Churchill, R. v. - Fourier Series and Boundary Value Problems

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FOURIERSERIESAND

BOUNDARYVALUE

PROBLEMSTHIRD EDlTION

RUEL v. CHURCHILLPRüFESSOR EMERITUS OF MA THEMA TlCS

THE UNIVERSITY OF MICHIGAN

JAMES WARD BROWNPROFESSOR OF MA THEMA TICS

THE UNIVERSITY OF MICHIGAN-DEARBORN

INTERNATIONAL STUDENT EDITION

McGRA W-HILL INTERNA TIONAL BOOK COMP ANY

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1..,

,,FOURIER SERIES AND BOUNDARY VALUE PROBLEMSlNTERNATIONALSTUDENTEDITlON

Copyright @ 1978

Exclusive rights by McGraw-HiII Book Co - Singapore, for

manufacture and exportoThis book cannot be re-exported from thecountry to which it is consigned by McGraw-HiII.

7th Printing 1985

Copyright @ 1978, 1963, by McGraw-HiII, Inc.

AU rights reserved.

Copyright 1941 by Mc-Graw-HiII, Inc. AIl rights reserved.

Copyright renewed 1959 by Ruel V. Churchill.No part of this publication may be reproduced, stored

in a retrieval system, or transmitted, in any form or byany means, electronic, mechanical, photocopying,

recording, or otherwise, without the prior

written permission of the publisher.

This book was set in Times Roman

The editors was A. Anthony Arthur~The production supervisor was David Damstra;

The cover was designed by John Hite.

Library 01 Congress Cata10ging in Pub6cation DataChurchill, Ruel Vance, date

Fourier series and boundary valueproblems

Bibliography: p.Includes index.

l. Fourier series 2. Functions, Orthogonal.

3. Boundary value problems. 1. Brown, James Ward,

joint author. 11.Title

QA404.C6 1978 515'.2432 77-12779ISBN 0-07-010843-9

When ordering this title use ISBN O.07-Y85100-X

PRINTED AND BOUND BY B& JO ENTERPRISE PTE LTD. S'PORE.

CONTENTS

PREFACE

1. PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS

Two Related Problems. Linear Boundary Value Problems. The VibratingString. Modifications and End Conditions. Other Examples or Wave Equa-tions. Conduction of Heat. Discussion or the Heat Equation. Laplace'sEquation. Cylindrical and Spherical Coordinates. Types of Equations andConditions.

2. THE METHOD OF SEPARATION OF VARIABLES

Linear Combinations. Examples. Series of Solutions. Uniform Conver-gence and Example. Separation of Variables. The Fourier Sine Series. APlucked String. Ordinary Differential Equations. General Solutions ofPartial DifferentialEquations. On Other Methods. Historical Development.

3. ORTHOGONAL SETS OF FUNCTIONS

The Inner Product orTwo Vectors. Orthonormal Sets of Vectors. Piecewise

Continuous Functions. Orthogonality of Functions. Generalized FourierSeries. Approximation in the Mean. Closed and Complete Sets. Complex-valued Functions. Other Types ofOrthogonality. Sturm-Liouville Problems.Orthogonality of the Eigenrunctions. Uniqueness or Eigenfunctions. An-other Example.

4. FOURIER SERIES

The Basic Series. Example. Fourier Cosine and Sine Series. Further Ex-amples. One-sided Derivatives. Preliminary Theory. A Fourier Theorem.Discussion of the Theorem. Other Forms or Fourier Series. The Ortho-normal Trigonometric Functions.

5. FURTHER PROPERTIES OF FOURIER SERIES

Uniform Convergence. Observations. Differentiation ofFourier Series. In-tegration of Fourier Series. More General Conditions.

6. BOUNDARY VALUE PROBLEMS

Formal and Rigorous Solutions. The Vibrating String Initially Displaced.Discussion of the Solution. Prescribed Initial Velocity. NonhomogeneousDifferentialEquations. ElasticBar. Temperaturesina Bar. Other BoundaryConditions. Dirichlet Problems. Fourier Series in Two Variables. Obser-vations and Further Examples.

.,

vii

1

23

46

78

104

116

)

vi CONTENTS I

I

7. FOURIER INTEGRALS AND APPLlCATIONS

The Fourier Integral Formula. . Preliminary Lemmas. A Fourier IntegralTheorem. The Cosine and Sine Integrals. The Exponential Form. FourierTransforms. More on Superposition of Solutions. Temperatures in a Semi-infinite Solid. Temperatures in an Unlimited Medium.

8. BESSEL FUNCTIONS AND APPLlCATIONS

Bessel's Equation. Bessel Functions J.. Some Other Bessel Functions.RecurrenceRelations. Bessel'sIntegral Form ofJ.. Consequencesof Bessel'sIntegral Form. The Zeros of Jo(x). Zeros of Related Functions. Or-thogonal Sets of Bessel Functions. The Orthonormal Functions. Fourier-Bessel Series. Temperatures in a Long Cylinder. Heat Transfer at theSurfacc of the Cylinder. Vibration of a Circular Membrane.

9. LEGENDRE POL YNOMIALS AND APPLlCATIONS

Solutions of Legendre's Equation. Orthogonality of Legendre Polynomials.Rodrigues' Formula and Norms. Laplace's Integral Form of p.. FurtherOrder Properties. Legendre Series. Convergencc of the Series. DirichletProblems in Spherical Regions. Steady Temperatures in a Hemisphere.Other Orthogonal Sets.

10. UNIQUENESS OF SOLUTIONS

Cauchy Criterion for Uniform Convergence. Abel's Test for Uniform Conver-gence. Uniqueness of Solutions of the Heat Equation. Example. Solutionsof Laplacc's or Poisson's Equation. An Application. Solutions of a WaveEquation.

BIBLlOGRAPHY

INDEX

149

174

218

II

248

263

267

"

PREFACE

This is an introductory treatment of Fourier series and their applicationsto boundary value problems in partial differential equations of engineeringand physics.It is designed for students who have completed the equivalentof one term of advanced calculus. The physical applications, explained insome detail, are kept on a fairly elementary level.

The first objective is to introduce the concept of orthogonal sets offunctions and representations of arbitrary functions in series of functionsfromsuch sets.The most prominent specialcases,representations of functionsby trigonometric Fourier series,are given special attention. Fourier integralrepresentations and expansions in series of Besselfunctions and Legendrepolynomials are also treated.

The secondobjective is a clear presentation of the classical method ofseparation of variables used in solvingboundary value problems with the aidof those representations. Some attention is given to the verification ofsolutions and to uniqueness of solutions; for the method cannot be presentedproperly without such considerations. Other methods are treated in the firstauthor's books "Operational Mathematics" and "Complex Variables andApplications:"

This edition is a revision of the 1963edition of the book, the originalone having been published in 1941.Considerable attention has been"givenhere to improving the exposition and including new problems.To note a fewof the specificchanges in this third edition, the motivation of the conceptof inner products of functions has been amplified, the proofs of theconvergencetheorems for Fourier series and integrals have been completelyrewritten, and there is more detail in the discussion of Besseland Legendrefunctions of the second kind. There are more footnotes referring to othertexts which give proofs of the more delicate results in advanced calculusthat are occasionally needed, and the Bibliography has been updated. Somerearrangement oftopics was founddesirable.For example,the chapter on theapplications of Fourier series to the solution of boundary value problemsnow precedes the chapter on Fourier integrals, thus allowing the studentto reach those applications earlier.

The chapters on Besselfunctionsand Legendre polynomials,Chapters 8and 9, are essentially independent of each other and can be taken up ineither order. Much of Chapter 5, on further properties of Fourier series,andChapter 10,on uniquenessof solutions, can be omitted to shorten the course;this also applies to some sections of other chapters.

vii

)

viii PkEFACE ",'

In preparing this edition, the authors have benefited from the commentsof a variety of people, inc1udingstudents. A number of specificsuggestionswere made by Craig Comstock and George Springer. Thanks are also dueR. P. Boas, Jr. for providing the reference to Kronecker's extension of themethod of integration by parts, which is often useful in evaluating integralsneeded in finding Fourier coefficients;and the derivation of the laplacian inspherical coordinates that is given was suggested by a note of R. P. Agnew'sin the AmericanMathematical Monthly, vol. 60 (1953).Finally, the authorsare indebted to Catherine A. Rader for her careful typing of the manuscript.

RUEL V. CHURCHILL

JAMESWARD BROWN

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CHAPTER

ONE

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8JBflol. Two Related Problems j;ECA- DE LA -.. FA!ULTA[)

We shall be concemed hedaEWt!~A~~r"~~R~s of problems. One typedeals with the representation of arbitrarily given'1l1nctionsby infinite seriesof functions from a prescribed setoThe other consists of boundary valueproblems in partial differential equations, with emphasis on equations thatare prominent in physics and engineering.

Representations by series are encountered iQsolving boundary valueproblems. The theories of those representations can be presented independ-ently. They have such attractive features as the extension of concepts ofgeometry, vector analysis, and algebra into the field of mathematicalanalysis. Their mathematical precision is also pleasing. But they gain inunity and interest when presented in connection with boundary valueproblems.

The set of functions that make up the terms in the series representationis determined by the boundary value problem. Representations or expan-sions in Fourier series,which are certain types of series of sine and cosinefunctions, are associated with the more common boundary value problems.We shall give special attention to the theory and application of Fourierseries. But we shall also consider extensions and generalizations of suchseries, incJuding Fourier integrals and series of Bessel functions andLegendre polynomials.

A boundary value problem is correctly set if it has one and only onesolution. Physical problems associated with partial differential equationsoften suggest boundary conditions under which a problem may be correctlysetoIn fact, it is sometimes helpful to interpret a problem physicallyin orderto judge whether the boundary conditions may be adequate. This is aprominent reason for associating such problems with their physical applica-tions, aside from the opportunity to display interesting and importantcontacts between mathematical analysis and the physical sciences.

EQUAT~ONS OFPHYSICS

PARTIAL DIFFERENTIAL

,

2 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS [SECo2

The theory of partial differential equations gives results on the existenceof solutions of boundary value problerns. But such results are necessarilylirnited and cornplicated by the great variety of features: types of equationsand conditions and types of dornains. Instead of appealing to general theoryin treating a specific problern, our approach will be to actually find asofution, which can often be shown to be the only possible solution.

2. Linear Boundary Value Problems

Theory and applications of ordinary or partial differential equations in adependent variable, or function, u usually require that u satisfy not only thedifferential equation throughout sorne dornain of its independent variable orvariables but also sorne conditions on boundaries of that dornain. The equa-tions that represent those boundary conditions rnay involve values of deriva-tives of u, as well as u itself, at points on the boundaries. In addition, sorneconditions on the continuity of u and its derivatives within the dornain andat the boundaries are required.

Such a set of requirernents constitutes a boundary value problem in thefunction u. We apply that terrn whenever the differential equation is accorn-panied by sorne boundary conditions, even though the conditions rnay notbe adequate to ensure a unique solution of the problern.

The three equations

(1)u"(x) - u(x) = -1

u'(O)==O, u(l) = O,(O< x < 1),

for exarnple, constitute a boundary value problern in ordinary differentialequations. The dornain of the independent variable x is the intervalO < x < 1 whose boundaries consist of the two points x = Oand x = 1.Thesolution of this problern which, together with each of its derivatives, iscontinuous everywhere is found to be

(2) u(x) = 1 - cosh xcosh 1 .

Frequently it is convenient to indicate partial differentiation by writingindependent variables as subscripts. Ir, for instance,u is a function of x andy, we rnay write

ouux or ux(x,y)for ox'

oZuUxxfor oxz'

oZu

Uxyfor oy ox '

and so on. We shall always assurne that the partial derivativesof u satisfyconditions allowing us to write UXy= Uyx'

sEc.2] PARTIAL DlFFERENTIAL EQUATIONS OF PHYSICS 3

AIso,weshall be free to use the syrnbols ux(xo,Y)and u"",,(xo,Y)to denotethe values of the functions oujox and o2ujox2, respectively, on the line x = xc;and corresponding syrnbols will be used for boundary values of otherderivatives.

The problernconsistingof the partialdifferentialequation .

(3) u"",,(x,y)+ Uyy(x,y)= O (x> O,y> O)

and the two boundary conditions

u(O,y)= ux(O,y) (y> O),

u(x,O)= sin x + cos x (x> O)

is an example of a boundary value problem in partial differential equations.The dornain is the first quadrant of the xy planeoThe reader can verify thatthe function

(4)

(5) u(x,y) = e-Y(sin x + cos x)

is a solution of that problem. This function and its partial derivativ~sareeverywherecontinuous in the two variables x,y together and bounded in thedornain x > O,Y > O.

A differentialequation in a function u, or a boundary condition on u, islinearif it is an equation oJthefirst degree in u and derivativesoJu. Thus theterrns of the equation are either prescribed functions of the independentvariables alone, including constants, or such functionsmultiplied by either uor one of the derivatives of u. -

The differential equations and boundary conditions (1), (3), and (4)above are alllinear. The differential equation

(6) zu"""+ xy2uyy - e"'u.= J(y,z)

in u(x,y,z) is linear. But the equation

u"""+ UUy= x

. is nonlinear in u(x,y) because the term UUyis not of the first degree as analgebraic expression in the two variables u and Uy.

Let the letters A through G denote either constants or functions of theindependent variables x and y only. Then the general linear partial differen-tial equation oJ the second order in u(x,y) has the form

(7) Auxx+ Buxy+ CUyy+ Dux+ Euy+ Fu = G.A boundary value problem is linearif its differentialequation and all its

boundary conditións are linear. Problem (1) and the problem consisting ofequations (3) and (4) are examples of linear boundary value problerns.

Methods of solution presented in this book do not apply to nonlinearboundary value problems.

-----

4 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS [SEC. 3

A linear differential equation or boundary condition is homogeneousifeachof its terms, other than zero itself,is of the firstdegree in the function uand its derivatives.

Equation (7) is homogeneous in a domain if and only if G = Othrough-out that domain. Equation (6) is nonhomogeneous, unlessf(y,z) = OforaIlvalues of y and z. Equation (3) and the first of conditions (4) are homogene-ous, but the second of those conditions is notoHomogeneous equations willplaya central role in our treatment of linear boundary value problems.

3. The Vibrating String

A tightly stretched string, whose position of equilibrium is some interval onthe x axis, is vibrating in the xy plane. Each point of the string, with coordi-nates (x,O)in the equilibrium position, has a transversedisplacementy(x,t) attime t. We assume that the displacementsy are smaIl relative to the length ofthe string, that slopes are smaIl, and that other conditions are such that themovement of each point is essentiaIlyin the direction of the y axis.Then attime t the point has coordinates (x,y).

Let the tension T of the string be great enough that the string behavesasif it were perfectlyflexible.That is, at each point (x,y) on the string the partof the string on the left of that point exerts a force of magnitude T in thetangential direction upon the part on the right, and the effect of bendingmoments at the point can be neglected.The magnitude of the x componentof the tensile force is denoted by H. See Fig. 1,where that x component hasthe same positive sense ;1sthe x axis. Our final assumption is that H isconstant; that is, the variation of H with x and t can be neglected.

Those idealizing assumptions are severe,but they are justified in manyapplications. They are adequately satisfied, for instance,by strings of musi-cal instruments under ordinary conditions of operation. MathematicaIly,the

y

-H

T ""..."

"......

V(x,t)

o~(x, O) (x + ~x, O) x

Figure 1

SECo 3] PARTIAL DlFFERENTIAL EQUATIONS OF PHYSICS S

assumptions willlead us to a partial differential equation in y(x,t) which islinear.

Now let V(x,t)denote the y component ofthe tensile forceexerted by theleft-hand portion of the string on the right-hand portion at the point (x,y).We take the positive sense of Vas that of the y axis. If ()(is the slope angle ofthe string at the point (x,y) at time t, then

- V(x,t) = tan ()(= yx(x,t),H

as indicated in Fig. 1.Thus the y component V(x,t) of theforce exerted at timet by the part of the string on the left of a point (x,y) upon the part on the rightis given by the equation

(1) V(x,t) = -HyAx,t) (H > O),

which is basic for deriving the equation ofmotion ofthe string.Equation (1)is also used in setting up certain types of boundary conditions.

Suppose that all external forces such as the weight of the string andresistance forces, other than forces at the end points, can be neglected.Consider a segment of the striñg not containing an end point and whoseprojection on the x axis has length ~x. Sincex components of displacementsare negligible, the mass of the segment is (j ~x where the constant (j is themass of the string per unit length. At time t the y component of the forceexerted by the string on the segmentat the left-hand end (x,y) is V(x,t), givenby equation (1).The tangential forceexerted on the other end of the segmentby the part of the string on the right ofthat end is also indicated in Fig. 1.Itsy component is evidently

(2) H tan f3= HyAx + ~x, t)

where f3is the slope angle of the string at the right-hand end ofthe segment.Note that it is, in fact, - V(x + ~x, t); the negative sign signifies that theforce is exerted upon the part of the string on the left by the part on the right.The acceleration of the end (x,y) in the y direction is Ytt(x,t). From Newton'ssecond law of motion (mass times acceleration equals force), it follows that

(3) (j ~x Ytt(x,t)= -HyAx,t) + HyAx + ~x, t),

approximately, when ~x is small. Hence

Ytt(x,t)= H lim yix + ~x, t) - yAx,t) H(j 4.>;-0 ~x = b yxx(x,t)

at each point where the partial derivatives exist.

~/


Thus the function y(x,t), representing the transverse displacements in astretched string under the conditions stated above, satisfies the one-dimensional wave equation

(4) yu(x,t) = a2Yxx(x,t) (a2 = H/{))

at points where no external forces act on the string. The constant a has thephysical dimensions of velocity.

4. Modifications and End Conditions

JWhen external forces parallel to the y axis act along the string, let F denotethe force per unit length of string, the positive sense of F being that of the yaxis. Then a term F tu must be added on the right-hand sideof equation (3),Seco3, and the equation of motion is

(1) yu(x,t) = a2YxAx,t)+~.

In particular, with the y axis vertical and its positive sense upward,suppose that the external force consists of the weight of the string. ThenF ~x = - ()~ g, where the positive constant g is the acceleration of grav-ity; and equation (1) becomes the linear nonhomogeneous equation

(2) yu(x,t) = a2YxAx,t) - g.

In equation (1), F may be a function of x, t, y, or derivativesof y. If theexternal forceper unit length is a damping forceproportional to the velocityin the y direétion, for example, F is replaced by - By" where the positiveconstant B is a damping coefficient.Then the equation of motion is linearand homogeneous:

(3) yu(x,t) = a2YxAx,t) - by,(x,t) (b = B/{)).

If one end x = O of the string is kept fixed at the origin at all times t ~ O,the boundary condition there is clearly

(4) y(O,t)= O (t ~ O).

But if that end is permitted to slide albng the y axis and if the end is movedalong that axis with a displacementf(t), the boundary condition is the linearnonhomogeneous condition

(5) y(O,t)=f(t) (t ~ O).

When the left-hand end is looped around the y axis and a force g(t)(t > O)in the y direction is applied to that end, g(t) is the limit of the force

SECo 5] PARTlAL DlFFERENTlAL EQUATlONS OF PHYSICS 7

V(x,t) described in Seco3 as x tends to zero through positive values. Theboundary condition is then

(6) -HyAO,t) = g(t) (t > O).

The negative sign disappears, however, if x = O is the right-hand endbecause g(t) is then the force exerted on the part of the string to tne left ofthat end.

5. Other Examples of Wave Equations

We can present further functions in physics and engineering which satisfywave equations and still limit our attention to fairly simple physicalphenomena.

(a) Longitudinal Vibrationsof Bars. Let the coordinate x denote dis-tances from one end of an elastic bar in the shape of a cylinder or prism toother cross sections when the bar is unstrained. Displacementsof the'endsorinitial displacementsor velocitiesofthe bar, all directed lengthwisealong thebar and uniform over each cross section involved~cause the sections of thebar to move in the direction ofthe x axis.At time t the longitudinal displace-ment of the section labeled x is denoted by y(x,t). Thus the origin ofthe displacement y of that section is fixed outside the bar, in the plane ofthe original referenceposition of that section (Fig. 2).

At the same time a neighboring section labeled x + Llx, to the right ofsection x, has a displacement y(x + Llx, t); thus the element ofthe bar withnatural length Llx is stretched by the amount y(x + Llx, t) - y(x,t). Weassume that such an extension or compression of the element satisfiesHooke's law, so that the force exerted upon the element over its left-handend is, except for the etTectof the inertia of the moving element,

- AE y(x + Llx, t) - y(x,t)Llx

whereA istheareaofa crosssectionand E is the modulusof elasticity of thematerial in tension and compression.When Llx tends to zero, it then followsthat the totallongitudinal forceg(x,t) exerted on the section x by the part ofthe bar on the left of that section is given by the basic equation

(1) g(x,t) = -AEYx(x,t).

1-- y(x.t)I

I I óx ! \O x x+óx x Figure 2

8 FOURIER SERIES AND BOUNDARY YALUE PROBLEMS [SECo 5

Let 1Jdenote the mass of the material per unit yolume. Then, applyingNewton's second law to the motion of an element of the bar of length ~x,write

(2) 1JA~x y,,(x,t) = -AEyAx,t) + AEyAx + ~x, t),

where the last term represents the force on the element at the end x + ~x.We find, after diyiding by 1JA~x and letting ~x tend to zero, that

(3) y,,(x,t) = a2YxAx,t) (a2 = E/1J).

Thus the longitudinal displacements y(x,t) in an elastic bar satisfy thewave equation (3) when no externallongitudinal forcesact on the bar otherthan at the ends. We have assumed only that displacements are smallenough thatlHooke's law applies and that sections remain planar after beingdisplaced. The elastic bar here may be replaced by a' column of air; thenequation (3) has applications in the theory of sound.

The boundary condition y(O,t)= ° signifiesthat the end x = ° ofthe baris heId fixed. If, instead, the end x = ° is free when t > O,then no forceactsacross that end; that is, g(O,t)= ° and, in view of equation (1),

(4) yAO,t)=° (t > O).

(b) Transverse Vibrationsof Membranes. Let z denote small displace-ments in the z direction, at time t, ofpoints ofa flexiblemembrane stretchedtightly oyer a frame in the xy plane. The tensile stress T, the tension per unit

.Iength across any line on the membrane, is large; and the magnitude H of itscomponent parallel to the xy plane is assumed to be constant. Then theinternal force in the z direction at a section x = xo, per unit length of thatline, is -HzAxo,y,t), corresponding to the force V (Sec.3) in the yibratingstring. The force in the z direction at a section y = Yo, perunit length, is- Hzy(x,yo,t).

Consider an element of the membrane whose projection onto the xyplane is a rectangle with opposite vertices (x,y) and (x + ~x, Y + ~y). WhenNewton's second law is applied to the motion of that element in the zdirection, we find that z(x,y,t) satisfiesthe two-dimensional waye equation

(5) z" = a2(zxx + Zy¡.) (a2 = H/1J).

Here 1Jis the mass of the membrane per unit area. Oetails of the derivationare left to the problems.

If an external transverse force F(x,y,t) per unit area act~ oyer themembrane, the equation of motion takes the form

(6) z" = a2(zxx+ ZyJ+~.

SECo 5] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 9

PROBLEMS

1. Give details in the derivation of equation (1), Seco4, for the forced vibrations of astretched string.

2. A tightly stretched string with its ends tixed at the points (0,0) and (2e,0) hangs at restunder its own weight. The y axis points vertically upward. State why the static displacementsy(x) of points on the string satisfy the boundary value problem

a2y"(x) - 9 = O

y(O) = y(2e) = O.

(O < x < 2e),

By solving this problem, show that the string hangs in the parabolic are

2a2

(ge2

)(x-e)2=- y+-9 2a2 (O::; x::; 2e)

tiand that the depth of the vertex of the are varies directly with b, the mass of the string per unitlength, and with e2 and inversely with H.

3. Use express ion (!). Seco3, for the vertical force V and the equation of the are in whichthe string in Problem 2 lies to show that the vertical force exerted on that string by eithersupport is gbe, half the weigh t of the string.

4. A strand of wire 1 ft long, stretched between the origin and the point (1,0), weighs0.032 lb (gb = 0.032, 9 = 32 ft/sec2) and H = 10 lb. At the instant t = Othe strand líes along thex axis but has a velocity of 1 ft/sec in the direction of the y axis, perhaps beca use the supportswere in motion and were brought to rest at that instant. Assuming that no external forces actalong the wire, show why the displacements y(x,t) should satisfy this boundary value problem:

y,,(x,t) = 10'yx.(x,t) (O < x < 1, t > O),

y(O,t) = y(l,t) = O, y(x,O)= O, y,(x,O)=1.

5. The physical dimensions of H, the magnitude of the x component ofthe tensile force in

a string. are those of mass times acceleration: M L T- 2 where M denotes mass. L length. andT time. Show that since a2 = H/b, the quantity a has the dimen'sionsofvelocity LT-1.

6. The end x = Oof a cylindrical elastic bar is kept tixed, and a constant compressive forceof magnitude Fa units per unit area is exerted at all times t > O over the end x = e. The bar isinitially unstrained and at rest with no external forces acting along it. Verify that the functiony(x,t) representing the longitudinal displacements of cross sections should satisfy this boundaryvalue problem:

y,,(x,t) = a2Yxx(x,t) (O < x < e, r > O; a2 = E/b),

y(O,r)= O, Ey.(e,r) = - Fa, y(x,O) = y,(x,O) = O.

7. The left-hand end x = O of an elastic bar is elastically supported in such a way that the

longitudina! force per unit arca exerted on the bar at that end is proportional to the displace-ment of the end, but opposite in sign. Show that the end condition there has the form

Ey.(O,r)= Ky(O,r) (K>O).

8. Deriveequation (6), Seco5. Also show that the sratíe transverse displacements =(x,y) of amembrane, over which a transverse force F(x,y) per unit area acts, satisfy Poisson's equation

Zxx + z,.,. + f = O (f= F/H).

10 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS [SECo 6

6. Conductionof "eat

Thermal energy is transferred from warmer to cooler regions interior to asolid body by conduction. It is convenient to refer to that transfer as thefiowof heat, as if heat were a fluid or gas which diffusesthrough the body fromregions of high concentration into regions of low concentration.

Let Po denote a point (xo,yo,zo)interior to the body and S aplane orsmooth curved surface through Po.At a time to thefiux ofheat <1>(xo,yo,zo,to)across S at Po is the quantity ofheat per unit area per unit time that is beingconducted across S at that point. Flux is measured in such units as caloriesper square centimeter per second.

Ir u(x,y,z,t)denotes the temperatures at points of the solidat time t and ifn is a coordinate that represents distance normal to the surface S at the

"

interior point Po (Fig. 3), the flux across S will be in the positive direction ofn if dufdn is negative and in the negative direction if dufdn is positive. Theflux is accordingly positive or negative, or zero; and a fundamental postlllateof the mathematical theory of heat conduction states that there is a thermalcoefficient K of the material such that

(1) <1>= - K dudn (K > O).

That is, the flux<1>(xo,Yo,zo,to)across S is proportional to the value at Po, toof the directional derivative of the temperature function u in a directionnormal to S at Po. The coefficientK is called the thermalconductivityof thematerial.

In particular, the flux across each of the planes x = Xo, y = Yo, andz = Zothrough the point Po has the values

(2) <1>1= -Kux, <1>2= -Ku)" <1>3=-Ku:,

respectively, where the partial derivatives are evaluated at Po, to. For a fixedvalue of t, the derivatives ux, 11).,and u: are the components ofthe gradient ofU. Since the direction of the vector grad u is that in which u increases mostrapidly, the vector

(3) J = - K grad u

Figure3

SECo6] PARTIAL DlFFERENTIAL EQUATlONS OF PHYSICS 11

is caIled the current of heat fiow, or total flux, at the point Po.t It meas-ures the flux across the isothermal surface

u(x,y,z,to) = u(xo,yo,zo,to)

through Po. The projection J(n)of J onto the normal to the arbitrary surfaceS is the flux <1>given by equation (1):

J(n) = <1> = - K du/dn.

Another thermal coefficient of the material is the heat capacity Coperunit volume. This is the quantity ofheat required to raise the temperature of aunit volume of the material by a unit on the temperature scale.

Unless otherwise stated, we shall always assume that the coefficients Coand K are constants. vlith that assumption, a second postulate of the math-ematical theory is that conduction leads to a temperature function u which,together with its derivative u/ as weIl as those of the first and second orderwith respect to x, y, and z, is a continuous function throughout each doma ininterior to the solid in which no heat is generated.

To derive the differential equation satisfied by u, first consider a spheri-cal surface So with center at Po, smaIl enough that aIl points within and onSo are interior to the solid body. The integral of Cou(x,y,z,t) over the regionRo bounded by So is a measure ofthe instantaneous heat content Qo(t) of Ro.Hence

Qó(t) = fr J.fto cou dV = CoJJto ~~dV,

where dV is written for the volume element in the integral.Heat enters Ro only by conduction through its boundary surface So.Ir n

is distance normal to that surface, positive in the outward direction, and ifdA denotes the element of area of So,then the time rate Qó(t)of conductionof heat into the sphere has the alternative expression

(4)

(5).. du ..

Qó(t) = LisoK dndA = -.ItJ(n) dA.

According to Gauss's divergence theorem for transforming surface inte-grals into volume integrals,

(6) rr J(n)dA = rrr div J dV... So ... Ro

t Results from Ihree-dimensional veClor calculus Ihal are used in Ihis seclion are developed

in, for example, Kaplan (1973, pp. 204 ITand 337 IT), listed in the Bibliography.


Thus expression (5) can be written

(7)Qó(t) = - JJJRO div J dV = K JJto div (grad u) dV.

Now div (grad u) is the laplacian

V2u = Uxx + uy). + u%%;

so it followsfrom expressions (4) and (7) that

(8) fff (cour- K V2u) dV = O.... Ro

Equation (8) holds for each spherical region Ro about Po, provided onlythat all points of Ro are interior to the solid and that Ro js free from sources.The integrand is a continuous function of x, y, and z in Ro at time t. If theintegrand has a positive value at Po, its continuity requires that there be aspherical region R I about Po, where R I is interior to Ro, such thatcou,- K V2u > O throughout R.. The integral in equation (8), with Roreplaced by R¡, would then have a positive value rather than the value zero.The corresponding contradiction arises ifwe assume that the integrand has anegative value at Po. Consequently, at Po

Cou, - K V2u = O.

Since Po represents any interior point of the body in a domain free fromsources, the temperature function u(x,y,z,t) in every such doma in satisfies theheat equation

(9) u, = k(uxx + uY)'+ u%%),

where

Kk=-., Co

The coefficient k is called the thermal diffusivity of the material.

(10)

7. Discussion of the "eat Equation

Suppose that the temperatures within a solid are independent of z; that is,there is no flowof heat in the z direction. The heat equation then reduces tothe equation for two-dimensional flow parallel to the xy plane:

(1) ur= k(uxx+u)').). ,For one-dimensional flow parallel to the x axis the equation becomes

(2) u,(x,t) = kUxAx,t).

SECo 7] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 13

Ir temperatures are in a steady state, where u does not vary with time, theheat equation becomes Laplace's equation V2u = O:

(3) u:c:c+ Uyy + Un = O.

We have assumedso far that heat is neither generated nor lost within thesolid but only transferred by conduction. Ir there is a uniform sourcethroughout the solid that generates heat at a constant rate q, where q de-notes the quantity of heat generated per unit volume per unit time, the heatequation takes the form

(4) cou, = K V2u + q.

The derivation of the heat equation in Seco6 can be modified to obtainequation (4) if the postulate on the continuity of u and its derivatives isextended to inc1udecases in which a constant uniform source is presento

Boundary conditions which describe the thermal conditions on the sur-faces of the solid and the initial temperature distribution throughout thesolid must accompany the heat equation ifwe are to determine the tempera-ture function U. The conditions on the surfaces may be other than justprescribed temperatures. Suppose, for example, that a surface is perfectlyinsulated. From Seco6 we know that if du/dn is the directional derivative of uin the direction of the outward normal at any point on the surface, the fluxout of the solid through that point is the value of - K du/dnthere. Since theflux is zero through any point on an insulated surface, it follows thatdu/dn=Oat sucha point.

On the other hand, there may be surface heat transfer from a boundarysurface into a medium whose temperature is a constant T. We assume thatthe flux of heat through that surface at each point is proportional to thedilTerencebetween the temperature at the point and the temperature T.According to this approximate linear law of cooling, sometimes calledNewton's law,there is a positive constant H, known as the surface conduct-ance of the material, such that

(5) \. du-K - = H(u - T)dn

(6)

at points on the surface. Condition (5) is often writtendu- = h(T - u)dn (h. = H/K > O).

For an iIIustration, consider a slab occupying the region O~ x ~ b. Ifthe surface x = Ois insulated and if there is surfaceheat transfer across thesurface x = b into a medium at temperature zero, the desired boundaryconditions are

uAO,y,z,t) = Oand

ujb,)',z,t)= -hu(b,)',z,t).

Finally, it should be pointed out that the postulate = - K du/dn alsoapplies to simple difTusion to represent the flux of a substance which isdifTusing within a porous solido In this case the function u denotes eoneentra-tion (the mass of the difTusing substance per unit volume of the solid) and Kis the eoefficient of dijJusion. Since the content of the substance in a givenregion is the integral of u over that region, we can replace eo by unity in thederivation in Seco6 to see that the concentration u satisfies the equation ofdijJusion

(7) u, = K V2u.

PROBLEMS

1. Lel u(x) denole Ihe sleady-slale lemperalures in a slab bounded by Ihe planes x = Oand x = b when those boundaries are kepl al fixed lemperalures u = Oand u = Uo. respeclively.

Sel up Ihe boundary value problem for u(x) and solve ¡llo show Ihal

Uou(x)=¡;X,

<II=-K~b

where <11is Ihe flux of heal across each plane x = XO (O S Xo S b).

2. A slab occupies Ihe region O S x S b. There is a conslant flux of heal <11= 4>0inlo Iheslab Ihrough Ihe face x = O. The face x = b is kepl allemperalUre u = O. Sel up and solve Iheboundary value problem for Ihe sleady-slale lemperatures u(x) in Ihe slab.

Ans. u(x) = ~ (b - x).

3. Lellhe slab O S x S b be subjecled lo surface heallransfer according to Newton's lawal ils faces x = Oand x = b, Ihe surface conduclance H being Ihe same on each face. Verify Ihalif the medium x < Ohas temperalure zero and Ihe medium x > b has Ihe constanl temperalureT, Ihen Ihe boundary value problem for sleady-slale lemperalures in Ihe slab is

Ku'(O) = Hu(O).

u"(x) =0

- Ku'(b) = H[u(b) - T],

(O< x < b).

where K is Ihe Ihermal conduclivily oflhe malerial in Ihe slab. Wrile h = HIK and derive Iheexpression

Tu(x)= ---(hx+l)bh + 2

for Ihose lemperalures.

4. Lel heat be generaled al a rale q per unit volume per unil lime Ihroughoul a sol id.Poinl oul why expression (5), Seco 6, should be modified lo read

Qó(r)= - rr J(n)dA + ¡Off q dV...So . . .Ro

"

SEC. 8] PAR TI AL DlFFERENTlAL EQUATIONS OF PHYSICS 15

Assume that q is a constant or a continuous function of x, y, Z. Then postulate the continuityof u and its derivatives as before. Complete the derivation of the form (4), Seco 7, of the heatequation where q is such a function.

S. Let u(x,t) denote temperatures within a sol id which is free from sources. Consider acylindrical element interior to the so lid such that the axis of the cylinder ís parallel to the x axisand of length Óx. Letting A denote the area of the base of the cylinder, show why

A[ -Kux(x,t) + Ku.(x + Óx, t)] = coA Óx u,(x,t).

approximately. if Óx is small. Thus obtain equation (2), Seco 7.

6. In Problem 5 let heat be generated at a constant rate q per unit volume per unít timethroughout the solid and derive the equation

cou, = Kuxx + q.

7. Let u(x,t) denote temperatures in a slender wire Iying along the x axis when surface heattransfer takes place along the wire into the surrounding medium at fixed temperature T.Assuming that the time rate of transfer per unit length is proportional to u(x,t) - T, use theprocedure indicated in Problem 5 to derive the equation

u, = kuxx - h(u- T),

where h is a positive constan t.

8. Suppose that the thermal coefficients Coand K are functions of x, y, z, and t. Modify thederivation in Seco6 to show that the heat equation takes the form

(cou), = (Kux)x + (Ku).). + (Ku,),

in a domain where all functions and deriva tives involved in that equation are c~ntinuous.

9. It is sometimes convenient to change the unit of time so that the constant k or a2 inthe heat or wave equation may be considered as unity. Show, for example, that the substitutionT = kt may be used to write the two-d.mensional heat equation (1), Sec.7, in the form

Ut = Uu + u),)""

8. Laplace's Equation

A functionu(x,y,z}that iscontinuous,togetherwithitspartialderivativesofthe first and second order, and satisfies Laplace's equation

(1) V2u = O,

where V2u is the laplacian

(2) V2u = div(grad u) = Uxx+ Uyy+ Uzz'

is called a harmonic function.We have seen that the steady-state temperatures at points interior to a

solid in which no heat is generated are represented by a harmonic function u.In fact, the temperature function u serves as a potential function for the flowof heat in the sense that - K grad u represents the current of heat flow byconduction, and so - K du/dn represents the flux of heat' in the direction ofthe coordinate n.


The steady-state concentration of a diffusingsubstance (Sec. 7) is like-wise represented by a harmonic function. From equation (5),Seco5, we cansee that the static transverse displacements z(x,y) of a stretched membranesatisfyLaplace's equation in two dimensions. Here the displacementsare theresult of displacements perpendicular to the xy plane of parts of the framethat supports the membrane when no external forces are exerted except atthe boundary.

Among the many physical examples of harmonic functions, the velocitypotential for the steady-state irrotational motion of an incompressible fluidis prominent in hydrodynamics and aerodynamics.t

An important harmonic function in electrical field theory is the elec-trostatic potential V(x,y,z) in a region of space that is free from electriccharges.The potential may be produced by any static distribution of electriccharges outside that region. The vector - grad V at (x,y,z) represents theelectrostatic intensity, the electrical force that would be exerted upon a unitpositive charge placed at that point. The fact that V is harmonic is a con-sequence of the inverse-square law of attraction or repulsion betweenelec-tric charges.

Likewise,gravitational potential is a harmonic function in domains ofspace not occupied by matter.

The physical problems treated in this book are limited to those forwhich the differential equations were ¡derivedin this chapter. A few otherproblems in the physical sciencesare noted in order to indicate the greatvariety of applications of partial differential equations. Derivations of thedifferential equations for such problems can be found in books on hydrody-namics, elasticity, vibrations and sound, electrical field theory, potentialtheory, and other branches of continuum mechanics.

9. Cylindrical and Spherical Coordinates

The cylindrical coordinates (p,eP,z),shown in Fig.4, determine a point Pwhose rectangular cartesian coordinates are

(1) x = p cos eP, y = p sin eP, z = Z.

Thus p and ePare the polar coordinates in the xy plane ofthe point Q whereQ is the projection of Ponto that plane.

The relations (1) can be written

p = .JX2 + y2, eP= tan-l(f),(2) z = z,

t See Churchill. Brown, and Verhey (1974, Chap. 9). lisled in Ihe Bibliography.

SEc.9] PARTIAL DlFFERENTIAL EQUATIONS OF PHYSICS 17

z

P(P. <P.z)

y

x Figure 4

where the quadrant to which the angle epbelongs is determined by the signsof X and y, not by the ratio y/x alone.

Let 1/denote a function of x, y, and z. Then, in view of equations (1), it isalso a function of the three independent variables p, ep,and z. If 1/and itsderivatives of the first and second order with respect to those variables are'continuous functions, we can obtain the laplacian of 1/in terms of them byusing the chain rule for differentiating .composite functions.

In view of equations (2), I

01/ Ol/Op 01/oep 01/ X 01/ Y+ ---OX- opox oepox - op-JX2+ y2 oep X2 + y2'

where y and z are kept fixed in the differentiation %x, epand z are keptfixed in the differentiation Ojop, and so on, and ep is measured in radians.Similarly,

(3)

(4)01/ 01/ . Y 01/ X--- +--ay- op -JX2 + y2 oep X2 + lO

From expression (3) we can write

021/ 01/ o

(X

)

01/ o

(

y

)X o

(01/

)Y o

(01/

)ox2= opox p - oepOX p2 + P ox op - p2 OX oep o

The chain rule applies to the last two indicated derivatives, giving

o(

01/

)

021/ X 021/ Y

OX op = Op2P - oepop p2 '

o(

01/

)

021/ X 021/ Y

ox oep = iJp oep P - ('ep2 p2 o

Substituting and simplifying, we find that

021/ y201/ 2xy 01/ X2 021/ 2x)' 021/ 'l 021/+--+ +--OX2 - p3 op p4 oep p20p2 p3 op oep p4 Oep2.

(5)

I\:1

t

¡

(6)

In like manner, we find from equation (4) that

02U X20U 2xy OU y202u 2xy 02U X202U+--+--+--oy2 - p3 op p4 o<p p20p2 p3 op o<p p4 o<p2o

According to equations (5) and (6), then,

(7)02U 02U 02U 1 OU 1 02U-'+---+__+__0OX2 oy2 - Op2 P op p2 o<p2'

and so the laplacian of u in cylindrical coordinates is

(8)02U 1 OU 1 02U 02U

V2u = Op2 + P op + p2 o<p2+ OZ2 .

We may group the first two terms and use the subscript notation for partialderivatives to write this in the form

2 1 ( )1

V u = - PUp p + 2 uq,q,+ Uz:'P p

The spherical coordinates (r,<p,lJ)of a point (Fig. 5) are related to x, y,and z as follows:

(9)

(10) x = r sin lJ cos <p, y = r sin lJ sin <p, z = r cos lJo

The coordinate <pis common to cylindrical and spherical coordinates,and the coordinates in the two systems are related by the equations

(11) z = r cbs lJ, p = r sin lJ, <p= <po

Expression (8) can be transformed into spherical coordinates quitereadily by means of the proper interchange of letters, without any furtherapplicationof the chain rule.This is accomplished in three steps, describedbelow.

y

Figure 5

SECo 9] PAR TI AL DIFFERENTIAL EQUATIONS OF PHYSICS 19

First, we observe that, except for the names ofthe variables involved,transformation (11) is the same as transformation (1). Since transformation(1) gave us equation (7), it follows that .

02U 02U 02U 1OU 1 02U-+---+--+--OZ2 Op2- or2 r or r2 002 .

Next, we note that when transformation (11) is used, the counterpart ofequation (4) is

(12)

OU p ou Z OU+--op - r or r200.

With this and the first two of equations .(11),we are able to write

1 ou 1 02U 1OU cot OOU 1 02U--+--=--+--+P op p20q.} r or r2 00 r2 sin2 O04J2.

Finally, by grouping the first and last terms in expression (8),and alsothe second and third terms there, we see that, according to equations (12)and (13), the lapladan of u in sphericalcoordinatesis,

2 02U 20u 102u 1 02U cot OOUV U=-+--+ +--+--.or2 r or r2 sin2 O04J2 r2 002 r2 00

(13)

(14)

Other forms of this expression are

V2u =! (ru)rr+ 2 .12 Ou",,,,+ ~ (Ugsin e)gr r SIn r SIn u(15)

and

(16)2 1

(2

)1 1

(.

)V u ="2 r u, ,+ 2 . 2 eu",,,, + ~ Ug SIn eg.r r SIn r SIn u

The two-dimensionallaplacian is sOq1etimes needed in the polar coor-dinates (p,4J).That form, already given j¡f equation (7), is of cóurse a specialcase of expression (8) when u is independent of the coordinate z.

PROBLEMS

l. Derive expressions (4) and (6) in Seco 9.

2. Show that in Seco 9 expressions (15) and (16) are the same as (14).

3. Obtain equation (7). Seco9. by starting with its right-hand side and transforming eachterm there by means of the chain rule. That is. find

OU ouox ouo)'-, = - - + - --. and so on.op oxop oyop

4. Let u(r) denote the steady-state temperatures in a solid bounded by two concentricspheres r = a and r = b (a < b) when u = Oon the inner surface r = a and u = Uo on the outer

surface r = b, where Uois a constan!. Show why the heat equation for u(r) reduces to

d22 (ru) = O,dr

and derive the expression

buo(

a

)u(r)=- 1--b -a r (a 5; r 5; b).

Sketch the graph of u(r) versus r.

5. In Problem 4 replace the condition on the outer sphere r = b with the condition that

there is surface heat transfer into a medium at constant temperature T according to Newton'slaw. Then obtain the expression

hb2T

(a

)u(r) = I - -a + hb(b - a) r

(a 5; r 5;b)

for the steady-state temperatures, jwhere h is the ratio of the surface conductance H to thethermal conductivity K of the material.

6. Let z(p) represent the static transverse displacements of a membrane stretched betweentwo circ1es P = 1 and P = Po (1 < Po) in the plan e z = O after the outer support P = Po isdisplaced by a distance z = Zo. State why the boundary value problem in z(p) can be written

~ (pt!!..)=0dp dp

(1 < p < Po),

z(I)= O, z(Po)= Zo,and obtain the expression

log pz(p)=Zo-

log Po

7. Show that the steady-state temperatures u(p) in a hollow cylinder I 5;p 5;Po,- 00 < z < 00 also satisfy the boundary value problem written in Problem 6 if u = Oon theinner cylindricalsurfaceand u=Zoon the surface p =Po'Thus showthat Problem6 is amembrane analogy for this heat problem. Soap films have been used to display such analogies.

8. A uniform transverse force of F o units per unit area acts over a membrane stretchedbetween the two circ1es p = 1, z = O and p = Po, z = O, where I < Po. From Problem 8, Seco 5,show that the static transverse displacements z(p) satisfy the equation

(15; p 5; Po).

(pz')' + fop = O (fo = Fo/H),

and derive the expression

- h. 2 - I (log p - ~--J)z(p) - 4 (Po) log Po P5- 1(1 5; p 5; Po).

10. Types of Equations and Conditions

The second-order linear partial differential equation

(1) Auxx + Buxy+ CUyy+ Dux + Euy+ Fu = G,

where A, B, ..., G are constants or functions of x and y only, is of elliptic,

SECo 10] PARTlAL DIFFERENTlAL EQUATIONS OF PHYSICS 21

parabolic,or hyperbolic type in a domain of the xy plane if the quantityB2 - 4AC is negative,zero, or positive,respectively,throughout the domain.The three types require differentkinds of boundary conditions to determinea solution.

Poisson's equation V2u = G and Laplace's equation V2u = Oin u(x,y)are special cases of equation (1) in which A = C = 1 and B, D, E, and Fvanish. Hence those equations are elliptic in every domain. The heat equa-tion kuxx - Uy = O is parabolic, and the wave equation a2uxx- Uyy = Oishyperbolic. .

In the theory of partial differential equations it is shown that equation(1) can be reduced to generalized forms of Laplace's equation or the heatequation or the wave equation, accoJding to whether equation (1)is elliptic,parabolic, or hyperbolic.t The reductions are made by substituting newindependent variables for x and y.

Another special case of equation (1) is the telegraphequation

(2) Vxx= KLv"+ (KR+ LS)v,+ RSv.Here v(x,t) represents either the electric potential or current at time t at apoint x units from one end of a transmission line or cable which has elec-trostatic capacity K, self-inductanceL, resistance R, and leakage conduct-ance S, all per unit length. The equation is hyperbolic if KL > O. It isparabolic if either K or L is zero.

Let u denote the unknown function in a boundary value problem. Acondition that prescribes the values of u itselfalong a boundary isknown asa boundary condition of thefirst type, ora Dirichletcondition.A problem ofdetermining a harmonic function interior to a region so that the functionassumes prescribed values over the boundaries of that region is a Dirichletproblem.In this case the values of the function can be interpreted as steady-state temperatures, and such a physical interpretation leads us to expect thata Dirichlet problem may have a unique solution if the functionsconsideredsatisfy certain requirements as to their regularity.

A boundary con"ition of the second type, also called a Neumanncondi-tion, prescribes the values of the normal derivative dujdn.of the function atthe boundaries. Among other kinds of boundary conditions are those of thethird type in which values of hu + dujdn are prescribed at the boundaries,where h is either a constant or a function of the independent variables.

If the partial differentialequation in u is of second order with respect toone of the independent variables t and if the values of both u and u, areprescribed on a boundary t = O,the boundary condition is one of Cauchytype with respect to t. When the differential equation is the wave equation

t See. for instance. the books on partial differential equations by Courant and Hilbert (vol. 2.1962). Greenspan (1961). or Weinberger (1965) listed in the Bibliography.

22 FOURIER SERIES AND ROUNDARY VALUE PROBLEMS [SEC'. 10

l/u = a2l/xx, such a condition corresponds physically to that of prescribingthe initial values of both the transverse displacements 11and velocities lit of astretched string. 80th conditions appear to be needed if the displacementsII(X,I) are to be determined.

When the equation is Laplace's equation IIxx= - 11, or the heat equa-lion kl/xx = lit, however, conditions C)fCauchy type with respect to x cannotbe imposed without severe restrictions. This again is suggested by interpret-ing 11physically a~ a temperature function. When the temperatures 11 of aslab O :5:x :5:h are prescribed on the boundary x = O, for example, the fluxKllx to the left through that boundary is ordinarily determined by the valuesof u there and by other conditions in the problem. Conversely, ¡flhe flux Kuxis prescribed at x = O,specific temperatures are needed there to produce thatflux.

CHAPTER

TWO

THE METHOD OF SEPARATION OFVARIABLES

11. Linear Combinations

If CI and C2are constants and UI and U2are functions, then the function

CI UI + C2U2

is called a linear combination of UI and U2.Note that CIU¡,C2U2,and UI :!:U2are special cases.

Suppose, for c!xample, that uI and U2are both functions of the independ-ent variables x and y. According to elementary properties of derivatives, aderivative of any linear combination of the two functions can be written asthe same linear combination of the individual derivatives. Thus

o OUI OU2OX(CIUI +C2U2)=Cla;+C2a;'

provided OUI/ox and OU2/0XexistoA linear space of functions, or function space, is a class of functions, with

a common domain of definition, such that each linear combination of anytwo functions in that class remains in it; that is, if UI and U2are in the class,then so is c¡ U¡ + C2U2.

A linear operator on a function space is an operator L that transformseach function u ofthat space into a function Lu and has the property that, foreach pair of functions U¡ and U2'

(2) L(C¡UI+ C2U2) = clLu¡ + c2Lu2

whenever c¡ and C2are constants. In particular, then,

(1)

L(u¡ + U2)= Lu¡ + Lu2,

If U3is a third function in the space,

L(c¡ u¡) = c¡ Lu¡, L(O)= o.

L(cI U¡ + C2U2 + C3U3) = L(c¡ U¡ + C2U2) + L(C31-l3)

= c¡Lu¡ + c2Lu2 + c3Lu3.

23


Proceeding by induction, we find that L transforms linear combinations of mfunctions in this manner:

(3)LC~ICnUn)= JICnÚln'

In viewof equation (1),%x is a linear operator on functions of x and ythat have derivatives of the first order with respect to x. It is naturallyc1assifiedas a linear differentialoperator. The operator L which multiplieseach function u(x,y) by a fixed functionf(x,y) is an example of a still moreelementary linear operator. Let us now point out why such composites as02/0X2,f %x, and i'2/0X2+ f C¡(lXare also linear operators.

Ir linear operators L and M, distinct or not, are such that M transformseach function u of some function space into a function Mu to which Lapplies and if UIand U2are functions of that space, it followsfrom equation(2) that

(4) LM(CI UI + C2U2)= L(cI MUI + c2Mu2) = CILMuI + C2LMU2'

That is, the produceLM of linear operators is a linear operator. Ir L and Mboth represent iJ/ox,it followsthat (¡2/(lX2is a linearoperator on the space ofall functions u whose derivatives Uxand UxxexistoSimilarly,ifL = f(x,y) andM = %x, it follows thatf %x is a linear operator.

The sum of two linear operators is defined by the equation

(5) (L + M)u = Úl + Mu.

Ir we replace u here by CIUI + C2U2,we can see that the sum L + M is alinear operator and hence that the sum of any finite number of linear opera-tors is linear. Thus 02/ax2 + f a/ax is a linear operator.

Each term of a linear homogeneous differential equation in u consists ofa product of a function of the independent variables by one of the derivativesof u or by u itself. Hence every linear homogeneous differential equation hasthe form

(6) Úl=O

where L is a linear differential operator. For example, if

a2 a2 a2 o aL= Aa 2 + B O a + C0-2 + Da + E_o + F,

x y x y x y(7)

where the letters A through F denote functions of x and y only, equation (6)is the linear homogeneous partial differential equation

(8)

in u(x,y).

Auxx + Bux)'+ Cu).y+ Dux + Eu).+ Fu = O

SEC'. 12] THE METHOD OF SEPARATION OF VARIABLES25

Linear homogeneous boundary conditions also have the form (6). Thenthe variables appearing as arguments of u and as arguments of functions thatserve as coefficients in the linear operator L are restricted so that theyrepresent points on the boundaries.

Now let u. denote functions that satisfy equation (6); that is, Lu. = O(n = 1,2,..., m). It follows from equation (3) that each linear combination of

. those functions also satisfies equation (6). We state that principie 01superpo-sition ofsolutions, which is fundamental to one ofthe most powerful methodsof solving linear boundary value problems, as follows.

Theorem 1. 11each 01 m lunctions u l' U2' ..., l/m satisfies a linear ho-mogeneol/s differential eql/ation LI/ = O, then ever,l' linear combination

(9) u = CI U1 + C2U2 + ... + cmum'

where the c's are arbitrary constallts, satisfies that differential equation. 11each

oltlre mfimctions satisfies a linear homogeneol/s bOl/ndary condition Lu = O,then every linear combination (9) satisfies that bOl/ndary condition.

12. Examples

The principie of superposition is useful in ordinary differential equations.For example, from the two solutions y = eXand y = e-X of the linearho-mogeneous equation y" - y = O,the general solution y = clex + c2e-x canbe written.

To iIIustrate the application of Theorem I to partial differential equa-tions, consider the linear homogeneous heat equation

(1) Ut(x,t) = uxx(x,t)

and the two linear homogeneous boundary conditions

(2) uAO,t) == O, uAI,t) = O.

It is easy to verify that each of the functions

l/o = 1, u. = exp (-n2n2t) cos I11tX (n = 1,2, oo.)

satisfies equation (1) and conditions (2).Thus every linear combination

(3)m

u:: Co+ I c. exp (- n2n2t)cos nnxn=1

of those functions satisfiesequation (1) and conditions (2).A principie of superposition for nonhomogeneous linear differential

equations should be noted. Suppose that L is a given linear differential op-erator and that "(4) Lu. =/. and Lu2 = 12'

26 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS [SEC'. 13

wheref1 andf2 are functions ofthe independent variables only. The functionu = U1 + U2 cIearly satisfies the linear differential equation

(5) Lu = fl + f2,

where Lis the same operator as in equations (4).A special case of this principIe is the fact that if

It is also tme that if functions u and U2satisfy equations (9) and (7), respec-tively, the function U1= u - U2 satisfiesequation(6).Thus the totalityofsolutions u of equation (9) consists of all sums of the form (8) where UIdenotes solutions of the corresponding homogeneous equation (6) and U2isany particular solution of equation (9) itself.

The above discussion of nonhomogeneous linear differential equationsis also valid when L is used to represent linear boundary conditions.

13. Series of Solutions

In order to extend our results on linear combinations of solutions 'to an

infinite set of functions U1' U2, ..., we must deal with convergence anddifferentiability of infinite series of functions.

Let the constants Cnand the functions Unbe such that the infinite serieswith terms CnUnconverges throughout some doma in of the independentvariables. The sum of that series is a function

(1)".

u = L CnU..n:::l

Let x represent one of the independent variables. The series is differentiable,

or termwise differentiable, with respect to x if the derivatives oun/ox andou/ox exist and the series of fun!}tions CnoUn/ox converges to ou/ox:

OU 'X Olln

--= LC .--ox . n= 1 n ex .

(2)

(6) Lul =0

and

(7) Lu2 = f,

then the function

(8) u = UI+ U2

satisfies the equation

(9) Lu =1

SECo 13] THE METHOD OF SEPARATION OF VARIABLES 27

Note that a series must necessarily be convergent to be difTerentiable.Sufficient conditions for difTerentiability will be noted in Seco 14.

If, in addition, series (2) is difTerentiable with respect to x, then series (1)is difTerentiable twice with respect to x, and so on for other derivatives.

Let L be a linear operator consisting of a functionf of the independentvariables times a derivative, or a sum of a finite number of such terms. Wenow show that if series (1) is difTerentiable for all the derivatives involved inL and if each of the functions Unin series (1) satisfies the linear homogeneousdifTerential equation LUn= O, then so does U; that is, Lu = O.

To accomplish this, we first note that according to the definition of thesum of an infinite series,.

fou

f l. ~ oUn l.

f o ~,,= 1m L- cn,,- = 1m "L- enUnuX m-cx.' n=l uX m-O) uX n=1

when series (1) is differentiablewith respect to X.Here the operator %x canbe replaced by other derivatives if the series is so difTerentiable.Then, byadding corresponding sides of equations similar to equation (3),we find that

Lu = lim L(I:CnUn)

.m"" 00 n=1

The sum on the right-hand side ofequation (4) is a linear combination ofthefunctions UI,U2"'" um;and since Lun = O(n = 1,2,.. .), Theorem 1 allows usto write

(3)

(4)

L(JI CnUn) = O

for every m. Hence frem equation (4) we have the desired result Lu = O.A linear homogeneous boundary condition may also be represented by

an equation Lu = O.In that case we may require the function Lu to satisfyacondition of continuity at points on the boundary in order that its valuesthere will represent limiting values as those points are approached from theinterior of the domain.

The followinggeneralization of Theorem 1 is now established.

Theorem 2. /f eaeh function of an infinite set Ul' U2,. .. satisfies a linearhomogeneous differential equation 01' boundary condition Lu = O, then the

funetion

(5)'"

u = I cnunn=1

also satisfies Lu = O, provided the eonstants en and the functions Unare suehthat series (5) converges ami is differentiahle fol' all derivatives involved in Lami pl'ovided the I'equil'edeontinuity condition at the houndal'Y is satisfied by Luwhen Lu = O is a boundary eondition.

28 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS SECo 14]

14. Uniform Convergence and Example

We recall some facts about uniformly convergent series of functions. tLet S(x) denote the sum of an infiniteseries of functionsfn(x),where the

series is convergent for all x in some interval a :::;x :::;b. That is,

(1)oc'

S(x) = ¿ fn(x) = lim Sm(x)n=1 m ((.t

(a:::; x :::; b)

where Sm(x) is the partial sum consisting of the first m terms of the series. Theseries converges uniformly with respect to x if the absolute value of itsremainder Rm(x) = S(x) - Sm(x) can be made arbitrarily small for all x inthe interval by taking m sufficiently large; that is, for each positive number e,there exists an integer m" independent of x, such that

(2) ¡S(x) - Sm(x) I < e whenever m > mE (a:::;x:::;b).

A useful sufficient condition for uniform convergence is given by theWeierstrassM-test: if there is a convergentseries

00

¿M n"=1

of positive constants such that for each n

(3) Ifn(x)1 :::;Mn (a :::;x:::; b),

then series (1) is uniformly convergent on the stated interval.Because S(x) = Sm(x)+ Rm(x)and IRm(x) Iis uniformly small when m is

sufficiently large, the following properties ofuniformly convergent series canbe established.

If the functionsfn(x) are continuous and if series (1) is uniformly conver-gent, then the sum S(x) ofthat series is a continuous function. Also, the seriescan be integrated term by term over the interval to give the integral of S(x)over the interval.

If the functionsfn and their derivativesf~ are continuous, if series (1)converges, and if the series whose terms aref~(x) is uniformly convergent,then series (1) is differentiable with respect to x.

Corresponding results hold for series of functions of two or moreindependent variables.

Let us use the above results and Theorem 2 to write a more generalsolution for the example on the heat equation introduced in Seco 12. To bespecific, we now require the function !I to satisfy the equation

(4) u,(x,t) = uxAx,t)

t See. for instance. Kaplan (1973. pp. 407 ff) or Taylor and Mann (1972. pp. 644 ff). listedin the Bibliography.

[SECo 14 THE METHOD OF SEPARATION OF VARIABLES 29

in the domain O< x < 1, t> to whenever to > O and these two Neumannconditions at the boundaries x = Oand x = 1:

(5) uAO,t) = O, uAI,t) = O.

Also, uAx,t) is to be a continuous function of x on the cIosed intervalO:;; x:;; 1 for each fixed t (t ~ tojo Then uAO,t) is the same as the limit ofuAx,t) as x -+ Ofrom the interior ofthe interval, and similarly for uAl,t); thatis, conditions (5) then imply that

lim uAh,t) = O,h-Oh>O

limuAI-h,t)=Oh-Oh>O

(t ~ to > O).

In Seco 12 we noted that each function of the infinite sequence

Uo = 1, Un = exp (-n2n2t) cos nnx (n = 1,2, oo.)

satisfies the linear homogeneous equations (4) and (5). Those functions andtheir partial derivatives are continuous everywhere. According to Theorem 2,the function

(6)'"

u(x,t) = Co+ LCnexp (-n2n2t) cos nnxn=l

also satisfies equations (4) and (5) if the constant coefficients Cnin the infiniteseries can be restricted so that the series is differentiable twice with respect tox and once with respect to t and so that Uxis continuous in x (O:;; x :;; 1).Wenow show that those conditions are satisfied when the sequence of constantsCn is bounded; thus we assume that a number B. independent of n,exists such that

(7) ICnl:;; B (n = 0,1,2, .oo). .

(8)

The terms CnUnof series (6) satisfy the condition

ICnexp (- n2n2t) cos nnx I :;; B exp (- n2n2to) (t ~ to),

and the ratio test shows the convergence of the series whose terms are theconstants B exp (-n2n2to) when to> O. Series (6) is therefore absolutelyconvergent by the comparison test and also uniformly convergent with re-spect to x and t according to the Weierstrass M-test. In like manner we seethat

I J~ (enun) 1:;; Bnn exp (-n2n2to).

The series of constants nk exp (- n2n2to), where k is any integer, convergesaccording to the ratio test. Hence the series whose terms are (cnun)x isuniformly convergent with respect to x and, in fact, with respect to x and ttogether. Series (6) is therefore differentiable with respect lo x, and the sum

30 FOURIER SERIES AND BOUNDARY VALUE PRORLEMS [SECo 14

ux(x,t) of the dilTerentiated series is continuous with respect to x for all xwhen t ~ to.

The uniform convergence of the series of derivatives (cnun)xx, which arethe same as (cnun)" is also evident. Thus series (6) is dilTerentiable twice withrespect to x and once with respect to t. Its sum 11therefore satisfies equa-tions (4) and (5), and it involves an infinite set of constants Cnwhich arearbitrary except for the boundedness condition (7).

We shall see later on that those constants can be determined so that u

will satisfy a nonhomogeneous initial condition

(9) U(x,O)= f(x) (O< x < 1).

Then, physically, u will represent temperatures in a slab ° ::;x ::; 1 withinsulated faces, in viewof conditions (5),and with initial temperatures givenby condition (9). Note that condition (9) is satisfiedby the function (6) ifCn(11= 0,1,2, ...) can be found such thatfhas the representation

(10)""

f(x) = Co + I c" COS I11tXn= 1

(O< x < 1).

PROBLEMS

1. Verify that each of the products "n = COS/IX sinh /Ir (/1= 1. 2. .. .). as well as thefunction Uo= y, satisfies Laplace's equation Uxx+ un' = Oand the three boundary conditions

u.(O.y) = u..(1t,y) = u(x,O) = o.

Then apply Theorem I to show that the linear combinationm

U= CoY + ¿:Cn COS /IX sinh n.rtI= I

also satisfies that difTerential equation and those boundary conditions. Find values of thecoefficients Cn(n = O. 1. 2, ...) for which that linear combination also satisfies 'the nonhomoge-neous boundary condition

u(x,2) = 4 + 3 cos X - cos 2x.

3 -1Ans. Co= 2, C,= -.- ,C2=~, cn= Owhen/1~ 3.smh 2 smh 4

2. Show that each of the functions

"n = exp [-(n - W1t2t] sin [(/1- !)1tx] (n = 1,2, oo.)

satisfies the heat equation

u/(x,t) = uxx(x,t)

and the boundary conditions

u(O,t)=0, u.(l,t) = O.

SECo 14] THE METHOD OF SEPARATlON OF VARIABLES 31

Then apply Theorem I to show that the Function

m

11= LCn exp [- (n - W1t2t] sin [In -!)1tx]n= I

also satisfies those three equations. Find values of the constants cn such that this function alsosatisfies the nonhomogeneous boundary condition

. 1tX . 51txu(x,O)=2 Sin - Sin - .2 2

Ans. e, = 2, C2 = O,e3 = -1, en = O when n ~ 4.

3. Verify that each of the functions

IImn= sin mx cos n,l'exp (-=)m2 + n2)

satisfies Laplace's partial differential equation

(m = 1,2, oo.; n = 0,1,2, oo.)

u..(x,y,z) + u).).(x,y,z) + u"(x,,I',z) = O


u(O,y,z) = u(1t,,I',z)= U,.(x,O,z)= u).(x,1t.z) = O.

Use superposition to obtain a function that satisfies not only Laplace's equation and thoseboundary conditions, but also the condition

u,(x,y,O) = (- 6 + 5 cos 4.1')sin 3x.

Ans. u = (2e- 3, - e- 5, cos 4.1')sin 3x.

4. Let u and l' denote two functions that satisfy the one-dimensional heat equation in xand t. where the thermal diffusivity k need not be constant; that is. u, = kuxx and v, = kv...where k may be a function of x and t. Multiply each sjde of those two equations by constants C1and C2' respectively, and add to show that the linear combination c, u + e2v also satisfies theheat equation. This iIIustrates a variation in the proof of Theorem 1.

S. Show that if an operator L has the two properties

L(u, + U2)= Lu, + Lu2' L(eu) = cLu.

for all functions u,. U2'and u and for every constant c, then Lis linear; that is, show that it hasproperty (2). Seco 11. .

6. Let u, and U2satisfy a linear nonhomogeneous differential equation Lu = f, wherefis afunction of Ihe independent variables only. Pro ve that the linear combination c, u, + c 2u2failsto satisfy that equation when C1 + C2 # 1.

7. Show that each of the functions ,1', = l/x and .1'2= 1/( l + x) satisfies the nonlineardifferential equation y' + ,1'2=O.Then showthat ifc isa constant (c '" O.e '" 1).neither ey, noreY2 satisfies that equation. Also show that the function y, + .1'2 does not satisfy it.

8. Use special cases of linear operators. such as L = x and M = c'/c'x, to show that theoperators LM and M L are not always the same.

9. Assuming that en is a bounded sequence of constants and Yo> O, prove Ihat thefunction

x

u(x,y) = L cne-n,. sin nxn=1

(.1'~Yo)

is twice differentiable with respect lo x and .r and satisfies Laplace's equation in the doma iny> Yo.

32 FOURIER SERIES AND BOUNDARY VALUE PROBUMS [SECo 15

10.. Prove that if n41c. I ::; B (n = 1, 2, .. .). where Bis a constant, the function'"

y(x,l)= L: c. sin nx cos nI,,=1

satisfies the wave equation y" = y.. for al! x and l.

15. Separation of Variables

Let us find an expression for the transverse displacements y(x,t) of a stringstretched between the points (0,0)and (e,O)if the string is initially displacedinto a position y =f(x) and released at rest from that position.

At this stage some of our steps in solving the problem must beformal, ormanipulative. Their validity cannot be established until we have developedfurther theory.

We assume that no external forces act along the string. Then the fune-tion y must satisfy the wave equation (Sec. 3)

(1) yu(x,t) = a2YxAx,t)

It must also satisfy these boundary conditions:

(O< x < e, t > O).

. (2)

(3)

y(O,t)= O, y(e,t)= O,

y(x,O) = f(x)

y,(x,O) = O,

(O~ x ~ e),

where the prescribed displacement functionf is continuous on tbe intervalO ~ x ~ e andf(O) = f(e) = O.

