Partial Differential Equations

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© Cambridge University Press 1975 First published 1975 Printed in Great Britain at the University Printing House. Cambridge (Euan Phillips, University Printer)

Library of Congres8 Cataloguing in Publication Data Copson, Edward Thomas, 1901Partial differential equations. Bibliography: p. 277 Includes index. 1. Differential equation, Partial.!. Title. QA377.Cn 515'.353 74--12965 ISBN 0 521 205832 hard covers ISBN 0 521098939 paperback

CONTENTS

Preface

page vii

1 Partial differential equations of the first order

1

2 Characteristics of equations of the second order

24

3 Boundary value and initial value problems

44

4 Equations of hyperbolic type

54

5 Riemann's method

77

6 The equation of wave motions

90

7 Marcel Riesz's method

107

8 Potential theory in the plane

131

9 Subharmonic functions and the problem of Dirichlet

175

10 Equations of elliptic type in the plane

186

11 Equations of elliptic type in space

207

12 The equation of heat

238

Appendix

271

Books for further reading

277

Index

279

PREFACE

This book has been written in memory of my father-in-law the late Professor Sir Edmund Whittaker, F.R.S., in gratitude for all the help and encouragement he gave me for over thirty years. Today is the hundredth anniversary of his birth. When I went to Edinburgh as a young lecturer in 1922, I was surprised to find how different the curriculum was from that in Oxford. It included such topics as Lebesgue integration, matrix theory, numerical analysis, Riemannian geometry, of which I knew nothing. I was particularly impressed by Whittaker's lectures on partial differential equations to undergraduate and postgraduate students, far different from the standard English textbooks of the time. This book is not based on Whittaker's lectures; yet without his inspiration it would never have been written. I have frequently given courses of lectures on partial differential equations and have always regretted that there was no book to which I could refer my stud~nts. Friends told me that the remedy was to write one myself; and here it is, a presentation of some of the theory by the methods of classical analysis. There are few references to original sources. Mter lecturing on the subject for so many years, I could not now say whence the material came. On page 277 will be found a list of the books which I have read with profit, many of them more advanced than this.

E.T.C. St Andrews 24 October 1973

[ vii]

1

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

1.1 Lagrange's equation Lagrange's partial differential equation ofthe first order is ofthe form Pp+Qq = R,

(1)

where p = ou/ox, q = ou/oy and P, Q, R are functions of x, y and u; it is sometimes called a quasi-linear equation since it is linear in the derivatives. If P, Q and R do not involve u, Lagrange's equation is said to be linear; if only R involves u it is said to be semi-linear. By a solution of (1), is meant a function u(x, y) which satisfies the differential equation; but we often have to be content with a solution defined implicitly by a relation !(x, y, u) = O. If we regard (x, y, u) as rectangular Cartesian coordinates, !(x, y, u) = 0 is the equation of a surface; if! = 0 provides a solution of (1), the surface is called an integral surface. The fundamental problem is: given a regular arct r in space, is there a unique integral surface through r? Alternatively, given a regular arc l' in the xy-plane, is there a solution u(x,y) of (1) which takes given values on 1'? Let the parametric equations of r be x = xo(t),

y = yo(t),

u = uo(t)·

On any surface, du = pdx+qdy. Hence, if there is an integral surface through r, the values Po(t), qo(t) of p and q on the integral surface at the point of parameter t of r satisfy (2)

where dots denote differentiation with respect to t. If we denote by Po, Qo,R othe values ofP, Q,R at the pointofr ofparametert, we have PoPo + Qoqo = R o·

(3)

Hence if XoQo - Yo Po is not zero, Po and qo are determined. It is conventional to denote the second derivatives u=, u Zll ' u llll by r, 8, t; the fact that we have also used t to denote the parameter of r t The term regular arc is defined [ 1 ]

in Note 3 of the Appendix.

2]

EQUATIONS OF THE FIRST ORDER

[1.1

will not cause any confusion. If we differentiate (1) with respect to x, weget Pr+Qs

=

F(x,y,u,p,q),

so that, at the point of r of parameter t, Poro+Qos o = Fo·

Since dp = rdx+sdy, If xoQo-YoPo is not zero, Po and qo are determined on r, and hence so also are ro and So' Similarly we can find all the partial derivatives of u on r. Thus we get a formal solution as a Taylor series u = uo+{Po(x-xo)+qo(y-yo)} + l-{ro(x - XO)2 + 2s o(x-x o) (y- Yo) + to(Y _YO)2} + ....

Under suitable conditions, it can be shown that the series converges in a neighbourhood of (x o' Yo' u o) of r, provided that x oQo - Yo Po is not zero. Now drop the suffix zero which has served its purpose. At a point of r, an integral surface satisfies Pp+Qq = R,

Hence

px+qy = u.

(Qx-Py)q = Rx-Pu,

and similarly for p. If Qx - Py vanishes at every point of r, this equation is impossible unless the transport equation Pu

(or equivalently)

=

Rx

Qu =Ry

°

is satisfied. Hence, if Qx-Py = on r, there is no integral surface through runless u satisfies the transport equation; and then there are an infinite number ofintegral surfaces since q can be chosen arbitrarily. An arc r which has this property is called a characteristic. There is one characteristic through each point of space at which P, Q, R are not all zero; a charact.eristic satisfies

x y P

u

Q "k

A characteristic is the curve of intersection of two integral surfaces. For ifu = u1(x,y), u = u 2(x,y) are two intersecting integral surfaces, P1dx+q1dy-du

=

0,

P2d:r+q2dy-du = 0,

EQUATIONS OF 1.1J in an obvious notation, and so

But

PPI +Qql = R,

so that

Pp2+Qq'J = R

~=_Q_= ql-q2

[3

THE FIRST ORDER

P2-Pl

R P2ql -Plq2

Therefore on the curve of intersection of two integral surfaces,

The differential equations for a characteristic can be written as

x=

Y = Q,

P,

1i = R

by a change of parameter. The solution of these equations contains three constants of integration; two of these can be the coordinates of the point where the characteristic cuts, say, the plane u = 0, and the third can be fixed by measuring t from that point - the differential equations are unaltered if we replace t by t + c. The characteristics then form a two-parameter family. If C is a noncharacteristic are, it can be shown that the unique integral surface through C is generated by the one-parameter family of characteristics which intersect C. Again, ifthe two-parameter family ofcharacteristics is given by ¢(x, y, u)

=

a,

'!fr(x, y, u) = b,

we can construct a one-parameter family by setting up a relation between a and b, say b = F(a). This one-parameter family generates the integral surface '!fr(x,y,u) = F{¢(x,y,u)}.

°

The projection'}' of a characteristic r on the plane u = is called a characteristic base-curve. If P and Q do not involve u, the characteristic base-curves satisfy

x=P,

y=Q.

In order that there may be a solution which takes given values on '}', the data must satisfy the equation

Pu =Rx.

4]

EQUATIONS OF THE FIRST ORDER

[1.2

1.2 Two examples We know that, if u is a homogeneous function of x and y of degree n then ou ou x ox + Y oy = nu. We now prove the converse. The subsidiary equations of Lagrange are dx x

dy y

du nu'

-=-=-

From these equations we get '!f...=a x '

Hence the general solution is

;n =f(~). As a second example, let. us find the integral surface of (y-u)p+ (u-x)q = x-y,

which goes through the curve u = 0, xy = 1. The characteristics are given by x = y-u,

which give

y = u-x,

x+y+u = 0,

U = x-y,

xx+yy+uu = 0.

Hence the characteristics are circles, x+y+u = a,

X2+y2+ U2 = b.

We have to choose the one-parameter family which goes through u = O,xy = 1. When u = O,xy = 1, a 2 = (X+y)2 = X2+y2+ 2xy

:z

b + 2.

The required integral surface is therefore (X+y+U)2

or

=

X2+y2+ U2+ 2 1-xy x+y'

u=--

1.3] 1.3

[5

EQUATIONS OF THE FIRST ORDER

The general first order equation

We now ask the same question concerning the general first order equation F(x, y, U,p, q) = 0. (1) Does there exist an integral surface through a given regular arc

x

= xo(t),

Y

= yo(t),

r

U = uo(t)?

The method is to try to construct a Taylor series which satisfies (1) and converges in a neighbourhood of an arbitrary point of r. This involves calculating at that point all the partial derivatives of u. The first derivatives of u satisfy the condition du = pdx+qdy, so their values Po and qo at the point of r of parameter t are given by

F(x o, Yo, Uo,Po, qo) xoPo+Yoqo

=

=

0,

uo'

We suppose that we can find a real pair (Po, qo) which satisfies these equations; if we cannot, there is no real integral surface. Next denote the partial derivatives of F with respect to x, y, U,p, q by X, Y, U, P, Q. Then, if we differentiate (1) partially with respect to x, the variables U,p, q now being functions of x and y, we get

Pr+Qs+X + Up = 0. By hypothesis, p is now known on

r. Using the condition

dp = rdx+sdy, the values of the second derivatives of rand s on

r

satisfy

~ro+Qo8o+Xo+UoPo = 0,

xoro+Yo8o = Po· (2)

Hence

(3)

Since where

Xdx+ Ydy+ Udu+Pdp+Qdq = 0, du

=

pdx+qdy,

(2) and (3) are in fact the same equa.tion. IfQoxo-~yo is not zero, the values ro,8o,to of the second derivatives are determined on r, and similarly for the derivatives of higher orders. Thus we again get a formal solution as a double Taylor

6]

EQUATIONS OF THE FIRST ORDER

[1.3

series which can, under suitable conditions, be shown to converge in some neighbourhood of the chosen point of r, provided that Qoxo-Poyo does not vanish there. Now drop the suffix zero. At a point ofr, an integral surface satisfies F(x,y,u,p,q) = 0, u = px+qy, Pr+Qs

and

=

-X-pU,

Ps+Qt = - Y -qU, rx+siJ = p,

where

(Qx-PiJ) s

Hence

=

sx+tiJ = q. -

(X +pU)x-Pp,

(Qx-PiJ)s = (Y +qU)iJ+Qq.

and

If Qx-PiJ vanishes at every point of r, there is no integral surface

through r unless the expressions on the right of the last two equations vanish. This means that there is no integral surface unless u,p, q are appropriately chosen on r. Thus we have now not an arc but a strip; a sort of narrow ribbon formed by the arc r and the associated surface elements specified by p and q. Such a ribbon is called a characteristic strip. The arc carrying the strip may be called a characteristic. The differential equations of a characteristic strip are

x'

u

p =~= pP+qQ

X+pU

F(x,y,u,p,q)

and also

=

Y+qU

0,

regarded, not as a differential equation, but as an equation in five variables. By a change in the parameter t, we can write the equations as x=P, where

iJ=Q,

u=pP+qQ,

p= -X-pU,

q= -Y-qU,

F(x, y, U,p, q) = 0.

The characteristic strips form a three-parameter family. There are five constants of integration; one of these is fixed by the identity F = 0, the second by choice of the origin of t. The unique integral surface which passes through a non-characteristic arc C is generated by a one-parameter family of characteristic strips. The first step is to construct an initial integral strip by asso-

1.3]

EQUATIONS OF THE FIRST ORDER

[7

ciating with each point of C a surface element whose normal is in the direction p:q: -1. If the parametric equations of Care x = xo(s),

y = yo(s),

u = uo(s),

where s need not be the arc-length, we choose p = Po(s), duo ds

so that

q = qo(s) dx o

8yo

= Po di + qo ds

and Through each surface element of the initial strip, there passes a unique characteristic strip. The one-parameter family of characteristic strips so formed generates the required integral surface, as illustrated by the example of the next section. This method is usually called the method of Lagrange and Charpit. It will be noticed that although the quasi-linear equation Pp+Qq = R

does possess characteristic strips no use is made of them in solving such an equation. This is because of an important geometrical difference between Lagrange's equation and the general equation F(x,y,u,p,q) = 0. If (x o, Yo' u o) is a point on an integral surface of F = 0, the direction ratios Po:qo: -1 of the normal there satisfy F(x o, Yo, uo,Po' qo) = 0.

Hence the normals to all possible integral surfaces through the point generate a cone N whose equation is F(X o, YO' u O' - x-xo , _y-Yo) u-u o u-u o

=

0.

The tangent planes at (x o, Yo, u o) to all possible integral surfaces through this point envelope another cone T whose equation is obtained by eliminating Po and qo from the equations u-

Uo

= Po(x - x o) + qo(y - Yo)'

(x-xo)Qo-(Y-yo)~

= 0,

F(x o,Yo' uo,Po, qo) = 0,

where ~ and Qo denote the values of 8F/8p and 8F/8q at

8]

EQUATIONS OF THE FIRST ORDER

[1.3

The tangent plane to a particular integral surface at (x o, Yo, u o) goes through a generator ofthe cone T; the normal there lies on the cone N. In the case of Lagrange's equation, the cone N degenerates into the plane the cone T becomes the straight line x-x o

u-u o

p;-

Jr' o

1.4 An example of the Lagrange-Charpit method We find by the method of characteristics the integral surface of pq = xy which goes through the curve u = x, y = O. The characteristic strips are given by the differential equations

x=

if

q,

=

u=

p,

P=

2pq,

q= x

y,

= xy. It turns out that t Aet+Be- , y = Cet+De- t, u = ACe 2t -BDe- 2t +E, p = Cet-De- t , q = Aet-Be-t ,

and the relation pq

x

=

where the constants of integration are connected by

AD+BC= O. x

On the initial curve

s,

=

y

=

u

0,

s.

=

On the initial integral strip, the equations du = pdx+qdy, pq = xy give p = 1, q = O. Let t be measured from the initial curve. Then when t = 0, we have

A+B

=

s,

C+D

C-D

1,

=

AC-BD+E = s,

A-B

=

0

AD + BC = O.

where These give

0,

=

A

=

B

=

!-s,

C = -D

=

i,

E

=

is,

the condition AD + BC = 0 being satisfied automatically since the initial strip is an integral strip. The characteristics through the initial integral strip are therefore

x

=

s cosht,

y

=

P = cosht,

sinht,

u

=

s cosh 2 t,

q = ssinht.

1.4]

EQUATIONS OF THE FIRST ORDER

[9

Eliminating 8 and t from the first three equations, we obtain u 2 = x2(1 +y2) as the equation of the required integral surface.

1.5 An initial value problemt In this section we prove that the equation P =f(x,y,u,q) (1) has, under certain conditions, a unique analytic solution which satisfies the initial conditions u(xo, y) = ¢(y), q(xo, y) = ¢'(y), (2) where ¢(y) is analytic. The result we obtain is a local result; we show, by the method of dominant functions, that there is a solution U = u(x, y) regular in a neighbourhood of any point (xo,Yo) of the initial line x = x o' It is convenient to write U o for ¢(Yo), qo for ¢'(Yo)' And we make the assumption that f(x, y, u, q) is an analytic function of four independent variables, regular in a neighbourhood of (xo,Yo, uo,qo)' The problem can be transformed into one involving three quasilinear equations with three dependent variables U,P, q. If there is an analytic solution, then oqjox = opjoy. From equation (1) we have oq op ox =fx+fup +fqox ' Hence U,P, q satisfy the equations OU -=P ox ' op op ox =fx+fup+fq Oy'

(3)

oq op ox = oy' under the initial conditions u(xo'y) = ¢(y), p(xo,Y) =f(xo,y,¢(y),¢'(y)), q(xo,Y) = ¢'(y). This system of three equations is equivalent to equation (1) under the initial conditions (2). From the first and last of equations (3),

~(q_ OU) ox

oy

= 0

t See Notes 1 and 2 in the Appendix.

10]

EQUATIONS OF THE FIRST ORDER

OU q- oy =

so that

[1.5

W 1 (Y)·

By the initial conditions, w1 (y) is identically zero, so that q = oujoy. The second equation of (3) gives

op oq ox =fx+fup+fq ox since

=

of ox

oq 02U 02U op ox = ox oy = oy ox = oy'

when u is analytic. Hence

P = f(x, y, u, q) + w2 (y)· Again, by the initial conditions, w2 (y) is identically zero, and so p = f(x, y, U, q). The coefficients in the second equation of (3) involve the independent variables x and y. We can get rid of this restriction by introducing two additional dependent variables £ and r; defined by

o£ or; ox = oy'

orJ

ox

=

°

under the initial conditions

£=

0,

r; = y-Yo,

when x = x o' Since r; is independent of x, r; = y - Yo for all x. Then o£lox = 1 so that £ = x-xo' If we put x = x o+£' y = yo+r; in f(x, y, U,p, q) we get an analytic function g(£, r;, U,p, q) of five variables regular in a neighbourhood of (0,0, uo'Po' qo)' We now have a system of five equations OU or; ox = P oy'

op Or; Op ox = {g,+gup} Oy +gq oy' oq op ox = oy' o£ or; ox = oy'

°

or; = ox '

1.5]

EQUATIONS OF THE FIRST ORDER

[11

with the initial conditions u

= ¢(y), P = f(x o, y, ¢(y), ¢'(y)), q = ¢'(y),

£=

0,

1J

= V-Yo

when x = x o' This system is of the form

i;

oUi _ OUi ox - j=1 (Iii oy

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

with the initial conditions Ui

= ¢i(Y)

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

when x = X o' The coefficients (Iii are analytic functions of the dependent variables only, regular in a neighbourhood of (¢l(YO)' ¢2(YO)' ... , ¢n(Yo))

and the data ¢i(Y) regular in a neighbourhood of Yo' Here n = 5; later, we shall need the case n = 8. The method of proof does not depend on the number of variables. The easiest case is when n = 2. The general case is discussed in Courant and Hilbert's book. Consider, then, the simplest case of two equations OU ox

= A OU +B oV

OV ox

= C OU +D OV

oy oy

oy' oy'

under the initial conditions U

= ¢(y), v = 1fr(y),

when x = X o, where ¢ and 1fr are analytic functions regular in a neighbourhood of Yo' For simplicity in writing, shift the origin so that X o and Yo are zero. The coefficients A, B, C, D are analytic functions of U and v, regularinaneighbourhoodof(u o, vo),whereu o = ¢(O),vo = ljr(O). Again, by a shift of origin in the uv-plane, we can take U o and Vo to be zero. Since ¢,1fr and the coefficients are analytic, we can find constants M and R so that ¢(y) and 1fr(y) are dominated by My/(l-y/R) in Iyl
12]

EQUATIONS OF THE FIRST ORDER

[1.5

Formal solutions of the initial value problem as double power series 00 U U= ~ ~n, xmyn, m,n=O m.n. v=

00 v "" -!!!:!!:....Xmyn m,n=O kJ m.n. " '

can be found by calculating the coefficients Umn =

(:::;;)0'

Vmn =

(::;;L

where the suffix zero denotes that the derivatives are evaluated at the origin. Since 9 and lfr are analytic, they have convergent expansions

where

90 = lfro =

0. Hence

From the differential equations we have u lO = A(0,0)91 +B(O,O)lfrv

vlO = 0(0,0)91 +D(O,O)lfrl'

Next, 2

2

2

8u 8u 8v (8U) 2 8u 8v (8V) 2 8x8y = A 8y 2 + V 8y 2 +A u 8y + (A,,+B u) 8y 8y +B" 8y so that Un

= A(O, 0) 92 + B(O, 0) lfr2 +Au(O, 0) 9i + {A,,(O, 0) +Bu(O, O)} 91lfrl +B,,(O, 0) lfri.

Similarly we can calculate successively the coefficients U 1n and V 1 nNext, by considering 82uj8x 2 , ()3uj8x 28y, ... we can calculate U20 , U2V .... In a similar way, the coefficients Umn and Vmn in the formal expansions for U and v can all be determined. If we replace A,B, O,D, 9 and lfr by their majorants, we obtain the system of equations 8U 8x

= 1-

M (8U 8V) U - V 8y + 8y ,

8V M (8U 8D 8x = 1- U - V 8y + 8yJ'

1.5]

[13

EQUATIONS OF THE FIRST ORDER

under the conditions

U = V = My/(l-y),

when x = 0, Iyl < 1, the factor y occurring because ¢ and 1fr vanish when y = 0. We show that this initial value problem has a unique solution regular near the origin, and the functions U and V will be majorants of u and v. o Since

°'

-(U - V) = ox

U - V is a function of y alone, and by the initial condition is identically zero. Hence U = V and oU 2J.lI oU ox = 1-2U oy'

Since the Jacobian of U and 2Mx + (1- 2U) Y vanishes identically, the general solution of this equation is 2Mx+(1-2U)y = F(U),

and, by the initial condition, F(U) = U(1-2U)/(M + U). Hence the required solution satisfies the quadratic equation 2J.Wx(M + U) + (1- 2U) (M + U) y = U(l- 2U).

°

This equation has two solutions; one takes the value at the origin, the other the value l-. It is the former which is needed; it can be written down explicitly, and is evidently an analytic function, regular in a neighbourhood of the origin. If we write down the expansion of U near the origin as U =

00

~

U. -!!!!!..

m,n=omln!

x m yn

'

the coefficients can be calculated successively just as in the original problem. Putting x = 0, we get Uoo

=

0,

UonI =M. n.

From the differential equation, UIO = 2M2.

Next, so that

2M 02U 4M (O~2 02U oxoy= 1-2U oy2+(1-2U)2 oyJ

14]

EQU ATIONS OF THE FIRST ORDER

[1.5

By repeated differentiations and respect to y, we can calculate all the coefficients U1n ; each is positive. Again 82U 2111 82U 4M 8U 8U 2 8x = 1-2U 8x8y +(1-2U)2 Ox Oy

so that

U20 = 2MUn

+ 4MUlO U01 = 4M3 + 8..il14 •

By repeated differentiations with respect to y, 'we can calculate all the coefficients U 2n; again, each is positive. Similarly all the coefficients Umn can be found, and each is positive. Lastly, by this iterative process, we see that

lumnl

< Umn ,

Ivmnl < Umn so that the formal series u and v are both dominated by U. Therefore there is a unique solution, regular near (x o, Yo), of the initial value problem with which we started. The result proved in this section is not as general as might appear at first sight. \Ve do not need to assume the data are analytic. For example, if we assume that ¢(y) is everywhere differentiable, the equation xp - yq + u = 0, under the initial conditions u(x o, y) = ¢(y),

q(xo, y) = ¢'(y),

where x o 9= 0, has the solution u =

:0 ¢(:~),

which is differentiable but not necessarily regular in the neighbourhood of any point (x o, Yo)'

1.6 Systems of semi-linear equations of the first order A system of semi-linear equations of the first order is of the form an U x + a l2 V x + bn u y + bl2 vy = hI'} a 21 U x + a 22 V x + b21 u y + b22 vy = h2,

(1)

where the coefficients aij' bii depend only on x and y, but hI and h2 may also involve u and v. The coefficients ai i ' bii are analytic functions, regular in some domain (open connected set) D and hI and h 2 are regular when (x, y) belongs to D and lui < K, Ivl < K for some constant K. The problem is to see whether there exists a unique analytic solution, regular in a neighbourhood of a regular arc y, which takes given values on y. The first step in a solution by Taylor

1.6]

EQUATIONS OF THE FIRST ORDER

[15

series is to show that, at any point of y, the partial derivatives of U and v of all orders are determinate. Let the parametric equations of y be x = :l:o(t), y = yo(t); on y, we are given that u = uo(t), v = vo(t). As we shall need derivatives ofthese four functions of all orders, we assume that they are analytic functions of t, regular near any given value of t. Dropping the suffix zero, we have

on y, where dots denote derivatives with respect to t. By (1) we then have (bn ~ - an~) u y + (bI2~ - aI2~) v y = hI ~ - an ~ - a l2 ~,} (2) (b 21 x - a 21 y) u y + (b 22 X - a 22 y) Vy = h 2x -a21 u - a 22 v. Usually these equations determine the first derivatives of u and vat any point of y. The six second derivatives satisfy four similar equations, namely an Uxx + a l2 vxx + bn UXy + bl2 Vxy : kl'l a21uXX+a22vXx+b21Uxy+b22Vxy - k2, an Uxy + a l2 VXy + bn u yy + bl2 Vyy = ka,

(3)

a21uxy+a22vxy+b21Uyy+b22Vyy = k4,

where kv k 2, ka, k4 involve, not only x, y, U, v, but also the first derivatives now supposed known on y. Since x, y, ii, ii are assumed to exist, we have two other equations ii = UxiX2 + 2uxyXfJ + uyyJ? + uxx + uyY, ii

= vxxx2+2vxyiy+vyyJj2+vxx+VyY

on y. These six equations usually determine the six second derivatives on y. A similar method determines the partial derivatives of all orders. We can simplify the calculation by using only the first two of equations (3) and the relations Ux = uxxx+uxyY,

Vx = vxxx+vxyy.

We then obtain a pair of equations similar to (2), namely (b n ~- an~) Uxy + (bI2~ - aI2~) VXy = kl ~ -an ~X-aI2~X'} (b 21 X - a 21 y) Uxy + (b 22 X - a 22 y) VXy = k 2x -a21 u x -a22 v x.

(4)

Again, these usually determine UXy and Vxy on y, and hence the other second derivatives there.

16]

EQUATIONS OF THE FIRST ORDER

[1.6

Equations (2) determine u y and v y uniquely only if the determinant

D. = Ibnx-any b21 X - a 21 y

bl2 x-al2 y I b22 X- a 22 y

is not zero. If the determinant is zero, the equations are usually inconsistent. If D. is zero, and if the determinants

hI X- an U- a 12 V I h2x-a2Iu-a22v' hI X- an U- al2 V I h 2x - a 21 u- a 22 v are both zero, equations (2) are identical; we can assign, say, u y as we please and calculate v y , but the equations do not determine u y and v y uniquely. The equation D. = 0 is a quadratic equation in x:y. At any point (x, y) it determines two directions, called the characteristic directions. If the characteristic directions at (x, y) are real and distinct, the system is said to be of hyperbolic type there. If the quadratic equation has equal roots, the system is said to be of parabolic type. But if the roots of D. = 0 are complex at (x, y), the system is said to be of elliptic type there. A real arc whose tangent at every point is in a characteristic direction is called a characteristic base curve, or, briefly, a characteristic. It does not depend on u or v. If the system (1) is of elliptic type everywhere in D, there are no characteristics in D. If it is of hyperbolic type there, there are two distinct families of characteristics. But if it is everywhere of parabolic type, there is only one family. If y is a characteristic of (1) and equations (2) are consistent, there exists a function j(t) such that the second equation of (2) becomes identical with the first if we multiply through by j(t). Hence, if there is a solution .of our problem, the values of u and v on y satisfy the ordinary differential equation h1x -an U - a l2 v = j{h 2x - a 21 u- a 22 v}.

This is called a transport equation. A hyperbolic system has two transport equations, one corresponding to each system of characteristics. To illustrate this, consider the simple example

1.6]

EQUATIONS OF THE FIRST ORDER

[17

Equations (2) then become YUy+XVy = u-h1x, iuy + YV y

= V- h2 x.

The characteristic directions are given by x2 -l? = 0. The system is ofhyperbolic type, the two families of characteristics being x ± y = constant. On a characteristic x + y = constant, the transport equation is on x-y

= constant,

u+v

=

x(h l +h 2 );

u-v

= x(h l -h 2 ).

In the simplest case when hI and h 2 are identically zero, the transport equations can be solved immediately. On each characteristic x+y = constant, u+v is constant; on x-y = constant, u-v is constant. Therefore

u+v = 2F(x+y),

u-v = 2G(x-y),

where F and G are arbitrary functions. Hence

u = F(x+y) +G(x-y),

v = F(x+y) -G(x-y).

Again, for the Cauchy-Riemann equations

ux-v y = 0,

vx+u y = 0,

the characteristic directions are given by x2+ y2 = 0, so that the system is of elliptic type. Lastly, the system

is of parabolic type, with one system of characteristics, y = constant. A system of semi-linear equations with two independent variables x and y and n dependent variables Uv U2 ' ••• , Un can be written in vector form A OU +B oU = h (5)

ox

oy

,

where A = {aij} and B = {b ij } are non-singular n x n matrices and and h are column vectors

U

Each coefficient aij' bij is a function of x and y alone, regular in some domain D. Each function hi depends on x, y and the components of u; each is assumed regular when (x, y) is in D and luil < K for all i.

18]

EQUATIONS OF THE FIRST ORDER

[1.6

Let y be a regular arc in D, with parametric equations x = x(t), Y = y(t). If we wish to show that there is a unique vector u which satisfies (5) and takes given values on y, then we must first show that we can calculate uniquely the values of all the partial derivatives of each component U i at any point of y. The procedure is just the same as in the case of two dependent variables. On y, u is known and . .ou .ou u=x-+y-. ox oy Hence

(xB-yA)

~~ = xh-Au.

(6)

The first derivatives are thus determined provided that the direction of y at the point considered does not make xB - yA singular. Ifu mn denote om+nujoxmoyn, it follows from (5) that

Au 20 + BUn = k v AU n

+ BU 02 =

k 2,

where k 1 and k 2 depend on x, y, u and the first derivatives of u, now known on y. But U = X2U20 + 2xyu n + y 2u 02 + xu lO + fju ov so that we have three equations to determine the three partial derivatives. Since we are assuming that A and B are not singular,

u 20

= A-1 k 1 -A-IBu 11l

U02

= B-1 k 2-B-IAu n ·

Therefore un satisfies on y the equation

(x 2A-IB-2xy+y2B-IA)u n = k a, where the vector k a is known on y. This equation can be written as

B-l(xB-yA)A-l(xB-yA) un = k a. Henee, if xB - yA is not singular, un = (xB-YA)-lA(xB-YA)-lBk a· on y. The second derivatives are thus determined provided that the direction of y, at the point considered, does not make xB - yA singular. A shorter proof can be given which depends on the equation

OV

OV

A-+Box oy = k 1 ,

1.6]

[19

EQUATIONS OF THE FIRST ORDER

where v = ou/ox. It follows that u20 = oV/2x, and un = determined on y if xB - yA is not singular. Similarly from A

ow

ow

ox + B oy

ov/oy

are

= k2,

where w = ou/oy, we can get the values ofu u and U 02 when xB-yA is not singular. And similarly for the derivatives of all orders. We shall not discuss the convergence of the resulting series. If however the matrix xB - yA is singular at the point of y under consideration, equation (6) either has no solution or an infinity of solutions, depending on the rank of the extended matrix [xB -yA,xh -Au].

In the latter case, the components of the vector xh -Au are not linearly independent; there then exists a transport equation n

~ il.ixh i -

i=l

~ il.iaijU j = 0 i,j

satisfied by the data on y. Since

det (xB - yA) = x n det B + ... +(- 1 )nyn det A,

the equation det (xB -yA) = 0 is of degree n in x: y. At any point (x, y) it determines n directions, called the characteristic directions. If the characteristic directions at (x, y) are all real and distinct, the system (5) is said to be of hyperbolic type there; if they are all complex, the system is said to be of elliptic type there. These are the two extreme cases. A regular arc y is called a characteristic if its tangent at every point is in a characteristic direction. If there are n distinct families of real characteristics, the system is said to be of hyperbolic type; if there are no real characteristics, it is of elliptic type. If y is a characteristic, there is usually no solution which takes given values on y. But if u satisfies the transport equation on y, there are an infinity of solutions. It will be recalled that we assumed that A and B are not singular. We can always rotate the axes about the point of y considered so that x = 0 and y = 0 are not characteristic directions; after the rotation, detA and detB are not zero. Hence the assumption that A and B are not singular at the point considered is no restriction. We could have dealt in much the same way with the quasi-linear case in which the matrices A and B also depend on the components of u. An example is given at the end of the chapter.

20]

EQUATIONS OF THE FIRST ORDER

[1.7

1.7 An application of the method of characteristics The linear system is of hyperbolic type, with characteristics x ± y = constant. On each characteristic x-y = constant, u-v is constant; on each characteristic x+y = constant, u+v is constant. Let us consider the problem of finding solutions U and v in the strip ~ x ~ 1 given that U = when x = and when x = l, and that

°

°

U = ¢(x),

°

v = Jjr(x)

°

when y = 0, ~ x ~ l. We assume that the data are continuous and continuously differentiable; in particular, ¢(o) = ¢(l) = 0. Consider the rectangle with vertices P(x,y),

where

°

~

x

~

Q(l,l+y-x),

R(l-x,l+y),

S(O,y+x),

l; the sides are characteristics. Then

UR+VR = uQ +vQ'

uR-VR = us-vs.

Since uQ and Us are zero,

that is

u(l-x,l+y) = -u(x,y),

v(l-x,l+y) = v(x,y).

From this it follows that U(x, y) = u(x, y + 2l),

v(x, y) = v(x, y + 2l)

so that U and v are periodic functions of period 2l. Hence we only need to solve the boundary value problem for ~ y ~ l. In fig. 1, AB is x = l, and the inclined lines are characteristics. There are four regions to consider, the triangles OAK, ABK, OKe, and the square KBLC. In the triangle OAK,

°

u(x,y) -v(x,y) = u(x-y, 0) -v(x-y, 0)

= ¢(x-y) -

Jjr(x-y) = 2F(x-y),

u(x,y)+v(x,y) = u(x+y,O)+v(x+y,O)

= ¢(x+y)+Jjr(x+y) =

2G(x+y),

1.7]

[21

EQUATIONS OF THE FIRST ORDER

y

B

o

.A

x

Fig. 1 since u + v and u - v are constant on characteristics. Therefore u(x,y) = F(x-y)+G(x+y),

v(x,y) = -F(x-y)+G(x+y),

where ¢-lfr = 2F, ¢+lfr = 2G. Next, in the triangle OKC, u(x,y)-v(x,y) = u(O,y-x)-v(O,y-x) = -v(O,y-x), u(x,y) +v(x,y) = u(x+y, 0) +v(x+Y, 0) = 2G(x+y).

But u(O,y-x)+v(O,y-x) = u(y-x,O)+v(y-x,O) = 2G(y-x),

so that

v(O,y-x) = 2G(y-x).

Hence, in the triangle OKC, u(x,y) = -G(y-x)+G(y+x),

In particular,

v(O, y)

v(x,y) = G(y-x)+G(y+x).

=

2G(y).

Note, that in the triangle OKC, u and v do not depend on F. Similarly, in the triangle ABK, u(x,y) = F(x-y)-F(21-x-y),

v(x,y) = -F(x-y)-F(21-x-y)

22]

[1.7

EQU A TIONS OF THE FIRST ORDER

are independent of G. In particular v(l,y) = -2F(l-y).

Lastly, in the square KBLC, u(x,y)-v(x,y) = u(O,y-x)-v(O,y-x) = -2G(y-x), u(x,y)+v(x,y) = u(l,x+y-l)+v(l,x+y-l) = -2F(2l-x-y).

Therefore

u(x,y) = -F(2l-x-y)-G(y-x), v(x,y) = -F(2l-x-y)+G(y-x).

Exercises 1. Solve the problem of §1.7 when u

x

= kv on x = l,

U

= -kv when

= 0, where k =l= 1. 2. Of what type is the system ou

ov

~+f(x,y)-;:;- = uX oy

0,

ou ov ~+f(x,y)~ = O? oy uX

Find the characteristics and the corresponding transport equations. 3. Find_the characteristics and corresponding transport equations of the system

4. Show that the system xyu",+v lI = 0,

ulI-v", =

°

is of elliptic type when xy > 0, of hyperbolic type when xy < parabolic type on the axes.

°and of

5. A semi-linear system has two distinct families of characteristics ;(x, y) = constant, 'YJ(x, y) = constant. Prove that, if; and 'YJ are taken as

independent variables, the equations of the system are OU OV alo; +a20; = hI'

OU ov blO'YJ +b 2 O'YJ = h 2·

(; and 'YJ are called characteristic variables.)

6. In the equation aw",,,, + 2bw"'l1 + CW IIII = h, the coefficients a, b, C depend only on x and y, but h is a function of x, y, W"', w lI " Show that it is equivalent to the half-linear system au", + bv",+ bUll + cV II = h(x,y,u,v),

with characteristics given by

v",-u ll =

°

ay2-2bXfj+cx2 = 0.

EQUATIONS OF THE FIRST ORDER

[23

7. Find the integral surface of yp -xq = 1 + n 2 which goes through the circle u = 0, x2 +y2 = 1. 8. Find the integral surface of yp +xq = u which goes through the curve r,y = 0.

u =

9. Find the integral surface of xpq+ yq2 = 1 which goes through the curve u = x, y = 0. 10. Find the characteristic strips of xp +yq = pq. Deduce the equation of the integral surface through u = lx, y = 0. 11. u and v satisfy the system of quasi-linear equations

ou ou ov -+u-+2v- =0, ox oy oy ov ov ou 2 -+2u-+v- = 0. ox oy oy If u and v are given on a regular arc y, prove that the~partial derivatives of u and v on y can be found uniquely if Y =l= (u ± v) x. Showthat,ify = (u+v)xon y, values can be found for the first derivatives, though not uniquely, if u+2v is constant on y. Find the corresponding result if x = (u-v) yon y. If there is a family ofstraight lines x = at, y = bt which are characteristic base curves on which if = (u + v) X, prove that u and v are functions of yjx and hence that 2Y Y U=3 x + A , v=--A, 3x where A is an arbitrary constant. 12. If, in Ex. 11, u and v are taken as independent variables and x and y as dependent variables, show that the equations become

ox ox oy 2v--u-+- =0, OU ov ov

ox Ox .oy 2u--v--2- =0. ou ov ou

Prove that this system has characteristics u ± 2v = constant. Show that, if ; = u+2v, 'Y/ = u-2v,

ox (;+3'Y/) 0;

oy

= 4 0;'

ox oy = 4-. 0'Y/ 0'Y/

(3s+'Y/)-

2

CHARACTERISTICS OF EQUATIONS OF THE SECOND ORDER

2.1

The general equation of the second order

The general equation of the second order with two independent variables is (1) F(x,y,u,p,q,r,s,t) = 0, where, in the usual notation, p = u x ' q = u lI ' r = u xx ' S = U Xll ' t = U lIlI " Given a regular arc r in space, (x, y, u) being rectangular Cartesian coordinates, does there exist a unique solution of (1), forwhichp and q take given values on n This is called the Problem of Cauchy; the existence theorem under appropriate conditions is the theorem of Cauchy and Kowalewsky. There are two other ways offormulatingthe problem. If D. is a developable surface containing r, does there exist a unique integral surface which touches D. along r? Alternatively, if y is a regular arc in the x, y-plane, does there exist a unique solution of (1) for which u,p,q take given values on y? The data must satisfy the strip condition du = pdx+qdy on y. The Cauchy-Kowalewsky theorem is a local theorem. It is assumed that F is an analytic function of the eight variables, and that the data U,p, q are analytic functions of the parameter, say the arc, which defines position on y. The result is that, if (x o, Yo) is any point of y, there exists a unique solution of (1) in the form of a convergent Taylor series co (x-x )m(y_y )n U

=

U

00

+ ~u 1 mn

0

m!n!

0

where U mn is the value of om+nujoxmoyn at (x o, Yo). The first step is to show that u has unique partial derivatives of all orders at every point ofy. The result is that the problem has a unique solution in a domain containing y. On y, we have u, p and q, analytic functions of the arc length 1say. Denote differentiation with respect to the arc by dashes. Then on y p' = rx' +sy',

q' =

SX'

+ty'.

(2)

We have three equations (1) and (2) to determine r, s, t on y. There are three possibilities. Firstly, there may be no set of real values of (r,s,t) which satisfies (1) and (2); in this case the problem has no [ 24 ]

2.1]

[25

CHARACTERISTICS

solution. Secondly, there may be just one real set (r, s, t) which satisfies (1) and (2); in this case, ifthere is a solution, it is unique. Thirdly, there may be several real sets, in which case we may get a solution corresponding to each set. Suppose that there is just one set. The next step is to calculate the third derivatives. If we differentiate (1) partially with respect to x, we obtain

Rrx+Ssx+Tt x =

-x - Up-Pr-Qs

(3)

in the obvious notation (R = of/or, ... ). In this equation, everything is known on y except rx , sx' t x. Sincedr = rxdx + sxdy, ds = sxdx+txdy, because r y = sx' Sy = t x' we have on y x'rx+Y'sx = r', . x'sx + y't x = s'.

(4)

vVe now have three equations to determine rx ' Sx and tx on y, namely (3) and (4). The three derivatives are determined uniquely provided that the determinant of (3) and (4) does not vanish. That is, provided that Ry'2 - Sx'y' + TX'2 does not vanish. Similarly we can calculate all the partial derivatives of u on y provided that this condition is satisfied. IfRy'2 - Sx'y' + TX'2 vanishes everywhere on y, y is called a characteristic base curve. The condition Ry'2_Sx 'y' + TX'2 = 0 for a curve y to be a characteristic usually depends on the values of u, p and q given on y. The following examples make this clearer. The curve y is a characteristic of p2r - 3pqs + 2q 2t = 0 py'2 - 3pqx'y' + 2q2x'2 =

if

o.

If y = x is a characteristic, p2 - 3pq + 2q2 = o. The data must satisfy p = q or p = 2q on the line; otherwise it is not a characteristic. The curve y is a characteristic of r-3q+2t

=

y'2 - 3x'y' + 2X'2

if

0

= 0,

a condition which does not depend on the data. Either y' = x' or y' = 2x' so that there are two real families of characteristics y = x + a and y = 2x + f3 where a and f3 are constants. Lastly, the curve y is a pr - qs = 0 characteristic of if

py'2 -qx'y' =

o.

There is one family, y = constant, of characteristics independent of the data; the otherfamily, given bypy' -qx' = O,depends on the data. 2

CPO

26]

2.2

[2.2

CHARACTERISTICS

The Cauchy-Kowalewsky theorem

If F(x,y,u,p,q,r,s,t) getting

= 0 involves

r explicitly, we can solve for r, (1)

r =j(x,y,u,p,q,s,t).

