Basic Theory of Ordinary Differential Equations Universitext

BASIC THEORY

OF ORDINARY

DIFFERENTIAL

EQUATIONS

Universitext

Editorial Board(North America):

S. AxlerF.W. Gehring

K.A. Ribet

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(cnnrinued after index)

Po-Fang Hsieh Yasutaka Sibuya

Basic Theory of OrdinaryDifferential Equations

With 114 Illustrations

Springer

Po-Fang Hsieh Yasutaka SibuyaDepartment of Mathematics School of Mathematics

and Statistics University of MinnesotaWestern Michigan University 206 Church Street SEKalamazoo, MI 49008 Minneapolis, MN 55455USA [email protected] sibuya 0 math. umn.edu

Editorial Board(North America):

S. Axler F.W. Gehring K.A. Ribet

Mathematics Department Mathematics Department Department ofSan Francisco State East Hall Mathematics

University University of Michigan University of CaliforniaSan Francisco, CA 94132 Ann Arbor. Ml 48109- at BerkeleyUSA 1109 Berkeley, CA 94720-3840

USA USA

Mathematics Subject Classification (1991): 34-01

Library of Congress Cataloging-in-Publication DataHsieh, Po-Fang.

Basic theory of ordinary differential equations I Po-Fang Hsieh,Yasutaka Sibuya.

p. cm. - (Unrversitext)Includes bibliographical references and index.

ISBN 0-387-98699-5 (alk. paper)1. Differential equations. I. Sibuya. Yasutaka. 1930-

II. Title. III. SeriesOA372_H84 1999515'.35-dc2l 99-18392

Printed on acid-free paper.

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9 8 7 6 5 4 3 2 1

ISBN 0-387-98699-5 Springer-Verlag New York Berlin Heidelberg SPIN 10707353

To Emmy and Yasuko

PREFACE

This graduate level textbook is developed from courses in ordinary differentialequations taught by the authors in several universities in the past 40 years orso. Prerequisite of this book is a knowledge of elementary linear algebra, realmultivariable calculus, and elementary manipulation with power series in severalcomplex variables. It is hoped that this book would provide the reader with thevery basic knowledge necessary to begin research on ordinary differential equations.To this purpose, materials are selected so that this book would provide the readerwith methods and results which are applicable to many problems in various fields.In order to accomplish this purpose, the book Theory of Differential Equations byE. A. Coddington and Norman Levinson is used as a role model. Also, the teachingof Masuo Hukuhara and Mitio Nagumo can be found either explicitly or in spiritin many chapters. This book is useful for both pure mathematician and user ofmathematics.

This book may be divided into four parts. The first part consists of Chapters I,II, and III and covers fundamental existence, uniqueness, smoothness with respectto data, and nonuniqueness. The second part consists of Chapters IV, VI, and VIIand covers the basic results concerning linear differential equations. The third partconsists of Chapters VIII, IX, and X and covers nonlinear differential equations.Finally, Chapters V, XI, XII, and XIII cover the basic results concerning powerseries solutions.

The particular contents of each chapter are as follows. The fundamental exis-tence and uniqueness theorems and smoothness in data of an initial problem areexplained in Chapters I and II, whereas the results concerning nonuniqueness areexplained in Chapter III. Topics in Chapter III include the Kneser theorem andmaximal and minimal solutions. Also, utilizing comparison theorems, some suf-ficient conditions for uniqueness are studied. In Chapter IV, the basic theoremsconcerning linear differential equations are explained. In particular, systems withconstant or periodic coefficients are treated in detail. In this study, the S-N de-composition of a matrix is used instead of the Jordan canonical form. The S-Ndecomposition is equivalent to the block-diagonalization separating distinct eigen-values. Computation of the S-N decomposition is easier than that of the Jordancanonical form. A detailed explanation of linear Hamiltonian systems with constantor periodic coefficients is also given. In Chapter V, formal power series solutionsand their convergence are explained. The main topic is singularities of the firstkind. The convergence of formal power series solutions is proven for nonlinearsystems. Also, the transformation of a linear system to a standard form at a sin-gular point of the first kind is explained as the S-N decomposition of a lineardifferential operator. The main idea is originally due to R. Gerard and A. H. M.Levelt. The Gerard-Levelt theorem is presented as the S-N decomposition of amatrix of infinite order. At the end of Chapter V, the classification of the singu-

vii

viii PREFACE

larities of linear differential equations is given. In Chapter VI, the main topics arethe basic results concerning boundary-value problems of the second-order lineardifferential equations. The comparison theorems, oscillation and nonoscillation ofsolutions, eigenvalue problems for the Sturm-Liouville boundary conditions, scat-tering problems (in the case of reflectionless potentials), and periodic potentials arestudied. The authors learned much about the scattering problems from the bookby S. Tanaka and E. Date [TD]. In Chapter VII, asymptotic behaviors of solu-tions of linear systems as the independent variable approaches infinity are treated.Topics include the Liapounoff numbers and the Levinson theorem together withits various improvements. In Chapter VIII, some fundamental theorems concern-ing stability, asymptotic stability, and perturbations of 2 x 2 linear systems areexplained, whereas in Chapter IX, results on autonomous systems which includethe LaSalle-Lefschetz theorem concerning behavior of solutions (or orbits) as theindependent variable tends to infinity, the basic properties of limit-invariant setsincluding the Poincar6-Bendixson theorem, and applications of indices of simpleclosed curves are studied. Those theorems are applied to some nonlinear oscillationproblems in Chapter X. In particular, the van der Pot equation is treated as botha problem of regular perturbations and a problem of singular perturbations. InChapters XII and XIII, asymptotic solutions of nonlinear differential equations asa parameter or the independent variable tends to its singularity are explained. Inthese chapters, the asymptotic expansions in the sense of Poincare are used mostof time. However, asymptotic solutions in the sense of the Gevrey asymptotics areexplained briefly. The basic properties of asymptotic expansions in the sense ofPoincare as well as of the Gevrey asymptotics are explained in Chapter XI.

At the beginning of each chapter, the contents and their history are discussedbriefly. Also, at the end of each chapter, many problems are given as exercises. Thepurposes of the exercises are (i) to help the reader to understand the materials ineach chapter, (ii) to encourage the reader to read research papers, and (iii) to helpthe reader to develop his (or her) ability to do research. Hints and comments formany exercises are provided.

The authors are indebted to many colleagues and former students for their valu-able suggestions, corrections, and assistance at the various stages of writing thisbook. In particular, the authors express their sincere gratitude to Mrs. Susan Cod-dington and Mrs. Zipporah Levinson for allowing the authors to use the materialsin the book Theory of Differential Equations by E. A. Coddington and NormanLevinson.

Finally, the authors could not have carried out their work all these years withoutthe support of their wives and children. Their contribution is immeasurable. Wethank them wholeheartedly.

PFHYS

March, 1999

CONTENTS

Preface vii

Chapter I. Fundamental Theorems of Ordinary Differential Equations 1

I-1. Existence and uniqueness with the Lipschitz condition 1

1-2. Existence without the Lipschitz condition 81-3. Some global properties of solutions 15

1-4. Analytic differential equations 20Exercises I 23

Chapter II. Dependence on Data 28II-1. Continuity with respect to initial data and parameters 2811-2. Differentiability 32Exercises II 35

Chapter III. Nonuniqueness 41

III-1. Examples 41111-2. The Kneser theorem 45111-3. Solution curves on the boundary of R(A) 49111-4. Maximal and minimal solutions 52111-5. A comparison theorem 58111-6. Sufficient conditions for uniqueness 61Exercises III 66

Chapter IV. General Theory of Linear Systems 69IV-1. Some basic results concerning matrices 69IV-2. Homogeneous systems of linear differential equations 78IV-3. Homogeneous systems with constant coefficients 81

IV-4. Systems with periodic coefficients 87IV-5. Linear Hamiltonian systems with periodic coefficients 90IV-6. Nonhomogeneous equations 96IV-7. Higher-order scalar equations 98Exercises IV 102

Chapter V. Singularities of the First Kind 108V-1. Formal solutions of an algebraic differential equation 109V-2. Convergence of formal solutions of a system of the first kind 113V-3. The S-N decomposition of a matrix of infinite order 118V-4. The S-N decomposition of a differential operator 120V-5. A normal form of a differential operator 121

V-6. Calculation of the normal form of a differential operator 130V-7. Classification of singularities of homogeneous linear systems 132Exercises V 137

ix

x CONTENTS

Chapter VI. Boundary-Value Problems of Linear DifferentialEquations of the Second-Order

VI- 1. Zeros of solutionsVI- 2. Sturm-Liouville problemsVI- 3. Eigenvalue problemsVI- 4. Eigenfunction expansionsVI- 5. Jost solutionsVI- 6. Scattering dataVI- 7. Reflectionless potentialsVI- 8. Construction of a potential for given dataVI- 9. Differential equations satisfied by reflectionless potentialsVI-10. Periodic potentialsExercises VI

Chapter VII. Asymptotic Behavior of Solutions of Linear SystemsVII-1. Liapounoff's type numbersVII-2. Liapounoff's type numbers of a homogeneous linear systemVII-3. Calculation of Liapounoff's type numbers of solutionsVII-4. A diagonalization theoremVII-5. Systems with asymptotically constant coefficientsVII-6. An application of the Floquet theoremExercises VII

Chapter VIII. StabilityVIII- 1. Basic definitionsVIII- 2. A sufficient condition for asymptotic stabilityVIII- 3. Stable manifoldsVIII- 4. Analytic structure of stable manifoldsVIII- 5. Two-dimensional linear systems with constant coefficientsVIII- 6. Analytic systems in R2VIII- 7. Perturbations of an improper node and a saddle pointVIII- 8. Perturbations of a proper nodeVIII- 9. Perturbation of a spiral pointVIII-10. Perturbation of a centerExercises VIII

Chapter IX. Autonomous SystemsIX-1. Limit-invariant setsIX-2. Liapounoff's direct methodIX-3. Orbital stabilityIX-4. The Poincar6-Bendixson theoremIX-5. Indices of Jordan curvesExercises IX

CONTENTS xi

Chapter X. The Second-Order Differential Equation

d-t2 + h(x) dt + g(x) = 0 304

X-1. Two-point boundary-value problems 305

X-2. Applications of the Liapounoff functions 309

X-3. Existence and uniqueness of periodic orbits 313

X-4. Multipliers of the periodic orbit of the van der Pol equation 318

X-5. The van der Pol equation for a small e > 0 319

X-6. The van der Pol equation for a large parameter 322

X-7. A theorem due to M. Nagumo 327

X-8. A singular perturbation problem 330

Exercises X 334

Chapter XI. Asymptotic Expansions 342

XI-1. Asymptotic expansions in the sense of Poincare 342

XI-2. Gevrey asymptotics 353

XI-3. Flat functions in the Gevrey asymptotics 357

XI-4. Basic properties of Gevrey asymptotic expansions 360

XI-5. Proof of Lemma XI-2-6 363

Exercises XI 365

Chapter XII. Asymptotic Expansions in a Parameter 372

XII-1. An existence theorem 372

XII-2. Basic estimates 374

XII-3. Proof of Theorem XII-1-2 378

XII-4. A block-diagonalization theorem 380

XII-5. Gevrey asymptotic solutions in a parameter 385

XII-6. Analytic simplification in a parameter 390

Exercises XII 395

Chapter XIII. Singularities of the Second Kind 403

XIII-1. An existence theorem 403

XIII-2. Basic estimates 406

XIII-3. Proof of Theorem XIII-1-2 417

XIII-4. A block-diagonalization theorem 420

XIII-5. Cyclic vectors (A lemma of P. Deligne) 424

XIII-6. The Hukuhara-T rrittin theorem 428

XIII-7. An n-th-order linear differential equation at a singular pointof the second kind 436

XIII-8. Gevrey property of asymptotic solutions at an irregularsingular point 441

Exercises XIII 443

References 453

Index 462

CHAPTER I

FUNDAMENTAL THEOREMS OFORDINARY DIFFERENTIAL EQUATIONS

In this chapter, we explain the fundamental problems of the existence and unique-ness of the initial-value problem

(P)dy

= y(to) _ 4odt

in the case when the entries of f (t, y) are real-valued and continuous in the variable(t, y), where t is a real independent variable and y is an unknown quantity in Rn.Here, R is the real line and R" is the set of all n-column vectors with real entries.In §I-1, we treat the problem when f (t, y) satisfies the Lipschitz condition in W. Themain tools are successive approximations and Gronwall's inequality (LemmaIn §1-2, we treat the problem without the Lipschitz condition. In this case, ap-proximating f (t, y) by smooth functions, f-approximate solutions are constructed.In order to find a convergent sequence of approximate solutions, we use Arzeli -Ascoli's lemma concerning a bounded and equicontinuous set of functions (Lemma1-2-3). The existence Theorem 1-2-5 is due to A. L. Cauchy and G. Peano [Peal]and the existence and uniqueness Theorem 1-1-4 is due to E. Picard [Pi1 and E.Lindelof [Lindl, Lind2J. The extension of these local solutions to a larger intervalis explained in §1-3, assuming some basic requirement.,, for such an extension. In§I-4, using successive approximations, we explain the power series expansion of asolution in the case when f (t, y) is analytic in (t, y). lit each section, examples andremarks are given for the benefit of the reader. In particular, remarks concerningother methods of proving these fundamental theorems are given at the end of §1-2.

I-1. Existence and uniqueness with the Lipschitz condition

We shall use the following notations throughout this book:

y=

Y1, f,

y2 f2

yn fn

!y = max{jy,j 1Y21-.. ItYn11.

In §§I-1, 1-2, and 1-3 we consider problem (P) under the following assumption.

Assumption 1. The entries of f (t, y) are real-valued and continuous on a rectan-gular region:

R = R((to,6),a,b) = {(t,yl: It - tot <a, ly - e< b},

where a and b are two positive numbers.

I

2 I. FUNDAMENTAL THEOREMS OF ODES

Since f (t, yam) is continuous on R, I f (t, yul is bounded. Let us denote by M themaximum value of I f (t, y-) I on R, i.e., M = mac I f (t, y) I. We define a positive

number a bya if M = 0,

amin Ca, Ml if M > 0.

To locate a solution in a neighborhood of its initial point, the number or plays animportant role as shown in the following lemma.

Lemma I-1-1. If ff = ¢(t) is a solution of problem (P) in an interval it - tol <a < a, then I'(t) - 41 < b in It - tol < a, i.e., (t,5(t)) E R((to,co),a,b) for.It - tol <a.

Proof

Assume that Lemma 1-1-1 is not true. Then, by the continuity of fi(t), thereexists a positive number $ such that

(i) < a,(ii)

1¢(t) - 41 < b for It - tol <'6,

I4(to+0) -QoI =b or 10(to-fl)-qol

The assumption a < a < a implies (3 < a. Hence, (t,j(t)) E R for It - tot < p.Therefore, I f (t, (t))I 5 M for it - to! < 0. From 4'(t) = f (t, ¢(t)) and 4(to) = 4o,it follows that

(1.1.2) fi(t) = co + / t f(s, 0(s))ds for It -tot < fi.io

Hence,

I4(t) - 4)1 = I1 f(s, i(s))ds < MIt - tol for It - tot < P.r

This implies that

(1.1.4)1 (t)-ZbI<-MQ<Ma<Ma<MM=b for It - tol<f.

In particular, 0) - cot < b. This contradicts assumption (ii).Similarly, we obtain the following result.

Lemma 1-1-2. If y' = 4(t) is a solution of problem (P) in an interval it - tol <a < a, then I¢(t,) - (t2)I < Mlt1 - t2l whenever t1 and t2 are in the intervalit - tot < a.

1. EXISTENCE AND UNIQUENESS WITH THE LIPSCHITZ CONDITION 3

Remark 1-1-3. Using Lemma 1-1-2, it is concluded that if y = j(t) is a solutionof problem (P) on an interval It - to I < it < a, then (to + & - 0) and (to - ar + 0)exist.

Now, consider problem (P) with Assumption 1 and the following assumption.

Assumption 2. The function f satisfies a Lipschitz condition on R: i.e., themexists a positive constant L /such that

If(t, 91) - f (t, W2)1:5 LIlh - 1121

whenever (t,yi) and (t, y2) are on the region R (cf. [Lip]).

Note that L is dependent on the region R. The constant L is called the Lipschitzconstant of f with respect to R.

The following theorem is the main conclusion of this section.

Theorem 1-1-4. Under Assumptions I and 2, them exists a unique solution ofproblem (P) on the interval It - to] < a.

In order to prove this theorem, the following lemma is useful.

Lemma 1-1-5 (T. H. Gronwall [Cr]). If(i) g(t) is continuous on to < t < ti,

(ii) g(t) satisfies the inequality

(1.1.5) 0 < g(t) < K + L g(s)ds on to < t < ti,to

then

(I.1.6)

onto<t< ti.Proof.

0 < g(t) < Kexp[L(t - to))

Set v(t) = J g(s)ds. Then, by (1.1.5), ddtt) < K+Lv(t) and v(to) = 0. Hence,io

(1.1.7) {exp[-L(t - to)]v(t)} < Kexp[-L(t - to)]

and, consequently,

exp[-L(t - to)]v(t) <

L

{1 - exp(-L(t - to)]).

This, in turn, implies that Lv(t) < K{exp[L(t - to)] - 1). Therefore, g(t) <K + Lv(t) < K exp[L(t - to)]. 0Proof of Theorem 1-1-4.

We shall prove this theorem in six steps by using successive approximationswhich are defined as follows:

(1.1.9)

t

Wi(t) _ 4o,

k=0,1,2,...$k+1(t) = Cq + f (s,Ok(s))ds,:o

Step 1. Each function 0k is well defined and (t,45k(t)) E R for It - tol < a(k = 0,1,2,...).

Proof.

We use mathematical induction in Steps 1 and 2. The statement above is evidentfor k = 0. Assume that (t, Ok(t)) E R on It - tol < a. Then, I f (t,Ok(t))I < M onit - toI < a. Hence,

IOk+I(t) -Q =t

f (s, Qsk(s))dsl < MIt - tol < Ma < MM = b.4 1 -

Thus, (t.0k+1(t)) E R for It - tol < a.

Step 2. The successive approximations given by (1.1.9) satisfy the estimates

MLkIlk+1(t) - mk(t)I < (k 1)1It - toIk+1

Proof.

for It - tol < a, k =0,1,2,... .

For k = 0, we have 1m1(t) - y5o(t)I = I J r f(s,co)ds` < MIt - tol. Assume thatco I

I&(t)(t)I < MLk1-1 It - tolk for it - tol < a. Then,

14k+I (t) - Qk(t)l = Jo`f (s, k(s)) - f (s k-1(s))} ds

< L I J r

tot kAkL I r It - toIkdsl = (ML1)r It - tolk+1.

+oo

Step 3. The series E (&(t) - 4k-1(t)) is uniformly convergent on the intervalk=1

It - tol<a.

Proof.

By virtue of the result of Step 2, for a given c > 0, there exists a positive integerN such that

°°/

M °o (LIt - tol)k M °O (La)kE I&(t) - 0k-I(t)I < L k1 < L k!

<k=N k=N k=N

Step 4. The sequence {k(t) : k = 0, 1, 2.... } converges to

- k-1(t))do(t) + F, I(&(t)k=1

uniformly on it - tol < a as k -+ +oo.

Proof.

Use the identity ¢N (t) = do (t) + E ($k (t) - $k_1(t)) and the result of Step 3.k=1

Step 5. fi(t) given by (I.1.10) is a solution of problem (P) on It - toI < a.

Proof.

The definitiontr

&+1(t) = e +J

AS, ('1k(s))ds,ca

the continuity of f and the results of Steps 3 and 4 imply that

fi(t) = 6o + f f(s, (s))ds.a

Step 6. fi(t) is the unique solution of problem (P) on It - tol < a.

Proof.

Suppose that y"= fi(t) is another solution of problem (P) on an interval it -toI <& < a for some &. Lemma I-1-1 and Remark 1-1-3 imply that (t, ti(t)) E R onIt - tol :5&. Note that

fi(t) + f (s, i1(s))ds

on It - tol < &. Hence,

At) - lL(t)I = f' {f(s,.(s)) - f(s,iG(S))} dsl <- L I fo Id(s) - i (s)Idsl

on it - tol < a. Now, by applying Lemma 1-1-5 with K = 0 to the inequality

L f Id(s) - +L(s)Idso

on to < t < to+&, we conclude that I¢(t) -tG(t)I = 0 on the interval to < t < to+&.Similarly, the same conclusion is obtained on the interval to - & < t < to. Steps 1through 6 complete the proof of Theorem 1-1-4. 0

The following lemma gives a sufficient condition that f (t, y") satisfies Assumption2 (i.e., the Lipschitz condition).

Lemma I-1-6. If Of (t' y-) (j = 1, 2, ... , n) exist, are continuous, and satisfy the8yj

condition: I of (t, y j

1

< Lo (j = 1, 2, .. - , n) for (t, y) E I x D, where I is an

interval, D is a region, and Lo is a constant, and if V is a convex open set, thenI f(t, y,) - f(t, y2)1 < nLollit - y"2I for t E z, gi E D, and y2 E D.

Proof.

Fix t E 1, and yl and #2 in V. Then, By, + (1 - O)y"2 E D for O < 6 < 1 since Dis convex. Setting g(9) = f (t, 09, + (1 - O)y2), we derive

9(O) = At, 92), 9(1) = f(t,91)-

44n

_ (yia - y2 j)' (t, 8y"1 +do 1:1=1

(1 - B)y'2),

where yh,1, Yh,2, ... , Yh.n are entries of yh (h = 1,2). Hence,

If(t,91)-f(t,y2)1 =II @(O)dO <nLolii-u21-

This is true for all t E Z, yl E D, and y2 E D. Thus, the lemma is proved.The following two simple problems illustrate Theorem 1-1-4.

Problem 1. Using Theorem 1-1-4, show that on the interval It - 11 < V"2-, theinitial-value problem

dy-

1

dt (3 - (t - 1)2)(9 - (y -5)2)'

y(1) - 5

has one and only one solution.

Answer.

The standard strategy is to find a suitable region

R = {(t,y): It-11 < a, ly-51 < b}

and find the maximum value M of if (t, y) I in R. Then, a = min (a, M)- Ifa > f, the problem is solved. In other words, the main idea is to trap thesolution curve in a suitable region R. To do this, for example, use the fact that thefunction

f ,y (3 - (t - 1)2)(9 - (y - 5)2)is continuous and satisfies the Lipschitz condition

If(t,Yi) - f(t,y2)1 < Iy' -Y21

5

on the region R= {(t, y) : It-11 < f, Iy-51 < 2} (cf. Exercise I-2). Furthermore,

If(t,y)I <- 1 = 1(3_(/)2)(9_22) 5

1Set a = V25, b = 2, and M =

5. Then,

a = min ( a, ) =min f, 2 min(v ,10) = f.

Therefore, the given initial-value problem has one and only one solution on theinterval It - 11 < f.

Problem 2. Using Theorem I-1-4, show that the initial-value problem

(P) dy = 1 4dt (1 + (t - 4)2)(5 + (y - 3)2)'

y() = 3

has one and only one solution on the interval -oo < t < +oc.

Answer.

The function1

,y(1 + (t - 4)2)(5 + (y - 3)2)

is continuous and satisfies the Lipschitz condition everywhere in the (t, y)-plane.

Also, If (t, y)I < everywhere. This implies that a = min(a, 5b), if Problem (P)is considered in a region

R = {(t, y) : It - 41 < a, ly - 31 < b}.

Hence, if a < 5b, Problem (P) has one and only one solution on the interval it - 41 <a. Now, this solution is independent of a due to the uniqueness. Therefore, we canconclude that (P) has one and only one solution on the interval -oo < t < +oo.

Remark 1-1-7. It is usually claimed that the general solution of the systemdt = f (t, y-) depends on n independent arbitrary constants. Theorem 1-1-4 verifies

this claim if we define the general solution very carefully, since the initial data (to, Clrepresent n arbitrary constants for a fixed to. However, if the reader looks into thisdefinition a little deeper, he will find various complicated situations. It is importantfor the reader to know that the number of independent arbitrary constants con-tained in the general solution depends strongly on the space of functions to whichthe general solution should belong. For example, if the differential equation

(E) tae = 1

is considered on the set j = {t E R : t A 0), the general solution is y(t) = cl +In Jtjfor t > 0 and y(t) = c2 + In JtJ for t < 0, where ct and c2 are independent arbitraryconstants. If the Heaviside function

H(t) ={1-I

if t > 0,if t < 0

is used, this general solution is given by

y(t) =c1 - c2 H(t)

+ c' + c2 + In JtI.2 2

Equation (E) may be looked at in terms of the generalized functions (or the distri-butions of L. Schwartz [Sc]). The transformation y = u + In Jti, changes equation(E) to

(E') t- = 0.

As a differential equation on the generalized functions, (E') is reduced to

(E") j = cb(t),where c is an arbitrary constant and b(t) is the Dirac delta function. Integrating(E"), we obtain u(t) = cH(t) + c, where c is another arbitrary constant. Therefore,the general solution of (E) is y(t) = cH(t) + c + in Jtt. For more information ondifferential equations on the generalized functions, see [Ko] for examples.

The following example might be a little strange. Let us construct solutions ofthe differential equation

(E)

as a formal Fourier series

(S)

L + 2y(cost + cos(2t)) = 0

+'y(t) _ am eimc

m=-oo

Inserting (S) into (E) we obtain

(R) imam + am-1 + am+1 + am-2 + am+2 = 0 (m E Z),where Z is the set of all integers. Equation (R) is a fourth-order difference equa-tion on am. Therefore, solution (S) contains four independent arbitrary constants.Of course, if (S) is restricted to be a convergent series, three of these four con-stants should be eliminated. A similar phenomenon may be observed if solutionsof a certain differential equation are expanded in positive and negative powers of

+oo

independent variables such as F, amxm. For such an example, see [Dw].m=-oo

1-2. Existence without the Lipschitz conditionTo treat the initial-value problem (P) without the Lipschitz condition, we need

some preparation.

2. EXISTENCE WITHOUT THE LIPSCHITZ CONDITION 9

Definition 1-2-1. A set.F of functions f (t) defined on an interval I is said to beequicontinuous on I if for every e > 0, there exists a b(e) > 0 such that ] f(tl) -f (t2)I < e for every f E .F whenever it, - t2I < b(E), tl E :r and t2 E Z.

For example, if the set F consists of all functions f such that f is continuouson a closed interval It - tol < a, f is continuous in it - to, < a, and If (t)l < L init - toy < a, then.F is erquioontinuous on the interval I = {t : It - tol < a). In fact,

since At 0 - f(t2) = J f'(s)ds for t1 E T and t2 E Z, we have ] f (tl) - f (t2)l < Et lt2

whenever It1 - t2l < b(c) = L.

Definition 1-2-2. A set F of functions is said to be bounded on the interval Z,if for every t E I there exists a non-negative number M(t) such that lf (t)I < M(t)for every t E Z and f E.F.

Lemma 1-2-3 (C. Arzela [Ar]-G. Ascoli [As]). On a bounded interval Z, let.F bean infinite, bounded, and equicontinuous set of functions. Then, F contains aninfinite sequence which is uniformly convergent on Z.

Proof.

Let Q be the set of all rational numbers contained in the interval Z. Then,(1) Q is dense in Z; i.e., for every 6 > 0 and every t E Z, there exists a rational

number r(t, b) E Q such that it - r(t, b)1 < b,(2) Q is an enumerable set; i.e., Q = {r, : j = 1, 2, 3, ... }.

Note that the choice of the rational number r(t, 6) is not unique. We prove thislemma in two steps.

Step 1. The set .F contains an infinite sequence { fh : h = 1, 2.... } such thatlim fh(r) exists for every rational number r E Q.

ho

Proof.

Choose from F a sequence of subsequences F. 1, 2.... }, j =1, 2, ... , such that

(i) .Fj +1 C .Fj for every j,(ii) t lim o f',,e(rl) = c, exists for every j.

To do this, first look at functions in F at t = r1. Using the boundedness of theset {f(r1) : f E .F}, we choose an infinite subsequence F1 = {f 1.t : t = 1, 2,... }from .F so that lim f1,t(r1) = cl exists. Next, look at the sequence F1 at t = r2.

1 +00

Using the boundedness of the set { f l,t(r2) : t = 1, 2.... }, we choose an infinitesubsequence F2 = {j 2 t : t = 1,2.... }from F1 so that lim h,t(r2) = c2 exists.

Continuing this way, we can choose subsequences .F,+1 from Y. successively.

Set fh = fh,h (h = 1, 2, 3.... ). Then, ft, E fl for every h >) since fh = fh,h E.Fh C .Fj if h > j. Therefore, lim fh(rj) = c, exists for every j, i.e., lira fh(r)- h-+oo h-+ooexists for every rational number r E Q.

Step 2. The sequence { fh : h = 1, 2.... } of Step 1 converges uniformly on theinterval I.

Proof

For a given positive number f and a rational number r E Q, there exist a positivenumber b(e) and a positive integer N(r, e) such that

lfh(t) - fh(r)I <c whenever it - rl < b(e),

Ifh (r) - ft(r)I < E whenever h > N(r, E) and I > N(r, E),

1J(r) - f (t)I < E whenever it - rl < b(e).

Now, observe that

Ifh(t) - ft(t)1 <- LAW - fh(r)I + Ifh(r) - ft(r)I + I h(r) - f!(t)I,

where t E I and r r= Q. Therefore, by choosing r = r (t, b ( 3)) , we obtain

(1.2.1) Ifh(t) - ft(t)I < E

whenever h > N (r I t, 6(3 bigg)) 3 I and t > N (r (t,b (3 ). 3)

Since the interval I is bounded, we can cover I by a ite number of openintervals 11, I2, ... ,1,(,) in such a way that the length of each interval I1 is not

greater than b (11) , where p(E) is a positive integer depending on E. For every t,

/E \choose a rational number rt(e) E 1t f1 Q. For all t E It, set r t,b1 3) I = rt(E)

and set

(r,(.E),1

N(e) = 1 maxE N 3 )Then, (1.2.1) is satisfied whenever h > N(e) and I > N(E). 0

To construct a solution of the initial-value problem (P) without the Lipschitz con-dition, approximate the given differential equation by another equation which sat-isfies the Lipschitz condition. The unique solution of such an approximate problemis called an E-approximate solution. Using the following lemma, an E-approximatesolution is found.

Lemma 1-2-4. Suppose that f (t, y) is continuous on a rectangular region

R={(t,y): It - tol a, Iy-col<b}.Then, for every positive number f, there exists a function FF(t, y) such that

(i) F( is continuous for It - toI < a and all g,(ii) 1 has continuous partial derivatives of all orders with respect to yl, y2,... , y

f o r It - tot < a and all y",(iii) I FE(t,y)I <c** max If(r,r1)I for It -to l<aand all y",

r=

X.(iv) IFe(t,y)-f(t,y)I <e on

Proof.

We prove this lemma in three steps.

Step 1. There exists a function P(t, yj such that(1) F is continuous for It - tot < a and all y,(2) IF(t, y)I < c >SERI If (-r, for It - tot < a and all f,

(3)

{f(tF(t, yJ =

0

on R,for It-tot <a and If-cal>b+1.

The construction of such an f is left to the reader as an exercise. Since f isuniformly continuous on R, properties (1), (2), and (3) of f imply that

(4) for every positive number e, there exists a positive number b(e) such that

IF(t,f1)-F(t,f2)i <e whenever It - tot < a, Ig, - f2l < b(e).

Step 2. For every positive number b, there exists a real-valued function p(t, 6) ofa real variable t such that

(a) p has continuous derivatives of all orders with respect to for -cc < t < +oo,(b) p(4, b) > 0 for -co < t < +oc,(c) p(.,b)=0forI{I>b,

(d) f(.5)de = 1.,o

The construction of such a function p is left to the reader as an exercise.

Set 0(g,6) = p(yi, b)p(y2, b) ... p(y,,, b). Then(a') has continuous partial derivatives of all orders with respect to y1, y2,.,. , y

for ally",(b') t(#, 6) 0 for all(c') 0(f, b) = 0 for Iyj > 6,

+00 +00(d') r ... f '(y, 6)dy1dy2 ... d1M = 1.fStep 3. Set

4(t, YJ = 1-00+- ...J+oo (y

- 1, b(E))F(t, ff)d ... dnn,00

where b(e) is defined in (4) of Step 1. Then, F,(t, y-) satisfies all of the requirements(i), (ii), (iii), and (iv) of Lemma 1-2-4.

Proof

Properties (i) and (ii) are evident. Property (iii) follows from the estimate

I e(t,M1<max{IF(t,10i: all ;t} j00

< iTMaExRIf(T,nl1.

Property (iv) follows from the estimate

I A, (t, yi - F'(t, Y) Ir+oo +

00

=l j (y-;),b(E)){F(t,n7)-F(t,)}dg1...dgl

r (9 - n,b(E)) {F(t,nj - F(t,y-) } dm ... dn._ ...

E tI ... r- n, b(E))drh ... dnn l = E.Il ff JW- l1 <_a(E)

This completes the proof of Lemma I-2-4.Now, we prove the fundamental existence theorem without the Lipschitz condi-

tion.

Theorem 1-2-5. If f (t, y-) is continuous on a region

P- = {(t,y): It-tol < a, I#-col < b}and M = ma c I f (t, )l, there exists a function fi(t) such that

(i) (to)(ii) 4'(t) = f (t, 0(t)) on It - toy < a,

whena

/if M = 0,

minla,il) if M > 0.

Proof.We prove this theorem in four steps.

Step 1. For every positive number e, construct a function FE (t, y1 which satisfiesall of the requirements given in Lemma 1-2-4. Using Theorem I-1-4, we constructa function 0 (t) such that

(1)S(,E(to) =_4o,

(2) P. (t, E(t)) on It - toI < o.Furthermore,

(3) (t,4(t)) E 1 on It - toj < a (cf. Lemma 1-1-1 and Remark 1-1-3).

Step 2. The set F = {¢E : E > 0} is bounded and equicontinuous on It - toI < a.

ProofProperty (3) of functions ¢E implies that F is bounded on It - tol < a and that

lFE(t, jE(t))I < M on It - tot < a. Hence, property (2) of ¢ implies that

I4(t1) - 4(t2)1:5 ljtl iE(s)dsl <Mlt1-

t21t2

if Iti - tol < a and It2 - t0I < a (cf. Lemma 1-1-2). Therefore, for a given positivenumber p, we have I0E(t0 - 0E(t2)l < µ whenever Itl - toI < a, It2 - toI < a, andIt1 - t21 < -E. This shows that Jr is equicontinuous on It - tol < a.

Step 3. Using Lemma 1-2-3 (Arzeli -Ascoli) choose a sequence {ee : I = 1, 2,. .. } of}positive numbers such that lim e, = 0 and that the sequence l = 1, 2....

r+ooconverges uniformly on It - tot < or as t -- +oo. Then, set

,(t) = lim ,,(t) on It - to, < a.+oo

Step 4. Observe that

(t) = c"o + fP, (s, ,(s))ds

and that

=ca+ f(s,0,(s))ds+J {f (s,t (s))-f(s,4(s))}dseo to

Ifca(s)) - f(s,Z (s))Ids < e It - tol on It - to; < a.

This is true for all e > 0. Letting a - 0, we obtain

(t} = c"o + f L f (s, 0(s))ds.u

This completes the proof of Theorem 1-2-5. 0

Example 1-2-6. Theorem 1-2-5 applies to the initial-value problem

(P)dy

dxxyl/s, y(3) = 0.

However, Theorem I-1-4 does not apply to (P). In fact y(x) = 0 is a solution ofProblem (P) and also

0 for x < 3,

y(x) _ (2(x2_ 9)1545

for x > 3

is a solution of Problem (P). Note that the right-hand side of the differentialequation (P) does not satisfy the Lipschitz condition on any y-interval containingy=0.

Two other methods of proving Theorem 1-2-5 are summarized in the followingtwo remarks.

Remark 1-2-7. Let every entry of an n-dimensional vector f (t, y-) be a real-valuedcontinuous function of n + 1 independent variables t and y = (yl, ... , y,,) on arectangular region R = {(t, y-) : It - tot < a, Iy"-cod < b}. Assume that I f(t,yj 1 < M

on R, where M is a positive number. Set a = min (a, M ) . Since f (t, yI is

uniformly continuous on R, there exists a positive n umber p(e) for every givenpositive number a such that I f (tl, y1) - f (t2,1%2)I < E Whenever Itl - t2[ < p(c) andIyl - 1%2I 5 p(e) for (xl, WI) E R and (x2, y2) E R. Let 0(e) : to = ro < r1 <

< rn(c) = to + a be a subdivision of the interval to < t < to + a such thatn(c)-1 p(E)m ax (I rj - rj+ll) < nun P(E),

M. Set

co for t = to,

91(t) = 1/e(rj-1) + J (r)-1,1/c(rj_1))(t - 7-j-1) for rj_1 < t < TI,(j = 1,2,... ,n(6)).

Then, we can show that(i) every entry of y, (t) is piecewise linear and continuous in t and (t, &(t)) E R

on the interval to < t < to + a,(ii) the set {yi : E > 0} is bounded and equicontinuous on the interval to < t <

to +a,(iii) if we choose a decreasing sequence {c) : j = 1, 2,... } such that Jim E) = 0

-+ooin a suitable way, the sequence {y',, (x) j = 1,2,. .. } converges to a solutionof the initial-value problem

(1.2.2)dt = f (t, J- , i(to) = 0o

uniformly on to < t < to + a as j -+ +oo.For more details, see [CL, pp. 3-5].

Remark 1-2-8. We use the same assumption, M, and a as in Remark 1-2-7. Also,let i1(t) be a continuously differentiable function of t such that #(to) = co and(t, #(t)) E R on the interval to - r < t < to, where r is a positive number. Forevery e such that 0 < e < r, define

q(t) for to - r < t < to,1%c(t) = c

clo+ff(s,il(s_6))ds for to < t < to + a.o

Then, we can show that(i) every entry of g, (t) is continuous in t and (t, &(t)) E R on the interval to-r <

t<to+a,(ii) every entry of 17, (t) is continuous in t on to - r < t < to +a except a jump at

x = to,(iii) the set {y'E : 0 < c < r} is bounded and equicontinuous on the interval

to -r<t<to +a,(iv) if we choose a decreasing sequence {c) : j = 1, 2, ... } such that lim e) = 0)+no

in a suitable way, the sequence {yEI (x) j = 1, 2, ... } converges to a solutionof the initial-value problem (1.2.2) uniformly on xo < x < xo+a as j - +oo.

For more details, see [CL, pp. 43-44] and [Hart, pp. 10-11].

3. SOME GLOBAL PROPERTIES OF SOLUTIONS 15

Remark 1-2-9. If Ax, y-) is assumed to be measurable for each fixed y, continuousin y for each fixed t, and I f (t, yj I is bounded by a Lebesgue-integrable function when(t, y") E R, a result similar to Theorem 1-2-5 (Caratheodory's existence theorem)[Ca, pp. 665-688) is obtained. Similarly, if, in addition, it satisfies an inequalitysimilar to Lipschitz condition with the Lipschitz constant replaced by a Lebesgue-integrable function L(t) when (t, y-) E R, we obtain also a result similar to Theorem1-1-4. (See [CL, p. 43), [Har2, p. 10], and [SC, p. 15).)

1-3. Some global properties of solutions

We can construct a solution to an initial-value problem (P) in a neighborhoodof the initial point by using Theorem 1-1-4 or 1-2-5. To extend such a local solutionto a larger interval of the independent variable t, the following lemma is useful.

Lemma 1-3-1. Assume that(i) a function f(t, yi) is continuous in an open set D in the (t, y-) -space,

(ii) a function fi(t) satisfies the condition ;s (t) = f(t,a(t)), and (t, Q(t)) E 'D, inan open interval T = {t : ri < t < r2}.

Under this assumption, if lim (t,, d(t,)) = (r1, ij) E D for some sequence {t,)tooj = 1,2.... } of points in the interval I, then lim (t, (t)) = (r1, ij). Similarly, if

lim (t,, 0(t,)) _ (T2,771 E D for some sequence {t. : j = 1,2,... } of points in thej+oointerval Z, then lim (t, fi(t)) = (r2i

Prof.

We shall prove the part of the lemma which concerns the left endpoint ri of theinterval Z. The other part can be proven similarly.

Let U be an open neighborhood of (ri, rl. We show that (t, 0(t)) E U in aninterval ri < t < r(U) for some r(U) which is determined by U. Assume that theclosure of U is contained in V and that I f (t, y-) I < M in U for some positive numberM. For every positive integer j and every positive number e, consider a rectangularregion R,(e) = {(t,y-) : it - t,I < e, Iy - 4(t,)j < Me}. Then, there exist an e > 0

1 E Rj (e) C U. Note that e =min Ce,!4fe

and a j such that (7-1,771/

MI and t, - e < rt .

Applying Lemma I-1-1 to the solution y" = d(t) of the initial-value problem

dy = f(t,y-), y(tj) = (tj), we obtain (t,y5(t)) E Rj(E) C U on the intervalr1 < t < tj. Since U is an arbitrary open neighborhood of (r1, i)), we conclude that

lim (ti,0(ti)) _ (ri,7-7) E D.-+oo

From Lemma 1-3-1, we derive the following result concerning the maximal inter-val on which a solution can be extended.

Theorem 1-3-2. If(i) a function f (t, yj is continuous in an open set V in the (t, y-')-space,

(ii) a function fi(t) satisfies the condition 4 (t) = f(t, ¢(t)) and (t, fi(t)) E D, inan open interval I = {t : r1 < t < r2},

(iii) 4(t) cannot be extended to the left of rl(or, respectively, to the right of r2)with property (ii),

(iv) jlimo(tj,0(ti)) = (TI,fl) (or, respectively, (r2, r))) exists for some sequence

{tj :j = 1,2....l of points in the interval Z,then the limit point (Ti, n (or, respectively, (T2, i))) must be on the boundary of V.

Proof

We shall prove this result by deriving a contradiction from the assumption that(r1,,7) E E) (or, respectively, (,r2, i-11 E D). Applying Lemma 1-3-1 to this situation,we obtain lim(t,¢(t)) = (rl,r) (or, respectively, (r2ir))). Hence, by applyingcrTheorem 1-2-5 to the initial-value problem

dy= f (t, y), y(rl) = ndt

(or, respectively, ff(r2)

the solution 45(t) can be extended to the left of rl (or, respectively, to the right ofr2). This is a contradiction. 0

The following example illustrates how to use Lemma 1-3-1.

Problem 1-3-3. Show that the solution of the initial-value problem

d2 y2

- 2xyL = y3 - y2 y(xo) = 17, Y (xo) = (,

exists at least on the interval 0 x < xo if xo > 0, q > 0, and (is any real number.

Answer.

Assume that the solution y = 4(x) of the given initial-value problem exists onan interval I = {x : £ < x < xo} for some positive number . Observe that

L2(.0')2+ 1.63 + 20-2] = 2xo(th')2 > 0

on Z. Hence,

(O'(x))2 + 14(x)3 + 14(x)-2 < 1(0'(xo))2 + 14(x0)3 + 14(x0)-2 (= Al > 0)3 2 2 3 2

on Z. Therefore, we have

z < 2M, 4(x)3 < 3M, 4(x)2(4 (x)) >21Lt

Now, apply Lemma 1-3-1.The proof of the following result is left to the reader as an exercise.

Corollary 1-3-4. Assume that f (t, yam) is continuous fort I < t < t2 and all y E R".Assume also that a function fi(t) satisfies the following conditions:

(a) and d+' are continuous in a subinterval I of the interval tl < t < t2,(b) (t) = f (t, ¢(t)) in Z.

Then, either(i) Qi(t) can be extended to the entire interval t1 < t < t2 as a solution of the

differential equation j = f (t, y), or

(ii) limO(t) = oo for some r in the interval ti < r < t2.t-rUsing Corollary I-3-4, we obtain the following important result concerning a

linear nonhomogeneous differential equation

dt = A(t)yy + bb(t),

where the entries of the n x n matrix A(t) and the entries of the R"-valued functionb(t) are continuous in an open interval I = It : ti < t < t2}.

Theorem 1-3-5. Every solution of differential equation (1.3.1) which is defined ina subinterval of the interval I can be extended uniquely to the entire interval I asa solution of (1.3.1).

Proof

Suppose that a solution y = d(t) of (1.3.1) exists in a subinterval Z' _ It : r1 <t < r2} of the interval I such that tj < 71 < r2 < t2. Then,

At)I <- Ato)I + jo

{A(s)4(s) + b(s)} dsl

in Z', where to is any point in the interval T. Setting

K = (r2 - ri) max Ib(t)I,'15'572

L = IA(t)I,

we obtain

in T.o

Thus, the estimate

Kexp[Llt - toll < Kexp[L(r2 - ri)[ <00 in Z'

is derived by using Lemma 1-1-5. Hence, case (ii) of Corollary 1-3-4 is eliminated.Since the right-hand side of (1.3.1) satisfies the Lipschitz condition, the extensionof the solution y5(t) is unique.

For nonlinear ordinary differential equations, we cannot expect to obtain thesame result as in Theorem 1-3-5. The following example shows the local blowup ofsolutions of the differential equation of Problem 1-3-3.

Problem 1-3-6. Show that for any fixed t > 0, there exists a real-valued function0(x) such that

(1) 0(x) is continuous on an interval 0 < C-x <- 6 for some small positive number6 which depends on (,

(2) fix) is continuous on an interval 0 < - z < 6,

(3) the function

y(x) -1 + (: - x)4(x)

F( - x)

satisfies the differential equation

(Eq) dxy - 2xyL y2y3

ono<x<6.Answer.

Step 1. Ifwesett=£-randyto

(1)

=t

, the given differential equation is changed

t2u" - 2tu' + 2u + 9(tu' - u)u = 2t(tu' - u)u + tsu-3 - tut,

where f =dt

1+u'1Step2. If wesetu=

(2)

and tu' differential equation (1) becomes

=w2,

_ - 2(1 + wl) + 3w2 - 2(w2 - (1 + wl))(1 + w1)

+ 2r1t(w2 - (1 +wl))(1 + w1)

+ t6 (1 + w1)-3 - (1 + w1)2.

If we set t = 0, wl = 0, and w2 = 0, the right-hand side of (2) vanishes. Now, (2)can be written in the form

td!W--'dt

dw2t

(3) dt

= W2,

=2w1+w2-2w1(w2-wl)

+ 2{-lt(w2 - (I + w1))(1 + w1)

+ t6t4(1 + w1)-3 - tt-1(I + wl)2.

Step 3. Write (3) in the following form:

(4) te = f2w +'Pp) + tG(t,w"),

where

_ wl _ 1 0w- [w2]' ['2 1] F(w) -[2w1(w2-wi)J'

with

(5)

G(t,w) = 22-'(w2 - (1 +w1))(1 +w1)

+ t5e(1 + w1)-3 - t:-'(1 + wl)2.

Two eigenvalues of f2 are -1 and 2, which are distinct. Let us diagonalize SZ by thetransformation

w" = Pv", where P =

Then, (4) becomes

(6) tv = Av + P-' F(PV) + tP-1G(t, Pv), A = [ O1

2} .

Set ii= ti. Then, t9 = ti+t2z and F(Pv-) = t2F(Pz). Therefore, (6) becomes

(7) ti' = Aoz + At' zl,

where Ao = [_20 I and f (t, z) = tP-'F(Pz) + P-'C(t, tPa). If t > 0 and

if tl1 and tll are small, there exist two positive constants Ko and L such thatIf(t, z-) <- tLli - {I and If(t,z)1 <- Ko + tL17.

Step 4. We change (7) to the following system of integral equations:

(8)

tzi(t) = t'2 fo tfi(r,z(t))dt

tz2(t) = d + t f r 2f2(t, t(t)o

)dt,

where to 1is a sufficiently small positive number and c is an arbitrary constant, and

z' = (21J

and f = 2J .

We want to construct a bounded solution of (8) on

0 < t < to. To do this, first note that, if IZ(t)I < K on 0 < t < to, we obtain

ttIzt(t)I < t-2 fo rI f1(r, zlr))Idr < t-2 r(Ko + rLK)dr

0

20+3LK < 20+ 3LK,

(9)

to

Izs(t)I < Iclt + t f r-21 f2(r, z(r))Idrtto

< Icit + tf

r-2(Ko + rLK )drt

= Iclt+t (' - 1 )Ko+t (In (12)) LK

< Iclto + K0 + e-1 t0LK.

a positive number co, fix another positive number K so that co + Ko < 2 andFor

then fix t0 > 0 so that to < 1, toL < 1, and toKIPI < 1. Here, IPI = max{IP.kl1 < j, k < 2}, where P = (Pjkj? k=1. Define successive approximations by

t_2Joi (t, C) =

dm > 0-

r-2f2(r,-im(r,c))dr+t JJJ(to

It can be shown that the successive approximations converges uniformly for Icl < Coand 0 < t < to. This establishes the existence of a bounded solution of the integralequation (8). Thus, we can complete the construction of solution (Sol) of equation(Eq).

1-4. Analytic differential equations

So far, it has been assumed that the independent variable t and the unknownquantity y are real. Since a complex number is a two-dimensional real vector, everyresult in the previous sections can be extended to the case when the entries of g'and of the function f (t, g) are complex. Furthermore, if the function f is analyticwith respect to (t, yam), we can construct analytic solutions by means of successiveapproximations. Denote by C the set of all complex numbers. The set of all n-column vectors with complex entries is denoted by C".

4. ANALYTIC DIFFERENTIAL EQUATIONS 21

Theorem 1-4-1. Assume that each entry of a Cn-valued function f (z, yam) is givenby a power series in (z, yl, y2, ... , yn) which converges in a polydisk:

D= ((z, y-) : Iz - zol < a, ly - col < b},

where z, yl, y2, , y" are complex quantities. Assume also that there exist positivenumbers M and L such that

If(z,-)I < /M in D,

l 0z, 91) - 1(z, g2)1 5 Lull - y21 whenever (z,y,) E D (j = 1,2).

Define successive approximations by

1m0(z) = 4o,

om(Z)=co+J f((,0-m-1( ))d(, m>1.0

where the path of integration may be taken to 1be the line segment zaz joining zo to

z in the complex z-plane. Set a = min I a,M

}. Then,

(i) For every m, the entries of the C"-valued))) function 4m(z) are power series inz - zo which are convergent in the open disk A = {z : iz - zoj < a},

(ii) the sequence { ,,,(z) m = 0,1, ... } satisfies the estimatesII ,c,4n+l(Z) - om(Z)I

!5 m

(nl + 1),IZ - zojm+l on 0, m = 0, 1, ... ,

(iii) the sequence {¢m(z) m = 0, 1,... } converges to

¢(z) = o(z) + (mk(z) - (Z))

k=1

uniformly in 0 as m -+ +oo,(iv) the entries of the C"-valued function b(z) are power series in z - zo which

are convergent in 0,(v) the function Q(z) is the unique solution to the initial-value problem

d = f(z,y), y(zo)

The proof of this theorem is left to the reader as an exercise. The reader must noticethat even if a sequence of analytic functions fn(t) of a real variable t converges to afunction f (t) uniformly on an interval Z, the limit function f (t) may not be analyticat all. For example, any continuous function can be uniformly approximated bypolynomials on a bounded closed interval. In order to derive analyticity of f (t),uniform convergence must be proved on a domain in the complex t-plane.Observation 1-4-2. The estimates

m+1(Z) - & (Z) 1 :5MLm

(m+ 1)IIZ - ZOlm+1 in A, n1 = 0, 1,...

+00

imply that there exists a power series Eck(z - zo)k such thatk=0

M

&(z) = E ck(z - z,)' +O(Iz - zol"'+1),k=O

Observation 1-4-3. Since

m

m=0,1,2,....

&(Z) = $O(Z) + E (&(Z)k=1

00

we obtain (z) - ¢,,,(z) = O(Iz - zol'"+1) and, hence, (z) _ >2Ck(Z - ZO)k.k=0

Observation 1-4-4. Set

m 00

(Z) = Ck(Z - ZO)k + 1: Cm.k(Z - Zo)k.k=0 k=m+1

Since cb1 < b in the disk 0, it follows that

1Cm,kI aNote that

(k = m + 1, m + 2,...).

1 0(z) - cak+14.,k =

2?ri Izo1=P (z - zo)

for any positive number p < a. This, in turn, implies that

1-IZZIDI

in the disk A.Example 1-4-5. In the case of the initial-value problem dzy = (1 - z)y2, y(0) = 2,we obtain

O(z) = 2 E(k + 1)zk =2

(1 - Z)2k=O

,n oo

Ck(Z-ZO)kl :5 b IZ aZOlkk=O k=m+1

1=blzzDI

a

EXERCISES I

by using the successive approximations

23

Oo(z) = 2 ,

01(z) = 2 + 4z - 2z2,

02(x) = 2 + 4z + 6x2 -83

- 6x4 + 4x523

¢3(z) = 2 + 4z + 6z2 + 8x3 _2x4 2825 82zs

3 3 9

256z7 124z8 68z9 46z10 56z"

+ 63 + 9 9 9 + 922z12 4z13 2z14

9 + 9 - 63

Combining Theorem 1-3-5 and Theorem 1-4-1, we obtain the following theorem:

Theorem I-4-6. If all the entries of an n x n matrix A(t) and C"-valued functionb(t) are analytic on an interval Z, then every solution of linear differential equation(1.3.1) is analytic on Z.

EXERCISES I

I-1. Solve the initial-value problem

d22 + 2y L = 0, y(r) = ,lo, b (r) = 171

The reader must consider various cases concerning the initial data (r, i , m)1).

I-2 Show that the function f (x y) _ 1

' (3 - (x - 1)2)(9 - (y - 5)2)

Lipschitz condition

If(x,yl)-f(x,y2)I < IY1Y2I

ifIx-1I 5 f, Iyi-5I < 2,

1-3. Show that the initial-value problem

(P)dy _ x3 + 3x + 1dx - sm

101-y2)

lye-51 < 2.

y(5) = 3,

satisfies the

has one and only one solution on the interval Ix - 51 < 7.

1-4. Suppose that u(x) is continuous and satisfies the integral equationx

u(x) = J sin(u(t))u(t)pdt0

on the interval 0 < x < 1. Show that u(x) = 0 on this interval if p > 0. Whatwould happen if p < 0?

-pHint. In case p < 0, the problem is reduced to the initial-value problem

udu =

sin(u)dx, u(O) = 0. Thus, the solution u(x) is not identically zero and has a branch pointat x = 0.

1-5. Show that if real-valued continuous functions f (x), g(x), and h(x) satisfy theinequalities

ff(x) ? 0, g(x) < h(x) +J0*'C

on an interval 0 < x < xo, then

ono < x <

g(x) < h(x) + f=

exp[JZ

f(n)dn, } di

xo.

Hint. If we put y(x) = J f then LY < f(x)h(x) + f (x)y and y(O) = 0.

1-6. Let a(t) be a real-valued function which is continuous on the interval I = It:0 < t < 11. For every positive integer rn, set

fm (t) = I + ta(0) for 0 < t < 1-mand

fm(t)= L=0(1+ma(m))]( l+lt-m/a\m)j

fork < t < k + 1and k = 1, 2, ... , rn - 1. Show that the sequence

m m1, 2, ... } converges uniformly on the interval Z. Also, find lim f,,,.m-oo

Hint. For large m, we have 1 + m a (m/

> 0 and

{fm

h m .exp [ma(m/ - m21 < 1 + ma \m! < exp [M1 a(m)]for some positive number K. Also

` lfta(r)dr-Lmn(mj-(t-mja(ml)E,,,

M -0

for some positive e,,, such that

lira. Em =

0.m-» too

:M=

EXERCISES I 25

1-7. Let f (t, y) be a real-valued function of two independent variables t and y whichis continuous on a region:

D = {(t, y) : 91(t) 5 y <- 92(t), r1 < t < r2}.

Suppose that(i) 91, g'1, 92, and g2 are continuous on rl < t

(ii) 91(t) < 92(t) on rl < t < T2,(iii)

< T2,

fgi(t) < f(t,9i(t)), rl < t < rz,9t> f(t, 92(t),z

(iv) Tl < t0 < r2 and g1(to) < CO < 92(to)-

Show that there exists a function 0(t) such that(1) 4 and 0' are continuous on to < t < T2 and (t,4i(t)) E D on to < t(2) Q'(t) = f (t, O(t)) on to < t < 72 and 0(to) = co.

1-8. Set0 for - oo < y 5 0,

{ f for 0 < y < +oo.f(u) =

Also, let rt(x) be an arbitrary continuous function of x on an interval 0 < x < xo,where xo is a positive number. Define successive approximations by

rt(r) for in = 0,tim(x) =

foxf (ym_l(t))dt for in = 1, 2,

Show that lim ym(x) exists uniformly on the interval 0 < x < xo.m-+oo

Hint. For this problem, see [Na3, pp. 143-160; in particular Bemerkung 2].

1-9. Let B be a Banach space over the field R of real numbers. Also, let f (t, x)be a B-valued function of the variables (t, x) E R x B. Consider an initial-valueproblem

(IP) d = f (t, x), X(r) _ .

where (r, {) is a given point in R x B. Assume that f satisfies the following condi-tions:

(i) f(t,x)iscontinuous onaregion R={(t,x)ERxB:It-rl <a, IIx-t:ll<b},where a and b are positive numbers and II II is the norm of B,

(ii) f(t.x) also satisfies the Lipschitz condition IIf(t,xl) -f(t,x2)II 5 LIIX1 -X211whenever (t, xl) E R and (t, x2) E R, where L is a positive constant.

Show that(1) there exists a positive number Af such that Ilf(t,x)II < M for (t,x) E R,(2) problem (IP) has one and only one solution on the interval It - rI < or, where\

a= min a,b

bI.

Hint. See [Huk7].

26 1. FUNDAMENTAL THEOREMS OF ODES

I-10. Let B be a Banach space over the field R of real numbers and let C be anonempty compact subset of B. Also, on a bounded interval Z, let F be an infiniteand equicontinuous set of C-valued functions. Show that F contains an infinitesequence which is uniformly convergent on Z.

I-11. Let B be a Banach space over the field R of real numbers and let C bea nonempty compact and convex subset of B. Assume that a C-valued functionf (t, x) is continuous on a region

R={(t,x)ERxB:It-rl <-a,llx-t1l <-b},

where r E R, t E B, a > 0, b > 0, and 11 11 is the norm of B. Show that(1) there exists a positive number M such that IIf(t,z)115 M for (t,x) E R,

(2) the initial-value problem - = f (t, x), x(r) has a solution on the interval

it - rl < or, where a = min ( a, f ).

Hint. Use a method similar to that of Remark 1-2-7. See, also, [Huk7l.

1-12. Let f (t, x") be Re-valued and continuous for (t, x") E Rr+t. Assume also thatE is a nonempty closed set in R"+t. Show that in order that for every (r,)) E E,there exists a real number a(r, depending on (r, ) such that r < a(-r, {) and

that the intial-value problemdi

= f (t, xF), x(r) _ has a solution s = ¢(t, r,

satisfying the condition (t, ¢(t, r, ) E E on an interval r < t < o(r), it is necessaryand sufficient that for every (r, f E E and every positive number e, there exists apoint (a, S) E E such that

0<a-r<e and < E,

where (a, ) generaly depends on (r, {, e).

Hint. The necessity is easy to prove. For sufficiency, use an idea similar to that ofRemark 1-2-7. See, also, [Huk6].

1-13. Assume that f (x, yt, y2,... , y,,) is real-valued, continuous, and continuouslydifferentiable function of (x, yt, ... , on a region R. = {(x, yl,... , a < x <b, lyj I < r (j = 1,... , n)), where a and b are real numbers such that a < b, and r isa positive number. Assume, also, that (1) c is a real number such that a < c < b,l(2) 6(x) is a real-valued solution of !° = f (x, y, Ly, ... , fn-Y) on an interval

lz - cl < p, where p is a positive number such that a < c - p < c + p < b, (3)

r (k = 0,1,... n - 1). and (4) f(x,0,... ,0) = 0 on a < x < b.dx*

Show that if there eixsts a sequence {ch : h = 1, 2, ... } of real numbers such thatUm ch = c and that ¢(ch) = 0 (h = 1, 2, ... ), then O(x) = 0 identically on

h_+00a<x<b.

EXERCISES I 27

1-14. Let u(x, y) be a continuously differentiable function of two independent vari-ables (x, y) in a domain D = {(x, y) : y > 0, a + Ay < x < b - Ay}, where A is apositive number and the two quantities a and b are real numbers such that a < b.Assume that u(x, y) is continuous on the closure f) of D and satisfies the condition

< AI e I +BluF+C

on D, where B and C are positive numbers. Set M = max(Iu(x,0)(: a < r < b}.Show that

u(x, y) 1 5 Me 11 +

B

(eBV - 1)

on D.

Hint. The last inequality is the Haar inequality. (See [Haa1, [Na5, pp. 51-561, and[Har2, pp. 140-141]).

1-15. Let H(u, t, x, q) be a function defined on an open set in R4 containing thepoint (u, t, x, q) = (0, 0, 0, 0) and satisfy a uniform Lipschitz condition with respectto (u, q). Let O(x) be a function of class C' satisfying 0(0) = 0 and 0=(0) = 0.Show that the initial-value problem

ut + H(u, t, x, u=) = 0, u(x, 0) = Q(x)

has at most one solution of class C1 in a neighborhood of x = 0.

Hint. This is Haar's uniqueness theorem of the given partial differential equation(cf. the references in Exercise 1-14). Upon applying Exercise 1-14, with M = C = 0,to the difference ti(t, x) = ul (t, x) - U2 (t, x) of two solutions ul(t, x) and U2 (t, x),the proof follows immediately.

CHAPTER it

DEPENDENCE ON DATA

The initial-value problem

(P) = fit, yam), YAto) = 4

is equivalent to the integral equation

(E) 11(t) = 4 + f (s, y(s))ds.t

The right-hand side of (E) is regarded as a function of (t, to, ca, f, yj. In this chap-ter, we explain some basic properties of solutions with respect to (t, to, ca, f). Aparameter a is used to represent the variable f. In §11-1, we explain the continuityof the solution of (P) with respect to (t, to, c"o, e) at (t, r, e, co) when the solution of(P) is unique for the initial data (r , c', ea) (Theorem 11-1-2). Theorem 11-1-2 wasfirst proved by I. Bendixson [Benl] for the scalar equation and later by G. Peano(Pea2] and E. Lindelof [Lind2] for a system of equations. In §11-2, we explain thedifferentiability with respect to initial data (to,ca) (Theorem 11-2-1) and with re-spect to a parameter a (Theorem 11-2-2). The discussion of these topics can befound in [CL, pp. 22-28, 57-60] and [Har2, pp. 94-100].

II-1. Continuity with respect to initial data and parametersLet e be a variable which takes values in a set with a separable topology. We

consider RI-valued functions f (t, y, e) and fi(t) under the following assumption.

Assumption 1. For the initial-value problem (P), assume that(i) f(t, y", e) is continuous in (t, y, e) in an open set Do in the (t, y, e)-space,

(ii) ¢'(t) = f (t, 6(t), co) and (t, ¢(t), eo) E Do on an interval Zo = {t : tt < t <t2}, where eo is a fixed value of the variable e,

(iii) y ' = ¢(t) is unique in the sense that i f 1r1 = f (t, tl'(t), co) and co) EDo on a subinterval I of the interval Zo and if y(r) = fi(r) at a point r ofthe interval Z, then L,(t) = ¢(t) identically on the interval Z.

We start our discussion with the following lemma.

Lemma II-1-1. Conditions (i) and (ii) of Assumption 1 imply that there exist anopen set A in the (t, y")-space and an open neighborhood llo of co in the a-space suchthat(a) A x Uo C Do,(b) (t, q',(t)) E A on the interval Zo,

28

1. CONTINUITY WRT INITITIAL DATA AND PARAMETERS 29

(c) I t) I < M on 0 x Uo for some positive number M.

Proof.

Let r E Zo. Since f is continuous in the open set Do and (r, ¢r(r), co) E Do,there exist an open neighborhood 0(r) of (r, m(r)) in the (t, yl-space and anopen neighborhood U(r) of to in the a-space such that i(r) x U(r) C Do andI f (t, y, t) - AT, fi(r), to) I < 1 on 0(r) x U(7-). This implies that I f (t, y e) I < 1 +Mo

on U 0(r) x U(r), where M o = m ax I f (r, fi(r), to)I. Since (t, fi(t)) E U 0(r)rEZ0 TEZo

on the interval Zo and the interval Zo is compact, there exists a finite set (r, : j =1, 2, ... , N) of points on the interval Zo such that (t, m(t)) E U A(rt) for all

N>1>1t E Zo. Set

N N

= U,&(r)), Uo = nU(r)),3=1 J=1

M = 1+Mo.

Then, all the requirements of Lemma II-1-1 are satisfied by 0, Uo, and M. 0The main purpose of this section is to prove the following theorem under As-

sumption 1.

Theorem II-1-2. Let E be an open neighborhood of to and let Z = {t : s1 < t < s2}be a subinterval of Zo. Assume that

(II.1.1) ,'(t, e) = f (t, >¢(t, t), e) and (t, ir(t, t), e) E Do on Z x E.

Assume also that there exists a real-valued function t(e) such that(1) t(e) E I fore E E,(2) lim t(e) = to,t-w(3) hm 0(t(e), e) = (to)

Then, toThen:-lim i/i(t, t)

it to

uniformly on the interval Z.

Proof.

We shall prove this theorem in two steps. In Step 1, we prove Theorem 11-1-2assuming that there exists an open neighborhood U of Co such that (t, tb(t, e)) E 0on Z x (E n U), where 0 is the open set in the (t, y)-space which was determinedby Lemma II-1-1. The existence of such an open neighborhood U of to is proved inStep 2.

Step 1. We can assume without any loss of generality that U is contained in theneighborhood Uo of to which was determined by Lemma 11-1-1. Furthermore, using

condition (3) of Theorem 11-1-2, we choose U in such a way that It (t((),e)I isbounded on U.

30

Since

II. DEPENDENCE ON DATA

(II.1.2) tjJ(t, E) = tp(t(E), E) + / t As, j(s, e), E)ds on Z x £,JJJt(E)

I

AT,we obtain I!(t(E),E)I+MIt-t(E)I onIx(£fU) and Ii(t,E) - E)I

MI t - rI if (t, c) and (r, E) E Z x (£ n U), where M is the positive constant whichwas determined by Lemma II-1-1. Therefore, the set F _ E) : E E £ n U} isbounded and equicontinuous on the interval Z.

Let us derive a contradiction from the assumption that ,JJ(t, e) does not con-verge to fi(t) uniformly on the interval I as e -+ co. This assumption means thatmax I does not tend to zero as e --+ co; i.e., there exists a positive

number p and a sequence (E. : j = 1,2,. .. } such that

(II.1.3)

Ej E£nU and lim E; = Eo,

m Zx 1 (t,Ej) - v(t)I > p.Here, use was made of the assumption that the topology of the E-space is separable.

Observe that the sequence { (-, j) : j = 1, 2.... } is a subset of Y. Hence,this sequence is bounded and equicontinuous on Z. Therefore, b y Lemma 1-2-3 (Arzela-Ascoli), there exists a subsequence v = 1, 2, ... } such thatj., -+ +oo as v - +oo and that tp(t, Ej converges uniformly on I as v -+ +oo.Set l(t) = lim r%(t, on Z. Since (11.1.2) holds for c = Ef for all v, we obtain

J(t) = $(to) + f f(s,rl'(s),EO)ds on I.to

Hence, f (t, J(t), Eo) on I and 1J(to) = ¢(to). Now, condition (iii) of As-sumption 1 implies that j(t) = fi(t) identically on the interval Z. This, in turn,implies that lim max I (t, E j ) - $(t) I = 0. This contradicts (1I.1.3).

v-+oo LET

Step 2. We shall prove that there exists an open neighborhood U of co such thatUClloand (t,1i(t,E))EAon Ix(EnU).

Choose no > 0 so that the rectangular region

Ro(r) = {(t.: It - rl < ao, (y - (r)I <-Mao}is contained in A for every r E I (cf. Figure 1).

FIGURE 1.

1. CONTINUITY WRT INITITIAL DATA AND PARAMETERS 31

For every positive number a less than ao, set

R((T, ti7); a) _ {(t, yl: It-.I < a, Il! - i,-I «sa).

Then, ?Z((T, f ); a) C Ro(r) if IT - T 15 ao - a, I i - (r)I < M(ao - a). Fixing apositive number a less than ao, choose TI, T2,. .. , TN so that

s1 = T1 < 72 < ... < TN = s2,to = rjo for some jo,

0 < T2+l-T2 < a (j=1,2,...,N-1).

Since lira t(e) = to and lim e) = (to), there exists a neighborhood U1 of

co such that It(e) - tol < ao - a and I>%,(t(e), e) - 4(to)I < M(ao - a) fore E E nU1.

Hence, R((t(e), t%i(t((), e)); a) C Ro(to) C 0 fore E E n U1. Now, using Lemma I-1-1, we can verify that (t, fl(t, e)) E R((t(e),

i_(t(e),e)); a) C i for It - t(e)I < a and

e E EnU1. Also, since lim t(e) = to = rjo, we have It-t(e)I < a for rjo_1 < t < rjo+i

t- !oand e E E n U1, if the neighborhood U1 of co is sufficiently small.

Using the same argument as in Step 1 for t (t, e) on the interval T,._ j 5 t < T3.+1and e E E n U1, we obtain lim t(t, e) = fi(t) uniformly on rjo_ 1 < t < 7-,,.+,. In

.1 toparticular,

11m1L(T)o_1ie) _ 0(r)o-1) and lim V(T,o+1, e) = d(rjo+1)f to (-to

Using the same method in the neighborhoods of and (rio+1,4(Tjo+i )), we obtain lim 1G(t, e) = Q(t) uniformly on rjo-2 t < rjo and r <e-co - -t < rjo+2. Hence, lim ti(t, e) = q(t) uniformly on To-2 < t <- Tjo+2. Contin-(5puing this process, we find an open neighborhood U of co such that U C Uo and(t,>G(t, e)) E i on I x (E nU) in a finite number of steps. 0Remark 11-1-3. If f (t, y, () is independent of e, Theorem 11-1-2 yields continuity

of solutions of the initial-value problem f = f (t, y-), 9(to) = 4 with respect tothe initial data (to, Q.

In Theorem II-1-2, it was assumed that (t, y(t, e), e) E Do on Z x E (cf. (II.1.1)).With regard to this assumption, we can improve Theorem 11-1-2 as follows.

Theorem II-1-4. Assume that Assumption 1 is satisfied. Let r be an open neigh-borhood of cc). Assume further that there exist functions t(e) and c(e) defined on Esuch that

(i) t(e) E Zo and (t(e), c 1c), e) E Do fore E E,(ii) llm t(e) = to and lim dde) = 0(to).`-'a cco

Then, we can choose a neighborhood U of eo sufficiently small so that the initial-value problem

dt -f (t,

y, e), e c)

32 II. DEPENDENCE ON DATA

has on the region lo x (E f1 Ll) a solution y = J(t, e) such that (t, 111(t, E), E) E Do.

Theorem II-1-4 can be proved by using an argument similar to Step 2 of the proofof Theorem 11-1-2 and the local existence theorem (cf. Theorem 1-2-5). Details areleft to the reader as an exercise.

11-2. Differentiability

In this section, we explain the differentiability of solutions of an initial-valueproblem with respect to the initial data under the following assumption.

Assumption 2. For the initial-value problem (P), assume that

(i) f (t, y-) and Lf' (t, y-) (j = 1, 2, ... , n) are continuous in (t, y-) on an open set

A in the (t, yj-space,

(ii) Q'(t) = f (t, and (t, ¢(t)) E ., on an interval To = {t : tl < t < t2}.

Using Lemma 1-1-6, Theorems 1-1-4, 11-1-2, II-1-4, and Remark 11-1-3, we canshow that there exists a positive number p such that the initial-value problem

(11.2.1) y(r) = rlf y = Y - )dt

has a unique solution y' = O(t, r, q') on the interval lo if r E Zo and fr) - ¢(r)j <p. The solution y = m(t, r, ri") is continuous in (t, r, q-) and satisfies the condition(t. (t, r, E 0 if t E Zo, r E lo, and (r)I < p. The solution t(t, r, IIIsatisfies the integral equation

m(t, r, q + f t As , (s, r, gi))ds.

If ¢(t, r, is differentiable with respect to (r, q-), then

- 1('r, q) + ft (s,( s, r, )) ds,

e) + (s, $(s, r, ij)) 80(s, 7,1-7) ds,8f

I" &R1

where q, is the j-th entry of i), eJ is the vector in IR" with entries 0 except for I at

the j-th entry, and Lf is the n x n matrix whose j-th column is . From this0Yj

speculation, we obtain the following result.

2. DIFFERENTIABILITY 33

Theorem 11-2-1. Partial derivativesO-r

and ft(j = 1, 2,... , n) exist and are

continuous in (t, r, i)) in the domain

(11.2.2) f) : t E lo, t1 < r < t2, Ii - 4(r) I < p}.

Furthermore, 8rproblem

(11.2.3)

Proof.

respectively 0is the unique solutionBRA

=(t,(t, r, n)) z,

1AT) = - AT, 4) ( respectively eej).

of the initial-value

We prove the existence and continuity of by showing that they are the unique

solution of initial-value problem (11.2.3) with the initial-value z-(-r) f(7-, r'). The

existence and continuity of L can be proved similarly.d7),

Let +;1(t, r, ij) be the unique solution of (11.2.3) with the initial condition z(r)- f (r, r)). From Theorems 1-3-5 and 11-1-2 and Remark 11-1-3, it follows that1i(t, r, n) exists and is continuous in (t, T,i) on the closure D of domain (11.2.2).Furthermore,

(I1 .2.4) i (t,r,,) _ -f(r,,) + (s, 0(s,r,rl)) 1'(s,r,q)dsJtOf

on '75. To show that 8 r, rl'), we first deriver

(11.2.5)(t,r+h,ff) = ri + f t f(s,G(s,r+h,i))ds,

- rh

d(t, r, i) = n f (s, (s, r, i))ds,+ fr

from (11.2.1) on D assuming that h 54 0 is sufficiently small. Using (11.2.4) and(H.2-5), we compute

(11.2.6) h) = h [(t, r + h, r1) - 4(t, r, 17)] - +G(t, r,

as follows.

Observation 1. For a fixed (t, r, fl) E D, we have

T j) = I t I f (s, $(s, r + h, AS- J(s, 7,70)] dsr l`

- 17,+hf ( s, $6, r + h,

if h 96 0 is sufficiently small.

Observation 2. Using an idea similar to the proof of Lemma 1-1-6, for a sufficientlysmall h, we obtain

f(s,j(s,r+h,n)) -

[101 L(s,0 (s,r+h,fl)+(1 -9)¢(s,r,n1)dOl (t,r,rr))

f o r s E 7 o and a fixed (r, i f ) such that (s, r, n E D. Note that (s, O (s, r + h, r') +(1- 0)4(s, r, n ) ) E A for s E 10 and a f i x e d (r, n) such that (s, r, n') E D, if h 0 0is sufficiently small.

Finally, applying Observations I and 2 to (11.2.4), (11.2.5), and (I1.2.6), we obtain

g(t, r, h)

1 rr+h

_ -h

Jl f(r,qj

rrr

(11.2.7)+JtIJ1

(s, r + h, J-7) + (1- 0)0(s, r, f )dO'If -

i9f-p.:+Observation 3. The first and the third terms of the right-hand side of (11.2.7)

8tend to zero as h - 0. Also, since 4r, r, J-7) = n and (s, d(s, r, rte} is bounded09

on D, there exist a positive constant L and a non-negative function K(h) such thatlimh_o K(h) = 0 and that

t1g'(t,r,if,h){ <K(h) + LI J1g(s,r,il,h)Ids for tEZo.INow, using Lemma 1-1-5, we obtain the estimate

19(t, r, n, h)l 5 K(h) exp[LIt - rIJ on To.

EXERCISES II 35

Thus, we conclude that g(t, r, il, h) - 0 ash -+ 0, i.e., (t, T, t) = rfi(t, r,'.

Upon applying Theorem 11-2-1 to the initial-value problem

d =f (t,

y, u),du

= 0, y{ r) = i1, u(r) = e,

we can prove the following theorem.

aTheorem 11-2-2. Let a be a real variable. Assume that At, y, e), (t, y, e) (,j =

1, 2, ... , n), and 8e (t, y', e) are continuous on an open set Do in the (t, y, e)-space.

Let y = y(t, T, i), e) be the unique solution of the initial-value problem

dy= f (t, y, e), y(r) _ *!

dt

Then, there exists an open set 11 in the (t, T, 77, e)-space such that the function00 00are continuous on fl. Furthermore, i = (t, r, 4J, e) is the(t, T, ii, e) and

unique solution of the initial-value problem

di Ofzi (t, (t, r, 1/, e), e)l + 8E (t, 4(t, r, n, E). E), Z(r) = 0.dt =

EXERCISES II

II-1. Given an interval I = {x : a < x < b}, show that if f ei is sufficiently small,the initial-value problem

dy

dxe cos(x(y - x)), y(a) = a + e

has the unique solution y = O(x, e) on Z. Show also that Ui m O(x, e) = x uniformlyon Z.

Hint. The function cos(x(y - x)) is continuously differentiable on the entire (x, y)-plane. Furthermore, I oos(x(y - x))I < 1. This shows the existence of the uniquesolution. Note that y = x is the unique solution in case e = 0. Complete the proofby using Theorem 11-1-2.


11-2. Let y = 4 1(x, A) and y =¢2(x,.1) be tvm solutions of the differential equation

2 + A(1 + x2)y = 0

determined respectively by the initial conditions

01(O,A) = 1, as(0, A) = 0 and 02(0, A) = 0, a (0, A) = 1.

Show that(i) 01 and 02 are analytic in (x, A) everywhere in the complex (x, A)-space (i.e.,

in C2),

(ii) aA (1, A0) # 0, if 02(1, 1\0) = 0.

Hint. Theorem II-2-2 implies that z (x, A) is the solution of the initial-value

problem

Z + A(1 + x2)z = -(1 + x2}02(x, A), z(0) = 0, z'(0) = 0.

The method of variation of parameters (cf. (IV.7.9) in Remark IV-7-2 of ChapterIV; also see, for example, [Cod, pp. 67-68, 122-123] or [Rab, pp. 241-2441) yields

8 A)(x,

01(X"\) J0 oHence, we obtain

020,\0)2(1+t2)dC 0.8 (1,A0) = 01\1,Ao)fo

Note that 01 and 02 are linearly independent and hence 01(1,Ao) 0.

11-3. Show that if IEI is small, the following boundary-value problem has the uniquesolution which is continuous in e:

2 = Esin (100-y2

y(-1) = 0, y(1) = 0.

Hint. Determine O(x, c, E) by the initial-value problem

d2y xy(-1) =E 5111 0, y'(-l) = c.

dZ2= ( 100 y2)

Then, the given boundary-value problem is reduced to the equation

(E) 0(1,c,e) = 0.First of all, 0(1, c, c) is analytic for small itI and #c(. Also, 0(1,c,0) = 2c and, hence,80(1, c, 0)

= 2 # 0. Therefore, equation (E) has the unique solution c = c(E) suchac

that c(0) = 0. Furthermore, c(E) is analytic for small lei.

EXERCISES II 37

11-4. Let Ax, y-) be an RI-valued function whose entries are continuously differ-entiable with respect to (x, yam) in a domain D C IIPn+1 Also, let y = (x, 171 be a

solution of the system fy = f (x, y) such that p(0) and that (x, 5(x, rtj) E Dfor 0 < x < 1 and it E Ao, where Do is a domain in R. Note that we must have(0, rl E D for rj E Ao. Set A = {(1, 1 : if E Do}. Show that

(a) Al is open in lR',(b) the mapping 4(1,r) : Ao - Al is one-to-one, onto, and differentiable with

respect to i in Do,(c) if we denote the inverse of the mapping ¢(1,i) by AI -. Ao, then

is also differentiable with respect to S.

11-5. For the differential equation

(1 + x2) LY -_ 2xy = 2.sy2,

find(i) the unique solution y = O(x, e, ri) satisfying the initial-condition y(£) = q;

(ii) the partial derivative 84(x, , n),

(iii) the artial derivative

ofam(x, £, r7)

(iv) 190(x, , n)t at x =8

a0(x, 4, rl)(v) at x =

Show also that z = 00(. '77) and z = 8¢(,17) both satisfy the differential8 Bin

equation:

(I+ x2) dz

dz- 2xz = 4xO(x, {, q)z .

II-6. Assume that f (t, x1, x2i ... , xn) is a real-valued, continuous, and continu-ously differentiable function of (t, xl,... , x,) on an open set A in the (t, x1.... , xn)-space. Assume also that o(t) is a real-valued solution of the n-th order differentialequation x(") = f(t,x,x',... ,x(n-1)) and (t,Oo(t),e0(t),... ,0On-1)(t))

E A on aninterval Zo = It : t1 < t < t2}. Show that there exists a positive number p suchthat

(i) the initial-value problem

x(n) = f (t, x, x , ... , x(n-1) ), x(r) = SI, x (r) = 52, ... , x(n-1) (r) = n

has a unique solution x = {n) on the interval 20 if r E Zo andt 3-O0-1)(r)I<p (j=1,2,...,n);

(ii) the solution x = qi(t, r, ti, ... , {n) is continuous in (t, r, t1, ... , n) and sat-isfies the condition t t 'r, 6, W) e(t r ) O("-1)(t r

(j=1,2,...,n);

(iii) the partial derivative t8tn exists and is continuous in (t, r, Ft,

n) on the domain D = {(t, r, 1, ... , {n) : t E To, t1 < T < t2,-')(T)I < p (1 = 1,2,...,n)};

(iv) u = Lo (t, r, is the unique solution of the initial-value problem

rn f ` tt-1

dnU - `dtn J= axf (t,7,6,... ,Sn)) dtj_1

U(T) = -6, u'(T) ,u(n-2)(7) u(n_1) = ,Ln).

11-7. Assume that f (t, X1, x2) is a real-valued, continuous, and continuously dif-ferentiable function of (t, xl, x2) on an open set A in the (t, x1, x2}space. Assumealso that O0(t) is a real-valued solution of the second-order differential equationx" = f (t, x, x') and (t, Oo(t), 00'(t)) E A on the interval Zo = (t : 0 < t < 1).Set ¢0(0) = a and 0o(0) = b. Denote by t(t,/3) the unique solution of the initialvalue-problem x" = f (t, x, x'), x(0) = a, x'(0) = ,B, where lb - 81 is sufficiently

small. Show that (1,b) > 0 if of 0 for t C-013

11-8. Let g(t) be a real-valued and continuous function oft on the interval 0 < t <1. Also, let A be a real parameter.(1) Show that, if Q(t, A) is a real-valued solution of the boundary-value problem

Ez + (g(t) + A)u = 0, u(0) = 0, u(1) = 0,

and if 8 &2 (t, A) is continuous on the region A = {(t, A) : 0 < t < 1, a <A < b}, then d(t, A) is identically equal to zero on A, where a and b are realnumbers.

(2) Does the same conclusion hold if 0(t,,\) is merely continuous on A?

11-9. Let a(x, y) and b(x, y) be two continuously differentiable functions of twovariables (x, y) in a domain Do = {(x, y) : IxI < a:, IyI < Q} and let F(x, y, z)be a continuously differentiable function of three variables (x, y, z) in a domainDo = {(x, y, z) : Ixl < a:, IyI < 3, Izl < 7}. Also, let x = f (t, i7), y = 9(t, ij), andz = h(t,17) be the unique solution of the initial-value problem

dt= a(x,y), L = b(x,y), = F(x,y,z),

x(0) = 0, y(0) = n, z(0) = c(+1).where c(q) is a differentiable function of 11 in the domain 14 = (q: In1 < p). Assumethat (t, rl) = (d(x, y), O(x, y)) is the inverse of the relation (x, y) = (f (t,17), g(t, rt)),where we assume that O(z, y) and i,1(x, y) are continuously differentiable with re-spect to (x, y) in a domain Al = {(x, y) : IxI < r, IyI < p}. Set H(x,y) =h(p(x, y), tp(x, y)). Show that the function H(x, y) satisfies the partial differentialequation

a(x, y)

8+ b(x, y) M = F(x, y, H)

and the initial-condition H(0, y) = c(y).

EXERCISES II

Hint. Differentiate H(x,y) with respect to (x,y).

11-10. Let F(x, y, z, p, q) be a twice continuously differentiable function(x,y,u,p,q) for (x,y,u,p,q) E 1R5. Also, let

x = x(t, s), y = y(t, s), z = z(t, s), p = P(t, s), q = q(t, s)

be the solution of the following system:

dx OF- (x+ y, z, p, q),

dy _ OFdt flq(T, y, z, p, q),

dz OF OFdt = pij (x,y,z,P,q) + gaq(x,y,z,P,q),

dp OF OFdt 8x (T, y, z, P, q) - P 8z (x, y, z, p, q),

dq OF OFdt 8y (x, y, z, P, q) - q az (x, y, z. P, q)

satisfying the initial condition

x(O,s) = xo(s), y(O,s) = yo(s), z(O,s) = zo(s),

p(O,s) = Po(s), q(O,s) = qo(s),

39

of

where x0 (s), yo(s), zo(s), po(s), and qo(s) are differentiable functions of s on R suchthat

F(xo(s),yo(s),zo(s),Po(s),go(s))=0,dzo(s)=Po(s)- (s)+go(s)dyo(s)

ds ds ds

on R. Show that

F(x(t, s), y(t, s), z(t, s), p(t, s), q(t, s)) = 0,

at (t, s) = At, s) 5 (t, s) + q(t, s) (t, s),

flz(t,s)= P(t,s)L(t,s) + q(t,s)ay(t,8)fls as as

as long as the solution (x, y, z, p, q) exists.

Comment. This is a traditional way of solving the partial differential equation

F(x, y, z, p, q) = 0, where p = 8 and q = For more details, see (Har2, pp.

131-143].

40 H. DEPENDENCE ON DATA

II-11. Let H(t, x, y, p, q) be a twice continuously differentiable function of(t, x, y, p, q) in RI. Show that we can solve the partial differential equation

Oz Z 8z\ =0+ H t, x, y,-,

by using the system of ordinary differential equations

dx 8H dy 8Hdt = 8p' t = q

,

dp 8H (LL dq 8Hdt 8x ' dt

_ _W,

OHdz 8H 8H du== u + p-p + q8q, dt_ --6T.

CHAPTER III

NONUNIQUENESS

We consider, in this chapter, an initial-value problem

(P)dy

= f(t,y1,dt

without assuming the uniqueness of solutions. Some examples of nonuniqueness aregiven in §III-1. Topological properties of a set covered by solution curves of problem(P) are explained in §§III-2 and 111-3. The main result is the Kneser theorem(Theorem III-2-4, cf. [Kn] ). In §1II-4, we explain maximal and minimal solutionsand their continuity with respect to data. In §§III-5 and 111-6, using differentialinequalities, we derive a comparison theorem to estimate solutions of (P) and alsosome sufficient conditions for the uniqueness of solutions of (P). An application ofthe Kneser theorem to a second-order nonlinear boundary-value problem will begiven in Chapter X (cf. §X-1).

111-1. Examples

In this section, four examples are given to illustrate the nonuniqueness of so-lutions of initial-value problems. As already known, problem (P) has the uniquesolution if f'(t, y) satisfies a Lipschitz condition (cf. Theorem I-1-4). Therefore, inorder to create nonuniqueness, f (t, y-) must be chosen so that the Lipschitz condi-tion is not satisfied.Example III-1-1. The initial-value problem

(III.1.1)

has at least three solutions

dy= yt/3dt y(to) = 0

(S.1.1) y(t) = 0 (-oo < t < +oc),

(S.1.2)

and

0, t < to,

23/2

(S.1.3) y(t) [3(t - to),

0,

I

3/2t - to) , t > to,y(t) _ [3 ]

t > to,t < to

41

42 111. NONUNIQUENESS

(cf. Figure 1). Actually, the region bounded by two solution curves (S.1.2) and(S.1.3) is covered by solution curves of problem (I11.1.1). Note that, in this case,solution (S.1.1) is the unique solution of problem (P) for t < to. Solutions are notunique only fort ? to.Example 111-1-2. Consider a curve defined by

(111.1.2) y = sin t (-oo < t < +oo)

and translate (111.1.2) along a straight line of slope 1. In other words, consider afamily of curves

(111.1.3) y = sin(t - c) + c,

where c is a real parameter. By eliminating c from the relations

dt = c os(t - c), y - t = sin(t - c) - (t - c),

we can derive the differential equation for family (111.1.3). In fact, since sin u - uis strictly decreasing, the relation v = sinu - u can be solved with respect to uto obtain u = G(v) - v, where G(v) is continuous and periodic of period 27r in v,G(2n7r) = 0 for every integer n, and G(v) is differentiable except at v = 2n7r forevery integer n. The differential equation for family (111.1.3) is given by

(II1.1.4) = 1OS[G(y-t)-(y-t)1.dy

dt

Since G(2n7r) = 0 and cos(-2n7r) = 1 for every integer n, differential equation(111. 1.4) has singular solutions y = t + 2mr, where n is an arbitrary integer. Theselines are envelopes of family (111.1.3) (cf. Figure 2).

FIGURE 1.

Example 111-1-3. The initial-value problem

(111.1.5)

has at least two solutions

FIGURE 2.

dt = y(to) = 0

(S.3.1) y(t) = 0 (-oo < t < +oo),

1. EXAMPLES

and

4 (t -t,0)2, t > to

(S.3.2) y(t),

- 4(t - to)2, t < to

43

(cf. Figure 3). The region bounded by two solution curves (S.3.1) and (S.3.2) iscovered by solution curves of problem (111.1.5).

FIGURE 3.

Consider the following two perturbations of problem

dy= + E, y(to) = 0

dy _ y2

y2 + C2 lyl y(to) = 0,

where a is a real positive parameter. Each of these two differential equations satisfiesthe Lipschitz condition. In particular, the unique solution of problem (111. 1.6) isgiven by

(S.3.3) y(t) =-E)2 - E, t > toJ4(t-to+2VI

-4(t-to-2f)2 + E, t < to

(cf. Figure 4). On the other hand, (S.3.1) is the unique solution of problem (111.1.7).Figure 5 shows shapes of solution curves of differential equation (111.1.7). Note thatnontrivial solution of (111.1.7) is an increasing function of t, but it does not reachy = 0 due to the uniqueness.

44 III. NONUNIQUENESS

y

FIGURE 4. FIGURE 5.

Generally speaking, starting from a differential equation which does not satisfyany uniqueness condition, we can create two drastically different families of curvesby utilizing two different smooth perturbations. In other words, a differential equa-tion without uniqueness condition can be regarded as a branch point in the spaceof differential equations (cf. [KS]).Example 111-1-4. The general solution of the differential equation

(1II.1.8)

is given by

d)2

+y2=1

(111.1.9) y = sin(t + c),

where c is a real arbitrary constant. Also, y = 1 and y = -1 are two singularsolutions. Two solution curves (111.1.9) with two different values of c intersect eachother. Hence, the uniqueness of solutions is violated (cf. Figure 6).

This phenomenon may be explained by observing that (111. 1.8) actually consistsof two differential equations:

(III.1.10) dy = 1 - yzdt

anddy=- l-y2.dt

Each of these two differential equations satisfies the Lipschitz condition for lyI <1. Figures 6-A and 6-B show solution curves of these two differential equations,respectively.

y=I

y=-I

FIGURE 6. FIGURE 6-A. FIGURE 6-B.

Observe that each of these two pictures gives only a partial information of thecomplete picture (Figure 6).

2. THE KNESER THEOREM 45

We can regard differential equation (111.1.8) as a differential equation

(111.1.11)

on the circle

(111.1.12)

dy

dt=w

w2 + y2 = 1.

If circle (111.1.12) is parameterized as y = sinu, w = cosu, differential equation(III.1.11) becomes

(III.1.13) d = 1 or cos u = 0 .

Solution curves u = t of (111.1.13) can be regarded as a curve on the cylinder

{(t, y, w) : y =sin u, w = cos u, -oc < u < +oo (mod 2w), -oo < t < +oo}

(cf. Figure 7). Figure 6 is the projection of this curve onto (t,y)-plane.

FIGURE 7.

In a case such as this example, a differential equation on a manifold would givea better explanation. To study a differential equation on a manifold, we generallyuse a covering of the manifold by open sets. We first study the differential equationon each open set (locally). Putting those local informations together, we obtain aglobal result. Each of Figures 6-A and 6-B is a local picture. If these two picturesare put together, the complete picture (Figure 6) is obtained.

111-2. The Kneser theoremWe consider a differential equation

(III.2.1) dt = f(t,y-)

under the assumption that the R"-valued function f is continuous and bounded ona region

(111.2.2) S2 = {(t, y-) : a:5 t < b , Iy1 < +oo }.

Under this assumption, every solution of differential equation (III.2.1) exists on theinterval Zfl = {t : a < t < b} if (to, y"(to)) E f2 for some to E Zo (cf. Theorem 1-3-2and Corollary I-3-4). The main concern in this section is to investigate topologi-cal properties of a set which is covered by solution curves of differential equation(111.2.1).

2. THE KNESER THEOREM 47

Theorem 111-2-4 ((Kn]). If A is compact and connected, then SS(A) is alsocompact and connected for every c E To.

Proof.

The compactness of SS(A) was already explained. So, we prove the connectednessonly.

Case 1. Suppose that A consists of a point (r, t), where we assume without anyloss of generality that r < c. A contradiction will be derived from the assumptionthat there exist two nonempty compact sets F1 and F2 such that

(111.2.3) SS(A)=F1uF2, F1nF2=0.

If t'1 E F1 and 2 E F2, there exist two solutions 1 and 4'2 of (111.2. 1) such that01(r) _ , 01(c) _ 1, and 42(r) _ ., 02(c) _ 6, (cf. Figure 9).Set

h(µ) _1(r+11)

{ &T + JJUJ)

for 05µ5c-r,f o r - , r )

Let { fk(t, y-) : k = 1, 2,... } be a sequence of R"-valued functions such that(a) the functions fk (k = 1, 2, ...) are continuously differentiable on 0,(b) I fk(t, y1I < M for (t, y) E f2, where M is a positive number independent of

(t, yl and k,(c)

k Elra+A = f'uniformly on each compact set in Q

oo

(cf. Lemma I-2-4). For each k, let &(t,µ) be the unique solution of the initial-

value problem = fk(t, y-), y(-r+ Iµ4) = h(µ). The solution 15k(t, µ) is continuous

for t E Zo and iµl < c - r. It is easy to show that the family k =1, 2,... ; jµj 5 c - r} is bounded and equicontinuous on the interval Io.

Note that the functions Z(rk(c, µ) (k = 1, 2, ...) are continuous in µ for 1µI <c - r and that r'k(c,c - r) = ¢'1(c) E F1 and t k(c,-(c - r)) = &c) E F2.Let d be the distance between two compact sets F1 and F2. Since &(c,µ) iscontinuous in p, there exists, for each k, a real number µk such that IµkI < c-r and

distance(!& (c, µk), F1) =d.

Since the family ilk) : k = 1, 2, ... } is bounded

and equicontinuous on the interval Io, there exists a subsequencej = 1,2.... } such that (i) lim k, _ +oo, (ii) lim µk, = po exists, and (iii)

i--+00 )-.+oourn 1Pk, (t, µk1) = 0(t) exists uniformly on To. It is easy to show that

s-+oo

tfi(t) = K(AO) + f f (s, ¢(s))ds,+ {µol

lµo1 < c - r, distance(¢(c), F1) = 2

Hence, (c) E $.(A) but * (c) 4 F1 U F2. This is a contradiction (cf. Figure 10).

!=c t=c

FIGURE 9. FIGURE 10.

Case 2 (general case). Assume (111.2.3) as in Case 1, set AI = AnR((c) x.Fl) andA2 = A n R({c} x F2), where {c} x F; = {(c, yl : y" E Fj } (j = 1, 2). Then, thetwo sets A, (j = 1, 2) are compact and not empty. Note that A = AI U A2. SinceA is connected, we must have AI n A2 0 0. Choose a point (r, {) E AI n A2. Then,

Sc((T,6) = {Ss((T,S)) nF1}U{S,.((T,6) n12} andS,,((r, ))n.Fi 36 0 (j = 1,2).This is a contradiction (cf. Case 1). 0

In order to apply the Kneser theorem (i.e., Theorem III-2-4), it is desirable toremove the boundedness of f from the assumption. To obtain such a refinementof Theorem 111-2-4, consider differential equation (111.2.1) under the following as-sumptions.

Assumption 1. A set Ao is a compact and connected subset of the region Q suchthat if ¢(t) is a solution of (111.2.1) satisfying

(111.2.4) (to, 0(to)) E Ao for some to E To,

then fi(t) exists on To.

As in Definition 111-2-1, denote by Ro the set of all points (t,yr) E Q such thaty = fi(t) for some solution 4 of differential equation (111.2.1) satisfying condition(111.2.4). Also, set $S = (9: (c, y-) E Ro) for c E Io.

Assumption 2. The set Ro is bounded.

Now, we prove the following theorem.

Theorem 111-2-5. If a compact and connected subset Ao of 12 satisfies Assump-tions 1 and 2, then the set Sc is also compact and connected for every c E I.

Proof.

Since Ro is bounded, there exists a positive number M such that

Ro c {(t,yj: tETo, Iyl SM) .

Setf(t,yl if tETo, Iyj<2M,

t9(, 0 = f ( t,j- y)

if t E To, ly7 ? 2M.

3. SOLUTION CURVES ON THE BOUNDARY OF R(A) 49

Then, g(t, g) is continuous and bounded on Q. Using the differential equation

dt = g(t, y), define R(Ao) and SS(Ao) by Definition 111-2-1. Then, we must have

Ro = 1Z(Ao). Otherwise, there would exist a solution ¢o(t) of differential equation(111.2.1) such that (to,&(to)) E Ao for to E Zo and (tt,do(tl)) V Ro for tl E -To-This is a contradiction. The theorem follows from Theorem 111-2-4. 0Remark 111-2-6. Any kind of shapes can be made by a wire. Therefore, the setSS(A) needs not to be a convex set even if A is convex.

Using Theorem 111-2-4, we can prove the following theorem, which is a refinementof Theorem II-1-4.

Theorem 111-2-7. Assume that the entries of an R"-valued function At' In arecontinuous in (t, yam) in a domain Do E Rn+l. Also, let fi(t) be a solution of system(111.2.1) on an interval a < t < b. Assume that (t,¢(t)) E Do on the intervala < t < b. Then, for any given positive number e, there exists another solution '(t)of (111.2.1) such that

(i) (t, e,1(t)) E Do on the interval a < t < b,(ii) JO(t) - v1'(t)l < e on the interval a < t < b,

(iii) a Y(r) # fi(r) for some r on the interval a < t < b.

Proof.

Using an idea similar to Step 2 of the proof of Theorem 11-1-2, the local existencetheorem (Theorem 1-2-5), and the connectedness of SS(A), we can complete theproof of Theorem 111-2-7. In fact, subdivide the interval a < t < b (i.e., a = to <tl < < tk = b) in such a way that [y(t) - ¢(t)1 < e for t, < t :5 tj+l if g(t)satisfies (111.2. 1) and Ib"(t,) - i(t,)I < 6. Set A. = {EE E R" : l e - (tj)I < b} andBl = {y-(t1) : y(to) E Ao}, where g(t) denotes any solution of (111.2.1) with initial-value y-(to). Since Bl is connected and contains ¢(t1), we must have Bl =if Bl n AI contains only (t1). In such a case, the proof is finished.

If Bl n Al contains more than one point, then B1 n A, contains a connected setSl containing 0(t1) and more than one point. Set B2 = {9(t2) : g(ti) E Si). SinceB2 is connected and contains (t2), we must have B2 = if B2 n A2 containsonly (t2). In such a case, the proof is finished. In this way, the proof is completedin a finite number of steps. 0

III-3. Solution curves on the boundary of 1Z(A)We still consider a differential equation

(III.3.1) dt =

under the assumption that the entries of the R°-valued function fare continuousand bounded on a region

(111.3.2) n = {(t, y-) : a < t < b , lyi < +oc}.

50 M. NONUNIQUENESS

As mentioned in §111-2, under this assumption, every solution of differential equa-tion (111.3.1) exists on the interval To = {t : a < t < b} if (to,y-(to)) E ft for someto E I. Define the sets 1(A) and SS(A) for a subset A of ) by Definition III-2-1.The main concern of this section is to prove the existence of solution curves on theboundary of R(A). We start with the following basic lemma.

Lemma 111-3-1. Suppose that (1) the set A consists of one point (c1, t) (i. e.,A = {(c,, f)}), (ii) a < cl < co < c2 < b, and (iii) ij is on the boundary of Sc2(A).Let B = Then, SI(B) contains at least a boundary point of S0(A).

Proof

A contradiction will be derived from the assumption that SI(B) does not containany boundary points of SS (A). Note that SI(B) n 4 (A) 54 0. Set S, = 4 (B) nS,,. (A), S2 = {y : y' E Se. (B) and Ii 0 SS(A)}. Then, S., (B) = 81 U S2 andS, n S2 = 0. It is known that S1 is a nonempty compact set. Also, $2 is closedand bounded, since S,,, (5) is compact and does not contain any boundary points ofSS (A). If S2 is not empty, the set Sc.(B) contains a point in Sc,, (A) and anotherpoint which does not belong to S,, (A). Then, S,.(8) contains a boundary point ofSc. (A), since S,, (B) is connected. This is a contradiction. Therefore, if it is provedthat S2 is not empty, the proof of Lemma 111-3-1 will be completed (cf. Figure 11).

To prove that S2 is not empty, observe first that Sam({(a2,()}) nS.,,(A) = 0 if0 SS,(A) (cf. Figure 12).


Let rj be on the boundary of $, (A). Then, there exists a sequence of points{Sk V Sc, (A) : k = 1, 2, ... ) and a sequence (¢k : k = 1, 2.... ) of solutions of(111.3.1) such that lira (k = '1, O&2) = (k, and mk(co) 0 SS(A). The familyk+oo

k = 1,2.... } is bounded and equicontinuous on the interval Zo. Therefore,there exists a subsequence { Jk, j = 1, 2, ...) such that lirn k, = +oo andj+oolim bk, (t) = Q(t) exists uniformly on Zo. The limit function is a solution of++oo

(111.3.1) such that ¢(c2) = tj. Hence, d(co) E Sc,(B). Since S, (B) does not containany boundary points of S,(A), we must have ¢'(co) if $,,(A). This implies thatS2 is not empty. 0

The following theorem is the main result in this section which is due to M.Hukuhara (Hukl( (see also (Huk2J and [HN31).

3. SOLUTION CURVES ON THE BOUNDARY OF R(A) 51

Theorem 111-3-2. Suppose that(1) a<cl <c2 <b,(2) is on the boundary of

Then, there exists a solution of differential equation (M.3.1) such that

(i) (Cl) = E, (C2) = 6,(ii) is on the boundary of R({(ci,l;))) for cl < t < c2.

Proof.

For a subdivision A : CI = To < Tl < . < Tm-1 < Tm = c2 of the intervalcl < t < c2, there exists a solution ¢o of (111.3.1) such that

(a) O0(CI) _ ., 4fiG(C2)(p) is on the boundary of R({(cl,O)}), where e = 1,2,... ,m - 1.

(Cf. Lemma 111-3-1 and Figure 13.)

FIGURE 13.

Choose a sequence {ok : k = 1, 2, ... } of subdivisions of the interval cl < t < c2such that

f Ok : C1 = Tk,O < Tk.1 < ... < Tk,2k-I < Tk,2" = C2,

e1 Tk,t = C1+2k(c2-cl), e=0,1,...,2k, k=1,2,....

Since Tk+1,21 = Tk,l, we have {Tk,t : e = 0, 1,2,... , 2k} C {Tkrl,e e

2k+1}. Set k =tt jok (k = 1,2,...). Then,

(ak) &(C1) = S, &(C2) = 1,(Qk) (Tk,t,&(Tk,t)) is on the boundary of where e = 1,2,... ,

2k - 1.

The family {4k k = 1, 2,... } is bounded and equicontinuous on 10. Hence,there exists a subsequence {&, : j = 1, 2,... } such that lim k, = +oo and thatj +00

lira ¢k, exists uniformly on 10. Then, is a solution of (111.3.1) such thati-.+oo(ao) (Cl) (C2)(Qo) (Tk,t,. k(Tk,t)) is on the boundary of R({(c1i,)}), where f

2k - 1 and k = 1,2,....Since the set {Tk,t : e = 1,2,... 2k - 1; k = 1,2.... } is dense on the interval

Cl < t < c2, the curve (t,$(t)) is on the boundary of for cl < t <C2- 0

Remark 111-3-3. In general, the conclusion of Theorem 111-3-2 does not holdon an interval larger than cl < t < c2.

111-4. Maximal and minimal solutions

Consider a scalar differential equation

(111.4.1) dt = f (t, y),

where t and y are real variables, and assume that f is real-valued, bounded, andcontinuous on a region H = {(t,y) : a < t < b, -oo < y < +oo}. Choose c inthe interval I = (t : a < t < b}. Then, Sr ({ (c, r,) }) is compact and connectedfor every r E I and every real number r) (cf. Theorem III-2-4). This impliesthat there exist two numbers 41(r) and ¢2(r) such that ST({(c,17)}) = { y ER : bl(r) < y < &2(r)}. Hence, R({(c,n)}) = {(t,y) E R2 : ¢1(t) < y <02(t), a < t < b} (cf. Figure 14). The two boundary curves (t,4i(t)) and (t,02(t))of R({(c,rl)}) are solution curves of differential equation (111.4.1) (cf. Theorem111-3-2). Every solution ¢(t) of (111.4.1) such that 0(c) = n satisfies the inequalities01(t) < ¢(t) < ¢2(t) on Z. The solution 02(t) (respectively 01(t)) is called themaximal (respectively minimal) solution of the initial-value problem

(111.4.2)dy

= f (t, y), y(c) = 77dt

on the interval Z. In this section, we explain the basic properties of the maximaland minimal solutions. Before we define the maximal and minimal solutions moreprecisely, let us make some observations.

FIGURE 14.

Observation 111-4-1. Assume that f (t, y) is real-valued and continuous on adomain Din the (t,y)-plane. Set lo = {t : a < t < b}. Let 01(t) and ¢2(t) be twosolutions of differential equation (111.4.1) such that (t, ¢1(t)) E D and (t, 02(t)) E Dfor t E Zo. Note that we do not assume boundedness of f on D. Set 0(t) =max {01(t), 02(t)} for t E Zo. Then, Q(t) is also a solution of (111.4.1) on theinterval Zo.

4. MAXIMAL AND MINIMAL SOLUTIONS 53

Proof.

For a fixed to E To, we prove that

(111.4.3) lim do(t) - O(to) = f(to,¢(to)).t-.to t - to

If ¢1(to) > -02(to), then there exists a positive number 6 such that 4(t) = 01(t) forIt - tol < J. Therefore, (111.4.3) holds. Similarly, (111.4.3) holds if 02(to) > &(to).Hence, we consider only the case when '(to) _ .01(to) = 02(to). In this case,

f o r each fixed t E To, we have0(t) - 0(to) = 0'(t) - 0'(to), where j = 1 or

t - to t- toj = 2, depending on t. Hence, by the Mean Value Theorem on O,(t), there exists

r E To such that r --r to as t - to and 0(t) - ¢(to) = f (r, 0,(r)). Also, 03 (r) _t - to

¢j(to) + f (a,-Oj (a)) (7 - to) for some a E To such that a to as r - to. Since f isbounded on the two curves (t, 01(t)) and (t, 02(t)), (111.4.3) follows immediately. 0Observation 111-4-2. Let f (t, y), D, and To be the same as in Observation111-4-1. Consider a set F of solutions of differential equation (111.4.1) such that(t, 0(t)) E D on To for every ¢ E F. Assuming that there exists a real number Ksuch that 0(t) < K on lo for every 0 E F, set 00 (t) = sup{Q(t) : ¢ E F} for t E To.Assume also that (t, 00(t)) e D on To. Then, 4'o(t) is a solution of differentialequation (111.4.1) on To.

Proof.

As in Observation 111-4-1, we prove that

(111.4.4) lim 00(t) - Oo(to) = f (to, bo(to))t-»to t - to

for each fixed to E To. Choose three positive numbers po, p, and S so that(i) A = {(t, y) : t E To, Iy - 4o(t)I < po} C D,

(ii) we have (t, 4(t)) E A on the intervalI

t - to I < 6, if d(t) is a solution of(III.4.1) such that 0 < 0o(r) - 4i(r) < p for some r in the interval It - tol < b.

There exists a positive number Af such that I f (t, y) I < M on A.Let us fix a point r on the interval It - tol < b. First, we prove the existence of

a solution ty(t; r) of (III.4.1) such that

(111.4.5) P(r; r) = 0o (r) and yJ(t; r) < 4o(t) for it - tol < b.

To do this, select a sequence {4'k : k = 1, 2,... } from the family F so thatlira thk(r) = 40(r). We may assume that (t, hk(t)) E A on It - tol < S for

o0Y+

k (cf. (ii) above). Then, the sequence {4'k : k = 1, 2, ... )is bounded andequicontinuous on It - tot < J. Hence, we may assume that lim 4'k(t) = 0(t; r)

k-»+ooexists uniformly on the interval It - tol < 6. It is easy to show that V'(t; r) is asolution of (111.4.1) and that (111.4.5) is satisfied.

Set ty(t) = max{t/,(t; r), tJ'(t; to)} for Jt - tol < b. Then, v is a solution of (111.4.1)such that (1) V)(r) = oo(r) and ty(to) = 00(to), (2) iP(t) < 4'o(t) for It - tol < b,and (3) (t, v/ (t)) E A for It - tol < 6 (cf. Figure I5).

Y= #o(')

Y s V(1)

FIGURE 15.

Using this solution tp(t), we obtainO0(T) - ¢040) = t.ti(T) - 0(t0) = f(a,tp(a))

T - to T- tofor some a such that Ja - tol < 6 and that or to as r -" to. Since 10(a) - tt'(td)J <Mba - t01, (111.4.4) can be derived immediately. 0

Let us define the maximal (respectively minimal) solution of an initial-valueproblem

{III.4.6 ) dt = f (t, ii), y(r)

where f is real-valued and continuous on a domain V in the (t, y)-plane and theinitial point (T, 1:) is fixed in D.

