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Algorithmic Lie Theoryfor Solving OrdinaryDifferential Equations

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Contents

1 Introduction 1

2 Linear Differential Equations 112.1 Linear Ordinary Differential Equations . . . . . . . . . . . . 112.2 Janet’s Algorithm . . . . . . . . . . . . . . . . . . . . . . . . 402.3 Properties of Janet Bases . . . . . . . . . . . . . . . . . . . . 552.4 Solving Partial Differential Equations . . . . . . . . . . . . . 77

3 Lie Transformation Groups 1053.1 Lie Groups and Transformation Groups . . . . . . . . . . . . 1053.2 Algebraic Properties of Vector Fields . . . . . . . . . . . . . 1153.3 Group Actions in the Plane . . . . . . . . . . . . . . . . . . . 1213.4 Classification of Lie Algebras and Lie Groups . . . . . . . . . 1353.5 Lie Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

4 Equivalence and Invariants of Differential Equations 1594.1 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . 1624.2 Nonlinear First Order Equations . . . . . . . . . . . . . . . . 1764.3 Nonlinear Equations of Second and Higher Order . . . . . . 186

5 Symmetries of Differential Equations 1935.1 Transformation of Differential Equations . . . . . . . . . . . 1945.2 Symmetries of First Order Equations . . . . . . . . . . . . . 2065.3 Symmetries of Second Order Equations . . . . . . . . . . . . 2105.4 Symmetries of Nonlinear Third Order Equations . . . . . . . 2245.5 Symmetries of Linearizable Equations . . . . . . . . . . . . . 236

6 Transformation to Canonical Form 2476.1 First Order Equations . . . . . . . . . . . . . . . . . . . . . . 2476.2 Second Order Equations . . . . . . . . . . . . . . . . . . . . 2496.3 Nonlinear Third Order Equations . . . . . . . . . . . . . . . 2796.4 Linearizable Third Order Equations . . . . . . . . . . . . . . 298

7 Solution Algorithms 3117.1 First Order Equations . . . . . . . . . . . . . . . . . . . . . . 3127.2 Second Order Equations . . . . . . . . . . . . . . . . . . . . 3197.3 Nonlinear Equations of Third Order . . . . . . . . . . . . . . 3287.4 Linearizable Third Order Equations . . . . . . . . . . . . . . 345

ix

x

8 Concluding Remarks 351

A Solutions to Selected Problems 355

B Collection of Useful Formulas 377

C Algebra of Monomials 383

D Loewy Decompositions of Kamke’s Collection 387

E Symmetries of Kamke’s Collection 403

F ALLTYPES Userinterface 417

References 419

Index 431

Chapter 1

Introduction

“...Es geht mir vorwarts, aber langsam, und es kostetunendliche und außerst langweilige Rechnungen...”1

Sophus Lie, Letter to A. Mayer of April 4, 1874.

Solving equations has been one of the most important driving forces in thehistory of mathematics. Particularly well known is the problem of solvingalgebraic equations of order higher than four and its eventual solution by La-grange, Ruffini, Abel and Galois. The complete answer given by the latterauthor provided not only the solution for the problem at hand, but also estab-lished a new field in mathematics, the theory of groups, which in turn laid thefoundation of modern algebra. A knowledgeable and detailed review of thissubject and various other topics discussed below may be found in the bookby Wussing [190], see also the Historical Remarks by Bourbaki [16].

Much less known is the fact that the theory of differential equations tooka similar course in the second half of the 19th century. At that time, solvingordinary differential equations (ode’s) had become one of the most importantproblems in applied mathematics, about 200 years after Leibniz and New-ton introduced the concept of the derivative and the integral of a function.Numerous phenomena in the physical sciences were described by formulas in-volving differentiation and integration, and the need arose to determine thefunctional dependencies between the variables involved. In other words, theproblem of solving a differential equation was born. In this book the phrasesolution of an ode means an expression for the general solution in some func-tion field, e.g., in terms of elementary or Liouvillian functions. The differentialequation should vanish identically upon its substitution. If n is the order ofthe equation the general solution involving n constant parameters is searchedfor. It should not involve derivatives, infinite series or products. Furthermore,any numerical or graphical representation is excluded.

1Translation by the author: “...I proceed, but slowly, and it takes me infinite, extremelyboring calculations...”

1

2

In the course of time, several ad hoc integration methods had been devel-oped for special classes of ode’s which occurred in the description of practicalproblems. It was quickly realized that a common feature of the various so-lution procedures consists in introducing new variables such that the givenequation is transformed into an equation with known solution, or for whichan integration method is available. Due to its importance, the special termequivalence problem was created for the task of recognizing whether two equa-tions may be transformed into each other, and the entirety of equations sharingthis property is called an equivalence class. In this general form, however, it ishardly justified to speak of a solution method because there is no indication ofhow to discover such a transformation in a concrete case, although infinitelymany may exist. In particular, for nonlinear equations, efforts to find a so-lution along these lines very often terminate in the worst case possible: Theattempts are abandoned without any evidence whether or not a solution inclosed form does exist. It is therefore highly desirable to develop algorithmicsolution schemes for the largest possible classes of equations.

The importance of equivalence problems had been realized before in al-gebraic geometry where the behavior of algebraic forms under the action ofcertain transformation groups, and in particular its invariants w.r.t. to thesetransformations, had been studied. The analogy with these methods turnedout to be extremely successful in the realm of differential equations. In hisarticle on invariants, Forsyth [47] expresses this as follows: Similarity in prop-erties of differential equations and of algebraic equations has long been of greatvalue, both in the development of the theory and in the indication of meth-ods of practical solution of the former equations. Based on preceding work byCockle [31], Laguerre [99], Brioschi [19] and Halphen [64], Forsyth applied thisprinciple in particular for calculating invariants of linear differential equationsunder predefined classes of transformations.

This was essentially the state of affairs when Sophus Lie got interested inthis problem in the sixties of the 19th century. The most important guide forhim was Galois’ theory for solving algebraic equations that had become widelyknown due to Liouville’s efforts around 1850. The most significant recognitionof Sophus Lie was that the transformation properties of an ode under certaingroups of continuous transformations play a fundamental role in answeringthis question, very much like the permutations of the solutions of an algebraicequation furnish the key to understanding its solution behavior. This is bestexplained by his own words, which are taken from the first chapter of his book[113] on differential equations: 2

2Translation by the author: “Previous investigations on ordinary differential equations asthey may be found in the customary textbooks do not form a systematic entirety. Specialintegration theories have been developed, e.g., for homogeneous differential equations, forlinear differential equations and other special forms of integrable differential equations. Themathematicians failed to observe, however, that the special theories may be subordinatedto a general method. The foundation of this method is the concept of an infinitesimaltransformation and closely related to it the concept of a one-parameter group.”

Introduction 3

“Die alteren Untersuchungen uber gewohnliche Differentialgleichungen,wie man sie in den gebrauchlichen Lehrbuchern findet, bilden kein sys-tematisches Ganzes. Man entwickelte specielle Integrationstheorien z. B.fur homogene Differentialgleichungen, fur die linearen Differentialglei-chungen und andere specielle Formen von Differentialgleichungen. Eswar aber den Mathematikern entgangen, dass diese speciellen Theoriensich unter eine allgemeine Methode unterordnen lassen. Das Funda-ment dieser Methode ist der Begriff der infinitesimalen Transformationund der damit aufs engste zusammenhangende Begriff der eingliedrigenGruppe.”

In order to provide an idea of how Lie acquired the intuition that finallyled him to create his theory, and to understand the basic principle underlyingit, a few simple but typical examples will be presented. Consider first theequation

y′′(y + x) + y′(y′ − 1) = 0 (1.1)

from the collection of solved equations by Kamke [85], equation 6.133 in hisenumeration. If new variables u and v are introduced by x = u+ v, y = u− vwith v ≡ v(u), equation (1.1) assumes the form

2v′′u+ v′2 − 1 = 0.

On the other hand, the transformation x = u+ 2v, y = 3u− 4v yields

v′′u− 12v′′v + 2

5v′3 − 9

10v′2 + 1

20v′ + 3

10 = 0.

Finally x = 12 (v2 + u), y = 1

2 (v2 − u) gives the particularly simple equation

v′′v3 + 14 = 0. (1.2)

From this latter equation the solution is readily obtained in the form

C1v2 = C2

1 (u+ C2)2 − 14

by means of two quadratures as may easily be verified by substitution. 3 Thusfinding among the infinity of possible variable transformations the particularone yielding (1.2) amounts to solving equation (1.1). Equation (1.2) has an-other remarkable property. Any of the following substitutions to new variablesz and w ≡ w(z),

u = −1− 1z , v = −wz , u = −4z + 4

8z + 7 , v = − w8z + 7 or

u = −9z − 9z − 10 , v = − w

z − 10 ,(1.3)

does not change its form, i.e., the result in all three cases is w′′w3 + 14 = 0.

A transformation with this particular property is called a symmetry of the

3The details of this example may be found on page 323.

4

differential equation. The three substitutions (1.3) may be subsumed underthe class of transformations of the form

u =(z + a)b2

1− (z + a)cb2, v =

wb

1− (z + a)cb2(1.4)

corresponding to the parameter values (a, b, c) = (1, 1, 1), (1, 2, 2) and (−1, 3, 1)respectively. Most remarkable, the entirety of transformations (1.4) forms agroup whose elements are parametrized in terms of a, b and c. It turns outthat invariance of (1.2) under the group of transformations (1.4) is a charac-teristic feature not only of eq. (1.2) but of all equations obtained from it by avariable transformation, i.e., for the full equivalence class determined by (1.2)which may be taken as a canonical representative. In general, a canonicalform is not uniquely determined by the symmetry type. The allowed freedommay be used in order to obtain an especially simple solution. An invariancetransformation of any member of the equivalence class determined by (1.2) isobtained from (1.3) or (1.4) if the same variable transformation is applied toit that leads to the transformed equation. The first transformation of (1.3)for example yields the symmetry

x =v

(u− v)2− 1

2, y =

u

(u− v)2+

12

of (1.1). The entirety of invariance transformations related by a variablechange is called a symmetry type. In general a group of transformations maybe a symmetry group for equations that are not equivalent to each other,e.g., if they have not the same order. The entirety of equations allowing thesame symmetry type is called a symmetry class, it is the union of equivalenceclasses.

With the insight gained by examples similar to those just described, Liedeveloped a solution scheme for ode’s that may be traced back to the followingquestion: Are there any nontrivial transformations leaving the form of thegiven equation invariant? If the answer is affirmative like for eq. (1.2) above,they determine a canonical form of the given equation. If the solution of thecanonical equation may be obtained and moreover it is possible to determinethe transformation of the given equation to a canonical form, the originalproblem is solved.

Due to the crucial role of various kinds of transformations in Lie’s theoryof differential equations, a major part of his work is devoted to establishing asystematic theory of groups of continuous transformations. Although it wasonly a device for his main goal of solving ode’s, the theory of Lie groups andthe closely related Lie algebras as they have been named by Hermann Weylbecame well-established fields in mathematics, independent from differentialequations.

Contrary to that, Lie’s theory for solving ordinary differential equations hasbeen almost forgotten after his death. Most textbooks on differential equa-tions do not even mention his name or the concept of a symmetry. For a long

Introduction 5

time the practical work of solving ode’s has remained almost unchanged fromthe way it is described in the above quotation by Lie. Basically it consists ofa set of heuristics that are applied one after another, usually accompanied bydatabase like tools in the form of collections of solved examples as the one byKamke [85], or the more recent collections by Polyanin [149] or Sachdev [157].This fact certainly needs some explanation. The most important obstacle forapplying his methods is the fact that they require huge amounts of analyticalcalculations that hardly may be performed by pencil and paper for almost anynontrivial example. The obvious remedy is to employ one of the computeralgebra systems that have become available lately to carry them through.

There is another more fundamental reason. It is related to the questionto what extent is his theory constructive, i.e., how much effort is needed todesign algorithms from it that may be applied for solving problems? This isparticularly important if the goal is a piece of computer algebra software thataccepts an ode as input and returns the best possible answer according to theunderlying theory. Lie himself did not attach a high value to this goal becausehe probably realized the practical difficulties due to the size of calculations.

After Lie had recognized that the symmetry of an ode is a fundamentalconcept that allows finding its solutions in closed form, he described severalsolution procedures based on it. On the one hand, if the infinitesimal genera-tors for the symmetry group of a given ode may be determined explicitly, thecanonical form and possibly the solution may be constructed from it. This isdescribed in detail in Lie’s book [113] on the subject; see also the books byOlver [140], Stephani [174] and Bluman [13]. In a second version the vectorfields generating the symmetry transformations are also determined explic-itly. The proper integration problem is reformulated and solved in terms ofan associated linear partial differential equation (pde), a subject that Lie hadconsidered in detail about ten years earlier. This procedure is also described inLie’s book quoted above. There is still another approach of solving equationswith symmetries. Its symmetry type is determined by suitable manipuationsof the determining equations of the symmetries without solving them explic-itly. It is shown that in principle the transformations to canonical form maybe obtained from it. Lie [109] outlined this proceeding in a series of articlesin the Archiv for Mathematik but never came back to it. The reason prob-ably is that it does not seem possible to set up the equations for the finitetransformations in terms of the given ode to be solved.

In order to make this latter approach into a working method, two funda-mental discoveries that were made within about twenty years after Lie’s deathturned out to be of fundamental importance. The first is Loewy’s theory oflinear ode’s, Loewy [127], and its generalization to certain systems of linearpde’s [105]. This was another effort to use the analogies between algebraicequations and differential equations, i.e., the reducibility of a linear ode andits representation in terms of irreducible equations of lowest degree. Loewy’smain result is a theorem that guarantees the existence of a unique decompo-sition in terms of largest so-called completely reducible factors. The second is

6

Janet’s theory of linear pde’s, Janet [83], and in particular the canonical formhe introduced for any such system which is known today as a Janet basis.This concept is closely related to Buchberger’s Grobner basis in polynomialideal theory, and turned out to be fundamental in the realm of differentialequations.

The importance of these results for Lie’s theory originates from the fact thatthe symmetries of any differential equation are obtained from its so-called de-termining system, a linear homogeneous system of pde’s. Representing it asa Janet basis allows one to identify its symmetry type from the coefficients.Furthermore, Loewy’s and Janet’s results allow to break down the solutionalgorithms for nonlinear ode’s based on its symmetries uniquely and system-atically into a small number of basic problems for which solution algorithmsmay be designed. At each step and in particular for the final result, thefunction field necessary for representing the answer is obtained.

Sophus Lie was not the only person who had the idea of generalizing Galois’results to differential equations. Almost at the same time as he developed histheories in Leipzig, Picard and Vessiot [146] in Paris initiated another effortwith a similar objective. Their approach is usually referred to as Picard-Vessiot-theory or differential Galois theory. Their goal is more limited thanLie’s because they consider only linear ode’s. Very much like ordinary Galoistheory deals with the field extensions generated by the solutions of univariatealgebraic equations, differential Galois theory investigates the differential fieldextensions over a base field generated by the solutions of linear ode’s. Theyobtain a theory that establishes a one-to-one correspondence between thestructure of the differential Galois group of an equation and the function fieldgenerated by its solutions. Therefore the theory of Picard and Vessiot mayvery well claim to be the proper analogue of Galois’ theory for linear ode’s.

Although Lie aimed for a general and complete theory for solving nonlinearode’s as well, this goal has only partially been achieved. In Drach’s [38] thesisthat has been supervised by E. Picard and J. Tannery, the state of affairs isexpressed as follows: ...Lie’s theorie des groupes a l’integration des equationsn’est pas la veritable generalization de la methode employee par Galois pourles equations algebriques.4 Drach’s main objections are that Lie’s theory isnot complete in the following sense: An equation may have a symmetry groupthat cannot be employed for solving it. This holds for first order equations,and may occur for equations of higher order as well. On the other hand, Lie’stheorems are not reciprocal, i.e., an equation may have a closed form solutionyet may not have a nontrivial symmetry group. An example is the equationgiven by Gonzales-Lopez [54]

y′′y − y′2 − y′y2(x2 + 1)− 2y3x = 0. (1.5)

4Translation by the author: “...Lie’s group theory for the integration of equations is notthe true generalization of the method applied by Galois for algebraic equations.”

Introduction 7

This equation has a trivial symmetry group, yet its general solution is

y =C3

1

(C1x+ 1)2 + C21 + 1 + C2e

C1x. (1.6)

In other words, Lie’s theory does not establish a unique relation between theexistence of symmetries of an equation and its solution behavior.

In this context the following question is frequently asked: What is therelationship between Lie’s theory and the theory of Picard and Vessiot? Al-though their point of departure is basically the same and group theory is anessential part of both theories, the answer is that there is virtually no con-nection between them. They are different approaches to the general problemof solving ordinary differential equations in closed form. Consequently, therearises a second question: What is the proper generalization of Galois’ theoryfor solving nonlinear differential equations? There does not seem to exist asatisfactory answer at present.

Another important aspect is raised by the following question: To whatextent do these theories support solving concrete problems? Although thetheory of Picard and Vessiot is quite satisfactory from a theoretical pointof view, it is only of limited value for obtaining explicit solutions. On theother hand, as it will turn out later on in this book, for solving nonlinearordinary differential equations Lie’s theory is a powerful tool. The solutionsfor the nonlinear second order equations in Chapter 6 of Kamke’s collection,for example, are almost exclusively due to Lie symmetries. Furthermore, anyequation obtained from them by a variable transformation may also be solvedby applying Lie’s theory, i.e., instead of a finite number of examples their fullequivalence classes become amenable to the solution procedure.

Sometimes the objection is raised that most equations do not have non-trivial Lie symmetries and therefore Lie’s theory does not apply. Althoughthis is true, one should keep in mind that a similar argument applies to lin-ear ode’s and their differential Galois groups, and is also true for algebraicequations. The equations that actually occur in applications very often havea special structure that expresses itself in the existence of symmetries. It is ofutmost importance to recognize these situations because it leads to a betterunderstanding of the underlying problem, and to take advantage of them forthe solution procedure. In this way Lie’s theory may finally lead to a betterunderstanding of the distinctive features that make an equation solvable, andmay even help to raise the intuition that is necessary in order to establish amore complete theory.

As mentioned before, the proceeding described in Lie [109] is the basis formost of the solution algorithms for second and third order ode’s in this book.In this approach, the symmetry type of an ode, being an invariant underpoint transformations, serves the purpose of narrowing down the possibleequivalence classes to which it may belong. The totality of equations from acertain family sharing the same symmetry type is called a symmetry class; it

8

is a union of equivalence classes. In this form the close relation between thesymmetry analysis and equivalence problems becomes particularly obvious.This point has been discussed in the preface of the book by Olver [141].

After the symmetry class of an equation and a possible canonical form areknown, the function field of admitted equivalence transformations must bespecified in order to obtain a well-defined problem. Furthermore, it should bepossible to solve equivalence problems within this function field constructivelyand to design solution algorithms. In this book most ode’s are polynomial inthe derivatives with coefficients that are rational functions of the dependentand the independent variables. The smallest field containing the coefficientsis called the base field. More general function fields are extensions of thebase field, e.g., elementary or Liouvillian extensions. The overall scheme forsolving second- or higher order ode’s along these lines decomposes in threemajor steps.

Determine the symmetry class of the given equation. This is achievedin terms of the Janet basis coefficients for the determining system of itssymmetries; it proceeds completely in the base field.

Transform the equation to a canonical form corresponding to its equiv-alence class which is contained in the symmetry class. To this end asystem of linear or Riccati-like pde’s will be set up for which solutionsin well-defined function fields may be determined algorithmically.

Solve the canonical form and determine the solution of the given equa-tion from it.

In addition to developing solution schemes for individual equations, equa-tions containing undetermined elements like parameters or functions depend-ing on a specified set of indeterminates may be considered. Examples are lin-ear equations, Riccati equations or Abel equations with coefficients that areundetermined functions of the independent variable. In other words, wholefamilies of equations are considered that are identified by a certain structure.The problem then is to decompose the totality of equations with a given struc-ture into symmetry or equivalence classes. The importance of this proceedingarises from the fact that many equations occuring in practical problems arespecified in terms of its structure.

The scope of this book has been restricted deliberately to ordinary differen-tial equations of low order. The aim is to obtain solution algorithms for well-defined classes of equations, leading to the best result including the possibleanswer that a solution of a certain type does not exist. Any extension of thisscope would lead to a significant increase of the size of the book. The simplestpde’s for a scalar function depending on two variables, or the simplest sys-tems of ode’s for two functions, require a classification of the finite Lie groupsof three-space comprising several hundred entries. It remains an interesting

Introduction 9

and challenging problem to develop solution schemes for these cases along thelines described in this book.

The goals of this book are twofold. On the one hand, it intends to be asource of reference for people working actively in the field. To this end, thedetailed discussion of many subjects and the extensive tabulations in the maintext and especially in the appendices should prove to be an indispensible tool.Beyond that it might serve as a textbook if supplemented by some additionalreading material, e.g., on the theory of ode’s in general or on Lie groups andLie algebras. Some bibliographic hints are given in the introductory remarksto the individual chapters.

The subsequent Chapter 2 provides the mathematical foundation for therest of the book. It deals mainly with linear differential equations. For linearode’s the most important topic dealt with is Loewy’s theory for the uniquedecomposition into completely reducible components. This is the basis forhandling more general problems later on. For linear pde’s a fairly completedescription of Janet’s work is given. Of particular importance are those sys-tems of pde’s that allow a finite-dimensional solution space. The dimensionof this space is called the order of the system. This generalizes the notion oforder of a linear ode. For values up to three, a complete classification of Janetbases is given. Loewy’s decomposition of linear ode’s is generalized to suchsystems. In Chapter 3, those results from the theory of continuous groups ofa two-dimensional manifold R2 are presented that are relevant for the mainsubject of this book, i.e., solving ode’s. The close relation between Lie’s sym-metry analysis and the equivalence problem is the subject of Chapter 4. Thetwo subsequent Chapters are the core of the book. The subject of Chapter 5 isto identify the symmetry class to which a given quasilinear equation of ordertwo or three belongs. To this end it is sufficient to know a Janet basis for itsdetermining system, it is not necessary to determine the symmetry generatorsexplicitly. The same is true for solving a given ode as is shown in Chapter6. The crucial problem of transforming a given ode with a nontrivial symme-try to canonical form is achieved by solving a system of linear or Riccati-likepde’s. As mentioned above, the symmetry class serves essentially the pur-pose of identifying a canonical form. Lie’s second approach, i.e., to determinethe Lie algebra of symmetry generators explicitly and to generate the canon-ical form of the differential equations from it, is the subject of Section 3 inChapter 6.

Various topics are postponed to the appendices A through F. The solutionsof selected exercises are given in Appendix A. A collection of useful formulaeis listed in Appendix B. Various properties of ideals of monomials that areneeded for Janet’s algorithm are presented in Appendix C. In Appendix Dfor selected linear ode’s from Kamke’s collection the Loewy decomposition isgiven explicitly. In Appendix E the symmetries of the equations of second andthird order of Kamke’s collection are listed. They serve as a reference in manyplaces of this book. Appendix F contains a description of the user interfaceof the software system ALLTYPES which is an important part of this book.

10

It provides a great many functions for carrying out the calculations that arenecessary in order to apply the theory to practical problems. Most of thesecalculations are too voluminous to be performed by pencil and paper. Thissoftware may be accessed through the website www.alltypes.de.

The most important sources of information for this book were Lie’s originalwritings on the subject, edited in seven volumes of the Gesammelte Abhand-lungen by Friedrich Engel, and also Engels’ Anmerkungen in the supplementsto these volumes. Furthermore, there are the books written under Lie’s guid-ance as quoted in the references at the end of this book. A lot of interestingbackground information may be found in the biography of Sophus Lie byStubhaug [175].

The author would like to thank several colleagues and students for theirhelpful comments. In particular, Gunter Czichowski, the late Eckehart Hotzeland several anonymous referees. The abiding encouragement by Carl-AdamPetri and the excellent working environment at the Fraunhofer Institute ofUlrich Trottenberg are also gratefully acknowledged.

Chapter 2

Linear Differential Equations

This chapter provides the fundamental algorithms for working with linear or-dinary and partial differential equations that are the building blocks for thesolution algorithms given later in this book. Systems of linear pde’s are char-acterized in the first place by the number m of dependent and the numbern of independent variables. Additional quantities of interest are the numberN of equations, the order of the highest derivatives that may occur and thesmallest function field in which the coefficients are contained, called the basefield. Without further specification this is assumed to be the field of ratio-nal functions in the independent variables with rational number coefficients.The special case m = n = N = 1 corresponds to a single linear ode. Thefundamental concepts described in this chapter are the Loewy decomposition,Loewy [127], and the Janet basis, Janet [83]. The latter term is chosen tohonour the French mathematician Maurice Janet who described this conceptand gave an algorithm to obtain it. After it had been forgotten for aboutfifty years, it was rediscovered [163] and utilized in various applications as de-scribed in this book. A good survey is also given in the article by Plesken andRobertz [148]. Janet bases are the differential counterpart of Grobner basesthat have been introduced by Bruno Buchberger and are a well-establishedtool in polynomial ideal theory and algebraic geometry now. The relevance ofa Janet basis for the main subject of this book, i.e., solving ordinary differen-tial equations or ode’s for short, originates from the fact that the symmetriesof any such equation are determined by a system of linear homogeneous pde’s.General references for this chapter are the book by Ince [80] or the two vol-umes on linear equations by Schlesinger [160]. For Chapter 2.1 the first 150pages of the book by van der Put and Singer [185], or the article by Buiumand Cassidy [23] are highly recommended.

2.1 Linear Ordinary Differential Equations

The theory of linear ode’s is important for several reasons. In the firstplace, many problems in connection with pde’s may be traced back to certainstandard problems involving ode’s. Secondly, the theory of linear ode’s is

11

12

more complete than the theory of pde’s and therefore serves as a guide forwhat kind of results may be expected for the latter. Many of the conceptsand the methods introduced in this section will be needed later on. Symmetryproperties of linear ode’s will be discussed in detail in Chapter 5.5.Solving Linear ODE’s. A linear homogeneous ode for a differential in-

determinate y, for which y′ = dydx

and more generally y(n) = dy(n−1)

dxis

defined for any natural number n, will be written as

L(y) ≡ y(n) + q1y(n−1) + q2y

(n−2) + . . .+ qn−1y′ + qny = 0. (2.1)

If not specified otherwise the coefficients qk, k = 1, . . . , n, are rational func-tions in the independent variable x with rational number coefficients, i.e.,qk ∈ Q(x) which is called the base field.

Equation (2.1) allows the trivial solution y ≡ 0. The general solution con-tains n constants C1, . . . , Cn. Due to its linearity, these constants appear inthe form y = C1y1 + . . .+Cnyn with the yk linearly independent over the fieldof constants, e. g. Ci ∈ Q in this case. They form a so-called fundamentalsystem and generate a n-dimensional vector space over the field of constantsQ. The coefficients qk of (2.1) may be rationally expressed in terms of afundamental system and its derivatives. To this end the Wronskian

W (n)(y1, . . . , yn) =

∣∣∣∣∣∣∣∣∣∣∣∣

y1 y2 . . . yn

y′1 y′2 . . . y′n

. . . . . . . . .

y(n−1)1 y

(n−1)2 . . . y

(n−1)n

∣∣∣∣∣∣∣∣∣∣∣∣(2.2)

is defined. It is different from zero iff the yk(x) are independent over theconstants. More generally, the determinants

W(n)k (y1, . . . , yn) =

∂

∂y(n−k)

∣∣∣∣∣∣∣∣∣∣∣∣

y y1 y2 . . . yn

y′ y′1 y′2 . . . y′n

. . . . . . . . . . . .

y(n) y(n)1 y

(n)2 . . . y

(n)n

∣∣∣∣∣∣∣∣∣∣∣∣(2.3)

are defined for k = 0, 1, . . . , n. There are the obvious relations W(n)0 =

(−1)n−1W (n) and dW (n)

dx= (−1)n−1W

(n)1 . The determinant at the right

hand side of (2.3) vanishes for y = yk, k = 1, . . . , n. Therefore, given a fun-damental system y1, . . . , yn, (2.3) is another way of writing the left hand sideof (2.1). This yields the representation

qk = (−1)kW

(n)k (y1, . . . , yn)

W (n)(y1, . . . , yn)(2.4)

Linear Differential Equations 13

for the coefficients qk, k = 1, . . . , n in terms of a fundamental system, i.e.,(2.1) is uniquely determined by its solution space. For k = 1 there followsW

(n)1 + q1W

(n) = W (n)′ + q1W(n) = 0, i.e., W (n) = exp (−

∫q1dx).

It is always possible to replace equation (2.1) by another equation over thesame base field, its so called rational normal form

y(n) + q2y(n−2) + . . .+ qn−1y

′ + qny = 0,

by introducing a new dependent variable y(x) by y = exp(− 1n

∫q1(x)dx

)y.

More general transformations with the same effect are discussed in Chap-ter 4.1.

Solving an equation (2.1) amounts to determining n independent elementsof a fundamental system, called a full set of solutions. In order to make thisa rigorous concept, the function field containing the yk’s has to be specified.After that an algorithm has to be designed in order to obtain the solutionexplicitly.

The simplest problem is to ask for solutions in the base field. If it is Q(x),this amounts to searching for the rational solutions of (2.1). In general anyrational solution has the form

y = C1y1 + . . .+ Ckyk (2.5)

with 0 ≤ k ≤ n, yk ∈ Q(x), Ck constant and y1, . . . , yk linearly independentover constants. It may occur that all elements of a fundamental system arerational, i.e., k = n, or that the equation does not have a rational solution atall. The most general rational solution is the one with the highest possiblevalue for k. If k < n it is an important problem to find the minimal extensionof the base field containing a full set of solutions.

Determining the most general solution in the base field occurs as a sub-problem of many other problems. It is therefore important that an efficientalgorithm is available for this purpose. Such an algorithm is designed accord-ing to the following general principles that are valid for more general functionfields as well. At first the problem has to be made finite, i.e., the possiblesolutions have to be parametrized such that a finite number of candidatesis obtained. Secondly, a search procedure must be provided that identifiesthe desired answer among the candidates, or assures that a solution does notexist.

The following observation yields an important constraint for the possibleform of any rational solution of (2.1). Assume a rational solution has a poleof order k at the finite position x = x0. If it is substituted into (2.1), a termproportional to 1

(x− x0)k+nis generated from the highest derivative y(n).

Furthermore, it is assumed that both the coefficients and the solution areuniquely represented as partial fractions. Then in order to make the left handside vanish, at least one more term of order k + n must occur. Because thelower derivatives of y cannot produce such a term, one of the coefficients must

14

contain a pole at x0. As a consequence, the only possible poles of a rationalsolution of (2.1) are those that occur in the coefficients qi(x). Alternatively, if(2.1) is considered as an equation over Q[x], by multiplication with the leastcommon multiple of the coefficient denominators, the possible poles are thezeros of the coefficient of the highest derivative y(n).

For any of these poles the coefficient of the highest order term must vanish.This leads to a nonlinear equation for the possible highest order whose largestsolution in natural numbers is the desired bound. Similar arguments apply forthe behavior at infinity. A polynomial ansatz with undetermined coefficientswithin these limits leads to a system of linear algebraic equations the solutionsof which determine the rational solutions. The following algorithm is basedon these considerations. More details may be found in Schwarz [162] and thereferences given there.

Algorithm 2.1 RationalSolutionsLode(L(y)). Given a linear homoge-neous ode

L(y) ≡ y(n) + q1y(n−1) + q2y

(n−2) + . . .+ qn−1y′ + qny = 0

with qk(x) ∈ Q(x), the maximal number of linearly independent rationalsolutions is returned.S1 : Singularities and bounds. Identify the possible positions of poles anddetermine an upper bound for any of them.S2 : Solve linear system. Set up the linear system for the undeterminedcoefficients for an ansatz within the limits found in S1 and determine itssolution. If only the trivial solution exists, return an empty list .S3 : Return result. For each solution found in S2 generate a solution ofL(y) and return the list y1, . . . , yk.

Example 2.1 Consider the equation y′′ +(2 + 1

x)y′ − 4

x2 y = 0. It hasno. 2.201 in Kamke’s collection. For the order N of a singularity at infinitythere follows immediately N = 0. For the order M of the only other pole atx = 0 there follows M(M + 1) −M − 4 = 0, i.e., M = 2. In step S2 theansatz a2

x2 + a1x + a0 yields the system 3a1 + 4a2 = 0, 2a0 + a1 = 0 with

the solution a1 = − 43a2, a0 = 2

3a2. With the normalization a2 = 1 the onlyrational solution obtained in step S3 has the form 1

x2 − 43x + 2

3 , i.e., k = 1 inthis case. The second element of a fundamental system is not rational, it willbe obtained in Example 2.14 below by a different method.

Solving a linear inhomogeneous equation

y(n) + q1y(n−1) + q2y

(n−2) + . . .+ qn−1y′ + qny = r (2.6)

may be traced back to the corresponding homogeneous problem with r = 0and integrations by the method of variation of constants. To this end, the Ciin the general solution y = C1y1 + . . . + Cnyn of the homogeneous equation


are considered as functions of x. Substituting this expression into (2.6) andimposing the constraints

C ′1y(k)1 + C ′2y

(k)2 + . . .+ C ′ny

(k)n = 0 (2.7)

for k = 0, . . . , n− 2 yields the additional condition

C ′1y(n−1)1 + C ′2y

(n−1)2 + . . .+ C ′ny

(n−1)n = r. (2.8)

The linear system (2.7), (2.8) for the C ′k has always a solution due to the non-vanishing determinant of its coefficient matrix which is the Wronskian W (n).Consequently, solving the inhomogeneous equation (2.6) requires only inte-grations if a fundamental system for the corresponding homogeneous problemis known.

Example 2.2 If y1 and y2 is a fundamental system for y′′+ q1y′+ q2y = 0,

the general solution of y′′ + q1y′ + q2y = r may be written as

y = C1y1 + C2y2 + y1

∫ry2W

dx− y2∫ry1W

dx (2.9)

where W = y′1y2 − y′2y1. Let the equation

y′′ − 4x2x− 1

y′ +4

2x− 1y =

e−2x

2x− 1

be given. A fundamental system for the homogeneous equation is y1 = x,

y2 = e2x. Therefore r = e−2x

2x− 1 and W = (2x− 1)e2x. Substituting this intothe above equation leads to

y = C1e2x + C2x+ e2x

∫xe−4x

(2x− 1)2dx− x

∫e−2x

(2x− 1)2dx.

Inhomogeneous problems will occur frequently later on if a linear homoge-neous equation of third order is reducible but not completely reducible.

If a linear ode (2.1) does not have a full set of solutions in its base field F ,i.e., if k < n in (2.5), the dimension of its solution space may increase ifan enlarged function field is admitted for a fundamental system. This isachieved by adjoining solutions of algebraic or differential equations of ordernot higher than n to the base field. If θ is the element to be adjoined, theenlarged differential extension field is denoted by F〈θ〉, extensions by morethan a single element are denoted by F〈θ1, . . . , θk〉. Frequently the newlyadjoined elements are required to obey special equations over the previouslydefined field. If these equations are algebraic or of first order, a special termis introduced for the corresponding extension.

16

Definition 2.1 (Liouvillian extensions). Let F be a differential fieldwith derivation d

dx≡ ′. A simple Liouvillian extension F〈θ〉 is obtained by

adjoining an element θ such that one of the following holds.

i) Algebraic extension, θ satisfies a polynomial equation with coefficientsin F .

ii) Extension by the exponential of an integral, θ′ − aθ = 0, a ∈ F .

iii) Extension by an integral, θ′ − a = 0, a ∈ F .

If this process is repeated and any finite number of elements θ1, . . . , θk is ad-joined, a Liouvillian extension F〈θ1, . . . , θk〉 is obtained.

The special case θ = ex, θ = log x or θ algebraic is also called an ele-mentary extension. If a full set of solutions is not obtained in a Liouvillianextension, solutions of higher order equations have to be adjoined to the basefield which are called Picard-Vessiot extensions. They may be considered asthe equivalent of the splitting field of an algebraic equation.

A systematic study of the possible field extensions is based on the conceptof an irreducible linear ode, and the factorization of reducible equations intoirreducible components. A basic prerequisite for this discussion is a theoryfor solving certain nonlinear equations due to Riccati that are discussed next.Solving Riccati Equations. Any n-th order linear homogeneous ode (2.1)may be transformed into a nonlinear equation of order n − 1 for a functionz(x) by means of the variable change y′ = zy. The higher order derivativesof y are then y(ν) = φν(z)y where

φν(z) =dφν−1(z)

dx+ zφν−1(z), ν ≥ 1 and φ0 = 1. (2.10)

For ν ≤ 4 their explicit form is

φ0 = 1, φ1 = z, φ2 = z′ + z2, φ3 = z′′ + 3zz′ + z3,

φ4 = z′′′ + 3z′2 + 4zz′′ + 6z′z2 + z4.

Substitution into (2.1) yields the Riccati equation associated to (2.1)

Rnx(z) ≡ q0φn + q1φn−1 + q2φn−2 + . . .+ qn−1φ1 + qnφ0 = 0 (2.11)

of order n − 1 for the unknown function z(x); the qk(x) for k = 1, . . . , n arethe same as in (2.1), q0 = 1. For n = 2, 3 and 4 these equations are explicitlygiven by

z′ + z2 + q1z + q2 = 0, (2.12)

z′′ + 3zz′ + z3 + q1(z′ + z2) + q2z + q3 = 0, (2.13)

z′′′+4zz′′+3z′2 +6z′z2 + z4 + q1(z′′+3zz′+ z3)+ q2(z′+ z2)+ q3z+ q4 = 0.(2.14)


Similar to linear equations, the existence of rational solutions of any Riccatiequation is of fundamental interest. If the general solution of an equation oforder n is rational it contains n constants. If this is not true, rational solutionsinvolving fewer constants or no constants at all may exist. The latter solutionsare called special rational solutions.

In order to make this more precise first it has to be made clear when twosolutions are considered as essentially different. To this end the followingconcept of equivalence of rational functions is introduced. Two rational func-tions p, q ∈ Q(x) are called equivalent if there exists another rational function

r ∈ Q(x) such that p − q = r′r holds, i.e., if p and q differ only by the log-

arithmic derivative of another rational function. It is easily seen that thisdefines an equivalence relation on Q(x). Applying this concept, the followingestimate of essentially different rational solutions of a Riccati equation maybe obtained.

Lemma 2.1 A generalized Riccati equation of order n has at most n + 1pairwise non equivalent rational solutions.

The proof of this result may be found in the article by Li and Schwarz [104].For equations of order one and two a detailed description of all possible ra-tional solutions is given next.

Theorem 2.1 If a first order Riccati equation z′ + z2 + az + b = 0 witha, b ∈ Q(x) has rational solutions, one of the following cases applies:

i) The general solution is rational and has the form

z =r′

r + C+ p (2.15)

where p, r ∈ Q(x), Q is a suitable algebraic extension of Q, and C is aconstant.

ii) There is only one, or there are two inequivalent special rational solu-tions.

Proof Let y1, y2 be a fundamental system for the linear homogeneousequation y′′ + ay′ + by = 0 corresponding to the given Riccati equation

z′ + z2 + az + b = 0. If both y′1y1 and y′2

y2 are not rational, a rational solu-

tion does not exist. If y′1y1 is rational but y′2

y2 is not, y′1y1 is the only rational

solution. If both y′1y1 and y′2

y2 are rational but

C1y′1 + C2y

′2

C1y1 + C2y2(2.16)

18

is not, y′1y1 and y′2

y2 are two special rational solutions of the Riccati equation.

This covers case ii). If y′1y1 , y

′2y2 and the ratio (2.16) are rational, it may be

rewritten as

C1y′1 + C2y

′2

C1y1 + C2y2=Cy′1/y1 + y′2/y1C + y2/y1

=(y2/y1)′

C + y2/y1+y′1y1

(2.17)

with C = C1C2

. The assignments p = y′1y1 and r = y2

y1 yield the expressions

given for case i).

For first order Riccati equations there exists an extension of the precedingtheorem for which there is no analogue if the order is higher than one.

Corollary 2.1 The general first order Riccati equation y′+py2+qy+r = 0may be transformed into normal form z′+z2+az+b = 0 by the variable change

y = zp with the result a = q − p′

p , b = pr. The following three cases may bedistinguished, depending on the number of special solutions of the normal formequation that are already known.

i) If a single special solution z1 is known, the general solution is z = z1+ 1w

where w is the general solution of w′ − (2z1 + a)w = 1.

ii) If two special solutions z1 and z2 are known, the general solution is

z = r′

C + r + z1 with r = − exp (∫

(z1 − z2)dx).

iii) If three special solutions z1, z2 and z3 are known, the general solutionmay be written as z − z2z − z1 : z3 − z2z3 − z1 = C or

z =(z1 − z2)z1 − z3z2 − z3C − z1 − z3

z2 − z3+ z1.

From these representations it follows in particular that two special rationalsolutions yield the general rational solution if z1 and z2 are equivalent, i.e.,if the difference z1 − z2 is a logarithmic derivative. If three special rationalsolutions are known, the general solution is always rational.

The proof is obtained by substituting the various solutions into the givenequation. It is instructive to consider various examples corresponding to thealternatives of the above results.

Example 2.3 The equation considered in Example 2.5 below has the twospecial rational solutions z = x and z = x+ 2

x . Because the difference of thetwo special solutions may be written as a logarithmic derivative in the form2x = (x2)′

x2 , the preceding discussion shows that the general solution must be


rational. With z1 = x and z2 = x + 2x , the same expression for the general

solution is obtained as before. If z1 = x+ 2x and z2 = x is chosen, its form is

z =2x3

C − 1x2

+ x+2x.

The two constants are related by CC + 1 = 0.

Example 2.4 The equation z′+z2−(1+ 1

x)z+ 1

x = 0 has the two specialrational solutions z1 = 1, z2 = 1

x+ 1. The difference z1 − z2 = xx+ 1 is not a

logarithmic derivative. As a consequence, the general solution

z =xe−x

C − (x+ 1)e−x+ 1

is not rational.

Theorem 2.2 If a second order Riccati equation

z′′ + 3zz′ + z3 + a(z′ + z2) + bz + c = 0

with a, b, c ∈ Q(x) has rational solutions, one of the following cases applies.


z =C2u

′ + v′

C1 + C2u+ v+ p (2.18)

where p, u, v ∈ Q(x), Q is a suitable algebraic extension of Q, and C1

and C2 are constants.

ii) There is a single rational solution containing a constant; it has the form(2.15).

iii) There is a rational solution containing a single constant as in the pre-ceding case, and in addition a single special rational solution.

iv) There is only a single one, or there are two or three special rationalsolutions that are pairwise inequivalent.

The proof for the second order equation proceeds along similar lines. There-fore it is left as Exercise 2.6.

For Riccati equations of order higher than two the general structure ofrational solutions has been described by Li and Schwarz [104]. Their result isgiven here without proof.

Theorem 2.3 The rational solutions of the Riccati equation (2.11) as-sociated to (2.1) may be described as follows. Let ri ∈ Q(x) and Pi ≡

20

pi,1, . . . , pi,k, pi,j ∈ Q[x], Ci,j constants, the sets Pi linearly independentover constants for i = 1, . . . ,m and j = 1, . . . , k, and define

Sri

Pi≡

ri +

Ci,1p′i,1 + . . .+ Ci,kp

′i,k

Ci,1pi,1 + . . .+ Ci,kpi,k

.

The rational solutions of (2.11) are the disjoint union of Sri

Pi, i = 1, . . . ,m.

Moreover,∑mi=1 |Pi| is not greater than n.

Due to its connection to many problems treated later on, a constructive pro-cedure for determining the rational solutions of a Riccati equation is needed.Similar to linear equations, the first step is to find the position of its singu-larities. For this purpose the explicit expression

φν(z) =∑

k0+2k1+...+νkν−1=ν

ν!k0!k1! . . . kν−1!

(z1

)k0(z′2!

)k1. . .

(z(ν−1)

ν!

)kν−1

(2.19)for the functions φν(z) in (2.11) is useful. It follows from a formal analogy tothe iterated chain rule of di Bruno [22]. For the subsequent discussion severalof its properties are required, e. g. the order of a pole of φν which it exhibits asa consequence of a pole of z. These properties are discussed in Exercise 2.13and are taken for granted from now on without mentioning it.

Let x0 be the position of a finite pole of order M > 1 in a possible solutionz(x). Due to (2.19) it generates exactly one term proportional to 1

(x− x0)νMin φν(z). In order to match it with some other term, the same pole mustoccur at least in a single coefficient qk(x) at the left hand side of equation(2.11). From this it follows that poles of order higher than one may occuronly at the pole positions of the coefficients. The same is true for a pole ofany order N at infinity. A bound for the order at any of these singularitiesis obtained by determining the growth for the various terms and looking forthe largest integer where at least two terms have the same value. Due to thenonlinearity of the Riccati equation, the equations for the coefficients of thevarious singularities in a partial fraction expansion obey nonlinear algebraicequations that may have nonrational numbers as solutions. Therefore rationalsolutions of Riccati equations are searched in appropriate algebraic extensionsQ of the rational numbers.

These arguments do not apply for finite poles of order one. Any term in(2.19) generates a pole of order ν in this case. Nonleading terms originatingfrom coefficient singularities may interfere with them. As a consequence, thefinite first order poles are not confined to those poles occuring in the coeffi-cients. It is a new problem to find their positions. According to Markoff [130]or Singer [169] one may proceed as follows. Let z be the desired solution of annth order Riccati equation. Due to the fact that the residues of the additionalfirst order poles are natural numbers, their contribution appears as a polyno-mial factor p in the solution of the corresponding linear equation. If the factor


originating from the higher order coefficient singularities is already known bythe methods described above, a linear homogeneous equation for p may beobtained; and its possible polynomial solutions determine the additional firstorder poles of z. These polynomial solutions may be obtained by specializingthe algorithm RationalSolutionsLode. The subsequent algorithm is designedaccordingly.

Algorithm 2.2 RationalSolutionsRiccati(R). Given a Riccati equation Rof order n, return the rational solutions in Q(x).S1 : Higher order poles and bounds. Determine the possible positions ofpoles of order higher than one that may occur in a solution of R anddetermine an upper bound for each of them.S2 : Solve algebraic system. Set up an algebraic system for the undeter-mined coefficients of an ansatz within the limits found in S1, determinethe coefficients from it and construct the corresponding solutions of R.If a solution does not exist, return an empty list.S3 : First order poles. For each solution obtained in S2 generate the cor-responding linear equation for a function p and determine its polynomialsolutions.S4 : Construct solution. From each polynomial solution obtained in S3and the corresponding solution obtained in S2 construct a solution of thegiven Riccati equation.S5 : Return result. Discard those solutions obtained in S4 that are sub-cases of some other solution, and return a list with the remaining ones.

Example 2.5 Consider the equation z′ + z2 −(2x + 1

x)z + x2 = 0. A

power xN in a possible solution generates terms of order N − 1, 2N , N + 1,N − 1 and 2, i.e., N = 1 is the largest value where at least two terms matcheach other. A pole at zero of the form 1

xMleads to singular terms of order

M + 1, 2M , M − 1, M + 1 and M − 2, therefore again M = 1 is the desiredbound. The ansatz z = ax+ b+ c

x yields the algebraic system

a2 − 2a+ 1 = 0, ab− b = 0, b2 + 2ac− 2c = 0, 2bc− b = 0, c2 − 2c = 0

with the two solutions a = 1, b = c = 0 and a = 1, b = 0, c = 2, i.e., two specialrational solutions are z = x and z = x+ 2

x . The corresponding equations fora polynomial factor are p′′ + 3

xp′ = 0 and p′′ − 1

xp′ = 0. The former equation

does not have any polynomial solution while a fundamental system for thelatter is 1, x2. This yields the general solution z = 2x

x2 + C+ x.

Gcrd, Lclm and Factorization of Linear ODE’s. Typical questionsthat arise when dealing with linear differential equations are the following:Is it possible to obtain some or all elements of a fundamental system for agiven linear ode as solutions of equations that are simpler than the given one?Without further specification this means that only equations of the same type,

22

i.e., linear homogeneous ode’s over the same base field, but of lower order,are admitted. More special problems are to decide whether all solutions of agiven equation are also solutions of some other equation, or whether two givenequations have some solutions in common. Certainly one would like to decidethese questions without determining the solutions explicitly. For algebraicequations there are well-known algorithms for answering these questions.

For the subsequent discussion a more algebraic language is appropriate. Tothis end, let Q(x)[D] with D = d

dxbe the ring of ordinary differential oper-

ators in the indeterminate x with rational function coefficients. Its elementshave the form

A ≡ a0Dn + a1D

n−1 + . . .+ an−1D + an

with ak ∈ Q(x). If a0 6= 0 the order of A is n. The linear ode (2.1) is obtainedby applying such an operator with a0 = 1, ai = qi for i = 1, . . . , n, to adifferential indeterminate y, i.e., by writing Ay = 0. Let

B ≡ bmDm + b1Dm−1 + . . .+ bm−1D + bm

be another operator. The sum A+B of two operators is defined by termwiseaddition in the obvious way. The product AB is defined by the prescriptionDbj(x) = bj(x)D + b′j(x) from which the general expression

ai(x)Dibj(x)Dj =i∑

k=0

(i

k

)ai(x)b

(k)j Di−k+j

follows. These definitions make the set of differential operators Q(x)[D] intoa ring. In general the product AB is different from BA, i.e., this ring is notcommutative. Consequently, one has to distinguish between left and right fac-tors in a product. In general, for differential operators A, B and C, C = A Bimplies ker(B) ⊂ ker(C), i.e., the solution space of B is a subspace of thesolution space of C. Usually differential operators or linear differential equa-tions are assumed to be primitive, i.e., the coefficient of the highest derivativeis unity.

For the problems mentioned above, the existence of a division algorithm isessential. It is defined in complete analogy to the corresponding divison inthe ring of univariate algebraic polynomials. As a consequence, if A and Bare operators of order n and m respectively with m ≤ n, there are uniqueoperators Q and R such that A = QB + R where the order of R is strictlyless than m. This guarantees that the ring Q(x)[D] is Noetherian as shownby Ore [143]. When R = 0, B is a right factor of A and Q is called the exactquotient of A and B.

Example 2.6 Let A = D2 + 1xD−

14 −

12x and B = D− 1

2 be given. Thecomplete division scheme for A and B is

Linear Differential Equations 23(D2 + 1

xD −14 −

12x

):(D − 1

2)

= D + 12 + 1

x

−(D2 − 1

2D)

(12 + 1

x)D −

(12 + 1

x)12

−(12 + 1

x)D +

(12 + 1

x)12

0This implies D2 + 1

xD −14 −

12x =

(D + 1

2 + 1x

)(D − 1

2). In this particular

case the remainder vanishes, i.e., the operator A may be written as a productwith the right factor B. Consequently, the equation y′′+ 1

xy′−

(14 + 1

2x)y = 0

has the solution exp (x2 ) in common with the equation y′ − 12y = 0.

Example 2.7 Let A = D3− 3xD

2 +(1+ 8

x2

)D− 2

x −10x3 and B = D− 2

x .Proceeding as above, the representation A = QB + R, with the operatorsQ = D2 − 1

xD + 1 + 2x2 and R = 1

x3 is obtained. This is easily verified byexpanding the right hand side, or by going through the steps of the divisionscheme.

The division algorithm for differential operators allows to calculate thegreatest common right divisor or Gcrd(A,B) of two given operators A andB by the Euclidian algorithm, exactly like for univariate polynomials, by uti-lizing Gcrd(A,B) = Gcrd(B,R) if A = QB + R. Because this definitionapplies division on the right, it is called greatest common right divisor. Inthis book the greatest common left divisor which may be defined similarly willnot occur. The solution space of the differential equation corresponding tothe Gcrd is the intersection of the solution spaces of the differential equationscorresponding to its arguments.

Example 2.8 Let A = D2 + 1xD−

14 −

12x and B = D2 +

(x− 1

2)D− x

2 .

A first division yields A = QB+R with Q = 1 and R =( 1x −x+ 1

2)(D− 1

2).

Dividing B by R yields a vanishing remainder, i.e., Gcrd(A,B) = D− 12 . This

result means that the two second order linear ode’s Ay = 0 and By = 0 havethe solution exp (x2 ) of y′ − 1

2y = 0 in common.

Similarly the least common left multiple or Lclm(A,B) for two differentialoperators A and B may be defined. It is the operator of lowest order thatis divided exactly by either argument from the right. The equation corre-sponding to the Lclm is the equation of lowest order with solution space thesum of its components. Its order is obviously at most the sum of the ordersof the two arguments, and it is strictly less than that if the two argumentshave a nontrivial Gcrd. Both the Gcrd and the Lclm are commutative andassociative.

Given any two operators, its Lclm may be determined as follows. Apply-ing the preceding remarks, determine its order and generate an ansatz with

24

undetermined coefficients. The conditions of divisibility generate a system oflinear algebraic equations, the solution of which determines the desired Lclm.

Example 2.9 Let A and B as in Example 2.8. Due to the first order Gcrd,the Lclm is of order three. Dividing D3 + c1D

2 + c2D + c3 by A and B, andequating the coefficients of the resulting remainder to zero leads to a systemof linear algebraic equations for c1, c2 and c3. Substituting its solution intothe third order ansatz finally yields

Lclm(A,B) = D3 +(x+ 2

x −2x− 1

2

x2 − 12x− 1

)D2

+(

34 −

1x +

12x−

94

x2 − 12x− 1

)D − 1

4x−12 +

14x+ 1

x2 − 12x− 1

.

From this construction it follows that exp(x2)

is one solution of the corre-sponding third order ode, the two remaining ones are independent solutionsof Ay = 0 and By = 0 respectively.

The general question of reducibility of a given operator or its correspondingequation, and how to obtain its factorization into irreducible components isconsidered next. An operator or an equation is called irreducible if a factor-ization into lower order operators over the same base field does not exist. Infull generality this problem has been solved at the end of the 19th centuryby Beke [9] and Schlesinger [160], vol. II, it requires to solve the so-calledassociated equations, their results are given next.

Theorem 2.4 (Beke 1894, Schlesinger 1897) Let the nth order linear ode

y(n) +Q1y(n−1) + . . .+Qny = 0 (2.20)

for y(x) be given, and let

y(m) + q1y(m−1) + . . .+ qmy = 0 (2.21)

be a possible right factor with 1 ≤ m ≤ n − 1, and N ≡(nm

). Further define

w0 ≡ W (m)(y1, . . . , ym) to be the Wronskian of (2.21), and wk ≡ W(m)k as

defined by (2.3) such that qk = (−1)kwkw0; w0 obeys the associated equation

w(N)0 + r1w

(N−1)0 + . . .+ rNw0 = 0 (2.22)

of order N ≤ N with ri ∈ Q〈Q1, . . . , Qn〉 for all i. The wk may be expressedas

wk = lk,N−1w(N−1)0 + lk,N−2w

(N−2)0 + . . .+ lk,0w0

with li,j ∈ Q〈Q1, . . . , Qn〉 for all i and j. A factor of order m of (2.20) exists

iff (2.22) has a hyperexponential solution, i.e., a solution such that w′0

w0∈ Q(x).


The proof of this theorem and more details may be found in the above

where factorization algorithms are described.

Example 2.10 Let n = 3 and m = 2, i.e., y′′′+Q1y′′+Q2y

′+Q3y = 0 andy1, y2 be a fundamental system of y′′ + q1y

′ + q2y = 0. If this latter equationis a factor of the former, the determinants

w0 =∣∣∣∣ y1 y2y′1 y

′2

∣∣∣∣ , w1 =∣∣∣∣ y1 y2y′′1 y

′′2

∣∣∣∣ and w2 =∣∣∣∣ y′1 y′2y′′1 y

′′2

∣∣∣∣obey w′0 = w1, w′1 = w2−Q1w1−Q2w0 and w′2 = −Q1w2+Q3w0. Reorderingthis system in a lex order with w2 > w1 > w0 yields

w′′′0 + 2Q1w′′0 + (Q′1 +Q2

1 +Q2)w′0 + (Q′2 +Q1Q2 −Q3)w0 = 0

and w1 = w′0, w2 = w′′0 + Q1w′0 + Q2w0, i.e., explicitly q1 = −w

′0

w0and

q2 = w′′0w0

+Q1w′0w0

+Q2.

In this book the general Beke-Schlesinger scheme is not considered further.Rather factorizations of second and third order equations are obtained by adirect method that leads to a more detailed description of the possible fac-torizations and the corresponding solutions. The subsequent lemma reducesthese factorization problems to solving Riccati equations.

Lemma 2.2 Determining the right irreducible factors of a linear homo-geneous ode up to order three with rational function coefficients amounts tofinding rational solutions of Riccati equations.

i) A second order ode y′′ + Ay′ + By = 0 has right factors of the formy′ + ay = 0 with A,B ∈ Q(x) if and only if a ∈ Q(x) satisfies

a′ − a2 +Aa−B = 0.

ii) A third order ode y′′′ +Ay′′ +By′ + Cy = 0 with A,B,C ∈ Q(x) has aright factor of the form y′ + ay = 0 if and only if a ∈ Q(x) satisfies

a′′ − 3aa′ + a3 +A(a′ − a2) +Ba− C = 0.

It has a right factor of the form y′′+by′+cy = 0 if and only if b, c ∈ Q(x)are solutions of

a′′ − 3aa′ + a3 + 2A(a′ − a2) + (A′ +A2 +B)a−B′ −AB − C = 0,

b = −(a′ − a2 +Aa−B).

Proof A first order factor y′ + ay = 0 implies

y′′ + (a′ − a2)y = 0, y′′′ + (a′′ − 3aa′ + a3)y = 0.

quoted references, see also Schwarz [162], Bronstein [21] and van Hoeij [73]

26

Reduction w.r.t. to these constraints yields immediately the given conditions.A second order factor y′′ + ay′ + by = 0 implies

y′′′ + (a′ − a2 + b)y′ + (b′ − ab)y = 0.

Reduction of the third order equation leads to the two conditions

a′ − a2 + aA+ b−B = 0, b′ − ab+ bA− C = 0.

The former one may be solved for b. If it is substituted into the latter the twoconditions of the theorem are obtained.

Example 2.11 Let the fourth order linear ode

L ≡ y(4) −(6x+

1x

)y′′′ + 12x2y′′ − (9x3 − 7x)y′ + (2x4 − 6x2)y = 0 (2.23)

be given. The corresponding third order Riccati equation has the two rationalsolutions z = x and z = 2x. They yield the two first order factors

L1 ≡ y′ − xy and L2 ≡ y′ − 2xy

with

Lclm(L1, L2) = y′′ − 3x2 + 1x

y′ + 2x2y.

Dividing it out from L yields

L3 ≡ y′′ − 3xy′ +(x2 + 3− 2

x2

)y.

This equation is irreducible. Therefore equation (2.23) has the representationL = L3Lclm(L1, L2) in terms of largest completely reducible right factors.

There is an important difference between the factorization of polynomialsand the factorization of differential operators, i.e., the latter is not unique asthe following example shows.

Example 2.12 (Landau 1902) Consider D2 − 2xD + 2

x2 . Two possiblefactorizations are(

D − 1x

)2

and(D − 1

x(1 + x)

)(D − 1 + 2x

x(1 + x)

).

More general, the factorization(D − 1

x(1 + ax)

)(D − 1 + 2ax

x(1 + ax)

)depends on a constant parameter a.


Example 2.13 Another aspect of the non-uniqueness of factorizationsmay be seen from the following representation of Landau’s operator as leastcommon multiple, e. g. in the form

Lclm(D − 2

x,D −

12

x−

32

x+ 2

)or Lclm

(D − 1

x− 1x+ 1

, D − 1x− 1x− 1

).

In general, the representation Lclm(D− 1

x−1

C + x

)with C = C1 and C = C2,

C1 6= C2 is valid. This means that the least common left multiple of anypair of first order operators corresponding to two different values C1 6= C2

yields Landau’s operator. The explanation of these phenomena is providedby Loewy’s results that are given next.

Loewy’s Decomposition. The factorization of differential operators intoirreducible factors is not unique, as Landau’s example has shown. In order toobtain a unique decomposition of a differential operator or the correspondinglinear differential equation in terms of lower order components, accordingto Loewy [127] or Ore [143] one has to proceed as follows. By applyinga factorization algorithm, determine all irreducible right factors beginningwith lowest order. The Lclm of these factors is called the largest completelyreducible part, it is uniquely determined by the given operator. If its orderequals n, the order of the differential operator, the operator is called completelyreducible and the procedure terminates. If this is not the case, the Lclm isdivided out and the same procedure is repeated with the exact quotient. Aftera finite number of steps the desired decomposition is obtained. By the abovedefinition an irreducible operator is completely reducible. This proceeding isbased on the following theorem by Loewy.

Theorem 2.5 (Loewy 1906) Let D = ddx

a derivative and ai ∈ F somedifferential field with derivative D. A differential operator

L ≡ Dn + a1Dn−1 + . . .+ an−1D + an

of order n may be written uniquely as the product of largest completely re-ducible factors L(dk)

k of order dk over the same base field in the form

L = L(dm)m L

(dm−1)m−1 . . . L

(d1)1

with d1 + . . .+ dm = n. The factors L(dk)k are unique.

The proof of this fundamental theorem may be found in the above quotedarticle by Loewy. In honor of Loewy the L(k)

j are called Loewy factors. Therightmost of them is simply called Loewy factor, it represents the largestcompletely reducible part.

In order to apply Theorem 2.5 for generating a fundamental system of thecorresponding linear ode, the further decomposition into irreducible factors isrequired as is described next.

28

Corollary 2.2 Any factor L(dk)k , k = 1, . . . ,m in Theorem 2.5 may be

written as the least common left multiple of irreducible operators l(ei)ji

withe1 + . . .+ ek = dk in the form

L(dk)k = Lclm(l(e1)j1

, l(e2)j2

, . . . , l(ek)jk

).

This result provides a detailed description of the function spaces containingthe solution of a reducible linear differential equation. A generalization ofthis result to certain systems of pde’s will be given later in this chapter.More general field extensions are studied by differential Galois theory. Goodintroductions are the above quoted articles by Kolchin, Singer and Bronstein,and the lecture by Magid [128].

For a fixed value of the order of a linear ode there is a finite set of possibilitiesdiffering by the number and the order of factors, some of which may containparameters. Each alternative is called a type of Loewy decomposition. Thecomplete answer for the most important cases n = 2 and n = 3 is listed inthe following corollaries.

Corollary 2.3 For n=2 the possible types of Loewy decompositions are

L21 : l(2), L2

2 : l(1)1 l(1)2 , L2

3 : Lclm(l(1)1 , l(1)2 ), L2

4 : Lclm(l(1)(C))

where C is a constant. All types except L22 are completely reducible.

The proof follows immediately from Lemma 2.2 and the classification ofsolutions of Riccati equations given in Theorem 2.1. A factor containing aconstant corresponds to a factorization that is not unique, any special valuefor C generates an irreducible factor. Because the originally given operatorhas order 2, two different special values must be chosen in order to representit as Lclm.

Example 2.14 The examples for the various types of Loewy decomposi-tions of second order equations are taken from Kamke’s collection, Chapter 2,they correspond to equations (2.162), (2.136), (2.201) and (2.146) respectively;as usual D = d

dx.

L21 : y′′ + 1

xy′ +

(1− ν2

x2

)y =

(D2 + 1

xD + 1− ν2

x2

)y = 0,

L22 : y′′ + 1

xy′ −

(14 + 1

2x

)y =

(D + 1

2 + 1x

)(D − 1

2

)y = 0,

L23 : y′′ +

(2 +

1x

)y′ − 4

x2 y

= Lclm(D +

2x− 2x− 2x2 − 2x+ 3

2

, D + 2 +2x− 1x+ 3

2

)y = 0,

L24 : y′′ − 6

x2 y = Lclm(D − 5x4

x5 + C− 2x

)y = 0.


These examples show clearly how a Loewy decomposition makes the so-lution space in the various functions fields completely explicit. Due to theirreducibility of all factors this representation is unique, except for type L2

4

decompositions which allow linear transformations over constants. The sameresult for equations of order three is given next.

Corollary 2.4 For n=3 the possible decompositions are

L31 : l(3), L3

2 : l(2)l(1), L33 : l(1)1 l

(1)2 l

(1)3 , L3

4 : Lclm(l(1)1 , l(1)2 )l(1)3 ,

L35 : Lclm(l(1)1 (C))l(1)2 , L3

6 : l(1)l(2), L37 : l(1)1 Lclm(l(1)2 , l

(1)3 ),

L38 : l(1)1 Lclm(l(1)2 (C)), L3

9 : Lclm(l(1)1 , l(1)2 , l

(1)3 ), L3

10 : Lclm(l(2), l(1)),

L311 : Lclm(l(1)1 (C), l(1)2 ), L3

12 : Lclm(l(1)(C1, C2))

where C, C1 and C2 are constants. The types L31, and L3

9 through L312 are

completely reducible. In the remaining cases the Loewy factor is of order lessthan three.

Whenever a constant C appears in a factor, two different special valuesmay be chosen such that the corresponding Lclm generates a second orderoperator. Similarly for the two constants in the last decomposition, threedifferent pairs of special values may be chosen such that the Lclm generatesa third order operator.

Example 2.15 Most of the examples of the various types of Loewy decom-positions of third order equations are again taken from Kamke’s collection,Chapter 3. The first three of them correspond to equations 3.6, 3.76 and 3.73respectively.

L31 : y′′′ + 2xy′ + y = 0,

L32 : y′′′ + 1

x4 y′′ − 2

x6 y =(D2 +

( 2x + 1

x4

)D + 2

x5

)(D − 2

x)y = 0,

L33 : y′′′ −

( 2x+ 1 + 2

x)y′′ +

( 6x + 4

x2 − 6x+ 1

)y′ +

( 8x −

8x2 − 4

x3 − 8x+ 1

)y

=(D − 2

x+ 1 −1x

)(D − 1

x)(D − 2

x)y = 0.

There is no type L34 decomposition in Kamke’s collection. An example is

L34 : y′′′ +

(x− 1

x+ 1 −1

x− 1 + 2x

)y′′ +

(1 + 1

x+ 1 −1

x− 1)y′

= Lclm(D + x,D + 1

x)Dy = 0.

The next four examples are equations 3.58, 3.45, 3.37 and 3.74 respectively

30

from Kamke’s collection.

L35 : y′′′ +

(14

+72x− 1

4x2

)y′′ +

( 1x

+1x2

)y′ +

12x2 y

= Lclm(D − 1

x+ C+

2x

) (D +

14− 1

2x− 1

4x2

)y = 0,

L36 : y′′′ + 4

xy′′ +

(1 + 2

x2

)y′ + 3

xy =(D + 3

x)(D2 + 1

xD + 1)y = 0,

L37 : y′′′ − y′′ −

( 1x− 2

− 1x

)y′ +

( 1x− 2

− 1x

)y

=(D +

1x− 2

+1x

)Lclm

(D − 1, D − 2

x

)y = 0,

L38 : y′′′ − 1

xy′′ + 1

x2 y′ = DLclm

(D − 2x

x2 + C

)y = 0.

The decompositions of type L39 in Kamke’s collection are more complicated.

A simple example is

L39 : y′′′ +

(x+ 1− 2x

x2 − 2− 2x+ 1x2 + x+ 1

)y′′ +

(x− 2− 2x− 1

x2 + x+ 1

)y′

−(2− 2x

x2 − 2− 2x2 + x+ 1

)y = Lclm

(D + 1, D + x,D − 2

x

)y = 0.

The next example is Kamke’s equation 3.29.

L310 : y′′′ +

3xy′′ + y = Lclm

(D2 −

(1− 2

x

)D + 1− 1

x,D + 1 +

1x

)y = 0.

There is no type L311 decomposition in Kamke’s collection. An example is

L311 : y′′′ −

(1− 3

x+

2x+ 1x2 + x− 1

)y′′ − x+ 3

x2 + x− 1y′

−( 3x− 3x+ 4x2 + x− 1

)y = Lclm

(D +

1x− 2xx2 + C

,D − 1)y = 0.

Finally there is equation 3.71 from Kamke’s collection.

L312 : y′′′−

(1

x+ 3+ 2x

)y′′+

(43x+ 2

x2− 43(x+ 3)

)y′+

(23x−

2x2− 2

3(x+ 3)

)y

= Lclm(D − 3x2 + 2C1x+ C2

x3 + C1x2 + C2x+ C2

)y = 0.

In Appendix D a fairly complete list of Loewy decompositions of the linearequations listed in Chapters 2 and 3 of Kamke’s collection is given.


Example 2.16 The type L36 decomposition considered in the above ex-

ample will be derived now by applying the procedure of Example 2.10. Thecoefficients of the given third order equation are Q1 = 4

x , Q2 = 1 + 2x2 and

Q3 = 3x . They lead to the associated equation

w′′′0 +8xw′′0 +

(1 +

14x2

)w′0 +

( 1x

+4x3

)w0 = 0

with the solution w0 = 1x . It yields q1 = 1

x and q2 = 1, i.e., the second orderfactor y′′ + 1

xy′ + y = 0.

There remains to be discussed how the solution procedure for a linear odewith a nontrivial Loewy decomposition is simplified. The general principle isdescribed in the following corollary the proof of which is obvious.

Corollary 2.5 Let a linear differential operator P of order n factor intoP = QR with R of order m and Q of order n−m. Further let y1, . . . , ym bea fundamental system for R(y) = 0, and y1, . . . , yn−m a fundamental systemfor Q(y) = 0. Then a fundamental system for P (y) = 0 is given by the unionof y1, . . . , ym and special solutions of R(y) = yi for i = 1, . . . , n−m.

Consequently, solving an inhomogeneous equation of any order with knownfundamental system for the homogeneous part requires only integrations. Atfirst this result will be applied to reducible second order equations. WithD = d

dx, y ≡ y(x) and ai ≡ ai(x), for the three nontrivial decompositions the

following fundamental systems y1 and y2 are obtained. In decomposition L23

a1 is not equivalent to a2 as defined on page 17.

L22 : (D + a2)(D + a1)y = 0,

y1 = exp (−∫a1dx), y2 = y1

∫exp (

∫(a1 − a2)dx)dx.

L23: Lclm(D + a2, D + a1)y = 0, yi = exp (−

∫aidx), i = 1, 2,

L24: Lclm(D + a(C))y = 0, yi = exp (−

∫a(Ci)dx), i = 1, 2.

There is an important difference between the last two cases. A fundamentalsystem corresponding to a L2

3 decomposition is linearly independent over thebase field, whereas this is not true for the last decomposition type L2

4. It isleft as Exercise 2.5 to derive these fundamental systems and their properties.An example for each decomposition type is given next.

Example 2.17 The decompositions of Example 2.14 are considered again.The first equation is the well-known Bessel equation with a fundamental sys-tem Jν(x) and Iν(x) in terms of series expansions. For the type L2

2 equationwith a2 = 1

2 + 1x and a1 = − 1

2 , a fundamental system is y1 = exp(x2)

and

32

y2 = exp(x2) ∫

e−x dxx . For the type L23 decomposition,

a2 = 2x −

4x− 42x2 − 4x+ 3

and a1 = 2 + 2x −

22x+ 3 .

Two independent integrations yield y1 = 23 −

43x + 1

x2 and y2 = 2x + 3

x2 e−2x.

Finally in the last case a(C) = − 5x4

x5 + C+ 2x , integration yields y = (x5+C) 1

x2

from which the fundamental system y1 = x3 and y2 = 1x2 corresponding to

C = 0 and C →∞ follows.

For differential equations of order three there are four decompositions intofirst order factors with no constants involved. Fundamental systems for themmay be obtained as follows.L3

3 : (D + a3)(D + a2)(D + a1)y = 0, y1 = exp (−∫a1dx),

y2 = y1∫

exp (∫

(a1 − a2)dx)dx, y3 = y1y2∫

exp (∫

(a2 − a3)dx)dx.

L34 : Lclm(D + a3, D + a2)(D + a1)y = 0, y1 = exp (−

∫a1dx),

y2 = y1∫

exp (∫

(a1 − a2)dx)dx, y3 = y1∫

exp (∫

(a1 − a3)dx)dx.

L37 : (D + a3)Lclm(D + a2, D + a1)y = 0, yi = exp (−

∫aidx), i = 1, 2,

y3 = y1∫

exp (∫

(a1 − a3)dx) dxa2 − a1

− y2∫

exp (∫

(a2 − a3)dx) dxa2 − a1

.

L39 : Lclm(D + a3, D + a2, D + a1)y = 0, yi = exp (−

∫aidx), i = 1, 2, 3.

The proof of these formulas is elementary, it is based on the above Corol-lary 2.5 and the expression for the solution of an inhomogeneous second orderlinear ode given in Example 2.2.

Example 2.18 The type L33 decomposition considered in Example 2.15 has

coefficients a1 = − 2x , a2 = − 1

x and a3 = 1x −

2x+ 1. If they are substituted

in the above expressions, the fundamental system y1 = x2, y2 = x2 log x andy3 = x+ x3 + x2 log (x)2 is obtained.

Example 2.19 The type L37 decomposition considered in Example 2.15 has

coefficients a1 = − 2x , a2 = −1 and a3 = 1

x + 1x− 2 . If they are substituted

in the above expressions, the fundamental system y1 = x2, y2 = ex and

y3 =x(x2 − 2)4(x− 2)

+x2

4log

x− 2x

+ ex∫e−x

dx

(x− 2)2

is obtained.

There are various decompositions into first order factors involving one ortwo constants. Their fundamental systems may be obtained analogous to thepreceding cases. The difference is that there may be dependencies over the


base field or an extension of it. All three elements of a fundamental systemcorresponding to a decomposition of type L3

12 are dependent over the basefield. Two elements are dependent over the base field for decompositions oftype L3

8 and L311, and two elements of a type L3

5 decomposition are dependentover the simple field extension of the base field generated by its first orderright factor.

There are two decomposition types involving second order irreducible fac-tors.L3

2 : (D2 +a3D+a2)(D+a1)y = 0. Let y2 and y3 be a fundamental systemof the left factor. Then

y1 = exp(−

∫a1dx

), y2 = y1

∫y2y1dx, y3 = y1

∫y3y1dx.

L36 : (D+a3)(D2 +a2D+a1)y = 0. Let y1 and y2 be a fundamental system

of the right factor, W = y′1y2 − y′2y1 its Wronskian. Then

y3 = y1∫

exp(−

∫a3dx

) y2W dx− y2

∫exp

(−

∫a3dx

) y1W dx.

Example 2.20 Consider the following equation with the type L32 decom-

position.

y′′′ +x2 + 1x

y′′ +4x2 − 4x2 y′ +

x2 − 3x

y =(D2 +

1xD +

x2 − 4x2

)(D + x)y = 0.

The left factor corresponds to the Bessel equation for n = 2 with the funda-mental system J2(x) and Y2(x). Applying the above formulas, the fundamen-tal system

y1 = exp(− 1

2x2), y2 = y1

∫J2(x) exp

(12x

2)dx, y3 = y1

∫Y2(x) exp

(12x

2)dx

in terms of integrals over Bessel functions and exponentials is obtained.

Symmetric Products and Symmetric Powers. The equations that areobeyed by the powerproducts of the elements of fundamental systems of givenlinear ode’s are important tools for understanding the field extensions gen-erated by the solutions of certain irreducible equations. They are called thesymmetric product and the symmetric power. The discussion in this subsectionis based on results by Singer and Ulmer [170, 172].

Definition 2.2 (Symmetric product, symmetric power) Let M = 0 andN = 0 be linear differential equations with fundamental systems y1, . . . , ymand z1, . . . , zn respectively. The equation L = 0 for which the productsy1z1, . . . , y1zn, . . . , ymzn generate the solution space is called the symmet-ric product and is written L = M ©s N . If M = N , this is also called thesymmetric power and written M©s 2 . The k-th power M©s k is defined corre-spondingly. Sometimes the same terminology is applied to the operators thatgenerate the respective equations.

34

Singer [170] has shown that if M has order n, the order of M©s k is nothigher than

(k+n−1n−1

). If n = 2, the order of M©s k is k + 1.

The symmetric product is associative, i.e., there holds (L1©s L2)©s L3 =L1©s (L2©s L3). Whenever a linear ode of order three or more is the sym-metric power of some linear second order ode, its solution is reduced to thislatter equation.

Example 2.21 Let L3 ≡ y′′′ − 3x2 y

′ + 3x3 y = 0 with solutions y1 = 1

x ,

y2 = x and y3 = x3 and L2 ≡ z′′ − 34x2 z = 0 with solutions z1 = 1√

xand

z2 = x√x. Then y1 = z2

1 , y2 = z1z2 and y3 = z22 . Consequently, L3 = L©s 2

2 .

In order to be applicable for a solution procedure it is necessary to know apriori that a given linear ode is the symmetric power of some other equation.In Theorem 5.21 it will be seen that the Lie symmetries of a linear equationprovide this information.

In some applications the symmetric powers of a given linear ode have to bedetermined. According to the above mentioned references one may proceedas follows. In order to construct M©s N , assume y and z are solutions of Mand N respectively. Differentiate their product w = yz such that

w′ = y′z + yz′, w′′ = y′′z + 2y′z′ + yz′′, . . . , w(k) =k∑j=0

(k

j

)y(j)z(k−j)

and reduce the result w.r.t. M and N . In the resulting expressions thereoccur at most mn different terms y(i)z(j) with i < m, j < n. Consequently,for some k these products are linearly dependent over constants, and a linearrelation for the derivatives of w results, it is the desired symmetric productequation. Its coefficients are in the base field of the original equations.

Example 2.22 (Kovacic 1986) Consider the equation y′′ +( 316x2 − 1

x)y = 0.

If y1 and y2 are two solutions one obtains for the product z = y1y2 the fol-lowing linear system for y(i)

1 y(j)2 , i, j = 0, 1.

y1y2 = z, y′1y2 + y1y′2 = z′, 2y′1y

′2 +

( 2x −

38x2

)y1y2 = z′′,( 4

x −3

4x2

)(y′1y2 + y1y

′2)−

( 2x2 − 3

4x3

)y1y2 = z′′′.

Transforming it to triangular form leads to the second symmetric power

z′′′ −( 4x −

34x2

)z′ +

( 2x2 − 3

4x3

)z = 0.

Computing symmetric powers may become fairly complex if the value of thepower increases and the coefficients of the given equation are complicated.Therefore an explicit list of symmetric powers of a second order equationL ≡ y′′ + ry = 0 is given in Appendix B on page 379. An example is thesecond symmetric power z′′′ + 4rz′ + 2r′z = 0 of y′′ + ry = 0.


Galois Theory for Second Order Equations. The structure of functionfields containing the solutions of a reducible linear second order ode is obviousfrom its Loewy decomposition L2

2, L23 or L2

4. A complete answer embracingalso irreducible equations is provided by the theory of Picard and Vessiotwhich is also called differential Galois theory. For second order equations itis explicitly described by Singer and Ulmer [172]. In the differential case,the Galois groups are subgroups of the general linear group GL2(C) overthe complex numbers C. According to Kaplansky [88], page 41, the Galoisgroup is a unimodular group contained in SL2(C) if the equation is in rationalnormal form y′′ + ry = 0. In order to obtain the possible Galois groups ofsuch equations the algebraic subgroups of SL2(C) must be known, they aredescribed next.

Theorem 2.6 The algebraic subgroups of SL2(C) may be classified, up toconjugacy, as follows.

i) The Borel group B =(

a b0 a−1

), a, b ∈ C, a 6= 0

.

ii) For m ∈ N, the groups Bm =(

a b0 a−1

), a, b ∈ C, am = 1

.

iii) The group Z =(

a 00 a−1

), a ∈ C, a 6= 0

.

iv) For m ∈ N, the cyclic groups Zm =(

a 00 a−1

), a ∈ C, am = 1

.

v) The group D =(

a 00 a−1

),

(0 1−1 0

), a ∈ C, a 6= 0

.

vi) For m ∈ N, the dihedral groups

Dm =(

a 00 a−1

),

(0 1−1 0

), a ∈ C, am = 1

.

vii) The tetrahedral group of order 24 which is generated by (θ8 = 1)

ASL24 =

⟨(i 00 −i

),

(0 1−1 0

),

1√2

(θ7 θ7

θ5 θ

)⟩.

viii) The octahedral group of order 48 which is generated by (θ8 = 1)

SSL24 =

⟨(i 00 −i

),

(0 1−1 0

),

(θ 00 θ7

),

1√2

(θ7 θ7

θ5 θ

)⟩.

36

ix) The icosahedral group of order 120 which is generated by (η5 = 1)

ASL25 =

⟨−

(η3 00 η2

),

1η2 − η3

(η + η4 1−1 −η + η4

)⟩.

x) The full unimodular group SL2(C).

A good reference for this result, including an introduction to the literature,is the article by Martins [131]. At first they are applied to reducible equations.

Theorem 2.7 (Singer and Ulmer 1993) Let L(y) ≡ y′′ + ry = 0 be areducible second order ode with r ∈ Q(x), L©s k its k−th symmetric power fork ≥ 1, and sk the number of independent nontrivial rational solutions of L©s k .The following Galois groups G(L) have to be distinguished.

i) G(L) is conjugate to the Borel group B or one of its subgroups Bm if

and only if there is a single nontrivial solution y such that y′

y ∈ Q(x).Furthermore, G(L) is conjugate to Bm if and only if sk = 0 for k =1, . . . ,m− 1 and sm = 1.

ii) G(L) is conjugate to Z or one of its finite subgroups Zm if and only if

there are two independent nontrivial solutions yi such that y′iyi ∈ Q(x)

for i = 1, 2. There holds y1y2 ∈ Q(x). Furthermore, G(L) is conjugateto Zm if and only if for k < m there holds sk = 0 for k odd, sk = 1 fork even and sm = 2 for m odd or sm = 3 for m even.

The proof of this theorem may be found in the article of Singer and Ulmer[172] on page 24. The criteria for identifying Bm and Zm are based on the tab-ulations given in Section 6.1 and 6.2 of [131]. The theorem covers the Loewydecompositions different from L2

1. The decomposition type L22 corresponds

to case i), the types L23 and L2

4 correspond to case ii).Applying this theorem, the following result for a reducible linear ode may

always be obtained: If the criteria for a fixed value of m are fulfilled, the groupis either Bm or Zm. If not, the Galois group may be either of them for a valuem greater than that for which the number of rational solutions of symmetricpowers can be determined, or it is B or Z. In order to obtain a completeanswer, a bound for the possible value of m is needed. Such a bound does notseem to be known at present. A more complete discussion of these questionsmay be found in the above quoted article by Martins [131]. For equationswith type L2

4 Loewy decomposition a complete answer is always possible.

Corollary 2.6 An equation with Galois group Z1 or Z2 has a type L24

Loewy decomposition.

The proof of this corollary is left as Exercise 2.7.


Example 2.23 The equation

L(y) ≡ y′′ +( 3

16x2 +1

4(x− 1)2− 1

4x(x− 1)

)y = 0

is discussed by Ulmer and Weil [182], page 192. Its type L22 Loewy decompo-

sition(D + 1

2(x− 1) + 14x

)(D − 1

2(x− 1) −14x

)y = 0 entails case i) of

the above theorem. A fundamental system is

y1 = (x− 1)1/2x1/4, y2 = (x− 1)1/2x1/4 log√x− 1√x+ 1

with y′1y1 = 3x− 1

4x(x− 1) ∈ Q(x) and y′2y2 not in the base field. Because y4

1 ∈ Q(x)

the Galois group is B4.

Example 2.24 The equation L(y) ≡ y′′+ 316x2 y = 0 is discussed by Ulmer

and Weil [182], page 182 and page 191. Its type L23 Loewy decomposition

Lclm(D − 1

4x,D −34x

)y = 0 yields the fundamental system y1 = x1/4 and

y2 = x3/4. There holds y1y2 = x and y41 = x. Consequently, according to the

above theorem, case ii), the Galois group is the cyclic group of order 4.

G(L) =(

1 00 1

),

(−1 00 −1

),

(i 00 −i

),

(−i 00 i

).

The next example is interesting because its solution requires the extensionof the field of constants.

Example 2.25 The equation L(y) ≡ y′′ + 716x2 y = 0 with L2

3 Loewy

decomposition Lclm(D − 2−

√−3

4x ,D − 2 +√−3

4x

)y = 0 is discussed by

Ulmer and Weil [182], page 197. A fundamental system is y1 = x(2+i√

3)/4

and y2 = x(2−i√

3)/4. For k ≤ 48, the symmetric powers L©s k have no rationalsolution for k odd, and a single rational solution for k even. By the aboveTheorem 2.7, case ii), the conditions for Zm are not satisfied for m ≤ 48.Consequently, the Galois group is either Zm with m > 48 or Z.

From the listing in Appendix D it is obvious that most second order linearode’s given there fall under the category covered by the above theorem. TheGalois groups of irreducible second order equations with type L2

1 decomposi-tion are discussed next. As far as possible, the conditions for the various typesof Galois groups are expressed in terms of solutions of symmetric powers.

Theorem 2.8 (Singer and Ulmer 1993, Ulmer and Weil 1996) Let L(y) ≡y′′ + ry = 0 be an irreducible second order ode with r ∈ Q(x). The followingGalois groups G(L) have to be distinguished; sk is the number of independentnontrivial rational solutions of L©s k = 0.

38

i) G(L) is conjugate to D if and only if there holds sk = 0 for k 6= 0 mod(4)and sk = 1 for k = 0 mod(4) for all k. It is conjugate to Dm if and onlyif for k < 2m the same conditions hold as in the previous case, s2m = 1for m even and s2m+2 = 2 for m odd.

ii) G(L) is conjugate to the tetrahedral group ASL24 if and only if sk = 0 for

k ≤ 5, s6 = 1.

iii) G(L) is conjugate to the octahedral group SSL24 if and only if sk = 0 for

k ≤ 7, s8 = 1.

iv) G(L) is conjugate to the icosahedral group ASL25 if and only if sk = 0

for k ≤ 11, s6 = 12.

v) G(L) is the full group SL2(C) if none of the preceding cases holds.

These criteria allow determining a finite Galois group of an irreducible sec-ond order equation algorithmically. For the imprimitive groups D and Dm

of case i) the same remarks on page 36 apply as for the groups of reducibleequations covered by Theorem 2.7.

In order to obtain algebraic solutions of a second order equation explicitily,the following result of Ulmer and Weil [182] is useful.

Theorem 2.9 (Ulmer and Weil 1996) Let L(y) = y′′ + r(x)y = 0 bean irreducible second order ode with r ∈ k, k some differential field, and let

u ≡ y′

y . Then the zeroes of P (u) = um +∑m−1i=0 biu

i with bi ∈ k are solutionsof the Riccati equation u′+u2 +r = 0 if and only if the coefficient bm−1 is thenegative logarithmic derivative of an exponential solution of L©s m = 0. Thecoefficients bi of P (u) are determined from the system

bm = 1, bi−1 =1

m+ 1− i[(i+ 1)rbi+1 − b′i + bm−1bi

]for 0 < i ≤ m− 1. In particular for m = 2 there holds b0 = r − 1

2 (b′1 − b21).

The proof may be found in the article by Ulmer and Weil [182]. The specialcase m = 2 is considered in Exercise 2.8.

Example 2.26 The equation considered in Example 2.22 is irreduciblewhereas its second symmetric power is reducible. Consequently, its Galoisgroup is imprimitive. The right factor z′ − 1

2xz = 0 of the second symmetric

power yields the solution z =√x from which b1 = 1

2x and b0 = 116x2 − 1

x are

obtained, i.e., P (u) = u2 − 12xu + 1

16x2 − 1x with solutions u1,2 = 1± 4

√x

4x .

They yield the fundamental system y1,2 = x1/4e±2√x.


Example 2.27 Equation 2.290 from Kamke’s collection

y′′ +27x

27x2 + 4y′ − 3

27x2 + 4y = 0

is irreducible with rational normal form y′′ + 14

405x2 − 264(27x2 + 4)2

y = 0. Its second

symmetric power is reducible, its type L310 Loewy decomposition has a first

order right factor z′ − 27x27x2 + 4

z = 0. Consequently, its Galois group is

imprimitive. By Theorem 2.9

P (u) = u2 +27x

27x2 + 4u+

34

135x2 − 1627x2 + 4

is obtained. Its zeros u1,2 = 12 (x2+ 4

27 )−1(

32x±

√x2 + 4

27

)yield the solutions

y1,2 = (x2 + 427 )1/4

(x +

√x2 + 4

27

)±1/3

of the equation in rational normal

form, and finally the solutions y1,2 =(x +

√x2 + 4

27

)±1/3

of the original

equation are obtained.

Example 2.28 Airy’s equation y′′ − xy = 0 is discussed by Singer andUlmer [172]. Its 6th symmetric power

y(7)−56xy(5)−140y(4) +784x2y′′′+2352y′′−4(576x3−295)y′−3456x2y = 0

is irreducible as shown by these authors. By Theorem 2.8, case v) the Galoisgroup of Airy’s equation is SL2(C).

At this point the discussion of linear ode’s is complete. For a second orderequation a nontrivial Loewy decomposition provides the complete answer forits solutions in terms of integrations. This implies that the Galois group of theequation is a proper subgroup of the respective linear group. If the equation isirreducible there may be a finite Galois group leading to algebraic solutions.If the Galois group is SL2(C), there may be some interesting informationon the solutions in terms of transcendents that have been introduced in theliterature in terms of so-called special functions like e. g. Bessel functions,Legendre functions etc. To recognize them amounts to solving an equivalenceproblem which is considered in Section 4.1 on page 170. If none of thesealternatives applies, its solutions have to be investigated numerically or bygraphical methods or, if it is interesting enough for large classes of problems,new transcendents may be introduced. In principle these remarks may begeneralized to linear ode’s of any order.

The collection of solved ode’s by Kamke [87] contains more than 400 linearequations of second order in Chapter C.2, and about 80 linear equation ofthird order in Chapter C.3. Most of them are listed in Appendix D of thisbook, and their Loewy decomposition is given if it is nontrivial, i.e., if it isnot of type L2

1 or L31.

40

2.2 Janet’s Algorithm

The theory of systems of linear homogeneous partial differential equations(pde’s) is of interest in its own right, independent of its applications for findingsymmetries and invariants of differential equations. Any such system maybe transformed in infinitely many ways by adding linear combinations of itsmembers or derivatives thereof without changing its solution space. In generalit is a difficult question whether there exist nontrivial solutions at all, or whatthe degree of arbitrariness of the general solution is. It may consist of a finiteset of constants, in which case there is a finite dimensional solution space overthe constants. This is true for the systems considered in this book originatingfrom symmetry analysis of ode’s. On the other hand, the general solution mayinvolve one or more functions depending on various numbers of arguments.The largest of these numbers is called the differential type, the number offunctions depending on them is called the typical differential dimension byKolchin [91], combined they have been baptized the gauge [62].

These questions were the starting point for Maurice Janet. He introduceda unique representation for these systems nowadays known as a Janet basis,similar to a Grobner basis representation of a system of algebraic equations.There are several advantages to using a Janet basis representation for a sys-tem of linear homogeneous pde’s. The degree of arbitrariness of its generalsolution, i.e., its gauge, is easily obtained from it. Due to the uniqueness of thecoefficients of a Janet basis, additional invariants may be obtained providingfurther information on the solutions.

Both Grobner bases and Janet bases have an important feature in common.Except in very simple cases it is virtually impossible to calculate any of themby pencil-and-paper, i.e., an efficient computer algebra implementation is cru-cial for handling concrete problems. A first example will make some of thesepoints clear.

Example 2.29 Consider the following linear homogeneous systems for twodifferential indeterminates z and w depending on x and y. The first systemis defined by fi = 0, i = 1, . . . , 5 where

f1 ≡ wy + x2y(x2 + y)

zy − 1yw, f2 ≡ zxy + y

xwy + 2yx zx,

f3 ≡ wxy − 2xy zxx −

xy2wx,

f4 ≡ wxy + zxy + 12ywy −

1ywx + x

y zy −1

2y2w,

f5 ≡ wyy + zxy − 1ywy + 1

y2w.

(2.24)


And the second one is gi = 0, i = 1, . . . , 4 where

g1 ≡ zyy + 12y zy, g2 ≡ wxx + 4y2wy − 8y2zx − 8yw,

g3 ≡ wxy − 12zxx −

12ywx − 6y2zy,

g4 ≡ wyy − 2zxy − 12ywy + 1

2y2w.

(2.25)

A priori, there does not seem to be any connection between systems (2.24) and(2.25). In particular it is not obvious whether there is any relation betweenthe space of solutions they describe. The answer is easily obtained, however,if they are written as

f1 = x2y(x2 + y)

e2 + e4, f2 = ∂∂xe2 + y

xe4 + 2yx e1,

f3 = ∂∂ye3 − 2x

y∂∂xe1,

f4 = ∂∂xe2 + ∂

∂ye3 + 1

2y e4 −1y e3 + x

y e2, f5 = ∂∂xe2 + ∂

∂ye4

(2.26)

andg1 = ∂

∂ye2 + 1

2y e2, g2 = ∂∂xe3 + 4y2e4 − 8y2e1,

g3 = ∂∂ye3 − 1

2∂∂xe1 − 6y2e2 − 1

4y e3,

g4 = ∂∂ye4 − 2 ∂

∂xe2 + 1

2y e4

(2.27)

where

e1 ≡ zx + 12yw, e2 ≡ zy, e3 ≡ wx, e4 ≡ wy − 1

yw. (2.28)

From this representation it is clear that both systems (2.24) and (2.25) allowall solutions of the system e1 = 0, e2 = 0, e3 = 0, e4 = 0. Explicitly, they arez = C1+C2x and w = −2C2y, where C1 and C2 are the integration constants,i.e., they form a two-dimensional vector space over the constants. It will beseen later on that there are no more solutions of (2.24) and (2.25).

Modules over Rings of Differential Operators. From the above exampleit is apparent that the object of primary interest is not an individual systemof pde’s, but the module that is generated from it over an appropriate ringof differential operators; they are called D-modules. This situation is similarto commutative algebra where the ideals generated by sets of polynomials areclosely related to the algebraic manifolds described by the corresponding setof algebraic equations. It turns out that the language of differential algebraapplied in the books by Kaplansky [88] and Kolchin [91] is the proper settingfor this discussion. Good introductions to D-modules are the articles by Oakuand Shimoyama [139] and Quadrat [150]. A few basic results from thesereferences will be given next.

42

An operator δ on a ring R is called a derivation operator if δ(a + b) =δ(a) + δ(b) and δ(ab) = δ(a)b + aδ(b) for all elements a, b ∈ R. A ring witha single derivation operator is called an ordinary differential ring. If there isa finite set ∆ of derivation operators such that δδ′(a) = δ′δ(a) for a ∈ R,δ, δ′ ∈ ∆ the ring is called a partial differential ring. Analogous definitionsapply for an ordinary differential field and a partial differential field if the ringR bis replaced by a field F .

Let Θ denote the commutative semigroup generated by the elements of∆. If it is written multiplicatively, its elements can be expressed in the formθ =

∏δ∈∆ δ

e(δ), where each exponent e(δ) is a natural number; e(δ) = 0defines the identity. The sum

∑δ∈∆ e(δ) is called the order of θ. The elements

of Θ are called derivative operators or simply derivatives.In this book most frequently rings of differential operators with derivation

operators ∂i = ∂∂xi

and rational function coefficients will be applied, they aredenoted by D = Q(x1, . . . , xn)[∂1, . . . , ∂n]. Its elements have the form

d =∑

ri1,...,in(x1, . . . , xn)∂i11 . . . ∂inn (2.29)

where ri1,...,in ∈ Q(x1, . . . , xn), almost all ri1,...,in equal to zero. The ring Dis noncommutative with commutation rules ∂ia = a∂i + ∂a

∂xifor 1 ≤ i ≤ n

and a ∈ Q(x1, . . . , xn).An m-dimensional left vector module Dm over D has elements (d1, . . . , dm)

with di ∈ D for all i. The sum of two elements of Dm is defined by com-ponentwise addition, multiplication with ring elements d by d(d1, . . . , dm) =(dd1, . . . , ddm).

The relation between the submodules of Dm and systems of linear pde’s isestablished as follows. Let (u1, . . . , um)T be an m-dimensional column vectorof differential indeterminates. Then the product

(d1, . . . , dm)(u1, . . . , um)T = d1u1 + d2u

2 + . . .+ dmum (2.30)

defines a linear differential polynomial in the u’s that may be considered asthe left hand side of a partial differential equation; u1, . . . , um are called thedependent variables or functions, depending on the independent variablesx1, . . . , xn.

A N × m matrix ci,j, i = 1, . . . , N , j = 1, . . . ,m, ci,j ∈ D, defines asystem of N linear homogeneous pde’s

ci,1u1 + . . .+ ci,mu

m = 0, i = 1, . . . , N. (2.31)

The ith equation of (2.31) corresponds to the vector

(ci,1, ci,2, . . . , ci,m) ∈ Dm for i = 1, . . . , N. (2.32)

This correspondence between the elements of Dm, the differential polynomials(2.30) and its associated pde’s(2.31) allows turning from one representation to


the other whenever it is appropriate. The module generated by these vectorsdetermines the solutions of the system.

Example 2.30 In Example 2.29 there holds n = m = 2. The elements(∂y −

1y,

x

2y(x2 + y)∂y

),

( y

x∂y, ∂xy +

2yx∂x

),

(∂xy −

x

y2 ∂x,2xy∂2x

),

(∂xy +

12y∂y −

1y∂x −

12y2 , ∂xy +

x

y∂y

),

(∂2y −

1y∂y +

1y2 , ∂xy

)generate the submodule of D2 corresponding to system (2.24). The samesubmodule is generated by( 1

2y, ∂x

), (0, ∂y), (∂x, 0),

(∂y −

1y, 0

)corresponding to system (2.28). The respective equations are obtained byapplying the module elements to

(wz

)with w, z differential indeterminates

with nonvanishing derivatives w.r.t. x and y.

Given any module M ⊂ Dm, there are two fundamental problems associ-ated with it.

Membership in M. Establish a procedure for deciding membership inM. This problem will be solved by constructing a Janet basis forM.

Structure of M. Determine all maximal submodules N with the prop-erty M⊂ N ⊆ Dm.

Answering these questions will be the subject of the remaining part of thissection.Term Orders and Rankings. Just like an ordering of monomials is afundamental prerequisite for defining a Grobner basis in commutative algebra,an ordering of partial derivatives is necessary for defining canonical forms fordifferential operators and differential equations. At first it will be given in thelanguage of differential polynomials and pde’s.

Definition 2.3 (Ranking) A ranking of derivatives is a total ordering suchthat for any derivatives δ, δ1 and δ2, and any derivation operator θ there holdsδ ≤ θδ, δ1 ≤ δ2 ⇒ δδ1 ≤ δδ2.

A term ti,j in a differential polynomial (2.31) is a differential indeterminateuα or a derivative of it, multiplied by an element of the base field. The lefthand side of a linear homogeneous pde is a sum of terms. For such a pde,the ordering of partial derivatives defined above generates a unique order forits terms. A system of linear homogeneous pde’s with terms ti,j is alwaysarranged in the form

44

t1,1 > t1,2 > . . . > t1,k1∧t2,1 > t2,2 > . . . > t2,k2∧...∧tN,1 > tN,2 > . . . > tN,kN

such that the ordering relations are valid as indicated. Each line of thisscheme corresponds to a differential polynomial or a differential equation ofthe system, its terms are arranged in decreasing order from left to right. Inorder to save space, sometimes several equations are arranged into a single line.In these cases, in any line the leading terms increase from left to right. Theterms in the rectangular boxes are the leading terms containing the leadingderivative, i.e., the leading term is that term preceding any other term in therespective differential polynomial or equation w.r.t. the given ordering. Forany given system of pde’s there is a finite number of term orderings leadingto different arrangements of terms, a trivial upper bound being the numberof permutations of its terms. In any term ordering there is a lowest termwhich is always one of the dependent variables corresponding to e(δ) = 0. ByDickson’s lemma [33], Chapter 2, $ 4, any decreasing sequence of derivativesmust terminate.

In order to generate the above mentioned form of a system of pde’s effec-tively it is advantageous to have a predicate to decide whether or not any pairof derivatives is in correct order. To this end, a matrix of weights (matrice decotes) has been introduced by Riquier [153]. The subsequent discussion fol-lows closely his original work and a more complete and systematic treatmentof the subject by Thomas [176].

First of all a fixed enumeration (u1, . . . , um) is chosen for the dependent and(x1, . . . , xn) for the independent variables. To any derivative of a function uα

a matrix I of weights is attached comprising a single column as follows.

∂uα

∂xi11 ∂xi22 . . . ∂xinn

←→ I ≡ (i1, i2, . . . , in, 0, . . . , 1, . . . , 0)T . (2.33)

The position of the element 1 is the (n + α)th row. The function uα itselfcorresponds to a matrix with i1 = i2 = . . . = in = 0. For a partial derivative(2.33), (i1, . . . , in) is called the derivative vector. Let J be the matrix ofweights associated with some other derivative of a function uβ . Riquier’smethod of placing these derivatives in a definite order is based on the followingrules. Let there be given a matrix M with n +m columns corresponding tothe independent and the dependent variables, and any number of rows. Bydefinition the derivative represented by I is higher than the one represented


by J if the first nonzero element in the column M(I − J) is positive, and viceversa if it is negative.

The particular form of the matrix M determines the type of ordering.Thomas [153] discussed the possible selections in full generality. He describedthe most general matrix leading to a given ordering and gave a canonical formfor a single representative. For the applications in this book, basically threetypes of orderings are applied, a lexicographic ordering, a graded lexicographicordering and a graded reverse lexicographic ordering abbreviated by lex, grlexand grevlex respectively.

In the lexicographic ordering, derivatives are ordered first by functions suchthat ui > uj if i < j in the arrangement (u1, . . . , um). If the functionscontained in two derivatives are identical to each other, the order of its partialderivatives is determined like in the lexicographic term order of two algebraicmonomials, i.e., xi > xj if i < j in the arrangement (x1, . . . , xn). The powerof an individual variable in the algebraic case corresponds to the order ofthe derivative w.r.t. this variable. The ordering according to these rules isachieved by the matrix

Mlex =

0 0 . . . 0 m m− 1 . . . 11 0 . . . 0 0 0 . . . 00 1 . . . 0 0 0 . . . 0

......

... . . .0 0 . . . 1 0 0 . . . 0

whereupon the lower left n× n corner is the n-dimensional unit matrix.

The graded lexicographic ordering grlex is obtained if the total ordersof the two derivatives are compared first. If they are different from eachother, the higher one precedes the other. If not, the above lex order is ap-plied. The corresponding matrix Mgrlex is obtained from Mlex if the first line1 1 . . . 1 0 0 . . . 0 is included, i.e., if it has the form

Mgrlex =

1 1 . . . 1 0 0 . . . 00 0 . . . 0 m m− 1 . . . 11 0 . . . 0 0 0 . . . 00 1 . . . 0 0 0 . . . 0

......

... . . .0 0 . . . 1 0 0 . . . 0

.

Finally, the graded reverse lexicographic ordering grevlex at first comparesthe total order of the derivatives like in the grlex ordering. If they are equalto each other, they are ordered by the functions. If they are identical, deriva-tive orders w.r.t. individual variables are compared, beginning with the lastvariable. The first pair with a different value decides the order, the deriva-tive with the lower value is higher than the other one. The following matrixgenerates this ordering.

46

Mgrevlex =

1 1 . . . 1 0 0 . . . 00 0 . . . 0 m m− 1 . . . 10 0 . . . −1 0 0 . . . 0

......

... . . .0 −1 . . . 0 0 0 . . . 0−1 0 . . . 1 0 0 . . . 0

.

There are numerous variations of these orderings in addition to the obviouspermutations of dependent and independent variables among themselves. Inany lexicographic ordering, for example, the independent variables may becompared ahead of the dependent ones, or a combination of lex and grlexorderings may be applied.

The generation of orderings by weight matrices provides an easy algorithmicway for establishing it for any number of derivatives. The following algorithmHigher? establishes a reflexive ordering of partial derivatives if a matrix ofweights is given. Because in a linear homogeneous system each term contains aunique derivative, it may also be applied for ordering the terms of its members.

Algorithm 2.3 Higher?(d1, d2,M). Given two derivatives d1 and d2, anda matrix M of weights, true is returned if d2 does not precede d1.S1 : Equality? If d1 = d2 return true.S2 : Generate cotes. Set up matrices I1 and I2 for d1 and d2 accordingto (2.33).S3 : Apply M. Generate the column M(I1− I2). If the first nonvanishingelement is positive, return true and false otherwise.

Example 2.31 Consider a problem with m = n = 2, comprising twofunctions (w, z) depending on (y, x). For the lexicographic orderings w > z,y > x and derivatives not higher than two, (2.33) yields

z zx zy zxx zxy zyy0001

1001

0101

2001

1101

0201

for the derivatives of z. The corresponding vectors for w are obtained if thelast two elements 0, 1 in each column are replaced by 1, 0. Then the matrix

M =

1 1 0 00 0 2 11 0 0 00 1 0 0

generates the graded lexicographic ordering

wyy > wxy > wxx > zyy > zxy > zxx > wy > wx > zy > zx > w > z.


The matrix

M =

0 0 2 11 0 0 00 1 0 0

generates the ordering

wyy > wxy > wxx > wy > wx > w > zyy > zxy > zxx > zy > zx > z

which is a lexicographic ordering.

The orderings described so far may be uniquely translated into moduleorderings for elements of Dm if the transformation rules described in thepreceding subsection on pages 41-43 are applied.

Example 2.32 The orderings of the preceding example are reconsidered.Applying the above rules, for w > z, y > x, w = (1, 0) and z = (0, 1) thegraded lexicographic ordering leads to

(∂yy, 0) > (∂xy, 0) > (∂xx, 0) > (0, ∂yy) > (0, ∂xy) > (0, ∂xx) >

(∂y, 0) > (∂x, 0) > (0, ∂y) > (0, ∂x) > (1, 0) > (0, 1)

for the elements ofD2 of order not higher than two. Similarly the lexicographicordering yields

(∂yy, 0) > (∂xy, 0) > (∂xx, 0) > (∂y, 0) > (∂x, 0) > (1, 0) >

(0, ∂yy) > (0, ∂xy) > (0, ∂xx) > (0, ∂y) > (0, ∂x) > (0, 1).

The latter order considered in this example is the POT order which hasbeen introduced by Adams and Loustaunau [3] for modules over polynomialrings. Their TOP order does not have an exact counterpart among the ordersdiscussed above.

Reduction and Autoreduction. The first fundamental operation of Janet’salgorithm is the reduction. Given any pair of linear differential polynomials e1and e2, it may occur that the leading derivative of e2, or of a suitable derivative∂e2 of it, equals some derivative in e1. This coincidence may be applied toremove the respective term from e1 by an operation which is called a reductionstep. To this end, multiply the appropriate derivative ∂e2 by the coefficientof its leading derivative in the term containing it in e1, and subtract it frome1 multiplied by the leading coefficient of e2. If e2 is monic which is usuallythe case, this latter operation may be skipped, otherwise the result has to bemade monic. In general, reductions may occur several times. However, due tothe properties of the term orderings described above and the genuine loweringof terms in any reduction step, by Dickson’s lemma the following algorithmalways terminates after a finite number of iterations. After its completion nofurther reductions are possible. In this case e1 is called completely reduced

48

with respect to e2. For any linear differential polynomial e the abbreviations

Lder(e), Lterm(e), Lcoef(e), Lfun(e)

are applied to denote the leading derivative, the leading term, the leading co-efficient or the leading function of e. Coefficient(e, t) denotes the coefficientof the term t contained in e.

Algorithm 2.4 Reduce(e1, e2). Given two monic linear differential poly-nomials e1 and e2, e1 is returned completely reduced w.r.t. e2.S1 : Reduction possible? Search for a term t in e1 the derivative of whichmay be obtained from the leading derivative of e2 by applying someappropriate ∂. If none is found, return e1.

S2 : Reduction step. Set e1 := e1 − Coefficient(e1, t)∂e2 and e1 :=Monic e1, then goto S1.

Example 2.33 Let e1 ≡ zy−x2

y2 zx−x− yy2 z, f1 ≡ zx+ 1

xz and f2 ≡ zy+1y z.

Reduction of e1 w.r.t. f1 yields in a single step zy + 1y z, whereas reduction

w.r.t. f2 yields −x2

y2 zx − xy2 z.

For some applications, a slightly more general algorithm for reduction isdesirable, reducing a given equation w.r.t. to a system of pde’s instead of asingle one.

Algorithm 2.5 Reduce(e, S). Given a linear differential polynomial e anda system of linear monic polynomials S = f1, f2, . . ., e is returned reducedw.r.t. S.S1 : Reduction possible? Search for a polynomial fi ∈ S w.r.t. which emay be reduced. If none is found, return e.

S2 : Reduction performed. Set e := Reduce(e, fi) and goto to S1.

It is obvious that its action is achieved by repeated application of the precedingalgorithm Reduce. Termination follows from the finiteness of the system Sand the termination of the former algorithm. If a differential polynomial hasbeen reduced w.r.t. to a given system, it is called in normal form w.r.t. tothis system. In general the result is not unique.

Example 2.34 Let e1, f1 and f2 be defined as in Example 2.33. If now e1is reduced w.r.t. to the system f1, f2, the result is zero, independent of theorder in which the reductions are performed.

Example 2.35 Consider f4 in (2.24) and its reduction w.r.t. the systemcomprising f1 and f2. Reduction w.r.t. to f1 is achieved in two steps. At firstthe leading derivative wxy is removed by subtraction of ∂f1

∂x. After division


by the leading coefficient the result is

f4a ≡ zxy +x2 + y

2x2y − x+ 2y2wy

+2x5 + 4x3y + x2 + 2xy2 − y(2x2y − x+ 2y2)(x2 + y)

zy −x2 + y

(2x2y − x+ 2y2)yw.

In the second step the term proportional to wy is removed by subtraction ofa proper multiple of f1. This leads to

f4b ≡ zxy +4x5y + 8x3y2 − x3 + 2x2y + 4xy3 − xy − 2y2

(2x2y − x+ 2y2)(x2 + y)yzy.

The result f4b may be further reduced w.r.t. to f2. This yields

f4c ≡ wy −4x6y + 8x4y2 − x4 + 2x3y + 4x2y3 − x2y − 2xy2

(2x2y − x+ 2y2)(x2 + y)y2 zy + 2zx.

Now there is again a reduction possible w.r.t. f1, which yields the answer

f4d ≡ zy−4(2x2y − x+ 2y2)(x2 + y)y2

(4x5y + 8x3y2 − x3 + 2x2y2 + 2x2y + 4xy3 − 2xy + 2y3 − 2y2)xzx

− (2x2y − x+ 2y2)(x2 + y)y(4x5y + 8x3y2 − x3 + 2x2y2 + 2x2y + 4xy3 − 2xy + 2y3 − 2y2)x

w = 0.

At this point f4d cannot be further reduced by any other equation.

For any given system, in general it is possible that some elements may bereduced w.r.t. the remaining ones. Janet’s algorithm requires that at first allsuch reductions be performed. A system with the property that there is noreduction possible of any element w.r.t. to any other in the system is calledan autoreduced system.

The following algorithm accepts any ordered system of linear differentialpolynomials as input and returns the corresponding autoreduced system bycalling the algorithm Reduce(ei, ej) repeatedly. These calls are organized asfollows. The lowest element of the given system which may be applied forany reduction is selected and all reductions w.r.t. this element are performed.The resulting system is treated in the same way until no further reduction ispossible in the full system, and the result is returned.

Algorithm 2.6 AutoReduce(S). Given an ordered system of linear monicdifferential polynomials S = e1, e2, . . ., a new system is returned, such thateach element is completely reduced w.r.t. to any other member of the system.S1 : Reduction possible? Find the first element ek in S w.r.t. which anyreduction may be performed in ek+1, ek+2 . . .. If none is found, returnthe system S unchanged.

50

S2 : Perform reductions. SetS := e1, e2, . . . ek ∪ Reduce(ej , ek)|j = k + 1, k + 2, . . .,

reorder S properly and return to S1.

Termination is assured due to the finiteness of the system and the fact thatany call to Reduce in step S2 generates a result with at least a single termreplaced by a lower one.

Example 2.36 Again system (2.24) is considered. In the first iteration ofAutoReduce the lowest element f1 is applied for removing all partial derivativesof w w.r.t. y in the remaining elements with the result

wy + x2x2y + 2y2 zy − 1

yw, zxy − 12x2 + 2y

zy + 2yx zx + 1

xw,

zxy + 4x5y + 8x3y2 − x3 + 2x2y + 4xy3 − xy − 2y2

4x4y2 − 2x3y + 8x2y3 − 2xy2 + 4y4 zy,

zxy + (4x2 + 4y)zxx + 2x3 − 2x2y + 2xy − 2y2

xy wx − x2 − yx3 + xy

zy,

zyy − 2x2y + 2y2

x zxy − x2 + 2yx2y + y2 zy.

The derivatives of zxy are removed by means of the second element next.

wy + x2x2y + 2y2 zy − 1

yw,

zy −8x4y3 − 4x3y2 + 16x2y4 − 4xy3 + 8y5

4x6y + 8x4y2 − x4 + 2x3y2 + 2x3y + 4x2y3 − 2x2y + 2xy3 − 2xy2 zx

− 4x4y2 − 2x3y + 8x2y3 − 2xy2 + 4y4

4x6y + 8x4y2 − x4 + 2x3y2 + 2x3y + 4x2y3 − 2x2y + 2xy3 − 2xy2w,

zxx + x− y2xy wx − 2x2 − x− 2y

8x5 + 16x3y + 8xy2 zy −y

2x3 + 2xyzx − 1

4x3 + 4xyw,

zxy − 12x2 + 2y

zy + 2yx zx + 1

xw,

zyy − x3 + x2y2 + 2xy + y3

x3y + xy2 zy + 4x2y2 + 4y3

x2 zx + 2x2y + 2y2

x2 w.

The next six iterations generate systems that are too large to be reproducedhere. After one more iteration, the lowest element becomes zx+ 1

2yw, and anadditional iteration generates the final result

zx + 12yw, zy, wx, wy + 1

yw.

Completion and Integrability Conditions. Although autoreduction al-ready removes a lot of arbitrariness from any sytem of differential polynomials,


there are still some shortcomings which disqualify them from being consideredas canonical form. In the first place, the outcome of autoreduction is highlynon-unique as the result of Example 2.36 and its comparison with the autore-duced system (2.25) shows. Furthermore, differentiating in (2.25) g2 w.r.t. yand g3 w.r.t. x, and reducing the result w.r.t. to the remaining elements ofthe system, the following result is obtained.

∂g2∂y≡ wxxy + 2ywy − 16yzx − 10w,

∂g3∂y≡ wxxy − 1

2zxxx − 6y2zxy − 14y zxx −

14y2wx − 3yzy.

Either of them may be considered as a representation of the leading deriva-tive wxxy in terms of its reductum which do not coincide. It is obvious thatinfinitely many instances like this may be generated by suitable differentia-tions. These apparent inconsistencies already hold the clue on how to pro-ceed further, i.e., an autoreduced system must be supplemented by additionalequations such that these inconsistencies disappear.

In order to obtain a finite problem, according to Janet [83] certain ele-ments which make it into a complete system have to be included. To thisend, at first the full system of pde’s is partitioned into subsystems with thesame function in the leading derivative. By definition, a system of lineardifferential polynomials of this kind is called complete if the derivative vec-tors of the leading terms in any subsystem form a complete set in the sensethat it is defined in Appendix C for degree vectors of monomials. Becausedifferentiation corresponds to multiplication by the respective variable, thesubsequently described completion algorithm is a straightforward generaliza-tion of the completion process for systems of monomials that is described inAppendix C.

Algorithm 2.7 CompleteSystem(S). Given an ordered autoreduced sys-tem of linear differential polynomials S with leading derivative vectors D, thecomplete system corresponding to S is returned.S1 : Separate. Represent S in the form S := S1 ∪ S2 ∪ . . .∪ Sk such thatthe elements in each Si contain the same leading function with leadingderivatives Di.S2 : Completion. For each i set Di := Complete(Di).S3 : Generate systems. For each Si generate the elements correspondingto a Di and assign it to Si.S4 : Generate complete system. Collect the Si, assign them toS := S1, S2, . . . Sk and return Reorder(S)

Termination of this process is assured by the finiteness of the completionalgorithm Complete(u) for systems of monomials described in Appendix C.

Consider any linear homogeneous system, choose two elements with thesame leading function and solve both with respect to its leading term. By

52

suitable cross differentiation it is always possible to obtain two new polyno-mials such that the derivatives on the left hand sides are identical to eachother. Intuitively it is expected that the same should hold true for the righthand sides if all reductions w.r.t. to the remaining elements of the system areperformed. This consideration justifies the following definition.

Definition 2.4 (Integrability condition). Let e1 and e2 be two elementsof a system of linear differential polynomials. Let ∂ and ∂′ be derivativesof minimal order that make their leading derivatives identical to each other.Then the constraint

Lcoef(∂′e2)∂e1 − Lcoef(∂e1)∂′e2 = 0is called an integrability condition for the system S.

The integrability conditions correspond to the S-pairs in a Grobner basiscalculation. A priori there are infinitely many of them in any given systemof pde’s. It is one of the achievements of Janet to identify a finite numberof them such that the remaining ones follow. This is the subject of the nexttheorem.

Theorem 2.10 (Janet 1920) Let S = e1, e2, . . . be an autoreduced com-plete system of linear differential polynomials. In order to satisfy all integra-bility conditions of this system it is sufficient to satisfy

∂ei∂xk− ∂m1+...+mpej

∂xm1i1

. . . ∂xmp

ip

= 0 (2.34)

provided that the leading derivatives of the two expressions at the left handside are identical to each other, xk is a nonmultiplier for ei and xi1 , . . . , xipare multipliers for ej.

Proof In a complete system the leading derivatives of a function parame-trize the classes of all principal derivatives. Derivatives w.r.t. the multipliervariables reproduce the derivatives of each class. Therefore in order to ob-tain any principal derivative in two different ways, at least a single derivationw.r.t. to a nonmultiplier variable must be involved. Let xk be a nonmulti-plier variable for ei. The leading derivatives of the derived element must becontained in some other class. Let the leading derivative of ej parametrizethis class. Therefore it must be possible by deriving it w.r.t. to its multipliervariables a suitable number of times to generate the same leading derivativeas in ∂ei

∂xk. Let this requirement be fulfilled for (2.34). Due to the uniqueness

of the representation of principal derivatives, any other element satisfying thiscondition has the form

∂n1+...+npej∂xn1

i1. . . ∂x

np

ip

(∂ei∂xk− ∂m1+...+mpej∂xm1

i1. . . ∂x

mp

ip

)= 0.

The above condition (2.34) assures their vanishing.


It may be said somewhat loosely that a system of pde’s for which all integra-bility conditions are satisfied contains all elements explicitly that are essentialfor determining the general solution. This property is of such a fundamentalimportance that a special term is introduced for it.

Definition 2.5 (Janet basis). For a given term ordering an autoreducedsystem of linear differential polynomials is called a Janet basis if all integra-bility conditions reduce to zero.

An autoreduced system satisfying all integrability conditions is also calledcoherent. The following algorithm that accepts any system of linear differen-tial polynomials and transforms it into a new system such that all integrabilityconditions are satisfied is due to Janet [83].

Algorithm 2.8 JanetBasis(S). Given a system of linear differential poly-nomials S = e1, e2, . . ., the Janet basis corresponding to S is returned.S1 : Autoreduction. Set S := Autoreduce(S).

S2 : Completion. Set S := CompleteSystem(S).S3 : Find integrability conditions. Find all pairs of leading terms vi ofei and vj of ej such that differentiation w.r.t. a nonmultiplier xik and

multipliers xj1 , . . . , xjl leads to ∂vi∂xik

= ∂p1+...+plvj∂xp1j1 . . . ∂x

pl

jl

and determine the

integrability conditions

ci,j = Lcoef(ej) · ∂ei∂xik− Lcoef(ei) ·

∂p1+...+plej∂xp1j1 . . . ∂x

pl

jl

.

S4 : Reduce integrability conditions. For all ci,j setci,j := Reduce(ci,j , S).

S5 : Termination? If all ci,j are zero return S, otherwise make theassignment S := S ∪ ci,j |ci,j 6= 0, reorder S properly and goto S1.

Proof Steps S1, S2 and S4 are finite due to the finiteness of the algo-rithms called there. Applying Theorem 2.10 to all subsystems correspondingto like leading functions leads to the integrability conditions in step S3. Uponreduction in step S4, the leading derivatives of any nonvanishing integrabil-ity condition are lower than any leading derivative in the current system.Therefore this process must terminate after a finite number of iterations.

The above definition of a Janet basis has the advantage of being construc-tive, i.e., it may be applied in straightforward manner to decide whether agiven system of linear differential polynomials is a Janet basis and, if this isnot the case, to transform it into one. The most important property of Janetbases is established in the next theorem. Sometimes this property of a Janetbasis is used as the definition.

Theorem 2.11 Let J be a Janet basis generating a module M ⊂ Dm.Then an element e ∈ Dm is contained in M if and only if it may be reducedto zero by J.

54

By means of this theorem it will be shown next that the result obtainedfrom the algorithm JanetBasis is unique.

Theorem 2.12 For a given module M and a given term ordering thereexists a unique Janet basis.

Proof Assume for a given moduleM there are two different Janet basesJ = j1, . . . , jk and H = h1, . . . , hl. Let their elements be monic andordered such that the leading terms increase with increasing index as explainedon page 44. By Theorem 2.11 there holds

Reduce(ji,H) = 0 for i = 1, . . . , k andReduce(hi, J) = 0 for i = 1, . . . , l.

In the first place this requires that the leading terms of j1 and h1 are identical,because if e. g. Lterm(j1) < Lterm(h1) holds, j1 cannot be reduced by anyelement of H. Consequently, the leading term of Reduce(j1, h1) = j1 − h1 islower than the common leading term of both j1 and h1 and cannot be reducedany further. By Theorem 2.11 it follows that j1 − h1 = 0, i.e., j1 = h1. Dueto this equality, both j1 and h1 cannot take part in any further reduction.Therefore the same argument may be repeated with j1 and h1 omitted andso forth until the complete bases are covered.

Example 2.37 As a first example consider the system (2.25) from above.It is already autoreduced. Integrability conditions can only arise from thesubsystem comprising g2 with multipliers x and y, and g3 and g4 with multi-plier y each. As it is already complete, there is no action in step S2. Theirmultipliers lead in step S3 to the integrability conditions

zxxx + 8y2wyy +1ywxx− 4y2zxy − 32yzx− 16w = 0, zxxy − 4y2zyy − 8yzy = 0.

In step S4 their left-hand sides are reduced to

g5 ≡ zxxx + 8y2wyy + 1ywxx − 4y2zxy − 32yzx − 16w,

g6 ≡ zxxy − 4y2zyy − 8yzy.

Because the result does not vanish, the algorithm is repeated with the enlargedsystem comprising equations g1, . . . , g6. The subsystem comprising g1, g5, g6is completed in step S2 to

zyy + 12y zy, zxxx + 12y2zxy − 24yzx − 12w,

zxxy − 6yzy = 0, zxyy + 12y zxy.

The first element has multiplier variables x and y, for the remaining ones y isthe only multiplier variable. They yield

zxxy + 12y2zyy + 12yzy, zxyy +12yzxy −

1y2wy −

2y2 zx.


In step S4 they are reduced to zy, wy + 2zx. The enlarged system generatedin S5 is autoreduced to

zx +12yw, zy, wx, wy −

1yw

at the beginning of the third pass. Because the only two integrability condi-tions reduce to zero, this is the final answer.

2.3 Properties of Janet Bases

From now on the emphasis will be shifted from linear differential polynomi-als to the systems of lpde’s that are associated with them, and their solutionsin suitable function fields. Furthermore, it is assumed frequently that thedimension of these solution spaces is finite, and that both the number ofdependent and of independent variables is not greater than two. These con-straints are a priori known to be true for the applications later in this book.It turns out that systems with these properties allow a complete classification;and explicit solution procedures may be designed for them.Type and Order of a Janet Basis. It turns out that the leading derivativesare of extraordinary importance for any Janet basis. They are applied fordefining the type of a Janet basis.

Definition 2.6 (Type of a Janet basis) Let f = f1, . . . , fp be a Janetbasis for a fixed term order. Its leading derivatives Lder(f1), . . . , Lder(fp)define the type of the Janet basis f .

The leading derivatives divide the totality of derivatives into two classes forwhich a special name is introduced.

Definition 2.7 (Parametric and principal derivatives. Rank). Thosederivatives that may be obtained from the leading derivatives of a Janet basisby suitable differentiation are called principal, the remaining ones are calledparametric. The number of parametric derivatives is called its rank.

The distinctive feature of a Janet basis as compared to an arbitrary systemof pde’s may be expressed in terms of its parametric derivatives as follows:A Janet basis does not generate any relation between the parametric deriva-tives by differentiation and elimination, i.e., the parametric derivatives areindependent of each other.

Another quantity of importance for any system of pde’s is the degree ofarbitrariness of its general solution. As a matter of fact, this question was thestarting point for Janet’s work. In general it may consist of any number offunctions with varying numbers of arguments. A special case is represented by

56

those systems the general solution of which contains only a finite number ofconstants. Due to the linearity, their general solution is a linear combinationof a fundamental system, i.e., it generates a finite dimensional vector space. Astraightforward extension of this property which is true for linear ode’s leadsto the following definition.

Definition 2.8 (Order). If the general solution of a linear homogeneoussystem of pde’s contains a finite number of constants, this number is calledthe order of the system. It is identical to the dimension of its solution space.

The importance of a Janet basis originates from the fact that the dimensionof the solution space is obvious from it due to the following result. Its proofmay be found in the book by Kolchin [91], Chapter IV, Section 5.

Theorem 2.13 (Kolchin 1973) If the rank of a Janet basis is finite, thenit is identical to its order.

Example 2.38 The type of the Janet basis generated in Example 2.37 isdetermined by zx, zy, wx, wy. The only parametric derivatives are z and w;therefore it allows a two-dimensional solution space as obtained previously.This information is available without determining the solution explicitly.

Theorem 2.14 If the dimension of the solution space of a system of linearhomogeneous pde’s is finite, any Janet basis in lexicographic term ordering hasa subsystem containing only the lowest dependent variable. This subsystemalso has a finite-dimensional solution space. Its lowest equation is a linearode in the lowest independent variable.

