Problems of Mathematical Physics (Selected Russian Publications in the Mathematical Sciences)

PROBLEMS OF MATHEMATICAL PHYSICS

\-¥SELECTED RUSSIAN PUBLICATIONS IN THE MATHEMATICAL SCIENCES Richard A. Silverman, Editor

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PRENTICE-HALL INTERNATIONAL, INC.,

London

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Sydney

I'RENTICE-liALL OF CANADA, LTD., Torollto l'RJ::NTICE-HALL OF INDIA (PRIVATE) LTD., PRENTICE-HAI.L OF JAPAN, INC.,

Tokyo

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PROBLEMS OF MATHEMATICAL PHYSICS

N. N. LEBEDEV I. P . SKALSKA YA Y. S. UFLYAND A. F. Joffe Physico-Technical Institute Academy of Sciences, U. S. S. R. Revised, Enlarged and Corrected English Edition Translated and Edited by

Richard A. Silverman

With a Supplement by

Edward L. Reiss Courant Institute of Mathematical Sciences New York University

PRENTICE-HALL, INC. Englewood Cliffs, N.J.

© 1965 by PRENTICE-HALL, lNC. Englewood Cliffs, N.J.

All rights reserved . No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher.

...

Library of Congress Catalog Card Number 65-26925 Printed in the United States of America 71767-C

AUTHORS' PREFACE

The aim of the present book is to ·help the reader acquire the proficiency needed to successfully apply the methods of mathematical physics to a variety of problems drawn from mechanics, the theory of heat conduction, and the theory of electric and magnetic phenomena. A wide range of topics is covered, including not only problems of the simpler sort, but also problems of a more complicated nature involving such things as curvilinear coordinates, integral transforms, certain kinds of integral equations, etc. The book is intended both for students concomitantly studying the corresponding topics in courses of mathematical physics, and for research scientists who in their work find it necessary to carry out calculations using the methods described here. We also think that quite apart from its value as a tool for acquiring technique, the book can also serve as a handbook, especially in view of the fact that answers to the problems are included. A rather solid background in applied mathematics is needed to profit from the book in its entirety. However, most of the problems appearing in Chapters 2 to 5 will be access~ble to those who have taken only the usual first course in methods of mathematical physics. Chapters 6 to 8 are more specialized, and presuppose some familiarity with special functions , integral transforms, integral equations, and so on. To make the book easier to use, each section begins with a brief introduction describing its contents and presenting a certain amount of relevant background information. However, it is not claimed that this information is complete in any sense, and the reader v

Vi

AUTHORS'PREFACE

desiring further details must consult the literature, e.g., the books and monographs cited at the end of each chapter. The majority of problems in this collection are accompanied by hints, facilitating the choice of meaningful methods of solution. In addition, certain problems, whose numbers are equipped with asterisks (e.g., *52, * 148, etc.), are solved in detail in a special section at the end of the book. The problems singled out in this way have been selected either because they illustrate the application of certain specific methods, or because of their special difficulty or particular importance in the applications. Because of the applied character of the book, we restrict ourselves to formal solutions, whose rigorous justification can be supplied by the interested reader. In compiling the collection, we have consulted not only the classic works on mathematical physics, but also a number of journal articles. Material accumulated during years of teaching and research in the Department of Mathematical Physics at the Leningrad Polytechnic Institute, as well as work done in connection with industrial projects, plays a role in the material presented here. It would be impractical, and in many cases impossible, to cite the original source where a given problem was solved for the first time. fhus references to the literature have been confined to cases we find particularly relevant.

We would like to take this opportunity to thank Prof. G. A. Grinberg for many valuable suggestions made in the course of writing the book. N. N. L. I.

P. S.

Y. S. U.

TRANSLATOR'S PREFACE

The present edition differs from the Russian original in various respects, of which three merit particular mention: 1. The Bibliography has been expanded and updated. For example, the original sources of works translated into Russian have been tracked down, all references have been equipped with titles, further references (especially, later editions and English translations) have been added, and so on. As in other volumes of this series, the system of references is in "letter-number form." Thus LlO refers to the tenth paper (or book) whose {first) author's surname begins with the letter L, where the entire Bibliography is arranged in lexicographic order, and chronological order as well, whenever there are several papers by the same author. 2. Working from an extensive list of errata sent me by the authors, I have corrected numerous misprints and mistakes present in the Russian edition. I am particularly grateful for their help, since the task of eliminating errors from a book of this type (consisting primarily of problems and answers) is both imperative and one which only the authors themselves can perform in finite time! The authors have also been kind enough to answer a number of specific questions that arose in the course of the translation. 3. It was felt that the English-language edition would benefit greatly by the addition of material on the approximate solution of problems of mathematical physics, since the emphasis of the Russian vii

Viii

TRANSLATOR'S PREFACE

edition is on exact solutions. This led to the writing of a Supplement on variational and related methods by Professor Edward L. Reiss of the Courant Institute of Mathematical Sciences of New York University. The Supplement is independent of the rest of the book, even to the extent of having its own references. Although the expression "mathematical physics" was preserved in the title of the book as entirely fitting and perhaps the most descriptive, one could just as well have substituted "applied physics and engineering" or "applied mathematics," at least in the sense in which these terms are used in the United States. R. A. S.

I

'

CONTENTS

PART

1

1

PROBLEMS,

Page 1.

.

DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS, Page 3. 1. Mechanics, 3. 2. Heat Conduction, 9. 3. Electricity and Magnetism, 11.

2

SOME SPECIAL METHODS FOR SOLVING HYPERBOLIC AND ELLIPTIC EQUATIONS, Page 20. 1. Hyperbolic Equations, 20. 2. E ll iptic Equations: The Green's Function Method, 27. 3. E ll iptic Equations: The Method of Conformal Mapping, 33.

3

STEADY-STATE HARMONIC OSCILLATIONS, Page 42. 1. Elastic Bodies: Free Oscillations, 43. 2. Elastic Bodies: Forced Oscillations, 46. 3. Electromagnetic Oscillations, 49.

4

THE FOURIER METHOD,

Page 55.

I. Mechanics : Vibrating Systems, Acoustics, 60. 2. Mechanics: Statics of Deformable Media, Fluid Dynamics, 73. ix

X

CONTENTS

4

THE FOURIER METHOD-Continued 3. Heat Conduction: Nonstationary Problems, 77. 4. Heat Conduction: Stationary Problems, 83. 5. Electricity and Magnetism, 91.

5

THE EIGENFUNCTION METHOD FOR SOLVING INHOMOGENEOUS PROBLEMS, Page 103. I. Mechanics: Vibrating Systems, 107.

2. 3. 4. 5.

6

Mechanics : Statics of Deformable Media, 114. Heat Conduction: Nonstationary Problems, 119. Heat Conduction: Stationary Problems, 124. Electricity and Magnetism, 131.

INTEGRAL TRANSFORMS,

Page 143.

1. The Fourier Transform, 146.

2. 3. 4. 5.

7

CURVILINEAR COORDINATES, 1. 2. 3. 4. 5. 6. 7. 8.

8

The Hankel Transform, 160. The Laplace Transform, 169. The Mellin Transform, 189. Integral Transforms Involving Cylinder Functions of Imaginary Order, 194.

Page 203.

Elliptic Coordinates, 204. Parabolic Coordinates, 210. Two-Dimensional Bipolar Coordinates, 212. Spheroidal Coordinates, 219. Paraboloidal Coordinates; 231. Toroidal Coordinates, 233. Three-Dimensional Bipolar Coordinates, 242. Some General Problems on Separation of Variables, 247.

INTEGRAL EQUATIONS, 1. Diffraction Theory, 254. 2. Electrostatics, 259.

Page 253.

CONTENTS PART

2

SOLUTIONS,

Xi

Page 273.

MATHEMATICAL APPENDIX,

Page 381.

I. Special Functions Appearing in the Text, 381. 2. Expansions in Series of Orthogonal Functions, 384. 3. Some Definite Integrals Frequently Encountered in the Applications, 386. 4. Expansion of Some Differential Operators in Orthogonal Curvilinear Coordinates, 388.

Supplement. VARIATIONAL AND RELATED METHODS, Page 391.

1. Variational Methods, 392. 1.1. Formulation of Variational Problems, 392. 1.2. The Ritz Method, 396. 1.3. Kantorovich's Method, 401. 2. Related Methods, 404. 2.1. Galerkin's Method, 404. 2.2. Collocation, 407. 2.3. Least Squares, 411. 3. References, 412. BIBLIOGRAPHY, NAME INDEX,

Page 415. Page 423.

SUBJECT INDEX,

Page 427.

Part

1

PROBLEMS

'Ii''

'

I,J

": I

''

I

1 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS

Chapter 1 is devoted to problem material on the derivation of the equations of mathematical physics and the formulation of appropriate initial and boundary conditions. It also serves as a convenient place to list the basic equations appearing later in the book. Throughout, we assume that the reader is familiar with the physical Jaws underlying the mathematical formulation of the problems which arise in various branches of physics. The chapter consists of three sections devoted in turn to problems of mechanics, heat conduction and the theory of electric and magnetic phenomena. Each section starts with the basic equations governjng the corresponding set of problems, with appropriate references to sources where the derivations can be found. Special attention is dev.oted to the formulation of problems of electrodynamics, since this subject is inadequately covered in the available Jiterature.1

I. Mechanics This section contains problems on the derivation of equations of motion and formulation of initial and boundary conditions for vibrating strings, membranes, rods and plates, as well as some examples pertaining to the statics of deformable media. It will be assumed that the reader has already 1 Those particularly in terested in mathematical aspects of the formulation of physical problems can find relevant material in CS, GJ, LJ, P2, Sl and SJ 3. (The reference scheme is explained in the Translator's Preface.)

3

4

DERIVATION OF EQUATIONS AKD FORMULATION OF PROBLEMS

encountered the basic equations in a f1rst course on mathematical physics. 2 Thus we shall merely list the equations concisely, at the same time explaining the notation to be used in the book. I. The equation of a vibrating string is

au

1

au

2

2

2

2

Ox

2

v Ot

_ q(x, t) T

where u(x, t) is the displacement of the point of the string with abscissa x at the timet, q(x, t) is the external load per unit length, Tis the tension, and p is the linear density. 2. The equation for longitudinal oscillations of a rod of constant cross section is v,........

J£

-'

p

where u(x, t) is the displacement of the cross section of the rod with abscissa x at the time 1, E is Young's modulus, and pis the density.

3. The equation for transverse oscillations of a rod (beam) is

where u(x, t) is the displacement of the points along the midline of the rod, q(x, t) is the external load per unit length, E is Young's modulus, J is the moment of inertia of a transverse cross section, pis the density, and Sis the cross-sectional area. 4. The equation of a vibrating membrane is 2

0u

ox'

+ o'u - ! o'u ~ oy 2 v' ot'

q(x,

y, t)

T

v~Jf;. p

where u(x, y, 1) is the displacement of the point (x, y) of the membrane at the timet, q(x, y, t) is the external load per unit area, Tis the tension per unit length of the boundary of the membrane, and p is the surface density. 5. The equation for transverse oscillations of a thin elastic plate is

!l.'u

+

_1_ b'

o'u ot'

=

q(x,

y, t)

D

'

b'~

JD

ph,

2 See 56 (Vol. II), S14, Tl and T2. Concerning the derivation of the equations of vibrating plates, see T4.

I' ll! 111.

I

DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS

5

where u(x, t) is the displacement of the point (x, y) of the midplane of the plate at the timet, q(x,y, t) is the density of the external load, D is the flexural rigidity, h is the thickness, p is the density, and

02

t::. = -

02

ox2

+- 2 oy

is the two-dimensional Laplacian operator. The above equations lead to corresponding equations for static deflections, if we regard the external load q and the unknown displacement u as independent of the time t. For example, the equilibrium equation for the membrane is

6. 02U

+0

2 U

ox 2 o/

= _ q(x, y) T

the static deflection of the plate satisfies the equation

7. and so on. Among the other equations governing the statics of elastic bodies which will figure in this book, we cite the familiar equation 8.

for twisting of a prismatic rod, where u(x, y) is the torsion function. We now give some problems on the formulation of initial and boundary conditions for these equations, and also some problems on the derivation of o ther differential equations. 1. Describe the initial and boundary conditions for a vibrating string with fixed ends (0 < x < /), which is stretched at the point x = c and time t = 0 to a heig\lt h, and then released without initial velocity.

Ans.

hx

uJ,_o = f (x)

=

c ' ( h(l - x) l - c '

0

< X< C,

C< X< l,

uJ:t=O= uJx=l = 0.

auot

I - o· t- o -

'

6

2

PROB.

DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS

2. A concentrated load of mass m 0 is fastened at the point x = c of a string 0 < x < I of length I. Find the equations describing vibrations of the string with arbitrary initial conditions, assuming that the ends of the string are fastened.

Ans. 0 - oo. However, these conditions need not be included in the fo rmu lation of the problem, since the uniqueness of the solution is guaranteed by the above requirement th at the potential u a pproach zero uniformly as R-->- oo.

+ +

18. A charge Q is placed at the point M 0 = (x 0 , y 0 , z0) near a conductor at potential V, bounded by a surface S. Formula te the corresponding pro blem of electrostatics. 8

The quantities A and u a re no t independent, but are connected by the relation . dJ V

A

ou + 41ti.L + E:!J. II = 0. c ot c

14

DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS

PROB.

19

Ans. The potential u satisfies Laplace's equation at every point of the region surrounding the conductor, except at the point M 0 , near which

q_ + a regular function,

u=

R

(R = IM0 MI is the distance between the points M and M 0 ). Moreover, the = V and the condition that potential satisfies the boundary condition --+ 0 uniformly in all directions.

uis

ui

0'.)

19. A thin charged wire of charge q per unit

length is placed inside a grounded cylindrical shell whose generators are parallel to the wire (see Figure 1). Formulate the corresponding twodimensional electrostatic problem. Ans. The potential u satisfies the twodimensional Laplace equation

0

r FIGURE

iJ2u

l:iu = - 2 OX

1

iJ2u

+ -oy

2

= 0

in the whole region D except at the point M 0 , where the potential has a logarithmic singularity u = - 2q ln R

The boundary condition is

+ a regular function.

ulr = 0.

20. Reformulate the preceding problem fo r the case where the charged wire is placed outside the conductor, and the total charge per unit length of the conductor is specified instead of its potential. Ans. T he boundary condition is now

auos

I=o r

(ofos denotes differentiation along the tangent to the contour

r), while the

condition at infinity becomes

uloo =

- 2(Q

+ q) ln R +a bounded function,

where Q is the charge per unit length of the conductor. 21. Show that the problem of the current distribution in a thin conducting · shell (see Figure 2) reduces to integration of the equation ·(e)

l:iu = .1..!!_' ah

PROB.

22

DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS

15

where u is the potential of the current distribution in the shell (j = - a grad u is the current density vector), cr is the conductivity and h the thickness of the shell, j ;,•> is the normal component of the density of current applied to the shell by using suitable electrodes, and tl is the appropriate two-dimensional Laplacian, i.e.,

1

tlu _ _ [~(H!l H<XH[3 HIX

oa

ou) + ~(HIX ou) ] oa

()~ H!l ()~

(ds is the element of arc length on the s urface of the shell).

Hint. Average equation (J ), p. 12 (giving the volume distribution of current) over the thickness of the shell. n

-

FIGURE

-

2

FIGURE

3

22. Suppose an object of arbitrary shape, made of magnetic material of permeability fL, is magnetized by being introduced into a homogeneous magnetic field H 0 (see F igure 3). Formulate mathematicaUy the corresponding problem of magnetostatics. Ans. If u is the potential of the magnetic field H (i.e., H = - grad u), the problem reduces to integration of the equation

,

tlu;

=

0,

i

= 1, 2,

with the boundary conditions u1is

oull

= Uzl s•

on s

=

j.L

ou21

on s

and the following conditions at infinity -grad u1 ioo

Ho,

=

Hint. It helps to keep in mind that in the source-free part of space, the magnetic field H satisfies the equations

curl H

=

0,

divH = 0,

which imply H = -grad u,

tlu

=

0.

16

DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS

PROB.

23

23. The differential equations for wave propagation down a long transmission line, with self-inductance L, capacitance C, resistance R and leakage conductance G per unit length, take the form

au ax

of + RI,

au at

of

- - ·C-+Gu,

- - = L-

ax

at

where u(x, t) and I(x, t) arc the values of the voltage and current at the point x at the time 1 (see Tl, p. 18). Formulate initial and FIGURE 4 boundary conditions for wave propagation along such a line, if at one end a constant voltage E is switched on in series with a lumped resistance R 0 , while the other end is terminated by a coil of self-inductance L (see Figure 4).

Ans. uJ,~o ~ 0,

uJx-0 + R,lJx~O• Show that if i''' ~ 0 and pJ,~o ·E ·.:

24. 0, then the differential equations for the electromagnetic potentials A and u can be satisfied by setting

u = -· div II, A ___ "-~-' an + 4"~-'"' n, c c where II is the Hertz vector satisfying the equation

at

Ll.II _ "fl c

2

a'II _ 4T:ay. an

at

2

c

2

at

~

0

'

Derive expressions for the vectors E and H in terms of the vector II.

Ans.

" ('ail + -4r.a II) ,

E ~ curl curl II. c Hint. According to footnote 7, p. 12, it follows from pJ,~o = 0 that p ~ 0 for arbitrary I. H = - curl -

at

"

i'''

= 0 and

25. Verify that ifj''' ~ 0 and pj 1 .• 0 ~ 0, then the vectors E and H satisfy the same differential equation as the Hertz vector, i.e., 9

11

In some prohlcms it is more convenient to start from these equations than from the

CljLJH!ions for the clct.:tromagnctic potentials or for the Hcrt:t: vector.

!'ROB.

27

DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS

(7

26. Show that for steady-state harmonic oscillations of frequency w, in which the time dependence of the quantities defining the electromagnetic lield (i.e., the vectors E and H, the charge and current densities which are the sources of the field ~tc.) is characterized by a factor ei"'t, Maxwell's equations (see p . 12) take the form

.1.A*

+ eA* = - 4itfL i el*, c

.! curl A* ,

H* =

fL

4itfL E* = - -iw2 ( curl curl A* - - J·

X

O< t < -,

0, v 2T

U =

V

For x

v

r a, 36. In a gas initially at rest, a condensations = s0 localized in the volume bounded by a surface a is created at the time t = 0. Show that the condensation at the point M = (x, y, z) at an arbitrary timet is given by the expression s(x, y, z, t)

=

s0 i

ot

(t .5!..!.), s., 1

where S.,, is the sphere of radius vt with center at the point M, and a 1 is the part cut out of S.,1 by the surface a. Hint. Use the general solution of the homogeneous wave equation !:,.u - _!_ o2u = 2 v2

ot

0

for arbitrary initial conditions (see Gl, p. 197). 37. The solution of the Cauchy problem for the three-dimensional inhomogeneous wave equation I 2u !:,.u - 2 -2 = - 4np(x, y, z, t), v with initial conditions

o ot

ult=o = 0, y > 0.

Ans. G(M, Mo)

1 [

= 4n

1 1 JMM0 1- JMM 2 l

1

J

1

+ JMM l- JMM 3

4

l '

where M = (x,y, z),M0 = (x0 ,y0 , z0),M2 = (-x0 ,y 0 , z0 ), M 3 = ( - x 0 ,-y0 , z0) and M 4 = (x0 , - y 0 , z0) .

Hint. Use the method of images. 43. Using the method of images, construct the Green's functi on for the Dirichlet problem in the case where the region D is the part of space lying between two parallel planes z = ±1/2. • See Tl , and in particular G5, which contains a number or interesting applications of the method of inversion to problems of electrostatics.

30

SOME SPECIAL METHODS

PROD.

44

Ans.

GM M - -1 ( ' o)-

00

47t 71~00

{

+ [z -

J R2

1

z0

-

2nlf 1

- J R + [z + z 0 2

where

R = Jcx - xo)2

}

(2n

+lW '

+ (y- Yo?·

44. Use the method of inversion to deduce the Green's function for the Dirichlet problem in the case where Dis a sphere of radius a with its center at the origin 0, assuming that the expression for the Green's function of a halfspace is known.

Ans.

1[ 1

a

1

J

G(M, Mo) = 47t IMM0 1 - IOM0 I IMMtl ' where M 1 is the image of th e point M 0 in the sphere.

45. Find the Green's function for a hemisphere of radius a.

Ans.

1[ 1 a 1 I G(M, Mo) = 47t IMM0 1- IOM 0 IIMM1I - IMM2i

a

1

J

+ IOMoiiMM31

'

where M 1 is the image of M 0 in the corresponding full sphere, M 2 is the image of M 0 in the diamctral plane of the hemisphere, and Ma is the image of M2 in the full sphere. 46. The Green's function G

= G(M, M 0 ) for the Neumann problem 5

flu = - F(M),

ou

I = f(P)

on; s

is defined by the conditions l. G is harmonic in D except at the fixed point M 0 , near which G is of the form G

=

1 47t IMM0 1

+ a regular function ;

5 Here M is a point of the three-dimensional region D, P is a point of the surface S bounding D, and the functions f and F satisfy the condition

Js fdcr = JD Fd-r

for the solvability of the Neumann problem. If the Green's function is known, the solutio n is given by the formula u(Mo)

= -

f

8

J(P)G dcr

+JD F(M)G d-r + const.