This is no restriction; for if it does not involve r explicitly, we can always make a change of independent variables to (x',y') say, so that 82u/oX'2 does occur explicitly. The Cauchy-Kowalewsky Theorem asserts that, if j is an analytic function of the seven variables, there is a unique analytic solution of (1) which is regular in a neighbourhood of (x o, Yo) and which satisfies the conditions u

=

¢(y),

p

=

(2)

Jjr(y),

when x = x o, provided that ¢(y) and Jjr(y) are regular in a neighbourhood of Yo. If these conditions are satisfied at eyery point (x o, Yo) of a finite interval y of the initial line x = x o' the Problem of Cauchy posed in (1) and (2) has a unique solution regular near y. The case when y is a regular arc x = w(y) can be reduced to this case by a change of independent variable. The values of u and its first and second derivatives at (x o, Yo) are Uo

=

¢(Yo), So

=

Po

=

Jjr'(yo),

Jjr(yo), to

=

qo

=

¢'(Yo),

¢H(yO),

r o = j(xo,Yo' uo,Po' qo, so' to)·

And all the other partial derivatives are uniquely determined at (x o,Yo). The existence of a unique analytic solution of (1) is proved by transforming the problem into one of solving a system of six quasilinear equations with six dependent variables U,p, q, r, s, t. The argument is like that of § 1.5. We do not now assume that p, q, r, s, tare partial derivatives of u; that they are, will be a consequence of the system of equations. The system is Ux

= p,

r, qx = Py, Sx = ry, tx = Sy,} r x = X + Up+Pr+Qpy+Sry+Tsy, Px

=

(3)

where X = oj/ox, etc.,j(x,y,u,p,q,s,t) being regarded as a function of seven independent variables. Note that in each of equations (3), the partial derivatives with respect to x occur on the left but not on the right. The conditions when x = X o are u

and

=

¢(y),

p

=

Jjr(y) ,

q

=

¢'(y),

S = Jjr'(y),

t

r = j(xo, y, ¢(y), Jjr(y) , ¢'(y), Jjr'(y), ¢JH(y)).

=

¢H(y),

(4)

2.2J

CHARACTERISTICS

[27

The first and second equations of (3) ensure that p and r have their usual meanings. The first and third give

~ ox

(q_ OU) = 0, oY

so that q-u y is independent of x. Using (4), we find that q = u y. The second and fourth equations give

!.(8- OP) ox oy

= 0

,

so that 8-py is independent ofx. Using (4), we find that 8 = Py = u xv ' Similarly for t and r. Thus system (3) with conditions (4) is equivalent to equation (1) with conditions (2). The coefficients in the last equation of (3) involve x and Y as well as the five dependent variables u,p,q,8,t. By introducing two more variables ~ and 1J, we get a system in which the coefficients do not involve x and y. We define 1J by 1Jx = 0 with the condition that 17=Y-Yo when x=x o; then 1J=Y-Yo for all x. We define ~ by ~x = 1Jy, where ~ = 0 when x = x o' This gives ~x = 1 and so ~ = X-X o' Now write f(XO+~'YO+1J,U,p,q,8,t)= g(~,1J,U,p,q,8,t).

vVe now have the system

rx = (gf,+guP+gpr)1Jy +gq py+gsry+gt 8y, ~x = 1Jy,

under the given conditions when x

i

OUi oU; ox = ;=lgii OY'

1Jx = 0 = X o.

This system is of the form

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

with initial conditions when x = x o, which was discussed in § 1.5. The coefficients gii are functions of the Uz only, and are regular in a neighbourhood of {
28]

2.3

CHARACTERISTICS

[2.3

The linear equation

The simplest second order equation with two independent variables is the linear equation ar+2h8+bt+2gp+2fq+cu+d = 0, where the coefficients are all analytic functions of x and y alone, regular in some open connected set (domain) D. Write this equation as ar+2hs+bt+F = O.

(1)

Let y be a regular arc in the xy-plane. Then, by the CauchyKowalewsky theorem, there exists a unique solution of (1) for which U,p, q assume given analytic values on y, and this solution is regular in a neighbourhood of y, provided always that y is not a characteristic base curve. On y, the data satisfy the strip condition u' = px' +qy' where dashes denote differentiation with respect to the arc length. In solving (1) by a Taylor series, we have to find all the derivatives of u on y. In particular, we have p' = rx' +8Y',

q' = 8X' +ty',

so that the value of 8 on y satisfies (ay'2- 2hx'y' + bx'2) 8 = ap'y' +bq'x' + Fx'y'.

(2)

Thus 8, and hence rand t, are determined uniquely on y provided that y is not a characteristic base curve. The characteristic base curves are given by ay'2 - 2hx'y' + bX'2 = O. As a, h, b depend only on x and y, the characteristic base curves do not depend on the data; the characteristics on an integral surface do not depend on the integral surface. The characteristic base curves are therefore frequently called the characteristics. If ab-h 2 > Oin D, there are no real characteristics, and the equation is of elliptic type. If ab - h2 < 0 in D, there are two distinct families of real characteristics, and the equation is of hyperbolic type. If ab - h2 = 0 everywhere in D, there is but one family of characteristics and the equation is of parabolic type. Laplace's equation r + t = 0 is of elliptic type; the equation of wave motions r - t = 0 is of hyperbolic type, and the equation of conduction of heat r - q = 0 is of parabolic type. There are also equations of mixed type; yr+t = 0 is of elliptic type when y > 0, but of hyperbolic type when y < o.

2.3]

CHARACTERISTICS

[29

Return now to equation (2). If y is a characteristic, (2) gives, in general, an infinite value for 8, and so there is no integral surface satisfying the prescribed conditions. But if the data on y satisfy ap'y' +bq'x' + Fx'y' = 0

(3)

we can assign 8 arbitrarily on y. This corresponds to the transport relation we found in the theory of linear systems of the first order. In fact, if (1) does not involve u explicitly, we can replace (1) by such a system, viz.

where F is linear in v and w, the second equation being the condition that vdx + wdy is an exact differential duo If y is a characteristic, there is no solution unless the data on y satisfy the transport condition (3); and, if the condition is satisfied, the solution is not unique. The simplest equation of hyperbolic type is the equation of wave motions r - t = O. The characteristics are given by y' = ± x', and so are straight lines x-y = constant and x+y = constant. The data on a characteristic must satisfy p'y' -q'x' = O. On x-y = constant, this shows that p - q is constant. Hence p-q = 2¢(x-y).

On x + y = constant, p

+ q is constant, and so p+q = 21fr(x+y).

Therefore p

= ¢(x-y)+1fr(x+y), q = -¢(x-y)+1fr(x+y),

and so

u = (x-y)+'¥(x+y),

the well-known solution. From this, the solution of the initial value problem u = f(x), U v = g(x) when y = 0, viz. U

= l{f(x-y)

+ f(x+y)}+l

f

X+V

x-v g(r)dr

follows. The problem of the vibrations of a string with fixed ends is to find the solution of r - t = Oin 0 ~ x ~ 1, y ~ 0 given thatu = f(x), U v = g(x) when y = 0, 0 ~ x ~ 1 and u = 0 when x = 0 or 1 and y ~ O. It can be discussed by the method of characteristics as in § 1.7. The problem of Cauchy is of little importance for equations of

[2.3

CHARACTERISTICS

30]

elliptic type. The fundamental problem for Laplace's equation is the problem of Dirichlet, to find a solution which takes given values on a simple closed regular curve C and which is analytic inside C. For example, if C is t,he circle x 2+ y2 = a 2, the solution which takes values cos 8 in polar coordinates on C and is analytic inside C is 'U = x/a; and the values of p and q on C follow. It is possible to ask whether there is a solution which satisfies Cauchy conditions on C. For example, if u = cos 8, ou/er = on C, the solution is r+t =

°

°

r

a

a

r

u = i-cos8+t-cos8.

This is regular except at the origin. This Cauchy problem has a solution regular near C, but no solution regular everywhere inside C.

2.4 The quasi-linear equation The quasi-linear equation of the second order is of the form Ar+2Hs+Bt+F = 0,

(1)

where A, H, B, F are analytic functions of five variables (x, y, v.,p, q) regular in some appropriate domain. Given a regular arc r in xyuspace, the theorem of Cauchy and Kowalewsky shows that there is a unique analytic solution, regular near r, on which p and q take given analytic values satisfying the strip condition du = pdx+qdy, provided that r is not a characteristic. Since the coefficients involve u, p and q, there is no unique family of characteristic base curves, so that the theory is more complicated than for the linear equation. On r, p' = rx' +sy', q' = sx' +ty', where dashes denote derivatives with respect to the arc length. Eliminating rand t from (1) we have (Ay'2 - 2Hx'y' + BX'2) s = Ap'y' + Bq'x' + Fx'y'

just as for the linear equation. Usually this determines s and so also rand t on r. But if Ay'2_2Hx'y'+Bx'2 = 0, (2) there is no solution unless Ap'y' + Bq'x' + Fx'y'

=

0.

(3)

If this last condition is satisfied, we can choose s arbitrarily and get an infinite member of solutions. A strip which satisfies equations (2) and (3) is called a characteristic strip. Suppose that A and B are not both zero. If they are, we make a

2.4]

[31

CHARACTERISTICS

linear change of variable. If A is zero, we interchange the parts played by x and y in what follows. Equation (2), regarded as a quadratic in y'lx', has two roots p and cr. If p and cr are complex, the equation is of elliptic type, and no characteristic strips exist. If p and cr are real and distinct, the equation is of hyperbolic type, and there are two families of characteristic strips. On the first family, y' =px', u' = px' + qy' , Ap'y' + Bq'x' + Fx'y' = 0.

Since B = pcrA, these equations become dy dx =p, du dx =p+qp,

(4)

dp dq A dx +Acr dx +F = 0.

The equations of the second system are obtained by interchanging p and cr. In equations (4), A, F, p and cr are functions of x, y, u, p and q. So we have three equations to determine y, u, p, q on a characteristic strip. If we put one of these variables equal to some arbitrary function of x, we have three equations to determine the other three. So the equations of a characteristic strip involve an arbitrary function. There are, in general, two families of characteristic strips. It can be shown that a surface generated by a one-parameter family of characteristic strips is an integral surface, and conversely. In practice, the method is complicated and of little use. We sometimes can make use of the differential equations of characteristic strips in a different way to find intermediate integrals, which are partial differential equations of the first order. The differential equations of one family of characteristic strips are dy-pdx

=

0,

du-pdx-qdy = 0, pAdp+Bdq+Fpdx

=

0.

If we can find functions a, fl, y such that a(dy-pdx) +fl(du-pdx-qdy) +y(Apdp +Bdq+Fpdx)

32]

CHARACTERISTICS

[2.4

is the exact differential dV of a function of five independent valuables x, y, U,p, q, the equation V = constant is called a first integral. It is a first order partial differential equation, and it can be shown that every solution of it satisfies Ar + 2Hs + Bt + F = O. If we can find two first integrals T~ = constant and T~ = constant, then
°

x 2 r-y 2t = O.

On any regular arc, and so

p' = rx' +sy', q' = sx' +ty', (X 2y'2 - y2x '2) S = x 2y'p' - y2x 'q'.

The characteristics are given by xy' = ± yx', and on any characteristic strip x 2y'p' _y2x 'q' = O.

If xy' = yx', y = kx is the equation of one family of characteristics, k being constant. The transport equation becomes

xp'-yq' = 0

or

p' -kq' = O.

Therefore p - kq is constant on y = kx, and so an intermediate integral is xp -yq = xf(yjx), where f is an arbitrary function. This is an equation of Lagrange's type; its general solution is U

= F(xy) +xO(yjx)

where P is an arbitrary function, and 0(0) is connected withf(O) by 0(0) - 200'(0) = f(O).

Thus, from a knowledge of an intermediate integral, we have found a general solution of x 2r-y 2t = 0 involving two arbitrary functions. The second. family of characteristics, given by xy' = -yx', is xy = k, for arty constant k. The transport equation is xp'+yq' = 0,

2.4]

[33

CHARACTERISTICS

satisfied when xy

= k. But

d(u-xp-yq) = (du-pdx-qdy) -(xdp+ydq) The first term on the right vanishes by the strip condition; the second term vanishes on a characteristic xy = k. Therefore u-xp-yq is constant on a characteristic xy = k; and so we have another intermediate integral u-xp-yq = g(xy), where q is an arbitrary function. This again is a first order equation of Lagrange's type. A better method of solving x 2r - y 2t = is to reduce the equation to normal form by a change of independent variables, as in the next section.

°

2.5 The normal form of a half-linear equation By a half-linear equation is meant an equation of the form L(u) == auxx + 2huxv + buyy +F

=

0,

(1 )

where F is a function of x, y, u and the first derivatives of u, but a, h, b depend only on x,y. An equation of this type can be reduced to a simple normal form by the use of characteristic variables. The transformation may be only a local one; for example YUxx + u yy = 0, which is of elliptic type in y > but hyperbolic type in y < 0, has different normal forms in the elliptic and hyperbolic half-planes. vVe suppose that a, h, b are analytic functions of x, y regular in a domain D and that L(u) = is of fixed type in D. Let

°

°

; = ;(x, y), 1J = 1J(x, y) be a bijection which maps D onto a domain~. Every point of D has a unique image in~; every point of ~ has a unique inverse image in D. We suppose also that; and 1J are real and have continuous second derivatives. The Jacobian J = 8(;,1J) 8(x, y)

does not vanish on D. This mapping gives L(u) == aUg + 2KU{" + flu"" + 20,

(2)

where 0 does not involve the second derivatives of u. A straightforward calculation shows that

[2.5

CHARACTERISTICS

34]

where Q is the quadratic form Q(X, Y) = aX2+2hXY +bP,

and TI is the polar form of Q, namely TI(X, Y;X', Y')

=

aXX'+h(XY'+X'Y)+bYY'.

At any fixed point (x, y) there, are three cases. If ab - h 2 > 0, the equation Q(X, Y) = ± 1, where the sign is + or - according as a > or a < 0, represents an ellipse in the XY-plane; the equation L(u) = is then of elliptic type at (x,y). If ab-h 2 < 0, Q(X, Y) = 1 is the equation of a hyperbola; and L(u) = is of hyperbolic type at (x, y). Lastly, if ab - h 2 = 0, aQ(X, Y) = 1 is the equation of a pair of parallel lines; and L(u) = is of parabolic type at (x, y). Suppose that L(u) = is of hyperbolic type at every point of some domain D. ",Ve can choose ~ and 1J so that a and fJ both vanish everywhere. Then ~ and 1J both satisfy the equation

°

°

°

°

°

a¢~ + 2h¢x ¢y + b¢~ = 0.

The curves ~ = constant and 11 = constant are therefore the characteristics of L(u) = since on them

°

ady 2 - 2hdydx + bdx 2 = 0.

(3)

Since a, h, b are analytic in D, so also are ~ and 1J. ",Vhen chosen in this way, we obtain the normal form K

8 2u O~(}rl +0 = 0,

~

and 1J are

(4)

the form of the original equation when a and b are both zero. If a is not zero, (3) gives dy dx a so that the curves ~ = constant and 1J = constant are nowhere tangent, and the Jacobian J does not vanish. If a is zero, ~ is equal to x, and the changes are straightforward. A different normal form is obtained by putting ~ + 11 = A, ~ -1] = fL in (4), namely,

If L(u) is of elliptic type in D, the functions complex functions. In equation (4) ,we put ~ +1J

~

=

and 1J are conjugate A, ~ -1J = iv, where

2.5]

[35

CHARACTERISTICS

A and v are real. This gives the normal form K

021£ 02n} {o/p + ov 2 + G =

°

for an equation of elliptic type with two independent variables. Lastly, if rib - h 2 = 0, we can have a = 0, h = 0, and L(l£) = comes bl£ yy +F = 0,

°be-

which is already of normal form. A similar result follows if b h = 0. If a, b, h are not all zero, equation (3) becomes ady-hdx

=

=

0,

0.

If this has solution 1J = constant, the coefficient j3 vanishes. If we take any analytic function 1J(x, y) such that the Jacobian .J is not zero, we have an(~x, ~y; 1Jx' 1Jy) = a~x(a1Jx + h1Jy) + h~y(a1Jx + h1Jy) = 0,

so that then

K

= 0. The normal form for an equation of parabolic type is al£~s+G = 0.

The linear equation with constant coefficients can be still further simplified. The transformation from (x,y) to (A,,u) in the hyperbolic case is a non-singular linear transformation, and the transformed equation becomes 1£"" -1£1'1' + 2gl£).. + 2fu" + el£ = 0, where g,j, e are constants. If we put 1£ = vexp ( - gA + l,u), we obtain

v,u-vl',,+(e+j2-g2)v = 0. And similarly for the equation of elliptic type.

2.6 The half-linear equation with three independent variables The half-linear equation ofthe second order with independent variables x, y, z is of the form auxx + buyy + CU zz + 2/1£yz + 2gl£xx + 2hl£xy + F = 0,

(1)

where F is a function of x, y, Z, 1£ and the first derivatives of 1£, but a, b, e,j, g, h depend only on x, y and z. The problem of Cauchy for thi!! . equation is to show that, under appropriate conditions, there is a unique solution of (1) for which 1£ and its normal derivative ol£/oN take given values on a given regular cape. t If the coefficients in (1) t

See Note 6.

36]

CHARACTERISTICS

[2.6

and the data on C are analytic functions regular in a neighbourhood of every point of C, the proof of the existence of the solution goes very like that for two independent variables; we do not go into this in detail. The proof depends on showing that, in a neighbourhood of any point of C, a formal Taylor series satisfying the equation can usually be constructed and its convergence proved by the method of dominant functions. Let N be the unit vector normal to C. Since u and au/oN are given on C, all the first derivatives of u are known on C. Now, if w is a scalar, gradw can be resolved into a component N.gradw in the direction of Nand N x grad w in a direction perpendicular to N. Hence if w is known on C, N x grad w, being a tangential derivative, is known on C. In particular, the three vectors N x grad u x '

N x grad u y,

N x grad U z

are known on C. They are not independent, since each is perpendicular to N. In this way, we obtain nine expressions, each linear in the second derivatives of u, which are lmown on C. If (l, m, n) are the components of N, we can choose five of these expressions which are linearly independent. Such a set is nuxx -luxz =

v

luyy - muXY = <1>2' muzz - nuyZ = <1>3' muxz - nuxy luyz - muzx where

=

<1>4'

=

<1>5'

(2)

auxx + buyy + cUzz + 2fuyz + 2guzx + 2huxy = - F.

The expressions on the right are all known on C. Hence we can find all the second derivatives on C, provided that the determinant a n 0 0 0 0

b 0

2f 2g 2h 0 -l 0 1 0 0 0 -m 0 m -nO 0 0 0 0 1)1, -n 0 0 1 -m 0 C

0

is not zero. This determinant is equal to lmn(al 2 + bm 2 + cn 2 + 2fmn + 2gnl + 2hlm).

CHARACTERISTICS [37 2.6] If 1mn is not zero, all the second derivatives of u are determined on the cap C, provided that the quadratic form

Q = a1 2 + bm 2 + cn 2 + 2fmn + 2gm1 + 2h1m does not vanish. This condition is also sufficient in the exceptional case 1mn = 0. If 1 = m = 0, n = 1, C lies on a plane z = constant. The data determine all the second derivatives except u zz ; the differential equation then gives u zz provided that c 4= 0. If n = 0, 1m 4= 0, u xz' u yZ' U zz are all determined on C. We then have

1uyy - muxy = <1>2' 1uxv - muxx = <1>6' which determine U xx' U yy , uxvprovided thata1 2 +bm2 + 2h1m 4= 0. Thus, in any case, unless Q = 0, the second derivatives of u are determined onC. We could go on to discuss higher derivatives, and prove the CauchyKowalewsky theorem for the half-linear equation. But what happens when Q = is more interesting; for if Q = everywhere on C, the problem proposed does not have a unique solution. C is said to be a characteristic. Suppose that the equation of Cis ¢(x, y, z) = 0. Since

°

°

l:m:n

=

¢x:¢y:¢z'

we have the partial differential equation a¢~+b¢~+c¢~+2f¢y¢z+2g¢z¢x+2h¢x¢y = 0,

(3)

satisfied by the characteristic surfaces of (1). Since the coefficients were assumed to be functions of x, y, z alone in the semi-linear equation (1), the family of characteristics is independent of the data in the Cauchy problem. We can arrive at the condition (3) in a somewhat different way. We can ask what conditions ¢ must satisfy if the problem of Cauchy does not have a unique solution with data on the surface ¢(x, y, z) = 0. In other words, can there exist two solutions u = ul(x,y,z) and u = u 2 (x, y, z) of (1) with the properties

on ¢ = o?

8ul

8u 2

8z

8z

[2.6

CHARACTERISTICS

38]

'\Trite u l

u2

-

= V,

so that v and its first derivatives vanish on

9 = 0. Since F has the same values for U I and U 2 on 9 = 0, (4)

there. Let y be a regular arc x = x(t), Y = y(t),

89. 89. 89;, _

8x x+ 8y y+ 8z Since 8vJ8x vanishes on

w

-

Z

= z(t) on 9 = 0; then

° •

9 = 0,

z, we have

Eliminating

(9x'cxz-9zvxx)X+(9yVxz-9zVxy)if = 0. But since y is an arbitrary regular are, we can choose X, if as we please and so Hence on

vxx:vxy:vxz

9 = 0. Similarl:,-

Therefore on

9 = 0,

•V • •• • V xx· yyoL ZZ ·

=

9x:9y:9z

Vyx:Vyy:VyZ = 9x:9y:9z' vzx:VZy:vzz = 9x:9y:9z· v yzovzx· .•, . vxy

_,1,2 • ,1,2 • ,1,2.,1, ,I, .,1, ,I, .,1, ,I, ¥'X"¥'y"¥'z·Yy't'Z"¥'Z¥'x·Yx't'y·

-

Hence, by (4)

a9;,+b¢;+C9;+2j¢y9z+2g9z9x+2h9x9y = 0,

°

so that if; = is a characteristic. If we solve if;(x, y, z) = for z, and if we denote the partial derivatives of Z by p and q, the differential equation for the characteristics becomes ap2+bq2+C- 2fq- 2gp+ 2hpq = 0. (5)

°

The characteristics can be built up from the characteristics of this first order equation, as we saw in eh. 1. The characteristics of (5) are called the bicharacteristics of equation (1).

2.7

The half-linear equation in general The half-linear equation with n independent variables (Xl. x 2 ' is of the form n 82u L: aij 8-;::- +F = 0, I xiox j

••• ,

x n ), (1)

2.7]

CHARACTERISTICS

[39

where a ij are functions of the independent variables alone, but F can also involve U and its first derivatiyes; it can be treated by the method of §2.6. Its characteristics satisfy the first-order equation n 8¢ 8¢ L:I a ij uX ;- " = 0. uX i

(2)

j

Consider the quadratic form

where the coefficients are evaluated at any fixed point Po. By a suitable non-singular linear transformation, we can express Q as a sum of squares m Q = L:JLi1J~, (3) i=l

where m ~ n. The number m does not depend on the particular linear transformation. If m is less than n, the equation (1) is said to be of parabolic type at Po. If m = n, the number of positive coefficients fLi does not depend on the particular linear transformation. If the transformation is orthogonal, the fLi are the latent roots (eigenvalues) of the matrix (a ij ). If all the coefficients fLi are of the same sign, (1) is said to be of elliptic type at -Po. An equation which is everywhere of elliptic type has no characteristics. If all but one of the fLi are of the same sign, the equation is said to be of hyperbolic type at Po. There are intermediate cases, such as Uxx+Uyy-Uzz-Utt

= 0,

which are of neither type; about these, little is known.

2.8

The half-linear equation with constant coefficients The half-linear equation with n independent variables (Xl' X 2 ' ••• , x n )

is of the form

n

82u

I

UXiUXj

L:aij~+F

= 0,

where F is a function of the independent variables and possibly of U and its first derivatives. If we introduce new independent variables £i

= £i(X I , x 2 '

••• ,

xn )

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

and try to choose the functions £i so that the resulting equation is of the form n 82u ~bi 8£;+0 = 0,

40]

CHARACTERISTICS

[2.8

we find that the n functions ~i have to satisfy tn(n-1) differential equations of the first order. This is, in general, impossible if n > 3; if n = 3, there are just enough conditions to determine the functions ~;. There are special cases when it is possible to make a transformation of this kind when n > 3; a particular case is when the coefficients a ij are constants. ",",Vhen the coefficients are constants, the differential operator

L(u)

=

13 2u

2:a·.-t] 13x 13x i

j

can be written in vector form

L(u) = STASu, where A is the real symmetric matrix (a ij ), ST is the row-vector operator

and S is the transpose of ST. Let x' = M x be a real non-singular linear transformation from the variables (X V x 2 , •.. ,xn ) to (x~,x~, ... ,x~), using the obvious vector notation. Then where S' is the transpose of the row-vector operator (S' ) T -_

(

8

8

uX!

uX 2

8 )

",",,,,", ... '<'l!, uXn

and MT is the transpose of M. Then L(u) = (S')TMAMTS'U = (S')TBS'u,

B=MAMT.

where

Since A is a real symmetric matrix, we can find an orthogonal matrix Jl so that JlAMT is a diagonal matrix. The diagonal elements of the resulting matrix B will be the non-zero latent roots !Lv !L2' ... ,!Lr of A, where r is the rank of A, and n- r zero elements if r < n; all the latent roots are real. "We have thus shown that

where each

Vi

is ± 1.

2.8]

[41

CHARACTERISTICS

This reduction of a half-linear equation with constant coefficients can be done in an elementary way by using Hermite's reduction of a quadratic form to a sum of squares, which is merely the process of 'completing the square', instead of an orthogonal transformation. This is illustrated below. Suppose that L(u)

= U xx

+ 3~tVY + 84uzz + 28uyZ + 16uzx + 2uXY

= (81 + 38~ + 848~ + 2882 83 + 1683 81 + 281 82 ) U, where 81' 82, 83 are the three components of the column-vector operator S. Now 81 + 38~ + 848~ + 2882 83 + 1683 81 + 281 82

= (81 + 82 + 883 )2 + 28~ + 1282 83 + 2085 = (81 + 82 + 883)2 + 2( 82 + 383 )2 + 28~

where

8~ =

83 •

If the linear transformation is x = Nx', then S' = NTS and so

NT~ [~

;l

1 1

0

Therefore

x = x',

y = x'+y',

z = 8x' + 3y' +z',

or

x' =x,

y' = y-x,

z' = -5x-3y+z.

With this transformation 82u 82u 82u L(u) = uX '" '2 + 2 <'li2 + 2 uX '" '2. uX 1

2

3

The process has to be modified slightly if the associated quadratic form contains no squares. For example, if (Lu) = 4uxy+2uyz +2uzx '

the quadratic form is

481 82 + 282 83 + 283 81 •

42]

[2.8

CHARACTERISTICS

If we put 81 = rx- (3,8 2 = rx+ (3, this becomes 4rx2 - 4(32 + 41X83 =

(2rx + 83 )2- 4(32-

=

(81 + 82+ 83 )2 - (81 - 82)2 -

8~

8~

= 8?- 8;2- 8~2, 8~

where

= 81 + 82+ 83 , 8; = 81 - 82,

x=x'+y',

Then

y=x'-y',

8~

= 83 .

z=x'+z'.

In terms of these new variables, 82u 82u 82u L(u) = -;r2-~-~. uX 0y oz

Exercises 1. Find the types ofthe following differential equations, and reduce them to normal form: (i) y2 r -t = 0, (ii) y2 r +2ys+t = p, (iii) y2 r + t = 0, (iv) r-2s+3t-q-u = 0, (v) (1+X 2)2r -t = 0, (vi) x 2 r-2xys+y2t = 0, (vii) x 2r+2xys+y 2t = 0.

2. Reduce the following equations to normal form: 82u 82u 82u 82u 82u 82u 82u (i) <">2+2<">2+3~ +4"'2 +2 ;-;:-+2~+2~ uX uy uZ ut uxoy OXuz uxut 82u 82u 82 u + 4 -;:-;::- + 4 -;:-;::- + 6 '" '" oyoz oyot uxut .. 8x2 82u 82u 82 u 82 u cPu (11) <">2+-;-2+2~+2~+2~+2~ uX uZ uxoy uXuZ uXOt ozut ...

82u uX

82u oxoy

82 u uXuz

= 0.

= 0.

82u uXut

(lll) <">2 + ;::-;::- + ~ + ~ = 0.

3. Prove that

has an intermediate integral p = qf(u), where f is an arbitrary function. Hence show that y+xf(u) = g(u), where g is a second arbitrary function.

CHARACTERISTICS

[43

4. Transform the equation 82u 82u x2- =y2_ 8x 2

8y 2

into one with characteristic variables. Hence show that

u

f(xy) +xg (~),

=

where f and g are arbitrary functions. 5. Find the characteristics of r -yt form when y > 0.

= 0. Reduce this equation to normal

6. Find the characteristics of (1+x 2 )r-(1+y2)t = 0, and reduce the equation to normal form. 7. Prove that x 2r+2xys+y 2t

=

°has an intermediate integral

px+qy-U+f(~) =

0,

where f is an arbitrary function. Hence show that

where g is also an arbitrary function. 8. Show that qr+ (uq-p) s-upt

°

= has an intermediate integral

p+qu =f(u), where f is an arbitrary function. Hence solve the !lquation. 9. Prove that the two families of characteristic strips of

s = F(x,y,u,p,q) are given by (i) dx = 0,

du-qdy = 0,

dp-Fdy = 0,

and (ii) dy = 0,

du-pdx = 0,

Hence solve the equation s

=

dq-Fdx = 0. pq.

3

BOUNDARY VALUE AND INITIAL VALUE PROBLEMS

3.1 Laplace's equation The theorem of Cauchy and Kowalewsky states that, under certain conditions, a second-order partial differential equation has a unique solution. But this result is of little importance in many problems of mathematical physics. Firstly, the solution in the CauchyKowalewsky theorem is only a local solution, valid near the manifold carrying the Cauchy type data; whereas a global solution is often needed. Secondly, the Cauchy type data are analytic functions, not necessarily the case in practice. Thirdly, in some problems, we have too much data for the existence of a global solution. In short, are we asking the right question? The simplest equation of elliptic type is Laplace's equation 2

'V u

_

82u ox

82u uy

82u_ - 0. oz

= ~+"2+~

It can be shown that there is but one solution which is regular in the sphere r ~ a, in spherical polar coordinates, and which takes the value cose on r = a, namely (ria) cose. But there isno solution, regular in r ~ a, which satisfies the Cauchy conditions

u = cose,

on r

=

au

°

Or = '

-

a. There is the solution u

2a

1 a2

= -3 -r cos e+-3 -r 2 cos e,

which is regular when r > 0, but has a singularity at the origin. Thus, when we ask for a solution regular in r ~ G, and satisfying Cauchy conditions on r = a, we are asking the wrong question. The right questions are suggested by physical problems. Let D be a spherical cavity inside an earthed perfectly conducting body; let S be the boundary of D. If there is a point charge e at a point Po inside S, a charge is induced on S, and the total electrostatic potential in D is

3.1]

BOUNDARY AND INITIAL VALUE PROBLEMS

[45

where R ois the distance from Po. The function u, which is the potential of the induced charge, has no singularities in D and satisfies Laplace's equation. On S, U = -ejRo. It is obvious to the physicist that such a potential u is uniquely determined; and a knowledge of u determines the normal derivative ofu on S. We cannot assign the normal derivative arbitrarily. The problem of finding u is a particular case of the problem of Dirichlet, a boundary value problem. Next let D be an earthed perfectly conducting spherical conductor with surface S. If there is a point charge at a point Po outside S, a charge is induced on S, and the total electrostatic potential outside Sis e 1> = u+ R' o

where R o is the distance from ~. The potential u of the induced charge satisfies Laplace's equation, has no singularities outside Sand has a gradient which vanishes at infinity. On S, u = -ejRo. Again it is 'obvious' that u is uniquely determined. Finding u is a particular case of the external problem of Dirichlet. Laplace's equation is also satisfied by the velocity potential defining the irrotational motion of an incompressible perfect fluid. Suppose that the motion of such a fluid in a spherical cavity D with rigid wall S is caused for a source m at ~ and an equal sink at Pl. If the distances of a point of the cavity from Po and PI are R o and R I respectively, the velocity potential is

m m 1>=u+---

Ro R I '

where u is a solution of Laplace's equation with no singularity in D. Since there is no flow across S, the normal velocity 81>j8N vanishes on S. Hence u satisfies the condition

:; =

8~ (;: - ;:)

on S. Obviously u is uniquely determined, up to an additive constant. This is a particular case of the problem of Neumann. Lastly suppose that we have a rigid spherical obstacle D fixed in a perfect fluid whose flow at infinity is uniform with velocity V. The velocity potential outside D is

1> = u+ Vx. The function u is a solution of Laplace's equation with no singularity

46]

BOUNDARY AND INITIAL VALUE PROBLEMS

[3.1

outside D. On the boundary of D, 8u 8 8N = - 8N(Vx)

and at infinity grad u vanishes. This is a particular case of the external problem of Neumann. Naturally, in all these problems, there is no special merit in considering only spherical boundaries except that it saves a lengthy explanation of the nature of the boundary. The important problems for Laplace's equations are thus boundary value problems, in "Which only one condition is satisfied on the boundary.

3.2 The equation of wave motions The simplest equation of hyperbolic type is the equation of "Wave motions 2 Y'2 Zt

1 8u

= c2 <'l2' ut

where c is a constant. For example, suppose that we have an infinitely long uniform string of density p, which is taut with tension T. For small vibrations, the displacement u at the point of abscissa x at time t satisfies 82u 1 82u 2 8x = C2 8t 2 '

(1)

where The displacement at any instant is determined if we are given the initial displacement and initial velocity of each point of the string, that is, if we are given u(x,O)

= (x), ut(x,O) = ':Y(x).

In this case, the data are of Cauchy type, and the Cauchy-Kowalewsky theorem shows that there is a unique solution analytic near every point of the x-axis if the data are analytic. But there is no need to assume analytic data. In terms of the characteristic variables; = x + ct, 1J = x - ct, equation (1) is Hence we have u = F(;) + G(1J) = F(x+ct)+G(x-ct) so that the general solution represents the sum of two disturbances

3.2]

BOUNDARY AND INITIAL VALUE PROBLEMS

[47

propagated with velocities ± c. If we use the initial conditions, we get the well-known solution u = t(x-ct)

1 fx+ct

+ !(x+ ct) +-2

c x-ct

'P'(r)dr

when t ~ 0. This solution has continuous derivatives of the second order, provided that "(r) and 'P"(r) are continuous. In some physical problems, the data are not everywhere continuously differentiable. For example, if (x)

= max (I-lxi, 0), 'P'(x) = 0,

the graph of u for any fixed value of t (~ 0) consists of straight segments. At the corners, the partial differential equation does not hold; u is not differentiable. Suppose next that the string is 'stopped' at x = ± I. The motion of the portion of the string when Ixl ~ I then satisfies the conditions

= (x), ut(x,O) = 'P'(x), u(l, t) = 0, u( -I, t) = 0, where <1>( ± I) = 0, 'P'( ± I) = 0. This is a mixed problem; initial conditions have to be satisfied on t = 0, boundary conditions on x = ± I. u(x,O)

The equation

82u

82u y

1 82u

+-=-8x 2 8 2 c2 8t 2 occurs in the theory of small vibrations of a uniform stretched membrane. If the membrane covers the whole xy-plane, the relevant problem is an initial value problem. But if the membrane is like the top of a drum, we have to solve a mixed problem - an initial value problem and a boundary value problem resulting from the fact that the membrane is fixed at its rim. The equation 82u 82u 82u 1 82u + - + - =c2-8t-2 8x 2 8y2 8z 2 appears in the theory of sound waves of small amplitude, and in other branches of applied mathematics.

3.3 Characteristics as wave fronts The characteristics of 82u 82u 1 82u

+-=-8x 2 8y 2 c2 8t 2

(1)

satisfy the first-order equation p2+ q2 =

1

_

c2 '

(2)

48]

BOUNDARY AND INITIAL VALUE PROBLEMS

[3.3

where p and q denote ot/ox and ot/oy respectively. The characteristics of (2) are called the bicharacteristics of (1) and satisfy the LagrangeCharpit equations dx ds = pc,

dy ds = qc,

dt ds

dp ds = 0,

1

c'

dq ds

=

°'

where s is a parametric variable. Hence on a bicharacteristic strip, P=

where

IX

1

Ccos IX,

1 .

q = - sIn IX, c

is a constant, and x = xo+SCOSIX,

Y = Yo+s sin IX,

ct = cto+s.

The bicharacteristic lines are straight lines making a constant angle cos-1 1!-J(1 + c2 ) with Ot; the bicharacteristic strips lie on planes (x-x o) COSIX+(y-Yo) sin IX = c(t-to). (3) Any characteristic of (1) is the envelope of a one-parametric family of such planes x cos IX +Y sin IX = ct + f(IX).

If x o, Yo, to are constants, the envelope of (3) is the characteristic cone Similarly, for

o2u 02U 02U 1 02U -2 + - + - = 2- ox oy2 OZ2 c ot 2 '

(4)

the characteristic cone with vertex (x o, Yo, zo, to) is (X-X O)2+(Y_YO)2+(Z-ZO)2 = c2(t-t O)2, the light-cone of special relativity. If we transform (4) to spherical polar coordinates (r, e, 1», a solution independent of e and 1> satisfies 02U 20u

or 2 +;; or and, if we put U =

1 o2u

= C2 8f2:

vir, we obtain the one-dimensional wave equation 02V

or 2

1 02V

= C2 ot 2 •

Hence a solution of (4) which represents expanding waves with spherical symmetry is of the form 1

U

= - f(ct- r). r

3.3]

BOUNDARY AND INITIAL VALUE PROBLEMS

Let us suppose thatf(r) andf'(r) vanish when r 1

U

= - f(ct-r),

r

~

ct;

= 0,

r

~

ct,

r

~

[49

O. Then

is the velocity potential of sound waves of small amplitude due to a source at the origin expanding into the gas at rest. The disturbance at a point distance r from the origin is zero until the instant t = ric. The plane sections t = constant of the characteristic cone x 2+ y2 + Z2 = c2t2 give the successive positions of the moving wave front. :More generally, a disturbance with wave front f(x, y, z) = t propagated into the gas initially at rest is specified by a velocity potential u which is zero on one side of the wave front and non-zero on the other. If the motion is not a shock wave, u and its first derivatives vanish on the wave front. Hence we have two solutions, U = U 1 representing the propagated waves and U = 0 representing the state of rest, which are equal and have equal first derivatives on the wave front, which is therefore a characteristic. Suppose that the normal at (x, y, z) onf(x, y, z) = t cutsf(x, y, z) = t' where t' > tat (x',y',z'). If the distance from (x,y,z) to (x',y',z') is J,

f(x+lJ,y+mJ,z+nJ) = t', where (l,m,n) are the direction cosines of the normal. Then

t' -t =f(x+lJ,y+mJ,z+nJ)-f(x,y,z) = J{lfx + mfy + nfz}, where fx,fy,fz are evaluated at (x + etJ, y + 8mJ, z + 8nJ) where 0<8<1.

Hence

lim.,i- = 1 . t'-+It -t lfx(X,y,z) +mfy(x,y,z) +nfz(x,y,z)

But

l:m:n =fx:fy:fz,

where

2 f2 f2 _ 1 f x+ y+ z - C2'

sincef(x,y,z) = t is a characteristic of (4). Thereforefx = lie, etc. and so J . 11m - - = c 1'-+1 t' -t . The wave front is therefore advancing with normal velocity c.

50]

BOUNDARY AND INITIAL VALUE PROBLEMS

[3.4

3.4

The equation of telegraphy If a uniform telegraph cable has self-inductance L, resistance R and capacity C, all per unit length, the electric potential V and the current i at a distance ::e from an end satisfy the equations 8V 8x

=

-L~ -Ri

8t'

Eliminating i, "We have Heaviside's equation of telegraphy 182V

C 8x2 =

82V 8V L 8t2 + Rat·

The physical problem to be solved for this equation is a mixed one. 'We are given V and 8Vj8t when t = 0, x ~ 0 and also we are given V at the end point x = 0 for all t.

3.5 The equation of heat The best known example of an equation of parabolic type is the diffusion equation, the equation governing the conduction of heat in a uniform solid of conductivity k, specific heat c and density p. The temperature u satisfies 82u

82u

82u

1 8u

+-+-=-8x 2 8y2 8z 2 a 2 8t

(a

2 =

!:....) cp ,

when there is no internal source of heat. If the body fills all space, a knowledge of the initial temperature distribution determines the temperature everywhere at any subsequent instant, assuming that the temperature at infinity remains finite as it obviously must. If, however, the body in which heat is flowing is bounded, there are also boundary conditions to be satisfied. There are three forms of boundary condition: (i) The temperature on the boundary may be given. This may be constant or a given function of the time. (ii) The rate of flow of heat, k8uj8N may be given. If there is no flow of heat, 8uj8...Y = O. (iii) The rate of flow of heat at the surface may be proportional to the difference between the temperature at the boundary and the temperature U o (not necessarily constant) of the medium in which the

3.5]

[51

BOUNDARY AND INITIAL VALUE PROBLEMS

body is immersed. Then on the boundary ou k oN

= -h(u-uo),

when %N denotes differentiation along the outward normal and h is non-negative. There are various particular cases. For a uniform straight rod whose surface is thermally insulated, the equation of heat reduces to a

202U

OU

ox

ot'

- 2= -

provided that the temperature is uniform over every cross section. But if the rod loses heat from its surface at a rate proportional to the temperature, the conduction of heat is governed by 202u _ OU

a ox 2

-

ot + vu,

where v is a positive constant depending on the size, shape and material of the rod. If the telegraph cable of § 3.4 has negligible self-inductance the equation of telegraphy reduces to the diffusion equation 02V

ox 2

=

RCoV

ot'

°

but now the data are simply that V is known when x ~ 0, t = and when x = 0, t ~ 0. We do not need to assign oV/ot on t = 0; it is determined from the differential equation.