Proof Let the dependent variables be ordered as z1 > z2 . . . > zm, theindependent variables as x1 > x2 > . . . > xn. The finite dimension of thesolution space requires that the Janet basis contains equations with leading

terms ∂kizm∂xki

i

for i = 1, . . . , n and ki ≥ 0. This assures the finite-dimensional

solution space for the lowest dependent variable. The applied term orderingguarantees that in these equations there does not occur a term containing afunction zi with i < m. Due to the ordering of the independent variables, the

equation with leading term ∂knzm∂xkn

n

cannot contain a derivative w.r.t. to xk

and k < n, i.e., it is an ode w.r.t. to xn.

Example 2.39 Consider the Janet basisz1,x +

1y − 2

z2 −1xz1, z1,y, z2,x, z2,y −

1y − 2

z2

(2.35)

in grlex order with z2 > z1 and x > y. There are four possible lex termorderings with the following result.z1,xx −

1xz1,x +

1x2 z1, z1,y, z2 + (y − 2)z1,x −

y − 2x

z1

for z2 > z1, y > x.

Linear Differential Equations 57z2,x, z2,y −

1y − 2

z2, z1,x −1xz1 +

1y − 2

z2, z1,y

for z1 > z2, y > x.

z1,y, z1,xx −1xz1,x +

1x2 z1, z2 + (y − 2)z1,x −

y − 2x

z1

for z2 > z1, x > y

and finallyz2,y −

1y − 2

z2, z2,x, z1,y, z1,x −1xz1 +

1y − 2

z2

for z1 > z2, x > y.

In each of these Janet bases the two lowest equations generate a Janet basisfor a single function. If y > x the lowest equation is a linear ode in x over thebase field Q(y) and vice versa.

J (1,2)1 J (1,2)

2,1 J (1,2)2,2

zx z zy zx, z zx z

zy z zxx zx, z zyy zy, z

J (1,2)3,1 J (1,2)

3,2 J (1,2)3,3

zy zx, z zxx zy, zx, z zx z

zxxx zxx, zx, z zxy zy, zx, z zyyy zyy, zy, z

zyy zy, zx, z

TABLE 2.1: The types of Janet bases for m = 1, n = 2, i.e., for a singlefunction z(x, y) of order up to 3. The term ordering is grlex with y > x. Anytype is uniquely characterized by its leading derivatives in the square boxes.

Classification for m=1, n=2. The simplest generalization of a linear ho-mogeneous ode is a system of lpde’s of finite order for a single function. Forany fixed value of the order and a fixed term ordering there is a finite numberof possible types of Janet bases. They are obtained by the following consid-erations. In the grlex term ordering with y > x the derivatives of order nothigher than three are

zyyy > zxyy > zxxy > zxxx > zyy > zxy > zxx > zy > zx > z.

According to the definition of a parametric derivative, for an r-dimensionalsolution space, principal derivatives must be selected from this arrangementsuch that none of it may be obtained by differentiation from any of the re-maining ones and that they allow r parametric derivatives. For r = 1, 2 or 3the possible selections are as follows.

r = 1 : zx, zy,r = 2 : zy, zxx, zx, zyy,r = 3 : zy, zxxx, zxx, zyx, zyy, zx, zyyy.

58

In Table 2.1 the structure of the corresponding Janet bases is indicated;and a unique notation J (m,n)

r,k for any of them is introduced. Here m andn denote the number of dependent and independent variables respectively, rdenotes the order, and k is a consecutive enumeration within this set. Thesymbol J (m,n) without the lower indices denotes collectively all Janet basistypes corresponding to the given values of m and n. Each line correspondsto an element of the Janet basis, its leading derivative is enclosed in a box,the parametric derivatives that may occur are listed to the right in decreasingorder. In a concrete Janet basis they are multiplied by a coefficient from thebase field.

The possible arrangements of parametric and leading derivatives may alsobe visualized geometrically. To this end, for any dependent variable its partialderivatives are assigned to a point of a rectangular lattice. Parametric deriva-tives are indicated by open dots, principal derivatives by heavy dots. To thesix types of Janet bases defined in Table 2.1 there correspond the followinggraphs.

-

6

c ssJ (1,2)1 :

x

y

1 2

1

2

-

6

cs c sJ (1,2)2,1 :

x

y

1 2

1

2

-

6

c scsJ (1,2)2,2 :

x

y

1 2

1

2

-

6

c c c ssJ (1,2)3,1 :

x

y

1 2 3

1

2

-

6

ccs

c ssJ (1,2)3,2 :

x

y

1 2

1

2

-

6

c sccs

J (1,2)3,3 :

x

y

1 2

1

2

3

The classification of Janet bases given above provides a complete surveyof all possible Janet bases for m = 1, n = 2 if the dimension of the solutionspace is not higher than three. If it is a priori known for a particular systemthat the dimension of its solution space is at most three, it is guaranteed thatthe resulting Janet basis type is included in this listing.

Next to the classification problem there occurs the following question: Whatare the most general coefficients for any given type such that the Janet basisproperty is assured? This is a new phenomenon that does not have a coun-terpart for linear ode’s of the form (2.1). Its coefficients qi may be chosen ar-bitrarily from the base field, yet the resulting equation has an n−dimensionalsolution space, only the function field in which a fundamental system is con-tained is determined by this choice. This is not true for Janet bases forsystems of lpde’s. The constraints for the coefficients guaranteeing the Janetbasis property are extremely important, particularly if they are combined with


additional restrictions expressing further properties of the solutions as theywill occur in later applications. Only if the former constraints are completelyknown, the latter ones may be identified as new features originating from thespecial problem.

The coefficient constraints for a given Janet basis type are essentially ob-tained in steps S2, S3 and S4 of the algorithm JanetBasis given above. Be-cause the type of the Janet basis is fixed now, upon completion of step S3 thenonvanishing integrability conditions are not added to the system. Rather inorder to preserve the type of the Janet basis, the coefficients of the variousparametric derivatives are supposed to vanish identically. The conditions ob-tained in this way represent a system of pde’s for the coefficient functions. Ifit is satisfied, the full system is guaranteed to be the most general Janet basisof the specified type. This is achieved by insertion of the additional step S5aas follows.

S5a : Separate Integrability Conditions. Assign

C :=⋃ci,j

Coefficients(ci,j)and return C.

In general the integrability conditions obtained in step S5a are a system ofnonlinear pde’s that may comprise redundant equations. However, the fol-lowing important property makes it amenable for further simplification.

Lemma 2.3 The integrability conditions obtained in step S5a are quasilin-ear, i.e., they are linear in the highest derivatives.

Proof In step S2 and the first part of S3 only differentiations of thegiven linear homogeneous equations are performed, i.e., the resulting systemis such that the reductum of any equation is linear and homogeneous in thecoefficients. The same is true for the integrability conditions ci,j obtained inthe second part of S3 because the leading terms cancel each other. For anygiven ci,j two kinds of reductions in step S4 may occur. In the first place thereare those applying an equation ek with an index different from i and j. Theycan never generate a quadratic term because there is no intersection betweenthe coefficient variables of any equation involved. On the other hand, anyreduction w.r.t. equation ei or ej involves only derivatives that are lowerthan those involved in forming the integrability condition. As a consequenceany derivative of maximal order can only occur linearly.

Applying the modified version of algorithm JanetBasis to the entries ofTable 2.1 yields the following answer.

Theorem 2.15 For m = 1, n = 2 and order r ≤ 3 the coherence conditionsor integrability conditions (IC’s for short) are explicitly given by the followingconstraints.

J (1,2)1 : zx + az, zy + bz. IC : ay − bx = 0.

60

J (1,2)2,1 : zy + a1zx + a2z, zxx + b1zx + b2z.

IC ′s :a1,xx − b1,xa1 − a1,xb1 + 2a2,x − b1,y = 0,a2,xx + a2,xb1 − 2a1,xb2 − b2,xa1 − b2,y = 0.

J (1,2)2,2 : zx + a1z, zyy + b1zy + b2z.

IC ′s : b1,x − 2a1,y = 0, a1,yy + a1,yb1 − b2,x = 0.

J (1,2)3,1 : zy + a1zx + a2z, zxxx + b1zxx + b2zx + b3z.

IC ′s :

a1,xx − 13b1,y −

13b1,xa1 + a2,x − 1

3a1,xb1 = 0,

a2,xxx + a2,xxb1 − b3,y − b3,xa1 + a2,xb2 − 3a1,xb3 = 0,

b1,xy + b1,xxa1 + 6a2,xx − 3b2,y − 3b2,xa1 + 43b1,yb1

+2b1,xa1,x + 2a2,xb1 − 6a1,xb2 + 43b1(a1b1)x = 0.

J (1,2)3,2 :

zxx + a1zy + a2zx + a3z, zxy + b1zy + b2zx + b3z,

zyy + c1zy + c2zx + c3z.

IC ′s :

a1,y − b1,x + a1b2 − a1c1 − a2b1 + a3 + b21 = 0,

a2,y − b2,x − a1c2 + b1b2 − b3 = 0,

a3,y − b3,x − a1c3 − a2b3 + a3b2 + b1b3 = 0,

b1,y − c1,x + a1c2 − b1b2 + b3 = 0,

b2,y − c2,x + a2c2 − b1c2 − b22 + b2c1 − c3 = 0,

b3,y − c3,x + a3c2 − b1c3 − b2b3 + b3c1 = 0.

J (1,2)3,3 : zx + a1z, zyyy + b1zyy + b2zy + b3z.

IC ′s :b1,x − 3a1,y = 0, a1,yy − 1

3b2,x + 23a1,yb1 = 0,

b2,xy − 3b3,x + 13b2,xb1 − 2b1,ya1,y + 3a1,yb2 − 2

3a1,yb21 = 0.

Proof For the J (1,2)1 Janet basis the only IC follows by subtracting

zxy + (ay − ab)z and zxy + (bx − ab)

from each other. These expressions are obtained by deriving the two elementsof the Janet basis w.r.t. y and x respectively and reducing the result. Thecondition obtained assures the path independence of integrals involving thegradient vector field of z.

As a second example, the IC’s for the type J (1,2)2,1 Janet basis will be ob-

tained. To this end, the first element has to be derived twice w.r.t. to the


multiplier x and all possible reductions have to be performed. The result is

zxy + (a1,x + a2 − a1b1)zx + (a2,x − a1b2)z,

zxxy + (a1,xx + 2a2,x − 2a1,xb1 − a1b1,x − a1b2 − a2b1 + a1b21)zx

+(a2,xx − 2a1,xb2 − a1b2,x − a2b2 + a1b1b2)z.

The second element has to be derived w.r.t. to its nonmultiplier y. After allreductions are performed, the expression

zxxy + (b1,y − a1,xb1 − a2b1 − a1b2 + a1b21)zx

+ (b2,y − a2,xb1 − a2b2 + a1b1b2)z

is obtained. If the two elements with the leading derivative zxxy are subtractedfrom each other, the resulting expression must vanish identically. Because theparametric derivatives zx and z are independent, its coefficients must vanish.After some simplification this yields the given IC’s. The calculations for theremaining cases are similar.

Example 2.40 Consider the set of differential polynomialszxx −

y

x(x+ y)zy +

1xzx, zxy +

1x+ y

zy, zyy +1

x+ yzy

.

Its leading terms suggest that it might be a type J (1,2)3,2 Janet basis. The

nonvanishing coefficients

a1 = − y

x(x+ y), a2 =

1x

and b1 = c1 =1

x+ y

satisfy the conditions given in the above theorem, i.e., the Janet basis propertyis assured without running the Janet basis algorithm. As is easily shown, abasis for its three-dimensional solution space is 1, log x, log (x+ y).

In general, any change in the coefficients destroys the Janet basis property,e.g. if the sign of the last term in the first equation is reversed. Runningthis new system through the algorithm JanetBasis yields the type J (1,2)

2,1

Janet basis zy, zxx − 1xzx. A basis for its two-dimensional solutions space

is1, x2

.

Classification for m=n=2. The second case to be discussed in full detailinvolves two dependent variables, they are denoted by z1 and z2. In grlexterm ordering with z2 > z1 and y > x, the derivatives of order not higherthan two are

. . . z2,yy > z2,yx > z2,xx > z1,yy > z1,yx > z1,xx >

z2,y > z2,x > z1,y > z1,x > z2 > z1.

62

J (2,2)1,1 J (2,2)

1,2

z1 z2 , z1

z2,x , z2 z1,x , z1

z2,y , z2 z1,y , z1

J (2,2)2,1 J (2,2)

2,2 J (2,2)2,3 J (2,2)

2,4 J (2,2)2,5

z1 z1 z1,x , z2, z1 z2 , z1 z2 , z1

z2,y , z2,x, z2 z2,x , z2 z1,y , z2, z1 z1,y , z1,x, z1 z1,x , z1,

z2,xx , z2,x, z2 z2,yy , z2,y, z2 z2,x , z2, z1 z1,xx , z1,x, z1, z1,yy , z1,y, z1

z2,y , z2, z1

TABLE 2.2: The Janet bases of types J (2,2)1,k and J (2,2)

2,k for linear ho-mogeneous systems in z1(x, y) and z2(x, y). The term ordering is grlex withz2 > z1 and y > x. Any type is uniquely characterized by its leading deriva-tives in the square boxes.

For an r-dimensional solution space, principal derivatives must be selectedfrom this arrangement such that none of it may be obtained by differenti-ation from any of the remaining ones and such that there are r parametricderivatives. For r = 1, 2 or 3 the possible selections are as follows.

r = 1 : z2, z1

r = 2 : z2,x, z2, z2,y, z2, z2, z1, z1,y, z1, z1,x, z1

r = 3 : z2,xx, z2,x, z2, z2,y, z2,x, z2, z2,yy, z2,y, z2, z2,x, z2, z1,

z2,y, z2, z1, z1,x, z2, z1, z1,y, z2, z1, z1,yy, z1,y, z1,

z1,y, z1,x, z1, z1,xx, z1,x, z1.

In Tables 2.2 and 2.3 the structure of the corresponding Janet bases isindicated and a unique notation for any of them is introduced. Each linecorresponds to an equation of the Janet basis, its leading derivative is enclosedin a box, the parametric derivatives that may occur in any equation are listedto the right of it in decreasing order. In a concrete Janet basis they aremultiplied by a coefficient from the base field. Similar schemes may be setup for higher dimensional solution spaces or for values of m and n differentfrom 2. The number of alternatives increases quickly if these values increase.

In the graphical representation there correspond to each Janet basis asmany drawings as there are dependent variables. For the five possible typesof Janet bases J (2,2)

2,k these graphs are as follows.


J (2,2)3,1 J (2,2)

3,2 J (2,2)3,3 J (2,2)

3,4

z1 z1 z1 z1,x , z2, z1

z2,y , z2,x, z2 z2,xx , z2,x, z2 z2,x , z2 z1,y , z2, z1

z2,xxx , z2,xx, z2,x, z2 z2,yx , z2,y, z2,x, z2 z2,yyy , z2,yy, z2,y, z2 z2,y , z2,x, z2, z1,

z2,yy , z2,y, z2,x, z2 z2,xx , z2,x, z2, z1

J (2,2)3,5 J (2,2)

3,6 J (2,2)3,7 J (2,2)

3,8

z1,x , z2, z1 z1,y , z1,x, z2, z1 z1,x , z2, z1 z2 , z1

z1,y , z2, z1 z2,x , z1,x, z2, z1 z2,x , z1,y, z2, z1 z1,x , z1

z2,x , z2, z1 z2,y , z1,x, z2, z1, z2,y , z1,y, z2, z1 z1,yyy , z1,yy, z1,y, z1

z2,yy , z2,y, z2, z1 z1,xx , z1,x, z2, z1 z1,yy , z1,y, z2, z1

J (2,2)3,9 J (2,2)

3,10

z2 , z1 z2 , z1z1,xx , z1,y, z1,x, z1 z1,y , z1,x, z1

z1,yx , z1,y, z1,x, z1 z1,xxx , z1,xx, z1,x, z1

z1,yy , z1,y, z1,x, z1

TABLE 2.3: The Janet bases of types J (2,2)3,k , k = 1, . . . , 10 for linear

homogeneous systems in z1(x, y) and z2(x, y). The term ordering is grlexwith z2 > z1 and y > x. Any type is uniquely characterized by its leadingderivatives in the square boxes.

-

6

sJ (2,2)2,1 :

x

y z1

1 2

1

2

-

6

c c ssx

y z2

1 2

1

2

z2,x > z2

-

6

sJ (2,2)2,2 :

x

y z1

1 2

1

2

-

6

c scsx

y z2

1 2

1

2

z2,y > z2

-

6

cs sJ (2,2)2,3 :

x

y z1

1 2

1

2

-

6

c ssx

y z2

1 2

1

2

z2 > z1

64

-

6

cs c sJ (2,2)2,4 :

x

y z1

1 2

1

2

-

6

sx

y z2

1 2

1

2

z1,x > z1

-

6

ccs

sJ (2,2)2,5 :

x

y z1

1 2

1

2

-

6

sx

y z2

1 2

1

2

z1,y > z1

From this representation the distinctive properties of parametric and non-parametric derivatives are obvious.

The analogue of Theorem 2.15 for Janet bases of type J (2,2) of order notgreater than three is given next. Due to the size of the result it is split intotwo cases, at first order one and two Janet bases are considered, and the thirdorder case thereafter.

Theorem 2.16 For m = n = 2 and order 1 or 2, the coherence conditionsor integrability conditions (IC’s for short) are explicitly given by the followingconstraints.

J (2,2)1,1 : z1, z2,x + az2, z2,y + bz2. IC : ay − bx = 0.

J (2,2)1,2 : z2 + az1, z1,x + bz1, z1,y + cz1. IC : by − cx = 0.

J (2,2)2,1 : z1, z2,y + a1z2,x + a2z2, z2,xx + b1z2,x + b2z2.

IC ′s :a1,xx − b1,y − b1,xa1 + 2a2,x − a1,xb1 = 0,

a2,xx − b2,y − b2,xa1 + a2,xb1 − 2a1,xb2 = 0.

J (2,2)2,2 : z1, z2,x + a1z2, z2,yy + b1z2,y + b2z2.IC ′s : b1,x − 2a1,y = 0, a1,yy − b2,x + a1,yb1 = 0.

J (2,2)2,3 :

z1,x + a1z2 + a2z1, z1,y + b1z2 + b2z1,

z2,x + c1z2 + c2z1, z2,y + d1z2 + d2z1.

IC ′s :

b1,x − a1,y + a1d1 − b1c1 − a1b2 + a2b1 = 0,

b2,x − a2,y + a1d2 − b1c2 = 0,

d1,x − c1,y − a1d2 + b1c2 = 0,

d2,x − c2,y + c1d2 − a2d2 − c2d1 + b2c2 = 0.

J (2,2)2,4 : z2 + a1z1, z1,y + b1z1,x + b2z1, z1,xx + c1z1,x + c2z1.

IC ′s :b1,xx − c1,y − c1,xb1 + 2b2,x − b1,xc1 = 0,

b2,xx − c2,y − c2,xb1 + b2,xc1 − 2b1,xc2 = 0.


J (2,2)2,5 : z2 + a1z1, z1,x + b1z1, z1,yy + c1z1,y + c2z1.IC ′s : c1,x − 2b1,y = 0, b1,yy − c2,x + b1,yc1 = 0.

Theorem 2.17 For m = n = 2 and order 3 the coherence conditions orintegrability conditions (IC’s for short) are explicitly given by the followingconstraints.

J (2,2)3,1 : z1, z2,y + a1z2,x + a2z2, z2,xxx + b1z2,xx + b2z2,x + b3z2.

IC ′s :

a1,xx − 13b1,y −

13b1,xa1 + a2,x − 1

3a1,xb1 = 0,

b1,xy + b1,xxa1 + 6a2,xx − 3b2,y − 3b2,xa1 + 43b1,yb1 + 2b1,xa1,x

+ 43b1,xa1b1 + 2a2,xb1 − 6a1,xb2 + 4

3a1,xb21 = 0,

a2,xxx + a2,xxb1 − b3,y − b3,xa1 + a2,xb2 − 3a1,xb3 = 0.

J (2,2)3,2 :

z1, z2,xx + a1z2,y + a2z2,x + a3z2, z2,xy + b1z2,y + b2z2,x + b3z2,

z2,yy + c1z2,y + c2z2,x + c3z2.

IC ′s :

b1,x − a1,y + a1c1 − a1b2 − b21 + a2b1 − a3 = 0,

b2,x − a2,y + a1c2 + b3 − b1b2 = 0,

b3,x − a3,y + a1c3 − b1b3 + a2b3 − a3b2 = 0,

c1,x − b1,y − a1c2 − b3 + b1b2 = 0,

c2,x − b2,y + c3 + b1c2 − a2c2 − b2c1 + b22 = 0,

c3,x − b3,y + b1c3 − a3c2 − b3c1 + b2b3 = 0.

J (2,2)3,3 : z1, z2,x + a1z2, z2,yyy + b1z2,yy + b2z2,y + b3z2.

IC ′s :b1,x − 3a1,y = 0, a1,yy − 1

3b2,x + 23a1,yb1 = 0,

b2,xy − 3b3,x + 13b2,xb1 − 2a1,yb1,y + 3a1,yb2 − 2

3a1,yb21 = 0.

J (2,2)3,4 :

z1,x + a1z2 + a2z1, z1,y + b1z2 + b2z1,

z2,y + c1z2,x + c2z2 + c3z1, z2,xx + d1z1,x + d2z2 + d3z1.

IC ′s :

a1c1 + b1 = 0, b2,x − a2,y + a1c3 = 0,b1,x − a1,y + a1c2 − a1b2 + a2b1 = 0,c1,xx − d1,y − d1,xc1 + 2c2,x − c1,xd1a1c3 = 0,c2,xx − d2,y − d2,xc1 − 2c3,xa1 + c2,xd1 − 2c1,xd2

−a1,xc3 − a1c3d1 + a1a2c3 = 0,c3,xx − d3,y − d3,xc1 + c3,xd1 − 2c3,xa2 − 2c1,xd3 − a2,xc3

−c2d3 + a2c1d3 + b2d3 + c3d2 − a2c3d1 + a22c3 = 0.

66

J (2,2)3,5 :

z1,x + a1z2 + a2z1, z1,y + b1z2 + b2z1,

z2,x + c1z2 + c2z1, z2,yy + d1z2,y + d2z2 + d3z1.

IC ′s :

a1 = 0, b1,x − b1c1 + a2b1 = 0,b2,x − a2,y − b1c2 = 0, d1,x − 2c1,y + b1c2 = 0,c1,yy − d2,x − 2c2,yb1 + c1,yd1 − b1,yc2 − b1c2d1 + b1b2c2 = 0,c2,yy − d3,x + c2,yd1 − 2c2,yb2 − b2,yc2

−c1d3 + a2d3 + c2d2 − b2c2d1 + c2b22 = 0.

J (2,2)3,6 :

z1,y + a1z1,x + a2z2 + a3z1, z2,x + b1z1,x + b2z2 + b3z1,

z2,y + c1z1,x + c2z2 + c3z1, z1,xx + d1z1,x + d2z2 + d3z1.

IC ′s :

c1,x − b1,y + a1,xb1 − c1d1 − a1b1d1 + c3

−b1c2 + b2c1 + a1d3 − b21a2 + a3b1 = 0,c2,x − b2,y + a2,xb1 − c1d2 − a1b1d2 + a2d3 − a2b1b2 = 0,c3,x − d3,y + a3,xb1 − c1d3 − a2b1d3

+b2c3 − c2d3 − a1b2d3 + a3d3 = 0,a1,xx − d1,y − d1,xa1 − b1,xa2 + 2a3,x − 2a2,xb1

−a1,xd1 + c1d1 + a1b1d2 − a2d3 + a2b1b2 = 0,a2,xx − d2,y − d2,xa1 − b2,xa2 + a2,xd1 − 2a2,xb2 − 2a1,xd2

+a2d3 + c2d2 + a1b2d2 + a2b1d2 − a3d2 − a2b2d1 + b21a2 = 0,a3,xx − d3,y − d3,xa1 − d3,xa2 + a3,xd1 − 2a2,xd3

−2a1,xd3 + a2b1d3 + c3d2 + a1d2d3 − a2d1d3 + a2b2d3 = 0.

J (2,2)3,7 :

z1,x + a1z2 + a2z1, z2,x + b1z1,y + b2z2 + b3z1,

z2,y + c1z1,y + c2z2 + c3z1, z1,yy + d1z1,y + d2z2 + d3z1.

IC ′s :

c1,x − b1,y + b1d1 − b1c2 + c21a1 − a2c1 + b2c1 − d3 = 0,c2,x − a1,yc1 + b1d2 − a1c3 + a1c1c2 − b2,y = 0,c3,x − a2,yc1 + b1d3 + a1c1c3 − a2c3 + b2c3 − c2d3 = 0,d1,x + c1,ya1 − 2a2,y + 2a1,yc1 − b1d2 + a1c3 − a1c1c2 = 0,a1,yy − d2,x − c2,ya1 + a1,yd1 − 2a1,yc2

+a1d3 + a1c1d2 − a2d2 + b2d2 − a1c2d1 + c22a1 = 0,a2,yy − d3,x − c3,ya1 + a2,yd1 − 2a1,yc3

+a1c1d3 + d2d3 − a1c3d1 + a1c2c3 = 0.

J (2,2)3,8 : z2 + a1z1, z1,x + b1z1, z1,yyy + c1z1,yy + c2z1,y + c3z1.

IC ′s :c1,x − 3b1,y = 0, b1,yy − 1

3c2,x + 23b1,yc1 = 0,

c2,xy − 3c3,x + 13c3,xc1 − 2c1,yb1,y + 3b1,yc2 − 2

3b1,yc21 = 0.


J (2,2)3,9 :

z2 + a1z1, z1,xx + b1z1,y + b2z1,x + b3z1,

z1,xy + c1z1,y + c2z1,x + c3z1, z1,yy + d1z1,y + d2z1,x + d3z1.

IC ′s :c3 − a1c2 = 0, d3 − a1d2 = 0, b2,y − b1,ya1 − a1,yb1 = 0,c1,x − c1,y + c1d1 − c21 − b2 + a1b1 = 0, d1,x − c1,y = 0.

J (2,2)3,10 : z2 + a1z1, z1,y + b1z1,x + b2z1, z1,xxx + c1z1,xx + c2z1,x + c3z1.

IC ′s :

b1,xx − 13c1,y −

13c1,xb1 + b2,x − 1

3b1,xc1 = 0,c1,xy + c1,xxb1 + 6b2,xx − 3c2,y − 3c2,xb1 + 4

3c1,yc1 + 2b1,xc1,x+ 4

3c1,xb1c1 + 2b2,xc1 − 6b1,xc2 + 43b1,xc

22 = 0,

b2,xxx + b2,xxc1 − c3,y − c3,xb1 + b2,xc2 − 3b1,xc3 = 0.

This explicit form provides a complete understanding of the features of agiven system of linear homogeneous pde’s. Of particular importance is theclear separation of all correlations for the coefficients due to the coherenceconditions given in the above theorem, and additional properties that may bedue to the structure of a specific problem. For example, the lowest equationin any Janet basis of type J (2,2)

3,5 has always the form z1,x + a2z1 = 0, thevanishing of the term proportional to z2 is not a characteristic feature of aspecific problem. The importance of this classification will become clear in thenext chapter where Janet bases of these types occur as determining systemsfor certain symmetry groups.

The quasilinearity of the constraints generated in step S5a (see page 59)allows application of the algorithm JanetBasis to the system C defined therewith the following modification. The algorithm is interrupted if an equationis generated that is not quasilinear. It turns out that this does not occur forthe types listed in Table 2.1, Table 2.2 or 2.3. The term ordering applied forthe integrability conditions is always grlex with

a1 < a2 < . . . < b1 < b2 < . . . < c1 < c2 < . . .

and x < y. In most cases the Janet basis property is either obtained uponproper reordering or by autoreduction.

Example 2.41 A typical case is type J (2,2)3,3 for which in the first place

the integrability conditions

b1,x − 3a1,y = 0, a1,yy − 13b2,x + 2

3a1,yb1 = 0,

a1,yyy + a1,yyb1 − b3,x + a1,yb2 = 0

are obtained. Autoreduction generates the Janet basis from it by replacingthe third equation with

b2,xy − 3b3,x + 13b2,xb1 − 2b1,ya1,y + 3a1,yb2 − 2

3a1,yb22 = 0.

68

Syzygies of Janet Bases. The coherence conditions of Theorems 2.15, 2.16and 2.17 imply certain relations between the elements of a Janet basis whichare known as syzygies. They will be required later on in this chapter forcomputing quotients. Let the Janet basis be f = f1, . . . , fp where fi ∈ Dm,D = Q(x, y)[∂x, ∂y]. Syzygies of f are relations of the form

dk,1f1 + . . .+ dk,pfp = 0

where dk,i ∈ D, i = 1, . . . p, k = 1, 2, . . .. The (dk,1, . . . , dk,p) may be con-sidered as elements of the module Dp. The totality of syzygies generates asubmodule.

Example 2.42 Consider the Janet basis f1 ≡ zx+ az, f2 ≡ zy + bz withthe constraint ay = bx. The coherence condition for zx − az − f1 = 0 andzy + bz − f2 = 0 yields azy + ayz − f1,y − bzx − bxz + f2,x = 0. Reductionw.r.t. to the given equations and some simplification yields the single syzygy(∂y + b)f1 − (∂x + a)f2 = 0.

Example 2.43 Consider the Janet basisf1 ≡ zxx +

4xzx +

2x2 z, f2 ≡ zxy +

1xzy, f3 ≡ zyy +

1yzy −

x

y2 zx −2y2 z

.

The integrability condition for zxx+ 4xzx+ 2

x2 z−f1 = 0 and zxy+ 1xzy−f2 = 0

yields upon reduction and simplification f1,y +f2,x− 3xf2 = 0. Similarly from

the last two elements f1 − y2

x f2,y −yxf2 + y2

x f3,x + y2

x2 f3 = 0 is obtained.Autoreduction of these two equations yields the final answer(

∂yy + 3y ∂y −

xy2 ∂x − 2

y2

)f2 −

(∂xy + 1

x∂y + 2y ∂x + 2

xy)f3 = 0,

f1 −(y2

x ∂y + yx

)f2 +

(y2

x ∂x + y2

x2

)f3 = 0.

Gcrd and Lclm of Modules. The greatest common right divisor (Gcrd) orsum of two modules I ≡< f1, . . . , fp > and J ≡< g1, . . . , gq >, fi, gj ∈ Dmfor all i and j, D = Q(x, y)[∂x, ∂y], is generated by the union of the generatorsof I and J (Cox et al. [34], page 191). The solution space of the equationscorresponding to Gcrd(I, J) is the intersection of the solution spaces of itsarguments. From this definition the following algorithm is self-explanatory.

Algorithm 2.9 Gcrd(I, J). Given two modules I ≡< f1, . . . , fp > andJ ≡< g1, . . . , gq > in Dm, the Janet basis for its greatest common divisor isreturned.S1 : Join input generators. Set h := f1, . . . , fp ∪ g1, . . . , gq .


S2 : Create Janet basis. Transform h in a Janet basis and return theresult.

For m = 1 the Gcrd of the ideals I, J ∈ D is obtained. For m = n = 1 theEuclidean algorithm for computing the Gcrd of two linear ode’s is subsumedas a special case under the above algorithm.

Example 2.44 Consider the ideals I =< ∂x + yx∂y, ∂xxx + 3

x∂xx > and

J =< ∂xx +1x∂x −

1x2 , ∂xy +

1x∂y +

1y∂x +

1xy, ∂yy +

1y∂y −

1y2 > .

Applying the above algorithm, the greatest common right divisor is obtainedas Gcrd(I, J) =< ∂x + y

x∂y, ∂yy + 1y ∂y −

1y2 >. In terms of solution spaces

this result may be understood as follows. For I a basis of the solution space is1, xy ,

yx

, and for J a basis is

1xy ,

xy ,yx

. A basis for their two-dimensional

intersection space isxy ,yx

, it is the solution space of Gcrd(I, J).

The Lclm of two modules I and J is the left intersection module of its argu-ments. The solution space of the equations corresponding to Lclm(I, J) is thesmallest space containing the solution spaces of its arguments. An algorithmfor generating the Lclm requires a more detailed analysis. Properly adjustingthe corresponding algorithm from commutative algebra, see for example Coxet al. [34], Proposition 3.11 on page 218 and the article by Grigoriev andSchwarz [61], Section 3, the following algorithm is obtained.

Algorithm 2.10 Lclm(I, J). Given the two modules I ≡< f1, . . . , fp >and J ≡< g1, . . . , gq > in Dm, the Janet basis for its least common leftmultiple is returned.S1 : Auxiliary system. Introducing two differential indeterminates wand z, generate the system f1w, . . . , fpw, g1w − g1z, . . . , gqw − gqz.S2 : Generate Janet basis. For the system obtained in S1 generate aJanet basis in lex term order with w > z.S3 : Return result. From the Janet basis obtained in S2 collect elementsdepending on z only and return the respective operators generating theresult.

Example 2.45 Consider the two ideals

I =< ∂y +1y, ∂x +

1x> and J =< ∂y +

1x+ y

, ∂x +1

x+ y> .

In step S1 the system

wy + 1yw, wx + 1

xw,

wy + 1x+ yw − zy −

1x+ y z, wx + 1

x+ yw − zx −1

x+ y z

70

is generated. The Janet basis obtained in step S2 is

zx − y2

x2 zy + x− yx2 z, zyy + 2x+ 4y

xy + y2 zy + 2xy + y2 z,

w + x2 + xyy zx + x

y z.

Consequently, the intersection ideal is

Lclm(I, J) =< ∂x − y2

x2 ∂y + x− yx2 , ∂yy + 2x+ 4y

xy + y2 ∂y + 2xy + y2 > .

In some places later in this book a shortened notation for the Lclm isapplied, e. g. Lclm(zy + az, zx + bz) = Lclm(∂y + a, ∂x + b)z.

Exact Quotient and Relative Syzygies. For ordinary differential oper-ators the exact quotient has been defined on page 22. In order to generatea full system of solutions for a system of linear pde’s which is reducible butnot completely reducible, a proper generalization of this exact quotient isrequired. It is given next.

Definition 2.9 (Exact quotient, relative syzygies) Let I ≡< f1, . . . , fp >∈Dm and J ≡< g1, . . . , gq >∈ Dm be such that I ⊆ J . The exact quotientmodule is generated by

(ei,1, . . . , ei,q) ∈ Dq|ei,1g1 + . . .+ ei,qgq = fi, i = 1, . . . , p.

The relative syzygies module Syz(I, J) is generated by

h = (h1, . . . , hq) ∈ Dq|h1g1 + . . .+ hqgq ∈ I.

If for two modules there holds I ⊆ J , J is also called a divisor or a componentof I. Obviously the relative syzygies module extends the syzygy module ofJ , the latter is obtained for the special choice I = 0. The following algorithmgenerates the relative syzygies module as the sum of the exact quotient andthe syzygies module.

Algorithm 2.11 RelativeSyzygies(I, J). Given two modules I ⊂ J ⊂ Dmby I ≡< f1, . . . , fp > and J ≡< g1, . . . , gq >, the Janet basis for its relativesyzygies module Syz(I, J) ∈ Dq is returned.S1 : Exact quotient. For i = 1, . . . , p determine ei = (ei,1, . . . , ei,q) fromei,1g1 + . . .+ ei,qgq = fi.S2 : Include syzygies. Generate the syzygies sk = sk,1g1 + . . .+ sk,qgq fork = 1, 2, . . ..S3 : Return Janet basis. Join the quotients qi and the syzygies sk, trans-form them into a Janet basis and return it.


Example 2.46 Consider the ideals

I =< ∂y −x2

y2 ∂x −x− yy2 , ∂xx +

2x

2x+ y

x+ y∂x +

2x(x+ y)

>

and J =< ∂x+ 1x, ∂y+ 1

y >. There holds I ⊂ J . Step S1 generates the matrix(− x

2

y2 , 1),(∂x+ 3x+ y

x(x+ y) , 0)

. In step S2 the single syzygy(∂y+ 1

y , ∂x+ 1x

)is obtained. Step S3 generates the Janet basis for the relative syzygies module

Syz(I, J) =<(− x2

y2 , 1),(∂x + 3x+ y

x(x+ y) , 0),(∂y − x− y

y(x+ y) , 0)>⊂ D2.

Example 2.47 (Li, Schwarz and Tsarev [105]) Consider the ideal I =< ∂y− xy ∂x, ∂xx−

xy − 1x ∂x− y

x > and its right factor J =< ∂x−y, ∂y−x >.

Division in step S1 yields(− xy , 1

),(∂x + 1

x, 0)

. There is a single syzygy(−∂y+x, ∂x−y). From these three generators step S3 generates a Janet basisfor the relative syzygies module Syz(I, J) =< (−xy , 1), (∂x+1

x, 0), (∂y, 0) >.

Example 2.48 Consider the ideal I =< ∂xx+ 1x∂x, ∂xy, ∂yy+

1y ∂y >. There

is a single maximal ideal J =< ∂x, ∂y > containing it. It yields the relativesyzygies module Syz(I, J) =<

(∂x + 1

x, 0), (∂y, 0), (0, ∂x),

(0, ∂y + 1

y)>. It

may be represented as the intersection of two maximal modules of order 1,i.e., there holds

Syz(I, J) = Lclm(< (1, 0), (0, ∂x)

(0, ∂y + 1

y)>,

<(xy , 1

),(∂x + 1

x, 0), (∂y, 0) >

).

Similar as for ordinary differential equations there remains to be discussedhow lower order right factors of a module of partial differential operatorssimplify the solution procedure, compare the discussion on page 31. Thefollowing result is due to Grigoriev and Schwarz [62].

Theorem 2.18 Let I ≡< f1, . . . , fp > and J ≡< g1, . . . , gq > be twosubmodules of Dm such that I ⊆ J , with solution space VI and VJ of finitedimension dI and dJ respectively. Let the relative syzygies module Syz(I, J) ≡K ⊂ Dq be generated by < h1, . . . , hr >, and z = (z1, . . . , zm) and z =(z1, . . . , zq) be differential indeterminates. A basis for the solution space VImay be constructed as follows.

i) Determine a basis of VJ from g1z = 0, . . . , gqz = 0, let its elements bevk = (vk,1, . . . , vk,m), k = 1, . . . , dJ .

72

ii) Determine a basis of VK from h1z = 0, . . . , hr z = 0, let its elements bewk = (wk,1, . . . , wk,q), k = 1, . . . , dK = dI − dJ .

iii) For each inhomogeneous system g1z = wk,1, . . . , gq z = wk,q, k = 1, . . . , dKdetermine a special solution vk = (vk,1, . . . , vk,m).

A basis of the solution space VI is v1, . . . , vdJ, v1, . . . , vdK

.

Example 2.49 In Example 2.46, the right factor J corresponds to thesystem zx + 1

xz = 0, zy + 1y z = 0, its solution yields the basis element

v1 = 1xy . The relative syzygies lead to the system

z2 −x2

y2 z1 = 0, z1,x +3x+ y

x(x+ y)z1 = 0, z1,y −

x− yy(x+ y)

z1 = 0,

it yields the basis element z1 = yx(x+ y)2

, z2 = xy(x+ y)2

. Substituting it

into the inhomogeneous system results in the special solution v1 = 1x+ y .

Consequently, a basis for the originally given system is 1xy ,

1x+ y

.

Example 2.50 In Example 2.47 the system of pde’s corresponding to Jis zx − yz = 0, zy − xz = 0, it yields the basis exy for VJ . The relativesyzygies module leads to the system z2 − x

y z1 = 0, z1,x + 1xz1 = 0 and

z1,y = 0; its one-dimensional solution space is generated by z1 = 1x , z2 = 1

y .

There is a single inhomogeneous system zx − yz = 1x , zy − xz = 1

y withthe special solution v1 = exyEi(−xy). The exponential integral is definedby Ei(ax) =

∫eax dxx . A basis for the solution space of the originally given

system is exy, exyEi(−xy).

Example 2.51 In Example 2.48, to the first order right factor correspondsthe system zx = 0, zy = 0, it yields the basis 1 vor VJ . The two argumentsof the Lclm lead to the systems z1 = 0, z2,x = 0, z2,y + 1

y z2 = 0 with the

solution z1,1 = 0, z1,2 = 0, and z2 + xy z1 = 0, z1,x + 1

xz1 = 0, z1,y = 0 with

the solution z2,1 = 1x , z2,2 = −1

y . Substituting them into the inhomogeneoussystem yields the two special solutions log y and log xy . Therefore a basis for

the given system is1, log y, log xy

.

For efficiency reasons it is advantageous to have closed forms available forthose relative syzygies that occur frequently in applications. In this way theJanet basis calculation which occurs in step S3 of the above algorithm maybe avoided. Because these applications usually involve solving differentialequations, they are given in terms of differential polynomials.


Lemma 2.4 A Janet basis zy + A1zx + A2z, zxx + B1zx + B2z of typeJ (1,2)

2,1 has a component z1 ≡ zx + az, z2 ≡ zy + bz of type J (1,2)1 iff there

holds b−A2 +A1a = 0,

ax − a2 +B1a−B2 = 0, ay +A1a2 + (A1,x −A1B1)a−A2,x +A1B2 = 0.

The relative syzygies module is of type J (2,2)1,2 and is explicitly given by

z2 +A1z1, z1,x + (B1 − a)z1, z1,y + (A1,x −A1B1 + 2aA1 + b)z1.

Proof Reduction of the type J (1,2)2,1 Janet basis w.r.t. z1 ≡ zx + az

and z2 ≡ zy + bz yields the first two generators of the quotient under theassumption that b is determined from the first constraint and the equationwith leading term ax is valid. The third element defining the quotient followsif in the integrability condition ay−bx = 0 the variable b is eliminated applyingthe first two constraints. In Example 2.42 it has been shown that the typeJ (1,2)

1 Janet basis obeys the syzygy z1,y − z2,x + bz1 − az2 = 0. Reducing itw.r.t. to the two previous relations, the last generator is obtained.

Example 2.52 Consider again the Janet bases from Example 2.46. Bythe above lemma, the coefficients

A1 = −x2

y2 , A2 = −x− yy2 , B1 =

2x

2x+ y

x+ yand B2 =

2x(x+ y)

yield immediately the Janet basis for the relative syzygies module

<(− x2

y2 , 1),

(∂x + 3x+ y

x(x+ y) , 0),

(∂y − x− y

y(x+ y) , 0)> .

Lemma 2.5 A Janet basis zx+A1z, zyy +B1zy +B2z of type J (1,2)2,2 has

a component z1 ≡ zx + az, z2 ≡ zy + bz of type J (1,2)1 iff there holds

A1 − a = 0, by − b2 −B1b−B2 = 0, bx + ay −B1,x = 0.

Its relative syzygies module is of type J (2,2)1,1 and is explicitly given by

z1, z2,x + az2, z2,y + (B1 − b)z2.

The proof is similar as for Lemma 2.4 and is therefore omitted.

Example 2.53 Consider the Janet basiszx + 1

xz, zyy + 1y zy −

1y2 z

.

Dividing outz1 ≡ zx+ 1

xz, z2 ≡ zy−1y z

yields

z1, z2,x+ 1

xz2, z2,y+ 2y z2

.

The original system has the solutionsyx,

1xy

whereas yx is the only solution

74

of its first order component z1, z2. The second member is obtained bysolving the quotient system with the result z1 = 0, z2 = − 2

xy2 , and then the

inhomogeneous system zx + 1xz = 0, zy − 1

y z = − 2xy2 with the result 1

xy .

Explicit forms for the possible quotients of Janet bases of order three aregiven next without proof.

Lemma 2.6 If a Janet basis zy+A1zx+A2z, zxxx+B1zxx+B2zx+B3zof type J (1,2)

3,1 has a component z1 ≡ zx + az, z2 ≡ zy + bz of type J (1,2)1 ,

its quotient is of type J (2,2)2,4 and is explicitly given by

z2 +A1z1, z1,y +A1z1,x + (A1,x +A2)z1,

z1,xx + (B1 − a)z1,x + (B2 − aB1 − 2ax + a2)z1.

If it has a component z1 ≡ zy + a1zx + a2z, z2 ≡ zxx + b1zx + b2z of typeJ (1,2)

2,1 , its quotient is of type J (2,2)1,1 and is explicitly given by

z1, z2,x + (B1 − b1)z2, z2,y + (2A1,x −A1B1 + b1A1 +A2)z2.

Example 2.54 Consider the Janet basiszy+x

y zx, zxxx+3xzxx

. Dividing

out the type J (1,2)1 component

zx− 1

xz, zy+1y z

yields the type J (2,2)

1,1 Janetbasis

z2 + xy z1, z1,y + x

y z1,x + 1y z1, z1,xx + 4

xz1,x + 2x2 z1

.

Lemma 2.7 If a Janet basiszxx +A1zy +A2zx +A3z, zxy +B1zy +B2zx +B3z,

zyy + C1zy + C2zx + C3z

of type J (1,2)3,2 has a component z1 ≡ zx + az, z2 ≡ zy + bz of type J (1,2)

1 ,

its quotient is of type J (2,2)2,3 and is explicitly given by

z1,x +A1z2 + (A2 − a)z1, z1,y + (B1 − a)z2 +B2z1,

z2,x +B1z2 + (B2 − b)z1, z2,y + (C1 − b)z2 + C2z1.

If it has a component z1 ≡ zy + a1zx + a2z, z2 ≡ zxx + b1zx + b2z of typeJ (1,2)

2,1 , its quotient is of type J (2,2)1,2 and is explicitly given by

z2 +A1z1, z1,x + (a1A1 +B1)z1, z1,y + (C1 − 2a21A1 − a2)z1.


If it has a component z1 ≡ zx+a1z, z2 ≡ zyy + b1zy + b2z of type J (1,2)2,2 , its

quotient is of type J (2,2)1,2 and is explicitly given by

z2 + C2z1, z1,x + (A2 − a1)z1, z1,y +B2z1.

Example 2.55 Consider the type J (1,2)3,2 Janet basis

zxx +4xzx +

2x2 z, zxy +

1xzy, zyy +

1yzy −

x

y2 zx −2y2 z

.

Dividing out the type J (1,2)1 Janet basis

z1 ≡ zx+ 1

xz, z2 ≡ zy + 1y z

yields

the type J (2,2)2,3 Janet basis

z1,x +3xz1, z1,y, z2,x +

1xz2 −

1yz1, z2,y −

x

y2 z1

.

Dividing out the type J (1,2)2,2 Janet basis z1 ≡ zx+ 1

xz, z2 ≡ zyy+ 1y zy−

1y2 z

yields the type J (2,2)1,2 Janet basis

z2 − x

y2 z1, z1,x + 3xz1, z1,y

.

Lemma 2.8 If a Janet basis zx + A1z, zyyy + B1zyy + B2zy + B3z oftype J (1,2)

3,3 has a component z1 ≡ zx + az, z2 ≡ zy + bz of type J (1,2)1 , its


z1, z2,x +A1z2, z2,yy + (B1 − b)z2,y + (B2 − bB1 − 2by + b2)z2.

If it has a component z1 ≡ zx+a1z, z2 ≡ zyy + b1zy + b2z of type J (1,2)2,2 , its


z1, z2,x +A1z2, z2,y + (B1 − b1)z2.

Construction of Janet Bases. It is difficult to construct examples forthe various types of Janet bases by prescribing the coefficients. The reasonis that finding sets of coefficients satisfying the coherence conditions comesdown to solving systems of nonlinear pde’s for which solution algorithms arenot known.

A different approach starts from the solutions. A basis for the solutionspace is given, and the problem is to generate a Janet basis with this solutionspace. For a single linear ode the answer has been given in (2.4), providingexplicit expressions for the coefficients in terms of a fundamental system. Forthe two second order Janet bases of type J (1,2) the answer is to be obtainedin Exercises 2.9 and 2.10.

It should be emphasized, however, that proceeding in this way only spe-cial classes of Janet bases may be obtained, i.e., those that are completely

76

reducible with first order right factors. Due to the requirement that the Janetbasis to be constructed has the rational function field of the independent vari-ables as base field, the fundamental system has to be hyperexponential. Anyother choice in general will lead to coefficients that are not in this base field.This problem exists already for linear ode’s.

An algorithm will be described now which is based on the constructive proofof a theorem due to Lie [112], vol. I, page 183. In the special case where thedependency on the integration constants is linear, the desired system is alsolinear and homogeneous. Assume the following system of equations is given.

z1 ≡ C1z1,1 + . . .+ Crz1,r,

......

zm ≡ C1zm,1 + . . .+ Crzm,r

(2.36)

where zj,k, j = 1, . . . ,m, k = 1, . . . , r are given functions of x and y, z1, . . . , zmare undetermined functions of x and y, and the Ck are constants to beeliminated. To this end, by a suitable number of differentiations, an en-larged system of rank r is generated from which the Ck may be eliminated.Upon resubstitution, the desired system of pde’s is obtained. The algorithmConstructJanetBasis takes expressions (2.36) as input and returns the cor-responding Janet basis in the specified term ordering.

Algorithm 2.12 ConstructJanetBasis(z1, . . . zm, O). Given a systemof expressions of the form (2.36), the Janet basis in the term ordering O isreturned such that (2.36) is its general solution.S1 : Set up linear system. Determine d such that (d+ 1)(d+ 2) > r.S2 : Rank? Generate linear system (L) for C1, . . . , Cr by differentiatingw.r.t. x and y up to order d. It has the form with j = 1, . . . ,m.

C1zj,1 + . . .+ Crzj,r = zj ,

C1zj,1,x + . . .+ Crzj,r,x = zj,x,

C1zj,1,y + . . .+ Crzj,r,y = zj,y, (L)...

......

C1zj,1,yy... + . . .+ Crzj,r,yy... = zj,yy...,

S3 : Rank? If the rank of the coefficient matrix is less than r, set d := d+1and goto S2.S4 : Determine C’s. Take a subsystem of (L) of rank r and determineC1, . . . , Cr from it.S5 : Generate Janet basis. Substitute the values for the C’s obtained instep S4 into the remaining equations and generate a Janet basis. If itssolution space has dimension r, return it, otherwise set d := d + 1 andgoto S2.


Example 2.56 For m = 1, r = 2 let z1 = z = C1(x2− y2)+C2xy. In stepS1, d = 1 is found. Step S2 generates the system of equations

C1(x2 − y2) + C2xy = z, 2C1x+ C2y = zx, −2C1y + C2x = zy.

The rank of the coefficient matrix is two. In step S4 the last two equationsare employed for determining C1 and C2 with the result

C1 =x

2(x2 + y2)zx −

y

2(x2 + y2)zy, C2 =

y

x2 + y2 zx +x

x2 + y2 zy.

Substitution into the first equation yields a single first order equation forz. Therefore in step S5 the value of d is increased to 2 and the algorithmcontinues with step S2 where the system 2C1 = zxx, C2 = zxy, −2C1 = zyy isgenerated. Substituting C1 and C2 and joining the previously obtained firstorder equation leads in step S5 to the Janet basis

zx +y

xzy −

2xz = 0, zyy −

2yx2 + y2 zy +

2x2 + y2 z = 0 (2.37)

which is the final answer.

2.4 Solving Partial Differential EquationsUntil now the goal has been to obtain information on the solutions of linear

pde’s without determining them explicitly. Furthermore, it has been shownhow given pde’s may be replaced by simpler ones, e. g. by decomposing theminto lower order equations. A common feature of these efforts was that thenecessary calculations took their course within the base field of the origi-nally given equations. In this section, the goal is to determine solutions inclosed form. This may require to enlarge the function field. Various casescorresponding to special values of n and m, the number of independent anddependent variables respectively, and the number N of equations will be con-sidered. Except in the last subsection, all equations are of first order.Linear Equations for a Single Function. First order systems containinga single dependent variable and any number of independent variables, i.e.,m = 1 and n arbitrary, have been studied extensively in the literature. Thediscussion in this section follows closely the Encyklopadie article by von Weber[187], or Chapter II in Goursat’s book [57], see also Kamke [87], page 45 ff.At first, N = 1 is assumed. Let the function z(x1, . . . , xn) be determined bythe homogeneous equation

Az ≡ a1∂z

∂x1+ a2

∂z

∂x2+ . . .+ an

∂z

∂xn= 0. (2.38)

78

In particular there is no term proportional to z, and ak ≡ ak(x1, . . . , xn)belongs to Q(x1, x2, . . . , xn) or an extension of it. A fundamental system orintegral basis for this equation is a set of n− 1 independent functions

ψ1(x1, . . . , xn), . . . , ψn−1(x1, . . . , xn)

such that Aψk = 0 for k = 1, . . . , n − 1. The following theorem reduces thesolution of (2.38) to solving a system of ode’s. Its proof may be found in theliterature quoted above.

Theorem 2.19 The general solution of equation (2.38) may be describedas follows.

i) A fundamental system for (2.38) is given by n-1 first integrals of theso-called characteristic system

dx1

dt= a1(x1, . . . , xn), . . . ,

dxndt

= an(x1, . . . , xn).

ii) The general solution of (2.38) is given by an undetermined functionΨ(ψ1, . . . , ψn−1) of the first integrals.

It should be noted, however, that there is no guarantee that the fundamentalsystem may always be explicitly determined. The reason is that solving thecharacteristic system amounts to solving general first order ode’s for whichan algorithm does not exist.

There is a remarkable generalization of the above theorem to inhomogeneousquasilinear equations by Kamke [86] II, §3, no. 15.

Theorem 2.20 A solution of the quasilinear first order equation

a1∂z

∂x1+ a2

∂z

∂x2+ . . .+ an

∂z

∂xn= r (2.39)

where ak ≡ ak(x1, . . . , xn, z) for k = 1, . . . , n and r ≡ r(x1, . . . , xn, z) may beobtained as follows. Determine any integral w = ψ(x1, . . . , xn, z) of

a1∂w

∂x1+ a2

∂w

∂x2+ . . .+ an

∂w

∂xn+ r

∂w

∂z= 0

by Theorem 2.19 and solve ψ(x1, . . . , xn, z) = C for z; C 6= 0 is a constant.

The special case n = 2 of the above theorem will occur frequently later onin this book. Therefore it is described in more detail in the following corollary.

Corollary 2.7 The general solution of the linear partial differential equa-tion

zx + a1zy + a2z + a3 = 0 (2.40)


with ak ≡ ak(x, y) ∈ Q(x, y), k = 1, . . . , 3 may be described as follows. Letφ(x, y) = const be the first integral obtained from the solution of the first order

ode dydx

= a1(x, y). Let y = φ(x, y) and assume that the inverse y = φ(x, y)exists. Then the general solution of (2.40) is

z(x, y) =(Φ(φ)−

∫a3(x, φ) exp

( ∫a2(x, φ)dx

)dx

)exp

(−

∫a2(x, φ)dx

) ∣∣∣y=φ

(2.41)with Φ an undetermined function of its argument.

Proof It is based on the procedure described in Kamke [87], Section 4.2. Ahomogeneous equation corresponding to (2.40) for a new functionw(x, y, z) is wx + a1wy − (a2z + a3)wz = 0. A first integral φ is obtained

from dydx

= a1. If it is applied for introducing a new variable y, the equationwx − (a2z + a3)|y=φwz = 0 for w is obtained with the first integral

ψ = z exp∫a2(x, φ)dx+

∫a3(x, φ) exp

( ∫a2(x, φ)dx

)dx.

Consequently, the general solution w = Ψ(φ, ψ) is obtained with an undeter-mined function Ψ. Resubstituting y = φ(x, y) and solving for z yields (2.41).

This corollary is particularly useful if a special solution of the homoge-neous equation is already known, then solving the inhomogeneous equation isreduced to an integration.

Example 2.57 Let 2xzx + 3yzy = 1 be the given equation. With the

notation of Corollary 2.7, a1 = 3y2x , a2 = 0 and a3 = − 1

2x . The first order

equation dydx

= 3y2x leads to φ(x, y) = y2

x3 . From (2.41) the general solution is

obtained in the form z(x, y) = Φ(y2

x3

)−1

2 log x with Φ an undetermined

function of its argument.

Now, let N > 1. It is interesting to see how several concepts like integra-bility conditions and reductions have evolved in the course of time for thisparticular case. The subsequent theorem shows how the Jacobian normalform (Kamke [87], Chapter 6.2) is obtained naturally as a Janet basis.

Theorem 2.21 Let a first order system for a function z(x1, . . . , xn) begiven as

a1,1∂z∂x1

+ a1,2∂z∂x2

+ . . .+ a1,n∂z∂xn

= 0,

a2,1∂z∂x1

+ a2,2∂z∂x2

+ . . .+ a2,n∂z∂xn

= 0,

. . . . . .

ak,1∂z∂x1

+ ak,2∂z∂x2

+ . . .+ ak,n∂z∂xn

= 0.

(2.42)

80

The coefficients ai,j(x1, . . . , xn) belong to some function field F . For the vari-able ordering x1 > x2 > . . . > xn, the Janet basis for this system is identicalto the Jacobian normal form, possibly after a new enumeration of the xi,

∂z∂xr

+ br,r+1∂z

∂xr+1+ . . .+ br,n

∂z∂xn

= 0,

∂z∂xr−1

+ br−1,r+1∂z

∂xr+1+ . . .+ br−1,n

∂z∂xn

= 0,

. . .

∂z∂x1

+ b1,r+1∂z

∂xr+1+ . . .+ b1,n

∂z∂xn

= 0

(2.43)

with 1 ≤ r ≤ n and bi,j ∈ F .

Proof By algebraic reduction of the system (2.42), the following autore-duced form may always be achieved, possibly after a new enumeration of theindependent variables.

∂z∂xp

+ ap,p+1∂z

∂xp+1+ . . .+ ap,n

∂z∂xn

= 0,

∂z∂xp−1

+ ap−1,p+1∂z

∂xp+1+ . . .+ ap−1,n

∂z∂xn

= 0,

. . . . . .

∂z∂x1

+ a1,p+1∂z

∂xp+1+ . . .+ a1,n

∂z∂xn

= 0

(2.44)

with 1 ≤ p ≤ n. The integrability conditions have the form

∂∂xj

(∂z∂xi

+ ai,p+1∂z

∂xp+1+ . . .+ ai,n

∂z∂xn

)− ∂∂xi

(∂z∂xj

+ aj,p+1∂z

∂xp+1+ . . .+ aj,n

∂z∂xn

)= 0

for 1 ≤ i < j ≤ p. All second order terms in these expressions cancel eachother identically. The remaining first order equations are reduced w.r.t. to(2.44). If the result is always zero, (2.44) is the desired Janet basis. Otherwisethose that do not vanish are added to the system (2.44) and the process isrepeated. The results described in the preceding section guarantee that thisprocess terminates after a finite number of iterations.

Example 2.58 The following system of pde’s

zx2 + (x4x3 − x2)zx3 + (x4x3x1 − x2x1 + x2)zx4 = 0,

zx1 + (x4 + x2 − 3x1)zx3 + (x4x1 + x3 + x2x1)zx4 = 0

for a function z ≡ z(x1, x2, x3, x4) has been discussed by Imschenetzky [79];as usual zxi

≡ ∂z∂xi

. The Janet basis for x1 > x2 > x3 > x4 iszx3 + x1zx4 , zx2 + x2zx4 , zx1 + (x3 + 3x2

1)zx4


and for x4 > x3 > x2 > x1 it iszx2 −

x2

x3 + 3x21

zx1 , zx3 −x1

x3 + 3x21

zx1 , zx4 +1

x3 + 3x21

zx1

.

From each of these Janet bases it is obvious that the general solution dependson an undetermined function of a single argument. Explicitly it is given byφ(x4 − x1x3 − 1

2x22 − x3

1

).

Linear Equations with Symmetries. In some applications it is advanta-geous to transform a linear equation for a function f(x1, x1, . . . , xn)

Af ≡ a1∂f

∂x1+ a2

∂f

∂x2+ . . .+ an

∂f

∂xn= 0 (2.45)

to a new set of variables y1, . . . , yn. Let them be defined by yk ≡ ψk(x1, . . . , xn)with the inversions xk ≡ φk(y1, . . . , yn). In these variables (2.45) becomes

B1∂g

∂y1+B2

∂g

∂y2+ . . .+Bn

∂g

∂yn= 0 (2.46)

where Bk = Aψk|x1=φ1,...,xn=φnfor 1 ≤ k ≤ n and g ≡ f(φ1, . . . , φn). The

proof of this expression is considered in Exercise 2.21.

Example 2.59 Let A = x∂x + y∂y by given and new variables u and v beintroduced by x = uv, y = v

u with the inverse u =√xy , v =

√xy. Applying

the above rule yields the transformed operator v∂v.

Of particular interest are those nontrivial transformations leaving the equa-tion Af = 0 unchanged. Due to its importance a special term is introducedfor them.

Definition 2.10 (Symmetry of a linear pde) If a linear homogeneous pde(2.45) is invariant w.r.t. a variable transformation, it is called a symmetry of(2.45). This is also expressed by saying that the transformation is admittedby (2.45).

It may occur that the totality of transformations admitted by a given equa-tion (2.45) make up a group depending on one or more parameters. For thisconcept the reader is referred to Section 3.1. A convenient criterion for theinvariance w.r.t. a group may be given in terms of its infinitesimal generatorsas follows.

Theorem 2.22 A linear homogeneous pde (2.45) is invariant under thegroup with the infinitesimal generator X = ξ1

∂∂x1

+ . . .+ ξn∂∂xn

if and onlyif there holds [A,X] = ρ(x1, . . . , xn)A.

The commutator [A,X] is defined by equation (3.10) in the next chapter.The solution procedure for an equation (2.45) varies with the type of the

82

symmetry group that it admits. A detailed discussion may be found in theoriginal work of Lie [108], § 10. Two special cases that are of particularimportance later on in this book are described next. The first one deals witha solvable symmetry group, for its definition see page 118.

Theorem 2.23 (Lie 1877) Let the linear homogeneous pde to be solved begiven in the form

Af ≡ a1∂f

∂x1+ a2

∂f

∂x2+ . . .+ an

∂f

∂xn= 0.

Assume that it admits

Xk ≡ ξk,1∂

∂x1+ ξk,2

∂

∂x2+ . . .+ ξk,n

∂

∂xn

for k = 1, . . . , n − 1, and that the relations [Xi, Xk] =∑k−1ρ=1 c

ρi,kXρ for

i < k and i = 1, . . . , n − 1, k = 2, . . . , n − 1, cρi,k constant are valid. Thena fundamental system may be obtained as follows. Let the determinant ∆ bedefined by

∆ =

∣∣∣∣∣∣∣∣∣a1 a2 . . . anξ1,1 ξ1,2 . . . ξ1,n...

...ξn−1,1 ξn−1,2 . . . ξn−1,n

∣∣∣∣∣∣∣∣∣ 6= 0.

Then an integral Φ1 of Af = 0 is

Φ1 =∫

1∆

∣∣∣∣∣∣∣∣∣∣∣

dx1 dx2 . . . dxna1 a2 . . . anξ1,1 ξ1,2 . . . ξ1,n...

...ξn−2,1 ξn−2,2 . . . ξn−2,n

∣∣∣∣∣∣∣∣∣∣∣. (2.47)

If new variables x1 = Φ1, xk = xk for k = 2, . . . , n are introduced, derivativesw.r.t. to x1 do not occur any more, Af = 0 and Xkf = 0 for k = 1, . . . , n− 3are a complete system allowing Xn−2. This problem is of the same type asthe original one with the number of variables diminished by one for which thesame procedure may be repeated until n− 1 integrals are obtained.

This theorem will be employed in Chapter 7 for solving ode’s with symme-tries. In order to apply it, the knowledge of a sufficient number of symmetriesof a certain kind is required. Because there is no algorithmic method for de-termining them, this knowledge must come from the application from whichthe equation Af = 0 originates. There are other methods for taking advan-tage of symmetries for solving first order pde’s or systems of pde’s. A fewcases are described in the subsequent examples.


Example 2.60 Let A ≡ ∂x + yz ∂y + y

z ∂z, X1 ≡ ∂x, X2 ≡ x∂x + y∂y + z∂z.Due to [X1, A] = 0, [X2, A] = −A, [X1, X2] = X1 the above theorem can beapplied. In the first step ∆ = y

x (y − z) and the integral for (2.47) is∫z

y

1y − z

(yzdy − y

zdz

)= 2 log (y − z).

Therefore Φ1 = y − z may be chosen. Introducing new variables x = x,y = y− z, z = z yields A = x∂x and X1 = ∂x. Integration and resubstitutiongives Φ2 = x− z + (y − z) log y.

Theorem 2.24 (Lie 1874) Let the linear homogeneous pde Af = 0 as inthe preceding theorem allow q infinitesimal generators B1, . . . , Bq. Assumethat there are exactly m relations of the form

Bk = βkm+1Bm+1 + . . .+ βkqBq + αA, k = 1, . . . ,m.

Then the coefficients βkj obey Aβkj = 0, i.e., they are integrals of the givenlinear pde Af = 0.

The proof may be found in Lie[107], §8, Theorem VII. A simple applicationis given next.

Example 2.61 Let A = ∂x + z∂y −z(z2 + 1)

2x ∂z. It allows the threesymmetry generators U1 = ∂y, U2 = x∂x+y∂y and U3 = xy∂x+ 1

2y2∂y−xz2∂z

because [A,U1] = 0, [A,U2] = A, [A,U3] = (xz + y)A. The relation

U3 +( 2xzz2 + 1

− y)U2 +

(2xz(xz − y)z2 + 1

+y2

2

)U1 −

2x2z

z2 + 1A = 0

yields the integrals Φ1 = 2xzz2 + 1

− y and Φ2 = 2xz(xz − y)z2 + 1

+ y2

2 .

Linear Systems for m=1, n=2 and N=2. In this subsection various kindsof linear pde’s are considered that will occur frequently in the applicationslater on in this book. Usually its solution may be expressed in terms of pathintegrals in the x− y−plane. If p ≡ p(x, y) and q ≡ q(x, y) are two functionsof x and y obeying py = qx, it is defined by∮pdx+ qdy =

∫ x

x0

p(ξ, y0)dξ +∫ y

y0

q(x, η)dη =∫ y

y0

q(x0, η)dη +∫ x

x0

p(ξ, y)dξ.

(2.48)The lower integration limit of the first integral contributes the integrationconstant. As a rule, any integration problem will be considered as solved assoon as it has been reduced to a form that is covered by the lemmata of thissubsection. A straightforward generalization of a linear homogeneous ode isconsidered first.

84

Lemma 2.9 The general solution of the homogeneous system

zx + az = 0, zy + bz = 0

where a ≡ a(x, y), b ≡ b(x, y) and ay = bx is z = exp (−∮adx+ bdy).

Example 2.62 Consider zx + xy2z = 0 and zy + x2yz = 0. The inte-gral (2.48) yields∮

xy2dx+ x2ydy =∫ξy2

0dξ +∫x2ηdη

= 12y

20(x2 − x2

0) + 12x

2(y2 − y20) = 1

2x2y2 + C

where C is a constant, and finally z = C exp (− 12x

2y2).

Lemma 2.10 The general solution of the inhomogeneous system

zx + az = p, zy + bz = q

where a, b, p and q depend on x and y and obey the integrability conditionsay = bx and py − qx = aq − bp may be written in the form

z = z0

( ∮pdx+ qdy

z0+ C

)where z0 = exp

(−

∮adx+ bdy

)is the solution of the homogeneous system with p = q = 0 and C is a constant.

Proof Assume that the desired solution has the form z = C(x, y)z0 withz0 a solution of the homogeneous system. Differentiation and substitutioninto the given system leads to Cx = p

z0 and Cy = qz0 from which the given

result is obvious.

Example 2.63 Consider the system zx + yz = y, zx + xz = x. ByLemma 2.9 the homogeneous system has the solution z0 = Ce−xy. Substitut-ing it into the integral of the preceding lemma leads to

z = e−xy(∮

exy(ydx+ xdy) + C)

= Ce−xy + 1.

A special second order system corresponding to Janet basis type J (1,2)2,1 that

will occur frequently later on is treated next.

Lemma 2.11 Let the function z(x, y) be determined by the system of equa-tions zy + azx = 0, zxx + czx = 0 with (ax − ac)x = cy. The nonconstantmember of a fundamental system is given by

z =∮

exp(−

∮cdx+ (ax − ac)dy

)(dx− ady).


The proof of this lemma is the subject of Exercise 2.27.

Example 2.64 An example of such a system is

zy +x

(x− 1)yzx = 0, zxx −

x2 − 2x+ 2x(x− 1)

zx = 0.

Applying the formulas given above yields for the nonconstant member of afundamental system z = 1

xy ex. In this particular case the same answer may

be obtained from the componentzx − x− 1

x z, zy + 1y z

because the given

system is completely reducible. In general, however, this is not true and theexpression in the lemma must be applied.

Quasilinear Systems for m=1, n=2 and N≥2. Systems comprising morethan two equations are the next topic. Moreover they do not need to be linearin the dependent variable any more.

Lemma 2.12 Let a system of first order quasilinear equations

aizx + bizy = ri, 2 ≤ i ≤ N

for z ≡ z(x, y) be given where ai ≡ ai(x, y), bi ≡ bi(x, y) and ri ≡ ri(x, y, z),a1b2 − a2b1 6= 0. Then z obeys

zx = − b1r2 − b2r1a1b2 − a2b1

, zy = a1r2 − a2r1a1b2 − a2b1

,

(a1b2 − a2b1)rj + (a2bj − ajb2)r1 − (a1bj − ajb1)r2 = 0 for j ≥ 3.(2.49)

Proof The expressions for zx and zy follow by algebraic elimination fromthe first two equations. The remaining conditions either assure the coherenceof the full system, or they reduce the originally given differential equations toan elimination problem.

Depending on the form of the right hand sides ri, various alternatives haveto be distinguished. If the ri do not contain z, or if only r1 and r2 dependlinearly on z and the remaining conditions are identically satisfied, the firsttwo equations lead to a representation of z as a path integral by Lemma 2.9or 2.10. If any ri contains z, and if the conditions for j ≥ 3 given in theabove lemma may be solved for z, the general solution does not contain aconstant, and the first two equations reduce to consistency conditions. Ifthey are violated the full system is inconsistent. Systems of this kind occurin the next chapter when groups acting on a two-dimensional manifold aretransformed into canonical form.Partial Riccati-Like Systems: A Single Function. The decompositionof Janet bases into irreducible components has been shown to come down to

86

solving various kinds of partial differential equations. Their structure suggeststo call them partial Riccati-like equations. In this section, those systemsoccuring in the decomposition of Janet bases of type J (1,2) of order not higherthan three will be considered.

First of all a few definitions are introduced. Two rational functions f, g ∈Q(x, y) are defined to be an integrable pair (f, g) if ∂yf = ∂xg. Two integrablepairs (f, g) and (p, q) are said to be equivalent, denoted by ” ∼ ”, if there exists

a nonzero h ∈ Q(x, y) such that f − p = ∂xhh

and g − q = ∂yhh

; ∼ defines anequivalence relation on the set of integrable pairs.

Let a and b belong to Q(x, y). If they are not an integrable pair it may bepossible that there exist two polynomials p, q ∈ Q[x, y] such that(a + ∂xp

p , b + ∂yqq

)is an integrable pair. In the above mentioned article

by Li and Schwarz [104] an algorithm is described that allows determining apair (p, q) for any given (a, b) if it does exist.

An element a of some larger function field comprising Q(x, y) is said to be

a hyperexponential if both ∂xaa and ∂ya

a belong to Q(x, y). If a is a hyperex-

ponential,(∂xaa ,

∂yaa

)is an integrable pair. Conversely, for an integrable pair

(f, g), the hyperexponential exp (∮fdx+ gdy) may be constructed. If (a, b)

and (p, q) are two integrable pairs, then (f, g) ∼ (p, q) if and only if the ratioof their hyperexponentials is in Q(x, y), possibly multiplied by some constant.

The simplest Riccati-like system involving a single unknown function is thesubject of the next theorem.

Theorem 2.25 The first order Riccati-like system of pde’s

e1 ≡ zx +A1z2 +A2z +A3 = 0, e2 ≡ zy +B1z

2 +B2z +B3 = 0 (2.50)

is coherent and its general solution depends on a single constant if and onlyif its coefficients satisfy the constraints

A1,y −B1,x −A1B2 +A2B1 = 0,

A2,y −B2,x − 2A1B3 + 2B1A3 = 0,

A3,y −B3,x −A2B3 +A3B2 = 0.

(2.51)

Let Ak, Bk ∈ Q(x, y), A1 6= 0, B1 6= 0 and system (2.51) be satsified. If(2.50) has a rational solution, one of the following alternatives applies.


1A1

rxr + C

+ p =1B1

ryr + C

+ p (2.52)

where p, r ∈ Q(x, y) and C is the integration constant.

ii) There is a single one, or there are two inequivalent special rational so-lutions.


Proof The coefficient constraints are the integrability conditions for sys-tem (2.50). Substituing z = u

w into the given equations yields

ux +u(A1u− wx)

w+A2u+A3w = 0, uy +

u(B1u− wy)w

+B2u+B3w = 0.

In order to obtain a linear system for u and w, the constraints

wx −A1u+ Pw = 0, wy −B1u+Qw = 0 (2.53)

are necessary where P,Q ∈ Q(x, y) still have to be determined. Substitutingthese conditions into the above system for u and w yields

ux + (A2 + P )u+A3w = 0, uy + (B2 +Q)u+B3w = 0. (2.54)

The consistency of the above substitution requires that the system comprisingthe equations (2.53) and (2.54) be a Janet basis. This is assured if

Py −Qx +A1B3 −A3B1 = Py −Qx + 12 (A2,y −B2,x) = 0.

The second form is obtained by applying (2.51). A possible choice for P andQ is P = − 1

2A2, Q = − 12B2 respectively. It leads to the linear system

ux + 12A2u+A3w = 0, uy + 1

2B2u+B3w = 0,

wx −A1u− 12A2w = 0, wy −B1u− 1

2B2w = 0.

Let u = C1u1 + C2u2, w = C1w1 + C2w2 be its general solution. If the ratiouw is rational, it may be rewritten as

z =u

w=Cu1

w1+u2

w2

C +w2

w1

=

u2

w1− u1

w1

w2

w1

C +w2

w1

+u1

w1.

Applying (w2w1

)x

= w2,xw1− w1w1,x

w21

= A1

(u2w2− u1w2

w21

),(

w2w1

)y

= w2,yw1− w1w1,y

w21

= B1

(u2w2− u1w2

w21

)the assignment r = w2

w1yields the representation of case i). On the other

hand, if u1w1

is rational and u2w1

is not or vice versa, (2.50) has exactly onerational solution. If neither are rational, there is none. If both are rationalbut the ratio u

w is not, there are exactly two rational solutions. These casescover alternative ii).

Based on this theorem, an algorithm for determining rational solutions ofsystems (2.50) may be designed as follows. The solution procedure is initi-ated with the first equation. If its general solution is rational, it contains a

88

constant w.r.t. x and has the general form rxr + C(y) + p with r, p ∈ Q(x, y).

Substituting this expression into the second equation yields the first orderequation

rxCy − (py +B1p2 +B2p+B3)C2

−[rxy + 2pyr + 2B1p(rx + pr) +B2(rx + 2pr) + 2B3r

]C

−(rxy + pyr)r + rxry −B1(rx + pr)2 −B2(rx + pr)r −B3r2 = 0

(2.55)for the y−dependence of C. Because the dependence on x is completelyexplicit, it may be decomposed w.r.t. powers of x. The coefficient of C2

has the form of the second equation of (2.50) for p. Therefore, dependingon whether or not p satisfies this equation, this leads to a system of linearor Riccati ode’s. If its general solution is rational, the same is true for thefull system. Otherwise special solutions may exist leading to special rationalsolutions of the full system. If the first equation does not allow a generalrational solution, special rational solutions may exist. Those also fullfillingthe second equation are special solutions of the full system. The followingalgorithm is organized according to these steps.

Algorithm 2.13 RationalSolutionPartialRiccati1(e1, e2). Given two par-tial Riccati-like equations (2.50), its general rational solution, or the maximalnumber of special rational solutions are returned, or failed if none exists.S1 : Rational solutions of first equation. Determine the rational solutionsof e1. If there are none return failed.S2 : Special rational solutions. If the rational solutions obtained in S1are special, substitute them into e2 an return those that satisfy it.S3 : Solve equation for C. Substitute p and r as obtained in S1 into(2.55) and solve the resulting system of ode’s for C. If it does not haveany rational solutions return failed, otherwise substitute them into zand return the result.

Example 2.65 Consider the equations

zx + z2 +2xz

x2 − 2y

−

2y(

x2 − 2y

)2= 0, zy +

x

yz2 +

( 2

y − 2x2

− 1y

)z = 0.

Its coefficients obey the coherence conditions (2.51). In step S1 of the abovealgorithm the general solution of the first equation is obtained in the form

z(x, y, C(y)

)=

1C(y) + x

− xy

x2y − 2.

In step S3 the single equation Cy + C = 0 for the y-dependence of C isobtained. Its solution C = C

y with C a constant yields the general rational

solution z = yxy + C −

xyx2y − 2

of the full system.


The system (2.50) is invariant w.r.t. exchange of x and y. Therefore itmakes no difference if the algorithm is started by solving the equation deter-mining the y-derivative, this is the subject of Exercise 2.25.

The special case B1 = 0 of (2.50), i.e., a Riccati equation for the x-dependence and a linear equation for the y-dependence is of some interestin later applications. In Exercise 2.27 it is shown that the solution to anysuch system may be written as

z =1

2A1

(A1,x

A1−A2

)+

1A1

C (2.56)

where C ≡ C(x) satisfies the Riccati equation

Cx + C2 +12

( A1,x

A1−A2

)x

+A1A3 −14

( A1,x

A1−A2

)2

= 0. (2.57)

The terms not containing C are independent of y. This property makes it intoa genuine ode problem.

A system closely related to the preceding one is

zxx + z2x +A1zx +A2 = 0, zy +B1zx +B2 = 0 (2.58)

which occurs in the analysis of several symmetry classes later on in this book.The coherence conditions for its coefficients are

B1,xx + 2B2,x − (A1B1)x = A1,y,

B2,xx +A1B2,x −A2B1,x − (A2B1)x = A2,y.

The first equation (2.58) is a first order Riccati equation for zx. If its generalsolution is rational, z is obtained from the system

zx =rx

C + r+ p, zy = − B1rx

C + r−B1p−B2 (2.59)

with C ≡ C(y). The integrability condition for its right hand sides leads tothe first order ode

rxCy −((B1p+B2)x + py

)C2

−[2r((B1p+B2)x + py) + (B1rx + ry)x

]C

+rxry − (B1rx + ry)xr +B1r2x = 0

(2.60)

for the y-dependence of C. If (B1p + B2)x + py = 0 it is a linear equation,otherwise a Riccati equation. Substituting the solution for C into (2.59), thesolution for z is obtained by Lemma 2.10. If the first equation (2.58) has onlyspecial rational solutions, they are substituted into the second one and z isagain obtained by Lemma 2.10.Partial Riccati-Like Systems: Two Functions. The next Riccati-likesystem to be considered contains two unknown functions z1 and z2, depending

90

on x and y. Originally it occured in Lie’s investigation of certain second orderode’s. As will be seen below, the decomposition of various types of Janetbases leads to systems of this kind as well. In order to make the algorithmsto be designed more generally applicable, the following class of systems willbe considered.

e1 ≡ z1,x + a0z21 + a1z1 + a2z2 + a3 = 0,

e2 ≡ z1,y + b0z1z2 + b1z1 + b2z2 + b3 = 0,

e3 ≡ z2,x + c0z1z2 + c1z1 + c2z2 + c3 = 0,

e4 ≡ z2,y + d0z22 + d1z1 + d2z2 + d3 = 0.

(2.61)

If a0d0 6= 0 is assumed, this system is coherent if either of the following setsof constraints is satisfied.

b0 − d0 = 0, a0 − c0 = 0, d0,x + b2c0 − c2d0 = 0,

c3,y − d3,x + a3d1 − b3c1 − c2d3 + c3d2 = 0,

c2,y − d2,x + a2d1 − b2c1 − b3c0 + 2c3d0 = 0,

c1,y − d1,x + a1d1 − b1c1 − c0d3 + c1d2 − c2d1 = 0,

c0,y − b1c0 + c1d0 = 0,

a3,y − b3,x − a1b3 − a2d3 + a3b1 + b2c3 = 0,

a2,y − b2,x − a1b2 + a2b1 − a2d2 + a3d0 + b2c2 = 0,

a1,y − b1,x − a2d1 + b2c1 − 2b3c0 + c3d0 = 0

(2.62)

or

d1 = 0, c1 = 0, c0 = 0, b2 = 0, b0 = 0, a2 = 0,

d0,x − c2d0 = 0, c3,y − d3,x − c2d3 + c3d2 = 0, c2,y − d2,x + 2c3d0 = 0,

a3,y − b3,x − a1b3 + a3b1 = 0, a1,y − b1,x − 2a0b3 = 0, a0,y − a0b1 = 0.(2.63)

If the first alternative (2.62) applies, new dependent variables z1 and z2 areintroduced by

z1 = 1a0z1 − 1

3a0

(a1 + c2 −

∂x(a0d0)a0d0

),

z2 = 1d0z2 − 1

3d0

(b1 + d2 −

∂y(a0d0)a0d0

).

(2.64)

The system (2.61) corresponding to conditions (2.62) is transformed into thenormalized coherent system

e1 ≡ z1,x + z21 +A1z2 +A2z1 +A3 = 0,

e2 ≡ z1,y + z1z2 +B1z2 +B2z1 +B3 = 0,

e3 ≡ z2,x + z1z2 +B1z2 +B2z1 +B3 = 0,

e4 ≡ z2,y + z22 +D1z2 +D2z1 +D3 = 0.

(2.65)


The new coefficients A1, . . . , D3 are nonlinear expressions of the old onesa0, . . . , d3. Due to the coherence they satisfy

A1,y −B1,x −A2B1 +A1B2 −A1D1 +A3 +B21 = 0,

A2,y −B2,x −A1D2 +B1B2 −B3 = 0,A3,y −B3,x −A2B3 −A1D3 +A3B2 +B1B3 = 0

B1,y −D1,x +A1D2 −B1B2 +B3 = 0,B2,y −D2,x +A2D2 −B2

2 +B2D1 −B1D2 −D3 = 0,B3,y −D3,x +A3D2 −B2B3 −B1D3 +B3D1 = 0.

(2.66)

Example 2.66 Consider the equations

z1,x + z21 + 3y2

x3 z1 + y3

x4 z2 −3x2y2 − 3y4

x6 = 0, z1,y + z1z2 + 2x2y − y3

x5 = 0,

z2,x + z1z2 − 2x2y + y3

x5 = 0, z2,y + z22 + 1

xz1 + 3yx2 z2 + x2 + 3y2

x4 = 0.

Its coefficients satisfy the conditions (2.62), but z1,y 6= z2,x. Introducing new

variables by z1 = z1 − y2

x3 , z2 = z2 − yx2 yields

z1,x + z21 + y2

x3 z1 + y3

x4 z2 = 0 z1,y + z1z2 − yx2 z1 −

y2

x3 z2 = 0,

z2,x + z1z2 − yx2 z1 −

y2

x3 z2 = 0, z2,y + z22 + 1

xz1 + yx2 z2 = 0,

i.e., the normal form (2.65).

If the second alternative (2.63) applies, new dependent variables z1 and z2are introduced by

z1 =1a0z1, z2 =

1a0z2.

In this case the system (2.61) corresponding to conditions (2.63) is trans-formed into

z1,x + z21 +A1z1 +A2 = 0, z1,y +B = 0,

z2,x + z22 + C = 0, z2,y + z2

2 +D1z2 +D2 = 0,

i.e., it decouples into two individual systems of the type (2.50) for z1 and z2and may be treated by the methods of the preceding subsection.

The subsequent theorem describes the structure of the various types ofrational solutions of system (2.65).

Theorem 2.26 If the coherent system (2.65) has a rational solution, oneof the following alternatives applies with r, s, p, q ∈ Q(x, y) and py = qx.

92

i) The general solution is rational and contains two constants. It may bewritten in the form

z1 =C2rx + sx

C1 + C2r + s+ p, z2 =

C2ry + syC1 + C2r + s

+ q.

ii) There is a rational solution containing a single constant, it may be writ-ten in the form

z1 =rx

C + r+ p, z2 =

ryC + r

+ q.

iii) There is a solution as described in the preceding case, and in additionthere is a special rational solution not equivalent to it.

iv) There is a single one, or there are two or three special rational solutionswhich are pairwise inequivalent.

Proof The coefficient constraints are the integrability conditions for thesystem (2.65). Substituting z1 = u

w , z2 = vw yields

ux + u(u− wx)w +A1u+A2v +A3w = 0,

uy + u(v − wy)w +B1u+B2v +B3w = 0,

vx + v(u− wx)w +B1u+B2v +B3w = 0,

vy + v(v − wy)w +D1u+D2v +D3w = 0.

In order to obtain a linear system for u, v and w, the constraints

wx − u+ Pw = 0, wy − v +Qw = 0 (2.67)

are necessary where P,Q ∈ Q(x, y) still have to be determined. Substitutingthese conditions into the above system yields

ux + (A1 + P )u+A2v +A3w = 0, uy + (B1 +Q)u+B2v +B3w = 0,vx +B1u+ (B2 + P )v +B3w = 0, vy +D1u+ (D2 +Q)v +D3w = 0.

(2.68)The consistency of the substitutions for z1 and z2 requires that the systemcomprising the equations (2.68) and (2.67) be a Janet basis. This is assuredif there holds

Py −Qx +B3 −B4 = Py −Qx + 13 (A1,y −B1,x +B2,y −D2,x) = 0.

The second form has been obtained by applying equations (2.66). Theseconditions are satisfied if P = − 1

3 (A1 + B2) and Q = − 13 (B1 + D2) are


chosen. It yields the linear system

ux + ( 23A1 − 1

3B2)u+A2v +A3w = 0,

uy + ( 23B1 − 1

3D2)u+B2v +B3w = 0,

vx +B1u− ( 13A1 − 2

3B2)v +B3w = 0,

vy +D1u− ( 13B1 − 2

3D2)v +D3w = 0,

wx − u− 13 (A1 +B2)w = 0, wy − v − 1

3 (B1 +D2)w = 0

(2.69)

that is equivalent to (2.61). Let its general solution be

u = C1u1 +C2u2 +C3u3, v = C1v1 +C2v2 +C3v3, w = C1w1 +C2w2 +C3w3

where wk 6= 0 for all k. The corresponding solutions of (2.61) are

z1 =u

w=

C1u1 + C2u2 + C3u3

C1w1 + C2w2 + C3w3, z2 =

v

w=

C1v1 + C2v2 + C3v3C1w1 + C2w2 + C3w3

.

If exactly one or two of the ratios ukwk and vk

wk are rational and the remainingones are not, system (2.61) has the corresponding number of rational solutions.If all three ratios are rational but u

w or vw are not, there are exactly three

rational solutions. These alternatives combined make up case iv).If not both u

w and vw are rational it may occur that they become rational

if one of the integration constants vanishes, e. g. C3. This yields

z1 =C1u1 + C2u2

C1w1 + C2w2=

u2

w1− u1

w2

w21

C +w2

w1

+u1

w1, z2 =

C1v1 + C2v2C1w1 + C2w2

=

v2w1− v1

w2

w21

C +w2

w1

+v1w1.

Applying the relations(wkw1

)x

=wk,xw1− wkw1,x

w21

=ukw1− u1

wk

w21

,

(wkw1

)y

=wk,yw1− wkw1,y

w21

=vkw1− v1

wk

w21

which are a consequence of the last two equations of (2.69), the solutions ofcase ii) are obtained. If in addition the ratios u3

w3and v3

w3are rational, this

is case iii).Assume now that u

w and vw both are rational. These quotients may be

written as

z1 =u

w=C1

u1

w1+ C2

u2

w1+u3

w1

C1 + C2w2

w1+w3

w1

=C2

( u2

w1− u1

w2

w21

)+u3

w1− u1

w3

w21

C1 + C2w2

w1+w3

w1

+u1

w1,

94

z2 =u

w=C1

v1w1

+ C2v2w1

+v3w1

C1 + C2w2

w1+w3

w1

=C2

( v2w1− v1

w2

w21

)+v3w1− v1

w3

w21

C1 + C2w2

w1+w3

w1

+v1w1

where C1 and C2 have been appropriately redefined. Applying again the aboveexpressions for the derivatives of wkw1

, the representation of case i) follow.

In order to determine the rational solutions of system (2.65) a more generalproblem will be considered first. Let u and v be two functions of x and y.Furthermore, let the x-dependence of u be determined by a Riccati ode in xof order n, and the y-dependence of v by a Riccati ode in y of order m, bothm,n ≥ 1. In addition uy = vx is required, i.e., the full Riccati-like system is

Rnx(u) = 0, Rmy (v) = 0, uy − vx = 0. (2.70)

The structure of its rational solutions is described next, it is the analogue ofTheorem 2.3 for pde’s in two dependent and two independent variables.

Theorem 2.27 Let ri, si ∈ Q(x, y) and Pi = pi,1, . . . , pi,k, pi,j ∈ Q[x, y],Ci,j ∈ Q, the sets Pi linearly independent over Q, j = 1, . . . , k, i = 1, . . . ,mand define

Sri,si

Pi≡ ri +

∂x(∑kj=1 Ci,jpi,j)∑kj=1 Ci,jpi,j

, si +∂y(

∑kj=1 Ci,jpi,j)∑kj=1 Ci,jpi,j

.

The rational solutions of (2.70) are the disjoint union of Sri,si

Pi, i = 1, . . . ,m

for some natural numbers k and m.

The proof of this result may be found in the article by Li and Schwarz [104].The main steps for designing an algorithm that determines the rational so-

lutions as described by the above theorem will be given next. At first thetwo given Riccati ode’s are solved by applying the algorithm RationalSolu-tionRiccati on page 21. If rational solutions do exist, it returns expressionsof the form given in Theorem 2.3. The Ci,j are constants w.r.t. x or y now.Their dependence on y and x respectively has to be determined such that thestructure described in Theorem 2.27 is achieved.

The last equation of the system (2.70) requires that its solutions be inte-grable pairs. In particular this is true for the terms ri and si occuring in Sri,si

Pi

because the terms involving the sums over the pi,j are already integrable pairsby themselves. If (ri, si) is not an integrable pair, but two polynomials havebeen found that make them into an integrable pair, the obvious relation

∂x(∑Ci,jhpi,j)∑Ci,jhpi,j

=∂x(

∑Ci,jpi,j)∑Ci,jpi,j

+∂xh

h(2.71)

which is valid for any h ∈ Q[x, y] may be applied to move over the logarithmicderivative of h to ri and si without changing the full expression Sri,si

Pi. If for


a given value of i it is not possible to make (ri, si) into an integrable pair bythis procedure, the respective Sri,si

Pihas to be discarded.

In order to make the integrable pairs obtained in this way into a genuinesolution of the full system (2.65), the y- and the x- dependence of the constantsw.r.t. x and y respectively have to be determined in a final step. Due to thefact that the x-dependence is completely explicit in the solution of the Riccatiequation in x, a bound may be obtained for the polynomials in x representingthe constants in y occuring in the solution of the Riccati equation in y, andvice versa. This leads to a finite linear problem over the constants, againdetails may be found in [104]. These steps are organized in terms of thesubsequent algorithm.

Algorithm 2.14 RationalSolutionPartialRiccati(Rnx ,Rmy ). Given a partial

Riccati-like system of the form (2.70), its rational solutions are returned, orfailed if none exists.S1 : Solve Riccati ode’s. Determine the rational solutions of Rnx andRmy by applying the algorithm RationalSolutionRiccati. If none existsneither for Rnx nor for Rmy , return failed.

S2 : Generate solution candidates. From the solutions obtained in S1form all possible sets Sri,si

Piand retain those that can be made into an

integrable pair. If none contains a constant, return failed.S3 : Generate solutions. For each integrable pair obtained in step S2,solve the linear problem for determining the integration constants. If aconsistent solution exists, construct a solution from it. Return the set ofall solutions or failed if none has been found.

Example 2.67 Consider the Riccati-like system

R1x(a) ≡ ax + a2 − 2

3xa+ 23x2 = 0, R1

y(b) ≡ by + b2 + 43y b = 0, (2.72)

supplemented by ay − bx = 0. In step S1, the special rational solutions 23x

and 1x are obtained for a, and − 1

3y and 0 for b. They yield the integrable

pairs (a, b) =( 23x,−

13y

),( 23x, 0

),( 1x,−

13y

)and

( 1x, 0

)in step S2. Because

they do not contain any undetermined elements, they are returned as result.

Example 2.68 Let the Riccati-like system

R2x(a) ≡ axx + 3axa+ a3 = 0 and R1

y(b) ≡ by + b2 +2

y − 1b = 0

supplemented by ay − bx = 0 be given. In step S1 the rational solutions

a =2C1(y)x+ C2(y)

C1(y)x2 + C2(y)x+ C3(y)and b =

C4(x)C4(x)y + C5(x)

− 1y − 1

96

are obtained. The single pair made up from them is left unchanged in stepS2 because the parts not depending on the C ′s, 0 and − 1

y − 1 , are an inte-grable pair. In step S3 the bounds 2 and 1 for the x- and the y-dependenciesrespectively lead to a linear system over the constants from which the finalanswer is obtained in the form

a = 1x+ y + C , b = 1

x+ y + C −1

y − 1 .

Rational and Hyperexponential Solutions of Linear PDE’s. In thissubsection the equations under consideration are linear and may be of orderhigher than one. The fundamental constraint now is that their solution spacebe finite-dimensional. This may easily be decided if a Janet basis is gener-ated first. Therefore it will be assumed from now on that the systems underconsideration are already in Janet basis form.

The general system comprises m functions z1 > . . . > zm depending onn independent variables x1 > . . . > xn in lex order, with coefficients fromthe base field Q(x1, . . . , xn). Similar as for linear ode’s, the solution algo-rithm depends crucially on the function field in which the desired solutionsare searched for. At first solutions in the base field are considered.

Two important simplifications may be performed according to Theorem 2.14.First of all, starting from a system involving m functions, it allows to gen-erate m subsystems for a single function zk, k = 1, . . . ,m, each of whichhas a finite-dimensional solution space as well. Secondly, any such subsys-tem involving a single unknown function, contains as lowest equation a linearode in the lowest variable xn. Its rational solutions may be determined byRationalSolutionsLode on page 14. As a result, the single unknown functionmay be expressed as a rational function of this lowest independent variablewith coefficients depending on the higher variables. Upon substitution intothe remaining equations of the system, a linear system for these coefficientsis obtained. It is of the same type as the original system with the number ofindependent variables diminished by one. Therefore, repeating this proceduren times, all dependencies for the coefficients are removed. In a final step alinear algebra problem has to be solved in order to make the solutions for them subsystems into a genuine solution of the originally given system for all mfunctions. The following algorithm is based on this scheme.

Algorithm 2.15 RationalSolutionLinearSystem(L). Given a linear homo-geneous system L for m functions z1, . . . , zm, depending on x1, . . . , xn withcoefficients from Q(x1, . . . , xn), with finite-dimensional solution space. Itsrational solutions are returned, or failed if none exists.

S1 : Systems for a single function. Generate a system for each individualfunction z1, . . . , zm by taking it in turn as lowest dependent variable inlex order.


S2 : Solve linear ode. For each system generated in S1, determine thegeneral rational solution of the linear ode for the lowest independentvariable, and store it on an auxiliary list. If all of them are trivial, returnfailed.S3 : Generate systems for coefficients. Substitute the rational solutionsobtained in step S2 into the remaining equations of the respective system.If the coefficients still carry a dependency with them, go to S1.S4 : Solve linear algebraic system. Substitute the relations obtained in S2and S3 into the original system L. Solve the resulting algebraic system,construct a solution from it and return the result.

The following example applies this algorithm, it originates from the symmetryanalysis of equation 6.91 from Kamke’s collection.

Example 2.69 Consider the system

z2,x − 2y3x2 z1 + 2

3xz2 = 0, z2,y − 23xz1 −

13y z2 = 0,

z1,x − 43xz1 + 1

3y z2 = 0, z1,y − 13y z1 + x

3y2 z2 = 0.(2.73)

In step S1 the two systems for z1 and z2 are generated.

z1,xx −23xz1,x +

23x2 z1 = 0, z1,y −

x

yz1,x +

1yz1 = 0. (2.74)

z2,xx +43xz2,x = 0, z2,y −

x

yz2,x −

1yz1 = 0. (2.75)

In step S2, the lowest equations for z1 and z2 yield the rational solutionsz1 = C1(y)x and z2 = C2(y). Substituting them into the remaining equationof the respective system in S3 yields C1,y = 0 and C2,y − 1

yC2 = 0. Dueto their dependencies, the algorithm returns to S1. There is nothing to dohere because only a single equation is left in either case. Step S2 generatesthe solutions C1(y) = C3 and C2 = C4y with C3, C4 constants. There is noaction in step S3. The linear algebraic system in step S4 reduces to the singleequation C3−C4 = 0. So the final answer is z1 = Cx, z2 = Cy, it generates aone-dimensional solution space. Because the general solution of (2.73) allowsa two-dimensional solution space, this result shows that the general solutionis not rational, i.e., there must be an additional basis element which is notrational, it will be determined in the next example.

The last problem to be considered in this chapter is to find hyperexponentialsolutions of linear homogeneous systems for z1, . . . , zm. Here the number ofindependent variables is restricted to two, usually x and y. This is sufficientfor the applications in the symmetry analysis in later chapters. The discussionfollows closely the article by Li and Schwarz [104]. At first m = 1 is assumed,i.e., a system for a single function z depending on x and y with coefficients from

98

Q(x, y) is considered. By Theorem 2.14 two linear ode’s w.r.t. x and y maybe generated. For both of them the associated Riccati equation for a functionu ≡ zx

z and v ≡ zyz is constructed as explained on page 16. These equations

are supplemented by the obvious integrability condition uy = vx which makesu and v into an integrable pair. This subsystem is of type (2.70), its rationalsolutions may be determined by the algorithm RationalSolutionPartialRiccati.They correspond to hyperexponential solutions of the original linear system.The remaining equations of the system for z generate linear relations betweenthe constants such that the solution space of the full system is generated. Thefollowing algorithm applies these steps.

Algorithm 2.16 HyperexponentialSolutionLinearSystem(L). Given a lin-ear homogeneous system L for a function z, depending on x and y, witha finite-dimensional solution space, its general hyperexponential solution isreturned, or failed if none exists.S1 : Generate Riccati-like system. Generate two Janet bases in lex orderfrom L with x > y and y > x. From the two lowest equations generateassociated Riccati equations in x and y respectivey.S2 : Solve Riccati-like system. Determine the rational solutions of theRiccati-like system corresponding to the equations obtained in step S1by algorithm RationalSolutionPartialRiccati. If none is found, returnfailed.S3 : Generate systems for coefficients. From the solutions obtained instep S2 construct the hyperexponential solutions, substitute them intothe remaining equations of the linear system and generate the linearsystem for the coefficients.S4 : Solve linear algebraic system. Solve the linear algebraic systemobtained in step S3, construct a solution from it and return the result.

If in a system m functions z1, . . . , zm are involved, at first m subsystemsfor each variable zk, k = 1, . . . ,m are generated. To each of them the abovealgorithm is applied, and in a final step an additional linear system for thecoefficients has to be solved in order to generate the solution of the full system.

Example 2.70 In order to find the hyperexponential solutions of (2.73),system (2.74) for z1 will be considered first. In step S1, from the two lowestequations

z1,xx −23xz1,x +

23x2 z1 = 0, z1,yy +

43yz1,y = 0 (2.76)

the Riccati equations (2.72) are generated. Its rational solutions have beendetermined in Example 2.67. The four integrable pairs obtained there yieldsthe general solution

z = C1x2/3

y1/3+ C2x

2/3 + C3x

y1/3+ C4x.


Substitution of this expression into (2.74) leads to the relations C1 = C4 = 0,C2 − C3 = 0; consequently, the full two-dimensional solution space of (2.74)

is generated by x and x2/3

y1/3 .

It turns out that system (2.75) leads to two equations for z2 which may beobtained from (2.76) by exchange of x and y, i.e., two basis elements for its

solution space are y and y2/3

x1/3 . Finally, if a general linear combination of these

basis elements is substituted into (2.73), a linear system for the coefficients isobtained, the solution of which yields the final answer

z1 = C1x+ C2x2/3

y1/3, z2 = C1y + 2C2

y2/3

x1/3.

This is the full solution space of system (2.73).

Decomposing Janet Bases. The hyperexponential solutions discussed inthe preceding section correspond to first order right factors. For Janet basesof order d three or higher, right factors of any order less than d may occur. Acomplete understanding of the solution space determined by such a Janet basisis only possible if these right factors may be found. To this end the analogueof the Loewy decomposition of linear ode’s is required which in turn requiresa generalized Beke-Schlesinger scheme for these pde’s. For the applicationslater in this book the general case is not needed. Therefore only a few resultsfor a certain third order Janet basis which occurs frequently in the symmetryanalysis will be described. They are based to a large extent on a recent articleby Li et al. [105].

For a nontrivial exact quotient q of two linear ode’s l1 and l2 to exist, i.e., fora relation l1 = ql2 to be valid, it is necessary that the order of l2 be strictly lessthan that of l1. For Janet bases the situation is more complicated. The partialorder defined in the subsequent lemma describes the possible candidates forexact division for the Janet basis types J (1,2) and order not higher thanthree. The following shortened notation is introduced for describing theseorder relations:

a ≺ b, c, d, . . . means a ≺ b, a ≺ c, a ≺ d . . . .

Lemma 2.13 For Janet bases of type J (1,2) and order not higher thanthree, the relations

J (1,2)1 ≺ J (1,2)

2,1 , J (1,2)2,2 ,

J (1,2)2,1 ≺ J (1,2)

3,1 , J (1,2)3,2 , J

(1,2)2,2 ≺ J (1,2)

3,2 , J (1,2)3,3

define a partial order that may be represented by the diagram

100

J (1,2)3,1 J (1,2)

3,2 J (1,2)3,3

J (1,2)2,1 J (1,2)

2,2

J (1,2)1

@@

@R

@@

@R

@@

@R

The exact quotient of two Janet bases e and f of these types may exist onlyif f is lower than e in this partial order.

Proof The partial order property is obvious. The leading derivative zyin the type J (1,2)

3,1 Janet basis cannot be reduced by any element of a type

J (1,2)2,2 Janet basis, therefore the latter can never divide the former. The same

is true for the leading derivative zx of a type J (1,2)3,3 Janet basis and a type

J (1,2)2,1 Janet basis. Similar arguments apply for the remaining cases.

The Janet basis of type J (1,2)3,2 in the upper line of the above diagram occurs

frequently in the symmetry analysis of ode’s. Therefore the complete answerfor its decomposition involving all possible Janet bases of lower order is givenexplicitly.

Theorem 2.28 A Janet basis

zxx+A1zy+A2zx+A3z, zxy+B1zy+B2zx+B3z, zyy+C1zy+C2zx+C3z

of type J (1,2)3,2 may have the following components.

i) A type J (1,2)1 component zx + az, zy + bz iff its coefficients a and b

are rational solutions of

ax − a2 +A2a+A1b−A3 = 0, ay − ab+B2a+B1b−B3 = 0,

bx − ab+B1b+B2a−B3 = 0, by − b2 + C1b+ C2a− C3 = 0.(2.77)

ii) A type J (1,2)2,1 component zy + a1zx + a2z, zxx + b1zx + b2z with

a1 = −w1w0, a2 = w2

w0, b1 = A1

w1w0

+A2, b2 = −A1w2w0

+A3, (2.78)

w1 = − 1A1

w0,x −A2

A1w0, w2 = −w0,y −

B1

A1w0,x −

(B2 −

A2

A1B1

)w0

(2.79)if A1 6= 0 and w0 is a hyperexponential solution of a coherent third ordersystem of type w0,xx, w0,xy, w0,yy.


iii) A component zx+a1z, zyy+ b1zy+ b2z of type J (1,2)2,2 if A1 = 0. Then

the coefficients are a1 = B1, b1 = C1 and b2 = C3 −B1C2.

Proof The coefficient constraints for case i) and case iii) are obtained bydirect reduction. The system (2.77) has been discussed in detail on page 90.

For case ii) let z1 and z2 be a fundamental system for a possible factor oftype J (1,2)

2,1 and define

w0 =∣∣∣∣ z1,x z2,xz1 z2

∣∣∣∣ , w1 =∣∣∣∣ z1,y z2,yz1 z2

∣∣∣∣ , w2 =∣∣∣∣ z1,y z2,yz1,x z2,x

∣∣∣∣ . (2.80)

The expressions (2.78) for the coefficients are obtained by substituting z1 andz2 into the type J (1,2)

2,1 system, using the given type J (1,2)3,2 system for reduction

and solving the respective linear algebraic systems. By differentiation andreduction w.r.t. to the given type J (1,2)

3,2 Janet basis the following first ordersystem for the determinants w0, w1 and w2 is obtained.

w0,x +A1w1 +A2w0 = 0, w0,y + w2 +B1w1 +B2w0 = 0,

w1,x − w2 +B1w1 +B2w0 = 0, w1,y + C1w1 + C2w0 = 0,

w2,x + (A2 +B1)w2 +A3w1 −B3w0 = 0,

w2,y + (B2 + C1)w2 +B3w1 − C3w0 = 0.

(2.81)

The coherence follows from the integrability conditions given in Theorem 2.15.The expressions (2.79) for w1 and w2 are obtained from the first two equa-tions. If they are applied for elimination, due to the coherence the resultingsystem for w0 may be transformed into a third order type J (1,2)

3,2 Janet basisin grlex term ordering with y > x. The details are discussed in Exercise 2.12.As a result the type J (1,2)

3,2 system for w0 is obtained explicitly for generalcoefficients A1, A2, . . . , C3.

Although the system for w0 obtained above appears quite cumbersome, itis useful for implementing decomposition algorithms because it is obtainedwithout any Janet basis calculation.

Example 2.71 If the type J (1,2)3,2 Janet basis considered in Example 2.55

has a type J (1,2)1 component, according to (2.77) its coefficients are solutions

ofax − a2 + 4

xa−2x2 = 0, ay − ab+ 1

xb = 0,

bx − ab+ 1xb = 0, by − b2 + 1

y b−xy2 a+ 2

y2 = 0.(2.82)

The transformation (2.64) which is a = −a + 53x , b = −b + 1

3y in this case

102

leads to

ax + a2 + 23xa−

29x2 = 0, ay + ab− 1

3y a−23x b+ 2

9xy = 0,

bx + ab− 13y a−

23x b+ 2

9xy = 0, by + b2 − xy2 a+ 1

3y b−2

9y2 = 0(2.83)

with the general solution

a =C2

1y + y

C1 + C2xy + xy

− 13x, b = −

C2xy2 − x

C1 + C2xy + xy

+13y.

The special choice C1 = 0, C2 −→ ∞ yields the solution a = − 23x , b = 2

3yand finally a = 1

x , b = 1y . Due to the fact that the general solution of the

Riccati-like system for a and b is rational, the originally given type J (1,2)3,2

Janet basis is completely reducible. It is left as an exercise to determine theremaining components of its Loewy factor.

Exercises

Exercise 2.1 In the following system of algebraic equations

x3 + a4x2 + a3y + a2x+ a1 = 0,

xy + b4x2 + b3y + b2x+ b1 = 0,

y2 + c4x2 + c3y + c2x+ c1 = 0

x and y are considered as indeterminates, the ai as parameters. Find theconstraints for the parameters ak, bk and ck such that the left hand sides ofthe above equations are a Grobner basis in grlex ordering with y > x.

Exercise 2.2 Let a third order ode y′′′+ py′+ 12p′y = 0 with p ≡ p(x) be

given. Show that a fundamental system yi, i = 1, . . . , 3 of the form y1 = z21 ,

y2 = z1z2, y3 = z22 exists where z1,2 is a fundamental system of z′′+ 1

4pz = 0.Later on it will be seen that an analogous property holds for the n− th orderlinear homogeneous equation with suitable constraints for its coefficients andis a consequence of its symmetry structure.

Exercise 2.3 Generalize Example 2.2 to a third order lode y′′′ + q1y′′ +

q2y′ + q3y = r and apply the result to the special equation y′′′ = r.

Exercise 2.4 Let a second order linear homogeneous ode be given. If itis completely reducible it may be represented in the form

y′′ + Py′ +Qy = Lclm(D + p,D + q)y = 0 with p 6= q.


Show that P and Q have the representation

P = p+ q − p′ − q′

p− q, Q = pq − p′q − q′p

p− q.

Give an explicit expression for the general solution of the inhomogeneousequation y′′ + Py′ +Q = R.

Exercise 2.5 Determine the solutions corresponding to the various typesof nontrivial Loewy decompositions of second order equations as given afterCorollary 2.5.

Exercise 2.6 Prove Theorem 2.1 for second order Riccati equations.

Exercise 2.7 Prove Corollary 2.6.

Exercise 2.8 Prove Theorem 2.9 for m = 2.

Exercise 2.9 Express the coefficients of the type J (1,2)2,1 Janet basis

zy + a1zx+ a2z, zxx+ b1zx+ b2z in terms of a fundamental system z1, z2.

Exercise 2.10 The same problem for the type J (1,2)2,2 Janet basis

zx + a1z, zyy + b1zy + b2z with fundamental system z1, z2.

Exercise 2.11 Determine by direct reduction a system of pde’s for thecoefficients of a type J (1,2)

2,1 component for a type J (1,2)3,2 Janet basis and

compare the result with Theorem 2.28.

Exercise 2.12 Determine the system for w0 of case ii) in Theorem 2.28explicitly.

Exercise 2.13 Determine the order of a pole in function (2.19) if q(x) hasa pole of order M at x = x0. Distinguish the cases M > 1 and M = 1.

Exercise 2.14 Assume the type J (1,2)2,1 Janet basis

zy +A1zx +A2z = 0, zxx +B1zx +B2z = 0

has the first order Loewy factor zx+ az = 0, zy + bz = 0. Express the generalsolution in terms of integrals.

Exercise 2.15 The same problem for the type J (2,2)2,3 Janet basis

z1,x +A1z2 +A2z1 = 0, z1,y +B1z2 +B2z1 = 0,z2,x + C1z2 + C2z1 = 0, z2,y +D1z2 +D2z1 = 0

with the first order Loewy factor z2 + az1 = 0, z1,x + bz1 = 0, z1,y + cz1 = 0.

Exercise 2.16 Let zx + aiz = 0, zy + biz = 0, i = 1, 2 be two first ordersystems. Determine its Lclm.

Exercise 2.17 Apply the results of Example 2.47 to obtain a basis for thesolution space of the corresponding system of pde’s.

104

Exercise 2.18 Let the system zy + a1zx + a2z = v, zxx + b1zx + b2z = wof inhomogeneous pde’s for z(x, y) be given where a1, a2, b1, b2, v, w ∈ Q(x, y)and the left hand sides form a type J (1,2)

2,1 Janet basis. Express a specialsolution in terms of an integral if a fundamental system z1 and z2 of thecorresponding homogeneous system is known.

Exercise 2.19 The same problem for zx + a1z = v, zyy + b1zy + b2z = w

where the left hand sides form a type J (1,2)2,2 Janet basis.

Exercise 2.20 The same problem for zy + a1zx + a2z = v, zxxx + b1zxx +b2zx + b3z = w where the left hand sides form a type J (1,2)

3,1 Janet basis andz1, z2 and z3 are a fundamental system of the corresponding homogeneoussystem.

Exercise 2.21 Show how (2.46) is obtained from (2.45).

Exercise 2.22 Apply Theorem 2.20 to solve the equation zy + azx = rfor z(x, y) where a ≡ a(x, y), r = b(x, y)z + c(x, y).

Exercise 2.23 The equation ∂I∂x

+ A∂I∂y

+ 2I ∂A∂y

+ ∂3A∂y3 = 0 has been

considered by Drach [38], page 364. Use the result of the preceding exerciseto describe its solution.

Exercise 2.24 Prove Theorem 2.24.

Exercise 2.25 Determine the general rational solution of the system con-sidered in Example 2.65 by solving the second equation first.

Exercise 2.26 Assume a special solution z1 for system (2.50) is known.Show that a second solution may be obtained by solving a linear system.

Exercise 2.27 Prove the representation (2.56) for the rational solutionsof zx +A1z

2 +A2z +A3 = 0, zy +B2z +B3 = 0.

Exercise 2.28 Prove Lemma 2.11.

Chapter 3

Lie Transformation Groups

This chapter describes the basic facts of Lie’s theory of continuous groupsin a form that is suited for the applications to differential equations in laterparts of this book. Originally Lie considered this only as an auxiliary subjectfor his main goal, i.e., solving differential equations in closed form by analogywith Galois’ theory of algebraic equations. After recognizing the fundamentalimportance of these groups for this latter problem he developed a fairly com-plete theory for them. Its original objective, solving differential equations,was almost completely forgotten, most textbooks on differential equations donot even mention it at all. Later on the theory of continuous groups became afield of independent interest under the name Lie groups. The same is true forthe algebraic objects introduced by Killing that were baptized Lie algebras byHermann Weyl. They were obtained by abstraction from the commutators ofvector fields occurring in Lie’s theory.

In the first Section 3.1, basic concepts of the theory of continuous groupsare put together for later reference, usually without proof. In Section 3.2below, some properties of abstract Lie algebras are discussed. Section 3.3applies the previous results to transformation groups of a two-dimensionalmanifold because these groups are most important for later parts of the book.Classifications of low-dimensional Lie algebras and transformation groups of atwo-dimensional manifold are given in Section 3.4. The subject of Section 3.5are so called Lie systems. These are special systems of linear homogeneouspde’s with the additional feature that their solutions define the coefficientsof infinitesimal generators of Lie transformation groups. These results areapplied later in the symmetry analysis of differential equations.

3.1 Lie Groups and Transformation Groups

This section follows closely the article by Gorbatsevich and Onishchik invol. 20 of the Encyclopaedia of Mathematical Sciences [43], part I; see alsoChapter 1 of the book by Olver [140], or Chapters 5, 6 and 7 of Ibragimov’sbook [78].

105

106

Lie Groups. The main ingredients for this section are the algebraic conceptof a group and the differential-geometric notion of a smooth manifold. Theterm smooth constrains the overlap functions of any coordinate chart to beC∞ functions. These concepts are merged by the following definition whichis the foundation for the subject of this chapter.

Definition 3.1 (Lie group) A set G is called a Lie group if there is givena structure on G satisfying the following three axioms.

i) G is a group.

ii) G is a smooth manifold.

iii) The group operations G×G→ G, (x, y)→ xy and G→ G, x→ x−1aresmooth functions.

If r is the dimension of the manifold, the group is called an r-parameter Liegroup and is denoted by Gr.

For the applications in this book usually one is not interested in the fullLie group, but only in that part which is connected to the identity element.Starting with any global Lie group, a local Lie group may be constructed bytaking a coordinate chart for the group manifold containing the identity, andretaining only those elements corresponding to parameter values contained inthis identity neighborhood. From now on, the term Lie group will always beused in this sense, i.e., meaning the local group just described.

Let G and H be Lie groups. A smooth map f : G→ H is a homomorphismif it is simultaneously a homomorphism of abstract groups and a smooth mapof the group manifolds. A homomorphism is called an isomorphism if thereexists a smooth inverse f−1 : H → G. Two Lie groups are called isomorphicif an isomorphism between them exists. Due to the restriction to local groups,it is sufficient that these maps be defined for a neighborhood of the identities.

A subsetH of a Lie group G is called a Lie subgroup ifH is a subgroup of theabstract group G and a submanifold of the manifold G. If H is even a normalsubgroup of the abstract group G, it is called a Lie normal subgroup. A Liegroup G is called simple if the only normal Lie subgroups are the identity andG itself. The commutator group G(1) = G′ = [G,G] is the group generatedby all commutators aba−1b−1 with a, b ∈ G. The higher commutator groupsare defined recursively by G(ν+1) = G(ν)′ for ν = 1, 2, . . .. A Lie group Gr iscalled integrable or solvable if its series of commutator groups G(ν) terminatesafter a finite number of steps with the identity.

A Lie group is called semi-simple if it does not contain a solvable normalsubgroup. Any Lie group has a maximal solvable normal subroup.

Let Gr be a r-parameter Lie group and a, b ∈ Gr be two elements with therepresentation a = (a1, . . . , ar) and b = (b1, . . . , br) in a coordinate chart. Forthe coordinates of the product c = ab there holds

cj = φj(a1, . . . , ar, b1, . . . , br) for j = 1, . . . , r (3.1)

Lie Transformation Groups 107

where φj are smooth functions of all arguments. Assume further now thatthe coordinate chart is such that the identity is represented by e = (0, . . . , 0).Then the lowest terms of (3.1) are

cj = aj + bj + 12

∑k,l γ

jklakbl + . . . (3.2)

with constants γjkl. The omitted terms are of order three or higher withrespect to ak and bl. The numbers

cjkl = γjkl − γjlk (3.3)

are anti-symmetric in k and l, they are called the structure constants of theLie group.Lie Transformation Groups. Let M be a n-dimensional smooth manifoldand G a Lie group. An action T of the group G on M is a smooth mapping

T : G×M →M, T (g, x) ≡ gx→ x (3.4)

with the following properties:

T (e, x) = x, T (a, T (b, x)) = T (ab, x) (3.5)

for any x ∈ M , g, a, b ∈ G, e ∈ G the unit element. Then G is called a Lietransformation group of the manifold M . For a r-parameter group Gr, in localcoordinates the action (3.4) has the special form

xi = fi(x1, . . . , xn, a1, . . . , ar) (3.6)

with smooth functions fi, i = 1, . . . , n. In this form Lie originally introducedhis continuous groups. Specializations of (3.6) are the basis for the symmetryanalysis in later parts of this book.

An additional assumption is that the number r of parameters is minimal,i.e., that it is not possible to introduce a smaller number of parameters gener-ating the same set of transformations. If this is true they are called essentialparameters. It is proved in Lie [112], vol. I, pages 13-14, that this is guaran-teed if there do not exist equations of the form

r∑k=1

χk(a1, . . . , ar)∂fi(x1, . . . , xn, a1, . . . , ar)

∂ak= 0

for i = 1, . . . , n. Furthermore, it is shown there how a constructive criterionmay be obtained from these requirements. From now on it is always assumedthat group parameters are essential without explicitly mentioning it.

The elements g ∈ G leaving a given element x ∈ M invariant form thestabilizer of x. If T is an action of a group G on M , one can define anequivalence relation on M by

x ∼ y ⇐⇒ x = gy for some g ∈ G.

108

The equivalence classes are called the orbits of the action. Every point ofx ∈M is contained in a unique orbit. The totality of orbits generate aninvariant decomposition or simply decomposition of the manifold.

The action of a transformation group G on M is called transitive if forarbitrary x, x ∈M there is a group element g ∈ G such that x = gx, otherwiseit is called intransitive. If the action is transitive there is only a single orbit ofif for x1, . . . , xk ∈ M , xi 6= xj for i 6= j, there is a g ∈ G such that xi = gxi,but the same is not true for k + 1, i.e., there are k orbits.

A transitive transformation group G of a manifold M is called imprimitiveif M is the union of submanifolds that are permuted by the group. Thesesubmanifolds are called systems of imprimitivity of G. A group that is notimprimitive is called primitive.

Let Gr be a r-parameter Lie group with coordinates a = (a1, . . . , ar) andsuch that e = (0, . . . , 0). Defining the quantities

ξk,i ≡∂fi∂ak

∣∣∣a1=...=ar=0

(3.7)

for k = 1, . . . , r and i = 1, . . . , n, expanding (3.6) at the identity and omittingterms of order higher than one in the group parameters there follows

xi = xi + ξ1,ia1 + . . .+ ξr,iar + . . . . (3.8)

The vector fields

Xk = ξk,1∂

∂x1+ . . .+ ξk,n

∂

∂xn, k = 1, . . . , r, (3.9)

are called the infinitesimal generators of the Lie transformation group. Theyare linearly independent over constants and obey the commutation relations

[Xk, Xl] = XkXl −XlXk =r∑j=1

cjklXj . (3.10)

Mathematical objects obeying these relations have been baptized a Lie alge-bra. They occur in various branches of mathematics and form an independentsubfield of algebra. Some of these algebraic properties are discussed in thenext Section 3.2. If X = [Xk, Xl] is the commutator of the two generators Xk

and Xl, its coefficients may be expressed as

ξi =n∑j=1

(ξk,i

∂ξl,i∂xj

− ξl,i∂ξk,i∂xj

)for i = 1, . . . , n. (3.11)

As usual, if q generators X1, X2, . . . , Xq do not obey a relation

λ1X1 + . . .+ λqXq ≡ 0 (3.12)


with λ1, . . . , λq constants, not all zero, they are said to be linearly independentover constants or simply independent. A generator Xk that may be expressedin the form

Xk = λ1X1 + . . .+ λqXq

is said to be dependent on X1, . . . , Xq. If q generators X1, X2, . . . , Xq do notobey a relation

φ1(x1, . . . , xn)X1 + . . .+ φq(x1, . . . , xn)Xq ≡ 0 (3.13)

with not all φi(x1, . . . , xn) identically zero they are said to be unconnected.Any generator Xk that may be expressed in the form

Xk = φ1(x1, . . . , xn)X1 + . . .+ φq(x1, . . . , xn)Xq

is said to be connected withX1, . . . , Xq. Unconnected generators are obviouslyindependent, but independent generators are not necessarily unconnected. Fora transformation group of a n-dimensional manifold there cannot be more thann unconnected generators, though there may be any number of independentgenerators. If a set of unconnected generators is such that the commutator ofany pair is connected with them, they are said to form a complete system.

The following rules for manipulating expressions involving a commutatorfollow immediately from the definition. Let φ(x1, . . . , xn) and ψ(x1, . . . , xn)be two functions of x1, . . . , xn, and X, X1 and X2 arbitrary vector fields.There holds

X(φ+ ψ) = Xφ+Xψ, X(φψ) = ψXφ+ φXψ,

[φX1, X2] = φ[X1, X2]−X2φX1, [X1, φX2] = φ[X1, X2] +X1φX2.

Let X1, . . . , Xr be the infinitesimal generators of an r-parameter Lie trans-formation group G with group parameters a1, . . . , ar. The vector fields

Zi = bi1(a1, . . . , ar)∂

∂e1+ . . .+ bir(a1, . . . , ar)

∂

∂er

where bik =∑rj=1 c

jikej for i = 1, . . . , r and k = 1, . . . , r are the generators

of a linear homogeneous transformation group of an r-dimensional manifoldwith coordinates (e1, . . . , er). It is called the adjoint group of G.Lie’s Fundamental Theorems. Lie himself summarized several importantresults of his theory in terms of three fundamental theorems that are givennext.

Theorem 3.1 (Lie’s First Theorem) If the equations (3.6) define anr-parameter group, then the xi, considered as functions of x1, . . . , xn anda1, . . . , ar, satisfy a system of pde’s of the special form

∂xi∂ak

=r∑j=1

ψj,k(a1, . . . , ar)ξj,i(x1, . . . , xn)

110

for k = 1, . . . , r. The determinant of the ψj,k does not vanish identically.On the other hand, if functions fi(x1, . . . , xn, a1, . . . , ar) satisfy a system ofdifferential equations of this form, they define a transformation group.

The proof of this theorem may be found in Lie [112], vol. I, Kapitel 2, pages27-34 and vol. III, Kapitel 25, pages 545-564.

The importance of the infinitesimal generators of a group arises from thefact that a group, due to the restriction to local groups as explained above, isuniquely determined by them. This relation is established in the next theorem.

Theorem 3.2 (Lie’s Second Theorem) Any r-parameter Lie group (3.6)defines r infinitesimal generators (3.9) satisfying r(r−1) relations of the form(3.10). Conversely, r independent generators (3.9) satisfying the relations(3.10) define a Lie group.

The proof of this theorem may be found in Lie [112], vol. III, Kapitel 25,pages 575-590. It is based on the system

dxidt

= λ1ξ1,i(x1, . . . , xn) + . . .+ λrξr,i(x1, . . . , xn)

for the general one-parameter group contained in the r-parameter group (3.6);the λ1, . . . , λr are constants. Together with the initial conditions xi = xi fort = 0, i = 1, . . . , n, they determine the one-parameter groups uniquely. If forany fixed k, 1 ≤ j ≤ r, λk = ak, λj = 0 for j 6= k are chosen, the system

dxidak

= ξk,i(x1, . . . , xn) (3.14)

for the one-parameter group corresponding to the generator Xk is obtained.Whenever the phrase “the group of the Xk” for a system of infinitesimalgenerators is used, it is based on Lie’s second theorem and the restriction tothe local group determined by them.

Theorem 3.3 (Lie’s Third Theorem) If a system of constants cjkl satisfiesthe relations cjkl + cjlk = 0 and

r∑s=1

(cjkscsli + cjlsc

sik + cjisc

skl) = 0

for i, j, k, l = 1, . . . , r, there exists a Lie group with these structure constants.

The proof of this theorem may be found in Lie [112], vol. III, pages 597 ff.Similarity of Transformation Groups. If any transformation group isgiven by relations (3.6), an apparently different group may be obtained byintroducing new coordinates of both the underlying manifoldM and the groupmanifold G. Two groups related to each other in this way are not essentially


different. A special term for this relation has been introduced by Lie [112],vol. I, page 24.

Definition 3.2 (Similarity of transformation groups) Let two r-parametertransformation groups be given by

xi = fi(x1, . . . , xn, a1, . . . , ar) and yi = gi(y1, . . . , yn, b1, . . . , br).

If there exist diffeomorphisms xi = ϕi(y1, . . . , yn) for i = 1, . . . , n andak = ψ(b1, . . . , br) for k = 1, . . . , r such that the two definitions become iden-tical, the groups are called similar.

The consequences of a simultaneous change of both coordinates are dis-cussed in detail by Lie [112], vol. I, Kapitel 2 and by Eisenhart [41], §7. Thetransformation of the infinitesimal generators under a coordinate change of Moccurs frequently in later applications. Let the two coordinates (x1, . . . , xn)and (y1, . . . , yn) be related by xk = φk(y1, . . . , yn) and yk = ψk(x1, . . . , xn).Defining ηi = Xkψi(x1, . . . , xn)

∣∣∣x1=φ1,...,xn=φn

for i = 1, . . . , n, the trans-

formed generator of (3.9) is

Yk = η1(y1, . . . , yn)∂

∂y1+ ηn(y1, . . . , yn)

∂

∂yn.