Definition 111-4-3. A solution tp(t) of problem (111.4.6-() is called the maximal(respectively minimal) solution of problem (111.4.6-() on an interval I = it : T <t < r' } if

(a) tp(t) is defined on I and (t, {1(t)) E D on Z,(b) if 4(t) is a solution of problem (II14.6-{) on a subinterval r < t r" of Z,

then

0(t) < y(t) (respectively ¢(t) > P(t)) on r < t < r".

The following two theorems are stated in terms of maximal solutions. Similarresults can be stated also in terms of minimal solutions. Such details are left tothe reader as an exercise. In the first of the two theorems, we consider anotherinitial-value problem

(III.4.6-n) !LYdt = f(t,y), y(r) = r)

together with problem (111-4.6-{)-

Theorem III-4-4. Suppose that the maximal solution tpt of Problem (IIL4.6-{)exists on an interval Z = {t : r < t < r'}. Then, there exists a positive number Sosuch that the maximal solution 0,1 of problem (III 4.6-r1) exists on I for t; < rl <t + b0. Fjrthermore, tpf(t) < ip,,(t) on I for £ < 17< + 60 and limipR(t) = tb,(t)

uniformly on Z.

Proof

Set 0(p) = {(t, y) : t E Z, yt(t) < y < ot(t) + p}. For a sufficiently smallpositive number p, we have A(p) C 1), and, hence, if (t, y)I is bounded on A(p) fora sufficiently small positive number p.

First, we prove that for a given p > 0, there exists a positive number 6 such that,if f < q < { + b, every solution 0(t) of problem (111.4.6-1?) defined on a subintervalr < t < r" of I satisfies

(III.4.7) 0(t) < ,'(t)±p for r < t <7".

Otherwise, there exist two sequences {qk : k = 1 , 2, ... ) and {rk : k = 1, 2, ... } ofreal numbers such that (1) qk > {,

klim qk = t;, and r < rk < r', (2) 0&-(-r) = nk-+oo

and Ok(Tk) = v((rk) + p, and (3) Ok(t) < vf(t) + p for r < t < rk. Furthermore,there exists a real number r(p) such that rk > r(p) > r (cf. Figure 16).

Y=W4+P

Set

Y Wt

t=r t= TA 1=',

FIGURE 16.

I

I i

- 1 I

max(0k(t), '' (t)),f (t)+p

,

T<t<Tk,Tk St ST.

It is easy to show that the sequence {0k W : k = 1, 2, ... } is bounded and equicon-tinuous on the interval Z. Hence, we can assume without any loss of generalitythat

(1) tim Tk = rp exists,k-.+oo

(2) tim 10k(t) = 0(t) exists uniformly on Z.

Note that(3) d(r) _ and O(ro) _ 'af(ro) + p,(4) ro?r(p)>r.

It is also easy to show that 0(t) is a solution of (III.4.6-t) on the subintervalr < t < ro of Z. Since 104 is the maximal solution of (III.4.6-1:) on 7, we musthave 0(t) < af(t) on the interval r < t < To. This is a contradiction, since0(ro) = i (ro) + p. Thus, (111.4.7) holds.

Now, for a given positive number p, there exists another positive number 6(p)such that (111.4.7) holds for any solution ¢(t) of problem (III.4.6-17) if < ' <t; + 6(p). Hence, using max(4(t), 0((t)), we can extend 6(t) on the interval Z insuch a way that

11)k (t )

(111.4.8) ?Gf(t) 5 y(t) < p on Z.

Let F be the set of all solutions 6(t) of (III.4.6-71) which satisfy condition (111.4.8).Set t[i,r(t) = sup{¢(t) : 0 E F j for t E I. Then, ty,, is the maximal solution of(III.4.6-r7) on I. Furthermore, 104(t) < 1,1(t) < ot(t) + p if t: < rt < t + 6(p).Letting p - 0, we complete the proof of the theorem.

Assuming again that f is continuous on a domain D in the (t, y)-plane, considerinitial-value problem (III.4.6-{) together with another initial-value problem

(III.4.9-E) dt = f (t, y) + E, y(r)

where (t, 1;) E V and f is a positive number. We prove the following theorem.

Theorem III-4-5. If the maximal solution tt'(t) of problem (111.4.6-,) on the in-terval I = {t : r < t < r'} exists, then, for any positive number p, there existsanother positive number e(p) such that for 0 < E < E(p), every solution d(t) of(11!.4.9-E) exists on I and tp(t) < d(t) < y',(t) + p on the interval I. In particu-lar, for every sufficiently small positive number f, then exists the maximal solutiontf=, (t) of problem (111.4.9-E) on I and lim tL, = tV uniformly on I.

Proof

The maximal solution tI'(t) satisfies the condition (t,>l,(t)) E D on I (cf. Defini-tion 111-4-3). Define a function 9(t, y) by

f(t,ty+(t)+p), y>V'(t)+p, t EI,9(t, y) = f (t, y), do(t) < y < 0(t) + p, t E I,

f (t, d,(t)), y < rif(t), t E I.

Then, g is continuous and bounded on the domain {t E I, -oo < y < +oo}. Hence,every solution ¢(t, e) of the initial-value problem

= 9(t, y) + E, y(r) _

exists on the interval I.Note that

(III.4.10) d0(t) = f (t, 0(t)) = 9(t, tl'(t)) < 9(t, trdt

(t)} + E

on I (cf. Figure 17). Hence,

(111.4.11) t'(t) < p(t, E) on t E I.

We prove that for a given positive number p, there exists another positive numbere(p) such that if 0 < e <_ E(p), we have 0(t, e) < t1(t)+p on I. Otherwise, there exista real number r(p) and two sequences {Ek : k = 1, 2.... ) and {rk : k = 1, 2, ... }of real numbers such that(1) ek > O and lim Ek = 0,

k +oo

(2) T' > rk ? T(p) > T and lim Tk = To exists,k+00(3) 0(7k, ek) = IP(rk) + p,(4) 0(t, (k) < V,(t) + p for r < t < Tk,(5) lim 0(t, ek) = 0(t) exists uniformly on 1.

Then, O(t) is a solution of problem on the interval r <_ t < ro < T'. Since0(ro) = Ty(ro) + p, this is a contradiction. Thus, it was proved that 0(t) S 0(t, e) <,y(t) + p on I for 0 < e < e(p). Therefore, Theorem 111-4-5 follows immediately. 0

In the proof of Theorem 111-4-5, (111.4. 11) was derived from (111.4.10) (cf. Figure17). In a similar manner, we can prove the following result.

Lemma 111-4-6. Assume that f (t, y) is continuous on a region 12 = {(t, y)w_(t) < y < w+(t), t E TO}, where l0 = {t : a < t < b} and

(i) w+ and w_ are real-valued, continuous, and differentiable on T0,(ii) w- (t) < w+ (t) on lo.

Assume also that

J

dw+(t)> f(t,w+(t)) on Zo,dt

dw

dt(t) < Pt"'-(O) on TO

and w_(a) w+(a). Then, every solution 0(t) of the initial-value problem

(111.4.12)dy

= f (t, y), y(a) _dt

exists on To and w_ (t) < 0(t) < w+(t) on 10 (i.e. (t, y5(t)) E Q). In particular, themaximal solution 01(t) and the minimal solution 02(t) of problem (111.4.12) on loexist and w_ (t) < 02 (t) < ¢, (t) < w+ (t) on Zo

Proof of this lemma is left to the reader as an exercise (cf. Figure 18).

Y = vKW)

Y=v= #(z)


Using Lemma 111-4-6, we prove the following theorem due to O. Perron (Per2j.

Theorem 111-4-7. Assume that f (t, y) is continuous on a region

fl = {(t, y) (t) < y < w+(t), t E Zo},

whereto = (t:a<t <b) and(i) w, and w_ are teal-valued, continuous, and differentiable on 4,

(ii) j- (t) 5 w+ (t) on 4.

Assume also thatdw+(t)

> f(t (t ) 1,W+ )dton 0,

dw (t) < f(t t ) 4,w-( )dton

-

and w_ (a) < 4 < w+(a). Then, there exists a solution Q(t) of initial-value problem(111.4.12) on To such that

w_ (t) < Q(t) < W+(t) on To, i.e., (t, ¢(t)) E Q .

Prof.

Let us define a function g(t, y) by

f(t,W+(t)), y ? w+(t), t E To,

g(t, y) = f(t, y), W - W+ (t), t E Zo,

f (t, W_ (t)), y < W- (t), t E lo.

Then, g is bounded and continuous on the region {(t, y) : t E 10, -00 < y < +oo).Hence, every solution ¢(t) of the initial-value problem

j = g(t, y), y(a)

exists on Za.

Set "I+(t,E) = W.}. (t) + E(t - a) and w_(t,e) = w_(t) - e(t - a), where c is apositive number and t E To. Then,

+(t, E) dW+(t)dt dt

dw_(t,E) _ d -(t)dt dt

+ ( > f(t,w?(t)} + E > g(t,w+(t,e)),

- E < f(t,W-(t)) - E < g(t,W_(t,E))-

Hence, w_ (t, c) < ¢(t) < w+ (t, e) on Zo (cf. Lemma 111-4-6). Letting e 0, wecomplete the proof of Theorem 111-4-7.

Comment III-4-8. The maximal and minimal solutions given in this section arealso defined and explained in [CL, pp. 45-48] and f Har2, p. 251.

111-5. A comparison theorem

In this section, we derive an estimate for solutions of a differential equation bymeans of differential inequalities. Let us introduce the basic assumption.

5. A COMPARISON THEOREM 59

Assumption 1. Let t and u be two real variables and let g(t,u) be a real-valuedand continuous function of (t, u) on a domain D in the (t, u)-plane. Also let .y(t)be the maximal solution of the initial-value problem

du= g(t,u), u(a) = uo

on an interval lo = It : a < t < b}, where (a, uo) E D and (t, i,b(t)) E D on To.

We first prove the basic theorem given below.

Theorem 111-5-1. Assume that(1) g(t, u) > 0 on D and uo > 0,

=(2) an R"-valued function ¢'(t) is continuously differentiable on a subinterval Iit: a< t <,r) of 4,

(3) uo,(4) (t, 14(t) I) E V on Z,(5)

Then,

do(t)dt

< g on Z.

P(t) on 1.

Proof.

Let a be a positive number and let 0(t, e) be any solution of the initial-valueproblem

dudt = 9(t, u) + e, u(a) = u0.

If e > 0 is sufficiently small, i/i(t, e) exists on Zo and lim 0(t, e) = P(t) uniformly on

To (cf. Theorem III-4-5). Let us make the following observations:

(I) At)I - Im(s)I <-

and

(II) 0(t, e) - t'(s, e)

Jt doI ds < f g(o, I m(o) I )d

t4Y(r,t(a,e)) + e1da,

where t > s. Suppose that l (s)I = r'(s, e) for some s E Z. Then, there exists apositive number b(e) such that

Ig(a,+G(o, e)) - 9(o, I$(a)I)I 2 for Iv - sl < b(e).

This implies that

{

Id(t)I 

0(t, e)

0(s, e)

for

for

s < t < s + b(e),

s - b(e) < t <s

60 111. NONUNIQUENESS

(cf. Figure 19).

FIGURE 19.

Note that

t'(t,e) - WWI > f [g(a,'p(a,e)) - g(a, j¢(o)j))do + e(t - s) for t > sta

and

[g(o, v(o, )) - g{O, (o )I)do - e(s - t) for tt)! C jt

Thus, it is concluded that I (t) < t'(t,f) on Z. Letting a 0, we complete theproof of Theorem III-5-1. 0

Example 111-5-2. If an R"-valued function 0(t) satisfies the condition

dj(t)dt

< C + MI¢(t)t for a < t < b and uo,

we can apply Theorem 111-5-1 to fi(t) with g(u) = C + Mu. Note that the initial-

value problem = C + Mu, u(a) = uo has the unique solution

+G{t) = uoeM(t-a) + M (eai(t-a) - i) .

Hence,

W(t)l <_ M (em('-4) - 1) for a < t < b.

We still assume that Assumption 1 is satisfied by the function g(t, u) and O(t)and that uo > 0 and g(t, u) > 0 on D. Consider a differential equation

(111.5.1)dy

= f (t, yIdt

under the following assumption:(a) the function f (t, yj is continuous on a domain f2 in the (t, y7)-space,(0) (t, Iyj) E D if (t, y-) E S2,

6. SUFFICIENT CONDITIONS FOR UNIQUENESS 61

(-v) If (t, y-) I 5 g(t, I yl) for (t, yj E 0.Then, the following result is a corollary of Theorem 111-5-1.

Corollary 111-5-3. If fi(t) is a solution of differential equation (111.5.1) on asubinterval a < t < r of the interval Zo such that (t, fi(t)) E 12 for a < t < T andthat jd(a)I S uo, then Im(t)I < v' (t) for a < t < T.

Proof of this result is left to the reader as an exercise.

111-6. Sufficient conditions for uniqueness

In this section, we derive some sufficient conditions for uniqueness by means ofdifferential inequalities. The basic assumption is given below.

Assumption 1. Let t and u be two real variables and let r(t), w(t), and g(t, u) bereal-valued functions such that

(1) r(t) and w(t) are continuous on an interval I = {t : a < t < b},(2) r(t)>0 andw(t)>0 on1.(3) g(t, u) is continuous on the set A = {(t, u) : 0 0 on A and g(t, 0) = 0 on Z,(5) the problem

(111.6.1)

du(t)dt

= g(t,u(t)),

aim u(t) = 0

(t,u(t)) E A on Z,

has only the trivial solution u = 0 on Z.

Let us consider a problem

(111.6.2)

d9(t)(t, y"(t)) E Q on Z,

dt

lim 19(01 = 0,t-.a r(t)

where S1 = {(t, y) : jyj < w(t), t E Z}. The main result of this section is thefollowing theorem.

Theorem 111-6-1. If the function f (t, y-) is continuous on 1 and if

I f(t, y-)I :5 g(t, Iyi) on Q,

then problem (111.6.2) has only the trivial solution y" = 6 on Z.

Proof

Suppose that problem (III.6.2) has a nontrivial solution j(t) on Z. This meansthat (a) ,-a< 6. for some a E Z. Choose a positive number 6 so that f <

min aminbw(t) }.

Let us make the following two observations.Observation 1. Note that the set A(i3) = {(t, u) : a < t < b, 0 < u < B} is asubset ofd and that u = 0 is a solution of the differential equation

(111.6.3)dudt = g(t, u)

on the interval a < t < b. Using Theorem 111-2-7, we can construct a nontrivialsolution uo(t) of (111.6.3) on the interval a < t < b so that (t, uo(t)) E A(f3) ona<t<b(cf. Figure 20).Observation 2. The nontrivial solution uo(t) of (111.6.3) which was constructedin Observation 1 can be continued on the interval a < t < a so that

(111.6.4) 0 < uo(t) < on a < t < a.

To show this, we first remark that for a given positive number e, any solution t(i(t, c)of the initial-value problem

(III.6.5) = g(t, u) + e, u(a) = uo(a)

satisfies the condition tfi(t, e) < on any subinterval a < t < a of I if it existson this subinterval (cf. Figure 21).

r=a r=b

I I u= 1 ;(r)1

I I u ='Kr' E)I 1 L -_Ju=P

TV, El u uo(r)u= u0(t) IU

U=0 u=0r=b r=a

FIGURE 20.

Now, define a real-valued function G(t, u) by

FIGURE 21.

(g(t, u), u > 0, a < t <a,1G(t, u) to, u < 0, a < t < a.

Then, any solution ty(t, e) of problem (III.6.5) can be continued on the interval

a < t < a as a solution of the differential equation = G(t, u) + e. Since the set

{t(i(-, e) : 0 < e < 1) is bounded and equicontinuous on every closed subinterval ofa < t < or, we can select a sequence {Ek : k = 1,2.... } of positive numbers suchthat limEk = 0 and IirW(t,ek) = fi(t) exists uniformly on each closed subinterval

k--0 k-0of a < t < a. It is easy to show that j)(t) is a solution of the initial-value problemdu

= g(t, u), u(a) = uo(a) which satisfies condition (111.6.4).

Thus, we constructed a nontrivial solution u0(t) of differential equation (111.6.3)such that (t,uo(t)) E 0 on I and 0 < uo(t) < ]5(t)l on a < t < a. Since

Urn ±L = 0, this contradicts condition (5) of Assumption 1. 0t-a r(t)From Theorem III-6-1, we derive the following result concerning uniqueness of

solutions.

Theorem 111-6-2. Suppose that Assumption 1 is satisfied and that(i) an 1R"-valued function f (t, y-) is continuous on a domain D in the (t, yl -space,

(ii) f satisfies the condition J f (t, 91) - f (f, 92)1 < g(t, Iyi - 92{) whenever t E 1,Iy, - 921:5 w(t), (t, yi) E V, and (t, y2) E D.

Let Q(t) be a solution of the differential equation

dt = f (t, y) on I = {t : a < t < b}

such that the set Do = {(t, y-) 4(t)1 < w(t), t E 1} is contained in D. Then,the problem

(111.6.6)

has only the

Proof.

dylt) = f(t,b(t)),dt

lun y(t)r0t-a ()

solution (t) itself.

(t, y'(t)) E Do on 1,

Set l(t) = z(t) + .(t). Then, problem (111-6.6) becomes

c-aHill r(t) = 0.

on Z,

Since I1 (t,y-Q(t))j < g(t, on Do, Theorem 111-6-2 follows from TheoremII1.6.1. 0

Let us apply Theorem III-6-2 to some initial-value problems.Example 1 (The Osgood condition (cf. [Os])). Consider a real-valued functiong(t, u) = h(t)p(u) in the case when

(1) h(t) is continuous and h(t) > 0 on an interval 1 = {t : a < t < b},

di(t)dt = F(t, i(t)) = A t ' z"(t) + fi(t)) - At' fi(t))

!AL)!(t 1="(t)!) E o on Z, and

(2) p(u) is continuous on an interval 0 ,.5 u < K, where K is a positive number,(3) p(0) = 0 and p(u) > O for 0 < u < K,

(4) r h(t)dt < +oc,a+

(5)(} jo+ p(u) = +oo.

Assume that(1) an R"-valued function f (t, y) is continuous on a domain V in the (t, yam)-space,

(ii) f satisfies the condition If (t,1%1) - f (t, il2)1 < 9(t, Iii - Y21) = -1721)whenever t E Z, I171 - g2 l < K, (t, y"1) E D, and (t, y2) E D.

Let ¢1(t) and 42(t) be two solutions of an initial-value problem = f (t, M, y"(a) _i such that

(a, T11 E D and (t, p1(t)) E D, (t, 2(t)} E D

on the interval Zo = {t : a < t < b}. Then, p1(t) = ¢'2(t) on Zo.

Proof.

It is sufficient to show that Assumption 1 is satisfied by three functions r(t) = 1,

w(t) = K, and g(t, u) = h(t)p(u), and that lim 1&t)(t)

1(t)) = 0. This limitt-acondition is evidently satisfied, since ¢2(a) = 41(a) and r(t) = 1. Conditions (1),(2), (3), and (4) of Assumption 1 are also evidently satisfied. Therefore, it sufficesto prove that condition (5) is also satisfied.

Let u(t) be a real-valued function such that(A) 0<u(t) <Kon asubinterval r<t<aof I={t:a<t<b},(B)

tlira+ u(t) = 0,

(C) dutt) = h(t)p(u(t)) on r < t < o.

Then,

r(1

FU (t))d t)dt = J o h(t) dt.

I

This contradicts condition (5).

Example 2 (The Lipschitz condition). If we choose h(t) = L and p(u) = u, whereL is a positive constant, then the Osgood condition becomes the Lipschitz condition(cf. Assumption 2 of §I-1).

Example 3 (The Nagumo condition). Define r(t), w(t). and g(t, u) by

r(t) = (t -a)', w(t) = K, 9(t, u) = t L a u,

where A is a non-negative constant and K is a positive constant. Assume also that

u > 0 and t > a. Then, the general solution of the differential equation = g(t, u)

is given by u(t) = c(t - a)', where c is an arbitrary constant. This implies thatAssumption 1 is satisfied by r(t), w(t), and g(t,u). In particular, choosing A = 1,we derive the following result due to M. Nagumo (Nal and Na2J. Assume that

(i) f (t, y') is continuous on a domain V in the (t, y-)-space,

(ii) f satisfies the condition j f(t, 91) - f (t, 92)1 < 191 _

a-921 whenever a < t < b,

III, -Y2I<K,(t,fi)EDand (t,Y2)EV.dY

Suppose that 1(t) and y+,2(t) are two solutions of the initial-value problemdt

=

f (t, y), g(a) = # such that

(a, 1 E V and (t, 41(t)) E V. (t, fi2(t)) E V

on the interval I= {t: a<t<b}. Then, $1(t) = &t) on 1.

= andjNote that, since 1(a) = $2(a) i

_WT= (a) = f (a, then

lim 1*1(t) - 02(t)] = 0.t-a t-aRemark 111-6-3. The sufficient conditions of Osgood's [Os] and M. Nagumo's[Nal and Na21 for uniqueness are also explained in [CL. pp. 48-601 and [Hart, pp.31-351.

Problem 4. Show that the initial-value problem

dy _ {tsin(f. if t 1 0,dt 0 if t = 0, y(r) = n

has one and only one solution.

Answer./ \

Note that t sin I tz I - t sin (L2) = yl t cos (fi) for some y on the interval

between y1 and y2. Therefore, we can use the Lipschitz condition for r - 0, whereaswe can use the Nagumo condition at r = 0.

Remark 111-6-4. In order to apply Theorem 111-6-2 to the initial-value problemsin Examples 1 and 3, it was assumed that f (t, y-) is continuous at the initial point(a,'qj. However, the reader must notice that problem (111.6.6) is more generalthan an initial-value problem. Actually, we can apply Theorem 111-6-2 even whenf (t, yr) is not continuous at t = a. For example, the Nagumo condition (ii) of

Example 3 is satisfied by the function sin 1 t I at a = 0. Therefore, the problem

{!) = sin (i), llm - = 0 I has only the trivial solution y(t) = 0.t-0 t

EXERCISES III

III-1.-Prove the Kneser theorem under the assumption that(i) f'(t, yj is continuous on R = {(t, yj : a < t < b, Myj < +oo),

(ii) 111(t, y-)l < K + Llyl on R for some positive numbers K and L.

Hint. Note that if I f (t, yj I < K + LI yj on R for some positive numbers K and

L, any solution of the system 9 = f (t, yJ) exists on a < t < b and satisfies the

estimate jy(t)l < (jy(a)I + K(t - a))eL(I -a) for a < t < b.

111-2. Find the maximal solution and the minimal solution of the initial-valueproblem _, y(0) _ -1 on the interval 0 < t < 10.

III-3. Set

H(v) =

and

h(u) _for u > 0,

- i for u < 0,

of Lh \v n) hn1J for v > 0,

-H(-v) for v < 0,

=0 for -1<t<-1+b,>0 for -I+6<t<0

1-4t),

0 for t = ,0

- c(-t) for 0 < t < 1,where 0 < d < 1. Show that

(a) v = 0 and v = ± 1 (n = 1,2.... ) are solutions of the differential equationn

(E)dv c(t)a

_H'(v) ,

where H'(v) = dHdv '

(b) solutions of (E) with the initial-values v(-1) = 0 and v(-1) = f 1 are notn

unique;(c) for any positive constant c,, the differential equation

dv172 ISin (r)

__ c(t)I

v2 sin +c LH'v) J(-")vl

satisfies the Lipschitz condition in v for jvi < +oo and Itt < 1.Also, find the maximal solution and the minimal solution of the solutions of (E)satisfying the initial-condition v(-1) = 0 on the interval -1 < t < 1.

Hint. See (KSI.

EXERCISES III 67

111-4. Show that if a continuously differentiable R"-valued function b(t) satisfiesan inequality

do(t)dt

< I40 for t > 0 and o(0) = 0,

then 1¢(t)1 < t2 for t > 0.

Hint. Use Theorem 111-5-1.

111-5. Assuming that an R'-valued function j(t) is continuously differentiable for0 < t < 1, show that if ¢(t) satisfies the condition

dd(t)dt

s ia(t)12

then ,(t) = 0 for 0 < t < 1.

Hint. Use Theorem III-5-1.

for 0 < t < 1 and lim 1d(t)I = 0,9 0+

111-6. Assuming that an R"-valued function fi(t) is continuously differentiable for0 < t < +oo, show that if a(t) satisfies the condition

d9(t)

l 10(t)i eXpft21 = 0,

then (t) = 0 for 0 < t < +oo.

Hint. Use Theorem 111-5-1.

111-7. Show that the problem

dt i

< 2tlm(t)I for 0 <- t < +oo,

dy(t) = sinIt1

y(t) lim y(t) = 0-(-0

has only the trivial solution y(t) = 0.

Hint. Note that

/yt

-sin(/

1M l = Y'- Y2cos1/

!Isin( It1 t1) ltl`t1 j,

for some y between yl and y2.

68 M. NONUNIQUENESS

111-8. Let f (t, i, y) be an R"-valued function of (t, x, y-) E R x R" x R'". Assumethat

(1) the entries of f'(t, i, yl are continuous in the region A 0 < t <a, jxrj < a, jyj < b}, where a, a, and b are fixed positive numbers,

(2) there exists a positive number K such that j f (t, x"j, y- )- f (t, x"2, y -)l < Kjxt -x"2 jif (t,i),y-) E A (j = 1, 2).

Let U denote the set of all R"'-valued functions u(t) such that ju(t)j < b for 0 <t < a and that JiZ(t) - u(r)j < Lit - rj if 0 _< t < a and 0 < r < a, where Lis a positive constant independent of u E U. Also, let ¢(t; u+) denote the unique

solution of the initial-value problem u(t)), x(0) = 0', where u" E U. It is'jjknown that there exists a positive number ao such that for all u E U, the solution(t, u') exists and u)j < a for 0 < t < aa. Denote by R the subset of Rn+'which is the union of solution curves {(t, 0(t, 65)) : 0 < t < ao) for all u E U, i.e.,R = {(t, ¢(t, u)) : 0 < t < aa, u E U}. Show that R is a closed set in R"+t

Hint. See [LM1, Theorem 2, pp. 44-47) and [LM2, Problem 6, pp. 282-283).

111-9. Let f (t, 2, yy be an R"-valued function of (t, a, y) E R x R" x R'". Assumethat

(1) the entries of f (t, f, y-) are continuous in the region 0 = {(t, 2, yj : 0 < t <a, Jx"j < a, jy"j < b}, where a, a, and b are fixed positive numbers,

(2) there exists a positive number K such that yl- f (t, x2, y-)J < Kjit-i2jif (t,i1,y)EA (j=1,2).

Let U denote the set of all R'"-valued functions g(t) such that I i(t)J < b for 0 <t < a and that the entries of u are piecewise continuous on the interval 0 <t < a. Also, let (t; u) denote the unique solution of the initial-value problemdidt

= f (t, x, u"(t)), x(0) = 0, where u" E U. It is known that there exists a positive

number ao such that for all u' E U, the solution (t, u) exists and Jd(t, u)j < a for0 <- t < ao. Denote by 1Z the subset of R"' which is the union of solution curves{(t, ¢(t, ul)) : 0 < t _< ao) for all u E U, i.e., R = {(t, (t, u)) : 0 < t < ao, u E U).Assume that a point (r, ¢(r, uo)) is on the boundary of R, where 0 < r < 00 and!!o E U. Show that the solution curve 0 <- t < r} is also on theboundary of R.

Hint. jLM2, Theorem 3 of Chapter 4 and its remark on pp. 254-257, and Problem2 on p. 258).

III-10. Let A(t, x) and f (t, x") be respectively an n x n matrix-valued and R"-valued functions whose entries are continuous and bounded in (t,x-) E R"+' ona domain 0 = { (t, x) : a < t < b, x" E R"), where a and b are real numbers.Also, assume that (r, {) E A. Show that every solution of the initial-value problemdx = A(t, x'a + f (t, i), i(r) = t exists on the interval a < t < b.dt

CHAPTER IV

GENERAL THEORY OF LINEAR SYSTEMS

The main topic of this chapter is the structure of solutions of a linear system

(LP) dt = A(t)f + b(t),

where entries of the n x n matrix A(t) are complex-valued (i.e., C-valued) contin-uous functions of a real independent variable t, and the Cn-valued function b(t) iscontinuous in t. The existence and uniqueness of solutions of problem (LP) weregiven by Theorem 1-3-5. In §IV-1, we explain some basic results concerning n x nmatrices whose entries are complex numbers. In particular, we explain the S-Ndecomposition (or the Jordan-Chevalley decomposition) of a matrix (cf. DefinitionIV-1-12; also see [Bou, Chapter 7], [HirS, Chapter 6], and [Hum, pp. 17-18]). TheS-N decomposition is equivalent to the block-diagonalization which separates dis-tinct eigenvalues. It is simpler than the Jordan canonical form. The basic toolsfor achieving this decomposition are the Cayley-Hamilton theorem (cf. TheoremIV-1-5) and the partial fraction decomposition of reciprocal of the characteristicpolynomial. It is relatively easy to obtain this decomposition with an elementarycalculation if all eigenvalues of a given matrix are known (cf. Examples IV-1-18and IV-1-19). In §IV-2, we explain the general aspect of linear homogeneous sys-tems. Homogeneous systems with constant coefficients are treated in §IV-3. Moreprecisely speaking, we define e'A and discuss its properties. In §IV-4, we explainthe structure of solutions of a homogeneous system with periodic coefficients. Themain result is the Floquet theorem (cf. Theorem IV-4-1 and [Fl]). The Hamilton-ian systems with periodic coefficients are the main subject of §IV-5. The Floquettheorem is extended to this case using canonical linear transformations (cf. [Si4]and [Marl). Also, we go through an elementary part of the theory of symplecticgroups. Finally, nonhomogeneous systems and scalar higher-order equations aretreated in §1V-6 and §IV-7, respectively. The topics of §§IV-2-IV-4, IV-6, andIV-7 are found also, for example, in [CL, Chapter 3] and [Har2, Chapter IV]. Forsymplectic groups, see, for example, [Ja, Chapter 6] and [We, Chapters 6 and 8].

IV-1. Some basic results concerning matrices

In this section, we explain the basic results concerning constant square matrices.Let Mn(C) denote the set of all n x n matrices whose entries are complex numbers.The set of all invertible matrices with entries in C is denoted by GL(n,C), whichstands for the general linear group of order n. We define a topology in Mn (C) bythe norm [A] = max [ask[ for A E Mn(C), where a3k is the entry of A on the j-th

1<j,k<nall a12 ... aln

row and the k-th column; i.e., A = all a22 a . A matrix A E Mn(C)

and an2 "' ann

69

70 IV. GENERAL THEORY OF LINEAR SYSTEMS

is said to be upper-triangular if ajk = 0 for j > k. The following lemma is a basicresult in the theory of matrices.

Lemma IV-1-1. For each A E Mn(C), there exists a matrix P E GL(n,C) suchthat P-1 AP is upper- triangular.

Proof.

Let A be an eigenvalue of A and pt be an eigenvector of A associated with theeigenvalue A. Then, Apt = A P t and P 1 # 0. Choose n - 1 vectors p", (j = 2, ... , n)so that Q = [#1A p"nJ E GL(n, C), where the p"j are column vectors of the matrix

A

Q. Then, the first column vector of Q-1AQ is0

Hence, ae can complete the

0proof of this lemma by induction on n.

A matrix A E Mn(C) is said to be diagonal if a,k = 0 for j k. We denoteby diag(di, d2, ... , d, the diagonal matrix with entries dl, d2, ... , do on the maindiagonal (i.e., d_, = aj.,). A matrix A E Mn(C) is said to be diagonalizable (orsemisimple) if there exists a matrix P E GL(n,C) such that P-1AP is diagonal.Denote by Sn the set of all diagonalizable matrices in Mn(C). The following lemmais another basic result in the theory of matrices.

Lemma IV-1-2. A matrix A E Mn(C) is diagonalizable if and only if A has nlinearly independent eigenvectors pt, p2, ... , PnProof

If A has n linearly independent eigenvectors pt, p'2, ... , p,, set P = [P'tpa ... pnJE GL(n,C). Then, P'1AP is diagonal. Conversely, if PAP is diagonal forP = [ 1 6 t h... fl, E GL(n, C), then p1, A, ... , pn 7are n linearly independenteigenvectors of A.In particular, if a matrix A E Mn(C) has n distinct eigenvalues, then n eigenvectorscorresponding to these n eigenvalues, respectively, are linearly independent (cf.[Rab, p. 1861). Therefore, we obtain the following corollary of Lemma IV-1-2.

Corollary IV-1-3. If a matrix A E Mn(C) has n distinct eigenvalues, then A ESn.

The set Mn(C) is a noncommutative C-algebra. This means that Mn(C) is avector space over C and a noncommutative ring. The set Sn is not a subalgebra ofMn(C). However, the following lemma shows an important topological property ofSn as a subset of Mn(C).

Lemma IV-1-4. The set Sn is dense in Mn(C).

Prioof.

It must be shown that, for each matrix A E Mn(C), there exists a sequence{Bk : k = 1,2,... } of matrices in Sn such that lim Bk = A. To do this, we

k +Wmay assume without any loss of generality that A is an upper-triangular matrixwith the eigenvalues At, ... , An on the main diagonal (cf. Lemma IV-1-1). Set

1. SOME BASIC RESULTS CONCERNING MATRICES 71

B k = A + diag[Ek.1, Ek,2, , Ek,n], where the quantities ek,,, (v = 1, 2, ... , n) arechosen in such a way that n numbers Al + Ek,1, A2 + 4,2, ... , An + Ek,n are distinctand that lim Ek,&, = 0 for v = 1, 2,... , n. Then, by Corollary IV-1-3, we obtain

Bk E Sn and lim Bk = A.k-r+oo

For a matrix A E Mn(C), denote by pA(A) the characteristic polynomial of Awith the expansion

n

(IV.1.1) pA(A) = det(AIn - A] = An + ph(A)An-''

h=1

where In denotes the n x n identity matrix. Note that

PA(A) = An + Eph(A)An-h, A° = In.h=1

Now, let us prove the Cayley-Hamilton theorem (see, for example, [Be13, pp. 200-201 and 220], (Cu, p. 220], and (Rab, p. 198]).

Theorem IV-1-5 (A. Cayley-W. R. Hamilton). If A E Mn(C), then its charac-teristic polynomial satisfies PA(A) = O, where 0 is the zero matrix of appropriatesize.

Remark IV-1-6. The coefficients ph(A) of pA(A) are polynomials in entries ap,of the matrix A with integer coefficients.

Proof of Theorem IV-1-5.

Since the entries of pA(A) are polynomials of entries a,k of the matrix A, theyare continuous in the entries of A. Therefore, if pA(A) = 0 for A E Sn, it isalso true for every A E Mn(C), since Sn is dense in Mn(C) (cf. Lemma IV-1-4).Note also that if B = P'1 AP for some P E GL(n, C), then pB(A) = PA(A) andpB(B) = P-'pA(A)P. Therefore, it suffices to prove Theorem IV-1-5 for diagonalmatrices. Set A = diag(A1, A2, ... , A.J. Then, pA(A) = (A - A1)(A - A2) ... (A - An)and pA(A) = diag[PA(A1),PA(A2),... ,PA(A.)] = 0.

It is an important application of Theorem IV-1-5 that an n x it matrix N satisfiesthe condition N" = 0 if its characteristic polynomial pN(A) is equal to A". IfN" = O, N is said to be nilpotent.

Lemma IV-1-7. A matrix N E Mn(C) is nilpotent if and only if all eigenvaluesof N are zero.

Proof.

If IV is an eigenvector of N associated with an eigenvalue A of N, then Nkp"= Akp'for every positive integer k. In particular, N"p = A"p Hence, if N" = 0, thenA = 0. On the other hand, if all eigenvalues of N are 0, the characteristic polynomialpN(A) is equal to A". Hence, N is nilpotent. 0

Applying Lemma IV-1-1 to a nilpotent matrix N, we obtain the following result.


Lemma IV-1-8. A matrix N E .Mn(C) is nilpotent if and only if then exists amatrix P E GL(n,C) such that P-I NP is upper-triangular and the entries on themain diagonal of N are all zero. Furthermore, if N is a real matrix, then thereexists a real matrix P that satisfies the requirement given above.

To verify the last statement of this lemma, use a method similar to the proof ofLemma IV-1-1 together with the fact that if an eigenvalue of a real matrix is real,then there exists a real eigenvector associated with this real eigenvalue. Details areleft to the reader as an exercise.

The main concern of this section is to explain the S-N decomposition of a matrixA E Mn(C) (cf. Theorem IV-1-11). Before introducing the S-N decomposition,we need some preparation.

Let A. (j = 1, 2,... , k) be the distinct eigenvalues of A and let m., (j =1,2.... , k) be their respective multiplicities. Then, the characteristic polynomialof the matrix A is given by pA(A) = (A - '\0M'(1\ - A2)m2 ... (A - Ak)m'. Decom-

pose 1 into partial fractions 1 = QU(A, where, for every j, the

pA(A) pA(A) (A - A,),n,

quantity Q, is a nonzero polynomial in A of degree not greater than m, - 1. Hence,k

I = EQ,(A) J1 (A - Ah)m". Setting)=1 h¢j

P,(A) = Qj(A) 1I (A - A,)meh#1

i = P2(A}.

J=1

Now that this is an identity in A. Therefore, setting

(IV.1.4) Pj(A) = Qj(A)fl(A-AhIn)m" (y = 1,2,....k),hv&1

we obtain

k

(IV.1.5) I. _ > Ph(A).h=1

In the following two lemmas, we show that (IV.1.5) is a resolution of the identity interms of projections Ph (A) onto invariant subspaces of A associated with eigenvaluesAh, respectively.

Lemma IV-1-9. The k matrices P, (A) (j = 1,2,... , k) given by (W-1.4) satisfythe following conditions:

(i) A and P, (A) (y = 1, 2, ... , k) commute.


(ii) (A-),In)'n'Pi(A) =0 (j = 1,2,... ,k),(iii) P,(A)Ph(A) = 0 ifj h,

k

(iv) >Ph(A) = In,h=1

(v) Pi(A)2=P,(A) (j=1.2,...,k),( v : ) P, (A) 0 (j = 1, 2, ... , k).

Proof.

Since P,(A) is a polynomial of A, we obtain (i). Using Theorem IV-1-5, wederive (ii) and (iii) from (IV.1.4) and (i). Statement (iv) is the same as (IV.1.5).Multiplying the both sides of (IV.1.5) by P,(A), we obtain

k

(IV.1.6) P,(A) _ >P,(A)Ph(A).h=1

Then, (v) follows from (IV.1.6) and (iii). To prove (vi), let IT, be an eigenvector ofA associated with the eigenvalue Al. Note that (IV.1.2) implies Ph(A)) = 0 if h 0 j.Therefore, we derive P,(A,) = 1 from (IV.1.3). Now, since P. (A)#, = P,(A3)p' #we obtain (vi).

Lemma IV-1-10. Denote by V. the image of the mapping P,(A) : C" -. Cn.Then,(1) p'E Cn belongs to V. if and only if P,(A)p= p(2) Pj(A)p"=0 for all fl E Vh if j 0 h.(3) Cn = V1 Ei3 V2 e e Vk (a direct sum).(4) for each j, V, is an invariant subspace of A.(5) the restriction of A on V, has a coordinates-wise representation:

(IV.1.7) Alv, : AjIj + A,,

where I. is the identity matrix and Nj is a nilpotent matrix.(6) dime V, = m, .

Proof

Each part of this lemma follows from Lemma IV-1-9 as follows.A vector IT E V3 if and only if p" = P, (A)q" for some q' E Cn. If p" = P, (A) q, weobtain P,(A)p= Pj(A)2q'= Pj(A)q =p""from (v) of Lemma IV-1-9.A vector p" E Vh if and only if ff = Ph (A),y for some q" E Cn. Hence, from (iii)of Lemma IV-1-9 we obtain P, (A)IF = Pj (A)Ph(A)q"= 0 if 0 h.(iv) of Lemma IV-1-9 implies p = P, (A)15 + + Pk- (A)p" for every p3 E C",while (1) implies that P, (A)p E Vj. On the other hand, if p" = )51 + + pkfor some g,E V2 (j = 1,2,... , k), then, by (1) and (2), we obtain P,(A)p" =

Pi(A)p1+...+PJ(A)pk =pj

Ap"= AP,(A)p = P,(A)Ap E V3 for every 15E Vj.Let n, be the dimension of the space V, over C and let {ff,,t : 1 = 1, 2, ... , n. }be a basis for V,. Then, there exists an n. X n, matrix N,, such that


(A-)'1In)V1,1PJ,2...p'1.n,J = [Pi,1PJ,2...p3i.nsJNj

as the coordinates-wise representation relative to this basis. This implies that

(A - A,1.)'Pi(A)(P1,1Pi,2...p),nj) _ I

IP1,191,2...Pj,Nj for (t = 1,2,... ).

In particular, from (ii) of Lemma IV-1-9, we derive N, O . Thus, weobtain

(IV-1-8) ...P = [l ,1P'r,2 ... p'f n,](A) I, + N,),

where 1, is the n, x n1 identity matrix. This proves (IV.1.7).(6) Let { " l , t : e = 1, 2,... , n, } be a basis for V, (j = 1, 2,... , k). Set

(IV.1.9) Po = (p1,1 . p'2...,

Then, Po E GL(n, C) and (IV.1.8) implies

([V.1.10) Po'AP0 = diagjAllt +N1,A212+N2,...,Aklk+Nk],

where the right-hand side of (IV.1.10) is a matrix in a block-diagonal form withentries Al h +N1, A212+N2, ... , Aklk+Nk on the main diagonal blocks. Hence,pA(A) A 2 ) ' 2 Also, PA(A) _ (A - A1)m'(A -X2 )12 (A - Ak)mk. Therefore, dimC V) = n, = m, (j = 1,2, ... , k). 0

The following theorem defines the S-N decomposition of a matrix A E Mn(C).

Theorem IV-1-11. Let A be an n x n matrix whose entries are complex numbers.Then, there exist two n x n matrices S and N such that(a) S is diagonalizable,(b) N is nilpotent,(c) A = S + N,(d) SN = NS.

The two matrices S and N are uniquely determined by these four conditions. IfA is real, then S and N are also real. Furthermore, they are polynomials in Awith coefficients in the smallest field Q(a,k, A,1) containing the field Q of rationalnumbers, the entries ajk of A, and the eigenvalues Al, A2 ... , Ak of A.

Proof

We prove this theorem in three steps.

Step 1. Existence of S and N. Using the projections P,(A) given by (IV.1.4), defineS and N by

S = A1P1(A)+A2Pz(A)+...+At Pk(A), N=A - S.

If P0 is given by (IV.1.9), then

(IV.1.11) Po 1SP0 = diag[A111, A212, ... , Aklk)


and

(IV.1.12) Po 1 NPo = diag[N1 i N2, ... , Nk]

from Lemmas IV-1-9 and IV-I-10 and (IV.1.10). Hence, S is diagonalizable and Nis nilpotent. Furthermore, NS = SN since S and N are polynomials in A. Thisshows the existence of S and N satisfying (a), (b), (c), and (d). Moreover, from(IV.1.4), it follows that two matrices S and N are polynomials in A with coefficientsin the field Q(ajk, Ah).

Step 2. Uniqueness of S and N. Assume that there exists another pair (S, N)of n x n matrices satisfying conditions (a), (b), (c), and (d). Then, (c) and (d)imply that SA = AS and NA = AN. Hence, SS = SS, NS = SN, SN = NS,and NN = NN since S and N are polynomials in A. This implies that S - S isdiagonalizable and N - N is nilpotent. Therefore, from S - S = N - N, it followsthat S-S=N-N=O.Step 3. The case when S and N are real. In case when A is real, let 5 and N bethe complex conjugates of S and N, respectively. Then, A = S + N = 3° + N.Hence, the uniqueness of S and N implies that S = 3 and N = N.

This completes the proof of Theorem IV-1-11.

Definition IV-1-12. The decomposition A = S + N of Theorem IV-1-11 is calledthe S-N decomposition of A.

Remark IV-1-13. From (IV.1.11), it follows immediately that S and A have thesame eigenvalues, counting their multiplicities. Therefore, S is invertible if and onlyif A is invertible.

Observation IV-1-14. Let A be an n x n matrix whose distinct eigenvalues areA = S + N be the S-N decomposition of A. It can be shown

that n x n matrices P1, P2, ... , Pk are uniquely determined by the following threeconditions:

(i)(ii) P,P1 = O if j 36 t,(iii) S = A11P1 + A2P2 + ... + AkPk.

Proof.

Note that

{In = P1(A) + P2(A) + ... + Pk(A),Pj(A)Ph(A) = O if j h,

S = A1P1(A) + A2P2(A) + ... + AkPk(A).

k

First, derive that P, S = SP; = \j P,. Then, this implies that .X P1 = >ahPj P1. (A).h=1

Hence, \jPjPA(A) = \h PiPh(A). Thus, PiPh(A) = 0 whenever j h. Therefore,it follows that P1 = P1(A) = P2P}(A).


Observation IV-1-15. Let A = S + N be the S-N decomposition of an n x nmatrix A. Let T be an n x n invertible matrix such that if we set A = T-1ST,then A = diag[A1I1, A2I2, ... , AkIk], Where A1, A2, ... , Ak are distinct eigenvaluesof S (and also of A), I, is the m3 x mj identity matrix, and m3 is the multiplicityof the eigenvalue A,. It is easy to show that

(i) if we set M = T-'NT, then M is nilpotent, MA = AM, and M =diag[M1i M2,... , Mk}, where Mj are mj x m j nilpotent matrices,

(ii) if we set Pj = Tdiag[Ej1iE.,2,... ,E,k]T-1, where E,1 = 0 if j 1, whileEjj = Ij, we obtain

(I =P1+P2+...+Pk, PjPh =0 (.1 h),

. S = A1P1 + A2P2 + ... + AkPk.Therefore, P. = P, (S) = P? (A) (j = 1, 2, ... , k) (cf. Observation IV-1-14).

The following two remarks concern real diagonalizable matrices.

Remark IV-1-16. Let A be a real nxn diagonalizable matrix and let A1, A2 ... , Anbe the eigenvalues of A. Then, there exists a real n x n invertible matrix P suchthat

(1) in the case when all eigenvalues A j (j = 1,2,... , n) are real, then P-1 AP isa real diagonal matrix whose entries on the main diagonal are A1, A2, ... , An,

(2) in the case when all eigenvalues are not real, then n is an even integer 2m andP-1A fP = diag[D1, D2, ... , DmJ, where A23_1 = a, + ibl, A2J = a) - ibj, and

Djabj a,

(3) in other cases, P1AP = diag[D1, D2i ... , Dhj, where A23_1 = a)+ib,, A2.1 =

a, - ib,, and D. = I b,a j

for j = 1, 2, ... , h - 1 and Dh is a real diagonal

matrix whose entries on the main diagonal are A, (j = 2h - 1,... , n).

Remark IV-1-17. For any given real n x n matrix A, there exists a sequence{Bk : k = 1, 2,. ..} of real n x n diagonalizable matrices such that lim Bk = A.

This can be proved in the following way:(i) let A = S + N be S-N decomposition of A,

(ii) using Remark IV-1-16, assume that S = diag[D1, D2, ... , Dhj, as in (3) ofRemark IV-1-16,

(iii) find the form of N by SN = NS,(iv) triangularize N without changing S,(v) use a method similar to the proof of Lemma IV-1-4.

Details of proofs of Remark IV-1-16 and IV-1-17 are left to the reader as exercises.Now, we give two examples of calculation of the S-N decomposition.

Example IV-1-18. The matrix A

252 498 4134 698-234 -465 -3885 -656

15 30 252 42

-10 -20 -166 -25

has two

distinct eigenvalues 3 and 4, and1

PA(A) = (A - 4)2(A - 3)2,PA(A) (A 14)2 A

2

4 + (A 13)2 + A2

3


Set

P1(A) = (A - 3)2-2(A-4)(A-3)2' P2(A) _ (A - 4)2 + 2(A - 3)(A - 4)2.

Then,

P1(A) =

Therefore,

-1 -2 134 198

1 2 -125 -1860 0 9 12

0 0 -6 -8

P2(A) =

2 2 -134 -198-1 -1 125 186

0 0 -8 -120 0 6 9

2 -2 134 198

S = 4P1(A) + 3P2(A) =1 5 -125 -1860 0 12 12 I'

10 0 -6 -5250 500 4000 500

-235 -470 -3760 -470N = A - S =

15 30 240 30

-10 -20 -160 -20

3 4 3

Example IV-1-19. The matrix A = 2 7 4 has two distinct eigenvalues-4 8 3

A1= 11, A2=1,and

PA(A)1 ) 21 1 ) ,

Hence,

Set

Then,

1 _ 1 _ (A+9)PA(A) 100(A - 11) 100(A - 1)2*

1 - (A - 1)2 - (A + 9)(A - 11)

100 100

P1(A) _ (A - 1)2P2(A) - -

(A + 9)(A - 11).

100 100

1 0 56 28 1 1100 -56 -28P1(A) = - 0 76 38 , P2(A) 0 24 -38

1000 48 24

1000 -48 76

Therefore,

10 56 281

S = 11P1(A) + P2(A) = 10 0 86 380 48 34

20 -16 2N=A - S= 10 20 -16 2

-40 32 -4

1

In this case, SN = NS = N and N2 = O. Let Vj = Image (PI(A)) (j = 1,2).Then, by virtue of (1) of Lemma IV-1-10, Pj(A)p" = p" for all IT E Vj (j=1,2).

14 1 0Furthermore, V1 is spanned by 19 and V2 is spanned by 0 and 1 Set

12 0 -214 1 0

Pa = 19 0 1 . Then,12 0 -2

Po 1 SPo =11 0 0

[000 1 0 Pp 1 NPo = 2 1 -1 .

0 0 1 0 1 -1

It is noteworthy that there is only one linearly independent eigenvector x =

for the eigenvalue A2 = 1.

It is not difficult to make a program for calculation of S and N with a computer.For more examples of calculation of S and N, see [HKSI.

IV-2. Homogeneous systems of linear differential equationsIn this section, we explain the basic results concerning the structure of solutions

of a homogeneous system of linear differential equations given by

(IV.2.1) dt = A(t)y',

where the entries of the n x n matrix A(t) are continuous on an interval I = It :

a < t < b). Let us prove the following basic theorem.

Theorem 1V-2-1. The solutions of (1V2. 1) forms an n-dimensional vector spaceover C.

We break the entire proof into three observations.

Observation IV-2-2. Any linear combination of a finite number of solutions of(IV.2.1) is also a solution of (IV.2.1). We can prove the existence of n linearlyindependent solutions of (IV.2.1) on the interval Z by using Theorem I-3-5 with nlinearly independent initial conditions at t = to. Notice that each column vector ofa solution Y of the differential equation

(IV.2.2) dt = A(t)Y

on an n x n unknown matrix Y is a solution of system (IV.2. 1). Therefore, construct-ing an invertible solution Y of (IV.2.2), we can construct n linearly independentsolutions of (IV.2.1) all at once. If an n x n matrix Y(t) is a solution of equation(IV.2.2)on an interval I={t:a<t<b}and Y(t)EGL(n,C)for all tE1,thenY(t) is called a fundamental matrix solution of system (1V.2.1) on.T. Furthermore,n columns of a fundamental matrix solution Y(t) of (IV.2.2) are said to form afundamental set of n linearly independent solutions of (IV.2.1) on the interval T.

2. HOMOGENEOUS SYSTEMS OF LINEAR DIFF. EQUATIONS 79

Observation IV-2-3. Let 4(t) be a solution of (IV.2.2) on Z. Also, let *(t) be asolution of the adjoint equation of (IV.2.2):

(IV.2.3)dZdt

= -ZA(t)

on the interval Z, where Z is an n x n unknown matrix. Then,

dt-W(t)A(t)4s(t) + $(t)A(t)4;(t) = 0.

This implies that the matrix %P(t)4i(t) is independent of t. Therefore, W(t)4i(t) =%P(r)t(r) for any fixed point r E Z and for all t E Z. Note that the initial values44(r) and %P(r) at t = r can be prescribed arbitrarily. In particular, in the casewhen 4?(r) E GL(n,C), by choosing %P(r) = we obtain WY(t)4i(t) = Infor all t E Z. Thus, we proved the following lemma.

Lemma IV-2-4. Let an n x n matrix 4i(t) be a solution of (IV.2.2) on the intervalZ. Then, 45(t) is invertible for all t E I (i.e., a fundamental matrix solution of(IV.2.1)) if 4i(r) is invertible for some r E Z. Furthermore, 4 (t)-1 is the uniquesolution of (IV.2.8) on Z satisfying the initial condition Z(r) = 4i(r)-1.

Observation IV-2-5. Denote by 4?(t; r) the unique solution of the initial-valueproblem

(IV.2.4)- = A(t)Y, Y(r) = In,dt

where r E Z. Then, 1(t; r) E GL(n, C) for all t E Z. The general structure ofsolutions of (IV.2.1) and (IV.2.2) are given by the following theorem, which can beeasily verified.

Theorem IV-2-6. The C"-valued functwn y(t) = 44(t; r)it is the unique solutionof the initial-value problem

dtA(t)y", y(r) =1,

where it E C. Also, the n x n matrix Y = 4i(t; r)r is the unique solution of theinitial-value problem

dY = A(t)Y, Y(r) = Idt

where r E

Theorem IV-2-1 is a corollary of Theorem IV-2-6. 0Remark IV-2-7.

(1) The general form of a fundamental matrix solution of (IV.2.1) is given byY(t) = 4i(t; r)r, where r E GL(n, C).

(2) If a fundamental matrix solution is given by Y(t) = 4'(t; r)r, then Y(r) = F.Hence,

(IV.2.5) 0(t; ,r) = Y(t)Y(r)-1 (t, r E Z)

80 W. GENERAL THEORY OF LINEAR SYSTEMS

for any fundamental matrix solution Y(t). In particular,

(IV.2.6) 1(t;r) _ for t, r, r1 E Z.

(3) In the case when A(t) is a scalar (i.e., n = 1), we obtain easily

t(IV.2.7) 1(t; r) = exp IJA(s)ds)]

V,

t 1

In the general case, we define exp I J A(s)ds] byr

+Mexp[B(t)] = In B(t)"', where

m=1T11.

B(t) = Jt A(s)ds.

However, generally speaking, (IV.2.7) holds only in the case when B(t) andB'(t) = A(t) commute. In particular, 4(t; r) = exp[(t - r )AJ if A = A(t) isindependent of t. In §IV-3, we shall explain how to calculate exp((t - r)AJ,using the S-N decomposition of A. Also, (IV.2.7) holds in the case whenA(t) is diagonal on the interval I. A less trivial case is given in Exer-cise IV-9. It is easy to see that B(t) and A(t) commute if A(t) is a 2 x 2upper-triangular matrix with an eigenvalue of multiplicity 2. For exam-

ple, the matrix A(t) = [coSttI satisfies the requirement. In this

case, 4b(t; r) = exp I J t A(s)ds = exp (jsin t o sin rsint - sin r, )

exp(sin t -sin r) 101

1 t 1 rJ1

ift

(4) det Y(t) = det Y(r) exp trA(s)ds) if Y(t) satisfies (IV.2.2), where det A

and trA are the determinant and trace of the matrix A. This formula isknown as Abel's formula (cf. (CL, p. 28]).Proof.

Regarding detY(t) as a function of n column vectors {y'1(t),... ofY(t), set det Y(t) = 7(g, (t), ... , y (t)). Then,

(IV.2.8)d det Y(t) - [: P(...

, A(t)ym(t), ... ).dt L

M=1

Denote the right-hand side of (IV.2.8) by G(y'1(t),... Then, 9 ismultilinear and alternating in y1(t),... ,j,,(t). Furthermore, 9 = trA(t) ifY(t) = I,,. Therefore, 9 = tr A(t) det Y(t). Solving the differential equationddet Y(t) = tr A(t) det Y(t), we obtain Abel's formula. 0

dt

3. HOMOGENEOUS SYSTEMS WITH CONSTANT COEFFICIENTS 81

IV-3. Homogeneous systems with constant coefficients

For an n x n matrix A, we define exp[A] by

(IV.3.1) exp[A] = In + E h Ah.h=1

It is easy to show that the matrix exp[A] satisfies the condition exp[A + B] =exp[A] exp[B] if A and B commute. This implies that exp[A] is invertible and

(exp[A])-1 = exp[-A]. Thus, we obtain a fundamental matrix solution Y = exp[tA]

of the system d = Ay with a constant matrix A by solving the initial-valueproblem

(IV.3.2) = AY, Y(0) = In.

This, in turn, implies that the unique solution of the initial-value problem

(IV.3.3) d = Ay, y(T) = P

is given by y = exp[(t - r)A]p, where p E C' is a constant vector.In this section, we explain how to calculate exp[tA] for a given constant matrix

A, using the S-N decomposition of A. Assume that an n x n matrix A has k distincteigenvalues Al, A2, ... , Ak. Let A = S + N be the S-N decomposition of A. Also,let Pj (A) (,j = 1, 2,... , k) be the projections defined in §IV-1 (cf. (IV.1.4)). Then,

k k

(IV.3.4) In = > PJ(A), S = AjPJ(A), N = A - S,J=1 )=1

and

(IV.3.5) PJ(A)Ph(A) 0 (A) if h

h

= jj

(3, h = 1, 2, .... k).36

The two matrices S and N commute.

Denote by V, the image of the mapping PJ(A) : C" Cn (cf. Lemma IV-I-10).It is known that Sp = \,)5 for pin VJ . Hence, Sl p" = \1)5 and

+00

exp[tS]p' = 1 + E (A2t)n }ic earh=1n-i h

On the other hand, exp(tN] = In + A Nh since N is nilpotent. Therefore,h=1

exp[tA)# = exp[t(S + N)]p' = expitN] exp[tS]p" = e'\'` exp[tN]P

(IV.3.6)= eA'`

[In+

n- i thp" for fl E Vj.

n=1

Applying (IV.3.4) and (IV.3.6) to a general p E C", we derive

C.Njt

n-10(IV.3.7) exp[tA]p = In + E hl Nh P, (A)p' for 9E C".

j=1 h=I

Thus, we proved the following theorem.

Theorem IV-3-1. The matrix exp[tA] is calculated by formula

k n-1 th(IV.3.8) exp[tA] _ > ea+` In + F,

T!Nh Pj (A).

=l h=1

Since the general solution of the differential equation

(IV.3.9)

is given by (IV.3.7), the following important result is obtained.

Theorem IV-3-2.(i) If R(,\,) < 0 for j = 1, 2, ... , k, then every solution of (IV. 3.9) tends to 06 as

t -' +00,(ii) if R(Aj) > 0 for some j, some solutions of (IV.3.9) tend to 00 as t - +00,

(iii) every solution of (IV.3.9) is bounded for t > 0 if and only if R(AE) < 0 forj = 1, 2, ... , k and NP, (A) = 0 if R(A)) = 0.

Now, we illustrate calculation of exp[tA] in two examples. Note that in the casewhen A has nonreal eigenvalues, we must use complex numbers in our calculation.Nevertheless, if A is a real matrix, then exp[tA] is also real. Hence, at the end ofour calculation, we obtain real-valued solutions of (IV.3.9) if A is real.

-2 1 0Example IV-3-3. Consider the matrix A = 0 -2 0 . The characteristic

3 2 1

polynomial of A is pA(A) _ (A -1)(A+2)2. By using the partial fraction decomposi-

tion of1 (A + 2)2 - (A + 5)(A - 1) (A + 2)2

PA(A), we derive 1 = 9 . Setting P1(A) =9

and+ 5)(A - 1),

we obtainP2(A)9

0 0 0 1 0 0P1(A) = 0 0 0, P2(A) = 0 1 0

1 1 1 -1 -1 0

Set

-2 0 0 0 1 0S = PI(A) - 2P2(A) = 0 -2 0 , N = A - S = 0 0 0.

3 3 1 0 -1 0


Note that N2 = 0. Hence,

exp[tA] = e'[ 13 + tN ] PI(A) + e-21 [ 13 + tN ] P2(A)e-2t to-2t 00 e-2c 0

et - e-2t et - (1 + t)e-2t et

The solution of the initial-value problem dt = Ay, y(0) = y is y(t) = exp[tA]i).

To find a solution satisfying the condition Jim y(0) = 0, we must choose it so thatt +mP1(A)ij = 0. Such an iJ is given by it = P2(A)c", where c is an arbitrary constantvector in C3.

0 -1 1

Example IV-3-4. Next, consider the matrix A = 1 0 -1 . The charac--1 1 0

teristic polynomial of A is PA(A) = A(A2 + 3) = A(A - if)(A + i\/3-). Using the1

partial fraction decomposition ofPA (A)

, we obtain

1 = (A- if)(A+ if) - A(A+ if)- sA(A- if}.

Setting

P1(A) = 3(A2 + 3), P2(A) IA(A + if), P3(A) A(A - if),

we obtain

1 1 1 1 1 -2 1-if 1+ifPi(A) = - 1 1 1 , P2(A) = -- 1+if -2 1-if

3 1 1 1 6 1-if 1+iv -2

and P3(A) is the complex conjugate of P2(A). If we set

S = (if)P2(A) - (if)P3(A),

then S = A. This implies that N = 0. Thus, we obtain

exp(tA] = P1(A) + e,.''P2(A) + e-;J1tP3(A)

= P1(A) + 23t (e'3tP2(A)) .

Using(e'' t(1 + if)) = cos(ft) - \/3-sin(ft),

2 (e'-1t(1 - if)) wa(ft) + f sin(ft),


we find

exp[tA] = 13

a(t) b(t) c(t)c(t) a(t) b(t)b(t) c(t) a(t)

wherea(t) = I + 2cos(ft),b(t) = 1 - {cos(ft) + f sin(ft)} ,

c(t) = 1 - {wa(ft) - f sin(ft))

Remark IV-3-5. Fhnctions of a matnx In this remark, we explain how to definefunctions of a matrix A.

I. A particular case: Let A0, I,,, and N be a number, the n x n identity matrix,and an n x n nilpotent matrix, respectively. Also, consider a function f (X) in aneighborhood of A0. Assume that f (A) has the Taylor series expansion (i.e., f isanalytic at A0)

(h)

f(A) = f(A0) + 1f h?Ao)(A

- Ao)h.

h=1

In this case, define f (,\o1 + N) by

f(AoI. + N) = f(Ao)I. +

=f(h)PLO)

A.A = f(Ao)II +

n-1 f(h)(AO) NA.t

h=1h.

h=1h.

n-1 (h)

Since N is nilpotent, the matrix >2f is also nilpotent. Therefore, theh=1

characteristic polynomial pf(A0,+N)(A) of f(AoI + N) is

Pf(aol-N)(A) = (A - 1(A0))".

II. The general case: Assume that the characteristic polynomial PA(A) of an n x nmatrix A is

PA(A) = (A - .l1)m1(A - A 2 ) ' - 2 ... (A - Ak)mr.

where A1,... , Ak are distinct eigenvalues of A. Construct P, (A) (j = 1, ... , k), S,and N as above. Then,

A = (A1In + N)P1(A) + (A21n + N)P2(A) + ... + 0k1n + N)Pk(A).

Therefore,

A' = (A11n + N)1P1(A) + (A2In + N)'P2(A) 4- ... + (Ak1n + N)'Pk(A)

for every integer P.


Assuming that a function f (A) has the Taylor series expansion

f(A) = f(A3) 000f(h)(A,)(A-A,)''h=1 h!

at A = A., for every j = 1, ... , k, we define f (A) by

(IV.3.10)f(A) = f (A11, + N)Pk(A) + f (A21n + N)P2(A)

+ ... + f (Akin + N)Pk(A).

Since P2(S) = P,(A) (cf. Observation IV-1-15), this definition applied to S yields

f(S) = f(AIIn)P1(A) + f(A21.)P2(A) + -' + f(AkII)Pk(A)

and f (A) - f (S) has a form N x (a polynomial in S and N). Therefore, f (A) - f (S)is nilpotent. Furthermore, f (S) and f (A) commute. This implies that

f(A) = f(S) + (f(A) - f(S))

is the S-N decomposition of f (A). Thus,

Pf(A)(A) = pf(s)(A) = (A - f(A1))m' ... (A - f(Ak))m

Example IV-3-6. In the case when f (A) = log(A), define log(A) by

log(A) = log(A1In+N)Pk(A) + log(A21n+N)Pk(A) + ... + log(Akln+N)Pk(A),

where we must assume that A is invertible so that A, # 0 for all eigenvalues of A.Let us look at log(AoI,, + N) more closely, assuming that A0 0 0. Since

/ +O° m+1log(Ao + u) = 1000) + log t 1 + -o) = logl o) + (-I)m I o

\\ m=1

we obtain

n-1log(.1oIn + N) = log(Ao)ln +

m=1

(-1)m+1

m

It is not difficult to show that exp[log(A)J = A. In fact, since

(log(A))m = (log(A11n + N))mPk(A) + (log(A21n + N))mp2(A)+ ... + (log(Akln + N))mPk(A),

it is sufficient to show that exp[1og(Aoln + N)J = A01,, + N. This can be proved byusing exp[log(Ao +,u)] =A0 +,u.


Observation IV-3-7. In the definition of log(A) in Example N-3-6, we usedlog(A,). The function log(A) is not single-valued. Therefore, the definition oflog(A) is not unique.

Observation IV-3-8. Let A = S + N be the S-N decomposition of A. If A isinvertible, S is also invertible. Therefore, we can write A as A = S(II + M), whereM = S' N = NS-1. Since S and N commute, two matrices S and M commute.Furthermore, M is nilpotent. Using this form, we can define log(A) by

log(A) = log(S) + log(Ih + M),

where

log(S) = Iog(A1)P1(A) + log(A2)P2(A) + . + log(Ak)Pk(A)

and(1)m+1

log(IR + Al) = E - Mm.m

M=1

This definition and the previous definition give the same function log(A) if the samedefinition of log(A,) is used.

3 4 3

Example N-3-9. Let us calculate sin(A) for A = 2 7 4 (cf. Example-4 8 3

IV-1-19). The matrix A has two eigenvalues 11 and 1. The corresponding projec-tions are

0 14 7 -14 -725 25 25 25

P1(A) = 0

25

L9 L9

50

P2(A) = 02s 50

12 6 -12 190

25 250

5 25

Define S = 11P1(A) + P2(A) and N = A - S. Then N2 = 0. Also,

( sin(11 + x) = -0.99999 + 0.0044257x + 0.499995x2 + 0(x3 ),

t sin(1 + x) = 0.841471+0.540302x-0.420735x 2 + 0(x3).

Therefore,

sin(A) _ (-0.9999913 + 0.0044257N)P1(A) + (0.84147113 + 0.540302N)P2(A)

1.92207 -1.8957 -0.4075491.0806 -1.42252 -0.591695

-2.1612 0.845065 0.1834

It is known that sin x has the series expansion

+00

sin x = 1 (2h +)1)IT2h+1

4. SYSTEMS WITH PERIODIC COEFFICIENTS 87

Therefore, we can also define sin(A) by

- + (-1)h A2h+1sin(A) - - (2h + 1)!

However, this approximation is not quite satisfactory if we notice that

sin(11) = -0.99999 and_1h

(2h + 1)!112h+1 = -117.147.

h=O

IV-4. Systems with periodic coefficients

In this section, we explain how to construct a fundamental matrix solution of asystem

(IV.4.1)dy = A(t)ydt

in the case when the n x n matrix A(t) satisfies the following conditions:(1) entries of A(t) are continuous on the entire real line R,(2) entries of A(t) are periodic in t of a (positive) period w, i.e.,

(IV.4.2) A(t + w) = A(t) for t E R.

Look at the unique n x n fundamental matrix solution 4'(t) defined by the initial-value problem

(IV.4.3)dY = A(t)Y, Y(0) = In.it

Since 4i '(t + w) = A(t + w)4S(t + w) = A(t)4'(t + w) and +(0 + w) = 4t(w), thematrix 41(t +w) is also a fundamental matrix solution of (IV.4.3). As mentioned in(1) of Remark IV-2-7, there exists a constant matrix r such that 44(t +w) = 4i(t)rand, consequently, r = 4?(w). Thus,

(IV.4.4) 41(t + w) = 4(t)t(w) for t E R.

Setting B = w-' log(4?(w)J (cf. Example IV-3-6), define an n x n matrix P(t) by

(IV.4.5) P(t) = d'(t) exp(-tBJ.

Then, P(t + w) = ((t + w) exp(-(t + w)BI = 4i(t)0(w) exp(-wB) exp(-tBJ =4'(t) exp(-tBJ = P(t). This shows that P(t) is periodic in t of period w. Thus, weproved the following theorem.

Theorem IV-4-1 (G. Floquet [Fl]). Under assumptions (1) and (2), the funda-mental matrix solution 4i(t) of (IV.4.1) defined by the initial-value problem (IV.4.3)has the form

(IV.4.6) 4'(t) = P(t) exp[tB],where P(t) and B are n x n matrices such that(a) P(t) is invertible, continuous, and periodic of period w in t,(Q) B is a constant matrix such that 4'(w) = exp[wB].

Observation IV-4-2. As was explained in Observation IV-3-8, letting 4'(w) _S + N = S(I + M) be the S-N decomposition of 4(w), we define log($(w)) bylog(4?(w)) = log(S) + log(1 + M), where

log(S) = log(A1)P1(4'(w)) + log(a2)P2(4'(w)) + - - + log(Ak)Pk('6(w))

andn- 1 l)m+l

log(!! + M) = E (-m

Mm.M=1

In the case when A(t) is a real matrix, the unique solution 4 (t) of problem (IV.4.3)is also a real matrix. Therefore, the entries of 4'(w) are real. Since S and N are realmatrices, the matrix M = S-'N is real. Therefore, log(I + M) is also real. Letus look at log(S) more closely. If \j is a complex eigenvalue of 4'(w), its complexconjugate A3 is also an eigenvalue of 4'(w). In this case, set A,+1 = A3. It is easyto see that the projection Pj+1(4?(w)) is also the complex conjugate of P,(4'(w)).However, if some eigenvalues of 4'(w) are negative, log[S] is not real. To see thismore clearly, rewrite log[s] in the form

log[S] _ E log[A,]Pi(4?(w)) + E log1A3]P'(4'(w))-A, <0 other j

The matrix log(A,]P,(4'(w)) is a real matrix, while the matrixother j

Elog[aj]Pis

not real. Therefore, log[S] is not real. To rectify this sit-A, <0

uation, let us look at S2. By virtue of the relations given in Lemma IV-1-9, weobtain

S2 =(,\J)2P+ _y2P

A, <0 other j

Notice that

log(S2] _ E log[(Ai)2]Pj(4'(w)) + 2 E log[a3]P,(4(w))af<0 other 7

is a real matrix. Therefore, we can find a real matrix

log[4'(w)2] = log[S2J + 2log(In + MI.

Now, observe that 4'(,)2 =' (2w) and 4'(t + 2w) = 4'(t). (2w). Thus, setting

(IV.4.7) B = _L log{4'(w)2] and P(t) = 4'(t) exp[-tB],

we obtain the following theorem.

4. SYSTEMS WITH PERIODIC COEFFICIENTS 89

Theorem IV-4-3. In the case when the matrix A(t) is real, the matrix 4s(t) hasthe form

P(t) exp[tB],

where P(t) and B are n x n real matrices such that(a) P(t) is invertible, continuous, and periodic of period 2w in t,(p) B is a constant matrix such that 4?(w)2 = exp[2wB].

Remark IV-4-4. In the case when 4?(t) = P(t) exp[tB], weA(t)P(t) - P(t)B. Therefore, (IV.4.1) is changed to

(IV.4.8)dz-- = Bzdt

havedP(t)

dt

by the transformation y" = P(t)!. As noted above, P(t) is periodic in t of periodw (or, respectively, 2w when A(t) is real), the behavior of solutions of (IV.4.1) isderived from the behavior of solutions of (IV.4.8).

Definition IV-4-5. The eigenvalues p1.... , p of the matrix B are called thecharacteristic exponents of equation (IV.4.1). The eigenvalues pi = exp[wpjl, ... ,p = of the matrix 4?(w) (or, respectively, when A(t) is real, the eigen-values pi = exp(2wpi],... , P = exp[2wµn] of 4?(w)2) are called the multipliers ofequation (IV.4.1).

The following basic result is a corollary of Theorem IV-3-1.

Corollary IV-4-6. All solutions of equation (IV.4.1) tend to 0' as t -. +oo if andonly if Ipj I < 1 for all j (or R(pj) < 0 for all j).

Observation IV-4-7. Let be an eigenvector of 45(w) associated with a multiplierp. Then, the solution d(t) = t(t)p" of (IV.4.1) satisfies the condition ¢(t + w) =pi(t). The terminology multiplier came from this fact. (In the case when A(t) isreal, choosing an eigenvector p" of 4;(w)2 associated with a multiplier p, we obtain(t + 2w) = pi(t) if d(t) = 4i(t)p.)Example IV-4-8. Consider the initial-value problem

(IV.4.9)

where

dY-it-

= A(t)Y, Y(0) = 12,

A(t) =

The solution of (IV.4.9) is given by

cost 1

0 cost]

r

fo

41(t) = exp I A(s)dsexp(sin t) I tt ]l L

(cf. (3) of Remark IV-2-7). Therefore,

$(27r) = I0

211 1

B = 2-. log[4i(2a)].


Since p.(\) = (a - 1) 2, Pt(4i) = 12, we obtain

S = 12i M = [ 020

J, and log[4s(27r)j = log[I2 + M) = M = I 0

2,r0

This gives B= 1 M= I 0 01. Therefore,

P(t) = i(t) exp [0 Ot, = 4i(t) l

It I = exp(sin t)I2

and fi(t) = P(t) exp[tBj = exp(sin t) exp I 0 t ]. Moreover, the characteristic

exponents are 0, 0, and the multipliers of (IV.4.9) are 1, 1. Note that in this case,as both of two multipliers are positive, it is not necessary to take 4i(2a)2 and period41r to find B.

IV-5. Linear Harniltonian systems with periodic coefficients

This section is divided into two parts. In Part 1, we explain the structure ofsolutions of a linear Hamiltonian system. The basic facts are found in the theoryof real symplectic groups and their Lie algebra. If we denote by J the real (2n) x (2n)

matrix I 0 O 1, where In is the n x n identity matrix, the Lie algebra of the

real symplectic group Sp(2n, R) of order 2n is the set of all real 2n x 2n matricesof the form JH, where H is a real (2n) x (2n) symmetric matrix. A linear system

T = A(t)f on y' E R2" is a linear Hamiltonian system, if there exists a real

(2n) x (2n) symmetric matrix H(t) such that A(t) = JH(t). Consequently, thefundamental matrix 0(t) of this system belongs to Sp(2n,R) if 0(0) = I. Weexplain all these elementary facts of Hamiltonian systems in Part 1. We also provethat if G E Sp(2n, R) and G = S+N = S(I2"+M) is the S-N decomposition of G,then S and 12" + M belong to Sp(2n, R). In Part 2, we refine the Floquet theorem(cf. Theorem IV-4-1) for the linear periodic Hamiltonian systems in such a waythat the periodic transformation y = P(t)i given in Remark IV-4-4 is canonical.This means that P(t) belongs to Sp(2n, R). The important consequence is that the

linear systemdx

= BE of Remark IV-4-4 is a Hamiltonian system.Part 1

Consider a linear Hamiltonian systemdp

= b-KH(t, y')

dq= -!2c (t, y-), where

dt 8q" dt 8Pand j are unknown vectors in R", y = [q1, ?{(t, y7 = 2y"TH(t)y with a real

on 8H

ON Bpi an Ni(2n) x (2n) symmetric matrix H(t), and Here,

8p 8q.

5. LINEAR HAMILTONIAN SYSTEMS 91

yT is the transpose of the column vector Y. Since

8 (t, y-) = [0, In[ H(t)y, (t, y-) = [1n, O[ H(t)y,

the Hamiltonian system can be written in the form

(IV.5.1) = JH(t)y,

where y is a unknown vector in R2n, H(t) is a real (2n) x (2n) symmetric matrix,and

(IV.5.2) J = O In[ -In O

The matrix J has the following important properties:

(IV.5.3) JT = -J, J-1 = -J,where JT is the transpose of J.

Let fi(t) be the unique real (2n) x (2n) matrix such that

do(t) = JH(t))(t) 4(0) = 12n.,

Lemma IV-5-1. The matrix 4(t) satisfies the condition 4(t)T J4(t) = J, wheret(t)T is the transpose of t(t).

Proof

Differentiate O(t)TJ4(t) to derive

a I(t)TH(t)(-J)J4(t) +

I(t)T H(t)41(t) - O.

Then, 1(t)TJ4(t) = 4(0)T Jo(0) = J. 0The converse of Lemma IV-5-1 is shown in the following lemma.

Lemma IV-5-2. Assume that a real (2n) x (2n) matrix G(t) satisfies the followingconditions:

(i) the entries of G(t) are continuously differentiable in t on an interval 1,(ii) G(t)T JG(t) = J for t E Z.

Then, G(t) is invertible and H(t) _ -Jd d(t)G(t)-I is symmetric for t E I.

Proof.

Computing the determinant of both sides of condition (ii), we obtain [det G(t)12= 1. Therefore, G(t) is invertible. To show that H(t) is symmetric, differentiatingboth sides of G(t)T JG(t) = J, we obtain

[dG(t)]TJG(t) + G(t)T J fi(t) = O.


Since [G(t)-']T = [G(t)T]-1, we further obtain

[G(t)-']T [dG(t)]T(- J) = JdG(t )G(t)_1.

Observing that the left-hand side of this relation is the transpose of the right-handside, we conclude that H(t) is symmetric. 0Observation 1. If GT JG = J, then (G-' )T JG-' = J and GJGT = J. In fact,GT JG = J implies that (G-' )T JG-' = J. Then, taking the inverse of both sidesof the first relation, we obtain the second relation (cf. (IV.5.3)).

Definition IV-5-3. A (2n) x (2n) real invertible matrix G such that GT JG = J iscalled a real symplectic matrix of order 2n. The set of all real symplectic matricesof order 2n is denoted by Sp(2n, R).

It is easy to show that Sp(2n, R) is a subgroup of GL(2n, R) (cf. Observation 1).Hence, Sp(2n, R) is called the real symplectic group of order 2n.Observation 2. If we change Hamiltonian system (IV.5.1) by a linear transfor-mation

(rV.5.1) becomes

dt - {P(t)-1JH(t)P(t) - P(t)-1 dt)J i.

Furthermore, if P(t) E Sp(2n, R) for t E Z, this system becomes

[p(t)TH(t)P(t)(IV.5.5) d J - JdP( 'P(t) i.

Since P-(t)P(t) = 12 and P(t)T E Sp(2n,R), we obtain

IdP(t) dP(t) dlz

P(t)_1dt + di P(t) = d = 0,

1 P(t)-'J = JP(t)T.

tObserve that -J

dP(t)P(t) is symmetric since P(t) E Sp(2n, R) (cf. Lemma

IV-5-2). This implies that (IV.5.5) is a Hamiltonian system.

Definition IV-5-4. Transformation (IV.5.4) is called a canonical transformationif P(t) E Sp(2n, R).

Observation 3. For a given constant matrix G in Sp(2n, R), let us construct theS-N decomposition G = S + N, where S is diagonalizable, N is nilpotent, andSN = NS. Since G is invertible, S is invertible (cf. Remark IV-1-13). If we setM = S-' N, then M is also nilpotent, and we derive the unique decomposition

(IV.5.6) G = S(12 + M), SM = MS.


The identity GTJG = J implies JGJ-' = (GT)-l. Using (IV.5.6), we can writethis last relation in the form

(JSJ-') (J(I2n + M)J-1) = (ST)-1 (I2, + MT)-'.

Set Si = JSJ-1, M1 = J(I2n + M)J-' - 12,,, S2 = (ST)-', and M2 =(12n + MT) -1 - 12n. It is not difficult to show that S. = 1, 2) are diagonal-izable and that Af. (j = 1,2) are nilpotent. Furthermore, S,llf, = MSS, (j = 1, 2).Hence, the uniqueness of S-N decomposition implies that S1 = S2 and 12n + M1 =12n + M2. Therefore,

(IV.5.7) JSJ-' = (ST)-', J(I2n + Af)J-' = (12n + AfT)-1.

Thus, we proved the following lemma.

Lemma IV-5-5. If we write a symplectic matrix G of order 2n in form (IV.5.6)by using the unique S-N decomposition of G (where Af = S- IN), the two matricesS and 12n + M are also symplectic matrices of order 2n.

Remark IV-5-6. Let J be the 2n x 2n matrix defined by (111.5.2) and let H be a2n x 2n symmetric matrix. Then, it is important to know that if A is an eigenvalueof JH with multiplicity m, then -A is also an eigenvalue of JH with multiplicitym. In fact, (JH)T = -HJ = HJ-'. Hence, J-'(JH)TJ = -JH, where ATdenotes the transpose of A.

Remark IV-5-7. Let G be a 2n x 2n symplectic matrix. Then, it is importantto know also that if A is an eigenvalue of G with multiplicity m. then is also an

eigenvalue of G with multiplicity m. In fact, JGJ-' = (CT)-'.

Remark IV-5-8. Assuming that the set of 2n x 2n symplectic matrices with 2ndistinct eigenvalues is dense in the symplectic group Sp(2n, R), it can be shownthat if I (and/or -1) is an eigenvalue of a symplectic matrix G, its multiplicity is

even. Also, det G = 1. In fact, slim, 1. It is also known that det A is equal

to the product of all eigenvalues of A.

Remark IV-5-9. Assuming that Sp(2n, R) is a connected set, we can show thatdet G = 1 for all G E Sp(2n, R). In fact, (det G)2 = 1 and the 2n x 2n identitymatrix is in Sp(2n, R).Part 2

Now, assume that the real symmetric matrix H(t) of system (IV.5.1) is periodicin t of a positive period w, i.e., H(t + w) = H(t). Then, applying Theorem IV-4-3and Remark IV-4-4 to system (IV.5.1), we conclude that the unique (2n) x (2n)

matrix solution 0(t) of the initial-value problem d dtt) = JH(t)4y(t), 4i(O) _12n can be written in the form 0(t) = P(t) exp[tB], where P(t) and B are real(2n) x (2n) matrices such that P(t) is invertible, P(t+2w) = P(t) for all t E R, andB is a constant matrix such that 4i(w)2 = exp[2wB]. Furthermore, system (IV.5.1)is changed to

dzF(IV.5.8)

dt= Bi

94 W. GENERAL THEORY OF LINEAR SYSTEMS

by the transformation

(IV.5.9) y = P(t)f.

The main concern of Part 2 is to show that a matrix P(t) can be chosen sothat the transformation (IV.5.9) is canonical. Note that Y = exp[tB] is the unique

solution of the initial-value problem = BY, Y(O) = I2,,. Therefore, if JB issymmetric, the matrix exp[tB) is symplectic. Hence, if JB is symmetric, the matrixP(t) = 4i(t) exp[-tB] is symplectic, since 4i(t) is symplectic. This implies that inorder to show that the transformation (IV.5.9) is canonical, it suffices to show thatwe can choose P(t) so that JB is symmetric.

Hereafter we use notations and results given in §IV-4. For examples,

(1V.5.10)

k

I2. = E P,('t(w)), 0 (j # t),,=1

Pj(4'(w))2 = P

Taking the transpose of (IV.5.10), we obtain

(IV.5.11)

k

l2n = yP,(.6(w))T, 0 U 36t),,=1

\P(.t(w))T )2 = Pj(.0(W))T.

Note that pAr(A) = PA(A) for any square matrix A. Also,

log[4?(w)2] = log[S2] + 2Iog[I2r, + MI,

log[S2] _ log[(,\,)2]Pi(I(w)) + 2 E log[Aj]Pj('(w)),ai <0 other ,

1 m+l(-) Mm,1og[I2n + M] = Em=1

and

B = log[4i(w)2].

In order to prove that JB is symmetric, it suffices to prove that J log[S2] andJ log[I2i + M] are symmetric. To do this, let us first prove the following lemma.

Lemma IV-5-10. For the symplectic matrix 4i(w), we have

(IV.5.12) Pj(0(w)T)JPe(0(w)) = 0 if A,Ae 34 1,

where Aj (j = 1, 2,... , k) are distinct eigenvalues of 4i(w).


Proof.

Upon applying (IV.5.6) to G = 0(w), write 4'(w) in the form -O(w) = S(I2n+M).Then, S is symplectic, i.e., ST JS = J = 12,,JI2n (cf. Lemma IV-5-5). Using theformulas

k k

(IV.5.13)

12. = > Pt(4 (w)), 12n = Pt=1=1k k

S = ST = EA,P,(,p(w))T,r=1 J=1

rewrite the identity ST JS = J = I21 JI2,, in the following form:

k k

(IV.5.14) A31 Pi(qD(w))TJPt(.D(w))

,,t=1 ,.t=1

Fixing a pair of indices j and t and multiplying both sides of (IV.5.14) by Pj (4ti(w))Tfrom the left and by Pt(4i(w)) from the right, we derive

A, AtP,(4 (w))T JPt(4 (w)) = P3(4'(w))TJPt(4 (w))

Hence, (IV.5.12) follows. 0

Observation 4. Let us prove that J log[S2] is symmetric. To do this, by using(IV.5.12) and (IV.5.13), we write J log[S2] and Iog[(S2)T](-J) in the followingform:

I2nJ log[S2] = log[(At)2]Pi(.O(w))TJPA,ae=1

log[(S2)T](-J)I2n = log[(A,)21A,A =1

Now, choose 1og[(A, )2] for j = 1, 2,... , k in such a way that

log[(At)2] = - log[(A,)2] if A,A, = 1.

Then,

1og[(S2)Tj(-J) _ log[(A1)2]P,(.t(w))TJPt(+(w)) = J log[S2].

Note that for any square matrix A, we have [log(A)IT = log(AT). Therefore,[J log(S2)]T = - log[(S2)T]J. Hence, J log[S2] is symmetric.Observation 5. Let us prove that J log[I2 + M) is symmetric. To do this, wewrite first [12n + M]TJ[I2n + M] = J in the form J[I2n + M] = (12n + MT)-'J.

Then, write [12n + MT J -1 in the form [12,, + MT J-1 = 12. + M. Using (IV.5.7), itis not difficult to show that M is nilpotent and JM = MJ. Hence,

JMm = (Kf)mJ for m= 1,2,....

Therefore,

00(IV.5.15)

J=

(-lmm+l MM _ (-lmm+lkm

/J

m=1 m=1

and

J 1og[I2,, + MI = (log(12n + 1GIJ) J

= (log [(I2 + MT)-1]) J = - (log [12" + Mn)]J.Thus,

[J log(I2 + M)]T = log(I2i + 111T)(-J) = J log(I2n + M).

This proves that J log[I2,, + M] is symmetric. Here, use was made of the fact that

if1 _+X =

1 + y(x), then

log I1 + xJ = log[1 + y(x)J = - log(1 + x]

as a power series in x. L

Thus, we arrived at the following conclusion.

Theorem IV-5-11. In the case when the matrix H(t) of system (IV.5.1) is real,symmetric, and periodic in t of a positive period w, there exist (2n) x (2n) realmatrices P(t) and B such that

(a) P(t) E Sp(2n, R) for all t E R,(b) P(t+2w)=P(t) for alit ER,(c) B is a constant matrix such that JB is symmetric,

(d) the canonical transformation y" = P(t): changes system (IV.5. 1) to L = BE.

For more information of exponentials in algebraic matrix groups, see (Mar] and(Si4].

IV-6. Nonhomogeneous equations

In this section, we explain how to solve an initial-value problem

(IV.6.1) di = A(t)9 + b'(t), y(r) = ij,

where the entries of the n x n matrix A(t) and the C"-valued function b(t) arecontinuous on an interval Z = {t : a < t < b} and r E Z, if E C". As it was

6. NONHOMOGENEOUS EQUATIONS 97

explained in §IV-2, let the n x n matrix 4>(t; r) be the unique fundamental matrixsolution of the homogeneous system

(IV.6.2)

dY = A(t)Y,Y(r) = In,

dt

determined by the initial-value problem

(IV.6.3)

where r E I. System (IV.6.2) is called the associated homogeneous equation of(IV.6.1). We treat the solution of (IV.6.1) by using the knowledge of the funda-mental matrix solution 4)(t; r) of (IV.6.2) (cf. §§IV-2 and IV-3).

Consider the transformation

(IV.6.4) y' = 4)(t; r)z.

Then, the problem (IV.6.1) is changed to

(IV.6.5)dz _ r)z(r) = 4.dt

dy= A(t)y

dt

In fact, differentiating both sides of (IV.6.4), we obtain

A(t)4)(t; r)l + b(t) = r)z""+ 4 (t; r)di*

.

Since the unique solution of problem (IV.6.5) is given by

z= i + jt(s;rY1(s)ds,t

(TV.6.6)

y" _ (t;r)Iit + j4(s;T)_1(s)ds]l r_

(t; r)r1 +J

t $(t; s)b(s)ds.r

Here, use was made of the fact that 0(t; s) = 4'(t; r)O(s; r)-1 (cf. (2) of RemarkIV-2-7; in particular (IV.2.6)).

In a similar way, using (IV.6.6), we can change a nonlinear initial-value problem

d = A(t)17 + 9(t y), y(r) = i

to an integral equation

(IV.6.7) y = 4 (t; 7-)17 +J

'(t; s)g(s,7(s))ds.tr

The function g(t, y-) is the nonlinear term which satisfies some suitable condition(s).In the case when the matrix A is independent of t, formulas (N.6.6) and (IV.6.7)

become

(IV.6.8) g" = exp[(t - r)A]it + / exp[(t - s)A]b(s)dstr

and

exp[(t - s)A]9(s,y1s))ds,(IV.6.9) exp[(t - r)A]>; +Jr

respectively.Example IV-6-1. Let us solve the initial-value problem

(IV.6.10) dt = Ay + 6(t), y(0)

where-2 1 0 [2] 1

A= 0 -2 0 , b(t)= 0 ,

#p=1

3 2 1 t 0

The matrix A was given in Example IV-3-3. As computed in §IV-3, a fundamentalmatrix solution of the associated homogeneous equation is

e-2t to-21 00) = exp(tA) = 0 e-2t 0

et - e-2t et - (1 + t)e-2t et

Therefore, the solution of (IV.6.10) is

r

fo

t 1 1 + to-2texp[tA] j i0 + exp[-sA]b(s)ds } e-2t

111 111 -9-t+5et-(l+t)e

IV-7. Higher-order scalar equations

In this section, we explain how to solve the initial-value problem of an n-th orderlinear ordinary differential equation

(IV.7.1)ao(t)ucn) + al(t)u(n-1) + ... + an-1(t)u + an(t)u = b(t),

u(r) = 'h, u'(r) = Q2, ... ,U(n-1)(,r)

=

where the coefficients a°(t), a1(t), ... , and the nonhomogeneous term b(t) arecontinuous on an interval I = {t : a < t < b}. In order to reduce (IV.7.1) to

7. HIGHER-ORDER SCALAR EQUATIONS 99

a system, setting yj = u and 112 = u', ...I yn = u(n-1), we introduce a vector

ylI. Then, problem (IV.7.1) is equivalent to

tin

(IV.7.2)

where

A(t) =

b(t) =

d11 = A(t)f + b(t),dt

0 1 0 0 ... 0 1

0 0 1 0 ... 0

an(t) an-1(t)an0

-3(t) _a 2(t) al(t)ao(t) ao(t) do(t) ao(t) ao(t)

0

0

b(t)

0t)

ao(

Assuming ao(t) 54 0 on Z, let 4?(t) = [ 1(t) 2(t) ... hn(t)J be a fundamentalmatrix solution of the associated homogeneous equation

(IV.7.3)d

= A(t)yat

of (IV.7.2). Using 4i(t), we can solve problem (IV.7.2) (cf. §IV-6). The firstcomponent of the solution of (IV.7.2) is the solution of problem (IV.7.1). Also,the first components of n column vectors of the matrix $(t) give the n linearlyindependent solutions of the associated homogeneous equation

(IV.7.4) ao(t)u(n) + al (t)u(n-1) +... + an(t)u = 0

of (IV.7.1).Example IV-7-1. Let us solve the initial-value problem

(IV.7.5) u"' - 2u" - 5u' + 6u = 3t, u(0) = 1, u'(0) = 2, u"(0) = 0.

To start with, set

111

111=u, 112=u', y2=u", and y= 1122

IY31


Then, problem (IV.7.5) is equivalent to

(IV-7.6)

where

(IV.7.7)

dy" = Ay" + b'(t),dt y(0) = #I,

10 1 0 0 1

A = 0 0 1, b(t) = 0 I, and [2].0-6 5 2 3t

Three eigenvalues of the matrix A are 1, -2, and 3. The corresponding projectionsare

-6P1(A) _ -6 -6

-6

4D(s;0)-1 6(s) ds

1 1 - (t + 1)e-t

1-2 1 1

P3 (A)= 10

-6 3 3-18 9 9

and N = 0. Therefore,

4)(t; 0) = exp(tAJ = etP1(A) + e-2tP2(A) + e3tP3(A)

et -6 -1=-6 -6 -1

-6 -1

Thus,

t

1-(t+1)e-`1 - (t + 1)e-t

and, consequently,

1

0

-1 1 1 3 -4 1

-1 1 I , P2(A) -6 8 -2-1 1

1512 -16 4

1 _2t 3 -4 1 3t -2 1 1

1 e -6 8 -2 + e -6 3 3.1

1512 -16 4

10 -18 9 9

1 1 + (2t - 1)e2t 1 1 - (3t + 1)e '3t+- -2(1+(2t-1)e2tJ +- 3(1-(3t+1)e-39 ,

204[1 + (2t - 1)e2t] 911 - (3t + 1)e-3t1

1 [1_(t+1)e_tl 1 1 + (2t - 1)e2t- 1 - (t + 1)e' + - -2(1 + (2t - 1)e2CJ

1 - (t + 1)e-t 20 4[1 + (2t - 1)e2L]

1 1 - (3t + 1)e-3t 30t + 25 - 17e'2t + 50et + 2e3L+ 3[1 - (3t + 1)e-3t] = 2(15 + 17e-2t + 25et + 3e3t)-

9(1 - (3t + 1)e-3t] 601 2(-34e-2t + 25et + 9e3t)

The first component of g(t) gives the solution of (IV.7.5).

7. HIGHER-ORDER SCALAR EQUATIONS 101

Remark IV-7-2. In the case of a second-order linear differential equation

(IV.7.8) ao(t)u" + al (t)u' + a2(t)u = b(t),

,,a fundamental matrix solution of (IV.7.3) has the form 4P(t) = 0' (t) ¢2(t).01(t) 02(t)

where 01 (t) and ¢2 (t) are two linearly independent solutions of the associated ho-

mogeneous equation (IV.7.4). Since 4i(t)-1 =1 .02(t) -02(t) where

W(t) -011(t) 01(t)W(t) = det(4i(t)), the first component of the formula

ty"(t) _ (t}it + -t(t)J

4>(s)-16(s)ds

gives the general solution of (IV.7.8) in the form

u(t) = r)1451(t) + 77202(t)(IV.7.9) 42(s) t 01(s)ao(s)W(s)b(s)ds + 02(t) r ao(s)W(s)b{s)ds,- 41(t) It

where ill and q2 are two arbitrary constants. This is known as the formula ofvariation of parameters (see, for example, [Rab, pp. 241-2461). Moreover,

(IV.7.10) W(t) =W(r)exp f dsdo (s) J

as in (4) of Remark IV-2-7.

Remark N-7-3. Write (IV.7.10) in the form

(IV.7.11) 01(t)02(t) - dl (t)t2(t) = c2 exp [ ftt s)ds

,L

-ao(s) I

where C2 is a constant. Regarding (IV.7.11) as a differential equation for 02(t), weobtain

1 al(s)02(t) = C101(t) +C201(t)

J t 01(T)2 [ - J t ao(s)ds }dry

fwhere cris anoth1er constant. Therefore, 1(t) and ¢1(t) 1(r)2

x exp [ - J t _(s)ds] d7- form a fundamental set of solutions of ao(t)u"+al(t)u'+s

a2(t)u = 0.

Remark IV-7-4. Let 01 (t) and 02(t) be linearly independent solutions of a third-order linear homogeneous differential equation

(IV.7.12) ao(t)u"' + a1(t)u" + a2(t)u' + a3(t)u = 0.

102 IV. GENERAL THEORY OF LINEAR. SYSTEMS

Also, let 0(t) be any solution of (IV.7.12). Then, (4) of Remark IV-2-7 implies that

(IV.7.13)0 01 0210' 401

452

'P 4i oz=c3exp

[ftao(s)ds]

where c3 is a constant. Write (IV.7.13) in the form

(IV.7.14) Au(t)o" + A1(t)O' + A2(t)4 = c3exp c al(s)ds ,[-I ao(s) ]

where

461 02 01

A1(t) = 101 ¢2 4 A2(t) - 1,iCOI 0'202

As(t)'02" ' 40 oz

A fundamental set of solutions of the homogeneous part of (IV.7.14) is given by{01,02}. Therefore, using (IV.7.9), we obtain

eXp I -J

T ao(s) I dT0(t) = C101 (t) + c202(t) + C3 I-01(t)I 02Ao(T3

+ 02 (t) I ` Ao(r)2 e L

- J ao(s)ds]dr}

where cl and c2 are constants. This implies that 01(t), 02(t), and

02(r) i) 1(r) ) ds] drao(s)dsj

dr - -02(t) I Ao(r)2 exp [ - f ao(s)Aa(T)2exp

[f'

form a fundamental set of solutions of (IV.7.12).

EXERCISES IV

IV-1. Let A bean n x n matrix, and let A1, A2, ... , Ak be all the distinct eigenvaluesof A. On the complex A-plane, consider k small closed disks Of = {A : IA - A2 I <r)} (j = 1,... , k}. Assume that Aj n At = 0 for j # 1. Set

f P] 2'rf dail=rj(AI,, -

A)-1dA (j = 1,... ,k),

and N=A-S,

where the integrals are taken in the counterclockwise orientation. Show that(i) Pl + ... + Pk = In,

(ii) Pi Pt = O if U'41),(iii) A = S + N is the S-N decomposition of A.

EXERCISES IV 103

Hint. Note that

r-o-1 -1-11 -1 .(rln - A)} = (oln - A){(u1n - A) - (7-In - A)

Also, if p(A) is a polynomial in A with constant coefficients, then

k

p(A) = 1 j p(A)(Aln - A)'1dA.J=12rri

For general information, see [Ka, pp. 38-43).

IV-2. Under the same assumptions as in Exercise IV-1, show that if f (A) is analyticin a neighborhood of each eigenvalue AJ, and an n x n matrix B is in a sufficientlysmall neighborhood of A, then f (B) is well defined and Biim f (B) = f (A).

Hint. If f (A) is analytic in a neighborhood of each eigenvalue A,, then

f (A) _ 1J=12rri

IV-3. Let A = S + N be the S-N decomposition of an n x n matrix A. Show that,as functions of A, the matrices S and N are not continuous.

Hint. Use Lemma IV-1-4.

IV-4. Let A be an n x n matrix with n eigenvalues Al, ... , An. Fix a real numberr satisfying the condition r > R[AJ] (j = 1,... , n). Show that

(F) exp[tA] =2T

j+00io)In - A)-'do

00

for t E R.

Remark. This result means that exp[tA] is the inverse Laplace transform of (sln-A) 1. For example, if A = [ U1

], then (s12 - A) -1 =s2 + 1 [_i 1 ]. Hence,

taking the inverse Laplace transform, we obtain exp(tA) = cos t sin tI-sint cost

I.

For-

mula (F) is useful in the study of exp[tA] of an unbounded operator A defined ona Banach space (cf. [HUP, Chapter XII, pp. 356-386]).

J(A2IV-5. Show that +T21" )-'dr = A-1 arctan(tA-1) if every eigenvalue of

an n x n matrix A is a nonzero real number.m

IV-6. For an n x n matrix A, show that lim (In + A) = eA.In +00 M


IV-7. Find the solution of the following initial-value problem

2 - 2A + A21r = etAe b7(0) = n, V(0) = S,

where A is a constant n x n matrix, {6, ,j,S are three constant vectors in C", andy E C' is the unknown quantity.

Hint. If we set y = e`AU, the given problem is changed to

Al =E, u"(0) =r1', u(0)dt2

IV-8. Find the solution of the following initial-value problem

2 +A21 =D, 9(0) = 40 , !Ly (0) = 41

where E C" is an unknown quantity, A is a constant n x n matrix such thatdet A = 0, and {rjo, >7j } are two constant vectors in C".

Hint. Note that cos(xA) and sin(xA) are not linearly independent when det A = 0.

If we define F(u) =sin u

then the general solution of the given differential equationu

isj(x) = cos(xA)c"1 + xF(xA)c2.

IV-9. Find explicitly a fundamental matrix solution of the system dy = A(t)yif there exists a constant matrix P E GL(n,C) such that P-"A(t)P is in Jordancanonical form, i.e.,

P-1A(t)P = diag[A, (t)I1 + N1iA2(t)12 + N2,... ,.1k(t)Ik +Nkj,

where for each j, Ap(t) is a C-valued continuous function on the interval a S t S 6,I1 is the n, x n j identity matrix, and N, is an nl x n, matrix whose entries are 1on the superdiagonal and zero everywhere else.

Hint. See [GH].

/r 3 0 -1j).aniiIV-10. Find log (I 11 ]).cos([ 1 2 -1

\` ` 0 1 1

252 498 4134 698 1).

arctan -234 -465 -3885 -65615 30 252 42

-10 -20 -166 -25IV-11. In the case when two invertible matrices A and B commute, find

log(AB) - log(A) - log B

if these three logarithms are defined as in Example IV-3-6.

EXERCISES IV 105

IV-12. Given that

0 0 1 0 yi

0 0 0 1 I/2A =

2 3 0 0 ' 1/3

4 -2 0 0 1/4

find a nonzero constant vector ii E C4 in such a way that the solution l(t) of theinitial-value problem

d9A9, 9(0)

satisfies the condition lim y"(t) = 0.t -+oo

IV-13. Assume that A(t), B(t), and F(t) are n x n, m x m, and n x m matrices,respectively, whose entries are continuous on the interval a < t < b. Let 4i(t) be

an n x n fundamental matrix of - = A(t)4 and W(t) be an in x in fundamental

matrix of Z = B(t)u. Show that the general solution of the differential equationon an ii x m matrix Y

(E)

is given by

dY = A(t)Y - YB(t) F(t),_WT

Y(t) = 4;(t) C 4,(t)-1 r`

where C is an arbitrary constant n x m matrix.

IV-14. Given the n x n linear system

dY = A(t)Y - YB(t),dt

where A(t) and B(t) are n x n matrices continuous in the interval a < t < b,(1) show that, if Y(to) -I exists at some point to in the interval a < t < b, then

Y(t)-i exists for all points of the interval a < t < b,(2) show that Z = Y-1 satisfies the differential equation

dZ = B(t)Z - ZA(t).dt

IV-15. Find the multipliers of the periodic system dt = A(t)f, where A(t) _[cost simt-sint cost

Hint. A(t) = exp It 101 0]]IV-16. Let A(t) and

l`B(t) be n x n and n x m matrices whose entries are real-

valued and continuous on the interval 0 < t < +oo. Denote by U the set of allR'-valued measurable functions u(t) such that ,u(t)j < 1 for 0 < t < +oo. Fix avector { E R". Denote by ¢(t, u) the unique R"-valued function which satisfies the

initial-value problemat

= A(t)i + B(t)u(t), i(0) where u E U. Also, set

R = {(t, ¢(t, u)) : 0:5 t < +oo, u E R}. Show that R is closed in Rn+i.

Hint. See ILM2, Theorem 1 of Chapter 2 on pp. 69-72 and Lemma 3A of Appendixof Chapter 2 on pp. 161-163j.

IV-17. Let u(t) be a real-valued, continuous, and periodic of period w > 0 in t onR. Also, for every real r, let ¢(t, r) and t1'(t, r) be two solutions of the differentialequation

(Eq)

such that

d2y - u(t) y = 0dt2

o(r, r) = 1, 0'(r, r) = 0, and 0(7-,T) = 0, t;J'(T,T) = 1,

where the prime denotes d Set

C(r) = 10'(0.'r)0(0, r) V(0, r) 1,i,I (0,r) J

M(r) _ 0(r+w,r) Vl(r+w,r)0(r+w,r) tG'(r+w,r)

(I) Show thatM(r) = C(r)'1M(O)C(r).

(11) Let A+ and A_ be two eigenvalues of the matrix M(r). Let I K+(r)J

K

and

be eigenvectors of M(r) corresponding to A+ and A_`, respectively._(r)how that

(i) K±(r) are two periodic solutions of period w of the differential equation

dK ± K2 + u(r) = 0,dT

(ii) if we set

Qf{t,r) = exp [J t Kt(() d(lr

these two functions satisfy the differential equation (Eq) and the conditions

13f(t+w,r) = A*O*(t,r).

EXERCISES IV 107

Hint for (I). Set 4'(t,r) _ (tmatrix solution of the system

Then, ((t, r) is a fundamental(t' )JJJ

I

-

and y = 4'(t, r)c is the solution satisfying the initial condition y(r) = c'. Note thatC(r) = 4'(0, r) and M(r) = 4'(r + w, r). Therefore, 4'(t, r) = 4'(t, 0)C(r) and4'(t + r) = 4'(t, r)M(r). Now, it is not difficult to see 4'(w, r) = 4'(w, 0)C(r) _4'(0, r)M(r) = C(r)M(r) and 4'(w, 0) = M(O).

Hint for (II). Since eigenvalues of M(r) and M(0) are the same, the eigenvalues ofM(r) is independent of r. Solutions 77+ (t) of (Eq) satisfying the condition Y7+ (t +

w) = A+n+(t) are linearly dependent on each other. Hence, W+(t) = K+(t) is

independent of any particular choice of such solutions q+(t). In particular K+(t +W) = 17'+ (t + w)

= K+(t). Problem (II) claims that the quantity K+(r) can be17+(t + w)

found by calculating eigenvectors of M(r) corresponding to A+. Furthermore,

A+ = exp I o K+(r)drJJ I I. The same remark applies to A_. The solutions 0±(x, r)o

of (Eq) are called the Block solutions.

IV-18. Show that(a) A real 2 x 2 matrix A is symplectic (i.e., A E Sp(2, R)) if and only if det A = 1;

(b) the matrixl

010 J

N is symmetric for any 2 x 2 real nilpotent matrix N.

Hint for (b). Note that det(etNJ = I for any real 2 x 2 nilpotent matrix N.

IV-19. Let G, H, and J are real (21n ) x (2n) matrices such that G is symplectic,

H is symmetric, and J = I 0n . Show that G-1JHGJ is symmetric.

IV-20. Suppose that the (2n) x (2n) matrix 4'(t) is the unique solution of the

initial-value problem L'P = JH(t)4', 4'(0) = 12n, where J = I 0n n and H(t)

l T Jis symmetric. Set L(t) = 4'(t)-1JH(t)4'(t)J. Show that d4'dt) = JL(t)4'(t)T,where 4'(t)T is the transpose of 4'(t).

CHAPTER V

SINGULARITIES OF THE FIRST KIND

In this chapter, we consider a system of differential equations

(E) xfy = AX, Y-),

assuming that the entries of the C"-valued function f are convergent power seriesin complex variables (x, y-) E Cn+I with coefficients in C, where x is a complex inde-pendent variable and y E C" is an unknown quantity. The main tool is calculationwith power series in x. In §I-4, using successive approximations, we constructedpower series solutions. However, generally speaking, in order to construct a power

00

series solution y"(x) = E xmd n, this expression is inserted into the given differ-M=1

ential equation to find relationships among the coefficients d, , and the coefficientsd,,, are calculated by using these relationships. In this stage of the calculation, wedo not pay any attention to the convergence of the series. This process leads usto the concept of formal power series solutions (cf. §V-1). Having found a formalpower series solution, we estimate Ia',,, i to test its convergence. As the functionx-I f (x, y) is not analytic at x = 0, Theorem 1-4-1 does not apply to system (E).Furthermore, the existence of formal power series solutions of (E) is not alwaysguaranteed. Nevertheless, it is known that if a formal power series solution of (E)exists, then the series is always convergent. This basic result is explained in §V-2(cf. [CL, Theorem 3.1, pp. 117-119) and [Wasl, Theorem 5.3, pp. 22-251 for thecase of linear differential equations]). In §V-3, we define the S-N decompositionfor a lower block-triangular matrix of infinite order. Using such a matrix, we canrepresent a linear differential operator

(LDO) G[Yj = x + f2(x)lj,

where f2(x) is an n x n matrix whose entries are formal power series in x withcoefficients in Cn. In this way, we derive the S-N decomposition of L in §V-4 and anormal form of L in §V-5 (cf. [HKS]). The S-N decomposition of L was originallydefined in [GerL]. In §V-6, we calculate the normal form of a given operator C byusing a method due to M. Hukuhara (cf. [Sill, §3.9, pp. 85-891). We explain theclassification of singularities of homogeneous linear differential equations in §V-7.Some basic results concerning linear differential equations given in this chapter arealso found in [CL, Chapter 4].

108

1. FORMAL SOLUTIONS OF AN ALGEBRAIC DE 109

V-1. Formal solutions of an algebraic differential equation

We denote by C[[x]] the set of all formal power series in x with coefficients inw 00

C. For two formal power series f = E amxm and g = b,,,x', we definem=0 m=0

f = g by the condition an = b", for all m > 0. Also, the sum f + g and the00 00 m

product f g are defined by f +g = (am +b,,,)xm and f g = (F am-nbn )x"',m=0 m=0 n=0

00

respectively. Furthermore, for c E C and f = >2 a,,,xm E C[[x]], we define cfm=0

00

by cf = >camx'. With these three operations, C[[x]] is a eommutattve algebram=0

over C with the identity element given by E bmxm, where bo = 1 and b,,, = 0m=0

if m > 1. (For commutative algebra, see, for example, [AM].) Also, we define the

derivative of f with respect to r by dx _ E(m+ 1)a,,,+,x' and the integralm=0

X r "0 00r

( f (x)dx by f f (x)dx = airi x"' for f = E a,,,xm E C[[xj]. Then, C[[x]]0 o m=1 m=0

is a commutative differential algebra over C with the identity element. There aresome subalgebras of C[[x]j that are useful in applications. For example, denote byC{x) the set of all power series in C[]x]] that have nonzero radii of convergence.Also, denote by C[x] the set of all polynomials in x with coefficients in C . Then,C{x} is a subalgebra of C[[x]] and C]x] is a subalgebra of C{x}. Consequently, C[x]is also a subalgebra of C[[x]].