In determining nontrivial (y ;f= O)solutions of all homogeneousequations(1) and (2) in the above boundary value problem, using ordinary difTerentialequations, we seek functions of the form

(4) y(x,t) = X(x)T(t)

wbich satisfy those equations. Note that X is a function of x alone and T afunction of t alone. Note, too, tbat X and T must be nontrivial (X ;f:.O,T;f=O).

If y = XT satisfies equation (1), then

X(x)T"(t) = a2X"(x)T(t);

and we can divide by a2XT to separate the variables:

(5)X"(x) T'(t)X(x) = a2T(t)'

Since the left-hand side here is a function of x alone, it does not vary with t.However, it is equal to a function of t alone, and so it cannot vary with x.


Hence both sides must have some constant value, which we write as - A, incommon; that is,

(6) X"(x) = -AX(X), T"(t) = -Aa2T(t).

While it may seem more natural to denote the separation constant by A,rather than - A,the choice is of course a minor matter of notation. It is onlyfor convenience later on (Secs. 31 through 33) that we have written - A.

If XT is to satisfy the first of conditions (2), then X(O)T(t) must vanishfor all t (t > O).With our requirement that T;f. O,it foIlowsthat X(O)= O.Likewise, the last two of conditions (2) are satisfied by XT if X(c) = OandT(O) = O.

Thus XT satisfiesequations (1) and (2) when X and T satisfy these twohomogeneous problems:

(7)

(8)

X"(x) + AX(X) = O, X(O)= O, X(c) = O,

T"(t)+ Aa2T(t)= O, T(O)= O,

where the parameter Ahas the samevafue in both problems. To find nontriv-ial solutions ofthis pair of problems, we first note that problem (8) has onlyone boundary condition and therefore many solutions for each value of A.Since problem (7) has two boundary condltions, it may have nontrivialsolutions for particular values of A. As we shaIl see in the next chapter,problem (7) is a so-caIled Sturm-Liouville problem, and A must be real-valued if the problem is to have a nontrivial solution.

If A = O,the ditTerentialequation in problem (7)becomesX" = Oand itsgeneral solution is X(x) = Ax + B. Since B = Oand Ac + B = Oif X(O)= Oand X(c) = O,it follows that A = B = O;therefore this problem has just thetrivialsolutionX(x) == OwhenA= O.

If A > O,we may write A = (X2(x > O).The ditTerential equation in prob-lem (7) then takes the form X" + (X2X = O, its general solution being

X(x) = C1 sin (xx+ C2 cos (Xx.

The condition X(O) = O implies that Cl = O;and ifthe condition X(c) = Oisto hold, C 1 sin (xc = O.In order for there to be a non trivial solution, then, (xmust satisfy the equation sin (Xc= O.That is,

(9)mr

(X=-c (n = 1,2, ...).

Thus,exceptfor the constantfactorCl'

(10) X(x) = sin mrxc (n = 1,2, ...).

The numbers A = (X2= n2n2/c2for which problem (7) has nontrivial solu-tions are caIled eigenvaluesof that problem, and the functions (10) are thecorresponding eigenfunctions.

34 FOURIER SERIESAND BOUNDARY VALUE PROBLEMS [SECo 15

When A.< O, let us write A.= - p2 (p > O). Then

X(x) = DI sinh px

is the solution of X" - p2X = O that satisfies the condition X(O) = O.Sincesinh pe =/=O, DI = O if X(e) = O. Thus problem (7) has no negativeeigenvalues.

When A.= n2n2/e2,problem (7) is a distinct problem for each clifferentpositive integer n. For a fixed integer n it has the solution (10), and problem(8) becomes

(nna

)2

T'(t) + e T(t)= O, T'(O)= O.

Except for a constant factor, then,

T( t) = cos nnate .

Therefore each function of the infinite set

(11). nnx nnat

Yn(x,t)= SIn-COS-e e (n = 1,2, ...)

satisfies all the homogeneous equations (1) and (2).The procedure here illustrates a fundamental method of obtaining non-

trivial solutions of the homogeneous equations in boundary value problems.It is called the method of separation of variables.

A linear combination of a finite number of the functions (11) alsosatisfies the homogeneous equations, according to Theorem 1. But whent = O, it reduces to a linear combination of a finite number of the functions

sin (nnx/c). Thus it will not satisfy the nonhomogeneous condition (3) unlessf(x) has that special character.

According to Theorem 2, the function

(12)00

y(x,t) = ¿ bn sin nnx cos nnatn=1 e e

also satisfies the homogeneous equations (1) and (2), provided thecoefficients bn can be restricted so that the infinite series is suitably con ver-gent and differentiable. That function will satisfy the remaining condition (3)iff can be represented in the form

(13)00

f(x ) = " b. nnx

'-- n Sln-n= I e

(O:::;x :::;e).

SECo 16] THE METHOD OF SEPARATlON OF VARIABLES 3S

In the following section we shall indicate why the coefficients bn should havethe values

2 ' e mrxbn= I f(x) sin - dxe 'o e

if representation (13) is to be valid.The formal solution of our boundary value problem in the displace-

ments of the string is, then, expression (12) with coefficients (14).

(n = 1,2, oo.)( 14)

16. The Fourier Sine Series

Recall from trigonometry the identities

2 sin A sin B = cos (A - B) - cos (A + B),

2 sin2 A = I - cos 2A.

Ir e denotes any positive number and ni and n are distinct positive integers, itis then easy to see that

I

'c . m1tx . n1tx di

I

'c

r

(m - n)1tx (m + n)1txj

d Osm - sm - x =_2 cos - cos x = ;'0 e e '0, e e

also,

. c I11tX I ,e

f

2n1tx

j

eI sin2-dx=-1 I-cos- dx=-.'0 e 2'0 e 2

That is, when m and n are positive integers,

c lo1

, . nl1tX . n1tXd ism-sm- x = <

'o e e ¡ ~

Using terminology to be developed in the next chapter, we thus find that theset of functions sin (n1txjc) (n = 1, 2, ...) is orthogonal on the interval(O,e); that is, the integral over that interval of the product of any two dis-tinct functions of the set is zero.

In Seco 15 we needed to determine the coefficients bn so that a series ofthose sine functions would converge to a prescribed function f(x) on theinterval ° ~ x ~ C. Assuming that an expansion

if m '" n,

(1)ifm = n,

(2) f( ) b. 1tX

b. 21tx

b. n1tx

x = I sm - + 2 sm - + ... + n sm - + ...e e e

is possible and that the series can be integrated term by term after being

36 fOURIER SERIES AND BOUNDARY VALUE PROBLEMS [SEC'. 17

multiplied by sin (nnx/c), for any fixed value of the positive integer n, wecan use the orthogonality property (1) .to find a formula for h..

Let all terms in equation (2) be multiplied by sin (nnx/c) and integratedfrom x = Oto x = c. The first term on the right becomes

.c . nx . nnxb¡ I Sin -sln-~-dx.'0 e e

This is zero unless n = 1,according to the orthogonality property. Likewise,all other terms on the right, except the nth one, become zero. In view ofproperty (1), then,

,e nnx ,e nnx e

I f(x) sin - dx = bn I sin2.' dx = 2 bno'o e 'o e

Hence the coefficients in representation (2) have the values

2 ,e nnxbn =. I f(x) sin - . dxe 'o e

The sine series (2) with coefficients (3) is called the Fourier sine series

corresponding to the function f on the intervalO < x < c. Using the tildesymbol - to indica te correspondence, we can write

(3) (n = 1, 2, .. o).

(4)2 OCJ nnx ,e nns

f(x) -.- I sin- I f(s) sin o dse n=1 c.o e(O< x < c)o

In Chap. 4 we shall establish general conditions onfunder which this seriesconverges to the values f(x), so that the correspondence (4) becomes anequalityo

17. A Plucked String

To treat a special case of the boundary value problem considered in Seco 15,let the string be stretched between the points (0,0) and (2,0) and suppose tha!its mid-point is raised to a height h above the x axiso The string is thenreleased from rest in that position. consisting of two line segments (Fig. '6).

r~ (1,hlj

'

Y = (x)

o (2.0) x Figure 6

SECo 17] THE METHOD OF SEPARATlON OF VARIABLES 37

In this case the function f, which describes the initial position of thestring, is given by the equations

(1)jhx

f(x)=\-hx+2h

when O ::; x ::; 1,when 1 ::; x ::; 2.

The coefficients bn in the Fourier sine series corresponding to that functionon the intervalO < x < 2 can be written

.2 nnxbn = I f(x) sin _2

e/x'o

1 nnx ' 2 nnx

= h r x sin_2 e/x + h I (-x + 2) sin- 2e/x.

'o '1

After integrating and simplifying, we find that

(2)8h . nn

b = - 2 2Sin_2.

n n n

Our formal solution (12), Seco15, for the displacements of the stringthen becomes

8h ro 1 . nn . nnx nnaty(x,t)= 2 L 2 Sin - Sin - COS--- .n n=1n 2 2 2

Note that sin (nn/2) = O when n is even. Series (3) can therefore be writtenmore efficiently by summing only the terms that occur when n is odd. This isaccomplished by replacing n by 2n - l wherever n appears on the right ofthe summationsymbol.The result is

(3)

( ) - 8h ~ (- 1)"+ 1 . (2n - 1)nx (2n - 1)naty x,t - n2 n';-I(2n - 1)2Sin 2 COS 2 '

where we have also used the fact that

(4)

. (2n- l)n . (n

)

. n( )

+ 1

Sin 2 = SIn nn - 2 = - cos nn Sin 2 = - In.

The infiniteseries in equations (3)and (4)is-notdifTerentiabletwicewithrespect to x or t. But in Seco51 we shall show how the seriescan be repre-sented in a form simple enough to see that the sum y(x,t) of the series isindeed a solution of the boundary value problem.

PROBLEMS

1, As pointed out in Seco 12. the linear homogeneous differential 1/,= l/xx andboundary conditions I/x(O,r)= O, I/.(I,t) = Oare all satisfied by these functions oftype X(x)T(t):

110= 1, 11. = exp (-n27t2t) cos n7tx (n = 1,2, ...).

Use the method of separation ofvariables (Sec. 15) to derive that set of solutions. [The solutionUo = 1 corresponds to the eigenvalue A = O for the problem X" + AX = O, X'(O) = X'(I) = O.]

2. Use the method of separation of variables to obtain the set of functions ". =exp [- (n - Wlr2t] sin [(n --!)lrx] (n = 1,2, ...) used in Problem 2, Seco 14.

3. In Problem 1, Seco 14, it was pointed out that each function of the set

Uo= y, U. = cos nx sinh ny (n = 1,2, Oo.)

satisfies Laplace's equation and the homogeneous boundary conditions

U.(O,y) = u.(Ir,y) = u(x,O) = O.

Use the method of separation of variables to derive that se!.

4. Suppose that equation (5)" Seco 15, had been written in the form

2X"(X) T'(t)

a X(x) = T(I)'

Set each side here equal to - A and show that the desired result (11), Seco 15, still follows.

5. Show that the Fourier sine seriescorresponding to the functionf(x) =3 sin 4lrxon theintervalO < x < 1 reduces to 3 sin 4lrx, the function itself.

6. Show that the Fourier sine seriescorresponding to thefunctionf(x) = I on the intervalO<x<lr~ .

4 '" sin (2n - l)x 4(

. 1 .

)- I -- =.- sm x + - sm 3x + ... .Ir .= I 2n - I Ir 3

7. Find the Fourier sine series corresponding to the function f(x) = x on the interval0< x< 1.

Au 2 " (_1)..' .I sm nlrx.11 n= I n

8. Give an example to show that linear combinations of functions of the type X(x )T(t) arenot necessarily functions of that restricted type.

18. Ordinary Differential Equations

In Seco 15 we found that thé homogeneous boundary value problem

{1) X"(x) + AX(X) = O, X~=~ X(c)= O

has many solutions X(x) = e 1 sin (nnxjc), one for each value of el, whenA = n2n2jc2 where n is a fixed positive integer. When A =1=n2n2jc2, the prob-

lem has the unique solution X(x) == O.This iIIustratesthe factthat unique-ness of solutions of boundary value problems may depend on the coefficientsin the differential equation and the kind of boundary conditions, not solelyon the order of the equation and the number of boundary conditions.This is also tme of the existenceof solutions, as we shall point out below.

SECo 18] THE METHOD OF SEPARATION OF VARIABLES39

The theory of ordinary differential equations ensures the existence anduniqueness of solutions of certain types of initial value problems, problems inwhich all boundary data are ~iven at one point. We state, without proof, atheorem for linear equations of the second order.

Theorem 3. út A, B, and C denote continuousfunctions ofx on an inter-val a ::;;x ::;;b.lfxo is afixed point in that interval and Yoand Yo are prescribedconstants. then there is one and only one function y, continuous together with

its derivative y' when a ::;;x ::;;b, which satisfies the differential equation

(2) y"(x) + A(x)y'(x) + B(x)y(x) = C(x) (a < x < b)

and the two initial conditions

(3) y(xo) = Yo, y'(xo) = Yo.

Proofs of this and corresponding theorems for equations of other orders,by methods of successive approximations, will be found in books on thetheory of ordinary differential equations.t According to equation (2),y" = C - Ay' - By; thus y" is also continuous. That the equation has ageneral solution involvingtwo arbitrary constants followsfrom the fact thatarbitrary values can be assigned to the constants Yo and Yo.

As an example, the equation

(4) x"(t) + k2x(t) = O,

where k is a constant and k +O, is satisfied by

(5) x(t) = CI sin kt + C2 cos kt

where C I and C2 are constants. On each interval of the taxis that containsthe point t = O the function (5) is the solution of equation (4) that satisfiesthe conditions x(O) = C 2 and x'(O) = kC l'

Suppose that x(t) satisfies the two boundary conditions

(6) x(O)= O, x(1)= 1

along with equation (4) and the requirement that x and x' be continuous onthe intervalO::;; t ::;;1. Since x(O) = O,it is necessary that

x(t) = CI sin kt;

and since x( 1)= 1,a value of C I must exist such that

(7) C 1 sin k = 1.

t See pp. 73 ff of Ihe book by Inee (1956) listed in the Bibliography.


This last equation determines el provided sin k =1= O.Thus

(8) x(t) = sin ktsin k (k =1=:!:nn, n = 1,2, ...).

But when k = :!:nn, the two-point boundary value problem consistingof equations (4) and (6) has no so/ution beca use there is no value of el, orx'(O)jk, that satisfies equation (7).

It is interesting to note a physical interpretation of the case k = n. Sincex" = -n2x, the function x represents the displacement ofa unit mass alongthe x axis under a force -n2x proportional to the displacement from theorigin x = O. The mas s is initially at the origin and is required to have thedisplacement x = 1 at time t = 1. But the number n is the natural frequencyk ofthis system; consequently, the mass returns to the origin at the instantst = 1, 2, ..., regardless of its initial velocity. Therefore x( 1) = O, and thecondition x( 1) = 1 cannot be satisfied.

19. General Solútions of Partial Differential Equations

Some boundary value problems in partial differential equations can besolved by a method corresponding to the one usually used to solve suchproblems in ordinary differential equations, the method of first finding ageneral solution of the differential equation.

Examp/e 1. Let us solve the boundary value problem

(1) uxAx,y) = O, u(O,y) = y2, u(l,y)= 1.

Successive integrations of the equation Uxx = O with respect to x, with ykept fixed,give the equations Ux= f(y) and

(2) u = xf(y) + g(y),

where f and 9 are arbitrary functions. The boundary conditions in prob-lem (1) require that

g(y) = y2, f(y) + g(y) = 1;

thusf(y) = 1- y2, and the solution of the problem is

(3) ¡¡(x,y)= x(l - y2) + y2.

(4)

Examp/e 2. We next solve the wave equation

y,,(x,t)= a2YxAx,t)

SEL. 19] THE METHOD OF SEPARATION OF VARIABLES 41

in the domain - 00 < x < 00, t > 0, subject to the boundary conditions

(5) y(x,O)= f(x), y,(x,O)= ° (-00 < x < 00),

in terms of the constant a and the functionjThe differential equation (4) can be simplified by introducing new

independent variables r and s:

r = x + at, s = x - ato

According to the chain rule for differentiation,

y, = y,r, + Yss, = ay, - ays.

Continuing with that rule, we find that

y" = a2(y" - 2yrs + Yss)' Yxx = y" + 2yrs + Yss.

Equation (4) therefore reduces to the form

Yrs = O,

which can be solved by successive integrations to give y, = g'(r) and

y = g(r) + h(s)

where 9 and h are arbitrary differentiable functions.A general solution of the wave equation (4) is therefore

(6) y = g(x + at) + h(x - at).

In this example the boundary conditions are simple enough that we candetermine the functions g and h. To be precise, the function (6) satisfiesconditions (5) when

g(x) + h(x) = f(x) and ag'(x) - ah'(x) = O.

Thus g(x) - h(x) = c, where c is a constant, and it foIlowsthat

2g(x) =f(x) + c and 2h(x) = f(x) - c.

Consequently,

(7)l

y(x,t) ="2 [f(x + at) + f(x - at)].

The solution (7) of the boundary value problem consisting of equations(4) and (5) is known as c/'Alembert's solutioll. It is easily verified under theassumption that f'(x) and .f"(x) exist for all x.


The method iIlustrated in the two examples here has severe limitations.The general solutions (2) and (6), solutions involving arbitrary functions,were obtained by successiveintegrations, a procedure that applies to rela-tively fewtypes of partial differentialequations. But, even in the exceptionalcases where such general solutions can be found, the determination of thearbitrary functions directly from the boundary conditions is often too dif-ticult.

20. 00 Other Methods

We have stresseda processbased on superposition for solving linear bound-ary value problems in partial differentialequations. The process, illustratedin Seco15,consists tirst of tinding solutions of all homogeneousequations inthe problem by the method of separation of variables. A generalized linearcombination, or superposition, of those solutions which will also satisfyanonhomogeneous boundary condition is then sought.

That process used in Seco15, which is sometimes called the FOl/riermethod, is a classical and powerful one. The treatment of its theory andapplications is the object of this book. Limitations of the method will benoted also.

There are other important methods of solving linear boundary valueproblems. The procedures of using Laplace transforms, Fourier transforms,or other integral transforms, all included in the subject of operational math-ematics, are especially elTective.tThe classical method of conformal map-ping in the theory of functions of a complex variable applies to a prominentclass of problems involving Laplace's equation in two dimensions.; Thereare still other ways of reducing or solving such problems, including appli-cations of so-called Green's functions and numerical or computationalmethods.

Even when a problem yields to more than one method, however,dilTer-ent methods sOl')etimesproduce dilTerentforms of the solution; and eachform may have its own desirable features.On the other hand, some problemsrequire successiveapplications of two or more methods. Others, includingsome linear problems of fairly simple types, have detied alJ known exactmethods. The development of new methods is an activity in present-daymathematical research.

t See Churchill (1972), listed in the Bibliography.~See Churchill, Brown. and Verhey (1974), listed in the Bibliography.

SECo 20] THE METHOD OF SEPARATION OF VARIABLES43

PROBLEMS

Use general solutions of the partial differentialequations 10solve the boundary value problemsin Problems 1 through 4: .

l. u..(x.y) = 6xy. u(O.y)= y. u.(I,y) = O.Ans. u = (xJ - 3x + I)y.

2. u.,.(x.y) = 2x. u(O,y)= O,u(x,O) = x2.

3. y,,(x,t) = a2Yu(x,t), y(x,O) = O,y,(x,O) = 1/(1 + X2).

Ans. u = x2(1 + y).

1

Ans. .1'=2a [Ian-I (x + al) - tan-' (x - at)].

4.y,,(x,t) = a2Yu(x,t), y(x,O) = O,y,(x,O) =g(x).

1 .X+~AB .1'= I gW~

2a'x_GI

S. Use superposition ofthe solution (7), Seco 19,and the solution of Problem 4 to write thesolution

1 1 .X+GI

y = _2(f(x + at)+f(x - al)]+ 2 J gIs) d.~a Jt-Ol

of the boundary value problem

y,,(x,l) = a2Yu(x,t), y(x,O) =f(x), )',(x,O) = g(x).

6. In Example 2, Sec. 19, Y(X,I) represents transverse displacements of a stretched string ofinfinite length, initially released at rest from a position y = f(x) ( - el:)< x < eI:)).From thesolution )' = !f(x + at) + !f(x - at) show how the instantaneous position ofthe string at time

t can be described as the curve obtained by adding ordinates of two curves, one obtained bytranslating the curve y = !f(x) to the left through the distance at, the other by translating it 10the right through the same distance. As t varies, the curve y = !f(x) moves as a wave withve10city a. Show graphically some instantaneous positions ofthe string whenf(x) is zero excepton a small interval about the origino

7. Derive the general solution

u =x + e-"g(x) + h(y)

of the partial differential equation l/.). + l/. = 1.

8. The boundary value problem

y,,(x,t) = a2Yu(x,t)

)'(x.O)= e-',

y.(O,t) = y(O,t),

y,(x,O)=0

limy(x.t)=O

(x > O,t > O).

(x> O). .

(t >0)

can be solved with the aid of Laplace transforms. lis solution is given by the equations

¡e-' cosh atv=. le-x cosh at + sinh (x - at) + (x - at) exp (x - at)

when x ~ ato

when x S ato

Show that this is a special case of the general solution (6). Seco 19. and hence that .1'satisfiesIhe wave equation in the domain x > O. t > O except at points on the line x = al. where the


derivalives of Y do nol exist. Poinl oul Ihal when I = Oand x> O.)' = e-X since x> al. ShowIhal y salisfies the resl of the boundary conditions. [Even Ihough Ihe general Solulion.of Ihepartial differenlial equation here is known, this fairly simple boundary value problem is poorlyadapted to the method (Sec. 19) of using the general Solulion.)

9. Consider Ihe partial difTerential equation

Ay" + By" + Cy" =O

where the coefficienls A, B. ahd C are constants.

(a) Use the Iransformalion

(A # O. C # O),

r = x + CXI, s = x + pr,

where cxand pare distincI conslants, lo oblain a new differenlial equation in .r wilh independenlvariables r and s.

(b) Assuming Ihal the given difTerential equation is of hyperbolic Iype (Sec. 10). so IhatB2 - 4AC > O, find values CXoand Po of cxand p, respectively, such that the Iransformed equa-lion is

y.. = O.

Then show Ihal Ihe general solulion of Ihe given difTerenlial equalion is

y =g(x + cxot) + h(x + PoI)

where 9 and h are arbilrary difTerenliab]e funclions.(e) By applying Ihe solulion oblained in parl (b) lo Ihe wave equalion -a2)'xx + )'" = O,

show how Ihe solulion (6), Seco ]9, of Ihal equalion follows as a special case.

-B:t JB2 - 4ACAns. (b) CXo.Po= .

21. Historical Development

Mathematical sciences experienced a burst of activity following the inven-tion ofcaIculus by Newton (1642-1727)and Leibnitz (1646-1716).Amongtopics in mathematical physics that attracted the attention of great scientistsduring that period were boundary value problems in vibrations of stringsstretched between fixed points and vibrations of bars or columns of air, allassociated with mathematical theories of musicalvibrations. Early contribu-tors to the theory of vibrating strings incIuded the English mathematicianBrook Taylor (1685-1731), the Swiss mathematicians Daniel Bernoulli(1700-1782)and Leonhard Euler (1707-1783),and lean d'Alembert (1717-1783)in France.

By the 1750'sd'Alembert, Bernoulli,and Euler had advanced the theoryof vibrating strings to the stage where the partial differential equationy" = a2yxxwas known and a solution of the boundary value problem forstrings had been found from the general solution of that equation. Also, theconcept of fundamental modes of vibration led those men to notions ofsuperposition of solutions, to a solution of the form (12),Seco15,where aseries of trigonometric functions appears, and thus to the matter of rep-resenting an arbitrary function by a trigonometric series. Later on, Euler


gave the formulas for the coefficientsin the series.But the general concept ofa function had not been elarified,and a lengthy controversy took place overthe question of representing arbitrary functions on a bounded interval byseriesof sine functions.This question of representation was finally settled bythe German mathematician P. G. Lejeune Dirichlet (1805-1859) about70 years later.

The French mathematical physicist Jean Baptiste Joseph Fourier(1768-1830)presented many instructive examples of expansions in trigono-metric series in connection with boundary value problems in the conduc-tion of heat. His book "Théorie analytique de la chaleur," published in1822,is a elassic on the theory of heat conduction. It was actually the thirdversion of a monograph that he originally submitted to the Instituí deFrance on December 21, 1807.t He etTectivelyillustrated the basicprocedures of separation of variables and superposition, and his work didmuch toward arousing interest in trigonometric series representations.

But Fourier's contributions to the representation problem did not in-elude conditions of validity of representations; he was interested in applica-tions and methods. As noted above, Dirichlet was the first to give suchconditions. In 1829he firmly established general conditions on a functionsufficient to ensure the convergence of its Fourier series to values of the-function.t

Representation theory has been refined and greatly extended sinceDirichlet's time. It is still growing.

t Freeman's early translation of Fourier's book into English was reprinted by Dover, New

York. in 1955. The original 1807 monograph itself remained unpublished until 1972. whenthe critical edition by Grattan-Guinness that is listed in the Bibliography appeared.

t For supplementary reading on the history of Fourier series, see the articles by Langer(1947) and Van Vleck (1914) listed in the Bibliography.

CHAPTER.

TOREE

ORTHOGONAL SETS OF FUNCTIONS

22. The Inner Product of Two Vectors

The concept of an orthogonal set of functions is analogous to the concept ofan orthogonal, or mutually perpendicular, set of vectors in ordinary three-dimensional space. Fundamental properties of the set of functions are sug-gested by corresponding properties of the set of vectors. In fact, manyaspects of orthogonal functions and three-dimensional vectors can betreated simultaneously within the context of generalized vector spaces. Weshalllimit ourselves,however,to analogieswhich willmotívate the terminol-ogy and results needed in our brief introduction to orthogonal functions.Our notation for three-dimensionalvectors in this and the followingsectionsis chosen in ordeno point up those analogies.

Let 1 denote a vector in three-dimensional space whose rectangularcomponents are the three real numbers al, a2,and a3'It can be thought ofasthe radius vector of the point having those three numbers as rectangularcartesian coordinates. When1 is the zero vector, the vector whose compo-nents are all zero, the radius vector reduces of course to a single point, theorigino Ir a vector g has components bl, b2, and b3, the inner, or scalar,product of1and g is given by the equation

(1)3

(f,g) = ¿ akbk.k;1

The norm, or length, of 1can then be written

(2) 11II1 = (JJ)I/2,

the number (f,f) being nonnegative.Although we have introduced the inner product first,and then the norm,

the norm is usually defined first by writing

(3)(

3

)1/2

11II1= k~1af ,

46

SECo 23] ORTHOGONAL SETS Of fUNCTIONS 47

which is the same as equation (2). When neither fnor g is the zero vector, theangle O (Os () S1t) between the radius vectors representing them is well-défihed; and the inner product (f,g) is introduced in terms of norms bymeans of the equation

(4) (f,g) = Ilfll Ilgll cos ().

Expression (1) is then obtained from equation (4). Ireitherfor gis the zeroveCtor, the angle () in equation (4) is undefined; but it is understood that(.f;g) = O then.

That approach has considerable advantage from a geometric point ofview. Equation (3) is simply a statement that the norm offis the length ofthedirected line segment representing it. Also, the length

(5)f

3

J

1/2

Ilf- gil= k~1 (ak - bk)2

of the vectorf - g, whosecomponents are al - bl, a2 - b2,and a3 - b3,canbe thought of as the distance between the tips of the vectors f and g, in-terpreted as radius vectors. Note, too, that if IIgll= 1, then, according toequation (4), (f,g) is the projection off onto the direction of g. As we shallsee, the analogous norms and inner products offunctions do not have suchimmediate geometric interpretations, and it will be more convenient to startwith inner products when we develop.the analogy in Seco25.

Finally, we note that whenf and g are orthogonal,or perpendicular, toeach other, the value of cos () in equation (4) is zero; that is,

(6) (f,g) = O.

Ir 11 f 11 = 1, then f is a unit vector, also called a normalized vector. In the nextsection we combine the concepts of orthogonal and normalized vectors.. I

23. Orthonormal Sets of Vectors

Let !/In (n = 1, 2, 3) denote nonzero vectors in three-dimensional spacewhich form an orthogonal set; that is, they are mutually orthogonal, so that(!/1m'!/In)= Owhen m '# n. A set of unit vectors clJn(n = 1,2,3) having the samedirections can be formed by dividing each vector !/Inby its length:

!/In

clJn= ~ (n =' 1, 2, 3).

This orthogonal set of normalized vectors clJnis called an orthonormal setoSuch a set is described by means of inner products by writing

(1) (clJm,clJn) = t5mn (m, n = 1, 2, 3)

where Ómnis Kronecker' s Ó:

Ómn= 1~

if m =1=n,

if m = n.

Condition (1) states that each of the vectors <P¡,<P2,<P3is perpendicular tothe other two and that each has unit length.

The symbol {<Pn}will be used to denote such orthonormal sets. A simpleexample is the set consisting of unit vectors along the three coordinate axesof the rectangular cartesian coordinate system.

Every vector f in three-dimensional space can be expressed as a linearcombination of the vectors <p¡,<P2,and <P3.That is, three numbers cl, C2,andC3can always be found such that

(2) f= CI <PI + C2<P2+ C3<P3

when the vector f is given. To find the number Cl, take the inner product ofboth sides of the vector equation (2) by <Pl.This gives

(f,<p¡)= (c¡ <p¡+ C2<P2 + C3<P3, <p¡)

= c¡(<P¡,<PI)+ C2(<P2,<PI)+ C3(<P3'<P¡)= Cl

since (<p¡,<pd= 1 and (<P2,<pd= (<P3,<pd= 0, according to condition (1). Sim-ilarly, C2 and C3 are found by taking th\,: inner products of botlt sides ofequation (2) by <P2and <P3,respectively. Thus

(3) Cn = (f,<Pn) (n = 1,2, 3),

Cnbeing the projection of f onto <Pn.The representation (2) can then bewritten

(4)

3

f = (f,<P¡)<P1+ (f,<P2)<P2+ (f,<P3)<P3= I (f,<Pn)<Pn.n= 1

The representation (2), or (4), is an expansion of the arbitrary vector finto a finite series of the orthonormal reference vectors. Those orthogonalreference vectors were assumed to be normalized only as a matter of con ven-ience in order to obtain the simple formulas (3) for the coefficients in theexpansion.

The definitions and results just given can be extended immediately tovectors in N-dimensional space, where a vector fhas N components insteadof 3. It is natural to define the inner product oftwo such vectorsfand g to bethe number

(5)N

(f,g)= I akbkk=l

when f and g have components al' a2, ..., af'{ and b¡, b2, ..., bf'{,respectively.

SECo 24] ORTHOGONAL SETSOF FUNCTIONS 49

A generalization of another sort is also possible. The units of length onthe rectangular coordinate axes, with respect to which the components ofvectors are measured, may vary from one axis to another. In such a case aninner product of two vectorsfand 9 in three-dimensional space has the form

3

(f,g) = I Pkakbk,k= I

where the positive weight numbers PI, P2, and P3 depend upon the units oflength used along the three axes.

24. Piecewise Continuous Functions

Before introducing inner products and norms of functions, we need tospecify the function space (Sec. 11)for which this is to be done.

Let a function f be continuous at all points of a bounded intervala ::;x ::;b except possibly for some finite set of points a, x h x 2, ..., Xn- 1, bwhere a < XI < X2 < ... < Xn-I < b. Then f is continuous on each of theopen intervals

a < x < XI' XI < X < X2, Xn- I < X < b.

It is not necessarily continuous or even defined at their end points. But if ineach of those subintervals f has finite limits as x approaches the end pointsfrom the interior,fis said to be piecewise continuous on the interval (a,b). Tobe precise, the one-sided limits f(a+), f(xI -), f(xI +), ..., f(b-) arerequired to exist, where such limits from the right and left at a point Xoaredefined, respectively, as folJows:

(1) f(xo+) = limf(xo + h),h-Oh>O

f(xo-) = limf(xo - h).h-Oh>O

The step function illustrated in Fig. 7 is piecewisecontinuous on theinterval (0,4), aJthough it is not continuous there. Note that a function ispiecewise continuous on (a,b) if it is continuous on the c/osed interval

¡(xl9 9 Q, I 'I I rI I II , rI I I

: , :Io '1 ;2, ,

I I6 6

3 4 x

Figure 7

a :::;x :::;b. Continuity on the open interval a < x < b does not, however,imply piecewise continuity there. The function I(x) = l/x defined on theintervalO < x < lis, for example, continuous but not piecewisecontinuouson that interval sinceI(O+) fails to existo

The integral of a piecewisecontinuous function exists. It is the sum ofthe integrals over the subintervals:

(2).b ."1 . "2 .bI I(x) dx = I Idx + I Idx +... + I I ~x.'a °a 'XI . Xn-l

The first integral on the right exists beca use it is the integral of a continuousfunction over the interval a :::;x :::;XI if we merely assign the values I(a + )andI(xI -) toIat a and X¡, respectively. Likewise, the other integrals existas integrals of continuous functions.

If two functions II and I2 are each piecewise continuous on an interval(a,b), t~ere is a subdivision of the interval such that both functions arecontinuous on each cIosed subinterval when the functions are given theirlimiting values from the interior at the two end points. Hence a linearcombination el}! + ed2, or the productId2, has that continuity on eachsubinterval and is therefore piecewise continuous on (a,b). Consequently, theintegral of the funttions edI + ed2, Id2' and [II (x)Y all exist on theinterval (a,b). Also, one-sided limits of those functions exist as the corre-sponding combinations of one-sided limits of I1 and I2'

Except when otherwise stated, in this book we shall restriet our attentiontu Iunctions thal are piecewise continuous on aIl bounded intervals underconsideration. When it is stated that a function is piecewise continuous onan interval, it is to be understood that the interval is bounded; and thenotion of piecewise continuity cIearly applies regardless of whether theinterval is open or cIosed.

Since, as noted above, any linear combination of functions that arepiecewise continuous also has that property, we may use the terminology ofSeco ll and refer to the cIass of aIl piecewise continuous functions defined pnthe interval (a,b) as a function space; we denote it by Cp(a,b). It is analogousto three-dimensional space, where linear combinations of vectors are weIl-defined vectors in that space. In the next section we shaIl extend the analogyby developing the concept of orthogonal functions.

Other function spaces can be used in the theory of orthogonal functions.The subspace consisting of all continuous functions on the interval

a :::;x :::;b is simpler but more restricted than the space Cp(a,b).The space ofaIl integrable functionsIon (a,b) whose products, incIuding squares [I(x)Y,are integrable is used in the general theory of functional analysis where amore general type of integral, known as the Lebesgue integral, is also used.Our special case involves more elementary concepts in mathematicalanalysis.

SEC. 25] ORTHOGONAL SETS OF FUNCTlONS 51

25. Orthogonality of Functions

We now introduce an inner product which can be applied to any two func-tionsf and g in the function space Cp(a,b)defined in the previous section. Inorder to simplifyour initial remarks, we assume for the moment thatfand gare actually cOl1til1uouSon the interval a ~ x ~ b.

Dividing that interval into subintervals of equallength Óx = (b - a)/N,where N is any positive integer, and letting Xkdenote any point in the kthsubinterval, we recall from elementary calculus that when N is large,

,b N

I f(x)g(x) dx ,; I f(xdg(Xk) óx,'a k; I

the symbol ,; here denoting approximate equality. That is,

(1).b N

I f(x)g(x) dx ,; I akbk'a k; I

where

ak = f(Xk)j/S:X and bk = g(Xk)j/S:X.

In view of equation (5), Seco23, the left-hand side of expression (1) is thenapproximately the inner product of two vectors in N-dimensional space,when N is large. The approximate equality becomes exact in the limit as thenumber of components considered tends to infinity.t

This suggests defining the il1l1erproduct of the fUl1ctiol1sf al1dg as thenumber

(2).b

(f,g)= I f(x)g(x) dx.'a

The inner product here is, of course, also well-defined when f and g areallowed to be piecewise continuous on (a,b).

The function space Cp(a,b) with inner product (2) is analogous to ordi-nary three-dimensional space. Indeed, the following counterparts of familiarproperties of vectors in three-dimensional space hold for any functionsf, g,and h in Cp(a,b):

(3)

(4)

(5)

U,g) = (gj),

U,g + h) = (f,g) + (f,h),

(cf,g) = c(f,g),

t Seepp. 210 rr of Ihe book by Lanczos (1966).listed in the Bibliography. for an elaboralionof this idea.

52 FOURIER SERIES AND BOUNDARY YALUE PROBLEMS [SECo 25

where e is any real number. and

(6) (ff) ;;::O.

The analogy is carried further with the introduction of the norm

(7) Ilfll = (ff)1/2

of a functionf in e p(a.b).It is evident from equation (2) that the norm offcan be written

(8){

b \1/2

Ilfll = t[J(xW dx¡ .

Two functions f and g in e p(a.b) are orthogonal when

(J,g) = O.

or

(9)b

f f(x)g(x) dx = O.'a

Also, if Ilfll = 1, the functionfis said to be normalized.We have carried our analogy too far to preserve the original meaning of

our geometric terminology.The norm of a functionfhas no interpretation asa length associated with f Geometrically, it is simply the square root of thearea under the graph of [J (x W.The orthogonality of two functions f and g

signifies nothing about perpendicularity, but instead that the product fgassumes both negative and positive values on the interval in such a mannerthat equation (9)holds. The so-calleddistancebetweentwofimctions f ami g,

I b \1/2

Ilf- gil = \t[J(x) - g(xW dx¡ ,(10)

is a measure of the mean distance between their graphs.A set offunctions {I/I.(x)} (n = 1.2, . ..) is orthogonal on an interval (a,b)

if (I/Im.I/I.)= O whenever m + n. If none of the functions 1/1.have zero norms.each function 1/1.can be normalized by dividing it by the positive constant111/1.11.The new set {<p.(x)}so formed. where

I/I.(x)

<P.(x ) = ~J(11) (n = 1,2, ...),

is orthonormal on the interval; that is.

( 12) (<Pm,<P.)= bm. (m,n = 1,2, ...),

where bm. is Kronecker's b (Sec. 23). Written in full, the characterization (12)of an orthonormal set becomes

(13),b 10

.la <Pm(x)<p.(x) dx = \ 1

if m + n,if m = n.

SECo 25] ORTHOGONAL SETS OF FUNCTIONS 53

The interval (a,b)over which the functions and their inner products aredefined is called the fundamental intervalo The same terminology is usedwhen the interval under consideration is cIosed, such as when the innerproduct is defined on the space offunctions that are continuous on the entireinterval a ::; x ::; b.

An example of an orthogonal set of functions was cited in Seco 16, whereit was noted that the set

¡

. mrx

}

SIn-e

is orthogonal on the interval (O,e). According to that section, the norm ofeach function is JCii; so the corresponding orthonormal set on (O?e)is

(14) (n = 1,2, ...)

(15) ¡A. mrx

}

-SIn-e e (n = 1,2, oo.).

The set (14) is also orthogonal on the interval (- e,e); in that case thenormalizing factor is l/Jc.

PROBLEMS

1. Show that the set

)1 f2 mrx\\je' -J ~cos71

(n = 1,2, oo.)

is orthonormal on the interval (O,e).

Suggesrion: Use the trigonometric identities (compare Seco 16)

2 cos A cos B = eos (A - B) + cos (A + B),

2 COS2 A = 1 + cos 2A,

2. (o) Use the fact that the set (15), Seco25, is orthonormal on the interval (O,e) to showthat the set

I 1 . mrx\ljesm 71

(n = 1,2, oo.)

is orthonormal on the interval (- e,e).(b) Use the fact that the set in Problem 1is orthonormal on the interval (O,e)to show that

the set

I 1 1 mr.::!1J2c' jecos7/

is orthonormal on the interval (- e,e).(c) Using the results in parts (o) and (b), show that the set

(n = 1,2, .oo)

I 1 1 III¡¡X 1 . n¡¡xl

1J2c' jecos-¡-, jesm~1 .

is orthonormal on the interval (- e,e).

(lII,n = 1,2, oo.)


Suggestion: Observe Ihal iff( -x) =f(x) for al! x in the interval (- e,e), its graph y = f(x)for that interval is symmetric with respect lo the y axis and

( f(x) e/x =2 [' f(x) e/x,.-~ 'o

provided f is integrable rrom x =O to x = e. Likewise, iff( -x) = -f(x). !he graph is sym-metric with respect to the origin and

(f(X) e/x=O.

3. Show Ihat the functions "',(x) = I and "'2(X) = x are orthogonal on Ihe interval (-1,1),and determine conslants A and B such that the function "'3(X)= I + Ax + BX2is orthogonal toboth "', and '" 2on thal interval.

Ans. A=O,B=-3.

4. Two continuous functionsf(x) and ",,(x) are linearly independent on an interval (a,b);that is,one is not a constanl times the other. Determine the linearcombinationf + A'" I of thosefunctionswhich is orthogonal to "', on !he interval. and thus obtain Ihe orthogonal pair ""''''2where

"'2(X)=f(x) - ~~~II~"',(x).

AIso, give lhe geometric inlerpretation or this expression for '" 2 when/. "',' and '" 2 represenlvectors in lhree-dimensional space.

5. If a functionfis continuous on a closed bounded interval, it is bounded there; that is, a

positive number M exists such that If(x) I ~ M for al! points x in the inlerval. Using this fact,

state why a piecewise continuous function is bounded on the set of points in ils interval a! whichit is defined.

6. Write f(x) = l/jX and (a) state why f is not piecewise con!inuous on the interval(0,1); also, (b) show that

.'I f(x) e/x'0

exists as an improper integral but that (f(x)j2 is not inlegrable over the interval (0,1).

7. Suppose tha! f(x)'= O except at a finite number of points on an interval a ~ x ~ b.State why f is a piecewise continuous function with zero norm (11fll = O) on that interval.

8. Given that the integral of a nonnegative continuous function has a positive value ifthefunction has a positive value somewhere in the interval of integration, show that if a functionfispiecewise continuous and Ilfll = O on an interval (a,b) of the x axis, thenf(x) = Oeverywhere,except possibly for a finite number or points, in the interval.

9. Verify thal

I .b .b

2.1. .1.(.f(x)g(,r) - g(x)f(,rW clx clJ'= IIfl1211gll' - Cf.g)2.

assuming Ihat f and 9 are piecewise continuous. Thus establish the Sehwarzinequalit,r

1(/.g)l ~ Ilfllllgll,

which is also valid when f and 9 denote vectors in three-dimensional space. In Ihat case it isknown as Cauchy's inequality and is an immediate consequence or definition (4). Seco 22. orthe inner product of two vectors.

SECo 26] ORTHOGONAL SETS OF FUNCTlONS SS

10. Prove that if 1 and 9 are piecewise continuous functions on a fundamental interval(a.h) and if eithcr has norm zero. say 11.r11 = O. then C{.g) = O (use Problem 9).

11. Prove that if1 and 9 are functions in the space e p(a.b). then

111+ gil :5 l/.fll + Ilgll.

\f 1 and 9 denote. instead, vectors in three-dimensional space. this is the familiar triangleinequality, which states that the length of one side of a triangle is less than or equal to the sum ofthe lengths of the other two sides.

Sugyesrion: Start the proof by showing that

111 + gl12 = 111112 + 2(f,g) + Ilg112.

and then use the Schwarz inequality (Problem 9).

26. Generalized Fourier Series

Let f be an arbitrarily given function in Cp(a,b), the space of piecewisecontinuous functions definedon the interval (a,b).When an orthonormal setoffunctions 4>n(x)(n = 1,2, oo.) in Cp(a,b)is specified,it maybe possibletorepresent f by a linear combination of those functions, generalized to aninfinite series which converges to f(x) at all but possibly a finite numberof points x in the fundamental interval (a,b):

(1) f(x) = Ct4>¡(x) + C24>2(X)+.oo + cn4>n(x) +oo. (a < x < b).

This corresponds to the representation (2), Seco23, of any vector inthree-dimensional space in terms of the three vectors of an orthonormal se\.We recall from Seco23 that for three-dimensional space only three vectors 4>nare required in any orthonormal reference set, three being the number ofcomponents of a vector in that space. However, as suggested by the discus-sion at the beginning of Seco25 leading up to the inner product (2) in thatsection, the number of functions 4>n(x)in the above orthonormal set will ingeneral need to be countably infinite; that is, there is a one-to-one corre-spondence between those functions and the set of all positive integers.

Ir the series in equation (1) does converge to f(x) and if, after we mul-tiply all terms in the equation by 4>n(x),the resulting series is integrable, wecan obtain the coefficients Cnas inner products by the process used forvectors. Upon multiplying through by 4>n(x)and integrating over the inter-val (a,b), we see that

(f,4>n) = c¡{4>¡,4>n)+ C2(4)2,4>n)+ ... + cn(4)n,4>n)+ ....

Since (4)m,4>n)= Dmn'it follows that

(2) (f,4>n) = Cn (n = 1, 2, oo.).

56 FOURIER SERIESAND BOUNDARY VAlUE PROBlEMS [SECo27

The numbers Cn are called the Fourier constants for f corresponding to

the orthonormal set {<Pn(x)}; they can be written

,b

Cn= I f(x)<Pn(x)dx'Q, (3) (n = 1,2, ...).

The series in equation (1) with those coefficients is the generalized Fourierseries corresponding to the function f, and we write

(4)00 oo.b

f(x) - I cn<Pn(x)= I <Pn(x) I f(s)<Pn(s)dsn==1 n= 1 °0

(a<x<b).

When {<Pn(x)}is the orthonormal set of functions (Sec. 25)

~ . mrx<Pn(x)= -Sln-

C C

in the function space ep(O,c),for example, the coirespondence (4) becomes

2 00 mrx ,e mrsf(x) -- I sin - I f(s) sin - ds

Cn=1 C'O C

which was obtained earlit:r in Seco 16.

The correspondence (4) betweenf(x) and its series will not always be anequality, even at all but a finite number of points. We may anticipate thislimitation by considering the case of vectors in three-dimensional space. Inthat case, if only two vectors <PIand <P2make up the orthonormal set, anyvector not in the plane ofthose two fails to have a representation ofthe formCI<PI+ C2<P2'The orthonormal reference system here is not complete in thesense that there are vectors in three-dimensional space which are perpen-dicular to both <PIand <P2'

Likewise in the correspondence (4), iff(x) is orthogonal to every func-tion <Pn(x)in the orthonormal set, then every term in the series is zero; andso the series does not represent,r. unless 11.r11= O (see Problem 8, Seco25).

An orthonormal set {<Pn(x)}is complete in the function space being con-sidered if there is no function in that space, with positive norm, which isorthogonal to each of the functions <Pn(x).We have just noted that the setmust necessarily be complete if the correspondence (4) is to be an equalityfor each function f of the space.

(n = 1,2, ...)

(O< x < c),

27. Approximation in the Mean

In three-dimensional space, a linear combination

K = YI<PI + Y2<P2

of two of the vectors of an orthonormal set {<PI,<P2,<P3}is a vector in the plane

SECo 28] ORTHOGONAl SETS OF FUNCTIONS 57

,,, ," / --V K = (',<1>, + ('2<1>2

:::~/~ = ;',<1>,+ ;'2<1>2

<1>,

Figure 8

of cP¡ and CP2'To make K the best approximation to a given vector / inthree dimensions, in the sense that the distance d = 11/- KII between thetips of/ and K is to be as small as possible when/ and K are thought of asradius vectors, we can see geometrically that K must be the projection offonto the plan e of cp¡ and CP2(Fig. 8). Consequently, )'¡ has the value(f,cP¡) = C¡, the projection of/onto cp¡,and)'2 = (f,CP2)= C2'Similarly, wesee that the coefficients Cn= (f,CPn)are those for which a linear combinationof any one, two, or three of the vectors CPnbest approximates f

There is a corresponding characterization ofthe Fourier constants Cnfora function / (x).

Consider functions in the space of piecewise continuous functions on an.interval (a,b). Let cp¡(x), CP2(X),..., CPm(x)denote m functions of an orthonor-mal set {CPn(x)}(n = 1, 2, ...) on that interval, and let Km(x) be a linearcombination of them:

(1) Km(x) = )'¡cp.(x) + }'2CP2(X)+... + )'mCPm(x).

We shall determine the constants )'n so that Km is the best approximationinthe mean to a given function/ in the sense thAt the value of the integral

.bE = I [J(x) - Km(x)Jl l/x,'.(2)

a measure of the error, is to be as small as possible. This is approximation off by least squares. We note that E is the square of the distance (Sec. 25)11/- Km 11between the functions / and Km'

Let Cnbe the Fourier constants Cn= (f,cP,,)for f; then

.b

E = I [J(x) - i'.CP.(X) - i'2CP2(X)-'" - i'mCPm(X)Pl/x'..b

= I [J(x)P l/x'.+ ,,2 + ,,2 + ... + ",2 - 2"' C - 2"' C -'" - 2", C

I 1 . I 2 , m fel I I 2 2 I m m"

We add and subtract d, d, c;, to complete the squares in this lastline; thus

.bE = I [J(xW dx - d - d - ... - e;,'Q

+ (YI - e.)2 + (1'2 - C2)2 +... + (Ym - cnY

It is cIear from equation (2) that E ¿ O.and so it folIows from equation(3) that E has its least value when ,'. = c.. }'2= e2. Ym= Cm,The resultcan be stated as folIows.

(3)

Theorem 1. 1fcl' C2. Cmare the Fourier eonstants ofajimctionf(x)with respect to the functions <P.(x), <P2(X)...., <Pm(x)of an orthonormal set,then, of all possible linear combinations qf' those mjimctions, fhe eombination

cl <PI(x) + C2<P2(X)+ ... + em<Pm(x)

is the best approximation in the mean to f(x) on the jimdamental intervalo

(4)

Write Yn= Cnin equation (3). Then. since E ¿ O..b

d + d +... + c;,::;I [J(xW dx = IIf1l2;'Q

this is known as Bessel's inequality. The term 11.1'112on the right is independ-ent ofm;and.as the number m ofelements from the set {<pl'<)}(n = 1.2, ...)increases, the sums on the left form a sequence which is bounded and nonde-creasing.That sequencetherefore convergesto a limit not greater than IIf112.

That is, the infinite series of squares of the Fourier constants for each func-tion fin our function space converges, and

(5)00 .bL c; ::; I [J(xW dx.

n= 1 "a

The convergence ofthis serjes implies that the general term tends to zeroas n becomes infinite. Hence the Fourier eO/lstants always approach zero asn --+C():

(6) lim Cn= lim (f,<Pn)= O.n-oo n-'"

Note that since the set

IJ

2. \, 7t Sin /IXI

is orthonormalon (O,n)(Sec.25),it folIowsthat

(n = l. 2. .. .)

(7) lim (f(x) sin nx dx = O.n-oc.o

The set

is also orthonormal onconsequently,

I 1 f2 \ -\)n' V ;reos nXI (n- 1,2, ...)

(O,1t), according to Problem 1, Seco 25; and

(8) lim (J(x) cos nx dx = O.n-+~'.0

28, Closed and Complete Sets

A sequence Sm (m = 1, 2, ...) of functionsdetined on a fundamental interval(a,b) is said to converge in the mean, or in the norm, to a functionJover thatinterval if 11 J - Sm 11 -> O as m -> 00; that is, it converges in the mean to J if

b .

lim r (f(x) - Sm(xW dx = O.m-'X¡ "a(1)

Condition (1) is also written

l.i.m. Sm(x) = J(x),m-~

where the abbreviation l.i.m. stands for limit in the mean.

Suppose now that the functions Sm are the partial sums of a gener-alized Fourier series corresponding to J on the fundamental interval (a,b):

(2)m

Sm(x) = I Cn<Pn(x).n=1

This is the linear combination Km(x) in Seco 27 when 'Yn= Cn.

U condition (1) is satistied by each function J in our function space, wesay that the orthonormal set {<Pn(x)}is c/osed in the sense of mean con ver-gence. Thus each function J can be approximated arbitrarily cIosely in themeaJi by some linear combination of functions <Pn(x)of a cIosed set, namelythe linear combination (2) when m is large enough. t

By expanding the integrand in equation (1) and keeping the detinition ofCnin mind, we can write that equation in the form

(3)~~: !((f(xW dx - 2nt¡c; + "t¡c;) = O.

t In the mathematicalliterature, the terms c10sed and complete are sometimes applied tosets which we have called complete (Sec. 26) and c1osed, respectively.


Hence for a cJosedset {4>n(X)}it is true that00 b

L c; = f[J(xW dx.n= 1 "o

This is known as Parsevafs equation.When writtenin the form

(4)

(5)00

L (f,4>n)2 = Ilfll2,n=!

it identifies the sum of the squares ofthe components off, with respect to thegeneralized reference set {4>n(x)},as the square of the norm off

Conversely, if each function f of the space satisfies Parseval's equation,the set {4>n(x)}is cJosed in the sense of mean convergence. This is truebecause equations (3) and (4) are merely alternative forms of equation (1).

Suppose that a function e(x) in the space, where Ilell f O,is orthogonalto each function 4>n(x)ofa cJosed orthonormal setoThen equation (5), withfreplaced by e, gives the contradiction Ilell= O.Thus the set is complete(Sec.26), and we have the followingtheorem.

Theorem 2. lf an orthonormal set {4>n(x)}is c/osed, it is complete.

This bare introduction to the theory of orthogonal functions, based onconvergence in the mean, will not be continued here. Convergencein themean does not ensurepointwiseconvergence;that is, the statement (1) is notthe same as the statement

(6) lim Sm(x)= f(x).,

m-oo

for each point x of the interval (a,b), even if some points may be excepted. tWe shall be concerned with pointwise convergence.

An orthonormal set is c/osed in the sense of pointwise convergence, let ussay, if its generalized Fourier series for each functionf of our function spaceconverges pointwise to f(x), except possibly at a finite number of points, onthe fundamental interval. At the end of Seco26 we indicated why a set that iscJosed in that sense must be complete. Thus Theorem 2 is true for suchcJosed sets in the space of piecewise continuous functions defined on thefundamental interval.

The direct application of Theorem 2 is limited, however, to denying thata set is cJosed when there is a function with positive norm which is orthog-onal to each function of the seto It is our representation theorems which willshow that certain sets are cJosed, and hence complete, in specified functionspaces.

t A simple example of a sequence of functions which converges in !he mean lo zero bu!which diverges ae each poi/!/ of the interval is given by Franklin (1964. p. 408). listed in theBibliography.

PROBLEMS

.1. According to Problem 1.Seco25. the set {j2/c cos ("'IX/")} (11= 1.2 ) is orthonormalon the interval O ~ x ~ c. Show that without Ihe inclusion of some constant function, corre-

sponding to the case 11= O. Ihe set is not complete even in the space of continuous functionswith continuous derivatives on that interval. We shall see later (Sec. 44) why the larger set iscomplete in the space.

2. From Problem 2(a), Seco 25, we know that the ser {sin n1tx} (11= 1,2, ...) is orthonor-mal on the interval - 1 ~ x ~ 1. Show that it is not complete even in the space of continuousfunctions on thal interval.

3. Suppose that a function f has a representation

f(x) = LA."'.(x)..,

on a fundamental interval (a,b) on which Ihe ser {"'.(x)} is orthogonal but not normalized(O< 11"'.11f 1). Use inner products to show formally that

A = (M.). 11"'.IP.

Also, point out how the above series with Ihese coefficients is the generalized Fourier series (4),

Seco 26, where <P.(x)= "'.(x)/II"'. 11.

4. Consider continuous funclions on a fundamental interval a ~ x ~ b. The norm of such

a function vanishes ifand only ifthe function vanishes at every point ofthe interval. Show that iftwo functions f and 9 have the same set of Fourier constants with respecl lo a completeorthonormal set {<p.(x):[c. = (f.<p.)= (g.<p.).11= 1.2 ]. then Ihe runctions must be identical;that is,f(x) = g(x) for all x in Ihe interval a ~ x ~ b. This is a uniqueness property ror functionswith Fourier constants c..

Suggesrion: Write l1(x) =f(x) - g(x) and show that l1(x) = O.

S. Consider continuous functions on a fundamental interval a S x ~ b and let:<p.(x)} be acOlllp/ele orthogonal sc!. Prove that if lhe corresponding gcneralized Fourier series ror afunctionfconverges uniformly over the interval. its sum S(x) is identicallof(x). (See Problem4.) Also, poinl out why the conclusion holds if the condition of completeness is replaced by thecondition that the set is cIosed in the sense of mean convergence.

6. .Show thatthe sequence of runctions Sm (m = 1.2, ...) where Sm(x) = O when O S x ~

l/m, Sm(x) = )/11 when l/m < x < 2/m. and Sm(x) = O when 2/111~ x S 1 converges to zero ateach point x of the intervalOs x S 1 but thalthe sequence does not converge in the mean tothe runctionf(x) = O over (0.1).

7. Consider the sequence of functions S", (111= 1,2. ...) defined on the intervalOs x ~ 1as follows:

10

Sm(x)= 11

whenx = 1.t. ~.

whenx '" l. t. ~.

Show that it converges in the mean to the functionf(x) = 1 over (0.1) but that. for each positive

inleger N. Sm(I/N) -> O as 111-> '$...

Suggestioll: Observe Ihat Sm(I/N) = O when 111~ N.


29. Complex-valued Functions

We shall have a few occasions to use complex-valued functions of a realvariable, functions of the type

(1) w(t) = u(t) + iv(t)

where u and v are real-valued functions of a real variable t.tThe derivative of w is detined in a natural way as

(2) w'(t) = u'(t) + iv'(t),

provided u and v are differentiable.Similarly,

(3).b .b .b

I w(t) dt = I u(t) dt + i I r(t) dt.-o "o °0

The operation W= u-iv of taking the complex conjugate of w.com-mutes with the operations of differentiation and integration; thus, for differ-entiation, [w(t)]'= w'(t). From detinition (2) one can show that theelementary rules of differentiation for real-valued functions, such as theformula for the derivative of a product, apply to these complex-valuedfunc-tions. Also, from detinitions (2) and (3) it follows that if u(t) and v(t) arecontinuous on the interval a ~ t ~ b and if W(t) = U(t) + iV(t) is a functionsuch that W(t) = w(t),then

.bI w(t) dt = W(b) - W(a).'0

The exponential function e=,or exp z, is detined by

(4) e%= eX(cos y + i sin y),

where z = x + iy, x and y are real, and the angle y means y radians. Fromthis detinition it can be seen that e=satisties the usual laws of exponents.Also, if x and y are functions of a real variable t, then z and e%are functionsof type (1); and, with the aid of equation (2), one can show that

d % % dz--e=e--.dt dt

Thus, for example, the function w = exp (ict) satisties the differentialequa-tion w"(t) + C2W(t)= Owhen c is a complex number.

From the detinitions

(5)l. .

sin z = - (el% - e-'=)2i '

l. .)cos z = - (el%+ e-'=2

t See Churchill. Brown. and Verhey (1974).listed in the Bibliography. for a more completetreatment of points raised in this section.

SECo 29]

and definition (4), we find that

(6) Isin zl2 = sin2 x + sinh2 y,

and, when z = z(t),

d . dzdt SinZ= cos z dt '

(7)

The hyperbolic functions

sinhz = ~ (e:- e-:),

ORTHOGONAL SETS OF FUNCTlONS 63

1 cos Z 12 = cos2 x + sinh2 y;

d . dz1

cos z = - Sin z -d- .

el t

1cosh z = (e:+ e-:)2

are cIearly related to the trigonometric functions as follows:

sin iz = i sinh z,

sinh iz = i sin z,

The identity

cos iz = cosh z,

cosh iz = cos z.

(8) IZ"=z(l-zm)"=1 1- z(z 4=1)

gives the sum of a finite number of terms in a geometric progression ofcomplex numbers, and its validity is cIear upon observing that

m m

(I-z)¿z"= ¿(z"_z"+1)=z_zm+1.n=1 "=1

It enables us to write, for example,m m m

2 ¿ cos nO= ¿ (eiO)" + ¿ (e-iO)""=1 "=1 "=1

ei0(1 - eimO)e-iO/2 e-iO(1 - e-imO)eiO/2- -. + --- 1 - eiO e-iO/2 1 - e-iO eiO/2

exp [i(m + 1)0]- exp [- i(m+ 1)0]=-1+ -- ----exp (iO/2)- exp (- iO/2) ,

or

m sin [(m + t)O]2"~1 cos nO= - I + ---sin (0/2) ,

where () is any angle such that 0#-0, :t27t, :t47t, Identity (9) isLagrange'strigonometricidentity,which wiIIbe used later on (Sec.40) in thetheory of Fourier Series.

(9)

30. Other Types of Orthogonality

Extensions of the concept of orthogonal sets of functions should be noted.(a) A set {ljIn(x)} is orthogonal on an interval (a,b) with respect to a

weight function p(x), which is positive when a < x < b, if

(1).b

I p(x)ljIm(x)ljIn(x)e/x = O'a

when m +- n.

The integral here represents the inner product (IjIm,ljIlI)with respect to theweight function, corresponding to the ioner product of vectors wheo weightnumbers are used (Sec. 23). The set is normalized by dividing IjIn(x)by IlljIn11,where

.bIIIjInl12= (IjIn,ljIn)= I p(x)[ljIn(xWe/x'a

and where it is assumed that IlljIn11+-O.

This jpe df orthogonality is reduced to the ordinary type by using theproducts p(x)ljIn(x) as the functions of the se!.Weight functions other than unity will arise in orthogonal sets ofBessel

functions and in other sets generated by Sturm-LiouvilIe problems. A fur-ther example is the set of Tchebyshe.ff po/ynomia/s

(2) T,,(x)= cos (n cos-I x) (n = O, 1,2, .. .),

which is orthogonal on the interval (- 1,1) with respect to the weightfunction

(3)1

p(x)=- .fi=7

(b) A set {wn(t)}ofcomplex-valued functions (Sec. 29) ofa real variable tis orthogonal in the hermitian sense on an interval (a,b) if

(4).b

I wm(t)wn(t) dt = O'a(m +- n).

The integral here is the hermitian inner product (wm,wn).The square of thenorm of Wnis real and nonnegative since

(5).b .b

Ilwnl12=1 wnwndt=1 (U;+ r;)e/t"a .a

if Wn= Un+ iVnwhere Unand Vnare real-valued functions of t.Certain complex-valued exponential functions furnish the most promi-

nent examples of such sets. The functions

(6) énl = cos nt + i sin nt (n = O, :t 1, :t2, . ..),

SECo 30] ORTHOGONAL SETS OF FUNCTlONS 65

for example, constitute a set with hermitian orthogonality on the interval(-n,n). The proof is left to the problems.

(e) For sets of functions of two independent variables the fundamentalinterval is replaced by a region in the xy plane, and the integrations aremade over that region. Similar extensions apply to functions of three ormore variables.

PROBLEMS

l. Verify that solutions of the difTerential equation

w"'(t) + e2w(t) = O,

in which e is any nonzero complex constant, may be written in each of the forms

w=C, exp (iet) + C2 exp(-ict)

= C 3 sin et + C4 cos et

= -tCs{exp [¡(et + C.)] + exp [ - i(et + C.)]}

=Cs cos (et + C.),

where C \, C 2' ..., C. are arbitrary complex constants.

2. Use equations (6), Seco29, to show (a) that the zeros ofthe functions sin z and cos z are

all real; (b) that Isin zl and Icos zl are unbounded as Iyl-+ oo.

3. Euler's formula

e;. = cos O + j sin O,

where O is any angle, is a special case of equation (4), Seco 29. Use it to write

11 ." .f(

r eX cos x dx + j I eX sin x d.x = I e(l +;¡x dx.'0 '0 'o

and then evaluate the two integrals on the left-hand side here by evaluating the single integralon the right.

Ans. - I + e" 1 + e"2 ,~.

4. Derive the identity

m

2 L cos (2n - 1)0 = sin 2mO"=1 ~

(OfO, :tlt, :t21t,...)

by noting that its left-hand side can be written

m m

e-l. L(eI2.)" + e;. L (e-;20)"11""I n= I

and then using equation (8), Seco 29.5. Show that

a cos O- a2

Ia"cosnO= l-2acosO+a-n=I(-1 < a < 1)

by using equation (8), Seco 29, to find an expression for the sum of the first 111terms of the series

on the left here and then letting 111tend to infinity. Do this by first writing

m m m

2 ¿ a. cos uO= ¿ (aew)"+ ¿ (ae-;")".no: 1 n= I n::::I

6. Establish the hermitian orthogonality of the exponential functions (6), Seco 30, on theinterval (-n,n). AIso show that the norm of each function is J2ii.

7. Show that the set

I I (2unr

)1\ .-. exp i~ b-al(u = O, :!:I, :!:2, ...)

is orthonormal in the hermitian sense on the interval (a,b).

8. Substitute O= cos- 1 X to. establish the orthogonality of the Tchebysheff polynomials

(2), Seco 30, with weight function (3) in that section.

9. Write x = cos O in expression (2), Seco 30, for the functions T.(x). Then, in de Moivre'sformula

(coi; O + i sin O). = cos uO + i sin uO

use the binomial expansion and equate real parts to show Ihat cos uOis a polynomial of degreen in cos O and hence that T,,(x) is actually a polynomial of degree n in X.

10. Let R be the square region O < x < n, O < Y < n in the xy plane. Show that ift/>m.(x,y)= sin mx sin ny (m,u = 1,2,.. .), the set (t/>m.(x,y)}is orthogonal over R in the sense that

rr t/>m.(x,y)t/>j'(x,y) dA = O"R

if 111f j or n f k.

31. Sturm-Liouville Problems

In Seco15 the method of separation of variables, followed by superposition offunctions of the type X(x)T(t) that satisfy the homogeneous equations, wasused to write a formal solution of a boundary value problem in the displace-ments y(x,t) of a stretched string. The process required the function X to be asolution of the homogeneous problem

(1) X"(x) + AX(X) = O, X(O)= O, X(c) = O.

For a discrete set of values of the parameter, il = n2n2/c2 (n = 1,2, .. ,), wefound that problem (1) has nontrivial solutions X = sin (nnx/c) and (Sec. 16)that those functions are orthogonal on the interval (O,c).

When applied to more general boundary value problems in partial dif-ferential equations, the process frequently leads to a homogeneous ordinarydifferential equation of the type

(2) X"(x) + R(x)X'(x) + (Q(x) + ilP(x)]X(x) = O,


involvinga parameter A in the manner indicated, and to a pair of homoge-neous boundary conditions of the type

(3) al X(a) + a2X'(a) = O, bl X(b) + b2X'(b) = O.

The functions P, Q, and R and the constants a, b, al, a2' bl, and b2 areprescribedby theprobleminpartialdifferentialequations;AandX(x) aretobe determined.

When its terms are niultiplied by a function

r(x) = exp [J R(x) e/xJ,

which is an integrating factor for X" + RX', equation (2) takes the form

d'

[

dX

Jdx r(x) dx + [q(x) + Ap(x)]X = O.(4)

Other types of homogeneous two-point boundary value problems in-volving a para meter may arise instead of the problem consisting of equation(4) with boundary conditions (3). But that problem, called a Sturm-Liouvilleprob/em, is of fundamental importance.t

Under rather general conditions on the functions p, q, and r it can beshown that there is always a countable infinity of values Al, ..12,... of thepara meter ), for each of which the above Sturm-LiouvilIe problem has asolution that is not identically zero. The numbers Anare the eigenvalues, orcharacteristic numbers, of the problem; and the corresponding solutionsX n(x) are the eigenfunctions, or characteristic functions. Note that eX n(x) isalso an eigenfunction, where e is any constant other than zero. In the specialcase (1), An= IJ2rt2/C2and Xn(x) = sin (IJrtx/c).

The orthogonality of the eigenfunctions with weight function p, on theinterval (a,b), is established in the following section. With respect to the setof normalized eigenfunctions

4>n(x)= 'f~~l' where IIXn 112= .(pX; dx,

the generalized Fourier series for a functionf is

(5)7.