PROB.48

SOME SPECIAL METHODS

31

2.

;~Is~~· where S 0 is the area of the surface ..~·;

3.

L

G dr;- 0.

Verify that in the special case where D is a sphere of radius a with center at the origin 0, the Green's function is G(M, M,)

1

a

4" IMM,I

4" IOMoiiM M,l

== - - - ' - - -

1 I - - In --=2- - 4rra 1 .. IOQI + IMM,I

IOM 1 1

21ta'

IOM,I

where M 1 is the image of the point M, in the sphere, and Q is the foot of the perpendicular dropped from the point M onto the line OM1 • 47. A conductor bounded by a closed surface S and held at a given potential V is introduced into an arbitrary external field with potential u0 • Suppose we know the charge density p(P, M,) at the point P of the surfaceS in the case where the surface is grounded and the external field is due to a unit charge at an arbitrary point M0 outside the conductor. Show that the potential distribution in the general case is given by the formula u(M,) == V

+ u0(M 0) +

Lu (P)p(P, M 0

0)

dcr.

Hint. The formula

w(M 0)

~i

s

f(P) a_G_ da

an

represents the solution of the boundary value problem Llw == 0 outside S,

Wls == j(P),

WI~-

0

in terms of the Green's function. Apply this formula to the function w == u- u0 , bearing in mind the electrostatic interpretation of the Green's function. 48. Find the Green's function for the two-dimensional Dirichlet problem, assuming that we know the function ~ • ~(z) mapping a given region in the z-plane conformally onto the upper half of the ~-plane (lm ~ > 0). Use the result to construct the Green's function for the half-strip x > 0, 0 < y ,;;; n.

32

SOME SPECIAL METHODS

PROB.49

Ans.

_!_ Re {In~

G(M, Mo) = -

- ~0},

~ - ~0

27t

where ~ and ~ 0 are the points of the half-plane corresponding to M = (x, y) and M 0 = (x0 , y 0). In the case of the half-strip,

G(M, M 0 ) = _ _!_In [cosh (x 4TI [cosh (x

+x

cos (y + Yo)][cosh (x - x 0 ) - cos (y - y 0 ) ] -Yo)][cosh (x - x 0 ) - cos (y + y 0 ) ] •

0) -

+ x 0 ) - cos (y

Hint. The conformal mapping of the half-strip onto the half-plane is accomplished by the function ~ = cosh z. 49. The boundary value problem !).2u

=

F(M),

ou on

uis = 0,

I= 0

s

(f). is the two-dimensional Laplacian, and M is a point of a planar region D bounded by a contour S) is encountered in the theory of bending of th in elastic plates. The solution of this problem can be written in the form u(M0 ) = fv F(M)G d-.

(d-r is an element of area), where the Green's function G = G(M, M 0 ) is defined by the conditions I. G is the solution of the biharmonic equation /). 2u = 0 which is regular (i.e., free of singularities) in D , except at the fixed point M 0 , near which G is of the form G =

2.

8~ IMM j ln jMMol + 2

0

a regular function ;

oGIs = o.

Gis = 0,

on

Verify that in the special case where Dis a disk of radius a with its center at the origi n 0, the Green's function is given by

G(M, Mo) =

~[IMMol l61t

a2

2

2

In

2

a IMMol 2 + (1 IOM 0 I2 IMM1I

2

_ IO~o1 ) (I a

_ IOM1 a2

where M 1 is the image of the point M 0 in the circle bo unding D.

2 )] . '

SOME SPECIAL METHODS

33

3. Elliptic Equations: The Method of Conformal Mapping In mathematical physics, one often encounters the problem of finding a function which is harmonic in a two-dimensional region D and satisfies the boundary condition (Sa) f or ~J; (5b)

ulsau

I

on s

wherefis a given function and n is the normal to the contourS bounding D. For example, such problems arise in studying electrostatics, magnetostatics, heat conduction, flow of ideal fluids, filtration phenomena, and so on. An effective method of solving problems of this kind is to construct a function of a complex variable ~ ~ F(z) such that F(z) is analytic in D and maps D conformally onto a region D* (with boundary S*) of a special form for which the solution of the given problem is either known or can be found more simply than for the original region D. Here it is assumed that F'(z) is nonzero in the region D, a condition which guarantees that the mapping is one-to-one. In asserting that this method leads to a solution of the boundary value problem, we rely on the fact that the Laplacian and the boundary conditions (Sa) and (5b) preserve their form' under the transformation from the variables x andy to the new variahles ~ and 1J defined by the relation ~

+ ;~~ ~ F(x + iy).

The method of conformal mapping can also be used to deal with more complicated boundary value problems, e.g., problems where the value of the unknown function u is specified on parts of the contour S while the value of oufon is specified on the rest of S, problems of jet flow of an ideal flow where the form of S is not known in advance but is determined in the course of solving the problem, and so on. In many cases, the construction of the function ~ ~ F(z) mapping the region D onto the region D* can be accomplished by consecutive application of several mappings which involve elementary functions. Of particular importance in applied work is the case where D is a polygon and D* is the upper half-plane. Then the function effecting the mapping can be found by using the familiar Schwarz-Christoffel transformation (see WI). The use of conformal mapping to solve problems of mathematical physics, involving 1 ; In the case of the boundary condition (5b), the value taken by the normal derivative on the contour S* is

34

PROB.

SOME SPECIAL METHODS

50

the biharmonic equation as well as Laplace's equation, is amply discussed in books on complex variable theory and in special monographs (see B3, FlO, MlO, S7, etc.). Hence we confine ourselves here to a few typical problems which illustrate the technique of the method, assuming that the reader is already familiar with the elementary theory of conformal mapping. In most of the problems, the required conformal mapping can be found by using the Schwarz-Christoffel transformation. Problems 51, 57 and 59 require knowledge of the properties of elliptic integrals and Jaco bian elliptic functions. In connection with Probs. 50-54, the following remarks will be found helpful : If cp is the potential of a stationary plane-parallel flow of an ideal fluid , described by the velocity field v = - grad cp, then by the complex potential w = w(z) is meant a function of the complex variable z = x iy wh ose real part eq uals cp. In other words, w = cp ilji, where 1j; is related to cp by the Cauchy-Riemann equations

+

+

oy

OX

ox

The lines of flow or streamlines are described by the family of curves 1j; = const, and hence the function x is called the stream function. The amount of fluid Q flowing per unit time between two streamlines lj; = o/ 1 and 1j; = o/ 2 (in a slab of unit thickness parallel to the xy-plane) is given by

The components of the velocity vector v = Vz tive of the complex potential by the formula

.

Vz- IV11

+ iv

11

are related to the deriva-

dw

=- -,

dz

The complex potential is a valuable tool for studying plane-parallel flows. 7 50. An ideal fluid, whose velocity at infinity equals v., flows past an obstacle in the shape of an elliptical cylinder

=

V 00 ,

v11 = 0,

Use the method of conformal mapping to find the complex potential of the flow. 1 Similarly, in the theory of stationary heat flow and in electrostatics, one can introduce complex potentials, with the role of v and

a, 0 < arg t
)A Y -k-''

(0 < arg ~

< 7t),

where

ak• A

=

--E(--k-:---)..:_:.:.:k---:;,,;-K-(k:-:-) '

and the modulus of the elliptic integrals is determined from the condition b a

E(k') - k 2 K(k') E(k)- k' 2 K(k) •

Hint. Use the Schwarz-Christoffcl transformation to map the region occupied by the ftow and lying above the axis of symmetry y = 0 onto the half-plane in such a way that the vertices =a, ±a ib go into the points

±1/k,

=

1_1.

*52. Study the two-dimensional motion of an ideal fluid in the channel of variable cross section shown in Figure 7, assuming that at infinity the direction of the flow coincides with the x-axis and has the values

(av.

=

bv0 ).

y

----,-------jC 0 ' 8

-

A

+--- 2~--E__,

2a ------"0

; FIGURE 7

36

PROD. 53

SOME SPECIAL METHODS

Find the distribution of velocity along the axis of symmetry of the flow.8

Ans. The velocity distribution in parametric form is given by the equations

=. = I[ln t - 1 + ~In 1 + a rr t + 1 a 1-

(b/a)t] (b/a)l

Hint. Using the Schwarz-Christoffel transformation, map the domain ABCDE onto the upper half-plane of the complex variable~ = l; iYJ, requirf A ing the points B , C and D to go into the points ~ = - I , ~ = - A and ~ = 0,

+

where A is a number between 0 a nd 1 which subsequent calculations show to be equal to the ratio b2 fa 2 •

c

53. Solve the preceding problem for the case where the channel has the form shown in Figure 8, assuming that FIGURE

8

v:tl:t""-oo =

Va.

Ans.

J

ta. - d 1 t -X = -1 [ ln l;+ ll~ a rr o t- 1

(l;

>

0).

Hint. Transform the region bounded by the wall of the cha nnel and the axis of symmetry of the flow into the upper half-plane of the complex variable ~. making the vertices B and C go into the points - I and 0. *54. Investigate the j et flow of a liquid through a slot of width 2a in a plane wa ll (see Figure 9), assuming that the amount of fluid flowing through the slot per unit time (in a slab of unit thickness parallel to the xy-plane) equals Q. Find th e form of the jet.

, 2b,

:I: Ans. In parametric form , the equatio n of the curve bounding the jet is 8

B

F IGURE

9

In Probs. 52- 54, where the flow is symmetric with respect to an axis, it is convenient to assume that ~ = 0 along this axis. The value of ~ along any other streamline can be found by using the formula Q = 1~ 1 - ~.J.

SO~l:. SPECIAL METHODS

PROB. 56

given by X

a

Ii + . mtz sm l • a sm mq>

. n1tz E'I' = B.! J m ( Ymn !.) sin m q> Sill • r a cos mq> I r ) cos m q>

Ez = CJm ( Ymn-

.

a sm m q>

H r __ M

H'l'

.! J m ( Ymn r

!.) sincos a

n1tz

COS- ,

I

mq> n1tz COS , mq> l

mq> n1tz = NJm, ( Ymn-r) cos . COS - .- ,

l

a sm m q>

H z -- p!J'm ( Ymn !.) sin mq> Sl.ll mtz a cos m q>

r

I

'

where m and n a re integers, the constants A, B, C, M , Nand Pare connected by the relations A Ymn

a

+ C n1t = l

A Ymn

±

B _!_

a M

= -

N = -

p =-

'

= 0,

m

l(=Fcm ik

_!_ (A ik

0

n'lt -

l

B n1t), l

Y

C mn) ' a

~(B Ymn ±Am), 1k

a

nnd the Ymn are consecutive roots of the equation J m(Y) = 0.

94. A high-frequency current I sin wt flows along a cyl indrical conductor of radius a, made of ma terial of conductivity cr and magnetic permeability fl-· Find the distribution of current density alo ng the cross section of the wire, nnd calculate the active resistance of the conductor at the frequency w (the .1•kin effect problem). Ans. The complex amplitude of the current density is given by the formula j(r) = J 0 (kr) j(a)

J 0(ka)'

54

STEADY-STATE HARMONiC OSCILLATIONS

PROB.

95

where j(a)

= .!.!£_ J 0(k a) . 2na J 1(ka)

The resistance per unit length of the conductor is

lkl

=

R

2

"'

2nacr(k

2

/?)

-

[k 1 0 (ka) _ kJ 0(ka )] J 1(ka) 1 1(ka) '

(the overbar denotes the complex conjugate), where k

J

=

4ncrcu[Li

- - -2c

=

1 ~

- ...; 2ncrcu[L

c

(1 - i).

Taking account of the asymptotic behavior of the Bessel functions for Large values of the argument, we find that

where

0-

c

- .j2ncrCU[L

and

Ro =

1 -2na cr

is the d-e resistance. 95. Solve the skin effect problem for a conductor whose cross section is a strip of width 2a. Find the corresponding current distribution and resistance. Ans.

J·c a) = -Ik cot k a, 2

j(x) cos kx = -- , j(a) cos ka

.

R, =

lkl 2

-

_

2 2cr(k 2 - k)

(k cot ka - k cot ka),

where I is the amplitude of the total current. For high frequencies, j(x)

Ij(a) where

I~

o is the same as

e-11

where the v,. are consecutive roots of the equation P~,.(cos u) 0, /',,( 1) • ~ 1111 Legendre function and H~!}y, (x) the second Hankel function. In lhl· ~ (ll'lllli case, ~ (,. II) I ja sin [wl - k(r - a)]- 2ka cos [wt . )_ 2 v0a U ( I, I 2 2 r 1 + 4k a 130. Solve the problem of diffraction or a plane sound by a spherical obstacle of radius a.

Ans. u

=

u 0 ei(wt-kz) -

X ~ (211 L..,

u0

W:IVl' ""'''(•·• I II)

J

rr e1"' 1 2kr

+ l )e- inn;z

.,_ 0

2kaJ'n-i y,(ka) - J., +y,(ka) II (~l, (k ·)/' (l ~~II) (2)' ( 2) 11 I 2 I 11 I o 2kaHn.,_ y,(ka) - H ,.+y,(ka)

where Jn.,_ ~,.(x) is the Bessel function of the first kind, H(x)~;~ , 1 the '>l'tond Hankel function , and P,.(x) the Legendre polynomial.

Hint. If the velocity potential is written as a sum

then solving the problem reduces to integrating Helmholtz's equation 6.u 1

+ku 2

1

= 0

with the boundary condition

_ au I or 1

r -a

= .£._ ce- ikzUo) 1r -a>

or

where u 1 must satisfy the radiation condition at infinity.

2. Mechanics: Statics of Deformable Media, Fluid Dynami cs 131. Find the equilibrium s hape of a rectangular membrane with 1-Hks

2a and 2b under the action of a uniformly distributed load q, choo!>i n!-l lhl· origin at the center of the membrane. Calculate the deflection of the l' l'llh'l of the membrane, assuming that the ratio b/a takes the values I , 2 and \,

Ans. u(x, y)

=

{.! (

2

2

qa 1 _ x 2) T 2 a

+ .!.§_1 (-

1

1t+ cosh [(2n rr3 .,_ 0 (211 1) 3 cosh [(2n

+

+ l )rryf2a] cos (211 + l)rrb/2a)

I I

211

)It\}.

74

132

PROD.

THE FOURIER METHOD

where Tis the tension per unit length of the boundary. Numerical calculations show that u(O, 0) = kQ{T, where Q = qab is the total load, and kjbfa~2 = 0.228,

klb/a=l = 0.295,

Hint. Make the problem homogeneous by subtracting out the particular solution of the equation for equilibrium of the membrane which depends only on the coordinate x and satisfies the boundary conditions on the sides

x =±a. *132. Find the equilibrium shape of a semicircular membrane of radius a (see Figure 20) under a uniformly distributed load q.

Ans. 2

u(r, cp)

= qa

2T

{.![(.!:E!) sin a r a cos 2cp - -1 ((r 2 +

Ans.

+ l)nb tanh (2n + l)nb]

2a 2a 2 cosh (2n ~a l)nb

,

where D is the flexural rigidity of the plate. Hint. Subtract out the particular solution of the deflection equation which depends only on the coordinate x and satisfies the boundary conditions tiw x =±a. 135. Solve the preceding problem, assuming that the boundaries x = ± a nrc simply supported, while the boundaries y = ±b are free. Calculate the deflection at the center of the plate. Ans.

11(0, 0)

= qa4 D

[2 + 24

!

64v (- I)" 5 n n=o (2n + 1)5

+

+

1 - v S111 . I1 (2n l)nb 1- v 2a

-

X

( 3 + v) sinh (2n

-

(2n

+ l)nb cosl1 (2n + l)nb 2a

+ l )nb cosh (2n + l)nb _ 2a

]

.o____;_----'0_

2a

(t _ v) (2n

+ l)nb

2a

2a where v is Poisson's ratio. *136. A semicircular plate of radius a is clamped along the semicircular llll' and simply supported along its recti linear edge. Find the deflection of lhl• plate under a uniform load. Write a formu la for the deflection of the ll~ i~ of symmetry of the plate, and represent the result in the form of a graph. Ans. U ( r, n) -

2

4

= -qa-

4 [r 4 - -1 (6

24D a

Sn

2

4

2

4

+ 12 -r + 5 -r a

a

2 4 2ar - 4 -a 2 - 3 -a 4 ) arc tan - .,----2 1·

- _!_(s ra 4n

Figure 21).

a3

r

11 !. a

a -

r2

+ 3 q_ + 3 aa)J r

r3

76

PROB.

THE FOURIER METHOD

137

u(r, f)

o':;~

0.08

r

0

0.5 fiGURE

75

1 21

Hint. Make the problem homogeneous by s ubtracting out the particular

solution _q_ r 4 sin 4 24D

m T '

of the deflection equation satisfying the boundary conditions on the rectilinear edge. 137. An infinite cylinder of radius a an ideal y ch oosing (r,cp) cylinder opposite ure 22).

is placed in a plane-parallel flow of fluid . Find the velocity potential, the origin at the center of the and the direction of the x-axis to the direction of flow (see Fig-

Ans.

~. ) 2

u(r, cp) = v00 (r + FIGURE

cos cp + const,

where v00 is the value of the flow velocity far from the cylinder.

22

138. F ind the velocity potential for flow of a n ideal fluid emanating from a source of strength m and flowing past an infinite cylinder of radius a, where the configuration of the cylinder and the source is shown in Figure 23. Ans.

m rb u(r, cp) = -In--:. + const, 27t

y

pp

where bb = a2 and the meaning of the various symbols is indicated in the figure. Hint. Subtract the so urce potential m

u1 = - - In p + const 27t

from the solution.

FIGURE

23

139. Solve the problem of plane-parallel flow of an ideal fluid past a sphere of radius a, choosing the origin of a system of spherical 900rdinates

PROD.

]43

77

THE FOURIER METHOD

r, 6, cp at the center of the sphere, with the direction of the z-axis opposite to that of the flow. Ans.

u(r, 0)

=

I

(r + ;:

v00

) 2

cos 6 + const,

where v 00 is the value of the flow velocity far from the sphere. * 140. Solve the problem of flow past a sphere of radius a due to a source of strength m at a distance b from the center. Ans.

u(r, O)

=

m[_! +

4TI p

a_ _

.! In _

bp

a

r(l + cos 6) -] •

p + r cos 6 - b

wi th the same notation as in Figure 23, except that the x-axis now becomes the z-axis.

3. Heat Conduction: Non stationary Problems 141. A slab of thickness 2a, thermal conductivity k, specific heat c and density p is heated to temperature T0 , and its faces are then held at temperature T0 , starting from the time t = 0 (see Figure 24). Find the temperature distribution l'(x, t) in the slab. Ans. 1'(-c, t)

= 4To 1t

I , _0

( - l )n

2n I 1 - (Z n H ) 2n 2T/4a 2

X e

where -r

=

cos (2n I 1) 7tX ,

2a

F IG UR E

ktfc p.

24

142. Describe the equalization of a given initial temperature distribution 0) = J(x ) in a slab whose faces x = 0 and x = a do not tra nsmi t heat.

/'( 1',

Ans.

T(x, t) =

.! laf(~) d~ ao

I

~

I

an

1

e- ,.•n•Tta• COS 1tn X l af(~) a o

COS

n1t~ d~. a

143. Starting fro m the time t = 0, a slab - a < x < a of thickness 2a Wllh n given initial Jemperature distribution T(x, 0) = J(x) radiates heat into

78

THE FOURIER METHOD

PROB.

144

the surrounding medium, whose temperature is taken to be zero. Assuming that the radiation obeys Newton's law, find the temperature distribution in the slab for arbitrary time t. Ans.

T(x t)

'

=

yx cos - "121 oo -1 a e- O. 181. Find the electrostatic potentia l u(x, y) ~ I ween two infinite parall el sheets if one

ahrl'l y

=

0 is at potential zero, while a given

0

u=O F IGURE

' X

34

92

182

PROS.

THE FOURIER METHOD

periodic potential ui 11 ~b

= f(x)

is maintained on the other sheet (wherefis a function with a given period 2a). Ans. nrtxi af(t") 1 ~ sinh (n.rty fa) [ y ) = - L, COS , where w is the frequency of the oscillations and k = w/c is the wave number.

References Bateman (B2), Frank and von Mises (F6), Franklin (F7), Gray and Mathews (G2), Grinberg (G5), Jackson (J I), Jeffreys and Jeffreys (14), Lebedev (L9, Chaps. 6 and 8), McLachlan (MS), Morse and Feshbach (M9), Tikhonov and Samarski (Tl), Tolstov (T7), Webster (WS). For further problems, see Budak, Samarski and Tikhonov (B6), Gyunter and Kuzmin (G7, Chap. 15), Smirnov (SS).

5 THE EIGENFUNCTION METHOD FOR SOLVING INHOMOGENEOUS PROBLEMS

In this chapter we study various inhomogeneous problems of mathematical physics leading to integrati on of the eq ua tion

!__

1'(:

y.