Of particular importance is the question of how the commutator of twovector fields is transformed under a change of coordinates. The answer isgiven in the subsequent lemma due to Lie [112], part I, page 84.

Lemma 3.1 If two vector fields Xk and Xj in coordinates (x1, . . . , xn)are transformed into Y1 and Y2 in coordinates (y1, . . . , yn), its commutator[Xk, Xj ] is transformed into [Yk, Yj ].

As a consequence of this lemma the structure constants of a Lie algebraof vector fields are invariant under a coordinate change of the transformedmanifold M . This is important because it decouples the algebraic propertiesof Lie algebras generated by vector fields from its appearance depending onthe actual coordinates of M .

Given any two transformation groups the problem arises how to decidesimilarity and, in case the answer is affirmative, how to actually determinethe transformation functions between them. The following theorem due toLie essentially provides the answer.

Theorem 3.4 (Lie 1888). In order that two groups with infinitesimal gen-erators X1, . . . , Xr in coordinates (x1, . . . , xn) and Y1, . . . , Yr in coordinates(y1, . . . , yn) respectively be similar, the following conditions are necessary andsufficient.

1. The two groups must have the same structure, i.e., if the relations[Xi, Xj ] =

∑rk=1 c

kijXk are valid it must be possible to choose suitable

112

linear combinations Yi =∑rj=1 γijYj with constant γ’s such that the Yi

satisfy relations [Yi, Yj ] =∑rk=1 c

kij Yk with the same structure constants

as for the Xi.

2. If X1, . . . , Xk are unconnected whereas Xk+i =∑kj=1 φi,j(x1, . . . , xn)Xj

for k ≤ r and i = 1, . . . , r − k, the corresponding generators Y1, . . . , Ykare also unconnected whereas the relations Yk+i =

∑kj=1 ψi,j(y1, . . . , yn)Yj

are valid such that the equations φi,j(x1, . . . , xn) = ψi,j(y1, . . . , yn) areconsistent. In particular they should not generate a relation amongx1, . . . , xn on the one hand or y1, . . . , yn on the other.

The proof may be found in Lie [112], part I, pages 353-355. It shows thatthe question of similarity of two groups may be divided into two independentparts. This separation is based on Lemma 3.1. In the first place, in orderto be similar the infinitesimal generators must generate isomorphic Lie alge-bras. This is a purely algebraic problem that will be discussed in detail in theSection 3.2. Only if both Lie algebras are isomorphic and a basis transfor-mation has been found such that their structure constants are identical, thesecond part dealing exclusively with the coordinate change of the transformedmanifold applies.

A special kind of similarity is involved in defining the so-called canoni-cal variables for a one-parameter group. Let the group equations in coordi-nates (x1, . . . , xn) be xi = fi(x1, . . . , xn, a) with the single group parametera. There exist always new coordinates (u1, . . . , un) such that the group equa-tions are ui = ui for i = 1, . . . , n− 1 and un = un + b, i.e., any one-parametergroup is similar to a group of translations in a single coordinate. In theselatter variables the infinitesimal generator is ∂

∂un. In Section 3.3 finding a

transformation to canonical form for groups acting in the plane will be dis-cussed. Furthermore, canonical forms for groups involving more than a singleparameter will be discussed.

Defining Equations of a Transformation Group. Starting with theinfinitesimal generators of a group, a third representation in terms of thedefining equations, as Lie [112], vol. I, page 185 called them, may be obtained.By definition, these equations are such that their general solution generatesthe vector space of the coefficients of the given generators. This representationis particularly important if the groups to be considered occur as symmetrygroups of differential equations because they are usually defined in terms ofsystems of pde’s.

Let the infinitesimal generators Xk of a group be given in the form (3.9).The general generator may be written as

ξ1∂

∂x1+ . . .+ ξn

∂

∂xn=

r∑k=1

ek

(ξk,1

∂

∂x1+ . . .+ ξk,n

∂

∂xn

)(3.15)


with constants ek. This defines the ξi by means of ξi =∑rk=1 ekξk,i(x1, . . . , xn).

The ξk,i are given functions of x1, . . . , xn. It has been shown by Lie [112],vol. I, § 48, page 181 that for the functions ξi a system of linear pde’s alwaysmay be constructed such that they form a fundamental system. Due to itsimportance a special name has been given to it.

Definition 3.3 (Defining equations of a group). The coefficients ξ1, . . . , ξnof the general infinitesimal generator of a Lie transformation group definedby (3.15) satisfy a system of linear homogeneous pde’s. They are called thedefining equations of the group.

Altogether there are three means now to describe a Lie transformationgroup. The finite equations (3.6), the infinitesimal generators (3.9) and thedefining equations just described. Due to the restriction to local Lie groupsthey are essentially equivalent. The infinitesimal generators are obtained fromthe finite equations by differentiation. The converse requires solving linearpde’s by Lie’s second theorem. The same is true for determining the infinites-imal generators if the defining equations are given. The relation between theserepresentations of a group and how to turn from one to another is illustratedby the following diagram. The operations corresponding to the downwardarrows at the left can always be performed algorithmically. Contrary to that,there is no generally valid algorithm for the operations at the right corre-sponding to the upward arrows.

Group Equations

?

6

DifferentiationSolving 1st OrderOde Systems

Infinitesimal Generators

?

6Solving Linear

Algebraic Equations Solving Linear Pde’s

Defining Equations

Invariants and Differential Invariants. A function F (x1, . . . , xn) is calledan invariant of the r-parameter Lie transformation group Gr defined by (3.6)if there holds

F (x1, . . . , xn) = F (x1, . . . , xn).

Invariants of this kind are also called absolute invariants in order to distinguishthem from another kind of invariants to be defined in Chapter 4. A criterionfor invariance may be given in terms of the infinitesimal generators of thegroup defined by (3.9). A function F (x1, . . . , xn) is an invariant of the groupif and only if there holds XkF (x1, . . . , xn) = 0 for k = 1, . . . , r. If F isan invariant of a group, then also Φ(F ) with Φ an undetermined function

114

is an invariant. If a group allows more than a single invariant, only thefunctionally independent invariants are relevant. If the generic rank q for thegroup generated by X1, . . . , Xr is defined by

rank

ξ1,1, ξ1,2 ξ1,nξ2,1, ξ2,2 ξ2,n...

...ξr,1, ξr,2 ξr,n

≡ qthere are n − q independent invariants. If q = n there are no invariants atall and the group is transitive. The proof may be found in Lie [112], vol. I,Kapitel 13 or in the book by Eisenhart [41], §21.

The concept of a differential invariant emerges from the fact that a groupacts on a manifold the coordinate variables of which may carry some de-pendencies with them. Whenever this occurs, the vector fields generatingthe group may be prolonged up to a predetermined order of the respectivederivatives. These prolonged generators of the original group are consideredas vector fields acting on a higher-dimensional manifold to which the aboveresults may be applied. If different dependencies are assumed, there are dif-ferent prolongations and consequently diverse types of differential invariants.In the remaining part of this subsection summation over repeated indices isalways understood.

Definition 3.4 (Prolongation of a vector field) Let the vector fieldU = ξi(x, u)∂xi +η

α(x, u)∂uα be the infinitesimal generator of a finite transfor-mation group on a manifold with coordinates x = (x1, . . . , xn) andu = (u1, . . . , um) with dependencies uα ≡ uα(x1, . . . , xn) for α = 1, . . . ,m.

Partial derivatives are denoted by uαi1,...,in ≡∂i1+...+inuα

∂xi11 . . . xinn. By definition the

k-th prolongation of U is

U (k) = U + ζαi∂

∂uαi+ . . .+ ζαi1,...,ik

∂

uαi1,...,ik.

The functions ζαi1,...,ik are recursively defined by ζαi = Di(ηα)− uαsDi(ξs) and

ζαi1,...,ik = Dik(ζαi1,...,ik−1)− uαi1,...,ik−1s

Dik(ξs).

Di is the operator of total differentiation w.r.t. xi.

Di =∂

∂xi+ uαi

∂

∂uα+ uαk,i

∂

∂uαk+ uαk,l,i

∂

∂uαk,l. . .

If the vector fields generating a group are prolonged there arises the questionas to what the algebraic relations between the prolonged vector fields are. The


subsequent theorem provides the simple answer that the structure constantsfor the prolonged generators remain unchanged.

Theorem 3.5 Let X1, . . . , Xr be vector fields generating a group acting ona manifold of arbitrary dimension obeying [Xi, Xj ] =

∑rk=1 c

kijXk. Further-

more, let there be some sort of dependencies declared between these variables.The prolongations X(l)

i of any order obey the same commutation relations asthe X ′s, i.e., there holds [X(l)

i , X(l)j ] =

∑rk=1 c

kijX

(l)k .

The proof may be found in Lie [112], vol. I, §130 or Eisenhart [41], §27.The next theorem due to Lie [114], page 760, assures the existence of an

infinite number of differential invariants. Furthermore, it shows that only afinite number of them has to be determined by integrating a complete systemof pde’s.

Theorem 3.6 (Lie 1893) For any continuous group, acting on a manifoldin any number of variables carrying certain dependencies, there exists an infi-nite number of differential invariants. They may be obtained by differentiationfrom a finite subset of them, it is called a full system of invariants.

The first part of this result is obvious. By differentiating a sufficient numberof times the number of variables increases until the condition for the existenceof invariants mentioned before is obeyed.

3.2 Algebraic Properties of Vector Fields

While Lie developed his theory of continuous groups and of integratingdifferential equations in Leipzig, the high school teacher Wilhelm Killing,apparently not knowing Lie’s work for a long time, developed a theory ofcertain algebraic structures that turned out to comprise the commutators ofthe infinitesimal generators occurring in Lie’s theory as special cases. On Lie’ssuggestion, Elie Cartan in Paris took up the subject in his thesis, correctingand completing Killing’s results, and continued for many years working in thisfield to become one of its most important contributors. Those topics of theresulting theory that will be applied later on are summarized in this section.In addition to Lie’s publications, good references are the thesis of Cartan[25] and the book by Jacobson [82]. For algorithmic questions the article byRand [151] or the thesis by de Graaf [37] is recommended. A comprehensivetreatment including a survey of the more recent literature may be found inthe three volumes of the Springer Encyclopaedia [43], entitled Lie Groups andLie Algebras. A lot of interesting background information is covered in partsII and III of the book by Hawkins [68].

116

General Properties of Lie Algebras. By abstraction from its geometricorigin, the algebraic objects obeying commutation relations (3.10) are studiedas a subject of interest of its own right, they are called Lie algebras. First ofall they have to be defined.

Definition 3.5 (Lie algebra). A Lie algebra L is a vector space over afield F equipped with a binary operation, called commutator, satisfying thefollowing constraints for any U, V,W ∈ L and α, β ∈ F .

i) [U, V ] = −[V,U ] (anti-commutativity)ii) [αU + βV,W ] = α[U,W ] + β[V,W ] (bilinearity)iii) [U, [V,W ]] + [V, [W,U ]] + [W, [U, V ]] = 0 (Jacobi identity)

In this book only finite dimensional Lie algebras are considered; and it isalways assumed that the field F has characteristic zero. If not stated otherwiseexplicitly it will be the complex number field C.

A Lie algebra is usually defined in terms of the commutators of basis ele-ments in the form (3.10). The numerical values of the structure constants ckijdepend on the basis of the Lie algebra. If two Lie algebras are such that bya suitable change of the basis their structure constants become identical theyare considered to be not essentialy different, clearly they are isomorphic. Asusual in algebra, this notion of isomorphism establishes an equivalence rela-tion between Lie algebras. As a consequence, a classification of Lie algebrasmay be performed by associating them to a canonical form which is given bya special basis. For algebras of low dimension canonical forms are given inSection 3.4 on page 135 ff.

Lie [112] has shown that deciding isomorphy of Lie algebras is a purelyalgebraic problem. To this end a suitable basis transformation has to be foundsuch that in the new basis the commutators coincide with a canonical one. Leta basis of an r-dimensional Lie algebra be U1, . . . , Ur with [Ui, Uj ] =

∑ckijUk.

The desired canonical basis elements U1, . . . , Ur with commutation relations[Ui, Uj ] =

∑ckijUk are expressed in terms of the given ones as Ui =

∑αijUj

with undetermined coefficients αij ∈ F . Substituting these expressions intothe canonical commutation relations leads to the system of algebraic equations

r∑l,m=1

cnlmαilαjm =r∑

k=1

ckijαkn, 1 ≤ i < j ≤ r, 1 ≤ n ≤ r (3.16)

with the constraint detαij 6= 0. This system has been given already byLie [112], vol. I, page 290, equations (3). Its solution yields the desired trans-formation. In general it is not unique. It has been discussed in detail byGerdt [52] applying Grobner basis methods for determining its solution.

Example 3.1 The vector fields considered in Example 3.3 below will beshown to generate a Lie algebra of type l3,2. In order to determine the transfor-mation to canonical generators explicitly, the new basis elements are writtenas Ui = αi1∂x + αi2∂y + αi3(y∂x − x∂y). The commutation relations of l3,2


require that the coefficients satisfy

α12α23 − α13α22 = 0, α11α23 − α13α21 = 0,α11α33 − α13α31 + α12 = 0, α12α33 − α13α32 − α11 = 0,α21α33 − α23α31 − α22 = 0, α22α33 − α23α32 + α21 = 0

and α13 = α23 = 0. Its general solution is α13 = α23 = 0, α33 = i,α12 = −iα11 and α22 = iα21. The coefficients α11, α21, α31 and α32 re-main undetermined. Choosing α11 = α21 = 1, α31 = α32 = 0 the final answerU1 = ∂x − i∂y, U2 = ∂x + i∂y and U3 = i(y∂x − x∂y) is obtained.

Although a priori any Lie algebra with equal dimension is a candidate for thecanonical form of any given Lie algebra, it is more efficient to narrow downthe possible types in advance by calculating several invariants. This pointhas been emphasized by de Graaf [37], Section 7.1. The following example istaken from the above mentioned publication by Gerdt [52]. The characteristicpolynomial which occurs there is explained below, see eq. (3.17).

Example 3.2 Consider the Lie algebra of vector fields v1 = ∂x, v2 = ∂t,v3 = t∂x+∂u, v4 = x∂x+3t∂t−2u∂u with nonvanishing commutators [v1, v4] =[v2, v3] = v1, [v2, v4] = 3v2 and [v3, v4] = −2v3. According to definition (3.17)its characteristic polynomial ω(ω − E4)(ω − 3E4)(ω + 2E4) is squarefree anddecomposes into linear factors. Consequently, the Lie algebra l4,5 defined onpage 136 is identified as the proper type.

A sub-vectorspace M ⊂ L is a subalgebra if the commutator of any twoelements fromM is again inM. This is written [M,M] ⊂M. An invariantsubalgebra or ideal is such that any commutator with elements from L is inM, i.e., [M,L] ⊂ M. The normalizer of M is the set of all elements l ∈ Lsuch that [l,M] ∈M. The normalizer of M is a subalgebra. If S is a subsetof L the centralizer of S is the set of all elements c ∈ L such that [c, s] = 0for all s ∈ S. The centralizer of L is called the center.

Fundamental problems are to classify all types of composition leading tononisomorphic Lie algebras, and to determine simpler substructures within agiven Lie algebra, i.e., to determine its invariant subalgebras. The character-istic polynomial

∆(ω) ≡ |∑Eic

kij − δjkω| =

ωr − ψ1(E)ωr−1 + ψ2(E)ωr−2 + . . .+ (−1)r−1ψr−1(E)ω(3.17)

or correspondingly the characteristic equation ∆(ω) = 0 where E1, . . . , Erare r indeterminates turns out to be an important tool for dealing with theseproblems. It has been introduced by Killing [89] in order to answer the ques-tion in which two-dimensional subalgebras a given element is contained. Thenumber of independent coefficients of the characteristic polynomial is calledthe rank of the Lie algebra. Important properties of the characteristic poly-nomial are given next.

118

Theorem 3.7 Let ∆(ω) and the ψk(E1, . . . , Er) be defined by (3.17).

i) The coefficient ψk(E1, . . . , Er) is a homogeneous polynomial of degree kin E1, . . . , Er.

ii) If new basis elements are introduced by Ui =∑αi,jUj, the ψk are

changed according to ψk(E1, . . . , Er) = ψk(∑αi,1Ei, . . . ,

∑αi,rEi).

iii) The coefficients ψk(E1, . . . , Er) satisfy the differential equations

r∑σ,τ=1

cτρσEσ∂ψ

∂Eτ= 0,

i.e. they are invariants of the adjoint group.

iv) The rank of a Lie algebra is not higher than the multiplicity of the rootω = 0 of the characteristic equation.

The proof may be found in Cartan’s Thesis [25], Chapitre II. Property ii)shows that the structure of the coefficients ψk is invariant under a basis changeof the corresponding Lie algebra. This property makes the characteristicpolynomial into an important tool for determining the type of a Lie algebraas it is shown in the following example.

Example 3.3 The Lie algebra generated by the vector fields ∂x, ∂y andy∂x − x∂y has the nonvanishing structure constants c132 = −1 and c233 = 1.They yield the characteristic polynomial (ω2 +E2

3)ω. By Theorem 3.7 the Liealgebra is of type l3,2 with c = −1 in the classification given in Section 3.4.This follows from the fact that it is the only algebra which has a third orderand a first order term in its characteristic polynomial, and the dimension ofits derived algebra (see below) has dimension 2.

This example is of particular importance because it applies exactly the kindof reasoning that is used in Chapter 5 for identifying the symmetry types ofdifferential equations.Solvable Lie Algebras. The so-called integrable or solvable groups and thecorresponding Lie algebras play an important role in the symmetry analysisof differential equations. This term is closely connected with their applicationfor solving linear pde’s as described in Theorem 2.23. The special structureof the commutators of the vector fields admitted by the given linear pde inthis theorem is generalized as follows. The derived Lie algebra L′ ≡ L(1) isgenerated by all commutators of L, i.e., L′ = [L,L]. The derived series is thesequence L,L′, . . . ,L(k) where L(j) = [L(j−1),L(j−1)].

Definition 3.6 (Solvable Lie algebra) A Lie algebra of dimension r iscalled integrable or solvable if L(k) = 0 for some positive integer k ≤ r.


The following theorem due to Lie and Engel provides a convenient means foridentifying solvable Lie algebras if they are given by basis vectors and structureconstants. Its proof may be found in Cartan’s Thesis [25], Theoreme IV onpage 45.

Theorem 3.8 Let L be an r-dimensional Lie algebra with structure con-stants ckij. Either of the following two conditions is necessary and sufficientfor L to be solvable.

i) The derived algebra of L has rank zero.ii) For i, j, k = 1, . . . , r there holds

∑rλ,µ,ν=1(c

µiλc

νjµc

λkν − cλiµc

µjνc

νkλ) = 0.

Example 3.4 Consider the three- and four-dimensional algebras listed onpage 135. The first and the second derived algebras of l3,1 and l4,1 havedimension 3, therefore by definition they are not solvable. The second or thethird derived algebras of the remaining algebras l3,2 to l3,6 and l4,2 to l4,17have zero dimension, consequently they are solvable.

Important properties of solvable algebras are described next, they may befound in Cartan’s Thesis [25] on pages 45-48.

Theorem 3.9 Solvable Lie algebras have the following important proper-ties.

i) Any solvable Lie algebra of dimension r contains an invariant subalgebraof dimension 1, which is contained in an invariant subalgebra of dimen-sion 2 etc., i.e., r independent elements U1, U2, . . . , Ur may be foundsuch that

[Ui, Uk] =j=i∑j=1

cji,kUj for i = 1, . . . , r, k = i+ 1, . . . , r.

ii) The characteristic polynomial of a solvable algebra decomposes into lin-ear factors, i.e.,

∆(ω) = −ωi=r−1∏i=1

( ρ=r∑ρ=i−1

Eρciρi − ω

).

Semi-simple and simple Lie algebras. If an algebra L is not solv-able it may have invariant subalgebras that are solvable. There is always aunique largest solvable ideal which contains every solvable ideal, it is called theradical R. If R = L the algebra is solvable, if R = 0 the algebra does nothave any nontrivial solvable ideal and the following definition applies.

Definition 3.7 (Semi-simple and simple) If a nonsolvable Lie algebra doesnot have any nontrivial solvable invariant subalgebra it is called semi-simple.If it does not have any invariant subalgebra at all it is called simple.

120

From the above discussion it follows that determining the radical of anyLie algebra is an important step for identifying its structure. The followingresult which is due to Cartan [25], Chapter III, Section 5, gives a constructiveanswer to this problem.

Theorem 3.10 (Cartan 1894) If L is a Lie algebra of dimension r withstructure constants ckij and characteristic polynomial (3.17), its radical is de-termined by a linear combination of the basis elements the coefficients of which

obey the constraints∑rρ=1 c

ρij∂ψ2(E)∂Eρ

= 0.

Because ψ2 is a quadratic form, the constraints given in this theorem gen-erate a system of linear homogeneous equations in the Ei, its solution de-termines the radical. The next theorem provides a constructive means fordeciding semi-simplicity.

Theorem 3.11 A Lie algebra is semi-simple if and only if the discriminantof the coefficient ψ2(E) of the characteristic polynomial is different from zero.

One of the major achievements of Killing and Cartan was to provide an ex-plicit and complete classification of simple complex Lie algebras. It comprisesfour series of so-called classical algebras An, Bn, Cn and Dn for n ≥ 1, 2, 3or 4 respectively, and the exceptional algebras E6, E7, E8, F4 and G2. Thesubindex denotes the rank of the respective algebra. Details may be found inthe literature quoted at the beginning of this section. The importance of thisclassification becomes obvious from the following theorem which is also dueto Killing and Cartan.

Theorem 3.12 Any semi-simple Lie algebra is the direct sum of invariantsimple subalgebras.

The decomposition ensured by this theorem is unique. The simple com-ponents may be transformed into canonical form. In order to obtain thisrepresentation explicitly for any given semi-simple Lie algebra suitable algo-rithms are required. A lot of information on these algorithmic problems maybe found in the thesis of de Graaf [37] and the references given there; see alsothe article by Cartan [26] on this subject.

Example 3.5 (Cartan 1896) Let a six-dimensional Lie algebra given bythe vector fields

U1 = ∂x + x(x∂x + y∂y + z∂z), U2 = ∂y + x(x∂x + y∂y + z∂z),

U3 = ∂z + x(x∂x + y∂y + z∂z),

U4 = z∂y − y∂x, U5 = x∂z − z∂z, U6 = y∂x − x∂y.

(3.18)

The coefficient ψ2(E) = E21 + E2

2 + E23 + E2

4 + E25 + E2

6 has nonvanishingdiscriminant, consequently by Theorem 3.11 the Lie algebra is semi-simple.


It is easy to see that U1 = U1 + U4, U2 = U2 + U5, U3 = U3 + U6 andU4 = U1 − U4, U5 = U2 − U5, U6 = U3 − U6 generate two type l3,1 algebraswith [Ui, Ui+3] = 0 for i = 1, 2, 3, i.e., the vector fields (3.18) generate asemi-simple algebra which is the direct sum of two three-dimensional simplealgebras of type l3,1.

The Levi Decomposition. A Lie algebra L having a nontrivial properradical may be represented as a semidirect sum R > S where R is theradical and S is semi-simple. The radical is unique and basis independent, thecomplement S is ismorphic to L/R. The formulation given in the subsequenttheorem is taken from Rand [151].

Theorem 3.13 (Rand 1988) Let L be a Lie algebra with basis l1, . . . , lr,radical R and Levi decomposition R > S. There is an algorithm that findsa new basis r1, . . . , rρ, s1, . . . , sσ with ρ + σ = r such that R is generated bythe ri, S by the sk and the commutation relations [R,R] ⊂ R, [S,S] = S and[R,S] ⊆ R are valid.

The proof of Rand [151] provides a constructive method for obtaining sucha basis explicitly. It is not given here because the examples that are relevantlater on in this book are simple enough such that it may be obtained byinspection, and the general method is not required.

Example 3.6 Consider the Lie algebra g23(r = 3) defined on page 139. Itis generated by the vector fields

U1 = ∂x, U2 = ∂y, U3 = x∂x, x∂y, U4 = x∂y, U5 = y∂y,

U6 = x2∂y, U7 = x2∂x + 2xy∂y

from which the nonvanishing structure constants c113 = 1, c214 = 1, c317 = 2,c517 = 2, c225 = 1, c427 = 2, c636 = 2, c737 = 1, c445 = 1, c647 = 1, c656 = −1and the coefficient ψ2 = 8E1E7 + 3E2

5 − 6E3E5 +E23 are obtained. Applying

Theorem 3.10 yields the constraints E1 = E3 = E7 = 0, i.e., the radical isgenerated by U2, U4, U5 and U6. It is easy to see that U1, U3 + U5 and U7

generate a simple three-dimensional algebra. Consequently, the Levi decom-position has the form l4,6(a = b = 1) >l3,1 or, its equivalent in terms ofvector fields

∂y, x∂y, x2∂y, y∂y >∂x, x∂x + y∂y, x2∂x + 2xy∂y.

3.3 Group Actions in the Plane

For the main subject of this book Lie transformation groups acting on atwo-dimensional manifold are most important. If not stated otherwise this will

122

be the complex plane C. Therefore the results described in the two precedingsections are specialized to this case and illustrated by numerous examples,usually taken from the collection in Section 3.4. Of particular importance forlater applications are constructive methods for dealing with these groups andits Lie algebras.Finite Groups and Their Infinitesimal Generators. The action of an rparameter Lie transformation group in coordinates x and y is written as

x = f(x, y, a1, . . . , ar), y = g(x, y, a1, . . . , ar) (3.19)

with f(x, y, 0, . . . , 0) = x and g(x, y, 0, . . . , 0) = y. If there is only a singleparameter, it is denoted by a without a subindex. Conforming with (3.7),the coefficients of the infinitesimal generators Uk = ξk(x, y)∂x+ηk(x, y)∂y aredefined by

ξk(x, y) ≡∂f

∂ak

∣∣∣a1=...=ar=0

and ηk(x, y) ≡∂g

∂ak

∣∣∣a1=...=ar=0

for k = 1, . . . , r. According to Lie’s First Theorem 3.1 there holds

∂x

∂ak=

r∑j=1

ϕj,k(a1, . . . , ar)ξj(x, y),∂y

∂ak=

r∑j=1

ψj,k(a1, . . . , ar)ηj(x, y).

(3.20)

Example 3.7 The rotations of the plane are parametrized by

x = x cosα− y sinα, y = x sinα+ y cosα (3.21)

with 0 ≤ α < 2π. An additional second transformation of this type withparameter β

¯x = x cosβ − y sinβ, ¯y = x sinβ + y cosβ (3.22)

may be combined into a single one corresponding to a parameter value γ =α + β, the reverse operation of (3.21) corresponds to −α, and no action toα = 0. The infinitesimal generator is U = −y∂x + x∂y. The group equations

(3.20) are dxdα

= −y and dydα

= x. A Janet basis for the defining equations in

grlex, η > ξ, y > x term order isη + x

y ξ = 0, ξx = 0, ξy + 1y ξ = 0

.

Example 3.8 Assume that the points with coordinates (x, y) are trans-formed into (x, y) according to x = x + a1 and y = y + a2. A second trans-formation of the same type with parameters b1 and b2 yields ¯x = x+ b1 and¯y = y + b2. Obviously the same effect may be achieved by a single transfor-mation with parameter values c1 = a1 +b1, c2 = a2 +b2. The group equations(3.20) are ∂x

∂a1= 1, ∂x

∂a2= 0, ∂y

∂a1= 0 and ∂y

∂a2= 1. For the infinitesimal

generators there follows U1 = ∂x and U2 = ∂y with [U1, U2] = 0, i.e., theygenerate the Lie algebra l2,2 defined on page 135.


In general a pair of substitution functions f and g does not define a groupas the following example shows.

Example 3.9 The transformations

x = (a+ 1)x, y = y + a and ¯x = (b+ 1)x, ¯y = y + b (3.23)

have the combined action

¯x = (a+ 1)(b+ 1)x, ¯y = y + a+ b.

In this case it is obvious from the finite group equations that there does notexist a parameter c that may be expressed in the form c = φ(a, b).

Example 3.10 An example of a group depending on three parameters isgenerated by the motions in the plane. If two subsequent motions are definedby

x = x cosα− y sinα+ a, y = x sinα+ y cosα+ b,

¯x = x cosβ − y sinβ + c, ¯y = x sinβ + y cosβ + d

a simple calculation shows that the combined motion is of the same form andis generated by the parameter values

α+ β, a cosβ + b sinβ + c, a sinβ + b cosβ + d

for the rotation angle and the two translations. The infinitesimal generatorscorresponding to the parameters a, b and α respectively are U1 = ∂x, U2 = ∂yand U3 = y∂x − x∂y with the nonvanishing commutators [U1, U3] = −U2 and[U2, U3] = U1.

Example 3.11 Consider the equations x = xea3 + a1, y = yea3 + a2. Bydifferentiation there follows ∂x

∂a1= 1, ∂x

∂a3= xea3 , ∂y

∂a2= 1 and ∂y

∂a3= yea3

for the nonvanishing derivatives. Consequently, corresponding to (3.20)

φ = ψ =

1 0 00 1 00 0 ea3

and ξ1 = 1, ξ2 = 0, ξ3 = x, η1 = 0, η2 = 1 and η3 = y. They satisfy (3.20), i.e.the above expressions for x and y define a three parameter group. Substitutinga2 = a3 = 0 yields x = x + a1, y = y and U1 = ∂x, substituting a1 = a3 = 0yields x = x, y = y + a2 and U2 = ∂y, finally substituting a1 = a2 = 0 yieldsx = xea3 , y = yea3 and U3 = x∂x + y∂y with the commutators [U1, U2] = 0,[U1, U3] = U1 and [U2, U3] = U2, i.e., they generate a Lie algebra of typel3,2(c = 1).

The finite transformations corresponding to an infinitesimal generatorU = ξ(x, y)∂x + η(x, y)∂y may be obtained from Lie’s equations (3.14). In

124

coordinates x and y they have the form

dx

da= ξ(x, y),

dy

da= η(x, y). (3.24)

This is a system of ode’s for the two functions x(x, y, a) and y(x, y, a) w.r.t. tothe group parameter a, complemented by the initial conditions x(a = 0) = x,y(a = 0) = y. The coordinates of the manifold M appear only as parameters.Although a solution for such a system always exists, in general it cannot beobtained in closed form. This is possible only if ξ(x, y) and η(x, y) have aspecial structure. The following examples illustrate these possibilites.

Example 3.12 Let U = x2∂x + xy∂y, i.e., dxda

= x2, dyda

= xy. Integration

of the first equation yields x = − 1a+ φ(x, y) where φ is an undetermined func-

tion of its arguments. The requirement x(a = 0) = x leads to φ(x, y) = − 1x

and x = x1− ax . Substituting this value into the second equation yields

dyda

= xy1− ax from which there follows y = ψ(x, y)

ax− 1 with ψ again an undeter-mined function of x and y. The condition y(a = 0) = y leads to ψ(x, y) = −yand finally y = y

1− ax .

Example 3.13 Let U = (x + y)∂x, i.e., dxda

= x + y, dyda

= 0. The lastequation leads immediately to y = y. Substitution into the first equationyields the linear first order equation dx

da− x = y with the general solution x =

ea(yea +φ(x, y)). The initial condition x(a = 0) = x leads to φ(x, y) = x− y,and the final answer x = eax+ ea(ea − 1)y.

Example 3.14 Let U = (y2 + 1)∂x + (x2 + 1)∂y, i.e., dxda

= y2 + 1, dyda

=x2 + 1. Eliminating x, the second order equation y′′2 − 4(y′ − 1)(y2 + 1) = 0is obtained for which a closed-form solution cannot be found.

For a selection of frequently occurring generators the corresponding finitetransformation is given in Table 3.1.

In order to find the finite transformations of an r-parameter group if thetransformations corresponding to the infinitesimal generators are known, amethod described in Lie [114], Kapitel 7, §3 will be applied.

Lemma 3.2 (Lie 1893) Let r one-parameter groups be determined by thetransformation functions fi(x, y, ai) and gi(x, y, ai) for i = 1, . . . , r. Then thegeneral transformation of an r-parameter group is obtained recursively by

x = fr(fr−1(. . . f2(f1(x, y, a1), a2), . . . , ar−1), ar),y = gr(gr−1(. . . g2(g1(x, y, a1), a2), . . . , ar−1), ar).

(3.25)

Example 3.15 Let a three-parameter group be determined by the in-finitesimal generators U1 = ∂x, U2 = ∂y, U3 = x∂x + y∂y. By means of the


Generator Transformation Generator Transformation

∂x x = x+ a, y = y ∂y x = x, y = y + a

x∂x x = xea, y = y y∂x x = x+ ay, y = y

x∂y x = x, y = y + ax y∂y x = x, y = yea

y∂x − x∂y x = x cos a+ y sin a, y∂x + x∂y x = x cosh a+ y sinh a,y = −x sin a+ y cos a y = x sinh a+ y cosh a

x2∂x + xy∂y x = x1− ax, y = y

1− ax xy∂x + y2∂y x = x1− ay , y = y

1− ayx2∂x + 2xy∂y x = x

1− ax, y = y(1− ax)2 (x+ y)∂x x = eax+ (ea − 1)y, y = y

x∂x + (x+ y)∂y x = eax, y = ea(ax+ y) x2∂x + y2∂y x = x1− ax, y = y

1− ayex∂y x = x, y = y + aex mx∂x + ny∂y x = xema, y = yena

yn∂x x = x+ ayn, y = y xn∂y x = x, y = y + axn

TABLE 3.1: The finite transformations for a selection of frequently oc-curring infinitesimal generators are listed; a is the group parameter, the valuea = 0 corresponds to the identity.

finite transformations given in Table 3.1 one gets x = x + a1, y = y for U1,¯x = x, ¯y = y + a2 from U2 and ¯x = ea3 ¯x, ¯y = ea3 ¯y from U3. Combining thethree transformations and making an obvious change of variables, the finitetransformations of the three-parameter group are obtained in the form

x = (x+ a1)ea3 , y = (y + a2)ea3 .

The final result must be independent of the order in which the infinitesimalgenerators are applied. Taking for example U1 = x∂x + y∂y, U2 = ∂x andU3 = ∂y one obtains by a similar calculation

x = eb1x+ b2, y = eb1y + b3.

This is the same group with a different set of parameters. They are relatedby the equations b1 = a3, b2 = a1e

a3 , b3 = a2ea3 .

On page 112 a third representation for a Lie transformation group in termsof the so-called defining equations has been introduced. Solving these linearhomogeneous systems of pde’s is the main issue of Chapter 2. The reverseprocess, i.e., setting up the defining equations from a given set of infinitesimalgenerators, has also been discussed there on page 75. The algorithm Con-structJanetBasis has been designed for this purpose. The following simpleexample may be performed by pencil and paper, yet it shows all basic steps.

Example 3.16 Let a group be given in terms of three vector fields by∂x, ∂y, y∂x−x∂y, consequently ξ = c1+c3y and η = c2−c3x. Differentiation

126

leads to ξx = 0, ξy = c3, ηx = −c3 and ηy = 0. Eliminating c3 there followsξx = 0, ξy + ηx = 0, ηy = 0. It is not a Janet basis. In a grlex term orderingwith η > ξ, y > x the integrability condition ξyy = 0 is obtained. It has to beadded to this system, i.e., the desired Janet basis is ξx, ηx + ξy, ηy, ξyy.

The importance of the defining equations arises from the fact that the Liesymmetries of an ode are always obtained in this form in the first place,therefore these equations are called determining system in this context. InSection 3.5 it will be shown how many important properties of Lie groupsmay be obtained from these defining equations without solving them.Geometric Properties. Let x = f(x, y, a) and y = g(x, y, a) be the finiteequations of a one-parameter group with coordinates (x, y) and infinitesimalgenerator U = ξ(x, y)∂x + η(x, y)∂y. Under the action of this group an arbi-trarily chosen point describes a curve the points of which are parametrized bya. Any two points on an individual curve may be transformed into each otherby a suitable group element, they are equivalent w.r.t. this group. If theyare located on different curves such a transformation does not exist. Thesecurves, generated by the action of a one-parameter group on an arbitrarypoint of the plane, are called its path curve. They are special cases of theorbits defined on page 108. The path curves generate an invariant decompo-sition of the plane. By definition, the path curves corresponding to U obeyequations (3.24). According to Theorem 2.19 these ode’s with first integralϕ(x, y) = const are equivalent to the pde Uϕ = 0. It shows that the orbits ofa one-parameter group of the plane are essentially determined by the invariantof the pde corresponding to its generator.

Example 3.17 Consider the one-parameter group x = x1− ax and y =

y1− ax with generator U = x2∂x + xy∂y. The invariant of U is φ = y

x .

Consequently, its orbits are yx = C, i.e., the straight lines through the origin.

Example 3.18 The group considered in the preceding example had theinvariant φ = y

x . Substituting the transformed variables x and y yields yx =y

1− ax1− axx = y

x , i.e., the invariance of φ is explicit.

Example 3.19 The invariant of the one-parameter group with generatorU = ax∂x + by∂y is determined by 1

adxx = 1

bdyy with the result φ = ya

xb.

If the finite transformations of the group are known, deciding transitivitycomes down to eliminating the group parameters a1, . . . , ar from equations(3.19) for given (x, y) and (x, y). Consider for example the group x = x+ a1,y = y + a2. If (x, y) is considered to be chosen at will, the parameter valuesa1 = x− x, a2 = y− y will perform the desired transformation. On the otherhand, if the group is x = x, y = y+ a1x+ a2, such a transformation does not


exist because the transformations of the group leave x unchanged and a pointwith x 6= x can never be reached. The above question may be generalized toany number of points sharing this property.

Example 3.20 The two-parameter groups g4 and g15 defined on page 138are intransitive because the first coordinate is transformed as x = x, i.e., ifx is chosen different from x there is no group element that brings about thetransformation. Group g25 has finite transformations x = ea2 , y = (y+a1)ea2

from which the expressions a1 = xy − xyx and a2 = log xx for the group

parameters are obtained, i.e., the group is transitive. The same is true for thegroup g26 that has been considered before.

Example 3.21 Consider the group g6 on page 138 with finite transforma-tions x = (x+ a1)ea3 , y = (y + a2)ea4 . Let two different points P1 = (x1, y1)and P2(x2, y2) be arbitrarily given. Applying a transformation of the groupto them yields xi = (xi+a1)ea3 , yi = (yi+a2)ea4 for i = 1, 2. The four groupparameters may be eliminated with the result

a1 =x2x1 − x1x2

x2 − x1, a2 =

y2y1 − y1y2y2 − y1

, ea3 =x1 − x2

x1 − x2, ea4 =

y1 − y2y1 − y2

.

This means, for any given pairs P1, P2 and P1 = (x1, y1), P2 = (x2, y2) atransformation of g6 may be determined moving Pi to Pi for i = 1, 2. Thesame procedure for three arbitrarily chosen points leads to an inconsistent setof equations. Consequently, g6 is 2-fold transitive.

If the finite transformations of the group are not known, or if their ex-pressions are rather complicated, the criterion for transitivity in terms of theinfinitesimal generators of the group given on page 114 may be applied. Itallows deciding transitivity for all groups that are listed on page 138 fromits vector field coefficients with the following answer. Transitive groups: Allprimitive groups g1, g2 and g3, the groups g5, g6, g7 and g9 to g12 allowingtwo systems of imprimitivity, the groups g13, g14 and g17 to g23 with a singlesystem of imprimitivity and finally the groups g24, g25 and g26 allowing aone-parameter family of systems of imprimitivity. Intransitive groups: g4, g8,g15, g16 and g27.

The possible types of transitivity of the transformation groups of the planeare completely described by the following result due to Lie [112], part I,page 632.

Theorem 3.14 A finite continuous group of a two-dimensional manifoldis at most four-fold transitive. The only group with maximal transitivity isthe eight-parameter projective group g3.

This discussion shows that a group on a two-dimensional manifold may al-low a single invariant, and this occurs if and only if all generators are multiplesof a single one. In particular this is true for any single-parameter group.

128

The transitive two-parameter group x = x + a1, y = y + a2 consideredpreviously transforms any line y = αx+ const. into y = αx+ const′. with thesame value of α, i.e., the straight lines with fixed slope α are permuted amongthemselves by the group. According to the definition given on page 108 theyform a system of imprimitivity.

The number of systems of imprimitivity is an important distinguishing fea-ture of any transformation group. For a Lie transformation group of the planethe complete answer has been obtained by Lie [114], page 342, as follows.

Theorem 3.15 A Lie transformation group of the plane is either primitive,or it has exactly one or two systems of imprimitivity, or there is a system ofimprimitivity depending on a parameter or a function of one variable.

In general it is difficult to identify a system of imprimitivity by means ofthe finite transformations of a group. A criterion for primitivity in terms ofthe infinitesimal generators of the group turns out to be more convenient.It is obtained by the following considerations. A system of imprimitivityin the plane may be regarded as a set of curves depending on a parameterwhich in turn may be described as the solution set of a first order ode. If anequation with this property does exist, the group under consideration mustbe a symmetry of it. Analytically this is assured (see Chapter 5) if for thegeneric rank of the prolonged coefficient matrix there holds

rank

ξ1, η1, ζ1ξ2, η2, ζ2

......

ξr, ηr, ζr

= 3

where ζk = ηk,x + y′(ηk,y − ξk,x) − y′2ξk,y, Lie [114], page 208. The ζ ′s arediscussed in detail in Chapter 5.

In later applications the following provides a particularly well-suited crite-rion in terms of the Janet basis of a group (Schwarz [164], see also Neumer [135]).

Theorem 3.16 Let the infinitesimal generators of a symmetry group ofthe x − y−plane be of the form U = ξ∂x + η∂y where the coefficients solvethe corresponding determining system. In order for a system of imprimitivitydifferent from x = const. or y = const. to exist, an equation

ηyω + ηx − ηωy − ξyω2 − ξxω − ξωx = 0 (3.26)

with a suitable ω ≡ ω(x, y) must be valid as a consequence of the determiningsystem of this symmetry group. If x=const. or y=const. is a system of impri-mitivity, (3.26) reduces to ξy = 0 or ηx = 0 respectively.

Proof Let the desired systems of imprimitivity be described by the firstorder ode y′ = ω(x, y). By definition, the solutions of this first order equationare permuted by the original symmetry group. Therefore it must allow it as


a symmetry group. Analytically this is expressed in terms of the requirementU (1)(y′ − ω) = 0 on the solution set of y′ = ω which is equivalent to (3.26),see Theorem 5.2; ω = 0 corresponds to y′ = 0 or y = const., by exchange ofx and y the condition for x = const. is obtained.

For Lie the number and type of systems of imprimitivity was an importanttool in order to obtain a classification of the transformation groups allowedby various manifolds. The classification of the groups of C2 given later in thischapter is organized by its systems of imprimitivity. The following examplesare taken from this listing on page 138.

Example 3.22 The group g1 has the Janet basis ηy+ξx, ξxx, ξxy, ξyy, ηxxfor its defining equations. Reducing the condition (3.26) w.r.t. to it yieldsthe equations ω = 0, ω2 = 0, ωx = 0, ωy = 0 and 1 = 0 which is obviouslyinconsistent. Therefore g1 is primitive.

Example 3.23 The group g7 has the Janet basis ξy, ηx, ηy− cξx, ξxx forits defining equations. From the first two equations x = const. and y = const.are obvious systems of imprimitivity according to the above lemma. The samegroup occurs in Example 5.16 as symmetry group. Reducing (3.26) w.r.t. tothe second Janet basis given there leads to the equations ωx = 0, ωy = 0 and(ω−6)(ω+1) = 0 from which the two systems of imprimitivity y−6x = constand y + x = const are obtained.

Example 3.24 The group g13 has the Janet basisξx−2

y η, ξy, ηy−1y η, ηxx

for its defining equations. The single system of imprimitivity x = const isobvious from the second equation, reduction of (3.26) w.r.t. to this Janetbasis leads to an inconsistent system. In Example 5.14 the same group typegenerates the Janet basis in some transformed variables. Reduction of (3.26)w.r.t. the Janet basis given there leads to the system

ωx +1

x+ yω2 − 2

x+ yω +

1x+ y

= 0, ωy +1

x+ yω2 − 2

x+ yω +

1x+ y

= 0

and (ω − 1)2 = 0 with the unique solution ω = 1, i.e., the single system ofimprimitivity is y − x = const.

Similarity. Transformation to Canonical Form. The appearance of anyLie transformation group may be changed in two ways: Either by a changeof the group parameters, leading essentially to a new basis of the Lie algebragenerated by the infinitesimal generators, or by changing the coordinates ofthe transformed manifold, leaving the commutation relations between theinfinitesimal generators unchanged according to Lemma 3.1, but changing theform of the generators as has been shown on page 111 for a n-dimensionalmanifold. Due to its importance for the rest of this chapter, it is given hereexplicitly for a two-dimensional manifold.

Lemma 3.3 Let U = ξ(x, y)∂x + η(x, y)∂y be a vector field in the plane

130

with coordinates x and y. Furthermore, let new coordinates u and v be definedby u = σ(x, y) and v = ρ(x, y) with the inverse x = φ(u, v) and y = ψ(u, v).Then the coefficients of the transformed vector field V = α(u, v)∂u+β(u, v)∂vare

α(u, v) = Uσ∣∣x=φ,y=ψ

, β(u, v) = Uρ∣∣x=φ,y=ψ

.

A constructive method for deciding similarity of two transformation groupsdefined in terms of their infinitesimal generators has been given in Theorem 2.It is illustrated next by a simple example.

Example 3.25 Consider the generators U1 = ∂u, U2 = ∂v, U3 = u∂u+v∂vin coordinates u and v with commutators [U1, U2] = 0, [U1, U3] = U1 and[U2, U3] = U2. They form a Lie algebra of type l3,2(c = 1). On the other

hand, there are the generators V1 = 1y ∂x + 1

x∂y, V2 = −x2

y ∂x + (x+ y)∂y and

V3 = 1y ∂x+

(y+ 1

x)∂y in coordinates x and y with commutators [V1, V2] = V1,

[V1, V3] = V1 and [V2, V3] = V2 − V3. The linear combinations V1 = 12V1,

V2 = 12(V1 + V2 − V3) and V3 = V3 − V1 obey the same commutation as the

generators Ui. Explicitly they are

V1 =12y∂x +

12x∂y, V2 = −x

2

2y∂x +

x

2∂y, V3 = y∂y.

The relations U3 = uU1 + vU2 and V3 = xyV1 + yxV2 lead to the coordinate

transformation u = xy and v = yx .

A special kind of similarity problem consists of the following question:Given the infinitesimal generators of a transformation group in some set ofcoordinates, is this group similar to another one which is also given in terms ofits infinitesimal generators? This latter group is considered to be a canonicalform chosen from a classification of groups like those described in Section 3.5.If similarity is assured, the transformation functions between the two coordi-nate sets may be determined.

For a one-parameter group there are no commutation relations. The canon-ical form of the group is the translation in one of the coordinates as explainedon page 112. A constructive solution to this problem is given next.

Lemma 3.4 Let ξ(x, y)∂x + η(x, y)∂y be a vector field with coordinates xand y, and ∂v its canonical form in coordinates u = σ(x, y), v = ρ(x, y).

i) If ξ 6= 0 and η 6= 0, the transformation functions are

σ(x, y) = Φ(φ(x, y)), ρ(x, y) = Ψ(φ(x, y)) +∫

dx

ξ(x, y = φ)|y=φ

where φ(x, y) = const is the first integral of dydx

= ηξ, y = φ(x, y) with

the inverse y = φ(x, y).


ii) If ξ = 0, η 6= 0 there holds σ = Φ(x), ρ = Ψ(x) +∫

dy

η(x, y).

iii) If ξ 6= 0, η = 0 there holds σ = Φ(y), ρ = Ψ(y)+∫

dx

ξ(x, y). In all cases

Φ and Ψ are undetermined functions of its argument.

Proof According to Lemma 3.3, because α = 1 and β = 0 now, σ andρ satisfy the system ξσx + ησy = 0 and ξρx + ηρy = 1. The solution of thissystem is obtained from Corollary 2.7 on page 78.

This lemma shows that in general it is not guaranteed that the canonicalform ∂v for an arbitrary vector field may be obtained in closed form. Thecrucial step is to solve the first order differential equation dy

dx= ηξ. This is

only possible if the coefficients ξ(x, y) and η(x, y) are sufficiently simple. Thisremark will be important in Chapter 5 when symmetries of first order ode’sare searched for.

Example 3.26 Let 2x∂x + 3y∂y be the given vector field. The equation

for σ is 2xσx + 3yσy = 0. Its invariant y2

x3 yields σ = Φ(y2

x3

)with Φ an

undetermined function. The equation for ρ is 2xρx + 3yρy = 1, it has beenconsidered in Example 2.57. If Φ = 0 is chosen in the solution for z given

there, ρ = − 12 log x is obtained. Consequently, canonical variables are u = y2

x3 ,

v = − 12 log x or x = e−2v, y =

√ue−3v.

Given an infinitesimal generator, the corresponding finite transformationsmay be obtained also by transforming it to canonical form first, and thensubstituting the backtransformation into the corresponding translation. Inthe next example this is shown for the group considered previously in Exam-ple 3.12.

Example 3.27 Let the infinitesimal generator be U = x2∂x + xy∂y. Thetransformation functions σ(x, y) and ρ(x, y) satisfy x2σx + xyσy = 0 and

x2ρx + xyρy = 1. By Corollary 2.7, the former equation leads to dyy = dx

xwith the special solution φ ≡ y

x = C. Introducing x = φ as new variable, theintegral for ρ yields

ρ =∫

dy

xy= x

∫dy

y2 = − xy

= − 1x.

The integration constant has been omitted. Substituting u = yx and v = − 1

xinto the canonical transformation finally leads to

yx = y

x, −1x = − 1

x + a or x = x1 + ax, y = y

1 + ax.

132

Generator σ(x, y) ρ(x, y) φ(u, v) ψ(u, v)x∂x y log x ev u

y∂x y xy uv u

x∂x + y∂yyx log x ev uev

x∂x − y∂y xy − log y uev e−v

ax∂x + by∂yya

xb1b

log y u−1/beav ebv

−y∂x + x∂y√x2 + y2 arctan yx u cos v u sin v

TABLE 3.2: For a selection of vector fields ξ(x, y)∂x+η(x, y)∂y the trans-formation functions x = φ(u, v), y = ψ(u, v) to canonical form ∂v, and theinverse functions u = σ(x, y), v = ρ(x, y) are given.

It is useful to have the transformation functions to canonical form for aselection of vector fields occurring frequently in applications easily available.They are listed in Table 3.2.

For groups with more than a single parameter, certain normal forms havebeen introduced by Lie. Let the generators be given as

Ui = ξi(x, y)∂x + ηi(x, y)∂y for i = 1, . . . , r. (3.27)

The determinants ∆ij = ξiηj − ξjηi and especially ∆12 ≡ ∆ will be usedfrequently. The canonical coordinates are u and v. The solution to the trans-formation problem is based on the expressions for the coefficients of the trans-formed generators given in Lemma 3.3, and the solution procedure for quasi-linear systems of pde’s described in Lemma 2.12 of the preceding chapter.The operations required are determined above all by the dimension r of thegroup. For r = 2 the complete answer is given next.

Lemma 3.5 Let ξi∂x + ηi∂y, i = 1, 2 be the generators of a two-parametergroup satisfying the proper commutator for one of the groups below. The twotransformation functions u = σ(x, y) and v = ρ(x, y) to canonical variablesare determined by the following systems of equations where ∆ = ξ1η2 − ξ2η1.

Group g4, ∂v, v∂v : Constraint ∆ = 0.

dx

ξ1=dy

η1, solution φ(x, y) = C, σ ≡ σ(φ(x, y)), ρ =

ξ2ξ1.

Group g15, r = 2, ∂v, u∂v : Constraint ∆ = 0.

σ =ξ2ξ1, ρ =

∫dy

η1(σ(x, y), y).

Group g26, ∂u, ∂v: Constraint ∆ 6= 0.

σx =η2∆, σy = −ξ2

∆, ρx = −η1

∆, ρy =

ξ1∆.


Group g25, ∂u, u∂u + v∂v: Constraint ∆ 6= 0.

σx +η1∆σ =

η2∆, σy −

ξ1∆σ = −ξ2

∆, ρx +

η1∆ρ = 0, ρy −

ξ1∆ρ = 0.

Proof For the group g4, Lemma 3.3 leads to the system

ξ1σx + η1σy = 0, ξ1ρx + η1ρy = 1, ξ2σx + η2σy = 0, ξ2ρx + η2ρy = ρ.

The existence of a nontrivial solution for σ requires the constraint ∆ = 0.If it is satisfied, σ is obtained from the first equation. The expression for ρfollows from the consistency of the two equations for ρ. For the second groupg15 the same system is obtained except that ρ at the right hand side of thelast equation is replaced by σ. Again ∆ = 0 is required. The consistency forthe equations determining ρx and ρy now leads to the expressions for σ; ρhas to be determined from ξ1ρx + η1ρy = 1. The corresponding homogeneousequation has σ as its solution which is already known. Consequently, theintegral for ρ follows by Lemma 3.4. For g26 the system for σ and ρ is

ξ1σx + η1σy = 1, ξ1ρx + η1ρy = 0, ξ2σx + η2σy = 0, ξ1ρx + η2ρy = 1,

and for g25

ξ1σx + η1σy = 1, ξ1ρx + η1ρy = 0, ξ2σx + η2σy = σ, ξ2ρx + η2ρy = ρ.

In either case algebraic elimination of the first partial derivatives yields theequations given above.

Example 3.28 Consider the vector fields U1 = ∂x + ∂y, U2 = y∂x + x∂ysatisfying [U1, U2] = U1 generating a group of type g25. According to theabove lemma, the transformation functions σ and ρ obey

σx +1

x− yσ =

x

x− y, σy −

1x− y

σ = − y

x− y,

ρx +1

x− yρ = 0, ρy −

1x− y

ρ = 0.

Its solution is obtained by Lemma 2.9 and Lemma 2.10 with the result σ =C1x− y + x+ y

2 and ρ = C2x− y . Choosing C1 = C2 = 1 yields u ≡ σ(x, y) =

1x− y + x+ y

2 and v ≡ ρ(x, y) = 1x− y , and the inverse transformations

x ≡ φ(u, v) = u− v + 12v and y ≡ ψ(u, v) = u− v − 1

2v .

Differential Invariants. In the plane there is only a single dependencypossible between the coordinates if no additional variables are introduced.Let the coordinates again be x and y and assume that y depends on x, i.e.,

134

y ≡ y(x). This case covers ordinary differential equations in a single dependentvariable. The next two examples deal with invariants of this type.

Example 3.29 Consider the vector fields U1 = ∂x, U2 = 2x∂x + y∂y andU3 = x2∂x + xy∂y obeying [U1, U2] = 2U1, [U1, U3] = U2 and [U2, U3] = 2U3.If y ≡ y(x) is assumed, its second prolongations are

U(2)1 = U1, U

(2)2 = U2− y′∂y′ − 3y′′∂y′′ , U

(2)3 = U3− (xy′− y)∂y′ − 3xy′′∂y′′ .

A simple calculation shows that the first two commutators of the prolongedvector fields remain unchanged. The last one is obtained from

U(2)2 U

(2)3 = 4x2∂x + 3xy∂y − (xy′ − y)∂y′ + 3xy′′∂y′′

andU

(2)3 U

(2)2 = 2x2∂x + xy∂y + (xy′ − y)∂y′ + 9xy′′∂y′′

with the result U (2)2 U

(2)3 − U (2)

3 U(2)2 = 2U (2)

3 .

Example 3.30 Consider the four-parameter group generated by ∂x, ∂y,x∂y and x∂x + y∂y. If it is assumed that y ≡ y(x), the fifth prolongations are∂x, ∂y, x∂y + ∂y′ and x∂x + y∂y − y′′∂y′′ − 2y′′′∂y′′′ − 3y(4)∂y(4) − 4y(5)∂y(5) .

By the methods explained in Chapter 2 the invariants Φ1 = y′′2

y′′′, Φ2 = y′′3

y(4)

and Φ3 = y′′4

y(5) are obtained. The invariants Φ1 and Φ2 are a full system.

More general types of differential invariants for transformation groups inthe plane may be obtained if dependencies on variables different from thecoordinates are assumed. In the next example the group is the same as in thepreceding one, the coordinates are now v and w, they are assumed to dependon x and y.

Example 3.31 Consider the four-parameter group generated by ∂v, ∂w,v∂w and v∂v + w∂w. It is assumed now that both v and w depend on xand y. These latter variables are not transformed by the group. The secondprolongations w.r.t. x and y are

∂v, ∂w, v∂w + vx∂wx+ vy∂wy

+ vxx∂wxx+ vxy∂wxy

+ vyy∂wyy,

v∂v + w∂w + vx∂vx+ wx∂wx

+ vy∂vy+ wy∂wy

+vxx∂vxx+ wxx∂wxx

+ vxy∂vxy+ wxy∂wxy

+ vyy∂vyy+ wyy∂wyy

.

There are two first order invariants Φ1 = vyvx and Φ2 = vxwy − vywx

v2x

, and

three second order invariants are

Φ6 = vxxwx − vxwxxv2x

, Φ7 = vyywy − vywyyv2y

, Φ8 = vxywx − vxwxyv2x

.


3.4 Classification of Lie Algebras and Lie Groups

Lie’s classification of groups of the complex plane and of the complex Liealgebras of low dimension is described in this section. It is not explained howLie arrived at these classifications, only references to the literature are given.Lie Algebras of Low Dimensions. For dimension not higher than fourthe listing of complex Lie algebras is complete. In some cases there occurparameters. They have to be constrained suitably in order to avoid dupli-cation. A few algebras of higher dimension are also given because they areneeded for the symmetry analysis later on. In addition the following generalrule is applied: Whenever an essential algebraic property like for example thedimension of the derived algebra changes for a special value of a parameter,the algebra corresponding to these parameter values is listed as a separateentry. More details may be found in Lie’s original work [112], vol. III, or invol. 41, Chapter 7 of the Encyclopedia of Mathematical Sciences.

Each entry of the subsequent listing is a representative of a full isomorphismclass. In order to identify a given Lie algebra, membership to a particular iso-morphism class has to be decided. Secondly, the transformation to canonicalform may be determined.

Various properties of Lie algebras may be read off easily from this listing.For example, all two-dimensional algebras, all three-dimensional algebras ex-cept l3,1 and all four-dimensional algebras except l4,1 are solvable; l3,1 is sim-ple; l4,1 has the Levi decomposition l1 >l3,1 where the radical is generatedby U4 and the simple three-dimensional algebra by the remaining generators.In the lisiting below only nonvanishing commutators are given.One-dimensional algebras

l1: commutative, Characteristic equation: ω = 0.Two-dimensional algebras

l2,1: [U1, U2] = U1, dim l′2,1 = 1, dim l′′2,1 = 0.Characteristic equation: ω2 − E2ω = ω(ω − E2) = 0.l2,2: commutative. Characteristic equation: ω2 = 0.

Three-dimensional algebrasl3,1: [U1, U2] = U1, [U1, U3] = 2U2, [U2, U3] = U3, dim l′3,1 = dim l′′3,1 = 3.Characteristic equation: ω3 + (4E3E1 − E2

2)ω = ω(ω2 + 4E3E1 − E22) = 0.

l3,2(c): [U1, U3] = U1, [U2, U3] = cU2, c 6= 0, dim l′3,2 = 2, dim l′′3,2 = 0.Characteristic equation: ω3−(c+1)E3ω

2+cE23ω = ω(ω−E3)(ω−cE3) = 0.

l3,3: [U1, U3] = U1, [U2, U3] = U1 + U2, dim l′3,3 = 2, dim l′′3,3 = 0.Characteristic equation: ω3 − 2E3ω

2 + E23ω = ω(ω − E3)2 = 0.

l3,4: [U1, U3] = U1, dim l′3,4 = 1, dim l′′3,4 = 0.Characteristic equation: ω3 − E3ω

2 = ω2(ω − E3) = 0.l3,5: [U2, U3] = U1, dim l′3,5 = 1, dim l′′3,5 = 0.Characteristic equation: ω3 = 0.l3,6: commutative. Characteristic equation: ω3 = 0.

136

Four-dimensional algebrasl4,1: [U1, U2] = U1, [U1, U3] = 2U2, [U2, U3] = U3, dim l′4,1 = dim l′′4,1 = 3.Characteristic equation: ω4 +(4E3E1−E2

2)ω2 = (ω2 +4E3E1−E22)ω2 = 0.

l4,2: [U1, U4] = U1, [U2, U4] = U1 + U2, [U3, U4] = U2 + U3,dim l′4,2 = 3, dim l′′4,2 = 0.Characteristic equation: ω4 − 3E4ω

3 + 3E24ω

2 − E34ω = ω(ω − E4)3 = 0.

l4,3(c): [U1, U4] = cU1, [U2, U4] = (c+ 1)U2, [U3, U4] = U1 + cU3,dim l′4,3 = 3, dim l′′4,3 = 0. Characteristic equation:ω4 − (3c+ 1)E4ω

3 + c(3c+ 2)E24ω

2 − c2(c+ 1)E34ω

= ω(ω − cE4)2(ω − (c+ 1)E4) = 0.l4,4: [U1, U4] = 2U1, [U2, U3] = U1, [U2, U4] = U2, [U3, U4] = U2 + U3,dim l′4,4 = 3, dim l′′4,4 = 1, dim l′′′4,4 = 0. Characteristic equation:ω4 − 4E4ω

3 + 5E24ω

2 − 2E34ω = −ω(ω − 2E4)(ω − E4)2 = 0.

l4,5(c): [U1, U4] = cU1, [U2, U3] = U1, [U2, U4] = U2, [U3, U4] = (c− 1)U3,c 6= 1, dim l′4,5 = 3, dim l′′4,5 = 1, dim l′′4,4 = 0. Characteristic equation:ω4 − 2cE4ω

3 + (c2 + c− 1)E24ω

2 − c(c− 1)E34ω

= ω(ω−E4)(ω−cE4)(ω− (c−1)E4)) = 0.l4,6(a, b): [U1, U4] = U1, [U2, U4] = aU2, [U3, U4] = bU3, ab 6= 0,dim l′4,6 = 3, dim l′′4,6 = 0.Characteristic equation: ω4 − 2E2ω

3 + E24ω

2 = ω2(ω − E4)2 = 0.l4,7: [U1, U4] = U1, [U2, U4] = U2, [U3, U4] = U2 + U3,dim l′4,7 = 2, dim l′′4,7 = 0.Characteristic equation: ω4 − 3E4ω

3 + 3E24ω

2 − E34ω = ω(ω − E4)3 = 0.

l4,8: [U1, U2] = U2, [U1, U3] = U2 + U3, dim l′4,8 = 2, dim l′′4,8 = 0.Characteristic equation: ω4 + 2E1ω

3 + E21ω

2 = ω2(ω + E1)2 = 0.l4,9(a): [U1, U2] = U2, [U1, U3] = aU3, a 6= 0, dim l′4,9 = 2, dim l′′4,9 = 0.Characteristic equation:

ω4 − (a+ 1)E4ω3 + aE2

4ω2 = ω2(ω − E4)(ω − aE4) = 0.

l4,10: [U1, U2] = U2, [U3, U4] = U4, dim l′4,10 = 2, dim l′′4,10 = 0.Characteristic equation:

ω4 + (E3 + e1)ω3 + E3E1ω2 = ω2(ω + E1)(ω + E3) = 0.

l4,11: [U2, U4] = U2, [U3, U4] = U1, dim l′4,11 = 2, dim l′′4,11 = 0.Characteristic equation: ω4 − E4ω

3 = ω3(ω − E4) = 0.l4,12: [U1, U4] = U1, [U2, U3] = U1, [U2, U4] = U2,dim l′4,12 = 2, dim l′′4,12 = 0.Characteristic equation: ω4 − 2E4ω

3 + E24ω

2 = ω2(ω − E4)2 = 0.l4,13: [U1, U4] = U2, [U3, U4] = U1, dim l′4,13 = 2, dim l′′4,13 = 0.Characteristic equation: ω4 = 0.l4,14: [U1, U4] = U1, [U2, U4] = U2, dim l′4,14 = 2, dim l′′4,14 = 0.Characteristic equation: ω4 − 2E2ω

3 + E24ω

2 = ω2(ω − E4)2 = 0.l4,15: [U1, U4] = U1, dim l′4,15 = 1, dim l′′4,15 = 0.Characteristic equation: ω4 − E4ω

3 = ω3(ω − E4).l4,16: [U1, U2] = U3, dim l′4,16 = 1, dim l′′4,16 = 0.Characteristic equation: ω4 = 0.


l4,17: commutative. Characteristic equation: ω4 = 0.

In order to faciliate the comparison with the classification as given by Lie[112], vol. III, §136 and §137, in the subsequent table the Lie algebras l4,i andthe corresponding equation number of Lie are given without the parameterconstraints.

i in l4,i 1 2 3 4 5 6 7 10 12 13 16 17

Lie′s enumeration 58 70 68 63 62 67 64 65 69 71 72 73

Algebra l4,8 is obtained from l4,3 for the parameter value c = −1, the permu-tations U2 and U4, U1 and U2 and the replacement U2 by −U2. Algebra l4,9is obtained from l4,6 for the parameter value b = 0. Algebra l4,11 is obtainedfrom l4,3 for the parameter value c = 0. Algebra l4,14 is obtained from l4,6 forthe parameter values a = 1, b = 0. Algebra l4,15 is obtained from l4,6 for theparameter values a = b = 0.Some algebras with more than four dimensions

l8: [U1, U3] = U1, [U1, U5] = U2, [U1, U7] = U4, [U1, U8] = U6 + 2U3,[U2, U4] = U1, [U2, U6] = U2, [U2, U7] = 2U6 + U3, [U2, U8] = U5,[U3, U4] = −U4, [U3, U5] = U5, [U3, U8] = U8, [U4, U5] = U6 − U3,[U4, U8] = U7, [U5, U6] = U5, [U6, U7] = U7.Characteristic equation: The two leading coefficients of (3.17) are

ψ1 = −2E6, ψ2 = 3E2E7 + 6E1E8 + E26 + E6E3 − 2E2

3 .

Finite Transformation Groups. These are the most important groupsoccurring in this book because any group of Lie symmetries of an ode mustbe similar to one of them. As it will turn out later on, however, not all of themdo actually occur as a symmetry group of an ode of low order. These questionswill be considered in Chapter 5. The subsequent classifications correspond toa manifold C2 with coordinates x and y as given by Lie [112], vol. III, Kapitel3, pages 28-78. In the real plane R2 the classification of primitive groups isdifferent, it is given in the same book on pages 360-392. A more completediscussion may be found in Gonzales-Lopez [54], and in Chapter 6 of vol. I of[43]. A partial ordering for these groups has been given by Krause [96], page260.

There are certain special classes of groups that have played an importantrole for the development of the subject, the so-called linear groups. They aresubgroups of the groupGL(V ) of linear transformations of a finite-dimensionalvector space over the field R or C. On the one hand they were the firstexamples of such groups investigated by Lie. Secondly, their structure turnedout to be fundamental in the sense that other groups may be composed outof them by standard operations like products or semidirect products. Theseclasses are described first.

138

i) The n(n+ 2)-parameter projective group

xk =ak,1x1 + ak,2x2 + . . .+ ak,nxn + ak,n+1xn+1

an+1,1x1 + an+1,2x2 + . . .+ an+1,nxn+1

with generators ∂xi, xi∂xk

, xi∑xj∂xj

, i, j, k = 1, . . . , n.

ii) The n(n+ 1)-parameter general linear group

xk = ak,1x1 + ak,2x2 + . . .+ ak,nxn + ak,n+1

with generators ∂xi , xi∂xk.

iii) The n2-parameter general linear homogeneous group

xk = ak,1x1 + ak,2x2 + . . .+ ak,nxn

with generators xi∂xk, i, k = 1, . . . , n.

iv) The n(n+1)−1-parameter special linear group of those transformationsin ii) with determinant equal to 1. Its generators are ∂xi

, xi∂xkand

xi∂xi− xk∂xk

, i, k = 1, . . . , n, i 6= k.

v) The n2 − 1-parameter special linear homogeneous group of those trans-formations in iii) with determinant equal to 1. Its generators are xi∂xk

,xi∂xi

− xk∂xk, i, k = 1, . . . , n, i 6= k.

A systematic notation for the groups defined in ii), iii), iv) and v) is GLn,SLn, GLHn and SLHn. However, as van der Waerden [186], page 5, haspointed out, it became popular to use the notation GLn and gln, SLn andsln for the homogeneous groups iii) and v) that will be applied in this book.For n = 2, using the variables x and y, their explicit form is

gl2 ≡ x∂x, y∂x, x∂y, y∂y, type l4,1,

sl2 ≡ x∂y, y∂x, x∂x − y∂y, type l3,1.

The problem of determining a basis for gl2 and sl2 such that the commutatortable given in the above listing applies is considered in Exercise 3.11.

Each entry of the subsequent listing is a representative of a group type, i.e.,the totality of all groups that are similar to this entry by a diffeomorphism.In order to identify any given group, its type has to be found. After that thetransformation to the representative in this list may be determined.

Primitive groupsg1: ∂x, ∂y, x∂y, x∂x − y∂y, y∂x.g2: ∂x, ∂y, x∂y, y∂y, x∂x, y∂x.g3: ∂x, ∂y, x∂y, y∂y, x∂x, y∂x, x2∂x + xy∂y, xy∂x + y2∂y.

Groups with two systems of imprimitivity x = const and y = const.g4: ∂y, y∂y.


g5: ∂x, ∂y, y∂y.g6: ∂x, ∂y, y∂y, x∂x.g7: ∂x, ∂y, x∂x + cy∂y with c 6= 0, 1.g8: ∂y, y∂y, y2∂y.g9: ∂x, ∂y, y∂y, y2∂y.g10: ∂x + ∂y, x∂x + y∂y, x

2∂x + y2∂y.g11: ∂x, ∂y, y∂y, x∂x, y2∂y.g12: ∂x, ∂y, y∂y, x∂x, y2∂y, x

2∂x.

Groups with the system of imprimitivity x = const

g13: ∂x, 2x∂x + y∂y, x2∂x + xy∂y.

g14: y∂y, ∂x, x∂x, x2∂x + xy∂y.g15: φ1(x)∂y, φ2(x)∂y, . . . , φr(x)∂y with r ≥ 2. Size: r ≥ 2.g16: φ1(x)∂y, φ2(x)∂y, . . . , φr(x)∂y, y∂y with r ≥ 2. Size: r + 1 ≥ 3.g17: eαkx∂y, xe

αkx∂y, . . . , xρkeαkx∂y, ∂x with l ≥ 1, 1 ≤ k ≤ l,∑

ρk + l ≥ 2, α1(α1 − 1) = 0. Size:∑ρk + l + 1 ≥ 3.

g18: eαkx∂y, xeαkx∂y, . . . , x

ρkeαkx∂y, y∂y, ∂x with l ≥ 1, 1 ≤ k ≤ l,l +

∑ρk ≥ 2, α1 = 0, α2 = 1. Size:

∑ρk + l + 2 ≥ 4.

g19: ∂y, x∂y, . . . , xr−1∂y, ∂x, x∂x + cy∂y with r ≥ 2. Size: r + 2 ≥ 4.g20: ∂y, x∂y, . . . , xr−1∂y, ∂x, x∂x+(ry+xr)∂y with r ≥ 1. Size: r+2 ≥ 3.g21: ∂y, x∂y, . . . , xr−1∂y, y∂y, ∂x, x∂x with r ≥ 2. Size: r + 3 ≥ 5.g22: ∂y, x∂y, . . . , xr−1∂y, ∂x, 2x∂x + (r − 1)y∂y, x2∂x + (r − 1)xy∂y

with r ≥ 2. Size: r + 3 ≥ 5.g23: ∂y, x∂y, . . . , xr−1∂y, y∂y, ∂x, x∂x, x

2∂x + (r − 1)xy∂y with r ≥ 2.Size: r + 4 ≥ 6.

Groups with the system of imprimitivity y = αx+ constg24: ∂x, ∂y, x∂x + y∂y.g25: ∂y, x∂x + y∂y.g26: ∂x, ∂y.

Groups with the system of imprimitivity y = φ(x) + constg27: ∂y.

This listing of the finite groups of the complex plane follows most closelythe one given in Lie [112], vol. III. For the indicated range of parameters thelisting is complete without any duplication. In earlier publications Lie gavesome of these groups in slightly different form, compare the discussion in Lie[112], vol. III, page 76. Group g7 with c 6= 1 is listed separately from groupg24 because the value c = 1 in the latter group generates a different systemof imprimitivity than for c 6= 1, and for the value c = 0 it is similar to g5.

A further distinction made by Lie concerns the multiplicity of a system ofimprimitivity. It generates the additional distinctions as follows. For groupg22 with r = 3 the system x = const. counts twofold (last line on page 71),for r = 2 and r ≥ 4 it counts as a single system (first item after the headingon page 72: b) Die invariante Schar zahlt doppelt). For group g20 and r = 1

140

the system is counted twofold (last line on page 72), all remaining cases r > 1are single counted (item six from top of page 72). Finally the imprimitivitysystem for group g19 with c = 1 counts twofold (second line from bottom ofpage 72) and only single for the remaining cases c 6= 1 (line five from topof page 72). These remarks allow it to identify each group in Lie’s listinguniquely.

Some applications require the knowledge of all group types with a givennumber of parameters without duplication. For most groups of the abovelisting it is easy to identify them. The coefficients of ∂y for the groups g17

and g18 are obtained by the following observations. The desired coefficientsare solutions of a linear homogeneous ode with constant coefficients of theproper order (Lie [112], vol. III, page 54ff). As a consequence they have thegeneral form xkeαx.

Subsequently the infinitesimal generators of various groups of a given size rup to r ≤ 7 are listed. For r ≤ 4 this listing is complete. For values higherthan four only those groups are listed that occur later on as symmetry groupsof differential equations. In addition the Janet bases for the defining equationsare given in grex term ordering with η > ξ and y > x. The coefficients φkfor k = 1, . . . , r occurring in the group types g15 and g16 are undeterminedfunctions of x. The coefficients qk occurring in the Janet basis for these groupsare defined by

qk = −W

(2)k (φ1, φ2)

W (2)(φ1, φ2)

for k = 1, . . . , r, see eq. (2.4). The algebraic properties of groups with r ≤ 4may be obtained from their Lie algebra type which is always given.One-parameter group

g27: ∂y. Type l1. Janet basis ξ, ηx, ηy.Two-parameter groups

g4: ∂y, y∂y. Type l2,1. Janet basis ξ, ηx, ηyy.g15(r = 2): φ1∂y, φ2∂y. Type l2,2. Janet basis ξ, ηy, ηxx + q1ηx + q2η.g25: ∂x, x∂x + y∂y. Type l2,1. Janet basis

ξx − 1

y η, ξy, ηx, ηy −1y η

.

g26: ∂x, ∂y. Type l2,2. Janet basis ξx, ξy, ηx, ηy.Three-parameter groups

g5: ∂x, ∂y, y∂y. Type l3,4. Janet basis ξx, ξy, ηx, ηyy.g7: ∂x, ∂y, x∂x + cy∂y with c 6= 0, 1. Type l3,2.Janet basis ξy, ηx, ηy − cξx, ξxx.g8: ∂y, y∂y, y2∂y. Type l3,1. Janet basis ξ, ηx, ηyyy.g10: ∂x + ∂y, x∂x + y∂y, x

2∂x + y2∂y. Type l3,1. Janet basisξy, ηx, ηy + ξx + 2

x− y (η − ξ), ξxx − 2x− y ξx −

2(x− y)2 (η − ξ)

.

g13: ∂x, 2x∂x + y∂y, x2∂x + xy∂y. Type l3,1.

Janet basisξx − 2

y η, ξy, ηy −1y η, ηxx

.

g15(r = 3): φ1∂y, φ2∂y, φ3∂y. Type l3,6.Janet basis ξ, ηy, ηxxx + q1ηxx + q2ηx + q3η.


g16(r = 2): φ1∂y, φ2∂y, y∂y. Type l3,2 with c = 1.Janet basis ξ, ηxx + q1ηx + q2η, ηxy, ηyy.g17(l = 1, ρ1 = 1, α1 = 0): ∂x, ∂y, x∂y. Type l3,5.Janet basis ξx, ξy, ηy, ηxx.g17(l = 1, ρ1 = 1, α1 = 1): ∂x, ex∂y, xex∂y. Type l3,3.Janet basis ξx, ξy, ηy, ηxx − 2ηx + η.g17(l = 2, ρ1 = ρ2 = 0, α1 = 0, α2 = 1): ∂x, ∂y, ex∂y. Type l3,4.Janet basis ξx, ξy, ηy, ηxx − ηx.g17(l = 2, ρ1 = ρ2 = 0, α1 = 1, α2 = c): ∂x, ex∂y, ecx∂y, c 6= 0, 1.Type l3,2. Janet basis ξx, ξy, ηy, ηxx − (c+ 1)ηx + cη.g20(r = 1): ∂x, ∂y, x∂x + (x+ y)∂y. Type l3,3.Janet basis ξy, ηx − ξx, ηy − ξx, ξxx.g24: ∂x, ∂y, x∂x + y∂y. Type l3,2 with c = 1.Janet basis ξy, ηx, ηy − ξx, ξxx.

Four-parameter groupsg6: ∂x, ∂y, x∂x, y∂y. Type l4,10. Janet basis ξy, ηx, ξxx, ηyy.g9: ∂x, ∂y, y∂y, y2∂y. Type l4,1. Janet basis ξx, ξy, ηx, ηyyy.g14: ∂x, x∂x, x2∂x + xy∂y, y∂y. Type l4,1.Janet basis

ξy, ηy − 1

y η, ξxx −2y ηx, ηxx

.

g15 with r = 4: φ1∂y, φ2∂y, φ3∂y, φ4∂y. Type l4,17.Janet basis ξ, ηy, η4x + q1ηxxx + q2ηxx + q3ηx + q4η.g16 with r = 3: φ1∂y, φ2∂y, φ3∂y, y∂y. Type l4,6 with a = b = 1.Janet basis ξ, ηxy, ηyy, ηxxx + q1ηxx + q2ηx + q3η.g17 with l = 1, ρ1 = 2, α1 = 0: ∂x, ∂y, x∂y, x2∂y. Type l4,13.Janet basis ξx, ξy, ηy, ηxxx.g17 with l = 1, ρ1 = 2, α1 = 1: ∂x, ex∂y, xex∂y, x2ex∂y. Type l4,2.Janet basis ξx, ξy, ηy, ηxxx − 3ηxx + 3ηx − η.g17 with l = 2, ρ1 = 0, ρ2 = 1, α1 = 0, α2 = 1: ∂x, ∂y, ex∂y, xex∂y.Type l4,8. Janet basis ξx, ξy, ηy, ηxxx − 2ηxx + ηx.g17 with l = 2, ρ1 = 1, ρ2 = 0, α1 = 0, α2 = 1: ∂x, ∂y, x∂y, ex∂y.Type l4,11. Janet basis ξx, ξy, ηy, ηxxx − ηxx.g17 with l = 2, ρ1 = 1, ρ2 = 0, α1 = 1, α2 = α: ∂x, ex∂y, xex∂y, eαx∂y.Type l4,3. Janet basis ξx, ξy, ηy, ηxxx − (α+ 1)ηxx + (2α+ 1)ηx − αη.g17 with l = 3, ρ1 = ρ2 = ρ3 = 0, α1 = 0, α2 = 1, α3 = α:∂x, ∂y, ex∂y, eαx∂y. Type l4,9.Janet basis ξx, ξy, ηy, ηxxx − (α+ 1)ηxx + αηx.g17 with l = 3, ρ1 = ρ2 = ρ3 = 0, α1 = 1, α2 = α, α3 = β : ∂x, ∂y, eαx∂y, eβx∂y.Type l4,6. Janet basis ξx, ξy, ηy, ηxxx − (α+ β)ηxx + αβηx.g18 with l = 1, ρ1 = 1, α1 = 0: ∂y, x∂y, y∂y, ∂x. Type l4,12.Janet basis ξx, ηy, ξyy, ηxx.g18 with l = 2, ρ1 = ρ2 = 0, α1 = 0, α2 = 1: ∂y, ex∂y, y∂y, ∂x.Type l4,10, Janet basis ξx, ξy, ηxx − ηx, ηxy, ηyy.g19 with r = 2: ∂y, x∂y, ∂x, x∂x + cy∂y. Type l4,5(c).Janet basis ξy, ηy − cξx, ξxx, ηxx.

142

g20 with r = 2: ∂x, ∂y, x∂y, x∂x + (x2 + 2y)∂y. Type l4,4.Janet basis ξy, ηy − 2ξx, ξxx, ηxx − 2ξx.

Selected Five-parameter groupsg1: ∂x, ∂y, x∂y, x∂x − y∂y, y∂x.Janet basis ηy + ξx, ξxx, ξxy, ξyy, ηxx.g11: ∂x, ∂y, y∂y, x∂x, y2∂x.Janet basis ηy, ξxx, ξxy, ξyy − 1

y ξy, ηxx.g15(r = 5): φ1∂y, φ2∂y, φ3∂y, φ4∂y, φ5∂y.Janet basis ξ, ηy, η5x + q1η4x + q2ηxxx + q3ηxx + q4ηx + q5η.g16(r = 4): φ1∂y, φ2∂y, φ3∂y, φ4∂y, y∂y.Janet basis ξ, ηxy, ηyy, η4x + q1ηxxx + q2ηxx + q3ηx + q4η.g18(l = 1, ρ1 = 3, α1 = 0): ∂y, x∂y, x2∂y, y∂y, ∂x. Janet basis ξx, ξy, ηxy, ηyy, ηxxx.g18(l = 2, ρ1 = 1, ρ2 = 0, α1 = 0, α2 = 1): ∂y, x∂y, ex∂y, y∂y, ∂x.Janet basis ξx, ξy, ηxy, ηyy, ηxxx − ηxx.g18(l = 3, ρ1 = ρ2 = ρ3 = 0, α1 = 0, α2 = 1, α3 = c), c 6= 0, 1:∂y, x∂y, ecx∂y, y∂y, ∂x.Janet basis ξx, ξy, ηxy, ηyy, ηxxx − (c+ 1)ηxx + cηx.g19(r = 3): ∂y, x∂y, x2∂y, ∂x, x∂x+cy∂y. Janet basis ξy, ηy−cξx, ξxx, ηxxx.g20(r = 3): ∂y, x∂y, x2∂y, ∂x, x∂x + (3y + x3)∂y.Janet basis ξy, ηy − 3ξx, ξxx, ηxxx − 6ξx.g21(r = 2): ∂x, ∂y, x∂y, y∂y, x∂x. Janet basis ξy, ξxx, ηxx, ηxy, ηyy.g22(r = 2): ∂x, ∂y, x∂y, 2x∂x + y∂y, x

2∂x + xy∂y.Janet basis ξy, ηy − 1

2ξx, ηxx, ξxxx.Selected Six-parameter groups

g2: ∂x, ∂y, x∂y, y∂y, x∂x, y∂x.Janet basis ξxx, ξxy, ξyy, ηxx, ηxy, ηyy.g12: ∂x, ∂y, y∂y, x∂x, y2∂y, x

2∂x.Janet basis ξy, ηx, ξxxx, ηyyy.g15(r = 6): φ1∂y, φ2∂y, φ3∂y, φ4∂y, φ5∂y, φ6∂y.Janet basis ξ, ηy, η6x + q1η5x + q2η4x + q3ηxxx + q4ηxx + q5ηx + q6η.g16(r = 5): φ1∂y, φ2∂y, φ3∂y, φ4∂y, φ5∂y, y∂y.Janet basis ξ, ηxy, ηyy, η4x + q1ηxxx + q2ηxx + q3ηx + q4η.g19(r = 4): ∂y, x∂y, x2∂y, x

3∂y, ∂x, x∂x + cy∂y.Janet basis ξy, ξxx, ηxy, ηyy, ηxxxx.

Selected Seven- and Eight-parameter groupsg23(r = 3) = ∂y, x∂y, x2∂y, y∂y, ∂x, x∂x, x

2∂x + 2xy∂y.Janet basis ξy, ηxy − ξxx, ηyy, ξxxx, ηxxx.g3: ∂x, ∂y, x∂y, y∂y, x∂x, y∂x, x2∂x + xy∂y, xy∂x + y2∂y.Janet basis ξyy, ηxx, ηxy, ηyy, ξxxx, ηxxy.

In addition to this classification of the finite groups of a two-dimensionalmanifold, Lie [110] derived a classification of its infinite groups. This listingis not given here because it is not required for the applications in this book.The only exception is group G, no. XXXV in Lie’s notation, which occurs inthe symmetry analysis of first order equations.


G: ξ(x, y)∂x, η(y)∂y. Janet basis ηx.

3.5 Lie Systems

In Chapter 2 general systems of linear homogeneous pde’s have been dis-cussed without any assumptions on their origin or special properties of theirsolutions. The uniqueness of the Janet basis representation makes it possi-ble to express additional knowledge in terms of the Janet basis coefficients.For the applications in this book, determining the symmetries of ordinarydifferential equations, this additional knowledge is of threefold origin.

Lie’s relations must always be true because they express the generalproperties of commutators of vector fields.

Geometric relations express geometric properties of particular grouptypes, e. g. the number of systems of imprimitivity.

Algebraic relations express algebraic properties of particular Lie algebratypes, e. g. in terms of the structure of its characteristic polynomial.

The importance of knowing the explicit form of the coherence conditions givenin Theorems 2.15, 2.16 and 2.17 becomes obvious at this point because onlythis knowledge, combined with the additional constraints, allows it to identifythe symmetry classes of ode’s.Constraints from Lie’s Relations. Let a Janet basis with k ≥ 2 parametricderivatives be given, whose solutions are the coefficients ξ(x, y) and η(x, y) ofk vector fields generating a finite group of the plane with coordinates x and y.Let (ξ1, η1) and (ξ2, η2) be any pair of solutions. Then according to (3.11) athird solution (ξ3, η3) is given by

ξ3 = ξ1ξ2,x − ξ1,xξ2 + η1ξ2,y − ξ1,yη2,

η3 = ξ1η2,x − η1,xξ2 + η1η2,y − η1,yη2.(3.28)

This property leads to severe additional restrictions for the coefficients of thecorresponding Janet basis. In order to obtain them explicitly, the conditionthat ξ3 and η3 as defined by (3.28) be a solution would have to be expressedin terms of these coefficients. To this end at first the right hand sides of (3.28)are reduced w.r.t. the Janet basis at hand. From the resulting expressionsthose derivatives of ξ3 and η3 that occur in the Janet basis are calculated andare again reduced w.r.t. to the Janet basis. The result is substituted into theJanet basis. Its elements become linear and homogeneous in 1

2k(k−1) bilinearexpressions of the form piqj − pjqi where the pj , qj traverse the k parametricderivatives involved. Their coefficients yield the desired constraints. Due to

144

its importance in later parts of this book, a special term for systems of pde’sobeying Lie’s relations is introduced.

Definition 3.8 (Lie system). A system of linear homogeneous pde’s withthe property that (ξ3, η3) defined by (3.28) is a solution if this is true for(ξ1, η1) and (ξ2, η2), is called a Lie system.

In the remaining part of this subsection the complete answer for severalJanet basis types is given explicitly. Obviously this concept makes sense onlyfor Janet bases of type J (2,2) of order not less than two. For Janet bases oforder two the complete answer is given next.

Theorem 3.17 For type J (2,2)2,1 , . . . ,J (2,2)

2,5 Janet bases Lie’s relations areas follows. The Janet basis

J (2,2)2,1 : ξ = 0, ηy + a1ηx + a2η = 0, ηxx + b1ηx + b2η = 0 (3.29)

is a Lie system if its coefficients satisfy the system of pde’s

a2,x − 12b1,y −

12a1b2 = 0, a1,y − a1a2 = 0,

b1,xy − 2a1,xb2 + b1,yb1 − 2b2,y − b2,xa1 + a1b1b2 = 0,a1,xx − a1,xb1 − b1,xa1 + a1b2 = 0.

The Janet basis

J (2,2)2,2 : ξ = 0, ηx + a1η = 0, ηyy + b1ηy + b2η = 0 (3.30)


a1 = 0, b2,x = 0, b1,x = 0, b1,y − b2 = 0.

The Janet basis

J (2,2)2,3 :

ξx + a1η + a2ξ = 0, ξy + b1η + b2ξ = 0,ηx + c1η + c2ξ = 0, ηy + d1η + d2ξ = 0

(3.31)


d1,x − d2,y + b1c2 − b2c1 = 0, c1,y − d2,y + a1d2 − b2c1 = 0,c2,y − d2,x + a2d2 − b2c2 − c1d2 + c2d1 = 0,c1,x − d2,x + a1c2 − a2c1 + a2d2 − b2c2 = 0,b1,x − b2,y − a1b2 + a2b1 − b1d2 + b2d1 = 0,

a2,y − b2,x − a1d2 + b1c2 = 0, a1,x − b2,x − a1d2 + b2c1 = 0,a1,y − b2,y − a1d1 + b1c1 − b1d2 + b2d1 = 0.

(3.32)

The Janet basis

J (2,2)2,4 : η + a2ξ = 0, ξy + b1ξx + b2ξ = 0, ξxx + c1ξx + c2ξ = 0 (3.33)



a2,yb1 + a2,xb21 + b1,ya2 − b1,x − a2b1b2 − b2,

b2,xx − 2b1,xc2 + b2,xc1 − c2,y − c2,xb1 = 0,b1,xx − b1,xc1 + 2b2,x − c1,y − c1,xb1 = 0,

a2,xx + a2,x(2b1,x − b1c1)− (2b2,x − c1,y)a2 − c1,x + (a2b1 + 1)c2 = 0.

The Janet basis

J (2,2)2,5 : η + a1ξ = 0, ξx + b1ξ = 0, ξyy + c1ξy + c2ξ = 0 (3.34)


b1,y − 12c1,x = 0, a1,x − a1b1 = 0,

c1,xy + c1,xc1 − 2c2,x = 0, a1,yy − a1,yc1 − c1,ya1 + a1c2 = 0.

Proof The proof is similar for all five cases. Therefore it is given onlyfor type J (2,2)

2,3 Janet bases. Reduction of the right hand sides of (3.28) w.r.t.(3.31) where ξ and η are replaced by (ξ1, η1) or (ξ2, η2), yields

ξ3 = (b2 − a1)B, η3 = (d2 − c1)B where B ≡ ξ1η2 − η1ξ2 (3.35)

and the useful relations Bx = −(a2 + c1)B, By = −(b2 + d1)B. The reducedfirst order partial derivatives of ξ3 and η3 are

ξ3,x =[b2,x − a1,x − (b2 − a1)(a2 + c1)

]B,

ξ3,y =[b2,y − a1,y − (b2 − a1)(b2 + d1)

]B,

η3,x =[d2,x − c1,x − (d2 − c1)(a2 + c1)]B,

η3,y =[d2,y − c1,y − (d2 − c1)(b2 + d1)

]B

(3.36)

with B as above. Substituting (3.35) and (3.36) into the Janet basis (3.31)and equating the coefficients of the parametric derivatives to zero yields thefollowing relations for the coefficients a1, a2, . . . , d2 and its first derivatives.

b2,x − a1,x − (b2 − a1)c1 + (d2 − c1)a1 = 0,

b2,y − a1,y − (b2 − a1)d1 + (d2 − c1)b1 = 0,

d2,x − c1,x + (b2 − a1)c2 − (d2 − c1)c1 = 0,

d1,y − c1,y + (b2 − a1)d2 − (d2 − c1)b2 = 0.

These conditions are necessary and sufficient for (3.31) to be a Lie system.They have to be supplemented by the integrability conditions of Theorem 2.17.After some final reductions have been performed, the above system (3.32) ofpde’s for the coefficients is obtained.

146

Example 3.32 Consider the type J (2,2)2,3 system

z1,x + 1y(y2 − 2)

z2 −2(y2 − 1)x(y2 − 2)

z1 = 0, z1,y + xy2(y2 − 2)

z2 − 2y(y2 − 2)

z1 = 0,

z2,x + 2x(y2 − 2)

z2 − 2y3

x2(y2 − 2)z1 = 0, z2,y −

2(y2 − 1)y(y2 − 2)

z2 + 2y2

x(y2 − 2)z1 = 0.

Because its coefficients satisfy the coherence conditions of Theorem 2.16 it isa Janet basis and has a two-dimensional solution space, a basis is (x2, 2xy)and

(xy , y

2). Because the constraints (3.32) are violated it is not a Lie system.

Consequently, the commutator of U1 = x2∂x + 2xy∂y and U2 = xy ∂x + y2∂y

which is [U1, U2] = −3x2

y ∂x + 2x(y2 − 1)∂y is not a linear combination of U1

and U2 over the constants, i.e., U1 and U2 do not generate a Lie algebra.


ξx−1xξ = 0, ξy = 0, ηx−

1x+ y

η+y

x(x+ y)ξ = 0, ηy−

1x+ y

η− 1x+ y

ξ = 0.

It satisfies the constraints (3.32) that include the coherence conditions ofTheorem 2.16, consequently it is a Lie system. A solution basis is (x, y) and(x,−x) from which the generators U1 = x∂x+y∂y and U2 = x(∂x−∂y) followwith [U1, U2] = 0.


ξx −3x2 + 1x2 + x

ξ = 0, ξy = 0, ηx +2y

x2 + 1ξ = 0, ηy −

1yη = 0.

It satisfies the constraints (3.32) that include the coherence conditions ofTheorem 2.16, consequently it is a Lie system. A solution basis is (0, y) and(x3 + x,−x2y). It is obtained from the preceding example by the variabletransformation x = xy and y = y

x and replacing the barred variables by x

and y.

As these examples show the main advantage of applying Theorem 3.17 isthat it does not require one to know the solutions; it uses only the coefficientsof a given system and arithmetic operations and differentiations.

For higher order systems Lie’s relations become increasingly more compli-cated due to the larger number of coefficients involved, therefore they cannotbe given explicitly in this book. Only a single third order system will betreated in detail because it occurs frequently in later applications.

Theorem 3.18 The type J (2,2)3,4 Janet basis

ξx + a1η + a2ξ = 0, ξy + b1η + b2ξ = 0,ηy + c1ηx + c2η + c3ξ = 0, ηxx + d1ηx + d2η + d3ξ = 0

(3.37)


is a Lie system if its coefficients satisfy the following system of pde’s.

b2 = 0, b3 = 0, a2c1 = 0, d2,x − d3,y + (2a2 − c2)d3 − (2a3 − c3)d2 = 0,

c3,xc1 + a2(a3 − 2c3 + d1)− c1(c23 + c1d3 − c3d1 + d2) = 0,

c3,y − 12d1,y + 1

2a2(a3 − 3c3 + d1)− 12c1d2 = 0,

c2,x − 12d1,y + 1

2a2(a3 − 3c3 + d1)− 12c1d2 = 0,

a3,x − 2c3,x + d1,x + a3(2c3 − a3 − d1) + c1d3 = 0, a3,y − a2c3 = 0,

c1,x + c1(a3 − c3) = 0, c1,y − c1c2 = 0, a2,y − a2c2 = 0,

d1,xy − 3c3,xa2 + d1,yd1 − 2d2,y − d3,yc1 − a2a3(a3 − 3c3 + 2d1)

+c1d2(2c3 + a3 + d1) + a2c23 − c1c2d3 − a2d

21 = 0,

c3,xx + c3,x(d1 − 2c3 − 2a3) + d1,xc3 − d3,y − d3,xc1

+2a3c3(c3 − d1) + c1d3(3a3 − c3)− c2d3 + c3d2 = 0.(3.38)

Proof Proceeding similarly to the preceding theorem, at first ξ3 and η3are obtained in the form

ξ3 = (a2 − b3)B3, η3 = c1B1 −B2 − c3B3 (3.39)

where the three bilinear expressions B1, B2 and B3 are defined in terms ofthe parametric derivatives ηx, η, ξ with the respective index, and

B1 = η1,xη2 − η1η2,x, B2 = η1,xξ2 − ξ1η2,x, B3 = η1ξ2 − ξ1η2. (3.40)

The following relations are useful for the subsequent calculations.

B1,x = −d1B1 + d3B3,

B1,y = −(c1,x − 2c2 − c1d1)B1 − c3B2 + (c3,x − a3c3 − c1d3)B3,

B2,x = −a2B1 − (a3 + d1)B2 − d2B3

B2,y = −b2B1 − (c1,x + b3 + c2 − c1d1)B2 − (c2,x − a1c3 − c1d2)B3,

B3,x = B2 − a3B3, B3,y = −c1B2 − (b3 + c2)B3.

(3.41)

Applying these expressions, the necessary derivatives of ξ3 and η3 may becomputed from (3.39). If they are substituted into (3.37), the coefficients ofthe B′s yield a system of equations for the coefficients a1, a2, . . . , d3. Com-bined with the integrability conditions of Theorem 2.17 they represent thenecessary and sufficient conditions (3.38) such that the general solution of(3.37) generates a three-dimensional Lie algebra of vector fields.

The system of pde’s (3.38) suggests that it may be split into several alter-natives due to algebraic factorization of some of its members. It turns outthat altogether three systems are generated in this way as is shown next.

148

Corollary 3.1 The system (3.38) splits into the following alternatives.

i) c1 = 0, b3 = 0, b2 = 0, a3 − 2c3 + d1 = 0,d2,x − d3,y + d3(2a2 − c2) + d2(2d1 − 3c3) = 0,

c3,y − 12d1,y − 1

2a2c3 = 0, c2,x − 12d1,y − 1

2a2c3 = 0,

a2,x − a2c3 = 0, a2,y − a2c2 = 0,

d1,xy − 3a2c3,x + d1,yd1 − 2d2,y + 3a2c3(c1 − d1) = 0,

c3,xx − 3c3,x(2c3 − d1) + d1,xc3 − d3,y

+2c23(2c3 − 3d1) + 2c3d21 − c2d3 + c3d2 = 0.

ii) c1 = 0, b3 = 0, b2 = 0, a2 = 0,

d2,x − d3,y − 2a3d2 − c2d3 + c3d2 = 0,

c3,y − 12d1,y = 0, c2,x − 1

2d1,y = 0,

a3,x − 2c3,x + d1,x − a23 + 2a3c3 − a3d1 = 0, a3,y = 0,

d1,xy + d1d1,y − 2d2,y = 0,

c3,xx − c3,x(2a3 + 2c3 − d1) + d1,xc3 − d3,y

+2a3c3(c3 − d1)− c2d3 + c3d2 = 0.

iii) b3 = 0, b2 = 0, a2 = 0,

d2,x − d3,y − 2a3d2 − c2d3 + c3d2 = 0,

c3,x − c1d3 − c23 + c3d1 − d2 = 0, c3,y − 12d1,y − 1

2c1d2 = 0,

c2,x − 12d1,y − 1

2c1d2 = 0,

c1,x + c1(a3 − c3) = 0, c1,y − c1c2 = 0,

a3,x + d1,x + (a3 + d1)(2c3 − a3) = 0, a3,y = 0,

d1,xy + d1d1,y − 2d2,y − d3,yc1 + c1d2(a3 + d1 − 2c3)− c1c2d3 = 0.

The next example shows how additional information beyond Lie’s conditionsis necessary in order to distinguish certain group types.

Example 3.35 Three type J (2,2)3,4 Lie systems are defined by nonvanishing

coefficients of (3.37) as follows.a) a3 = − 2

x , d1 = 2x , b) a3 = − 2

x , d1 = 2x + 2

x2 , d2 = 1x4 ,

c) a3 = − 2x , d1 = 2

x + 1x2 .

All three cases satisfy system (3.38). Case a) satisfies only alternative i)of Corollary 3.1, case b) satisfies only alternative ii) whereas case c) satisfiesalternatives ii) and iii).


Geometric Constraints. In the symmetry analysis of ode’s the most im-portant distinguishing feature for the type of a symmetry group which is ofgeometric origin is its system of imprimitivity. Therefore for those Janet basistypes that occur in the symmetry analysis the constraints determining theirsystems of imprimitivity are described in this subsection. They are based onTheorem 3.16. For the two first order bases the answer is rather simple.

Theorem 3.19 The Lie systems of types

J (2,2)1,1 : ξ = 0, ηx + aη = 0, ηy + bη = 0,

J (2,2)1,2 : η + aξ = 0, ξx + bξ = 0, ξy + cξ = 0

define one-parameter groups allowing a system of imprimitivity depending onan undetermined function.

Proof The systems of imprimitivity are determined by

ωy + bω + a = 0 or ωx + aωy + cω2 + (ay − b− c)ω + ax − ab = 0

respectively. The general solution of either of these pde’s for ω(x, y) containsan undetermined function of a single argument. Consequently, they determinesystems of imprimitivity depending on a function.

This result was to be expected because one-parameter groups according tothe tabulation on page 139 always allow a system of imprimitivity of this kind.

Theorem 3.20 The Lie systems of types J (2,2)2,1 , . . . ,J (2,2)

2,5 define vectorfields for two-parameter groups with the following systems of imprimitivity.

J (2,2)2,1 : System (3.29) does not allow a one-parameter system of imprimitivity.

It allows two systems of imprimitivity if a1 6= 0, and a single one ifa1 = 0.

J (2,2)2,2 : System (3.30) always allows two systems of imprimitivity.

J (2,2)2,3 : System (3.31) always defines a one-parameter system of imprimitivity.

J (2,2)2,4 : System (3.33) always allows two systems of imprimitivity.

J (2,2)2,5 : System (3.34) does not allow a one-parameter system of imprimitivity.

It allows two systems of imprimitivity if a1 6= 0 and a single one ifa1 = 0.

Proof Type J (2,2)2,1 : There is always the system of imprimitivity corre-

sponding to ξy = 0. Any other system of imprimitivity is determined by theequations a1ω − 1 = 0, ωy + a2ω = 0. If a1 6= 0 there is one solution, ifa1 = 0 the first equation becomes inconsistent, consequently a solution doesnot exist.

150

Type J (2,2)2,2 : There are always the systems of imprimitivity corresponding

to ξy = 0, and a second one that follows from the equations ω = 0, ωy+a1 = 0with the constraint a1 = 0 according to Theorem 3.17.

Type J (2,2)2,3 : Any system of imprimitivity is determined by the equations

ωx − b2ω2 − (a2 − d2)ω + c2 = 0, ωy − b1ω2 − (a1 − d1)ω + c1 = 0.

Due to Theorem 3.17 it satisfies the coherence conditions (2.51). Conse-quently, it determines always a system of imprimitivity depending on a pa-rameter.


b1ω2 + (a2b1 − 1)ω = 0, a2ωy − ωx + b2ω

2 − (a2,y − a2b2)ω − a2,x = 0.

If b1 6= 0, the first equation determines two systems of imprimitivity becausea1b1 + 1 6= 0. If b1 = 0, by Theorem 3.17 it follows b2 = 0, i.e., there is onesystem of imprimitivity corresponding to ξy = 0, and a second one followsfrom ω = 0.


ω(ω + a1) = 0, a1ωy − ωx − (a1,y + b1)ω − a1,x + a1b1 = 0.

There is always the solution ω = 0. If a1 6= 0 there is a second system ofimprimitivity corresponding to ω = −a1.

There are three Janet basis types that occur in the symmetry analysis ofsecond and third order ode’s in Chapter 5, their systems of imprimitivity areconsidered next.

Theorem 3.21 The Lie systems of types J (2,2)3,4 , J (2,2)

3,6 and J (2,2)3,7 define

vector fields for three-parameter groups with the following systems of imprim-itivity.

J (2,2)3,4 : The Lie system

ξx + a2η + a3ξ = 0, ξy + b2η + b3ξ = 0,

ηy + c1ηx + c2η + c3ξ = 0, ηxx + d1ηx + d2η + d3ξ = 0(3.42)

defines a three-parameter group allowing two systems of imprimitivityif c1 6= 0 and a single one if c1 = 0. There cannot be a system ofimprimitivity depending on a parameter.


ξy + a1ξx + a2η + a3ξ = 0, ηx + b1ξx + b2η + b3ξ = 0,

ηy + c1ξx + c2η + c3ξ = 0, ξxx + d1ξx + d2η + d3ξ = 0(3.43)


defines a three-parameter group allowing a system of imprimitivity de-pending on a parameter if a1 = b1 = 0 and c1 = −1, two systems of im-primitivity if either ai = bi = 0, i = 1, 2, 3 or D ≡ 4a1b1 +(c1 +1)2 6= 0,and a single one if D = 0 and a1 6= 0 or b1 6= 0 or c1 6= −1.


ξx + a2η + a3ξ = 0, ηx + b1ξy + b2η + b3ξ = 0,

ηy + c1ξy + c2η + c3ξ = 0, ξyy + d1ξy + d2η + d3ξ = 0(3.44)

defines a two-parameter group which does not allow a system of imprim-itivity depending on a parameter. It allows two systems of imprimitivityif D ≡ b1 − 1

4c21 6= 0 and a single one if D = 0.

Proof Type J (2,2)3,4 : According to Theorem 3.18, for any Janet basis

of type J (2,2)3,4 there holds b2 = b3 = 0, i.e., there is always a system of

imprimitivity corresponding to ξy = 0. A possible second one is determinedby the equations

c1ω − 1 = 0, ωx + (c3 − a3)ω = 0, ωy + (c2 − a2)ω = 0.

The coherence conditions for the system of first order pde’s are satisfied due toTheorem 3.18. Consequently, it allows an additional system of imprimitivityif and only if c1 6= 0.J (2,2)

3,6 : Any system of imprimitivity is determined by the equations

a1ω2−(c1+1)ω−b1 = 0, ωx−a3ω

2+c3ω+b3 = 0, ωy−a2ω2+c2ω+b2 = 0.

If the first equation vanishes identically the remaining system of pde’s which iscoherent defines a system of imprimitivity depending on a parameter. If not,the discriminant of the first equation determines the number of different so-lutions and thereby the systems of imprimitivity. If the first equation reducesto ξy = 0 or the second one to ηx = 0, an additional system of imprimitivityis obtained.J (2,2)

3,7 : Any system of imprimitivity is determined by the equations

ω2 + c1ω + b1 = 0,ωx + (c3 − a3)ω + b3 = 0, ωy + (c2 − a2)ω + b2 = 0.

Because the quadratic equation for ω cannot vanish identically and the sys-tem of pde’s for ω is coherent, the discriminant D of the quadratic equationdetermines the number of imprimitivity systems.

Example 3.36 For all three alternatives of Corollary 3.1 there holdsb2 = b3 = 0, i.e., by the above lemma there is always one system of im-primitivity corresponding to ξy = 0. Reducing condition (3.26) by (3.42) and

152

separating the result w.r.t. the parametric derivatives leads to the system

c1ω − 1 = 0, ωy − b2ω2 + (c2 − a2)ω = 0, ωx − b3ω2 + (c3 − a3)ω = 0.

If c1 = 0 the first equation becomes inconsistent and a second system ofimprimitivity does not exist. If c1 6= 0, ω = 1

c1 is the unique solution ofthe full system and generates a second system of imprimitivity. Summingup this means, alternatives i) and ii) allow a single system of imprimitivity,alternative iii) allows two of them.

There occur two Lie systems of order four which have to be analyzed w.r.t.to their group properties. Their systems of imprimitivity are considered next.

Theorem 3.22 The Lie systems of types J (2,2)4,9 and J (2,2)

4,14 define vectorfields for four-parameter groups with the following systems of imprimitivity.


ξy + a2ξx + a3η + a4ξ = 0, ηy + b1ηx + b2ξx + b3η + b4ξ = 0,

ξxx + c1ηx + c2ξx + c3η + c4ξ = 0, ηxx + d1ηx + d2ξx + d3η + d4ξ = 0(3.45)

defines a four-parameter group which does not allow a system of imprim-itivity depending on a parameter. It allows two systems of imprimitivityif a1 = 0, a2 = a3 = a4 = 0, b2 + 1 = 0 and b1 6= 0, and a single one ifa2 = a3 = a4 = b1 = 0.


ηx + a1ξy + a2ξx + a3η + a4ξ = 0, ηy + b1ξy + b2ξx + b3η + b4ξ = 0,

ξxx + c1ξy + c2ξx + c3η + c4ξ = 0, ξxy + d1ξy + d2ξx + d3η + d4ξ = 0,

ξyy + e1ξy + e2ξx + e3η + e4ξ = 0(3.46)

defines a four-parameter group which does not allow a system of imprimi-tivity depending on a parameter. It allows two systems of imprimitivityif a1 = 0, a2 = a3 = a4 = 0, b2 +1 = 0 and b1 6= 0 or a2 = 0, b2 +1 = 0,b21 − 4a1 6= 0 and a1 6= 0 or a3 6= 0 or a4 6= 0.

The proof is similar as for the preceding theorems and is therefore skipped.

Algebraic Relations. The third set of constraints for the coefficients ofJanet bases occurring in the symmetry analysis of ode’s originates from thealgebraic properties of their Lie algebras. They are based on Theorem 3.7, inparticular on part ii) describing the properties of characteristic polynomials.The Lie algebra types li,j have been defined in Section 3.4 on page 135 ff.

Theorem 3.23 The Lie system (3.31) of type J (2,2)2,3 may define a Lie

algebra of vector fields of the following types:

l2,1 if a1 6= b2 or c1 6= d2. l2,2 if a1 = b2 and c1 = d2.


Proof For the general Lie system of type J (2,2)2,3 the characteristic poly-

nomial has the form ω2 − [(c1 − d2)E2 − (a1 − b2)E1]ω. The above coefficientconstraints generate the structure of the two-dimensional Lie algebras l2,1 orl2,2 respectively listed on page 135.

Theorem 3.24 The Lie system (3.42) of type J (2,2)3,4 may define a Lie

algebra of vector fields of the following types:

l3,1 if c1 = 0, a3 − 2c3 + d1 = 0 and either

a2 6= 0 or c3,x − c23 + c3d1 − d2 6= 0.

l3,2(c) if c3,x + 14a3(a3 − 4c3 + 2d1) + 1

4d21 − d2 6= 0, c 6= 0,

c2 + 2c3,x + 2c23 − 2c3d1 − 4a3c3 + a21 + d2

1 − 2d2

c3,x − c23 + c3d1 − d2c+ 1 = 0.

l3,3 if a1 = 0, a2 = 0, c3,x + 14a3(a3 − 4c3 + 2d1 + 1

4d21 − d2 = 0,

a3 − 2c3 + d1 6= 0.

l3,4 if a2 = 0, c3,x − c23 + c3d1 − c1d3 − d2 = 0 and either

c1 6= 0 or a3 − 2c3 + d1 6= 0.

l3,5 if c1 = 0, a2 = 0, a3 − 2c3 + d1 = 0, c3,x − c23 + c3d1 − c1d3 − d2 = 0.

l3,6 not possible.

Proof According to the definition (3.17) the characteristic polynomial forany three-dimensional Lie algebra has the form ω3 + C2ω

2 + C1ω. In termsof a Lie system of type J (2,2)

3,4 the coefficients may be expressed as follows.

C2 = c1E1 + (a3 − c3 + d1)c1E2 − (a3 − 2c3 + d1)E3,

C1 = a2c1E1E2 − a2(a2 − a3c1 + c1c3 − c1d1)E22 − 2a2E1E3

+[c3,x − c1d3 − c23 + c3d1 − d2)c1 − a2(a3 + d1)

]E2E3

−(c3,x − c1d3 − c23 + c3d1 − d2)E23 .

Imposing the constraints on C1 and C2 that follow from the structure givenin the listing on page 135, the above conditions are obtained after some sim-plifications. The condition for l3,6 follows from the fact that some structureconstants are numbers different from zero.

154

Example 3.37 The three alternatives obtained in Corollary 3.1 yield thecharacteristic equations

ω3 −[(c3,x − c23 + c3d1 − d2)E2

3 + 2a2c3E3E2 + 2a2E3E1 + a22E

22

]ω = 0,

ω3 − (a3 − 2c3 + d1)E3ω2 − (c3,x − c23 + c3d1 − d2)E2

3ω = 0,

ω3 −[(a3 − 2c3 + d1)E3 − (a3c1 − c1c3 + c1d1)E2 − c1E1

]ω2 = 0

respectively. The first alternative does not contain a quadratic term but alinear one, therefore generically it corresponds to a type l3,1 Lie algebra. Atype l3,2(c = −1) Lie algebra is excluded because the coefficient of ω is not acomplete square of a linear form in the E′s. If c3,x − c23 + c2d1 − d2 = 0 anda2 = 0, a type l3,5 or type l3,6 Lie algebra is possible.

For the second alternative, define C1 ≡ c3,x−c23+c3d1−d2, C2 ≡ a3−2c3+d1

and the discriminant D ≡ C22 − 4C1. The coefficient of the first order term is

proportional to a complete square of a linear form in the E′s. As a consequencetype l3,1 is excluded. If C1 6= 0 and C2 6= 0, the condition D 6= 0 correspondsto type l3,2(c 6= −1) and D = 0 to type l3,3. Finally C1 = 0, C2 6= 0corresponds to the type l3,4, and C1 = C2 = 0 to types l3,5 or l3,6.

Due to the constraint c1 6= 0 for the last alternative, there is always aquadratic term and no linear term. Consequently it corresponds uniquely toa Lie algebra of type l3,4.

There are four additional Janet basis types the geometric properties ofwhich have been determined in Theorems 3.21, 3.22 and 3.22. Their Liealgebra types are described in the following two theorems without proof.

Example 3.38 The Lie algebra types corresponding to the three Lie sys-tems considered in Example 3.35 are easiliy identified by the above theoremas follows: a) type l3,5, b) type l3,3 and c) type l3,4.

Theorem 3.25 A Lie system (3.43) of type J (2,2)3,6 may define a Lie algebra

of vector fields of the following types:

l3,1 if c1 = 1, a3b1 − a1b3 + b2 − c3 + d1 = 0 and either

a1b1 + 1 6= 0 or a1(b1 − c3) + a3 6= 0 or a3b1 − b2 + c3 6= 0

or a3,x − a3(a1b3 + c3 − d1)− a1d3 − d2 6= 0

or a3,xb1 + (a1b3 + c3 − d1)(c3 − b2)− b1d2 + d3 6= 0

or a3,x + 12 (a1b2 − a1c3 − a3)(a1b3 + c3 − d1) 6= 0.

l3,2(c) if c2 + 2a1b1 + c21 + 1a2b1 + c1

c+ 1 = 0, c1 6= 1.


l3,3 if any of the following set of conditions is satisfied.

c1 + 1 = 0, b1 = 0, a1b3 + b2 − c3 − d1 = 0,

a3,x + a1(b2 − c3)2 − a3b2 − a1d3 − d2 = 0

or c1 + 1 = 0, b1 = 0,

a3,xb1 − 14a

23b

21 − 1

4 (b2 − c1 − d1)2 − 12a3b1(b2 + c3 − d1)− b1d1 = 0

or 4a1b1 + (c1 + 1)2 = 0, b1 6= 1,

(c1 + 3)(b2 − c3) + (c1 − 1)(d1 − a1b3) + (c1 − 5)a1b1 = 0,

a3(c1 + 1)2(c1 − 5) + 4a1(a1b3 − d1)(c1 − 1)− 4a1(c1 + 3)(b2 − c3) = 0,

a3,xb1 − 18 (a3b3 − 2d3)(c1 + 1)2 − 1

4 (b2 − d1)2 − 14 (a2

1b23 + a2

3b21)− b1d2

− 12a3b1(b2 + c3 + d1)− 1

2a1b3(b2 − c3 − d1) + 14c3(2b2 − c3 − 2d1) = 0.

l3,4 if any of the following set of conditions is satisfied.

a1 = 0, c1 = 0, a3b1 = b2 − c3, a3,x = a3(c3 − d1) + d2, c3 − b2 6= 12d1

or a1b1 + 1 = 0, a3c1 + a1(b2 − c3) = 0, a3b1 = b2 − c3,a3,x − a3(a1b3 + c3 − d1)− a1d3 − d2 = 0,

a1b3 − 2(b2 − c3) 6= d1, a3 + a1(a1b3 + c3 − b2 − d1) 6= 0.

l3,5 if c1 = 1, a1b1 + 1 = 0, a1b3 − 2(b2 − c3)− d1 = 0, a3 + a1(b2 − c3) = 0,

(b2,x − c3,x)a1 + (b2 − c3)(a1d1 − a3c1) + a1d3 + d2 = 0.

l3,6 not possible.



l3,1 if c1 = 0, a2 − 12 (c2 + d1) = 0 and either

b1 6= 0 or c3 6= 0 or c2,y + d1,y − 12c

22 + 1

2d21 − 2d3 6= 0.

l3,2(c) if c2 + 2b1 − c21b1

c+ 1 = 0, c1 6= 1.

l3,3 if b1 − 14c

21 = 0, c3 − 1

2c1(a2 + 2c2 + 2d1) = 0,

c2,y + d1,y + c1d2 − 12c

22 + 1

2d21 − 2d3 = 0.

l3,4 if c2,y + d1,y − 2a2(a2 + c2 + d1) + c1d2 − c2d1 − c22 − 2d3 = 0,

b1 = 0, c3 − a2c1 = 0 and either c1 6= 0 or a2 − 12 (c2 + d1) 6= 0.

l3,5 if b1 = 0, c1 = 0, c3 = 0, a2 − 12 (c2 + d1) = 0,

c2,y + d1,y − 12c

22 + 1

2d21 − 2d3 = 0.

l3,6 not possible.



156

l4,1 if a2 = b1, b2 = 0, b1 6= 0.

l4,4 if b2 = −2, b1 = a2 = 0, a4 = 12c1.

l4,5(c) if b2 6= 0, b2 + 1 6= 0. Then c = −b2.



l4,4 if b2 = − 12 , l4,5(c) if b2 = −1

c , c 6= 0, c 6= −2.

The proofs of the preceding three theorems are similar to those of Theo-rems 3.23 and 3.24 and are therefore omitted. The theorems of this subsectionprovide the interface between geometric and algebraic properties of Lie trans-formation groups of the plane on the one hand, and the symmetries of ode’sexpressed in terms of the Janet bases of its determining systems on the other.They will be applied frequently later in this book.

It turns out that not only the Lie algebra type of the vector fields defined bya given Lie system may be obtained from its coefficients, but even its structureconstants ckij w.r.t. some basis. The subsequent discussion is based on articlesby Neumer [135] and Reid et al. [152].

Assume a Lie system for two functions ξ(x, y) and η(x, y) is given as aJanet basis. Let this system be of finite rank r, i.e., its general solution de-termines the infinitesimal generators of a group with r parameters. A specialsolution of this system is uniquely determined by assigning fixed values to theparametric derivatives p1, . . . , pr. Let two special solutions ξ1, η1 and ξ2, η2be determined by p1,1, . . . , p1,r and p2,1, . . . , p2,r respectively. The Lie sys-tem property assures that there exists a third solution ξ3, η3 determined byp3,1, . . . , p3,r corresponding to the commutator of the vector fields ξ1∂x+η1∂yand ξ2∂x + η2∂y and is explicitly given by (3.28). Its parametric derivativesmay be computed by deriving the right hand sides of (3.28) and reducing themw.r.t. to the given Lie system. Due to the linearity of all operations involved,they are alternating bilinear forms in the parametric derivatives p1,1, . . . , p1,r

and p2,1, . . . , p2,r and may be expressed in terms of Bk = p1,ip2,j −p1,jp2,i fork = 1, . . . , 1

2r(r − 1), i, j = 1, . . . , r, i < j. This proceeding is illustrated by athree-parameter group.

Example 3.39 Consider again the Janet basis ξx, ηx + ξy, ηy, ξyy of thepreceding example. The parametric derivatives ξy, η and ξ lead to

B1 = ξ1,yη2 − η1ξ2,y, B2 = ξ1,yξ2 − ξ2,yξ1, B3 = η1ξ2 − ξ1η2.

Differentiating and reducing the right hand sides yields ξ3,y = 0, η3 = −B2,ξ3 = B1, i.e., the only nonvanishing structure constants are c132 = −1 andc233 = 1. The same answer is of course obtained if they are computed fromthe generators ∂x, ∂y and y∂x − x∂y.


Exercises

Exercise 3.1 Show that the transformations x = x+y+a, y = x−y+ bdo not define a group. Which part of the First Fundamental Theorem isviolated? (Lie [112], vol. I, page 72).

Exercise 3.2 Determine the finite transformations of the four-parametergroup generated by U1 = ∂x, U2 = x∂x, U3 = y∂y, U4 = x2∂x + xy∂y.

Exercise 3.3 Determine the order of transitivity of the group g2.

Exercise 3.4 Show that any group of F1(x)∂y, . . . , Fr(x)∂y with r ≥ 3 ofthe type g15 is similar to a group of the form ∂v, u∂v, G3(u)∂v, . . . , Gr(u)∂v.

Exercise 3.5 If the Lie algebra type of the vector fields in Example 3.1is not a priori known, one may start with some other algebra of the listing onpage 135, e. g. with type l3,1. How do the calculations proceed in this case?

Exercise 3.6 Applying (3.17) determine the characteristic polynomial forr = 2 and r = 3 explicitly.

Exercise 3.7 Describe the groups g17 for l = 1, 2 or 3 in detail, i.e., listall possible groups, its generators and put up its commutator table.

Exercise 3.8 For the three-parameter groups listed above determine theassignment of vector fields to the generators U1, U2 and U3 such that thecommutation relations of the respective Lie algebra have the canonical formas given in the tabulation on page 135.

Exercise 3.9 The same problem for the four-parameter groups.

Exercise 3.10 The special structure of the characteristic equation entailsalgebraic relations for its coefficients in expanded form that are invariantw.r.t. basis transformations of the Lie algebra. Determine these relations forthe algebras l4,4 and l4,5.

Exercise 3.11 Determine the assignment of vector fields of gl2 and sl2 tothe generators Uk such that the commutation relations of the respective Liealgebra l4,1 and l3,1 in canonical form are obeyed.

Chapter 4

Equivalence and Invariants ofDifferential Equations

The two main topics of this chapter, equivalence and invariance of differen-tial equations w.r.t. certain transformation groups, have their origin in thetheory of algebraic forms as developed in the 19th century. Comprehensiveintroductions to the latter may be found in one of the classical text bookslike Clebsch [30] or Gordan [55], or in the expository article by Kung andRota [97]. There was a general belief around the middle of the 19th centurythat many concepts in algebra must have an important meaning for differen-tial equations if they were appropriately generalized. Following these ideasCockle [31] introduced a notion that came fairly close to an invariant of adifferential equation in terms of his so-called criticoids. He obtained themfor linear second order ode’s by elementary methods without applying thenotion of a group of transformations. Laguerre [98] was the first to realizethe close connection between the type of transformations admitted by a lin-ear second order equation and the invariants belonging to it. Brioschi [19]and Halphen [65] generalized these results to equations of third and fourthorder. For quasilinear equations of first and second order Liouville [117, 120]determined various classes of invariants and distinguished absolute and rela-tive invariants. Later, Tresse [181] in his Preisschrift presented an extensivediscussion of the invariants of the equation y′′ = F (x, y, y′) and solved thecorresponding equivalence problem. Some of his results have been generalizedto third order equations of the form y′′′ = F (x, y, y′, y′′) by Leja [103]. Asurvey on the history of the subject, including some comments and reprints ofrelevant articles may be found in the booklet by Czichowski and Fritzsche [36].A more modern treatement of the subject is given in the book by Olver [141].

In the context of differential equations the diffeomorphisms between coor-dinates are usually called point transformations in order to distinguish themfrom so-called contact transformations. The latter involve certain transfor-mations of the first derivatives in addition to the variables themselves. Thisterminology will frequently be applied in the remaining part of this book.

Whenever dealing with differential equations, it is essential to know whethertwo given equations may be transformed into each other by a suitable variablechange. A special name for this fundamental concept is introduced next.

159

160

Definition 4.1 (Equivalence problem). Given two differential equations,decide whether there exists a transformation which takes one of them into theother. If the answer is affirmative the two equations are said to be equivalentunder the type of transformations admitted.

This definition creates a partition of all ode’s into equivalence classes. Solv-ing an individual equation may be generalized to solving equations of a fullequivalence class. To this end, in the first place it must be possible to decidemembership in any given equivalence class algorithmically. As shown below,the invariants of a differential equation are a useful device for answering thisquestion. Secondly, within a given equivalence class, a canonical representa-tive is defined which may be applied to characterize this class. An importantsubproblem of this latter step is to identify the smallest function field in whichthe transformation functions are contained. In Chapter 5 it will be seen howthe symmetry type of an equation narrows down the possible equivalenceclasses to which it may belong.

In order to make the invariants of a differential equation into a meaningfulconcept, it is necessary to identify those transformations leading to a well-defined behavior of its coefficients. For linear equations this has been noticedfirst by Stackel [173] who obtained the complete answer in this case. Fora general equation Goursat [56] gave a definition of the invariants based onthe behavior of its coefficients under certain groups of transformations. Thesubsequent discussion follows closely this latter reference.

Let an n−th order ode ω(x, y, y′, . . . , y(n)) = 0 be given with indepen-dent variable x and dependent variable y. Assume that it is polynomial inthe derivatives, with coefficients a1, a2, . . . , aN depending on x and y. Let atransformation x = φ(x, y; θ) and y = ψ(x, y; θ) of a certain group be appliedto it where θ denotes collectively the parameters or functions specifying thegroup element. They are limited by the requirement that the structure of thedifferential equation remains unchanged under its action. The structure of adifferential equation may be expressed in terms of certain restrictions on itsorder and the power to which the various derivatives may occur. A linearequation, for example, is determined by the requirement that the dependentvariable and its derivatives occur only to the first power in each term. For thelargest group leaving the structure unchanged a proper name is introducedby the following definition.

Definition 4.2 (Structure invariance group). Let a class of differentialequations be defined in terms of certain restrictions on its structure. Thelargest group of transformations leaving this structure unchanged is called thestructure invariance group of this class. In general it will be an infinite di-mensional Lie group.

Originally the concept of a structure invariance group goes back to Ovsian-nikov [144], Chapter II, Section 4; he called them equivalence transformations.The name structure invariance group, however, appears to be more suggestive

Equivalence and Invariants 161

of the basic idea behind it. By definition the elements of the structure invari-ance group operate only on the unspecified elements of an equation, i.e., theparameters and undetermined functions that it may contain. If an equationdoes not contain any of them, its structure invariance group specializes to itssymmetry group to be explained in the next chapter.