I cn4>n(x)n=1

where.b

Cn= I Pf4>ndx.'Q

For prominent special cases of the Sturm-LiouvilIe problem we shall estab-lish the convergence of this series tof(x) on the interval (a,b).

t Papers by 1. C. F. Slurm and J. Liouville giving the first exlensive development or thetheory or this problem appeared in vols. 1-3 or JOllrnal d~ matlll!matiqll~. 1836-1838.

68 FOURIER SERIES AND nOUNDARY VALUE PROnLEMS [SEC'. 32

The convergence of the series in the general case is not treated in thisbook.t For the general case, proofs that series (5) converges tof(x) usuallyemploy either the theory of residues of functions of a complex variable or acomparison of the series with an ordinary Fourier series that representsfProofs are complicated by the fact that explicit solutions of the Sturm-Liouville equation with arbitrary coefficientscannot be written. Propertiesof solutions are found, by interesting and usefuldevices,from the differentialequation itself.

The adjoint of a second-order linear differential operator M, where

(6) M[X(x)) = A(x)X"(x) + B(x)X'(x) + C(x)X(x),

is the operator M* such that

(7) M*[X(¡x))= [A(x)X(x)]" - [B(x)X(x))' + C(x)X(x).

The operator L defined by the equation

(8) L[X(x)) = [r(x)X'(x)]' + q(x)X(x)

is self-adjoint; that is, L and L*are the same since

rX" + r'X' + qX = (rX)" - (r'X)' + qX.

Form (4) of the Sturm-Liouville equation turns out to be !!speciallyusefulbecause it is in self-adjoint form:

(9) L[X(x)) + J.p(x)X(x) = O.

Some properties of L wii'ibe noted explicitly in the problems.

32. Orthogonality of the Eigenfunctions

A few results in the general theory can be established here. They will beuseful in the chapters to follow. In special cases the eigenvaluesand eigen-functions will be found, and so their existence will not be in doubt.

We assume,ul1/essothenvisestated,that the coefficientsin the Sturm-Liouville problem

(1)

(2)

(rX')' + (q + J.p)X= O,

al X(a) + a2X'(a) = O, bl X(b) + b2X'(b) = O

satisfy these conditions: p, q, r, and r' are rea/-ra/ued contil1uousful1ctiol1sof

t The general case is trealed in Churchih11972, Chap. 9). For other trealmenlS or SlUrm-Liouville theory, some quile extensive, see Ihe book by Ince (1956), the one by Coddington andLevinson (1955). and thc two volumes by Titchmarsh (1962. 1958). These are alllislcd in IheBibliography.

the real variable x (a ::; x ::; b) and are independentof .,1.;also, p(x) > Oandr(x) > O when a < x < b, and the constantsal' a2, bl> and b2 are real andindependentof .,1..It is usually understood that al and a2 are not both zero,and the same is true of the constants bl and b2. Eigenfunctions Xn are tosatisfy the regularity conditions usually required of solutions of differentialequations of the second order (Sec. 18);namely, Xn amiX~areto becontin-uous on the imerval a ::; x ::; b.

Theorem 3. Let A.mand A.nbe any two distinct eigenvalues of the Sturm-Liouvi/le problem (1) and (2), with correspondingeigerifunctionsXm andXn'

Then Xm and Xn are orthogonal with weight function p on the interval (a,b).The orthogonality also holds in the following cases:

(a) when r(a) = O ami thefi.rst ofboundary conditions (2) is droppedfrom

the problem;(b) when r(b) = O and the second of conditions (2) is dropped;

(c) when r(a) = r(b) and conditions (2) are replaced by the conditions

(3) X(a) = X(b), X'(a) = X'(b).

In cases (a) and (b) the Sturm-Liouville problem is said to be singular.This terminology is used when either p(x)or r(x) vanishesat an end point orq(x) is discontinuous there.1t also applies when the interval is replacedby anunbounded one. Note that the dropping of the first of conditions (2) incase (a) is the same as letting both al and a2 be zero; a similar remarkregarding bl and b2 applies to the dropping of the second of thoseconditions in case (b).Conditions (3) in case (c) are called periodicboundaryconditions.Theycommonlyarisewhenx representsthe angle4Jincylindrical coordinates, on the interval (-n,n).

To prove the theorem, we first note that

(rX;"Y+ qXm= -A.mpXm, (rX~Y+ qXn= -A.npXn

since each eigenfunction satisfies equation (1) when A;is the correspondingeigenvalue. We multiply each side'of these two equations by Xn and Xm,respectively, and subtract to get the relation

(4) (A.m- An)pXmXn = Xm(rX~Y - Xn(rX;"Y = ~[r(XmX~- X;"Xn)].

While the final reduction here to an ex1tctderivative is elementary, it is madepossible by the special nature of the Sturm-Liouville operator L defined inequation (8), Seco 31. Details regarding this point are left to the problems.

Our continuity conditions now permit us to write.b

(Am- A.n)I pX m X n dx = [r(x) ~(x )]~'0(5)

"


where Ll(x) is the determinant

(6)I

Xm(X)

Ll(x) = Xn(X)

X;"(X)

l

.X~(X) ,

that is,

(7),b

(Am- An)I pXmXn dx = r(b) Ll(b) - r(a) Ll(a).'Q .

The first of boundary conditions (2) requires that

aIXm(a) + a2X;"(a) = O, aIXn(a) + a2X~(a) = O;

and for these simultaneous equations in al and a2 to be satisfied by numbersal and a2, not both zero, it is necessary that the determinant Ll(a) be zero.SimilarIy, from the second boundary condition, where bl and b2 are notboth zero, we see that Ll(b) = O.Then, according to equation (7),

,b

(A..- An)I pX..Xn dx = O'Q

(8)

and, since Am#- An,the desired orthogonality property follows:

(9),b

I p(x)X..(x)Xn(x) dx = O'Q

(Am =1=An).

Ir r(a) = O, property (9) follows from equation (7) even when Ll(a) #- O,or when al = a2 = O, in which case the first of boundary conditions (2)disappears. Similarly, if r(b) = O, the second of those conditions is notneeded.

When r(a) = r(b) and the periodic conditions (3) are used in place ofconditions (2), then

r(b) Ll(b) = r(a) Ll(a)

and, again, equation (9) follows.This completes the proof of Theorem 3.Suppose that X is an eigenfunction corresponding to an eigenvalue

A= (X+ ip, where (Xand p are real numbers. Then X(x), which may becomplex-valued, satisfiesequatjons (1) and (2).Taking complexconjugatesof all terms in those equations and recalling which coefficientsare real-valued, we see that (Sec.29)

(rX')' + (q + Xp)X = O,

al X(a) + a2X'(a) = O, bl X(b) + b2X'(b) = O.

Thus X is an eigenfunction corresponding to the eigenvalue X; and, accord-ing to equation (8),

,b

(A- X) I p(x)X(x)X(x) dx = O.'Q

But p(x) > Owhena < x < b. and XX = IX12.So the integralherehas apositive value. Since A - A = 2iP. it follows that p = O;that is. A is real.

The argument also applies to cases (a). (b).and (c) in Theorem 3.

Theorem 4. For the Sturm-Liouville problem (1) ami (2), and itslIIodifications(a), (b), ami (c) citell in Theorem 3, each eigenvalueis real.

33. Uniqueness of Eigenfunctions

Let X and Y denote two eigenfunctions ofthe Sturm-Liouville problem (1)and (2), Seco32, corresponding to the same real eigenvalue A.Suppose that rdoes not vanish at one end point ofthe interval. say at x = a. where the valueof r is positive.

The linear combination

(1) W(x) = Y'(a)X(x) - X'(a)Y(x)

satisfies the linear homogeneous differential equation

(2) . (rW)' + (q + Ap)W = O;

also W(a) = O.Since X and Y satisfy the equations

(3) al X(a) + a2X'(a) = O, al Y(a) + a2 Y'(a) = O,

where al and a2 are not both zero, and since W(a) is the determin~mt of thatpair of equations in al and a2, then W(a) = O.According to our uniquenesstheorem for solutions of linear differential equations (Sec. 18), W(x) == Oisthe only solution of equation (2) for which W(a) = W(a) = O.Thus

(4) Y'(a)X(x) - X'(a)Y(x) = O.

Unless Y'(a) = X'(a) = O,equation (4) states that the functions X and Yare linearly dependent; that is, one is a constant times the other. Recall thatzero is not an eigenfunction. Ir Y'(a) = X'(a) = O, then al = O in equations(3). In that case we use the ,linear combination

(5) Z(x) = Y(a)X(x) - X(a)Y(x)

to provein likemannerthat Z(x) == Oand hencethat X and Yare linearlydependent.

Suppose that X = u + iv is a complex-valued eigenfunction correspond-ing to a real eigenvalue A..When we substitute u + ir for X in the Sturm-Liouville problem and separate real and imaginary parts, we see that u andv are each eigenfunctions corresponding to A.. Hence l' = ku andX = (1 + ik )u, where k is a constant. That is, X is real except possibly for animaginary constant factor.

We collect our results as follows.


Theorem 5. Under the additiona/ condition that either r(a) > ° orr(b) > O, the Sturm- Liouvil/e prob/em (1) and (2) in Seco 32 Ca/1/lOthave tIVOlinear/y independent eigef!functions that correspond to the same eigenva/ue;a/so, each eigenfunction can be made rea/-va/ued by mu/tip/ying it by an appro-priate nonzero constant.

,1

That Theorem 5 does not apply when the conditions (2), Seco32, arereplaced by periodic boundary conditions is shown by this important specialcase:

(6) X" + AX = 0, X( -n) = X(n), X'( -n) = X'(n).

When A > 0, or A = (X2((X> O), the solution of the differential equation hereis

X(x)=C1 sin(Xx+C2 cos(xx.

If it is to satisfyboth boundary conditions, we find that

(7) e I sin (Xn= ° and C 2 sin (Xn= 00

Since el and C2 must not both vanish if X is to be an eigenfunction, itfollows that A= n2 (n = 1, 2, ...), while the constants CI and C2 areotherwise arbitrary.

In particular, for any constant Bn, the tIVOfunctions

(8) Xn(X) = sin nx, y"(x) = Bn sin nx + cos nx

are linear/y independenteigenfunctionscorrespondingto the same eigenva/ueA.= n2.Other linear combinations of sin nx and cos nx can be used to formsuch pairs.

When two functions are linearly independent, there is always a linearcombination of the two which is orthogonal to one of the functions (Prob-lem 4, Seco25). If Bn= ° in the second of equations (8), Y"is orthogonal toXn on the interval (-n,n). [See Problem 2(c),Seco25.]

A constant is the only eigenfunctionofproblem (6)corresponding to theeigenvalue A = 0, and there are no negative eigenvalues. According toTheorem 3, eigenfunctions corresponding to distinct eigenvalues are or-thogonal.Herep(x)= 1.Consequently,the set

(9) {1,cos x, cos 2x, ..., sin x, sin 2x, .. .},

a set of eigenfunctionsof problem (6), is orthogonal on the interval (-n,n).Note that this ineludes the orthogonality of sin nx and cos nx, each of whichcorresponds to the same eigenvalue A= n2.

SECo34] ORTHOGONAL SETS OF FUNCTlONS 73

34. Another Example

Graphical techniques can often be used to demonstrate the existence of aninfinite number of eigenvalues.For an iIIustration, we consider here theproblem

(1)

(2)

X"(x) + AX(X) = O,

X'(O) = O, hX(I) + X'(I) = O .

where h is a positive constant. .

The case A= Oleads to the trivial solution X(x) == O,and zero is there-fore not an eigenvalue.

When A > O, or A= rx2(rx> O), the general solution of the dilTerentialequation together with the first boundary condition is X(x) = e2 cos rxx.Imposing the secoild boundary condition, we find that in order for e 2 to benonzero rxmust be a (positive) root of the equation

htan rx= - .

rx(3)

Figure 9, where the graphs of y = tan rxand y = h/rxare plotted, shows thatequation (3) has an infinite number of consecutive positive roots rxl' rx2,...;they are the positive values of rx for which those graphs intersect.t The

t Roots orlhis and a related equation. ror certain values or h. are tabulated. ror example. on

pp. 224-225 or the handbook edited by Abramowitz and Stegun (1965) which is listed in theBibliography. .

)'

:x

Figure 9

corresponding eigenvalues are then An= a; (11= 1, 2, .. .), with eigenfunc-tions X n(x)= cos anX.

Finally, when A < O.or A = - /32(/3> O),it is straightforward to showthat in order for a nontrivial solution to occur /3must satisfy the equation

htanh /3= - - .

.f3(4)

This equation, however. has no real roots since the graphs of the functionsy = tanh /3 and y = -h//3 do not intersect. .

In summary, then. the complete solution of the given Sturm-LiouvilIeproblem is

An=a;, Xn(x)=cosanx (11= 1.2 )

where al. a2, ... are the consecutive positive roots of equation (3).Keeping in mind the fact that an= (h cos an)/sina.. wecan readily show

that

( X;(x) dx = h + ~~2 an

Thus the normalized eigenfunctionsare

(

2h

)1/2

c/>n(X)= h . 2 COSanx+ Sin an

(5) (11 = 1,2. ...).

(6) (11= 1,2, ...).

PROBLEMS

Find all eigenvalues and eigenfunctions ofthe Slurm-Liouville problems in Problems 1 through7 below. Also state what the interval and weight function p are in the orthogonality relationensured by Theorem 3.

1. X" + ).X = O,X(O) = O,X'(n/2)= O.A/1s. X. = sin (2/1- I)x (/1= 1,2, ...).

2. X" + ).X = O, X'(O) = O, X'(e) = O.

/1nxAlIs. Xo = 1,X. = cos- e (/1= 1,2, ...).

3. X" + ).X = O, X'(O) = O, X(e) = O.

AliS. X. = cos (211- l)nx2e (/1= 1,2, ...).

4. X" + ).X = O, X( -n) = O, X'(n) = O.SlIggestioll: Make the substitution x = s-n and transrorm this problem into a special

case of Problem 2.

A/1s. Xo = 1, X. = cos /1(n + x)2 (/1= 1,2, ...).

S. X" +).X = O,X(O) = O, I1x(1) + X'(I) = Owhere 11is a positive constant.Am. X. = sin :x.x (11= 1,2,...) where :x. are the consecutive positive roots of the equa-

tion tan :x= -:x/I1.6. X" +).X = O,X(O) = O,X(I) - X'(I) = O.Alls. Xo = x, X. = sin 0:.X (11= 1,2, ...) where:x. are the consecutive positive roots of

the equation tan o:= 0:.7, X" +).X = O,X(O) = O,I1X(I) + X'(I) = Owhere 11is a constant and 11< -1.A liS. X o = sinh 0:0x where 0:0is the positive root of the equation tanh :x= - :x/11and

X. = sin IX.x (11= 1,2, ...) where IX. are the consecutive positive roots of the equationtan o:= -0:/11.

8. Give the steps that lead to equations (7). Seco33. Also, show that X (x) = 1 is theeigenfunction of the Sturm-Liouville problem (6), Seco33, corresponding to the eigenvalue). = Oand that there are no negative eigenvalues.

9. In the Sturm-Liouville problem (6). Seco33,(a) point out why it is true that if 11is a positive integer and IX(radians) is a constant angle

such that sin IX +-O, then the two linearly independent functions sin IIX and sin (/IX + 0:) are

eigenfunctions corresponding to the same eigenvalue ). = /12;(b) verify that

sin (/IX + IX)cos IX- COS(/IX + IX)sin IX= sin /IX

and hence that cos (/IX + 0:) is a linear combination of the two eigenfunctions in part (a);(e) show that cos (/IX + 0:) is orthogonal to sin (/IX + 0:)and, with the aid of parts (a)

and (b), show that the functions

1, cos (x + IX),cos (2x + IX),..., sin (x + IX),sin (2x + 0:),...

constitute an orthogonal set of eigenfunctions on the interval (-1t,1t).

10. For the eigenvalue problem

X" + ).X= O, XI-e) = X(e), X'(-e) = X'(e),

obtain this set of eigenfunctions, which is orthogonal on the interval (- e,e) (see'Sec.33):

I m1tx. /l1tx\\1, cos~, sln~1

(m,lI= 1,2,...).

11. Use the function Z defined by equation (5), Seco33, to prove that the functions X andy there are linearly dependentwhena, = O.

12. Show that the eigenvalues and normalized eigenfunctions of the Sturm-Liouvilleproblem

X" + ).X = O, X'(O)=0, I1X(e) + X'(e) =O (11>0),

which is a generalization of the one solved in Seco 34, are

An= ~;. (211

)1/2

<P.(x)= 11 . 2 COSo:.XC + Sin o:.C(11 = 1,2, ...)

where 0:. are the consecutive positive roots of the equation tan o:c= 11/:x.Do this (a) by usingthe method of Seco34 and (b) by substituting s = x/e and writing the new Sturm-Liouvilleproblem

d2X

--¡;;2 + ¡IX = O,X'(O)=O, kX(1) + X'(I) = O (¡I = ÁC2. k = Ilc)

in the ¡ndependent variable s and then using the results already obtained in Seco 34.

76 FOURIER SERIES ANO. BOUNDARY VALUE PROBLEMS [SEC. 34

I

\13. Ir A, B, and C are constants, the dilTerential equation

Ax2X" + BxX' + CX = O

is caIled a Cauchy-Euler equation.(a) Show that with the substitution x = e' this dilTerential equation can be put into the

form

d2X dXA-+(B-A)-+CX=O.ds2 ds

(b) Use the result in part (a) and the solution ofthe Sturm-Liouville problem (1), Seco 31,to solve the related problem

(xX')' + l(~)X = O,X(I)=O, X(b)= O,

where b is a constant and b > 1. Normalize ¡he eigenfunctions.

Ans. l.= (10; b)2,f2 (

log X

),p.(x)= V10gb sin n1t log b(n = 1,2, ...).

14. Solve the Sturm-Liouville problem

(X3X')' + lxX = O, X(I)=O, X(e)=O

by noting that the dilTerential equation can be written as a Cauchy-Euler equation and applyingthe transformation x = e' [~ee Problem 13(a)). State the weight function for the eigenfunctions.

1 .Ans. X. = -Sin (n1tlog x) (n = 1,2, ...);x p(.x)= x.

15. Solve the Sturm-Liouville problem

(xX')' + l(~)X = O,X'(I) =0, X(b)=O (b> 1)

by noting that the dilTerentialequation can be written in Cauchy-Euler form (see Problem 13),using the transformation x = e', and then referring to the solution of Problem 3. Normalize theeigenfunctions.

Ans. l. =[

(2n - 1)Ir

]

2

2 log b ' - J 2 [

(2n - 1)1r log x

],p.(x) - log bcos 2 log h(n= 1,2, ...).

16. Note that the dilTerential equation in the Sturm-LiouvilIe problem

(xX')' + l(~)X = O,X'(I)=O, hX(b) + X'(b) = O (h>O,b> 1)

can be written as a Cauchy-Euler equation (Problem 13). Then, with the aid ofthe transforma-tion x = e' and using the solu!ion of Problem 12, find the eigenvalues and normalizedeigenfunctions.

Ans. l. = :x~. ,p.(x)=[

_~bh1

'/2

bh log b + sin2 (a. log b) cos (a. log x)(n = 1,2, ...)

where a. are the consecutive positive roots or the equation tan (a log b) = bh/a.

SECo34] ORTHOGONAL SETSOF FUNCTIONS 77

17. (a) Prove tbat tbe self-adjoint Sturm-Liouville operator L defined by equation (8),Seco 31, satisfies lAgrange's identity

dXL(Y]- YL(X] = -[r(XY' - X'Y)]dx

for eacb pair of functions X and Y, assuming tbat all derivatives involved exist.(b) Sbow tbal Ibis identily can also be written

dX(rY')' - Y(rX')' = -[r(XY' - X'Y)].dx

18. Let tbe operalor L and Ibe functions X and Y be tbose in Problem 17. Witb tbe aid ofLagrange's identity, oblained in tbat problem, sbow tbat if X and Y satisfy tbe Sturm-Liouville

boundary conditions (2) or (3) in Seco 32, tben

(X, L(Y]) = (Y, L(X]),

wbere tbese inner products on tbe interval (a,b) are witb weigbt function unity.

19. Sbow tbat if N is tbe operator d4Jdx4, Iben

d

XN[Y] - YN[X] = dx (XY'" - YX'" - X'Y" + Y'X").

Tbus sbow Ibal if XI and X z are eigenfunclions of Ibe fourtb-order eigenvalue problem

N[X] + A.X= 0, X(O)= X"(O)= 0, X(e) = X"(e) = 0,

corresponding lO distincI eigenvalues Al and Az' Iben XI is orlbogonal lo X z on Ibe inlerval(O,e).

"CHAPTER

FOURFOURIER SERIES

35. The Basic Series

The trigonometric series

(1)1 Q)

2ao + "~1(a" cos nx + b" sin nx)

is a Fourier series if its coefficients are given by the formulas

1 "a" = - f f(x) cos nx dxn -"

1 "b"= - r f(x) sin nx dxn . -"

where f is some function defined on the interval (-n,n).In Seco33 we saw that the functions 1,cos x, cos 2x, . . ., sin x, sin 2x, . . .

constitute a set of eigenfunctions,orthogonal on the interval (-n,n), for theeigenvalue problem

(n = O,1,2, oo.),

(2)

(n=1,2,oo,)

X"(x) + AX(X) = O,

X(-n)=X(n), X'(-n)=X'(n).

The normalized orthogonal set is [Problem 2(c), Seco25]

(4){

~, cos mx, sin nx}-/iTc v0c v0c

and the generalizedFourierseries(4),Seco26,correspondingto a functionfwith respectto that orthonormalset,is

(3)

(m, n = 1,2, .oo);

1" 1r->= r f(s) r->= ds

v' 2n . -" v' 2n

+ Ireos nx r" f(s) cos ns ds + sin nx r" f(s) sin nsds

J

."=1 v0c '-" v0c v0c '-" v0c

This is series (1) with coefficients (2),.'

78

SECo36] FOURIER SERIES 79

The correspondence can be written

1 ft

(5) f(x) '" 2nL/(s) ds1 co

[

ft K

]+ ~ "~1 cos nx L/(s) cos ns ds + sin nx LKf(S) sin ns ds .A morecompactforrois

(6)1 ft 1 co .K

f(x) '" 2n L/(s) ds + ~ "~1 L/(s) cos (n(s - x)] ds.

Note that the constant term in the series. or the term tao in series (1). is themean va/ueoff (x) over the interva/ (-n,n).

Each term of series (1) is periodic in x with period 2n. Consequently,when the series converges to f(x) on the fundamental interval (-n,n), itrepresents a periodicextensionoff; that is, it convergesto a periodic functionof period 2n which coincides with f on the fundamental interval. Note, too,that iffis detined as a periodic function with period 2n, so thatf(x + 2n) =f(x), series (1) representsf outside the interval (-n,n) when the representa-tion is valid on it.

Thus the Fourierseries (1) serveseitherof two importantpurposes:(a)torepresent a function detined on the interval (-n,n), for values of x in thatinterval, or (b) to represent a periodic function,with period 2n, for all valuesof x. Clearly, it cannot represent a function for all x if the function is notperiodic.

In this chapter we shall establish the convergence of the basic Fourierseries (1) to f(x) under fairly general conditions on f. Representations byFourier series for fundamental intervalsother than (-n,n) willfolloweasily.

36. Example

Let us write the Fourier series corresponding to the functionfdetined on theinterval -n < x < n as follows:

(1)f(x) = {~

when -n < x ~ O,when O< x < n.

The graph of the function is indicated by bold line segments in Fig. 10.Note that f is piecewisecontinuous on (-n,n) and the existenceof the

integrals in formulas (2), Seco35, is therefore ensured. The Fouriercoefficientsobtained from those formulas are

1 .K 1

(r

O

r

K

)

n

ao = ~L/ (x) dx = ~ . - KO dx + .o x dx = 2"

y

p~,,, .,,,

1< l',.,,. . ,, ,, ,~ ,~ o L----

21t 31t 41t X-21t -1t o 1t

Figure 10

and

- 1 fn

d - nx sin nx + cos nxJ

n - (- 1t - 1an-- xcosnx x- 2 - 2'

1t o 1tn o 1tn

b1f

n.d

- nx cos nx + sin nx

J

n (- 1tn= - x sin nx x = 2 = - -

1t o 1tn o n

where n = 1,2, Therefore, on the interval (-1t,1t),

1t ex:>

[

( - l)n - 1 (- l)n]

f(x)-- 4 + I 2 cosnx--sinnx.n=1 1tn n

We shall soon see (Sec.42) why this Fourier series convergesto f(x)when -1t < x < 1t.Then it will follow that the series also represents theperiodic extension of the functionfindicated by the dotted line segments inFig. 10.The periodic extension has jumps at the points x = :!::1t, :!::31t,.....Our theory will show that the sum of the series at any such peint must be1t/2,the mean value of the one-sided limits at the point. With the dotted linesegments and the solid dots inc1uded,Fig. 10shows the graph of the func-tion that series (2) represents for all X.

As an illustration of the convergenceof the series to the function (1) onthe interval -1t < X< 1t,a few of its terms may be summed by addition ofordinates. It will be found, for instance, that the graph of the function

1t2 .1'2Y = - - - cos x + Sinx - - Sin X4 1t 2

(2)

is a wavy approximation to the graph of y = f(x) shown in Fig. 10.

37. Fourier Cosine and Sine Series

Irf (- x) = f (x) for all values of x for which f (x) is defined,f is an evenfunction. The graph of y = f(x) is symmetricwith respect to the y axisand, iff is integrable from x = Oto x = c,

(1)e e

f f(x) dx = 2 f f(x) dx.-e O

SECo 37] FOURIER SERIES 81

An odd functionf is one such that f (- X) = - f (X). Its graph is sym-metric with respect to the origin, and

(2) rC f(x) dx = O.. -c

As already indicated in Problem 2, Seco25, properties (1) and (2) are oftenuseful in dealing with Fourier series.

The functions 1, X2, cos nx, and x sin nx are even, while x, X3,sin nx,and X2sin nx are odd functions. Although most functions are neither evennor odd, each function definedon an interval (- c,c)is expressedas a sum ofan even and an odd function by means of the identity

(3)1 1

f(x) = 2[J(x) + f( -x)) + 2 [J(x) - f( -x)].

When f is an even function on the interval (-x,x), the productsf(x) cos nx (n = O,1,2, oo.) are even.The productsf(x) sin nx (n = 1,2, oo.)are, on the other hand, odd. Consequently, when the functionfin integrals(2), Seco 35, is even, the Fourier coefficients have the values

2 na. = - f f(x) cos nx dx

x o

b. = O(n = 1,2, .. .); and the correspondence (5), Seco35, reduces to

1 n 2 00 nf(x) -- r f(s) ds + - I cos nx r f(s) cos ns ds.X'o X .=.1 '0

(4) (n = 0,1,2, oo.),

(5)

This Fourier cosine series is the generalized Fourier series for fwith respectto the set {1/J1t, -/ffir. cos nx} (n = 1,2,...) ofnormalized eigenfunctions ofthe Sturm-Liouville problem

(6) X"(x) + AX(X) = O, X'(O) = O, X'(x) = O,

(O,x). (See Problems 2, Seco34, and 1,the fundamental interval beingSeco25.)

Since the eigenfunctions are even, series (5) represents an even functionon the interval (-x,x) if it convergeswhenO< x < x. Whenfis definedonlyon the interval (O,x),the series is the Fourier series for the even periodicextension off, with period 2x.

When f is an odd function, a. = O(n = O, 1,2, oo.)and

2 nb. = - r f(x) sin nx dxx .o(7) (n = 1,2, .oo);

thus

(8)2 00 .n

f(x) -- I sin nx I f(s) sin ns ds.x .=1 '0

82 FOURIER SERIESAND BOUNDARY VALUE PROBLEMS [SEC.38

This Fourier sine series(compareSeco16)willserveto representodd func-tions defined on the interval (-n,n) or odd periodic functions of period 2n.It will also serve to represent functions defined only on the interval (O,n).Infact, it is the generalized Fourier series forfwith respect to the orthonormalset {J2/ir sin nx} (n = 1, 2, .. .), which is the set of normalized (Sec. 25)eigenfunctionsof the Sturm-Liouville problem

(9) X"(x) + ..1.X(x)= O, X(O)= O, X(n) = Othat arose in Seco15 with e instead of n.

38. Further Examples

To write the cosine series (5), Seco37, for the functionf(x) = sin x on theinterval (O,n),we observe that

2 " 4ao = - f sin x dx = - ,

non

and, when n = 2,3, ..., that

2 "al = - r sin x cos x dx = On '0

2 ."an= - J

sin x cos nx dxn o

=! ('[sin (1 + n)x + sin (1 - n)x) dxn.o

= !r

- cos (1 + n)x - cos (1 - n)xJ

" = ~ 1 + ( - !t .n l+n 1-n o n 1-n

Thus

. 2 2 '" l+(-l)nSIn x -- + - L 1 . cos nxn n n=2 - n

(O< x < n).

Observe that in this particular series the terms occurring when n is anodd integer vanish; that is,

a2n+l = O and4 1

a2n = ; 1 - 4n2 (n = 1,2, ...).

Hence we may replace n by 2n after the summation symbol and write (com-pare Seco17)

. 2 4 '" cos 2nxSIn x ---- - L---Y-n n n= l 4n - 1

Let us assume that the correspondence here is an equality for each value of xin the intervalO::; x ::; n, as we shall see later on. Then for all values of x

(1) (O< x < n).


~-n O n 2n x

y = Isinxl.

Figure 11

outside that interval the seriesconvergesto the even periodic extension, withperiod 21t,of sin x (O::::;x ::::;1t).That extension, shown in Fig. 11, is thefunction y = Isin x l.

Note that since sin x is orthogonal to sin nx when n = 2, 3, ..., on theinterval (0,1t),the Fourier sine series (8), Seco37, for the function f(x) =sin x on (0,1t)consists of a single term, naI1)elysin X.

In the evaluation of integrals representing Fourier coefficients, it issometimes necessary to apply integration by parts more than once.We nowgive an example where this can be accomplished by means of a single for-mula due to L. Kronecker (1823-1891).t Let p(x) be a polynomial of degreem and suppose thatf(x) is continuous. Then, except for an arbitrary additiveconstant,

(2) J p(x)f(x) dx = pF1- p'F2 + p"F3- ... + (-ltp(m)Fm+1

where p is successively differentiated, F 1 denotes an indefinite integral off,F2an indefinite integral of Fl' and soon,andalternatingsignsare affixedtothe terms. Note that the differentiation of p begins with the second term,while the integration off begins with the firse termoThe formula is readilyproved by differentiating its right-hand side to obtain p(x)f(x).

To illustrate the usefulnessof formula (2), let us write the Fourier sineseries (8), Seco37, for the particular functionf(x) = X3 on the interval (0,1t).With the aid of that formula, we find that

2 .nbn= - I X3 sin nx dx1t .o

= ~ r(X3)( - CO~IIIX)- (3X2)( - Si:2I1X)+ (6x)(CO:3I1X) - (6)(Si:4I1X) J:2 2 6

= 2(_1)n+11I 1t 3--11 (11= 1,2, ...).

t Kronecker actually treated the problem more extensively in papers which originally ap-peared in the Berlill Sil=lIllgsherichle. 1885 and 1889.

Hence

00 n2n2 - 6X3 -2 I (~1)n+ 1 3 sin nxn=1 n

Since X3 is an odd function, the serieshere is also the basic Fourier seriesforf(x) = X3 on the interval (-n,n).

(3) (O< x < n).

PROBLEMS

Write the Fourier series on the interval (-It,It) for the funclions described in Problems 1Ihrough 6:

1. f(x) = x when -It < x < It. Nole Ihat Ihe sum of Ihe series is zero when x = :tlt.

'" (_1)0.1Ans. 2 L -- sin nx.

n= 1 n

2. f(x) = j-It when -It < x <O,\ Owhen O< x < It.

It ;. sin (2n - I)xAns. - - + 2 L- .2 0=1 2n - 1

3. f(x) is Ihe funclion such Ihal Ihe graph of y = f(x) consisls of the Iwo ¡ine segmenlsshown in Fig. 12.

y

(1t,2)

o x Figure 12

Ans.3 '"

[

1-(-1)0 (-1)0.'

1- + 2 L ()2 cos nx + - sin /IX .2 0= 1 /lit nlt

4. f(x) = O when -It ::; x::; O,f(x) = sin x when O < x ::; It. Also, given Ihal Ihe seriesconverges lo f(x) when -It ::; x ::; It, show graphically Ihe funclion represenled by Ihe series forall x.

Suggestion: To find Ihe series, wrile Ihe funclion in Ihe form

( )sin x +

,

Isinxlfx=- (-It::;x::; It)

and Ihen use Ihe Fourier cosine series for sin x found in Seco 38.

Ans. 1 1. 2~cos2nx- + -SIn x - - L---'-'It 2 It o- I 4n - I

SECo38] FOUIUliN. SERIES 85

S. f(x) = eox(a # O)when -n < x < n.Suggestion: Using Euler's formula (see Problem 3, Seco30), write

l' .ao+ ibo= - r f(x) e'" dx

1t. -/1(n = 1,2, ...)

and evaluate the integral here.

Ans.2 sinh an

[~ + f ~-1)02 (a ¿os nx - n sin nx) J.n 2a 0= 1 a + n

6.f(x) = sinh x when -n < x < n.Suggestion: Use the series found in Problem 5.

Ans.2

.h ., ",Sin n '" (-1 )0+1 SUlI/<.

- L. 1~,,21t n= I

7. Write the Fourier cosine series for the function f(x) = x (O::; x::; n) on the interval(O,n).Given that the series converges to f(x) when O::; x ::;n, show graphically the functionrepresented by it for all x.

Ans. ~- ~ f cos (2/1 - 1)x2 n 0= 1 (21/ - 1)2 .

8. Prove that the cosine series found in Problem 7 converges uniformly (Sec. 14) withrespect to x for all x and hence that the series really does represent an even periodic runctionwhich is everywhere continuous.

Write (a) the Fourier cosine series and (b) the Fourier sine series for the functions dcfinedon the interval (O,n) in Problems 9 through 11:

9. f(x) = 1 (O< x < n).

Ans. (a) 1; (b) ~ f sin (2n - I)xn 0= I 2n - 1 .

10.f(x) = n-x (O< x < n).

( ) ~ ~ ~ cos (2n - 1)x . (b) ~ sin nxAns. a + L.(2 1)2 ' 2 L. .2 1[n=1 n- n=1 n

11. f(x) = 1 when O < x < n/2,f(x) = O when n/2 < x < n.

2 .,(Ans. (b) - ¿: 1- cos /In)

sin /IX

11 0= I 2 /l'

Write the Fourier cosine series for the functions defined on the interval (O,n) in Problems12 through 14 below. Note that they are also the Fourier series for those functions defincd onthe interval (-n,n).

12. f(x) = X2 (O< X < n).

Ans. n2 '" (- 1)"- + 4 ¿:---y- COS /IX.3 0= I n

13.f(x) = x4 (O< X < n).

Ans.n4 " /l2n2 - 6- + 8 ¿: ( - 1)0 COS/IX.5 0=I /1

0'.

86 FOURIER SERIES ANO BOUNOARY VALUE PROBLEMS [SECo 39

14. f(x) = cos ax (O < x < 1t), where a *O, :!: 1, :!:2,....

Suggestion:The idenlity cos ax = (iax + e-iU)/2 (see Seco29) and the seriesobtained inProblem 5 can be used here.

2a sin a1t

[

1 "( _I)n+ I jAns. -- -.. + ¿ -cosnx.1t 2a' n=I ,,' - a'

15. Using the series in Problems 1 and 12, find the Fourier series for the function

1

f(x) =x + ;¡:x' (-1t<x <1t).

1t' 00

Ans. - + '\' (-I )n (~OS_~X - 2 sin nx.)12 '- ,-----n= 1 n n'

16. Prove Kronecker's formula (2), Seco 38.

39. One-sided Derivatives

Letf denote a function whose right-hand limitf(xo + ) exists at a point Xo(see Seco24). The right-handderivative,or derivative from the right, of fat Xo is defined as follows:

(1) f'( ) - 1' f(xo+h)-f(xo+)

R Xo - 1m h 'h-Oh>O

provided the limit here exists. Note that although f(xo) need not exist,f(xo + ) must exist iff~(xo) does. When the ordinary, or two-sided, deriva-tive f'(xo) exists, it is obvious thatfR(xO) = f'(xo).

Similarly, iff(xo -) exists, the left-hand derivative offat Xo is given bythe equation

(2) f~(xo) = lim f(xo + h) - f(xo - )h-O hh<O

when this limit exists; andf~(xo) = f'(xo) whenf'(xo) exists.A usefulalter-native form of definition (2),

f'( ) - 1' f(xo-)-f(xo-h)

L Xo - 1m h'

h-Oh>O

(3)

is obtained by replacing h by - h there.To iIIustrate, consider first the continuous function

¡X2 when x ::;;O,f(x) = \ sin x when x > O.

We find thatfR(O) = 1 andf~(O) = O, as the graph offindicates;f'(O) doesnot exist.

SEC. 39] FOURIER SERIES 87

For the step function

f(x) = 1011

when x < O,

when x > O,

1'(0) does not exist; but both f~(O) and f~(O) exist and have the commonvalue zero.

The function f(x) = Jx (x ~ O) is an example of a function that hasno right-hand derivative at x = Oa1though it is continuous there.

A number of properties of ordinary derivatives are also valid for one-sided derivatives. If, for example.each bf two functionsf and g has a right-hand derivative at a point Xo, then so does their productoA direct proof isleft to the problems. But a proof can be based on the corresponding propertyof ordinary derivatives in the followingway. We usef(xo + )and g(xo + )asthe valúes off and g at Xo;we also definethose functionswhen x ::s;Xoas thelinear functions represented by the tangent lines at the points (xo, f(xo + ))and (xo, g(xo + )) with slopesf~(xo) and g~(xo), respectively.Those exten-sions off and g are differentiableat xo, with derivatives equal to the right-hand derivatives.Thus the derivativeoftheir product exists there; its value isthe right-hand derivative off(x)g(x) at Xo.

Likewise,iffí.(xo) and g~(xo)exist, the left-hand derivative of the prod-uctf(x)g(x) exists at Xo.

A further property of one-sidedderivativeswill be usefulin our theory ofFourier series and integrals.

Suppose that bothfand its derivativef' are piecewisecontinuous func-tions on some c10sedinterval, and let the interval a ::s;x ::s;b be any one ofthe subintervals interior to which both f and l' are continuous and haveone-sided limits from the interior at the end points. If we define f(a) asf(a+) and f(b) as f(b-), then f is continuous on the c10sed intervala ::s;x ::s;b. Also,f' exists on the open interval a < x < b; so the law of themean applies to any interval a ::s;x ::s;a + h where O< h < b - a. That is,there exists a number Oh(O< Oh< 1) such that

(4) f(a + h) - .D~~.! = f'(a + Ohh).

Since f'(a + ) exists, the limit off'(a + Ohh) as h -+ Oexists and has that value.The difference quotient on the left in equation (4) therefore has the samelimit; that is,f~(a) = f'(a+). Similarly,f~(b) = f'(b-).

Thus at each point Xo of a c/osed interval on which both f and f' arepiecewise continuous the one-sided derivatives off, from the interior of theinterval, exist and are the same as the corresponding one-sided limits off':

(5) f~(xo) = f'(xo + ), f~(xo) = f'(xo - ).

The continuous function

(6) f(x) =2 . 1x sm-

x when x =1=O,

O when x = O

illustrates the distinction between one-sidedderivativesand one-sided lirnitsof derivatives.Heref~(O) =f~(O) = O,while the one-sided Iimitsf'(O+ )and1'(0- ) do not existoThe verification of this is left as a problem.

40. Preliminary Theory

We begin our discussion of the convergence of Fourier series with twopreliminary theorems, or lemmas. The first is often referred to as theRiemann-Lebesguelemma,and we present it in,somewhat greater generalitythan we actually need in order that it can be used as well in Chap. 7, wherethe convergence of Fourier integrals is treated, and also in Chap. 8.

Lernrna 1. lf afunetion G(u) is pieeewise eontinuous on an interval (O,e),then

(1)\ e e

lim f G(u) sin ru du = lim r G(u) cos ru du = O.r-oo o r-oo.o

We shall verify only the first limit here and leave verification of thesecond, which is similar, to thryproblems.

To verifythe first limit,it is sufficientto showthat if G(u) is contin-uous at each point of an interval a ~ u ~ b, then

b

lim r G(u) sin ru du = O.r- 00.a(2)

For, in viewofthe discussionof integrals of piecewisecontiriuousfunctions inSeco24, the integral in the first of limits (1) can be expressedas the sum of afinite number of integrals of the type appearing in equation (2).

Assuming, then, that G(u)is continuous on the closed bounded intervala ~ u ~ b, we note that it must also be uniformlycontinuous there. That is,for each positive number e there exists a positive number b such thatIG(u) - G(v)l < e whenever u and v lie in the interval and satisfy the inequal-ity lu - vi < b.t Writing

eoe = 2(b - a)

t See, for example, Taylor and Mann (1972, pp. 558-561) or Buck (1978, Seco2.3),listed inthe Bibliography.

where EOis an arbitrary positive number, we are thus assured that there is apositive number b sueh thílt

(3)1 G(u)- G(v)1 < "In Eo _\ whenever 1 U - vi < b.

To obtain the limit (2), divide the interval a :s;u :s;b into N subintervalsof equallength (b - a)/N by means of the points a = Uo, U¡, U2' oo., UN = b,where Uo< U¡ < U2< ... < UN'and let N be so large that the length of eaehsubinterval is less than the number b in eondition (3). Then write

.b N ,Un

I G(u) sin ru du = ¿ I G(u) sin ru du"a "=I'Un-l

N .Un N Un

= ¿ I [G(U) - G(u.)]sin ru du + I G(u.) r sin ru du,"=1 '"n-l n=1 'Un-l

or

(4) 1,( G(u) sin ru du I

:s;.t¡ .C. IG(u) - G(u.)llsin rul du + J11G(U.)II.r:.~, sin ru du l.

In view of eor.dition (3)and the faet that 1 sin ruI :s;1, it is easy to seethat

EO b - a Eo(n IG(u)- G(!l.) 11 sin ru 1 du < 2(b =-¡;j ---¡¡- = 2N'Un-' (n=1,2,...,N).

Also, sinee G(u) is eontinuous on the closed interval a :s;U:s;b, it is boundedthere; that is, there is a positivenumber M sueh that 1 G(u)I:s;M for all Ubetween a and b inclusive.Furthermore,

Il

,un .I

I

leos ru.1 + leos ru.-¡I 2SIn ru 1U :s; :s;-'Un-l r r

(n = 1,2, .. ., N),

where it is understood that r > O. With these observations, we find thatinequality (4) yields the stiltement

1,( G(u) sin ru du I < i + ~~N.

90 FOURIER SERIES AND BOUNDARY VALUE 'PROBLEMS [SEC. 40

Now Write R = 4M N leo and observe that if r > R, then 2M N Ir < eo12.Consequently,

I

r

b .l

eo eo

'Q G(u)SIn ru du < 2 + 2 = eo

and limit (2) is established.Note that the lemma is, in particular, valid if ,. denotéJ only positive

integers, of half integers, and tends to infinity through thoSe values ratherthan continuously.

Qur second lemma involves the Dirichlet kernel

whenever r > R;

(5)1 m

Dm(u) = 2 + n~l cos nu,

which is a continuou!i function defined for all u and where ni is allowed to be

any positive integer. Note that Dm(u) is even and periodic with period 2n.The Dirichlet kernel plays a central role in our theory, and the following twoproperties will be useful:

(6) rn Dm(u) du =?:'0 2'

Dm(u) = sin [(ni + !)u]2 sin (uI2) -

(u + O, :t2n, :t4n, .. .).(7)

Property (6) is obvious upon integrating each side of equation (5), andproperty (7) is identity (9), Seco29, in only slightly differentnotation.

Lernrna2. Suppose that a function F(u) is piecewise continuous on theinterval (O,n) and that the right-hand derivative F~(O) exists. Then

,n n

lim I F(u)Dm(u)du= 2 F(O+),m-co '0

I

where Dm(u) is defined by equation (5).

(8)

Qur proof is based on the fact that, for each positive integer ni, we maywrite

(9)n

r. F(u)Dm(u) du = 1m+ Jm'0

wheren

1m = f [F(u) - F(O+ )]Dm(u) du,'o Jm = (F(O+ )Dm(u) duo'0


In view of expression (7), the first of these integrals can be put into theform

(10) rnF(U)-F(O+).

f(

1

) J1m= .O 2 sin (u/2) SIn m + 2 u duo

Observe that the function

G(u) = F(u) - F(O+)2 sin (u/2)

is the quotient of two functions which are piecewise continuous on theinterval (O,n).Although the denominator vanishes at the point u = O, theexistence of F~(O)ensures the existenceof G(O+ ):

lim G(O+ h) = lim F(O+ h) - F(O+) lim .(h/2) = F~(O).h~O h-O h h-O SIn (h/2)h>O h>O h>O

Hence G(u) is itself piecewise continuous on the interval (O,n). ApplyingLemma 1 to integral (10), we therefore conclude that

(ll) lim 1m= O.m~co

With property (6) of the Dirichlet kernel, we know that Jm= jF(O+), or

(12) lim Jm= ~ F(O+).m~oo

The desired result (8) now follows from equation (9) and limits (11) and (12).

41. A Fourier Theorem

A theorem that gives conditions under which a Fourier series converges toits function is called a Fourier theorem. One such theorem will now be

established. Although it is stated for periodic functions of period 2n, itapplies as well to functions defined only orf the fundamental interval ( - n,n);for we need only consider the periodic extensions of those functions, withperiod 2n.

Theorem 1. !.et f (x) be a function which is piecewise continuous on theinterval ( -n,n) and periodic with period 2n. 1ts Fourier series

(1)1 .n 1 co .n

2n L/(s) ds + ~ n~l L/(s) cos [n(s- x)] ds

converges to the va/ue

(2)1

"2[J(x+)+f(x-)]

at each point x (- 00 < x < OC!)where the one-sided derivatives f~(x) andf~(x) both existo

As already pointed out in Seco35, series (1) is an alternative form ofthebasic Fourier series

1 00

"2ao + n~1 (an cos nx + bn sin nx).

Note that the piecewisecontinuity off ensures the existenceof integrals (2),Seco35, which define the coefficientsanand bn.AIsonote that the quantity(2) is simply the mean value of the one-sided limits offat x and is actuallyf(x) iffis continuous at the point X.

It should be emphasized that the conditions given in the theorem areonly sufficient,and there is no claim that they are necessary conditions.Indeed, there are well-known examples of functions which even becomeunbounded at certain points but which nevertheless have valid Fourierseries representations.t The conditions we giveare, however,broad enoughfor use in many physical applications and are suitable for solving theboundary value problems treated in this book.

Turning now to the proof of the theorem, we let Sm(x)denote the sum ofthe first m + 1 terms, where m ~ 1, in series (1):

1 .n 1 m ,n

Sm(x)= 27tL/(s) ds +;; n~1L/(s) cos (n(s - x)] ds.

Using the Dirichlet kernel, defined in equation (5) of the previous section, wecan express Sm(x) more compactly in the form

1 .1t

Sm(x) = - I f(s)Dm(s - x) ds.7t . -n

Observe that the periodicity of the integrand in this last integral allows us tochange the interval of integration to any inthval of length 27twithout alter-ing the value of the integral (see Problem 15, Seco42). Thus

(3)I .x+n

Sm(x) = - I f(s)Dm(s - x) e/s,1t . :c-n

t See, ror instance, Tolstov (1976. pp. 91-94), listed in the Bibliography.

T

SEc.41] FOURIER SERIES 93

where the point x is at the center of the interval we have chosen; and,according to equation (3),

(4)1

Sm(x)= - [/m(x)+ Jm(x)]1t

where

(5).x+n

Im(x) = I f(s)Dm(s - x) ds'x

and

(6) Jm(X) = ( f(s)Dm(s - x) ds.. x-n

If we replace the variable of integration s in integral (5) by the newvariable u = s - x, that integral becomes

Im(x)= (f(x + u)Dm(u)duo'0

Assuming that f~(x) exists, we can now write F(l/) = f(x + u) and applyLemma 2 (see Problem 13,Seco42). The result is

(7) lim Im(x)= ~f(x+).m-O(.

If, on the other hand, we substitute the variable u = x - s into integral(6)and recallthat Dm(u)is an evenfunctionof u,wefindthat

Jm(x)= (f(x - u)Dm(u)du.'0

This time wewrite F(u) =f(x - u) and note that F~(O)= -fí.(x) whenfí.(x) exists.. Lemma 2 then reveals that

(8) lim Jm(x) = ~2f(x-).m-x. i

We now conclude from equations (4). (7), and (8) that when f~(x) andfí.(x) exist,

lim Sm(x)= ~ [J(x+) + f(x-)];m-x.

and the proof of the theorem is complete.

42. Discussion of the Theorem

Suppose that a functionfis defined only on the interval (-n,n) and that it ispiecewise continuous there. As pointed out just prior to the statement ofTheorem 1,that theorem then applies to the periodic extension off Thus ateach interior point x of the interval -n ::;;x ::;;n where both one-sided de-rivatives exist the Fourier series forf converges to the mean value

l

2 (f(x+) +/(x-)] (-n < x < n).

But at both end points x = In it conl'erges to the mllle

(1)l .2 (f( -n+) +.1 (n- )],

provided /~( -n) and f~Jn) exist, because that is the mean value of theons-sided limits of the periodic extension at each of thos points. The meanvalue (l) reducls to/( -n + ), or to f(n - ), if and only iff( -n + ) = f(n - ).

Note that the theorem assumes the existenceof both one-sided deriva-tives of the function only at those points where it ensuresconvergenceoftheseries to the mean value of the function. The function f(x) =X2/3, forinstance, is continuous on the interval -n::;; x ::;;n, and it has one-sidedderivatives there except at the point x = O; also /( -n) = f(n). HenceTheorem 1 shows that the Fourier series for f converges to f(x) when-n ::;;x < O and when O< x ::;;n; it does not ensure convergence whenx=O.

If in Theorem 1 both / ami r are piecewisecontinl/ol/son the interval-n ::;;x ::;;n, then the one-sided derivatives of the periodic extension of /exist at every point (see Seco39); so the series cOlll'ergesererywhere to themean value of the limits from the right and left for the periodic extension.

In Seco36 we wrote the Fourier series for the function

~) f~)=¡OIxwhen -n < x ::;;O,when O< x < n.

Both f and f' are piecewise con tinuous on the interval - n ::;;x ::;;n; so theone-sided derivatives of the periodic extension of f, with period 2n, existeverywhere. Also,f( -n+) = Oandf(n-) = n. Theorem l therefore showsthat the Fourier series found in Seco36 converges to/(x) when -n < x < nand to (O+ n)/2, or n/2, when x = In. The periodic graph in Fig. lO, withthe points (I n,n/2), (I 3n,n/2), ... included, represents the sum of the seriesfor all X.

Representations of piecewise continuous functions on the interval (O,n)by their Fourier cosine series and Fourier sine series are often ensured by

SEC o 42] FOURIER SERIES 9S

Theorem 1 beca use it applies to the even and odd extensions of those func-tions to functions defined on the interval (-n,n). The series for such exten-sions reduce to the cosine series and the sine series, respectively (Sec. 37). Westate the special Fourier theorems as a corolIary.

Corollary 1. Let a .limetion f(x) he pieeewise eol1til1llOUSon the inter-va/ O < x < n a/1(/,for eOI1l'e/1ienee,/et the va/ue of f he defined at each pointwhere it is discol1tinllousas the mean ra/ue ofits right-hand anc/left-hand limitsthere. Then at eaeh point x of that il1terra/ where the one-sided derivativesfÍ1(x) and fí.(x) exist,f is represel1ted hy its Fourier cosine series

1 '<

f(x) = "2ao+ J¡an cos nx(3) (O< x < n),

where

2 .nan= - I f(x) cos nx dxn.o

it is also representedby its Fouriersine series

(4) (/1 = O, 1,2, .. .);

(5):x.

f(x) = I bn sin /1X"=1

(O< x < n),

where

(6)2 .n

bn= I f(x) sin I1Xdxn.o(n = 1,2, ...).

In view of the even periodic function represented by the cosine series (3),that series converges at the point x = O to f(O+ ) iffÍ1(O) exists; at x = n itconverges to f(n-) iff~Jn) existsoThe sum of the sine series (5) is cIearlyzero when x = Oand when x = n.

Broaderconditions than those givenin Theorem 1,under whicha Fouri-er seriesconverges to its function,are stated in Chap. 5 and in a number ofthe referenceslisted in the Bibliography.

PROBLEMS

1. Use Theorem 1 and series (1). Seco 38. to show that. for every value of x.

Isin x I= ~- ~ I co~ 2/1x o

1[ 1[ 0=,4/1 - I

By writing x = O and x = 1[/2 in this expansion. obtain the following summations:

, J 1L: 4/12- 1 = 2'.n=I

, (-1)" J 1[~4/12 - I = 2' - '4 o0-'

96 FOURIER 'SERIES AND BOUNDARY VALUE PROBLEMS [SECo 42

2. With the aid of Theorem 1, sta te why the series found in Problem 4, Seco 38, for thefunction described by the equations J(x) = O when -n::; x::; O and J(x) = sin x whenO < x ::; n must converge to J(x) everywhere in the interval -n ::; x ::; n.

3. Show that the functions described in Problems 2 and 5 o( Seco 38 satisfy condi-

tions under which the series found there actually converge on the interval -n ::; x ::; n. Give thesum of the first series when x = O, :t n and the sum of the second when x = :t n.

An". Prob.2:x=0. :tn.sum= -n/2. Prob.5:x= :tn.sum=coshan.

4. State why the function Ixl is represented by its Fourier series everywhere in theinterval -n ::; x ::; n. From that series, found in Problem 7, Seco 38, show that

x I n2

~ (2n - 1)2= 8 .n-'

5. Describe graphically the function to which the sine series found in Problem 11,Seco 38,must converge for all values of x ( - 00 < x < 00).

6. Show that the odd functionJ(x) = X"3 (-n < x < n) does not have one-sided deriva-tives at the point x = O. Without finding its Fourier series. state why that series converges to

J(x) on the interval -n < x < n, incIuding the point x = O. Thus illustrate the fact that theexistence of one-sided deriva ti ves is not a necessary condition for convergence.

7. Use the series in Problem 12. Seco38. to show that

x (_l)n+1 n2¿-, =-

n=1 n 12'

. 1 n2¿ n2= 6'n=I

8. Write x = n in the series in Problem 13, Seco 38. and then. with the aid of the secondresult in Problem 7 above, show that

4x I n

¿~"=90',,=1

9. With the aid of the series in Problem 14. Seco 38. obtain the expansion

on x (_l)n+1~-- = I + 202¿ -¡-,-sm Qn ,,=I " - a

(o *O. :t 1, :t2, .. .).

10. Show that the function

J(x) = ¡ : si<

when x *O,

when x = O

is continuous at x = O but that neither of the one-sided derivatives exists at that point.

11. Letl be the function defined by equations (6). Seco 39: namely,f(x) = X2 sin (l/x)

when x * O andJ(O) = O. Show thatf'R(O) = Obut thatf'(O + ) fails to exist. Likewise, show thatJiJO) = O but thatf'(O-) does not exist.

12. With the aid of I'Hospital's rule, findJ(O+) andJ~(O) for the function

J(x) = eX- Ix (x *O).

An". 1.1.13. Suppose that a function f(x) has a right-hand derivative i'R(XO) at a point Xo. Show

that the function F(x) =J(xo + x) has a right-hand derivativeat x =Oand that F'R(O)=f'R(XO)'

Likewise,suppose thatf'L(xo) exists and show that if F(x) =J(xo - x), then F'R(O)= -f'L(XO)'


14. Lemma I in Seco40 was needed in order to prove Lemma 2 in that section; but only

the first limit in Lemma 1 occurring when e = 1t and r = IIJ+ 1/2, where IIJdenotes positiveintegers, was used. Obtain that special case of Lemma I from limits (7) and (8), Seco 27; that is,use those limits to show that if G(u) is piecewise continuous on the interval (0,1t), then

lim (G(u) sin [(m + ~)uJdu = O.

m-C(-o

Suggestion: Start by rewriting the integrand in the above integral according to the trigo-nometric identity

sin (A + B) = sin A cos B + cos A sin B.

15, Let f denote a function which is piecewise continuQus on an interval (- e,e) andpetiodic with period 2e. Show that, for any number a,

(f(X) dx = .C.,'f(x) dx.

Suggestion: Write

[' f(x) dx = (' f(x) dx + [' f(x) dx. -e . -e 'a+c

and then substitute the variable s = x - 2e into the second integral on the right-hand side ofthis equation.

16. Prove property (7), Seco40, of the Dirichlet kernel by writing A = nu and B = u/2 inthe trigonometric identity

2 cos A sin B = sin (A + B) - sin (A - B)

and then summing each side of the result from n = 1 to n = m.

17, Given that the right-hand derivatives of two functionsfand 9 exist at a point Xo,provethat the product fg of those functions has a right-hand derivative there by inserting the termf(xo + h)g(xo + ) and its negative into the numerator of the difTerence quotient

f(xo + h)g(xo + h) - f(xo + )g(xo + )+-- ~- .-

h

18. Verify the second limit in Lemma 1, Seco40.

43. Other Forms of Fourier Series

Let e denote any positive number andfa periodic functionofperiod 2e thatis piecewisecontinuous on the interval (- e,e).We detinef(x) at each pointofdiscontinuity xas the mean value off(x+ )andf(x-). A Fourier theoremfor f can be obtained from Theorem 1 by changing the unit of length onthe x axis. .

To do this, we introduce a new independent variable

(1)¡r;x

S=-.e

98 FOURIER SERIES ANO BOUNOARY VALUE PROBLEMS [SEC. 43

Then -n < s < n when -e < x < e, andf(x) =f(cs/n); we write F(s) =f(es/n)o The function F(s) is periodic with period 2n, and it is continuous at apoint s,iff(x) is continuous at the corresponding point XoFrom the piece-wise continuity off(x) on the interval (- e,e) we can conc1udethat F(s) ispiecewisecontinuous on (-n,n); and, from the definition off(x) at points ofdiscontinuity, we find that the value F(s) is the mean of the values F(s+ ) andF(s-) at points s of discontinuityoMoreover, the existence of one-sidedderivatives off (x) at a point x ensures the existenceofthose derivativesofF(s) at the corresponding point so

According to Theorem 1,F(s) is represented by its Fourier seriesat eachpoint s where F~(s)and F~(s)both existoThat is,

(es

)1 ex)

f - = _2 ao + L (a" cos ns + b" sin ns)n ,,=1

(2)

where

(3)

1r

"

(es

)a,,=- f -- cosnsdsno_" n

1 o"

(

es

)b" = - J f . sin ns dsn -" n

(n = O, 1, 2, ooo),

(n=I,2,ooo)0

With the substitution (1), equation (2) becomes1 ""

(

nnx o nnx

)(4) f(x) = -2 ao + L a" cos- + b" sm- .,,=1 e e

Formulas (3) for the coefficientscan be written as follows,by making thesubstitution (1) for the variable of integration:

1r

c nnxa" = - f(x) cos-dx

e o -c e(n = O, 1,2, oo o),

(5) 1r

c o nnxb,,=- f(x)sm-dx

e .-c e

Note that the right-hand side of equation (4) can then be written

(n = 1,2, 000)0

(6)1

r

c 1 ex)

r

c

[

nn

J2e o -/(s) ds + ~ "~I o -/(s) cos ~ (s- x) ds,

where here s is a new variable of integration and is not to be confused withthe variable s defined in equation (1)0[Compare series (6), Seco 350]

The series in equation (4) with coefficients (5), or the series (6), is theFourier series for periodic funetions of period 2eoThe set of functions

(7) I mnx. nnx\\1, cos~, sm-~I

(m, n = 1,2, oo.)


used in expansion (4) is the orthogonal set of eigenfunctions of theeigenvalueproblem(Problem10,Seco34)

(8) X" + AX = O, X( -e) = X(e), X'( -e) = X'(e).

The representation (4) is used for functions defined only on the interval(- e,e),as wellas for periodic functions.But, as in the special case e = 1t,westate the Fourier theorem as follows in terms of periodic extensions of thefunctions.

Corollary 2. If afunetionf(x) is periodie with period 2e, pieeewise eontin-uous on the interval (- e,e), and defined as its mean valuefrom the right and leftat its points of diseontinuity, then at each point x wheref'¡¡(x) andfí(x) existf(x) has the Fourier series representation (4) with the eoefficients (5).

Irfis an even function, expansion (4) reduces to the Fourier eosine seriesrepresentation

(9)I 'J:' n1tX

f(x) = -2ao+ L an cos -n=\ e

where

(10)2 .c n1tX

an = -1 f(x) cos--dxe.o e(n = 0,1,2, oo.).

This is the representation of a functionf on the interval (O,e)in terms of theeigenfunctions of the Sturm-Liouville problem (Problem 2, Seco34)

(11) X" + AX = O, X'(O)= O, X'(e) = O.

Whenfis odd, the series in representation (4) becomes the Fourier sineseries for the interval (O,e), introduced in Seco 16.

Ir we express cos (n1tx/e) and sin '(n1tx/e) in terms of complex exponen-tial functions (Sec. 29) and write

(12)I

Yo = 2ao,I

Yn = 2 (an- ibn),

IY-n = 2 (an + ibn)

(n = 1,2, oo.),

representation (4) takes the form

m-x.n=-m

m ." II11tXL. Yn exp--. e(13) f(x) = lim

~

100 FOURIER SERIESAND BOUNDARY VALUE PROBLEMS [SECo44

This is the exponentialform of the Fourier series expansion of a periodicfunction with period 2e. Equations (12)can be written

1r

e

(- imCX

)')In=- 2 . f(x)exp - dx

e .-e e

(n = O, :t 1,,:t2, . oo).

Note that the sum in equation (13) includes the term }'o, correspondingto n = O.The limit there is the principalvalue of the series

(14)

(15)<1:) imcx <1:) innx 00 - innxL ')In exp-- =')lo+ L')In exp- + L')I-n exp ,

n=-<1:) e n=I e n=I e

obtained by grouping ')lowith the first m terms of each of t~e last two seriesand then taking the limit, as m -+ 00, of the sum of terms in that group. Theprincipal value of a seriessummed from n = - 00 to n = 00 sometimes existswhen the seriesdiverges;but ifthe seriesconverges,its sum is the same as theprincipal value. Details are left to the problems.

44. The Orthonormal Trigonometric Functions

We use the symbol C~(a,b)to denote the set, or function space, of all func-tionsf(x) such thatf andf' arebothpieeewiseeontinuouson the interval(a,b).At every point x in the interval a :s;x :s;b the one-sidedderivativesoff, fromthe interior of that interval, therefore exist; also, the number of discontinui-ties of f in the interval is finite.

Corollary 2 applies to the periodic extension, with period 2e, of eachfunction f in the space C~(- e,e) to show that f(x) has the Fourier seriesrepresen tation (4), Seco43, at all poin ts x (- e < x < e) where f is contin-uous. That series is the generalized Fourier series with respect to the set

¡

1 1 mnx 1 . nnx\M:. ' í:. cos -, í:. SIn- /v 2e v e e v e e

which is orthonormal on the interval (- e,e) [seeProblem 2(e),Seco25] andis contained in the space C~(-e,e). Since the representation is ensured

-except for at most a finite number of points in the interval (- e,e), we canstate the following result in the terminology introduced in Chap. 3.

(1) (m, n = 1,2, oo.),

Corollary 3. In the funetion spaee C~(- e,e) the orthonormal set offune-tions (1) is c/osed in the sense of pointwise eonvergenee. 1t is also complete.

It was pointed out in Seco28 that such a set must be complete if it isclosed.Note that the corollary is stated for functionswhose one-sidedderiv-atives exist everywhere in the interval.

,.


From the representations of functions on the interval (O,e)by Fouriercosine and sine series, corresponding statements follow for the orthonormalsets

(2) I 1 f2 n1tX

}\JC' V e cos c-(n = 1,2, ...)

(Problem 1, Seco25) and

(3) {~. mr:x

}

-SIn-e e (n = 1,2, ...),

(see Seco25) on that interval.

Corollary 4. In the funetion spaee C~(O,e)eaeh ofthe orthonormal sets (2)and (3) is closed in the sense of pointwise eonvergenee, and eaeh is complete.

In particular, the set (2) is complete in the space of continuous functionswith continuous derivatives on the intervalO::; x ::;e (see Problem 1,Seco28).

PROBLEMS

l. Show that iff(x + 2e) = f(x) for all x andf(x) = -1 when -e < x < O,f(x) = 1whenO< x < e, andf(O) = f(e) = O,then

f(x) = ~ I ~sin (2n - 1)7tx7to=.2n-1 e

2. Suppose thatf(x) =Owhen -2 < x < 1 andf(x) = 1when 1 < x < 2; also,f( -2) =f(l) = f(2) = l Show that', for all x in the interval -2::; x::; 2,

(-00 < x < 00).

1 1 '" I

[

n7t n7tx

(n7t

). n7tx

]f(x)=--- ¿ - sin-cos-+ cosn7t-cos- Sln-.

4 7t 0= 1n 2 2 2 2

3. Prove that when O ::; x ::; e,

2 e2 4e2 ",. (-1)0 n7txx =-+2 ¿ --¡-cos-.

3 7t 0= 1 n e

Suggestion: The expansion in Problem 12, Seco38, can be used here, and no new integralsneed be evalua ted.

4. Show that

2 2 '"

[

(_1)0.1 1- (-1)0

]

. n7txx = 2e ¿ - - 2 3 3 Sln-

n= I nn n 1t e(O::;x < e).

5. Show that iff(x) = O when -~ < x < O,f(x) = 1 when O < x < 3, andf(O) =!, then

1 2 "- 1 . (2n - 1)7tXf(x)=-+- ¿-Sin (-3<x<3).

2 7t0=12n - 1 3

Suggestion: The expansion in Problem 1 can be used here.

6. Show that when - 1 < x < 1,

1 2 '"

(2 1

)X+X2= + I(-J)O 2-cosmrx- sinllnx.3 n 0=' 11n 11

7. Suppose that f(x) =cos nx when O< x < 1, f(x) = O when 1 < x < 2, f(O) = 1,f(l) = -1, and f(x + 2) = f(x) ror all x. Write the Fourier series ror f and sta te why it mustconverge everywhere to f(x).

Alls.1 4 'X 11

cos nx + I -2 sin 2mrx.2 n 0= I 411 - I

8. Obtain the Fourier sine series representation

8 'X 11

cos nx = I '-'2---' sin 211nxn 0= 1411 - I

(O<x < 1).

9. Suppose that

I ~ - x/(X)='

,~

x - 4

ewhen O < x < - .- -2

when : < x < e.2 -

Establish the Fourier cosine series representation

2e" 1 (411- 2)nxf(x) ="2 I zcos---

n 0=1 (211- 1) e

10. State why the set of cosine runctions {cos (IInx/e)} (11= 1,2,.. .), excludinga constantfunction, is closed in the sense or pointwise convergence in the subspace of the function space

C~(O,e) consisting or all functions f of that space ror which

(O::; x::; e).

(f(x) dx =O.'0

j

I

11. Letf be piecewise continuous on an interval (O,e)and let bo denote the coefficients inthe corresponding Fourier sine series for! (Sec. 16). Use Bessel's inequality (Sec. 27) to provethat the series of terms b; converges and that

«' 2 ,eI b;::; - I[f(x))' dx.

"= 1 e '0

Thus deduce that bo-> Oas 11-> 00, and show how this conclusion also follows rrom Lemma 1.

12, Let! be piecewise continuous on an interval (O,e)and pro ve that the coefficients (10).Seco 43, satisfy the condition

1 'X 2 .'-a~ + Ia;::; -1 [f(x))' dx.2 0=' e.o

(Compare Problem 1l.) Deduce that ao -> O as 11->oo.

13. Let! be piecewise continuous on an interval (- e.e) and show that the coefficients (5).Seco 43, satisfy the condition

1" 1 .'

2 at + o~, (a; + b;) ::; ~.I }!(x))' dx.