+ const,

260. Two faces of a rectangular bar are thermally insulated, and the other two are held at temperature zero (see F igure 53). Find the stationary tempera ture distribution, assuming that heat is produced with density Q inside the bar.

126

PROB.

THE EIGENFUNCTION METHOD

261

Ans.

16Qa 2 ~ ( - l)" { cosh [(2n T (x y) = -3- L1' 1t k n~o (2n + 1)3 cosh [(2n

+ l)7ty/2a ]} cos (2n + l)1tx . + 1)7tb/2a] 2a

y /.

FIGURE

53

FIGURE

54

*261. Heat is produced with density Q in the bar shown in Figure 54. Find the stationary temperature distribution, assuming that a heat current of constant density q leaves the bar thro ugh the section Jxl < c of the upper face, while the rest of the surface of the bar is thermally insulated . Ans.

T(x y) = _ ~[l ' k 2

+ 2a2 b 7t2C

I n- 1

sin (n7tc/a) cosh (nTiy fa) cos n1tx] n2 sinh (n1tb{a) a

+ const, (14) where k is the thermal conductivity. Another form of the solution, sui ta ble for afb ~ J, is

T(x, y) . .

{a -

= -Q - -c x 2 - -a y 2 2

kc

2

2

+

2ab ~(- l )"sinh [n7t(a - c)/b) h 177tX 117tJI} - Lcos cos 2 2 7t n~1 II Sinh (n1ta{b) b b

-

JxJ < c,

x T(x, y) = Q{ k a Jxl - 2

2

2

2ab ~( - l ts inh (n1tc{b) h n1t(a - -2- Lcos 2

7t

C ,._ 1

+ const,

n sinh (n1ta{b)

-lxl) cos -n1ty} -!

b

b

const,

JxJ >c. 262. Find the stationary temperature distribution in a conductor of rectangular cross section heated by a d-e current producing heat with density Q, if the surface of the conductor gives off heat according to Newto n's law.

PROB.

264

THE EIGENFUNCTION METHOO

127

Ans. 2

= 4Qa ~

T(x y) '

k

sin y,. y~(2y, + sin 2y,)

1l= t

X

ah J [ 1 - -y,. tanh (y,b fa)

cosh (y,.yfa)J y"x cos + (a hfy , ) cosh (y, bfa) a '

where h is the heat excha nge coefficient, and the y,. are consecutive positive roots of the equation ah tan y = - . y 263. Find the statio nary temperature distribution in a rectang ular paral lelepiped 0 < x -~ a, 0 C" y ,.., b, 0 .-- z .- c, if the faces x = 0, y '--= 0, z = 0 a re held at temperature zero, while the o ther faces have the tempera ture distribution

T/.,=a= fa(y, z) , Ans. T( x,y,z) = - 4 L~ L~ {sinh - . y.,.,.z ab m - 1 , _ 1 smh y,.,.c I { - -Ymn Jo Gmn(~, z) c

[

i"ib

r (!:

Jc'->•"IJ

0

) sm . m7t~ . nrrYJ d!: d - sm .., "I) a

0

(- I)

b

- J,{0 fa.( "f),~) sm. -nrrYJ d""f)

m m7t

a o

b

"1r b(!:c..,, >"') sm . 1717t~ . nmx . n7ty I) 117tl - d"'] c, d>"'} sm - su1 ,

-! ( -

11

~.,

b o

~.,

a

b

a

where

G,.,.(~, z) = .

I X {s~nh Ym,.~ s~nh Ymn(c - z), smh y,.,c smh Ymnz smh y,.,.(c - ~),

~

~

< >

z, z.

264. A heat current Q enters a bar of semicirc ular cross section through its plane face a nd leaves through the curved face (see Figure 55). F ind the statio na ry temperature distributio n in the bar, assuming that the incoming a nd o utgoing currents have co n!.tant density.

Ans. T (r, cp)

~[ 1 - 2!(-

=

7t/W

X

I

l )"

,._ l

( l /2n)(r/a) • 4n· - I

2

"-

1

cos 2n cp

where k is the thermal conductivity.

J+ const, F IGURE

55

128

PROB.

THE EIGENfUNCTION METHOD

265

265. Find the temperature distribution in a bar whose cross section is the "curvilinear rectangle"" a < r < b, 0 < q> < a, given the following temperature distribution on the faces of the bar: T[,~a ~ 0,

Ans. _ T.

T ( r, q>)-

0

[InIn (rfa) + (bfa) X~ L

~

l

1t

(-1)" sinh

n-•

nn(~cp) +sinh __ll7t5l_ In (b/a) In (bja) . mt In (rja) sm . h mw. In (h/a) nsm - - -

.

In (bja)

Another form of the solution is

'!'

T(r, q>) ~ T0 [

0.

l

1t

(-')' "/". - [.1 -. (-1) . n(h)nnfa] (a)"nfa [1 - (-1) "(a)""/"] .b •L, n [ (b)""i" - (a' - )""l"] ~

X

+ ~?

a

a

~

a

.r

. nr.rp

Sill----;- .

b

266. Find the stationary temperature distribution T(r, z) in a cylinder 0 < r < a, 0 .:.:; z (~) sin___.::, d~ I ,~ 1 1 0 (nr.af/) I o I

L

\.

nrrz [(-l)"f,(p)- j 0 (p)]pG,.(r, p) dpJ sm-,

/

0

/

p




r,

PROB.

268

THE LIGENFUNCTION METHOD

129

where ! 0(x) and K 0 (x) are cylinder functions of imaginary argument. Another form of the solution is

i .

T(r, z) = 2• 1 2 {sinh YnZ {l'f t( p)plo(' YnP) dp a" ,.=1 smh (y,.lfa)J 1(yn) a a

Jo

+ sinh y,.(l a- z) f!o(P)PJ o ( Y:P) dp

L

+ aJl(Yn) cp(~)G,.(z, Q a~}lo(y~r). sinh

'Yn~ sinh y,.(/ - z), a

a

G,.(z, ~) = { · h YnZ S111 a

. h SIO

y,.( l a

0

,

~

< z,

~

> z,

where the y,. are consecutive p ositive roots of the equa tion J 0 (y)

=

0.

267. A heat current Q enters a cylinder 0 < r < a , 0 < z < I through its ends and leaves thro ugh the lateral surface. Find the temperature d istribut ion in the cylinde r, assuming that the incoming a nd o utgoing cu rrents have constant density. Ans.

T(r z) '

2 [z2- - -z - ,. ] = -QJ2 2 2 2na k /

l

2[

+ const '

where k is the thermal co nduc tivity. 268. Find the tempera ture distributi on in a cylinder 0 < r < a, - I < z < I inside whic h heat is produced with density Q, if the surface radiates heat into the surrou nding medium according to Newton's law. Ans.

T(r z) '

=

2

2Ql k

~

sin y,.

,.=1 y;Jl + (sin 2y.,J2y,.)) 1-

X

[

hl 'Yn

l1(y,.aj /) + .!!..! 10(y,.a fl) Yn

]

fo(Ynl'/ l) Io(Ynafl)

COS

y,.z

l '

130

PROB.

THE EIGENFUNCTION METHOD

269

where h is the heat exchange coefficient, 10(x) and 11 (x) are Bessel functions of imaginary argument, and the y n are consecutive positive roots of the equation tan y

hl y

=- .

Another form of the solution is

T (r, z )

=

1

2Qha ~ L k n=l 3

(ah jy,) cosh (y,z/a)

_

+ (ahfy,) ] cosh (y, lfa) 1 o(y,r) , 4 + (a h/ y , )2 ]y,J a 0(y.,)

[tanh (y,lfa)

[1

where the Yn are consecutive positive roots of the equation

*269. A thin wire heated by a d-e current p roduc ing Joule heat Q per unit length is placed inside a cylindrical object (see Figure 56). Find the temperature distribution in the object, assum ing that the lateral surface of the cylinder is held at temperature zero, wh ile t he e nds radiate heat into the surrounding medium according to Newton's law. I

Ans.

I I

21

ol

T(r, z)

I I I

Q

l L Tik Y, J l Y..)

=-

2

2

n= l

--1--X FIGURE

<X>

56

[1

_

J (y,.r)

ah cosh (ynz fa) 1 y, sinh (y,L/a) + ah cosh (y, Lfa) · 0

a

'

where the y,. are consecutive posit ive roots of the eq uation lo(Y) = 0.

Hinl. Replace the line so urce by a sou rce di stributed over a cyli nder of small radiu s c:, and then take the limit as c: ~ 0.

270. Fi nd the stationary temperature d istrib ution T(r, 6) in a sphere of radius a, assuming that heat is produced with de nsity Q inside the sphere, while the bo unda ry conditio n

( a~ a, + hr) I

= f(6),

r=a

involving a g iven function f(6), is satisfied on the surface of the sphere.

PROS. 272

THE EIGENFUNCflON METHOD

131

Ans.

T(r, fJ) = Q (a 2 . 6k

-

r 2)

+ Qa + ..!._ ('~<j(fJ) sin f) dfJ 3kh 2h Jo 2n + + -a L -1 ( -r)"P ,.(cos fJ) 2 n~l ah + a «>

i"'

j(fJ)P,.(cos fJ) sin

ll

e dfJ.

0

5. Electricity and Magnetism 271. Calculate the two-dimensional electrostatic field due to the electrodes shown in Figures 57(a) a nd 57(b).

0

b v=

-v

v=V

0

X

(b)

(o) FIGURE 57

Ans.

J2 v + y) a

a) u(x, y) = - - (x

{c

- 4 v! - lr' sinh 117t[a -

7t

. h n1t[a + sm

b) u(x, y)

= V [~ + ~ a

1t

!

,.~ 1

.ji(y -

x)]

2a

n- t

+ J2(y - x)]} 2a

. n1t[a - J2(y sm 2a n sinh 117t

+ x)]

cosh (1myja) sin (n1txja)J, cosh (1mbja) n

where u(x, y) is the electrostatic potential. *272. Find t he electrostatic field in the electron-optical d evice shown in Figure 58. 11 What is the distribution of potential along the axis of symmetry? 11

By an electron-optical device (for example, a lens), we mean a system of conductors

at given potentials producing an electrostatic field used to govern the trajectories of charged

rnrticles.

132

THE EIGENFUNCTION METHOD

PROB.

273

Ans.

}'

uiii=O,:. c,

,

where the y,. are consecutive positive roots of the equation J 0 (y) = 0. Another form of the solution is

u( r, z)

4q~ l 0(mrafl)K0 (n7tr/l)

- K 0(n7ta/l)l 0(nm"/l) . n1tc . n1tz sm stn . l 0(n1tajl) l l

=-

L l ,_1

278. F ind the electrostatic field inside a cylind rical shell 0 < r < a, 0 < z < l whose ends and lateral surface are at the potentials V0 , V 1 and V, respectively. Ans. The electrostatic potential is u(r, z)

=

V(1 - f) +

V1

0

T

~ [1 - (+ -1t2 L.

lt]V

+ (-

1)"' V1 -

n

n-1

V0 l 0 (n7tr/l) . n1tZ sm-. l 0(n1ta/l) l

Another form of the solution is u( r, z ) = V

+ 2( V

~sinh

1 -

V) L.

,._ 1

(y.,z/a) J 0(y 11r/a)

sinh (y,.lfa) Ynlt(y.,) \-

(V. _ V)~sin h [y,(l - z)/a] J 0 (ynr/a) 2 0 L. . , ,~ 1 smh (y,.l/a) y,)1 (y,.)

where J0(x), J 1(x) and T0(x) are cylinder functions, and they,. a re consecutive positive roots of the equation 10 (y) = 0. 279. Find the electrostatic potential along the axis of a cylindrical shell 0 < r < a, 0 < z < I if the lateral surface is held at a given potential uJr=a = f(z),

while the ends are held at potentials V0 Ans. uJr=o = Vz l

+ ?! [ V l (I n = l 1t

tY' n

=

0 and V 1 = V.

+ (f(~) sin n1t~ d~J sin (n7t z/l), Jo l 1 (n1ta/l)

where l 0(x) is the Bessel function of imaginary a rgument.

0

PRO D.

282

T HE EIGENFUNCTION METHOD

135

280. Examine the specia l cases of t he precedi ng p roblem which correspond to the fo llowi ng potentia l distributions on t he latera l surface of the cylinder :

b) f (z) =

vk

ulr o =

b) u

N

k

(k = l ,2, .. . ,N);

< z < -

N

< z < c, c < z < I.

0, c) f(z ) = { V, Ans. a)

(k - 1)

for

0

Vz l

+ ~ ~ [(- J)"(V1t n - 1

Va) + Val sin (n1tz/l) ; nlo(n1tafl)

Vz 2 I = -+I r

0

1t

X~

[(ka)H~1 >(kr 0)

CCX

-

n- 1

X

A 0(r, cp) =

2

CCX

J,.(kr)...,: nrtcp0 J,.(ka) ex

•

nrtcp ex

--~m-- sm-- ,

r


(kr)H~1>(1w)]

rt J 0 i ! [H~2 >(ka)H~1 >(kr)

2

H~2 >(kr0)m;>(ka)]

n= l

J ,.(kr0 ) • nrtcp0 . nrtcp x --sm - - sm - J"(ka) ex ex '

k

=

wf c is the wave number, and J,.(x), H,~ll(x), H~2 >(x) are cylinder functions.

*296. Find the e lectromagne tic oscillations in a cylindrical resonator 0 < r < a, - I < z _.. I excited by a dipole of moment P locat ed at the origin of coordinates a nd directed along the z-axis.

Ans. The complex a mplitude of the z-component of the vector potential is given by the Fourier expansion A(r, z) = rtP[a0 (r) 2!a,.(r)cos~]. lc , _1 l where the coefficients a,. have the values

+

1

a,.=--- [/0 (cx,.a)K0(cx,.r) - K 0(cx,.a)J0(cx,.r)], J0(cx,a) k is the wave number, and J0(x), K 0(x) are Bessel functions of imaginary nrgument.

297. Find the electromagnetic field in an infinite cylindrical waveguide with perfectly conducting walls, ass uming that the source of the oscilla tio ns is a current J sin wt in a coil of given dimensions, with a single uniformly wound layer (see F igure 67).

Ans. The complex amplitudes of the components of the electro magnetic field are

Er = E. = 0,

1\

= _ 8rtiw bJN ~ J~(y11b/a) J

'P

22

ca h

JX

{

L22 )1 ,.~ 1 0(y,. 2

e-rx·"'

f--0 -

II

a

cosh cx,.z,

0

< x < h/2,

z

> h/2,

c 1 () II = ~ oErp H . = - -:- - (rE'~')' r iw oz ' I W ror

-

b t-

l

(y,.r)

cx, J

sinh cx,h e-«•z, 2

z

H'~'

= 0,

0

F IGURE 67

r

142

THE EIGENFUNCTION METHOD

PROB.

297

where

J0 (x) and J 1(x) arc Bessel functions, k is the wave number, c the velocity of light, N the number of turns in the coil, and the y n are consecutive positive roots of the equation J 1(y) ~ 0.

References Grinberg (04, 05, 06), Morse and Feshbach (M9), Tikhonov and Samarski (Tl).

6 INTEGRAL TRANSFORMS

If application of the Fourier method to a given problem leads to a set of particular solutions depending continuously on some real or complex parameter, we say that the problem has a continuous spectrum. 1 Characteristically, the soluti on of a problem of this kind is constructed from appropriate particular solutions by integrating with respect to the parameter, i.e., the solution takes the form of an integral expansion involving the eigenfunctions (the continuous analogue of the series expansions considered in Chaps. 4 und 5). 2 P roblems with continuous spectra are encountered in all branches of mathematical physics, and can often be solved by the method of integral transforms, to which the present chapter is devoted. We begin by reminding the reader of the necessa ry background information. By an integral transform of a functionf(x), defined in an interval (a, oo), we mean an expression of the form j(T)

=

f''

f(x)K(x, T) dx,

C


0

Ans.

T(x, y)

=

To[l - 2hioo ~

Tl .,..o =

0

'Ay] d"A,

sin A.(A. + h)

e- Az

:0 [(~- Si (yh))cos yh + Ci (yh)sin yh].

2

where Si (x) and Ci (x) are the sine and cosine integrals.

PROS.

307

INTEGRAL TRANSFORMS

(5(

Hint. Take the Fourier sine transform of the required function T(x, y), i.e., mu ltiply the relevant differential equation by sin )..y and integrate with respect toy from 0 to oo. 305. The end of a semi-infinite cylinder 0 < r < a, 0 < z < ex:> is held at constant temperature T 0 , while the lateral surface is held at temperature zero. Find the stationary temperature distribution in the cylinder, by expanding the required function in a Fourier sine integral with respect to z. Ans.

T=

To[l _ ~

( a:> 10(Ar) sin

n Jo / 0()-.a)

)..z d)..],

A

where J0 (x) is the Bessel function of imaginary argument.

Hint. Introduce a new unknown function u = T- T 0 , and use the integral l

= ~ { a:> sin AX c/)..,

X> 0. n Jo A 306. Solve the preceding problem, assuming that a given temperature distribution Tl=~o = f(r) is maintained on the end of the cylinder, while the luteral surface radiates heat into the surrounding medium according to Newton's law. Ans. where

{[AK1(Aa) - hK 0()..a)]/ 0()..r)

-1 [)../ 1(Aa) + hI 0()..a )] K0(Ar)} I o(A p) , P __. r, D(A)

{[AK1 (Aa) - h K0(Aa)]1 0(Ap)

1- [),Ii)..a) I hI 0()..a)] K0(Ap)} I o(Ar) , D(A)

p -;,. r,

D(A) = )..11()-.a) + h/0()-.a), l,.(z) and Kn(z) are Bessel functions of imag inary argument, and h is the heat rxchange coefficient.

307. Find the statio nary temperature distribu tion in a semi-infinite l'ylinder 0 < r < a, 0 ..-- z < ex:> if the lateral surface is maintained at the ll'lllperature Tlr- a = f(z), while the end rad iates hea t into t he surrounding med ium. Ans. 2 { <X> I (Ar) d).. { <X> T(r, z) Jo q>"(z) I:(Aa) h2 )..2 Jo f(~)q>~.(~) d~,

=;

+ q>~.(z) = A cos )..z + h sin AZ.

152

INTEGRAL TRANSFORMS

PROB.

308

Hint. To solve the problem, use the following generalization of the Fourier integral theorem (see Ll3, p. 79):

f(x) =

~ { "" 2rp~.(x) 2 dA {""f(~)cp,_(~) d~,

1t

Jo

h

0 <X
0, if the potential distribution uiv- o = f(x) is maintained on the plane y = 0. Ans.

= E J +oo

f(~)2

d~ = ! Jn/

2

f(x + y tan f>) d6. x) y 1t -n/2 Hint. To reduce the solution to final form, use the integral u(x, y)

1t

-oo (~ -

+

2

oo e-a"' cos bx dx =

J o

a

a2 + b2

a > 0.

,

y P=(x,yl

--~--~--~~0~~--~-o--~r--x

u=£'(, u=v;

u=~- 1 u=~ FIGURE

u=~- 1 u=~

71

3~9. Examine the special case of the preceding problem corresponding to the piecewise constant potential distribution in the plane y = 0 shown in Figure 71. Ans.

1 .. u(x, y) = v.,~k·

L:

1t k- 0

where vk is the value of the potential in the interval (xk-1> xJ and ~k is tlw angle subtended by (xk-1 , xJ at the point P = (x, y). 310. Find the distribution of electrostatic potential in the planar electron optical lens shown in Figure 12 (cf. Prob. 282). 8 Ans.

u (x, y) = 8

.

10

v2 + vl + v2 - vl f "" cosh "Ay sin Ax d - A. 2

1t

o cosh Ah

A

Note that the integrals representing the solutions of Probs. 310- 311 can be exprcssnl terms of elementary functions.

I'ROB. 312

INTEGRAL TRANSFORMS

Hint. Subtract out the particular solution !{V2 equation, and then use the expansion

1=

~ f oo sin Ax dA 1t

x

>

0. y

u =V2

u: VI

J

of Laplace's

A

o

y

-2h

+ VJ

I 53

-

X

0

F tGURE

I

I I

j

10

I

I

I I

"2

72

Vz I

t

2h

u =V2

u=V1

v;

Vz

L2~~ FIGURE

v2 73

311. Find the distribution of electrostatic potential o n the axis of the planar electron-optical lens shown in Figure 73. Ans.

(

U X,

J) =

sin ),a cosh AY ~ a~ v.2 + 2 vl - v2f<Xl COS AX A

o

1t

A.

cosh Ah

312. A thin charged wire of charge q per unit length is placed between lwo parallel conducting planes (see Figy ure 74). Find the res ulting distribution ol' electrostatic potentia l, a nd also the u=O dl·nsity of charge o n the planes y = 0 nndy = h. -r q 0 Ans. The potential distri bution is 11 (x,y)

sinh A(h - a) o A sinh Ah x sinh AY cos AX dA, y

= 4q

i

oo

F IGURE

< a,

74

where the correspo nding form ula for y > a is obtained by permuti ng y and a. The charge density on the planes is q

. 1ta

cr(x) = - -Sin 2h

h

1 cosh (1txj h) - cos (1t0jh)'

X {

· --

1

- -- - - - ,

cosh (1txjh)

+ cos (1tajh)

y

=

0,

y = h.