Let F(x, yo, y, , ... , yn) be a polynomial in yo, y,..... yn with coefficients inC[[x]]. Then, a differential equation

(V.1.1) F y,,d"y =0dy,...dxn

is called an algebrutc differential equation, where y is the unknown quantity and x

is the independent variable. If F ` x, f, ... , f = 0 for some f in C[[x]], thendxn

such an f is called a formal solution of equation (V.1.1). In this definition, it is notnecessary to assume that the coefficients of F are in C{x}.

Example V-1-1. To find a formal solution of

(V.1.2) xLY +y-x=0,set y = Famx'. Then, it follows from (V.1.2) that ao = 0, 2a, = 1, and

m>0

(m + 1)am = 0 ( m > 2 ). Hence, y = 2x is a formal solution of equation (V.1.2).

110 V. SINGULARITIES OF THE FIRST KIND

In general, if f E C{x} is a formal solution of (V.1.1), then the sum of f as aconvergent series is an actual solution of (V.1.1) if all coefficients of the polynomialFareinC{x}.

Denote by x°C([x]j the set of formal series x°f (x), where f (x) E C([x]], ais a complex number, and x° = exp[a logx]. For 0 = x°f E x°C[[x]], 0 =x°g E x°Cj[x]j, and h E C(jx]], define 0 + iP = x(f + g), hO = x°hf, and

xL = ox°f + x°x . Then, x°C([xfl is a commutative differential module over

the algebra C((x]]. Similarly, let x°C{x} denote the set of convergent series x°f (x),where f (x) E C{x}. The set x°C{x} is a commutative differential module overthe algebra C{x}. Furthermore, if a is a non-negative integer, then x°C([x]] CC[[x]]. If F(x, yo,... , y,) is a formal power series in (x, yo,... , yn) and if fo(x) ExC[[x]j, ... , fn(x) E xC([xj], then F(x, fo(x),... , fn(x)) E C[[xJj. Also, ifF(x, yo,. .. , yn) is a convergent power series in (x, yo,... , yn) and if fa(x) E xC{x},... , fn(x) E xC{x}, then F(x, fo(x), ... , f,,(x)) E C{x}. Therefore, using the no-tation x°C[[x]]n to denote the set of all vectors with n entries in x°C[(x]], we can de-

fine a formal solution fi(x) E xC[[x]]n of system (E) by the condition x = Ax, i)in C[[x]j . (Similarly, we define x°C{x}n.) Also, in the case of a homogeneous sys-

tem of linear differential equations xfy A(x)g, a series d(x) E x°C[[xj]n is said

to be a formal solution if x = A(x)e in x°C[[x]jn. Now, let us prove the followingtheorem.

Theorem V-1-2. Suppose that A(x) is an n x n matrix whose entries are formalpower series in x and that A is an eigenvalue of A(O). Assume also that A + k arenot eigenvalues of A(O) for all positive integers k. Then, the differential equation

(V.1.3) x!Ly = A(x)g

has a nontrivial formal solution fi(x) = xa f (x) E x"C[[x]]n

Proof.00

Insert g = xa E x'nam into (V.1.3). Setting A(x) _ y2 xmAm, where An Em=0 m=0

Mn(C) and Ao = A(O), we obtain

00 Co

xa (A + m)xm= xa LtAm_i.Therefore, in order to construct a formal solution, the coefficients am must bedetermined by the equations

m-h,1ao = Aoa"o and (A + m)am = Aodd,n + E Am-hah (m > 1 ).

h=0

1. FORMAL SOLUTIONS OF AN ALGEBRAIC DE 111

Hence, ado must be an eigenvector of Ao associated with the eigenvalue A, whereasm-h

ii',,, = ((A + m)1. - A0)-1 Am-hah form? 1. 0h

For an eigenvalue Ao of A(O), let h be the maximum integer such that Ao + his also an eigenvalue of A(O). Then, Theorem V-1-2 applies to A = A0 + h. Theconvergence of the formal solution ¢(x) of Theorem V-1-2 will be proved in §V-2,assuming that the entries of the matrix A(x) are in C(x) (cf. Remark V-2-9).

n

Let P = Eah (x)&h (b = x d /be a differential operator with the coefficients

h=0ah(x) in C([x]]. Assume that n > 1 and an(x) 54 0. Set

ab

P[x'] = x' fm(s) xm,m=np

where the coefficients f,,, (s) are polynomials in s and no is a non-negative integersuch that 0. Then, we can prove the following theorem.

Theorem V-1-3.(i) The degree of f,,,, in s is not greater than n.

(ii) If zeros of fno do not differ by integers, then, for each zero r of f,,, thereexists a formal series x' (x) E xr+IC[[x]] such that P[xr(1 +O(x))] = 0.

Proof

(i) Since 6h(x'] = shx', it is evident that the degree of f,,,, in s is not greater thann.

+o0

(ii) For a formal power series d(x) = E ckxk, we havek=1

P[x'(I + b(x))] = P(x'] +

xa+no AK(+ + E Ckfno+m-k(s + k)) x"'J

.

M=1 k=1

+oo

L. Ck P[x'+k J

k=1+oo +oo

(frn(s+oo

= x' (s)xm + > Ckxk + k)xm/{mnof" k =1 m=no }

+oo m( -no= x' fno(s)x + (,fm(s) + Ckfm-k(s + k)) x"`m=no+1

k=1 J1

In order that y = x''(1 +b(x)) be a formal solution of P[y] = 0, it is necessary andsufficient that the coefficients cm and r satisfy the equations

fno (r) = 0,m

fno+m (r) + E Ckfno+m-k(r + k) = 0 (m > 1).k=1

It is assumed that f,,,, (r + m) 54 0 for nonzero integer m if r is a zero ofTherefore, r and c,,, are determined by

1m-1

(r) = 0 and cm = -7,(r + m) ckfn+m-k(r + k) (m > 1).

Convergence of the formal solution x'(1+¢(x)) of Theorem V-1-3 will be provedat the end of §V-7, assuming that the degree of in s is n and the coefficientsah(x) of the operator P belong to C{x}. The polynomial f.,, (s) is called the indicialpolynomial of the operator P.

Remark V-1-4. Formal solutions of algebraic differential equation (V.1.1) as de-fined above are not necessarily convergent, even if all coefficients of the polynomial

00

F are in C{x}. For example, y = Y (-1)m(m!)xm+i is a formal solution of them=0

differential equation 2d2y

2+ y - x = 0. Also, the formal solution x' (I + fi(x)) of

Theorem V- 1-3 is not necessarily convergent if the degree of f,,. (s) in s is less than

n. In order that f = >a,,,xm be convergent, it is necessary and sufficient thatm=0

la,,, I < KA' for all non-negative integers m, where K and A are non-negative num-00

bers. Also, it is known that some power series such as (m!)mxm do not satisfy'n--Rany algebraic differential equation. The following result gives a reasonable neces-

sary condition that a power series be a formal solution of an algebraic differentialequation.

Theorem V-1-5 (E. Maillet [Mail). Let F(x, yo, yi, ... , yn) be a nonzero polyno-

mial in yo, yi, ... , y, with coefficients in C{x}, and let f = E amxm E CI[x]] bem=0

a formal solution of the differential equation F(x, y, Ly, ... , dny / = 0. Then,dxn

there exist non-negative numbers K, p, and A such that

(V.1.4) I am I < K(m!)PA' (m > 0).

We shall return to this result later in §XIII-8.

Remark V-1-6. In various applications, including some problems in analytic num-ber theory, sharp estimates of lower and upper bounds of coefficients am of a formal

solution f = F, axm of an algebraic differential equations are very important.00

m =9For those results, details are found, for example, in [Mah], [Pop], [SS1], (SS2], and[SS3]. The book [GerT] contains many informations concerning upper estimates ofcoefficients lam I.

2. CONVERGENCE OF FORMAL SOLUTIONS 113

V-2. Convergence of formal solutions of a system of the first kind

In this section, we prove convergence of formal solutions of a system of differentialequations

(V.2.1) Xfy = AX, Y-),

where y E C' is an unknown quantity and the entries of the C"-valued function fare convergent power series in (x, y7) with coefficients in C. A formal power series

00

(V.2.2) E x' , ,,, E xC([xfl" (c,,, E C" )rn-t

is a formal solution of system (V.2.1) if

(V.2.3) X dx= f(x,0)

as a formal power series.To achieve our main goal, we need some preparations.

Observation V-2-1. In order that a formal power series (V.2.2) satisfy condition(V.2.3), it is necessary that f (O, 0) = 0. Therefore, write f in the form

f (x, y-) = o(2-) + A(x)g + F, y~'fv(x),Iel>2

where(1) p = (pl,... ,p") and the p, are non-negative integers,(2) jpj = pi + + pn and yam' = yi' yn^, where yi,...are the entries

of g,

(3) fo E xC{x}" and f, E C{x}",(4) A(x) is an n x n matrix with the entries in C{x}.

Note that

fo(x) = Ax, 6) and A(x) _ Lf(x,0).

Setting

A(x) = F, x'A"s Ao = 0,0) ILZM=0

where the coefficients A", are in M"(C), write condition (V.2.3) in the form

xd'o = Ao + f (x, ¢) - A0 .dx

Then,

(V.2.4) "t,, = A0E + rym for m = 1, 2, ... ,


where

f (x, 0) - AoO = o(x) + IA(x) - Ao1 fi(x) +(J(x))p

fp(x)IpI>2

00

1: xm7mM=1

and Im E C n. Note that ry'm is determined when c1 i ... , c,,,_ 1 are determined andthat the matrices mI" - A0 are invertible if positive integers m are sufficientlylarge. This implies that there exists a positive integers mo such that if 61, ... , C,,,a

are determined, then 4. is uniquely determined for all integers m greater than mo.Therefore, the system of a finite number of equations

(V.2.5) mEm = A0 + (m = 1,2,... ,mo)

decides whether a formal solution ¢(x) exists. If system (V.2.5) has a solution{c1 i ... , c,,,a }, those mo constants vectors determine a formal solution (x) uniquely.

Observation V-2-2. Supposing that formal power series (V.2.2) is a formal solu-N

tion of (V.2.1), set ¢N (x) _ x'6,. Sincem=1

Ax, (x)) - f (x, N(x)) = A(x)(d(x) - N(x)) + F, [d(x)p - N(x)D] ff(x),Ipl>2

it follows that AX, $(x)) - f (X, dN(x)) E xN+C[Ix]]". Also,

xd¢N(x) = xdd(x) - xdoN(x) E xN+1C((x((n.d

Hence, AX, 4V W)

dx x dx

- xdON(x) E xN+1C{x}". Setdx

(V.2.6) 9N,0(x) = zddN(x)dx

Now, by means of the transformation y" = z1+ y5N(x), change system (V.2.1) to thesystem

(V.2.7)

on i E C", where

dzxdx = JN(x,z

9N(x, l = f (x, z + dN(x)) -xdjN(x)

dx

= 9N,O(x) + f(x,Z+dN(x))f r- f(x,dN(x))

= 9N,0(x) + A(x)z" + L. [(+ $N(x))" - N(x)'] f,,(x)InI?2


As in Observation V-2-1, write gN in the form

9N(x, = 9N,O(x) + BN(x)z + L xr 9N,p(x),Ip1?2

where(1) 9N,o E xN+1C{x}" and g""N,p E C{x}",(2) BN(x) is an n x n matrix with the entries in C{x},(3) the entries of the matrix BN(X) - Ao are contained in xC{x}.

Observation V-2-3. System (V.2.7) has a formal solution00

(V.2.8) ON(x) = O(x) - ON(x) _ x'"cn, E xN+1C1jx11 n.m=N+1

The coefficients cm are determined recursively by

me,, = Aocm + '7m for m = N + 1, N + 2, ... ,

where

9N(x,ILN(x)) - AO1GN(x) = f(x,$(x)) - Ao1 N(x) - dx

= f (x, fi(x)) - AOcN(x) - xdVx)

= 9N,O(x) + [BN(x) - Ao1 i'N(x) + E (ZGN(x))p 9N,p(x)Ip1>2

ou

= E xm-'1'm

m=N+1

Note that the matrices min - Ao are invertible for m = N + 1, N + 2, ... , if N issufficiently large.

Observation V-2-4. Suppose that system (V.2.1) has an actual solution q(x)such that the entries of i(x) are analytic at x = 0 and that ij(0) = 0. Then, the

Taylor expansion ¢(x) _ j dim(0) of q(x) at x = 0 is a formal solution of

(V.2.1). Furthermore, is convergent and E xC{x}".Keeping these observations in mind, let us prove the following theorem.

Theorem V-2-5. Suppose that fo(x) = f (x, 0) E xN+1C{x}n and that the ma-

trices mIn - A0 (m > N + 1) are invertible, where Ao = (0, 0). Then, system69

(V.2.1) has a unique formal solution

(V.2.9) t(x) =00

E X'n4n E x^'+1C((x}]n.m=N+1

Furthermore, ¢ E xN+1C{x}n.

Remark V-2-6. Under the assumptions of Theorem V-2-5, system (V.2.1) pos-sibly has many formal solutions in xC[[x)J". However, Theorem V-2-5 states thatthere is only one formal solution in xN+1C[[xJJn

Proof of Theorem V-2-5.

We prove this theorem in six steps.

Step 1. Using the argument of Observation V-2-1, we can prove the existence anduniqueness of formal solution (V.2.9). In fact,

f (x, 4) - Ao¢ = o(x) + JA(x) - Ao}fi(x) + E (_)P_)IpI?2

00

xmrym.m=N+1

This implies that rym = 0 for m = 1, 2, ... , N. Hence, (m > N + 1) areuniquely determined by

mc,,, = Ao6,,, + rym for m=N+1,N+2,....

Step 2. Suppose that system (V.2.1) has an actual solution q(x) satisfying thefollowing conditions:

(i) the entries of rl(x) are analytic at x = 0,(ii) there exist two positive numbers K and 6 such that

jy(x)I < KIxjN+1 for IxI < 6.00

xm(0) of #(x) at x = 0 is a formal solutionThen, the Taylor expansion E ml a.,.7m'1

m=N+1of (V.2.1) (cf. Observation V-2-4). Since such a formal solution is unique, it follows

xm d1ijthat $(x) = E -(0). Because the Taylor expansion of if(x) at x = 0 is

m=N+1convergent, the formal solution ¢ is convergent and E xN+1C{x}".

Step 3. Hereafter, we shall construct an actual solution f(x) of (V.2.1) that sat-isfies conditions (i) and (ii) of Step 2. To do this, first notice that there exist threepositive numbers H, b, and p such that

(I)

and

If(x,o)I 5 HIxIN+1 for IxI < 5

(II) If(x,91) - f(x,y2)I 5 (IAoI + 1) 1111 - y2Ifor Ix) < b and Iy, I < p (j = 1, 2).

Hence,

(III) If(x, y-)i <_ HIxl "+1 + (IAoI + 1)Iyl for Ixl < b and Iyl <_ p.

Using the transformation of Observation V-2-2, N can be made as large as wewant without changing the matrix A0. Hence, assume without loss of any generalitythat

(V.2.10)IAoi + 1 1

N+1 < 2Also, fix two positive numbers K and b so that

(V.2.11) K > H + NAol i1]K and K6N+1 < p

Step 4. Change system (V.2.1) to an integral equation

(V.2.12)J

:f n+(f ))

Define successive approximations

)b(x) = 0 and 1?k+1(x) = IZ for k=0,1,2,....

Now, we shall show that

lnk(x)I < KIxIN+1 for jxl <b and k=0,1,2....Since this is true for k = 0, we show this recursively with respect to k as follows.First if this is true for k, then

I i k(x)1 < Kjxl N+l < K&N+I < p for jxj < b.Hence,

I'.F(c nk (o) < {H + IIAol + 1jK}j£VN for 6.

Therefore,

II7k+1(x)I <H + IAoI

1

1]K IxIN+I < K 1xIN+1 for jxj < S.

Step 5. Set

Then, since

we obtain

nkll = maxiInk+iIZI ++1k(x)I InI < b}.I1'Rk+1 --

I ik+1(x) - ik(x)I = fo

IInk+1 - llkIIj

N ++1 171k - '1k-11I 2lI'7k - 1k-l II

This implies that

lim ilk(x) _ #e+1(x) - #((X)k-.+00 xN+1 xN+1

e=0

exists uniformly for Ixl < J.

Step 6. Setting

#(x) = xN+1 ("liM xJ`'+1) = ku>nik(x),

it is easy to show that n'(x) satisfies integral equation (V.2.12). It is also evidentthat i7(x) is analytic for IxI < 6. Thus, the proof of Theorem V-2-5 is completed. 0

Now, finally, by using the argument given in Observations V-2-2 and V-2-3, weobtain the following theorem.

Theorem V-2-7. Every formal solution E xC[[zjJ" of system (V.2.1) is conver-gent, i.e., E xC{x}".

Remark V-2-8. In general, (V.2.1) may not have any formal solutions. However,Theorem V-2-7 states that if (V.2.1) has formal solutions, then every formal solutionis convergent.

Remark V-2-9. If yi(x) = xA f (x) E XAC[[xfj" is a formal solution of a linear

system X!LY = A(x)y, the formal power series f is a formal solution of xd _[A(x) - A!, }i. Therefore, f E C{x}" by virtue of Theorem V-2-7. This provesconvergence of the formal solution constructed in Theorem V-1-2.

V-3. The S-N decomposition of a matrix of infinite orderIn §V-4, we shall define the S-N decomposition of a linear differential operator.

As a linear differential operator will be represented by a lower block-triangularmatrix of infinite order, we derive, in this section, the S-N decomposition of sucha matrix

All 0 0 ... ... ...A21 A22 0 ... ... ...

Amt Am2 ... Am,n 0 . .

where for each (j, k), the quantity AJk is an n, x nk constant matrix. Set

Al = All,

Am =

AllA21

Am l

0A22

O .. 0O .. 0

Am2 ... ... Am,

(m > 2).

Form > 2, we write Am - Am-! 0 Set Nm = m nt. Then, A. is an[ B. Amm J l-1

N", x Nm matrix, while Bm is an nm x N,"_ 1 matrix. Also, let Amm = Smm +Nmm

3. THE S-N DECOMPOSITION OF A MATRIX OF INFINITE ORDER 119

and Am = Sm + Aim be the S-N decompositions of Amm and Am, respectively,where Smm is an nm x nm diagonalizable matrix, Sm is an Nm x Nm diagonalizablematrix, Nmm is an nm x nm nilpotent matrix, Aim is an N. x Nn nilpotent matrix,SmmNmm = NmmSmm, and SmAm = NmSm. The following lemma shows howthe two matrices Sm and Aim look.

Lemma V-3-1. The matrices Sm and Aim have the following forms:

JS1 = s11, S"'

=rI Sm_1 0 l

(m > 2),L Cm Smm

N, = Ni1, Aim =[Aim_i

O 1

(m > 2),fm Nmm

where Cm and fm are am x N,_1 matrices.

Proof

Consider the case m > 2. Since the matrices Sm and Aim are polynomials in Amwith constant coefficients, it follows that

Sm = [B m1ILM0

and Aim = Ifm1 01,

where 13m-1 and Dm-1 are Nm-1 x Nm_1 matrices, Cm and fm are nm x Nm_1matrices, and µm and vm are nm x nm matrices. Furthermore, Dm_1 and Vm arenilpotent. Also, Bm-1Dm-1 = Dm-1Bm-i and tlmvm = 1.m µm. Hence, it sufficesto show that Bm-1 and µm are diagonalizable.

Note that since Sm is diagonalizable, Sm has Nm linearly independent eigen

vectors. An eigenvector of Sm has one of two forms kP J

and [?.] , where p is an

eigenvector of Bm_1i whereas q is an eigenvector of µm. Therefore, if we countthose independent eigenvectors, it can be shown that Bm-1 has N,, -I linearly in-dependent eigenvectors, while Pm has nm linearly independent eigenvectors. Notethat Nm = Nm_ 1 + nm. This completes the proof of Lemma V-3-1. 0

Lemma V-3-1 implies that the matrices Sm and Nn have the following forms:

S1 = SI1,S11 0 0 ... 0C21 S22 0 ... 0

Sm = (m>2)

Cmi Cm2 ... ... Smm

andN1 = N11,

Nil 0 D ... 0Y21 JN22 0 0

Aim =

Fm 1 Fm2 ... ... Nmm

(m>2),


where C,k and Fjk are n3 x nk matrices.Set

S11 0 0C21 S22 0

S =

Cml Cm2 "' Smm 0

fN11 0 0f21 N22 0 "'

N=1m 1 Fm2 ... .Wmm 0

Then,

(V.3.1) A=S+N and SN=NS.

We call (V.3.1) the S-N decomposition of the matrix A.

V-4. The S-N decomposition of a differential operator

Consider a differential operator

(V.4.1) C(yl = x + f2(x)Y,

wwhere f2(x) _ > xt1 and 1 E Let us identify a formal power series

t=oao

xt dt with the vector pa1

= a2 E C. Then, the operator C is representedt>o

by the matrix

flo 0 0 0 ...f2l In + f2o 0 0 ...f12 l1 21, + f2o 0 ...

A =

11m f4.-l fZn-2 ... f21 mI, + fZa 0 .

5. A NORMAL FORM OF A DIFFERENTIAL OPERATOR 121

where I is the n x n identity matrix. Let A = S + N be the S-N decomposition

11

of A. Since A =L

j) A1, where (i = R2 and I,> is the oo x oo identityL

matrix, the matrices S and N have the forms

So 0 0 O ... ...

al I. + So 0 0 ... ...

a2 al 21n + So 0 ... ...

(V.4.2) S =am

and

(V.4.3) N =

am-l am-2

No O OV1 No 0V2 V1 No

at min + SO 0

Vm Vm-1 Vm-2 ... VI No 0 ...

respectively, where the a. and v, are n x n matrices and S2o = So +No is the S-Ndecomposition of the matrix S2o.

00 00

Set a(x) = So + E xtat and v(x) = No + trot. Then, S represents a

t_1 tvl

differential operator Lo[y1 = x + a(x)y, while N represents multiplication byv(x) (i.e., the operator: y" -+ v(x)y-). Since A = S + N and SN = NS, it followsthat

(V.4.4) L[gf = Lo[Yj + v(x)y" and Lo[v(x)y1 = v(x)Lo[yl-

We call (V.4.4) the S-N decomposition of operator (V.4.1). We shall show in thenext section that Lo[yj is diagonalizable and v(x)n = O.

V-5. A normal form of a differential operatorLet us again consider the differential operator

(V.5.1) L[yj = Xf + Q(x)U,

00

where f2(x) = Ex1fle and Sgt E WC). In §V-4, we derived the S-N decomposi-t=o

tion

(V.5.2) L[y-] = Lo[yl + v(x)y" and Co[v(x)yl = v(x)Lo[yj.

122

where

(V.5.3)

and

V. SINGULARITIES OF THE FIRST KIND

Co[YI = x + a(x)y

00

Co

a(x) = So + xtat and v(x) = No + xtvt

t=1 t_1

(cf. (V.4.4)). Notice that N = So + No is the S-N decomposition of S2o andthat the operator Go[ul and multiplication by v(x) are represented respectively bymatrices (V.4.2) and (V.4.3).

Set

Sm =

0 0 0I. +,% 0 0

al 21 + So 0

am_ 1 am-2

Since So is diagonalizable, there exist an invertible n x n matrix P0 and a diagonalmatrix Ao such that

SoPo = PoAo, NO = diag[al, A2, ... , . \n)-

Note that A, (j = 1, 2,... , n) are eigenvalues of So. Hence, A. (j = 1, 2,... , n) arealso eigenvalues of flo. For every positive integer m, Sm is diagonalizable. Hence,

Po

there exists an (m + 1)n x n matrix P.. =Pi

such that SmPm = PmA0. If

Pmm is sufficiently large, we can further determine matrices Pt for t > m + 1 by theequations

t(V.5.4) (tl + S0)Pt + E ahP1_h = PtAo.

h=1

Equation (V.5.4) can be solved with respect to Pt, since the linear operator Pt -+(tl + So)P1 - PtAo is invertible if m is sufficiently large. In this way, we can find

an oo x n matrix P = I such that SP = PAa. Set P(x) = >x'P,. Then,

AP1

Pm t=D

entries of P(x)-1 are also formal power series in x and

(V.5.5) Co[P(x)j = P(x)Ao.


Define two differential operators and an n x n matrix by

X [U1 = P(x)-iC(P(x)uZ, /Co(u1 = P(x)-'Co(P(x)Uj,{ vo(x) = P(x)-'v(x)P(x),

respectively. Then 1C(u'j = X 0(u'] + vo(x)u'. Observe that

.- - _. _.. _, f _. , l du' \ _ . _. 1

(V.5.6)du=x +Aou".

This shows that the operator Co(y1 is diagonalizable. Furthermore,

dvo

(V.5.7) x dxx) + Aovo(x) = P(x)-'Lo(P(x)vo(x)1 = P(x)`Co[v(x)P(x)1

= P(x)-'v(x)Lo(P(x)1 = vo(x)P(x)-'Co(P(x)1 = vo(x)Ao

This shows that the entry i,k(x) on the j-th row and k-th column of the matrixvo(x) must have the form v,k(x) = 7jkxAk-a', where ryjk is a constant. Since Lt(x)is a formal power series in x, it follows that

(V.5.8) vjk(x) = 0 if Ak - )j is not a non-negative integer.

Observe further that the matrix &. (x) can be written in the form

vo(x) = x-^°rx^°, where r ='711 'Yin

'7ni Inn

Hence, for any non-negative integer p, we have vo(x)P = x-A0rPxAo, where

z^O = exp ((log x)Ao] = diag(za' , Z112'... , za^ I.

On the other hand, since No is nilpotent, vo(x)P can be written in a form i (x)P =X'pQP(x), where mp is a non-negative integer such that lim mP = +oo andP+oothe entries of the matrix Qp(x) are power series in x with constant coefficients.Therefore, L0(x)P = 0 if p is sufficiently large. This implies that the matrix r isnilpotent. Since v(x) = P(x)vo(x)P(x)-1, we obtain v(x)^ = O.

Thus, we arrive at the following conclusion.

Theorem V-5-1. For a given differential operator (V.5.1), let 0o = So +No bethe S-N decomposition of the matrix N. Then, there exists an n x n matrix P(x)such that

(1) the entries of P(x) are formal power series in x with constant coefficients,(2) P(O) is invertible and SoP(0) = P(0)Ao, where A0 is a diagonal matrix whose

diagonal entries are eigenvalues of Q0r


(5) the transformation

(V.5.9) P(x)u

changes the differential operator (V.5.1) to another differential operator

(V.5.10) P(x)-1G(P(x)ui = KOK + vo(x)u,

where

and

dilKo(61 = x + Aou, vo(x) = x-^Orx^0,

r=

with constants "ljk such that

(V.5.11) Vj k = 0 if

Ynl Inn

Ak - Aj is not a non-negative integer.

Furthermore, the matrix r as nilpotent.

Remark V-5-2. It is easily verified that the matrix P(x) is a formal solution ofthe system

xd) = P(x)(Ao + vo(x)) - Sl(x)P(x).

Since the entries of vo(x) are polynomials in x, the power series P(x) is convergentif f2(x) is convergent (cf. Theorem V-2-7). Therefore, in such a case, v(x) isconvergent and, hence, o(x) is convergent.

Observation V-5-3. Choose integers l1, ... , to so that A.,+f j = Ak +fk if Aj -Akis an integer. Then, vo(x) = x4rx-L, where L = diag(e1, f2, ... , en]. If we set7t{V1 = x'! 1C(xf and ho{v") = x-LICo(x' i , it follows that

?{o(vl = x'L fXLTf + zLLv + Aoz' 7}

Hence,

(V.5.12)

Note that

dv'= xdx

R{v = x + (Ao + L + r)v.

+ (Ao + L)v

Ao + L = diag(A1 + P1, A2 + e2, ... , An + enj

and

(V.5.13) (Ao + L)r = r(A0 + L).

It was already shower in §V-4 that r is nilpotent. Thus, the following theorem isproved.


Theorem V-5-4. The transformation y = P(x)xLiT changes the system

(V.5.14) G[yi = xdy + fl(x)yl = 0

to

d1U(V.5.15) VV! = x + (Ao + L + r)u = 0.

Observation V-5-5. The matrixn-1 h

o{x) = x-Ao-L-r = x-Ao-L I + (-1)h[logx[ rh

h-1 h!

is a fundamental matrix solution of system (V.5.15). Note that if (A3+tj)-(Ak+tk)is an integer, then A, + ej = Ak + tk in Ao + L.

Remark V-5-6. A fundamental matrix solution 4D(x) of system (V.5.14) is givenby 4>(x) = P(x)x4o(x), which can be written in the form

n-1 (lo x h45(x) = P(x)xLx-M-L In F+ {-1)h h! rhIL h=1

Since L - A0 - L = -A0, the matrix 44(x) can be also written in the form

n-1 h

4i(x) = P(x)x-ne In + (-1)h (lohx) rh

h=1

n-i hHowever, x-Ao and

II nr >(-1)h [lohx[ rh do not commute, whereas x-A°-L

h=1n-I h

and In + E(-1)h[log x[

rh commute.h=1

h! IRemark V-5-7. The methods used in §§V-3, V-4, and V-5 are based on the

original idea given in [GerL[.

Example V-5-8.In order to illustrate the results of this section, consider the differential equation

of the Bessel functions

(V.5.16) za (zT) + (z2 - a2)y = 0,

where a is a non-negative integer. If we change the independent variable z byx = z2, (V.5.16) becomes

dx lxdx

2y=0.


This equation is equivalent to the system

(V.5.17) x + Q(x)y = 0, y =

0 -1 0 -1 0 0

where 11(x) = x - a2 =10+x11i c o =a2 and 0i =

1

40

40

4 0

To begin with, let us remark that the S-N decomposition of the matrix 11 isgiven by S2o = So + No, where

80 =O if a = 0, N if a = 0,

{Q if a > 0, and N0 - to if a>0.a 0

Also, set Ao = 2 a Note that two eigenvalues of Q are

20

Now, we calculate in the following steps:

Step 1. Fix a non-negative integer m so that m > 2 -(_a)

= a.

Step 2. Find three 2(m + 1) x 2(m + 1) matrices A,,,, Sm, and Nm (cf. §§V-4 andV-5).

PoP1

Step 3. Find a 2(m + 1) x 2 matrix Pm = , where the Pr are 2 x 2 matrices

Pmsuch that SmPm = Pn&.

m in

Step 4. Find two 2 x 2 matrices Mm(x) = No + >2xtvt and Qm(x) _ >xtPtt=1 t-o

(cf. §§V-4 and V-5).

Step 5. We must obtain Qm(x)-1Mm(x)Qm(x) = vo(x) +O(xm+i), where

Po InoP0

vo(x) r 0 a(a)xn0 0 J

if a = 0,

if a>0,

where a(a) is a real constant depending on a (cf. Theorem V-5-1).After these calculations have been completed, we come to the following conclu-

sion.


+00

Conclusion V-5-9. There exists a unique 2 x 2 matrix P(x) = >xtPe such that1=o

(1) the matrices Pl for t = 0,1, ... , m are given in Step 8,(2) the power series P(x) converges for every x,(8) the transformation y = P(x)u" changes (V.5.17) to

dii + (Ao + vo(x)) i = 0'.(V.5.18) xaj

We illustrate the scheme given above in the case when a = 0. First we fix m = 2.Then,

0 -1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 -1 0 01 1

1 0 0 04 2

A2 =1 0 0 1 0 0

S2 = 1 10 1 0 0

4 4 4

0 0 0 0 2 -1 3 3 1 12 0

32 32 4 2

0 04

0 0 2 - 1 3 1 10 2

16 32 4 4

and

0 -1 0 0 0 0

0 0 0 0 0 0

1 1

0 -1 0 04 2

N2= 10 0 0 0 0

3 3 1 1

0 -132 32 4 2

16 T2-0

90 0

1192 -320

Calculating eigenvectors of S2, we find a 6 x 2 matrix P2 =

that S2P2 = 0. Note that, in this case, Ao = 0.

64 -128-16 16

such-32 480 1

1 0


Set

and

1 1 3 3

4 2 32 32

0 1 1 3

4 16 32-M2 (x) = 1 o + xvl + x2v2

01(V.5.19)

-320 -16 16 1192PO = [ 64 -18] '

P1

= [ -32 48]

,P2

=[0

'

Q2(r) = P° + xPl + x2P2.

Then, Qz-' 11f2(x)Q2(x) -1 2] + O(x3). Note that

64 64

Thus, we arrived at the following conclusion.

Conclusion V-5-10. There exists a unique 2 x 2 matrix P(x)

that(1)

(2)

(3)

(V.5.20)

the matrices Pt for i = 0, 1, 2 are given by (V.5.19).the power series P(x) conueryes for every x,the transformation y" = P(x)u" changes the system

to

dy+xdx

0 -1

X 0 Y = 0'4

d6

-+ [_1 2, u" = 0.(V.5.21) x3i

00

Zx'Pr suche=o

Furthermore, the transformation y" = P(x)Po iv changes (V.5.20) to

1 5

41,° 12 634

[001]

[19264 -128] =

[--212].

"(V.5.22) x + 1 0 0= 0.


In the case when a=1, wefixm=2. Then

1 0 -1 0 0 0 0

1

4

A2 =

and

132=

0 -1 0 0 0 0

0 0 0 0 0

11 -1 0 0

8 4

3 1 1 1 0 016 8 4

1 1 1 12 -1

16 16 4

3 1 3 1 1 2

64 16 16 8 4

0 0 0 0 0 01

0 0 0 0 0 0

1 10 0 0 0

8 4Ar2 = 1 1

16 8 0 0 0 0

- 1 1 1 1 0 016 16 8 4

-3 1 1 10 0

64 16 16 8

384 32192 -16

Calculating eigenvectors of S2, we find a 6 x 2 matrix P2 = _72 -12 such4

2 0

5 1

01that S2P2 = P2Ao. Note that, in this case, A0 = 2 1

1

. SetL0

and

21 1 1 1

8 4 16 16vl = 1 1 v2 =

3

1

16 8 64 16

M2(x) = xl/1 + x2v2,

Po[19

84 -48 -123

62

J' Pl - [-72 -14] , 1'2 = [2 1j ,(V.5.23)

Q2(x) = Po + xP1 + T2 p2.

0 0 0 0 0

0 0 1 -1 0 0

0 -4 1 0 0

0 0 0 0 2 -1

0 0

4

0 -4 2 J

1


0x

Then, Q2-'M2(X)Q2(X) = 48 + Q(x3). Thus, we arrived at the following0 0

conclusion.+oo

Conclusion V-5-11. There exists a unique 2 x 2 matrix P(x) = >xtPt suche=o

that(1) the matrices Pt for P = 0, 1, 2 are given by (V.5.23),(2) the power series P(x) convet es for every x,(3) the transformation y = P(x)u changes the system

(V.5.24) xdx +

-1

0

V = 0

to

(V.5.25)2 48I u = U.x- +

102

For a further discussion, see IHKS[. A computer might help the reader to calculateS2i N2, and P2 in the cases when a = 0 and a = 1. Such a calculation is not difficultin these cases since eigenvalues of A2 are found easily (cf. §IV-1).

V-6. Calculation of the normal form of a differential operator

In this section, we present another proof of Theorem V-5-1. The main idea is toconstruct a power series P(x) as a formal solution of the system

dP(x)= P(x)(Ao + vo(x)) - Q(x)P(x)

(cf. Remark V-5-2).

Another proof of Theorem V-5-1.

To simplify the presentation, we assume that So = A0 = diag[,ulI,, µ2I2, ... ,Aklk[, where pi, p2, ... , µk are distinct eigenvalues of no with multiplicities m1,respectively, and the matrix I, is m, x m j identity matrix. Since AOA(o = NoAo,the matrix No must have the form No = diag(N01,No2, ... ,Nok[, Where Nor is an

00

mi x m1 nilpotent matrix. Let us determine two matrices P(x) = I,a + E xmP,,,M=1

6. CALCULATION OF THE NORMAL FORM 131

00

and B(x) = Sto + E xmBm by the equationm=1

dxa

This equation is equivalent to

(in+mm) (Oo+ *n=1 xmBm)

Cep + m=1 (In + m 'nPm)

m-1mPm = PmS2o - S20Pm + Bm - 1m + ( PhBm-h - fl.-hPh) (m > 1).

[=1

Therefore, it suffices to solve the equation

(V.6.1) mX + (A0 +N4)X - X(Ao+No) - Y = H,

where X and Y are n x n unknown matrices, whereas the matrix H is given. If wewrite X, Y, and H in the block-form

X11 ... Xlk [Y11 ... Xlk H11 ... Xlk

Xkl Xkk Ykl ... Ykk Hkl ... Hkk

where XXh, YYh, and H,h are m, x mh matrices, equation (V.6.1) becomes

(m + P) - µh)X)h + NO,Xjh - X,hNOh - Yjh = Hjh,

where j, h = 1,... , k. We can determine XJh and Y,h by setting Y,;h = 0 ifm + u, - Ah 4 0, and X,,h = 0 if m + p, - µh = 0. More precisely speaking, ifm + p - Ph 96 0, we determine X3h uniquely by solving

(m + p, - µh)Xlh + NojXjh - X3hNoh = H)h.

If M + Aj - Ah = 0, we set Y,h = H,h. In this way, we can determine P(z) andB(x). In particular, go + B(x) has the form x ^°I'xA0, where r is a constantn x n nilpotent matrix. Furthermore, the operator x d

+A0 and the multiplicationoperator by No + B(x) commute. D

ds

The idea of this proof is due to M. Hukuhara (cf. [Si17, §3.9, pp. 85-891).

Remark V-6-1. In the case of a second-order linear homogeneous differentialequation at a regular singular point x = a, there exists a solution of the form

+00

01(z) = (x - a)aE c, (x - a)n, where the coefficients c,, are constants, co 0 0,n=O

and the power series is convergent. If there is no other linearly independent solu-tion of this form, a second solution can be constructed by using the idea explainedin Remark IV-7-3. This second solution contains a logarithmic term. Similarly,a third-order linear homogeneous differential equation has a solution of the form

+00

01(x) _ (x - a)° c, (x - a)" at a regular singular point x = a. Using this solu-n=0

tion, the given equation can be reduced to a second-order equation. In particular, ifthere exists another solution ¢(x) of this kind such that ¢1 and ¢2 are linearly in-dependent, then the idea given in Remark IV-7-4 can be used to find a fundamentalset of solutions.

In general, a fundamental matrix solution of system (V.5.14) can be constructedif the transformation y" = P(x)xLV of Theorem V-5-4 is found. In fact, if thedefinition of fo(x) of Observation V-5-5 is used, P(x)xL2-no-L-r is a fundamentalmatrix solution of (V.5.14). The matrix P(x) can be calculated by using the methodof Hukuhara, which was explained earlier.

V-7. Classification of singularities of homogeneous linear systems

In this chapter, so far we have studied a system

(V.7.1) x dt = A(x)y, y E C",

where the entries of the n x n matrix A(r) are convergent power series in x withcomplex coefficients. In this case, the singularity at x = 0 is said to be of the firstkind. If a system has the form

(V.7.2) xk+1 = A(x)y", y" E C",

where k is a positive integer and the entries of the n x n matrix A(x) are convergentpower series in x with complex coefficients, then the singularity at x = 0 is said tobe of the second kind

In §V-5, we proved the following theorem (cf. Theorem V-5-4).

Theorem V-7-1. For system (V.7.1), there exist a constant n x n matrix A0 andan n x n invertible matrix P(x) such that

(i) the entries of P(x) and P(x)-1 are analytic and single-valued in a domain0 < jxi < r and have, at worst, a pole at x = 0, where r is a positive number,

(ii) the transformation

(V.7.3) y' = P(x)i7

changes (V.7.1) to a system

(V.7.4) du = Aou".

Theorem V-7-1 can be generalized to system (V.7.2) as follows.

7. CLASSIFICATION OF SINGULARITIES OF LINEAR SYSTEMS 133

Theorem V-7-2. For system (V.7.2), there exist a constant n x n matrix A0 andan n x n invertible matrix P(x) such that

(i) the entries of P(x) and P(x)-1 are analytic and single-valued in a domainV = {x : 0 < lxi < r}, where r is a positive number,


(V.7.5) y = P(x)i

changes (V.7.2) to (V.7.4).

Proof.

Let 4i(x) be a fundamental matrix of (V.7.2) in D. Since A(x) is analytic andsingle-valued in D, fi(x) _ I (xe2"t) is also a fundamental matrix of (V.7.2). There-fore, there exists an invertible constant matrix C such that 46(x) = 4i(x)C (cf. (1)of Remark IV-2-7). Choose a constant matrix Ao so that C. = exp[2rriAo] (cf. Ex-ample IV-3-6) and let P(x) = 4(x)exp[-(logx)Ao]. Then, P(x) and P(x)-1 areanalytic and single-valued in D. Furthermore,

dP(x) = dam) exp(_(logx)Ao] - P(x)(x-'Ao)

= x-(k+1)A(x)P(x) - P(x)(x-'Ao).

This completes the proof of the theorem. 0An important difference between Theorems V-7-1 and V-7-2 is the fact that thematrix P(x) in Theorem V-7-2 possibly has an essential singularity at x = 0.

The proof of Theorem V-7-2 immediately suggests that Theorem V-7-2 can beextended to a system

(V.7.6)dg

dx= F(x)y,

where every entry of the n x n matrix F(x) is analytic and single-valued on thedomain V even if such an entry of F(x) possibly has an essential singularity atx = 0. More precisely speaking, for system (V.7.6), there exist a constant n x nmatrix Ao and an n x n invertible matrix P(x) satisfying conditions (i) and (ii) ofTheorem V-7-2 such that transformation (V.7.5) changes (V.7.6) to (V.7.4).

Now, we state a definition of regular singularity of (V.7.6) at x = 0 as follows.

Definition V-7-3. Let P(x) be a matrix satisfying conditions (i) and (ii) suchthat transformation (V.7.5) changes (V.7.6) to (V.7.4). Then, the singularity of(V.7.6) at x = 0 is said to be regular if every entry of P(x) has, at worst, a poleatx=0.Remark V-7-4. Theorem V-7-1 implies that a singularity of the first kind is aregular singularity. The converse is not true. However, it can be proved easily thata regular singularity is, at worst, a singularity of the second kind. Furthermore, if(V.7.2) has a regular singularity at x = 0, then the matrix A(0) is nilpotent. Thisis a consequence of the following theorem.

Theorem V-7-5. Let A(x) and B(x) be two n x n matrices whose entries areformal power series in x with constant coefficients. Also, let r and s be two positiveintegers. Suppose that there exists an n x n matrix P(x) such that(a) the entries of P(x) are formal power series in x with constant coefficients,(b) det(P(x)] ,E 0 as a formal power series in x,

16 -1

(c) the transformation y = P(x)ii changes the system x' = A(x)y to x' 2i _X

B(x)il.Suppose also that s > r. Then, the matrix B(O) must be nilpotent.

Proof

Step 1. Applying to the matrix P(x) suitable elementary row and column opera-tions successively, we can prove the following lemma.

Lemma V-7-6. There exist two n x n matrices

+oo +oo

T(x) _ xmTm and S(x) _ x"'Sm,m=0 m=o

and n integers 1\1,,\2, ... , an such that(i) the entries of n x n matrices T,,, and Sm are constants.

(ii) det To 0 0 and det So 0 0,(iii)

T(x)P(x)S(x) = A(z) = diag(x''',zA2,...(iv)A1<A2<...<An .

The proof of this lemma is left to the reader as an exercise.

Step 2. Change the two systems x'd = A(x)y and x'd6= B(x)u, respectively,

to xrd = C(x)a and x'd = D(x)ii by the transformations z' = T(x)y andii = S(x)v. Then, F = A(x)v. Furthermore, if the matrix D(0) is nilpotent, thematrix B(0) is also nilpotent.

Step 3. Look at D(x) = x'-'A(x)-1C(x)A(x) - x'A(x)-ld ). This shows

clearly that if s > r, the matrix D(0) is nilpotent.

The following corollary of Theorem V-7-5 is important.

Corollary V-7-7. Assume that conditions (a), (b), and (c) of Theorem V-7-5 aresatisfied. Also, assume that A(O) and B(O) are not nilpotent. Then, r = s.

Definition V-7-8. If a singularity of the second kind is not a regular singularity,this singularity is said to be irregular. In particular, if the matrix A(O) of (V.7.2)is not nilpotent, the singularity of (V.7.2) at x = 0 is said to be irregular of orderk. Also, a regular singularity is said to be of order zero.

In order to define the order of singularity at r = 0 for all systems (V.7.2), theindependent variable x must be replaced by x'1P with a suitable positive integer p.

7. CLASSIFICATION OF SINGULARITIES OF LINEAR SYSTEMS 135

For example, for a differential equation

(V.7.7) ah(x)bhl/ = 0,h=0

assume that the coefficients ah are convergent power series in x and that an (x) # 0.Set y; = 6'-1y (j = 0,... , n - 1). Then, (V.7.7) becomes the system

f 0 1 0 0

(V.7.8) 6y =0 0 0 ... 0

0 0 0 . 1

a0 an-1an as

/

yl

91 i! =yn

If we change (V.7.8) by the transformation y" = diag[l, x-O,then(V.7.9)x°6u

0 1 0 ... 0

0 0 0 0

0 0 0 ... 1

-xna aoa. -x'an-1

an

'.T-(n-1)0 Iii,

+ x°diag[l, a, 2a,... , (n - 1)a) u'.

ah(x)x(n-h)OSet bh(x) =

an(x) . Choose a non-negative rational number a so that bh(X)

(h = 0,... , n - 1) are bounded in a neighborhood of x = 0. Also, if a must bepositive, choose a so that bh(0) 0 0 for some h. Then, the matrix on the right-handside of (V.7.9) is not nilpotent at x = 0 if or > 0. Hence, we define a as the order ofthe singularity of (V.7.7) at x = 0. System (V.7.2) can be reduced to an equation(V.7.7) (cf. §XIII-5).

The following theorem due to J. Moser [Mo) concerns the order of a given sin-gularity-

Theorem V-7-9. Let A(x) = x-a E x'A be an n x n matriz, where p is an=o

integer, A are n x n constant matrices, A0 y6 0, and the power series is convergent.Set

Im(A) = max (P_ 1 + n,0) ,

µ(A) = minlm(T-1AT-T-1dx)J

Here, r = rank(Ao) and min is taken over all n x n invertible matrices T of the

form T(x) = x-9 E x°T,,, where q is an integer, T are constant n x n matrices,

and the power series is convergent. Note that det T(x) 0 0, but det To may be 0.Assume that m(A) > 1. Then, we have m(A) > p(A) if and only if the polynomial

P(A) =

vanishes identically in A, where In is the n x n identity matrix.

The main idea is to find out p(A) in a finite number of steps. Using the quantityp(A), we can calculate the order of the singularity at x = 0. The criterion form(A) > p(A) is given in terms of a finite number of conditions on the coefficientsof A(x). There is a computer program to work with those conditions. In particular,in this way, we can decide, in a finite number of steps, whether a given singularityat x = 0 is regular.

Notice that IP(x)I can be estimated at z = 0 for the matrix P(x) of Theorem V-7-2, using similar estimates for fundamental matrix solutions of (V.7.2) and (V.7.4).Therefore, an analytic criterion that a singularity of second kind at x = 0 be aregular singularity is that in any sectorial domain V with the vertex at x = 0,every solution fi(x) satisfies an estimate

c K[xj' (x E V)

for some positive number K and a real number m. These two numbers may dependon 0 and V. In fact, Theorem V-7-2 and its remark imply that a fundamental matrixsolution of (V.7.6) is P(x)xA0. The matrix P(x) has, at worst, a pole at x = 0 ifand only if !P(x)I < Kjxjr in a neighborhood of x = 0 for some positive numberK and a real number p (cf. [CL, §2 of Chapter 4, pp. 111-1411). Another criterionwhich depends only on a finite number of coefficients of power series expansion ofthe matrix A(x) of (V.7.2) was given in a very concrete form by W. Jurkat and D.A. Lutz [JL] (see also 15117, Chapter V, pp. 115-1411).

Now, let us look into the problem of convergence of the formal solution whichwas constructed in Theorem V-1-3. Let

n

P = Eah(x)Jhh=0

be a differential operator with coefficients ah(x) in C{x). Assume that n > I anda,(x) 0 0. Defining the indicial polynomial of the operator P as in §V-1,we prove the following theorem.

Theorem V-7-10.(i) In the case when the degree of f b in s is equal to n, if we change the equation

P[y1 = 0 to a system by setting yt = y and yj = 61-1 [y) (j = 2, ... , n), thesystem has a singularity of the first kind at x = 0.

EXERCISES V 137

(ii) In the case when the degree of fn ins is less than n, if we change the equationP[y] = 0 by setting y1 = y and y, = b'-1 [y] (j = 2,... , n), the system has anirregular singularity at x = 0.

Proofn

Ltahi00

Look at P(xd] _ >ah(x)bh[x°] = xe= x" fm(S)xm and setTn=r+o

ah(x) = xnobh(x). Then, the functions bh(x) are analytic at x = 0. Furthermore,if the degree of is n, we must have bn(0) # 0. Therefore, in this case we can

n-1write the equation P(y) = 0 in the form bo[y] = -bn(Ebh(x)bh(yJ. Claim (i)

h=0follows immediately from this form of the equation.

If the degree of fn0 is less than n, we must have bn(0) = 0 and b,(0) - 0 forsome j such that 0 < j < n. To show (ii), change Yh further by zh = x(h-1)ayh

with a suitable positive rational number a so that the system for (z1, ... , zn) hasa singularity of the second kind of a positive order (cf. the arguments given rightafter Definition V-7-8, and also §XIII-7).

From Theorem V-7-10, we conclude that the formal solution x''(1 + O(x)) ofTheorem V-1-3 is convergent if the degree of f,, (s) in s is n. The following corollaryof Theorem V-7-10 is a basic result due to L. Fuchs [Fu].

Corollary V-7-11. The differential equation Ply] = 0 has, at worst, a regular

singularity at x = 0 if and only if the functions ah(x) (h = 0, ... , n - 1) arex

analytic at x = 0.

Some of the results of this section are also found in [CL, Chapter 4, pp. 108-137].

EXERCISES V

V-1. Show that if A is a nonzero constant, H is a constant n x m matrix, and N1and N2 are n x n and m x in nilpotent matrices, respectively, then there exists oneand only one n x m matrix X satisfying the equation AX + NIX - XN2 = H.

V-2. Show that the convergent power series

(a+1)...(a+m-1Q(Q+1) Q+m-1)y = F(a,x) = 1+a x,,

m=1


x(1-x)!f Y2 + +1)J - c3y = 0,

where a, /3, and -y are complex constants.


Comment. The series F(a, Q, ry, x) is called the hypergeometric series (see, forexample, [CL; p. 1351, 101; p. 159], and [IKSYJ).

V-3. For each of the following differential equations, find all formal solutions ofc

the form x'' I 1 + > cmxm] . Examine also if they are convergent.L m=1

(i) xb2y + aby + /3y = 0,

(ii) b2y + aby + $y = rxmy,

where b = xa, the quantities a, 3, and 7 are nonzero complex constants and mis a positive integer.

V-4. Given the system

(E) dt = (N + R(t))il, N = [0 O] ,and R(t) = t-3 I 0 0J ,

show that(i) (E) has two linearly independent solutions

and

, where (t) om!(m+1)!'

J2(t) = 1612 (t , where 02(t) = t + bmt-'n - ¢1(t)logt,+00(t))1

m-1

with the constants bm determined by b_1 = 1, bo = 0, and

m(m + 1)bm - b,,,-,2m + 1

= m!(m+ 1)!(m = 1,2,... ),

(ii) 1 lim t-1(Y(t)-e IN) = O for the fundamental matrix solution Y(t) = [if1(t)12(t)J,

(iii) the limit of e-'NY(t) as t -+ +oo does not exists.

V-5. Let y be a column vector with n entries and let Ax, y-) be a vector with nentries which are formal power series in n + 1 variables (x, yj with coefficients inC. Also, let u' be a vector with n entries and let P(x, u) be a vector with n entrieswhich are formal power series in n + 1 variables (x, u") with coefficients in C. Findthe most general P(x, u") such that the transformation u + xP(x, u7 changes

dy" du"the differential equation ds

= f (X, y-) to = 0.

Hint. Expand xP(x, u) and f (x, u+xP(x, u)) as power series in V. Identify coeffi-

cients ofd[xP(x, V-)]

with those of Ax, u'+zP(x, iX)) to derive differential equations

which are satisfied by coefficients of xP(x, u).

EXERCISES V 139

V-6. Suppose that three n x n matrices A, B, and P(x) satisfy the followingconditions:

(a) the entries of A and B are constants,(b) the entries of P(x) are analytic and single-valued in 0 < [xj < r for some

positive number r,dii

(c) the transformation y' = P(x)u changes the system xfy = Ay" to xjj = Bu.Show that there exists an integer p such that the entries of xPP(x) are polynomialsin x.

Hint. The three matrices P(x), A, and B satisfy the equation xd) = AP(x) -+"0

P(x)B. Setting P(x) = >2 xmPm, we must have mPm = APm - PmB for allm=-oo

integers m. Hence, there exists a large positive integer p such that Pm = 0 forImI ? P.

V-7. Let ff be a column vector with n entries {y,... , and let A(ye) be an n x nmatrix whose entries are convergent power series in {yj, ... , yn} with coefficientsin C. Assume that A(0) has an eigenvalue A such that mA is not an eigenvalue ofA(0 for any positive integer m. Show that there exists a nontrivial vector O(x)

with n entries in C{x} such that y" = b(exp[At]) satisfies the system L = A(y-)y.

Hint. Calculate the derivative of (exp(At]) to derive the system Axe =

for .

N

V-8. Consider a nonzero differential operator P = >2 ak(x)Dk with coefficientsk=O

ak(x) E C([x]], where D = dx. Regarding C([x]] as a vector space over C, de-

fine a homomorphism P : C[[x]] -. C[[x]j. Show that C((xjj/P(C[[x]j] is a finite-dimensional vector space over C.

Hint. Show that the equation P(y) = x"`'4(x) has a solution in C([xjj for anyO(x) E C[[x]] if a positive integer N is sufficiently large. To do this, use the indicialpolynomial of P.

NV-9. Consider a nonzero differential operator P = >2 ak(x)dk with coefficients

k=0

ak(x) E CI[x]j, where b = x2j, n > 1, and 54 0. Define the indicial poly-

nomial as in Theorem V-1-3. Show that if the degree of is n andif n zeros {A1, ... , A, } of do not differ by integers, we can factor P in thefollowing form:

P = a+,(x)(b - &1(x))(b - 42(x))...(6 - 0n(x)),

where all the functions ¢, (x) (j = 1, 2, ... , n) are convergent power series in x and4f(O) = Aj.

Hint. Without any loss of generality, we can assume that an(x) = 1. Then, An +n-1Eak(O)Ak = (A - A1)... (A - An). Define constants {yo... , In-2} by A' ' +k=0n-2

,YkAk = (A - A2) ... (A - An). For v] (X) E xC[[xI] and ch(x) E xC([x]l (h =k=O0,. . . , n - 2), solve the equation

( n-2

(C) P = (6 - Al - 2(7h+Ch(x))6h

h=0

If we eliminate O(x) by Al + V(x) = 1`n-2 + ct-2(x) - an-1(x), condition (C)becomes the differential equations

6(CO(x)1 = a0(x) + (-Yn-2 + Cn-2(x) -a. -1(x))(1'0 +40(x)),

1 b[ch(x)] = ah(x) + (1'n-2 + Cn-2(x) - an-1(x))(7h +Ch(x)) - (1h-1 + Ch-1(x)),

where h=I,...,n-2.n

V-10. Consider a linear differential operator P = E atbt, where ao,... , an aret=o

complex numbers and 6 = x. The differential equation

(P) P[y1= 0

is called the Cauchy-Euler differential equation. Find a fundamental set of solutionsof equation (P).

Hint. If we set t = log x, then b = dt

V-11. Find a fundamental set of solutions of the differential equation

(6-06-8)(6-a-8)[yl =xn'Y'

where 6 = xjj and m is a positive integer, whereas a and /3 are complex numbers

such that they are not integers and R[al < 0 < 8t[01.

V-12. Find the fundamental set of solutions of the differential equation

(LGE) f(1-x2)LJ +a(a+1)y=0

at x = 0, where a is a complex parameter.

EXERCISES V 141

Remark. Differential equation (LGE) is called the Legendre equation (ef. [AS, pp.331-338] or [01, pp. 161-189]).

V-13. Show that for a non-negative integer n, the polynomial Pn(x) = 12nn!

x do [(x2 - 1)'] satisfies the Legendre equation

d[(1-x2) ] +n(n+1)Pn=0.

Show also that these polynomials satisfy the following conditions:

(1) Pn(-x) = (-1)nPn(x), (2) Pn(1) = 1, (3) JPn(t)j ? 1 for {xj > 1,+00

(4) 1 - 2xt + t =E Pn(x)tn, and (5) lPn(x)I < 1 for Jxj < 1.n=o

Hint. Set g(x) = (1 - x2)n. Then, (1 - x2)g'(x) + 2nxg(x) = 0. Differentiate thisrelation (n + 1) times with respect to x to obtain

(1,- x2)g(n+2)(x) - 2(n + 1)xg(n+1)(x) - n(n + 1)g(n)(x)

+ 2nxg(n+1) (x) + 2n(n + 1)g(")(x) = 0

or

(1 -X 2)g(n+2)(x) - 2ng(n+1)(x) + n(n + 1)g(n)(x) = 0.

Statements (1), (2), (3), and (4) can be proved with straight forward calculations.Statement (5) also can be proved similarly by using (4) (cf. [WhW, Chapter XV,Example 2 on p. 303]). However, the following proof is shorter. To begin with, set

F(x) = Pn(x)2n(n + 1){1 -x2)((1 - x2)P''(x))2.

2

Then, F(±1) = Pn(±1)2 = 1, p(X) _ (P"(x))n(n + 1)

,and F(x) = Pn(x)2 if (x)0. Therefore, for 0 < x < 1, local maximal values of Pn(x)2 are less than F(1) = 1,whereas, for -1 < x < 0, local maximal values of Pn(x)2 are less than F(-1) = 1.Hence, Pn(x)I < 1 for JxI < 1 (cf. [Sz, §§7.2-7.3, pp. 161-172; in particular,Theorems 7.2 and 7.3, pp. 161-162]. See also (NU, pp. 45-46].)

The polynomials Pn(x) are called the Legendne polynomials.

V-14. Find a fundamental set of solutions of (LGE) at x = oo. In particular, showwhat happends when a is a non-negative integer.

V-15. Show that the differential equation b"y = xy has a fundamental set ofsolutions consisting of n solutions of the following form:

(logx)k-1-A

A-o (k - 1 - h)!OA(x) (k = 1, ... , n),

where the functions Oh(x) (h = 0,... , n -1) are entire in x, 00(0) = 1, and Oh(O) _0 (h = 1, ... , n -1). Also, find O0 (x).


V-16. Find the order of singularity at x = 0 of each of the following three equa-tions.

(i) x5{66y - 362y + 4y} = y,(ii) x5{56y - 3b2y + 4y} + x7{b3y - 55y} = y,

(iii) x5y"' + 5x2y" + dy + 20y = 0.

V-17. Find a fundamental matrix solution of the system

x2dbi = xyi + y2,dx

x2dy2

= 2xy2 + 2y3,dx

x2 dx3 = x3y! + 3y3.

V-18. Let A(x) be an n x n matrix whose entries are holomorphic at x = 0. Also,let A be an n x n diagonal matrix whose entries are non-negative integers. Showthat for every sufficiently large positive integer N and every C"-valued functionfi(x) whose entries are polynomials in x such that those of xA¢(x) are of degree N,there exists a C"-valued function f (x; A, ¢, N) such that

(a) the entries off are polynomials in x of degree N-1 with coefficients dependingon A, N, and ¢,

(b) f is linear and homogeneous in tb,(c) the linear system

xAdd9 =z

A(x)3l +

has a solution #(x) whose entries are holomorphic at x = 0 and il(x) is linearand homogeneous in ',

(d) fi(x)) = O(xN+i) as x 0.

Hint. See [HSW].

V-19. Let A(x) and A be the same as in Exercise V-18. Assuming that n >

trace(A), show that the system xA dz = A(x)y has at least n - trace(A) linearlyindependent solutions holomorphic at x = 0.

Hint. This result is due to F. Lettenmeyer [Let]. To solve Exercise V-19, calculatef (x; A, ¢, N) of Exercise V-18 and solve f(x; A, 4, N) = 0 to determine a suitablefunction 0.

00

V-20. Suppose that for a formal power series ¢(x) _ > cx' E C[]x]], there ex-m=0/dl d

ist two nonzero differential operators P = E ak (x) ` I k and Q = > bk (x) (d) kk=0 rk=10

with coefficients ak(x) and bk(x) in C{x} such that P[0] = 0 and Q IJ

= 0. Show

that 0 is convergent. l

EXERCISES V 143

Hint. The main ideas are(a) derive two algebraic (nonlinear) ordinary differential equations

(E) F(x,v,v',... ,v(")) = 0 and G(x,v,v',... ,v(' ) = 0

for v = from the given equations P[¢) = 0 and Q ['01 = 0.

(b) eliminate all derivatives of v from (E) to derive a nontrivial purely algebraicequation H(x, v) = 0 on v. See [HaS1] and [HaS2[ for details.

CHAPTER VI

BOUNDARY-VALUE PROBLEMS OF LINEARDIFFERENTIAL EQUATIONS OF THE SECOND-ORDER

In this chapter, we explain (1) oscillation of solutions of a homogeneous second-order linear differential equation (§VI-1), (2) the Sturm-Liouville problems (§§VI-2-VI-4, topics including Green's functions, self-adjointness, distribution of eigen-values, and eigenfunction expansion), (3) scattering problems (§§VI-5-VI-9, mostlyfocusing on reflectionless potentials), and (4) periodic potentials (§VI-10). Thematerials concerning these topics are also found in [CL, Chapters 7, 8, and 11],[Hart Chapter XI], [Copl, Chapter 1], [Be12, Chapter 61, and [TD]. Singular self-adjoint boundary-value problems (in particular continuous spectrum, limit-pointand limit-circle cases) are not explained in this book. For these topics, see [CL,Chapter 9].

VI-1. Zeros of solutions

It is known that real-valued solutions of the differential equation L2 + y = 0 are

linear combinations of sin x and cos x. These solutions have infinitely many zeros onthe real line P. It is also known that real-valued solutions of the differential equation

2 - y = 0 are linear combination of e= and a-x. Therefore, nontrivial solutions

has at most one zero on the real line R. Furthermore, solutions of the differential

equation2 + 2y = 0 have more zeros than solutions of L2 + y = 0. In this

section, keeping these examples in mind, we explain the basic results concerningzeros of solutions of the second-order homogeneous linear differential equations. In§§VI-1-VI-4, every quantity is supposed to take real values only.

We start with the most well-known comparison theorem concerning a homoge-neous second-order linear differential equation

(VI.1.1)dx2

+ 9(x)y = 0.

Theorem VI-1-1. Suppose that(i) g, (x) and g2(x) are continuous and g2(x) > 91(x) on an interval a < x < b,

(ii)d 2 01 (x)

+ 91(x)o1(x) = 0 and d2622x) + g2(x)02(x) = 0 on a < x < b,

(iii) S1 and 6 are successive zeros of 01(x) on a < x < b.q2(x) must vanish at some point £3 between t:l and C2.Then.

144

1. ZEROS OF SOLUTIONS 145

Proof.

Assume without any loss of generality that Sl < 1;2 and 01(x) > 0 on 1;1 < x < £2.

Notice that 41(1::1) = 0, 1 (l:l) > 0, 01(S2) = 0, and (1:2) < 0. A contradiction

will be derived from the assumption that 02(x) > 0 on S1 < z < 1;2. In fact,assumption (ii) implies

d 20, (7)VI 2 ) d202 (x)(

=(x)0 (x)(x)10[ (x) -02(x)

.1. _ 01 x 2192 91

and, hence,

VI 1 3 1 L ( ) ) =0 V )!!LV rE2(x)dx(x)j¢ (x)0(x) -( - - ) 0202) 6 - 2 1 1

.2191E1

The left-hand side of (VI.1.3) is nonpositive, but the right-hand side of (VI.1.3) ispositive. This is a contradiction. 0

A similar argument yields the following theorem.

Theorem VI-1-2. Suppose that(i) g(x) is continuous on an interval a < x < b,(ii) 0i(x) and 02(x) are two linearly independent solutions of (VI.1.1),

(iii) 1;1 and 6 are successive zeros of 01 (x) on a < x < b.Then, 02(x) must vanish at some point 3 between t;1 and 1;2.

Proof.

Assume without any loss of generality that Sl < £2 and .01 (x) > 0 on S1 < x < 1;2.

Notice that 01(11) = 0, 1 (S1) > 0, 01(12) = 0, and X1(1;2) < 0. Note also that

if 02(1) = 0 or 02(6) = 0, then 01 and 02 are linearly dependent. Now, acontradiction will be derived from the assumption

(x)assumption that 02(x) > 0 on fi < x < 2

(x)In fact, assumption (u) implies 02(x) a-

dS2- p1(z) d2

d02

x2= 0 and, hence,

(VI.1.4) 02(2) 1(6) - 4'201) 1(W =0.

Since the left-hand side of (VI.1.4) is positive, this is a contradiction. 0The following result is a simple consequence of Theorem VI.1.2.

Corollary VI-1-3. Let g(x) be a real-valued and continuous function on the in-terval Zo = {x : 0 < x < +oo}. Then,

(a) if a nontrivial solution of the differential equation (VI.1.1) has infinitely manyzeros on I , then every solution of (VI.1.1) has an infinitely many zeros onZo,

(b) if a solution of (VI.1.1) has m zeros on an open subinterval Z = {x : a < x <3} of Zo, then every nontrivial solution of (VI.1.1) has at most m + 1 zeroson Z.

Denote by W(x) = W(x;01,02) =02(x)

I the Wronskian of the set of

functions (01(x), 02(x)}. For further discussion, we need the following lemma.

146 VI. BOUNDARY-VALUE PROBLEMS

Lemma VI-1-4. Let g(x) be a real-valued and continuous function on the intervalTo = {x : 0 < x < +oo}, and let 771(x) and M(x) be two solutions of differentialequation (VI 1.1). Then,

(a) W (x; 772 , ri2 ), the Wronskian of {rli (x), 712(x)}, is independent of x,

(b) if we set t;(x) =771(x),

then4(x) = c , where c is a constant.

772(x) dx 772(x)2

Also, if 77(x) is a nontrivial solution of (VI.1.1) and if we set w(x) = then

we obtain

(VI.1.5)dw(x)

+ w(x)2 + g(x) = 0 .

Proof.

(a) It can be easily shown that

d 771(x) 772(x) 771(x) 772(x)

dx I77i(x) 772(x)=

77i'(x) 772(x)-g(x) !

771(x)712(x) = 0712(x)

(b) Note thatdl;(x} = - 1 I rl' (x) 712(x) ! and that dw(x) if'(x)

1

dx 712(x)2 771(x) 772(x)1 dx :l(x)

((xx))12 = -g(x) - w(x)2.

The following theorem shows the structure of solutions in the case when everynontrivial solution of differential equation (VI.1.1) has only a finite number of zeroson To= {x:0<x<+oo}.

Theorem VI-1-5. Let g(x) be a real-valued and continuous function on the inter-val I. Assume that every nontrivial solution of differential equation (VI.1.1) hasonly a finite number of zeros on T. Then, (VI.1.1) has two linearly independentsolutions 777(x) and 712(x) such that

(a) lim 771(x) = 0,x-+oo 712(x)

(b)

1.0

+00 dx= +oo and(VI.1.6) x)2

771(

where xo is a sufficiently large positive number,(c) the solution 712(x) is unbounded on To.

Proof.

1+00 dx

Jxo 7r (x)2 < +00,

(a) Let (1(x) and (2(x) be two linearly independent solutions of (VI.1.1) such that(, (x) > 0 (j = 1, 2) for x > xo > 0. Using (b) of Lemma VI-1-4, we can derive the

1. ZEROS OF SOLUTIONS 147

(1(x) =following three possibilities (1) =limo(2(x) 0, (ii) =U +oo, and (iii)

Urn ( 1( x )>0 . Setx-+oo (2 (x)

(i(x) in case (i), (2(x) in case (i),

711(x) _ (2(x) in case (ii), 772(x) = (i(x) in cage (ii),

(i(x) - 7'(2(x) in case (iii), 1(2(x) in case (iii).

Then, (a) follows immediately.(b) Conclusion (b) of Lemma VI-1-4 implies that

1 _Id(Y12(x)and

1 _-1d >)1(x)

ql (x)2 ^ c dx . q1(x)) 712(x)2 c dx \ 2(x)

where c is the Wronskian of q1 and rte. Hence, (VI.1.6) follows.+oc

(c) If q2 is bounded, we must have j()2 = +00. 0-

,

The following theorem gives a simple sufficient condition that solutions of (V1.1.1)have infinitely many zeros on the interval Zo.

Theorem VI-1-6. Let g(x) be a real-valued and continuous function on the in-

terval lo. If f g(x)dx = +oo, then every solution of the differential equation0

(VI.1.1) has infinitely many zeros on Zo.

Proof.

Suppose that a solution 17(x) satisfies the condition that 77(x) > 0 for x >

xo > 0. Set w(x) = !L.(x) Then, w(x) satisfies differential equation (VI.1.5).77(x)

Hence, lien w(x) = -oo. This implies that lim 17(x) = 0 since 17(x) = r7(xo):-+co :-+oo

This contradicts (c) of Theorem VI-1-5. Ox exp VoJ

The converse of Theorem VI-1-6 is not true, as shown by the following example.

Example VI-1-7. Solutions of the differential equation

(VI.1.7)2 + 1A

z2= 0,

where A is a constant, have infinitely many zeros on the interval -oo < x < +oo if

and only if A > 4

Proof

Look at (VI.1.7) near x = oo. To do this, set t = to change (VI.1.7) to

r

x

(VL1.8)L

462 + 26 +1 + t, 0,

where b = t. Note that t = 0 is a regular singular point of (VI.1.8) and the

indicial equation is 4s2 + 2s + A = 0 whose roots are s = - 2 f 4 - A. These

roots are real if and only if A < 4 This verifies the claim. 0+00

Note that J1 + x2

dx2*0

The following theorem shows some structure of solutions in the case when

+00

ig(x)ldx < +oo.0

Theorem VI-1-8. Let g(x) be a real-valued and continuous function on the inter-f+C0

valA. If 1 jg(x)ldx < +oo, then there exist unbounded solutions of differential0

equation (VI. 1. 1) on Ia.

Proof.

The assumption of this theorem and (VI.1.1) imply that lim dq(x) = -Y exists:-+oo dxfor any bounded solution n(x) of (VI.1.1). If 7 0, then i(x) is unbounded. Hence,

zlim odd) = 0. Therefore, calculating the Wronskian of two bounded solutions

of (VI.1.1), we find that those bounded solutions are linearly dependent on eachother. This implies that there must be unbounded solutions. 0

Remark VI-1-9. Theorems VI-1-1 and VI-1-2 are also explained in ]CL, §1 ofChapter 8, pp. 208-211] and [Hart, Chapter XI, pp. 322-403]. For details concern-ing other results in this section, see also [Cop2, Chapter 1, pp. 4-33] and [Be12,Chapter 6, pp. 107-142].

VI-2. Sturm-Liouville problems

A Sturm-Liouville problem is a boundary-value problem

(BP)dd-x (p(x) Ly) + u(x)y = f (.T),

y(a) cos a -Ip(a)y'(a) sin or = 0,

y(b) cos Q - P(b)y(b) sin )3 = 0,

under the assumptions:(i) the quantities a, b, a, and $ are real numbers such that a < b,

(ii) two functions u(x) and f (x) are real-valued and continuous on the intervalZ(a,b) = {x: a < x < b},

(iii) the function p(x) is real-valued and continuously differentiable, and p(x) > 0on 1(a, b).

In this section, we explain some basic results concerning problem (BP).

2. STRUM-LIOUVILLE PROBLEMS 149

Let 4(x) and O(z) be two solutions of the homogeneous linear differential equa-tion

(VI-2.1)-dx

(p(x)A) + u()y = 0

such that

(VI.2.2) m(a) = sins, p(a)d'(a) = cosa, ,p(b) = sin f3, p(b)t'(b) = cosfi,

respectively. Then, these two solutions satisfy the boundary conditions

(VI.2.3) ¢(a) cos a - p(a)4,'(a) sin a = 0, i (b) cos /3 - p(b)0'(b) sin /3 = 0.

The two solutions 4(x) and /;(x) are linearly independent if and only if

(VI.2.4) 4(b) cos Q - p(b)©'(b) sin /3 # 0 or '(a) cos a - p(a)rb'(a) sin a # 0.

The first basic result of this section is the following theorem, which concerns theexistence and uniqueness of solution of (BP).

Theorem VI-2-1. If the two solutions 4(x) and ip(x) of (VI.2.1) are linearlyindependent, then problem (BP) has one and only one solutwn on the intervalI(a, b).

Proof

Using the method of variation of parameters (cf. Remark IV-7-2), write thegeneral solution y(x) of the differential equation of (BP) and its derivative y'(x)respectively in the following form:(VI.2.5)

VW = CIOW + C20 W + 4(x)J

6'j'(_)f (_) 4 + i,&(x)1= o(f)fP(f)WW a

Y' (X) = Clo'(x) + C21k'(x) + 0'(x) I 4 (x) J A,s P(A)W JJJu P(A)W V)

where cl and c2 are arbitrary constants and W (x) denotes the Wronskian ofNow, using (VI.2.3) and (VI.2.4), it can be shown that solution

(VI.2.5) satisfies the boundary conditions of (BP) if and only if cl = 0 and c2 =0. 0Observation VI-2-2. It can be shown easily that p(x)W(x) is independent of x.

Observation VI-2-3. Under the assumption that the two solutions 40(x) and tfi(x)of (VI.2.1) are linearly independent, the unique solution of (BP) is given by

y(x) = d(x)rb

10(x) I (- ))- W(f )(VI.2.6) Jx P(A)W (O iY 'W = O(x)

rb owfwdC + V,'(x) r= 4(W W 4.1r W Ja P(OW (O

Setting

4(x)1() if < <bCx ,

(VI.2.7) G(x, ) = p(OW (c)O WOW if < <p(f)W(c)

ta x,

we can write (VI.2.6) in the form

b b aG

y(x) =J

G(x, )f y '(x) =J 8x (x, Of (Ode.

The function G(x,l;) is called Green's function of problem (BP). The followingtheorem gives the characterization of Green's function.

Theorem VI-2-4. The function G(x, C) given by (VI. 2.7) satisfies the followingconditions:

(i) G(x, l:) is continuous with respect to (x, e) on the region D = {(x,) a < x <b, a< £<b},

(ii) 8 (x, l:) is continuous with respect to (x, C) on the region D - a <

C < b},(iii) at ({, l;), we have

8 (tr + 0. 0 - ( - 0, 0 =p(0

for a < < b,

(iv) as a function of x, G(x, l:) satisfies homogeneous linear differential equation(VI. 2. 1) if x 0 4,

(v) as a function of x, G(x, 4) satisfies the boundary conditions of (BP), i. e.,

G(b,Z:)cosf3-p(b)8G(b,{)sinf3=0

for a:5 < b.Furthermore, the function G(x, f) is uniquely determined by these five conditions.

Proof.

It is evident that the function G(x, l;) satisfies these five conditions. SupposeH(x,t) satisfies the five conditions. Conditions (iv) and (v) imply that

Cl(l;)4(x) for a < x < l;,{ C2(t;)V(x) for { < x < b

for (x,l;') E D, where Ci(1;) and C2(C) must be determined by conditions (i), (ii),and (iii). This means that C,({) and must be determined by

0 and-PW

2. STRUM-LIOUVILLE PROBLEMS 151

This OW and C2({) = 4(f) Since {(x,e) : a < x < }yields Cl(a;) = p(OW(O p(OW(O - -x<t<b}and {(x,t): 1;<x< b}{(x,£): a<t<x}for

(x, ) E V, we obtain H(x,1;) = G(x, l;). 0The second basic result co//ncerns self-adjointness of problem (BP). Define a dif-

ferential operator L[y] _ (p(x) I + u(z)y and a vector space V(a, b) over the

real number field 1R by \ /

V(a, b) _ { f r C2(a, b) : f (a) cos a - p(a) f'(a) sin a = 0,

f(b)cos5 - p(b)f'(b) sin,3 = 01,

where C2(a, b) denotes the set of all real-valued functions which are twice contin-uously differentiable on the interval I(a, b) = {x : a < x < b). Define also aninner product (f, g) for two real-valued continuous functions f and g on I(a, b) by

(f,9) = f Sincea

(f, 4191) = J6f()

[(())

+4

b

p(b)f (b)9 (b) - p(a)f (a)9'(a) + f dl;=a

for f E V(a, b) and g E V(a, b), we obtain

(f,L[g) - (L(f1,9)={p(b)f(b)9'(b) - p(b)g(b)f'(b)} - {p(a)f(a)9 (a) - p(a)9(a)f'(a)}.

Also, since

( f (a) cos a - p(a)f'(a) sin a = 0, f (b) cos Q - p(b) f'(b) sin O= 0,

t 9(a) cos a - p(a)9'(a) sin a = 0, g(b) cos, - p(6)9'(b) sin /3 = 0

for f E V(a. b) and g E V(a, b), we obtain

p(a)f(a)9'(a) - p(a)g(a)f(a) = 0, p(b)f(b)9'(b) - p(b)g(b)f(b) = 0.

Thus, we proved the following theorem.

Theorem VI-2-5 (self-adjointness). The operator L has the following property:

(f, 0191) = (L(f], 9) for f E V (a, b) and g E V(a, b).


Observation VI-2-6. In Theorem VI.2.1, it was assumed that the two solutions¢(x) and '(x) are linearly independent. Consider the case when this assumptionis not satisfied. So, assume that ¢(x) and ?P(x) are linearly dependent. This meansthat

0(a) cos a - p(a)0'(a) sin a = 0 and ¢(b) cos ji - p(b)5'(b) sin fi = 0.

In other words,

(VI.2.8) lr[o) = 0 and 0 E V(a, b).

The boundary-value problem (BP) can be written in the form

(VI.2.9) G[yl = f and y E V(a, b).Using (VI.2.8) and (VI.2.9), we obtain

(f, 0) = (C [y], 0) = (y, x[01) = 0,if there exists a solution y of problem (VI.2.9). The converse is also true, as shownin the following theorem.

Theorem VI-2-7. Assume that O(x) satisfies condition (VI.2.8). Then, if a real-valued continuous function f (x) on T (a, b) satisfies the condition

(VI.2.10) (f. 0) = 01

problem (BP) has solutions depending on an arbitrary constant.

Proof.

Using the method of variation of parameters, write the general solution y(x) ofthe differential equation of (BP) and its derivative y(x) respectively in the followingform:(VI.2.11)

b__ 11

y(x) = cjW(x) + c2p(X) + 0(x)j + (x)JP(A)W f ) o P(A)W (E)

?% (z) = J p _ { + F' (r) 1-vr P(f)ww u P(t)

1-v

where u(x) is a solution of the linear homogeneous differential equation (VI.2.1)such that 0 and p are linearly independent, two quantities ca and c2 are arbitraryconstants, and W(x) denotes the Wronskian of {0(x),p(x)}. Note that p(x)W(z)is independent of x and that

(VI.2.12) p(a)cosa-p(a)u'(a)sina,-40, p(b)cosQ-p(b)p'(b)sinfl,40.From (VI.2.10) and (VI.2.11), we derive

y(a) = ctm(a) + c2p(a) + 0(a)J

b

c P(4)W(E)

y'(a) = ca0'(a) + c2 (a) + 0'(a) r° A,a -N)

y(b) = ciO(b) + c2p(b), y'(b) = c,O'(b) + c2p'(b)The condition that y E V(a,b) and (VI.2.12) imply that ca is arbitrary and c2 =

_),

0. 0

3. EIGENVALUE PROBLEMS 153

Remark VI-2-8. The materials of this section are also found in (CL, Chapters 7and 11] and [Har2, Chapter XI].

VI-3. Eigenvalue problemsIn this section, we consider the eigenvalue problem

(EP)(P(x) dx ) + u(x)y = Ay,

y(a)coso - p(a)y'(a)sina = 0,y(b) cos /3 - p(b)y'(b) sin /3 = 0,

under the assumptions:(i) the quantities a, b, a, and 0 are real numbers such that a < b,(ii) u(x) is a real-valued and continuous function on the interval Z(a, b) = {x :

a<x<b},(iii) the function p(x) is real-valued and continuously differentiable, and p(x) > 0

on I(a, b).The quantity A is the eigenvalue parameter.

Let 6(x,,\) and ti(x, A) be two solutions of the homogeneous linear differentialequation

(VI.3.1)

such that

(VI.3.2)

(px) + u(x)y = Ay

{¢(a, A) = sin a, p(a)4'(a, A) = cos a,,O(b, A) = sin /3, p(b)t,V(b, A) = cos f3,

respectively. Then,

(VI.3.3)-O(a, A) cos a - p(a)6'(a, A) sin a = 0,tp(b, A) cos /3 - p(b) 0'(b, A) sin /3 = 0.

These two solutions 6(x, A) and tp(x, A) are analytic in A everywhere in C. Also,they are linearly independent if and only if

(VI.3.4) f 0(b, A) cos (3 - p(b)¢'(b, A) sin /3 i4 0 or

t'(a,A)cosa - p(a)ty'(a,A)sina # 0.

Therefore, we obtain the following result.

Theorem VI-3-1. In order that A be an eigenvalue of (EP), it is necessary andsufficient OW ,\ satisfies the equation

(VI.3.5) ¢(b, A) cos i3 - p(b)cp'(b, A) sin 0 = 0

or, equivalently,+'(a, A) cos a - p(a)0'(a, A) sin a = 0.


Observation VI-3-2. All roots of (VI.3.5) are simple, since, if A is a root of(VI.3.5), we can prove

80(b, A) 8¢'(b, A)

8Acos j3 - P(b)

8Asun f # 0

in the following way. Let A be a root of (VI.3.5). Then, notice that z =is a solution of the differential equation

CP(x)L1 + u(x)z = Az + ¢(x,A).

84(x, A)

8A

This implies that(VI.3.6) r_

Z(X) = c16(x, A) + c2p(x) - O(x, A)J

A) dC + p(x) f = ' , A)2 de,

1 n Ja P(O1't'(Oz'(x) = clOx, A) + C21L (x) - 4'(x, A) f pW¢(f, A)dC + p '(x) = w(), A)2

dd,PwW (f) j P(O-N)

where p(x) is a solution of homogeneous linear differential equation (VI.3.1) suchthat 4(x, A) and p(x) are linearly independent, two quantities ct and c2 are con-stants, and W(x) denotes the Wronskian of {¢(x,A),p(x)}. Note that p(x)W(x) isindependent of x and that

(VI.3.7) p(a) cos a - p(a)p'(a) sin a 0, p(b) cos,3 - p(b)1 (b) sin Q # 0.

Formulas (VI.3.6) imply that(VI.3.8)

z(a) = clo(a,A) + c2p(a), z'(a) = ci4'(a,A) + c2,u (a),

z(b) = cj¢(b, A) + c2p(b) - 4(b, A)J

p(b)fb

n Pwwwz'(b) = c1 ¢'(b, A) + c2p'(b) - 0'(b.,\) j + p'(b) rP(u)W

w o P(4)W (0

Since (VI.3.3) is true for all values of A, it follows that z(a) cos a - p(a)z(a) sin a0. Hence, from (VI.3.7) and (VI.3.8), we conclude that c2 = 0. Therefore,

z(b) cos - P(b) z'(b) sin,3 = (,u(b) cos$ - P(b) (b) sin 0) I° 0(Td # 0. 0

a a P(A)W w

Remark VI-3-3. If A = 0 is not an eigenvalue of (EP), we can define Green'sfunction G(x,t) of problem (BP of §V-2 so that (EP) is changed equivalently to

the integral equation y(x) = A G(x,

To prove that all eigenvalues of (EP) are real, it is convenient to extend TheoremVI-2-5 (self-adjointness) to complex-valued functions f and g. Define a differential

operator C[y] = ((x)) + u(x)y and a vector space U(a, b) over the complex

number field C by

U(a,b) _ If E C2 (a, b; C) : f (a) cos a - p(a) f'(a) sin a = 0,

f (b) cos a - p(b) f'(b) sin,3 = 0},

where C2(a, b; C) denotes the set of all complex-valued functions which are twicecontinuously differentiable on the interval I(a, b) = {x : a < x < b}. Also, define aninner product (f, g) for two complex-valued continuous functions f and g on 2(a, b)by

b

(f, 9) =J

f (09(041n

where g(1;) denotes the complex conjugate of g(f ). Since

(f,'CWi) = 1b

f \p(e) <J + u(C)9(C)] dCa

= P(b)f(b)9'(b) - p(a)f(a)9'(a) +Je [-Pwfv)9'(o + uh)fw9(-j <'

for f E U(a, b) and g E U(a, b), we obtain

(f, C(91) - (C[f ], 9)

= {p(b)f(b)9'(b) - p(b)9(b)f'(b)} - {p(a)f(a)g(a) - p(a)9(a)f'(a)}.

Also, since

f (a) cos a - p(a) f'(a) sin a = 0, f (b) cos Q - p(b)f'(b) sin /3 = 0,

. 9(a) cos a - p(a)9'(a) sina = 0, 9(b) cosQ - p(b)91(b) sin p = 0

for f E U(a,b) and g E U(a,b), we obtain

p(a)f(a)9'(a) - p(a)9(a)f'(a) = 0, p(b)f(b)9'(b) - p(b)9(b)f'(b) = 0.

Thus, we extended Theorem VI-2-5 as follows.

Theorem VI-3-4 (self-adjointness). The operator C has the following property:

(f, C[9]) = (C[f], 9) for f E U(a, b) and g E U(a, b).

Theorem VI-3-4 implies the following conclusion.


Theorem VI-3-5. All eigenvalues of (EP) are meal.

Proof.

Let A be an eigenvalue of (EP) and let y(x) be an eigenfunction correspondingto A. We may assume that (y, y) = 1. Then,

(y,Cjy]) = (y, Ay) = X(y,y) = X,

(hjyl,y) = (Ay,y) = A(yy) = A,and Theorem VI-3-4 imply that J = A. 0

In order to explain distribution of eigenvalues on the real line IR, look at the basiccomparison theorem (Theorem VI-1-1) more closely. Let us rewrite the differentialequation

(VI.3.9)

in the form

and

(VI.3.10)

by setting

(P(X) dy) + g(x)y = 0

p(x) = (1 - p(x)g(x))rsin(O)cos(9)

p(x)dO = I + (p(x)g(x) - 1) sin2(a)

y = r sin(O), p(x) Li = r cos(B).

The right-hand side of (VI.3.10) is independent of r. Therefore, (VI.3.10) is a first-order nonlinear differential equation on 0. Our main concern is the behavior ofsolutions of (VI.3.10).

Remark VI-3-6. Set w = ! ((x) L) . Then, differential equation (VI.3.9) be-

comes p(x) dx + w2 + p(x)g(x) = 0. This equation is further changed to (VI.3.10)by the transformation w = cot(8).

Remark VI-3-7. If the function g(x) is continuous on the interval 2(a, b), thenevery solution 0(x) of (VI.3.10) exists on T(a,b).

Now, we prove the following basic lemma.

Lemma VI-3-8. Assume that(1) four functions gl(x), g2(x), pi (x), and p2(x) are continuous on the interval

2(a, b),(2) 92(x) ? gi(x) and pi(x) ? P2(x) > 0 on T(a,b),

(3) 01(x) (j = 1,2) are solutions of the differential equations

Pi (x)! = 1 + [pi (x)gg (x) - 1J sin2(01) (j = 1, 2),

respectively.

Then,02(x) > 61(x) on I(a,b) if 02(a) > 01(a).

Proof.

Set

9(x) - 1 stn2(02) - sin2(0)l(1x,61,62()1p()

/ l 02 01

h(x,02) _ (92(x) - 91(x))Sin2(62) +1

P2(x)1

WS2(02)

P1 (x)

Then, the function w(x) = 02(x) - 01(x) satisfies the differential equation

dw- A(x,01(x),02(x)) w = h(x,02(x)) > 0.

dx -

Therefore,

w(x)eXPL-J IL a

rw(a) + J x h(s,02(s))eXP [- f ds > 0

a

on T(a, b). 0Remark VI-3-9. The proof given above also showed that if pi (x) > p2(x) > 0 onZ(a, b), then we obtain(a) 02(x) > 61(x) on I(a,b) if 02(a) > 01(a),

02(x) > 01(x) for a < x 01(a) and g2(x) > g1(x) on a < x < b,( y) in the case when g2(x) > g1(x) on a < x < b, if 01(a) = 0 and 61(b) = 7r and

if 0<02(a)<rr,then 02(b)>a.

Also, y=rsin(0)>0 if and only if 0<0<r(mod 2rr)and p(x) = 1 > 0 at0 = 0 (mod jr). Theorem VI-1-1 follows from (y).

Setting 9(x,,\) = u(x) - A, apply Lemma VI-3-8 to problem (EP).

Lemma VI-3-10. Let 0(x, A) be the unique solution to the initial-value problem

Ax) = 1 + (P(x)(u(x) - A) - 1)sin2(0), 0(a,A) = a

on the interval Z(a, b), where u(x) is continuous on Z(a, b), p(x) is continuous andpositive on I(a, b), A is a real parameter, and at is a fixed real number such that0 < a < rr. Then, for every c E 2(a, b) such that c > a,

(i) 0(c, A) is a continuous and strictly decreasing function of A for -oo < A <+00,

(ii)a

lira 0(c, A) = +oo,

(iii) lim 0(c, A) = 0.A +00

Proof.

We prove this lemma in three steps.

Step 1. To prove (i), apply Lemma VI-3-8 to pi(x) = 12(x) = p(x), 91(x) =g(x, A1), and 92(x) = g(x, A2). If A2 < A1, then g2(x) > 91(x) on Z(a, b). Hence,9(x, A2) > 9(x, A1) on a < x < b, since 9(a, A2) = 9(a, Al) = a.

Step 2. To prove (ii), choosing a real number m and a positive number P so thatu(x) > m and p(x) m -A and P > p(x) > 0 for x E I(a, b), -oo < A < +ooand a > 0, it follows from Lemma VI-3-8 that

9(c, A) > 9o(c, A) for a < c < b and - oo < A </+oo.

Observe that v = P tan(Oo) satisfies the differential equation = 1 + (MP A ) v2.

Hence, tan(9o(x, A)) =P(m - A)

tan ( mP A (x - a) f for A < m. Note that

tan(9o(x, A)) = 0 if and only if tan I VmP A (x - a)) = 0, i.e.,

m - Atan(8o(x,A)) = 0 at

P(x - a) = na for n=1,2....

and that 9o(x, A) is strictly decreasing with respect to A if x > a (cf. Step 1).

Furthermore, since tan(9o(x, m)) = x P a, we must have 0 < 9o(x, m) < 2 fora < x. Therefore, ()2)

9 m - P = ntr for n = 1, 2, ... .

Hence,r

9 I x, m -(.)2)

> na for n = 1, 2, ... .x - a

Thus, we conclude that lim 9(c, A) = +oa if c > a.accStep 3. To prove (iii, first notice that 0(c, A) > 0 for -oo < A < +oo, since

0(a, A) = a > 0 and = P(IX)> 0 if 0 = 0. Choose a positive number M so that

n

2(e)I < M on I(a, b).lu(x) sin2(9) + 1

siAx)

Then,

dO < M - A sin2(0) on Z(a, b).

For any positive number 6 such that 0 < a < r - 6, fix another positive numberA(6) so that

M - A sin2(6) < - c-10afor 6<0< r - b and A > A(6).- - -

This implies that dO < 0 if 9 = 6. Hence, 9(x, A) > 6 for a < x < c if 9(c, A) > 6,

whereas9(c,A) <a-10<r-10<0if 9(x, A)>6fora<x<c. Thisisacontradiction. Therefore, 0 < 9(c, A) < 6 for A > A(6). Thus, we conclude that

lim 9(c, A) = 0. 0+00Now, using Lemma VI-3-10, we prove the following theorem concerning problem

(EP).

Theorem VI-3-11. Assume that u(x) is continuous on the interval I(a.b) = (x:a < x < b} and that p(x) is continuously differentiable and positive on I(a,b).Then,(1) problem (EP) has infinitely many eigenvalues:

AI >A2>...>An >...,

(2)n

lira An = -oo,

(3) every eigenfunction Q,,(x) corresponding to the eigenvalue An has exactly n-1zeros on the open interval a < x < b.

Proof.

Assume without any loss of generality that 0 < a < r and 0 < 0 < r. Define9(x, A) by the initial-value problem

(VI.3.11) p(x) = 1 + (p(x)(u(x) - A) - 1)sin2(9), 0(a, A) = a,Tx-

and, then, define r(x, A) by

(VI.3.12) p(x)d. = (1 - p(x)(u(x) - A))rsin(9) cos(9), r(a,A) = 1.

Then,

(VI.3.13) y = i(x, A) = r(x, A) sin(9(x, A))

is a nontrivial solution of the differential equation

(VI.3.14)d

+ u(x)y = Ayds (P(x)±y)d.T

that satisfies the condition y(a)cos (a) - p(a)LY(a)sin (a) = 0. Note that from

(VI.3.11), (VI.3.12), and (VI.3.13), it follows that p(x) = r(x, A) cos(9)(x, A).A)

The eigenvalues of (EP) are determined by the condition 0(b,,\) = f3 ( mod sr). Since8(b, A) is strictly decreasing as A +oo and since 8(b, A) takes all positive values(cf. Lemma VI-3-10), the eigenvalues A. are determined by

0(b,,\.) = Q + (n - 1)a, n = 1,2,... .

Observe that

(I) ¢(x, A) = 0 if and only if 8(x, A) = 0 (mod a),

(II) 2 =p(x) >0if8=0(mod

a).

This implies that 4(x, A) has exactly k zeros on the open interval a < x < b if andonly if

kar < 0(b,,\) < (k + 1)7r.

Observe also that(n - 1)-,r < ,3 + (n - I)a < nn.

Hence, the eigenfunction 45(x, has n - I zeros on the open intervala<x<b. 0Observation VI-3-12. As shown earlier, if u(x) > m and p(x) n'rr for n = 1, 2, ... .

/(VI.3.15) an>m - P( n,)z

b - an = 1, 2,....

Observation VI-3-13. If u(x) p > 0 on I(a.b), determine 81(x, A)by the initial-value problem

p dx = 1 + (p(p sin2(81), 81(a, A) _ ?r.

Then, since u(x) - A p > 0 for x E T(a, b), -oo < A < +oo,and a < s, it follows from Lemma VI-3-8 that

0(c,,\) < Oi(c,A) for a < c < b and -00 < A < +oo.

Observe that v = p tan(81) satisfies the differential equation L = 1 + (1f -P

A) v2.

Hence,

tan(81(x, A)) = 1 tanP

(x - a) for a < µ.Pµ-

(x - a) = nrr for n = 1,2,... andNote that tan(91(x,A)) = 0 at AP2that 01(2, A) is strictly decreasing with respect to A if x > a. Furthermore,

7r < 91(x, p) <2

for a < x since tan(01(x, {e)) =x p a . Therefore,

na \291 (X, p (- I = (n + 1)ir for n = 1, 2, ... .xa

Hence,

P ( nTr )2)< (n + 1);r for n = 1, 2,...

x - a

This implies that

(VI.3.16) An+2 < p - p nrr 2

for n = 1,2,... .G-aIf Al and A2 are determined by

(VI.3.17) 01(b,A1) = Q and 01(b,A2) = 0 + it,

then Al < Al and A2 < A2. From (VI.3.15) and (VI.3.16) we derive the followingresult concerning distribution of eigenvalues of (EP).

Theorem VI-3-14. For n > 3, eigenvalues An satisfy the following estimates:

(i)

and

if

n2 +P(b- a)

2

< Iml if An -P Wan2

where in, p, P and p are real numbers such that

P > p(x) > p > 0 and p > u(x) > m for x E Z(a, b).

Remark VI-3-15. Lemma VI-3-8, Lemma VI-10, and Theorem VI-3-11 are alsofound in (CL, §§1 and 2 of Chapter 8] and [Hart, Chapter XI].

VI-4. Eigenfunction expansions

!LV + u(x)y and the vectorLet us define a differential operator L[y] = dx (p(z))space V(a, b) over the real number field R in the same way as in §VI-2. Also, asin §VI-2, define the inner product (f, g) for two real-valued continuous functions f

b

and g on Z(a, b) _ {x : a < x < b} by (f, g) = J It is known that the

operator L has the following property:

(f,L[g]) = (L[f],g) for f E V(a,b) and g E V(a,b)

(cf. Theorem VI-2-5). Define a norm of a continuous function on X(a, b) by 11f 11 =(f, f). Then, 1(f, g)f < Ilf II llgll. The following theorem is a basic result in the

theory of self-adjoint boundary-value problems.

Theorem VI-4-1. Let Al and A2 be two distinct eigenvalues of the problem

(EP) L[y] = Ay, y E V(a, b).

Let r11(x) and r}2(x) be eigenfunctions corresponding to Al and A2, respectively.Then, (rll,rh) = 0.

Proof.

Note that (r71,L[r12]) = A2(rh,g2) and (Ljrll1,rh) = Al(rl1,rn). This implies thatA2(r11, r12) = AI (r11, r12). Therefore, (r1i, r12) = 0 since Al 0 A2 0

It is known that problem (EP) has real eigenvalues

Al > A2 > ... > An > ... ( lien An = -co).n-++oo

Let r)1(x),-. .*2(x), , be the eigenfunctions corresponding to Al, A2,...such that 11,11 = 1 (n = 1, 2, ... ). Theorem VI-4-1 implies that (' h,''m) = 0 ifh # k. From these properties of the eigenfunctions rjn, we obtain

k k11f - >(f,vh)T1hII2 = Ilfll2 - E(f.rlh)2 > 0

h=1 h=1

and hence+oo

E(f,7,lh)2 < IIffl2 (the Bessel inequality)h=1

for any continuous function f on Z(a, b).In this section, we explain the generalized Fourier expansion of a function f (x)

in terms of the orthonormal sequence {rl (x) : n = 1,2,... }. As a preparation, weprove the following theorem.

4. EIGENFUNCTION EXPANSIONS 163

Theorem VI-4-2. Let µo not be an eigenvalue of (EP) and let f (x) be a contin-uous function on Z(a, b). Then, the series

(f, 1h)t1h(x)

h=1 Ah - Jb

is uniformly convergent on the interval Z(a, b).

Proof

Define a linear differential operator Go by £o[yl = G[yl - poy. Then, there existsGreen's function of the boundary-value problem

(BP) Go [y1 = f (x), y E V (a, b)

such that the unique solution of problem (BP) is given by

b

y(x) =JJ.

(cf. §VI-2; in particular, Observation VI-2-3). Define the operator G[f[ by

b

G(f 1 = (G(x, ), f) =J

G(x, Of WA

for any continuous function f(x) on 1(a, b). Since Go[g[f]l = f, we obtain

(I,G[.1) = (4[9[f)), 9191) = (9[f], 4191911) = (9[fl,9)

for any continuous functions f (z) and g(x) on Z(a, b). Also, since Go[rlh1 = c[nh1-

A01% = (Ah - l'o)llh, we obtain 9['7h] =Ahr1hUO . Hence,

(f, rlh)ghI

< Klz lz

(f, '1017h = K (f, ih)2,h=l, h=11

where K is a positive constant such that J!G(x, ) Il < K on 1(a, b). Finally,

12

the Bessel inequality implies that JimI

(f' nh)nh(x) l = 0 uniformly onli,ls-+oo h=l j

- 110,

T(a, b). 0Now, we claim that the limit of the uniformly convergent series of Theorem

VI-4-2 is equal to 9[f].

Theorem VI-4-3. For any continuous function f (x) on Z(a, b), the sum+00

(h,nh) is equal to 9[f].

Proof

Set 91 (f ] = 9 If, - 1: (f, TO 17h . Then, it suffices to showh=1=1

Ah - PO

(VIA. 1) lim IIGe[f111 = 0

for every continuous function f (x) on Z(a, b). We prove (VIA.1) in four steps.Without any loss of generality, we assume that Ah - uo < 0 for h > t . Also, notethat

(VI.4.2) (f,9e[9]) = (94f1,9)

for any real-valued continuous functions f and 9 on Z(a, b).Step 1. It can be shown easily that Gr[>jh] = chrjh, where

Ch =

for h = 1,2,... ,e,

for h > t.

Let y be an eigenvalue of 9t and let ¢(r) be an eigenfunction of 9t associated with

y. Then, ¢ E V(a,b), 11011 # 0, and y¢ = 9[41 - (0, 17017h . Also, (VI.4.2)h=1 An - go

implies that (0,9471h]) = (91'101,170. Hence, ch(¢, nh) = y(¢, nh) Consequently,(¢, rjh) = 0 if y 0 ch. Therefore, (¢, rjh) = 0 for h = 1, ... , I and y¢ = G[¢] ify 96 0. Hence, either y = O or y = ch (h > P).Step 2. In this step, we prove that

(VI.4.3) sup 1190JI1 = sup 1(911f1.f)I-uni=1 nfa=1

In fact, the inequality I(9e[f),f)1 5 11Ge[f]IIIIfII implies sup 1I9t[f111 >-ufu=1

sup 1(91[f],f)I Also, since (Ge[f ±9],f ±9) = (911f],f)+(9t[9],9)±2(9t[f],9),Of H=1we obtain

4(91[f],9) = (91 [f+9) ,f+9) - (9e[f-9],f-9)

< ( sup l(Ge[w], w)I (IIf + 9II2 + If - 9112)lH'°t=1

= 2 sup I(9e[wj,w)l (11f112+119112).N++ N=1

Suppose that 9t[f] 0 and IIfIII = 1, and set g = Gt[f]Then, IIGe[f]II -

IIGt[f]II'sup I (Gt[w], w)I. This shows that sup 119t[f]II sup I (Gt[f]+f )I Thus, the

IIwIl=1 11!11=1 I1I11=1

proof of (VI.4.3) is completed.Step 3. In this step, we prove that - sup I (Gt If], f) I is an eigenvalue of Gt. To

IlfI1=1do this, set c = sup I(Ge[f], f)I. Note first that c > I(Gt[ne+I],ne+i)I = Ice+II >

Ill 11=1

0. Also, note that c = sup (911f], f) or -c = inf (Gt[f], f). Suppose thatIif11=1 [if U=1

c = sup (9t If], f ). Then, there exists a sequence f (j = 1, 2.... ) of continuous11111=1

functions on T(a, b) such that lim (Gt[ fi], fi) = c and IIf, II = 1(j = 1, 2, . . . )

Since {Gt[ fj] : j = 1,2.... } is bounded and equicontinuous on T(a, b), assumewithout any loss of generality that lim Gt[ f)] = g E G(a, b) uniformly on T(a, b).

-.+00Let us look at

(VI-4.4) IIGt[fjI - cf, II2 = I1Gt[fj]II2 + c2 - 2c(Gt[f,], f,) <- 2c2 - 2c(Ge[fjl , fi).

This implies that lim 11 9t [f, I - cfj II = 0 and, hence, Um 119 - c f j II = 0. Thus,)+oo j- +00Gt[g] = cg. Also, from (VI.4.4), it follows that IIgII2 = c2 > 0. Therefore, c must bean eigenvalue of Gt. However, since all eigenvalues of Gt are nonpositive, this is acontradiction. Therefore, -c = inf (91 [fit f ). We can now prove in a similar way

Illil=lthat

-c = ct+1 =1

),ttl - µ0

Step 4. Since sup IIGt[f]II = 1 , we conclude that lien sup I191[f]II = 011111=1 !Lo - t+1 t-'+0°1!11!=l

and the proof of (VI.4.1) is completed.The following theorem is the basic result concerning eigenfunction expansion.

+00

Theorem VI-4-4. For every f E V(a, b), the series f = 1: (f, nh)nh converges toh=1

f uniformly on T(a, b).

Proof.

Set g = .Co[f], then f = g[g]. Therefore,

+00 +C0f = (9+17 h)17h = 1: (GO[J]+nh)nh

h=1 Ah - {b h=1 Ah - 110

+00(1 £ofrlh])r)h

+00

_ ah _ _ >(f, rlh)rlhh=1 h=1

For every continuous function f on 1(a, b) and a positive number e, there existsan f, E V(a, b) such that 11f - fe II <_ e. Since

t

/

t t

f - (f, 77h)rh = f - ff + fe - 1: (ff,17h)nh + 1: (ff - f, 17077h,h=1 h=1 h=1

C

we obtain lim I f - (f , 7nh )T7h l = 0. Therefore, f = 0 if (f, Tjh) = 0 for h =t-.+00

h=11, 2,.... This proves the following theorem.

Theorem VI-4-5. If a continuous function f on 1(a, b) satisfies the conditionU, 17h) = 0 f o r h = 1, 2, ... , then f is identically equal to zero on the interval1(a, b).

Remark VI-4-6. Also, we have

+00

j11112 = >(f, rth)2 (the Parseval equality).h=1

Furthermore, this, in turn, implies that if f (x) and g(x) are continuous on 1(a, b),then

h=1

Example VI-4-7. Using the eigenvalue problem = Ay, y'(0) = 0, y(a) = 0,dX2

we construct an orthogonal sequence {cos(nx) : n = 0, 1, 2.... } of eigenfunctions.00

2 Ean cos(nx) of a function f (x), whereThis yields the Fourier cosine series a-° +n=1

r*an = 2

J f(x) cos(nx)ds. By virtue of Theorem V1-4-4, this series convergesa ouniformly to f (x) on 1(0, 7r) if f (x) is twice continuously differentiable on 1(0, zr)and f'(0) = 0 and f'(Tr) = 0.

Example VI-4-8. Using the eigenvalue problem d = Ay, y(O) = 0, y(7r)

= 0, we construct an orthogonal sequence {sin(nx) : n = 1,2,...l of eigenfunc-00

tions. This yields the Fourier sine series >bn sin(nx) of a function f (x), wheren=1

/bn = f f (x) sin(nx)ds. By virtue of Theorem VI-4-4, this series convergesa ouniformly to f (x) on 1(0, ir) if f (x) is twice continuously differentiable on 1(0, ir)and f (0) = 0 and f (a) = 0.Example VI-4-9. If f (x) is twice continuously differentiable on Z(-ir, ir), f (-vr) _

f (ir), and f'(1r) = f'(-7r), set fe(x) = 2 if (x)+f (-x)] and fo(x) [f (x)-f (-x))

Then, fe satisfies the conditions of Example VI-4-7, while fo(x) satisfies the condi-tions of Example VI-4-8. Thus, we obtain the Fourier series

cc

ao + [an cos(nx) + bn sin(nx)],n=1

r

J* f (x) oos(nx)ds and bn = ! J f (x) sin(nx)ds. The Fourierwhere an =

7r

series converges uniformly to f (x) on Z(-7r, 7r).

Remark VI-4-10. The Fourier series of Example VI-4-9 can be constructed also

from the eigenvalue problem L2 = Ay, y(-7r) = y(7r), y'(-7r) = y'(ir). Including

this eigenvalue problem, more general cases are systematically explained in [CL,Chapters 7 and 11]. For the uniform convergence of the Fourier series, it is notnecessary to assume that f (x) be twice continuously differentiable. For example,it suffices to assume that f (x) is continuous and f'(x) is piecewise continuous on1(-7r, 7r), and f (7r) = f(-7r). For those informations, see [Z].

Remark VI-4-11. The results in §§V1-2, VI-3, and VI-4 can be extended to thecase when p(x) is continuous on 1(a, b), p(x) > 0 on a < x < b, and p(a)p(b) =0 under some suitable assumptions and with some suitable boundary conditions.Although we do not go into these cases in this book, the following example illustratessuch a case.

Example VI-4-12. Let us consider the boundary-value problem

d f_dy . _

(VI.4.5)y(x) is bounded in a neighborhood of x = 0, and y(l) = 0,

where u(x) and f (x) are real-valued and continuous on the interval 1(0, 1).

Step 1. Consider the linear homogeneous equation

(VI.4.6)d

(X2idv)+ u(x)v = 0

on the interval 1(0,1). Since 1 and logs form a fundamental set of solutions of

the differential equation ± (xdx = 0, a general solution of (VI.4.6) can be

constructed by solving the intergral)equation

v(x) = c1 + C2 109X +1

(logs)10 u(C)V(4)dC,0 0

where c1 and c2 are arbitrary constants. In this way, we find two solutions O(x) and-i(x) of (VI.4.6) such that O(x) and 4'(x) are continuous on 1(0,1), lim O(x) = 1,

and lien x¢'(x) = 0, whereas tI'(z) and tY(x) are continuous for 0 < x < 1,r_0+

lim (O(z) - logs) = 0, and urn (xo'(x) - 1) = 0. In general, this step of analy-0+ z-+0+

sis is very important. The behavior of solutions at x = 0 determines the nature ofeigenvalues of the given problem. Denote also by p(x) the unique soltion of (VI.4.6)such that p(1)=0andp'(1)=1.

Step 2. Assuming that 0(l) # 0, set

¢(x AO f 0< < <1or x e ,

G(x,WW¢(F)p(x) for

where W(x) is the Wronskian of 10,p) and xW(x) = 0(1). Then, G(x,t) isGreen's function of problem (VI.4.5). The unique solution of (VI.4.5) is given

i

by y(x) = 1

G(x, )2dxd{ < +oo.Jo 0

Step 3. It is easy to prove the self-adjointness of problem (VI.4.5). Hence, alleigenvalues are real, and the orthogonality of eigenfunctions follows. By virtue of(VI.4.7), eigenfunction expansions can be derived in the exactly same way as in§VI-4.

Step 4. We can also derive Theorem VI-3-11 in the exactly same way as in §VI-3.To do this, consider the differential equation

(VI.4.8) 0-0 + u(x)y = )y.