(Compare Problems 11 and 12.)

r

I

14. (a) The coefficients b. in the Fourier sine series for a piecewise continuous functionf

on an inlerval (O,e) are those for which a finite linear combination ofthe sine functions becomes

the best approximation in the mean tof(x) over the interval (O,e). Show how Ihis follows fromTheorem 1 in Seco 27.

(h) Find Ihe values of C ,. C 2' and C 3 such that Ihe funclion

. nx . 2nx . 3nx

.r = C, SIDT + C2 SID 2 + CJ SID Tis Ihe bestapproximation in Ihe mean 10Ihe functionf(x) = 1over Ihe interval (0,2).Also, drawthe graph of y versus x using the coefficients found, and compare it to the graph of y = f(x).

4 4Am. C,=-,C2=0,C3=-.

n 3n

15. Theorem I ensures the convergence of Fourier series wilh corresponding cosine andsine lerms grouped as a single lerm

(a. cos nx + b. sin nx).

Show Ihal whenfsalisfies the conditionsstated in that theorem and when x is a point such Ihatf'(x) andf'(-x) bolh exist, then

1 1 ""- [J(x) + f( -x)] = -ao + ¿a. cos nx.2 2 .-,

Thus the cosine terms themselves form a convergenl series at Ihe poin\. Why must the series of

sine terms converge there also?

16. In Seco 43, write the Fourier representation (4) in the form

f( )I

1. ~ (nnx . nnx

)x = - ao + 1m L.. a. cos- + b. SID- ;2 m-x. ,,-1 e e

then obtain the exponential form (13) of that representation and expression (14) for thecoefficients Y.. Also, show how those samevalues ofy. can be found formally from equation (13)by using the hermitian orthogonality (seeProblem 7, Seco30) of the functions exp (innx/e) onthe interval (- e,e).

17. (a) The serieson the left-hand side of equation (15),Seco43, is said to converge if eachof the individual series on the righl-hand side converges. Prove Ihal if Ihe left.hand sideconverges, Ihen its principal value exisls.

(b) Use the exponential form (13), Seco43, of Fourier seriesrepresenlalions, when e = n,lo show that

~e'= " (-1)..smh 7l L -elllX.=_"I-in

(-n <x < n).

where Ihe principal value of Ihe series is laken.(e) Show how il follows Ihal Ihe principal value of Ihe seriesin parl (b) exisls, wilh value

n colh n, when x = n. Show Ihal when x = n, Ihe series does not, however, converge in Ihesensedescribed in parl (a).t The converse of Ihe slalemenl proved in parl (a) is Iherefore nOIIrue.

t For delails regarding Ihe convergence of series with complex lerms, see, for example,Churchill, Brown, and Verhey (1974, Chap. 6), lisled in Ihe Bibliography.

,~-.

CHAPTER

FIVEFURTHER PROPERTIES OFFOURIER SERIES

j

II

45. Uniform Convergence

-~

In establishing the uniqueness of solutions of boundary value problems inpartial differential equations (Chapo 10) it is sometimes helpful to knowconditions on a function under which its Fourier series will converge uni-formly over the fundamental interval. We present such conditions in thissectionoThe development in Secso46 through 49, dealing mostlywith differ-entiation and integration of Fourier series, will be used only occasionallylater onoThe reader can skip those sections at this time without disruptionand refer back to results that will be specificallycited as neededo

Our treatment of the uniform convergenceof Fourier serieswilldependon an important inequality, and wenow present a simplebut indirect deriva-tion of it.

Let A. and B. (n= 1,2,o o o, m)denoterealnumberswhereat leastoneofthe A., say AN' is nQnzeroo Ir the quadratic equation

m m m

X2 I A;+ 2x I A.B. + I B;= O.=1 .=1 .=1

in the variable x has a real root xo, we need only write that equation in theform

m

I (A.x + B.)2 = O.= 1

toseethat A.xo + B. = Oforeach n,in particularwhen n = NoConsequently,the only possiblevalue ofXois Xo= - BN/AN'and wefindthat our quadrat-ic equation cannot have two distinct real rootsoIts discriminant is thereforenegative or zero; that is,

(1)(JI A.B. r ~ (JI A;)(JI B;).

This result is obviously valid even when all of the numbers A. are zero.

104

SECo 45] FURTHER PROPERTIES OF FOURIER SERIES 105

Condition (1) is known as Cauchy's inequality.When m = 3, it simplystates that the square of the inner product of two vectors in three-dimensional space does not exceedthe product ofthe squares oftheir norms.The corresponding property for inner products of functions is the Schwarzinequality (Problem 9, Seco25).We shall use Cauchy's inequality in provingthe following theorem.

Theorem l. Let f be a continuous function on the interval -n =:;;x =:;;nsuch that f( -n) = f(n), and let its derivative l' be piecewise continuous onthat intervalo If a. and b. are the Fourier coefficients

l' ."a. = - I f(x) cos nx dx,n._"

1 "b. = - r f(x) sin nx dx,n . -"

the series

(2)CIO

IJa; + b;n= 1

converges.

From the comparison test we note that each of the series

(3)CIO

Ila.i,.;1

CIO

Ilb.1n=1

converges as a consequence of the convergence of series (2).We begin the proof ofthe theorem with the observation that the Fourier

coefficients off',

1 ,,'IX.= - r .f'(x) cos nx dx,n . -"

(4)1 "

P. = - r f'(x) sin nx dx,n .-"exist because of the piecewise continuity of 1'. Sincef is continuous andf( -n) = f(n),

(5)1 ." 1

1X0=-1 f'(x) dx=-[J(n)-f(-n)] =0.n._" n

Also, when n = 1,2, ..., integration by parts reveals that

(6)n ." 1

IX. = - I f(x) sin nx dx + - [J(x) cos nx]':"n._" n

cos nn= lib. + - [J(n) - f( -n)] = nb.,n

(7)11." 1

P. = - I f(x) cos nx dx + -[J(x) sin nx]':" = -na..n._" n.


Note that the conditionf( -n) = f(n), under which the periodic exten-sion off is also continuous, is necessary if expression (6) is to reduce to theform OCn= nbn.

Now let Smdenote the sum of the first m terms of the infinite series (2). Inview of relations (6) and (7),

S - ~ ~ b2 - ~ I ) 2{32

m - L, Van + UII- L, - OCn+ n'n;1 n;ln,

Since Sm;:::O,it follows from Cauchy's inequality (1) that

f

m 1 m

J

1/2

(8) Sm:::;; n~1 n2 n~1 (OC;+ {3;) .

The first sum on the right here is bounded for all m because the infiniteseriesof positive terms l/n2 converges.

Using the piecewise continuous functionJ' in Bessel's inequality (seeSeco27 and Problem 13, Seco44) with respect to the orthonormal set

j~ cosx cos2x sinx sin2x I\~' fi' fi 'oo.,fi' fi ""j

on the interval (-n,n), we find that for every m it is true thatmi"L (oc; + {3;) :::;;- f (J'(x)Y dx,n;1 n -n

the coefficient OCobeing zero. The right-hand side of inequality (8) is thereforebounded for all m, and so is Sm.

Since Sm is a bounded nondecreasing sequence, its limit as m tends toinfinity exists; that IS, series (2) converges, as stated in Theorem 1.

The next theorem is cIosely related to the first.

Theorem 2. Under the conditions stated in Theorem 1 the convergence ofthe Fourier series

1 00

2ao + n~1(an cos nx + bn sin nx)

to f (x) on the interval -n :::;;x :::;;n is absolute and uniform with respect to x onthat interval.

(9)

The conditions onfandJ' ensure the continuity, and also the existenceof one-sided derivatives, of the periodic extension off for all x. It followsfrom the Fourier theorem in Seco41 that series (9) convergestof(x) every-where in the interval -n :::;;x :::;;n. Now, to prove Theorem 2, we observethat

Ian cos nx + bn sin nx I :::;;IanI + IbnI

I

r

II

J

j

SECo46] FURTHER PROPERTlES OF FOURIER SERIES 107

and also that the series of constants lanl + Ibnl converges because eachof series (3) converges. The comparison test and the Weierstrass M-test(Sec. 14)therefore apply to show that the convergence of series (9) is indeedabsolute and uniformo

The tests apply as well to show the convergenceof the seriesof cosine orsine terms only; in fact, the convergence is absolute and uniformoThereforeseries (9) is the sum of those series; that is,

10000

(10) f(x) = 2ao + n~lan COS/IX+ n~,bn sin /IX

and both series here converge absolutely and uniformly.

(-11: ::; X ::; 11:),

46. Observations

A Fourier series cannot converge uniformly on an interval that contains adiscontinuity of its sum because a uniformly convergent series of continuousfunctions always converges to a continuous function. Hence some continuityrequirement onJ, such as the continuity assumed in Theorem 2, is necessaryin order to ensure uniform convergence of the Fourier series to f(x).

Modifications of Theorems I and 2 for cosine series or sine series, or forFourier series on an interval (- e,e), are apparent. For instance, it followsfrom Theorem 2 that the Fourier cosine series for a continuous functionJ, onthe intervalO::; x::; 11:,converges uniformly tof(x) on that interval if!, ispiecewise continuous on the interval. For the sine series, however, the fur-ther conditions f(O) = f(1I:) = O are needed.

Consider the space of functions satisfying the conditions stated inTheorem 1. Parseval's equation (Sec. 28) with respect to the orthonormaltrigonometric functions on the interval (-11:,11:)is satisfiedbyeach functionfof that space. To see this, we multiply the Fourier series expansion offbyf(x), thus leaving the series still uniformly convergent, and then integrate:

r[J(xW dx-n

1 n 00

[

n .n

J= 2ao L/(x) dx+ n~1 an L/(x) cos /IXdx + bn L/(x) sin /IXc/x .

This is Parseval's equation

1 n . 1 00

n L}f(xW dx = 2 a5 + n~1(a; + b;).

From it we conclude (Sec.28) that the orthonormal set of trigonometricfunctions is cIosed in the sense of mean convergence in the function space.

(1)

47. Differentiation oí Fourier Series

We have seen why the Fourier series for the functionf(x) = x on the interval-1t < X< 1tconvergestof(x) at each point of the interval; that is, (Problem1, Seco38)

c:o

X = 2 ¿ (- 1t + 1 sin nx.=1 n

(- 1t < X < 1t).4

But the series here is not differentiable.The differentiated seriesc:o

2¿(-1).+1 cosnx.=1

does not converge since its nth term fails to approach zero as n tends toinfinity.The periodic extension off with period 21t,represented by the firstseries for alI x, has discontinuities at the points x = :t 1t,:t 31t,....

Continuity of the periodic functions is an important condition for differ-entiability of Fourier series. Sufficientconditions can be stated as folIows.

Theorem 3. Let f be a continuous function on the interval -1t :5:x :5:7tsuch that f (-1t) = f(1t), and let f' be piecewise continuous on that intervaloThen the Fourier series in the representation

(1)1 c:o

f(x) = 2: ao + .~1 (a. cos nx + b. sin nx)(-1t :5:x :5:1t),

where

1 n

a. = - r f(x) cos nx dx,1t '-n

1 n .

b. = ~ r f(x) sin nx dx,1t '-n

is differentiable at each point x in the interval -1t < X < 1twheref"(x) exists:

(2)c:o

f'(x) = ¿ n(-a. sin nx + b. cos nx)..=1

To prove this, we note first that since the Fourier theorem in Seco41 canbe applied to the functionf', that function is represented by its Fourier seriesat each point x (-1t < X < 1t) where its derivativef"(x) exists. At such apoint!, is continuous, and so

1 '"

(3) f'(x) = 2:lXo+ .~1 (IX.cos nx + P. sin nx)

where IX.and P. are the coefficients (4), Seco45. But when f satisfies theconditions stated in Theorem 3, we know from Seco45 that

(4) IXO= O, IX.= nb., P. = -nao (n = 1,2, ...).

SECo 48] FURTHER PROPERTIES OF FOVRIER SERIES 109

When these substitutions are made, equation (3) takes the form (2). Thiscompletes the proof of the theorem.

At a point x wherej"(x) does not exist,but wheref' has derivatives fromthe right and left,the differentiationis still valid in the sensethat the series inequation (2) converges to the mean of the valuesf'(x+ ) andf'(x- ). This isalso true for the periodic extension off

Theorem 3 applies with natural changes to other formsof Fourier series.For instance, iffis continuous andf' is piecewisecontinuous on an intervalO:::;x :::;c, then the Fourier cosine series for f on that interval is differen-tiable at each point wherej"(x) exists.

48. Integration of Fourier Series

Integration of a Fourier series is possible under much more general condi-tions than those for differentiation.This is to be expectedbecausean integra-tion introduces a factor n in the denominator of the general termoIn thefollowingtheorem it is not even essential that the original seriesconvergetoits function in order that the integrated series converge to the integral ofthefunction.The integrated seriesis not, however,a Fourier series if ao * O;forit contains a term aox/2.

Theorem 4. Let f be piecewise continuous on the interval (-n,n). Then,whether the Fourier series corresponding to f,

(1)1 '"

f(x) - 2 ao + .~1 (a. cos nx + b. sin nx),

converges or not, the following equality is true when -n :::;x :::;n:

(2)XI'" 1

L/(s) ds = 2ao(x + n) + .~1 ~[a. sin nx - b.(cos nx - cos nn)].

The latter series is obtainedby integratingtheformer term by termo

Our proof starts with the fact that sincef is piecewisecontinuous, thefunction

(3).X 1

F(x) = L/(s) ds - 2aox

is continuous; moreover,

1F'(x) = f(x) - 2 ao


except at points wheref is discontinuous. Therefore F' is piecewisecontin-uous on the interval (-1t,1t). Also,

xII 1F(1t) = f f(s) ds - "2ao1t= ao1t - "2ao1t= "2ao1t-x

and F(-1t)=ao1t/2; hence F(1t)=F(-1t). According to our Fouriertheorem (Sec.41), then, for all x in the interval -1t :;;X :;;1tit is true that

1.

1 ex>

F(x) = "2Ao + n~1 (An cos nx + Bn sin nx)j

1

where

1 x

An = - f F(x) cos nx dx,1t -x

1 x

Bn= - f F(x) sin nx dx..1t - xj1I

.When n =1=o, we integrate the last two integrals by parts, using the fact

that F is continuous and F' is piecewise continuous. Thus

An = ~[F(x) sin nx]~x - ~ rF'(x) sin nx dxn1t n1t -x

1 x

r

1

J=--f f(x)--aon1t - x 2

sin nx dx = - !bn.n~j

I

~

Similarly, Bn = an/n; hence

1 ex>1F(x) = _2 Ao + I - (an sin nx - bn cos nx)

n= 1 n(4)

when -1t :;; x :;; 1t. Out, since F(1t) = ao 1t/2,

1 1 ex>1- ao1t= - Ao - I - bn cos n1t.2 2 n=1 n

I~

With the value of Ao given here, equation (4) becomes

1 ex>1F(x) = _2 ao 1t+ I - [ansin nx - bn(cosnx - cos n1t)].

n= 1 n

In view of equation (3), equation (2) followsat once.The theorem can be written for the integral from Xo to x, where

-1t :;; Xo :;; 1t and -1t :;; X :;; 1t, by noting that

x .x xo

f f(s) ds = I f(5) ds - r f(s) ds..%0 . -n '-n

I

I

II

SECo49] FURTHER PROPERTIES OF FOURIER SERIES 111

49. More General Conditions

A few of the many more general results in the theory of Fourier series wiII benoted here. They are stated without proof; our purpose is only to inform thereader of the existence of such theorems. We first introduce some conceptsinvolved in the theorems. .

A function g defined on a c10sed interval is monotone nondecreasing thereif its value g(x) never decreases when x is increased. One example, on theinterval -n :s;x :s;n, is the step function

n

when -n :s;x < 2'o

(1) ~nG(x)= \ n + 1

n n + 1when-n <x <-n

11+1 - 11+2 (n = 1,2, ...),

n when x = n.

This function is no! piecewise continuous on the interval; graphically, it hassteps up to the line y = x at the infinite set of points x = nn/(n + 1).Thefunction - G(x) is monotone nonincreasing.

A functionf ofboundedvariationon an interval a :s;x :s;b can be definedas one that is the sum of two monotone functions g and h,

f(x) = g(x) + h(x) (a :s; x :s; b),

>

where g is nondecreasing and h is nonincreasing. Each such functionf hasthe following properties.t The one-sided limitsf(x+) andf(x-) from theinterior of the interval exist at each point;fhas at most a countable infinityof discontinuities in the interval, and f is bounded and integrable over theinterval.

(a) Another FourierTheorem. Letfdenote a periodic functionofperiod2n whose integral from -n to n exists.If that integral is improper, let it beabsolutely convergent. There is a theorem which states that at each point xwhich is interior to an interval on which f is of bounded variation, theFourier series for the function converges to the value

12 (f(x+) + f(x-)).

The periodic extension of the function G defined on the interval-n :s;x :s;n by equations (1), for example, satisfies the conditions in thistheorem, but not those in our Fourier theorem in Seco41. For that periodic

t See, for inslance, Franklin (1964. pp. 255 ff) or Apostol (1974, pp. 127 ff), listed in the

Bibliography.

A

function the series converges at every point to the mean value of the limitsfrom the right and left.

The cosine series for the function JX on the intervalO:;; x :;;n con-verges to JX over that interval, incIuding the point x = Owhere the right-hand derivative fails to existoAlso, the series for the unbounded functionX-1/3on the interval (-n,n) represents the function over the interval exceptat the point x = O. .

(b) Uniform Convergence.Let the periodic integrable function fdescribed in (a) above satisfy the additional condition that on some intervala :;;x :;;b it is contínuous and of bounded variation. Then its Fourier seriesconvergesuniformly to f(x) on each cIosed interval interior to the interval(a,b).The series for the functionf(x) = x on the interval (-n,n), for exam-pIe, converges uniformly to x on the interval - 3 :;; x :;; 3.

We noted earlier (Sec.46) that a Fourier series for.a periodic functionfcannot converge uniformlytof(x) on an interval that contains a point wherefis discontinuous. The nature ofthe deviation ofthe partial sums Sm(x)fromf(x) on such intervals is known as the Gibbs phenomenon.t Suppose, forexample, thatf(x) = -n when -n < x < Oandf(x) = Owhen O< x < n.The sequence of sums Sm(x)of the first m + I nonzero terms in the Fourierseries forf on the interval (-n,n) converges,of course, to zero for all x in thesubinterval O< x < n. It can be shown, however, that there exists a fixedpositive number u such that for large m there are alwayspositivenumbers xin that subinterval for which the number Sm(x) is cIose to u. See Fig. 13,which shows how "spikes" in the graphs of the partial sums form near thepoint x = O,wherefis discontinuous, as m increases.The tips ofthese spikestend toward the point (O,u)on the vertical axis.The behavior of the partíalsums is similar on the subinterval -n < x < O.This ilIustrates how special

t See Carslaw (1930, Chap. IX) for a detailed analysis of this phenomenon.

Sm(X)

(O. (1)

7[ x7[

Figure 13

\

1:1,

SECo 49] FURTHER PROPERTlES OF FOURIER SERIES 113

care must be taken when a function is approximated by the partial sums ofits Fourier series near a discontinuity.

(e) 1ntegration. Parseval's equation

(2)1 "" 1 n

2 a5+ L (a; + b;) = - f (f(x)P dxn=1 n'-n

is satisfied wheneverf is bounded and integrable over the interval (-n,n).That is, this series of squares of the Fourier coefficientsoff converges to

11 f 112In.

Proofs of the results stated in (a), (b), and (e) above are given in some ofthe books listed in the Bibliography.t

Let a function Falso satisfy the conditions onf stated under (e), and letAn and Bn denote its Fourier coefficients. Then an + An and bn + Bn are thecoefficients for f + F and, according to Parseval's equation,

1 nI""

; Ln(f(x) + F(x)P dx = 2(ao + AO)2+ n~Yan + An)2+ (bn+ Bn)2].

When we subtract the corresponding equation for f - F, we obtainParsevafs equationfor the inner product:

(3)l n l co

; L/(X)F(X) dx = 2aoAo + Jl(anAn + bnBn).

In equation (3) let us write

F x - jg(x) when - n < x < t,( ) - \0 when t < x < n (- n ~ t ~ n),

where g is bounded and integrable on the interval (-n,n). Then equation (3)takes the form

.'(4) I f(x)g(x) dx'-n

= ~ ao (.g(x) dx + Jl fan (/(x) cos nx dx + bn (ng(X) sin nx dx l.

So it follows from statement (e) that if the Fourier series correspondingto any bounded integrable functionfis multiplied by any other function g ofthe same c1ass and then integrated term by term, the resulting series con-verges to the integral of the productf(x)g(x). When g(x) = 1, we have ageneral theorem for integration of a Fourier series.

t See. in parlicular. Chap. 9 of the book by Whittaker and Walson (1963) Ihal is lisledIhere.

.ti


According to Seco28, statement (e) implies that, in the space ofboundedintegrable functions on the interval (-n,n), the orthonormal set of trigo-nometric functions (4), Seco 35, is c\osed in the sense of mean convergenceo

PROBLEMS

l. Show that the function

f(x) = I~in xwhen -It $; x $; O,

when O < x $; It

satisfies all the conditions in Theorems I and 2. Verify direclly from Ihe Fourier series for f,found in Problem 4, Seco38, Ihal Ihe series converges uniformly for all X. Also, slate why thatseries is difTerentiabIe on the inlerval -It $; x$; It except allhe poinls x = O,;tlt, and describeIhe funclion represenled by Ihe difTerentialed series for all x.

20 DifTerentiale the Fourier cosine series for Ihe funclion f(x) = x on Ihe inlervalO < x < It (Problem 7, Seco 38) lo obtain Ihe Fourier sine series expansion of the functionf'(x) = I on Ihat interval. Stale why the procedure is reliable in Ihis case.

30 Slate Theorem 3 as il applies lo Fourier sine series on Ihe inlerval (0,1t). Point out, inparlicular, why the conditionsf(O) =f(lt) = O are present in this case.

4. Prove that the Fourier coefficienls a. and b. for the functionfin Theorem I satisfy theconditions

lim nao = O, lim nb. = O.

5. Integrate from s = O to s = x (-It $; X $; n) Ihe Fourier series

'" (-Ir+l2 L -sinns

n::::1 "

and

It ",.

- 2 + 2 L Sin (2n - l)s.=1 2n-t '"

oblained in Problems 1 and 2, respeclively, of Seco 38. In each case describe Ihe functionrepresenled by the new series.

6. Let f be Ihe function described by the equations f (x) = - It when -n < x < O andf(x) = O when O < x < It (Fig. 13).

(a) Using Ihe results in Problem 2, Seco 38, and Problem 4, Seco30, show Ihallhe sumSm(x) of.lhe firsl m + I (m ;:: 1) nonzero lerms of Ihe F ourier series ror f on Ihe interval ( - n,n)can be wrillen -

It .' sin 2muSm(x)= - - + I -:-- duo2 .o Sin U

(b) Write

1

9lfsin s 1[u= -ds--,

'0 s 2

\

¡

{SECo 49] FURTHER PROPERTIESOF FOURIER SERIES 115

the value of <1being approximately O.28.t Then, using the integral representation for Sm(x)found in part (a), show that

(11

)..,(2m)

(U - sin U

)Sm - -<1= I ~ sin 2mudu.2m '0 U Sin U

I

(e) Show that the function (u - sin u)/(u sin u) is piecewise continuous on the intervalO~ u ::;11/2and is therefore bounded on that interval (see Problem 5, Seco25). Then showhow it followsfrom the result in part (b) that

lim Sm(~ )= <1.111-00 2m

(d) Suppose that the sequence Sm(x) converges uniformly to O on the intervalO < x < 11.There is then a positive integer m. such that, for all x in the interval, Sm(x) < <1/2wheneverm> mi' <1being the number defined in part (b). By using the main result in part (e), show thatthere is a positive integer m2 such that

s (~ )>'!.m 2m 2

whenever m > m2 ,

and reach a contradiction. The convergence is therefore 1101uniformo

t The integral here occurs as a particular value of the so-called sine integral function Si(x).which is tabulated, for instance, on p. 244 of the handbook edited by Abramowitz and Stegun(1965) that is listed in the Bibliography. Approximation methods for evaluating definite in te-grals can also be used lo find <1.

CHAPTER

SIX

BOUNDARY VALUE PROBLEMS

SO. Formal and Rigorous Solutions

The foregoing theory of representing prescribed functions by Fourier seriesenables us to use the method of separation of variables to solve importanttypes of boundary value problems in partial differential equations. Themethod is illustrated in this chapter for a variety of boundary value pro!?-lems which are mathematical formulations of problems in physics.

We illustrate ways of proving that the function we find truly satisfiesthepartial differential equation and all boundary conditions and continuityrequirements. When that is done, our function is rigorously established as asolution of the problem. The physical problem may indicate that thereshould be only onesolution. In Chap. 10we shall givesome attention to thatquestion of uniqueness of solutions.

Even for some of the simpler problems, the full treatment, consistingofverifyinga solution and proving that it is the only solution, may be lengthyor difficult.We shall give the full treatment in only a fewcases. Most of theproblems will be solvedformally in the sense that we may not verify oursolution fully'or that we may not prove that our solution is unique.

51. The Vibrating String Initially Displaced

In Seco15 we found, but did not verify, an expression for the transversedisplacements y(x,t) of a string, stretched between the origin and the fixedpoint' (e,O),after the string is released at rest from a position y = f(x) in thexy plane (Fig. 14).The function y was therefore required to satisfyall condi-tions of the boundary value problem

(1) Yrr(x,t)= a2y'".,(x,t)

(2)

(3)

y(O,t)= O,

y(x,O) = f(x),

y(e,t)= O

y,(x,O)= O

(O < x < e, t > O),

(t ~ O),

(O :s; x :s; e).

116

SECo 51] BOUNDARY VALUE PROBLEMS 117

y

,

(c,Of' ,.,'y = F(x)

x(-c, 0)"",,,,,,-__"'''''''''''0

Ji

Figure 14

)

I

We used the method of separation of variables, and also superpositionand the orthogonality of the functions sin (nnx/c), to arrive at the formalsolution

(4)00

Y(x t) = '\' b. nnx nnat

, f.., n sm-cos-n= I C C

II

{

I

where

(5)2f

e. nnxbn =- f(x) slO-dx

c. O C(n = 1,2, ," .).

j

The given function f was assumed to be continuous on the intervalO~ x ~ c; also,f(O) = f(c) = o. Assuming further that f' is at least piece-wise continuous on the interval, we now know that f is represented by itsFourier sine series there. The coefficientsin that series are the numbers bngiven by equation (5). Hence when t = O, the series in expresslon (4)converges to f(x); that is, y(x,O)= f(x) when O~ x ~ c.

The nature of the problem calls for a solution y(x,t) that is continuous inx and t when O~ x ~ c and t ~ Oand is such that Yt(x,t)is continuous in t att = o. Thus the prescribed boundary valuesare limitingvalues at the bound-aries of the domain O< x < c, t> O:y(O,t)= y(O+,t), y(c,t) = y(c-,t), andso on.

To verify that expression (4) represents a solution, we must prove thatthe series there converges to a continuous function y(x,t) which satisfiesthewaveequation (1)and all the bóundary conditions. But the seriesmay not betwice differentiable with respect to x and t even though it has a sum y(x,t)that may satisfy the wave equation. In the case of the plucked stringin Seco 17, for instance, the coefficients bn are proportional to(l/n2) sin (nn/2); so after the series for y(x,t) is differentiated twice withrespect to either x or t, the resulting series fails to converge.

It is possible to sum the series in expression (4); that is, we can write ItSsum without using infinite series. This will simplify the verification of thesolution.

Since

)

~

I 2 . nnx /1nat . nn(x + at) . nn(x - at)SIO- cos - = sm + SIO ,e e c c

equation (4) can be written

(6) ,( ) -! ~ b . nn(x + at) ! ~ b . nn(x - at)j X,¡ - 2 1..- n SIn + 2 1..- n SIn .

n=1 e n=1 e

Let F(x) be defined for all real x by the sine series for f:

00 nnxF(x) = I bn sin-

n=1 e(7) ( - 00 < x < 00).

Then F(x) is the odd periodic extension, with period 2e, off(x); that is,

(8)F(x) = f(x)

F( -x) = -F(x), F(x + 2e) = F(x)

when O ~ x ~ e,

for all X.

In viewof equation (7), expression (6) can be written

(9)1

y(x,t) ="2 [F(x + at) + F(x - at)].

Thus series (4) issummed with the aid ofthe function F definedby equations(8). The convergence of series (6) and (4) followsfrom the convergenceofseries (7).

So/ution Verified. Our conditions on f are such that its extension F iscontinuous for all x (Fig. 14).Hence F(x + at) and F(x - at), and thereforethe function y(x,t) given by equation (9), are continuous functionsof x and tfor all x and t. While it isevident from expression (4)that our functiony(x,t)satisfiesthe conditions y(O,t)= O,y(e,t) = O,and y(x,O)= f(x), expression (9)can also be used to verify this. Note, for example, that when x = e in expres-sion (9), we can write F(e - at) = -F(at - e) = -F(at + e); thereforey(c,t) =' O.

Since F(- x) = - F(x), then - F'(-x) = - F'(x) whenever F'(x) exists,where the prime denotes the derivative with respect to the argument of F.That is, F' is an even function. Likewise,F" is an odd function.

If f' andf" are eontinuouswhen O~ x ~ e and

1"(0) = f"(e) = O,

then F'(x) and F"(x) are continuous for all x, as indicated in Fig. 15.Thus

y,(x,t) = ~[F'(X + at) - F'(x - at)),

y, is continuous for all x and t, and y,(x,O)= O. According to Seco19,F(x + at) and F(x - at) satisfy the wave equation (1); therefore y(x,t)satisfiesthat equation, as wellas all boundary conditions. The functiony(x,t)given by equation (9) is then fully verified as a solution 01'our boundary

I

lJI

t

J

I

I

1

{II

rrr,

(I,~

I

1


F'(x)

(-e,O) "..------,...-----,I

I"I,

I..'x

F"(x)

,'-.....................,.......,

x(-e. O)o:

Figure 15

value problem. In Chap. 10 we shall show why it is the only possible solu-tion which, together with its deriva tives of the first and second order, iseverywhere continuous.

Ir the conditions on!' andf" are relaxed by merely requiring those twofunctions to be piecewise continuous, we find that at each instant t there maybe a finite number of points x (O =::;;x =::;;e) where the partial derivatives of yfail to exist. Except at those points, our function satisfies the wave equationand the condition y,(x,O)= O.The other boundary conditions are satisfiedasbefore. In this case we have a solution of our boundary value problem in abroader sense.

52. Discussion of the Solution

From either equation (4) or its alternate form (9) in the previous section wecan see that for each fixedx the displacement y(x,t) is a periodic functionoftime t, with period

(1) To= 2c.a'

The period is independent of the initial displacementj(x). Since a2 = HID,where H is the magnitude of the horizontal component of the tensile forceand D is the mass of the string per unit leng!h (Sec. 3), the period variesdirectly with e and ~ and inversely with JH.

From either equation (4) or (9) in Seco51 it is also evident that, for agiven length e and initial displacementf(x), the displacement y depends onlyon the value of x and the value of the product at. That is, y = </J(x,at)wherethe function </Jis the same function regardless of the value of the constant a.Let al and a2 denote dilTerent values of that constant, and let YI(x,t) andY2(X,t) be the corresponding displacements. Then

(2) YI(X,tl) = Y2(X,t2) if al ti = a2t2 (O =::;; x =::;; e).

...-...-


In particular, suppose that only the constant H differs,with values H 1

and H 2. The same set of instantaneous positions is then taken by the stringwhen H = H 1 and when H = H 2' but the times ti and t 2 required to reachany one position have the ratio

(3) ~ = [H;..t2 V H;

Approximate Solutions. Except for the nonhomogeneous condition

(4) .y(x,O) = f(x),

our boundary value problem is satisfied by any partial sum

(5)m

Ym(x,t)= I bn sin mrx cos mratn=1 e e

ofthe infinite series (4), Seco 51. Instead ofmeeting requirement (4), however,it satisfies the condition

(6) Ym(x,O)= i: bn sin mrx .n= 1 e

The sum on the right-hand side of equation (6) is, of course, the partial sumconsisting of the first m terms of the Fourier sine series for f(x) on theinterval (O,e).Since the odd periodic extension off is continuous and f' ispiecewisecontinuous, that seriesconverges uniformly tof(x) on the intervalO::;;x ::;;e (Chap. 5). Hence, if m is taken sufficientlylarge, the sum Ym(x,O)can be made to approximatef(x) arbitrarily closelyfor al! valuesof x in thatinterval. .

The function Ym(x,t),which is everywhere continuous together with al!its partial derivatives, is therefore established as a solution of the approxi-mating problem obtained by replacing condition (4) in the original problemby condition (6).

Cprresponding approximations can be made to other problems. But aremarkable feature in the present case is that Ym(x,t)never deviatesfrom theactual displacement y(x,t) by more than the greatest deviation of Ym(x,O)fromf(x). This is true because

( )1

[

~ b . mr(x + at) ~ b . mr(x - at)J

Ymx,t = _2 '-- n Sin + '-- n Sinn=l e n=l e

and two sums here are those of the first m terms ofthe sine series for the oddperiodic extension F offwith arguments x + at and x - atoBut the greatestdeviation of the first sum from F(x + at), or of the second from F(x - at), isthe same as the greatest deviation of Ym(x,O)fromf(x).

SECo53] BOUNDARY VALUE PROBLEMS 12l

PROBLEMS

I,fIII¡tt

1. A string is stretched between the fixed points (0,0) and (1,0) and released at rest fromthe position y = A sin ItX, where A is a constan!. Find an expression for its subsequent displace-ments y(X,I), and verify the result fully. Sketch the position of the string at severa1 instants oftime.

Ans. y = A sin ItX cos Itat.

2. Solve Problem 1 when the initial displacement there is changed to y = B sin 2ltx, whereB is a constan!.

Ans. y = B sin 2ltx cos 2Mt.

3. Show why the sum of the two functions y(x,t) found in Problems 1and 2 represents thedisplacements after the string is released at rest from the position y = A sin ItX + B sin 2ltx.

4. For the initially displaced string oflength e considered in Secs. 51 and 52, show why thefrequency v of vibration in cycles per unit time has the value

v =~ =;C\h'

¡Il'

I

Show that if H = 200 lb, the weight per foot is 0.01 lb (go = 0.01, 9 = 32), and the Jengthis 2 ft. then v = 200 cycles/sec.

5. In Seco 51 the position ofthe string at each instant can be shown graphically by movingthe graph of the periodic function ~F(x) to the right with velocity a and an identical curve to theleft at the same rate and adding ordinates, on the intervalO::; x ::; e, of the two curves soobtained at the instant l. Show how this follows from expression (9), Seco 51.

6. Plot some positions of the plucked string considered in Seco 17 by the method describedin Problem 5 to verify that the string assumes such positions as indicated by the bold line

segments in Fig. 16.II

I

II1I

\II,~

!

J

J

y

(1, h)

.,,,...-...-...............

o (2,0) xFigure 16

7. Write the boundary value problem (1) through (3), Seco 51, in terms ofthe two independ-

ent variables x and . = al to show that the problem in y as a function of x and . does not

involve the constant a. Thus, without solving the problem, deduce that the solution has the form

y = 4>(x,.) = 4>(x,al) and hence that the relation (2), Seco 52, follows.

53. Prescribed Initial Velocity

When, initially, the string ,hassome prescribed distribution ofvelocities g(x)parallel to the y axis in its position of equilibrium y = O,the boundary valueproblem for the displacements y(x,t) becomes

(1)

(2)

(3)

YII(x,t) = a2Yxx(x,t)

y(O,t) = O,

Y(X,O) = O,

y(e,t) = O

y,(X,O)= g(X)

(O< x < e, t > O),

(t ~ O),

(O :::; x :::; e).

If the xy plane, with the string Iying on the x axis, is moving parallel tothe y axis and is brought to rest at the instant t = O,the function g(x) is aconstant. The hammer action in a piano may produce approximately auniform initial velocity over a short span of a piano wire, in which caseg(x) may be considered to be a step function.

As in Seco15, we find the functions of type X(x)T(t) that satisfy allhomogeneousequations in 'theboundary value problem. The Sturm-Liouvilleproblem in X is the sameas the one found in Seco15; thus). = n2n2je2andX = sin (nnxje) where n = 1,2, The conditions on T become

T"(t) + (l1;a) 2T(t) = O,T(O)= O;

hence T = sin (nnatje). The homogeneous equations are then formallysatisfied by the function

00

y(x,t ) = "B sl'

n nnx . nnatL.. - SIn-

.= 1 e e'

where the constants B. are to be determined fram the second of condi-tions (3):

(4)00 nna. I1nxL B.- SIn-- = g(x)

.=1 e e(O :::; x :::; e).

Under the assumption that 9 and g' are piecewise continuous, the seriesin equation (4) is the Fourier sine series for g(x) on the interval (O,e) ifB.(nnajc) = b. where

(5)2

re . nnx

b. = - g(x) SIn - dx.e.o e

Thus B. = b.(cjn1ta) and

(6) ( ) e ~ b. . nnx . l11tatyx,t =- L. -Sln-Sln-.1ta .= 1 n e e

We can sum the series here by first writing

00 . nnx nnat 1 1y,(x,t) = Lb. SIn - COS - =_2 G(x+ at) + _2 G(x - at)

.= 1 e e

, ----..

I

Ir

¡I

where G is the odd periodic extension, with period 2e, ofthe given function g(compare Seco 51). Then, since y(x,O) = O,

1 ' 1 '

y(x,t) = 2 fo G(x + at) dt + 2 tG(x - at) dt

1 x+a, 1 x-a'= _2 f G(s) ds - _2 f G(s) ds;a x a x

and, in terms of the function

(7) H(x) = ( G(s) ds'0

1y(x,t) = 2a [H(x + at) - H(x - at)).

(-00 < x < 00),

(8)I

!

~

The verification of the solution in the form (8) is left to the problems.Superpositionof Solutions. If the string is given both a nonzero initial

disp1acementand a nonzero initial velocity,

(9) y(x,O) = f(x), y,(x,O) = g(x),

the displacements y(x,t) can be written as a superposition of the solution(9), Seco51, and the solution (8) above:

1 1(10) y(x,t) = 2 [F(x + at) + F(x - at)] + 2a[H(x + at) - H(x - at)).

Note that both terms satisfy the homogeneous equations (1) and (2),whiletheir sum cIearly satisfies the nonhomogeneous conditions (9). (CompareProblem 5, Seco20.)

In general the solution of a linear problem containing more than onenonhomogeneous condition can be written as a sum of solutions of prob-lems each of which contains only one nonhomogeneous condition. Theresolution of the original problem in this way, though not an essential step,often simplifiesthe process of solving the problem.

54. Nonhomogeneous Differential Equations

The substitution of a new unknown function sometimes reduces a nonho-

mogeneous partial differential equation to one that is homogeneous. Then, ifvariables can be separated and if the new two-point boundary conditionscan be made homogeneous by properly selecting the new function so that aSturm-Liouville problem arises, our method involving superposition may beeffective. .

To iIIustrate, consider the displacements in a stretched string uponwhich an external force per unit length acts, paraIlel to the y axis. Let thatforce be proportional to the distance from one end, and let the initialdisplacement and velocity be zero. Units for x and t can be chosen so thatthe problem becomes

(1)

(2)

y,,(x,t) = yxx(x,t) + Ax (O< x < 1, t > O),

y(O,t) = y(1,t) = O,y(x,O) = y,(x,O) = O,

where A is a constant.In terms of a new function Y(x,t) where

y(x,t) = Y(x,t) + cf¡(x)

and cf¡(x)is yet to be determined, equation (1) becomes

Yrt(x,t) = Yxx(x,t) + cf¡"(x)+ Ax.

This will be homogeneous if

(3) cf¡"(x)= - Ax.

The two-point boundary conditions on Y are

Y(O,t) + cf¡(0) = O,

they are homogeneous if

Y(1,t) + cf¡(1)= O;

(4) cf¡(0)= O, cf¡(1)= O.

From equations (3) and (4) we find that

(5)A

cf¡(x)= -x(l - X2)6 (O =::;x =::; 1).

The boundary value problem inYrt= Yxx, Y(O,t)= Y(l,t) = O,and

Y(x,O) = -cf¡(x),

y now consists of the equations

Yr(x,O) = O.

This is a special case of the problem solved in Seco51. The solution of ourproblem can therefore be written

(6)1

y(x,t) = cf¡(x)- 2 [<II(x+ t) + <II(x- t)]

where <II(x)is defined for aIl finite x as the odd periodic extension of cf¡(x),with period 2. {..

Analogous substitutions can sometimes be made in order to transformnonhomogeneous two-point boundary conditions into homogeneous ones,so that a Sturm-Liouville problem arises. (See Problem 11, Seco55, andalso Seco 57.)

J

1

t

J

J

I

~

J

¡

I

fIt


55. Elastic Bar

A cylindrical bar of naturallength e is initialIy stretched by an amount be(Figo17)and at rest. The initiallongitudinal displacementsof its sections arethen proportional to the distance from the fixed end x = 00 At the instantt = Oboth ends are released and kept freeoThe boundary value probleminthe longitudinal displaeements y(x,t) is (Seco5)

(1)

(2)

(3)

Ytt(x,t) = a2YxAx,t) (O< x < e, t> O;a2 = E/(j),

yAO,t)= O, yAe,t) = O,

y(x,O)= bx, y,(x,O)= 00

The homogeneous two-point boundary conditions (2) state that the forceper unit area E oy/ox on the end sections is zeroo

Functions X(x)T(t) satisfy aIl the homogeneous equations above whenX is an eigenfunction of the problem

(4) X"(x) + AX(X) = O, X'(O)= 0, X'(e)= O

and when, for the same eigenvalue A,

(5) T"(t) + Aa2T(t) = O, r(o) = 00

The Sturm-Liouville problem (4) generates the eigenfunctions used in Fou-rier cosine series on the interval (O,e)oThe eigenvalues, all real, are Ao = O,An= n2n2/e2(n = 1,2,o oo); also Xo(x) = 1,To(t)= 1and

Xn(x) = cos nnxe ' 1;.(t) = cos nnate o

(n = 1,2, 0")0

1

FormaIly, then, the generalized linear combination

1 co nnx nnaty(x,t) = -2 ao + Lan cos - cos -

n=1 e e

satisfies aIl the equations (1) through (3) provided that

(6)1 co nnx

bx = _2ao + L an COS,----

n=1 e(O< x < e)o

r-y(Xot) --~(~-,

IX! j L~ L-;xO Figure 17

The functionbx is such that it is represented by the Fourier cosine series (6)on the intervalO :5:x :5:e, where

2b .< mrxQn= - I x cos - dxe '0 e

(n = O, 1,2, 00 0)0

We find that Qo= be, Qn= -:-2be[l - (- 1)n]/(mr)2(n = 1,2, .00);hence

(7) y(x,t)=!be- 4b2e I -~--2cosPn-l)7txcos(2n-l)1tQto2 7t n = I (2n - 1) e e

Ir P(x) denotes the e~enperiodic extension, with°period2e,of the func-tion bx (O:5:x :5:e), equation (7) can be written

1y(x,t) = 2 [P(x + Qt)+ P(x - Qt)].(8)

This reduction and its verification are left as a problemo

PROBLEMS

l. Show that, for each fixed x. the displacements y given by equation (10), Seco 53. areperiodic functions of l with period 2c/a.

2. Show that the motion of each cross section of the elastic bar treated in Seco 55 is

periodic in l with period 2c/a.

J. A string stretched between the points (0.0) and (n,O) is initially straight with velocityy,(x,O) = b sin x, where b is a constan!. Write the boundary value problem in y(x,t), solve it.

. and verify the solution:b

Ans. y = sin x sin al.a

40 A wire stretched between the points (0,0) and (1,0) of a horizontal x axis is initially atrest along that axis, its subsequent motion being caused by the force of gravity. 1f the y axis ispositive upwards, the equation of motion is y" = a2yu - g, where g is the acceleration due togravity. Write the boundary value problem in the vertical displacements y(x,l) and solve i!.Show that

1y(x,l) = q,(x) - i(<tI(x + al) + <tI(x- at))

where the function <tiis the odd periodic extension, with period 2, of the function

gq,(x) = 2X(X - 1)2a

5. Display graphically the periodic functions G(x) and H(x) in Seco53 whenall points ofthe string there have the same initial velocity g(x) = ¡'o. Then, using expression (8), Seco 53,indicate some instantaneous :->ositionsof the slring by means of bold line segments, which willbe similar lo those in Fig. 16.

6. VerifyIhe solution (8),Seco53.Show thal the function H definedby equation (7)in thatsection is continuous, even, and periodic and Ihal H'(x) = G(x) wherever G is conlinuous.

(O~x ~ 1).

...-.......

7. In Seco 53 we noted thatl' n1ts

G(s) = L: b. sin-n=I e

(-00 < s < 00).

Integrate that series (Sec. 48) to show that

e " 1

(nllx

)H(x)=- L: -b. I-cos--7t n= In e

(-oo<x<oo)

»

and hence that the function (8) in Seco 53 is represented by the series (6) there.

8. Reduce the representation (7). Seco55. ofthe function y(x.e) to the form (8) in that sectionand verify the solution in that formo

9. From expression (8), Seco 55, show that y(O,e) = P(ae) and hence that the end x = ° ofthe bar moveswithconstantvelocityabduringthehalfperiod° < e< e/aand withvelocity- ab during the next half periodo

10. Let the function y(x,e) satisfying conditions (1) through (3). Seco 55, be interpreted asrepresenting transverse displacements in a stretched string, where a2 = H/o. Describe the initialand end conditions for the string. [Note that conditions (2), Seco 55, imply that no vertical forceacts at the ends, as if the ends were looped about perfectlysmooth rods along the lines x =°andx = e.]

11. The end x = ° of an elastic bar is-free, and a constant longitud inal force F o per unítarea is applied at the end x = e. The bar is initially unstrained and at rest. Set up the boundaryvalue problem for the longitud inal displacements y(x,e), the condition at the end x = e beingy.(e,e)= Fo/E (Sec. 5).

(a) After noting that the method of separation of variables cannot be applied directly tofind y(x,e), make the substitution

y(x,e) = Y(x,e) + .p(x) + I/I(e)

.,

in the above boundary value problem and determine .p(x) and I/I(e)so that Y satisfies the wave

equation and the boundary conditions

Y.(O,e)= O. YAe,e)= 0,

Y(x,O)= -.p(x), Y,(x,O)= O.Then show that

y(x,e) = ~o (X2 + a2(2) - 1 [<t>(x + ae) + <t>(x- ae)]2eE 2

where <t>(;x-)is the even periodic extension, with period 2e, of the function

Fo.p(:<)= - :<2.2cE

(b) Use the result in part (a) to show that the end x = ° of the bar remains at rest untiltime e = e/a and then moves with velocity Vo = 2aFo/E when e/a < e < 3e/a, with velocity 2vo

when 3e/a < e < Se/a, and so on.

56. Temperatures in a Bar

The lateral surface of a solid right cylinder of length 1tis insulated. The initialtemperature distribution within the bar is a prescribed function f of thedistance x from the end x = O.At the instant t = Othe temperature of both

7/////////////////////////////////////////////////////////////////////,

, -01 .', Oi=MI" -ooW//###7//$/##///////#//#//////////.IW/////////////////////kX=71

xFigure 18

ends X = Oand X = rris brought to zero and kept at that value (Fig. 18).Ifno heat is generated in the solid, the temperatures should be the values of afunction u(x,t) that satisfies the heat equation

(1) u,(x,t) = kUxAx,t) (O< X< rr, t > O)


u(O+,t)=O, u(rr-,t)=0 (t>O),

u(x,O+) = f(x) (O< x < rr).

The constant k is the thermal diffusivityofthe material (Sec. 6).We havewritten the boundary conditions as limits in order to indicate continuityproperties that should be satisfiedby u.The problem is also that ofdetermin-ing temperatures u(x,t) in a slab bounded by the planes x = O and x = rrand initialIy at temperature f(x), with its faces kept at temperature zero.

By separation of variables we find that X(x)T(t) satisfies the homoge-neous equations (1) and (2) if

(4) X"(x) + AX(X) = O, X(O)= O,

(5) T(t) + AkT(t) = O.

The Sturm-Liouville problem (4) has eigenvaluesA = n2and eigenfunctionsX = sin nx (n = 1,2,.. .).The corresponding functions fromequation (5)areT = exp (- n2kt). FormalIy, then, the function

(2)

(3)

X(n) = O,

(6)ex>

u(x,t) = Lb" exp (- n2kt) sin nx"=1

satisfies all equations in the problem, inc1uding(3), if

(7)ex>

f(x) = Lb" sin nx"=1

(O< x < rr).

We assume thatfandf' are piecewisecontinuous. Thenfis represented byits Fourier sine series (7),where

2 "b" = - f f(x) sin nx dx

n o

Expression (6),with coefficientsb"definedby equation (8),is our formalsolution of the boundary value problem.

(8) (n = 1,2, oo.).

{SECo 56] BOUNDARY VALUE PROBLEMS 129 /

t

(a) Verification. According to Seco27, b. -> O as n -> oo. Hence thosecoefficients are bounded for all n; that is, lb. I < B where B is some posi-tive constant. So whenever t :2:to, where to is a positive constant,

lb. exp (-n2kt) sin nx 1< B exp (-n2kto).

Since the infinite series with constant terms exp (- n2kto) converges, accord-ing to the ratio test, the Weierstrass M-test ensures the uniform convergenceof series (6) with respect to x and t when O :$ x S n, t :2:to > O.The terms ofthat series are continuous functions; hence series (6) converges to a contin-uous function u(x,t) when t :2:to, or whenever t > O since to is an arbitrarypositive number. In particular, u(O+,t) = u(O,t) when t > O; and sinceu(O,t) = O, the first of conditions (2) is satisfied by the function u defined byequation (6). Similarly, the second of those conditions is satisfied.

The series with terms n exp (- n2kto), or .n2 exp (- n2kto), also con-verges. Hence series (6) can be differentiated twice with respect to x and oncewith respect to t, when t > O, because the series of derivatives convergeuniformly when t :2:to. But the terms of series (6) satisfy the heat equation(1), and so the sum u(x,t) of the series satisfies that linear homogeneous differ-ential equation.

It remains to show that u satisfies the initial condition (3). This can bedone as follows, with the aid of a convergence test due to Abelt and to bederived in Chap. 10. For each fixed x (O< x < n) the series with termsb. sin nx converges tof(x); at a point of discontinuity we definef(x) as themean ofthe valuesf(x+ ) andf(x- ). According to Abel's test, the new seriesformed by multiplying the terms of a convergent series by correspondingterms of a bounded sequence of functions of t whose values never increasewith n, such as e~p (-n2kt), is uniformly convergent with respcct to t. Ourseries (6) therefore converges uniformly with respect to t when t :2:O.t

It followsthat our function u is continuous in t (t :2:O);thus u(x,O+) =u(x,O).Since u(x,O)= f(x), condition (3) is satisfied.The function u definedby equation (6) is now verifiedas a solution.

(b) Uniqueness.Consider the somewhat special case in whichfis con-tinuous (OS x ~ n),j(O) = f(n) = O,andf' is piecewisecontinuous, so thatthe Fourier sine series (7) converges uniformly to f (x). Then Abel's testshows uniform convergence of series (6) with respect to x and t together inthe region Os x S n, t :2:O.ConsequentIy, OUTfunction u is continuous inthat region. As before, u" ux, and Uxxare continuous when t > O andOs x S n.

A simple uniqueness theorem for solutions of such boundary valueproblems is proved in Chap. 10.It shows that not more than one functionU(x,y,z,t) can satisfy the heat equation throughout the interior of a c10sedbounded region R in space, such as our cylindrical region, and meet the

¡

t Niels H. Abel (pronounced AIÍ-bel), Norwegian, 1802-1829.

following requirements: V is to be a continuous function of x, y, z, and ttogether when t ¿ Oand when (x,y,z) is in R, where R incIudesits boundarysurface; the derivatives of V that appear in the heat equation are continuousin R when t > O; V is prescribed on part of the boundary and the normalderivative of V is prescribed on the rest when t > O.

Our function u(x,t), which is independent of y and z, satisfies all thoserequiremen~s, incIuding conditions (1) through (3). For each fixed t thegradient of u is paraIlel to the x axis; so the normal derivative of u at thelateral surface of the cylinder,or the fluxof heat across that surface,is zero.Hence expression (6) represents the solution of our problem that satisfiesthe continuity conditions stated. We have now justified the assumption thatu is a function of x and t only.

57. Other Boundary Conditions

We point out modifications of the procedure that are useful when the slabO:5:x :5:n, or the bar with insulated lateral surface, is subjected to othersimple thermal conditions at its boundary surfaces x = Oand x = n. In eachcase the temperature function is to satisfythe heat equation in x and t withinthe slab:

(1) u,(x,t) = kuxx(x,t) (O< x < n, t > O).

(a) One Face at TemperatureUo.If the slab is initially at temperaturezero throughout and the facex = Ois kept at that temperature whilethe facex = n is kept at a constant temperature Uowhen t > O,then

(2) I u(O,t)= O, .) u(n,t)= 110, u(x,O)= O.. I

The boundary value problem consistingof equations (1)and (2) isnot inproper form for the method of separation of variables because one of thetwo-point boundary conditions is not homogeneous. If we write (compareSeco54)

u(x,t) = V(x,t) + 4>(x),

however, those equations become

V,(x,t) = kVxAx,t) + k4>"(x)

and

V(O,t) + 4>(0)= O, U(n,t) + 4>(n)= uo, V(x,O) + 4>(x) = o.We thus find that if 4>(x)= uox/n, the function V satisfies the conditions

(3) V, = kVxx' V(O,t)= V(n,t)=0, V(x,O)= -4>(~.

This is a special case ofthe problem solved in Seco 56, wheref(x) = - Uox/no

Hence we obtain the result

(4) u(x,t)= ~~[X + 2 f (-I)n exp (-n2kt) sin nx

J

.1t n=1 n

Note that the function 4>(x)represents the steady-state temperatures in theslab.

(b) lnsulated Faces. The flux of heat across the faces x = Oand x = 1t is

the value of - Kux at those faces, where K is the thermal conductivity(Sec.6).If both facesofthe slabare insulatedand the initialtemperatureisf (x), the boundary value problem consistsof equation (1)and the conditions

(5) ux(O,t) = O, uA1t,t) = O, u(x,O) = f(x) (O< x < 1t).

Separation of variables leads to the Sturm-Liouville problem associatedwith the Fourier cosine series,

X"(x) + lX(x) = O, X'(O)= O, X'(1t)= O,

and to the equation T'(t) + lkT(t) = O~The temperature function is thenfound to be

(6)1 co

u(x,t) = 2ao + n~lan exp (-n2kt) cos nx

2 nan = - f f (x) cos nx dx

1t o. (n = 0,1,2, .,,).where

(c) One Face lnsulated. If the face x = 1t is insulated while the facex = Ois kept at temperature zero (Fig. 19),then

(7) u(O,t)= O, uA1t,t) = O, u(x,O) = f(x) (O< x < 1t),

wheref(x) is the initial temperature. Upon separating variables, we obtainthe Sturm-LiouvilIe problem

X"(x) + lX(x) = O, X(O) = O, X'(1t) = O

o

~'-"''''-''''''''''''-'1~ I~ I~ 1~I/(X, O) 11/ = O~ =f(2n - x):~ I~ 1~ -4~ I~ I~ 1~ 1~ 1~ I!'t J

x

x=1t x = 2n Figure 19

and the equation T'(t) + AkT(t) = O.The eigenvalues and normalized eigen-functions of that problem are found to be

(8) An = (2n - 1)24 ' rPn(X)= ~ sin (2n - l)xV~ 2

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

Note that the orthogonal sine functions here are not those used in theFourier sine series for the interval (0,7t).

Qur generalized linear combination of functions of type XT, written interms of the functions rPn,is

( ) ~l

(2n - 1fkt

Jft. (2n - I)x

u x,t = t Cnexp - -SIO--.n=1 4 7t 2

It satisfies the condition ~(x,O)= f(x) if

(9)

(10) f(x) = f Cn ~sin (2n - l)xn=1 V ~ 2

(O< x < 7t).

Series (10) is the generalized Fourier series corresponding tofwith respect tothe eigenfunctions rPn if

(11) I"

ft r" (2n - l)x

Cn= f(x)rPn(x)dx = - f(x) sin - dx.o 7t.o

Ir we assume that the representation (10) is va lid, as proved in references toSturm-Liouville theory cited in Seco 31, expression (9) with coefficients (11)represents a formal solution of our problem. .

But we can actually establish the special representation (10)with the aidof Fourier sine series on the larger interval (0,27t).Let the given functionfand its derivativef' be oit.;ewisecontinuous on the interval (0,7t).Then letfbe defined on (7t,27t)so that its graph is symmetric with respect to the linex = 7t; that is, write

(12) f(x) = f(27t - x) when 7t< x < 27t.

This procedure is suggested by considering the temperature problem in theslab extended to the plane x = 27twith that plane kept at temperature zero(Fig. 19). When the initial temperature distribution has the symmetryproperty (12), no heat should flow across the midsection x = 7t. So thetemperature function u should be the one sought when OS;x S;7t.The newproblem is essentially the one solved in Seco56 except that now the sineseries on the interval (0,27t),for the symmetric extension off, is to be used.

That sine series representation on (0,27t)is

(13)00

f(x) = Lbn sin nxn= 1 2

(O< x < 27t)

( 1f2!r nx

where b. =; o f(x) sin 2 dx

1f

n nx 1r

2n ns= - f(x) sin- dx + - f(2x - s)sin- ds.

Xo 2 X'n 2

In the last integral we substitute x = 2x - s and note that sin (ns/2) thenbecomes - (-1). sin (nx/2). Thus

1f

n nx

b.=;[1-(-l}"] of(x)sin2dx;that is,

2f

n . (2n- l)x(14) b2.-1=-;¡ of(x)sm 2 dx, b2.=O (n=1,2,...).

The representation (13) with coefficients (14) is therefore valid for the givenfunctionf on the interval (O,x). It is the same as the representation (10) withcoefficien ts (11).

PROBLEMSt

1. The faces x = O and x = e of a slab which is initially at temperaluresf(x) are kept attemperature lero (compare Seco56). Derive the temperature formula

2 '"

(n27t2kt

)n7tx' n7ts

u(x,t) = - L exp - sin - f f(s) sin-ds.e ,,-1 e e o e

2. Let the initial temperature distribution be uniform over the slab in Problem 1, so that

f(x) = Uo. Write the formula for u(x,t) and find the flux - Ku.(xo,t) across a plane x = Xo(O :s; Xo :s; e, t > O). Show that no heat fIows across the center plane x = e/2.

3. Suppose thatf(x) = sin (7tx/e) in Problem 1. Find u(x,t) and verify the result fully.

(7t2kt

). 7tX

Ans. u(x,t) = exp -7 SIO-;.

4. (a) Show that if A is a constant and f(x) = A when O < x < e/2 and f(x) = O whenc/2 < x < e, the temperature formula in Problem 1 becomes

( ) 4A ~ sin2 (n7t/4) (n27t2kt

). n7tx

u x,t = - ~ exp - SIO-.7t n=I n e e

(b) Two slabs o(¡ron (k = 0.15 centimeter-gram-second unit), each 20 cm thick, are suchthat one is at temperature 100°C and the other at O°C throughout. They are placed face to facein perfect contact, and their outer faces are kept at O°c. Compute the temperature at theircommon face 10 min after contact has been made to show that it is then approximately 36°C.

t Only formal solutions of the boundary value problems here and in the sets of problems tofollow are expected. unless the problem states that the solution found is to be fully verified.Partial verification is often easy and helpful.

134 FOURIER SERIES AND ROUNDARY VALUE PRORLEMS [SECo 57

(e) Show that ifthe slabs in part (b) are made of concrete (k = 0.005 cgs unít), it takes 5 hr

for the common face to reach that temperature of 36°C. [Note that U(X,I)depends on theproduct kl.)

S. The facesx = Oand x =e of a slab which is initially at temperaturesf(x) are insulated[compare Seco57(b)).Derive the temperature formula

I '

(n21t2kl

)n1tx

U(X,I)= - aQ+ L:a. exp - r- cos-2 .- 1 e e

where2 .' n1tX

a.=-I f(x)cos-cIxe.o e(n = 0,1,2, ...).

Show how ít follows that II(X.I)= 110whenf(x) = 110.where "0 is a constan!.

6. Verify the result in Problem 5 [compare Seco 56(a)).

. 7. Let V(X,I) denote temperatures in a slender wire Iying along the x axis. Variations ofthetemperature over each cross section are to be neglected. At the lateral surface the linear law ofsurface heat transTer belween the wire and its surroundings is assumed to apply (see Problem 7,Seco 7). Let the surroundings be at temperature zero: then

V,(X,I) = kvx,(x,l) -lw(x,l) (h constant, h > O).

Ot t t

~I I~-O + + + x

0° x = e Figure 20

The ends x = Oand x = e of the wire are insulated (Fig. 20) and the initial temperature distribu-tion isf(x). Solve the boundary value problem for v and then show that

V(X,I) = e-.'u(x,l)

where u is the temperature function found in Problem 5.

8. Use the substitution V(X,I) = e-.'u(x,l) to reduce the boundary value problem in Prob-lelO 7 to the one in Problem 5.

9. Assuming that the ends of the wire in Problem 7 are not insulated, but kept at tempera-ture zero instead, find the temperature function.

10. Heat is generated at a constant rate uniformly throughout a slab which is initially at

temperaturesf(x) and whose faces x = Oand x = 1tare kept at temperature zero. Then the heatequation has the form (Sec. 7)

U,(X,I) = kll,,(X,I) + C

where C is a positive constan!. Solve the boundary value problem to obtain the temperatureformula

Cx(1t-x) 00 .u(x,t) = + Lb. exp (-n2kl) Sin nx

2k .=1

2 f'

[

CX(X-1t)

]b, = - + f(x) sin nx e/x.

1t.o 2k

In particular, find u(x,t) whenf(x) = Cx(1t- x)j(2k).

where

11. Show that whenf(x) = O (O < x < 71)in Problem 10, the temperature formula therecan be written

4C '" 1 - exp[-(2n - 1)2kt].u(x,t)= - ¿ (2 " Sin (2n - I)x.tnk 0=1 n - 1

Suggestion: Note that whenf(x) = O (O < x < 71), the constants bo in Problem 10 are the

coefficients in the Fourier sine series representation of Cx(x - n)/(2k)on the interval (0.71). That

representalion is readily oblained by combining the series found in Problem 1, Seco38, and

Problem 4, Seco 44. ~ ~ ~ -f)) I'4:.. Solve this boundary value problem ror u(x,t): l ¡

u,(x,t) = u..(x,t) - hu(x,t) (O < x < 71, t > O; h constant, h > O).

u(O.t)= O, u(n,t) = 1 (t > O),

u(x,O)= O (O< x < 71).

sinh (xjh) 2 x (-I)On .

AliS. u(x,t)= jh +-exp(-ht)¿ ~ h exP(-II2t)slnnx.sinh(n h) 71 0-1 n +

13. Use 'superposition of known solutions (Secs. 56 and 57) to write a formula for temper-alures in a slab whose faces x = O and x = 71 are kept at temperatures O and uo' respectively, and

whose initial temperature distribution isf(x).

14. The race x = O of a slab is kept at lemperalure rero while heat is supplied or extractedat the olher face x = 71al a constant rate per unit area so that u.(n,t) = A. The slab is initially atlemperature zero and the unit of time is chosen so that we can write k = 1. Derive Ihe tempera-ture formula

( ) 8A ~ (-1)0[

(2n - 1)2tJ

. (2n - I)xuxt =Ax+- t... -exp - Sln-

, 71 0=.(2n-1)2 4 2

with the aid of the series representalion (10),Seco57.IS. Let f and f' be piecewise conlinuous on an inlerval (O,e).By Ihe process used in

Seco57(e),establish the representation

( ) 2 ~ . (2n - I)nxr

'( )

. (2n- l)nsdf x = - t... SIO f S SIO Se 0=1 2e. o 2e

(O< x < e)

offin lerms of the eigenrunctions of the Slurm-Liouville problem

X"(x) + U(x) = O, X(O)=O, X'(e) =0.

58. Dirichlet Problems

As already noted in Seco10,a boundary value problem in u is said to be aDirichlet problem when it consists of Laplace's equation V2u = O,whichstates that u is harmonic in the interior of a given region, together with

t This result oecurs, for example, in Ihe theory of gluing of wood with the aid of radio-frequency heating. See G. H. Brown, Proc.lnst. Radio Engrs., vol. 31, no. 10.pp. 537-548,1943, .where operational methods are used.

>'1 u = O (a. b)

,.{~]_o01 u=f(x) x

,Figure 21

prescribed values of Uon the boundaries. We now iIIustrate the use of themethod of separation of variables in solving such problems for regions in theplane.

(a) Cartesiancoordinates.Let u be harmonic in the interior of a rectan-gular region O~ x ~ a, O~ Y ~ b, so that

(1) uxx(x,y)+ Uyy(x,y)= O (O< x < a, O< Y < b),

with these values prescribed on the boundary (Fig. 21):

(2)

(3)

u(O,y) = O,

u(x,O) = f(x),

u(a,y)= O

u(x,b)= O

(O< y < b),

(O< x < a).

Separation of variables leads to the Sturm-Liouville problem

(4) X"(x) + lX(x) = O, X(O)= O, X(a) = O,

where l = n2Tt2/a2and)( = sin (nTtx/a)(n = 1,2, ...), and to the conditions

Y"(y) - (n:) 2Y(y) = O,

Thus Y can be written Y = sinh [nTt(b- y)/a], and the function

(5) Y(b) = O.

(6) u(x,y) = I B. sinh nTt(b- y) sin nTtx.= 1 a a

formally satisfies all the equations (1) through (3) if

(7)00

f(x) = LB. sinh nTtbsin nTtx.= 1 a a

(O< x < a).

We assume that f and f' are piecewise continuous. Then series (7) is theFourier sine series representation of f on the interval (O,a) providedB. sinh (nTtb/a)= b. where

2J

G . nTtXb. = - f(x) Sin-dx.

a o a

The function defined by equation (6) with coefficients

(8) B2 ,0

" =I

f(nnx

a sinh (nnb/a).o x) sin-;-dx

is therefore our formal solution. The result can be verified by the methodused in Seco56. We defer the verification, however, till Chap. 10, whereuniqueness is also considered.

Ir y is replaced by the new variable b - Y in the above problem andsolution, and if f(x) is replaced by g(x), the nonhomogeneous conditionsatisfied by u becomes u(x,b) = g(x). An interchange of x and y places thenonhomogeneous conditions at the boundaries x = O or x = a. Superposi-tion of the four solutions then gives the harmonic function whose valuesareprescribed as functions of position along the entire boundary of therectangle.

From the equations (1) through (3) we note that u(x,y) represents thesteady-state temperaturesin a rectangular plate with insulated faces whenu = f(x) on the edge y = Oand u = Oon the other three edges.The functionu also represents the electrostatic potential in the space bounded by theplanes x = O,x = a, y = O,and y = b when the space is free of charges andthe planar boundaries are kept at potentials given by conditions (2)and (3).Irf is continuous and f(O) = f(a) = O,u(x,y) represents the static transversedisplacementsin a membranestretched over a rectangular frame after thesupporting side along y = O is displaced according to the equationII(X,O)=/(x).

(b) Cylindrical coordinates.The same Sturm-Liouville problem canarise in seemingly unrelated boundary value problems. Problem (4) above,for example, first arose in finding transverse displacements of a string(Sec. 15);and a special case of it arose in finding time-dependent tempera-tures in a bar (Sec.56). As we shall now see, the same Sturm-Liouvilleproblem can arise even when a different coordinate system is used.

Let u(p,</J)denote a function of the cylindrical coordinates p and </J

which is harmonic in the region 1 < p < b, O< </J< 7t of the plane z = O(Fig. 22). That is (Sec.9),

(9) p2Upp(p,</J)+ pup(p,</J) + u¡p¡p(p,</J)= O (1 < p < b, O < </J< n).

u =Uo

u=o Figure 22

CUI I Vn.I,",L "UM. '"' .