1/int. F irst ass ume that the charge is uniformly distributed over the ngle - ~ < x < a- o: < y < a + o:, and then take the limit as • 0. To solve the corresponding Poisson equation, take the Fourier cosine

o,

o,

154

PROB.

I NTEGRAL TRANSFORMS

313

transform of the unknown function, by multiplying the equation by cos A.x and integrating with respect to x from 0 to oo.

*313. Find the electrostatic field of a t hin charged wire of charge q per unit length located near the plane interface between two dielectric slabs (see Figure 75). Ans.

where 75

FIGURE

Hint. To avo id any difficulties associated with t he behavior of the l ogarithmic pote ntial at infinity, set up a system o f equatio ns for the components of the electrostatic field.

314. Find the potential distribution in the electron-optical lens shown in Figure 76. Ans. VI+ ur ) ( z=

'

2

v2 + v2 - vl f oo ----(). Io(A.r) sin AZ I o I 0 (A.a)

1t

A

'

where J0(x) is t he Bessel function of imaginary argum ent.

0

I

I

u= V. 0\ u=Vz ' {- ---- ..... r- -----I

I

1

z

I

FIGURE 76

Hint. Reduce the problem to one with bounda ry conditions which a1r odd in the variable z, and then make a sine ex pa nsion, using the formula

J _ ~ n

f co sin AZ df.., o

A

z > 0.

315. Find the potential of the electrostatic field due to a point ch a rgl' 'I placed on the axis of a n infini te conducting cylinder of radius a.

PROB.

317

INTEGRAL TRANSFORMS

155

Ans.

u(r, z) =

.J

q r

2

+z

2

2qf<X) K 0()..a) I 0()..r) cos ).,z d).., 1t o I 0(Aa)

-

where / 0(x) and K0(x) are Bessel functions of imaginary argument.

Hint. ln the co urse of the calculations, use the following integral representation of Macdonald's function:

CX) cos AZ K 0 (Aa) =

Jo

.Ja + z 2

2

dz.

316. Find the distribution of electrostatic potential inside a conducting cone 0 < r < oo, 0 < < 00 due to a point c ha rge q on its axis (see Figure 77).

e

Ans. u(r 6)

'

=

q -;======== 2 .Ja - 2a r cos 0 r2

+

q J CX) P -~l-iT( - -=

-cos 6

0)

P- ~t-;'t(cos0 0)

.Jra o

P-~+iT(cos 6) cos

(

T

r)

dT

In - - - - , a cosh7tT

where P.(x) is the Legendre function of the first k ind.

Hint . Introduce new variables x

=

In !..

u

a'

=

r- 112 v.

J'o expand the so urce, u se the following integral representation of the Legendre function:

i <XJ

2

P- ~+;T(cos oc) = -cosh 1tT

COSTX

o v12 cosh x - 2 cos oc

1t

dx.

317. A point cu rrent source is p laced on the axis of a t ube filled with a medium of cond uctivity cr1 and "mounded by a medium of conductivity cr2. Find the potential of the current field in each medium. 9

~·y lind rical

Ans. z)

11 1(r,

= _!_ [ 1 47tcr1 .j,.2 z2

+

+ ~ (cr 1t

1

77

J

- cr2) r <X) Ko(Aa)Kl(Aa)Io(Ar) cos AZ dA Jo cr1 K0(Aa)I1(Aa) cr2 l 0(Aa)Ki)..a) '

+

_ _!_ i <X) (r, z ) 2

t/2

0 FIGURE

K 0 ()..r) cos AZ d).. 27t a o cr1 K0()..a)I 1(Aa) -r cr2/0(Aa)K1(Aa) A '

• This is the problem of "electrical coring" (see Fock's paper F2).

156

PROB.

INTEGRAL TRANSFORMS

318

where I .,(x) and Kn(x) are Bessel functions of imaginary argument, J is the current emanating from the electrode, and a is the rad ius of the tube. 318. A line current J is p laced b etween the boundary p lanes of two massive bodies made from iron of magnetic permeability fL (see Figure 78). Find the magnetic field in the air space.

Ans.

f

21

H "' = -

---- 1t

P>-(r)

o Jg(Aa)

2

cosh Az dA],

+ Y (/..a) cosh Ah 0

A

where cp1.(r) has the same meaning as in the preceding pro blem.

FtOURE

83

FtOURE

84

ll7. The walls of a cylindrical hole terminating a t the plane surface of an body (see Figure 84) a re held at a given temperature T0 • Find the unary temperature distribution in the b ody, assuming that it radiates

164

INTEGRAL TRANSFORMS

PROB.

338

heat from its surface into the surrounding medium according to Newton's law. Ans. T("1 ) _ T. [l _ 2hf"' cp"(r)e- "• d"A 0 1t o A.(A. + h)[J~(A.a) + Y~(A.a)] ' ' z where cp"(r) = J 0(A.a)Y0(A.r) - Y0(A.a)J0(A.r), and h is the heat exchange coefficient.

J

338. Find the distribution of electrostatic potential in the space between two grounded plane electrodes z = a, due to a point charge q at the point r = 0, z = 0. Ans. q coshA.z J (A.r) dA., u(r, z) = - q e-Aa 0 .Jr 2 + z2 o cosh A.a in terms of the Bessel function J 0(x).

J"'

Hint. Use the formula

.J

r2

1

+ z2

=J"'o e-"•Jo("Ar) d"A,

z

>

0.

339. Find the electrostatic field due to a point charge q located near the plane interface between two media with different dielectric constants (see Figure 85). Ans. q 1 q £1 - £2 1 u1(r, z) = - - + - -- - , £1

u 2(r, z)

R1

= ___3_L ..!_ , £1

+ c:2 R1

£1 £1

+ £2 R2

R 1, 2 = ...;I r 2

+ (z =F a) . 2

Hint. To represent the solution in closed form, use the hint to Prob. 338.

FIGURE

85

FIGURE

86

PROB.

343

INTEGRAL TRANSFORMS

340. Find the electrostatic field produced by two point charges

165

+q and

- q, between which there is a slab of material of dielectric constant e (see Figure 86). Calculate the field on the line j oining the charges. Ans. · - 2q roo :Ae-A(a-b) cosh AZ d:A, Jo sinh :Ab + e cosh :Ab

E.lr=O = {

q

+ q

(lz J- a?

roo sinh :Ab - e cosh :Ab e- A(a- 2b+l•l> Ad:A, Jo sinh :Ab+ecosh :Ab

Jzl < b, JzJ > b.

341. A d-e current J enters the ground through a n electrode mak ing contact with the earth's surface (z = 0) over the area of a disk of radius a. Find the current distribution in the earth, and examine the limiting case of a point contact. Ans. The potential of the current field is

u(r, z) = - J

l ao e- J.• Jl:Aa)J ()..r)-, d:A

z > 0, :A where J 0(x) and J 1(x) are Bessel functions, and cr is the conductivity of the t•nrth. In the limiting case, J u(r,z)= ~ · 2~crv r + z· 342. A po int electrode carrying current J is placed on terrain consisting of two layers of different conductivities (see Figure 87). Calculate the potential of the c urrent field on the earth's s urface. Ans. J J e- :l.alo(:Ar) ul .~o = - (crl- cr2) d)... 27tcr1 r 2~cr1 o cr1 sinh )..a + cr2 cosh )..a ~acr

i oo

+-

FIGURE

0

o

87

F IGURE

88

.\43. Determine the electromagnetic field of a vertical radiator (antenna)

plnn·d at height h over the plane surface of the earth, ass umed to be perfectly mu lucting (see Figure 88).

166

INTEGRAL TRANSFORMS

PROB.

344

Ans. The z-co mponent of the vector potential of the electromagnetic field is p ( e-ikll e- ikll) A(r, z) = ~ R ,

R +

(all other components vanish), where

R = ~(z- 17) 2 -!

R. =

r 2,

J(= + 11) + r 2, 2

cu is the frequency of the oscillations, c is the velocity of light, P is the moment of the radiating dipole, and k = cufc. Hint. Use the expansion e-ikVr +•'

i..,

e-Vt..'-k =Jo("Ar)A.

J r2 + 2 2

0

~)...2 - k2

2

--- =

2

dA..

344. Solve the preceding problem, assuming that the earth has finite conductivity. Ans. The vector potential of the electromagnetic field is

2

,-.-.

Pk2ioo

A (2) = _ _2

c

2

,...----..

A.e- I•V /.. - kt -tz v /.. - 1..-. I 2 2 ~ I •

2

ok2vA-k1 + kiv"A" - k2

J o("Ar) d"A,

where

cu J(r.cu - 47tcr i)cu , kl = - , k2 = c c in terms of the earth's dielectric constant r. and conductivity cr. 11 345. Using the solution of Prob. 344, find an expression for the normal component of the electric field on the earth's surface for the case where the dipole is placed directly on the surface itself (h--+ 0). Ans. 2 E.j,_o= 2~wi (k2) k 1k 2 { ( •k 1 k~ • _ ik2 _ k2_!_) e- iA·,r c·r k1 k1 - k~ kj t- k;. r k1 r2

_ (

k~k2 ki + ki

k3k32

- ir

I

(ki 11

_ ikl _ k1 .!..)e- ik,r

+ k~)

r k2 r2 J ...;k,'·t ko'lk, 312

- .ds } e- irk,kz&/Yk," fkt 2 - - 2 Yk,1Hs1 1ko Js - 1

Details on the transformatio n of these expressions into a form suitable for calculatiuu as well as an analysis of the corresponding physical picture of wave propagation, can Ill' found in the specialized literature (see e.g., F6, Chap. 23, Sec. 1 and Sl4, Sees. 31-32).

J'ROB.

348

167

I NTEGRAL TRANSFORMS

Hint. Use the Van der Pol substitution .

k~JA 2 - k~ + k~.JA2 - k~

rv

_, -.-

=

1

k1k2

k~ - k~

.

k t.+kt/kl

Vlq 1+ kt21k•

1

d

JA

.J s 2 - 1

k~k;

2 _

ki 346. Determine the electromagnetic field of a ho rizontal radiator located at height h above the plane surface of the earth, nssumed to be a perfect conductor (see Figure 89). Ans. The vector potential of the electromagnetic field has the components A.,

=

A(r, z)

p (e-ikll

= ~

R -

+ k~

2 5

(I)

(2)

e-ikR) R

, FIGURE

A u = A.= 0.

89

347. Solve the preceding problem, assuming that the earth has finite nmductivity. Ans.

P{e-;k,n

,t= - - ~

c

R

+J oo 0

2P

,t ~l) = -

c

A .JA2 .JA2 - ki .JA2 -

ki - .JA2- k~ e- vA•-k,•(=+h)Jo(Ar) dA}, k~ .JA2- k~

+

(k'f - ki) cos cp

X

.y-.- • ...;-.- . A -.t, 11+ A -k• •J (Ar)

oo

A2e-

J (k~JA - ki + ki.JA2 2

1

+

2 k';J(.JA2 - ki .JA I ur further deta ils, see F6, Chap. 23, Sec. 2 and Sl4, Sec. 33 . 0

~ -

k';)

.148. Find the magnetic field of a horizontal radiator lying on the plane rli1ce of the earth, assumed to have dielectric constant e and conductivity a.

168

PROB.

INTEGRAL TRANSFORMS

349

Show that the magnetic field on the earth's surface can be expressed .in terms of elementary functions. Ans.

H.l.-o=

2P sin 0.

1hl'Il the Laplace transform of f(t) is defined by the formula (16)

+

Ythrrc p = cr h: is any complex number in the half-plane Rep > cr1 . 16 If h 1- also assumed that f(t) is of bounded variation in every finite subinterval ' ' S1.-e the relevant books cited at the eod of this chapter (p. 202). '' I aplace transforms can also be defined for functions satisfying weaker conditions. that the function/ is an analytic function of p in the domain Rep > u 1• The values /In the rest of the complex plane can be determined by analytic continuation.

J7Q

INTEGRAL TRANSFORMS

[a, TJ,17 then formula (16) can be inverted by using the Fourier-Mel/In theorem f(t) = ___!_. f(p)e., 1 dp, (17)

f

27tl r where r is a straight line parallel to the imaginary axis lying to the right of the line Rep = cr1 (see Figure 90). Conversely, (17) implies (16) if /(p) satisfies appropriate conditions. The application of the Laplace transform r p method is called for in nonstationary problems r leading to integration of the equation

-----0~~~~,-;------~

a at

1 2u Lu- 2 -2 a

-

au at

b - = f(t , .. .),

where L is a linear differential operator which does not contain t, a and bare given constants, and f(t, . . .) is a given function. Its use allows FIGURE 90 us to eliminate the time t, thereby reducing the problem to the determination of a function 11 satisfying a simpler equation. In particu lar, if the unknown function u depends only on one spatia l variable (in addition to the time), the equation for ii will be an ordinary differential equation. After finding ii, the problem can be solved by using the inversion formula (17), where the path of integration r must be chosen in such a way that all the singular points of ii lie to the left of r . The actual calculation of the complex integral (17) can be carried out by various methods, the most important of which involve the use of Cauchy's theorem and residue theory, expansion in series, application of the convolution theorem, use of appropriate tables,18 etc. The varietY. of available methods makes it possible to obtain the solution of the problem quickly, in the form most suitable for understand· ing the physics of the situation and making subsequent numerical calculations. This constitutes the great advantage of the Laplace transform method, which is particularly suitable for studying wave propagation along transmission lines, physical problems with boundary conditions involving time derivative~ (see Probs. 365, 367, 370), and so on. This section contains a variety of nonstationary problems, dealing first with heat conduction, then with electricity and magnetism, and finally with mechanics. Because of the abundance of specialized literature on Laplace transforms, we have omitted the simplest problems belonging to these categories. At the end of the section, we give a few problems of a more " In particular, this condition is satisfied if f(t) is piecewise smooth in [a, T], or if f(t) satisfies Dirichlet conditions in [a, T}. 1 6 The tables in E3 are particularly complete.

l' tlOil.

INTEGRAL TRANSFORMS

355

171

ttlmplicated nature (e.g., Probs. 391, 405, 406), to be solved by combining tht• Laplace transform with some other integral transform (e.g., the Fourier t111nsform or the Hankel transform). 353. Starting from the time t = 0, the plane boundary of a semi-infinite hody of thermal conductivity k, specific heat c and density p is maintained nl the temperatureT j.,_0 = f(-r), where -r = ktjcp. Find the subsequent tempt·rature distribution in the body, assuming that the initial temperature is zero. Ans.

=

T(x, t)

2

I-""•. r • ( 1 yTt

2_ _.,_e-"f -r- -x

4u

.v.,-

2

)

X> 0.

du,

354. Consider the following special cases of the preceding problem:

a) f(-r) = T0 ; 0 < -r < -ro, C)!(~). __ {To, 0, -r > -ro;

b) f(-r)

=

d ) f( -r)

. w-r. = "~'~ 0 s1n

A-r;

Ans.

a) T(x, t)

=

To[1 - (~;r)

+

b) T(x, t) = A-r{ ( 1

=

;J [

1 -

0

< -r
0, y > 0), whose s urface is held at temperature T0 starting from the timet = 0 (the initial temperature is assum ed to be zero). Plo t the corresponding isotherms.

2

Ans.

T (x,

X

OL-----~-----2 ~-2/T F IGURE

92

y, t) = ro[ t - cJ> CJ=.)$(~:J J.

The result of the calculations is shown in F ig ure 92. Hint. Look for a solution of the form T = To[l + u(x, t)v(y, t)], and then reduce th e problem to Prob. 354, Case a.

362. F ind the temperature distribution inside a body shaped like an odant (x > 0, y > 0, z > 0), whose surface is held at temperature T0

174

INTEGRAL TRANSFORMS

PROB.

363

= 0 (the initial temperature is assumed to be

starting from the time t zero).

Ans.

363. Find the temperature T(x, t) in a slab of finite thickness, if one face x = 0 is held at temperature T 0 starting from the time 1 = 0 , while the other x = a is held at temperature zero. lt is assumed that the whole slab is initially at temperature zero. Give two forms of the solution, one suitable for large t, the other for smaU t.

Ans.

=

T(x, t)

To[t- ~ - ~ a

T(x, t) = To

I n=O

Isin (n7txja) n

e- n"n"-r!""],

7t n = I

[(2a- x + 2na) 2.J-r

(x + ~na)J. 2.J-r

_

Hint. To obtain the second form of the solution, expand the Laplace transform of the desired function in ascending powers of the quantity e_.,.y;. 364. Solve the preceding problem, assuming that a thermal current of constant density q is incident on the face x = a, whi le the face x = 0 radiates heat according to Newton's law.

Ans.

{1. +

T(x, t) = qa ~- 2(ah)2I cos [y.,.(a - x)/~] e;Y~2-r~a• }, k ah a .,._ 1 [ah(l + ah) + y;;]y,. Sll1 Yn where they.,. are consecutive positive roots of the equation coty

y

= -, ah

and h is th e heat exchange coefficient. 365. Solve Prob. 363 assuming that the face x = 0 is held at constant temperature T0 , while the face x = a is connected to a thermal capacitance. 20 Derive expressions for the density of heat current o n the faces of t he slab.

Ans.

I

_ 2kT L.. ~

q .t- o -

qIx- a

0

a

11= 1

+ o:\~ e- Y.'-r/"2' 1 + o: + o: y;, 1

2

_ 2akT0 ~

-

a

L.. (

n- 1

•

-y. 2-r:fr?

"( 11

1

+ 0: + 0: y.,) S ill y, 2

2

•

e

'

1

20 By a "thermal capacitance" we mean a body in which any temperature drop can be neglected. Jn Probs. 365, 367, etc., C0 denotes the amount of heat needed to raise the temperature of the body by I degree, referred to uni t area, unit length, etc.

I' IIOB.

367

INTEGRAL TRANSFORMS

175

where the Yn are consecutive positive roots of the equation cot y

Co

= ay,

Ct.=-.

cpa

Hint. The boundary condition a t x =a has the form

where C0 is the thermal capacitance per unit area and k is the thermal \'onductivity.

366. Find the temperature distribu- a < x < a in wh ich, ~l arting from the time t = 0, there is 11 process periodically producing heat uccording to the law shown in Figure 93. 'l'he temperature of the faces of the •,lub and the initial temperature are n ~sumed to be zero.

1ion in a slab

FIGURE

93

Ans. ,,

2

Q {a - x 16a ~ (-1)"cos(T0 A~,x/a) --- •~.• a• t) - - 0 - - - L. e '' 0 " 1 .' - k 4 it3 n=o(2n -J- 1)3[1 -J- e~o•l.n•ta•] 2

2

I (X

2 To~ 1 2 rc ~(2n + I?

[( 1

_

4

cosh A.,(x + a) cos ),.,(x - a) + cosh A,(x - a) cos A,.(x + a)) cos 2 'A!T cosh 2A,a cos 2A,.a

+

I sinh )..,(x - a) sin A11(x +a)+ si nh A,.(x + a) sin A,(x - c1) sin 2 A~TJ ), 4

cosh 2)..,.a

+ cos 2)..,a

where An

= J(2n -1-

I)TC.

2T0

367. A t hin cyl indrical rod (probe), heated to temperature T0 , is inserted the ground, in order to measure the ground's thermal properties. Ikscribe how the temperature of the probe varies with time, assuming that I he temperature drop inside the rod can be neglected (see T JO). 1111.0

Ans. 4 T0a ( ""

T Jr=a =

7

e-1-'~!a"

Jo [AcxJ 0(A) - Jl("-)f

+ [f..ocYo("-)-

dA Ylf..)]

2

):'

176

INTEGRAL TRANSFORMS

PROB.

368

where Jn(x) and Yn(x) are Bessel func tions of the first and second kinds, = C0 /2na2cp, C0 is the thermal capacitance of the probe per unit length, t1 is the radius of the probe, k is the thermal conductivity, c the specific heat and p the density of the ground, and -r = ktfcp. IX

Hint. Solve the heat conduction problem for the domain r boundary condition C0 ()T

ot

I= r- a

()T 2nak-

or

I

t

, r-a

>

>

a with the

0.

368. Use the Laplace transform to solve Prob. 335, and then show that the two answers are equivalent. Ans. T( r, t)

=

T.

o

[1 - ~ l ao _

1t

2

rob. 335 follows from the expansion

1=

3 roo 1t

Jo

0 through an opening in the impermeable wall z = 0 (see Figure 96). F ind the amou nt of substance inside the tube as a function of time, ass umi ng that the fl ow of current is constant over a cross section of the tube a nd that the initial values of the concentratio n of the substance in the tube a nd in the half-space equal ('0 (per u nit length) and 0, respectively. AfiS.

M (t)

=

2Mo

fro

1t

Jo 1

(1 + Sin.

2

X -

ax) e-Dt~••

cos l

. 2 Sin

( si n

II

. (1

X Sin

+ Ia)

X

x)2dx,

x

I'ROB.