3.6 Well-posed problems It is essential to know whether a boundary value or initial value problem has a solution and whether the solution is unique. But more than this is needed in mathematical physics. The data in a physical problem, being derived from experiment, are necessarily approximate. If the problem is to have any physical meaning, a small change in the data must give rise to a small change in the solution. Again, the numerical analyst, solving such a problem numerically, is forced to approximate his data in order to apply numerical methods. It is not enough to know that the problem has a unique solution. It must also be known that the process of making an approximation to the data produces only a slight change in the solution.

52]

BOUNDARY AND INITIAL VALUE PROBLEMS

[3.6

The same difficulty arises in connexion with the Cauchy-Kowalewsky theorem, in which the initial data are analytic. 'We can approximate to any continuous function as accurately as we please by polynomials. So we could apply the Cauchy-Kowalewsky theorem with continuous data by using polynomial approximations only if a small change in the analytic data produces a small change in the solution. An initial value problem, boundary value problem or mixed problem which has a unique solution is said to be well posed if a small change in the data produces a small change in the solution. If (x), '(x), '¥(x) are differentiable, the initial value problem 82u

82u

8x Z = 8t Z ' u(x, 0) = (x),

where

ut(x,O) = '¥(x)

has the d'Alembert solution u = !{(x+t) + (x-t)}+'21 JX-H . '¥(r) dr. x-I

If we replace by + 8, '¥ by '¥ + 8'¥ on an interval (a, b) where 18<1>1 < 10, 18'f"1 < 10, the absolute value of the change in u is less than €+t€(b-a). Hence the problem is well posed. Next suppose thatJ(8) is cont,inuous. Then the problem of Dirichlet for

u(R cos8,Rsin8) =J(8)

with

has a unique solution, regular in formula u(rcos8,rsin8)

=

1

-2 7T

J

2 "

0

X

(0

~

8

~

27T),

Z+ yZ < RZ, given by Poisson's

RZ-r z RZ - 2R'1 cos (8 - ¢) +r zJ(¢)d¢.

If we replace J by J + 8J, the absolute value of the change in u is less than or equal to 1 27T

J 2

SUP

18J(¢)1

0

"

R2_ r Z RZ-2Rrcos(8-¢)+r zd ¢

= sup 18J(9)1. Hence if 18J(9)1 < 10, the change in the absolute value of the solution u is less than 10. This problem is thus well posed.

3.6]

BOUNDARY AND INITIAL VALUE PROBLEMS

[53

The problem of Cauchy for 82u 82u 8x 2 + 8y 2

=0

with data on y = 0 is not well posed. If U (x,O)

then

= 0, =

u(x, y)

uy(x,O)

1 . smnx,

=-

n

~ sin nxsinh ny.

n

The data can be made as small as we please by choosing n large enough. But if y =l= 0, the solution is unbounded as n -+00. Now consider the two Cauchy initial value problems: (i)

u1(x,0) = x 2 ,

=

x 2,

8u 1(x,0)j8y = 0;

(ii)

u 2 (x, 0)

Then

U1

= x 2_y2,

u2

=

X2 _

8u 2 (x, 0)j8y

y 2+

\

n

= ~ sin nx.

sin nx sinh ny.

Although the data for u 2 tend to the data for tend to U 1 •

U1

as n-+oo, u 2 does not

4

EQUATIONS OF HYPERBOLIC TYPE

4.1 The half-linear equation of hyperbolic type The general half-linear equation with two independent variables is of the form ar+ 2hs+bt = J(x, y, u,p, q), where a, h, b are functions of x and y alone. As we saw in §2.5, if this equation is of hyperbolic type, it can be reduced to the normal form 8

=J(x,y,u,p,q)

(1)

by using characteristic variables, though the transformation to characteristic variables may be only a local one. The characteristics are x = constant on which du dy = q,

dp -=J, dy

and y = constant on which du dx =p,

dq =J

dx

.

In particular, ifJ is identically zero, p is constant on each characteristic x = constant and q is constant on each characteristic y = constant. If we rotate the axes in (1) so that the characteristics are x ± y = constant,

we obtain a second normal form r-

t = J(x, y, u,p, q).

(2)

If J is identically zero, p + q is constant on every characteristic x + y = constant, p - q is constant on every characteristic x - y = constant.

4.2 The equation U XY = 0 The general solution of u xy = 0 is U = F(x) + G(y), so that p is constant on every characteristic x = oonstant, q is constant on every characteristic y = constant. If u has continuous second derivatives, F" and G" are continuous. [54 ]

4.2]

EQUATIONS OF HYBERPOLIC TYPE

[55

The problem of Cauchy cannot be solved with Cauchy data on a characteristic, for if we try to solve the Cauchy problem with data u = J(X),

p

=:

on a characteristic y = b, where have to find F and G to satisfy F(x)+G(b) =J(x),

g(X),

q = h(x),

l'

by the strip condition, we

= g

F'(x) = g(x) =1'(x),

G'(b) = h(x).

The third equation is impossible for arbitrary h(x); it can only be satisfied if h(x) is a constant C and G(y) = Cy. The solution then is u = J(x) + C(y-b).

In what is called the characteristic boundary value problem, u is given on two characteristics x = a and y = b. On x = a, U = (y); on y = b, u = '¥(x) where '¥(a) =
= '¥(x) + (y) -

'Y(a).

The most interesting case of the problem of Cauchy for U XY = 0 is that which corresponds to the initial value problem for the wave equation; the Cauchy data u = J(x),

are given on the line x + y =

p = g(x),

q = h(x),

o. The strip condition is then

1'(x) = y(x) - h(x).

It is readily verified that the solution is

u = J(x) + J:y h(r) dr.

In order that this solution may have continuous derivatives of the second order, it is necessary and sufficient that 1", g' and hi be continuous; and p = g(x), q = h( -y) everywhere. If, however, h(x) has a finite jump at x = a, so that the limits h(a + 0) and h(a - 0) exist but are not equal, then q(x, -a+e)-q(x, -a-c) = h(a-e)-h(a+e)

does not tend to zero as e-+O. Hence q(x,y) is discontinuous across y = -a. A discontinuity in the given value of q at (a, -a) persists across the whole characteristic y = - a. Similarly a discontinuity in

56]

EQUATIONS OF HYPERBOLIC TYPE

[4.2

the given value ofp at (a, -a) is propagated along the whole characteristic x = a. Next consider the case when 'U

= f(x), p = g(x), q = h(x)

on x+y = 0, x> 0, wherej'

g-h, and

=

u = F(x),

p = G(x),

q = H(x)

on x+y = 0, x < 0, where F' = G-H. Suppose thatj",g',h' are continuous in x ~ 0, and that F", G', H' are continuous in x:( O. \Ve assume in addition that f( + 0) = F( - 0); for, if not, the solution u, if it existed, would be discontinuous at the origin. The charact€ristics x = 0, y = 0 divide the half-plane x + y ~ 0 into three regions D I , D 2 , D s where (x~O,y:(O),

respectively. In D lI

11

(x:(O,y~O),

=

(x~O,y~O)

f(x) +fx h(7) d7. -y

u = F(x) + f:y H(7) d7.

In D 2 , Hence when x

~

0, y

= 0, u(x, 0) =j(x)+ f; h(7)d7,

but when x = O,y > 0, u(O,y) = F(O)+ f~y H(7)d7.

To get u in D s, we have to solve a characteristic boundary value problem, the data being continuous sincef(O) = F(O). The solution is u(x,y) =f(X)+fxh(7)d7+fO H(7)d7, o -y

which is just what we should have got if we had forgotten that the given values of q on x+y = 0 are not necessarily continuous at the origin. The first derivatives are given by p

=

j'(x) +h(x),

q = h( -y),

inD lI

p

=

F'(x) +H(x),

q == H( -y),

inD 2 ,

q == H( -y),

inDs'

P = j'(x) + h(x),

4.2]

[57

EQUATIONS OF HYPERBOLIC TYPE

In D 1 UD a, P is continuous, but q is discontinuous across the axis of x, the jump being H(O) -h(O). In D 2 UD a, q is continuous, but p is discontinuous across the axis of y, the jump being 17(0) - G(O). If, however, G(O) = g(O), H(O) = h(O), the first derivatives are continuous in the half plane x+y > O. If the first derivatives are continuous in x + y > 0, we have in D1UDa r = F"(x) +H'(x) = G'(x) in D 2 • Hence, if g'(O) ,.. G'(O), r is discontinuous across Oy in the half-plane x+y> o. A discontinuity in the data for p produces a discontinuity in r along Oy. Similarly a discontinuity in the data for q produces a discontinuity in t along Ox. The argument can be extended by a change of variable to the case when the Cauchy data are given on a curve y = ¢(x), where ¢(x) is strictly monotonic and is such that no characteristic touches y = ¢(x) or cuts it in more than one point. Such a curve is said to be duly r=!,,(x)+h'(x)=g'(x)

inclined.

4.3 The uniqueness theorem for u xy = 0 In § 4.2 we found a solution of u xy == 0 satisfying the Cauchy conditions U = f(x), p == g(x), q = h(x) on x+y = 0, where!, = g-h. If AB is a closed segment ofx+y = 0 on which!", g', h' are continuous, the solution we found has continuous second derivatives on the square with AB as diagonal. If there were two such solutions U 1 and U 2 , U = U 1 - U 2 would also be a solution of Uxy = 0 satisfying the conditions U = P = q = 0 on AB. To prove uniqueness, we have to show that U 1 - u 2 is identically zero on the square. Suppose, then, that U is a solution of Uxy = 0 with zero Cauchy data on the segment AB of x+ y = 0, and that U has continuous second derivatives on the squal'e with AB as diagonal. Let the characteristics through any point P(x, y) of the square cut AB in Q( - y, y) and R(x, - x). If D is the triangle PQR, r its perimeter, we have, by Green's theorem given in Note 5 of the Appendix,

fr

(lu;+muVds

= 2 ff»

(u'I U 'Ig+ u gu g'l)dgd1]

= 0,

land m being the direction cosines of the outward normal to Therefore _y u~(g, y) dg +, -x u;(x, 1]) d1] = 0

fX

3

IY

r.

58]

[4.3

EQUATIONS OF HYPERBOLIC TYPE

since the first derivatives of 11 vanish on QR. Hence uf,(;,y) = O,(-y <; < x);

un(x,y) = O,(-x < 17 < y).

It follows that P and q are zr:,ro everywhere on the square, and so U is constant. But 11 vanishes on AB. Hence U is identically zero on the square.

4.4

The Cauchy problem for the half-linear equation of hyperbolic type The Cauchy-Kowalewsky theorem asserts that the half-linear equation 8=F(x,y,u,p,q) (1)

has a unique solution analytic in a neighbourhood of a duly-inclined arc carrying the anal~ytic Cauchy data. \Ve now turn to a case when the data are not analytic; suppose that on the line x+y = we are given that U = f(x), P = g(x), q = h(x),

°

where the derivativesf", g' and h' are continuous, and!, the strip condition. If F == 0, so that the equation is 8 = 0, the solution is U

=

g-h by

o=f(x)+ f:y h(7)d7.

If we put U = U o + U in (1) we obtain an equation of the same form, but U, u~ vanish on x+y = 0. Thus we need only consider equation (1) in the case when

ux,

U

= 0, P = 0, q =

°

on x+y = 0. We show that this problem has a unique solution on any square jxj :( y, jyj :( y provided that F is a continuous function of (x,y,u,p,q) when jxj :( y, jyj :( y,U 2+p2+ q2:( 02, where it satisfies a Lipschitz condition, jF(x, y, U1: PI> ql) - F(x, y, U 2,P2' q2) j

:( M{jU1-U2j+jpI-P2j+jql-q2j}, M being a constant. Such a condition is certainly satisfied when F is linear in u,p, q and has continuous coefficients. Let, P (x, y) be any point ofthe sqmare j;j < y, j17j < Y and let D be the triangle bounded by ; + 17 = 0, ; ;:= X,17 = y. If u xy is continuous,

EQUATIONS OF HYPERBOLIC TYPE 4.4] since u q vanishes when; = -rj. Integrating again, we have

u(x, y)

and so

u(x, y) =

ff ff =

D

D

[59

u gn (;, rj) d; d'!,

F(;, rj, u,p, q) d;drj,

(2)

it being understood that u,p, q are functions ofthe coordinates (;, rj) of the integration point. Thus, if there is a solution with continuous second derivatives, it satisfies the integro-differential equation (2). Conversely, suppose that (2) has a continuously differentiable solution. Evidently it vanishes on x + y = O. Since the integrand is continuous, we may write the integral as a repeated integral in two ways and differentiate to obtain p(x,y)

=

q(x,y)

=

f: f

x

F(X,rj,1t(X,rj),P(X,rj),q(X,rj))drj,} (3)

x

_y

F(;,y,u(;,y),p(s,y),q(;,y))d;,

so that P and q vanish on x+y = O. Since the integrands in (3) are also continuous, we see that P and q are also differentiable with respect to y and x respectively, giving op oq oy = ox = F(x,y,u,p,q).

Therefore 8 is continuous, and u satisfies the partial differential equation. The integro-differential equation can be solved by Picard's method of successive approximations, by constructing a sequence of functions {un(x, y)} connected by the recurrence formula un+l(x, y)

=

ff

D

F(;, rj, u n(;, rj),Pn(;, rj), qn(;, rj)) dgdrj,

the first term uo(x, y) being any continuously differential function which, together with its first deriva.tives, vanishes on x + y = O. By induction, all the functions of the sequence are continuously differentiable and

f:x = f:y

Pn+l(x,y) =

F(x,rj,un(x,rj),Pn(x,rj),qn(x,rj))drj

qn+l(x,y)

F(;,y,un(t,y),Pn(;,y),qn(;,y))d;.

un,Pn,qn all vanish on x+y

=

O.

60]

[4.4

EQUATIONS OF HYPERBOLIC TYPE

We show that l1 n,Pn' qn tend to limits uniformly on by proving that the series

Ixl : : ;

y,

lyl : : ;

y

00

~

o

(q,,+1 - qn),

are uniformly and absolutely convergent. \Ve may suppose x+y for, if not, we have merely to replace x and y by - x and - y. By the Lipschitz condition,

IU n+1- Unl ::::; M

ff

D

~

0;

+ IPn -Pn-ll + Iqn-qn-lJ)d;d1J,

(Iun- un-II

IPn+1-Pnl ::::; M f:x (lun -un- 1 1+ IPn-Pn-ll

+ Iqn-qn-ll)g=x d1/,

with a similar expression for Iqn+1 - gn I· Since U o and U 1 are continuously differentiable, they and their first derivatives are bounded on Ixl : : ; y, Iyl : : ; y. Hence there exists a constant K such that

lu 2(x,y)-u 1 (x,Y)1 ::::; MK

ffn d;d1J = llJIK(x+y)2,

Ip2(X,Y)-Pl(X,y)1 ::::; MK f:x d1J = MK(x+y), Iq2(X,y)-ql(X,y)1 ::::; MK(x+y). Then

lu 2-u1 1+ Ip2-Pll

+ Iq2-qll ::::; MK(y+2) (x+y),

since 0 ::::; x+Y ::::; 2y. Next

lu 3(x,y)-u 2(x,y)1 ::::; M2K(y+2) ffn(;+1J)d;d 1J _ M2K (y + 2) ( )3 31 x+y, y

Ip3(X,y)-P2(X,y)1 ::::; M2K(y+2) -x (x+1J)d1J = f Iq3(X,y)-q3(X,y)1 ::::;

M2K(y+2) 2!

1I12K~;+2) (X+y)2, ~.

and so

lu3-u 21+ Ip3-P21

+ Iq3-q21 ::::;

M2K(y+2)2 2! (X+y)2.

(X+y)2,

4.4]

EQUATIONS OF HYPERBOLIC TYPE

By induction,

~

IUn+l-Unl

[61

1J-lnK(y+ 2)n-l(X + y)n+l (n+ 1)! '

, MnK(y+ 2)n-l(X+y)n IPp+l-Pnl ~ n! '

By comparison with the exponential series, the series 1;(un+l - un) and the series obtained by term-by-term differentiation are uniformly and absolutely convergent on Ixl ~ y, Iyl ~ 'Y. Hence the integrodifferential equation has a continuously differentiable solution. Since u xy = F(x,y,u,p,q), uX1/ is also continuous. To prove the solution is unique, we have to show that the limit does not depend on the initial member of the sequence. If we construct two sequences {un(x,y)} and {u~(x,y)}, we have IUn(x, y) -u~(x, y) I ~

II

)F(;, 1], u n(;, 1]),Pn(;, 1]), qn(;, 1]))

- F(;, 1], u~(;, 1]) ,p~(;, 1]), q~(;, 1])) I d;d1]

~M

IID (IUn-u~1 + IPn-p~1 + Iqn-q~l)d;d1],

with similar formulae for IPn(x,y)-p~(x,y)1 and Iqn(x,y)-q~(x,y)l. By induction, we can show that, with some constant Kl>

so that un -u~ tends to zero as n-?-oo. Therefore the sequences {un(x, y)} and {u~(x, y)} converge to the same limit whatever uo(x, y) and u~(x, y) may be. Lastly, to show that the problem is well posed, we revert to the original form. The unique solution of 8 = F(x, y, U,P, q) which satisfies on a segment AB of x+Y = 0 the Cauchy conditions U = f(x),

P

= g(x), q = h(x),

where!, = g -h, satisfies U(x,y) =f(x)+

I:y

h(7)d7+

IID

F(;,1],u,p,q)d;d1].

62]

[4.4

EQUATIONS OF HYPERBOLIC TYPE

Let u 1 be another unique solution which satisfies the conditions u 1 = fv PI = gv q1 = hI wherefi = gl - hI' \Ve have to show that if If1 - fl < €, Ig1 - gl < €, Ih 1- hi < e, where € is small, then U1- U is small, at any rate near the segment AB. Consider the strip 0 ::::; x+y ::::; 0 in the square with diagonal AB. Let the least upper bound of lUI - ul + Ip1- pi + Iq1- ql in this strip be K. Then

lu1(x,y)-u(x,Y)1 ::::; If1(X)-f(x)1

+ I:y Ih1(T)-h(T)1 dT

+ II)F(;, 1], UvPv q1) -F(;, 1], U,P, q)1 d;d1] < €+€(x+y)+MK

IID d;d1]

::::; €+€0+tMK 02. Next

Ip1(X,y) - p(x,y) I:: :; Ig1(X) - g(x) 1+ I:)F(x,1],U1,P1' q1) - F(x,1],u,P, q) Id1] < €+MK I:y d1] ::::; €+MKo;

and similarly for Iq1-ql. Hence we have, on O::::;x+y::::;o,

lu 1-ul + Ip1-pl

+ Iq1-ql

< 3€+€0+ 2MKo+lL11K02 =3€+€0+tK

if we choose 0 (> 0) so that M 02+ 4Mo= 1. Since the least upper bound of the expression on the left is K, we have

tK::::; 3€+€0, and so U1- U, PI - P, q1 - q are all of the order of o ::::; x+Y ::::; 0, "Where 0 does not depend on €.

€

in the strip

4.5 Two other applications of Picard's method In the characteristic boundary value problem for 8

= F(x, y, U,p, q)

(1 )

is given on two intersecting charal:lteristics which we may take to be the axes. ,Ye are given that

11

U = ¢(x) (y = O,x where ¢(O)

=

~

¥f(O), and ¢ and

0),

U = 1/f(y) (x = O,y

~

0),

1/f have continuous second derivatives.

4.5]

EQUATIONS OF HYPERBOLIC TYPE

[63

°

The problem is to show that there is a unique solution in x ~ 0, y ~ which has continuous second derivatives. If F is identically zero, the solution is Uo

If we put u =

'1IO

+ v,

=

¢(x) + 1/f(y) - ¢(O).

we find that v satisfies an equation V ru

=

F1(x, y, v, v x , v y )

and vanishes on the positive parts of the axes. Thus it suffices to deal with (1) when ¢ and 1/f are identioally zero. It can be shown that this simplified problem is equivalent to solving the integro-differential equation u(x, y) =

II

n F(;, 1], u(f" 1]),p(;, 1]), q(;, 1])) d;d1]

(x > 0, y > 0),

where D is now the rectangle with the origin and the point (x, y) as opposite corners. The existence and uniqueness of a solution of this problem can be demonstrated by Picard's method when F satisfies a Lipschitz condition. The second problem is the problem of Goursat when u is given on a characteristic and on a duly-inclined curve, which we may take to be y = x. The data are u

=

¢(x) (y

=

O,x ~ 0),

u

=

1/f(x) (y

=

x,x ~ 0)

when ¢(O) = 1/f(0), and ¢ and 1/f have continuous second derivatives. We wish to prove that the problem has a unique solution in the angle x ~ y ~ 0. If F is identically zero, the problem has a unique solution Uo

=

¢(x)~ ¢(y)

+ 1/f(y).

If we put u = U o+ v, we find that v satisfies a half-linear equation of the form (1) and vanishes on y = O,x ~ and on y = x,x ~ 0. Thus it suffices to deal with (1) when ¢ and 1/f are identically zero. This simplified problem is equivalent to showing that

°

u(x,y)

=

IIn

F(;,1],u(;,1]),p(;,1]),q(;,1]))d;,

whereDistherectanglewithsidesg = x,; = y,1] = 0,1] = y(x ~ y ~ 0), has a unique solution with continuous second derivatives. This can be done by Picard's method when F satisfies a Lipschitz condition.

4.6 Duly inclined initial lines A simple example shows why the curve carrying the Cauchy data for a hyperbolic equation with two independent variables must be duly

64]

EQUATIONS OF HYPERBOLIC TYPE

[4.6

inclined. Consider the Cauchy problem for u xy = 0 with data u

= j(x), p = g(x), q = h(x)

on the parabola y = x 2 ; we assume that 1", g', h' are continuous and that the strip condition l' = g + 2xh is satisfied. At a point (x, y) in the region y > x 2 , a unique solution with continuous second derivatives is determined by the Cauchy data on the arc from (x, x 2 ) to (,,/y, y). Another such solution is determined by the data on the arc from (--!y,y) to (x,x 2 ). These two solutions are not necessarily the same, as j, g, h are arbitrary functions which merely have to satisfy the condition that 1", g', h' are continuous. The Cauchy problem in this case does not have a unique solution. The curve y = x 2 is not duly inclined; it touches the characteristic y = 0 and cuts the characteristic y = a2 in two points ( ± a, a2 ). Ify = ¢(x) is a duly inclined curve, the change of variable Xl = ¢(x) does not alter the form of the equation 8 =

F(x,y,u,p,q),

and the Cauchy data, originally carried by y = ¢(x), are now carried by the straight line y = Xl'

4.7 The equation of wave motions The equation of wave motions in one dimension 82u 1 82u 8x2 -c2 8t 2

=

0,

where c is a constant becomes 82u 82u 8x 2- 8y2 = 0

(1)

if we put y for ct. This is a hyperbolic equation with characteristics ;(; ± y = constant. It is equivalent to the system Px-qy = 0,

qx-Py =.0

discussed in § 1.7, since the second equation shows that p dx + qdy is an exact differential duo The general solution of (1) is 11

= F(x+y)+G(x-y),

from which it follows that p + q is 'Constant on each characteristic x + y = constant, p - q is constant on each characteristic x - y = constant.

4.7]

[65

EQUATIONS OF HYPERBOLIC TYPE

y

o

x

D

Fig. 2. The solution of the initial value problem u

= f(x),

q = h(x)

when y = 0, where 1", g' are continuous, is U

= l{j(x+y)+f(x-y)}+l

x+ y x-v h(T)dT.

f

(2)

vVe can relax the conditions to continuity of f and piece-wise continuity of1',j", h, h', that is, these four functions are continuous on closed segments. A discontinuity of l' or h at x = a produces discontinuities of p and q across the characteristics x ± y = a. A discontinuity of 1" or h' at x = /3 produces discontinuities of r, 8 and t across x ± y = /3. The discontinuities in the data are propagated along characteristics. Iff and g are given only along the segment from A(a, 0) to B(b, 0), equation (2) determines the solution only if x ± y lie between a and b. Let 0 be (l(a+ b), l(b - a)), D(l(a +b), l(a - b)) as in fig. 2; the characteristics through A and B meet at 0 and D. Then u is determined at P(x, y) by the data on AB if and only if P lies in the square ADBO. And we can then assign whatever data we please outside AB without altering the value of u at P. The square ADBO is called the domain of influence of the data on AB. Let P be any point of the square, and let the characteristics through P cut Ox in Q and R. The value of u at P depends only on the data on QR. QR is called the domain of dependence associated with the point P. The ideas of domains of influence and dependence occur in the theory of hyperbolic equations because data on an initial line or curve are propagated along characteristics.

66]

EQUATIONS OF HYPERBOLIC TYPE

[4.7

The violin string problem, the problem of determining the small transverse vibrations of a taut unifmtm string with fixed end points, depends on finding the solution of a mixed initial value and boundary value problem for the equation of wave motion. ,Ve are given the initial displacement and velocity ofevery point oHhe string. Replacing ct by y in the equation of wave motions, we have to solve 82u ;2U 8x 2 - 8y 2 = 0,

= f(x), q = h(x) (y = 0), wherefvanishes when x = or l, and given that

U

°

°

U= (X"" 0, l). 'We assume that 1" and h' are continuous; if we merely assumed piece-wise continuity of j',1",h,h' the solution would have discontinuities propagated along characteristics. We could solve this problem by translating the results for the

system

p x - qy

=

° ,

qit _ p y

=

°

obtained in § 1. 7, but prefer to base the discussion on the use of Green's theorem, preparing for a discussion of Riemann's method. Let C be a regular closed curve bounding a domain D whose closure is D. Then if U,p, q are continuous in D and rand t are continuous in D,

IIn

(u=-uyy)dxdy

=

Ie

uydx+uxdy

I

where C is described in the positive sense. Hence if u satisfies the wave equation e uydx+uxdy = 0. Take C to be the rectangle PQRS of fig. 3, where AQ is the line x = l. If Pis (x,y) ,then R is (l-x,l+y). On PQ and RS, dy = dx; 011 PS and QR, dy = -dx. Then 0= =

Ie

uydx+uxdy

I -f

uxdx+uydy-

PQ

SP

f

uxdx+uydy+

QR

l.l :r dx + U ydy

= [uJ~- [uJZ + [uJ:k- [uJf = - 2up - 21.lR'

I

uxdx+uydy

RS.

4.7]

[67

EQUATIONS OF HYPERBOLIC TYPE y

R S

Q p

0

A

x

Fig. 3.

since U vanishes at Q and S. Therefore UR = - Up, or u(l-x,l+y) = -u(x,y)

°

If we can find U for ~ y ~ l, this formula gives U for l Moreover, the equation gives

~

y

~

2l.

u(x, y + 2l) = u(x, y)

so that U is periodic in y of period 2l. It suffices to find u in the square ~ x ~ l,O ~ y ~ l.DividethesquareintofourtrianglesDv D 2 ,Ds,D4 by the characteristics through 0 and A, as in Fig. 4. When P is in D v let the characteristics through P cut Ox in Q and R. Then, taking 0 to be PQR,

°

o=f

OR

=

so that or

f

OR

uydx-f

RP

(UxdX+Uydy)+f

PO

(uxdx+uydy)

h(x)dx+UR-Up+UO-U P ' .

l(UO+UR)+~foR h(x)dx 1 fX+lI u(x,y) = 1{J(x-y)+!(x+Y)}+2 X-ll h(T)dT. Up =

When P is in D4 , take 0 to be PQRS, as in Fig. 5; the inclined lines are characteristics. Then 0= fRS uydx- fsp du+ fpo du- fOR du =

f RS

h(x)dx+us -2up-UR'

68]

[4.7

EQUATIONS OF HYPERBOLIC TYPE

11

x

Fig. 4.

11

o

x

Fig. 5.

so that

or

u(x,y)

=

lfV+X h(T)dT.

H!(y+x)-!(y-x)}+2

Similarly if P(x, y) is in D 2 ,

u(x,y)

= H!(x-y)-t!(2l-x-y)}+21

II-X

f

2Z

-

X-II

X -

V

h(T)dT.

4.7]

[69

EQUATIONS OF HYPERBOLIC TYPE

T

Fig. 6. Lastly if P(x,y) is in D a, take C to be PQRST of Fig. 6, where the inclined lines are characteristics. Then 0= f =

so that or

RS

uydx+f

ST

du-f

TP

du+f

po

du-f

OR

du

fRS h(X)dx-u s -2u P -u R, up

= -l(uR+us ) +~fRS h(x)dx

u(x,y) = -1{f(y-x)+f(2l-x-y)}+21

f

2Z X 1I - 1I-X h(T)dT.

All these formulae can be put into one rule by extending the range of definition off and h by the following rules: (i) f(x) and g(x) are periodic of period 2l, (ii) f( -x) = -f(x),h( -x) = -h(x), (iii) f(2l-x) = -f(x),h(2l-x) = -h(x).

The solution then is, for all x and y, lfx+1I X-1I h(T)dT.

u(x,y) = Hf(x-y)+f(x+Y)}+2

70]

[4.8

EQUATIONS OF HYPERBOLIC TYPE

This corresponds to the problem of an infinite string stopped at the points 0, ± l, ± 2l, ... with antisymmetry of the initial state about each node; the resulting motion is antisymmetrical about each node.

4.8 The uniqueness theorem It is evident physically that the problem of §4. 7 has a unique solution. To prove it, we have to show that the only solution when f and h are identically zero is u == o. "\Ve prove here a more general uniqueness theorem which covers the case when the end conditions may be either u = 0 or U x = 0 and which does not assume continuity of the second derivatives. Let u(x, y), Ux and u ll be continuous in the closed rectangle 6.:0 ~ x ~ l,O ~ y ~ Y. Let the second derivatives u xx ' u xll ' u llz , u llll exist and be bounded in the open rectangle .6.:0<x
(ii) If f(x, t) is integrable in Lebesgue's sense over a ~ x ~ b for each fixed t in a ~ t ~ fl, and if ft(x, t) exists and is bounded in a ~ x ~ b, a ~ t ~ fl, then ~f:f(x,t)dX = f:ft(x,t)dX. Consider

I(a'~lY) = •

fba (u~+u~)dx,

where a and b are constants such that 0 < a < b < l. This integral is a continuous function of a, b, y and tends to J(y) =

f~(U;,+u~)dX.

when a-+ + 0, b -+ l- o. J(y) is a continuous function of yin 0 and vanishes when y = o.

~

y

~

Y,

4.8] By (ii), since

[71

EQUATIONS OF HYPERBOLIC TYPE

:yI(a,b,y) =

f~:y(U~+1t~)dX,

a (2 2 = 2u U + 2u y1t ay Ux + Uy) x Xll w

is bounded in

o< a

,,; y

~

fJ

< Y,

0 < a ,,;

.1: ~

b < l.

Hence, by (i)

= 2u x(b, y) ulI(b, y) - 2u x(a, y) ulI(a, y).

But ux(x, y) uy(x, y) is uniformly continuous in ~ and so tends to zero uniformly with respect to y as x -+ + 0 or x -+ l- O. Hence

d dy 1(a" b, y) -+ 0 as a -+ + 0, b -+ l- 0, uniformly with respect to yin 0 < a ,,; y ~ I t follows that d -J(y) = 0 dy

fJ

< Y.

for 0 < y < Y, and so J(y) is a constant. But J(y) is continuous on o ,,; y ~ Y and vanishes when y = O. Hence

f~ (u~+u~)dx =

0

for 0 ,,; y ,,; Y. This implies that U x and u ll vanish on ~, and so u is constant on~. But, by hypothesis, u = 0 for 0 ~ x ~ l, y = O. Hence u is identically zero on ~.

4.9 The use of Fourier series The violin string problem of §4.7 can also be solved by the method of separation of variables, the Solliltion being in the form of Fourier series.

72]

[4.9

EQUATIONS OF HYPERBOLIC TYPE

If we put u = XY, where X is a function of x alone, Y a function of y alone, in u xx - u llll = 0, we get X" X

y"

=Y'

dashes denoting differentiation with respect to the relevant variable. Hence X" +k2 X = 0, when k 2 is the separation constant. This gives a particular solution

u = (A cos kx + B sin kx) (a cos ky + b sin ky)

°

if k =l= 0. If u vanishes when x = and when x = Zfor all values of y, A is zero and k = nrT/Z where n is a positive integer. If k = the solution is u = (A +Bx) (a+by)

°

and the end conditions give A = B = 0, so that u == 0. As the equation is linear, we get a formal solution

f

co (

u =

n7TY b . n7TY) . n7TX an cos -Z- + n sm -Z- sm -Z-·

(1)

If the initial conditions are u = f(x), u ll = g(x) (0 ~ x ~ Z), we should expect that co • n7TX co n7T . n7TX f(x) = fansm -Z-, h(x) = ~ Tbn sm-- ' (2) Z where an and bn are given by Fourier's rule

an =

12

II

nrrt /(t) sin -Z- dt,

2

bn = n7T

II

0 h(t)

n7Tt sin -Z- dt.

There are no simple conditions to ensure that the series (1) and (2) are convergent. This is purely formal; but with special initial data, it may happen that the series (1) is convergent and gives the desired solution.

4.10

The equation of telegraphy

The equation satisfied by the potential V in the propagation of signals along a uniform telegraph cable with self-inductance L, resistance R and capacity per unit length is

°

02V

oV

L0 ai2 +R° ai

02V =

ox

2•

4.10]

EQUATIONS OF HYPERBOLIC TYPE

[73

We may take LO = 1, RO = 2 by changing the units of x and t. So we consider 2 2

a~-~-2T=0. v a v av

If we put V =

ux uy uy ue- we get a2u a2u ax 2 - ay 2 +u = O. lI ,

We treat this equation in an unrigorous manner as an example of the Fourier integral method of solving the problem of Cauchy. We want the solution which satisfies for all x the conditions u = f(x), u ll = h(x) when y = O. If we represent u by a complex Fourier integral

u(x, y)

1

= ,J(27T)

foo

-00

U(;, y) eixsd;,

the differential equation becomes

f:oo {~:~ +(;2_1) U}eiXEd; = 0, which can be satisfied by U = Fm cos{y,J(;2-1)}+Hm sin{y,J(;2-1)} ,J(;2-1) , where F(;) and H(;) are arbitrary functions. This gives the formal solution

u(x,y) =

,J(~7TJ:oo [F(;)cos{y,J(;2- 1)} + -vI(~(~ 1) sin {y ,J(;2 - 1)}] eixsd;,

where F and G have to be determined from the initial conditions, which give

Using the inversion formula, we get

74]

EQUATIONS OF HYPERBOLIC TYPE

1

= 2

7T

where

fOC;

Q(X-T,y)h(T)dT+

-00

-foo

Q( x, y ) -

1 : 2IT uy

foo

[4.10

Q(X-T,y)f(T)dT,

-00

ix,;sin{y,\/(~2-1)}dl: -J (~2 _ 1) s·

_ e 00

It can be shown that, if y > 0, Q(x,y) = 7T10 {-J(y2- x2)} when x lies between ± y, and is zero elsewhere. 10 denotes the Bessel function of imaginary argument and order zero. In this way, we get formally the solution

'We obtain this solution by other methods in the next chapter.

Exercises 1. If u(x, y) satisfies u xv

= 0, prove that

f

c U g(£, 1/) d~

taken round any simple closed curve is zero. Deduce that, if u = f(x), ou/ox = g(x) when x + y = 0, then u(x,y) =f(-y)+

f:/(~)df'

2. If u(:r, t) satisfies the wave equation 02U 1 02U ---=0 ' 8x 2 c2 2

ot

where c is a constant, show that

round any simple closed curve is zero. Deduce that the solution which

[75

EQUATIONS OF HYPERBOLIC TYPE

satisfies the conditions U

u(x,t)

= f(x),

Ut

°

= g(x) when t = is 1 fx+ct

= t/(x+ct)+t!(x-ct)+-2

c x-ct

g(f;)df;.

Find the particular solution when f(x) = a sin x, g(x) = flsinx, where a and fl are constants. 82u 82u -2 -=0

3. Find the solution of

8x

8y2

= 3x2, u1/ = 2x when y = 0.

given that u

82u 82u -2 -=0

4. Find the solution of

8x

8y 2

inx ~ O,y ~ O,given thatu = 3x2,8uj8y = 2xwheny = O,X ~ O,and u when x = 0, y ~ 0. Examine the continuity of u and its derivatives.

=

°

2

5. Find the solution of

82u_8 u=0 8x2 8y 2

°

°

in the half-plane y > given that, on y = 0, u = Sin1TX when ::s;; x::s;; 1, u = when x > 1 and when x < 0, and u ll = for all x. Solve the same equation given that, on y = 0, u = for all x, U1/ = sin 1TX when ::s;; x ::s;; 1, U v = when x > 1 and when x ::s;; 0.

°

°

°

6. Find the solution in x> 0, y >

°

°

°of

8 u -~ =0 8x 2 8y2 2

°

°

given that u = on the axes and that, On y = 0, u1/ = x when ::s;; x ::s;; 1 and u1/ = 1 when x > 1. Examine the continuity of u and its derivatives. 7. Find by using Fourier series the solution in -l ::s;; x ::s;; l, t

~

°of

82u_..!..~=0 8x2 c2 8t a

given that, when t = 0, and that u =

u=

h

f (l-Ixi),

°for all t when x

=

Ut

= 0,

± l.

8. Show that the solution of the equation of telegraphy 82u 82u ---+'11,=0 8x 2 8t 2

is

u=

f:",

eiq", [O(q) cos{t.j(q2-1)}+Ol(q) SinJ(~~~~l)}] dq

76] if

EQUATIONS OF HYPERBOLIC TYPE

u= J~oc e(q)eiqxdq,

'lit::;:

J:oo

e 1 (q)e iqx dq

when t = O.

9. u(x, y) is the solution of u xx - U yy = 0 with continuous first and second derivatiyes in x;?; 0, y ;?; O. It satisfies the Cauchy conditions 'U = f(x), u y = g(x) on y = O,x ;?; 0 and u = F(y), U x = H(y) on x = O,y ;?; 0, where f(O) = F(O), g(O) = F'(O),j'(O) = H(O). Since the boundary is not duly inclined, the data are not independent. Prove that j'(x)+g(x)

when x ;?; O.

= F'(x)+H(x),

5 RIEMANN'S METHOD

5.1 Adjoint linear operators To a linear operator L, defined by L(u) = au zz + 2huzll + bU llil + 2guz + 2full + CU,

(1)

where the coefficients are continuously differentiable functions of x and y alone, there corresponds a unique linear operator L*, called the adjoint of L, such that vL(u) - uL*(v) is a divergence 8H/8x+8K/8y.

If D is a domain whose boundary is a regular closed curve C, it follows by Green's theorem that IID{vL(u) - uL*(v)} dxdy = Ie (lH + mK) ds,

where (1, m) are the direction cosines of the outward normal to C. Since L* does not depend on the particular domain D, we may find it by considering the particular case when D is a rectangle with sides x = Xo, x = Xl' Y = Yo, Y = YI" By integration by parts, IID vau=dxdy = I::

[avuz]i~dy-IID (av)zuzdxdy

= I:"[avuz-(av)zu]i~dY+ IID u(av)zzdxdy so that I ID [vauzz-u(av)zz] dxdy = I:: [avu z -

(av)zu]i~dy.

Similarly IID[VhUzy-U(hV)zy]dXdY =

I:"[hUVll]~dY-I:'[(hv)zU]~~dX

IID[Vbullz-U(bV)lIlI]dXd y = [ 77 ]

I::[bVUll-(bV)lIU]~~dX,

78]

[5.1

RIEMANN'S METHOD

IID[gvux + u(gV)xJ dxdy = I:" [gvuJ~~dy, IID [fvuv+u(fv)yJdxdy I:' [fvuJt~dx. =

It follows that and that

L*(v) = (av)xx+ 2(hv)xy+ (bv)yy- 2(gv)x- 2(fv)v+cv,

(2)

H = avux-u(av)x+hvuy-u(hv)y + 2guv,

(3)

K = hvux - (hv)x + bvu y - u(bv)y +2fuv.

(4)

Hand K are not unique; we can add to them 08/oyand - 08/ox respectively; but L* is unique. And the adjoint of L* is L. L is said to be self-adjoint if L* = L. The condition for this is ax+hv = 2g,

hx+b y = 2f

so that a self-adjoint operator can be written in the form

o (au


uX

0 0 0 (hu y ) + -;;- (hu x ) +
5.2 Riemann's method Riemann's method of solving the problem of Cauchy for a linear equation L(u) = F(x, y) of hyperbolic type first appeared in his memoirt on the propagation of sound waves of finite amplitude. It depends on finding a particular solution, known as the RiemannGreen function, of the characteristic boundary value problem. Suppose that we use characteristic variables. Then we have to solve L(u)

= 2uxv +2gux +2fuv +cu =

F(x,y)

given u, u x ' U v on some duly-inclined regular arc C. The data satisfy the strip condition du= uxdx + uvdy on C. The functions g,f, c, Fare assumed to be continuously differentiable functions of x and y alone. Let the characteristics y = Yo and x = X o through P(x o, Yo) cut C in Q and R respectively. Let D be the domain bounded by PQ, PR and C; let r be its boundary. Then, if L(u) = F, L*(r) = 0, we have

IID vFdxdy IID {vL(u)-uL*(v)}dxdy IID (~~ + ~~) dxdy =

=

= { (-Kdx+Hdy) • r

t

Gott. Abh. 8 (1860). p, 43. This paper will be found in Riemann's Gesammeltc Mathematischc H"erke (Dover Press reprint, New York, 1953), pp. 156-78.