For every value of A, there exists a unique solution O(x, A) such that lim O(x, A) = 1xo+

and lim x¢'(x, A) = 0 (cf. Step 1). It is easy to see that p(x, AT and x*'(x, A)X--o+

are continuous in (x, A) for 0 < x r(( )) =

4(x,A) B(0, A) = . Using the same method as in

§VI-3, we can verify that ¢(1,A) is strictly decreasing and takes all values be-tween 0 and -oo. Eigenvalues of problem (VI.4.8) are determined by the equations9(1, A) = mar (m = 1, 2.... ).

Theorem VI-3-14 cannot be extended to the present case, since there is no posi-tive lower bound of x on 1(0,1).

VI-5. Jost solutions

So far, we have studied boundary-value problems on a bounded interval on thereal line R. Hereafter, we consider the scattering problem, which is a problem onthe entire real line. To explain the essential part of the problem, let us consider

a homogeneous linear differential equation Ly + ((I - u(x))y = 0, where ( is a

complex parameter and u(x) is a real-valued continuous function of x such thatu(x) = 0 for all sufficiently large values of ext. It is evident that e_ttx and e'(' aretwo linearly independent solutions of this equation if 1xI is large. Therefore, if a

5. JOST SOLUTIONS 169

positive number M is sufficiently large, the solution y = e-'(= for x < -M becomesa linear combination a(()e-'t=+b(()e't= of two solutions a-K= and eK' for x > M.The main problems are (i) the properties of a(() and b(() as functions of ( and (ii)construction of u(x) for given data {a((), b(()). Keeping this introduction in mind,let us consider a differential operator

(VI.5.1) G = - D2 + u(x),

where D =

ajand u(x) is real-valued and belongs to C°°(-oo, +oo) such that

+o°(VI.5.2)

J(1 + jxj)lu(x)jdx < +oc.

We study the differential equation

(VI.5.3) Gy = (2y,where (is a complex parameter. First, we construct two basic solutions which arecalled the Jost solutions of (VI.5.3).

Theorem VI-5-1. There exist two solutions f+(x, () and f_ (x, () of (VI.5.S) suchthat

(i) ft are continuous for

(VI.5.4) -oc < x < +oo, 0,

(ii) ft are analytic in ( for £ (() > 0,(iii)

-°jf+(x, () - e'`=I <_ C(x)P(x)e:

1 + 1(I

f- (x, () - e-K=I < C(-x)P(x)1 +e"x

I(1

for (VI.5.4),, where q = Q3'((). C(x) is non-negative and nonincmasing,t00

P(x) = I (1 + IrI)Iu(r)Idr, and O(x) =J

(1 + Irj)ju(r)jdr.o0

Proof.

Step 1. Construction of f+: Solve the integral equation

y(x, () = e"zsin(((s - r)) u(r)y(r,

()dr

by setting

yo(x,() = eK=,

ym+1(x, ()+00 sin((((- r))u(r)ym(r, ()dr (m > 0),

x

and00

f+(x, () = E Ym(x, ()m-o

Step 2. Proof of (i) and (ii): To prove (i) and (ii), it suffices to prove that if

e(x) - ( 21x1 + 2 for x < 0,2 for x > 0,

then

(VI.5.6) Iym(x> }l < mj(C(x)P(x))me-1, m= 0,1,2....

for (V1.5.4).Inequality (VI.5.6) is true for m = 0. Assume that (VI.5.6) is true for an m;

then,

Iym+i(x,C)J < mir+ Isin(((x - r)) I

Z

Note that

sin(((x - r)) a (1 - e-2i(z-7)() = eft(Z-r)

z-re-2it=dz.

2i( 1Hence,

sln(((x - r)) I < e-q(z-T)(r - x)C

for r1 > 0.

Therefore,

Since

I ym+1(x,C)I <(C(x)me-nx f

+00

(r - x)tu(r)IP(r)mdr.z

j +00 +00 (1+ I rI )I u(r)I p(r)mdr(T - x)Iu(T)IP(r)mdr < C(x)f.

m + 1 C(x)P(x)m+

we obtainI ym+1(x, ()I <

(m 1)!(C(x)P(x))m+ie-nZ

Remark VI-5-2. At the last estimate, the following argument was used:(a) r-x<r<1+rforx>O,(b)

+00

TI u(r)I P(T)mdTJ

= ( +J

< 2Ixl1 I u(r){P(r)mdr + 2fzz z z z

for x < 0.

5. JOST SOLUTIONS 171

Step 3. Proof of (iii): First, the estimate

(VI.5.7)

'o II f+(x, 0- I <- E n,, (O(x)v(x))me-''z

m=100

< C(x)p(x) I li (C(x)p(x))m-1 I e-"z

m=1

follows from (VI.5.6). However, this estimate is not enough to prove (iii). So, letus derive another estimate for large I(I. To do this, we shall prove that

(VI.5.8) I ym(x, ()I rn!(k)

m e-?=,m=0,1,2....

or(x) = f"

iu(r)jdr.x

Inequality (VI.5.8) is true for m = 0. Assume that (VI.5.8) is true for an m. Then,

Iym+1(x, ()1 -M! I(I 1 J

+aoI sin(((x - r))IIu(r)I e-"'a(r)mdT.

z

Note that

sin(((x - r))e-+,zl = e 2 11 - e-2it(z-7)1 e-'

Hence,

if 3(() 2! 0.

1 e-'7= +00 1 m+1

Iym+1(x,()I m! KIm+1I iu(r)io(r)mdr = (m+ 1)!(!!-(X)

1(i}a-' .

J.

This establishes (VI.5.8). Therefore,

(VI.5.9) If+(x, () - ei(zl < v(x)= 1

OWC-1-))M-,l e-qz.(i M! I

The proof of Theorem VI-5-1 for f+ (x, () can be completed by using (VI.5.7) and(VI.5.9).

For f-(x,(), change x by -x to derive

(VI.5.10) -LY + u(-x)y = (2y.

The solution f+ for (VI.5.10) is the solution f- for (VI.5.3). 0

Remark VI-5-3. The solutions f± are uniquely determined by the conditionsgiven in Theorem VI-5-1. Also,

(VI.5.11)d

(x,j+00 mi(((x

- r))u(r)f+(r, ()dr

and

Id'

(x,() - i(e{C2 + f cos(((x - r))u(r)eK=dr

(VI.5.12)-

< C(x)p(x)a(x) 1+I(Ifor (VI.5.4).

Remark VI-5-4. If u(x) = 0 identically on the real line, then f f (x, () = et't'.If u(x) = 0 for Ixl _> M for some positive constant M, then f+(x, () = e'(= forx > M, while f- (x, () = e-K= for x < -M.

VI-6. Scattering dataFor real (, 1+ (x, () = f+ (x, -(), where f denotes the conjugate complex of f .

Both f+ and f+ are solutions of Gy = (2y and

I f+(x, f) - e Z I < C(x)P(x)1 + ICI

(x)(x, + i(e-+

r+r))u(r)e(rdrl C() +

16

dldx

for -oo < x < +oo and -oo < f < +oo. Let W(f,gj denotes the Vlwronskian of{f,g}. Then,

W If+, f+l - I f+(x, () f+, (x, () J

-2i( (-oo < < +oo).

This implies that I f+, f+} is linearly independent if 36 0. This, in turn, impliesthat the solution f_ is a linear combination of I f+, 1+}. Set

(VI.6.1) f-(x,() = a(()f+(x,() + b(f)f+(x,()It is easy to see that

i i f+(x+ f-(x,()(VI.6.2) a(() =

and

(VI.6.3) b(() = Wif+,f-j =I f+(x,() f .(x,()

I.

The function a(() is analytic for %() > 0 and

(VI.6.4) a(() = 1 + 0((-1) as (-' 00 on `3(() > 0,

whereas the function b(() is defined and continuous on -oo < ( < +oo. Further-more, b(() = O(Ifl-1) as I(I ---. +oo. To simplify the situation, we introduce thefollowing assumption.

6. SCATTERING DATA

Assumption VI-6-1. We assume that

Iu(x)I < Ae-kj=j (-oo < x < +oo)

for some positive numbers A and k.

The boundary-value problem

(VI.6.5) Gy = Ay, y E L2(-oo,+oo)

173

is self-adjoint, where L2(-00, +oo) denotes the set of all complex-valued func-r+00

tions f (x) satisfying the condition J `f (x) I2dx < +oo. The self-adjointness

of problem (VI.6.5) can be proved by using an inner product in the vector spaceL2(-oo, +oo) in the same way as the proof of Theorem VI-3-4. Therefore, eigenval-ues of problem (VI.6.5) are real. Furthermore, all eigenvalues are negative. In fact,if t 96 0 is real, then A = £2 > 0 and the general solution ci f+ + c2f+ is asymptoti-cally equal to cle`4=+c2e-'4= as x --. +oc. If t = 0, then f+r(x,0) is asymptotically

equal to 1 as x +oc. Moreover, another solution f+J z is asymptoticallyf+

equal to x as x +oc. Therefore, all eigenvalues are A = (ir7)2 < 0. Furthermore,all eigenvalues of (VI.6.5) are determined by

(VI.6.6) a(iry) = 0.

This implies that all zeros of a(() for 3(() > 0 are purely imaginary.

Under Assumption VI-6-1, ft(x, () are analytic for 9'(() > - 2. Hence, a(()has only a finite number of zeros for 3(() > 0 (cf. (VI.6.4)). Let S = irlj 0 =1, 2, ... , N) be the zeros of a(() for (() > 0. Then, f+(x, ir73) are real-valued andf+(x, ins) E L2(-oo, +oo). Set

(VI.6.7)

c) _ r+ao 1 (j = 1, 2, ... , N),f+(x,n,)2dr

(-oc < { < +oo).

Observation VI-6-2. Every eigenvalue of (VI.6.5) is simple, i.e.,

da(C) ;f 0 if a(() = 0.4Proof.

From a(t;) = 0, it follows that

`ia(() i d W[ f+, f-[ - 2 W [f+, f-I = r (u'[f+(, f-] + N'[f+, f-<l),d( 2S d( 2 2Z

where ft denotes 4f . Also, two relations

`f+S + of+C = (2f+C + uf+

imply that

ddw(f+(, f-1 = 2-f+f-

and

andd

W(f+, f-cl = -2f+f_

Since a(() = 0, there exists a constant d(() such that f_(x,() = d(()f+(x,C).

Therefore,

W(f+t, f-) =-2S+00

f+f-dr and W(f+, f-tl = -2Cf f+f-dr.+o0

Thus, we obtain

da(()

d(-i +oo

f+f-dr = -id(()f+oo f+dr = -id(()

# 0.00 oo Ci

Observation VI-6-3. The quantities a(() and b({) satisfy the following relation:

(VI.6.8) Ia(t)12 - Ib(E)12 = 1 for - oo < +00.

Proof.

From f_(x, -{), it follows that

aW = a(-4), b(() = b(-(), and W(f-,f-l = 2i,.

Therefore, (VI.6.8) follows from

f- = af+ + b(af- - bf_) = a(f+ + bf-) - Ib12f__ aaf- - Ibl2f_ = {1a12 - Ibl2}f .

Observation VI-6-4. The formula

(VI.6.9) a(() _ (- iq, exp [-L f+°° log(1 -i1h 2ri ao (- S J

for %() > 0 shows that the quantities q, and r({) determine a(().

Proof

Set f (() = a(()j Then,

7. REFLECTIONLESS POTENTIALS

(i) f (() is analytic for $(() > -2 and (94 0,

175

(ii) ((f(() - 1) is bounded for 3(() > -2,(iii) f (() & 0 for Qr(() >- 0 and (54 0.

Set also F(() = log(f (()). Then, O(log(()) near = 0 and F(() _0((-1) as ( -, 00 on 3(() > 0. Observe that f(() - 1 = O((-'). From thisobservation, it follows without any complication that

1 /'}Oc 2log if (01 for £(() > 0F(() - Trif_. - ((cf. Exercise VI-18). Now, (VI.6.9) follows from If(t) = Ia(()j and

loBja(()f2 = log(Ta(f)22/

Definition V1-6-5. The set {r((), (n,,n2,... ,nN), (cl,c2,... ,CN)} is called thescattering data associated with the potential u(x).

VI-7. Reflectionless potentials

The function is called the reflection coefficient. If this coefficient is zero, thepotential u(x) becomes a function of simple form. Let us look into this situation.

Observation VI-7-1. If r(() = 0, then b(() = 0. This means that f_(x,t:) =a(t) f+(x, £) = a(t) f+(x, -t) (cf. (§VI-6) ). Therefore, using the relation

f+(x,0) =f-(x, -C)

for 3(() !5 0,a(-()we can extend f+ for all ( as a meromorphic function in (. Furthermore, if irlj (y =1, 2,... , N) are zeros of a(() in 3'(() > 0, then -inj (j = 1, 2,... , N) are simplepoles of f+(x, () in ( and

Ftesidueoff+at -inj = - f-(x,ir,) = C)f-(x,irli) -= -icj f+(x, ins).a<(inj) id(irtj)

Observation VI-7-2. Set gJ (x) = e-" cj f+(x, iii ). Then, from the fact thate_,<=f+(x, () - I as I(I - +oo, it follows that

(VI.7.1) f+(x,0) = e'S= 1 - i 9,(x)

=1 (+ ins

Setting ( = int in (VI.7.1), we obtain

e2ne= N 1

(VI.7.2) 9t(x) + nt+n'gj(x) = 1 (t=1,2,...,N).C-1 J=1

Observation VI-7-3. If we set F(x,() = e-'t= f+(x, (), then -F" - 2i(F'+uF =` 91 (x) , it follows that'0. Since F(x, ) = 1 - i-1 +

(VI.7.3)N

u(x) = 2E gj,(x).

.7_1

Observation VI-7-4. Let us solve (VI.7.2) for the g,(x). First, set

hi = gie'''= (j = 1,2,... ,N).

Then, (VI.7.2) becomes

(VI.7.4) ht + c h = cte-''i: (tJ=1 rJt + I?j

Write the coefficient matrix of (VI.7.4) in the form IN + C(x), where IN is theN x N identity matrix. Since

N +op N 2

E 717t E 7 e-n,,= dx)t°1 'b + 171

=L i p

and 77, (j = 1, 2,... , N) are distinct, the matrix IN + C(x) is invertible for -oo <x < +oo. Set L(x) = det(IN + C(x)). Then, manipulating with Cramer's rule, we

can write g, (j = 1, 2,... , N) anddLda

in the following forms:

(VI.7.5)

and

(VI.7.6)dL(x)

dx

N

J=1

Thus finally, from (VI.7.3), it follows that

or

(j = 1, 2,... , N)9j (x) = A(X)

N

u(x) =2E g,.;(x)

= -2 a(x)

-2z(log(0(x))}.(VI.7.7) u(x) =

7. REFLECTIONLESS POTENTIALS 177

This implies that u(x) is a rational function of exponential functions and satisfiesAssumption VI-6-1 of §VI-6. To see this, the following remarks are also useful:

(a) we have the identityN N

gg(x) 2 ?1 e2n'x9s(x)2,j=1 j=1 i

(b) g j (-oo) = = limo g . (x) ( j = 1, 2, ... , N) exist and

N 1

E g,(-oo) = 11=1171+n,

(e = 1, 2,... , N).

To show (a), derive

e2nez N 2cr 9t(x) +

1h + ng, (x) = - cl[ e2arxgt(x) ((= 1, 2, ... , N)

=1

from (VI.7.2). Multiplying both sides by ge, adding them up over f, and interchang-ing the orders of summation, we obtain (a). To show (b), calculate the inverse ofthe coefficient matrix of (VI.7.2). Using Cramer's rule on (VI.7.4), it can be shownthat he(x) -, 0 exponentially as x -+ oo. Also, (b) implies that e2°,, xg,(x)2 is expo-

nentially small as x -oo. Therefore, (a) implies that u(x) satisfies AssumptionVI-6.1.

gz1

Example VI-7-5. In the case N = 1, if we determine g(x) bye2+

/g(x) _

g

1, then u(x) = 2g'(x). Since g(x) = c' we obtain u(x) = -- e2'rzg(x)2.C

QxAlso,

2ce+7x + FF17e-nx g(x) = 1 implies that g(x) = 2 e-n=

(211

xsech(11(x + p)), where p =In

` c ,211

and, hence,

(VI.7.8) u(x) = -2112 sech2(g(x + p)).

4

In particular, formula (VI.7.8) yields u(x) = -8 sech (2 (x + In (s) )I in the

case when N = 1, n1 = 2, and c1 = 5. On the other hand, a\straightforward calcu-

lation using formula (VI.7.7) yields u(x) _ (5+4)2. Also, u(s) = -2sech(x)

is the reflectionless potential corresponding to the data r1= 1 and c = 2.

Example VI-7-6. If N = 2, ill = 2, rh = 3, c1 = 5, and c2 = 2, we obtain

40e42(16 + 135e2-- + 600e6x + 2160e10s + 3600e12=)U(X)

(1 + 20e4x + 75e6. + 60ebox)2

by calculating (VI.7.7).

Remark VI-7-7. If the reflection coefficient r(t) = 0 and if there is no eigenvaluefor the problem

d2y-2

+ u(x)y = Ay, y E L2(-oo, +oo),

then u(x) = 0 identically for -oo < x < +oo. In fact, f+(x,() is analytic ineverywhere in the (-plane. Furthermore, l e-`<= f+(x, () -11 -+ 0 as - oo. Hence,f+(x,() = e'(=. This shows that u(x) = 0.

Remark VI-7-8. If u(x) 0 but u(x) = 0 for }xI _> M, where M is a positivenumber, then the reflection coefficient r(S) # 0. In fact, if r(C) = 0, then u(x)is analytic for -oo < x < +oo. Hence, u(x) = 0. This is a contradiction. Thisremark reveals that the problem posed at the beginning of §VI-5 is not as simpleas it looks.

VI-8. Construction of a potential for given dataIf scattering data {0, (71,... 7x), (cl,... ,cN)} are given, formula (VI.7.3)

gives the corresponding reflectionless potential u(x) as it is shown in Examples VI-7-5 and VI-7-6. However, in these two examples, we assumed implicitly that sucha potential exist. Since the existence of u(x) has not been shown yet, it must beproved. To do this, for given data

7i>0, 72>0, ..., 7N>0 and c1>0, c2>0, ..., cN>0,

define g, (j = 1, 2,... , N) by (VI.7.2) and define f+ and u by (VI.7.1) and (VI.7.3).Hereafter, the first thing that we have to do is to show that y = f+ is one of theJost solutions of Ly = (2y.

Observation VI-8-1. Let gi (t = 1, 2,... , N) be determined by (VI.7.2) and u(z)be defined by (VI.7.3). Then,

(VI.8.1) gi + 27r9t - ugt = 0 (=1,2,... , N).

Proof

Write (VI.7.2) in the form

N

(VI.8.2) E ae, (x)9, (x) = 1 (t = 1, 2,... , N)3=1

and differentiate both sides with reaspect to z. Then,

(VI.8.3) Eat, (x)9f (x) + 2t, 9j(x) = 0 (t = 1, 2,... , N)}=1

and, hence,

N N

Eata(x)9j'(x) + 27t 11 - 1 g, (x) = 0 (t = 1,2,... ,N).1=1 i=1 7e+7)

8. CONSTRUCTION OF A POTENTIAL FOR A GIVEN DATA 179

From (VI.8.2), it follows that

N NE ajj(x){g,(x) + 2gegj(x)} - 237tE g, (x) = 0

j=1 j=1 ge +T7j

(e = 1, 2, ... , N)

or

N N N

Eatj(x){gf(x) + 2gtgj(x)} - 2>9,(x)+ 2E g, 92(x) = 03=1 j=1 j=1

711+17.,

(e = 1, 2, ... , N).

Differentiating again, we obtain

N 2o[sE ata (x){gj"(x) + 2r7eg, (x) } + 2ge (9e(x) + 2gtgt(x)) - u(x),=1

N+ 2E gj 9'(x) = 0 (e = 1,2,... N)

3=1 ge+g,

or

N N

Eat,j(x){g,"(x) + 217eg,(x)) - Eaea(x)u(x)9J(x),=1 f=1

N

+ aea(x)2%g,'(x) + 2171 C27jte2nas

) = ... ,9e(x)0 (e = 1,2,N).ct

,=1

Thus, we derive

N

F'ae,,(x){9,(x) + 217,g,(x) - u(x)g,(x)}j=1

N

2171 E atj (x)9, (x) + 217t ct 91(x) = 0 (t = 1, 2, ... , N),,=1

and (VI.8.3) implies that

N

F'at,1(x){9;'(x) + 2s7,g,,(x) - u(x)g,(x)} = 0 (e= 1,2,... ,N).,=1

Therefore, (VI.8.1) follows.

Observation VI-8-2. If we further define f+ by (VI.7.1), then £f+ = (2f+.

Proof.

(e-"=f+)11 + u(e_" f+)

Observation VI-8-3. Since

N-i g1' + 2i(g - ttgJ -+ i

29J

J=1 ( ,=1g;,

+ 2rligl' - ugJ = 0.-i N

Ej=1 (+ tnJ

N xen,"

f+(x,it7t) = e-fez 1 - g., (x) _ -gt(x) E L2(-oo,+oo),J=1 711+17.; ct

N numbers -1l,2 (j = 1, 2,... , N) are eigenvalues.N

Observation VI-8-4. Set a (cf. (VI.6.9) with r 0) andJ=1

N

f-(x,() = a(()f+(x, -() =a(S)e-Cz 1 + iE g3 (x)

J=1(-ig,

(cf. Observation VI-7-1). Then, £f_ _ (2f_ . Furthermore, since

E gJ (-oo) = 1 (e = 1, 2, ... , N)J=1171+17J

(cf. (VI.7.2)), we obtain

(VI.8.4) a(() = 1 - i1 ____ )

j=1(+nJIn fact, this follows from the fact that both sides of (VI.8.4) are rational functionsin ( with the same zeros, the same poles, and the same limits as ( , oo.

Observation VI-8-5. The functions f±(.x, () are the Jost solutions of Ly = (2y.

Proof.

Note first that lim f+(x. ()e-'t= = 1 (cf. (VI.7.1)). Also, we haves+oo

lim f+(x,()e-`<x = 1 - gg(-o) = a(() (cf. (VI-7.1) and (VI.8.4)).i=1

This implies that

lim f-(x,()e'S= = a(() lim f+(x,-()e"x = a(()a(-() = 1. Ot--oo x---oo

9. DIFF. EQS. SATISFIED BY REFLECTIONLESS POTENTIALS 181

Observation VI-8-6. Note that f_ (x, ) = a(() f+(x, -{) = f+(x, ) for ( =C real. This means that 0. Hence, r(C) = 0. Thus, we conclude that u(x) isreflectionless.

Remark VI-8-7. For the general r((), the potential u(x) can be constructed bysolving the integral equation of Gel'fand-Levitan:

K(x, r) + F(x + r) ++00

F(x + r + s)K(x, s)ds = 0,+0

where

F(x) = 1J + 2cL` cle-2n,=R oo J=1

Find K(x, r) by this equation. Then, the potential is given by

u(x) 8x(x,0).

If we set r({) = 0 in this integral equation, we can derive (VI.7.2) and (VI.7.3).Details are left to the reader as exercises (cf. (Ge1LJ ).

V1-9. Differential equations satisfied by reflectionless potentials

Suppose that u(x) is a reflectionless potential whose associated scattering dataare given by 0, (r11, r12, ... flN ), (cl, c2,... , cN) },where 0 < ril < ]2 <

< rily. It was proven that one of the Jost solutions is given by

N

f+(T,() = e`t= 1 - i 9j (T)j_1 (+

Furthermore,

f+(x, -() = e`= 1 -N

g1 (r)

-( + trj7

is also a solution of

dX2 - (u(x) - A)y = 0, where J = (2.

Therefore,

(VI.9.1)N

P(x, A) = 11 ! 1+11- f+(T,C)f+(x, -()3=1 A )


3

(E)2 dz3

+ 2(u(x) - A) dP + ddx)P = 0.

182

Let

VI. BOUNDARY-VALUE PROBLEMS

+oo

(S) P = E an (PO 3 0)

n=0

be a formal solution in powers of A-1 of differential equation (E). Then,

0 = P [-- + 2(u(x) - A)P + u'(x)P

P2, + (4)2 + (u(x)

- A)P2J

and, hence,

A(P2)' = I - PZ it + (4)2 + u(x)P2 }, .

Therefore,(i) the coefficients pn can be determined successively by

po = ca, where co is a constant,n n n

pip.-t -nPtPn+1-t

r2C Pn+l = 2 E PtFri-1 + 4E Ptpn-t + uEt=o t=o c=o 1=1

(n = 0,1,2,...),

(ii) the coefficients pn are polynomials in u and its derivatives with constantcoefficients,

(iii) the formal power series (S) is uniquely determined by the condition P = 1 foru = 0,

+00lv f we denote the unique P of by G u, A G"( u) Go = 1), then the() i q (iii) Y ( ) =

An(

n=0general formal solution of (S) is given by P(u, A) = P(0, A)G(u, A).

Example VI-9-1. A straight forward calculation yields

u 3u2 - u" 10u3 - lOuu" - 5(u')2 + u(4)Gl = Z' G2 = 8 G3 = 32

Observe that P(x, A) of (VIA 1) is a polynomial in of degree N. Therefore,

the potential u(x) satisfies a differential equation

(VI.9.2) GN+1(u) + a1GN(u) + a2GN-1(u) + ... + QNG1(u) = 0

for some suitable constants a,, a2, ... , 0N. More precisely speaking, it is shownabove that P(u, A) = P(0, A)G(u, A). Compute the coefficients of A-(N+1) on bothsides of this identity. In fact,

N N

u(x) = 2 g}(x) 4 1` 171e2,7ixgj(x)2

C.,i=1 1=1

10. PERIODIC POTENTIALS 183

implies that u = 0 if and only if gl = 0 (1 = 1, 2, - , N). This, in turn, impliesthat

and

P(0, A) = 11 C1 +A

G(u, A) = f+(x, C)f+(x, where

Hence, ao = 1 and

1 +

N N 2E =+

3=1 =11

A=C2.

Example VI-9-2. The function u(x) = -2q2 sech2(17(x /+ p)) is a reflectionless

potential corresponding to the data {r7, c}, where p = 2-

In 1c) (cf. Example VI-

7-5). In this case, G2(u)+172G1(u) = 0. Also, G3(u)+(n1+7)2 2(u)+rI14Gi(u) _0 in the case when N = 2 and {r(C) = 0, (i71, n2), (cl, c2)} are the scattering data.In particular, G3 (U) + 13G2(u) +36G1(u) = 0 if i1 = 2 and '12 = 3.

Remark VI-9-3. The materials of §§VI-5-VI-9 are also found in (TDJ.

VI-10. Periodic potentials

In this section, we consider the differential equation

(VI.10.1) 2 + (A - u(x))y = 0

under the assumption that(I) u(x) is continuous for -co < x < +oo,

(II) u(x) is periodic of period 1, i.e., u(x + e) = u(x) for -oo < x < +oo.The period a is a positive number and A is a real parameter. Denote by 01(x, A)and 02(x, A) the two linearly independent solutions of (VI.10.1) such that

(VI.10.2) 01(0,A) = 1, 1(o A) = 0, 02 (0, A) = 0, d2(0,A) = 1.

Set01(e, A) 02(t, A)

.D(A)= O1 (t"\) (e A)

The two eigenvalues Z1(A) and Z2 (A) of the matrix are the multipliers of theperiodic system

(VI.10.3)dxd

[Y2J = [u(--O-A 0, [y21

(cf. Definition IV-4-5). It is easy to show that det [1(A)] = 1 for -oo < A <+oo. Hence, the two multipliers Z1(A) and Z2(A) are determined by the equationZ2 - f (,\)Z + 1 = 0, where

(VI.10.4) f(A) = 2(t,A)

Note that f (A) is continuous for -oo < A < oo (cf. Theorem 11-1-2). We derivefirst the following conclusion.

Lemma VI-10-1. The two multipliers Z1(A) and Z2(A) of system (VI.10.3) are

2 lf(,\) + f (A)2 - 4, ,

2 l f (A) - f (A)2 - 4, ,

where f (A) is given by M. 10.4). Therefore,(i) if I f (A) I > 2, two multipliers are real and distinct, i.e..

ZI(A) > 1, 0 < Z2(A) < 1 if f(A) > 2,

-1 <Z1(A) < 0, Z2(A) < -1 if f(A) < -2,

(ii) if I f (A) I < 2, two multipliers are complex and distinct, and

I Zt (A) I= 1 and I Z2(A) 1 = 1,

(iii) if f (A) = 2, then Z1(A) = Z2(A) = 1.(iv) if f(A) _ -2, then Z1(A) = Z2(A) = -1.

Observation VI-10-2. If f (A) = 2, let ICCl2

Jbe an eigenvector of 4;(A) associated

with the eigenvalue 1. This means that cl and c2 are two real numbers not bothzero and that

c141(t, A) + C202(1, A) = c1, Cl d ' (t, A) + c2 d-i (t, A) = C2 -

Set 6(x,,\) = cl o1(x, A) + c202(x, A). Then, d(x, A) is a nontrivial solution of(VI.10.1) such that

.0(t, A) = 0(0,A) and LO (t, A) = (0, A).

This implies that O(x, A) is a nontrivial periodic solution of (VI.10.1) of period t.

Observation VI-10-3. If f (A) = -2, it can be shown that equation (VI.10.1) hasa nontrivial solution O(x, A) such that

0(t, A) = -0(0,A) and (t, A) = - (O, A)

This implies that ¢(x, A) satisfies the condition

¢(x + e, A) = -&, A) for - oo < x < +oo.Observation VI-10-4. Cconsider the eigenvalue problem

(VI.10.5) 2 + (A - tL(x) y = 0, y(o) = 0, y(e) = 0.

Applying Theorem VI-3-11 (with -A instead of A) to problem (VI.10.5), it can beshown that (VI.10.5) has infinitely many eigenvalues

(VI.10.6) µI < i2 < < µ <which are determined by the equation(VI.10.7) t2(£, A) = 0.The eigenfunction 02 (x, has exactly n -1 zeros on the open interval 0 < x < e.

Since .2 (0, A) = 1 > 0, it follows that

dO2 < 0 for n is odd,

dx (e'µ") I > 0 for n is even.

Furthermore, condition (VI.10.7) implies that det t I (A)] = A) = 1.

Thus, f (A) = 01(e, A) +1

follows from (VI.10.4). Therefore,Z V. A)

f(A) > 2 if (e, A) >0 and f(A) :5-2 if (e, A) < 0,

> 2 if n is even,f (µ") { < -2 if n is odd

(cf. (VI.10.7)). Here, use was made of the fact that jaj+ 1a1,

ifa76 0.Observation VI-10-5. If(VI.10.8) A < min{u(x) : -oo < x < +oo},

then f (A) > 2. In fact, 1 (x, A) = (u(x) - A)01 (x, A) > 0 as long as 01(x, A) > 0dX2

and, hence,dol

(x, A) > 0. Thus, we obtain s1(£, A) > 1. Similarly, 2 (P, A) > 1

if (VI.10.8) is satisfied. In this way, a rough picture of the graph of the functionf (A) is obtained (cf. Figure 1).

At A2 3 44 A3=T

-F3 I l4W lA

Ul U2 U3=N3U4 u5 U6

f=2

f= -2

FIGURE 1.Actually, we can prove the following theorem.

Theorem VI-10-6. Assuming that u(x) is continuous and periodic of period t > 0on the entire real line R, consider three boundary-value problems:

(A) 2 + (A - u(x)) y = 0, y(0) = 0, y(t) = 0,

($)2 + (A - u(x)) y = 0, y(0),W) = y(0), dx (t) _

(C) d2 + (A - u(x)) y = 0, y(t) = -y(0), dY(t) _ L(0).

Then, each of these three problems has infinitely many eigenvalues:

µI <µ2</A3...<An<...,

(VI.10.9) A0<\1<A2<A3<a4 <...<A2n_I<\2n <...,V1 <v2 <v3 :5 L14 <... <V2n-1 <V2n <... ,

respectively. Furthermore, if we denote by an(x), and 7n(x) the eigenfunc-tions associated with the eigenvalues µn, An, and vn, respectively, then(1) A0 < v1,(2) v2n-1 :5 92n-I 1'2n < A2n-1 !5 µ2n < A2n < V2n+1 < µ2n+1 < i/2(n+l) for

n = 1,2,...,(3) 02n_1(x) and /32n(x) are linearly independent if A2,,-1 = A2n,(4) 72n-1(x) and 72n (x) are linearly independent if v2n-1 = v2n,(5) 3D(x) does not have any zero on the interval 0 < x < t,(6) 02._1(x) and /32,,(x) have exactly 2n zeros on the interval 0< x < t,(7) 72n-1(x) and 72n (x) have exactly 2n - 1 zeros on the interval 0 < x < t.

Proof.

We prove this theorem in five steps. Note first that (VI.10.9) is obtained fromFigure 1.

Step 1. If ¢(x, A) is a solution of differential equation (VI.10.1), then (x, A) is

a solution of the initial-value problem(VI.10.10)

d2w

dx2+ (A - u(x)) w + O(x, A) = 0,

w(0, A) = (0, )L),

i i (0,,\) =TA_ (dx

(cf. §lI-2). Therefore, using the variation of parameters method, we obtain(VI.10.11) _

-1(x,A) _ (x, A) fT 01(t, A)02(t, A)dt ¢2(x, A)

f01(t"\)2dt'

d z (x,,\) = 01(x"\) 102(t, A)2dt - 02(X,,\) f -01(t, A) 02(t, A)dt

and, hence,

da (dd-x (x, A)) = 1(x, A) f ; m2(t, A)2dt - (_, A)la

mi (t, A)yi2(t, A)dt.

Therefore,df r

dA= r Q(01 (t, A), ¢2(t, A))dt follows from (VI. 10.4), where

0

Q(Y1,Y2) = -02(e,A)Y2 + dTl(e,A)Y2 + [1(t1A) - (e,A)Jr1Y2.

Step 2. The discriminant of the quadratic form Q is f(,\)2 - 4. In fact,

2

1(t"\)A)]

+ 4xd

l (e, A)o2(t, A)

2

2 (e, A)] + 401(e, A)dO2 (e, A) - 4 = f(,\)2 - 4.

Note that

41(e, ) (e, A) - ' (e, A) = det [ I (A)] = 1.

Thus, Q(01 (X, A), ¢2(x, A)) does not change sign for 0 < x < e if If (A)f < 2. Itfollows that

(I) df(A) 0dA

Step 3. Also, it can be proven that

df(A)

if If(A)f < 2.

< 0 if A < pl and f(,\)2 = 4,d,\

(-1)1 d('\) < 0 if pl < A < pi+1 and f(A)2 = 4.

To prove this, notice that Q is a perfect square if f(,\)2 = 4 and that 02(1,.\) # 0

if A 0 pj for every j. Hence, dA\) ¢2(e, A) < 0. Also, using Lemma VI-3-11 (with-A instead of A), we obtain

02 (1, A) > 0 if A < p1,

A) > 0 if p., < A < pj+1.

Thus, (1) is verified.

Step 4. Let us prove that

dzf(a)J< 0 if f (a) = 2 and d(\) = 0,

dal df(A)> 0 if f (A) _ -2 andd,\

= 0.

First observe that Q is a perfect square if (f(A)l = 2. Hence,

t

71K (A)= ± f [c101(t, A) + c202(t, \)12 dt = 0,

0

where cl and c2 are some real numbers. Therefore, cl = 0 and c2 = 0, since 01 and02 are linearly independent. This means that

(IV) 02(f,,\)=0, ' (e, A) = 0,UX ax

Since det 1, we conclude that

(V) fi(A) =d

12 if f (a) = 2 and d (a) = 0,

f- I2 if f (A) = -2 and -(A) = 0,

where 12 is the 2 x 2 identity matrix. Furthermore,

z e2 (a) =

Jo- (t, A)0 (t, A)2 + da ( ') (e, a)2

d

+ d ) (e, A) } m (t, A)m2(t, A)1 dt.

Also, if f (A) = 2 andd

(A) = 0, it follows from (VI.10.11) and (IV) that

Id (e, a) =

z {e, ) = j42(s,A)2ds,

td0i (s, A)2d

0

da ( f t 0r (s, A)00

Hence,

s,

2(s, A)ds.

0d (A) _ - f t ft(0f(t,

A)02{s, A) - 0e(s, A)02(t,A)j2dtds <

if f (A) = 2 and I (A) = 0. Similarly,

d2f t tf2 (A) = f [Ol (t, A)02(s, A) - 41(s, A)4 2(t, A)J2dtds > 0

if f (A) = -2 and df (A) = 0. Note that, if A2,,-1 = A2,,, it follows from (VI. 10.2),

(IV), and (V) that 01(x, A) and 02(x, A) are two linearly independent solutions forproblem (B). Similarly, if w2r_1 = v2,,, it follows from (VI.10.2), (IV), and (V) that01(x, A) and 62(x, A) are two linearly independent solutions for problem (C). Thus,(3) and (4) are verified.

Step 5. The functions (respectively -yn(x)) have an even (respectively odd)number of zeros on the interval 0 < x < 1, since /3, (0) = Qn(1) and 7n(0) Yn(1)As can be seen in Figure 1,

112n-i < A2n-1 A2n < 12n+l-

Therefore, by virtue of Theorems VI-3-11 and VI-1-1, we conclude that fit, 1 andper have more than 2n - 1, and less than 2n + 2, zeros on the interval 0 < x < 1.Hence, they have exactly 2n zeros there. Similarly, since

(VII) /12n-2 < V2n-1 V2n < 112n,

and -t2n have exactly 2n - I zeros on the interval 0 < x < e. The function,8a does not have any zero on the interval 0 < x < t since Ao < 111. Thus, (5), (6),and (7) are verified. Finally, (2) follows from (VI.10.9), (VI), and (VII).

Definition VI-10-7.(I) The set {A : f(A)2 < 4} is called the stability region of the differential

equation (VI.10.1).(11) The set {A : f(A)2 > 4} is called the instability region of the differential

equation (6210.1).

Example VI-10-8. A periodic function u(x) is called a finite-zone potential if the

function - 4 of the differential d2yf (A)' equation

dx2+ (A - u(x))y = 0 has a finite

number of simple zeros (i.e., all other zeros are double). For example, considerthe case when u(x) = a, where a is a constant. The differential equation becomesd? 2

dx+ (A - a)y = 0 and, hence,

cos(v1'"T-_ax) if A > a,

01 (X, A) = 1 if A =a,

cosh( a --Ax) if A < a

and( sin(v1_T_-ax)

A-a02(x, A) = x

sinh(x)

if A>a,

if A =a,

if A < a.

Therefore,

AA) = A) + (e, A) _

12 cos(\/-), --at) if A >a,2 if A=a,2 cosh( a --At) if A < a.

Thus, we conclude that in this case, f (A)2 -4 has only one simple zero a (cf. Figure2).

FicURE 2.

The materials in this section are also found in [CL, Chapter 8j.

EXERCISES VI

VI-1. Assume that u(x) is a real-valued continuous function on the interval 20 ={x : 0 < x < +oo} such that u(x) > rno for x >_ xo for some positive numbers "toand x0. Show that

(1) every nontrivial solution of the differential equation

(E)day - u(x)y = 0&2

has at most a finite number of zeros on Z0,(2) the differential equation (E) has a nontrivial solution 1)(x) such that

hm q(x) = 0.

Hint.(1) Note that if y(xo) > 0, then y"(xo) > 0. Hence, y(x) > 0 for x > x0 if

1/(xo) > 0.(2) It is sufficient to find a solution 0(x) such that 0(x) > 0 and 0'(x) < 0 for

x>xo.

EXERCISES VI 191

VI-2. For the eigenvalue-problem

(EP) L2 + u(x)y = Ay, Y(()) = y(1), y(0) = y'(1),

where u(x) is real-valued and continuous on the interval 0 < x < 1,(1) construct Green's function,(2) show that (EP) is self-adjoint,(3) show that (EP) has infinitely many eigenvalues.

VI-3. Let A, > A2 > - > An > - be eigenvalues of the boundary-value problem

d2y+ u(x)y = Ay, y(a) = 0, y '(b) = 0,

where u(x) is real-valued and continuous on the interval a < x < b. Show thatthere exists a positive number K such that

2irn +

n2 b - aKn

for n = 1, 2, 3, ... .

VI-4. Assuming that u(x) is real-valued and continuous on the interval 0 < x <+oo and that lim u(x) = +oo, consider the eigenvalue-problemr+oo

dx2 - u(x)y = Ay, y(0) cos a - y'(0) sin a = 0,z

lim y(x) = 0.

where a is a non-negative constant. Show that(a) there exist infinitely many real eigenvalues AI > A2 > . . . such that lim An =n-.ioo

-00,(b) eigenfunctions corresponding to the eigenvalue An have exactly n -1 zeros on

the interval 0 < x < +oo.

Hint. Let A1(b) > A2(b) > . . . be the eigenvalues of

dx2 - u(x)y = Ay, y(0) cos a - y'(0) sin a = 0, y(b) = 0,

where b > 0. Define Am by lim A,(b). See JCL, Problem 1 on p. 2541.6 +oo

VI-5. Show that if a function O(x) is real-valued, twice continuously differen-tiable, and ¢"(x) + e-'O(x) = 0 on the interval 10 = {x : 0 < x < +oo} and

if J O(x)2dx < +oo, then O(x) is identically equal to zero on I .J0

VI-6. Using the notations and definitions of §VI-4, show that

+oo

(f,C(f)) = j>n(f,17n)2n=1

jb

= {u(x)f(x)- P(x)f'(x)2}dx + P(b)f(b)f'(b) - P(a)f(a)f'(a),

if f E V(a, b).

VI-7. Assume that u(x) is real-valued and continuous and u(x) < 0 on the intervalI(a, b), where a < b. Denote by yh(x, A) the unique solution of the initial-value

problem 2 + u(x)y = Ay, y(a) = 0, y(a) = 1, where A is a comlex parameter.Show that

(i) O(b, A) is an entire function of A,(ii) ¢(b, A) 54 0 if A is a positive real number,(iii) ¢(b, A) has infinitely many zeros A,, such that 0 > A0 > Al > A2 > and-2

.

\r l

)llm A" - - \ /n +oo n2 b - a

VI-8. Find the unique solution O(x) of the differential equation

I x ) = y such that is analytic at .x = 0 and 0(0) = 1. Also, show that(i)\\\ m(x)

is an entire function of x,(ii) ¢(x) A 0 if x is a positive real number,

(iii) ¢(x) has infinitely many zeros an such that 0 > \o > a1 > A2 > . andlim 1n = -oc,n-+00

(iv) J nx)dx = 0 if n 96 m.0

VI-9. Show that the Legendre polynomials

Pn(x)=

2"n1 d"!dx"((x2 - 1)'J (n = 0,1,2,...)

satisfy the following conditions:(i) deg P,, (x) = n (n = 0,1,2,...),

(ii)J

1 Pn(x)Pm(x)dx = 0 if n 0 m,

(iii) j P n (x)2dx = 1 (n = 0,1, 2, ... ),n z

J(iv) xkPn(x)dx = 0 for k = 0,... , n - 1,

(v) Pn(x) (n > 1) has n simple zeros in the interval Ixl < 1,

EXERCISES VI 193

(vi) if f (x) is real-valued and continuous on the interval jxj < 1, then

1 Nlim f f (x) - E (n+

N-+oo 1n=O

2

Z) (f, PP)Pn(x)) dx = 0,

1

where (f, Pn) = f f (x)PP(x)dx,

(vii) the series F, (n + 21 (f, PP)Pn(x) converges to f uniformly on the interval+00n=0

jxj < 1 if f, f', and f" are continuous on the interval jxj < 1.

Hint. See Exercise V-13. Also, note that if f (x) is continuous on the intervaljxj < 1, then f (x) can be approximated on this interval uniformly by a polynomialin x. To prove (vii), construct the Green function G(x,t) for the boundary-valueproblem

((I -x2)d/ +aoy=f(W),

y(x) is bounded in the neighborhood of x = ±1,

1 1

where ao is not a non-negative integer. Show that f f G(x, )2dx< < +oo.J! 1J 1

Then, we can use a method similar to that of §VI-4.

VI-10. Assume that (1) p(x) and p'(x) are continuous on an interval Zo(a,b) _{x : a < x < b}, (2) p(x) > 0 on Zo, and (3) u(x, A) is a real-valued and continuousfunction of (x, A) on the region Zo x ll = {(x, A) : x E Zo, A E It} such that

lim u(x, A) = boo uniformly for x E Zo. Assume also that u(x, A) is strictlyA-toodecreasing in A E R for each fixed x on I. Denote by O(x, A) the unique solutionof the intial-value problem

(P(x)L ) + u(x,.\)y = 0 y(a) = 0, y(a) = 1.

Show that there exists a sequence {pn : n = 0, 1, 2, ... } of real numbers such that(i) pn < 1An-1 (n = 1, 2, ... ), and lim pn = -oo,n .+oo(ii) 4(b, pn) = 0 (n = 0,1, 2, ... ),

(iii) ¢(x, .\) ,- 0 on a < x po, and q(x, A) (n > 1) has n simple zeroson a<x<bforpn<\<,un-1,

(iv) strictly decreases from +oo to -oo as A decreases from pn_1

to An-

VI-11. Assume that p(x) and u(x) are real-valued and continuous on the interval1(0,1) = {x : 0 <- x < 1} and that p(x) is also continuous on 7(0,1), p(x) > 0for 0 < x < 1, p(0) = 0, and p'(0) < 0. Show that the differential equation

dx(px )d +u(x)y = 0 has a fundamental set {0, ii} of solutions on the interval

dx)0 < x < 1 such that

(i) slim 4(x) = 1 and Zli o p(x)O'(x) = 0,

(ii)slo

(i(z) - 11 F )0 and =li m- p(x)ii (x) = 1.

VI-12. Setfor 0<x<

for 2 < x 1.

Denote by 01(x, A) and 42(x, A) the two unique solutions of the differential equationy

+ (A - u(x))y = 0 satisfying the initial conditions 4i(0, A) = 1, X1(0,A) = 0,

42(0, A) = 0, and (0, A) = 1, where A is a real parameter. Sketch the graph of

the function f (A) = 4(1, A) + (1, A).

VI-13. For the scattering data {r({) = 0, (1,2,3), (1,1,1)}, find the potential u(x)and the Jost solutions ft(x,().

VI-14. Calculate the scattering data for each of u(x + 1) and u(-x), assumingthat {r({), (271, ... ,T)N), (cl, ... , cN)} are the scattering data for u(x) and thatu(x) satisfies a condition Eu(x)I < Ae-k1=I for some positive numbers A and k.

Hint. Let f f(x, () be the Jost solutions for u(x). Two quantities a(() andare given by

a(S) =2C I f+(x,C) f_(x,<) I

and(X,0 f-' (x:0

1

respectively.

The Jost solutions f o r u(x + 1 ) are e ' ft (x + 1, (). Therefore, the scatteringdata for u(x + 1) are

(171,... ,nN), (e-2"'c1,... ,e-2"r'cN)}.

The Jost solutions for u(-x) are ff(-x, (). Hence, the scattering data for u(-x)are

b(l)'(171,...,17N)I 1 ,...,- 1 l}

a(C) ( cia<(iv71)2 CNa<(LT,N)21 J

Note thatf-(x,iil3) = i a<(iul,)f+(x,iu,)

EXERCISES VI 195

VI-15. Let u(x) be real-valued, continuous, and periodic of period t > 0. Also,for every real , let O(x, {, A) be the solution of the differential equation

(Eq) 2 + (A - u(x))y = 0

satisfying the initial conditions 0({, {,A) = 0 and {G'(£, t, A) = 1. Let

A = Al (0 < A2W < µ3(f) < ...

be all roots of ii({ + t, , A) = 0 with respect to A. Show that

( + [', , A) = QAm(t) - A

+00 2m=1

e

Hint.Step 1. Let us construct O(x, , A) for negative A. To do this, change (Eq) to theintegral equation

sinh(p(x - )) + 1 rsinh(µ(x -

p p f

where it = v > 0. Since e-;`(=-() sinh(µ(x - )) is bounded as p - +oo, it canbe shown that

e-v(x-E)+G(x, , A) = e-al=-E)such{µ x- f )) + O( A 1

on the interval 0 < x - < t as A -e -oo. Thus, we derive

lim t'(4+f,.,A) = 1.(sinh(e))

Step 2. There exists an entire function R(A) of A such that

sin(Pf)J aA) =R(

sinh(CX)

for A > 0,

for A <0.

This function R(A) has the following factorization:

R(A) = t-_,

m=1 \1

mrwhere Cm = tStep 3. If the function V)((+1, 1, A) is entire in .\ and ''(+t, 1,,\) = O(exp(1 JaJ))as A -+ oo, we can write 1'(( + 1, t,,\) in the following form:

lp(f + t, (, a) = c[i(A]

where c is independent of A. Here, we used the fact that cn,J is boundedas m +oc (cf. Theorem VI-3-14).

Step 4. Note that

(1)

Note also that

_((+t,t,X) - 'c' + [0-W_- aR(A) ( ) H cam, - A ]

c -a cm -AThis implies the uniform convergence of the infinite product on the right-hand side

of (1) for -oo < A < 0. Thus, limZ'((+ 1, t, _ = 1 = c

a-.-OO R(a) lRemark. For expressions of analytic functions in infinite products, see, for exam-ple, (Pa, pp. 490-5041.

VI-16. The functions are defined in §VI-9. Calculate G4(u) and G5(u).I

VI-17. Use the same notations as in Theorem VI-10-6. Show that if J u(t)dt = 0,0

ft

then Ao < 0. Also show that if / u(t)dt = 0 and = 0, then u(x) = 0 identicallyJoon-oc<x<+00.

Hint. Since j3o(z) # 0 on -oc < x < +oo, set w(z) = !Lx). Then, w'(x) +

w(x)Z + A - u(x) = 0 (cf. 1A'IW, Theorem 4.4, p. 621).

V1-18. Set F(() = log f (() for `a( > 0, ( # 0, where(i) f (() is continuous for QJ( > 0, ( 0,

(ii) f (() is analytic for `£( > 0,(iii) ((f (() - 1) is bounded for !a( > 0,(iv) f (() # 0 for !'( > 0, (0 0.

Denote by SR the semicircle {(: 1(I = R,0 < arg(< r) which is oriented counter-clockwise, where R is a positive number. Show that

(a) IR1 1 F(z) dz = f F(() (`1( > 0,1(1 < R),

2Ri2ri 0 (`'(<0,1(1<R),

(b) F(() = tai j F'(() d{ + 7 f-g for a(> 0,

where fl() is the complex conjugate of F().

CHAPTER VII

ASYMPTOTIC BEHAVIOR OFSOLUTIONS OF LINEAR SYSTEMS

In this chapter, we explain the behavior of solutions of a homogeneous linear

system = A(t)y as t +oo in the case when the coefficient matrix A(t)has a limit A0 as t -+ +oo. The purpose is to show how much information wecan glean from the limit matrix Ao. We are interested in the exponential growthof solutions and the asymptotic behavior of solutions. In order to measure theexponential growth of a function, we use Liapounoff's type numbers which wasoriginally introduced by A. Liapounoff in [Lia]. Liapounoff's type numbers areexplained in §VII-1. Also, we explain Liapounoff's type numbers of solutions of ahomogeneous linear system in §VII-2. (Liapounoff's type numbers are also foundin L. Cesari [Ce, pp. 50-551.) In §VII-3, assuming that lim A(t) = Ao, wet-toocalculate Liapotimoff's type numbers of solutions in terms of the eigenvalues of A0(cf. [Huk3j, [Hart, Chapter X1, and [Si9]). In §VII-4. we explain how to derive theasymptotic behavior of solutions by diagonalizing the given system. A theorem ofM. Hukuhara and N. Nagumo gives the original motivation (cf. Theorem VII-4-1).The main result is Theorem of N. Levinson (cf. Theorem VII-4-2). (Generalizationsand refinements of Theorem of Levinson are found, for example, in [Bell], [Bel2],[CK], [Cop1]. [Dev], [DK], [El], [E2], [Gi], [GHS], [HarL1], [HarL2], [HarL3], [HW1],[HW2], [HX1], [HX2], and [HX3].) In §VII-5, the Theorem of Levinson is appliedto a system whose matrix has a limit as t -+ +oc and its derivative is absolutelyintegrable on the interval 0 _< t < oo. The topics of §§VII-4 and VII-5 are alsofound in [CL, §8 of Chapter 3, pp. 91-97]. In §VII-6, we explain how we can reduce

some problems such as the differential equation dt2 + {X + h(t) sin(at)}rj = 0 to

the Theorem of Levinson, even if the derivative of h(t)sin(at) is small but notabsolutely integrable on the interval 0 < t < oo. The main idea is to apply the

Floquet theorem (Theorem IV-4-1) to z + {1 + e sin(ot)}rt = 0 to eliminatethe periodic parts of coefficients so that we can use the Theorem of Levinson (cf.[HaS3]; see also [HarLl], [HarL2], [HarL3]).

VII-1. Liapounoff's type numbers

In order to measure the exponential growth of a function, let us introduce Lia-pounof''s type numbers.

Definition VII-1-1. Let f (t) be a C"-valued function whose entries are continu-ous on an interval Z = {t : to < t < +oo}. Let us denote by A the set of all real

197

198 VII. ASYMPTOTIC BEHAVIOR OF LINEAR SYSTEMS

numbers a such that exp[-at] f (t) is bounded on the interval Z. Set

+oo ifA =0,(VII.1.1) 1 (f) = inf{o : a E A} i f A j4R and A 3 40 ,

-00 ifA=R.

The quantity A (f) is called Liapounoff's type number off at t = +oo.

Note that if a E A and a <,6, then a E A.

Example VII-1-2.

J+ (0) = -00, A (exp[-t2j) _ -00, A (tm) = 0 (for all constants m),

{ A (exp[ort]) = a, A (exp[t2j) _ +00.

Here, it was assumed implicitly that to > 0 if m < 0.The following lemma can be proved easily.

Lemma VII-1-3. Let A (f) be Liapounoff's type number of a C' -valued function

f (t) whose entries are continuous on an interval Z = It: to < t < +oo}. Then,

(i) exp I - (. (f) + e) t] f(t) is bounded on I if e > 0, and unbounded if e < 0,

whenever A (f) # -oo,

(ii) A fi < max{A (f,) : j = 1,2,... m

(tii) A f, = A(ft), tfA(f1) > A(fi)forj=2....,m,1<f<m

(iv) f1, f2, ... , f, are linearly independent on the interval if A if 1) , ... , A (fm)are mutually distinct,

m

(V) A(f1f2...fm) <- FA(f,) in the case when f1i... , fm are C-valued junc-

tions.(vi) A (P(t)f) = A (f), if the entries of an n x n matrix P(t) and the entries

of its inverse P(t)-1 are bounded on the interval Z.

The following lemma characterizes Liapounoff's type number.

Lemma VII-1-4. If A(f) is Liapounoff 's type number of a function f(t) at t =+oo, then

a = lim sup log If (s) I(f) t-.+oo lt<s<oo s

2. LIAPOUNOFF'S TYPE NUMBERS OF A LINEAR SYSTEM 199

Proof.

Since there exists a positive constant K such that I f (s)I < Keia(f)+`i' for e > 0and large values of s, it follows that

logIf(s)I <A(f)+e + logK 5.1U) +e+logtK

Hence,

lim suplog If(3)1 < a(f)

t-+mIt<5<+m s

for t < s.

Also, for a fixed positive number a and any positive integer in, there exists a large

value of sm such that If (sm)1 and lim sm = +oo. Hence,log m +

e < log If (sm)1 Therefore, we obtainSm

lim sup log If(s) I > (f )t-+oo<a<+ao s

Thus, Lemma VII-1-4 is proved.

sm

VII-2. Liapounoff's type numbers of a homogeneous linear system

In this section, we explain Liapounoff's type numbers of solutions of a homoge-neous linear system

(VII.2.1)dy

dt= A(t)9

under the assumption that the entries of the n x n matrix A(t) are continuous andbounded on an interval T = It : to < t < +oc}. Let us start with the followingfundamental result.

Theorem VII-2-1. If y" = fi(t) is a nontrivial solution of system (VII. 2. 1) and iff A(t)e < K on the interval I = It : to < t < +oo} for some non-negative numberK, then

(VII.2.2)

Proof.

Let the n x n invertible matrix 4P(t) be the unique solution of the initial-valuedYproblem 7 = A(t)Y, Y(to) = In, where In is the n x n identity matrix. Then,

dZ4i(t)-1 is the unique solution of the initial-value problem

dt= -ZA(t), Z(to) _

I (cf. Lemma IV-2-4). Since 1, we obtain

1 + KJ I4(t)Idt,to

t

1 + K /ITherefore,

)4i(t)I < exp[K(t - to)) and I < exp[K(t - to)] for t E T

(cf. Lemma 1-1-5). Observe that fi(t) = 4D(t)j(to) and b(to) Fromthe definition of the norm of a matrix, it follows that

exp[-K(t - to)] < 1 (t)I < 1 (to)j exp[K(t - to)] for t E Z.

Therefore, (VII.2.2) follows, since I¢(to)I 0 0.

Set A = {a Q is a nontrivial solution of (VII.2.1)}. Then, (iv) of LemmaVII-1-3 implies that A is a nonempty subset of R which contains at most n numbers.Let Al > A2 > ... > Am (1 < m < n) be all of the distinct numbers in A. Set

(VI1.2.3) Vj = {j : is a solution of (VII.2.1) such that A (p) < A-}

for j = 1,2.... Tn. Then, (ii) and (vi) of Lemma VII-1-3 imply that V, is a vectorspace over C. Set

(VII.2.4) 1i = dim Vj (j = 1,2,... ,m).The following lemma states that there exists a particular basis for each space V,which consists of y, solutions whose type numbers are equal to A,.

Lemma VII-2-2. For system (VI1.2.1) and y, given by (VII. 2-4), it holds that(_) yi = n,(ii) ym < -Ym-1 < ... < y,

(ii:) for each j, there exists a basis for V, that consists of y, linearly independentsolutions d,,- (v = 1,2,... , y,) of (VIL2.1) such that A (p,,°) = A3.

Proof

It is easy to derive (i) and (ii) from the definition of V, and the definition ofthe numbers at, ... , am. To prove (iii), let y,,,, (v = 1, 2, ... . y,) be a basis for Vj.Assume that

for t r- T., I dt

( A, for v = 1,...t,for v=f+1...... ).

It follows from (ii) of Lemma VII-1-3 and definitions of V, and A, that t > 1. Set

for v = 1.... , t,+ for v = t + l.... , y2.

Then, , , (v = 1, ... , y,) satisfy all the requirements of (iii).

2. LIAPOUNOFF'S TYPE NUMBERS OF A LINEAR SYSTEM 201

Observation VII-2-3. The maximum number of linearly independent solutionsof (VII.2.1) having Liapounoff's type number A,, is rye.

Definition VII-2-4. The numbers Al, A2, ... , Am are called Liapounoff's typenumbers of system (VIL2.1) at t = +oc. For every j = 1, 2,..., m, the multiplicityof Liapounoff's type number A, is defined by

(VII.2.5) hj _7m

for j=1,2,... in- 1,for j = in.

The structure of solutions of (VI1.2.1) according with their type numbers is givenin the following result.

Theorem VII-2-5. Let {(1, ¢2r ... , iin} be a fundamental set of n linearly in-dependent solutions of system (VII.2.1). Then, the following four conditions aremutually equivalent:

(1) for every j, the total number of those 4t such that A (y5t) = A, is h, (cf.(VII. 2.5)),

(2) for every j, the subset {¢t : A (Qt) < A, } is a basis for V,,

(3) A ctt > A) if the constants ct are not all zero,

(4) A = max {A (dt) : ct 0} for every nontrwzal linear combi-

n

nation FC, it of e Wn}t=1

Proof.

Assume that (1) is satisfied. Then, the total number of those ¢t such thatm

A (3t) < A, is Fht = ryr = dims Vj. Hence, (2) is also satisfied. Conversely,t=j

assume that (2) is satisfied. Then, the total number of those t such that A (fit) <Aj is equal to dime V) = -,. Hence, (1) is also satisfied (cf. (VII.2.5)).

Assume that (2) is satisfied. Then, if A ct4 A, for some ct, it

follows that ctc E V,, and hence

clot = tt

for some c e. Since {¢j : t = 1,2,... , n} is linearly independent, all constants ccmust be zero. Thus, (3) is satisfied.

n

Assume that (3) is satisfied. Write a linear combination I:ct4t in the formt=1

mcj¢t. Then, A CWI = A), if ci4r 96 0. Hence, (4)

j=1 a(ft)=A, a(mt)=a1 a(Ot}=a,follows from (iii) of Lemma VII-1-3.

Finally, assume that (4) is satisfied. Then, every solution d of (VII.2.1) withA (j) < A, must be a linear combination of the subset {¢t : A ((rt) S Aj }. Hence,(2) is satisfied. 0

Definition VII-2-6. A fundamental set (01,02, ... , On } of n linearly independentsolutions of system (VIL2.1) is said to be normal if one of four conditions (1) - (4)of Theorem VII-2-5 is satisfied.

Since V. C Vm-1 c . C V2 C V1, it is easy to construct a fundamental setof (VII.2.1) that satisfies condition (1) of Theorem VII-2-5. Thus, we obtain thefollowing theorem.

Theorem VII-2-7. If the entries of the matrix A(t) are continuous and boundedon an interval Z = It : to < t < +oo}, system (VIL2.1) has a normal fundamentalset of n linearly independent solutions on the interval Z.

Example VII-2-8. For a system = Ay" with a constant matrix A, Liapounoff's type numbers A1, A2, ... , Am at t = +oo and their respective multiplicitiesh1, h2, ... , hm are determined in the following way.

Let IL I , p2, ... , Pk be the distinct eigenvalues of A and M1, m2, ... , Mk be theirrespective multiplicities. Set v, = R(pj) (j = 1, 2,... , k). Let Al > A2 > > Am

be the distinct real numbers in the set {v1i v2, ... , vk }. Set h, _ mr fort=at

j = 1, 2, ... , m. Then, A1, A2, .... Am are Liapounoff's type numbers ofdy

= Ay"

at t = +oc and h1, h2, ... , hm are their respective multiplicities. The prom of thisresult is left to the reader as an exercise.Example VII-2-9. For a system

(VII.2.6)dy

= A(t)ydt

with a matrix A(t) whose entries are continuous and periodic of a positive period c..'on the entire real line R, Liapounoff's type numbers )k1, A2, ... , Am at t = +oo andtheir respective multiplicities h1, h2, ... , hm are determined in the following way:

There exists an n x n matrix P(t) such that(i) the entries of P(t) are continuous and periodic of period w,(ii) P(t) is invertible for all t E R.

3. CALCULATION OF LIAPOUNOFF'S TYPE NUMBERS 203

(iii) the transformation y = P(t)z changes system (VII.2.6) to

(VII.2.7)

with a constant matrix B.Therefore, (vi) of Lemma VII-1-3 implies that systems (VII.2.6) and (VII.2.7) havethe same Liapounoff's type numbers at t = +oo with the same respective multiplic-ities. Liapounoff's type numbers of system (VII.2.7) can be determined by usingExample VII-2-8. Note that if pl, p2,. .. , p" are the multipliers of system (VII.2.6),

then p, log[p,] (j = 1, 2,... , n) are the characteristic exponents of system(VII.2.6), i.e., the eigenvalues of B, if we choose log[p3] in a suitable way. Hence,

R(µ1) log[[p2I] (j = 1, 2,... , n).

Those numbers are independent of the choice of branches of log[p,].

{etExample VHI-2-10. For the system = IJ

the fundamental set[b],

eu [0] I is normal, but the fundamental set {[] , [J} is not normal.I 2t -e

VII-3. Calculation of Liapunoff's type numbers of solutions

The main concern of this section is to show that Liapounoff's type numbers of

a system J = B(t)y" at t = +oc and their respective multiplicities are exactly the

same as those of the system = Ay" with a constant matrix A if iimoB(t) = A.dt t-+0

It is known that any constant matrix A is similar to a block-diagonal form

diag[pllm, + W.421,., + Rf2i ... , µkl,,, + Mk],

where µl, ... , Pk are distinct eigenvalues of A whose respective multiplicities arem1, ... , mk, is the mj x mj identity matrix, and M1 is an mj x m, nilpotentmatrix (cf. (IV.1.10)).

Consider a system of the form

(VII.3.1) 10 = A2y1 + EBjt(t)Ut (j = 1,2,...t=1

where y2 E C'-,, A. is an n, x n) constant matrix, and B2t(t) is an n, x nt matrixwhose entries are continuous on the interval Za = {t : 0 < t < +oo}, under thefollowing assumption.

Assumption 1.(i) For each j, the matrix A. has the form

(VII-3.2) Aj = )1In, + E. + N., (j = 1,2,... ,m),

where Al is a real number, In, is the nJ x nl identity matrix, E. is an n. x njconstant diagonal matrix whose entries on the main diagonal are all purelyimaginary, and Nj is an nj x n, nilpotent matrix,

(ii) Am < )1m-1 < ... < 1\2 < Al,(iii) N,E, = E?NJ (j = 1,2,... ,m),(iv) t limp B,t(t) = 0 (j,t = 1,2,... ,m).

The following result is a basic block-diagonalization theorem.

Theorem VII-3-1. Under Assumption 1, there exist a non-negative numberand a linear transformation

(VII-3.3) y , = z ' , + I: T , (t)zt (j = 1,2,... ,m)tj

to

with ni x nt matrices T,t(t) such that

(1) for every pair (j, t) such that j f, the derivative dt tt (t) exists and the

entries of Tat and dj tt are continuous on the intervalI = (t : to < t < +oo)(2) lim T;t(t) = O (j # e),t-+00(3) transformation (VII.3.3) changes system (VI1.3.1) to

(VII.3.4) t Lzj A,+B,,(t)+EBjh(t)Th,(t)I (j=1,2,...,m).

Proof.

We prove this theorem in eight steps.

Step 1. Differentiating both sides of (VII.3.3), we obtain

dzl+

T,tdz"t dTit5tdt dt t#? df

(VII.3.5) = A) % + E Tj tatt0r

m

Bjt t + Tt Vt=1#tAj 4' Bjj + BjhThj i, + AjTjt + Bjt + BjhTht

h* t hitzt,


where j = 1, 2,... , m. Note thatr

in n+ 1

2, = F BjnTnv zve=1 Vol v=1 h#v

Rewrite (VII.3.5) in the form

{-[At+Fj }+ET,t{ Lz' -[At+Fe]zt}

JJJJJJ t#i l 1

_ [AJTJt + Bit + BlhThe - Tlt (At + Ft) dttt#, h#t

where j = 1, 2,... , m, and

F t = B» + B)hTh) M )

h#j

Define the Tit by the following system of differential equations:

Ztf

(.l r e)(VII.3.6)d t t

Th#t

Then,

{ -[A,+F,]x",}+Tj dt -[At+Ft]zt}=Q (j=1,2,...,m).

This implies that we can derive (VII.3.4) on the interval I if the Tit satisfy (VII.3.6)and condition (2) of Theorem V1l-3-1, and to is sufficiently large.

Step 2. Let us find a solution T of system (VII.3.6) that satisfies condition (2) ofTheorem VII-3-1. To do this, change (VII.3.6) to a system of nonlinear integralequations

J(VII.3.7) T,t(t) = exp[(t - s)A,]U,t(s,T(s))exp[-(t - s)A1]ds,

for j 0 e, where the initial points rat are to be specified and

U,t(t,T) = B11(t) + E Bjh(t)Tht - Tjt Btt(t) + E Bth(t)Thth961 h#t

Using conditions given in Assumption 1, rewrite (V11.3.7) in the form

rt 1

(VII.3.7') Tit(t) = J exp 2(a1 - t)(t - s)] s,s,T(s})ds,

where j t andWjt(r,s,T)

(VII.3.8)[(A,= exp - At)r] exp[rEj j exp[rNj[UJt(s, T) exp[-rEtj exp[-rN,].

On the right-hand side of (VII.3.7'), choose the initial points rat in the followingway

(VII.3.9) r tat

( +00 if Aj > At (i.e., j < t),j`

o if A < A (ie j > t)1t..,.Step 3. Let us prove the following lemma.

Lemma VII-3-2. Let a be a positive number and let f (t, s) be continuous in (t, s)on the region Z x Z, where Z = {t : to < t < +oo}. Then,

(VII.3.10)ft"

t exp[-a(t - s)] f (t, s)dsl < - max f (t, s)to s < tIII

yl - J

for each fixed t E T. Also, if I f (t, s)' is bounded for to < t < s < +oo, then

(VII.3.11)I

rt exp[a(t - s)j f (t, s)dsl <_ 1 sup { I f (t, $)I : to < t < s < +oo}J+00 a

for each fixed t E Z.

Proof

In fact, estimates (VII.3.10) and (VII.3.11) follow respectively fromrt

J exp[-a(t - s)]ds = a {1 - exp[-a(t - to)]} <to

t

exp[a(t - s)]ds = O- 1 .

a+ooStep 4. Let us estimate s, T) fort < s < +oo if A j > at, and for to < s < tif A_, < at. If A, > At, r < 0, and s > t, it follows that

i1',t(r,s,T)I < Ko(l + Irlm'+mi-2) exp 12(A, - A,)7-1 IVit(s,T)I

for some positive number 1Co. It can be shown `easily that there exist a positivenumber X and a non-negative valued function p(t) such that

] X, for r < 0,I + Ir1m'*m`_2 exp [(A1 - At)r <

B, (s) Q(t) for t < s < +oo and all pairs (p, q),lim p(t) = 0.t-+00

Hence, if Aj > At, r < 0, and s > t, we obtain IW,t(r,a,T)I < /C2,3(t) {1 + ITI2}for some positive number 1C2, where ITI = maxIT,t[. Similarly, the estimate

t#jI Wjt(r, s, T) I < 1C219(to) { 1 + ITI2) is obtained by choosing a positive number 1C2sufficiently large, if A., < at, r > 0, and to < s < t.

Step 5. In a manner similar to Step 4, we can derive the following estimates:

IWj1(T,s,T) - Wj1(r,s,T)I

K3$(t) {1 +ITI + ITI} IT - TI if A, > A1, T < 0, s > t,

1 K3$(to){1+IT{+ITI}IT-T{ if A<A1, T>0, to<s<t

for some positive number K.

Step 6. Define successive approximations on the interval I as follows:

TO ye(t) = 0,

TP1t;)t(t) = f[ exp [(A, - A,)(t - s)j 9,1(t - s, s, Tp(s))ds,i..

where j 54 a and p = 1, 2, .... Suppose that

(VU.3.12) I Tp;j,(t) j < C on the interval I = ft: to < t < +oo}

for some positive number C. Then, Lemma VII-3-2 and Step 4 imply that

2 K2i3(t){1 + C2} if A) > A,,

I TP .iat(t) I <_ A' - Al

Al2

AjK21300){I +C2} if A, < '\1

on the interval Z. Hence, choosing to so large that

we obtain

2 K2/3(t){1 + C2} < C on Z,IA, - A,I

ITp+1,je(t)j < C on the interval I

from (VII.3.12).

Step 7. Suppose that (VII.3.12) holds for p = 0, 1, 2,.... Then,

+i.t(t) - Tp,1(t)I S 2 aup ITP(s) - TP-1(s)I <_ C on Z,WPlTp

where ITpI = i ITpael if to is so large that K30(t){1 + 2C} < on Z. SinceP

Tp.e(t) _ {TQae(t) - TQ_lae(t)}, it can be shown easily that lim Tpe(t) _

Tje(t) exist for all (j, 1) such that j # f uniformly on the interval T. The limitTj1(t) satisfies integral equation (VII.3.7).

Step 8. In this final step, we prove that the bounded solution Tjt of (VII.3.7)satisfies condition (2) of Theorem VII-3-1. It easily follows from Steps 6 and 4 that

[(A3 - at)(t - s)1 ir4jt(t - s, s,T(s))ds = 01+

t lim o exp1

00

if > X1.For (j, f) such that ), < at, write the right-hand side of (VII.3.7') in the following

form:

(A s)] IV1t(t - s, s,T(s))dsJ t exp 12to11

fexpQ

L(A - a)(t - s)] - s, s,T(s))dsor

r 1

+J

t exp 12(At - at)(t - s)] bt%tt(t - s, s, T(s) )dso l

for any a such that to < or < t. Observe that

fexp - At)(t - s)] Wjt(t - s,s,T(s))dso

< exp L2 (.1j - '\t)(t - a)]I 1'. a exp

12 (a) - at)(a - s)J 14 t(t -- s, s, T(s))ds`

<2KO(t°)

{l + C2} exp 1(aI - AE)(t - a)At - A) 1:5

J ,

if JT(t)[ < C on I (cf. (VII.3.10)). Observe also that

jf exp L2(,\j - at)(t - s)]ji)t(t - s,s,T(s))dsl A i3(a) {1 +C2}

if T(t)j < C on I (cf. (VII.3.10)). Hence,

jexpt 1(A3 - a)(t - s)] - s, s, T(s))dso

<2K1\1 211-C2} {(to)exP {(A, - At)(t - )1 +

J

f T(t) I < C on T. Therefore, letting a and t tend to +oo, we obtaini

rtlim I exp 12 (A, - At)(t - s)] Yt,jt(t - s, s, T(s))ds = 0.t-+3o to L

This completes the proof of Theorem VII-3-1.In order to find Liapounoff's type numbers of system (VII.3.4), it is necessary

to establish the following lemma.

Lemma VII-3-3. If N is a nilpotent matrix, then for any positive number e suchthat 0 < e < 1, there exists an invertible matrix P(e) such that

P(e)-INP(e)I < Xe

for some positive number IC independent of e.

Proof.

Assume without any loss of generality that N is in an upper-triangular form withzeros on the main diagonal (cf. Lemma IV-1-8). Set A(e) = diag[1, e, ... , e"-1].Then, A(e)-'NA(e) = [ ek-Jz,k ] (a shearing transformation). Hence, as 0 < e <1 and j < k, we obtain IA(e)-1NA(e)I < e(N(. 0

Let us find Liapounoff's type numbers of system (VII.3.4), i.e.,

dzja. = IA + B,,(t) + z, (7 = 1,2,... ,m).

h*j

Theorem VII-3-4. A system of the form

dz = (AIn+E+N+B(t)Jadt

has only one Liapounoff's type number ,\ at t = +oo if(i) A is a real number,

(ii) In is the n x n identity matrix, E is an n x n constant diagonal matrix whoseentries on the main diagonal are all purely imaginary, N is an n x n constantnilpotent matrix, and EN = NE,

(iii) the entries of the n x n matrix B(t) are continuous on the interval I = {t :to < t < +oo},

(iv) t lim B(t) = 0.+00

Proof.

Set i = eaèÈU. Then,

(VII.3.13) dt = [N + C(t)) u, C(t) = e-ÈB(t)etE

Let p be any Liapounoff's type number of (VII.3.13) at t = +oo. Then, (µ( <sup(N + C(t)( (cf. Theorem VII-2-1). This is true even if to -+ +oo. Sincetezlim C(t) = 0, we conclude that 1µ1 < (N(. Using Lemma VII-3-3, it can bet+00

shown that any Liapounoff's type number p of (VII.3.13) at t = +oo is zero. This,in turn, completes the proof of Theorem VII-3-4. 0

Remark VII-3-5. For a nilpotent matrix N, the two matrices N and eN aresimilar to each other for any nonzero number e. To prove this, use the Jordancanonical form of N.

From the argument given above, we conclude that Liapounoff's type numbers ofsystem (VII.3.4) are at, ... ,.1,,, and their respective multiplicities are n1, ... , r4,,.Thus, the following theorem was proved.

Theorem VII-3-6. Liapounof's type numbers of a system = B(t)y at t =

+oo and their respective multiplicities are exactly the same as those of the systemdy

dt= Ay with a constant matrix A if the entries of the matrix B(t) are continuous

on the interval Io = {t : 0 < t < +oo} and lim B(t) = A.t-.+ao

Applying Lemma VII-1-4 to the solutions of the system = B(t)y', we obtainthe following corollary of Theorem VII-3-6.

Corollary VII-3-7. If an n x n matrix B(t) satisfies the conditions(i) the entries of B(t) are continuous on the interval To = {t : 0 < t < +oo},

(ii) lim B(t) = A exists,t-.

then, for every nontrivial solution ¢(t) of the systemdy

= B(t)y, limlog 1-0(t)I

dt t--+oo tp exists and p is the real part of an eigenvalue of the matrix A.

Remark VII-3-8. The conclusion of Theorem VII-3-1 still holds even if condition(iv) of Assumption 1 of this section is replaced by IB)k(t)j < f (t) (j, k = 1, 2,... , m)on the interval Zo, where f (t) satisfies

rP(VII.3.14) sup(1 +p- t)-1 J f(s)ds - 0 as t +oo.

p>t t

To see this, let us assume that a positive-valued function f (t) satisfies conditiontP

(VII.3.14). Set h(t) = sup ((1 + p - t)-1 ( f (s)ds) for a fixed positive numberp>t \\ ft

t. Also, set E(t) = suph(r). Then, lim E(t) = 0, and E(t2) < E(t1) if t1 < t2.r> t t--+oo

Now, it is sufficient to prove the following lemma.

Lemma VII-3-9. For any positive numbers t, to, and c, it holds that

rt e-c(t-a) f(s)ds < (I + -) E(to),(a)to

ec(t-s) f(s)ds < 1 + cE(t).f+

lim f t e-`(t-`) f (s)ds = 0.t-.+oo

o

Proof of (a).

tSet p(t) = f e-`(t-°)f (s)ds. Then, 0'(t) = -cb(t) + f (t) and, hence,I.0(r) - 0(o) = -c / T th(s)ds + / T f (s)ds

0 0

for to < a < r. Suppose that there exists a positive number 6 such that

b(r) = C1 + - + 6 E(to), 0(a) = C- + 6 E(to),

and

(!±0(s) > b E(to) for a < s < r.

Then,

E(to) + c l! + 6 E(to)(r - a) < E(a)(1 + (r - o)).

This is a contradiction. \Proof of (b).

fSet ,L(t) = et-f (s)ds for 0 t T for a fixed T > 0. Then, "(t) _

r

rcO(t) - f (t) and, hence, w(r) - ?P(t) = cJ ;(s)ds J f (s)ds for t < r < T.Suppose that there exists a positive number 6 such that t

><i(r) = I ! + 6) E(t), y(t) = (1 + 1 +6) E(t),

and

Then,

1l'(s) > (!+o)E(t) for t < s < r.

E(t) + c (1+ 6) E(t)(r - t) < E(t)(1 + (r - t)).

This is a contradiction. This, in turn, proves that

JT

f (s)ds < C1 + c j E(t) for T > t.

Since lim E(t) = 0, the integral+430

+ e'('-")f (s)ds exists andt-+o ftj+00

ec(t-e)f(s)ds < 1+ 1 f E(t).e \ c

The proof of (c) is left to the reader as an exercise. 0For the argument given above, see [Harl).

VII-4. A diagonalization theorem

Liapounoff's type number of a solution is useful information since it providessome idea about the behavior of the solution as t --. +oo. However, it is not quiteenough when we look for a more specific information. For example, Liapounoff'stype number of the second-order differential equation

x

dt2 + {1 + R(t)}r, = 0

is 0 and its multiplicity is 2 if the function R(t) is continuous on the interval 0 < t <+oc and lim R(t) = 0 (cf. Theorem VII-3-6). However, this does not imply thet-+"0boundedness of solutions as t --+ oc. 0. Perron [Per3[ constructed a function R(t)satisfying the two conditions given above in such a way that solutions of (VII.4.1)are not bounded as t +oc. Looking for better information concerning equation(VII.4.1), M. Hukuhara and M. Nagumo [HNI[ proved the following theorem.

Theorem VII-4-1. Every solution of differentud equation (VII.4.1) is bounded as1

+t +x, if+C0

jR(t)ldt < +oc.0

Proof.

r+xFirst, fix to > 0 in such a way that a jR(t)ldt < 1. Write a solution 0(t)

SU

of (VII.4.1) in the form

b(t) = q(to) cost - to) + 0'(to) sin(t - to) - J t R(s)¢(s) sin(t - s)ds.to

Choose a positive number K so that 4(to)j + kd'(to)j < K and choose another

positive number At so that M > 1KQ

> K. Then,

K + At J t JR(s)jds < At for to < t < ttto

if jb(t)j < At for to < t < t1. Hence, 14(t)I < M for to < t < +oc. 0In this section, we explain the behavior of solutions of a system of linear dif-

ferential equations under a condition similar to the Hukuhara-Nagumo condition.Precisely speaking, we consider a system of the form

(VII.9.2) d [A(t) + R(t))1i

under the following assumptions.

4. A DIAGONALIZATION THEOREM 213

Assumption 2. Assume that A(t) is an n x n diagonal matrix

(VII.4.3) A(t) = diag(A1(t), A2(t), ... , An (t)],

R(t) is an n x n matrix whose entries are continuous on the interval Zo = {t : 0 <t < +00)1 and

(VII.4.4) J IR(t)Idt < +oo.

Set

(VII.4.5) A,k(t) = al(t) -)tk(t) and D-,k(t) = R(AJk(t)) (j,k = 1,2,... ,n).Concerning the functions A. (t) (j = 1, 2,... , n), the following is the main assump-tion.

Assumption 3. The functions Al (t), A2(t), ... , A,, (t) are continuous on the inter-val 4. Furthermore, for each fixed j, the set of all positive integers not greaterthan n is the union of two disjoint subsets P.t and P32, where

(i) kEP) I if

lim I Djk(r)dr = -oo andt-+oo 0

for some positive number K,(ii) k E Pj2 if

`t

f/a

LiDik(r)dr < K for 0 < s < t

Dik(r)dr < K for s > t > 0

for some positive number K.

Remark VII-4-2. Assume that the functions A1(t), ... , An(t) are continuous onthe interval Z o and that lim A, (t) = pj (j = 1, 2, ... , n) exist. Then, the functions

t +ooA1(t),... ,An(t) satisfy Assumption 3 if the real parts of pl,... p, are mutuallydistinct. The proof of this fact is left to the reader as an exercise.

The main concern in this section is to prove the following theorem due to N.Levinson.

Theorem VII-4-3 ([Levil]). Under Assumptions 2 and 3, there exists an n x nmatrix Q(t) such that

(1) the derivative dQ(t) exists and the entries of Q and Q are continuous ond

the interval Z0,

(2) t1 1 Q(t)(3) the transformation

(VII.4.6) y' = V. + Q(t)] Echanges system (VII.4.2) to

(VII.4.7)d"

dt= A(t)i

on the interval Z0, where I is the n x n identity matrix.Prool

We prove this theorem in six steps.

Step 1. Differentiating both sides of (VII.4.6), derive

a + (In + Qj dr = [A(t) + R(t)J (I + Q) z.

Then, it follows from (VII.4.7) that Q should satisfy the linear differential equation

(VII.4.8) dt _ (A(t) + R(t)1[I,, + Q] - (In + Q] A(t)

or, equivalently,

(VII.4.9) dQ = A(t)Q - Q A(t) + R(t) (In + Qf .

The general solution Q(t) of (VII.4.9) can be written in the form

(VII.4.9') Q(t) ='Nt)C41(t)-1 + f0(t)O(s)-1R(s)`I,(s),I,(t)-1ds,

where C is an arbitrary constant matrix, 44(t) is an n x n fundamental matrix of

dt = [A(t) + R(t)]4>, and 'F(t) is an n x n fundamental matrix of Al = (cf.

Exercise IV-13). Thus, the general solution Q(t) of (VII.4.9) exists and satisfiescondition (1) of Theorem VII-4-3 on 10. Therefore, the proof of Theorem VII-4-3will be completed if we prove the existence of a solution of (V1I.4.9) which satisfiescondition (2) of Theorem VII-4-3 on an interval I = {t : to < t < +oo} for alarge to.

Step 2. Now, construct Q(t) by using equation (VII.4.9) and condition (2) ofTheorem VII-4-3. To do this, let 4?(t. s) be the unique solution of the initial-valueproblem

dYdt

= A(t)Y, Y(s) = In.

Then, (VII.4.9) is equivalent to the following linear integral equation:

1(VII.4.10) Q(t) = J (t, s)R(s) f 1 + Q(s)145(t, s)'1ds,

where4?(t, s) = diag(Fl (t, s), F2(t,1s), ... , Fn(t, s)],

F, (t, s) = exp [ I A,(r)d; { (j = 1,2,.JJJ

.. ,n).

Step 3. Letting q,k(t) and rik(t) be the entries on the j-th row and the k-thcolumn of Q(t) and R(t), respectively, write the integral equation (VII.4.10) in theform

(VII.4.10') ggk(t) = Jexp Aik(r)drr)k(s) + F ds,

[ft11 A=1

where j, k = 1, 2,... , n and the .1jk(t) are defined by (VII.4.5). Note that

t t

lexp(I

J(Ajk(r)dr}I = exp I J Djk(T)dr] < eK

ll a

if

0< s< t for k E Pj1 and t< s< +oo for k E Pj2,

where j, k = 1, 2,... , n and the Djk(t) are defined by (VII.4.5). The initial pointsrjk are chosen as follows:

rjk =

for some to > 0.

to if k E P1I,{ +00 if k E Pj 2

Step 4. Define successive approximations by

gojk(t) = 0,

lgPJk(t) = j t expLJ

t Ajk(r)dTJ rk(S) + rja(S)gp-I;hk(s) ds,h=1

where p = 1, 2, .... Then, we obtain I gpuk(t) I < eK (1 + nC]J r(s)ds on theto

interval I={t:to <t<+oc} if

(VII.4.11) Igp_I,jk(t)I < C on the interval 2,

where r(t) = max(j,k) Irjk(t)I. Using assumption (VII.4.4), choose to so large that

r+ooe'11 + nCJ J

r(s)ds < C.to

Then, from (VII.4.11), it follows that Igpjk(t)I < C on the interval Z.

Step 5. Similarly, if (VII.4.11) holds for p = 1, 2, ... , we obtain

P

Igp+Iuk(t) - gpjk(t)l < (2} C

on the interval I = {t : to < t < +oo} if to is chosen sufficiently large. Hence,plim gpUk(t) = gjk(t) exists uniformly on the interval Z, and the limit satisfies

integral equation (VII.4.10').

Step 6. Let us prove that the bounded solution qjk(t) satisfies condition (2) ofTheorem VII-4-3. If k E Pj2, then Tjk = +oo. Hence,

lim q)k(t)

f f1im f expLJ

AJk(T)d-rl [rik(s)+rih(s)chk(s)](is = 0.h=1

In the case when k E PJ1, note thatt

[f>tik(r)dr]q)k(t) = f exp [r)k(s)+r)h(s)hk(s)} ds

h=1

and

exp IJIrJ//

ft \Jk(r)dr] rik(S) + r)h(S)hk(S)v L s l1 A=1

ll

= fo

exp

[ft.\)k(T)dr] [r)k(S) +> rJh(S)hk(s)J ds

h=1

+ \.k(T)dr] {r)k(S) + n rJh(S)ghk(S)dss h-1

for any a such that to <l1a

f

< t. Observe that

1

exp [ft ak(T)dr] Irk(s)+ rJh(s)hk(s)1

dsl[fC.Jk(r)dr] [rJk(s)

= eXp Lf t Dk(r)dTJ f exP + r)h(s)hk(s)l1 dsh=1 JI

( t +oo

< exp J f D)k(r)dr] eK (1 + nCj ft) r(s)dsl o

f (q)k(t)(< C on Z. Observe also thati

I

t tf exp I f AJk(T)dr] [r)k(s) + rJh(s)hk(s) dsh=t

+co< eK(1 + tnC} f r(s)ds

0if lq)k(t)E < C on I. For a given positive number fix a positive number a so large

that eK (1 + nCjft-

r(s)ds < Since

lim f t D)k(r)dT = -oo if k E P)t-»+0for a fixed a, we obtain

t 1 +e

expL

f DJk(r)dr] eK [1 + nC] f r(s)ds <o to

for large t. Therfore, for any positive number e, there exists such that Iq,k(t)( <e for t > t(e). This complete the proof of Theorem IV-4-3.


Remark VII-4-4. The n x n matrix W(t, s) = [In + Q(t)j$(t, s) is the uniquesolution of the initial-value problem

dYdt

= [A(t) + R(t)]Y, Y(s) = In + Q(s),

where di(t, s) is the diagonal matrix defined in Step 2 of the proof of TheoremVII-4-3 (cf. (VII.4.10)). Since det[%P(t,s)] i4 0 for large t, the matrix WY(t,s) isinvertible for all (t, s) in Yo x Yo (cf. Exercises IV-8). This, in turn, implies thatthe matrix In + Q(t) is invertible for all t in Io.

Remark VII-4-5. Theorem VII-4-3 has been shown to be the basis for manyresults concerning asymptotic integration (cf. [El, E2] and [HarLl, HarL2, HarL3]).

Remark VII-4-6. Using the results of globally analytic simplifications of matrixfunctions in [GH[, H. Gingold, et al. [GHS] shows some results similar to TheoremVII-4-3 with Q(t) analytic on the entire interval Io under suitable conditions.

Remark VII-4-7. Instead of (VII.4.4), [HX1] and [HX2] assume only the inte-grability at t = oc for above (or below) the diagonal entries of R(t) and obtain theresults similar to Theorem VII-4-3. More results were obtained in [HX3] applyinga result of [Si9j.

The following example illustrates applications of Theorem VII-4-3.

Example VII-4-8. Let us look at a second-order linear differential equation

(VII.4.12) d 22 + p(t)rt = 0.

If we set

dt

equation (VII.4.12) becomes the system

(VII.4.13)

1 = , A(t) = 0 1ldo -P(t) 0J

dy = A(t)ydt

The two eigenvalues and corresponding eigenvectors of the matrix A(t) are

'\t(t)= i p(t)112,

A2(t) = -t p(t)'/

Set Po(t) _ [i P(t)'/2 -ip(t)'/2J. Then,

911 _ [i p(t)1/2]1pct) _ [-i pt)1/2J

dPo(t) - ip'(t) [0 0 1POW-1 =

-2p(t)

i [ip(t)'/2 1 1dt 2p(t)'/2 1 1 ' '/2 jP(t)1/2

1 ,

and

Po(t)-'A(t)Po(t) _[i

P(to1/2

0_i

pt)ti21

The transformation y = Po (t) i changes system (VII.4.13) to

- P0(t)-1 {A(t)Po(t) - dP t)jz.

Using the computations given above, we can write this system in the form

(VII.4.14) .ii = {t P(t) 1/ 2 [0 01] - 49t) [ 11 11

]

Suppose that(1) a function p(t) is continuous on the interval Zo = {t : 0 < t < +oo},(2) there exists a positive number c such that p(t) c > 0 on the interval Zo,(3) the derivative p'(t) of p(t) is absolutely integrable on Io,

Then, Theorem VII-4-3 applies to system (VII.4.14) and yields the following theo-rem (cf. IHN2]).

Theorem VII-4-9. If a function p(t) satisfies the conditions (1), (2), and (3)given above, every solution of equation (IV.4.12) and its derivative are bounded onthe interval Zo.

The proof of this theorem is left to the reader as an exercise. Note that condition(3) implies the boundedness of p(t) on the interval 1 .

If p(t) is not absolutely integrable, set

(VII.4.15) z = [I2 + q(t)E]t,

where I2 is the 2 x 2 identity matrix, q(t) is a unknown complex-valued function,and E is a constant 2 x 2 unknown matrix. Then, the transformation (VII.4.15)changes system (VI I.4.14) to(VII.4.16)

dt = [Iz + q(t)E]-1 { [ip(t)h12A0 -4p(t) A11

[12 + q(t)E] dtt) E} u,J

where

Ao = Al =

Anticipating that(i) q(t)I and Ip'(t)I are of the same size,

(ii) Jq'(t) I and Ip"(t)I are of the same size,choose q(t) and E so that two off-diagonal entries of the matrix on the right-handside of (VII.4.16) become as small as Ip'(t)I2 + [p"(t)I. In fact, choosing

q(t) = -i p'(t) , E = [0 -118 p(t)312 1 0 J

S. SYSTEMS WITH ALMOST CONSTANT COEFFICIENTS 219

we obtain

and

where

I 2lp(t)1/2Ao- p,(t)12+E(p,p,p')-q(t)dq(t)12 ,

1 -r q(t) I 4p(t) dt

+g(t}AoE+E(p,p',p") - q(t) dtt)12}

1

2jip(t)1/2A0 _ p'(t) EAo - p'(t)

Al1 + q(t) 8p(t) 4p(t)

E2 = -12, EAo = -AoE = 0 Il11 0

and E(p, p', p") is the sum of a finite number of terms of the form

a (P (t) 12 + $ p"(t)p(t)h/2

with some rational numbers or and /3 and some positive integers h. Applying The-orem VII-4-3 to system (VII.4.16), we can prove the following theorem.

Theorem VII-410. Suppose that(1) a function p(t) is continuous on the interval Zo = {t : 0 < t < +oo},(2) there exists a positive number c such that p(t) > c > 0 on the interval ID,(3)

+00

(Ip(t)120

(4) limop'(t) = 0.

+ Ip'(t)I } dt < +oo.

Then, every solution of equation (VII.4.12) is bounded on the interval 7o.

The proof of this theorem is left to the reader. Condition (3) of Theorem VII-4-10does not imply the boundedness of p(t) on the interval Io. For example, TheoremVII-4-10 applies to equation (VII.4.12) with p(t) = log(2 + t).

VII-5. Systems with asymptotically constant coefficientsIn this section, we apply Theorem VII-4-3 to a system of the form

(VII-5.1)d9

where A is a constant n x n matrix and V(t) is an n x n matrix whose entries andtheir derivatives are continuous in t on the interval Zo = It : 0 < t < +oo} underthe following assumption.

(12 + q(t)E]' = 1 + I9(t)2 (12 - q(t)E(

(12 + q(t)E]-1 { [1(t)1/2Ao - 4p(t) A11 (12 + q(t)E1 - dt) E}

Assumption 4. The matrix A has n mutually distinct eigenvalues p 1, u2, ... , An,and the matrix V(t) satisfies the conditions

lim V(t) = 0 and+00+ [V'(t)[dt < +oo.

t + Cc o

Let A1(t), 1\2 (t).... , an(t) be the eigenvalues of the matrix A+V(t). Then, theseare continuous on the interval I. Furthermore, it can be assumed that

lim ),(t) = µj (j = 1,2,... ,n).t-+,Choose to > 0 so that A) (t),.\2(t), ... , an(t) are mutually distinct on the intervalI={t:to<t<+oo}. Set

F(t, A) = det[AII - A - V(t)].

Then, F(t, \j(t)) = 0 on II for 1 = 1,2,... ,n, and

j (t, A, (t)) + as (t, \,(t)) A (t) = 0 (j =1,2,... n) on I.

Also,OF8A

(t, \j (t)) # 0 (j = 1,2,... ,n),

(since III, µ2i ... , p, are mutually distinct. Observe that .X (t)

8F (t,A, t ))

aF is

8a (t, af(t))linear homogeneous in the entries of the matrix V'(t). In this way, we obtain thefollowing lemma.

Lemma VII-5-1. Under Assumption 4, the derivatives of the eigenvalues of thematrix A + V (t) are absolutely integrable over the interval I = {t : to < t < +oo},i. e.,

JA(t)dtl < +oo (j = 1,2,... ,n).w

An eigenvector p, (t) of the matrix A + V(t) associated with the eigenvalue as(t)can be constructed in the following manner. Observe that the characteristic poly-nomial of A + V(t) can be factored as

F(t, \) = (A - \1(t))(A -1\2(t)) ... (A - \.(t))

Hence, by virtue of Theorem IV-1-5 (Cayley-Hamilton), we obtain

(VII.5.2) (A+V(t)-.11(t)In)(A+V(t)-.\2(t)In)...(A+V(t)-an(t)IJ) = 0

5. SYSTEMS WITH ALMOST CONSTANT COEFFICIENTS 221

on Zo. Furthermore, if t > to,

J1 (A+V(t) - Ah(t)In) # 0,h 0j

lim fl(A + V(t) - Ah(t)In) = H(A - phln) 36 0,t-+oo h#l h#j

for j = 1, 2,... , n. From (VII.5.2), it follows that

[A + V(t)] 1(A+ V(t) - Ah(t)In) = A,(t) [I(A + V(t) - Ah(t)In)h#J h#)

Hence, choosing a suitable column vector j,(t) of H(A+V(t)-Ah(t)In), we obtainh#)

p3(t) 1-1 0 and [A+V(t)]p,(t) = \j(t)jYj(t)

on the interval I = It : to < t < +oo} if to > 0 is sufficiently large. Furthermore,

lim j5, (t) = 4, (.7 = 1,2,... n)t+oo

are eigenvectors of the matrix A associated with the eigenvalues p. (j = 1, 2,... , n),respectively. Observe that the entries of the vectors pF,(t) (j = 1,2,... , n) arepolynomials in the entries of V(t) and A1(t),... ,An(t) with constant coefficients.Hence,

+00

Ip"j'(t)I dt < +oc (j = 1,2,... ,n).Jta

Thus, we proved the following lemma.

Lemma VII-5-2. Under Assumption 4, them exists a non-negative number tosuch that

(1) the matrix A + V (t) has n mutually distinct eigenvalues A1(t)...... n (t) onthe interval I = It : to < t < +oo},

(2) the eigenvalues Al(t), ... , An(t) are continuously differentiable on I andlim Aj(t) = p, (j = 1,2,... ,n),

(3) the matrix A + V(t) has n eigenvectors p1(t),... ,#n(t) associated with theeigenvalues \I(t) .... ,An(t), respectively, such that lim 73 (t) = 4j (j =t-+oo1, 2, ... , n) are eigenvectors of the matrix A associated with the eigenvaluesp) (j = 1,2,... ,n),

(4) the derivatives of the eigenvalues A,(t) (j = 1,2,... , n) and the derivativeso f the eigenvectors p ' , (t) (j = 1, 2, ... , n) with respect to t are absolutely inte-grable on the interval Z.

Set

Po(t) _ [ 91 (t) #2 (t) ... ( t ) [ ,[ , Q = [ 91 92 ... 4n [ ,A(t) = diag[A1(t), A2(t), ... , An (t)[, M = diag[pl, p2, ... , pn]-

Then,

t-+ooPo(t)-'[A + V(t)]Po(t) = A(t), Q-'AQ = M,

+oof I Po(t)_ 1 dP°(t) I dt < +oo.to dt

lim Po(t) = Q, lim A(t) = M,

Observe that the transformation y"= P°(t)z changes system (V11.5. 1) to

dt = POW-1 [[A + V(t)]P0(t) -dl t)1 i= [A(t) - Po(t)-'d t)] Y.

Applying Theorem VII-4-3 to this system, we obtain the following theorem.

Theorem V11-5-3. Under Assumption 4, if the eigenvalues A1(t)...... n (t) ofthe matrix A + V (t) satisfy all requirements given in Assumption 3 (cf. § VII--4) onthe interval .7 = {t : to < t < +oo}, a fundamental matrix solution of (VII.5.1) willbe given by

1(t) = P°(t)[In + H(t)]exp I / ( A(s)ds],

where H(t) is an n x n matrix whose entries are continuously differentiable on theinterval I and lim H(t) = O.t+00Remark VII-5-4.

r: 1

(a) As t -. +oo, the matrix exp [ - J A(s)ds] 4i(t) approaches the matrix Q,

(b) we can prove a result similar to Theorem VII-5-3 even if system (VII.5.1) isreplaced by

d = (A+V(t)+R(t)]y,where the matrix A + V (t) satisfies all the requirements of Assumption 4 andthe entries of the matrix R(t) are absolutely integrable for t > 0, i.e.,

f+00 JR(t)jdt < +oo.0

Observation VII-5-5. Let us look into the case when the matrix A has multi-ple eigenvalues. To do this, consider system (VII.3.1) under Assumption 1 givenin §VII-3. By virtue of Theorem VII-3-1, system (VII.3.1) is changed to system(VII.3.4) by transformation (VII.3.3). Furthermore, (VII.3.4) is changed to

(VII.5.3) t [ N 3 + R 3 (t)) u 3 (j = 1, 2, ... m),dwhere

R3(t) = e-'E, B33(t)+E B3h(t)Ttj(t) etE, (j=1,2,...,m)h #)


z3 = explt(A3I., + E, )]u3 (.) = 1, 2, ... , m).

Therefore, we look into the following two cases.

5. SYSTEMS WITH ALMOST CONSTANT COEFFICIENTS 223

Case VII-5-5-1. If N is an n x n nilpotent matrix such that N' = 0 and R(t)is an n x n matrix whose entries are continuous on the interval 0 < t < +oo andsatisfy the condition

+(VII.5.4) jt2(T_1)IR(t)Idt < +oo,

we can construct a fundamental matrix solution 45(t) of the system

(VII.5.5)dy

= (N + R(t))y'dt

such that lim e-' N4(t) = I, , where I is the n x n identity matrix. In fact, thet -.+00

diitransformation y" = etNii changes system (VII.5.5) to = e-tNR(t)e1NU. Since

leftNI < Kjtjr-1 for some positive number K, the integral equation X(t) = Ine-R(r)erNX(r)dT can be solved easily.j

Case VII-5-5-2. Let N be an n x n nilpotent matrix such that Nr = 0 and letR(t) be an n x n matrix whose entries are continuous on the interval 0 < t < +ooand satisfy the condition

+00(VII.5.6)

10tr-1IR(t)jdt < +oo.

Exercise V-4 shows that this case is different from Case VII-5-5-1. As a matter offact, in this case, we can construct a fundamental matrix solution 'P(t) of system(VII.5.5) such that

(VII.5.7) lira t-(k-t) (41(t) - etN) c" = 0', whenever Nkc_ = 0,t-+00

where k < r. We prove this result in three steps.Step 1. First of all, if an n x n matrix T(t) satisfies condition (VII-5.7), then

c' = 0 if 41(t)c = 0 for large t. In fact, noice first that limeWe N'5

trf!

if

NPe = 0 for p = F + 1, ... , r - 1. Also, note that limetNt: = 0 if a(t)e = 0

t-+oo tktkland Nkcc = 0. Therefore, N'lE = 0 if k = r. Hence, Nr-2c' = 0. In this way,we obtain c" = 0. Thus, we showed that if the matrix %P(t) satisfies the differential

equation (N + R(t))* and condition (VII.5.7), then %P(t) is a fundamentalmatrix solution of (VII.5.5).Step 2. To prove the existence of such a matrix W(t), we estimate integrals ofske(t-")NR(s) with respect to s, where 0 < k < r - 1. Observe that

ske(t_e)N =r1 Sk(t - S)hNh

h1h=0

Let us look at the quantity sk(t - s)h. Note that 0 5 k 5 r - 1 and 0 5 h < r - 1.

Case 1. k + h < r - 1. In this case, since sk(t - s)h = sk+h C t _ 11, define- s

JtSk(th'

8)h N°R(s)ds = / t 8k(th! S)h NhR(s)ds.

Sk+pth-pCase 2. k + h > r. In this case, look at sk(t - 8)h = D-1)p (hp)

p_Q

Subcase 2(i). k + p < r - 1. In this case, sk+pth-p = Sr-I

\1 t", where

sr-1-u=k+p,u+v=h-p. Since

v = h - p - p = (h-p) + (k+p) - (r-1) = (k+h) - (r-1) = k - (r-1-h),

we obtain 0 < v < k. Now, in this case, define

Irt

ht

j Sk+pth-PNhR(S)dS= J+ao p) sk+Pth-PNhR(s)ds.

Subcase 2(ii). k + p > r. In this case,

Sk+pth-p = Sr-1(tlµt", where r-l+p=k+p, v -p=h -p.

Since

v = h - p + p = (h-p) + (k+p) - (r-1) = (k+h) - (r-1) = k - (r-i-h),

we obtain 0 < v < k. Also, note that h-p<(r-1)+(k-r)=k-1. Now, inthis case, define

t=

k ) k+P h PNhR d kh-3 t - {s) s = -r;o

From the definition given above, it follows that

where t > to.

f',t Ske(t-s)NR(S)d8 = etN / t Ske sNR(s)ds,

vrt

lim I / ske(`'")NR(s)ds = O (k = 0,... , r - 1),t+00 F. q

h

k+h-P h$ Pt R(s)d$,N

Settingr

U(t) = f t ske(`-e)NR(s)ds,

6. AN APPLICATION OF THE FLOQUET THEOREM

we obtaindU(t)

225

= NU(t) + tkR(t).dt

Step 3. Let us construct n x n matrices Uk (t) (k = 0, 1, 2, ... , r - 1) by the integralequations

Then,

Uk(t) = Nk +

lira Uk(t) = Nkt-++00

Observe that

Hence, setting

we obtain

This implies that

tkUk(t) tkNk

k! k!

1(t) = eIN +

do(t)dt

It can be easily shown that

1

ftske(t-")NR(s)Ui(s)ds

rt e(t_s)NR(s)skUk(s)ds.n k!

tkUk(t)k!

k=0

r-1t k r

k=o nY' r

e(t-,)NR(s) sds.

k!!

_ (N + R(t))W(t).

limL(t)c = Urn

etNc' - Nk"t-+oo tk t-+00 tic

if Nk+16= 0. Thus, the construction of W(t) is completed.Case VII-5-5-2 was given as Exercise 35 in [CL, p. 1061.

VII-6. An application of the Floquet theoremThe method discussed in §VII-5 does not apply directly to the scalar differential

equation

d-t2 + {1 + h(t)sin(at)} ri = 0

when h(t) is a small function such as i ,t112, l01 t, andlog t

sin(tt/2) since theg g

derivative of h(t) sin(at) is not absolutely integrable over the interval 0 < t < +oo.In this section, using the Floquet theorem (cf. Theorem IV-4-1), we eliminateperiodic parts of coefficients so that Theorem VII-4-3 applies. Keeping this schemein mind, consider a system of the form

djj = A(t,h(t))y(VII-6-1)

under the following assumption.

Assumption VII-6-1.(1) The entries of an n x n matrix A(t, ej are continuous in (t, e) E R x A(r)

and analytic in f E A(r) for each fixed t E LR, where FE C'" unth the entrieset, ... , A(r) = IF: Ie < r}, and r is a positive number.

(2) The entries of A(t, e) are periodic in t of a positive period w.(3) The entrees hl,... , h,,, of a C"t -valued function h(t) are continuous on the

interval T(to) = {t : to < t < +oo} for some non-negative number to andlim h(t) = 6.t+00

Let us consider the following two cases.

Case 1. The function h(t) is supposed to be continuously differentiable on T(to)and satisfy the condition

+00(VII.6.2) J Ih(t)Idt = +oo and

I.f + Ih'(t)Idt < +oo.

Case 2. The function h(t) is supposed to be twice continuously differentiable onT(to) and satisfy the condition

(VII.6.3)

limt -+00

+00 +00

L00I;(t)ldtr= +00, / = +oo,

0

JIh"(t)Idt < +oo, J + Ih'(t)I2dt < +oo

0 0

For example, the function h(t) _ I satisfies (VI1.6.2), whereas h(t) =sin(f)f

satisfies (VII.6.3).

Observation in Case 1. In Case 1, we use the following lemma.

Lemma VII-6-2. If a matrix A(x, ej satisfies conditions (1) and (2) of Assump-tion VII-6-1, them exist n x n matrices P(t,e) and H(t") such that

(i) the entries of P(t, ) are continuous in (t, e) E R x A(f) and analytic inFE A(f) for each fixed t E R, where f is a suitable positive number.

(ii) P(t + w, ej = P(t, t-) for (t, F) E R x ©(f),(iii) P(t, fj is invertible for every (t, t) E R x A(f),(iv) the entries of H(t) are analytic in FE A(f),

2ai(v) any two distinct eigenvalues of H(0) do not differ by integral multiples of -,

(vi) a P(t, a exists for (t, e) E R x L(f) and given byw

(VII.6.4) P(t, eM = A(t, t)P(t, e) - P(t, e)H(e)

for (t, e) E R x L(f).

6. AN APPLICATION OF THE FLOQUET THEOREM 227

Proof

In order to prove this lemma, construct a fundamental matrix solution $(t, e)

of the differential equation d = A(t, e)y by solving the initial-value problemdX = A(t, e-) X, X (O) = I, where In is the n x n identity matrix. The en-dt bog( w( , andtries of 4!(w, ej are analytic in A(r). Define H(e) by H(ej =

P(t,') = 4i(t, cl exp(-tH(E)]. Then, (VII.6.4) follows.The most delicate part of this proof is the definition of H(). Details are left forthe reader as an exercise (cf. [Sill).

Changing system (VI I.6.1) by the transformation

(VII-6-5) W = P(t, i(t))u

we obtain the following theorem.

Theorem VII-6-3. Transformation (VII.6.5) changes system (VII.6.1) to

(VII.6.6) di =H(i(t)) - P(t,h(t))-1 hj(t)aP(t,h(t)) }u.

Proof

In fact,

(VII.6.7)

1<j<m

dil= P(t, h(t))-1 {A(t(t))P(t&(t)) - dt [P(t, h(t))] } u"

iiT

= P(t, i(t))-1( A(t, h(t))P(t, h(t)) - (t, K (t))

1 < <m

Since H(e) is given by (VII.6.4), equation (VII.6.6) follows from (VII.6.7).Observe that

(IV.6.8)r+oo

froP(t, h(t))-' h(t))

1<j<m jdt < +oo

under assumption (VII-6.2). Also, observe that H(i(t)) does not contain any pe-

riodic quantities anddH(i(t))

dt is absolutely integrable. Therefore, if eigenvalues

of H(6) satisfy suitable conditions, the argument given in §VII-5 applies to system(VII.6.6).

Observation in Case 2. Set

Bo (f) =

(VII-6-9) 8PB1(t, E, iZ) _ -P(t, E)`1 u8e

(t, E1,I<j<m

where µ E C' and the entries of µ are u1, P2, ... , µm. Then, differential equation(VII.6.6) can be written in the form

(VII.6.10) dt = {Bo(h(t)) + B,

Notice that the entries of the matrix B, (t, E, µ) are periodic in t of period tv andthat B1 (t, E, 6) = 0. Furthermore, any two distinct eigenvalues of Bo (6) do not

27ndiffer by integral multiples of -. Set D(r) = IE + lµI < r}.

In Case 2, the following lemma is used.