If the structure remains unchanged under a certain transformation, it gen-erates a new differential equation of the form ω(x, y, y′, . . . , y(n)) = 0 withcoefficients

ak = fk(a1, . . . , aN , a1,x, a1,y, . . . , aN,x, aN,y, . . . , θ)

for k = 1, . . . , N where the functions fk are uniquely determined by the trans-formation corresponding to θ. Consequently, it generates a group of trans-formations among the coefficients a1, . . . , aN that may be prolonged to itsderivatives. According to Lie [112], vol. I, pages 212 and 523, any such groupallows an infinite series of differential invariants.

Definition 4.3 (Invariants of an ode) Let ω(x, y, y′, . . . , y(n)) = 0 be anode which is polynomial in y, y′, . . . , y(n) with coefficients a1, . . . , aN dependingon x and y. If a transformation of the structure invariance group of ω withgroup parameters θ is applied, a new equation with transformed coefficientsa1, . . . , aN is obtained. An expression Φ that is transformed like

Φ(a1, a2, . . . , aN ) = w(θ) · Φ(a1, a2, . . . , aN ).

is called an invariant of the ode ω with weight w. More precisely, if w is equalto 1, Φ is an absolute invariant or simply invariant, otherwise it is a relativeinvariant with weight w.

The meaning of this relation is as follows. If the left hand side Φ is evaluatedwith arguments ak, an expression depending on x and y is obtained. If at theright hand side the same function Φ is evaluated with ak = fk, the sameexpession is obtained, possibly multiplied with a factor depending only on thegroup parameters θ. Sometimes also a dependence on x and y is allowed.

Subsequently the term invariant of a differential equation without furtherspecification means always an invariant w.r.t. the structure invariance groupof the equation under consideration. These invariants will usually be denotedby Φ. Any other invariants are denoted by small Greek letters φ, ψ, etc. Thesubgroup to which they correspond will be specified explicitly, or it will beclear from the context. Phrases like semi-invariants will not be applied.

The problem is now to develop a set of criteria for deciding equivalence forgiven pairs of differential equations. It turns out that an answer may only beobtained for special classes of equations, e. g. linear equations or special typesof first or second order equations. Moreover, if two equations have been shownto be equivalent, it is still an additional task to determine a transformationbetween them. It turns out that the necessary computations quickly lead tounmanageable equations. Therefore the following observation is vital in many

162

cases in order to replace a given problem by an equivalent one which is mucheasier to handle.

The complexity of dealing with any class of ode’s is determined above allby the number of undetermined elements, and by the function field in whichthey are contained. This suggests using the undetermined elements in theadmitted transformations to eliminate as many coefficients in the equation aspossible, without enlarging the function field determined by the initially givenequation. For the result of this transformation a special term is introduced inthe following definition.

Definition 4.4 (Rational normal form) A rational normal form of an or-dinary differential equation that is polynomial in the dependent variable andits derivatives is an equation with the minimal number of nonconstant coeffi-cients that may be obtained from it by a point transformation in the dependentand the independent variable.

In general, a rational normal form is not unique, but defines a new structurewith a structure invariance group of its own. Applying these concepts, a fairlycomplete discussion of linear equations is given in Section 4.1. Nonlinearequations of first order are the subject of Section 4.2. In the last Section 4.3quasilinear equations of second order are considered.

4.1 Linear EquationsAs mentioned in the introduction to this chapter, linear ode’s were histor-

ically the first class of differential equations for which a systematic study ofthe invariants corresponding to a group of transformations had been made inanalogy to algebraic invariants. As a consequence the results are more com-plete for linear equations than for nonlinear ones. As usual, the general linearhomogeneous equation will be written in the form

y(n) + q1y(n−1) + . . .+ qn−1y

′ + qny = 0. (4.1)

Structure Invariance, Normal Forms and Invariants. The subsequenttheorem is originally due to Stackel [173]. The proof follows essentially anarticle of Neumer [137].

Theorem 4.1 (Stackel 1893, Neumer 1937) The structure invariancegroup of a linear homogenous equation (4.1) consists of transformations ofthe form x = F (u), y = G(u)v + L(x) with F and G undetermined functionsand L the general solution.

Proof Let x = φ(u, v) and y = ψ(u, v) be a general point transforma-tion. An undetermined dependence on v in the coefficients of the transformed


equation requires φv = 0, i.e., x = F (u). Under this constraint the expres-sion (B.9) for y(k) shows that an unmatched term proportional to v′v(k−1)

can only be avoided if ψvv = 0, i.e., ψ(u, v) = G(u)v + H(u). Furthermore,the expression (B.9) shows that the homogeneity requires H = 0 in order toavoid any term not proportional to v or any derivative of it. L(x) is a linearcombination of a fundamental system, therefore its contribution vanishes.

Although the term L(x) leaves the equation invariant and therefore maybe considered as trivial, it is important to recognize its existence because insome applications the maximal number of parameters involved in the structureinvariance group is an important invariant. For later reference the followingspecial case due to Neumer is formulated as a corollary.

Corollary 4.1 (Neumer 1937) The structure invariance group of a linearhomogeneous equation with constant coefficients is x = αu, y = eβuv + L(x)with α and β constants.

Theorem 4.2 Equation (4.1) has the rational normal form

y(n) + q2y(n−2) + . . .+ qn−1y

′ + qny = 0. (4.2)

It is obtained from (4.1) by the transformations

x = F (x), y = CF ′(n−1)/2 exp(− 1n

∫q1(F )F ′dx

)y (4.3)

where C is a constant and F (x) an undetermined function. The structureinvariance group of (4.2) is

x = F (u), y = CF ′(u)(n−1)/2v + L(u) (4.4)

where C is a constant and L is a linear combination of a fundamental system.

Proof Transforming equation (4.1) according to its structure invariancegroup by x = F (x), y = G(x)y leads to a new equation (see equation (B.16)of Appendix B) with the coefficient of y(n−1) equal to

nG

F ′n

( G′

G− n− 1

n

F ′′

F ′

)+q1

G

F ′n−1 .

It vanishes ifG = CF ′(n−1)/2 exp

(− 1n

∫q1F

′dx)

with C a constant. This yields (4.3).Any transformation of the structure invariance group of (4.2) must have

the form x = F (u), y = G(u)v. Applying it to (4.2) yields an equation for vcontaining the term

nG

F ′n

(G′G− n− 1

2F ′′

F ′

)v(n−1). (4.5)

164

It vanishes if G is proportional to F ′(n−1)/2. Any contribution originatingfrom L obviously drops in the final result. This yields (4.4).

The transformation (4.3) is the most general one leading to rational nor-mal form. Different choices of F (x) lead to rational normal forms that arerelated by a transformation of its structure invariance group (4.4). Oftenthe special choice x = x is made with the corresponding transformationy = exp (− 1

n∫q1(x)dx)y as in Section 2.1.

If the coefficients of the transformed equations are allowed to be in an ex-tension of the base field, a more special canonical form, the so-called Laguerre-Forsyth canonical form (Laguerre [99], Forsyth [47]), of equation (4.1) maybe obtained.

Theorem 4.3 (Laguerre 1879, Forsyth 1888) Any linear homogeneousequation in rational normal form

y(n) + q2y(n−2) + . . .+ qn−1y

′ + qny = 0 (4.6)

is equivalent to an equation in Laguerre-Forsyth canonical form

v(n) + p3v(n−3) + . . .+ pn−1v

′ + pnv = 0 (4.7)

with v ≡ v(u) and pk ≡ pk(u). It is obtained by the transformation x = F (u),y = F ′(u)(n−1)/2v where the inverse function u = F (x) is

F (x) = C1 exp∫z(x)dx+ C2 with z′ − 1

2z2 − 12q2(x)

n(n2 − 1)= 0. (4.8)

The structure invariance group of (4.7) is

u =a1u+ a2

a3u+ a4, v =

v

(a3u+ a4)n−1 + L(u)

where a1, . . . , a4 are constants and L(u) is a linear combination of a funda-mental system.

Proof Due to Theorem 4.2, the desired transformation must be of theform x = F (u), y = F ′(u)(n−1)/2v. Substituting it into (4.6) (for the requiredderivatives see (B.16) of Appendix B) and imposing the constraint p2(u) = 0yields the equation( F ′′

F ′

)′−1

2

( F ′′

F ′

)2

+12q2(F )F ′2

n(n2 − 1)= 0

for F (u) with F ′ = dFdu

. For the inverse function u = F (x) there follows

( F ′′

F ′

)′−1

2

( F ′′

F ′

)2

− 12q2(x)n(n2 − 1)

= 0 (4.9)


where F ′ = dFdx

. This first order Riccati equation for z ≡¯F ′′F ′

has always asolution.

Any transformation of the structure invariance group of (4.7) is obtainedby further specialization of (4.4). Substituting the corresponding relationG′

G = n− 1n

F ′′

F ′into the coefficient of y(n−2) (see (B.16) of Appendix B) leads

after some simplifications to( F ′′

F ′

)′−1

2

( F ′′

F ′

)2

= 0.

From its general solution and after a further integration for obtaining G thefinal answer is

F =C1u+ C2

u+ C3, G =

C4

(u+ C3)n−1.

Introducing new constants by

C1 =a1

a3, C2 =

a1

a3, C3 =

a4

a3, C4 =

1an3

the above form of the structure invariance group corresponding to parametersa1, . . . , a4 is obtained.

The structure invariance group of the Laguerre-Forsyth canonical form hasbeen given by Bouton [18], Chapter 5, equation (46). This is the first ex-ample in this chapter where the structure invariance group is a finite dimen-sional group. In order to determine the transformation to Laguerre-Forsythcanonical form explicitly, the Riccati equation (4.8) for z has to be solved.Substituting F ′ = 1

θ(x)2the linear second order equation

θ′′ +6q2

n(n2 − 1)θ = 0 (4.10)

for θ is obtained. Then the transformation function F (u) is the inverse of

u = F (x) =∫

dx

θ(x)2.

The above equation (4.10) for θ shows that determining the Laguerre-Forsythnormal form is equivalent to solving a general second order linear equation as ithas already been shown by Schlesinger [160], vol. II, page 198. For n = 2 this isequivalent to solving the original equation and no simplification of the solutionprocedure is obtained. This was to be expected because otherwise the generalsecond order equation would be solvable which is known to be impossible.For equations of order higher than the second, however, the transformation tocanonical form is a problem of more minor complexity than solving the givenequation itself.

166

Example 4.1 Consider y′′′ + 1xy = 0. Applying a general transformation

of its structure invariance group yields

y′′′ +(a1a4 − a2a3)3

(a1x+ a2)(a3x+ a4)5y = 0.

In addition there are the invariance transformations x = x, y = a5y, andthose corresponding to L(x).

The third subject to be treated in this subsection are invariants of linearode’s. A first result due to Laguerre is given next.

Theorem 4.4 (Laguerre 1879) Any third order linear homogeneous equa-tion

y′′′ + p1y′′ + p2y

′ + p3y = 0, pk ≡ pk(x) (4.11)

allows the relative second order invariant

Θ3 = p′′1 + 2p′1p1 − 3p′2 + 49p

31 − 2p1p2 + 6p3 (4.12)

w.r.t. to the structure invariance group x = F (u), y = G(u)v. It is trans-formed according to Θ3(x)|x=F (u) · F ′(u)3 = Θ3(u).

Proof The infinitesimal variations corresponding to the group F = u +f · δt, G = 1 are δx = f · δt, δy = 0, δy′ = −f ′y′ · δt, δy′′ = −f ′′y′ − 2f ′y′′

and δy′′′ = −f ′′′y′ − 3f ′′y′′ − 3f ′y′′′. These values are substituted into theequation

δy′′′ + p1δy′′ + p2δy

′ + p3δy + δp1y′′ + δp2y

′ + δp3y = 0.

The third derivative y′′′ is substituted by those of lower order applying thegiven ode. The resulting expression can only vanish if the coefficient of eachderivative including y itself vanishes. This leads to a system of linear equationsfrom which the variations of the coeffcients may be determined. The result is

δp1 = 3f ′′ − p1f′, δp2 = f ′′′ + p1f

′′ − 2p2f′, δp3 = −3p3f

′.

The variations of p′k and p′′k are obtained from δpk by repeated application ofLie’s relation (Lie [114], page 670)

δφ′ =d

dxδφ− φ′ d

dxδx (4.13)

which is valid for any φ. If invariants Φ of second order are to be determined,these variations have to be substituted into the equation∑

k=1,2,3

( ∂Φ∂pk

δpk +∂Φ∂p′k

δp′k +∂Φ∂p′′k

δp′′k

)= αΦ.

The coefficients of the various derivatives of f yield a linear first order systemfor Φ.


For the group generated by G = (1 + g · δt)v, F = u, the infinitesimalvariations are δx = 0, δy = gy ·δt, δy′ = (g′y+gy′) ·δt, δy′′ = g′′y+2g′y′+gy′′

and δy′′′ = g′′′y + 3g′′y′ + 3g′y′′ + gy′′′. By a similar analysis as above thevariations

δp1 = −3g′, δp2 = −4g′′ − 2p1g′, δp3 = −g′′′ − p1g

′′ − p2g′

are obtained. Together with its first and second derivatives they yield anotherlinear first order system for Φ. The two systems have to be combined. Aftera lengthy calculation the following Janet basis is obtained.

Φp′′3 = 0, Φp′′2 = 0, Θ3Φp′′1 = Φ,

Φp′3 = 0, Θ3Φp′2 = −3Φ, Φp′1 = 2p1Φ,

Θ3Φp3 = 6Φ, Θ3Φp2 = 2p1Φ, Θ3Φp1 =(2P ′1 + 4

3p21 − 2p2

)Φ

with Θ3 given by (4.12). This system has Φ = Θ3 as solution which is obtainedby integration.

Example 4.2 Consider y′′′+( 1x + 1

x2

)y′′− 1

x2 y = 0 which is equation 3.41

from the collection by Kamke with Θ3(x) = − 1x6

(6x4− 4

9x3− 4

3x2 + 8

3x−49

).

In new variables x = 1u , y = uv with v ≡ v(u) the equation becomes

v′′′ −(1− 8

u

)v′′ −

( 4u− 14u2

)v′ −

( 2u2 −

4u3 −

1u4

)v = 0

with Θ3(u) = − 1u4

(49u

4 − 83u

3 + 43u

2 + 49u − 6

). A simple calculation shows

that the two invariants obey the relation (4.12).

Laguerre’s Theorem 4.4 does not require the equation to be in rationalnormal form. As explained in the introduction to this chapter however, thecomplexity of any calculation involving differential equations increases signifi-cantly if the number of nonvanishing coefficients increases. This is particularlytrue for any kind of invariants. Therefore from now on in this section any lin-ear ode will be assumed in rational normal form.

Theorem 4.5 (Schlesinger 1897) Any linear homogeneous equation inrational normal form

y(n) + q2y(n−2) + . . .+ qn−1y

′ + qny = 0

allows a series of n−2 relative invariants Θk, k = 3, . . . , n w.r.t. to the groupx = F (u), y = F ′(u)(n−1)/2v transforming like Θk(u) = F ′(u)kΘk(x)|x=F .Explicitly the three lowest invariants are

Θ3 = q3 −n− 2

2q′2,

168

Θ4 = q4 −n− 3

2q′3 +

(n− 2)(n− 3)10

q′′2 −110

5n+ 7n+ 1

(n− 2)(n− 3)n(n− 1)

q22 , (4.14)

Θ5 = q5 − n− 42 q′4 + 3(n− 3)(n− 4)

28 q′′3 −(n− 2)(n− 3)(n− 4)

84 q′′2

−17

7n+ 13n+ 1

(n− 3)(n− 4)n(n− 1) q2q3 + 1

147n+ 13n+ 1

(n− 2)(n− 3)(n− 4)n(n− 1) q2q

′2.

The proof may be found in Schlesinger [160], vol. II, page 147. If p1 = 0,p2 = q2 and p3 = q3 are substituted into (4.12), it is identical to Θ3 defined in(4.12), up to a constant factor. This example shows the considerable simplifi-cation due to rational normal form. For higher order invariants the differencein size is even more prominent.

Theorem 4.6 (Forsyth 1888) Any linear homogeneous equation in rationalnormal form

y(n) + q2y(n−2) + . . .+ qn−1y

′ + qny = 0

allows a series of n−2 relative invariants Φk, k = 3, . . . , n w.r.t. to the groupx = F (u), y = F ′(u)(n−1)/2v transforming like Φk(u) = F ′(u)2Φk(x)|x=F . Interms of the invariants Θk they are

Φk(x) = 2kΘ′′kΘk− (2k + 1)

(Θ′kΘk

)2

− 24k2q2

n(n2 − 1).

Combining the Θk and the Φk, the following series of absolute invariants

Θj,k ≡Θjk

Θkj

and Ψk ≡ΦkkΘ2k

is obtained.

The invariants Φk are obtained by differentiation and substitution from theinvariants Θk. The details may be found in the original article by Forsyth[47], a good account is also given in the book by Schlesinger [160], part II,Chapter 6, page 200.Equivalence of Second Order Equations. In order to proceed furtherwith identifying equivalence classes of linear ode’s, equations of fixed orderare considered. The complete answer for second order linear ode’s is givennext.

Theorem 4.7 Any linear second order ode y′′+p1y+p2y = 0 is equivalentto v′′ = 0. As a consequence all second order linear ode’s form a singleequivalence class with canonical representative v′′ = 0.

Proof At first the given equation is transformed into its rational normalform y′′ + py = 0 with p = p2 − 1

2p′1 − 1

4p21. It remains to be shown that


v′′ = 0 may be transformed into it by a transformation u = f(x), v = g(x)y.According to (B.15) the second derivative is

v′′ =g

f ′2

[y′′ +

(2g′

g− f ′′

f ′

)y′ +

(g′′g− f ′′

f ′g′

g

)y].

It yields the equation y′′ + py = 0 if

2g′

g− f ′′

f ′= 0,

g′′

g− f ′′

f ′g′

g= p.

Substituting g′

g = 2f′′

f ′and g′′

g =(g′g

)′ + (g′g

)2 into the latter equation, the

Riccati equation z′ − z2 − p2 + 12p′1 + 1

4p21 = 0 for z ≡ g′

g is obtained. Solvingfor f and g the expressions

f = C1

∫exp

(2

∫zdx

)dx+ C2, g = C3 exp

( ∫zdx

)follow; z is a solution of the above Riccati equation.

Example 4.3 Let 1, u and y1, y2 be fundamental systems for v′′ = 0and y′′ + p1y

′ + p2y = 0 respectively. The transformation u = σ ≡ y1y2 and

v = ρ ≡ yy2 transforms the latter into the former. Correspondingly v′′ = 0 is

transformed into the latter equation in the form (2.3).

This result has been known for a long time; see for example Schlesinger[160], vol. II, page 184, equation (8). It is mainly of theoretical interestbecause it does not help in solving any second order equation. Solving theRiccati equation that occurs at the end of the proof of Theorem 4.7 comesdown to solving the original linear second order ode.

Numerous higher transcendents or special functions are defined in terms oflinear second order ode’s, e. g. Bessel functions, hypergeometric functions,etc. In applications there often arises the problem of recognizing whetherthe solution of a given differential equation may be expressed in terms ofparticular special functions. The answer may be obtained by the followingresult which generalizes the above theorem.

Theorem 4.8 A linear second order ode y′′ + p(x)y = 0 is transformedinto v′′ + q(u)v = 0 by x = f(u), y = f(u)′1/2v if a function f(u) may befound that obeys ( f ′′

f ′

)′−1

2

( f ′′

f ′

)2

+2f ′2p(f) = 2q(u). (4.15)

This third order ode has always a six-parameter group of symmetries.

Proof By Theorem 4.2 the desired equivalence transformation must havethe form (4.4) for n = 2. Substituting it into y′′ + p(x)y = 0, applying (B.15)

170

and equating the result to v′′ + q(u)v = 0 leads immediately to the aboveequation for f . A simple calculation shows that the conditions for case ii) ofTheorem 5.15 are satisfied, i.e., (4.15) belongs to symmetry class S3

6 .

In order to determine an equivalence transformation to any equation defin-ing a special function by Theorem 4.8, a rational normal form must be knownfor it. The most important ones are given next. The defining equation iswritten in terms of the independent variable z and dependent variable w. Forthe rational normal forms the dependent variable is w. Further details maybe found in Abramowitz [2]. It should be emphasized again that the givenrational normal forms are not unique. The Bessel equation

w′′ +1zw′ +

(1− ν2

z2

)w = 0 (4.16)

with fundamental system J±ν(z). A rational normal form is

w′′ +(1−

ν2 − 14

z2

)w = 0 with w =

√zw. (4.17)

The hypergeometrical equation

w′′ +(α+ β + 1)z − γ

z(z − 1)w′ +

αβ

z(z − 1)w = 0. (4.18)

A rational normal form is

w′′ − [(α− β)2 − 1]z2 + 2[(α+ β − 1)γ − 2αβ]z − γ(γ − 2)4z2(z − 1)2

w = 0 (4.19)

with w = zγ/2(z − 1)(α+β+1−γ)/2w. The Legendre equation

w′′ +2z

z2 − 1w′ − ν(ν + 1)

z2 − 1w = 0. (4.20)


w′′ − ν(ν − 1)(z2 − 1)− 1(z2 − 1)2

w = 0 with w =√z2 − 1w. (4.21)

Weber’s equationw′′ − zw′ − aw = 0. (4.22)


w′′ − 14 [z2 + 2(a− 1)]w = 0 with w = exp

(− 1

4z2)w. (4.23)

The confluent hypergeometric equation or Whittaker’s equation is already inrational normal form.

y′′ −(1

4− k

x+

4m2 − 14x2

)y = 0. (4.24)


Despite the large symmetry group of equation (4.15), the solution algorithm

The reason is that it requires solving the second order linear ode’s for which theequivalence transformation is desired. Details are discussed in Exercise 4.8.

In order to proceed, the problem has to be further specialized. It is basedon the observation that equivalence to an equation defining a special functionis only of interest if the transformation is simple in some sense, e. g. it shouldbe rational or algebraic. For equivalence to Bessel’s equation the followingresult is useful.

Lemma 4.1 The transformations z = axα, w = (αaxα−1)1/2y generate theequations

y′′+[α2a2x2α−2 +

(14 − α

2ν2) 1x2

]y = 0 (4.25)

from the rational normal form (4.17) of Bessel’s equation.

It is easily proved by applying the formulas (B.15) given in Appendix B tothe suggested transformations. Let y′′ + q1y

′ + q2y = 0 be a given equation,y′′ + q(x)y = 0 its rational normal form obtained by y = exp

(− 1

2

∫q1dx

)y.

If q(x) has the form of the square bracket in (4.25), a fundamental system ofthe original equation in y(x) may be written as

√x exp (− 1

2

∫q1dx)w1,2(axα)

if w1,2(z) is a fundamental system of (4.16).It turns out that the class of transformations admitted in Lemma 4.1 cov-

ers virtually all equivalence problems for Bessel’s equation listed in Kamke’scollection. A few examples are given next.

Example 4.4 Equation 2.347 from Kamke’s collection y′′+ 1xy′+ 1

x4 y = 0

has the rational normal form y′′ +( 1x4 + 1

4x2

)y = 0 if y = 1√

xy is chosen.

The above lemma yields α = −1, a = ±1 and ν = 0. With a = −1 therefollows z = 1

x , w = yx . Finally the full transformation from the originally

given equation to (4.16) with ν = 0 is x = 1z , y = w, consequently, it has a

fundamental system J0

( 1x

)and Y0

( 1x

).

Example 4.5 Let the irreducible Airy equation y′′ + xy = 0 be given. Itis already in rational normal form. By Lemma 4.1 α = 3

2 , a = 23 and ν = 1

3is obtained. Consequently, a fundamental system for the original equation is√xJ±1/3

(23x

3/2).

For Weber’s equation the equivalent of Lemma 4.1 is given next. Again thisclass of transformations covers all cases listed in Kamke’s collection.

Lemma 4.2 The transformation z = axα, w = (αaxα−1)1/2y generates theequation

y′′+[

14α

2a4x4α−2 + 12α

2a2(k − 1)x2α−2 + α2 − 14x2

]y = 0 (4.26)

from the rational normal form (4.17) of Bessel’s equation.

for this symmetry type described in Chapter 7 on page 344 is of no help here.

172

Equivalence of Third Order Equations. For linear equations of thirdorder it occurs for the first time that there is more than a single equivalenceclass. This is because a linear homogeneous equation of order n contains ncoefficient functions. Only two of them may be transformed into each other bythe two functions F (u) and G(u) contained in the structure invariance groupgiven in Theorem 4.1. Obviously this is only possible for n = 2. The nexttheorem applies the invariants defined in the preceding subsection in order toidentify all equivalence classes of linear third order equations.

Theorem 4.9 Let two third order linear homogeneous ode’s

y′′′ +A(x)y′ +B(x)y = 0 and v′′′ + P (u)v′ +Q(u)v = 0 (4.27)

in rational normal form be given. The four alternatives below define equiva-lence classes, canonical forms and give the equations that are obeyed by thetransformation function x = F (u), y = F ′(u)v to canonical form, or its in-verse u = F (x), v = F ′(x)y. The required invariants

Θ3 = B − 12A

′, Φ3 = 6Θ′′3Θ3− 7

( Θ′3Θ3

)2

−9A and Ψ3 = Φ33

Θ23

are obtained from the preceding Theorems 4.14 and 4.6 for n = 3.

E31 : Θ3 = 0, canonical form v′′′ = 0, F (x) obeys( F ′′

F ′

)′− 3

2

(F ′′

F ′

)2

− 12A = 0. (4.28)

E32 : Θ3 6= 0, Φ3 = 0, canonical form v′′′ + v = 0, F obeys F ′ = Θ3(x)

13 .

E33 (c) : Θ3 6= 0, Φ3 6= 0, Ψ3 = const, canonical form v′′′+cv′+v = 0, c 6= 0. The

constant c parametrizes equivalence classes, while F (u) is determined byintegration from F ′2 · Φ3(x)|x=F = −9c.

E34 (Q) : If none of the preceding three cases applies, the two equations (4.27) are

equivalent to each other if both

Θ3(x)|x=FF ′(u)3 = Θ3(u) and Ψ3(x)|x=F = Ψ3(u)

are valid. The equation for y is equivalent to v′′′ + Q(u)v = 0 if F (x)is determined from (4.28), and its inverse F (u) is substituted into thesecond equation of (4.29) from which Q is obtained.

Proof Applying a general transformation of the structure invariance groupx = F (u) and y = F ′(u)v to the above equation for y(x) yields

v′′′+[

2F′′′

F ′− 3

(F ′′

F ′

)2

+A(F )F ′2]v′

+[F (4)

F ′+ 3

(F ′′

F ′

)3

−4F′′′F ′′

F ′2+A(F )F ′F ′′ +B(F )F ′3

]v = 0.


In order to generate the equation for v(u), F (u) has to obey the constraints

F ′′′

F ′− 3

2

(F ′′

F ′

)2

+ 12A(F )F ′2 = 1

2P,

F (4)

F ′+ 3

(F ′′

F ′

)3

−4F′′′F ′′

F ′2+A(F )F ′F ′′ +B(F )F ′3 = Q.

(4.29)

Reduction of the second equation w.r.t. the first one yields[B(F )− 1

2A′(F )

]F ′3 = Q− 1

2P′ or Θ3(x)

∣∣∣x=F

F ′(u)3 = Θ3(u)

(4.30)

where A′(F ) = dA(F )dF

. If Θ3(x) = Θ3(u) = 0, the second equation of (4.29)is the derivative of the first and the system reduces to a single third order odefor F which has always a solution. If P = Q = 0 is chosen for the canonicalform equation, F (u) satisfies(

F ′′

F ′

)′− 1

2

(F ′′

F ′

)2

+ 12A(F )F ′2 = 0

from which the Riccati equation (4.28) for the inverse function F (x) is ob-tained. If Θ3(x) = 0 and Θ3(u) 6= 0 or Θ3(x) 6= 0 and Θ3(u) = 0, the abovesystem (4.29) for F is inconsistent and the two equations for y and v cannotbe equivalent to each other. Consequently, the condition Θ3 = 0 characterizesthe single equivalence class E3

1 .If both Θ3(x) 6= 0 and Θ3(u) 6= 0, reducing the first equation of (4.29) w.r.t.

(4.30) yields

6Θ′′3(u)Θ(u)

− 7( Θ′3(u)

Θ3(u)

)2

−9P =[

6Θ′′3(F )Θ(F )

− 7( Θ′3(F )

Θ3(F )

)2

−9A(F )]F ′2

where Θ′3(F ) = dΘ3(F )dF

, or

Φ3(u) = Φ3(x)|x=F · F ′(u)2. (4.31)

If Φ3(x) = Φ3(u) = 0, F is determined by the first order ode (4.30) which hasalways a solution. If P = 0, Q = 1 is chosen for the canonical form equation,θ3(u) = 1 and the equation F (x)′ = Θ3(x)

13 is obtained. If Φ3(x) = 0

and Φ3(u) 6= 0 or Φ3(x) 6= 0 and Φ3(u) = 0, the system for F is againinconsistent and the two equations are not equivalent. Consequently, Θ3 = 0,Φ3 = 0 characterize another equivalence class E3

2 .If both Φ3(x) 6= 0 and Φ3(u) 6= 0, consistency of (4.30) and (4.31) requires

Ψ3(u) = Ψ3(x)|x=F . If both Ψ3(x) and Ψ3(u) are constant, a possible choicefor the canonical form isQ = 1 and P = c = const, the value of c is determinedby Ψ3(u) = −729c3 = Ψ3(x). In general, two canonical equations with c 6= care not equivalent, i.e., any given value of c determines an equivalence class

174

E33 (c). Substituting the corresponding value for Φ3(u) = −9c into (4.31) leads

to the equation given for F given above.If both invariants Ψ3(x) and Ψ3(u) are not constant, the transformation

function F (u) has to be determined as a solution of the system (4.29) orequivalently (4.30) and (4.31). Its consistency condition Ψ3(u) = Ψ3(x)|x=Fis necessary for equivalence. If it is satisfied, the validity of (4.30) is sufficientfor equivalence because in this case (4.31) is a consequence of it. ChoosingP = 0 for the canonical form equation, the first equation determines F (u); asa consequence F (x) satisfies the same equation as in case i). Q is obtainedby substituting the solution for F (u) into the second equation of (4.29).

The canonical form for E34 (Q) is actually the Laguerre-Forsyth canonical

form of Theorem 4.3, the Riccati equation of the above theorem correspondsto the equation (4.8) for z given there. In order to decide equivalence of twogiven equations in E3

3 (Q) and E33 (Q), it is not necessary to solve a Riccati

equation. Rather a possible transformation function is determined from thecondition Ψ3(u) = Ψ3(x)|x=F for the invariants which is a purely algebraicproblem. If the resulting function F (u) satisfies (4.30) and (4.31), equivalenceis established, otherwise the two equations are not equivalent to each other.The following examples illustrate this case.

Example 4.6 Let the two equations y′′′ + xy′ + y = 0 with Θ3(x) = 12 ,

Φ3(x) = −9x, Ψ3(x) = −2916x3 and v′′′+ 1u5 v

′− 3u6 v = 0 with Θ3(u) = − 1

2u6 ,

Φ3(u) = − 9u5 and Ψ3(u) = −2916

u3 be given, i.e., they belong to E34 (Q). By

Theorem 4.9 the differential equation F ′ = −Fu with the solution F = Cu is

obtained. The condition for the invariants Ψ3 leads to F = 1u , i.e., the two

equations are in the same equivalence class; x = 1u and y = − v

u2 establish

the transformation.

Example 4.7 Let the equations y′′′ + xy′ + y = 0 like in the preceding

example and v′′′ + v′ + uv = 0 with Θ3(u) = u, Φ3(u) = −9u2 + 7u2 and

Ψ3(u) = − (9u2 + 7)3

u8 be given. By Theorem 4.9 the constraints

F 3 =(9u2 + 7)3

2916u8, F ′(u) =

18u3F

9u2 + 7−→ F 9 =

Ce9u2

(9u2 + 7)7

are obtained which are inconsistent for any value of the integration constant C.As a consequence the two equations are not equivalent to each other.

Example 4.8 The equation y′′′ − y′′ = 0 with rational normal formy′′′− 1

3y′− 2

27y = 0 has invariants Θ3 = − 227 , Φ3 = 3 and Ψ3 = 19683

4 . Transfor-mation to canonical form v′′′− 3

23√

2v′+ v = 0 is achieved by F (u) = − 32

3√

4u,i.e., it belongs to E3

3 (− 32

3√

2). Similarly, y′′′ − 1xy′′ = 0 with rational normal


form y′′′− 43x2 y

′+ 1627x3 y = 0 and invariants Θ3 = − 20

27x3 , Φ3 = 21x2 and Ψ3 =

137781400 belongs to equivalence class E33 (− 21

2 3√50), because it is transformed

to canonical form v′′′ − 212 3√50

v′ + v = 0 by F (u) = exp (− 310

3√

50u).

In the next chapter it will be shown how the various equivalence classesdescribed in Theorem 4.9 combine into symmetry classes. In this contextequations with constant coefficients play an important role. They are thesubject of the next example.

Example 4.9 For linear ode’s with constant coefficients the results ofTheorem 4.9 may be described more explicitly. The generic equation ofthe equivalence class E3

1 is y′′′ + ay′ = 0 with a nonzero constant a. Byx = u

i√a

log i√a, y = v

u it is transformed into the canonical form v′′′ = 0.

The generic equation for the equivalence class E32 is y′′′ + by = 0 where b 6= 0

is again a constant. By x =(qb

)1/3u, y =

(qb

)1/3v it is transformed into

v′′′+ qv = 0 where q is an arbitrary constant. The canonical form v′′′+ v = 0is obtained by x =

(1a)1/3

u, y =(1a)1/3

v. The generic equation for case E33 (c)

has the form y′′′ + ay′ + by = 0 with both a, b nonvanishing constants. Itcannot be transformed into an arbitrary equation of the same kind, only oneof its coefficients may be chosen at will, e. g. the coefficient of v may be madeto unity as in the preceding case with the result v′′′ + a

b2/3v′ + v = 0.

In later chapters different canonical forms will be applied for equivalenceclasses E3

2 and E33 (c). They are described in the subsequent lemma.

Lemma 4.3 A third order equation y′′′ + cy′ + y = 0 with c 6= 0 andc 6= − 3

23√

2 is equivalent to v′′′ − (a+ 1)v′′ + av′ = 0, if the constants a and care related to each other by

(a− 12 )2(a+ 1)2(a− 2)2c3 + 27

4 (a2 − a+ 1)3 = 0. (4.32)

The transformation between the two equations is achieved by x = αu andy = eβuv where

α = 19

(a+ 1)(a− 2)(2a− 1)ca2 − a+ 1

, β = − 13 (a+ 1).

If c = 0 it is equivalent to v′′′−(a+1)v′′+av′ = 0 with a = 12 (1±i

√3) whereby

α = 13 (a + 1), β = − 1

3 (a + 1). If c = − 32

3√

2 it is equivalent to v′′′ − v′′ = 0whereby α = − 1

33√

2, β = − 13 .

Proof By Corollary 4.1 the transformation between the two linear ode’smust have the form x = αu, y = eβuv. Substitution into the first equationyields the algebraic system

3β + a+ 1 = 0, cα2 + 3β2 − a = 0, α3 + cα2β + β3 = 0

176

from which α and β are obtained by elimination if the constraint (4.32) issatisfied. The calculations in the remaining cases are similar.

The relation between equivalence classes and the value of c in the canonicalform representative y′′′ + cy′ + y = 0 is not one-to-one. The same is truefor the canonical form y′′′ − (a + 1)y′′ + ay′ = 0 in the above lemma. Theserelations are discussed in more detail in the Exercises 4.6 and 4.7.

4.2 Nonlinear First Order Equations

The first result in this section is mainly of theoretical interest. It shows thatpoint symmetries will be of limited help in solving first order ode’s becausethey are members of a single equivalence class. Consequently, they all sharethe same symmetry type.

Theorem 4.10 Any quasilinear first order equation y′ + r(x, y) = 0 isequivalent to v′ = 0, i.e., all first order equations of this kind form a singleequivalence class under point transformations for which v′ = 0 may be chosenas a canonical form.

Proof The point transformation x = φ(u, v), y = ψ(u, v) transforms the

given equation into v′ + ψu + r(φ, ψ)φuψv + r(φ, ψ)φv

= 0. The desired canonical form is

obtained if ψu + r(φ, ψ)φu = 0. This equation always has a solution.

Although this result guarantees that an equivalence transformation to canon-ical form v′ = 0 does exist for any given first order ode, it does not providea method for finding such a transformation. In general it comes down tosolving the given equation as is obvious from rewriting the condition in theproof as dψ

dφ+ r(φ, ψ) = 0. In Exercise 4.2 the reverse transformation from

the canonical form to a general first order equation is considered.

Example 4.10 The Riccati equation y′+y2 + 4xy+ 2

x2 = 0 has the general

solution y = 1x+ C −

2x . It yields the first integral x(xy + 1)

xy + 2 = C. According

to Exercise 4.2 this yields the canonical variables v ≡ ρ(x, y) = x(xy + 1)xy + 2 and

u ≡ σ(x, y) where σ(x, y) may be chosen arbitrarily. For u = 1x one obtains

the transformation x = 1u , y = u(1− 2uv)

uv − 1 . For u = x the transformation is

x = u, y = 2v − uu(u− v) .

From this result follows that in order to proceed, more special classes of


equations have to be considered, e. g. by limiting the admitted transforma-tions to the corresponding structure invariance groups. The two major classesare Riccati’s equation containing a quadratic term in the dependent variable,and Abel’s equation with a cubic term.Riccati’s Equation. Algorithms for finding special solutions of these equa-tions have been described already in Chapter 2. Here they will be reconsideredin the context which is the subject of this chapter. Its structure invariance isdetermined first.

Theorem 4.11 The structure invariance group of Riccati’s equation

y′ + a2y2 + a1y + a0 = 0, ak ≡ ak(x) for k = 0, 1, 2 (4.33)

is x = F (u), y = G(u)v + H(u) where v ≡ v(u) and F , G and H areundetermined functions of its argument.

Proof A general point transformation from x and y to new variables uand v with v ≡ v(u)

x = φ(u, v), y = ψ(u, v), y′ =ψu + ψvv

′

φu + φvv′

generates the equation

v′ +(b2ψ2 + b1ψ + b0)φu + ψu

(b2ψ2 + b1ψ + b0)φv + ψv= 0 (4.34)

where bk ≡ ak(φ). In order to avoid the occurrence of an unspecified depen-dence on v via the coefficients ak, φv = 0 is required, i.e., φ = F (u). Thedenominator must be independent of v, this requires ψv to be a function of ualone, i.e., it must be linear in v and has the form ψ = G(u)v+H(u). Underthese constraints (4.34) simplifies to

v′ + b2F′Gv2+

[ G′G

+ (2b2H + b1)F ′]v + (b2H2 + b1H + b0)

F ′

G+H ′

G= 0,

(4.35)i.e., it has the desired structure.

Theorem 4.12 The rational normal form of Riccati’s equation (4.33) isy′ + y2 +A(x) = 0. It has the structure invariance group

x = F (u), y =1F ′

(v +

12F ′′

F ′

). (4.36)

Proof By Theorem 4.11 the desired transformation must have the generalform x = F (u), y = G(u)v +H(u). From (4.35) the constraints

b2F′G = 1,

G′

G+ (2b2H + b1)F ′ = 0

178

are obtained with bk ≡ ak(F ). Eliminating G and H yields for the transfor-mation to rational normal form

x = F, y =v

b2F′ +

b2F′′ + b′2F

′ − b1b2F ′2

2b22F′2 . (4.37)

If F ≡ F (u) is in the base field of (4.33), the same is true for y and y′ obtainedfrom it. On the other hand, any additional constraint on the term which is freeof the dependent variable in order to obtain a particular form A(x) requiressolving a Riccati equation for H which in general is not possible within thebase field. This proves the first part.

Any transformation of the structure invariance group of y′ + y2 + A = 0must have the general form (4.35) with the additional constraints F ′G = 1,G′

G + 2HF ′ = 0. This yields (4.36).

Sometimes it occurs that the rational normal form allows one to obtain thesolution immediately as the following examples show.

Example 4.11 The two equations from Kamke’s collection

no.1.140 : y′ + y2 + 4xy + 2

x2 = 0 and

no.1.165 : y′ + 12x2 − xy

2 − 4x+ 12x2 − xy + 4

2x− 1 = 0

have rational normal form v′ + v2 = 0 with general solution v = 1u+ C .

If the rational normal form does not contain the independent variable ex-plicitly, the solution may be obtained as explained on page 18. Two examplesare given next.

Example 4.12 Equation 1.15 y′+y2−2x2y+x4−2x−1 = 0 from Kamke’scollection has rational normal form v′ + v2 − 1 = 0 with a special solutionv = 1. Equation 1.140 y′ + y2 + 4

xy + 2x2 = 0 with rational normal form

v′+ v2 = 0 has the special solution v = 0. In either case, the general solutionmay be obtained by Corollary 2.1 or by direct integration.

Corollary 4.2 Any Riccati equations y′ + y2 + A(x) = 0 and v′ + v2 +P (u) = 0 are equivalent w.r.t. to the structure invariance group (4.36).

Proof Applying a transformation (4.36) to the first equation yields

v′ + v2 +12

( F ′′

F ′

)′−1

4

( F ′′

F ′

)2

+A(F )F ′2 = 0.

This represents the second equation if 12

(F ′′

F ′

)′−1

4

(F ′′

F ′

)2

+A(F )F ′2 =P (u). Because this third order ode always has a solution, equivalence is alwaysassured.


This result is an immediate consequence of Theorem 4.10. It shows thatalso for Riccati equations an equivalence transformation in general cannot befound.Abel’s Equation. The simplest first order equation that cannot be trans-formed easily into a linear one as is true for Riccati’s equation is the equa-tion introduced about 200 years ago by the famous Norwegian mathematicianAbel. It has the form

y′ + a3y3 + a2y

2 + a1y + a0 = 0 (4.38)

with ak ≡ ak(x) for k = 0, . . . , 3, a3 6= 0. The more general equation

y′ +f3y

3 + f2y2 + f1y + f0

y + g= 0 (4.39)

which is usually called Abel’s equation of second kind may be reduced to(4.38) by changing the dependent variable y = 1

v(x) − g with the result

a3 = f3g3 − f2g2 + f1g − f0, a2 = g′ − 3f3g2 + 2f2g − f1,

a1 = 3f3g − f2, a0 = −f3.(4.40)

If in (4.39) the numerator is of degree two in y, i.e., if f3 = 0, it follows thata0 = 0.

As usual by now, the structure invariance group of (4.38) will be identifiedfirst. Although the result has been known for a long time, for example toAppell and Painleve, it has never really been proved. Therefore it is derivednext.

Theorem 4.13 The structure invariance group of Abel’s equation (4.38)is

x = F (u), y = G(u)v +H(u) (4.41)

where v ≡ v(u) and F , G and H are undetermined functions of its argument.

Proof The first part of the proof is identical to the proof of Theorem 4.11with the result φ = F (u) and ψ = G(u)v+H(u). Applying this transformationto (4.38) yields

v′ + b3F′G2v3 + (3b3H + b2)F ′Gv2+

(3b3H2 + 2b2H + b1 + G′

F ′G

)F ′v

+(b3H3 + b2H2 + b1H + b0)F

′

G + H ′

G = 0(4.42)

with bk(u) ≡ ak(F ), i.e., it has the desired structure.

Abel’s equation (4.38) contains four coefficient functions, whereas there areonly three in the structure invariance group (4.41). This suggests that Abel’s

180

equation may be transformed into an equation with only a single nonconstantcoefficient involved. This transformation was performed by Appell [6] withthe result y′ + y3 + J(x) = 0. However, in general J(x) is not contained inthe base field; it is left as an exercise to determine it explicitly. The optimalresult that may be achieved without leaving it is obtained next.

Theorem 4.14 There are two different rational normal forms of Abel’sequation (4.38). Let A ≡ A(x) and B ≡ B(x) be rational functions ofa0, . . . , a3 and its derivatives, A 6= 0.

i) y′+Ay3+By = 0 with structure invariance group x = F (u), y = G(u)v.

ii) y′ + Ay3 + By + 1 = 0 with structure invariance group x = F (u),y = F ′(u)v.

Proof If any transformation (4.41) is applied to (4.38), it assumes theform (4.42). The choice H = − b2

3b3entails the vanishing of the coefficient of

v2. As a consequence applying the transformation

y = v − a2

3a3, y′ = v′ − 1

3

( a2

a3

)′which leaves the independent variable unchanged, Abel’s equation assumesthe form v′ + b3v

3 + b1v + b0 = 0 with

b3 = a3, b1 = a1 −a22

3a3, b0 = a0 −

a1a2

3a3+

2a32

27a23

− 13

( a2

a3

)′. (4.43)

If b0 = 0, the first alternative is obtained. Otherwise introducing again a newfunction w by v = b0w, v′ = b0w

′ + b′0w leads to

w′ +Aw3 +Bw + 1 = 0 with A = b20b3, B = b1 +b′0b0.

This is the canonical form for case ii). For the structure invariance groupof either case the terms independent of v in (4.42) lead immediately to theconstraint H = 0. If b0 = 1 the additional constraint G = F ′ is necessary inorder not to change this value.

The rational normal form of case i) is actually a Bernoulli equation. It isa pleasant feature that this special case is identified without any additionaleffort. As is well known, if the new function v is replaced by 1

v2 the linearequation

v′ − 2Bv − 2A = 0

is obtained that is easily solved in terms of quadratures with the result

v = exp(

2∫Bdx)

[2

∫A exp (−2

∫Bdx

)dx+ C

]. (4.44)


As will be seen later on, this is a consequence of a two-parameter symmetrygroup that may always be determined explicitly for a Bernoulli equation.

Example 4.13 Equation 1.146 y′ − x2y3 + 3y2 − 1x2 y + 2x− 1

x4 = 0 from

Kamke’s collection has rational normal form v′ − u2v3 + 2u2 v = 0, i.e., it is

a Bernoulli equation the solution of which may be obtained by integration asexplained above.

The invariants of Abel’s equation with respect to the group (4.41) weregiven for the first time by Liouville [119]. He did not mention, however, howhe obtained them. For the applications in this book only the two lowestinvariants derived in the next theorem are needed.

Theorem 4.15 (Liouville 1887) For Abel’s equation

y′ + a3y3 + a2y

2 + a1y + a0 = 0 (4.45)

where ak ≡ ak(x) a relative invariant w.r.t. its structure invariance groupx = F (u), y = G(u)v +H(u) with v ≡ v(u) is

Φ3 = a2a′3 − a′2a3 + 3a0a

23 − a1a2a3 + 2

9a32. (4.46)

By differentiation the series of higher order relative invariants

Φ2m+1 = a3Φ′2m−1 − (2m− 1)[a′3 + 13a

22 − a1a3]Φ2m−1 (4.47)

for m = 2, 3, . . . is obtained. To the invariant Φk the weight k is assigned. De-noting the coefficients of the original and the transformed equation collectivelyby a and b, the transformation law between the invariants is

Φk(u) = F (u)′k ·G(u)k · Φk(x)|x=F (u).

Rational expressions of these relative invariants with the property that the sumof the products of their order and the corresponding exponent are the same inthe numerator and the denominator are absolute invariants.

Proof The expression (4.46) for the first order invariant Φ3 will be derivedin detail. The infinitesimal variations corresponding to the group F = u+f ·δt,G = 1, H = 0 are δx = f · δt, δy = 0, δy′ = −f ′y′ · δt. Substituting thesevalues into the variation of Abel’s equation yields

δy′ + (3a3y2 + 2a2y + a1)δy + δa3y

3 + δa2y2 + δa1y + δa0 =

−f ′y′ + δa3y3 + δa2y

2 + δa1y + δa0 = 0.

If in the latter expression y′ is eliminated by means of Abel’s equation and theresult is separated w.r.t. to y, a system of equations is obtained from whichthe variations of the coefficients may be determined as δak = −f ′ak · δt. Thevariations of the derivatives a′k follow from these expressions by means of the

182

identity (4.13) with the result δa′k = −(akf ′′ + 2akf ′) · δt. Accordingly arelative invariant Φ for this group has to obey the system of equations

a3Φa′3 + a2Φa′2 + a1Φa′1 + a0Φa′0 = 0,

2a′3Φa′3 + 2a′2Φa′2 + 2a′1Φa′1 + 2a′0Φa′0 + a3Φa3 + a2Φa2 + a1Φa1 + a0Φa0 = 3Φ.(4.48)

The term 3Φ at the right hand side originates from the variation of the factorF ′m that yields the contribution mf ′ · δt. It turns out that only m = 3 allowsa nontrivial solution for Φ.

For the group generated by G = (1 + g · δt)v, F = u and H = 0, theinfinitesimal variations are δx = 0, δy = gy · δt, δy′ = (g′y + gy′) · δt. By asimilar analysis to the above the variations

δa0 = ga0, δa1 = −g′, δa2 = −ga2, δa3 = −2ga3,

δa′0 = g′a0 + ga′0, δa′1 = −g′′, δa′2 = −ga′2 − g′a2, δa3 = −2g′a3 − 2ga′3

of the coefficients and its derivatives are obtained. They yield the system

Φa′1 = 0, 2a3Φa′3 + a2Φa′2 − a0Φa′0 + Φa1 = 0,

2a′3Φa′3 + a′2Φa′2 − a′0Φa′0 + 2a3Φa3 + a2Φa2 − a0Φa0 = 3Φ

(4.49)

for the invariants. Finally the group F = u, G = 1 and H = h · δt generatesthe variations δx = 0, δy = h · δt and δy′ = h′ · δt of the variable and the firstderivative. The variations of the coefficients and its derivatives are

δa0 = ha0, δa1 = −h′, δa2 = −a2h, δa3 = −2ha3,

δa′0 = h′a0 + ha′0, δa′1 = −h′′, δa′2 = −a′2h− a2h′, δa3 = −2h′a3 − 2ha′3.

They yield the system

Φa′0 = 0, 3a3Φa′2 + a2Φa′1 + Φa0 = 0,

3a′3Φa′2 + 2a′2Φa′1 + 3a3Φa2 + 2a2Φa1 + a1Φa0 = 0.(4.50)

The full system comprising (4.48), (4.49) and (4.50) determines the desiredinvariants. It is arranged in grlex term order with ak > aj for k > j. Due tothe group properties of the transformations of x and y it is a complete system.As a consequence, a Janet basis is obtained by purely algebraic autoreductionsteps with the result

Φa0 −3a2

3Φ3

Φ = 0, Φa1 + a2a3Φ3

Φ = 0,

Φa2 −3a1a3 − 2a2

2 − 3a′33Φ3

Φ = 0, Φa3 −6a0a3 − a1a2 − a′2

Φ3Φ = 0,

Φa′0 = 0, Φa′1 = 0, Φa′2 + a3Φ3

Φ = 0, Φa′3 −a2Φ3

Φ = 0


with Φ3 given by (4.46). It is easily seen that Φ ≡ Φ3 solves this system.The higher order invariants (4.47) may be verified by differentiation and

elimination.If in a rational expression involving the invariants the powers of F (u)′ in

the numerators and the denominator are the same, F (u)′ drops from thisexpression. The same is true for G(u), i.e., a rational expression with thisproperty is an absolute invariant.

Due to the fact that the structure invariance group of Abel’s equationsin rational normal form is a subgroup of (4.41), their invariants may be ob-tained by specialization of the invariants determined in the above theorem.Of particular interest for solving equivalence problems are the lowest absoluteinvariants.

Corollary 4.3 Two absolute invariants for Abel’s equation in rationalnormal form case ii) are

K(A,B) =1A

(3B − A′

A

)3

and (4.51)

J(A,B) =1A

(3B − A′

A

)[(3B − A′

A

)′+

(5B − 2

A′

A

)(3B − A′

A

)].

(4.52)For the rational normal form in case i) J and K are not defined.

Proof For case ii), a3 = A, a2 = 0, a1 = B and a0 = 1 have to besubstituted into (4.46) with the result Φ3 = 3A2, and into (4.47) with theresult

Φ5 = 3A3(

3B − A′

A

)and

Φ7 = 3A4[(

3B − A′

A

)′+

(5B − 2A

′

A

)(3B − A′

A

)].

The ratio Φ35

Φ53

is the above invariant K(A,B) up to an irrelevant factor 9 which

is due to a different normalization. The quotient Φ5Φ7

Φ43

is J(A,B). For the

rational normal form of case i), a2 = a0 = 0 leads to Φ3 = Φ5 = 0, i.e., J andK are not defined.

The absolute invariants (4.51) and (4.52) are the decisive tool for solvingequivalence problems of case ii).

Theorem 4.16 The two rational normal forms have to be considered sep-arately.

i) Any two Abel’s equations in rational normal form of first kind

y′ +A(x)y3 +B(x)y = 0 and v′ + P (u)v3 +Q(u)v = 0

are equivalent to each other w.r.t. the group x = F (u), y = G(u)v.

184

ii) Any two equations in rational normal form of second kind

y′ +A(x)y3 +B(x)y + 1 = 0 and v′ + P (u)v3 +Q(u)v + 1 = 0

are equivalent to each other w.r.t. the group x = F (u), y = F ′(u)v ifand only if for the absolut invariants (4.51) and (4.52) there holds

K(A,B)|x=F (u) = K(P,Q) (4.53)

and

[3J(A,B)− 5K(A,B)]3

K(A,B)

∣∣∣x=F (u)

=[3J(P,Q)− 5K(P,Q)]3

K(P,Q). (4.54)

Proof In case i) the first equation is transformed by x = F (u) andy = G(u)v with the result

v′ +A(F )F ′G2v3+[B(F )F ′ +

G′

G

]v = 0

with ′ = ddu

. The conditions for equivalence are therefore

A(F )F ′G2 = P, B(F )F ′ +G′

G= Q

or

F ′ =P

A(F )G2 , G′ = QG− B(F )A(F )

P

G.

This first order system for the two undetermined functions F and G alwayshas a solution depending on two constants.

In case ii) the first equation is transformed by x = F (u) and y = F ′(u)v.This leads to

v′ +A(F )F ′3v3+[B(F )F ′ +

F ′′

F ′

]v + 1 = 0

from which the conditions for equivalence

A(F )F ′3 = P, B(F )F ′ +F ′′

F ′= Q (4.55)

follow. This is a system of ode’s for the unknown function F . It may betransformed into a Janet basis by two reductions with the result

1A

(3B − A′

A

)3∣∣∣x=F (u)

=1P

(3Q− P ′

P

)3

,


1A2/3

[(3B − A′

A

)′−1

3A′

A

(3B − A′

A

)]∣∣∣x=F (u)

=

1P 2/3

[(3Q− P ′

P

)′−1

3P ′

P

(3Q− Q′

Q

)].

Expressing these equations in terms of the invariants K and J , the conditions(4.53) and (4.54) are obtained.

Up to now the following results have been obtained. The totality of Abel’sequations (4.45) comprises the following equivalence classes w.r.t. the group(4.41). Bernoulli’s equations form a single class, all other classes are parame-trized by two rational functions, the absolute invariants (4.51) and (4.52).

Example 4.14 Consider the Abel equations y′ + xy3 + y + 1 = 0 withinvariants

K(A,B) = 27x4

(x− 1

3

)3 and J(A,B) = 45x4

(x2 − 11

15x+ 15

)(x− 1

3

),

and v′ − 1u7 v

3 − 2u+ 1u2 v + 1 = 0 with invariants

K(P,Q) = −u(u− 3)3 and J(P,Q) = −3u(u2 − 11

3 u+ 5)(u− 3)

in the notation of the above theorem. The constraints (4.53) and (4.54) leadto

u(u− 3)3F 4 + 27F 3 − 27F 2 + 9F − 1 = 0,

u5(4u− 3)3F 8 + 27F 3 − 108F 2 + 144F − 64 = 0.

The greatest common divisor of its left hand sides is uF−1; it yields the trans-formation function F = 1

u . Consequently, the two equations are equivalent,the transformation is achieved by x = 1

u and y = − vu2 .

Example 4.15 In the preceding example the first equation is modified toy′ + xy3 + 2y + 1 = 0 with the absolute invariants

K(A,B) = 216x4

(x− 1

6

)3 and J(A,B) = 360x4

(x2 − 11

30x+ 120

)(x− 1

6

).

The second equation remains unchanged. The constraints (4.53) and (4.54)lead to

u(u− 3)3F 4 + 216F 3 − 108F 2 + 18F − 1 = 0,

u5(4u− 3)3F 8 + 216F 3 − 432F 2 + 288F − 64 = 0.

The gcd of the left hand sides is 1 now. Consequently, the two equations arenot equivalent.

186

4.3 Nonlinear Equations of Second and Higher Order

For nonlinear equations of any order the results are far less complete thanfor linear or first order equations. They are basically limited to quasilinearequations of order two and three, and equations that are homogeneous in thedependent variable and its derivatives.Lie’s Equation. An important class of second order equations is

y′′ +Ay′3 +By′2 + Cy′ +D = 0 (4.56)

where A ≡ A(x, y), . . . , D ≡ D(x, y); almost all equations in Chapter 6 ofKamke’s collection are of this type. It is suggested that it is called Lie’sequation because Lie was the first to investigate it in detail in connection withhis symmetry analysis. These aspects will be discussed in the subsequentChapter 5. Most of the results in this subsection have been given for the firsttime by Roger Liouville in a series of articles between 1885 and 1900.

Theorem 4.17 The structure invariance group of (4.56) is x = F (u, v),y = G(u, v) where v ≡ v(u), i.e., the full group of point-transformations ofthe plane.

Proof From (B.7) it is obvious that the numerator of y′′ is linear in v′′

and a third order polynomial in v′, (B.6) shows that y′ is a fraction withnumerator and denominator linear in v′. Consequently, upon substitutioninto (4.56) an equation of the same structure is generated.

A special case of this result is considered in Exercise 4.5.

Theorem 4.18 (Liouville 1887) Lie’s equation (4.56) is equivalent tov′′ = 0 if and only if L1 = L2 = 0 where

L1 ≡ Dyy +BDy −ADx + (By − 2Ax)D + 13Bxx −

23Cxy + 1

3C(Bx − 2Cy),

L2 ≡ 2ADy +AyD + 13Cyy −

23Bxy +Axx − 1

3BCy + 23BBx −AxC −ACx.

This important result singles out a single equivalence class of Lie’s equa-tions. Later on in Theorem 5.11 it will turn out that the conditions L1 =L2 = 0 assure the maximal eight-parameter symmetry group for second orderequations, i.e., they determine the symmetry class S2

8 . If they are not satisfiedanother set of invariants may be obtained.

Theorem 4.19 (Liouville 1889) Lie’s equation allows a series of relativeinvariants w.r.t. point transformations which are determined by

ν5 = L1(L2L1,y − L1L2,y) + L2(L1L2,x − L2L1,x)

−AL31 +BL2

1L2 − CL1L22 +DL3

2,

νk+2 = L1νk,y − L2νk,x + kνk(L2,x − L1,y), k = 5, 7, . . .

(4.57)


If x = φ(u, v), y = ψ(u, v) and ∆ ≡ ∂uφ∂vψ − ∂vφ∂uψ, these invariants aretransformed according to νk(x, y)|x=φ,y=ψ = ∆k · νk(u, v). From these relative

invariants the series of absolute invariants tk = ν5k

νk5is obtained.

The invariants νk and tk are derived in Liouville [125], pages 19-23. Liouvilledescribes a procedure of how to decide equivalence from the absolute invariantst7 and t9. To this end, they must be defined and moreover they must befunctionally independent. It turns out that this is not true for most equationsthat occur in actual problems, e.g. for the equations listed in Chapter 6of Kamke’s collection. In these cases another series of invariants may bedetermined.

Theorem 4.20 (Liouville 1889) If L1L2 6= 0 and ν5 = 0, Lie’s equationallows the following series of invariants. At first the invariant

w1 = 1L4

1

[L31(α1L1 − α2L2) +R1(L2

1)x − L21R1,x + L1R1( 1

3CL1 −DL2)] or

w1 = 1L4

2

[L32(α1L2 − α0L1)−R2(L2

2)y + L22R2,y − L2R2(AL1 − 1

3BL2)]

is defined for L1 6= 0 or L2 6= 0 respectively. R1, R2, α0, α1 and α2 are

R1 = L1L2,x − L2L1,x + 13BL

21 − 2

3CL1L2 +DL22,

R2 = L1L2,y − L2L1,y +AL21 − 2

3BL1L2 + 13CL

22,

α0 = 13By −Ax + 2

3 (AC − 13B

2), α1 = 13 (Cy −Bx) +AD − 1

9BC,

α2 = Dy − 13Cx + 2

3 (BD − 13C

2).

From w1 the series of relative invariants

wk+2 = L1wk,y − L2wk,x + k(L2,x − L1,y)wk

is obtained from which the absolute invariants uk = wkwk1

for k = 3, 5, . . . follow

if w1 6= 0.

The invariants wk and uk are also derived in Liouville [125], pages 38-42.Finally, there is a third series of invariants if in addition there holds w1 = 0.

Theorem 4.21 (Liouville 1889) If L1 6= 0, ν5 = 0 and w1 = 0, Lie’sequation allows the absolute invariant

i2 =3R1

L1+ L2,x − L1,y

and the series of relative invariants

i2k+2 = L1i2k,y − L2i2k,x + 2ki2k(L2,x − L1,y), k ≥ 1.

188

They yield the series of absolute invariants j2k = i2kik2

if i2 6= 0.

These invariants may be found in Liouville [125], page 50. The case ν5 = 0,w1 = 0 deserves special attention because any Lie equation with this propertyis equivalent to a generalized Emden-Fowler equation.

Theorem 4.22 (Babich and Bordag 1997) Any Lie equation (4.56) withν5 = w1 = 0 is equivalent to an equation of the form v′′ + r(u, v) = 0. If inaddition L2 = 0, the transformation to this form is determined by u = σ(x, y),v = ρ(x, y) with

σ(x, y) =∫

exp(

2∫Bdy −

∫Cdx

)dx and ρ(x, y) =

∫exp

( ∫Bdy

)dy.

The proof may found in the article by Babich [8]. They discuss in detailthe application of this result to the six types of Painleve’s equations.

Example 4.16 Painleve’s equation III has the form

y′′ =y′2

y− y′

x+αy2 + β

x+ γy3 +

δ

y.

With B = −1y and C = 1

x one obtains σ = log x, ρ = log y and the canonicalform

v′′ = αeu+v + βeu−v + γe2(u+v) + δe2(u−v).

Example 4.17 Equation 6.159 in Kamke’s collection y′′− 34y y

′2−3y2 = 0

fullfills the assumptions of the above theorem. Its coeffients B = − 34y and

C = 0 yield σ = x, ρ = 1256y

4 and the canonical form v′′ − 31024v

5 = 0. Itshould be compared with Lie’s canonical form obtained in Example 6.9.

The Emden-Fowler canonical form obtained according to the above theoremis not unique as the next result shows.

Theorem 4.23 (Babich and Bordag 1997) The structure invariance groupof the Emden-Fowler equation y′′ + r(x, y) = 0 is

x = c1

∫a(x)2dx+ c2, y = a(x)y + b(x)

where a and b are undetermined functions of its argument, c1 and c2 areconstants, and y depends on x.

Quasilinear Equations of Second Order. In his award-winning articlefor the Furstlich Jablonowski’sche Gesellschaft Tresse [181] (see also [180])applied Lie’s theory of differential invariants to the equivalence problem ofsecond order quasilinear ode’s of the form y′′ = ω(x, y, z) with z ≡ y′. In order


to explain his results some notation has to be introduced first. If ψ ≡ ψ(x, y, z)is any function of x, y and z, define

ψ10 =∂ψ

∂x+ z

∂ψ

∂y, ψ01 =

∂ψ

∂yand ψk =

∂kψ

∂zk.

Due to the commutativity of the first two differential operators, the definitionψkij ≡ (ψk)ij is meaningful. To any such symbol the two weights r = i− k+2and s = j + k − 1 are assigned. Furthermore, three differential parameters∆x, ∆y and ∆z are defined in terms of their action on any invariant ψ withweights r and s. They return an invariant the order of which is increased byone. For ω 6= 0 they are defined as follows.

∆xψ = ψ10 + ω∆zψ+[

(3r + 2s)(ω1 +

3ωω5

5ω4

)+(2r + s)

ω410

ω4

]ψ

returns an invariant with weights r + 1 and s,

∆yψ = ψ01 +ω5

5ω4∆xψ +

ω410 + ωω5 + 2ω1ω4

ω4∆zψ

−

(3r + 2s)[ ω2

8+

320ω5(ω4

10 + ωω5 + 2ω1ω4)ω4ω4

]+r + 2s

4ω4

01

ω4

ψ

returns an invariant with weights r and s+ 1 and

∆zψ = ψ1 + (r − s) ω5

5ω4ψ

returns an invariant with weights r− 1 and s+ 1. Applying this notation theinvariants of y′′ = ω may be described as follows.

Theorem 4.24 (Tresse 1896) A quasilinear ode y′′ = ω(x, y, z) with z ≡ y′and ∂4ω/∂z4 6= 0 has the following relative invariants w.r.t. point transfor-mations x = σ(x, y) and y = ρ(x, y).

i) Two invariants of order four: ω4 with r = −2, s = 3 and

H ≡ ω220 − 4ω1

11 + 6ω02 − ω(3ω201 − 2ω3

10) + ω1(4ω101 − ω2

10)

−3ω2ω01 + ω3ω10 + ωωω4 with r = 2, s = 1.

ii) Three invariants of order five:

H10 ≡ ∆xH with r = 3, s = 1, H01 ≡ ∆yH with r = s = 2,

K ≡ ∆zH with r = 1, s = 2.

iii) Eleven invariants of order six:

Ω6 ≡ ω6 − 65ω5ω5

ω4with r = −4, s = 5,

190

Ω510 ≡

5ω4

24H(∆zxH −∆xzH −∆yH) with r = −2, s = 4,

Ω501 = 4

9 (∆zΩ510 −∆xΩ6) with r = −3, s = 5,

Ω420 = ∆zzH −

Ω6

5ω4H with r = 0, s = 3,

Ω411 = 4

3 (∆zΩ420 −∆xΩ5

10) with r = −1, s = 4,

Ω402 ≡

45

(∆yΩ5

10 −∆xΩ501 +

Ω6Ω420

ω4+

Ω510Ω

510

5ω4

)with r = −2, s = 5,

∆xzH with r = s = 2, ∆yzH with r = 1, s = 3,

∆xxH with r = 4, s = 1,

∆xyH with r = 3, s = 2, ∆yyH with r = 2, s = 3.

Any of these invariants I transforms according to

I

I=

(σx + y′σy)s−r

(σxρy − σyρx)s.

The proof involves rather lengthy calculations. They may be found in theabove quoted article by Tresse. By inspection of the invariants it is seen thatall invariants may be obtained by application of the differential parameters toω4, H, Ω6 and Ω5

10.The transformation law for the invariants given in Theorem 4.24 implies

that absolute invariants may be formed from suitable powers of three properlychosen relative invariants. If ω4H 6= 0 and ψ is any relative invariant withweights r and s, an absolute invariant

ψ =ψ

(ω4)α(H)β

is obtained if α = 18 (2s − r) and β = 1

8 (2s + 3r) are chosen. In order toobtain rational expressions, they are raised to the power of the least commonmultiple of the denominators of α and β. For the invariants up to order sixthe complete answer is given next. Absolute invariants are denoted by a haton top of the corresponding symbol for the relative invariant.

Corollary 4.4 With the same notation as in Theorem 4.24 and assumingω4H 6= 0, the absolute invariants for y′′ = ω(x, y, z) of order 5 are

K =K8

(ω4)3(H)7, H10 =

(H10)8ω4

(H)11, H01 =

(H01)4

ω4(H)5.


The invariants of order 6 are

Ω6 =(Ω6)4H(ω4)7

, Ω510 =

(Ω510)

4

(ω4)5H, Ω5

01 =(Ω5

01)8

(ω4)13H,

Ω420 =

(Ω420)

4

(ω4)3(H)3, Ω4

11 =(Ω4

11)8

(ω4)9(H)5, Ω4

02 =(Ω4

02)2

(ω4)3H,

(∆xzH)4

ω4(H)5,

(∆yzH)8

(ω4)5(H)9,

(∆xxH)4ω4

(H)7,

(∆xyH)8

ω4(H13),

(∆yyH)2

ω4(H3).

Some of the results just described will be applied in Chapter 5.3 for deter-mining symmetry classes of second order ode’s.

Exercises

Exercise 4.1 Let a linear homogeneous ode be given in the form

p0y(n) + p1y

(n−1) + . . .+ pn−1y′ + pny = 0

with pk ≡ pk(x) polynomials in x for all k. Find a transformation thatgenerates a new equation of the same type from it such that the constraintp1 = −p′0 is valid. Discuss the result (Hirsch [72]).

Exercise 4.2 Determine a transformation from the equation v′ = 0 to agiven equation y′ + r(x, y) = 0. Compare the result with Theorem 4.10.

Exercise 4.3 Derive the rational normal form obtained by choosingF (u) = u in (4.37).

Exercise 4.4 Determine the structure invariance group of an Abel equa-tion in Appel’s normal form y′ + y3 + r(x) = 0.

Exercise 4.5 How does Lie’s equation (4.56) change if the dependent andthe independent variable are exchanged by each other?

Exercise 4.6 Show that the three equations y′′ + cy′ + y = 0 andy′′ + 1

2 (3± i√

3)cy′ + y = 0 are pairwise equivalent to each other.

Exercise 4.7 Show that the six equations y′′ − (a + 1)y′′ + ay′ = 0 andy′′ − (a + 1)y′′ + ay′ = 0 with a = 1

a , a = 1 − a, a = 11− a , a = 1 − 1

a and

a = 1 + 1a− 1 are pairwise equivalent to each other.

Exercise 4.8 Apply the algorithm LieSolve3.6 described on page 344to equation (4.15). Why is it not effective for the particular application ofTheorem 4.8 ?

192

Exercise 4.9 Design an algorithm that accepts a linear second order odeas input and returns a rational transformation to Weber’s equation (4.22) ifthere exists one or failed otherwise.

Exercise 4.10 The same problem for the Bessel equation (4.16).

Chapter 5

Symmetries of Differential Equations

This chapter deals with the most important topic of this book, i.e., the non-trivial symmetries that a given ordinary differential equation (ode) may admit.Let two coordinate sets of the plane be defined with the additional assump-tion that in either of them one coordinate variable is dependent on the other.Roughly speaking a symmetry of an ode is a diffeomorphism connecting thesecoordinates for which this ode is an invariant. In the literature on symmetriesthese transformations are often called point transformations or variable trans-formations in order to distinguish them from more general transformationsalso involving the first derivative. This notation will frequently be used fromnow on whenever symmetries of an ode are the main topic.

It is obvious that the entirety of symmetries of any given ode forms agroup. The term symmetry group of a differential equation is applied to thelargest group of transformations sharing this property. The Lie algebra of itsinfinitesimal generators forms the corresponding symmetry algebra. If a vari-able transformation is applied to a given differential equation, the symmetrygroup of the transformed equation is similar to the original one according toDefinition 3.2. The equivalence class to which the symmetry group of a par-ticular ode belongs is called its symmetry type. Consequently, all equationscontained in an equivalence class have the same symmetry type. The reverseis not true. As a consequence, the entirety of all differential equations allowingthe same type of symmetry group is the union of equivalence classes. Krauseand Michel [96] called it the stratum of ode’s corresponding to a symmetrytype. In this book it will be called the symmetry class.

The symmetries of a differential equation occur in different connections. Onthe one hand, there is the classification problem. Its aim is to determine allpossible symmetry types for a family of ode’s, e. g. ode’s of a fixed order. Thestarting point for this approach is the listing of groups given in Section 3.4;its differential invariants determine the general form of an ode that may beinvariant under the respective group. On the other hand, if any particularode is given, its symmetry type has to be determined if it is to be applied forfinding its solutions.

The symmetry problem for equations of order one is significantly differentfrom that of equations of order two or higher. The main difference is thefact that for first order equations in general there is no algorithm availablefor determining any symmetry generator; only heuristics or insight into the

193

194

problem from which the equation originates may allow finding them. Whatmakes the problem especially difficult is the fact that any first order odeallows infinitely many symmetries as discussed at the beginning of Section 5.2.Contrary to this, for equations of order two or higher, the symmetry type mayalways be identified algorithmically, and there is a well-defined procedure fortransforming it to a canonical form. The key for this is the Janet basis forthe determining system and the theorems derived in Section 3.5.

In the subsequent Section 5.1, general properties of the behavior of differen-tial equations under a change of variables are discussed and various conceptsrelated to its symmetries are defined. Section 5.2 discusses the symmetrystructure of first order equations. The most important field of application forLie himself was equations of second order; they are the subject of Section 5.3.In Section 5.4 a complete discussion of equations of order three is given follow-ing a similar approach to the second order equations. Finally in Section 5.5the symmetries of linear equations of any order are discussed.

5.1 Transformation of Differential Equations

Introducing new variables into a given ode has been a popular method inorder to faciliate the solution procedure all the time. Usually this is donein an ad hoc manner without guaranteed success. In particular there is nocriterion for deciding whether a certain class of transformations will lead to anintegrable equation or not. A critical examination of these methods was thestarting point for Lie’s symmetry analysis. Therefore the behavior of ode’sunder various kinds of transformations will be investigated first.Properties of Point Transformations. Let an ode of order n in the inde-pendent variable x and the dependent variable y ≡ y(x) be given as

ω(x, y, y′, . . . , y(n)) = 0. (5.1)

If not stated explicitly, it will be assumed that ω is a polynomial in thederivatives with coefficients in some base field which is usually the field ofrational functions in x and y, i.e., ω ∈ Q(x, y)[y′, . . . , y(n)]. Any other fieldthat occurs later on during the solution procedure is an extension of this basefield.

A point-transformation between two planes with coordinates (x, y) and(u, v), and dependencies y ≡ y(x) and v ≡ v(u) respectively, is consideredin the form

u = σ(x, y), v = ρ(x, y). (5.2)

The term point-transformation expresses the fact that only the coordinatesoccur as arguments of the transformation functions. If, in addition, the first

Symmetries of Differential Equations 195

derivative is transformed by an independent function of x, y and y′ of a certainform it is called a contact-transformation. They are not dicussed in this book.In general equations (5.2) define a diffeomorphism between the two coordinateplanes. Depending on the particular situation, the function field in which σand ρ are contained has to be specified.

Let a curve in the x− y-plane described by y = f(x) be transformed under

(5.2) into v = g(u). There arises the question of how the derivative y′ = dfdx

corresponds to v′ = dgdu

under this transformation. A simple calculation leadsto the first prolongation

v′ =dv

du=ρx + ρyy

′

σx + σyy′ ≡ χ1(x, y, y′). (5.3)

It is remarkable that the knowledge of (x, y, y′) and the equations of the pointtransformation (5.2) are sufficient for computing the derivative v′, knowingthe equation of the curve is not required. This is also expressed by sayingthat the line element (x, y, y′) is transformed into the line element (u, v, v′)under the action of a point transformation. Similarly the transformation lawfor derivatives of second order is obtained as

v′′ =dv′

du=χ1,x + χ1,yy

′ + χ1,y′y′′

σx + σyy′ ≡ χ2(x, y, y′, y′′). (5.4)

Explicitly in terms of σ and ρ it is

v′′ = 1(σx + σyy

′)3(σxρy − σyρx)y′′ + (σyρyy − σyyρy)y′3

+[σxρyy − σyyρx + 2(σyρxy − σxyρy)]y′2

+[σyρxx − σxxρy + 2(σxρxy − σxyρx)]y′ + σxρxx − σxxρx.

(5.5)

In general there holds

v(n) =dv(n−1)

du≡ χn(x, y, y′, . . . , y(n)) (5.6)

and

v(n+1) =dv(n)

du=χn,x + χn,yy

′ + . . .+ χn,y(n−1)y(n)

σx + σyy′ .

The form of a differential equation is extremely sensitive to a variablechange. The following definition introduces a particular term for the excep-tional cases where this is not true.

Definition 5.1 (Symmetry of an ode) Equation (5.1) is said to be invari-ant under the transformation

x = φ(u, v), y = ψ(u, v) with v ≡ v(u) (5.7)

196

if the functional dependence on u, v, v′, . . . , v(n) of the transformed equation isthe same as that of the original equation (5.1) on x, y, y′, . . . , y(n), i.e., if it isa differential invariant for (5.7). Such a transformation is called a symmetryof the differential equation.

The general solution of equation (5.1) is a family of curves

φ(x, y, C1, . . . , Cn) = 0 (5.8)

in the x−y-plane depending on n constant parameters C1, . . . , Cn. The trans-formation (5.7) acts on the curves (5.8) just as on the differential equation.If it is a symmetry, the functional dependence of the transformed curves on uand v is the same as the dependence on x and y in (5.8). This is not necessarilytrue for the parameters because they do not occur in the differential equationitself. This means, the entirety of curves described by the two equations is thesame; to any fixed values of the constants, however, there may correspond adifferent curve in either set. In other words, the solution curves are permutedamong themselves by a symmetry transformation.

Example 5.1 Consider the ode 2yy′′ − y′2 = 0 with the general solutiony = (C1x + C2)2. Its symmetry group is x = a1u

1− a2u+ a3, y = a1v

(1− a2u)2where a1, a2 and a3 are the group parameters. In the transformed variablesthe solution is v = (C1u+ C2)2 with

C1 = 1√a1

[C1(a1 − a2a3)− C2a2], C2 = a3 + 1√a1

C2.

From the above Definition 5.1 of a symmetry of an ode and the geometricconsiderations on its solution curves it is obvious that all symmetry transfor-mations of a given ode form a group. This important concept is introducednext.

Definition 5.2 (Symmetry group, symmetry type) The totality of sym-metry transformations of an ode forms a continuous group; it is called thesymmetry group of that equation. The type of this group is called its symme-try type.

At this point the connection to Chapter 3 of this book becomes obvious.All that has been said about groups of transformations of a two-dimensionalmanifold applies to the symmetry groups of ode’s. In particular this applies tothe classification of these groups and the resulting constraints for the possiblesymmetries of a differential equation. In order to establish a systematic theoryof symmetries of ode’s, it must be known how they behave under variabletransformations. The answer is given in the following theorem.

Theorem 5.1 (Lie 1891) Equivalent equations have the same symmetrytype. The reverse is not true. In general to any symmetry type, there corre-sponds the union of one or more equivalence classes.


Proof Let ω(x, y, y′, . . . , y(n)) = 0 be a differential equation in x andy ≡ y(x), and ω(x, y, y′, . . . , y(n)) = 0 the transformed equation under thevariable change Φ : x = α(x, y), y = β(x, y) with y ≡ y(x) and inverse Φ−1.Furthermore, let g : x = φ(u, v), y = ψ(u, v) with v ≡ v(u) be a symmetrytransformation of ω with the result ω(u, v, v′, . . . , v(n)) = 0. Then g ≡ Φ−1gΦtransforms ω(x, y, y′, . . . , y(n)) = 0 into ω(u, v, v′, . . . , v(n)) = 0, i.e., it definesa symmetry of ω of the same type as g.

On the other hand, by Theorems 6.4 and 6.16, the equations y′′+y′4+1 = 0and y′′′+ y′′y′+ 1 both belong to symmetry classes S2

2,1 and S32,1 respectively

with the same symmetry type, yet due to its different order they are obviouslynot equivalent to each other.

In order to utilize the symmetries of an ode for the solution procedure, thegeometrical considerations above must be expressed in analytical terms. Ofparticular importance is the fact that this may be achieved in terms of theinfinitesimal generators of a symmetry, the expressions for its finite transfor-mations are not required. Due to its importance for the rest of this bookthe prolongation of a vector field in x and y(x) is introduced next. It is aspecialization of Definition 3.4 in Chapter 3.

Definition 5.3 (Prolongation in the x-y-plane) Let an infinitesimal gen-erator U = ξ(x, y)∂x + η(x, y)∂y be given and y ≡ y(x) depend on x. Its nthprolongation is

U (n) = U + ζ(1) ∂

∂y′+ ζ(2) ∂

∂y′′+ . . .+ ζ(n) ∂

∂y(n). (5.9)

The functions ζ(k) are recursively defined by

ζ(1) = D(η)− y′D(ξ), ζ(k) = D(ζ(k−1))− y(k)D(ξ) (5.10)

for k ≥ 2, D = ∂x + y′∂y + y′′∂y′ . . . is the operator of total differentiationwith respect to x.

From (5.9) it is obvious that the prolonged generators operate on the spaceof derivatives up to order n. A few remarkable properties of the functions ζ(k)

are formulated subsequently as a lemma. They follow immediately from itsdefinition and are quite useful for practical calculations.

Lemma 5.1 The functions ζ(k) defined by (5.10) have the following prop-erties.

1. They are linear and homogeneous in ξ(x, y), η(x, y) and its derivativesup to order k.

2. Its coefficients depend explicitly on y′, y′′, . . . , y(k); for k > 1, y(k) oc-curs linearly and y′ with power k+1. In particular they do not dependexplicitly on x or y.

198

From this lemma it is obvious that the amount of computations growsenormously with the order of the differential equation due to the increase inthe number of terms in ζ(k). A rough estimate for this number is 2k. Fork = 1, . . . , 10 the exact figure is given in the subsequent table.

k 1 2 3 4 5 6 7 8 9 10Number of terms 4 9 17 29 47 73 110 162 234 332

The three lowest ζ ′s are

ζ(1) = ηx + (ηy − ξx)y′ − ξyy′2,

ζ(2) = ηxx + (2ηxy − ξxx)y′ + (ηyy − 2ξxy)y′2 − ξyyy

′3

+(ηy − 2ξx)y′′ − 3ξyy′y′′,

ζ(3) = ηxxx + (3ηxxy − ξxxx)y′ + 3(ηxyy − ξxxy)y′2 + (ηyyy − 3ξxyy)y

′3

−ξyyyy′4 + 3(ηxy − ξxx)y′′ + 3(ηyy − 3ξxy)y′y′′ − 6ξyyy

′2y′′

+(ηy − 3ξx)y′′′ − 4ξyy′y′′′ − 3ξyy′′2.(5.11)

It is convenient to have the prolongations for some simple vector fields explic-itly available. For linear coefficients ξ = ax, η = by or ξ = ay, η = bx, a andb constant, the third prolongations are

U (3) = ax∂x + by∂y + (b− a)y′∂y′ + (b− 2a)y′′∂y′′ + (b− 3a)y′′′∂y′′′

U (3) = ay∂x + bx∂y + (b− ay′2)∂y′ − 3ay′y′′∂y′′ − a(4y′y′′′ + 3y′′2)∂y′′′ .

For ξ = x2, η = y2 it is

U (3) = x2∂x + y2∂y + 2(y − x)y′∂y′ + 2[y′(y′ − 1) + (y − 2x)y′′]∂y′′

+2[3y′′(y′ − 1) + (y − 3x)y′′′]∂y′′′ .

For later use a relation between applying the prolongation operator andtaking the total derivative is given next.

Lemma 5.2 The prolongation (5.9) of U = ξ∂x + η∂y and the operatorof total differentiation D obey U (n+1)D(φ) = D(U (n)φ)−D(ξ)D(φ); φ is anundetermined function of x, y(x) and its derivatives up to any order.

The proof is based on expanding the left and the right hand side of thisrelation; it is left as Exercise 5.1.

The next result which is due to Lie [113], Kapitel 16, is fundamental forthe further proceeding because it is the basis of a constructive method fordetermining the symmetries of large classes of ode’s. It is an immediate con-sequence of Definition 5.2.


Theorem 5.2 (Lie 1891) A differential equation ω(x, y, y′, . . . , y(n)) = 0with n ≥ 1 is invariant under a transformation with infinitesimal generatorU = ξ(x, y)∂x + η(x, y)∂y if and only if

U (n)ω(x, y, y′, . . . , y(n)) = 0 mod (ω = 0) (5.12)

in the space of variables x, y, y′, . . . , y(n).

There is a remarkable constraint for the number of possible symmetries ofan ode if the order is not less than two which is also due to Lie. In additionto the theoretical interest in this result it has the important consequence thatthe symmetry classification becomes a finite problem.

Theorem 5.3 (Lie 1893) The maximal number of symmetry generators ofan ode of order n ≥ 2 is finite. For n = 2 it is 8, for n ≥ 3 it is n+ 4.

The proof which is based some geometric considerations may be found inLie [114], Kapitel 12, Satz 3.

For reasons that will become clear shortly the further discussion is limitedto n ≥ 2. The case n = 1 will be considered in Section 5.2. In order tomake the symmetry problem for ode’s manageable the functional dependenceof ω on its arguments is restricted as follows. If not stated otherwise it isassumed to be quasilinear, i.e., linear in the highest derivative, and rationalin its remaining arguments. Consequently, equations of the form

y(n) + r(x, y, y′, . . . , y(n−1)) = 0 (5.13)

are considered with r ∈ Q(x, y, y′, . . . , y(n−1)). This class covers a wide rangeof interesting equations as it may be seen from the respective chapters ofthe collection by Kamke [85]. The quasilinearity and the rationality in thederivatives have another important property, they are invariant under general,i.e., not necessarily rational point transformations as it may be seen from eqs.(5.3) to (5.4).

Based on Theorem 5.2 a procedure for determining the infinitesimal sym-metry generators of an ode (5.13) may be designed as follows. Due to theassumptions on the form of ω, the expression at the left hand side of (5.12) isrational in the derivatives with coefficients depending on x and y, the unknownfunctions ξ(x, y) and η(x, y) and derivatives thereof. Because the dependenceon the derivatives y(k) is completely explicit, it can only vanish if each coeffi-cient of the monomials in the derivatives occuring in the numerator vanishes.As a consequence of Lemma 5.1, the resulting constraints form a system of lin-ear homogeneous partial differential equations for ξ and η in which derivativesy(k) do not occur. It is called the determining system. Its general solutiondefines the symmetries of the original differential equation. The algorithmsdescribed in Chapter 2 are applied in the following algorithm for determininglarge classes of symmetry generators.

200

Algorithm 5.1 Symmetries(ω). Given a quasilinear ode ω of ordern ≥ 2, its infinitesimal symmetry generators with Liouvillian coefficients arereturned.S1 : Set up determining system. Reduce the n−th prolongation of ωw.r.t. ω, collect the coefficients of the monomials in the derivatives andequate them to zero.S2 : Janet basis. By means of the algorithm given in Chapter 2 transformthe system obtained in S1 into a Janet basis.S3 : Determine coefficients. Decompose the Janet basis obtained in S2applying the algorithms of Chapter 2 and determine the Liouvillian so-lutions from it. If there is no nontrivial solution, return failed.S4 : Return vector fields. From the solutions obtained in S3 constructthe infinitesimal generators and return them.

This algorithm will be applied frequently in later parts of this chapter, bothfor determining the symmetry generators explicitly, or the symmetry classto which a given equation belongs. It should be emphasized, however, thatsymmetry generators with non Liouvillian coefficients in general cannot beobtained from it.

Example 5.2 Equation 6.159 of Kamke’s collection 4y′′y−3y′2−12y3 = 0generates the determining system

ξyy + 34y ξy = 0, ηxx + 3y2ηy − 6y2ξx − 6yη = 0,

ηxy − 12ξxx −

34y ηx −

92y

2ξy = 0, ηyy − 2ξxy − 34y ηy + 3

4y2 η = 0.

This is the result of step S1 of the above algorithm. The Janet basis

ξx +12yη = 0, ξy = 0, ηx = 0, ηy −

1yη = 0

is obtained in step S2 from which the solution ξ = − 12C1x+ C2, η = C1y in

step S3 follows. Finally in step S4 the two symmetry generators U1 = ∂x andU2 = x∂x− 2y∂y are obtained. Its commutator [U1, U2] = U1 shows that theyare canonical generators of the group g25.

Classification of Differential Invariants. In order to obtain a classifi-cation of possible symmetry types of ode’s, the proper differential invariantsof all finite transformation groups of a two-dimensional manifold have to bedetermined. The starting point for this classification is the listing of transfor-mation groups as described in Chapter 3. Coordinates are x and y, y dependson x. For an r-parameter group with infinitesimal generators ξip + ηiq fori = 1, . . . , r, the differential invariants of order m are solutions of the linear


homogeneous system

ξ1∂Φ∂x

+ η1∂Φ∂y

+ ζ(1)1

∂Φ∂y′

+ . . .+ ζ(m)1

∂Φ∂y(m) = 0,

ξ2∂Φ∂x

+ η2∂Φ∂y

+ ζ(1)2

∂Φ∂y′

+ . . .+ ζ(m)2

∂Φ∂y(m) = 0,

......

ξr∂Φ∂x

+ ηr∂Φ∂y

+ ζ(1)r

∂Φ∂y′

+ . . .+ ζ(m)r

∂Φ∂y(m) = 0.

(5.14)

In this subsection, all systems of pde’s are represented in grlex term orderwith x > y > y′ > y′′ > y′′′. The group property guarantees that this is acomplete system for Φ with m + 2 − r solutions for 1 ≤ r ≤ m + 2. Due toLemma 5.1 their dependencies may be chosen such that

Φ1 ≡ Φ1(x, y, y′, . . . , y(r−1)),Φ2 ≡ Φ2(x, y, y′, . . . , y(r)),

......

Φm−r+2 ≡ Φm−r+2(x, y, y′, . . . , y(m)).

The two lowest invariants Φ1 and Φ2 are called fundamental invariants.They are determined by the methods described in Section 3.2 for solvingJacobian systems. The Jacobian scheme breaks down if the determinant ofthe coefficient matrix of the first r − 2 rows vanishes. Correspondingly thelower equations of the group are obtained from the irreducible factors of theLie determinant

∆ =

∣∣∣∣∣∣∣∣∣∣∣

ξ1 η1 ζ(1)1 . . . ζ

(r−2)1

ξ2 η2 ζ(1)2 . . . ζ

(r−2)2

......

......

ξr ηr ζ(1)r . . . ζ

(r−2)r

∣∣∣∣∣∣∣∣∣∣∣. (5.15)

Its irreducible factors may determine additional invariants. According to Lie[109], part I, page 247, the higher invariants may be obtained by differentiationas it is shown next. Lie gave some intuitive geometric arguments in favor ofthis result; see also the remarks by Krause and Michel [96] on page 254,footnote 4. This important result is proved next.

Theorem 5.4 The higher invariants Φj for j ≥ 3 may be obtained by

differentiation Φj+1 = dΦjdΦ1

.

Proof The above relation may be written as Φj+1 = DΦjDΦ1

because by

definition there holds dΦjdx

= DΦj and dΦ1dx

= DΦ1. Let Φj be an invariant oforder n. If Φj+1 is an invariant of order n+1, it has to obey U (n+1)Φj+1 = 0.

202

The numerator of the left hand side is

D(Φ1)U (n+1)D(Φj)−D(Φj)U (n+1)D(Φ1) =

D(Φ1)[D(U (n)Φj)−D(ξ)D(Φj)]−D(Φj)[D(U (n)Φ1) +D(ξ)D(φ1)] = 0.

In the last line Lemma 5.2 has been applied. Due to the assumption that Φ1

and Φj are invariants there holds U (n)Φ1 = 0 and U (n)Φj = 0. The remainingterms cancel each other.

Example 5.3 Extending the three generators of the group g13 definedon page 139 up to third order yields the following system of pde’s for anyinvariant Φ.

Φx = 0, 2xΦx + yΦy − y′Φy′ − 3y′′Φy′′ − 5y′′′Φy′′′ = 0,

x2Φx + xyΦy − (y′x− y)Φy′ − 3y′′xΦy′′ − (5y′′′x+ 3y′′)Φy′′′ = 0

with ∆ = y2. Applying the algorithm JanetBasis of the previous chaptergenerates the Jacobian form

X1 ≡ Φx = 0, X2 ≡ Φy′ − 3y′′y Φy′′′ = 0,

X3 ≡ Φy − 3y′′y Φy′′ − 5y′′′y + 3y′′y′

y2 Φy′′′ = 0.

The first equation implies that the invariants do not depend explicitly on x.In the last equation, the third derivative y′′′ does not occur in the coefficientsof Φy and Φy′′ . Therefore a solution independent of y′′′ may be obtained

from 3dyy + dy′′

y′′= 0 with the result Φ1 = y′′y3. Because also X1Φ1 = 0 and

X2Φ1 = 0, Φ1 is the first invariant searched for.

For the second equation a solution φ is obtained from 3dy′

y + dy′′′

y′′= 0

in the form φ = y′′′y + 3y′′y′. Because X3φ = −4φy , the extension of φ toa solution of the full system must have the form Φ(φ, y) and satisfies the

equation ∂Φ∂φ

4φy + ∂Φ

∂y= 0 or equivalently dφ

φ+ 4dyy = 0, i.e., φy4 = C.

Therefore Φ = φy4 is a solution of the complete system, the second invariantis Φ2 = y′′′y5 + 3y′′y′y4.

Example 5.4 Consider the group g25 with generators ∂y and x∂x + y∂y.The pde’s for invariants up to order n are xΦx−

∑nk=2(k−1)y(k)Φy(k) = 0 and

Φy = 0, with invariants Φ1 = y′, Φk = xk−1y(k) for k = 2, . . . , n. Applying

Theorem 5.4 yields dΦ2dΦ1

= 1 + Φ3Φ2

and dΦ3dΦ1

= 2Φ3 + Φ4Φ2

whereas the above

formula leads to Φ2 = xy′′, Φ3 = x2y′′′ and Φ4 = x3y(4). It shows that theinvariants obtained by differentiation may be not in simplest form.


The listing below contains the complete information on the possible groupsof point symmetries of ordinary differential equations of arbitrary order. It isdue to Lie [109], part I. Some minor corrections have been included that maybe found in the Anmerkungen by Engel in vol. V of the Gesammelte Abhand-lungen, pages 676-679. Furthermore, various corrections and improvementsdescribed in the dissertation of Heineck [69] have been added. In general thenecessary calculations proceed similarly to the above examples.

In the subsequent tabulation the fundamental invariants Φ(k)1 and Φ(k)

2 ,and the Lie determinants ∆(k) for the groups gk, k = 1, . . . , 27, are explicitlygiven. Different from Lie’s listing, all invariants are rational. This faciliatesthe explicit calculations that will be performed later on. In various cases theSchwarzian derivative of y is denoted by the abbreviation

w = y′′′

y′− 3

2

(y′′

y′

)2

. (5.16)

Group g1: Φ(1)1 = [3y′′y(4) − 5y′′′2]3

y′′8, Φ(1)

2 = 9y′′2y(5) − 45y′′y′′′y(4) + 40y′′′3

3y′′4,

∆(1) = 9y′′3.

Group g2: Φ(2)1 = [9y′′2y(5) − 45y′′y′′′y(4) + 40y′′′3]2

3(3y′′y(4) − 5y′′′2)3,

Φ(2)2 = 9y′′3y(6) − 63y′′2y′′′y(5) + 105y′′y′′′2y(4) − 35y′′′4

3(3y′′y(4) − 5y′′′2)2,

∆(2) = 2y′′2(5y′′′2 − 3y′′y(4)).

Group g3: Φ(3)1 = u3

ρ85

, Φ(3)2 = v

ρ35

where

ρ4 = 3y′′y(4) − 4y′′′3, ρ5 = 3y′′2y(5) − 15y′′y′′′y(4) + 403 y

′′′3,

ρ6 = 3y′′2y(6) − 24y′′2y′′′y(5) + 60y′′y′′′2y(4) − 40y′′′4,

ρ7 = 9y′′4y(7) − 105y′′3y′′′y(6) + 420y′′2y′′′2y(5) − 7000y′′y′′′3y(4) + 11203 y′′′5,

ρ8 = 27y′′5y(8) − 48y′′′ρ7 − 840y′′′2ρ6 − 2240y′′′3ρ5 − 2800y′′′4ρ4 − 22403 y′′′6,

u = 2ρ5ρ7 − 35ρ4ρ25 − 7(ρ6 − 5

3ρ24)

2,

v = ρ5(ρ8 − 84ρ4ρ6 + 2453 ρ3

4)− 12(ρ7 − 352 ρ4ρ5)(ρ6 − 5

3ρ24) + 28

ρ5(ρ6 − 5

3ρ24)

3,

∆(3) = −2y′′(9y′′2y(5) − 45y′′y′′′y(4) + 40y′′′3)2.

Group g4: Φ(4)1 = x, Φ(4)

2 = y′′

y′, Φ(4)

3 = y′′′

y′, ∆(4) = 0.

Group g5: Φ(5)1 = y′′

y′, Φ(5)

2 = y′′′

y′, ∆(5) = y′.

Group g6: Φ(6)1 = y′y′′′

y′′2, Φ(6)

2 = y′2y(4)

y′′3, ∆(6) = −y′y′′.

204

Group g7: Φ(7)1 = y′′y′(2−c)(c−1), Φ(7)

2 = y′′′y′(3−c)(c−1),∆(7) = (c− 1)y′, c 6= 1.

Group g8: Φ(8)1 = x, Φ(8)

2 = w, ∆(8) = 0.

Group g9: Φ(9)1 = w, Φ(9)

2 = w′, ∆(9) = 2y′3.

Group g10: Φ(10)1 = [(x− y)y′′ + 2y′(y′ + 1)]2

y′3,

Φ(10)2 = (x− y)2y′′′ + 6(x− y)(y′ + 1)y′′

y′2+ 6(y′2 + 4y′ + 1)

y′,

∆(10) = 2(y − x)2y′.

Group g11: Φ(11)1 = w′2

w3 , Φ(11)2 = w′′

w2 , ∆(11) = 4y′2(y′y′′′ − 32y′′2).

Group g12: Φ(12)1 = 4ww′′ − 5w′2

w3 , Φ(12)2 = (4w2w′′′ − 18ww′w′′ + 15w′3)2

w9 ,

∆(12) = 4y′2(y′y′′′ − 32y′′2).

Group g13: Φ(13)1 = y′′y3, Φ(13)

2 = y′′′y5 + 3y′′y′y4, ∆(13) = 2y2.

Group g14: Φ(14)1 = (yy′′′ + 3y′y′′)2

yy′′3, Φ(14)

2 = 3yy′′y(4) − 4yy′′′2

y′′3,

∆(14) = 2y2y′′.

Group g15: Let the determinants D and Di, for i = 1, 2, . . . be defined by

D =

∣∣∣∣∣∣∣∣∣∣φ1 φ

′1 . . . φ

(r)1

......

...φr φ

′r . . . φ

(r)r

y y′ . . . y(r)

∣∣∣∣∣∣∣∣∣∣, Di =

∣∣∣∣∣∣∣∣∣∣φ1 φ

′1 . . . φ

(r−1)1 φ

(r+i)1

......

......

φr φ′r . . . φ

(r−1)r φ

(r+i)r

y y′ . . . y(r−1) y(r+i)

∣∣∣∣∣∣∣∣∣∣.

Then Φ(15)1 = x, Φ(15)

2 = D, Φ(15)3 = D1, ∆(15) = 0.

Group g16: Φ(16)1 = x, Φ(16)

2 = (logD)′ = D1D with D as above, ∆(16) = 0.

Group g17: Φ(17)1 = c0y + c1y

′ + . . .+ cry(r), Φ(17)

2 = Φ(17)′

1 ,

∆(17) =

∣∣∣∣∣∣∣∣φ1 φ

′1 . . . φ

(r−1)1

......

...φr φ

′r . . . φ

(r−1)r

∣∣∣∣∣∣∣∣ .Group g18: In terms of the previously defined determinants D, D1 and D2

they may be written as

Φ(18)1 =

D1

D, Φ(18)

2 =D2

D


if the proper values for the φk are substituted. They are of order l+∑ρk +1

and l +∑ρk + 2 respectively, ∆(18) = D.

Group g19: For c arbitrary: ∆(19) = (c− r)y(r),

for c 6= r: Φ(19)1 = y(r+1)c−r

y(r)c−r−1 , Φ(19)2 = y(r+2)c−r

y(r)c−r−2 ,

for c = r: Φ(19)1 = y(r), Φ(19)

2 = y(r+2)

y(r+1)2.

Group g20: Φ(20)1 = y(r+1)ey

(r)/r!, Φ(20)2 = y(r+2)e2y

(r)/r!, ∆(20) =∏rk=1 k!.

Group g21: Φ(21)1 = y(r)y(r+2)

y(r+1)2, Φ(21)

2 = y(r)2y(r+3)

y(r+1)3, ∆(21) = y(r)y(r+1).

Group g22: Φ(22)1 = vr+1

1

y(r)2(r+3) , Φ(22)2 = vr+1

2

y(r)3(r+3) , ∆(22) = y(r)2

where

v1 = (r + 1)y(r)y(r+2) − (r + 2)y(r+1)2,

v2 = (r + 1)2y(r)2y(r+3) − 3(r + 1)(r + 3)y(r)y(r+1)y(r+2)

+2(r + 2)(r + 3)y(r+1)3.

Group g23: Φ(23)1 = v2

2

v31

, Φ(23)2 = v3

v21

where v1 and v2 are the same as for the preceding group and

v3 = (r + 1)3y(r)3y(r+4) − 4(r + 1)2(r + 4)y(r)2y(r+1)y(r+3)

+6(r + 1)(r + 3)(r + 4)y(r)y(r+1)2y(r+2) − 3(r + 2)(r + 3)(r + 4)y(r+1)4 ,

∆(23) = y(r)[(r + 2)y(r+1)2 − (r + 1)y(r)y(r+2)

].

Group g24: Φ(24)1 = y′, Φ(24)

2 = y′′′

y′′2, ∆(24) = 0.

Group g25: Φ(25)1 = y′, Φ(25)

2 = xy′′, Φ(25)3 = x2y′′′, ∆(25) = y.

Group g26: Φ(26)k = y(k) for k ≥ 1, ∆(26) = 1.

Group g27: Φ(27)1 = x, Φ(27)

k = y(k) for k ≥ 1, ∆(27) not defined.

There is an important distinction on whether an invariant equation origi-nates from the invariant of a group or from a lower equation. In the lattercase, the left hand side of an invariant equation in normal form is identicalto the factor of the lower equation as given in the above listing. Invariantequations originating from an invariant may be any function of them.

206

5.2 Symmetries of First Order EquationsAlthough first order quasilinear equations ω ≡ y′ + r(x, y) = 0 are simpler

than their higher order counterparts, applying its symmetries for solving themis limited by the fact that there is no algorithmic procedure for obtaining thesymmetry generators explicitly. This is obvious from the expression at theleft hand side of (5.12) for n = 1; it is a single linear homogeneous pde offirst order for the two unknown functions ξ(x, y) and η(x, y). With ω as givenabove it is

ξrx + ηry + ηx − (ηy − ξx)r − ξyr2 = 0. (5.17)

There is no solution algorithm available for this pde. What makes thingseven worse, there are a lot of trivial solutions that cannot be utilized forthe solution procedure as it will turn out later on. They yield the trivialsymmetries ξ = φ(x, y) and η = −rφ(x, y) with φ(x, y) an undeterminedfunction of its arguments. The problem is to find the nontrivial solutions thatmay be applied for the solution procedure.Classification of Symmetries. According to Theorem 4.10 the quasilinearfirst oder equations form a single equivalence class for which v′ = 0 is acanonical form. Consequently, there is only a single symmetry class, i.e. allequations of this kind share the same symmetries up to similarity. They aredescribed in the subsequent theorem.Theorem 5.5 Any symmetry generator of a first order quasilinear ode

v’=0 in canonical variables u and v ≡ v(u) is an operator α(u, v)∂u + β(v)∂vof the infinite-dimensional Lie algebra corresponding to the transformationgroup S1 ≡ G with Janet basis βu; α and β undetermined functions of itsarguments.

Proof The determining system for the coefficients α(u, v) and β(u, v) incanonical variables is immediately obtained from (5.12) in the form βu = 0with the solution α(u, v) and β(v).

This theorem describes the symmetries of first order ode’s in full generality.Any finite group of invariance transformations of such an equation must beup to similarity contained in the pseudogroup given there. From the listingof groups and invariants in the preceding section it follows that the group g12

with generators ∂u, u∂u, u2∂u, ∂v, v∂v, v2∂v leaves v′ = 0 invariant.

An arbitrary first order equation y′ + r(x, y) = 0 not in canonical variablesallows the same invariance transformations; its symmetry generators are sim-ilar to those just mentioned.Corollary 5.1 Any symmetry generator of a first order quasilinear odey′ + r(x, y) = 0 obtained from v′ = 0 by the transformation u = σ(x, y),v = ρ(x, y) with the inverse x = φ(u, v), y = ψ(u, v) has the form

[α(u, v)(φu∂x + ψu∂y) + β(v)(φv∂x + ψv∂y)]u=σ(x,y),v=ρ(x,y)

with α(u, v) and β(v) undetermined functions of its arguments.


This is a straightforward consequence of Theorem 5.5 and the transforma-tion formulas of Lemma 3.3. The above corollary may be applied to obtainfinite groups of invariance transformations for first order ode’s as the followingexample shows.

Example 5.5 For the Riccati equation of Example 4.10 the transformation

functions to canonical variables were σ = 1x and ρ = x(xy + 1)

xy + 2 with the

inverse φ = 1u and ψ = u(1− 2uv)

uv − 1 . From the expression in the corollary, anysymmetry generator of the Riccati equation has form

α( 1x,x(xy + 1)xy + 2

) [−x2∂x+(x2y2+4xy+2)∂y

]+β

( x(xy + 1)xy + 2

) (xy + 2)2

x2∂y.

The generators of the group g12 in noncanonical variables are obtained fromthis expression upon substitution of the proper values for α and β with thefollowing result.

∂u : −x2∂x + (x2y2 + 4xy + 2)∂y,

u∂u : −x∂x + (xy2 + 4y + 2x )∂y, u2∂u : −∂x +

(y2 + 4y

x + 2x

)∂y,

∂v :(y + 2

x)2∂y, v∂v : 1

x (xy + 1)(xy + 1)∂y, v2∂v : (xy + 1)2∂y.

Lie gave another characterization of the most general symmetry generatorof a first order ode; it is described next.

Theorem 5.6 (Lie 1891) Any symmetry generator of y′+ r(x, y) = 0 maybe written in the form ρ(x, y)A + Ω(ω(x, y))U where A = ∂x − r(x, y)∂y,Aω = 0, ρ(x, y) and Ω(ω) are undetermined functions of its arguments.U is a special symmetry generator.

The proof of this result may be found in Lie [113], Kapitel 7. Both Lie’s rep-resentation given in this theorem and the expression in Corollary 5.1 containan undetermined function depending on two arguments, i.e., α or ρ respec-tively, and one undetermined function depending on a single argument, i.e.,β and Ω. Furthermore, in order to obtain a generator explicitly, either thetransformation to canonical form must be known, or the general solution ofthe equation at issue.

Example 5.6 Coming back to the Riccati equation of Example 5.5, with

r(x, y) = y2 + 4xy + 2

x2 , ω(x, y) = x(xy + 1)xy + 2 and U = ξ(x, y)∂x + η(x, y)∂y a

given symmetry generator for the Riccati equation at issue, the representationfor an arbitrary generator is

ρ(x, y)[∂x−

(y2 +

4yx

+2x2

)∂y

]+Ω

( x(xy + 1)xy + 2

) [ξ(x, y)∂x+η(x, y)∂y

].

This result should be compared with the representation given in the precedingExample 5.5

208

Classification of Invariant Equations. Due to the fact that invariancetransformations for first order equations in general cannot be obtained algo-rithmically, it is of some interest to consider the inverse problem, i.e., to startfrom any generator and to determine the corresponding invariant equation.Such a tabulation may be applied to raise the intuition in order to guess agenerator for a given first order ode. The subsequent discussion follows Lie[113], Kapitel 8; see also Cohen [32], §18 and §19 and the listing given inTable 2 on pages 231-235, and Emanuel [42], Appendix 5.

The type of calculation required to obtain these tabulations is similar to thecomputation of invariants in the preceding section. However, now the startingpoints are certain simple symmetry generators for which there is a good chanceof occurrence in a significant number of problems as previous experience hasshown, in particular those on which the usual heuristic methods of solvingfirst order equations are based. Its first order invariants determine the typeof equation taken into account. A few examples of symmetry generators andcorresponding invariant equations are given next.

U = ∂x : y′ = F (y), U = ∂y : y′ = F (x),

U = x∂x + y∂y : y′ = F(yx

), U = x∂x − y∂y : y′ =

y

xF (xy),

U = x∂x + ky∂y : y′ =y

xF

( y

xk

),

U = φ(x)(x∂x + ky∂y) : y′ =y

x

[k +

1φ(x)

F( y

xk

)].

The latter result is obtained from the prolongation

U (1) = xφ∂x + kyφ∂y +[kyφ′ + (kφ− φ− xφ′)y′

]∂y′

with the two invariants yxk

and xy′ − kyxk

φ. For F linear or quadratic in y itcomprises certain classes of Riccati or Abel equations. They will be discussedin more detail in Chapter 7 on solution algorithms.Abel’s Equation. In this subsection the symmetry properties of Abel’sequation are considered. It turns out that case i) and case ii) of Theorem 4.14have a significantly different behavior that is described first.

Theorem 5.7 The two rational normal forms of Theorem 4.14 are con-sidered separately.

i) Equation y′+Ay3+By = 0 with A 6= 0 allows the two symmetry generators

U1 = BA (∂x −By∂y), U2 = B

A

∫A

Bdx(∂x −By∂y)−

12y∂y (5.18)

with B = exp (2∫Bdx) and the commutator [U1, U2] = U1.


ii) Equation y′+Ay3 +By+1 = 0 with A 6= 0 allows the symmetry generator

U =1

A1/3

(∂x −

13A′

Ay∂y

)(5.19)

if and only if its absolute invariant K = 1A

(3B− A′

A

)3

is a constant,i.e., if there holds

dK(A,B)dx

= 0. (5.20)

Proof In general a first simplification is obtained by the following obser-vation. Substituting an infinitesimal generator with coefficients

ξ = ξ(x), η = φ1(x) + φ2(x)y

into (5.17) and equating the coefficients of y in the resulting polynomial withzero, the coefficient of y2 yields the equation φ1A = 0, i.e., in general φ1 = 0.Let the simplified form of η be η = φ(x)y. For the determining system of thesesymmetries the lexicographic term ordering with φ > ξ is always applied, i.e.,any term involving φ is considered to be higher than any term involving ξ.

For case i) substitution of these values for ξ and η into (5.17) leads to

φ+1

2ξ′ +

1

2

A′

Aξ = 0, (φ+Bξ)′ = 0 (5.21)

where ′ = ddx

. Reduction with respect to the first equation yields

[ξ′+

( A′

A− 2B

)ξ

]′= 0, φ+

1

2ξ′ +

1

2

A′

Aξ = 0.

The former equation for ξ has the general solution

ξ =1A

exp(2

∫Bdx

) [C1

∫A exp

(− 2

∫Bdx

)+ C2

].

The coefficients of the generators (5.18) are obtained by choosing C1 = 0,C2 = 1 or C1 = 1, C2 = 0 respectively, applying the latter equation fordetermining φ and defining B = exp (2

∫Bdx).

For case ii) there is the additional equation φ − ξ′ = 0. It is used foreliminating φ from system (5.21) with the result

ξ′ +1

3

A′

Aξ = 0, (ξ′ +Bξ)′ = 0, φ− ξ′ = 0. (5.22)

Two further reductions with respect to the first equation yield[(B − 1

3

A′

A

)ξ

]′= 0, ξ′ +

1

3

A′

Aξ = 0, φ+

1

3

A′

Aξ = 0. (5.23)

210

The latter two equations yield ξ = CA1/3 and φ = 1

3CA1/3

A′

A , i.e., the generator

(5.19), if and only if (5.20) is satisfied. This condition is obtained from thefirst equation (5.22) upon substituting ξ.

The basic achievement of this theorem is the fact that certain symmetrygenerators with Liouvillian coefficients may be determined algorithmically.Condition (5.20) of this theorem is convenient in applications for deciding theexistence of such a symmetry. It is interesting to note that it allows an explicitparametrization of all Abel equations with rational normal form of case ii)with a symmetry (5.19) in terms of an unconstrained coefficent A. To thisend it may be explicitly solved for B with the result

B =13A′

A+ const. ·A1/3.

If condition (5.20) is not satisfied the above theorem does not apply. It ispossible that different types of symmetries may exist.

Example 5.7 Equation 1.46 of Kamke’s collection

y′ − x2y3 + 3y2 − 1x2y +

2x− 1x4

= 0 (5.24)

assumes the rational normal form z′ − x2z3 + 2x2 z = 0 by the transformation

y = z + 1x2 , i.e., a Bernoulli equation belonging to case i).

Example 5.8 Equation 1.41 of Kamke’s collection is y′ + axy3 + by2 = 0with constants a and b, and has rational normal form

y′ +(9a+ 2b2)2b2

729x3y3 − 6a+ b2

3axy + 1 = 0, (5.25)

i.e., it belongs to case ii). Its invariant K = −729b2

(3a+ b2)3

(9a+ 2b2)2entails that

there is a symmetry generator

U = 81x(9a+ 2b2)2/3b2/3

(6a+ b2ax ∂x + 3y

x ∂y

).

5.3 Symmetries of Second Order Equations

In this section the objects of primary interest are not the various groupsany more, but the differential equations and the symmetries attached to them.The link between these two concepts is provided by the tabulation of differ-ential invariants given in the Section 5.1. Although it contains the complete


information on all possible symmetries of any ode, the information given thereis somehow organized by the wrong key for the applications that are the sub-ject of this book. Here the symmetry problem occurs in the following form:for any given ode, determine its symmetries, and if there are any, utilize themfor finding its closed form solutions.Classification of Symmetries. As a first partial answer to the symmetryproblem all possible symmetry types of a quasilinear ode of order two willbe determined. This is an important constraint that will turn out to be ofconsiderable help later on for identifying the symmetry class to which a givenode belongs.

An equation which is a function of the invariants of a certain group isguaranteed to allow a group of invariance transformations that comprises thisgroup; it is possible, however, that its symmetry group is actually larger.It turns out that this does occur in a few cases. Taking these observationsinto account, the subsequent theorem describes the possible point symmetrygroups for quasilinear ode’s of order two. The explicit form of the ode’sallowing the various symmetry groups which also follows from the listing inSection 5.1 is the subject of Chapter 6.

Theorem 5.8 (Lie 1883). Any symmetry algebra of a second order quasi-linear ode is similar to one in canonical variables u and v ≡ v(u) as givenin the following listing. In addition the corresponding Janet basis for the de-termining system is given where α(u, v) and β(u, v) are the coefficients of ∂uand ∂v respectively.One-parameter groupS2

1 : g27 = ∂v. Janet basis α, βu, βv.Two-parameter groupsS2

2,1 : g26 = ∂u, ∂v. Janet basis αu, αv, βu, βv.S2

2,2 : g25 = ∂v, u∂u + v∂v. Janet basis αv, βu, βv − 1vβ, αuu.

Three-parameter groupsS2

3,1 : g10 = ∂u + ∂v, u∂u + v∂v, u2∂u + v2∂v. Janet basis

αv, βu, βv + αu + 2u− v (β − α), αuu − 2

u− vαu −2

(u− v)2 (β − α).

S23,2 : g13 = ∂u, 2u∂u + v∂v, u

2∂u + uv∂v. Janet basis

αu − 2vβ, αv, βv −

1vβ, βuu.

S23,3(c) : g7(c) = ∂u, ∂v, u∂u + cv∂v, c 6= 1.

Janet basis αv, βu, βv − cαu, αuu.S2

3,4 : g20(r = 1) = ∂u, ∂v, u∂u + (u+ v)∂v.Janet basis αv, βu − αu, βv − αu, αuu.

Eight-parameter group:S2

8 : g3 = ∂u, ∂v, u∂v, v∂v, u∂u, v∂u, u2∂u + uv∂v, uv∂u + v2∂v.Janet basis αvv, βuu, βuv, βvv, αuuu, βuuv.

This listing shows in particular that there does not exist any second order odeallowing a group of point symmetries with 4, 5, 6 or 7 parameters.

212

Proof At first invariant differential equations originating from the invari-ants listed in Section 5.1 will be considered. For the groups gk with k = 1,2, 3, 6, 8, 9, 11, 12, 14 and 24 an invariant of order two obviously does notexist, a second order ode with these symmetry groups is therefore excluded.For the groups with k = 16, 18, 21, 22 or 23 the parameter r is constrained tor ≥ 2 as may be seen from the listing of Section 3.4. Therefore an invariantof order two is excluded for these groups as well. The same is true for k = 19and c 6= r.

For the groups g4 and g5 the only invariant of order not higher than two has

the form y′′

y′, and there is no invariant of order one or zero, i.e., depending on

y or y′. As a consequence, any quasilinear equation that may be constructedfrom it must be linear. By Lemma 4.7 its symmetry group is the eight-parameter projective group g3. A similar argument applies to the group g15

for r = 2 with the lowest invariant of nonvanishing order

D ≡ (φ1φ′2 − φ′1φ2)y′′ − (φ1φ

′3 − φ′1φ3)y′ + (φ′1φ

′′2 − φ′′1φ2)y

and the group g17 where D is a linear and homogeneous in y, y′, . . . , y(r) withconstant coefficients.

For the remaining values k = 7, 10, 13, 20 with r = 1, 25, 26 and 27, equa-tions may be constructed from the given invariants with the respective sym-metry groups.

The only lower equation of order two is of the form y′′ = 0 with the eight-parameter projective group g3 as symmetry group. This is also true for theremaining case of the group g19 with to c = r.

Case 7:σ(x, y), ρ(x, y)

)

@@

@@R

PPPPPPPPPPPqCase 3:σ(x), ρ(x, y)

Case 4:σ(y), ρ(x, y)

Case 5:σ(x, y), ρ(x)

Case 6:σ(x, y), ρ(y)

@@

@@

@@R

@@@R

Case 2: σ(y), ρ(x)

Case 1: σ(x), ρ(y)

FIGURE 5.1: Arrows indicate a possible specializations. The notationCase k → Case j means that Case j is obtained by specialization from Case k.

Determining the Symmetry Class. In Theorem 5.8 the symmetry groupsof a second order ode are characterized in terms of canonical variables. Inapplications, however, they occur in actual variables. In order to apply the


results of this theorem, the relation between these two descriptions has tobe found. In particular this applies to the Janet bases for the determiningsystem because it can be determined algorithmically from the given ode. Thisis based on the following observation.

Take the Janet basis for any group defined by Theorem 5.8 in canonicalvariables u and v, and transform it into actual variables by means of (5.2). Ingeneral this change of variables will destroy the Janet basis property. In orderto reestablish it, the algorithm JanetBasis has to be applied. During thisprocess it may occur that a leading coefficient of an equation that is appliedfor reduction vanishes due to a special choice of the transformation (5.2). Thishas the consequence that alternatives may occur leading to different types ofJanet bases. In order to obtain the complete answer, each of these alternativeshas to be investigated separately. Finally all Janet bases have to be combinedsuch that a minimal number of generic cases is retained, and all those thatmay be obtained from them by specialization are discarded. It turns outthat up to a few exceptional cases all alternatives are due to vanishing firstorder partial derivatives of the transformation functions σ and ρ. Takinginto account the constraint ∆ ≡ σxρy − σyρx 6= 0, there are altogether sevenpossible combinations. There exists a partial order between them describedby the diagram shown in Figure 5.1. If there is only a single generic case, itmust obviously be Case 7, all others may be obtained from it. Due to the sizeof the expressions involved, the details of the necessary calculations can onlybe given for the simplest groups.

In order to identify a particular group type of an ode from its Janet basis,the results of Chapter 3 relating various group properties to its determiningsystem are applied. In general this proceeds in three steps.

The first distinguishing property is the number of group parameters thatis obvious from the Janet basis type.

The second distinguishing property is the number of imprimitivity sys-tems, it is purely geometric.

Finally the algebraic structure of the corresponding Lie algebra ex-pressed in terms of invariant properties of the coefficients of its char-acteristic polynomial as described in Theorem 3.7 is applied.

It is part of the problem to assure that a set of criteria is obtained that willlead to a unique answer for the symmetry type of the ode at hand. For secondorder quasilinear ode’s the complete answer may be described as follows.

Theorem 5.9 (Schwarz 1996) The following criteria provide a decisionprocedure for the symmetry type of a second order ode if its Janet basis in agrlex term ordering with η > ξ, y > x is given.

One-parameter group

S21 : Group g27, Janet basis type J (2,2)

1,1 or J (2,2)1,2 .

214

Two-parameter groups

S22,1 : Group g26, Lie algebra l2,2, Janet basis type J (2,2)

2,3 .

S22,2 : Group g27, Lie algebra l2,1, Janet basis type J (2,2)

2,3 .

Three-parameter groups

S23,1 : Group g10 with two systems of imprimitivity, Lie algebra l3,1, Janet

basis type J (2,2)3,6 or J (2,2)

3,7 .

S23,2 : Group g13 with one system of imprimitivity, Lie algebra l3,1, Janet basis

type J (2,2)3,4 , J (2,2)

3,6 or J (2,2)3,7 .

S23,3(c) : Group g7(c) with two systems of imprimitivity, Lie algebra l3,2(c), Janet


3,7 .

S23,4 : Group g20(r = 1) with one system of imprimitivity, Lie algebra l3,3,

Janet basis type J (2,2)3,6 or J (2,2)

3,7 .

Eight-parameter group

S28 : Group g3, Janet basis type ξyy, ηxx, ηxy, ηyy, ξxxx, ξxxy.

Proof One-parameter group. A one-parameter symmetry is uniquely de-termined by the respective Janet basis type.

Two-parameter groups. The two groups are identified by their Lie algebravia Lemma 3.23.

Three-parameter groups. At first the groups are distinguished by the num-ber of imprimitivity systems they allow by Theorem 3.21. The two pairscorresponding to one or two systems of imprimitivity are separated w.r.t.their Lie algebras by Theorem 3.24, 3.25 and 3.26.

The projective group. The eight-parameter symmetry is uniquly identifiedby its Janet basis type.


4x2y′′ − x4y′2 + 4y = 0 (5.26)

has the Janet basis

η +2yxξ = 0, ξx −

1xξ = 0, ξy = 0 (5.27)

for its point symmetries. By Theorem 5.9 it follows that it has a one-parametersymmetry group.



x4y′′ − x2y′2 − x3y′ + 4y2 = 0 (5.28)

generates the type J (2,2)1,2 Janet basis

η − 2yxξ = 0, ξx −

1xξ = 0, ξy = 0. (5.29)

Consequently, it has a one-parameter symmetry. The equation

y′′− 2x(x− 1)x+ 1

y′2− 4(x− 1)x+ 1

yy′+4x2 + 5x+ 2x(x+ 1)2

− 2(x− 1)x(x+ 1)

y2+2x+ 1x(x+ 1)2

y = 0

(5.30)generates the type J (2,2)

1,1 Janet basis ξ = 0, ηx + 1xη = 0, ηy = 0 and

therefore also a one-parameter symmetry. The close relation between the twoequations (5.28) and (5.30) will become clear later on.


(x− y′)y′′ + 4y′2 = 0


ξx −1xξ = 0, ξy = 0, ηx = 0, ηy −

1yη = 0.

By Theorem 3.23 its Lie algebra is l2,2, therefore by Theorem 5.9 its symmetryclass is S2

2,1.


4y′′y − 3y′2 − 12y3 = 0

has the structure (5.34) with φ1 = − 154 , φ2 = 0, i.e., the projective group is

excluded as symmetry group. The Janet basis

ξx +12yη = 0, ξy = 0, ηx = 0, ηy −

1yη = 0

is of type J (2,2)2,3 . By Theorem 3.23 its Lie algebra is l2,1, therefore by Theo-

rem 5.9 its symmetry class is S22,2.


x2(x2y − 2)2y′′2 − 4x4(x2y − 2)y′′y′2 − 8x(x2y − 2)y′′y′

+4x6y′4 + 24x3y′3 + 24x2y′2y + 16y′2 + 24xy′y2 + 8y3 = 0

216

has been written in rational form, therefore it is not quasilinear. Its Janetbasis

ξy = 0, ηx + 2yx ξx −

x2y + 2x(x2y − 2)

η − 4y2

x2y − 2ξ = 0,

ηy + ξx − 2x2

x2y − 2η − 3x2y + 2

x(x2y − 2)ξ = 0,

ξxx − 2xyx2y − 2

ξx + 4x(x2y − 2)2

η + 2y(x2y + 2)(x2y − 2)2

ξ = 0

is of type J (2,2)3,6 . By Theorem 3.21 it allows two systems of imprimitivity. By

Theorem 3.25 its Lie algebra type is l3,1. Consequently by Theorem 5.9 itssymmetry class is S2

3,1.


y′′y + xy′′ + y′2 − y′ = 0

generates the Janet basis

ξy + ξx − 1x+ y η −

1x+ y ξ = 0, ηx − ξx + 1

x+ y η + 1x+ y ξ = 0,

ηy + ξx − 2x+ y η −

2x+ y ξ = 0, ξxx − 3

x+ y ξx + 3(x+ y)2

η + 3(x+ y)2

ξ = 0

of type J (2,2)3,6 . By Theorem 3.21 it allows a single system of imprimitivity.

By Theorem 3.25 its Lie algebra type is l3,1. Consequently by Theorem 5.9its symmetry class is S2

3,2.


y′′y′yx6 − 2y′3x6 + 2y′2yx5 + y5 = 0 (5.31)


ξy = 0, ηx = 0, ηy −32ξx −

2yη +

3xξ = 0, ξxx −

2xξx +

2x2 ξ = 0. (5.32)

By Theorem 3.25 its Lie algebra type is l3,2( 32 ); therefore by Theorem 5.9 its

symmetry class is S23,3(

32 ). The equation

y′y′′ + 2y′′ − y′4 − 12y′3 − 54y′2 − 108y′ − 81 = 0

generates the Janet basis

ξx = 0, ηx + 6ξy = 0, ηy + 5ξy = 0, ξyy = 0

of type J (2,2)3,7 . By a similar reasoning as in the preceding case the symmetry

class is again S23,3(

32 ).


Example 5.16 Any equation of symmetry class S23,4 must contain a term

proportional to an exponential in y′. An example is

y′′y + y′2 − 1xy′y − 1

2x= 2x2 exp

(− 1 + 2y′y

2x

)(5.33)

with type J (2,2)3,6 Janet basis

ξy = 0, ηx−x

yξx−

2x+ 12xy

ξ = 0, ηy−ξx+1yη− 1

xξ = 0, ξxx+

1xξx−

1x2 ξ = 0.