5.2]

RIEMANN'S METHOD

[79

by Green's theorem, assuming that U and v have continuous second derivatives. Using equations (3) and (4) of §5.1, the integral round r is equal to

-f

PQ

{(vu)z- 2u(vz - fv)}dx +f (-Kdu+Hdy) +f {(vu)1I- 21t (v ll -gv)}dy QR RP

provided that

vz-fv = 0

when

vll-gv = 0

when

y = Yo, x = x o,

that is, provided that v(x,Yo) =

expfxf(~'Yo)d~, x.

where we have taken v(xo, Yo) to be unity, as we may without loss of generality. Hence v is given by a characteristic boundary value problem for L*(v) = O. This problem has a solution, the Riemann-Green function, which we denote by v(x, y; x o, Yo). If Q is (xv Yo), and R is (xo, Yl)' it follows that u(xo, Yo) = iu(xv Yo) v(xv Yo; x o' Yo)

+ iu(xo, Yl) v(xo, Yl; x o' Yo)

+~fQR (Kdx-Hdy) +~ffD v(x, y;xo,Yo) F(x, y)dxdy. This is the required solution, since H = vull -uvll +2guv,

K = uVz -uvz +2fuv

are given on QR by the Cauchy data. If v*(x, y; Xv Yl) is the Riemann-Green function for the adjoint equation L*(v) = 0,

80]

[5.2

RIEMANN'S METHOD

In particular if L is self-adjoint.

To prove this symmetry property, let r be the rectangle with opposite corners Po (x o, Yo), PI (Xl> YI)' Then

Ir (vv~

- v*v y + 2gvv*) dy-

(vv~ -

PoQ~R

v*v x + 2fvv*) dx = O.

On y = yo,vx =fv; on x = XO,v y = gv. On y = YI'V~ = -fv*; on Xl' v: = - gv*. The contribution of the line from Po to Q (Xl' Yo) is

x =

Similarly the contributions of QPI , PI R, RPo are respectively (vv*)Q - v PI '

(VV*)R- vI'l'

Hence the value of the integral round

r

Vf,o - (vv*)R'

is 2vf,o - 2v PI' and so

which was to be proved.

5.1 Another form of Riemann's method Riemann's method also applies when the equation is given in the form with characteristics X L*(v)

Then

± Y = constant. The adjoint operator is

= v=-vyy-2gvx-2fVy+(c-gx-fy)v. oH

oK

vL(u)-uL*(v) = -+~ ox oy'

where If D is a domain bounded by a regular closed curve

IID

vFdxdy =

Ir

r,

(-Kdx+Hdy).

Suppose that the Cauchy data are carried by a duly-inclined regular arc C. Let the characteristic y - X = Yo - X o through Po (x o' Yo) cut C in Q; let y+x = Yo+x o cut C in R. Take r to be the segment PoQ, the arc QR and the segment RPo'

5.3]

RIEMANN'S METHOD

Since dy = dx on f

[81

PoQ

(-Kdx+Hdy) p~

=f Hdx-Kdy =fp.Q vdu-udv+2guvdx-2fuvdy ~Q

=f

p.Q

vdu+udv-

2fP.Q u(vz +Yll+fv -gv ) dx

= (uv)Q - (uv)p o'

provided that v z +vll = (g-f)v on y = Yo. Again, on RPo, dx = -dy. Hence f RP.

-K~+H~=-f

RP.

= =

-f -f

RP. RP.

H~-K~ vdu - udv + 2g1Wdx - 2fuvdy VdU+UdV+2f u(vz -vll -fv-gv)dx RP.

= (UV)R- (uv)Po'

provided that V z - v = (g + f) v ll on x+Y = xo+Yo. We again have a characteristic boundary value problem, which determines a Riemann-Green function v(x, y; x o' Yo), the constant of integration being chosen so that v(xo, Yo; x o' Yo) = 1. The resulting solution is u(xo, Yo) = !(uv)Q +!(UV)R+~fQR(-Kdx+Hdy)

-~ffDv(x,y;xo, Yo) F(x, y)dxdy. 5.4 Determination of the Riemann-Green function The difficulty in Riemann's solution is the determination of the Riemann-Green function. The method replaces a Cauchy problem by a characteristic boundary value problem. It suffices to consider the case when the independent variables are characteristic variables, so that the linear operator is L(u) = 2uzy + 2guz + 2full + c.

82J

[5.4

RIEMANN'S METHOD

If we make the change of dependent variable through by ¢, we get a linear operator

11,

= ¢U,

and divide

= 2Uxv + 2GUx + 2FUv + CU,

M( U)

where G = g+ ~\

¢'

=1+ ¢.'t

F

x'

The Riemann-Green function for L(u), v(x, Y; x o' Yo), satisfies the adjoint equation L*(u) = 0 and the conditions

and It easily follows that the Riemann-Green function for M(U) = 0 is T..

_

¢(x, y)

T (x,y,xo,Yo) - '""( 'f'

xo,Yo

.

)v(x,y,xo,Yo)'

For example, if we put u = (x+y) U in u xv Riemann-Green function is constant, we get

= 0,

for which the

Ux+Uv_O

--xv +x+y

U

for which the Riemann-Green function is therefore x+y V(x,y;xo,Yo) = - - . xo+Yo

The change of dependent variable is useful if we can choose ¢ so that M(U) = 0 is self-adjoint. This occurs if

I

= -

8

ox log ¢,

g= -

0

oy log ¢ ;

such a transformation is thus possible when gx = IvThe Riemann-Green function is the solution of a characteristic boundary value problem, and does not depend in any way on the arc ca,rrying the Cauchy data. If it is possible to solve by some other method the problem of Cauchy with a simple curve carrying the data, a comparison of this solution and the Riemann solution should give the Riemann-Green function. In the case of the two equations discussed by Riemann, it was possible to do this; he solved the problem of Cauchy with data on a straight line by using Fourier cosine transforms.

5.4J

[83

RIEMANN'S METHOD

The method given below is suggested by Hadamard's observation in his Lectures on Cauchy's Problem that the Riemann-Green function is the coefficient of the logarithmic term in his elementary solution.

5.5

A series formula for the Riemann-Green function

The Riemann-Green function for L(u) = u:cy+gu z +full +cu = 0 satisfies

L*(v) = V:cy-gv z -fvlI + (c-g z -fll ) V = 0

under the conditions Vz = fv

on

and

y = Yo,

vlI = gv

on

x = x o,

v(xo, Yo; x o, Yo) = 1.

Following Hadamard, we try v

=

v Ti

ex>

~~,

(1)

o J.J.

where v j are functions of x and y to be determined and

r = (x -

xo)(y - Yo).

Assuming that it is legitimate to differentiate term-by-term, we have j ex> ri ex> r -I { ov OV o} L*(v) = ~ DL*(Vj)+ ~ .'(. -1)' (x-xo)~+(Y-Yoh..2 o J oJ. I J. J . uX uy 0

ex>

ri-I

ex>

rj-I

j + ~ ("-1)'( ·_l)'V j . -~I J..'( J'_l),{(x-xo)f+(y-yo)g}v . I l . J . Since L*(v) = 0, equating to zero the coefficients of powers of obtain jL*(v j_I ) +jv j + (x-x o) ~;. + (Y-Yo)

~~. -

r,

we

f(x-x o)v j - g(y-yo) v j = 0 (2)

for j = 1,2, .... The coefficient V o is at our disposal; once it is fixed, equation (2) gives successively the coefficients VI' V 2, •••• On the characteristic y = Yo, V = V o and Ov/ox = ov%x. Hence

o

ox vo(x, Yo) = f(x, Yo) vo(x, Yo)· If we take Vo = 1 at (xo, Yo), we get logvo(x, Yo) =

IXx. f(~, Yo) df

(3)

84]

[5.5

RIEMANN'S METHOD

J

log vo(x o, y) =

Similarly

1/

1/.

g(x o, 1J) d1J.

(4)

Thus we know '/,'o(x, y) on the two characteristics x = x o' y = Yo and have to assign V o elsewhere. There a,re many ways of doing this; each gives rise to a different formula for the Riemann-Green function. We follow Hadamard and define (X,V)

log vo(x, y) =

J

f(g, 17) d~ + g(g, 1J) d1J,

(5)

(x., v.)

where integration is along the straight line from (x o, Yo) to (x, y). This function reduces to that given in (3) on the characteristic y = Yo and to that in (4) on x = X o' The integral in (5) would be changed if we altered the path, since f(g, 1J) dg + g(g, 1J) d1J is not necessarily an exact differential. Ifx = xo+1'cosO,y = Yo+1'sinO, we have

°

g = xo+scosO,

1J

where is constant and s varies from log vo(x, y) =

f:

= yo+ssinO

°to 1'. Equation (5) then becomes

{(g - xo)f(g, 1J) + (1J - Yo) g(~, 1J)} :s,

If we denote the value of V o given by (6) by

(6)

n, -Ie get

on = -;:{(x-xo)f(x,y) n a;: + (Y-Yo' g(x,y)}. Using polar coordinates, (2) becomes

oV j

l'

on

.

1'a;- n a; v j +Jv j

= -J'L* (v j _ I ),

or Hence since v j is finite when l' = 0, v·] =

jn fT sj-I -L*(v·I)ds 0 n ]- ,

-~ 1'J

(7)

where, in L*(v j _ I ),; = xo+scosO,1J = yo+ssinO. Equation (7) determines successively all the coefficients Vj' The convergence of the resulting series can be proved by the method of dominant functions, as Hadamard does for his elementary solution. But it may not be possible to carry out the integrations. A knowledge of the first few terms sometimes suggests the form of the RiemannGreen function which can then be found by elementary methods.

5.6J

[85

RIEMANN'S METHOD

5.6 The equation of telegraphy The equation of telegraphy a2u

a2u

L(u) = - 2 - - 2 - ox at

+U

(1)

= 0

is self-adjoint; and

a

vL(u) - uL(v) = ox (V1£x - nvx )

-ata (vu t -

uvt )·

Hence, if U and v are two solutions with continuous second derivatives within and on a regular closed curve r

fr

(vUt-uvt)dx+(vuz-uv:zJdt =

To solve the initial value problem, to find U = f(x),

1£

o.

(2)

in t > 0 given that

Ut = h(x),

when t = 0, we take r to be the triangle with vertices Po(xo, to), Q(xo - to, 0), R(xo + to, 0). The inclined sides of the triangle are characteristics. On PoQ,dx = dt; on PoR,dx = -dt. Hence equation (2) becomes

f

QR

(VUt-UVt)dX-f (VdU-Udv)+f (vdu-udv) = 0, RP. p.Q

whence

f

QR

(VUt-UVt)dX-[UV]~o+2f

RP,

UdV+[UV]~o-2f

~Q

udv

=

O.

If v is constant and equal to unity on PoQ and PoR, this gives up

o

=

1

!UQ+tuR+-2 f

QR

(VUt-uvt)dx.

(3)

The function v is the Riemann-Green function which we have to determine in order to complete the solution. We try to find the Riemann-Green function as a series of the form

where The conditions on the characteristics PoQ and PoR are satisfied by

86]

[5.6

RIEMANN'S METHOD

taking V o == 1. The differential equation L(v) = 0 becomes 00

fi

fi-1

00

2:o JDJ . L(v j ) + 4 2:1 (J.-1)' J., 0

0

0

{

OVo Ovo} (x- x o) -;-l + (t- to) "'t] uX u

4 [ (x-x ) -",-+(t-t OVj+l OV j ..1-1] Rence L(v j )+-.' +4V j ..1-1 = o o)--",

J+1

uX

ut

'

o.

This is satisfied when v j is constant and jj+l

=-

!L(v j )

::;: -

i vj

o

v j = (- !)i vo = (- i-)i.

Hence

Thus the Riemann-Green function is 00

v(x,t;xo,t o) =

(-1)ifi

2: -.-,-.,- 4i o J.J.

,

= Jo(.v f ),

where Jo is the Bessel function of order zero. In our initial value problem, f is negative, and so v = 1 0 [,J{(t-t o)2- (X-X O)2}],

where 10 is the Bessel function 'of imaginary argument'. This result can be easily verified. The solution (3) then becomes U(X o, to)

=

1fx'+to H(x o-to)+i!(xo +to)+2 zo_t,h(x)10 [,J{t 5 -(x-xo)2}]dx 1fx'+to 0 +2 !(x)fj10 [,/{t5- (X-X O)2}] dx. z,-t, to

5.7 More examples of the Riemann-Green function The Riemann-Green function v(x, y; x o, Yo) for the self-adjoint equation 02U v(1- v) --+---u=O oxoy (X+y)2 '

(1)

where v is a constant, is constant on the characteristics x = x o, y = Yo' and so is equal to unity when f

=

(x- xo)(y- Yo)

vanishes. Hence we try

with V o = 1.

v Ti

TI. o J.J.

00

v(x, y; x o, Yo)

= 2:

5.7]

RIEMANN'S METHOD

[87

The coefficients vi satisfy the recurrence relation ov i ov i · .02vi _ 1 .v(l-v) -0 (x-xo)
Hence, if vi = Ki(x+y)i where K i is a constant, K i = -Ki _1 (V+j-1) (1-v+j-1)!(x o+Yo)'

Since K 0 = 1,

_(~v~+.;:-j:-;)r:;-;(7"l-~v:-+,,-,j) K . = ( - l ) i. r r (v) r (1- v) } (x o + yo)1 .

(. .

(x-xo) (y-Yo)) )( ) , x+y xo+Yo

Therefore v(x,y,xo,yo)=F v,l-v,l,-( which can also be written as

(x-xo) (y-Yo)) v(x, y; x O' Yo) = P-v ( 1 + 2 ( )( . ). x+Y xo+Yo

If, in the equation 02u +_v_ 01£ +_v_ 01£ = 0 oxoy x+yox x+y oy ,

(2)

where v is a constant, we put u = U(x+y)-V, we find that U satisfies equation (1) and hence the Riemann-Green function for (2) is v(x, y; Xo, Yo) = (

(x+y)V ( (x-x o) (y-Yo)) )V F v, 1 - v; 1; - ( )( ). xo+Yo x+y xo+Yo

(3)

With a different choice of V O' we can obtain an alternative form (X+y)2v ( (x-x o) (y-yo)) v(x,y;xo,Yo) = (x+Yoy(xo+y)V F v,v;l;(x+yo)(xo+Y) .

(4)

If the constants It and v in a2u + --.!!:..- au + _v_ au = 0 axay x+yay (x+Y) ay

are distinct, we cannot reduce it to self-adjoint form. If we try to find a Riemann-Green function of the form

a possible form for V o is

88]

RIEMANN'S METHOD

[5.7

Comparing this with the formula (4) for the case fl = v, we substitute in the adjoint equation (x +y)fl+' V = (x + YO)fl (x + y)" F(t), o t

where

=

(x-x o) (y-Yo) (x+Yo) (xo+y)'

It turns out that d 2F dF t(l-t) dt 2 +{1-(fl+V+1)t}llt-flVF= 0

of which the solution which takes the value 1 at t = 0 is F(j.L, v; 1; t). Hence v(x y'X y) ,

,

0'

0

-

(x + y)fl+' F(U V· l' (x-x o)(y-Yo)) (X+YO)fl(XO+Y)' r'" '(x+yo)(xo+y) ,

which reduces to (4) when ,U = v. 'With a different choice of vo, we could get the alternative form v(x y'X y) = (x+Y)'(XO+y)fl-' F(fl 1 - V' 1 ' _(x-xo) (y-Yo)) , , 0' 0 (x o+ YO)fl '" (x+y) (xo+Yo) corresponding to (3).

Exercises 1. Verify that the Riemann-Green function for 82u

2

-----u=o 8x8y (X+y)2 is

. ) _ (x + Yo) (xo+y)+(x-x o) (Y-Yo) (x+y) (xo+Yo) . v (x,y,xo,Yo -

Use Riemann's method to show that the solution which satisfies the con· ditions U = 0, U'" = x 2 on y = x is U = i(x-y) (X+y)2.

2. Prove that the Riemann-Green function for u"'l/+u = 0 is Jo(2.jr) , where r = (x-x o) (y-yo). 3. Show that the Riemann-Green function for 2 8 u +_2_ (8U + 8U) = 0 8x8y x+y 8x 8y

is

[89

RIEMANN'S METHOD

Use Riemann's method to show that the solution which satisfies the conditions U = 0, ou/ox = 3x2 on Y = x is U = 2x3-3x2y+3xy2-y3 • 4. Show that the Riemann-Green function for o2u o2u 20u - 2- - - - - =0 ox oy2 X oX is

v(x,y;xo' Yo) =

X2+X~-

(Y-YO)2 2x 2 .

Hence find the solution in x > y > 0 which satisfies the conditions U Ul/ = h(x) when y = o.

= f(x),

5. Prove that the Riemann-Green function for 02U

o2u

v(1- v)

---+--u=O ox2 oy2 x2 v(x,y; x O' Yo)

is

= F(v, 1-v; 1; f/(4xx o))'

where Hence find the solution which satisfies the conditions wheny = O.

U

= f(x),

ul/

= h(x)

6. Prove that the Riemann-Green function for o2u a OU b OU -+--+--=0 oXOy

x ox

x oy

,

where a and b are constants is v(x,y;xo, Yo) = ( XX)b exp {a(y-yo)} X IFI (1-b; 1; -af/(xxo)), o o f = (x-x o) (Y-Yo), where and

IFI (a;f3;

r(f3) ex> r(a+n) tn t) = r(a) ~ r(f3+n) n t"

7 v(x,y;xo,Yo) is the Riemann-Green function of

L(u) == u"'l/+gu",+ful/+cu = O. u(x, y; Xv YI) is the solution of L( u) u'" = - fu on y = Yv Ul/ = -gy on x = tion to

=0 Xl'

whi ch satisfies the conditions

By applying Green's transforma-

IID (vL(u)-~L*(v))dxdY,

where D is the rectangle with opposite corners (xo' Yo) and (xv Yo) and (Xl' YI)' prove that 4

CPD

6

THE EQUATION OF WAVE MOTIONS

6.1 Spherical waves The equation of wave motions in space of n dimensions is 82 U

,12U-= 0 8t 2

with an appropriate choice of units, \72 being Laplace's operator. In spherical polar coordinates, this is 2 2 8 U + n-1 8U +.!.A(U)- 8 U = 0, 81'2 l' 81' 1'2 8t 2

where A is a linear operator containing only derivatives with respect to the angle variables. A solution U with spherical symmetry about the origin does not involve the angle variables, and so satisfies 82U n-18U 82U (1) 81'2 +-1'- a;:- 8t2 = O. This can be reduced to the self-adjoint form 82 u

821l

01'2 - 8t2 -

(n- 1) (n- 3) 41'2 U = 0

(2)

by putting U = ur-(n-l)!2. When n = 3, this is the one-dimensional wave equation. Hence the general solution of the wave equation with spherical symmetry in three dimensional space is U =

~{
(3)

The term involving 0 by 1 { (r-t)f(r-t)+(r+t)f(r+t)+ JT+t F = 21' T-t sh(s)ds},

provided that

l'

;?; t.

(4)

But when t > 1', this solution fails because f(1') [ 90 ]

6.1]

[91

WAVE MOTIONS

and h(r) are not defined for r < O. This is what we should expect; Cauchy data on t = 0, r ;?; 0 determine U only up to the characteristic r = t in the rt-plane. To determine U when t > r another condition must be satisfied on r = O. If we require U to be finite at r = 0 for all t ;?; 0, U must vanish for r = O. Now lfT+t U = !(r-t)f(r-t)+!(r+t)f(r+t)+2 T-t sh(s)ds,

when 0 in 0

~

r

~ ~

t

~

r, and

lfr'(t+r)

Ur

=

+ Ut is constant

on every characteristic r + t =

(r+t)f'(r+t)+f(r+t)+(r+t)h(r+t), lfr'(t) = tf'(t) + f(t) + th(t).

or Therefore and

(5)

¢(t) + lfr(t) = 0,

t, where

for t ;?; O. But since constant, we have

u = !¢(t-r)+ilfr(t+r)

lfr(t) = tf(t) + f~ sh(s) ds + lfr(O) ¢(t)

= - tf(t) - f~ sh(s) ds- lfr(O).

Inserting these values in (5), we find that, when t

;?;

r,

1 { -(t-r)f(t-r)+(t+r)f(t+r)+ ft+T U = 2r t-T sh(s)ds} .

Iff(s) and h(s) are zero for s ;?; a, then, for a fixed r ;?; a, U remains zero until t = r - a and again when t > r + a. Thd initial bounded disturbance is thus propagated with a sharp head and tail.

6.2 Cylindrical waves The equation of wave motions in two-dimensional space is 82U 1 8U 1 a2u 82U 8r2 802 - 8t2 = 0,

+r -ar+rz

in polar coordinates. This equation can be solved by the method of separation of variables. In particular, if U is independent of 0, it is of the form V(r) eit'T where T is a constant and

d2 V

1 dV

-+ - - + T 2 V= 0 dr 2 r dr '

w AVE

92]

[6.2

MOTIONS

which is Bessel's equation of order zero. There are two independent solutions Jo(r1') and :Yo(n); but it is preferable to use the Hankel functionst H~l)(r1')

= .lo(r1') + i:Yo(r1') ,

H~2)(r1')

= Jo(r1') -iYo(r1')

because of their simpler behaviour when r1' is large. Vile have, then, two solutions symmetrical about the origin U1 =

eiTtH~l)(n),

U2 =

eiTtH~2)(r1').

'Vhen r is real and r1' large, U1

(~)t exp(ir(t+1')-!7Ti), 7Tr1'

,....,

U2 ,...., (~)\ t exp (ir(t - 1') + !7Ti), 7Tr1'

so that the real parts of ~ and U2 represent contracting and expanding cylindrical waves respectively which vary simply-harmonically with the time. Using the integral formulae for the Hankel functions, we find that U1

= ~foo exp (itr +irr cosh¢» d¢>,

U2 =

7TZ

0

-~foo exp(itT-irrcosh¢»d¢>. 0 7T~

It follows that a general solution of the two-dimensional wave equation with symmetry about the origin is

oo

U

= f~oo dr fo d¢>exp (ir(t-1'cosh¢»)f(r)

oo

oo

or

U

+ f~oo dr fo

d¢>exp(ir(t+1'cosh¢»)g(r),

oo

= fo F(t-1'cosh¢»d¢>+ fo G(t+ l' cosh¢» d¢>.

(1)

This result, which corresponds to the solution (3) for spherical waves, has been obtained by a purely formal argument. To verify that, if F" is continuous, U =

oo fo F(t-1'cosh¢)d¢

(2)

t See, for example, Copson, Functions of a Complex Variable (Oxford, 1935), p. 323.

6.2]

[93

WAVE MOTIONS

satisfies the equation 02U

loU

02U

-+ ----= 0 dr 2 r or ot 2 is quite straightforward, provided that we can justify differentiation under the sign of integration and provided sinh ¢F'(t-r cosh¢) ~ as

¢~

+00.

°

The function (2) can be regarded as the cylindrical wave function due to a source of strength 27TF(t) at r = 0. A peculiarity of the twodimensional propagation of waves which it describes is the existence of a 'tail' to the disturbance. For if F(t) = when t > to, we have

°

U

=

!soo F(t-rcosh¢)d¢

when t > to+r, where rcoshs = t-t o; this expression does not generally vanish. In this respect, waves in two dimension differ essentially from those in spaces of one or three dimensions. A different integral formula for a cylindrical wave function U = f:f(t-rcOS¢)d¢ was given by Poisson; this is certainly valid iff" is continuous. None of these formulae is of much use in solving the initial value problem for the equation of cylindrical waves.

6.3 Poisson's mean value solution If p(x, y, z) has continuous partial derivatives of the first and second orders, the equation of wave motions (1)

has the solution

u

= tvll(p; x, y, z; t),

(2)

when t > 0, where vii denotes the mean value of p over the sphere with centre (x, y, z) and radius t; this solution satisfies the initial conditions u = 0, U t = P when t = and has continuous derivatives of the first and second orders. The proof uses Boussinesq's spherical potential

°

94]

[6.3

WAVE MOTIONS

where S is the sphere with centre (x, y, z) and radius t. If (l, m, n) are the direction cosines of the outward normal to S,

V(X,y,z,t) = t IInP(X+lt,y+mt,z+nt)dQ,

(3)

where Q is the unit sphere l2+ m2+ n 2 = 1. Evidently V has continuous deriYatives of the first and second orders, and vanishes when t = O. Now \72V(X,y,z,t) = tIIn \J2p(x+lt,y+mt,z+nt)dQ

If If ( = -+- If ( Iff 1 =t

eV V -+t ott

Also

V t

=

S

{02P 02p 02p } 0;2 + 0'17 2+ OS2 dS.

op op OP) dQ l-+m-+nnOx oy OZ

1

t

op op OP) dS l-+m-+n0; 0'17 os

s

V 1 = T+t

D

(02 p 02p 02p ) 0;2+ 0'172+ OS2 df,d1J d S,

(4)

where D is the interior of S. This may be written as

1ft dT Ifn T2\J2p(x+lT,y+mT,z+nr)dQ.

oV = -+V -d t t t

0

~

Hence 2

ae-

0ot2 V = t1 0V t2 V -i2 1

02V ot 2

By (4)

=t

If

Iff

+t

n

D

\J2p d;d1J dS

JIn \J2p(x+ It, y+rnt, z+

\J2p(x+lt,y+mt,z+nt)dQ

nt) dQ.

= \J2V,

(5)

so that V satisfies the equation of wave motions. Moreover

o-;,T' = ut so that

If

n

p(x+lt,y+mt,z+ nt) dQ+tff

n

(l~P +m 0; +n ~P) dQ, ox uy oz

T~(x,y,z,O)=47Tp(X,y,z).

(6)

Since V = 47TtJ/, it follows that

u = tvlt(p;x,y,z;t) is a solution of the equation of wave motions which has the required differentiability properties and satisfies the initial conditionE! 11 = 0

6.3] Ut

[95

WAVE MOTIONS

= p. We see later that it is unique. Moreover, if Ipl

< e, we have < te ~ Te in 0 ~ t ~ T, so that the problem is well posed. Ifu = V;(x,y,Z;t)!47T it follows from (6) and (5) that

InI

u(x, y, z; 0) = p(x, y, z),

ut(x, y, z, 0) =

o.

8

Hence

U

= at {tvK(p; x, y, z; t)}

satisfies the equation of wave motions under the initial conditions = p, U t = O. In order that this solution should have continuous derivatives of the second order, it suffices that p has continuous derivatives of the third order. Again, the solution is unique. And if Ipl < e, Igradpl < e, we have lui < e+te ~ (1 +T)e in 0 ~ t ~ T, so that the problem is well posed.

U

To sum up, if f(x, y, z) has continuous derivatives of the third order and h(x, y, z) of the second order, a solution of the wave equation in t > 0 satisfying the initial conditions u = f, U t = h is U

=

8

tvK(h;x,y,z;t)+8i{tvK(f;x,y,z;t)}.

(7)

We show later that this solution is unique. The expression on the right of (7) depends only on the initial data on the sphere with centre (x, y, z) and radius t. If f and h are zero except on some bounded closed set F, and if the greatest and least distances from F of some point (x, y, z) not in Fare R 1 and R 2 , then U = 0 when t > R 1 and when 0 < t < R 2 , since in these cases the sphere over which the mean is taken does not then cut F. The solution represents a wave motion with a sharply defined front and rear; there is no residual after-effect.

6.4 The method of descent If we use the method of §6.3 with the initial conditions u(x,y,z,O) =f(x,y),

ut(x,y,z,O)

=

h(x,y),

(1)

the two mean values in (7) do not involve z, and so the solution satisfies 82u 82u 8x 2 + 8y 2

82u

= 8i2 .

This device was called by Hadamard the method of descent.

(2)

96]

[6.4

WAVE MOTIONS

With this sort of data,

tJI(f;x,y;t) =

4~tffDf(x+£,y+1J)dS,

where integration is over the sphere £2 + 1J2 + S2 = t2 • Hence 1 tJI(f;x,y;t) = 27T

If

dsd1J r-.!(X+ S,Y+1J) \/(t2_S2_1J2)'

where ~o is the disc S2 + 1J2 ~ t 2, since dS = tdsd1JIISI; a factor 2 arises since there are two hemispheres in S ~ and S ~ whose projections on S = are both the disc ~o. The solution in t > of the initial value problem (1) for the equation of wave motions in the plane is thus

°

° U

1

= 27T

If

r-

° °

dsd1J h(s,1J) .J{t2_(X-S)2_(y-1J)2} 1 0 + 27T Ft

If/(s,

dsd1J 1J) \/{t2_ (X-S)2- (Y-1J)2}'

(3)

where integration is over the disc ~ on which (X-S)2+(Y-1J)2 ~ t 2. Suppose that f and h are zero except on a bounded closed set F. If (x, y) is a point outside F at a distance R from F, both integrals in (3) vanish when ~ t < R, but not when t > R. As t.-+ +00, the integrals tend to integrals over the whole plane, which are usually not zero. The solution (3) represents a wave motion which has a sharply defined start but leaves a residual after-effect. We obtain formula (3) again by a different method in §7.4 where we show that, if h has continuous second order derivatives and f has continuous third order derivatives, then (3) gives the solution of (2) which satisfies the initial conditions and has continuous second order derivatives. If f and h are defined only on some domain D of the initial plane t = 0, formula (3) determines u(x, y; t) when t is positive only if the disc ~ lies in D. This determines the domain of influence of D. For example, if D is the disc S2 + 1J2 ~ a 2 on the plane t =- 0, the domain of influence in t ~ is X2+y2 ~ (a-:t)2, ~ t ~ a.

°

°

6.5

°

The uniqueness theorem

To prove that the solution of the equation 02U 82u 82u

- =ot-2 ox2+8y:

found in § 6.4 is unique, we have to show that, if u = 0, = 0, then u = for all t > 0.

t

°

Ut

=

°when

WAVE MOTIONS [97 6.5] Regard (x, y,.t) as Cartesian coordinates in a three-dimensional Euclidean space. Let D be the volume defined by

(X-X O)2+(Y_YO)2

~

(t-t O)2,

0 ~ t ~ to.

Then if U has continuous second order derivatives, we have, by Green's theorem,

Iff 8t D

=

=

OU (02u o2u 02u) ot2 - ox2 - oy2 dxdydt

II {:e (u;,+u~ ~IIs {n(u;,+u~ +u~)

~I

D

+Un - 2 :x (uxut ) - 2 :y (UyUt )} dxdydt - 21U x Ut- 2muyu t }dS,

where (1, m, n) are the direction cosines of the outward normal to the boundary S of D. If U satisfies the wave equation, this surface integral vanishes. S consists of two parts, a disc in the plane t = 0 and a portion So of the cone (X-X O)2+(Y_YO)2 = (t-t O)2

cut off by the plane t = O. On So, n = 1/")2 and so 12+ m 2 = n 2. Ifwe multiply through by n, we get

or

IIs. {(l2+m2)u~+n2u;+n2y~-21nuxut-2mnuyut}dS IIs. o.

= 0,

{(lUt- nu x)2+ (mUt- nu y)2}dS =

The integrand must be zero, and so

on So. Denote each fraction by v. Then, on a generator of So, du = uxdx+uydy+uzdz = v(ldx+mdy+ndz) = 0

since the normal with direction cosines (l,m,n) is perpendicular to the generator. Therefore u is constant on every generator, and so is zero since U = 0 when t = o. In particular, u(x o, Yo' to) is zero, which was to be proved. There is a vanant of this proof in which we take D to be the truncated cone (X-X O)2+(Y_YO)2 ~ (t-t O)2, 0 ~ t ~ t l ,

98]

[6.5

WAVE MOTIONS

where t1 < to. As before, we get

ffs{n(U;'+U~+Un-2lUxUt-21nUyUt}d8=

0,

where 8 is now the boundary of the truncated cone. On the base, the integrand vanishes. It follows that

~ff

n

~

{(lUt- nUx)2+(mut -nU y )2}d8+ff

~

(u;,+u~+uF)dxdy=

where n = 1N2 on the curved boundary 8 1 of D, and in which t = t1 cuts D. Therefore

ff1:.

(u;,+u~+undxdy =

~1

0,

is the disc

0,

so that u x' u y,Ut all vanish on ~1. As t1 is arbitrary, Ux' uy, Ut vanish everywhere in D, and so u is a constant. As u vanishes when t = 0, u is zero everywhere in D. These proofs also apply to the equation of wave mot,ions with any number of space variables, though the details are more complicated.

6.6 The Euler-Poisson-Darboux equation In § 6.1 we saw that spherically symmetrical solutions of the equation of wave motions in space of m + 1 dimensions satisfy the equation

02U m oU 02U -+ ----=0. or 2 r or ot 2

(1)

If we introduce the characteristic variables x = r + t, Y = r - t, we obtain the equation of Euler, Poisson and Darboux

o2u N (Ou Ou) -0 --+-oxoy x+Y -+ox oy - ,

(2)

where N = im. 'Ve do not restrict our discussion to integral values of 2N; the equation arises in some physical problems when 2N is not an integer. Nor shall we restrict our attention to the case when x and yare real. Laplace's equation

has axially symmetrical solutions, which are functions of y and

6.6]

[99

WAVE MOTIONS

m+ 1

r=

Jf 4·

These solutions satisfy oZu + m ou + oZu = 0 or 2 r or oy2 .

The characteristic variables are now conjugate complex numbers z = r+iy,

z = r-iy.

In terms of these variables, the equation becomes

OZU N (OU OU) ozoz+ z+z OZ + oz =

o.

The self-adjoint form of (2), 02V + N(1-N)v = 0 oxoy (X+y)2

(3)

is obtained by putting U = v(X+y)-N. Equation (3) is unaltered if we replace N by 1- N. It follows that, if U is a solution of (2),

U = u(x + y)2N-l is a solution of

02U oxoy

(OU + o~ + 1-N x+Y ox oy -

0

.

Again, if we put ux+u y = (x+y)v in (2), we have uxy+Nv =

Hence

o.

(ux+uy)xy+N(vx+v y ) = 0,

or {(x+Y) v}xy +N(vx+vy ) = o. Therefore v satisfies o2v + N + 1 + OV) = 0 oxoy x+Y ox oy .

(Ov

These recurrence formulae enable us to deduce from a solution of (2) a solution ofthe same equation with N replaced by 1 +N or 1-N. When N is not an integer, we need only consider the case when N lies between 0 and 1. Considerable use has been made of these formulae by Weinstein, to whom they are due. The general solution of (2) when N = 0 is By the first relation,

Uo

=

(x) + '¥(y).

1

u1 = {(x) + '¥(y)} x+y

100]

WAVE MOTIONS

[6.6

is the general solution when N = 1. In particular (x) U1

is a solution when

.J..\T =

= x+y

1. Therefore

is a solution when N = 2; so also is o 'Y(x) oy (X+y)2·

Similarly

1 (OU2 OU2) 02 (x) 3 U = x+Y OX + oy = ox 2 (X+y)3

is a solution when N = 3. And so on. For any integer N, the general solution of (2) is ON-l (x) ON-l 'Y(y) +-1 N OX - (X+y)N oyN-l (X+y)N'

U=--

(4)

where and 'Yare arbitrary functions, and N is a positive integer. The first term in (4) is (N -l)!f 2rri

(s)ds

o(s- X)N (s + y)N

if C is a simple closed contour surrounding x but not -y, provided that (S) is an analytic function of the complex variable S, regular in and on C. By the first recurrence formula (x + y)l-2Nf

¢(S) o (s _X)l-N (s + y)l-.l\°

is also a solution when N is zero or a negative integer. This suggests that we should consider complex integrals of the form (5)

for general values of the constant N. Suppose that (S) is an analytic function regular in a domain which contains the disc lsi ~ R. Let x and y be two points of the disc. The integrand is not one-valued in the disc if N is not an integer: it has branch points at x and - y. If we choose a particular branch at, say, R and take C to be a figure of eight starting at R, enclosing x in the posit,ive sense, -y in the negative sense, and returning to R,

6.6]

[101

WAVE MOTIONS

the integrand returns to its original value. \Ve can deform C, so long as it remains a figure of eight enclosing x in the positive sense, - y in the negative sense, without altering the value ofthe integral. Moreover, - the integral is an analytic function of x and of y; its partial derivatives can be calculated by differentiation under the sign of integration. Since (S-X)-N(S+y)-N satisfies the Euler-Poisson-Darboux equation, so also does the integral. Alternatively, if (S) is an integral function, we can choose a particular branch and take C to be a path which starts at infinity, passes round x in the positive sense and returns to infinity.

6.7 Poisson's solutions Suppose that x and yare real, and that C is the figure of eight contour of §6.6. Choose the branch of(s-x)N (s +y)N which is real and positive when S is real and greater than max (lxi, Iyl). Consider the case when x > - y. "Ve can deform C into a small circle of radius e and centre x, described in the positive sense, the real axis from x - e to - y + 8, a small circle of radius 8 and centre - y, described in the negative sense, the real axis from - y + 8 to x-e.

f

In U

1 -

(S)

O(S-X)N(s+y)N

dt: !,,,

the integrals round the small circles tend to zero as e and 8 tend to zero if N < 1. Then U1 = U1

2isinN1T

f:y (X-~)-N(y+~)-N(~)df

vanishes if N is zero or a negative integer. If we put

S = !{(x-y) + (x+y) cos y,} and drop a factor which is unimportant if 0 < N < 1, we obtain U1

)l-2Nftr {!(x-y)+!(x+y) cos y,} sin 1-2Ny,dy,.

X+ Y = ( -2-

0

Restoring the original variables r = 4(x+y),

t = !(x-y),

we find that a solution of oZu + ~ OU _ oZu = 0

or 2

is

U1

f:

= r 1- m

r or

ot 2

(t+rcosy,)sin1-my,dy"

102]

[6.7

WAVE MOTIONS

when 0 < m < 2. 'Ve assumed that x+y = 2r is positive; but the argument can be modified when x + y is negative, with the same result. Similarly, from - ()1-22Vf 'Y(S) dY U2 x+y O(S-X)l-N(S+y)l-l, ~ we find the solution U2

= f: 'Y(t + r cos ljr) sinm-1ljr dljr,

when 0 < m < 2. The two solutions are identical when m = 1. The proof assumed that and 'Yare analytic, but the formulae give solutions which have continuous derivatives of the first and second orders if <1>" is continuous. Whenm = 1,

To get a second solution, consider the limit as m -+ 1 of

_1_f1T 'Y(t + r cos ljr) (sinm-1ljr- r

m-1

1-

m sin 1-

mljr) dljr.

0

This gives the other solution U3

=

f:'Y(t +r cos ljr) log (rsin 2ljr) dljr

when m = 1. All these formulae are due to Poisson, who did not derive them in this way.

6.8

The formulae of Volterra and Hobson

In the solution where x> -y, take C to be the path starting at +00 on the real axis encircling x in the positive sense and returning to + 00. Start with (S-X)N(S+y)N real and positive when S is real and larger than Ixl and Iyl. If (S) is analytic, we may suppose N is not zero or a negative integer, since the integral then vanishes. Deform C into the real axis from + 00 to x + t, followed by Is- x I = t described in the positive sense, and then the real axis from x + t to +00. The integral round the small circle tends to zero if N < 1, which we assume to be the case. Then

6.8]

[103

WAVE MOTIONS

!(x-y)+!(x+y)cosh V

~ =

If we put

and drop the non-zero numerical factor, we get U

=

I:

{l(x + yj}l-2N

{t(x - y) + !(x + y) cosh lfr} sinhl - 2N 1frd1fr.

Restoring the original variables, we obtain Hobson's solution of 02U + m OU _ o2u _ 0 or 2 r or ot 2 - ,

namely

u 1 = r 1- mIooo (t+rcosh1fr)sinh1-m1frd1fr.

This solution in the special case m = 1 was found by Volterra. t We do not need to assume that is analytic. The result is true if m < 2, <1>" is continuous and behaves suitably at infinity. Since the equation is unaltered if we replace r by -r, a second solution is U2

= r1-m

Iooo (t -

r cosh 1fr) sinh1 - m1fr d1fr.

The first solution u 1 represents converging waves, the second expanding waves. Using the first recurrence formula, we get solutions

U2

ua= Iooo (t+rcosh1fr)sinhm-l1frd1fr and

U4

=

I:

(t-rcosh1fr)sinhm-1ifrd1fr

valid when m > o. When m = 1, u 2 and u 4 are the same. A second solution is then lim

_1_Ioo (t -

m-+lm-1

r cosh 1fr) (sinhm -l

0

=

I:

1fr -

r 1- msinh1 - m1fr) d1fr

(t - r cosh 1fr) log (r sinh2 1fr) d1fr.

Similarly, since u 1 and u a are the same when m = 1, the second solution is then (t + r cosh 1fr) log (r sinh2 1fr) d1fr.

I:

t

Volterra, Acta math. 18 (1894), 161. Hobson, Proc. London Math. Soc. (1) 12 (1891), 431-449.

104]

WAVE MOTIONS

Exercises 1. Use Poisson's formula to find the solution of the equation

given that

U = 0,

ut

= x 2+xY+Z2

when

t = 0.

2. Use Poisson's formula to solve the wave equation, given that, when t = 0, U = 0, and Ut = 1 for X2+y2+ Z2 ~ a 2, Ut = for X2+y2+ Z2 ~ a 2. Examine the discontinuities of the solution.