379

lNTEGRAL TRANSFORMS

179

where a is the radius of the tube and M 0 is the initial amount of substance inside the tube (M0 = C0/). 376. The end x = 0 of an infinite transmission line, with self-inductance Land capacitance C per unit length, is joined at the timet = 0 to a source of c.m.f. E = f(t). F ind the voltage u(x, t) at every point of the line. Ans. X

t

< -' v

t

> -.

X

v

where v = J(J LC is the velocity of wave p ropagation along the line. 377. Solve the preceding problem for the case of self-inductance L, capacitance C, resistance R and leakage conductance G per unit length, chosen to satisfy the relation RC = LG (a distortionless line). Ans. X

t

-'

x

v

.JL/C is the wave resistance of the line, and a =

1/C0 Z .

379. The end x = 0 of an infinite line with self-i nductance L , capacitance ( 'and resistance R per unit length is connected at the timet= 0 to a source ol' constant e.m.f. Study the res ulting process of propagation of a voltage wave along the line (see C2, p. 202). Ans.

t

X

< -, v

X

t

>-v '

180

INTEGRAL TRANSFORMS

PROB.

380

where E is the size of the applied e.m.f., ()( = R /2L and I 1 (x) is t he Bessel functi o n o f imaginary a rgument.

x=O

f

380. A line of length I with parameters L and Cis terminated at the end x = I by a resistance R 0 (see Figure 97). Find the subsequent FIGURE 97 voltage in the load R 0 , ass uming that the end x = 0 is suddenly connected a t the time t = 0 to a source of consta nt e.m.f. E. Under what conditions is there no reflection of waves from the end of the line? Ans.

O < t < T, (2n - l )T

< t
truck at t he timet = 0 by a mass M 0 moving with velocity v0 • Ans.

I

u x=o =

where

IX

Vo{21X.Jl .j7r. -

IX2

= 2../2 paSfM and (x) is

1

+e

2

a.

t[ L - ""( r;) } ' l ' IXy t ] ,

the probability integral.

400. Find the transverse oscillations of a beam - / < x < !, simply supported at both ends, due to an impulse P acting at the center of the beam. Write an expression for the deflection of the center of the beam. · Ans.

ui .,=o = 4~/a ~sin [(2n + lin a t/4l 2 7t EJ n = O (2n + 1) 2

2 2

2 ].

401. Find the deflection of an infinite elastic plat e, if at the time t force Q is applied to the point x = y = 0 (see Ll6, p. 424).

11 constant

Ans. u(r, t) 22

= -Q-r- [7t- 4nD 2

(r2) -

Si 4-r

r2 + -rz Ci (r2 - )] ,

sin 4-r

Problems 397- 399 are treated in Lurye's book Ll6.

4T

4T

= 0

188

INTEGRAL TRANSFORMS

PROB.

402

where D is the flexural rigidity and p the surface density of the plate, -r = t,J D/ p and Si(x), Ci(x) are the sine and cosine integrals. Hint. At the point where the force is applied, uir=o must be bounded, and moreover

0

r-

or

Q

6.ulr=O= - - .

21tD *402. Solve the preceding problem, assuming that an impulse P (rather than a force Q) is applied to the point x = y = 0.

Ans. u(r, t)

p [7t 1t,JPI5 4

l-

=

Si

(~"2)] -r .

4 403. At the time t = 0 an impulse with components P., = 0, P 11 = P is applied to the point x = y = 0 of an infinite elastic plate. Describe the resulting process of wave propagation. Ans. The elastic potentials are given by the formulas r

0, cp(r, t)

=

t

< -' a

t - a2 '

t

>-'

0,

t

, and is the appropriate tool to use for solving problems of two-dimensional elasticity theory and potential theory involving angular regions. The required technique can easily be acquired by working through the following small sci of problems. 24 FIGURE 101

*407. Find the stationary temperature distribution inside the dihedml angle 0 < r < cc, 0 < t:p < ex ~< rr, if one boundary is held at temperature zero, while the temperature distribution

0

< r a is maintained on the other boundary. Ans. T(r, rp)

~

(a)";" .

"'P smT. r, a ___!) arc tan ---'-''------"'--"' 1 + ({l_)nfacos 7t_rp r. oc

•

408. Solve the preceding problem, assuming that a given distribution of heat current

I

qrprp=:X

23

qo,

=

(0

a-s< r ) 2(rr 1-2-cos .a 3 1- 2(_.':_)'cos 2(rp 'ro· ,,

(' r)'i" +(:)· 0)

0

< r
- 7t -

l ao

fw

21~ d K;~(Aa)K;~(A.r) sinh 2(7t - a}r d cos AZ A T, 1t 3cr1 o o sinh TIT + ~ sinh (1t - 2a)T

l ao

l ao

41 d K;~(Aa)K;-r(Ar) sinh m d cos AZ A T, 1t3(cr1 + cr 2) o o sinh 1tT + ~sinh (1t - 2a)T

This problem was first solved by Macdonald (Ml).

PROB.

425

INTEGRAL TRANSFORMS

199

where g ,....- cr2- crt 'cr2+cr1' J is the current and KJx) is Macdonald's function (see S4).

425. Show that the solution of the preceding problem can be reduced to the form

" vI (r +at+ z2

I - ~'j'~ sinh 3~ d~ ) 2 2 2 sin 0 o sinh r=~ ~i(r 1 a) -1-- z + 4ar sinh h~~ '

uio~n ~

\(-;==='=1= =

J

11:(cr,

+

cr,)

.Jcr +a)'+ z'

3 if oc

~

~

arc

cos~

2

Jt/4, and to the form

j''"

1+ ~ 2 cos ~0 o cosh r=~ ,,..i(r

cosh 3~ d~

\

1+~ a= arccos--

2

if CJ.

~

Jt/3.

200

PROB.

INTEGRAL TRANSFORMS

426

426. Solve the problem of diffraction of a plane electromagnetic wave

E., = E" = 0,

FIGURE

Ez = Eoei(wt- kx)

(where k = wfcis the wave number) incident on a thin perfectly conducting sheet (see Figure 108) making an angle rx with the direction of wave propagation (Sommerfeld's problem).

108

Ans. The complex amplitude of the z-component of the total field is

E

. [1 + --=

eirt/4i

= Eoe-•"• cos cp

-

2

.J7t

v

!-iq>

2kr sin

.2

e- " ds

J

0

+ Eoe-'lk-r. cos. -~·-> [

-

(see K3).

4

+ eirr1 2 .fit 1

-

iV

J

2krsln ~(cp-21x) e_,s. , ds

0

427. Using the result of the p receding problem, find the current distribution on each side of the sheet. Consider the special cases where a) rx = 0; b) (J. = 7t/2. Ans. The requi red densities are determined by the system of linear equations jl

+ j2 =j =

E : 2

cj2 [ e-ik..-

rx

2e in/4

.

{

~ .Jikr cos2. + .J"i"t sin rxe- •krcosa.Jo

v 2kr sin !-ia.

.,

J

e- " ds ,

j l - j2 = Eoc sin rxe- ikrcosa.

27t where h and }2 are the densities on the upper and lower sides of the sheet, · respectively. In the special cases, a) b)

jl

j1

'

2

= j2 =

E c [ 1 e- ikr 47t 7t .J i kr

:fo ~~::. ;

2

2eml4l .,;;.;; . ' e-u ds 7t o

= - 0 .J--= + .J-

±

J

1 .

428. A line source of a-c current J = J 0eiwt is placed parallel to the edge of a thin conducting sheet 0 < x < oo, - oo < y < oo. Find the distribution of induced currents if the source lies in the plane of the sheet at a distance a from its edge. Ans. The complex amplitude of the current density is .

Jo

]= -

27t

Ja X

e- ikCx+a> .

X+

a

20 I

INTEGRAL TRANSFORMS

PROB.431

429. Find the electromagnetic field of a dipole of moment P located on the axis of a perfectly conducting conical reflector of vertex angle 2o:, if the dipole lies at a distance a from the vertex of the cone (see LJO).

Ans. If r series


a is obtained by permuting the symbols r and a in the general term of the series. 430. A plane acoustic wave u 0 eiC 0, cp = <X. F ind the wave reflected from the screen.

Ans. The complex amplitude of the velocity potential at an arbitrary point is

_

.

Uoe-•krcos (c;>--2«)

[

_

1

_

+ eirt/_4i v

2

-y11t

2kr sin

~(cp-2ot)

. •

J

e-u d s .

0

431. A point so urce of so und, radiating a spherical wave

u = u0

sin (wt - kR)

R

,

is placed on the axis of a conical resonator 0 < 0 < o: with perfectly reflecting walls. Find the velocity potential inside the cone.

Ans. T he complex amplitude of the veloci_ty potential is

I

_ Uo1t21.~( L.,

U r O

Ans.

=

ds2

L'l.u _ -

c2(0:.2

a

y

"'h"~ :x0 Il

=a~ Sill

tJ

1--l ~ La11e

(nl

13)

(cosh oc -i- cos

2

'Y2)(a-ao)sin (n -I ~)[3' )

a

7t n~... o

where

. smh a 0

h(2a-h) 2a(a - h) The rigidity is

(h

:::::--: - - - - · -

C = Ga' sinh'




0.

(5)

Here P .(z) is the Legendre function of the first kind, and the plus sign pertains to the interior region 0 < ~ < ~ 0 and the minus sign to the exterior region ~ 0 < ~ < 11:. The general solution is now constructed by integrating (5) with respect to -r. To determine M"', we use the Mehler-Fock theorem, 8 instead of the theory of expansions in series of spherical harmonics. Next we consider oblate spheroidal coordinates a, ~. q> related to the rectangular coordinates x, y, z by the formulas x = c cosh rx sin

~cos q>,

y

= c

cosh a sin

~

sin q>,

z = c sinh a cos

~.

where 0


different from the conductivity cr2 of the rest of the ground (see Figure 131).

226

PROB.476

CURVIUNEAR COORDINATES

Find the current distribution in the ground, assuming that the boundary between the two media is the prolate spheroid with equation

z

>

0.

Ans. The potentials of the current field in the two media are given by u1

1 2rtcrlR

= -

+

J(cr2 - cr1) !(4n 2rtcrf. J a 2 - b2 n- o

X

U2

=

Q2n(cosh exo)Q~n(cosh exo)P2 ,.(cosh cx)P2,.(cos ~) cr1Q2n(cosh cx0)P~,.(cosh cx0) - cr2P 2,.(cosh cx0)Q; 11(cosh cx0) ' ----~~--~~~~--~~~~~~~~~---

J 00 2C4n 2 2 2 2rtV a - b sinb cx0 n- o 1

X

+ 1)P2n{O)

+ 1)P2n(O) Q 211(cosh cx)P211(COs ~)

------------~~----~~--~~----------­

cr1Q2,.(cosh exo)P~n(cosh exo) -

cr2P2 ,.(cosh exo)Q~n(cosh exo)'

where R is the distance from the source to the field point, tanh cx0 = bfa, P 11(x) and Q,.(x) a re Legendre functions, and

Po(O)

= 1, P2..(0) = (- 1)" 1. 3 . 5 ... (2n - 1),

n = 1, 2, ...

2 · 4 · 6 · · · 2n

476. A d-e c urrent enters ground of conductivity cr through a grounding plate in the form of a disk of radius a (see Figure 132). Find the distribution of current under the plate, and calculate the resistance of the plate.

Ans. The potential of the c urre nt field is 2V . u = -arc cot smh ex,

I

z

F IO URE 132

1t

where Vis the potential of the plate. The resistance is

R = - 1-. 4cra

Hint. Introduce a system of spheroidal coordinates (0 0

< ~


0, b

>

0) .

491. Find the stationary temperature distribution in a body shaped like a paraboloid of revolution ~ = ~ 0 , if a given ax ially symmetric temperature distributio n

T(rx, ~)la=!lo is maintained on its surface.

= f(rx)

Ans.

~) = J "' Io(f.~) J 0 (f.rx)f. dt..J J(C:)lo(t..C:)C: dC:.

T(cx.,

o I 0 (/..~o)

o

492. Solve the Dirichlet problem for the domain bounded by the paraboloid of revol ution ~ = ~ 0 , assuming that the boundary condition is of the form cos n

0), and are of the form

+

u

=

uT = ,)2 cosh ~ -

2 cos ex [M -r cos -r~

+N

-r

sin -r~]P-~+;-.( ± cos ex.), T

> 0, (13)

where the plus sign corresponds to the exterior problem (0 < ex < ex0 ) and the minus sign to the interior problem (ex0 < ex < rc). Jn this case, the general solution is obtained by integrating (13) with respect to -r, and the factors M-r and N.., are determined by taking Fourier cosine and sine transforms with respect to ~· This section contains problems from various branches of mathematical physics which can be solved by using three-dimensional bipolar coordinates. The last three problems (Probs. 512514) involve limiting cases of bipolar and toroidal coordinates, and lead to a= e legant formulas for the capacitance ::......:'-R'~~~-=t.:.:._-~~~:F-....._z of such objects as a pair of spheres in contact or t he surface obtained by rotating a circle about a tangent line.

507. Find the e lectrostatic field in a spa rk gap consisting of two conducting spheres of radius a, with centers a distance 2/ apart, if the spheres are at potentials V1 and V2 respectively (see Figure 144). FIGURE 144

Ans. The electrostatic potential is

u( ex,

A) I-'

= v12 cos h I-'A

-

+ i)~ + k>~o sinh (n + !)~j e-ln+l--•Jf>op,.(cos ex.) , + v2 -2 vi sinh ( n + n~

~ [-=V•c....: 2 cos oc L. - +~Vl cos h (n n- o 2 cosh (n

in terms of the Legendre polynomials P,.(x), where

I cosh ~ 0 = -. a

Hint. Use the expansion 1

-;;;==;::::;::=:::;o===

,}2 cosh

~ - 2 cos ex

00

= L:e-< 11 +~>flp11 (cosoc). I

,.=0

*508. Find the capacitances Cu. C 12 and C22 of a system of conductors consisting of two spheres of radii a 1 and a 2 , with centers a distance 2/ apart. 20 •• Concerning the meaning of C11 , C,! a nd Ce:. see the solution, p. 370.

I'ROB.

509

245

CURVILINEAR COORDINATES

Assuming that the radii are equal (a, = a, = a), tabulate C12 as a function of the ratio /fa. Ans. 1 + ~ [e- 0, z = 0. Denoting the components of the resulting electric field (the sum of the incident and reflected waves) by 0, 0, E and setting E

=f(t - ~) - u,

show that the reflected wave u can be represented in the form u

=f"l


YJ

=

cp(s)

- s + (2afv) t -

r v a'

ds

r =

.Jx2 + l )'

where cp(s) is the solution of the integra l equation

f ~;

- oo .J~

cp(s)

- s + (2afv)

ds = {(~). ·

Hint. Look for a solution of the wave equationdepending only on ~ and YJ·

256

PROB.

INTEGRAL EQUATIONS

527

527. Solve the preceding problem of diffraction theory, assuming that the wave makes contact with the screen at the time t ~ 0 and is continuous along its front, 1 i.e.,

/(~)

(g(~).

=

g(O) •o 0.

0, Describe the diffraction process graphically. Ans.

u

=

{f.

,

ds,

'f'(s)

- s + (2ajv)

o J~

0,

1J

< 0,

where

9(s) Here

g and K are

~ - 1 f..~ 2r.i

r

K

e"·'dp.

the Laplace transforms of the functions g(O and

• ~ (·c, ·r. -;;2a)· ''', K( 0, f(~) = {g(~), g(O) = 0. 0, ~ < 0,

> 0),

Ans. In the excited zone (YJ

1

o~ - s

where cp(s)

cp(s)

1)

u =

=a- 1 -

+ (2afv) ds'

f.

pg

v 2-rri r 1 - (a fv)pK

e.,• dp.

In the last formula, g and K a re the Laplace transforms of g(~) and the kernel K(" ) c; - ~

1 + (2a fv) '

and the path of integration r is a straight line parallel to the imaginary axis lying to the right of the singular points of the integrand. Note that

K=

e- 2a.,fvEi ( -

2:P),

in terms of the exponential integral Ei(x). 533. Consider the problem of diffraction of a plane wave by a paraboloid of revolution r = z + 2a with homogeneous bo undary conditio ns of the first kind. Show that. the reflected wave has the representation u

=I~ -

z + a ( ~ = t - -v- ,

<X)

~

-

s

rp(s) ds + (2a fv) '

r- a

= t - -v-,

1)

r

=

.Jx2

where cp(s) is the solution of the integral equation

f

r;

-<X)

~

-

s

cp(s) ds + (2afv)

= !(~).

+ / + z2) ,

536

I'I(()U.

INTEGRAL EQUATIONS

259

534. Using the Laplace transform, solve the integral eq uatio n of Prob. 'd3 for the case of a wave of the form

/(~) = {g(~).

~ ~

0,

> 0, < 0,

g(O)

=

0.

Ans. In the notation of Prob. 532,

= -1

a) from the center of the sphere. 0

Ans. cr(N) =

~ - !L b2 - .at'

4-rr aR 3 where R is the distance from the charge to the given po int N of the surface of the sphere. 4-rra

538. Solve the preceding problem, given the total charge Q of the sphere (rather than its potential). Use the formula so obtained to solve t he problem of the charge distribution on the surface of a n initially uncharged insulated sphere introduced into a homogeneous external field E 0 •

Ans. cr(N) = E?,(N) --[~ ii - u\N)], 2-rr 4-rra a where ii 0 is the average over the sphere of the potential of the external field: 1

+

+

uo = _ 1_2 ( uo(N) dS. 4-rra

Jr:

In the special case

3E cr(N) = -cos !.l, 4-rr where 0 is the a ngle between the direction of the external field E 0 and the radius vector drawn from the cente r of the sphere to the point N. 539. A cylindrical conductor with cross section bounded by an arbitrary contour r (see Figure 152) is introduced into a given plane-parallel field E 0 • Show that the density of charge satisfies the integral n equation E?,(N)

cr(N) = -

-

2TI

+ -1 TI

J

cr(M) - - cos (r 111N, n) ds,

(2)

r lrMNI

•"

F IGURE

152

where M and N a re two arbitrary points of the contour r, ds iS the element of arc length, r MN is the VeCtOr joining M to N, n is the u nit exterior normal to r at the point N, and E?, = E 0 • n is the projecti on of E 0 onto n.

1'~08.

542

INTEGRAL EQUATIONS

261

540. Suppose a conductor shaped like an infinite circular cylinder of radius a, carrying charge Q per unit length, is introduced into an external plane-parallel field E 0 • Show that the density of induced charge on the surface of the conductor is given by

=

a(N)

JL + E~(N) . 21ta

27t

Consider the special case where a) The extern al field is homogeneous; b) The source of the external field is a line charge with c harge q per unit length, placed o utside the cylinder at the distance b from its axis.

Ans. a( N)

a)

+ E cos 0 ,

= _Q_ 21ta

27t

where 0 is the angle between the direction of the homogeneous field (of strength E) and the vector drawn from the center of the cylinder to the given point No n the surface of the conductor; b) where R is the distance from the line charge to N. 541. Find the distribution of charge density on the inner surface of a grounded cylindrical shell of radius a, assumi ng that the external field is produced by line charges parallel to the axis of the cylinder passing through the points M k = (ak, rpk), k = I , 2, ... , n.

Ans.

1

1t0

a a;. 2

n

L_q,,

a(N) = - -2

-

R2

k- 1

,

k

where qk is the charge per unit length of the line charge passing through the point Mk, and Rk is the distance between the points M k a nd N. *542. The e lectrostatic field in the region 0 < y < lz between two grounded parallel planes is due to line so urces whose free-space field is E 0 • Show that the densities a 0(x) and a,.(x) of induced charge on the planes y = 0 andy = lz sati sfy the system of integral equations a (x) - _!_ Eoi o - 2 7t 11 v- o

a"(x)

=-

~ 7t

7t

and then solve this system.

-oo

h

I

ol -2 EY v - Ii

f oo

-

-

7t

ah(~) X )2

("C, -

Joo -oo

+ h2

dr:

"'' (3)

ao(~) )2 + 12 dt,, X l v

("

C, -

262

INTEGRAL EQUATIONS

PROS.

543

Ans.

(

cro x )

= _1 J oo j'o + e- IJ..II'j'" -iJ..x d'A

-oo 1 - e- 211-lh e ' = _ _1 J oo .j'h + e- IJ..Ih'j'0 - i'AX d~ cr,.(X ) "' 47t2 -oo 1 - e- 21>-lh e 47t2

where I is the Fourier transform of f(x), i.e.,4

I =L :

f(x )ei>-x dx,

and

fo(x) = E~jy=O• f,.(x) = E~jY=h· Hint. Take the Fourier transform of each of the equations (3). 543. Solve the preceding problem for the special case where the field E0 is due to a line source with charge q per unit length, passing through the point M 0 = (0, b). Ans. . 1tb . 1tb sm Si n cr .(x) = - !L ____ h _ __ 1 cro(x) = - ;h 1tX h 1tb ' 2fl 1tX 1tb cosh- - cos cosh - + cos-

h

h

h

h

Hint. To obtain the solution in closed form, use formula 15, p. 385. 544. Suppose a system of line sources, whose free-space field E0 has components E~, E:, 0 in cylindrical coordinates, is placed inside a di hedral angle 0 < q> < ex with grounded conducting walls. Show that the charge densities cr0 (r) and cra(r) on the walls satisfy the system of integral equations

·) = _!_ Eoj'P rp= O _

O'o(t

27t

sin 1t

exl ""

pcr..(p) d p, • • o p- + r· - 2rp cos ex

(t·) = _ _!_ Eoj 'P

_ sin exl oo 2 pcr0 (p) d ROB.