Suppose further that

(10)

(11)

u(p,O) = .o,

u(l,4»= O,

u(p,n)= O

u(b,4» = Uo

(l<p<b),

(O < 4>< n),

where Uo is a constant.Substituting u = R(p )<1>(4))into the homogeneous equations and sepa-

rating variables, we find that

(12) p2R"(p) + pR'(p) - 2R(p) = O, R(l) = O

and

(13) <1>''(4)) + 2<1>(4)) = O, <1>(0)= O, (n)= O.

Evidently, then, 2 = n2 and <1>= sin n4>(n = 1,2, .. .). The difTerentialequa-tion in R is a Cauchy-Euler equation [see Problem 13(a),Seco34], and itfollowsfrom equations (12) that R = pn - p-n when 2 = n2.Thus, formally,

00

u(p,4»= L Bn(pn - p-n) sin n4>n=1

where, according to the second of conditions (11), the constants Bn are suchthat

00

Uo = L Bn(bn - b-n) sin n4>n=1

(O < 4>< n).

It readily follows from the Fourier sine series expansion [see Problem9(b), Seco 38]

Uo= 4uo I sin (2n - 1)4>n n= 1 2n - 1

(O< 4> < n)

that

B2nWn - b-2n) = O (n = 1,2, ...)and

B2n-l[b2n-1- b-(2n-l)]= 4uo~n 2n - 1 (n = 1,2, ...).

The solution of this Dirichlet problem is therefore

_4uo 00

[

p2n-l_p-(2n-l)

]

sin(2n-1)4>u(p,4»- L b2n-1 b-(2n-1) - . .n n= 1 -(14)

SECo59] BOUNDARY VALUE PROBLEMS 139

59. Fourier Series in Two Variables

Let z(x,y,t) denote the transverse displacement at each point (x,y) at time t ina membrane stretched across a rigid square frame in the xy planeo To sim-plify the notation, we select the origin and the point (n,n) as ends of adiagonal of the frameo Ir the membrane is released at rest with a given initialdisplacement f(x,y) which is continuous and vanishes on the boundary ofthe square, then

(1)

(2)

(3)

Ztt= a2(zxx + Zyy) (O< x < n, O< Y < n, t > O),

z(O,y,t)= z(n,y,t) = z(x,O,t)= z(x,n,t) = O,

z(x,y,O)= f(x,y), z,(x,y,O)= O (OS;x S; n, OS;Y S;n)o

Let us assume that the partial derivatives fAx,y) and fy(x,y) are alsocontinuous.

Functions of the type X(x)Y(y)T(t) satisfyequation (1) if

X"(x) T'(t) Y"(y)X(x) = a2T(t)- Y(y)'

Thus X"/X = -..1..and Y"/Y = T"/(a2T) + A.,where l is a constant; alsoY"/Y = - Jl and T'/(a2T) + l = - Jl,where Jl is a constant independent of l.Separation ofvariables in the homogeneous equations therefore leads to thetwo Sturm-Liouville problems

X"(x) + lX(x) = O,

Y"(y) + JlY(Y)= O,

X(O)= O,

Y(O)= O,

X(n) = O,

Y(n) = O

and to the conditions

T'(t) + a2(l + Jl)T(t) = O, T(O) = 00

Consequentlyl = m2, X = sin mx m = 1, 2, .o.)and Jl = n2,y = sin ny(n = 1,2, o o o); also, T = cos (at m2+ n ).

The formal solution of our problem is therefore

(4)00 00

z(x,y,t) = L L Bmn sin mx sin ny cos (atJm2 +112)m=1 n=1

where the coefficients Bmnneed to be determined so that

(5)00 00

f(x,y) = L L Bmn sin mx sin nym= 1 n= 1

when OS;x S;n and OS;Y S;noBy grouping terms in this iterated sine series

1""""" "" ,""'.""'. ..".


so as to display the total coefficient of sin mx for each m, we can write,formally,

f(x,y) = mt (Jl Bmnsin ny) sin mx.

For each fixed y (O~ Y ~ 1t) equation (6) is the Fourier sine seriesrepresentation of the functionf(x,y) of the variable x (O~ x ~ 1t),providedthat

(6)

co 2 n

L Bmn sin ny = - J f(x,y) sin mx dxn=1 1t o

The right-hand side here is a sequence offunctions Fm(Y)(m = 1,2,. o.), eachrepresented by its Fourier sine series (7) on the interval (0,1t) when

(7) (m = 1, 2, 0")0

2 nBmn=- J Fm(y)sinnydy

1t o

The coefficients Bmntherefore have the values

(n = 1, 2, o..).

o 4 n nBmn =""2 J sin ny r f(x,y) sin mx dx dy.1t o '0

A formal solution of our membrane problem is given by equation (4)with the coefficientsdefined by equation (8).

Since the numbers .Jm2 + n2 do not change by integral multiples ofsome fixed number as m and n vary through integral values, the cosinefunctions in equation (4) have no common period in the variable t; so thedisplacement z is not generally a periodic function of t. Consequently thevibrating membrane, in contrast to the vibrating string, generally does notproduce a musical note. It can be made to do so, however, by givingit theproper initial displacement. Ir, for example,

z(x,y,O)= A sin x sin y

where A is a constant, the displacements (4) are given by a single term

z(x,y,t) = A sin x sin y cos (aJ2 t);

then z is periodic in t with period 1tJ2/a.

(8)

PROBLEMS

l. One edge of a square plate with insulated faces is kept at a uniform temperature "o. andthe other three edges are kept at temperature zero. Without solving a boundary value problem,but by superposition of solutions of like problems to obtain the trivial case in which all fouredges are at temperature "o. show why the steady lemperaiUre at Ihe cenler of.lhe given platemust be "0/4.

CUI I '-Inl""L ~, "'" .

2. A square plate has its faces and its edges x =Oand x =1t (O < Y < 1t) insulated. lisedges y = O and y = 1t (O < X < 1t) are kept at temperatures zero and f(x), respectively. Letu(x,y) denote its steady temperatures and derive the expression

1 '" sinh nyu(x,y)= -aoy + L a'--:-h

cos nx21t .=1 Sin n1t

where2 .

a.= - r f(x) cos nx dx1t '0

(n = 0,1,2, ...).

Find u(x,y) whenf(x) = uo, where Uois a constan!.

3. Find the harmonic function u(x,y) in the semi-infinite strip O < x < 1t,Y > O such that

u(O,y) = u(1t,y) = O

u(x,O)= 1

(y> O),

(O<x<1t),

and Iu(x,y) I < M where M is some constan!. (This boundedness requirement serves as aeondition at the missing upper boundary of the strip.)

4 00 sin (2n - l)xAns. u(x,y)=- L exp[-(2n-l)y] 2 .1t .= 1 n - 1

4. Given that the imaginary part of log (a + ib) is tan-I (b/o) when o and b are realnumbers and that

1 + z '" 1log-=2L -Z2.-1

1-z .=12n-1

write z = e-'e;X(y > O)and equate imaginary parts in the above expansion to show that theanswer to Problem 3 can be written

(lzl < 1),t

2 _,(SinX

)u(x,y)= ;- tan sinhy(O < x < 1t, Y > O).

Verify the answer in this formo

S. Derive an expression for the bounded steady temperatures u(x,y) in a semi-infinite wallO < x < e, y> O whose planar faces x = O and x = e are insulated and where u(x,O) = f(x).

Assume thatfandf' are pieeewise eontinuous on (O,e).

6. Let p, </>,z be eylindrical coordinates. Find the harmonie function u(p,</»in the region1 < 1t/2 of the plane z = O when u = O and u =f(</» on the ares p = 1,0< 4>< 1t/2 and p = b, 0<4> < 1t/2, respeetively, and u~(p,O) = u~(p,1t/2) = O (1 

og b .= 1

4 .,200 = - r f(</» d</>

1t.o

4 012

O. =- r f(4)) cos 2n</>d</>1t.o

(n = 1,2, ...).and

7. Let the faces of a plate in the shape of a wedge O :5 P :5 o, O :5 </>:5 a: be insulated. Find

the steady temperatures u(p,</» in the plate when u = O on the two rays 4> = O,</>= a: (O < p < o)

t See, for example, Churchill, Brown, and Verhey (1974), listed in the Bibliography.

142 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS [SEC.60

and u =f(q,) on the are p = a, O < q, < ce.Assume thatfandf' are piecewise continuous andthat u is bounded.

2 '"

(P

)"" nxq,' nxO

Ans. u(p,q,)=- L - sin- r f(O) sin-dO.IX,,-1 a IX '0 ex

8. AIl four faces of an infinilely long rectangular prism bounded by the planes x = O,x = a, y = O, and y = b are kept at temperature zero. Let the initial temperature distribution bef(x,y) and derive this expression for the temperatures u(x,y,e) in the prism:

00 00

[ (m2 n2

)J

mxx nx)'u(x,y,e)= L L Bm'exp -x2ke 2" + -b2 sin- sin-bM-l,.-1 a a

where 4 b nxy' mxxBm'= -b r sin -b r f(x,y) sin - dx d)'.

a '0 '0 a

9. Writef(x,y) = g(x)h(y) in Problem 8 and show that the double series obtained there foru reduces to the product

u(x,y,e) = v(x,e)w(y,e)

of two series where vand w represent temperatures in the slabs O S x S a and O S )' S b withtheir faces at temperature zero and with initial temperatures g(x) and h()'), respectively.

10. Let the functions v(x,t) and w(y,t) satisfy the heat equation for one-dimensional now:

v,= kv.., w,= kw"..Showbydifferentiationthat their product 11=1"Wsatisfies the heat equa-lion u, = k(uu + u".). Use this fact lO arrive althe solution in Problem 9.

60. Observations and Further Examples

Not all linear homogeneous partial difTerential equations yield to themethod of separation of variables.Conditions for the existenceof solutionsV = X(x)Y(y) of the equation

V",,(x,y) + (x + y)2Yyy(X,y) = O,

for instance, do not lead to difTerential equations ofthe Sturm-Liouville typein either X or Y.

Although the separation process does apply to Laplace's equation inrectangular, cylindrical, or spherical coordinates, to certain wave equationsand heat equations involving the laplacian operator in those coordinates,and to many other equations, the general test consists of trying out theprocess on each equation.

(a) Periodic Boundary Conditions. The followingexample willillustrateproblems that involve expansions in the basic Fourier series c~ntaining bothsine and cosine terms.

Let u(p,4» denote the steady temperature~in a long solid cylinderp 5 1,- 00 < Z < 00 when the temperature of the surface p = l is a given function

1(4)); the variables p, 4>and Z are, of course, cylindrical coordinates. Then

(1) p2Upp(p,4» + pUp(p,4» + u.p.p(p,4» = O (p < 1, - 7t < 4> 57t),

CUII VI'\I"'L "''''''' 1'" .

(2) u(l,cf» = f(cf» ( - n < cf>< n)o

Also, u and its partial derivatives of the first order are to be continuousinterior to the cylindero

Iffunctions ofthe type R(p )(cf»are to satisfy equation (1) and the con-tinuity requirements, we and that

(3)

(4)

pZR"(p) + pR'(p) - AR(p) = O (O:s;p < 1),

"(cf»+ A(cf»= O, <1>(-n) = (n), <1>'(-n) = '(n).

The eigenvalues of problem (4), whose boundary conditions are of peri-odie type,are the numbersA= Oand A= nZ(n= 1,2, o. .); and the set

{1, cos x, cos 2x, .. o, sin x, sin 2x, .. o}

is the set of corresponding eigenfunctions, orthogonal on the interval(-n,n). (See Seco 330)Equation (3) is ofthe Cauchy-Euler typeoWhen A= O,R = A 10gp + B; and when A= nZ (n = 1, 2, o. o), the general solution isR "" CI pn + Czp-'. Ir R is to be continuous on the axis p = O, thenA = CZ = O.

The generalized linear combination of our continuous functions Ristherefore

(5)1 00

u(p,cf» = "2ao + .~/n(a. cos ncf> + b. sin ncf»o

This satisfies boundary condition (2) if a. and b., including ao, are thecoefficients in the Fourier series forf on the interval (-n,n):

(6)J n

Q. = - r f(cf» cos ncf>dcf>,n'-n

1 n

b. = - r f(cf» sin ncf>dcf>on'-n

We assume thatfsatisfies the conditions in our Fourier theorem (Seco41)0(b) Sturm-LiouvilleSeriesoEven when separation of variables applies,

the superposition process may lead to a generalizedFourier series represen-tation of a given functiono Our formal procedure may then include anassumption that the representation is valido

To illustrate, let u(x,t) denote temperatures in a slab, bounded by twoplanesx = O,x = 1and initially at temperatures f(x), when u = Oon the facex = Oand surface heat transfer takes place at the face x = 1 into a mediumat temperature zero (Fig. 23)0The rate of heat transfer per unit area intothat medium is assumed to be proportional to u(1,t) - O,so that uA1,t) =-/11I(1,t) where h is a positive constant (Seco7)0Ir the thermal diffusivity k isunity,. then

(7)

(8) u(O,t)= O,

ut(x,t) = uxAx,t)

uA1,t) = -hu(1,t),

(O< x < 1, t > O),

u(x,O) = f(x)o

EDITORIALAGATA.

u = 01 U(X,O) = ¡(X) O.

o X.

X = I Figure 23

Ir we let functions of the type X(x)T(t) satisfy the homogeneous equa-tions here, we have the Sturm-Liouville problem

X"(x) + lX(x) = O, X(O)= O, hX(l) + X'(l) = O,

along with the condition T'(t) + lT(t) = O.The full set of eigenvalues,.ob-tained in Problem 5,S~c.34,is l. = IX;(n = 1,2, ...) where IX.are the consec-utive positive roots of the equation tan IX= -IX/h; the normalized eigen-functions are fouhd to be

(9) l/J.(x)=(

2h

)

1/2

h + COS2IX. sin IX.X(n=1,2,...).

Hence the formal solution of our problem is

(10)oc

u(x,t)= Le. exp (-IX;t)l/J.(x),.=1

where, in order that u(x,O)= f(x) when O< x < 1,

(11). 1

(

2h

)1/2. 1

c. = I f(x)l/J.(x) dx = h 2 I f(x) sin IX.Xdx.'0. + cos IX. '0

We have assumed that our function f is represented by its generalizedFourier series with respect to the orthonormal eigenfunctions(9), so that

x

f(x) = L c.l/J.(x).=1(O< x < 1)

where c. = (f,l/J.). Representation theorems in Sturm-Liouville theory,proved in referencescited in Seco31, show that the series does converge tof(x) at each point x (O< x < 1)wherefis continuous, provided thatfandf'are piecewisecontinuous on the interval (0,1).

.-. JlI' .KIAAI M~."'I AA.

PROBLEMS

1. Show that when f(x) = 1 (O< x < 1) in the boundary value problem consisting ofequations (7) and (8) in Seco 60, the solution obtained there reduces to

~ 1-cosa. ( 2)

.u(x,t) = 2h L. (h 2' exp -a.t SIOa.x.11-1 a" + cos tX"

2. Solve the boundary value problem

u,(x,t) = kuxx(x,t)

u(-n,t) = u(n,t), ux(-n,t) = u.(n,t),

(-n<x<n,t>O),

u(x,O) =f(x).

The solution u(x,t) represents, for example, temperatures in an insulated wire oflength 2n whichis bent into a unit circ1e and which has a given temperature distribulion along it. For con ven-ience, Ihe wire is Ihoughl of as being cut al one poinl and laid on Ihe x axis between Ihe poinlsx = -n and x = n. The variable x then measures the distance along the wíre starting at Ihe

point x = -n, the points x = -n and x = 1tdenote the same point on the circ1e, and the firsltwo boundary conditions in the problem state that the temperalures and the lIux musl be thesame for each of those values of x. This problem was of considerable interesllo Fourier himself,and the wire has come to be known as Fourier's ring. .

Ans. u(x,t) = ~ao + f: a.e-""(a. cos nx + b. sin nx)2 .= 1

where1 .

a.= - r f(x) cos nx dx1t.-e

(n = 0,1,2, ...)

and1 "

b. = - r f(x) sin nx dx1t. -11:(n = 1,2, ...).

3. A string stretched between the poinls (0,0) and (1t,O)is released at rest from an inilialdisplacement y = f(x). lis molion is opposed by air resistance proportionallo the velocity at

each poin 1. Le! the unit of time be chosen so that Ihe equation of motion becomes

y,,(x,t) = yxx(x,t) - 2hy,(x,t),

where h is a constant. Assurning that O < h < 1, derive the expression

y(x,t) = e-h' i:b.(COS K.t + ~ sin K.t )sin nx,,,""1 Kn

where K. = Jn2 - h2,2 ."

b. = - I f(x) sin nx dx,1t.o

for the transverse displacements.

4. Show forrnally that the expression obtained in Seco 60 for the steady temperatures inthe cylinder p ::; 1, -:c < ; < x whose surface p = 1 is kept al lemperatures f(4)) can bewrilten

1 ." 1 - 2

u(p"f» = 2;;L /(0) p . dO(p < 1).

This is known as Poisson's integral formula. .Suggeslion: Afler replacing the variable of inlegration </>by O in expressions (6), Seco60,

substitute those expressions into equation (5) of that section. Then use the surnrnation formuladerived in Problern 5. Seco 30. wilh a and O Ihere replaced by p and O-</>.respeclively.

~nlTngllU .o.r.:.o.T.o.

S. Using cylindrical coordinates p, cf>,and z, let u(p,cf» denote steady temperatures in a

long hollow cylinder a ::; p ::; b. - 00 < z < 00 when the temperatures on the inner surface p = aaref (cf»and lemperature of the outer surface p =b is zero.

(a) Derive the temperature formula

I10g (b/p) "' (b/p). - (p/b). .

u(p,cf» = 210g (b/a) ao + .~IILI_\. UL\. (a. cos ncf> + b. sm ncf»

where the coefficients a. and b., including ao, are given by equations (6), Seco 60.

(b) Use the result in part (a) to show Ihat if, in particular,f(cf» = A + B sin cf>where A andB are constants, then

log (b/p) abB(b P

),

u(p,cf»=A-+ - --- smcf>.log (b/a) b2 - a2 p b

6. Find the harmonic function u(p,cf» in the region I < 11of the z planesuch Ihat u = f (p) on the part of the boundary where cf>= 11 and u = O on the rest of the

b?undary. [The Sturm-Liouville problem that arises is the one in Problem 13(b). Seco34.]"'

(Sinh CX.cf»

(n1l

)Ans. u(p,cf»= L C. --:-- sin (cx.logp) cx.=-.=, smh cx.11 logb

2 b 1

Jere C. = - r - f(p) sin (cx.log p) dp10gb ., p

: 7. Show that (a) whenf(p) = p in Problem 6,

"' n[1- b(-Ir](

Sinh cx.cf» ,

u(p,cf» = 211 L (1 W ( )2 --:--hsm (cx. log p)

.=1 og + n1l S1l1 CX.1I

..

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

(cx. = 10:1Ib);

(b) when Uo is a constant andf(p) = Uo in that problem,

4U"'

('h P

u(P.cf»=~ L ~ )Sin (P. logp)

11 .. 1 smh P.1I 2n - 1where P. = (2n - 1)11

log b '

Suggestioll: Make the substitution s = log p in the integrals needed lO find Ihe coeffi-cients C.'

~A bounded harmonic function u(x,y) in the semi-infinile strip x> O, O < Y < 1 is tosatisfy the boundary conditions

u(O,y) = Uo-~

"(O<y< 1),

where Uo is a constanl, and

u,(x,O) = O, u,(x,l) = -hu(x,l) (x > O),

where h is a positive constan\. Assuming the validily of the Sturm-Liouville expansion thatarises. derive the e~pression

CO sin (x"

u(x,y) = 2huo L (h - 2 \ exp (-cx.x) cos cx.yn.z:1 IX" + sin /XII

where cx. are the consecutive positive roots of the equation lan cx= h/a (see Seco 34). Give a

physical interpretation. of the function u(x,y) here.

-- - - - - oo--- - _L- --- -

.9. Solve the boundary value problem

uxx(x.y) + u,.,.(x.y)= O

u.(O.y) = O. u.(a.y) = - hu(a,y)

u(x,O) = O. u(x.b) =f(x)

(O< x < a, O< Y < b).

(O< Y < b).

(O< x < a).

where h is a positive constant. and interpret u(x.y) physically. (The Sturm-Liouville problemthat arises is treated in Problem 12, Seco 34.)

'"

(cos oc_x

)(Sinh OC_Y

),.

Ans. u(x.y)= 2hI . 2 --: b I f(s) cos oc_sds_~I ah + Sin oc_a SInh oc- 'o

where oc-are the consecutive positive roots of the equation tan oca= hloc.

10. Let u(r.e) denote eemperatures in a solid sphere bounded by the surface r = e. where r is

the spherical coordinate, when the sol id is initially at a uniform temperature Uoand the surfaceis then kept at temperature zero. The function u(r,e) satisfies the conditions

ou k 02

a¡= ~or2 (ru), u(e.e)= O. u(r.O) = Uo (O S; r < e).

Introduce the new unknown function v(r.e) = ru(r.e) and note that v(O.e) = O because u is contin-

uous at the center r = O. Thus derive the expression

2u e '"

(n21[2ke

)I n1[r

u(r.()=~ I(-I)n+'exp ---y- -sin-on n=1 e nr e

11. A spherical body 40 cm in diameter. initially at l00.C throughout. is cooled bykeeping its surface at O.C (see Problem 10). Find the approximate temperature of its center10 min after cooling begins when the material is (a) iron, for which k = 0.15 cgs unit;

(b) concrete. for which k = 0.005 cgs uni!.

A liS. (a) 22°C; (b) 100°C.12. The boundary r = 1 of a solid sphere is kept insulated. Initially the solid has tempera-

tures f(r). Assuming the validity of the Sturm-Liouville expansion that arises (see Problem 6,Seco 34). derive this expression for the temperatures in the sphere:

1 C() sin o: ru(r,e) = 3 r s2f(s) ds + I Cne~p (-oc;ke) ~'0 n=1 r

where OCnare the consecutive positive roots of the equation tan oc=ocand

2 ,1

J Cn=--:--y-.I s/(s)sinocnsdsocnsds. (11=1.2 ).Sin ocn.o

( I~ Let P. tJ>,z denote cylindrical coordinates and solve the following boundary valueproblem in the region 1 S;p S;b, OS; tJ> S; 71of the plane z = O:

p2Upp(p.tJ» + pup(P.tJ» + u~~(P.tJ» = O

up(I.tJ» = o. up(b.tJ»= -hu(b,tJ»

u(P.O)= O. u(P.1t)= f(p)

(1 < 1[),

(O < tJ>< 71),

(1 <p<b).

where h is a positive constan!. Interpret the function u(P.<I» physically. (The Sturm-Liou"ille

problem that arises is the one solved in Problem 16. Seco 34.)

'"

(Sinh!X. <1»

Ans. u(P.<I»= L C. --, cos (!X.log p).=1 smh !X.1t

where ex. are the consecutive positive roots of the equation tan (ex log b) = bit/ex and

C = 2bh . I. .. 'ogb+ sin2 (ex.log b) J. p J(p) cos (ex.log p) dp.

\

CHAPTER

SEVENFOURIER INTEGRALS AND

APPLICA TIONO'i.

61. The Fourier Integral Formula BIBI:IÓ:r,-ECA- DE lA -From Seco43we knowconditionsunderwhichthe Fourierseries FAS:UlTAD

1 e 1 ex> e . f)E CIBNC'1ASQUIMICAS

(1) 2eL/(s) ds+~JI L/(s) cos [n; (s - x)J ds

converges to f(x) when -e < x < e. To be precise, it is sufficient thatfbepiecewisecontinuous on the interval (- e,e), that f have one-sided deriva-tives at each point x in that interval, and that the value offat x be the meanvalue of the limitsf(x + ) and f(x - ).

Suppose that f satisfies such conditions on every bounded interval(-e,e). Then e may be given any fixed positive value, arbitrarily large butfinite, and series (1) will representf(x) over the large segment - e < x < e ofthe x axis. But that series representation cannot apply over the rest of the xaxis unless f is periodic with period 2e because the sum of the series has thatperiodicity.

In seeking a representation that is va lid for all real x when f is notperiodic, it is natural to try to extend the above representation by letting etend to infinity.The first term in series (1)would then vanish, provided thatfis such that the improper integral

,ex>

I f(s) ds, -ex>

exists. We write ~cx= n/e, and the remaining terms take the form

(2)1 ex> e

- L ~cxr f(s) cos [n ~cx(s- x)] dsn n=1 '-e(e = n/~cx),

or

(3)1 ex>- Lgc(n~a, x) ~an n=l

(~cx= n/e)

149

ISO FOURIER SERIES AND BOUNDARY VALUE PROBLEMS [SECo 61

where

(4), e

ge(O('X) = I f(s) cos [O((S- x)] ds.'-e

Let the value of X be fixed and e be large, so that ~O(is a small positivenumber. The points I! ~oc (I! = 1,2, ...) are equally spaced along the entirepositive O(axis; and because of the resemblance of the series in expression (3)to a sum of the type used in defining definite integrals: one might expect thatthe sum of that series tends to

(5),'"I ge(O('x)dO(,'o

or possibly

(6),'"

.lo g",(oc,x)dO(,

as ~octends to zero. As integral (6) indicates, however, the function ge(O(,x)changeswith ~O(becausee = 1t/~0(.AIso, the limit of the seriesin expression(3)as~octendsto zero isnot, in fact, the definition of the improper integral (5)even if e could be kept fixed.

The above manipulations merely suggestthat under appropriate condi-tions on f the function may have the representation

(7)1 ,'" ,'"

f(x) = - I I f(s) cos [oc(s- x)] ds dO(7t 'o '- '"(-00 < x < 00).

This is the Fourier integralformula for the functionf, to beestablish~din thefollowing two sections.

The formula can be written in terms of separatecosine and sine func-tions as follows:

(8),ore

f(x) = I [A(O() cos o(x + 8(0() sin !Xx]dO('0(-oo<x<oo)

where

(9)

1 ,'"A(O()=-1 f(x) cos ocxdx,1t.-x,

1 '"B(O()= - r f(x) sin o(xdx.1t ' - '"

Formulas (8) and (9) bear a resemblanceto Fourier series representationsand formulas for the coefficientsa. and b., statedat the beginning of Seco35.

SIi(" 62] FOURJER INTEGRALS AND APPLICATIONS 151

62. Prelirninary Lernrnas

Just as we prefaced our Fourier theorem in Sec, 41 with some preliminarytheory, we digress here with two lemmas that are essential to our proof of aFourier integral theoremwhich gives conditions under which representation(7) in the previous section is valid, The statement and the proof of thetheorem appear in the next section.

Our first lemma is an evaluation of an improper integral that occursoften in applied mathematics.

Lernrna1.1

,00 sin x d - n:x -2

,t'0 x

Before evaluating the integral, we show that it actually converges, Notethat the integrand (sin x)/x is piecewise continuous on any bounded inter-val; and since

(1) ,

., sin x ' 1 sin x " sin x- dx = I dx + I -. dx,'o x 'o X '1 X

where e is any positive number, it suffices to show that the integral

"

ex. sin xdo_o X

'1 X(2)

converges. To accomplish this, we apply the method of integration by partsto the function (sin x)/x = x' 1(d/dx)( -cos x) and write

(3),'sin x cos e "cos x

I --_o dx = cos l - - - I '2- dx,'1 X e, 1 X

Since I(cos e)/eI :::;;l/e, the second term on the right-hand side here tends tozero as e tends to infinity,AIso,since I(cos x )/x21 :::;;l/x2, the integral

1

, 00 cos xd~- X

'1 X(4)

is (absolutely) convergen!. !he limit of the left-hand side of equation (3) as etends to infinity therefore exists; that is, integral (2) converges,

Now that we have established that the given integral converges to somenumber L, or that

l.

r

'sin x1m -=L,

e-C().O X

t For another proof of Lemma 1 that is fairly standard, see, for example, Buck (1978,Sec, 6.4). A method of proof involving complex variables is pointed out in Churchill, Brown,and Verhey (1974, p, 193). Both books are listed in the Bibliography-

we note that, in particular,

r<m+1/21" sin x

(5) Jim -dx = Lm-(Q .o X

as m passes through positive integers. That is,

(6) lim f" sin [(m + !)u] du = L,m-co o U

where the substitution x = (m + !)u has been made for the variable of inte-gration. Observe that equation (6) can be written

(7)"

lim f F(u)Dm(u)du = Lm-co o

where

(8) F(u) = 2 sin (u/2)u

and Dm(u)is the Oirichlet kernel (Sec.40)

D ( ) =sin [(m+!)u]m U 2 sin (u/2) .

The function F(u), moreover, satisfies the conditions in Lemma 2, Seco40,and F(O+) = 1(seeProblem 1,Seco63). So, by that lemma, limit (7) has thevalue n/2; and, by uniqueness of limits, L = n/2. The proof of Lemma 1 inthis section is now complete.

Our second lemma makes direct use of the first one.

Lemma2. Suppcse that afunction F(u) is piecewisecontinuouson everybounded interval of the positive u axis and that the right-hand derivative F~(O)exists. If the improper integral

(9).""

I IF(u)j du'0 .

converges, then

(10).00 sin ru n

Jim I F(u)- du= _2 F(O+).,-x .0 U

Observe that the integrand appearing in equation (10) is piecewise con-tinuous on the same intervals as F(u) and that when u ¿ 1,

IF(U)sinuru 1:::; IF(u)l.

SECo 62J FOURIER INTEGRALS AND APPLlCATIONS 153

Thus the convergence of integral (9) ensures the existence of the integralin equation (10).

We begin the proof of the lemma by demonstrating its validity when therange of integration is replaced by any bounded interval (O,e).That is, wefirst show that if a function F(u) is piecewise continuous on a boundedinterval (O,e)and F~(O)exists, then

c .l. fF( )

SIn rud

1tF(O )1m u - u = _

2,..

r-'X) .0 U(11)

To prove this, we write

.c . sin ruI F(u)- du = I(r) + J(r)'o u

where

and

,c

f

F(U)-F(O+)

JI(r) = I sin ru du'o u

f)E

,C sin ru

J(r) = I F(O+ )-du.'o u

Since the function G(u) = [F(u)- F(O+)JIuis piecewisecontinuous on theinterval (O,e),where G(O+) = F~(O),we need only refer to Lemma 1 inSeco40 to see that

(12) lim I(r) = O.,~""

On the other hand, ifwe substitute x = ru into the integral representingJ(r)and apply Lemma 1of the present section, we find that

,cr sin x 1tlim J(r) = F(O+) lim I -dx = F(O+ )-.r~ ex. ,~ "- . o x 2

Limit (11) is evidently now a consequence of limits (12) and (13).To actualIy obtain limit (10),we note that

I ( F(u) sinuru du I ~ ( IF(u) I du,

where we assume that e ~ 1.We then write

(13)

l

. "" sin ru 1t

I(14) .lo F(u)-¡¡-du.- 2"F(O+)

1

, c sin ru 1t

l

' oc

~ .loF(u)-¡¡-du-2"F(O+) + t IF(u)I,lu,

choosing c to be so large that the second absolute value on the right, or theabsolute value of the remainder of integral (9), is less than e/2 where e is anarbitrary positive number independent ofthe value ofr. In view oflimit (11),there exists a positive number R such that whenever r > R, the first absolutevalue on the right-hand side of inequality (14) is also less than e/2. It thenfollows that

If

00 sin ru 1t

l

eeF(u)-du--F(O+) <-+-=e

o u 2 2 2

whenever r > R, and this is the same as statement (10).

63. A Fourier Integral Theorem

The following theorem gives conditions under which the Fourier integralrepresentation (7), Seco61, is valid.t

Theorem 1. út f denote afunction which is piecewise continuous on everybounded interval of the x axis and suppose that it is absolutely integrable overthat axis.. that is, the improper integral

00

f If(x)1 dx-00

converges. The Fourier integral

(1)1 00 00

- r f f(s) cos [a(s- x)] ds da1t.o -00

then converges to the value

(2)12[J(x+) + f(x-)]

at each point x( - 00 < x < 00) where the one-sided derivativesfR(x) andfí.(x)both existo

We begin our proof with the observation that integral (1) represents the'limit as r tends to infinity of the integral

(3)1 r 00 1- f f f(s) cos [a(s - x)] ds da = -[I(r,x) + J(r,x)]1t o -00 . 1t

t For other conditions see, for instance, Carslaw (1930, pp. 315 ff) and Titchmarsh (1967,pp. 13 ff), both listed in the Bibliography.

SECo63] FOURIER INTEGRALS AND APPLlCATlONS 155

where, 00

l(r,x) = f f f(s) cos [a(s- x)] ds dao x

and

J(r,x)= r ( f(s) cos [a(s- x)] ds da.o -00

We now show that the individual integrals l(r,x) and J(r,x) exist; and,assuming that f~(x) and f~(x) exist, we examine the behavior of these inte-grals as r tends to infinity.

Turning to l(r,x) first, we introduce the new variable of integrationu = s - x and write that integral in the form

(4), 00

l(r,x) = f f f(x + u) cos au du da.o o

Since

If(x + u) cos aul ~ If(x + u)1and because

00 00 00

f If(x + u)1 du = f If(s)1 ds ~ f If(s)1 ds,o x -00

the Weierstrass M-test for improper integrals applies to show that the integral00

f f(x + u) cos au duo

converges uniformly with respect to the variable a. Consequently, not onlydoes the iterated integral (4) exist, but the order ofintegration there can bereversed:

00

f' 00 sin ru

l(r,x) = f f(x + u) cos au da du = f f(x + u) -- du.to o o u

Observe now that the function F(u) = f(x + u) satisfies all of the conditionsin Lemma 2, Seco62. Hence, if we apply that lemma to this last integral, wefind that

(5) lim l(r,x) = if(x+).'-00

t Theorems on improper integrals used here are developed in Buck's "Advanced Calculus"

(1978, Seco 6.4), listed in the Bibliography, as well as in most other texts with such titles. Thetheorems are usually given for integrals with continuous integrands, but they are also validwhen the integrands are piecewise continuous.

The limit of J(r,x) as r tends to infinity is treated similarly. Here wemake the substitution u = x - s and write

fr

foo 00 sin ruJ(r,x)= f(x - u) cos cxudu dcx=f f(x - u)-du.

(j o o u

When F(u)=f(x - u), the limit

(6) lim J(r,x) = ~2 f(x-)r-oo

also folIowsfrom Lemma 2 in Seco62.FinalIy, in viewof limits (5) and (6),we see that the limit ofthe left-hand

side of equation (3) as r -+00 has the value (2), which is then the value ofintegral (1) at any point where the one-sided derivatives of f existo

Note that since the in\egrals in formulas (9),Seco61, for the coefficientsA(cx)and B(cx)exist whenfsatisfies the conditions stated in the theorem, form(8), Seco61, of the Fourier integral formula is also justified.

PROBLEMS

1, Show Ihat the function

F(u) = 2 sin (u/2)u '

used in equation (7), Seco 62, satisfies the conditions in Lemma 2, Seco40. To be precise, showthat it is piecewise continuous on the interval (O,x) and that F~(O) = O.

Suggestion: To obtain F~(O), show that

F~(O) = lim 2 sin (h(l) - h>-0 h2'>0

and apply I'Hospital's rule twice.

2. Verify that all the conditions in Theorem I are satisfied by the function

f(x) =<!when Ixl < 1,

when Ixl = 1,

o when Ixl > 1.

Thus show that for every x (- 00 < x < 00)

f(x) =~JO) sin [a(l + x)) + sin [a(l- x)] da =~JO> sin a cos ax da.

n o a. x o a

SECo63] FOURIER INTEGRALS AND APPLICATlONS 157

3. Show how it foIlows from Lemma l. Seco 62. that

j when k > O.

f<X>sin ka

-da=o a

o when k = O,

-j when k < O.

Then use this result to evaluate the. first integral written in Problem 2 when Ixl < 1, whenIxl = 1, and when Ixl > 1, thereby making a direct verification of the integral representationfound for the function f there.

4. Show that the function

f(X)J:

je-X

when x < O,

when x = O,

when x > O

satisfies the conditions in Theorem 1 and hence that

f( )1f

<X>cos ax + a sin axx = - 1 2 da1t o + el

(-00 < x < (0).

Verify this representation directly at the point x = O.5. Prove that

2f

"'cosaxexp (-Ixi)=- ¡da

. 1t o 1 + a(-00 < x < (0).

6. Prove that iff(x) = sin x when O S; x S; 1tandf(x) = O when x < O and when x > 1t,then

f(x) = ~f <X> cos ax + cos [~(1t - x)) da1t o 1 - a

(-00 < x < (0).

In particular, write x = 1t/2 to show that

f'" cos (a1t/2) 1t

2 da=- 2'

o 1- a

7. Show why the Fourier integral formula fails to represent the functionf(x) = I (- 00 <

x < (0). Also, point out which condition in Theorem 1 is not satisfied by that function.

8. Give the details showing that the integral J(r,x) in Seco63 actuaIly exists and that limit(6) in that section holds.

9. Let f be a nonzero function which is periodic, with period 2c, for example. Point outwhy the integrals

( f(x) dx-<X>

and L: If(x)1 dx

fail to exist.

64. The Cosine and Sine Integrals

Letf(x) be an even function that satisfies the conditions in Theorem 1. Sincef(x) cos ax andf(x) sin ax are even and odd functions of x, respectively, thecoefficients A(a) and B(a) given by formulas (9), Seco 61, are

1 .C() 2 .C(). A(a)= - I f(x) cosax dx =- I f(x) cosax dxn._C() n.o

and

1 .""B(a) = - I f(x) sin ax dx = 00n . - C()

The Fourier integral formula

(1)C() .

f(x) = r [A(a) cos ax + B(a) sin ax] da'0(-oo<x<oo)

thus reduces to the Fourier cosine integral formula

(2)2 .C() .C()

f(x) = - I cos ax I f(s) cos as ds da.n '0 . o

Ir, instead, f is an odd function, then

A(a)= 0,2 .C()

B(a) = - I f(x) sin (Xxdx;n .o

and formula (l) becomes the Fourier sine integralformula

(3)2 ."- .""

f(x) = -1 sin ax I f(s) sin as ds da.n .o oo

Irf(x) is definedonly on the halfline x> 0, formulas (2)and (3)apply tothe even and odd extensionsoff, respectively,and representf(x) whenx> Ounder the following conditions.

Corollary 1. Let a function f be piecewise continuous on el'ery boundedinterval of the positive x axis.. and,for convenience, let the value of f be definedat each point where it is discontinuous as the mean value of its one-sided limitsthere. Suppose further that the improper integral

."-

I If(x)1 dx'0

converges. Then, at each point x(x > O)where the one-sided derivativesfR(x)and fÍ.(x) exist, f is represented hy its Fourier cosine integral (2) and its Fouriersine integral (3)0

'¡

SECo 65] FOURIER lNTEGRALS AND APPLlCATIONS 159

Under those conditions the cosine integral formula (2)representsf(O+ )whenx = OiffR(O)exists,as weseeby consideringthe evenextensionoffThe sine integral clearly has the value zero when x = O.

The eigenvalue problem

X"(x) + AX(X) = O

X(9)=0, ¡X(x)1 <Mwhere M is some positive' constant, is singular (Sec.32) because its fun-damental interval x> O is unbounded. If A= O, we have only the trivialsolution. If A is a real number such that A > Oand we write A = ex2.(ex> O),we readily find that, except for a constant factor, the eigenfunctions areX = sin exxwhere extakes on all positive values. The eigenvalues A = ex2 are'

continuous rather than discreteoIf A < O,or A= - p2 (P> O),the solutionof the differential equation and the bound~ry condition at x = O isX = DI sinh px. This is, however, unbounded on the half line x > OunlessDI =O.So the caseA< Oyieldsno neweigenfunctions.Casesother thanthose in which A is real need not be considered since they yield unboundedsolutions of the differential equation (seeProblem 12,Seco66).Although theeigenfunctionssin exxhave no orthogonality property, the Fourier sine inte-gral formula (3) gives representations of functionsf on the interval x> Owhich are generalized linear combinations of those eigenfunctions.

Likewise,A= ex2(ex~ O)and X = cos exxare the eigenvaluesand eigen-functions of the singular problem

(x > O),

(x > O),(4)

(5)(x > O),

(x > O);

X"(x) + AX(X) = O

X(O) = O, ¡X(x)1 < M

and formula (2) represents functionsf in terms of cos exX.

65. The Exponential Forro

Under the conditions stated in Theorem l the Fourier integral formula canbe written '

(1)1 .r .'"

f(x) = 2~ !~.lo L", 2f(s) cos [ex(s- x)] ds dex.

Let the cosine function be expressed in terms of exponential functions. Sincef is absolutely integrable, the integrals

.cx,

I . f(s)ei2Sds,.- x.

"-

r f(s)e-¡«s ds._"

converge uniformly with respect to exfor all positive ex,according to theWeierstrass M-test for improper integrals. That uniform convergence

together with the piecewisecontinuity offimplies that those integrals repre-sent continuous functions of IX(see Problem 9, Seco66). Thus the iteratedintegral in equation (1) can be written as the sum

(2).r .'X .r . "X

I e-i.x I f(s)ei.. ds dlX+ I é'x I f(s)e-¡U ds dlX'0 . -ac '0 . -O<'

since the integrals with respect to a exist as integrals of continuous functionsof IX.(The improper integrals from IX= Oto a = 00 may not exist, however.)

With the substitution f3= -IX, the second term of the sum (2) takes theform

.0 . CI()

I e-iPx I f(s)eiPS ds df3."-r "-00

Evidently, then, the sum (2) can be written

( e-iox f'" f(s)ei«s ds dlX."-r O-cc

Thus the exponential forrn of the Fourier integral formula (1) is

(3) f(x) = 2~ !~~ .C e-iox ("",f(s)ei«S ds da,

where - 00 < x < oo.The minus sign in the exponent - iax can be shifted tothe exponent ias by using - a, instead of IX,as a variable of integration.

The limit in formula (3) is called the Cauchy principal value of theimpropér integral from -00 to 00with respect to IX.The principal value mayexist when the improper integral does not. The integral

co

f a dlX,.-co

for example, fails to exist; but its principal value does exist, and it is equal to. zero because

,

f a da = O-,

for each value of r. When the improper integral does exist, it is equal to itsprincipal value.

Unless further conditions are imposed on f, the principal value used informula (3) cannot be replaced by the improper integral itself,as the follow-ing example will show. The function

(4) f(x) = e¡:-x

when x < O,

when x = O,

when x > O

SECo66] FOURIER INTEGRALS AND APPLlCATIONS 161

satisfies all the conditions. in our Fourier integral theorem. It is then repre-sented by formula (3) at everypoint x, and at x = Oin particular. In this case

(5) r

C() C()

. f(s)eia. ds = r e-.(1-ia) ds = ~ = ~+ ia.-C() '0 1-ia. 1+0:2'

Thus when x = O in formula (3), the iterated iÍ1tegral becomes

, 1 + io: f i

J

'f -1 2 do: = tan - I o: +-2 log (1 + 0:2) = 2 tan- 1 r,-, + o: -,

which has the limit 1tasr -+ oo.The right-hand sideof equation (3) thereforehasthe value 1/2,which isf(O). But the improper integral of the function (5)from - 00 to 00doesnot exist since the function 0:/(1+ 0:2)is not integrablefrom Oto oo.

The functions X = e-iax (- 00 < o:< 00) are eigenfunctions of the sin-gular eigenvalue problem

(6) X"(x) + ÁX(x) = O, ¡X(x)1 < M (-00 <x<oo),

where M is some positive constant. The eigenvaluesÁ= 0:2 consist of all realnonnegative numbers, and representations of functions f in terms of theeigenfunctions are given by formula (3).

66. Fourier Transforms

The Fourier sine integral formula (3),Seco64, can be written in the form

Ir f is a given function, then equation (1) is an integralequation in theunknown function F. appearing in the integrand there. It is a singularinte-gral equation because the integral is improper. Equation (2)givesa solutionwhenf satisfiesthe conditions stated in Corollary 1,as weseeby substitutingthe right-hand side of that equation, with the variable of integration xreplaced by s, into equation (1).

The functionF. givenby equation (2) is the Fouriersinetransformofthe functionj Transformation (2),which we may abbreviate as

(3) F.(a.) = Sa{f},

establishes a correspondence between functionsfand F.. Functionsfsatisfy-ing the conditions in Corollary 1have transforms F. such that equation (1)

(1)2 C()

(x > O)f(x) = - f F.(o:) sin o:x do:1t o

whereC()

(2) F.(o:) = f f(x) sin o:x dx (o:> O).o

gives f in terms of its transform; that is, equation (1) defines the inversetransformation.

Suppose thatf,f', andf" are continuous when x ~ O,thatf(x) andf'(x)tend to zero as x tends to infinity, and that

00

r If(x)1 dx'0

exists. Then successive integrations by parts show that

(4) S.{f"(x)} = -a2Fs(a) + af(O)

(see Problem 7 of this section). This is the basic operational property of thesine transformation. It replaces the difTerential form f"(x) by an algebraicform in the transform Ps(ex)and involves the initial value f(O).

Property (4), together with other operational properties of the transfor-mation, may be used to reduce certain types of boundary value problems tosimpler problems in the transforms of the unknown functions. The develop-ment of this operational method, using the various Fourier transforms andthe related Laplace transform, is treated elsewhere.t

The Fourier cosine transform Fe of a function f is

(5)<X>

Fc(ex)= f f(x) cos ax dx = e.{f}o(ex> O).

The inverse of the transformation e.{f} is given by the Fourier cosineintegral formula (2), Seco64; that is,

(6)2 '"

f(x) = - f Fc(a) cos exx dex1t o

(x > O).

The basic operational property, valid under the conditions given above forproperty (4), involves1'(0) rather than f(O):

(7) e.{f"(x)} = -ex2FAa) - 1'(0).

Note that S. and e. are linear operators (Sec. 11).The exponential Fourier transformis the function

(8)'"

F(a) = f f(x)eiOX dx = E.{f}-'"(-co<ex<co)

whose inverse is obtained from formula (3), Seco 65. There are also certainintegral transformations over bounded intervals calIed finite Fouriertransforms.

t See. for example. Churchill (1972). listed in the Bibliography.

SECo 66] FOURIER INTEGRALS AND APPLICATIONS 163

PROBLEMS

l. By applying the Fourier sine integral formula to the function

f(x) = ( !O

when O < x < k.

whenx = k.

when x > k,

obtain the representation

2 .' I - cos kO!f(x) = - 1 sin o!x dO!

1t.o O!(x> O).

2. Use Corollary I and the Fourier sine integral formula to show that

2 .' o!' sin o!xe-xcosx=-I -dO!1t.o :x4+ 4

(x> O).

3. Use the Fourier cosine integral formula to pro ve that

2 .>- 0!2 + 2e-x cosx =-1 coso!xdO!

1t.o O! + 4(x ~O).

4. Apply the operalional property (4), Seco 66, lo the function e-kx, where k is a positiveconstant, to show that the sine transform of that function is :x/(0!2+ k2). Thus obtain therepresen tation

- b 2 l. oc' a: sin o:x

e =- -dO!1t.o 0!2 + k'

(k > O. x > O).

S. Use the operational property (7), Seco66. to prove Ihat the cosine transform of e-kx.where k is a positive constanl, is k/(0!2 + P). Thus show that

-kx 2kr

'" cos o!xe =- -dO!

1t . o 0!2+ k2(k > O. x ~ O).

6. By regarding the positive constant k in the equation obtained in Problem 5 as avariable and then differentiatingeach side of that equation with respect to k, showformally that

4r

"'~dO!(1 + x)e-X =;'0 (0!2+ 1)2

(x ~O).

7. Assuming that f, 1', and f" are continuous (x ~ O), use integralion by parts to provethat, for each positive number c,

f f"(x) sin ax dx = f'(e) sin o!c- af(e) cos O!e+ af(O) - a2 (f(x) sin ax dx.o o

[The continuity off" here can be replaced by the condition thatf" be piecewise continuous oneach interval (O.e).] Assuming also thalf(x) andf'(x) tend to zero as x tends to infinity and thatthe Fourier sine transform F,(O!)offexists, show that the right-hand sideoftheaboveequationhas the limit O!f(O) - 0!2F,(O!) as e tends to infinity. Deduce that the sine transform off" exists

and that it satisfies the operational property (4), Seco 66.

8. Derive the operational properly (7), Seco 66, for the cosine transform off". (CompareProblem 7.)

9. Let f(s) be piecewise continuous on each interval (-e.e) of the s axis. State why thefunction

(a)= .Cf(s)e'" ds

is continuous for all real a. If. in additionJ is absolutely integrable along the entire s axis. thenthe integral

'"

-,"-, f(s)e'" ds = F(a)

is uniformly convergent with respect to a. Write

F(a) = (a) + r -, f(s)e'" tls+ f' f(s)e'" ,/s'-~ 'c

and preve that F(a + Lla) - F(a) O as Lla O by first mak ing e large. independent of IXandLla. and ¡hen taking Llrxsmall to make (IX+ LlIX)- (IX)small. Thus establish the continuity ofF(a) that was used at the beginning of Seco65.

10. Establish the following operational property of the exponential Fourier transform

F(IX)= f'" f(x)e'U tlx. -,(- ex)< IX<XJ):

ifIJ'. andf" are everywhere continuous. ifthe transform F(IX) off exists. and iff(x) andf'(x)both tend to zero as Ixl tends to oo. then

E,{f"(x)} = -a2F(a) (-O:<IX<012).

11. Verify the Fourier sine integral representation

x 2.' ."'ssinIXs---z--2= . I sin IXXI ~ k2 ds dIXx + k n.o '0 ., +

by observing that according to Problem 4. the inner integral here has the value (n/2)e-h andthe outer integral is then the sine transform ofthat exponential function of IX.Point out why thefunction X/(X2 + k2) is no! absolutely integrable from x = O to x = w.

12. Using the fact that ¡he function Isin(x + iy)l is unbounded as Iyl x (see Problem 2.Seco 30), show that the eigenvalues of ¡he singular Sturm-Liouville problem (4), Seco 64. must bereal.

(k> O)

67. More on Superposition of Solutions

In Seco 11 we showed that linear combinations of solutions of linear homo-

geneous differential equations and boundary conditions are also solutions.In Seco 13 we extended that result to inelude infinite series of solutions, thusproviding the basis of the technique for solving boundary value problemsin Chap. 6. Another useful extension is illustrated by the following example,where superposition consists of integration with respect to a parameter ainstead of summation with respect to a\1 index 11.It will enable us to solvecertain boundary value problems where Fourier integrals, rather than Fou-rier series, are required.

SECo67] FOURIER INTEGRALS AND APPLlCATIONS 165

The functionsofthe set exp (-ay) sin ax, where the parameter a (a > O)is independent of x and y, satisfy Laplace's equation

(1) uxx(x,y) + Uyy(x,y) = O (x > O,Y > O)

and the boundary condition

(2) u(O,y)= O (y> O).

Those functions are bounded in the dornain x > O,Y > O; they are obtainedfrorn conditions (1) and (2) by the rnethod of separation of variables whenthat boundedness condition is incIuded (Problern 1, Seco69).

We now show that their cornbination of the type

(3)00

u(x,y) = f g(a)e-«Y sin ax da'0(x > O,Y > O)

also represents a solution of the homogeneous equations (1) and (2) whichis bounded in the dornain x> O.y> Ofor each function g(a) that is boundedand continuous on the half line a > Oand absolutely integrable over it.

To accomplish this, we use tests for improper integrals that are analo-gous to those for infinite series.t The integral in equation (3) convergesabsolutely and uniformly with respect'to x and y because

(4) Ig(a)e-«Y sin ax I ~ Ig(a) I (x ~ O,Y ~ O)

and g is independent of x and y and absolutely integrable from zero toinfinity'with respect to a. Also

(5)00 00

Iu(x,y) I ~ f Ig(a)e-«Y sin ax I da ~ f Ig(a) Ida,o o

so that u is bounded; it is also a continuous function of x and y (x ~ O,Y ~ O) because of the uniforrn convergence ofthe integral and the continuityof the integrando Clearly u = O when x = O.

When y > O,

aau= aa 1'00g(a)e-«Y sin ax da = f00 aa [g(a)e-«Y sin ax] da;

x X'o '0 x(6)

for if Ig(a) I ~ go and y ~ Yo where Yo is sorne small positive nurnber, thenthe absolute value of the integrand of the integral on the far right does notexceed goa. exp (- a.Yo),which is independent of x and y and integrable frorna = Oto IX = oo. Hence that integral is uniforrnlyconvergent. Integral (3) is

t See KapJan (1973. pp. 447 fT) or TayJor and Mann (1972. pp. 722 fT). listcd in theBibliography.

then differentiable with respect to x, and similarly for the other derivativesinvolved in the laplacian operator V2 = (}2/0X2 + 02/oy2. Therefore

(7).00

V2u = I g(a)V2(e-o)'sin ax) da = O'0

(x > O, Y > O).

Suppose now that the function (3) is also required to satisfy the nonho-mogeneous boundary condition

(8) u(x,O) = f(x) (x > O),

where f is a given function satisfying the conditions stated in Corollary'1. Weneed to determine the function g(a) in equation (3) so that

(9).ro

f(x) = I g(a) sin ax da'0

(x > O).

This is easily done since representation (9) is the Fourier sine integral for-mula (3), Seco64, when

2 .<X'g(a) = I f(x) sin ax dx1[ .o

(10) (a > O).

We have shown here that the function (3) with g(a) given by equation(10) is a solution of the boundary value problem in the quadrant x ~ O,Y ~ Oconsisting ofequations (1), (2), and (8) together with the requirementthat u be bounded.

68. Temperatures in a Semi-infiniteSolid

The face x = O of a semi-infinite solid x ~ O is kept at temperature zero(Fig. 24). Let us find the temperatures u(x,t) in the solid when the initialtemperature distribution isf(x), assuming at present thatfandf' are piece-wise continuous on each bounded interval of the positive x axis and that-f is bounded and absolutely integrable from x = O to x = oo.

If the so lid is considered as a limiting case of a slab O :::;x :::;e as eincreases, some condition corresponding to a thermal condition on the facex = e seems to be needed; otherwise the temperatures on that face may be

u=o

xoFigure 24

SECo 68] FOURIER INTEGRALS AND APPLlCATIONS 167

increased in any manner as e increases.We require that our function u bebounded; that condition also impliesthat there isno instantaneous source ofheat on the face x = O at the instant t = O.Then

(1) u,(x,t) = kU",,(x,t)

u(O,t)= O

lu(x,t)I < M

(x > O,t > O),

(t > O),

(x> O,t > O),(2)

where M is some positive constant, and

(3) u(x,O) = f(x) (x > O).

Linear combinations of functions XT will not ordinarily be boundedunless X and T themselvesare bounded. Upon separating variables,we thenhave the conditions

(4)

X"(x) + )'X(x) = O

X(O) = O, IX(x)1 < M¡,

T'(t) + )'kT(t) = O, IT(t) I< M2

(x > O),

and (t > O),

where MI and M 2 are positive constants. As pointed out in Seco 64, thesingular eigenvalue problem (4) has continuous eigenvalues). = a2 where a

. represents all positive real numbe"rs;X = sin ax are the eigenfunctions. Inthis case the corresponding functions T = exp (-a2kt) are bounded. Thegener3¡lizedlinear combination of the functions XT for all positive a,

(5)00

u(x,t) = r g(a) exp (-a2kt) sin ax da,'0

will formally satisfy all conditions of the boundary value problem if thefunction g can be determined so that

(6),00

f(x) = I g(a)sin ax da'0(x> O).

As in the previous section, we note that representation (6) is the Fouriersine integral formula (3), Seco64, for our function f if

(7)2 00

g(a) = - J f(x) sin ax dxn o

(a> O).

Our formal solution is therefore

(8)2 ,00

r

oo

u(x,t)= -1 exp (-a2kt) sin ax f(s) sin as ds da.n 'o .o

I

We can simplify this result by formally reversing the order of integra-tion, replacing 2 sin a.ssin a.x by cos [a.(s- x)] - cos [a.(s+ x)], and thenapplying the integration formula (Problem 20, Seco69)

'" 1 (it (b2

)(9) t exp (-a.2a) cos a.bda.=2"Va exp - 4a

Equation (8) then becomes

(10) u(x,t)= 2~ .C f(s){exp[ - (s~k:)2J - exp[-(s;k;)2J} ds

(a > O).

when t > O.An alternati.!,¡eform of equation (10),obtained by introducingnew variables of integration, is

(11)1 '"

u(x,t) = r.: f f(x + 2rJkt) exp (-r2) drv 7t - x¡(2.ji(f¡1 '"

- c f f( -x + 2r,jkt) exp (-r2) dr.v 7t x¡(2M

Our use of the Fourier sine integral formula in obtaining the solution(8) suggests that we apply Corollary 1 in verifying that solution. Theforms (10)and (11)suggest,however, that the condition in Corollary 1 thatIf(x)1 be integrable from Oto 00 can be relaxed in verifying u(x,t). To .beprecise, when s is kept fixed and t > O,the functions

, 1

[

- (s :t X)2

]Jt exp 4kt

satisfy the heat equation (1).Then, under the assumption thatfis continuousand bounded when x ~ O, it is possible to show that. the function (10)satisfies the heat equation when Xo< x < Xl and to < t < ti' where Xo,Xl'to,and ti are any positivenumbers. Conditions (2)and (3)can beverifiedbyusing expression (11).By adding step functions (seeProblem 4, Seco69) tof,we can permit f to have a finite number of jumps on the half line X > O.Except for special cases,details in the verification of formal solutions of thisproblem are, however, tedious.

Whenf(x) = 1, it followsfrom equation (11) that

1

[

'" '"

]u(x,t) =-c f exp (-r2) dr - f exp (-r2) dr .V 7t -x¡(2.¡Ifi x¡(2.¡kt)(12)

In terms of the error function

(13)2 x

erf (x) = r.:f e!'p (-r2) dr,v 7t o

where erf (x) -+ 1 as x -+ 00 (see Problem 19, Seco69), equation (12) can bewritten

(14) u(x,t)= erfC~).The fullverificationof thisresult is not difficult.

69. Temperatures in an Unlimited Medium

For an application of the general Fourier integral formula, we now deriveexpressions for the temperatures u(x,t) in a medium that occupies all space,where the initial temperature distribution is f(x). We assume that f isbounded and, for the present, that it satisfies conditions under which it isrepresented by its Fourier integral formula. The boundary value.problemconsists of a boundedness condition Iu(x,t) I < M and the conditions

(1) u,(x,t) = ku",,(x,t) (- 00 < x < 00, t > O),

(2) u(x,O)= f(x) (- 00 < x < (0).

Separation of variables leads to the singular eigenvalue problem

X"(x) + ,tX(x) = O, IX(x)1 < MI (-00 < X < (0),

whose eigenvalues are ,t = a2 where a ;e::O,and to two linearly independenteigenfunctions cos ax and sin ax corresponding to each nonzero value of a.

Our generalized linear combination of functions XT ist

.a;,

u(x,t) = I exp (-a2kt)[A(a) cos ax + B(a) sin ax] da.'0

The coefficients A(a) and B(a) are to be determined so that the integral hererepresentsf(x) (- 00 < x < (0) when t =O.According to equations (8) and(9)ofSec.61and our Fourierintegraltheorem(Sec.63),the representationis validif

1 .OG

A(a) = - J f(s) cos as ds,n -a;,


(3)

1 a;,

B(a) =- f f(s) sin as ds.n_a;,

(4)1 a;, a;,

u(x,t) = - f exp (-a2kt) f f(s) cos [a(s- x)] ds da.n o -a;,

Ir we formally reverse the order of integration here, the integration for-mula (9) of Seco68 can be used to write equation (4) in the form

1 a;,

[

(s X)2

]u(x,t) = r.:;;;;f f(s) exp - ~k ds2 ; nkt -a;, t(5) (t > O).

170 J10URIER SERIESAND BOUNDARY VALUE PROBLEMS [SEC. 69

An alternative form of this is

1 .oc

u(x,t) = Jñ L 00f(x + 2rJkt) exp (-r2) dr.

Forms (5) and (6) can be verified by assuming only that f is piecewisecontinuous over some bounded interval Ixl < e and continuous andbounded oyer the rest of the x axis, when Ixl ~ c. If f is an odd function,u(x,t) becomes the function found in the preceding Section for positiyeyalues of x.

(6)

PROBLEMS

l. (a) Give details showing how the functions exp (-IXY) sin IXX(IX> O)arise by means ofseparation ofvariables from equations (1) and (2). Seco67. and the condition that the functionu(x,y) there is to be bounded when x > O, Y > O.

(b) Show that if the function u(x.y) in Seco67 satisfies the nonhomogeneous boundarycondition

u(x.O)= e-X - e-Ix (x ~O).

then

u(x.y) = J'" g(lX)e-'Y sin IXX dlXo

(x ~ O. y ~ O)

where g(oc)is the absolutely integrable function

g(lX)= 6 oc (o:~O).

Suggestion: Note that the singular eigenvalue problem that arises in pan (a) has alreadybeen solved in Seco 64. In part (b), refer to the result in Problem 4. Seco66. to write

e-X-e-Ix=~r'" IXsin IXX do:7['0 (0:2 + 1)(1X2+ 4)

(x ~O)

and then use this representation.

2. (a) Substitute expression (10). Seco67. for the function g(o:) into equation (3) of thatsection. Then. by formally reversing the order of integration, show that the solution of theboundary value problem treated in Seco 67 can be written

\,.'"

[

I 1

]~~~=~ fN ~

. 7[.10 (s - X)2 + .1'2 (s + X)2 + .1'2

(b) Show that when f(x) = 1.the form of the solution obtained in part (a) can bewritten in terms of the inverse tangent function as

2'

(

X

)I/(x.y) = - tan- I -.

7[ y

3. Verify that the function u = erf[xj(2jkt)] in Seco 68 satisfies the heat equation u, = kuxx

when x > O,t > Oas well as the conditions

u(O+,t) = O

u(x,O+) = 1

lu(x,t)I < 1

(t > O).

(x> O),

(x > O. t > O).

4. Show that iff(x) = Owhen O < x < e andf(x) = 1when x > e, expression (11). Seco68,reducesto

1

(e + X

)1

(e - X

)u(x,t)= Zerf 2.jkc - lerf 2jkt .

Verify this solution of the boundary value problem in Seco 68 when f is this function.

5. The face x = O of a semi-infinite so lid x ;:: O is kept at constant temperature Uoafter its

interior x > Ois initially at temperature zero throughout. Obtain an expression for the tempera-tures u(x,t) in the body.

Ans. u(x.t) = Uo [1 - erf(2,;dj.6. The face x = O of a semi-infinite solid x ;:: O is insulated, and the initial temperature

distribution isf(x). Derive the temperature formula

1 '"u(x,t)= 1- r f(x + 2rJki) exp (-r2) dr

y 7[.-xf(2jiÍ)

. 1 '"+ 1- r f( -x + 2r.jkc) exp (-r2) dr.

y7[ 'x/(2M

7. In Problem 6, writef(x) = 1 when O < x < e andf(x) = Owhen x > e, Then show that

1

(e + X

) 1 f (E...=...!. )u(x,t) ='2erf 2jkt + '2er 2.jkc'

8. Let the initial temperature distribution f(x) in the unlimited medium in Seco 69 bef(x) = O when x < O andf(x) = 1 when x > O. Show that

1 1

(X

)u(x,t)= '2+ Zerf 2.jki .

Verify this solution of the boundary value problem in Seco69 when f is this function.

9. Derive this solution of the wave equation y" = a2)'xx (- OCJ< x < OCJ.t > O) that

satisfies the conditions y(x,O)= f(x), y,(x,O)= Owhen - OCJ< X < OCJ:

1 "" "

y(x,l)= - f cos aal f f(s) cos [a(s - x)] ds da.no' - 00

AIso, reduce the solution to the form obtained in Example 2. Seco 19:

1y(x.t) = '2(f(x + at) + f(x - al)].

,10. A semi-infinite string with one end fixed at the origin is stretched along the positive

half of the x axis and released at rest from a position y = f(x) (x ;:: O). Derive the expression

2 x x.

y(x.t) = -,' COSO/atsin O/xf f(s) sin o<sds d:xn 'o '0

for ils Iransverse displacemenls. Lel F(x) (-00 < x < (0) be Ihe odd exlension off(x) andshow how Ihis resull reduces lo Ihe form

1y(x,l) = 2 [F(x + al) + F(x - al)].

[Compare SolUlion (9~ Sec. 51, of the string problem treated in Ihat section.]

11. Find Ihe bounded harmonic funclion V(x,y) in Ihe semi-infinite Slrip x > O,O < Y < 1Ihat satisfies the boundary conditions (Fig. 25)

V.(O,y) = O, V,(x,O) = O, V(x,I)=f(x)

(O J V =f(x)

v.~ol "V-O \-01 v,=0 x Figure25

where (a)f(x) = 1 when O < x < 1 andf(x) = O when x > 1; (b)f(x) = e-' (x > O).

Ans. (a) V(x,y) = ~f'" sin a cos ax cosh ay

n o a cosh a da;

2f " cos ax cosh ay da.

(b) V(x,y) =;; o (1 + (2) cosh a

12. Find V(x,y) when Ihe boundary condilions in Problem 11 are replaced by Ihecondilions

V.(O,y) = O, V,(x,l) = - V(x,I), V(x,O)=f(x)

wheref(x) = 1 when O < x < 1 andf(x) = O when x > 1. Inlerpret this problem physically.

( )2f

'" a cosh [a(1 - y)] + sinh [a(1 - y)] . dAns. V x,y = - 2 . h 510a cosax a.n o a cosha+as1O a

13. Find the bounded harmonic function V(x,y) in Ihe semi-infinile slrip O < x < 1, y> Owhich satisfies the conditions

V,(x,O) = O, V(O,y)= O, V.(I,y) =f(y).

2 f'" sinh ax cos ayf

'"

Ans. V(x,y) = - h f(s) cos as ds da.no acosa o

14. Find V(x,y) when the boundary conditions in Problem 13 are replaced by theconditions

V,(x,O) = O, V(O,y) = e-", V(I,y)=O.

( ) - ~f'" sinh [a(1 - x)) cos ay

Ans. V x,y - (2) . h da.n o 1 +a 510 a

15. Find the bounded h~rmonic function V(x,y)in the strip -00 < x < 00, O< Y < bsuch that V(x,O)= Oand V(x,b) = f(x) (- 00 < x < 00~ wherefis bounded and represented byits Fourier integral.

1 '" sinh ay '"Ans. V(x,y)= - f -:--b f f(s) cos [a(s - x)] ds da.n o s10ha -'"

16. Let a semi-infinite solid x ~ O,which is initially at a uniform temperature, be cooled orheated by keeping its boundary at a uniform constant temperature (Sec. 68). Show that thetimes required for two interior points to reach the same temperature are proportional to thesquares of the distances of those points from the boundary plane.

17. Solve the following boundary value problem for steady temperatures u(x,y) in a thinplate in the shape of a semi-infinite strip when heat transfer to the surroundings at temperaturezero takes place at the faces of the plate:

u..(x,y) + Uyy(x,y) - hu(x,y) = O

ux(O,y)= O

u(x,O)= O, u(x,l)=f(x)

(x>O,O<y< 1),

(O<y< 1),

(x >0)

where h is a positive constant andf(x) = I when O < x < c andf(x) = O when x > C.

( ) - 2J""sin ac cos ax sinh (yP+h)Ans. u x,y -- ~ da.

1t o a sinh v a + h

18. Verifythat, for any constant C. the function

~ (_X2

)v(x t) = Cxt-312 exp -, 4kt

satisfies the heat equation V, = kv.. when x > O and t > O. Also verify that v(O+,t) = O whent > Oand that v(x,O + ) = Owhen x > O. Thus v can be added to the function u found in Seco68to form other solutions of the problem there if the temperature function is not required to bebounded. But note that vis unbounded as x and t tend to zero, as can be seen by letting;'; vanishwhilet=x2. .

19. Let / denote the integral of exp (_X2) from O to r::t:)and write

ex) co tO c:o

/2 = J e-x' dxJ e-Y' dy =J f e-(x'+Y') dx dy.o o o o

Evaluate the iterated integral by using polar coordinates and show that 1 = j;j2. Thusverify that erf (x), defined in equation (13), Seco68, tends to unity as x tends to infinity.