°

3. U is a solution of utt-u",,,,-u'j/'j/-uzz=O of the form v(r,t) cosO in spherical polar coordinates. Prove that

02V 20v 2 o2v -+----v= 8r2 r or r 2 ot 2 • Hence show that

U = cosO ~ ¢(r-t):1fr(r+t),

where ¢ and 1fr are arbitrary functions. Obtain the same result from the fact that, if U is a wave function, so also is u"'. 4. Show that (1-XW)-N(1 +yW)-N satisfies the equation

02U N (OU OU) oxoy + (x+y) ox + oy = 0. By considering its expansion in ascending powers of w, prove that the equation has homogeneous polynomial solutions

where (N)" = N(N +1) ". (N +p-1). Prove that, if K =

IN! and

R = max (\x\, !y\) !F,,(x,y)!

then

~

(2K)"

-,_Rn. n.

°

5. Prove that Tricomi's equation of mixed type yU",,,,+u'j/'j/ = has imaginary characteristics x ± iiyf = constant in the elliptic half-plane y > and real characteristics x ± i( - y)f = constant in the hyperbolic half-plane y < 0. Prove that, if y < 0, the equation can be written as

°

02U 1 OU o2u or2 + 3r or - 8t2 = 0, where r =

!( _y)f, t =

x.

WAVE MOTIONS

6. If u satisfies

[105

oZu Mou Nou -+--+---=0 oxoy x+yox x+y oy ,

where M and N are real constants, prove that v = (X+y),u+N-lU satisfies the same equation with M and N replaced by 1 -:v and 1- M respectively 7. Show that u = ('_X)-N('+y)-M satisfies the equation of Ex. 6. By considering

where C is the double circuit contour of fig. 7, deduce the solution

Fig. 7 (after Whittaker & Watson, A Course of Modern Analysis). when 0 < ffI < 1,0 < N < 1. Obtain also the solution

u 2 = (x + y)l-M-N f:y <1>(;) (X_;)M-l(y+;)N-1d;. These solutions are the same when M + N = 1. Obtain the second solution whenM+N= 1. 8. If u satisfies the equation of Ex. 6, show that

MU:zJ+Null x+y

V= -..;;;,._--:

satisfies

o2v M + 1 ov N + 1 ov --+---+---=0. oXOy x+y ox x+y oy

9. v(x, y, z; t; r) is the solution of the equation of wave motions L(v)

02V o2v o2v 02V =------=0 0&2 ox 2 oy2 OZ2

which satisfies the initial conditions

v(x,y,z;r;r) = 0, for all r. If

vt(x,y,z;r;r) = h(x,y,z;r)

u(x,y,z;t) =

f~V(X,y,z;t;r)dr

106]

WAVE MOTIONS

prove that U satisfies the equation L(v) = h(x, y, z; t) under the initial conditions v(x,y,z;O) = 0, ut(x,Y,z;O) = 0. By using Poisson's mean value solution, show that

where 10. v(x, t) is the solution of Utt - U xx = h(x, t) which satisfies the initial conditions U = 0, U t = when t = 0. By considering the integral of utt-u xx over a suitable triangle, prove that

°

u(x,t)

1ft dT fX+7 d~h(~,t-T). =2

o X-7 Obtain this result also by a modification of the method of the previous example.

7

MARCEL RIESZ'S METHOD

7.1 A comparison with potential theory Let u be a solution of Laplace's equation V'2u = which has continuous derivatives of the second order inside and on a simple regular closed surface ~, and let Po(xo,yo,zo) be any point inside ~. The elementary solution which plays an important part in potential theory is

°

v=

1

....,.,..--~----o------::-:

..){(x - XO)2 + (y - YO)2 + (z - ZO)2} .

If we apply Green's transformation to

I I I V (UV'2V - vV'2u) dxdydz = 0, where V is bounded externally by ~ and internally by a small sphere S with centre (xo,Yo' zo), v being this elementary solution, we obtain

II~ (u :;-v :;)dS+ IIs (u :;-v :;)dS =

0,

where %N denotes differentiation along the normal drawn out of V. [f we now make the radius of S tend to zero, we find that

u(xo,Yo,zo)

=

4~II~ (v:;-U:;)dS,

which expresses u at a point Po inside ~ in terms of the values taken by u and au/oN on ~. It does not solve the problem of Cauchy, since we cannot assign u and ou/oN arbitrarily on ~; given u on ~, ou/oN is determined there. If we replace x and y by ix and iy respectively and z by t, Laplace's equation becomes the two-dimensional equation of wave motions

L(u)

~2u

o2u =o2u ot2 - ox 2- oy2 = 0.

We might expect that an application of Green's transformation to

IIIv{UL(v)-VL(U)}dXdYdt with

=

°

1

v = ..){(t-tO)2_(X-XO)2_(Y_YO)2} , [ 107 ]

108]

MARCEL RIESZ'S METHOD

[7.1

would enable us to find u in terms of the boundary values of 1t and ou/oN on a surface ~ in the space with (x, y, t) as Cartesian coordinates. But we immediately meet difficulties which do not appear in potential theory. In the first place, v is real only when (X-X O)2+(y_yo)2

~

(t-t O)2,

so we take V to be bounded in part by the characteristic cone (X-X O)2+ (y_YO)2

=

(t-t O)2.

Next, the surface S on which we have the Cauchy data must be duly inclined. The characteristic cone then cuts S in a simple closed curve, bounding an area ~ on S. The boundary of V is ~ and the characteristic cone. For example, if S is the plane t = 0 so that we are dealing with an initial value problem, V is given by (X-XO)2+ (y_YO)2

~

(t- t o)2,

0 ~ t ~ to.

The next difficulty is more serious. On the characteristic cone, v and its derivatives are infinite. There are two classical methods of getting over this difficulty. Volterra used, instead of v, its integral cosh- 1

It-tol ,J{(x - XO)2 + (y - YO)2} ,

which behaves satisfactorily on the characteristic cone but has a line of singularities on the axis of the cone. By cutting out of V the singularities on the axis by means of a small coaxial cylinder and then applying Green's transformation, Volterra obtained a formula for

and hence a formula for u. The values of u and ou/oN on the characteristic cone do not appear in the solution because of the properties of Volterra's function. Hadamard, in his Lectures on Cauchy's Problem, on the other hand, did not try to avoid the occurrence of divergent integrals, but developed a method of picking out the' finite part' of a divergent integral. Marcel Riesz has shown how the difficulties of Hadamard's method disappear if we introduce a complex parameter a whose real part can be chosen so large that all the integrals converge. The solution is then obtained by the analytic continuation of a function of a complex variable. Moreover Riesz's method is applicable to the wave equation

7.1]

MARCEL RIESZ'S METHOD

[109

in a Euclidean space of any number of dimensions and more generally to the wave equation in a Riemannian space with positive definite metric. What Riesz does is to introduce a generalization of the Riemann-Liouville integral of fractional order of a function of one variable·t

7.2 The Riesz integral of functional order The problem of Cauchy for the equation L(u)

02U

= -ot2 -

n

~

k=l

o2u -

o~

=

°

(i)

is to find u, given the values taken by u and its first derivatives on some duly-inclined n-dimensional manifold S in the (n+ i)-dimensional Euclidean space with rectangular Cartesian coordinates

In the case of the initial value problem, Sis t = 0. As t plays a special part in equation (1), we treat it separately and denote a point in the space-time manifold by (x;t) where x is the vector (xl>x 2 , ••• ,xn ). The equation of the characteristic cone with vertex P(x; t) is

r = (t-7)2_lx-;12

=

°

where (;;7) is a variable point in the space-time manifold. This cone divides space-time into three domains, viz. r > 0,7 > t; r > 0,7 < t; r < 0. The region in which r < lies outside the cone. r = is a double cone; the part for which 7 ~ t is called the retrograde cone. Since, by hypothesis, S is duly inclined, the retrograde cone cuts S is a simple closed curve, if t is large enough. We denote by V(P, S) the domain bounded by S and the retrograde cone. In the particular case of the initial value problem, t is positive and V(P, S) is the set of points (;; 7) for which r ~ 0, ~ 7 ~ t. The Riesz integral offractional order a of a continuous function f is defined to be

°

°

°

where t is chosen so that V(P, S) is not an empty set. The constant H(a, n) is chosen so that ]lZe t = et when S is 7 = Then

t Acta math.

81 (1949), pp. 1-223.

-00.

MARCEL RIESZ'S METHOD

110]

[7.2

where V is the region r ~ 0, r ~ t. By a change of origin, we may assume that x = 0; and if we put r = t-r', we get

I

H(a, n) =

e- T'r<",-n-l)/2d;dr',

TV

where W is defined by

r

= r'2 - r 2 ~ 0,

r'

~

0,

r = ~i + ~~ + ... + g. 2

and

If we replace (~l> ~2' ., ., ~n) by spherical polar coordinates and drop the dashes, we have H(a,n)

=I'"

drfT drf

o

0

dQne-T(r2-r2)<",-n-l)/2rn-1,

On

where dQn is the element of solid angle in n-dimensional Euclidean space, and so H(a,

n) = Q n Io'" dr I: dre-Trn-l(r2_r2)(",-n-l)/2.

Since the total solid angle

Qn

is 27T n /2 /r(!n), this gives

H(a,n) = 2",-1rr
With this choice of H(a,n),]"'j(x;t) is an analytic function of the complex variable a, regular when rea> n-1. The index notation is used because ]"'1fJ j(x; t) = ]"'+fJj(x; t).

'Ve prove this first when the real parts of a and fJ exceed n + 1, and then use analytic continuation. 'When we do this, all the integrands are continuous. If the real parts of a and fJ exceed n + 1, H(a, n) H(fJ, n) ]"']fJj( x; t) =

I

V(P,S)

d;' dT'j(;'; r') r<",-n-l)!2 rfO- n- 1)/2,

d;drf V(Q,S)

where Q is the point (;;r) and r 1 = (r-r')2-1;-;'12. When j is continuous we can invert the order of integration to get H(a, n) H(fJ, n) ]aJfJj( x; t)

=

I

V(P,S)

where

j(;'; r') g( x, t; ;', r') d;' dr',

7.2]

[111

MARCEL RIESZ'S METHOD

The domain W is defined by

r

> 0,

rI

> 0, r' < r < t. But

g(x,t;;',r') =a(x-;',t-r';O,O), a( x, t; 0, 0) = f

and

w,

r(
where

m

and is defined by r > 0, r o > 0, 0< r < t. By a notation of axes, we can replace x in this integral by (1',0,0, ... ,0),

where

l'

=

1x I· Then, by a Lorentz transformation Xl = xi cosh y + t' sinh y, t = xi sinh y + t' cosh y

we can replace l' by zero, and t by 8 = ,J(t2 -1' 2 ). Hence

g( x, t; 0, 0) = g(O, 8; 0, 0). But g(O, 8; 0,0) = f

w,

{(8- r)2 - ;2}(
where on

ni. Therefore

and so where Hence

g(0,8; 0,0) = K(a,f3)8
r2 =

(t-r')2_lx_;'12.

H(a, n)H(f3, n)J"'1Pf(x; t) = K(a,f3)f

f(;', r') V(P,Sl

r~
= K(a,f3)H(a+f3;n)J
H(a,n)H(fJ,n) = K(a,f3)H(a+f3,n), and so J
112]

[7.2

MARCEL RIESZ'S METHOD

Lastly, when the real part of a is sufficiently large, we can differentiate under the sign of integration in the equation Ja.+2f( x; t) =

1

H(a + 2, n)

f

fe;, r) r
despite the fact that the variables xl' X 2, ••• , X n ' t occur not only in the integrand but also in the equations defining the boundary of V(P, S). For, if h > 0, we have H(a ~ 2, n) {Ja.+2f( x; t + h) _ Ja.+2f( x; t)}

=f V(P,S)

f(;; r) {r~-n+l)f2 - r
where

r+ =

r~-n+I)/2 d~dr,

(t+h-r)2-lx-;12,

°

and Er is the conical shell bounded by r = 0, r + = and S. The volume of the shell is O(h) and r + vanishes on the outer boundary of the shell. Hence the second term tends to zero as h-+ ifre a > n - 1. A similar argument applies when his negative. Hence, when rea> n+ 1,

°

8 H(a + 2, n) 8t Ja.+2f( x; t) =

(a-n+ 1)f

f(;; r) r
V(P.S)

Similarly, when re a > n + 3,

82 H(a+2,n) 8t2Ja.+2f(x;t)

= (a - n + 1) (a- n -1)f

f(;; r) r
+ (a - n + 1)f

f(;, r) r
and

82 H(a + 2, n) ';"2 Ja.+2f( x; t) uX I

=

(a - n + 1) (a - n-1)f

f(;; r) r
- (a-n-1)f

f(;; r) r
7.2]

[113

MARCEL RIESZ'S METHOD

Therefore, when rea> n+ 3, H(a+ 2, n) L]a.+2f(x; t) = a(a -n + 1)f

fe;; 7) pa.-n-l)/2d;dT V(P,S)

= a(a-n+ 1)H(a,n)]a.f(x;t) or

Lra.+2f(x;t) = ]a.f(x;t).

This result holds when re a > n - 1, since the expressions on both sides of the equation are analytic functions, regular when re a > n-1. The operators Land] do not commute. For L]a.+2f(x;t)_]a.+2Lf(x;t)

=

1, f H(a+2,n)

(fAv-vAf)d;dT, V(P,S)

where

.

and

0'1

Af(;, 7) = 07 2 -

f

n

0'1

o~r

The integral can be transformed into a integral over the boundary of V(P, S). The integral over the part of the retrograde cone vanishes when rea> n+ 1, but the integral over the relevant part of S does not in general vanish. In what follows we write H(a) for H(a, n), the dimension n being evident from the context.

7.3 The analytical continuation of Riesz's integral In the initial value problem for the equation of wave motions, the Cauchy conditions are satisfied on t = 0 and the solution is required for t > O. In this case, ]a.f(x;t)

= H~a)fvf(;,7)r(a.-n-l)f2dTa;,

(1)

where V is the region on which r ~ 0, 0 ~ 7 ~ t. If we write ~o = ,Jr, this becomes 1 f ~a.-n ]a.f(x;t) = H(a) wf(;;t-r)7d~oa;, (2) r2

where

n

= ~+ ~ (~k-Xk)2, 1

and W is the hemisphere 0 ~ r ~ t, ~o ~ 0 in the (n+ 1)-dimensional space with Cartesian coordinates (~o, ~l> ~2' ... , ~n)' Sometimes it is more convenient to use polar coordinates ~o =

rcos(},

~k =

xk+lkrsin(},

114]

MARCEL RIESZ'S METHOD

" where 2:

n=

[7.3

1. In polar coordinates, (2) becomes

1

]af(x;t)=H

1 (a)

IldrIt"dof 0

0

On

dQnf(x+lrsinO;t-r) X r a- 1sin n- 1 0 cos",-n 0,

(3)

where 1 is the vector (l1l 12 , ••• , In) and dQ n is the element of solid angle in n-dimensional space. From this it follows at once that ]"'f(x; t) is an analytic function of a, regular in rea> max(0,n-1), on the assumption that f( x; t) is continuous. To extend the region of regularity, we assume that f is continuously differentiable to a suitable order. Since (a- n+ 1) H(a) = H(a+ 2)fa, we may write (2) in the form a I 1 df,,,,-n+l ]af( x; t) = H(a + 2) wfC;; t - r);: ~£o d£oa;· Integrating by parts, we obtain, when rea> n-1, a I £",-n+2 ]af(x;t) = H(a+2) w[f(;;t-r)+rft(;;t-r)]Td£od;

f

a f,,,,-n+l +H(a+2) ~f(;;O)-O-t-a;, (4) n

£5+2:(£k- X k)2=t 2,

where 2: is

O~£o~t.

1

On 2:, ; = x+lp, where p = t sin O. Then, on 2:, d; = pn-1dpdQn = t n sinn-lOcos OdOdQ. The last term in (4) is

J.

=

3

at'" H(a+2)

It" 0

dof dQ f(x+ltsinO' 0) sin n- 1 0 cos",-n+20 On n ' .

When t > 0, this is an analytic function regular when re a > n - 3. The term involving ft> which by hypothesis is continuous, is J2 = H(aa+ 2)

I~ dr I:" dO x

f

dQnft( x + lr sin 0; t- r) r" sin n- 1 0 cos",-n+20

On

which is an analytic function ofa, regularwhenrea > max (-1, n- 3). The remaining term is J1 =

H(aa+2)I~ drI:" dO IOn dQnf(x+ll'sinO;t-r) X

r"-l sin"-l 0 cos",-n+2 0

which is regular when rea> max (0, n- 3).

7.3]

MARCEL RIESZ'S METHOD

[115

(i) When n = 2, 21TJ3 is equal to

1T

at:t. It I~7T r(a+1) dOsinOcosC1.0 0 d¢f(x+ltsinO;O),

°

where x = (x, y), 1 = (cos ¢, sin ¢) so that dQ 2 = d¢. This is an analytic function of a, when t > 0, regular when re a > - 1. Hence J3 is zero when a = 0. Next, 21TJ2 is equal to a ItodrrC1. r(a+1)

fi°1T dO sin 0 cosJ. 0 I °1T d¢ft(x+lrsinO;t-r). 2

This again is regular when re a > - 1; and vanishes when a = 0, Lastly, when re a > 0, 21T~ is equal to

fi°1T dOsinOcos:t.O I °21T d¢f(x+lrsinO;t-r).

a It drrC1.-1 r(a+1)

°

Integration by parts is valid since f is continuously differentiable. Hence 21TJ1 is equal to

tC1. I r(a+1)

2

i

°1T dOsinOcosC1.0 I °1T d¢f(x+ltsinO;O)

1 1) It drrC1. I - r(a+

°

i

2

0 °1T dOsinOcos:t.O I °1T d¢ o/(x+lrsinO;t-r)

which is regular when rea> -1. When a is zero, 21TJ1 becomes i

I

1T

o dO sin 0

I21T ° d¢f(x+ltsinOsinO;O) i

0 - I °1T dOsinO I21T ° d¢ It°dr arf(x+lrsinO;t-r) i

=

°1T dOsinO I21T ° d¢f(x,t) = I

21Tf(x;t).

Therefore, when n = 2 and f(x, y; t) is continuously differentiable, Riesz's integral ]J.f(x,y;t) is regular in rea> -1, and

]Of(x, y; t) = f(x, y; t).

116]

[7.3

MARCEL RIESZ'S METHOD

(ii) When n = 3 and rea> 0,

t

J3 = " r(1. ") r(1. )f!1T dof do'3sin20cos"-10f(x+ltsinO; 0), 2

7T

2a

0

2a

0,

where listhe vector (sin ¢ cos ljf, sin ¢sin ljf, cos¢) anddo' 3 = sin¢d¢dljf. If f is continuously differentiable, we may integrate by parts with respect to 0 and obtain

1T J3 = 2"H7T r(1: t a ) r(ta) f: dO fo, do' 3cos" 0

x :0 {sin Of( x+ ltsin 0; O)}. Hence J3 is regular when re a > - 1 and vanishes when a = 0. To deal with Jl and J2 we make the additional assumption that f has continuous derivatives of the second order. When re a > 0, we integrate by parts in J2 =

2"'7Tr(t~) r(ta)f~ dr f:

1T

dO fo, do' 3 x ft( x + lrsin 0; t- r) r(X sin 2 0 COS,,-l 0

and obtain J. = 2

1

2"H7T r(1 +ta) r(ta ) X

ft drf!1T dof dO, 0

0

0,

3

ret cos" 0 :oft( x + lr sin 0; t- r).

Hence J2 is also regular when re a > - 1 and vanishes when a = 0. Lastly, when re a > 0, we may again integrate by parts with respect to in

e

J = l

2"7Tr(t~) r(ta)f~ dr f:

1T

dO fo, do' 3 X f( X + lr sin 0; t - r) r,,-l sin 2 0 COS,,-l 0

and obtain

~ = 2"H7Tr(1 ~ta) r(ta)f~ dr f: where

1T dO fo, do'3r"-1 cos" OF

. F = :0 (f(x+lrsinO;t-r)sinO),

since, by h~ypothesis,· F is continuous. In fact, F is continuously differentiable, so we may integrate by parts with respect to r, to get

.11

= 2"+27T r(1 +

~a) r(1 + ta)

U:

1T dO fo, do' 3cos" Ot"[FJr=t

1T - f: dO fo.

d~3f~ drr" cos" 0 ~~}.

7.3]

[117

MARCEL RIESZ'S METHOD

Thus J1 is regular when rea> -1. When a = 0,

Ii" I = Ii" f = 4~ In. I:"

1 J1 = -4

IT

1 -4

IT

0

0

de

1 dQ3 [FJr=t--4

~

de

n.

dQ 3

IT

Ii" f It 0

de

~

dQ 3

0

r8F T

ur

dQ 3 [FJr=o de :e {f( x; t) sin e}

=f(x;t). Thus when f(x;t) has continuous derivatives of the second order, ](l.f( x; t) is an analytic function of a, regular in re a> -1, and

]Of(x;t) =f(x;t). (iii) The argument can be repeated for larger values of n. ](1. f( x; t) is regular in rea> -1 provided that we assume that f(x;t) has continuous partial derivatives of a sufficiently high order. The cases n = 2 and n = 3 suffice for our applications. The simple case n = 1 occurs in Ex. 2 at the end of the chapter.

7.4 Cauchy's problem for the non-homogeneous wave equation in two dimensions If

82u

Lu

=

Lu

the problem is to solve when t > 0, given that

82u

82u

8£2- 8x2- 8y 2' =

u = f(x, y),

F(x,y;t), Ut

(1)

= h(x, y),

when t = 0. Heref, hand F are continuous functions, with continuous partial derivatives of orders which emerge in the sequel. The solution is to have continuous second order derivatives. If v is a function of ;, 1], r, it is convenient to write 82v

Av

=

82v

82v

8r2 - 8;2 - 81]2·

Let P be any fixed point (x,y; t) in space-time with t > 0, and let V be the region r ~ 0, ~ r ~ t where

°

118]

[7.4

MARCEL RIESZ'S METHOD

By Green's transformation

I

{U(g,1;: r) Av(;, 17; r) - v(;, 17; r)" \U(g, 1J; r)} df,d1Jdr

r

Is {v (u~~ - v:~) - A( :~ v:;) - # ( u:;- v:;)}dS,

=

U

-

where S is the boundary of V and (A,#, v) are the direction cosines of the normal to S drawn out of V. The transformation is valid when u and v have continuous derivatives of the second order. Let u(x,y;t) be the desired solution of (1), and let V= P,,-1)/2. Then Au = F(;,1J; r), and Av = a(a...£ 1) pcx-3)!2. Then, when rea> 3,

Iv

{a(a-1)uP"-3l/2_FP"-ll/2}d;d1J dr

Is [(a- 1)

=

p,,-3)f2 U{A(f, -

+ v( r- t)} + p"-1)f2{AUg +#u1J -IJUT}J dS.

x) + #(1J - y)

Since re a > 3, the integral over the retrograde cone part of S vanishes; the rest of Sis L, specified by r

On L, A = #

I

=

=

0, v

0,

r o == t 2_(X-;)2_(Y-1J)2 ~

= -

1. Hence

0.

{a(a-1) uPcx-3l!2_FP"-ll/2}d;d1Jdr

v =

t

{Q,,-ll/2uT + (a-1) r&,,-3 l/2tu}d;d1J.

Using the initial conditions, this is, in Riesz's notation, ]"u-]"..,.2F=

I

1 r<,,-ll/2h(;.1J)d;d1J 27Tr(a+ 1) ~ 0 .

f

1 (3 r<,,-ll!2!(1: ) dl:d +27Tr(a+1)8t ~ 0 ,:>,1J ':> 1J.

Now the expressions on each side of this equation are analytic functions of a, regular when re a > - 1; although we have prO'-ed the result only for re a > 3, it is true in this larger domain. In particular, taking a = 0, we have u(x, y; t)

=

]2F(x, y; t)

+ 2~ I~ ro~h(g, 17) d;d1J 1

cI

-~

+27Tat ~ro !(;,1J)d; d17,

7.4] or

MARCEL RIESZ'S METHOD

'll(X,y;t)

[119

217TIvF(;,1];r){(t-r)2-(x-~)2_(Y-1])2}-~d;d1]dr

=

+ 2~ IE h(;, 1]){t 2- (x- ;)2 - (y -1])2}-~d;d1] 2+~ ~I f(;iJ){t 27T 8t E '

(x- ;)2_ (Y-1])2}-~d;d1].

(2)

When F is identically zero, this reduces to the solution found in §6.4. If we put ; = x+pcos¢., 1] = y+psin¢, the first term in (2) becomes 2

1 It0 dr It-r 1(X, y; t) = 27T 0 pdp I 0 1T d¢ x F(x+p cos ¢, y +psin ¢; r){(t- r)2_p2}-~

1

= 27T I~ dr I: pdp I:1T d¢ x F(x+p cos ¢, y +psin ¢; t- r) {r2_p2}-~. If we then put p = tp', r = tr', we get

1(X, y; t)

=

~: I: dr I: pdp I:1T d¢ x F(x+tp cos¢' Y +tp sin ¢; t-tr)(r2_p2)-~ (3)

dropping the dashes. This representation as a repeated integral is valid since F is, by hypothesis, continuous. The expression on the right of (3) is continuous in t ~ o. If F(x, y; t) has continuous derivatives of the second order, so also has 1(X, y; t); and <1>1 and 8<1>1!8t vanish when t =0. The second term in (2) is 2

2(X,y;t)

=

t II0 pdp I 0 1T d¢h(x+tpcos¢,y+tpsin¢) (1-p2)-~, 27T

which vanishes when t = O. Since h is continuous and

t1 {$2(X, y; t) -

2(X, y; O)} 2

= -1 II pdp I 1T d¢h(x+tpcos¢, y+ tpsin ¢) (1_p2)-~ 27T

0

0

120]

MARCEL RIESZ'S METHOD

82!8t exists and is equal to 1 27T h (X,y)

II 0

pdp

~(l_p2)

I

2

0

1T

d¢

[7.4

= h(x,y).

If h has continuous derivatives of the second order, so also has 2(X, y; t).

The third term in (2) is

II I II I

2

8 3(X,y;t) = - 1 7:;t 27T ut

pdp

0

0

1T d¢f(x+tpcos¢,y+tpsin¢) (l_p2)-~

2

1 oPdp = 27T

0 1T d¢f(x+tpcos¢,y+tpsin¢)(l-p2)-t

~ fl d f21T d",8f(x+tpcoS¢,y+tpsin¢)(l_ 2)-t +27TtOPPO'fJ 8t p,

differentiation under the sign of integration being certainly valid when the partial derivatives fx and f v are continuous; and then 3(X, y; 0) = f(x, y). In order that 3(X, y; t) may have continuous partial derivatives of the second order, it suffices that f should have continuous derivatives of the third order. We have thus proved that equation (2) solves the Cauchy problem for

when F(x, y; t) and h(x, y) have continuous partial derivatives of the second order,f(x, y) ofthe third order, the solution (2) has continuous second order derivatives.

7.5 The equation of wave motions in three dimensions The Riesz solution for 82u

82u

;2.u

82u

Lu == 8t 2 - 8x 2- 8y 2- 8z 2 = F(x, y; z, t)

(1)

in three spatial dimensions starts like the corresponding solution in two dimensions. The initial conditions are u

= f(x, y, z),

Ut

= h(x, y, z),

(2)

when t = o. Here f, hand F are continuous functions, differentiable to orders which suffice to ensure the existence of a solution ·with continuous second derivatives.

7.5]

[121

MARCEL RIESZ'S METHOD

With any fixed point (x, y, z; t) in space-time, let V be the region in r = 0 and r = 0, where

r ~ 0 bounded by

r

(t-r)2_(x-;)2_(Y-1])2_(z-S')2.

=

If u and v have continuous second derivatives, Green's transformation gives

Ir.

{u(;, 'i],

S'; r)Av(;, 1], S'; r) -v(;, 1], S'; r) Au(;, 1], S'; r)}d;d1]dS'dr =

f {Til s

(u ov -v oU) -;\(u ov -v oU) or or 0; o~ - p (u

:~ -

v :;) - v (u

:~ - v :~) } dB,

where B is the boundary of V and (;\,p, v, ro) are the direction cosines of the outward normal. As before, Au = u 77 -uf;;-u'l'l-u". Let u be the desired solution of our problem, and let v = r
Then

vVe assume in the first instance that re a > 4, and then use analytical continuation. We have

I

v {a(a- 2) u(;, 1], S'; r) r<,,-4)/2 - F(;, 1], S'; r) r<"-2)/2}d;d1]dS'dr =

Is [(a- 2) Ur<,,-4)/2{;\(; - x) + p(1] - y) + v(S' -z) +ro(r - t)} oU}] dB. ou ou 8u +r<,,-2)/2 {;\-+p-+v--m0; 01] oS' or

If we divide through by

= 2"+1n'r(ta) r(ta+ 1),

H(a+2)

we obtain I"u(x, y, z; t) -1,,+2F(x, y, z; t) =

1 H(a+ 2)

I

1:

{r
+ (a- 2) r&O:-4)/2tu(;, 1], S'; O)} d;d1]dS', where

ro=

t 2_(X-;)2_(Y-1])2_(Z-S')2

and 2: is the region of the plane r = 0 on which 0 ~ r 0 ~ t 2 • The reason for this is that when re a > 4, the integrand vanishes on the part of B belonging to the retrograde cone. On 2:, the rest of B, 5

CPO

122]

[7.5

MARCEL RIESZ'S METHOD

we have T = 0, A = It = )J = 0 and Til = -1. Using the initial conditions, ]"u(x, y, z; t) -]"+2F(x, y, z; t) 1

= H (a + 2) f

~ q,,-2)!2 h(f" 1j, t,) df, d1j dt, Q"-2)!2!(f" rj, t,)df,d1jdt,. a,_ ~tf 0 E

+ H( 1,

(3)

'»

The expression on the left is an analytic function of a, regular when re a > -1 when u has continuous derivatives of the second order; its value when a = 0 is

u(x, y, z; t) -]2F(x, y, z; t). If we introduce spherical polar coordinates f,

= x+lr, 1j = y+mr,

1= sin 0 cos ep,

where

t, = z+nr,

m=sinOsinep,

n=cosO,

the first term on the right of (3) is J,

1

=

1

H (a + 2)

ft r2drf dO. (t 2-r2)
0

3

" ,

do. 3 = sinOdOdep.

where

o ::;

and

0 ::::; 7T,

0::::; ep ::::; 27T.

Integration by parts gives

~ = - aH(~+ 2)fn, do. f~ drrh(x + lr, y +mr, z+nr) :r (t 23

=

aH(~+ 2)

fn.

r 2 )a12

do. 3 f: dr (t 2_r 2)a!2 {h(x+lr, y+mr, z +nr) + r :r h(x + lr, y + mr, z + nr)},

which is valid if h is continuously differentiable. Now lim aH(a+ 2) = lim 2a+27Tr(ta+ 1) r(ta+ 1) = 47T.

Hence

a-O

a-O

. ~ = -4 1 f do. ft dr~{rh(x+lr,y+mr,z+nr)} (3 hm 3 7T n, 0 ur

a --+ -;- 0

1

= 4

f

7T n,

th(x+lt,y+mt,z+nt) do. 3

= tJl(h; x, y, z; t),

7.5]

[123

MARCEL RIESZ'S METHOD

where vIt(h; x, y, z; t) is the mean value of h(f" 17,?;) over the sphere

(f,-X)2+(1J_y)2+(?;-Z)2 = t 2 • A similar expression follows for the second term on the right of (3). Lastly, by §7.3 (2),

IzHF(x,y,z;t)

a

=

f

r,,-1

2"+2n{r(1+ta)}2 w F (f,,1J,?;;t-R)T d f,d1J d?;dr,

where with a slight change of notation. Here W is 0 ~ R ~ t, r ~ O. When we integrate by parts with respect to r, r varies from 0 to ,J(t 2 -r2 ). We then have 1 f {[ F(f, 1J ?;.t_R)]v'(I'-r') r"+2F(x, y; Z, t) = 2"+2n{r(1 + t a )}2 v T"" " 0

Ii

- f:W-r') r":r F(f,,1J'i;t-R) dr}df,d1J d?;, where Vis

(X-f,)2+(Y-1J)2+(Z-?;)2

~

t 2.

We have transformed 1"+2F(x, y, z; t) into a function regular when rea> -1. Hence

1 2F(x,y,z;t) = ~f {[F(f,'1J,?;;t-R] 47T v R r=v'(l'-r') _fv'(l'-r') ~ F(f" 1J, ?;; t - R) dr} df,d1Jd?; o or R =

It follows that

-!...f F(f,,1J,?;;t-r)df,d d?;. 47T v r 1J

o

u(x,y,z;t) = tJ(h;x,y,z;t)+8itJ(f;x,y,z;t) +--!..f F(f,,1J,?;;t-r)dl:d d Y 47T v r S 1J S

(4)

is the required solution. It has continuous derivatives of the second order if h(x,y,z) and F(x,y,z;t) have continuous derivatives of the second order, and !(x,y,z) has continuous derivatives of the third order. Where F is identically zero, this reduces to the formula (7) of § 6.3, which we obtained by using Boussinesq's spherical potential. 5-2

124]

MARCEL RIESZ'S METHOD

The solution of (1) which satisfies the conditions u = 0, U t = -a is u(x,y,z;t) = 41 F(f"1j,s;t-r)df,d1jdl;,, 7T Va r where T'~ is now (f,-X)2+(1j_y)2+(l;,-Z)2::::; (t+a)2.

t

f

=

If we make a -+

[7.5

°on

+ 00, we get the well-known retarded potential

F(f,,1j,s;t-7')dl:d1jd Y u (x,y,z,·t) = ~f 4 s ~, 7T V r

where integration is through all space.

7.6 Babha's equation In his theory of the meson, Bhabha found it necessary to solve the differential equation 82u 82u 82u 82u 8t 2 - 8x 2 - 8 2- 8z 2 +k2u = F(x, y, z; t). (1) y The theory of this equation is well knovm, but rather difficult. vVe show how the use of Riesz's operator leads to the solution very simply. For brevity, we omit the rigorous details. The equation (1) is, in the usual notation, Lu+k2u = F. If u and

Ut

yanish when t = 0, 1 2Lu = L1 2u = u. Hence u + k 21 2u = 1 2F. k2i12iu + k2i+212i+2u

This implies that

=

1 2i+2F.

Omitting conyergence considerations, But 2r 1 F(x, y, z; t)

where

=

1 227 - 17T(r_ 1)! (r- 2) JvF(f" 1j,

r

=

s; r) rr-2df,d1jdl;,dr,

(t- r)2_ (X-f,)2- (Y_1j)2_ (z- S)2

and V is the region on which r r ~ 2. Hence

~

0,

°:: :;

r ::::;

t. This formula holds when

u(x,y,z;t) = 1 2F(x,y,z;t)

f

00' ( 1 )nk2n +2:22n+I ( -1)1 I F(f,,17, S;r) r n- 1 df,d1jdSdr. 1 7T n . n. v

7.6]

[125

MARCEL RIESZ'S METHOD

r 2 = (X-~)2+(Y-1])2+(Z-S)2,

If we write

R2

this becomes 1 u(x,y,z;t) = 4 7T

f

w

=

r

= (t-r)2- r 2,

F(;,1],s;t-r) d;d1] d s

r

-~f F(;,1],s,r)J,(kR)dl:d dYd 47T v R 1 S 1] S r,

°:: ;

where W is the sphere r ::::; t, and J1(kR) is the Bessel function of order 1. The form ofthe solution used by Bhabha differs slightly from this; his result can be obtained by assuming that U and 14 vanish on t = -a and making a-+ +00.

7.7 A mixed boundary and initial value problem The problem is to find the solution of 82u

82u

82u

Lu == 8t 2 - 8x 2- 8y 2 = F(x,y;t)

(1)

°

in y > O,t > 0, given that U =!(x,y),Ut = h(x,y) on t = O,y;:::. and = ¢(x, t) on y = 0, t ;:::. 0. The data are assumed to be continuous, and, in particular,j(x, 0) = ¢(x, 0), h(x, 0) = ¢t(x, 0). We may assume that ¢(x, t) == 0. For, if not, the substitution U = v + ¢(x, t) leads to the same problem with different F, ! and h and with v zero on y = O,t > 0. We wish to find u(x, y; t) when y > 0, t > 0. The retrograde characteristic cone

U

cuts the plane r =

°in the circle (;-X)2+(1]_y)2

=

t 2.

°

If t ::::; y, the disc with this circle as boundary lies in 1] ;:::. on which ~l and u t are given, so that the solution is that given in §7.4. But if t> y, the circle crosses 1] = and bounds a disc on part of which U and 14 are unknown. From now on, assume that t > Y > 0. As in §7.4,

°

f

v (a(a -1)

pa-3l/2 u

-

pa- 1l/2F) d;d1]dr

f

= s [(a - 1) pa-3l/2 u {A(; - x) + ,a (1] - y) + v( r - t)}

+ r
vu T } ] dB,

126]

MARCEL RIESZ'S METHOD

°

[7.7

where V is the region bounded by S on which r = 0, 1] = and r = 0, and (A, #, v) are the direction cosines of the normal to S drawn out of V. Since t > Y > 0, S consists of three parts. The part of S belonging to the retrograde cone contributes nothing when rea> 3. Next there is a segment 2: 0 of a disc in r = 0, defined by {;-X)2+{1]_y)2 ~ t 2,

Lastly there is a region 2: 1 in 1]

=

1] ~

0.

°defined by °

{t-r)2_{;-x)2 ~ y2,

~

r ~ t-y.

On 2:0 , A = # = O,V = -1; on 2:1> A = v = 0,# = -1. Using the conditions U = !(;, 1]), u t = h{;,1]) on r = 0, 1] ~ 0, and u = on 1] = 0, r ~ 0, we have

°

f

1

r<"-3l(2 u {;,1];r)d;d1]dr

27Tr{a-1) v

=

f

1 r<"-1 l/2F{;,1];r)d;d1]dr 27Tr{a-1) v

1l/2h{;, 1]) d;d1] {f~o r
1 27Tr{a+ 1)

+ :ef~, r~"-1)12!{;, 1]) d~d1] - f~1 rf"-1lI2~{;, r) d;dr} , r o = {r)7=0,

where

r 1 = (r)'1=o

and ~(x, t) is the unknown value of U v at (x, 0; t). We get over the difficulty that U v is unknovi'll on y of reflection. Let r = {t- r)2 - {x- ;)2_ {y+1])2.

°

(2)

=

°by a method

Then r = is the characteristic cone with vertex (x, - y; t). The retrograde part of r = cuts off from Va region li whose boundary is part of the retrograde cone, the same area 2: 1 on y = 0, and a segment 2: 2 defined by

°

(;-X)2+{1]+y)2 ~ t 2,

Y

~

0.

On 2: 2 , u = !, u t = h. Repeating the argument, we find that, when rea> 3,

f-

1 27Tr{a-1) r r<"-3lt2u{~,17;r)d;d1]dr l

-

is equal to

1

---:::=-:---

27Tr{a+ 1)

f-r
l/2F{£ -'

1]' r) d; d1]dr ,

7.7]

[127

MARCEL RIESZ'S METHOD

1 {f 1'(a-lJ/2h(; 1j)d;d1j+~f 1'
-f~1 q%-I)/2~(;,T)d;dT}, where

1'0

=

(1')T=O'

(1'),,=0

=

(r),,=o

=

(3)

r 1·

If we subtract (3) from (2) the terms involving ~ cancel. The expressions which appear in (2) and (3) are analytic functions of a regular when re a > - 1. Let r; andVsbe the parts of Von which t-y ::::; T::::; tandO::::; T::::; t-y. Then 1 f r,'!:>

is Riesz's integral of order a corresponding to the initial plane T

its limit as

a-.+O

= t-y;

is u(x,y;t). But the limit of

1 f r" S

is zero since (x, y; t) is outside

r;. Similarly the limit of

is zero. It follows from the limiting forms of (2) and (3) that .t)=.2-f F(;,1j;T)dl:d d -~f F(;,1j;T)dl:d d 27T v ,.,jr S "I T 27T v, .jr S "I T

u (x, y,

+.2-f h(;,1j)d;d1j-..l-f h(;!-1j)d;d1j 27T 1:, ,.,jr0 21T I:, ,.,jr0

+..!- ~f !(;, "I) d;d1j _..l- ~f 27T 8t

1:,

,.,jr0

21T 8t

1:,

!(;!-1j) d;d1j. ,.,jr0

This is precisely the result we should have got if we had extended the range of definition of F,f and h to y < 0 by requiring them to be odd functions of y. If we are given Cauchy data on t = 0,0 ::::; y ::::; 1, and if u is zero on y = 0 and on y = 1, we can solve the problem by requiring F,! and h to be odd functions of y, periodic in u of period 21.

128]

[7.8

MARCEL RIESZ'S METHOD

7.8 A note The problems discussed in this chapter seem to have a rather restricted scope. The Riesz integral has proved of use in some branches of mathematical physics, such as the solution of Maxwell's equations, the theory of the meson field, quantum electrodynamics. Riesz has also developed integrals of fractional order associated with Laplace's operator and Viith the parabolic operator of the equation of heat. For the connexion of Riesz's theory with semi-groups, see Einar Hille's American Matheplatical Society's Colloquium Publication Functional Analysis and Semi-Groups, of which the first edition appeared in 1948. Exercises 1. Iff(x) is continuous, prove that l"f(x)

=-1

r(a)

IX f(t) (x-t)"-ldt 0

is an analytic function of a, regular in rea> O. If f(x) has a continuous derivativef'(x), show that l"f(x)

=

IX

x" 1 r(a+1/(OH r (a+1) /'(t) (x-t)"dt

is regular in re a > -1, and hence that 1°f(x) = f(x). Prove also that 1"I/lf(x) = 1"+/lf(x) and that d dx 1"+lf(x)

=

l"f(x).