546

INTEGRAL EQUATIONS

263

where j is the Mellin transform of f(r), i.e.,

j=

L"

1

f(r)r 11- dr,

and

fo(r)

=

E~l~=o•

fr,_(r)

= E~lq>=cx·

Hint.

( «>

Jo

=

t ··dt

t

2

-

2t cos a f- l

~sin

sin a

(1t - a)s si n 1tS

- 1 < Res < 1.

545. Solve the preceding problem, assuming that the field E 0 is due to a line source with charge q per unit length, passi ng through the point M 0 = (r0 , cp0). Use the formula so obtained to find t he electrostatic field due to a charged line placed al distance a from the edge of a conducting half-plane (a = 27t, r 0 = a, cp0 = 1t) or near a right-angular corner (a = 37t/2, r 0 = a, 'Po = 7t). Ans. sin 7tCflo

O"o(r)

= - .!L _____...:.a.:..___ __

-

+ (ro)n/cx - -

ar ( r )n/cx

'o

O"cx(r)

r

7t<po ' 2cosa

sin 7tCflo ...:.ex::___ _ _

= - .!L _____

)"'"'+(r )n/cx+ 2 cos--'! 7t

+ h2J'

a,.(~) d~

(x _ ~?

([z)

e:2

+ e:2) X [Eol . + 2hf"' ~ v=h 21t(e:l

-«>

+ h2J.

Go(~) d~

(x _ ~)2

(Ez)

Solve this system of equations. FIGURE

Ans. cro(x) = a11(x) =

~ -2

47t

~ -2

47t

f "" -«>

~e-l).lhlh - fo 2 - 21:1.111 e 1- ~ e

f "" ~~~ -«>

1-

~e-P·I"fo ~

L:

- i1.:t

2 - 21'-1'• e

e

where J is the Fourier transform ofJ(x), i.e.,

I=

iA:t

f(x )ei'-z dx,

d"A, d"A,

158

270

PROD.

INTEGRAL EQUATIONS

564

and

fo(x)

=

E~lv-O• fix)

=

~lv=IP ~

=

1 E

2 -

El

E

+ E2

•

564. Solve the preceding problem for the special case where t he external field E 0 is due to a line source with cha rge q per unit length passing through the point x = 0, y = h/2.

Ans. cr0(x)

= a,.(x) = -q~

i"

e-M/ 2

ne:1 o 1 -

~e-

cos "Ax d"A.

IJ

'

*565. A perfectly conducting ha lf-plane x > 0, y an external electromagnetic field with components

E"'

= 0 is introduced

into

= Eu = 0.

+

Show that the sum j = j 1 j 2 of the current densities flowing in the upper and lower sides of the half-plane satisfy the integral equation E(x)

= n~ {""H~2>[k lx - ~l]j(~) d~, c

Jo

where E(x)

= E0(x, 0),

and e, !L and cr are t he dielectric constant, the magnetic permeability and the conductivity of the medium, while the difference between the current densities is . . c2e-i"f2i)Eo(x, y) )1 - ]2 = - . (9) 2m.u y v- 0

a

I

Solve the integral equation by using the transfo rm (27), p. 196.

Ans. j(x)

= ~ f""{~ [E(x) - E(O)e- ika:] -

J

2nwx o 2

_E(O) }"" si nh2 n-r e"(kx) d-r, sm h n-r cosh n-r

where

[it is assumed that f(x) approaches zero as x integral converges at its lower limit].

- >-

0 in such a way that the •

*566. A plane electromagnetic wave with components

I'ROB.

566

INTEGRAL EQUATIONS

271

is incident on a perfectly conducting half-plane r > 0, cp = <X. Using the result of the preceding problem, find the distribution of current density on the half-plane.

Ans.

References Books: Grinberg (G5), Kupradze (K5), Mikhlin (M7), Morse and Feshbach (M9), Muskhclishvili (M 10, M ll ), Noble (N 1), Smirnov (S6, Vol. IV), Titchmarsh (T5), Trico mi (T9). Papers: Fock (FJ, F5), Lebedev (L5), Wiener and Hopf (WlO).

Part

2

SOLUTIONS

SOLUTIONS

52. The solution of the problem reduces to the determination of the complex flow potential w = q> + i~, whose imaginary part is a harmonic function which equals zero on the axis of symmetry and takes the value •.V = -v0 a on the walls of the channel. To determine w, we need only find a conformal mapping of the region ABCDE onto the upper half-plane of the variable ?,:= ~ + iYJ. Suppose that in applying the Schwarz-Cbristoffel transformation, we make the points of the z and ~-planes correspond in the way suggested in the hint to the problem. Then the relation between z and ~ is obtaine9 by integrating the equation

~~ = M(~ + 1)112(~ + A)-112~-1, where M is a constant to be determined later. Bearing in mind that z if~ = - 1, we find that z

= ib

= MJ~ (~ + 1Y'2 d~ + ib, - 1

~ + )J

~

where the integration is along any path joining the point = - 1 to a given point ~ in the upper half-plane. It follows from the condition lim [zl~-· e-o

- zlc=•] =

ia

that M = aVA/rt, and hence it only remains to determine the value of the parameter A. This is done by using the correspondence between the points ;: = ia and ~ = - A. Since in evaluating the integral with ~ = -A as its upper limit, we can integrate along the line segment joining the points ~= -I and~ = - A, on which ~

+1=

1 - s,

~

+ A=

e;"(s - A),

275

~ = - s

(A

<s
[lo(:') Yo(~p) - Yo(:')lo(~p) J

pdp.

The constant B equals zero because of the requirement that the solution be bounded at the point r = 0. The constant A is determined from the condition w(a) = 0, which gives

A =

rt

2TJ 0 (wa fv)

lji(p)[ro(wa)lo(wp) - J (wa) Yo(wp)J pdp. v v v v

(a

Jo

0

After some manipulation, the desired expression for the amplitude takes the fo rm

w(r) =

J:lji(p)G(p, r)p dp,

(4)

282

PROB. 91

SOLUTIONS

where

G(p, r)

=

Jo(wpfv)[Jo(wr) Y0 (wa) J 0(wafv) v v

1t

2

T

Y0 (~) 1 0 (~)] , v v

p


r.

(5) Substituting (3) and (5) into (4), and using the formu las

l

o

I:

p

p

plo(l-p) c/p = - J1(l-p), !.

p

- Y(t.p) A 1

p Y0('- p) d p

2 +1t),2

and the familiar expression

J0(x)Y~(x) - Y0(x)J~(x) = 2 TCX for the Wronskian of the Bessel functions, we finally obtain the answer o n

p. 48. 91. If the z-axis is parallel to the generators of the wave guide, then the only component of the electric field of the TM-wave is

( w is the frequency of the oscillations and v is the propagation constant),

whose amplitude satisfies Helmholtz's equation

~. ~( 1_(JE)

ror or

2

+ _!_ oE + (k2 _ r2

ocp

2

v2)E = O

(k = wfc = 2TC/l., where!. is the wavelength) a nd the homogeneous boundary conditions

Elq;- o = Eq;

·a

= 0.

These equations have infinitely many non zero solutions of the form

E

= Emn =

(Ymn~") ·

J mrrfa -a-

Sin

mTCcp - a- '

m

=

J, 2, ... ,

where they"'" are the roots of the equation

J .,,rrla_(y) = 0, and the value of the propagation constant corresponding to Ymn is 'I

=

'~mn

=

Jk2 - (y""')2· a

11ROB.

96

283

SOLUTIONS

A wave with an imaginary value of v,,.11 falls off exponentially in the z-direction and is essentially unable to propagate in the wave guide, i.e., a wave can propagate in the guide only if v,.,. is real. This leads to the inequality ,

21ta

A< - . Ynm

The maximum wavelength which can propagate in the guide is given by the formula 21ta

Anwx = - ,

Yo where 'Yo is the smallest positive root of the equations

lmn/a('Y)

=

0,

=

m

1, 2, . . .

96. The problem reduces to integration of the equation 2

2

d u - .!_ d u = 0 dx 2 v2 d/ 2 ' with initial conditions

hx

< X < C,

0

•I•-' ~ f(x) ~ { h~l ·- x) ' l- c and boundary conditions

uj.,_0 = uj,_1 =

0.

Setting u(x, t) ·= X(x)T(t) and separating variables, we arrive at the equations

X" -l :AX = 0,

T"

+ :Av T = 2

0.

Solving the first of these equations with the boundary conditions X(O) X(l) = 0, we find the corresponding eigenvalues and eigenfunctions

X = Xn(x)

.

=

1l1tX

sm - 1

(n

=

1, 2, . ..).

The solutio n of the second equation satisfying the conditions T'(O) given by T

=

T,.(t)

=

=

0 is

t11tVI

= c,. cos - - . l

Therefore the set of particular solutions of the equation of the vibrating string satisfying all the homogeneous conditions is U

=

U

n

=

.

C

n

1l1tX

Stn -

l

1l1tVI

COS - -

l

'

tl

= 1, 2, ...

284

PROB.

SOLUTIONS

108

According to the basic idea of the Fourier method, we now look for a solution of the given problem in the form of a series ~ . mtx n1tvt u(x, t) =L en Sin cos--, n- 1 I I

where the coefficients c,. are determined from the condition u 1i=o = f(x), i.e., coincide with the coefficients of the expansion of the function f(x) in a Fourier series

f(x)

"" . n1tx = 2: c,. Sin - ,

< X
=1t

vir-a=

sin4 q>,

. 4 (~)]z d~ .

Evaluating the integrals in (21), we obtain 2 To tan y n (COS y n

c.,

=

yn

I

(Y,.a) (k

0 -

-

hi

1

k2

-

. 2 y,.h2 sm - h!

~~ 2)

cos Y

+ -h2 sm . hl

2

yn

)

•

Substitution of these coefficients into (20) gives the answer on p . 89.

(21)

304

SOLUTIONS

PROB.

176

176. The problem reduces to integration of Laplace's equation

ar) + - 1- ~(sin 6 ar) = o or sin 6 o6 o6 '

~(r2 or

with boundary condition

1

T /r=a = j(6)

o < e < oc, oc < e < rr.

= {~0 '

The required harmonic function is constructed as a series

(r)"

oo T(r,6) = 2cnP ,.(cos 6), n= O

(22)

a

where P ,.(x) is the Legendre polynomial of degree n. Because of the boundary condition, the coefficients c,. must coincide with the expansion coefficients of the function f(6) with respect to the Legendre polynomials, i.e., 00

/(6) which implies

=

0


= A(r 6) 'I'

Setting

'

i = 1,2,3 .

'

where

Ao(r, 6)

= 2Jr0 { "' c

cos

Jo .Jr

2

-

tp

dtp

2rro sin 6 cos

tp

(23)

+ r~

is the vector potential o f the loop, we reduce the problem to determination of the functions A 1 satisfying the differential equations4

~(,. 2 oA 1)

or

or

+ _1_ .E._ ( sin 6 oA1) _ ~ = 0

ior (r oA2) + or 2

.E..(r 2

or 'Note that

sin 6 o6

sin 2 6

oA2) _ __6_ = 0

_ 1_ i(sin 6 sin 6 o6 o6

oA 3) + _ 1_ or

o6

.E._ ( sin 6 oA3)

sin 6 oe

o6

sin 2 6

_ -...6_ = 0 sin2 6

' ' '

0

< r < a,

a < r < b, b < r < oo

308

SOLUTIONS

PROB.

201

and the boundary conditions

~ (rA)l r-- a = !~ a (rA12l)l r=a '

ar Al2ll r= b

fL r

= Al3)1 r= b ' Al3l i _,_ 0 .!. ~ (rAt2l)i = ar ~ (rA)!. r- 00 ' fL ar r=b r -b

Looking for solutions of these equations of the form A = R(r)0(6), we obtain the equations 1 1 - )0 = -.- (sin 6 · 0 ')' + (24) 2

(A - -.- 6

6

Stn

0.

Stn

T he permissible values of the parameter A are determined from the condition that the second of the equations (24) have solutions which are regular in the closed interval 0 < 6 < 7t. This requirement leads to the eigenvalues and corresponding eigenfunctions A = A11 = n(n

+ 1),

(n = 0, 1, 2, ...),

0 = 0 11 = P!(cos6)

where the P!(x) a re associated Legendre functions of the first kind. The general solution of the first of the equations (24) is

T aking account of the behavior of the functions A i near r = 0 and r = oo, we find that they can be represented as series of the form 00

A 1 (r, 6) = LA 11 r 11 P~(cos 6),

L D 11r_ _ P~(cos 6). 00

A 3 (r, 6) =

11 1

n- o

The vector potential of the source can also be represented as an expansion in terms of Legendre functions, by starting from the formula 00

1== = = = -;=====

Jr

2

-

2rr0 sin 6 cos (/)

+ r~

= -1

+ 2:P,.(sin 6 cos(/))

r

)11,

(_!! '

r

n- 1

r

>

r 0•

(25)

Using the addition fo rmula for spherical harmonics P nCsin

ecos (/)) =

P 11(0)P 11(cos 6)

+2i m= l

r(n - m r(n m

+ 1)

+ + 1)

X

P;:'(O)P:;'(cos 6) cos m(/),

PROD.

309

SOLUTIONS

206

and substituting (25) into the integral (23) for A 0 , we find after some simple calculations that 21LJ .., 1 = P .,(O)P..(cos 6) . c n-o n(n 1) r

Aolr>ro - 2

1 1

+

('o)n+1

Then the boundary conditions lead to the following system of equations for determining the coefficients A .. , B m C., and D .. :

a"A - a"B " "

(n

+ 1)a"A.. -

n

a-n-1c

21LJ P~(O) (~)n+l c n(n + 1) a '

= -

"

(!:.2)"+1,

+ 1 a"B,. +

!!.a-n- te,.= 21LJ P!.(O) fl. cn + la

fl.

b"B .. + b- "- 1 C.. - b-"-1 D., 11

=

0,

+ 1 b"B.. -

~ b- "- 1 C., + nb-"-1 D .. = 0.

fl.

fl.

Solving this system we obtain D 11

=

21Llf.l. cn(n 1)

+

X

[n(f.l.

+ 1) +

(2n ..L 1) 2P~(O)r;+ 1 1][n(f.l. + 1) + !J.l - (afb)211+1n(n

+ 1)(!J. -

2

1)

'

which leads to the solution in the region outside the shield given on p. 99, if we bear in mind that Pik(O) = 0. 206. The magnetic field in the spherical resonator has only a cp-component with complex amplitude H~ = H(r, 6). Writing H = H 0 + Hi> where5

H0

= -p sin - 2 -6 (1 +

"k r) e-ikr

1

cr

is the magnetic field of the source, we find that H 1 satisfies the equation 6.Hl

+

(k

2

-

r

2

~

SIO

2

)

6

HI

Next we introduce a new unknown function u H _ 1 5

-

=

0.

= u(r, 6) such that

ou ae

This expression can be obtained from the relations Ho

p e - ikr = (curl A101)'1', A(O) = ---cos e. r c r

A(O)

0

p e -ikr = - - - - s in c r

e.

3 I0

P ROB.

SOLUTIONS

210

Then u is the solution of Helmholtz's equation

l 2 ~(r2 du) + -2 r dr

1 .E_(sin 0 r sin 0 ()0

•dr

()u) + k u= 0 ae 2

which is regular inside the sphere. Since the tangential component of the electric field i

()

Eo = - - (rH) kr dr

must vanish on the surface of the sphere, it follows that

() (r -()0au) I

-

dr

r=a

a

= - - (rH 0 )

dr

I

•

r=a

Using the Fourier method to solve the differential equation for u, we find that (26) in terms of the Legendre polynomials P ,(x) and the Bessel function s of halfintegral order J n+J-i(x) . Using the boundary condition and the familiar relation Ja;lz)

=

J

2 (sin ;; -z-z - cos z ) ,

we find that the coefficients en equal

J2--; 1tk Pk

1

e =

- ika

e

1 + ika - k 2a 2 (l - k 2a 2) sin ka - ka cos ka ·

Substituting these values of en into (26) and differentiating with respect to 6, we arrive at the expression for H~ = H(r, 0) given on p. 101. 210. The problem reduces to solving the equation

(27) of the vibrating string, with zero initial conditions

I

u t- o =

dudt I

and homogeneous boundary conditions

!= 0

=

0

PROD.

210

SOLUTIONS

3I I

of the first kind. We look for a solution in the form of an expansion

with respect to the eigenfunctions X.,.(x) of the corresponding homogeneous problem, where the weight r equals 1 and ii.,.

= J; 11 X .,.(x) dx.

The func tions X (x) are the nontrivial solutions of the equation 11

X" + )..X = O

satisfying the homogeneous boundary conditions X(O) = X(!) = 0.

Such solutions exist for

n= 1, 2 .. . , and are of the form

=

X

X (x) n

=

sin n1tx . l

To determine the coefficients ii,., we multiply (27) by X ,.(x) and integrate with respect to x from 0 to /. 6 Integrating by parts twice and taking account of the boundary conditions, we obtain

_, + (n7tv) v Jo q(x, t) sm -n1tx - .- u,. = T - dx . 2

u,.

2

_

{

1

.

1

1

The solution of this equation can be found by variation of constants: - = A cos n7tvt n n l

'U

+B

n7tVt sm l 0

n

+ -vlll1tT

it

l'

)

n1t~ d~"' sm n7tv(t- •t') d 't' q c~"' 't' sml 0 ' l . 0

0

0

To calculate the constants A .,. and B,., we use the initial conditions for the function ii.,., which are obtained by multiplying the original initial conditions by X,.(x) and integrating with respect to x from 0 to l. The result is ii,.(O)

=

ii~(O)

= 0,

which implies A ,.= B ,. = 0. In this way, we arrive at the answer on p. 108. 6 In the interest of using a unified approach, we fo llow the general scheme on p. 105. For problems of the type under consideration, this method is entirely equivalent to that described on p. 104.

3 12

SOLUTfONS

PROB.

217

217. We have to solve the inhomogeneous equation

o u + __!_ ou = ox a ot 4

2

4

4

2

q sin wt EJ

(28)

for transverse oscillations of the beam, with zero initial conditions ~

ouat It=o=

u/t=o =

0

and homogeneous boundary conditions

u,:r-±, = au I =

ox

0.

>:=±1

A feature of this problem is that it involves an expansion in terms of eigenfunctions of a fourth-order differential operator. Following the usual m ethod, we represent the solution as an expansion

'! J' r

=

u(x, t)

n=l

- I

ii,.

X n(x)

X~(x) dx

with respect to the eigenfunctions X n(x) of the homogeneous problem, where iin

=J'

- !

ruX,.(x ) d x .

In the present case, the weight r = 1, and the functions Xn(x) are the solutions of the equation x< i~ )-

A.X = o

satisfying the boundary conditions X(± l)

= X'(±l) = 0.

Simple calculations show that7 n

=

1, 2, ... ,

X,.(x) = cosh Yn cos y.,.x - cos Yn cosh YnX,

l l where the y n are consecutive positive roots of the equation tan y

+ tan h y =

0.

To determine the functions iin, we multiply (28) by Xn(x) and integrate with respect to x from - 1 to I. Integrating by parts four times and taking account of the boundary conditions we obtain ii~ 7

4

+ ( y l a) iin = __!!__

4

qa - sin wt EJ

JHX nCx) dx. - L

Concerning the orthogonality of the functions Xn(x), see the solution of Prob. 11 8.

PROBo

222

SOLUTIONS

3 13

The solution of this equation satisfying the zero initial conditions

u,.(O) =

u~(O)

= o

obtained by multiplying the original initial conditions by ing with respect to x from - l to /, is give n by a y~,t l2

y~a - 2l

2

0

W Sin qa 2 l 2

_

u11

= --

Ely~

-

(

2

(J.)-

2 0

SIO Wt

f!

2) 2

2

x ..(x) and integrat-

a Yn

-z

--

X,.(x) dxo

12

T he final form of the solution, as given in the answer on po 110, is found by taking account of the easily verified for mulas

J

Xn

f X~(x)

dx

z

( )d X

_ 41 sin Yn cosh y,. , Yn

X -

-z

1

-z

= .J. X~2(l) = 2

21 cosh 2 y n cos2 y no

222. The problem reduces to integrating the equation

o4 u

1 02u a

(29)

- 4+ -2 = 0 4

ox

ot

for the oscillating beam, with zero initial conditions

uit=O= OU.l

at t- o=

0

and inhomogeneous boundary conditions

ui

= z- O

au I OX

a2u I OX

=

2

z= O

ou I 3

= 0

z= Z

'

OX3

= - P(t)

z= !

0

EJ

Applying Grinberg's method, we look for a solution in the form of an expansion u(x, t)

=

1r

n=l

0

u,.