Lemma VII-6-4. There exist n x n matrices P, (t, e, {'t) and HI (E, µ) such that(:) the entrees of P1 (t, E, ui) are continuous in (t, E, u') E R x D(r) and analytic in

(E, )i) E D(r) for each fixed t E Ii£, where r is a suitable positive number,(i:) P, (t + w, F,)!) = P, (t, e, it) for (t, F, tt+) E 1 x D{r

(iii) P1(t, E, 6) = O for (t, i) E R x 0(r),(iv) the entries of H1 (E, u') are analytic in (E, Et) E D(r) and H1 (E, 6) = 0 for

E E O(r),

(v) at P, (t, e, µ') exists for (t, E, p) E R x D(r) and is given by

19

(VII.6.11)Pi(t,e,u) ={Bo(e) + B1(t,e,µ)) (In + Pi(t,E,ls)}

+ P1 (t, i,#)) { Bo(E) + Hi (E, µ) }

for (t, E, µ) E IR x D(r-), where I, is the n x it identity matrix.

Remark VII-6-5. Equation (V1I.6.11) can be simplified as

(VII.6.12) a P1(t,Ej) = Bo(E)P1(t,E,i) - Pi(t,E., )Bo(Ej

+ {B1(t,E,N)P:(t,E,Ji) - +B1(t,,rA-)} -Hl(E,u-)-

Proof of Lemma VII-6-4.M

Given p = (pi,... ,p,,,), where the p, are non-negative integers, denote EIpjI=1

and 4' um by Ipl and 91, respectively. Set

Pi (t, E, u') _ gP1,p(t, E), B, (t, e,

ipl>1y

$pl?1

Ip1?1

6. AN APPLICATION OF THE FLOQUET THEOREM 229

Then, equation (VII.6.12) is equivalent to

(VII.6.13)8

p = B0(E)P1,p - P1,pBo(E) + Q1,p(t, E) -

where

(VII.6.14) Q1,p(t,E') = E (B1,ptP1,p2 - Pi,p2Hl,ps) + B1,p.P1+p2=p, 0p1IJ{r21>_1)

Hence,

(VII.6.15)Pi,p(t,e-) = {c( + 1 exp[-sBo(0

x 1 Q1,p(s, ) - H1,$,(e)) exP[sBo(E))ds} exp[-t B0(Z)1,

where C(e) and H1,p(e) are n x n matrices to be determined by the condition thatP1,p(t,e) is periodic in t of period w, i.e.,

exp[wBoQ-)1C(E) - C(i) exp[wBo(E)1

(VII.6.16)- exp[wBo(E)1

Jexp[-sBo(e)1H1,p(i) exp[sBo(i)1ds

0

_ - exp[wBo(E)1 exp[-sBo(e)1Ql.p(s, e) exp[sBo(t1ds.0

It is not difficult to see that condition (VII.6.16) determines the matrices C(ep)and Hl,p(ej. Then, the matrix P1,p(t, E) is determined by (VII.6.15). The conver-gence of power series P1 and H1 can be shown by using suitable majorant series. 0

In the same way as the proof of Theorem VII-6-3, the following theorem can beproven.

Theorem VII-6-6. The transformation

(VII.6.17) it = {In + Pi (t,h(t),h'(t)))v"

changes differential equation (VII.6.10) to

dv = fH(i(t)) + H1(h(t), h'(t))dt l

(VII.6.18) - [In + P1(t, h(t), hh'(t)),'1 I hj(t) aP1(t, h(t), h'(t})l

+ h (t) a_ (t, h{t), h'(t))J j v ,

where P1(t, E, µ) and H1(E, µ) are those two matrices given in Lemma VII-6-4.

Observe that under assumption (VII.6.3), we obtain

f0+00

(VII.6.19)

In + P1(t,h(t),h'(t))j_18Ej

+ hJ (t) a. (t, h(t), h'(t))J dt < +oo.

Observe also that the matrix H(h(t))+H1(h(t), h'(t)) does not contain any periodicterms. Furthermore, H(6) + H1(6, 6) = H(0). It is clear that the derivative ofH(h(t)) is not necessarily absolutely integrable over to < t < +oo. Therefore, inorder to apply the argument of MI-5, the matrix H(h(t)) must be examined moreclosely in each application. Details are left to the reader for further observation.

EXERCISES VII

VII-l. Find the Liapounoff's type number of each of the following four functionsf (t) at t = +00:

sint(dt](3) inin(exp[3t],exp[5t]sin67rt]),(1) exp [t2sin (1L1) ]' (2) exp Lfo

(4) the solution of the initial-value problem y"-y'-6y = eat, y(O) = 1, y'(0) = 4.

VII-2. Find a normal fundamental set of four linearly independent solutions of

the system dy = Ay" on the interval 0 < t < +oo, where

252 498 4134 698

A = -234 -465 -3885 -65615 30 252 42

-10 -20 -166 -25

Hint. See Example IV-1-18.

VII-3. Let 4i(x) be a fundamental matrix solution of the system

log(l+x) e= expx2

= x3 1+ 1+ x sin x

exp(e') cosx arctanx

on the interval 0 < x < +oo. Find Liapunoff's type number of det 4i(x) at x = +oo.

EXERCISES VII 231

VII-4. Assuming that the entries of an n x n matrix A(t) are convergent power

series in t-1, calculate lim p log I (t) for a nontrivial solution ¢(t) of the differential

equation t = A(t)9 at t = +oo.

Hint. Set t = e".

VII-5. Let dY = xP-'A(x)i be a system of linear differential equations such thaty E C" is an unknown quantity, p is a positive integer, the entries of an n x n matrixA(x) are convergent power series in x-1, and lim A(x) is not nilpotent. Show that

there exists a solution g(x) of this system such that limln(l y-(Teie)j)

is a positiver-.+oo rpnumber for some real number 6.

Hint. Set x = re!e for a fixed 0 and t = rp. Then, the given system becomesd# e+'0 e' PO e'= -A(re's)y'. Note that lim -A(re'B) _ -A(oo). The eigenvalues

P r-.+oo P Pei>' ei_)Iof

PA(oo) are p ' , where the a, are the eigenvalues of A(oo). Now, apply

Corollary VII-3-7.

VII-6. For each of two matrices A(t) given below, find a unitary matrix U(t)analytic on (-oo, oo) for each matrix such that U-' (t)A(t)U(t) is diagonal or upper-triangular.

r 111 t 0

(a) A(t) =L

Ot 0J, (b) A(t) = t 1 + 2t 0i

0 0 1+t2

Hint. See [GH].

VII-7. Show that if a function p(t) is continuous on the interval 0 < t < +oo and

lim t-Pp(t) = I for some positive integer p, the differential equation d2 +p(t)y =tt +oo0 has two linearly independent solutions rlf(t) such that

t77.k(t) = P(t)-114(1 +o(1))exp [±ij )ds(,JJ

77, (t) = P(t)114(+1 +o(1))exp [f iJ d4]tto

,

as t +oo, where to is a sufficiently large positive number and o(1) denotes aquantity which tends to zero as t - +oo.

Hint. Use an argument similar to Example VII-4-8.m-1

VII-8. Let Q(x) = x' + E ahxh and P(x, e) _ ,Tn+l + Q(x), where in is ah=o

positive integer, x is an independent variable, and am-1) are complex

parameters. Set A(x, f) = {(0)11. Also, let Ro, po, and ao be arbitrary but

fixed positive numbers. Suppose that 0 < po < 2. Show that the system

d:V= A(x,f)y, y = 17121

has two solutions y, (x, f) (j = 1, 2) satisfying the following conditions:(a) the entries of y, (x, f) are holomorphic with respect to (x, f, ao,... , am_ 1) in

the domainD={(x,f,ao,...,am_,): rEC, 0<Ifl<ao, Iargfl<Po,

laol + ... + Iam-I I < Ro},

(b) y', (x, f) (j = 1, 2) possess the asymptotic representations

exp P(t,f)1/2dtl111 +o(1)JP(x,f)

L fro J L it - o(1)]P(x,f)1/4

expl- Jro P(t,E) 1/2dt

11 +o(1)] P(x,E)-1/4, f -l + o(1)]P(x, f)'/4,

respectively as x - +oc on the positive real line in the x-plane, where xois a positive number depending on (Ra.po.ao) and o(1) denotes a quantitywhich tends to 0 as x ---. +oo on the positive real line uniformly with respectto (f, ao,... , am-I) for IEI < ao, I arg EI < po, and Iaol + ... + lam-1l < Ro.

Hint. Use a method similar to that for Exercise V1I-7. See, also, [Nful and [Si13,Chapter 3].

VII-9. Let A1(t)...... (t) be n continuous functions of t on the interval Zo ={tER:0<t<+oo}such that R.eja1(t)-aJ+1(t))>1(j=1,...,n-1) onAlso, let A(t) be an n x n matrix such that the entries are continuous in t on Zoand that

sup(1 + p - t) 1 1 I A(s) l ds - 0 as t -+ +oo.pp>t 1

Show that there exist a non-negative number to and an n x n matrix T(t) such that

(1) the derivative

E(t) exists and the entries of T(x) and . (x) are continuous

in t on the interval 2 = It : to < t < +oo},(2) t limo T(t) = 0,(3) transformation (I + T(x)) z changes the system

dt = (diag[A1(t),A2(t),... +T(t))yto

d= diag(A1(t) + bl (t), A2(t) + b2(t), ... , An(t) +

dtwhere y E C, i E C, n functions b1(t),... ,b,,(t) are complex-valued andcontinuous in t on T, and lim b,(t) = 0 (j = 1, ... , n).

t +00

EXERCISES VII 233

VII-10. Show that if R(t) is a real-valued and continuous function on Zo = {t E It :+00

0 < t < +oc} satisfying the condition J JR(t)j < +oo, the differential equation0

d2y+ (1 + R(t))y = 0 has a solution q5(t) such that lim (O (t) - sint) = 0. Also,

dt2 c-+OQ

show that 0(t) has infinitely many positive zeros an such that lim n = 7r.n+00 n

VII-11. Show that every solution of a differetial equation tz + R(t)y = 0 has

at most a finite number of zeros on the interval Zo = {t E Lea : 0 <_ t < +oo}if R(t) is a real-valued and continuous function on Zo satisfying the condition

L+a,

tIR(t)j < +oc.

+00tjR(t)j = +oo,Remark. See (CL, Problem 28 on p. 1031. For the case when f

of+a

but J IR(t)l < +oo, see Example VI-1-i.0

VII-12. Suppose that u(t) is a real-valued, continuous, and bounded function of tr+00

on the interval 0 < t < +oo. Also, assume that J ju(t)jdt < +oc. Show that if0

X is an eigenvalue of the eigenvalue problem

f+00d'y7t2 + u(t)y = Ay, y(O) = 0,

Jy(t)2dt < +oo,

0

then 0 < A < sup Ja(i)l.o<:<+c*

VII-13. Let A(t) be an n x it matrix whose entries are real-valued, continuous,and periodic of period 1 on R. Show that there exist two n x n matrices P(t, c) andB(c) such that

(a) the entries of P(t, c) and B(c) are power series in c which are uniformly con-vergent for -oc < t < +oo and small JEJ,

(b) P(t, 0) = I, (-oo < t < +oo), where In is then x n identity matrix,(c) the entries of P(t, c) is periodic in t of period 1,(d) P(t, c) and B(E) satisfy the equation

eP8t = cIA(t)P(t, e) - P(t, c)B(c)j.

VII-14. Apply Lemma VII-6-2 to the following system:

dy =dt

[nL + (a + 8cos(2t))(K - L)Jg, g _[Y2J

where n is a positive integer, the two quantities a and,3 are real parameters, and

K = f0U]

and L = [ 01 0]


VII-15. Using Theorem VII-6-3, find the asymptotic behavior of solutions of thedifferential equation

2 + {1 + h(t) sin(at)} rl = 0, h(t) = ln(2 + t)

as t -+ +oo, where a is a real parameter.

CHAPTER VIII

STABILITY

In the previous chapter, we explained the behavior of solutions of linear systemsas t -. +oo. In this chapter, we look into similar problems for nonlinear systems.To start with, in §VIII-1, we introduce the concepts of stability and asymptoticstability of a given particular solution as t +oo. We illustrate those conceptswith simple examples. Reducing the given solution to the trivial solution by asimple transformation, we concentrate our explanation on the stability propertyof the trivial solution. It is well known that the trivial solution is asymptoticallystable as t +oo if real parts of eigenvalues of the leading matrix of the givensystem are all negative. This basic result is given as Theorem VIII-2-1 in §VIII-2.The case when some of those real parts are not negative is treated in §VIII-3. Inparticular, we discuss the stable and unstable manifolds. In §VIII-4, we look intothe structure of stable manifolds more closely for analytic differential equations.First we change a given system by an analytic transformation to a simple standardform. By virtue of such a simplification, we can construct the stable manifold in asimple analytic form. This idea is applied to analytic systems in IR2 in §VIII-6. In§§VIII-7-VIII-10, using the polar coordinates, we explain continuous perturbationsof linear systems in J2. In §VIII-5, we summarize some known facts concerninglinear systems with constant coefficients in JR2. The topics discussed in this chapterare also found in [CL, pp. 371-388], [Har2, pp. 160-161, 220-227], and [SC, pp.49-96]. The materials in §§VIII-4 and VIII-6 are also found in [Du], [Huk5], and[Si2].

VIII-1. Basic definitions

Let us consider a system of differential equations

(VIII.1.1)dyd t=

where y E lR" and the R'-valued function At, M is continuous on a region

A(ro) = Zo x D (ro) = {(t, y1 E J"+I : 0 < t < +oo, 1171 < ro}.

Also, assume that a solution mo(t) of (VIII.1.1) is defined on the entire interval 10and that (t, mo(t)) E A(ro) on Yo. The main topic in this chapter is the behaviorof solutions of the initial-value problem

(VIII.1.2) dt = f (t, o'

as t +oo. To start, we introduce the concept of stability.

235

236 VIII. STABILITY

Definition VIII-1-1. The solution do(t) is said to be stable as t - +oo if, forany given positive number c, there exists another positive number b(c) such thatwhenever I&(0) - 7 < 5(c), every solution 4(t) of initial-value problem (VIII.1.2)exists on the entire interval Zo and satisfies the condition

(VIII.1.3) I$o(t) - (t)I < E on 10.

Remark VIII-1-2. Suppose that the initial-value problem

(VIII.1.2.r)dt

= At' y), y(r) = it

has the unique solution y = (t, r, i) if (r, rl) E 0(ro). Then, qs(t, r, 771 is continuouswith respect to (t, r, '). Therefore, for any r E Zo and any given positive numbersT and f, there exists a positive number p(r,T,e) such that whenever 100(r) - )1 <p(r, T, c), the solution 0(t, r, 71) exists on the interval 0 < t < T and I0o(t) -¢(t, r, Y-7)1 < E on the interval 0 < t < T (cf. §11-1). This implies that if the solution0o(t) is stable as t +oo, then for any r E Zo and any positive number c, thereexists another positive number b(r, c) such that whenever loo (,r) - n7 < b(r, E), thesolution (t, r, rt) exists on the entire interval 10 and (t, r, J) satisfies the conditionIV'0(t) - (¢(t, r, T-D I < E on I.

We also introduce the concept of asymptotic stability.

Definition VIII-1-3. The solution do(t) is said to be asymptotically stable ast-4 +00 if

(i) the solution do(t) is stable as t - +oo,(ii) there exists a positive number r such that whenever I o(0) - i7 < r, every

solution fi(t) of initial-value problem (VIII. 1.2) satisfies the condition

(VIII.1.4) lim I&(t) - (t)I = 0.

Remark VIII-1-4. Set g= E+ ¢0(t). Then, system (VIII.1.1) is changed to

(VIII.1.5) dt = At' z + do(t)) - f'(t, do(t)).

Hence, the study of the solution o(t) of (VIII.1.1) is reduced to that of the trivialsolution z1(t) = 0 of (VIII.1.5). Thus, the solution .o(t) of (VIII.1.1) is stable(respectively asymptotically stable) as t -+ +oo if and only if the trivial solution of(VIII.1.5) is stable (respectively asymptotically stable) as t -- +co. In the followingsections, we shall study stability and asymptotic stability of the trivial solution.

The following three examples illustrate stability and asymptotic stability.Example VIII-1-5. The first example is the system given by

(VIII.1.6) ddtl = -yi, dt = (yl - y2)y2

1. BASIC DEFINITIONS

To find the general solution of (VIII.1.6), solve the first-order equation

(VIII.1.7) dye = -y2 + .dyi y1

237

The transformation u = b2 changes this equation to a first-order linear differential

du 2 vt a-2nequation - = 2u - -. Thus, we obtain u(yl) = ezvi c - 2'I dq . Since

7Y1 yl rl

u = yz > 0, the quantity c must be positive. It is evident that y1(t) = y e-tsatisfies the first equation of system (VIII.1.6) with the initial value yt(0) = y.Therefore,

(VIII.1.8) yl(t) = -( e-t, 112(t) = ±u(yi(t))

is a solution of (VIII.1.6). Observe that

y<yl(t)<0 if y<0 and 0<y1(t)<y if y>0

for t > 0. Therefore, Jry, (t)-2

dt) < 0 for t > 0. Thus, y2(t)2 <e-2Yt(t)

J. 77

C

e2171

c= e21',1e2'ry2(0)2 for t > 0. This proves that the trivial solution of system

(VIII.1.6) is stable as t +oc. Furthermore, since

fvt(t) a-2q a-2nlim J -dt7 = lim

fYidi = -oo,t-+o0 r) Y1 -0 11

the trivial solution of system (VIII.1.6) is asymptotically stable as t -- +oo. Thisresult is shown also by Figure 1.

Y2

FIGURE 1.

Example VIII-1-6. The second example is a second-order differential equation

(VIII.1.9)d-t2

+ g(t7) = 0,

1

238 VIII. STABILITY

where 9(17) is a real-valued and continuously differentiable function of g on the realline R. Equation (VIII.1.9) is equivalent to the system

(VIII.1.10) dti = Y2,d

= -9(yi)

If g(a) = 0, then system (VIII.1.10) has a constant solution yt = a, Y2 = 0. Setyi = zi + a and y2 = z2. Then, system (VIII.1.10) becomes

(VIII.1.11)

where

diiii = Az + #(z-),

z = lz2J , A = j-9'(a) 0] [-[g(21 + a) -9()z1)]

The eigenvalues of A are ±(-g'(a))I/2. Note also that

lim g(zl + a) - g'(a)zt = 0.z,-0 zi

Set

17

(VIII.1.12) G(i7) = / g(s)ds.0

Then, G(n) takes a local minimum (respectively a local maximum) at n = a ifg(a) = 0 and g'(a) > 0 (respectively g'(a) < 0). Also, set

(VIII.1.13) y2) = G(yi) + y2H(yi ,2

Then,dH(yi(t),y2(t)) dyl(t) dy2(t)(t)

( (t)) +=

Therefore,

Y1 Y29dt dt dt

dH(yi(t),y2(t))= 9(YI(t))y2(t) - y2(t)9(yI(t)) = 0

dt

if (yi(t),y2(t)) is a solution of (VIII.1.10). This means that

(VIII.1.14) G(zi(t) + a) + z2(t)2 =

G(zi(0) + a) + z2 0)2

2 2for all t

if z(t) is a solution of (VIII.1.11).Case when g'(a) < 0: If g'(a) < 0, there exists a positive number PO such that

G(rl) < G(a) for 0 < 177 - al < Po Let t;(t) be the unique solution of (VIII.1.11)satisfying the initial condition z1(0) = 0, z2(0) = Co > 0. Since t;2(t) = (1(t),

1. BASIC DEFINITIONS 239

we obtain ('(t) = 2(G(a) - G((1(t) + a)) + co >- Co as long as 0 < (1(t) < po.Hence, there must be a positive number to depending on (o such that (I (to) = Pano matter how small to may be. This implies that the trivial solution of (VIII.1.11)is not stable.

Case when g'(a) > 0: For a given positive number p, let V (p) be the connectedcomponent containing 0 of the set {[; : G(( +a) < G(a)+p}. Then, V (pl) C V (P2)if p1 < p2. Furthermore, if g(a) = 0, g(a) > 0, and if a positive number p issufficiently small, there exist two positive numbers e1(p) and e2(p) such that(1) e2(P) < e1(P),

(2) Plmoe1(P) = 0,

(3) {( : ISO < C2 W} C 7) (P) C < el(p)} (cf. Figure 2).S=o

G= G(S+a)

G=G(a)+p

G=G(a)

1 P)

FIGURE 2.

Observe also that G(a) < G(( + a) < G(a) + p for [; E V (p) if p is a sufficientlysmall positive number.

For a given positive number e, choose another positive number p so that ei(p) < e

and p:5 e. Choose also the initial value z(0) so that Jz1(0)j < e2 (2) and z2(0)2 < p.

Then, G(z1(0)+a) < G(a)+

2

since z1(0) EP (2). Hence, (VIII.1.14) implies that

G(zi(t) + a) < G(a) + p for all t. Observe that the set {z1(t) : all t} is connectedand zi(0) E D (2) C V (p). Therefore, zi(t) E D(p) for all t. Hence, Iz1(t) <ei (p) < e. On the other hand, G(a) < G(z1(t) + a) < G(a) + p since z1(t) E D (p)

2(tfor all t. Hence, (VIII.1.14) implies that z2< G(z1(0) +a) - G(a) +

2

< p < e.

This proves that the trivial solution of system (VIII.1.11) is stable as t -+ +oo.The analysis given in Example VIII-1-6 is an example of an application of the

Liapounoff functions to which we will return in Chapters IX and X.The third example is the following result.

Theorem VIII-1-7. If f (x, y) and g(x, y) are real-valued continuously differen-tiable functions such that

(i) f (0, 0) = 0 and g(0, 0) = 0,(ii) (f (x, y), g(x, y)) 0 (0, 0) if (x, Y) # (0, 0),

(x, y) = 0,(iii) 8x (x, y) +49Y

then, the trivial solution (x, y) = (0, 0) of the system

(S) dt = f(x,y), dt = g(x,y)

240 VIII. STABILITY

is not asymptotically stable as t -- +oo.

Proof.

A contradiction will be derived from the assumption that the trivial solution

is stable. Let t CI, C2)x(t, ct, c2) be the unique solution ofasymptotically (, t, 2) y(t,cl,c2)

system (S) satisfying the initial condition 0(0.cl, c2) =[ci]. If the trivial solution

is asymptotically stable as t -. +oo, the trivial solution is also stable as t -. +oo.Therefore, for every positive number c, there exists another positive number 8(e)such that it (t,cl,c2)1 < e for 0 < t < +oo whenever max{jcj1,lc21} < b(e). Sincef and g are independent of t, O(t - r, cl, c2) is also a solution of (S) and satisfiesthe initial condition (x(r), y(r)) = (Cl, c2). This implies that I¢(t, cl, c2)1 < e forr < t < +oo if I¢(r,c1,c2)I < 8(e). Denote by 0(r) the disk {(c1,c2) E 1R2 :

Ic112 + 1c212 < r}. Also, for a fixed value r of t, let us denote by D(r, r) theset {0r,C1,c2) : (cl,c2) E A(r)}. Then, the mapping (C1, C2) --+ (r,Cl,C2) is ahomeomorphism of 0(r) onto D(r, r) (cf. Exercise 11-4). Fix a sufficiently smallpositive number ro. Since (0, 0) is asymptotically stable as t +00 and the disk

ro lA(ro) is compact, there exists a positive number ro such that D(ro,ro) C 02 ).

This implies that the area of V(ro, ro) is definitely smaller than the area of A(ro).Observe that(VIII.1.15)

area of D(ro, ro) =fA(r0)

det tO(roc1c2) a$(ro, C1, C2)8c1 8c2

It is known that the matrix 1L(t,cl,c2) =

unique solution of the initial-value problem

1of ofdX ax 8y

dc 1 de,.

aj(ro,c1,c2)is

L ac, 0Ce

dt a9 09ax Dy tr,y)=Oit.c,,cz)

X. X (0) = I2.

where 12 is the 2 x 2 identity matrix (cf. Theorem 11-2-1). Therefore,

r /'zJ

(det '(t, cl, c2) = expLI 8f (x, y) + ag (x, y)1 I dtJ = 1ft=,v>=ecl.c,,cz1o l ax Oly

the

(cf. (4) of Remark IV-2-7). Now, it follows from (VIII.1.15) that the area ofD(ro, ro) is equal to the area of 0(ro). This is a contradiction. 0

2. A SUFFICIENT CONDITION FOR ASYMPTOTIC STABILITY 241

VIII-2. A sufficient condition for asymptotic stabilityIn this section, we prove a basic sufficient condition for asymptotic stability. Let

us consider a system of differential equations

(VIII.2.1) = Ag + g(t,yj

under the assumptions that(i) A is a constant n x n matrix,

(ii) the entries of the 1R"-valued function §(t, y-) are continuous and satisfy theestimate

(VIII.2.2) I9(t,y)l < k(t)lyy + colylt+w for (t, y)) E A(ro),

where, vo > O, co > O, ro > O, k(t) > O and bounded fort > 0, lim k(t) = 0,t+xand A(ro) = {(t, yj : 0 < t < +oo, Iy1 < ro).

The following theorem is the main result in this section.

Theorem VIII-2-1. If the real part of every eigenvalue of A is negative, the tnvialsolution g = 0 of (VIII. 2.1) is asymptotically stable as t -i +oo.

Proof.

We prove this theorem in four steps.

Step 1. Let A, (j = 1, 2,... , n) be the eigenvalues of A. Then, RIA,I (j = 1, 2, ... ,

n) are Liapounoff's type numbers of the system d= Ay (cf. Example VII-2-8).t

Therefore, choosing a positive number p so that 0 < p < -9RIA,I f o r j = 1, 2, ... , n,we obtain

IeXpIAtll < Ke-"t on the interval To = {t : 0 < t < +oc}for some positive constant K.

Step 2. Fixing a non-negative number T, change system (VIII.2.1) to the integralequation

g(t) = e(t-T)Ag(T) + fte(t-')Ag(s,g(s))ds.T

If Ig(t)I < b for some positive number b on an interval T < t < T1, then

ly(t)l <- Ke-"(t-T)Ig(T)I + Kff e-"(t-d) {k(s) + cabv0} Ig(s)Ids

on the same interval T < t < Ti. Setting p(T) = sup k(s), change this inequalitya>T

to the forme"(t-T)Ig(t)I <- Klg(T)I + K {p(T) + coa"o}

T

e"('-T)Iy(s)Ids

for T:5 t < Tt. Then,e"(t-T)lg(t)I <- Klg(T)I exp[K(p(T) + cob&°)(t -T)I

and, hence,

(VIII.2.3) lg(t)I <- KI g(T)I exp I- (u - K(p(T) + cotVO))(t - T)Jfor T:5 t < Tl (cf. Lemma 1-1-5).

242 VIII. STABILITY

Step 3. Fix a non-negative number T and two positive numbers 6 and 61 in sucha way that

p-K(p(T)+co6'o) > 0, 0<61 <6, K61 <6.

Assume that 19(T)J < 61. Then, inequality (VIII.2.3) holds for T < t < T1 as longas Iy(t)I < bon the interval T < t < T1. This, in turn, implies that

Jg(t)J < K61 < 6 for T < t < T1.

This is true for all T1 not less than T. Hence, inequality (VIII.2.3) holds for t > T.

Step 4. If Jy(t)J < b < 1 for 0 < t < T, there exists a positive number is such

that I d t)I < n19(t)j, for 0 < t < T. This implies that Jy-(t)J < Jy-(0)JeK' as long

as Jy(t)I < 6 T if Iy(0)J is small. This completes theproof of Theorem VIII-2-1. O

Example VIII-2-2. For the following system of differential equations d = A#+where

( 1y = [ yj A = r -0.4--2

.21 (Y2

+ y2) I 11JY

the trivial solution y = 0 is asymptotically stable as t -' +oc. In f act, the char-acteristic polynomial of the matrix A is PA(a) _ (A + 0.3)2 + 1.99. Therefore, thereal part of two eigenvalues are negative.

Remark VIII-2-3. The same conclusion as Theorem VIII-2-1 can be proven, evenif (VIII.2.2) is replaced by

(VIII.2.4) 19(t, y-)J < (k(t) + h(t)Jyj")Jy7 for (t, y) E A(ro),

where p is a positive number, and two functions h(t) and k(t) are continuous fort > 0 such that k(t) > 0 for t > 0, lim k(t) = 0, h(t) > 0 fort > 0, andLiapounoff's type number of h(t) at t = +oo is not positive (cf. [CL, Theorem 1.3on pp. 318-319]).

Remark VIII-2-4. The converse of Theorem VIII-2-1 is not true, as clearly shownin Example VIII-1-5. In that example, the eigenvalues of the matrix A are -1 and 0,but the trivial solution is asymptotically stable as t -+ +oo. Also, in that example,solutions starting in a neighborhood of 0 do not tend to 0 exponentially as t -+ +oo.

Remark VIII-2-5. Even if the matrix A is diagonalizable and its eigenvalues areall purely imaginary, the trivial solution is not necessarily stable as t -+ +oo. Insuch a case, we must frequently go through tedious analysis to decide if the trivialsolution is stable as t -+ +oo (cf. the case when g'(a) > 0 in Example VIII-1-6).We shall return to such cases in IR2 later in §§VIII-6 and VIII-10 .

3. STABLE MANIFOLDS 243

Remark VIII-2-6. If the real part of an eigenvalue of A is positive, then thetrivial solution is not stable as it is claimed in the following theorem.

Theorem VIII-2-7. Assume that(i) A is a constant n x n matrix,

(ii) the entries of y(t, yam) are continuous and satisfy the estimate

I9(t,y)l 5 e(t,yyly7 for (t,y-) E o(ro) _ {(t,y-) : 0 5 t < +oo, Iy-I <_ ro},

where ro > 0, e(t, y-) > 0 for (t, y) E i(ro), and lim e(t, y) = 0,t-1+19{o

(iii) the real part of an eigenvalue of the matrix A is positive.

Then, the trivial solution of the systemdydt

= Ay + §(t, y-) is not stable as t -. +oo.

A proof of this theorem is given in (CL, Theorem 1.2, pp. 317-3181. We shall provethis theorem for a particular case in the next section (cf. (vi) of Remark VIII-3-2).An example of instability covered by this theorem is the case when g'(a) < 0 ofExample VIII-1-6. The converse of Theorem VIII-2-7 is not true. In fact, Figure 3shows that the trivial solution of the system

I= -yI, dY2

= (yi + yi)112

is not stable as t - +oo. Note that, in this case, eigenvalues of the matrix A are-1 and 0.

Y2

Yt

FIGURE 3.

VIII-3. Stable manifoldsA stable manifold of the trivial solution is a set of points such that solutions

starting from them approach the trivial solution as t --. +oo. In order to illustratesuch a manifold, consider a system of the form

(VIII-3-1)dx = AIi + LY = Alb + 92(40,

where Y E iR", 11 E R n, entries of R"-valued function gl and RI-valued function g2are continuous in (1, y-) for max(IiI, Iyj) 5 po and satisfy the Lipschitz condition

19't(_,y)- 9,(f,nil 5 L(P)max(Ii-.1, Iv-ffl) (1=1,2)

244 VIII. STABILITY

for max(jil, Iy1) < po and max(11 i, Ii1) < po, where po is a positive number andu oL(p) = 0. Furthermore, assume that gf(0, 0) = 0 (j = 1, 2). Two matrices AI

and A2 are respectively constant n x n and m x m matrices satisfying the followingcondition:

(VIII.3.2)

where K, and of (j = 1,2) are positive constants. Condition (VIII.3.2) impliesthat the real parts of eigenvalues of AI are not greater than -01, whereas the realparts of eigenvalues of A2 are not less than -02. Assume that

(VIII.3.3) a1 > 02.

Let us change (VIII.3.1) to the following system of integral equations:

(VIII.3.4)

The main result in this section is the following theorem.

Theorem VIII-3-1. Fix a positive number c so that a1 > 02 + E. Then, thereexists another positive number p(() such that if an arbitrary constant vector c' in

R' satisfies the condition KI I c1 < 2 , a solution (i(t, c), y"(t, cam)) of (VIII.3.1) can

be constructed so that

(VIII.3.5) 1(0,61=6 and max(Ii(t, c)I, I y(t, c)I) < p(e)e-(Ol -t)t

for 0 < t < +oo. Furthermore, this solution (i(t, cl, #(t, c)) is uniquely determinedby condition (VIII.3.5).

Proof

Observe that if

max(Ix(t,c)I,Iy(t,c)I) < pe-(°'-`)tfor 0 < t < +00,

le(t-9)A' I < Kie-o1(t-a) for t > s,Je(t-s)A21 < K2e-02(t-8) for t < s,t

I

x(t, c) = e1AI C + J e(1-a)A' gi (Y(s, c-), y(s, c))ds,0

At, c ) = J e(t 8)A2 2(x(s, c), y(s, c))ds.tt o0

then

I fot e(t-s)A'91(Y(s, c), y(s, c))ds1 < K1 L(P)P ft e-`(t-8)e-(" -`)sds

0

KiL(P)Pe-0,t (e" - I) < Ki L(P)Pe-(o, -c)t

and

t

00

+00

IJ e(t-a)A2 2(j(s, c), y(s, c))dsl < K2L(p)p f+ t

e-02(t-8)e-(0 $ -')8ds

K2L(P)P e

al -o2-E

3. STABLE MANIFOLDS 245

This implies that if a positive number p is chosen so small thatK2L(p) < 1

01-o2-e - 2and K1L(p) <

2

and if the arbitrary vector c' in R" satisfies the condition

Kl I i <2

, then, using successive approximations, a solution (i(t, cl , y(t, c)) of

(VIII.3.4) can be constructed so that

x(O, c-)= c and max(Ii(t, c1I, Iy(t, c)I) < pe-(c for 0 < t < +oo.

Details are left to the reader as an exercise.

Remark VIII-3-2.(i) The positive number a is given to start with and the choice of p depends on

e. However, since this solution approaches the trivial solution, the constant emay be eventually replaced by any smaller number, since the right-hand sideof (VIII.3.1) is independent of t. This implies that the curve (i(t, c), y'(t, c-))is independent of a as t -. +oo. More precisely, if a solution (x"(t), y"(t)) of(VIII.3.1) satisfies a condition

max(Ii(t)I, ly(t)I) < Ke-(°'-'0)t for 0 < t < +00

with some positive constants K and eo such that of -o2-eo > 0, then for everypositive a smaller than co, there exists to > 0 such that (i(to + t), y(to + t)) _(i(t, ), y(t, cam)), where c = ?(to).

(ii) The initial value of y(t, c-) is given by

0

00, cl = f e-sA292(i(s, c_), y(s, c_))ds.+oo

exist and are continuous in a neighborhood of (6,6) and if(iii) Ifay

and09

(0, 0) = 0', then i(t, c) and y"(t, cl are continuously8 (6, U = 0' and89

differentiable with respect to c.(iv) If the real parts of all eigenvalues of the matrix Al of (VIII.3.1) are negative

and the real parts of all eigenvalues of the matrix A2 are positive, then thestable manifold of the trivial solution of system (VIII.3.1) is given by S =((6, 9(0, c-)) : 161 < p}, where p is a sufficiently small positive number.

(v) Consider a system

(VIII.3.6)dy-

Ay +dt 9(y

in the following situation:(a) A is a constant n x n matrix,(b) the entries of §(y1 are continuous for Iy1 < po and satisfy the Lipschitz

condition 19(y-) - L(p)ly - r11 for ly-I < po and i31 < po,where PO is a positive number and lim L(p) = 0,

p-0

246 VIII. STABILITY

(c) 9(0,6) = 6.

Suppose further that A has an eigenvalue with positive real part. Then, applying

Theorem VIII-3-1 to the system = -Ay' - g(y-), we can construct the stable

manifold U of the trivial solution. This means that if a solution fi(t) of (VIII.3.6)starts from a point on U, then urn Q(t) = 6. This shows that the trivial solution of

t - +oo, and Theorem VIII-2-7 is proved for (VIII.3.6).The set U is called the unstable manifold of the trivial solution of (VIII.3.6).

The materials in this section are also found in [CL, §§4 and 5 of Chapter 81 and[Hart, Chapter IX; in particular Theorem 6.1 on p. 2421.

VIII-4. Analytic structure of stable manifolds

In order to look closely into the structure of the stable manifold of the trivialsolution of a system of analytic differential equations

(VIII.4.1)dy

dt= Ay + f (y),

let us construct a formal simplification of system (VIII.4.1). To do this, considersystem (VIII.4.1) under the following assumption.

Assumption VIII-4-1. The unknown quantity g is a vector in C" with entries{y1, ... , y" }, A is a constant n x n matrix, and

(VIII.4.2) f (Y-) = E #vfajpI>2

is a formal power series with coefficients fp E C', where p = (pl,... , p") withn

non-negative integers pl, ... , p", > Ph, and yam' = yl' yP,,-.

h=1

The following theorem is a basic result concerning formal simplifications of sys-tem (VIII.4.1).

Theorem VIII-4-2. Under Assumption VIII-4-1, there exists a formal power se-ries

(VIII.4.3) P'(u) = Pou" + EPP,,

IpI>2

in a vector u E C" with entries {ul,... ,u"} such that(i) Po is an invertible constant n x n matrix and PP, E C",

(ii) the formal transformation y' = P(u") reduces system (VIII 4. 1) to

(VIII.4.4) j = Boii + g"(u)

4. ANALYTIC STRUCTURE OF STABLE MANIFOLDS 247

with a constant n x n matrix Bo and the formal power series g(ui) _ E upgpIpi>2

with coeficients gp in Cn such that(iia) the matrix B0 is lower triangular with the diagonal entries Al.... , An,

and the entry bo(j, k) on the j-th row and k-th column of B0 is zerowhenever A, j4 Ak,

(iib) for p with IpI > 2, the j-th entry gp, of the vector gp is zero whenever

n

(VIII.4.5)A1

1: PhAhh=1

Proof

Observe that if y" = P(d), then

d9 = p p Pp

dtPo + 1: [L Pp P22 Pp ... un Pp Boi + F 4Lpgp

W>2 Ipl>2

and

Furthermore,

Ay + f(y) = A Pod + F u-'pPp + E P(u)p fp.IpI>2 IPI>_2

[Plup,5, P2upP- poop P_pBo

J

u'Po +ul u2

p ...

UnIpl>2

= PoBou" +n

uhphAh u"PPp + plp),A-lpl>2 h_1 1<h<35n

ul

where 03,k is the entry of B0 on the 7-th row and k-th column.Let us introduce a linear order pi -< p2 for p. = (p,',... , pin) (j = 1, 2) by the

relationPlh = p2h for (h < he) and P1ho < P2ho

Now, calculating the coefficients of iip 1) on both sides of (VIII.4.1), weobtain

(VIII.4.6) PoBo = APo

and

(VIII.4.7)(PhAh)P + Po9p - APp

h=1

=j (I5,' p -<p) + yp(Pp',ff : 10 <ItI)

248 VIII. STABILITY

for Jpj > 2. From (VIII.4.6), it is concluded that the diagonal entries A1, ... , A,, ofBo are eigenvalues of A and that this allows us to set A = B0 and P0 = In, whereI,, is the n x n identity matrix. Then, (VIII.4.7) becomes

(VIII.4.8)(PhAh)PP + 9p - Bo,6,

h=1

= ff(PP, p) + 9$'05P" 9P, : 10 < 1p1)

Solve (VIII.4.8) by solving equations of the form

(VIII.4.9) (PhAh - A7J

PP,1 + 9P.Jh-

= F'p,J

successively, where PP,, and gp,, are the j-th entries of the vector P. and

respectively, and FP,, are known quantities. If >phAh - Aj 36 0, set 9,j = 0 andh=1

solve (VIII.4.9). If >phAh - A? = 0, then set 9P,, = Fr,, and choose PPJ in anyh=1

way. This completes the proof of Theorem VIII.4.2.

Observation VIII-4-3. Assume that R(AE) < 0 for j = 1, ... , r and R(A3) > 0for j 1,... , r. In this case, if uh = 0 for h I, ... , r, the a-th entry of thevector Bou + g"(u) is zero for t ¢ I, ... , r. In fact, look at g"(u"). Then, the f-

n

th entry of the coefficient g"p is zero if Al 96 >phAh. Note that, if t > r andh=1

n

At = > Ph Ah, then (Pr+ 1, . Pn) 0 (0, ... , 0). Hence, in such a case, uP = 0 ifh=1

(ur+1, ... , u,) = (0, ... , 0). Therefore, the 1-th entry of the vector Boti + 9"(uiszero forI>rif(ur+1,...,u,,)=(0,....0).Observation VIII-4-4. Under the same assumption on the Aj as in ObservationVIII-4-3, set (ur+1, ... , un) = (0,... , 0). Then, the system of differential equationson (u1,... has the form

(VIII.4.10)

dud = A3u3+ QJ,huhdt

+ 9(P,.A,=p,a,+ +p.a.,lai>2

.P04ul1 ... u p , ( i=1 , - - -... , r)-

r rObserve that since A. - ph Ah # 0 if ph is sufficiently large, the right-hand

h=1 h=lmembers of (VIII.4.10) are polynomials in (ul,... ,ur).

4. ANALYTIC STRUCTURE OF STABLE MANIFOLDS 249

Observation VIII-4-5. Assume that R[Ah+1J <_ R[Ah]. Then, (V111.4.10) can bewritten in the form

dul du

dt = Aiu1 and = A)u) + u_,-i) for j=2,... ,r.

Hence, system (VIII.4.10) can be solved with an elementary method. To see thestructure of solutions of (VIII.4.10) more clearly, change (ul, ... , ur) by u2 _ea'ivi (j = 1,... , r). This transformation changes (VIII.4.10) to

dv)

dt Q),hvhan

g(P,.. ...vf (.i

A, =P, a,+...+prAr,Ipl>2

This, in turn, shows that the general solution of (VIII.4.10) has the form u) _ea'ii) (t, cl, ... , cr) (j = 1.... , r), where (cl,... , c,.) are arbitrary constants and0,(t,c1,....c,-) are polynomials in (t.cl,... ,G)

An analytic justification of the formal series 15(ii) is given by the following the-orem.

Theorem VIII-4-6. In the case when the entries of the Z"-valued function f'(y-)on the right-hand side of (VI11.4.1) are analytic in a neighborhood of 0, under thesame assumption on the A) as in Observation VIII-4-3, the power series P(d) isconvergent if (ur+i, ... , u") = (0,... , 0).

The proof of this theorem is straight forward but lengthy (cf. [Si2J). The keyfact is the inequality

A) - >PhAhh=1 h=1

r

for some positive number a if E ph is large. In this case, a iajorant series for Ph=1

can be constructed.

The construction of such a majorant series is illustrated for a simple case of asystem

dy=dt Ay + f (y),

where A is a nonzero complex number and AM is a convergent power series in ygiven by (VIII.4.2). According to Theorem VIII-4-2, in this case there exists a

ddformal transfomation ff = d + Q(u) such that

dt= Ad, where

IpI?2

This implies that E AIpJi1PQg, = AQ(u) + f (u" + Set f (ii + Q(u)) _IeI>2

250 VIII. STABILITY

ui"A,. Then,lal>2

Set

Apfor all p (IpJ > 2).a= AOPI-1)

1

' (y1= > IfP11FIpI>2 1

Then, F(y-) is a majorant series for ft y-). Determine a power series V (u-) = uupii

IpI>2

by the equation v = ICI 0(u" + v'). Set F(u + V(U)) _ iBa. Then,Ipl>2

vp =B

Ii for all p (JpJ > 2).

It can be shown easily that Y(ul) is a convergent majorant series of Q(ur). Thisproves the convergence of Q(ii).

Putting P(u") and the general solution of (VIII.4.10) together, we obtain a par-ticular solution y"

=

P(#(t, cl) of (VIII.4.1), where

(t,cj = (e'\It7Pi(t,cl,... ,cr),... ,e*\1 'Wr(t,cl,... ,cr),0,... ,0).

This particular solution is depending on r arbitrary constants c = (cl,... , c,.).Furthermore, this solution represents the stable manifold of the trivial solution of(VIII.4.1)if W(Aj)>0for j#1,...,r.

Remark VIII-4.7. In the case when y, A, and AM are real, but A has someeigenvalues which are not real, then P(yl must be constructed carefully so that the

particular solution P(i(t, c)) is also real-valued. For example, if A = a b

b a ,

the eigenvalues of A are a ± ib. If IV = ['] is changed by ul

= - iy, system (VIII.4.1) becomes

(VIII.4.11)

= 1/i + iy2i and

dul= (a + ib)ul + 9P,,p2u 'uZ

OFFPl+P2>2

due_ (a - ib)u2 +

d ,t

where gP,,- is the complex conjugate of g,,,p,. If a 54 0, using Theorem VIII-4-2,simplify (VIII.4.11) to

(VIII.4.12) dtl = (a + ib)vl,ddt

= (a - ib)v2

S. TWO-DIMENSIONAL LINEAR SYSTEMS 251


(VIII.4.13)

VI +

V2 +

PP,.P2'UP11t?221

p, +p2>_2

7P,.p2V2P1VP13.

P,+p2>2

Now, system (VIII.4.12), in turn, is changed back to - = At by wI = V 2 V2

VI - V2and w2 =

2i , where (w1, w2) are the entries of the vector tu. Observe that

U l + u22

uI - u22i

w1 + Ep,+p2>2

U'2 + g2.P,,pzu'P1 u"p, +p3 22

where ql,p,,p2 and g2,p,,p2 are real numbers. Similar arguments can be used ingeneral cases to construct real-valued solutions. (For complexification, see, forexample, [HirS, pp. 64-65].)

For classical works related with the materials in this section as well as moregeneral problems, see, for example, [Du).

VIII-5. Two-dimensional linear systems with constant coefficients

Throughout the rest of this chapter, we shall study the behavior of solutions ofnonlinear systems in 1R2. The R2-plane is called the phase plane and a solutioncurve projected to the phase plane is called an orbit of the system of equations. Adiagram that shows the orbits in the phase plane is called a phase portrait of theorbits of the system of equations. As a preparation, in this section, we summarizethe basic facts concerning linear systems with constant coefficients in R2.

Consider a linear system

(VIII.5.1)dy =dt Ay,

where E R2 and A is a real, constant, and invertible 2 x 2 matrix. Set p = trace(A)and q = det(A), where q ¢ 0. Then, the characteristic equation of the matrix A is1\2 - pA + q = 0. Hence, two eigenvalues of A are given by

AI = + 4 q and A2 = 2 - - q.2 4

It is known that

p = AI +A2, q = AIA2, and Al - A2 =2 P2- q.

4

252 VIII. STABILITY

Also, let t and ij be two eigenvectors of A associated with the eigenvalues Al andA2, respectively, i.e., At = All;, l 0 0, and Au = A2ij, ij 0 0. Observe that, ify(t) is a solution of (VIII.5.1), then cy(t +r) is also a solution of (VIII.5.1) for anyconstants c and T. This fact is useful in order to find orbits of equation (VIII.5.1)in the phase plane.

Case 1. Assume that two eigenvalues Al and A2 are real and distinct. In thiscase, two eigenvectors { and ij are linearly independent and the general solution ofdifferential equation (VIII.5.1) is given by

1!(t) = cl eAI t + c2ea'tti = e)lt (cl + c2e(1\2-,,,)t11= ea2t[cie(AI -a2)t (+ c2i17,

where cl and c2 are arbitrary constants and -oo < t < +oo.2

la: In the case when Al > A2 > 0 (i.e., p > 0, q > 0 and 4 > q), the phase portrait

of orbits of (VIII.5.1) is shown by Figure 4. The arrow indicates the direction inwhich t increases. The trivial solution 0 is unstable as t +oo. Note that ast -oo , the solutions y(t) tends to 0 in one of the four directions of and-ij. The point (0, 0) is called an unstable improper node.

2

lb: In the case when 0 > Al > A2 (i.e., p < 0, q > 0 and 4 > q), the phaseportrait of orbits of (VIII.5.1) is shown by Figure 5. The trivial solution 0 is stableas t --i +oo. The point (0, 0) is called a stable improper node.

4

FIGURE 4. FIGURE 5.

1c: In the case when Al > 0 > A2 (i.e., q < 0), the phase portrait of orbits of(VIII.5.1) is shown by Figure 6. The trivial solution 0 is unstable as Itl +oo.Note that as t -+ -oo (or +oo), only two orbits of (VIII.5.1) tend to 0. The point(0, 0) is called a saddle point.

2

Case 2. Assume that two eigenvalues Al and A2 are equal. Then, q = 4 and

Al=A2=29& 0.

2a: Assume furhter that A is diagonalizable; i.e., A = 212i where 12 is the 2 x 2identity matrix. Then, every nonzero vector c is an eigenvector of A, and the general

solution of (VIII.5.1) is given by y"(t) = exp [t]e. In this case, the phase portrait

of orbits of (VIII.5.1) is shown by Figures 7-1 and 7-2. As t -+ +oo, the trivialsolution 6 is unstable (respectively stable) if p > 0 (respectively p < 0). Note that,

5. TWO-DIMENSIONAL LINEAR SYSTEMS 253

for every direction n, there exists an orbit which tends to 6 in the direction n" as ttends to -oo (respectively +oo). The point (0, 0) is called an unstable (respectivelystable) proper node if p > 0 (respectively p < 0).

FIGURE 6.

p>0

FIGURE 7-1.p

p<0

FIGURE 7-2.

2b: Assume that A is not diagonalizable; i.e., A = 212 + N, where 12 is the 2 x 2

identity matrix and Nris a 2 x 2 nilpotent matrix. Note that N 0 and N2 = O.

Hence, exp[tA] = exp 12t] {I2 + tN}. Observe also that a nonzero vector c' is an

eigenvector of A if and only if NcE = 0. Since N(Nc) = 0, the vector NcE iseither 6 or an eigenvector of A. Hence, NE = ct(-) , where is the eigenvectorof A which was given at the beginning of this section and a(c) is a real-valuedlinear homogeneous function of F. Observe also that at(c) = 0 if and only if c' is aconstant multiple of the eigenvector . The general solution of (VIII.5.1) is given

by y(t) = exp [t] {c + ta(c7}, where c" is an arbitrary constant vector. In this

case, the phase portrait of orbits of (VIII.5.1) is shown by Figures 8-1 and 8-2. Thetrivial solution 6 is unstable (respectively stable) as t - +oo if p > 0 (respectivelyp < 0). The point (0, 0) is called an unstable (respectively stable) improper node ifp > 0 (respectively p < 0).

p>0

FIGURE 8-1.

p<0

FIGURE 8-2.

Case 3. If two eigenvalues \I and .12 are not real, then q >

4and

Al =a+ib, A2=a-ib, a= 2,

4

Note that b > 0. Set A = a12 + B. Then, B2 = -b212, since two eigenvalues of B

254 VIII. STABILITY

are ib and -ib. Therefore, exp[tB] = (cos(bt))I2 +Smbbt)

B.

3a: Assume that a = 0 (i.e., p = 0). Then, the general solution of (VIII.5.1) is

given by '(t) = exp[tB]c = (cos(bt))c"+slnbbt)

Bc, which is periodic of period 2b

in t. The phase portrait of orbits of (VIII.5.1) is shown by Figures 9-1 and 9-2.The trivial solution 0 is stable as It 4 +oo. The point (0, 0) is called a center. It

is important to notice that every orbit y(t) is invariant by the operator

b. In fact,

sin bt +r,

B j(t) = cos(bt) Bc-(sin(bt))c = (cos (bt + ))e+ 2 Be= 17 (t +j

)T b b 2b

In other words, dy(t) = by (t + . Note that "" is 1 of the period2r

dt 2b) 2b 4 b

FIGURE 9-1. FIGURE 9-2.

3b: Assume that a 0 0 (i.e., p 0 0). Then, the general solution of (VIII.5.1) is

given by y(t) = exp[at]il(t), where u(t) is the general solution of dt = Bu. The

solution 0 is unstable (respectively stable) as t - +oo if p > 0 (respectively p < 0).The orbits, as shown by Figures 10-1 and 10-2, go around the point (0, 0) infinitelymany times as y(t) -- 0. The point (0, 0) is called an unstable (respectively stable)spiral point if p > 0 (respectively p < 0).

(0.0)

p>0 p<o


Let us summarize the results given above by using Figure 11:(1) (0, 0) is an improper node.(2) (0, 0) is a saddle point.

6. ANALYTIC SYSTEMS IN R2 255

(3) (0, 0) is a proper or improper node.(4) (0, 0) is a center.(5) (0, 0) is a spiral point.

Stability of the trivial solution 6 is summarized as follows:(1) If q < 0, the trivial solution 0' is unstable as Itl -+ +oo.

(II) If q > 0 and p > 0, the trivial solution 0 is unstable as t +oo.(III) If q > 0 and p < 0, the trivial solution 0 is stable as t -+ +00.(IV) If p = 0 and q > 0, the trivial solution 0 is stable as It( - +oo.(Cf. Figure 12).

q(P0) q(P=0)

P2(5) (5) 4

(4) (IV)(3) (3) (111) (Ill

-------- -------


VIII-6. Analytic systems in R2

In this section, we apply Theorems VIII-4-2 and VIII-4-6 to an analytic systemin R2. Consider a system

(VIII.6.1)dt Ay" + E ypfp,

tp!>2

where y' E LR2 with the entries yj and y2, A is a real, invertible, and constant2 x 2 matrix. p = (pi, p2) with two non-negative integers pi and p2i Ipl = pi + p2,yam' = y'' y2 , the entries of vectors fp E R2 are real constants independent of t, andthe series on the right-hand side of (VIII.6.1) is uniformly convergent in a domainA(po) = (y E R2 : fyj < po) for some positive number po. Let us look into thestructure of solutions of (VIII.6.1) in the following five cases.

Case 1. If the point (0, 0) is a stable proper node of the linear system = Ay, thenA = )J2i where A is a negative number and 12 is the 2 x 2 identity matrix. To applyTheorem VIII-4-2 to this case, look at the equation A = pi.1 + p2A on non-negativeintegers pi and p2 such that pl + p2 > 2. Since no such (pi, p2) exists, there existsan R2-valued function P(u) whose entries are convergent power series in a vector

it E R2 with real coefficients such that 5 () = 12 and that the transformation

= P( u reduces system (VIII.6.1) to dt` = Ail. This, in turn, implies that the

256 VIII. STABILITY

point (0, 0) is also a stable proper node of (VIII.6.1) and that the general solutionof (VIII.6.1) is y = P(eJ1°cl, where c is an arbitrary constant vector in R2.

Case 2. If the point (0, 0) is a stable improper node of the linear system dt = Ay,

then we may assume that either (1) A = AI2+N, where 1A is a negative number and

N 4 0 is a 2 x 2 nilpotent matrix, or (2) A = I 1 a2J ,

where Al and A2 are real

negative numbers such that Al > A. In case (1), the same conclusion is obtainedconcerning the existence of an 12-valued function P(i7). Therefore, the point (0,0)is a stable improper node of (VIII.6.1), and the general solution of (VIII.6.I) isy" = P(eAt(12 + tN)c), where c is an arbitrary constant vector in 1R2. In case (2),looking at the equations Al = p1 A 1 + p2A2 and A2 = p1 At + p2A2 on non-negativeintegers pi and p2 such that pl + p2 ? 2, it is concluded that there exists an 1R2-valued function P(ui) whose entries are convergent power series in a vector ii E R2

with real coefficients such thatdP

(d) = 12 and that the transformation y" = P(u)

reduces system (VIII.6.1) to dt = ['Ju.&Ajul2u21, where (u1, u2) are the entries of

1

i , M is a positive integer such that A2 = MA1, and -y is a real constant which mustbe 0 if A2 54 MA1 for any positive integer M. Therefore, in case (2), the general

A,:solution of (VIII.1.6) is y" = PAlt , where cl and c2 are arbitrary

Ct 7t + C2)e )([(''constants. This, in turn, implies that the point (0, 0) is a stable improper node of(VIII.6.1).

Case 3. If the point (0, 0) is a stable spiral point of the linear systemdy

= Ay",

then we may assume that A = I b ab 1, where a and b are real numbers such that

a < 0 and b # 0. The eigenvalues of A are At = a ± ib. This implies that there areno non-negative integers pi and p2 satisfying the condition A = p1 A+ +P2A_ andpl + > 2. Therefore, there exists an R 2-valued function P(g) whose entries are

convergent power series in a vector u" E R2 with real coefficients such that LP (0) _

at

8u

12, and the transformation y = P(u) reduces system (VIII.6.1) to dt = Au (cf.Remark VIII.4.7). This, in turn, implies that the point (0, 0) is also a stable spiralpoint of (VIII.6.1) and that the general solution of (VIII.6.1) is y = P(eAC1, wherec" is an arbitrary constant vector in 1R2.

Case 4. If the point (0, 0) is a saddle point of the linear system = Ay, theeigenvalues At and A2 of A are real and At < 0 < A2. Construct two nontrivialand real-valued convergent power series fi(x) and rG(x) in a variable x so that¢(eA'tcl) (t > 0) and ,G(eA2tc2) (t < 0) are solutions of (VIII.6.1), where cl and c2are arbitrary constants (cf. Exercise V-7). The solution y' = *(eAt tct) represents thestable manifold of the trivial solution of (VIII.6.1), while the solution y" =1 (eA2tc2)represents the unstable manifold of the trivial solution of (VIII.6.1). The point (0, 0)


is a saddle point of (VIII.6.1). In the next section, we shall explain the behavior ofsolutions in a neighborhood of a saddle point in a more general case.

dyCase 5. If the point (0, 0) is a center of the linear system - = Ay, both eigenvalues

of A are purely imaginary. Assume that they are ±i and A = [ 001

J .Set

y" = 2L

1i1

Jv, where v = [t1].

Then, the given system (VIII.6.1) is changedL J 2

to

(VIII.6.2)

where

du Ig(vl,v2)ldt - 9(27) 012, v1)

g(tvl, v2) = ivl + 9P'9111 2,p+9=2

+00

012,V0 = -iv2 + p.9v2v1p+9=2

Here, a denotes the complex conjugate of a complex number a. Let us applyTheorem VIII-4-2 to system (VIII.6.2).

Observation VIII-6-1. First, setting Al = i and A2 = -i, look at Al = p1A1 +p2A2 and A2 = Q1A1 +g2A2, where p1, p2, q1, and q2 are non-negative integers suchthat p1 + P2 > 2 and ql + q2 > 2. Then, p1 - 1 = p2 and q2 - 1 = q1. This impliesthat system (VIII.6.2) can be changed to

(VIII 6 3)du1 due

( (= -iC= i ) ). .d

W ulu2 u1U2u1, 0 U2,dt

where w(z) = 1 ++oo

E Wmxm, by a formal transformationm=1

(T) v = f(17) = ul + h(ul,u2)

Here,

u2 + h(u2,u1)

+00 +0 _h(ul,u2) _ hp.qupU2, h(u2,u1) _ hP,9u'lU2

p+q=2 p+9=2

In particular, h(ul, u2) can be construced so that

(VIII.6.4) the quantities hp+1,p are real for all positive integers p.

Observation VIII-6-2. We can show that if one of the Wm is not real, then y = 0is a spiral point (cf. Exercises VIII-14). Hence, let us look into the case when the

+00

Wm are all real. Furthermore, if a formal power series a(t) = 1 + withm=1

real coefficients am is chosen in a suitable way, the transformation

at

(VIII.6.5) u1 = a(B11j2)131, u2 = W10002

258 VIII. STABILITY

changes (VIII.6.3) to another system

d,31= in(AiQ2)Qi, 2 _ -0#10002,

where f2(() is a polynomial in C with real coefficients which has one of the followingtwo forms:

1, Case IQ(o - 1 + coS"'°, Case II

where co is a nonzero real number and tno is a positive integer.

Observation VIII-6-3. Let us look into Case I. Assume that system (VIII.6.3)is

(VIII.6.3.1)dul

dt = iu1,due

dt = -iu2.

Note that transformation (VIII.6.5) does not change (VIII.6.3.1). Using a trans-formation of form (VIII.6.5), change h(ul, u2) so that

(VIII.6.6) 1,P = 0 for all positive integers p.

Since (u1,u2) = (ce`t,de-'t) is a solution of system (VIII.6.3.1) with a complexarbitrary constant c, the formal series

(FS-1) v = f(-) =cell + h(eett, cue-,t)

ce-u + h(ce-tt cett)

is a formal solution of system (VIII.6.2) which depends on two real arbitrary con-stants.

Let

Vt + H(t,(S) u(t, , ) _

fe-'t

be the solution of system (VIII.6.2) satisfying the initial condition

(C) t'(0, f, 6 = III].

Note that

H(t, , ) _ HP,q(t)ef°,D+9=2

+«0

H(tto = EP+v=2

are power series in { and which are convergent uniformly on any fixed boundedinterval on the real t line.


Using (VIII.6.6), we obtain

IZx 2*

0 h(ceic, ee ")e-'tdt = 0, r h(Ce-'t, cett)e'Ldt = 0.0 0

Fix c and c by the equations

T7r/ 2x H(s, t:, )e-"ds and e = { + 2 J2w H(s, {, {)e'sds.c = £ + -0 0

Then,= c + -(c, c) and t; = e + ?(c, c),

where =(c, e) and c(e, c) are convergent power series in (c, e).Now, we can prove that two formal solutions

ce`t +h(ce",Ce-'t)

v'(t,c+F(c,e),e+ =2(c' i!)) andee-tt + h(&-'t, cent ) j

of system (VIII.6.2) are identical. The first of these two is convergent; hence, thesecond is also convergent. This finishes Case I.

Observation VIII-6-4. Let us look into Case II. Assume that system (VIII.6.3)has the form

(VIII.6.3.2) dtl = iul(l + co(u1u2)'0), dt2 = -iu2(1 + co(uiu2)"'°)

Note that we have (VIII.6.4).

Since (ul,u2) = (ce't(1+co(ce)'"°) ee-u(1+c0(ct)'"0)) is a solution of system(VIII.6.3.2) with a complex arbitrary constant c, the formal series

(FS-2) v = f (u') =ce't(I+co(ce)'"0) + h(ceit(l+co(ct)"'O),

ee-'t(1+c0(ct)_0))

ee-'t(1+c0(ce)'"0) +h(ce-tt(1+co(ce)'"° ), celt(1+co(ct)"O) )

is a formal solution of system (VIII.6.2) which depends on two real arbitrary con-stants.

Again, as in Observation VIII-6-3, let (S) be the solution of system (VIII.6.2)satisfying the initial condition (C). Set

(VIII.6.7) t; = ?(c, cJ = c + h(c, c) and = =(c, c) = e + h(e, c).

Then,

I ce1t(1+co(ct)"°)

+h(ceit(1+co(ce)'"°),ee-'t(l+co(ct)"'0)

ee-tt(1+c0(ct)'"0) + h(ee-it(t+co(ct)'"O),ceut(l+co(et)"O))

260 VIII. STABILITY

as formal power series in (c, e). This series has a formal period

T(cc) =27r

with respect to t, i.e.,1 + co(ce)m0

E(c, c)(VIII.6.8) v(T(cc), =[c, e], E[c, c])

E(e, C)

Solving (VIII.6.7) with respect to (c, e), we obtain two power series in

c = and e = e( , £).

Set t) = T(c(., t)e(, £)). Then, (VIII.6.8) becomes

(VIII.6.9)

Using (VIII.6.9) as equations for P({,£), it can be shown that the formal powerseries is convergent. This implies that

C(S,S)C(CS) [(CO) 1/ J

1/7+o

is convergent. (Here, some details are left to the reader as an exercise.)On the other hand,

T(ct){{ese + H(s,e-.(1+c0(c4),"O)ds

0

LT(cz)

1Cesa(1+co(ct)'"°) + h(ceps(1+Co(cr)m0),ce-1e(1+co(Ce)m°))J

and

P=1

`T(cC) +ooJ/ { e + H(s, , ) }

e'8(1+co(cc)"° )ds 1 + t he+1,P(cz)p0

P=1

since the quantity hP+1,P are real (cf. (VIII.6.4)). This proves that C(Cr/ ) is conver-C(l, 7)

gent. Hence, c(l;, ) and 4) are convergent. Thus, the proof of the convergenceof A g) in Case II is completed.

Thus, it was proved that if all of the coefficients w,,, on the right-hand side ofsystem (VIII.6.3) are real, the point (0,0) is a center of system (VIII.6.1). Thegeneral solution of (VIII.6.1) can be constructed by using various transformationsof (VII.6.1) which bring the system to either (VIII.6.3.1) or (VIII.6.3.2). Periodsof solutions in t are independent of each solution in Case I, but depend on eachsolution in Case H.

+00

x e-fs(1+G)(cz)m°)ds = c 1 + Ehp+l,p(cz)P

7. PERTURBATIONS OF AN IMPROPER NODE AND A SADDLE PT. 261

Remark VIII-6-5. The argument given above does not apply to the case whenthe right-hand side of (VIII.6.1) is of C(O°) but not analytic. A counterexample isgiven by

2 _

dt [h(1) h(r) y' r2 = Y1 + y2+

where h(r2) = e-1jr2 sin I I. Using the polar coordinates (r,O), this system isrchanged to

dt = rh(r2) and dand periodic solutions are given by r2 = 1 with any integers nt. This is an

MRexample of centers in the sense of Bendixson (cf. [Ben2, p. 26] and §VIII-10).

For more information concerning analytic systems in JR2, see [Huk5].

VIII-7. Perturbations of an improper node and a saddle pointIn the previous section, the structure of solutions of an analytic system in JR2

was showen by means of the method with power series. Hereafter, in this chapter,we consider a system

(VIII.7.1)dy _

Ay + §(y-),dt

under the assumption that(i) A is a real constant 2 x 2 matrix,(ii) the entries of the R2-valued function §(y) are continuous in g E JR2 near 0' and

satisfies the condition

(VIII.7.2) lim 9(y) = 0.y-o Iy1

In this section, it is also assumed that the initial-value problem

(VIII.7.3) df = Ay + §(y-), g(0) = g

has one and only one solution as long as g belongs to a sufficiently smallneighborhood of 0.

Remark VIII-7-1. In the case, when the entries of an 1R2-valued function f (Z-)are continuously differentiable with respect to the entries z1 and z2 of i in a neigh-

borhood of i = 0 and f (0) = 0, write the differential equation dt = f(E) in form(VIII.7.1) by setting

aft eftA = of(o) - 0z'

(o)19x2

(o)

i9Lof2

aft8x1 az2 ( )

and g(y)=f(y)-Ay.

262 VIII. STABILITY

Using thepolar coordinates (r, 0) in the g-plane, write the vector g in the form

y = r [c?}. Then, system (VIII.7.1) is written in terms of (r, B) as follows:

d=r[a, i cos2(B) + a22 sin2(9) + (a12 + a21) sin(g) cos(9)]

(VIII.7.4)

where A = alla21

+ 91(-1 COs(0) + 92() sin(g),

rdte =r[-a12 sin2(0) + a21 COS2(0) + (a22 - all) sin(e) cos(0)]

- 91(y-) sin(g) + 92 M cos(e),

a12 J and g"(y) = [iW)l Here, use was made of the formula22 92 M

dr = dy1 (9) + dye sin(6),r de

= -dy1 sin(g) + dye cos(9).dt dt

cosdt dt dt dt

In this section, we consider the case when the two eigenvalues Ai and A2 of thematrix A are real, distinct, and at least one of them is negative, i.e., A2 > Aland Ai < 0. Throughout this section, assume that all = Al, a22 = A2, anda12 = a21 = 0. Then, in terms of the polar coordinates (r, 0), system (VIII.7.1) canbe writen in the form(VIII.7.5)

[Ai(cos(8))2 1

dt = r + A2(sin (e))2 + (e) + g2(g)sin (e))]

d9= (A2 - Ai )sin (0) cos (0) + r (e) + 92(y-)cos (e))

Observation VIII-7-2. The point (0,0) is not a proper node of (VIII.7.1). Infact, if (r(t), 9(t)) is a solution of (VIII.7.5) such that r(t) 0 and e(t) --+ w as

t - +oo or -oo, then d8 --* (A2 - AI) sin w cos w as t -+ +oo or -oo. Hence,

sin w cos w = 0. This implies that w = -7r, -7r,

0, or2

Observation VIII-7-3. For any given positive number e > 0, there exists anotherpositive number p(e) > 0 such that

> 0 for - 7r + e < 0 < - 2 - e,

de < 0 for - 2 + e << 0 < - e,dt > 0 for e < e < -

IT-+e<0<ir-e,I < 0 for2

if 0 < r < p(e) (cf. Figure 13).Observation VIII-7-4. If 0 > A2 > Al, there exists a positive number ro > 0

such that dt < Air for 0 < r < ro.

Observation VIII-7-5. In the case when 1\2 > 0 > AI, find a real number wo

such that 0 < wo < 2 and tanwo = -1. Then, for any given positive number2

E > 0, there exists another positive number p(E) > 0 such that

> 0 for (wo - 7r) + E < 0 < -"lo - e,dr < 0 for - wo + f < 0 < wo - E,dt > 0 for wo + E <0< (7r - wo) - E,

< 0 for (7r - wo) + f < 8 < (7r + wo) E,

if 0 < r < p(E) (cf. Figure 14).


Observation VIII-7-6. If two positive numbers f and p are sufficiently small,Observations VIII-7-2 and VIII-7-3 show that the behavior of orbits of (VIII.7.1)in the sectorial region S = {(r cos 0, r sin 0) : $6 < E, 0 < r < p} looks like Figure15.

Denote by fl the set of all real numbers w such that(i) IwI 5 E,(ii) the orbit (r(t), 8(t)) of (VIII.7.1) such that r(0) = p and 0(0) = w leaves the

sectorial region S through the boundary 0 = -E.Set a = sup(w : w E 0). Then, ja$ < E. Furthermore, the orbit (r(t), B(t)) of(VIII.7.1) such that r(0) = p and 0(0) = a satisfies the condition lim r(t) = 0and Iim 6(t) = 0. This can be shown by using the continuity of orbits with respect+00

c-.+ooto initial data (cf. Figures 16-1 and 16-2). Here, use was made of the uniquenessof solutions of initial-value problem (VIII.7.3).

FIGURE 15. FIGURE 16-1. FIGURE 16-2.

264 VIII. STABILITY

Observation VIII-7-7. In the case when 0 > A2 > A1, the point (0, 0) is a stableimproper node of (VIII.7.1) as t -+ +oo (cf. Figure 17).

Observation VIII-7-8. In the case when A2 > 0 > A1, the point (0,0) is a saddlepoint of (VIII.7.1) (cf. Figure 18).


Observation VIII-7-9. Figures 17 and 18 indicate that there are many orbitstend to the point (0,0) in the direction 0 = 0 as t +oc. This is the generalsituation, as the following example shows.

Fixing two positive numbers c and v, define two regions Do and DI in the y"-planeas follows (cf. Figure 19):

'D0 = (!7: Iy2I < yi+f. U < yI},

DI = -2 - 19: Iy2I 5 (1 + E)yi+U, 0 < yl}.

Choose three real numbers A1, A2, and A3 so that Al < 0, A2 > Al, and

(VIII.7.6) A3 > 1 +51 v.

Let µ(y) be a real-valued function such that (a) u is bounded on tit, (b) p is continu-ouslv differentiable in y" on e - {(0.0)}. (c)U(y) = A3 on Do. and

(d) µ(y') = A2 on DI. Set 9(y) = 0J

Then, 9(y) = 0 if y" E Dl.(l4(y) - A2) Y21-

Furthermore, 0 < yl. 11121 5 (1 + ()y1+" if y"" D1, and

I9'(y)I = I (µ(y) - A2)y2I 5 KIy2I < K(1 + E)yi"" 5 K(1 + e)Iyj1+

where K is a suitable positive constant.

Consider the system

(VIII.7.7) ddll = -\01- dt = AM + (µ(y) - A21y2

If y"(0) E Do, then db2 = (.!) 2 Hence, y2 = y1 ' . Now, condition (VIII.7.6)dy1 ,\I 1/i

implies that g(t) E Do for t > 0, lim r(t) = 0, and lim 9(t) = 0. (cf. Figure 20).t-+oo t-+oo

(I +E)Yj.V

(I +e)Yi


Observation VIII-7-10. If we assume some condition on smoothness of §(y-),there is only one orbit of (VIII.7. 1) that tends to the point (0, 0) in the direction9 = 0 as t - +oc.Theorem VIII-7-11. Assume that all = A1, a22 = A2, and a12 = a2I = 0, andthat(1) A2 > \1 and Al < 0,(2) g(j) is continuous in a neighborhood of

(3) lim " =y-.o lye

a9(y(4) exists2

Then, there exists

i

0,

and is continuous in a neighborhood of 6.

a unique orbit y(t) _ [01(1)1 such that02(t)

01(t) > 0 for t > 0,

b2(t)urn 01(t) = 0, lim 02(t) = 0, lim = 0.t-.+oo t--+co 01(t)

Proof.

The existence of such orbit is seen in previous observations. To show the unique-ness of such orbit, rewrite (VIII.7.1) in the form

I

(VIII.7.8) dye = 1\2Y2 + 92(Y) _ All yz + a y192(y - (Aty1)291(Y)

dy1 AIyI+91(91 \Y1 1+ 1 91()J)AIYI

Next, introduce a new unknown u by y2 = y1u 1or u = y2yI. Then, (VIII.7.8)

becomes

VIII.7.9 du = (L2)( ) y1- + u u+

dy1

1 JI2u1y192(YIU) - \I-91(y1u)

1

1 +'\

Iy1gI(y1u)

266 VIII. STABILITY

where u = [:1]. Observe that (a) 161 = 1 if Jul < 1, (b) -L (d) = U g(0, y2)-0 Y2

= d, and (c) - 1 < 0.i

Write system (VIII.7.9) in the forml

- 1 I u + G(yi.u).Q1 A2u

- 92(yiu) - a 91(yl UG(yi, u) =

11 + -9i(y1iZ)aiyi

It is easy to show that limac

v5:u (y1, u) = 0 uniformly on Jul < 1. Hence, there exists

--oa non-negative valued function K(yj) such that

!G(yi, u) - G(y1, v)! < K(yi )lu - vlwhenever Jul < 1, !v! < 1 and !y1! is sufficiently small. Furthermore, slim K(yt)

=0.Let u = 01(yi) and u = ip2(yi) be two solutions of (VIII.7.9) such that

(1) ti i (yi) and .'2(y1) are defined for 0 < yi < r) for some sufficiently smallpositive number 'I,

(2) lim i'1(yi) = 0 and lim i2(yl) = 0.vl o+ v,

Set '(yi) = 01 (y1) - s (yi ). By virtue of uniqueness, assume without any lossof generality that y(yi) > 0 for 0 < yi < ri. Then, for a sufficiently small positive r?,

y1d

-)tt'(yi) for 0 < y1 < i , where 7 = 2 1 I < 0. This implies that

dJ

[y1 "V,(yi)j < 0 for 0 < yi < rj. Hence, 0 < y1 lim yi 'W(yi) = 0 for0dyi Y:-

0 < y1 < 17. Therefore, ;1(yi) = 0 for 0 < y1 < r).The materials of this section are also found in (CL, §§5 and 6 of Chapter 151.

VIII-8. Perturbations of a proper nodeIn this section, we consider a system of the form

(VIII.8.1)dt

= All + where y" E R2,

under the assumptions that(i) A is a negative number,

(ii) the entries of the R2-valued function g(y} are continuous in a neighborhoodof 0,

(iii) 1§(y -)l < K!y"l1+i in a neighborhood of 6, where K is a nor-negative numberand v is a positive number.

The main result is the following theorem.

8. PERTURBATIONS OF A PROPER NODE 267

Theorem VIII-8-1. Under the assumption given above, the point (0, 0) is a stableproper node of system (VIII.8.1) as t --, +oo.

Before we prove this theorem, we illustrate the general situation by an example.

Example VIII-8-2. Look at the system

(VIII.8.2) dt = -y + 9(yj, 9(y _ (yi + y2) [y

where y E R2 with the entries yl and y2 It is known that the point (0,0) is a

stable proper node of the linear system dt = -y. To find the general solution of(VIII.8.2), change (VIII.8.2) to

(VIII.8.3)

by the transformation y' = e-'Z. Next, introduce a new independent variable s by

s = 2e-2t. Then, system (VIII.8.3) becomes

(VIII.8.4)

Observe that

ds -g(z)

If a solution y(t) of system (VIII.8.2) satisfies an initial condition

(VIII.8.5) 9(o) = n = 171

?12 '

the corresponding solution z(s) = et y'(t) of system (VIII.8.4) is the solution of the

2initial-value problem ds = r(7)2 f zz l

z" C 2 /= rj, where

(VIII.8.6) r(i) = (n,)2 + (m)2.

Hence,

zl(s) = r(ir)sin(r(i)2s + 6(rl)), z2(s) = r(i)7)cos(r(1-7)2s + 0(r1)),

where

2

z, (0) = r(rl sin 111 = r(i) sin( 0(rl

{ Z2(0) = r(n7) cosoos (i = r( 2n n

268 VIII. STABILITY

(cf. Figure 21).The general solution of (VIII.8.2) is

(L( 2

yI(t) = e-tr(7f)sin 7 a-2t

y2(0 = e-tr(771 cos (!2e_2t

where y(0) and r(fl) are given by (VIII.8.5) and (VIII.8.6), respectively. Every orbitof (VIII.8.2) goes around the point (0, 0) only a finite number of times. Figure 22shows that the point (0, 0) is a stable proper node as t -+ +oo.


Remark VIII-8-3. If condition (iii) of the assumption given above is relaxed, thepoint (0, 0) may become a stable spiral point (cf. Exercise VIII-10).

Proof of Theorem VIII-8-1.

We prove Theorem VIII-8-1 in two steps. To start with, using the polar coordi-nates, write system (VIII.8.1) in the form

(VIII.8.7) J

dr= Ar + gI (y) cos(9) + g2(y) sin(g),

dt

r dO= -g1(y) sin(g) + g2(y COs(9)t

(cf. (VIII.7.4)). By virtue of the assumption given above, there exists a positivenumber ro such that

(VIII.8.8) 2Ar < Ar + gi (y-)cos(9) + g2(y) sin(g) < 2r < 0

for0<r<ro.Step 1. We show first that the point (0, 0) is a stable node. To do this, let(r(t), 9(t)) be a solution of (VIII.8.7) satisfying the initial condition

(VIII.8.9) r(0) = p, 9(0) = w,

where p is a positive number and w is a real number. This solution exists fort > 0. Furthermore, estimate (VIII.8.8) implies that the function r(t) satisfies

8. PERTURBATIONS OF A PROPER NODE 269

the following two conditions: (a) r(t) is strictly decreasing as t -+ +oo and (b)lim r(t) = 0. Therefore, we obtain (c) 0 < r(t) 0._ _t-. +00Since r(t) is strictly decreasing for t _> 0, the variable t can be regarded as a

function of r, i.e., t = t(r), where t(p) = 0 and lim t(r) = +oo. Use r as theindependent variable. Then, (VIII.8.7) becomes

V1II 8 10 d9-gi (y-)sin(g) + g2(Y)cas(e)

0F( . . )Tr

(r, ) _r[Ar + g1(y) cos(0) + g2(y) sin(0)J

Let 0 = 4D(r) = 0(t(r)) (0 < r < ro) be the solution of (VIII.8.10) satisfying theinitial condition (b(p) = w. Then,

fi(r) = w +J

r F(s, I(s))ds (0 < r < ro).P

Estimate (VIII.8.8) implies that

2Ar2 I - gt (y sin(0) + 92(Y) cos(0) I F(r, 0) < ar2 (I g1(y)I + Ig2(y1I)

0 < r < ro.for

Jto

F(s, $(s))ds =Pr(t)

F(s, (s))dsP

Now, by virtue of condition (iii), the improper integral

limJ

r F(s, 4(s))ds exists. Observe that 0(t) = 4(r(t)) = w +r-.0 P

and that lim r(t) = 0. Therefore, lim 0(t) = w +t--+oo t-'+oo

0

J Thus, we conclude that the point (0,0) is a stable node ofP

(VIII.8.1) as t +oc.

Step 2. In this step, we prove that the point (0,0) is a proper node. To dothis, for a given real number c, find a solution (r(t), 0(t)) of (VIII.8.7) such thatrim 0(t) = c and lim r(t) = 0. Selecting a sequence {pm : m = 1, 2, ... } oft-+oo t -+oo

positive numbers such that 0 < p,,, < ro and lim pm = 0, define a sequence of

functions {cm(r) : in = 1, 2.... } by the initial-value problems

d0),

WT= F(r, 0), 0(Pm) = c (m = 1, 2....

respectively. Those functions are defined for 0 < r < ro, and

Set

c + / r F(s, 4m(s))ds (m = 1, 2, ... ).P

o F(s, Om(s))ds.Cm = lim 4)m(r) = c + jP1.- +«;

270 VIII. STABILITY

Then, 4b,,,(0) = c,,,, lira c,,, = c, andm=+oo

(r) = cm + J r F(s, 4,,,(s))ds (m = 1,2.... )0

for 0 < r < ro. Since the sequence {4D m(r) : m = 1, 2.... } is bounded andequicontinuous on the interval 0 < r < ro, assume without any loss of generalitythat lim fi(r) exists uniformly on the interval 0 < r < ro (otherwise

M +00choose a subsequence). It can be shown easily that

fi(r) = c + fF(s,(s))ds (0 r ro).

Therefore, 7(r) is a solution of (VIII.8.10) such that rim 1(r) = c.

Define r(t) by the initial-value problem

dr = )r + 92(17(r))sm(4'(r)), r(0) = ro,dt

where g(r) = r [c?1]. The function r(t) is defined for t > 0, is decreasing,

and tends to 0 as t -4 +oo. Set 0(t) = $(r(t)). Then, it is easily shown that(r(t), B(t)) is a solution of (VIII.8.7) such that limo 0(t) = 1i mt(r) = c. 0

t +0

The materials of this section are also found in (CL, §3 of Chapter 151.

VIII-9. Perturbation of a spiral point

In this section, we show that (0,0) is a stable spiral point of (VIII.7.1) as t -, +oo

if (0, 0) is a stable spiral point of the linear system = Ay". The main result isthe following theorem.

Theorem VIII-9-1. If(i) two eigenvalues of A are not real and their real parts are negative,(ii) the entries of the L2-valued function ff(y-) is continuous in a neighborhood of

the point (0,0),

(iii) lira MI = 6,g-alyl

then the point (0, 0) is a stable spiral point of (VIII. 7.1) as t +oo.

Proof.

Assume that

(VIII.9.1) all = a22 = a < 0 and a12 = -a21 = b 0 0.

10. PERTURBATIONS OF A CENTER 271

Then, from system (VIII.7.4), it follows that

dr

dt

r t

ar + 91(yjcos(9) + 92(ylsin(g),

= -br - 91(y-) sin(g) + g2(Y)cos(0).

Therefore, a positive number ro can be fouind so that r(t) < r(0) exp1at

J for t > 0

if 0 < r(0) < ro. This, in turn, implies that lim r(t) = 0 and that dt < -

2

ift -+oo

b > 0, while de > - 2 if b < 0, whenever 0 < r(0) < ro. Therefore, lim00(t) = -oo

if b > 0 and lim 0(t) = +oo if b < 0, whenever 0 < r(0) < ro. Thus, we concludet too -

that (0, 0) is a stable spiral point of system (VIII.7.1) as t -+ +oo. 0The materials of this section are also found in fCL, §3 of Chapter 151-

VIII-10. Perturbation of a centerWe still consider a system (VIII.7.1) under the assumption that the entries of the

RI-valued function §(y) is continuous in a neighborhood of 0 and satisfies condition(VIII.7.2). In this section, we show that if the 2 x 2 matrix A has two purelyimaginary eigenvalues Al = ib and A2 = -ib, where b is a nonzero real number,then the point (0,0) is either a center or a spiral point of

[°b

(VIAssumewithout any loss of generality that the matrix A has the form b

Jand system

(VIII.7.4) becomes

(VIII 10 1). .dB = -b + 92(y')cos (B)).(B) +rdt

Observe that system (VI I I.10.1) can be written in the form

(VIII.10.2)dr = 91(y')cos (0) + 92(y-)sin (0)d8 -b + r (-91(yjsin (0) + g2(y-')cos (0))

If a positive number p is sufficiently small, the solution r(0, p) of (VIII.10.2) satis-fying the initial condition r(0, p) = p exists for 0 < 0 < 27r.

Case 1. If there exists a sequence (pm : rn = 1, 2, ...) of positive numbers suchthat

lim p,,, = 0 and r(2xr, Pm) = Pm (m. = 1, 2, ... ),m-.+00

dr _dt = 91Mcos (0) + 92(y-)sin (0) ,

1

then the point (0,0) is a center of (VIII.7.1) in the sense of Bendixson (Bent, p.261 (cf. Figure 23).

272 VIII. STABILITY

Case 2. Assume that b < 0. If there exists a positive number po such that

r(2ir, p) > p (respectively < p)

whenever 0 < p < po, then the point (0, 0) is an unstable (respectively stable)spiral point as t - +oo (cf. Figures 24-1 and 24-2).


Example VIII-10-1. The point (0,0) is a center of the system

d

dt [y2J [ yuj + yi +y2 I yy1J

drsince

d8= 0.

Example VIII-10-2. The point (0, 0) is an unstable spiral point of the system

d = [ Y2dt l Y2 l -Yi J

+ ` y 2 + 2 yIIY2

Y2

as t - -oc, since dr = r2, which has the solution r = 1 r(0) = 1and

dt ro - t ror +oo when t ro. (Note: t < ro.)

Example VIII-10-3. For the system

dt = - y, d y = x + x2 - xy + cry2,

the point (0, 0) is (a) a stable spiral point if a < -1, (b) a center if a = -1, and(c) an unstable spiral point if a > -1.

Proof.

Use the polar coordinates (r, 0) for (x, y) to write the given system in the form

(1+rcos8(cos20-cos0sin8+asin28)) = r2sin8(cos28-cos0sin8+asin20).

10. PERTURBATIONS OF A CENTER 273

Settingtoo

r(9, c) _ E r,,, (9)c"', where r(0, c) = c,m=1

and comparing r(2r, c) and c for sufficiently small positive c, it can be shown that

r1(t) = 1, r2(2r) = 0, r3(2r) =(1 + Or

4

Thus, r(27r, c) < c if a < -1. Therefore, (a) follows. Similarly, r(2r, c) > c ifa > -1. Therefore, (c) follows.

In the case when a x2 - xy - y2 = (x - +2 ) (x - (1-2 ). Set

W = - (1 Then. a = (1 Therefore, changing x and y by u =

X - (1 +2,/)y and v = x - (1 -

2)y, the given system is changed to

du

dtWv(1 + u),

dv= ---u(1 + v).

Hence, on any solution curve, the function

W2(v - ln(1 + v)) + (u - ln(1 + v)) = u2t2W2v2

+

is independent of t. This implies that in the neighborhood of (0, 0), orbits are closedcurves. This shows that (0, 0) is a center. 0

Example VIII-10-4. The point (0,0) is a center of the system

dy = bA ' +dt y 9(y), y= y'Jy2

if (1) b is a nonzero real number, (2) A = 101 1J, (3) the entries of the R2-

valued function g"(yj is continuous and continuously differentiable in a neighborhood

of 0, (4) lim " = 6, and (5) there exists a function M(y) such that M is positiveg-6 1Y1

valued, continuous, and continuously differentiable in a neighborhood of 0 and that8(Mf1)(y) + 8(M 2)(y..) = 0 in a neighborhood of 0, where f, (y-) and f2 (y-) are

the entries of the vector MY + §(y).

Proof

The point (0, 0) is either a center or a spiral point. Look at the system

dt =bAy + 9(y) and 7t- = M(y-

274 VIII. STABILITY

Using s as the independent variable, change the given system to ds = M(y)(bAy+g"(y)). Upon applying Theorem VIII-1-7, we conclude that (0, 0) is not asymptoti-cally stable as t - ±oo. Hence, (0, 0) is a center.

The materials of this section are also found in [CL, §4 of Chapter 15] and [Sai,§21 of Chapter 3, pp. 89-1001.

EXERCISES VIII

VIII-I. For each of the following five matrices A, find a phase portrait of orbitsd:i

of the system = AY.

I 14 1]' [ 11 31 ' [ 15 11]' [1

5 1

5

VIII-2. Find all nontrivial solutions (x, y) = (d(t), P (t)). if any, of the system

dt = -xy2, _ -x4y(1 + y)

satisfying the condition lim (th(t),tb(t)) = (0,0).c-.+00

VIII-3. Let f (x, y) and g(x, y) be continuously differentiable functions of (x, y)such that

(a) f (0, 0) = 0 and g(0, 0) = 0,(b) (f(x,y),g(x,y)) # (0,0) if (x,y) 0 (0,0),(c) f and g are homogeneous of degree m in (x, y), i.e., f (rx, ry) = rm f (x, y) and

g(rx, ry) = rmg(x, y). Let (r, 8) be polar coordinates of (x, y), i.e., x = r cos 8and y = r sin 0. Set F(8) = f (cos 0, sin 8) and G(O) = g(cos 8, sin 8).

(I) Show that

dr= rm(F(8) cos 8 + G(8) sin 8),

(S) dtd8

dt = rm-1(-F(8)sin8 + G(O)cosO).

(II) Using system (S), discuss the stability property of the trivial solution of eachof the following three systems:

x y = 2xy;(i) = 2 - 2

dt4ii

= x3 (x2 + y2) - 2x(x2 + y2)2, = -y3(x2 + y2);

(iii) d = x4 - 6x2y2 + y4, dy = 4x3y - 4xy3.

EXERCISES VIII 275

VIII-4. Let J be the 2n x 2n matrix defined by (IV.5.2) and let H be a real constant2n x 2n symmetric matrix. Show that the trivial solution of the Hamiltonian system

a = JHy is not asymptotically stable as t +oo.

V11I-5. Show that the trivial solution of the system

dy'd = Ay" + 9(t, y-)

is asymptotically stable as t +oo if the following conditions are satisfied:(i) A is a real constant n x it matrix,

(ii) the real part of every eigenvalue of A is negative,(iii) the entries of the IRI-valued function g(t, y-) are continuous in the region

0(ro) = To x D(ro) = {(t, yl : 0 < t < +oo, Iyj < ro) for some ro,

(iv) g(t, yj satisfies the estimate

19(t, y) 1 _< eo(t, y')Iyj for (t, y) E o(ro),

where Eo(t, y) is continuous, and positive, andfun Eo(t, y) = 0 in L(ro).+Ii7 -.o

VIII-6. Show that the point (0, 0) is a stable improper node of the system

dy = Ay + Ng + 9(y), y = Iyil,

where(1) A is a negative number,(2) N is a real constant nilpotent 2 x 2 matrix and N # 0,(3) the 1R2-valued function y'(yj is continuous in a neighborhood of(4) 19(y-) 11-- cly-1I+" for some positive numbers c and v in a neighborhood of

CA

Hint. Suppose N = 00 0

2 . If we set yj = r cos 0 and y2 = r sin 6, we obtain

d = r IA _ EA sin2 cos 0 + 9i (y-)cos 9 + 92(y-)sin O,

where 9"(y'i = {9i]. Hence, if r(0) is small, r(t) is bounded by r(0) and92(V

lim r(t) = 0 for t > 0. This implies that if 1#(0)I is small, g(t) is boundedt+0 -and lim y(t) = 0'. Look att-+o

y(t) = eAtetN {(o) + -sN9(y($))d$]

276 VIII. STABILITY

If we set y(t) = e' e'Nu, we obtain

d(t) = y-(0) + r e-aae-aN9(g(s))ds.0

Choose e > 0 so that 1 - e > 1

+

Then, condition (4) implies that

+00

6(+ = 9(0) ++00

e-aae_a'

0

exists.Now,

(1) if d(+oo) = 0, then d(t) = 0 identically, since, in this case,

d(t) = f e-X$e-ajV9(y(s))ds;

+a

(2) if t7(+oo) = 0 0, then t line-a`y"(t) = i"(+00);

(3) if iZ(+oc) =L

RZJ with 2 # 0, then limo ( e _) 0 2

Hence, the point (0, 0) is a stable improper node of the given system.

VIII-7. Determine whether the point (0,0) is a center or a spiral point of thesystem

dt = -y, dt = 2x + r3 - x2(2 - x)y.

VIII-8. Show that the point (0, 0) is the center of the system

dt = y + xy3 - y7, _ -x + xy2 - yg.

Hint. This system does not change even if (t, y) is replaced by (-t, -y).

VIII-9. For the system

dt = 3y + 5x(x2 + y2), dty = 27x + 5y(x2 + y2),

find an approximation for the orbit(s) approaching (0,0) as t - +oo.

VIII-IO. Show that the point (0, 0) is a stable spiral point of the system

d + 1 y2

--[

dt[y,]

y2 -[yjyIn yi + y2 yi

as t-++oc.

EXERCISES VIII 277

Hint. If we set yl = r cos 0 and y2 = r sin 0, the given system becomes

dr _ dO 1

dt -r' dt = In r'

VIII-11. Suppose that a solution ¢(t) of a system

dy _dt

AJ + §(y-)

satisfies the conditionlim 4(t) = 0t-r

for a positive number r. Show that if(1) A is an n x n constant matrix,(2) the n-dimensional vector W(if) is continuous in a neighborhood of 0,(3)

lien9(Y) = 0,

V-6 !y1then ¢(t) = 6 for all values oft.(Note that the uniqueness of solutions of initial-value problems is not assumed.)

VIII-12. Assume that(i) AI.... .. 1n are complex numbers that are in the interior of a half-plane in the

complex A-plane whose boundary contains A = 0,(ii) there are no relations \, = plat + P2 1\2 + + pnan for j = 1, ... , n and

non-negative integers p,,... , pn such that pi + - - + pn >_ 2,(iii) f(y' = yam' ft, is a convergent power series in ff E Un with coefficients

jp1>2

fpEC'.Show that there exists a convergent power series ((u) _ ul'Qp with coefficients

Inl>2 _dgQp E Cn such that the transformation if = u" + Q(ur) changes the system - _

A#+ f (y-) to j = Au, where A = diag[.1t, A2,4, ... , A ].

VIII-13. Show that the trivial solution of the system

d d

dt

x r- -st(x +X22) + x2ezl+ 2,dt

-22(22 + 22) _ Xlez'+ 2

is aymptotically stable as t +oo. Sketch a phase portrait of the orbits of thissystem.

VIII-14. In Observation VIII-6-2, it is stated that if one of the is not real,then y" is a spiral point. Verify this statement.

VIII-15. Discuss the stability of the trivial solution of the system

j1 = x21 722 = xi(I - x1)

as t - +00.

278 VIII. STABILITY

VIII-16. Find the general solution of the system

= 223 + 2j + X122 + 22,

= 3x4 + + X122 t 2122 + 21X3 + x223,

explicitly.

CHAPTER IX

AUTONOMOUS SYSTEMS

In this chapter, we explain the behavior of solutions of an autonomous systemdg

= f (y). We look at solution curves in the y-space rather than the (t, y-)-space.dtSuch curves are called orbits of the given system. In general, an orbit does nottend to a limit point as t -+ +oo. However, a bounded orbit accumulates to a setas t -4 +oo. Such a set is called a limit-invariant set. In §IX-1, we explain thebasic properties of limit-invariant sets. In §IX-2, using the Liapounoff functions,we explain how to locate limit-invariant sets. The main tool is a theorem due toJ. LaSalle and S. Lefschetz [LaL, Chapter 2, §13, pp. 56-71] (cf. Theorem IX-2-1).The topic of §IX-4 is the Poincare-Bendixson theorem which characterizes limit-sets in the plane (cf. Theorem IX-4-1). In §IX-3, orbital stability and orbitallyasymptotic stability are explained. In §IX-5, we explain how to use the indices ofthe Jordan curves to the study of autonomous systems in the plane. Most of topicsdiscussed in this chapter are also in [CL, Chapters 13 and 16], [Har2, Chapter 7],and [SC, Chapter 4, pp. 159-171].

IX-1. Limit-invariant sets

In this chapter, we explain behavior of solutions of a system of differential equa-tions of the form

(IX.1.1)dg

fW

where y" E Ill;" and the entries of the RI-valued function f (y) are continuous in theentire y-space II8". We also assume that every initial-value problem

(IX.1.2)dy

= Ay), y(0) = ifdt

has a unique solution y = p(t, y7). System (IX.1.1) is called an autonomous sys-tem since the right-hand side f (y) does not depend on the independent vari-able t.

Observe that p(t + r, r) is also a solution of (IX.1.1) for every real number T.Furthermore, At + r, rl) = p(r, il) at t = 0. Hence, uniqueness of the solution ofinitial-value problem (IX.1.2) implies that p(t +r, rt) = p(t, p(-r, rt)) whenever bothsides are defined. For each il, let T(rl) be the maximal t-interval on which thesolution p(t, r) is defined. Set C(rt) = {y" = p(t, rl) : t E Z(rl)}. The curve C(rl) iscalled the orbit passing through the point il. Two orbits C(iji) and C(rj2) do notintersect unless they are identical as a curve. In fact,

(IX.1.3) C(ijl) = C(i@ if and only if 7)2 E C(il, ).

279

280 IX. AUTONOMOUS SYSTEMS

If f (n) = 0, the point ij is called a stationary point. If , is a stationary point, the

orbit consists of a point, i.e., C(17) = Generalizing property (IX.1.3) oforbits, we introduce the concept of invariant sets which play a central role in thestudy of autonomous system (IX.1.1).

Definition IX-1-1. A set M C R is said to be invariant if i E M impliesC(n) C M.

For example, every orbit is an invariant set.

Remark IX-1-2. If M1 and M2 are invariant sets, then M1 UM2 and M1 f1M2are also invariant. For a given set f2 C 1R", let Ma (A E A, an index set) be allinvariant subsets of S2, then U M>, is the largest invariant subset of Q.

AEA

Hereafter, assume that every solution p(t, nJ) is defined for t > 0, i.e., (t : t >0) C Z(n). In general, lim Pit, nom) may not exist. However, if p(t, n) is bounded for

t > 0, the orbit C(171 accumulates to a set as t -+ +oo. This set is very importantin the study of behavior of C(n) as t -+ +oo.

Definition IX-1-3. Let C+(i7-7) denote the set of all y" E II8" such thatlim p(tk, n ) = y f o r some increasing sequence {tk : k = 1, 2, ... } of real numbers

k-++oosuch that lim tk = +oo. The set C+(n) is called the limit-invariant set for the

k-+ooinitial point r7.

The basic properties of are given in the following theorem.

Theorem IX-1-4. If p(t, n) is bounded fort > 0, then C+(n) is nonempty, bounded,closed, connected, and invariant.

Proof.

(1) C+ (1-71 is nonempty: In fact, let {sk : k = 1,2,... } be an increasing sequenceof real numbers such that lim sk = +oo. Since p(t, n) is bounded for t > 0,

k-.+oothere exists a subsequence {tk : k = 1, 2.... } such that lim tk = +oo and that

lim p(tk, exists. This limit belongs to f_+ (17).k-.+oo

(2) The boundedness of C+(n7) follows from the boundedness of p(t,i7) immedi-ately.

(3) is closed: To prove this, suppose that lim yk = y for 9k E L+(1-7).k-+oo

It must be shown that y" E L+(771. Since g k E it follows that ilk =lira p(tk,t, r 7 ) for some {tk,t : I = 1, 2, ... } such that lim tk,l = +00. Choose tk

t-.+oo t-+oo

so that lim tk,t,, = +oo and 19k - P(tk,t,,,'Th !5 ! Then, since ly - P(tk,t4, n)I -k-.+oo

k+ ly - ykl, we obtain

klim oP(tk,l,,, n) = y E C+(1_7)

(4) C+(777) is connected: Otherwise, there must be two nonempty, bounded, andclosed sets Sl and S2 in R" such that

2. LIAPOUNOFF'S DIRECT METHOD 281

(1) S1 n S2 = 0,(2) S1 U S2 L+(rf)

Set d = distance(S{{l, S2) = min{lyl - y21:

yl E S1, y2 E S2}. Note that d >

0. Set also So = g: distance(y, SI) =

2

1. Then, So is not empty, bounded,

and closed. Furthermore, So n C+(i) = 0. Choose two points y"I E SI andy ' 2 E S 2 and t w o sequences {tk : k = 1, 2, ... } and {sk : k = 1,2,...) ofreal numbers so that tk < Sk (k = 1, 2, ... ), lim tk = +oo, lim sk = +oo,

k-+oo k-+oo

klimp P(tk, y"1, and k lim G p1sk, r7) = 92. Assume that r), SO <

2 and distance(p(sk, r), SI) > 2. Then, there exists a Tk for each k such that

tk < Tk < sk and p-(Tk, rl) E So. Choose a subsequence (at : e = 1,2.... ) of{rk : k = 1, 2, ... } so that lim at = +oo and lim p(at, r) = # exist. Then,

Y E So n C+ (Y-)) = 0. This is a contradiction.(5) C+ (1-7) is invariant: It must be shown that if ff E C+ (r), then p1t, y) E L+(1-7)f o r all t E 11(y). In fact, there exists a sequence {tk : k = 1, 2, ... } of real numbers

lim P-14, r) = 17. From (IX.1.1) and the continuitysuch thatkIii o tk = +oo and k o

of P(t,y) of y", it follows that for each fixed t.

+cotk, n) =

kUM P(t, Pptk, r1)) = Pit, y) E L+(7). El

kli

The materials of this section are also found in [CL, Chapter 16, §1, pp. 389-3911and [Har2, Chapter VII, §1, pp. 144-1461.

IX-2. Liapounoff's direct methodIn order to find the behavior of pit, r)) as t +oo, it is important to locate

L+(rl). As a matter of fact, if p(t, r)) is bounded for t _> 0 and if a set M contains,C+ (1-7), then p(t. r)) tends to M as t -+ +oo, i.e., lim inf(1p(t, r)) - yj : y E M) = 0.

t +a:Otherwise, there must be a positive number co and a sequence {tk . k = 1, 2.... } ofreal numbers such that

klim tk = +00,

klimo P140 7) exists, and ipjtk, >)) -y"1 eo

for all y E Jul. Hence, lim -1) V M. This is a contradiction. Keeping thisk-.too

fact in mind, let us prove the following theorem (cf. [LaL, Chapter 2, §13, pp.56-71]).

Theorem IX-2-1. Let V(y) be a real-valued, continuous, and continuously differ-entiable function for Jyl < ro, where 0 < ro < +oo. Set

fDt = J#: V (y-) < e}, St = {y" E DI : V-(y) J(y) = 0}.Mt = the largest invariant set in St,

F' a`, yl fl(where Vi(y) ' f(y) _ ayJ fJ(y), y , and f( y) _ . Suppose

J-1 yn fn (y)

that there exist a mat number t and a positive number r such that 0 < r < ro,Dt C {y : Iy1 < r}, and Vg(y-) f (y) < 0 on De. Then, L+ (1-7) C Me for all n E Vt.

Proof.

Set u(t) = V(p(t,rl")). Then,

dot)= V9(r(t,r7))' dtp(t,i) =

Therefore,du(t) < 0 as long as p(t, t) E Vt. This implies that u(t) < V(1-7) < e

dtfor r) E Vt as long as p(t, n-) E Vt. Hence, p(t, rl E Dt for t > 0 if ij E Vt.Consequently, Ip(t, rlI < r for t > 0. It is known that C+ (171 is nonempty, bounded,closed, connected, and invariant (cf. Theorem IX-1-4). Furthermore, C Vt.

Let po be the minimum of V(rl') for Iy7 < r. Then, po < u(t) andd t)

< 0 if

it E 1)t. Hence, lim u(t) = uo exists and no > po. This implies that V(y) = uo

for all y E C+(t)). Since C+(71 is invariant, V(p(t, y)) = uo for all y E L+(171 andt > 0. Therefore, VV(p(t, y)) &I t, y)) = 0 for all y E C+ (q) and t > 0. Settingt = 0, we obtain Vc(y) f (y) = 0 if y E C+(rl, i.e., C St if i E Vt. Hence,C+() C Mt for if E Vt.

The following theorem is useful in many situations and it can be proved in a waysimilar to the proof of Theorem IX-2-1.

Theorem IX-2-2. Let V (y) be a real-valued and continuously differentiable func-tion for ally E lR' such that VV(y) f (y) < 0 for all y" E R". Assume also that theorbit p(t, rte) is bounded fort > 0. Set S = 1g: Vi(y) f(y) = 0) and let M be thelargest invariant set in S. Then, L+ (7-1) C M.

The proof of this theorem is left to the reader as an exercise.In order to use Theorem IX-2-2, the boundedness of p(t, must be shown in

advance. To do this, the following theorem is useful.

Theorem IX-2-3. If V (y) is a real-valued and continuously differentiable functionfor all y" E lR^ such that VV(y) f(y) < 0 for Iyi > ro, where ro is a positive number,and that lim V (y) = +oo, then all solutions p(t, r)7) are bounded for t > 0.

1Q1+00

Proof.

It suffices to consider the case when p(to,rl > ro for some to > 0. There are twopossibilities:

(1) IP-(t,r1)I > ro fort > to,(2) IP(t,n')I > ro for to < t < t, and Ip1ti,17)1 = ro for some tl > to.

Case (1). In this case, dt V (p( t, r)1) = Vj(p(t, rl) f (p(t, 4-7)) < 0 fort > 0. Therefore,

V (r(t, V (r(to, t))) for t > to. Hence, p(t, tt") is bounded for t > 0.

Case (2). In this case, it can be shown that

V(p(t, 71)) <_ max IV(P(to,' )), max(V(y) : 1yi <_ ro)1

3. ORBITAL STABILITY 283

in a way similar to Case (1). The details are left to the reader as an exercise. 0

IX-3. Orbital stabilityIn this section, we introduce a concept of stability (respectively asymptotic sta-

bility) in a sense different from the stability (respectively the asymptotic stability)of Chapter VIII (cf. Definitions VIII-1-1 and VIII-1-3). We denote again by C(rfthe orbit passing through fl. Also, assume that every solution p(t, if) is defined fort>0.Definition IX-3-1. An orbit C(sjo) is said to be orbitally stable as t -+ +00 if,for any given positive number e, there exists another positive number b(e) such that

distanee(p"(t, rl"), C(rlo)) < c for t > 0 whenever 14 - rlo1 < b(e).

This definition is independent of the choice of the point ijo on the orbit C(ilo). IfC(i-lo) = C(#,) and if the condition of this definition is satisfied in terms of ilo, thenthe same condition is satisfied also in terms of , with a different choice of b(e).The following example illustrates these new concepts.

Example IX-3-2. The orbit of the system

d. [y2] = yl +y2 [ -2]

passing through the point ij = I ] is given by

yj = r(f[)cos{r(i))t + 9(t)}, y2 = r(ij)sin{r(if)t + 9(f))),

where it = [c?]. It is easy to show that if r(n) - r(q) = b > 0 and

9(ili) = 9(ij2), then distance(p(t, f2), C(#,)) = b. Hence, the orbit C(ij) is orbitallystable as t -+ +oo for every i ) . However, i h ) - p (b ff) I = r(il') + r(')2 )since 1r(ffi)t+9(ili)1-1r(il2)t+9(il2)j = 6t (cf. Figure 1). Therefore, every nontrivialsolution p"(t, i)) is not stable as t +oc in the sense of Definition VIII-1-1.

Definition IX-3-3. An orbit C(fo) is said to be orbitally asymptotically stableas t - +00 if

(i) C(ilo) is orbitally stable as t +oo,(ii) there exists a positive number 6o such that lim C(i )) = 0

whenever {il - iloj < 60.

This definition is independent of the choice of the point 4 on the orbit. However,the choice of 6o depends on rjo.

Consider a system of differential equations

(IX.3.1) = f(y,

where the entries of the R"-valued function f is continuously differentiable on theentire y-space W1. Assume also that the solution At, rjo) is periodic in t of period1 (i.e., #(t + 1, ito) = p1t, i"Jo) for -oo < t < +oo) and that f'(p1t, ijo)) 54 0. Thesystem of linear differential equations

(IX.3.2)di7 Of

=

is called the first variation of system (IX.3.1) with respect to the solution pit, rlo).The coefficients matrix of (IX.3.2) is periodic in t of period 1.

Let pi, p2, ... , p be the multipliers of (IX.3.2) (cf. Definition IV-4-5). Since2 p-(t, (pit, o)) d p1t, ijo), linear system (IX.3.2) has a nontrivial periodic

solution fjt, im). This implies that one of the multipliers must be 1. Set pl = 1.The following theorem gives a basic sufficient condition for orbitally asymptoticstability.

Theorem IX-3-4. If n - 1 multipliers p2,... , p, satisfy the condition ipiI < 1(j = 2,... , n), then the periodic orbit C(rjo) is orbitally asymptotically stable ast-4+oo.Prof.

We prove this theorem in five steps. It suffices to find an (n - 1)-dimensionalmanifold M in a neighborhood of the point so that

(1) M is transversal to the orbit C(i) at ilo; i.e., the tangent of C(yo) is not inthe tangent space of M at ilo,

(2) there exist two positive numbers K and r such that

Iu(t,n - p1t,i )I <- KIn - iole-°: for i EM and t>0

(cf. Figure 2).

FIGURE 1. FIGURE 2.

Step 1. There exists an invertible n x n matrix P(t) whose entries are real-valued,continuously differentiable, and periodic in t of period 1 (or 2) such that the trans-formation

(1X.3.3) 6 = P(t)v"

3. ORBITAL STABILITY

changes (IX.3.2) to

(IX.3.4) dt = AV, A = [ 0 B,

285

where B is a constant (n - 1) x (n - 1) matrix (cf. Theorems IV-4-1 and IV-4-3).Since the absolute values of n -1 multipliers are less than 1, there exist two positivenumbers Co and a such that(IX.3.5) Iexp[tB] I < Coe-tot for t > 0.

A fundamental matrix solution of (IX.3.2) is given by

1/(t) = P(t) exp[tA], exp[tA] = [1 0 1

o exp[tBj J

This implies that the first column vector of P(t) is a periodic solution of (IX.3.2)and all the other columns of W(t) are not periodic. Hence, assume that

(IX.3.6) P(t) _ [d AtA), Q(t),

where Q(t) is an n x (ii - 1) matrix. Note that det[P(t)] 56 0.

Step 2. Change (IX.3.1) by(IX.3.7)

to

(IX.3.8)

where

y = z' + At, rlo)

Tt 8 ay(p1t,no))z + h(t, zdt =

h(t,z-) = f(z + pit,t)o)) - I(plt,no)) - y(pit,qo))z

It is easy to show that h(t, 6) = 0 and az (t, 6) = 0.