2. Marcel Riesz's operator associated with the differential operator L, where and Cauchy data on t = 0 is l"u(x·t) ,

I

= 2"-1(r(la)}2 1 u(s'r)rl"-2)/2dsdr v ' ,

where r = (t-r)2_(x-s)2 and V is defined by r ~ 0, 0 ,,;;; r ,,;;; t. Prove the characteristic properties of l"u(x;t). u(x;t) is the solution of Lu-u = F(x;t) which satisfies the initial conditions u = f(x), u t = h(x) when t = O. Prove that , 1 2"u-1 2"+2u "

fx+t r"h(s)ds x-t ° 1 cIx+tt rH(s)ds, + 2 2"+l{r(a + 1)}2 8t 1

= 1 2"+2F+ 2 2 "+l{r(a+1)}2

x-

[129

MARCEL RIESZ'S METHOD

where f °is the value of f when r

= O. Deduce that

3. u(x, y; t) is the solution of 82 u

02U

82u

8t 2 - 8x 2 - 8

y

2 -It

= F(x, y; t)

with continuous derivatives of the second order. When t = 0, u =f(x,y), = h(x, y). The functions F and h have continuous derivatives of the second order,f of the third order. Prove by Riesz's method that

Ut

u(x,y;t)

1 = 7)

f

.1T

v

F(;,7J;r)

cosh .jf 1 .jf d;d7Jdr+7)

f

~

.1T

h(;,7J)

cosh.jf° .jf d;d7J 0

~ ~f f(C ) cosh.jf odCd

+27Tot

where

f

~ <",7J

.jfo

<" 7J,

= (t-r)2_(x-;)2_(Y-7J)2,

f o = t 2- (X-;)2- (Y-7J)2,

and where V is f

~

0, 0 ,,;;; r ,,;;; t, and

~

is f ° ~ O.

4. u(x; t) is a function of the variables x = (Xl' x 2' ... , X 2m - l ) and t with continuous partial derivatives of order m-1. The function U(x; t) is defined by U(x; t)

where, on V,

f

=

f

= v u(!;; r) d!;dr,

(t-r)2-lx-!;12 ~ 0,

0,,;;; r ,,;;; t.

If where V'2 is Laplace's operator in the x-space, prove that LmU(x; t) = 22m -

Ut

I

rrm- l (m -1) ! u(x; t).

If u is a solution of Lu = 0 which satisfies the initial conditions u =f(x), = h(x) when t = 0, prove by applying Green's transformation to

that

130J where

MARCEL RIESZ'S METHOD

r 0 is the value of r n(x; t)

=

2 2m +l

when T

:-.1

7T

I Lm

m.

= 0 and on TV, r 0 = O. Deduce that

{fIf' rog(!;) d!; +-(}~tf TV r o!(!;) d!;}

and hence that n(x; t)

=

22m-17Tm-~ (m-l) ! Lm-1 {fw g(!;) d!; +~ fw!(!;) d!;}.

To what does this reduce when m = 1 ?

5. D is the triangle bounded by the lines; x+y> O. An operator Ja is defined by Jau(x, y)

= {~}2f r(a)

D

= x,

"I

= y, ; +"1 = 0 where

u(;, "I) (x_;)a-1(Y_?J)a- 1d;d1J.

Prove that, if u is continuous, Jau is an analy--tic function of a regular in re:x. > 0, and that, if u has continuous first derivatives, Jau tends to u as :x. -+ + O. Show also that JaJfJu(x, y) = Ja+fJu(x, y),

82

8x8y Jau(x, y)

=

Ja-1 u (x, y).

u is the solution of U"y + u = F(x, yj which satisfies the conditions u = 0, u" = 0 orr x+y = O. By applying Green's transformation to

f

D (uvf,'1- VUf,'1)

where

v

=

d;d?J,

(x-;)a(Y-?J)a,

show that, when re a > 0, Jau+Ja+lu

=

Ja+1F.

Deduce that

where J o is Bessel's function of order zero.

8

POTENTIAL THEORY IN THE PLANE

8.1 Gravitation The simplest equation of elliptic type is Laplace's equation where

V'2U

= 0,

This equation with n = 2 or 3 is of frequent occurrence in mathematical physics, notably in the theory of gravitation. N ewton's law of universal gravitation asserts that every particle of matter in the universe attracts every other particle with a force whose direction is that ofthe line joining them, and whose magnitude varies directly as the product of their masses and inversely as the square of their distance apart. With an appropriate choice of units, the attractive force is mm'Ir 2 where m and m' are the masses of the particles and r their distance apart. If a particle of unit mass moves under the attraction of a particle of mass m fixed at, say, the origin, the increase in the kinetic energy of the particle of unit mass as it moves from a position Po to a position Pis

m

The expression

u=-

OP

is called the gravitational potential of m. If the coordinates of the variable point Pare (x,y,z), 82u

82u

82u

8x 2 + 8y 2 + 8z 2

=

0;

and the force is grad (mlr). By a generalisation of Newton's law, the attractive force due to a continuous distribution of matter of density m(x, y, z) in a volume V acting on a particle of unit mass at a point (x,y,z) outside V is equal to gradu, where u= v m(;, 1], ~d;d1]dS'

fff

s)

[ 131 ]

132]

[8.1

POTENTIAL THEORY

R2 = (X-;)2+(Y-1])2+(Z-{;')2.

where

This potential satisfies Laplace's equation outside V, but in V it satisfies Poisson's equation V'2U = - 47Tm. Similar formulae hold for the gravitational potential of distributions of matter on curves or surfaces. For example, an infinitely long uniform straight thin wire of mass m per unit length located on the z-axis attracts a particle of unit mass at (x, y, z) with a force normal to the ·wire of magnitude IOO r d {;', -00 m RS r 2 = X2+y2,

where

R2

= r 2+(z_{;')2.

This is equal to 2mjr and is independent of z. If the unit particle moves from a position Po to a position P under the attraction of the wire, its increase in kinetic energy is r 2m log.J!, r where r o and r are the distances of Po and P from the wire. The expression 1 2m logr is the gravitational potential of the wire; its gradient is the attractive force. More generally, if we have an infinitely long straight rod of cross section D parallel to the z-axis, the density O"(x, y) being independent of z, it attracts a unit particle at (x, y, z) outside the rod with a force 2 grad u, where u = IIn O"(;,1])log~d;d1], R2

where

=

(X-;)2+(Y-1])2.

Such a function u is called a logarithmic potential. It is independent of z and satisfies Laplace's equation outside the rod; but inside the rod, V'2u = - 27T0"(X, y). There is one important difference between the Newtonian potential and the logarithmic potential. If V is bounded, u == I I Iv

m(~,.1], {;') d;~d{;' '" ~(X2+~2+Z2)II

Ir,

m(;, 1],

{;')d~d1]d{;'

at great distances. But if D is bounded, u = IIn

O"(;,1])log~d;d{;''" log ~(X2~y2)IIn 0"(;, 17)d;d1].

8.1]

[133

POTENTIAL THEORY

The Newtonian potential behaves like the potential of a particle, and vanishes at infinity. The logarithmic potential behaves like the potential of a thin wire, and is infinite at infinity. In this chapter, we deal with the solution of Laplace's equation and Poisson's equation in the plane. Fundamental tools in the theory are the first and second identities of Green, which are consequences of Green's theorem enunciated in Note 5 of the Appendix. Let D be a domain bounded by a regular closed curve C. Let ¢, ljr and their first partial derivatives be continuous in the closure of D. Then, if we put u = ¢ljrx' V = ¢ljrll in Green's theorem, we have

IID (¢V ljr+¢xljrx + ¢lIljrlI)dxdy = Ia ¢ :t ds , 2

where 8ljrJ8N is the derivative along the outward normal, provided that ljrxx and ljrllll exist and are bounded in D and the double integrals of ¢ljrxx and ¢ljrllll over D exist. This is Green's first identity. It holds when ¢ and ljr are interchanged, provided that ¢xx and ¢VII exist and are bounded in D and the double integrals of ljr¢xx and ljr¢lIlI over D exist. By subtraction, we have Green's second identity

I IfD a m (¢V2ljr_ljrV2¢) dxdy =

(¢ 8ljr -ljr 8¢) ds.

m

Simple conditions for the truth of these results are that ¢ and ljr should have continuous second derivatives in 15, the closure of D. D Il?-ay be a multiply-connected domain bounded by several nonintersecting regular closed curves. These results still hold, the integral over C being replaced by the sum of the integrals over the boundary curves; N is still the unit normal vector drawn out of the domain D.

8.2 Green's equivalent layer It is convenient to use (;,1]) as the coordinates of the integration point and (x, y) as the coordinates of a fixed point, usually the coordinates of the point at which the solution is desired; and we use ~ to denote Laplace's operator with (;,1]) as variables. In Green's second identity, put 1Jr = log l/R where R2 = (X-;)2+(Y-1])2.

Then if C is a regular closed curve bounding a domain D, we have

134]

[8.2

POTENTIAL THEORY

This is certainly true if ¢ has continuous second derivatives in D, provided that (x, y) is not a point of D. If (x,y) is a point of D, the region R ::::; e lies in D for all sufficiently small values of e. Let Do be the domain D when the region R ::::; e has been deleted; and let 0 0 be R = e. Then

Ie {¢ o~Tlog~ - :tTIog~}dS+ Ie, {¢ o~log~ - :tlog~}dS = - IIn, log~ t!.¢ df,d1J. On 0 0 , %N = - %R. The integral over 0 0 is then

I {Yl+ I:" e,

=

¢

8¢

1}

oR log Yl ds

{¢(X +e cose, y +e sin e) +e¢.(x+e cos e, y+esin e) log~} de

-+ 27T¢(X,

y)

as e -+ 0, since ¢ is continuously differentiable. Therefore

1 ¢(x, y) = - 27T

IInlOg~ .t!.¢ .df,d1J .

1

I (0

1

1

o¢)

- 27T e ¢ oN log Yl -log Yl oN ds,

(1 )

when (x, y) is a point of D. But when (x, y) is outside the curve 0, the value of the expression on the right of (1) is zero. If ¢ is a solution of Laplace's equation, we have 1

I

1

o¢

1

f

0

1

¢(x,y) = 27T e logYl oN ds - 27T e ¢ oN log Yl ds .

(2)

The first term is the logarithmic potential of a distribution of matter of line density on o. The second term is the logarithmic potential of a distribution of normally directed doublets of linear strength 1

27T ¢. These two distributions on 0 form Green's equivalent layer. It must be emphasised that the formula does not provide a solution of the

8.2]

[135

POTENTIAL THEORY

problem of Cauchy; we cannot assign ¢ and

c. If ¢ satisfies Poisson's equation 'i/ 2¢ = -

o¢/oN independently on

21T(j,

there is an additional

term

in (1). This is the potential of a distribution of matter of density (J' on D. If ¢ satisfies Poisson's equation everywhere, we may take C to be a circle of large radius. If the integral over C in (1) tends to zero as the radius tends to infinity, then

holds everywhere, integration being over the whole plane.

8.3

Properties of the logarithmic potentials

If (J'(x, y) is bounded and integrable over a domain D, the logarithmic potential

when R2 = (X-;")2+(Y-1])2, is continuous everywhere. Its first derivatives are also continuous, and can be found by differentiating under the sign of integration. If (J'(x, y) has continuous first derivatives on D, the second derivatives of u are continuous on D and satisfy Poisson's equation 'i/2u = - 27T(J'. If we differentiate formally, we obtain the expressions

We have to showthatu, X, Yare continuous and that grad u = (X, Y). If we introduce polar coordinates, ; = x+Rcose,1] = y+Rsine, we obtain u(x,y) =

ffn (J'(x+Rcose,y+Rsine)Rlog~dRde,

with corresponding results for X and Y. Since (J'(x, y) is bounded and integrable, the integrals defining u, X and Y converge everywhere. On the exterior of D, u has continuous derivatives of all orders, obtained by differentiation under the sign of integration; it satisfies Laplace's equation there.

136]

POTENTIAL THEORY

To deal with the case when function

1

= 10g:n

(x,y)

[8.3

is a point of D, introduce the

(R ~ 8),

where 8 is so small that the disc R < 8 lies in D; denote the disc by Do. ThenJ(R; 8) is continuous and continuously differentiable everywhere. If

then On Do, the expression in square brackets is not negative; and there exists a constant K such that 10'"1 ~ K on D. Then

lu-u81~

2rrKJ: [ -RlogR-!R+Rlog 8+ ~:2] dR = t rrK82 .

Hence U 8 tends to u uniformly as 8-+ also is u. Similarly,

+ 0. Since u 8 is continuous, so

X

x - ~:8 = JJDo 0'"(;,71) [;';2 -;~2X] d;dlj

8~-;:2 (;-x)d;d7J. Do 2 X _ ou8! ~ 2 KJ8 8 - R 2dR = 4rrK8 ox "rr 0 82 3 . ! = JJ O'"(;,7J)

and so

Hence, as 8-+ + 0, OU8/0X converges uniformly to X. Since OU8/0X is continuous, X is continuous everywhere, and ou/ox = X. In the same way, Y is continuous, and ou/oy = Y. On the exterior of D, that is, on the complement ofthe closure of D, u has continuous derivatives of all orders and satisfies Laplace's equation. In order to deal with second derivatives on D, we assume that 0'" has continuous first order derivatives. 'Ve could have assumed that 0'" has piece-wise-continuous first derivatives, or, even that it satisfies a Heilder condition; these weaker conditions make the analysis more difficult.

8.3]

[137

POTENTIAL THEORY

Let D 1 be an open disc contained in D, and let D 2 be the rest of D. Write

u 1 (x, y) =

ff

D, <7(;, 1]) log~ d; d1] , 'u 2(x, y) =

ff

D, <7(;, 1]) log ~ d;d1].

The function u 2 has continuous derivatives of all orders on D 1 and satisfies Laplace's equation there. If (x, y) is any point of D v

Since <7 has continuous first derivatives, we have

=-

Jo,A<710g~dS+ JJD,:;log~d;d1],

where 0 1 is the boundary of the disc D 1 and (A, fl) are the direction cosines of the outward normal. This expresses o~/ox as the sum of the logarithmic potential of a continuous distribution of density O<7/ox over D 1 and the logarithmic potential of a continuous linear distribution over 01' Hence ou1/ox has at each point of D 1 a continuous derivative with respect to x or y; and

V'2u = V'~

0

1

= J a, <7 oN log Il ds +

If

(O<7;-X O<71]-y) D, 0; R2 + 01] R2 d;d1].

This holds when (x,y) is any point of any disc D 1 contained in D. Now take 0 1 to be the circle R = € with centre (x, y), € being chosen so that D 1 is contained in D. Then 0 1 Ja, <7 oNlogIlds

= -

J2" <7(x+€cos8,y+€sin8)d8, 0

138]

POTENTIAL THEORY

which tends to - 2mr(x, y) as € -+ double integral over D 1 is equal to

J: J:" dR

°since

[8.3

CT is continuous. Also the

de{coseCTx(x+Rcose,y+Rsine)

+ sin eCTy(X +Rcos e, y+Rsin e)}, which tends to zero as € -+ 0, since CTx and CTyare continuous. But V'2 U does not depend on €, and so is equal to the limit of\i 2 u1 as € -+ 0. Therefore V'2 U

= - 21TCT(X, y).

8.4 Some other logarithmic potentials We have already met two other forms of logarithmic potential which are of importance in mathematical physics. As we shall not make much use of them, their properties will be stated without proof. Let C be a regular arc with parametric equations x = !(8), y = g(8), 8 being the arc length. If fl(8) is continuous, the logarithmic potential u(X, y)

where

J

= Ofl(8) log~d8,

R2 = (X-;)2+(Y-1J)2,

(;,1J) being the coordinates of the integration point, is continuous everywhere. At points not on C, u has continuous derivatives of all orders and satisfies Laplace's Equation. Suppose, in addition, that fl(8) has a bounded derivative and that C has bounded curvature. Let the unit tangent and normal vectors at a typical point Po(x o' Yo) of C be To and No. If P crosses the surface along the normal at Po, ou(x, y)/oTo is continuous, but ou(x, y)/ol\~ is not. As P moves to Po along the normal,

J

ou(x,y) lj-yo oNo -+ ±fl(Po) + Ofl(8) (;-X )2+(1J-YO)2 d8 , O

the sign being + or - according as P approaches Po in the direction No or - No. Thus the tangential component of grad u is continuous, but the normal component is not. The logarithmic potential of a particle of mass -111 at the origin and a particle of mass +m at (h, 0), where h > 0, is m log r - m log (r 2 - 2h,'J,; + h 2 )1z

8.4]

[139

POTENTIAL THEORY

at the point (x,y). If we use polar coordinates, this is 2

2

h h ) = -cosO+mO mh (h0- ) -!mlog ( 1-2-cosO+r r2 r rif hlr is small. If we make h-+ 0, mh-+fl, we obtain

cosO fl- -, r which is the potential of a doublet at 0, directed in the positive direction of the x-axis; it can be written as

If the doublet is in the direction of the unit vector N, its potential is

8 fl8Nlogr.

The second type of logarithmic potential in Green's equivalent layer is that due to a normally directed distribution of doublets. If we have a normally directed distribution of doublets on a regular arc C, the strength per unit length beingfl(8), the resulting potential is u(x,y) =

J o

fl(8)

8~logRd8.

The gradient of u is continuous across C, but u itself has a jump 27Tfl. The value of u at a point of C is the mean of the limits as we approach C from the two sides.

8.5 Harmonic functions A function u(x, y) is said to be harmonic in a domain D if it has continuous second derivatives and satisfies Laplace's equation. In this section, we discuss the properties of harmonic functions by means of Green's theorem. We assume that D is bounded but not necessarily simply connected. The boundary 8D of the domain is assumed to consist of a finite number of non-intersecting regular closed curves, in the terminology of Note 4 of the Appendix. If D is simply connected, 8D consists of one regular closed curve; D is then the domain inside 8D. The closure of D, denoted by l5, is the union of D and 8D.

140]

POTENTIAL THEORY

[8.5

If u is harmonic in a bounded domain D I containing 15, then

J

oD

8u 81\7 ds =

o.

a/eN denotes differentiation along the normal to aD drawn out of D. If the direction cosines of the normal are (l, m), the integral is equal to Green's theorem is applicable because 'II has continuous second derivatives in D I . If 'II has continuous second derivatives in a bounded domain D and if

J

Co

:;Tds = 0

for every regular closed curve 0 0 bounding a domain Do contained in D, then 'II is harmonic in D. The condition implies that

JJDo "V 2udxdy = 0 for every domain Do. Hence 'II satisfies Lapl~ce's equation everywhere in D. If 'II is harmonic in a disc D with centre (xo, Yo) and radius R and if 'II is continuous in 15, then 1

u(xo,Yo) = 27T If 0 <

r

J211 . 0 u(xo+Ecose,Yo+Esme)de.

Jc:;dS =

< E,

where 0 is the circle ordinates, we have "211 8

0,

(X-X O)2+(Y_YO)2

= 1'2. Hence, using polar co-

Jo -;or:- u(xo+rcose,yo+rsine)de = 0,

and so, if 0 < R' < E,

J:' J:" dr

:r u(x o+ r cos e, Yo + r sin e) de = O.

Since 'II is continuous in 15, we may invert the order of integration and obtain

8.5]

POTENTIAL THE OR Y

[141

Therefore u(x o, Yo)

="..7T f211 u(xo+R'cos8,Yo+R'sin8)d8 1

0

for every R' < R. Since u(xo+rcos8,yo+rsin8) is uniformly continuous when 0 ~ r ~ R, we may proceed to the limit as R' -+ R - 0 under the sign of integration and obtain the desired result, which is called Gauss's mean value theorem. A corollary of this result is obtained as follows. If 0 < R' < R, u(x o, Yo) and so

J

R. rdr

o

=" JR' 1 ..7T

u(x o, Yo)

0

=

rdr

J211 u(xo+rcos8,yo+rsin8)d8, 0

7T~'2 JJu(x, y) dxdy,

where integration is over the disc (x - xo)2 + (y - YO)2 < R'2. Making R' -+R-O, we obtain u(xo, Yo) =

7T~2 JJD u(x, y) dxdy,

where D is the disc with centre (xo' Yo) and radius R. Let u be harmonic in a bounded domain D and continuously differentiable in Jj. If u vanishes everywhere on oD, u vanishes identically in D. If ou/oN vanishes everywhere on oD, u is constant in D. This follows from the identity

If u vanishes everywhere on oD, or if ou/oN vanishes everywhere on oD, the double integral vanishes. Since U x and u y are real and continuous, U x and u y vanish everywhere on D, and so u is constant. The two results follow. Let U l and U2 be harmonic in a bounded domain D and continuously differentiable in D.If U l = U2 everywhere on oD, U l = U2 everywhere on D. If oul/oN = oU2/oN everywhere on oD, U l differs from U2 by a constant. This result, known as the uniqueness theorem, follows from the previous one if we put u = U l -1t 2 . A different proof of this result, which uses the maximum principle, is given at the end of this section. If u is harmonic in a bounded domain D, it possesses derivatives of all orders, themselves harmonic in D.

142]

[8.5

POTENTIAL THEORY

If (Xo' Yo) is a point of D, it is the centre of a disc K whose closure lies in D. If (x,y) is any point of K, by formula (2) of §8.2,

lf

lf

lou

0

1

u(X,y)=27T oKlog:RoNds-27T oKueNlog:Rds,

where

R2

=

(1)

(X-~)2+(Y-l1)2,

(G,1]) being a typical point of oK. Since any number of differentiations can be carried out under the sign ofintegration, u has derivatives of all orders on K, and thence throughout D. Since

Ux

is harmonic in D - similarly for all the derivat,ives. Let U mn be the value of om+nujoxmoyn at (xo,Yo)' Then

is the formal Taylor series expansion of u. It can be shown that the series is uniformly and absolutely convergent on a square Ix-xol + IY-Yol < k, and hence that a junction harmonic on a bounded domain D is analytic there. Suppose that oK has radius a, so that G = xo+acos¢,

On

oK,

u =j(¢),

1] = yo+asin¢.

OU

oN = g(¢),

wherej(¢) and g(¢) are differentiable as often as we please. If x

= X o+

r cos 8,

y

=

Yo + r sin 8

we have Hence

00 rn logE = loga- ~ -cosn(8-¢), 1 nan

the series being uniformly and absolutely convergent when

o :(

r :( a' < a.

8.5]

POTENTIAL THEORY

[143

Substituting these series in (1) and integrating term-by-term as we may when r ~ a', we have 00 rn u(x,y) =ao+~ (axcosnO+bnsinnO)-, (2) 1 an where

ao =

2~J:"j(ep)dep, 1

an = 2 J211 j(ep) cos nepdep +" a J211 g(ep) cos nep dep , .rrn 0

rr 0

1 J211 a J211 g(ep)sinnepdep. bn =-2 j(ep)sinnepdep+-,,rr 0 .rrn 0

The coefficient a o does not involve g(ep) since

J:" g(ep)dep Now

=

O.

rncosnO = (x-x O)n_nC2(x-x o)n-2(Y_Yo)2+ ... , rnsin nO = nC1(x-xO)n- nC3(x _xo)n-3(Y_Yo)3+ ....

The general term of (2) then has the absolute value rn

Ian cos ne + bn sin nO I an ~

M

n a

n

~ ncp Ix-xol n- p IY-Yol p

p=O

M

= an {Ix-xol + IY-Yol}n. We can therefore substitute in (2) for rncosnO and rnsinnO in terms of x and y, and obtain a double series which converges uniformly and absolutely with respect to x and Y in any square

Ix-xol + IY-Yol ~ k, whenever k < a. Hence u(x, y) is analytic in D. Ij u is harmonic on a bounded domain D and continuous on D (the closure oj D), it is bounded on D and attains its supremum on the boundary aD. Ij it also attains its supremum at a point oj D, it is constant. (The maximum principle). Let 1.11 be the supremum of u on D. Since u is continuous, there is at least one point of D at which u = M. We have to show that if u = M at a point inside aD, then u is a constant; it will then follow that, ifu is not constant, it must attain its supremum only at a point of aD.

144]

POTENTIAL THEORY

[8.5

Suppose that u attains the supremum lJ1 at a point (xo, Yo) of the bounded open connected set D. Since u(x o, Yo)

=

1 27T

J211 0 u(xo+Rcos8,Yo+Rsin8)d8

for every circle (X-X O)2+ (Y_YO)2 = R2 lying in D, u is equal to lJ1 on every such circle. It follows that u = 1.11 on the largest disc with centre (x o, Yo) contained in D. LetFbetheset of points of D at which u = lJ1. Since uis continuous, F is a closed set. 'We have to show that F is D. If this is not so, there is at least one point P of of which belongs to D. But since P belongs to F, u is equal to M at P. Hence there is a disc with centre P which lies in D and on which u is equal to M. This disc is therefore part of F, and P is an interior point of F and so does not belong to of, a contradiction. Hence F is D, and u is constant on D. In particular, if u is harmonic on a bounded domain D and continuous on D, and if u is identically zero on oD, then u is zero everywhere on D. For the suprema of u and - u are both zero on D. From this follows a uniqueness theorem. If U 1 and U 2 w'e harmonic in a bounded domain D and continuous on D, and if U 1 = U 2 everywhere on oD, then U 1 = U 2 everywhere on D. This is a consequence of the preceding result; take u = U 1 - U2 .

8.6

Dirichlet's principle

Let D be a bounded domain whose boundary oD consists of a finite number of non-intersecting regular closed curves. The problem of Dirichlet is to show that there exists a function u(x, y), harmonic in D and continuous in D, which satisfies on oD the boundary condition u = f, where f is a given continuous function. If there is such a function, it is, as we have seen, unique. Let 5:2t be the family of all functions u, which are continuous in D and have continuoul;! second derivatives in D, and which satisfy the boundary condition u = f on oD. If u belongs to 5:2t ,

'I(u)

=

JJn (u;+u~)dX~y

exists and is not negative. The set of all such integrals has a nonnegative infimum m. But it is not necessarily the Case that there exists a function U, belonging to 5:2t for which I(U) = m. Assuming that there is such a function, let us consider u = U +eV where V belongs to 5:20 and e is a constant. Since V = 0 on oD, U + e F

8.6]

POTENTIAL THE OR Y

[145

belongs to I!2t , and I(U +eV) has the minimum value m when e = 0, no matter what function V of I!2o we use. Then, by the ordinary calculus rule, But

fan V:~d8 = ffn (Uxv~+Uy~)dxdy+ ffn VV 2Udxdy. The integral on the left is zero. Hence

ffn VV2Udxdy = 0 for every function V belonging to I!2o. This implies that V2U = 0 throughout D. For suppose V2U is positive, say, at some point Po of D. By continuity, there exists a disc K wIth centre Po on which V2U is positive. But we can choose a function V of I!2o which is zero outside K and positive inside K; and then

ffn VV2Udxdy > o. So V2U is not positive anywhere on D; similarly it is not negative. Hence All we have proved is that, if there is a function U belonging to I!2f for which l(u) attains the infimum m, then U is harmonic. The assumption that there is a function which minimises the integral1(u) was called by Riemann Dirichlet' 8 principle. The principle remained suspect until, in 1899, Hilbert showed that it can be used under proper conditions on the domain, the boundary values and the class of admissable functions.

8.7 A problem in electrostatics Consider an infinity long perfectly conducting cylinder with generators parallel to the z-axis. Suppose that the plane z = 0 cuts the cylinder in a regular closed curve bounding a domain D. An infinitely long line charge of density e per unit length parallel to the z-axis through the point (x o, Yo' 0) of D has electrostatic potential 2elog where

1

R,

146]

[8.7

POTENTIAL THEORY

If the conducting cylinder is earthed, a surface charge of density (T is induced on the conductor. This surface charge produces an electrostatic field of potential u inside the cylinder; u does not depend on z, is harmonic in D and continuous in D. The total field inside the cylinder has potential 1 U = u+2elog:n'

The electric force is grad U. On first sight, we should expect on physical grounds, that U would haye continuous first derivatives on D. But since U is zero on the conductor, the electric force is normal to the conductor and has magnitude aU/aN. As we go round aD, the direction of the normal vector changes suddenly at each corner, if such exist. The direction of the electric force would then seem to change suddenly at a corner. This is impossible unless aU/aN either tends to zero or to infinity as we approach the corner. Since the charge density (T is equal to (1/47T) aU/aN, the total charge per unit length is

J au

1 d 47T oD aN s

and this is equal to - e. Thus, although aU/aN may be infinite at a finite number of points of aD, it is integrable. The potential U, with e = t, is called the Green's Junction, and is denoted usually by G(x,y;xo,yo)' No doubt it is physically obvious that Green's function exists; to prove mathematically that it exists is equivalent to proving the existence of the solution of a particular case of the problem of Dirichlet. The same physical argument can be applied when aD consists of a finite number of non-intersecting regular closed curves.

8.8 Green's function and the problem of Dirichlet In this section, we show how a knowledge of Green's function enables one to solve the problem of Dirichlet for a bounded domain D. 'We assume that D is simply connected and that its frontier is a regular closed curve aD. If u is continuously differentiable in D, and if u has continuous second derivatives in D, then 1 ( ( u aloer au ) dS+-1 u(x,y)=? ,":~T R -"I\TlogR 27T _7T. oJ) 0..1.\ u1

If 01

D

Llu(~,11)logRd~(11/

when (x, y) is a point of D, provided that the double integral exists: here

POTENTIAL THEOR Y 8.8] If v is harmonic in D and continuously differentiable in

J

have

1 (av au) 1 O=21T aD uaN-v aN +21T

If

[147

D,

we also

D(j.u(;,rJ).v(;,1J)dgd1J.

Subtracting, u(x, y) = -

2~ JOD {u a~v (v-logR) -

(v-logR) :;} ds

1 - 21T I ID(j.u(;, 1J) {I:(;, 1J) -logR}d;d1J.

Now suppose that D has a Green's function G(;,1J;x,y) = v(;,1J;x,y)-logR

which vanishes when (;,1J) is a point of aD. vVith this value of v, u(x, y)

= -

1 21T JOD u(;, 1J) :;'dS-

2~ IID (j.u(;, rJ) G(;, 1J; x, y)d;d1J. (1 )

This holds if G is continuously differentiable with respect to ; and 1J in D, or, more generally, if aG/aN is integrable round aD. If u(x,y) satisfies Poisson's equation V'2u = -21TCT(X,y), this becomes u(x, y)

J

= - ~1T ..} aD u(;, 1J) ~~T ds +JJ 0"(;, 1J) G(;, 1J; x, y) d;d1J. Ulv D

Presumably this would give the solution of Poisson's equation in D when u is known on aD. It is necessary to prove the existence of Green's function and to show that 1 u(x, y) = - 21T JOD!(;' 1J) :;. ds + J J D 0"(;, 1J) G(;, 1J; x, y) d;d1J (2)

tends to !(xo' Yo) as (x, y) tends, on D, to any point (xo, Yo) of aD. The corresponding solution for Laplace's equation is (3)

8.9 Properties of Green's function Let D be a bounded domain whose frontier consists of a finite number of non-interesting regular closed curves and which possesses a Green's function G(x,y;xo,yo)

where

= v(x,y;xo,yo)-logR,

148]

[8.9

POTENTIAL THEORY

(x o, Yo) being a fixed point of D. Regarded as a function of the coordinates (x, y) of a variable point of D, v is assumed to be harmonic in D, continuous in D and continuously differentiable in D. The integral of 8Gj8N round 8D is - 27T. This follows from equation (1) of §8.8 if we put u == 1. The physical argument suggests that it holds even if 8D has corners. For every pair of points (x, y) and (xo, Yo) of D, G(x, y; x o, Yo) > o. For values of e which are not too large, the closed disc R ~ e lies in D. Let 00 be the circle R = e, Do the domain obtained by deleting the disc from D. Regarded as a function of (x, y), G is harmonic in Do, continuous in Do. On 8D, G is zero; on 00' G is positive since G-+ +00 as R-+O. By the maximum principle, G attains its supremum and infimum in Do on the boundary of Do. Hence G is strictly positive when (x, y) is any point of Do. Since e can be as small as we please, this proves the result. The Green's function possesses the symmetry property G(x, y; x o, Yo)

= G(x o, Yo; x, y)

for every pair of points (x, y) and (xo' Yo) of D. Write G(x,y;x o, Yo) = Go(x,y), G(x,y;XVYl) = Gl(x,y).

Let R o and R l be the distances of (x, y) from (x o, Yo) and (xv Yl) respectively. Let 00 be the circle R o = 8, 01 the circle R l = e, where 8 and e are such that 0 1 and 8D do not intersect; then the discs R o < 8, R l < e lie in D. Since Go and Gl are harmonic in the domain bounded by 00' 0 1 and 8D,

°,°

J (a°8N CO

8Gl

G 8Go) l8N ds+ +

J( J( C,

oD

8Gl G 8Go) G08N - l8N ds 8Gl 8Go) G08N -G18N ds =

o.

The third integral vanishes since Go and Gl vanish when (x, y) is on 8D. Near (xo,Yo), G1(x,y) is continuous, but Go is of the form v(x, y; x O' Yo) -logRo'

where v is harmonic. It follows that the first integral tends to - 27TG1(x o, Yo)

as 8-+0; similarly the second integral tends to 27TG o(X V Yl) as e-+O. Hence G1 (xo' Yo) = Go(x v Yl)'

8.10] 8.10

POTENTIAL THEORY

[149

The case of polynomial data

In continuation of §8.8, let f(x,y) be any polynomial. If there is a function u(x, y) harmonic in D and continuously differentiable in D, which takes on the boundary aD the same values as f(x, V), then U

=

u-f(x,y)

vanishes on aD and satisfies Poisson's equation

V'2U = - 27TCT(X, V), where CT

=

V'2ff47T is also a polynomial. By (2) of §8.8, we have U(x,y)

= JJn CT(;, 1]) G(;, 1]; x, y) d;d1].

(1)

Hence, instead of (3) of §8.8, we have a different formula u(x, y)

= f(x, y) +J J n CT(;, 1]) G(;, 1]; x, y) d;d1]

(2)

for the harmonic function. Since G(;, 1]; x, y) vanishes when (x, y) is a point of aD by the symmetry property, the function defined by (2) does take the prescribed valuef(x, y) when (x, y) is a point of D. This is not enough; we have to show that (1) gives a function harmonic in D and continuous in D; for, if this were so, u(x, y) would tend to f(x, y) as (.1:, y) moves up to the boundary. The Green's function G(;,1];x,y)

we write as

=

G(;, 1]; x, y) = 10g

v(;,1];x,y)-logR

a

B + {v(;, 1];x,y) -loga}

where a is a constant greater than the diameter of D. On aD, v-logR is zero, and so v -log a is negative. By the maximum principle, v -log a is negative on D. Therefore o ~ G(;, 1]; x, y) ~ 10g a

B

for all points (;,1]) of D. Let (xo' Yo) be any point of aD. Let Do be the part of D for which (; - XO)2 + (1] - YO)2 < e2

and let D 1 be the rest of D. Then U(x,y)

=

ffno CT(;, 1]) G(;, 1]; x,y) d;d1] + ffn, CT(;,1])G(;,rj;x,y)d;d1].

150]

POTENTIAL THEORY

[8.10

Since er is a polynomial, it is bounded on I5. There exists a constant K such that !er! ~ K there, and so

!U(x,y)!

~ K ffDoG(;,1J;x,Y)d;d1J+K ffD,G(;,17;X,Y)d~d1J.

Suppose that (x, y) is a point of Do. The integral over D1 tends to zero as (x, y) moves to (x o' Yo) on D. Hence limsup !U(x,y)! ~ limsuPKJJ G(;,1J;x,y)d;d1J. (X,Y)-(Xo,Yo) Do But, if D 2 is the disc R < 2e, D 2 contains Do and so

ffDoG(;'1J; x,y) d;d1J

~ fJDologid;d1J
= 21Te2 (1+210ga-210g2e). As e can be as small as we please, U(x,y) tends to zero as (x,y) moves to (xo' Yo) on aD. By (2), u(x, y) tends to f(x o, Yo)· We can write (1) in the form

U(x,y)

=

ffD er(;'1J)log~d;d1J+ffD er(;,1J)v(;,1J;x,y)d;d1J.

The first term is a logarithmic potential of the type considered in §8.2. It is continuous and continuously differentiable everywhere. Moreover, if er is continuously differentiable (as is certainly the case here since it is polynomial), the first term has continuous second derivatives on D and satisfies Poisson's equation there. By the symmetry relation v(;, 1J; x, y) is a harmonic function of (x,y) in D. Regarded as a function of (;,1J) it is harmonic in D and continuous differentiable in I5 save possibly at corners of aD. So the second term is harmonic in D. Hence V 2 U = - 21Ter. The restrictions onf(x, y) could be lightened. For example, it would suffice iff(x, y) had continuous derivatives of the third order on some domain containing D.

8.11 Some examples of Green's function The first example is Green's function for a disc whose boundary is the circle C, with equation r = a in polar coordinates. If Po is the fixed point of the disc which is the singularity of the Green's function and P is a variable point of the disc, G(P; Po)

1

=

log Po P o

+ v,

8.11]

POTENTIAL THEORY

[151

where v is a harmonic function. Guided by known results in electrostatics, we expect that G(P; Po)

1

log P. P -log

=

o

1 D

P + constant,

"'1

where PI is the point inverse to Po with respect to the circle C. Now if Q is any point on the circle C, the triangles OPoQ and OQP1 are similar, and so POQ/P1 Q = OPo/OQ.

Therefore

G(Q; Po) =

log~~~ + constant,

which is zero if the constant is log OPo/OQ. Hence G(P' P.) - I ,

0

-

PIP .OPo og PoP. OQ .

Let P, Po have polar coordinates (p, ¢), (r, 0); then PI has polar coordinates (a 2 /r, 0). It follows that a4 - 2a 2prcos (O_A) + p2 1·2 G(P' P.) = t log -;:-;--::--'---'-,--::-..:.-'jJ"':----'--:-:, 0 a 2{r 2-2prcos (O-¢) +p2}" It can be readily checked that this function has all the required properties. On C, oG/oNis the limit ofoG/op asp-+a. Hence at the pointQ(a, ¢)

o G(Q' P.)

oN

'0

-

-

-

a 2 -r2 a(a 2 -2arcos(0-¢)+r2)"

The function, harmonic on the disc, which takes the value j(¢) at the point (a cos ¢, a sin ¢) of C, is therefore . 1 2 2J211 j(¢)d¢ u(rcosO,rsmO) = -2 (a -r ) 0 a 2 - 2ar cos (0 - ¢) + r 2' 7T

This is known as Poisson's integral. It reduces to Gauss's mean value formula when r = O. Now suppose that Po lies in the semicircle 0 < r < a,O < 0 < 7T. As before, let PI be the point inverse to Po with respect to the circle C; and let P~, P~ be the images of --Po and PI with respect to the x-axis. The Green's function for the semicircle is

152]

[8.11

POTENTIAL THEORY

or, in polar coordinates,

G(p cos ¢,psin ¢; rcos 0, rsin 0) =

lJo {r 2_ 2rp cos (0 + ¢) +p2}{a4_ 2a 2rp cos (0- ¢) +r 2p2} "2 g {r2_ 2rpcos (o-¢) +p2}{a4 _ 2a 2rpcos (0+ ¢) +r2p 2}"

This function is a harmonic function of (pcos¢,psin¢) in any domain which contains none of the points Po, P~, PI> P~, and so is continuously differentiable. In particular, oGjoN is continuous on the boundary of the semicircle and vanishes at the corners. The Green's function for the half-plane 1J > with singularity (x, y) where y > 0, can be found by the method of images; it is

°

From this, we should expect that, if u is harmonic in y > 0, u(x, y)

= -1JOO u(;, 0) ( 7T

-00

x-

Y ;)2

+Y

2d;

under suitable conditions. The discussion must be deferred until after we have considered functions harmonic on an unbounded domain, Similarly, the Green's function for the quadrant; > 0,1J > with singularity (x, y) is . -.1 (;-X)2+(1J+y)2 (;+X)2+(1J_y)2 G(;, 1/,X,y) - 2 log (; _X)2+ (1J_y)2' (; +X)2+ (1J +y)2'

°

again by the method of images. G is a harmonic function of (;,1/) on any domain which does not contain any of the points (± x, ± V), and so is continuously differentiable. oGjoN is continuous on the boundary of the quadrant and vanishes at the origin which is a corner of the boundary. The method of images can be used to find the Green's function for other domains, such as an infinite strip or a rectangle, but the results are complicated since an infinity of images arise. Care must be taken that the process of taking images does not introduce unwanted singularities. A simple instance of this is when the domain D is the whole plane cut along the positive part of the x-axis. If we simply took the image of (x, y) in the x-axis, we should get a singularity at (x, - y) which lies in D; we should get the Green's function for the half-plane. In the cut plane, the angle variables are restricted to lie between (I and 27T. It can be shown, by using Sommerfield's multiple-valued

8.11]

[153

POTENTIAL THEORY

potential, t that the Green's function for the cut plane is

'. . _ p + r - 2-J(pr) cos t(O + cp) G(pcoscp,psmcp,rcosO,rsmO) - tlog 9-J( 1(0 cpr p + r - ~ pr) cos ~ This vanishes when cp = 0 and when cp = 21T. It has the correct logarithmic singularity when the point of polar coordinates (p, cp) moves to (r,O) since

{p +r - 2-J(pr) cos t(O - cp)}{p + r + 2-J(pr) cos t(O -cp)} = p2+r2-2prcos(O-cp). It also has alogarithmicsingularitywhenp = r,cost(O+cp) = 1. Now cos t( 0 + cp) = 1 when cp = 4n1T - 0 where n is an integer or zero. As and cp are restricted to lie between 0 and 21T, this is impossible; the Green's function has but one singularity in the cut plane. The origin is a 'corner' of the boundary of angle 21T. When p is small, oG/op and p-l oG/OO are both of the order O(p-t) and so tend to infinity as the point of polar coordinates (p, cp) moves up to the corner.

o

8.12 Poisson's integral Let u(x,y) be harmonic in a domain D. If K is an open disc with centre (xo, Yo) and radius a, which is contained in D, then if (x, y) is any point of K, u(x,y) where and

=

1 21T

J2" u(xo+acoscp,yo+asincp)P(r,O-cp)dcp, 0

x = xo+rcosO, P(r,O)

=

y = yo+rsinO

(r < a)

a 2 -r2 2 a - 2ar cos O + r·

(1)

2

Here u is known to be harmonic in a domain containing K; the formula merely states a relation connecting the known value of u at any point of the disc K and the known values of u on oK. We now ask what properties 1 u(x,y) = 21T 0 f(cp)P(r,O-cp)dcp (2)

J2"

has whenf is an arbitrary function. By a shift of origin, we may suppose X o and Yo to be zero. The simplest case is whenf(cp) is a continuous periodic function of t

A brief account of Sommerfeld's multiple.valued potentials is given in Jeans, The Mathematical Theory of Electricity and Magneti8m, 5th edn (Cambridge, 1925), pp. 279-283. 6

CPD

154]

[8.12

POTENTIAL THEORY

period 27T. 'Ve need the following properties of the Poisson kernel P(r,O): (i) for every value of ep, P(r, 0- ep) is harmonic in r < a, (ii) (iii )

P(r, 0) > 0 when r < a, 217 o P(r, 0 - ep) dep = 27T.