X,.(x)

rX~(x) dx

with respect to the eigenfu nctions X,.(x) of the homogeneous problem, where U11

=

J: ru X .,(x) dxo

Explicit expressions for the functions X ..(x) are obtained by solving the equation x~v>- 'AX = 0,

3I 4

PROB.

SOLUTIONS

223

with homogeneous boundary conditions

,

X(O)

= X'(O} =

XH(l)

=

=

x m(l )

0.

This gives 4

'A = 'A,. =

Y:•,

n = 1, 2, ... ,

l

X ..(x) = (sin Yn + sinh y ..)( cos y~x

- cosh y,;x)

,.x),

. hy . y ,.x - ( cos Yn + cosh y.,) ( sm - - - sm - 1 1

where they 11 are consecutive positive roots of the equation cosy cosh y + l = 0. The functions X.,(x) are orthogonal on the interval (0, /) with weight r = 1 (see the solution to Prob. ll8), and the integral of X!(x) is

{ X~(x) dx = 4!. X~(l) = /(sinh y., + sin y ..?. Jo To determine the coefficients u.., we multiply (29) by X,.(x) and integrate 1

with respect to x from 0 to I. After a bit of manipulation, we arrive at the equation

u~ + (y~aJu.. = ;~P(t)X ..(l). The solution of this equation satisfying the initial conditions

u..(O)

=

u~(O) =

o,

obtained by multiplying the original initial conditions by X ,.(x) and integrating with respect to x from 0 to l, is given by

- =

u..

a2[2

2

EJy..

X (l) J tP( ) . y~a2(t - -r) l n 'r sm 2 l 'r' l

o

and immediately leads to the answer on p. 112. 223. Clearly we can express the dependence of the external load on the coordinate x and the time t in the form A sin wt q(x, t)

where e:

>

=

(

for vt - e:

0 2e:


----'-- - - ' - 2b and taking the limit as c; ~ 0, we obtain 2pb\ - 1}" 1t 2 T(2n + 1) 2

. h (2n sm cosh

+ l )7t(a -

Ixi)

2b (2n

+ 1)1ta 2b

We also take account of the boundary cond ition uix=a = 0 and the relation lx=O= 0 implied by the symmetry of the problem.

(oufox)

320

SOLUTIONS

PROB.

240

which implies formula (13), p. 115. Which form of the solution to use in making calculations depends on which series converges more rapidly (this depends primarily on the ratio afb of the sides of the rectangle).

240. The problem reduces to solving the biharmonic equation

a4u

ox4

+ 2 ~ + a4u = 0, OX 2 ol

ol

with boundary conditions

uix=O= uix=a = 0 , u/u=±b/2

= o.

= au 1

oy

v- ±b/2

It is easy to construct a function u*

=

mx(a - x) 2D

satisfy ing both the differential equation and the boundary conditions at x and x = a. If we set u = u* v,

=

0

+

f

then the new unknown function v must be a solution of the homogeneous biharmonic equation satisfying homogeneous boundary conditions in x :

vlx=O=

vl:t=

=

2 02 v I = -o2v I = 0. OX :t=O OX :t=a

-2

This enables us to use the Fourier method, where the boundary conditions in the variables y take t he form vlv= ± b/ 2

=

av I

u*(x) ,

oy

= 0

·v =±b/2

.

Taking account of the boundary conditions in the variable x, we look for particular solutions of the biharmonic equation f:!. 2 v = 0 of the form V

=

. mtx

Vn (y ) S i n - ,

a

n

=

1, 2, . ..

The amplitude vn must then be a solution of the differential equation

(n7t)

2

(iv ) Vn -

-

a

+ (n7t) Vn = 4

, Vn

O

a

which is even in y , and hence n1ty

V11

.

n1ty

= An cosh- + Bny smh a

a

.

PROB.

241

SOLUTIONS

321

Writing v as a series

v = !(An cosh n1ty n= l a

+ BnY sinh n1ty) sin mtx, a

a

we determine the coefficients An and Bn from the remaining conditions

ov'y= b/2 = ay

vf11= b/ 2 = u*(x),

0 ·

This requires expanding the known function u*(x) in a Fourier series with respect to sin (mufa). In this way, we eventually arrive at the answer on p. 119. 241. Suppose the line load p is replaced by a load uniformly distributed over the sector - e: < cp < e:, 0 < r < a with central angle 2e:, where e: > 0 is arbitrarily small. Then the problem reduces to solving the inhomogeneous biharmonic equation

a(

a) + ~1 o<pa J[1~ora(

1 [ ~or r or

2

2

ou] {_p_ , + ~ ocp = ~~D

au)

1

2

2

r or

with homogeneous boundary conditions

ufr=a =

au I

= 0.

or r=a . With our way of measuring angles, u is an even function of cp and hence can be written as a cosine series

u (r, cp) = -1_ u0 + -2~L. U n cos ncp, 1t n=l

7t

where

un = fo7t u cos ncp dcp. To find un, we multiply the equation for u by cos ncp and integrate with respect to cp from 0 to 1t. Then, integrating by parts four times, we find that

+ .!. !!_ _ n (~ dr r dr r

2 2 )

2

2

un=

p sin ne:, e:aDn

where the right-hand side can be replaced by pfaD after taking the limit as e:-+ 0. We are interested in the solution of this equation which is regular for r = 0, i.e., iin = A.,r" Bnr"+2 where

+

+ u:,

322

PROB.

SOLUTIONS

except for the cases n

=

=

2 and n

242

4:

p r In r u-* - -2 aD 48 '

4

u-* -

4

p r In-r - -·aD 96 ·

4 -

The constants A n and B n are determined from the conditions

=

iin(a)

ii~(a)

= 0.

242. The problem reduces to integration of the heat conduction equation

a T aT 2

(32)

with the zero initial condition and inhomogeneous boundary conditions

oTi

- kOX

z~o

= q,

It is easy to see that the linear function T*

=

T0

+ q_ (a k

x)

is a solution of (32) satisfying both inhomogeneous boundary condit!ions in the variable x. Therefore, writing

T = T* - u, we find t hat u satisfies the differential equati on

0 2U

-

ox2

- OU OT

with initial condition

ul~~o = T*(x) = T0 + q_ (a - x) k

and homogeneous boundary conditions

au I

OX

= 0,

uj.,=a=

0.

z=O

Application of the Fourier method gives

where the en are the coefficients of the Fourier expansion of T*(x) with respect to the functions cos [(2n l )7txf2a].

+

SOLUTIONS

PROB. 247

323

247. We have to solve the inhomogeneous equation

.!.~(rar) r or

=

ar _ ~ , ih

or

(33)

k

with the zero initial condition

Ti.,.~o

=

0

and homogeneous boundary condition of the third kind:

+ hr)\ = o. (ar or r=a Suppose the solution is of the form

T =

L""

r

f

n

,._1JoarR!(r) dr

Ri r),

where

T., = tTRn(r)r dr, in terms of the eigenfunctions R n(r) of the homogeneous problem. The latter must satisfy the equation

.!_ (rR')' r the boundary condition R'(a)

+ "AR = 0,

+ hR(a) =

(34)

0

and the requirement that R(O) be bounded. Solutions of the required type exist if

n = 1,2, . . . , where the y,, are consecutive positive roots of the equation

The corresponding sol utions of (34) are R

y,.r) · = R n(r) = lo ( -;;

These functions are orthogonal with weight ron the interval (0, a), and moreover

l

a rR~(r)

0

2

dr = -a [J~(Yn) 2

+ J~(yn)l =

a2 2

-

J~(y,.) [ 1

(ha)1 ·

+ -

y ..

324

SOLUTIO~S

PROD.

261

To find the functions T, we multiply (33) by R,(r) and integrate from 0 to a. Then, integrating by parts twice and taking account of the boundary conditions, we find that

T' n

+

(y, a. )'r

~ Qa' J,(y,) k

1
I,

which implies T0

= - Qal - - + A 0 + B0y , 2k

n1ty T,. = A,. cosh a

. n1ty + B,. smh ,

n

a

> 1.

Using the boundary conditions in y , we find that

I _0

dT,. dy v=o

-dt., I = '

dy v=b

J"!( ) X

n1tx d X

COS - -

o

2

= -

a

Qa b . n7tc --Sin n1tkc a '

which leads to the following values of the constants :10 A ,.

= -

Qa 3 b sin (n1tcja) n21t2 kc sinh (n7tbfa)'

B,.

=

O.

Substituting A ,. and B ,. into (36), we obtain formula (14), p. 126. To obtain the other form of the solution, we set

T(x, y)

~ -

n1ty L T,. cos , b n~ L b

= -1 T- 0 + -2 b

where

t,. =

f

b

n7ty Tcos dy. b

0

Then, by the same procedure as before, we obtain the differential equation

r~-

(:7trr. =

C.

2 ,

After some manipulation, we find that Qax 2 Qx2

lxl < c,

2kc '

T(x,y) = - -+ 2k { Qa lxl - k-,

+

+ const lxl > c,

2Qab2 "'

nn:y (- 1)" cosb

-2 -----=-h nn:a

-2

1t

kc

n=l

9

•

n-sm -

b h nn:x . h nn:(a - c) COS b b

Stn

X

. h nn:c h nn:(a ( - sm -cos b b

-

. h nn:a b '

Stn

lxl) ,

lxl < c, lxl > c.

The form of the solution given in the answer on p. 126 is obtained if we improve the convergence by using the form ula

2 <X)

n= l

(

- 1)n+l

. n-

cos nx

7t2

x2

12

4

=- - - ,

to carry out partial summation of the series.

- 7t < X < 7t

PROB.

272

SOLUTIONS

327

269. To solve the problem, we assume that heat is produced with uniform density Q/rte 2 inside a cylinder of arbitrarily small radius e. Then the problem reduces to integrating Poisson's equation

1o ( ror) r or or

2

+ o- T

= f(r)

OZ 2

=

~

{- rtke , 0

e

'

with boundary conditions

( oT OZ

±

h

r) I

=

z=±l

< r < a,

o.

Expa nding the solution in a series of eigenfunctions of the corresponding homogent:ous problem depending on the variable r, we find that

T(r,z) = 2z! 2Tn a n=l Jl(y,.)

lo(yn r), a

T n=i oaTJo(Yanr)rdr,

where the Yn are consecutive positive roots of the equation J 0 (y) = 0. Multiplying the original equation by rJ0 (y nrfa) and integrating with respect to r from 0 to a, we obtain

or

(Yn )2rn = - JL a 2rtk

T" n

after taking the limit as e--+ 0. The solution of t his equation satisfying the boundary co nditions

( dTn ± hTn) dz is 2

T. _ Qa n - 2rtky~

[t _

I

=

0

z=±t

ah cosh (ynzfa)

J

Yn sinh (ynlfa) + ah cosh (ynlfa)

'

which leads to the answer given on p. 130. 272. T his problem of electrostatics reduces to finding a solution of Laplace's equation

o2u +-= o2u 0

-

OX 2

ol

satisfying the following inhomogeneous bo undary conditions of the first kind : 0 < x < a,

a< x < oo.

328

PROD.

SOLUTIONS

272

Following Grinberg's method, we look for a solution in t he form of an expansion with respect to the eigenfunctions of the corresponding homogeneous problem,12 i.e., 2 ~ _ (2n 1)1ry u(x, y) = - L.., U n cos , b 1!= 0 2b where

+

- ib

un =

u

COS

(2n

+ 1)7ty d y . 2b

0

To determine the unknown quantities ii"' we multiply Laplace's equation by 1)7ty/2b] and integrate with respect to y from 0 to b. Taking cos [(2n account of the boundary conditions, we obtain

+

u~ -

+ l)7t] 2ii'll =

[(2n

( - l ) n+ l (2n + 1)7t f(x).

(38)

2b

2b

We want the solution of (38) wh ich is bounded at infinity and satisfies the condition

uI

=ib

V cos (2n

" :t=O

+ l)7ty d 2b

o

= 2bV(-W y

(2 n

+ 1)7t

It is easy to see that this solution can be written in the form

Un

-(l) _ Un -

= {

!1~2)

=

B

n

. h (2n

Sin

+ l)1tx + 2bV(- 1)n , 2b (2n + l)7t

Cne- (2 n+l)rrx/2b'

x < a, x > a,

where the values of the constants B n and Cn are d etermined from the "contact conditions" which imply

B = 2b V( - 1)"+1 en (2 n l)7t

+

en = 2bV(-

C2n+1Jrrat 2b

'

(2n

Substitution of these values of the coefficients into series solution of t he problem :

I

_ 4V ~ L..,

I

~ ( - 1)" h (2n = 4V -L.., - - - COS

U xa

7t

n- 0

( - l)n [ _ 1 e 2n + 1

1t n- o 2n

+1

(2n+ l)Ttb/ 2a

.

SIO

h (2n

It cosh (2n

+ 1)7t

+

l)1ta.

2b

u,. leads to the following

+ 1)7tXJ COS (2 n + l)1ty , 2a

2b

+ 1)1ta e- c n+l)nx/ 2b COS (2n + l )7ty 2

2b

.

-

2b

12 C hoosing the other form of the solution leads to an expansio n in a Fourier sine integral over the integral (0, co).

I'ROB.

277

SOLUTIONS

329

To obtain the final form of the solution, we improve the convergence by using the formula

(- Dn 2: --cos(2n + 2n + 1 ~

K

K

lxl < -

l)x = - , 4

n- O

2

to sum the slowly convergent part of the first series. It would be noted that the solution can also be written in closed form. 277. To solve the problem, we first assume that the charge q is uniformly distributed with density p over an arbitrarily small cylinder 0 < r < 8, c - tt:: < z < c + !t::, i.e., we reduce the problem to integration of Poisson's equation

a( au) ou ror r -or + -oz = 2

-1 -

2

- 4xp(r, z),

where for

0

< r < 8,

c-

te < z < c + !t::,

otherwise, subject to the boundary conditions

uj,....., = ui z~o = ulz=l = 0. One of the two possible forms of the solution is an expansion with respect to the function s J 0(y ,.rfa), which are the eigenfunctions of the corresponding homogeneous problem, i.e. ,

u ( r, z )

~ = 22 £.. a

n• l

u,.

- 2--

Jly,.)

(y"')

Jo -

a

'

where the y n are consecutive positive roots of the equation J 0 (y) = 0. Multiplying the original equation by rJ0(y nrfa) and integrating with respect to r from 0 to a, we arrive at the equation

u~ -

(:"rii,. =

f

- 4x

(39)

p(t, z)Jo(y; t) t dt,

which is to be so lved with zero boundary conditions

The general solution of (39) satisfying the first of these conditions is iin = A n s~nh (ynz fa) - 4xai"sinh Yn(z smh (y,.lfa) y,. o a

~) d~ia p(t, ~)Jo(y,.t) t dt. o

a

330

PROB.

SOLUTIONS

282

Using the other boundary condition to calculate A n, and then passing to the limit 3, e: - 0, we obtain

=

2aq sinh YnCl- c). Yn a Thus the coefficients iin are equal to A

Un

2aq

= ----'-.

. hyn{l - c). hy..z sm sm - , a a

0

< z < c,

{ . b Ync . h Yn(l - c) sm . -sm , a a

c

< z < I,

X

Ynl

Yn smh -

n

a

which immediately leads to the answer on p. 134. The other form of the solutio n can be obtained by expanding u(r, z) in a series with respect to the eigenfunctions in the variable z, i.e., 2 Lao _

nnz

.

u( r z ) = u sm' l n- 1 n l '

fin =

f

. nnz u sm--dz.

t

l

0

282. Since the potential distribution must be an odd function of the coordinate z, the problem reduces to solving Laplace's equation

! ~(r au) + ()2u

,. or or

= 0,

Oz 2

with boundary conditions

ulz-0=

0,

uJz~t =

v, ulr=a=

f(z),

where

v, f(z)

< z < l, 0 < z < 3. ~

= ( V sm. nz

23'

To obtain homogeneous boundary conditions in the variable z, we set u

=

v.:I -

v.

Then the function v(r, z) will be the solution of Laplace's equation satisfying the homogeneous conditions vJz~o = vJz~t = 0 and the following boundary condition on the lateral surface

v(f - t), vfr=a = cp(z) = ) { (.:_-sin nz l 23 ' V

r.

p

Then, passing to the limit a rrive at the expression

o, e ---+ 0 and bearing in mind that lim Je =

0, we

•-o

nP 1 An = - - - Uo(ocna)Ko(ocnr) - Ko(ocna)l0 (ocnr)] , c ! 0 (ocna)

which immediately implies the a n swer o n p. 141.

303. We want the solution of Laplace's equation o2T

o2T

- +ox ol 2

= 0

(0


(i = 1, 2). A convenient way of solving (41) is to use the method of integral transforms, by taking the sine transform of E;il and the cosine transform of E;il (i = 1, 2). 14 Thus we multiply the first of the equations (41) and the second of each pair of boundary conditions (42) by cos AX, and the second of the equations (41) and the first of each pair of boundary conditions by sin AX. Then, integrating from 0 to co, we find that E., •

£11

11-++oo -+

-(i )

dEv dy

•

a; •

Ell

+ AE =

!1-+-co -+

d

0

X

-(i)

E., dy

'

[e:2£-(2) 11 -

•

+ A£.

e:l

+ e:2

Note that the cosine transform of £~ 11 and the sine transform of £!11 vanish, because of the symmetry of the problem. 14

336

PROB.

SOLUTIONS

321

Using the inversion formulas E(i) =

"'

~ i co.E(i) sin AX dA 7t

0

"'

E(i) 11

'

= ~ i co.£!il cos AX dA 11 ' 7t

0

and making a few simple calculatio ns, we arri ve at the expressions for the components of the electric field given in the answer on p . 154.

321. The electric field has only a z-co mponent, whose complex amplitude we denote by E(x, y) . If we regard the current as distributed over an arbitrarily sm all rectangle a- I> < x < a + 1>, lyl < e, then the solution of the problem reduces to integration of the inhomogeneo us Helmholtz equation (44) where

j(x, y)

~

for

a -

I>




cr.,,

0"

11

themselVeS.

328. Replacing the concentrated force P by a load uniformly distributed over the arbitrarily small rectangle

a

a

2

2'

- -<x 0 and can be written as an integral (54) while the second quantity is biharmonic in the region z written in the form

>

0 and can be

(55) (note that the integ rand is biharmonic). Comparing the result of differentiating (54) twice with respect to z with the result of applying the operator

342

SOLUTIONS

PROB.

355

to (55), we find that CA = -t:AAA. To determine the remaining constants, we have to use the boundary conditions

which can be written as conditions on the fu nction u 1 : 18

~[(2- v)~ul - ()2u21JI

OZ

OZ

[ (1 -

=z=O

v)~ul - o2u21J I = OZ

_£_[(2 - v)~uo - o2u2oJI OZ OZ v)~uo - O\oJI

- [(1 -

OZ

z-0

z~o

'

. z=O

Performing the differentiations on the right, expanding t he resu lts in Hankel integrals and substituting from (54) and {55), we obtain a system of linear equations determining t he constants A'), and B~.. T he formu la given in t he answer on p. 168 is obtained after evaluating certain integrals of a familia r type. 355. The problem reduces to integra tion of the one-dimensiona l heat conduction equ ation

with the initial condition the boundary condition

aT !

= q(-r)

- k-

OX

o:-0

a nd the condition at infinity

Ti.,-.oo--+ 0. Introducing the Laplace transform

T = f"" Te-"1: d-r

Jo

'

we m ultiply the diffe rential equation and boundary co nditions by e -vT and integrate with respect to -r from 0 to oo. If we take account of the initial condition, this gives

f"- pT = 0, - kf''l.,=o = ij, Tj.,_, 00 --+ 0, which implies

"" = _1_ e_.y;.,

kJp

.1 18

'

Re Jp> 0.

The second of these equations foll ows from the formula T

rz

= -o[(1

or

-

o•u] v)!::.u - -oz•

0

PROD.

SOLUTIONS

355

343

The problem is now solved by using the Fourier-Mellin inversion theorem

I. -J p

l ,~_;;., q~, dp T = -- e 21tik r

where r is a straight line parallel to the imaginary axis lying to the right of all the sing ular points of the integrand. In Case a, where q = q0 = const, we have T = Cf..9 - 1 ev•-.Y;., dp_ . (56) T k 21ti r p =tJ+ir As the next step, we calculate the derivative r aT = _ ~-~-I. e"'_.y;., dp. k 21ti r p Applying Cauchy's integral theo rem to the contour shown in Figure 159, and then taking the limit as e: - >- 0, R -+ oo, we obtain19

J.

pJ p

ox

-aT = - q- 0 [ 1 -

ox

k

-2

i ooe- n sm. J-r x dr] r

1t o

FIGURE

159

where O

0 .

Taking Laplace transforms and using the initial condition, we obtain the equation

df)

1d (rr dr 2 °For this branch, grand.

-

dr

Vp =F - h, and hence p =

-

pT = O 0 is the only singular point of t he inte-

PROB.

375

SOLUTIONS

345

and the condition

which together imply

T=

rY.aT0 lo(.J p r) .jp[r1.aJP 10(./P a) + 11(J p a)].

The solution of the problem is given by the inversion formula

T

=

r1.aT0 27ti

f. J r

l 0(Jp r)em dp p[r1.a.{P 10(./P a) + 11 (./P a)].