Step 3. Change (IX.3.8) by the transformation i = P(t)w". Since A =

P(t)-1 18 (p-(t, jo))P(t) - d (,t) , it follows thatL

(IX.3.9) dt = Aw + #(t, tip),

where g(t, w) = P(t)-'h(t, P(t)w). Note that g(t, 6) = 0 and 8w (t, 0') = 0. Thisimplies that for any given positive number r, there exists another positive numberK(r) such that(IX.3.10) 19(t, w1) - g"(t, w2)I < K(r)[w1 - t62I for t > 0whenever kw"II < r and I w2I < r and that

(IX.3.11) lim K(r) = 0.r-.0

Set ti = ti and w2 = 0. Then, (IX.3.10) becomes(IX.3.12) jg"(t,w)I < K(r)Iti'j for t > 0.

Step 4. Let us write (1X.3.9) in the form

(IX.3.13)dt

= 9i(t,19),

tt9i (s,+u(s,&ds

JL0

By + g2(t, w),

where v and g2 are (n - 1)-dimensional vectors w" _ VJ

and g = I _J .

System

(IX.3.13) can be changed to the system of integral equations111

(IX.3.14)

v(t, ) = p (

t

C))ds,u(t, J00+

exp [(t - s)BI g2(s, w(s, ))ds,tB1 + f0

where 1; is an (n -1)-dimensional arbitrary constant vector. In this step, a solutionw(t, t) of (IX.3.14) will be constructed in such a way that

(IX.3.15) ko 1 ie-ot

for

(IX.3.16) t > 0 and 1t1 < 60,

where ko and bo are suitable positive numbers.First fix three positive numbers ko, ro, and 6o so that Ico > 2, kobo < ro, and

K(r)< 2 for 0 < r < ro. Then, from (IX.3.5), (1X.3.10), (IX.3.11), (IX.3.14), and

a(IX.3.15), it follows that

(I)

and

e-2otlrj + f eo"dstS0

S e of I 1 + K( (1 + ) < ko1 1e-OL

Next, set lIJt ({) = sup (eot 1 (t, 4)1 : t > 0) if the entries of an R"-valued func-

tion t(i(t, £) are continuous for (IX.3.16) and l}} < kolle-ot for (1X.3.16).Then,

I

J+9i(s,t,ii(s,())ds

dv _dt

+< K(ko[ it

e-O°dst

K(aIf1)

exp (tBJ ( + jexE(t -

l- ft 9i(s,tG2(s,&ds+oo

< K(ko at < 2IItLI - 1211(&-ot

and

f exp[(t-s)B]92(s,))ds - f+ exp((t-s)B](s,(IV)

<

t

<Q

- 2[[i1 - &IOe-oc

for (IX.3.16) if the entries of R'-valued functions 1(t, )) and (t, t; are continuousfor (1X.3.16) and that J ,(t,£)I < ko1&-OC (j = 1,2) for (IX.3.16).

Let us define successive approximations as follows:

15m(t, =L gm(t,

where %ao(t, = 0 and

f= g1(s,(s,))ds,ao

exp[tB) +J

exp[(t-s)B]92(s, ' mt0

Then, it can be shown without any difficulties that lim >im(t, existsm --+oc

uniformly for (1X.3.16) and the limit %i(t, { is a solution of integral equations

(1X.3.14) satisfying condition (IX.3.15) for (IX.3.16). Note that i(O,t;)

0

where a(t) = f gl (s, z/i(s, })ds. This implies that00

Step 5. Set00,6 = Plt,vio)

Then, is a solution of system (IX.3.1) such that

(IX.3.18) At, f) - pjt, i o)[ 5 Ko{f [e-°` for (IX.3.16),

where Ko is a positive constant. Define an (n - 1)-dimensional manifold M by

M = (g=4(0'6: 01<<601-Using (IX.3.6), (IX.3.17), (IX.3.18), and (1X.3.11), we obtain

(0, f) = P(0)W (0, a + i?o = dP (0, 40) + Q(O) + rlo,

det dp(0, 'b), Q(0)] 0.-76

Thus, the manifold M is transversal to the orbit C(fjo) at rp. Note that (t, tAt, ¢(0, 6).

Moreover, by continuity of the solutions of (IX.3.1) with respect to the initialconditions, there exist two positive numbers 61 and ro such that for i satisfying thecondition

11 -401

there is a real number r(q such that 1r(q-)l < ro and p(r(g71,'7) E M. Therefore,(IX.3.18) implies that

1pit+r(17),q - P-(t,il0){ <Koboe-°` for t > 0.

Thus, C(ilo) is orbitally asymptotically stable as t --+ +oc.

Remark IX-3-5. If the entries of an R'-valued function f (t,1/) is continuouslydifferentiable on the entire f.-space lRn and system (IX.3.1) has a periodic solutionp ( t , i l o ) of period 1 such that f (p(t, i )) j4 0, there exist n - 1 vectors 4j(t) CRn (j = 2, ... , n) such that

(i) the entries of vectors 9'J (t) (j = 2,... , n) are continuously differentiable andperiodic in t of period 1,

(ii) n vectors dpig0), q"2(t), ..., form an orthogonal system.

Proof.

If n = 2, it is easy to find 42(t). For n > 3, there is the recipe.

Note first thatdpdtilo} - - f(At, 40))

f(p1t,ilo))00. Set g1(t)=f (pit, no)) f (pit, ilo))

where u" b denotes the usual dot product. Since fl(t) is smooth, there exists aconstant vector i71 such that nl ill = 1 and that ili + qi (t) A 0 for 0 < t < 1.Choosing an orthonormal set {ill.... , in I, where n, are constant vectors, define

Q'n(t) = T)n - on(t) (rJi + ji(t)) (1 = 2, ... , n),+ai(t)

where ah(t) _'h - ji(t). 0For this construction, see, for example, (U).

Example IX-3-6. The orthogonal system of Remark IX-3-5 is useful to studysolutions of (IX.3.1) in a neighborhood of a periodic solution. To illustrate thisapplication, let us look at a system

dy = fdt

where f is an 1R3-valued function whose entries are continuously differentiable onthe entire. y-space R3. Assume that (SI) has a periodic solution p(t, irjo) of periodI such that f(plt,r'b)) 54 0. In this case, there exist two vectors j2(t) E IR3 and4fi(t) E L3 such that


(i) the entries of vectors 4"4(t) (j = 2,3) are continuously differentiable and peri-odic in t of period 1,

, 4'2(t), and 43(t) form an orthogonal system.(ii) three vectors dQ

Note thatdp(ttrlo)

= f (plt, no)).

For any fixed real non-negative number r, the set

P(T) = {l7-,m) + u142(r) + u2Q3(T) (ul,u2) E R2}

is the plane which is perpendicular to the orbit at piT,, o). Letting t, u1, andu2 be three functions of r to be determined, set

(IX.3.19) f(t) = p( r, no) + u1Q2(r) +

From (IX.3.19) and the given system (S1), we derive

u243(T)

dt {/,-/ du1 due _ dge(r) dQ3(r)f(f(t))dT = f(P(r>io)) + dT 42(r) + dT 93(r) + u1

d-,+ u2

dT

This yields(IX.3.20)

42(T)df11 dt ( 1 d93(r)1W - 2(T) f(y(t))dr - 2(t)' dr l u1 - ((t) dT J u2,

and

due dt ((t) dQ2(T)\ dg3(r))y' dr u1 - 93(t) dr

(IX.3.21)= [f(PIT'770)) f(PIT,, )) + (fcvlr,no)) u1

-',*lo}).N(t))}+ (f(p1r,TIo))d9d(T)) uz] x [f(Ar, f(

where a" b denotes the usual dot product and we assumed that q"1(r} = 1 (j =2, 3).

Note that

(PIT,r, ))42(r)+ u2 f(YlT,r/o))93(T)+O(ItII +Iu2I)AM)) = RAT, 7lo))+ul89 09

Hence, from(IX.3.20) and (IX.3.21), we derive

dt77.

= 1 + 00U11 + Iu21)

290

and(IX.3.22)

dul

dr

IX. AUTONOMOUS SYSTEMS

42(r) ' l ul + [i(T). (r))1 U2

- (o) . _d_r) Jut - (0) -

d9drr))u2 + O(lul l + 1U21),

d7-93(T) ' W (FFIr, 1I0))42(r) 1 ul + [i(r). r, 90))93(r) 11 U2

(i(t). ddrr)) ul - (t(t). d (r) J U2 + O(lulI + 1U21)

Let Q(t) be the 3 x 3 matrix whose column vectors are (at, qo)), fi(t), and,fi(t), i.e., Q(t) = [ f (p1t, rlo)) 6(t) q"3(t)! . The transformation w = Q(t)v changesthe linear system

(S2)

to

dtr,_ L(PIt,rl *6

dv' 0 /31(t) 32(t)= 0 atl(t) a12(t) 16,

0 a21(t) a22(t)

Using these notations, write (IX.3.22) in the form

(IX.3.23)

where

dul= all(r)ul + a12(r)u2 + 91(T,ul,u2),

d7-

1 dug= a21(r)u1 + a22(r)u2 + 92(T,u1,u2),

dr

191(r,u1,u2) - K(r)(lul-V11+1u2-V21) (i = 1,2)

with K(r) > 0 and limK(r) = 0 for lull + lull < r and Ivll + 1v21 < r. Observer-O

that the two multipliers of the linear system

du' = [all(t) a12(t)j a21(t) a22(t)U

are also multipliers of system (S2). Therefore, using system (IX.3.23), TheoremIX-3-4 can be proven. In general, we obtain more precise information concerningthe behavior of solutions in a neighborhood of a periodic solution in this way.

The materials of this section are also found in 1CL, Chapter 13, §2, pp. 321-3271.

4. THE POINCARE-BENDIXSON THEOREM 291

IX-4. The Poincare-Bendixson theorem

In this section, we explain the structure of C+(i-) on the plane. Consider anR2-valued function f (y) of y E R2 such that the entries of f (y7) are continuouslydifferentiable on the entire g-plane R2. Denote again by pit, ill the unique solution

of the initial-value problem dg = f (y-), y(0) = i. The main result of this section

is the following theorem due to H. Poincare [Poll] and 1. Bendixson [Ben2].

Theorem IX-4-1. Suppose that the solution pit, qo) is bounded for t > 0 andthat C+(ilo) contains only a finite number of stationary points. Then, there are thefollowing three possibilities:

(i) C+(i-Io) is a periodic orbit,(ii) C+(i o) consists of a stationary point,

(iii) C+(i)) consists of a finite number of stationary points and a set of orbits eachof which tends to one of these stationary points as I ti tends to +oo.

To prove this theorem, we need some preparation.

Definition IX-4-2. A finite closed segment t of a straight line in R2 is called atransversal with respect to f if f (y1 0 0 at every point on t and if the vector f (y)is not parallel to t at every point on t (cf. Figure 3).

Observation IX-4-3. For every transversal t and every point i , the set tnC+(contains at most one point (cf. Figures 4-1 and 4-2).

I

PU,

FIGURE 3. FIGURE 4-1. FIGURE 4-2.The following lemma is the main part of the proof of Theorem IX-4-1.

Lemma IX-4-4. If pit, qo) is bounded fort > 0 and if there exists a point i7 EC+(7) such that C+ (11) contains a nonstationary point, then C+(i ) is a periodicorbit.

Proof.

We prove this lemma in three steps. Since C+(qo) is invariant and rj1 E C+(7 ),it follows that p1t,i7l) E C+(jo) for all t. Furthermore, C+(ijt) C C+(io) sinceC+(ilo) is closed.

Step 1. Let rj be a nonstationary point on C+(t7 ). Also, let t be a transversal withrespect to f that passes through I. Then, t n C+(10) = {77} since rj E G+(ijl) CC+(no).

Step 2. Since tj E C+(r)l ), there exists a sequence {tk : k = 1,2.... } of realnumbers such that lira tk = +oo and p1tk, ill) E t (k = 1, 2, ... ). Note that

m+oop(tk, ill) E C+(ilo). Therefore, p(tk, m) = #(k = 1,2.... ). This implies thatthere exist two distinct real numbers rl and T2 such that it = p(rl, ill) = (r2, ill)and hence p(t, rl = p(t, p(ri , ili )) = p1t, p(r2i ill )). This, in turn, implies thatp(t + T1, ill) = p(t + r2i ill) for t > 0. Therefore, the orbit C(ill) is periodic in t ofperiod Irl - r21. Furthermore, C(ill) C C+(rjo). Note that there is no stationarypoint on C(qj ).

Step 3. Since C+(ilo) is connected, it follows that distance(C(ijl), C+(ilo)-C(ni))}= 0. Hence, if C(ijl) # L+ (10), there exists a sequence { k E C+(i o) : k = 1, 2....

such that tk C(rj,) (k = 1, 2, ...) and lim £k = { E C(ill ). Assume thatk-.+oo

there exists a transversal t such that E e and lk E t (k = 1, 2.... ). Note thatl; E C(ill) C C+(ilo). Then, k = t c C(ill) (k = 1, 2, ... ). This is a contradiction.Thus, it is concluded that C+(rlo) = C(iji).

Now, we complete the proof of Theorem IX-4-1 as follows.

Proof of Theorem IX-4-1-

If C+ (, ) does not contain any stationary points, then (i) follows (cf. LemmaIX-4-4). If C+(no) consists of stationary points only, we obtain (ii), since C+(ilo) isconnected. If f-+ (Q contains stationary and nonstationary points, then C+(7-) for

any point it E C+(ilo) does not contain nonstationary points (cf. Lemma IX-4-4).This is true also for t < 0. Hence, (iii) follows.

Observation IX-4-5. In cases (i) and (iii), the set 1R2 - C+(ilo) is not connected.Furthermore, if an orbit C(i)) is contained in C+(i o), two sides of the curve p1t, tl")belong to two different connected components of R2 - C+(t ). In fact, if we considera simple Jordan curve C which intersects with the orbit C(tl at n transversally, thenthe curve p(t,,o) intersects with C in a neighborhood of two distinct points on Cinfinitely many times (cf. Figure 5).

Theorem IX-4-6. If C(ijo) n C+(ilo) 0 0, then C+(ijo) = C(rjo) and either no isa stationary point or the orbit C(ilo) is periodic.

Proof

In this case, C(ilo) C C+(ilo). If C+ (8o) contains nonstationary points, the orbitC(i o) consists of nonstationary points. Choose a transversal a at ilo. Then, it canbe shown that C(ilo) is periodic in a way similar to the proof of Lemma IX-4-4,since r1o E C+(tlo).

Observation IX-4-7. If C(ilo) n L+(no) = 0, then lim distance(plt,ilo) andt-+oo,C+(Q) = 0. It follows that if C+(ilo) consists of a stationary point then

lim p(t, irjo) If C+(ijo) is a periodic orbit, then C+(rjo) is called a limitt+oocycle (cf. Figures 6-1 and 6-2).

5. INDICES OF JORDAN CURVES 293


The materials of this section are also found in [CL, Chapter 16, §§1 and 2, pp.391-3981 and (Har2, Chapter VII, §§4 and 5, pp. 151-1581. In [Hart[, the Poincare-Bendixson Theorem was proved without the uniqueness of solutions of initial-valueproblems.

IX-5. Indices of Jordan curvesIn this section, we explain applications of index of a Jordan curve in the plane to

the study of solutions of an autonomous system in 1R2. Consider again an R'-valuedfunction f (y-) of y E 7,k2 whose entries are continuously differentiable on the entirey-plane R. Denote also by pit, J) the unique solution of the initial-value problemdg

= f (y1, yi(0)dt

To begin with, let us introduce the concept of indices of Jordan curves. Let Cbe a Jordan curve y = il(s) (0 < s < 1) with the counterclockwise orientation (cf.

Figure 7). Assume that &(s)) # 0 for 0 < s < 1. Set u(s) = f ('l(s)) (0 <If(s))1

s < 1). Then, il(s) is the unit vector in the direction of f'(il(s)). There exists areal-valued continuous function 0(s) defined on the interval 0 < s < I such that

a (s)(s) _ Isin0(s)I'

Definition IX-5-1. The index of the Jordan curve C with respect to the vector

field f (y) is given by I f{C) =0(1) - 8(0) .

2r,

This definition is independent of the choice of a parameterization rj(s) of C and thefunction 0(s). Let us denote by D the domain bounded by C (cf. Figure 7).

Observation IX-5-2. If the domain D is divided into two domains DI and D2by a simple curve £, then 1 f{C) = l f{",) + If-(8D2) if f (rl) 0 on £ U C, where8D? (j = 1, 2) denote the boundaries of domains DI and D2, respectively, as theportions of I f{8Di) and I j{d Dt) along £ canceled each other (cf. Figure 8).

Observation IX-5-3. If All) A 0 on C U D, then I f{C) = 0. To show this, divideV into sufficiently small subdomains, use the fact that the vector field 19(s) has nochange on a small subdomain, and apply Observation IX-5-2.

Observation IX-5-4. Assume that a point d E V is a stationary point, i.e.,f(d) = 0. Assume also that f(,) 96 0 on C U V except at a. Then, If(C) dependsonly on aa. Therefore, we define the index of an isolated stationary point d withrespect to f by Ilea) = I f(OV), where V is a neighborhood of as such that there areno stationary points in V other than a.

Observation IX-5-5. If D contains only a finite number of stationary pointsN

dl, d2, ... , dN, then I j<C) = FI J(a"k). Proof of this result is similar to that ofk=1

Observation IX-5-3.

Observation IX-5-6. If f (y) # 0 on C and if f(,7(s)) is tangent to C at everypoint i(s) (0 < s < 1) of C, then I1{C) = 1 (cf. (CL, Chapter 16, §4, Theorem4.31).

Proof.

It is evident that the index of C with respect to f and the index of C with respect

to the tangent vector to C are the same, i.e.,

If4C) = hn)(C)-Assume without any loss of generalities that

# ( 0 )=# ( I ) ) = 0 and i (s) > 0 f o r 0 < s < 1,where rig (s) (j = 1, 2) are the entries of the vector r"(s) (cf. Figure 9).

- __11 n2=0

FIGURE 7.

For 0<r<s<1,set

V(T,s) =

FIGURE 8.

(0.0)

FIGURE 9.

if T = s,

if T < s (mod 1),

1 (T, s) _ (0, 1).

Then, v(T, s) is continuous in (T, s) for 0 <_ r < s < 1. Hence, studying how 16(0, s)changes from s = 0 to s = 1 and how i6(T,1) changes from T = 0 to T = 1, weobtain Ij.(.)(C) = 1. 13

Observation IX-5-7. If C is a periodic orbit of the autonomous system

(IX.5.1)didt f(+1+

then 1 f-(C) = 1.

Observation IX-5-8. If C is a periodic orbit of autonomous system (IX.5.1), thenthe domain D contains at least a stationary point (cf. Observations IX-5-3 andIX-5-7).

Observation IX-5-9. Suppose that f(it(s)) is tangent to C only at a finite numberof points ij(rl ), #(7-2),. .. ,1t(TN) on C. Assume also that at each point it(rk), either

(1) p(t,it(rk)) V V for ItI < 5k, or(2) p(t, it(rk)) E D for 0 < ItI < bk,

for some sufficiently small positive numbers 51, ... , SN. Let us call rj(rk) an exterior(respectively interior) contact point in case (1) (respectively (2)) (cf. Figure 10).Let us denote by E (respectively H) the total number of interior (respectivelyexterior) contact points among N points it(-r1), ij(r2), ... , #(T-N). Then E+H = Nand

(IX.5.2) 1 f-(C) =E-H+2

2

Sketch of proof.

Let rj(r,) and fj(rk) be two consecutive contact points such that T. < rk. Then,(i) if both of these two points are exterior contact points, then the tangent to C

changes 7r in angle more than the vector field f does from the point #(T.) tothe point fj(rk) (cf. Figure 11-1),

(ii) if both of these two points are interior contact points, then the vector field fchanges it in angle more than the tangent to C does from the point #(T,) tothe point it(rk) (cf. Figure 11-2),

(iii) if these two points are an exterior contact point and an interior contact point,then the tangent to C and the vector field f change in the same amount inangle (cf. Figures 12-1 and 12-2).

Since the total amount of change of the tangent to C in angle is 27r (cf. ObservationIX-5-6), we arrive at formula (IX.5.2). 0

q(

Observation IX-5-10. Let d be an isolated stationary point. Assume that aneighborhood V of d is divided into a finite number of sectorial regions S1, S2,... , SNby a finite number of orbits C(rji ), C(il3),... , C(iv) in such a way that(a) rli, Tt2, - , nN E 8V,(Q) either p1r,rjk) E V fort > 0 and tends to 99 as t -+ +oo, or p5r,ilk) E V for

t < 0 and tends to d as t - -cc (cf. Figures 13-1 and 13-2).

FIGURE 12-1. FIGURE 12-2. FIGURE 13-1. FIGURE 13-2.

Let us assume also that each sectorial region Sk satisfies one of the followingfour conditions:

(I) if it E Sk and {il - d] is sufficiently small, then C(r)) C Sk and pit, q-) tends tod as Itl +oo (cf. Figure 14),

(II) if i) E Sk and Jil - dl is sufficiently small, then pit, i)) E Sk only on a finitet-interval ak < t < Yk (cf. Figure 15),

FIGURE 14. FIGURE 15.(111) if it E Sk and ail - dl is sufficiently small, then p1t,i) tends to dd in Sk as

t +oc, but plt, q") does not tend to dd in Sk as t -+ -oo (cf. Figure 16-1),(IV) if ij E Sk and Jrj - al is sufficiently small, then p1t,71) tends to ad in Sk as

t -. -oc, but plt, q+) does not tend to dd in Sk as t - +oo (cf. Figure 16-2).

The sectorial region Sk is said to be elliptic (respectively hyperbolic) in case (I)(respectively (II)). In cases (III) and (IV), the sectorial region Sk is said to beparabolic. Let us denote by E (respectively H) the total number of elliptic sectorialregions (respectively hyperbolic sectorial regions) among St, ... , SN. Then, it isknown that

(IX.5.3)E--H+2

2

This result is similar to formula (1X.5.2) of Observation IX-5-9. For a proof indetail, see (Har2, Chapter VII, §9, pp. 166-172).

Observation IX-5-11. Suppose that the vector field f (y e) depends on a param-eter e E 0 continuously, where 0 is a connected set. In this case, if f (if(s), e) # 0at every point if(s) on the Jordan curve C for all e E A, then the index I f(.,()(C) is

a constant independent of e. In fact, in this case, the integer I f( E)(C) depends on

e continuously, and hence it is a constant.

Example IX-5-12. If an isolated stationary point d is a node, a spiral point, ora center, then E = H = 0. Hence It{a") = I.

Example IX-5-13. If an isolated stationary point d is a saddle point, then E = 0and H = 4. Hence -1 (cf. Figure 17).

FIGURE 16-1. FIGURE 16-2. FIGURE 17.

Example IX-5-14. More general cases of isolated stationary points a are shownby Figures 18-1, 18-2, and 18-3. In fact,

1E = 0, H = 2 and hence II-(a") = 0 in the case of Figure 18-1,E = 2, H = 0 and hence It-(d) = 2 in the case of Figure 18-2,

E = 1, H = 1 and hence l1-(d) = 1 in the case of Figure 18-3.

Example IX-5-15. Let p(z) be a polynomial in a complex variable z with complexcoefficients. Set y' = t(zj and y2 = !3(zJ (i.e., z = yl + iy2) and regard the

differential equationdz

= p(z) as a system of two differential equations

(IX.5.4) ddt' _ 9(p(yl + iy2)], dtY2 = (p(yl + iy2)]

on the y"-plane. It is easy to see that, if zo = 171 +ir12 is a zero of p(z) of multiplicity

m, then if#?) = m, where AM _ R(p(yl + iy2)] and i) = 111

%P(YI + iy2)] [112, .

if p(z) = iz2, system (IX.5.4) becomes

dtl -2y1Y2, ddt2 = y1 - y2

The point 0 is an isolated stationary point and II<O) = 2.

For example,

If we define a vector field f (y, e) by f (W, e) = [!1], then, 1- (0) = 2 for

e > 0 (cf. Observation IX-5-11).

Example IX-5-16. Assume that f ( = fl (pl'Y2) satisfies the condition f (arty)f2(y142)

= aP f (rte, where A is a real variable and p is an integer. Set y = r os0 I. Then,

the autonomous system = f(it) can be written in the form

dr dO

dt = rPFI (0), rdt

rPF2(0),

whereF1(0) = f 1(cos 8, sin o) cos 0 + f2(cos 0, sin 0) sin 0,

Sl F2(0) = -f1(cos0,sin6)sin0 + f2(cos0,sin0)cos0

(cf. Exercise VIII-3). If a ray to is defined by F2(0) = 0, then to is an orbit of the

system dff = f (r))). Thus, the entire y"-plane can be divided into sectorial regionsby those orbits QB determined by equation F2(0) = 0. For example, in the case ofsystem (IX.5.5), we obtain Fl (0) = - sin 0 and F2(0) = cos 0. Observe that

>o for -2 <0<2,>0 for -r<0<0,dr = 0 for 0=-r and 0, and =0 f o r 0 = - and .

'dt dt 2 2<0 for 0<0<r

3< 0 for 2 < 0 < 1r

Hence, E = 2 and H = 0 (cf. Figure 19). Therefore, I f-{0) = 2 (cf. ExampleIX-5-15).

FIGURE 18-1. FIGURE 18-2. FIGURE 18-3. FIGURE 19.

The materials of this section are also found in [CL, Chapter 16, §§4 and 5, pp.398-402] and [Har2, Chapter VII, §§2 and 3, pp. 146-151, and §§8 and 9, pp.161-1741. For a geometric and topological treatment of indices, see (Mil.

EXERCISES IX

IX-1. For each of the following three systems, using the given function V (x, y),show that all orbits are bounded as t -. +oo .

EXERCISES IX

(1) _ -xy2 - 4y, _ -yx2 + 3x, V(x,y) = 3x2 +4Y2;

(2) = y, = -x3 - y, V(x,y) = x4 +2Y2;

299

= y,dy

= - (xs - 3x4 + 2x3 + 120x2 - 23x + 5) - (1 + x2)y,

V(x,y) =2

+ J (ss - 3x4 + 2x3 + 120x2 - 23s + 5)ds.0

IX-2. Show that every solution y(t) and its derivative L(t)(t) of the differential

tend to 0 as t -i +oo.equation j2 + dt + y3 =0

IX-3. Consider an autonomous system

(S1)yj __ fj(11j,112)

Wt [y2] [f2(yi,y2) '

where f j and f2 are continuously differentiable on the entire (yj, y2)-plane. Assumethat

(1)

fl(yl,y2)>0 for y2>0 and -oo<yl<+oo,fi(yi,y2) < 0 for y2<0 and - oo < yl < +oo,

(2) fj(yi,1) > 1 and fi(yi, -1) < -1 for -oo < yl < +00,(3) f2(yi,1) = 0 and f2(yi, -1) = 0 for -oo < y1 < +oo,(4)

2 - 1) for jy2j < 1 and 0 < yi < +oo,1 f2(11j,y2) <_ YI(112

f2(Y1,Y2) ? Y1 (Y22 - 1) for jy21 < 1 and - oo < yi < 0,

Find G+((n1,n2)) for (771,772) such that In2t < 1.

Hint. There are two possibilities:

Case 1. The solution (yi(t),y2(t)) of (Si) that satisfies the initial condition

(C) yi(0) = 771 and

is bounded as t -+ +oo.

112(0) = 172

Case 2. The solution (y1(t),y2(t)) of (SI) that satisfies the initial condition (C) isunbounded as t +oo.

In Case 1, G+((nj, n2)) is either {(0, 0)} or a periodic orbit. In Case 2, G+((nj, 92))is{(x,±1):-oo<x<+oo}.

Examples.(a) Every orbit in Iy21 < 1 of the system with fl(yl,y2) = y2 and f2(yi,y2) _

yl(y2 - 1) is periodic.(b) The stationary point (0,0) is a stable spiral point if fl(yl,y2) = y2(2 -

sin(yly2)) and f2(yj,y2) = 2yi(y2 - 1).(c) The stationary point (0, 0) is an unstable spiral point if ff(yi, y2) = y2(2 +

sin(yiy2)) and f2(yi,y2) = 2y, (y22 - 1).Verify these statements by using the function V(yl, y2) = yi - ln(1 - y2).

IX-4. Let us consider a system

(52) dt - W= [y2 ] , f(U _ [f2(Y-)

where f, (y-) and f2(y-) are continuously differentiable with respect toy in a domainDo C R2. Assume that

(i) system (S2) has a periodic orbit p-(t, ik) of period 1 which is contained in thedomain Do,

(ii) RA t' v) 6,

(iii) the integral (1 ( i (pjt, qo)) + (5t, o))) dt is negative.,9y 0

Show that the periodic orbit p(t, jo) is orbitally asymptotically stable as t -' +00.

Hint. This is called Poincnre's criterion. Look at

dzV 49f

dt =

Then, I det Y(1)I < 1 for the fundamental matrix solution Y(t) of this system suchthat Y(0) = 12 (cf. (4) of Remark IV-2-7). Hence, an eigenvalue p of Y(1) mustsatisfy the condition dpi < 1. (The other eigenvalue of Y(1) is 1.) Now, use TheoremIX-3-4.

IX-5. Assume that two functions f (x) and g(x) are continuously differentiable in

x E R. Assume also that the differential equation dt2 + f (x) drt+ g(x) = 0 has

a nontrivial periodic solution x(t) of period I such that f f (x(t))dt > 0. Show0

that (x(t), x'(t)) is an orbitally asymptotically stable orbit as t +00 in the (x, x')phase plane.

Hint. Use Exercise IX-4-

IX-6. Show that there exists a nontrivial periodic orbit of the system

dI7=f (y), y= Ly2J,

Lf2h

W)(Y-)

where(1) the entries of the 1R2-valued function f (y-) is continuously differentiable on the

entire y-plane,

EXERCISES IX 301

(2) f(6) and f(y) ifil96 ,

(3) of (O) = i, eft (0) = 8, 2Z-2 (0) = -2 and (p)= 1,

ay1 OW ay1

(4) Ivll+lliM +oo(ft(Y1,Y2) + yr) and 1Y11+liM +oo(f2(yt,N2) +

IX-7. Given that

yz) exist.

Y1 - y2 y2

find the total number E of elliptic sectorial regions, the total number H of hyper-bolic sectorial regions, and the total number of parabolic sectorial regions in the

neighborhood of the isolated stationary point 0 of the systemdt

= f (y, E) in the

following two cases: (i) E = 2 and (ii) E _ 2. Also, find I y(0) for e 96 0.

IX-8. Let us consider a system

(S3) dt = f (y),

where the entries of the l 3-valued functionf is continuously differentiable on theentire y-space R3. Assume that (S3) has a periodic orbit pit,ip) of period 1 suchthat f (r'(t, o)) 0 0. Assume also that the first variation of system (S3) with respectto the solution p(t,' o), i.e.,

d9 Ofdt

(r(t,io))u,

has three multipliers p1 = 1, p2i and p3 satisfying the condition: 1P21 < 1 and jp3j >1, respectively. Construct the general orbits pit, r) of (53) such that distance(At,17),C(s'7o)) tends to0ast-+too.

IX-9. Show that the differential equation d-t2 + (x + 3)(x + 2)drt

+ x(x + 1) = 0does not have nontrivial periodic solutions.

Hint. Two stationary points are a node (0,0) and a saddle (-1,0). Furthermore,

setting f(x1,x2) = x2 , we obtain divf(x1,x2) _I -(xt + 3)(x1 + 2)x2 - xt(xt + 1) J

-(x1 + 3)(x1 + 2) < 0 if x1 > -1. Also, use index in §IX-5.

IX-10. For the system

dt= f(x,y) = x(1 - x2 - y2) - 3y, = g(x,y) = y(1 - x2 - y2) + 3x,

(1) find and classify all critical points,(2) find

dt = f (x, y)

.-+ 9(x, y) 5


for the function V (x, y) = x2 + y2,

{(x): dV(3) find the set S =

dt= 01,

(4) examine if S is an invariant set,(5) find the phase-portrait of orbits.

IX-11. For the systemdx

dt =f(x,y,x) = - x(1 - x2 _ y2)2 + 3xz + y,

dt =9(x,y,z) = - y(1 - x2 - y2)2 + 3yz - x,

dz

dt =h(x,y,z) = - z - 3(x2 + y2),

(1) find all critical points and determine if they are asymptotically stable,(2) find

(3)

(4)

(5)

dV= f (x, y, x) 8V + g(x, y, z)

8V + h(x, y, x) 8Vdt 8x 8y 8z

for the function V(x, y, z) = x2 + y21+ z2,

find thesetS=((x,y,z): dV =0},find the maximal invariant set M in S,find the phase portrait of orbits.

IX-12. Find the a phase portrait of orbits of the system

dt = y, dt = y + (1 - x2)x2(4 - x2).

IX-13. Let f (x, y) and g(x, y) be real-valued, continuous, and continuously dif-ferentiable functions of two real variables (x, y) in an open, connected, and sim-

ply connected set D in the (x, y)-plane such that 8f (x, y) + Lf (x, y) j4 0 for all

(x, y) E D. Show that the systemdxt

= f (x, y), d = g(x, y) does not have anynontrivial periodic orbit that is contained entirely in D.

IX-14. Let p(z) be a polynomial in a complex variable z and deg p(z) > 0. Setx = 3t(z] and y = Q [z]. Verify the following statements.

(i) In the neighborhood of each of stationary points of system

(A) d = R[p(z)], dt = `3`[6 (x)],

there are no hyperbolic sectors. Also, system (A) does not have any isolatednontrivial periodic orbit.

(ii) In the neighborhood of each of stationary points of system

(B)dt = 3R[p(z)], L = -`3'[p(x)],

there are no elliptic sectors. Also, system (B) does not have any nontrivialperiodic orbits.

EXERCISES IX 303

IX-15. Find the phase portrait of orbits of the system

d = x2 - y2 -3x+2, d _ -2xy + 3y.

Hint. See Exercise IX-14 with p(z) = (z - 1)(z - 2).dx dy

IX-16. Find explicitly a two-dimensional system dt = f (x, y),d t= 9(x, y) so

that it has exactly five stationary points and all of them are centers.

IX-17. Consider a system

(S4)dy

= f (y1,dt

where the entries of the R'-valued function f are analytic with respect toy in adomain Do C R'. Assume that

(i) system (S2) has a periodic orbit pit, ijo) of period 1 which is contained in thedomain Do,

(ii) AP-Tt,no)) 0(iii) for any open subset V of Do which contains the periodic orbits p(t, i"p), there

exists an open subset U of V which also contains p'(t, o) such that for anypoint i in U, the orbits p'(t, i of (S4) is contained in V and periodic in t.

Show that if U is sufficiently small, for any point i in U, we can fix a positive periodof p1t, y) so that is bounded and analytic with respect to i in any simply

connected bounded open subset of U.

Hint. Apply the following observation.

Observation. Let Do be a connected, simply connected, open, and bounded set in1Rk and let T2 (j = 1, 2, ...) be analytic mappings of Do to l . Suppose that, forany pointy E Do, there exists a j such that T3 [yl = y, where j may depends on Y.Then, there exists a jo such that Tjo [yj = y for all y E Do.

Proof.

SetE j _ {y E D o : T j [ y 1 = Y - ) . j = 1, 2, ... .

Then,(1) E. is closed in Do,

+00

(2) Do = U E,,3=1

(3) Do is of the second category in the sense of Baire.Hence, for some jo, the set Ego contains a nonempty open subset (cf. Baire'sTheorem). Since Tea is analytic, we obtain Tao [ 7 = y" for all y" E Do. O

For Baire's Theorem, see, for example, [Bar, pp. 91-921.

CHAPTER X

THE SECOND-ORDER DIFFERENTIAL

EQUATION dt2 + h(x) + g(x) = 0

In this chapter, we explain the basic results concerning the behavior of solutionsof a system

112

[-h(yi)112 - 9(111)

as t -- +oo. In §X-2, using results given in §IX-2, we show the boundedness ofsolutions and apply these results to the van der Pol equation

(E) x + E(x2 - 1) d + x = 0

(cf. Example X-2-5). The boundedness of solutions and the instability of the uniquestationary point imply that the van der Pol equation has a nontrivial periodicsolution. This is a consequence of the Poincar&$endixson Theorem (cf. TheoremIX-4-1). In §X-3, we prove the uniqueness of periodic orbits in such a way that itcan be applied to equation (E). In §X-4, we show that the absolute value of oneof the two multipliers of the unique periodic solution of (E) is less than 1. Theargument in §X-4 gives another proof of the uniqueness of periodic orbit of (E). In§X-5, we explain how to approximate the unique periodic solution of (E) in the casewhen a is positive and small. This is a typical problem of regular perturbations. In§X-6, we explain how to locate the unique periodic solution of (E) geometrically ase - +oo. In §X-8, we explain how to find an approximation of the periodic solutionof (E) analytically as a +oc. This is a typical problem of singular perturbations.Concerning singular perturbations, we also explain a basic result due to M. Nagumo[Na6] in §X-7. In §X-1, we look at a boundary-value problem

y = F (t, y, d i ) , y(a) = o, y(b) = 8.

Using the Kneser Theorems (cf. Theorems 1II-2-4 and III-2-5), we show the exis-tence of solutions for this problem in the case when F(t, y, u) is bounded on theentire (y, u)-space. Also, we explain a basic theorem due to M. Nagumo [Na4] (cf.Theorem X-1-3) which we can use in more general situations including singularperturbation problems (cf. [How]).

For more singular perturbation problems, see, for example, [Levi2], [LeL], (FL],[HabL], [Si5], [How], [Wasl], and [O'M].

304

1. TWO-POINT BOUNDARY-VALUE PROBLEMS 305

X-1. Two-point boundary-value problemsIn this section, first as an application of Theorems 111-2-4 and 111-2-5 (cf. [Kn]),

we prove the following theorem concerning a boundary-value problem

(X.1.1) d 2 = F t, y, dt) , y(a) = a, y(b) = Q

Theorem X-1-1. If the function F(t, yt, y2) is continuous and bounded on a region11 = {(t, Y1, y2) : a < t < b, lyt I < +oo, Iy2I < +oo}, then problem (X.1.1) has asolution (or solutions).

Proof.

For any positive number K, the set Aa = {(a, or, y2) : Iy21 < K} is a compactand connected subset of ft We shall show that A0 satisfies Assumptions 1 and 2of §111-2 for every positive number K. In fact, writing the second-order equation(X.1.1) as a system

(X.1.2)

we derive

dyt _ dyedt - Y2, dt = F(t, yt, y2),

y1(t) = y1(a) + J y2(s)ds,a

y2(t) = y2(a) + JF(s,yi(s),y2(s))ds.`

Hence, if (a, y1(a), Y2 (a)) E A0, we obtain

Jy2(t)I < K + M(b - a),ly1(t)I < lal + [K + M(b - a)](b - a),

where I F(t, yi, Y2)1 < Al on Q. Therefore, Ao satisfies Assumptions 1 and 2 of§111-2. Thus, Theorem 111-2-5 implies that SS is also compact and connected forevery c on the interval Te = {t : a < t < b}.

We shall prove that if K > 0 is sufficiently large, the set Sb contains two points(771,772) and (t;t, (2) such that

(X.1.3) n, < 0 < (1.

In fact, by using the Taylor series at t = a, write yt (b) in the form

yt(b) = a + y2(a)(b - a) +2 dt

(c) (b - a)2,

where c is a certain point in the interval 10. Since I ddt2 (c) M, the quantity

yI(b)I can be made as large as we wish by choosing Iy2(a)I sufficiently large. Thus,there are two points (nt, n2) and ((1, (2) in Sb such that (X.1.3) is satisfied.

Since the set Sb is compact and connected, there must be a point (Q, () in theset Sb. This implies the existence of a solution of problem (X.1.1). 0

306 X. THE SECOND-ORDER DIFFERENTIAL EQUATION

Example X-1-2. Theorem X-1-1 applies to the following two problems:

(X.1.4) dt2 + sin y = 0, y(a) = a, y(b)

and

(X.1.5)dt2 + p-+-, =

0, y(a) = a, y(b) = Q

However, Theorem X-1-1 does not apply to

(X.1.6) d 22 + y = 0, y(a) = a, y(b) = Q.

For more general cases, the following theorem due to M. Nagumo (Na4] is useful.

Theorem X-1-3. Assume that09 f

(i) a real-valued function f (t, x, y) and its derivatives az and are continuous

in a region V = {(t, x, y) : (t, x) E A, -oo < y < +oo}, where 0 is a boundedand closed set in the (t, x) -space;

(ii) in the region D, the function f satisfies the condition

(I) 1f(t,x,y)1 : 0(IyI),where 0(u) is a positive-valued function on the interval 0 < u < +oo suchthat

(II)+°° udu

+00;

(iii) two real-valued functions wi (t) and w2(t) are twice continuously differentiableon an interval a < t f 1 t, wi (t), d t) )(IV)

2 dt

d2dt(t) < f t, W2 (t),

ddt t)) ,2

(iv) two real numbers A and B satisfy the condition

for a < t <b;

(V) wi(a) < A < w2(a), and wi(b) < B < w2(b).

Then, the boundary-value problem

d 22 = f I t, x, ') , x(a) = A, x(b) = B,

has a solution x(t) such that (t, x(t)) E Ao for a < t < b, i.e.,

wi(t) < x(t) < w2(t), for a < t < b.Proof.

The main tools are the following two lemmas.

1. TWO-POINT BOUNDARY-VALUE PROBLEMS 307

Lemma X-1-4. Let x(t, to,rl) be the solution of the initial-value problem

d2x = f (t, x,d .)

, x(to) x'(to) = n,

where a < to < b, (to, l;) E Ao. Then, for any given positive number M, there existsa positive number a(M) such that l x'(t, to, l;, il) j < a(M) if

(X.1.7) jr71 < M and (T, x(r, to, e, rl)) E Do for to < r < t or t:5 r < to.

Proof.

Letting L be a positive number such that

(X.1.8) w2(t) - w1(t) < L for a < t < b,

choose a(M) > 0 for any given positive number M in such a way that a(M) > 111and

(X.1.9)a(M) udu

u)> L.

+! q5(

Suppose that there exist ri and r2 such that to < Ti < r2 < t and that

x'(r1) = M < x'(r) < x'(r2) = a(M) for r1 < r < T2,

where x(T) = x(r, to, t ,17). Then, since x'(r) > 0 for ri < r < r2, it follows that

xxO(X'

(T))) < x/ (7-) for T1 < -r < -r2.

Hence,

fa(M) u/du

0u) -This contradicts the choice of L by (X.1.9). Therefore, Lenuna X-1-4 is true forto < r < t. We can treat the case t < r < to similarly, since if we change t by -t,the differential equation x" = f (t, x, x') becomes x" = f (-t, x, -x'). 0

Lemma X-1-5. Set

7 = rim {a(ba)I wi(a)+e, -w2' (a)-eI,

where e is an arbitrarily fixed positive number. Let also x(t, c) be the solution to theinitial-value problem

d2x = f (t, x,)

, x(a) = A, i (a) =

Then,(1) two curves x = x(t, c) and x = w2(t) meet for some t on the interval a < t 'Y;(2) two curves x = x(t, c) and x = wl (t) meet for some ton the interval a < t La

if (r,x(t,c)) E Ao for

a < r < t. The proof of part (2) is similar.

Proof of Theorem X-1-3.

Now, let us complete the proof of Theorem X-1-3. The main point is that whentwo curves x = x(t,c) and x = w2(t) or two curves x = x(t,c) and x = w, (t) meet,they cut through each other. So look at Figure 1.

(b, B)

(a. A)

t=a

FIGURE 1.

r=b

Example X-1-6. Theorem X-1-3 applies to the boundary-value problem

(X.1.10)d2x _dt2 Ax'

x(0) = A, x(1) = B

if A is a positive number. In fact, assume that 0(u) is a suitable positive constant. Ifwi(t) = sinh(ft) -a and w2(t) = sinh(ft)+$ with two positive numbers a and,3 such that -a < A < j3 and sinh(/) - a < B < sinh(VrA_) +,Q, all requirementsof Theorem X-1-3 are satisfied.

If A is negative, Theorem X-1-3 does not apply to problem (X.1.10). Details areleft to the reader as an exercise.

2. APPLICATIONS OF THE LIAPOUNOFF FUNCTIONS 309

X-2. Applications of the Liapounoff functions

In this section, using the results of §IX-2, we explain the behavior of orbits of asystem

(X.2.1)d __ Y2

dt[yjy[-h(yi)y2 - 9(yi)

as t +oo. Set y = I ylJ

and f (yl = [_h(yi JLet us assume thatl)y2- 9(y1)h(x), g(x), and dd(x) are continuous with respect to x on the entire real line R.

Also, we denote by p(t,r) the solution of (X.2.1) satisfying the initial conditiony(0) = n

Fi g(s)ds. Then,rst set V (y) =2Y2

+ G(yl ), where G(x) = fox

09= [9(yi ), Y21, 8y'

- f (y-) = -h(yi) y22.

Set also, S = { g : R Y) = 0 y. Then, U E S if and only if either h(yl) = 0 or

Y2 = 0.l JJJ

Observation X-2-1. Denote by M the set of all stationary points of system(X.2.1), i.e., M = {9: g(yl) = 0, y2 = 01. Then, M is the largest invariant set inS if the following three conditions are satisfied:

(1) h(x) > 0 for -oo < x < +oc,(2) h(x) has only isolated zeros on the entire real line IR,(3) g(x) has only isolated zeros on the entire real line R.

The proof of this result is left to the reader as an exercise (cf. Figure 2, where0, 0 and 0).

By using Theorem IX-2-2, we conclude that lim p(t, y) = 17 E M if conditionst-+00(1), (2), and (3) are satisfied and if the solution p(t,g) is bounded for t _> 0. Notethat G+(rt) is a connected subset of M.

In Observation X-2-1, the boundedness of the solution p(t, i) for t _> 0 wasassumed. In the following three observations, we explore the boundedness of allsolutions of (X.2.1). Set

X

G(x) = o g(s)ds and H(x) = f h(s)ds.Jo o

Observation X-2-2. Every solution p(t, q) of (X.2.1) is bounded for t > 0 if(i) h(x) > 0 for -oc < x < +oo and (ii) lim G(x) = +oo.

1x1 +00This is a simple consequence of Theorem IX-2-3. In fact, lim V (Y-) = +oo.

191-.+00

Observation X-2-3. Every solution pit, of (X.2.1) is bounded for t > 0, if(i) h(x) > 0 for -oo < x < +oc, (ii) urn IH(x)l = +oo, and (iii) xg(x) > 0

I=t-+o0for -oc < x < +oo.

Proof

Change system (X.2.1) to

(X.2.2)dt 22

d [zl] Z2 - H(z1)-g(z1)


Y1 = Z1, Y2 = z2 - H(zi).

Denote by qqt,() the solution of (X.2.2) such that z(0) _ . Set V, (z-) = zz2 +G(z1). Then,

L9 V,

ay - [g(zl), z21,8z"

,P = -g(z1)H(zl)

Note that g(x)H(x) > 0 for -oo < x < +oo and that

dt V1(ggt,C)) - z(g-(t,C)) F(glt,S)) S 0 for t > 0.

Hence, setting q"(t, S) = z1(t, and V, = c > 0, we obtain

2[z2{t,S)]2 < c for t>0,

since G(x) > 0 for -oo < x < +oo. Therefore, Figure 3 clearly shows that q1 t, C)is bounded for t > 0. This implies that all solutions of (X.2.2) are also bounded fort>0.

{s2, 0) (43,0) (41.0)Y2=0

-

di-1 z1=0

z2=0

FtcuRE 2. F1cuRE 3.

2. APPLICATIONS OF THE LIAPOUNOFF FUNCTIONS 311

Observation X-2-4. Every solution p(t, >)1 of (X.2.2) is bounded for t > 0 if(i) = lim H(x) = +oo, (ii)

=lim H(x) = -oo, (iii) g(x) > a for x > as > 0,

+00 --Ccand (iv) g(x) < -a for x < -ao < 0, where a and ao are some positive numbers.

Proof.

In Observation X-2-3, the Liapounof function

Vi(z) = G(zi) = 2[y2 + H(yl)]2 + G(yi)

was used. Now, let us modify Vl to a form

V2(yl = 2[y2 + H(yi) - k(yi)J2 + G(yl)

The

OV2

n,

1

09_ [[y2 + H(yi) - k(yi )J {h(yi) - dyyt )1 + 9(yi ), Y2 + H(1h) - k(yi) I

and J J

a z (Y ') {y22 + [H(yi) - k(yi)J y2}dy,

- 9(yi) [H(yi) - k(yl)JUsing (i) and (ii), three positive numbers M, a, and c can be chosen so that

fa [H(x) - cJ > M for x > a > ao,-a [H(x) + c] > M for x < -a < -ao.

Also choose a function k(x) so that

(1) k(x) ={ c for x > 2a,

-C for x!5 -2a,

(II} Jk(x)J < c anddk(x) > 0

and

for - on < x - +oo ,

dx

dk(x) >m > 0 for JxJ < a

dx

for some positive number m (cf. Figure 4).k

k=c

x

x=-2a x=-a x=O x=a x=2ak=-c

FIGURE 4.

If a positive number b is chosen sufficiently large,

OV2 .f 0 for lyil ? 2a,

In fact,

or 1y21 > b.

(A) -g(yl) [H(bi) - k(yl))- a IH(yl) - cI < -M < 0 for yl > a,

a IH(y1) + cj < -M < 0 for yl < -a

and

(B) Y2 + IH(yj) - k(yi )11/2 ? 1 for 1yi 15 2a, 1y21 > b

if b > 0 is sufficiently large. Therefore,

av2(i) ' f = -g(yi) IH(yi) - k(yi)1 < 0 for

(u)

and

Iyi I >- 2a, 11/21 < +00,

f < -- (y1 < 0 for a < 1yi I G 2a, 1y21 ? b,09

OV2(iii)

f < - m {y22 + IH(yi) - k(yi)1 y2}

g(yt)IH(yi) - k(yi)I < 0 forlyil < a, 1y21 ? b

if b > 0 is sufficiently large.

Since lim G(x) = +oo, Theorem IX-2-3 implies that every solution of (X.2.2)1XI +00

is bounded.

Example X-2-5. For the van der Pol equation

2 + e(x2 - 1) dt + x= 0,

where a is a positive number, h(x) = e(x2 - 1) and g(r) = x. Hence,

G(x) = 2x2 and H(x) = e (3x3 - x)

Therefore, conditions (i), (ii), (iii), and (iv) of Observation X-2-4 are satisfied. Thisimplies that every solution of the van der Pol equation

X.2.3 yl y2( )

ddt y2 -E(Y - 1)y2 - yi,

is bounded for t > 0. System (X.2.3) has only one stationary point 6. It is easyto see that 0 is an unstable stationary point as t -+ +oo. Therefore, using thePoincaare-Bendixson Theorem (cf. Theorem IX-4-1), we conclude that there existsat least one limit cycle. In §X-3, it will be shown that system (X.2.3) has exactlyone periodic solution.

3. EXISTENCE AND UNIQUENESS OF PERIODIC ORBITS 313

Example X-2-6. For a given positive number a, the system

d[yi

(X.2.4) _ y2

Y2 J [ -aye - sin (yl )

satisfies three conditions (1), (2), and (3) of Observation X-2-1. But, (X.2.4) doesnot satisfy conditions of Observations X-2-2, X-2-3, and X-2-4. Therefore, in orderto prove the boundedness of solutions of (X.2.4), we must use some other methods.

ev.In fact, using the Liapounoff function V (y) = -yZ - cos (yl ), we obtain f =

WY=

-aye < 0, where f (y = [ij.2Since V (y-)

lim at +oo

exists for every solution p-(t, ii) of (X.2.4). Now, observe that(1) We must have a < 1. Otherwise we would have y2(t,,)2 > 2(a+cos(yi(t,rte))

for t > 0. This implies that dt V (pl t, r7-')) < -2a(o - 1) < 0. This contradicts

(X.2.5).(2) If -1 < a < 1, the solution p(t, 7) must stay in one of connected components

of the set {y" : V(y) < a + e < 1} for large positive t. Those connectedcomponents are bounded sets.

(3) In case a = 1, we can show the boundedness of p(t, t) by investigating thebehavior of solutions of (X.2.4) on the boundary of the set {g: V(y-) < 1).

X-3. Existence and uniqueness of periodic orbits

In this section, we prove the following theorem (cf. (CL, p. 402, Problem 51).

Theorem X-3-1. Assume that(i) two real-valued functions h(x) and g(x), and

dg(x) are continuous for -oo <

X < +00,(ii) g(-x) = -g(x) and h(-x) = h(x) for -oo < x < +oo,

(iii) g(x) > 0 for x > 0,(iv) h(0) < 0,

(v) H(x) = f h(s)ds has only one positive zero at x = a,0

(vi) h(x) > 0 for x > a,(vii) H(x) tends to +oo as x -- +oo.Then, the system

(X.3.1)dtd [Yyj

[-h(yi)y22- 9(yi)

has exactly one nontrivial periodic orbit and all the other orbits (except for thestationary point 0) tend asymptotically to this periodic orbit as t +oo.

Proof.

Change system (X.3.1) to

(X.3.2)


Setting

where

d

dt Lz2J Lz2-zi)

Y2J Lz2 -zH(zi),

V(z) =

2+ G(z1),

G(x) = J z g(s)ds and z =0 [z2]

look at the way in which the function V(z) changes along an orbit of (X.3.2). Forexample,

dt

along an orbit of (X.3.2). Hence,

dz2 g(zl) dzi z2 - H(zi)

dzi z2 - H(zl)' dz2 9(z1)dV 9(z1)H(z1) dVdz1 z2 - H(z1)' dz2

_H(zi)

along an orbit of (X.3.2).

z1(t' a)be the orbit of (X.3.2) such that z(0, a)Observation 1. Let z(t, a) = IZ24,a)]

[0]. Then, V (i(0, a)) = 2 a2. There exists exactly one positive number ao suchh

that

V (zI = 9(zl)[z2 - H(zi)] - z29(z1) = -g(z1)H(z1)

z1(ao, ao) = [0 ] and z2 (t, ao) >0 for 0:5t < oo

for some positive number co, where a is the unique positive zero of H(x) given incondition (v) (cf. Figure 5).

Observation 2. Since 0 when z2 = H(z1), there exist two positive numbers

r(a) and Q(a)such that

z(r(a), a) 0

-Q(a)and 0 < zi (t, a) :5a for 0< t < r(a)

if 0 < a < ao. Also, since H(x) < 0 for 0 < x < a, we obtain

dtV(z'(t,a)) = -g(zi(t,a))H(zi(t,a)) > 0 for 0 < t < r(a)

except for a = ao and t = oo. Therefore,

(A) V(z"(r(a),a)) - V(z(0,a)) > 0 for 0 < a < a0 (cf. Figure 6).

(0.ao)

zI = 0

FIGURE 5. FIGURE 6.

Observation 3. If ao < a, there exists a positive number r0(a) such that

J z1(ro(a), a) = a, 0 < z1(t, a) < a for 0 < t < ro(a),10 < z2(t,a) - H(zl(t,a)) for 0 < t < ro(a)

(cf. Figure 7). In particular ro(ao) = Co (cf. Observation 1).If the variable t is restricted to the interval 0 < t < 7-0 (a), the quantity z2(t, a)

can be regarded as a function of z1(t,a), i.e.,

z2(t,a) = Z(zl(t,a),a),

where Z(x, a) is a continuous function of (x, a) for 0 < x < a and a > a0, andcontinuously differentiable for 0 < x < a and a > ao except for x = a and a = a0.Furthermore, Z(x, al) < Z(x, a2) for 0:5 x:5 a if a0 < a1 < 02 (cf. Figure 7).

Set 2(x, a) = I Z(z, a)J .

Then,

d g(x)H(x),

dxV(Z(x,a)) - Z(x,a) - H(x) > 0 for 0<x<a

and, hence,

3jV(Z(x,a1)) > ±V(i(x,a2)) for 0 < x < a

if 00 < 01 < 02. Thus, we obtain

(I) V(zro(al),al)) - V(z(O,al)) > V(E(ro(a2),a2)) - V(E(O,a2)) > 0

for go<a1<a2.

Observation 4. If a0 < a, there exists a positive number ri(a) such that ri(a) >ro(a), zl (r1(a), a) = a, and zi(t, a) >a for ro(a) <t <7-1 (a) (cf. Figure 8).

(0. a)(a, z2(rp(a), a))

O,a(a. z2(r0(a), a))

z2=H(zt) ( ) z2=H(z0(O.a0) I (O.ao)

zi =0 zi=a

z2=0

Z, =0

=0

FIGURE 7. FIGURE 8.Note that z2(ri(a),a) < 0 and that H(zi(t,a)) > 0 for ro(a) < t < 7-1(a) sincezi (t, a) > a on this t-interval. Regarding zi (t, a) as a function of z2(t, a) forz2(r1(a), a) < z2 < z2(ro(a), a), we obtain

(II) 0 > V(zi(ri(ai),ai))-V(zlro(ai),ai)) > V(z(ri(a2),a2))-V(z(ro(a2),a2))

for ao < al < a2 in a way similar to Observation 3 (cf. Figure 8).

Observation 5. If ao < a, there exists a positive number r(a) such that r(a) >71(a), z1(r(a), a) = 0, and 0 < zi (t, a) < a for r1(a) < t < r(a) (cf. Figure 9).Note that z2(t, a) < H(zi (t, a)) < 0 for r2 (a) < t < r(a). Again, regarding Z2 (t, a)as a function of zi (t, a) in the same way as in Observation 3, we can derive

(III) V(z(r(ai),al))-V(z-'(ri(ai),ai)) > V(i(r(a2),a2))-V4flri(a2),a2)) > 0

if a0 < aI < a2 (cf. Figure 9).

Observation 6. Thus, by adding (I), (II), and (III), we obtain

(B) V(z(r(aI),at)) - V(z(0,a1)) > V(zr(a2),a2)) - V(z0,a2)) > 0

if a0 < al < a2. This implies that the function G(a) defined by

g(a) = V((r(a),a)) - V(z(0,0 = Zz2(r(a),a)2 - 2a2

is strictly decreasing for a > ao as a -+ +oo. Also, 9(a) > 0 for 0 < a 5(A)).

Observation 7. Since

limdV = 0 uniformly for 0 < z1 < a,

jz21-+oo dzi

z1lim00 a = +oo uniformly for - oo < z2 < +oo,

a0 (cf.

it follows that

Therefore,

(C)

lim (V(i(ro(a),a)) - V(z(0,0J = 0,a+oolim (V(z(rl(a),a)) - V(z(ro(a),a))J = -oo,a+oolim JV(z(r(a),a)) - V(i(ri(a),a))J = 0.

a-.+00

lim Cg(a) = -oo.a-'+00

Thus, we conclude that g has exactly one positive zero a+, i.e.,

(X.3.3) 9(a)> 0,

Zz2(r(a), a)2 - 2a2= 0,< 0,

0<a<a+,a = a+,a>a+

FIGURE 9. FIGURE 10.From (X.3.3) and symmetric properties (ii) of the functions h(x) and g(x), we

conclude that a(t, a+) is the only periodic orbit, and all the other orbits tends toz"(t, a+) asymptotically since

Iz2(r(a), a)I > a if 0<0<a+,Iz2(r(a),a)I < Of if a> a+.

Thus, we complete the proof of Theorem X-3-1. 0Remark X-3-2. Condition (iv) of Theorem X-3-1 can be replaced by the followingcondition:(iv') there exists a positive number 6 such that H(x) < 0 for 0 < x < S.

X-4. Multipliers of the periodic orbit of the van der Pol equationIn Example X-2-5, we looked at the van der Pol equation

(X.4.1) LX + e(x2 - 1) d + x = 0,where a is a positive number. Using Observation X-2-4, it was shown that everyorbit of the system

i /(X.4.2) ddt

[y112

] [ -E(y1 -112

1)112 - 111

is bounded for t > 0. It was also remarked that 0 is the only stationary point ofsystem (X.4.2) and that the stationary point 0 is not stable as t - +oo. In fact, the

linear part of the right-hand side of (X.4.2) at y = 0 is Ay, where A = [ 01 and

[i]. Since trace[A]=f and det (A) = 1, the stationary point is an unstableY21-

node for e > 2, while 0 is an unstable spiral point for 0 < e < 2. Therefore, usingthe Poincare-Bendixson Theorem (cf. Theorem IX-4-1), we conclude that thereexists at least one limit cycle. Now, Theorem X-3-1 implies that system (X.4.2)has exactly one limit cycle and all the other orbits except for the stationary point 6approach this limit cycle asymptotically as t - +oo. In fact, since h(x) = e(x2 -1),

3

H(x) = e I3

- x] , and g(x) = x, the seven conditions (i) - (vii) of Theorem X-3-1

are satisfied. In particular, the positive zero of H(x) is a = J > 1.In this section, we prove the following theorem concerning the multipliers of the

unique periodic solution x = x(t, e) of the van der Pol equation (X.4.1).

Theorem X-4-1. The multipliers of the periodic solution x(t, e) of (X.4.1) at 1

and p such that )pI < 1.

Proof.z z

If we! set v =x

211 , wwhere 11 = d , then dt = -e(x2 - 1)y2. This implies

that eJ (x(t)2 - 1)dt = - / Now. Now look at Figure 11.

Y

FIGURE 11.

5. THE VAN DER POL EQUATION FOR A SMALL PARAMETER 319

On each of two curves Cl and C2, let us denote y as a function of v by yl (v) and y2(v)respectively. Note that x2 > 1 on C1, but x2 < 1 on C2. Hence, yl(v)2 < y2(v)2.This implies that

efT

(x(t)2 - 1)dt > 0,

where T is a period. Therefore, we can complete the proof by using the followinglemma.

Lemma X-4-2. Assume that two functions f (x) and g(x) are continuously differ-entiable in R. Assume also that the differential equation

d 22 + f(x) dt + g(x) = 0

r1has a nontrivial periodic solution x(t) of period 1 such that J f (x(t))dt > 0. Then,

0the multipliers of the periodic solution x(t) are 1 and p such that jpI < 1.

Remark X-4-3. If system (X.4.2) has more than one periodic orbit, then at leastone of them must be orbitally unstable. Therefore, the proof of Theorem X-4-1 isanother proof of the uniqueness of periodic orbit of (X.4.2).

X-5. The van der Pol equation for a small e > 0In this section, we explain a method to locate the unique periodic orbit of dif-

ferential equation (X.4.1) (or system (X.4.2)) for a small e > 0.

Set e = 0. Then, system (X.4.2) becomes

(X.5.1) dt[y2J LyYi

Every orbit of (X.5.1) is a circle and of period 2n in t. We expect that the periodicorbit of (X.4.2) must be approximated by one of those circles as e -+ 0. The mainproblem is to find the radius of the circle which approximates the periodic orbit of(X.4.2) for a sufficiently small e > 0.

Since the periodic orbit and its period are functions of e, we normalize theindependent variable t by the change of independent variable

(X.5.2) t = (I +eW)T

so that the period of the periodic orbit becomes 27r for every e > 0. Transformation(X.5.2) changes differential equation (X.4.1) to

x722+ e(1+eW)(x2-1)d + (1+&,d)2x = 0

that can be written in the form

(X.5.3)d 2X

22

+ x = e(1 + ew)(1 - x2) - e(2W + ew2)x.


Observation X-5-1. In the case when a real-valued function f (r) is continuousand periodic of period 27r in r, the general solution of the linear nonhomogeneousdifferential equation

(X.5.4)

is given by

(X.5.5)

d2x

d-T' + x = f(r)

/'r+2nx(r) = K cos(r + m) + a- J sf(s)sin(r - s)ds

r

as it is shown with a straight forward calculation, where K and 0 are arbitraryconstants. This solution is periodic in r of period 2ir if and only if

(X.5.6)10

f (s) cos(s)ds = 0.2w

f (s) sin(s)ds = 0 and J2w

0

Observation X-5-2. In the case when condition (X.5.6) is not satisfied, set

27r

f(s) sin(s)ds and C[J] = 1 J2, f (S)

cos(s)ds.(X.5.7) S[ f ] = 1 foi 7t o

Then,

v(T) = Kcos(r + 0)

(X.5.8) 1

+ 2ajr+21r

s f (s) - S[ f ] sin(s) - C[ f ] cos(s)Jsin(r - s)ds

is the general solution of the differential equation

dr2 + x = f(r) - S[f]sin(r) -

Furthermore, v(r) is periodic of period 27r.

Observation X-5-3. Set

f ( X, ,W, E) _ (I + (w)(I - x2) - (2W + EW2)x.

Letting K be a parameter and a function v(r, K, w, f) be periodic in r of period 27r,set

j

2w1

S(K,w,) _ f (v(sKw,_(s,K,WE)WJf

2w

I 7rC(K,, E) = - f (1T cos(s)ds.

5. THE VAN DER POL EQUATION FOR A SMALL PARAMETER 321

Now, let us consider an integral equation

(X.5.9) fr+2 a

v(r, K, w, E) = K cos(t) +2n

s I f lV, dr, w, EL

- S(K, w, c) sin(s) - C(K, w, f) tos(s)] sin(T - s)ds.

Solutions of (X.5.9) satisfy the differential equation

(X.5.10) dd 2V

-,r2+ V = E [f( V,

Va

, w, E I - S(K, w, c) sin(r) - C(K, w, E) cos(r)

as long as v is periodic in r of period 2ir. This integral equation can be solved byusing successive approximations in such a way that the solution v(r, K, W, E) is aconvergent power series in c:

(X.5.11) v(r,K,w,E) _ EmVm(T,fi,w) = Rcos(T)+E vl(r,K,w)+ ,

m=0

where vm(r, K, w) are polynomials in (K, w) with coefficients periodic in r of period2n. For any given positive numbers Ko and wo, there exists another positive numberEo(Ko,wo) such that series (X.5.11) is uniformly convergent for

I K1 < K0, IwI <- wo, IEI <- Eo(Ko,wo), -00<T<+00.

This implies that S(K, w, c) and C(K, w, E) are also convergent power series in c:

00 0C

(X.5.12) S(K,w,E) = E EmS,,,(K,w) and C(K,w,E) _ F, E"'Cm(K,w),m=0 m=0

where Sm(K,w) and C,,,(K,w) are polynomials in (K,w) with constant coefficients.These two power series also converge uniformly for

I K1 <- Ko, IwI < wo, IEI Eo(Ko,wo)

Observation X-5-4. Inserting series (X.5.11) and (X.5.12) into system (X.5.10),we obtain

d2vi

dr2+ vi = (1 - K2cos2(r))(-Ksin(r)) - 2wKcos(r)

- So(K,w) sin(r) - Co(K,w) cos(r)= K3 cos2(r) sin(r) - K sin(r) - 2wK cos(r)

- So(K,w) sin(1T) - Co(K,w) cos(r)1

= 4 K sin(3r) +4

- K - So(K, w)J sin(r)

- [2wK + Co(K, w)1 cos(r).


Since v1 is periodic in r of period 2,r, we must have

So(K,w) = I K3 - K and Co(K,w) _ -2wK.

This implies that S(2, 0, 0) = 0 , C(2, 0, 0) = 0, and

8K(2,0,0) x(2,0,0)

2 0 --8.200 8C200

_

-10 41

Therefore, the system of equations S(K, w, e) = 0, C(K, w, e) = 0 has a solutionK(e) = 2 + 0(e), w(e) = 0(e). The functions K(E) and w(e) are power series in ewhich converge if Iej is sufficiently small.

Observation X-5-5. Set

fx(t, e) = v (T+ f1w(E)IK(E), w(E), e/

Then, x(t, e) is a periodic solution of (X.5.3) and

x(t, e) = K(e) cos t + 0(e) as a --+ 0.(1+Ew(e)

From the fact that

ty2(t, E) _ (t, f) = -1 K(f) sin C 1 +

ew(e)) + 0(e) as e -+ 0,

we obtain the following conclusion.

Conclusion X-5-6. The unique periodic orbit of (X.4.2) tends to the circle yl +Y22 = 4 ase-.0+.

X-6. The van der Pol equation for a large parameterIn this section, we consider the van der Pol equation (X.4.1) for a large e.

us write (X.4.1) in the form

(X.6.1)(`1_

dt [z21

3

by setting x=z1 and =z2-ef 3 - zl).

Set {z2]21 - I Eu'I

and t = er. Then, system (X.6.1) becomes

(X.6.2)ddr

[$2 Il

-z1

Let

1where a = -

E

6. THE VAN DER POL EQUATION FOR A LARGE PARAMETER 323

Observation X-6-1. Note that, along an orbit of (X.6.2),

(X.6.3)dud

Q2w1dwl 01

3 - wtlof any orbit is small at aTherefore, if 3 > 0 is sufficiently small, the slope dw,

point (wl, w2) far away from the curve C : w2 = 3 - w!. This implies that every

orbit moves toward the curve C almost horizontally (cf. Figure 12).

Observation X-6-2. From Observation X-6-1 a rough picture of orbits of (X.6.2)is obtained (cf. Figure 13).


Actually, defining the curve Co by Figure 14, we prove the following theorem.

Theorem X-6-3. For a sufficiently small positive number J3, the unique periodicorbit of (X.6.2) is located in an open set V(/3) such that the closure of V(f3) containsthe curve Co and shrinks to CD as 0 -+ 0+.

Proof of Theorem X-6-8. In eight steps, we construct an open set V(13) so that(1) the closure of V(O) contains the curve Co,(2) the closure of V(J3) shrinks to the curve Co as O 0+,(3) if an orbit of (X.6.2) enters in the open set V(!3) at r = ro, then the orbit

stays in V(/3) for r > ro.

Step 1. Fixing a number a(3) > 2, we use the line segment

C1(a) = j wi, a(3)3 - a((3) I : 0 < wi <_ a(/3)}

as a part of the boundary 8V(i3) of V(/3) (cf. Figure 15).

Note that the slope of an orbit given by (X.6.3) is negative on C1 (O) and decreasing

to -oo as wt increases to a(f3). Also,dw2

is 0 on C1(f3).

Step 2. Let us consider a curve

CO) = j (wi, ws) : w2 = 3' - wl - Op(wi, 0), 1 < wl < a(fl)} ,

where(i) µ(w1,3) is continuous for 1 < wl < a(0) and continuously differentiable for

1 < wl < a($),(ii) µ(w1, 0) > 0 for 1 < w1 < a(0), and (iii) µ(a(,3), 0) = 0.

On the curve C2(f3),

\

3

31 - wl = -0µ(w1,13)

and, hence,_ /32w1 Owl

W2 _ f wl _ wl l µ(w1, 0)3 /f

On the other hand, the slope of the curve C2($) is

dw2 _ 2 dEc(w1,f)dwl - w1 - 1 - ,8

dwl

This implies that if µ is fixed by the initial-value problem

(X.6.4)dwl 1, µ(a(0),,8) = 0,

we obtain

/32w1

1dw2

C-

w'/dwl

W2 -3

for 1 < wl < a($),

where dw2 denotes the slope of C2(0). The unique solution to problem (X.6.4) isi

given by

a(/3)2 - wl.Thus, we choose the curve

C2(O) _ {(wt, wz) : u,2 =

3- wl - O a(0)2 - wi, 1 < w1 < a(0) }

as a part of the boundary OV(#) of V(f3) (cf. Figure 16). Note that on the curveC2(,8),

w2 = -3 - Q a(f3)2 - 1 at w1 = 1.

6. THE VAN DER POL EQUATION FOR A LARGE PARAMETER 325

Step 3. We choose the curve

C3(/3) = {(wIw2): W2 = -3 - Q a(N)2 - w1, 0 < wl < 11

as a part of the boundary OV(/3) of V(/3) (cf. Figure 17).

FIGURE 16.

On the curve C3(/3),

1w3W2

31

and, hence,

/32w1(,w3

31

a(Q)2 - wl

wl)

wl/

CC1(p)

FIGURE 17.

23 - 31 - wl l - /3 a(/3)2 - wi

0'w1

/ a(Q)2 - w1 + 3+

(31 - wl

On the other hand, the slope of the curve C3(/3) idw2

sdw,1

implies that02W, <dm2 for 0<w1<1,(u, ) dw,

w2-3

-wI

This

denotes the slope of C3(0) (cf. Figure 17). Note that on the curvewhereawl

C3(3), W2 = - 3 - Qa(Q) at wI = 0.

Step 4. Now, let us fix the positive number a(/3) > 2 by the equation

3

(X.6.5)2

+ 3a(3) = a(3) - a(13).

Denote by C+(/3) the curve consisting of Cl(/3), C2(13), and C3()3), i.e., C+(Q) _C1(Q) UC2(/3) UC3($). Let C_(0) be the symmetric image of C+(Q) with respect tothe point (0, 0). Then, C+(3) U C_ (/3) is a closed curve if a(,Q) satisfies condition(X.6.5). We use the closed curve C+(/3) UC_(/3) as a part of the boundary 8V(/3)of V(/3) (cf. Figure 18).

3

Step 5. Fixing a positive number b($) < 2 so that 3 =b(3)

- b(j3) +,A(0), wechoose the curve

r!(,3) = {twiw2) : w2 =b(3)3-

b(j3) +,6b($)2 - w;, 0 < w, < b(13)}

as a part of the boundary 8V(j3) of V($) (cf. Figure 19). Note that, on r,(j3),2w2=3 at w, = 0.

FIGURE 18.

On the curve r, (,B),

/32w1

w2

FIGURE 19.

02w1

Q b(B)2 - wl +b(3) 3

- b(8) - (3'Q wl> -

b(3) - w,

On the other hand, the slope of the curve 171(3) isdullw2

ap)2'- wi

Step 6. We use the two curves

r2(0) = {(wiw2): w2

r3($) = {(wltw2): U2

3 -wi, 1 <w, <b(S)3

ll

-2, 0 < wl <3

JJ

w,)

as parts of the boundary 8V(j3) of V(8) (cf. Figure 20). Note that on r3(Q), the

slope of an orbit given by (X.6.3) is positive and dd is negative.

Step 7. Denote by r+(j3) the curve consisting of r,(8), r2(8), and r3(8), i.e.,

r+w) = r1(Q) u r2(a) u r3(8).

Let r_ (j3) be the symmetric image of r+(j3) with respect to the point (0, 0). Then,r+(p) U r_(8) is a closed curve. We use the closed curve r+(j3) U r_ (f3) as a partof the boundary OV(0) of V(8) (cf. Figure 21).

(iii) the function Y(x) is a solution of the differential equation

F (x, Y, I = 0,

andd'Yx(x)

M for a positive constant M on the interval 0 < x < t.

Let y(x) be any solution of the differential equation

AL2 + f (x,y,LY) = 0 (A>0)

satisfying the conditions y(O) = Y(0) and ly'(0) - Y'(0)l l< p. Then,

ly(x) - Y(x)I < {h, + a(L + k)}exp[ Lx]

on the interval 0 < x < if c and A are sufficiently small.

Proof.

We prove this theorem in four steps.

Step 1. Setting

y=u+Y(x) and -__v+dY),change the equation

to the systern

(X.7.1)

where

ad.z + f (x, y, ) = 0 (A > 0)d-T

du dvdx = v, F(X, u, v, a),

F (x, Y(x), v + d ) I + F(x, u, v,.)l < (e + AM) + Kf ul,

as long as (x, u, v) is in the regi/on l

Do = {(x, u, v) : 0 < x < t, Jul <- a(x), lv) < pe-`= + b(x)}.Also,

F (riY(x)v + d()) > Lv for v > 0,

F x, Y(x). v + d()) < Lv for v < 0,

in Do. Hence, in Do,

F(x, u, v, A) < -Lv + Klul + (e + AM) for v > 0,(X.7.2))

F(x, u, v, A) > -Lv - Klul - (E + AM) for v 0.

8. A SINGULAR PERTURBATION PROBLEM 329

Step 2. Suppose that two functions wl (x, .A) and w2(x, A) satisfy the followingconditions:(X.7.3)

10 < wl(x,.A) < a(x), 0 < w2(x,A) < pe-' + b(x),wl(0,A) > 0, w2(0,.1) > p,

wi (x, A) > W&, A), Aw2(x, A) > -Lw2(x, A) + Kwl (x, A) + (E + AM)

on the interval 0 < x < e. Then, as long as (x, u, v) is in the region

Dl = {(x,u,v): 0 <_ x < e, IuI < wi(x,A), IvI < w2(x,A)},

it holds thatIvI < wi (x, A),.F(x, u, w2(x, A), A) < Aw2(x, )1),

F(x. u, -w2(x, A), .A) > -aw2(x, I\).

Look at the right-hand side of (X.7.1) on the boundary of V. Then, it can beeasily seen that if a solution of (X.7.1) starts from Dl, it will stay in Dl on theinterval 0 < x < e.

Step 3. Show that two functions

Jwi(x,A) = J x A)l + 62(1+1),0

w2(x A) = pe-Lx/A + Alex:/L

ApA e + AM + K62(e + 1)where bl = L2 + L , satisfy the requirements (X.7.3) if f, A,

and a positive constant 62 are sufficiently small. Observe that two roots of AX2 +

LX -K=0 are -L -(o and (o = L +O(A).

Step 4. Note that

wl (x, A) = (L /(1 - e-Lx/A) + bK I (eKx/L - 1) + 82(x + 1).

To complete the proof, look at Jul < wl (x, A) as 62 -4 0. 0

The inequality v< wr a, as 62 0,d

I VI 2( ) yields the following estimate of

dy dY(x)dx dx

Note that

< pe-Lx/A + f E + (h + M1 ] ex'/L.L L L J

lim a-Lx/A = 0 if x > 0.

X-8. A singular perturbation problemIn this section, we look at behavior of solutions of the van der Pol equation

(X.4.1) as a --+ +oo more closely. Setting t = Er and A = c2, let us change (X.4.1)to

d2x (x2 1)dx0.J1dr2 + - dr + x =

Set A = 0. Then, (X.8.1) becomes

(X.8.2) (x2 - 1) d + x = 0.

Solving (X.8.2) with an initial value x(O) = xo < -1, we obtain x = 0(r), where

2 )2 - In I0(r) I = -r + 2 - In(-xo).

Observe that

d0(r)

dr0(r) > 0 if d(r) < -1.0(7)2-1

2

Note also that setting ro = 2 - In(-xo) -- 2 > 0, we obtain ,(ro) = -1 and45(7-)2 - 1 > 0 for 0:5 r < ro. The graph of 0(r) is given in Figure 22.

FIGURE 22.

(i) Behavior for 0 < r < 7-o -bo, where do > 0: Let us denote by x(r, A) the solutionto the initial-value problem

d2z 2 dx40,A)(X.8.3)

a dr2 + (x -1) dr- + x = 0, x(O, A) = xo, = 7-7,

where the prime is d7- and q is a fixed constant. Using Theorem X-7-1 (Na6J, wederive

IIx(r, A) - O(T)I 5 AK,Ix'(r, A) - 4'(r)l 5 I7-7 - 0'(0)le-P'1x + AK,

for 0 < T < To - 60, where K and a are suitable positive constants.

(11) Behavior for lx+1l< o: First note that lim p0'(T) _ +oo. Set ¢'(ro-2bo) _

1 > 0. Then, there exists T1(A) for sufficiently small A > 0 such thatPi

llm T1(A) _ 7.0 - 280,O<ri(A)<ro_So,{X

x(T1(A),A) _ O(Tp - 26p), x'(T,

Set p =1A)

Then,

(X.8.4) ALP = p2(x2 - 1) + p3x,

regarding p = p(x, A) as a function of x and A. Note that

0 < p(x, A) < 2p' for ¢(ro -26o) < x < x(ro-b0, A) < -1, 0 < A < AO,

where A0 is a sufficiently small positive number. It is important to notice that ifwe make 60 small and if we make A0 also small accordingly, we can make p1 small.

Set

(X.8.5) o = -1 - ¢(T0 - 26o) and M0 = max Ix2 - 11.I1+xI<a

Then,

(X.8.6) o > 0, lien o = 0, and lim Mp = 0.0-+0+

Furthermore, 0 < p1 < Mo since 1 - ((TO - 25p)2P1 =0(TO - 260)

Therefore,

0 < p(z, A) < 2Mo for ¢(r0-25o) < x < x(ro - 50, A), 0 < A < A0.

Now, we shall show that

(X.8.7) 0 < p(x, A) < 2Mo for Ii + xt < o, 0 < A < A0.

If (X.8.7) is not true, there must exist a such that

p({, A) = 2M0, 0 < p(x, A) < 2Mo for -1-a:5 x<

Then, this implies that

5 A)2(Mo + A)2(1 + g) < 0.

A) >2P1

for r1(A) < T < TO - 60.

This is impossible.

(iii) Behavior for -1 + or < x < 2 - al: To begin with, it should be remarked that

(X.8.8)

J(2_1)d

1)d. =

Fix a positive number al so that

[..-_]2 i=

8-2- ( -3+1J

C3-x)-3<0,

(X.8.9) j(2 - 1)4 < - y

= 0,

for -1<x<2.

for -1+a< x < 2-al,

where y is a sufficiently small positive constant. Note that ry --+ 0 as do 0+ andal -0+-

Integrating (X.8.4), it follows that

p(z ) = P( I,A) - J(C2 - 1) - f 1 agd

(I)P( x<0. Hence, if we choose \0>0x,,\-so that

A 7

< 4Kyfor 0 < A < A0, where

2

K = f0

w e obtain 0 < p(x, A) < 9K for -1+a <x<0, 0<A <A0.

(II)a

p(x A) > y -K max A) for 0 < x < 2 - al. Suppose that

0 <2K

for 0 < < x, 0 < A < A0.

Then, 0 < p(x, A) <2

<2KL. Thus, we proved that

(X.8.10) 0 < p(x,A) < 2K for -1+a < x < 2 - al, 0 < A < A0.

(iv) Behavior for 2 - al < x < 2 + a2: First look at

p(2 - at, A) 2&0), A) -1) + p(, )ld

Notice that p(2 - al, A) is independent of 5o. Then, it can be shown that

A0+ as a, -+ 0+ and A 0+.

p(2 - a,, A)

For a fixed positive number a2, assume that

0 < p(x, A) < 1 for 2 - al < x < 2 + a2i 0 < A < Ao.

Then,

2+02 g2 - 1) + P( A)fl > Q.P(2 + 2, A) p(2 -- o , A) 12-,,

Hence,

J2+0a g2 - 1) + < A for 2 - al <x < 2 + a2.-o, p(2 - al, A)

This is a contradiction if al and A are small. Therefore, if Ao > 0 is sufficientlysmall, there exists an x(A) such that

2 - al < x(A) < 2 + a2 and p(x(A),A) = 1 for 0 < A < Ao.

It can be shown that Urn x(A) = 2.ao*Setting r(x(A)) = r(A), it can be shown that liM +r(A) = ro. Now again, apply

Theorem X-7-1 (Na6) to the initial-value problem

AT2 + (x2 - 1)d + x = 0, x(r(A)) = x(A), x'(r(A)) = 1.

Figure 23 shows the general behavior of x(r, A) for small positive A.

TO

FIGURE 23.

EXERCISES X

X-1. (1) Show that if F(t,yl,y2) is continuous and bounded on

Il = {(t, Y1, Y2)

the boundary-value problem

a < t < b, [yl[ < +oo, 1y2[ < +oo},

dt2 = F (t, y, &) !LY (a) = a, y(b)

has a solution.(2) Does the boundary-value problem

dtY =dt2

F() dY (b) _ 3t, y,dtdY

dtdy(a) = a' dt

have a solution?(3) Show that the boundary-value problem

d2x

dt2= tx + x3, x(-2) = A, x(3) = B

has a solution for any real numbers A and B.

Hint. A counterexample for (2) is

d2 = 0, (n) = 0, (b) = 1.

For (3), apply Theorem X-1-3 [Na4j with w,(t) = a and w2(t) where -aand 3 are sufficiently large positive constants.

X-2. Find the global phase portrait of each of the following two differential equa-tions:

(i)d2x

2+ x2(1 - x)2 dt +

d2x dxdx2

+dt

-

(x - 1)2(x + 1)x = 0;

1

l + x2= 0.

X-3. Suppose that(i) f (x, y, t) is continuously differentiable everywhere in the (x, y, t)-space (i.e.,

in p3),(ii) g(x) is continuously differentiable in -oo < x < +oo,

(iii) e(t) is continuous on 0 < t < +oo,(iv) f (x, y, t)

r>0 for x2 + y2 > r2 and t > 0, where r is a positive number r,

(v) G(x) = J g(t)dC --' +oo as jx[ +oo,0

(vi) 1 je(r)Idr is bounded for 0 < t < +oo.`0

Show that every solution (x(t), y(t)) of the systemdx _ dy =

' df-f(x,y,t)Y - g(x) + e(t)

Ydt -is well defined and bounded for 0 < t < +oo.

EXERCISES X 335

Hint. Set V (x, y) = 2 + G(x). Then, d = - f (x, y, t)yz + e(t)y. There exists a

positive number ro > r such that V(x, y) > 4 for x2 + y2 > ro. Hence, if an orbit

(x(t), y(t)) satisfies conditions that x(t)2 + y(t)2 > ro for to < t < ti, we obtain

d V(x(t),y(t)) < Ie(t)Iiy(t)I <2(e(t)j V(x(t),y(t)) for to <t <t,.

This yields

V(x(t),y(t)) < V(x(to),y(to)) + ft., (e(r)!dr for to <t < t,.

X-4. Show that the differential equation

E.1d yi _ ill

( ) dt [y2 ] [ -E(yi - 1)y2 - i/i - Eyihas exactly one periodic orbit, where a is a positive parameter-

X-5. Find an approximation for the unique periodic orbit of (E.1) for small e > 0and for large e.

X-6. Considering the system of two differential equations

yi y2(E.2)dtd

[ y2] - [ -e(yi - 1)y2 + yi - byl

where a and 6 are positive numbers,(a) show that every orbit is bounded for t > 0,(b) determine whether each stationary point is stable or unstable, examining every

possibility,(c) using the function

!-y2 + E(-yi

y,+ 6YNY2 + J (-s + bs3J[1 + E2(si)]ds

0

and Theorem IX-2-2, show that if 0 < 6< 3, every orbit of system (E.2) tends toone of the stationary points as t -' +oo,

(d) discuss the uniqueness of periodic orbits.

X-7. Show that the differential equationd2xd z

+ f (x) d+ 9(x) = 0has at most one nontrivial periodic solution if fix) and g(x) are continuously dif-ferentiable in !R and satisfy the following conditions

< 0 for 1x3 < 1,(i)

f (x) { > 0 for IzI > 1,

<0 for -2<x<0,(ii) 9(x)

> 0 for (x + 2)x > 0,

(iii) J g(x)dx = 0.

Hint. See [Sat].The main ideas are as follows:

(1) If there are more than one nontrivial periodic orbits, then at least one of themshould not be orbitally asymptotically stable.

(2) For a nontrivial periodic orbit (x(t), x'(t)) of period T > 0, the inequalityfT

of (x(t))dt > 0 implies that this orbit is orbitally asymptotically stable

(cf. Exercise IX-5).(3) Set

(

x(t)

a(t) = V(x(t), x'(t)) x'(t))2= +2

fThen,

da(t) _ -f(x(t))(x'(t))2

Therefore, we obtain

I

.

dt

)dt = - r dAJ t( ) f(x( ) 2' )J (x

Now, investigate the quantity (x'(t))2 as a function of A along the periodic orbit(x(t), x'(t)). For a fixed value of A, compare (x'(t))2 for different values of x(t),using the assumptions on f and g.

Step 1. First the following remarks are very important.(a) If we set y = x', the given differential equation is reduced to the system

(S) dt = y, dt = -f (x)y -A careful observation shows that the index of the critical point (0, 0) is 1,while the index of the critical point (-2,0) is -1. There are only two criticalpoints of (S).

(c) The periodic orbit and any line {x = a constant) intersect each other at mosttwice.

(d) The critical point (-2,0) should not be contained in the domain bounded bythe periodic orbit. To show this, use the fact that the index of any periodicorbit is 1.

These facts imply that the periodic orbit should be confined in the half-plane x >-2.

2 r:Step 2. A level curve of V (x, y) = 2 + J is a closed curve if the curve is

0totally confined in the strip jxj < 1. In particular, (1, 0) and (-1, 0) are on the same

level curve V (x, y) = / 1

I_ 1 g(C)d{. Since A(t) is increasing as long as

0 o

jx(t)I < 1, the periodic orbit cannot intersect the level curve V(x,y) = / ,0

Therefore, the periodic orbit must intersect the lines x = 1 (as well as the linex = -1) twice. Denote these four points by (1,s1(A)), (1,-rt(B)), (-1,-q(C)),and (-1, 77(D)), where r)(A), n(B), q(C), and q(D) are positive numbers.

EXERCISES X 337

Step 3. Set

AA = V(1, rl(A)), AB = V (1, -n(B)),

Ac = V(-1, -n(C)), AD = V (-1, rl(D))

Then,AA > AB, AC > AB, AC > AD, AA > AD-

SetZ(A) = { A : AA > A > max(AB, AD) },Z(B) = { A : AB < A < min(AA, )'C) },Z(C) = { A : Ac > A > max(AB, AD) },Z(D) = { A: AD < A < min(AA, Ac) }.

Note that the interval lo = JA: min(AA, Ac) > A > max(AB, AD)} is contained inZ(A) n Z(B) n Z(C) n I(D).

Now,(a) on the arc between (1, rl(A)) and (1, -rl(B)) of the periodic orbit, regard y as

a function of A and denote it by yi (A), where AB < A 5 AA,(Q) on the arc between (1, -rl(B)) and (-1, -r1(C)) of the periodic orbit, regard

y as a function of A and denote it by y2 (A), where AB < A < AC,(ry) on the are between (-1, -rl(C)) and (-1, rl(D)) of the periodic orbit, regard

y as a function of A and denote it by yi (A), where AD < A < AC,(6) on the arc between (-1, rl(D)) and (1, rl(A)) of the periodic orbit, regard y as

a function of A and denote it by yz (A), where AD :5 A < AA.Then, it can be shown that

yi (A)2

yi (A)2

yi (A)2

yi (A)2

< y2 (A)2 on I(A),< y2 (A)2 on Z(B),

< y2 (A)2 on I(C),< y2 (A)2 on Z(D).

Step 4. Now, fixing a AO E Z(A) n Z(B) n 1(C) n T(D), evaluate the integral (I)as follows:

J = -{JA8rT

f(s(t))dtdA rc dA rAD dA

IA

AA dA 1

w yi (A)2 + J y2 (A)2 +'Ac yl (A)2 + a y2 (A)2 T ,

where

J\A

dA dA r" dA

yi (A)2 as yi (A)2 lao yi (A)2,

rac dA _ ra° dA AC dA

Joe y2 (A)2 = Jaa y2 (A)2+

Jiby2(A)2,

I" dA ra° dA rAc dA

yi (A)2 = - JaD yi (A)2 - ao yi(A)2,

ao ys (A)2 aD y2 (A) 2 Jao y2(A)2.

Thus, we conclude that

JT f (x(t))dt > 0.

This implies that every periodic orbit is orbitally asymptotically stable. Hence,there exists at most one periodic orbit.

X-8. For a nonzero real number c and a real-valued, continuous, and periodicfunction f (t) of period T which is defined on -oo < t < +00, find a uniquesolution 0(t, E) of the differential equation

Edt = -y + f (t)

which is periodic of period T. Also, find the uniform limit of 0(t, c) on the interval-oo<t<+ooasE-+0.X-9. Assuming that X(t) satisfies the condition

(X(t)2 - X(t) = 0 and IX(t)I > 1

on an interval 0 < t < t, where a is a positive constant, and that x(t, A) is theunique solution to the initial-value problem

22 + (x2 -1) dt + x = 0, x(O, A) = X (O), x'(O, A) = X'(O),

find lim d2x(2' as a function of X (t) for 0 < t < Q.a-.o+ dt

X-10. Assuming that y(t, E) is the solution to the initial-value problem

d2Yt2 + 2ydy = 0, y(0) = a, y'(0) = Qdt2 dt

where e is a positive parameter, or is a real number, and ,8 is a positive number,find lim y(t, e) for any fixed t > 0.

f 0+

Hint. Look at E dt = c - y2. Then, c = Q + a2 > a2. Hence, y2 increases to c

very quickly. Then, use Theorem X-7-1 [Na6], or find the solution explicitly (cf.Exercise I-1).

X-11. Find, if any, solution(s) y(t, c) of the boundary-value problem

4 + 2ydty = 0, y(0) = A, y(1) = B,dt2

in the following six cases: (1) 0 IAA, (4) B = -A > 0,(5) CBI < -A, and (6) B < 0 < A, assuming that e is a positive parameter. Also,find lim y(t, e) for 0 < t < 1 in each of the six cases.

e 0+

EXERCISES X 339

Hint. Use explicit solutions together with the Nagumo Theorems (Theorems X-1-3 and X-7-1) on boundary-value and singular perturbation problems. See alsoExercises X-10 and X-12, and [How].

X-12. Let f (x, y, t, e) be a real-valued funcion of four real variables (x, y, t, e).

Assume that(i) 0(t) is a real-valued, continuous, twice-continuously differentiable function

on the interval Zo = {t : 0 < t < 1) and satisfies the conditions 0 =f (¢(t), 4'(t), t, 0) and 4(1) = B on Zo, where B is a given real number,

(ii) the function f (x, y, t, e) and its partial derivatives with respect to (x, y) arecontinuous in (x, y, t, e) on a region R = {(x, y, t, e) : Ix - ¢(t)I < rl, IyI <+00, t E Zo, 0 < E < r2}, where r1 and r2 are positive nmbers,

(iii) If (4(t), 4'(t), t, e) I < Ke for t E Zo, where K is a positive number,

(iv) there exists a positive number µ such that Of (x, y, t, e) < -p on R,

(v) there is a positive-valued and continuous function '(s) defined on the interval

0 < s < +oo such thatJ+00

+oo and that If (x, y, t, e)I < +'(IyI) on

R,(vi) A is a given real number.

Then, there exists a positive number co such that for each positive f not greaterthan co, there exists a solution x(t, E) of the boundary-value problem

d2xf d#2 = f (X, d , t, E) , x(0, e) = A, x(1, c) = B

such that

Ix(t,E) - di(t)I < IA - h(0)led=te-ft/` +C2e on Zo,

x'(t, e) - 4'(t) C3e + co for 0 < c5e <_ t < 1,e

where c1 i c2, c3, c4, and cs are positive numbers.

Hint. Cf. Theorem X-1-3. See also [How].

X-13. Let a(t), b(t), and f (t) be real-valued, continuous, and continuously differ-entiable functions of t which are defined on the interval 0 < t < 1. Also, let 4o(t)

be a real-valued solution of the differential equation a(t) dt + b(t)z = f (t).For a positive number e, denote by y(t, e) the unique solution of the initial-valueproblem

+ b(t)y = f (t),d3y

3

+ d-t2 + a(t)dy

y(0) ='7o(E), y'(0) = 171 W, y"(0) = 12(E)-

Show that ly(t, e) -00(t)I+ Iy'(t, e) -40 (t)I +Iy"(t, e) -40 (t) I tends to zero uniformlyon the interval 0:5 t < 1 if 4o(0)I + Iti1(E) - 40(0)I + Irn(E) - 0011(0)1 --+ 0 ase-0+.

X-14. Let Y E lR", y E Rm, t E R, and e E R. Also, let ft f, y, t, e) and g(x", y, t, e)be respectively R'-valued and 1R'"-valued functions of (x, y, t, e). Assume that

(i) an R"-valued function fi(t) and an R'-valued function t%i(t) are continuousand continuously differentiable on an interval Zo = {t : a < t < b} and satisfythe system of equations

dx"

dt= f (Y, i, t, 0), 0 = g(x, y, t, 0) on Zo,

(ii) f (Y, y, t, e), g(x', y, t, e), and their partial derivatives with respect to (x, y-) arecontinuous on a region R = {(:e, y, t, e) : I x - m(t) I < rl, Iy - j(t) I < r2, t Elo, 0 < e < r3 }, where rl, r2, and r3 are positive numbers,

(iii) all eigenvalues of the matrix (1(t), ti(t), t, 0) are less than a negative num-

ber M on .70.Show that the initial-value problem

d:5YT = f (: , y, t, E), a dt = §(Y, 9, t, f), x(a) = ((e), y(a) = #(f)

has a unique solution (2(t, e), y(t, f)) for t E Zo and 0 < e < r4 if e + IC(e) -

(a)I + Ir(e) - t'(a)I is sufficiently small on 0 < e < r4i where r4 is a positivenumber. Also, show that Ii(t, e) - ¢(t)I + I y(t, e) - tj,(t)I - 0 uniformly on Zo ase + IC(E) - (a)I + I#(E) - (a)I - 0.

Hint. See [LeL].

X-15. Let x E lR", y E R"', t E R, and e E R. Also, let t, e) and g(i, y, t, e)be respectively R"-valued and R'-valued functions of (i, y", t, e) which are periodicin t of period T. Assume that

(i) an R"-valued function fi(t) and an R"'-valued function t%'(t) are continuous,continuously differentiable, and periodic of period T on the interval Zo = it :

-00 < t < +oo} and satisfy the system of equations d = f (:F, y", t, 0), 0 =

g"(x, y, t, 0) on Za,

(ii) f(i, y", t, e), g(i, y", t, e), and their partial derivatives with respect to (x, y') arecontinuous on a region R = {(i, y, t, e) : Ix - m(t)I <- ri, I y - 7G(t)I < r2, t EZo, IEI < r3}, where rl, r2, and r3 are positive numbers,

(iii) there exists a real m x m matrix P(t) such that(iiia) the entries of P(t) and P(t)-1 are real-valued, continuous, continuouslydifferentiable, and periodic of period T[Baton Zo,

(iiib) P(t)8b(m(t),1(t),t,0)P(t) = B2(t)], where Bi(t) is a real

m1 x ml matrix with eigenvaules havingnegative real parts on 10i while B2(t)is a real (m - ml) x (m - ml) matrix with eigenvalues having positive realparts on Zo,

(iv) the system

EXERCISES X 341

d,F= 8iq(t), (t),t,0).iF

- Of (fi(t), G(t), t, 0) [ w(t), +G(t), t,0)J

ON(* it), t, 0) z

does not have nontrivial periodic solutions of period T.Show that the system

di dy"

dt - f(s,y,t,f), adt =9'(2,y,t,e)

has a unique periodic solution (a(t, e), y(t, e)) of period T on To if 0 < Iej is suffi-ciently small. Also, show that jT(t, e) - (t)I + lc(t, e) - iJJ(t)j 0 uniformly on To

Hint. See [FL]

CHAPTER XI

ASYMPTOTIC EXPANSIONS

In §§V-1 and V-2, we defined formal solutions of a system of analytic dif-ferential equations. Formal solutions are not necessarily convergent. For ex-ample, as we mentioned it in Remark V-1-4, the divergent formal power series

00

f = (-1)m (m! )x'"+1 is a formal solution of x2dy

+ y - x = 0. This equationm=0 r_

has an actual solution f (x) = el/=J t-1e-hIldt for x > 0. Integrating by parts, we0

obtain f = (-1)m(m!)xm+1 + (- 1)N+1((N + 1)!)e1/x tNe-l'tdt. Since 0 <N

foxM=0sel/=10,7 tNe-l/tdt = xN+2 - (N + 2)el/= r tN+1e-l/tdt < xN+2, we conclude that

0N

f (X) - (_l)m(m!)xm+l < ((N + 1)!)xN+2 for x > 0. This is an example ofM=0

an asymptotic representation of an actual solution by means of a formal solution.In this chapter, we explain the asymptotic expansions of functions in the sense ofPoincare and in the sense of the Gevrey asymptotics. In the Poincare asymptotics,

flat functions are characterized by the condition lim E f) = 0 for all positive inte-xm

gers m, whereas in the Gevrey asymptotic, flat functions are characterized by thecondition If (x) I exp(cIxI-k) < M as x -i 0, where c, k, and M are some positivenumbers. Generally speaking, the Poincare asymptotics is too general for the studyof ordinary differential equations. A motivation of the Gevrey asymptotics is alsogiven by the Maillet Theorem (cf. Theorem V-1-5). In §XI-1, we summarize thebasic properties of asymptotic expansions of functions in the sense of Poincare. TheGevrey asymptotics is explained in §§XI-2-XI-5.

For more information concerning the Poincare asymptotics, see, for example,[Wasl]. The Gevrey asymptotics was originally introduced in [Wat] and furtherdeveloped in [Nell. To understand the materials concerning the Gevrey asymptoticsof this chapter, [Ram 1], (Ram 21, [Ram3], [Si17, Appendices], [Si18], and (Si19] arehelpful.

XI-1. Asymptotic expansions in the sense of Poincare

In this section, we explain the asymptotic expansions in the sense of Poincare.Let x = a be a point on the extended complex x-plane. Consider a formal powerseries

(XI.1.1)00

P(x) = E c,(x - a)m.M=0

342

1. ASYMPTOTIC EXPANSIONS IN THE SENSE OF POINCAR$ 343

Let D be a sector in the x-plane with vertex at x = a and Do be a neighborhoodof x = a in D. Assume that f (x) is defined and continuous in Do.

Definition XI-1-1. The formal series (XI.1.1) is said to bean asymptotic (series)expansion of f (x) as x -k a in V if for every non-negative integer N, there existsa constant KN such that

(XI.1.2)N

f(x)->2cm(x-a)m < KN Lx - aIN+r, N = 0, 1, 2,- ..M-0

for all x in Do.

Such an asymptotic relation is denoted by

(XI.1.3) f(x) p(x) as x a in D.

This definition of an asymptotic expansion of a function was originally given by H.Poincare [Poi2J.

Before we explain some basic properties of asymptotic expansions, it is worth-while to make the following remarks.Remark XI-1-2. The vertex x = a can be x = oo. In that case, the asymptotic

on

series is in the form C nx-"'.m=0

Remark XI-1-3. Assume that f (x, t) is a function defined and continuous in (x, t)for x in D and t a parameter in a domain H in the t-plane. A formal power series

N

F c,,,(t)(x-a)'", where c,(t) is a function oft, is said to bean asymptotic (series)M=0expansion of f (x, t) as x 0 in V if for every non-negative integer N, there existsa function KN(t), independent of x, such that

N

f(x, t) - Cm(t)(x - a)m

M=0

<K1v(t)Ix-a1N+1, N=0,1,2,...

for all x in Do. If, moreover, KN(t) are independent of t, then the asymptoticexpansion is said to be uniform with respect to t.Remark XI-1-4. If f (x) is holomorphic (i.e., analytic and single-valued) in aneighborhood of x = a, by virtue of the Taylor's Remainder Theorem, f (x) admitsits Taylor's series expansion as its asymptotic expansion in any sector with vertexat x = a.Theorem XI-1-5. For a continuous function f (x), there is at most one asymptoticexpansion as x -+ a in a sector with vertex at x = a.

ProofAssume that there are two asymptotic expansions of f (x) at x = a

00

(XI.1.4) f (x) ^_- c,,(x - a)14 as x -a in DM=0

344 XI. ASYMPTOTIC EXPANSIONS

and

f(x)^, 7m(x-a)' as x -i a in D.M=0

Then, for every non-negative integer N, there exist two constants KN and LN suchthat

N

f (x) - > cm(x - a)mm=0

N

.f (x) - ,1.(x - a)mm=0

< KN Ix - alN+i , N = 0, 1, 2, .. .

<LNIx-aIN+1 N=0,1,2,...

for all x in a neighborhood Do of x = a in D. For N = 0, we have Ieo - 7ol <(Ko + Lo)lx - al for x in Do. Let x a in D; then, we obtain co = 70

Now, assume that Ck = 7k for k = 0, 1,... , N - 1. Then, from (XI.1.5) and(XI.1.6), it follows that ICN -7NIIx-aIN < (KN+LN)Is-alN+l for x in Do. Letx -+ a in D. Then, cN = 7N. Thus, cm = 7m is true for every non-negative integerm. OTheorem XI-1-6. Assume that f (x) and g(x) are two continuous functions suchthat(XI.1.7)

00

f (x) > cm(x - a)' and g(x) ^-' E 7m(x - a)"' as x --+ a in D.M=0 m=0

f (x) ± g(x) (cm ± 7.)(x - a)m as x -+ a in D00M=0

Proof.

f(x)g(x) 1 Ch7k) (x - a)m

m=0 h+k=mas x -+a in D.

N

Fix a non-negative integer N and put f (x) _ > cm(x - a)'+Ei(x)(x-a)N+iM=0

N

and g(x) _ > 'ym(x - a)m + E2(x)(x - a)N+1 Then, there exists two constantsm=0

KN and LN such that

(XI.1.10) IE1(x)I <_ KN, IE2(x)I <_ LN

for x in Do. Therefore,

(XI.1.11)N

If (x) ± 9(x)] - E I(Cm ± 1'.](x - a)m < (KN + LN)I x - aIN+1M=0

for x in Do. Thus, (XI.1.8) holds. Also, it is easy to see that there exists a positiveconstant k such that

f(x)9(x) - ( Ch7k) (x - a)m

m=0 h+k=m< KIx - aIN+1

for x in Do. Thus, (XI.1.9) holds.

Theorem XI-1-7. Suppose that f (x) is a function holomorphic in 0 < Ix - al < 8and admits an asymptotic expansion (XL1.4) with V = (0 < Ix - at < b}. Then,the asymptotic series actually converges to f (x) in 0 < Ix - at < S.

Proof.

Since lim f (x) = co, x = a is a removable singularity. Thus, f (x) is holomorphicrain Ix - al < b. Therefore, the asymptotic series agrees with the Taylor's series (cf.Remark XI-1-4 and Theorem XI-1-5). Thus, the asymptotic series converges inIx - aI < 8.

For a vector z E cm with entries (z1,z2,... zm) and p = (p1,p2,... Pm) withnon-negative integers p, (j = 1, 2,... , m), we define I6oI = P1 + p2 + + p,,,z' = zr z2 ... zm , and Iz7 = max{Iz1I, IZ21, ... , Ixml}.

Theorem XI-1-8. Suppose that F(x, z) is a function with power series expansion00

F(x, ) = F.(x)zY, which converges uniformly forx E D, I zl < bo, where Fp(x)Ip1=o

00

is continuous in D and admits an asymptotic expansion Fp(x) c 1: Fpk(x - a)kk=0

as x -, a E D. Define the formal power series in (x,

00

(XI.1.12) 4i(x, 2) = E 4ip(x)iV,lpI=0

where

0(XI.1.13) 4ip(x) _ E Fpk(x - a)k.

k=0

If f (x) is a continuous Cm-valued function with entrywise asymptotic expansions

00

f (x) ^-- > f (x - a)' __ p"(x) as x -' a in D,1=0

then 45(x, pjx) - fo) defines a formal power series of (x - a) and

(XI.1.14) F(x, f (x) - fo) = 4'(x, p(x) - fo) as x -+ a in D.

Proof.

Choose p so small that I f (x) - fol < 6o for Ix - al < p. Fix a positive integer Nand put

N 00

F(x,f(x) - fo) = E F,(x)(f(x) - fo)p + > Fp(x)(f(x) - fo)p.

Ip1=0 Ipl=N+1

Then, by virtue of Theorem X1-1-6, we obtain

N N/.-y

F,,(x)(f(x) - f0)p o)'.

Ip1=0 1p1=0

00

Set

Then,

(XI.1.17)

and, hence,

(XI.1.18)

E Fp(x)(f(x) - fo)p = O(Ix - alN+1).

Ipi=N+1

N(x) =,D(x,Plx) - fo) = >2 Hm(x - a).M=0

N N

op(x)Q3(x) - f0)p = Hm(x - a)m +O((x - a)N+1)1p1=0 m=0

N N

E Fp(x)(f(x) - fo)p = E Hm(x - a)m +O(Ix - aIN+1).Ip1=0 m=o

Since this is true for all positive integers N, the theorem follows immediately. 0

Theorem XI-1-9. Suppose that f (x) is a continuous function with asymptotic ex-pansion (XI.1.4) and co 34 0. Let p(x) denote the formal power series(X7.1.1). Then, Pox) defines a formal power series in (x - a) and f(x) ox) asx -+a in D.

Proof.

Consider F(z) = I . Then, the theorem follows from Theorem X1-1-8. 0co+z

Theorem XI-1-10. Suppose that f (x) is a continuous function with asymptoticexpansion (XL 1.4) as x -' a in D. Let p(x) denote the formal power series (XI. 1.1).

Ten,1

Jf(t)dt m+1cm(x-a)m+1 as x --* a in D,

where the path of integration is the line segment joining a and x, i.e.,

f aZ f(t)dt f =P(t)dta a

Proof

as x -a in D.

N

Set pN(x) _ E c, (x - a)'. From (XI.1.4) and (XI.1.5), it follows thatm-0

If (t) -PN(t)Idtl 5 J x_al If(C) PN(t)l ds0

Js-al Ix - al N+2KN J sN+1 = KN

N + 20

for every non-negative integer N, where s = It - al. O

Theorem XI-1-11. Suppose that f (x) is a holomorphic function with asymptoticexpansion (X1.1.4). Let p(x) denote the formal power series (XI.1.1). Let D be a

proper subsector of D with vertex x = a. Then, I ax as x -- a in D, where

00 .dp = E mr,,,(x - a)m-1

m=1

Proof.

For a non-negative integer N, put f (x) = pN(x) + EN(x). Then, there exists apositive constant KN such that

(XI.1.19) EN(x)I 5 KNIx - alN+'

for x in a neighborhood of x = a in D (cf. Definition XI-1-1). Let the radialboundaries of D be arg(x - a) = 6t and those of D be arg(x - a) = 0± (9_ <0_ < 0+ < 9+). Let 0 = min{6+ - 0+, 0_ - 9_ }. Let x be a point of D. Consider

a circle I' _ {t1 It - xI = 2Ix - al sin 9 }. Then, r and its interior are contained

in D. Using Cauchy's integral formula, we obtain

df (x) 1 r f 1 r PN (t) EN (C)I=dx 2rri Jr (t - x)2

__

2rri Jr (t - x)2I

+ 2rri r (t - x)2dt.

ei4' ( e'4'2Ix-al sin 0.Then, r_(rll>I= 2Ix-aIsin0,0<0<27r

and (XI.1.19) impliesl 11

1 1 EN () I fEN()dtai r (x)2 172 rl

_ 1 12x

EN(e)e"'dcb = 0 (Ix - aIN) .7

Thus, the theorem is proved.