20. Derive the integration formula (9~ Seco 68, by first writing

y(x) = (' exp (-a2a) cos ax da'0

(a> O)

and differentiating the integral to find y'(x). Then integrate the new integral by parts to showthat 2ay'(x) = -xy(x), point out why

l

Ay(O)=- -2 a

(see Problem 19), and solve for y(x). The desired result is the value of y when x = b.t

t Another derivation is indicated in Churchill, Brown, and Verhey (1974, p. 191), listed inthe Bibliography.

CHAPTER

EIGHTBESSEL FUNCTIONS AND APPLICATIONS

70. Bessel's Equation

In boundary value problems that involve the laplacian V2u expressed incylindrical coordinates, the process of separating variables often producesan equation of the form

d2 d

p2 dp;' + Pd; + (2p2 - V2)y= O,

where y is a function of the cylindricalcoordinate p. In such problems,as weshall observe in examples presented in this chapter, - 2 is the sep,arationconstant and the values of 2 are eigenvalues associated with equation (1).The parameter v isa real number determined by other aspects of the bound-ary value problem. Since v is squared where it appears in equation (1), it issufficient to solve that equation with the agreement that v ~ o. Usually v iseither zero or a positive integer.

In our applications it turns out that 2 ~ O; and, when 2 > O, thesubstitution

(1)

x = flp

can be used to transform equation (1) into a form that is free of 2:

(2) x2y"(X) + xy'(x) + (x2 - V2)y(X)= o.This linear homogeneous differential equation is known as Bessel's equation.Its solutions are called Bessel functions, or sometimes cylindrical functions.

Upon comparing equation (2) with the standard form

y"(x) + A(x)y'(x) + B(x)y(x) = O,

we see that A(x) = l/x and B(x) = 1 - (v/xf Those coefficients are bothcontinuous except at the origin, which is a singular point of Bessel'sequa-tion. The existence and uniqueness theorem stated in Seco18applies to theequation for any elosed bounded interval that does not inelude the origino

174

SECo71] BESSELFUNCTIONS AND APPLICATlONS 175

But for boundary value problems in regions interior to circles or cylindersp = e,the originx = 0,correspondingto thecenteror axisp = 0,is interiorto the region. The interval for the variable x then has the origin as an endpoint.

We limit our attention primarily to the casesv = n where n = 0, 1,2,....In those cases we shall see that there is a solution of Bessel'sequation whichis ana/ytic for all va/ues of x, including the origin; that is, one solution isrepresented by a power series in x which is convergent for every x. Thatsolution,.denoted by Jn(x), and its derivatives of all orders are thereforeeverywhere continuous functions. In referring to any power series, weagree that we shall mean a Maclaurin series, or a Taylor series aboutthe origino

71. Bessel Functions J n

We let n denote any fixed nonnegative integer and seek a solution ofBessel's'equation

(1) xZy"(x) + xy'(x) + (XZ- nZ)y(x) = ° (n = 0,1,2, oo.)

in the form of a power"seriesmultiplied by xP,where the first term in thatseries is nonzero and p is some constant. That is, we attempt to determine pand the coefficientsaj so that the function

(2)00 00

y = xP I ajxj = I ajxp+jj=O j=O

(aO'" O)

satisfies equation (1).tAssume for the present that the series is differentiable. Then, upon sub-

stituting the function (2) and its derivatives into equation (1), we obtain theequation

00

I [(p + j)(p + j - 1) + (p + j) + (XZ- nZ)]ajxp+j = 0,j=O

or00 00

I [(p+ N - nZ)ajxj + I ajxj+ Z = o.j=O j=O

The last sum can be written00

Iaj-zxj;j= Z

J .

t The series method used here lo solve equation (1) is often referred lO as the method ofFrobenius and is Irealed in introductory texts on ordinary dilTerential equations. such as theone by Boyce and DiPrima (1977) or the one by Rainville and Bedient (1974). Both ofthese arelisted in the Bibliography.

\ .


thus

(3) (p - n)(p + n )aO + (p - n + 1)(p + n + l)a 1x

co

+ I [(p - n + j)(p + n + j)aj + aj-2]xj = O.j=2

Equation (3) is an identity in x if the coefficient of each power of xvanishes.This condition is satisfied if p = n or p = - n, so that the constantterm vanishes,and if al = Oand

(p - n + j)(p+ n + j)aj + aj- 2 = O

We make the choice p = n. Then the recurrence relation

(j = 2,3, ...).

(4) a.- -1J - .J(2n + j) aj- 2

(j = 2, 3, ...)

is obtained, giving eachcoefficient in terms of the secondone precedingit inthe series. Note that 2n + j ..¡:O for alI j in relation (4); when n"¡: O, thechoice p = - n fails to give such a recurrence relation.

Since al = O,relation (4) requires that a3 = O; then as = O,and so on.That is,

(5) a2k+1 = O (k = O,1, 2, ...).

To obtain the remaining coefficients,we let k denote any positive integerand use relation (4) to write the following k equations:

-1

a2 = l(n + 1)22ao,

-1

a4 = 2(n + 2)22a2,

-1

a2k= k(n + k)22a2k-2'

Upon equating the product of the left-hand sidesof theseequations to theproduct of their right-hand sides,we find that

(6)( -l)ka - a

2k- k!(n + l)(n + 2)'" (n + k)22k o(k = 1,2, ...).

The coefficient ao may have any nonzero value. To simplify expression (6),we write

1

ao = n! 2n .

SECo71] BESSEL FUNCTIONS AND APPLlCATIONS 177

The nonvanishing coefficients are then

(7)(-l)k 1

a2k= k!(n+k)!2n+2k(k = 0,1,2, o..),

where we use the convention that O! = 1.

Our proposed solution (2) ofBessel's equation (1) can now be written, inview of expressions (5) and (7), as

y = Jn(x)

where Jn is the Besse/function of the first kind of order, or index, n (n = O,1, 2, .. .):

~ (-l)k (x

)

n+2k

Jn(x?= k~ok!(n+k)! "2

~

[

X2 0-- 1- +-2nn! 22l!(n+l) 242!(n+l)(n+2)

X6

]

- .. +... .

The power series (8) is absolutely convergent for all x (- 00 < x < (0),according to the ratio test. It therefore represents a continuous functionandis dilTerentiablewith respect to x any number of times. Since it is dilTeren-tiable and its coefficientssatisfy the recurrence relation needed to make itssum satisfyBessel'sequation (1), Y= Jn(x) is indeed a solution of thatequation.

(8)

Theorem l. For all x, the ana/yticfunction Jn(x) is a so/ution of Bessefsequation (1).

Since Bessel's equation is linear and homogeneous, the function CJ n(x),where C is any constant, is also a solution.

From expression (8) we note that

(9) Jn(-x) = (-Ir Jn(X);

that is, J n is an even function if n = O,2, 4, ..., bu! odd ifThe series representation

. X2. X4 X6

Jo(x) = 1 - 22+ 2242 - 224262 +...

n = 1,3, 5, ....

(10)

bears some resemblance to the power series for cos x. There is also a similar-ity between the power series representations of the odd functionsJ 1(x) andsin x. Similarities between the properties of those functions inelude, as we

178 FOURIER SERIESAND BOUNDARY VALUE PROBLEMS [SEC. 72

shall see, the differentiation formula Jó(x) = -J dx), corresponding to theformula for the derivative of cos x. Graphs of Jo and J I will be shown inSeco 77.

72. Some Other Bessel Functions

Functions linearly independent of J n that satisfy Bessel's equation

(1) x2y"(X) + xy'(x) + (X2 - n2)y(x) = O (n = 0,1,2, ...)

can be obtained by various methods of a fairly elementary nature.The singular point x = Oof equation (1) belongs to the class of such

points known as regular singular points. The power series procedure, ex-tended so as to givegeneral solutions near regular singular points, applies toBessel's equation. We do not give further details here but only state theresults.

When n = O,the general s'olution is found to be

(2) y = AJo(x)

[

X2 X4

(

1

)X6

(1 1

) J+ B Jo(x ) log x + - - - l + - + - l + - + - -."

22 2242 2 224262 2 3 '

where A and B are arbitrary constants and x > O.Observe that as long asB =fO, any choice of A and B yields a solution which is unbounded as xtends to zero through positive values. Such a solution cannot therefore beexpressed as a constant times Jo(x),which tends to unity as x tends to zero.So Jo(x) and the solution (2)are linearly independent when B =fO.It is mostcommon to use Euler'sconstanty = 0.5772..., which is defined as the limitof the sequence

(3)1 1 1

s = 1 + - + - + ... + - - lognn 2 3 n (n = 1,2, ...),

(4)

and write A = (2/n)()'- log 2), B = 2/n. When A and B are assigned thosevalues, the second solution that arises is Weber'sBesselfunctionofthe secondkind %rder zero:

Yo(x) = ~ [(IOgi + Y)Jo(X)

x2 X4

(

1

)X6

(

1 1

) ]+ 22 - 2242 1 + 2 + 224262 1 + 2 +"3 -'" .t

t There are other Bessel functions, and the notation varies widely throughout the literature.

The treatise by Watson (1952) that is listed in the Bibliography is, however, usually regarded asthe standard reference.

SECo 72] BESSEL FUNCTIONS AND APPLlCATIONS 179

More generally, when n has any one of the values n = O,1,2, . .., equa-tion (1)has a solution y"(x)which is valid when x > Oand is unbounded as xtends to zero. SinceJn(x) is continuous at x = O,Jn(x)and y"(x)are linearlyindependent; and when x > O,the general solution of equation (1) is

(5) y = CJn(x) + DY,,(x) (n = 0,1,2, ...),

where e and D are arbitrary constants. The theory of the second solutionY"(x) is considerably more involved than that of Jn(x),and we shalllimit ourapplications to problems where it is only necessary to know that Y,,(x)isunbounded on any interval (O,e).

To write the general solution of Bessel'sequation

(6) x2y"(X) + xy'(x) + (X2- V2)y(X)= O (v> O, 1'=1= 1,2, ...),

where v is any positive number other than 1, 2, . . ., it is necessary to knowsome elementary properties of the gammafimetion, defined when v > Obymeans of the equation

(7.)~

f(1')= r e-'tV-1 dt'0(v> O).

An integration by parts shows that

(8) f(1' + 1) = vf(v)

when v > O.That property is assigned to the function when v < O,so thatf(v) = f(v + 1)/1'when -1 < v < O, -2 < v < -1, and so on. Thus equa-tions (7) and (8) together define f( v) for alll' except v = O,- 1, - 2, . . .. Wefind from equation (7) that f( 1)= 1; also, it can be shown that r is contin-uous when v> O.It then follows from property (8) that f(0+) = 00 andconsequently that If( v) Ibecomes infinite as l' -> - n (n = 1,2, ...).

When 1'= 1,2, 3, ..., f(1') reduces to a factorial; specifically,

(9) f(n + 1) = n! (n = O, 1,2, ...).

\!.{¡

The proof of property (9) and the further property that

rH) = fi

I¡I

is left to the problems.We define the Besselfunetion of thefirst kind of order v (v;;:::O) as

~ (-l)k (X

)"+2k

(10) Jv(x) = Jo k!f(v + k + 1) 2 '

and we note that this becomes expression (8), Sec.71, for Jn(x) when1'= n = O, 1,2, The Bessel funetion J -v(x) (v> O) is also well-definedwhen v is replaced by - v in equation (10); if v is a positive integer, the termsin the express ion for J - v(x) where the argument (- v + k + 1) of r is


zero or a negative integer are to be dropped from the series.It is not difficultto verify by direct substitution into equation (6) that J v and J - v are solu-tions of that equatioq. These solutions are arrived at by a modification,involving property (8) of the gamma function, of the procedure used inSeco71.

When v > Oand v #- 1,2, oo.,the Bessel function J - \.(x) is the productof l/xv and a power series in x whose initial term (k = O)is nonzero; henceJ - v(x) is unbounded as x -+O. Since J v(x) tends to zero as x -+O, it isevident that J,. and J -v are linearly independent functions. The generalsolution of Bessel'sequation (6) can therefore be written

(11) y = CJv(x) + DJ -v(x) (v> O, v'" 1, 2, oo.),

where C and D are arbitrary constants. [Contrast this with solution (5) ofequation (1).]

It can be shown that Jn and J -n are linearly dependent because

(12) J -n(x) = (-I)nJn(x) (n = 0,1,2, oo.)

(see Problem 7, Seco73).So if v = n = O,1,2,..., solution (11) cannot be thegeneral solution of equation (6).

73. Recurrence Relations

Starting with the equation

-n 1 co (-lt (X

)2k

X Jn(x)=2n k~ok!(n+k)! 2(n = 0,1,2, oo.),

let us write

d -n 1 co k( -1)k (X

)2k-1

dx [x Jn(x)] = 2n k~l k(k - 1)!(n + k)! 2" .

Ir we replace k by k + 1here, so that k runs from Oto OCJagain, it follows that

d -n -n(x

)n co (-I)k+ 1

(X

)2k+ 1

dx[X Jn(x)]=x 2" k~ok!(n+k+l)! 2"

-n CO (-I)k (x

)n+l+2k

=-x k~ok!(n+l+k)! 2" '

or

(1)d

dx [x-nJn(x)] = -x-nJn+ I(X) (n = O,1,2,.. .).

SECo73] BESSELFUNCTIONS AND APPLICATIONS lSI

The special case

(2) Jó(X)= -J¡(x)

was mentioned at the end of Seco 71.

Similarly, from the power series representation of xnJn(x), we can showthat

(3)ddx (xnJn(x)]= xnJn-l(X) (n = 1,2, oo.).

Relations (1) and (3), which are called recurrence relations, can bewritten

xJ~(x) = nJn(x) - XJn+l(X),

xJ~(x) = -nJn(x) + xJn-¡(x).

Eliminating J~(x) from these equations, we find that

(4) xJn+ ¡(x) = 2nJn(x) - xJn-¡(x) (n = 1,2, oo.).

This recurrence relation expresses J n + 1 in terms of the functions J n and J n- ¡

of lower orders.From equation (3) we get the integration formula

(5) (5"Jn-l(S) ds = xnJn(x)o(n = 1,2, ...).

An important special case is"

f sJo(s)ds=xJ1(x).o

Relations (1), (3),and (4)are valid when n is replaced by the unrestrictedparameter v. Modifications of the derivations simply consist of writingr(v + k + 1) or (v + k)r(v + k) in place of (n + k)!.

(6)

PROBLEMS

1. Show ¡hat Jo(O) = 1,J.(O) = o (n = 1, 2, o o 0)0

2. From ¡he series representa¡ion (8), Seco 71, for J. show that

'" 1

(X

)'+ 2k

i-'J.(ix)= t~ok!(n+k)! '2(n = O, 1,2, ...).

The function I.(x) = i-'J.(ix) is the modified Bessellunction olthe first kind olorder noShowthat the series here converges for all x, that I.(x) > Owhen x> O,and that 1.( -.t) = (-1).1 .(x~Also show that since J.(x) satisfies Bessel's equation (1), Seco 71, I.(x) is a solution of themodified equation

xly"(X) + xy'(x) - (XI + nl)y(x) = O.

182 FOURIF.R SERIES AND BOUNDARY VALUE PROBLEMS [SEC. 73

3. (a) Derive the factorial property r(v + 1) = vr(v) or the gamma function when v> O.(b) Show that r(1) = 1 and hence that r(n + 1)= n! when n = O,1,2, ..oo

4. Rcrcr to the rcsult obtained in Problem 19, Seco 69, and show that

rH) = 2 (' e-x' dx= JIt.o

5. Veriry that the runction J, (v ~ O) defined by equation (10), Seco 72, satisfies Bessel'sequation (6) in that section. Point out how it follows that J -, is also a solution.

6. From definition (10), Seco 72, or J, show that

ff .(a) J'/2(X)= -SIOX;11:X (b) J -1/2(X) = fT cos X.V11:X

7. Using series (10), Seco 72, and recalling that certain terms are to be dropped, show that

J -.(x) = (-I)"J.(x) (n = 0,1,2, oo.)

and hence that the runctions J. and J -. are linearly dependen!.8. Derive relation (3), Seco73.9, Establish the dilTerentiationformula

x2J:(x) = (n2 - n - X2)J.(X)+ xJ.. ,(x) (n = 0,1,2, oo.).

lO. Derive the diITerentiation formula

d

dx [x-'J,(x)] = -x-"J,.. ,(x),

where v ~ O, iind point out why it is also valid when v is replaced by - v (v > O). [Comparerelation (1), Seco 73.]

11. Use results in Problems 6 and 10 to show that

J2(Sinx

)J J/2(X) = 11:X -:;- - cos x .

12. Let y(x) denote a solution or Bessel's equation (1), Seco 72, when 11= O.

(a) Put that special case orBessel's equation in self-adjoint rorm L[y(x)] = O (see Seco31)and, by writing X = Jo and Y = .1' in Lagrange's identity [Problem 17(a), Seco 34], show that

<1

r

y(x)

J

B

<Ix Jo(x) = x(Jo(xW

where B is some constan!.

(b) Assuming that the function 1/[Jo(xW has a Maclaurin series expansion of the rorm

1 "

[Jo(xW= 1+ LC,x2't,-,

t This valid assumption is most easily justified by methods from the theory orrunctions or acomplex variable. See. ror instance. Churchill, Brown. and Verhey (1974. pp. 164-165).listed inthe Bibliography.

I

ISECo73] BESSELFUNCTfONS AND APPLICATlONS 183

~

fand that the resulting expansion obtained by multiplying each side by l/x can be integratedterm by termo use the result in part (a) to show formally that y can be written in the form

y = AJo(x) + B [Jo(X) log x + .t/.X2. J

JJ

where A, B. and d. (k = 1, 2. ...) are constants. [Compare equation (2), Seco72.]

13. Since the two Bessel functions Jo(x) and Yo(x) are linearly independent, theirWronskian

I

Jo

W[Jo(x), Yo(x)] = J'o

Yo

I= Jo Y'o - J'o Yo

Y'o

never vanishes when x> O.t Find a simple expression for that Wronskian in thefollowing way.(a) By writing y(x) ==Yo(x) in the result obtained in Problem 12(a). show that

eW[Jo(x). Yo(x)] = ~ (x >0)

where e is some constant.

(b) Use expression (4). Seco72, for Yo(x) to show that

2 2W[Jo(x), Yo(x)] = - [Jo(xW + - [Jo(x)S'(x) - J'o(x)S(x)]1tX 1t

where

X2 X4

(1

)x.

(l 1

)S(x)= 22- 2W l + 2 + 224262l + '2+ 3 -... (x > O).

Thus show that

. 2hm xW[Jo(x). Yo(x)] =-.x-o nx>o

(e) Conclude from part (b) that the constant e in part (a) has the value e = 2/1tand hencethat

2W[Jo(x). Yo(x)] =- 1tX (x> O).

14, Let -'R(n = 1,2. ...) be the sequence defined in equation (3). Seco 72. Show that SR> O

and SR- SR. 1 > O for each n. Thus show that the sequence is bounded and decreasing and that

it therefore converges to some number y. Also. point out how it follows that O S; y < 1.

Suggestion: Observe from the graph of the function y = l/x that

,

(

R-1 l .RdxL: - > I - = log n

.= 1k . I X(n ;;: 2)

and

1 .R. I dx-< I -= log(n+ I)-Iognn + 1 'R X

(n;;: 1).

t See. for example. one of the texts mentioned in the footnote in Seco 71 for a discussion of

the Wronskian. which plays an important role in the theory of ordinary difTerential equations.


15. (a) Derive the retluctionformula

(~Jo(s)ds=x'J,(x)+(n-1)x"-'Jo(x)-(n-1)2 (s"-2JO(s)tI.~'0 '0

(n = 2.3. ...)

by using integration by parts twice and using the relations (Sec. 73)

ti- [sJ.(s)) = sJo('~)'ds

d

~Jo(s)= -J.(s)

in the !irst and second of those integrations, respectively.

. (b) Note that. in view of equation (6). Seco 73. the identity obtained in part (a) can beapplied successively to eva1uate the integral on the left-hand side of that identity when theinteger n is odd. t lIIustrate this by showing that

( SSJo(s)ds = X(X2 - 8)[4xJ o(x) + (X2 - 8)1I(x)).o

74. Bessel's Integral Form of J n

Consider two power series converging for every real number x, with sumsf(x) and g(x):

(1)00

I amxm=f(x),m=O

00

I bmxm = g(x).m=O

The Cauchy product

aobo + (aobt + at bo)x + (aobz + at b¡ + azbo)xZ + ...

of those series also converges for every x and represents thef(x)g(x); that is,

product

(2)00

I cmxm= f(x)g(x)m=O

m

where Cm= I ajbm- j'j=O

This is true, moreover, when the coefficients am and bm are complexnumbers.t

In particular, since00

(i/2)m

(iX .s)m~o~eimSxm = exp Te'

t Note, too, that when n is even, the reduction formula can be used to transform theproblem of evaluating J~s"Jo(s) ds into that of evaluating J~Jo(s) ds. which is tabulated forvarious values of x on, for example, pp. 492-493 of the book edited by Abramowitzand Stegun(1965) that is listed in the Bibliography. Further references are given on pp. 490-491 of thatbook.

t See, for example, Taylor and Mann (1972, pp. 635 and 662) and a1so Churchil~ Brown.

and Verhey (1974, pp. 164-165), both listed in the Bibliography.

rSECo74] BESSEL FUNCTIONS AND APPLICATIONS 185

and

f (i/2t ei(n-m)Oxm= exp (iX e-W)einO,

m=O m. 2.'

where Ois independent of x and where n is a fixed nonnegative integer, itfollows that

(3)<Xc

I cm(O)xm = exp (ix cos O)einOm=O

r where

(4)'m m l

C (O)=~ I ei(n-m+2j)Om 2mj=oj!(m-j)! .

We shall now show that, except for a multiplicative constant, the inte-gral from - n to n with respect to O of the right-hand side of equation (3) isthe Bessel function Jn(x). To do this, we observe first that the series on theleft-hand side of that equation is uniformly convergent with respect to Oonthe interval -n:::; O :::;n. For expression (4) and the binomial expansionformula allow us to write

ICm(O)xml:::;(M)m~ I m! =

(M )m(l + 1)m_Ixl m

2 m! j=oj!(m-j)! 2 m! ---;;;!'

and the Weierstrass M-test can be applied. Term by term integration of theseries in equation (3) over the interval -TC:::;O:::;n is therefore justified.That is,

(5)'X. .ft . n

I I cm(O)dOxm= I exp (ix cos O)énOdO.m=O "-n "-1(

The integrals on the left-hand side of this last equation can be evaluatedas follows. Using expression (4), write

(6).n im m l n

l e (O)dO =- I rei(n-m+2j)O dO

'-nm 2mj=oj!(m-j)!'-n .

Since

n

(

O

f éN9 dO=- n 2n

if N = :t 1, :t2, ...,if N=O,

it is evident that all the terms on the right-hand side of equation (6) vanishunless m = n + 2j. So the only integers m for which it is possible that theintegral on the left-hand side of eqtlation (6) isnonzero are those ofthe form


m = n + 2k (k = O,1,2, . ..). When m is such an integer, all the terms on theright-hand side of equation (6) vanish except when j = k. To be precise,

fn 21t

(i

)n+ 2k

(7) -nCn+2k(0)dO=kT6~'+k)! 2 (k=0,1,2,...).

Equation (5) therefore reduces to

co (-1t(

x

)n+2k n .

21tinI kl( k)1 '2' = r exp(ixcosO)ern8dO,k=O . n + . . - n

or

Jn(X)= i2-n rn exp (ix co~ 0)ein8dO (n = 0,1,2, ...).1t . - n

Now i = exp (i1t/2). Hence

Jn(x)= 2~(n exp (ix cosO)exp fin(o - n J dO.

Ir we make the substitution 4>= (1t/2) - Oand note that the resulting inte-grand is periodic in 4>with period 21t (see Problem 15, Seco42), we find that

1 .n .Jn(x) = 21t Lnexp [i(x sin 4>- n4»] d4>.

(8)

(9)

The imaginary part of this integral vanishes, of course, since Jn(x) is realwhen x is real. Therefore

1 .n

J n(x) = 21tL n cos (x sin 4>- 114» d4>.

Finally, since the integrand here is an even function of 4>,

1 .n

Jn(x) = - I cos (n4>- x sin 4» d4>1t.0

This is Bessel's integral form of J n(x).

(10) (n = 0,1,2, oo.).

75. Consequences of Bessel's Integral Form

From the integral representation

(1)1 .n

Jn(x) = -1 cos (n4> - x sin 4» d4>1['0

(n = O. l. 2, .00)

just obtained, it follows that

J~(x) = ! ( sin (n4>- x sin 4» sin 4>d4>.1t.o

Continued differentiation yields integral representations for J~(x), and so on.In each case the absolute value of the integrand that arises does not exceedunity. The following boundedness properties are then consequences of Bes-sel's integral form (1):

(2) IJn(x)1~ 1,I

dk

dXk J n(x) I ~ 1(-00 < x < 00, k = 1,2, oo.).

For example, the first inequality is true because

1 " 1 "I1n(x)I ~ - J Icos (n4>- x sin 4»1 d4>~ - J d4>= 1.

no no

The integrand in expression (1) can be written

cos (n4>- x sin 4» = fn(x,4» + 9n(X,4»

where

j,,(x,4» = cos n4>cos (x sin 4», 9n(X,4» = sin n4>sin (x sin 4».

For each fixed x the graphs ofj" and 9nhave these properties of symmetrywith respect to the line 4>= n/2:

(3)j,,(x,n - 4» = (- 1)nj,,(x,4»,

9n(x,n - 4» = - ( - 1)"9n(X,4».

Consequently." .~2

I f2n(X, 4» d4>= 2 I f2.(X, 4» d4>,.o .o (g2n(X, 4» d4>= o,.o

and therefore

1 ."J 2n(X)= - I cos 2n4>cos (x sin 4» d4>,n .o

or

(4)2 .,,/2

J 2n(X)= - I cos 2n4> cos (x sin 4» d4>n .o (n = 0,1,2, oo.).

In like manner, we see that

J 2n- ¡(x) = ~ 1"" sin (2n - 1)4>sin (x sin 4» d4>,n .oor

(5)2 .",2

J 2n- 1(X) = - I sin (2n - 1)4>sin (x sin 4» d4>1t .o(n = 1,2, oo.).

(6)

Note the special case2 n/2

Jo(x) = - r cos (x sin cf» dcf>11:.o

of representation (4), which can also be written

2 n/2Jo(x) = - r cos (x cos O) dO11:.o

by means of the substitution O= (11:/2)- cf>.For each fixed x the Riemann-Lebesgue lemma (Lemma 1, Seco 40)

applies to the integrals in expressions (4) and (5) to show that J2.(x) andJ 2.- .(x) tend to zero as n increases. Thus

(7)

(8) lim J.(x) = O (n = O, 1, 2, oo.).'-0<>

Another important property is that, for each fixed n,

(9) lim J.(x) = O (n = 0,1,2, oo.).X-a)

To indicate a method of proving this, we consider the special case n = O.Wesubstitute u = sin cf>in equation (6) to write

e I11:

( ) r

cos xud r

cos xud- J Ox = u + u,

2 .O Jl=-? . e Jl=-?where O < e < 1.The second integral here is improper but uniformly conver-gent with respect to X.Corresponding to any positive number e, the absolutevalue of that integral can be made lessthan e/2, uniformly for all x, byselecting e so that 1 - e is sufficientIy small and positive. The Riemann-Lebesgue lemma then applies to the first integral with that value of C.Thatis, there is a number x, such that the absolute value of the first integral is lessthan e/2 whenever x > x,. Therefore

11:

"2 iJo(x) I <e whenever x > x"

and this pro ves property (9) when n = O.The proof for the other cases is leftto the problems.

PROBLEMS

1. Use integral representations for J. to verify that (a) Jo(O)= 1; (b) J.(O) = O whenn = 1.2 ; (e) Jó(x) = -J,(x).

2. Deduce from expression (4). Seco75, that

2 ./2

J 1.(X) = (-1)' - f cos 2n9 cos (x cos O) dO1t o

w.hen n = O, 1, 2. ....

1

tSEC. 75] BESSEL FUNCTIONS AND APPLICATIONS 189

3. Deduce from expression (5). Seco 75. that

2 . ./2J20- ,(x) = -( -1)"-1 cos(2n- 1)0sin(xcosO)dO11'0

when n = 1.2. ....

4. Complete the proof of property (9). Seco 75. that

lim Jn(x) = O (n = O, \,2. ...).

S. Verify directly from the representation

2 ../2

Jo(X)= -1 cos (x sin eJ»deJ>11'0

that J o(x) satisfies Bessel's equation (1). Seco 71. when n =Othere.6. Apply integration by parts to representations (4) and (5) in Seco75 and then use the

Riemann-Lebesgue lemma (Sec.40) to show that

lim nJn(x) = O (n = O, \,2, ...)

[ for each fixed X.

7. Recall from equation (4). Seco 75. that when n = O, 1. 2. ....

2 . ./2J 2n(X)= -1 COs(x sin eJ»cos 2neJ>deJ>.11'0

Then, by using this expression and Corollary 2 in Seco 43 applied to Fourier cosine serieson the intervalO < eJ>< 11/2.prove that

cos (x sin eJ»=Jo(x) + 2 I J 2n(X) cos 2neJ>,,=1

f

~

for all real x and eJ>.Observe that this is also a Fourier cosine series on the intervalO < eJ>< 11.

8. Show that when n = \,2. ...,

[" sin (x sin eJ» sin 2nq, deJ>= O..o

Then, using the Fourier sine series for the function sin (x sin q,) on the intervalO < eJ>< 11,prove that

fAsin (x sin q,) = 2 I J 2n- ,(x) sin (2n - 1)q,

,,==1

for all real x and eJ>.

9. According to Seco 46. if a function / and its deriva ti ve .f' are both continuous on theinterval -11 ~ X ~ 11and if/(-1I) =/(11). then Parseval's equation

I .' 1 '"-1 (f(x)j2 dx = -a5 + I (a; + b;)1t. -'1( 2 n=1

holds, where the numbers un (n = O, 1,2, ...) and bn (n = \,2, ...) are the Fourier coefficients

\ .'an = -:-1 [(x) cos l/X dx,

x. -'1(

\ .b. = - f [(x) sin nx dx.x. -ft:

f¡,

190 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS [sEc.75

(a) By applying that resulI to f(¡P) = cos (x sin ¡P).an even function of ¡P.and referring tothe Fourier (cosine) series for f(¡P) in Problem 7. show that

1 ,/1 J.

-. I COS2 (x sin ¡P) tl¡p = (Jo(xW + 2 ¿: [l2.(XWn. o n=I(-x < x < x).

(b) Similarly. by writing f(¡P) = sin (x sin ¡P) and referring to the Fourier (sine) seriesexpansion in Problem 8. show that

1 .11 X

- I sin2 (x sin ¡P)tl¡p= 2 ¿:[l2.- [(xW1t.o tI=l

(-::c<x<oc).

(e) Combine the results in parts (a) and (h) to show that

(Jo(xW + 2 ¿: [l.(xW = In=1

(-x <x<oo).

and point out how it follows from this identity that

1.2. ...) for all x.

10. Since the series in the cxpansions

Ilo(x)! S l and ¡l.(x)! S I/.fi (n =

(XI

) 'xi iexp-2 =.&i!2il.)-0.

exp (-X

)= i!.=.I)'x'_,21 '=0 k!2' l

are absolutely convergent when x isany real number and l *O.the product ofthe two functionsrepresented is itself represented by the series formed by multiplying each term in one series byevery term in the other and then arranging the resulting terms in allY oI'Cler.tBy noting that eachof the terms in the resulting series can be written as a function of x times either l. (n = O. 1.2. ...)or 1-. (n = 1.2. ...) and arranging those terms according to powers of l. show that

exp[

: (1 - ~)j= 1 o(x) + i: [l .(x )1. + ( - 1).1 .(X)I -.]2 l .=[

(1,.0).

Also. write this result in the form

eXP[~ (I-~ )j = lim ~l.(X)I.... t "'-1. rI"-m

(1 #0).

The exponential function here is said to be a generalillg .I;/II('liollfor the Bessel functions l.(x)(n = O. :!: l. :!:2. ...).

Suggeslioll:Observe that the coefficientof 1"(11= n. 1.2. ...) can be obtainedby multi-plying the kth term of the second series by ¡he term of the first series whose index is j = 11+ kand then summing over k from O to x. Similarly. ¡he coefficient of 1-. (11= 1. 2. ...) can beobtained by multiplying thejth term of the first series by the term ofthe second series with indexk = n + j and summing over j from O to x.

11. By writing l = ei~ in the identity found in Problem 10and noting that its left-hand sidebecomes exp (ix sin ¡P).obtain the Fourier series expansions in Problems 7 and 8.

t For a justification of this procedure. see. for example. Taylor and Mann (1972.pp. 635-636). listed in theBibliography.

j(

I

1

t

I

IIf

)

/'

SECo76] BESSELFUNCTIONS AND APPLICATIONS 191

76. The Zeros oí Jo(x)

A modified form of Bessel'sequation

(1) x2y"(X) + xy'(x) + (x2 - 1,2)y(X)= °in which the term containing the first derivative is absent is sometimesuseful.That form is easily found (Problem 1, Seco78) by making the sub-stitution y(x) = XCu(x)in equation (1) and observing that the coefficientofu'(x) in the resulting differential equation in u,

(2) X2U"(X)+ (1 + 2c)xu'(x) + (X2 - y2 + C2)U(X) = 0,

is zero if c = -1. The desired modified form of equation (1) is then

(3)X2U"(X)+ (X2 - y2 + ~)II(X) = 0,

and the function 11= JxJ,.(x) is evidently a solution of it. In particular,when y = 0, we see that the function 11=JxJ o(x)satisfiesthe differentialequation

(4)1I"(X) + (1 + 4~2)II(X)=0.

Observe now that the differential operator L = ¡P/dX2 is self-adjoint(Sec.31) and that Lagrange's identity [Problem 17(0),Seco34] for that op-erator is

(5) U(x)V"(x) - V(x)U"(x) = _id (UV' - U'V),¡X

where U(x) and V(x) are any functions whose first and second derivativesexist. We write

(6) U(x) = JXJo(x)

and note that if V(x) is chosen so that V"(x) = - V(x), identity (5) becomes

-[U"(x) + U(x)]V(x) = ¡;~ (UV' - U'V).

But, since the function 11= U(x) satisfies equation (4), we know that- [U"(x) + U(x)] = U(X)/(4X2).Hence if V satisfiesthe condition V" = - Vand if ° < o < h, then

(7) (U(.:.~~(x) dx = [U(x)V'(x)- U'(x)V(x)]:.a

With the aid of identity (7), we shall now show that the positive zeros ofJ o(x), or roots q{ the equation J o(x) = OJorm an infinite sequence of numbersXj (j = 1,2, ...) such that Xj-+ 00 as j -+ oo.t -

First we show that our function U(x) = JxJo(x), and hence Jo(x), hasat least one zero in each interval 2kn ::; x ::; 2kn + n (k = 1, 2, .. .). We dothis by assuming that U(x) =/= Oanywhere in the interval2kn ::;x ::;2kn + nand obtaining a contradiction. According to that assumption, eitherU(x) > O for all x in the interval or U(x) < O for all such x, since U(x) iscontinuous and it cannot therefore change sign without having a zero valueat some point in the interval.

Suppose that U(x) > O when 2kn::; x ::;2kn + n. In identity (7) writea = 2kn, b = 2kn + n, and V(x) = sin x. Since V(a) = V(b) = O, V'(a) = 1,and V'(b)= -1, that identity becomes

.2kn+n sin xI U(X)-4 2 dx = -[U(2kn) + U(2kn + n)].. 2kn X

(8)

The integrand here is positive when 2kn < x < 2kn + n. Hence the left-handside of this equation has a positive value while the right-hand side is nega-tive, giving a contradiction.

Ir, on the other hand, U(x) < O when 2kn ::; x ::; 2kn + n, those twosides of equation (8) are negative and positive, respectively. This is again acontradiction. Thus Jo(x) has at least one zero in each interval 2kn ::;x ::;2kn + n (k = 1, 2, .. .).

Actually, J o(x) can have at most a finite number of zeros in any cIosedbounded interval a ::; x ::; b. To see that this is so, we assume that theinterval a::; x::; b does contain an infinite number ofzeros. From advancedcalculus we know that if a given infinite set of points lies in a cIosed boundedinterval, there is always a sequence of distinct points in that set whichconverges to a point in the interval.t In particular, then, our assumption thatthe interval a ::; x ::; b contains an infinite number of zeros of J o(x) impliesthat there exists a sequence xm (m = 1, 2, ...) of distinct zeros such thatxm -+ e as m -+ 00, where e is a point which also lies in the interval. Since thefunetion Jo(x) is eontinuous, Jo(c)= O;and, by the definition of the limit ofa sequenee, every interval eentered about e eontains other zeros of J o(x). Butthe faet that Jo(x) is not identieally zero and has a MacIaurin series rep-resentation whieh is va lid for all x means that there exists some interval

centered at e whieh contains no other zeros.§ Sinee this is eontrary to what

tOur method is a modification of the one used by A. Czarnecki, Amer. Math. Month/y,vol. 71, no. 4, pp. 403-404, 1964, who considers Bessel functions J ,,(x) where -1 ~ v ~ 1-

~ See, for example, Taylor and Mann (1972, pp. 542-547), listed in the Bibliography.

§, That is, the zeros of such afunction are iso/ated. The argument for this is given in asomewhat more general setting by Churchill, Brown, and Verhey (1974, p. 167), listed in theBibliography.

SECo77] BESSEL FUNCTIONS AND APPLICATIONS 193

Table 1. Jo(Xj) = O

I

I

has just been shown, the number of zeros in the interval a ~ x ~ b cannotthen be infinite.

It is now evident that the positive zeros of J o(x) can, in fact, be arrangedas an infinite sequence of numbers tending to infinity. Table 1 gives thevalues, correct to four significant figures, of the first five zeros of J o(x) andthe corresponding values of J I(x). Extensive tables of numerical values ofBessel and related functions, together with their zeros, will be found in bookslisted in the Bibliography. t

77. Zeros of Related Functions

If for some pair of positive numbers a and b it is true that J n(a)= OandJn(b)= O,then x-n Jn(x)also vanisheswhen x = a and when x = b. It followsfrom Rolle's theorem that the derivative of x-nJ n(x)vanishesfor at least onevalue of x between a and b. But (Sec. 73)

dd)x-nJn(x)] = -x-nJn+I(X) (n = 0,1,2, oo.);

and so, whenn = O,1,2, .. ., there is at leastone zeroof Jn+ I(x) betweentwo positive zeros of J n(x). Also, just as in the case of J o(x) (Sec. 76), J n+ I (x)can have at most a finite number of zeros on each bounded interval.

We have already shown that the positive zeros of J o(x) constitute anunbounded infinite sequence of numbers. It now follows that the zeros ofJ I(X) must form such a sel. The same is then tfUe for J 2(X),and so on. Thatis, for each fixed nonnegative integer n, the set of all positive roots of theequation J n(x) = O forms an infinite sequence x = Xnj (j = 1, 2, ...) whereXnj- 00 asj- oo.

Graphs of Jo(x) and J I (x) are shown in Fig. 26.The function y = Jn(x) satisfies Bessel's equation, which is a linear ho-

mogeneous differential equation of the second order with the origin as asingular poinl. According to the uniqueness theorem stated in Seco 18,thereis just one solution that satisfies conditions y(c) = y'(c) = O where e> O;

t See the book edited by Abramowitz and Slegun (1965) and the ones by Jahnke, Emde, andLosch (1960~Oray and Mathews (1966),and Watson (1952),all1isted there.

j 1 2 3 4 5

Xj 2.405 5.520 8.654 11.79 14.93J,(Xj) 0.5191 - 0.3403 0.2715 -0.2325 0.2065

x

-0.5

Fi~ure 26

tbat solution is identically zero. Consequently, tbere is no positive number csuch that J n(C)= J~(c)= O.That is, J~(x)cannot vanish at a positivezero ofJn(X); thus Jn(X) must change its sign at that point.

Let a and b (O< a < b) be two consecutive zeros of Jn(x). If J~(a) >0,tben Jn(x) > O when a < x O.So the values of J~ alternatein sign at consecutive positive zeros of J no

We now consider the function hJn(x) + xJ~(x), where h;;::O andconstant. Its zeros will also arise in certain boundáry value problems. If aand b are consecutive positive zeros of J n(x), the function hJn(x) + xJ~(x)has the values aJ~(a) and bJ~(b) at tbe points x = a and x = b, respectively.Since one of tbose values is positive and the other negative, the functionvanishes at some point, or at some finite number of points, between a and b.It therefore has an infinite sequence of positive zeros. t We collect our princi-pal results as follows.

Theorem 2. For eachfixed n (n = O,1,2, . . .), the set 01all positive roots 01the equation

(1) Jn(x) = O

consists 01an infinite sequence x = Xnj (j = 1, 2, ...) such that Xnj-+ 00 asj -+ 00; also, the set 01all positive roots 01the equation

(2) hJn(x) + xJ~(x) = O (h ;;::O),

where h is a constant, is always a sequence 01 that type.

Observe that x = O is a root of botb equations (1) and (2) if n is apositive integer. It is also a root of the equation Jó(x) = O.

t In the important special case when n = O. the first six zeros are tabulated for various

positive values or h in. ror example. Appendix IV or the book on heat conduction by Carslawand Jaeger (1959) that is listed in the Bibliography.

1

¡

1

)

~

{

I

SECo 78] BESSELFuNcrlONS AND APPLlCATIONS 19S

Ir x = e is a root of equation (1), then x = - e is also a root sinceJ .( - e) = (- 1).J .(e). That statement is true of equation (2) as well; for, inview of the recurrence relation (Sec. 73)

xJ~(x) = nJ.(x) - xJ.+ I(X),

equation (2) can be written

(3)and we note that

(h + n)J.(-x) - (-X)J.+I(-X) = (-I).[(h + n)J.(x)- xJ.+¡(x)].

(h + n)J.(x) - xJ.+ ¡(x) = O,

Finally, although our discussion leading up to Theorem 2 need nothave excluded the possibility that h be negative, those values of h willnot arise in our applications.

,

II

I

I

I

)

78. Orthogonal Sets of Bessel Functions

As indicated in Seco70, where somewhat different notation was used, thephysical applications in this chapter will.involvesolutions of the differentialequation

(1)2 d2X dX

(2 2

x --+x-+ A.x -n )X=OdX2 dx (n = O, 1,2, ...).

Equation (1) is readily expressed in the self-adjoint form (Sec. 31)d

(dX

) (

n2

)- x -- + AX- - X = O.dx dx x(2)

For each fixed n, this form of equation (1) is a special case of the Sturm-Liouville equation (4), Seco31, where r(x) = p(x) = x and q(x) = -n2/x.

In particular, we shall need to solve the singular (Sec.32) Sturm-Liouville problem, on an interval (O,e),consisting of equation (1) and aboundary condition

(3) b¡ X(e) + b2X'(e) = O,\

\

\

J

!

where it is understood that X and X' are lO be eontinuous on the closed

intervalO::::;x::::;e. The constants b1 and b2 are real and not both zero. Theend point x = Ois the singular point of the differential equation; also, q isdiscontinuous there unless n = O.

In the important special case when b2 = O,the boundary condition (3)is

(4) X(e) = O.

When b2 +- O,we may multiply through condition (3)by e/b2and write it as

(5) hX(e) + eX'(e) = O

196 FOURIER SERIESAND BOUNDARY VALUE PROBLEMS .[SECo78

where h = cb¡/b2. In solving our Sturm-Liouville problem, we shall find itconvenient to use the boundary condition (3) in its separa te forms (4) and(5); and, when using condition (5), we shall always aSSl/me that h ~ O.

Cases (a) of Theorems 3 and 4 in Seco 32 give conditions ensuring thatour singular Sturm-Liouville problem has orthogonal eigenfunctions andreal eigenvall/es A..We note that the proofs of those theorems do not dependon the continuity of q(x) at the end point x = a (a = O in the present case) aslong as X and X' are continuous on the cIosed interval a :::;x :::;b. We nowconsider the three possibilities that A.be zero, positive, or negative.

When A.= O, equation (1) is a Cauchy-Euler equation (see Problem 13,Seco 34):

(6) 2d2X dX 2X dX2 + x dx - n X = O.

To solve it, we write x = eSand put it into the form

(7) d2 X - n2X = O.ds2

If n = 1,2, ..., it followsthat X = Ae'lS+ Be-nS,or X(x) = Axn + Bx-n,where A and B are constants. Since our solution must be continuous, andtherefore bounded, on the intervalO:::;x:::;c, we require that B = O;that is,X(x) = Axn. It is now easy to see that A = Oif either of conditions (4)or (5)is to be satisfied,and we arrive at only the trivial solution X(x) ;: O.Thuszero is not an eigenvalue if n = 1,2, ....

If, on the other hand, n = O, equation (7) has the general solutionX = As + B; and the general solution of equation (6) when n = O is there-fore X(x) = A log x + B. According to the continuity requirements, then,X(x) = B. When condition'14fis imposed, B = O;the same is true of condi-tion (5) when h> O. But when h = O, condition (5) becomes simplyX'(c) = O,and B can remain arbitrary. So if n = O and condition (5) is l/sedwhen h = O, we have the eigenjimction

X(x) = 1 corresponding lO A.= O.

This is the only case in which A.= O is an eigenvall/e.We consider next the case when A.> O and write A.= 0:2(o:> O). Equa-

tion (1) is then

(8)d2X dX

X2 dX2 + x dx + (0:2X2- n2)X= O.

Except for the notation used, it has airead y been pointed out in Seco70 that

SECo 78] BESSEL FUNCTIONS AND APPLlCATlONS 197II

the substitution s = rx.xmay be used to transform equation (8) into Bessel'sequation

(9)d2X dX

S2- + s- + (S2 - n2)X = O.ds2 ds '

and the general solution of equation (8) is evidentiy (Sec. 72)

(10) X(x) = C1Jn(rx.x) + C2 Y,,(rx.x).

Qur continuity requirements imply that C2 = O,since y"(rx.x)is unboundedon the interval (O,c). Hence any non trivial solution of equation (8) whichmeets those requirements must be a nonzero constant multiple of the func-tion X(x) = Jn(rx.x).

In applying the boundary condition at x = c, let us agree thaÚhe symbolJ~(rx.x)stands for the derivative of J n with respect to the argument of J n' Thatis, J~(rx.x)= dlds(Jn(s)]where dlds(Jn(s)]is to be evaluated at s == rx.x. Then

dldx[Jn(rx.x)] = rx.J~(rx.x);and conditions (4) and (5) require that

(11) J n(rx.c)= Oand

(12) hJn(rx.c)+ (rx.c)J~(rx.c) = O,

respectively. Note that since equation (2), Seco77, can be written in the form(3) in that section, equation (12) can also be written

(h + n)Jn(rx.c) - (rx.C)Jn+l(rx.C) = O.

According to Theorem 2, there is an infinite sequence of positive valuesof rx.csatisfyingequation (11)or (12). If we designate such a sequence by xj

(j = 1,2, .. .), but keep in mind that the numbersXj dependon the value of nand also on the value of h in the case of equation (12), it follows thatequation (11) or (12) is satisfied when rx.c= Xj' That is, the positive roots ofeither of those equations have the form

(13) rx.._XjJ-- c (j = 1,2,...).

Qur Sturm-Liouville problem thus has eigenvalues Aj = rx.i (j = 1,2,.. .),andthe corresponding eigenfunctions are

(14) XAx) = Jn(rx.jx) (j = 1,2, ...).

We note that if the numbers rx.jare the positive roots of equation (12)when n = h = O, which is the only case when A = O was found to be aneigenvalue, that equation can be wr.itten

(15) J 1(rx.c)= O.

The numbers exj are then given more directly as the positive roots of equation(15). Also, making a minor exception in our notation, we let the subscriptjrange over the values j = 2, 3, oo.,instead of starting from unity. The sub-script j = 1 is reserved for writing ex¡ = Oand Al = exi= O.This aIlowsus toinc1ude the eigenvalue A¡ = O together with the eigenfunction X ¡(x) =) o (ex¡ x) = 1, obtained earlier for the case n = h = O. Note that it is alsopossible to describe the numbers exj(j = 1,2, oo.) here as the nonnegativeroots of equation (15).

FinaIly, we consider the case when A < O,or A = - {l2 ({l > O),and writeequation (1) as

(16)d2X dX

X2- + x- - ({l2x2+ n2)X= O.dX2 dx

The substitution s = {lx,whichisliketheoneusedto transformequation(8)into Bessel's equation (9), can be used here to put equation (16) into theform

(17)2 d2 X dX 2 2

S - + s - - (s + n )X = O.ds2 ds

From Problem 2, Seco73, we know that the modified Bessel functionX = 1.(s) = i-.).(is) satisfiesequation (17); and, since it has a power seriesrepresentation which converges for aIl x, it satisfies the continuity require-ments in our problem. As was the casewith equation (9),equation (17)has asecond solution which is discontinuous at s = O, that solution being analo-gous to Y".tThus we know that, except for an arbitrary constant factor,X(x) = 1,,({lx).

We now show that for each positive value of {l the function X(x) =1.({lx) fails to satisfyeither ofthe conditions (4)or (5).In each caseour proofrests on the fact that 1.(x) > Owhen x > O,as demonstrated in Problem 2,Seco73.

Since 1.({le)> O when {l> O, it is obvious that condition (4), whichrequires that 1.({lc)= O,fails to be satisfiedby any positive number {l.Also,in view of the alternative form (3),Seco77, of equation (2) in that section,condition (5), when applied to our function X(x) = 1.({lx)= i-.).(i{lx),becomes

(h + n)i-.).(i{lc) + fJci-(.+ 1)).+ ¡(ifJc) = O,or

(18) (h + n)I.({lc) + fJcI.+¡(fJc) = O.

t For a detailed discussion of this, see, for example, pp. 16 ff of the book by Tranter (1969)tisted in the Bibliography.

SECo 78] BESSELFUNCTIONS AND APPLlCATIONS 199

Since {J> O,the left-hand side of this last equation is positive; and, onceagain, no positive values of (Jcan occur as roots. We conclude, then, thatthere are no negative eigenvalues.

We have now completely solved our singular Sturm-Liouville problemconsisting of equations (1) and (3), where, since we have assumed that theconstant h = ebtlb2 is nonnegative, it is understood that the constants b1and b2 in equation (3) have the same sign when neither is zero.

The eigenvaluesare aIl represented by the numbers ..1.)=aJwherethe a)are given by equation (13) and where ..1.}> O,except that ..1.1= Oin the casen = h = O. Let us agree to arrange the numbers IX}in ascendingorder 01magnitudeso that a) < a)+I, and therelore..1.}< ..1.}+1'

Theorem 3, Sec. 32, gives the orthogonality propertye

r xJn(a}X)Jn(akx) dx = O'0(19) U :f=k).

Note that this orthogonality ofthe eigenfunctionswith weight functionx, onthe interval (O,e),is the same as ordinary orthogonality of the functionsJxJn(a}x) on that same intervaI. Also note that many orthogonal sets arerepresented here, depending on the values of n, e, and ho

We summarize our results in the foIlowingtheoremo

Theorem 3. Let n have one of the values n = O, 1, 2, . . oo The set {J.(:x} x)}

(j = 1,2,oo.) is orthogonalon the interval(O,e)withweightfimetionx whellthenumhersa = a} U= 1,2, ...) are either

(a) the positive roots of the equation

J n(ae) = O,

(b) the positive roots of the equatioll

hJ n(ac) + (ae )J~(ae) = O (h ¿ O,h + n > O),

01'(e) (whellll = O) zero alld the positive roots olthe equatioll

Jó(ae) = O.

The numbers A} = a] are the eigenvalues, and Xj = Jn(ajx) are the eorre-spolldillg eigerifulletiolls, of the singular Sturm-Liouvil/e problem eOllsistillgofthe equatioll

d2X dXX2 - + x - + (Ax2- n2)X= O,dX2 dx

whose self-adjoint form is

(20)

d

(dX

) (n2

)- x - + Ax- - X = O,dx dx x

together with the boundary condition

(21)

in case (a), the condition

X(c) = O

(22) hX(c) + cX'(c) = O (h ~ O,h + n > O)

in case (b), or the condition

(23) X'(c) = O

in case (e), the value of n in equation (20) for this last case being zero.

PROBLEMS

1. By means of the substitution y(x) = x'u(x), transform Bessel's equation

x2y"(X) + xy'(x) + (x2 - V2)y(X) = O

into the differential equation

X2U"(X)+ (1 + 2e)xu'(x) + (X2- V2 + e2)u(x)= O,

which becomes equation (3), Seco 76, when e = -1.

z. Use equation (3), Seco 76, to obtain a general solution of Bessel's equation when v = 1.

Then, using the expressions in Problem 6, Seco 73, point out how J 1/2(X) and J -112(X) arespecial cases of that solution.

3. By referring to Theorem 3, show that the eigenvalues of the singular Sturm-Liouvilleproblem

(xX')' + AXX = O

X(2)=0

(O< x < 2),

are the numbers Aj = /X; (j = 1,2, ...) where /Xjare the positive roots ofthe equation J 0(2/X) = O

and that the corresponding eigenfunctions are X j = J o (/XjX) (j = 1,2,. ..). With the aid ofTable

1 in Seco 76, obtain the numerical values /XI = 1.2, /X2 = 2.8, /X3 = 4.3, valid to one decimal place.

4. Write U(x) = jXJ.(/Xx) where n has any one of the values n = O, 1,2, ... and /Xis apositive constan!.

(a) Use equation (3), Seco 76, to show that

(1- 4n2

)U"(x)+ /X2+ 4T U(x)= O.

(b) Let e denote any fixed positive number and write Ujx) = jXJ.(/Xjx) (j = 1,2, ...)where /Xiare the positive roots of the equation J .(/Xe)= O.Use the result in part (a) to show that

(/X] - /Xf)U¡(x)u.(x) = UiU; - U. Uj.

(e) Use the result in part (b) and Lagrange's identity [Problem 17(a), Seco 34] for theself-adjoint operator L = d2fdx2 to show that the set (U¡(x)} (j = 1, 2, ...) is orthogonal on(O,e) with weight function unity. Thus give a new proof that the set {J ,(<Xjx)} (j = 1,2, ...)in case (a) of Theorem 3 is orthogonal on (O,e)with weight function x.


5. Lel 11have any one of Ihe fixed values 11= O. l. 2. ....(a) Suppose Ihal Jn(ib) = O (b #- O) and use resulls in Problem 2. Seco 73. 10 reach a

contradiclion. Thus show Ihal Ihe funclion J n(z) has no pure imaginary zeros z = ib (b #- O).(b) Since our series represenlalion of J n(x) converges when x is replaced by any complex

number z and since Ihe coefficienls oflhe powers of z in Ihal represenlalion are all real. il is IrueIhal Jn(:) = Jn(z}. where z denoles Ihe complex conjugale x - iy of Ihe number z = x + iy.t

Also. Ihe proof of orthogonalily in Problem 4 above remains valid when exis a nonzero complex

number and Ihe sel of rOOISexj Ihere is allowed 10 include any nonzero complex rOOISof Iheequalion Jn(exc)= OIhal may occur. Use Ihese facIs lo show Ihal iflhe complex number a + ib(a +- o. b +- O) is a zero of J .(z). Ihen a - ib is also a zero and Ihal

I I

0= r xJn[(a + ib)x]Jn[(a - ib)x] dx = r xlJn[(a + ib)x]12 dx.. o . o

Poinl oul why Ihe value of Ihe inlegral on Ihe far righl is aClually posilive ando wilh Ihiscontradiclion. deduce Ihal J .(z) has no zeros of Ihe form a + ib (a +- O. b +- O). Conclude Ihal if

z = x j (j = 1.2. ...) are Ihe posilive zeros of J n(z). Ihe only olher zeros, real or complex. are Ihe

numbers z = -Xj U = 1,2. ...). and also z = O when ti 2: 1.

79. The Orthonormal Functions

From the previous section we know that if exis a positive constant, thefunction

X(x) = Jn(exx) (n = 0,1,2, ...)

satisfies the equation

(1) (xX')' + (ex2X-:)X=0

Wemultiplyeachsideby 2xX' and write

(n = 0,1,2, ...).

d d- (xX'f + (ex2x2- n2)-(X2) = O.dx dx

After integrating both terms here and using integration by parts in thesecond term, we find that

[(XX')2 + (ex2x2- n2)X2]o- 2ex2(XX2 dx = O,o

where e is any positivenumber. When n = O,the quantity inside the bracketscIearly vanishes at x = O; and the same is tme when n = 1, 2, ..., sinceX(O) = Jn(O)= Othen. We thus arrive at the expression

.c2ex2 I x[Jn(exxW dx = ex2C2[J~(exC)P + (ex2c2 - n2)[Jn(exc)P,'0

(2)

t The ralío test applies even when Ihe lerms in a series are complex numbers. Also. seeChurchill. Brown. and Verhey (1974.p. 291). lisled in Ihe Bibliography. .


which we now use to find the norms of all the eigenfunctions in Theorem 3except for the one corresponding to the zero eigenvalue in case (c) there.That norm is treated separately.

(a) When I'J.j(j = 1,2, ...) arethepositiverootso[the equatio/1

(3) Jn(I'J.C) = O,

equation (2) becomese

.2 f x[Jn(l'J.jxW dx = C2[J~(l'J.jCW,O

The integral here is the square of the norm of Jn(Ct.jx)on the interval (O,c)with weight function x. AIso (Sec. 73)

xJ~(x) = nln(x) - XJn+I(X),

and therefore Ct.jcJ~(Ct.jc)= -Ct.jCJn+I(l'J.jC).Hence

(4)2 .

IIJn(l'J.jX) 112 = ~ [Jn+¡(c.tjC)y (j = 1,2, ...).

(b) When Ct.j(j = 1,2, ...) are the positive roots o[ the equation

(5) hJn(l'J.c) + (Ct.c)J~((;(c) = O (h ~ O,h + n > O),

we find fram equation (2) that

(6)1'J.2C2- n2 + h2

I/1n(l'J.jx}lll = J ''-.2 [Jn(Ct.jCWj

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

(C) Suppose now that Ct.¡= O and that Ct.j(j = 2, 3, ...) are the positiveroots o[ the equation

(7)

Since Jo(Ct.¡x) = 1,

Jó(I'J.C) = O.

.c C2

IIJo(Ct.¡x}lll=.10 x dx =2'

Expressions for 1/10(Ct.j c)II 2 (j = 2,3, . . .) are obtained by writing /1= h = Oinequation (6):

(8)

(9)C2

1/10(lXjx)112 = 2 [Jo(l'J.jcW (j = 2, 3, ...).

For equation (7) is simply equatiQn (5) when 11= h = O,and the restrictionh + /1> Ois not actually needed in deriving expression (6).

SECo80] BESSEL FUNCTlONS AND APPLICATIONS 203

The orthogonal eigenfunctions Xj(x) = Jn(CXjx)of our singular Sturm-Liouville problem can now be written in normalized form as

(10) 4>j(X)= .Jn(CXjx) (j = 1, 2, ...).

The norms here are given for the eigenfunctions in cases (a) and (b) ofTheorem 3 by equations (4) and (6), respectively.In case (c) they are givenby equations (8) and (9). The fact that the set (10) is orthonormal on theinterval (O,c)with weight functionsx is,of course,expressedby the equations

e

{

O

fo X4>ix}cPk(X)dx = 1if j :f=k,

ifj = k.

80. Fourier-Bessel Series

Since the functions 4>j(x)constitute al! the normalized eigenfunctions ofoursingular Sturm-Liouville problem, we can anticipate representations offunctions f(x) on the interval (O,c)by means of generalized Fourier series(Sec. 26) involving those orthonormal functions.

Let Cj (j = 1,2, ...) denote the Fourier constants of a functionf(x) withrespect to the set {4>j(x)}(j = 1,2, ...) on the interval (O,c).Then

.e 1r

e

Cj = t xf(x)4>ix) dx = IIJn(cx¡x)11'0 xf(x)Jn(CXjx) dx,

and the generalized Fourier series corresponding to f is

(1)ex> ex> Jn(CXjx) e

.L Cj4>ix) = .L IIJ ( . )112r sf(sVn(CXjS) ds.)=1 )=1 nCX)X '0

In view of the expressions for the norms found in the preceding section,we thus have the correspondence

(2)ex>

f(x) -L AjJn(cxjx)j= 1

(O< x < c)

where the coefficients Aj have the fol!owing values.(a) When CXj(j = 1,2, ...) are the positiveroots of the equation

(3)

@

Jn(CXC) = O,

2 .c .

c2Un+ 1 (cxAY .loxf(x)Jn(CXjx) dx(j = 1,2, .. .).Aj

(5)

(6)

(b) Whenaj (j = 1,2, ...) arethe positiverootsof the equation

hJ.(ac) + (ac)J~(ac)= O (h ~ O,h + n> O),

2a~ .e

Aj=(22 2 h2)[J( wl Xf(X)J.((Xjx)dx (j=1,2,...).ajC -n + .ajc.O

(C) When n = O in series (2) and when a¡ = Oand (Xj(j = 2, 3, ...) are thepositive roots of the equation

(7)

then

J{¡((Xc) = O,

(8)2 .c

A¡=21 xf(x)dxc '0

and

(9) A - 2 r

e

j - c2fJo(ajcW '0 xf(x)Jo(ajx) dx(j = 2, 3, ...).

Note that equation (9) reduces to equation (8) ifj = 1,since (X¡= O incase (c).

Proofs that correspondence (2) is actualIy an equality, under conditionssimilar to those used to ensure the representation of a functionby its Fouriercosine or sine series, usually involve the theory of functions of a complexvariable. We state without proof one form of such a representation theoremand refer the reader to the Bibliography.t

Theorem 4. If f is a function which, together with its derivative 1', ispiecewise continuous on the intervalO < x < c, then series (2) converges to themean value of the one-sided limits of f at each point in that interval. That is,

1 ex)

(10) 2[J(x+)+f(x-)]= J¡AjJ.(ajx) (O<x<c)

where the coefficients Aj are defined by equation (4) or (6) or the pair ofequations (8) and (9), depending on the particular equation that determines theroots aj .

The expansion (10) is called a Fourier-Besselseries representation offAn example is the expansion ofthe functionf(x) = 1 (O< x < c) into a

series of the functions Jo(ajx) (j = 1,2, ...) when aj are the positive roots of

t This theorem is proved in the book by Watson (1952). Also see pp. 84-86 in part IoCthework by Titchmarsh (1962). as well as the books by Gray and Mathews (1966) and Bowman(1958). These are alllisted in the Bibliography.

the equation Jo(cxc)= O.In view of the integration formula (Sec.73)

(sJo(s) ds = xJ l(X),'0

it is easy to evaluate the integral in expression (4) that is needed:

r

C 1 ajC exJo(CXjx)dx = 2" J sJo(s) ds = -Jl(CXjc).

'0 CXj o CXj

Consequently,

1=~ I Jo(CXjx)e j= 1 cxJ 1(CXjc)

The results stated in Theorems 2, 3, and 4 are also va lid when n isreplaced by an arbitrary positive number v, although we have not developedproperties of the functions J v far enough to establish this fact.

For functions on the unbounded interval x> O there is an integralrepresentation in terms of J" corresponding to the Fourier cosine or sineintegral formula.t The representation, for a fixed v (v ;?:-!), is

(11) (O< x < c).

00 00

f(x) = J cxJv(cxx) f sf(s)lv(cxs)ds dcxo o(x > O)

and is known as Hankefs integralformula.It is valid iffandf' are piecewisecontinuous on each bounded interval and ifjXf(x) is absolutely integrablefrom zero to infinity andf(x) is defined as its mean value at each point ofdiscontinuity.

Ir the interval (O,c)is replaced by some interval (a,b)where O< a < b,the Sturm-Liouville problem treated in Seco78 is no longer singular whenthe same differentialequation is used and boundary conditions of type (3) inthat section are applied at each end point. The eigenfunctions involvebothof the Bessel functions J n and y".

PROBLEMS

1. Find lhe coefficienls Aj U = 1.2, ...) in lhe expansion

.J.,

1 = L AJo(cxjx)j= I

(O< x < c)

when <XI= O and CXjU = 2.3. ...) are lhe positive roots of the equation Jo(<Xc)= O.Ans. AI=I,Aj=O U=2,3 )

t See Chap. 2 of the book by Sneddon (1951) that is listed in the Bibliography. For asurnrnary ofrepresentations in terrns ofBessel functions. see"Chap. 7 in vol. 2 ofthe work editedby Erdélyi (1953) which is also listed.

2. (a) Oblain Ihe represenlalion

l = 2c t rxJJ¡(IXJc)Jo(O!Jx)J-I (O!;CZ + IIZ)[Jo(O!JcJ]Z

(O< x < c)

wherc rxJU - l. 2,...) are the posilive roots of the equation

hJo(rxc) + (IXC)Jó(rxc) = O (h >0).

[Contrast this representation with Ihe representation (11). Seco 80. and also Iheone in Problem l.]

(b) Show how the result in part (a) can be written in Ihe form

1 =~r(~) J ¡(IXJC)Jo(rxJx)

c J-I. rxJ [Jo(IXJCJ]Z+ [J1(rxJCJ]Z(O< x < c).

3. Show that if

f(x) = I~

when O < x < 1,

when l < x < 2,

and f(l) =! then

1 ~ J1(rxJ)¡(x) =-2 L.. [J (2 m Jo(rx¡x)J-. aj 1 aj(O< x < 2)

where rx¡(j = 1,2, oo.) are Ihe positive roots ofthe equation Jo(2rx) = O.

4. Let rx¡(j = 1,2. oo.)denote the positive roots ofthe equation Jo(rxc) = O, where c is afixed positive number.

(a) With Ihe aid of Ihe reduction formula found in Problem 15. Seco 73, obtain Iheexpansion

Z 2 ~ (rxjc)Z- 4 ( )x =- L.. 3 Jo rx¡xc j-' rxjJ¡(rx¡c)

(O< x < c).

(b) Combine expansion (11), Seco 80. with Ihe one in parl (a) lo show Iha!

8 " Jo(rx¡x)

CZ - XZ = -¡ ¡~, rxJJ1(rxjc)(O< x < c).

5. Show Ihal

ao

[

1 .J

1

J ( )x=2¿ 1-- Jo(s)ds ~

¡=I rx;JI(rx¡)fo IXjJ1(rx¡)

where rxj (j = 1,2, oo.) are Ihe positive roots oflhe equation Jo(rx) = O.t

6. Lel n have any one of the posilive values n = 1, 2, . ... Wilh Ihe aid of Ihe integral ionformula (5), Seco 73, show Ihal

(0< x < 1)

xn=2I rx¡Jn+l(rxJ)¡-, (IX;- nZ)[Jn(rx/uJ.(IX¡X)

(O< x < 1)

where IX¡(j = 1, 2, ...) are Ihe posilive rools of Ihe equalion J~(rx)= O.

t See Ihe footnole wilh Problem 15(b) Seco 73.

,

SECo81] BESSEL FUNCTIONS AND APPLlCATIONS 207

7. Point out why ¡he eigenvalues of the singular Sturm-Liouville problem

(xX')'+ (h -~)X =0 (O<x < 1),

X(I)=O

are the numbers Aj= IXJ(j= 1,2, o o o) where IXjare the l10sitive roots of ¡he equation J ¡(IX)=Oand why the corresponding eigenfunctions are Xj = J ,(IX jX) (j = l. 2. . ..). Then obtain therepresen tation

x=2I J,(ajx)j=llXjJ2(1X)

(O::;x < 1)

in terms of ¡hose eigenfunctionso

81. Temperatures in a Long Cylinder

Let the lateral surface p = e of an infinitely long circular cylinder (Fig. 27),or a circular cylinder of finite altitude with insulated bases, be kept attemperature zero; and let the initial temperature distribution be a givenfunction of only the variable p, the distance from the axis ofthe cylindér. Weshall derive an expression for the temperatures u(p,t) in the cylinder, assum-ing that the material of that solid is homogeneous.