The contour integral can be evaluated by residues, since the integrand is single-valued . T he singular points of the integrand consist of poles at the points p = 0 a nd p = Pn = -y!/a2, where t hey,. are consecutive positive roots of t he equation J1(y) r:t..lo(Y) = 0.

+

Calculating the residues at these points, we immediately find the answer on p. 177.21 As in other problems with boundary conditions involving time derivatives, the solution of this problem is greatly simplified by the use of Laplace tran sforms. 375. In the fi rst region 0 C1 (r, z, t) satisfies the equation



lllx=l

=

f 2Jx=1

+ Rii2 1x=l· 0

These equations can be solved for iih ii2, 11 and 12 . In particular, for ii2 we obtain the expression Eo e- r>Cx- t>/ v u = -2 p oc cosh pT [1 (Z/R 0)) sinh pT'

+

+ +

where v = 1/J LC is the propagation velocity, T = lfv is the time it takes a wave to traverse the part of the line going from x = 0 to x =I, and Z = J L/C is the wave resistance. Then the Fourier-Mellin inversion formula leads to the following representation of u2 as a contour integral:

i

Eo ert-(o:tv)+T] dp u (x t) = -~ ' 27ti r cosh pT + [1 + (Z/R 0)] sinh pT p + oc'

X > [.

The most interesting form of the solution can be obtained by using the expansion 1 2R 0e-P7 ' Z cosh pT [1 (Z/R0)] sinh pT = 2R0 Z n = O 2R0 + Z e

+ +

+

I(

)n -2n,7'

348

SOLUTIONS

PROB.402

and then integrating term by term. This gives u 2(x t ) =

'

f.

)n

2RoEo ~ Z v [t-(xfv)-2n7' J dp L. ( -l e -- . 2R0 + Z n = O 2R0 + Z 2rci r p +a

According to the formula

f.

T

1 e m -dp - - (0, 2ni r p + a e-"'~

< 0, 0,

-r >

all the terms of this series vanish for fixed x and t, starting from some value of n. In particular, we have

_

I

U2 xfv

ll·

t


=O acp "'="'

Multiplying (68) by r P+ 2 (where pis a suitably chosen complex number), and integrating from 0 to oo, we find that 26 2 !:iu - (p {r"+ ~ ar

+ (p -

+ l )rP+l Au + (p + 1)2r" au ar

1)\ p

+ 1)2u + [(p -

1)2

(p - l)(p

+ 1)2r,_ u} I 1

<X)

0

+ (p + 1)2] d2~ + d4~ dcp

dcp

1 l <X)q(r, cp)rv+2 dr, D o

=-

(69)

where (70)

Suppose the function u is such that the quantities r- 1 u, aujar, r Au and r 2(a!:iujar) are all O(r• 1) as r --+ 0 and all 0(,-••) as r --+ oo, where s1 > 0, s2 > 0. Then the integrated term 1{. . .}1: in (69) vanishes if - s1 < Rep < s2 , thereby reducing (69) to the ordinary differential eq uation 27

d4~ + ((p -

dcp

1)2

+ (p + llJ d2~ + (p dcp

1)2(p

+ 1)2ii

1 00

= -1

D

q(r, cp)r:v-+-2 dr.

o

26 In problems of elasticity theory involving integration of the biharmonic equation, it is best to use a modification of the Mellin transform, in which the exponent p is replaced byp - I. 2 ' By the same token, the integral (70) is analytic in the strip - s 1 < Rep < s 2 , being uniformly convergent in every closed subset of the strip.

354

SOLUTIONS

PROB.

418

Using the method of variation of constants, we find that ii ~A cos (p - l):p ;· B sin (p- 1)rp ·!- C cos (p .L

0? I E sin (I' ·I 1)1'

1)] dtj(x q(p, t)p

_1_1."[sin (p -- l)(rp- t) _sin (p + l)(q> 4Dp .o p- 1 p~ I

. dp,

1 1 )'

0

where the boundary conditions

•J

u9-o

diil

~d''l ~ utt-··-o: •J ~ dr:p drp 'f-0

serve to determine the coefficients A, B, e ···~ 0 and solving for these coefficients, function with poles at the points \Vherc vanishes, and moreover that the number equation

~0 Q-'X

a,

C and E. Passing to the limit we find that it is a merom orphic the expression p2 sin 2 C< sin:,. p'Y. s1 =· s2 is the smallest root of the

The bending moment M and the shear stress N along the edge '? determined from the relations ---,

2

d u dt.(

Mr 19~o= --D-2

I

0 can be

'

r;=O

Using the inversion formula for the Mellin transform, and choosing the imaginary axis as the path of integration, we find, after a certain amount of ca1culation, that !vi and N are the same as in the answer on p. 193. 418. Following the Fourier method, we look for particular solutions of Laplace's equation of the form . n~:z T o R(r)(•p) sm-. I

Separating variables and integrating the resulting equations, we find that

T= [AI.V--A r. (n-;r;rfl.)

+ BKJ-:: (n-;r;rfi,)] [C cosh ,il- rp -L D sinh ,i~ co] sin nr.z, ~·A

'

[

where Tv(x) and K,.(x) arc cylinder functions of imaginary argument. Because of the behavior of l,(x) and K,(x) as r ~ 0 and r > :JC, the bounded ness of the solutions T requires that A c~ 0 and i, > 0. Thus the particular solutions needed to solve :he boundary value problem, which has a continuous spectrum (0, co), arc of the form

T= T .,

=.

'f h T)) -A-. ' ' sm ·. h ·rrt)Ki--: (_ll7t/') . II7;Z ('"-:-cos -, sm-,

1

1

0



where R is the distance from the source to an arbitrary poin t of space, we reduce the problem to integration of Laplace's equation

with the boundary condition

ullcx~o =

-

q_ I

=

-

R cx- o

J q c sin h2 cx.0

+ sin

2

~

. h cx. = -d sm 0

'

a

and the condition at infinity In keepi ng with the discussion on p. in the form of a series

222, we look for a secondary potential

00

u1

= 2: A,.Q,.(i sinh cx.)P,.(cos ~), n-o

where the coefficients A,. are determined from the boundary condition. Using the theorem on expansion of an arbi trary function in a series of Legendre polynomials, we find that 2n

A,.

f ~ J

+ 1q

= - 2Q,.(O)

1

- 1

P ,.(x) dx cosh2 cx - x 2

(81)

0

for even n, while A,.= 0 for odd n. To evaluate (81) for even n, we use the · integral

b

>

1,

PROB.

471

363

SOLUTIONS

which can be evaluated by expanding (b 2 - x 2)-1' 2 in a power series and then integrating term by term. Using well-known formulas, we find that J

1

=l 1

f

= _! ~

1

rem + t ) _1_ x2mp 2 ex) dx 2 n b tn= O rmrem + 1) b "' - 1 n 22n + l ~ rem + t)r(2m + l)r(m + n + 1) 1 = - b-,. n rmrem + l)rem - n + l)re2m + 2n + 2) b 2m = r 2en + t) (n + tMn + ~)k ._fo b2 n+1 re2n + ~) k =o k! e2n + ·D,, b 2 ' P 2n{x) dx b J_l .J1 - exfb)2

I

(l)k

where ("A)

=

k

+

r("A k) . r("A)

The result can be expressed in terms of the hypergeometric function

Fe<X,

fl..

1'-' '

.

y' z

)

= L. ~ (<XM~)k zk ' k = O k!

i.e.,

r 2(n + t) .Jrt b2n+lre2n + ~) F(n + 1

J -

er)k n

Z>

n -

...!.. 1· I Z>

2n

+

a

.l...) b2 •

Y>

U sing the familiar formula Fe<X,

~; y; z) = (1 -

z)- "F(<X, y -

~; y; _z_), z- 1

we find that

r en + t ) 1 F( n + J. n + J ,. 2n + .a. . _ " - __!_ .J-1t re2n + v sinh n+ <Xo 2

J _

2

1

2,

2'

_

1_) 2

sinh <Xo

'

or 1n

=

2iP2,.(0) Q2n(i sinh <Xo),

because of the definition of the Legendre function of the second kind . Thus the required values of the coefficients A n are A 2n

=

2

q e4n

+ l)Q n(i sinh <X 2

0),

7t:C

which leads to the potential distribution u

= 9.. + ~ q_ ~ e4n + 1)Q2n(i sinh <X0)Q 2,.(i sinh <X)P2necos ~), R

1t

c n- o

if we note that P2,.(0) = 2i Q2n(O) 1t

364

PROB.

SOLUTIONS

481

The distribution of charge on the surface of the disk is now fou nd by differentiation , according to the formula

(J

1(

=-

47t

I

1 au) c.Jsinh 2 ex + cos 2 ~ aex tx=~

481. The problem reduces to solving the system of equations ~u

aex cx~txo

= au(2) I aex

.

<X= txo

Setting u ex0) , we have 00

u1


0), and the constants Nand D

+

SOLUTIONS

PROB.483

367

must be set eq ual to zero.34 Thus we arrive at the particular solutions -r

> 0.

To construct the general solution, we integrate over the parameter -r, obtaining

u1 = foaa CTP-~+;~(cosh ot)P-~+iT(cos [3) d-r, where the coefficients C" must satisfy the boundary condition

ull(>-ao =

=

-

c(cosh ot

q

cos [30)

-

L.,

C"P- !A+iT(cos (30 )P-!A+iT(cosh ot) d-r,

ot

>

0.

Using the inversion formula implied by the Mehler-Fock theorem, we find that

CT =

-

l oa

q-r tanh 1t-r sinh ot p ( h )d -l-f+iT COS ()( ()( c P l-f+i-r(cos [30 ) o cosh ot - cos [30

= _ f{

T tanh 1tT

c P-l-f+;-r(cos [30)

roo

J

1

p -~+iT(~) d~. ~ - cos [30

Evaluating t he integral, we arrive at the formula for the electrostatic potentia l given in the answer on p. 229.35 34

Jn particular, we use the formula

Pv(cosh ot) =

r(v _

+ J)

vrr rev + ~)

e- lv+llot tan rrvF(v

+ J, t; v + '~; e-

1

"')

+ ret + v) e•"' F( v'rr r(l + v)

v, t; t - v; e-'"'),

which shows that a bounded solution in the interval (0, co) exists on ly if - 1 < Rev The extra requirement that u1 be real compels us to set v = - t + i-r. 35 To prove the formula J

=

f

1

aa

P -!-f+l/1;) dl; I; - cos ~o

(see L7), use the integral representation

< 0.

rr cosh rr-r

= - - - P- ~-i+IT( - cos ~o)

21"'

cos -rsds P _!-f ..,T(coshot) = , rr o v'2 cosh ex - 2 cosh s and then reverse the order of integration with respect toot and s. After evaluat ing the inner integral, this gives cos -rs ds rr 1 = 2 = - - P- !-ftt-r( - cos ~o), o V2 cosh s - 2 cos ~ 0 cosh rr-r

l

oa

where we have used another integral representation of P- \-f+iT(x).

368

SOLUTIONS

PROB.

49.4

494. Bearing in mind that the potential u can be represented in the form E0z uh where u1 is harmonic outside the torus and goes to zero ut infinity, we look for a solution of the form

u

=

+

~ P -~(cosh ex) . 2 cos ~ L. A,. " sm n~. n=l P n- !A(cosh ex0) The coefficients A ,. are found from the boundary condition a nd coincide with the coefficients of the function

u

=

E0 z

+ .J2 cosh ex -

-2E0 c sin ~(2 cosh ex0 - 2 cos ~)- 3 ' 2 when expanded in a Fourier sine series in the interval (0, 1t). Thus we find that 4E0c {"' sin ~ sin n~ d~ A,. = - -;-Jo (2 cosh ex0 - 2 cos f3) 312 • Integrating by parts and using the formula given in the hint to Pro b. 493, we arrive at the answer given on p. 237. 498. Setting

T =

Qrz

- - + u,

4k we reduce the problem to finding the function u. This function is harmonic outside the torus (0 < ex < ex0) and goes to zero at infinity (i.e., as ex -+ 0, f3-+ 0). We look fo r a solution of the form

cos n~, Q,_!-1l(cosh ex0) where the coefficients A,. are determined from the boundary condition u

=

.J2 cosh ex - 2

cos~ ~A,. Qn-~i(cosh ex) n= O

uia=ao

=

~r

2

la=ao·

It follows from the theory of Fourier series that A

2

= Qc sinh 2 ex0 { "'

df3

Jo (2 cosh ex0 -

krt

J"'

2 cos

~) 51 2 '

2Qc sinh ex0 cos n~ df3 , n = 1, 2, ... k1t o (2 cosh ex0 - 2 cos f3) 51 ~ Evaluating these integrals, we eventually arrive at the answer on p. 238. An

2

=

2

502. If we subtract out the singularity at the point r u

=

q

..J r2 + z2

= z=

0 by setting

+ ul>

the potential u1 of the secondary field is harmonic in the region 0 < ex < oo, ~ 0 < ~ < 27t ~ 0 outside the conductor, and vanishes as ex-+ 0, f3 -+ 27t.

+

I' ROB.

502

SOLUTION S

369

The function u1 can be represented in the form of an integral q~

u1 = -

c

J 2 cosh

2 cos~

IX -

i"" o

+

~ 0 - ~) P- l-2+i-r(cosh a) dT. M -r cosh (n cosh n't'

It follows from the boundary condition uil3-13o = uil3- 2n+l3o = 0 that M-r coincides with the coefficients of the expansion of the function

(2 cosh

IX

+ 2 cos ~ 0)- '

1 2

in a Mehler-Fock integral with respect to the functions P-!-2+i-r(cosh a), i.e.,

ex > 0. In the present case, we cannot determine M " directly by using the inversion formula implied by the Mehler-Fock theorem, since the function being expanded does not belong to the class for which the theorem holds (see L9, p. 228). However, it can be shown without recourse to the Mehler-Fo 0.

PROB.

523

SOLUTIONS

371

The solution is then constructed in the form

U sing the well-known formula

--=== = 1

-Jo:2+~2

1 00

e-~ 0,

and taking account of the boundary condition ul~=f!o N 1, = e - 1'f!o, and hence u

=

~2 rooe-~(k'r)[l • 2 •• ..,

7tLT

1 . H~2>(k~) Sinh {7tT

= H(2l(k~)[ 0 t;.

2 ' h 7tT

LT Stn

+ 2 L. ~

m - 1 In

2

+

" 2

T

J

2

~ enT/ 2 coth 7tT Hi;>(k~). LT

1 ]· 'ht2 7tT

.

LT Sin

Thus we fina lly have

( "' H~z>(k lx - ~I) - e-ik"'Hb2l(Jc~) enT/2Hg\kx) dx

Jo

X

2 . . [H~2>(k~) '" smh 7tT

errT/

2

coth nT

H!!l(k~)].

If we now introduce the integral transform

-( ) 1oo cpxe· () (kx) ,

0

X

n-/ 2

0




The sine integral

t

Si (z)

=

Ci (z)

=f· a)

arg z


dx =

a > 0. '

a > 0.

2a

o

2

a > 0.

2ab,

a > 0,

2a

= r oo cos x 2 dx = !J~

Jo

2

.

2

b

>

.

. Ttr . 7tp smh - s m -

smh px . d 7t 2a 2a sm rx x = - - - - - - - o cosh qx q rtp cos h7tr - + cos«>

0.

q

0

< p < q.

q

Ttr

14.

i

00

o

15.

16.

sinh-

cosh px sm . rx d x = -------''---7t q sinh qx 2q h Ttr + 7tp cos cosq q

. oo smh px ---cos rx dx o sinh qx

i

l

ao

o

cosh px

d cos rx x

coshqx

17. r oo cosh px dx Jo (cosh x)Q

7t

= -

2q

q

= 2q- 2


0. Rev > - 1.

b > 0,

20.

r <Xl x•+lJ,(bx) a•- IJ.biJ. (x2 + ~2)1J.+l dx = 2'LI'(!L + 1} K._IJ.(ab) ,

Jo

a

x2;

21. f <Xl KIJ.(a J l) o (x2 + YY' 2

Jv(bx)x·~

>

0,

>

0,

- l


= d"ufdx". Determine the Euler equation of the functional J[u]

= J:F(x, u, u', ... , u) dx,

assuming as admissibility conditions that u, u', ... , u 1"> are continuous in [a, b] and have prescribed values at x = a, b. Ans.

-aF - (a -

au

F)'+

ou'

(oF)" )(">= 0. - ··· + (- 1)" ( - oF ou" ou 1")

1.1.4. Determine the Eu ler equation of the functional J[u]

= JJru! + u! + 2f(x,

y)u] dx dy

D

for the functions u(x, y) defined on the domain Din the xy-plane bounded by the contour C (the subscripts denote the corresponding partial derivatives, e.g., u., == oufox). The ad missibil ity conditions are that u and its first and second partial derivatives be continuous and that u satisfy the boundary condition u = cp(s) on C, (6) where sis arc length along C and cp is a specified function on C. The function is prescribed and continuous on C.

f

Ans. 5

D.u

=

u.,., + u11v = f(x, y).

For a discussion of more general boundary conditions, see [1), pp. 203- 207.

PROB.

1.1.8

SUPPLEMENT

395

1.1.5. Tn Prob. 1.1.4, alter the admissibility conditions so that (6) is satisfied on a subarc C1 of C. On C2 - C - C1o there are no specified boundary conditions. Determine the Euler equation and the natural boundary condition that must be satisfied on C 2 in order that u minimize I.

tl.u

Ans.

= f(x,y) ,

where the subscript n denotes differentiation with respect to the unit outward normal n to D . 1.1.6. Determine the Euler equation of the functional

I =

JJ[a(x, y)u! + b(x, y)u! + c(x, y)u + 2f(x, y)u] dx dy, 2

D

using the admissibility conditions of Prob. 1.1.4. Here a, b, c and fare prescribed continuous functions on D, and a and b have continuous first partial derivatives.

Ans.

(au,.),,

+ (bu~)11

-

cu =

f

1.1.7. Determine the Euler equation and natural boundary condition for the functional

I

=

ff[a(x, y)(u! + uz) + c(x, y)u

2

+ 2f(x, y)u] dx dy

D

+fc a[x(s), y(s)][A(s)u

2

-

2cp(s)u] ds,

where C is the contour bounding D , the functions a, c and fare prescribed and continuous on D, and a has continuous first partial derivatives. The prescribed functions A(s) and cp(s) are continuous on C.

Ans.

u,. +Au = cp. 1.1.8. Determine the Euler equation of the functional

l[u]

= JJ[(6.u)2 -

2f(x, y)u] dx dy,

D

where f is a prescribed continuous function on D. The admissible functions u(x, y) have continuous partial derivatives up to and including the fourth order, and satisfy the boundary conditions

u = cp(s) ,

Ans. 6.2 u

Hint. Let ii =

u.. =

=

u=,

u,.

=

+ 2u,.uu +

0 for x, y on C.

on C.

tjJ(s) y

U 11 1111

= f(x, y).

(7)

396

SUPPLEMENT

PROB.

1.1 .9

1.1.9. Determine the Euler equation of the functional I[u] = I I [(6-u? - 2(1 -

v)(u.,.,u~~ -

u! 11)

-

2fu] dx dy,

D

where the admissible functions have the same continuity properties as in Prob. 1.1.8 and the condition (7) is satisfied on the subarc C1 of C. The remaining part of the boundary c2 = c - cl is "free," i.e., no conditions are specified on C2 • Determine the natural boundary conditions on C2 • When c2 = 0, compare the results with those obtained in the previous problem. The constant vis a specified number in the range 0 < v < ~ · Ans. 2

6. u = f,

(6-u)n

v6.u +( I - v)(u.,.,n; + 2u.,un1 n 2 + Ll 11 un~)

+ (1 -

= 0,

+

v)((uvu - u.,.,)n 1 n 2 u.,y(n; - 11;)]., where n1 and n 2 are the x andy-components of the outward unit normal to D, and the subscripts denotes differentiation with respect to arc le ngths a long C.

Hint. On C 2 , ii and iin are arbitrary. 1.1.10. Determine admissibility conditions and a functional whose Euler equation and natural boundary co ndition yield the following boundary value problem for the region D in the xu-plane with contour C: 6. 2u

u

= f(x, y)

for x , yin D ,

0, v 6-u -1 (1 - v)(u.,.,n~ + 2u.,11n1 n2 f- ll 11 vn~) = 0 for x, y on C. Ans. The functional is given in the preceding problem. The adm issible

=

functions have the same continuity properties as in Prob. l.l.8, and in addition, u = 0 on C.

1.2.

THE RITZ METHOD [14]

The minimum property of the solutions of boundary va lue problems suggests a method for their approximate determination. Suppose that a sequence of admissible functions is constructed whose limit minimizes an appropriate functional. Then the function obtained by truncating the sequence after a finite number of terms may provide an approximation to the minimizing function. The approximation is presumably more accu rate when more terms in the sequence are retained. Specifically, we select a family of admissible functions u = U(x;c) (8) depending on n (unknown) parameters c = (cl> c 2 , •• • , en)· Inserting these functions into the functiona l and performing the necessary integrations, we obtain

/[ U(x; c)] =