Theorem XI-1-12. Suppose that {fn(x)In = 1,2,3,...) is a sequence of contin-uous functions such that they admit unifonn asymptotic expansions

00

(XI.1.20) fn(x) ^-- > a)' (n = 1, 2, 3, ...) as x --+ a in D.M=O

Assume that {f(x)In = 1,2,3.... } converges uniformly to a function f (x) in asubsector Do of D with vertex x = a. Then,

lira Cnm = Cm

00

f (x) ^- E cm(x - a)' as x -+ a in Do.M=0

ProofThe assumption implies that for each non-negative integer N, there exists a

positive constant KN, independent of n, such that

N

(XI.1.23)I

a ) ' 1 : 5(n = 1,2,3,... )

M=0

for x in a neighborhood of a in D. Furthermore, for each pair of positive integers(j, k), there exists a positive constant bjk such that

(XI.1.24)

for x in Do, where

(XI.1.25)

Put

If,(x) - fk(x)I S bjk

bjk-+0 as j, k -+ oo.

N

(XI.1.26) pjN(x) =E cjm(x - a)n' (j = 1, 2, 3, ... ).M=0

1. ASYMPTOTIC EXPANSIONS IN THE SENSE OF POINCARE 349

Then,

(XI.1.27) Ifj(x) -P)N(x)I 5 KNIx-alN+1 (j = 1,2,3,...).

Thus,

(XI.1.28) IP) N(x)-PkN(x)I 5 5)k+2KNIx-alN+l (j,k=1,2,3,...).

In particular,

(XI.1.29) Ic)o - ckol < b)k+2Kolx-al (j,k= 1,2,3,...)

for N = 0 and x in Do. Therefore, (XI.1.25) and (XI.1.29) imply that lim c)o = CO

exists.

Assume that lim c),,, = c,,, exist for m < N. Then, from (XI.1.28), it follows1 00

thatN-1

E (C),,, - Ckm)(x - a)m + (CjN - CkN)(x - a)NM=0

< S)k + 2KNIx - alN+1 (j,k = 1,2,3,...)

for x in Do. Thus,

IC)N - CkNI Ix - aIN < e)k + 61k + 2KNIx - alN+1 k=1 ,2 , 3 . . . .

where e)k 0 as j, k -. oo. Hence,

Ic)N -CkNl :5-11

IzaIN +2KNIx-al (j,k= 1,2,3,...).

Therefore, lim c)N = cN. Consequently, lim c),, = c,,, for all m.) --00 l-00

Furthermore, since (XI.1.27) holds independently of j, we obtain

N

f(x) - > cm(x - a)mM=0

for x in Do. Thus, (XI.1.22) holds.

< KNlx - alN+1

The following basic theorem is due to E. Borel and J. F. Ritt.

Theorem XI-1-13 ((Bor( and [Ril). For a given formal power series in (x - a)

(XI.1.30) P(x) = > c,,,(x - a)mm=0

and a sector with vertex x = a

(XI.1.31) D={xIO<Ix -al <ry, (31<arg(x-a)<p2},

them exists a function f (x) which is continuous in D and holomorphic in the inte-rior of D, and

00

(XI.1.32) f (x) ^_- E c,,,(x - a)' as x - a in D.M=0

Proof.

Without loss of generality, assume that the sector D contains the ray arg(x -a) = 0. Construct functions a,,, (x) for m = 1, 2.... such that f (x) = co +

00

E ca,,,(x)(x - a)m is continuous in D, holomorphic in the interior of D, andm=1satisfies (XI.1.32). Let bm and 0 be positive numbers, where bm are to be specifiedlater, whereas 0 is chosen to satisfy 0 < 0 < 1 and

(XI.1.33) 01 arg(x - a)I < 2 for x E D.

Consider

r 1

(XI.1.34) am(x) = 1 - exp l - 2m1,m(x - a)0 J , m = 1, 2, ... .

Then, am(x) are holomorphic in a `sector containing D. Also, from (XI.1.33) andthe fact that

it follows that

(XI 1 35)

I1-esl=l jetdt

bm<

<Izi for z<0,

1 2. . lam(x)I - 2mymlx - aIe' m= , ,...,

in D. Put

(XI.1.36)

Then,

Hence,

bm = Icml-' if c,,, 0 0,I0 if Cm=0.

00 00

Icmam(x) (x - a)mI < EM=1 m=1

Ix - alm-e2m.ym for x E D.

Oo

f (x) = co + E c,nam(x) (x - a)mm=1

converges uniformly in D. Consequently, f (x) is continuous in D and holomorphicin the interior of D.

1. ASYMPTOTIC EXPANSIONS IN THE SENSE OF POINCARE 351

To show that f (x) satisfies (XI.1.32), let N be a positive integer and observethat

N+1 N+1

AX) -0D - E cm(x - a)m = E C exp2--y-(Z - ax - a)m

7)01(m=1 m=100

+ (x - a)N+1 i cmam(x)(x -a) m-N-1].

Lm=N+2

Since (XI.1.33) implies that

iexp [ 2'n7m(x - a)e ] Ix - aj-"I, r = 0. 1, 2,... , N,

are bounded for x E D, there exists a positive constant H1 such that

,,=1m-1

<Hljx-aIN+1 for xED.

Also, (XI.I.35) implies that there exists a positive constant H2 such that

00

E cmam(x) (x - a)m-N-1

m=N+2

1 [00 Ix -aim-N-1

(27)N+1Ix - aj0 (2'r)m-N-1

m=N+2

IN+2- i (2y) +2 {x - aI - H2 for X E D.

2y

Thus, there exists a positive constant KN such that

N

f(x)- c,n(x-a)m <KNIx-aIN+1 for xED.

Therefore, (XI.1.32) is satisfied. 0Similarly to Theorem XI-1-13, we can prove the following theorem.

Theorem XI-1-14. For a given formal series p(x, t) _ cm(t)(x - a)m in pour00

mers of (x - a and a sector D given by (XI.1.31), where q. (t) (m = 0, 1, ...) are

holomorphic and are bounded in a domain S2, there exists a function f (z, t)which is holomorphic with respect to (x, t) in D x S2 and satisfies the conditions

N r _ bmc-exp I 2mrym(x - a)e] (x - a)m

(XI.1.37) A x , t) c,,,(t)(x - a)m as x -+ a in DM=0

and

(XI.1.38)( X ,

00

as x -. a in Vdt m=o dt

uniformly fort in Il.

Proof.

AsI dcd't(t) are bounded in Sl (m = 0,1, ... ), choose

6m ={ I ax 1cm(t)I] + I

I d d (t) J }-1if c -(t) 0,

t r:

0 L ` ifC,,,(t)-0.Set

00

f (x, t) = co(t) + Cm(t)am(x)(x - a)m,m=1

where a,,, (x) are given in (XI.1.34). Then, it can be shown that f (x, t) is holo-morphic with respect to (x, t) in D x 1 and satisfies (XI.1.37) and (XI.1.38) in amanner similar to the proof of Theorem XI-1-13. O

Examples XI-1-15. The followings are two examples of functions admitting as-ymptotic expansions.

1. Let r(z) denote the Gamma function and Log z denote the principal branch ofthe complex natural logarithm of z. If 0 = arg z satisfies IOI < 7r, a real number wsatisfies Iw I < 2 , and 10 + w l < 2 , then

rN m-1Log[r(z)] _ (z - Logz - x +

2m(2mM=1

+ eN(z) BN+1 -(2N+1)2(N+1)(2N+1)z

where IeN(z) < ( ( ))2N+1I ( )I and the B,,, are the Bernoulli numberscos 8 + w sin 2w

(see, for example, [01, p. 294]).

2. The exponential integral function Ei(z) =J

ettdt has the following form:zo0

N Z

Ei(z) = er J(k - 1)!z-k + N!100 ett-N-ldt.k=1

From this observation, it follows thatN

e-ZEi(x) - E (k - 1)!z-kI < {N! + (N + 1)!(7r + (N + 1)-1)) IzI-N-1

k=1

for I arg(-z)I < 7r. The asymptotic expansion is valid for I arg(-z)I <32

(see, for

example, (Wasl, p. 31[).

2. GEVREY ASYMPTOTICS 353

XI-2. Gevrey asymptotics

The Gevrey asymptotics is based on the following two definitions.

Definition XI-2-1. Let s be a non-negative number. A formal power series p =+ooE anx' E C[[x]) is said to be of Gevrey order s if there exist two non-negativem=0numbers C and A such that

(XI.2.1) [am[ < C(m!)'Am for m = 0,1,2,... .

We denote by C([x]], the set of all power series of Gevrey order s.

Definition XI-2-2. A function 0 of x is said to admit an asymptotic expansionof Gevrey orders (s > 0) as x -+ 0 on a sector

D(r, a, b) = {x : 0 < fix[ < r, a < arg(x) < b},

where a and b are two real numbers such that a a,,,x"' E G[[x]], such that an in-m

equality

(XI.2.2)N-1

0(x) -E amxmM=0

< Kv,a,A(N!)'(Bv,a,3)8[x[N

holds on D(p, a, 0) for every positive integer N and every (p, a, 03) satisfyingthe inequalities 0 < p < r and a < a <,3 < b, when KP,Q,o and Bp,,,,o arenon-negative numbers determined by (p, a, /3).

We denote by A, (r, a, b) the set of all functions admitting asymptotic expansionsof Gevrey order s as x -' 0 on the sector D(r, a, b). We also set J(4) = p for0 E A. (r, a, b). In §XI-3, we shall explain the basic properties of ar[[x]] A. (r, a, b),and the map J : A. (r, a, b) -+ G[[x]],.

In the example given at the beginning of this chapter, the formal solution p =00

E (-1)m(m!)x"+1 of X2!Ly +y-x = 0 is of Gevrey order 1 and the solution f (x) =M=0

eL/z Jxt-1e-lltdt admits an asymptotic expansion of Gevrey order 1. Furthermore,JJa

J(f) = p. Also, the Maillet Theorem (cf. Theorem V-1-5) states that any formalsolutions of an algebraic differential equation belong to C[[x]], for some s, where sdepends on each solution.

It is well known that if a complex-valued function 0(x) is holomorphic andbounded on a domain 0 < [xj < r, where r is a positive number, then 0 is repre-sented by a convergent power series in x. The Gevrey asymptotic expansions arise

354 XI. ASYMPTOTICS EXPANSIONS

in a similar but more general situation. To explain such a situation, let us considerN sectors

St={x: 0<Ixl <r, at<arg(x)<b1} (1=1,2,...,N)

N

which satisfy the condition USt = (x: 0 < IxI < r}. The set {S11S2,... SN} ist=1

called a covering at x = 0. Also, a covering {S1, S2, ... , SN } at x = 0 is said to begood if

(i) at < at+l (t = 1, 2, ... , N), where aN+l = al + 27r,(ii) bt - at < x (1 = 1, 2,... , N),

( ) S t n St+1 0 (1 = 1, 2, ... , N) and Stn Sk = 0 otherwise if 1,,E k, whereSN+1 = S1.

The following theorem is the most basic fact in the Gevrey asymptotics.

Theorem XI-2-3. Assume that a covering {Si, S2, ... , S1 %r) at x = 0 is good andthat N functions 01(x), ¢2 (x), ... -ON (x) satisfy the conditions

(1) 4t(x) is holomorphic on St,(2) dt(z) is bounded on St,(3) it holds that

(XI.2.3) I¢e(x) - 01+1(x)I < 'Yexp [-

Set

on St n St+1,

where -y > 0, A > 0, and k > 0 are suitable numbers independent of 1.

(X1.2.4)

+00

Then, there ex,sts a formal power series p = E a ..xm E C[[x)]. such that ¢t EM=0

A. (r, at, bt) and J(41) = p for each 1.

There are various situations in which Gevrey asymptotic expansions arise. Toillustrate such a situation, let ¢(x) be a convergent power series in x with coefficientsin C. For two positive numbers r and k, set

k rr

J 0(t)e-('1x)}tk- 1 dt.x 0

This integral is called an incomplete Leroy transform ofd of order k.+oo

Theorem XI-2-4. For every ¢ = cx"j E C{x}, it holds thatm=0

R if4,k(0) E Al/k P,-2k'2k

2. GEVREY ASYMPTOTICS 355

and+00

J(tr,k(4)) _ r 1 1 + k Cm2'",M=0

where k and p are any positive numbers and r is any positive number smaller thanthe radius of convergence of 0.

Proof.

The following proof is suggested by B. L. J. Braaksma. For an arbitrary smallpositive number e, let {Sl (e), S2(e), ... , SN(c)} be a good covering at x = 0, where

St(E) _ {x : I argx - dtI < 2k - e, 0 < jxj < r}

with real numbers dt such that -ir < d 1 < . - - < dN ir, and set

J OtW _ (t = 1, 2, ... , N).I. 4t(x) = tr,k(0t)(xe-d`)

In particular, choose dt = 0 for some t. Then, {Sj(e),S2(e),... ,SN(e)) and{01, 02, . - . , ON} satisfy conditions (1), (2), and (3) of Theorem XI-2-3. In fact, (1)and (2) are evident. To prove (3), note that

k / re'd' rQe(x) =

zkJ 0(a)e-(o/s) o 'da,

0

where the path of integration is the line segment connecting 0 to re{d'. Therefore,

of (x) - bt+1(x) = kj 0(a)e-(ol=)'ak-1d,

X t

where the path ryt of integration is the circular arc connecting re'd'+1 to re'd'.Statement (3) follows, since

/e-(°/=)` exp - t fx cos[k(arg a - arg x)J

and kj arga-argxj < 2 -ke for a E ?Y and x E St(e)f1St+1(E) Since a is arbitrary,it follows that

61 E Allk I P, de - r,-, dt +2k

(t = 1, 2, ... , N).

Furthermore, J(01) can be computed easily (cf. Exercise XI-13).+00

Observe that a power series p = 1: a,nxm E C[[xjj belongs to C[[xJ), if and onlym=0

+aoif d(x) = I'(l

+sm)xm

belongs to C{x}. Therefore, we obtain an important

corollary of Theorem XI-2-4.

Corollary XI-2-5. For any p E G[[x]], and any real number d, there exists afunction ¢(x) E A. (r, d - 2 , d +

2) such that J(O) = p.

This corollary corresponds to the Borel-Ritt Theorem (cf. Theorem XI-1-13) ofthe Poincar6 asymptotics. Also, this corollary implies that the map JA8 (r, d -

2s7r , d + 2) -+ c[[x]], is onto.

Theorem XI-2-3 is a corollary of the following lemma.

Lemma XI-2-6. Assume that a covering {St : e = 1, 2,... , N} at x = 0 is goodand that N functions 61(x), 62(x), ... , 6v (x) satisfy the following conditions:

(i) 6t is holomorphic on St n St+1,(ii) I6t(x)I 7exp[-Alxl-k] on St n St+1, where -y > 0, A > 0, and k > 0 are

suitable numbers independent of e.Define s by (XI. 2-4). Then, there exist N functions ), (x), t.b2(x), ... ,'IPN(x) and a

+oo

formal power series p = > E C[[x]], such thatm=o

(a) 4,t E A. (r, at, bt) and J(4't) = p, where St = {x : 0 < I x[ < r, at < arg(x) <bt} (e = 1, 2, ... , N),

(b) 6t(x) _ t(x) - i/'e+l(x) onSt n St+1.

Let us prove Theorem XI-2-3 by using Lemma XI-2-6.

Proof.

Set

5t(x) = 4t(x) - 41+1(x) (e = 1, 2, ... , N).

Then, there exist N functions 4111(x), 02(z), ... , ON (x) satisfying conditions (a) and(b) of Lemma XI-2-6. In particular, (b) implies that

Oe(x) - 4,t+1(x) = t t(x) - 4Gt+1(x) (e = 1, 2,... , N)

on St n St+1. This, in turn, implies that

01(x) - 4't(x) = 4,1+1(x) - 4Gt+1(x) (e = 1, 2,... , N)

on St n St+1. Define a function 0 by

4,(x) = 4t(x) - 4Vt(x) on St (e = 1, 2, ... , N).

Then, 0 is holomorphic and bounded for 0 < IxI < r. Therefore, 0 is representedby a convergent power series. Since .01 = -01 + 0, Theorem XI-2-3 follows immedi-ately. 0

We shall prove Lemma XI-2-6 in §XI-5.

Because the Gevrey asymptotics of functions containing parameters will be usedlater, we state the following two definitions.

3. FLAT FUNCTIONS IN THE GEVREY ASYMPTOTICS 357

00

Definition XI-2-7. Lets be a positive number. A formal power series am(u-')emm=0

is said to be of Gevrey order s uniformly on a domain V in the u-space if thereexist two non-negative numbers Co and C, such that

(XI.2.5) Iam(i )I < Co(m!)sCr

for u E D and rn = 0, 1, 2, ....

Set V = D(60, a0, Qo) = {e : 0 < Ie] < 5o, ao < arg e < (30} and W = D(b, a, Q).

Definition XI-2-8. Let s be a positive number. A function f (u e) is said to admit00

an asymptotic expansion am(u")em of Gevrey order s as c 0 in V uniformlym=0

on D if

(i) > am(i )em is of Gevrey order s uniformly on D,m=0

(ii) for each W such that a0 < a < 0 < X30 and 0 am('tL)cm < 1)!]8LM=0

fori ED,eEIV andN=1, 2, ....Theorem XI-2-3, Theorem XI-2-4, Corollary XI-2-5, and Lemma XI-2-6 can be

extended in a natural way so that we can use them for functions containing param-eters. We leave such details to the reader as an exercise.

The materials of this section are also found in [Rain 11, [Ram 21, [Si17, Appen-dices], and [Si18).

XI-3. Flat functions in the Gevrey asymptoticsIn the next section, we shall show that C[[x)), and A. (r, a, b) are differential

algebras over C and the map J : A, (r, a, b) C[[x)), is a homomorphism ofdifferential algebras over C. In §XI-2, it was shown that the map J is onto ifb - a < sir (cf. Corollary XI-2-5). In this section, we explain the basic resultsconcerning the nullspace of J. To begin with, we introduce the following definition.

Definition XI-3-1. A function f (ii, e) is said to be flat of Gevrey order s as e - 0in a sector

V = D(r,a,b)={e: 0<je]<r, a<arg(e)<b)uniformly on a domain D in the u'-space if f (ii, e) admits an asymptotic expansion

+00

p = E am (u")em of Gevrey orders as f 0 in V uniformly on D and the expansionm=0

off is 0, i.e., all the coefficients a, (ii) of p are equal to zero.

The following theorem characterizes flat functions of Gevrey order s.

Theorem XI-3-2. A function f (u", c) is flat of Gevrey order s as e -- 0 in asector D(r, a, b) uniformly on a domain D in the ii-space if and only if for eachW = D(p, a, Q) such that 0 < p < r and a < c< < ft < b, there exist non-negativenumbers KW and Aw such that

(XI.3.1)

where

(XI.3.2) k =I.

s

Proof.

Suppose that for a fixed W, there exist two non-negative numbers C and A suchthat

I f ( u " , e)j < C(m!)"Am jelm for (11,f) E D x W and m=0,1,2,....

Then, by virtue of Stirling's formula

(XI-3.3)

If(u,e)I<KtyexpI-ikI for (i,e)EDxW,

r`k) 2m

lm! = 2nm mme'm [i +0(01,

If(u,e)I < Com"'Ag'jem for (u, e) E D x W and m = 0, 1, 2, .. .

For a given e E W, choose a non-negative integer m so that

k

me < (;i)< (m+1)e.

Then,

1

I f(u' e) 5 Comm/k (me) = Coe-"'/k = Co exp Ike

{e - (m + 1)e}]

Hence, I f (u", e) I < K exp[-A ej-k] for every (6,e) ED x W, where K = Coexp[s]and A =

e. The converse is evident. 0

0

3. FLAT FUNCTIONS IN THE GEVREY ASYMPTOTICS 359

Remark XI-3-3. It is known that a function f (11, e) is identically zero on D x Vif

(i) f is holomorphic in a on V for each fixed ii in D,(ii) f is flat of Gevrey order s as e --+ 0 in V uniformly on D,

This implies that the homomorphism J :A, (r, a, b) - C[[xBJ, is one-to-one ifb - a > sa. Remark XI-3-3 is a consequence of the following lemma.

Lemma XI-3-4 ([Wat]). Let V = D A - T2k

,2k

), when p is a positive number.

If f (x) is holomorphic in V and If (x)1 < Cexp[-Blxl-kJ for x E V for somepositive numbers k, B, and C, then f = 0 identically in V.

Proof

Fora>k,set

F(x,a) = f(x) P cos(kB

x_k,2a)

Then, for each fixed value of a, F is holomorphic in V and

(k7r))IF(x,a)l < Cexp - B C1 - coss((k9)IxE-k I for x E V,

where 0 = argx. In particular,

IF(x,a)l < C for argx = ±7r, x E V,

IF(x,a)1 < Cexp IB ( () - 1 r-kcos-E 1

for Ixj = r <pand IargxI <

Therefore, by virtue of the Phragmen-Lindelof Theorem (cf. Lemma XI-3-5), weconclude that

IF(x,a)I < Cexp [B (COS Icos(U')

for I arg x1 < , x E V.

In particular,

If (x)I < C exp [BI i)p-k - ) sk] for argx = 0, x E V.COS

Letting a -+ k, we derive f (x) = 0 if argx = 0 and x < p. This completes the proofof Lemma 111-3-5.

Since use was made of the Phragmdn-Lindelof Theorem, we shall state and provethe theorem precisely (cf. [Ne2, pp. 43-44J).

Lemma XI-3-5 (Phragmen-Lindelof). Let W be a closed sector in C given by

W = {x : a < arg x < ,Q, 0 < IxI < p(arg x) },

where p(w) is a positive-valued and continuous function on the interval a < w < Q.Assume that(1) Q - a < A , where k is a positive number,(2) f is continuous on W,

0 0

(3) f is holomorphic in W, where W is the interior of W,(4) I f(x)I < Kexp[AIxI-k] for x E W, where A and K are positive numbers,(5) If (x) I < M for x on the boundary of W except for x = 0 , where M is a

positive number.Then,

(XI.3.4) If (x)I < M for x E W.

Proof.

Choose I > k so that Q - a <a

< 7r. Set g(x) = f (x) exp[-e(e-'sx)-r], whereP 1

0= a2

Q and e> 0. Then,.

I9(x)I = If(x)Iexp[-ecos(t(argx - 0))Ixl-`]< Kexp[-IxI-r(ecos(f(argx - 0)) - AIxIt'k)]

for x E W. Note that eI arg x - 01 < f(3-2

a) < 2 for x E W. Therefore,J

lim g(x) = 0 and I9(x)I 5 If(x)I for x E W. Since g is holomorphic in W, itIzl-ufollows that Ig(x)I < M for x E W. Therefore,

If (x)I < 11f lexp(e(e-sex)-t)I for x E W .

Letting e - 0, we derive (XI.3.4).We can also prove the following theorem without any complication.

Theorem XI-3-6. If f is holomorphic in e on D(r, a, b) for each fixed u" in V andf is flat of Gevrey orders as e - 0 in D(r, a, b) uniformly on a domain D in the

u"-space, then f (u, a)do and of (a, e) are also flat of Cevney orders as e -' 0 inja, b)

uniform'

Ofly on D.

The proof of this theorem is left to the reader as an exercise.For more information see [R.aml], [Ram2] and [Si17, Appendices].

XI-4. Basic properties of Gevrey asymptotic expansionsIn this section, we summarize the basic properties of the Gevrey asymptotic

expansions. First we prove the following theorem.

4. BASIC PROPERTIES OF GEVREY ASYMPTOTIC EXPANSIONS 361

Theorem XI-4-1. Assume that(i) a C-valued function F(ul,... , un, e) is holomorphic in (ul,... , un, e) in a do-

main D(rl,... ,rn) x D(r,a,b), where D(rl,... ,rn) = {(u1,... ,un) : Iu3f <r,, j = 1, ... n},

(ii) flu l, ... , un, e) admits an asymptotic expansion p(ul, ... , tin, E) _+oo

Ean,(u1 i ... , un)E' of Gevrey order s as a -' 0 in D(r, a, b) uniformlymc0on D(rl,... , r.), where the coefficients a,n(u1 i ... , un) are holomorphic inD(ri, . . . , rn),

(iii) n C-valued functions 61 (v", e), .... On(v, e) are holomorphic in a domain D xD(r, a, b), where D is a domain in the 6-space,

(iv) for each j, the function 03 (v, e) admits an asymptotic expansion p, (v, e) _+ooE b3,,n(v')e'n of Gevrey order si as a -+ 0 in D(r, a, b) uniformly on V, wherem=1the coefficients bj,,n(v) are holomorphic in D.

Regarding the coefficients a,n(u1,... , un) as power series in ( u1 , , u.), define aformal power series in a by

P(16, c) = p(p1(i', e).... Pi-(Of"M=0

where the coefficients P,n(v") are holomorphic in D. Then, for any (a, 0) satisfyingthe condition a < a < Q < b, there exists a positive number p(a, 3) such that0 < p(a, (3) < r and that the function F(Oi (6, e), ... , thin (6,t), c) is holomorphic inD x D(p(a, 0), a, Q) and admits the formal power series P(U, e) as its asymptoticexpansion of Gevrey order max(s, si , ... , sn) as e 0 in V(p(a, 03), a, 03) uniformlyon the domain D.

Remark XI-4-2. As a consequence of this theorem. we conclude that under thesame assumptions as in Theorem XI-4-1, the formal power series P(v, e) in a is ofGevrey order max(s, s1, ... , sn) uniformly on D.

Proof of Theorem XI-4-1.

Fixing a pair (a, (3) satisfying the condition a < a < 0 < b, choose a suitablegood covering {S1, ... , SN } ate = 0 and N(n + 1) functions Ft(u1, ... , un, e),¢1,e(vU, e), ... , QSn,t(U, e) (e = 1,... , N) such that(1) for each e, the function Ft(ui,... ,un,e) is holomorphic in the domain

D(r1 i ... , r,,) x St and admits the formal power series p(u1,... , Unit) asits asymptotic expansion of Gevrey order s as e -' 0 in St uniformly onD(r1,... , rn),

(2) for each (j, e), the function 6,,e(17, e) is holomorphic in the domain V x St andadmits the formal power series pj (u", e) as its asymptotic expansion of Gevreyorder sJ as e -+ 0 in St uniformly on V.

(3) there exist two positive numbers K and A such that

a,un,E) - Ft+1(ul,... Un,()! < Kexp - Iefk

on D(rl,... , rn) x (St n St+l ), where k = 1,

(4) for each j = 1, ... , n, it holds that

101,1 (V,e) -o,,e+1(V,e)I Kexp -JEIk on D x (St n St+1),

where k, =1

,

Si(5) $t = D(p, a, E3), Ft. = F, and O,,& n) for some fo.

We can accomplish this by virtue of Corollary XI-2-5 and Theorem XI-3-2.

Observe that

Fe+1(&,e+1(v,e),...

tl

Fl(41,t(V, E), ... , ¢n,[(ve E), E) - (v, E), ... , 41n,,+1(v, e), e) I

+ JFt(¢1,1+1(Lt,E), +dn,[+1(v,e),E) - F1+1(dl.[+1(6,e), ,1+1(V,e), e)1n

< LKexp [-IIk

+ Kexp [-(IC`J'-1 E

E

for (6, e) E V x (St n S[+1), where L, K, and A are suitable positive numbers.Therefore, using Theorem XI-2-3, we complete the proof of Theorem XI-4-1. 0

As a corollary of Theorem X1-4-1, the following result is obtained without anycomplication.

Theorem XI-4-3. C([x]], and A. (r, a, b) are commutative differential algebraover C, t.e.,

(i) f + g E C((x]], (respectively A. (r, a, b)) if f and g E C((x]], (respectivelyA, (r, a, b)),

(ii) f9 E C([xJ], (respectively A. (r, a, b)) if f and g E C((xi], (respectivelyA. (r, a, b)),

(iii) cf E C[(x]], (respectively A. (r, a, b)) if c E C and f E C((x]J, (respectivelyA. (r, a, b)),

(iv) 1 E C((xJ], (respectively A, (r, a. b)) if f E C((x)J, (respectively A. (r, a, b)),

(v) / Z f dx E C((x]], (respectively A, (r, a, b)) if f E C((xf, (respectively

A. (r, a, b))Furthermore, the map J : A. (r, a, b) --t C((xi], is a homomorphism of differentialalgebra over C.

Remark XI-4-4. Using Theorem XI-3-6, we can prove (iv) and (v) of TheoremXI-4-3 in a way similar to the proof of Theorem X1-4-1. Also, it can be shown

+"Q1

that if f (x) E A, (r, a, b) with J(f) = > a,,,x' and ao 54 0, we obtain

fE

m=o

5. PROOF OF LEMMA XI-2-6 363

A. (p(a, 8), a, 0) and J\ I / = J{f) , where a < a < /3 < b and p(a, /3) is a

suitable positive number depending on (a,fl).The materials in this section are also found in [Ram1], [Ram2], and [Si17, Ap-

pendices].

XI-5. Proof of Lemma XI-2-6

In order to prove Lemma XI-2-6, choose N line segments C1, C2, ... , CN sothat Ct E St n St+1 (t = 1, 2,... , N), i.e., Ct : z = teie1 (0 < t < r), wherefor each t, Bt is a fixed number satisfying the condition at+1 < 8t < bt. These Nline segments divide the open punctured disk V = {x : 0 < IzI < r} into N opensectors S1, S2, ... , SN, where

St = {x : 9t_ 1 < arg(x) < 01, 0 < txI < r}.

Set

lrZt(x) = 1

N

21rth=1

for x E St, t = 1, 2,... , N. The functions +Gt can be continued analytically ontoSt by deforming Ct_1 and Ct without moving any of their endpoints. In doing this,we do not change other line segments Ch (h i4 t - 1, t). Thus, the function tyt isholomorphic on St, t = 1, 2,... , N, respectively.

Now, assuming that x E St n St+1, compute t/t(x) - tPt+1(x). To do this, write'Pt and tLt+1 in the following forms respectively:

Ot+1 {x) = 27ri e", t - x + 2Ri c, - x

tt't(x) _ {2 =) / (2 =) E C" Ln2ft

h#lJ

where the paths Ct and Ct+1 of integration are obtained by deforming C, withoutmoving either of its endpoints so that

(1) Ct C St n St+l and Ct+l c St n St+1,(2) -Ct + Ct+1 is a simple closed curve whose interior contains x.

Thus, (b) follows, i.e.,

0&) - 01+1(x) = fe,tai + e«,at(x) on St nSt,.1.

Let us derive asymptotic properties of ib1. To do this, fix an t and a closed sectorW contained in St. Let Ct (respectively Ct_1) be a path obtained by deforming Ct(respectively CL_1) without moving the endpoints so that W is contained in the

interior of the simple closed curve Ct_ 1 + 7t - C1, where 1't is a circular arcjoining two points reist-' and rest. For x E W, it holds that

={-1) sew (-1) bt-1(1) (-1)

ot (x) - 27ri f x + 2?7 f xd{

+ 27ri f x

It may be assumed that the path Ct is given as a union of a line segment L1 and acurve rt defined by

Lt : x = te'" (0 < t < r1 < r),

rt : x = Ftt(T) (0 <T < 1),

where we is a constant such that 01 < we < bt, and

pt(0) = r1e'", pt(i) = re'B', r1 < litt(r)I < r (0 < 7- < 1).

It can be also assumed that there exists a positive number a < 1 such that JxI < ar1

for x E W. Then, 1 f8t(-)

dC is represented by a convergent power series in x21ri r, c - x

whose radius of convergence is not less than r1.

Let us estimate the integral 1J bt() Since27ri L, c - x

1

x

we obtain

where

1 1x Q)m +

mM+1

x,

11 6t{f) M

[Ymx"' xM+lEM+1(x),27ri L - x m=0

__ 1 ram5() (m > 0) and EAi+I(x) = 1 f 61W <.

21ri JL, Cm+1 - 27ri £M+1({ - x)

Now, estimate the a,,, as follows:

lQm= 1 I fr' ebt(Teu.")eudJ I /r'

J 27r f Tm+1 dT

7 +oor_m-1e_ar- dr = 7 A-Ilk m r( l

l ! \ k 127r f 2k-,r

-I- \A-1lk\m( 27)1-k (v') -m/k (m!)l",I \_7r

where the Stirling's formula (XI.3.3) was used as m -' +co.

EXERCISES XI 365

To estimate EN+i(x), note that there exists a positive number 0 depending on

W such thatit - xJ > Jtj sin(0) for t E L1, x E W.

Hence,

1 r ^ye-ar '

JEN+i(x)J.ya-(N+1)/k N+ 1

2a JorN+2sin(0)dr <

2kirsin(0) r ( k

<

7/N+llity+l)/k

k sin(O) ` )eke )

We can estimate

-112ri c,_, f - x and

-12iri Jc - x (h94 Q,t-1)

in a similar manner. Thus, the proof of Lemma XI-2-6 is completed.The material of this section is also found in [Si18].

EXERCISES XI

XI-1. Let S be an open sector D(ro, a, b), i.e.,

S = {x : 0 < JxI < ro, a < argx < b},

where a and b are real numbers and ro is a positive number. Denote by(a) A(S) the set of all functions which are holomorphic in S and admit asymptotic

expansions in powers of x as x -. 0 in every open subsector

W = {x : 0 < JxJ < r < ro, a < a < argx < Q < b}

of S,(b) J[f] the asymptotic expansion for f E A(S),(c) Ao(S) the set of all those functions f (x) in A(S) such that f (x) ^-- 0 as x 0

in every W (i.e. AO(S) = If E A(S) : J[f] = 0}).Show that C[Jxll. A(S), and Ao(S) are differential algebras over the field C andthat the map J : A(S) C[[x] is a homomorphism of differential algebras overthe field C.

XI-2. Using the same notations as in Exercise XI-1 and considering a linear dif-N

ferential operator P = E ak(x)Dk with coefficients ak(x) which are holomorphick-o

in a disk JxJ < r containing S, where D = d, show that

(i) P defines three homomorphisms

P1 = P : Ao(S) - Ao(S), P2 = P : A(S) -- A(S), P3 = P : C[[x]] C[[x]]

of vector spaces over the field C,(ii) two vector spaces A(S)/Ao(S) and C[[x]] are isomorphic,

(iii) dimensions of three vector spaces Nullspace ofP1i Nullspace ofP2, andNullspace ofP3 over C are not greater than the order of P,

(iv) two vector spaces {Nullspace of P2}/{Nullspace of P1} and J(Nullspace of P2]are isomorphic,

(v) there exists a homomorphism T : Nullspace of P3 .A0(S)/P1 [A0(S)] of vec-tor spaces over C such that Nullspace of T = J[Nullspace of P2] andT[Nullspace of P3] = {P2[A(S)] nAo(S)}/P1[Ao(S)]

Hint.

(ii) J is onto (cf. Theorem X1-1-13) and .Ao(S) is Nullspace of J.(iii) Use Theorem IV-2-1 for Nullspace of P1 and Nullspace of P2; use an idea

similar to the proof of Theorem V-I-3 for Nullspace of P3.(iv) Nullspace of the homomorphism J : Nullspace of P2 -+ J[Nulspace of P2] is

Nullspace of P1. Also, this homomorphism is onto.(v) Assume that P3[¢] = 0 in C{[xJI. There exists a b E A(S) such that J[4] = 4

(cf. Theorem XI-1-13). Such b is not unique. Actually, ¢ + also satisfiesthe condition J[4, + '] = 4, if and only if iP E A0(S). So define T by T[4,] =P2[4J (mod P1[Ao(S)]). Note that J[P2[4,]] = P3[4'] = 0. This implies thatP2[O] E Ao(S).

X1-3. For each of the following three power series, find s such that f E C[[x]],.cc

(a) f(x)

(c) f (x)

00 r(1+ )_ (mx)m, (b) f (x) = E 2m

,n=o m_or 1 + 300 2m

_ Exm 11 (5+ V5-h).m=o h=m

XI-4. For two real numbers s > 0 and A > 0 and for a formal power series+00 +00

f = 1: a,nxm E C[[x]1, set I I f 11 &,A = E ( Denote by E(s, A) the set of

7=0M=0

all f E C[[x]] satisfying the condition IIf II..A < +oc. Show that(i) E(s, A) is a Banach space with the norm IIf II.,A,(ii) IIf9II.,A <_ IIf II..A I19II..A if f E E(s, A) and g E E(s, A),

(iii) if B > A, then E(s, A) C E(s, B), and II f II,,a < IIf II,,A for f E E(s, A),

(iv) if B > A, then Il f II.,e - If (0)I o a. xm,(v) for f E E(s, A) and an integer q such that sq > 1, it holds that

df11 5 -LIIfII.,AeA

EXERCISES XI 367

Comment. The formal power series f E C([x]] is of Gevrey order s if and only iff E E(s, A) for some positive number A.

XI-5. Show that f (e) = 0 identically in a sector S if f is holomorphic and flat ofGevrey order 0 as e --+ 0 in S.

Hint. I f (e)I < KA' IeIr' fore E Sand N = 0,1,2,..., where K and A are positivenumbers independent of N.

XI-6. Show that C[[x]]o = C{x}.

),Hint. f = E E C[[x]]o if and only if Ic,.. I < KAt If I' (m = 0, 1, 2....00

m=G

where K and A are positive numbers independent of m.

X1-7. Consider a system of differential equations

(S)xy+t dy = f0 + bg +

Ip!>_o

where q is a positive integer, fo and fp E C[(x]]n, 4> is an n x n matrix whose entriesare formal power series in x, p = (pl, ... , p,>) with n non-negative integers pt, ... ,Pn, IpI = Pt + - - - + p,,, and yam' = yP' - - - VP-. Assume that sq > 1 and that

(i) fo and f. E E(s, A)n, and the entries of 0 belongs to E(s, A) for some A > 0,(ii) the power series F, IIfpIIs,A9P is a convergent power series in y", where

Ipl>2IIY1Ia,A = max{IIy,II5,A :1 < j < n},

(iii) fo(0) = 0,(iv) (3(0) is invertible.

Show that there exists a unique power series E C[[x]]n satisfying system (S) andthe condition ¢(0) = 0. Furthermore, E (C[[x]],)n.

Hint. Write system (S) in a form y" = go + x9+1 1P

dx

+ E y'gp and use theipl>2

Banach fixed-point theorem in terms of the norm II IIs.A-

XI-8. Let D(r) = {z : IzI < r}, S(r, p) = {z : 0 < IzI < r, I arg zI < p}, anda power+w

series f (z, e) = E fn(e)zn is convergent uniformly in a domain D(ro) x S(r, p),n=o

where ro, r, and p are positive numbers. Assume that the coefficients fn(e) of theseries f are holomorphic in S(r, p) and admit asymptotic expansions in powers ofe as e -' 0 in S(r, p). Suppose also that limo fo(e) = 0 and limo f, (c) 0 0. Show

C- 1-that there exist a positive number rt and a function O(e) such that (1) O(e) isholomorphic in S(rt, p) and admits an asymptotic expansion in powers of a ase -+ 0 in S(rt, p), (2) l ¢(e) = 0, and (3) f (c(e), e) = 0 identically in S(rt, p).

X1-9. Consider a formal power series

(P) F(x, y, _ip,l+IpI>o

#,y fa,,n2,

yi z1

"where %' _ , fn:,as E C[[x]] , pi = (pll,... ,pln), and Pz =yn xm

(p21, ... , p2",) with non-negative integers P1k and p2k. Denote by Fg(x, y z) theJacobian matrix of ! with respect to W. Assume that power series (P) satisfies theconditions (1) f;(0,6,6) = 0, (2) Fy(0, 0, 0') is invertible, (3) fn, a, E E(s, A)" forsome s > 0 and A > 0, and (4) the power series i[ fn, a, Ij, Aye' is a

ia,I+?P2I?0convergent power series in y and F. Show that there exists a unique power series¢(x, z) 92 in (x, z) such that (i) drr,, E C[[x]]", (ii) b(0,0) = 0, and (iii)

lack>o

F(x, (x, z), z) = 0. Also, show that (a, E E(s, B)" for some B > 0 and the powerseries E II( a' i" is a convergent power series in z.

IP3i>o

Hint. This is an implicit function theorem. Write f in the form f = fo + fiy +F'gv' z"P2 fn,,r,, , where fo and fn,,n, are in E(s, A)", $ E M"(E(s, A)), and >'is the sum over all (pi i p) such that either jell = 0 and [g-,j = I or Jial j +jp2j > 2. Here, M.(R) denotes the set of all n x it matrices with entries in a ringR. Assumption (1) implies that fo(0) = 0. Assumption (2) implies that -t(0) isinvertible. Therefore, I ' exists in M"(C[[x]],), and hence II4 'jI, A < +oo forsome A > 0, where III-' IIB,A = sup{jj4V' f Ij, A : f E E(s, A)n, Ij f [j,.A = 11. Since

IIfIIe,B 5 11A .,A for f E (C[[x]],)" if B > A, assume without any loss of generalitythat A = A. Then, t' IF = 90 + y" + zP2§p., p,, where g"o E E(s, A)",9a,,a, E E(s, A)", 90(0) = 0, and 'II9n,,a2I1a,A9" i is a convergent power series

ulin y" and F. Consider the equation 0 = u"+ y"+ F,'#P- z g"n, ,n, , u" = . . Then,

unthere exists a unique power series 5(x, u", _ lip ' z'na"n, such that

fa,l+lpiI>Idn,,a, E E(s, A)" and 0 = u'+ a'+ identically. The unique powerseries satisfying conditions (i), (ii), and (iii) is given by

(S) (x, z_) = 5(x, 90, zfl

Now, consider the equation

&'$121 Is,A

(R) y'=u".+z-V Gn,,p, where Ga,,i7 = ER".9P% ,P2 Il .w

EXERCISES XI 369

Equation (R) has a unique solution y = E 9P'z-r'pp,,p, such that Qp,,p, E1p1I+1P21>>-1

]R" , the entries of pP,,p, are non-negative, and the series is a convergent powerseries in u and z1 . Use this series as a majorant to show that defined by (S)satisfies conditions (iv) and (v).

XI-10. Assume that a covering (S,,$2,... , SN} at x = 0 is good and that Nfunctions 01(x), 02(x),... , ON (x) satisfy the conditions:

(1) 0t(x) is holomorphic in St,(2) of(x)^_-Oasx-.0 in St,(3) 10e(x) - Ot+1(x)l < -yexp[-AIxI-kJ on St n St+1i where -y > 0, A > 0 and

k > 0 are suitable numbers independent of e.Show that there exists a positive number H such that

[4t(x)I < Hexp[-AIxI-kJ in St.

Hint. See [Si15J.

XI-11. Assume that a covering {S1, S2, ... , SN } at x = 0 is good and that Nfunctions 01(x), 02(x),... , y` N (x) satisfy the conditions

(1) 4,(x) is holomorphic on St,(2) 4c(x) is bounded on St,(3) we have

IOt(x) - 01+1(x)l < Knlxl' (n = 1,2,...) on St n St+1,

where K are positive numbers.

Show that there exists a formal power series p = > a,,,xm E d[[xJJ such that form=0

each t, we obtain ¢t E A(S1) and J(¢1) = p, where the notations A(S) and J aredefined in Exercise XI-1.

XI-12. Assume that a covering {S1, S2.... , SN } at z = 0 is good and that n x nmatrices 4i1(x),t2(x),... ,4N(x) satisfy the following conditions: (1) the entriesof tt(x) belong to A(S1 n St+1) and (2) J(tt) = I,. Show that there exists a

formal power series Q = I,, + E xmQm having constants n x n matrices Qm asm=1

coefficients, and n x n matrices P1(x), P2(x), ... , PN(x) such that(i) for each e, the entries of P1(x) and P1(x)-1 belong to A(S1),

(ii) J(P1) = Q (t = 1, 2,... , N),(iii) It(x) = PI(x)-1Pt+1(x)(e= 1,2,... ,N),

where the notation A(S) and J are defined in Exercise XI-1. Also, show that ifthe entries of (t(x) belong to A.(St), respectively, then the entries of P,(x) andP, 1(x) also belong to A,(Se), respectively.

Hint. See [Si17, Theorem 6.4.1 on p. 150, its proof on pp. 152-161, and §A.2.4 onpp. 207-2081.

XI-13. Prove the formula

00

M=0

which is given in Therorem XI-2-4.

Hint. Assume that j argx1 < 2k - b, where b is a small positive number. Then,

k

k rT tme-(t/x)ktk-Idt = xm (r'/) v"'/ke °doJ0 0

Jr°Or (1 + m) xm - xmJ

Om/ke-`do.T (t/z)k

Hence,

N

1,,k(6) _ r (i + m ) Cmxm - E cmxm Om/ke °d0M=0

k m-0 (t/k)k

/r +00

+ k J cmtm e-(t/x)ktk-ldt.xk 0 m=N+1

XI-14. Let a be a positive number larger than 1. Also, let

SJ={x: a, <argx<bj, 0<jxj <p) (1=1,2,... v)

be a good covering of the sector S = {x : j argxj < 'r,0 < jxj < p}, where a1 =

, b = , and p is a positive number. Assume that v functions &1(x), 4(x),satisfy the following conditions:

(i) Oj(x) is holomorphic in Sj and continuous on the closed sector19, ={x:aj5argx<bj,O<lxj<p},

(ii) 103(x)1 < Aexp[clxl-1] in S, for some positive numbers A and c,(iii) 10,+1(x) - Oj(x)j < Mo in S, fl S,t1 for some positive number M0,(iv) 1O1(x)j < M0 on the line segment argx = a1, 0 < lxj < p,(v) < M0 on the line segment argx = b,,, 0 < lxj < p.

Show that there exists a positive number M such that

IOj(x)1 < M in SS (j=1,2....,v).

Hint. This is a generalization of the Phragmen-Lindelof Theorem (Lemma XI-3-5).See ]Lin].

EXERCISES XI 371

XI-15. Let k be a positive number. A formal power series f E C[[x]] is said to bek-summable in a direction arg x = 0 if

E I m a g e J : AIT 7r \f { 1/k (Ap> e - 2k - E, 0+

2k+ EJ C{[x]]1/k

for some positive numbers po and E. Show that if a formal power series f E C[[x]]is k-summable in a direction arg x = 0, there exists one and only one functionF E Al/k (po, 9 - 2k - e, 0 + 2 + EJ such that J[F] = f.

Hint. Use Remark XI-3-3. For more informations concerning summability, see, forexample, [Ba13], (Ram2), [Ram3], [Si17, Appendices], and [Si19].

r e_t-sXI-16. Consider the integral f (x) = f - x dl;, where -1 is a line segment 0 <

x < 1 in the sectorial domain D\

1, - 4 27r + 7r ! . Using the argument given in

§XI-5, show that /(i) f(x) admits an ``asymptotic expansion J[f] E C[[x])112 as x -. 0 in

DI 1,-427x+ 4),(ii) J[f] is 2-summable in the direction argx = 0 if 0 < 0 < 27r.

CHAPTER XII

ASYMPTOTIC SOLUTIONS IN A PARAMETER

In this chapter, we explain asymptotic solutions of a system of differential equa-

tions E°fy = f (x, y, f) as a -i 0. In §§XII-1, XII-2, and XII-3, existence of such as-ymptotic solutions in the sense of Poincar6 is proved in detill. In §XII-4, this result

is used to prove a block-diagonalization theorem of a linear system E° d = A(x, e)y.

In §XII-5, we explain similar results in the Gevrey asymptotics. In §XII-6, we ex-plain how much we can simplify a linear system by means of a linear transformationwith a coefficient matrix whose entries are convergent power series in the param-eter. This result is given in [Hs1] and similar to a theorem due to G. D. Birkhoff[Bi] concerning singularity with respect to the independent variable x (cf. TheoremXII-6-1 and (Si17, Chapter III]). The materials in §§XII-1- XII-5 are also found in[Wasl], [Si3], [Si7], and [Si22].

XII-1. An existence theorem

In §§XII-1, XII-2, and XII-3, we consider a system of differential equations

dv(XII.1.1)

E° V - = f3 (x, V 1 , V2, ... , Vn, E) (j = 1, 2, ... n),

where a is a positive integer and f J (x, V1, v2, ... , vn, f) are holomorphic with respectto complex variables (x, v1, v2, ... , vn, E) in a domain

(XII.1.2) Ix[ < do, 0< IEI < Po, I arg EI < ao, Iv. I < 'ro (j = 1, 2, ... , n),

6o, po, ao, and 'Yo being positive constants. Set

n

(XII.1.3) f j (x, V,

f)= fJO(x,

()a,h(x, e)Vh + 1 fi, (x, E)iJ ,

h=1 Icl?2

where i I E Cn with the entries (V1, V2, ... , vn).

We look at (XII.1.1) under the following three assumptions.

Assumption I. Each function f j (x, v, c) (j = 1, 2,... , n) has an asymptotic ex-pansion

(XII.1.4) fi (x, v, E) "' E f v)E"00

L=o

in the sense of Poincare as c 0 in the sector

(XII.1.5) 0 < IEI < po, I arg EI < ao,

372

1. AN EXISTENCE THEOREM

where coefficients f j,, (x, v) are holomorphic in the domain

(XII.1.6) W < 6o, IV1 < 7o.

Furthermore, we assume that

(XII.1.7) fjo(x,) = 0 (j = 1, 2, ... , n) for 1x1 < do.

373

Observation XII-1-1. Under Assumption I, fjo(x, e), a jh(x, e), and f jp(x, e) ad-mit asymptotic expansions

00 00

(XI1.1.8) fjo(x, () E f o (x)e", fjp(x, e) ' E fJp.(x)f"=1 v=0

and

(XII.1.9) aj h(x, e) '- > (j, h = 1, 2, ... , n)--o

as e - 0 in sector (XII.1.5) with coefficients holomorphic in the domain

(XII.1.10) 1x(< 60-

Let A(x, e) be the n x n matrix whose (j, k)-entry is a.,&, e), respectively (i.e.,A(x, e) = (ajk(x, e)). Then, A(x, c) admits an asymptotic expansion

(XII.1.11) A(x, e) ^_- E e"A,,(x)=o

as a -+ 0 in sector (XII.1.5), where the entries of coefficient matrices(ajk,,(x)) are holomorphic in domain (XII.1.10). The following second assumptionis technical and we do not lose any generality with it.

Assumption II. The matrix Ao(0) has the following S-N decomposition

Ao(0) = diag [µh,02,... ,p1 + N,

where Al, µ2f ... , An are eigenvalues of Ao(0) and .,V is a lower-triangular nilpotentmatrix.

Note that [Ni can be made as small as we wish (cf. Lemma VII-3-3).The following third assumption plays a key roll.

Assumption III. The matrix Ao(0) is invertible, i.e.,

t<J#0 (j = 1, 2,. .. , n).

In §§XII-2 and XII-3, we shall prove the following theorem.

374 XII. ASYMPTOTIC SOLUTIONS IN A PARAMETER

Theorem XII-1-2. Under Assumptions I, II, and III, system (X11.1.1) has asolution

(XII.1.12) vJ = pj (x, c) (j = 1,2,... ,n)

such that(r) pi(x,e) are holomorphie in a domain

(XII.1.13) jxl < b, 0 < lei < p, l arg el < a,

where b, p, and a are suitable positive constants such that 0 < 6 < bo, 0 <p<po, and 0<a<ao,

(ii) p, (x, e) admit asymptotic expansions

00

(XII.1.14) p3(x,e) - (j = 1,2,... n)

as e - 0 in the sector

(XII.1.15) 0 < Iej < p, urge) < a,

where coefficients of (XII.1.14) are holomorphic with respect to x inthe domain {x : lxi < 6}.

XII-2. Basic estimates

In order to prove Theorem XII-1-2, let us change system (XII-1-1) to a systemof integral equations.

Observation XII-2-1. Expansion (XII.1.14) of the solution p3(x,e)

00

(XII.2.1) vJ = Epi,(x)e" (j = 1,2,... n)1-1

must be a formal solution of system (XII.1.1). The existence of such a formalsolution (XII.2.1) of system (XII.1.1) follows immediately from Assumptions I andIII. The proof of this fact is left to the reader as an exercise.

Observation XII-2-2. For each j = 1, 2.... , n, using Theorem XI-1-14, let usconstruct a function g1(x,e) such that

(i) q, is holomorphic in a domain

(XI1.2.2) lxi < 6', 0 < jej < p', l arg el < a',

where 0<6'<6o,0<p'<po,and 0<a'<ao,

2. BASIC ESTIMATES 375

d-zadmit asymptotic expansions(ii) qj and !j

(XII 2 )3) (x > (x) " d d9, dp x),E) ". . ,Egj p,v E an E_ E=i

,,V=1

as f -. 0 in the sector

(XII.2.4) 0 < IEI < p', I argel < o .

Consider the change of variables

v, = u, + q,(x,E) (j = 1,2,...,n).

Denote (ql, q2, ... , qn) and (UI, U2,. .. , un) by q" and u, respectively. Then, ii sat-isfies the system of differential equations

(XII.2.5) E°dd

xj = gj(x,19, c) (j = 1,2,... ,n),

where

93

Set

(XII.2.6)

In particular,

(XII.2.7)

and

(j = 1, 2,. .. , n).

n 00

g)(x,u,E) = g2o(x,E) + bjk(x,E)uk + E bJp(x,E)u-P

k=1 IpI?2

(j = 1, 2, ... , n).

g,o(x,E) = fi(x,9,E) -E°dq)'E) ^ 0 (j = 1,2,... n)

b,k(x, E) - a,k(x, c) = O(E) k = 1, 2,... , n)

as c - 0 in sector (XII.2.4). Therefore,

(XII.2.8) bik(x, E) = ajko(x) + O(e) (j, k = 1, 2,... , n)

asE-+0insector (XII.2.4).Set

(XII.2.9) g, (x, ",E) = p, u, +R.(x,u,E) (j = 1,2,... ,n).

Then, for sufficiently small positive numbers 6, p, and -y, there exists a positiveconstant c, independent of j, such that for every positive integer N, the estimates

(x,u,E) = fj(x,U + q' f) -E° dq,(x,E)

dx

(XII.2.10) IR7(x,u,E)I CIUI+BNIEIN (j=1,2,...,n)

and

(XII.2.11) eIu - u I (j = 1, 2,... , n)

hold whenever (x, u", f) and (x, u~', e) are in the domain

(XII.2.12) lxI <8, 0<lEH<p, largfl <a', lull <-y (j=1,2,... n).

Here, BN is a positive constant depending on N and 1191.From (XII.2.5) and (XII.2.9), it follows that

(XII.2.13) E° uj = µ,u, + R.(x,u,E) (j = 1,2,... ,n).

Change system (XII.2.13) to the system of integral equations

(XII.2.14)u, = I

exp of (t - x) R, (t, u, E)dtE

(

where the paths of integration must be chosen carefully so that uniformly convergentsuccessive approximations can be defined.

Hereafter in this section, we explain how to choose paths of integration on theright-hand side of (XII.2.14).

Observation XII-2-3. Set

(XII.2.15) w,=argµ, (j=1,2,...,n)

and suppose that

(XII.2.16) - 27r < wj < 27r (j = 1, 2,... , n).

Then, there exists a positive number a less than a such that

7r and w, - e -2ir ,n).-2rr <w, - e < 21

Without loss of generality, suppose that n real numbers w, are divided into thefollowing two groups:

(I) <w,-e<2ir (j =1,2,...,m')

and

(II) -27r<w,-6<-27r (3=m'+1,...,n).

Choose two positive numbers a and /3 sufficiently small such that

-2ir+aa+3 <wl -6 < 7 r - (Qa+Q) (.7 = m'),

3 12"+oa+13<wj -A<-27r-(aa+p) (j=m'+1,...,n).

Set

x(1) = be-i°, x(2) = ix(1) tan [3, x(3) = -x(1), and x(4) _ -x(2),

where b is a sufficiently small positive number. Then, a rhombus is defined by itsfour vertices x( j) (j = 1,2,3,4) (cf. Figure 1). Note that the angle at x(1) is 20.

FIGURE 1.

Denote the interior of this rhombus by D(5). It is noteworthy that the domainD(b) contains a small open neighborhood of x = 0 and is contained in the domain{x : IxI < b}.

The basic estimates for the proof of Theorem XII-1-2 are given by the followinglemma.

Lemma XII-2-4. For each j = 1.... , n, consider the function

J=exp - µi(t_ dt

(XII.2.17) U3(x,E) =[z, L J

where the path of integration is the straight line 77! and

( x(1) (j = 1, 2,... , m'),xj=51x(3) (j = m' + 1, ... , m).

Then, there exists a positive number c such that

(XII.2.18) I Ul (x, E)1 :5 c1cl, lexpL

'1l(x - xl)

JE°

for

(XII.2.19) x E D(8), I argel <a, IEI > 0.

Proof

We prove this lemma for the cases j = 1, 2,... , m'. The cases j = m' + 1,... , mcan be treated in a similar manner. Set f = It - x(l)I and 0 = arg(t - x(')) fort E D(b) and set w = arg e. Then,(XII.2.20)

11

I U j (x, e) I s Ix- (l, l exp [-19'J' £ cos(w, + 0 - aw)] d { (j = 1, 2, ... m').

From the definition of D(8), it follows that 7r - e - f3 < 0 < 7r - e +,6 for x E D(8).Thus, if a is in the sector I arg eI < a, the inequalities

1 32a<wj +7r-(aa+(3) <wj +0-aw<wj + ir - 0 + (aa + 0) < 2a

hold f o r x E D(b) and j = 1, 2, ... , m'. Therefore, there exists a positive constantc' such that

(XII.2.21) -cos(wj +0-aw)>c (j=1,2,...,m')for (x, e) in (XII.2.19). By virtue of (XII.2.20) and (XII.2.21), we obtain

IUj(x,E)I < leXp [µ,( xj) (j = 1,2,... ,m')

for (x, e) in (XII.2.19). Thus, Lemma XII-2-4 is proved. DIn the next section, we shall consider system (XII-2-14) of integral equations

assuming that x E 1)(6) and the paths of integration are chosen in the same wayas in Lemma XII-2-4.

XII-3. Proof of Theorem XII-1-2Let us construct a solution u, _ 0, (x, e) of (XII.2.14) so that

(XII.3.1) Oj(x,e) 0 (j = 1,2,... n)as e-'0in (XII.1.15).

Now, by virtue of Assumption II, three positive quantities 6, p, and (Nj can bechosen so small that c in (Xl.2.10) and (XII.2.11) satisfies the condition

(XII.3.2) cc < 1,

where c is the constant given in (XII.2.18).Define successive approximations of a solution of (XII.2.14) in the following way:

I°)u (x, e) 0,

[__i.(t1i (Z

(XII.3.3) Uh(xe) _ exp - x)} Ri (t, u"("-1)(t, e), e)dt

(j = 1,2,... ,n; h=1,2,...)1

3. PROOF OF THEOREM XII-1-2 379

where x E D(b) and the integration is taken over the straight line . For a givenpositive integer N, it will be shown that

(i) for each j, the sequence I (h) (x, E) : h = 0,1, ... } is well defined,(ii) for the given integer N, there exist positive constants KN and PN (0 < PN

p) such that

(XII.3.4) Iu(h)(x,E)I<KNIEIN (j=1.2,...,n; h=0,1,...),

uniformly for (x, e) in the domain

(XII.3.5) x E D(b), 0 < IEl < pN, I argel < a',

(iii) the sequence {u(') (x, f) : h = 0,1, ...) converges uniformly to (x, E) _(01(x, E), ¢2(x, E), ... , 0n (x, E)) in (XII.3.5), where iZ (x, f) is the C"-valuedfunction with the entries (u(h) (x, E), ... , u,(th) (x, E)).

The limit function (x, e) is independent of N since the successive approximationsare independent of N.

If (i), (ii), and (iii) are proved, it follows that

(XII.3.6) 0l(x,E) = llim u(jh)(x,E) ^-0 (j = 1,2,... ,n)

as e - 0 in the sector S = E : 0 < lei < sup(pN), I argel < a' }, and the functionsN

(x, c) are holomorphicinll

the domain D(b) x S and satisfy (XII.2.14). Settingpf(x,E) = j(x,E) +g2(x,E)(j = 1,... ,n), we obtain a solution v. = p., (x,E)(j =1,2,... , n) of (XII.1.1) which satisfies all of the requirements of Theorem XII-1-2.Thus, the proof of Theorem XII-1-2 will be completed.

To show (i) and (ii), choose two constants KN and pN so that KN > iBNOC and

PNNKN < ry. This is possible since condition (XII.3.2) is satisfied. Now, assumingthat (i) and (ii) are true for ugh-1)(x,E) in (XII.3.5), let us prove thatalso satisfy conditions (i) and (ii). First from (XII.3.3), it follows that

..(h)i_ _' _ I _-._ I uJ I_ _ ,l

(XII.3.7)x j:exp { (t -x.,)]Ri(tu(h-1)E)dt (J = 1,2,....n).

Then, from Lemma XII-2-4, (XII.2.10), (XII.3.4), and the inductive assumption,we conclude that

E)I < C{CKN + BN}IEI N < KNIEIN

for (x, e) in (XII.3.5). Thus, (i) and (ii) are true for ujh) (x, E) (j = 1, 2, ... , n).

To show (iii), from (XII.2.11), (XII.3.3), and Lemma XII-2-4, we obtain

Iu-(h+l)(x,E)-u-(h)(x,E)I :5c8 sup Iiz(h)(t,E)-tL(h-1)(t,E)ItED(6)

in (XII.3.5) for h = 1 , 2, 3, .... Hence, the sequence (u -1 h ) (x, E) : h = 0,1, 2, ... }is convergent to (x, f) uniformly in (XII.3.5). Furthermore, uj = Oj (x, e) (j =1, 2,... , n) satisfy system (XII.2.14). Thus, the proof of Theorem XII-1-2 is com-pleted. 0

XII-4. A block-diagonalization theoremConsider a system of linear differential equations

(XII.4.1) E° fy = A(x, E)y,

where a is a positive integer, y E C", and A(x, c) is an n x n matrix. The entriesof A(x, E) are holomorphic with respect to a complex variable x and a complexparameter c in a domain

(XII.4.2) lxi < ao, 0 < IEI < po, I argEl < no,

where bo, po, and ao are positive numbers. Assume that the matrix A(x, e) admitsa uniform asymptotic expansion in the sense of Poincar6,

00

(XII.4.3) A(x, e) E E"A, (x),V=o

in domain (XII.4.2) as c --# 0 in the sector

(XII.4.4) 0< IEI < po,I

ao,

where the entries of coefficients are holomorphic with respect to x in thedomain

(XII.4.5) Ixl < do.

Suppose that Ao(O) has a distinct eigenvalues A1, A2i ... , At with multiplicitiesn1, n2i ... , nt, respectively (n1 + n2 + - - + ne = n). Without loss of general-ity, assume that Ao(0) is in a block-diagonal form

(XII.4.6) Ao(0) = diag [Al, A21 ... , At] ,

where Aj are n, x n, matrices in the form

(XII.4.7) Aj =A,I",+N, (j=1,2,...,t).Here, I,, is the nj x n, identity matrix and Yj is an n, x nj lower-triangularnilpotent matrix. The main result of this section is the following theorem.

4. A BLOCK-DIAGONALIZATION THEOREM 381

Theorem XII-4-1 ([Si7]). Under assumptions (X11.4.8) and (X11.4.6), thenexists an n x n matrix P(x, e) such that

(i) the entries of P(x, e) are holomorphic with respect to (x, e) in a domain

(XII.4.8) Ix j < 6, 0< kkI < p, l argeI < a,

where 6, p, and a are positive numbers such that 0 < 5 < 60, 0 < p < po and0<or <ao,

(ii) P(x, e) admits a uniform asymptotic expansion

(XII.4.9)00

P(x, e) etPP(x) (P0(0) = I,,)V=0

in domain (XII.4.8) as e - 0 in the sector

(XII.4.10) 0< IkI < p, argel < a,

where the entries of coefficients P,(x) are holomorphic with respect to x inthe domain

(XII.4.11) fix, < 6,

(iii) the transformation

(XII.4.12) y" = P(x, e)zi

reduces system (X11.4.1) to a system

(XII.4.13) e° d = B(x, e)z

with the coefficient matrix B(x, e) in a block-diagonal form

(XII.4.14) B(x, e) = diag[Bj (x, e), B2(x, e), ... , B,(x, e)],

where B , (x, e) is an n, x n, matrix (j = 1, 2, ... , e),(iv) the matrix B, (x, e) admits a uniform asymptotic expansion

00(XII.4.15) Bj (x, e) ^- 1:

e in sector (XII.4.10), where the entries of coef-ficients are holomorphic with respect to x in domain (X11.4.11).

Remark XII-4-2.(a) Set

(XII.4.16) diag [B1v(x), B2 ,(x), ... , Bl,(x)] -

Then, the coefficient matrix B(x, e) of system (XII.4.13) admits a uniform0asymptotic expansion B(x, e)

'=a(b) When a fundamental matrix solution Z(x, e) of (XII.4.13) is known, a funda-

mental matrix solution Y(x, e) of (XII.4.1) is given by Y(x, e) = P(x, e)Z(x, e).(c) In the case when the matrix Ao(O) has n distinct eigenvalues, by Theorem

XII-4-1, we can diagonalize system (XII.4.1).(d) In the case when eigenvalues of the matrix Ao(O) are not completely distinct,

the point x = 0 is, in general, a so-called transition point. In order to studybehavior of solutions in the neighborhood of a transition point, we need amuch deeper analysis of solutions of system (XII.4.1). For these informations,see, for example, [Wash], [Was2], and [Si12].

Proof of Theorem XII-4-1.

We prove this theorem in two steps. The proof is similar to that of TheoremVII-3-1.

Step 1. We show that there exist a positive number 6 (< oo) and an n x n matrixPo(x) such that

(i) the entries of Po(x) and Po(x)-1 are holomorphic in the domain {x : JxI < b}and Po(0) = In,

(ii) the matrix Co(x) = Po(x)-1Ao(x)Po(x) is in a block-diagonal form

(XIL4.17) Co(x) = diag [Col)(x),Co2)(x),... ,C(c)(x)]

where Ca) (x) is an n, x nj matrix such that

Co)(0) = Aj=a,ln,+Arj (J=1,2,...,e).

In fact, two matrices Po(x) and Co(x) must be determined by the equation

(XII.4.18) Ao(x)Po(x) - Po(x)Co(x) = 0.

Set 4(11)(x) 4(12)(x)' 4(13)(x) ... 40(2)

2 1) ) 2) u)

nd

Ao(x)Ao (x

Aat1)(x)

Poll)(x)

Pa21)(x)

4 (x)

Aotz}(x)

Po12)(x)

P )(x)

AA(x) ...o

Ao)(x)

Po's)(x)

Po2)(x)

Ao (x)

Ao r)(x)

Pol() (x)

Po2'(x)Po(x) =

Poti)(x)

oP ' (x) Po )(x)

Poti)(x) 1

,


where A4(jk) (x) and POk) (x) are nj x nk matrices. Furthermore, set PO l) (x) _In, (j = 1,2,... , t). Then,

(XIL4.19) Co ) (x) = Ao(jj) (x) + 1` 4h) (x) p(hj)(x) (j = 1, 2,... , f)h#3

and

(XII.4.20)

Ao(jj)(x)pok)(x) - pok)(x)Cc(k)(x)

+ +AD(jk)(x) = 0h#j,k

(j i4 k)

from ( XII.4.18). Combining (XII.4.19) we obtain

41) (x)p0 k) (X)-

p(1k) (x)("Okk)(x)

+4h)

(x)pOhk)(x))

(XII.4.21)hjAk

"vh)(x) p(hk)(x) + 0 (j # k).h#j,k

Upon applying the implicit function theorem to (XII.4.21), matrices POk)(x) (j Lk) can be constructed. Then, Co(x) is given by (XII.4.19) and (XII.4.17).

Step 2. Now, assume without loss of generality that Ao(x) is in a block-diagonalform

(XII.4.22) Ao(x) = diag [4)(x),4)(x),... ,Afl[)(x)] .

To prove Theorem XII-4-1, it suffices to solve the differential equation

(XII.4.23) EadP(x, e)

= A(x, E)P(x, E) - P(x, E)B(x, E),

where

(XII.4.24)

Set

A(x, E) = Ao(x) + EA(x, E), P(x, E) = In + EP(x, E) and

B(x, E) = Ao(x) + EB(x, e).

A(11)(x,E) A(12)(x,E) ... A(11)(x,E)A(21)(,, ) A(22) (x, E) ... A(21) ('T' E)

A(x, c) _

A([I) (x, E) A(M) (x, E) ... A(t[) (x, E)

P(11)(x,E) P(12)(x,E) ... P(I[)(x'E)P(21) (x, E) P(22) (x, E) ... P(2[) (x, E)

P([1)(x E)p(t2)(x, E) ... PIP (x,


and

B(x, e) = diag [b(') (X, E), b(2) (X, E), . . . , B(') (x, E)] ,

where A(jk) (x) and P(Jk) (x) are nj x nk matrices and BJ (x, c) is an nj x nJ matrix(j, k = 1,... , e). Furthermore, set

(XII.4.25) P(JJ)(x,E) = 0 (j = 1,2,...

Then, equation (XII.4.23) becomes

(XII.4.26)

lB(.)(x, f) = A(Jj)(x,E) + Ey: A(Jh)(x,E)P(hJ)(x,e)

h=1

(j = 1,2,...,t)

and(XII.4.27)

EodP(jk)(x,E) = A(J)(x)P(Jk)(x,E) - P(Jk)(x,E)A(k)(x) + A(jk)(x,E)

+ E E A(jh)(x f)P(hk)(x,E) - P(jk)(x,f)P(k)(x,E) (i k).1hjAk

Combining (XII.4.26) and (XII.4.27), we obtain

Eo dP(J c) (x, E) 4J) x P(Jk) x E - P(Jk) x, e A(k) x + A(Jk) x, E

I

+ E L.:A(Jh)(x, E)P(hk)(x, E) - Ep(Jk)(x, E)A(kk)(x, E)

h=1I

- (2P(Jk) (x, c) A(kh) (x, E)P(hk) (x, E)h=1

(j0k).

Replacing P(jk)(x,E) by an nJ x nk matrix X(Jk), we derive a system of nonlineardifferential equations

o dX (2k) (J) ( ) (jk) - (jk) (k) (X)( ,E

dx= A x X X A x+ A(Jk) x E)

(XII.4.28)

I+ E E A(Jh)(x, C)X(hk) - EX (Jk)A(kk)(x, f)

h=1I

- E2X(jk) FA(kh)(x,e)X(hk)

h=1(j0k)

Consider the entries of X (jk) (j, k = 1, 2, ... , e; j # k) altogether to formsystem of nonlinear differential equations. Upon applying Theorem XII-1-2, wea

5. GEVREY ASYMPTOTIC SOLUTIONS IN A PARAMETER 385

construct an analytic solution of this nonlinear system in a domain (XII.4.8) admit-ting an asymptotic expansion in e as described in Theorem XII-4-1. This completesthe proof of Theorem Xll-4-1. 0

XII-5. Gevrey asymptotic solutions in a parameter

In this section, using Theorems XI-2-3 and the proof of Theorem XII-1-2, weconstruct Gevrey asymptotic solutions of a system of differential equations

(XII.5.1)Ea dx = Ax, y, E),

where x is a complex variable, y E Cn, E is a complex parameter, a is a positiveinteger, and Ax, if, e) is a Ca-valued function of (x, y', f). Define three domains by

(XII.5.2)

A(So) = {x E C : IxI < bo}.

S(ro, no) = {E E , arg E

Also, define two matrices A(x, E) and Ao(x) by

(XII.5.3) A(x, e)

and

(XII.5.4)

respectively.

< ao, 0 < H < ro}.

89

8y1(x,E)

Ao(x) = limo A(x,e),

Oyn (x, 0, f)

We first prove the following lemma.

Lemma XII-5-1. Assume that(i) Ax, y", e) is holomorphic with respect to (x, f, c) in a domain A(bo) x Q(po) x

S(ro, no), where 6o, po, ro, and no are positive numbers,(ii) f (x, y", c) is bounded on A(60) x i2(po) x S(ro, no),

(iii) the matrix Ao(x) defined by (XII. 5.4) exists as e - 0 in S(ro, no) uniformlyin x E A(6o) and Ao(0) is invertible,

(iv) f (x, 0, e) is flat of Gevrey order -r as e -+ 0 in S(ro, no) uniformly in x EA(60), where r is a non-negative number.

Then, there exist three positive numbers b, r, and or such that system (XIL5.1) hasa solution O(x, e) which is holomorphic in (x, t) E A(6) x S(r, a) and that (x, c)is flat of Gevrey order -r as e -' 0 in S(r,a) uniformly in x E L1(6).

Proof.

Note first that

(XII.5.5) I exp1- k1I = expJ-cJfJ-'cos(k(arge))]

for any positive numbers c and k. Note next that assumptions (i) and (iv) imply thatin the case when r = 0, Ax, 0, e) is identically equal to 0 for (t, f) E A(b) x S(r, a).Hence, in this case, 0 is a solution of (XII.5.1). In the case when r > 0, it holdsthat

(XII.5.6) If(x,o,e)I < h exp[-2cIEl-k1

for some positive numbers K and c if (x, e) E A(b) x S(r, a) for sufficiently smallpositive numbers 6, r, and a and if

(XII.5.7)k

(cf. Theorem X1-3-2). Therefore, (XII.5.5) and (XII.5.6) imply that

I exp[ce-k] I If (x, O, E)I = if (x, o, e)I expf clkl-k cos(k(arg e))](XII.S.8)

<K exp(-c{E1,'k]

for (x, e) E A(6) x S(r, a). Note also that

(XII.5.9) cos(k(arge)) > cos(ka) > 0 for f E S(r,a) if ka < 2.

Let us change l/ in (XII.5.1) by

(XII.5.10) Ii = exp(-ce-'16.

Then, (XII.5.1) is reduced to

(XII.5.11) el" = exp[m-kJf(x,eXP[-ce-k]u,e).

Set

f (x+ FJ, e) = f (x, 0, e) + A(x, e)il + E yPfp(x, e).JpI>2

Then,

exp[cE-k1f(x, eXp[-cE-k]u, e)

= eXPfce-kJf (x, 0, e) + A(x, e)U + explc(1 - lpl)E-k]ul fp(x, e)'ip1>2

Using a method similar to the proof of Theorem XII-1-2, we construct a boundedsolution u = tlr(x, e) of (XII.5.11). Therefore, system (XII.5.1) has a solution ofthe form y" _ ¢(x,e) = expl-ce-k]y,(x,e). This completes the proof of LemmaXII.5.1.

The main result of this section is the following theorem, which was originallyconjectured by J: P. Ramis.

Theorem XII-5-2. Assume that(i) f (x, y', e) is holomorphic in (x, y, e) on a domain 0(6o) x f2(po) x S(ro, ao),

where bo, po, ro, and ao are positive numbers,(ii) f (x, y, e) admits an asymptotic expansion F(x, y, e) of Cevmy orders as a --+ 0

in S(ro, ao) uniformly in (x, y-) E A(bo) x f2(po), where s is a non-negativenumber,

(iii) the matrix Ao(x) given by (XIL5.4) is invertible on 0(bo),(iv) it holds that

(XII.5.12) lim f (x, 0, e) = 0 on 0(bo).

Then,(1) (XIL5.1) has a unique formal solution

(X1.5.13) Plx, e) _ eep'e(x)c-

with coefficients p"t(x) which are holomorphic on A(bo),(2) there exist three positive numbers 6, r, and a such that (XII. 5.1) has an actual

solution ¢(x, e) which is holomorphic in (x, e) E A(5) x S(r, a) and that ¢(x, e)admits the formal solution P(x, e) as its asymptotic expansion of Gevrey order

max { 1, s } as e - 0 in S(r, a) uniformly in x E L1(b).a

J11

Proof.

If a positive number & is sufficiently small, for every real number 6, there existsa C"-valued function fe(x,y,e) such that

(a) fe(x, y', e) is holomorphic in (x, y", e) on a domain A(60) x f2(po) x Se(ro, &),where

(XIL5.14) Se(ro,&) = {e: jarge - 61 < &, 0 < fej < ro},

(b) fe(x, y, e) admits an asymptotic expansion F(x, y e) of Gevrey order s ase - 0 in Se(ro, &) uniformly in (x, y) E 0(bo) x f2(po), where s is the non-negative number given in Theorem XII.5.2.

Such a function fe(x, y, e) exists if & is sufficiently small (cf. Corollary XI-2-5). Inparticular, set

(XII.5.15) fo(x, b, e) = f (x, v, e)

Let y = & (x, e) be a solution of the system

(XII.5.16) e° = fe(x, y, e)

such that &(x, e) is holomorphic and bounded in (x, e) E i (b1) x Seri, al) forsuitable positive numbers 61, r1, and al. Using Theorem XII-1-2, it can be shownthat such a solution of (XII.5.16) exists.

Suppose that So,(ri,al) nSo,(ri,al) 0 0. Set

(XII.5.17) /i(x, E) = 4 (x, E) - 4 (x, E)

for (x, E) E A(5j) X {So,(r1,a1)nSo.(rl,a,)}. Then, obi satisfies the following linearsystem

(XII.5.18)

where

E°L = B(x, E)/+ 6(x, E),

(XII.5.19)B(x,E)W(x,E) = fet(x,iez(x,E),E),

6(x,0 =

Note that

(XII.5.20) B(x, c) _

and that, if s > 0, then

JlOfJ (x,tiet(x,E) + (1 -

(XII.5.21) jb(x,E)I <pexp{-vlEl-k}

for (x,E) E 0(6) x {Se,(rl,a,)nS02(rl,a,)}, where

(XII.5.22)k

and u and v are suitable positive numbers (cf. Theorem XI-3-2). From (XII.5.20),it follows that

(XII.5.23) lim B(x, E) = Ao(x),

where the matrix Ao(x) is given by (XII.5.4). If s = 0, the power series f isconvergent in c (cf. Exercise XI-6). Hence, b(x, E) = 0.

If

(XII.5.24) Se(r2,a2) C Se,(ri,a'i)nSo,(ri,ai)

and if positive numbers 62, r2, and a2 are sufficiently small, applying Lemma XII-5-1 to (XII.5.18), the functions j'(x, f) can be written in the form

(XII.5.25) i(x, E) = I (x, E)y'(E) + w(x, E),

where

(XII526) 1.-.f .)l.. 1=0 if s=0,

for (x, e) E A(S2) x Se (r2, a2), K and c are suitable positive numbers, t(x, e) isa fundamental matrix solution of the homogeneous system

(XII.5.27) to dV = B(x, E)v,

and I(e) is a C"-valued function of a independent of x.Assuming that Ao(0) has distinct eigenvalues A1,.12, ... , Al with multiplicities

nl, n2i ... , ne, respectively (n1 + n2 + + ne = n) and that AO(O) is in a block-diagonal form (XII.4.6) with (XII.4.7), where N, (j = 1, 2,... , l) are n, x nj nilpo-tent matrices, respectively, apply Theorem XII-4-1 to block-diagonalize (XII.5.27).More precisely speaking, there exist three positive numbers 53, r3, 03, and an n x nmatrix P(x, e) which is holomorphic in (x, e) E A(53) x So (r3, a3) such that

(i) P(x, e) admits an asymptotic expansion (XII.4.9) in the sense of Poincart ase -+ 0 in S(r3,a3) uniformly in x E A(53),


(XII.5.28) V = P(x, E)z

reduces (XII.5.27) to

(XII.5.29) E° Tdi'

= E(z, E) L,

where E(x, e) is in the block-diagonal form

(XII.5.30) E(x, e) = diag [El (z, e), E2(z, c),. . . , Et(x, E)J ,

with an n1 x n, matrix E J (x, c) for each j = 1, 2, ... , t, and E(x, e) admits

an asymptotic expansion in the sense of Poincar6 as t -+ 0 in

SB(rs,a3) uniformly in x E A(63), where Eo(0) = Ao(0).Let us construct a rhombus 7)(64) with vertices x{'), x(2), xi3i, and xt4i as it is

defined in §XII-2, in such a way that(a) 7)(54) C a(53),(b) it holds that

(XII.5.31)

(Ai(x_r(I))\< _i4> HIx--x""I

E° J IEI°

E° IEI°(A,(z_z(3))\ <i4 A,1 Ix - x3l

for 3=1,...,m',

for j=m'+1,...,1

on the domain V(54) x So(r4,Q4), where 64i r4, and a4 are suitable positivenumbers.

El

Write fin the form z" _ I where z"J E C"' (j = 1, 2, ... , f). Then, (XII.5.29)ztJ

can be written in the form

(XII.5.32) E° L = (j = 1,2... ,f).

For each j, let 4ij(x, e) be a fundamental matrix solution of cvThen, the general solution of (XII.5.29) is given by

= Ej(x, e)z j.

(XII.5.33) %(x,E) _

where

(XII.5.34)xl for j = 1,... , m',x3 for j = m+ I,... , 1.

In this way, we can prove that t 5(x, e) - w'(x, e) is flat of Gevrey order 1 as e - O inS e, (r1, a i) n Sez (ri , a 1) uniformly for x E A(S) if a positive number 6 is sufficientlysmall (cf. [Si6; Proof of Lemma 1 on pp. 377-379]). Therefore, by using TheoremX1-2-3, we can complete the proof of Theorem XII-5-2.

Materials of this section are also found in JSi22J.

XII-6. Analytic simplification in a parameterFor a system of linear differential equations of the form

(E) dx = xkAl ly,

where k is an integer, y E C', and A(t) is an n x n matrix whose entries areholomorphic in a domain

(D) { t : it( < }, (t = x; Ro : positive constant),

G. D. Birkhoff proved the following theorem.

Theorem XII-6-1 ((Bil). There exists an n x n matrix P(t) such that the entriesof P(t) and P(t)-' are holomorphic in domain (D) and that the transformation

(T) y = PMg

reduces system (E) to a system of the form

(Eo) = xkB(z}v,

where B(t) is an n x n matrix whose entries are polynomials in t. Moreover, thematrix P(t) can be chosen so that x = 0 is, at worst, a regular singular pointof (E).

Observe that if we set t = x-3, (E) becomes

dg=

_t-k-2A(t)9.{E1)

dt

6. ANALYTIC SIMPLIFICATION IN A PARAMETER 391

If k < -2, (E1) has the coefficient holomorphic in (D). Hence, there is a funda-mental matrix solution V(t) of (E1) whose entries are holomorphic in (D). Then,

the transformation v = V (1) i reduces (E) to the system f = 6. Therefore, the

main claim of Theorem XII-6-1 concerns the case when k > -1. In this theoremof G. D. Birkhoff, the entries of the matrix B(t) are polynomials in t. However,even though we can choose P(t) so that x = 0 is, at worst, a regular singular pointof (Eo), the degree of B(t) with respect to t may be very large. In order that thedegree of B(t) with respect to t is at most k + 1 so that x = 0 is a singularityof the first kind of system (Eo), we must impose a certain condition on A(t) (cf.Exercises XII-8, XII-9, and XII-10). For interesting discussions on this matter, see,for example, [u2], [JLP], [Ball], and [Ba12j. A complete proof of Theorem XII-6-1is found, for example, in [Si17, Chapter 3].

In this section, we prove a result similar to Theorem XII-6-1 for a system oflinear differential equations

(XII.6.1) f '!Ldx

= A(x,e)y,

under the assumption that a is a positive integer, y" E C", and A(x, e) is an n x nmatrix whose entries are holomorphic with respect to complex variables (x, e) in adomain

(XII.6.2) X E Do, [e[ < 60,

where Do is a domain in the x-plane containing x = 0, and 6o is a positive constant.Let

00

(XII.6.3) A(x,e) = E ekAk(x)k=0

be the expansion of A(x, e) in powers of e, where the entries of coefficients Ak(x) are00

holomorphic in Do. We assume that the series >26o (Ak(x)I is convergent uniformlyk=0

in Do. The main result of this section is the following theorem, which was originallyproved in [Hsl].

Theorem XII-6-2 ([Hsl]). For each non-negative integer m, there is an n x nmatrix P(x, e) satisfying the following conditions:

(i) the entries of P(x, e) are holomorphic in (x, e) in a domain

(XII.6.4) x E D1i jeJ < 60,

where V1 is a subdomain of Do containing x = 0,(ii) P(x, 0) = In for X E V1 and P(0, e) = In for je[ < 6o,

(iii) the system (XII.6.1) is reduced to a system of the form

m o-1dii(XII.6.5) e° _ E{kA(X) + em+1 r` c'Bk(: W

) uk=0 k--O



(XII.6.6) P(x, e)u,

where Bk(x) (k = 1, 2, ... , or - 1) are n x n matrices whose entries are holo-morphic for x E D1.

Remark XII-6-3. In case when a = 1, (XII.6.5) becomes

diled nE CkAk(x)+em+IB0(x)I!.

k=0

In particular, if a = I and m = 0, (XII.6.5) has the form

dile = {Ao(x) + eB0(x)} u.

It should be noticed that in Theorem XII-6-2, if we put m = 0, then the right-handside of (XII.6.5) is a polynomial in a of degree at most a without any restrictionson A(x, e). We prove Theorem XII-6-2 by a direct method based on the theory ofordinary differential equations in a Banach space. (See also [Si8j.)

Proof of Theorem XII-6-2.

The proof is given in three steps.

Step 1. Put

(XII.6.7)

where

(XII.6.8)

a m+oP(x, e) = I + Em+1 [-` ekPk(x) and B(x, e) = E EkBk(x),

k=0 k--0

Bk(x) _ J A&. (x) (k = 0,1, ... , m),Bk-m-1(x) (k = m+ 1,m+2,... ,m+a).

From (XII.6.1), (XII.6.5), and (XII.6.6), it follows that the matrices P(x,e) andB(x, e) must satisfy the equation

(XII.6.9)QdP

e = A(.x, e)P - PB(x, e).

From (XII.6.3), (XII.6.7), (XII.6.8), and (XII.6.9), we obtain

k

(XII.6.10)Am+1+k(x) - Bm+1+k(x)+ E{Ak-h(x)Ph(x) - Ph(x)Bk-h(x)}

h=O

(k=0,1,...,a-1)

6. ANALYTIC SIMPLIFICATION IN A PARAMETER 393

and

(XII.6.11)

where

o+k

E Ph(x)Bo+k-h(x) (k = 0,1,2,...),

h=k-m

(XIL6.12) Ph(x) = 0 if h < 0.

It should be noted that the formal power series P and B that satisfy the equation(XII.6.9) are not convergent in general. In order to construct P as a convergentpower series in e, we must choose a suitable B. To do this, first solve equation(XII.6.10) for B,,,+1+k(x) to derive

(XII.6.13)

o+kdPk(x)

+k(x) + Ad+k-h(x)Ph(x)+1+=A m odx

h=o

Bm+1+k(x) = Am+I+k(x)+Hm+1+k(x;P0,P1,... Pk)(k=0,1, ..,Q-1),

where H, are defined by(XII.6.14)

H , = 0, (k = 0,1, ... , m),k k

Hm+I+k(x; Po, PI, ... , Pk) = E{Ak-h(x)Ph - PhAk-h(x)) - E PhHk-h,h=0 h=0

(k = 0,1,...,a - 1).

Denote by P an infinite-dimensional vector {Pk : k = 0, 1, 2,... }. Then, by substi-tuting (XII.6.13) into (XII.6.11), we obtain

(XII.6.15)

where

dPk(x)dx

= fk(x; P) (k = 0,1, 2, ... ),

o+k o+k

fk(x; P) = Am+1+o+k(x) + E Ao+k-h(x)Ph - 1: PhA,,+k-h(x)h=0 h=k-m

(XII.6.16)a+k

- E PhHo+k-h(x;P) (k=0,1,2,...).h=k-m

Denote by -F(x; P) the infinite-dimensional vector { fk(x; P) : k = 0,1, 2,... }.Then, equation (XII.6.15) can be written in the form

(XII.6.17)dPdx

= F(x; P).

Solve this differential equation in a suitable Banach space with the initial conditionP(0) = 0. Actually, we solve the integral equation

(XII.6.18) P(x) = f0F(;P())de.This is equivalent to the system

(XII.6.19) Pk(x) = ffk(;P())de (k = 0,1, 2, ... ).

If P is determined, then the matrices Bk are determined by (XII.6.13) and (XI1.6.14).

Step 2. We still assume that A(x, c) is holomorphic in domain (XII.6.2). Denoteby B the set of all infinite-dimensional vectors P = {Pk : k = 0, 1, 2.... } such that

(i) Pk are n x n matrices of complex entries,00

(ii) j:boilPkll < oo, where IlPkll is the sum of the absolute values of entriesk=0

of Pk.

For each P, define a norm 11P11 by

(XII.6.20)00

IIPII = Ebo IlPkllk=0

Then, we can regard B as a Banach space over the field of complex numbers.Now, we can establish the following lemma.

Lemma XII-6-4. Let F(x; P) be the infinite-dimensional vector whose entriesfk(x,P) are given by (XII.6.16). Then, for each positive numbers R, there existtwo positive numbers G(R) and K(R) such that

(XII.6.21)

and

(XII.6.22)

by

flF(x; P) II 5 G(R) for IIPII <- R

II.(x; P) - F(x; P)ll 5 K(R)IIP - PII for IIPII 5 R, IIPI{ 5 R.

In order to prove this lemma, consider a formal power series of e which is defined

(XII.6.23) .r(x,P,E) _ Ekfk(x;P)-k

Then, using (XII.6.16), we obtain(XII.6.24)

00

.F(x, P, e) _ >2 EkA,,,+1+o+k(x) + 1 \ EkAk(x)/(e'Pk)

k=0 l k=0

r

k=0

a-1 kC* M+V

Ek [Ak(x) +Ik(x; PA)- E Ek hE Ak-h Ph k=o Ekpk/ k ok=O

0-1 kl- > ek E Ph[Ak-h(x) + Hk-h(x;P)1 }.

k=0 h=O

Hence, Lemma XII-6-4 follows immediately.

EXERCISES XII 395

Step 3. We construct the matrix P(x, e), solving the integral equation (XII.6.18)by the method of successive approximations similar to that given in Chapter I. Byvirtue of Lemma XII-6-4, we can construct a solution P(x) in a subdomain Dl ofDo containing x = 0 in its interior. Since (XII.6.18) is equivalent to differentialequation (XII.6.15) with the initial condition P(0) = 0, the solution P(x) givesthe desired P(x, e). The matrix B(x, c) is given by (XII.6.13) and (XII.6.14). 0

EXERCISES XII

00

XII-1. Find a formal power series solution y = F, e'd,,,(x) of the system ofm=0

differential equations

LY = Ay" +CO

> Embm(x),m=0

where y E C", A is an invertible constant n x n matrix, and 5,n (x) and bm(x) areC"-valued functions whose entries are holomorphic in x in a neighborhood of x = 0.

XII-2. Using Theorem MI-4-1, diagonalize the system

dy =11-X0 1+x1

e dx ex y'where [1.

XII-3. Using Theorem XII-4.1, find two linearly independent formal solutions ofeach of the following two differential equations which do not involve any fractionalpowers of e.

(1)E2d2

Y2 + y = eq(x)y, (2) a2 + y = eq(x)y,

where q(x) is holomorphic in x for small jxj.

Hint. If we set '62 = e, differential equation (2) has two linearly independentsolutions e4=/00(x, 0) and a-1/190(x, -/3). The two solutions

= Q reiz/°4(x, Q) - -Q),

do not involve any fractional powers of e.

XII-4. Let x be a complex independent variable, y" E C", z" E C, e be a complexparameter, A(x, y, z, e) be an n x n matrix whose entries are holomorphic withrespect to (x, y, z, e) in a domain Do = {(x, y, X, e) : jxj < ro, jyi < Pi, Izl < p2, 0 <jej < ao, j arg ej < flo}, f (x, y, z, e) be a C-valued function whose entries areholomorphic with respect to (x, y, 1, e) in Do, and #(x, ,F, e) be a C"-valued functionwhose entries are holomorphic with respect to (x, z, e) in the domain Uo = ((x, zl, e) :jxj < ro, jzj < p2,0 < jej < ao, j argej < /3o}. Assume that the entries of the matrix

A(x, if, z, c), the functions f (x, y, z, e), and §(x, .F, e) admit asymptotic expansions asf - 0 in the sector So = {E : 0 < IEI < ao, I arg EI < /30} uniformly in (x, y, z) in thedomain Ao = {(x, y, z) : IxI < ro, Iv'I < pl, I1 < p2} or (x, z-) in Vo = {(x, z) : I xI <ro, Izl < p2}. The coefficients of those expansions are holomorphic with respect to(x, y", I) in Do or (x, z-) in V0. Assume also that det A(0, 0, 0, 0) 0 0. Show that thesystem

f dy = A(x, y, z, E)y + E9(x, Z, f), = Ef (x, g, Z, f)

has one and only one solution (y, z) = (¢(x, f), e)) satisfying the followingconditions:

(a) the entries of functions O(x, E) and '(x, f) are holomorphic with respect to(x, e) in a domain {(x, f) : I xI < r, 0 < lei < a, I arg EI < /3} for some positivenumbers r, a, and 0 such that r < ro, a < ao and /3 < /io,

(b) the entries of functions (x, c) and i/ (x, c) admit asymptotic expansions inpowers of f as c - 0 in the sector if : 0 < lei < a, I argel < (3} uniformlyin x in the disk (x : Ixi < r), where coefficients of these expansions areholomorphic with respect to x in the disk {x: Ixi < r},

(c) 0(0,c) = O for f E {E : 0 < IEI < a, I arg EI < ,0}.

Hint. Use a method similar to that of §§X11-2 and XII-3.

rXII-5. Find the following limits: (1) lim

1

/ f (t) sin 1at dt and (2) limc-0 0 \ E

f (t) sine I I dt, where a is a nonzero real number and f (t) is continuousjand continuously differentiable on the interval 0 < t < 1.

XII-6. Discuss the behavior of real-valued solutions of the system

Edg = rEE -1 +Ex1 y as e -40, where y"_ [y2J

XII-7. Discuss the behavior of real-valued solutions of the following two differential

equations as f O+: (1) E2d3 y + 2 + xty = O, (2) E3 1 (1 x)y = O.

XII-8. Assume that(i) the entries of a C"-valued function Ax, y", e) are holomorphic with respect to

(x, y", e) in a domain A1(Eo) X II(po) X O2(ro), where b0, po, and ro are positivenumbers and

0l(6o)={xE(C:IxI <bo}, I(po)={iEC":Iv7<po},A2(ro) = {E E C : IEI < ro},

Of -(ii) the matrix Ao(x) = (x, 0, 0) is invertible on 0(bo),

(iii) f(x, 0, 0) = 0 on A(bo),(iv) a is a positive integer.

EXERCISES XII 397

Denote by S(r, a, 0) the sector (EEC : 0 < IEI < r, I arg e - 01 < a). Show thaty

(1) the system (S) c'±dx = f (x, y, E) has a unique formal solution p(x, E) _+00

with coefficients p"l(x), which are holomorphic in A(50),c=1

(2) for any real number 0, there exist three positive numbers b, r, and a such that(S) has an actual solution ¢(x, e), which is holomorphic in (x, E) E 0(b) xS(r, a, 0), and that ¢(x, c) has the formal solution p(x, e) as its asymptotic

expansion of Gevrey order 1 as c -. 0 in S(r, a, 0) uniformly in x E 0(b)or

Hint. This is a special case of Theorem XII-5-2.+M

XII-9. Assume that p(x,E) _ E Emp',,,(x) is a formal solution of a system (S)M=1

dyca

dx= f (x, 17, f), where a is a positive integer, y" E C, Ax, y", e) is a C"-valued

function whose entries are holomorphic in a neighborhood of (x, y, f) = (0, 0, 0),and the entries of pm(x) (m = 0.1, ...) are holomorphic in a disk Ixi < ro, wherero is a positive ember. Assume also that p(x, E) E {G[[E]]3}" uniformly for Ixi < ro

and that 0 < s < 1. Show that if S is an open sector in the E-plane with vertex ate = 0 and whose opening is smaller than sir, there exists two positive numbers riand r2 and a solution $(x, e) of (S) such that the entries of q,(x, e) are holomorphicin (x, E) for Ixt < r1, IEI < r2,( E S, and that (x, E) admits the formal solutionp(x, c) as the asymptotic expansion of Gevrey order s as e -+ 0 in S n {I-El < r2}uniformly for IxI < r1.

Note. No additional conditions on the linear part of Ax, y", c) are assumed.+00

XJI-10. Assume that p(x, c) = F E"`p""m(x) is a formal solution of a systemm--O

e = f (z, y, E), where y E C", f (x, y, e) is a C"-valued function whose entries are

holomorphic in a neighborhood of (x, y, c) = (0, 6, 0), p".. E C[(x]]" (m = 0,1,... ),p"o(0) = 6, and p(O, e) is convergent. Show that p(x, e) is convegent for ]x) < r andIEI < p for some positive numbers r and p.

Hint. Regard p(ET, e) as the solution of the initial-value problem y = f (.ET, y, E),d

00) = P(0. E.

XII-11. Let t and e be complex variables, y" E G", and the entries of a G"-valued function f(t, y, e) be holomorphic with respect to (t, y. E) in a domain 1)o =

E) : ICI < do, -oo < tt < oo, Jyj < po, 0 < IEI < ro, I arg E) < ao). As-00

some that f(t, y, E) E E'f,,,(t, y-) as e -+ 0 in the sector So = {E : 0 < IEI <m=0

ro, I arg EI < ao } uniformly with respect to (t, y-) in the domain Do = {(t, y) : J 3'tI <

do, -oo < Iftt < +oc, Iyl < po}, where the coefficients fm(t,y-) are holomorphic in

Ao. Assume also that f(t, y, e) and the coefficients fm(t, y) are periodic in t of

period 1. Show that the differential equationdydt

= 27riy + e[y + ef (t, y, e)] has a

periodic solution U = e) of period 1 such that(a) the entries of (t, e) are holomorphic with respect to (t, e) in a domain D =

{ (t, e) : I `3'tJ < d, -oo < Rt < +oo, 0 < JeJ < r, I arg el < a} for some (d, r, a)such that0<d<do,0<r<r0,and0<a<ao,

00

(b) j(t, e) ^ - - E em&m(t) as a -+ 0 in the sector S = If : 0 < Jej < r, J arg el <M=0

a}, where coefficients pm(t) are holomorphic and periodic of period 1 in thedomain D' = It : J:33tI < d, -oo < Rt < +oo}.

Hint.Step 1. Construct the solution +li(t, c", e) to the initial-value problem

(IP) dt = 27riy+ e[y+ ef (t, y, e)], 9(0) = c.

Substep 1. First, construct a formal solution+00

+G(t, , e) _ E em'+Gm(t, l = e2x1cd +M=0

ete2attC + O(e2).

The coefficients 1& (t, a) are holomorphic in a domain

S2o = {(t, cl : 13'tI < do, -oo < Rt < +oo, Icl < ryo),

where -yo is a positive number. To prove this, set y = e2i't(c + CZ-) to change (IP)to

(IP') d = 6 + e[z + e-2i't f (t, e2"`t(c + ez_), e)J, z(0) = 0.

The function f (t, e2x't (c" + ez-),e) admits an asymptotic expansion in powers of e(cf. Proof of Theorem XI-1-8) whose coefficients are polynomials in the entries ofthe vector z. Coefficients ci can be calculated successively.

Substep 2. Show that iP(t, c", e) V;(t, c, e) as e -. 0 in a sector S = {e : 0 < je1 <r, I argel < a} uniformly in a domain 1Y = {(t,c) : J)tj < d, IRtI < T, Icl < ry},where T is an arbitrary positive number and -y > 0 depends on T. In this argument,the Gronwall's inequality (Lemma 1-1-5) is useful.

Step 2. Solve the equation 1 (1, cE, e) = E. This equation has the form

(E) c" = eh(c, e),

where h(c", e) admits an asymptotic expansion as a -' 0 in the sector S uniformlyfor Icd < y. To solve this, we can use successive approximations as follows:

4W = 0, Ch(e) = e46y_1(e),e).

Then, the approximations 64(e) converge to a limit c(e). Using Theorem XI-1-12,we can conclude that ee) admits an asymptotic expansion in powers of e.

EXERCISES XII 399

Step 3. The particular solution ¢(t, e) = j(t, cue), e) is the periodic solution satis-fying all of the requirements.

XII-12. Consider a system of differential equations

(s)dy

dx= xka(x,e)y,

where x is a complex independent variable, a is a complex parameter, k is a non-negative integer, and y is an unknown element in a Banach algebra U over the fieldC of complex numbers with a unit element I . Assume that

+00

(a) a(x, e) _ E x-ma,.,(e),m=o

where am(e) E U and these quantities are holomorphic and bounded with respecttoe in a sector S = {e : 0 < je' < bo, I argel < bl}, and the series (a) is convergentin norm for lx[ > Ra uniformly for e in S. Also, assume that a(x, e) admits anasymptotic expansion in powers of a uniformly for x in jxl > Ro as e -. 0 in S.Show that if a positive integer N is sufficiently large, there exist elements p(x, e),bo(e ), ... , bM (e) of U such that

(i) p(x, e) is holomorphic and bounded in S and large [x[,(ii) p(x, e) admits an asymptotic expansion in powers of e uniformly for large [xi

a positive integer and the quantities bo(e), ..., bkf(e) are holomorphicin S and admit asymptotic expansions in powers of e as a -+ 0 in S,

(iv) m>k+l,(v) the transformation y = [I + x-(N+1)p(x, e)Ju changes (s) to

(S')du

dx

N ME x-mam(e) + x- (N+1) E x-mb,,,(e)I U.xk

Comment and Hint. See [Sill; in particular Theorem 2 on p. 157]. The pointx = 0 is not necessarily a regular singular point of (s') as in Theorem XII-6-1.

Step 1. The main idea is, assuming that N > k, to solve equations of the followingforms:

-k d((x) +-(k+l)Q(x) -(N + 1) )Q( ) Q( B(( ) ( ) )(1)

x x a x -x x a x - x

+ x-(N+1) [a(x)Q(x) - Q(x)B(x)] = F(x)

and

(N + 1)x-(k+l)Q(x) - x-kdQ(x) + a(x)Q( ) ( ) - B( )) - Q((II)

x a x xx

+ x-(N+1) [a(x)Q(x) - Q(x)$(x) - 'Y(x)B(x)) = F(x),


where(1) Q(x) is an unknown quantity which should be a convergent power series

in x-1,(2) B(x) is an unknown quantity which should be a polynomial in x-1 of

degree k,(3) a(x) is a given polynomial in x-1 of degree N,(4) a(x) is a given convergent power series in x-1,(5) 0 is a given polynomial in x-1 of degree k,(6) -r(x) is a given convergent power series in x-1,(7) F(x) is a given convergent power series in x-1.

To solve these equations, first express B in terms of Q. In fact, settingtoo k N

Q(x) _ E x-'Q,,,. B(x) _ F, x-mBm, a(x) _ E X -0m,m=0 m=0 m=0

we obtain

Bm =M

1:fam-hQh - Qham-hl,h=1

Then, we can write (I) and (II) in the form1

m = 0,1,..., k.

(III) QM = (m>0),

where 9,,, is either quadratic or linear in Q (n > 0).

Step 2. If N > 0 is sufficiently large, we can solve (III) by defining a norm for a+oo +oo

convergent power series P(x) = E x-'P,,, by 11P11 = E p'11 P,,, 11, where p is am0 m=0

sufficiently small positive number.

Step 3. Using Steps 1 and 2, we can construct a formal power series q(x, e) _+00E e'ge(x) such that the coefficients qt(x) are holomorphic and bounded in a diskt=0A = {x : Ix] > R > 0} and that the formal transformation y = [I+x-(N+1)4(x, e)Juchanges (s) to (s').

Step 4. Find a function q(x, e) such that q is holomorphic and bounded in 0 x Sand that q(x, e) q(x, e) as e --+ 0 in S uniformly for x E A. Then, transformationY= [I + x-(N+1)q(x, e)Iv changes (s) to(s")dv

dx= xk

where

and

FM=

x-mam(E) + x- (N+1) E x-mbm(E) + Em=0 m=k+1

k +oo

bm(E) ^_- b,1 (E) as E 0 in S

x-mbm(0)1 u,

x-mam(E) ^- 0m=k+1

as 0 in S uniformly for x E A.

EXERCISES XII 401

Step 5. Using again Steps 1 and 2, find a function r(x, e) such that(a) r(x, e) is holomorphic and bounded for jxj > R' > 0 and e E S,(b) r(x, e) = 0 as a - 0 in S uniformly for jx[ > R',(c) for a sufficiently large M' > 0, the transformation v = [I +x-(h1'+1)r(x,e)jw

changes (s") to

dw = xk ` x-mam(f) + xN M

-(N+1) E ? bm(e) w,Lm=oM-

whereb,,,(e)-0ase-0 inSform>k+1.:XII-13. Let A(t) be a 2 x 2 matrix whose entries are holomorphic in a disk It

it[ < po} such that the matrix

(M) P ()' {kA(!)P(1) -P (-X1)1

is not triangular for any 2 x 2 matrix P(t) such that the entries of P(t) and P(t)-1are holomorphic in the disk D = It : jtj < po}. Show that there exists such a 2 x 2

matrix P(x) for which matrix (M) has the form xk B Gl)

with a 2 x 2 matrix B(t)

wh ose entries are polynomials in t and whose degree in t is at most k + 1.

Hint. See [JLP]. This result can be generalized to the case when A(t) is a 3 x 3matrix (cf. [Ball) and [Bal2J).

XII-14. Let P(t) be a 2 x 2 matrix such that the entries of P(t) and P(t)-1 are

holomorphic in a disk V = it : jtj < po} and that the transformation y" = P (!) u"X

changes the system

dil [ l +x-1 0 1

dxI.

x-3 1-x'1Jy'

to

dilds Blx/u, u =

y =

u1

112

[ i12,

with a 2x 2 matrix B(t) whose entries are polynomials in t. Show that the degreeof the polynomial B(t) is not less than 2. Also, show that there exists P(t) suchthat the degree of the polynomial B(t) is equal to 2.

Hint. To prove that there exists P(t) such that the degree of the polynomial B(t)

is equal to 2, apply Theorem V-5-1 to the system t d'r =L

-17+ at1 +tbt ] v with.ii t

suitable constants a, Q, y, and b. To show that the degree of the polynomial B(t)is not less than 2, assuming that the degree of the polynomial B(t) is less than 2,derive a contradiction from the following fact:

The transformation y = 0 0 u changes the system

dy rl+x'1 0 1

dx l x-3 xtJ',to

dil= [lx-I

1-x-11u, u = IuaJ.

dx

XII-15. Let A(t) be a 2 x 2 triangular matrix whose entries are holomorphic in adisk it : Its < po}. Show that there exists a 2 x 2 matrix Q(x) such that

(i) the entries of Q(x) and Q(x)-I are holomorphic for jxi > o and meromorphic

at x = oo,

(ii) the transformation y" = Q(x)t changes the system dx = xkA\ x/ y to _

xkB 1 f u" with a 2 x 2 matrix B(t) whose entries are polynomials in t and

whose degree in t is at most k + 1.

Hint. See IJLP]. This result can be generalized to the case when A(t) is a 3 x 3matrix (cf. [Ball] and [Bal2j).

CHAPTER XIII

SINGULARITIES OF THE SECOND KIND

In this chapter, we explain the structure of asymptotic solutions of a system ofdifferential equations at a singular point of the second kind. In §§XIII-1, XIII-2, andXIII-3, a basic existence theorem of asymptotic solutions in the sense of Poincar6is proved in detail. In §XII-4, this result is used to prove a block-diagonalizationtheorem of a linear system. The materials in §§XIII-I-XIII-4 are also found in[Si7J. The main topic of §XIII-5 is the equivalence between a system of lineardifferential equations and an n-th-order linear differential equation. The equivalenceis based on the existence of a cyclic vector for a linear differential operator. Theexistence of cyclic vectors was originally proved in [De1J. In §XIII-6, we explain abasic theorem concerning the structure of solutions of a linear system at a singularpoint of the second kind. This theorem was proved independently in [Huk4] and(Tull. In §XIII-7, the Newton polygon of a linear differential operator is defined.This polygon is useful when we calculate formal solutions of an n-tb-order lineardifferential equation (cf. [Stj). In §XIII-8, we explain asymptotic solutions in theGevrey asymptotics. To understand materials in §XIII-8, the expository paper[Ram3J is very helpful. In §§XIII-I-XIII-4, the singularity is at x = oo, but from§XIII-5 through §XIII-8, the singularity is at x = 0. Any singularity at x = oo is

changed to a singularity of the same kind at = 0 by the transformation x = I

XIII-1. An existence theorem

In §§XIII-1, XUI-2, and XIII-3, we consider a system of differential equations

(XIII.1.1) = x'f,(x,L'i,v2,... ,un) (.7 = 1,2,... n),

where r is a non-negative integer and ff (x, v1, t,2,... , vn) are holomorphic withrespect to complex variables (x, v1, v2, .... vn) in a domain

(XIII.1.2) JxJ>No, JargxI<ao, Jv21<6o (j=1,2,...,n),

No, oo, and 6a being positive constants. Set

n

(XIII.1.3) f2 (x, ffo(x) + E ain(x)va + f, ,(x)v +

h=1 Ip1?2

where v" E C" with the entries (vl, V2.... , vn).We look at (XIII.1.1) under the following three assumptions.

403

404 XIII. SINGULARITIES OF THE SECOND KIND

Assumption I. Each function f j (x, V-) (j = 1, 2,... , n) admits a uniform asymp-totic expansion

(XIII.1.4) fi(x,v1 = >fjk(la k00

k=0

in the sense of Poincare as x - oo in the sector

(XIII.1.5) Jx j > No, J argxj < cto,

where coefficients fjk(v-) are holomorphic in the domain

(XIII.1.6) jvl < bo.

Furthermore. we assume that

(XIII.1.7) fio(d) = 0 (j = 1, 2,... , n).

Observation XIII-1-1. Under Assumption I, fo(x), ajh(x), and fl,(x) admitasymptotic expansions

oe

f}0(x) f70kx-k,

k=1

a,h(x) ti Eajhkx-

00

k=0

00

fjp(x) ti 1: fjpkx-k,k=0

(j, h = 1, 2,... , n)

as x oo in sector (XIII.1.5) with coefficients in C. Let A(x) be then x it matrixwhose (j, k)-entry is ajk(x), respectively (i.e., A(x) = (a,h(x)). Then, A(x) admitsan asymptotic expansion

00

(XIII.1.10) A(x) '=- E x-kAkk=0

as x - oo in sector (XIII.1.5), where Ak = (ajhk). The following assumption istechnical and we do not lose any generality with it.

Assumption II. The matrix A0 has the following S-N decomposition:

(XIII.1.11) Ao = diag [µ1,µ2, ... 44n] + N,

where µ1,µ,2, ... , µ are eigenvalues of Aa and N is a lower-triangular nilpotentmatrix.

Note that WI can be made as small as we wish (cf. Lemma VII-3-3).The following assumption plays a key roll.