J

It is evident that we may differentiate (2) under the sign of integration as often as we please with respect to x and y, provided that r < a. It follows, using (i) that the function u(x, y), defined by (2), is harmonic in r < a, and has partial derivatives of all orders, which are also harmonic in r < a. We next show that u(x,y) tends to f(ex) as (x, y) tends to (a cos ex, a sin ex) in any manner in r < a. Now

lu(r cos 0, r sin 0) - f(ex)1 ~ lu(r cos 0, r sin 0) - f(O)

1

+ If(O) - f(ex) I.

Since f is continuous, the second term can be made as small as we please by taking 0 near enough to ex. Hence it suffices to consider radial approach to the boundary. By a rotation, we can make ex equal to zero. Then 1 J217 u(r, 0) - f(O) = 27T 0 {f(ep) - f(O)}P(r, ep) dep, and so

1 J217

lu(r, 0) - f(O)1 ~ 27T

0

If(ep) - f(O)1 P(r, ep) dep.

8.12]

[155

POTENTIAL THEORY

Hence

lu(r,O)-f(O)I
Keeping 8 fixed,

2.LV(a 2 2 2 ~

r2 )

+ r - :.ar cos 8'

limsuplu(r,O)-f(O)1 ~c. r-.a-O

As c is arbitrary, u(r, 0) tends tof(O) as r-+a-O, which was to be proved. Another method of justifying Poisson's integral depends on the use ofFourier series. (See Notes 8 and 9.) Suppose that {an} and {b n} are bo unded se quences. Then if x = r cos 0, y = r sin 0, the series

is absolutely convergent in r ~ ka, for any positive value of k less than unity, and converges uniformly; so also do the series obtained by differentiating with respect to rand 0 (or x and y) any number of times. Since rn cos nO and rn sin nO satisfy Laplace's equation, it follows by differentiating under the sign of integration that u(x, y) is harmonic in r < a. Now suppose that an and bn are the coefficients in the Fourier series of a function f(O) which is integrable in Lebesgue's sense, so that 1 f2lT 1 f2lT an = f(¢)cosn¢d<jJ, bn = f(¢)sinn¢d¢. rr

rr

0

0

Since {an} and {b n} are null-sequences, they are certainly bounded. Hence u(x,y)

n 1 {f2lT = -2 f(¢)d¢+22:, r- f2lT f(¢)cosn(O-¢)d¢ } 00

rr

0

1

an

0

is harmonic in r < a. A Fourier series is summable by Abel's method; hence lim u(x, y) = f(O) r-.a-O

iff(¢) is continuous when ¢ = 0; the limit isf(O) almost everywhere. If ~ r ~ ka, where < k < 1, we may alter the order of integration and summation in the series solution to obtain

°

°

n

1 f2lT f(¢) {1+22:, n r u(x,y) =-2 -cosn(O-¢) } d¢ rr 0 1 an

6'2

156]

POTENTIAL THEORY

[8.12

which is Poisson's integral. Thus, if f(1;) is integrable in Lebesgue's sense, Poisson's integral gives a function u(r cos e, r sin e), harmonic in r < a, which tends tofte) as r-+a- 0, provided thatfis continuous when 1; = e. In particular, iff(¢) is continuous, u(r cos e, r sin e) tends to f(e) as r-+a- 0, for all values of e.

8.13 The problem of Neumann The problem ofKeumann is to find a function u harmonic in a bounded domain D given the values taken by ou/oN on the frontier of D. There are two points to note. Firstly, the problem does not have a unique solution, since the addition of a constant to u does not alter the boundary values of ou/oN; if there is a solution, it is unique only up to an additive constant. Secondly, the boundary values cannot be assigned completely arbitrarily, since

f

aD

OU

~1i' dB

°'

=

0.

For simplicity, we assume that the domain D is simply connected, bounded by a regular closed curve C. We have to try to find a function u which is harmonic in D and continuously differentiable in 15 and for which ou/oN takes given continuous values on C satisfying the above condition. If (x, y) is a point of D, we have, as in §8.8,

u(x,y)

1 = - 21T

f[

o

r u(S,7j) ~N{v(S,7j)-logR}

C

-

ou(S,7j) ] oN {v(s,7j)-logR} dB,

(1)

where v is any function, harmonic in D and continuously differentiable in 15, and 'We cannot eliminate the unknown boundary values of u from thif' equation by choosing v so that the normal derivative of v -log R vanishes every,,-here on D. For, if we could so so, we should have

f

OV dB = ~1iY

CO

f

~ 0",' log R dB, coJ.\

which is impossible since the left-hand side is equal to zero but the expression on the right has the value 21T. There are two ways of a voiding this difficulty suggested by the·

8.13]

POTENTIAL THEORY

[157

theory of two-dimensional flow of a perfect incompressible fluid. The function v = v-logR is the velocity potential a.t the point (~, "I) of irrotational flow due to a source at the point (x, y). If C is a rigid boundary a.cross which there is no flow of the fluid, there must be an equal sink. If there is no sink, there must be a flow across C. Suppose that there is no sink and that oVloN = con C, where c is a constant. Then, if l is the length of C,

The first term on the right is zero since v is harmonic, the second is - 21T. The constant is thus 21T c = - l- . With this choice of V, (1) becomes 1

1£(X,y) = 21T

f C V oNds+y 01£ 1fcuds, 1

or

u(x,y) = 21T

f

C

01£

V oN ds + K ,

where K is a constant. The function V, called the Green's function of the second kind, we denote by r(~, "I; x, y). Then (2)

If the problem of Neumann has a solution, it is given by (2) where K is an arbitrary constant. In the alternative method, we make C a rigid boundary and introduce an equal sink at the fixed point (x o,Yo) of D. If

R5 =

(~-XO)2+(7J-YO)2,

we have u(x,y)-u(xo,Yo) =

-2~fetu o~(V-logR+logRo) OU} ds. -(v-logR+logRo) oN

158]

[8.13

POTENTIAL THE OR Y

"\Ve now choose v so that 8v ;; R oN = eN log R o

on C, as we may since it gives fc:;7 dS = If we have

fco~7l0gRdS- fc8~,10gROdS = o.

N(s,11;x,y;x o,yo)

= v-logR+logRo,

where the constant K is u(x o, Yo).

8.14 Harnack's first theorem on convergence Let {un(x,y)} be a sequence of functions, each harmonic in a bounded domain D and continuous on 15, the closure of D. Let un(x, y) be equal to fn(x, y) when (x, y) is any point of aD. If the sequence {fn(x, y)} converges tof(x,y) uniformly on oD, the sequence {un(x,y)} converges uniformly on 15 to a function u(x,y) which is harmonic on D. ]}foreover, u(x, y) is continuous on 15 and is equal to f(x, y) on aD. Also any derivative of un(x, y) converges uniformly to the corresponding derivative of u on any closed subset of D. For any positive value of 1':, there exists an integer no such that, for all points (x, y) on aD, - I':

< fm(x, y) - fn(x, y) <

1':,

wherever m > n ;:: no. Hence, by the maximum principle, -I':

< um(x,y)-un(x,y) <

I':

for all (x, y) on 15, whenever m > n;:: no. Hence the sequence {un(x, y)} converges uniformly on 15 to a function u(x,y); u is continuous on 15 and is equal to f on aD. Use polar coordinates, with any point (x o, Yo) of D as origin. 'Ve can choose a so that r:::; a lies in D. If x = xo+rcos8,y = yo+rsin8 where r < a, un(x, y) is equal to 1

-2

r

21T

1T ~ 0

2

•

un(xo+acos¢,yo+asm¢)

2

a -

2

a _r2

ar cos

(8

-

¢)

+ r2

d ¢.

(1)

POTENTIAL 8.14] 'When m > n ): no, we have

[159

THEORY

I{um(xo+acos ¢, yo+asin ¢) - un(xo+a cos ¢, yo+asin¢)} a 2_r 2 I< a-r' a+r a -2arcos(8-¢)+r2 x

€--

2

and so the integrand in (1) converges uniformly with respect to 8 and ¢ for any fixed r '< a. Hence 1

u(x,y) = -2 1T

f2" u(xo+acos¢,yo+asm¢) . 0

2 2

2

a -

a _r

8 ¢) 2d ¢, ar cos ( - + r

2

and so u(x, y) is harmonic on the arbitrary disc r < a and hence everywhere on D. Next oun(x, y) 1 f2" ,/... ,/. 0 P 8 ,/.. ,/.. ox = 21T 0 un(xo+aCOS'f',yo+aSm'f') ox (r, -'f')d'f', (2)

where P is Poisson's kernel. It can be shown that

~p(

I ox

r,

8-,/..)1 ~ 2a(a+3r) ~ 2{1+3k) 'f'

'"

(a-r)3

"'a(1-k)3'

where r ~ ka < a. The integrand in (2) converges uniformly on the closed disc r ~ ka, and so oun/ox converges uniformly to ou/ox. Similarly for derivatives of any order. If F is any closed set contained in D, for every point (X1'Yl) of F we can find a closed disc (x - X1)2 + (y - Yl)2 ~ ai contained in D such that the sequence {oun/ox} converges uniformly to ou/ox on this disc. The corresponding set of open discs is an infinite open cover of F. By the Heine-Borel theorem, we can choose a finite number of these open discs which also cover F. Hence F is covered by a finite number of closed discs on which the convergence is uniform. Hence {oun/ox} converges uniformly to ou/ox on F.

8.15 Harnack's inequality Let u(x,y) be a non-negative function , harmonic in a bounded domain D. Let (X-XO)2+(Y_YO)2 ~ a 2 be a closed disc contained in D. If (x, y) is an interior point of the disc

at a distance r from (x o, Yo), then a-r a+r - - u(x o, Yo) ~ u(x,y) ~ --u(xo,Yo)· a+r a-r

(1)

160]

POTENTIAL THEORY

[8.15

This follows at once from Poisson's integral u(x,y) = -1

21T

since u

~

0 and

f2" u(xo+acos¢,Yo+asin¢)P(r,8-¢)d9, 0

a-r -a+r

~

P(r, 8)

~

a+r --. a-r

From this follows an analogue of Liouville's theorem. A nonnegative function, harmonic in every bounded domain, is a constant. For such a function is harmonic on every disc (X-X O)2+ (y- YO)2 < a 2. If we make a -+ 00 in (1), we get u(x,y) = u(xo,Yo), the desired result. A more general result is as follows. Let u(x, y) be harmonic and nonnegative in a domain D, and let F be a bounded closed subset of D. Ij (x o, yo) is a fixed point of F, there exist positive constants Co and c l , independent of u, such that cou(xo,Yo)

~

u(x,y)

~

clu(XO,Yo)

at every point (x, y) of F. Since aD and F are disjoint closed sets, they are at a positive distance 4R apart. Consider the set of all open discs of radius R with centres at points of F. By the Heine-Borel theorem, we can choose a finite number, n say, of these discs which cover F. We may SUppOSf: that the disc with centre (x o, Yo) belongs to this family, which Wf call I:. The function u(x, y) is harmonic and non-negative in the closed disc with centre (x o, Yo) and radius 4R, since every point of this disc: is an interior point of D. If K o is the open disc with centre (xo, Yo) and radius 2R,

everywhere on K o. There exists at least one point (Xl' YI)' other than (x o, Yo), whicb lies in K 0 and is the centre of a disc of the fami!y I:. Again, u is harmoni c and non-negative on the closed disc with centre (xv YI) and radius 4R. If K I is the open disc with centre (xv YI) and radius 2R, !u(x l , YI) ~ u(x, y) ~ 3u(xl , YI)

everywhere on K 1 . But, since (Xl' YI) is a point of K o, iu(xo, Yo) ~ u(x v YI) ~ 3u(xo, Yo)'

8.15]

[161

POTENTIAL THEORY

Hence, at every point (x, y) of K l , tu(xo, Yo) ~ u(x, y) ~ 9u(xo' Yo)·

This also holds on K o. Proceeding in this way, we find after n steps, that 1 3n u(xo, Yo) ~ u(x, y) ~ 3nu(xo, Yo)

on all the discs of the family I:, and hence everywhere on I:.

8.16

Harnach's second theorem on convergence

Let {1£n(X, y)} be a sequence of functions, each harmonic in a bounded domain D, such that at every point of D. If the sequence is bounded at one point (xo, Yo) of D, it converges on D to a harmonic function; and the convergence is uniform on every closed subset of D. The sequence {un(xo, Yo)} is monotonic and bounded, and so is

convergent. For every positive value of €, there exists, therefore, an integer no such that

o ~ um(xo, Yo) -

un(xo, Yo) <

€

whenever m > n ~ no. If F is a closed subset of D containing (xo, Yo), we have, by the extension of Harnack's inequality,

and so

o ~ um(x,y)-un(x,y) ~ Cl{Um(Xo,Yo)-un(xo,Yo)} o ~ um(x, y) - un(x, y) < Cl €

everywhere on F, whenever m > n ~ no. Hence the sequence {un(x, y)} converges uniformly on F; let its limit be u(x,y). By Harnack's first convergence theorem, u(x, y) is harmonic in the interior ofF. As F was any closed subset of D, u(x,y) is harmonic onD.

8.17 Functions harmonic in an annulus If u is harmonic in a domain containing an annulus 0 < a ~ r polar coordinates, its value at a point (x, y) of the annulus is u(x,y)

=

2~ {fe. + feJ (u Ol~;R -logR:;) ds,

where 0 1 is r = a, O2 is r = b, and

~

bin

162]

[8.17

POTENTIAL THEORY

On 02'

S=

b cos ¢, 17 = b sin ¢, and so R2 = r 2- 2brcos (¢- 8) +b 2,

where (r, 8) are the polar coordinates of (x, y). As in § 8.5, 1 -2

1T

But on 01'

f

c,

n

=

00 ao+ 2: (ancosn8+bnsinn8)br n ·

1

S=

acos¢, 17 = asin¢, and R2 = a 2-2arcos(¢-8)+r2, where r > a. Then 00

a:/l,

10gR = logr- 2: - n cosn(8-¢) 1 nr n 810gR 810gR 00 a - 1 and - - = - - - = 2: -cosl1(8-¢). 8N 8a 1 rn It follows that

f

1 = 21T c. Here

a~10gr+~(a;,cosn8+b~sinn8)an. n r

1

,I fc. 8N 8u d s;

ao =

21T

but this is not necessarily zero since we are not given that u is harmonic in r ~ a. A function u(x, y), harmonic in a domain containing the annulus 0 < a ~ r ~ b, is of the form 00 rn u(x, y) = ao +a~logr + 2: (all, cosn8 +bnsin n8)-b n

1

00

all,

1

r

+ 2: (a~ cosn8+b~sinn8)-.n The first infinite series is convergent when r < b and uniformly convergent in rand 8 when r ~ b' < b. The second is convergent when r > a and uniformly convergent when r ~ a' > a. This is the analogue of Laurent's theorem for harmonic functions. If the one-valued function u(x, y) is harmonic in a neighbourhood of a point Po, except at Po itself, and is bounded in the neighbourhood of Po, then we can define u at Po so that u is harmonic in the whole neighbourhood. "\Ve may take Po to be the origin and the neighbourhood to contain r ~ b. Then, changing the notation, 00

u(x,y) =

ao+a~logr+ 1',

(all, cosn8+b" sinn8)rn

1 00

+ 2: 1

(a~ cosn8+b~sin

n8) r- n ,

8.17]

[163

POTENTIAL THEORY

the series being uniformly convergent on any closed set contained in 0 < r < b. Suppose that lui ~ Kin r ~ b. By integrating term-byterm we have

21T(ao+a~10gr) =

f:"

u(rcos 0, rsinO) dO

lao+a~logrl ~ K.

and so

As this holds no matter how small r may be, we must have Similarly, if n > 0,

lanrn+a~r-nl

=

a~ =

O.

I~f:" u(rcoso,rsino)cosnOdol ~ 2K.

Again, as this holds no matter how small r may be, b~ = O. Hence

a~ =

O. Similarly

aJ

u(x, y) = a o +

L: (an cos nO + bn sin nO) rn 1

for 0 < r < b, and also for r = 0 if we define u(O, 0) to be ao' Since ~(ianl + Ibnll cn is convergent for c < b, u is harmonic in r < b. ~ is called a removable singularity. If the one-valued function u(x, y) is harmonic in a neighbourhood of a point Po except at Po itself, and u tends to + 00 as (x, y) tends to Po, then u = a~logr+v, where a~ is negative and v is harmonic in a neighbourhood of Po· Since u-.+ +00, for any positive value of K, no matter how large, we can choose r o so that u > K when 0 < r < roo Using the Laurent

expansion, we have 21T(ao +a~ logr) +1T(an r n +a~r-n)

= ~

since 1 + cos nO

~

O. Similarly

f

2"

0

K

u(rcos 0, rsinO) {1 + cos nO} dO 2" 0

f

(l+cosnO)dO = 21TK,

21T(ao +a~logr) -1T(an r n +a~r-n)

~

21TK.

These inequalities hold in 0 < r < ro, no matter how small r may be. If a~ =l= 0, these inequalities cannot both be true since ± 1Ta~r-n are the dominant terms on the left. Hence a~ = 0; similarly b~ = O. The inequalities also imply that a~logr-.+ +00 as r-.+O; hence a~ < O. Hence aJ u(x,y) = ao+~logr+.s (an cosnO+bnsin nO) r n 1

which was to be proved.

164]

[8.18

POTENTIAL THEORY

8.18 Unbounded domains So far, we have considered only bounded domains. If D is unbounded, we say that u(x, y) is harmonic on D if it is bounded and is harmonic on every bounded domain contained in D. 'We may suppose that the origin is not a point of D, so that D lies outside a circle X2+y2 = a 2. The point (x',y') inverse to (x,y) with respect to this circle has coordinates x' = a2xj(x 2+y2),

y' = a2yj(x 2+y2),

or, in polar coordinates, r' = a 2 jr,

8'

=

8.

The inverse of the domain D is a domain D', which lies inside the circle. The origin is not a point of D'; but D' does contain a' punctured' disc Since the function

(x

2

2

2

U 8 u) + Y2) (88x2 + 8y 2

2 _ ('2 '2) (8 U 82u ) - X +Y 8X'2 + 8y'2 '

a2x' a2y' ) u1(x', y') = U ( X'2+ '2' X'2+ '2 y y

is harmonic on the punctured disc, and is bounded there. The origin is therefore a removable singularity of u1(x',y'); as r' -+ 0, u1(x',y') then tends to a limit ao' The function which is equal to u1(x',y') on < r' < b and is equal to a o at the origin is harmonic on r' < b. Let C be a regular closed curve. The exterior problem of Dirichlet is to find a function which is harmonic on the domain D exterior to C and which takes given continuous values on C. By the definition, we require the function u(x,y) to be bounded and to be harmonic on every domain contained in D. If the boundedness condition is dropped, the problem does not have a unique solution. Choose as origin a point inside C. The inverse of D with respect to the circle r = a is a bounded domain, punctured at the origin. The inverse of C is a regular closed curve r. If u1(x',y') is the unique function which is harmonic inside r and takes the corresponding continuous boundary values on r, then

°

U

1

a2x a2y ) ( x2 + y2' x2 + y2

is the required solution of the exterior problem of Dirichlet. The existence and uniqueness ofthe solution ofthe interior problem ofDirichlet

POTENTIAL THEORY [165 8.18] implies the existence and uniqueness of the solution of the exterior problem of Dirichlet in the class of bounded functions. A simple example is the solution of the exterior problem of Dirichlet when the boundary is a circle r = a. The solution of the interior problem is 1 f2" a 2-r2 u(r,8) = 21T 0 a2-2arcos(8-¢;)+r 2 !(¢;)d¢;.

If we invert with respect to r = a, the interior domain becomes the unbounded domain r > a. Replacing r by a 2jr, we get 1

u(r,8) = 21T

f2" 0

r 2-a 2 a2-2arcos(8-¢;) + r2!(¢;) d¢;,

a result which can also be obtained from the fact that a function which is bounded and harmonic when r > a is of the form

:More generally, the solution of the exterior problem of Dirichlet in the class of bounded functions can be deduced from the Green's function solution of the interior problem obtained by inversion. This implies that we can define the Green's function in the ordinary way for an unbounded domain, provided that we require it to be bounded at infinity. For example, the Green's function for the halfplane 1] > 0 with singularity (x, y) where y > 0 is G(s, 1]; x, y) = !log

(s(S-X)2+(1]+y)2 r + ( r' -x~

1]-y~

This satisfies the condition of vanishing when 1] = 0; it has the correct logarithmic singularity at (x, y) and is bounded at infinity.

8.19

Connexion with complex variable theory Letw = !(z) be an analytic function of the complex variablez = x+iy, regular in a bounded domain D. If w = u + iv, where u and v are real, u and v are harmonic in D. For both have continuous derivatives of all orders; and since U x = vlJ' ulJ == - vx' both u and v satisfy Laplace's equation. Conversely, ifu is harmonic in a simply connected domain D, whose boundary is a regular closed curve C, then u is the real part of an analytic function, regular in D. By Green's theorem,

fr

(uxdy -ulJdx)

166]

[8.19

POTENTIAL THEORY

r

vanishes for every regular closed curve

lying in D. Hence

(X,y)

f

v(x,y) =

(uxdy-uydx)

(Xo,Yo)

is independent of the path of integration; v(x, y) is a one-valued function which has continuous derivatives inD, and V x = - u y, v y = u x ' Therefore u + iv is an analytic function of z, regular in D. The harmonic function v is called the conjugate of u. If (x o' Yo) is a point of D, u is the real part of an analytic function 1(z), regular in a disc Iz - zol < R contained in D. By Taylor's theorem, 00

1(z) = 2:cn (z-zo)n o

is convergent on the disc. Hence u can be expressed, either as 00

u = 2: o

where Cn =

Icnl eia.n and z =

Icnl rncos (n8+iX n ),

zo+re 8i , or as

m

U

=

2: amn(x-xo)m(Y_Yo)n. m,n=O

By the sort of argument used in § 8.5, we can show that this double series is absolutely convergent on the square

Ix-xol + IY-Yol

< R,

and hence that u(x, y) is analytic.

8.20

Conformal mapping

If S = F(z) is an analytic function, regular in a bounded domain D, itmapsD into adomain~inthe splane. Such a mapping is conformal, since the angle between the tangents at the point of intersection of two curves is unaltered by the mapping. If the mapping is a bijection, that is, to every point of D there corresponds one point of ~ and to every point of ~ there corresponds one point of D, S = F(z) is said to map D conformally on ~; such a mapping is sometimes said to be simple or schlicht. If S = F(z) maps D conformally onto ~, for any function u(x, y),

we have

82

(~2u 82 )

82

8x~.+ 8Y~ ~

IF'(z)12

~f,2 + 87J~

,

where z = x+ iy, s = f, + i7J. Hence ifu is harmonic on D, the conformal . mapping turns it into a function harmonic on ~.

POTENTIAL THEORY [167 8.20] Suppose that D is bounded by a regular closed curve C. Then, by Riemann's theorem on conformal mapping, there exists a unique analytic function F(z), regular in D, such that S = F(z) maps D conformally on lsi < 1, transforms any point a within C into the origin, and a given direction at a into the positive direction of the real axis. It would seem, therefore, that all we have to do to solve the problem of Dirichlet for D is to map D conformally on the unit circle lsi < 1 and solve the corresponding problem in the S plane. If we know an explicit formula for the mapping, this is satisfactory. The argument fails to prove the general existence theorem since Riemann's theorem is equivalent to the theorem concerning the existence and uniqueness of the solution of Dirichlet's problem. As an example, we deduce Poisson's formula from Gauss's mean value theorem. The conformal transformation

where Izol < a and 20 is the conjugate of zo, maps Izl ~ a onto lsi ~ 1, the image of Zo being the origin of the Splane. This mapping turns a function u(x, y) harmonic in the z plane into a function U(s,7J) harmonic in lsi ~ 1; and u(xo, Yo) = U(O, 0). But U(O, 0)

=

2~

5

r U(s, 71) dCT,

where CT is the arc length of the unit circle Hence

where 8 is the arc length of C, where r < a. Now

Ids I I

Izi

= a.

r whose equation is lsi

On C,

z

= ae
and

Zo

=

1.

= re 8i

I

2 2 2 dCT a(a -l zoI2) a _r 2 2 d8 = dz = a(a -zz o)2 = a(a - 2arcos (8-¢) +r2 )'

Therefore 1 U(Xo,Yo) = -2 1T

52" u(rcos¢,rsin¢) a - 2ar cos (8 - ¢) +r 2

0

2

a _r

2

2d

¢.

8.21 The problem of Neumann Let C be a regular closed curve bounding a domain D. Position on C is defined by the arc length 8, measured from some fixed point on C; 8 is chosen to increase as the point of parameter 8 describes C in the

168]

POTENTIAL THEORY

[8.21

positive sense. The problem of Neumann is to find a function u(x, Y), harmonic in D and continuously differentiable in 15, which is such that ou/oN takes given continuous values on C. Thus, on C,

ou

~"H

01.\' ~

= g(s),

where g(s) is periodic, of period l, where l is the length of C. This problem can be transformed into the Dirichlet problem, by using conjugate functions. Let v(x, y) be the harmonic function conjugate to u. By the CauchyRiemann conditions,

ou

on C. Hence, on C,

v

=

I:

ov

oN = os g(s)ds+A =j(s),

where A is an arbitrary constant. Since g(s) is periodic of period l and z+a a g(s)ds = 0

I

for all a, j(s) is a continuous function of period l. If v(x, y) is the solution of the problem of Dirichlet for D with boundary data j(s), the solution u of the problem of Neumann is the function whose conjugate is v, or, alternatively, is the conjugate of v with the sign changed. As an example, consider the problem of Neumann for the disc r < a. The solution is to be harmonic in D, continuously differentiable in 15; on r = a, ou/or = g(¢), with a change of parameter to ¢ = s/a. The conjugate function v takes on C the values defined by j(¢) = a

I:

g(¢)d¢,

dropping the constant of integration. Since 2 1 0 17 j(¢)P(r,8-¢)d¢, v(rcos8,rsin8) = 21T

I

the solution of the problem of Neumann is 1 u(rcos8,rsin8) = - 21T

I

2

0 17 j(¢)Q(r,8-¢)d¢,

where Q is the conjugate of the Poisson kernel. But ae~;i+reei

ae¢'i-re&i

= a2 -

a 2 _r 2 2iarsin (8-¢) 2arcos (8- ¢) +r 2 + a 2 - 2arc~s (8~ ¢) +r2 '

The second term on the right is iQ(r, 8 - ¢).

8.21]

POTENTIAL THE OR Y

We thus have .

1 1T

u(rcos8,rsm8) = --2

f2" j(¢) 0

[169

2arsin(8-¢) 2 (8 ¢) a - ar cos + r 2 d¢. 2

To express this in terms of the given function g(¢), we integrate by parts: this gives u(rcos8,rsin8)

=

1 21T

f2" j(¢) o¢log a (a 0

a =-21T

f2" g(¢) log (a 0

2

2

-2arcos(8-¢) +r 2 )d¢

-2arcos (8-¢)+r 2 )d¢.

An arbitrary constant can be added to this solution.

8.22 Green's function and conformal mapping In the complex z-plane, let D be a simply connected domain bounded by a regular closed curve C. By Riemann's mapping theorem, there exists a unique analytic function j(z) such that S = j(z) maps D conformally on the disc lsi < 1, maps a given point Zo of D on the origin and turns any given direction at Zo into the positive direction of the real axis in the Splane. When z is on C, lsi = 1. The function j(z) has a simple zero at Zo and so is of the form (z -zo) g(z) where g(z) is analytic and does not vanish on D; for only one point of D is mapped into the origin. If we put j(z) = e- U- iV , then

U

=

-loglj(z)1

where

= u-Ioglz-zol,

u = -relogg(z)

is harmonic on D. On C, U vanishes. Hence U is the Green's function for D. If we know a conformal mapping of D of the desired type, the Green's function can be found. An interesting case is the Green's function for a closed polygon. Suppose that the corners of the polygon are Zl,Z2' ""Zn taken in order in the positive sense. As we go through the corner Zk in the positive sense, the direction of the side turns through an angle 1TCi,k' At an ordinary corner 0 < Ci,k < 1, but at a re-entrant corner -1 <

Ci,k

< 0;

and I:Ci,k = 2. The Schwarz-Christoffel transformation which maps the polygon onto lsi < 1 and turns a given point Zo inside the polygon into the origin is given by z-zo

=

xf'

o

n(S-Sk)-akds. 1

170]

[8.22

POTENTIAL THEORY

Since

Z-Zl

=

KJS (~-~l)-alrr (~-~k)-akd~, S1

2

the mapping near the corner z, has the form 00

Z - Zl

= (~- ~l)l-al

2: Ap(~ o

~l)P,

where the infinite series has a finite radius of convergence. Hence, by the formula for the reversion of series

where the series again has a finite radius of convergence. If we write ~ = e- u - w , U is the Green's function for D. And Igrad UI Since

~1

= IUx- iUyl

=

lUx + ilS,1 = I~ ~~I·

is not zero, Igrad UI near the corner Zl' behaves like bl._1_lz_Z111/(l-al)-1 = K 1p1/(1-a1)-1, 1~11 1- (Xl

where K 1 is a non-zero constant and p is the distance of Z from the corner Zl. If Zl is an ordinary corner, 0 < (Xl < 1. Hence as Z moves up to Zl on the interior of the polygon, grad U tends to zero like a multiple of p a l/(l- al). But if Zl is a re-entrant corner, - 1 < (Xl < o. If we write (Xl = - 13, then as Z moves up to zIon the interior of the polygon, grad U tends to infinity like a multiple of p-fJ/(l+fJ). In either case, 1/(1- (Xl) is positive, and so grad U is integrable round the polygon. If we know the Green's function for a domain D, we can deduce the Green's function for a domain ~ if we can map D conformally on ~. For example, the Green's function for the half-plane 1] > 0 with singularity (x, y) where y > 0 is G(~, 1]; x, y)

If we write z = x + iy,

~

(x - ~)2 + (y + 1])2 = i log (x- ~)2 + (y-1] )2·

= ~ + i1], this becomes Iz-~I

G = log Iz _

~I'

where ~ is the conjugate of S. When z and ~ lie in the upper half-plane 0< phz < 1T,O < ph~ < 1T. If ~ lies on the boundary, ph~ = 0 or ph~ =

1T.

8.22]

[171

POTENTIAL THEORY

If 0 < IX ~ 2, the relation w = za: maps the half-plane 0 < phz < 1T onto the angle 0 < ph w < 1TIX (the cut plane if IX = 2), and the mapping is conformal except at the origin. Hence the Green's function for the angle is IWI/a: _ (ro)I/a:1 o -- log "'-c--:-:---'-"-:-:IWI/a: - WI/a: I If we put

where 0 <

e < 1TIX, 0 < if; < 1TIX, this becomes 1

r 2/a: _ 2rl /a:p l/a: cos - (e + if;) + p2/a: 0= ilog

IX

1

r 2/a: _ 2rl /a:p l/a: cos - (e _ if;) + p2/a:

'

IX

the result we stated in §8.11 for the case IX = 2. When p is small,

80 . 2 pl/a:-l .

e.

if;

-8p =. -IX --sm-smrl/er. IX iX'

e if; --sm-cosrl/a: IX IX

1 80 . 2 pl/a:-l .

-- = -

p 8if; .

IX

Hence, if 0 < IX < 1, when the angle is an ordinary acute or obtuse angle, 80j8p and p- 1 80j8if; tend to zero as p-+O like multiples of pl/a:-l. But if 1 < IX ~ 2, when the angle is a reflex angle, 80j8p and p- 1 80j8if; tend to infinity as p-+O. This agrees with the result obtained for a polygon. Since the problem of what happens at a corner is a local problem we should expect that the Green's function for a domain whose boundary consists of a finite number of regular arcs would behave in a similar way at corners.

Exercises 1. Prove that, if D is the disc ;2+ 17 2 < a2, the logarithmic potential

ffDIog~d;dl7 is equal to !1T(a2-r2)-1Ta2Ioga when r2 =x2+y2 a. 2. Prove that the logarithmic potential

172] where R2

POTENTIAL THEORY

= (X_;)2+ y 2, is equal to

t(x -a) log {(x - a)2+ y2} - t(x +a) log {(x +a)2 +y2} + 2a x+a x-a -ytan-1--+ytan-1- -

y

provided that (x,y) is not a point of the segment -G Show that, on the segment, the potential is

y

~

X

~

a, y

= O.

2a- (a-x)log (a-x) - (a+x) log (a +x).

Verify that this potential has the properties stated in §8.4. Show also that

where N is parallel to the axis of y, is equal to x+a y

x-a y ,

tan-1 - - -tan-1 - when y =F O. What happens when y

= O?

3. Prove that the potential at a point P of a normally directed distribution of doublets of uniform density a on a regular arc C is O'lfr where y is the angle subtended by C at P. 4. u(x,y) is harmonic on the disc X 2+y2 < a 2 and vanishes on the diameter y = O. Prove that u is an odd function of y.

5. Find the function u(x,y), harmonic on the disc r 2 = X 2+y2 < a 2 given that, on r = a, U = 1 when y > 0 and U = 0 when y < O. Find also the function, harmonic and bounded when r > a, which takes the same values on

r=a. 6. Find the function u(x, y), harmonic on r < a, which takes the

Iyl on r =

value~

a.

7. The functionf(;) is bounded and continuous. Prove that u(x, y)

= -1T1fOO

-00

f(;) (

x-

Y

;)2+ 2 d;

Y

is harmonic in y > 0 and that as (x, y) tends to (a,O) on any path in y > 0 u(x, y) tends to f(a). Show that u(x, y) is also harmonic in y < 0, but i~

f

discontinuous across the x-axis. Deduce that 1 u(x,y) = _ 21T

00

g(;)log{(x-;)2+ y 2}d;

-00

is the solution of the problem of Neumann for the half-plane y > O.

POTENTIAL THEORY

[173

8. u(x, y) is harmonic on the disc x 2+ y 2 < a 2 and is identically zero on the smaller disc X2+y2 < b2. Show that u is identically zero. Hence prove that, if U 1 and u 2 are harmonic on a domain D and are equal on a domain contained in D, then u 1 = U 2 everywhere on D.

9. u(x, y) is harmonic in a domain containing the closed disc K, X2+y2 ..,; a2. If

ff

K

{u(x, y)}2 dxdy = .JtI,

(M)! a-;

lu(O,O)1 ..,; 1

prove that

by using Gauss's mean value theorem. Deduce that if (x,y) is an interior point of K, lu(x,y)l..,; - 1 , a-r 1T where r2 = X2+y2. Hence show that, if u is harmonic everywhere and is not identically zero, the integral of u 2 over the whole plane is divergent.

(M)!

10. Deduce from Harnack's Inequality that if u(x, y) is non-negative and harmonic everywhere, it is constant.

11. Prove that, if

r(;, 71; x, y) is

the Green's function of the second kind,

r(Xo,Yo;Xvy1)+f fa r(;,7J;X 1'Yl)ds =

r(XVY1;XO,Yo)+}fa r(;, 71; xo,Yo) ds.

12. Prove that, for the disc X 2 +y 2 ..,; a 2 , the Green's function of the second kind is r(;, 7J;x,y) = - !log (r 2_ 2prcos (O_¢>)+p2) - !log (a4 -2a2pr cos (O_¢»p2r2),

where (p,¢» and (r,O) are the polar coordinates of (;,71) and (x,y). Deduce the solution of the problem of Neumann for the disc.

°..,;

13. Show that w = eZ maps the infinite strip Y ..,; 1T, where z = x+ iy, conformally on the half-plane im w ~ 0. Hence show that the Green's function for the strip is G(;. ) - ~ll e2"'-2eo:+Scos(Y+7J)+e~ , 7J,X, Y - ~ og e2"'-2e"'+S cos (Y-7J) +e~·

Deduce that, if u is harmonic in the strip,

U(;,1T)} d",.C SinYf<XI {U(;,O) ( ) =-ux,y + 21T -<XI cosh(;-x)-cosy cosh(;-x)+cosy

174]

POTENTIAL THEORY

14. ll(x, y) is bounded and harmonic in the whole plane cut along the positive part of the x-axis. As (x, y) moves to the point (s,O) where S > 0 by a path in the upper half plane, u(x, y) tends to f+(s) wheref+ is bounded and continuous. As (x, y) moves to the same point by a path in the lower half-plane, u(x, y) tends to f-(s) where f- is also bounded and continuous. Prove that if (x,y) has polar coordinates (r,8) when 0 < 8 < 21T,

15. ll(x, y) has continuous second derivatives on a bounded domain D where it satisfies y2U = - 21Ta(x, y) where a is non-negative. If (x o,Yo) is any point of D, there exists a disc with centre (x o,Yo) and radius a which lies in D. Prove that

J:"

u(xo+rcos8,yo+rsin8)d8

is a decreasing function of r when r < a. Hence show that

9

SUBHARMONIC FUNCTIONS AND THE PROBLEM OF DIRICHLET

9.1 The enunciation of the problem If D is a bounded domain, the problem is to show that there exists a function, harmonic in D and continuous in 15, which takes given continuous values on the frontier aD of D. A domain is an open connected set; its frontier is the intersection of its closure and the closure of its complement. There are several proofs of this existence theorem. vVe give here the proof, associated with the names of Perron and Remak, which uses the theory of subharmonic functions. No assumption is made initially about the connectivity of D or the structure of aD. A function is found which is harmonic in D and is specially related to the boundary values. Conditions are found on the structure of aD, under which this harmonic function is continuous in 15 and takes the given values on aD. These conditions are satisfied when D is simply connected and is bounded by a regular closed curve; more generally, when D is multiply connected and is bounded by a finite number of non-intersecting regular closed curves. Although the function which solves the problem of Dirichlet is 'found', the method does not provide a practical method of solving particular problems.

9.2 Subharmonic functions Afunctionu(x,y), continuous in a domain D, is said to be subharmonic in D if, for every closed disc (S-X)2+(1J_y)2 ~ r 2 contained in D U(x,y)

~ ~f21T u(x+rcos8,y+rsin8)d8. 21T

0

If U and - U are both subharmonic, U is harmonic, by the converse of Gauss's mean value theorem. IfulJu2' ... 'U n are subharmonic on a domain D, so also are

[ 175 ]

176J

[9.2

SUBHARHONIC FUNCTIONS

where c1 , c2 ,

••• , Cn

are positive constants, and

max (u1 , U 2'

•.. ,

un)'

The first result is obvious. The second can be proved by induction since where So it suffices to prove the second result in the case n Let

= 2.

This means that, at each point (x, y), the value of u(x, y) is the greater of u1(x, y) and u 2 (x, y). Since U U

= t(u1 +u2 + !U1 - U2 J)

is continuous on D. If U = u(x,y)

=

U1

at the point (x, y), we have

1

f2lT 0 u1(x+rcosO,y+rsinO)dO

1

f2lT 0 u(x+1·cosO,y+rsinO)dO.

u1(x,y) :::; 27T

:::; 27T

We get the same result if U = subharmonic.