With the cylinder situated as shown in Fig. 27, the heat equation andboundary conditions are, in cylindrical coordinates,

Also, when t > O,the function u is to be continuous throughout the cylinderand, in particular, on the axis p = O. We assume thatfandj' are piecewise

=1

p=c

y

u=O

Figure 27

(1)au = k(a2u + au)

(O O),at ap2 p ap

(2) u(c-, t) = O (t > O),

(3) u(p, 0+) =f(p) o (O< p < e).

continuous on the interval (O,e)and, for convenience, thatfis defined as themean value of its one-sided limits at each point of discontinuity.

Any solutions of the homogeneous equations (1) and (2) that are of thetype R(p )T(t) must satisfy the equations

RT' = kT(R" + ~Rl R(e)T(t) = O.

Separating variables in the first equation here, we have

:;=~(R"+ ~R') = -A,

where - A is some constant yet to be specified.Thus

(4)pR"(p) + R'(p) + ApR(p) = O

R(e) = O,

(O O).

The differential equation in R is BesseJ's equation with the parameter A,in which n = O. Problem (4), together with continuity conditions on R andR' on the intervalO:::; p :::;e, is a special case of the singular Sturm-Liouville problem consisting of equations (20) and (21) in Theorem 3 ofSeco 78. According to that theorem, the eigenvalues Aj ofproblem (4) are thenumbers Aj = a} (j = 1,2, ...) where a.jare the positive roots ofthe equation

(6) Jo(a.e)= O;

and R = Jo(a.jp) are the corresponding eigenfunctions.

When A = Aj, we see from equation (5) that T = exp (-a.Jkt). So, exceptfor a constant factor,

R(p)T(t) = Jo(a.jp)exp (-a.Jkt) (j = 1,2, ...).

The extended linear combination of those functions,

(7)'"

u(p,t) = L AjJo(a.jp) exp (-a.Jkt),j= I

formally satisfies the homogeneous equations (1) and (2) in our boundaryvalue problem. It also satisfies the nonhomogeneous initial condition (3) ifu(p, 0+ ) = u(p,O)and if the coefficientsAj can be determined so that

'"f(p) = LAjJo(a.jp)

j= 1(O < p < e).

sEc.81] BESSELFUNCfJONS AND APPLICATlONS 209

This is a valid Fourier-Bessel series representation (Theorem 4) if thecoefficients have the values

(8)2 .c

Aj = -2rr 1- -\121 pf(p)Jo(rxjp) dp'0(j = 1,2, ...),

obtained by writing n = Oin equation (4), Seco80.The formal solution of the boundary value problem is therefore given by

equation (7) with the coefficients (8), where rxjare the positive roots ofequation (6).That is, our temperature formula can be written as

( ) - 2 ~ Jo(rxjp) (2

fc

u p,t - 1: L...[J ( )]

2 exp -rxjkt) sf(s)Jo(rxjS)ds.e j= 1 1 rxje o

Verifieation. We can verify formula (7), with coefficients (8), as asolution of our problem by the procedure followedin Seco56 if we use twoadditional properties of the zeros of Bessel functions, namely thatrxj+l -- rxr-+n/easj -+ex)and that the sequence ofnumbers 1/[~á~J1(rxje)](j = 1,2, ...) is bounded.t

Since the sequence 1/[~J1(rxje)] (j = 1, 2, ...) is bounded and sinceJo and f are bounded functions, it follows from expression (8) that thesequence ofnumbers A)rxj (j = 1,2, ...) is also bounded. Therefore a posi-tive number B exists such that, for each positive n!lmber to, the absolutevalues of the terms in series (7) are less than the constant terms

(9) Brxjexp (-rxJkto)

when O :s;p :s;e and t ~ to. The series of those constant terms convergesbecause rxj+1 - rxj-+n/e as j -+ex)(see Problem 5, Seco82). Series (7) there-fore converges uniformly with respect to p and t (t ~ to), and its sum u(p,t) isa continuous function of its two variables when O :s;p :s; e and t > O. But

u(e,t) = O; hence condition (2) is satisfied.The derivatives of Jo(x) are also bounded (Sec. 75), and the series of

constants Erx} exp (-rxJkto), where 111= 2, 3, converge. So it follows in thesame way that the differentiated series converge uniformly when t ~ to, andhence that the function (7) satisfies the heat equation (1).

t These properties follow from an asymplotic represenlalion of J.(x). for large values of x.,

I (1 1

)1

..j""itXJ.(x) = J'I cos x - - nlt - -7t + -O.(x)2 4 x

where O.(x) is bounded as x lends lO infinily. For a derivalion. see vol. l. p. 526. oflhe work byCouranl and Hilbert (1953) or Ihe book by Walson (1952). bolh lisled in ¡he Bibliography.

FinalIy, in view ofthe convergence ofseries (7) tof(p) when t = O,Abel'stest, which is to be proved in Chap. 10, applies to show that u(p,O+ ) =u(p,O) when O < P < c. Condition (3) is therefore satisfied, and our solutionis verified.

82. "eat Transfer at the Surface of the Cylinder

Let us replace the condition that the surface of the infinite cylinder is attempera.ture zero by a condition that heat transfer takes place there intosurroundings at temperature zero. As in Seco7, where Newton's law for sur-face heat transfer was discussed, the flux through the surface is assumed tobe proportional to the differencebetween the temperature of the surface andthat of its surroundings. That is,

-Kup(c,t) = H[u(c,t) - O] (K > O,H > O),

where K is the thermal conductivity of the material of the cylinder and H isits surface conductance.

The boundary value problem for the temperature function u(p,t) is

u,(p,t) = k [upp(P,t) + ~ UAp,t)] (O < P'< c, t > O),

cup(c,t) = - hu(c,t) (t > O),

u(p,O) = f(p) (O< p < c).

(1)

(2)

(3)

We have written h = cH/K; and, for convenience, we alIow the possibilitythat the constant h be zero. In that case condition (2) simply states that thesurface p = c is insulated.

When u = R(p)T(t), separation of variables produces the eigenvalueproblem

(4))pR"(p) + R'(p) + ApR(p) = O

hR(c) + cR'(c) = O.

(O O,the eigenvalues are, according to Theorem 3, Aj= r:xJ(j = 1,2, ...) where r:xjare the positive roots of the equation

(5) hJo(r:xc) + (r:xc)Jó(r:xc) = O;

and, since T(t) + AkT(t) = O,we find that

R(p)T(t) = Jo(r:xjp)exp (-r:xJkt) (j = 1,2, ...).

The formal solution of our problem is then

(6)00

u(p,t)= ¿AjJo(r:xjp)exp (-r:xJkt)j= 1

SECo 82] BESSELfUNCTIONS AND APPLlCATIONS 211

where, according to the initial condition (3) and Theorem 4,

20:~f

c

A j = f..2-2 , 1.2\~' f.. _\12 pf(p)J O(O:jp) dpO

(7) (j = 1,2, ...).

I

II

}

(

J

If h = O,the boundary condition in our eigenvalue problem becomesR'(c) = O.In that case, Aj= 0:] (j = 1, 2, ...) where 0:1= O and O:j(j = 2,3, ...) are the positive roots of the equation J{¡(o:c)= O,or

(8) J I(o:c) = O.

Noting that JO(O:Ip) = 1,we may now write

(9)co

u(p,t) = Al + I AjJo(O:jp)exp (-o:]kt)j=2

where (Theorem 4)

(10)2 .c

Al =? t pf(p) dp,

A. = 2 ,<J c2rJO(O:jCw.lopf(p)Jo(O:jp) dp

(j = 2,3, ...).(11)

PROBLEMS

1. In Seco81, use expression (7). with coefficients (8), to find the lemperatures U(p,l) in aninfinite cylinder p ::; I under the conditions u(I,I) = O, u(p,O) = uo, where Uois a constan!. Giveapproximate numerical values of the first three coefficients in the series.

Ans. U(p,l) = 2uo[O.80Jo(2.4p) exp (-S.8kl) - 0.53Jo(5.5p) exp (-30kl)

+ 0.43Jo(8.7¡,) exp (- 76kl) - ...¡.

2. Over a long solid cylinder p ::; 1, at uniform temperature A, there is tightly fitted a longhollow cylinder 1 ::; p ::; 2 of the same material at temperature B. The outer surface p = 2 isthen kept at temperature B. Derive this expression for the temperatures io the cylinder of radius2 so formed:

A - B '" JI(a.)u(p,I)=B+- L [ (JWJo(ajp)exp(-aJkl)2 j=1 aj JI 2aj

where aj are the positive roots of J 0(201) = O.This is a temperature problem in shruoken fittings.

3, A function V(p,z) is harmonic ioterior to the cylinder bounded by the three surfacesl' = e, Z = O, and z = b. Assuming that V = O on the first two of those surfaces and fhat

V(p,b) = f(p) (O < l' < e), derive the express ion

'" sinh ajzV(p,z) = L AjJo(ajp) sinh a.bj= 1 J

where IXjare the positive roots of Jo(ae) =O and the coefficients Aj are given by equation (8),Seco 81.

(<f.\Derive an expression for the steady temperatures u(p,z) in the solid cylinder bounded

by !QJlhree surfaces p = 1, Z= O, and Z= 1 when u = O on the side, the bottom is in-sulaled, and u = 1 on Ihe 10p.

Ans. u(p,z) = 2 Í Jo(CXjl» cosh CXjZj=1 aJl(CXj) coshcx.. J

[Jo(CX) = O, 'J.j > O).

S. Apply Ihe ratio lest lo show Ihat Ihe series of constant lerms (9), Seco SI, converges.

Suggestion: Firsl show Ihal (CXj+,/CXj)- 1 -+.0, or (CXj+,/CXj)-+ 1, as j -+ oo.

6. A solid cylinder is bounded by Ihe Ihree surfaces p = l. Z = O, and Z = h. The side isinsulaled, Ihe bottom kepl al temperature zero, and the 10p at temperaluresf(p). Derive Ihisexpression for the sleady temperalures u(p,z) in Ihe cylinder: .

2zr

1 00 Jo(CXjp)sinh CXjZr

1

u(p,z) ='-b sf(s) ds + 2 L -

[J ( )]2 --;---h h sf(s)Jo(CXjs) ds

'0 j=2 o'J.j Sin CXj'o

where CX2'cx),... are Ihe posilive rools of J,(cx) = O.7. Show that whenf(p) = 1 (O< p < 1) in Problem 6, u(p,z) = z/b.

8. Find Ihe bounded sleady lemperalures u(p,z) in the semi-infinile cylinder p ;5;1, Z ~ Owhen u = 1 on Ihe base and Ihere is heal Iransfer into surroundings at temperalure zero,

according to Ihe linear law used in Seco S2, at the surface p = 1, Z > O.

Ans. u(p,z) = 2h i Jo(CXjp)exp (-CXjz). j=' JO(CXj)(cxJ+ h2)

[hJo("j) = CXjJ,(CXj),"j > O).

9. Solve Ihis boundary value problem for u(x,t):

xu, = (xu.). - (n2/x)1I

u(c,t) =0

u(x,O) = f(x)

(O < x < c, t > O),

(e >0),

(O< x < c),

where u is continuous when O ;5;x ;5;c, e > O and where n is a nonnegative integer.'"

An.'. u(x,e) = L AjJ.(cxj x) exp (-"Jt) where "j and Aj are given by equalions (3) and (4)j= 1

in Seco SO.

10. Lel V(p,z) denote a funclion which is harmonic interior lo the cylinder bounded by thethree surfaces p = c, Z = O, and Z = b. Given Ihat V = O on bolh the 10p and bottom of Ihecylinder and thal V(c,z) = f(z) (O < z < b), derive Ihe expression

V(p,z) = f b Io(mrp/b) . mtZ.=, .Io(n1[c/b)Sin ¡;-

where

2f

b n1[Z

b. = ¡; f(z) sinbdz.o

[See Problem 2, Seco 73, as well as the commenls immediately following equalion (I7), Seco 7S,regarding Ihe solulions of Ihat modified form of Bessel's equalion.]

(

I

,

IIII,t1

fI

t)JI

¡

f\ .(

/!

SECo 83] BESSEL FUNcrlONS AND APPLlCATIONS 213

83. Vibration of a Circular Membrane

A membrane, stretched over a fixedcircular frame p = c in the plane z = O,is given an initial displacement z =f(NP) and released from rest in thatposition. Its transverse displacements z(p,cf>,t),wherep, cf>,and z are cylindri-cal coordinates, will be found as the continuous function that satisfies thisboundary value problem:

(1) 2

(

1 1

)z" = a zpp + ¡;zp + p2 z~~ '

(2)

(3)

z(c,cf>,t)= O (-n S; cf>S; n, t ;;::O),

z(p,cf>,O)= f(p,cf», z,(p,cf>,O)= O (OS; p S; C, -1t S; cf>S; n),

where the function z(p,cf>,t)is periodic with period 2n in the variable cf>.A function z = R(p)«1>(cf»T(t)satisfies equation (1) if

T" = ~(R" + .!. R' )+ ~ «1>"= -).

a2T R p p2 «1> '(4)

where -). is any constant. We separate variables again in the second ofequations (4) and write «1>"/«1>= - JJ..Then we find that the function R«1>T

satisfies the homogeneous equations and has the necessary periodicity withrespect to cf> if R and «1>are eigenfunctions of the Sturm-Liouville problems

(5)

(6)

p2R"(p) + pR'(p) + ().p2- JJ.)R(p) = O, R(c)= O,

«1>"(cf»+ JJ.«1>(cf»= O, «1>(-n) = «1>(n), «1>'(-1t) = «1>'(n)

and T is such that

T'(t) + ).a2T(t) = O,

If JJ.has one of the values

T(O) = o.

JJ.= n2 (n= 0,1,2, ...),

Theorem 3 can be applied to problem (5); and if we consider problem (6)first, we see that the constant JJ.must, in fact, have one of those values. For,according to Seco33, they are eigenvalues of problem (6). To be precise,«1>(cf»= 1whenn = O;and whenn = 1,2,..., «1>(cf»can beany linearcombi-nation of cos ncf>and sin ncf>.From Theorem '3 we now see that theeigenvalues of problem (5) are 'the numbers Anj= CX;j(j = 1,2, ...) where CXnjare the positive roots of the equation

(7) Jn(CXc)= O (n = 0,1,2, ...),

the corresponding eigenfunctions being R(p) = Jn(cxnjP)'Then T(t) =cos (cxnj.at).

214 FOURIER SERIES ANO BOUNOARY VALUE PROBLEMS [SEC 83

The generalized linear combination of our functions Rct>T,

(8)00

z(p,cp,t)= ¿ AoJo(aojp) cos (aOjat)j=l

00 00

+ ¿ ¿ Jn(anjp)(Anj COSncp+ Bnj sin ncp)cos (anjat),n= 1 j= 1

formally satisfies all the homogeneous equations. It also satisfies the condi-tion z(p,cp,O)= f(p,cp) if the coefficientsAOj,Anj, and Bnjare such that

(9)00

f(p,cp) = ¿ AOjJo(aOjp)j= 1

+ JI ¡ [j~IAnjJn(anjP)J cos ncp + fJI BnjJn(anjP)J sin ncp}

when O :$; P :$; c, - 1t :$; cp :$; 1t.

For each fixedvalue of p, series (9)is the Fourier series forf(p,cp) on theinterval - 1t :$; cp :$; 1t if

00 1 ' .JI AOjJo(aOjp) = 21t L /(p,cp) dcp,

'" 1 ,.¿ AnjJn(anjP)= -1 f(p,cp) cos ncp dcp

j=l 1t,-.

oc 1 ,.¿ BnjJn(anjp) = -1 f(p,cp) sin I1CPdcp

j=1 1t,-.

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

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

For each fixed n, the equations here are Fourier-Bessel series representationsof the functions of p appearing on the right-hand sides of those equations, onthe interval (O,c),provided that (Theorem 4)

IcWyA - 1 ,e ,.

Oj - 1tC2fJ l(aOjcW .lopJo(IXOjp) L/(p,cp) dcp dp,

and

2 ,e ,.

(11) Anj= 2[J ( W I pJn(IXnjP) I f(p,</» cos 11</>dcp dp,1tC n + 1 IXnjC 'o ' - .2 ,e .1t

(12) Bnj= 2fJ ( W I pJn(IXnjP) I f(p,</» sin I1CPd</>dp1tC n + 1 rxnj C ' o ' - .when n = 1,2, ..,.

The displacements z(p,</>,t)are then given by equation (8) when thecoefficients have the values (10), (11), and (12). We assume, of course, thatthe functionfis such that the series in expression (8) has adequate propertiesof convergence and differentiability.

...

I

III

IIII

¡

1I1

I

),,I

\I

,}

)\

II

I

SECo 83] BESSELFUNCTIONS AND APPLlCATIONS 215

PROBLEMS

1. Suppose that in Seco83 the initial displacement functionf(p,4» is a linear combination

of a finitenumber of the functionsJo(OtOjp) and J .(Ot.jp) cos n4>,J .(Otnjp) sin n4>(n = 1, 2, .. .).Point out why the iterated series in expression (8) of that section then contains only a finitenumber of terms and represents a rigorous solution of the boundary value problem.

2. Let the initial displacement of the membrane in Seco 83 bef(p),.a function of p only,and derive the expression

2 '< Jo(Otjp) cos (Otjat) .'z(P.t) ="2 I [ Y I sf(.~)Jo(Otjs) ds,C j=' J,(Otjc) '0

where Otjare the positive roots of Jo(Otc) = O, for the displacements when t > O.

3. Show that if the initial displacement of the membrane in Seco 83 is AJ o(Ot.p), where A isa constant and Ot. is some positive rOOI of J o(Otc) = O.the subsequent displacements are

z(P.t) = AJo(Ot.p) cos (Ot.at).

Observe that these displacements are all periodic in t with a common period; thus the mem-brane gives a musical note.

4. Replace the initial conditions (3), Seco 83, by the conditions that z = Oand z, = 1whent = O. This is the case if the membrane and its frame are moving with unit velocity in the zdirection and the frame is brought to rest at the instant t = O. Derive the expression

2 ' sin (Otjat)z(p,c) =- I --y---( )

Jo(ajp),ac j= 1 OtjJ. OtjC

r whye Otjare the positive roots of J o(ac) = O, for the displacements when e > O.

V 5, Derive the following expression for the temperatures lI(p,4>,t) in an infinite cylinderp ::; C when 11= O on the surface p = C and 11= f(p.4» at time t = O:

>.

lI(p,4>,t)= I AOjJo(OtOjp) exp (-Ot~jkt)j=1

'< >.

+ I IJn(Otnjp)(Anj cos n4>+ B.j sin n4» exp (-Ot;jkt)n= I j= I

where Ot.j'AOj. A.j' and B.j are the numbers defined in Seco 83.

6. Derive an expression for the temperatures lI(p,z,t) in a solid cylinder p ::; c, O::; z ::; 1!whose entire surface is kept at temperature zero and whose initial temperature is a constant A.Show tha! it can be written

lI(p,z,t) = Ar(z,t)w(p,t)

where

4' sin (2n- I)zr(z,t) = - I exp[-(2n - 1)2kt]1!.=, 2n- l

and

2 ' Jo(Ot.p)w(p,t)=- I-

(J exp(-Ot;kt)Cj=,Ot¡l,Ot¡c)

Show that v(z,t) represents temperatures in a slab O::; z ::; 1! and w(p,t) temperatures in aninfinite cylinder p ::; c, both with zero boundary temperature and unit initial temperature (seeSecs. 56 and 81).

(Jo(OtjC) = O, aj > O].

7. Derive Ihe following expression for lemperalures u(p,c/>,t)in Ihe long righl-angled cylin-drical wedge formed by Ihe surface p = I and Ihe planes cf>= O and cf>= rt/2 when u = O on ilsenlire surface and 11=f(p,cf» al lime t = O:

'" x

lI(p,cf>,t)= L L BnjJ 2.(rxnjP) sin 2nc/>exp (-rx~jkt)11= Ij=1

where rx.j are Ihe posilive rools of J 2.(rx)= O and

8 .,2 ,B.;[J2n+,(rx.jJ]2= - r sin 2ncf>r pf(p,cf>)J2n(rx.jP)dp dcf>.

1t.o '0

8. Show Ihal if Ihe plane cf>= rt/2 in' Problem 7 is replaced by aplane cf>= cf>o, Ihe

expression for Ihe lemperalures in Ihe wedge will in general involve Bessel funclions J, ofnonintegral orders:

9. Solve Problem 7 when Ihe enlire surface of Ihe wedge is insulaled. inslead of being kepl

~I lemperalure zero.

tíO;I Solve Ihe following problem for lemperalures u(p,t) in a Ihin circular plale wilh healransfer from ils faces inlo surroundings al lemperalure zero:

I11,= u.. + - u. - hu

p (p < 1, t > O; h > O).

u(I,t) =0, u(p,O)= 1.

A ( ) 2 -., ~ Jo(rxjp) ( 2)ns. up,r = e. ... - ( )exp -rxjt

j=' rxjJ, rxj(Jo(rx;) = O, C/.j> O).

11. Solve Problem lO afler replacing Ihe condilion u(I,t) = Oby Ihis surface heal Iransfercondilion al Ihe edge:

u.(I,t)= -hou(I,t)

12. Solve Ihis Dirichlel problem ror V(p,z):

(ho >0).

v2V = O (p < 1,z > O),

(z> O),

(p < 1),

V(I,z)= O

V(p,O)=I

and V is lo be bounded in Ihe domain p < 1, z > O.

Ans. V(P.z)= 2 I Jo(rxj(

P)

)exp (-rxjz)

j=' rxjJ, rxj(Jo(rx;) = O, rxj> O].

13. Lel Ihe sleady lemperalures lI(p,Z) in a semi-infinile cylinder p :::;1, Z ;:: O whose baseis insulaled be such Ihal

u(I,z)= !~when O < z < 1,

when z > 1.

Derive Ihe express ion

2J00 /o(rxp) .u(p,z)=- -

( )cos rxz Sin rx drx

.rt o rx/o rx

ror Ihose temperalures. '.

SECo 83] BESSELFUNCTIONS AND APPLICATIONS 217

14. Given a functionj(z) that is represented by its Fourier integral formula for all real z,derive the following expression for the harmonic function V(P.z) inside the cylinder p = c suchthat V(c,z) =j(z) ( - 00 < z < 00):

l1

," lo(ap)r

"

V(p,z)= - -( )

j(s) cos [a(s - z)] ds da,n 'o lo ac '- '"

\

1

1

,

)

I

I

I

iI!I

r

CHAPTER

NINE

LEGENDRE POLYNOMIALSAND APPLICATIONS

84. Solutions of Legendre's Equation

As we shall see later on (Sec.91),separation of variables in Laplace's equa-tion written in terms of the spherical coordinates r and O leads, after thesubstitution x = cos O is made, to Legendre's equation

(1) (1 - X2)y"(X) - 2xy'(x) + AY(X)= O.

The points x = 1and x = - 1,corresponding to O= Oand O= n, are singu-lar points of that linear homogeneous differential equation. We show nowthat there is a certain infinite set of values of the parameter A for each ofwhich equafion (1) has a polynomial as a solution.

The point x = O is an ordinarypoint of equation (1); and so, to deter-mine a solution, we substitute

(2).CX)

)' = I ajxjj=O

into that equation.t This yields the identityro oc:

I j(j - l)ajxj-2 - I UU - 1) + 2j - A]ajxj= O.j=O j=O

Since the first two terms in the first series here are actually zero and sincej(j - 1) + 2j = j(j + 1), we may write

0:' x'

Ij(j-l)ajxj-2- IU(j+ 1)-A]ajxj=O.j=2 j=O

Modifying the first series once again, we write it in the form<L

I (j + 2)(j + l)aj+2xjj=O

t For a discussion of ordinary points and a justification for this substitution, see, forexample, the books referred to earlier in the footnote in Seco 7J.

218

SECo84] LEGENDRE POLYNOMIALS AND APPLlCATIONS 219

so that

(3)""

¿ {(j + 2)(j + 1)aj+2 - [j(j + 1) - 2]aj}xj= O.j=O

Equation (3) is an identify in x ifthe coefficients aj satisfy the recurrencerelation

~ (4)j(j + 1)- 2

aj+2 = (j + 2)(j + 1)aj(j = O, 1,2, ...).

The power series (2) then represents a solution of Legendre's equationwithin its interval of convergence if its coefficientssatisfy relation (4).Thisleaves ao and al as arbitrary constants.

Ir al = O,it follows from relation (4) that a3 = as = .oo = O.Thus onenontrivial solution of Legendre's equation is

(5)IX)

Yl = ao + ¿ aZkx2kk= 1

(aO -1=0)

where ao is an arbitrary nonzero constant and where the remainingcoefficientsaz, a4, ... are expressed in terms of ao by successiveapplicationsof relation (4). Another such solution is obtained by writing ao = O andletting al be arbitrary. To be precise, the series

(6)

IX)

Y2 = al x + ¿ a2k+ l X2k+ lk=1

(al -1= O)

satisfiesLegendre's equation for any nonzero value of al when a3'as,... arewritten in terms of al in accordance with relation (4).These two solutionsare, of course, linearly independent since they are not constant multiples ofeach other.

From relation (4)it is c1earthat the value of 2 alTectsthe values of all butthe first coefficientsin series (5) and (6).In particular, if A.has any one of theintegral values

(7) A.=n(n+ 1) (n = 0,1,2, ...),

so that

(8)j(j + 1) - n(n + 1)

aj+2 = (j + 2)(j + 1) aj(j = O, 1,2, oo.),

one of those series always terminates since an+z= O and, consequently,an+4= an+6=... = O.

220 FOURIER SERIES ANO BOUNOARY VALUE PROBLEMS [SEC. 84

Note that if n = O,then a2= a4= a6 = ... = O;and series (5) becomessimplyYI = ao.Ir,moreover,nisanyoneofthe evenintegers2,4,..., so thatn = 2m (m = 1,2, .. .),it is true that a2mf Oand a2(m+1)= a2(m+2) = ... = O.Series (5) thus reduces to a polynomial whose degree is 2m, or n. On theother hand, if n = 1, we see that Y2 = al x; and if n is any one of the oddintegers n = 2m + 1 (m = 1,2,...), then a2m+I ""O and we may writea2(m+ 1)+ I = a2(m+ 2)+ I = ... = O.Hence series (.6)becomes a polynomial ofdegree n if n is odd.

To summarize our discussion of solutions of Legendre's equation (1)when A.= n(n + 1) (n = 0,1,2,.. .), solution (5) reduces to the polynomial

(9) YI = ao + a2x2 + ... + anxn (anf O)

when n is even; and solution (6) becomes

(10) Y2 = alx + a3x3 +... + anxn (anf O)

when n is odd. The coefficients ao and al are arbitrary nonzero constants, andthe others are determined by successive applications of relation (8). Observethat when n is even, solution (6) remains an infinite series and that when n isodd, the same is true of solution (5).

Ir n is even, it is customary to assign a value to ao such that when thecoefficients a2' ..., an in expression (9) are determined by means of relation(8), the final coefficient an has the value

(11)(2n)!

-~.an - 2n(n!)

The precise value of ao that is required is not important to usoUsing theconvention that O!= 1,we note that ao = 1ifn = O.In thatcase,Yl = 1.Irnis odd, let us choose al so that the final coefficientin expression (10)is alsogiven by equation (11). Note that )"2= X if n = 1, since al = 1for that valueof n.

When n = 2, 3, ..., relation (8) can be used to write al! the coefficientsthat precede anin expressions (9) and (10)in terms of anoTo accomplishtbis,we first observe that the numerator on the right-hand sideofrelation (8)canbe written

j(j+ 1)-n(n+ 1)= -[(n2-l)+(n-j)]= -(n-j)(n+j+ 1).

We then solve for aj; the result is

(12) aj = - (j + 2)(j + 1)(n - j)(n + j + 1) aj+2"

,

SECo 84] LEGENDRE POLYNOMIALS AND APPLICATlONS 221

To express an-2k in terms of an, we now use relation (12) to write thefollowingk equations:

(n)(n-1)- - Qm

an-2 - (2)(2n - 1)

(n - 2)(n - 3) an-2,an-4 = - (4)(2n - 3)

(n-2k+2)(n-2k+ 1)an-2k= --(2k)(2n-2k+ 1) an-2k+2'

Equating the product of the left-hand sides of these equations to the productof their right-hand sides, we find that

(-IY n(n-l)'" (n-2k+ 1)an-2k = 2kk!- (2n - 1)(2n- 3) ... (2n - 2k + 1)ano

Then, upon substituting expression (11) for an into equation (13)and com-biningvarioustermsintotheappropriatefactorials(seeProblem 1,Seco86),we arrive that the desired expression:

(13)

(14)1 (-I)k (2n-2k)!

an-2k = 2n"k! (n - 2k)!(n - k)!'

As usual, we agree that O!= l.In viewof equation (14), the polynomials (9)and (10),with the values of

ao and al assigned in the manner described above, can be written

(15) 1 m (-I)k (2~=}kt~ xn- 2kPn(x)= 2nJo k! (n - 2k)!(n- k)! (n = O,1,2, ...)

where m = n/2 ifn is even and m = (n - 1)/2 ifn is odd.Another expressionfor Pn(x) will be given in Seco86.

The polynomial Pn(x) is called the Legendre polynomial 01degree n. Forthe first several values of n expression (15) becomes

Po(X) = 1, P¡{X) = x,1

P2(x)=2(3x2_1),

1P4(x) = 8 (35x4 - 30X2+ 3),P3(X)= ~ (5X3 - 3x),

1( S 3

)Ps(x)-= 8 63x - 70x + 15x .

We have just seen that Legendre's equation

(16) (1 - x2)y"(x) - 2xy'(x) + n(n + I)y(x) = O (n = 0,1,2, oo.)

always has the polynomial solution y = Pn(x), which is solution (9) (n even)or solution (10) (n odd) when appropriate values are assigned to the arbi-trary constants ao and al in those solutionso Details regarding the standardform of the accompanying series solution, which is denoted by Qn(x) and iscalled a Legendre function of the secolld kind, are left lo the problemso Theseries representation for Qn(x) is shown there to be convergent on the inter-val -1 < x < 1, and each Qn(x) is Iherefore continuous on that open inter-val. Although it is somewhat more difficult to show, the series diverges at theend points x = :t I.t It will, however, be sufficient for us to know that Qn(x)and Q~(x) faíl to be a pair of continuous functions on the closed interval-1 :::;;x :::;;1 (Problem 13, Seco90). Since Pn(x) and Qn(x) are linearlyindependent, the general solution of equation (16) is

(17) y = CPn(x) + DQn(x),

where C and D are arbitrary constan ts.Finally, we note that when A is unrestricted, two fundamental solutions

of Legendre's equation (1) can be written as power series in x by means ofrelation (4)0

85. Orthogonality of Legendre Polynomials

The self-adjoint form of Legendre's equation (1),Seco84, is

(1)d

f

2 dY

Jix (1 - x )dx + AY= 00

It is a speeial case of the Sturm-Liouville equation (4), Seco31, in whichr(x) = 1- X2, p(x) = 1, and q(x) = 00 Since r(-I) = O and r(l) = O, noboundary conditions at the ends of the interval (- 1,1)are needed in order tocomplete the singular Sturm-Liouvilleproblem on that interval (seeSeco32).The problem consists of equation (1) and the condition that y and y' becontinuous when - 1 :::;; x :::;; 1.

Equation (17) in the preceding section gives the general solution of. equation (1) here ifA = n(1I + 1)where n = O, 1,2, o o o o Since the polynomíalPn(x) and its derivative are conlinuous on the entire interval - 1 :::;; x:::;; 1and since this is nol true of the Legendre funclion Qn(x),it is cIear Ihat thecontinuity requirements on y and y' are met only when y is a constant

t See, for example, pp. 230-231 oflhe book by Bell (1968) Ihal is lisled in Ihe Bibliography.

SECo 85] LEGENDRE POLYNOMIALS AND APPLlCATIONS 223

multiple of Pn(x). Our singular Sturm-Liouville problem on the interval(- 1,1)therefore has eigenvalues

(2) A=n(n+l) (n = O, 1, 2, oo.)

and corresponding eigenfunctions

(3) y = Pn(x) (n = 0,1,2, ,oo).

Since these eigenfunctions all correspond to different eigenvalues,wemay conclude from Theorem 3,Seco32,that the set {Pn(x)}(n = 0,1, 2,.,.) ofLegendrepolynomialsisorthogonalon the interval(- 1,1)with weightfunctionp(x) = 1.That is,

(4).1I Pm(x)Pn(x)dx = O

, -1(m #: n).

In the notation used for inner products, property (4) reads (Pm,Pn)= O(m -+ n).

Later on (Secs. 89 and 90) we shall prove that whenever a functionfandits derivative f' are piecewise continuous on the interval -1 < x < 1, thegeneralized Fourier series for f with respect to the normalized form of theorthogonal set {Pn(x)} converges to f(x) at each point in the interval wherefis continuous. That orthonormal set is therefore cIosed in the sense of point-wise convergence, and this means that it is complete in the cIass of all suchfunctionsf(see Seco28). It follows that there can be no other eigenvalues thanthe set (2). For, suppose Yo(x) is an eigenfunction corresponding to anothereigenvalue Ao, Then (Yo,Pn) = O for each n; and the completeness of theset {Pn(x)} requires that Yo, which is continuous on the entire interval- 1 :::;;x :::;;1, have va\tie zero for each x in that interval. Hence Yo cannot bean eigenfunction.

Since Pn(x) is a polynomial containing only even powers of x if n is evenand only odd powers if n is odd, it is an even or an odd function, dependingon whether n is even or odd; that is,

(5) Pn(-x) = (-I)"Pn(x) (n = 0,1,2, oo.).

Hence if m and n are both even or both odd, the product Pm(x)Pn(x)is aneven function. It then followsfrom property (4) that

,1

I P2m(X)P2n(x) dx = O'0(6) (m #: n)

and

(7).1

I P2m+¡(x)P2n+I(X) dx = O'o

where m, n = 0,1,2, ....

(m #: n),


Condition (6) states that the set {P2n(x)}(n = 0, 1, 2, o..) of Legendrepolynomials of even degree is orthogol1al 011the il1terval (0,1). Since P~n(O)= 0,those polynomials are the eigenfunctions of the singular Sturm-Liouvilleproblem consisting of Legendre's equation (1) on the interval (0,1) and thecondition

(8) y'(O)= 0,

together with the condition that y and y' be continuous when ° ~ x ~ 1.Theeigenvalues are A.= 2n(2n + 1) where n = 0, 1, 2, . . . .

Similarly, according to condition (7), the set {P2n+¡(x)} (11= 0,1,2, ...)of Legel1dre polynomials of odd degree is orthogonal 0/1 the interval (0,1).Those polynomials are the eigenfunctions of the problem consisting of equa-tion (1), the boundary condition

(9) y(O)=0,

and the same continuity conditions on the c10sedinterval ° ~ x ~ 1. Theeigenvalues are A.= (2/1+ 1)(211 + 2) where /1= 0, 1,2, ....

The Fourier cosine series and sine series representations on half inter-vals followed from the basic Fourier series of sines and cosines (Sec. 37).Injust the same way, representations of functions on the interval (0,1) interms of either P2nor P2n+¡ follow from representations in terms of Pnon the interval (-1,1).

Further properties of Pn(x) must be developed before we can establishthe expansion theorem and certain order properties that are needed in ourapplications of those polynomialsot

86. Rodrigues' Formula and Norms

According to Seco84,.

( ) ( ) 1 ~ ( )k /1! (211- 2k)! n- 2k1 Pn X = 2n I L... - 1 k '( - k) l ( _A'" xn. k=O . 11 o 11 ,

where m = n/2 if n is even and m = (n - 1)/2 if /1 is odd. Since

dn 2n- 2k (2n - 2k)! n- 2k-x = xdxn (11-2k)!

and because of the linearity of the differentialoperator dn/dxn,expression(1)can be written

1 dn m k n! o 2n- 2kPn(X)=- 2n I d---¡¡ ¿(-1) k'( -k)'X ./1. X k=O . n .

(O~ k ~ m)

(2)

t Readers who wish 10 pass 10 Ihe applicalions al Ihis lime may read Theorem 1 (Sec. 88)and Seco 89 and Ihen proceed 10 Seco 91.


The powers of x in the sum here decrease in steps of two as the index kincreases ; and the lowest power is 2n - 2m, which is n if n is even and n + 1 ifn is odd. Evidently, then, the sum can be extended so that k ranges from Oton; for the additional polynomial that is introduced is of degree less than n,and its nth derivative is therefore zero. Since the resulting sum is the bino-mial expansion of (X2- 1)",it follows from eguation (2) that

1 d"P (x ) = - - (X2 - 1)

"" 2"n!dx"(3) (n = O,1, 2, . ..).

This is Rodrigues' formula for the Legendre polynomials.Various useful properties of Legendre polynomials are readily obtained

from Rodrigues' formula with the aid of Leibnitz' rule for the nth derivativeD"[f(x)g(x)]of the productof two functions: .

" I

D"(fg) = ¿ LO 'un. u, Dk(f)D"-k(g),k=O

(4)

where it is understood that aIl the required deriva tives exist and that thezero-order deriva tive of a function is the function itself.

We note, for example, that if we write u = X2- 1, 'so that

u" = (X2- 1)"= (x + 1)"(x- 1)",

it followsfrom Leibnitz' rule that

" niD"u"= ¿ . Dk[(X+ 1)"]D"-k[(x- 1)"].

k=Ok! (n - k)!

Since only the first term, occurring when k = O, in this sum is free of thefactor x - 1, we find that the value ofthe sum when x = 1 is 2"n!. Accordingto Rodrigues' formula, then,

(5) P"(I)= 1 (n = O,1, 2, .. .).

(See Problem 13of this section for another derivation of this result.) Notehow it follows from this and equation (5), Seco85, that p"( -1) = (-1)".

For another application of Rodrigues' formula, we write

2"+I(n+ 1)!P"+l(X)= D"+IU"+I= D"-I(D2u"+I),

where u = X2- 1. But

Du"+I = 2(n + l)xu.,

and so

D2U"+I = 2(n + 1)(u" + 2nx2u.- 1)

=2(n+ 1)[u"+2n(x2-1)u"-1 +2nu.-I]

= 2(n+ 1)[(2n+ l)u"+ 2nu"-I].


Consequently,

2nn! Pn+ I(X) = (211+ I)Dn-Illn + 2nDn-llIn-l.

Substituting 2n-l(n - 1)!Pn- ¡(x) for Dn-1un-1 here, we find that

(6)211+ 1

Pn+I(X) - Pn-¡(x)= - 2n , Dn-Iun.n.

On the other hand, Leibnitz' rule allows us to write

Dn(Dlln+l) Dn(xun) XDnlln+l1Dn-lllnP (x) - - - .n+l. - 2n+1(11+ 1)! - ~ - 2nl1! '

and since Dnun = 2nn! P n(;X),

(7)Pn+ I(X) - xPn(x) = 2n:! Dn-Illn.

Elimination of Dn-1un between this and equation (6) gives the reCllrrel1cerelatíon

(8) (n + I)Pn+I(X) + nPn- ¡(x) = (211+ l)xPn(x) (11= 1,2, ...).

Note, too, that the relation

(9) P~+I(X) - P~-I(X) = (211+ I)Pn(x) (n = 1,2, ...)

is an immediate consequence of equation (6).We now show how relation (8) and its form

(10) nPn(X) + (11- I)Pn-2(x) = (211- l)xPn-l(x) (n = 2,3, oo.),

obtained by replacing 11by 11- 1, can be used to find the norms IIPn11=(Pn,Pn)1/2 of the orthogonal polynomials Pn. Keeping in mind that(Pn+I'Pn- ¡) = O and (Pn- 2,Pn)= O, we find from equations (8) and (10),respectively, that

(11)

and

n(Pn-I,Pn- ¡) = (2n + 1)(xPn,Pn-.)

(12) n(Pn,Pn) = (211- l)(xPn-I,Pn).

The integrals representing (xPn,Pn- ¡) and (xPn- I'Pn) are identical, and weneed only eliminate those quantities from equations (11) and (12) to see that

(2n + 1)(Pn,Pn)= (211- 1)(Pn-I,Pn-I)'

or

(13) (211+ 1)IIPnI12=(211-1)IIPn-II/2 (11= 2,3, ...).

SECo 86] LEGENDRE POL YNOMIALS AND APPLlCATIONS 227

It is easy to verify directly that equation (13) is also valid when n = 1.Next let 11be any fixed positive integer and use equation (13) to write the

following 11equations:

(211+ 1)IIPn112 = (211- 1)IIPn-¡112,

(211- 1)IIPn-¡112= (211- 3)IIPn-2112,

(5)IIP2112= (3)IIP¡1I2,

(3)IIP ¡112 = (1 )/lPo 1120

Setting the product of the left-hand sides of these equations equal to theproduct of their right-hand sides and canceling appropriately, we arrive atthe result

(2/1 + 1)IIPn/l2 = l!Po112

Since I!P o 112= 2, this means that

(n = 1,2, oo.).

(14)IIP.II= ) 211: 1

(11 = O, 1, 2, . ..).

The set

(15)1)211; 1Pn(X)}

(11= 0,1,2, oo.)

is therefore orthOl1orma/011the il1terl'a/ ( - 1,1).

PROBLEMS

l. Obtain expression (14). Seco 840 for the coefficients an-l> from equations (11) and (13)in that sectiono

SlIggestioll: Observe that the factorials in equation (11). Seco 84. can be written

(211)! = (211)(211- 1)(211- 2) ..0 (211- 2k + 1)(211- 2k)!.

II! = 11(11- 1)'" (11- 2k + 1)(11- 2k)!.

1I!=/I(II-I)'" (II-k+ I)(II-k)!o

2. Establish these properties of Legendre polynomials:

(a) P2n+,(O)=O;(211)! .

(h) P2n(O)= (-1)" 22n(II!)2 .(e) P2n(O) = 00

3. Verify directly that the set of functions Po(x). P,(x). and P2(x) is orthogonal on theinterval (- 1.1). Draw the graphs of these functionso

4. Use induction on the integer 11to verify Leibnitz' rule (4). Seco 860

5. From relation (9). Seco 86, obtain the integration formula

1 1

J p.(s) ds= --[P.-,(x)- p.+,(x)]% 2n + 1

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

. 6. From the orthogonality of the set {p.(x)}. state why

1

(a) J p.(x) dx = Owhen n = 1,2, ...;-,1

(b) J (Ax + B)P.(x) dx = O when n = 2,3, ... (A, B constant).-1

7. Why is the set {J 4n + 1P 2.(X)} (n = O. 1, 2, ...) orthonormal on the interval (O,I)?

8. Why is the.set {J4n + 3 P20+I(X)} (n = 0,1,2, ...) orthonormal on rhe interval (O,I)?

9. With the aid of express ion (15), Seco 84. for p.(x), show that when n = 2. 3. theconstants ao and al in equations (9) and (10) in that section must have the following values inorder for the final constant a. to have the value specified in equation (11) there:

ao = (_1)"'2 (1)(31(~::_:J~ - 1)(2)(4)(6)... ¡;,¡

a, = (-1)<,-1)/2 (1)(3)(5t.. (n)(2)(4)(6)'" ¡;;'=I)

(n = 2. 4. ...),

(n = 3, 5. ...).

10. When the value ofthe constant Á in Legendre'sequation (1), Seco 84, is unrestricted. itis customary to write that equation as

(1 - X2)y"(X) - 2xy'(x) + v(v + I)y(x) =O,

where v is an unrestricted complex number.(a) Show that relation (4), Seco 84, takes the form

(v - j)(v + j + 1)aj+2=- (j+2)(j+-I)a)

(j = O, 1,2, .. .).

and then use it to obtain the following nontrivial solutions of Legendre's equation:

y, = a)1 + I (-I)'[v(v - 2)'" (v- 2k+ 2)][(v+ I)(v+ 3)'" (v+ 2k - 1)] nI\ ,=, (2k)! -- x ,

Y2 = a.J x + I: (-1)' [(v- I)(v - 3) ... (v - 2k + I)][(v+ 2)(v + 4) ... (v + 2k)] 2B ,)\ ,=, (2k+I)! -x ,

where ao and a, are arbitrary nonzero constants.(b) Apply the ratio test to the series obtained in part (a) to show that they are convergent

when-1 <x< 1.

11. Note that the solutions YI and Y2 obtained in Problem lO are solutions (5) and (6) inSeco 84 when Á = v(v + 1). They remain infinite series when v = n = 1.3,5,... and v = n = 0,2,4, ..., respectively. When v = n = 2m (m = 0,1,2,.. .), the Legendre function Q. of the secondkind is defined as Y2 where

( -1)m22m(m!)2

al = -(2m)! -;

SECo86] . LEGENDRE POLYNOMIALSAND APPLlCATIONS 229

and when v = n =2m + ) (m = 0,1,2,..o),il is definedas Y. where

( -lr2z'"(m!)Zao = - (2m + I)! o

Using Ihe facl Ihal

(1+ X

)"' XZH 1

log - =2¿-1- x .-02k + 1(-1 < x < 1),

show Ihal1

(1 + X

)Qo(x) = 210g 1 - xand

X

(1 + X

)QI(x)=-log - -1 =xQo(x)-1.2 I-x

12. Wrile

F(x,t) = (1- 2xt + tZ¡-./z,

where Ixl ~ 1 and t is as yel unreslricledo(a) Nole Ihat x = cos 8 for some uniquely delermined value of 8 between ° and 1tinclu-

sive, and show Ihat

F(x,t) = (1 - el6t¡-IIZ(1- e-l6t¡-IIZo

Then, using Ihe facI Ihat (1- Z¡-I/Z has a valid Maclaurin series expansion when Izl < 1,point oul why il is Irue thal Ihe funclions (1 - e:t;'t¡-.'Z, considered as funclions of t, can berepresented by Mac1aurin series which are valid when It I < 1. It follows that the product ofthose two functions also has such a reprcsentation when It I < 1. t That is, there are functions

J.(x) (n = 0, 1, 2, o..) such that"'

F(x,t)= ¿ I.(x)t'.-0(Itl < 1)0

(b) Show that the function F(x,t) satisfies the identity

aF(1 - 2xt + tZ)- = (x - t)F,at

and use this resul! to show that the functions I.(x) in part (a) salisfy the recurrence relation

(n + 1)/.. .(x) + nl.-.(x) = (2n + I)xl.(x) (n = 1,2, 0")0

(e) Show that the first two functions lo(x) and/.(x) in part (a) are 1 and x, respectively,and notice that the recurrence relation oblained in part (b) can then be used to delermineJ.(x)when n = 2, 3, o o o o Compare Ihat relalion with relation (8), Seco86, and conc1ude that thefunctions I.(x) are, in fact, the Legendre polynomials p.(x); that is, show that

"'(1 - 2xt + tZ¡- I/Z= ¿ P.(x)t".-0 (lxl SI, Itl < 1)0

The function F is thus a generating lunction for the Legendre polynomials.

130 Use the generating function obtained in Problem 12 to show that P.(I) = 1 (n = 0,1,2, .. 0)0

t For a discussion of this point, see, for example, Churchil~ Brown, and Verhey (1974,po 164~1istedin the Bibliographyo

87. Laplace's Integral Form of p"

Useful orders of magnitude of p"(x) with respect to x and n followfrom acertain integral representation for those polynomials.

To derive the representation, we start with the fact that

f" eiN4>d</>= {~1t-"when N = 1, 2, . . . ,when N = O.

Because of the nature of the binomial expansion of (x + yei4>)k(k = O,1,2, .. .),wherex and y (y # O)are independentof </>, it followsthat

1 "21t f (x + yei4»k d</>= Xk-"

(k = 0,1,2, ...)

and hence that any polynomial"

q(x) = I akxkk=O

satisfies the identity

(1)1 ."

q(x) = 21t .1-"q(x + yei4» d</>.

Details are left to the problemsoIn particular, let q(x) be the Legendre polynomial p"(x). In view of

Rodrigues' formula (3) in the previous section, this choice of q(x) can also bewritten as the nth derivative p("I(x) of the polynomial

(2)l

p(x) = 2" , (X2- 1)".n.


(3)1 ."

p"(x)= - I p(")(x + yei4» d</>.21t .- "

Now it is readily shown by means of integration by parts that

o" . . k + 1'" . .(4) I e-1k4>p("-kl(x+ yel4» d</>= - I e-I(H l)4>p("-k- II(x + yel4» d</>. -" y .-"

where k is any integer such that O::;;k ::;;n - 1 (see Problem 9, Seco90).Successiveapplications ofthis reduction formula enable us to writeequation(3) in the form

(5)II! o". .

p"(x)= - I e-"4>p(x+ yel4»d</>o2lty" 0_"

If at this point we require that x =1= :t 1 and write y = JX2=1,definition (2) of the polynomial p(x) and some elementary simplificationsreveal that

~ e-in</!p(x+ yei</!)= (x + JX2=1 cos </»n.

With the above choice of y, then, expression (5) reduces to

1 "

Pn(x)= 2n L}x + JX2=1 cos</»nd</>,

or

1 "Pn(x) = - f (x + p-=t cos </>)"d</>(n = O,1,2, . ..).

n o

This is called Laplace'sintegralform of Pn(x).SincePn(l) = 1and Pn(-1) =(-1)n (Sec.86), it is also valid when x = :t 1.Thus it is valid for all x.

Suppose that - 1 ~ x ~ 1. Since x = cos Ofor some value óf ObetweenOand n inclusive,equation (6) can be written

(6)

(7)1 "

Pn(cosO)= - r (cos 0+ i sin O cos </>)"d</>n.o(O~ O ~ n).

Furthermore,

Icos O+ i sin Ocos </>1 = (1- sin2 Osin2 </»1/2;

and so, if we also note that sin (n - </» = sin </>,we have the inequality

(8)2 ,,/2

IPn(cos0)1 ~ - r (1- sin2 O sin2 </»n/2d</>.. n.o

Since the nonnegative quantity 1 - sin2 O sin2 </>is always less than orequal to unity, it follows from inequality (8) that 1 Pn(cosO)1 ~ 1, or

(9) IPn(x) 1 ~ 1 when - 1 ~ x ~ 1 (n = 0,1,2, ...).

To obtain another important consequence of inequality (8), we observefrom the graph of sin </>that sin </>> 2</>/nwhen O < </>< n/2. Thus ifu = (2</>/n)2sin 2 O,it is true that

1 - sin2 Osin2 </>< 1 - u

when O< </>< n/2 and O< O< n. Also 1 - u ~ e-u, as can be seen graph-icallyor from the Maclaurin series for e-u. According to inequality (8), then,

2 .,,/2(

2n sin2 O

)IPn(cos0)1 < ~ I exp - 2 </>2 d</> (O < O < n, n = 1, 2, ...).n.o n

Changing the variable of integration in this last integral by means of thesubstitution r = (.J2ñsin 0)<f;/1tand noting that the resulting upper limit ofintegration can be replaced by 00, we find that

~ 1 00

IPn(cosO)1< ---:--of exp(-r2)drnsm o

FinalIy, since the value ofthe integral here is Jñ/2 (Problem 19,Seco69),wearrive at the inequality

(10) IPn{x)I< J2n(1~X2) (- 1< x < 1, n = 1,2, ...).

That is,for each fixed x where - 1 < x < 1, Pn(x) is 01 the order 01n- 1/2as nincreases.

(O< O < 1t, n = 1, 2, . . .).

88. Further Order Properties

Since Pk(X) is a polynomial of degree k containing only alternate powers ofx, weknowthat ,

Xk = CPk(X)+ Ck-2Xk-2 + Ck-4Xk-4 +...

where the coefficients are constants. Similarly, Xk- 2 is a linear combinationof Pk- 2(X)and a polynomial of degree k - 4, and so on. Thus Xk is a finitelinear combination of the polynomials Pk(X), Pk- 2(X), Pk-4(X), ....

The derivative P~(x) (n = 1,2,...) is of degree n - 1 and contains onlyalternate powers of X; and each of those powers can be written in the abovemanner as a linear combination of Legendre polynomials. Therefore P~(x)itself can be written as a linear combination

(1) P~(X)= Cn-1Pn-I(X) + Cn-3 Pn-3(X) + ... (n = 1,2, ...)

of Legendre polynomials. To find any one of the constants Cj (j = n - 1,n - 3, .. .), we take inner products of both sides of equation (1) with Pix):

1

f Pj(x)P~(x) dx = Cj(Pj,Pj)'-1

When integrated by parts, the integral on the left becomes1

[Pix)Pn(x)]~ 1 - f Pn(x)Pj(x)dx;-1

and this last integral vanishes because Pj(x) is a linear combination ofLegendre polynomials of degree less than n. Also Pk(l) = 1, Pk(-1) =(-1)\ and (Pj,Pj) = 2/(2j + 1).Therefore

Cj = 2j + 1 (j = n - 1, n - 3, ...).

SECo 89] LEGENDRE POLYNOMIALS AND APPLlCATlONS 233

Representation (1), which is valid for all x, then becomes

(2) P~(x)= (2n - l)Pn- ¡(x) + (2n - S)Pn-3(X)+.oo (n = 1,2, ..o),

ending with 3P1(x) if n is even and with Po(x) if n is oddoAssume now tbat -1::;; x::;;1.Then IPn(x)I ::;;1 (Seco87), and it fol-

lows from equation (2) that

IP2n(x)1::;;(4n-1)+(4n-S)+'''+3=n(2n+ 1),

IP2n+¡(x)1::;;(4n + 1) + (4n - 3) + 000 + 1 = (n + 1)(2n+ 1).

But n(2n + 1)::;;(2n)2and (n + 1)(2n+ 1)::;;(2n + 1)2;and so

(3) IP~(x)1 ::;;n2 when -1::;; x::;; 1 (n = 1,2, "0)0

Differentiating both sides of equation (2) and then noting thatIP~-I(X)l< n2, IP~- 3(x)1 < n2, and so on, when -1 ::;;x ::;; 1, we see by themethod used above that

(4) IP~(x)1 ::;;n4 when -1::;; x::;; 1

Similarly, IP~k)(X) I ::;; n2k when k = 3, 4, o" and - 1 ::;;x ::;; 1.We collect our order properties of Po as followso

(n = 1,2, o. ')0

Theorem l. For each positive integer n and at all points of the interval- 1 ::;;x ::;;1 the val¡¡esof each of the f¡¡nctions

IPn(x)l,1

n2 IP~(x) 1,

1n4IP~(x)l, o..

never exceed ¡¡nityo Also (Seco 87), for each fixed n¡¡mber x such that- 1 < x < 1 it is tr¡¡e that

M

IPn(x)I<-fiwhere the value of M depends only on the choice of X.

(n = 1,2, '0')'

89. Legendre Series

In Seco86 we saw that the set of polynomials

(1)~

cPn(x)= V y- Pn(x)(n = O, 1, 2, o..)

isorthonormalon theinterval(-1,1)0TheFourierconstantswithrespecttothat set, for a functionf defined on the interval (- 1,1),are

[2iI+fJ

1

Cn = (f,cPn)= V y- -1 f(x)Pn(x) dx;

and the generalized Fourier series corresponding to f is

00 00 2n + 1 1

I c.4>.(x)= I ~p.(x) f f(s)P.(s) ds..=0 .=0 -1That is,

(2)00

f(x) '" I A.P.(x).=0

(-l<x<l)

where

(3) - 2n + 1f 1 f(x)P.(x) dxA. - 2 - 1

(n = 0, 1,2, .. .).

Series (2) with coefficients (3) is a Legendre series. In the followingsection we shall prove that it converges to f(x) under conditions stated inthis theorem.

Theorem 2. Let f denote a piecewise continuousfunction on the interval- 1 < x < 1. At each point x in that interval wheref is contimlOusand hasderivativesfrom the right and left the Legendreseries (2) convergestof(x);that is,

(4)00

f(x) = I A.P.(x).=0(-I<x<l)

where the coefficients A. are given by equation (3).

Note that iffand its derivativej' are both piecewisecontinuous on theinterval - 1 < x < 1,representation (4) must then be valid at any point x inthat interval wheref is continuous.

The proof of the theorem can be extended to show that the series con-verges to the mean ofthe valuesf(x+) andf(x-) whenfhas ajump at thepoint x, provided both one-sidedderivativesexist there.t That result willnotbe needed here.

Irf is an evenfunction, the productf(x)P.(x)is even when n is even,andodd when n is odd. Hence A2.+1= ° (n = O, 1,2, ...) and

1

A2. = (4n + 1) r f(X)P2.(X) dx'0(5) (n = 0,1,2, ...).

Thus, if we apply Theorem 2 to the even extension of a function f which ispiecewise continuous on the intervalO < x < 1 and which has one-sidedderivativesat points of continuity x, wefind that

(6)00

f(x) = I A2.P2.(X).=0

(O< x < 1)

where the coefficients A2. have the values (5).

t See pp. 65 ff of lhe book by Jackson (1941) lhat is listed in the Bibliography.

SECo90] LEGENDRE POLYNOMIALS AND APPLlCATIONS 235

Similarly, when the odd extension off is taken, we may write

(7)00

f(x)= IA2n+IP2n+¡{X)n=O

(O< x < 1)

/"1

A2n+1 = (4n + 3) f f(X)P2n+1 dxo

Both of the sets {P2n} and {P2n+I}' when normalized on the interval(0,1), are therefore closed in the sense of pointwise convergence (Seco28)0

where

(8) (n = 0,1,2, '0')'

90. Convergence of the Series

To prove Theorem 2, let x (-1 < x < 1) denote a point at which the fune-tionfis eontinuous and has one-sided derivatives.At x the value ofthe sumof the first m + 1 terms of the Legendre series (2), Seco89, can be written

(1)m 1

Sm(X) = I AnPn(x)= f f(s)Km(s,x) dsn=O -1

where

(2)Km(s,x)= ~ [1 + .tyn + I)Pn(S)Pn(X)].

The recurrence relation

(3) (2n + l)xPn(x) = (n + l)Pn+l(x) + nPn-I(X) (n = 1,2, oo.),

whieh is equation (8), Seco 86, leads to a more compact expression for Km inthe following way. Multiply each side of relation (3) by - Pn(s)and each sideof the same relation in s,

(2n + l)sPn(s) = (n + l)pn+ I(S) + nPn- ¡(s),

by Pn(x) and then add corresponding sides. Using the notation

(4) Rn(s,x) = Pn(S)Pn-I(X) - Pn-I(S)Pn(x),

we find that

(2n + I)Pn(s)Pn(x) = (n + l)Rn+l(s,x) - nRn(s,x)s-x (s 4=x).

When this expression for (2n + l)Pn(s)Pn(x) is used in the sum (2), the termsof that sum cancel in pairs, giving

(5) Km(s,x) = m + 1 Pm+I(S)pm(X) - Pm(S)pm+I(X)2 s - x (s4=x)o

The numerator of the last fraction, which is Rm+,(S,X), is a polynomial in sthat vanisheswhen s = x; and so it contains s - x as a factor. Therefore thefractionhas a finitelimitas s-x. We haveassumedthat m ~ 1; but notethat equation(5)yieldsKo(s,x)=!, whichagreeswithexpression(1)whenm=O.

Accordingto equations(1),(4),and (5),

Sm(x)= m2+ 1

f ' f(s) Rm+ ,(s,x) ds,-, s - x

(6)

the integrand here being piecewise continuous. Consider for the moment the

special casef(s) = 1 = Po(s). Then Ao = 1 and An = O (n = 1,2,.. .); and, inview of equation (1),Sm(x)= 1.Thus, fram equation (6),

(7) 1 = m + 1f ' Rm+ ¡(s,x)ds.2 -, s-x

We now multiply each side of equation (7) byf(x) and subtract fram thecorresponding sides of equation (6) to write

m + 1f

' f(s) - f(x)Sm(x)- f(x) = ~ -, - u Rm+,(s,x) ds,

or

(8)m+ 1 '

Sm(x) - f(x) = ~ t, F(s)[Pm+,(s)Pm(x)- Pm(s)pm+,(x)] dswhere

F(s) = f(s) - f(x)s - x .

. Since F(x + ) and F(x - ) are the one-sided derivativesof the givenfunctionfat the point x, F(s) is piecewisecontinuous on the interval -1 < s < 1.LetCndenote the Fourier constants of F corresponding to the orthonormal setof functions <Pn(x)= ,j(2n + 1)/2Pn(x).Then equation (8) can be written

m+1 m+1

Sm(x)-f(x) = J;+lPm(X)Cm+, - J;+l Pm+,(x)Cm.2 m+i 2 m+!

But IPm(x)1< MI';; (m = 1,2, oo.),where M is independent of m(Theorem 1).Thus it followsfram equation (9) that

(9)

m+lISm(x)-f(x)1 <M~(lCm+,1 + ICm!) (m = 1,2, oo.);

and, since Cm- O as m - 00 (Sec. 27), it is true that

(10) lim Sm(x)=f(x) (- 1 < x < 1).m-C()

This completes the proof of Theorem 2.

PROBLEMS

1. Use Theorem 2 and results found in Problems 2 and 5, Seco 86, to show that iff(x) = Owhen -1 < x < O andf(x) = 1 when O < x < 1, then

1 1 .,f(x) = 2 Po(X) + 2 n~}P2n(0) - P2ndO)]P2n+ ,(x)

1 3 00 n(4n + 3

)(2n)!

= 2 + ¡ x + n~' (-1) 4n + 4 22n(n!)2 P2n+ ,(x)

when - 1 < x < 1 and x '" O. Show that iff(O) is defined as 1/2, the expansion is also valid when

x =0.

2. Suppose thatf(x) = Owhen -1 < x ::;;Oandf(x) = x when O < x < 1.(a) State whyf is represented by its Legendre series (4), Seco89, at each point of the

interval -1 < x < 1.(b) Show that A2n+,= O(n = 1,2, ...) in the series in part (a).(e) Find the first four nonvanishing terms of the series in part (a).

1 1 5 3Ans. (e) f(x) =.,. + -P,(x) +- 6 P2(X) - -Pix) +oo.4 2 1 32

(-1 < x < 1).

3. Prove that for all x

1 2 3 . 2(a) X2=-PO(X)+-P2(x); (b) X3=_P,(X)+-P3(X).3 3 5 5

4. Expand the functionf(x) = 1 (O < x < 1) into a series of Legendre polynomials of odddegree on the interval (0,1). What function does the same series represent on the interval- 1< x < O?

Ans. 1 = I (-I)n (4n + 3

) (2n)!

n~O 2n+2 22n(n!)2P2n+'(X)(O< x < 1).

S. Obtain the first three nonzero terms in the series of Legendre polynomials of even

degree representing the functionf(x) = x (O< x < 1) to show that

1 5 3x =-Po(x) +-P2(x)- -p.(x)+ oo.2 8 16

'.~'(O<x<:: 1).

Point out why this expansion remains valid when x = O,and state what function the seriesrepresen ts on the in terval - 1 < x < 1.

6. State why it is true that when x is a fixed number such that - 1 < x < 1, then

lim Pn(x) = O.

7. State why

,limJ4n+T f f(X)P2n(x)dx = On-ao o

whenfis piecewise continuous on the interval (0,1).

8. Give details in the discussion starting at the beginning of Seco 87 and leading up toidentity (1), Seco 87, which states that

1 .q(x) = 27tJ-.q(x + yei.) d</>

for any polynomial q(x).

9. Use integration by parts to obtain the reduction formula (4) in Seco 87.Suggestion: Write the first integral appearing in that formula as

r e-i(H1)4> !...[~p('-'-I)(X + yei.)

]d</>.-. d</>'Y

10. Using the fact that x' (k = 0, 1, 2, ...) can be written as a finite linear combination ofthe Legendre polynomials p.(x), p.-2(x), p.-.(x),... (Sec. 88), point out why it is true that

1

J p.(x)p(x) dx = °-1

. where p.(x) is a Legendre polynomial of degree n (n = 1, 2, ...) and p(x) is any polynomialwhose degree is less than n.

11. Give details in the derivation of expression (5~ Seco 90, for Km(s,x).

12. Let n have any one of the values n = 1, 2, . ...(a) By recalling the result in Problem 6(a), Seco 86, point out why it is true that p.(x) must

change sign at least once in the open interval - 1 < x < 1. Then let x l' x 2' ..., x. denote thetOlality of distinct points in that interval where P .(x) changes sign. Since any polynomial ofdegree n has at most n zeros, we know that 1 ~ k ~ n.

(b) Assume that the number of points x,, X2' ..., x. in part (a) is such that k < n, andconsider the polynomial

p(x) = (x - X,)(X - X2)'" (x - x.).

With the aid of the result in Problem 10, point out why the integral

1

r p.(x)p(x) dx. -1

has value zero; and, afler noting that P.(x) and p(x) change sign at precisely the same points inthe interval -1 < x < 1, state why the value ofthe integral cannot be zero. Having reached thiscontradiclion, conduce that k = n and hence that the zeros of a Legendre polynomial p.(x) areall real and distinct and lie in the open interval - 1 < x < 1.

13. Show in the following way that for each value of n (n = 0, 1, 2, ...) Ihe Legendrefunction of the second kind Q.(x) and its derivative Q~(x) fail to be a pair of continuousfunctions on Ihe dosed interval -1 ~ x ~ 1. Suppose that there is an inleger N such that QN(X)and Q~(x)are conlinuous on the staled interval.The functionsQN(X)and p.(x) (n f N) are theneigenfunctions corresponding to difTerent eigenvalues of the Sturm-Liouville problem (1).Seco 85. Point out how it follows that

.1I QN(X)P.(x) dx = °. -1

(nf N)

and then use Theorem 2 to show that QN(X)= ANPN(X) where AN is some constan!. This is,however. impossible since P ,(x) and QN(X) are linearly independent.

sEc.91] LEGENDRE POLYNOMIALS AND APPLlCATIONS 239

91. Dirichlet Problems in Spherical Regions

For our first application of Legendre series, we shall determine the harmonicfunction V in the regio n r < e such that V assumes prescribed values F(O)onthe spherical surface r = e (Fig. 28). Here r, 4>,and Oare spherical coordi-nates, and Vis independent of 4>.Thus V satisfies Laplace's equation

(1)02 1 o

(. OV

)r or2(rV) + sin 000 SinO00 = O (r < e, O< O< n)

and the condition

(2) lim V(r,O) = F(O) (O< O < n);r-cr<c

also, V and its partial derivatives of the first and second order are to becontinuous throughout the interior (O~ r < e, O~ O~ n) of the sphere.

Physically, the function V may denote steady temperatures in a solidsphere r ~ e whose surface temperatures depend only on O; that is, thesurface temperatures are uniform over each circle r = e, O= Oo.Also, Vrepresents electrostatic potential in the space r < e, which is free of charges,when V = F(O)on the boundary r = c.

Consider now a solution of equation (1) of the form V = R(r )e( O) thatsatisfies the continuity requirements. Separation of variables shows that, forsome constant A, the function R(r) must satisfy the differential equation

(3) r(rR)" - AR = O (r < e)

and be continuous when O~ r < c. Also, for the same constant A,

(4) ~ d(

. de~sin OdO Sin 0dil} + Ae = O

(O< O < n),

where e, e', and en are to be continuous on the cIosedjnterval O~ O~ n.

=1

Figure 28

If in equation (4) we make the substitution x = cos 8, so that

de 1 de desin 8- = (1 - COS28)-- = -(1 - X2)-d8 sin 8 d8 dx '

it followsreadily that

(5)d

[

de]

- (1 - X2)- + Ae = odx dx (-1 < x < 1),

where e and its first two derivativeswith respect to x are continuous on theentire closed interval -1 ~ x ~ 1. Equation (5) is Legendre's equation inself-adjoint form; and from Seco85 we know that, except when e is iden-tically zero, A must be one of the eigenvalues

A=n(n+1)

and that e = Pn(x).Then e(8) = Pn(cos8).Writing equation (3) in the form

r2R" + 2rR' - AR = O,

we see that it is a Cauchy-Euler equation, which reduces to a ditTerentialequation with constant coefficientsafter the substitution r = e' is made (seeProblem 13,Seco34). When A= n(n + 1),its general solution is

(6) R=C1r"+C2r-n-l,

(n = 0,1,2, .oo)

as is easily verified. The continuity of R at the origin r = Orequires thatC2 =0.