1. AN EXISTENCE THEOREM 405

Assumption III. The matrix A0 is invertible, i.e.,

(XIII.1.12) p,54 0 (j=1,2,...,n).

Set

(XIII.1.13) w = arg x, w , = arg p3 1 ,2 , .n ) .

and denote by D, (N,'y,q) the domain defined by

D3 (N, y, q) x : Ixi > N, jwj < ao,(XIII.1.14) l

(2q+)1r_v}(2q_)r+Y<wj+(r+1)w< ,whereq is an integer, N is a sufficiently large positive constant, and -y is a sufficiently

small positive constant. For each j, there exists at least one integer q, such thatthe real half-line defined by x > N is contained in the interior of V., (N, y, qj). Set

n

(XIIL1.15) 7)(N,7) = n V, (N,y,qj).=1

In §§XIII-2 and XIII-3, we shall prove the following theorem.

Theorem XIII-1-2. If N-1 and y are sufficiently small positive numbers, thenunder Assumptions 1, II, and III, system (XIII. 1.1) has a solution

(XHI.1.16) vj = pj (x) U = 1,2,... , m),

such that(i) p,(x) are holomorphic in D(N,-y),

(ii) p, (x) admit asymptotic expansions

00

(XIII.1.17) Pi (z) >2 P,kx-k (j = 1, 2, ... , m)k=1

as x - oo in 7)(N,7), where p,k E C.

To illustrate Theorem XIII-1-2, we prove the following corollary.

Corollary XIII-1-3. Let A(x) be an n x n matrix whose entries are holomorphicand bounded in a domain Ao = {x : lxi > Ro} and let f (x) be a C'-valued functionwhose entries are holomorphic and bounded in the domain Do. Also, let A1,A2,... An be eigenvalues of A(oo). Assume that A(oo) is invertible. Assume alsothat none of the quantities Aje1ka (j = 1, 2,... , n) are real and negative for a realnumber 9 and a positive integer k. Then, there exist a positive number e and a

solution y"= J(x) of the system d-.yi = xk-1A(x)y+x-1 f(x) such that the entries of

O(x) are holomorphic and admit asymptotic expansions in powers of x-1 as x -+ 00

in the sector S = {x : IxI > Ro, I argx - 01 < Zk + e}.

Proof.

The main claim of this corollary is that the asymptotic expansion of the solutionis valid in a sector I argx-8 < 2k +e whose opening is greater than k So, we look

at the sector D(N, y) of Theorem XIII-1-2. In the given case, ao = +oo, r = k -1,and µ, = Aj (j = 1,... , n). The assumptions given in the corollary imply that

w, + k6 -A (2p + 1)7r, where wj = arg Aj (j = 1, ... , n),

for any integer p. Therefore, for each 3, there exists an integer q j such that

either - it < w, - 2qjr, + k8 < 0 or 0 < w, - 2gjir + k8 < rr.

Therefore, either

3r. < wj - 2q, 7r + k8 - 22

- < w) - 2q,rr + kO + < 2

or

2<w) - 2 q , i r + k 8 - 2 < wj - 2gjrr + k8 + 2 < 32T.

This that a sector S = x : x > &,proves { , I argx - 81 <2nk

+ e is in D(N, y)

for a sufficiently large Ro > 0 and a sufficiently small e > 0 if we use xe-i8 as theindependent variable instead of x.

XIII-2. Basic estimates

In order to prove Theorem XIII-1-2, let us change system (XIII-1-1) to a systemof integral equations.

Observation XIII-2-1. Expansion (XIII.I.17) of the solution pj (x)

00

(XIII.2.1) uJ = Epjkx_k (j = 1,2,... n)k=1

must be a formal solution of system (XIII.1.1). The existence of such a formalsolution (XIII.2.1) of system (XIII.1.1) follows immediately from Assumptions Iand III. The proof of this fact is left to the reader as an exercise.

Observation XIII-2-2. For each j = 1,2,... , n, using Theorem XI-1-14, let usconstruct a function z3(x) such that

(i) z j (x) is holomorphic in a sector

(XIII.2.2) In l > No, I argxl < ao,

where No is a positive number not smaller than No,

(ii) zj (x) anddz)

admit asymptotic expansions

(XIII.2.3) z,(x) L.f,kx-k anddz)

E(-k)P,kx-k-1

k=1

as x - oo in sector (XIII.2.2), respectively.

Consider the change of variables

k=1

(XIII.2.4) v, = u, +z,(x) (j = 1,2,... ,n).

Denote (zl, z2, ... , z,a) and (u1, U2,. - - , by i and u, respectively. Then, u' sat-isfies the system of differential equations

(XIII.2.5)

where

du,= x'g,(x,U-) (.7 = 1,2,... ,n),

dx

Set

m

00(XIII.2.6) g,(x, u") = go(x) + >2 b,k(x)uk + E b,p(x)iI (j = 1, 2,... , n).k=1 jp1>2

In particular,

(XIII.2.7) 92o(x) = f, (x, a) - x-' d dxx)0 (j = 1, 2,... , n)

and

b3k(x)-ajk(x)=O(IxI-') (j,k=1,2,...,n)as x -+ oo in sector (XIII.2.2). Thus,

()III.2.8) b,k(x) = a,k(oc) +o(IxI-1) (j, k = 1,2,... , n)

as x -' oo in sector (XIII.2.2).

Observation XIII-2-3. Set

(XIII.2.9) g1(x,u") = u, uj +R1(x,u') (j = 1,2,... ,n).

Then, by virtue of Assumption III, (XIII.2.7), and (XIII.2.8), for sufficiently smallpositive numbers No 1 and 5, there exists a positive constant c, independent of j,such that for any positive integer h, the estimates

(XIII.2.10) lR3(x,u)I ( clui +bhlxl-(n+1) (j = 1,2,... n)

and

(XIII.2.11) IR,(x,u) - R, (x,1U') I < clu"- u l (j = 1,2,... ,n)

hold whenever (x, u") and (x, u"') are in the domain

(XIII.2.12) lxl > No, Iargxl <oo, luil <6 (j =1,2,... ,n).

Here, by, is a positive constant depending on h. Furthermore, by virtue of Assump-tion II, it can be assumed without loss of generality that the constant c satisfiesthe condition

(XIII.2.13)r+1c<

H,

where H is a positive constant to be specified later.From (XIII.2.5) and (XIII.2.9), it follows that

(XIII.2.14)dds = X'" [µuj + R, (x, u)]

Change system (XII/I.2.14) to a system of integral equations

(XIII.2.15) u, = J trR, (t, u) exp Ir - 1(tr+l _ xr+1)1 dt (j = 1, 2,... , n),t J

where the paths of integration L3 start from x. The paths of integration mustbe chosen carefully so that uniformly convergent successive approximations canbe defined in such a way that the limit is a solution uf(x) of (XIII.2.15) whichsatisfies the conditions (i) u , (x) (j = 1, 2, ... , n) are holomorphic in D(N,'y) and(ii) u., (x) ^ _ - 0 (j = 1, 2, ... , n) as x - oo in D(N, ry) for suitable positive numbersN and ry.

Hereafter in this section, we explain how to choose paths of integration on theright-hand side of (XIII.2.15).

Observation XIII-2-4. Since for each j, the domain Dj (N, y, q,) contains thereal half-line defined by x > N in its interior, their common part D(N, ry) is givenby the inequalities

(XIII.2.16) l xI > N, -t < argx < e',

where t and f are positive constants. It is noteworthy that if x E D(N, y), then xsatisfies the inequalities

(XIII.2.17)argxI <ao, - 2r+y<wj +(r+1)argx< 2r-y

(j=1,2,...,n),

provided that w1 = arg u, are chosen suitably. Therefore,

(XIII.2.18)

r <w2 - (r+1)f <wj +(r+1)argx2

2r-y<w, +(r+1)P'< 3

Moreover, since D(N,1) is the common part of Dj (N, y, qj), the equalities

(XIII.2.19)

and

(XIII.2.20)

-Zr+y =w3 -(r+ 1)f

3 r-y=Wh+(r+1)f

hold for some j and h. This implies that t and f depend on y. Thus, the quantityy can be chosen so that

(XIII.2.21) Wa - (r+ 1)f 36 f2r

and

(XIII.2.22) wj +(r+1)f 54± r

for all j.

Observation XIII-2-5. Define the paths of integration in each of the followingtwo cases.

Case 1. Consider first the case when

(XIII.2.23) (r+1)(f+f) <Tr.

In this case, the set of indices J = {j : j = 1, 2, ... , n} is divided into four groups:

(XIII.2.24) G1 = jj : 2r <W3 -(r+1)f <w, +(r+1)f' < 2r

(XIII.2.25) G2 {j: -2tr<w1-(r+1)f<wj +(r+1)t<-2r},

For every point E D(N), the lines Lj4 in D(N) (except possibly for theirstarting points) are defined by

I s = fo + t expji arg( - £o)], 0 < t < I - toI for j E G1 u G2,

(XIII.2.36) s = t + texp[i(r + 1)t'], 0 < t < oo for j E G3,s=C+texp[-i(r+1)t], 0<t<oo for jEG4.

Note that from (XIII.2.28) and (XIII.2.29), it follows that

Ir+ <w2 2 2 2

J-2r+ 2 <wj 2

Case 2. For the case when

(XIII.2.37) (r + 1)(t + t') > r,

the set of indices J is divided into three groups:

(XIII.2.38.1)

(XIII.2.38.2)

(XIII.2.38.3)-2r +'y' <W3 - (r + 1)t < -r - -y',

2r+

where -y' is a sufficiently small positive constant such that r - 2^y' > 0. Note that

(r + 1) (t + t') > r - y'. These inequalities imply that 2 r - 2 ry' > 0 and

-(r + 1)t < -(r + 1)t - 27' + 17r < (r + 1)t' + 27' - 2r < (r + 1)t'.

In D(N), let

1 = Nr+1 exp i{ (r + 1)t' + 12'Y' - Zr}J ,ifl

1,YC2Nr+lexp i -(r+1)t-+2r} ,

(j E G1),

(jEG2)

-2r+ry' <wj -(r+1)t <-r

-2-.+y <wj+(r+l)C<2r-ry

62= j: -2r+ry'<c3-(r+1)t<Zr

2r+-y' <w) +(r+1)t' < r-ry'},

For every t E D(N), the paths of integration in t(N) are defined in the followingmanner:(a) For j E G1, from (XIII.2.37) and (XIII.2.38.1), it follows that -27r + 27' <

w, -(r+1)t-21+rand w, -(r+l)t-2ry'+r<w1+(r+1)f'<2r-'y'and,hence,

-2r+2ry'<wj -(r+1)P-2ry'+r<w, +(r+1)e<2r-ry'.

The lines L14 in D(N) are defined by

(XIII.2.39) s = + texp(i(O({)) (0 < t < cc),

where 0({) is a real-valued and continuous function off such that

-(r+1)(- 2y'+r (r+1)t"

in D(N). This implies that - 2 + 2 < wJ + 8(t;) < 2 - 2 . Precisely speaking,

the function is defined in the following way. Note that (XIII.2.37) implies

1)e- lr<(r+1)(- ir+ lry`.-(r+1)t- 127 + <(r+

22 2 2

Let

6 =Nr+1expli{(r+1)f -2r}J.

Then, o is on arc (A-3). Let t;' be a point on arc (A-3) such that argt2 < argt' <arg {o. Then, the tangent T4, of the circle j = Nr+1 at the point t:' is givenby s = £' + t < +oo), where 0(l;') is continuous in {' and

-(r + 1)t - 2 + r < (r + 1)e'. Moreover, the part of T. in D(N) is givenby

(XIII.2.40) s + t exp(iQ(t:')) (0 < t < +oo).

If a point f in D(N) is on such a tangent (XIII.2.40), set 9(e) = Note thatthe tangent of the circle I{1 = Nr+' at to is parallel to the line (A-1). If a point

is in a domain between these two lines, then set O(ff) = (r + 1)Q'. For all points

other than those given above, set 9(e) = -(r + 1)t - try' + r.(b) For j E G2, (XIII.2.37) and (XIII.2.38.2) imply that

-2r+y'<wj -(r+1)Q<w,+(r+1)f +2'?'-r<2r-Zry'

The lines LSE is defined by (XIII.2.39) with 9(4) such that

-(r+1)e<B(t) <(r+1)P'+ I r

in )(N). This implies that - 2 + 2 < wj + O(l:) < 2 - 2 The function O(C) isdefined in a way similar to the previous case.(c) For j E C3, from (XIII.2.38.3), it follows that

-17r+ I < wj - (r+ 1)e- 'y'+1r < 17r - 32 2 2 2 2

lry'-7r< 7- 1>'2 2

Letl

o = N''+1 exp [i{(r+1)t'+.i'_ 32rr1 .

In the case when (r + 1)(t + e') >_ 2a - ry', the inequality

(XIII.2.41) arglo ? argf2,

holds, whereas in the case when (r + 1)(t + e') < 2a - ', it holds that

(XIII.2.42) arg o < arg t;2

In the case when (XIII.2.41) holds, the lines Lj( are defined by (XIII.2.39) with

-(r+1)e- 2y'+a (r+1)e'+ 2ry'-7r

in D(N). This implies that - 2 + 2 < wj + B({) < 2 - 2 . Since to is on arc

(A-3), O(l;) is defined as given above. In the case when (XIII.2.42) holds, the linesLjf are defined by (XIII.2.39) with

(r+1)e'+ try'-7r <9({) -(r+l)e- Zry'+7r

in D(N). This implies that - 2 + 32 < wj + 9({) < 2 - 2Y.In this case, o is

not on arc (A-3), but 8(e) is still defined in a way similar to the previous cases.For all of the cases considered above, we prove the following lemma.

Lemma XIII-2-6. There exists a positive constant H independent of h such thatthe inequalities

(XIII.2.43) I Isi-11 I exp[ -r+ 1

a11 Ids) <Hjl:I-h l exp 1r l](j = 1,2,... n)

hold in a domain D(Nh) for any positive number h, where Nh is a sufficiently largepositive constant depending only on h.

Proof

First let j E G1 U G2. Then,

r+1ji r+l{to+texP(iarg(t-to))}JIexpexp [ - I [-

- I exp [ r + 1 to] II exp I - r +tl exp(i arg(t - Wd

expL r "J 1 to]

I exp I +I1 cos(wj + g(t - to))]

for E D(N). By virtue of (XIII.2.28), (XIII.2.29), and (XIII.2.35), it holds that

27r+2Y'<w, (9EG1),

-rr + try' < W, + arg( - W) < -Zrr - 27 (j E G2)-2

1Therefore, - cos(w, + arg(C - co)) > sin (-') for E D(Nh) and j E G1 U G2 ..

Moreover, (XIII.2.34) implies

arg( - o) - 2(r + 1)(e+ e') + 2ry1

and (XIII.2.28) and (XIII.2.29) imply (r + 1)(t + t') < n - ry'. Hence,

(XIII.2.44) I arg( - {o) - ir - 27

Observe that idol = Nn+' implies 1812 = M2 + t2 + 2cMt, where M = Nh+1 andc= -cos(rr-0) with 0= (cf. Figure 4).

Let b = sin ('). Then,

(XIII.2.45) 0<b<6 <1

(cf. (XIII.2.44)). Set

Y(r) = (M2 + 7-2 + 2aMr)h/2 exp(M,r) or dt,(M2 + t2 + 26Mt)h/2

where M, = iµj I cos(Wj +arg( - co)) . Then,r

(XIII.2.46) dr = { Mj+ h M2 + T2+ 2f Mr

IY + 1

'r + amand Af, < - rµ+l

i< 0. From (XIII.2.45), it follows that

M2 + r + 2c"Mr

Mb (r > 0). If M satisfies an inequality M >2hµJIb21),

thenT- <

2(1

r + 1)Y+

1. Since Y = 0 for r = 0, we obtain Y(r) <2(r + 1)

(r > 0). This implies thatIpj lb

exp 1 112r+ ix J ISI-h exp I - r +tl exp{i arg(e - o)}J I Idsl < 2(lu

Ib

1)

G,c l

Therefore, (XIII.2.43) follows. In this case, H = 2(rub 1), where y = min{p,).

In other cases (i.e., G3, G4, G1, G2, and G3), the lines L, are given by (XIII.2.39),

where 9(£) satisfies -2 it + 27' < wj + 9(e) < 27r - in D(Nh). Therefore,

(XIII.2.47) cos(w, + 9(e)) > sin try' I =b

for E D(N),).

(i) Consider the case when 10(x) - a given point { E D(Nh). Then,

Isle =112 + t2 - cos(sr - IB(S) - argil) ? IC12.Therefore,

J,<exp

L p1 Sr+1 Idyl <f 1 exp [ - r±1 ( +

I dt0

I31-h

=141-h exp _ Iexp +tl (wj + dt

< exp l - Nj, 1I(r1}1 ` r+1J lµ)Ib

This implies (XIII.2.43).

(ii) Consider the case when 27r < 19(x) - a given point t E TD(Nh).

Since the lines Lg are in D(Nh), the distance D from the origin to L,t is not lessthan M = Nh+i. Then, IS12 can be written in the form 1x12 = D2+o2 (cf. Figure5).

FIGURE 4. FIGURE 5.

3. PROOF OF THEOREM XIII-1-2 417

Consider a function Y(r) = (D2 + r2)h/2exp(Mjr)100 xp+ M, ) da, where

M2 = 1'U2I `os(+ + OW). Note that b > 0. Hence, for r > 0, (XIII.2.47) implies

that Y(r) < M . On the other hand, Y(r) satisfies the differential equation dY =

M + h Y -1. Sincer

< 1 , it follows thatdY

>lit, Ib Y-1

{ D2 + r2 } D2+72 2D dr 2(r + 1)h(r + 1)

if M >1i2 1b

. Here, a use was made of the inequality M < D. Since Y(0) <-1 < (r + 1)

we obtain Y(r) < 2(r i 1) (r < 0). Therefore, (XIII.2.43) follows.M2 1A2 ib 1µ, b

2(r + 1)This completes the proof of Lemma XIII-2-6. In this case, again, H = µb ,

where µ =min{µ2 }. 02

We fix the paths of integration on the right-hand sides of (XIII.2.15) as explainedabove. The constant H in (XIII.2.13) is chosen to be equal to the constant H givenin Lemma XIII-2-6. Note that H is independent of h.

XIII-3. Proof of Theorem XIII-1-2

Let us construct a solution u2 = 02 (x) of (XIII.2.15) so that

(XIII.3.1) O2 (x) 0 (j = 1,2,... n)

as x oo in (XIII.1.15).

Observation XIII-3-1. As in the previous section, denote by Do(Nh) the domainin the x-plane which corresponds to the domain D(Nh) in the -pllane. The domain

Do(Nh) depends on h. Set o i = Nh+1 exp {i(r + 1)(e + and denote by

),xoh) the corresponding point in Do(Nh ). Also, setting h' = r + 1(h = 1,2,3....

consider the domain Do(Nh,). As mentioned in the previous section, the constantH in (XIII.2.13) is chosen to be equal to the constant H in Lemma XIII-2-6.

Choose a positive constant 8 so that r H1 { cb + Nh2 <_ S. Furthermore, by

111

virtue of (XIII.2.13), Nh, can be chosen so large that Nh, > No and b < d (cf.(XIII.2.12)). Fix Nh, in this way. Then, by a method similar to that in §XII-3 foreach j (j = 1, 2, ... , n), successive approximations can be defined to construct asolution u2 = O,(x) of (XIII.2.15) which is holomorphic in Do(Nh-) and 101(x)l <

(j = 1,2,... ,n), where = xT+'. In this way, the existence of a boundedsolution u2 = 02 (x) (j = 1, 2, ... , n) of system (XIII.2.15) is proved.

Observation XIII-3-2. In order to prove that Oj (x) 0 (j = 1, 2,... , n) asx -+ oo in Do(Nh'), use the fact that, for each positive integer k, the functions4j (x) (j = 1, 2, ... , n) also satisfy the integral equations

(XIII.3.2) uj(x) =

Oj(x(k))exp1r+1

.(xr+1 _ (xok))r+1)I

+ f tTRj(t, i (t))exp [r-JAI

(tr+l - xr+1) I dtLjc J

(j E G1 U G2),

trR1(t,, (t))exp [r+i(tr+1 _xr+1)ldtI JJ

VG1UG2)U

in the domain Do(Nk). Here, u(x) denotes the Cn-valued function whose entries are(u1(x),u2(x),... ,un(x)). Note that we can assume Do(Nk) C Do(Nh') withoutloss of generality.

Upon applying successive approximations similar to those of §XII-3 togetherwith Lemma XIII-2-6 to (XIII.3.2), it can be proved that (XIII.3.2) has a solutionu,(x) = 0., (x) (j = 1,2,... ,n) such that IiP, (x) I < CkIxI-k U = 1,2,... ,n) inDo(Nk), where Ck is a suitable positive number. Also, using Lemma XIII-2-6 and(XIII.2.13), it can be shown that ¢j (x) = ikj (x) (j = 1,2,... , n) in VO(Nk), since

Oj (x) - j (x) = j tr [RR (t, fi(t)) - Rj(t,,(t))I exp Ir +1(tr+1 _ xr+1)l dt,

J

where j = 1, 2,... , n, and ¢(x) and tG(x) denote en-valued functions whose entriesare (01(x), 02(x), ... , On(x)) and (ti1(x), ,11 (x)), respectively. Thus, weconclude that 0, (x) 0 (j = 1, 2,... , n) as x -' oo in Do(Nh,).

Now, let us return to Observation XIII-2-2. If we set pj(x) = Oj(x) + zj(x), thesolution v , = pj(x) (j = 1, 2, ... , n) of (XIII.1.1) satisfies all of the requirements ofTheorem XIII-1-2. 0

Remark XIII-3-3. Let Ax, y', µ, e) be a Cn-valued function of a variable (x, e)E C x Cn x Cn x C such that

(a) the entries of f are bounded and holomorphic in a domain

Do = { (x, b, µ, e) : IxI > Ro, Iu-1 < do, lµl < p o, IeI < Eo, I arg EI < po},

where R0, 60, µo, co, and po are some positive constants,(b) f admits a uniform asymptotic expansion in Do

s/+0

r/x,a/, e) ti Ehfh(x, . p)

h=0

as e - 0,

where coefficients fh (x, y, i) are single-valued, bounded, and holomorphic inthe domain

Ao = {(x, 9, #) : IxI > Ro, I vi < 6o, Iµl < po},

3. PROOF OF THEOREM XIII-1-2 419

(c) if f (x, y, µ', e) = fo (x, #,c) + A(x, µ, e)y + O(I y12), then

fo(oo,µ',e) = 0 and detA(oo,0,0) 0 0.

Let )q, A2, ... , )1n be n eigenvalues of A(oo, 0', 0) and set wf = arg A, (j _1, 2, ... , n). Note that the w, are not unique. Set also S(pl, p2) _ {x : Pi <argx < p2}. Let a and r be two non-negative integers. Assume that the sectorS(pl, p2) is contained in the sector

32 - min{wj} + apo < (r+ 1)argx < 32 - max{wj} - apo

f o r a suitable choice of {wj : j = 1, 2, ... , n}.Under these assumptions, we can prove the following theorem in a way similar

to the proof of Theorem XIII-1-2.

Theorem XIII-3-4. The system e° fy = x"f (x, c) has two solutions

¢(x, 91, e) and i (x, µ, e) such that(i) these two solutions are bounded and holomorphic in the domain

So = {(x, 9, e) : Ix1 > N, x E S(p1, p2), 1141 < ul, 0 < IEI < e1,1 arg el < p0},

if p1 and E1 are sufficiently small,

(ii) the solution ¢(x, µ, e) admits an uniform asymptotic expansion

+ooF, x-k&(µ, e) as x -. ook=1

in So, where coefficients &(µ,e) are bounded, holomorphic, and admit uni-form asymptotic expansions in powers of a as c -+ 0 in the domain {(µ, e)Iµ1 < {31, 0 < lei < el, I argel < po},

(iii) the solution 1'(x, N, e) admits an uniform asymptotic expansion:

+oo

(x, , e) ' ehWh(x, µ) as f -4 0

h=0

in So, where coefficients >Gh(x,#) are bounded, holomorphic, and admit uni-form asymptotic expansions in powers of x'1 as x - oo in the domain{(x,µ') : Ixf > N, x E S(p1,p2), Jill < p1}.

FLrther more, there exist functions tc(x, µ, e) for t = 0,1, ... such that(a) these functions ¢'c satisfy conditions (i) and (is) given above,(b)

h=0

A complete proof of Theorem XIII-3-4 is found in [Si7J and (Si101.

Remark XIII-3-5. In the proof of Theorem XHI-1-2, we used the assumption thatthe matrix Ao on the right-hand side of (XIII.1.10) is invertible (cf. AssumptionIII of §XIII-1). Without such an assumption, we can prove the following theorem.

Theorem XIII-3-6. Let F(x, yo, yr, ... , yn) be a nonzero polynomial in yo, yl, ,00

yn whose coefficients are convergent power series in x'1, and let p(x) = an,x-°`M--0

E C[(x-1)) be a formal solution/ of the differential equation

(XIII.3.3) F(x,y,Ly,..., fn)=0.

Then, for any given direction

\arg

x = 0, there exist two positive numbers 6 and aand a function 0(x) such that

(i) ¢(x) is holomorphic in the sector S = {x E C:0< [xj <6, (arg x - 0[ < a},(ii) 0(x) admits the formal solution p(x) as its asymptotic expansion as x o0

in S,(iii) 0(x) satisfies differential equation (XIII.3.3) in S.

The main idea of the proof is similar to the proof of Theorem XIII-1-2. However,we need the thorough knowledge of the structure of solutions of a linear system

xd = A(x)y that we shall explain in §XIII-6. Also, it is more difficult to define

the paths of integration (cf. [I] and (RS11). A complete proof of Theorem XIII-3-6is found in [RS1). See, also, §XIII-8.

XIII-4. A block-diagonalization theoremConsider a system of linear differential equations

(XIIi.4.1)dg

= x' A(x}y,dx

where r is a positive integer, y" E C', and A(x) is an n x n matrix. The entries ofA(x) are holomorphic with respect to a complex variable x in a sector

(XIII.4.2) jxl > No, f argx[ < ao,

where No and ao are positive numbers. Assume that the matrix A(x) admits anasymptotic expansion in the sense of Poincare

00

(X1II.4.3) A(x) E x'"A&'=o

as x - oo in sector (XIII.4.2), where coefficients A are constant n x n matrices.Suppose also that Ao = A(oo) has a distinct eigenvalues A1, A2, ... , J1t with mul-tiplicities n1, n2, ... , nt, respectively (nl + n2 + + nt = n). Without loss ofgenerality, assume that Ao is in a block-diagonal form:

()UII.4.4) AO = diag [Al, A2, ... , At) ,

where Aj are of x n, matrices in the form

(XIII.4.5) A, =A,In, +Nj (j=1,2,...,1).

Here, A , is a lower-triangular nilpotent n, x n, matrix. The main result of thissection is the following theorem (cf. (Si7J).

Theorem XIII-4-1. Under the assumption (XIIL4.3) and (XIII.4.4), there existsan n x n matrix P(x) whose entries are holomorphic in a sector

(XIII.4.6) IxI > N1, I argxj < al,

where Ni 1 and a1 are sufficiently small positive numbers, such that(i) P(x) admits an asymptotic expansion

00

(XIII.4.7) P(x) Ex-"p- (Pp = In),"=0

as x -' oo in sector (X111.4.6), where coefficients P" are constant n x nmatrices,


(XIII.4.8) y = P(x)zi

reduces system (X111.4.1) to

(XIII.4.9)di ,

= x B(x)i,

where B(x) is in a block-diagonal form

(XIII.4.10) B(x) = diag [BI (x), B2(x),... , Bt(x)J .

Hen, B,(x) an n, x n. matrices and admit asymptotic expanswns

00

(XIII.4.11) B, (x) ^-- E xBj""=o

as x oo in (XIII.4.6), where coefficients B,, are constant n) x n) matrices.

Proof.

From (XIII.4.1), (XIII.4.8), and (XIII.4.9), we derive the equation

(XIII.4.12)dP

= xr(A(x)P - PB(x)]

that determines the matrices P(x) and B(x). Set

(XIII.4.13) A(x) = Ao + E(x), B(x) = A0 + F(x), P(x) = In + Q(x).


Then, E(x) = O(x-1), F(x) = O(x'1), and Q(x) = O(x-1). Furthermore,(XIII.4.12) becomes

(XIII.4.14) dQ = x''[AoQ - QAo + E - F + EQ - QF].dx

Write each of three matrices E(x), F(x), and Q(x) in a block-matrix form accordingto that of Ao in (XIII.4.4), i.e.,(XIII.4.15)

F(x) = diag]Fl,F2,... ,Ft],E11 E12 ... EllE21 E22 ... E2t

E(x) _

Ell E 2 Eu

Q(x) =

Q11 Q12 ... Qlt

Q11 Q2t . Qttwhere EJk and QJk are nJ X nk matrices and F, are nJ x nJ matrices. Set

(XIII.4.16) QJJ = 0 (j = 1,2,... ,t).

From (XIII.4.4), (XIII.4.14), (XIII.4.15), and (XIII.4.16), it follows that

(XIII.4.17) F,=EJJ+FEJhQh, (j=1,2,...,t)h#J

and

(XIII.4.18)dQjk

= x' [43Qik - QJkAk + EJk + EJhQhk - QJkAk] (j j4 k).h*k

Substituting (XIII.4.17) into (XIrII.4.18), a system of nonlinear differential equations

dQJk= x' LAJQJk - QJkAk + EJhQhk

dx{ h*k

(XIII.4.19)

- QJk (Ekk + EkhQhk) + EJkJ (j # k)h*k

is obtained. Since it is assumed that .11i ... , Al are distinct eigenvalues of Ao andthat A0 is in the block-diagonal form (XIII.4.5), upon applying Theorem XIII-1-2to (XIII.4.19) we can construct a desired holomorphic solution Qjk(x) of (XIII.4.19)

00

which admit an asymptotic expansion Qjk(x) > x "Qjk" (j, k = 1, 2,... , 3; j"=1

k), where Qjk" are constant nj by nk matrices. Defining Fj by (XIII.4.17) and thenB(x) by (XIII.4.13), the proof of Theorem XIII-4-1 is completed. 0

Theorem XIII-4-1 concerns the behavior of solutions of system (XIII.4.1) nearx = oo. Since it is useful to give a similar result concerning behavior of solutionsnear x = 0, we consider, hereafter in this section, a system of differential equations

(XIII.4.20) xa+1 dY = A(x)y,

where d is a positive integer and the entries of n x n matrix A(x) are holomorphicin a neighborhood of x = 0. Also, assume that A(0) is in a block-diagonal form

()III.4.21) A(0) = diag[A11,,, +N2,... +JVt],

where a1, ... , .1t are distinct eigenvalues of A(0) with multiplicities n1, n2, ... , nt,respectively (n1 + n2 + + nt = n), and, for each j, .W is a lower-triangular andnilpotent nj x nj matrix.

Comparing the present situation with that of Theorem XIII-4-1, we notice thefollowing two differences:

(a) singularity is at x = 0 in the present situation, while singularity is at x = 00in Theorem XIII-4-1,

(b) The power series expansion of A(x) is convergent in the present situation,while A(x) in Theorem XIII-4-1 admits only an asymptotic expansion in asector containing the direction arg x = 0.

We can change any singularity at x = 0 to a singularity at x = oo by changing

the independent variable x by1. Also, any direction argx = 8 can be changedx

to the direction arg x = 0 by rotating the independent variable x. Furthermore,the asymptotic expansion P of Po(x) and the expansion b of Be are formal power

series satisfying the equation xd+i dP = AP - PB. This implies that two matrices

P and b are independent of P. Hence, using Corollary XIII-1-3, the following resultis obtained.

Theorem XIII-4-2. Let A(x) be an n x n matrix whose entries are holomorphicin a neighborhood of x = 0. Also, let d be a positive integer. Assume that the matrixA(0) is in block-diagonal form (XIII.4.21), where al, ... , at are distinct eigenvaluesof A(0) with multiplicities n1, n2, ... , nt, respectively (n1 + n2 +, + nt = n), andfor each j, JVj is a lower-triangular and nilpotent n, x nj matrix. Fix a real number0 so that (a, - Xk)e-utO V EP for j 96 k. Then, there exists two positive numbers beand ee and an n x n matrix Pe(x) such that

(a) the entries of Po(x) are holomorphic and admit asymptotic expansions in pow-

ers of x as x - 0 in the sector Se = Ix: 0 < IxI < be, I arg x - 01 < + to },00

(b) if >2 xmP,,, is the asymptotic expansion of Pe(x), then this expansion ism=0

independent of 8 and Po = I,,,(c) the transformation PB(x)u" changes system (XIIL4.20) to a system

(XIII.4.22) xd+1 = Be(x)u,

where the matrix Be(x) is in a block-diagonal form

(XIII.4.23) Be(x) = diag (B1e(x), B2e(x), ... , Bte(x)]

For each,, Bje is an n, xn2 matrix which admits also an asymptotic expansionin x as x--i0 in So.

The main claims of this theorem are

(i) the asymptotic expansion of Pe(x) is independent of 0,

(ii) the opening of the sector So is larger than k.

Proof of Theorem XIII-4-2 is left to the reader as an exercises.

Remark XIII-4-3. Using Theorem XIII-3-4, we can generalize Theorem XIII-

4-1 to the system e°dx = xr'A(x, µ, e)y", where r and o are non-negative integers,

y" E C^, and A(x, µ', e) is an n x n matrix with the entries that are bounded andholomorphic with respect to the variable (x, µ, e) E C x C'° x C in a domainDo = {(x, µ, e) : 1x1 > N, Iµ1 < µo, 0 < 1e1 < eo, 0 < 1 arg e1 < po}. Also, assume

00

that A admits a uniform asymptotic expansion A(x, µ, e) - > ehAh(x, µ) in Doh=0

as a -+ 0, where coefficients Ah (X, µ) are bounded and holomorphic in the domain{(x,µ) : 1x1 > N, 1µ1 < µo}. A complete proof of this result is found in [Si7] and[Hs2].

XIII-5. Cyclic vectors (A lemma of P. Deligne)

In the study of singularities, a single n-th-order differential equation is, in manycases, easier to treat than a system of differential equations. In this section, weexplain equivalence between a system of linear differential equations and a singlen-th-order linear differential equation.

Let us denote by 1C the field of fractions of the ring C[(x]] of formal power seriesin x, i.e.,

K=I

9:

p E C[Ix]J, 9 E C[Ix]j, 9 34 0}

Also, denote by V the set of all row vectors (cl(x), c2(x), ... , c,,(x)), where theentries are in the field K. The set V is an n-dimensional vector space over thefield 1C.

Define a linear differential operator C : V - V by G[vl = by + 7l(x) (v' E V ),

where b = xd and S2(x) is an n x n matrix whose entries are in the field 1C. Wefirst prove the following lemma.

Lemma XIII-5-1 (P. Deligne [Del]). There exists an element iio E V such that{v"o, Gv"o, G2v"o, ... , G"-lvo} is a basis for V as a vector space over IC.

Proof.

For each nonzero element v of V, denote by µ(v'' the largest integer t such that{ii, Cii, C2v, ... ,.CeV) is linearly independent over K. In two steps, we shall derivea contradiction from the assumption that max{µ(v') : v' E V} < n - 1.

5. CYCLIC VECTORS 425

Step 1. First, we introduce a criterion for linear dependence of a set of elements ofV. Consider a set {i11, ... vm } C V, where m is a positive integer not greater thann = dim, V. Let 6j = (c31,c12,... ,Cin) (j = 1,2,... ,m). Set .7 = {(jl,... ,jm)1 < it < j2 < . < jm < n}, and introduce a linear order .7 -' { 1, 2,... , (M-)) inthe set J. Let us now define a map

m (.)V ={(v1,V2,...,Vm): 61 EV (j=l,...,m)} -+ Id.)

(VI, ... , Vm) --+ V1 A v2 A ... A vm, where

1,AV2A ... A vm

f cI31 Cj32 ... clim

det : : : (jl,... ,jm) E .7

C-i 1 Cmj2 C-3-

It is easy to verify the following properties of V1 A v"2 A A v,,,:

Vk_1 A 1m_1 A Um

(1) = Vk_1 A Vm_1 A v"m

+ v'k_1 A Vm_1 A Vm,

(2)

V1 A ... A Vk_1 A (a Vk) A vk+I A ... A Vm_1 A Vm

= 6k_1 A irk A vm),

for all aEIC,

(3)VIA.-.A irk A ...A...AVjA ... A Vm= -(11A...A iY A A Vm),

(4) a sety{ {6j, 62,... , Vm } E V is linearly dependent if and only if v'1 AV2 A A 17m =

6 in K('^).

Step 2. Fix an element Vo of V such that µ(6o) = max{p(V) : v" E V} < n-1. Sincep(VO) < n-1, another element w of V can be chosen so that {VO, Cv"o, CZi3o, ... , CnOv',w"} is linearly independent, where no = max{µ(v") : v E V}. Set v' = vo+Ax"'O E V,where A E C and m is an integer. Then, Cwt = Cw"o + C'(Ax-ti) = Crvo +Axm(C + m)tw. Since {V, Cv", ... , Cn0+1V} is linearly dependent, it follows thatv A DU A A CnOv' A C"O+I V = 6 for all A E C and all integers in. Note thatv A,06 A A C"OiYA C"°+16 is a polynomial in A. Since this polynomial is iden-tically zero, each coefficients must be zero. For example, the constant term of thispolynomial is i6o A CVO A ... A 00+I V"o. This is zero since {iio, Ciio..... G"0+Iuo}is linearly dependent. Compute the coefficient of the linear term in A of the poly-nomial. Then,

w A C60 A ... A Cna+1v'o + v'Y A ... A 00 Vo A (C + m)n0,+1ti

no

+ 1:60A...ACi-IVoA (C+m)lw A CJ+'A A...ACnb+Ii = 0J=1


identically for all integers m. The left-hand side of this identity is a polynomial inin of degree no + I. Hence, each coefficient of this polynomial must be zero. Inparticular, computing the coefficient of mn0+1, we obtain vOAGvOA . A GnbtloAw' =0. This is a contradiction, since {vo, Gvo, ... , L"Ovo, w} is linearly independent.This completes the proof of Lemma XIII-5-1.

Definition XIII-5-2. An element vo E V is called a cyclic vector of G if {6o, Lu0,L2v'o, ... , Ln-1%) is a basis for V as a vector space over X.

Observation XI II-5-3. Let uo be a cyclic vector of Land let P(x) be the n x n ma-Uo

trix whose row vectors are {vo,G'o,... ,L' 'io}, i.e., P(x) =Gvo

. Then,Gn-lvo

nv"oL21 yo

=o

and, hence, setting A(x) = L[P(x)]P(x)-1 = bP(x)P(x)-l +

P(x)f2(x)P(x)-1, we obtain

0 1 0 0 .. 00 0 1 0 ... 0

A

0 0 0 0 ... 1

ap al a2 a3 ... an-1

with the entries a, E C. Thus, we proved the following theorem, which is the mainresult of this section.

Theorem XIII-5-4. The system of differential equations

(XIII.5.1) by' = fl(x)y", where g _y1

Y.

becomes

(XIII.5.2) bu" = A(x)u,

if y is changed by u = P(x)y". System (XIll.5.2) is equivalent to the n-th-orderdifferential equation

n-1

(XHI.5.3) bnq - arbrq = 0, where q = yl.1=o

5. CYCLIC VECTORS

Example XIII-5-5.(1) Let us consider the system

(a) by" = 0,

The transformation

(T)

changes system (a) to

(E)

y1

where y =yn

u = diag[l, X'... xn-11y,

bu = diag[0,1,... ,n - 1]u.

Further, the transformation

(r) --

changes (E) to the form

00

0

ao

1

2

22

2n-1

1 0 0 0

0 1 0 0

0 0 0 1

a1 a2 a3 ... an-1

U

w,

where ao, a,,.. . , an-1 are integers. Hence, the transformation

1 1 1 ... 1

0 1 2 .. n- 1

w" = 0 1 2 2 ... (n - 1)2 diag [1, x,... ,x°-1

0 1 2n-1 ... (n - 1)n-1

y

427

changes (a) to (E'). This implies that vo = (1, x, x2'. .. xn-1) is a cyclic vector inthis case.(II) Next, consider the system

(b) by = Ay,

where A is a constant diagonal n x n matrix. Choose a transformation similar to(T) of (a) to change system (b) to

(E") oiZ = A'u


so that A' is a diagonal matrix with n distinct diagonal entries. Then, a transfor-mation similar to (T') can be found so that (E") is changed to (E') with suitableconstants ao, al, ... , an_ I .

XIII-6. The Hukuhara-Turrittin theorem

In this section, we explain a theorem due to M. Hukuhara and H. L. TLrrittin thatclearly shows the structure of solutions of a system of linear differential equationsof the form

(XIII.6.1)

where the entries of the n x n matrix A are in K (cf. §XIII-5). In order to statethis theorem, we must introduce a field extension f- of K. To define L, we first set

+aE a.nxm1" : a.., E C and M E Z ,

M=M I+a

where 7L is the set of all integers. For any element a = F a,nxn'/° of K,,, weM=M

define xji by xda +00

= F \ v) amx'nl '. Then, K, is a differential field. The fieldM=M

+oo

L is given by L = U K which is also a differential field containing K as a subfield.L=1

Furthermore, L is algebraically closed. The Hukuhara-T rrittin theorem is givenas follows.

Theorem XIII-6-1 ({Huk4j and [Tu1j). There exists a transformation

(XIII.6.2) y' = UI

such that

(i) the entries of the matrix U are in L and det U ;j-1 0,(ii) transformation (XIII.6.2) changes system (X111.6.1) to

(XIII.6.3) xd = Bz',

where B is an n x n matrix in the Jordan canonical form

(XIII.6.4)B=diag[Bl,B2,...,Bpj, Bj=diag[BJi,Bj2i...,B,,n,],Bjk =AI In,,. + Jn,,,.

6. THE HUKUHARA-TURRITTIN THEOREM 429

Here, I,,,, is the njk x njk identity matrix, J.,,, is an n 1k x n3k nilpotentmatrix of the form

0 1 00 0 1

(XIII.6.5) Jn",0 0 0

0 0 0

and the Aj are polynomials in xfor some positive integer s, i.e.,d,

(XIII.6.6) Aj = > A x-'18 where all E C (j = 1, 2, ... , p)r=O

and(XII1.6.7)

A,d, 0 if dJ > 0 and A., - A, are not integers if i 0 j.

Proof.

Without loss of generality, assume that the matrix A of system (XIII.6.1) hasthe form

0 1 0 0 00 0 1 0 ... 0

A0 0 0 0 ... 1

00 a1 a2 a3 "' an-1

where ak E 1C (k = 0, 1, ... , n - 1) (cf. Theorem XIII-5-4). Set E= A -an- 1

n In.

Then,

(XIIi.6.8) trace [E] = 0.Consider the system

(XIII.6.9) X E = E.

Case 1. If there exists an n x n matrix S with the entries in X such that det S 0 0dii

and the transformation w = Sii changes (XIII.6.9) to x = A(x)u", where the

entries of A are in C[[x]], then there exists another n x n matrix S with the entriesin X such that det S 96 0 and the transformation

(XIII.6.10) w = Suu

digchanges (XIII.6.9) to x = Aoii, where the entries of the matrix A0 are in C.F irthermore, any two distinct eigenvalues of A0 do not differ by an integer (cf.Theorem V-5-4). Hence, in this case, system (XIII.6.1) is changed by transforma-tion (XIII.6.10) to

dilxaj = [an 1 In+Ao]u.

This proved Theorem XIII-6-1 in this case.


Case 2. Assume that there is no n x n matrix S with entries in X such thatdet.S 0 and the transformation w = Su' changes (XM.6.9) to xdu = A(x)u,

where the entries of A are in C([x]]. Since

E =

-lan_1 1n

0

0

ap al a2 a3 an-2 an-1 -- -an-1,

1 0 0

- lan_1 1 0n

0 0 0

a cyclic vector can be found for system (XIII.6.9) by using the matrix W definedby

w11 winwith

writ Wnn

[wll ... Win] = (10 ... 01,

[w,,1 ... wJn] = V1-1[1 0 01,

where V([c1 cn]) = x[c1 [c1 - - cn]E. The matrix W is lower-triangular and the diagonal entries are {1, ... ,1}, i.e.,

(XIII.6.11) W =

1

If (XIII.6.9) is changed by the transformation v" = Wtv", then

(XIU.6.12) X!LV _ x + WEJ W-Y.

It follows from (XIII.6.8) and (XIII.6.11) that

XIII 136 t WE]W- l 0( . . ) racel x + = .

Also,

0 1 0 0 00 0 1 0 0

+ WE] W-1 = V[ W]W-1x11 0 0 l0 0

W i31 / A32

2 2 An-1

1 0 0 01 0 0

where 1k E )C (k = 0, 1,... , n - 1). In particular, from (XIII.6.13), it follows thatOn-1 = 0. Under our assumption, not all Qt are in C[[x]]. Set 9 = (t: at V C[[x]]}

+00

and set Also, /3t = x-u" E Qtmxm (t E 3), where, for each t, the quantity µt is am=0

+oo

positive integer, 1: /3tmxm E C[[xJ] and #to 54 0. Setm=0

k=max(n µt tJ

:IE91.

Then, ut < k(n - t) for every t E J and µt = k(n - t) for some t E J. This impliesthat

(I) Of = L. Qt,mxm

m>-k(n-t)(t=0,1,... n- 1)

and

(11) Qt = x-k(n-t) (Ct + xqt)

for some t such that k(n - t) is a positive integer, ct is a nonzero number in C,qt E C[[x]], and Qt,m E C. We may assume without any loss of generality that

k = h for some positive integers h and q.9

Let us change system (XIII.6.12) by the transformation

v" = diag [1, x-k, ... , x-(n-1)kI u".

Then,

(XIII.6.14)

where

(XIII.6.15) F =

and

0 1 0 0

0 0 1 0

0 0 0 0

70 71 72 73

+ kxkdiag [0,1, ... , n - 1]

7t = xk(n-t)o, = : $,,mxm+k(n-t) E C[ [x'1911 (0 < t < n - 1).m>-k(n-t)

In particular, 7n-1 = 0.


+oo

Setting F = E xm/9Fm, where the entries of F,,, are in C, we obtainm=o

Fo=

0 1 0 0 0 01

COC1 C2 C3 ... Cn_2 0

where the constants co, cl ... , cn_2 are not all zero. This implies that the matrixF0 must have at least two distinct eigenvalues. Hence, there exists an n x n matrixT such that

(1) T = XM19Tm, where the entries of the matrices Tm are in C and To ism=o

invertible,(2) the transformation

(XIII.6.16) y = Ti

changes system (XIII.6.14) to a system xd = x-kGxi with a matrix G in

0a block-diagonal form G = [0G2 J

, where Gl and G2 are respectively

n1 x nl and n2 x n2 matrices with entries in C[[x1/91and that nl + n2 = nand n., > 0 (j = 1, 2) (cf. §XIII-5). Therefore, the proof of Theorem XIII-6-1can be completed recursively on n. 0

Observation XIII-6-2. In order to find a fundamental matrix solution of (XIII.6.1),let us construct a fundamental matrix solution of (XIII.6.3) in the following way:

Step 1. For each (j, k), set

4'1k = xA,0 eXp[AJ (x)] exp[(log x)Jf, ],

where

-t/s

if dj = 0,

if d1>0.

Step 2. For each j, set

Dj = diag ['j1, xAJ0 exp[A,(x)] exp[(logx)Jj],

where J. = diag [J,,,,, J,,72 , ... J,mJ ] .

Step 3. Set

Then,

C = diag [,\101., + J1, 1\20I., + J2, ... ,

A = ding [Al(x)In A2(x)I,,,, ... ,

(XIII.6.17) 4) = diag [451, 4i2, ... , 4ip] = xC exp[A]

is a fundamental matrix solution of (XIII.6.3), where

xC = diag [xA1OxJ, xl\2OxJ2' ... 'X aPOxJP] , xJ, = exp[(logx)JjJ.

The matrix

(XIII.6.18) U4i = Uxc exp[A]

is a formal fundamental matrix solution of system (XIII.6.1), where U is the matrixof transformation (XIII.6.2) of Theorem XIII-6-1. The two matrices exp[A] and xCcommute.

Observation XIII-6-3. Theorem XIII-6-1 is given totally in terms of formalpower series. However, even if the matrix A(x) of system (XIII.6.1) is given ana-lytically, the entries of U of transformation (XIII.6.2) are, in general, formal powerseries in xl1", since the entries of the matrix T(x) of transformation (XIII.6.16)are formal power series in general. Transformation (XIII.6.16) changes system(XIII.6.14) to a block-diagonal form. Therefore, in a situation to which Theo-rem XIII-4-1 applies, transformation (XIII.6.2) can be justified analytically. Thefollowing theorem gives such a result.

Theorem XIII-6-4. Assume that the entries of an n x it matrix A(x) are holo-morphic in a sector So = {x E C 0 < lxJ < ro, I argxi < ao} and admit asymptoticexpansions in powers of x as x 0 in So, where ro and ao are positive numbers.Assume also that d is a positive integer and y" E C'. Let S be a subsector of Sowhose opening is sufficiently small. Then, Theorem XIII-6-1 applies to the system

(XIII.6.19) xd+1dy = A(x)fdx

with transformation (XIII.6.2) such that the entries of the matrix U of (XIII.6.2)are holomorphic in S and each of them is in a form x,00(x) where p is a rationalnumber and O(x) admits an asymptotic expansion in powers of xl"° as x 0 in S,where s is a positive integer.

Observation XIII-6-5. In the case when the entries of the matrix A(x) on theright-hand side of (XIII.6.19) are in C[[x]l and A(0) has n distinct eigenvalues, thematrix A(x) also has n distinct eigenvalues A1(x), A2(x), ... , which are inC((xJJ. Furthermore, the corresponding eigenvectors p"1(x), p"2(x), ... ,15n(x) can beconstructed in such a way that their entries are in Chill and that p""1(0),7"2(0), ... ,

p",,(0) are n eigenvectors of A(O). Denote by P(x) the n x n matrix whose columnvectors are p1(x), p2(x), ... , p,+(x). Then, detP(O) 36 0 and P(x)-lA(x)P(x) =diag[A1(x),A2(x),... ,An(x)]. This implies that the transformation y = P(x)iichanges system (XIII.6.19) to

(XIII.6.20) xd+1 dx {diagiAi(x)A2(x).... , An(x)] - xd+1P(x)-1 d ( )

It is easy to construct another n x n matrix Q(x) so that (a) the entries of Q(x)are in C[(x]], (b) Q(0) = 4,, and (c) the transformation i = Q(x)v changes system(XIII.6.20) to

(XIII.6.21) xd+1 dv = diag [ft1(x), µ2(x), ... , µn(x)] v,dx

where µ1(x), µ2(x), ... , i (x) are polynomials in x of degree at most d such thatA, (x) = it) (x) + O(xd+1) ( j = 1, 2, ... , n). Therefore, in this case, the entries ofthe matrix U of transformation (XIII.6.2) are in 1C.

Observation XIII-6-6. Assume that the entries of A(x) of (XIII.6.19) are inC([x]]. Assume also that A(O) is invertible. Then, upon applying Theorem XIII-6-1

to system (XIII.6.19), we obtaind1

= d for all j. Using this fact, we can prove thefollowing theorem.

s

Theorem XIII-6-7. Let Q;i1(x) and Qi2(x) be two solutions of a system

(XIII.6.22) xd+1 dy = A(x)yf + x f (x),dx

where d is a positive integer, the entries of the n x n matrix A(x) and the C"-valuedfunction 1 *(x) are holomorphic in a neighborhood of x = 0, and A(O) is invertible.Assume that for each j = 1, 2, the solution ¢,(x) admits an asymptotic expansionin powers of x as x - 0 in a sector S3 = {x E C : lxl < ro, aj < arg x < b3 },where ro is a positive number, while aJ and bi are neat numbers. Suppose also thatS1 n S2 0. Then, there exist positive numbers K and A and a closed subsector S ={x : lxl < R,a < argx < b} of Sl nS2 such that K exp[-Alxl -d]in S.

Proof.

Since the matrix A(O) is invertible, the asymptotic expansions of1(x) and ¢2(x)are identical. Set 1 (x) = ¢1(x) -$2(x). Then, the C"-valued function >G(x) satisfies

system (XIII.6.19) in S, n$2 and ii(x) ^_- 6 as x 0 in S1nS2. By virtue of TheoremXIII-6-4, a constant vector 66 E C" can be found so that r%i(x) = U4D(x)6, wherelb(x) is given by (XIII.6.17). Now, using Observation XIII-6-6, we can completethe proof of Theorem XIII-6-7.


Observation XIII-6-8. The matrix A = diag [AI In,, A2In3, ... , API,y] on theright-hand side of (XIII.6.18) is unique in the following sense. Assume that an-other formal fundamental matrix solution Uxcexp[A] of system (XIII.6.1) is con-structed with three matrices U, C, and A similar to U, C, and A. Since thematrices Uxc exp[A] and UxO exp[A] are two formal fundamental matrices of sys-tem (XHI.6.1), there exists a constant n x n matrix r E GL(n, C) such thatUxc exp[A] = U5C exp[A]I' (cf. Remark IV-2-7(1)). Hence, exp[A]T exp[-A] =x-CU-IUxC. Using the fact that r is invertible, it can be easily shown that A = Aif the diagonal entries of A are arranged suitably. For more information concerningthe uniqueness of the Jordan form (XIII.6.3) and transformation (XIII.6.2), see, forexample, [BJL], [Ju], and [Leve].

Observation XIII-6-9. The quantities A.,(x) are polynomials in x1l'. Set w =2a[!] 1/a = 1/a 1/e Cexp and x wx Then, i = x. Therefore, if z in Ur exp[A] is

replaced by zI/', then another formal fundamental matrix of (XIII.6.1) is obtained.This implies that the two sets {Aj (i) : j = 1, 2,... , p} and (A.,(x) : j = 1, 2,... , p)are identical by virtue of Observation XIII-6-8.

Observation XIII-6-10. A power series p(x) in x1/' can be written in a forma-1

Ax) = Exh1'gh(x), where ql(x) E C[[x]] (j = 0, 1, ... , s - 1). Using this fact andh=0

Observation XIII-6-9, we can derive the following result from Theorem XIII-6-1.

Theorem XIII-6-11. There exist an integer q and an n x n matnx T(x) whoseentries are in C[[x]] such that

(a) det T (x) 96 0 as a formal power series in x,dil(b) the transformation y" = T(x)t changes system (XII1.6.1) to x = E(x)iZ

with an it x it matrix E(x) such that entries of x9E(x) are polynomials in x.

The main issue here is to construct, starting from Theorem XIII-6-1, a formaltransformation whose matrix does not involve any fractional powers of x in such away that the given system is reduced to another system with a matrix as simple aspossible. A proof of Theorem XIII-6-11 is found in [BJL].

Changing the independent variable x by x-1, we can apply Theorem XIII-6-1 tosingularities at x = oo. The following example illustrates such a case.

Example XIII-6-12. A system of the form

P(x) 0]b, y = lyd,

where P(x) = xm+ ahxin is a positive odd integer, and the ah are complexh=1

numbers, has a formal fundamental matrix solution of the form

1 0 f 1 1 e 0x F(x)0 x-1/2 ] l I -111 0 eE(t,a) I

where

+m

ak1/2 +00

bk(a)E xk + E xkk=1 k=1

2E(x, a) = (m +

2) x(m+2)/2 + i (.m + 2 - 2h) bh(a)x(m+2-2h)/2,

1<h<(m+2)/2

+00

and F(x) = xq> x-h Fr, with an integer q and 2x2 constant matrices Fh such thath--O

+00

Fh ,-6 0 as a formal power series in x-l. For details of construction,det h=O x-h

see [HsSj and [Sil3j.

XIII-7. An n-th-order linear differential equation at a singular point ofthe second kind

Let us look at the formal fundamental matrix solution (XIII.6.18) of system

(XIII.6.1). First, notice that if we set k = max { a :j = 1,2,... pthen k is thes

JJJ

order of singularity of system (XIII.6.1) at x = 0 (cf. Definition V-7-8). Let(XIII.7.1) 0<k1 <k2 <... <k9_1 <kq

be the distinct values among p non-negative rational numbersd1 d2 , ... ,. Also,s s s

set th = n, (h = 1, 2,... , q). It is easy to see that lh > 0 (h = 1, 2,... , q)d /s=kh

q P

and Elh = En, = n.h=1 7=1

Observation XIII-7-1. System (XIII.6.1) has th linearly independent formal so-lutions of the form(XIII.7.2) 'ih,v(x) = x'1h - exp[Qh,v(x)Wh,,,(x) (v = 1, ... ,;h = 1, ... , q),

where -yh,,, E C, Qh,,,(x) is either equal to 0 or a polynomial in x-1/' of the form

Qh.v(x) = I2h,vx-kh(I +O(x1/')) (11h,v E C and I2h,, # 0),

and the entries of lh,,,(x) are polynomials in logx with coefficients inDefine q + 1 points (Xh, Yh) (h = 0,1, ... , q) recursively by

(Xo,Y0)

=

(010),

(Xh, Yh) = (Xh-1 + eh, Yh-t + kheh) (h = 1, 2,... , q).

Let us denote by N the polygon whose vertices are q + 1 points (Xh, Yh) (h =0,1, ... , q). The polygon N has q distinct slopes kh given in (XIII.7.1).

7. AN N-TH-ORDER LINEAR DIFFERENTIAL EQUATION 437

Definition XIII-7-2. The polygon N is called the Newton polygon of system(XIII.6.1) at x = 0.

Observation XIII-7-3. In §XIII-5, it was shown that system (XIII.6.1) is equiv-alent to an n-th-order linear differential equation

n-1

(XIII.7.3) anb" r) + E atbtrl = 0,t=o

where b = x , at E C((xj), and an # 0 (cf. Theorem XIII 5 4). For the differentialoperator

n-1

(XIII.7.4) C(>,J = anb"r) + j:atdtrt.t=o

the Newton polygon N(C) is defined in the following way.

If a power series a = E c,,x'n E C((xj] is not 0, we set v(a) = min{m : cnm=0

0}. If a = 0, set v(0) = +oo. For operator (XIII.7.4), consider n + 1 points(Q, v(at)) (t = 0, 1, ... , n) on an (X, Y)-plane. Set

Pt = ((X, Y) : 0 < X < e, Y > v(at)}, and P = UP,.t=o

Definition XIII-7-4. The boundary curve C of the smallest convex set containingP is called the Newton polygon of the operator C at x = 0.

Denote by N(C) the Newton polygon of C at x = 0.

Definition XIII-7-5. Two Newton polygons are said to be identical if the twopolygons become the same by moving one or the other upward in the direction ofthe Y-axis.

Now, we prove the following theorem.

Theorem XIII-7-6. If system (XIII.6.1) and differential equation (X111.7.3) areequivalent in the sense of Theorem X111-5-4. then the two Newton polygons N andN(C) are identical.

Proof.

The proof of this theorem will be completed if the following three statements areverified:

(a) If./V(C) has only one nonvertical side with slope k, then differential equation

(XIII.7.3) is equivalent to a system xk+1 = A(x)g with a matrix A(x)

whose entries are power series in x11", and A(0) is invertible if k > 0, wheres is a positive integer such that sk is an integer.

(b) If (XIII.7.1) gives slopes of all nonvertical sides of N(L), then the operator Lis factored in the following way:

(XIII.7.5)

where, for each h, N(Lh) has only one nonvertical side with slope kh.(c) If L is factored as in (b) and each differential equation Lh[r)h] = 0 is equiv-

alent to a system xdih = Ah(x)uh, then the differential equation (XIII.7.3) is

equivalent to x !L = diag [A, (x), A2(x), ... , Aq(x)] y'.

Statement (a) can be proved by an idea similar to the argument which is used toreduce system (XIII.6.12) to (XIII.6.14) in Case 2 of the proof of Theorem XIII-6-1.A proof of Statement (b) is found in [Mall], [Si16] and [Si17, Appendix 1]. Lookat the system

Lq]u] = V, L1... L9_1[v] = 0.

Then, Statement (c) can be verified recursively on q without any complication. Theproof of Theorem XIII-7-6 in detail is left to the reader as an exercise.

Combining Observation XIII-7-1 and Theorem XIII-7-6, we obtain the followingtheorem.

Theorem XIII-7-7. If the distinct slopes of the nonvertical sides of N(L) aregiven by (XIII.7.1), then the n-th-order differential equation (XIIL7.S) has, at x =0, n linearly independent formal solutions of the form

(XIII.7.6) rlh,v(x) = xtin.,. exp[Qh.v(x)]Oh,v(x)

where(i) if

{(X, Yh-I + kh(X - Xh_1)) : Xh-1 C X < Xh}

is the nonvertical side of N(L) of slope kh, then

It, = Xh - Xh-1 (h = 1, 2,... , 9),

(ii) ryh,v E C, Qh,v(x) is either equal to 0 or a polynomial in x- 1/11 of the form

Qh,v(x) = lAh,vx-k"(1 + O(xi'°)) (I<h,v E C and uh,v 54 0),

and the quantities Oh,v(x) are polynomials in log x with coefficients in C[[x'/']]Here, s is a positive integer such that skh (h = 1,... , q) are integers, and PI + P2 +

+ Pq = n.

A complete proof of Theorem XIII-7-7 is found in [St].

We can construct formal solutions (XIII.7.6), using an effective method with theNewton polygon N(L). The following example illustrates such a method.

7. AN N-TH-ORDER LINEAR DIFFERENTIAL EQUATION 439

Example XIII-7-8. Consider the differential operator C = x63 - x62 - 5 - 1 orthe third-order differential equation C[q] = 0. The Newton polygon N(C) is given

by Figure 6. In this case, q = 2, kl = 0, and k2 = 1.

1 3

(i) For

Hence,

FIGURE 6.

(co 0). Then,

C[q] _ { (A + m)3 - (,\ + m)2} c,,,xa+m+1M=0

t o0

(A + m + 1)C,,,xa+m = 0.m=o

+00

k1 = 0, set i _ E c,,,xX+mM=0

+00

((A + 1)co = 0,1 (A+m+1)Cm = {(A+m-1)3-(A+m-1)2}C,,,_1

Therefore,

Thus,

for m > 1.

A = -1 and c,,, _ (m - 2)2(m -3)c,,,_1 for m > 1.

m -

A = -1, cl = -2co, c, ,, = 0 for m>2.

This implies that q = CO (x-1 - 2) is a solution of C[r7] = 0.

(ii) For k2 = 2, set n = exp[Ax-112](. Then,

).b"['1] = exp[Ax-112] \a -x_1 2)

[S] (n =0,1,2....

Therefore, C[n] = 0 is equivalent to

3 )2_ft Cb - Zx-1/2) - x I tS - -\X-112 (b - 2x-1/2) - 1 0.

440

Since

XIII. SINGULARITIES OF THE SECOND KIND

22x-1/21

2

- \x-112d +Ax-1/2

+ 42x-1,J

2x-1/23

/=d3 - tax-1/252 + (x_1/2

+4,\2x-I

/ d

- 18,\x-1/2 +3,\2X-1 + 8A3x-3/21

,

it follows that

/X I d - 2x-1/2/3 - x I d - 2x-1/2

2_(d - x-1/2) - 1

= x63 - (xhI2 + x 62 + (.x2 - 1 + 4Ax1/2) d2 4

+ 1

2- g31 x-1/2 - 1 + g-) + x1/2.

The Newton polygon of this operator has only one nonvertical side of slope 2 forarbitrary A (cf. Figure 7-1). However, if A is determined by the equation

A \3

2 8 0,

then the Newton polygon has a horizontal side (cf. Figure 7-2).


Hence, we can find two formal solutions of equation £[77] = 0:

exp [2x-1 12] xv, 11 + x1 /2f,J,

exp [ - 2x-1/2]xl2[1 + x1/2f2J,

where p1 and p2 are constants and f1 and f2 are formal power series in C[[x1/2JJ.

8. GEVREY PROPERTY OF ASYMPTOTIC SOLUTIONS 441

XIII-8. Gevrey property of asymptotic solutions at an irregular singularpoint

In this section we prove a result which is more precise than Theorem V-1-5([Mai]). In §XIII-3, we stated an existence theorem of asymptotic solutions for agiven formal solution of an algebraic differential equation (cf. Theorem XIII-3-6).If differential equation (XIII.3.3) has a formal solution, we can transform (XIII.3.3)to the form

(XIII.8.1) f_[y] = x"'G(x,y,by,... b"-1y) (b = x;),where

n

(XIII.8.2) G = F_ ah(x)bh,h=o

and G(x, yo, yl, ... , y.-I) is a convergent power series in (x, yo, yl ... y,,-,) (cf.(SS31 and [Mal2j). Here, it can be assumed without any loss of generality that

(i) ah (h = 0,1, ... , n) are convergent power series in x and a # 0,00

(ii) differential equation (XIII.8.1) has a formal solution p(x) = E a,,,x"' Em=0

CI[x]I,(iii) M is an integer such that for any differential operator At of order not greater

than n, the two Newton polygons N(C - xMIC) and N(C) are identical (cf.Definitions XIII-7-4 and XILI-7-5).

Using Theorem XIII-3-6, we can find(a) a good covering {S1, $2, ... , SN } at x = 0,(b) N solutions 01(x), 02(x), ... , ON (x) of (XIII.8.1) in S1, S2,... , SN, respec-

tively such that of are holomorphic and admit the formal solution p(x) astheir asymptotic expansions as x -+ 0 in St, respectively.

Set ui = 01 - 0t+1 on Se n Sjt+1. Then, ue are flat in the sense of Poincarein sectors Se n S(+1, respectively, where Sv+1 = S1. Furthermore, if we definedifferential operators IQ by

bh (tot + (1 - t)-Ol+l I.... )dt bh,Kt E

LJO 1 h-(x,...

O<h<n-1

then

(C - x"'K t)[ut] = 0 on se n St+l (f = 1, 2, .... N).

If the Newton polygon N(,C) has only one nonvertical side of slope 0, then x = 0is a regular singular point of C[iI] = 0. Therefore, in this case the formal solutionp(x) is convergent (cf. Theorem V-2-7). Let us assume that./V(C) has at least oneside of positive slope. In such a case, let

(XIII.8.3) 0 < kl < k2 < . . . < k9 < +00


be all of the positive slopes of the Newton polygon N(,C). Then, since N(1 C -xMJCe)and N(G) are identical, we must have

lut(x)l : -yeXp(-A Ixl-k] on Sr n Se+1

for a non-negative number -y and a positive number A, where k E {k1, k2 ... , ky}

(cf. Theorem XIII-7-7). Now, by virtue of Theorem XI-2-3, we obtain the followingtheorem.

Theorem XIII-8-1. Under the assumptions given above, the formal solution p(x)

is a formal power series of Gevrey order 1 and, for each t, the solution 0e admits

p(x) as its asymptotic expansion of Gevrey order k as x -. 0 in S1, where k E{k1ik2... ,kq}.

This theorem was originally prove in (Raml] for a linear system. For nonlinearcases, see, for example, ISi17, §A.2.4, pp. 207-211J.

Remark XIII-8-2. In Exercise XI-14, we gave the definition of a k-summablepower series. As stated in Exercise XI-14, if a formal power series f E C((x]]is k-summable in a direction argx = 0, there exists one and only one function

F E Al /k (Po, 0 - 2k-e,6+ 2k + e) such that J(F] = f , where Po and a are

positive unmbers. This function F is called the sum of f in the direction arg x = 0.If we use the idea of Corollary XIII-1-3 and Theorem XIII-6-7, we can prove thefollowing theorem concerning the k-summability of a formal solution of a nonlinearsystem

(XIII.8.4) xk+1 dy= A(x)y" + xb(x, yj.

dx

Theorem XIII-8-3. Under the assumptions(i) k is a positive integer,

(ii) A(x) is an n x n matrix whose entries are holomorphic in a neighborhood ofx = 0 and b(x, y-) is a C"-valued function whose entries are holomorphic in aneighborhood of (x, yl = (0,

(iii) A(0) is invertible,00

system (XIII.8.4) has one and only one formal solution y = p(x) xmpm andm=1

p(x) is k-summable in any direction arg x = 0 except a finite number of values of 6.Furthermore, the sum of p(x) in the direction arg x = 0 is a solution of (XIII.8.4).

To prove this theorem, it suffices to choose the good covering {S1, S2, ... , SN }at x = 0 in the proof of Theorem XIII-8-1 so that opening of each of sectors{S1, S2, ... , SN } is larger than k . We can prove a more general theorem.

EXERCISES XIII 443

Theorem XIII.8.4. Let a linear differential operator L = 6 - A(x) be given,

where 6 = , and A(x) is an n x n matrix whose entries are meromorphic in a

neighborhood of x = 0. Also, let (X111.8.3) be all the positive slopes of the Newtonpolygon N(C) of the operator C. Assume that

(1) k1 ?51

(2) C[ f] is meromorphic at x = 0 for a f (x) E C[[xiln.Then, there exist a finite number of directions arg x = 0 (1= 1,2,... , p) such thati f 0 34 Ot f o r I = 1 , 2, ... , p, there exist q formal power series j, (v = 1, 2, ... , q)satisfying the following conditions:

(a) for each v, the power series f is in the direction argx = 0,(b)f=fi+fz+...+fq.

A complete proof of this theorem is found in [BBRS]. In this case, f is said tobe {k1, k2 ... , kq }-multisummable in the direction arg x = 0. We can also provemultisummability of formal solutions of a nonlinear system. For those informations,see, for example, [Ram3], [Br), [RS21, and [Bal3j.

EXERCISES XIII

XIII-1. Using Observation XIII-6-5, diagonalize the following system:

`yildg _ x+ 5 x+ 8y where

dx 3 -x+ 1 l yz J

for large )x[.

XIII-2. Show that there is no rational function f (x) in x such that

(a"f)(r) + rzf(x) = x

Hint. One method is to show that the given equation has a unique power seriessolution in X-1 which is divergent at x = oo. Another method is to observe thatany solution of this equation has no singularity in ]x] < +oo except possibly atx = 0. Furthermore, if p(x) is a rational solution, then some inspection shows thatp(x) does not have any pole at x = 0. This implies that p(x) must be a polynomial.But, we can easily see that this equation does not have any polynomial solution(cf. {Si20J).

XIII-3. Find a cyclic vector for the differential operator £[ 3 = x;ii + 'A, with a

constant n x n matrix A of the form A = I At0 1, where for each j = 1, 2, the


quantity Aj is an of x n, matrix of the form

0 1 0 0 ... 00 0 1 0 ... 0

A?=0 0 0 0 ... 1

CYO, a 1,) a2,j C13,j ... ant -1j

with complex constants ah,j.

XIII-4. Find the Newton polygon and a complete set of linearly independent for-mal solutions for each of the following three differential equations:

(a) x(52y + 45y - y = 0, (b) x2(52y + xby - y = 0, (c) xb2y + x6y - y = 0,

dwhere b = x .

XIII-5. Find the Newton polygon of the following system:

1 1 0

x3 d = 0 x xdx

10 0 x2

Hint. For the given system, a cyclic vector is (1, 0, 0). This implies that the trans-formation

(T)

1 0 0X-2 x-2 0

X-4-2x-2 x-4 + x-3 - 2x-2 x -

changes the given system to

(E) (5 - 1)(x5 - 1)(x2(5- 1)u1 = 0, where b = x .

Note that transformation (T) is equivalent to u1 = yl, u2 = dyl, u3 = 52y1, andthe given system can be written in the form

1/2 = (x25 - 1/3 = (xb - 1)1/2, (5 - 1)1/3 = 0.

The Newton polygon of (E) has three sides with slopes 0, 1, and 2, respectively.On the other hand, The standard form of the given system in the sense of

Theorem XIII-6-1 is

3 dua(x) 0 0

X - = 0 b(x) 0 u,dx

0 0 x2

wherea(x) = 1 + 0(x), b(x) = x + 0(x2).

Hence, this also shows that the Newton polygon of the given system has three sideswith slopes 0, 1 and 2, respectively.

EXERCISES XIII 445

XIII-6. Let A(x) be an n x n matrix whose entries are holomorphic and boundedin a domain Ao = {x : 3x! < ro} and let j (x) be a Cn-valued function whoseentries are holomorphic and bounded in the domain Ao. Also, let Al, A2, ... , Anbe eigenvalues of A(O). Assume that det A(0) # 0. Assume further that two realnumbers 01 and 82 satisfy the following conditions:

(1) 01 < 02,(2) none of the quantities Aye-'ke (j = 1, 2, ... , n) are real and positive for a

positive integer k if 01 < 8 < 02,(3) Ape-ike' and Aqe-'k93 are real and positive for some p and q.

Show that there exist one and only one solution f = fi(x) of the system x'`+1 ds =

A(x)f+ x f (x) such that the entries of fi(x) are holomorphic and admit asymptoticexpansions in powers of x as x - 0 in the sector S = {x : 0 < !xj < r0, 81 - <

2k

argx<82+2k

Hint. If we use Corollary XIII-1-3 at x = 0, it can be shown that for each 8 in theinterval B1 -

2k< B < B2 + 7, a solution ¢'(x; 8) is found so that the entries of

O(x; 8) are holomorphic and admit asymptotic expansions in powers of x as x - 0in the sectorial domain S

'ra = x : 0 < jxj < ro, I arg x - 81 < 2k + ea }, where Ee is

a sufficiently small positive constant depending on 8. If 10 - 8'l is sufficiently small,then ¢(x; 0) = d(x; 8).

XIII-7.

(a) Show that j (x) _ (-1)m(m!)xr+L is a formal solution ofm=o

(E)2 dy

dx+y-x=0

and that f is not convergent except at x = 0.(b) For a given direction 8, find a solution fe(x) of (E) such that fe(x) ^- j (x) as

x 0 in the direction arg x = 8.(c) Calculate fe, (x) - fe, (x) for two given directions 81 and 02.

Hint. For Part (b), use the following steps:

Step 1. Apply Theorem XIII-1-2 to the given differential equation. To do this, we

must change x = 0 to t = oo by x = 1. Then, the differential equation becomesdu 1

= y - t . In this case, n = 1, r = 0, and the eigenvalue is ,u = 1. Set arg u = 0.

Then, the domain D(N, ry) (cf. (XI 11. 1. 15)) is

D(N,7) = {t:iti > N, I argt - 2gir! < 32 - f}

where q is an integer, N is a sufficiently large positive number, and -y is a sufficientlysmall positive number. In terms of x, the domain D(N, y) becomes

S(N, -t) = {x:O < IxI < N, I arg x - 2pirI < 2 - 'Y}

where p = -q. Since there is no singular point of the given differential equation inthe domain 0 = {x: 0 < Ixi < oo}, we can conclude that, for each fixed integer p,there exists a solution ¢p(x) of the given differential equation such that

(i) op(x) is analytic (but not single-valued) in !2,(ii) -ip(x) j (x) as x 0 in the domain

Dp = {x: 0 < xl < oc, argx - 2pl < 37r

_ 2

Step 2. It is easy to see that f (x) = el/=J 2t-1e-1l`dt (x > 0) satisfies the given0

differential equation and has the asymptotic property f (x) = j (x) as x 0. Since

¢p(x) - f (x) is a solution of the homogeneous differential equation x2dy + y = 0,

it follows s that ¢p(x) = f (x) + cell=, where c is a constant. From this, op(x) _

elI1 t-le-ll`dt follows for argx = 2pir.

Step 3. Using an argument similar to that of Step 2,

fe(x) =

rt4p(x) if 10 - 2p7ri <

21

Op(x) + cge,/, if 2 < 0 - 2pir < 2,

is obtained, where co is an arbitrary constant.

Step 4. Note that

ICt-le-1'`dt = 27ri,

where C is a counterclockwise oriented circle with the center at x = 0. Hence, usinganalytic continuation of f (x), we obtain

fi(x) = ¢p(x) + 2pirie' for argx = 2pir.

XIII-8. Show that the following differential equation has a nontrivial convergentpower series solution:

3d + dX+ y = Q.2x2

EXERCISES XIII 447

Hint.Step 1: The given differential equation has three linearly independent formal so-lutions

+oo +- +oo

1(x) = ell= 1 + amxm 2(x) = 1+ L. bmxm, and o3(x) = x+ amxm.M=1 m=2 m=3

In fact, Ol can be found through some calculation with the Newton polygon. Theother two can be found by solving the equations

f ao + 2a2 = 0, al + 603 = 0,(l a,, + (m + 1)m(m - 1)0,,,+1 + (m + 2)(m + 1)am+2 = 0 for m > 2

+00

for a formal solution E a,,,x"`.m=o

Step 2: The given differential equation has three linearly independent actual solu-

tions such that e-'/=O1(x) ^ as x 0 in the sector j argx + 7r1 <3ir

T,and 02(x) fi(x) and 03(x) ^_- 3(x) as x -. 0 in the sector I argxI < 32

Step 3: In the direction arg x = -7r,

02(x) -02(2e2x,)

= C201(x) and 03(x) -03(xe2ai)

= C301 (-T)

for some constants c2 and C3. Then, c3 (x) - c2.3(x) is a convergent power seriessolution of the given differential equation.Remark:

(1) See [HIJ.(2) This result was originally proved for a more general case in [Per1].(3) There is another proof based on Exercises V-18 and V-19 (cf. [HSWI).

XIII-9. Consider a system of differential equations

(E) xp+1du

= F(x, y, u),dx

where p is a positive integer, x is a complex independent variable, y is a complexparameter, it and F are n-dimensional vectors (i.e., E C'), and entries of F(x, y, u")are holomorphic with respect to (x, y, u-) in a neighborhood of (x, y, u) = (0, 0, 0).Assume that there exists a formal solution of system (E)

00

u = V, (X, y) _ yh+Gh(x),h=0

where coefficients zlih(x) are R'-valued functions whose entries are holomorphic ino9p

a neighborhood of x = 0. Assume also that O0(0) = 0 and det [(oo)] 54 0.

Show that i (x, y) is convergent in a neighborhood of (x, y) = (0, 0).


Hint. See ]Si14] and [Si21].

XIII-10. Consider a Pfaffian system

(S)J xl

aaxu= F(x, y, u,

alay4+t = G(x, y, u-),

where p and q are positive integers, x and y are two complex independent variables,iZ, F, and d are n-dimensional vectors (i.e., E C"), the entries of F and G areholomorphic with respect to (x, y, 11) in a neighborhood of (x, y, u) _ (0, 0, andsystem (S) is completely integrable, i.e., F and d satisfy the condition

yq+I5F (x, y, v') +

041(x, y, u)G(x, y, u = xp+l (x, y, u) + 5 (x, y, U-)F(x, y, u).

Assume that F(0, 0, 0) = 0, d(0, 0, 0) = 0, det [(oo)J 36 0, and

detL

(0, 0, 0)J j4 0. Show that system (S) has one and only one solution

11 = (x, y) such that i (0,O) = 0 and that entries of y) are holomorphicwith respect to (x, y) in a neighborhood of (x, y) = (0, 0).

Hint. This is an application of Exercise XIII-9.Step 1. Construct a formal power series solution

(FS)

of the system

11 = '1G(x, y) =+00

E yhrGh(x)h=o

yq+1 = G (x yay , ,

in such a way that coefficients t/ih(x) are holomorphic in a neighborhood of x = 0.

Step 2. For t _ l%i(x, y), note that

yl+9ay (i+ Pax - F(x,y,11))

0211= yl+qxl+p

8i-ax

8211= yi+qxl+p

0;-ax

8211= xl+pyl+q

5y-ax

= x1+p

y1+qaFay su ay

Y1+q aF

ay

x1+paGTX

ad ac 011ax + au ax

- OF'yl+ga11

OF -- &aG(x,y,ur)

- auF(x,y,u)

- +paG - adx

ax su F'(x, y, u)

811

(xl+Pax - F(x,y,i)

EXERCISES XIII 449

Using this result, it can be shown that (FS) also satisfies the system

(E) xP+i a = Ax, y, u)

Step 3. Upon applying Exercise XIII-9 to (E), the convergence of t%i(x, y) is proved.

XIII-11. Complete the proof of Theorem XIII-7-6 by verifying rigorously state-ments (a), (b), and (c) in the proof.

XIII-12. Show that the series(FS)

+00

y = P(x) = x-1/4 1 + >( -1)h(3)h xexp

[

E54hh!r(h +)

[_x3/2}3

is a formal solution of the differential equatond2-xZ

-xy = 0, where r is the Gamma-function.

XIII-13. Show that the differential equationd2-xZ

- xy= 0 has a unique solution

O(x) such that (1) b(x) is entire in x and (2) ¢(x)exp [3x3/21 admits the formal

series p(x) exp [x3I2J as its asymptotic exansion as x -oc in the sector I arg xj <

r, where p(x) is givenby (FS).

Remark. Ai(x) = 2(--X) is called the Airy function (cf. [AS, p. 446), [Wasl, pp.v/Fr

124-1261, and [01, pp. 392-394]).

XIII-14. Using the same notations in Exercises XIII-12 and XIII-13, show that if

w = elxp (2ril3 J then ¢(w'ix) and Q(wx) are two solutions of equation (S). Also,

(i) derive asymptotic expansions of ¢(w-lx) and m(wx),(ii) show that {Q(x),y5(w-lx)}, {0(w-1x),O(wx)}, and {d(wx),Q(x)} are three

fundamental sets of solutions of (S),1

(iii) show that if we set m(x) = c1O(w-1x)+c2y5(wx), then c2 = -w and [ctO

J3

w 11is equal to the 2 x 2 identity matrix,

(iv) using (iii), show that cl = -w-1.

XIII-15. Show that if O(x, A) is an eigenfunction of the eigenvalue problem

+00d2

(EP)ylda La

then

IT(i) Q(x) is entire in x and Q(x) exp

Jis a polynomial,

(ii) all negative odd integers are eigenvalues of (EP) and there is no other eigen-value,

(iii) for every non-negative integer n, Hn(x) = (-1)ne=2 Un (e-y2) is a polyno-r 2

mial, and -0n(x) = H.(x)exp I - 2J

is an eigenfunction of (EP) for the

eigenvalue -(2n + 1), L

r+00 +00

(iv) JOn(x)/m(x)dx = 0 if n 9k m, and / On(x)2dx = 2nn!y'.

0o J o00Remark. The polynomials Hn(x) are called the Hermit polynomials (cf. [AS, p.775] and [01, p. 49]).

XIII-16. Construct Green's function of the boundary-value problem

j- x2y = f (x), y(x)2dx < +00. Show also thato0

j+00 r+00(i)

JG(x,)2dxd< +oo,

00 00+

(ii) if f (x) is real-valued, f (x), f'(x), and f"(x) are continuous,+00

f (x)2dx <o0

+00 00

+00, and f{f"(x) - x2 f (x)}2dx < +oo, then the series Y2nn! (f, On)

00

xOn(x) converges to f(x) uniformly on the interval -oo < x < +oo, where

j+00

On (x) are defined in Exercise XIII-15, and (f, g) _ f (x)g(x)dx.00

Hint. See §VI-4.n-1 h

XIII-17. Consider a differential operator C[y] _ dxn + 1: an-h(x) h , whereh=0

+00

aj(x) = ajmx-'n E C[[x1]. Also, assume that aj,_m1 # 0 if a,(x) is notm=-mj

equal to zero identically, while mj = -oo if aj(x) is zero identically. In the (X, Y)-plane, consider the points Pj = (j, mj) (j = 1,... , n). Construct a convex polygonII whose vertices are (0, 0), pi, p2, ... , p, such that each pk is one of those pointsPj, and that all other points Pj are situated below the polygon. Set po = (0, 0)and Pk = (ak, Pk) (k = 1) ... , s), where an = n. Denote by pk the slope of thesegment k k+1 (k = 0, ... , s - 1). Then, Po > Pl > P2 > ... > p,-,. Assumethat pk>-1 (k<vo)and pk<-1 (k>vo). Show that

(i) the differential equation £[y] = 0 has n - at,,, linearly independent formal so-Me

lutions of the form ye(x) = xe" E Req(x)(logx)q (e = a,,. + 1,... , n), whereq=o

Rrq(x) E C[[x-1]], b E C, and the Mt are non-negative integers,(ii) if k < vo, then the differential equation £[y] = 0 has ak+1 - ak linearly

me

independent formal solutions of the form yr = e^<(x)x6, E Rrq(x)(log x)q (e =q-0

EXERCISES XIII 451

1 + ok,... , nk+1), where At(x) = Arxl+1 b + terms of lower degree, At E C,A, 96 0, bt E C, M, are non-negative integers, and Req E C[[x-lIP]], p being apositive integer.

Hint. Compare the polygon in this problem with the Newton polygon at x=0defined in §XIII-7.

XIII-18. Consider a differential operator C[yl = x9+1 dx + 1(x)y', where q is a00

positive integer and f2(x) _ > xlf2, with Hi being n x n constant matrices. As int=o

§V-4, the operator C can be represented by the matrix

01J,, +A

H1f12

where H = and [Oq]J_ _I...

Here, 0, is the nr x oo zero matrix and I,,,, is the oc x oo identity matrix. Let

Qo=So+No and A=S+N

be the S-N decomposition of 0o and that of A in the sense of §V-3. Also, letPo, P1, P2, ... , Pm, ... be n x n constants matrices and let AO be an n x n diagonalmatrix such that det Po 0 0 and that

A 1 0 0

0 A2 0

(1) SoPo = PoAo, Ao = 0 0 A3

0 0 0

and

(2) SP=PAo,

where

P1

P2P0P=

P.

0 0

0 0

0 0

0 An

Note that At are eigenvalues of So. Hence, A, are eigenvalues of 00. Show that(a) the matrix S represents a multiplication operation: y -+ o(x)y for an n x n

matrix o(x) = So + O(x) whose entries are formal power series in x,


(b) the matrix N represents a differential operator Lo[yM = x4+1 fy + v(x)y foran n x n matrix v(x) = No +O(x) whose entries are formal power series in x,

(c) C[f, = £o[yl + o(x)y and Go[o(x)yj = a(x)Go[yl,+00

(d) if we set P(x) _ F xmP,,,, then P(x)-1o(x)P(x) = A0,m=0

(e) if we set /C[uZ = P(x)-'C[P(x)uZ and Ko[u] = P(x)-1Go[P(x)ul, then K[u"j =Ko[ul + Aou,

(f) if we set

q+1 duVii

Ko[i ] = x + vo(x)u, vo(x) = .

vn1

then

vjk(x) = 0 if .\.7 # Ak

(cf. [HKS]).

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(Sil3] Y. Sibuya, Global Theory of a Second Order Linear Ordinary Differential Equa-tion with a Polynomial Coefficient, Math. Studies 18, North-Holland, Amster-dam, 1975.

[Si14] Y. Sibuya, Convergence of formal power series solutions of a system of nonlineardifferential equations at an irregular singular point, Lecture Notes in Math. No.810, Springer-Verlag, New York, 1980, pp. 135-142.

[Si15] Y. Sibuya, A theorem concerning uniform simplification at a transition pointand the problem of resonance, SIAM J. Math. Anal., 12 (1981), 653-668.

[Si16] Y. Sibuya, Differential modules and theorem of Hukuhara-Turrittin, Proc. ofInternational Conf. on Nonlinear Phenomena in Math. Sci., Academic Press,New York,1982, pp. 927-937.

[Si17] Y. Sibuya, Linear Differential Equations in the Complex Domain: Problems ofAnalytic Continuation, Translations of Math. Monographs 82, American Math-ematical Society, Providence, RI, 1990.

REFERENCES 461

[Si18] Y. Sibuya, Gevrey asymptotics, Ordinary and Delay Differential Equations, J.Wiener and J. K. Hale, ed., Pitman Research Notes in Math. Series 272, Pitman,London, 1991, pp. 207-217.

[Si19] Y. Sibuya, The Gevrey asypmtotics and exact asymptotics, Trends and Develop-ments in Ordinary Differential Equations, Y. Alavi and P. F. Hsieh, eds., WorldScientific, Singapore, 1993, pp. 325-335.

(Si20] Y. Sibuya, A remark on Bessel functions, Differential Equations, DynamicalSystems and Control Science, W. N. Everitt and B. Lee, eds., Lecture Notes inPure and Applied Math. No. 152, Marcel Dekker, New York, 1994, pp. 301-306.

[Si21] Y. Sibuya, Convergence of formal solutions of meromorphic differential equationscontaining parameters, Funk. Ekv., 37(1994), 395-400.

[Si22] Y. Sibuya, The Gevrey asymptotics in case of singular perturbations, Preprint,Univ. of Minnesota, 1997.

[SS1] Y. Sibuya and S. Sperber, Some new results on power series solutions of algebraicdifferential equations, Proc. of Advanced Seminar on Singular Perturbations andAsymptotics, May 1980, MRC, Univ. of Wisconsin-Madison, Academic Press,New York, 1980, pp. 379-404.

[SS2] Y. Sibuya and S. Sperber, Arithmetic properties of power series solutions ofalgebraic differential equations, Ann. Math., 113(1981), 111-157.

(SS3] Y. Sibuya and S. Sperber, Power Series Solutions of Algebraic Differential Equa-tions, Publication I.R.M.A. et Departement de Mathematique, Univ. Louis Pas-teur, Strasbourg, France, 1982.

[St] W. Sternberg, Uber die asymptotische Integration von Differentialgleichungen,Math. Ann., 81(1920), 119-186.

[Sz] G. Szego, Orthogonal Polynomials, Colloquium Publications XXIII, AmericanMathematical Society, Providence, RI, 1939.

[TD] S. Tanaka and E. Date, KdV Equation: An Introduction to Nonlinear Mathe-matical Physics, Kinokuniya, Tokyo, 1979 (in Japanese).

[1111] H. L. Turrittin, Convergent solutions of ordinary linear homogeneous differen-tial equations in the neighborhood of an irregular singular point, Acta Math.,93(1955), 27-66.

[Tu2] H. L. Turrittin, Reduction of ordinary differential equations to the Birkhoffcanonical form, Trans. Am. Math. Soc., 107(1963), 485-507.

[U] M. Urabe, Geometric study of nonlinear autonomous oscillations, Funk. Ekv.,1(1958), 1-83.

[Wasl] W. Wasow, Asymptotic Expansions for Ordinary Differential Equations, Dover,New York, 1987. (Also John Wiley, New York, 1965.)

[Was2] W. Wasow, Linear Turning Point Theory, Applied Math. Sci. 54, Springer-Verlag, New York, 1985.

[Wat] G. N. Watson, A theory of asymptotic series, Phil. Trans. Roy. Soc. of London(Ser. A), 211(1911), 279-313.

[We] H. Weyl, The Classical Groups, Princeton University Press, Princeton, NJ, 1946.[WhW] E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Cambridge

University Press, Cambridge, 1927.[Z] A. Zygmund, Trigonometric Series, Vol. 1, 2nd ed., Cambridge University Press,

Cambridge, 1959.

INDEX

,A.(r,a,b), 353Abel's formula, 80Abramowitz, M., 141, 449, 450Adjoint equation, 79Airy function, 449Algebraic differential equation, 109, 353Analytic differential equation, 20Analytic simplification of a system,

at a singular point, 421with a parameter, 391

Approximate solutions, c-, 10Approximations, successive, 3, 20, 21,

378, 418Arzelh, C., 9Arzelb-Ascoli Lemma, 9, 13, 30Ascoli, G., 9Associated homogeneous equation, 97, 99Asymptotics, Gevrey, 353

Poincar6, 342Asymptotic behavior of solutions, 197

of a linear second-orderequation, 226of linear systems, 209, 213, 219

Asymptotic (series) expansion, 343, 353definition, 343differentiation, 347integration, 347inverse of, 346of a holomorphic function, 345of Ei(z), 352of Gevrey order s, 353of Log(o(z)), 352of uniformly convergent sequences, 348product, 344sum, 344Taylor's series as, 345uniform with respect to a parameter,

343, 357, 361of Gevrey order s, 357uniqueness, 343

Asymptotic reduction of a linearsystem with singularity, 405of a singularly perturbed linearsystem, 381

Asymptotic solution, 374Asymptotic stability, 236, 300

a sufficient condition for, 241Atiyah, M. F., 109Autonomous system, 279

Balser, W., 371, 391, 401, 435, 443Baire's Theorem, 303Banach space, 25, 26, 103, 366, 394

differential equation in, 25, 394Bartle, R. G., 303Basis of the image space, 73Bellman, R., 71, 144, 148, 197Bendixson, 1., 27, 279, 291, 304Bendixson center, 261Bessel function, 125Bessel inequality, 162Birkhoff, G. D., 372, 390Block-diagonalization theorem, 190, 204,

380, 422, 423Block solution, 107Blowup of solutions, local, 17Borel, E., 349, 356Borel-Ritt Theorem, 349, 356Boundary-value problem, 144

of the second-order equation, 305Sturm-Liouville, 148

Bounded set of functions, 9Bourbaki, N., 69Braaksma, B. L. J., 359, 355, 443Branch point, 44

C, field of complex numbers, 20C", set of all n-column vectors in C, 20C[xJ, set of all polynomials, 109C{x}, set of all convergent power

series, 109C[[xJJ, set of all formal power series, 109C{x}", set of all convergent power

series with coefficients in C", 110C[[xfJ", set of all formal power series

with coefficients in C", 110C[[xJJ set of all power series of Gevrey

order s, 353

462

INDEX 463

C-algebra, 70

Cayley-Hamilton Theorem, 71, 220Canonical transformation of a

Hamiltonian system, 92, 94Cauchy, A. L., 1Cauchy-Euler differential equation, 140Caratheodory, C., 15Center, 255, 257, 261, 272

perturbation of, 271Cesari, L., 197Characteristic exponents of an

equation, 89, 203Characteristic polynomial of a matrix,

71, 72of a function of a matrix f(A), 84, 85partial fraction decomposition of theinverse of, 72, 76

Chevalley, Jordan-, decomposition, 69Chiba, K., 197Classification of singularity of a

homogeneous linear system, 132Coddington, E. A.,14, 28, 36, 58, 65, 69,

108, 137, 138, 144, 148, 153, 162,191, 197, 225, 233, 246, 266 274,281, 290, 293, 298, 313

Commutative algebra, 109Commutative differential algebra, 109,

362Commutative differential module, 110Comparison theorem, 58, 144Complexification, 252Contact point, exterior (interior), 295Conti, R., 15, 235, 279Continuity of a solution, 17, 29

with respect to the initial point andinitial condition, 29, 31with respect to a parameter, 31

Convergence of formal solution, 113, 118Coppel, W. A., 144, 148, 197Covering, good, 354Cullen, C. G., 71Cyclic vector, 424, 426

in S-N decomposition, 74, 119, 122Diagonalization theorem, Levinson's, 213Differentiability with respect to initial

values, 35Differential operator C, 120, 151, 169,

424, 437, 443S-N decomposition, 122normal form, 121

Dirac delta function, 8Distribution, of L. Schwartz, 8

of eigenvalues, 159Dulac, H., 235, 251Dwork, B., 8

E(s, A), 366E(s, A)", 367Eastham, M. S. P., 197, 217Eigenfunction, 162Eigenvalues of a matrix, 70, 71Eigenvalue-problem, 153, 154, 156

of a boundary-value problem, 153, 186Eigenvectors of a matrix, 70Elliptic sectorial domain, 296, 302Equicontinuity, 9

equicontinuous set, 47, 55, 63Euler, Cauchy-, differential equation, 140Existence, 1, 12

of solution of an initial-value problem,1, 3, 12, 21

with Lipschitz condition, 3without Lipschitz condition, 12of solution of a boundary-valueproblem, 148of S-N decomposition of a matrix, 74

Existence theorem for a nonlinearsystem at a singular point, 405

Existence theorem for a singularlyperturbed nonlinear system, 420

cxp[Al, exponential of a matrix A, 80exp[tAl with S-N decomposition, 82

Extension of a solution, 16of a nonhomogeneous equation, 17

Date, E., 144Deligne, P., 403, 424Denseness of diagonalizable matrices, 70Dependence on data, 28,Devinatz, A., 197Diagonalizable matrix, 70

Finite-zone, potentials, 189First variation of an equation, 284Flat of Gevrey order s, 342, 367, 385Flatto, L., 341Floquet, G., Theorem, 69, 80, 225Formal power series, 109, 343, 349

464

Formal power series (cont.)as an asymptotic series, 350of Gevrey order s, 353, 355with a parameter as an asymptoticseries, 351

Formal solution, 109, 342convergence, 115, 118Taylor's series as, 345uniqueness, 115

Fourier series, 165, 166, 167Fuchs, L., 137Fundamental matrix solution, 78, 105,

125, 122of a system with constant coefficient,

81

with periodic coefficient, 87, 88of an equation at the singularity ofthe first kind, 125of an equation at the singularity ofthe second kind, 136, 432, 433

Fundamental theorem of existence anduniqueness, 3, 12without Lipschitz condition, 8

Fundamental set of linearlyindependent solutions, 78normal, 202

Gel'fand, 1. M., 181Gel'fand-Levitan integral equation, 181General solution, 7, 105Gerard, R., 108, 112, 125Gevrey asymptotics, 353Gevrey order 8, 356, 367

asymptotic expansion of, 353flat of, 342, 367, 385formal power series, 354uniformly on a domain, 357, 361

Gevrey property of asymptoticsolutions at an irregular singularity,

441solutions in a parameter, 385

Gingold, H., 104, 197, 217, 231Global properties of solutions, 15GL(n, C), general linear group, 69Good covering, 354Green's function, 150, 154Gronwall, T. If., Lemma, 3, 398

Haber, S., 304

INDEX

Hamiltonian system with periodiccoefficients, 90

Harr inequality, 27Harr's uniqueness theorem, 27Harris, W. A., Jr., 142, 143, 197, 433, 447Hartman, P., 14, 27, 28, 39, 58, 69, 144,

148, 153, 162, 197, 246, 281, 293,298

Heaviside function, 8Hermit polynomials, 450Higher order scalar equations, 98Hille, E., 103Hirsch, M. W., 69, 251Holomorphic function asymptotic to a

formal series, 349, 356Homogeneous systems, 78Homomorphism of differential algebras,

359, 362, 365Howes, F. W., 304, 339Hsieh, P. F., 78, 104, 108, 130, 197, 217,

231, 372, 424, 436, 452Hukuhara, M, 25, 26, 50, 108, 131, 197,

212, 218, 261, 403, 428, 447Hukuhara-Nagumo condition, 218

for bounded solution of a secondorder equation, 212for bounded solution of a system, 218

Hukuhara-Turrittin theorem, 428Humphreys, J. E., 69Hyperbolic sectorial domain, 296, 302Hypergeometric series, 138

Improper node, 252, 254perturbation of, 261stable, 252, 253, 256unstable, 252, 253

Independent solutions, 78Index of isolated stationary point, 293Index of a Jordan curve, 293Indicial polynomial, 112Infinite-dimensional matrix, 118Infinite-dimensional vector, 120, 393Infinite product of analytic function, 196Initial-value problem, complex

variable, 21partial derivative of a solution as asolution of, 33

Initial-value problem, real variable, 1, 28,32, 39

INDEX 465

nonhomogeneous equation, 96nonlinear, 1, 16, 32, 39second order equation, 324, 333

Inner product, 151Instability, 243Instability region, 189Integral inequality, 377, 418Invariant set, 280, 299Irregular singular point, 134, 441Iwano, M., 447Iwasaki, 1., 138

J, map of A,(r,a,b) to C[[x]] 353one-to-one, 359onto, 356

Jacobson, N., 69Jordan-Chevalley decomposition, 69Jordan canonical form, 421, 426Jordan curve, index of, 293Jost solution, 168, 180, 194Jurkat, W., 136, 391, 401, 402, 435

K, the field of fractions of C[[x[], 424K[u'J, differential operator, 123Kaplan, J., 197Kato, J., 44, 66Kato, T., 103Kimura, II., 138Kimura, T., 197Kneser, H., theorem, 41, 47, 305Kohno, M., 78, 108, 130, 452Komatsu, H., 8

C, differential operator, 120, 151, 169,424, 437, 443

normal form, 121calculation of, 130

C+, limit-invariant set, 280Laplace transform, inverse, 103LaSalle, J., 279, 281Lebesgue-integrable function, 15Lee, E. B., 68, 106Lefschetz, S., 279, 281Legendre, equation, 141

polynomials, 141, 192Leroy transform, incomplete, 354Lettenmeyer, F., 142Levelt, A. H. M., 108, 125, 435Levin, J. J., 340

Levinson, N, 14, 28, 36. 65, 69, 108, 137,138, 144, 153, 162, 191, 197, 225,233, 246, 266, 274, 281, 290, 293,298, 304, 340, 341

Levinson's diagonalization theorem, 21:3Levitan, B. M., 181Levitan, Gel'fand-, integral equation, 181Liapounoff, A., 197Liapounoff function, 239, 309, 311Liapounoff's direct method, 281Liapounoff's type number, 198

calculation of, 203multiplicities, 202of a function, 198of a system at t = oc, 201, 204of a solution, 199properties, 198

Lie algebra, 90Limit cycle, 292Limit invariant set, 280Lin, C.-H., 370Lindelbf, E., 1, 28, 359Lipschitz condition, 3, 43, 64

constant, 3sufficient condition for, 5existence without, 8

Local blowup of solutions, 17Logarithm of a matrix, 86

log[ 1 + ltf ], 86, 88log[8(w)[, 88log[8(w)2], 88, 94log[s], 86, 88log[S2], 88

Lutz, D. A., 136, 197, 391, 401, 402, 435

M,,(C), set of all n x n matricesin C, 69

MacDonald, 1. C., 109Magnus, W., 196Mahler, K., 112Maillet, E., 112, 353, 441Malgrange, B., 438Manifold, 45

differential equation on, 45stable, 243unstable, 246

Markus, L., 68, 69, 96, 106Matrix, 69

functions of a matrix f (A), 84

466

Matrix (cont.)diagonalizable, 70in S-N decomposition, 74infinite-dimensional, 118nilpotent, 71in S-N decomposition, 74norm of a matrix A, 11A11, 69semisimple, 70symplectic, 93, 95upper-triangular, 70

Maximal interval, 16Maximal solution, 52, 54, 56, 66Minimal solution, 52, 54, 66Milnor, J. W., 298Moser, J., 135Mullen, F. E., 232Multiplicity of an eigenvalue of a

matrix, 72, 74Multiplicity of Liapounoff's type

number, 201of a system of equations with constantcoefficients, 202of a system of equations with periodiccoefficients, 202

Multipliers, of an autonomoussystem, 284of a system with periodiccoefficients, 89, 183, 203of periodic orbits, 318, 319

N(1), Newton polygon of G, 437, 439Nagumo, M., 27, 50, 197, 212, 218, 304,

306, 327, 330, 333, 334, 342Nagumo condition for uniqueness ofsolution, 64, 65

Nevanlinna, F., 342Nevanlinna, R., 359Newton polygon, 437, 439Nikiforov, A. F., 141Nilpotent matrix, 71

in S-N decomposition, 74Node, 253, 254

improper, 253, 255proper, 253, 256

Nonhomogeneous equation, 17Nonhomogeneous initial-value problem,

96Nonlinear equation, 18, 372, 384

INDEX

Nonlinear initial-value problem, 17, 2128, 32, 41

Nonstationary point, 291, 292Nonuniqueness of solution of an

initial-value problem, 41Norm, (in C[[x]], 366

in C[[x]l", 367in CI[x]j 366of a continuous function, 162in E(s, A)", 367of a matrix A, 1PAIl, 69of a vector, 1, 345of an infinite-dimensional vector, 394

Normal form of a differentialoperator, 121

Normal fundamental set, 202Null space of a homomorphism, 366

Olver, F. W. J., 138, 141, 449, 450O'Malley, R. E., 304Orbit, 251, 279

periodic, 304, 318, 319of van der Pol equation, 313, 318

Orbitally asymptotical stability, 283Orbital stability, 283Orthogonal sequence, 166Osgood, W. F., 63Osgood condition for uniqueness of

solution, 63, 65

Palka, B. P., 196Parabolic sectorial domain, 296Parseval inequality, 166Partial differential equation, 39Peano, G., 1, 28Periodic coefficients, system with, 87Periodic orbit, 291, 297, 318, 319

of van der Pol equation, 313, 318Periodic potentials, 183Periodic solution, 184, 288

of van der Pol equation, 313, 318Perron, 0., 57, 212, 443Perturbation, of initial-value problem, 43

of a center, 271of a proper node, 266of a saddle point, 263of a spiral point, 270of an improper node, 263

Peyerimhoff, A., 391, 401, 402

INDEX 467

Pfaffian system, 448Phase plane, 251Phase portrait of orbits, 251, 302Phillips, R. S., 103Phragmen-Lindelof theorem, 359, 370Picard, E., 1Poincarc, H., 279, 291, 304Poincarc asymptotics, 342Poincarc-Bendixson Theorem, 291, 293Poincarc's criterion, 300Popken, J., 112Potentials, 175, 178

finite-zone, 189,reflectionless, 175, 177, 178, 181periodic, 183

Projection, P,(A), 72properties, 73

Proper node, 253, 254perturbation of, 266stable, 253, 255, 267unstable, 253

R, real line, IR", set of all n-column vectors in R, 1Rabenstein, A. L., 36, 70, 101Ramis, J.-P., 342, 360, 363, 371, 386, 420,

443Reflection coefficient, 175Reflectionless potentials, 177, 178, 181

construction, 178Regular singular point, 131, 133, 134, 394Ritt, J. F., 349, 356

S,,, set of all diagonalizable matrices, 70S-N decomposition, 74, 75

existence, 74of a differential operator, 121of a function of a matrix f(A), 84of a matrix for a periodic equation, 88of infinite order, 118, 120of a real matrix, 75uniqueness, 75

Saddle point, 252, 255, 256perturbation of, 261

Sansone, G., 15, 235, 279Saito, T., 274Sato, Y., 336Scalar equations, higher order, 98Scattering data, 172, 175

Schwartz, L., 8Second-order equation, boundary-value

problem, 304, 308Sectorial region, 296

elliptic, 296, 302hyperbolic, 296, 302parabolic, 296

Self- adjoi nt ness, 151, 155Semisimple matrix, 70Shearing transformation, 209Shimomura, S., 138Sibuya, Y., 44, 66, 69, 78, 96, 108, 112,

130, 131, 136, 142, 143, 197, 217,227, 232, 304, 342, 360, 363, 365,369, 371, 372, 381, 382, 390, 391,392, 399, 419, 420, 424, 436, 438,447, 448, 452

Singular solution, 42Singular perturbation of van der Pol

equations, 330Singularity of a linear homogeneous

system, 132, 134of the first kind, 113, 132, 136of the second kind, 132, 134, 136, 403n-th-order linear equation, 436regular, 131, 133, 134, 394irregular, of order k, 134, 441

Smale, S., 69, 251Solution, asymptotic, 374

asymptotically stable, 236, 241independent, 78fundamental matrix, 78, 105, 125, 222periodic, 184, 288of van der Pol equation, 313, 318singular, 42trivial, 236, 243uniqueness, 1, 3, 21

Solution curves, 42, 49Spiral point, 254, 255

perturbation of, 270stable, 254, 256, 270, 272unstable, 254, 272

Sperber, S., 112Stability, 235Stability region, 189Stationary point, 280, 291, 312

isolated, 294Stable manifolds, 243

analytic structure, 246

468 INDEX

Stegun, 1. A., 141, 449, 450Sternberg, W., 403, 438Stirling formula, 358, 364Sturm-Liouville problem, 148Subalgebra, 109Successive approximations, 3, 20, 21, 378,

418Sufficient condition, for asymptotic

stability, 241for uniqueness, 61, 63for unstability, 243

Summability of a formal power series,371, 442, 443

Symplectic group Sp(2n, R), 92, 93matrix of order 2n, 92, 93 107

Tahara, H., 112Tanaka, S., 144Taylor's series, 345Transform, incomplete Leroy, 354Transformation, shearing, 209Transversal, to orbit, 291, 292

with respect to a vector, 291Trivial solution, 236, 243Turrittin, H. L., 391, 403, 428Turrittin, Hukuhara-, theorem, 428Two dimensional system with constant

coefficients, 251

Unbounded operators, 103Uniformly asymptotic series, 343, 357,

361Uniqueness, Lipschitz condition, 1, 3, 64

Nagumo condition, 64, 65of S-N decomposition of a matrix, 75of asymptotic expansion, 343of formal solution, 115of solution of a boundary-valueproblem, 150

468

of solution of an initial-valueproblem, 1, 3, 21, 79Osgood condition, 63, 65sufficient conditions for, 61, 63

Unstable manifold, 246Upper-triangular matrix, 70Urabe, M., 288Uvarov, V. Q., 141

V image of P, (A), 73direct sum for C", 73

van der Pol equation, 312for a large parameter, 322for a small parameter, 319singular perturbation of, 330

Variation of parameters, formula of, 101Vector, infinite-dimensional, 120, 393

norm of, 393Vector space of solutions, 78

Wasow, W., 108, 304, 342, 372, 382, 449Watson, G. N., 141, 342, 359Weinberg, L., 142, 447Weyl, H., 69Whittaker, E. T., 141Winkler, S., 196Wintner, A., 197Wronskian, 145, 149, 168, 172

Xie, F., 197, 217x"C{x}", 110x"C((Fx)l", 110

Yoshida, M., 138

Zeros, of eigenfunctions, 185of solutions, 144

Zygmund, A., 167

Universitext (continued )

Moise: Introductory Problems Course in Analysis and TopologyMorris: Introduction to Game TheoryPolster: A Geometrical Picture BookPorter/Woods: Extensions and Absolutes of Hausdorff SpacesRamsay/Rlchtmyer: Introduction to Hyperbolic GeometryReisel: Elementary Theory of Metric SpacesRickart: Natural Function AlgebrasRotman: Galois TheoryRubel/Colliander: Entire and Meromorphic FunctionsSagan: Space-Filling CursesSamelson: Notes on Lie AlgebrasSchif: Normal FamiliesShapiro: Composition Operators and Classical Function TheorySimonnet: Measures and ProhabilitSmith: Power Senes From a Computational Point of ViewSmorytiski: Self-Reference and Modal LogicStillwell: Geometry of SurfacesStroock: An Introduction to the Theory of Large DeviationsSunder: An Invitation to von Neumann AlgebrasTondeur: Foliations on Riemannian ManifoldsWong: Wcyl TransformsZhang: Matrix Theory Basic Results and TechniquesZong: Strange Phenomena in Convex and Discrete GeometryZong: Sphere Packing-,

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