By Theorem 3.25 its Lie algebra type is l3,3, therefore by Theorem 5.9 itssymmetry class is S2

3,3(32 ).

Symmetry Type and Invariants. Tresse’s invariants and their applicationto equivalence problems have been discussed in Section 4.3. In the theorembelow it will be shown how they may be applied for determining the symmetryclass of any second order quasilinear ode.

Theorem 5.10 (Tresse 1896) The following criteria provide a decisionprocedure for the type of symmetry group of a second order quasilinear ode interms of its invariants under point transformations.

i) If there are three functionally independent absolute invariants the equa-tion does not have a nontrivial symmetry.

ii) If there are two functionally independent absolute invariants the equationbelongs to symmetry class S2

1 .

iii) If there is a single functionally independent absolute invariant there isa two-parameter symmetry group. If for any two invariants ψ1 and ψ2

with weights r1, s1 and r2, s2, respectively, there holds

2∣∣∣∣∆xψ1 ∆zψ1

∆xψ2 ∆zψ2

∣∣∣∣ +∣∣∣∣∆yψ1 (3r1 + 2s1)ψ1

∆yψ2 (3r2 + 2s2)ψ2

∣∣∣∣ = 0

the symmetry class is S22,1, and S2

2,2 otherwise.

iv) If ω4H 6= 0 and all absolute invariants are constant there is a three-parameter symmetry group. If ω4 = 0 the symmetry class is S2

3,2. If forany invariant ψ with weight r and s there holds

∆zψ∆xψψ∆yψ

= 3241 (3r + 2s)

the symmetry class is S23,4. If there holds

∆xψ = 0, ∆zψ = 0, ∆yψψ

14√ω4H

6= 18 (3r + 2s)

the symmetry class is S23,1, otherwise the symmetry class is S2

3,3.

218

v) If ω4 = 0 and H = 0 the symmetry class is S28 .

The invariants used in this theorem have been defined in Theorem 4.24 onpage 189. The proof may again be found in Tresse’s thesis [181], page 83.In the following two examples the criteria of this theorem are applied foridentifying the symmetry class of an ode. For comparison the same answer isobtained by the Janet basis analysis of Theorem 5.9

Example 5.17 Consider the second order ode

8xy′′y6 − 9x5y′4 − 16xy′2y5 + 16y′y6 = 0.

Its relative invariants up to order five are

ω4 =27x4

y6, H =

218764

y′8x12

y18, K =

21878

y′7x12

y18,

H10 =98415

64y′

11x16

y24, H01 =

255879128

y′10x16

y24.

All absolute invariants are constant, therefore the symmetry group has threeparameters. Due to ω4 6= 0,

∆zH∆xH

H∆yH=

80y′x4 + 64y5

13y′2x4 + 8y56= 256

41

and∆xH =

9841564

(y′2x4 + 45y

5)y′9x12y12 6= 0

by Theorem 5.10, iv) the symmetry class S23,3 is identified. On the other hand

the Janet basis for the determining system is

ξy = 0, ηx = 0, ηy −23ξx −

2yη +

43xξ = 0, ξxx −

2xξx +

2x2 ξ = 0.

Applying Theorem 5.9 yields the same answer.

Example 5.18 As a second example for applying Tresse’s invariants con-sider the equation y′′y′2 + y′′y2 + y3 = 0 with the absolute invariants

K =6710886487890625

(y′8 − 214 y

′6y2 + 74y′4y4 − 23

4 y′2y6 + 9

4y8)8y′8

(y′4 − 2y′2y2 + 15y

4)10y8(y′2 + y′y + y2)6(y′2 − y′y + y2)6,

H10 =67108864

225(y′8 − 21

4 y′6y2 + 7

4y′4y4 − 23

4 y′2y6 + 9

4y8)8y′8

(y′4 − 2y′2y2 + 15y

4)10y8(y′2 + y′y + y2)6(y′2 − y′y + y2)6.

The remaining invariants cannot be given here due to its size. As a conse-quence it is difficult to apply Theorem 5.10 for determining the symmetry


class of this equation. The Janet basis ξx = 0, ξy = 0, ηx = 0, ηy − 1yη = 0

immediately yields the symmetry class S22,1.

If Tresse’s invariants are computed for a large collection of ode’s, the follow-ing general pattern appears. The calculations become extremely complex forsmall symmetry groups with less than three parameters. That means, despitethe theoretical interest in his results, for any practical purpose it is advanta-geous to determine the symmetry class from the Janet basis by Theorem 5.9.This is even more true because the subsequent step in the solution algorithm,i.e., determining the canonical form, requires the knowledge of the Janet basisas well.

Symmetries of Lie’s Equation. The case ∂4ω∂y′

4 = 0 has been excluded in

Theorem 4.24. Quasilinear equations which are third order polynomials in y′,are called Lie’s equation. They deserve special attention due to its importancein various application fields and when the linearizability of second order ode’sis discussed. Two of its symmetry classes may be identified by the conditionsgiven below.

Theorem 5.11 Let a second order ode of the form

y′′ +Ay′3 +By′2 + Cy′ +D = 0 (5.34)

be given where A ≡ A(x, y), . . . , D ≡ D(x, y). If it allows a nontrivial pointsymmetry the following alternatives may occur.

i) Its symmetry class is S28 iff the two relations L1 = 0, L2 = 0 are satisfied

where

L1 ≡ Dyy +BDy −ADx + (By − 2Ax)D

+ 13Bxx −

23Cxy + 1

3C(Bx − 2Cy),

L2 ≡ 2ADy +AyD + 13Cyy −

23Bxy +Axx

− 13BCy + 2

3BBx −AxC −ACx,

ii) Its symmetry class is S23,2 if L1 6= 0 or L2 6= 0 and Φ = Ψ = 0 where

Φ ≡ L2(L1L2,x − L1,xL2)− L1(L1L2,y − L1,yL2)

−AL31 +BL2

1L2 − CL1L22 +DL3

2,

Ψ ≡ L2(L1L2,yy − L1,yyL2)− (2L2,y + 3AL1 −BL2)(L1L2,y − L1,yL2)

−A2L31 + (AB +Ay)L2

1L2 − (AC −Ax +By)L1L22

+(AD − 13Bx + 2

3Cy)L32.

220

Proof The general expression for the transformation of a second orderderivative v′′(u) by the transformation u = σ(x, y), v = ρ(x, y) is

v′′ = 1(σx + σyy

′)3(σxρy − σyρx)y′′ + (σyρyy − σyyρy)y′3

+[σxρyy − σyyρx + 2(σyρxy − σxyρy)]y′2

+[σyρxx − σxxρy + 2(σxρxy − σxyρx)]y + σxρxx − σxxρx.

It has the structure of Lie’s equation. On the other hand, any equation withthe structure (5.34) leads to a determining system of the form

ξyy − 2Aηy −Bξy +Aξx −Ayη −Axξ = 0,

ηxx −Dηy + Cηx + 2Dξx +Dyη +Dxξ = 0,

ηxy − 12ξxx +Bηx + 3

2Dξy + 12Cξx + 1

2Cyη + 12Cxξ = 0,

ηyy − 2ξxy +Bηy + 3Aηx + 2Cξy +Byη +Bxξ = 0.

(5.35)

In order to transform it into a Janet basis, the algorithm JanetBasis describedin Chapter 1 has to be applied. In the first step it yields the integrabilityconditions

ξxxx − 3Dξxy − (2Bx − Cy − 6AD)ηx

−(Dx − CD)ξy − (2Cx − 4Dy − 4BD + C2)ξx

+(2ByD − Cxy − CyC + 2Dyy + 2DyB)η

+(2BxD − Cxx − CxC + 2Dxy + 2DxB)ξ = 0,

(5.36)

ξxxy − Cξxy − ( 23Bx −

13Cy)ηy − (2Ax − 2AC)ηx

−(Cx −Dy −BD)ξy − ( 23Bx −

13Cy − 3AD)ξx

+(AyD − 23Bxy + 1

3Cyy + 2DyA)η

+(AxD − 23Bxx + 1

3Cxy + 2DxA)ξ = 0.

(5.37)

If they are added to the original system (5.35) as the two highest equations,it has the same leading terms as the Janet basis for the projective symmetrygroup. In order to become a Janet basis for the group S2

8 , the two integrabilityconditions

2L2ηy + L1ξy + L2ξx + L2,yη + L2,xξ = 0,

L1ηy + L2ηx + 2L1ξx + L1,yη + L1,xξ = 0

for this enlarged system are obtained. It is easy to see that they are a conse-quence of L1 = L2 = 0. This proves part i). The calculations for part ii) aresimilar, although they are more lengthy due to the larger coefficients.

The first part of this theorem has been known already to Lie [109]. He gave arather implicit criterion for S2

8 symmetry in terms of a set of pde’s from which


the conditions L1 = L2 = 0 may be obtained as integrability conditions, seeeqs. (3) and (4) in Lie [109], part II. Liouville [118] gave these constraintsfor the first time exlicitly in terms of the invariants L1 and L2. Tresse [181],Theorem VII on page 56, rederived and extended this result. Since then theyhave been rediscovered by various authors, Grissom [63] e. g. obtained themas conditions for linearizability of a second order ode. However, the abovemethod of proving this theorem is more transparent and in addition it providesan explicit form of the Janet basis for any equation with S2

8 symmetry. Thisremarkable result is formulated as follows.

Corollary 5.2 For any second order equation (5.34) with S28 symmetry

the Janet basis in grlex term ordering with η > ξ and y > x is given by theunion of the systems (5.35) and (5.36), (5.37).

Example 5.19 Equation 6.98 of Kamke’s collection has been consideredpreviously in Example 5.10. It has the structure (5.34). From its coefficientsthe functions

φ1 =83

2x2 − 3yx6 , φ2 = − 4

3x5

are obtained. Therefore the projective group is excluded. Because

Φ = −4009

5x4 + 2x2 + x

(x2 + 1)10, Ψ = −128

81x2

(x2 + 1)11

the symmetry class S23,2 is excluded as well.

Example 5.20 Kamke’s equation 6.133 has been considered in Exam-ple 6.11. By Theorem 5.11 with φ1,2 = ± 3

(x+ y)3, the symmetry group of

this equation cannot be the projective group. However both Φ = 0 and Ψ = 0and the above theorem yields the symmetry class S2

3,2.

Example 5.21 Equation Kamke 6.180 is

y′′(y − 1)x2 − 2y′2x2 − 2y′(y − 1)x− 2y(y − 1)2 = 0. (5.38)

Its coefficients A = 0, B = − 2y − 1 , C = − 2

x and D = −2y(y − 1)x2 satisfy

the constraints φ1 = φ2 = 0 of Theorem 5.11, i.e., its symmetry group is theprojective group.

A special case of Lie’s equation is obtained for A = B = C = 0, sometimesit is called generalized Emden-Fowler equation. The subsequent theorem de-scribes its partition into symmetry classes.

Theorem 5.12 The symmetry classes of the generalized Emden-Fowlerequation y′′ + r(x, y) = 0 may be described as follows.

i) The symmetry class is S28 if ryy = 0.

222

ii) The symmetry class is S23,2 if ryyy 6= 0 and(

log ryyy − 65 log ryy

)y

= 0,(log rxyy − 6

5 log ryy)x

= 0,

rxxyyryyyryy − 65r

2xyyryyy − r2yyyryyry − 5ryyyr2yyry + 25

3 r4yy = 0.

iii) The symmetry class is S22,2 if ryyy = 0 and

rxxxxyyr3yy − 32

5 rxxxyyrxyyr2yy − 43

10r2xxyyr

2yy + 594

25 rxxyyr2xyyryy

+5rxxyyr3yyry − 5rxxyr4yy − 1782125 r

4xyy − 10r2xyyr

2yyry

+10rxyyrxyr3yy + 5r5yyr − 52r

4yyr

2y = 0.

iv) The symmetry class is S21 if r obeys a differential polynomial of order

eight.

Proof The determining system for y′′ + r(x, y) = 0 is

ξyy = 0, ηxx − rηy + 2rξx + ryη + rxξ = 0,

ηxy − 12ξxx + 3

2rξy = 0, ηyy − 2ξxy = 0(5.39)

with the integrability conditions

ξxxx − 3rξxy − rxξy + 4ryξx + 2ryyη + 2rxyξ = 0, ξxxy + ryξy = 0,

ryyξy = 0, 2ryyηy + 4ryyξx + 2ryyyη + 2rxyyξ = 0.

The first two conditions, combined with (5.39), form a Janet basis for thesymmetry class S2

8 if and only if ryy = 0 such that the last two conditions aresatisfied. This covers case i). If ryy 6= 0, these latter conditions have to beadded to the system. After two autoreduction steps it contains an equationwith the leading term ryyy

ryy ξx and a further branching occurs. If ryyy 6= 0

a type J (2,2)3,6 Janet basis is generated if the conditions ii) are satisfied. By

Theorem 5.9 it determines uniquely the symmetry class S23,2. If ryyy = 0, a

type J (2,2)2,3 is obtained under the constraints of case iii). By the criterion

given in Theorem 5.9 the symmetry class S22,2 is identified. Finally it may

belong to symmetry class S21 if r obeys a differential polynomial of order eight

that is too complicated to be given here.

Lie Algebra of Symmetry Generators. The methods for solving dif-ferential equations given by Lie [113] require the explicit knowledge of thesymmetry generators. By means of the algorithms of Chapter 2, all Liou-villian generators may be obtained from the Janet basis for its coefficients.Because the symmetry type is also known from the Janet basis, the only addi-tional step is to generate a set of generators satisfying canonical commutation


relations as it is described in Chapter 3 on page 116. These steps will beexplained now by some examples.

Example 5.22 The two equations 6.90 and 6.98 considered above havethe generators x∂x − 2y∂y and x∂x + 2y∂y respectively. This shows the closeconnection between the two equations although this is not obvious from theirappearance.

Example 5.23 Equation 6.228 has the Abelian two-parameter symmetryalgebra with generators U1 = x∂x + y∂y and U2 = (x + y)∂y. Example 5.12with equation 6.159 has the two parameter algebra U1 = ∂x, U2 = x∂x−2y∂ywith commutator [U1, U2] = U1 which is already in canonical form.

Example 5.24 Kamke’s equation 6.133 of Example 5.14 in symmetry classS2

3,2 has generators

U1 = ∂x − ∂y, U2 = x∂x + y∂y,

U3 =(x2 − 2

3xy −13y

2)∂x +

(13x

2 + 23xy − y

2)∂y.

In order to satisfy the commutation relations in canonical form, the linearcombinations U1 − 3

4U1, U2 = 2U2 and U3 = U3 have to be formed.

Example 5.25 The first equation in symmetry class S23,3 considered in

Example 5.15 has symmetry generators

U1 = x∂x + 32y∂y, U2 = y2∂y, U3 = x2∂x.

In order to obtain the canonical commutation relations of the l3,2 Lie algebrathe linear combinations U1 = U2, U2 = U3 and U3 = − 2

3U1 have to beformed. The second equation of this example in the same symmetry class hasgenerators U1 = ∂x, U2 = ∂y and U3 = y∂y − (6x+ 5y)∂y. A simple algebraiccomputation yields U1 = − 1

3U1+U2, U2 = − 12U1+U2 and U3 = U1+U2− 1

3U3.In either case there holds [U1, U2[= 0, [U1, U3] = U1 and [U2, U3] = 2

3U2.

Example 5.26 Equation 6.180 has been considered above in Example 5.21.Its Janet basis in grlex term ordering is

ξyy + 2y − 1ξy = 0,

ηxx + 2y2 − 2yx2 ηy − 2

xηx −4y2 − 4y

x2 ξx − 4y − 2x2 η + 4y2 − 4y

x3 ξ = 0,

ηxy − 12ξxx −

2y − 1ηx −

3y2 − 3yx2 ξy − 1

xξx + 1x2 ξ = 0,

ηyy − 2ξxy − 2y − 1ηy −

4xξy + 2

y2 − 2y + 1η = 0,

ξxxx + 6y2 − 6yx2 ξxy = 0, ξxxy + 2

xξxy = 0

224

in accordance with Corollary 5.2. The special structure of this system allowsit to be solved by an ad hoc method. It has a rational fundamental systemleading to the vector fields

U1 = x∂x, U2 = y(y − 1)∂y, U3 = x2∂x − x(y2 − 1)∂y,

U4 = x2∂x − 2xy(y − 1)∂y, U5 = x2(y − 1)2∂y,

U6 = yx(y − 1)∂x −

y2

x2 ∂y, U7 = y − 2y − 1∂x + 2y

x ∂y,

U8 = yy − 1∂x −

2y2

x ∂y.

5.4 Symmetries of Nonlinear Third Order Equations

The methods described in the preceding section may be extended to higherorder equations more or less straightforwardly, although the larger number ofsymmetry types leads to more alternatives that have to be considered. Forvarious reasons it appears to be appropriate to distinguish equations that areequivalent to a linear one from those that are not. These latter equations aresimply called genuinely nonlinear third order equations; they are the subjectof the current section whereas the linearizable equations are considered inSection 5.5.Classification of Symmetries. The main result of this section, the equiv-alent of Theorem 5.8 for nonlinear third order equations, is given first.

Theorem 5.13 Any symmetry algebra of a genuinely nonlinear third orderquasilinear ode is similar to one in canonical variables u and v ≡ v(u) asgiven in the following listing. In addition the corresponding Janet basis isgiven where α(u, v) and β(u, v) are the coefficients of ∂u and ∂v respectively.One-parameter groupS3

1 : g27 = ∂v. Janet basis α, βu, βv.Two-parameter groupsS3

2,1 : g26 = ∂u, ∂v. Janet basis αu, αv, βu, βv.S3

2,2 : g25 = u∂u + v∂v, ∂v. Janet basis αv, βu, βv − 1vβ, αuu.

S32,3 : g4 = v∂v, ∂v. Janet basis α, βu, βvv.S3

2,4 : g15(r = 2) = ∂v, u∂v. Janet basis α, βv, βuu.Three-parameter groupsS3

3,1 : g10 = ∂u + ∂v, u∂u + v∂v, u2∂u + v2∂v. Janet basis

αv, βu, βv + αu + 2u− v (β − α), αuu − 2

u− vαu −2

(u− v)2 (β − α).

S33,2 : g13 = ∂u, 2u∂u + v∂v, u

2∂u + uv∂v. Janet basisαu − 2

vβ, αv, βv −1vβ, βuu.


S33,3(c) : g7 = ∂u, ∂v, u∂u + cv∂v, c 6= 0, 1.

Janet basis αv, βu, βv − cαu, αuu.S3

3,4 : g20(r = 1) = ∂u, ∂v, u∂u + (u+ v)∂v. Janet basisαv, βu − αu, βv − αu, αuu.S3

3,5 : g5 = ∂u, ∂v, v∂v. Janet basis αu, αv, βu, βvv.S3

3,6 : g17(l = 1, ρ1 = 1, α1 = 0) = ∂u, ∂v, u∂v. Janet basis αu, αv, βv, βuu.S3

3,7 : g17(l = 1, ρ1 = 1, α1 = 1) = eu∂v, ueu∂v, ∂u.Janet basis αu, αv, βv, βuu − 2βu + β.S3

3,8 : g17(l = 2, ρ1 = ρ2 = 0, α1 = 0, α2 = 1) = eu∂v, ∂v, ∂u.Janet basis αu, αv, βv, βuu − βu.S3

3,9(c) : g17(l = 2, ρ1 = ρ2 = 0, α1 = 1, α2 = c) = eu∂v, ecu∂v, ∂u,c 6= 0, 1. Janet basis αu, αv, βv, βuu − (a+ 1)βu + aβ.S3

3,10 : g24 = ∂u, ∂v, u∂u + v∂v. Janet basis αv, βu, βv − αu, αuu.Four-parameter groupsS3

4,1 : g6 = ∂u, ∂v, v∂v, u∂u. Janet basis αv, βu, αuu, βvv.S3

4,2 : g14 = ∂u, u∂u, v∂v, u2∂u + uv∂v.Janet basis αv, βv − 1

vβ, αuu −2vβu, βuu.

S34,3(c) : g19(r = 2) = ∂u, ∂v, u∂v, u∂u + cv∂v, c 6= 2

Janet basis αv, βv − cαu, αuu, βuu.S3

4,4 : g20(r = 2) = ∂u, ∂v, u∂v, u∂u + (u2 + 2v)∂v.Janet basis αv, βv − 2αu, αuu, βuu − 2αu.

Groups with six parametersS3

6 : g12 = ∂u, ∂v, u∂u, v∂v, u2∂u, v2∂v. Janet basis αv, βu, αuuu, βvvv.

Proof At first invariant differential equations originating from the invari-ants listed in Section 5.1 will be considered. For the groups gk with k = 1, 2,3, 11 and 12 an invariant of order three obviously does not exist. A third orderode with these symmetry groups is therefore excluded. For the groups withk = 21, 22 or 23 the parameter r is constrained to r ≥ 2 as it may be seen fromthe listing of Section 3.4. Therefore, an invariant of order three is excludedfor these groups as well. The quasilinear equation y′′′y′ − 3

2y′′2 + r(x) = 0

corresponding to g8 is invariant under the larger group g12 ⊃ g8. Thereforeits symmetry group is g12. A similar argument applies to group g9.

The group g15 for r = 2 allows the three lowest invariants Φ(15)1 = x,

Φ(15)2 = D ≡

∣∣∣∣∣∣φ1 φ

′1 φ

′′1

φ2 φ′2 φ

′′2

y y′ y′′

∣∣∣∣∣∣ and Φ(15)3 = D1 ≡

∣∣∣∣∣∣φ1 φ

′1 φ

′′′1

φ2 φ′2 φ

′′′2

y y′ y′′′

∣∣∣∣∣∣as it may be seen from the listing in Section 5.1. The special choice φ1 = 1and φ2 = x yields the simplification Φ(15)

2 = y′′ and Φ(15)3 = y′′′ from which a

nonlinear third order equation may be constructed. For r = 3, again Φ(15)1 = x

and

Φ(15)2 =

∣∣∣∣∣∣φ1 φ

′1 φ

′′1 φ

′′′1

φ2 φ′2 φ

′′2 φ

′′′2

y y′ y′′ y′′′

∣∣∣∣∣∣ .

226

Choosing φ1 and φ2 as above and leaving φ3 undetermined yields now Φ(15)2 =

φ′′y′′′ − φ′′′y′′. It is not possible to construct a nonlinear third order equa-tion from it which is linear in y′′′ and is therefore excluded; see, however,Theorem 5.16.

The group g16 for r = 2 has the lowest invariants Φ(16)1 = x and Φ(16)

2 = D1D

with D and D1 as above. Applying the same simplification leads to Φ(16)2 =

y′′′

y′′. Again it is not possible to construct a nonlinear third order equation

from it which is linear in y′′′ and is therefore excluded.For the remaining values k = 4, . . . , 7, 9, 10, 12, 13, 14, 19, 20 and 24, . . . , 27

equations may be constructed from the given invariants with the respectivesymmetry groups as listed above.

The groups g17 have at least three parameters. There exist exactly fourof them for this minimal value; they have been listed in Section 3.4. Settingup the system (5.14) for these groups, the two lowest invariants of the formΦ, Φ′ are obtained. For the groups S3

3,6, . . . ,S33,9 one gets Φ = v′′, Φ =

v′′ − 2v′ + v, Φ = v′′ − v′ and Φ = v′′ − (α + 1)v′ + αv respectively. In eachcase, a nonlinear third order equation that is linear in v′′′ may be constructed.Proceeding in a similar way for the seven types of four-parameter groups g17

listed in Section 3.4 yields lowest invariants of the form v′′′+ c1v′′+ c2v

′+ c3vwith constants ci from which a nonlinear equation, linear in v′′′, cannot beconstructed. Therefore these groups are excluded.

The lower equations for k = 1, . . . , 10, 13, 14, 20, 23, . . . , 27 are excludedbecause its order is different from three. For k = 19, 21 and 22 the lower equa-tion is linear, they are covered by Theorem 5.16 below. The lower equationsfor g11 ⊂ g12 are already subsumed under the equation with g12 symmetryas listed above.

In his book on differential equations Lie [113] gives in Chapter 22, §3 acomplete listing of three parameter groups of a two-dimensional manifold.It is instructive to identify the symmetry groups of the above theorem inLie’s list. For symmetry classes S3

3,k, k = 1, 2, . . . , 6 this is fairly obvious,the corresponding groups are 1), 2), 4), 8), 10) and 12) in his enumerationrespectively. The symmetry type of class S3

3,7 is similar to Lie’s group 9), thetransformations functions are σ = −x, ρ = ye−x; the symmetry type of classS3

3,8 is similar to Lie’s group 11), the transformation functions are σ = − log x,ρ = y

x . Finally the symmetry type of S33,9 is similar to Lie’s group 5) with

transformation functions σ = 1c− 1 log x, ρ = yx1(c−1) and α = c.

In the subsequent Section 5.5 it will be shown that equations with a five-or seven-parameter symmetry group are always linearizable, therefore they donot occur in the above theorem.

Determining the Symmetry Class. The preceding theorem describes thepossible symmetry types of third order quasilinear ode’s in canonical coor-dinates. Based on this result, the next theorem allows it to determine the


symmetry type of any such equation in actual variables from the Janet basisof its determining system. In addition to the Janet bases that occurred forsecond order ode’s in the previous section, various Janet bases of order fourare required. They are given in Appendix B on page 380.

Theorem 5.14 The following criteria provide a decision procedure fordetermining the symmetry class of a third order ode if its Janet basis in agrlex term ordering with η > ξ, y > x is given.

One-parameter group

S31 : Janet basis of type J (2,2)

1,1 or J (2,2)1,2 .

Two-parameter groups

S32,1 : Group g26 with a one-parameter system of imprimitivity, Lie algebra

l2,2, Janet basis type J (2,2)2,3 .


l2,1, Janet basis type J (2,2)2,3 .

S32,3 : Group g4 with two systems of imprimitivity, Lie algebra l2,1, Janet basis

type J (2,2)2,1 , J (2,2)

2,2 or J (2,2)2,4 .



2,5 .

Three-parameter groups



3,7 .


type J (2,2)3,4 , J (2,2)

3,6 or J (2,2)3,7 .

S33,3(c) : Group g7(c) with two systems of imprimitivity, Lie algebra l3,2(c),


3,7 .


Janet basis type J (2,2)3,6 with c1 6= 1 or J (2,2)

3,7 with c1 6= 0.

S33,5 : Group g5 with two systems of imprimitivity, Lie algebra l3,4, Janet basis

type J (2,2)3,4 , J (2,2)

3,5 , J (2,2)3,6 or J (2,2)

3,7 .

S33,6 : Group g17(l = 1, ρ1 = 1, α1 = 0) with one system of imprimitivity, Lie

algebra l3,5, Janet basis type J (2,2)3,4 , J (2,2)

3,6 or J (2,2)3,7 .

228

S33,7 : Group g17(l = 1, ρ1 = 1, α1 = 1) with one system of imprimitivity, Lie

algebra l3,3, Janet basis type J (2,2)3,4 , J (2,2)

3,6 with c1 = 1 or J (2,2)3,7 with

c1 = 0.

S33,8 : Group g17(l = 2, ρ1 = ρ2 = 0, α1 = 0, α2 = 1) with one system of

imprimitivity, Lie algebra l3,4, Janet basis type J (2,2)3,4 , J (2,2)

3,6 or J (2,2)3,7 .

S33,9(c) : Group g17(l = 2, ρ1 = ρ2 = 0, α1 = 1, α2 = c) with one system of

imprimitivity, Lie algebra l3,2(c), Janet basis type J (2,2)3,4 , J (2,2)

3,6 or J (2,2)3,7 .


l3,2(c = 1), Janet basis type J (2,2)3,6 .

Four-parameter groups


basis type J (2,2)4,9 , J (2,2)

4,10 or J (2,2)4,14 .


type J (2,2)4,9 or J (2,2)

4,12 .

S34,3(c) : Group g19(r = 2) with one system of imprimitivity, Lie algebra

l4,5(c), Janet basis type J (2,2)4,9 or J (2,2)

4,14 .



4,14 .

Six-parameter group

S36 : Janet basis of type ξy, ηx, ξxxx, ηyyy, type ξy, ηy, ξxxx, ηxxx, typeηx, ηy, ξxy, ξxxx, ξyyy or type ηx, ηy, ξyy, ξxxx, ξxxy.

Proof One-parameter group. A one-parameter symmetry is uniquely de-termined by the Janet basis type.

Two-parameter groups. The first two symmetry classes are singled out bytheir one-parameter system of imprimitivity by Theorem 3.20. The furtherdistinction between the two groups is obtained through their Lie algebra byTheorem 3.23. For Janet basis type J (2,2)

2,1 the symmetry class is uniquelyidentified by Theorem 3.20 through their systems of imprimitivity. If theJanet basis type is J (2,2)

2,2 , J (2,2)2,4 or J (2,2)

2,5 the symmetry class is unique.

Three-parameter groups. If there is a type J (2,2)3,5 Janet basis, the symmetry

class S33,5 is uniquely identified. For Janet basis type J (2,2)

3,4 , by Theorem 3.21again symmetry class S3

3,5 is uniquely identified by its two systems of im-primitivity. The remaining cases allow a single system of imprimitivity; the


symmetry type is identified from its Lie algebra by means of Theorem 3.24.For Janet basis type J (2,2)

3,6 , Theorem 3.21 gives the unique answer S33,10 if

a one-parameter system of imprimitivity is found; otherwise it combines thesymmetry classes into those allowing two systems of imprimitivity, i.e., S3

3,1,S3

3,3(c) and S33,5, and the remaining ones. In the former case Theorem 3.25

leads to a unique identification by its Lie algebra. In the latter the sameis true except for the two symmetry classes S3

3,4 and S33,7, both with a l3,3

Lie algebra. They are distinguished by their different connectivity propertieswhich are identified by c1 6= 1 or c1 = 1 respectively. For Janet basis typeJ (2,2)

3,7 the discussion is identical; Theorems 3.21 and 3.26 are applied now.The distinction between symmetry classes S3

3,4 and S33,7 is obtained by the

condition c1 6= 0 or c1 = 0 respectively.Four-parameter groups. The symmetry classes S3

4,1 and S34,2 are uniquely

identified from the Janet basis types J (2,2)4,10 and J (2,2)

4,12 . For Janet basis type

J (2,2)4,9 , by Theorem 3.22, symmetry class S3

4,1 is uniquely identified by its twosystems of imprimitivity. The remaining three symmetry types are distin-guished by their Lie algebra applying Theorem 3.27. For Janet basis typeJ (2,2)

4,14 , by Theorem 3.22, again symmetry class S34,1 is uniquely identified

through its two systems of imprimitivity. The distinction between the tworemaining cases is obtained from Theorem 3.28.

Six-parameter group. The only symmetry class S36 comprising six parame-

ters is identified from any of the Janet bases of order six without constraintson its coefficients.

For later use some algebraic properties of the various symmetry algebrasare given next.

Corollary 5.3 All one- and two-dimensional symmetry algebras, allthree-dimensional symmetry algebras except those of symmetry classes S3

3,1

and S33,2 that are simple and of type l3,1, and all four-dimensional symmetry

algebras except that of symmetry class S34,1 of type l4,1 are solvable.

Proof The solvable cases follow immediately from the Definition 3.6of solvability. The nonsolvable algebra of S3

4,1 has the Levi decompositionR⊕S with radical R generated by y∂y and the simple Levi factor of typel3,1 generated by ∂x, x∂x + 1

2y∂y, x2∂x + xy∂y. The six-dimensional alge-

bra g12 is semi-simple with the direct-sum decomposition ∂u, u∂u, u2∂u ⊕∂v, v∂v, v2∂v into two simple type l3,1 algebras.

Example 5.27 Equation 7.2 in Kamke’s collection

x2y′′′ + xy′′(y − 1) + xy′2 − y′(y − 1) = 0


η, ξx− 1

xξ, ξy; consequently it belongs

to symmetry class S31 .

230


y′′′y′y − 3y′′2y + y′′y′(2xyy′ + 3y′ + 2y2)− y′3(y′2 + 2xy′ + 4y) = 0


ξx + 1

y η, ξy + 1y ξ, ηx, ηy

with nonva-

nishing coefficients a1 = b2 = 1y . By Theorem 3.23 it generates a l2,2 Lie

algebra, therefore by Theorem 5.14 its symmetry class is S32,1.

Example 5.29 Equation 7.7 from Kamke’s collection yy′′′−y′y′′+y2y′ = 0generates the type J (2,2)

2,3 Janet basisξx + 1

y η, ξy, ηx, ηy −1y η

with nonvan-

ishing coefficients a1 = 1y , d1 = −1

y . By Theorem 3.23 it generates a l2,1 Lie

algebra; consequently, by Theorem 5.14 its symmetry class is S32,2.


y′′′y′3x3 + x3y′′3 − 3x(2y′ + x)(xy′ − 2)y′y′′ − x3(y + 1)y′6 − 6xy′4 + 8y′3 = 0

generates the type J (2,2)2,4 Janet basis; consequently it belongs to symmetry

class S32,3.


y′′′y′2−(3y′+4x2 )y′′2+y′

(y′2− 6

xy′+

4yx3 y

′′+x2y′6+2xy′4− 6

x2 y′3+

4yx4 y

′2)

= 0

generates the type J (2,2)2,5 Janet basis; consequently it belongs to symmetry

class S32,4.


(y′′′y′ − 4y′′2)(y − x)2 + 10y′′y′(y′ + 1)(y − x)− y′2(10y′2 + 3y′ + 10) = 0

generates a type J (2,2)3,6 Janet basis for its symmetry generators with nonvan-

ishing coefficients c1 = 1, c2 = −c3 = 2x− y , d1 = − 2

x− y and d3 = −d2 =2

(x− y)2 . By Theorem 3.21 it allows two systems of imprimitivity. By Theo-

rem 3.25, its Lie algebra is l3,1. Consequently, by Theorem 5.14 its symmetryclass is S3

3,1.


y′′′y′y6−3y′′2y6+(6y′y+

2x2 y

′+3y2

x

)y′′y′y4− 1

x5 y′5−2

(3y+

2x2

)y′4y3−6y5

xy′3 = 0

generates a type J (2,2)3,7 Janet basis for its symmetry generators with nonvan-

ishing coefficients a3 = − 1x , c2 = −2

y , c3 = − 2x and d1 = 2

y . By Theorem 3.22


it allows a single system of imprimitivity, by Theorem 3.26 its Lie algebra isl3,1. Consequently, by Theorem 5.14 its symmetry class is S3

3,2.

Example 5.34 The equation(y′′′ +

3xy′′

)(y′3 − 3

xy′2y +

3x2 y

′y2 − 1x3 y

3)

+1x8 = 0

generates a type J (2,2)3,6 Janet basis for its symmetry generators with coeffi-

cients a1 = 0, b1 = y2x , c1 = − 3

2 . By Theorem 3.21 it allows two systems ofimprimitivity. By Theorem 3.25 its Lie algebra is l3,2( 3

2 ). Consequently, byTheorem 5.14 its symmetry class is S3

3,3(32 ).

Example 5.35 The equation y′′′y′2 − 3y′′2y′ − y′′2 − y′6e−2/y′ = 0 gen-erates a type J (2,2)

3,6 Janet basis for its symmetry generators with coefficientsa1 = c1 = −1. By Theorem 3.21 it allows a single system of imprimitivity.By Theorem 3.25 its Lie algebra is l3,3. Consequently, by Theorem 5.14 itssymmetry class is S3

3,4.


x2y′′′y′y4 − 4x2y′′2y4 + 10x2y′′y′

2y3 − 10xy′′y′y4

−10x2y′4y2 + 2x2y′

4 + 20xy′3y3 − 10y′2y4 = 0

generates a type J (2,2)3,6 Janet basis for its symmetry generators with nonva-

nishing coefficients c2 = −2y , d1 = − 2

x and d3 = 2x2 . By Theorem 3.21 it

allows two systems of imprimitivity. By Theorem 3.25 its Lie algebra is l3,4.Consequently, by Theorem 5.14 its symmetry class is S3

3,5.

Example 5.37 Consider the following four equations generating typeJ (2,2)

3,4 Janet bases for its symmetry generators, the nonvanishing coefficientsof which are also given.

x10y′′′y′+2x9y′′′y′+x8(6x−1)y′′2 +2x7(9x−2)y′′y′+4x6(3x−1)y′2−1 = 0

with nonvanishing coefficients a3 = − 2x and d1 = 2

x .

x10y′′′y′′ + 2x8(x+ 1)y′′′y′ + x6y′′′y + x8(6x+ 1)y′′2

+x6(18x2 + 16x+ 1)y′′y′ + 6x5y′′y

+2x4(x+ 1)(6x2 + 2x− 1)y′2 + 3x2(2x2 − 1)y′y − y2 − 1 = 0

with nonvanishing coefficients a3 = − 2x , d1 = 2

x + 2x2 and d2 = 1

x4 .

x10y′′′y′′ + x8(2x+ 1)y′′′y′ + 6x9y′′2 + x6(18x2 + 6x− 1)y′′y′

+x4(6x2 − 1)(2x+ 1)y′2 − 1 = 0

232

with nonvanishing coefficients a3 = − 2x and d1 = 2

x + 1x2 .

x10y′′′y′′ + 2x8(x+ 2)y′′′y′ + 3x6y′′′y + 3x8(2x+ 1)y′′2

+x6(18x2 + 36x+ 11)y′′y′ + 6x4(3x+ 1)y′′y

+2x4(x+ 2)(6x2 + 6x− 1)y′2 + 3x2(6x2 + 4x− 5)y′y − 9y2 − 1 = 0

with nonvanishing coefficients a3 = − 2x , d1 = 2

x + 4x2 and d2 = 2

x4 . There

are altogether six possible symmetry classes for the Janet basis type J (2,2)3,4 .

By Theorem 3.21 it follows that in all four cases there is a single systemof imprimitivity; consequently, symmetry class S3

3,5 is excluded. By Theo-rem 3.24 the Lie algebras l3,5, l3,3, l3,4 and l3,2( 1

2 ) are identified. From theclassification in Theorem 5.14 the symmetry classes are S3

3,6, S33,7, S3

3,8 andS3

3,9 respectively.

Example 5.38 Consider the equation

x2y′′′y′ + x6y′′2y′2 + x2y′′2 + 4x5y′′y′3 + 10xy′′y′ + 4x4y′4 + 10y′2 = 0.

It generates a type J (2,2)3,6 Janet basis for its symmetry generators with nonva-

nishing coefficients c1 = −1, c3 = 2x , d1 = − 2

x and d3 = 2x2 . By Theorem 3.21

it allows a one-parameter system of imprimitivity, by Theorem 5.14 its sym-metry class S3

3,10 is identified.


y′′′y′y3(y′ + 1

xy)− y′′2y3

(3y′ + 1

2xy)

−y′′y′2y2(y′ + 11

x y)

+ 4xy′6 + 16y′5y + 22x y

′4y2 = 0.

It generates a type J (2,2)4,14 Janet basis for its symmetry generators with non-

vanishing coefficients a1 = a2 = a3 = a4 = 0, b1 = yx and b2 = −1. By The-

orem 3.22 it allows two systems of imprimitivity, therefore by Theorem 5.14its symmetry class is S3

4,1.


y′′′ − x8y′′3 − 6x6(2xy′ + y)y′′2 − 3(16x6y′2 + 16x5y′y + 4x4y2 − 3

x)y′′

−64x5y′3 − 96x4y′2y − 6(8x3y2 − 3

x2

)y′ − 8x2y3 + 6

x3 y = 0

with Janet basis type J (2,2)4,9 for its symmetry generators with nonvanishing

coefficients b2 = − 32 , b4, c2, c4 and all d′is. By Theorem 3.22 it allows a

single system of imprimitivity, by Theorem 3.28 its Lie algebra is l4,5( 32 ).

Consequently, by Theorem 5.14 its symmetry class is S34,3(

32 ).


Equations with a Six-Parameter Symmetry. Although equations withthis symmetry type are covered by the above theorem, a different proceedingwill be described next. It does not require the Janet basis for the determin-ing system. Later on, for third order equations allowing a seven-parametersymmetry group, a similar method will be applied exclusively.

Theorem 5.15 If a quasilinear third order ode belongs to symmetry class S36 ,

either of the following two cases applies where Aj ≡ Aj(x, y) and Bk ≡Bk(x, y) for all j and k.

i) The equation has the structure

(A0y′2 +A1y

′ + 1)y′′′ − (3A0y′ + 3

2A1)y′′2 + (A2y′3 +A3y

′2 +A4y′ +A5)y′′

+B1y′6 +B2y

′5 +B3y′4 +B4y

′3 +B5y′2 +B6y

′ +B7 = 0.(5.40)

Necessary coefficient constraints for symmetry class S36 are as follows.

A0A1,x − 12A0A1A5 + 1

3A0A4 − 14A1,xA

21 + 1

12A31A5 − 1

6A1A3 + 13A2 = 0,

A20A5 + 3A0A1,y − 1

2A0A21A5 − 1

2A0A1A4 +A0A3 − 34A1,yA

21

+ 14A

31A4 − 1

2A21A3 + 1

2A1A2 = 0,

A20A1A5,x − 2A2

0A5,y − 32A0A1,yA1A5 +A0A1,yA4 − 1

4A0A31A5,x + 1

4A0A31A

25

− 12A0A

21A4,x − 1

4A0A21A4A5 +A0A

21A5,y +A0A1A3,x − 1

2A0A1A3A5

+ 16A0A1A

24 − 2A0A2,x +A0A2A5 + 3

8A1,yA31A5 − 1

4A1,yA21A4

+ 18A

41A4,x − 1

8A41A4A5 − 1

8A41A5,y − 1

4A31A3,x + 5

12A31A3A5 + 1

12A31A

24

+ 12A

21A2,x − 3

4A21A2A5 − 1

3A21A3A4 + 1

2A1A2A4 + 16A1A

23 − 1

3A2A3 = 0,

A30A5,x − 13

4 A20A

21A5,x − 1

2A20A1A4,x + 13

2 A20A1A5,y +A2

0A3,x

−A20A4,y + 1

6A20A

24 + 3

2A0A21,y + 33

8 A0A1,yA21A5 − 7

2A0A1,yA1A4

+3A0A1,yA3 + 34A0A

41A5,x − 17

24A0A41A

25 + 13

8 A0A31A4,x + 19

24A0A31A4A5

− 258 A0A

31A5,y − 13

4 A0A21A3,x + 7

6A0A21A3A5 + 1

4A0A21A4,y − 13

24A0A21A

24

+ 112 A0A1A2,x − 31

12A0A1A2A5 + 12A0A1A3,y − 1

3A0A1A3A4 −A0A2,y

+ 14A1,yA1A2 − 3

8A51A4,x + 17

48A51A4A5 + 3

8A51A5,y + 3

4A41A3,x

− 2924A

41A3A5 − 13

48A41A

24 − 11

8 A31A2,x + 13

6 A31A2A5 − 1

8A31A3,y

+ 3124A

31A3A4 + 1

4A21A2,y − 43

24A21A2A4 −A2

1A23 + 25

12A1A2A3 − 56A

22 = 0,

B5 − 52B6A1 −A0B7 + 4A2

1B7 + 12A1A5,x + 1

2A1A25 − 1

3A4A5 −A5,y = 0,

B4 − 2B6A0 − 2B6A21 + 4A0A1B7 + 1

3A0A5,x + 29A0A

25 + 4A3

1B7

+ 23A

21A5,x + 4

9A21A

25 + 1

6A1A4,x − 32A1A5,y − 1

3A3A5 − 13A4,y − 1

9A24 = 0,

234

B3 − 3B6A0A1 − 12B6A

31 +A2

0B7 + 7A0A21B7 + 13

12A0A1A5,x + 23A0A1A

25

+ 16A0A4,x − 2

9A0A4A5 − 53A0A5,y +A4

1B7 + 16A

31A5,x + 1

9A31A

25

+ 18A

21A4,x − 11

24A21A5,y + 1

12A1A3,x − 118A1A3A5 − 1

3A1A4,y

+ 118A1A

24 − 2

9A2A5 − 16A3,y − 2

9A3A4 = 0,

B2A0 − 14B2A

21 −B6A

30 − 3

4B6A20A

21 + 1

4B6A0A41 + 4A3

0A1B7 +A20A

31B7

+ 12A

20A1A4,x − 1

2A20A1A5,y − 1

3A20A3,x − 1

12A20A

24 + 3

4A0A21,y

− 316A0A1,yA

21A5 − 3

4A0A1,yA1A4 + 13A0A1,yA3 − 1

2A0A51B7

− 5144A0A

41A

25 + 1

24A0A31A4,x + 31

144A0A31A4A5 − 1

24A0A31A5,y

− 14A0A

21A3,x − 1

72A0A21A3A5 + 1

48A0A21A

24 +A0A1A2,x

− 4372A0A1A2A5 − 1

3A0A1A3,y − 19A0A1A3A4 + 1

3A0A2,y − 16A0A2A4

− 316A

21,yA

21 + 1

32A1,yA41A5 + 7

32A1,yA31A4 − 7

48A1,yA21A3 + 1

8A1,yA1A2

− 124A

51A4,x − 1

96A51A4A5 + 1

24A51A5,y + 1

12A41A3,x − 5

144A41A3A5

− 596A

41A

24 − 1

4A31A2,x + 25

144A31A2A5 + 1

12A31A3,y + 5

36A31A3A4

− 112A

21A2,y − 11

144A21A2A4 − 1

72A21A

23 − 11

72A1A2A3 + 1336A

22 = 0,

B1 − 12B2A1 + 1

2B6A0A31 +A3

0B7 −A20A

21B7 + 1

6A20A4,x − 1

2A20A5,y

− 18A0A1,yA1A5 + 1

12A0A1,yA4 −A0A41B7 − 3

16A0A31A5,x

− 13144A0A

31A

25 − 1

6A0A21A4,x − 1

48A0A21A4A5 + 13

24A0A21A5,y

+ 572A0A1A3A5 + 1

6A0A1A4,y + 172A0A1A

24 + 1

6A0A2,x − 536A0A2A5

− 16A0A3,y + 1

32A1,yA31A5 − 1

48A1,yA21A4 + 1

96A41A4,x − 1

96A41A4A5

− 196A

41A5,y − 1

48A31A3,x + 5

144A31A3A5 + 1

144A31A

24 − 1

24A21A2,x

+ 112A

21A3,y − 1

18A21A3A4 + 11

72A1A2A4 + 572A1A

23 − 5

36A2A3 = 0.

ii) The equation has the structure

(A0y′2 + y′)y′′′ − (3A0y

′ + 32 )y′′2 + (A2y

′3 +A3y′2)y′′

+B1y′6 +B2y

′5 +B3y′4 +B4y

′3 +B5y′2 = 0.

(5.41)

Its symmetry class is S36 if and only if its coefficients satisfy the following


constraints.

B1 −B2A0 +B3A20 −B4A

30 +B5A

40 − 1

6A23A

20 + 1

3A3A2A0 − 16A

22 = 0,

A0,x + 13A3 = 0, A0,y + 1

3A2 = 0,

A2,y − 2A2,xA0 − 3B2 + 6B3A0 − 8B4A20 + 8B5A

30 − 4

3A23A0 + 4

3A3A2 = 0,

A3,x −B4 + 4B5A0 = 0, A3,y −A2,x = 0,

B4,x − 3B5,y −B5,xA0 − 23B5A3 = 0, B3,x −B4,y − 2B5,yA0 − 4

3B5A2 = 0,

A2,x,x −B4,y + 4B5,yA0 − 43B5A2 = 0,

B2,xx − 13B4,yy − 8

3B5,yyA0 + 49B4,yA3 − 4

9B5,yA3A0 − 209 B5,yA2

− 109 A2,xB4 + 8

9A2,xB5A0 − 43B2B5 + 2

3B3B4

− 89B

24A0 + 8

9B4B5A20 − 4

27B4A23 + 8

27B5A3A2 = 0.

The proof of this theorem is similar to that of Theorem 5.11. Startingfrom a generic equation with structure (5.40) or (5.41) that is obtained fromthe canonical form v′′′v′ − 3

2v′′2 = 0 (see Theorem 6.16) by a general point

transformation, a Janet basis involving the undetermined functions Aj and Bkis generated. The above constraints assure that this Janet basis correspondsto a six-parameter symmetry. The intermediate calculations are very lengthyand are therefore omitted. The advantage of these criteria is that they arevery fast.


2x3y′′′y′2y − 2xy′′′y3 − 6x3y′′

2y′y + 6x3y′′y′

3 − 6x2y′′y′2y

+6xy′′y′y2 − 6y′′y3 + 6x2y′4 − 12xy′3y + 6y′2y2 = 0

with the structure (5.40). Its coefficients

A0 = −x2

y2 , A1 = 0, A2 = −3x2

y3 , A3 = 3xy2 , A4 = −3

y , A5 = 3x,

B1 = B2 = 0, B3 = −3xy3 , B4 = 6

y2 , B5 = − 3xy , B6 = B7 = 0

satisfy the necessary constraints of Theorem 5.15, case i). Furthermore, theJanet basis

ηx − y2

x2 ξy = 0, ηy − ξx − 1y η + 1

xξ = 0,

ξy,y − x2

y2 ξx,x + 1y ξy + x

y2 ξx − 1y2 ξ = 0, ξxxx = 0, ξxxy − 1

xξx,y = 0

is generated. Consequently, it belongs to symmetry class S36 . If the above ode

is slightly changed, e. g. by adding the term y, i.e., B7 = y, the symmetry

236

is destroyed. This may be discovered by testing the coefficient constraints ofTheorem 5.15 which is very fast, avoiding in this way the much more elaborateJanet basis calculation.

Example 5.42 Consider the equation[x

(x− 1)y y′2 + y′

]y′′′−

[3x

(x− 1)y y′ + 3

2

]y′′2

+[

3x(x− 1)y2 y

′3 + 3(x− 1)2y

y′2]y′′ + 3x(x− 2)

(x− 1)2y2 y′4

+2(x2 − 3x+ 3)(x− 1)2y

y′3 + x2 − 4x+ 62(x− 1)2

y′2 = 0


A0 = xxy − y , A2

3xxy2 − y2 , A3 = 3

x2y − 2xy + y, B1 = B2 = 0,

B3 = 3x2 − 6xx2y2 − 2xy2 + y2 , B4 = 2x2 − 6x+ 6

x2y − 2xy + y, B5 = x2 − 4x+ 6

2x2 − 4x+ 2

satisfy the constraints of Theorem 5.15, case ii). Consequently, it belongs tosymmetry class S3

6 .

5.5 Symmetries of Linearizable Equations

For order higher than two, linearizable equations comprise more than a sin-gle equivalence class as it has been shown in Theorem 4.9. Moreover, theycombine into several symmetry classes as it will be seen in this section. Be-cause the methods for solving linear equations are different from the nonlinearones, a special section is devoted to them. At first equations of third order areconsidered; the results for them are more detailed than for arbitrary order.Classification of Symmetries. At first a complete survey of all possiblesymmetry types is provided.

Theorem 5.16 Any symmetry algebra of a third order linearizable ode issimilar to one in canonical variables u and v ≡ v(u) as given in the followinglisting. In addition the corresponding Janet basis is given where α(u, v) andβ(u, v) are the coefficients of ∂u and ∂v respectively.Four-parameter symmetryS3

4,5 : g16(r = 3) = ∂v, u∂v, φ(u)∂v, v∂v.

Janet basisα, βuv, βvv, βuuu − φ′′′

φ′′βuu

.

Five-parameter symmetry


S35,1 : g18(l = 2, ρ1 = 1, ρ2 = 0, α1 = 0, α2 = 1) = ∂v, u∂v, eu∂v, v∂v, ∂u.

Janet basis: αu, αv, βuv, βvv, βuuu − βuu.S3

5,2(a) ≡ g18(l = 3, ρ1 = ρ2 = ρ3 = 0, α1 = 0, α2 = 1, α3 = a) =∂v, eu∂v, eau∂v, v∂v, ∂u, a 6= 0, 1.Janet basis: αu, αv, βuv, βvv, βuuu − (a+ 1)βuu + aβu.

Seven-parameter symmetryS3

7 : g23(r = 3) = ∂v, u∂v, u2∂v, v∂v, ∂u, u∂u, u2∂u + 2uv∂v.

Janet basis: αv, βuv − αuu, βvv, αuuu, βuuu.

Proof The equivalence classes of third order equations with a linearrepresentative have been determined in Theorem 4.9. The canonical equa-tion v′′′ = 0 for E3

1 generates the Janet basis of the seven-parameter groupg23(r = 3) given in the listing on page 142. It corresponds to symmetryclass S3

7 . The canonical equation v′′′ + v = 0 for E32 has been shown to be

equivalent to v′′′ − (a + 1)v′′ + av = 0 with a = 12 (1 ± i

√2) in Lemma 4.3.

Similarly, the canonical equations v′′′ + cv′ + v = 0 for c 6= 0, c 6= − 32

3√

2of E3

3 (c) have been shown to be equivalent to equations of the same formif c and a are related by (4.32). Equations of this latter form generate theJanet basis αu, αv, βuv, βvv, βuuu − (a + 1)βuu + aβu. From the listing onpage 142 the group g18(l = 3, ρ1 = ρ2 = ρ3 = 0, α1 = 0, α2 = 1, α3 = a) isidentified as a symmetry group. It corresponds to symmetry class S3

5,2(a). Ifc = − 3

23√

2 in the canonical form of E33 (c), it is equivalent to v′′′− v′′ = 0 with

Janet basis αu, αv, βuv, βvv, βuuu−βuu. From the listing on page 142 groupg18(l = 2, ρ1 = 1, ρ2 = 0, α1 = 0, α2 = 1) is identified as a symmetry groupcorresponding to symmetry class S3

5,1. Finally the canonical representativev′′′ +Q(x)v = 0 of E3

4 (Q) generates the Janet basis α, βuv, βvv, βuuu +Qβ.From the listing on page 142 the group g16(r = 3) for symmetry class S3

4,5 isobtained.

In Section 4.1 on page 172 the equivalence classes of linearizable third orderequations have been completely identified. The subsequent corollary combinesthese results with the symmetry classes obtained in the preceding theorem.

Corollary 5.4 The symmetry classes of linear third order equations arecomposed of equivalence classes as follows.

S37 ≡ E3

1 , S35,1 ≡ E3

3 (− 32

√2),

S35,2(a) ≡ E3

2 ∪ E33 (c) with c 6= − 3

2

√2, S3

4,5 ≡ E34 (Q).

The constants a and c for the five-parameter symmetries are related by (4.32).

The symmetries identified above will be applied in the next chapter forsolving the corresponding equations. To this end the algebraic structure ofthese symmetry algebras is described next.

238

Corollary 5.5 The symmetry algebras of the four- and five-dimensionalsymmetries are solvable. The seven-dimensional algebra has the Levi decom-position R >S with radical R = ∂v, u∂v, u2∂v, v∂v of type l4,6(a = b = 1)and the simple Levi factor S = ∂u, u∂u + v∂v, u

2∂u + 2uv∂v of type l3,1.

Proof The solvable cases follow immediately from the Definition 3.6 ofsolvability, the seven-dimensional algebra has been considered before in Ex-ample 3.6.

Determining the Symmetry Class. Based on the above classification ofsymmetries, the subsequent theorem provides algorithmic means for determin-ing the symmetry class of any given quasilinear linearizable third order ode.The Janet basis types required for its identification are listed in Appendix Bon page 380.

Theorem 5.17 The following criteria provide a decision procedure foridentifying the symmetry class of a linearizable third order ode if its Janetbasis in grlex term ordering with η > ξ, y > x is given.

Four-parameter symmetry

S34,5 : Janet basis of type J (2,2)

4,2 , type J (2,2)4,17 or type J (2,2)

4,19 .Five-parameter symmetries

S35,1 and S3

5,2(a) : Three Janet basis types may occur. In all three cases theparameter a determining the symmetry type is a solution of the equation

P 2(a2 − a+ 1)3 +Q2R(a− 12 )2(a+ 1)2(a− 2)2) = 0. (5.42)

If a 6= 0 the symmetry class is S35,2(a) and S3

5,1 otherwise. Its coefficients P ,Q and R for the various cases are given below.

i) Janet basis type J (2,2)5,1 .

P = − 92

(a1,xa1 + 1

3a1,xe1 + c3,x − a31 − 2

3a21e1 − 2a1c3

− 13a1e

21 + a1e3 + c1e2 − 2

27e31 + 1

3e1e3 − e4),

Q = a1,x − 2a21 − a1e1 − 3c3 − 1

3e21 + e3, R = 3Q.

ii) Janet basis type J (2,2)5,2 .

P = − 92

(b1,yb1 + 1

3b1,ye1 + d2,y − b31 − 23b

21e1 − 2b1d2

− 13b1e

21 + b1e2 + c1e3 − 2

27e31 + 1

3e1e2 − e5),

Q = b1,y − 2b21 − b1e1 − 3d2 − 13e

21 + e2, R = 3Q.


iii) Janet basis type J (2,2)5,3 .

P = 92

(a3,xa3c

31 − 1

3a3,xc21e1 + a3,xc1c3 + c5,x + a2a

23c

31 − 1

3a2a3c21e1

+a2a3c1c3 − a2c5 + 2a33c

41 − a2

3c31e1 + 3a2

3c21c3 + a3c

31e2

+ 13a3c

21e

21 − a3c

21e3 + 2a3c1c5 + a3c4 − 1

3c21e1e2

+c1c3e2 − 227c1e

31 + 1

3c2e1e3 − c1e5 − e4),

Q = a3,xc21 + a2a3c

21 + 3a2

3c31 − a3c

21e1 + 3a3c1c3 + c21e2

+ 13c1e

21 − c1e3 + 3c5, R = 3

c1Q.

Seven-parameter symmetryS3

7 : Janet basis of type ξy, ηxy, ηyy, ξxxx, ηxxx, type ηx, ξxx, ηyy, ξxyy, ξyyyor type ηy, ξyy, ηxx, ξxxx, ξxxy.

Proof A Janet basis of type J (2,2)4,2 , J (2,2)

4,17 or J (2,2)4,19 identifies the sym-

metry class S34,5 uniquely because these Janet basis types do not occur for

any other four-parameter symmetry as it may be seen from Theorem 5.14. Ifthe equations between the generic coefficients of a Janet basis of type J (2,2)

5,1 ,

J (2,2)5,2 or J (2,2)

5,3 and the transformation functions σ(x, y) and ρ(x, y) are trans-formed into a Janet basis, one of the equations is (5.42) from which the valueof a and thereby the symmetry class S3

5,2(a) may be obtained if a 6= 0, or S35,1

if a = 0. A Janet basis allowing a seven-parameter symmetry group identifiesuniquely symmetry class S3

7 .

In the first three examples equations in symmetry class S34,5 are discussed.

As it will turn out in the next chapter, they are all equivalent to each other.


y′′′y2 − 6y′′y′y +3xy′′y2 + 6y′3 − 6

xy′2y − 8x

2x− 1y′y2 = 0


ξ = 0, ηxy − 2y ηx = 0,

ηyy − 2y ηy + 2

y2 η = 0, ηxxx + 3xηxx −

8x2x− 1ηx = 0.

By case a) of the above theorem it belongs to symmetry class S34,5.


(y′′′y′ − 3y′′2)y3(y − 12 )− 3y′′y′2y2(y − 1

2 )

−y′5x(8y3 − 6y + 3) + y′4y(4y3 − 6y + 3) = 0

240


η = 0, ξxx = 0, ξxy = 0,

ξyyy − 3y ξyy −

4y3 − 6y + 3y3(y − 1

2 )ξy −

x(8y3 − 6y + 3)y3(y − 1

2 )ξy + 8y3 − 6y + 3

y3(y − 12 )

ξ = 0.

By case b) of the above theorem it belongs to symmetry class S34,5.


y′′′(y′ + 1)(y + x− 12 )− 3y′′2(y + x− 1

2 )− 4y′5x+ 4y′4(y − 4x)

+8y′3(2y − 3x) + 8y′2(3y − 2x) + 4y′(4y − x) + 4y = 0


η + ξ = 0, ξxy − ξxx = 0, ξyy − ξxx = 0,

ξxxx − 4yx+ y − 1

2

ξy − 4xx+ y − 1

2

ξx + 4x+ y − 1

2

ξ = 0.

By case c) of the above theorem it belongs to symmetry class S34,5.


y′′′y′y + 1xy′′′y2 − 3y′′2y − xy′′y′2y + 3y′′y′2 − 2y′′y′y2 − 12

x y′′y′y

− 1xy′′y3 − 3

x2 y′′y2 + 2xy′4 + 3y′3y + 15

x y′3 − 1

xy′2y2

− 3x2 y

′2y − 3x2 y

′y3 − 9x3 y

′y2 − 1x3 y

4 − 3x4 y

3 = 0

generates the type J 2,25,3 Janet basis

ηx + yxξx + 1

xη = 0, ηy + yxξy + 1

xξ = 0,

ξxy − xy ξxx −

1xξy + 3

y ξx + 1y2 η − 2

xy ξ = 0,

ξyy − x2

y2 ξxx + 3y ξy + 3x

y2 ξx + 2xy3 η − 1

y2 ξ = 0,

ξxxx − xy + 6x ξxx + xy2 + 3y

x3 ξy + 3xy + 15x2 ξx + 2xy + 9

x2yη − xy + 6

x3 ξ = 0.

By the above theorem, a five-parameter symmetry group is assured. Fromthe Janet basis coefficients P = − 1

3xy2, Q = − 1

3xy and R = xy are obtained.Substituting these values into (5.42) leads to a2(a−1) = 0 with the zero a = 0corresponding to the symmetry class S3

5,1.



y′′′y′y3 − 3y′′2y3 − 4y′′y′2y3 − 3y′′y′2y2 + 3xy′5y2

+8xy′5y + 6xy′5 − 3y′4y3 − 8y′4y2 − 6y′4y = 0

generates the type J 2,25,2 Janet basis

ηx = 0, ηy = 0, ξxx = 0, ξxy + 1y2 η = 0,

ξyyy − 4y + 3y ξyy + 3y2 + 8y + 6

y2 ξy + 3xy2 + 8xy + 6xy3 ξx

+3xy2 + 16xy + 18xy4 η − 3y2 + 8y + 6

y3 ξ = 0.

The Janet basis coefficients yield P = − 103 , Q = − 7

3xy and R = −7. Withthese values (5.42) leads to

(a+ 2)(a+ 12 )(a− 1

3 )(a− 23 )(a− 3

2 )(a− 3) = 0.

The proper choice for a will be determined in the subsequent chapter.

Third Order Equations with Maximal Symmetry. Similar to secondorder equations with projective symmetry, third order equations with sym-metry groups comprising seven parameters deserve special attention. Due tothe fact that their canonical form v′′′ = 0 does not contain parameters orundetermined functions, equations with this symmetry type have a specialstructure from which the S3

7 symmetry may be identified without generatinga Janet basis. The answer has been given essentially by Zorawski [192], astudent of Lie at the end of the 19th century. As usual, σ(x, y) and ρ(x, y)are the transformation functions from a canonical form to the actual variablesx and y.

Theorem 5.18 (Zorawski 1897, Neumer 1929) A quasilinear third orderode belongs to symmetry class S3

7 if and only if either of the following twocases applies where Aj ≡ Aj(x, y) and Bk ≡ Bk(x, y) for all j and k.

a) If σx 6= 0 the equation has the structure

(A0y′ + 1)y′′′ − 3A0y

′′2 + (A1y′2 +A2y

′ +A3)y′′

+B1y′5 +B2y

′4 +B3y′3 +B4y

′2 +B5y′ +B6 = 0.

(5.43)

The coefficients satisfy the constraints

A0,xx − 16A0,xA3 − 1

6A0A3,x + 16A2,x − 1

6A3,y = 0,

A0,y −A0,xA0 + 16A

20A3 − 1

6A0A2 + 16A1 = 0,

242

B5,x + 23B5A3 − 2B6,y − 3B6,xA0 − 5B6A0,x − 2B6A0A3

− 23B6A2 − 1

3A3,xx − 23A3,xA3 − 4

27A33 = 0,

B5,y − 2B5A0,x + 23B5A0A3 − 7B6,yA0 + 2B6,xA

20 + 5B6A0,xA0

− 76B6A

20A3 − 3

2B6A0A2 + 56B6A1 + 4

3A0,xA3,x

+ 1118A0,xA

23 + 1

3A0A3,xx − 118A0A3,xA3 − 4

27A0A33

+ 13A2,xx + 1

18A2,xA3 − 43A3,x,y − 13

18A3,yA3 = 0,

B4 − 4B5A0 + 10B6A20 +A0A3,x + 2

3A0A23 − 1

3A2A3 −A3,y = 0,

B3 − 6B5A20 + 20B6A

30 + 7

3A20A3,x + 23

18A20A

23 + 1

3A0A2,x

− 518A0A2A3 − 7

3A0A3,y − 518A1A3 − 1

3A2,y − 19A

22 = 0,

B2 − 4B5A30 + 15B6A

40 + 11

6 A30A3,x + 17

18A30A

23 + 1

2A20A2,x

− 536A

20A2A3 − 11

6 A20A3,y + 1

6A0A1,x − 16A0A1A3

− 12A0A2,y − 1

36A0A22 − 1

6A1,y − 736A1A2 = 0,

B1 −B5A40 + 4B6A

50 + 1

2A40A3,x + 1

4A40A

23 + 1

6A30A2,x

− 136A

30A2A3 − 1

2A30A3,y + 1

6A20A1,x − 1

18A20A1A3

− 16A

20A2,y − 1

6A0A1,y − 136A0A1A2 − 1

12A21 = 0.

b) If σx = 0 the equation has the structure

y′y′′′ − 3y′′2 + (A1y′2 +A2y

′)y′′ +B1y′5 +B2y

′4 +B3y′3 +B4y

′2 = 0. (5.44)

The coefficients satisfy the constraints

B4 + 13A2,x + 1

9A22 = 0, B3 +A1,x + 1

3A1A2 = 0,

A2,y −A1,x = 0, A1,xy + 23A1A1,x +B2,x = 0,

A1,yy + 3B2,y − 6B1,x + 2A1A1,y + 49A

31 + 2(A1B2 −A2B1) = 0.

Proof If the expressions for the derivatives up to third order (5.3), (5.5)and (5.6) (see also are into the canonical formv′′′ = 0, and the resulting expression is separated w.r.t. derivatives of y,the structure (5.43) or (5.44) follows. For any equation obtained in this way,equating the coefficients of the monomials in the derivatives to the coefficientsAj and Bk of (5.43) or (5.44) leads to a system of equations that is obeyed bythe transformation functions. The above constraints are obtained by trans-forming this system into a Janet basis. If they are satisfied it is guaranteedthat a general solution involving seven constants does exist.

In the subsequent chapter the systems of pde’s determining the transforma-tion functions to canonical form will be studied in more detail. In particular,its relation to certain invariants will become apparent.

(B.6), substituted(B.7))


Neumer (1929) has actually given more complete results on equations withthe structure (5.43) or (5.44). On pages 20-23 he determines criteria for iden-tifying a four-, five- or seven-dimensional symmetry in terms of its coefficients.However, in the former two cases they are considerably more complicated, andthe Janet bases are required anyhow for determining the transformation func-tions to the equivalent linear equation. Therefore they are not given here. Aspecial case of this type of results for identifying an enlarged symmetry groupof a linear equation is discussed in the Exercise 5.8.

Example 5.48 Equation 7.8 from Kamke’s collection

y′′′y2 − 92y′′y′y + 15

4 y′3 = 0

has structure (5.43) with nonvanishing coefficients A2 = − 92y and B3 = 15

4y2 .

They satisfy the necessary constraints of Theorem 5.18, case a); consequently,this equation belongs to symmetry class S3

7 . Alternatively, the symmetry classmay be read off from its Janet basis

ξy = 0, ηxy − ξxx −32yηx = 0,

ηyy − 32y ηy + 3

2y2 η = 0, ξxxx = 0, ηxxx = 0.


y′′′y′y + 1xy′′′y2 − 3y′′2y + 3y′′y′2 − 12

x y′′y′y − 3

x2 y′′y2

+15x y

′3 − 3x2 y

′2y − 9x3 y

′y2 − 3x4 y

3 = 0


A0 = xy , A1 = 3x

y2 , A2 = −12y , A3 = −3

x ,

B1 = B2 = 0, B3 = 15y2 , B4 = −3

xy B5 = −9x2 , B6 = −3y

x3

satisfy the necessary constraints of Theorem 5.18, case a); consequently itbelongs to symmetry class S3

7 .


y′′′y′ − 3y′′2 + 9y y′′y′

2 + 3x2 − 6x+ 6x(x− 1) y′′y′ − 6x

y3(x− 1)y′

5 − 18y2 y

′4

−9x2 − 18x+ 18xy(x− 1) y′

3 − x3 − 3x2 + 6x− 6x2(x− 1)

y′2 = 0

244


A1 = 9y , A2 = 3x2 − 6x+ 6

x(x− 1) , B1 = −6xy3(x− 1)

, B2 = −18y2 ,

B3 = −9x2 + 18x− 18xy(x− 1) , B4 = −x

3 + 3x2 − 6x+ 6x2(x− 1)

satisfy the constraints of Theorem 5.18, case b); consequently, its symmetryclass is S3

7 .

Linear Equations of Any Order. The special structure of these equationsentails that their symmetries may be described in much more detail than for ageneral quasilinear equation. Amazingly they have been studied only recentlyby Krause and Michel [95, 96], and Mahomed and Leach [129]. Let a linearhomogeneous equation of n-th order be given in rational normal form by

L ≡ y(n) + q2y(n−2) + q3y

(n−3) + . . .+ qn−1y′ + qny = 0. (5.45)

The following theorem describes the possible symmetries of these equations.

Theorem 5.19 (Mahomed and Leach 1990, Krause and Michel 1988,1991) The symmetry algebra of a linear nth order differential equation (5.45)with fundamental system y1, y2, . . . , yn has one of three possible forms.

i) A (n+4)-parameter symmetry algebra y1∂y, y2∂y, . . . , yn∂y, y∂y >sl2.This case applies if and only if (5.45) is equivalent to v(n) = 0.

ii) A (n + 2)-parameter symmetry algebra ∂x, y1∂y, y2∂y, . . . , yn∂y, y∂y.This case applies if and only if (5.45) is equivalent to a linear equationwith constant coefficients.

iii) A (n+ 1)-parameter symmetry algebra y1∂y, y2∂y, . . . , yn∂y, y∂y.

It cannot have a n+ 3-dimensional symmetry algebra.

The proof of this theorem is based on a direct computation of the vectorfields generating the symmetries. The details may be found in the articlesby Mahomed [129] or Krause and Michel [96]. If an equation has maximalsymmetry, the transformation to v(n) = 0 is actually achieved by the Laguerre-Forsyth transformation. The special case of third order equations has beentreated above in Theorem 5.17.

Linear equations with maximal symmetry group have a very special struc-ture, in fact they may be parametrized by a single independent coefficient asthe following result shows.

Theorem 5.20 (Mahomed and Leach 1990) For linear equations (5.45)with maximal symmetry, the coefficient p of y(n−2) determines the equation


completely. For n not higher than five these equations are

y′′′ + py′ + 12p′y = 0,

y(4) + py′′ + p′y′ +(

310p

′′ + 9100p

2)y = 0,

y(5) + py′′′ + 32p′y′′ +

(910p

′′ + 425p

2)y′ +

(15p′′′ + 4

25pp′)y = 0.

In the article by Mahomed [129] these equations are given up to ordereight. The severe constraints on the coefficients of an equation with maximalsymmetry described in this theorem have their counterpart in a very specialstructure of the solutions of such equations as it is shown next.

Theorem 5.21 (Krause and Michel 1988) The linear equations with max-imal symmetry of Theorem 5.20 are the symmetric power of the second ordersource equation M ≡ z′′ + 1(n+ 1

3

)pz = 0, i.e., L ≡ M©sn−1 ; a fundamental

system therefore has the form yk = zk1zn−1−k2 for k = 0, . . . , n−1 where z1, z2

is a fundamental system for M .

This result reduces the problem of solving a n-th order equation with max-imal symmetry to a second order problem.


y(4) −(5

2+

20x2

)y′′ +

40x3 y

′ +( 9

16+

9x2

)y = 0 (5.46)

has a maximal, i.e., an eight-parameter symmetry group. The source equationz′′−

(14 + 2

x2

)z = 0 has the two independent solutions z1,2 = (1± 2

x ) exp (∓x2 ).

By the above theorem a fundamental system for (5.46) is

y1 =(x− 2)3

x2 exp (3x2

), y2 =(x+ 2)(x− 2)2

x3 exp (x

2),

y3 =(x+ 2)2(x− 2)

x3 exp (−x2), y4 =

(x+ 2)3

x3 exp (−3x2

).

It is left as Exercise 5.9 to obtain the same answer from the Loewy decompo-sition of (5.46).

246

Exercises

Exercise 5.1 Prove the relation between U (n) and D given in Lemma 5.2.

Exercise 5.2 Let y′′ + r(x)y = 0 be a second order linear ode in rationalnormal form with the general solution y = C1y1 + C2y2. Determine its eightsymmetry generators in terms of this fundamental system.

Exercise 5.3 Show that any linear homogeneous ode of order n for y ≡y(x) with fundamental system y1, y2, . . . , yn has n + 1 symmetry generatorsyk(x)∂y, k = 1, . . . , n, and y∂y.

Exercise 5.4 The form of the two equations (2.67) is constrained by therequirements that the complete system for w is linear and homogeneous andthat no new integrability conditions are introduced. Determine their mostgeneral form in conformity with these requirements.

Exercise 5.5 S33,1, . . . ,S3

3,10 of Theorem 5.13 coincide with the listinggiven in Lie [113], pages 501-502. Why is this list only partially included inTheorem 5.8 ?

Exercise 5.6 Determine the structure invariance group for the canonicalform equation v′′′ + v′r

(v′′v′, u

)= 0 of the symmetry class S3

2,3.

Exercise 5.7 The same for v′′′ + r(u, v′′) = 0 and symmetry class S32,4.

Exercise 5.8 Show that y′′′+A(x)y′+B(x)y = 0 allows a five-parametersymmetry group if and only if one of the following two conditions is satisfied.

6Θ′′3Θ3−7

(Θ′3Θ3

)2

−9A = 0 orΘ′′′3Θ3−4

Θ′′3Θ3

Θ′3Θ3

+289

( Θ′3Θ3

)3

+AΘ′3Θ3− 3

2A′ = 0

where Θ3 is defined in Theorem 4.9.

Exercise 5.9 Determine the fundamental system of (5.46) from its Loewydecomposition.

Chapter 6

Transformation to Canonical Form

It is assumed now that a differential equation of order two or three allows anontrivial symmetry of a known type or, if its order is one, at least a singlesymmetry generator is explicitly known. This knowledge is utilized for trans-forming it into a canonical form corresponding to its symmetry class wheneverthis is possible. To this end a system of equations is constructed the solutionsof which are the desired transformation functions. These equations must besuch that algorithms are available for determining its solutions in well-definedfunction fields, e. g. rational or Liouvillian functions. A complete understand-ing of these equations is achieved by determining for each symmetry class thestructure invariance group of its canonical form equations, and the differentialinvariants that may be constructed from the transformation functions.

6.1 First Order Equations

Any first order equation y′ + r(x, y) = 0 is equivalent to the canonicalform v′ = 0 according to Theorem 4.10. However, in general there is noalgorithm available for determining the transformation functions that actuallygenerate it. Furthermore, it is not guaranteed that a symmetry generatormay be explicitly obtained despite the fact that infinitely many do exist ashas been explained in Section 5.2. As a consequence, in order to obtain amanageable problem it must be assumed that at least a single symmetrygenerator is a priori known. That means, the first order ode together withone or more symmetry generators is the input to the problem at issue. Thefollowing theorem, which is due to Lie [113], Kapitel 6, § 5, describes howa canonical form for the given ode may be obtained if a single nontrivialsymmetry generator is known.

Theorem 6.1 (Lie 1881) Let the first order ode y′ + r(x, y) = 0 allow thesymmetry generator U = ξ(x, y)∂x + η(x, y)∂y. If canonical variables u and vmay be determined such that it represents the translation ∂v, the ode assumesthe form v′ + s(u) = 0 in these variables.

247

248

Proof The first prolongation of the canonical generator ∂v allows theinvariant Φ(u, v′). From this the given form of the canonical ode followsimmediately.

The crucial step in this theorem is to solve the ode dxξ(x, y) = dy

η(x, y) as

required by Lemma 3.4. If it fails, a canonical form may not be obtained. Inthis case the symmetry may be applied to generate an integrating factor aswill be explained in Section 7.1.

Example 6.1 Equation 1.15 from the collection by Kamke

y′ + y2 − 2x2y + x4 − 2x− 1 = 0

allows the symmetry generator U = ∂x + 2x∂y. By Lemma 3.4 the transfor-

mation function σ(x, y) to canonical variables is obtained by solving dydx

= 2xwith the result u ≡ σ(x, y) = y − x2, then v ≡ ρ(x, y) follows from

ρ =∫

dy√y − u

∣∣∣u=σ

=√y − u

∣∣∣u=σ

= x.

The inverse functions x = v, y = u+ v2 yield y′ = 2v+ 1v′

and finally lead to

the canonical form v′ + 1u2 − 1

= 0.

If two generators corresponding to the groups g26 or g25 are known, thetransformation to canonical form is always possible as is shown next.

Theorem 6.2 Let the first order ode y′ + r(x, y) = 0 allow two symmetrygenerators Ui = ξi(x, y)∂x + ηi(x, y)∂y for i = 1, 2. If they generate the groupg25 or g26, a Liouvillian transformation according to Lemma 3.5 transformsit to v′ + a = 0 where a is a constant.

Proof The canonical form is an immediate consequence of the invariantsgiven in the listing in Chapter 5.1 on page 205.

The constant a introduced in this theorem is not an integration constant.Its value is uniquely tied up with the particular canonical form. The system ofequations for the transformation functions given in Lemma 3.5, and the solu-tion algorithms described in Lemma 2.9 and Lemma 2.10, imply that knowingtwo symmetry generators for a first order ode y′+ r(x, y) = 0 guarantees thata transformation to canonical form may be achieved by Liouvillian functionswhich may always be determined.

Example 6.2 The equation y′+ (x− y)2 + 1(x− y)2 − 1

= 0 allows the two symmetry

generators U1 = ∂x + ∂y and U2 = y∂x + x∂y of a type g25 group. Itstransformation to canonical form has been obtained in Example 3.28 with the

Transformation to Canonical Form 249

result x ≡ φ(u, v) = u − v + 12v and y ≡ ψ(u, v) = u − v − 1

2v . It yields

y′ = 2v2 + (1− 2v2)v′

2v2 − (1 + 2v2)v′and finally v′ = 2

3 .

6.2 Second Order Equations

Let a second order ode be given in actual variables x and y ≡ y(x). Ingeneral it will be assumed that it is polynomial in the derivatives, and rationalin the dependent and the independent variables. In addition the dependenceon y′′ is assumed to be linear if it is not stated otherwise explicitly. For anyequation of this kind with a nontrivial symmetry new variables u and v(u)are searched for such that the equation is transformed to a canonical formcorresponding to its symmetry class. This process consists of two steps.

i) A system of equations is constructed that is obeyed by the desired trans-formation functions.

ii) Solutions of these equations in well-defined function fields have to bedetermined.

A priori there is no guarantee that the desired transformation functions maybe found. However, due to the special structure of these equations, by themethods that have been described in Chapter 2 it is assured that large classesof transformation functions may be obtained algorithmically. A completelisting of possible equation types and the corresponding classes of solutionsthat may be obtained for them is given next, it is a partial summary of theresults of this section. In order to arrive at these results, Lemma 2.9, 2.10and 2.11 are applied expressing a solution in terms of integrals.

Theorem 6.3 In order to transform a second order quasilinear ode with anontrivial symmetry to canonical form, the type of equations and the functionfields for which solution algorithms are available may be described as follows.

S21 : First order ode, in general no algorithm available.

S22,1: Third order system

zxx+a1zy+a2zx = 0, zxy+b1zy+b2zx = 0, zyy+c1zy+c2zy = 0. (6.1)

Hyperexponential solutions may be determined.

S22,2: Liouvillian functions explicitly known.

S23,1: Partial Riccati-like system

zx + a1z2 + a2z + a3 = 0, zy + b1z

2 + b2z + b3 = 0.

Liouvillian functions in terms of integrals over rational solutions for z.

250

S23,2: Partial Riccati-like system

zxx + a1z2x + a2zx + a3 = 0, zy + b1zx + b2 = 0.

Liouvillian functions in terms of integrals over rational solutions for z.

S23,3 and S2

3,4: Liouvillian functions over the base field explicitly known.

S28 : Third order system (6.1) with coefficients that are rational solutions of

partial Riccati-like systems

z1,x + z21 + a1z1 + a2z2 + a3 = 0, z1,y + z1z2 + b1z1 + b2z2 + b3 = 0,

z2,x + z1z2 + c1z1 + c2z2 + c3 = 0, z2,y + z22 + d1z1 + d2z2 + d3 = 0.

Liouvillian solutions determined by first order Loewy components of(6.1).

These results show that for an equation in symmetry classes S22,2, S2

3,3 orS2

3,4 a canonical form may always be obtained from its Janet basis coefficients.If it belongs to symmetry class S2

2 , S23,1, S2

3,2 or S28 a canonical form may

be obtained for large classes of transformation functions. An equation insymmetry class S2

1 in general may not be transformed to canonical form.The details of these proceedings will be described now for the various sym-

metry classes one after another, proofs are given for the individual cases. Atfirst the corresponding structure invariance groups will be determined, andafter that its differential invariants up to second order. As usual the notation∆ = σxρy − σyρx is applied.Canonical Forms and Structure Invariance Groups. In general thecanonical forms of the invariant equations for the various symmetry classesare not unique. For the special case of quasilinear equations this freedom iscompletely described in Theorem 6.4 below. The structure invariance groupsgiven there have an important meaning for the respective differential equation.In the first place they describe the degree of arbitrariness for the transforma-tion functions to canonical form. This is an important information for solvingthe systems of pde’s describing these transformations as it will be seen later onin this chapter. Secondly, knowing the freedom that is allowed for the canon-ical form is a necessary prerequirement in order to obtain definite statementson the existence of exact solutions and for designing solution algorithms.

Theorem 6.4 In canonical variables u and v ≡ v(u) the second orderquasilinear equations with nontrivial symmetries have the following canonicalforms and corresponding structure invariance groups; f , g and r are unde-termined functions of their respective arguments, a, c and ai for all i areconstants.

One-parameter symmetry group

S21 : v

′′ + r(u, v′) = 0 allows the transformations u = f(x), v = g(x) + cy.


Two-parameter symmetry groups

S22,1: v

′′ + r(v′) = 0 allows u = a1x+ a2y + a3, v = a4x+ a5y + a6.

S22,2: v

′′u+ r(v′) = 0 allows u = a1x, v = a2x+ a3y + a4.

Three-parameter symmetry groups

S23,1: v

′′(u− v) + 2v′(v′ + a√v′ + 1) = 0 allows

u =a1 + (a2 − a1a3)x

1− a3x, v =

a1 + (a2 − a1a3)y1− a3y

.

S23,2: v

′′v3 + a = 0, a 6= 0, allows

u =a1(x+ a3)

1 + a4(x+ a3), v =

a2y

1 + a4(x+ a3).

S23,3(c): v

′′ + av′(c−2)/(c−1) = 0, c 6= 0, 1, allowsu = a1x+ a2, v = a3y + a4.

S23,4: v

′′ − ae−v′ = 0 allows u = a1x+ a2, v = a3x+ a1y + a4.

Eight-parameter symmetry group

S28 : v

′′ = 0, projective group of the plane.

Proof A general point transformation u = σ(x, y) and v = ρ(x, y) with

y ≡ y(x) generates the transformation v′ = ρx + ρyy′

σx + σyy′ of the first derivative.

For S21 , in order to avoid any dependence on y to be generated via r(u, v′),

σy = ρxy = ρyy = 0 is required. This yields the above structure with

v′ =1f ′

(g′ + cy′), v′′ =1f ′3

(cf ′y′′ − cf ′′y′ + f ′g′′ − f ′′g′).

The expression for v′′ shows that no further constraints are necessary.For the two parameter groups, any dependence on x and y in the trans-

formed function r is avoided if

x = a1u+ a2v + a3, y = a4u+ a5v + a6

where a1, . . . , a6 are constant. The second derivative is transformed accordingto

y′′ =a1a5 − a2a4

(a1 + a2v′)3v′′.

For S22,1 this assures the desired structure. For S2

2,2 the second derivativemust be proportional to the independent variable. This requires in additiona2 = a3 = 0.

252

For S23,1 the transformed first derivative must be proportional to y′; this

requires σy = ρx = 0. The condition for the invariance of the second orderinvariant leads to a fairly complicated system of pde’s for σ and ρ. A Janetbasis for the full system is

σy = 0, ρx = 0, ρyσx−( ρ− σx− y

)2

= 0, σxx +2σ2

x

ρ− σ+

2σxx− y

= 0.

The transformation functions given above, containing three constants, give itsgeneral solution.

For S23,2 the transformed second derivative must be proportional to y′′ and

must be independent of x and y′. This is assured if

σy = 0, ρyy = 0, 2σxρxy − σxxρy = 0, σxρxx − σxxρx = 0

holds. Then v′′ = ρyσ2x

y′′. In the transformed equation the coefficient of the

second derivative must be independent of x and proportional to y3. Thisrequires in addition( ρyρ

3

σx

)x= 0,

( ρyρ3

σx

)y

σ2x

ρyρ3 −

3y

= 0.

The combined constraints may be transformed into the Janet basis

σy = 0, yρy − ρ = 0, σxxρ− 2ρxσx = 0, ρxxρ− 2ρ2x = 0

with the general solution

σ =C1x+ C2

C3(C3x+ C4), ρ =

y

C3x+ C4.

By a suitable change of the integration constants Ck the above expressionsfor the transformation functions are obtained.

For the group S23,3 the transformed first derivative must be proportional

to y′; this requires σy = ρx = 0. The transformed second derivative cannotcontain a term proportional to a power of y′; this requires σxx = ρyy = 0.The general solution of these equations represent the transformation functionsgiven in the above theorem.

For the group S23,4 the transformed first derivative must have the form

−y′ + constant; this requires

σy = 0, ρy − σx = 0, σxρxy − σxyρx = 0, σxρxx − σxxρx = 0

or the equivalent Janet basis σy = 0, σxx = 0, ρy − σx = 0, ρxx = 0 with thesolution given above.

Finally the structure invariance group of v′′ = 0 is obviously identical to itssymmetry group S2

8 , i.e., the eight parameter projective group of the plane.


For later use the Lie algebras corresponding to the two- and three-parametersymmetries of the preceding theorem are given explicitly next.

Corollary 6.1Two-parameter symmetry groups

S22,1: ∂x, ∂y, x∂x, y∂x, x∂y, y∂y. S2

2,2: ∂y, x∂x, x∂y, y∂y


S23,1: ∂x + ∂y, x∂x + y∂y, x

2∂x + y2∂y. S23,2: ∂x, x∂x, y∂y, x2∂x + xy∂y.

S23,3(c): ∂x, ∂y, x∂x, y∂y. S2

3,4: ∂x, ∂y, x∂x + y∂y, x∂y.

The proof is a straightforward application of the definition of the infinites-imal generators given in Chapter 3.

A symmetry class may comprise only a single equivalence class. This is truee. g. for S2

8 . Because its canonical representative v′′ = 0 does not containany undetermined function or parameter, this symmetry class is at the sametime an equivalence class as has been mentioned before. In general, however,a symmetry class may comprise more than a single equivalence class. In thesecases there remains the question of how the individual equivalence classescontained in it may be uniquely parametrized. For the symmetry classescorresponding to three-parameter groups the answer is given next.

Corollary 6.2 The symmetry class S23,1 decomposes into equivalence

classes which are uniquely determined by the value of the parameter a in thecanonical form equation. The remaining symmetry classes S2

3,2, S23,3(c) and

S23,4 comprise a single equivalence class.

Proof The structure invariance group of S23,1 is identical to the symmetry

group. Therefore it cannot transform equations with different values of ainto each other. In the remaining cases, canonical equations with parametervalues a and b are transformed into each other if the group parameters of therespective structure invariance groups are determined by

b

a=

( a1

a22

)2

, andb

a=

( ac1a3

)1/(c−1)

orb

a=

1a1

exp(−a3

a1

).

That is, in all three cases there is a one-parameter set of transformations bywhich this transformation may be achieved.

For two-parameter symmetries the situation is more complicated. A partialanswer may be obtained from the following example dealing with symmetryclass S2

2,2.

Example 6.3 Applying a general transformation of the structure invari-ance group to the canonical form uv′′+r(v′) = 0 of symmetry class S2

2,2 yields

254

xy′′ + s(y′) = 0 with s(y′) = 1a3r(a2a1

+ a3a1y′). Its action consists essentially

of a linear transformation of the argument of r.

Example 6.4 Applying a general transformation of the structure invari-ance group to the canonical form v′′v3 + a = 0 for S2

3,2 symmetry yields

y′′y3 + b = 0 with b = aa21

a42

, i.e., any value for the constant b may be achieved,

e. g. b = 1.

The parameters and undetermined functions occurring in the canoncialforms of Theorem 6.4 have to obey certain constraints in order to guaran-tee the respective symmetry type. They are a consequence of the definitionthat the symmetry type of an equation is determined by the maximal groupleaving it invariant. Obvious examples of these constraints are a 6= 0 in the

canonical forms of S23,2, S2

3,3 and S23,4, or dr(v′)

dv′6= 0 in the canonical forms

of S22,1 and S2

2,2 assuring that the projective symmetry is excluded. For theapplications in this book, however, this has no importance because the sym-metry type of an equation is determined by its Janet basis for the symmetrieswhich always gives the correct answer for a given equation in actual variables.These questions are discussed in more detail in Exercise 6.1

The discussion in the remaining part of this chapter is based on a seriesof articles by Lie [109], especially part III. As Lie mentions at the beginningof page 371, he outlines the ample calculations only schematically. Here allnecessary steps and case distinctions are described in full detail such that thetheorems given below readily may be applied for solving concrete examples.

The differential equations satisfied by the transformation functions to canon-ical form have to obey severe constraints which are a consequence of the fol-lowing result due to Lie [109], part III, page 379ff.

Theorem 6.5 (Lie 1883) The differential equations satisfied by the trans-formation functions to canonical form allow the structure invariance group ofthe respective symmetry class as a symmetry group.

Lie’s proof uses the fact that the determining systems for the actual differen-tial equation and the corresponding equation in canonical form are related bythe desired transformation functions as well. Knowing their symmetry groupand its differential invariants yields a better understanding of their structure.On page 382, part III of [109] the following remarkable result is proved.

Theorem 6.6 (Lie 1883) Let Ik be the invariants for a particular symmetrytype, and σ(x, y) and ρ(x, y) be the transformation functions to canonicalform. There are relations of the form Ik = Bk(x, y) with known functions Bk.


After these preliminary remarks the equations determining the transforma-tion functions are determined explicitly.One-Parameter Symmetry. This is the simplest type of nontrivial invari-ance that may occur for any ode. The freedom involved in the canonicalform transformation, i.e., two unspecified functions of a single variable anda constant, correspond to the respective quantities generating the structureinvariance group.

Theorem 6.7 If a second order ode belongs to symmetry class S21 , two

types of Janet bases may occur; f , g, h and k are undetermined functions ofits argument.

i) Janet basis of type J (2,2)1,1 : σ = f(x), ρ =

∫exp

(F (x, y)

)dy+g(x) where

F (x, y) =∮a(x, y)dx+ b(x, y)dy.

ii) Janet basis of type J (2,2)1,2 : σ = h(φ) where φ(x, y) = C is the solution

of y′ + a = 0, ρ = C∫

exp(G(x, y)

)dx + k(y) where C is a constant,

and G(x, y) =∮b(x, y)dx+ c(x, y)dy.

Proof In case i) the Janet basis

ξ = 0, ηx + (log ρy)xη = 0, ηy + (log ρy)yη = 0

yields the relations (log ρy)x = a, (log ρy)y = b and σy = 0 with the solutiongiven above. In case ii) the Janet basis

η +σxσyξ = 0, ξx+

(log

∆σy

)xξ = 0, ξy+

(log

∆σy

)yξ = 0

leads to the system

σxσy

= a,(

log∆σy

)x= b,

(log

∆σy

)y= c

for σ and ρ. A few simplifications lead to

σy −1aσx = 0, ρy −

1aρx =

1a

exp G(x, y)

with G as defined above. By Theorem 2.19 the invariant φ(x, y) for thefirst equation is obtained. By Theorem 2.20 the function w(x, y, ρ) obeysthe homogeneous equation wy − 1

awx + 1ae

Gwρ = 0. The variable changey = φ(x, y) yields wx − eGwρ = 0 with the solution ρ as given above.

The arbitrariness due to the undetermined elements is obviously a conse-quence of the structure invariance group given in Theorem 6.4.

The crucial step of this proceeding is to solve the ode y′ + a(x, y) = 0 incase ii). If a solution cannot be found it fails. Consequently, the examples of

256

second order equations with a one-parameter symmetry given in the literaturegenerate a Janet basis with a simple expression for the coefficient a(x, y), e. g.linear in y. Many equations are in the form y′′ + r(y, y′) = 0. Interchangingthe dependent and the independent variable with each other by x = v, y = u

and applying (B.18) yields v′′−v′3r(u, 1v′

)= 0. This case is discussed in more

detail in Exercise 6.4.

Example 6.5 Equation 6.90 of Kamke’s collection has been shown in 5.9to belong to symmetry class S2

1 . Its type J (2,2)1,2 Janet basis yields a = 2y

x ,

b = − 1x and c = 0. According to case ii) y′ + 2

xy = 0 is obtained with thesolution x2y = C. It leads to φ = x2y and eG = C 1

x . Special solutionstherefore are σ = x2y and ρ = log x. Inverting these relations with the resultx = ev and y = ue−2v leads to the canonical form

v′′ + u(u+ 5)v′3 − (u+ 3)v′2 + 14v′ = 0.

Example 6.6 Equation 6.98 has been considered before in Example 5.10.Its type J (2,2)

1,2 Janet basis differs from the preceding case only by the value of

a; now it is a = −2yx . A similar calculation as above yields σ = y

x2 , ρ = log x

and x = ev, y = ue2v with the canoncial form

v′′ + 2(2u− 1)v′2 + v′ = 0. (6.2)

The second equation considered in Example 5.10 had a type J (2,2)1,1 Janet

basis with a = 1x and b = 0. Therefore case i) of the above theorem applies.

A few simple calculations yield the general transformation σ = f(x) andρ = Cxy + g(x). With the special choice f = 1

x+ 1, C = 1 and g = 0 thevariable transformations

u =1

x+ 1, v = xy and x =

1− uu

, y =uv

1− u

are obtained. It transforms equation (5.30) into the canonical form, too.

Two-Parameter Symmetries. According to Corollary 6.1 the structureinvariance group for the symmetry class S2

2,1 is generated by the six operators∂x, ∂y, x∂x, y∂x, x∂y, y∂y. Its invariants up to second order are given next.

Lemma 6.1 The pde’s for the canonical form transformations σ(x, y) andρ(x, y) for the symmetry class S2

2,1 allow the following invariants w.r.t. theinvariance group generated by ∂σ, ∂ρ, σ∂σ, ρ∂σ, σ∂ρ, ρ∂ρ. There are no firstorder invariants, and there are six second order invariants

I1 = σyρyy − σyyρyσxρy − σyρx , I2 = σxρyy − σyyρx

σxρy − σyρx , I3 = σyρxy − σxyρyσxρy − σyρx ,

I4 = σyρxx − σxxρyσxρy − σyρx , I5 = σxρxy − σxyρx

σxρy − σyρx , I6 = σxρxx − σxxρxσxρy − σyρx .


Proof The system of pde’s determining any invariant Φ of order nothigher than two comprises Φσ = Φρ = 0 and

σxΦσx+ σyΦσy

+ σxxΦσxx+ σxyΦσxy

+ σyyΦσyy= 0,

ρxΦσx+ ρyΦσy

+ ρxxΦσxx+ ρxyΦσxy

+ ρyyΦσyy= 0,

σxΦρx + σyΦρy + σxxΦρxx + σxyΦρxy + σyyΦρyy = 0,

ρxΦρx + ρyΦρy + ρxxΦρxx + ρxyΦρxy + ρyyΦρyy = 0.

The subsystem containing only derivatives w.r.t. σ, ρ and its first order deriva-tives contains six equations in six indeterminates with a nonvanishing coeffi-cient determinant σxρy −σyρx. Consequently, a first order invariant does notexist. The full system for the second order invariants comprises six equationsin twelve indeterminates σ, σx, σy, σxx, σxy and σyy, and the correspondingderivatives of ρ, i.e., six second order invariants may exist. They may beobtained by the methods described in Chapter 2.

Theorem 6.8 If a second order ode has a Janet basis of type J (2,2)2,3 and

a1 = b2 and c1 = d2, it belongs to symmetry class S22,1. The transformation

function σ is determined by

σxx− c2σy − a2σx = 0, σxy − c1σy − a1σx = 0, σyy − d1σy − b1σx = 0. (6.3)

There is an identical set of equations for ρ. The general solution of the fullsystem contains six constants.

Proof From the Janet basis given in equation (11) of Schwarz [165], thecoefficients a1, a2, . . . , d2 may be expressed in terms of the transformationfunctions σ and ρ as

σyρxy − σxyρy = a1∆, σyρxx − σxxρy = a2∆,σyρyy − σyyρy = b1∆, σyρxy − σxyρy = b2∆,

σxρxy − σxyρx = −c1∆, σxρxx − σxxρx = −c2∆,σxρyy − σyyρx = −d1∆, σxρxy − σxyρx = −d2∆

(6.4)

where as usual ∆ = σxρy − σyρx. This system may be transformed into aJanet basis in grlex term ordering with ρ > σ > d2 > d1 > . . . > a2 > a1 andy > x with the result

b2 − a1 = 0, d2 − c1 = 0,a2,y − a1,x + c2b1 − c1a1 = 0, b1,x − a1,y + d1a1 − b1c1 + b1a2 − a2

1 = 0,c2,y − c1,x + d1c2 − c2a1 − c21 + c1a2 = 0, d1,x − c1,y + b1c2 − c1a1 = 0,σxx − c2σy − a2σx = 0, σxy − c1σy − a1σx = 0, σyy − d1σy − b1σx = 0,ρxx − c2ρy − a2ρx = 0, ρxy − c1ρy − a1ρx = 0, ρyy − d1ρy − b1ρx = 0.

(6.5)

258

The lower equations not involving σ and ρ represent the coherence conditionsfor the Janet basis coefficients. The upper half of this Janet basis representsthe two identical linear systems for σ and ρ respectively as given above.

Example 6.7 The symmetry type of equation 6.227 has been identified asS2

2,1 in Example 5.11. By Theorem 6.8, case i) the type J (1,2)3,2 Janet basis for

the transformation function σ isσxx +

1xσx = 0, σxy = 0, σyy +

1yσy = 0

,

and an identical one for ρ. It is not completely reducible; its Loewy factorcomprises only σx, σy. Dividing it out yields a type J (2,2)

2,3 Janet basis forthe quotient in terms of σ1 ≡ σx and σ2 ≡ σy. It is completely reducible anddecomposes into type J (2,2)

1,1 and type J (2,2)1,2 Janet bases according to

σ1,x + 1xσ1 = 0, σ1,y = 0, σ2,x = 0, σ2,y + 1

yσ2 = 0

= Lclm(σ1 = 0, σ2,x = 0, σ2,y + 1

yσ2 = 0,

σ2 + xy σ1 = 0, σ1,x + 1

xσ1 = 0, σ1,y = 0)

from which the special solutions σ1 = 0, σ2 = 1y and σ1 = 1

x , σ2 = −1y ,

and finally σ = log y and σ = log xy are obtained. A possible choice ofthe transformation functions to new variables u ≡ σ and v ≡ ρ therefore isσ = log y and ρ = log xy with the inverse x = eu+v, y = eu. It yields the

canonical form v′′ + v′3 + 5v′2 + 8v′ + 4v′

= 0.

As has been shown in Theorem 6.4, the canonical form for the symmetryclasses corresponding to two-parameter groups is highly non-unique. Thisarbitrariness is discussed next.

Example 6.8 The preceding example is continued. A fundamental systemfor the transformation functions σ and ρ has been shown to be 1, log x, log y.Any linear combination leading to a nonvanishing functional determinant isappropriate, although it is not a priori clear whether there is a particularlyfavorable one for the subsequent solution procedure. The linear combinationsu = c1 log x + c2 log y, v = c3 log x + c4 log y with c1c4 − c2c3 6= 0 lead tothe most general transformation eu = xc1yc2 and ev = xc3yc4 . The special

choice c1 = c4 = 0, c2 = c3 = 1 leads to v′′ + v′(v′ + 1)2

v′ − 1= 0 for which r(v′)

decomposes into linear factors. The transformation between the two canonicalforms is discussed in Exercise 6.2.

The structure invariance group for the symmetry class S22,2 is generated by

the four operators ∂y, x∂x, x∂y, y∂y according to Corollary 6.1. Its invariantsare similar to the preceding case.



2,2 allow the following invariants w.r.t. theinvariance group that is generated by ∂ρ, σ∂σ, σ∂ρ, ρ∂ρ. There are two firstorder invariants K1 = σx

σ and K2 = σyσ . If σx 6= 0, ρx 6= 0 and ρy 6= 0,

there are the same six second order invariants that have been determined inLemma 6.1

Proof The system of pde’s determining the invariants Φ of order nothigher than two comprises Φρ = 0 and

σΦσ + σxΦσx+ σyΦσy


+ σyyΦσyy= 0,

σxΦρx+ σyΦρy

+ σxxΦρxx+ σxyΦρxy

+ σyyΦρyy= 0,

ρxΦρx+ ρyΦρy

+ ρxxΦρxx+ ρxyΦρxy

+ ρyyΦρyy= 0.

The subsystem containing only derivatives w.r.t. σ, ρ and its first derivativescomprises four independent equations in six variables; i.e., there are two firstorder invariants. The four independent equations in twelve variables of thefull system allow 12-4-2=6 second order invariants.

In Exercise 6.3 the Janet basis coefficients of the two parameter symmetrieswill be determined in terms of the respective invariants.

Theorem 6.9 If a second order ode has a Janet basis of type J (2,2)2,3 and

belongs to symmetry class S22,2, the transformation function σ is

σ = exp[ ∮

(c1 − d2)dx+ (b2 − a1)dy]. (6.6)

For the other transformation function ρ two alternatives may occur.

a) If c1 6= d2, ρ is explicitly given by

ρ =∮ρ1c2dx+ d2dy

d2 − c1− σ

∮ρ1

σ

c2dx+ c1dy

d2 − c1(6.7)

where

ρ1 = exp[ ∮ ( a1 − b2

c1 − d2c2 + c1

)dx+

( a1 − b2c1 − d2

c1 + b2 − a1 + d1

)dy

].

b) If c1 = d2, ρ is explicitly given by

ρ =∮ρ1a1dx− b1dya1 − b2

− σ∮ρ1

σ

b2dx+ b1dy

a1 − b2(6.8)

whereρ1 = exp

( ∮a2dx+ b2dy

).

260

Proof From the Janet basis given in equation (10) of Schwarz [165], thecoefficients a1, a2, . . . , d2 may be expressed in terms of the transformationfunctions σ and ρ as

(σyρxy − σxyρy)σ + σy∆ = −a1σ∆, (σyρxx − σxxρy)σ + σx∆ = −a2σ∆,σyρyy − σyyρy = −b1∆, σyρxy − σxyρy = −b2∆,σxρxy − σxyρx = c1∆, σxρxx − σxxρx = c2∆,

(σxρyy − σyyρx)σ − σy∆ = d1σ∆, (σxρxy − σxyρx)σ − σx∆ = d2σ∆.(6.9)

In the same term ordering as system (6.5) above the Janet basis

a2,y − a1,x + c2b1 − c1b2 = 0, b1,x − a1,y + a1d1 − b1c1 − a1b2 + a2b1 = 0,b2,x − a1,x + a1d2 − c1b2 = 0, b2,y − a1,y + b1d2 − b2d1 + a1d1 − c1b2 = 0,c2,y − c1,x − c1d2 + c2d1 − a1c2 + c1a2 = 0, d1,x − c1,y − a1d2 + b1c2 = 0,d2,x − c1,x − a2d2 + b2c2 − a1c2 + c1a2 = 0, d2,y − c1,y − a1d2 + c1b2 = 0,

σx + (d2 − c1)σ = 0, σy + (a1 − b2)σ = 0,ρxx − c2ρy + (d2 − c1 − a2)ρx = 0,

ρxy − c1ρy − b2ρx = 0,ρyy + (a1 − d1 − b2)ρy − cρx = 0

is obtained. The eight lowest equations, i.e., the first four lines, are the coher-ence conditions for the coefficients. The two subsequent equations determineσ in terms of the path integral (6.6). The three remaining equations for ρ al-low the solutions ρ = 1 and ρ = σ. This follows from the fact that they may bereduced to zero by both ρx, ρy and also by ρx+(d2−c1)ρ, ρy+(a1−b2)ρ,i.e., the same system that determines σ. If c1 6= d2, dividing out the leastcommon left multiple

Lclm(ρx, ρy, ρx + (d2 − c1)ρ, ρy + (a1 − b2)ρ

)=

ρ1 ≡ ρy + a1 − b2

c1 − d2ρx, ρ2 ≡ ρxx+

(a1 − b2c1 − d2

c2 + d2 − c1 − a2

)ρx

(6.10)

yields the exact quotientρ2−c2ρ1, ρ1,x−

( a1 − b2c1 − d2

c2+c1)ρ1, ρ1,y−

( a1 − b2c1 − d2

c1+b2−a1+d1

)ρ1

from which ρ1 and ρ2 may be be determined. Consequently, the second line of(6.10) forms an inhomogeneous system of pde’s with known solution of the ho-mogeneous part and known right hand sides. It has been shown in Excercise 5of Chapter 2, see also Appendix A, how to express a special solution in termsof integrals. Applying the results obtained there leads to the expression (6.7).If c1 = d2 it follows c1 = c2 = d2 = 0. The least common left multiple now is

Lclm(ρx, ρy, ρx, ρy + (a1 − b2

)ρ) =

ρ1 ≡ ρx, ρ2 ≡ ρyy + (a1 − b2 − d1)ρy(6.11)


and yields the exact quotient ρ2 − b1ρ1, ρ1,x − a2ρ1, ρ1,y − b2ρ1. Applyingthe results of Exercise 5 in Chapter 2 leads to the representation (6.8).

This result for the symmetry class S22,2 is more explicit than for symmetry

class S22,1 because the transformation functions are obtained in closed form

without solving any equation. Consequently, whenever a second order quasi-linear equation allows this symmetry type, it is guaranteed that it may betransformed into canonical form by a Liouvillian transformation.

Example 6.9 It has been shown in Example 5.12 that Equation 6.159of Kamke’s collection belongs to symmetry class S2

2,2. The integral (6.6) ofTheorem 6.9 yields σ = 1√

y. Because c1 = d2 = 0, the integral (6.8) yields

ρ = x. The inverse of this transformation x = v, y = 1u2 yields the canonical

form v′′u− 32v′(v′2 − 1) = 0.

Three-Parameter Symmetries. The general proceeding in this subsectionis the same as in the preceding one, i.e., at first the invariants of the structureinvariance groups for the various symmetry classes are determined. Subse-quently the equations for the transformation functions to canonical form areobtained. They turn out to be considerably more complicated than for thetwo-parameter symmetries.

Lemma 6.3 The pde’s for the canonical form transformations σ(x, y) andρ(x, y) for the symmetry type S2

3,1 allow the following invariants w.r.t. theinvariance group that is generated by ∂σ + ∂ρ, σ∂σ + ρ∂ρ, σ

2∂σ + ρ2∂ρ. Ifσx 6= 0 and ρx 6= 0 there are three first order invariants K1 = σy

σx , K2 = ρyρx

and K3 = σxρx(ρ− σ)2

, and six second order invariants

I1 =σxxσx

+2σxρ− σ

, I2 =σxyσx

+2σyρ− σ

, I3 =σyyσy

+2σyρ− σ

,

I4 =ρxxρx− 2ρxρ− σ

, I5 =ρxyρx− 2ρyρ− σ

, I6 =ρyyρy− 2ρyρ− σ

.

Proof The system of pde’s determining any invariant Φ of order nothigher than two comprises Φσ + Φρ = 0 and

σΦσ + ρΦρ + σxΦσx+ σyΦσy

+ ρxΦρx+ ρyΦρy

+σxxΦσxx+ σxyΦσxy

+ σyyΦσyy+ ρxxΦρxx

+ ρxyΦρxy+ ρyyΦρyy

= 0,

σ2Φσ + ρ2Φρ + 2σσxΦσx + 2σσyΦσy + 2ρρxΦρx + 2ρρyΦρy

+2(σσxx + σ2x)Φσxx

+ 2(σσxy + σxσy)Φσxy+ 2(σσyy + σ2

y)Φσyy

+2(ρρxx + ρ2x)Φρxx + 2(ρρxy + ρxρy)Φρxy + 2(ρρyy + ρ2

y)Φρyy = 0.

262

The subsystem containing only derivatives w.r.t. σ, ρ and its first derivativescomprises three independent equations in six variables, i.e., there are threefirst order invariants. The three independent equations in twelve variablesinvolving second derivatives allow 12-3-3=6 second order invariants.

Theorem 6.10 If a second order ode belongs to symmetry class S23,1 two

types of Janet bases may occur. The transformation functions σ and ρ aredetermined by the following systems of equations.

a) Janet basis type J (2,2)3,6 : Three cases have to be distinguished. If b1 6= 0

define z2 ≡ a1b1 + 1 and

P (x, y) ≡ 1(a1b1 + 1)b1

[(z + 1)(a1b1 + 1)b3 − z(a3b1 + c3 − b2)b1

],

Q(x, y) ≡ b14z2

[z2(a3,x − a1d3 − d2) + a3(a3b1 + 2c3 − 3b2)

−a1(c3 − b2)2 − a1a3b1b2].

(6.12)R is determined from

Rx +R2 − PR+Q = 0,

Ry + z + 1b1

R2−[z + 1b1

P+(z + 1b1

)x

]R+ z − 1

b1Q = 0.

(6.13)

The system for σ and ρ is

σy − z + 1b1

σx = 0, σxx + (2R− P )σx = 0, ρ = 1Rσx + σ = 0.

(6.14)If b1 = 0 define

P ≡ 12(a3,x − a1d3 − a1c

23 + 2a3c3 − d2), Q ≡ a1c3 + c2 − a3. (6.15)

R is either determined from

Rx + P = 0, Ry +R2 −QR− 12a1P = 0 (6.16)

and the system for σ(x, y) and ρ(x, y) is

σx = 0, σyy + (2R−Q)σy = 0, ρ =1Rσy + σ. (6.17)

Or R is determined from

Rx +R2 − c3R = 0, Ry + 12a1R

2 − (a1c3 − a3)R+ P = 0 (6.18)

and the system for σ and ρ is

σy + 12a1σx = 0, σxx + (2R− c3)σx = 0, ρ = 1

Rσx + σ = 0. (6.19)


b) Janet basis type J (2,2)3,7 : Defining b1 and B by

b1 ≡ −c2, B ≡ exp(

2∮a3dx+

cx + a3c

c2dy

),

at first R and S may be determined from

Rx +R2 − (a2c+ a3)R+B = 0,

Ry − 1cR

2 − cx − (a2c+ a3)cc2

R− 1cB = 0,

Sx − S2 + (a2c− a3)S −B = 0,

Sy + 1cS

2 − cx + (a2c− a3)cc2

S + 1cB = 0.

(6.20)

Then the system for σ and ρ is

σy − 1c σx = 0, σxx + (2R− a2c− a3)σx = 0,

ρy + 1c ρx = 0, ρxx − (2S − a2c+ a3)ρx = 0.

(6.21)

Proof At first case a) will be considered. If b1 6= 0, the Janet basis for σand ρ in grlex ordering with ρ > σ, y > x is

σy − z + 1b1

σx = 0, ρy + z − 1b1

ρx = 0,

σxx(ρ− σ) + 2σ2x − Pσx(ρ− σ) = 0, ρxσx −Q(ρ− σ)2 = 0.

The new function R ≡ σxρ− σ is introduced with the result

σx −R(ρ− σ) = 0, σy − z + 1b1

R(ρ− σ) = 0,

ρx − QP (ρ− σ) = 0, ρy − z − 1

b1QR (ρ− σ) = 0.

P , Q and z are defined as above. The integrability conditions of this systemare the two first equations of (6.14). If a lex term order with ρ > σ > R isapplied, the complete system (6.14) is obtained.

If b1 = 0, σx = 0, ρx 6= 0, the Janet basis for σ and ρ in grlex term orderingwith ρ > σ, x > y is

σx = 0, σyy(ρ− σ) + 2σ2y −Qσy(ρ− σ) = 0,

ρxσy − P (ρ− σ)2 = 0, ρyσy + 12a1P (ρ− σ)2 = 0

where P and Q are defined by (6.15). The new function R = σyρ− σ is intro-

duced. It yields the system

σx = 0, σyy + (2R−Q)σy = 0, ρyR+ 12a1P (ρ− σ) = 0, ρxR− P (ρ− σ) = 0.

264

Its integrability conditions are the equations (6.16). In order to show this, therelations Qx + 2P = 0, Px = c3P and Py = c2P are useful.

If b1 = 0, σx 6= 0 and ρx = 0, the Janet basis for σ and ρ in the same termordering as above is

σy + 12a1σx = 0, σxx(ρ− σ) + 2σ2

x − c3σx(ρ− σ) = 0,

ρx = 0, ρyσx − P (ρ− σ)2 = 0.

Introducing now R = σxρ− σ as a new function, the system

σx −R(ρ− σ) = 0, σy + 12a1R(ρ− σ) = 0, ρx = 0, ρyR− P (ρ− σ) = 0

for σ and ρ is obtained. A similar reasoning to the above leads to equations(6.18) and (6.19).

In case b), the type J (2,2)3,7 Janet basis for σ and ρ is

σy − 1c σx = 0, ρy + 1

c ρx = 0,

σxx(ρ− σ) + 2σ2x − (a2c+ a3)σx(ρ− σ) = 0,

ρxx(ρ− σ)− 2ρ2x + (a2c− a3)ρx(ρ− σ) = 0.

Defining R ≡ σxρ− σ and S ≡ ρx

ρ− σ leads to the equations

Rx +R2 +RS − (a2c+ a3)R = 0, Ry − 1cR

2 − 1cRS −

1c2

[cx − (a2c+ a3)c]R = 0,

Sx − S2 −RS + (a2c− a3)S = 0, Sy + 1cS

2 + 1cRS −

1c2

[cx + (a2c− a3)c]S = 0.

Combining the two equations determining the x-derivatives and the y-derivativesrespectively leads to

(RS)x − 2a3RS = 0, (RS)y −2c2

(cx + a3c)RS = 0

from which the above expression for B follows.

Example 6.10 The equation considered in Example 5.13 in symmetryclass S2

3,1 is continued. Due to b1 6= 0 the first alternative of case a) in theabove theorem applies. The Riccati system (6.13) has the form

Rx +R2 + 2x

x2 − 2y

R−

2y(

x2 − 2y

)2 = 0,

Ry + xyR

2 +(

2

y − 2x2

− 1y

)R = 0


with the rational solution R = −xyx2y − 2

+ yC + xy . Substituting the special

solution −xyx2y − 2

corresponding to C →∞ into (6.14) yields

σxx = 0, σy −x

yσx = 0, ρ+ (x− 2

xy)σx − σ = 0.

The Janet basis for σ is completely reducible, it may be written as

Lclm(σx −

1xσ, σy −

1yσ, σx, σy

).

The non-constant solution σ = xy yields ρ = 2x . The corresponding transfor-

mation to canonical variables u and v is x = 2v , y = 1

2uv. Substitution intothe original equation finally yields the equation in canonical form

v′′2(u− v)2 + 4v′′v′(v′ + 1)(u− v) + 4v′4 + 4v′3 + 4v′2 = 0.


3,2 allow the following invariants w.r.t. theinvariance group that is generated by ∂σ, σ∂σ, ρ∂ρ, σ2∂σ + σρ∂ρ. If σx 6= 0and ρx 6= 0 there are two first order invariants K1 = σy

σx , K2 = ρyρ −

ρxρσyσx ,

and six second order invariants

I1 =σxxσx− 2

ρxρ, I2 =

σxyσy− 2

ρxρ, I3 =

σyyσy− 2

ρyρ,

I4 =ρxxρ− ρx

ρ

σxxσx

, I5 =ρxyρ− 1

2ρxρ

σxyσx− 1

2ρyρ

σxyσy

, I6 =ρyyρ− ρy

ρ

σyyσy

.

Proof The system of pde’s determining any invariant Φ of order nothigher than two comprises Φσ = 0 and

σxΦσx+ σyΦσy


+ σyyΦσyy= 0,

ρΦρ + ρxΦρx+ ρyΦρy


+ ρyyΦρyy= 0,

σρΦρ + 2σσxΦσx+ 2σσyΦσy

+ (ρσx + σρx)Φρx+ (ρσy + σρy)Φρy

+2(σσxx + σ2x)Φσxx

+ 2(σσxy + σxσy)Φσxy+ 2(σσyy + σ2

y)Φσyy

+(ρσxx + σρxx + 2σxρx)Φρxx + (ρσxy + σρxy + σxρy + σyρx)Φρxy

+(ρσyy + σρyy + 2σyρy)Φρyy= 0.

The number of invariants is determined in the same way as in the proof ofLemma 6.2.

The four parameters of the structure invariance group for the symmetryclass S2

3,2 are obtained by combining the integration constant of a second orderRiccati-like system and a second order Janet basis for σ or ρ respectively.

266

Theorem 6.11 If a second order ode belongs to symmetry class S23,2, three

types of Janet bases may occur. The transformation functions σ and ρ aredetermined by the following systems of equations.

a) Janet basis of type J (2,2)3,4 .

Rxx +R2x − a3Rx − 1

4 (c3,x + c23 − a3c3 − d2) = 0, Ry − 12a2 = 0,

ρx +Rxρ = 0, ρy +Ryρ = 0, σy = 0, σxx − (2Rx − a3)σx = 0.(6.22)

b) Janet basis of type J (2,2)3,7 .

Ryy +R2y − c2Ry − 1

4 (a2,y + a22 − a2c2 − d3) = 0, Rx − 1

2c3 = 0,

ρx +Rxρ = 0, ρy +Ryρ = 0, σx = 0, σyy + (2Ry − c2)σy = 0.(6.23)

c) Janet basis of type J (2,2)3,6 .

Rxx +R2x + a1b3Rx − Q

P = 0, Ry + a1Rx − 12P = 0,

ρx +Rxρ = 0, ρy +Ryρ = 0, σy + a1σx = 0, σxx + (2Rx + a1b3)σx = 0(6.24)

where P (x, y) ≡ a1c3 − a1b2 − a3 ≡ −2 ∆σxρ 6= 0 and

Q(x, y) = 14

[(b1a3 + c3 − b2)(a3,x − d2) + a1(c3 − b2)3

−d3P (x, y) + 2b3P (x, y)2 + a3(b2c3 − a3b1b2 − c23)].

Proof The proof will be given in detail for case a) for Janet basis typeJ (2,2)

3,4 . If its coefficients are compared to the corresponding expressions interms of the transformation functions (see also Theorem 5.8), the followingsystem of pde’s is obtained.

(log ρ2)y = a2,(

log σxρ2

)x= a3,

(log ρyρ

)y= c1, c2 = 0,

(log ρyρ

)x= c3,(

logρ2yσx

)= d1, (log ρx)y(log σx)x − (log ρxx)y + 3(log ρ)x

(log σxρx

)x= d2,

ρxρy

[(log ρxxσxx

)x

+3(

log σxρx

)x

(log σxρ

)x

]= d3.


From this system of equations a Janet basis in total degree, then lexicographicterm order ρ > σ > d3 > d2 > . . . > a3 > a2 is generated.

d1 − 2c3 − a3 = 0, a2,x − a2c3 = 0, a2,y − a2c2 = 0, a3,y − c3a2 = 0,

c3,y − c2,x = 0, d3,y − d2,x + d3c2 − 2d3a2 − d2c3 + 2d2a3 = 0,

c2,xx − d2,y − 2c3,xa2 + 2c2,xc3 − c2,xa3 − 3c23a2 + 2a2a3c3 = 0,

c3,xx − d2,x + 2c3,xc3 − 3c3,xa3 − a3,xc3 − 2d3a2 + 2d2a3 − 2c23a3 + 2c3a23 = 0,

σy = 0, σxxρ− 2σxρx − σxρa3 = 0,

ρy + 12ρa2 = 0, ρxxρ− 2ρ2

x − ρxρa3 + 14ρ

2(c3,x − d2 + c23 − a3c3) = 0.

Introducing the new function R ≡ − log ρ into the last four equations yieldssystem (6.22). For case b) the four highest equations of the Janet basis forthe full system expressing σ and ρ in terms of the coefficients a2, . . . , d2, d3

areσx = 0, ρx + 1

2c3ρ = 0, σyyρ− 2ρyσy − c2σyρ = 0,

ρyyρ− 2ρ2y − ρyρc2 + 1

4 (a2,y + a22 − d3 − a2c2)ρ2 = 0.

Substituting R ≡ − log ρ yields (6.24). Finally for case c) a fairly complicatedsystem of pde’s is obtained for σ and ρ. Introducting again R ≡ − log ρ, itmay be written in the form (6.23).

The equations for R in the systems (6.22), (6.23) and (6.24) are of the form(2.58) on page 89. The solution procedure described there is always applied.

Example 6.11 The symmetry class of equation 6.133 of Kamke’s collectionhas been identified as S2

3,2 in Example 5.14. Because its Janet basis is of type

J (2,2)3,6 , case c) of the above theorem applies. The system (6.24) is

Rxx +R2x + 1

x+ yRx −14

(x+ y)2= 0, Ry +Rx + 1

x+ y = 0.

Solving the first equation leads to the system

Rx =1

x− y + C−

12

x+ y, Ry = − 1

x− y + C−

12

x+ y

for R with the integration constant C ≡ C(y). Because the equation for C(y)obtained from (2.60) is linear, the special solution corresponding to C → ∞may be chosen and the system for σ becomes σy + σx, σxx. It is completelyreducible and may be written as

Lclm(σx − 1

x− y + C1σ, σy + 1

x− y + C1σ)

=

Lclm(σx, σy,

σx + 1

x− yσ, σy −1

x− yσ)

= 0.

268

The two components of the latter representation correspond to the valuesC1 → ∞ and C1 = 0 respectively. A special solution is σ = x − y. Tothe chosen value of C there corresponds the system Rx = Ry = 1

2(x+ y) .

A special solution is R = 12 log (x+ y) and consequently ρ =

√x+ y. The

inverse of the transformation

u = x− y, v =√x+ y is x = 1

2 (v2 + u), y = 12 (v2 − u),

yields the canonical form v′′v3 + 14 = 0 for equation 6.133 of Kamke’s collec-

tion. The transformations of the structure invariance group may be appliedto generate any value for the constant term. Choosing a1 = 2, a2 = 1 anda3 = a4 for example yields v′′v3 + 1 = 0.

For both symmetry classes S23,3(c) and S2

3,4 the four-parameter structureinvariance originates from the second order Janet bases for the transformationfunctions σ and ρ as it is shown in the next lemma. Moreover, the sameremarks apply as for symmetry class S2

2,1 on page 261, i.e., no equations haveto be solved in order to determine the transformation functions.

Lemma 6.5 The pde’s for the canonical form transformations σ(x, y) andρ(x, y) for symmetry class S2

3,3(c) allow the following invariants w.r.t. theinvariance group that is generated by ∂σ, ∂ρ, σ∂σ, ρ∂ρ. If all first derivativesare different from zero there are two first order invariants K1 = σy

σx , K2 = ρyρx

and six second order invariants

I1 =σxxσx

, I2 =σxyσx

, I3 =σyyσy

, I4 =ρxxρx

, I5 =ρxyρx

, I6 =ρyyρy

.


σxΦσx+ σyΦσy


+ σyyΦσyy= 0,

ρxΦρx+ ρyΦρy


+ ρyyΦρyy= 0.


For the symmetry class S23,3(c) the parameter c has been determined in

Theorem 5.8 and may be considered as already known.

Theorem 6.12 If a second order ode belongs to symmetry class S23,3(c)

with c 6= 1, two types of Janet bases may occur. The transformation functionsσ and ρ are determined by the following systems of pde’s.

a) Janet basis of type J (2,2)3,6 and σxρy − cσyρx 6= 0. There are two alter-

natives. If b1 = 0 the system is either

σx = 0, σyy −(a1c3 + c2 + a3

c)σy = 0,

ρy + ca1c− 1ρx = 0, ρxx + cc3ρx = 0

(6.25)


orσy − a1

c− 1σx = 0, σxx − d1σx = 0,

ρx = 0, ρyy − (a1c3 + c2 + ca3)ρy = 0.(6.26)

If b1 6= 0 and c+ 1 6= 0, the system is

σy − c1 + c(c+ 1)b1

σx = 0, σxx + R(x, y)T (x, y)σx = 0,

ρy − cc1 + 1(c+ 1)b1

ρx = 0, ρxx + S(x, y)T (x, y)ρx = 0

(6.27)

where

R(x, y) ≡ c2[b1(a3b1 + c3 − b2 − a1b3)− b3

]+c

[b1(c1 + 2)(a3b1 − a1b3 + c3 − b2)− (2c1 + 1)b3

]+(c1 + 1)

[b1(a3b1 − a1b3 + c3 − b2)− c1b3

],

S(x, y) ≡ c3b1(a3b1 + c3 − b2)

+c2[b1(c1 + 2)(a3b1 + c3 − b2)− (a1b1 + 1)c1b3

]+c

[b1(c1 + 1)(a3b1 + c3 − b2)− (a1b1c1 + 1)b3

],

T (x, y) = b1[c1 + a1b1 + c(c+ 2)(a1b1 + 1)

].

b) Janet basis of type J (2,2)3,7 and σxρy − cσyρx = 0. The transformation

functions σ and ρ are determined by

σy − c+ 1cc1 σx = 0, σxx + a2c1 − (c+ 1)a3

c+ 1 σx = 0,

ρy − c+ 1c1 ρx = 0, ρxx + a2c1 + (c+ 1)a3

c+ 1 ρx = 0.(6.28)

Proof The original expressions of the Janet basis coefficients in terms of σand ρ are too voluminous to be given here explicitly. If they are transformedinto a Janet basis in total degree, then lexicographic term order with ρ > σand y > x, the above linear system for σ and ρ is obtained.