The functions r"Pn(cos8) (n = O, 1, 2, oo.)therefore satisfy Laplace'sequation (1) and the continuity conditions accompanying that equation.Formally, their generalized linear combination

(7)'00

V(r,8)= I Bnr"Pn(cos 8)n=O

(r ~ e)

is a solution of our boundary value problem if the constan ts Bnare such thatV(c,8) = F(8).

We find those constants by introducing a new functionjdefined on theinterval - 1 < x < 1. In terms of the principal values of the inverse cosine,we writej(x) = F(COS-lx), so that F(8) = j(cos 8). It is then evident thatthe constants Bn are related to the coefficients(Sec. 89)

2n + 1f

1

(8) An ='"""""2 - /(x)Pn(x) dx

in the Legendre series representation

(n = O, 1,2, oo.)

(9)

00

j(x) = I AnPn(x)n=O .

(-l<x<l)

SECo 91] LEGENDRE POLYNOMIALSAND APPLlCATlONS 241

by means of the equation An = Bnen.We assume thatfandf' are piecewisecontinuous on the interval (-1,1), so that representation (9) is valid at eachpoint in the interval at whichfis continuous (Theorem 2).

The harmonic function V(r,O)can then be written in terms of thecoefficients(8) as

(10)00

(r

)

n

V(r,O)= n~oAn e Pn(cos O)(r ~ e).

A full verification is given below. A compact form of expression (10) is, ofcourse,

00 2n + 1

(

r

)

n

f1

V(r,O)= Jo -y- e Pn(COSO) -1 f(s)Pn(s) ds(r ~ e).

Before proceeding to the verification, we note that the harmoniefunetionW in the unbounded region r > e, exterior to the spherical surface r = e,which assumes the valuesf(cos O) on that surface and which is bounded asr 00 can be found in like manner. Here el = o in equation (6) if R isto remain bounded as r 00, and our solutions of equation (1) arer-n-1Pn(cos O).Thus

00 B

(11) W(r,O) = n~o ,.:1 Pn(COSO) (r ;:::e)

where the Bn are this time related to the coefficients (8) by means of theequation An = Bn/c"+1. That is,

(12)00

(

e

)n+ 1

W(r,O)= n~oAn ~ Pn(cosO)(r;:::e).

Verifieation. For convenience, let us define the function v(r,x) which isrelated to V(r,O) by the equation x = cos O (O~ O ~ 7t).In equation (1) thesecond term, which is a dilTerential form in O,can be written

1 o

[

2 1 oV]

o[

2 ov

]sin 000 (1 - cos O)sin O00 = ox (1 - x )ox .Irwe write F(O)=f(cos 8), as we did earlier, then v(r,x) is required to satisfythe conditions

(13)

(14)

r(rv)rr + [(1 - X2)vx]x = O

v(e-,x) = f(x)

(r < e, - 1 < x < 1),

(-I<x<I);

and v and its deriva tives of the first and second order must be continuous

functions of r and x when O ~ r < e, -1 ~ x ~ 1. Solution (lO) becomes

(15)00

(

r

)

n

v(r,x) = n~oAn e Pn(x)(r ~ e);

and, by the method used in Seco56, we now verify it as a solution of thisalternative form of our boundary value problem.

To prave that v(r,x) satisfies boundary condition (14) for each fixed xwhere f(x) is continuous, we first note that, in view of the Legendre seriesrepresentation (9), series (15)convergestof(x) when r = e.But the sequenceof functions (r/et (n = O,1,2, ...) is bounded and monotonic with respect ton. Hence Abel's test, which is developed later on in Chap. 10, shows thatseries (15) is uniformly convergent with respect to r (O:::;;r :::;;e). Therefore vis a continuous function of r, and v(e-,x) = v(e,x) = f(x).

When n ;::=1, each term in series (15) can be written as the product ofthethree factors A,,/n, P,,(x), and lI(r/e)". The first two factors are bounded for alln and x involved here. The third is nonnegative and not greater than n(ro/e)"if O :::;;r :::;;1'0' When ro < e, the series with terms n(ro/et converges.There-fore series (15) converges uniformly with respect to l' and x when O:::;;r:::;;roand - 1 :::;; x :::;; 1.Thus v is continuouswhenr < e; and it is boundedwhenl' :::;;1'0 < e and - 1 :::;;x :::;;1.

But the series with terms nk(ro/e)" also converges whenever 0< ro < e,for each fixedpositive value of k. Since P~(x)/n2and P~(x)/n4(-1 :::;;x :::;;1,11= 1. 2, ...) are bounded, according to Theorem 1, it follows readily thatseries (15) is twice differentiable with respect to r and x when r < e andthat the derivatives of v are continuous when r < e. Since each term of theseries satisfiesequation (13), the sum of the series satisfies that equation.

This completes the verification of solution (10) in its alternative form(15). The solution (12) for the harmonic function W in the external regioncan be established in the same manner. By writing s/e for e/r in the alterna-tive form

00

(e

)"+ 1

w(r,x) = Jo A" -; P,,(x) (r ;::=e)

ofsolution (12), we find fram the above discussion ofseries (15) that rw(r,x)is bounded for large values of l' (s:::;;So < e) and for all x (-1 :::;;x :::;;1).

92. Steady Temperatures in a Hemisphere

The base l' < 1, O= n/2 of a solid hemisphere r :::;;1, O :::;;8 :::;;n/2, part ofwhich is shown in Fig. 29, is insulated. The flux of heat inward through thehemispherical surface is kept at prescribed values f(cos 8). In order thattemperatures can be steady, those values are such that the resultant rate offlow thraugh the hemispherical surface is zero. That is, f satisfies thecondition

,,/2

f f(cos 8)2n sin e de = O,()

SECo92] LEGENDRE POLYNOMIALS AND APPLICATIONS 243

Figure 29

which can also be written

(1),1

I f(x) dx = O.'0

If u denotes temperatures as a function of r and O,the condition that thebase be insulated is

The boundary value problem in u(r,O) consists of Laplace's equation

a2 1 o(

. OU

)(3) r or2(ru) + sin 000 SinO00 = O

condition (2), and the flux condition

(r < 1,O< O< n/2),

(4) Ku,(I,O) = f(cos O) (O< 8 < n/2),

whete K is thermal conductivity. Writing x = cos 8, we assume thatf(x) and

f'(x) are piecewise continuous on the interval (0,1) and thatf(x) satisfiescondition (1). Also, u is to satisfy the usual continuity conditions when r < 1and O::::;;8 ::::;;n/2.

Separating variables by replacing u in equations (2) and (3) by R(r)e(8),we obtain the equations

(5) r(rR)" - AR= Owhere R must be continuous when O ::::;;r < 1, and

(r < 1),

(6)

~ d(

. de

)sin OdO Sin 8 dO + Ae = O

e'(n/2) = O,

(O< O < n/2),

where e and e' are to be continuous when O ::::;;O ::::;;n/2.

)

! ou = O when O= ?!. .r 00 2'

that is,

(2) uo(r,n/2) = O (O< r < 1).

The substitution x = cos () transforms equations (6) into the singularSturm-Liouville problem consisting of Legendre's equation (see Seco91)

d

[

de]

- (l_X2)- +).e=odx dx (O< x < 1)

and the condition that

de =0dx when x = O,

where e and de/dx are to be continuous when O~ x ~ 1. According toSeco85, this problem has eigenvalues ). = 2n(2n + 1),where n = O,1,2, . . .,and eigenfunctions e = P2n(X);hence e(O) = P2n(cosO).The correspond-ing solution of the Cauchy-Euler equation (5) is R = r2n.

Formally, then,00

u(r,O)= L Bnr2np 2n(COS O)n=O

if the constants Bn are such that condition (4) is satisfied. That conditionrequires that

(7)00

2K L nBnP2n(X) = f(x)n= 1

(O< x < 1),

where x = cos O.This is the representation forf(x) on the interval (0,1) in aseries of Legendre polynomials of even degree (Sec. 89) if 2KnBn = A2nwhere

(8)1

A2n = (4n+ 1)f f(X)P2n(X) dxo(n = 1,2, ...)

and iff is such that Ao = O,which is preciselycondition (1).Thus Bo is leftarbitrary, and

(9)1 00 1

u(r,O) = Bo + 2K L - A2nr2nP2n(cos O)n= 1 n

(r ~ 1, O ~ O ~ n/2)

where the coefficients A2n have the values (8).The constant Bo is the temperature at the origin r = O. Solutions of

such problems with just Neumann conditions (Sec. 10)are determined onlyup to such an arbitrary additive constant because all the boundary condi-tioris prescribe values of derivatives of the unknown harmonic functions.

"-1

93. Other Orthogonal Sets

Laplace's equation for a function V of aIl three spherical coordinates r, </>,()IS

02 1 02V 1 o

(

o OV

)r or2 (rV) + sin2 () 0</>2+ sin () o(} SIn () o(} = 00

If V and its derivatives are to be continuous throughout the interiorr < e of a sphere, those functionsmust be periodic in </>with period 21toThenseparation of variables leads to the solutions

(1)

(2) r"(am cos m</>+ bm sin m</»P':(x) (m,n = 0,1,2, 0.0)

where x = cos () and where the functions y = P,:(x), called associatedLegendrefunctions, satisfy the difTerentialequation

d

[

dY

] [

m2

]dx (1 - X2) dx + n(n+ 1)- 1 - X2 y = 00

The functions P,:(x) are generalizations of Legendre polynomials andare related to them as foIlows:

(3)

(4)dm

P:;'(x) = (1 - x2r/2 dxm Pn(x) (m,n = O, 1,2, 0")0

For each fixed m the set P,:(x) (n = 0,1,2, oo.) is orthogonal on the interval(-1,1) with weight function unity.t Note that P~(x) = Pn(x) and thatP,:(x) = O if m > no

Other generalizations of Legendre polynomials indude the J acobi poly-nomials p~a,/i)(x)(n = O, 1, 2, 'oo), where (X> - 1 and p> -1; they aredefinedas

( l )n dn

p~a./i)(x)= ;n! (1 - xta(l + xt/i dxn[(1 - x)a+n(l+ x)/i+n].

When (Xand pare fixed, those polynomials are orthogonal on the interval(-1,1) with weight function (1 - x)a(l + x)/ioWhen a = p = O,they reduceto Pn(x) (see Seco 86)ot

t For Irealmenls of associated Legendre funClions, see Chap. 15 of Ihe book by Whiuakerand Walson (1963) or Chap. 3 oflhe one by Hobson (1955);applicalions are given in Secs. 9.11and 9.12 of Ihe book by Carslaw and Jaeger (1959) on heal conduclion. All oflhese referencesare lisled in Ihe Bibliography.

¡ An inlroduclion lo Jacobi polynomials and also lo Hermile and Laguerre polynomials,which are orlhogonal on unbounded inlervals, is given in Ihe book by Jackson (1941) Ihat islisted in Ihe Bibliography. Also see Ihe book on special funclions by Rainville (1972) and theone by Lebedev (1972), listed there.

PROBLEMS

1. Suppose that V is harmonie throughout the regions r < e and r > e, that V -+ O asr -+ 00, and that V = 1 on the spherical surfaee r = C. Show from results found in See. 91that V = 1 when r ~ e and V =c/r when r ;:: C.

2. Suppose that, for all <p,the steady temperatures u(r,lI) in a so lid sphere r ~ 1 are suehthat u(I,II) = 1 when 0< 11< n/2 and u(I,II) = O when n/2 < 11< n. Obtain the express ion

1 1 ~[ ] 20+ , )u(r,lI) = - + - l.. P20(0) - P 20+2(0) r P 20+' (eos 112 2 0=0

for those temperatures.

3. The base r < 1, 11= n/2 of a solid hemisphere r ~ 1, O ~ 11~ n/2 is kept at temperatureu = O, while u = 1 on the hemispherieal surfaee r = 1, O < 11< n/2. Derive the expression

00

(4n + 3

)(2n)'

u(r,lI) = O~I(-1)0 2n + 2 22o(n,')2r20+1P2ft+1(eos 11)

for the steady temperatures in that solid.

4. The base r < e, O = n/2 of a solid hemisphere r ~ e, O ~ O ~ n/2 is insulated. Thetemperature distribution on the hemispherieal surfaee is u =f(eos O). Derive the expression

00

(r

)20 1

u(r,O)= 0~0(4n+ 1) ~ P20(eosll) t f(S)P20(s)ds

for the steady temperatures in the solid. Also, show that u(r,O) = 1 whenf(eos O) = 1.

5. A funetion V is harmonie and bounded in the unbounded region r > c, O ~ <p~ 2n,O ~ O < n/2. Also, V = O everywhere on the lIat boundary r > c, O = n/2 and V = f(eos 11)onthe hemispherieal boundary r = e, O < 11< n/2. Derive the expression

00

(C

)20+2 ,

V(r,O)= Io (4n + 3) ~ P20+,(eos O)fo f(S)P20+,(s) ds.

6. The lIux of heat Ku,(l,O) into a solid sphere at its surfaee r = 1 is a preseribed funetion

f(eos 11),wheref is sueh that the net time rate of lIow of heat into the solid is zero. Thus (seeSee. 92)

1

f f(x) dx = O.-,Assuming that u = O at the eenter r = O, derive the expression

1 00 2n+ 1 'u(r,O)= - I - r"Po(eosO)f f(s)Po(s) ds2K 0=' n -1

for the steady temperatures throughout the en tire sphere O ~ r ~ 1.

7. Let u(r,O) denote steady temperatures in a hollow sphere a :s; r ~ b when u(a,O) =

f(eos O) and u(b,lI) = O (O< O < n). Derive the expression

00 b20+ 1 - r20+'

(a

)o+ 1

u(r,lI)= I Ao 20+' 20+1 - Po(eos11)0=0 b - a r

where

A = 2n + ~( f(x)Po(x) dx.o 2 -1

SECo 93] LEGENORE POLYNOMIALS ANO APPLlCATIONS 247

8. Let U(X,I) represent temperatures in a nonhomogeneous insulated bar -1 :5;x :5;1along the x axis, and suppose that the thermal conductivity is proportional to 1 - X2. The heatequation takes the form

ou o[

2 ou

]-=b-. (I-x)-ot ox ox (b >0).

Here b is constant since we assume that the thermal coefficient Co used in Seco 6 and in Problem

8, Seco 7, is constan!. Note that the ends x = :f: 1 are insulated because the conductivity vanishes

there. Assuming that u(x,O) = f(x) (-1 < x < 1), derive the expression

<X>2n + 1 I

u(x,t) = L - exp [-n(n + l)bt]P.(x) f f(s)P.(s) ds..=0 2 - I

9. Show that iff(x) = X2 (-1 < x < 1) in Problem 8. then

u(x,t)= ~ + (X2 -~) exp (-6bt).

10. Suppose that the initial temperature distribution in a solid sphere r :5;1 is a prescribedfunction f(cos O) which is independent of the spherical coordinates r and </>.If k = 1, thetemperature function u(r,O,t) satisfies the heat equation

2OU 02 1 o (OU

)r - = r-(ru) + sinIJ-01 or2 sin IJolJ oIJ

and the usual conditions of continuity. Given that u = O when r = 1,separate va;iables in theproblem to obtain solutions R(r)0(IJ)T(t) of the homogeneous equations, showing that 0,taken as a function.of the variable x = cos IJ, satisfies Legendre's equation where ). = n(n + 1)and that .

r2R" + 2rR' + Ú¡2,2 - n(n + I)]R = O, R(I) = O.

Verify that R = r-1/2 J.+ 1/2(/ljr) where /l = /lj are roots of the equation J.+ 1/2(/l) = O. Thus,without completing the solution for u, show that

R(r)0(IJ)T(t) = r-1/2 J.+ 1/2(/ljr)P.(cos IJ) exp (- /lft).

The functions J.. 1/2(X)are called spherical Besstrlfunctions.

CHAPTER

TEN;

UNIQUENESS OF SOLUTIONS

94. Cauchy Criterion for Uniform Convergence

In this chapter we examine in greater detail the question of verifying solu-tions of boundary value problems of certain types and, more particularly,the question of establishing that a solution of a given problem is the ónlypossible solution. A multiplicity of solutions may actually arise when thestatement ofthe problem does not demand adequate continuity or bounded-ness of t~e unknown function and its derivatives. This was illustrated inProblem 18, Seco69.

Abel's test for uniform convergence (Theorem 1) willenable us to deter-mine further continuity.properties of solutions obtained in tlie form of series,properties that are useful both in verifyingsolutions and in proving a solu-tion unique. The remaining theorems giveconditions under which a solutionis unique. They apply only to specific types of problems, and their applica-tions are further limited becausethey require a rather high degreeofregular-ity of the functions involved.

We begin with a brief discussion of the Cauchy criterion for unilormconvergence. Let sn(x) denote the sum of the first n terms of a series offunctions X¡(x) which converges to the sum s(x):

(1)n

sn(x) = LX¡(x),i=1

S(x) = lim sn(x).n-ex>

Suppose that the series converges uniformly with respect to x for all x insome interval. Then, as noted in Seco 14, for each positive number 8 thereexists a positive integer n.. independent of x, such that

8

1 s(x) - sn(x) 1< "2whenever n > n.

for every x in the interval. Let j denote any positive integer. Then

ISn+j-Snl = ISn+j-s+s-snl ~ Is-sn+jl + IS-Snl <8,

provided n > n..

248

SECo95] UNIQUENESS OF SOLUTlONS 249

Thus a necessary condition for uniform convergence of the series is that,for all positive integers j,

(2) ISn+ix) - sn(x)I< e whenever n > n,.

Condition (2) is a sufficientcondition, known as the Cauchy criterion,for convergenceof the series for each fixedx even if n, is not independent ofx. Hence it implies that the sum s(x) exists.Then, for any fixedn and x andfor the given number e, an integer j,(x) exists such that

(3) Is(x) - sn+ix)1 < e whenever j > j,(x).

To show that condition (2) implies uniform convergence,let n denote aparticular integer greater than n" where n, corresponds to the givennumbere in the sensethat condition (2)is satisfiedfor all x. Then, for each particularx, condition (3) is satisfied when j > j,(x); and, since

Is-snl = Is-sn+j+sn+j-snl ~ Is-sn+jl + ISn+j-snl,

it followsfrom conditions (3) and (2) that

(4) Is(x) - sn(x)I< 2e

when j > j.(x) and n > n,. Thus Is(x) - sn(x)1,which is independent ofj, isarbitrarily small for each x when n > n,. Since n, is independent of x, theuniform convergence is established.

Note that x here may equally well denote elements (x" X2' oo.,XN)ofsome set in N-dimensional space. The uniform convergence is then withrespect to all N variables x" X2' 00 ., xN.

95. Abel's Test for Uniform Convergence

We now establish a test for the uniform convergenceof infiniteserieswhose. terms are products of functions .9f certain classes. Applications of the test

were made earlier (Secs.56, 81, and 91) for the purpose of verifyingformal'solutions of boundary value problems. .

The functions in a sequence of functions 1;(t) (i = 1,2, ...) are uniformlybounded for all points t in an interval if a constant K, independent of i, existssuch that

(1) I1;(t) I<K (i=1,2,00.)

for all t in the interval. The sequence is monotonic with respect to i if for everyt in the interval either

(2)

or else, for every t,

1;+,(t) ~ 1;(t) (i = 1,2, oo.)

(3) 1;+¡(t) ~ 1;(t) (i = 1,2, oo.).

;. .;."-)

The following generalized form of a test known as Abefs test shows thatwhen the terms of a uniformly convergent series are multiplied by functionsT¡(t) of the type just described, the new series is uniformly convergent.

Theorem1. The series

(4)00

I X¡(x)T¡(t)¡=I

converges uniformly with respect to the two variables x and t together in aregion R of the xt plane provided that (a) the series

00

IX¡(x)i=1

converges uniformly with respect lOx for all x such that (x,t) is in R and (b)the functions T¡(t) are uniformly bounded and monolOnic with respect lOi (i = 1,2, ...)for all t such that (x,t) is in R.

~

To start the proof, we let S. denote partial sums of series (4):.S.(x,t)= IXi(x)T¡(t).i=1

As indicated in the preceding section, the uniform convergence ofthat serieswill be established if we prove that to each positive number &there corre-sponds an integer nE,independent of x and t, such that

ISm(x,t)-S.(x,t)1 <& whenever n > nE,

for all integers m = n + 1,n + 2, ... and for all points (x,t) in R.We write s. for the partial sum

s.(x) = X I(X) + X 2(X)+ ... + X.(x).

Then for each pair of integers m, n (m > n), Sm.- S. can be written

X.+l T,,+1+ X.+2 T,,+2+... + Xm Tm

= (s.+1 - s.)T,,+1 + (S.+2 - S.+¡)T,,+2 + ... + (Sm- Sm-l )Tm

= (s. +1 - S.)T,,+1 + (s.+ 2 - S.)T,,+2 - (s.+ 1 - S.)T,,+ 2

+ .. . + (Sm - s.)Tm - (Sm-1 - s.)Tm.

By pairing alternate terms here, we find that

(5) Sm- S. = (S.+1 - S.)(T,,+1 - T,,+2) + (S.+ 2 - S.)(T,,+ 2 'C""T,,+ 3)

+ .. . + (Sm-1 - s.)(Tm-1- Tm)+ (Sm- s.)Tm.

SECo 96] UNIQUENESS OF SOLUTIONS 251

Suppose now that the functions 1; are nonincreasing with respect to i, sothat they satisfy condition (2), and that they also satisfy the uniform bound-edness condition (1). Then the factors 1;.+1- 1;.+2' 1;.+2 - 1;.+3' and soon, in equation (5) are nonnegative and I1;(t)I< K. Since the series withtermsX¡(x)convergesuniformly,an integern, existssuch that

E

ISn+ix) - Sn(x) I < 3K whenever n > n"

for all positive integers j, where E is any given positive number and n, isindependent of x (Sec.94). Then if n > n, and m > n, it followsfrom equa-tion (5) that

E

ISm-Snl <3K[(1;.+1-1;.+2)+(1;.+2-1;.+3)+"'+ ITml]

E

= 3K (1;.+ 1 - Tm + ITm 1)E

::;; 3K ( I1;.+ 1 I + 21 Tm 1).

Therefore

ISm(x,t) - Sn(x,t) I < E whenever m > n > n"

and the uniform convergence of series (4) is established.The proof is similar when the functions 1; are nondecreasing with res-

pect to i. .

When x is kept fixed, the series with terms X ¡ is a series of constants; andthe only requirement placed on that series is that it be convergent. Then thetheorem shows that when 1; are bounded and monotonic, the series of termsX¡ 1;(t) is uniformly convergent with respect to t.

Extensions of the theorem to cases in which X¡ are functions of x and t,or where X¡ and 1; are functions of several variables, become evident when itis observed that our proof rests on the uniform convergence of the series ofterms X¡ and the bounded monotonic nature of the functions 1;.

96. Uniqueness of Solutions of the Heat Equation

Let D denote the domain consistingof all points interior to a c10sedsurfaceS; and let l5 be the'c1osureof that domain, consistingof all points in D andall points on S. We assume always that the c10sedsurface S is piecewisesmooth.That is, it is a continuous surface consisting of a finite number ofparts over each of which the outward-drawn unit normal vector váries con-tinuously from point to point. Then if W is a function of x, y, and z which is


continuous in J5, together with its partial derivatives of the first and secondorder, an extended form of Gauss' divergence theorem states that

(1) ff wddW dA = frr (W V2W + W; + W; + W;) dV.S n "D

Here dA is the area element on S, dV represents dx dy dz, and dW/dn is thederivative in the direction of the outward-drawn normal to S.t

Consider a homogeneous solid whose interior is the domain D andwhose temperatures at time t are denoted by u(x,y,z,t). A fairly generalproblem in heat conduction is the following:

(2)

(3)

(4)

u = k V2u + cP(x,y,z,t)

u(x,y,z,O) = j(x,y,z)

u = g(x,y,z,t)

..

[(x,y,z,t) in D, t > O],

[(x,y,z) in D],

[(x,y,z) on S, t ;:::O]. fThis is the problem of determining temperatures in a body, with p~escribedinitial temperatures j(x,y,z) and surface temperatures g(x,y,z,t), interior towhichheat may be generated continuously at arate per unit volume propor-tional to cP(x,y,z,t).

Suppose that the problem has two solutions

u = uI(x,y,z,t), u = U2(X,y,z,t)

where both UIand U2are continuous functions in the cJosedregion J5whent ;:::0, while their derivatives of the first order with respect to t and of thefirst and second order with respect to x, y, and z are continuous in 15whent > O. Since UI and U2satisfy the linear equations (2), (3), and (4), theirdifference

w(x,y,z,t) = uI(x,y,z,t) - U2(X,y,Z,t)

satisfies the homogeneous problem

(5)

(6)

(7)

w, = k V2w

w(x,y,z,O) = O

w=O

[(x,y,z) in D, t > O],

[(x,y,z) in 6],

[(x,y,z) on S, t ;:::O].

Moreover, w and its derivatives have the continuity properties of Ul and U2assumed above.

We shall show now that w = Oin D when t > O,so that the two solutions

UI and U2are identical. It will follow that not more than one solution oftheboundary value problem in u can exist if the solution is required to satisfythe continuity conditions stated.

' ,.

t Equation (1) is found by applying lhe basic divergence lheorem used in Seco6 10 thevector field W grad W.See Taylor and Mann (1972.pp. 520-521). listed in the Bibliography.

SECo 96] UNIQUENESS OF 'SOLUTIONS 253

The continuity of w with respect to x, y, z, and t together in the cIosedregion fj when t :2:Oimplies that the integral

(8)I(t) = ~ fff)w(x,y,z,tW dV

is a continuous function of t when t:2:O. According to equation (6),1(0)= O.In viewof the continuity of w,when t!> Oand equation (5),wecanwrite

l'(t) = fffww, dV = k fff W V2w dVD D(t > O).

The divergenceformula (1) applies to the last integral becauseof the contin-uity of the derivatives of w when t > O;thus

(9) fftwV2WdV=.ftw~:dA- fft(w;+w;+w;)dV

when t > O.But w = Oon S; therefore

l'(t) = -k fff (w; + w; + w;) dV:s; O.D

The mean value theorem applies to I(t). Thus for each positive t anumber tI (O< tI < t) exists such that

I(t) - 1(0) = tl'(t¡);

and, since 1(0)= O and l'(t¡):S; O, it follows that l(t):S; O.definition (8) of the integral shows that I(t) :2:O.Therefore

However,

I(t) = O (t :2:O),

and so the nonnegative integrand W2cannot have a positive value at anypoint in D; for if so, the continuity of W2requires that W2be positive through-out a neighborhood of the point and then I(t) > O. Consequently

w(x,y,z,t) = O [(x,y,z) in fj, t :2:O],

and the following theorem on uniqueness is established.

Theorem 2. Let u satisfy these conditions ofregularity: (a) it is a contin-uous function of the variables x, y, z, and t, taken together, when the point(x,y,z) is in the c/osed region fj and t :2:O; (b) those derivatives of u present inthe h!!at equation (2) are continuous in the same sense when t > O.Then if u is asolution of"the boundary value problem (2) through (4), it is the only possiblesolution satisfying conditions (a) and (b).

The condition that u be continuous in fj when t = O restricts theusefulnessof our theorem. It is cIearlynot satisfiedif the initial temperdture

I


functionfin condition (3) fails to be continuous throughout 15,or ifat somepoint on S the initial value g(x,y,z,O)of the prescribed surface temperaturediffers from the value f(x,y,z). The continuity requirement at t = O can berelaxed in some cases.t

When conditions (a) and (b) in Theorem 2 are added to the requirementthat u is to satisfy the heat equation and the boundary conditions, ourboundary value problem iscompletelystated, provided it has a solution; forthat will be the only possible solution.

The proof of Theorem 2 required that the integral

., dw1 1 w- .dA. 's dn

in equation (9) should either vanish or have a negative value. It vanishedbecause w = O on S. But it is never positive ifcondition (4) is replaced by theboundary condition I

(10)du-d-- + /tu = g(x,y,z,t)n

~

[(x,y,z) on S, t > O],

where /t ~ O. For in that case dw/dn = -/tw on S, and w dw/dn ~ O.Thusour theorert} can be modified as follows.

-11Ieorem 3. T/te cOIle/usionin T/teorem 2 is tl'ue ifboundary condition (4)is replaced by cOlldition (10), 01' if condition (4) is satisfied on part of t/tesurface S and condition (10) is satisfied on the resto

97. Example

In the problem of temperature distribution in a slab with insulated facesx = Oand x = n and initialtemperaturesf(x)[Seco57(b)],assume thatfiscontinuous and f' is piecewisecontinuous on the interval O~ x ~ n. Thenthe Fourier cosine series forf converges uniformly to f(x) on that iriterval.

Let u(x,t) denote the sum ofthe series

(1)1 x

2ao + n~1an exp (-n2kt) cos nx(O~ x ~ n, t ~ O),

obtained as a formal solution of the problem, aoand an(n = 1,2, ...) beingthe coefficientsin the Fourier cosine series'forf

t Integral transforms can sometimes be used to prove uniqueness of solutions of certaintypes of boundary value problems. This is iIIustrated in Seco79 ofChurchill (1972).listed in theBibliography.

)

SECo97] UNIQUENESS OF SOLUTIONS 255

We can see from Abel'stest (Theorem 1) that series (1) converges uni-formly with respect to x and t together in the region O:::;;x :::;;n, t ~ O;thus uis continuous there. When 1~ lo,where lo is any positive number, the seriesobtained by difTerentiatingseries (1) term by term,any number of times withrespect to x or 1, is uniformly convergent according to the WeierstrassM-test. It followsreadily that u not only satisfiesall equations in the bound-ary value problem (compare Seco56), but also that u/, ux, and Uxxarecontinuous functions in the region O:::;;x :::;;n, 1> O.Thus u satisfies theregularity conditions (a) and (b) in Theorem 2.

The temperature problem for the slab is the same as the problem for aprismatic bar with its bases in the planes x = Oand x = 7[and with its lateralsurface, parallel to the x axis, insulated (du/dn = O).Let the domain D con-sist of all interior points of the prism. Then Theorem 3 applies to show thatthe sum U(X,l)of series (1) is the only solution that satisfiesthe conditions (a)and (b).

PROBLEMS

l. In the problem oftemperature distribution in the hemisphere r S 1,O S 8 S 71/2,solvedformally in Seco 92, assume that f and f' are piecewise continuous. With the aid of Abel's test,prove that, for each fixed x (O < x < 1, x = cos O) where f is continuous, the series in thesolution (9), Seco 92, namely

1 .. 1Bo+ 2K L Az.rz.pz.(x)

,,-1 n(x = cos8),

converges uniformly with respect to r (OS r SI). Also prove that the series is difTerentiablewith respect to r and that

'"KU,(I-, O)=f(cos O)

where u(r,8) is the sum of the series.

2. In Problem lO, Seco 57, on temperatures u(x,t) in a slab initially at temperaturesf(x),throughout which heat is generated at a constant rate, assume that f is continuous and f' ispiecewise continuous (O S x S 71) and thatf(O) = f(7I) = O. Prove that the function u(x,t) given

there is the only solution of the problej!Lwhich satisfies the regularity conditions (a) and (b)stated in Theorem 2.

3. Verify the solution of Problem 14,Seco57, and prove that it is the only possible solutionsatisfying the regularity conditions (a) and (b) stated in Theorem 2. Note that in this case theWeierstrass M-test suffices for all proofs of uniform convergence.

4. In Seco 81, given that the initial temperature functionf(p) for the cylinder is continuouson the intervalOS p S e and is such that the F ourier-Bessel series used there convergesuniformly to f(p) on that interval. prove that the solution established there is the only one thatsatisfies our regularity conditions.

256 fOURIER SERIES AND BOUNDARY VALUE PROBLEMS [SEC. 98

98. Solutionsof Laplace's or Poisson's Equation

Let W be a harmonic function in a domain D of three-dimensional spacebounded by a continuous closed surface S that is piecewisesmooth. Assumealso that Wand its partial derivativesofthe firstorder are continuous in theclosure i5 of the domain. Then, since

(1) V2W(x,y,z) = O [(x,y,z) in D],

form (1), Seco96, of Gauss' divergence formula becomes

(2) ft wdd~dA = fft(w.~ + W; + W:) dV.

This equation is va lid for our function W even though we have required thederivatives of the second order to be continuous only in D, and not in theclosed region D. It is not difficult to prove that because V2W = O,modification of the usual conditions in the divergence theorem is possible.t

Suppose that W = Oat all points on the surface S. Then the first integralin equation (2), and therefore the second, vanishes. But the integrand ofthesecond is nonnegative and continuous in D. It must therefore vanishthroughout i5; that is,

(3) W"= »j = Jv.= O [(x,y,z)in 15].

Consequently, W(x,y,z) is constant; but it is zero on S and continuous in D,and therefore W = Othroughout D.

Suppose that dW/dn, instead of W, vanishes on S; or, to make thecondition more general, suppose that

(4) dW +hW=Odn [(x,y,x) on S]

where h ~ Oand h is either a constant or a function of x, y, and z. Then on S

dW 2W-= -hW <Odn - ,

so that the first integral in equati~n (2) is less than or equal to zero. But thesecond integral is greater than or equal to zero; hence it must vanish, andagain conditions (3) follow. So W is constant in D.

Ir W vanishes over part of S and satisfies condition (4) on the rest of thatsurface, our argument still shows that W is constant in D.In this case theconstantmustbe zero.

t See the book by Kellogg (1953, p. 119) that is listed in the Bibliography.

4

j

1

1

~

,

..


Now let U denote a function which is continuous in 15, together with itspartial derivatives of the first order. Suppose that it also has continuousderivatives of the second order in D and satisfies these conditions:

(5)

(6)

V2U(x,y,z) = f(x,y,z)

dUp-+hU=gdn

[(x,y,z) in D],

[(x,y,z) on S].

Heref, p, h, and g denote prescribed constants or functions of (x,y,z). Weassume that p ~ Oand h ~ O.

Equation (5) is known as Poisson'sequationand is a generalization ofLaplace's equation, which occurs whenf(x,y,z) is identically zero through-out D. Boundary condition (6) inc1udesimportant specialcases.When p = Oon S, or on part of S, the value of U is assigned there.When h = O,the valueof dU/dn is assigned. Of course p and h must not vanish simuItaneously.

Ir U = U1(x,y,z) and U = U2(X,y,Z)are two solutions of this problem,their dilTerence

W=U.-U2

satisfies Laplace's equation in D and the condition

dWp dn + hW = O

on S. Moreover, W satisfies the conditions of regularity required of U 1 andU2' Thus it is harmonic in D, and W and its derivatives of the first order arecontinuousin 15. It followsfromthe,resuItsestablishedaboveforharmonicfunctions that W must be constant throughout 15. ThusdW/dn= Oon S.Irh "#O at some pomt on S, then W vanishes there and W = O throughout D.For the harmonic function W vanishes over the c10sed surface S and hencecannot have values other than zero interior to S.

We have now established the following uniqueness theorem for prob-lems in electrostatic or gravitational potential, steady temperatures, or otherboundary value problems for Laplace's or Poisson's equation.

Theorem 4. Let U(x,y,z) satisf~ these conditions ofregularity in a domainD bounded by a closed surface S: (a) it is continuous, together with its deriva-tives of the first order, in 15,and (b) its derivatives of the second order arecontinuous in D. .Then if U is a solution ofthe boundary value problem consist-ing of equations (5) and (6), it is the only solution satisfying conditions (a) and(b), except possibly for U + e where e is an arbitrary constant. Unless h = Oat every point on S, e = o and the solution is unique.

It is possible to show that this theorem also applies when D is theunbounded domain exterior to the closed slirfaceS, provided U satisfiesthe I

, 1:

I

additional requirement that the absolute values of rU, r2Uxx' r2UYY' andr2Uzz be bounded for all r greater than some fixed number, where r is thedistance from (x,y,z) to the originot Then, since U vanishes as r -> 00, theconstant e is zero and the solution is unique. But note that S is a closedsurface, so that this extension of the theorem does not apply, for instance, tounbounded domains between two planes or inside a cylinder.

Condition (a) in Theorem 4 is severe because it requires U and itsderivatives of the first order to be continuous on the surface S. For problemsin which p = O on S, so that U is prescribed on the entire boundary, thecondition can be relaxed so as to require only the continuity of U itselfin Dif derivatives are continuous in D. This followsdirectly froro a fundamentalresult in potential theory: if afunction other than a constant is harmonicinD and continuousin D, its maximumandminimumvaluesareassumedat pointson S, and never in D.t

J99. An Application

To illustrate the use of Theorem 4, consider the problem in Seco58 of deter-mining steady temperatures u(x,y) in a rectangular plate with three edgesattemperature zero and an assigned temperature distribution on the fourthedge. The faces of the plate are insulated. For convenience, we shallconsider the plate to be square with edge length 1t. As long as du/dn = O onthe faces, the thickness of the plate does not affect the problem.

The domain D is the interior of the region bounded by the planes x = O,x = 1t,Y = O,Y = 1t,Z= z¡, and z = Z2,where z¡ and Z2are constants. ThenS is the boundary of that domain. The required function u is harmonic in D.It vanishes on the three parts x = O,x = 1t,and y = 1tof S; and u = f(x) onthe part y = O.AIso,Uz= Oon the parts z = z¡ and z = Z2'Thus Theorem 4applies if u and its derivatives of the first order are continuous in D.

First, suppose u is inde~ndent of Z.Then

(1)

(2)

(3)

uxx(x,y)+ Uyy(x,y)= O (O< x < 1t,O< Y < 1t),

u(O,y) = u(1t,y) = O (O :::; y :::; 1t),

u(x,O) = f(x), u(x,1t)= O (O:::; x :::; 1t).

t See the book by Kellogg (1953) referred to earlier in this section..t Physically, the theorem seems evident since it states that steady temperatures cannot have

maximum or minimum values interior to a solid in which no heat is generated. For a proof in

three dimensions, see the book by Kellogg (1953) cited earlier in this section and, for twodimensions, see Churchill, Brown, and Verhey (1974, Seco 54).

"

~

sEc.99] UNIQUENESS OF SOLUTlONS 259

The formal solution found in Seco 58 becomes

00 sinh n(n - y) .u(x,y)= I bn---;--- sm nx

.= 1 smhnn

whereb. are thecoefficientsin the Fourier sineseriesforf on the interval (O,n).To show that the function (4) satisfies the regularity conditions, let us

require thatfandf' be continuous andf" piecewisecontinuous (O~ x ~ n)and that

(4)

f(O) =f(n) = O.

Results found in Chap. 5 then show that

(5)00

f(x) = I bn sin nx,.= 1

ro

f'(x) = I nb. cos nx,n=1

and both series converge uniformly on the interval O~ x ~ n. The secondseries,obtained by differentiatingthe first, is the Fourier cosine series forf';and, sincef' is continuous and f" is piecewisecontinuous, not only is thatseries uniformly convergent but the series of absolute values 100.1of itscoefficientsconverges. It foIlowsfrom the Weierstrass M-test that the series

(6)00

I nbn sin nx.=1(O~ x ~ n)

also converges uniformly with respect to x.Let us show that for each fixedy the sequence of functions

(7)sinh n(n - y)-- - .-

sinh nn (n = 1,2, oo.; O~ Y ~ n)

appearing in series (4) is mopotonic and nonincreasing as n increases. This isevident when y = Oand when y = n. It is true when O < Y < n ifthe function

T(t) = ~inh ptsinh IXt (t > O, IX> p > O)

always decreases as t grows. To see that this is so, write

T(t) sinh2 IXt= P sinh IXtcosh pt - IXsinh pt cosh IXt

= - ~ (IX - P) sinh (IX+ P)t + ~ (IX + P) sinh (IX- P)t

'= - 1X2 - p2f

sinh (IX+jn~ - sinh (IX- P)t]2 IX+P IX-P

= - 1X2 - p2 I: (IX+ P)2. - (IX - P)2. t2.+ l.2 .=0 (2n + 1)!

The terms of this series are positive, so that T'(t) < O; therefore T(t)decreases as t grows.

Likewise, the positive-valued functions

cosh n(n - y)sinhnn(8) (O~ y ~ n),

arising in the series for u¡., never increase in value as n grows beca use theirsquares can be written

(9) _,_.12 - + rSin:i:~nn: y) r

and each term is nonincreasing.The values of functions (7) cIearlyvary only from zero to unity for all n

and y involved. The functions (8) are also uniformly bounded. Thereforethose functions can be used in Abel's test for uniform convergence.From theuniform convergence of the series in equations (5) and of series (6), on theinterval O~ x ~ n, we conclude not only that series (4)convergesuniformlywith respect to x and y together in the region O~ x ~ n, O~ Y ~ n but alsothat the uniform convergence holds true for the seriesobtained by dilTeren-tiating series (4) once termwise with respect to either x or y.

Consequently,series (4)is differentiablewith respect to x and y; also, itssum u(x,y) and Uxand u¡.are continuous in the cIosed region O~ x ~ n,O~ Y ~ n. Clearly u(x,y) satisfies bouridary conditions (2) and (3).

The derivatives of the second order, with respect to either x or y, of theterms in series (4) have absolute values not greater than

n21bnI sinh .n(n -Yo)smh nn(10) )

iI

¡

i

when O ~ x ~ n and Yo ~ Y ~ n, where Yo > O. Let B be chosen such thatIbnl < B for all n. Then, from the inequalities

2 sinh n(n - Yo) < exp n(n - Yo), 2 sinh nn ~ enn(1 - e- 2n)

it follows that the numbers (10) are less than

B ( 2 \ n2 exp (- nyo).- exp - n jThe series with these terms converges, according to the ratio test, sinceYo> O; and so the Weierstrass M-test ensures the uniform convergence ofthe series of second-order derivativesof terms in series (4)whenYo ~ Y ~ n.Thus series (4) is twicedifferentiable;also, uxxand Uyyare continuous in theregion O~ x ~ n, O< Y ~ n.

Since the terms in series (4) satisfyLaplace's equation, the sum u(x,y) ofthat series satisfies condition (1).Thus u is established as a solution of our

J

boundary value problem. Moreover, u satisfies our regularity conditions,even with respect to Z since it is independent of Z and Uz= Oeverywhere, onthe parts Z = z¡ and Z = Z2 of S in particular. According to Theorem 4, thefunction defined by series (4) is the only possible solution that satisfies theregularity conditions.

100. Solutions of a Wave Equation

Consider the following generalization of the problem solved in Seco51 forthe transverse displacements in a stretched string:

(1)

(2)

(3)

YI/(x,t)= a2YxAx,t)+ ¡P(x,t)

y(O,t)= p(t), y(c,t) = q(t)

y(x,O)= f(x), YI(X,O)= g(x)

(O< x < e, t > O),

(t ~ O),

(O :::;x :::;e).

But we now require y to be of cIass e2 in the region R: O:::;x :::;e, t ~ O,bywhich we shall mean that y and its derivatives of the first and second order,incIuding YXIand YIX'are to be continuous functions in R. As indicated byexamples in Chap. 6, the prescribed functions ¡P,p, q, f, and g must berestricted if the problem is to have a solution of cIass e2.

Suppose there are two solutions Yl(X,t) and Y2(X,t)of that cIass. Thenthe difference Z= YI - Y2 is of cIass e2 in R, and it satisfies the homoge-neous problem

(4)

(5)

(6)

ZI/(x,t) = a2zxx(x,t)

z(O,t) = O, z(c,t)= O

z(x,O)= O, ZI(X,O)= O

We shall prove that Z= Othroughout R; thus Yl = Y2'The integrand of the integral

1 C

(

1

)(7) !(t) = 2 fo z; + a2z; dx

satisfies conditions such that we can write

(8)!'(t) = ( (ZXZXI + :2 ZIZII) dx.

Since ZI/ = a2zxx,the integrand here can be writtena

ZXZIX + ZIZXX = OX (zx ZI)'

So in view of equations (5), from which it follows that

ZI(O,t) = O, ZI(C,t) = O,

(O< x < e, t > O),

(t ~ O),

(O :::;x :::;e).

(t ~ O)

we can write

(9) I'(t) = zx(c,t)Zt(c,t)- zAO,t)z,(O,t)= O.

Hence I(t) is a constant. But definition (7) shows that 1(0)= O becausez(x,O)= O,and so zx(x,O)= O;also z,(x,O)= O.Thus I(t) = O.The nonnega-tive continuous inlegrand of that integral must therefore vanish; that is,

zAx,t) = z,(x,t) = O (0:$ x :$ e, t ~ O).

So z is constant; in fact, z(x,t) = Obecause z(x,O)= O. .

Thus the boundary value problem consisting of equations (1), (2), and (3)cannot have more than one solution of class e2 in R.

If Yx' instead of y, is prescribed at the end point in either or both ofconditions (2), the proof of uniqueness is still valid beca use condition (9) isagain satisfied.

The requirement of continuity on derivatives of y is severe. Solutions ofmany simple problems in a wave equation have discontinuities in theirderivatives.

I

PROBLEMS

1. In the Dirichlet problem for a rectangle treated in Seco 58(a). letfandf' be piecewisecontinuous on the interval (O,a). Verify that the formal solution found there satisfies the

condition u(x,O+) = f(x) when O < x < a ifJ(x) is defined as the mean off(x+) andf(x-) atits points of discontinuity.

2. Formulate a complete statement of the boundary value problem for steady tempera-tures in a square plate with insulated faces when the edges x = O.x = n, and r = Oare insulatedand the edge y = n is kept at temperatures u = f(x). Show that if!.f'. andJ" are continuous onthe inter\!31 O ;5;x ;5;n and 1'(0) = f'(n) = O, the problem has the unique solution

I '" cosh nyu(x,y) = _2 ao+ La. - h

cos nx11=1 cos n7r

where

2 .'a. = -1 f(x) cosnx dxn.o

(n = O. 1.2. ...).

3. Replace the infinite cylinder in Seco 6O(a) by a cylinder bounded by the surfaces p = l.z = z.. and z = Z2' where u, = O on the last two parts. Also letf(,p) be a periodic function ofperiod 2n with a continuous derivative of the second order everywhere. Then show that thefunction u given by equation (5). Seco 60. is the unique solution. satisfying our conditions ofregularity. of the problem in steady temperatures.

4. Use the uniqueness that was established in Seco 100 to show that the solution

y = A sin nx cos nat of Problem l. Seco 52. is the only solution of c1ass C2 in the region0;5; x ;5; l. t ~ O.

5. Show that the solution of Problem 3. Seco 55, is unique in the c1ass C2.

6. In Seco 51. let f(x) be such that its odd periodic extension F(x) has a continuousderivative F"(x) for all x( - ex; < x < oc); then show that the solution (9) there is unique in thec1ass C2.

t

BIBLIOGRAPHY

The following list of books and articIes for supplementary study of thevarious topies that have been introdueed is far from exhaustive. Furtherrefereneesare given in the ones mentioned here.

Abramowitz, M., and 1. A. Stegun (Eds.): "Handbook of Mathematical Functions with For-mulas, Graphs, and Mathematical Tables," Dover Publications, Inc., New York, 1965.

Apostol, T. M.: "Mathematical Analysis," 2d ed., Addison-Wesley Publishing Company, Read-ins. Massachusetts, 1974.

Bell. W. W.: "Special Functions for Scienlists and Engineers," D. Van Nostrand Company,Ltd.. London, 1968.

Berg, P. W., and J. L. McGregor: "Elemenlary Parlial DifTerenlial Equalions," Holden-Day,Inc., San Francisco, 1966.

Bowman, F.: "Inlroduction lo Bessel Funclions," Dover Publications,lnc., New York, 1958.

Boyce, W. E., and R. C. DiPrima: "Elemenlary DifTerenlial Equalions," 3d ed., John Wiley &Sons, Inc., New York, 1977.

Buck, R. c.: ¡'Advanced Calculus," 3d ed., McGraw-HilI Book Company, New York, 1978.Budak, B. M., A. A. Samarskii, and A. N. Tikhonov: "A Collection of Problems on Malhemali-

cal Physics," Pergamon Press, LId., Oxford, 1964.Byerly, W. E.: "Fourier's Series and Spherical Harmonics,". Dover Publications, Inc., New

York, 1959.

Carslaw, H. S.: "Theory of Fourier's Series and Integrals," 3d ed.. Dover Publicalions, Inc.,New York, 1930.

-, and J. C. Jaeger: "Conduction of Heat in Solids," 2d ed., Oxford Universily Press,London, 1959.

Churchill, R. V.: "Operational Malhemalics," 3d ed., McGraw-HilI Book Company, NewYork, 1972.

-, J. W. Brown, and R. F. Verhey: "Complex Variables and Applications," 3d ed.,McGraw-HilI Book Company, New York, 1974.

Coddinglon, E. A.. and N. Levinson: "Theory ofOrdinary DifTerential Equalions," McGraw-HilI Book Company, Inc., New York, 1955.

Courant. R.. and D. Hilberl:" Methods of Mathemalical Physics,"lnterscience Publishers,lnc.,New York, vol. 1, 1953; "Partial DifTerenlial Equations," vol. 2, 1962.

Davis. H. F.:" Fourier Series and Orlhogonal Funclions," AlIyn and Bacon,lnc.. Boslon, 1963.Erdélyi, A. (Ed.): "Higher Transcendental Functions," McGraw-HilI Book Company. Inc..

New York. vols. 1,2, 1953; vol. 3, 1955.Farrell, O. J., and B. Ross:" Solved Problems in Analysis as applied lO Gamma, Beta, Legendre,

and Bessel Functions." Dover Publicalions. Inc., New York. 1971.

Fourier, J.: "The Analytical Theory of Heat," translated by A. Freeman. Dover Publications,Inc.. New York. 1955.

Franklin. P.: "A Trealise on Advanced Calculus," Dover Publications, Inc., New York, 1964.

263

264 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS

Grattan-Guinness, l.: "Joseph Fourier, 1768-1830," The MIT Press, Cambridge, Massa-chusetts and London, England, 1972.

Gray, A., and G. B. Mathews: "A Treatise on Bessel Functions and their Applications toPhysics," 2d ed. (with T. M. MacRobert~ Dover Publications, Inc., New York, 1966.

Greenspan, D.: "Introduction to Partial DilTerential Equations," McGraw-HiII BookCompany, Inc., New York, 1961.

Hobson, E. W.: "The Theory of Spherical and Ellipsoidal Harmonics," Chelsea PublishingCompany, Inc., Bronx, New York, 1955.

Hochstadt, H.: "The Functions of Mathematical Physics," John Wiley & Sons, New York,1971.

Inee, E. L.: "Ordinary DilTerential Equations," Dover Publications, Inc., New York, 1956.Jackson, D.: " F ourier Series and Orthogonal Polynomials," Carus Mathematical Monographs,

no. 6, Mathematical Association of America, 1941.Jahnke, E., F. Emde, and F. Losch: "Tables of Higher Functions," 6th ed., McGraw- HiII Book

Company, Inc., New York, 1960.

Kaplan, W.: "Advaneed Calculus," 2d ed., Addison-Wesley Publishing Company, Inc., Reading,Massachusetts, 1973.

Kellogg, O. D.: "F('1Jndations of Potential Theory," Dover Publications, Inc., New York, 1953.Lanczos, c.: "Discourse on Fourier Series," Hafner Publishing Company, New York, 1966.Langer, R. E.: Fourier's Series: The Genesis and Evolution of a Theory, Slaught Memorial

Papers, no. 1, Amer. Math. Monthly, vol. 54, no. 7, part 2, pp. 1-86, 1947.Lebedev, N. N.: "Special Functions and their Applications," Dover Publications, Inc., New

York, 1972.

-, 1. P. Skalskaya, and Y. S. Uftyand: "Problems of Mathematical Physics," Prentice-Hal~ Inc., Englewood ClilTs, New Jersey, 1965.

McLachlan, N. W.: " Bessel Functions for Engineers," 2d ed., Oxford University Press, London,1955.

MacRobert, T. M.: "Spherical Harmonics," 3d ed. (with l. N. Sneddon), Pergamon Press, Ltd.,Oxford, 1967.

Magnus, W., F. Oberhettinger, and R. P. Soni: "Formulas and Theorems for the SpecialFunctions of Mathematical Physics," 3d ed., Springer-Verlag, New York, Inc., 1966.

Oberhettinger, F.: "Fourier Expansions: A Collection of Formulas," Academic Press, NewYork and London, 1973.

Powers, D. L.: "Boundary Value Problems," Academic Press, Inc., New York, 1972.Rainville, E. D.: "Special Functions," Chelsea Publishing Company, Inc., Bronx, New York,

1972.

-, and P. E. Bedient: "Elementary DilTerential Equations," 5th ed., Macmillan PublishingCo., Inc., New York, 1974.

Rogosinski, W:: "Fourier Series," 2d ed., Chelsea Publishing Company, Inc., Bronx, New York,1959."

Sagan, H.: "Boundary and Eigenvalue Problems in Mathematical Physics," John Wiley &Sons, Inc., New York, 1961.

Sansone, G.: "Orthogonal Functions," rey. ed., Robert E. Krieger Publishing Co., Inc.,Huntington, New York, 1975.

Seeley, R. T.: "An Introduction to Fourier Series and Integrals," W. A. Benjamin, Inc., New. York, 1966. .

Smirnov, V. l.: "'yqmplex ,yariables-Special Functions," Addison-Wesley PublishingCompany, Inc., Reading, 'Massachusetts, 1964.

Sneddon, 1. N.: "Elements of Partial DilTerential Equations," McGraw-HiII Book Company,Inc., New York, 1957.

-: "Fourier ~ransforms," McGraw-HiII Book Company, Inc., New York, 1951.

~

BIBLlOGRAPHY 265

-: "Special Functions of Mathematical Physics and Chemistry," 2d ed., Oliver and Boyd,Ltd., Edinburgh, 1961.

-: "The Use of Integral Transforms," McGraw-HiII Rook Company, Inc., New York,1972.

Taylor, A. E., and W. R. Mann: "Advanced Calculus," 2d ed., Xerox College Publishing,Lexington, Massachusetts, 1972.

Titchmarsh, E. C.: "Eigenfunction Expansions Associated with Second-order Dift'erential Equa-tions," Oxford University Press, London, part 1, 2d ed., 1962; part 11, 1958.

-: "Theory of Fourier Integrals," 2d ed., Oxford University Press, London, 1967.Tolstov, G. P.: "Fourier Series," Dover Publications, Inc., New York, 1976.Tranter, C. J.: "Bessel Functions with Some Physical Applications," Hart Publishing

Company, Inc., New York, 1969.Van Vleck, E. B.: The Inlluence of Fourier's Series upon the Development of Mathematics,

Science, vol. 39, pp. 113-124, 1914.Watson, G. N.: "A Treatise on the Theory ofBessel Functions," 2d ed., Cambridge University

Press, London, 1952.Weinberger, H. F.: "A First Course in Partial DifTerential Equations," Xerox College Publish-

ing, Lexington, Massachusetts, 1965.Whittaker, E. T., and G. N. Watson: "A Course of Modero Analysis," 4th ed., Cambridge

University Press, London, 1963.Zygmund, A.: "Trigonometric Series," 2d ed., 2 vols. in one, Cambridge University Press,

London, 1968.

(

BIBU01;ECA-' DE CA -

fAf-UlTADDE CI~(1 ASQtl\M'Ct-,S

A

--

Abe~ N. H., 129

Abel's test for uniform convergence, 129,210,242

proof o~ 249Adjoint of operator, 68d'Alembert, J, 44d'Alembert's solution, 41Analytic function, 175, 177Approximate solutions, 120 ""-Approximation in meaR, 57Associated Legendre functions, 245Asymptotic representation of Bessel

functions, 209n.

Basic Fourier series, 78Bemoull~ D, 44Bessel functions, 174

of first kind, 171, 179

asymptotic representation of, 209n.boundedness of, 187, 190derivatives o~ 180, 181, 187generating function for, 190graphs o~ 194integral form of, 186integrals of, 181, 184modified, 181, 198norms of, 202

orthogomil seis of, 195orthonormal seis of, 201recurrence relations for, 180series of, 204zeros of, 191-195,201

of second kind, Weber's, 178-179spherica~ 247Wronskian of, 183

INDEX"1-/

Bessel's equation, 174modified, 181self-adjoint form of, 195

Bessel's inequality, 58. 102Bessel's integral form. 186Boundary conditions, 2

Cauchy type. 21Dirichlet type, 21linear, 3linear homogeneous, 4Neumann type. 21. 244periodic, 69, 142of third type, 21types o~ 20

Boundary value problems. 2, 116approximate solutions of, 120formal solutions of, 32, 116. 133

methods of solving, 42

solutions verified, 118, 12~09. 241Boundedness. of Bessel functions. 187, 190

of Legendre polynomials, 232, 233Brown, G. H.. 135n.

Cauchy criterion for uniform convergence,248

Cauchy principal value, 160Cauchy product, 184Cauchy-Euler equation, 76Cauchy's inequality, 105Characteristic functions. 67Characteristic numbers. 67Closed orthonormal sets, 59, 60

of Legendre polynomials. 223. 235of trigonometric functions. 100. 101. 107

267

268 INDEX

Complete orthonormal sets, 56, 60, 100, 101

Complex-valued functions, 62Conductance, surface, nConductivity, tberma~ 10Conformal mapping, 42Continuous eigenvalues, 159Continuous functions, piecewise, 49

uniformly, 88Convergence, of Fourier series, 91, 94, 106

of Fourier cosine series, 95, 99, 107of F ourier sine series, 95, 99, 107in mean, 59pointw¡s~,6(j

(See also Uniform convergence)Current of heat ftow, 11

Cylindrical coordinates, 16laplacian in, 18

Czarnecki, A., 192

Derivative, left-hand, 86

right-hand, 86Differential equations, Cauchy-Euler, 76

linear, 3linear homogeneous, 4nonhomogeneous, 25, 31, 123

ordinary,38 '-(See also Partial differential equations)

Differentiation of series, 26, 28, 108Diffusion, coefficient of, 14

equation of, 14Diffusivity, thermal, 12Dirichlet, P. G. L., 45Dirichlet condition, 21Dirichlet kernel, 90, 152Dirichlet problems, 21

in rectangle, 262in regions of plane, 135-138in spherical regions, 239

Distance between functions, 52, 57

Eigenfunctions, ~3, 67linear independence of, 72

Eigenvalue problems (see Sturm-Liouvilleproblems)

Eigenvalues, 33, 67continuous, 159

Elastic (vibrating) bar, 7, 125-127Elasticity, modulus of, 7Elliptic type, partial differential equations of,

20-21Error function, 168E uler, L., 44Euler's constant, 178

j

Euler's formula, 65

Even function, 80Exponential form of Fourier series, 100

Expo~ential function, 62Extension, periodic, 79

Flux, of heat, 10total, 11

Formal solutions, 32, 116, 133F ourier, J. B. J., 45Fourier constants, 56

Fourier cosine integral formula, 158Fourier cosine series, 81, 19

convergence of, 95, 107differentiation of, 109

Fourier integral formula, 150applications of, 166-173exponential form of, 160

F ourier integral theorem, 154Fourier method, 42F ourier series, basic, 78

convergence of, 91, 94, 106differentiation of, 108exponential form of, 100generalized, 55, 143integration of, 109, 113in two variables, 139uniform convergence of, 106, 112

Fourier sine integral formula, 158F ourier sine series, 36. 82, 99

convergence of, 95, 107

Fourier theorem, 91, 99, 111Fourier transforms, 42. 161Fourier-Bessel series, 203

in two variables, 214, 215, 216F ourier's ring, 145Frobenius, method of, 175n.

Function space, 23, 50, 100Functions, analytic, 175, 177

Bessel (seeBesselfunctions)characteristic, 67

complex-valued, 62error, 168even, 80

exponential, 62gamma, 179generating, for Bessel functions, 190

for Legendre polynomials, 229harmonic (see Harmonic functions)hyperbolic, 63inner product of, 51Legendre, associated, 245

of second k ind, 222, 228. 238

normalized, 52norms of, 52odd,81

piecewise continuous, 49potentia~ 15sine integral, 115n.trigonometric, orthonormal sets of, 53, 78

closed, 100, 101, 107uniformly continuous, 88weight,64

Fundamental interval, 53

Gamma function, 179Generalized Fourier series, 55, 143

Gibbs phenomenon, 112

Hankel's integral formula, 205Harmonic functions, 15

in circular regions, 137, 141in rectangular region, 136in spherical regions, 239-242, 246in strip, 141, 172

Heat conduction, postulates of, !O, 11Heat equation, 12

solutions of, product of, 142uniqueness of, 129, 251

Heat ftow, current of, 11 ----Heat transfer at surface, 13, 134, 143, 210

Hermite polynomials, 245n.Hermitian orthogonality, 64, 66Homogeneous equations, 4Hooke's law, 7Hyperbolic functions, 63Hyperbolic type, partial differential

equations of, 21, 44

Improper integral, principal value of, 160uniform convergence of, 155

Inequality, Bessel's, 58, 102Cauchy's, 105Schwarz. 54

Initial value problem, 39Inner prod uct, 46, 51Insulated surface, 13

Integral equation, 161Integral form, of Bessel functions, 186

of Legendre polynomials, 231I.ntegral formula, Hankel's, 205

(See a/so Fourier integral formula)Integral theorem, Fourier, 154Integratiol\ of series, 28, 109, 113

Jacobi polynomials, 245

I

INDEX 269

Kronecker, L., 83Kronecker's <5,48, 52

Lagrange's identity. 77trigonometric, 63

Laguerre polynomials, 245n.Laplace transforms, 42, 43, 162Laplace's equation, 13, 15 -Laplace's integral form, 231Laplacian operator, 12, 15

in cylindrical coordinates, 18in spherical coordinates, 19

Least squares approximation, 57

Left-hand derivative, 86 .Legendre functions, associated, 245

of second kind, 222, 228, 238

Legendre polynomials, 221boundedness of, 233closed sets of, 223, 235derivatives of, 226, 233

generating function for, 229integral form of, 231integrals of, 228norms of, 227

order property of, 232orthogonal sets of, 223, 224orthonormal sets of, 227, 228recurrence relations for, 226Rodrigues' formula for, 225series of, 234zeros of, 238

Legendre series, 233Legendre's equation, 218

self-adjoint form of, 222Leibnitz. 44Leibnitz' rule, 225Limits, in mean, 59

one-sided, 49Linear boundary condition, 3

homogeneous, 4Linear combinations, 23

extensions of, by integrals, 164by series, 26

Linear differential equation, 3Linear independence of eigenfunctions, 72Linear operators, 23, 162

adjoint of, 68product of, 24self-adjoint, 68sum of, 24

Liouville, J., 67n.

Mapping, conforma~ 42

)

270 INDEX

Mean. approximalion in. 57limit in. 59

Mean convergence. c10sed in sense of. 59Membrane. analogy, 20

partial differential equations foro 8static. 9, 16, 20, 137(See o/so Vibrating membrane)

Method of Frobenius, 175n.Modified Bessel functions. 181. 198

Modulus of Elasticity, 7de Moivre's formula, 66

M-test, Weierstrass, for uniform convergence,28,155

Neumann condition, 21, 244Newton, 44

Newton's law of cooling, 13, 14,210Newton's second law of motion, 5, 8

Nonhomogeneous partial differentialequations, 4, 25, 31, 123

Normalized functions, 52Normalized vectors, 47Norms, of Bessel functions, 202

of functions, 52of Legendre polynomials, 227 ---of vectors, 46

Odd function, 81One-sided derivatives, 86One-sided limits, 49

Operators, laplacian, 12, 15linear, 23, 162

adjoint of, 68product of, 24self-adjoint, 68sum of, 24

Ordinary point, 218Orthogona] sets, 35, 47, 52

of Bessel functions, 195

of Legendre polynomials, 223, 224Orthogonality, 35, 47, 52, 64

of eigenfunctions, 69hermitian, 64, 66

Orthonormal sets, 47, 52

of Bessel functions, .201c1osed, 59, 60

complete, 56,60, 100, 101of Legendre polynomials, 227, 228of Irigonometric functions, 53, 78, 100, 107

Parabolic type, partia] differential equationsof,21

Parseval's equation, 60, 107, lB, 189

')

~

~,Partial differential equations. of diffusion, ]4

for elastic bar. 8

genera] linear. of second order. 3general solutions of, 40for membrane, 8

for stretched string. 6types of, 20

Periodic boundary conditions, 69, 142Periodic extension, 79Piecewise continuous functions, 49

Pointwise convergence, c10sed in sense of, 60Poisson's equation, 9, 21

uniqueness of solution of, 257Poisson's integral formula. 145Polynomials. Hermite, 245n.

Jacobi. 245

Laguerre. 245n.

Legendre (see Legendre polynomials)Tchebysheff, 64

Postu]ates of heat conduction, lO, 11Potential, 15, 16

in space bounded by planes, 137in spherical region, 239

Principal value, of improper integral, 160of series, 100

Principie of superposition of solutions, 25

Recurrence relations, for Bessel functions. 180

for Legendre polynomials, 226Riemann-Lebesgue lemma, 88Right-hand derivative, 86Rodrigues' formula, 225

Schwarz inequality, 54Self-adjoint operator, 68Semi-infinite solid, temperatures in, ]66, 171,

173

Semi-infinite vibrating string. 171Separation constant, 33Separation of variables, method of, 34, 42Series, of Bessel functions, 204

differentiation of, 26, 28, 108Fourier-Bessel, 203

integration of, 28, 109, 113Legendre, 233of Legendre polynomials, 234principal value of, 100Sturm-Liouville, 143

superposition of solutions by, 26(See o/so Fourier cosine series; Fourier

series; F ourier sine series; U niformconvergence )

Sine integral function, 115n.

)

Singular point. 174. 178Singular Sturm-Liouville problems, 69. 159,. 161, 195, 222

Space ofTunctions, 23, 50, 100Spherical Bessel functions, 247Spherical coordinates, 18

laplacian in. 19Spherical regions, Dirichlet problems in. 239

harmonic functions in, 239-242. 246

Steady temperatures, 13in cylinder, 142,212,216,262in hemisphere, 242, 246in rectangular plate, 137, 140, 141,258,262in slab, 14

in sphere, 239, 246hollow, 20

in wall, 141

in wedge-shaped plate, 141String (see Vibrating string)Sturm, J. C. F., 67n.

Sturm-Liouville equation, 67, 68Sturm-Liouville problems, 33, 66-70-

singular, 69, 159, 161, 195, 222Sturm-Liouville series, 143Superposition of solutions, 25, 42, 123

by integrals, 164by series, 26

Surface, insulated, 13Surface conductance, 13

Taylor, B., 44Tchebysheff polynomials, 64Telegraph equation, 21Temperatures, in bar, 127, 247

in cylinder, 207-211, 215, 255in prism, 142in semi-infinite solid, 166, 171, 173in slab, 30, 127-135, 143, 254

in sphere, 147, 247in unlimited medium, 169in wedge, 216in wire, 15, 134(Seeo/so Steady temperatures)

Tension, in membrane, 8

in string, 4Thermal cJnductivity, 10Thermal diffusivity, 12Tilde symboL -,36Total flux, 11

INDEX 271

Transforms, Fourier, 42, 161

Laplace, 42, 43, 162. Trigonometric functions, orthonormal seIS

of, 53, 78. 100, 107

Trigonometric idenlity, Lagrange's, 63

Uniform convergence, Abel's test for, 129,210, 242, 249

Cauchy criterion for, 248of Fourier series, 106. 112of improper integrals, 155of series, 28

Uniformly continuous function, 88

Uniqueness of solutions, 248of heat equation, 129, 251of Laplace's equation, 257of ordinary differential equations, 39of Poisson 's equation, 257of wave equation, 262

Vectors, 46-49

Vibrating bar, 7, 125-127Vibrating membrane, 8

circular. 213-215rectangular, 137

Vibrating string, 4-7, 32-37, 116--124, 126,127

approximating problem for, 120end conditions for, 6

equation of motion of, 6initially displaced, 32-37, 43, 116--121, 123

with air resistance, 145semi-infinite, 171

Wave equation, 6-8, 44, 171,261solution of, general, 41

uniqueness of, 262Weber's Bessel function of second kind,

178-179

Weierstrass M-test for uniform convergence,28, 155

Weight function, 64Wronskian of Bessel functions, 183

Zeros, of Bessel functions, 191-195,201

of Legendre polynomials, 238

Churchill, R. v. - Fourier Series and Boundary Value Problems

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