Example 6.12 In Chapter 5, Example 5.15, two equations in symmetryclass S2

3,3(c) have been considered. For the first equation, due to b1 = 0 thefirst alternative of case a) applies. It is not a priori clear which choice of cleads to the correct canonical form. For the first alternative, c = 2

3 and thesystem

σx = 0, σyy +2yσy = 0, ρy = 0, ρxx = 0

is obtained. Its general solution is σ = C1y + C2, ρ = C3

x + C4. The specialchoice C1 = C3 = 1, C2 = C4 = 0 yields the canonical form v′′+ v′4 = 0. The

270

second alternative leads to y′y′′ − 1 = 0. Therefore the former choice is thecorrect one. For the second equation c is determined by c2− 13

6 c+1 = 0 withthe same solutions as above. The value c = 2

3 leads to the Janet basis

σy − 12σx = 0, σxx = 0, ρy − 1

3ρx = 0, ρxx = 0.

A special solution is σ = 2x+y, ρ = 3x+y. Substituting the inverse x = v−uand y = 3u− 2v into the latter equation of Example 5.15 yields v′′ + v′4 = 0,i.e., the same canonical form as the preceding equation.


3,4 allow the following invariants w.r.t. theinvariance group that is generated by ∂σ, ∂ρ, σ∂σ+ρ∂ρ, σ∂ρ. If σx 6= 0 there

are two first order invariants K1 = σyσx , K2 = σxρy − σyρx

σ2x

. If σx 6= 0 and

σy 6= 0 there are six second order invariants

I1 =σxxσx

, I2 =σxyσx

, I3 =σyyσy

,

I4 =σxxρx − σxρxx

σ2x

, I5 =σyyρy − σyρyy

σ2y

, I6 =σxyρx − σxρxy

σ2x

.


σxΦσx+ σyΦσy

+ ρxΦρx+ ρyΦρy


+σyyΦσyy+ ρxxΦρxx

+ ρxyΦρxy+ ρyyΦρyy

= 0,

σxΦρx + σyΦρy + σxxΦρxx + σxyΦρxy + σyyΦρyy = 0.


Theorem 6.13 If a second order ode belongs to symmetry class S23,4 two

types of Janet bases may occur.

a) If the Janet basis is of type J (2,2)3,6 three cases have to be distinguished. If

b1 6= 0, c1 + 1 6= 0 and T (x, y) ≡ 2a1b1 + c1 + 1 6= 0, the transformationfunctions σ and ρ are determined by

σy + 2a1c1 + 1σx = 0, σxx + R(x, y)

T (x, y)σx = 0,

ρy − c1 + 12b1

ρx − c1 − 12b1

σx = 0, ρxx + R(x, y)T (x, y)ρx + S(x, y)

b1T (x, y)σx = 0

(6.29)where

R(x, y) ≡ (c1 + 1)(a3b1 + c3 − 4)− (c1 − 1)a1b3,

S(x, y) ≡ (c1 + 1)[(a3b1 + c3 − b2)b1 + b3]− (c1 − 3)a1b1b3.


If b1 6= 0, c1 + 1 = T = 0 the system for σ and ρ is

σy = 0, σxx − (c3 − b2)σx = 0,

ρy + 1b1σx = 0, ρxx + (c3 − b2)ρx +

(c3 − b2 + f

b1

)σx = 0.

(6.30)

If b1 = c1 + 1 = T = 0 there is always a1 6= 0, the system for σ and ρ is

σx = 0, σyy − (a1c3 + a3 − c2)σy = 0,

ρx + 1a1σy = 0, ρyy − (a1c3 + a3 − c2)ρy −

(a1c3 + a3 − a2

a1

)σy = 0.

(6.31)

b) If the Janet basis is of type J (2,2)3,7 there holds c1 6= 0. The system for σ

and ρ is

σy − 2c1σx = 0, σxx +

(12a2c1 − a3

)σx = 0,

ρy − 2c1 ρx −

2c1σx = 0, ρxx

(12a2c1 − a3

)ρx + 1

2a2c1σx = 0.(6.32)

Proof At first case a) is considered. In terms of σ and ρ the coefficientsa1, b1 and c1 are

a1 = −σ2y

σxσy −∆, b1 =

σ2x

σxσy −∆, c1 =

σxσy + ∆σxσy −∆

where as usual ∆ = σxρy − σyρx. It follows that

c1 + 1 =2σxσy

σxσy −∆, T (x, y) = − 2σxσy∆

σxσy −∆.

The three alternatives correspond to σx 6= 0 and σy 6= 0, σx 6= 0, σy = 0and σx = 0, σy 6= 0 respectively. In the latter case there follows a1 6= 0. Incase b) there holds c1 = 2σxσy . Consequently, c1 = 0 entails σx = 0 which is

not possible due to the constraint σxσy = ∆ 6= 0, i.e., p 6= 0 is assured.

Example 6.13 An equation with this symmetry type has been consideredin Example 5.16. From the type J (2,2)

3,6 Janet basis given there it followsa1 = 0, b1 = −xy and c1 = −1. Therefore the second alternative of case a)applies. It yields

σy = 0, σxx +1xσx = 0, ρy −

y

xσx = 0, ρxx −

1xρx +

1xσx = 0.

A non-constant solution for σ is σ = x2. Substituting it in the equations forρ yields the independent solution ρ = x+ y2.

272

The Symmetry Class of the Projective Group. The largest group ofpoint symmetries that any second order ode may allow has been shown tobe the eight-parameter projective group of R2. Because there is no degreeof freedom in the canonical form v′′ = 0 corresponding to this symmetry,its structure invariance group is identical to the symmetry group itself. Theinvariants of order two of the projective group and its largest subgroup aregiven in the following lemma. The notation for the invariants is chosen suchthat its relation to the differential equations to be discussed subsequently isobvious.

Lemma 6.7 (Lie 1883) The projective group of the plane with generators∂σ, ∂ρ, σ∂ρ, ρ∂ρ, σ∂σ, ρ∂σ, σ

2∂σ + σρ∂ρ, σρ∂σ + ρ2∂ρ

(6.33)

does not have any first order invariants. A fundamental system of four secondorder invariants is

σyρyy − σyyρyσxρy − σyρx ≡ A,

σxρyy − σyyρxσxρy − σyρx + 2σyρxy − σxyρyσxρy − σyρx ≡ B,

σyρxx − σxxρyσxρy − σyρx + 2σxρxy − σxyρxσxρy − σyρx ≡ C,

σxρxx − σxxρxσxρy − σyρx ≡ D.

(6.34)

The largest subgroup of the projective group is generated by∂σ, ∂ρ, σ∂ρ, ρ∂ρ, σ∂σ, ρ∂σ

,

it does not have any first order invariant either. A fundamental system of sixsecond order invariants is

σyρyy − σyyρyσxρy − σyρx ≡ A,

σxρyy − σyyρxσxρy − σyρx ≡ B1,

σyρxy − σxyρyσxρy − σyρx ≡ B2,

σyρxx − σxxρyσxρy − σyρx ≡ C1,

σxρxy − σxyρxσxρy − σyρx ≡ C2,

σxρxx − σxxρxσxρy − σyρx ≡ D.

The proof may be found in the above quoted article by Lie [109], part III, §1.

Theorem 6.14 (Lie 1883) Any second order ode

y′′ +A(x, y)y′3 +B(x, y)y′2 + C(x, y)y′ +D(x, y) = 0 (6.35)

with projective symmetry group is similar to the equation v′′(u) = 0. Thetransformation functions u = σ(x, y) and v = ρ(x, y) are solutions of

σxx −Dσy + (C − 2C2)σx = 0,σxy − C2σy +B2σx = 0,

σyy − (B − 2B2)σy +Aσx = 0(6.36)


and an identical system for ρ such that σxρy − σyρx 6= 0. The coefficientsB2 = b and C2 = −a are determined by the Riccati system

ax + a2 + Ca−Db+Dy +BD = 0,ay + ab− 1

3Bx + 23Cy +AD = 0,

bx + ab− 23Bx + 1

3Cy +AD = 0,by + b2 −Bb+Aa−Ax +AC = 0

(6.37)

of the form (2.65). Its rational solutions may be determined by the algorithmdescribed in Chapter 2. Substituting them into (6.36) generates Janet bases oftype J (1,2)

3,2 for σ and ρ. They may be decomposed by the algorithms describedin Chapter 2 as well.

Proof A general point transformation u = σ(x, y), v = ρ(x, y) of v′′(u) = 0generates an equation of the form (6.35) where the coefficients A, B, C andD are given by the corresponding expressions in Lemma 6.7. These equationsmay be considered as a system of pde’s expressing σ and ρ in terms of the co-efficients of (6.35). Furthermore, there are the relations expressing B1, B2, C1

and C2 from the second part of this lemma in terms of σ and ρ. By means ofthe obvious relationsB = B1+2B2 and C = C1+2C2, B1 and C1 are expressedin terms of B2 and C2, and the resulting system is transformed into a Janetbasis with lexicographical term ordering ρ > σ > B2 > C2 > A > . . . > D.The full Janet basis comprises twelve equations, the upper half of which isgiven by (6.36), and an identical system for ρ. The two lowest equationsare the constraints for A, . . . ,D of Theorem 5.11 guaranteeing the projectivesymmetry. In between there is the system

C2,x − C22 +DB2 + CC2 −Dy −BD = 0,

C2,y +B2C2 + 13Bx −

23Cy −AD = 0,

B2,x −B2C2 − 23Bx + 1

3Cy +AD = 0,

B2,y +B22 −BB2 −AC2 −Ax +AC = 0.

(6.38)

Combined with the two lowest equations they may be considered as coherenceconditions for the linear homogeneous system (6.36). Equations (6.38) expressthe functions B2 and C2 in terms of the known coefficients A, B, C and D.Substituting C2 = −a and B2 = b yields the Riccati system (6.37) for aand b.

The above decomposition of the coefficients of equation (6.35) into B1,B2, C1 and C2 is not obvious. It seems to be more natural to start withthe coefficients A, B, C and D, expressed in terms of σ and ρ, comparethem to the actual coefficients in terms of x and y and generate a Janetbasis from it. In principle this leads to the same answer. However, theresulting system contains several fairly complicated equations for which asolution algorithm does not seem to exist. This is particularly obvious if B2

274

and C2 are substituted by their definition in terms of σ and ρ into the system(6.36). Lemma 6.7 shows clearly the group theoretical origin of the proceedingdescribed in the above proof, it follows closely the discussion by Lie [109], partIII, pages 364-370.

The existence of the crucial relations (6.38) may be understood as follows.In addition to the six second order invariants the group (6.34) has eight inde-pendent third order invariants. Due to the fact that the twelve first derivativesof the second order invariants are also invariants of third order, four relationsmust exist between them. For example, by differentiation and some rearrange-ment the difference C2,x −Dy may be written as

C2,x −Dy =2∆

(σxxρxy − σxyρxx

)− ∆x

∆C2 +

∆y

∆D

with ∆ = σxρy−σyρx. Using ∆x∆ = C2−C1 and ∆y

∆ = B1−B2 and applyingsome simplifications, the relation

C2,x −Dy = (B1 +B2)D − (C1 + C2)C2

is obtained which is the first equation of (6.38). Due to the group structurethey are obtained as integrability conditions during the Janet basis algorithms.

Example 6.14 Consider equation 6.180

x2(y − 1)y′′ − 2x2y′2 − 2x(y − 1)y′ − 2y(y − 1)2 = 0

of the collection by Kamke with A = 0, B = − 2y − 1 , C = − 2

x and

D = −2y(y − 1)x2 . By Theorem 5.34 it belongs to symmetry class S2

8 . The

system (6.37) for a and b is

ax + a2 − 2xa+ 2y(y − 1)

x2 b+ 2x2 = 0, ay + ab = 0,

bx + ab = 0, by + b2 + 2y − 1b = 0.

Its general solution is rational and may be written as

a =C2y − 1y

+2x(y − 1)

y

C1 + C2x(y − 1)

y+x2(y − 1)

y

+1x,

b =C2

x

y2 +x2

y2

C1 + C2x(y − 1)

y+x2(y − 1)

y

− 1y(y − 1)


where C1 and C2 are the integration constants. For the values C1 → ∞ andC2 = 0 the special solution a = 0 and b = − 1

y(y − 1) is obtained. They yield

the Janet basis

σxx + 2y(y − 1)x2 σy − 2

xσx = 0, σxy − 1y(y − 1)σx = 0, σyy + 2

yσy = 0

for σ and an identical one for ρ. It is completely reducible with the Loewydecomposition

Lclm(σx−

(C2 + 2x)(y − 1)C1y + (C2 + x)x(y − 1)

σ, σy−(C2 + x)x

C1y2 + (C2 + x)xy(y − 1)

σ)

= 0.

Choosing C1 → ∞, C2 = 0 or C1 = C2 = 0 or C1 = 0, C2 → ∞ leads to therepresentation

Lclm(σx, σy

,σx −

2xσ, σy −

1y2 − y

σ,σx −

1xσ, σy −

1y2 − y

σ)

= 0

from which the fundamental system1, x(y − 1)

y ,x2(y − 1)

y

is obtained.Therefore possible transformation functions are

u =x(y − 1)

y, v =

x2(y − 1)y

with inverse x =v

u, y =

v

v − u2 .

The projective symmetry entails that any transformation

σ =a1σ + a2ρ+ a3

a7σ + a8ρ+ a9, ρ =

a4σ + a5ρ+ a6

a7σ + a8ρ+ a9

generates the same equation 6.180 from the canonical form v′′(u) = 0. Forexample, choosing ak = 1 for k = 1, 3, 5, . . . , 9 and a2 = a4 = 2 yields thetransformation

u ≡ σ =x(2x+ 1)(y − 1)

y, v ≡ ρ =

x(x+ 2)(y − 1)y

and its reverse

x ≡ φ =v − 2uu− 2v

, y ≡ ψ =3(v − 2u

u2 − 4uv − 6u+ 4v2 + 3v.

If it is applied to equation 6.180 it generates

− (2u− v)3(u− 2v)6

3[(4v + 3)v + u2 − 2(2v + 3)u]3(uv′ − v)3v′′ = 0,

i.e., up to a lower order factor the canonical form v′′ = 0.

276

Example 6.15 As a second example consider equation 6.124

yy′′ − 3y′2 + 3yy′ − y2 = 0

of the collection by Kamke where A = 0, B = −3y , C = 3 and D = −y. By

Theorem 5.34 it belongs to symmetry class S28 . The system (6.37) is

ax + a2 + 3a+ by + 2 = 0, ay + ab = 0, bx + ab = 0, by + b2 +3yb = 0.

Its rational solutions are a = −1, b = 0 or a = −2, b = 0 or a = 0, b = −2y .

Choosing the first alternative yields B2 = 0, C2 = 1 and leads to the Janetbasis

σxx + yσy + σx = 0, σxy − σy = 0, σyy +3yσy = 0.

It is completely reducible and may be represented as

Lclm(σx, σy

,

σx + σ, σy

,σx − σ, σy +

2yσ)

= 0.

There is an identical one for ρ. They yield the fundamental system1, e−x, e

x

y2

.

A possible choice for the transformation functions is

u = e−x, v =ex

y2 with inverse x = log1u, y =

1√uv.

It is instructive to evaluate the same example after a transformation to a newset of variables has been performed. Introducing x and y by x = y

x , y = 1x

and renaming them x and y, the equation 6.124 becomes

y′′− 1xy′3 +

3x2 y

′2y− 3xy′2− 3

x3 y′y2 +

6x2 y

′y− 3xy′+

1x4 y

3− 3x3 y

2 +3x2 y = 0.

There is a third order equation for the x-dependence of w now with the Loewydecomposition

Lclm[wx, wxx+

( 4x− 2x− y

P

)wx+

( 3xy− 1x2 +

2yx3 −

y2

x4 −3x− 2yyP

)w

]= 0

where P ≡ x2 − xy + 13y

2. The first order factor yields the Janet basis

σxxx +6x2 − 9xy + 4y2

3x3 − 3x2y + xy2σxx −6x4 − 12x3y + 10x2y2 − 5xy3 + y4

3x6 − 3x5y + x4y2 σx = 0,

σy −13x

4

x2y − xy2 + 13y

3σx,x +

13x

3 − 23x

2y + 13xy

2

x2y − xy2 + 13y

3σx = 0.


It is completely reducible with the representation

Lclm(σxx − 1

x− yσx + 3x2y − 3xy2 + y3

x5 − x4yσ,

σy − xx− yσx + 1

x− yσ,

σx, σy

)= 0.

In some applications there occur equations of the general form (5.34) withA = 0. This may be due to the fact that the transformation to canonical formis achieved by a so-called fiber-preserving transformation, i.e., the transforma-tion function σ is independent of y. This specialization leads to a considerablesimplification of the the procedure described in the above Theorem 6.14. Thedetails are given in the subsequent corollary.

Corollary 6.3 (Hsu and Kamran 1989) A second order ode of the formy′′ + B(x, y)y′2 + C(x, y)y′ + D(x, y) = 0 is similar to v′′(u) = 0 by a fiber-preserving point transformation if and only if its coefficients satisfy

Cy − 2Bx = 0,(Cx + 1

2C2 − 2Dy − 2BD

)y

= 0.

The transformation function u = σ(x) and v = ρ(x, y) are determined by

ρxx −Dρy − σxxσx ρx = 0, ρxy − 1

2Cρy −12σxxσx ρx = 0, ρyy −Bρy = 0,(σxx

σx

)x− 1

2

(σxxσx

)2 + Cx + 12C

2 − 2Dy − 2BD = 0.

Proof Applying a general fiber-preserving transformation u = σ(x) andv = ρ(x, y) to v′′ = 0 yields

σxρyy′′ + σxρyyy

′2 + (2σxρxy − σxxρy)y′ + σxρxx − σxxρx = 0

from which the relationsρyyρy

= B, 2ρxyρy− σxx

σx= C,

ρxρy

(ρxxρx− σxx

σx

)= D

follow. Transforming this system of pde’s into a Janet basis in lexicographicterm ordering with ρ > σ > B > C > D and y > x yields the above equationsfor σ and ρ. The two lowest equations in the Janet basis are the coefficientconstraints.

It should be noticed that the Riccati equation for σxxσx does not depend ony due to the second coefficient constraint. In this form the above corollarysolves the equivalence problem for second order equations with a maximalnumber of fiber-preserving symmetries constructively, i.e., for all equationsthat occur Liouvillian solutions may be obtained by the algorithms describedin Chapter 2.Canonical Form Transformation and Infinitesimal Generators. In hismain publication on solving differential equations Lie determines at first the

278

symmetry generators for a given second order ode explicitly, and then appliesthem for determining its canonical form. The following theorem is based onLie [113], Kapitel 18, pages 412-425 and Kapitel 23, pages 503-531.

Theorem 6.15 (Lie 1891) Let ξi∂x+ηi∂y, i = 1, . . . , r, r = 1, 2 or 3, be theinfinitesimal generators for a second order ode obeying canonical commutationrelations. In terms of these generators, the transformation functions σ and ρare determined by the following systems of equations where ∆ij = ξiηj − ξjηi.

S21 : ξ1σx + η1σy = 0, ξ1ρx + η1ρy = 1.

S22,1 : σx =

η2∆12

, σy = − ξ2∆12

, ρx = − η1∆12

, ρy =ξ1

∆12.

S22,2 : σx+

η1∆12

σ = 0, σy−ξ1

∆12σ = 0, ρx+

η1∆12

ρ =η2

∆12, ρy−

ξ1∆12

ρ = − ξ2∆12

.

S23,1 : σ + ρ =

∆13

∆12, σρ =

∆23

∆12.

S23,2 : σ2 =

∆13

∆12, ρx +

η1∆12

ρ = 0, ρy −ξ1

∆12ρ = 0.

S23,3 : σ = −∆23

∆12, cρ =

∆13

∆12. S2

3,4 : σ =∆13

∆12, ρ = −∆13 + ∆23

∆12.

Proof For a one-parameter symmetry the solution is given by Lemma 3.4on page 130. For two-parameter symmetries the proof has been given inLemma 3.5 on page 132. For symmetry class S2

2,1 corresponding to group g26,the system

ξ1σx + η1σy = 1, ξ2σx + η2σy = 0, ξ1ρx + η1ρy = 0, ξ2ρx + η2ρy = 1

is obtained; and for symmetry class S22,2 corresponding to group g25 it is

ξ1σx + η1σy = 0, ξ2σx + η2σy = σ, ξ1ρx + η1ρy = 1, ξ2ρx + η2ρy = ρ.

In both cases the above result follows by algebraic elimination of the firstorder derivatives. The transformstion functions are obtained in terms of pathintegrals.

For the three-parameter groups, due to N = 3 in Lemma 2.12, two algebraicrelations for σ and ρ are obtained. For the symmetry S2

3,1 the systems for σand ρ are

ξ1σx + η1σy = 1, ξ2σx + η2σy = σ, ξ3ρx + η3ρy = σ2,

ξ1ρx + η1ρy = 1, ξ2ρx + η2ρy = ρ, ξ3ρx + η3ρy = ρ2


leading to

∆12σ2 −∆13σ + ∆23 = 0, ∆12ρ

2 −∆13ρ+ ∆23 = 0

with the solution as given above. For the symmetry class S23,2 the systems for

σ and ρ are

ξ1σx + η1σy = 1, ξ2σx + η2σy = 2σ, ξ3ρx + η3ρy = σ2,

ξ1ρx + η1ρy = 0, ξ2ρx + η2ρy = ρ, ξ3ρx + η3ρy = σρ2.

The two algebraic relations

∆12σ2 + 2∆23σ = 0, ∆12σρ−∆13ρ = 0

yield σ = ∆13∆12

and σ2 = ∆23∆12

. The second function ρ is not determined bythese relations. Therefore, the first two equations for ρ have to be solved forits first derivatives with the above result. For symmetry class S2

3,3 the systemfor σ and ρ is

ξ1σx + η1σy = 1, ξ2σx + η2σy = 0, ξ3ρx + η3ρy = σ,

ξ1ρx + η1ρy = 0, ξ2ρx + η2ρy = ρ, ξ3ρx + η3ρy = cρ2

leading to ∆12σ + ∆23 = 0 and c∆12 −∆13 = 0. Finally, for S23,4 the system

ξ1σx + η1σy = 0, ξ2σx + η2σy = 0, ξ3ρx + η3ρy = σ,

ξ1ρx + η1ρy = 1, ξ2ρx + η2ρy = 0, ξ3ρx + η3ρy = σ + ρ2

leads to ∆12σ − ∆13 = 0 and ∆12(σ + ρ) + ∆23 = 0 from which the aboverelations follow immediately.

This result shows clearly the decreasing complexity of finding the canon-ical variables with increasing number of symmetries. For the full problem,however, it must be taken into account that symmetry generators have to bedetermined first.

6.3 Nonlinear Third Order Equations

For second order equations there is only a single equivalence class with alinear canonical representative which may be chosen as v′′ = 0. This equi-valence class makes up the full symmetry class S2

8 . The situation is differentfor equations of order higher than two. There are infinitely many equivalenceclasses with linear canonical representatives that may be combined to form

280

several symmetry classes. As it has been shown in Chapter 2, there arespecial solution procedures for linear equations that do not exist for nonlinearones. Therefore, it appears reasonable to consider the two cases separately.The subject of this Section 6.3 is third order equations that are genuinelynonlinear, i.e., they are not equivalent to a linear equation. These latterequations will be treated in the subsequent Section 6.4.Canonical Forms and Structure Invariance Groups. Similar to equa-tions of order two, the tabulation of the groups of the plane including itsinvariants allows determining the possible forms of invariant equations. Theyare described in the subsequent theorem together with its structure invariancegroups.

Theorem 6.16 In canonical variables u and v ≡ v(u) the third orderquasilinear equations with nontrivial symmetries have the following canonicalforms and corresponding structure invariance groups of point transformationsu = σ(x, y) and v = ρ(x, y); f , g and r are undetermined functions of therespective arguments, a, ai for all i and c are constants.

One-parameter symmetry group

S31 : v

′′′ + r(u, v′, v′′) = 0 allows the pseudogroup u = f(x), v = g(x) + cy.

Two-parameter symmetry groups

S32,1: v

′′′ + r(v′′, v′) = 0 allows u = a1x+ a2y + a3, v = a4x+ a5y + a6.

S32,2: u

2v′′′ + r(uv′′, v′) = 0 allows u = a1x, v = a2x+ a3y + a4.

S32,3: v

′′′ + v′r(v′′v′, u

)= 0 allows u = f(x), v = a1y + a2.

S32,4: v

′′′ + r(v′′, u) = 0 allows

σ =(x+ a1)a2

1− (x+ a1)a2a3, ρ =

a3y

1− (x+ a1)a2a4+ f(x).


S33,1: Φ(10)

1 = [(u− v)v′′ + 2v′(v′ + 1)]2

v′3,

Φ(10)2 =

(u− v)2v′′′ + 6(u− v)(v′ + 1)v′′

v′2+

6(v′2 + 4v′ + 1)v′

.

ω(Φ(10)

1 ,Φ(10)2

)= 0 allows

u =a1(x+ a2)

1− a1a3(x+ a2), v =

a1(y + a2)1− a1a3(y + a2)

.


S33,2: v

′′′v5 + 3v′′v′v4 + r(v′′v3) = 0 allows

u =a1(x+ a3)

1 + a4(x+ a3), v =

a2y

1 + a4(x+ a3).

S33,3(c): v

′′′ + v′(c−3)/(c−1)r(v′′v′(2−c)/(c−1)

)= 0 allows u = a1x+ a2,

v = a3y + a4.

S33,4: v

′′′e2v′+ r(v′′ev

′) = 0 allows u = a1x+ a2, v = a3x+ a1y + a4.

S33,5: v

′′′ + v′r(v′′v′

)= 0 allows u = a1x+ a2, v = a3y + a4.

S33,6: v

′′′ + r(v′′) = 0 allows u = a1x+ a2, v = a3y + a4x2 + a5x+ a6.

S33,7: v

′′′ − 2v′′ + v′ + r(v′′ − 2v′ + v) = 0 allows u = x+ a1,

v = a2y + a3 + a4ex + a5xe

x.

S33,8: v

′′′ − v′′ + r(v′′ − v′) = 0 allows u = x+ a1, v = a2y + a3ex + a4x+ a5.

S33,9: v

′′′ − (α + 1)v′′ + αv′ + r[v′′ − (α + 1)v′ + αv] = 0 allows u = x + a1,v = a2y + a3 + a4e

x + a5eαx.

S33,10: v

′′′ + v′′2r(v′) = 0 allows u = a1x+ a2y + a3, v = a4x+ a5y + a6.

Four-parameter symmetry groups

S34,1: v

′′′v′ + av′′2 = 0 with a 6= 0,− 32 allows u = a1x+ a2, v = a3y + a4.

S34,2: v

′′′v + 3v′′v′ + av′′√v′′v = 0 allows u = a1x

1− a2x+ a3, v = a4y

1− a2x.

S34,3(c): v

′′′ + av′′(c−3)/(c−2) = 0 allows u = a1x+ a2, v = a3x+ a4y + a5.

S34,4: v

′′′ + ae−12v

′′= 0 allows u = a1x+ a2, v = a2

1y + a3x2 + a4x+ a5.

Six-parameter symmetry group

S36 : v

′′′v′ − 32v′′2 = 0 allows u = a1 + a2x+ a3x

2, v = a4 + a5y + a6y2.

The proof is similar to that for Theorem 6.4 and is not given. A fewsymmetry types are considered in the exercises.One- and Two-Parameter Symmetries. The symmetry class S3

1 andthe first two symmetry classes S3

2,1 and S32,2 described in Theorem 5.13 cor-

respond to the same groups as the respective symmetry classes for secondorder equations. Its transformation to canonical form is therefore known byTheorems 6.8 and 6.9.

282


1 generates a type J (2,2)1,2 Janet basis. Therefore case b) of Theorem 6.7

applies. The equation dydx

= 0 yields σ = y, and the path integral leads toρ = − log x. Substituting these two functions the canonical form

v′′′v′ − 3v′′2 − (u− 4)v′′v′2 + (2u− 4)v′4 + v′3 = 0

is obtained.

Example 6.17 The equation considered in Example 5.28 in symmetry classS3

2,1 generates a type J (2,2)2,3 Janet basis. Therefore case b) of Theorem 6.8

applies. It yields the system σxx = 0, σxy− 1yσx = 0, σyy = 0 and an identical

system for ρ with a solution basis 1, y, xy. The choice σ = y and ρ = xyleads to x = v

u , y = u and the canonical equation v′′′ + 2v′′v′ + 1 = 0.


2,2 generates a type J (2,2)2,3 Janet basis. The expressions (6.6) and (6.8)

of Theorem 6.9 yield σ = 1y and ρ = x from which the canonical form

u2v′′′ − 3v′u2v′′2 − 5uv′′ − v′3 − 4v′ = 0

is obtained.

The remaining symmetry classes S32,3 and S3

2,4 of this subsection are treatedin the following two theorems.

Theorem 6.17 If a third order ode belongs to symmetry class S32,3, the

transformation functions σ and ρ may be determined as follows; Ci are con-stants.

a) Janet basis type J (2,2)2,2 : σ = f(x), f is an undetermined function of x,

ρ obeysρx = 0, ρyy − b1ρy = 0. (6.39)

b) Janet basis type J (2,2)2,1 : σ is the same as in the preceding case, ρ obeys

ρy + a1ρx = 0, ρxx +(2a1,x

a1− b1

)ρx = 0. (6.40)

c) Janet basis of type J (2,2)2,5 : σ(x, y) = f(ϕ), ϕ(x, y) = C0 is the integral

ofdy

dx= −a1, (6.41)

ρ obeysρx = 0, ρyy +

(2a1,y

a1− c1

)ρy = 0. (6.42)


d) Janet basis type J (2,2)2,4 : σ is the same as in the preceding case, ρ obeys

ρy + b1ρx = 0, ρxx +[2(a1b1)xa1b1 + 1

− c1]ρx = 0. (6.43)

Proof In all four cases the equations for ρ are obtained as part of theJanet basis expressing the transformation functions in terms of the actualcoefficients. In case a) and b) σ obeys σy = 0; in case c) and d) it obeysσy − 1

a1σx = 0 from which the given expressions follow from Corollary 2.7 on

page 78.


2,3 generates a type J (2,2)2,4 Janet basis. The expressions (6.6) and (6.8)

of Theorem 6.9 yield σ = 1y and ρ = x from which the canonical form

v′′′ + v′[(v′′

v′

)3

+ u+ 1]

= 0

is obtained.

Theorem 6.18 If a third order ode belongs to symmetry class S32,4, three

types of Janet bases may occur; Ci are constants.

a) Janet basis type J (2,2)2,1 : σ = C1

∫exp (

∫S(x)dx)dx + C2, S 6= 0 is a

solution of

Sx − 12S

2 + b1,x + 12b

21 − 2b2 = 0, Sy = 0 (6.44)

or σ = C1x+ C2 if S = 0.

ρ = C3 exp ( 12

∫(S + b1)dx−

∫a2dy)

∫exp (

∫a2dy)dy + f(x)

where f(x) is an undetermined function of its argument.

b) Janet basis type J (2,2)2,5 : σ = C1 exp (

∫R(y)dy) + C2, R 6= 0 a solution

ofRy − 1

2R2 + c1,y + 1

2c21 − 2c2 = 0, Rx = 0 (6.45)

or σ = C1y + C2 if R = 0.

ρ = C3 exp ( 12

∫(R+ c1)dy −

∫b1dx)

∫exp (

∫b1dx)dx+ g(y)

where g(y) is an undetermined function of its argument.

c) Janet basis type J (2,2)2,4 : σ obeys the system σy+b1σx = 0, σxx−Sσx = 0

where S is a solution of

Sx − 12S

2 + c1,x + 12c

21 − 2c2 = 0. (6.46)

284

If σ is known, ∆ ≡ σxρy − σyρx is determined from

∆x−(

32σxxσx + a2,xb1 + 1

2c1

)∆ = 0,

∆y−(

32σyyσy + 1

2b1,x + 12a2b1,y + b2 − 1

2b1c1

)∆ = 0

and, finally,

ρ =∫

∆(φ(x, y), y)σx(φ(x, y), y)

dy∣∣∣x=σ(x,y)

(6.47)

where φ(x, y) is defined as the inverse of x = σ(x, y).

Proof In case a) for σ and ρ the Janet basis

σy = 0, ρxy − 12

(σxxσx + b1

)ρy = 0, ρyy − a2ρy = 0,

σxxxσx − 32σ

2xx + (b1,x + 1

2b21 − 2b2)σ2

x = 0

is obtained. Defining S ≡ σxxσx , the last equation becomes the Riccati equa-

tion for S given above with the general solution S ≡ S(x,C0), C0 the inte-gration constant. From the third equation ρ = C3(x)

∫exp (

∫a2dy)dy + f(x)

is obtained. Substituting the derivatives ρy and ρxy into the second equationyields the linear first order equation C3,x + (

∫a2,xdy − 1

2S −12b1)C3 = 0 for

C3. Consequently,

C3 = C3 exp(

12

∫(S + b1)dx−

∫a2,xdydx

)with C3 constant. Substituting this value into the expression for ρ completescase a). In case b) the equations for σ and ρ are

σx = 0, ρxx − b1ρx = 0, ρxy − 12

(σyyσy + c1

)ρx,

σyyyσy − 32σ

2yy + (c1,y + 1

2c21 − 2c2)σ2

y = 0.

A similar calculation as in case a) leads to the above result. In case c) theequations for σ are

σy + b1σx = 0, σxxxσx − 32σ

2xx +

(c1,x − 2c2 + 1

2c21

)σ2x = 0

from which the given representation follows if S ≡ σxxσx is defined. The system

for ρ is

ρxy + b1ρxx − 12

(σxxσx + 2a2,xb1 + c1

)ρy − 1

2b1

(σxxσx + c1

)ρx = 0,

ρyy − b21ρxx + b1

(σxxσx − a2,y + a2,xb1 + c1 + a2b2

)ρy

+b21(σxxσx + c1 + a2b2

)ρx = 0.


Defining ∆ as above and expressing the derivatives of ρ in terms of ∆, theabove system for ∆ is obtained. If σ and ∆ are known, ρ may be determinedfrom it applying part ii) of Corollary 2.7.


2,4 generates a type J (2,2)2,5 Janet basis. Consequently, case b) applies.

R = 0 leads to the special solution σ = y. The corresponding value for ρ is 1x .

From these transformation functions the canonical form v′′′+uv′′2+v′′+1 = 0follows.

Three-Parameter Symmetries. The first four symmetry classes in thissubsection are S3

3,1, . . . ,S33,4. They correspond to the respective symmetry

classes of second order equations, and it has been described in Theorems 6.10to 6.13 how the transformation functions to canonical form are determined.These results may be applied to third order equations without any change.

Example 6.21 The canonical form transformation of the ode in symmetryclass S3

3,1 considered in Example 5.32 is covered by Theorem 6.10, case a).Because b1 = 0, P and Q are determined from (6.15) with the result P =

1(x− y)2 , Q = 2

x− y . R is obtained from (6.16) with the result R = xy(x− y) .

The system for σ is σx = 0, σyy + 2yσy = 0 with the non-constant solution

σ = 1y from which ρ = 1

x follows. With these transformation functions thecanonical form

v′′′ + 6v′′(v′ + 1)u− v + 6v′

(u− v)2 (v′2 + 4v′ + 1)

+ v′2

(u− v)2[ (v′′(u− v) + 2v′(v′ + 1))2

v′3+ 1

]= 0

is obtained.


3,2 considered in Example 5.33 is covered by Theorem 6.11, case b).At first Ryy + R2

y + 2yRy = 0 yields the special solution Ry = −1

y . With

Rx = − 1x the system ρx − 1

xρ =, ρy − 1y ρ = is obtained with the solution

ρ = xy. Finally, σx = σyy = 0 with the special solution σ = y leads to thecanonical form v′′′v5 + 3v′′v′v4 + 2v′′v2 + 1 = 0.


3,3(c) considered in Example 5.34 is covered by Theorem 6.12. Because

b1 = y2x 6= 0 of its type J (2,2)

3,6 Janet basis, the last case of a) applies. The

systems for σ and ρ are σy = 0, σxx+ 2xσx = 0 and ρy+x

y ρx = 0, ρxx+xy ρx = 0

from which the transformation functions σ = 1x and ρ = y

x are obtained. Theyyield the canonical form v′′′v′3 + 1 = 0.

286


3,4 considered in Example 5.35 is covered by Theorem 6.13, case a).Due to b1 = c1 = −1, the system for σ is σx = σyy = 0 with the specialsolution σ = y. The remaining equations ρx = σy, ρyy = 0 yield ρ = x andthe the canonical form v′′′ + v′′2 + e−2v′ = 0 follows.

The remaining six three-parameter symmetries S33,5, . . . ,S3

3,10 do not havea counterpart for second order equations and are treated next. For the sym-metry class S3

3,9(c) the parameter c has been determined in Theorem 5.14 interms of Janet basis coefficients; i.e., the symmetry class may be considered asknown. The proof of Theorem 6.19 to 6.24 is essentially identical in all cases,the given equations are immediately obtained as part of a Janet basis for thepde’s expressing the general determining system in terms of the respectiveJanet basis coefficients.

Theorem 6.19 If a third order ode belongs to symmetry class S33,5, the

transformation functions σ and ρ may be determined by the following systemsof pde’s.

a) Janet basis type J (2,2)3,5 :

σy = 0, σxx − a2σx = 0, ρx = 0, ρyy − d1ρy = 0. (6.48)

b) Janet basis type J (2,2)3,4 :

σy = 0, σxx − a3σx = 0, ρy + c1ρx = 0, ρxx − (2c3 − 2a3 − d1)ρx = 0.(6.49)

c) Janet basis type J (2,2)3,7 : There holds always c1 6= 0.

σy−1c1σx = 0 = 0, σxx−a3σx = 0, ρx = 0, ρyy−(2a2−2c2−d1)ρy = 0.

(6.50)

d) Janet basis type J (2,2)3,6 : If b1 = 0, the system is

σx = 0, σyy − c2σy = 0, ρy + a1ρx = 0, ρxx − d1ρx = 0 (6.51)

and if b1 6= 0

σy − 1b1σx = 0, σxx − b3

b1σx = 0,

ρy + a1ρx = 0, ρxx + (2a1b3 − 2a3b1 − d1)ρx = 0.(6.52)

Example 6.25 The Janet basis coefficients given in Example 5.36 lead tothe systems σx = 0, σyy+ 2

yσy = 0 and ρy = 0, ρxx+ 2xρx = 0. A functionally


independent set of solutions is σ = 1y , ρ = 1

x . It yields the canonical form

v′′′v′ + v′′2 − 2v′2 = 0.

Theorem 6.20 If a second order ode belongs to symmetry class S33,6, the

transformation functions σ and ρ are determined by the following systems ofpde’s.


σy = 0, σxx − a3σx = 0, ρxy − c3ρy = 0, ρyy − c2ρy = 0,

ρxxx − 3a3ρxx − d3ρy − (a3,x − 2a23)ρx = 0.


σx = 0, σyy − c2σy = 0, ρxx − a3ρx = 0, ρxy − 12 (c2 + d1)ρx = 0,

ρyyy − 3c2ρyy + (d1,y + 32c

22 + 1

2d21 − 2d3)ρy − d2ρx = 0.

c) Janet basis type J (2,2)3,6 : There holds always a1 6= 0.

σy + a1σx = 0, σxx + a1b3σx = 0,

ρxy + a1ρxx − (a1b3 − a3b1 + b2)ρy − a3ρx = 0,

ρyy − a21ρxx + (a2

1b3 + a3 − c2)ρy + (a1a3 − a2)ρx = 0,

ρxxx + 3a1b3ρxx −(b3,x + 3a1b

23 + b2b3 + 2a3

a1b3 + d3

a1

)ρy

−(2a3b3 + d3)ρx = 0.

An example for a canonical form transformation in symmetry class S33,6

is given following Theorem 6.23 in conjunction with examples for symmetryclasses S3

3,7, S33,8 and S3

3,9(c) which are treated in the next three theorems.




σy = 0, σx = c3 − 12 (a3 + d1), ρxy − c3ρy = 0, ρyy − c2ρy = 0,

ρxxx + (d1 − 2a3 − 2c3)ρxx − d3ρy

−(a3,x − 54a

23 − 1

4d21 + d1c3 + 1

2a3d1 − c23 − a3c3)ρx = 0.

288


σx = 0, σy = a2 − 12 (c1 + d1), ρxx − a3ρx = 0, ρxy − a2ρx = 0,

ρyyy − (2a2 + 2c2 − d1)ρyy + (d1,y + a22 + a2c2 − a2d1

+ 34c

22 − 1

2c2d1 + 34d

21 − 2d3)ρy − d2ρx = 0.

c) Janet basis type J (2,2)3,6 : There holds always a2

1b3 − a1d1 + 2a3 6= 0 anda1 6= 0.

σx = a1b3 − 2a3b1 − d1, σy = a1d1 − a21b3 − 2a3,



1b3 + a3 − c2)ρy + (a1a3 − a2)ρx = 0,

ρxxx +P1(x, y)Q(x, y)

ρxx +P2(x, y)Q(x, y)

ρy +P3(x, y)4Q(x, y)

ρx = 0

where

P1(x, y) ≡ 2a31b

23 − a2

1b3d1 + 2a1a3b3 − a1d21 + 4a2

3b1 + 4a3d1, (6.53)

P2(x, y) ≡ −b3,xa21b3 + b3,xa1d1 − 2b3,xa3 − 2a3

1b33 − a2

1b2b23

+a21b

23d1 − 4a1a3b

23 + a1b2b3d1 + a1b3d

21 − a1b3d3

+2a3b1d3 − 2a3b2b3 − 2a3b3d1 + d1d3,

(6.54)

P3(x, y) ≡ a41b

33 − 3a3

1b23d1 − 2a2

1a3b23 + 3a2

1b3d21 − 4a2

1b3d3

−4a1a3b3d1 − a1d31 + 4a1d1d3 + 8a3

3b21

+12a23b1d1 − 4a2

3b3 + 6a3d21 − 8a3d3,

Q(x, y) ≡ a21b3 − a1d1 + 2a3. (6.55)




σx = 2c3 − a3 − d1, σy = 0, ρxy − c3ρy = 0, ρyy − c2ρy = 0,

ρxxx + (d1 − 2a3 − 2c3)ρxx − d3ρy

−(a3,x − a23 + a3d1 − 2a3c3)ρx = 0.



σx = 0, σy = 2a2 − c2 − d1, ρxx − a3ρx = 0, ρxy − a2ρx = 0,

ρyyy − (2a2 + 2c2 − d1)ρyy + (d1,y − 2a22 + 4a2c2

+2a2d1 − 2c2d1 − 2d3)ρy − d2ρx = 0.


σx = a1b3 − 2a3b1 − d1, σy = a1d1 − a21b3 − 2a3,



1b3 + a3 − c2)ρy + (a1a3 − a2)ρx = 0,

ρxxx + (2a1b3 + d1 − 2a3a1

)ρxx

−(b3,x + b3d1 + 2a1b23 + b2b3 + d3

a1)ρy − 2a3b3ρx = 0.

Theorem 6.23 If a second order ode belongs to symmetry class S33,9(c) the



σx =1

c+ 1(2c3 − a3 − d1), σy = 0, ρxy − c3ρy = 0, ρyy − c2ρy = 0,

ρxxx + (d1 − 2a3 − 2c3)ρxx − d3ρy + [a3,x − a23 − c

(c+ 1)2(a2

3 + d21)

− c2 + 1(c+ 1)2

a3(2c3 − d1)− 4c(c+ 1)2

c3(c3 − d1)]ρx = 0.


σx = 0, σy =1

c+ 1(2a2 − c2 − d1), ρxx − a3ρx = 0, ρxy − a2ρx = 0,

ρyyy + (d1 − 2a2 − 2c2)ρyy +[c2,y − c22 − c

(c+ 1)2(c22 + d2

1)

− c2 + 1(c+ 1)2

c2(2a2 − d1)− 4c(c+ 1)2

a2(a2 − d1)]ρx = 0.


σx =1

c+ 1(a1b3 − d1 + 2

a3

a1

), σy =

1c+ 1

(a1d1 − a21b3 − 2a3),

290



1b3 + a3 − c2)ρy + (a1a3 − a2)ρx = 0,

ρxxx +(2a1b3 + d1 − 2a3

a1

)ρxx −

(b3,x + 2a1b

23 + b2b3 + b3d1 + d3

a1

)ρy

−[d3 + 2 c2 + 1

(c+ 1)2a3b3 − c

(c+ 1)2(a1b3 − d1)2

− 4c(c+ 1)2

a3

a21

(a3 − a1d1)]ρx = 0.

Example 6.26 The type J (2,2)3,4 Janet bases for the ode’s that have been

considered in Example 5.37 are dealt with in case a) of Theorems 6.20 to6.23. In the first case, for σ the equations σy = 0, σxx− 2

xσx = 0 are obtainedwith the fundamental system

1, 1x

. For the remaining three cases σ obeys

σy = 0, σx = 1x2 with the special solution σ = − 1

x . For all four cases thesystems for ρ comprise the equations ρxy = 0 and ρyy = 0. In addition thereis a third order equation

ρxxx + 6xρxx + 6

x2 ρx = 0, ρxxx +(

6x + 2

x2

)ρxx +

(6x2 + 4

x3 + 3x4

)ρx = 0,

ρxxx +(

6x + 1

x2

)ρxx −

(2x −

2x3

)ρx = 0

orρxxx +

( 6x

+4x2

)ρxx −

( 2x2 −

8x3 +

6x4

)ρx = 0

It is not necessary to obtain a fundamental system for ρ, rather it is sufficientto find a single solution such that σ and ρ are functionally independent. Itturns out that ρ = y together with σ = − 1

x have this property and transformthe ode’s of Example 5.37 into a canonical form. In all four cases it has thegeneral structure dz

dxz+z2+1 = 0 where z is defined by z ≡ v′′, z ≡ v′′−2v′+v,

z ≡ v′′ − v′ or z ≡ v′′ − 4v′ + 3v.

Theorem 6.24 If a second order ode belongs to symmetry class S33,10 the

transformation function σ is determined by the system of pde’s

σxx − b3σy − (b2 − c3)σx = 0, σxy − b2σy − a3σx = 0,

σyy − (a3 + c2)σy − a2σx = 0.(6.56)

The function ρ satisfies an identical system.

Example 6.27 The Janet basis coefficients given in Example 5.38 leadto the system σxx + 2

xσx = 0, σxy = 0, σyy = 0 for σ and an identical onefor ρ with the solution basis

1, 1x, y

. According to the preceding theorem a

proper choice for the transformation functions is σ = 1x , ρ = y. It yields the

canonical form v′′′v′ + v′′2(v′2 + 1) = 0.


Four-Parameter Symmetries. They occur for the first time for equationsof order three. For the symmetry class S3

4,3(c) the parameter c has been deter-mined in Theorem 5.14 in terms of Janet basis coefficients; i.e., the symmetryclass may be considered as known.




σx = 0, σyy − d1σy = 0, ρy = 0, ρxx − c2ρx = 0. (6.57)


σy + b1σx = 0, σxx + (b1d2 − c2)σx, ρy = 0, ρxx − c2ρx = 0. (6.58)

c) Janet basis type J (2,2)4,14 : Two cases have to be distinguished. If a1 = 0

σy −1b1σx = 0, σxx − c2σx = 0, ρx = 0, ρyy − (2b3 − c1)ρy = 0 (6.59)

or an identical system with σ and ρ exchanged; and if a1 6= 0

σy − 1√−a1

σx = 0, σxx +(b4 + a1e2 − d2

√−a1

)σx = 0,

ρy + 1√−a1

ρx = 0, ρxx +(b4 + a1e2 + d2

√−a1

)ρx = 0.

(6.60)

Proof The given equations are immediately obtained as part of a Janetbasis for the pde’s expressing the general determining system in terms of therespective Janet basis coefficients. In case c) for a1 = 0 it cannot be decideda priori which alternative applies. If u = σ and v = ρ, as obtained by theabove system, does not yield the canonical form, the proper transformationis u = ρ and v = σ. The algebraic function that may occur for a1 6= 0 doesnot prevent the solution because only integrations are involved.

Example 6.28 The coefficients of the type J (2,2)4,14 Janet basis in Exam-

ple 5.39 single out the first alternative of case c). The system σy − yxσx = 0,

σxx = 0 yields σ = xy, and the system ρx = 0, ρyy + 2y ρy = 0 yields ρ = 1

y .Substitution into the equation of Example 5.39 yields the canonical formv′′′v′− 1

2v′′2 = 0. Exchanging σ and ρ, the canonical form v′′′v′− 5

2v′′2 = 0 is

obtained.

292

Theorem 6.26 If a second order belongs to symmetry class S34,2 the trans-

formation functions σ and ρ are determined by the following systems of pde’s.


Rx + 12d2 = 0, Ryy +R2

y − d1Ry − 14c1d2 = 0,

ρx +Rxρ = 0, ρy +Ryρ = 0, σx = 0, σyy − (2Ry + d1)σy = 0.(6.61)


Ry + b1Rx + 12b

21d2 + a4 − b1c2 − 1

2c1 = 0,

Rxx +R2x + (b1d2 − c2)Rx + 1

4d2(b21d2 + 2a4 − 2b1c2 − c1) = 0,

ρx +Rxρ = 0, ρy +Ryρ = 0,

σy + b1σx = 0, σxx + (2Rx + b1d2 − c2)σx = 0.(6.62)

Proof In case a) the system of pde’s expressing the transformation func-tions in terms of the Janet basis coefficients comprises the following equationscontaining σ and ρ.

σx = 0, σyy − (2ρyρ + d1)σy = 0,

ρx + 12d2ρ = 0, ρyyρ− 2ρ2

y − d1ρyρ+ 14c1d2 = 0.

Introducing the new function R by log ρ = −R leads to the equations givenabove. In case b) the equations obtained in the first place are

σy + b1σx = 0, σxx − (2ρxρ − b1d2c2)σx = 0,

ρy + b1ρx + (b1c2 + 12c1 − a4 − 1

2b21d2)ρ = 0,

ρxxρ− 2ρ2x + (b1d2 − c2)ρxρ

−( 12a4d2 + 1

4a4d2 + 14b

21b

22 − 1

2b1c2d2 − 14c1d2)ρ2 = 0.

The same substitution log ρ = −R yields the equations for case b).

Theorem 6.27 If a second order ode belongs to symmetry class S34,3(c) the



σy = 0, σxx − c2σx = 0, ρxx − d2b2 + 2ρy − c2ρx = 0,

ρxy + (b2c2 − b4)ρy = 0, ρyy − b3ρy = 0.(6.63)



σy − b2 + 1a2

σx = 0, σxx +[(b2 + 1)(a3 + d1 − 2a4

a2− c1a2

) + b2c2]σx = 0,(6.64)

ρxx + 12b2 + 1

[2a2(b2 + 1)(a3 + d1)− (2b2 + 1)(2a4 + c1) + 2a2b2c2

]ρy

1(b2 + 1)(2b2 + 1)

[(b2 + 1)(2b4 + d1 − a3) + a2b2(b3 + d2 + e1)

]ρx = 0,

ρxy − b1 + 1a2

ρxx + 12a2

[2a4(b2 + 1)− a2(a3 + 2b4 + c2 + d1)

]ρy

+ 12a2

(b2 + 1)(a3 + c2 + d1)ρx = 0,

ρyy +[

12b2(b3 + d2 + e1)− b3 − a2e2

(b1 + 1)(2b2 + 1)

]ρy + e2

2b2 + 1ρx = 0.

(6.65)

Proof The given equations are immediately obtained as part of a Janetbasis for the pde’s expressing the general determining system in terms of therespective Janet basis coefficients. In case a) the constraint c 6= −2 assuresthat the denominator b2 + 2 6= 0, see Theorem 5.13.

Example 6.29 The type J (2,2)4,9 Janet basis in Example 5.40 singles out

case a). The system σy = 0, σxx + 2xσx = 0 yields the non-constant solution

σ = 1x . The system

ρxx −2yx2 ρy +

2xρx = 0, ρxy −

1xρy = 0, ρyy = 0

yields the independent solution ρ = xy. Substitution into the equation ofExample 5.40 leads to the canonical form v′′′ + v′′3 = 0.



a) Janet basis type J (2,2)4,9 : There holds always d2 6= 0.

σy = 0, σxx − c2σx = 0,

ρy = −2σ2x

d2, ρxxx − 3c2ρxx + (2c22 − c4)ρx = 2

(c2 − d4

d2

)σ2x.

(6.66)

b) Janet basis type J (2,2)4,14 : There holds always e2 6= 0. Two cases have to

be distinguished. If b1 = 0 the system is

σx = 0, σyy − (b3 + d2)σy = 0, ρx = − 1e2σ

2y,

ρyyy − 3(b3 + d2)ρyy −(b3,y + 1

2e4

−2b23 − 3b3d2 + b4e2 − 3d22

)ρy =

(2d2 + e3

e2

)σ2y

(6.67)

294

and if b1 6= 0

σy − 32b1

σx = 0, σxx −(

49b

21e2 + 3a4

b1+ 3c1

2b1)σx = 0,

ρy − 32b1

e2ρx =( 32b1

)3σ2x,

(6.68)

ρxxx − 1b1

(9a4 + 4

3b31e2 + 9

2c1)ρxx

+ 1b21

(32c1,xb1 + 18a2

4 + 8a4b31e2 + 63

4 a4c1 + 6481b

61e

22

+ 83b

31c1e2 + 2

3b31c3 + 2b21c1d2 − b21c4 + 3

2b1b4c1 + 92c

21

)ρx

= σ2x

b41e2

(− 27

8 c1,xb1 + 24316 a4c1 + 3b31c1e2 − 3

2b31c3 − 9

2b21c1d2 − 27

8 b1b4c1).

(6.69)

Proof The given equations are immediately obtained as part of a Janetbasis for the pde’s expressing the general determining system in terms of therespective Janet basis coefficients.

Six-Parameter Symmetry. The only symmetry class is S36 with canoni-

cal form v′′′v′ − 32v′′2 = 0. It is treated along similar lines as the projective

symmetry S28 of second order equations. At first the invariants are deter-

mined from which the coefficients of an equation with this symmetry may beconstructed.

Lemma 6.8 The differential invariants up to order three for the group S36

generated by the vector fields ∂σ, ∂ρ, σ∂σ, ρ∂ρ, σ2∂σ, ρ2∂ρ and the transfor-

mation u = σ(x, y), v = ρ(x, y) may be described as follows.

a) For σx 6= 0 and ρx 6= 0 there are two first order invariants, four secondorder invariants

I1 =σyσx, I2 = I1,x, I3 = I1,y, K1 =

ρyρx, K2 = K1,x, K3 = K1,y

and eight third order invariants

I4 = σxxxσx −

32

(σxxσx

)2

, I5 = σyyyσy −

32

(σyyσy

)2

, I6 = I2,x, I7 = I2,y,

K4 = ρxxxρx −

32

(ρxxρx

)2

, K5 = ρyyyρy −

32

(ρyyρy

)2

, K6 = K2,x, K7 = K2,y.

b) For σx = 0, ρx 6= 0 there is no invariant involving derivatives of σ upto order two and one invariant

J1 = σyyyσy −

32

(σyyσy

)2

of order three with J1,x = 0. The invariants K1, . . . ,K7 involving ρ arethe same as in the preceding case.


c) For σx 6= 0, ρx = 0 the invariants I1, . . . , I7 involving σ are the same asin case a). There is no invariant involving derivatives of ρ up to ordertwo and one invariant

L1 = ρyyyρy −

32

(ρyyρy

)2

of order three with L1,x = 0.

Proof The prolongations of the group generators up to order three are

∂σ, σx∂σx+ σy∂σy

+ σxx∂σxx+ σxy∂σxy

+ σyy∂σyy

+σxxx∂σxxx+ σxxy∂σxxy

+ σxyy∂σxyy+ σyyy∂σyyy

,

σ2x∂σxx

+ σxσy∂σxy+ σ2

y∂σyy+ 3σxσxx∂σxxx

+(2σxσxy + σxxσy)∂σxxy+ (2σxyσy + σxσyy)∂σxyy

+ 3σyσyy∂σyyy

and the same set with σ replaced by ρ. Due to the occurrence ∂σ and ∂ρ, anyinvariant is independent of σ and ρ. There are two equations involving thefour first order derivatives. Consequently, there are two first order invariants.Up to order two there are twelve variables involved leading to six invariantsaltogether. Up to order three there are twenty variables involved leading tofourteen invariants up to this order. If the respective numbers of lower orderinvariants are taken into account the above figures are obtained. The func-tional independence is guaranteed due to the fact that each invariant dependson a different set of variables. The actual determination of the invariants isachieved by the methods described in Section 5 of Chapter 1. If σx = 0 andρx 6= 0, the three prolongations involving derivatives of σ are

∂σ, σy∂σy+ σyy∂σyy

+ σyyy∂σyyy, σy∂σyy

+ 3σyy∂σyyy

with the obvious solution J1. If σx 6= 0 and ρx = 0, the same prolongationsinvolving derivatives of ρ are obtained with the solution L1.

Lemma 6.9 The coefficients of an equation with S36 symmetry may be

expressed in terms of the invariants introduced in the above lemma.

a) For equation (5.40), σx 6= 0 and ρx 6= 0, the expressions for the coeffi-cients are:

A0 = I1K1, A1 = I1 +K1, A2 = 3I3K21 − I2

1K3

I1 −K1,

A3 = 3K21I2 − I2

1K2 + 2(K1I3 − I1K3)I1 −K1

,

A4 = 3I3 −K3 + 2(K1I2 − I1K2)I1 −K1

, A5 = 3I2 −K2I1 −K1

,

B1 = I21K

21I5 −K5I1 −K1

,

B2 = A0(A0 +A21)B6 − 2A0A1(2A0 +A1)B7,

296

B3 = A1(3A0 + 12A

21)B6 − (A2

0 + 7A0A21 +A4

1)B7,

B4 = 2(A0 +A21)(B6 − 2A1B7), B5 = 5

2A1B6 + (A0 − 4A21)B7,

B6 = 3(I6 −K6) + 2(K1I4 − I1K4) + I1I4 −K1K4

I1 −K1, B7 = I4 −K4

I1 −K1.

b) For equation (5.41), σx = 0 and ρx 6= 0 the expressions for the coeffi-cients are

A0 = K1, A2 = −3K3, A3 = 3K2, B1 = K21 (J1 −K5),

B2 = 2J1K1 − 3K31K4 − 3K2

1K6 + 32K1K

22 −K1K5

−3K1K7 + 3K2K3 + 32

K23

K1,

B3 = J1 − 6K21K4 − 6K1K6 + 3K2

2 − 3K7,

B4 = −4K1K4 − 3K6, B5 = −K4.

c) For equation (5.41), σx 6= 0 and ρx = 0 the expressions for the coeffi-cients are

A0 = I1, A2 = −3I3, A3 = −3I2, B1 = I21 (L1 − I5),

B2 = 2L1I1 − 3I31I4 − 3I2

1I6 + 32I1I

22 − I1I5

−3I1I7 + 3I2I3 + 32

I23I1,

B3 = L1 − 6I21I4 − 6I1I6 + 3I2

2 − 3I7,

B4 = −4I1I4 − 3I6, B5 = −I4.

The proof is essentially the same as for Lemma 6.9 except that the canonicalform now is v′′′v′ − 3

2v′′2 = 0. Therefore it is omitted.

Theorem 6.29 The transformation of an equation in symmetry class S36

to canonical form may be determined as follows.

a) If it has the form (5.40) the transformation functions σ and ρ are de-termined by the following system of pde’s.

Sx − 12S

2 − I4 = 0, Sy − 12I1S

2 − I1,xS − I1I4 − I1,xx = 0,

σy − I1σx = 0, σxx − Sσx = 0,(6.70)

Rx − 12R

2 −K4 = 0, Ry − 12K1R

2 −K1,xR−K1K4 −K1,xx = 0,

ρy −K1ρx = 0, ρxx −Rρx = 0.(6.71)

The required invariants may be expressed in terms of the coefficients of(5.40) as follows.

I1 = 12 (A1 + Z), I4 = 1

2B6 − 12 (3A1 − Z)B7 − 3

2ZxxZ .


K1 and K4 are obtained from the corresponding I ′s by replacing Z ≡√A2

1 − 4A0 with −Z. Z is contained in the same field as the transfor-mation functions.

b) If it has the form (5.41), two cases have to be distinguished. Either thetransformation function σ is determined by the system

Sx = 0, Sy − 12S

2 − J1 = 0, σx = 0, σyy − Sσy = 0 (6.72)

whereJ1 = B3 − 6A2

0B5 + 6A0A0,xx − 3A20,x + 3A0,xy,

and the transformation function ρ is obtained from system (6.71) wherenow K1 = A0, K4 = −B5. Or σ is determined by the system (6.70) andρ is determined by the system

Rx = 0, Ry − 12R

2 − J1 = 0, ρx = 0, ρyy −Rρy = 0 (6.73)

with J1 as above.

Proof Case a). The first equation of (6.70) for S ≡ σxxσx is essentially the

definition of I4 in Lemma 6.8. The second equation involving the y-derivativeof S is obtained in the same way as in Theorem 6.33. The equations for σ areessentially the definitions of I1 and S. Similar arguments lead to the system(6.71) for R ≡ ρxx

ρx .

Case b). The equation for S ≡ σyyσy in (6.72) is essentially the definition of

J1 in Lemma 6.8.In either case, the expressions for the invariants are obtained by elimination

from the relations given in Lemma 6.10. Writing Z as Z =√A2

1 − 4A0 =∆2

σ2xρ

2x

, ∆ = σxρy −σyρx, shows that it is contained in the field determined by

the transformation functions.

Example 6.30 An equation in symmetry class S36 , case a), has been

considered in Example 5.41 of the preceding chapter. From the coefficientsgiven there, the invariants I1 = x

y , K1 = −xy , I4 = K4 = 0 are obtained.

Consequently, the equations for S are Sx − 12S

2 = 0, Sy − x2yS

2 − 1yS = 0

with the general solution S = − 2yxy + C . Choosing C →∞ yields S = 0 and

finally σ = xy. The system for R is Rx− 12R

2 = 0, Ry + x2yR

2 + 1yR = 0 with

the general solution R = − 2Cy + x . Choosing now C = 0 yields R = − 2

x and

ρy + xy ρx = 0, ρxx + 2

xρx = 0 with the non-constant solution ρ = yx .

Example 6.31 An equation in symmetry class S36 , case b), has been con-

sidered in Example 5.42 of the preceding chapter. From the coefficients given

298

there, the invariants

J1 = 0, K1 =x

(x− 1)y, K4 = −1

2x2 − 4x+ 6

(x− 1)2

are obtained. Consequently, the equations for S and σ are the same as inExample 6.39, the special solution σ = 1

y is chosen again. R is determined bythe system

Rx − 12R

2 + 12x2 − 4x+ 6

(x− 1)2= 0,

Ry − 12

x(x− 1)yR

2 + 1(x− 1)2y

R+ 12

(x− 2)(x2 − 2x+ 2)(x− 1)3y

= 0.

It has the two rational solutions x2 − 2x+ 2x(x− 1) and −x− 2

x− 1 . Choosing the first

alternative, the system

ρy −x

xy − yρx = 0, ρxx −

x2 − 2x+ 2x2 − x

ρx = 0

for ρ is obtained with the non-constant solution ρ = yxe

x.

6.4 Linearizable Third Order Equations

The distinctive feature of the equations covered by this section is the factthat they allow canonical forms that are linear. Consequently, the solutionprocedures described in Chapter 2 for linear differential equations apply assoon as the transformation to canonical form has been found. At first thecanonical forms and its structure invariance groups are described, after thatthe transformation to these canonical forms for the various symmetry classesis considered.Canonical Forms and Structure Invariance Groups. The subsequenttheorem describes canoncial forms and structure invariance groups for thefour symmetry types that occur in this section.

Theorem 6.30 In canonical variables u and v ≡ v(u) the third orderequations that are equivalent to a linear one and allow nontrivial symmetrieshave the following canonical forms and corresponding structure invariancegroups of point transformations u = σ(x, y) and v = ρ(x, y); ai for all i areconstant.

Four-parameter symmetry group

S34,5: v

′′′ + P (u)v′ +Q(u)v = 0 allows u = a1x+ a2a3x+ a4

, v = y(a3x+ a4)2

.


Five-parameter symmetry groups

S35,1: v

′′′ − v′′ = 0 allows u = x+ a1, v = a2y + a3x+ a4ex + a5.

S35,2(a): v

′′′ − (a + 1)v′′ + av′ = 0 allows u = x + a1, v = a2y + a3ex +

a4aa1eax + a5. In addition five transformations u = αx, v = eβuv with

α = 1a , β = 0; α = β = −1; α = β = 1

a− 1 ; α = 1a , β = 1 and

α = 1a− 1 , β = a

a− 1 .

Seven-parameter symmetry groups

S37 : v

′′′ = 0 allows u = a1 + a2xx+ a3

, v = a4 + a5x+ a6x2 + a7y

x+ a3

2

.

Proof The structure invariance group for the symmetry class S34,5 has

been given in Theorem 4.3. For the symmetry classes S35,1 and S3

7 they areidentical to the respective symmetry groups as given in Theorem 5.16 becausethere are no parameters or undetermined functions involved.

Four-Parameter Symmetry. There is only a single symmetry class, thegeneric linear homogeneous third order ode belongs to it. The subsequenttheorem describes how to transform a general equation in this symmetry classto a linear equation in rational normal form.

Theorem 6.31 If a third order ode belongs to the symmetry class S34,5

three types of Janet bases may occur. The transformation functions σ and ρ torational normal form v′′′+P (u)v′+Q(u)v = 0 are determined as follows; f , gand h are undetermined functions of its arguments; A ≡ P (σ) and B ≡ Q(σ).

a) Janet basis of type J (2,2)4,2 : σ = f(x), the system for ρ is

ρxy−( σxxσx

+13c1

)ρy = 0, ρyy − b2ρy = 0,

ρxxx − 3σxxσx

ρxx + c2ρy

−[ 3

2

( σxxσx

)2

+12c1,x +

16c21 −

12c3 +

32σ2xA

]ρx + σ3

xBρ = 0. (6.74)

A and B are determined through

2( σxxσx

)x−

( σxxσx

)2

+c1,x +13c21 − c3 + σ2

xA = 0,

c1,xx+2c1c1,x−3c3,x−6b2c2 + 49c

31−2c1c3 +6c4 +3σ2

xAx−6σ3xB = 0.

(6.75)

300

b) Janet basis of type J (2,2)4,19 : σ = g(y), the system for ρ is

ρxx − b2ρx = 0, ρxy−( σyyσy

+ 13c1

)ρx = 0,

ρyy − 3σyyσy

ρyy−[

32

( σyyσy

)2

+ 12c1,y + 1

6c21 − 1

2c2 + 32σ

2yA

]ρy

+ c3ρx + σ3yBρ = 0. (6.76)


2( σyyσy

)y−

( σyyσy

)2

+c1,y + 13c

21 − c2 + σ2

yA = 0,

c1,yy+2c1c1,y−3c2,y−6b2c3 + 49c

31−2c1c2 +6c4 +3σ2

yAy−6σ3yB = 0.

(6.77)

c) Janet basis of type J (2,2)4,17 : σ = h(φ) where φ(x, y) = C is the solution

of y′ − 1b1

= 0, the system for ρ is

ρxy + b1ρxx−( σxxσx− b2 + 1

3d1

)ρy−

(b1σxxσx

+ 13b1d1

)ρx = 0,

ρyy − b21ρxx+(

2b1σxxσx

+ 23b1d1 + c2 − 2b3

)ρy

+(

2b21σxxσx

+ 23b

21d1 − b1b3

)ρx = 0,

ρxxx − 3σxxσx

ρxx + d2ρy−[

32

( σxxσx

)2

+ 12d1,x

+ 16d

21 + 3

2b1d2 − 12d3 + 3

2σ2xA

]ρx + σ3

xBρ = 0. (6.78)


2(σxxσx

)x−

(σxxσx

)2

+d1,x + 13d

21 − d3 + b1d2 + σ2

xA = 0,

d1,xx + 2d1d1,x − 3d3,x + 3b1d2,x + 2d1,xd1 + 6d4 − 2d1d3

+2b1d1d2 − 6b2d2 − 3b1b2d2 + 49d

31 + 3σ2

xAx − 6σ3xB = 0.

(6.79)

Proof In case a) the given equations are obtained from a Janet basisrelating σ and ρ to the coefficients of the given type J (2,2)

4,2 for the symmetrygenerators. A(x) ≡ P (σ) and B(x) ≡ Q(σ) are the transformed coefficientfunctions P and Q. Their actual shape is determined by the choice of σ.


This freedom is allowed by the structure invariance group for the canonicalform according to Theorem 6.30. The remaining four constants occur in thesolution for ρ. The discussion for case b) is similar.

Example 6.32 The equation considered in Example 5.43 generates thetype J (2,2)

4,2 Janet basis, therefore case a) of the above theorem applies. Ifσ = x is chosen, the system (6.74) yields the independent solution ρ = x

y andthe canonical form

v′′′ − 8u2u− 1

v′ +8

2u− 1v = 0. (6.80)

Choosing σ = 1x yields ρ = 1

xy , and the canonical form

v′′′ +8

u4(u− 2)v′ − 8

u5(u− 2)v = 0 (6.81)

is obtained.


4,19 Janet basis, therefore case b) of the above theorem applies. Ifσ = y is chosen, the system (6.74) yields the independent solution ρ = x

y and

the canonical form (6.80). Choosing σ = 1y instead yields ρ = 1

xy with the

canonical form (6.81).


4,17 Janet basis, therefore case c) of the above theorem applies. Thefirst order equation for σ is y′+1 = 0; i.e., φ(x, y) = x+y. Choosing σ = x+yyields a system for ρ with the fundamental system 0, x, y. Consequently, anylinear combination ρ = c1x+c2y with c1 6= c2 yields a possible transformationto a canonical form which turns out to be (6.80).

Five-Parameter Symmetries. The parameter a that defines the symmetryclass S3

5,2(a) has been determined in Theorem 5.17. The expressions for P ,Q and R occurring in the following theorem are also given there. The Janetbases of order five that are required are listed on page 382.

Theorem 6.32 If a third order ode belongs to the symmetry class S35,1 or

S35,2(a), three types of Janet bases may occur. The systems of pde’s for the

transformation functions σ and ρ are as follows.


σx =P

Q

a2 − a+ 1(a− 1

2 )(a+ 1)(a− 2), σy = 0, (6.82)

302

ρxy − 13

[(a+ 1)σx + 3a1 + e1

]ρy = 0, ρyy − c1ρy = 0,

ρxxx −[(a+ 1)σx + 3a1

]ρxx + e2ρy

+[aσ2

x + (a+ 1)a1σx + 2a21 − a1,x

]ρx = 0.

(6.83)


σx = 0, σy =P

Q

a2 − a+ 1(a− 1

2 )(a+ 1)(a− 2), (6.84)

ρxx − c1ρx = 0, ρxy − 13

[(a+ 1)σy + 3b1 + e1

]ρx = 0,

ρyyy −[(a+ 1)σy + 3b1

]ρyy

+[aσ2

y + (a+ 1)b1σy + 2b21 − b1,y]ρy + e3ρx = 0.

c) Janet basis type J (2,2)5,3 :

σx =P

Q

a2 − a+ 1(a− 1

2 )(a+ 1)(a− 2), σy = −c1σx, (6.85)

ρxy + c1ρxx − 13

[(a+ 1)σx + e1 + 3a2

]ρy

− 13

[c1e1 − 3c21a3 + c1(a+ 1)σx

]ρx = 0,

ρyy − c21ρxx +[23 (a+ 1)c1σx + 2

3e1c1 − c3 − c21a3 − b2

]ρy

+[23 (a+ 1)c1σx + 2

3e1c1 − c3 − 2c21a3

]c1ρx = 0,

ρxxx −[(a+ 1)σx − 3c1a3

]ρxx + e2ρy

+[aσ2

x − (a+ 1)c1a3σx + (a3,x + e2 + a3a2 + 3c1a23)c1

]ρx = 0.

(6.86)

Example 6.35 From the Janet basis coefficients in Example 5.46, thesystem for σ is σx = y, σy = x, i.e., σ = xy + C. For ρ the system

ρxy − xy ρxx + 1

xρy −2y ρx = 0,

ρyy − x2

y2 ρxx + 1y ρy −

xy2 ρx = 0, ρxxx − yρxx + xy2 + 3y

x3 ρy − xy + 3x2 ρx = 0

is obtained with the three rational solutions ρ = 1, ρ = xy and ρ = yx . The last

alternative yields a transformation with nonvanishing functional determinant;i.e., the desired transformation is x =

√uv , y =

√uv, and the canonical form

is v′′′ − v′′ = 0.


Example 6.36 Choosing a = 3 from Example 5.47, case b) of the abovetheorem leads to σx = 0, σy = 1, i.e., σ = y + C. The system

ρxx = 0, ρxy + 1y ρx = 0,

ρyyy − 4ρyy + 3ρy + 3xy2 + 8xy + 6xy3 ρx = 0

has the rational solution ρ = xy . Together with the special solution σ = y, it

yields the canonical form v′′′ − 4v′′ + 3v′ = 0 for the ode of Example 5.47.

Maximal Symmetry Group with Seven Parameters. The canonicalform corresponding to this group does not contain any unspecified elements;its structure invariance group is therefore identical to its symmetry group. Theproceeding is similar to the second order equations with projective symmetry.This case has been considered in detail by Zorawski [192].

Lemma 6.10 (Zorawski 1897) The differential invariants up to order threefor the group S3

7 generated by∂σ, ∂ρ, σ∂σ, ρ∂ρ, σ∂ρ, σ

2∂ρ, σ2∂σ+2σρ∂ρ

with

σ ≡ σ(x, y) and ρ ≡ ρ(x, y) may be described as follows.

a) For σx 6= 0 there is a single first order invariant I1 = σyσx , there are four

second order invariants I2 = I1,x, I3 = I1,y,

I4 =σxρxy − σxyρx + σxxρy − σyρxx

σxρy − σyρx− 2

σxxσx

,

I5 =σxρyy − σyyρx + σxyρy − σyρxy

σxρy − σyρx− 2

σxyσx

and there are eight third order invariants

I6 =σxxxσx− 3

2

(σxxσx

)2

, I7 =σxρxxx − σxxxρxσxρy − σyρx

− 3σxρxx − σxxρxσxρy − σyρx

σxxσx

,

I8 = I4,x, I9 = I4,y, I10 = I5,y, I11 = I2,x, I12 = I2,y, I13 = I3,y.

b) For σx = 0 there are no first order invariants, there are two secondorder invariants

J1 =ρxxρx

, J2 =ρxyρx− σyy

σy

and there are five third order invariants

J3 =σyyyσy− 3

2

( σyyσy

)2

, J5 = J1,x, J6 = J1,y, J7 = J2,y,

J4 =ρyyyσy − σyyyρy

σyρx− 3σyy

ρyyσy − σyyρyσ2yρx

.

304

Proof The prolongations of the group generators up to order three are,after some reductions have been performed, ∂σ, ∂ρ,

σx∂σx + σy∂σy + σxx∂σxx + σxy∂σxy + σyy∂σyy

+σxxx∂σxxx + σxxy∂σxxy + σxyy∂σxyy + σyyy∂σyyy ,

ρx∂ρx + ρy∂ρy + ρxx∂ρxx + ρxy∂ρxy + ρyy∂ρyy

+ρxxx∂ρxxx+ ρxxy∂ρxxy

+ ρxyy∂ρxyy+ ρyyy∂ρyyy

,

σx∂ρx + σy∂ρy + σxx∂ρxx + σxy∂ρxy + σyy∂ρyy

+σxxx∂ρxxx+ σxxy∂ρxxy

+ σxyy∂ρxyy+ σyyy∂ρyyy

,

σ2x∂ρxx + σxσy∂ρxy + σ2

y∂ρyy + 3σxσxxρσxxx

+(2σxσxy + σxxσy)∂ρxxy+ (2σxyσy + σxσyy)∂ρxyy

+ 3σyσyy∂ρyyy,

ρxσ∂ρx + ρyσ∂ρy + σ2x∂σxx + 2σxρx∂ρxx + σxσy∂σxy + (ρxσy + ρyσx)∂ρxy

+σ2y∂σyy

+ 2ρyσy∂ρyy+ 3σxxσx∂σxxx

+ 3(ρxxσx + ρxσxx)∂ρxxx

+(2σxyσx + σxxσy)∂σxxy+ (2σxyσy + σxσyy)∂σxyy

+(2ρxyσx + ρxxyσ + ρxxσy + 2ρxσxy + ρyσxx)∂ρxxy

+(2ρxyσy + σxσyy + ρyyσx + 2ρyσxy)∂ρxyy+ 3σyyσy∂σyyy

+3(ρyyσy + ρyσyy)∂ρyyy .

They generate a system of linear homogeneous pde’s in the derivatives of σand ρ for the desired invariants. There are 4, 10 or 18 derivatives up to order1, 2 or 3 respectively. If all derivatives are nonvanishing, the above system hasrank three w.r.t. to derivatives of first order, and rank five w.r.t. derivatives oforder higher than one. From these figures the number of invariants in case a)follows. Their explicit form is obtained by applying the methods describedin Section 5 of Chapter 2. The occurrence of the derivatives of the variousorders guarantees their functional independence. If some partial derivativesvanish as it occurs in case b), the corresponding terms in the system for theinvariants disappear.

Lemma 6.11 (Zorawski 1897) The coefficients of an equation with S37

symmetry may be expressed in terms of the invariants introduced in the abovelemma.

a) For equation (5.43) the expressions for the coefficients in terms of theI ′s are:

A0 = I1, A1 = 3I1I5 − 6I3, A2 = 3I1I4 + 3I5 − 6I2, A3 = 3I4,

B1 = I51I7 + I4

1 (2I6 + I11 + I24 ) + I3

1 (I12 + I8 + I4I5 − I2I4)

+ I21 (I13 + I2

3 − I3I4 − I2I5)− I1(I10 + 3I3I5 − I2I3) + 3I23 ,


B2 = 5I41I7 + I3

1 (8I6 + 5I11 + 5I24 ) + I2

1 (5I12 + 3I8 + 5I4I5 − 5I2I4)

+ I1(2I13 − 2I9 + 2I25 − 5I3I4 − 5I2I5 + I2

2 )− I10 − 3I3I5 + 7I2I3,

B3 = 10I31I7 + 2I2

1 (6I6 + 5I11 + 5I24 ) + I1(7I12 + 2I8 + 7I4I5− 10I2I4)

+ I13 − 2I9 + I23 − 4I3I4 − 4I2I5 + 4I2

2 ,

B4 = 10I21I7 + I1(8I6 + 9I11 + 9I2

4 ) + 3I12 + 3I4I5 − 6I2I4,

B5 = 5I1I7 + 2I6 + 3I11 + 3I24 , B6 = I7.

b) For equation (5.44) the expressions for the coefficients in terms of theJ ′s are:

A1 = 3J2, A2 = 3J1, B1 = −J4, B2 = −3(J22 + J7 + J3),

B3 = −3(J1J2 + J6), B4 = −J21 − J5.

Proof If the canonical form equation v′′′(u) = 0 is transformed by ageneral transformation u = σ(x, y) and v = ρ(x, y), the coefficients Aj and Bkof (5.43) or (5.44) are obtained in terms of σ, ρ and its derivatives dependingon whether σx 6= 0 or σx = 0. On the other hand, the invariants have beenexpressed in terms of them in Lemma 6.10. If σ, ρ and its derivatives areeliminated between the two sets of equations, the above expressions for thecoefficients are obtained.

Theorem 6.33 The transformation of an equation with S37 symmetry to

canonical form is determined by a system of partial differential equations thegeneral solution of which contains seven constants.

a) For equation (5.43), the transformation function σ is determined by thesystem

Sx − 12S

2 − I6 = 0, Sy − 12I1S

2 − I1,xS − I1I6 − I1,xx = 0, (6.87)

σy − I1σx = 0, σxx − Sσx = 0. (6.88)

The transformation function ρ is explicitly given by

ρ = −∮

12I7

( σ

σx

)2

ρ1dx

+[

12

(σσx

)2

(ρ1,xx − I1I7ρ1)− σσx

(12σσx

σxxσx + 1

)ρ1,x + ρ1

]dy

+σ∮I7σ

σ2x

dx+σ

σ2x

[ρ1,xx−

( σxxσx

+σxσ

)ρ1,x + I1I7ρ1

]dy

−σ2

∮1

2σ2x

[I7ρ1dx+

(ρ1,xx −

σxxσx

ρ1,x + I1I7ρ1

)]dy

(6.89)

306

where

ρ1 = exp∮ (

I4 +σxxσx

)dx+

(I5 + I1,x + I1

σxxσx

)dy.

The invariants occurring in these equations are expressed in terms ofthe coefficients of (5.43) by

I1 = A0, I4 = 13A3, I5 = 2A0,x + 1

3 (A2 −A0A3),

I6 = − 12A3,x + 1

2B5 − 16A

23 − 5

2A0B6, I7 = B6.

b) For equation (5.44), the transformation function σ is determined by thesystem

Sx = 0, Sy − 12S

2 − J3 = 0, σx = 0, σyy − Sσy = 0. (6.90)

The transformation function ρ is explicitly given by

ρ =∮ [

−12

( σ

σy

)2

ρ1,yy +σ

σy

( 12σ

σy

σyyσy

+ 1)ρ1,y − ρ1

]dx

−12J4

(σσy

)2

ρ1dy

+σ∮

σ

σ2y

[ρ1,yy−

( σyyσy

+σyσ

)ρ1,y

]dx+ J4

σ

σ2y

ρ1dy

−σ2

∮1

2σ2y

[(ρ1,yy −

σyyσy

ρ1,y

)dx+ J4ρ1dy

](6.91)

where

ρ1 = exp∮J2dx+

(J2 +

σyyσy

)dy.

The required invariants may be expressed in terms of the coefficients as

J1 = 13A2, J2 = 1

3A1, J3 = − 12B2 − 1

2A1,y − 16A

21, J3,x = 0, J4 = −B1.

Proof Case a). The first equation (6.87) for S ≡ σxxσx is essentially the

definition of I6 in Lemma 6.10. In order to obtain the second equation, therelation σy

σx = I1 is differentiated twice w.r.t. x. In the resulting expression,(σxxσx

)x

is substituted by means of the first equation of (6.87), and the obvious

relation(σxxσx

)y

= σxxyσx −

σxxσx

σxyσx is used to eliminate σxxy

σx .Eliminating the ρ-derivatives from the expressions for I4, I5 and I7 in

Lemma 6.10 and using the equations (6.87) for eliminating derivatives of σxxσx ,


the linear homogeneous system

ρxy − I1ρxx−(I4 +

σxxσx

)ρy+

(I1I4 − I1,x + I1

σxxσx

)ρx = 0,

ρyy − I21ρxx−

(I1I4 + I5 + I1,x + 2I1

σxxσx

)ρy

+(I21I4 + I1I5 − I1,y + 2I2

1

σxxσx

)ρx = 0,

ρxxx − 3σxxσx

ρxx − I7ρy+[I1I7 − I6 +

32

( σxxσx

)2]ρx = 0 (6.92)

for ρ is obtained. It has the solutions ρ = 1, ρ = σ and ρ = σ2 as it may beverified by substitution. Consequently, the right factor

Lclm(ρx, ρy

,ρx − σx

σ ρ, ρy −σyσ ρ

,ρx − 2σxσ ρ, ρy − 2σyσ ρ

)=

ρy − I1ρx ≡ ρ1, ρxxx − 3σxxσx ρxx−(I6 − 3

2σ2xx

σ2x

)ρx ≡ ρ2

(6.93)

may be divided out yielding the exact quotientρ2 − I7ρ1, ρ1,x−

(I4 + σxx

σx

)ρ1, ρ1,y − (I5 + I1,x + I1

σxxσx )ρ1

from which ρ1 and ρ2 may be determined by integration. Substituting itinto the right hand sides of the second line of (6.93) yields an inhomogeneoussystem for ρ with known solution of the homogeneous part. It has been shownin Exercise 6 of Chapter 2, see also Appendix A, how a special solution forit may be obtained in terms of integrals. Applying these results leads to therepresentation (6.89) for ρ.

Case b). The function σ follows from the expression for the invariant J3 inLemma 6.10, the substitution σyy

σy = − 12S and two subsequent integrations.

The expressions for J1, J2 and J4 in the same lemma yield the system

ρxx − J1ρx = 0, ρxy−(J2 + σyy

σy

)ρx = 0,

ρyyy − 3σyyσy ρyy−(σyyyσy − 3

σ2yy

σ2y

)ρy − J4ρx = 0

for ρ with the three solutions ρ = 1, ρ = σ and ρ = σ2, see Lie [109], part III,page 409. Consequently, the right factor

ρx ≡ ρ1, ρyyy − 3σyyσy

ρyy−( σyyy

σy− 3

σ2yy

σ2y

)ρy ≡ ρ2

may be divided out yielding the exact quotient

ρ2 − J4ρ1, ρ1,x − J1ρ1, ρ1,y−(J2 +

σyyσy

)ρ1

.

308

By similar methods as in case a) the representation (6.91) is obtained.The required invariants are obtained by elimination from the expressions

for A0, . . . , A3, B6 and B7 in Lemma 6.11.

In either case, determining σ requires solving a first order partial Riccati-like system and integrations. Although the expressions (6.89) and (6.91) forρ appear somewhat awkward, they are in fact very simple. Containing onlyarithmetic and integrations, they are guaranteed to be Liouvillian over thefield determined by σ and ρ1.

Example 6.37 Equation 7.8 of Kamke’s collection in symmetry class S37

has been considered in Example 5.48 of the preceding chapter. From itscoefficients the invariants I1 = I4 = I6 = I7 = 0 and I5 = − 3

2y are obtained.

They yield Sx − 12S

2 = 0, Sy = 0 for S. The special solution S = 0 leads toσy = σxx = 0 with the non-constant solution σ = x. Substituting it togetherwith ρ1 = 1

y√y

into (6.89) yields ρ = 2√y− 2.

Example 6.38 An equation in symmetry class S37 , case a), has been

considered in Example 5.49 of the preceding chapter. From the coefficientsgiven there the invariants

I1 =x

y, I4 = − 1

x, I5 = −1

y, I6 = 0, I7 = −3y

x3

are obtained. Substituting I1 and I6 into (6.87) yields Sx − 12S

2 = 0,Sy + x

2yS2 − 1

yS = 0 with the only rational solution S = 0. It leads to

σy − xy σx = 0, σxx = 0 a special solution of which is σ = xy. Substituting

this choice of σ and the corresponding value ρ1 = 1x into (6.89) finally yields

ρ = yx .

Example 6.39 An equation in symmetry class S37 , case b), has been con-

sidered in Example 5.50 of the preceding chapter. From the coefficients giventhere the invariants

J1 =x2 − 2x+ 2x(x− 1)

, J2 =3y, J3 = 0, J4 = − 6x

y3(x− 1)

are obtained. Substituting J3 into (6.90) yields Sy− 12S

2 = 0 with the solutionS = − 2

y + C . Choosing C = 0 leads to σyy + 2yσy = 0 a special solution of

which is σ = 1y . Substituting this choice of σ and the corresponding value

ρ1 = (x− 1)yx2 ex into (6.91) finally yields ρ = y

xex.


Exercises

Exercise 6.1 Determine the constraints for the parameters and undeter-mined functions occurring in the canonical forms of Theorem 6.4 guaranteeingthe correct symmetry type. Show that these constraints are related to thestructure of the symmetry algebras.

Exercise 6.2 Determine the transformation between the two canonicalforms given in Examples 6.7 and 6.8.

Exercise 6.3 Express the Janet basis coefficients of the two parametersymmetry groups in terms of the invariants determined in Lemma 6.1 and 6.2.Show that this is an application of Theorem 6.6.

Exercise 6.4 Apply Theorem 6.7 to equation y′′ + r(y, y′) = 0 in orderto generate the most general canonical form as given in Theorem 6.4.

Exercise 6.5 Consider again the equation discussed in Examples 5.15and 6.12. Assume now that only a two-parameter group of type g26 is known.Determine the canonical form under this assumption and discuss the result.

Exercise 6.6 Consider again equation 6.133 from Kamke’s collection dis-cussed in Example 6.11. Assume now that only a two-parameter group oftype g25 is known. Determine the canonical form under this assumption anddiscuss the result.

Exercise 6.7 The number of subgroups that may be applied for generatinga canonical form of a second order equation is maximal for symmetry class S2

8 .Reconsider equation 6.180 from Kamke’s collection that has been discussed inExample 6.14. Choose the two-parameter subgroups of type g25 and g26 andgenerate the respective canonical form. What about subgroups comprisingthree or four parameters?

Chapter 7

Solution Algorithms

It is assumed now that for a given ode of order not higher than three withnontrivial Lie symmetries the transformation functions from actual variablesx and y(x) to canonical variables u and v(u) are known. This implies that itsinfinitesimal symmetry generators are explicitly known in canonical variablesfrom the tabulation in Section 5.1 without further calculations. What remainsto be done is to solve the canonical equation, and to generate the solution inactual variables from it. The term solution always means general solution;i.e., a n-parameter family of solutions if n is the order of the equation. Thisexcludes the so-called singular solutions which may or may not be special-izations of the general solution. The relation between these two concepts isdiscussed in detail in the article by Buium and Cassidy [23], Section 1.8, andarticles by Ritt quoted there.

Before this proceeding is described in detail for the individual symmetryclasses, the underlying general principles will be outlined. The following dis-cussion is based on Engel [44], vol. V, Anmerkungen by Engel, pages 643-669and 682-687 where many more details may be found.

Let the canonical ode of order n have an r-parameter symmetry group.The further proceeding depends crucially on the relative values of r and n.At first it is assumed that r < n and ∆ 6= 0; the Lie determinant ∆ is definedby (5.15). The fundamental invariants Φ1 and Φ2 defined on page 201 areof order r − 1 and r respectively. In terms of these invariants the canonicalequation has the form

Ω(

Φr−1,Φr,dΦrdΦr−1

, . . . ,dn−rΦrdΦn−rr−1

)= 0 (7.1)

where Ω is an undetermined function. This is a general ode of order n − r.It has to be solved first. If this is not possible the main achievement of thesymmetry analysis consists in lowering the order of the originally given odeof order n to an equation of order n− r in the independent variable Φr−1 anddependent variable Φr. If the solution of (7.1) may be found, however, anequation of order r allowing an r-parameter symmetry group is obtained.

Consequently, the case n = r has to be considered next. It is discussed infull detail for arbitrary values of n in the above mentioned Anmerkungen byEngel. The complete answer for n = 2 and n = 3 that is relevant here is givenin the book by Lie [113] on differential equations, Theorem 45 on page 464and Kapitel 25, in particular Theorem 49 on page 555. It may be formulatedas follows.

311

312

Theorem 7.1 (Lie 1891) If a second order ode has a two-parameter sym-metry group and its symmetry generators are explicitly known, its solutioninvolves two integrations. If a third order ode has a three-parameter solvablesymmetry group with known generators, its solution involves three integra-tions. If the group is not solvable, the solution of a first order Riccati equationis required.

The proof is based on the representation of the given ode as an equivalentfirst order pde, and Theorem 2.23 dealing with its solution. The details maybe found in Lie’s book.

If n < r, i.e., if the number of symmetries is higher than the order of thedifferential equation, the solution process may be further simplified. In manycases the solution may be given explicitly without any integrations involved.The complete answer for arbitrary values of n and r including all exceptionsis due to Engel [44]; see in particular pages 651-665. Engel’s comprehensiveproof is based to a large extent on a case-by-case discussion. Therefore it isnot given here. Rather for order n = 2 and n = 3 all possible symmetry typeswith r ≥ 3 or r ≥ 4 respectively are treated in detail below.

For first order equations the situation is different because point symmetriescannot be determined algorithmically. Therefore, at least a single nontrivialsymmetry generator must be known a priori in order to apply the algorithmsdescribed in this chapter. Alternatively, the symmetry problem may be spe-cialized to certain classes of equations for which algorithmic methods are avail-able, e. g. for Abel’s equation. This is discussed in Section 7.1. Subsequently,equations of order two and nonlinear equations of order three are treated inSections 7.2 and 7.3 respectively. The subject of the final Section 7.4 areequations of order three that are equivalent to linear ones.

7.1 First Order Equations

In this section it is assumed that in addition to the first order equationto be solved at least a single symmetry generator is explicitly known; it isconsidered as part of the input to the solution algorithms.One-Parameter Symmetry. If for a quasilinear first order equation

ω ≡ y′ + r(x, y) = 0 (7.2)

a single nontrivial symmetry generator is known, it may be possible to in-troduce new variables u and v(u) according to Theorem 6.1 such that thesymmetry transformation becomes the translation in v. In these variablesequation (7.2) takes the form

v′ + s(u) = 0. (7.3)

Solution Algorithms 313

Integration yields the representation

v +∫s(u)du = C (7.4)

of its general solution. If the actual variables are resubstituted into (7.4), thegeneral solution of (7.2) is obtained. The function field for its representation isdetermined by the integral in (7.4) and the transformation functions betweenthe actual variables x, y and the variables u, v.

According to Lemma 3.4 it is not guaranteed that the canonical form (7.3)from the knowledge of a single symmetry generator can be obtained. If thisis not possible, the following lemma which is also due to Lie [113], Kapitel 6,§1, may be applied.

Lemma 7.1 If a first order ode y′ + r(x, y) = 0 allows the nontrivialsymmetry generator U = ξ(x, y)∂x + η(x, y)∂y, the expression 1

η + rξis an

integrating factor. Consequently, there holds

φ =∮rdx+ dy

rξ + η. (7.5)

Proof Let φ(x, y) = C be the general solution of y′ + r(x, y) = 0. Bydefinition of a symmetry generator, it obeys ξφx + ηφy = 1. Moreover, as asolution, φ has to satisfy φx + y′φy = φx − rφy = 0. Eliminating the firstorder derivatives from these two expressions yields φx = r

η + rξ, φy = 1

η + rξ.

Substituting them into the right hand side of dφ = φxdx+ φydy yields dφ =dy + rdxη + rξ

which is a total differential by construction. Consequently, 1η + rξ

is an integrating factor and the representation is obtained.

The various steps described above are applied in the subsequent algorithmfor solving equation (7.2) with a known symmetry generator.

Algorithm 7.1 LieSolve1.1(ω,U). Given a quasilinear first order ode ωin the form (7.2) with the known symmetry generator U , its general solutionis returned.S1 : Determine transformation functions. Applying Lemma 3.4 deter-mine the transformation of ω to canonical form (7.3). If it cannot bedetermined return goto S4.S2 : Canonical form. Applying the result of step S1, generate the canon-ical form (7.3).S3 : Return solution. Integrate the canonical form. Resubstitute theactual variables and return the result.S4 : Use integrating factor. Evaluate the integral in (7.5) and return theresult.

314

Example 7.1 For equation 1.15 from the collection by Kamke the canonicalform v′ + 1

u2 − 1= 0 has been obtained in Example 6.1 by the transforma-

tion x = v and y = u + v2. It may be integrated to v + 12 log u− 1

u+ 1 = C.Resubstituting the actual variables by u = y − x2 and v = x finally yields

y = C(x2 − 1)e−2x − x2 − 1Ce−2x − 1

.

Two-Parameter Symmetries. If for a first order equation (7.2) the gen-erators of a two-parameter symmetry group are explicitly known, its generalsolution may often be obtained by algebraic operations as the next resultshows.

Theorem 7.2 Let an equation (7.2) have two symmetries with generators

Ui = ξi(x, y)∂x + ηi(x, y)∂y for i = 1, 2, and define F ≡ η2 + rξ2η1 + rξ1

. If ∂F∂x6= 0

and ∂F∂y6= 0, its general solution is F (x, y) = C.

Proof The symmetry type g26 is considered first, i.e., [U1, U2] = 0. Letthe canonical generators in variables u and v ≡ v(u) be V1 = ∂u and V2 = ∂v.By Lemma 3.3 the transformation functions x = φ(u, v) and y = ψ(u, v) obeythe relations

V1φ = φu = ξ1, V2φ = φv = ξ2, V1ψ = ψu = η1, V2ψ = ψv = η2

from which the first derivative y′ = η1 + η2v′

ξ1 + ξ2v′ is obtained. Substitution into

(7.2) and some simplifications yields v′ + η2 + rξ2η1 + rξ1

= 0. According to The-

orem 6.2 the canoncial form for this symmetry class is v′ + a = 0. This isassured if the relation given above holds.

The symmetry type g25 is considered next, i.e., now [U1, U2] = U1. Fromthe canoncial generators V1 = ∂v and V2 = u∂u + v∂v, the relations

V1φ = φv = ξ1, V1ψ = ψv = η1,

V2φ = uφu + vφv = ξ2, V2ψ = uψu + vψv = η2

follow from which the first order derivatives

φu =ξ2 − vξ1

u, φv = ξ1, ψu =

η2 − vη1u

, ψv = η1

of φ and ψ are obtained. They lead to the expression y′ = η2 − η1v + η1uv′

ξ2 − ξ1v + ξ1uv′ for

the first derivative. Substitution into (7.2) and rearranging terms yields now

v′u− v+ η2 + rξ2η1 + rξ1

= 0. This represents the correct canonical form v′ + a = 0

if η2 + rξ2η1 + rξ1

− v is proportional to u; i.e., if there holds η2 + rξ2η1 + rξ1

= v+au = C.


If F defined in Theorem 7.2 is identically a constant, the above theoremcannot be applied for the solution procedure. In this case the canonical form

v′ + a = 0 (7.6)

has to be determined by Theorem 6.2 from which the solution may alwaysbe obtained. Lie [113] discusses the two alternatives in detail on page 124,Satz 1. On pages 126 and 127 he explains the different behavior by geometricalconsiderations. A solution scheme based on the preceding discussion may bedesigned as follows.

Algorithm 7.2 LieSolve1.2(ω,U1, U2). Given a quasilinear first orderode ω in the form (7.2) with two known symmetry generators U1 and U2, itsgeneral solution is returned.

S1 : Generate F. From the coefficients of U1 and U2 generate F definedin Theorem 7.2. If F depends explicitly on x and y, return it.S2 : Return result. Generate the canonical form (7.6). Determine thesolution in actual variables from it and return the result.It should be noted that due to the explicit knowledge of two symmetry

generators this algorithm never fails.

Example 7.2 Consider the equation y′ + (x− y)2 + 1(x− y)2 − 1

= 0 with the two

symmetry generators U1 = ∂x + ∂y and U2 = y∂x + x∂y of a type g25 group.Substituting the coefficients of these generators into the definition of F inTheorem 7.2 yields after some simplifications the general solution in the formof (x+ y + C)(x− y) = 1.

Example 7.3 The same equation as in the preceding example is con-sidered. If only the generator U = ∂x + ∂y is known, the solution scheme ofalgorithm LieSolve1 has to be applied. By Lemma 3.4 the canonical variables

are obtained as x = v − u, y = v with canonical equation v′ − u2 + 12u2 = 0.

Integration and resubstitution of the original variables yields the same solu-tion as above. However, knowing only a single symmetry generator requiresan additional integration.

Abel’s Equation. If one or two symmetry generators of an Abel equationmay be found, its solution may be obtained by applying the above results forgeneral first order equations. However, due to the special structure, the moreexplicit answer given in the subsequent theorem is usually more convenient.

Theorem 7.3 The two rational normal forms of Theorem 4.14 are con-sidered separately.

i) If the rational normal form is y′+Ay3 +By = 0, the general solution is

y =1√

2B(∫ABdx+ C)

(7.7)

316

where B = exp (2∫Bdx) and C is the integration constant.

ii) If the rational normal form is y′ + Ay3 + By + 1 = 0 and allows aone-parameter symmetry with generator

U =1

A1/3

(∂x −

13A′

Ay∂y

),

its canonical variables are u = A1/3y and v =∫A1/3dx. They lead to

the canonical form

v′ +1

u3 + ku+ 1= 0 (7.8)

where k = 3√K, and K is the absolute invariant (4.51). From (7.8) its

general solution is obtained by integration as

v +∫

du

u3 + ku+ 1= C. (7.9)

If k = 0 the integral may be performed with the result

v+∫

du

u3 + 1= v+

13

logu+ 1√u2 − u+ 1

+1√3

arctan2u− 1√

3= C. (7.10)

Proof The transformation functions u ≡ σ(x, y) and v ≡ ρ(x, y) obeythe system σxξ + σyη = 0 and ρxξ + ρyη = 1. With the above values for ξand η the solution σ = A1/3y and ρ =

∫A1/3dx follows. The first derivative

y′ is transformed according to y′ = −ρx − σxv′

ρy − σyv′(compare (B.6) and (B.3))

with the result y′ = 1v′− 1

3A′

A y. Substituting this expression into the rational

normal form yields after some simplifications (7.8).

The various alternatives that may occur for solving an Abel equation ac-cording to this theorem are combined in the following algorithm. The fun-damental difference compared to LieSolve1.1 and LieSolve1.2 is the fact thatit is not required to know any symmetry generator in advance. Rather it isautomatically determined whenever this is possible.

Algorithm 7.3 SolveAbel(ω). Given an Abel equation ω, its generalsolution is returned if a symmetry may be found, or failed otherwise.S1 : Second kind equation? If ω is an Abel equation of second kind, applythe transformation (4.40).S2 : Rational normal form. Transform ω to rational normal form.S3 : Bernoulli case? If the result of S2 is a Bernoulli equation, determineits solution from (7.7) and goto S5.S4 : Symmetry? If K(A,B) = constant, determine its solution from (7.9)or (7.10). Otherwise return failed.


S5 : Backtransformation. Apply the reverse transformations performedin S1 and S2 and return the result.

Next to Riccati’s equation, Abel’s equations are the most thoroughly stud-ied differential equations in the history of mathematics. In the course of timeseveral integrable cases have been found by some kind of heuristics, a shortdiscussion of them may be found in Kamke [85], pages 24-28. It turns outthat most of them may be subsumed under the symmetry analysis describedhere.

Scalizzi (1917) describes the integrable case of (4.38) for the constraint

a0 =13

( a2

a3

)′+

13a1a2

a3− 2

27a32

a23

.

From the last relation in (4.43) it is obvious that this condition identifies therational normal form allowing a two-parameter symmetry with the solution(7.7).

Chiellini (1931) found an integrable case of (4.38) for a0 = a1 = 0, (a3a2

)′ =const. · a2 and a2 6= 0, a3 6= 0. Under these constraints the coefficients

(4.43) are b3 = a3, b1 = − a22

3a3and b0 = const. · a

32

a23

; i.e., due to b0 6= 0 a

rational normal form corresponding to case ii) of Theorem 4.14 is obtained.

Applying the relation a′3a3− a′2a2

= const. · a22a3

which is a consequence of theabove constraints it follows that the absolute invariant K(A,B) is constant.Consequently, there is a one-parameter symmetry group with generator

U =a3

a22

[∂x−

(2a′2a2− a′3a3

)y∂y

].

Chini (1924) observed that the condition Φ3 = 0 for the invariant definedin Theorem 4.15 leads to a Bernoulli equation. Evaluating it for the equationwith coefficients bk in the proof of Theorem 4.14 yields Φ3 = 3b0b23. Becauseb3 6= 0 it follows b0 = 0, i.e., the Bernoulli rational normal form. For Φ3 6= 0Chini showed that under the condition

a3Φ′3 + (3a1a3 − a22 − 3a′3)Φ3 = 3αΦ5/3

3

the integral representations described in case ii) of Theorem 7.3 exist. Eval-uating the above constraint for the rational normal form of this case showsthat it is equivalent to the condition K(A,B) = const.; i.e., the correspondingequation allows a symmetry generator.

For the equation of the second kind (4.39) Abel himself described severalintegrable cases if certain relations between the coefficients are satisfied; e.g.,if f1 = 2f2 − g′ the transformed equation becomes

y′ + (f0 − f2g2 + gg′)y3 + f2y = 0

318

which is a Bernoulli equation.

Finally Todorov and Kristev [177] consider an Abel equation of the secondkind (4.39) and impose the constraints

f0 = (g3 − 1)f3 + gg′, f1 = 3f3g2 − g′, f2 = 3f3g.

The signs of the fk’s have been reversed in order to agree with the notation

in (4.39). They imply the rational normal form y′+f23 y

3 + f ′3f3y+1 = 0 with a

vanishing absolute invariant and the integral representation (7.10) (equation(14) in Todorov and Kristev [177]).

Example 7.4 For equation 1.46 of Kamke’s collection the rational normalform z′ − x2z3 + 2

x2 z = 0 has been obtained in Example 5.7. The solution of

this Bernoulli equation is obtained from (7.7) in step S3. Resubstituting theoriginal variable z = y − 1

x2 yields the final answer

y =e2/x√

−2∫x2e4/xdx+ C

+1x2.

It is not always advisable to generate the canonical form (7.8) because itmay lead to unnecessary algebraic numbers due to the cubic root in k = 3

√K.

The next example shows how they may be avoided.

Example 7.5 Equation 1.255(y − 3

x)y′ + 1

xy2 − 1

x2 y = 0 from Kamke’scollection is an Abel equation of the second kind. Introducing the new depen-dent variable z by y = 1

z+ 3x (compare equation (4.39) and the transformation

given there) leads to z′− 6x3 z

3− 2x2 z

2− 1xz = 0. It is transformed into rational

normal form w′− 3219683

w3

x3 − 79wx +1 = 0 by z = − 1

243 (4w+27x). Its absolute

invariant is K = − 7294 ; consequently, k = − 9

23√

2. In order to avoid the cubic

root, it is better to introduce canonical variables by u = 2y9x and v = − 2

9 log xwith inverse x = exp (− 9

2v) and y = 92u exp (− 9

2v). It yields

v′ +1

427u

3 − u− 1= v′ +

274

(u+ 32 )2(u− 3)

= 0

with the solution v = 13 log 2u+ 3

u− 3 −3

2u+ 3 +C. Resubstituting the originalvariables w, z and finally y yields the solution of the originally given equationin the form xy3 = C exp (xy).


7.2 Second Order Equations

The subjects of this section are second order equations that are linear inthe highest derivative. A cursory glance at the various collections of solvedexamples shows that they have been investigated more thoroughly than anyother family of equations. Lie [113] himself devoted to them almost completelyhis book on symmetries of differential equations. These results are discussedin detail in the current section.One-Parameter Symmetry. If a second order quasilinear ode belongs tosymmetry class S2

1 it is equivalent to an equation of the form

v′′ + r(u, v′) = 0 (7.11)

with r(u, v′) an undetermined function of its arguments. The substitutionw = v′ in (7.11) yields

w′ + r(u,w) = 0. (7.12)

Because there are no constraints on r(u,w), this is a general first order equa-tion. It is the first example of an equation of the form (7.1) because thenumber of symmetries is lower than the order of the equation. As explainedin the introduction to this chapter, solving it is a new problem that has tobe handled as described in the preceding Section 7.1. The further proceed-ing depends on whether or not this first order equation can be solved. Thealgorithm below applies these steps. The solution procedure may terminatein step S2 if σ cannot be obtained, or in step S3 if equation (7.12) cannot besolved.

Algorithm 7.4 LieSolve2.1(ω). Given a quasilinear second order ode insymmetry class S2

1 , its general solution or an equivalent first order equationis returned; or failed if neither may be found.S1 : Janet basis for symmetries. Generate a Janet basis for the deter-mining system of its Lie symmetries.S2 : Determine transformation functions. If case a) of Theorem 6.7applies choose σ. If case b) applies solve the first order equation for σ. Ifit cannot be obtained return failed. Determine ρ by integration.S3 : Canonical form. Applying σ and ρ obtained in S2 generate thecanonical form (7.12) and apply the solution algorithms of Section 7.1 toit. If a solution has not been found, return the canonical form equationtogether with the transformation functions σ and ρ.S4 : Return solution. From the solution obtained in step S3, constructthe solution of ω and return the result.

Example 7.6 For equation 6.90 of Kamke’s collection the canonical formhas been obtained in Example 6.5. Substituting v′ = w yields

w′ + u(u+ 5)w3 − (u+ 3)w2 + 14w = 0.

320

This is an Abel equation the solution of which cannot be obtained.

Example 7.7 The canonical form (6.2) for equation 6.98 of Kamke’s col-lection leads to w′ + 2(2u − 1)w2 + w = 0. This Bernoulli equation has the

general solution w = 1C1e

x − 4x− 2 and finally v =∫

du

C1eu − 4u− 2

+ C2.

If the actual variables u = yx2 and v = log x are resubstituted, an integral

representation of the general solution of equation 6.98 is obtained.

Two-Parameter Symmetries. The canonical form of a second order quasi-linear ode in the symmetry class S2

2,1 has been shown to be

v′′ + r(v′) = 0 (7.13)

where r(v′) is an undetermined function of v′. In order to solve this equationit is rewritten as

v′′ + r(v′) =dv′

du+ r(v′) = 0 or

dv′

r(v′)+ du = 0. (7.14)

Integration yields

R(v′) + u = C1 with R(v′) =∫

dv′

r(v′). (7.15)

If r(v′) is such that the latter integration may be performed, the originalequation is reduced to a first order equation. Depending on the form of R(v′)it may be possible to continue the solution procedure. In any case the generalsolution may always be represented in parameter form as

u+R(p) = C1, v + pR(p) =∫R(p)dp+ C2. (7.16)

This is a consequence of the fact that (7.15) is a first order differential equationthat does not contain the dependent variable explicitly; i.e., it allows thesymmetry ∂v. Details may be found in Kamke [86], vol. I, page 49; see alsoExercise 7.4. If u ≡ σ(x, y) and v ≡ ρ(x, y) are resubstituted, the solution inactual variables is obtained.

In the subsequent algorithm the various steps for solving an equation insymmetry class S2

2,1 are put together. Only step S2 may fail if the transfor-mation to canonical form is not Liouvillian.

Algorithm 7.5 LieSolve2.2.1(ω). Given a quasilinear second order odeω in symmetry class S2

2,1, its general solution is returned explicitly or inparameter form. Or failed if it cannot be determined.S1 : Equations for σ and ρ. Generate a Janet basis for the symmetrygenerators of ω and set up system (6.3).


S2 : Determine transformation functions. Determine an independent pairof Liouvillian solutions of the system obtained in step S1. If it cannot befound return failed.S3 : Generate canonical form. By means of the transformation functionsobtained in S2 generate the canonical form (7.13).S4 : Return solution. Using r(v′) from step S3 construct the solution(7.16), resubstitute actual variables and return the result.

Example 7.8 In the preceding chapter, a canonical form of Kamke’s equa-tion 6.227 has been determined in Example 6.7. Substituting the respectivevalue for r(v′) into (7.15) and (7.16) and performing the integrations yields

u+ logp+ 2p+ 1

+p

p+ 2= C1, v + log (p+ 1)− 2p

p+ 2= C2.

If the canonical variables are resubstituted, the parameter representation

x = C11

p+ 2exp

p

p+ 2, y = C2

p+ 1p+ 2

exp(− p

p+ 2

)for x and y is obtained.

In Exercise 7.6 the same answer will be obtained from the canonical formof Example 6.8.

The other two-parameter symmetry class is S22,2. A quasilinear second order

ode in this symmetry class has the canonical form

uv′′ + r(v′) = 0 (7.17)

where again r(v′) is an undetermined function of v′. Rewriting it as

v′′ +r(v′)u

=dv′

du+r(v′)u

= 0 ordv′

r(v′)+du

u= 0 (7.18)

it may be integrated once with the result

R(v′) + log u = C1 with R(v′) =∫

dv′

r(v′). (7.19)

The same general remarks apply to this equation as for (7.15). If a furtherintegration is possible, the general solution of the canonical equation followsexplicitly. Otherwise the parameter representation

u = C1e−R(p), v = C1(pe−R(p) −

∫e−R(p)dp) + C2 (7.20)

in canonical variables or, after substitution of σ(x, y) and ρ(x, y) in actualvariables is obtained. As in the previous case, details may be found in Kamke[86], vol. I, page 49.

322

A solution scheme for equations in symmetry class S22,2 is realized in terms

of the following algorithm.


2,2, its general solution is returned explicitly or inparameter form.S1 : Equations for σ and ρ. Generate a Janet basis for the symmetrygenerators of ω.S2 : Determine transformation functions. Determine the transformationfunctions σ from (6.6), and ρ from (6.7) or (6.8).S3 Generate canonical form. By means of the transformtion functionsobtained in S2 generate the canonical form (7.17).S4 : Return solution. Construct the solution (7.20), resubstitute theactual variables and return the result.According to Theorem 6.3, for any equation in this symmetry class a canon-

ical form may be obtained. Consequently, a solution in parameter form isguaranteed and the above algorithm LieSolve2.2.2(ω) does not have a failedexit. In those cases where (7.19) may be solved for v′, the solution may beexpressed as an integral as it is the case in the following example.

Example 7.9 The canonical form for Kamke’s equation 6.159 has beendetermined in Example 6.9. With the above notation there follows

r(v′) = − 32v′(v′2 − 1) and R(v′) = 2

3 log v′ − 13 log (v′2 − 1).

Substituting it into (7.20) leads to u3v′2 = C1(v′2 − 1). Solving for v′ andintegrating leads to

v = C1

∫du√

u3 + C21

+ C2.

If the actual variables u = 1√y

and v = x are resubstituted, and the integra-

tion constant C1 is replaced by its inverse, the solution of the original equationis obtained in the form

x+ 12

∫dy√

y√y√y√y + C1

= C2.

Three-Parameter Symmetries. In this subsection the number of symme-tries exceeds the order of the respective ode by one and the remarks of thesecond paragraph on page 312 apply. In the first symmetry class S2

3,1 to bediscussed the canonical form is

v′′(u− v) + 2v′(v′ + a

√v′ + 1

)= 0 (7.21)

where a is a constant, a 6= 0, or in rational form

v′′2(u− v)2 + 4v′′v′(v′ + 1)(u− v) + 4v′4 + 4(2− a3)v′3 + 4v′2 = 0. (7.22)


Two subsequent integrations of the canonical form (7.21) yield

(u− v)√v′

1 + a√v′ + v′

=1C1

and finally a third integration leads to the general solution

(C1v + C2)(C1u+ C2 + a) + 1 = 0. (7.23)

The various steps for solving an equation in this symmetry class are puttogether in the following algorithm.


3,1, its general solution is returned; or failed if it cannotbe determined.S1 : Equations for R and S. Generate Janet basis for the symmetrygenerators of u. Set up system (6.13), (6.16) or (6.18) respectively for R,or system (6.20) for R and S.S2 : Determine R and S. From the equations obtained in S1 determinerational solutions for R and S. If none can be found return failed.S3 : System for transformation functions. Applying the result of S2 setup the system (6.14), (6.17), (6.19) or (6.21) for σ and ρ.S4 : Determine transformation functions. Determine two independentLiouvillian solutions for the system obtained in S2.S5 : Return solution. Generate the canonical form and determine thevalue for a from it. Substitute a and the canonical variables into (7.23)and return the result.

Example 7.10 An equation in this symmetry class has been consideredbefore in Examples 5.13 and 6.10. Comparing the coefficients of the cubicpower of v′ in the canonical form of latter example and of (7.22) yields a = 1.Substituting this value and the expressions for the canonical variables u = xy

and v = 2x into (7.23) yields the solution, after some simplifications, in the

form

y = (C22 + C2 + 1)x+ 2C1(C2 + 1)

(2C1 + C2x)C1x.

A quasilinear second order ode in symmetry class S23,2 is equivalent to an

equation of the formv′′v3 + a = 0 (7.24)

where a is constant, a 6= 0. In order to obtain the solution, the canonicalequation (7.24) is written as (v′2)′ −

( av2

)′ = 0. In this form the integration

is obvious and leads to v′2 − av2 = C1. Solving for v′ yields v′ =

√C1v2 + av .

Separation of variables allows it to perform the second integration with theresult

vdv√C1v2 + a

= du or1C1

√C1v2 + a− u = C2.

324

The latter relation may be rewritten as

C1v2 − C2

1 (u+ C2)2 + a = 0. (7.25)

Similar to the preceding case the solution procedure comes down to de-termining the transformation functions of the actual to canonical variables,determining the constant a from the canonical form and substituting it to-gether with u ≡ σ and v ≡ ρ into the solution of the latter. Theorem 6.11explains how the transformation to canonical variables is achieved. Thesesteps are put together in the subsequent algorithm.


3,2, its general solution is returned; or failed if it cannotbe determined.S1 : Equations for R. Generate Janet basis for the symmetry generatorsof ω. If it is of type J (2,2)

3,4 set up system (6.22), of type J (2,2)3,7 system

(6.23) and of type J (2,2)3,6 system (6.24) for R.

S2 : Determine R. From the equations obtained in S1 determine a ra-tional solution for R. If it cannot be found return failed.S3 : System for transformation functions. Applying the result of S2 setup the systems (6.22), (6.23) or (6.24) respectively for σ and ρ.S4 : Determine transformation functions. Determine two independentLiouvillian solutions for the system obtained in S3.S5 : Return solution. Generate the canonical form and determine thevalue for a from it. Substitute a and the canonical variables into (7.25)and return the result.

Example 7.11 The solution procedure for equation 6.133 of Kamke’scollection may now be completed. Substituting u = x − y, v2 = x + y anda = 1

4 as obtained in Example 6.11 into (7.25) yields the general solution inactual variables

C1(x+ y) + C21 (x− y + C2)2 + 1

4 = 0.

For an equation in symmetry class S23,3(c) it is assumed that the parameter

c has already be determined. Its canonical form is

v′′ + av′(c−2)/(c−1) = 0 (7.26)

with a constant a 6= 0. In order to yield an integer power v′m, the parameterc must be determined by c = 1 + 1

1−m . For |m| ≤ 5 the resulting values ofc are given in the following table.

m -5 -4 -3 -2 -1 0 2 3 4 5c 7/6 6/5 5/4 4/3 3/2 2 0 1/2 2/3 3/4


Integrating the canonical form (7.26) once yields

(c− 1)v′1/(c−1) + au = C1.

The second integration gives the final result

v =(1− c)1−c

ac(au− C1)c + C2. (7.27)

This form is most suitable for applications. The value of the constant cdetermines the symmetry type completely. In Theorem 6.12 it is expressedin terms of the coefficients of the Janet basis for the symmetries. Combiningthese results the following algorithm is obtained.

Algorithm 7.9 LieSolve2.3.3(ω). Given a second order quasilinear or-dinary differential equation in symmetry class S2

3,3(c), the general solution isreturned.S1 : Janet basis for symmetries. Generate a Janet basis for the symmetrygenerators of ω.S2 : System for transformation functions. If the Janet basis obtained instep S1 has type J (2,2)

3,6 , set up system (6.25), (6.26) or (6.27) respectively.

If it has type J (2,2)3,7 set up system (6.28).

S3 : Determine transformation functions. Determine two independentLiouvillian solutions for the system obtained in S2.S4 : Return solution. Generate the canonical form (7.26), substitute aand the canonical variables into (7.27), and return the result.

Example 7.12 Two equations with this symmetry group have been con-sidered in Example 6.12. In both cases the canonical form is v′′ + v′4 = 0corresponding to γ = 2

3 and a = 1. Substituting these values into (7.27)yields

v =3

2 3√

3(u− C1)

23 + C2

for the solution in canonical variables. Inverting the substitution functionsx = 1

v and y = 1u for the first alternative leads to the solution[

(8C32 − 9C1)x3 + 24C2

2x2 + 24C2x+ 8

]y2 − 18C1x

3y − 9x3 = 0.

In the other case, inverting x = v − u and y = 3u− 2v yields the solution

y3 − 9(x+ 1

3C1 − 18

)y2 + 27

[x2 + ( 2

3C1 − 16 )x+ 1

9C21 − 1

12C2

]y

+27[x3 + (C1 − 1

6 )x2 + ( 13C

21 − 1

6C2)x− 124C

22

]= 0

for the second equation of Example 6.12.

An equation in symmetry class S23,4 has the canonical form

v′′ − ae−v′= 0 (7.28)

326

with constant a 6= 0. Two integrations lead immediately to the general solu-tion

v =au+ C1

a

[log (au+ C1)− 1

]+ C2. (7.29)

Similar to S23,1 and S2

3,2 there is a single constant that has to be determinedfrom the canonical form of the given equation. It is substituted together withthe transformation functions into (7.29) in order to generate the solution forthe given equation. The algorithm below performs these steps, it returns thesolution for any equation in this symmetry class.

Algorithm 7.10 LieSolve2.3.4(ω). Given a second order quasilinear or-dinary differential equation in symmetry class S2

3,4, the general solution isreturned.S1 : Janet basis for symmetries. Generate a Janet basis for the symmetrygenerators of ω.S2 : System for transformation functions. If the Janet basis obtained inS1 has type J (2,2)

3,6 , set up system (6.29), (6.30) or (6.31) respectively. If

it has type J (2,2)3,7 set up system (6.32).

S3 : Determine transformation functions. Determine two independentLiouvillian solutions for the system obtained in S2.S4 : Return solution. Generate the canonical form and determine thevalue for a from it. Substitute a and the canonical variables into (7.29)and return the result.

Example 7.13 The transformation functions for an equation in symmetryclass S2

3,4 have been obtained in Example 6.13 as σ = x2 and ρ = x + y2.Substituting the inverse x =

√u, y =

√v −√u leads to a canonical form

(7.28) with a = 1. Substituting a and the transformation functions into(7.29) yields finally the solution

y2 = (C1 + x2)[log (C1 + x2)− 1

]− x+ C2

for the equation considered in Example 5.16

Eight-Parameter Projective Symmetry. The canonical form v′′(u) = 0with the general solution v = C1 + C2u of an equation with this symmetryis particularly simple because it does not contain any unspecified elements.Solving such an equation therefore comes down to determining the transfor-mation functions σ and ρ. According to Theorem 6.14 this is a two-stepprocedure. At first a rational solution of the partial Riccati system (6.37) hasto be found. Secondly, an independent set of Liouvillian solutions σ and ρhas to be determined by solving (6.36). The various steps of this proceedingare put together in the subsequent algorithm.


Algorithm 7.11 LieSolve2.8(ω). Given a quasilinear second order ode ωin symmetry class S2

8 . Its general solution is returned; or failed if it cannotbe determined.S1 : Determine a and b. From the coefficients of ω generate the system(6.37) and determine a special rational solution. If none is found returnfailed.S2 : System for transformation functions. Apply the result of S1 andconstruct the Janet basis (6.36).S3 : Determine transformation functions. Find two independent Liou-villian solutions of the Janet basis constructed in S2. If none are foundreturn failed.S4 : Return solution. Substitute u = σ(x, y) and v = ρ(x, y) into thecanonical form v = C1 + C2u and return the result.

Example 7.14 Equation 6.180 has been considered already in Exam-ple 6.14. Substituting the values for u ≡ σ(x, y) and v ≡ ρ(x, y) obtainedthere into v = C1 +C2u yields after some simplifications the general solution

y = C2x2 − x

C1 + C2x2 − x.

Example 7.15 Equation 6.124 has been considered before in Example 6.15.Substituting the values for u ≡ σ(x, y) and v ≡ ρ(x, y) obtained there intov = C1 + C2u yields after some simplifications the general solution

y = 1√C1e−x + C2e−2x

.

Lie’s Second Integration Method. In his book on differential equationsLie [113], Kapitel 20, §4 and 5, a solution scheme for differential equationswith symmetries is described that he called second method of integration. It isbased on the observation that any second order quasilinear ode y′′ = ω(x, y, y′)

may be written as the system y′ = dydx

, y′′ = dy′

dxwhich in turn is equivalent

to the linear pde∂f

∂x+ y′

∂f

∂y+ ω(x, y, y′)

∂f

∂y′= 0.

The application of this second integration scheme is based on the followinglemma.

Lemma 7.2 (Lie 1891) The differential equation y′′+ r(x, y, y′) = 0 allowsthe symmetry with the generator U = ξ∂x + η∂y if and only if the linear pde

Af ≡ ∂f

∂x+ y′

∂f

∂y− r(x, y, y′) ∂f

∂y′= 0

allows the first extension of U , i.e., if [U,A] = ρ(x, y, y′)A for some function ρ.

328

The proof may be found in the above quoted book by Lie. Based on thislemma, the problem of solving an ode with symmetries is replaced by a linearpde allowing the once extended vector fields and the theory of Chapter 2,Section 5 applies. This scheme is applied in the following example.

Example 7.16 Equation 6.159 has been solved in Example 7.9. For thisequation r = −3y2 − 3

4y y′2, its symmetry generators are (see Example 5.23)

U1 = ∂x and U2 = x∂x − 2y∂y. Therefore the system

∂xf + y′∂yf +(3y2 − 3

4yy′2

)∂y′f = 0, ∂xf = 0, x∂xf − 2y∂yf − 3y′∂y′f = 0

is obtained. It leads to ∆ = 32y′2 − 6y3 and the integral (2.47) yields

Φ1 =∫ (

3y2 +34yy′2

)dy∆−

∫y′dy′

∆= log

y√y

y′2 − 4y3 .

Its exponential is the first integral u3v′2

v′2 − 1obtained in Example 7.9 written

in actual variables x and y.

7.3 Nonlinear Equations of Third Order

In this section a complete discussion of symmetry methods for solving gen-uinely nonlinear equations of third order is presented, similar to second orderequations above. In Lie’s book [113] on differential equations there is only ashort chapter on third order equations allowing certain three-parameter sym-metry groups.

One-Parameter Symmetry. A quasilinear third order ode in symmetryclass S3

1 has the canonical form

v′′′ + r(v′′, v′, u) = 0. (7.30)

Substituting v′ = w(u) yields

w′′ + r(w′, w, u) = 0. (7.31)

This is a general second order equation for w(u) because r(w′, w, u) is anundetermined function of its arguments as it has been explained in the intro-duction on page 311. If it has a symmetry, the results of the preceding sectionmay be applied for solving it, if not it is returned by the following algorithmperforming these steps.


Algorithm 7.12 LieSolve3.1(ω). Given a quasilinear third order ode ω insymmetry class S3

1 , its general solution or a lower order equation is returned,or failed if neither may be found.S1 : Janet basis for symmetries. Generate a Janet basis for the deter-mining system of the Lie symmetries of ω.S2 : Determine transformation functions. If case a) of Theorem 6.7applies choose σ. If case b) applies solve the first order equation for σ. Ifit cannot be obtained return failed. Determine ρ by integration.S3 : Canonical form. Applying the solutions obtained in S2 generatethe canonical form (7.31). If it does not allow any nontrivial symmetryreturn it.S4 : Return solution. Proceed with the proper solution procedure fora second order equation. If a solution is found, resubstitute the actualvariables and return the result.

Example 7.17 The canonical form obtained in Example 5.27 leads to thesecond order equation (7.30)

w′′w − 3w′2 − (u− 4)w′w2 + (2u− 4)w4 + w3 = 0.

Because it does not have any nontrivial symmetry, this is the best possibleanswer in the context of Lie’s theory.

Two-Parameter Symmetries. According to the discussion in the introduc-tion of this chapter, a third order equation with a two-parameter symmetryin general may be reduced to a first order ode of type (7.1). For the four two-parameter symmetry classes discussed there these are the equations (7.32),(7.35), (7.39) and (7.42) below. The first symmetry class to be discussed isS3

2,1. It has the canonical form

v′′′ + r(v′′, v′) = 0. (7.32)

Substituting v′′′ = v′′ dv′′

dv′yields

dv′′

dv′+r(v′′, v′)v′′

= 0. (7.33)

This is a general first order equation for v′′(v′) because r(v′′, v′) is an unde-termined function of its arguments. If v′′ = w(v′, C1) is its general solution,an integration yields ∫

dv′

w(v′, C1)= u+ C2. (7.34)

If the integration may be performed, the resulting first order ode allows theone-parameter symmetry ∂v. Therefore the desired solution of (7.32) may berepresented in terms of an integral.

330

The various steps are put together in the algorithm below. The solutionprocedure may fail in step S2 if the transformation functions are not Liouvil-lian, or in step S4 if equation (7.33) cannot be solved.

Algorithm 7.13 LieSolve3.2.1(ω). Given a quasilinear third order odeω in symmetry class S3

2,1, its general solution or some lower order equation isreturned.S1 : Equations for σ and ρ. Generate a Janet basis for the symmetrygenerators of ω and set up system (6.3).S2 : Determine transformation functions. Determine an independent pairof Liouvillian solutions of the system obtained in step S1. If it cannot befound return failed.S3 : Second order equation. By means of the transformation functionsobtained in S2 generate equation (7.33) and determine its solution. If itcannot be found return (7.33).S4 : Integral representation. Perform the integration at the left hand sideof (7.34). If it cannot be done return (7.34).S5 : Return solution. Solve the first order equation obtained in step S4and return the result. If a solution cannot be found return the first orderequation and the transformation functions.

Example 7.18 The canonical equation v′′′+2v′′v′+1 = 0 of Example 6.17has the form (7.32). With v′′ = w and v′ = z the Abel equation of secondkind ww′ + 2wz + 1 = 0 for w(z) is obtained. According to Chapter 4.2 itmay be transformed into an Abel equation of first kind for which a symmetrycannot be found.

A quasilinear third order ode in symmetry class S32,2 has the form

v′′′ +1u2r(uv′′, v′) = 0. (7.35)

Substituting first v′ = z yields the equation z′′+ 1u2 r(uz

′, z) = 0 allowing theone-parameter group u∂u. By Theorem 6.7 new variables w and t defined byu = ew, z = t with w ≡ w(t), and finally w′(t) = w(t) yield the first orderequation

w′ + w2 − w3r( 1w, t

)= 0. (7.36)

If its general solution is w = w0(t, C1), the relation

u = C2 exp( ∫

w0(v′, C1)dv′)

(7.37)

between v′ and u involving two constants follows. If the integration may beperformed, the solution procedure may be continued.

The subsequent algorithm performs these steps. The crucial step is S3. Ifit fails the first order equation (7.36) is the best possible answer.



2,2, its general solution or some lower order equation isreturned.S1 : Equations for σ and ρ. Generate a Janet basis for the symmetrygenerators of ω.S2 : Determine transformation functions. Determine the transformationfunctions σ from (6.6), and ρ from (6.7) or (6.8).S3 : First order equation. By means of the transformation functionsobtained in S2 generate equation (7.36) and determine its solution. If itcannot be found return (7.36).S4 : Integral representation. Perform the integration at the left hand sideof (7.34). If it cannot be done return (7.37).S5 : Return solution. Solve the first order equation obtained in step S4and return the result. If a solution cannot be found return the first orderequation and the transformation functions.

It may occur that the above second order equation for z allows a largersymmetry group than just the one-parameter generator u∂u as it is shownin the example below. In these cases it is advantageous to apply this largersymmetry for the further solution procedure.

Example 7.19 The canonical equation obtained in Example 6.18 has theform (7.35). It turns out that after substituting v′ = z the resulting equation

z′′ − 3zz′2 − 5

uz′ − z(z2 + 4)

u2= 0 (7.38)

belongs to symmetry class S28 . Consequently, algorithm LieSolve2.8 may be

applied. It yields the solution z = 4C1√C2u4 + u2 − 4C2

1

, and finally

x =∫

dy√C1y2 + C2 − 1

4y4

+ C3

in actual variables. In Exercise 7.5 the same result is obtained from the firstorder equation (7.36).


v′′′ + v′r( v′′

v′, u

)= 0. (7.39)

Substituting first v′ = z yields

z′′ + zr( z′

z, u

)= 0 or

z′′

z+ r

( z′

z, u

)= 0.

332

Finally substituting z′z = w leads to

w′ + w2 + r(w, u) = 0. (7.40)

If this general first order equation may be solved, and w = w0(u,C1) is itsgeneral solution, the solution of (7.39) is

v = C2

∫exp

( ∫w0(u,C1)du

)du+ C3. (7.41)

The subsequent algorithm may terminate in step S2 without generating thedesired solution if the first order ode for σ cannot be solved, or in step S3 ifequation (7.40) cannot be solved.


2,3, its general solution or some lower order equation isreturned.S1 : Equations for σ and ρ. Generate a Janet basis for the symmetrygenerators of ω, and set up equation (6.41) for σ and equations (6.39),...,(6.43) for ρ.S2 : Determine transformation functions. Choose σ or solve (6.41). If σ isnot known at this point return failed. Otherwise determine independentsolution for ρ.S3 : Second order equation. By means of the transformation functionsobtained in S2 generate equation (7.40) and determine its solution. If itcannot be found return (7.40).S4 : Integral representation. Perform the integration at the left hand sideof (7.41). If it cannot be done return (7.41).S5 : Return solution. Solve the first order equation obtained in step S4and return the result. If a solution cannot be found return the first orderequation and the transformation functions.

Example 7.20 The canonical equation obtained in Example 6.19 has theform (7.39) from which the Abel equation (7.40) w′ + w3 + w2 + u + 1 = 0follows. Because a symmetry cannot be found for it, this first order equationis the best possible answer in the realm of symmetry analysis.


v′′′ + r(v′′, u) = 0. (7.42)

Substituting v′′ = w the general first order equation

w′ + r(w, u) = 0 (7.43)

is obtained. If w = w0(u,C1) is its general solution, the solution of (7.42) is

v =∫ ∫

w0(u,C1)du du+ C2u+ C3. (7.44)


The subsequent algorithm applies these steps. It may fail in step S2, or afirst order equation may be returned in step S3 as the best possible answerof the symmetry analysis.


2,4, its general solution or some lower order equation isreturned.S1 : Equations for R or S. Generate a Janet basis for the symmetrygenerators of ω and set up equations (6.44), (6.45) or (6.46).S2 : Determine R or S. From the equations obtained in S1 determinerational solution for R or S. If none is found return failed.S2 : Determine transformation functions. Applying the solution obtainedin S2 determine σ and ρ by integration.S4 : First order equation. Generate the first order equation (7.43) withthe help of σ and ρ. If a solution cannot be found return this equation.S5 : Integral representation. Using the solution obtained in S4 returnthe integral representation (7.44) and the transformation functions.

Example 7.21 The canonical equation obtained in Example 6.20 has theform (7.42) from which the Riccati equation (7.43) w′ + uw2 + w + 1 = 0follows. Because a closed form solution cannot be found it is the best possibleresult of the symmetry analysis.

Three-Parameter Symmetries. The general solution procedure for theequations covered in this subsection has been described in Theorem 7.1 inthe introduction to this section. According to the classification of symmetriesgiven in Theorem 5.13 and the listing of groups and Lie algebras on page 135,the first two symmetry classes with nonsolvable Lie algebras of symmetrygenerators require solving a Riccati equation, i.e., equations (7.46) and (7.49)below. In the remaining cases with a solvable symmetry algebra the solutionreduces essentially to integrations.

The first symmetry class S33,1 has a canonical form

v′′′+6v′′(v′ + 1)u− v

+6v′(v′2 + 4v′ + 1)

(u− v)2+

v′2

(u− v)2r(v′′(u− v) + 2v′(v′ + 1)

v′√v′

)= 0.

(7.45)

Introducing w =√v′ and z = v′′(u− v) + 2v′(v′ + 1)

v′√v′

as new variables, the

Riccati equationdw

dz+

w2 − 12zw + 1

r(z)− 32z

2 − 12= 0 (7.46)

for w ≡ w(z) is obtained. If its general solution may be determined in theform Φ(v′, z) = C1, two more independent relations between u, v, v′ and zmay be obtained by differentiation, see Lie [109], page 288.

(v − u)v′Φv′ = C2 and (v − u)v′[(v − u)v′Φv′v′ + 2vΦv′

]= C3. (7.47)

334

From these relations v may be obtained by elimination. These steps areapplied in the subsequent algorithm. It may terminate in step S2 if a rationalsolution of the respective Riccati equations may not be found, or in step S4if equation (7.46) cannot be solved.

Algorithm 7.17 LieSolve3.3.1(ω). Given a quasilinear third order ode ωin symmetry class S3

3,1, its general solution is returned; or failed if it cannotbe determined.S1 : Equations for R and S. Generate Janet basis for the symmetrygenerators of u and set up system (6.13), (6.16) or (6.18) for R, or system(6.20) for R and S respectively.S2 : Determine R and S. From the equations obtained in S1 determinea rational solution for R and S. If none can be found return failed.S3 : Determine transformation functions. Applying the result of S2 setup the system (6.14), (6.17), (6.19) or (6.21) for σ and ρ respectively anddetermine two independent Liouvillian solutions.S4 : Riccati equation. Applying the transformation functions obtainedin step S3 generate equation (7.46). If its solution cannot be determinedreturn this equation.S5 : Return solution. Set up system (7.47), eliminate v, resubstitute theactual variables and return the result.

Example 7.22 The canonical equation obtained in Example 6.21 hasthe form (7.45) with r(z) = z2 + 1 and z as given above. It leads to the

Riccati equation (7.49) w′ − 2w2 − zw + 2z2 + 22

= 0 for w(z). Because a closed

form solution cannot be found it is the best possible result of the symmetryanalysis.

A quasilinear third order ode in symmetry class S33,2 has the canonical form

v′′′v5 + 3v′′v′v4 + r(v′′v3) = 0. (7.48)

If the two lowest invariants w ≡ vv′ and z ≡ v′′v3 of the two-parametersubgroup ∂u, 2u∂u+v∂v are introduced as new dependent and independentvariables and dw

du= vv′′ + v′2, dz

du= v′′′v3 + 3v′′v′v2 are applied, (7.48)

assumes the formdw

dz+

1r(z)

w2 +z

r(z)= 0. (7.49)

This Riccati equation for w(z) has to be solved. If w = w0(z, C) is its generalsolution, in the original variables u and v a second order equation of the form

w0(v′′v3, C)− vv′ = 0

is obtained allowing the above mentioned two-parameter symmetry group.Because it is not in the canonical form as applied in Theorem 5.8, this second


order ode is not in canonical form (7.17). Introducing new variables u andv(u) by u = v− u and v =

√u yields ∂v and u∂u + v∂v, and the second order

equation

w0

(−2uv′′ + v′ − 1

4(v′ − 1)3, C

)− 1

2(v′ − 1)= 0. (7.50)

Under suitable constraints for the function w0 it may be treated by the al-gorithm LieSolve2.2.2 on page 322. These steps are applied in the algorithmbelow. It may terminate in step S2 if a rational solution for R cannot befound, or in step S5 if (7.49) cannot be solved, or if (7.50) is too complicated.


3,2, its general solution is returned; or failed if it cannotbe found.S1 : Equations for R. Generate Janet basis for the symmetry generatorsof ω. Set up system (6.22), (6.23) or (6.24) for R from it.S2 : Determine R. From the equations obtained in S1 determine a ra-tional solution for R. If it cannot be found return failed.S3 : System for transformation functions. Applying the result of S2 setup the systems (6.22), (6.23) or (6.24) for σ and ρ.S4 : Determine transformation functions. Determine two independentLiouvillian solutions for the system obtained in S3.S5 : Riccati equation. Generate equation (7.49) and determine a rationalsolution. If it cannot be found return this equation.S6 : Return solution. Employing the solution obtained in S5, generate(7.50), apply algorithm LieSolve2.2.2 to it and return the result.

Example 7.23 The canonical equation obtained in Example 6.22 has theform (7.48) with r(v′′v3) = 2v′′v3 +1 and leads to the Riccati equation (7.49)dwdz

+ 12z + 1w

2 + z2z + 1 = 0. Because a closed form solution cannot be found

it is the best possible result of the symmetry analysis.


v′′′ + v′(c−3)/(c−1)r(v′′v′−(c−2)/(c−1)

)= 0. (7.51)

With v′′′ = dv′′

dv′v′′ it may be rewritten as a first order ode for v′′(v′)

dv′′

dv′v′′ + v′(c−3)/(c−1)r

(v′′v′−(c−2)/(c−1)

)= 0

with the symmetry (c − 1)v′∂v′ + (c − 2)v′′∂v′′ . Applying Theorem 6.1,new variables u and v(u) are introduced by v′ = u−(c−1)/(c−2)e(c−1)v andv′′ = e(c−2)v with the result

v =c− 1c− 2

∫1

(c− 1)r(u) + (c− 2)u2

du

u+ C1 ≡ R(u) + C1. (7.52)

336

If the original variables are resubstituted by u = v′′v′−(c−2)/(c−1) andv = 1

c− 2 log v′′, the equation

R( v′′

v′(c−2)/(c−1)

)− 1c− 2

log v′′ + C1 = 0 (7.53)

is obtained. If R does not involve integrals, this is a second order ode allowingthe two symmetries ∂u and ∂v. Under suitable constraints for R it may betreated by the algorithm LieSolve2.2.1 on page 322. The subsequent algorithmis designed according to this scheme. The only bottleneck where it may failis the integral (7.52).

Algorithm 7.19 LieSolve3.3.3(ω). Given a third order quasilinear or-dinary differential equation in symmetry class S3

3,3(c), its general solution isreturned, or equation (7.53) if it cannot be determined.S1 : Janet basis for symmetries. Generate a Janet basis for the sym-metry generators of ω and set up system (6.25), (6.26), (6.27) or (6.28)respectively.S2 : Determine transformation functions. Determine two independentLiouvillian solutions for the system obtained in S2.S3 : Integral representation. Generate (7.53). If the integration may notbe executed in closed form return this integral.S4 : Return solution. If (7.53) is in S2

2,2, apply algorithm LieSolve2.2.1to it and return the result. Otherwise return (7.53).

Example 7.24 If into the canonical form v′′′v′3 + 1 = 0 of Example 6.23a new dependent variable is introduced by v′ = w, the second order equationw′′w3+1 = 0 in symmetry class S2

3,2 is obtained. Applying (7.25) an additionalintegration yields its solution

v = 12 (u+ C2)

√C2

1 (u+ C2)2 − 1

− 12C1

log[C1(u+ C2) +

√C2

1 (u+ C2)2 − 1]+ C3.

Substituting the canonical variables determined in Example 6.23, the generalsolution

y = 12x (C2x+ 1)

√C2

1 (C2x+ 1)2 − x2 + 12C1

log x

− 12C1

logC1(C2x+ 1) +√C2

1 (C2x+ 1)2 − x2 + C3

of the equation in Example 5.34 is finally obtained.


v′′′ + e−2v′r(v′′ev

′)= 0. (7.54)


Writing v′′′ = dv′′

dv′v′′ yields the first order equation

dv′′

dv′+e−2v′

v′′r(v′′ev

′) = 0 (7.55)

for v′′(v′) with the symmetry ∂v′ − v′′∂v′′ . Applying Theorem 6.1, new vari-ables u and v(u) are introduced by v′ = log u + v and v′′ = e−v with theresult

v =∫

r(u)u2 − r(u)

du

u+ C1 ≡ R(u) + C1. (7.56)

Upon resubstitution of the original variables by u = v′′ev′

and v = − log v′′,the equation

R(v′′ev

′)+ log v′′ + C1 = 0 (7.57)

is obtained. If R does not involve integrals, this is a second order ode allow-ing the two symmetries ∂u and ∂v. Similar to the preceding case, the onlybottleneck where the algorithm below may fail is the integral (7.56).

Algorithm 7.20 LieSolve3.3.4(ω). Given a third order quasilinear or-dinary differential equation ω in symmetry class S3

3,4, the general solution isreturned, or equation (7.56) if it cannot be determined.S1 : Janet basis for symmetries. Generate a Janet basis for the sym-metry generators of ω and set up system (6.29), (6.30), (6.31) or (6.32)respectively.S2 : Determine transformation functions. Determine two independentLiouvillian solutions for the system obtained in S2.S3 : Integral representation. Generate (7.57). If the integration may notbe performed return this integral.S4 : Return solution. Apply algorithm LieSolve2.2.1 to the equationobtained in S3 and return the result.

Example 7.25 The canonical equation obtained in Example 6.24 has theform (7.54) with r(v′′ev

′) = v′′2e2v

′+ 1. In step S2 the equation

v′′ −√

2√v′ − C1e

−v′ = 0 is obtained. Finally, step S4 yields the param-eter representation u = f(p) + C2 and v =

∫f(p)dp − pf(p) + C3 where

f(p) = − 1√2

∫epdp√p− C1

.


v′′′ + v′r( v′′

v′

)= 0. (7.58)

Substituting v′′

v′= z yields the first order equation dz

du+ z2 + r(z) = 0 from

which ∫dz

z2 + r(z)+ u = C1 (7.59)

338

follows. If the integration may be performed and the result may be solved forz, i.e., z = z0(u,C1), the second order linear equation

v′′ − z0(u,C1)v′ = 0 (7.60)

is obtained with the general solution

v = C2

∫exp

( ∫z0(u,C1)du

)du+ C3. (7.61)

The algorithm below is based on this solution scheme. It may fail in stepS3 if the integration in (7.59) may not be executed, or if the result of theintegration may not be solved for z. In these cases an expression involving v′′

and v′ is returned. On the other hand, if equation (7.60) may be obtained,the general solution is returned.


3,5, its general solution is returned.S1 : Janet basis for symmetries. Generate Janet basis for symmetrygenerators of ω and set up equations (6.48), (6.49), (6.50), (6.51) or(6.52) respectively.S2 : Determine transformation functions. From the equations obtainedin S2 determine two independent Liouvillian solutions.S3 : Integral representation. Applying the solutions obtained in S3 gen-erate the equation (7.59). If z0(u,C1) cannot be obtained, return thisexpression.S4 : Return solution. Generate equation (7.60) and its solution (7.61),resubstitute actual variables and return the result.The symmetry class S3

3,5 is one of the few cases of third order equationsconsidered by Lie [113], pages 556-558. He uses canonical variables u = vand v = u such that the group generators have the form ∂u, ∂v and u∂u

with invariants v′′

v′2, v

′′′

v′3and the canonical equation v′′′ + v′′r

(v′′

v′3

)= 0.

Its integration leads finally to the same answer (7.61). Of particular interestare Lie’s remarks on page 556, line 7 to 10 from the bottom, concerning thecomparison of this proceeding with his second integration method describedon page 327.1

Example 7.26 The canonical form obtained in Example 6.25 corresponds

to r(z) = z2 − 2 and the solution z = Ce4u + 1Ce4u − 1

. Substituting it into the

second order equation v′′ − zv′ = 0 yields

v = C1+C2u2+

12

∫ [log

(C3e

u − 1)+log

(C3e

u + 1)+log

(C2

3e2u + 1

) ]du.

1Translation by the author: In certain cases this method requires simpler operations thanthe above... .


The four symmetry classes S33,6 to S3

3,8 and S33,9(c) arise from the groups

g17 in Chapter 3, page 139. Their common canonical form

dz

du+ r(z) = 0 where z(u) ≡ v′′ + c1v

′ + c2v (7.62)

is linear and homogeneous in v ≡ v(u), v′ and v′′, with constant coefficientsc1 and c2. This first order equation for z may always be integrated with theresult ∫

dz

r(z)+ u+ C ≡ R(z) + u+ C = 0. (7.63)

This relation determines z either explicitly in the form z = z0(u,C) or im-plicitly as function of u. In any case v is the solution of the linear equation

v′′ + c1v′ + c2v = z0(u,C). (7.64)

The left hand side of this equation is always completely reducible (see Theo-

S33,6 : v = C1 + C2u+

∫ ∫z0(u,C)dudu,

S33,7 : v = C1e

u + C2ueu − (u+ 1)e−uz0(u,C) +

∫(u+ 1)e−uz0(u,C)du,

S33,8 : v = C1 + C2e

u + z0(u,C) +∫z0(u,C)du− eu

∫e−uz′0(u,C)du,

S33,9(c) : v = C1e

u + C2ecu + eu

c∫e−uz0(u,C)du− ecu

c∫e−cuz0(u,C)du.

(7.65)Resubstituting the actual variables into these expressions finally yields thesolution of the given equation. Due to the similiar course of the solution pro-cedure for all four groups under consideration, the various steps have beenput together in the following algorithm. It may fail in step S3 if a hyperex-ponential solution for ρ cannot be found.

Algorithm 7.22 LieSolve3.3.6to9(ω). Given a quasilinear third order odeω in one of the symmetry classes S3

3,6 to S33,8 or S3

3,9(c), its general solutionis returned; or failed if it cannot be determined.S1 : Equations for transformation functions. Generate Janet basis forsymmetry generators of ω and set up equations for σ and ρ as describedin Theorems 6.20, 6.21, 6.22 or 6.23 respectively.S2 : Determine transformation functions. From the equations obtainedin S1 determine a pair of independent Liouvillian solutions for σ andhyperexponential solutions for ρ. If the latter may not be found returnfailed.S3 : Canonical form. Applying the solutions obtained in S2 generate thecanonical form (7.64).

rem 6.16). Therefore its general solution may be written as follows.

340

S4 : Return solution. Substitute the value of z0 and the actual variablesinto the appropriate expression (7.65) and return the result.

Example 7.27 The canonical form equation dzdx

+ r(z) = 0 with r(z) =

−z2 + 1z that has been obtained in Example 7.27 yields R(z) = 1

2 log z2 + 1and, consequently, z0(u,C) =

√Ce−2u − 1. This value for z0(u,C) has to

be substituted into the expressions for v given above. Finally resubstitutingthe actual variables by u = 1

x and v = y and renaming C into C3 yields thesolution

y = C1 + C21x

+∫ ( ∫

z0(x,C3)dx

x2

)dxx2

of the first ode’s in Example 5.37 with z0(x,C3) =√C3 exp (− 2

x )− 1. Pro-ceeding along the same lines, the answer for the remaining three ode’s is

v = C1 exp( 1x

)+ C2

1x

exp( 1x

)−

(1 +

1x

)exp

(− 1x

)z0(x,C3)

+∫ (

1 +1x

exp(− 1x

)z0(x,C3

) dx

x2 ,

v = C1 + C2 exp( 1x

)+ z0(x,C3)−

∫z0(x,C3)

dx

x2

+ exp( 1x

) ∫exp

(− 1x

)z′0(x,C3)dx,

v = C1 exp( 1x

)+ C1 exp

( c

x

)+

1c

[exp (

1x

)∫

exp (− 1x

)z0(x,C3)dx

x2

− exp (c

x)∫

exp(− cx

)z0(x,C3)

dx

x2

]respectively.

A quasilinear third order ode in symmetry class S33,10 has the canonical

formv′′′ + v′′2r(v′) = 0. (7.66)

Substituting v′′′ = v′′ dv′′

dv′yields the linear first order equation dv′′

dv′+ r(v′)v′′

= 0 for v′′(v′). Integration leads to

v′′ + C1 exp(−

∫r(v′)dv′

)= 0. (7.67)

If the integral may be performed, this is a second order ode in symmetry classS2

2,1. For special forms of r(v′) its solution may be obtained explicitly. In


any case, defining R(p) =∫

exp (∫r(p)dp)dp the expressions (7.16) yield the

parameter representation

u+ C1R(p) = C2, v + C1pR(p) = C1

∫R(p)dp+ C3. (7.68)

The following algorithm is designed according to this scheme. For any equa-tion in this symmetry class at least the parameter representation (7.68) isreturned.


3,10, its general solution is returned either explicitly oras a parameter representationS1 : Equations for transformation functions. Generate Janet basis forsymmetry generators of ω and set up equations (6.56) for σ and ρ.S2 : Determine transformation functions. From the equations obtainedin S1 determine a pair of independent Liouvillian solutions or failed if itmay not be found.S3 : Canonical form. Applying the solutions obtained in S3 generate theequation (7.67).S4 : Return solution. Apply algorithm LieSolve2.2.1 to the equationobtained in S3 and return the result.

Example 7.28 The canonical form of the ode considered in Example 6.27in symmetry class S3

3,10 corresponds to r(v′) = v′ + 1v′

. For this particularchoice of r the integrals at the left hand side of (7.67) may be executed inclosed form with the result v′2 = 2 log (C1u+ C2). It leads to the generalsolution

v =√

2∫ √

log (C1u+ C2)du+ C3.

Substituting u = 1x and v = y yields the solution

y = −√

2∫ √

log (C11x

+ C2)1x2 dx+ C3

of the ode in Example 5.38.

Four-Parameter Symmetries. In the remaining part of this section thenumber of symmetries exceeds the order of the given equation at least by one.Consequently, the solution procedure is simplified. It requires only algebraicoperations.

The first symmetry class S34,1 has a canonical form

v′′′v′ + av′′2 = 0 (7.69)

with constant a 6= 0, a 6= − 32 . Dividing by v′′v′ and performing three consec-

utive integrations leads to the general solution

(a+ 2)a+1(C1v + C2)a+1 = (a+ 1)a+2(C1u+ C3)a+2.

342

In this form the invariance under exchange of u and v combined with thereplacement of a by −(a+ 3) is obvious. Usually the explicit form

v = (C1u+ C2)(a+2)/(a+1) + C3 (7.70)

with new constants Ck is more convenient for solving equations. The constanta together with the transformation functions has to be substituted into (7.70)in order to generate the solution for the given equation.

The subsequent algorithm applies these steps for obtaining the general so-lution. For any equation in this symmetry class it is guaranteed that thesolution is found.


4,1, its general solution is returned.S1 : Equations for transformation functions. Generate Janet basis forsymmetry generators of ω and set up equations (6.57), (6.58), (6.59) or(6.60) respectively depending on the type of the Janet basis.S2 : Determine transformation functions. From the equations obtainedin S1 generate an independent pair of solutions.S3 : Canonical form. Applying the solutions obtained in S2 generate thecanonical form (7.69).S4 : Return solution. Substitute the value of a and the actual variablesinto (7.70) and return the result.

Example 7.29 Substituting the values a = − 12 and u = xy, v = 1

y into(7.70) ensures the solution (C1xy+C2)3y+C3y+12C1 = 0, i.e., a third orderalgebraic function. It is easy to see that a = − 5

2 and u = 1y , v = xy lead to

the same answer.

The canonical form of an ode in symmetry class S34,2 is

v′′′v + 3v′′v′ + av′′√v′′v = 0. (7.71)

Its symmetry group contains ∂u, 2u∂u + v∂v, u2∂u + uv∂v as a subgroup.

Writing (7.71) as v′′′v5 + 3v′′v′v4 + a(v′′v3)32 = 0 shows that it is a special

case of (7.48) with r(v′′v) = a(v′′v3)32 . This special form of r is a consequence

of the additional symmetry v∂v. Applying the same transformation which ledto (7.49) yields a special Riccati equation with an additional symmetry. It is

more efficient however to introduce the new dependent variable w = v′v which

has to satisfyw′′ + 6ww′ + 4w3 + a(w′ + w2)3/2 = 0.

Its general solution is rational and leads to the general solution

v = C1 exp∫

a2(x+ C2)dxa2x(x+ C3) + 4C2

2 − (a2 − 16)C2(C2 − C3)(7.72)


of (7.71). Another method for arriving at this result is discussed in Exer-cise 7.11.

The subsequent algorithm implements the various steps. The desired solu-tion is only returned if rational transformation functions in step S2 may befound.


4,2, its general solution is returned; or failed if it cannotbe determined.S1 : Equations for R. Generate Janet basis for symmetry generators ofω and set up equations for R according to (6.61) or (6.62) depending onthe type of the Janet basis.S2 : Determine R. From the equations obtained in S1 determine rationalsolution for R. If none may be found return failed.S3 : Equations for transformation functions. Applying the solutions ob-tained in S2 set up equations (6.61) or (6.61) for σ and ρ.S4 : Return solution. Applying the transformation function obtainedin S3 generate (7.72), resubstitute the actual variables and return theresult.For an ode in symmetry class S3

4,3(c) it is assumed that the value of theparameter c is already known. The canonical form of an equation in thissymmetry class is

v′′′ + av′′α = 0 with α =c− 3c− 2

. (7.73)

Substituting v′′ = w leads to w′ + awα = 0. Three successive integrationsyield the general solution

v =1

a2(2− α)(3− 2α)(α− 1)(3−2α)/(1−α)(au+ C1)(3−2α)/(1−α) + C2u+ C3.

(7.74)The subsequent algorithm implements the various steps. The desired solu-

tion is only returned if hyperexponential transformation functions in step S2may be found.


4,3(c), its general solution is returned; or failed if it cannotbe determined.S1 : Equations for transformation functions. Generate Janet basis for thesymmetry generators of ω and set up equations (6.63), (6.64) or (6.65)respectively depending on the type of Janet basis.S2 : Determine transformation functions. From the equations obtainedin S1 generate an independent pair of hyperexponential solutions. If noneis found return failed.S3 : Canonical form. Applying the solutions obtained in S3 generate thecanonical form (7.73).

344

S4 : Return solution. Substitute the value of a and the actual variablesinto (7.74) and return the result.

Example 7.30 Substituting the values a = 1, c = 32 , i.e., α = 3 and

u = 1x , v = xy into (7.74) ensures the general solution

y = 2√

23x

(C1 + 1

x)3/2 + C2

x2 + C3x .

Finally the canonical form of an ode in symmetry class S34,4 is

v′′′ + a exp(− 1

2v′′) = 0. (7.75)

Substituting v′′ = w leads to w′ + a exp (− 12w) = 0. Similar to the preceding

case, three succesive integrations lead to

v = (u− C1)2 log(u− C1) +(log (− 1

2a)−32

)u2 + C2u+ C3. (7.76)

The subsequent algorithm implements the various steps. The desired solu-tion is only returned if hyperexponential transformation functions in step S2may be found.


4,4, its general solution is returned; or failed if it cannotbe determined.S1 : Equations for transformation functions. Generate a Janet basis forthe symmetry generators of ω and set up equations (6.66), (6.67), (6.68)or (6.69).S2 : Determine transformation functions. From the equations obtainedin S1 generate an independent pair of hyperexponential solutions. If noneis found return failed.S3 : Canonical form. Applying the solutions obtained in S3 generate thecanonical form (7.75).S4 : Return solution. Substitute the value of a and the actual variablesinto (7.76) and return the result.

Six-Parameter Symmetry S36 . In canonical variables an ode in symmetry

class S36 has the form

v′′′v′ − 32v′′2 = 0. (7.77)

The subgroup ∂u, ∂v, u∂u, v∂v may be utilized for its integration. The spe-cial value a = − 3

2 in (7.70) leads to the general solution

v =C1

u+ C2+ C3

after a suitable change of the integration constants. The algorithm belowapplies these steps. It may fail if rational solutions for R and S cannot befound.


Algorithm 7.28 LieSolve3.6(ω). Given a quasilinear third order ode ωin symmetry class S3

6 , its general solution is returned; or failed if it cannot bedetermined.S1 : Equations for R and S. If u has the form (5.40) set up the equationsfrom (6.70) and (6.71). If ω has the form (5.41), set up equations from(6.72) and (6.73).S2 : Determine R and S. From the equations obtained in S1 determinea rational solution for R and S. If none may be found return failed.S3 : System for transformation functions. Applying the result of S2 setup the system from (6.70), (6.71) or (6.72) and (6.73) for σ and ρ.S4 : Determine transformation functions. Determine two independentLiouvillian solutions for the system obtained in S3.S5 : Return solution. Substitute the solutions from S4 into (7.77) andreturn the result.

Example 7.31 For the equation with symmetry type S36 and structure

(5.40) in Example 5.41 the transformation functions σ = xy and ρ = yx

have been obtained. Substituting these values into (7.77) leads to the generalsolution

x3y2 + (C1x2 + C2)y + C3x = 0.

Example 7.32 For the equation in symmetry class S36 and structure (5.41)

of Example 5.42 the transformation functions σ = 1y and ρ = y

xex have been

obtained. Substituting these values into (7.77) leads to the general solution

C1y2 + (C2xe

−x + 1)y + C3xe−x = 0.

7.4 Linearizable Third Order Equations

The equations considered in this section are assumed to be equivalent toa linear ode. Consequently, the methods described in the first section ofChapter 2 may be applied after the canonical form has been obtained.Four-Parameter Symmetry. The canonical form for any equation in theonly symmetry class S3

4,5 of this subsection is the generic linear third orderequation in rational normal form

v′′′ + P (u)v′ +Q(u)v = 0. (7.78)

The coefficients P (u) and Q(u) may assume quite different shapes by trans-formations of the structure invariance group u = F (u) and v = F ′(u)v. Thefurther proceeding depends on the structure of the differential Galois group of(7.78) which may allow it to obtain solutions explicitly, e. g. if it is reduciblewith a nontrivial Loewy decomposition. In general, however, it is irreducible

346

and the only information that is guaranteed for equations in this symmetryclass is its linearizability.

Algorithm 7.29 LieSolve3.45(ω). Given a quasilinear third order ode ωin symmetry class S3

4,5, its general solution is returned; or failed if it cannotbe determined.S1 : Identify case a), b) or c). Determine the Janet basis for the Liesymmetries of ω. By means of the criteria given in Theorem 6.31 identifythe case that applies for equation ω.S2 : Equations for σ and ρ. From the Janet basis obtained in step S1 con-struct the system for ρ according to (6.74), (6.76) or (6.78) respectively.In case c) set up the first order ode for σ.S3 : Determine σ and ρ. In case a) and b) choose σ, in case c) determineit from the first order ode. From the equations constructed in step S2find a hyperexponential solution for ρ, if none exists return failed.S4 : Canonical form equation. Applying the transformation functionsfrom step S3 generate a canonical form equation, determine its Loewydecomposition and construct the most general explicit solution from it.S5 : Find solution. Substitute u = σ(x, y) and v = ρ(x, y) into thecanonical form solution obtained in step S4 and return the result.

Example 7.33 The first canonical form (6.80) obtained in Example 6.32has the Loewy decomposition

(v′+ 4u

2u− 1v)Lclm

(v′− 2v, v′− 1

uv)

= 0 fromwhich the general solution

v = C1u+ C2e2u + C3

[e2u

∫ue−4u

(2u− 1)2du− u

∫e−2u

(2u− 1)2du

]follows. Substituting the transformation u = x and v = x

y given there yieldsthe solution of the equation in Example 5.43 in the form

y =x

C1x+ C2e2x + C3

[e2x

∫xe−4x

(2x− 1)2dx− x

∫e−2x

(2x− 1)2dx

] . (7.79)

The second canonical form (6.81) has the Loewy decomposition[v′+

( 1u− 2

+1u− 2u2

)v

]Lclm

(v′ − 1

uv, v′ − (

2u− 2u2 )v

)= 0

from which the solution

v = C1u+ C2u2 exp

2u

+ C3

[u2 exp

2u

∫exp (− 4

u )duu(u− 2)2

− u∫

exp (− 2u )du

(u− 2)2]

is obtained. Substituting u = 1x and v = x

y yields again (7.79).



4,19 Janet basis, therefore case b) applies. If σ = y is chosen, thesystem (6.74) yields the independent solution ρ = x

y and the canonical form

(6.80). Choosing σ = 1y instead yields ρ = 1

xy with the canonical form (6.81).


4,17 Janet basis, therefore case c) applies. The first order equation forσ is y′+ 1 = 0, i.e., φ(x, y) = x+ y. Choosing σ = x+ y yields a system for ρwith the fundamental system 0, x, y. Consequently, any linear combinationρ = c1x + c2y with c1 6= c2 yields a possible transformation to a canonicalform which turns out to be (6.80).

Five-Parameter Symmetries. The canonical form for the symmetry classesS3

5,1 and S35,2(a) is v′′′−(a+1)v′′+av′ = 0 with a some number. This canonical

form is particularly convenient because it is completely reducible for any valueof a without introducing new algebraic numbers. If a = 0 it decomposes as

v′′′ − v′′ = Lclm(v′, v′ − 1

uv, v′ − v

)= 0 (7.80)

with the general solution v = C1 + C2u+ C3eu, and if a 6= 0

v′′′ − (a+ 1)v′′ + av′ = Lclm(v′, v′ − v, v′ − av) = 0 (7.81)

with the general solution v = C1 +C2eu +C3e

au. Substituting ρ and σ for vand u immediately yields the desired solution of the actual equation. Thesesteps are organized in terms of the subsequent algorithm for both S3

5,1 andS3

5,2(c) symmetries.

Algorithm 7.30 LieSolve3.5(ω). Given a quasilinear third order odeω with any five-parameter group of Lie symmetries, its general solution isreturned; or failed if it cannot be determined.S1 : Identify case a), b) or c). Determine the Janet basis for the Liesymmetries of ω. By means of the criteria given in Theorem 6.11, identifythe case that applies for the equation at issue.S2 : Determine value of parameter a. Solve equation (5.42) and choosesome value of a from it.S3 : Equations for σ and ρ. From the Janet basis constructed in step S1and the value for a obtained in step S2 construct the systems for σ andρ according to (6.82) and (6.83), (6.84) and (6.32) or (6.85) and (6.86).S4 : Determine σ and ρ. From the equations constructed in step S3determine σ by integration. Find a hyperexponential solution for ρ, ifnone exists return failed.S5 : Return solution. Substitute u = σ(x, y) and v = ρ(x, y) into thecanonical form solution (7.80) or (7.81) and return the result.

348

Example 7.36 The transformation functions σ = xy and ρ = yx obtained

in Example 6.35 yield the general solution

(C3x2 + 1)y + C2e

xyx+ C1x = 0

for the differential equation considered in Example 5.46.

Example 7.37 The transformation functions σ = y and ρ = xy obtained

in Example 6.36 yield the general solution(C3e

3x + C2ey + C1

)y + x = 0

for the differential equation considered in Example 5.47.

Seven-Parameter Symmetry. In canonical variables u and v ≡ v(u), aquasilinear third order ode in symmetry class S3

7 has the form v′′′ = 0 with theobvious solution v = C1+C2u+C3u

2. Therefore, the whole solution procedurecomes down to generating the canonical form of the given equation, very muchlike for a second order equation with eight-parameter projective symmetrygroup. For third order equations the two alternatives of Theorems 5.18 and6.33 have to be distinguished. The various steps are put together in thesubsequent algorithm.

Algorithm 7.31 LieSolve3.7(ω). Given a quasilinear third order odeω with a seven-parameter group of Lie symmetries, its general solution isreturned; or failed if it cannot be determined.S1 : Identify case a) or case b). By means of the criteria given in Theo-rem 5.18 determine which case applies for equation ω.S2 : Equations for σ and ρ. Determine a solution for S from equations(6.87) or (6.87). If none is found return failed; otherwise construct thesystem for σ from it.S3 : Determine σ and ρ. From the equations constructed in step S2determine Liouvillian transformation functions σ and ρ.S4 : Return solution. Substitute u = σ(x, y) and v = ρ(x, y) into thecanonical form v = C1 + C2u+ C2u

2 and return the result.

Example 7.38 For equation 7.8 of Kamke’s collection in Example 6.37the transformation functions σ = x and ρ = 2√

y− 2 have been obtained.

Substituting them into the above solution for v′′′ = 0 yields the solution4 2√

y− 2 = C1 + C2x+ C3x

2 or equivalently

y =1

(C1 + C2x+ C3)2

after a redefinition of the integration constants.


(5.44) in Example 5.49 the transformation functions σ = xy and ρ = yx have


been obtained. Substituting these values into the above solution for v′′′ = 0leads to the general solution x3y2 + (C1x

2 + C2)y + C3)x = 0.


(5.44) in Example 5.50 the transformation functions σ = 1y and ρ = y

xex have

been obtained. Substituting these values into the above solution for v′′′ = 0leads to the general solution y3 + (C1y

2 + C2y + C3)xe−x = 0.

Exercises

Exercise 7.1 Apply Lemma 7.1 for solving the equation in Example 6.2.

Exercise 7.2 Solve the equation of Example 7.2 by transforming it intothe form v′ = C.

Exercise 7.3 Solve the equation of Example 7.2 under the assumptionthat only the second symmetry generator U2 = y∂x + x∂y is known.

Exercise 7.4 Derive the expression (7.16) for the solution of (7.13).

Exercise 7.5 Generate the solution of the equation considered in Exam-ple 7.19 from the first order equation (7.36).

Exercise 7.6 Determine the solution of equation Kamke 6.227 from thecanonical form obtained in Example 6.8.

Exercise 7.7 Solve the equation y′′ + r(y) = 0 by symmetry methods.

Exercise 7.8 Solve the equation y′′y+r(x)y′2 = 0 by symmetry methods.

Exercise 7.9 Determine the solution of (7.24) from the equivalent pdeand compare the result with (7.25).

Exercise 7.10 Derive a first order Riccati equation from (7.71) and de-termine its solution.

Exercise 7.11 Derive solution (7.72) from the equivalent pde of (7.71).

Chapter 8

Concluding Remarks

After discussing the various integration methods for ode’s developed by Lieit appears appropriate to conclude with a few words of comparison, assessits advantages and limitations, and also put them into proper perspective.Furthermore, several directions of future research are outlined.Comparison of Lie’s Methods. There are basically two alternative meth-ods for solving differential equations using its symmetries. Either an equationwith a certain nontrivial symmetry is transformed into a canonical form thesolution of which may be determined. Or the infinitesimal symmetry genera-tors are determined explicitly, and are used to determine a set of first integralsfrom which the solution may be constructed. The relation between these twoalternatives is most easily understood from the following drawing. The ordern of the given quasilinear ode is two or three.

Equation in actual variablesy(n) + r(x, y, y, . . . , y(n−1)) = 0

?h1

Generate Janet basis fordetermining system,

determine symmetry type

h2 HHHHHHj

h5Equation in canonical variablesv(n) + s(u, v, v, . . . , v(n−1)) = 0

Generators U1, . . . , UrUi = ξi(x, y)∂x + ηi(x, y)∂y

?

h3?

h6

h6a

Solution in canonical variablesh4Integrals from equivalent pdeh7HH

HHHHj

Solution in actual variables

Step 1 takes the given ode in actual variables as input, constructs the de-termining system from it and transforms it into a Janet basis. To this end

351

352

only differentiations and arithmetic in the base field are required. From theJanet basis coefficients the symmetry type may be obtained as described inTheorems 5.9 and 5.14. This step is guaranteed and it is always performedfirst.

After that there are two main alternatives. The left branch which is em-phasized in this book begins with step 2. At first a system of equations for thetransformation functions to canonical form is set up. If an independent pairof solutions for the transformation functions may be obtained, it is appliedfor generating the canonical form equation for the respective symmetry class.In step 3 its solution, or at least a simplified equation, is determined. In thefinal step 4 this result is transformed back to the actual variables x and y.

Alternatively the vector fields generating the symmetries may be deter-mined explicitly in step 5. They are applied in step 6 to determine a set offirst integrals for the pde which is equivalent to the given ode. In step 7 byelimination the desired solution or some lower order equation may be con-structed from the first integrals.

These two main branches are not completely disconnected. One may leavethe right branch with step 6a and then proceed along the left branch. Thereare several reasons why the left branch in the above drawing is preferred.

i) The equations which determine the transformations to canonical formare usually simpler than the equations determining the symmetry gen-erators. For all possible types of equations occurring in the symmetryanalysis of second and third order ode’s algorithms have been describedfor decomposing them into components of lowest order.

ii) If the symmetry generators are determined explicitly, they have to betransformed such that they obey the canonical commutators of the re-spective Lie algebra.

iii) The first integrals for the equivalent pde often lead to unnecessary com-plicated expressions for the solution of the given ode which are difficultto simplify.

There is an additional, more fundamental reason. In general, solving adifferential equation may be considered as an equivalence problem consist-ing of two parts. At first membership of the given equation to a particularequivalence class has to be decided. Secondly, within this equivalence class thetransformation of the given equation to a canonical form has to be determinedalgorithmically. This scheme is valid independent of whether an equation hassymmetries or not. In the former case, the symmetry class of the equationnarrows down the possible equivalence classes, and the canonical form for itis already known. These aspects are clearly discussed by Tresse [181] as it isexplained in Section 4.3 on page 188.

If there are no symmetries, candidates for equivalence classes are usuallyspecified in terms of differential equations defining certain special functions

Concluding Remarks 353

like e. g. Painleve transcendents, Bessel functions, etc. As soon as membershipto any of them is established, the transformation to the canonical equationhas to be found. These two steps may be combined into a single one. Thisapproach is dicussed in the Dissertation of Berth [10].Problems for Further Research. It would be highly desirable to admitmore general base fields, e. g. elementary extensions by one or more expo-nentials, angular functions or logarithms, or algebraic functions. Because thebase field occurs as coefficient field for the Janet basis of the determining sys-tem, and also for the equations determining the transformation to canonicalform, the decompositions of these Janet bases over more general coefficientfields must be known first if an algorithmic theory without heuristics is thegoal.

For various applications it might be of interest to have a similar schemeavailable for equations of order four comprising more than one hundred sym-metry classes.

Lie himself announced a book on point and contact symmetries of partialdifferential equations. Due to his early decease he has not been able to realizethis project; and so another field of possible research is left. Due to the largenumber of possible symmetry types that have to be considered, a completeclassification is rather voluminous and relies even more on appropriate com-puter algebra tools. Similar arguments apply to the symmetry analysis ofsystems of ode’s. There have been various attempts to extend Lie’s symmetryconcept in order to make more equations amenable to a solution procedure. Tothis end, Barbara Abraham-Shrauner et al. [1] applied nonlocal symmetries.Muriel et al. [134, 49] define the new concept of a potential symmetry.

A completely different approach has been pursued by Zaitsev and Polyanin[191]. As opposed to Lie, they apply discrete groups for solving ordinarydifferential equations. It would be highly interesting to compare their resultswith those obtained by Lie analysis.

Another comparison concerns the relation between Galois’ theory and Lie’ssymmetry analysis. For second order linear equations there is certainly nochance for any correlation because any such equation has the full eight-parameter projective group as its symmetry group. The situation is different,however, for equations of order three as may be seen from the listings inAppendix D and E. For example, all equations in symmetry class S3

7 havea Loewy decomposition of type L3

9. All equations in symmetry class S35,2(c)

have Loewy decompositions of type L39 or type L3

12. This suggests that forlinear equations of order higher than two there may be some correlations be-tween its symmetry type and the type of its Loewy decomposition, which inturn entails a special Galois group.

Appendix A

Solutions to Selected Problems

In this Appendix the solutions to most exercises are given, although in somecases only a few hints are provided. The reader is strongly advised to try firston his own to find the answers.

2. Linear Differential Equations

2.1 The constraints for the coefficients of the given system of algebraicequations are

a1 = 2b4a3b3 − b2a3 + b33 − b23a4 + b3a2 + c3a3,

b1 = b2b3 − a3b24 − c4a3 − b4b23,

c1 = a2c4 + a2b24 − b22 + b2c3 − b3c2 + a3b

34 + b4c4a3 + b24b

23,

c2 = a4c4 − 2b2b4 + a4b24 + b4c3 + b24b3 − c4b3.

2.2 If y1 = z21 is differentiated three times, and z′′1 + 1

4pz1 = 0 is appliedfor reduction whenever possible, one obtains y′1 = 2z1z′1, y

′′1 = − 1

2pz21 + 2z′21

and y′′′1 = − 12p′z2

1 −2pz1z′1. Substitution into the given third order ode showsthat y1 is a solution. Similarly differentiating and substituting y2 = z1z2,y′2 = z′1z2 + z1z

′2, y

′′2 = − 1

2pz1z2 + 2z′1z′2 and y′′′2 = − 1

2p′z2

1 − 2pz1z′1 showsthat y2 is a solution as well. The same is true for y3.

2.3 Let y1, y2 and y3 be a fundamental system for the given equation andW its Wronskian. The general solution may be written as

y = C1y1 + C2y2 + C3y3 +∫

r

W(y2y′3 − y′2y3)dx

−∫

r

W(y1y′3 − y′1y3)dx+

∫r

W(y1y′2 − y′1y2)dx

where C1, C2 and C3 are constants. For y′′′ = r with the fundamental systemy1 = 1, y2 = x and y3 = x2 there follows

y = C1 + C2x+ C3x2 + 1

2

∫x2rdx− x

∫xrdx+ 1

2x2∫rdx.

2.4 Reducing y′′ + Py′ +Qy = 0 w.r.t. y′ + py = 0 and y′ + qy = 0 yieldsQ − pP = p′ − p2 and Q − qP = q′ − q2. Solving for P and Q yields thegiven expressions. Substituting the fundamental system y1 = exp (−

∫pdx),

355

356

y2 = exp (−∫qdx) and the Wronskian W = (q − p) exp (−

∫(p+ q)dx) into

(2.9) yields

y = exp (−∫pdx)

( ∫exp (

∫pdx) Rdxq − p + C1

)+exp (−

∫qdx)

( ∫exp (

∫qdx) Rdxq − p + C2

).

This representation generalizes the solution of a first order equation in anobvious way.

2.5 For L22, the first solution y1 is obtained from (D+ a1)y = 0, the second

from (D + a1)y = y2 with y2 a solution of (D + a2)y = 0.For L2

3 the two solutions are obtained from (D + ai)y = 0. Linear de-pendence over the base field would imply a relation q1y1 + q2y2 = 0 withq1, q2 from the base field. Substituting the solutions this would entail q1q2 =− exp

∫(a2 − a1)dx. Due to the nonequivalence of a1 and a2, its difference

is not a logarithmic derivative. Consequently the right hand side cannot berational.

For L24 the equation (D+a(C))y = 0 has to be solved; then C is specialized

to C1 and C2. Substituting ai = r′

r + Ci+ p in the above quotient, the

integration may be performed with the result q1q2 = −r + C2r + C1

which is rational.2.6 Let y′′′ + ay′′ + by′ + cy = 0 with fundamental system y1, y2, y3 be the

third order linear equation corresponding to the second order Riccati equation

z′′ + 3zz′ + z3 + a(z′ + z2) + bz + c = 0. If y′iyi is rational for a single value

of i, or for two or three values of i, but for any pair i, j with 1 ≤ i < j ≤ 3,the ratios

Ciy′i + Cjy

′j

Ciyi + Cjyj(A.1)

are not, the corresponding number of rational solutions of the Riccati equationexists. This is case i) for the second order equation. If there is a pair ofindices such that the ratio (2.17) is rational, case ii) or iii) respectively appliesdepending on whether the logarithmic derivative of the remaining element ofthe fundamental system is rational or not. Finally, if the ratio

C1y′1 + C2y

′2 + C3y

′3

C1y1 + C2y2 + C3y3

is rational, it may be rewritten as

Cy′1/y1 + C2y′2/y1 + y′3/y1

C1 + C2y2/y1 + y3/y1=

C2(y2/y1)′ + (y3/y1)′

C1 + C2(y2/y1) + y3/y1+y′1y1

(A.2)

with the ratios C1C3

and C2C3

appropriately redefined and p = y′1y1 , q = y′2

y1 ,

r = y′3y1 , u = y2

y1 and v = y3y1 . This is case iv).

Solutions to Selected Problems 357

2.7 If the Galois group is Z1, by Theorem 2.7 the equation has a rationalfundamental system y1, y2; i.e., it is linearly dependent over the base field.By the result of Exercise 2.4 the Loewy decomposition is of type L2

4. If thegroup is Z2, its second symmetric power has a rational fundamental system

y21 , y1y2 and y2

2 . Consequently, the ratio y21

y1y2 = y1y2 is rational and the

decomposition L24 follows as above.

2.8 Substituting P (u) ≡ u2 + B1u + b0 = 0 into u′ + u2 + r = 0 leads tob0 = r − 1

2 (b′1 − b21) and b1b′1 − 2b′1 + 12b1(b

21 − 4b0). If the expression for b0 is

substituted into this latter relation, the constraint

b′′1 − 3b1b′1 + b31 + 4rb1 − 2r′ = 0

follows. With b1 = −z′z the third order equation z′′′+ 4rz′+ 2r′z = 0 for z is

obtained which is the the second symmetric power of y′′ + ry = 0.2.9 Substituting z1 and z2 into the equations of the given Janet basis leads

to a linear system for the coefficients a1, a2, b1 and b2. Defining

w0 =∣∣∣∣ z1,x z2,xz1 z2

∣∣∣∣ , w1 =∣∣∣∣ z1,y z2,yz1 z2

∣∣∣∣ , w2 =∣∣∣∣ z1,y z2,yz1,x z2,x

∣∣∣∣ ,w3 =

∣∣∣∣ z1,xx z2,xxz1 z2

∣∣∣∣ , w4 =∣∣∣∣ z1,yy z2,yyz1 z2

∣∣∣∣ ,w5 =

∣∣∣∣ z1,xx z2,xxz1,x z2,x

∣∣∣∣ , w6 =∣∣∣∣ z1,yy z2,yyz1,y z2,y

∣∣∣∣they may be expressed as

a1 = −w1

w0, a2 =

w2

w0, b1 = −w3

w0, b2 =

w5

w0.

2.10 With the same notation as in the previous exercise the representationsa1 = −z1,xz1 = −z2,xz2 , b1 = −w4

w1and b2 = w6

w1are obtained.

2.11 By direct reduction the following Riccati-like system of pde’s is ob-tained.

a1,x +A1a21 + (B1 −A2)a1 + a2 −B2 = 0,

a1,y +B1a21 − a1a2 + (C1 −B2)a1 − C2 = 0,

a2,x +A1a1a2 +B1a2 −A3a1 −B3 = 0,

a2,y − a22 +B1a1a2 + C1a2 −B3a1 − C3 = 0,

b1 +A1a1 −A2 = 0, b2 +A1a2 −A3 = 0.

(A.3)

358

2.12 Substituting (2.79) into system (2.81) and reordering yields after afew autoreduction steps the following result.

w0,xx −A1w0,y−(A1,x

A1−A2 − 2B1

)w0,x

−(A1,x

A1A2 −A2,x + 2A1B2 − 2A2B1

)w0 = 0,

w0,xy +A2w0,y − 1A1

(B1,x −A1B2 +A2B1 −A3 −B21)w0,x

+[B2,x +A2B2 −B1B2 +B3 − A2

A1(B1,x −B2

1 +A22B1 −A2A3)

]w0 = 0,

w0,yy + (2B2 + C1)w0,y − 1A1

(C1,x −A1C2 + 2B1B2 − 2B3)w0,x

+[C2,x + 2B2

2 + 2C3 − A2A1

(C1,x + 2B1B2 − 2B3)]w0 = 0.

2.13 Each term in (2.19) is uniquely determined by an index sequencek0, k1, . . . , kν−1. If the lexicographic ordering with respect to this sequence isassumed, the leading terms corresponding to these sequences are given in thesubsequent table. Furthermore, the sum of the indices is given.

No. k0 k1 k2 k3 . . . kν−2 kν−1

ν−1∑j=0

kj(M + j)

1 ν 0 0 0 . . . 0 0 νM

2 ν − 2 1 0 0 . . . 0 0 (ν − 1)M + 1

3 ν − 3 0 1 0 . . . 0 0 (ν − 2)M + 2

4 ν − 4 2 0 0 . . . 0 0 (ν − 2)M + 2

5 ν − 4 0 0 1 . . . 0 0 (ν − 3)M + 3

...... . . .

......

last 0 0 0 0 . . . 0 1 M + ν − 1

The rightmost column contains the order of a pole of order M for the termdetermined by the values in the preceding columns. Because ν ≥ 1, the valueνM in the first line is maximal if M > 1. For M = 1 they are all equal to ν.

2.14 The first member of a fundamental system is z1 = exp (−∮adx+ bdy).

Dividing out w1 ≡ zx + az and w2 ≡ zy + bz yields the system

w2 +A1w1 = 0, w1,x+(B1−a)w1 = 0, w1,y+(A1,x+A1 +A2−A1B1)w1 = 0

with the solution (see Lemma 2.9)

w1 = exp(−

∮(B1 − a)dx+ (A1,x +A1 +A2 −A1B1)dy

), w2 = −A1w1.


The corresponding inhomogeneous system zx + az = w1, zy + bz = −A1w1

finally yields the second member of a fundamental system (see Lemma 2.10)

z2 = z1

∮w1z1 (dx−A1dy).

2.15 The first member of a fundamental system is z1,1 = exp (−∮bdx+ cdy),

z2,1 = −az1,1. Dividing out the Loewy factor w1 ≡ z2 + az1, w2 ≡ z1,x + bz1and w3 ≡ z1,y + cz1 yields the system

w1,x + (aA1 + C1)w1 = 0, w1,y + (aB1 +D1)w1 = 0,w2 +A1w1 = 0, w3 +B1w1 = 0

with the solution (see Lemma 2.9)

w1 = exp(−

∮(aA1 + C1)dx+ (aB1 +D1)dy

), w2 = −A1w1, w3 = −B1w1.

The corresponding inhomogeneous system

z2 + az1 = w1, z1,x + bz1 = −A1w1, z1,y + cz1 = −B1w1

finally yields the second member of a fundamental system (see Lemma 2.10)

z1,2 = −z1,1∮

w1

z1,1(A1dx+B1dy), z2,2 = −az1,2.

2.16 Reduction w.r.t. the given two first order systems yields a linear alge-braic system for the coeffients of the desired Lclm. If a1 6= a2 , the result isof type J (1,2)

2,1 ,

zy − b1 − b2a1 − a2

zx + a1b2 − a2b1a1 − a2

z = 0,

zxx + (a1 + a2 −a1,x − a2,xa1 − a2

)zx + (a1a2 + a1a2,x − a2a1,xa1 − a2

)z = 0.

If a1 = a2 and b1 6= b2, the result is of type J (1,2)2,1 ,

zx+ a1z = 0, zyy +(b1 + b2−

b1,y − b2,yb1 − b2

)zy +

(b1b2−

b1b2,y − b1,yb2b1 − b2

)z = 0.

2.17 Introducing a differential indeterminate z, the system of linear pde’s

zy −x

yzx = 0, zxx −

xy − 1x

zx −y

xz = 0 (A.4)

is obtained. The right factor leads to zx − yz = 0, zy − xz = 0 with thesolution exy. It is the first basis element for the solution space of (A.4).Introducing coordinates z1 and z2 in D2, the exact quotient generates thepde’s z2 − x

y z1 = 0, z1,x + 1xz1 = 0 and z1,y = 0 with the solution z1 = 1

x ,

360

z2 = 1y . In order to obtain the second basis element for the solution space

of (A.4), the inhomogeneous system zx − yz = 1x , zy − xz = 1

y has to besolved with the result exyEi(−xy). The exponential integral is defined by

Ei(−x) = −∫ ∞

x

e−tdt

t.

2.18 A special solution of the inhomogeneous system is searched for in theform z = C1z1 + C2z2 where both C1 ≡ C1(x, y) and C2 ≡ C2(x, y) areundetermined functions of x and y. They will be obtained by an extension ofLagrange’s variation of constants. The last equation yields

C1,x = −wz2W

, C2,x =wz1W

where W =∣∣∣∣ z1, z2z1,x, z2,x

∣∣∣∣ .Substituting the expression for z into the first equation and the equation

zxy + (a1,x + a2 − a1b1)zx + (a2,x − a1b2)z = vx − a1w

that is obtained from it by derivation w.r.t. y yields

C1,yz1 + C2,yz2 = v, C1,yz1,x + C2,yz2,x = vx − a1w.

By elimination the expressions

C1,y =vz2,x − (vx − a1w)z2

W, C2,y = −vz1,x − (vx − a1w)z1

W

for the y-derivatives are obtained. Combining these results, both coefficientsC1 and C2, and consequently a special solution of the inhomogeneous system,may be obtained as a path integral.

2.19 A special solution of the inhomogeneous system is searched for in theform z = C1z1 + C2z2 with

C1,x =vz2,y − vyz2

W, C2,x =

vyz1 − vz1,yW

where W =∣∣∣∣ z1, z2z1,y, z2,y

∣∣∣∣ ,C1,y = −wz2

W, C2,y =

wz1W

.

It is most easily obtained from the preceding case by exchange of x and y andsubstituting a = 0.

2.20 A special solution of the inhomogeneous system is searched for in theform z = C1z1 +C2z2 +C3z3, Ck ≡ Ck(x, y) for k = 1, 2, 3 are undeterminedfunctions of x and y. Lagrange’s variation of constants leads to

C1,x =w

W

∣∣∣∣ z2, z3z2,x, z3,x

∣∣∣∣ , C1,y =1W

∣∣∣∣∣∣v, z2, z3vx, z2,x, z3,x

vxx − aw, z2,xx, z3,xx

∣∣∣∣∣∣ ,


C2,x = − w

W

∣∣∣∣ z1, z3z1,x, z3,x

∣∣∣∣ , C2,y =1W

∣∣∣∣∣∣z1, v, z3z1,x, vx, z3,xz1,xx, vxx − aw, z3,xx

∣∣∣∣∣∣ ,

C3,x =w

W

∣∣∣∣ z1, z2z1,x, z2,x

∣∣∣∣ , C3,y =1W

∣∣∣∣∣∣z1, z2, vz1,x, z2,x, vxz1,xx, z2,xx, vxx − aw

∣∣∣∣∣∣where

W =

∣∣∣∣∣∣z1, z2, z3z1,x, z2,x, z3,xz1,xx, z2,xx, z3,xx

∣∣∣∣∣∣ .From these expressions the Ck(x, y) may be expressed in terms of path inte-grals.

2.21 With the substitutions xk ≡ φk(y1, . . . , yn), yk ≡ ψ(x1, . . . , xn) andg(y1, . . . , yn) ≡ f(φ1, . . . , φn), the partial derivatives of f become

∂f

∂xi=

∂g

∂y1

∂y1∂xi

+ . . .+∂g

∂yn

∂yn∂xi

.

Substituting them into (2.45), rearranging terms and defining

Bk = a1∂ψk∂x1

+ . . .+ an∂ψk∂xn

for k = 1, . . . , n yields (2.46).2.22 With the notation of Theorem 2.20, the equation for w(x, y, z) is

wy + awx + rwz = 0. From dy = dxa there follows dy

dx= 1

a and the firstintegral ψ1(x, y). Introducing a new variable x = ψ1(x, y) with the inversex = φ(x, y), the equation for w becomes wy+(b(φ, y)z+c(φ, y))wz = 0. Fromdy = dz

bz + cthere follows dz

dy− b(φ, y)z = c(φ, y). It yields a second first

integral

ψ2(x, y, z) =[z exp

(−

∫b(φ, y)dy

)−

∫c(φ, y) exp

(−

∫b(φ, y)dy

)dy

]∣∣∣x=ψ1

.

The general integral of the originally given equation may be written asΦ(ψ1, ψ2) = C with a constant C. The particular choice of the functionΦ determines the actual form of the answer. As an example, consider the

equation zy + 2x3y zx = 1

3y z with the first integrals ψ1 = y2

x3 and ψ2 = zy1/3 .

Choosing Φ = ψ2

ψ1/31

yields z = Cyx .

2.24 Taking the commutator with A of both sides of the assumed relations

362

yields(A,Bk) = (A, βkm+1Bm+1) + . . .+ (A, βkqBq) + (A,αA)

= Aβkm+1 ·Bm+1 + . . .+Aβkq ·Bq+βkm+1(A,Bm+1) + . . .+ βkq (A,Bq) +Aα ·A.

By assumption the commutators involving the Bk in the last line are propor-tional to A; i.e., the complete expression is a relation of the same type givenin the Theorem which is a contradiction. Consequently, Aβkj = 0 for all kand j.

2.25 The second equation has the general solution

z(x, y, C(x)) =yx

C(x) + y− xy

x2y − 2.

Substitution into (2.55) yields Cxx + C = 0; i.e., C = Cx with C a constant

from which the general rational solution z = yxy + C −

xyx2y − 2

follows.

2.26 Substituting z = z1 + 1w into (2.50) yields the linear system

wx − (2A1z1 +A2)w = A1, wy − (2B1z1 +B2)w = B1

for w in complete analogy to a first order ordinary Riccati equation.2.27 If in the system zx+A1z

2 +A2z+A3 = 0, zy +B2z+B3 = 0 the newfunction w is introduced by z = w

A1, the following system for w is obtained.

wx +w2+(A2 −

A1,x

A1

)w+A1A3 = 0, wy+

(B2 −

A1,y

A1

)w+A1B3 = 0.

The coefficient of w in the latter equation vanishes identically due to the firstconstraint in (2.51). The second constraint of (2.51) yields for the last term

A1B3 =12

(A2 −

A1,x

A1

)y.

The y-integration may be performed with the result

w =12

( A1,x

A1−A2

)+C

where C ≡ C(x) is a function of x. Substituting this expression for w into thefirst equation yields the Riccati equation (2.57) for the x-dependence of C.If the terms not containing C are derived w.r.t. y and the constraints (2.51)are applied for eliminating all y-derivatives, zero is obtained. If the solutionfor C is substituted into the above expression for w and z = w

A1is used, the

representation (2.56) for the solution follows.


2.28 Dividing out the exact component zx ≡ z1, zy ≡ z2 yieldsz2 + az1, z1,x + cz1, z1,y + (ax − ac)z1. The corresponding linear systemhas the general solution

z1 = exp (−∫cdx+ (ax − ac)dy), z2 = −az1.

It leads to the system zx = z1, zy = −az1 with the solution

z =∫z1dx− az1dy =

∫exp (−

∫cdx+ (ax − ac)dy)(dx− ady)

which is the non-constant member of a fundamental system.

3. Lie Transformation Groups

3.1 The differential equations

∂x

∂a= 1,

∂x

∂b= 0,

∂y

∂a= 0,

∂y

∂b= 1

have the required form; however, there does not exist an identity element.3.2 Let ak correspond to the one-parameter group Uk. Then the finite

transformations are

x =(x+ a1)ea2

1− (x+ a1)ea2a4, y =

ea3y

1− (x+ a1)ea2a4

3.3 The finite transformations of g2 are x = a1x + a2y + a3, y = a4x +a5y + a6. Accordingly, three arbitrarily chosen points (x1, y1), (x2, y2) and(x3, y3) are transformed into xi = a1xi + a2yi + a3 and yi = a4xi + a5yi + a6

for i = 1, 2, 3. These relations may be solved for the group parameters asfollows.

a1 =1D

∣∣∣∣∣∣x1 y1 1x2 y2 1x3 y3 1

∣∣∣∣∣∣ , a2 =1D

∣∣∣∣∣∣x1 x1 1x2 x2 1x3 x3 1

∣∣∣∣∣∣ , a3 =1D

∣∣∣∣∣∣x1 y1 1x2 y2 1x1 x2 x3

∣∣∣∣∣∣ ,

a4 =1D

∣∣∣∣∣∣y1 y1 1y2 y2 1y3 y3 1

∣∣∣∣∣∣ , a5 =1D

∣∣∣∣∣∣x1 y1 1x2 y2 1x3 y3 1

∣∣∣∣∣∣ , a6 =1D

∣∣∣∣∣∣x1 y1 1x2 y2 1y1 y2 y3

∣∣∣∣∣∣ ,D is defined by

∣∣∣∣∣∣x1 y1 1x2 y2 1x3 y3 1

∣∣∣∣∣∣.3.4 Introducing new variables u = F2(x)

F1(x)≡ σ(x) and v = y

F1(x)≡ ρ(x, y),

Lemma 3.3 yields vanishing coefficients for all ∂u and Gk(u) = F2(x)F1(x)

with x

expressed by u for k = 1, . . . , r. Obviously G1 = 1 and G2(u) = u as required.

364

3.5 Applying the same notation as in Example 3.1, the following system ofequations for the undetermined coefficients is obtained.

α12α23 − α13α22 = α11, α11α23 − α13α21 = −α12,

α12α33 − α13α32 = 2α21 = 0, α11α33 − α13α31 = −2α22 = 0,α22α33 − α23α32 = α31 = 0, α21α33 − α23α31 = −α32 = 0

and α13 = α23 = α33 = 0. Obviously, it allows only the trivial solutionαij = 0 for all i and j.

3.6 For r = 2 there holds ∆(ω) = ω2 −(c212E1 − c112E2

)ω, for r = 3

∆(ω) = ω3 −[(c212 + c313)E1 − (c112 − c323)E2 − (c113 + c223)E3

]ω2

+[(c212c

313 − c213c312)E2

1 − (c112c323 − c123c312)E2

2 + (c113c223 − c123c213)E2

3

−(c112c313 − c113c312 − c212c323 + c223c

312)E1E2

+(c112c213 − c113c212 + c213c

323 − c223c313)E1E3

+(c112c223 − c113c323 − c123c212 + c123c

313)E2E3

]ω.

3.7 For l = 1, k = 1, ρ1 ≥ 1. For any value of ρ1, there are two groups ofsize ρ1 + 2 corresponding to α1 = 0 or α1 = 1, respectively. Their generatorsand non-vanishing commutators are

∂y, x∂y, . . . , xρ1∂y, ∂x with [∂x, xk∂y] = kxk−1∂y and

ex∂y, xex∂y, . . . , x

ρ1ex∂y, ∂x with [ex∂x, xk∂y] = kxk−1ex∂y + xkex∂y.

For l = 2, 1 ≤ k ≤ 2, the constraint ρ1 + ρ2 ≥ 0 becomes redundant. Forany pair of values ρ1 and ρ2, there are two groups of size ρ1 + ρ2 + 3 corre-sponding to α1 = 0 or α1 = 1 respectively. Their generators and nonvanishingcommutators are

∂y, x∂y, . . . , xρ1∂y, e

α2x∂y, xeα2x∂y, . . . , x

ρ2eα2x∂y, ∂x with

[∂x, xk∂y] = kxk−1∂y, [∂x, xkeα2x∂y] = kxk−1eα2x∂y + α2xkeα2x∂y

and

ex∂y, xex∂y, . . . , x

ρ1ex∂y, eα2x∂y, xe

α2x∂y, . . . , xρ2eα2x∂y, ∂x with

[∂x, xkex∂y] = kxk−1ex∂y + xkex∂y,

[∂x, xkeα2x∂y] = kxk−1eα2x∂y + α2xkeα2x∂y.

3.8 In general, the answer is not unique. For example, the most generalvalid choice for group g17(l = 1, ρ1 = 1, α1 = 0) is U1 = (β1γ3 − γ1β3)∂y,U2 = β1∂x + β2∂y + β3x∂y, U3 = γ1∂x + γ2∂y + γ3x∂y where β1, . . . , γ3 are


arbitrary constants. Choosing β1 = γ3 = 1 and 0 for the remaining coefficientsleads to special case given below.

g5: U1 = ∂y, U2 = ∂x, U3 = y∂y.g7: U1 = ∂x, U2 = ∂y, U3 = x∂x + cy∂y.g8: U1 = ∂y, U2 = y∂y, U3 = y2∂y.g10: U1 = ∂x + ∂y, U2 = x∂x + y∂y, U3 = x2∂x + y2∂y.g13: U1 = ∂x, U2 = x∂x + 1

2y∂y, U3 = x2∂x + xy∂y.g15(r = 3): U1 = φ1∂y, U2 = φ2∂y, U3 = φ3∂y.g16(r = 2): U1 = φ1∂y, U2 = φ2∂y, U3 = y∂y.g17(l = 1, ρ1 = 1, α1 = 0): U1 = ∂y, U2 = ∂x, U3 = x∂y.g17(l = 1, ρ1 = 1, α1 = 1): U1 = ex∂y, U2 = xex∂y, U3 = −∂x.g17(l = 2, ρ1 = ρ2 = 0, α1 = 0, α2 = 1): U1 = ex∂y, U2 = ∂y, U3 = −∂x.g17(l = 2, ρ1 = ρ2 = 0, α1 = 1, α2 = c): U1 = ex∂y, U2 = ecx∂y, U3 = −∂x.g20(r = 1): U1 = ∂x − ∂y, U2 = ∂x + ∂y, U3 = x∂x + (x+ y)∂y.g24: U1 = ∂x, U2 = ∂y, U3 = x∂x + y∂y.

3.9 The same remark applies as in the preceding exercise.g6: U1 = −x∂x, U2 = ∂x, U3 = −y∂y, U4 = ∂y.g9: U1 = ∂y, U2 = y∂y, U3 = y2∂y, U4 = ∂x.g14: U1 = ∂x, U2 = x∂x, U3 = x2∂x + xy∂y, U4 = y∂y.g15(r = 2): Uk = φk∂y for k = 1, . . . , 4.g16(r = 1): U1 = ∂y, U2 = x∂y, U3 = φ(x)∂y, U4 = y∂y.g17(l = 1, ρ1 = 2, α1 = 0): U1 = −2x∂y, U2 = 2∂y, U3 = x2∂y, U4 = ∂x.g17(l = 1, ρ1 = 2, α1 = 1):U1 = ex∂y, U2 = xex∂y, U3 = 1

2x2ex∂y, U4 = −∂x.

g17(l = 2, ρ1 = 0, ρ2 = 1, α1 = 0, α2 = 1):U1 = ∂x, U2 = ex∂y, U3 = xex∂y, U4 = ∂y.g17(l = 2, ρ1 = 1, ρ2 = 0, α1 = 0, α2 = 1):U1 = ∂x, U2 = x∂y, U3 = ∂y, U4 = ex∂y.g17(l = 2, ρ1 = 1, ρ2 = 0, α1 = 1, α2 = α):U1 = ∂x, U2 = ex∂y, U3 = xex∂y, U4 = eαx∂y.g17(l = 3, ρ1 = ρ2 = ρ3 = 0, α1 = 0, α2 = 1, α3 = c):U1 = ∂x, U2 = ex∂y, U3 = ecx∂y, U4 = ∂y.g17(l = 3, ρ1 = ρ2 = ρ3 = 0, α1 = 1, α2 = α, α3 = c):U1 = ∂x, U2 = ∂y, U3 = eαx∂y, U4 = ecx∂y.g18(l = 1, ρ1 = 1, α1 = 0): U1 = −∂y, U2 = x∂y, U3 = ∂x, U4 = y∂y.g18(l = 2, ρ1 = ρ2 = 0): U1 = ∂x, U2 = ex∂y, U3 = −y∂y, U4 = ∂y.g19(r = 2): U1 = ∂y, U2 = ∂x, U3 = x∂y, U4 = x∂x + cy∂y.g20(r = 2): U1 = −2∂y, U2 = 2x∂y, U3 = ∂x, U4 = x∂x + (x2 + 2y)∂y.3.10 Let the expanded form of the characteristic equation be

ω(ω3 +Aω2 +Bω + C) = 0.

For the algebra l4,4 the form of the roots entails the invariant relations

B = 516A

2, C = 132A

3

366

and for l4,5(c)

B =c2 + c− 1

4c2A2, C =

c− 18c2

A3.

3.11 For gl2 the answer is U1 = −x∂y, U2 = 12 (y∂y − x∂x), U3 = y∂x and

U4 = x∂x+y∂y. For sl2 the generators U1, U2 and U3 are the same as for gl2.

4. Equivalence and Invariants of Differential Equations

4.1 Let p0y(n) + p1y

(n−1) + . . .+ pn−1y′+ pny = 0 be a linear homogeneous

ode with polynomial coefficients, i.e., pk ∈ Q[x]. Applying (B.16), the generaltransformation leading to the desired form is x = F (u), y = G(u)v with

G = F ′n/2q−1/(n+1)0 exp

(− 1n+ 1

∫q1q0F ′du

).

F (u) is an undetermined function of u, qk(u) = pk(F ). According to Hirsch [72],p1 = −p′0 is a necessary condition such that there is a fundamental sys-tem of polynomial integrals which is by no means sufficient. An example isx2y′′− 3xy′+ 5y = 0 with the fundamental system y1 = x2 and y2 = x2 log x.Choosing F = u or F = 1

u yields

u7/3v′′ − 73u

4/3v′ − 119 u

1/3v = 0 or u−1/3v′′ + 13u−4/3v′ − 35

9 u−7/3v = 0

respectively. Although q1 = −q′0, neither of these equations has a polynomialfundamental system.

4.2 A general point transformation u = σ(x, y) and v = ρ(x, y) leads fromv′ = 0 to y′+ ρx

ρy = 0. This is the given equation y′+r(x, y) = 0 if there holdsρx− r(x, y)ρy = 0; i.e., ρ is the first integral of the given equation. The othertransformation function σ(x, y) may be chosen arbitrarily. This result makesthe degree of arbitrariness for the equation given in the proof of Theorem 4.10completely explicit.

4.3 If F (u) = u, the transformation (4.37) simplifies to

x = u, y = 1a2

(v + 1

2a′2a2− 1

2a1

), y′ = v′

a2− a′2a22

v + a′′22a2

2

− a′12a2

+ a1a′2

2a22

− a′22a32

.

Substitution into (4.33) and some simplification yields

v′ + v2 + 12z′ − 1

4z2 + a0a2 where z = a′2

a2− a1 and ak ≡ ak(u).

4.4 By Theorem 4.13 the general form of the desired transformation isx = F (u), y = G(u)v +H(u). Upon substitution into Appell’s normal formthe constraints F ′G2 = 1, G′ = 0 and H = 0 are obtained. They yield thetwo-parameter group x = eau+b, y = e−a/2v with Lie algebra ∂x, ∂x− 1

2∂y.


4.5 Applying formulas (B.18) to Lie’s equation (4.56) yields after somerearrangement

v′′ + A(u, v)v′3 + B(u, v)v′2 + C(u, v)v′ + D(u, v) = 0;

i.e., there holds A(u, v) = −D(x, y)|x=v,y=u, B(u, v) = −C(x, y)|x=v,y=u,C(u, v) = −B(x, y)|x=v,y=u and D(u, v) = −A(x, y)|x=v,y=u.

4.6 Applying the transformation x = αu, y = eβuv to y′′′ + cy′ + y = 0yields

v′′′ + 3βv′′ + (3β2 + cα2)v′ + (α3 + α2βc+ β3)v = 0.

Retaining the structure of the original equations requires β = 0 and α3 = 1.From the factorization α3 − 1 = (α− 1)(α2 − α− 1) the asserted equivalenceis obvious.

4.7 If y′′′ − (a + 1)y′′ + ay′ = 0 is transformed by x = αu, y = eβuv, theequation

v′′′+(3β−aα−α)v′′+(3β2+aα2−2aαβ−2αβ)v′+(β2+aα2−aαβ−αβ)βv = 0

is obtained. In order to retain the structure of the original equation, thecoefficient of v must vanish. This yields β(α − β)(aα − β) = 0; i.e., β = 0,β = α or β = aα. Substituting these values into the coefficients of v′′ and v′

and imposing the conditions for the structural invariance leads to the givenfive alternatives.

4.8 By Theorem 5.15, case ii), the only nonvanishing coefficients of equation(5.41) are B3 = 2p(f) and B5 = 2q(u). Consequently, it belongs to symmetryclass S3

6 . The invariant J1 = 2p(f) in equations (6.72) or (6.73) shows thatdetermining S or R comes down to solving a linear second order equation. Ifthe goal is to solve any given third order equation in S3

6 this is an importantsimplification. In the particular application at issue, however, it does not leadtoward the desired answer.

4.9 Let the given equation be in rational normal form y′′ + q(x)y = 0 withq(x) ∈ Q(x). In order to decide the desired equivalence, a rational solution ofthe equation ( f ′′

f ′

)′−1

2

( f ′′

f ′

)2

−12f ′2f2 − (a− 1)f ′2 = 2q(x) (A.5)

for f(x) has to be found. In a partial fraction expansion of f let the leadingterm for x→∞ be aNxN . The first two terms at the left hand side generate

a term 1−N2

x2 from it if N > 1, whereas the remaining two terms generatethe leading powers 4N − 2 and 2N − 2 for N ≥ 1. If n is the leading powerin the partial fraction expansion of q(x), the leading power 4N − 2 must bematched by it which leads to the boundN = n+ 1

4 for the behavior at infinity.For any pole of order M at x = x0, a similar consideration yields the term1−M2

(x− x0)2for M > 1 and two more terms of highest power 4M+2 and 2M for

368

M ≥ 1. The leading power 4M +2 must be matched by a singularity of orderm such that M = m− 2

4 . Consequently, all singularities of a rational solutionare determined by the function q(x) at the right hand side and its ordersare properly bounded. An ansatz with undetermined coefficients within thesebounds leads to a linear system any solution of which determines a rationalsolution for f .

4.10 With the same notation as above, a rational solution has to be foundfor

( f ′′

f ′

)′−1

2

( f ′′

f ′

)2

−2(ν2 − 1

4

)( f ′

f

)2

+2f ′2 = 2q(x). (A.6)

By a similar argumentation, the bounds for the polynomial part and any finitepole are now N = n

2 +1 and M = m2 −1 respectively leading again to a linear

system for the undetermined coefficients.

5. Symmetries of Differential Equations

5.1 By Definition 5.3 the left hand side of the relation at issue may bewritten as

U (n+1)D(φ) = ξ∂D(φ)∂x

+ η∂D(φ)∂y

+ ζ(1) ∂D(φ)∂y′

+ . . .

+ζ(n) ∂D(φ)∂y(n) + ζ(n+1) ∂D(φ)

∂y(n+1) .(A.7)

Expanding the right hand side yields

D(U (n))−D(ξ)D(φ)

= D(ξ∂φ∂x

+ η∂φ∂y

+ ζ(1) ∂φ∂y′

+ . . .+ ζ(n) ∂φ

∂y(n)

)−D(ξ)D(φ)

= D(ξ)∂φ∂x

+D(η)∂φ∂y

+D(ζ(1)) ∂φ∂y′

+ . . .+D(ζ(n)) ∂φ

∂y(n)

+ξD(∂φ∂x

)+ηD

(∂φ∂y

)+ζ(1)D

(∂φ∂y′

)+ . . .

+ζ(n)D(

∂φ

∂y(n)

)−D(ξ)D(φ).

Applying the relations

D(η) = ζ(1) + y′D(ξ), D(ζ(k)) = ζ(k+1) + y(k+1)D(ξ)


for k = 1, . . . , n that follow from (5.10) it may be rewritten as

D(ξ)∂φ∂x

+[ζ(1) + y′D(ξ)

]∂φ∂y

+[ζ(2) + y′′D(ξ)

] ∂φ∂y′

+ . . .

+[ζ(n+1) + y(n+1)D(ξ)

] ∂φ

∂y(n)

+ξD(∂φ∂x

)+ηD

(∂φ∂y

)+ζ(1)D

(∂φ∂y′

)+ . . .

+ζ(n)D(

∂φ

∂y(n)

)−D(ξ)D(φ).

The first term and the second term in the square brackets combined cancelthe last term with the result

ξD(∂φ∂x

)+ηD

(∂φ∂y

)+ζ(1)

[∂φ∂y

+D(∂φ∂y′

)]+ . . .

+ζ(n+1)[

∂φ

∂y(n) +D(

∂φ

∂y(n+1)

)].

The terms in the square brackets may be simplified using the obvious relations

∂xD = D∂x, ∂yD = D∂y, ∂y(k)D = ∂y(k−1) +D∂y(k)

for k = 1, . . . n such that they yield the right hand side of (A.7).5.2 Applying Theorem 5.2 to y′′ + r(x)y = 0 leads to the determining

systemξyy = 0, ηyy − 2ξxy = 0, ηxy − 1

2ξxx + 32ryξy = 0,

ηxx − ry(ηy − 2ξx) + rη + r′yξ = 0.

Its general solution contains eight constants and yields the generators

U1 = y1∂y, U2 = y2∂y, U3 = y∂y,

U4 = y1y∂x + y′1y2∂y, U5 = y2y∂x + y′2y

2∂y,

U6 = y21∂x + y1y

′1y∂y, U7 = y2

2∂x + y2y′2∂y,

U8 = 2y1y2∂x + (y′1y2 + y1y′2)y∂y.

5.3 If a symmetry generator is searched for in the form U = η(x)∂y, (5.10)

yields ζ(k) = ∂(k)η∂xk

. Consequently, the determining system for η(x) is

p0η + p1ηx + p2ηxx + . . .+∂(n)η

∂xn= 0;

i.e., it is identical to the given equation for y. Now if a symmetry generatoris searched for in the form U = η(y)∂y, from equations (5.10) it follows ζ(1) =

370

y′ηy and ζ(k) = D(ζ(k−1)) with D = y′∂y + y′′∂y′ + . . ., therefore the generalstructure of ζ(k) is

ζ(k) = y′∂kη

∂yk+ y(k)ηy + . . .

where the omitted terms are proportional to derivatives ∂iη∂yi

with 1 < i <

n − 1. From this the determining system ηyy = 0, pnη − pnyηy = 0 followswith the solution η = y.

5.4 The most general form that is linear and homogeneous in u, v and w is

wx + u+ Pv +Qw = 0, wy +Ru− v + Sw = 0.

The requirement that no new integrability conditions are introduced is assuredunder the condition Sx −Qy − 1

3Cy −13Bx = 0.

5.5 The subsequent listing allows one to associate Lie’s enumeration ofgroups 1), . . . , 13) with the notation in this book: 1) ≡ g10, 2) ≡ g13, 3) ≡ g8,4) ≡ g7, 5) ≡ g17 with l = 2, ρ1 = ρ2 = 0, α1 = 1, α2 = α 6= 0, 1, 6) ≡ g24,7) ≡ g16, 8) ≡ g20, 9) ≡ g17 with l = 1, ρ1 = 1, α1 = 1, 10) ≡ g5, 11) ≡ g17

with l = 2, ρ1 = ρ2 = 0, α1 = 0, α2 = 1, 12) ≡ g17 with l = 1, ρ1 = 1,α1 = 0, 13) ≡ g15. For three groups a change of variables has to be appliedin order to obtain the generators in Lie’s form: 9) is obtained from g17 bythe change x = −x, y = yex, 11) by the change x = − log x, y = y

x and 5)by x = 1

α− 1 log x and y = yx1/(α−1). Finally, the barred variables have tobe replaced by the original ones. In Theorem 5.8 there occur only groups 1),2), 4) and 8) because there are no second order ode’s with a symmetry groupdifferent from these. This is related to the definition of a symmetry group asthe maximal group of transformations leaving the equation invariant; see alsothe remarks by Lie [113] on page 538.

5.6 In order to avoid an unspecified dependence on y in the transformedfunction r, σy = 0 is required; i.e., σ(x, y) ≡ f(x). The quotient v′′

v′is

reproduced if the two derivatives are proportional to y′′ and y′ respectively.This is assured if ρx = 0 and ρyy = 0, i.e., ρ = a1y + a2. It yields

v′′

v′=

1f ′y′′

y′− f ′′

f ′2and v′′′ =

a1

f ′3y′′′ − 3a1f

′′

f ′4y′′ +

3a1(f ′′2 − f ′′′f ′)f ′5

y′.

If the second term of the expression for v′′′ is rewritten as −3a1f′′

f ′4y′′

y′y′ its

correct structure is obvious.5.7 In order to avoid an unspecified dependence on y in the transformed

function r, σy = 0 is required, i.e., σ = h(x). The independence of v′′ on y′ inthe first place requires ρyy = 0 due to the quadratic term in equation (5.5),i.e., it has the form ρ(x, y) = f(x)+ g(x)y. The coefficients of the linear term


in y′ and the independence on y vanish if f and g satisfy

2g′

g− h′′

h′= 0,

g′′

g′− h′′

h′= 0

with the general solution g = 1C1x+ C2

, h = C3g + C4, and, consequently,

σ =(C1x+ C2)C4 + C3

C1x+ C2, ρ =

y

C1x+ C2.

The function f(x) remains undetermined. Introducing new constants by

C1 = −a2a4

a3, C2 =

1− a1a2a4

a3, C3 =

1a4, C4 = −a3

a4

yields the required answer.5.8 Both conditions are subsumed under the constraint Ψ′3 = 0 with Ψ3

defined in Theorem 4.9 as a simple calculation shows. They may also beobtained from the Janet basis for the symmetry generators as it has beenshown by Schwarz [166].

5.9 Equation (5.46) is completely reducible into first order factors

Lclm(D − 32

3x− 2 + 3

x, D −12 −

1x+ 2 −

2x− 2 + 3

x,

D + 12 −

2x+ 2 −

1x− 2 + 3

x, D + 32 −

3x+ 2 + 3

x )y = 0

where D ≡ ddx

. From this representation the answer is obvious.

6. Transformation to Canonical Form

6.1 Denoting the symmetry algebra of symmetry class S21 by L2

1, symmetryclass S2

2,1 by L22,1, etc., the diagram

L2

1

L2

2,1 L2

2,2

L2

3,4 L2

3,3 L2

3,2 L2

3,1

L2

8

@@R

@@R

?PPPPPPq

)

describes the relationship between the various Lie algebras. An arrow fromLi to Lj means on the one hand that Li ⊂ Lj . On the other hand it implies

372

that the undetermined elements in the canonical form corresponding to theformer symmetry algebra may be specialized such that the symmetry algebraenlarges to the latter. Accordingly, the arrows originating from L2

1 implythe relations ru(u, v′) = 0 or uru(u, v′) + r(u, v′) = 0, respectively, for theundetermined function r(u, v′) in the canonical form for the symmetry classS2

1 as given in Theorem 6.4. The arrows originating from L22,1 imply that the

undetermined function r(v′) in the canonical form for symmetry class S22,1

has the special form required for symmetry class S23,3 or S2

3,4. The remainingarrows imply that the respective undetermined elements vanish such that thecanonical form v′′ = 0 for symmetry class S2

8 is obtained.6.2 The desired transformation must have the form u = a1u + a2v + a3,

v = a4u + a5v + a6. Substituting it into the canonical form of Example 6.7leads to the algebraic system

a1(4a31 + 8a2

1a4 + 5a1a24 + a3

4) = 0, a2(4a32 − 8a2

2a4 + 5a2a24 − a3

4) = 0,

(a1 + a2)a24 = 1.

There are 24 alternatives. A particularly simple one is a1 = 0, a2 = a4 = 1,a5 = −1, a3 and a6 arbitrary.

6.3 For symmetry class S22,1 the system (6.4) may be solved for the Janet

basis coefficients with the result

a1 = b2 = I3, a2 = I4, b1 = I1, c1 = d2 = −I5, c2 = −I6, d1 = −I2.

These relations have obviously the form given in Theorem 6.6. For symmetryclass S2

2,2, system (6.9) yields

a1 = −I3 −K2, a2 = −I4 −K1, b1 = −I1, b2 = −I3,

c1 = I5, c2 = I6, d1 = I2 −K2, d2 = I5 −K1.

They may be rewritten in the form given in Theorem 6.6 as follows.

I1 = −b1, I2 = d1 − a1 + b2, I3 = −b2, I4 = d2 − a2 − c1,

I5 = c1, I6 = c2, K1 = c1 − d2, K2 = b2 − a1.

6.4 Applying the algorithm Symmetries on page 199 to y′′ + r(y, y′) = 0generates the Janet basis η = ξx = ξy = 0; i.e., case b) of Theorem 6.7with a = b = c = 0. Consequently, φ = y and G = 0, and the generaltransformation may be written in the form x = Cv + f(u), y = g(u) withsuitably defined functions f and g. Together with

y′ =g′

Cv′ + f ′, y′′ =

−Cg′v′′ + Cg′′v′ + f ′g′′ − f ′′g′

(Cv′ + f ′)2

the desired transformation follows. The special values C = 1, f = 0 and g = ugenerate the variables exchange discussed after the proof of Theorem 6.7.


6.5 The vector fields determined by ξx − 2xξ = 0, ξy = 0, ηx = 0 and

ηy − 2y η = 0 generate a two-parameter subgroup of the full symmetry group

because it reduces the Janet basis (5.32) to zero. Applying the scheme de-scribed in Theorem 6.8, the system σxx+ 1

xσx = 0, σxy = 0, σyy + 2yσy = 0 is

obtained with the fundamental system1, 1x,

1y. The transformation x = 1

u

and y = 1v yields v′′v′−1 = 0; i.e., a canonical form v′′+r(v′) = 0 correspond-

ing to symmetry class S22,1. In order to obtain the special value r(v′) = v′4,

the genuine symmetry class has to be known.

6.6 The vector fields determined by ξx − η + ξx+ y = 0, ξy = 0, ηx = 0 and

ηy − η + ξx+ y η = 0 generate a two-parameter subgroup of the full symmetry

group because it reduces the Janet basis given in Example 5.14 to zero. Eval-uating the integrals given in Theorem 6.9 yields for the transformation func-tions σ = x + y and ρ = y. Substituting x = u − v and y = v into the givenequation yields the canonical form uv′′ − 2v′3 − 3v′2 − v′ = 0. The canonicalform given in Example 6.11 cannot be obtained by specialization from it.

6.7 Three generators of a sub-algebra of the full symmetry algebra areU1 = 1

2x∂x, U2 = x2(y − 1)2∂y and U3 = y(y − 1)∂y with commutators[U1, U2] = U2, [U1, U3] = 0 and [U2, U3] = −U2. A g26 subgroup is generatedby U1 and U3. From its Janet basis

ξx +1xξ = 0, ξy = 0, ηx = 0, ηy −

2y − 1y(y − 1)

η = 0

Theorem 6.8 yields σxx+ 1xσx = 0, σxy = 0, σyy+ 2y − 1

y(y − 1)σy = 0 with funda-

mental system1, log x, log y − 1

y. It leads to the transformation functions

u = log x and v = log y − 1y with the inverse x = eu, y = 1

1− ev from whichthe canonical form v′′ − v′2 − 3v′ − 1 = 0 is obtained.

Choosing the two-parameter group of type g25 generated by U1 and U2 withJanet basis ξx− 1

xξ = 0, ξy = 0, ηx− 2xη = 0, ηy− 2

y − 1η = 0, by Theorem 6.8

there follows u = 1x2 , v = 1

x2(y − 1). From its inverse x = 1√

uand y = u

v +1

the canonical form uv′′ + 12v′ + 1

2 = 0 follows, i.e., an inhomogeneous linearode.

The three generators U1, U2 and U3 generate a three-parameter group withLie algebra l3,4. This symmetry type is not included in the classificationgiven in Theorem 5.8 because by definition a symmetry class is defined by themaximal group leaving it invariant.

374

7. Solution Algorithms

7.1 For the ode in Example 6.2 the equation

dy + (y2 − 2x2y + x4 − 2x− 1)dxy2 − 2x2y + x4 − 1

= 0

is obtained. Upon integration it yields the same answer as given there. Inthis case, the integrations for obtaining the canonical form are simpler.

7.2 In Example 6.2 the transformation functions x ≡ φ(u, v) = u− v + 12v

and y ≡ ψ(u, v) = u − v − 12v have been shown to generate the canonical

form v′ = 13 . Substituting its inverse u = 1

x− y + x+ y2 , v = 1

x− y into the

canonical form solution v = 23u+ C yields the solution in actual variables as

given in Example 6.2.7.3 According to Lemma 3.4, canonical coordinates are u = y2−x2 and v =

12 log y + x

y − x . They lead to the canonical form v′ + 12u

u+ 1u− 1 = 0. Integrating

and resubstituting x and y yield the solution given in Example 6.2.7.4 Let x = g(y′) be a first order ode and consider y′ as a parameter, say

p, such that p = y′ = h(x) with h = g−1. It follows y =∫h(x)dx + C or

y =∫pg′(p)dp + C. Partial integration yields y = pg(p) −

∫g(p)dp + C.

Substituting u and v for x and y respectively, −R(p) + C1 for g, and C2 forC leads to (7.16).

7.5 Substituting u = e ¯w, z = t with ¯w ≡ ¯w(t), and then ¯w′ = w into (7.38)yields w′+ t(t2 +4)w3 +6w2 + 3

t w = 0. This is an Abel equation. Its rational

normal form w′ + t(t2 + 4)w3 + 3tt2 + 4

w = 0 is obtained by substituting (see2

t(t2 + 4)and leaving t unchanged. This

Bernoulli equation has the general solution

w =[C(t2 + 4)3 + 1

4 (t2 + 4)2t2]− 1

2 .

Resubstitution of the original variables and redefining the integration con-stants finally yields the solution given in Example 7.19.

7.6 Now r(p) = p(p+ 1)2p− 1 , R(p) = log p+ 1

p + 2pp+ 1 and

∫R(p)dp =

(p− 1) log (p+ 1)− p log p+ 2p. Substitution into (7.16) yields

u+ logp+ 1p

+2pp+ 1

= C1, v + log (p+ 1)− 2pp+ 1

= C2

and finally

x =C1

p+ 1exp

2pp+ 1

, y =C2p

p+ 1exp− 2p

p+ 1.

7.7 For generic r(y), a type J (2,2)1,2 Janet basis with a = b = c = 0 is

obtained for the determining system. By Theorem 6.7, case b), the trans-formation to canonical form is achieved by x = v, y = u with the result

the proof of Theorem 4.14) w = w−


v′′ − v′3r(u) = 0. Two integrations and resubstitution of the actual variablesleads to the integral representation∫

dy√C1 − 2

∫r(y)dy

= x+ C2

of the general solution.7.8 For generic r(x), a type J (2,2)

1,1 Janet basis with a = 0, b = −1y is

obtained for the determining system. By Theorem 6.7, case a), the trans-formation to canonical form is achieved by x = u, y = ev with the resultv′′ + r(u)v′2 + v′ = 0. This Riccati equation for v′ has the general solution

v′ =e−u∫

r(u)e−udu+ C1.

Integration and resubstitution leads to the general solution in actual variables

y = C2 exp∫

e−xdx∫r(x)e−xdx+ C1

.

7.9 The first integrals following from the equivalent pde may be found in Lie[113], page 540. In the notation applied in this book they yield the relations

u2 +v2 − 2uvv′

v′2 − a

v2

= C1, u− vv′

v′2 − a

v2

= C2.

Elimination of v′ leads to

v4 − au4 − 4C1u

3 + (4C21 − 2C2)u2 + 4C1C2u+ C2

2

C21 + C2

= 0.

This representation is equivalent to (7.25) if C21 = a

C2

1+ C2, C2 = −C1.

This result shows that the solutions obtained from the first integrals of theequivalent pde may be unnecessarily complicated.

Determine the solution of (7.24) from the equivalent pde and compare theresult with (7.25).

7.10 Write the equation to be solved as y′+ 1ax√xy2 + 1

a√x

= 0. The sym-

metry generator x∂x+ 12y∂y yields the canonical variables u = y√

x, v = 2 log y.

Substituting the inverse x = 1u2 e

v, y = ev/2 yields v′ = 4u

u2 + 12u2 + au+ 2

. An

integration leads to v = 2 log u+ 4ai√a2 − 16

arctan 4u+ a

i√a2 − 16

+C. Applying

the relation arctanx = − i2 log 1 + ix1− ix , it assumes the form

v = 2 log u+2a√a2 − 16

log

√a2 − 16 + a+ 4u√a2 − 16− a− 4u

+ C.

376

Resubstitution of the original variables and some simplifications lead to

y =a√x

2(c− 1

2 )xc − (c+ 12 )C

xc + C

where c =√a2 − 162a .

7.11 The pde for (7.71) is Af = 0 with A = ∂u + v′∂v + v′′∂v′ + ω∂v′′ and

ω = −v′′v (3v′ + a

√v′′v). The linear relation between A and the four vector

fields generating the symmetry group yields the first integrals

Φ1 = u−2vv′

D, Φ2 = 2u−4vv′ + av

√v′′v

D, Φ3 = u2+

2v2 − (4vv′ + av√v′′v)u

D

with D = 2vv′′ + 2v′2 + av′√v′′v. Solving the relations Φi = Ci, i = 1, 2, 3,

for v′v yields the integrand in (7.71).

Appendix B

Collection of Useful Formulas

For ease of reference in this appendix those formulas are put together thatoccur frequently in the main text. Moreover, they might be of general interestfor any person working in this field.Transformations of the Plane. Let a coordinate transformation from x, yto u, v and its inverse be defined by

x = φ(u, v), y = ψ(u, v) u = σ(x, y), v = ρ(x, y). (B.1)

They entail the following relations for the first order partial derivatives.(σx σyρx ρy

)=

(φu φvψu ψv

)−1

=1

φuψv − φvψu

(ψv −φv−ψu φu

)(B.2)

and (φu φvψu ψv

)=

(σx σyρx ρy

)−1

=1

σxρy − σyρx

(ρy −σy−ρx σx

). (B.3)

It is assumed that ∆ ≡ σxρy − σyρx 6= 0 and φuψv − φvψu 6= 0. The first andsecond order derivatives of φ and ψ are

φu =ρy∆, φv =

−σy∆

, ψu =−ρx∆

, ψv =σx∆, (B.4)

φuu = 1∆3 [ρy(ρxy∆− ρy∆x)− ρx(ρyy∆− ρy∆y)],

φuv = 1∆3 [−σy(ρxy∆− ρy∆x) + σx(ρyy∆− ρy∆y)],

φvv = 1∆3 [σy(σxy∆− σy∆x)− σx(σyy∆− σy∆y)],

ψuu = 1∆3 [−ρy(ρxx∆− ρx∆x) + ρx(ρxy∆− ρx∆y)],

ψuv = 1∆3 [σy(ρxx∆− ρx∆x)− σx(ρxy∆− ρx∆y)],

ψvv = 1∆3 [−σy(σxx∆− σx∆x) + σx(σxy∆− σx∆y)].

(B.5)

If y ≡ y(x) and v ≡ v(u), the substitutions (B.1) generate the followingtransformation of the derivatives y′, y′′ and y′′′.

y′ =ψu + ψvv

′

φu + φvv′ , (B.6)

377

378

y′′ = 1(φu + φvv

′)3(φuψv − φvψu)v′′ + (φvψvv − φvvψv)v′3

+[φuψvv − φvvψu + 2(φvψuv − φuvψv)

]v′2

+[φvψuu − φuuψv + 2(φuψuv − φuvψu)

]v′ + φuψuu − φuuψu

.

(B.7)Higher order derivatives are obtained from (5.6). In general the structure ofthe kth derivative is

y(k) = 1(φu + φvv

′)k+1

[(φuψv − φvψu)v(k) − mkφv(φuψv − φvψu)v′′v(k−1)

φu + φvv′

+P (v′)v(k−1) +Q(v′, . . . , v(k−2)

(φu + φvv′)k−2)

].

(B.8)P and Q are polynomial in their arguments and furthermore depend on u andv; mk is a natural number. If φv = 0, this expression simplifies to

y(k) = 1φku

[ψvy

(k) + kv′v(k−1) + F (u, v)v(k−1) +G(v′, . . . , v(k−2))φk−1u

]. (B.9)

The expressions (B.8) and (B.9) are due to Neumer [137].Various special cases that occur frequently in this book lead to simplifica-

tions of these expressions. If

x = F (u), y = G(v), F ′ =dF

duand G′ =

dG

dv(B.10)

the derivatives are

y′ = G′

F ′v′, y′′ = 1

F ′3(F ′G′v′′ + F ′G′′v′2 − F ′′G′v′),

y′′′ = 1F ′5

(F ′2G′v′′′ + 3F ′2G′′v′v′′ + F ′2G′′′v′3 − 3F ′′F ′G′v′′

−3F ′′F ′G′′v′2 + 3F ′′2G′v′ + F ′′′F ′G′v′).

(B.11)

If

x = F (u), y = G(u)v+H(u), F ′ =dF

du, G′ =

dG

duand H ′ =

dH

du(B.12)

the derivatives are

y′ = GF ′

(v′ + G′

G v +H),

y′′ = GF ′2

[v′′+

(2G

′

G F ′ − F ′′)F ′2v′

+(G′′

G F ′ − G′

G F ′′)F ′2v + (H ′′F ′ −H ′F ′′)F

′2

G

].

(B.13)

Collection of Useful Formulas 379

Ifx = F (u), y = G(u)v, F ′ =

dF

duand G′ =

dG

du(B.14)

the derivatives are

y′ = GF ′

(v′ + G′

G v),

y′′ = GF ′2

[v′′+

(2G

′

G −F ′′

F ′

)v′+

(G′′

G −F ′′

F ′G′

G

)v

],

y′′′ = GF ′3

v′′′ + 3

(G′

G −F ′′

F ′

)v′′

+[

3(F ′′

F ′

)2

+3G′′

G − 6F′′

F ′G′

G −F ′′′

F ′

]v′

+[

3(F ′′

F ′

)2G′

G + G′′′

G − F ′′′

F ′G′

G − 3F′′

F ′G′′

G

]v

.

(B.15)For some applications the explicit form of the coefficients of the three leadingderivatives are required. They may be seen from

y(n) = GF ′n

v(n) + n

[G′

G −n− 1

2F ′′

F ′

]v(n−1)

+n(n− 1)2

[G′′

G − (n− 1)F′′

F ′G′

G

−n− 23

F ′′′

F ′+ (n− 2)(n+ 1)

4

(F ′′

F ′

)2]v(n−2)

+ . . .

(B.16)Upon further specialization G = F ′, it simplifies to

y(n) = 1F ′n−1

v(n) − n(n− 3)

2F ′′

F ′v(n−1)

−n(n− 1)2

[n− 5

3F ′′′

F ′− n2 − 5n+ 2

4 (F′′

F ′)2

]v(n−2)

+ . . .

(B.17)The independent variable x and the dependent variable y ≡ y(x) are ex-

changed by the transformation x = v, y = u with v ≡ v(u). The derivativesup to order four are transformed by

y′ = 1v′, y′′ = − v

′′

v′3, y′′′ = −v

′′′

v′4+ 3v′′2

v′5,

y(4) = −v(4)

v′5+ 10v′′′v′′

v′6− 15v′′3

v′6.

(B.18)

Symmetric Powers of Second Order Equations. The listing below givesthe k-th symmetric powers of y′′ + ry = 0 from k = 2 to k = 6.

k = 2 : y′′′ + 4ry′ + 2r′y = 0,

380

k = 3 : y(4) + 10ry′′ + 10r′y′ + (3r′′ + 9r2)y = 0,

k = 4 : y(5) + 20ry′′′ + 30r′y′′ + (18r′′ + 64r2)y′ + (4r′′′ + 64r′r)y = 0,

k = 5 : y(6) + 35ry(4) + 70r′y′′′ + (63r′′ + 259r2)y′′ + (28r′′′ + 518r′r)y′

+(5r(4) + 155r′′r + 130r′2 + 225r3)y = 0,

k = 6 : y(7) + 56ry(5) + 140r′y(4) + (168r′′ + 784r2)y′′′ + (112r′′′ + 2352r′r)y′′

+(40r(4) + 1408r′′r + 1180r′2 + 2304r3)y′

+(6r(5) + 312r′′′r + 708r′′r′ + 3456r′r2)y = 0,

Janet Basis Types. The Janet basis types that are required in earlier partsof this book are listed, beginning with the Janet bases of type J (1,2) of order 1,2 and 3. Their coherence conditions may be found on page 59. The dependentvariable is z, the independent variables are x and y, the grlex term order withy > x is always applied.

J (1,2)1 : zx + az = 0, zy + bz = 0.

J (1,2)2,1 : zy + a1zx + a2z = 0, zxx + b1zx + b2z = 0.

J (1,2)2,2 : zx + a1z, zyy + b1zy + b2z = 0.

J (1,2)3,1 : zy + a1zx + a2z = 0, zxxx + b1zxx + b2zx + b3z = 0.

J (1,2)3,2 :

zxx + a1zy + a2zx + a3z = 0, zxy + b1zy + b2zx + b3z = 0,zyy + c1zy + c2zx + c3z = 0.

J (1,2)3,3 : zx + a1z = 0, zyyy + b1zyy + b2zy + b3z = 0.

The Janet bases of type J (2,2) of order 1, 2 and 3 are given next. Theircoherence conditions may be found on pages 64 to 67. The dependent variablesare z1 and z2. The term order is grlex with z2 > z1 and y > x.

J (2,2)1,1 : z1 = 0, z2,x + az2 = 0, z2,y + bz2 = 0.

J (2,2)1,2 : z2 + az1 = 0, z1,x + bz1 = 0, z1,y + cz1 = 0.

J (2,2)2,1 : z1 = 0, z2,y + a1z2,x + a2z2 = 0, z2,xx + b1z2,x + b2z2 = 0.

J (2,2)2,2 : z1 = 0, z2,x + a1z2 = 0, z2,yy + b1z2,y + b2z2 = 0.

J (2,2)2,3 :

z1,x + a1z2 + a2z1 = 0, z1,y + b1z2 + b2z1 = 0,z2,x + c1z2 + c2z1 = 0, z2,y + d1z2 + d2z1 = 0.

Collection of Useful Formulas 381

J (2,2)2,4 : z2 + a1z1 = 0, z1,y + b1z1,x + b2z1 = 0, z1,xx + c1z1,x + c2z1 = 0.

J (2,2)2,5 : z2 + a1z1 = 0, z1,x + b1z1 = 0, z1,yy + c1z1,y + c2z1 = 0.

J (2,2)3,1 : z1 = 0, z2,y +a1z2,x+a2z2 = 0, z2,xxx+ b1z2,xx+ b2z2,x+ b3z2 = 0.

J (2,2)3,2 :

z1 = 0, z2,xx + a1z2,y + a2z2,x + a3z2 = 0,z2,xy + b1z2,y + b2z2,x + b3z2 = 0,z2,yy + c1z2,y + c2z2,x + c3z2 = 0.

J (2,2)3,3 : z1 = 0, z2,x + a1z2 = 0, z2,yyy + b1z2,yy + b2z2,y + b3z2 = 0.

J (2,2)3,4 :

z1,x + a1z2 + a2z1 = 0, z1,y + b1z2 + b2z1 = 0,z2,y + c1z2,x + c2z2 + c3z1 = 0, z2,xx + d1z1,x + d2z2 + d3z1 = 0.

J (2,2)3,5 :

z1,x + a1z2 + a2z1 = 0, z1,y + b1z2 + b2z1 = 0,z2,x + c1z2 + c2z1 = 0, z2,yy + d1z2,y + d2z2 + d3z1 = 0.

J (2,2)3,6 :

z1,y + a1z1,x + a2z2 + a3z1 = 0, z2,x + b1z1,x + b2z2 + b3z1 = 0,z2,y + c1z1,x + c2z2 + c3z1 = 0, z1,xx + d1z1,x + d2z2 + d3z1 = 0.

J (2,2)3,7 :

z1,x + a1z2 + a2z1 = 0, z2,x + b1z1,y + b2z2 + b3z1 = 0,z2,y + c1z1,y + c2z2 + c3z1 = 0, z1,yy + d1z1,y + d2z2 + d3z1 = 0.

J (2,2)3,8 : z2 + a1z1 = 0, z1,x + b1z1 = 0, z1,yyy + c1z1,yy + c2z1,y + c3z1 = 0.

J (2,2)3,9 :

z2 + a1z1 = 0, z1,xx + b1z1,y + b2z1,x + b3z1 = 0,z1,xy + c1z1,y + c2z1,x + c3z1 = 0, z1,yy + d1z1,y + d2z1,x + d3z1 = 0.

J (2,2)3,10 : z2+a1z1 = 0, z1,y+b1z1,x+b2z1 = 0, z1,xxx+c1z1,xx+c2z1,x+c3z1 = 0.

A partial listing of Janet basis types of order four is given next; they arerequired for the symmetry class identification on page 227.

J (2,2)4,2 :

z1 = 0, z2,xy + a1z2,xx + a2z2,y + a3z2,x + a4z2 = 0,z2,yy + b1z2,xx + b2z2,y + b3z2,x + b4z2 = 0,z2,xxx + c1z2,xx + c2z2,y + c3z2,x + c4z2 = 0.

J (2,2)4,9 :

z1,y + a2z1,x + a3z2 + a4z1 = 0, z2,y + b1z2,x + b2z1,x + b3z2 + b4z1 = 0,z1,xx + c1z2,x + c2z1,x + c3z2 + c4z1 = 0,z2,xx + d1z2,x + d2z1,x + d3z2 + d4z1 = 0.

382

J (2,2)4,10 :

z1,y + a2z1,x + a3z2 + a4z1 = 0, z2,x + b2z1,x + b3z2 + b4z1 = 0,z1,xx + c1z2,y + c2z1,x + c3z2 + c4z1 = 0,z2,yy + d1z2,y + d2z1,x + d3z2 + d4z1 = 0.

J (2,2)4,12 :

z1,x + a3z2 + a4z1 = 0, z2,x + b2z1,y + b3z2 + b4z1 = 0,z1,yy + c1z2,y + c2z1,y + c3z2 + c4z1 = 0,z2,yy + d1z2,y + d2z1,y + d3z2 + d4z1 = 0.

J (2,2)4,14 :

z2,x + a1z1,y + a2z1,x + a3z2 + a4z1 = 0,z2,y + b1z1,y + b2z1,x + b3z2 + b4z1 = 0,z1,xx + c1z1,y + c2z1,x + c3z2 + c4z1 = 0,z1,yy + d1z1,y + d2z1,x + d3z2 + d4z1 = 0,z1,yy + e1z1,y + e2z1,x + e3z2 + e4z1 = 0.

J (2,2)4,17 :

z2 + a4z1 = 0, z1,xy + b1z1,xx + b2z1,y + b3z1,x + b4z1 = 0,z1,yy + c1z1,xx + c2z1,y + c3z1,x + c4z1 = 0,z1,xxx + d1z1,xx + d2z1,y + d3z1,x + d4z1 = 0.

J (2,2)4,19 :

z2 = 0, z1,xx + a2z1,y + a3z1,x + a4z1 = 0,z1,xy + b2z1,y + b3z1,x + b4z1 = 0,

z1,yyy + c1z1,yy + c2z1,y + c3z1,x + c4z1 = 0.

Finally, the following fifth order Janet bases are required for symmetry classidentification on page 238.

J (2,2)5,1 :

z1,x + a4z2 + a5z1 = 0, z1,y + b4z2 + b5z1 = 0,z2,xy + c1z2,xx + c2z2,y + c3z2,y + c4z2 + c5z1 = 0,z2,yy + d1z2,xx + d2z2,y + d3z2,y + d4z2 + d5z1 = 0,z2,xxx + e1z2,xx + e2z2,y + e3z2,y + e4z2 + e5z1 = 0.

J (2,2)5,2 :

z2,x + a2z1,y + a3z1,x + a4z2 + a5z1 = 0,z2,y + b2z1,y + b3z1,x + b4z2 + b5z1 = 0,

z1,xx + c1z1,yy + c2z1,y + c3z1,x + c4z2 + c5z1 = 0,z1,xy + d1z1,yy + d2z1,y + d3z1,x + d4z2 + d5z1 = 0,z1,yyy + e1z1,yy + e2z1,y + e3z1,x + e4z2 + e5z1 = 0.

J (2,2)5,3 :

z2,x + a2z1,y + a3z1,x + a4z2 + a5z1 = 0,z2,y + b2z1,y + b3z1,x + b4z2 + b5z1 = 0,

z1,xy + c1z1,xx + c2z1,y + c3z1,x + c4z2 + c5z1 = 0,z1,yy + d1z1,xx + d2z1,y + d3z1,x + d4z2 + d5z1 = 0,z1,xxx + e1z1,xx + e2z1,y + e3z1,x + e4z2 + e5z1 = 0.

Appendix C

Algebra of Monomials

In this appendix various properties of semigroups of monomials are described.Details and proofs may be found in the article by Janet [83]; see also Castroand Moreno [27]. The notation applied here is independent from the rest ofthe book.

Let x1, x2, . . . , xn be n indeterminates equipped with the lex orderingx1 < x2 < . . . < xn. Monomials in the xi are denoted by m = xα1

1 . . . xαnn

with (α1, . . . , αn) ∈ Nn or, if several monomials are to be distinguished, by

mi = xαi

11 . . . x

αinn . Because the product of two monomials is again a monomial,

and the product is commutative and associative, they form a commutativesemigroup under multiplication. Because (1, . . . , 1) is the identity element,they even form a monoid. The connection of the results described in this ap-pendix with partial differential equations is established by the identification

of the degree vector (α1, . . . , αn) with the partial derivative ∂α1+...+αn

∂xα11 . . . ∂xαn

n.

Let M be a finite set of such monomials. They generate an ideal in thismonoid. This ideal should not be mistaken for an ideal in the polynomial ringk[x1, . . . , xn] with k some number field. The totality of all monomials of themonoid decomposes into those which are contained in the ideal generated byM and those which are not.

For the applications in the main part of this book a unique representationof the members of both subsets is required. To this end for both subsets adecomposition into classes will be obtained. The classes are defined as theentirety of all multiples of any monomial of a complete set with respect itsmultipliers, i.e., a certain subset of its variables. It is the main subject of thisappendix to define a procedure for constructing such a decomposition intoclasses for any given set of monomials.

The finiteness of various algorithms dealing with monomials is based on thefollowing lemma which is proved in Janet’s article [83], pages 69-70.

Lemma C.1 A set of monomials such that none of its elements is a multipleof any other monomial is necessarily finite.

Let M = m1,m2, . . . be a finite set of monomials and a1 < . . . < akthe degrees in xn which do actually occur in M. Mα ⊂ M denotes thosemonomials in M which occur for a fixed value of α ∈ a1, . . . , ak. The firststep toward the decomposition mentioned above is the subdivision of variables

383

384

and monomials as described in the following definitions.

Definition C.1 (Multipliers and non-multipliers) The leading variable xnis called a multiplier for mi ∈ M if αin = ak; otherwise it is called a non-multiplier. If n > 1, the same procedure is applied to the monomials of Maj

,j = 1, . . . , k, w.r.t. n− 1 variables x1, . . . , xn−1.

Definition C.2 (Classes) The totality of all multiples of a monomial mi ∈M with respect to its multipliers is called the class defined by this monomialand is denoted by mi. The union of the classes of all monomials in M isdenoted by M.

It should be noted that this definition of multipliers is only meaningfulwith respect to a fixed ordering for the variables. In general there will bemonomials in the ideal generated by M which do not belong to any class atall.

Example C.1 Consider the following set M comprising four monomialsin three variables.

Monomial Multipliers Monomial Multipliers

x21x

22x

33 x1, x2, x3 x3

1x2x3 x1, x2

x31x

33 x1, x3 x2x3 x2

The monomial x1x2x3 obviously may not be represented as the product of anymonomial inM and some of its multipliers; and therefore does not belong toany class.

The special sets of monomials which have the property that any monomialin the ideal generated by it belongs at least to a single class are defined next.

Definition C.3 (Complete system) A set of monomialsM is called com-plete if any monomial in the ideal generated by M belongs to at least one ofits classes.

Due to this property of complete sets it is desirable to have a procedurewhich determines for any given set of monomials an equivalent complete set,i.e., a complete set generating the same ideal as the given one. The followingthree theorems due to Janet are the prerequisite for such an algorithm.

Theorem C.1 A set of monomials M is complete if and only if

i) The sets Ma1 , . . . ,Makare complete with respect to x1, . . . , xn−1.

ii) For any monomial m ∈ M there holds m/xαn ∈ Mα → xn ·m ∈Mα+1

for α < ak.

Theorem C.2 LetM be a set of monomials. If the product of any mono-mial in M with any of its non-multipliers is contained in M, then M iscomplete.

Algebra of Monomials 385

Theorem C.3 Any finite set of monomialsM may be extended to a com-plete set generating the same ideal as M.

The proof of these theorems may be found in Janet [83], page 74 ff.An obvious consequence of Theorem C.1 is the constraint ai+1 = ai + 1 for

1 ≤ i ≤ k−1 for complete sets. Based on these results the following algorithmComplete may be designed which is also due to Janet. The notation is thesame as above.

Algorithm C.1 Complete(M). Given a setM = m1,m2, . . . of mono-mials, the complete set corresponding to M is returned.S1 : Find Multipliers & Non-Multipliers. Determine multipliers and non-multipliers for all monomials inM.S2 : Products with Non-Multipliers. Generate the set M0 of monomialswhich are the product of an mi w.r.t. any one of its non-multipliers.S3 : In Classes? Remove all those elements fromM0 which are containedin M.S4 : Termination? If M0 is empty return M, else set M := M∪M0

and goto S1.

Example C.2 If this algorithm is applied to example C.1, the followingcomplete set of monomials is obtained.

Monomial Multipliers Monomial Multipliers

x21x

22x

33 x1, x2, x3 x3

1x2x23 x1, x2

x1x22x

33 x2, x3 x2

1x2x23 x2

x22x

33 x2, x3 x1x2x

23 x2

x31x2x

33 x1, x3 x2x

23 x2

x21x2x

33 x3 x3

1x2x3 x2

x1x2x33 x3 x2

1x2x3 x2

x2x33 x3 x1x2x3 x2

x31x

33 x3, x1 x2x3 x2

Complete sets of monomials have several properties that make them im-portant tools for analyzing sets of pde’s. They are collected in the followingtheorem.

Theorem C.4 A complete set of monomials M has the following proper-ties.

i) Any monomial in the ideal generated by M belongs exactly to a singleof its classes.

ii) The product of any monomial inM by one of its non-multipliers is equalto the product of some other monomial by its multipliers.

386

LetM denote a complete set of monomials. To obtain a description of themonomials which are not contained in the ideal generated by M, a new setof monomials, the complementary set N , will be constructed as follows. Ifan > 0 then xαn ∈ N for 0 ≤ α < a1 with multipliers x1, x2, . . . xn−1. For1 ≤ i ≤ k, considerMai . Let b1 < b2 < . . . bj be the exponents of xn−1 whichdo occur in Mai . If b1 > 0 then the monomials xai

n xβn−1 ∈ N for 0 ≤ β < b1.

Its multipliers are x1, x2 . . . xn−2 if i < k, and in addition xn if i = k. Thisproceeding is applied recursively all the way down until n = 1 is reached.

Appendix D

Loewy Decompositions of Kamke’sCollection

In this appendix the types of Loewy decompositions of linear second and thirdorder equations in x and y(x) with coefficients in Q from Kamke’s collectionare listed. Excluded are equations containing undetermined functions, morethan two parameters, or a parameter in an exponent. For these equationsusually a more sophisticated analysis is required in order to determine theconstraints leading possibly to different decomposition types. If there areparameters involved, the given decomposition applies to unconstrained pa-rameter values. Any relations between them may change the decompositiontype. For inhomogeneous equations only the left hand side is taken into ac-count. From the given nontrivial decompositions fundamental systems maybe obtained by integration according to the formulas on page 31.Equations of Second Order. These are the equations of Chapter 2 ofKamke’s collection. The tabulation below is organized by decomposition typeas described in Corollary 2.3. For types L2

k, k = 2, 3 or 4 the numbers of thecorresponding equations are listed.

For irreducible equations of type L21 a further distinction is made according

to the type of Galois group of its rational normal form as discussed in Chap-ter 2 on page 35. If the Galois group is SL2, the equivalence to an equationdefining a special function is given whenever possible. The following equationsdefining special functions w(z) are taken into account.

Bessel : w′′ + 1zw

′−(

1− ν2

z2

)w = 0,

Hypergeometrical : w′′ + (α+ β + 1)z − γz(z − 1) w′ + αβ

z(z − 1)w = 0.

Legendre : w′′ +2z

z2 − 1w′ − ν(ν + 1)

z2 − 1)w = 0.

Weber : w′′ − zw′ − kw = 0.

Whittaker : w′′−( 1

4− k

z+

4m2 − 14z2

)w = 0.

Confluent Hypergeometrical : w′′−(

1− a

z

)w′ − b

zw = 0.

387

388

L21 : Galois group SL2, equivalent to Bessel equation. 2.10, 2.14, 2.15, 2.86,

2.94, 2.95, 2.102, 2.103, 2.104, 2.105, 2.106, 2.123, 2.130, 2.149, 2.155,2.161, 2.162, 2.164, 2.165, 2.167, 2.169, 2.170, 2.172, 2.173, 2.180, 2.185,2.189, 2.197, 2.200, 2.206, 2.216, 2.272, 2.274, 2.347.

L21 : Galois group SL2, equivalent to Hypergeometrical equation. 2.12, 2.252,

2.258, 2.265, 2.287, 2.291, 2.293, 2.294, 2.406.

L21 : Galois group SL2, equivalent to Legendre equation. 2.226, 2.231, 2.239,

2.240, 2.241, 2.244, 2.245, 2.249, 2.256, 2.269, 2.311, 2.313.

L21 : Galois group SL2, equivalent to Weber equation. 2.42, 2.44, 2.46, 2.87,

2.131, 2.132, 2.139.

L21 : Galois group SL2, equivalent to Whittaker equation. 2.16, 2.37, 2.114,

2.154, 2.195, 2.273.

L21 : Galois group SL2, equivalent to Confluent Hypergeometrical equation.

2.12, 2.92, 2.96, 2.107, 2.110, 2.113, 2.134, 2.138, 2.190.

L21 : Imprimitive Galois group Dm with m > 24, or D. 2.130, 2.135, 2.222,

2.288, 2.289.

L21 : Imprimitive Galois group D3. 2.290, 2.292.

L22 : 2.11, 2.13, 2.39, 2.40, 2.41, 2.43, 2.45, 2.50, 2.56, 2.57, 2.58, 2.59, 2.93,

2.109, 2.121, 2.122, 2.133, 2.136, 2.141, 2.158, 2.166, 2.181, 2.182, 2.192,2.194, 2.196, 2.198, 2.199, 2.208, 2.209, 2.223, 2.225, 2.230, 2.234, 2.237,2.238, 2.242, 2.246, 2.251, 2.253, 2.254, 2.257, 2.262, 2.263, 2.264, 2.267,2.270, 2.271, 2.280, 2.299, 2.300, 2.304, 2.308, 2.310, 2.324, 2.326, 2.319,2.321, 2.328, 2.332, 2.334, 2.336, 2.338, 2.345, 2.347a, 2.351, 2.355, 2.358,2.365, 2.378, 2.379, 2.384, 2.386, 2.387, 2.399.

L23 : 2.2,... 2.7, 2.35, 2.47, 2.49, 2.51, 2.53, 2.91, 2.98, 2.100, 2.101, 2.111,

2.112, 2.115, 2.119, 2.125, 2.126, 2.129, 2.148, 2.150, 2.151, 2.152, 2.160,2.165a, 2.176, 2.179, 2.191, 2.193, 2.201, 2.202, 2.203, 2.204, 2.211, 2.243,2.247, 2.255, 2.276, 2.277, 2.279, 2.282, 2.320, 2.331, 2.337, 2.342, 2.350,2.353, 2.354, 2.361, 2.370, 2.380, 2.400, 2.401, 2.404, 2.405.

L24, Galois group Z1: 2.1, 2.146, 2.147, 2.174, 2.183, 2.184, 2.229, 2.250,

2.281, 2.312, 2.322, 2.323, 2.340, 2.346, 2.366.

L24, Galois group Z2: 2.159, 2.168, 2.175, 2.186, 2.227, 2.229, 2.266, 2.284,

2.286, 2.391.

In the subsequent listing the number of the equation and its decompositiontype is given. If the equation is irreducible and it is equivalent to any ofthe known equations defining a special function, the transformation to thisequation is given; as usual D = d

dx.

Loewy Decompositions 389

2.1: y′′ = Lclm(D − 1

C + x

)y = 0. Type L2

4.

2.2,...,2.5: y′′ + y = Lclm(D − cotx,D + tanx)y = 0. Type L23.

2.6: y′′ − y = Lclm(D + 1, D − 1)y = 0. Type L23.

2.7: y′′ − 2y = Lclm(D +√

2, D −√

2)y = 0. Type L23.

2.8: y′′ + a2y = Lclm(D + ia,D − ia)y = 0. Type L23.

2.9: y′′ + ay = Lclm(D + i√a,D − i

√a)y = 0. Type L2

3.2.11: y′′ − (x2 + 1)y = (D + x)(D − x)y = 0. Type L2

2.2.13: y′′ − (a2x2 + a)y = Lclm(D + ax,D − ax)y = 0. Type L2

2.

2.35: y′′ + ay′ + by = Lclm(D+ 1

2a+√

14a

2 − b,D+ 12a−

√14a

2 − b)y = 0.

Type L23.

2.39: y′′ + xy′ + y = D(D + x)y = 0. Type L22.

2.40: y′′ + xy′ − y =(D + x+ 1

x)(D − 1

x)y = 0. Type L2

2.

2.41: y′′+xy′+(n+1)y =(D+ p′n

pn

)(D+x− p

′npn

)y = 0. Type L2

2; pn monicpolynomial of degree n; p1 = x, p2 = x2−1, p3 = x3−3x, p4 = x4−6x2+3.

2.42: y′′ + xy′ − ny = 0. Type L21. Weber equation, k = −n, x = iz, y = w.

2.43: y′′ − xy′ + 2y =(D − x+ 1

x+ 1 + 1x− 1

)(D − 1

x+ 1 −1

x− 1)y = 0.

Type L22.

2.44: y′′ − xy′ − ay = 0. Type L21. Weber equation.

2.45: y′′ − xy′ + (x− 1)y = (D − x+ 1)(D − 1)y = 0. Type L22.

2.46: y′′ − 2xy′ + ay = 0. Type L21. Weber equation, k = − 1

2a, x = 1√2z,

y = w.

2.47: y′′ + 4xy′ + (4x2 + 2)y = Lclm(D + 2x− 1

x,D + 2x)y = 0. Type L2

3.2.49: y′′ − 4xy′ + (4x2 − 1)y = Lclm(D − 2x− θ) = expx2, θ2 + 1 = 0.

Type L23.

2.50: y′′ − 4xy′ + (4x2 − 2)y = (D − 2x)(D − 2x)y = 0. Type L22.

2.51: y′′ − 4xy′ + (4x2 − 3)y = Lclm(D − 2x+ 1, D − 2x− 1)y = 0.Type L2

3.2.53: y′′ + 2axy′ + a2x2y = Lclm(D + ax− θ) = 0, θ2 − a = 0. Type L2

3.

2.56: y′′ − x2y′ + xy =(D − x2 + 1

x)(D − 1

x)y = 0. Type L2

2.2.57: y′′ − x2y′ − (x2 + 2x+ 1)y = (D + 1)(D − x2 − 1)y = 0. Type L2

2.2.58: y′′ − (x3 + x2)y′ + (x5 − 2x)y = (D − x3)(D − x2)y = 0. Type L2

2.

2.59: y′′ + x4y′ − x3y =(D + x4 + 1

x)(D − 1

x)y = 0. Type L2

2.

2.86: y′′ + 94xy = 0. Type L2

1. Bessel equation, ν = 13 , x = z2/3, y = z1/3w.

2.87: y′′ − 14 (x2 + a)y = 0. Type L2

1. Weber equation, k = − 14 (a+ 1), x = z,

y = exp(− 14z

2)w.

390

2.91: y′′ + y = Lclm(D − θ)y = 0, θ2 + 1 = 0. Type L23.

2.93: y′′ + 1xy′ =

(D + 1

x)Dy = 0. Type L2

2.

2.94: y′′ + 1xy′ + a

xy = 0. Type L21. Bessel equation, ν = 0, x = z2

4a , y = w.

2.95: y′′ + 1xy′ + λy = 0. Type L2

1. Bessel equation, ν = 0, x = z√λ

, y = w.

2.96: y′′+ 1xy′+

(1+ c

x)y = 0. Type L2

1. Confluent Hypergeometric equation,a = 1, b = 1

2 (1 + ic), x = 12 iz, y = exp (− 1

2z)w.

2.98: y′′ − 1xy′ − ax2y = Lclm(D − θx) = 0, θ2 − a = 0. Type L2

3.

2.100: y′′ + 2xy′ − y = Lclm

(D + 1 + 1

x,D − 1 + 1x

)y = 0. Type L2

3.

2.101: y′′ + 2xy′ + ay = Lclm

(D − θ + 1

x)

= 0, θ2 + a = 0. Type L23.

2.104: y′′ + αxy

′ + βxy = 0. Type L2

1. Bessel equation, ν = α− 1,

x = z2

4β , y =( z2

4β)(1−α)/2

w.

2.105: y′′ + axy′ + by = 0. Type L2

1. Bessel equation, ν = 1− a2 ,

x = − z√b, y =

(− z√

b

)νw.

2.106: y′′ + axy′ + bxα−1y. Type L2

1. Bessel equation, ν = 1− a1 + α ,

x =( (α+ 1)2z2

4b

)1/(α+1)

, y =( (α+ 1)2z2

4b

)(1−a)/(2α+2)

w.

2.109: y′′ − y′ − 1xy =

(D + 1

x)(D − 1− 1

x)y = 0. Type L2

2.

2.111: y′′ −(1 + 1

x)y′ + 1

xy = Lclm(D − 1, D − 1 + 1

x+ 1)y = 0. Type L2

3.

2.112: y′′ −(1 + 1

x)y′ −

(2− 2

x)y = Lclm

(D + 1− 1

x+ 13

, D − 2)y = 0.

Type L23.

2.115: y′′ − (3− 2x )y′ +

(2− 3

x)y = Lclm

(D − 1 + 1

x,D − 2 + 1x

)y = 0.

Type L23.

2.119: y′′− 2(ax+ b)x y′+ a(ax+ 2b)

x y = Lclm(D− a+ 2b+ 1

x ,D− a)y = 0.

Type L23.

2.121: y′′ − (x− 1)y′ +(1− 1

x)y =

(D− x+ 1 + 1

x )(D− 1x

)y = 0. Type L2

2.

2.122: y′′ −(x− 1− 2

x)y′ − (x+ 3)y =

(D+ 1 + 1

x)(D− x)y = 0. Type L2

2.

2.125: y′′ +(4x− 1

x)y′ − 4x2y = Lclm(D − 2αx,D − 2βx)y = 0 where

α+ β + 2 = 0, αβ + 1 = 0. Type L23.

2.126: y′′+(2ax2− 1

x)y′+(ax4 +ax)y = Lclm

(D+ax2− 2

x,D+ax2)y = 0.

Type L23.

2.129: y′′ − 4x− 9x− 3 y

′ + 3x− 6x− 3 y =


Lclm(D − 1, D − 3− 12x2 − 28x+ 50

4x3 − 42x2 + 150x− 183

)y = 0. Type L2

3.

2.131: y′′ − x− 12x y′ + a

2xy = 0. Type L21. Weber equation, k = −2a,

x = z2, y = w.

2.133: y′′ − 3x− 42x− 1y

′ + x− 32x− 1y =

(D − 1

2 +52

x− 12

)(D − 1)y = 0. Type L2

2.

2.134: y′′ − x+ c4x y = 0. Type L2

1. Confluent Hypergeometric equation,a = 2, b = 1 + c

4 , x = z, y = z exp(− 1

2z)w.

2.136: y′′ + 1xy′ − x+ 2

4x y =(D + 1

2 + 1x

)(D − 1

2

)y = 0. Type L2

2.

2.139: y′′ + 12xy

′ − x+ a16x y = 0. Type L2

1. Weber equation, k = − 14 (a+ 2)

x = z2, y = exp(− 1

4z2)w.

2.141: y′′ + 3a+ bxax y′ + 3b

axy =(D + 3

x)(D + b

a)y = 0. Type L2

2.

2.146: y′′ − 6x2 y = Lclm

(D − 5x4

x5 + C+ 2x

)y = 0. Type L2

4.

2.147: y′′ − 12x2 y = Lclm

(D − 7x6

x7 + C+ 3x

)y = 0. Type L2

4.

2.148: y′′ + ax2 y = Lclm

(D − θ

x)y = 0, θ2 − θ + a = 0. Type L2

3.

2.149: y′′ + ax+ bx2 y = 0. Type L2

1. Bessel equation, ν =√

1− 4b,

x = z2

4a , y = z2√aw.

2.150: y′′ +(1− 2

x2

)y = Lclm

(D + θ − 1

x+ θ+ 1x

)y = 0. Type L2

3.

2.151: y′′ −(a+ 2

x2

)y = Lclm

(D + a−

√ax + 1

x2 , D − a−√ax −

1x2

)y = 0.

Type L23.

2.152: y′′ +(1− 6

x2

)y = Lclm

(D − θ + 2

x −2x+ 3θ

x2 + 3θx− 3

)y = 0. Type L2

3.

2.153: y′′ +(a− ν(ν − 1)

x2

)y = 0. Type L2

1. Bessel equation, ν + 12 ,

x = − z√a, y = i

√z

a1/4w.

2.155: y′′ +(axk−2 − b(b− 1)

x2

)y = 0. Type L2

1. Bessel equation,

ν2 = 1k

√1 + 4b(b− 1), x =

(k2z2

4a

)1/k

, y =(k2z2

4a

)1/(2k)

w.

2.158: y′′ + ax2 y

′ −(b2 + ab

x2

)y = Lclm

(D + b+ a

x2 , D − b)y = 0. Type L2

2.

2.159: y′′ − 1xy′ − y = Lclm

(D − 1

x2 + C+ 1x

)y = 0. Type L2

4.

2.160: y′′ + 1xy′ + a

x2 y = Lclm(D − θ − 1

2x

)y = 0, θ2 + a+ 1

4 = 0. Type L23.

392

2.161: y′′ + 1xy′ −

( 1x + a

x2

)y = 0. Type L2

1. Bessel equation, ν = 2√a,

x = − 14z

2, y = w.

2.164: y′′ + 1xy′ +

(λ− ν2

x2

)y. Type L2

1. Bessel equation, x = z√λ

, y = w.

2.165: y′′ + 1xy′ + 4

(x2 − ν2

x2

)y = 0. Type L2

1. Bessel equation,

x =√z, y = w.

2.165a: y′′ +( 1x + a

x2

)y′ − y = Lclm

(D − 1

x + ax2 , D − 1 + 1

x+ a)y = 0.

Type L23.

2.166: y′′ − 1xy′ + 1

x2 y = D(D − 1

x)y = 0. Type L2

2.

2.167: y′′ − 1xy′ +

(axm−2 + b

x2

)y = 0 Type L2

1. Bessel equation,

ν2 = 4(1− b)m2 , x =

(m2z2

4a

)1/m

, y =(m2z2

4a

)1/m

w.

2.168: y′′ + 2xy′ = Lclm

(D − 1

x+ C + 1x

)= 0. Type L2

4.

2.169: y′′ + 2xy′ +

(ax −

b2

x2

)y = 0. Type L2

1. Bessel equation, ν2 = 1 + 4b2,

x = z2

4a , y = 2√awz .

2.170: y′′ + 2xy′ +

(a+ b

x2

)y = 0. Type L2

1. Bessel equation, ν2 = 14 − b,

x = z√a, y = a1/4w√

z.

2.174: y′′ − 2xy′ + 2

x2 y = Lclm(D + 1− 1

x −1

x+ C

)y = 0. Type L2

4.

2.175: y′′ − 2xy′ − 4

x2 y = Lclm(D + 1

x −5x4

x5 + C

)y = 0. Type L2

4.

2.176: y′′ − 2xy′ +

(1 + 2

x2

)y = Lclm

(D− i− 1

x,D− i+1x

)y = 0. Type L2

3.

2.179: y′′− 2xy′+

(a+ 2

x2

)y = Lclm

(D− θ− 1

x)y = 0, θ2 + a = 0. Type L2

3.

2.180: y′′ + 3xy′ +

(1 + 1− k2

x2

)y = f(x). Type L2

1. Bessel equation, ν = 0,

x = z, y = −wz .

2.181: y′′ + 3x− 1x2 y′ + 1

x2 y =(D + 2

x)(D + 1

x −1x2

)y = 0. Type L2

2.

2.182: y′′ − 3xy′ + 4

x2 y =(D − 1

x)(D − 2

x)y = 5

x . Type L22.

2.183: y′′ − 3xy′ − 5

x2 y = Lclm(D − 6x5

C + x6 + 1x

)y = x2 log x. Type L2

4.

2.184: y′′ − 4xy′ + 6

x2 y = Lclm(D + 1

C + x + 2x

)y = x2 − 1. Type L2

4.

2.185: y′′ + 5xy′ −

(2x− 4

x2

)y = 0. Type L2



x = − 12 (3z)2/3, y = −4(3z)2/3w

9z2 .

2.186: y′′ − 5xy′ + 8

x2 y = Lclm(D − 2

x −2x

x2 + C

)y = 0. Type L2

4.

2.189: y′′ + axy′ +

(bxm−2 + c

4x2

)y = 0. Type L2

1. Bessel equation,

ν = 1m

√(1− a)2 − 4c, x =

(m2z2

4b

)1/m

, y =(m2z2

4b

)(1−a)/(2m)

w.

2.191: y′′− y′− 2x2 y = Lclm

(D+ 1− 1

2(x+ 2) + 12x,D−

1x− 2 + 1

x

)y = 0.

Type L23.

2.192: y′′ +(1− 1

x2

)y′ − 1

x2 y =(D − 1

x2

)(D + 1)y = 0. Type L2

2.

2.193: y′′+(1+ 1

x)y′+

( 1x−

9x2

)y = Lclm

(D+1+ 3

x−3x2 + 18x+ 36

x3 + 9x2 + 36x+ 60,

D + 3x −

2x− 8x2 − 8x+ 20

)y = 0. Type L2

3.

2.194: y′′+(1+ 1

x)y′+(3

x−1x2 )y =

(D+ 1

x− 3 + 2x

)(D+1− 1

x− 3−1x

)y = 0.

Type L22.

2.196: y′′ −(1− 1

x)y′ +

( 1x −

1x2

)y =

(D − 1 + 2

x)(D − 1

x)y = 0. Type L2

2.

2.198: y′′ −(1− 2

x)y′ −

( 3x + 2

x2

)y =

(D + 3

x)(D − 1− 1

x)y = 0. Type L2

2.

2.199: y′′ − (1 + 4x )y′ − 4

x2 y = D(D − 1− 4x )y = 0. Type L2

2.

2.201: y′′ +(2 + 1

x)y′ − 4

x2 y =

Lclm(D − 2(x− 1)

x2 − 2x+ 32

, D + 2− 1x+ 3

2

+ 2x

)y = 0. Type L2

3.

2.202: y′′−(2+ 2

x)y′+

( 2x+ 2

x2

)y = Lclm

(D− 1

x,D−2− 1x

)y = 0. Type L2

3.

2.203: y′′ + ay′ − 2x2 y = Lclm

(D − a

ax− 2 + 1x,D + a+ a

ax+ 2 + 1x

)y = 0.

Type L23.

2.204: y′′ +(a+ 2b)y′ + (ab+ b2 − 2

x2

)y =

Lclm(D + a+ b− 1

ax+ 2 + 1x,D + ab− 2b

x + 1x

)y = 0. Type L2

3.

2.208: y′′ + xy′ +(1− 2

x2

)y =

(D + x− 1

x)(D + 1

x)y = 0. Type L2

2.

2.209: y′′ + x2 + 2x y′ + x2 − 2

x2 y = D(D + x+ 2

x)y = 0. Type L2

2.

2.211: y′′+4xy′+(4x2 +2+ 1x2 )y = Lclm(D+2x− θ

x )y = 0, θ2− θ+1 = 0.

Type L23.

2.223: y′′ + xx2 + 1

y′ − 9x2 + 1

y =(D + 2x

x2 + 34

+ xx2 + 1

+ 1x

)(D − 2x

x2 + 34

+ 1x

)y = 0. Type L2

2.

394

2.225: y′′+ xx2 + 1

y′+ 1x2 + 1

y =(D+ 1

x −x

x2 + 1)(D− 1

x)y = 0. Type L2

2.

2.227: y′′− 2xx2 + 1

y′+ 2x2 + 1

y = Lclm(D− 1

x−1 +

1x2

C + x− 1x

)y = 0. Type L2

4.

2.229: y′′+ 4xx2 + 1

y′+ 2x2 + 1

y = Lclm(D− 1

C + x+ 2xx2 + 1

)y = 2 cosx− 2x

x2 + 1.

Type L24.

2.230: y′′ + axx2 + 1

y′ + a− 1x2 + 1

y =(D + 2x

x2 + 1

)(D + (a− 2)x

x2 + 1

)y = 0.

Type L22.

2.234: y′′ +( 1

2x+ 1 +

12

x− 1

)y′ =

(D +

12

x+ 1 +12

x− 1

)Dy = 0. Type L2

2.

2.237: y′′ +( 1x+ 1 + 1

x− 1)y′ =

(D + 1

x+ 1 + 1x− 1

)Dy = 0. Type L2

2.

2.238: y′′ + 2xx2 − 1

y′ =(D + 2x

x2 − 1)Dy = 0. Type L2

2.

2.242: y′′ − 3x+ 1x2 − 1

− xx+ 1y =

(D + 2

x− 1 −1

x+ 1)(D + 1− 2

x+ 1)y = 0.

Type L22.

2.243: y′′+ 4xx2 − 1

+ x2 + 1x2 − 1

y = Lclm(D−θ+ 1

x+ 1 + 1x− 1

)y = 0, θ2+1 = 0.

Type L23.

2.247: y′′ + a( 1x+ 1 + 1

x− 1)y′ − a(a− 1)2 ( 1

x+ 1 −1

x− 1)y =

Lclm(D + a− 1x+ 1 , D + a− 1

x− 1)y = 0. Type L23.

2.250: y′′ +( 4x+ a + 4

x− a)y′ − 1

a( 4x+ a + 4

x− a)

=

Lclm(D − 12C(a2 + x2)x+ 9a4 + 54a2x2 + 9x4

C(a4 + 2a2x2 − x4) + 3x(3a4 − 2a2x2 − x4)

)y = 0. Type L2

4.

2.251: y′′ −( 2x+ 1 −

1x

)y =

(D − 2

x+ 1 + 1x− 1 + 1

x)(D − 1

x− 1)y = 0.

Type L22.

2.253: y′′+( 2x+ 1 + 1

x)y =

(D+ 1

x+ 1 + 1x

)(D− 1

x+ 1 + 1x

)y = 0. Type L2

2.

2.254: y′′ +(1− 2

x+ 2)y′ −

(6− 3

x+ 2 + 3x− 1

)y =(

D + 3− 2x+ 2 + 1

x− 1)(D − 2 + 1

x− 1)y = 0. Type L2

2.

2.255: y′′ + a( 1x− 1 −

1x

)y′ −

( 2x− 1 −

2x

)y =

Lclm(D + a− 1

x− 1 −a+ 1x ,D − 2x+ 2a− 1

x2 + ax− 2 + a(a− 1)/2

)y = 0.

Type L23.

2.256: y′′ + (2x− 1x(x− 1)y

′ − ν(ν + 1)x(x− 1)y = 0. Type L2

1. Legendre equation,

x = 1− z2 , y = w.


2.257: y′′ + (a+ 1)x+ bx(x− 1) y′ =

(D + a+ b+ 1

x− 1 − bx

)Dy = 0. Type L2

2.

2.262: y′′+x2 + x− 1(x− 1)2

y′− x+ 2(x− 1)2

y =(D− 1

x+ 1−1

x+ 2x+ 1)(D+1)y = 0.

Type L22.

2.263: y′′ + 3x− 1(x2 + 3x)2

y′ + 1(x2 + 3x)

y =(D + 2x+ 3

x2 + 3x

)(D +

73

x+ 3 −43x

)y = 0. Type L2

2.

2.264: y′′ + x2 + x+ 1x2 + 3x+ 4

y′ − 2x+ 3x2 + 3x+ 1

y =(D + 2x+ 1

x2 + x+ 3− 2x+ 3x3 + 3x+ 4

)(D − 2x+ 1

x2 + x+ 3

)y = 0. Type L2

2.

2.265: y′′ − 2x− 3(x− 1)(x− 3)y

′ + 1(x− 1)(x− 3)y = 0. Type L2

1.

Hypergeometric equation, α = − 12 (3 +

√5), β = − 1

2 (3−√

5). γ = −1,x = z + 1, y = w.

2.266: y′′ − 1x− 2y

′ − 3(x− 2)2

y = Lclm(D − 4(x− 1)3)

C + (x− 2)4− 1x− 2

)y = 0.

Type L24.

2.267: y′′ −12ν + x2 − 5

2

x2 y′ +2x− 1

2

x2 y =(D + 2

x)(D − 1 + 1

2x −ν

2x2

)y = 0.

Type L22.

2.269: y′′ + (2ν + 5)x− 2ν − 32x(x− 1) y′ + ν + 1

2x(x− 1)y = 0. Type L21. Legendre

equation, inverse of z = x+ 12√x

and w = x(ν+1)/2y.

2.270: y′′ + 5x2 + 21/2x+ 4x2 + 3x+ 2

y′ + 6x2 + 17/2x+ 4x2 + 3x+ 2

y =(D + 4x2 + 11x+ 4

2x2 + 6x+ 4

)(D + 3− 4

x+ 2

)y = 0. Type L2

2.

2.271: y′′ + 14x2 y =

(D + 1

2x)(D − 1

2x)y = 0. Type L2

2.

2.272: y′′ +(a2 + 1

4x4

)y = 0. Type L2


x = za , y =

(za)1/2

w.

2.274: y′′ + 1xy′ + x− ν2

4x2 y = 0. Type L21. Bessel equation, x = z2, y = w.

2.276: y′′ + 1xy′ − 4x2 + 1

4x2 y = Lclm(D + 1 + 1

2x,D − 1 + 12x

)y =√xex

x .

Type L23.

2.277: y′′ + 1xy′ − ax2 + 1

4x2 y = Lclm(D +

√a

2 −12x,D −

√a

2 + 12x

)y = 0.

Type L23.

2.279: y′′ + 54xy

′ − 14x2 y = Lclm

(D − θ

x)y = 0, θ2 + 1

4θ −14 = 0. Type L2

3.

396

2.280: y′′+ 2xy′− x

2 + 3x+ 34

x2 y =(D+1+ 5

2x)(D−1− 1

2x)y = 0. Type L2

2.

2.281: y′′ − 2x− 1x y′ − 4x2 − 4x− 1

4x2 y = Lclm(D − 2x− 1

2x − 1C + x

)y = 0.

Type L24.

2.282: y′′ + xy′ +(

14x

2 + 12 − 6

)y = Lclm

(D + 1

2x+ 2x,D + 1

x −3x

)y = 0.

Type L23.

2.284: y′′ − 22x+ 1y

′ − 12(2x+ 1)2

y =

Lclm(D − 1

x+ 12

+4x3 + 6x2 + 3x+ 1

2

C + x4 + 2x3 + 32x

2 + 12x

)y = 0. Type L2

4.

2.286: y′′ + 33x− 1y

′ − 9(3x− 1)2

y =

Lclm(D − 2x− 2

3

C + x2 − 23x

+ 1x− 1

3

)y = 0. Type L2

4.

2.287: y′′ + 3(2x− 1)9x(x− 1)y

′ − 209x(x− 1)y = 0. Type L2

1. Hypergeometric

equation, α = 13 , β = − 20

3 .

2.290: y′′ + 27x27x2 + 4

y′ − 327x2 + 4

y = 0. Type L21. Galois group D3.

2.299: y′′ + 2a2xa2x2 − 1

y′ =(D + a

ax+ 1 + aax− 1

)Dy = 0. Type L2

2.

2.300: y′′+ 2a2xa2x2 − 1

y′− 2a2

a2x2 − 1y =

(D+ a

ax+ 1+ aax− 1+1

x)(D− 1

x)y = 0.

Type L22.

2.304: y′′ + 1x2 y

′ − 2x+ 3x3 y =

(D − 1

x,D + 1x + 1

x2

)y = 0. Type L2

2.

2.307: y′′ + x+ 1x2 y′ − 2

x3 y = Lclm(D − 1

x+ 1 , D + 1x + 1

x2 − 1x+ 1

)y = 0.

Type L22.

2.308: y′′ − 1xy′ + 1

x2 y = D(D − 1

x)y = log (x)3

x3 . Type L22.

2.310: y′′ + 3xy′ + 1

x2 y =(D + 2

x)(D + 1

x)y = 1. Type L2

2.

2.311: y′′ + 2x2 + 1x(x2 + 1)

y′ − ν(ν + 1)xx2 + 1

y = 0. Type L21. Legendre equation,

x =√z2 − 1, y = w.

2.312: y′′ + 2x2 − 2x3 + x

y′ − 2x2 + 1

y = Lclm(D − 3x2

C + x3 + 2xx2 + 1)

)y = 0.

Type L24.

2.319: y′′− 1x3 + x

y′− 6x2 + 1

y =(D+ 2x

x2 + 1+ 1x

)(D− x

x2 + 1− 2x

)y = 0.

Type L22.

2.320: y′′ − x3 + 3x2 − 2x− 1x3 − 2x

y′ + x4 + 4x+ 2x3 − 2x

y =


Lclm(D − 1− 2

x,D −1

x− 1)y = 0. Type L2

3.

2.321: y′′ − 2x+ 1x2 + x

y′ + 2x+ 1x3 + x2 y =

(D − 1

x+ 1)(D − 1

x)

= 0. Type L22.

2.322: y′′ + 6x+ 4x2 + x

y′ + 6x+ 2x3 + x2 y = Lclm

(D − 1

x+ C + 3x+ 2x(x+ 1)

)y = 0.

Type L24.

2.323: y′′+ 2x− 1x2 − xy

′− 2x+ 2x3 − x2 y = Lclm

(D− 3x2 − 6x+ 3

C + x3 − 3x2 + 3x+ 2x

)y = 0.

Type L24.

2.324: y′′− 5x− 4x2 − xy

′+ 9x− 6x3 − x2 y =

(D− 1

x− 1 −1x

)(D− 3

x)y = 0. Type L2

2.

2.326: y′′ + 1x+ 1y

′ + 1x3 + 2x2 + x

y =(D + 1

x)(D + 1

x+ 1 −1x

)y = 0.

Type L22.

2.328: y′′−(

2x−

2x− 1 + 2

(x− 1)2)y =

(D+ 1

x−1

x− 1)(D− 1

x+ 1x− 1

)y = 0.

Type L22.

2.331: y′′ +( 1

2x− 2 −

1x

)y′ −

( 18

x− 2 −18x −

34

x2

)y =

Lclm(D −12

x− 2 + 12x,D −

12x )y = 0. Type L2

3.

2.332: y′′ − 1x+ 1y

′ + 3x+ 14x3 + 4x2 y =

(D − 1

x+ 1 + 12x

)(D − 1

2x)y = 0.

Type L22.

2.333: y′ + 3x− 12x(x− 1)y

′ − ν(ν + 1)4x2 y = 0. Type L2


x = 2z2 − 1 + 2z√z2 − 1, y = w.

2.334: y′′− 5x− 4x2 − xy

′+ 9x− 6x3 − x2 y =

(D+ 1

x− 1 −1x

)(D− 3

x)y = 0. Type L2

2.

2.336: y′′ − 3x− 4(x− 1)(2x− 1)2

y =(D + 1

x− 1 −12

x− 12

)(D − 1

x− 1 +12

x− 12

)y = 0. Type L2

2.

2.337: y′′ + 3x+ a+ 2b2(x+ a)(x+ b)y

′ + a− b4(x+ a)2(x+ b)2

y =

Lclm(D + 1

2(x+ a) , D −a− b

2(x+ a)(x+ b)

)y = 0. Type L2

3.

2.338: y′′ − 18x− 39x2 − 2x

y′ − 39x3 − 2x2 y =

Lclm(D + 1

x +2x− 1

3

x2 − 13x+ 1

54

, D + 32x +

32

x− 29

)y = 0.

Type L23.

2.340: y′′− 2ax+ bax2 + bx

y′+ 2ax+ 6bax3 + bx2 y = Lclm

(D− 1

x+ C + ax+ bax2 + bx

)y = 0.

Type L24.

398

2.342: y′′ + ax4 y = Lclm

(D + 1

x + i√a

x2 , D + 1x −

i√a

x2

)y = 0. Type L2

3.

2.345: y′′ + 1x3 y

′ − 2x4 y =

(D + 1

x)(D − 1

x + 1x3

)y = 0. Type L2

2.

2.346: y′′ − 2x2 y

′ + 2x+ 1x4 y = Lclm

(D − 1

x2 − 1x+ C

)y = 0. Type L2

4.

2.347: y′′ + 1xy′ + 1

x4 y = 0. Type L21. Bessel equation, ν = 0, x = 1

z , y = w.

2.347a: y′′ + 1xy′ + x− 1

x4 y =(D + 1

x + 1x2

)(D − 1

x2

)y = 0. Type L2

2.

2.350: y′′ + 2xy′ + a2

x4 y = Lclm(D − θ

x2

)y = 0, θ2 + a2 = 0. Type L2

3.

2.351: y′′ + 2x2 + 1x3 y′ − 1

x4 y =(D + 2

x)(D + 1

x3

)y = 0. Type L2

2.

2.353: y′′ −( 2x −

1x3

)y′ + 2

x4 y =(D + 1

x −4x3 + 4x

x4 + 2x2 − 1

)(D +

121x

3 − 237 x

x4 + 3x2 + 3+

27

x+ 1 +27

x− 1 + 3x

)y = 0. Type L2

2.

2.354: y′′ −( 2x −

1x3

)y′ + 1

x4 y =(D − 4

x + 1x3 + 2x

x2 − 15

)(D + 2

x −2

x2 − 15

)y = 0. Type L2

2.

2.355: y′′+( 4

3x−23

x2 − x+ 1+

23

x+ 1 −1x

)y′ − x

x3 + 1y =(

D + 2x− 1x2 − x+ 1

+ 1x+ 1 −

1x

)(D − x2

x3 + 1

)y = 0. Type L2

2.

2.358: y′′ − x2 − 2x3 − xy

′ − x2 − 2x4 − xy =

(D +

12

x+ 1 +12

x− 1 −1x

)(D − 1

x)y = 0.

Type L22.

2.359: y′′ + 2x3

x2(x2 − 1)y′ + ν(ν + 1)

x2(x2 − 1)y = 0. Type L2


x = 1z , y = w.

2.361: y′′ − 2xx2 − 1

y′ − a2(x2 − 1) + a(3x2 − 1)x4 − x2 y =

Lclm(D + a

x,D −a+ 1x − 2x

x2 − (a+ 32 )/(a+ 1

2 )

)y = 0. Type L2

3.

2.365: y′′ + a(x2 + 1)2

y = D(D − 1

x+ 1 −1

x− 1 + xx2 + 1

)y = 0. Type L2

2

2.366: y′′ + 2xx2 + 1

y′ + 1(x2 + 1)2

y = Lclm(D − 1

x+ C + xx2 + 1

)y = 0.

Type L24

2.370: y′′ + 2xx2 − 1

y′ − a2

(x2 − 1)2y =

Lclm(D + a

2( 1x+ 1 −

1x− 1

), D − a

2( 1x+ 1 −

1x− 1

))y = 0. Type L2

3.


2.378: y′′+ 2(x+ 1)x(x− 1)y

′− 2(x2 − x− 1)x2(x− 1)2

y =(D+ 3

x− 1)(D+( 1

x− 1−2x

)y = 0.

Type L22.

2.379: y′′ − 12(x+ 1)2(x2 + 2x+ 3)

y =(D − 4

(x+ 1)(x2 + 2x+ 3)

)(D + 4

(x+ 1)(x2 + 2x+ 3)

)y = 0. Type L2

2.

2.380: y′′+m(m− 1)a2

x2(x− a)2 y = Lclm(D+ m

x− a−m− 1x ,D+m− 1

x− a −mx

)y = 0.

Type L23.

2.384: y′′ − (ax− b)2 − 6bx− x2

4x2 y =(D − (ax− b)2 − 2bx− x2

(2a+ 1)x3 − 2bx2

)(D + a+ 1

2x − b2x2 − 1

x− b/(a+ 1)

)y = 0. Type L2

2.

2.386: y′′ − 18(2x+ 1)2(x2 + x+ 1)

y =(D − 2x+ 1

x2 + x+ 1+ 2x+ 1

2

)(D + 2x+ 1

x2 + x+ 1− 2x+ 1

2

)y = 0. Type L2

2.

2.387: y′′ −34

(x2 + x+ 1)2y =

(D +

x+ 12

x2 + x+ 1

)(D − x+ 1

2

x2 + x+ 1

)y = 0.

Type L22.

2.390: y′′ +316

x2(x− 1)2y = Lclm

(D −

34

x− 1 −14x ,D −

34x −

14

x− 1

)y = 0.

Type L23.

2.391: y′′ − 7ax2 + 5ax3 + x

y′ + 15ax2 + 5ax4 + x2 y =

Lclm(D − (8ax3(ax2 + 1))/(2ax2 + 1)2

C + (2ax4)/(2ax2 + 1)+ 6ax2 + 1x(2ax2 + 1

)y = 0. Type L2

4.

2.397: y′′ + 1x4 y

′ − 1x5 y =

(D − 1

x −1x4

)(D − 1

x)y = 0. Type L2

2.

2.399: y′′ − 3x+ 1x2 − 1

y′ + 12(x+ 1)2

(x− 1)2(x+ 53 )2

y =

(D − 1

x+ 1 −12

x− 1 +12

x+ 53

)(D −

32

x− 1 −12

x+ 53

)y = 0. Type L2

2.

2.400: y′′ − 1xy′ + a

x6 y = Lclm(D− 2

x +√a

x3 , D− 2x −√a

x3

)y = 0. Type L2

3.

2.401: y′′ + 3x2 + ax3 y′ − a2m(m− 1)

x6 y =

Lclm(D − a(m− 1)

x3 , D + amx3

)y = 0. Type L2

3.

2.404: y′′ + 2x2 + 1x3 y′ − 2x2 − 1

4x6 y =

Lclm(D + 1

x + 12x3 , D + 1

2x3

)y = 0. Type L2

3.

400

2.405: y′′ + 2x2 + 1x3 y′ − 8x4 + 10x2 + 1

4x6 y =

Lclm(D − 1

x −1

2x3 , D − 2x −

12x3

)y = 0. Type L2

3.

2.406: y′′ + 2716

x(x3 − 1)2

y = Lclm(D + 1

x + 12x3 , D + 1

2x3

)y = 0. Type L2

1.

Equations of Third Order. These equations are from Chapter 3 of Kamke’scollection. The tabulation below is organized by decomposition type L3

k,k = 1, . . . , 12 for third order equations as described in Corollary 2.4.

L31 : 3.6, 3.41, 3.54, 3.57, 3.77, L3

2 : 3.54, 3.76,

L33 : 3.73, L3

5 : 3.33, 3.58,

L36 : 3.45, L3

7 : 3.37, 3.38, 3.39, 3.56, 3.65, 3.68, 3.70,

L38 : 3.27, 3.46, 3.63, 3.74, 3.75, L3

9 : 3.1, 3.4, 3.5, 3.16, 3.17, 3.29, 3.47, 3.64,

3.66, 3.78,

L310 : 3.32, 3.35, 3.42, L3

12 : 3.18, 3.71.

Similar to equations of order two, in the listing below the decomposition ofequations of Chapter 3 is given if its type is different from L3

1.3.1: y′′′ + λy = Lclm(D − θ)y = 0, θ3 + λ = 0. Type L3

9.3.4: y′′′ + 3y′ − 4y = Lclm(D − 1, D − θ)y = 0, θ2 + θ + 4 = 0. Type L3

9.3.5: y′′′ − a2y′ = Lclm(D + a,D − a,D)y = 0. Type L3

9.3.16: y′′′ − 2y′′ − 3y′ + 10y = Lclm(D + 2, D − θ)y = 0, θ2 + 4θ + 5 = 0.

Type L39.

3.17: y′′′ − 2y′′ − a2y′ + 2a2y = Lclm(D + a,D − a,D − 2)y = 0. Type L39.

3.18: y′′′ − 3ay′′ + 3a2y′ − a3y = Lclm(D − a− 2x+ C1

x2 + C1x+ C2

)y = 0.

Type L312.

3.27: y′′′ − 2y′′ − 114 y

′ − 34y = (D− 3)Lclm

(D+ 1

2 −1

x+ C

)y = 0. Type L3

8.

3.29: y′′′ + 3xy′′ + y = Lclm

(D + 1 + 1

x,D − θ + 1x

)y = 0, θ2 − θ + 1 = 0.

Type L39.

3.32: y′′′ − x+ 2νx y′′ − x− 2ν − 1

x )y′ − x− 1x y =

Lclm(D2 − 2ν + 1

x D − 1, D − 1)y = 0. Type L3

10.

3.33: y′′′+(x− 3

x)y′′+4y′+ 2

xy = Lclm(D− 1

x+ C + 1x )(D+x− 5

x)y = 0.

Type L35.

3.35: y′′′ − 2(x+ ν − 2)x y′′ +

(x+

x+3ν− 5

2

x)y′ − ν − 1

2x y =

Lclm(D2 − x+ 2ν − 1

x D +ν − 1

2x ,D − 1

)y = 0. Type L3

10.


3.37: y′′′ − y′′ − 2x2 − 2x

y′ + 2x2 − 2x

y =(D + 1

x− 2 + 1x )Lclm(D − 1, D − 2

x)y = 0. Type L3

7.

3.38: y′′′− 8x2x− 1y

′+ 82x− 1y =

(D+2+ 1

x− 1/2)Lclm(D−2, D− 1x

)y = 0.

Type L37.

3.39: y′′′ + x+ 42x− 1y

′′ + 22x− 1y

′ =(D + 1

x + 1x− 1

2

)Lclm

(D + 1

2 +14

x− 12

, D)y = 0. Type L3

7.

3.42: y′′′ − 1xy′′ + (x+ 1

x )y′ = Lclm(D2 − 1

xD + 1, D)y = 0. Type L3

10.

3.45: y′′′+ 4xy′′+

(x+ 2

x)y′+ 3

xy =(D+ 3

x)(D2 + 1

xD+ 1)y = 0. Type L3

6.

3.46: y′′′ + 5xy′′ + 4

x2 y′ =

(D + 3

x)Lclm

(D − 1

x+ C + 1x

)y = 0. Type L3

8.

3.47: y′′′ + 6xy′′ + 6

x2 )y′ = Lclm(D + 2

x,D + 1x,D

)y = 0. Type L3

9.

3.54: y′′′− x3 − 6x y′′− 2x3 − 6

x2 y′+2y =(D2−(x2− 4

x )D+ 2x2

)(D+ 2

x)y = 0.

Type L32.

3.55: y′′′ + 8xx2 + 1

y′′ + 10x2 + 1

y′ =

Lclm(D + 4x

x2 + 1− 4x3 + 3C2x

2 + 4x+ 3C2

x4 + C2x3 + 2x2 + 3C2x+ C1

)y = 0. Type L3

12.

3.56: y′′′+ 2xx2 + 2

y′′+ y′+ 2xx2 + 2

y = Lclm(D− 2x,D+ θ)y = 0, θ2 + 1 = 0.

Type L39.

3.58: y′′′ +(14 + 7

2x −1

4x2

)y′′ +

( 1x + 1

x2

)y′ + 1

2x2 y =

Lclm(D + 1

C + x −2x

)(D + 1

4 −12x −

14x2

)y = 0. Type L3

5.

3.63: y′′′ + 3xy′′ − 2

x2 y′ + 2

x3 y =(D + 1

x)Lclm

(D − 3x2

C + x3 + 2x

)y = 0.

Type L38.

3.64: y′′′ + 3xy′′ − a2 − 1

x2 )y = Lclm(D + a

x,D −1x,D

)y = 0. Type L3

9.

3.65: y′′′ − 4xy′′ + (1 + 8

x2 )y = Lclm(D − 2

x,D + θ + 1x

)y = 0, θ2 + 1 = 0.

Type L39.

3.66: y′′′+ 6xy′′+(a− 12

x3 )y = Lclm(D− αx + 2

x2

)y = 0, α3 +a = 0. Type L3

9.

3.68: y′′′ + x+ 3x y′′ + 5x− 30

x2 y′ + 4x+ 30x3 y =

(D + 1

x + 2x− 60x2 − 60x+ 450

)Lclm

(D+ 6

x−4x3 − 252x2 + 4032x− 20160

x4 − 84x3 + 2016x2 − 20160x+ 75600, D+1+ 6

x−pq

)y = 0.

p := 8x7 + 196x6 + 2700x5 + 25500x4 + 171600x3 + 801360x2 + 2358720x+3326400, q := x8 + 28x7 + 450x6 + 5100x5 + 42900x4 + 267120x3

402

+1179360x2 + 3326400x+ 4536000. Type L37.

3.70: y′′′ +(

3xx2 + 1

+ 3x

)y′′ −

(12xx2 + 1

− 12x

)y =(

D + 2xx2 + 1

+ 3x

)Lclm

(D − 2x

x2 + 12

, D − xx2 + 1

+ 1x

)y = 0. Type L3

7.

3.71: y′′′ −(

1x+ 3 + 2

x

)y′′ −

( 43

x+ 3 −43x −

2x2 )y′ − (

23

x+ 3 −23x + 2

x2 )y =

Lclm(D − C1 + 2C2x+ 3x2

C1(x+ 1) + C2x2 + x3

)y = 0. Type L3

12.

3.73: y′′′−(

2x+ 1 + 2

x

)y′′−

(6

x+ 1 + 6x+ 4

x2

)y′−

(8

x+ 1−8x+ 8

x2 + 4x3

)y =(

D − 2x+ 1 + 1

x

)(D − 1

x )(D − 2x

)y = 0. Type L3

3.

3.74: y′′′ − 1xy′′ + 1

x2 y′ = DLclm

(D − 2x

C + x2

)y = 0. Type L3

8.

3.75: y′′′ −(

2x + 2x

x2 + 1

)y′′ +

(4x2 + 6

x2 + 1

)y′ −

(8x + 4

x3 − 8xx2 + 1

)y =(

D− 2xx2 + 1

+ 1x+ 1 + 1

x− 1

)Lclm

(D− 3Cx2 + C + 2x

Cx3 + Cx+ x2

)y = 0. Type L3

8.

3.76: y′′′+ 1x4 y

′′− 2x6 y

′ =(D2 +

( 2x + 1

x4

)D+ 2

x5

)(D− 2

x

)y = 0. Type L3

2.

3.78: y′′′ −(2− 4

x −1x2 + 4x3 + 4x+ 2

x4 + 2x2 + 2x+ 1

)y′′

+(1− 8

x −2x2 + 8x3 + 8x+ 4

x4 + 2x2 + 2x+ 1

)y′+

(4x + 1

x2 − 4x3 + 4x+ 2x4 + 2x2 + 2x+ 1

)y =

Lclm(D − 1, D − 1− 1

x,D + 1x2

)y = 0. Type L3

9.

Appendix E

Symmetries of Kamke’s Collection

In this appendix the symmetry classes of a large number of quasilinear sec-ond order, and linear and quasilinear equations of third order from Kamke’scollection are listed. Equations that are not rational in all arguments, containundetermined functions, more than two parameters or parameters in an expo-nent, are excluded. If there are parameters involved it is assumed that theyare unconstrained, any relations between them may change the symmetryclass.Equations of Second Order. These are the second order nonlinear equa-tions of Chapter 6. The following tabulation is organized by symmetry class;for each class the numbers of the equations which it contains are given.

Trivial : 6.3, 6.5, 6.6, 6.8, 6.9, 6.27, 6.108, 6.142, 6.144, 6.145, 6.147, 6.171,6.211, 6.212.

S21 : 6.4, 6.10, 6.11, 6.21, 6.26, 6.31, 6.40, 6.45, 6.47, 6.50, 6.56, 6.73,

6.74, 6.79, 6.80, 6.82, 6.86, 6.87, 6.89, 6.90, 6.92, 6.94, 6.96, 6.98,6.105, 6.106, 6.118, 6.119, 6.120, 6.141, 6.143, 6.153, 6.156, 6.160,6.172, 6.189, 6.190, 6.219, 6.226, 6.233, 6.240, 6.245.

S22,1 : 6.227, 6.228, 6.229, 6.232, 6.239, 6.243.S2

2,2 : 6.1, 6.2, 6.7, 6.12, 6.30, 6.32, 6.42, 6.43, 6.57, 6.78, 6.97, 6.104,6.109, 6.110, 6.111, 6.127, 6.130, 6.137, 6.140, 6.146, 6.154, 6.155,6.159, 6.174, 6.188, 6.205, 6.237.

S23,1 : none.S2

3,2 : 6.57, 6.81, 6.133, 6.138, 6.158, 6.162, 6.163, 6.183, 6.209, 6.244S2

3,3 : 6.71.S2

3,4 : none.S2

8 : 6.93, 6.99, 6.107, 6.117, 6.124, 6.125, 6.126, 6.128, 6.134, 6.135,6.150, 6.151, 6.157, 6.164, 6.168, 6.169, 6.173, 6.175, 6.176, 6.178,6.179, 6.180, 6.181, 6.182, 6.184, 6.185, 6.186, 6.191, 6.192, 6.193,6.194, 6.195, 6.206, 6.208, 6.210, 6.214.

In the subsequent listing the number of the equation, its symmetry classand, if it is different from S2

8 , its symmetry generators are given. When-ever possible, the solution is given explicitly or it is indicated what type oftranscendentals for its representation are required.

403

404

6.1: y′′ − y2 = 0. Symmetry class S22,2. Generators ∂x, x∂x − 2y∂y.

Solution: x =∫

dy√23y

3 − C1

+ C2.

6.2: y′′ − 6y2 = 0. Symmetry class S22,2. Generators ∂x, x∂x − 2y∂y.

Solution: Weierstraß’ P-function.6.3: y′′ − 6y2 − x = 0. Trivial symmetry. Solution: Painleve’s Transcendent.6.4: y′′ − 6y2 + 4y = 0. Symmetry class S2

1 . Generator ∂x.Solution: Elliptic integral.

6.5: y′′+ ay2 + bx+ c = 0. Trivial symmetry. Solution: Painleve’s Transcen-dent.6.6: y′′− 2y3−xy+ a = 0. Trivial symmetry. Solution: Painleve’s Transcen-dent.6.7: y′′ − ay3 = 0. Symmetry class S2

2,2. Generators ∂x, x∂x − y∂y.Solution: Elliptic integral.

6.8: y′′−2ay3+2abxy−b = 0. Trivial symmetry. Reduction to y′+ay2−bx =0.6.9: y′′ + ay3 + bxy + cy + d = 0. Trivial symmetry. Solution: Painleve’sTranscendent.6.10: y′′ + ay3 + by2 + cy + d = 0. Symmetry class S2

1 . Generator ∂x.

Solution: x = −∫

dy√12ay

4 + 23by

3 + cy2 + 2dy + C1

+ C2.

6.11: y′′ + axνyn = 0. Symmetry class S21 . Generator ∂x − ν + 1

n− 1∂y.

6.12: y′′ + (n+ 1)a2ny2n+1 = 0. Symmetry class S22,2.

Generators∂x, x∂x − 1

ny∂y. Solution: x =

∫dy

y√

(ay)2n − 1 + C1

+ C2.

6.21: y′′ − 3y′ − y2 − 2y = 0. Symmetry class S21 . Generator ∂x.

6.23: y′′ + 5ay′ + 6ya2 − 6y2 = 0. Symmetry class S22,2.

Generators ∂x, eax(∂x − 2ay∂y.Solution: y = a2C2

1e−2axP(C1e

−ax + C2, 0,−1).6.24: y′′ + 3ay′ + 2a2y − 2y3 = 0. Symmetry class S2

2,2.Generators ∂x, eax(∂x − ay∂y).Solution: y = −iaC1e

−axsnk2=−1(C1e−ax + C2).

6.26: y′′ + ay′ + byn + a2 − 14 y = 0. Symmetry class S2

1 . Generator ∂x.

6.27: y′′ + ay′ + bxνyn = 0. Trivial symmetry.6.30: y′′ + yy′ − y3 = 0. Symmetry class S2

2,2. Generators ∂x, x∂x − y∂y).

Solution: y = C1P ′(u, 0, 1)P(u, 0, 1) where u = C1x+ C2.

6.31: y′′ + yy′ − y3 + ay = 0. Symmetry class S21 . Generator ∂x.

Symmetries of Kamke’s Collection 405

Solution: 12

√a3P ′(u, 12, C1)P(u, 12, C1)− 1 with u = x

2

√a3 + C2.

6.32: y′′ + yy′ + 2ay′ − y3 + ay2 + 2a2y = 0. Symmetry class S22,2.

Generators ∂x, eax(∂x − ay∂y).

Solution: y = C1e−axP ′(u, 0, 1)P(u, 0, 1) where u = C1

a e−ax + C2, a 6= 0.

6.40: y′′ − 3yy′ − (3ay2 + 4a2y + b) = 0. Symmetry class S21 . Generator ∂x.

6.42: y′′ − 2ayy′ = 0. Symmetry class S22,2. Generators ∂x, x∂x − y∂y.

6.43: y′′+ ayy′+ by3 = 0. Symmetry class S22,2. Generators ∂x, x∂x− y∂y.

Solution: x =∫

dy

uy2 log y+C2, u is defined by

∫udu

2u2 + au+ b+log y = C1.

6.45: y′′ + ay′2 + by = 0. Symmetry class S21 . Generator ∂x.

Solution: x = 2a2

∫dy

2a2C1e−2ay + b(1− 2ay)

+ C2.

6.47: y′′ + ay′2 + by′ + cy = 0. Symmetry class S21 . Generator ∂x.

Reduction to Abel’s equation.6.50: y′′ + ayy′2 + by = 0. Symmetry class S2

1 . Generator ∂x.

Solution: x =∫

dy√C1 exp (−ay2)− b/a

+ C2.

6.56: y′′ + ay(y′ + 1)2 = 0. Symmetry class S21 . Generator ∂x.

Solution: x =∫ √

ay2 + C1

1− ay2 − C2

dy + C2.

6.57 y′′ − a(xy′ − y)ν = 0. ν = 1: Symmetry class S28 .

Solution: y = C1x+ C2x

∫exp

(12x

2)dxx

.

ν = 3: Symmetry class S23,2. Generators y∂x, x∂y, x∂x − y∂y.

Solution: (C21C

22 + a)y2 − 2C2

1C2xy + C22x

2 − C1 = 0.ν = 2 or ν ≥ 4: Symmetry class S2

2,2. Generators x∂y, x∂x − 2ν − 1y∂y.

Solution: Parameter representation.6.58: y′′ − kxayby′c = 0. Symmetry class S2

1 . Generator ∂x.6.71: 8y′′ + 9y′4 = 0. Symmetry class S2

3,2. Generators ∂x, ∂y, x∂x + 23y∂y.

Solution: (C1 + y)3 = (C2 + x)2.

6.73: xy′′ + 2y′ − xyn = 0. Symmetry class S21 . Generator x∂x − 2

n− 1y∂y.

6.74: xy′′ + 2y′2 + axνyn = 0. Symmetry class S28 for n = 1; symmetry

class S21 for n ≥ 2, ν 6= n, generator x∂x − ν + 1

n− 1y∂y; symmetry class S22

for ν = n ≥ 2, additional generator ∂x − yx∂y.

6.78: xy′′ + yy′ − y′ = 0. Symmetry class S22,2.

Generators x∂x, x log x∂x+(y−2)∂y. Solution: y = 2−2 tan(C1(log x− C2)

).

6.79: xy′′ − x2y′2 + 2y′ + y2 = 0. Symmetry class S21 . Generator x∂x − y∂y.

406

Solution: x = C1 exp∫

dη

C2eη + 2η + 1

with η = xy.

6.80: xy′′ + a(xy′ − y)2 = b. Symmetry class S21 . Generator x∂y.

6.81: 2xy′′ + y′3 + y′ = 0. Symmetry class S23,2.

Generators ∂y, x∂x + y∂y, xy∂x + 12y

2∂y.Solution: (y + C1)2 = 2C2x− C2

2 .6.82: xy′′ = a(yn − y). Symmetry class S2

1 . Generator x∂x.6.86: x2y′′ + a(xy′ − y)2 = bx2. Symmetry class S2

1 . Generator x∂y.6.87: x2y′′ + ayy′2 + bx = 0. Symmetry class S2

1 . Generator x∂x + y∂y.6.89: y′′ + xy′′ + y′2 = 0. Symmetry class S2

1 . Generator ∂y.Solution: y = C1 + C2x+ (C2

2 ) log x− C2.6.90: 4x2y′′ − x4y′2 + 4y = 0. Symmetry class S2

1 . Generator x∂x − 2y∂y.6.91: 9x2y′′ + ay3 + 2y = 0. Symmetry class S2

2,2.Generators

x∂x, 3x2/3∂x− y

x1/3 ∂y. Solution in terms of elliptic functions.

6.92: 9x3(y′′ + yy′ − y3) + 12xy + 24 = 0. Symmetry class S21 .

Generator x∂x − y∂y. Solution: Parameter representation.6.93: 9x2y′′−a(xy′−y)2 = 0. Symmetry class S2

8 . Solution: y = xa log x

C1x+ C2.

6.94: 2x2y′′ + (2x3y + 9x2)y′ − 2x3y3 + 3x2y2 + axy + b = 0.Symmetry class S2

1 . Generator x∂x − y∂y.

6.96: x4y′′ + a2yν = 0. Symmetry class S21 . Generator x∂x + 2

ν − 1y∂y.

6.97: x4y′′ − 2xyy′ − x3y′ + 4y2 = 0. Symmetry class S22,2.

Generators x∂x + 2y∂y, x log x∂x + (2y log x+ x2 − y)∂y.Solution: y = [C2 tan(C1 log x+ C2)]x2.

6.98: x4y′′ − x2y′2 − x3y′ + 4y2 = 0. Symmetry class S21 .

Generator x∂x+2y∂y. Solution: x = exp∫

dy

x2(C1 exp (y

x2 )− 4y

x2 − 2+C2.

6.99: x4y′′ + (xy′ − y)3 = 0. Symmetry class S28 .

Solution: y = C1x+ x arcsin C2x .

6.104: yy′′ − a = 0. Symmetry class S22,2. Generators ∂x, x∂x + y∂y.

Solution: x =∫

dy√2a log y + C1

.

6.105: yy′′ − ax = 0. Symmetry class S21 . Generator x∂x + 3

2y∂y.6.106: yy′′ − ax2 = 0. Symmetry class S2

1 . Generator x∂x + 2y∂y.6.107: yy′′+y′2−a = 0. Symmetry class S2

8 . Solution y =√ax2 + C1x+ C2.

6.108: yy′′ + y′2 = ax+ b. Trivial symmetry.6.109: yy′′ + y′2 − y′ = 0. Symmetry class S2

2,2.Generators ∂x, x∂x + y∂y. Solution: x = y + C1 log y − C1 + C2.


6.110: yy′′ − y′2 + 1 = 0. Symmetry class S22,2.

Generators ∂x, x∂x + y∂y. Solution: C1y = sin (C1x) + C2.6.111: yy′′ − y′2 − 1 = 0. Symmetry class S2

2,2.Generators ∂x, x∂x + y∂y. Solution: C1y = cosh (C1x) + C2.

6.117: y′′y − y′2 + ay′y + by2 = 0. Symmetry class S28 .

Solution y = exp (C1e−ax − b

ax+ C2).6.118: y′′y− y′2 + ay′y+ by3 − 2ay2 = 0. Symmetry class S2

1 . Generator ∂x.6.119: y′′y − y′2 + (ay − 1)y′ − 2b2y3 + 2a2y2 + ay = 0. Symmetry class S2

1 .Generator ∂x.

6.120: yy′′ − y′2 + a(y − 1)y′ − y(y + 1)(b2y2 − a2) = 0. Symmetry class S21 .

Generator ∂x.6.124: y′′y − 3y′2 + 3y′y − y2 = 0. Symmetry class S2

8 .Solution: (C1e

x + 1)y2 + C2e2x = 0.

6.125: y′′y − ay′2 = 0. Symmetry class S28 .

Solution: y = (C1x+ C2)1/(1−a) for a 6= 1, y = C1eC2x for a = 1.

6.126: yy′′ + ay′2 + a = 0. Symmetry class S22,2.

Generators ∂x, x∂x + y∂y. Solution: x =∫

dy√C1y−2a − 1

+ C2.

6.127: yy′′ + ay′2 + by3 = 0. Symmetry class S22,2

Generators ∂x, x∂x − 2y∂y. Solution: Parameter representation.6.128: yy′′ + ay′2 + byy′ + cy2 + dy1−a = 0. Symmetry class S2

8 .6.130: yy′′ + ay′2 + by2y′ + cy4 = 0. Symmetry class S2

2,2.Generators ∂x, x∂x − y∂y.

6.133: y′′(y + x) + y′2 − y′ = 0. Symmetry class S23,2.

Generators∂x−∂y, x∂x+ y∂y, (x2− 2

3xy−13y

2)∂x+( 13x

2 + 23xy− y

2)∂y.

Solution: y2 − (3C2 + 94C1 + 2x)y + 9

4C22 + 3C2x− 9

4C1x+ 8164C1 + x2 = 0.

6.134: yy′′ − xy′′ − 2y′2 − 2y′ = 0. Symmetry class S28 .

Solution y = C1 + C2x− C1

.

6.135: yy′′ − xy′′ + y′3 + y′2 + y′ + 1 = 0. Symmetry class S28 .

Solution: (y + C1)2 = C22 − (x+ C1)2.

6.137: 2yy′′+ y′2 + 1 = 0. Symmetry class S22,2. Generators ∂x, x∂x + y∂y.

Solution: x = C1(t− sin t) + C2, y = C1(1− cos t).6.138: 2yy′′ − y′2 + a = 0. Symmetry class S2

3,2.Generators ∂x, x∂x + y∂y, x

2∂x + 2xy∂y.Solution: C1y = ( 1

2C1x+ C2)2 − a.6.140: 2y′′y − y′2 − 8y3 = 0. Symmetry class S2

2,2.

Generators ∂x, x∂x − 2y∂y. Solution: x+ 12

∫dy√

y(y2 − C1)= C2.

6.141: 2y′′y − y′2 − 8y3 − 4y2 = 0. Symmetry class S21 . Generator ∂x.

408

Solution: Elliptic functions.

6.142: 2y′′y − y′2 − 8y3 − 4xy2 = 0. Trivial symmetry.

6.143: 2y′′y − y′2 + ay3 + by2 = 0. Symmetry class S21 . Generator ∂x.

6.144: 2y′′y − y′2 + ay3 + 2xy2 + 1 = 0. Trivial symmetry.

6.145: 2y′′y − y′2 + ay3 + bxy2 = 0. Trivial symmetry.

6.146: 2yy′′ − y′2 + 3y4 = 0. Symmetry class S22,2.

Generators ∂x, x∂x − y∂y Solution: Elliptic function.

6.147: 2yy′′ − y′2 − 3y4 − 8xy3 − 4(x2 + a)y2 + b = 0. Trivial symmetry.Solution: Painleve’s Transcendent.

6.150: 2yy′′ − 3y′2 = 0. Symmetry class S28 . Solution: y = C1

x+ C2.

6.151: 2yy′′−3y′2−4y2 = 0. Symmetry class S28 . Solution: y = C1

cos (x+ C2)2 .

6.153: 2yy′′ − 6y′2 + ay5 + y2 = 0. Symmetry class S21 . Generator ∂x

Solution: x =∫

dy

y√C1y4 + ay3 + 1

4

+ C2.

6.154: 2yy′′−y′4−y′2 = 0. Symmetry class S22,2. Generators ∂x, x∂x+y∂y

Solution: x = C1t+ C2 sin t+ C2, y = C1(1− cos t).

6.155: 4yy′′ − 4ay′′ + 2y′2 + 2 = 0. Symmetry class S22,2.

Generators ∂x, x∂x + (y − a)∂y.

Solution: x = 12√

(y − a+ C2)(a− y)− C2 arctan√

a− yy − a+ C2

+ C1.

6.156: 3yy′′ − 2y′2 = ax2 + bx+ c. Symmetry class S21 .

Generator (ax2 + bx+ c)∂x +(3a+ 3b

2a)∂y.

6.157: 3yy′′ − 5y′2 = 0. Symmetry class S28 . Solution: y = 1

(C1x+ C2)3/2.

6.158: 4yy′′ − 3y′2 + 4y = 0. Symmetry class S23,2.

Generators ∂x, x∂x+2y∂y, x2∂x+4xy∂y. Solution x = 12

∫dy√

C1y +√y

+ C2.

6.159: 4yy′′ − 3y′2 − 12y3 = 0. Symmetry class S22,2.

Generators ∂x, x∂x − 2y∂y. Solution x =∫ √

C1 − 4y3

y√y

dy + C2.

6.160: 4yy′′ − 3y′2 + ay3 + by2 + cy = 0. Symmetry class S21 . Generator ∂x

6.162: 4yy′′ − 5y′2 + ay3 = 0.Symmetry class S2

3,2. Generators ∂x, x∂x − 2y∂y, x2∂x − 4xy∂y.

Solution: 2C21C2y

14 + (C2

1C2 − C1x2 − 1

16a)√y + C2

1 = 0.

6.163: 12yy′′ − 15y′2 + 8y3 = 0. Symmetry class S23,2.

Generators ∂x, x∂x − 2y∂y, x2∂x − 4xy∂y.


Solution: y = 6C2

[(C1 + x)2 + C2]2.

6.164: nyy′′−(n−1)y′2 = 0. Symmetry class S28 . Solution: y = (C1x+C2)n.

6.168: (ay + b)y′′ + cy′2 = 0. Symmetry class S28 .

Solution: y = (C1x+C2)a/(a+c)− ba if a+c 6= 0, y = C1eC2x− ba if a+c = 0.

6.169: xyy′′+xy′2−yy′ = 0. Symmetry class S28 . Solution: y =

√1

C1x2 + C2

.

6.171: xyy′′ − xy′2 + yy′ + axy4 + by3 + cy + d = 0. Trivial symmetry.6.172: xyy′′ − xy′2 + ayy′ + bxy3 = 0. Symmetry class S2

1 .Generator x∂x − 2y∂y.

6.173: xyy′′ + 2xy′2 + ayy′ = 0. Symmetry class S28 .

Solution: y =√

1C1x

1−a + C2.

6.174: xyy′′ − 2xy′2 + yy′ + y′ = 0. Symmetry class S22,2.

Generators x∂x, x log x∂x + y∂y.6.175: xyy′′ − 2xy′2 + ayy′ = 0. Symmetry class S2

8 .Solution: y = 1

C1x1−a + C2

for a 6= 1, y = 1C1 log x+ C2

for a = 1.

6.176: xyy′′ − 4xy′2 + 5yy′ + y′ = 0. Symmetry class S28 .

Solution: y = x(C1x

3 + C2)1/3.

6.178: x(y + x)y′′ + xy′2 − (y − x)y′ − y = 0. Symmetry class S28 .

Solution: y =√C1x2 + C2 − x.

6.179: 2y′′yx− y′2x+ y′y = 0. Symmetry class S28 .

Solution: y = C1(√x+ C2)2.

6.180: x2(y − 1)y′′ − 2x2y′2 − 2x(y − 1)y′ − 2y(y − 1)2 = 0.

Symmetry class S28 . Solution: y = C1x

2 + C2xC1x

2 + C2x− 1.

6.181: x2(y + x)y′′ − (xy′ − y)2 = 0. Symmetry class S28 .

Solution: y = C1x exp(C2x

)− x.

6.182: x2(y − x)y′′ − a(xy′ − y)2 = 0. Symmetry class S28 .

Solution: y = C1x( xC2x+ a− 1

)1/(a−1) + x.

6.183: 2x2yy′′ − x2y′2 + y2 − x2 = 0. Symmetry class S23,2. Generators

x∂x+y∂y, x log (x)∂x+(log (x)+1

)y∂y, x log (x)2∂x+

(log (x)+2

)log (x)y∂y.

Solution: y = x(C1 +

√4C1C2 − 1 log x+ C2 log x2

).

6.184: ax2yy′′ + bx2y′2 + cxyy′ + dy2 = 0 Symmetry class S28 .

6.185: x(x+ 1)2yy′′ − x(x+ 1)2y′ + 2(x+ 1)2yy′ − ax(x+ 2)y2 = 0.Symmetry class S2

8 . Solution: y = C1(x+ 1)a exp(C2x

).

6.186: 8(x3 − 1)yy′′ − 4(x3 − 1)y′2 + 12x2yy′ − 3x(x+ 2)y2 = 0Symmetry class S2

8 . Solution in terms of hypergeometric functions.

410

6.188: y2y′′ − a = 0. Symmetry class S22,2. Generators ∂x, x∂x + 2

3y∂y.

Solution x =∫ √

ydy√C1y − 2a

+ C2.

6.189: y2y′′yy′2 + ax = 0. Symmetry class S21 . Generator x∂x + y∂y.

6.190: y2y′′+yyy′2 = ax+b. Symmetry class S21 . Generator

(x+ b

a)∂x+y∂y.

First integral: y3y′3 − 3y2y′(ax+ b) + ay3 + 1a (ax+ b)3 = C.

6.191: y′′(y2 + 1)− y′2(2y − 1) = 0. Symmetry class S28 .

Solution: y = tan log(C1x+ C2)6.192: (y2 + 1)y′′ − 3yy′2 = 0. Symmetry class S2

8 .Solution: y = C1x+ C2√

C1x+ C2)2 − 16.193: (y2 + x)y′′ + 2(y2 − x)y′3 + 4yy′2 + y′ = 0. Symmetry class S2

8 .Solution: y = C1e

C2y − x.6.194: (y2 + x)y′′ − (y′2 + 1)(xy′ − y) = 0. Symmetry class S2

8 .Solution: y = C1 exp arctan

(C2

yx

).

6.195: (y2 + x2)y′′ − 2(y′2 + 1)(xy′ − y) = 0. Symmetry class S28 .

Solution: y2 + C1y + C2x+ x2 = 0.6.205: xy2y′′ − a = 0.

Symmetry class S22,2. Generators x∂x + 1

3y∂y, x2∂x + xy∂y.

Solution: Parameter representation.6.206: (x2 − a2)(y2 − a2)y′′ − (x2 − a2)yy′2 + x(y2 − a2)y′ = 0.

Symmetry class S28 . Solution

√y2 − a2 + y = C1(

√x2 − a2 + x) + C2.

6.208: x3y2y′′ + (xy′ − y)3(y + x) = 0. Symmetry class S28 .

6.209: y3y′′ − a = 0.Symmetry class S2

3,2. Generators ∂x, x∂x + 12y∂y, x

2∂x + xy∂y.Solution: (C2x− C2)2 + C1y

2 + a = 0.6.210: (y3 + y)y′′ − (3y2 − 1)y′2 = 0. Symmetry class S2

8 .

Solution: y =√

1− C1x− C2C1x+ C2

.

6.211: 2y3y′′ + y4 − a2xy2 = 1. Trivial symmetry.6.212: 2y3y′′ + y2y′2 = ax2 + bx+ c. Trivial symmetry.6.214: (4y3−g2y−g3)y′′−(6y2− 1

2g2)y′2 = 0; g2 and g3 constant. Symmetry

class S28 . Solution: y = P(C1x+ C2, g2, g3); P Weierstraß’ P-function.

6.219: (y2 + ax2 + 2bx+ c2)y′′ + dy = 0. Symmetry class S21 .

Generator (ax2 + 2bx+ c)∂x + (xy + bay)∂y.

6.226: y′y′′ − x2yy′ − xy2 = 0. Symmetry class S21 . Generator y∂y.

6.227: xy′y′′ − yy′′ + 4y′2 = 0. Symmetry class S22,1. Generators x∂x, y∂y.

Solution: x = C1(t− 1)e2t, y = C2te−2t.

6.228: xy′y′′ − yy′′ − y′4 − 2y′2 − 1 = 0.


Symmetry class S22,1. Generators x∂x + y∂y, (x+ y)∂y.

6.229: ay′′y′x3 + by2 = 0. Symmetry class S22,1. Generators x∂x, y∂y.

6.231: (2y2y′ + x2)y′′ + 2yy′3 + 3xy′ + y = 0. Symmetry class S21 .

Generator x∂x + y∂y. First integral: y2y′2 + x2y′ + xy = C.6.232: y′′y′2 + y′′y2 + y3 = 0. Symmetry class S2

1 . Generators ∂x, y∂y.

Solution: log y = 12 log | sin (x

√3 + C1)| ±

∫ [(1 + 3

4 cot2 (x√

3 + C1)]1/2

dx+ C2.

6.233: (y′2 + a(xy′ − y))y′′ = b. Symmetry class S21 . Generator ∂x − 1

2ax∂y.6.236: y′′2 + ay + b = 0.

Symmetry class S22,2. Generators

∂x, x∂x + 4

(y + b

a)∂y

.

6.237: a2y′′2 − 2axy′′ + y′ = 0.Symmetry class S2

2,2. Generators ∂y, x∂x + 3y∂y.6.238: 2(x2 + 1)y′′2 − x(4y′ + y)y′′ + 2(y′ + x)y′ − 2y = 0. Trivial symmetry.6.239: 3x2y′′2 − 2(3xy′ + y)y′′ + 4y′2 = 0. Symmetry class S2

2,1.Generators x∂x, y∂y. Solution y = C2

1x2 + C1C2x+ C2

2 .6.240: (9x3 − 2x2)y′′2 − 6x(6x− 1)y′y′′ − 6y′y′′ + 36xy′2 = 0.

Symmetry class S21 . Generator y∂y. Solution y = C2

1x3 + C1C2x+ C2

2 .6.243: (a2y2 − b2)y′′2 − 2a2yy′2y′′ + (a2y′2 − 1)y′2 = 0. Symmetry class S2

2,1.

Generator ∂x. Solution y = C1 exp (C2x)± 1a

√(b2 + 1

C22

).

6.244: (x2yy′′ − x2y′2 + y2)2 = 4xy(xy′ − y)3. Symmetry class S23,2.

Generators∂x + y

x∂y, x∂x + y log xy ∂y, ∂y. Solution y = C1x exp 1

C2 − x .

6.245: (2yy′′ − y′2)3 + 32y′′(xy′′ − y′3)3 = 0. Symmetry class S21 .

Generator x∂x + y∂y. Solution y = 1C1C

32

((C2

1x+ 1)2 + 2C22

).

Nonlinear Equations of Third Order. These equations are from Chapter 7of Kamke’s collection.

S31 : 7.2, 7.5, 7.6. S3

2,1 : 7.1. S32,2 : 7.3, 7.4, 7.7.

S34,4 : 7.12 S3

6 : 7.10, 7.11. S37 : 7.8, 7.9.

7.1: y′′′ − a2(y′5 − 2y′3 − y′) = 0. Symmetry class S32,1. Generators ∂x, ∂y.

Solution in parameter representation:

x =1a

∫dp√

C1 + 13 (p2 + 1)3

+ C2, y =1a

∫pdp√

C1 + 13 (p2 + 1)3

+ C3.

7.2: y′′′ + yy′′ − y′2 = 0. Symmetry class S31 . Generator ∂x.

7.3: y′′′ − yy′′ + y′2 = 0. Symmetry class S32,2. Generators ∂x, x∂x − y∂y.

Reduction to Abel equation.

412

7.4: y′′′ + ayy′′ = 0. Symmetry class S32,2. Generators ∂x, x∂x − y∂y.

Reduction to Abel equation.7.5: x2y′′′ + xy′′ + (2xy − 1)y′ + y2 = 0. Symmetry class S3

1 . Generatorx∂x − y∂y.

7.6: x2y′′′ + xyy′′ − xy′′ + xy′2 − yy′ + y′ = 0. Symmetry class S31 = x∂x.

First integral: xy′′ + (y − 1)y′ = Cx.7.7: yy′′′− y′y′′+ y3y′ = 0. Symmetry class S3

2,2. Generator ∂x, x∂x− y∂y.

Solution: x =∫

dy√C1y2 + C2 − 1

4y4

+ C3.

7.8: 4y2y′′′ − 18yy′y′′ + 15y′3 = 0. Symmetry class S37 .

Solution: y = 1(C1x

2 + C2x+ C3)2.

7.9: 9y2y′′′ − 45yy′y′′ + 40y′3 = 0.Symmetry class S3

7 . Generators ∂x, x∂x, x2∂x − 3xy∂y.Solution: y2 = 1

(C1x2 + C2x+ C3)3

.

7.10: 2y′y′′′ − 3y′′2 = 0. Symmetry class S36 .

Generators ∂x, ∂y, x∂x, y∂y, x2∂x, y2∂y. Solution: y = C1x+ C2

C3x+ C4.

7.11: y′2y′′′ + y′′′ − 3y′y′′2 = 0. Symmetry class S36 . Generators

∂x, ∂y, x∂x + y∂y, y∂x − x∂y, xy∂x − 12 (x2 − y2)∂y, (x2 − y2∂x + 2xy∂y.

Solution: (y − C1)2 + (x− C2)2 = C23 .

7.12: y′2y′′′ + y′′′ − 3y′y′′2 − ay′′2 = 0. Symmetry class S34,4.

Generators ∂x, ∂y, x∂x + y∂y, y∂x − x∂y.Solution: x = C2 + C1e

−at(a cos t− sin t), y = C3 + C1e−at(a sin t+ cos t).

Linear Equations of Third Order. These equations are from Chapter 3 ofKamke’s collection. There occur the three symmetry classes S3

4,5, S35,1 and S3

7 ,and the parametrized class S3

5,2(a). The symmetry generators are not givenexplicitly because they are closely related to the elements of a fundamentalsystem (see Exercise 5.3). Whenever a closed form for the latter exists it isalso given. It is usually obtained by means of a Loewy decomposition.

S34,5 : 3.2, 3.7, 3.30, 3.31, 3.32, 3.33, 3.34, 3.35, 3.37, 3.38, 3.39, 3.40,

3.41,3.42, 3.49, 3.50, 3.51, 3.53, 3.54, 3.56, 3.58, 3.60, 3.65, 3.66,3.68, 3.70, 3.73, 3.75, 3.76, 3.77.

S35,1 : 3.27, 3.63, 3.74. S3

5,2(a) : 3.1, 3.4, 3.16, 3.29, 3.71.S3

7 : 3.6, 3.21, 3.45, 3.47, 3.48, 3.57, 3.61, 3.64.

3.1: y′′′ + λy = 0. Symmetry class S35,2(θ), θ

2 − θ + 1 = 0. Fundamentalsystem: exp (−θ1x), exp ( 1

2θ1x) cos 12θ1√

3x), exp ( 12θ1x) sin 1

2θ1√

3x),θ1 real root of θ3 − λ = 0.

3.2: y′′′ + ax3y + bx = 0. Symmetry class S34,5.


3.3: y′′′ = axby. Symmetry class S34,5.

3.4: y′′′+3y′−4y = 0. Symmetry class S35,2(θ), θ

2− θ+ 25 = 0. Fundamental

system: exp(x), exp (− 12x) cos ( 1

2

√15x), exp (− 1

2x) sin ( 12

√15x).

3.5: y′′′−a2y′ = 0. Symmetry class S37 . Fundamental system: 1, exp (±ax).

3.6: y′′′ + 2axy′ + ay = 0. Symmetry class S37 . Fundamental system:

u2, uv, v2, u and v fundamental system for w′′ + 12axw = 0.

3.7: y′′′ − x2y′′ + (a+ b− 1)y′ − aby = 0. Symmetry class S34,5.

3.16: y′′′ − 2y′′ − 3y′ + 10y = 0. Symmetry class S35,2(θ), θ

2 − 417θ + 4

17 = 0.Fundamental system: exp (−2x), cos(x) exp (2x), sin(x) exp (2x).

3.17: y′′′ − 2y′′ − a2y′ + 2a2y = 0. Symmetry class: S35,2(3) for a = 1,

S35,1 for a = 2, S3

5,2(6) for a = 3, S35,2(4) for a = 4.

Fundamental system: exp (2x), exp (±ax).3.18: y′′′ − 3ay′′ + 3a2y′ − a3y = 0. Symmetry class S3

5,2(θ), θ2 − θ + 1 = 0.

Fundamental system: exp (ax), x exp (ax), x2 exp (ax).3.20: y′′′−6xy′′−(4a+8x2−2)y′−8axy = 0. Symmetry class S3

7 . Fundamen-tal system: u2, uv, v2; u and v fundamental system of w′′−2xw+aw = 0.

3.21: y′′′ + 3axy′′ + 3a2x2y′ + a3x3y = 0. Symmetry class S37 . Solution:

y = exp (− 12ax

2)[C1 + C2 cos (x

√3a) + C3 sin (x

√3a)

].

3.27: 4y′′′ − 8y′′ − 11y′ − 3y = 0. Symmetry class S35,1.

Fundamental system: exp (3x), exp (− 12x), x exp (− 1

2x).3.29: xy′′′ + 3y′′ + xy = 0. Symmetry class S3

5,2(θ), θ2 − θ + 1 = 0.

Fundamental system: 1xe−x, 1

x exp ( 12 (1± i

√3)x)

.

3.30: xy′′′ + 3y′′ − ax2y = 0. Symmetry class S34,5.

3.31: xy′′′ + (a+ b)y′′ − xy′ − ay = 0. Symmetry class S34,5.

3.32: xy′′′ + (a+ b)y′′ − xy′ − ay = 0. Symmetry class S34,5.

Solution: y = C1ex + xν+1[C2Jν+1(ix) + C3Yν+1(ix)].

3.33: xy′′′− (x+ 2ν)y′′− (x− 2ν− 1)y′+ (x− 1)y = 0. Symmetry class S34,5.

Fundamental system:x5 exp (− 1

2x2), x5 exp (− 1

2x2)

∫exp ( 1

2x2)dxx6

.

3.34: xy′′′ + 32y′′ + 1

2axy = 0. Symmetry class S34,5.

3.35: xy′′′ − 2(x+ ν − 1)y′′ + (x+ 3ν − 52 )y′ − (ν − 1

2 )y = 0.Symmetry class S3

4,5.Solution: y = C1e

x + xν exp ( 12x)

[C2Jν

(12 ix

)+ C3Yν

(12 ix

)].

3.37: (x− 2)xy′′′ − (x− 2)xy′′ − 2y′ + 2y = 0. Symmetry class S34,5.

Fundamental system:x2, ex, ex

∫e−x

[x(x− 2) log x

x− 2 − 2(x− 1)]dx

.

3.38: (2x−1)y′′′−8xy′+8y = 0. Symmetry class S34,5. Fundamental system:

exp (2x), x exp (2x), exp (2x)∫x exp (−4x)dx4x(x− 1) + 1

− x∫

exp (−2x)dx4x(x− 1) + 1

.

3.39: (2x− 1)y′′′ + (x+ 4)y′′ + 2y′ = 0. Symmetry class S34,5.

414

3.40: x2y′′′ − 6y′ + ax2y = 0. Symmetry class S34,5.

3.41: x2y′′′ + (x+ 1)y′′ − y = 0. Symmetry class S34,5.

3.42: x2y′′′ − xy′′ + (x2 + 1)y′ = 0. Symmetry class S34,5.

3.45: x2y′′′ + 4xy′′ + (x2 + 2)y′ + 3xy = 0. Symmetry class S34,5.

3.46: x2y′′′ + 5xy′′ + 4y = 0. Symmetry class: S35,1.

Fundamental system: 1, 1x,

log xx .

3.47: x2y′′′ + 6xy′′ + 6y′ = 0. Symmetry class S37 .

Fundamental system: 1, 1x,

1x2 .

3.48: x2y′′′ + 6xy′′ + 6y′ + ax2y = 0. Symmetry class S37 .

Fundamental system: 1x2 exp θx, θ3 + a = 0.

3.49: x2y′′′ − 3(p+ q)xy′′ + 3p(3q + 1)y′ − x2y = 0. Symmetry class S34,5.

3.50: x2y′′′ − 2(n+ 1)xy′′ + (ax2 + 6n)y′ − 2axy = 0, n ∈ N.Symmetry class S3

4,5.Fundamental system: ax2 +4n− 2, exp (x

√−a)P (x), exp (−x

√−a)Q(x),

P (x) and Q(x) polynomials of degree ≤ 2n+ 2.3.51: x2y′′′ − x(x− 2)y′′ + (x2 + ν2 − 1

4 )y′ − (x2 − 2x+ ν2 − 14 )y = 0.

Symmetry class S34,5.

3.52: x2y′′′ − (x+ ν)xy′′ + ν(2x+ 1)y′ − ν(x+ 1)y = 0.Symmetry class S3

4,5.

3.53: x2y′′′ − x(x− 1)y′′ + (x2 − 2x− ν2 + 14 )y′ − (ν2 − 1

4 )y = 0.Symmetry class S3

4,5.3.54: x3y′′′ − (x4 − 6x)y′′ − (2x3 − 6)y′ + 2x2y = 0.


3.55: (x2 + 1)y′′′ + 8xy′′ + 10y′ = 0. Symmetry class S34,5.

Fundamental system:

1(x2 + 1)2

,x(x2 + 3)(x2 + 1)2

,x2(x2 + 2)(x2 + 1)2

.

3.56: (x2 + 2)y′′′ − 2xy′′ + (x2 + 2)y′ − 2xy = 0. Symmetry class S34,5.

Fundamental system: x2, cosx, sinx.3.57: x(x− 1)y′′′ + 3(x− 1

2 )y′′ + (ax+ 12b)y

′ + 12ay = 0. Symmetry class S3

7 .Fundamental system: y2

1 , y1y2, y22, y1 and y2 hypergeometric functions.

3.58: 4x2y′′′ + (x2 + 14x− 1)y′′ + 4(x+ 1)y′ + 2y = 0.Symmetry class S3

4,5. Solution:

y =√x exp−x

2 + 14x

[C1 +

∫ (C2x

− 52 + C3x

− 32)exp

x2 + 14x

].

3.60: x3y′′′(ν2 − 1)xy′′ + (ax3 + ν2 − 1)y = 0. Symmetry class S34,5.

3.61: x2y′′′ + (4x3 − 4ν2x+ x)y′ + (4ν2 − 1)y = 0. Symmetry class S37 .

Fundamental system: xJ2ν (x), xJν(x)Yν(x), xY

2ν (x).

3.62: x3y′′′ + (ax2ν + 1− ν2)xy′ + [bx3ν + a(ν − 1)x2ν + ν2 − 1]y = 0.



3.63: x3y′′′ + 3x2y′′ − 2xy′ + 2y = 0. Symmetry class S35,1.

Fundamental system:x, 1x2 , x log x

.

3.64: x3y′′′ + 3x2y′′ + (1− a2)xy′ = 0. Symmetry class S37 .

Fundamental system: 1, xa, 1xa.

3.65: x3y′′′ − 4x2y′′ + (x2 + 8)xy′ − 2(x2 + 4)y = 0. Symmetry class S34,5.

Fundamental system: x2, x cosx, x sinx.3.66: x3y′′′ + 6x2y′′ + (ax3 − 12)y = 0. Symmetry class S3

4,5.

Fundamental system:θx− 2

x3 exp (θx), θ3 + a = 0.

3.68: x3y′′′ + (x+ 3)x2y′′ + 5(x− 6)xy′ + (4x+ 30)y = 0.Symmetry class S3

4,5.

Solution: y = x5 d6

dx6

x2 d2

dx2

[x−5e−x

( ∫ (C1 + C2 log x

)x4exdx+ C3

)].

3.70: (x2 + 1)xy′′′ + 3(2x2 + 1)y′′ − 12y = 0. Symmetry class S34,5.

Fundamental system: 2x2 + 1, xu, 2x+ 23x − xu log u+ 1

u− 1, with

u =√x2 + 1.

3.71: (x+3)x2y′′′−3(x+2)xy′′+6(x+1)y′−6y = 0. Symmetry class S35,2(3).

Fundamental system: x+ 1, x2, x3.3.73: (x+1)x3y′′′− (4x+2)x2y′′+(10x+4)xy′− 4(3x+1)y = 0. Symmetry

class S34,5. Fundamental system: x2, x2 log x, x+ x3 + x2 log (x)2).

3.74: 4x4y′′′ − 4x3y′′ + 4x2y′ = 1. Symmetry class S35,1.

Fundamental system: 1, x2, x2 log x− 136x.

3.75: (x2 + 1)x3y′′′ − (4x2 + 2)x2y′′ + (10x2 + 4)xy′ − 4(3x2 + 1)y = 0.Symmetry class S3

4,5. Fundamental system: x2, x3 + x, x2 log |x|.3.76: x6y′′′ + x2y′′ − 2y = 0. Symmetry class S3

4,5.3.77: x6y′′′ + 6x5y′′ + ay = 0. Symmetry class S3

4,5.

Appendix F

ALLTYPES Userinterface

To a large extent the calculations involved in the analysis of differential equa-tions described in this book can hardly be performed by pencil and paper.Therefore, the userinterface of the computer algebra system ALLTYPES hasbeen developed which is available on the website www.alltypes.de. It pro-vides a great number of functions that perform the laborious calculations in-volved automatically. A complete listing of the functions currently providedmay be obtained interactively by submittingUserFunctions();

after the ALLTYPES system has been loaded. A detailed description of anindividual function with the name FunctionName is obtained by callingUserFunction FunctionName;

In this Appendix the connection to the mathematical background for anyparticular function is provided. Subsequently, all user interface functions arelisted in alphabetical order and the page number where it is explained is given.If there occur several page numbers, they correspond to the various argumenttypes, i.e., the corresponding Methods in the ALLTYPES system.

AbsoluteInvariants(LODE|ABEL)Returns absolute invariants for linear and Abel’s equation.

CharacteristicPolynomial(LIEVEC|LIEALG)Returns characteristic polynomial of Lie algebra.

CommutatorTable(LIEVEC)Returns commutator table of Lie algebra.

DeterminingSystem(ODE|PDE)Returns determining system for Lie symmetries.

ExactQuotient(LODE|LDFMOD)Returns exact quotient or failed if it does not exist.

ForsythQuadrinvariants(LODE)Returns Forsyth’s quadrinvariants for equation of order up to three.

Gcd(LODE|LDFMOD)Returns greatest common divisor of the two arguments.

GroebnerBase(DPOLID)Returns Grobner basis in specified term order.

Integral(POLY|PARF|RATF)Returns integral of first argument w.r.t. second.

417

418

Invariants(LODE)Returns different kinds of invariants.

JacobianProlongation(LIEVEC)Returns k-th prolongation of first argument in Jacobian form.

JanetBase(ODE |LDFMOD|LFMOD)Returns Janet base in specified term order.

JanetResolution(LDFMOD)Returns Janet resolution.

LaguerreForsythNormalForm(LODE3)Returns Laguerre-Forsyth normal form.

Lclm(LODE|LDFMOD)Returns least common left multiple of the two arguments.

LieCanonicalForm(ODE2|ODE3)Returns canonical form corresponding to Lie symmetry class.

LiouvillianSolutions(LODE|LDFMOD)Returns Liouvillian solutions following from first order components.

LoewyDecomposition(LODE)Returns Loewy decomposition into irreducible components.

Prolongation(LDO |LIEVEC)Returns k-th prolongation of first argument.

RationalNormalForm(LODE| RICCATI|ABEL)Returns rational normal form.

RationalSolution(UPOLY|LODE|RICCATI )Returns most general rational solution including constants.

RationalSolutions(UPOLY|LODE|RICCATI)Returns all rational solutions.

RelativeInvariants(LODE|ABEL)Returns relative invariants.

SemiSimple?(LIEALG|LIEVEC)Returns t if Lie algebra is semi-simple.

Socle(LODE|LDFMOD)Returns largest completely reducible right component.

Solvable?(LIEALG| LIEVEC)Returns t if Lie algebra is solvable.

Solve(LODE|ODE)Returns solution or failed if it cannot be found.

Symmetries(ODE| PDE)Returns vector fields of infinitesimal generators.

SymmetryClass(LODE|ODE )Returns Lie symmetry class Sni,j.

TresseInvariants(ODE)Returns Tresse invariants of order not higher than four.

References

[1] Abraham-Shrauner, B., Govinder K. S., Leach, P. G. L., Integration of Se-cond Order Ordinary Differential Equations Not Possessing Lie Point Sym-metries, Physics Letters A 203 (1995), 169-174.

[2] Abramowitz, M., Stegun, I., Handbook of Mathematical Functions, Dover,New York, 1965.

[3] Adams, W., Loustaunau, P., An Introduction to Grobner Bases, AmericanMathematical Society, Providence, RI, 1996.

[4] Appell, P., Sur les equations differentielles algebriques et homogenes par rap-port a la fonction inconnus et a ses derivees, Comptes Rendus 104 (1887),1176–1779.

[5] Appell, P., Sur les invariants des equations differentielles, Comptes Rendus105 (1887), 55–58.

[6] Appell, P., Sur les invariants de quelques equations differentielles, Journal deMathematique 5 (1889), 361–423.

[7] Appell, P., Sur des equations differentielles lineaires transformable en elles-memes par un changement de fonction et de variable, Acta Mathematica 15(1891), 281–314.

[8] Babich, M. V., Bordag, L. A., Projective differential geometrical structure ofthe Painleve equations, Preprint NTZ 35/1997.

[9] Beke, E., Die Irreduzibilitat der homogenen linearen Differentialgleichungen,Mathematische Annalen 45 (1894), 278–294.

[10] Berth, M., Invariants of Ordinary Differential Equations, Dissertation,Greifswald, 1999.

[11] Berth, M., Czichowski, G., Using Invariants to Solve the Equivalence Problemfor Ordinary Differential Equations, Applied Algebra in Engineering, Com-munication and Computing 11 (1998), 359-376.

[12] Bianci, L., Lezioni sulla teoria dei gruppi continui finiti di transformazioni,Nicola Zanichelli, Bologna, 1928.

[13] Bluman, G. W., Kumei, S., Symmetries of Differential Equations, Springer,Berlin, 1990.

[14] Blyth, T. S., Module Theory, Clarendon Press, Oxford, 1977.

[15] Bordag, L. A., Dryuma, V. S., Investigation of Dynamical Systems UsingTools of the Theory of Invariants and Projective Geometry, Z. angew. Math.Phys. 48 (1997), 725-743.

419

420 References

[16] Bourbaki, N., Elements d’histoire des mathematiques, Masson Editeur, Paris,1984 [English translation, Elements of the History of Mathematics, Springer,Berlin, 1994].

[17] Borel, A., Essays in the History of Lie Groups and Algebraic Groups, Amer-ican Mathematical Society and London Mathematical Society, 2001.

[18] Bouton, C. L., Invariants of the General Linear Differential Equation andtheir Relation to the Theory of Continuous Groups, American Journal ofMathematics 21 (1899), 25–84.

[19] Brioschi, F., Les invariants des equations differentielles lineaires, ActaMathematica 14 (1890), 233–248.

[20] Bronstein, M., On Solutions of Linear Differential Equations in their Coeffi-cient Field, Journal of Symbolic Computation 13 (1992), 413-439.

[21] Bronstein, M., Linear Ordinary Differential Equations, Breaking through theOrder 2 Barrier, Proceedings of the ISSAC 1992, Berkeley, California, ACMPress, 1992; P. S. Wang, Ed., 42-48.

[22] di Bruno, Faa., Note sur une nouvelle formule de calcul differentielle, Quar-terly Journal of Pure and Applied Mathematics 1 (1857), 359-360.

[23] Buium A., Cassidy, Ph., Differential Algebraic Geometry and DifferentialAlgebraic Groups: From Algebraic Differential Equations to Diophantine Ge-ometry, in: Selected Works of Ellis Kolchin, AMS Press, Providence, RhodeIsland, 1999; H. Bass, A. Buium, Ph. Cassidy, Eds.

[24] Campbell, J. E., Introductory Treatise on Lie’s Theory of Finite ContinuousTransformation Groups, 1903 [Reprinted by Chelsea Publishing Company,New York, 1966].

[25] Cartan, E., Sur la structure des groupes de transformations finis et continus,Premiere These, Paris, 1894.

[26] Cartan, E., Sur la reduction a sa forme canonique de la structure d’un groupede transformations fini et continu, American Journal of Mathematics 18(1896), 1-61.

[27] Castro-Jimenez, F. J., Moreno-Frıas, M.A., Janet Bases and Grobner Bases,Prepublicaciones del Departamento de Algebra de la Universidad de Sevillano 1, 2000.

[28] Chiellini, A., Sull’ integrazione dell’equazione differentiale y′+Py2+Qy3 = 0,Bolletino Unione Mat. Italiana 10 (1931), 301-307.

[29] Chini, M., Sull’integrazione di alcune equazioni differenziali del primo ordine,Rendiconti Istituto Lombardo (2) 57 (1924), 506-511.

[30] Clebsch, A., Theorie der binaren algebraischen Formen, Teubner, Leipzig,1872.

[31] Cockle, J., Correlations of Analysis, Philosophical Magazine 24(IV)(1862), 24.

References 421

[32] Cohen, A., An Introduction to the Lie Theory of One-Parameter Groups,Hafner Publishing Company, New York, 1911.

[33] Cox, D., Little, J., O’Shea, D., Ideals, Varieties and Algorithms, Springer,Berlin, 1991.

[34] Cox, D., Little, J., O’Shea, D., Using Algebraic Geometry, Springer, Berlin,1998.

[35] Curtiss, D. R., On the Invariants of a Homogeneous Quadratic DifferentialEquation of the Second Order, American Journal of Mathematics 25 (1903),365–382.

[36] Czichowski, G., Fritzsche, B., Beitrage zur Theorie der Differentialinvari-anten, Teubner Verlagsgesellschaft, 1993.

[37] de Graaf, W. A., Algorithms for Finite-Dimensional Lie Algebras, Disserta-tion, Eindhoven, 1997.

[38] Drach, J., Essai sur une theorie general de l’integration et sur la classificationdes transcendentes, Gauthier-Villars et Cie, Paris, 1898.

[39] Dresner, L., Applications of Lie’s Theory of Ordinary and Partial DifferentialEquations, IOP Publishing, Bristol and Philadelphia, 1999.

[40] Emanuel, G., Solution of Ordinary Differential Equations by ContinuousGroups, Chapman & Hall/CRC, London, New York, 2001.

[41] Eisenhart, L. P., Continuous Groups of Transformations, Princeton Univer-sity Press, Princeton, 1933 [Reprinted by Dover Publications, New York,1961].

[42] Emanuel, G., Solution of Ordinary Differential Equations by ContinuousGroups, Chapman & Hall/CRC, 2000.

[43] Encyclopaedia of Mathematical Sciences, Lie Groups and Lie Algebras, vol. I,A. L. Onishchik, Ed., Springer, Berlin, 1993; vol. II, A. L. Onishchik,E. B. Vinberg, Eds., Springer, Berlin, 2000; vol. III, A. L. Onishchik,E. B. Vinberg, Eds.; Springer, Berlin, 1994.

[44] Engel, F., Gesammelte Abhandlungen Sophus Lie, vol. I to VII, Teubner,Leipzig, 1922–1960.

[45] Faber, K., Differentialgleichungen, die eine irreduzible Gruppe von Beruh-rungstransformationen gestatten, Mitteilungen des Mathematischen Seminarsder Universiat Giessen 13 (1924), 1-30.

[46] Faber, K., Differentialgleichungen, die durch eine Beruhrungstransformationauf die Form y′′′ = 0 gebracht werden konnen, Leipziger Berichte 78 (1926),9-22.

[47] Forsyth, A. R., Invariants, Covariants and Quotient-Derivatives Associatedwith Linear Differential Equations, Philosophical Transactions 179 (1888),377–489.

[48] Fritzsche, B., Sophus Lie, A Sketch of his Life and Work, Journal of LieTheory 9 (1999), 1-38.

422 References

[49] Gandarias, M. L., Medina, E., Muriel, C., New Symmetry Reductions forSome Ordinary Differential Equations, Journal of Nonlinear MathematicalPhysics 9 (2002), 47-58.

[50] Gasiorowski, L., Uber die Definitionsgleichungen der endlichen con-tinuierlichen Gruppen von Beruhrungstransformationen der Ebene, PraceMatematyczno-Fizyczne 26 (1914), 133-203.

[51] Gat, O., Symmetry algebras of third-order ordinary differential equations,Journal of Mathematical Physics, 33 (1992), 2966-2971.

[52] Gerdt, V., Lassner, W., Isomorphism Verification for Complex and real LieAlgebras by Grobner Base Technique, Proceedings of the International Work-shop Acireale, Catania, N. H. Ibragimov, N. Torrisi, A. Valenti, Eds.; KluwerAcademic Publishers, Dordrecht, 245-254, 1993.

[53] Gonzales-Lopez, A., Symmetry and Integrability by Quadratures of OrdinaryDifferential Equations, Physics Letters, A133 (1988), 190-194.

[54] Gonzales-Lopez, A., Kamran, N., Olver, P., Lie Algebras of Vector Fieldsin the Real Plane, Proceedings of the London Mathematical Society (3) 64(1992), 339–368.

[55] Gordan, P., Invariantentheorie, Teubner, Leipzig, 1888 [Reprinted by ChelseaPublishing Company, New York, 1987].

[56] Goursat, E., Sur les invariants des equations differentielles, Comptes Rendus107 (1888), 898–900.

[57] Goursat, E., Lecons sur l’integration des equations aux derivees partielles,Gauthier-Villars et Cie, Paris, 1921.

[58] Grebot, G., The Characterization of Third Order Ordinary Differential Equa-tions Admitting a Transitive Fiber-Preserving Point Symmetry Group, Jour-nal of Mathematical Analysis and Applications 206 (1997), 364-388.

[59] Grebot, G., The Third Order Ordinary Differential Equations Admitting thePainleve I and II as First Integrals, Journal of Mathematical Analysis andApplications 220 (1997), 110-124.

[60] Greuel, G. M., Pfister, G., A Singular Introduction to Commutative Algebra,Springer, Berlin, 2002.

[61] Grigoriev, D., Schwarz, F. Factoring and Solving Linear Partial DifferentialEquations, Computing 73 (2004), 179-197.

[62] Grigoriev, D., Schwarz, F. Generalized Loewy-Decomposition of D-Modules,Proceedings of the ISSAC’05, 163-170, ACM Press, 2005.

[63] Grissom, C., Thomson, G., Wilkens, G., Linearization of Second Order Or-dinary Differential Equations via Cartan’s Equivalence Method, Journal ofDifferential Equations 77 (1989), 1–15.

[64] Halphen, G. H., Sur les invariants differentielles, These presentee a la Facultede Sciences de Paris, 1878.

References 423

[65] Halphen, G. H., Memoire sur la reduction des equations differentielleslineaires aux formes integrable, Memoires Presentes par divers Savants 28(1878), 1–301.

[66] Halphen, G. H., Sur la reduction de certaines equations differentielles dupremier ordre a la forme lineaire, Comptes rendus 87 (1882), 741–743.

[67] Halphen, G. H., Sur les invariants des equations differentielles lineaires duquatrieme ordre, Acta Mathematica 3 (1883), 325–380.

[68] Hawkins, T., Emergence of the Theory of Lie Groups, Springer, Berlin, 2000.

[69] Heineck, C., Invariante Kurvenintegrale bei infinitesimalen Transformatio-nen in drei Veranderlichen x, y und z und deren Verwertung, Dissertation,Teubner, Leipzig, 1899.

[70] Hereman, W., Review of Symbolic Software for the Computation of Lie Sym-metries of Differential Equations, Euromath Bulletin 1 (1994), 45–97.

[71] Hill, J. M., Differential Equations and Group Methods, CRC Press, BocaRaton, Florida, 1992.

[72] Hirsch, A., Zur Theorie der linearen Differentialgleichungen mit rationalemIntegral, Dissertation, Konigsberg, 1892.

[73] van Hoeij, M., Factorization of Differential Operators with Rational FunctionCoefficients, Journal of Symbolic Computation 24 (1997), 537-561.

[74] Hsu, L., Kamran, N., Symmetries of Second-Order Ordinary DifferentialEquations and Elie Cartan’s Method of Equivalence, Letters in Mathemat-ical Physics 15 (1988), 91-99.

[75] Hsu, L., Kamran, N., Classification of Second-Order Ordinary DifferentialEquations Admitting Lie Groups of Fibre-Preserving Point Symmetries, Pro-ceedings of the London Mathematical Society (3) 58 (1989), 387-416.

[76] Hydon, P. E., Symmetry Methods for Differential Equations, Cambridge Uni-versity Press, 2000.

[77] Ibragimov, N. H., Transformation Groups Applied to Mathematical Physics,D. Reidel Publishing Company, Dordrecht, 1985.

[78] Ibragimov, N. H., Elementary Lie Group Analysis of Ordinary DifferentialEquations, John Wiley & Sons, New York, 1999.

[79] Imschenetzky, V. G., Sur l’integration des equations aux deriveees partiellesdu premier ordre, Grunerts Archiv fur Mathematik 50 (1869), 278.

[80] Ince, E. L., Ordinary Differential Equations, Longmans, Green and Co., 1926[Reprint by Dover Publications Inc., 1956].

[81] Jacobson, N., The Structure of Rings, American Mathematical Society, Prov-idence, Rhode Island, 1956.

[82] Jacobson, N., Lie Algebras, Interscience Publishers, New York, 1962.

[83] Janet, M., Les systemes d’equations aux derivees partielles, Journal demathematiques 83 (1920), 65–123.

424 References

[84] Janet, M., Lecons sur les systemes d’equations aux derivees partielles,Gauthier-Villars et Cie, Paris, 1929.

[85] Kamke, E., Differentialgleichungen, Losungsmethoden und Losungen, I.Gewohnliche Differentialgleichungen, Akademische Verlagsgesellschaft,Leipzig, 1961.

[86] Kamke, E., Differentialgleichungen I, Gewohnliche Differentialgleichungenund II, Partielle Differentialgleichungen, Akademische Verlagsgesellschaft,Leipzig, 1962.

[87] Kamke, E., Differentialgleichungen, Losungsmethoden und Losungen, II. Par-tielle Differentialgleichungen, Akademische Verlagsgesellschaft, Leipzig, 1965.

[88] Kaplansky, I., An Introduction to Differential Algebra, Hermann, Paris, 1957.

[89] Killing, W., Die Zusammensetzung der stetigen endlichen Transformations-gruppen, Mathematische Annalen 31 (1887), 252-290; 33, 1-48; 34, 57-122;36, 161-189.

[90] Kolchin, E. R., Algebraic Matrix Groups and the Picard-Vessiot Theory ofHomogeneous Linear Ordinary Differential Equations, Annals of Mathematics49 (1948), 1-42.

[91] Kolchin, E. R., Differential Algebra and Algebraic Groups, Academic Press,New York and London, 1973.

[92] Kovacic, J., An Algorithm for Solving Second Order Linear HomogeneousDifferential Equations, Journal of Symbolic Computation 2(1986), 3-43.

[93] Kowalewski, G., Einfuhrung in die Theorie der kontinuierlichen Gruppen,Akademische Verlagsgesellschaft, Leipzig, 1931.

[94] Kowalewski, G., Integrationsmethoden der Lie’schen Theorie, AkademischeVerlagsgesellschaft, Leipzig, 1933.

[95] Krause, J., Michel, L., Equations differentielles lineaires d’ordre n > 2 ayantune algeebre de Lie de symetrie de dimension n + 4, Comptes Rendus desAcademie des Sciences 307 (1988), 905–910.

[96] Krause, J., Michel, L., Classification of the Symmetries of Ordinary Differ-ential Equations, Proceedings of the Summer School in Group TheoreticalMethods in Physics, Moscow, 1990, Springer LNP 382, 1991.

[97] Kung, J. P. S., Rota, G. C., Theory of Binary Forms, Bulletin of the AmericanMathematical Society 10 (1984), 27-85.

[98] Laguerre, E., Sur les equations differentielles lineaires du troisieme ordre,Comptes Rendus 88 (1879), 116–119.

[99] Laguerre, E., Sur quelques invariants des equations differentielles, ComptesRendus 88 (1879), 224–227.

[100] Lambek, J., Lectures on Rings and Modules, Blaisdell Publishing Company,Waltham, 1966.

References 425

[101] Landau, E., Uber die Faktorisierung linearer Differentialgleichungen, Journalfur die reine und angewandte Mathematik 124 (1902), 115-120.

[102] Leja, F., Proprietes des equations differentielles ordinaires de 3-eme ordrepar rapport aux transformations tangentielles, Prace Matematyczno-Fizyczne29 (1918), 179-256.

[103] Leja, F., Bestimmung der Invarianten der gewohnlichen Differentialgleichun-gen 3. Ordnung in bezug auf Punkttransformationen, Monatshefte fur Math-ematik und Physik 29 (1918), 203-254.

[104] Li, Z., Schwarz, F., Rational Solutions of Riccati Like Systems of PartialDifferential Equations, Journal of Symbolic Computation 31 (2001), 691-716.

[105] Li, Z., Schwarz, F., Tsarev, S., Factoring Zero-dimensional Ideals of LinearPartial Differential Operators, Proceedings of the ISSAC’02, ACM Press,T. Mora, Ed., 168-175, 2002.

[106] Li, Z., Schwarz, F., Tsarev, S., Factoring Systems of Linear PDE’s withFinite-Dimensional Solution Spaces, Journal of Symbolic Computation 36(2002), 443-471.

[107] Lie, S., Verallgemeinerung und neue Verwertung des Jacobischen Multip-likators, Christ. Forb. (1874), 255-274 [Gesammelte Abhandlungen, vol. III,188-206].

[108] Lie, S., Allgemeine Theorie der partiellen Differentialgleichungen erster Ord-nung, Mathematische Annalen 11 (1877), 464–557 [Gesammelte Abhandlun-gen, vol. IV, 163-250].

[109] Lie, S., Klassifikation und Integration von gewohnlichen Differentialgleichun-gen zwischen x, y, die eine Gruppe von Transformationen gestatten I, II, IIIand IV, Archiv for Mathematik 8 (1883), 187-224, 249-288, 371-458 and 9,431-44. [Gesammelte Abhandlungen, vol. V, 240-281, 282-310, 362-427 and432-446].

[110] Lie, S., Uber unendliche kontinuierliche Gruppen, Christ. Forh. Aar, Nr. 12(1883) [Gesammelte Abhandlungen, vol. V, 314-361].

[111] Lie, S., Uber Differentialinvarianten, Mathematische Annalen 24 (1884), 537–578 [Gesammelte Abhandlungen, vol. VI, 95-138].

[112] Lie, S., Theorie der Transformationsgruppen I, II and III, Teubner, Leipzig,1888 [Reprinted by Chelsea Publishing Company, New York, 1970].

[113] Lie, S., Vorlesungen uber Differentialgleichungen mit bekannten infinitesi-malen Transformationen, Teubner, Leipzig, 1891 [Reprinted by Chelsea Pub-lishing Company, New York, 1967].

[114] Lie, S., Vorlesungen uber continuierliche Gruppen, Teubner, Leipzig, 1893[Reprinted by Chelsea Publishing Company, New York, 1971].

[115] Lie, S., Geometrie der Beruhrungstransformationen, Teubner, Leipzig, 1896[Reprinted by Chelsea Publishing Company, New York, 1977].

426 References

[116] Liouville, R., Sur quelques equations differentielle non lineaires, ComptesRendus 103 (1886), 457–460.

[117] Liouville, R., Sur certaines equations differentielles du premier ordre,Comptes Rendus 103 (1886), 476-479.

[118] Liouville, R., Sur une classe d’equations differentielles non lineaires, ComptesRendus 103 (1886), 520–523.

[119] Liouville, R., Sur une classe d’equations differentielles de premier ordre et surles formations invariantes qui s’y rapportent, Comptes Rendus 105 (1887),460–463.

[120] Liouville, R., Sur une classe d’equations differentielles, parmis lesquelles,en particulier, toutes celles des lignes geodesiques se trouvent comprises,Comptes Rendus 105 (1887), 1062–1064.

[121] Liouville, R., Sur quelques equations differentielles non lineaires, Journal del’Ecole Polytechnique 57 (1887), 189–250.

[122] Liouville, R., Sur certaines equations differentielles de premier ordre,Comptes Rendus 106 (1888), 1648–1651.

[123] Liouville, R., Sur les lignes geodesiques des surface a courbure constante,American Journal of Mathematics 10 (1888), 283–292.

[124] Liouville, R., Sur les invariants de certaines equations differentielles de pre-mier ordre et sur leurs applications, Comptes Rendus 109 (1889), 560–563.

[125] Liouville, R., Memoire sur les invariants de certaines equations differentielleset sur leurs applications, Journal de l’Ecole Polytechnique 59 (1889), 7–76.

[126] Liouville, R., Sur une equation differentielles du premier ordre, Acta Mathe-matica 27 (1903), 55–78.

[127] Loewy, A., Uber vollstandig reduzible lineare homogene Differentialgleichun-gen, Mathematische Annalen 56 (1906), 89–117.

[128] Magid, A. R., Lectures on Differential Galois Theory, American Mathema-tical Society, Providence, Rhode Island, 1991.

[129] Mahomed, F. M., Leach, P. G. L., Symmetry Lie Algebras of n-th OrderOrdinary Differential Equations, Journal of Mathematical Analysis and Ap-plications 151 (1990), 80–107.

[130] Markoff, A., Sur les equations differentielles lineaires, Comptes Rendus 113(1891), 685-688.

[131] Martins, J., Algebraic Subgroups of SL(2, C), private communication, 2003.

[132] Maurer, L., Burkhardt, H., Kontinuierliche Transformationsgruppen, Encyk-lopadie der Mathematischen Wissenschaften II.1., Teubner, Leipzig, 401-436,1916.

[133] Miller, G. A., Blichfeldt, H. F., Dickson, L. E., Finite Groups, Dover Publi-cations, New York, 1961.

References 427

[134] Muriel, C., Romero J. L, New Methods of Reduction for Ordinary DifferentialEquations, IMA Journal of Mathematics 66 (2001), 111-125.

[135] Neumer, W., Uber gewohnliche Differentialgleichungen, die lineare homogeneForm erhalten konnen, Mitteilungen des Mathematischen Seminars der Uni-versitat Giessen 16 (1929), 1-52.

[136] Neumer, W., Die allgemeinste mit der projektiven G8 der Ebene ahnlicheGruppe von Beruhrungstransformationen, Journal fur die reine und ange-wandte Mathematik 173 (1935), 125-159.

[137] Neumer, W., Die gewohnlichen Differentialgleichungen dritter und vierterOrdnung, die lineare homogene Form erhalten konnen, Journal fur die reineund angewandte Mathematik, I, 176 (1937), 224-249; II 177 (1937), 13-36;III 177 (1937), 65-81.

[138] Noth, G., Differentialinvarianten und invariante Differentialgleichungenzweier zehngliedriger Gruppen, Dissertation, Teubner, Leipzig, 1904.

[139] Oaku, T., Shimoyama, T., A Grobner Basis Method for Modules over Rings ofDifferential Operators, Journal of Symbolic Computation 18 (1994), 223-248.

[140] Olver, P., Application of Lie Groups to Differential Equations, Springer,Berlin, 1986.

[141] Olver, P., Equivalence, Invariants and Symmetries, Cambridge UniversityPress, Cambridge, UK, 1995.

[142] Olver, P., Classical Invariant Theory, Cambridge University Press, Cam-bridge, UK, 1999.

[143] Ore, O., Formale Theorie der linearen Differentialgleichungen, Journal furdie reine und angewandte Mathematik 167 (1932), 221–234, and 168 (1932),233-257.

[144] Ovsiannikov, L. V., Group Analysis of Differential Equations, AcademicPress, New York, 1982.

[145] Peyovitch, T., Sur les semi-invariants des equations differentielles lineaires,Bulletin de la Societe mathematique de France 53 (1923), 208–225.

[146] Picard, E., Sur les groupes de transformation des equations differentielleslineaires, Comptes Rendus 86 (1883), 1131–1134.

[147] Picard, E., Analogies entre la theorie des equations differentielles lineaires atla theorie des equations algebriques, Gauthier-Villars, Paris, 1936.

[148] Plesken W., Robertz, D., Janet’s Approach to Presentations and Resolutionsfor Polynomials and Linear PDEs, Arch. Math. 84 (2005), 22-37.

[149] Polyanin, A. D., Zaitsev, V. V., Handbuch der linearen Differentialgleichun-gen, Exakte Losungen, Spektrum Akademischer Verlag, Heidelberg, 1996[Translated from Russian].

[150] Quadrat, A., An Introduction to the Algebraic Theory of Linear Systems ofPartial Differential Equations,www-sop.inria.fr/cafe/Alban.Quadrat/Temporaire.html

428 References

[151] Rand D., Winternitz, P., Zassenhaus, H., On the Identification of a Lie Alge-bra Given by Its Structure Constants. I. Direct Decompositions, Levi Decom-positions, and Nilradicals, Linear Algebra and its Applications 109 (1988),197-246.

[152] Reid, G., Lisle, I. G., Boulton, A., Wittkopf, A. D., Algorithmic Determina-tion of Commutation Relations for Lie Symmetry Algebras of PDE’s, Pro-ceedings of the ISSAC’92, ACM Press, G. Gonnet, Ed., 63-68, 1992.

[153] Riquier, C., Les Systemes d’Equations aux Derivees Partielles, Gauthier-Villars, Paris, 1910.

[154] Riverau, P., Sur les invariants de certaines classes d’equations differentielleshomogenes par rapport a la fonction inconnue et a ses derivees, Thesespresentees a la Faculte des Sciences de Paris, Gauthiers-Villars, Paris, 1890.

[155] Riverau, P., Sur les invariants des equations differentielles lineaires, Annalesde la Faculte des Sciences de Toulouse 4 (1890), 1-5.

[156] Riverau, P., Sur les invariants de quelques equations differentielles, Journalde Mathematiques pures et appliquees 4(8) (1892), 233–268.

[157] Sachdev, P. L., A Compendium on Nonlinear Ordinary Differential Equations,Wiley-Interscience, New York, 1997.

[158] Scalizzi, P., Soluzione di alcune equazioni del tipo Abel, Atti Accad. Lincei26(5) (1917), 60-64.

[159] Schlesinger, L., Ein Beitrag zur Theorie der linearen homogenen Differen-tialgleichungen dritter Ordnung, Inauguraldissertation Kiel, Mayer & Muller,Berlin, 1897.

[160] Schlesinger, L., Handbuch der Theorie der linearen Differentialgleichungen,vol. I and II, Teubner, Leipzig, 1897.

[161] Schmucker, A., Czichowski, G., Symmetry Algebras and Normal Forms ofThird Order Ordinary Differential Equations, Journal of Lie Theory 8 (1998),129-137.

[162] Schwarz, F., A Factorization Algorithm for Linear Ordinary DifferentialEquations, Proceedings of the ISSAC’89, ACM Press, G. Gonnet, Ed., 17–25,1989

[163] Schwarz, F., An Algorithm for Determining the Size of Symmetry Groups,Computing 49 (1992), 95-115.

[164] Schwarz, F., Symmetries of 2nd and 3rd Order ODE’s, Proceedings of theISSAC’95, ACM Press, A. Levelt, Ed., 16-25, 1995.

[165] Schwarz, F., Janet Bases of 2nd Order Ordinary Differential Equations, Pro-ceedings of the ISSAC’96, ACM Press, Y. N. Lakshman, Ed., 179-187, 1996.

[166] Schwarz, F., On Third Order Ordinary Differential Equations with MaximalSymmetry Group, Computing 57 (1996), 273-280.

[167] Schwarz, F., ALLTYPES, An ALgebraic Language and TYPE System,Springer, LNAI 1476, 270-283, 1998.

References 429

[168] Schwarz, F., Solving Third Order Differential Equations with Maximal Sym-metry Group, Computing 65 (2000), 155-167.

[169] Singer, M., Liouvillian Solutions of n-th Order Homogeneous Linear Differ-ential Equations, American Journal of Mathematics 103 (1981), 661-682.

[170] Singer, M., Solving Homogeneous Linear Differential Equations in Terms ofSecond Order Linear Differential Equations, American Journal of Mathemat-ics 107 (1985), 663-696.

[171] Singer, M., Liouvillian Solutions of Linear Differential Equations, Journal ofSymbolic Computation 11 (1991), 251-273.

[172] Singer, M., Ulmer, F., Galois Groups of Second and Third Order Lin-ear Differential Equations, Journal of Symbolic Computation 16 (1993),9-36.

[173] Stackel, P., Uber Transformationen von Differentialgleichungen, Journal furreine und angewandte Mathematik 111 (1890), 290–302.

[174] Stephani, H., Differential equations. Their solution using symmetries, Cam-bridge University Press, Cambridge, UK, 1989.

[175] Stubhaug, A., The Mathematician Sophus Lie, Springer, Berlin, 2001.

[176] Thomas, J. M., Matrices of Integers Ordering Derivatives, Transactions ofthe American Mathematical Society 33 (1931), 389–410.

[177] Todorov, P. G., Kristev, G. A., Some Classes of the Reduction and Gen-eralizations of the Abel Equation, Differensialnye Uravneniya 28 (1992),2178-2179 (in Russian).

[178] Tresse, A., Sur les groupes infinis de transformations, Comptes Rendus 115(1892), 1003–1006.

[179] Tresse, A., Sur les invariants differentielles des groupes continus de transfor-mations, Acta Mathematica 18 (1894), 1–88.

[180] Tresse, A., Sur les invariants ponctuels de l’equation differentielle ordinairedu second ordre, Comptes Rendus 120 (1895), 429-431.

[181] Tresse, A., Determination des Invariants ponctuels de l’Equation differentielleordinaire du second ordre, y′′ = ω(x, y, y′), Preisschriften der FurstlichJablonowski’schen Gesellschaft, Teubner, Leipzig, 1896.

[182] Ulmer, F., Weil, J. A., Note on Kovacic’s Algorithm, Journal of SymbolicComputation 22 (1996), 179-200.

[183] Vivanti, G., Lecons elementaires sur la theorie des groupes de transforma-tions, Gauthiers-Villars, Paris, 1904.

[184] van der Put, M., Galois Theory of Differential Equations, Algebraic Groupsand Lie Algebras, Journal of Symbolic Computation 28 (1999), 441-473.

[185] van der Put, M., Singer, M., Galois Theory of Linear Differential Equations,Springer, Berlin, 2003.

430 References

[186] van der Waerden, B. L., Gruppen von linearen Transformationen, ChelseaPublishing Company, New York, 1948.

[187] von Weber, E., Partielle Differentialgleichungen, Encyklopadie der Mathe-matischen Wissenschaften II.1., Teubner, Leipzig, 296-321, 1916.

[188] Wallenberg, G., Anwendung der Theorie der Differentialinvarianten auf dieUntersuchung der algebraischen Integrierbarkeit der linearen homogenen Dif-ferentialgleichungen, Journal fur reine und angewandte Mathematik 113(1894), 1–41.

[189] Wilczynski, E. J., Projective Differential Geometry of Curves and Ruled Sur-faces, Teubner, Leipzig, 1906 [Reprinted by Chelsea Publishing Company,New York, 1961].

[190] Wussing, H., Die Genesis des abstrakten Gruppenbegriffes, VEB DeutscherVerlag der Wissenschaften, Berlin, 1969.

[191] Zaitsev, V. F., Polyanin, A. D., Discrete-Group Methods for Integrating Equa-tions of Nonlinear Mechanics, CRC Press, Boca Raton, Florida, 1994.

[192] Zorawski, K., Uber die Integration einer Klasse gewohnlicher Differential-gleichungen dritter Ordnung, Krakauer Abhandlungen 34 (1897), 141-205(Polish).

Index

Abel’s equation, 179absolute invariant, 113, 161adjoint group, 109Appell, 179associated equations, 24autoreduced system, 49autoreduction, 47

base field, 8, 11Bernoulli equation, 180Bouton, 165

center, 117centralizer, 117characteristic equation, 117characteristic polynomial, 117characteristic system, 78coherent, 53commutation relation, 108commutator group, 106complete set of monomials, 384complete system, 51, 109completely reducible, 27component, 70connected, 109contact transformation, 159, 195criticoid, 159

defining equations, 112dependent variables, 42derivation operator, 42derivative, 42derivative operator, 42derivative vector, 44derived Lie algebra, 118derived series, 118determining system, 199Dickson’s lemma, 44

differential Galois theoryfor second order equations, 35

differential invariant, 114differential type, 40divisor, 70Drach’s equation, 104

elementary extension, 16Emden-Fowler equation, generalized,

221equivalence

of first order ode’s, 176of linear second order ode’s,

168of linear third order ode’s, 172

equivalence class, 160equivalence problem, 2, 160equivalence transformation, 160equivalent rational solutions, 17essential group parameters, 107exact quotient, 22exact quotient module, 70

factorization, 21full set of solutions, 13full system of differential invariants,

115function, 42fundamental invariants, 201fundamental system, 12, 78

gauge, 40Gcd, 21Gcrd, 23general solution, 311Goursat, 160graded lexicographic ordering, 45

431

432 Index

graded reverse lexicographic order-ing, 45

greatest common divisor, 21greatest common left divisor, 23greatest common right divisor, 23group type, 138

homomorphism of Lie groups, 106hyperexponential solution, 86

imprimitive Lie group, 108independent variables, 42infinitesimal generator, 108inhomogeneous equation, 14integrability conditions, 50integrable Lie algebra, 118integrable Lie group, 106integrable pair, 86integral basis, 78integrating factor, 313, 349intersection module, 69intransitive Lie group, 108invariant

absolute, 161of a differential equation, 160of a Lie group, 113relative, 161semi-, 161

invariant decomposition, 108invariants

absolute of linear ode, 168relative of linear ode, 166–168

irreducible, 24isomorphism of Lie groups, 106

Jacobi normal form, 79Janet basis, 11Janet basis

algorithm for, 53classification of, 57, 61construction of from solutions,

75decomposition of, 99definition of, 53exact quotient of, 70

for Lie vector fields, 143type of, 55

k-fold transitive, 108Krause, 193

Laguerre, 166Laguerre-Forsyth canonical form, 164Landau, 26Lclm, 23Lcm, 21leading derivative, 44leading term, 44least common left multiple, 23least common multiple, 21Levi decomposition, 121lexicographic ordering, 45Lie, 10Lie algebra, 116Lie algebra

classical, 120classification of, 135definition of, 116derived, 118exceptional, 120integrable, 118semi-simple, 119solvable, 118

Lie group, 106Lie system, 144Lie transformation group, 107Lie’s equation, 219Lie’s first theorem, 109Lie’s relations, 143Lie’s second theorem, 110Lie’s third theorem, 110Lie-determinant, 201line element, 195linear ode, 12Liouville R., 181Liouville, R., 186Liouvillian extension, 16local Lie group, 106Loewy decomposition, 11Loewy decomposition

Index 433

type of, 28Loewy factor, 27lower equations, 201

matrice de cotes, 44matrix of weights, 44Michel, 193

Neumer, 162normalizer, 117

orbit, 108order of a derivative, 42ordinary differential field, 42ordinary differential ring, 42

parametric derivative, 55partial differential field, 42partial differential ring, 42path curve, 126Picard-Vessiot extension, 16point transformation, 159, 194primitive Lie group, 108principal derivative, 55prolongation

of a symmetry generator, 197of a vector field, 114

quotientof ordinary differential opera-

tors, 22

radical, 119rank

of a Janet basis, 55of a Lie algebra, 117

ranking, 43rational normal form

definition of, 162of Abel’s equation, 180of Riccati’s equation, 177

reduction, 47relative invariant, 161relative syzygies module, 70Riccati equation, 16Riccati equation

associated to a linear equation,16

Riccati-like partial differential equa-tions, 85

semi-invariant, 161semi-simple Lie algebra, 119semi-simple Lie group, 106similar Lie groups, 111simple Lie algebra, 119simple Lie group, 106singular solution, 311solvable Lie algebra, 118solvable Lie group, 106special functions, 169special rational solutions, 17Stackel, 162stabilizer, 107stratum, 193structure constants, 107structure invariance group

definition of, 160of Abel’s equation, 179of Lie’s equation, 186of linear differential equation,

162of Riccati’s equation, 177

symmetric powerand Lie symmetries, 245definition of, 33

symmetric product, 33symmetry, 3, 196symmetry

of an ode, 195symmetry algebra, 193symmetry class, 4, 7, 193symmetry group, 193symmetry type, 4, 193system of imprimitivity, 108

term, 43term ordering

definition of, 43grevlex, 45grlex, 45

434 Index

lex, 45transitive Lie group, 108Tresse, 188trivial symmetries, 206typical differential dimension, 40

unconnected, 109

variation of constants, 14

weight, 161Wronskian, 12

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