U2

at the point considered. Hence

U

is

If U is subharmonic on a bounded domain D and has a relative maximum at a point of D, theTe is a disc, which lies in D and has the point as centre, on which u is constant. Take the point as origin. Then there is a disc r < a whose closure

lies in D. Since u is subharmonic, 1

u(O,O) :::; 27T

f2lT 0 u(r cos 0, r sin 0) dO

for all r < a. Suppose that there is no disc with centre 0 on which 11 is constant. Since u has a relative maximum at 0, we can find a circle r = b on which the supremum M of u is less than u(O, 0). Let w(x, y) be the function, given by Poisson's integral, which takes on r = b the same values as u(x, y). Then w(O, 0)

and so

=

1

27T

f2lT 0 u(bcosO,bsinO)dO ~ w(O, 0) > M.

u(O,O),

SUBHARMONIC FUNCTIONS [177 9.2J This is impossible since w(x, y), being harmonic in r < b and continuous in r :::; b, cannot take at the origin a value greater than its supremum on r = b. It follows that there must be a disc with centre 0 on which u is constant. If u is subharm<Jnic on a bounded domain D and continuous on D, it attains its supremum on D at a point or points of aD. If it also attains this s~tpremum at a point of D, it is a constant. Since u is continuous on D, it is bounded and attains its supremum at a point of D. We have to show that, if u is not constant, it cannot attain its supremum at a point of D. Suppose this is not so. Let A be the set of points of D at which u = M, where M is the supremum of u on D. If (x, y) is a point of A, u has a relative maximum at that point, and so there is an open disc with centre (x, y) on which u = M. The set A is therefore the union of a family of open discs and hence is an open set. Let B be the set of points of D at which u < 3-'J. Since u is continuous, B is an open set. But D is the union of the disjoint open sets A and B, which is impossible since D, being a domain, is connected. Hence either A or B is an empty set. If B is empty, A is D and u = J.lJ everywhere on D. This is impossible if u is not constant. Hence A is empty, and u does not attain its supremum on D at any point of the interior of D. Let Ll be an open disc whose closure is contained in D. We can associate with any function u which is continuous in D a function Ut. with the following properties: (i) Ull. is continuous in D; (ii) Ull. = U on the complement of Ll with respect to D; (iii) ut. is harmonic on Ll. In fact, Ull. on Ll is given by Poisson's integral. If u :::; Ull. for every open disc Ll where closure is contained in D, then u is subharmonic in D. Conversely, if u is subharmonic in D, so also is Ut. for every disc Ll whose closure is contained in D; and u :::; Ut.. If (x,y) is any point of D, there exists a positive number a such that, if r < a, the closure of the disc Ll with centre (x, y) and radius r lies in D. Then, if r < a,

u(x,y) :::; ut.(x,y)

=

1

27T

J211' . 0 ut.(x+rcosO,y+rsinO)dO

1 J211' = 27T 0 u(x+rcosO,y+rsinO)dO,

since ull. = u on all. Hence u is subharmonic.

178J

SUB HARMONIC FUNCTIONS

[9.2

'We now consider the converse. Since U,:,. is harmonic on Ll, it is subharmonic on Ll. Since U,:,. = U on the part of D outside l, it is subharmonic there. It remains to discuss points on all. Since - u ~ is harmonic on Ll, it is subharmonic. Hence U - U,:,. is subharmonic on Ll, and so attains its supremum in Zi. at points of all. But u-u::, vanishes on all; hence u-uj. :::; 0 on l. But u,:,. = U on the part of D outside Ll. Therefore U :::; U,:,. everywhere on D. If (x, y) is a point of all,

u,:,.(x,y) = u(x,y):::;

f:

7T

u(x+pcosO,y+psinO)dO

for all sufficiently small values of p. It follows that

ut,.(x,y):::;

f

27T

0

u,:,.(x+pcosO,y+psinO)dO,

so that Ut,. is subharmonic at every point of all.

9.3 Lower functions Let f be continuous on the frontier aD of a bounded domain D. A function v, which is continuous on D and subharmonic on D and which satisfies v :::; f on aD, is called a lower function or subfunction associated v;rith f on D. 'We denote the family of all lower functions associated withf on D by SIj. The family is not empty; the infimum m off on aD belongs to SIj. By the maximum principle, every function v belonging to SIj satisfies on D the inequality v :::; JJf, where JJf is the supremum off on aD. If V 1 ,V2 ""'Vn belong to SIj, so also does max(vl ;v2 ,oo.,vn ). For max (VI' v 2 , ... , vn ) is continuous on D, subharmonic on D and does not exceedf on aD. If v belongs to SIj, then v,:,. belongs to SIj for every disc Ll whose closuTe lies in D. v,:,. is continuous in D, subharmonic on D and, on aD, v,:,. = v :::; f. 9.4

Perron's function If v belongs to SIj, v :::; JJf at every point of D. The values taken by the functions of the family of SIj at a given point (x, y) of D have, therefore, a finite supremum which we denote by u(x, y). Perron's function u(x,y) = supv(x,y) •• E!I'f

is therefore defined everywhere on D, where it satisfies u:::; JJ1. On aD, u:::;f. This is all we know about u(x,y); we prove that it is harmonic on D.

SUB HARMONIC FUNCTIONS [179 9.4J Let Ll be a disc whose closure is contained in D. If Po is a point of ~, there exist functions of 9f which take values as near as we please to u at Po. Hence we can find a function v~ of 9f such that

v~ (Po) ~ u(Po) - 1.

If we write

then VI also belongs to 9f and is harmonic on Ll. Since v~ ~ we have Again, there exists a function

VI

on 15,

v; of 9f such that

v;(Po) ~ u(Po) -

i·

The function also belongs to

9f and is harmonic on :l. Since

we have Proceeding in this way, we can construct two sequences

{v~}

and

{v n } offunctions of 9f such that

v~(Po) ~

u(Po)

-~, n

Since both sequences converge to u(Po) at Po. Now

Vn- I

~ max(vn_l>V~) ~ [max(vn_l>v~)]a = Vn ~]}[

everywhere on 15. Hence {v n }, being a bounded increasing sequence, converges everywhere on 15 to a function v. Since each function V n is harmonic on Ll, the limit V is harmonic on Ll, by Harnack's second convergence theorem. The convergence of {v n } to V is uniform on every closed subset of Ll. We know that V = u at Po; we shall now show that v = u everywhere on Ll. By definition, V n ~ u and so v ~ u on Ll. If v 9= u everywhere on Ll, there is a point PI of Ll such that v(PI ) < u(PI ). We carry out the same construction starting with the point PI instead of Po. We get

180J

[9.4

SUBHARMONIC FUNCTIONS

two sequences {u';,} and {wn } of functions ofthe family SIj such that u(-R.)

~ W~(Pl) ~

u(P 1)-!, 11

Both sequences converge to u(P1 ) at -R.. And, as before, {w n } converges to a function w which is harmonic in Ll. Now consider the sequences {W~}, {Jv,.}, where

TF;, = max (v n , w n ),

Jv,. =

[Wa~·

All these functions belong to the family SIj, and {W~} is a bounded increasing sequence since HY~

= max (v n , wn )

Tl~

:::; max (v nH , w nH )

=

is harmonic on Ll. W~H'

The sequence {Tl~} is also a bounded increasing sequence. For, outside Ll, Jv,. = W~, n~+l = W~+I' and so {Tl~} is increasing outside Ll. On Ll Tl~+l - Jv,. is harmonic and takes non-negative values W~+I - W~ on all; hence Jv,.+l - Tl~ ~ 0 on Ll. The sequence {Tl~} converges, therefore, to a function W which is harmonic on Ll. Moreover, W ~ v, W ~ w. Since u is.the supremum of the functions of the family 9;, Tl~(Po)

and so

:::; u(Po)

HT(PO) :::; u(po) = v(Po) :::; Tl1(Po).

Therefore the function W - v, which is harmonic on Ll, vanishes at Po. But W - v ~ O. Hence TV-v attains its minimum at an interior point of Ll, and so is identically zero on Ll. In particular, v(P1 )

=

W(P1 ) ~ w(-R.)

=

u(-R.).

This is a contradiction, since we assumed V(P1 ) < u(P1 ). Hence v = u everywhere on Ll, and v is harmonic on Ll. Hence Perron's function u is harmonic on any disc Ll where closure is contained in D. As D is an open set, the union of a family of open discs, it follows that Perron's function is harmonic in D.

9.5 .Barriers In this section, P denotes a point of D, but Q and Qo are points of aD. We have to find conditions under which Perron's function u(P) tends tof(Qo) as P tends to Qo by a path in D. To do this, we introduce

SUBHARMONIC FUNCTIONS [181 9.5J the concept of a barrier. A barrier is a function w(P; Q) u'hich is

continuous in D, harmonic in D, whose boundary-values w(Q; Qo) a1'e strictly positive except at the point Qo; and (o(Qo; Qo) = O. If there is such a barrier at Qo, u(P) does tend to f(Qo) as P tends to Qo' If

every point of aD has a barrier, Perron's function is the required solution of the problem of Dirichlet, which we know is unique. The boundary function f(Q) is continuous on the closed set aD and so is bounded; there exists a constant K such that I f(Q)1 :::; K. The result will follow if we can show that, for every positive value of e, limsupu(P) :::;f(Qo)+e, p-~

liminfu(P) ?:-f(Qo)-e, p-~

when there exists a barrier w(P;Q o)' Let Ll be a disc with centre Qo where radius is so small that Ll does not contain the whole of the domain D; ~' is the complement of U. For every positive value of e, we can choose the radius of 6. so that If(Q) -f(Qo)1 < e

for all points Q in Ll n aD. On D n Ll', w(P; Qo) is harmonic and strictly positive, and so has a positive minimum Wo which is attained on the boundary of D n Ll'. Consider the harmonic function W(P) = f(Qo) + e+ w(P; Qo) [K - f(Qo)]' Wo

Everywhere on

D,

W(P) ~ f(Qo) +e.

Sincef(Q) lies betweenf(Qo) ± e on aD

n Ll, we have

W(Q) > f(Q)

there. But on aD n 6.'

Thus W - f > 0 on aD. Ifv belongs to the family.9j, v - Wis continuous in D, subharmonic in D, and v-W:::;f-W
on aD. Hence, by the maximum principle, v(P) < W(P)

everywhere on D. Since u is the supremum of all functions v belonging to .9j, u(P) :::; W(P),

SUBHARMONIC FUNCTIONS

182J

[9.5

and so limsupu(P)

~

lim sup HT(P) = lim HT(P) =f(Qo)+e.

P-.Qo

P-.Qo

P-.Qo

Next consider the harmonic function w(P, Qo) [K +f(Qo)]' wo V(P) ~ f(Qo) - e,

V(P) =f(Qo)-e

Everywhere on

D,

and, in particular on aD n Ll, But on aD

V(Q)

n Ll', V(Q)

~f(Qo)-e


~f(Qo)-e-[K+f(Qo)] =

-K-e
Therefore V belongs to SIj, and so V(P)

Hence

liminfu(P) P-.Qo

~

~

u(P).

liminfV(P) = lim V(P) =f(Qo)-e. P-.Qo

p-.Qo

Vi7e have thus proved that, for every positive value of e,

f(Qo)-e

~

liminfu(P) p-.Qo

~

limsupu(P)

~f(Qo)+e.

P-.Qo

Hence u(P) tends tof(Qo) as P-+Qo, provided that there is a barrier at Qo'

9.6 Some examples of barriers The simplest case of a barrier arises when D lies in an open half-plane except for one point Qo of aD which lies on the line which is the frontier of the half-plane. Take Qo as origin, y = 0 as the line and y > 0 as the half-plane. Then w = y is a barrier for Qo' For example, if the frontier of D is an ellipse, this geometrical condition is satisfied at each point of aD. Hence the problem of Dirichlet has a solution for an ellipse. A somewhat similar test occurs when there exists a disc Ll such that Ll and D are disjoint, but Li and D have a single point Qo in common. If 0 is the centre of the disc, OP w(P) = logOQo :..:t.~

is [{' barrier at Qo' This condition is again satisfied at each point of aD if the' frontier of D is an ellipse. It follows again that the problem of Dirichlet has a unique solution for an ellipse.

9.6J

SUBHARMONIC FUNCTIONS

[183

In another simple example, the point Qo is the end-point of a straight segment QoA every point of which except Qo lies in the complement of 15. If we choose the axes so that Qo is the origin and A is (a,O) where a > 0, we can use polar coordinates (r, 0) and (r v 01 ) with Qo and A as origins, so that x = rcosO = a+ r 1 cos 01 ,

Y = rsinO = r l sin01 ,

where 0 and 01 lie between ± 7T. Then

J(~)

cos !(O- 01 )

is a barrier at Qo' It is harmonic, since it is the real part of ,J{zj(z -a)}, it vanishes on y = 0, x < a and is positive elsewhere. If this geometrical condition is satisfied at each point of aD, the problem of Dirichlet has a unique solution for D. A simple example is the doublyconnected domain bounded by two concentric circles. If D is a simply-connected domain bounded by a regular closed curve, aD may have a finite number of corners. If Qo is not a corner, we may take QoA along the outward normal. If Qo is a corner which is not re-entrant, we may take QoA along either of the outward normals at Qo to the two arCs which meet there. So, if aD is a regular closed curve with no re-entrant corners, the problem of Dirichlet has a solution for D. The situation is more complicated at are-entrant corner Qo' If the two arcs which join at Qo do not touch, we can still find a segment QoA there. Even if the arcs touch but form a keratoid cusp at which the two arcs lie on opposite sides of the tangent at Qo' the test is applicable. But if the two touching arcs form a ramphoid cusp at which the two arcs lie on the same side of the tangent at Qo' the test fails. There is another barrier which enables us to take account of reentrant corners. Suppose that, at a point Qo of aD, there is a regular arc r which lies, apart from its end point Qo, in 15', the complement of15. Let Ll be a disc with centre Qo of so small a radius that its circumference intersects aD and r. As we go along r from Qo, there will be a first point A which lies on all. Cut Ll along the arc QoA. Call the cut disc Ll o' Take Qo as origin, the tangent to r at Qo as axis of x, and write x = rcosO,Y = rsinO. Then w= _ log (ria) ( ) {log (rja)}2+ 02 r9=O,

°:: ;

=

°

(r

=

0)

184J

[9.6

SUB HARMONIC FUNCTIONS

is a barrier at Qo' (u is one-valued and continuous on [j and on Ll o' Since, when z 9= 0, (j)

= - re log (zja)'

w is harmonic on D. And, if we choose a large enough, w >

°

on 8D except at Qo' Hence the result. This enables us to complete the discussion where D is simplyconnected and 8D is a regular closed curve. If Qo is not are-entrant corner, we take as the regular arc the straight segments ofthe previous test. If Qo is a re-entrant corner, take Qo as origin and choose the xaxis so that, near Qo, lJ' lies in x > 0. Near Qo, the two arcs of 8D have equations y = ¢(x) and y = ¢(x) where ¢ and ljf are differentiable. Since the two arcs intersect only at Qo, there is an interval < x < a on which ¢(x) < ljf(x); and ¢(x) < y < ljf(x) belongs to lJ'. Then,

°

for

°.:; x < a,

y = H¢(x) + ljf(x)}

is a regular arc which lies, apart from its end point Qo, in [j'. The problem of Dirichlet thus has a unique solution when 8D is a regular closed curve. The same argument applies when D is a multiplyconnected domain bounded by a finite number of non-intersecting regular closed curves.

9.7 Discontinuous boundary data In the case ,,·hen the domain D is bounded by a regular closed curve, the requirement that the boundary data be continuous can be lightened; we can allow for a finite number of ordinary discontinuities. It suffices to consider one such discontinuity. Suppose that. the boundary function f has one ordinary discontinuity at a point (a, b) which is not a cusp of 8D. Since 8D is regular, there is either a tangent to 8D at (a,b) or the two regular arcs which join these have tangents in different directions. Vi7e may suppose that the axes are chosen so that the approach to (a, b) on the two arcs is as x-+a-O and as x-+a+O. As x-+a+O on 8D,

y-b

- - -+tana x-a and as x-+a- 0,

y-b

- - -+tanfJ x-a

(-!7T < a < !7T),

(t 7T < jJ

< i7T)·

If there is a unique tangent at (a,b),jJ=7T+a. As x->a+O on 8D,f-+f+; as x-+,a-O,f-+f_·

SUBHARMONIC FUNCTIONS

9.7J

If

[185

y-b F(x,y) =/(x,y)+Ktan-1- - , x-a

F-+I++Krx as x-+a+O, F-+I_+KfJ as x-+a-O. Hence F is continuous at (a,b) if K = (/+-I-)/(fJ-rx). Let U(x,y) be the function which is harmonic in D, continuous on D and which takes the continuous boundary values F on aD. Then b u(x,y) = U(x,y)-Ktan- 1y - , x-a

with an appropriate choice of the branch of the inverse tangent, is harmonic in D and takes the continuous boundary values 1 on aD except at (a, b). Since U is continuous on D, U tends to a limit L as (x,y) tends to (a, b) on a path in D not tangent to the regular arcs which meet at (a, b); and the inverse tangent tends to a limit y definitely between rx and fJ. Since F is continuous, L =1++Krx = 1-+KfJ.

Hence

u-+I+ -K(y-rx)

1

=

fJ-rx {(fJ-y)j+ + (y-rx)/_}·

Hence u tends to a limit, depending on the direction in which (x, y) approaches (a, b); this limit lies between 1+ and 1-.

7

CPD

10 EQUATIONS OF ELLIPTIC TYPE IN THE PLANE

10.1 The linear equation Suppose that a linear equation of the second order with two independent variables is ofelliptic type in some region in which the coordinates (x, y) can be chosen so that the characteristics are x ± iy = constant. Then the equation is of the form 82u 82 u 8u 8u 8x2 + 8y 2 + 2a 8x + 2b 8y + cu = F,

(1)

where a, b, c and F are functions of x and y alone. For brevity we write this equation as L(u) = F. Associated with L is the adjoint operator L* defined by

L*(u) Then

82u 8x

82u 8y 2

8 8x

8 8y

vL(u)-uL*(v)

= -~-+-

= -+--2-(au)-2-(bu)+cu. 2 8H ox

(2)

oK 8y'

H = vu x -uvx +2auv, K = VU y -uv y +2buv. L is said to be self-adjoint if the operators Land L* are identical; this happens if and only if a and b are zero. If a and b are constants, the operator M defined Q.y where

M(u) = eax+bYL(e-ax-byu) is self-adjoint. The problem of Dirichlet for the linear equation (1) is to find a solution which is continuously differentiable in the closure of a bounded domain D, which has continuous second derivatives in D and which takes given continuous boundary values On the frontier 8D of D. ViTe have seen that, in the particular case of Laplace's equation or Poisson's equation when a, b, c are identically zero, the problem of Dirichlet has a solution and that it is unique, provided that the boundary satisfies certain conditions. This is not necessarily the case for the general linear equation. [ 186 ]

10.lJ

ELLIPTIC EQUATIONS

[187

If the linear equation (1) had two solutions u 1 and u 2 , the function U 1 - U 2 would satisfy L(u) = and would vanish on 8D. For Laplace's equation, this implies that u is identically zero on the closure of D, so that, if the problem of Dirichlet has a solution, it is unique. It can happen that the problem of Dirichlet for L(u) = with zero boundary values on 8D possesses solutions which are not identically zero on 15, so that the problem of Dirichlet for L(u) = F does not necessarily possess a unique solution. The other problem associated with a bounded domain is the problem of Neumann, in which the normal derivative is assigned on 8D. In the case of an unbounded domain, we have to impose in both these problems conditions on the behaviour ofthe solution at infinity, just as we did for Laplace's equation.

u =

°

°

10.2

The reduced wave equation

In the theory of small vibrations of a uniform stretched plane membrane with a fixed rim, the small normal displacement U of the membrane at the point (x,y) at the instant t satisfies the equation of wave motions (1)

with an appropriate choice of units of length and time. Since the equation is linear, the general solution is the sum of solutions corresponding to the vibration~ in normal modes of oscillation. In a normal mode, U is of the form u(x, y) cos (kt+a) where k and a are constants, and u satisfies the reduced wave equation 82u 82u 8x2+ 8y2 + k2u = 0.

(2)

If the boundary of the membrane is a regular closed curve 0, we require u to satisfy the usual differentiability conditions inside and on 0 and to vanish on C. In the simplest case, 0 is the square with sides x = 0, x = a, y = 0, y = a. Then (2) has the solution . mrrx . nrry

u=sm--,;:-sm--,;-, which vanishes on 0 provided that m and n are positive integers and that (3)

In the problem ofthe vibrations of a square membrane, this equation 7-2

188J

ELLIPTIC EQUATIONS

[10.2

gives the frequencies of the normal modes of vibration, and the general solution of equation (1) is of the form ~ . mrrx . nrry cos (k'mn t + cx ), .£..J a SIn - smmn m,n mn a a

where Now forget the membrane problem and consider the reduced wave equation (2) as an elliptic equation with a given k. If a = ..)2rrjk, equation (2) has the solution . kx . ky (4) u = sm ")2 sm ..)2' which vanishes on C, but does not vanish identically inside C. Therefore for this square, the problem of Dirichlet does not have a unique solution; we can always add to any solution a multiple ofthe solution (4). But if a < ..)2rrjk, there is no non-zero solution of(2) which vanishes on C; hence, if the problem of Dirichlet for (2) has a solution, it seems likely that it is unique if the square is sufficiently small. The problem of the normal modes of vibration of a circular membrane with fixed rim l' = a, using polar coordinates, can be discussed in a similar way. In a normal mode of vibration, U is of the form ucos (kt+cx) where u satisfies the reduced wave equation. Now, in polar coordinates, equation (2) is 02U 1 ou 1 02u 2 01'2 +;- or +;2 082+ k u

= o.

By the method of separation of variables, we find that this equation has the particular solution u

=

Jm(kr) cos (m()+cm),

where m is a positive integer or zero, to ensure that the solution is finite and one-valued when l' ::::; a, and Jm is the Bessel function of order m. In the membrane problem, we get the normal modes by choosing k so that Jm(ka) vanishes, and the general solution is the sum of solution corresponding to the various normal modes. Now z-mJm(z) is an even function of z, whose zeros are all real and simple. Denote the positive zeros by jm,1,jm,2'''' in increasing order of magnitude. Then the norma,l modes of vibration of the circular membrane with fixed rim l' = a are given by U = JmUm,n1'ja) cos (m8+ €m,n) cos (km,nt+cxm,n),

where

10.2J

ELLIPTIC EQUATIONS

[189

Again, forget about the membrane problem and consider the reduced wave equation (2) in its own right. If k = jm,n/a, there exists a solution which vanishes on r = a. Hence for the disc r :::; a, the problem of Dirichlet does not have a unique solution. The least of the zerosjm,n isjo,l' If a < jo,l/k, there is no non-zero solution of (2) which vanishes on r = a. Hence, if the problem of Dirichlet for (2) has a solution, it seems likely that it is unique if the disc is sufficiently small. The argument in this section is not entirely rigorous, but it does suggest the sort of result we should expect.

10.3 The elementary solution The theory of harmonic functions in the plane depends on the existence of the solution log R where R is the distance from the singularity (xo, Yo) to the variable point (x, y). A homogeneous linear equation (1)

also possesses a solution with a similar logarithmic singularity; it was called by Hadamard the elementary solution. Its existence for V'2u+cu = 0 was first proved by Picard. In his Lectures on Cauchy's Problem, Hadamard deals with the general linear homogeneous equation which is not a parabolic type. He discusses the problem in a space of any number of dimensions; in n-dimensional space, the elementary solution behaves near the singularity like 1/Rn-2. The reduced wave equation u=+u YV +k2u

=

0

has two solutions which depend only on R, viz. Jo(kR) and Yo(kR). Jo(z) is an integral function of z in the complex plane, but :Yo(z) has a logarithmic singularity at z = 0; in fact, ~(z)

2

= -Jo(z)logz+ V(z), 7T

where V(z) is an integral function. Hence :Yo(kR) is the required elementary solution of the reduced wave equation. It is not uniquely determined; we could add to it any constant multiple of Jo(kR). The discussion here is based on Hadamard's proof and the method of Frobenius for solving by series an ordinary linear homogeneous differential equation.

190J

ELLIPTIC EQUATIONS

[10.3

'Ve choose the singular point as origin, and we assume that there is a neighbourhood of the origin in which the coefficients a, b, c in the equation [(1 )J

are analytic functions. Follow Hadamard, we "rrite f for r 2 , where (r, e) are the polar coordinates of the point (x, y). We ask whether L(u) = has a solution ofthe form

°

(2)

where v is a constant at our disposal and where the coefficients Un are analytic.

ffin~

L(UfV )

au = f L(U)+4vf"-1 ( ra;-+(v+'Y) U ), V

'Y = ax+by,

where we have L(u)

=

co

co

~fn+vL(Un)+4ffn+v-l(n+v)

(

au

)

r arn+(n+v+'Y)U" ,

assuming that term-by-term differentiation is valid. It follo,,-s that L(u)

= 4vfv- 1

(r aa~o + (v+ 'Y) Uo)

(3)

if the other coefficients satisfy the recurrence relation r

aa~n + (n + v + 'Y) Un = -

4(n

~ v) L(Un-l)'

(4)

If Uo is given, this differential equation gives the other coefficients successively; it is an ordinary differential equation in which the variable is only a parameter. If we choose Uo so that o r~+'YUo = 0,

e

au ur

,

(3) becomes

the resemblance to the case of equal exponents in Frobenius's method being evident. The equation satisfied by Uo is an ordinary differential equation with solution

Uo = A exp ( where A may depend on

e.

f:

\F ~~) ,

10.3J

[191

ELLIPTIC EQUATIONS

Since a and b are analytic near the origin, there is a disc on which '¥ = ax+by

=~' amnxmyn,

where the~' indicate summation over all non-negative integer~ except

o.

m = n = Since

rxmyn xmyn - - dr = cos me sin ne fr rm+n-1dr = -, fo r o m+n r dr xmyn ,¥-=~'a -mn m + n fo r

we have

on the disc where the series for '¥ is uniformly and absolutely convergent. Hence myn x -) . Uo = Aexp ( -~ , a mn m+n If, in (4), we substitute '¥ we obtain

r :r

(g;) +

(n + v)

=

-;0 °o~o,

g; = -

4(n: v) U L(Un_1)· o

It follows that Un = - 4(

U fr r n +v - 1 0) nv + L(Un_1) dr, n+vr 0 0

-u.

the lower limit of integration being taken to be zero to ensure that each coefficient Un is analytic near the origin. If we write U = r(v+l) u.w: n r( V + n + 1) 0 n' where Jfo = 1, we have r(v+ 1) v (5) u = uor r( 1) ~rn. o v+n+

z: (X)

The recurrence formula becomes

where with

02W 02W oW oW L1(W) = ox 2 + oy2 + 2a1"8X+ 2b 1ay +C1W, a1 = a+

1 oUo

Uo

ox'

b1 = b+

1 oUo

Uo

8Y'

c1 =

1 L

U.

llo (

0)·

Since Uo is analytic and does not vanish in a neighbourhood of the

192J

ELLIPTIC EQUATIONS

[10.3

ongm, these coefficients are all analyt,ic there. By the method of dominant functions, it can be shown that the series (5) is uniformly and absolutely convergent in a neighbourhood of the origin, and the formal method by which (5) was obtained is then valid. The details under more general conditions will be found in Hadamard's book. Since U defined by (5) satisfies L(u) = 4v 2

r 1 uo, v-

where rand Uo do not depend on v, we obtain a solution U 1 of L(u) = 0 by putting v = 0; this solution

COw

U1

where

Wo

= UoL:o -f rn, n.

= 1, w n = -

4~n

J:

L 1 (u'n_l)1· n - 1 dr,

is analytic in a neighbourhood of the origin. The series (5) and the series obtained by differentiation with respect to v can be shown to converge uniformly on any interval a ::::; l' ::::; jJ which contains no negative integer. Since L

(OU) ov

=

a second solution of L(u)

8vrv - 1 [1,0 + 4v 2 r v - 1 [1,0 log r' =

0 is

u2

=

OU) ' (IOv. v=o

or This is the required elementary solution. The first term contains the logarithmic singularity, the second is analytic in a neighbourhood ofthe origin; we could add to the second term any solution ofL(u) = 0, analytic near the origin. This discussion of the theory of Hadamard's elementary solution is incomplete - a discussion of the convergence of the series (5) is needed. All that is given here is a method of finding the form of an elementary solution. The difficulty in the method is the evaluation of the integrals which determine successively the coefficients IT:;'. In some problems it is possible to calculate explicitly the first few coefficients and to make an inspired guess at the form ofthe elementary solution and then to find it by direct substitution in the differential equation.

10AJ 10.4

[193

ELLIPTIC EQUATIONS

Boundary value problems

Let D be a bounded domain whose frontier aD consists of a finite number of non-intersecting regular closed curves. The most important boundary value problems for the linear equation L(u) == V2u-t2au x +2buy +cu = F(x,y),

where a, b, c, F are analytic, are the problems of Dirichlet and of Neumann. In the problem of Dirichlet, we have to find a solution with continuous first derivatives in 15 and continuous bounded second derivatives in D, given that u = f(x, y) on aD, f being continuous. In the problem of Neumann, au/oN is given on aD. We try to solve these problems by the use of Green's theorem. (See Note 5.) Let. L* be the operator adjoint to L. If the double integral exists ffn{VL(U)-uL*(v)}dxd Y = fan (lH+mK)ds,

where (I, m) are the direction cosines of the normal to aD drawn out of D, and H = vux -uvx +2auv, K = VU y -uvy +2buv. This is true if u and v have continuous bounded second derivatives on D and continuous first derivatives on 15. In particular, if L*(v) = 0, ffn vFdxdy = fan (v

:~-u ~~+2(la+mb)Uv)ds.

vVe assumed that a, b, c are analytic. Then L*(v) = mentary solution 1 v = vologR+vv

(1)

° has an ele-

where and

vo(xo' Yo) = 1.

If (x o' Yo) is a point of D, Vo and VI are analytic in D, but v has a logarithmic singularity at (x o' Yo)· Equation (1) no longer holds; it is replaced by u(xo,Yo) -=

2~fan {(v :~-u:~) +2(la+mb)Uv}ds-2~ffn vFdxdy,

by the argument we used in the theory of Laplace's equation. This formula contains the values of u and au/oN on aD. In order to solve the problem of Dirichlet, we must in some way eliminate au/oN. The elementary solution v is not uniquely determined; we

194J

[lOA

ELLIPTIC EQUATIONS

can add to VI any solution v 2 of L*(v) = 0 which has no singularitiei' on 15. Suppose that VI is fixed. Then we can take as the elementary solution 1 V = V olog 'R, + VI + v2 and choose v2 so that

v2

= -

1

V

olog 'R, -

VI

on aD, if this is possible. Assuming that this particular problem of Dirichlet for the adjoint equation is soluble, V vanishes on aD, and we get the required solution of L(u) = F. The elementary solution chosen in this way is the Green's function. In the same way, we can solve the problem of Neumann if we can find an elementary solution of L*(v) = 0 of the form 1

V

= volog'R, + VI +v s

where the solution Vs of the adjoint equation L*(v) = 0 satisfies the condition

1)

(1)

av 2 (la+mb)v s = - oN a (volog'R,+vI +2(la+mb) volog'R,+vI aNs

on aD. But V s is not a solution of the problem of Neumann for the adjoint equation; it has to satisfy a more general condition of the form av ()1I~ + hvs = g(x, y) on the boundary aD. Even if we succeed in finding a solution ofthe problem of Dirichlet for the equation L(u) = F(x, y), we still have to show that the solution is unique. If there were two solutions U I and u 2 , u = U I - U 2 would satisfy L(u) = 0 and would vanish on aD. To prove uniqueness, WE' have to show that the only solution of this last problem is identicall~ zero. As we have seen in the case of the reduced wave equation, thif: is not necessarily the case. Consider the self-adjoint equation

V'2U + CU

= 0,

where c is now not constant on D. Since

ffn {uV'2u+ui+u~}dxdy= fan u :;d8, v;,e have

ffn

(u~+u~)dxdy =

ffn cu 2 dxdy

10.4[

[195

ELLIPTIC EQUATIONS

if u vanishes on aD. If c ::::; 0 everywhere on D, ffn

(u;+u~)dxdy::::; o.

Therefore U x and u y vanish everywhere on D. Since we are considering solutions which are continuously differentiable on the closure of D, u is constant on D and so is identically zero on 15. It follows that, if the problem of Dirichlet for V'2U+ CU = F(;-,;,y), where c::::; 0 on D, has a solution, the solution is unique. This result can be extended. If c ::::; 0 on D, the equation L(u) == V'2U+ 2aux + 2bull +C1t

= F(x,y)

has at most one solution which is continuously differentiable on 15, which has continuous bounded second derivatives on D and which takes given continuous values on the boundary aD. This is equivalent to proving that if u satisfies the differentiability conditions and is a solution of L(u) = 0 which vanishes on aD, then u is identically zero on 15. Suppose firstly that c < 0 everywhere on D. Then no solution of L(u) = 0 has a positive maximum or a negative minimum at any point of D. For if such a solution had a relative maximum at a point P of D, U X and u y would vanish at P and we should have u xx ::::; 0, 'U yy ::::; 0 there. By the differential equation, CU ~ 0 at P and hence u ::::; 0 there. Similarly if it had a relative minimum at a point P of D, we should have u ~ 0 there. Since u is continuous, it attains its supremum and its infimum on 15. A supremum attained at a point of D (an interior point of 15) is a relative maximum and so cannot be positive. But u vanishes on aD, and therefore the supremum of u on 15 is zero. Similarly the infimum of u on 15 is zero. Hence u vanishes identically on 15. The case c ::::; 0 can be transformed into the case c < 0 by a change of dependent variable. If we put

u = (A _e aX ) U, where A and a are positive constants, the equation L(u) = 0 becomes V'2U + 2a'Ux + 2b'UlI +c'U = 0,

where

C

'

= Ac-eaX (a 2 +2aa+c)

A _e ax

•

Now a and c, being continuous, are bounded on 15. Choose a so large that

196]

ELLIPTIC EQUATIONS

[lOA

on 15. Then chouse A so that A > e where h is the supremum of x on 15. Then A _e IXX ~ A _e lXh > 0 lXh

e' = _Aiel +e

on 15. But

IXX

(a 2 +2aa+e) < 0

A_e lXx

on 15. It follows that U1 and hence u, are identically zero on 15. If the condition e ~ 0 does not hold everywhere on D, it suffices to require that the area of D is sufficienHy small. The result is stated by Petrovsky·t

10.5 The linear equation with constant coefficients The homogeneous linear equation of the second order with constant coefficients can be put into the form 82u 82u 8x 2 + 8 2 +AU = 0,

y

(1)

where A is a constant. If the problem of Dirichlet for this equation has a solution, it is unique when A ~ 0 but not necessarily unique when A> o. Let us consider this in more detail when the domain is a disc D with the circle C (equation r = a in the polar coordinates) as frontier. Try to find a solution which is continuously differentiable in r ~ a, which has continuous second derivatives in r < a, and which satisfies the condition U = 1(0) when r = a, 1 being of period 21T. The function 1(0) must then have a continuous derivative 1'(0). Then 1(0) has a Fourier series 00 tao + '1: (an cos nO + bnsin nO), 1

which is uniformly convergent on any finite interval, since nan and nb n tend to zero as n ....HX). If A = k 2 , (1) has a formal solution _ 1 Jo(h) u - ~ao Jo(ka)

.

+

t 00

(an COS n

0 'b T

. 0 In(kr) n S111 n ) In(ka) .

(2)

Now z-nJn(z) is an even function ofzwhich has only real simple zeros; the positive zeros arejn.l,jn,2,jn,3' .. , in increasing order.ofmagnitude. If Jm(ka) = 0, the series (2) becomes meaningless, except in the case when am = bm = O. The series (2) does then give a solution of the problem of Dirichlet; but it is not unique - we can always add to it a term of the form . (A cosmO+BsmmO)Jm(lcr).

t

Lectures on Partial DijJenntial Equations (Kew York, 1955), p. 232,

10.5]

[197

ELLIPTIC EQUATIONS

This difficulty does not arise if a is small enough. If jn,l is the least positive zero of z-nJn(z), the numbers Un,l:n = 0, 1,2, ...} form an increasing sequence. If < ka < jo, v none of the functions

°

vanishes, and the formal solution exists. If A is negative, say A = - P, equation (1) has solutions In(lcr) cos nO,

where

In(kr) sin nO,

• •

In(z) = ~-nJn(~z) = (iz)n

(-!z )2k

00

'1: k'( n+ k)". k=O l;.

The zeros of In(z), apart from the zero at the origin, are all purely imaginary, being of the form ± ijn.v ± ijn, 2' .... A formal solution is then . .1 Io(kr) 00 O ' 0 In(kr) a 2 O Io(ka) + ~ (an cos n + bn sm n ) In(ka)' (3) When r = a, its sum is 1(0). We have to discuss its convergence and continuity on r ~ a. When x ~ 0, the function I n (x) behaves like a multiple of x n when n is large, since _ (!x)n{ 00 n! 2 k} In(x) - - I 1+ '1: ( k)lk,(i x ) . n. k=l n+ . . (!x)n

= - , {1 + 1Jn(x)},

n.

where

1 1 1 °~ 1Jn(x) ~ n+ --1 '1: k' (!X 2)k < --1 exp (!x 2). k=l . n+ 00

Hence Since {nan} and {nb n} are null sequences, the series (3) and the series obtained by differentiation as often as we please are uniformly and absolutely convergent on every disc r ~ R when R < a. The sum of the series is therefore a solution of V'2u - k2u = continuously differentiable as often as we please on r < a. On r ~ a and on any finite interval of values of 0, the series (3) is of the form

°

where

198]

[10.5

ELLIPTIC EQUATIONS

The series ~un(8) is uniformly convergent. For any particular value of r in the interval [O,a], we have O~vn(r)~l;

and since where wn(r,a) tends to zero as n-+oo uniformly with respect to r, {vn(r)} is a non-increasing sequence on [0, aJ. It follows by Abel's test that the series (3) is uniformly convergent on X 2 +y2 ~ a 2 and so is continuous there. The series (3) thus gives a solution of 'V2u-k 2u = which is continuous on r ~ a, which is continuously differentiable as often as we please on r < a and which takes the given continuously differentiable valuesf(8) on r = a. The corresponding problem for V'2u + k 2u = can be discussed in a similar way.

°

°

10.6 The use of the elementary solution Let D be a bounded domain whose frontier consists of a finite number of non-intersecting regular closed curves. If L(u) == 'V2u+k2u where k is a positive constant, we have

ffD {uL(v)-vL(u)}dxdy = foD ('11 :;-v :;) ds, where OfoN denotes differentiation along the normal to oD drawnout of D. Sufficient conditions for this use of Green's theorem are that '11 and v be continuously differentiable in the closure of D and possess bounded continuous second derivatives in D. Ifu and v are solutions of L(w) = 0, the left-hand side vanishes and (1)

If R is the distance from a fixed point (xo, Yo) of D to a variable point (x, y), this result holds if v is Jo(kR), but not if v is the elementary solution 2 :fo(kR) = -logR·Jo(kR) + ..., 1T

where the terms omitted have no singularity in D. If v is the elementary solution, we have to exclude from D a disc with centre (xo,Yo) just as in the theory of Laplace's equation. The

10.6]

ELLIPTIC EQUATIONS

[199

result is that, if (x o' Yo) is a point of D, u(xo, Yo) =

~fiJD {u O~:Yo(kR) -

:Yo(kR) :;} ds.

(2)

This formula involves the boundary values of u and its normal derivative, and so does not provide a solution either of the problem of Dirichlet or of the problem of Neumann. To solve them, we have to construct Green's functions. If oD is a circle X 2+y2 = a 2, a typical point of the boundary has polar coordinates (a, e). If (x o' Yo) has polar coordinates (r o, eo), then R2 = a 2+ r~ - 2aro cos (e - eo)'

If we use the addition theorem <xl

:Yo(kR)

= :Yo(ka) Jo(kr o) + 2 '1: Yn(ka) In(kr o) cos n(e - eo), 1

we obtain a solution 00

u(ro' eo) = !aoJo(kro) + '1: (an cos neo + bn sin neo) In(kr o), 1

where the coefficients an and bn depend on the boundary values of u and ou/oN. In particular, (2) gives the value of u at the origin Uo

= !kY~(ka) fe uds-i:Yo(ka) fe:;ds,

where C is the circle r = a. But (1), with v = Jo(kR), gives

o = tkJ~(ka) fe uds-iJo(ka) fe:;ds. Hence

Jo(ka)u o =

~{Jo(ka) Y~(ka)-J~(ka):Yo(ka)} fe uds

in which the values of ou/oN on the boundary do not appear. The expression is called the Wronskian of Jo and:Yo; it has the value 2/(1TZ). We have thus arrived at the mean-value formula Uo

=

21T~(ka)fe uds.

Since J~(z) = -J1(z), we have another mean-value formula uo = -

f

1 -ou ds 21TaJ;.(ka) eoN

depending only on the values of the normal derivative on C.

200]

[10.7

ELLIPTIC EQUATIONS

10.7 Divergent waves So far, we have considered the solution of the reduced wave equation only on a bounded domain. When the domain is unbounded, a condition at infinity is necessary. An important case arises in the theory of the reflection of sound waves of small amplitude by a rigid obstacle. Ifthe incident waves have a velocity potential9i' the reflected waves a velocity potential ¢s' the condition on the boundary of the obstacle is

The condition at infinity is that ¢s should represent a divergent wave motion. 'Ve assume that the motion is 'monochromatic', that the velocity potentials vary in a 'simple-harmonic manner with the time. It is convenient to use complex velocity potentials of the form

¢ = ue-ikct where k > 0, and take real parts at the end. Then u satisfies the reduced wave equation V'2u+k 2u = 0. In this section we consider the case when the wave motion is cylindrical; the obstacle is a cylinder with generators parallel to the z-axis; ¢i' and hence also 9s' are independent of z. We saw in §6.1 that the velocity potential of cylindrical waves due to a source of strength 21TF(t) per unit length on the z-axis is

¢=

f:oo F(t-rcosha)da

where r is the distance from the axis of z. In particular, for monochromatic divergent cylindrical waves .

¢

=

exp ( -ikct)

f:

00

exp (ikr cosh a) da = 1TiH&l) (kr) exp ( - ikct),

where H&l) denotes the Hankel function of order zero. The function of order n is Using the asymptotic expansion for the Hankel function H&l)(kr), we find that

¢o

=

H&l) (kr) exp ( - ikct) '"

r

(1T~r

exp {i(kr - kct -

i 1T )}

as r -+ 00. The velocity potential ¢o represents expanding harmonic

10.7]

[201

ELLIPTIC EQUATIONS

cylindrical waves with amplitude decreasing like 1Nr. A more general velocity potential is

¢n

=

H~)(kr) C?S nO exp (-ikct), sm

which behaves like a multiple of exp {ik( r - ct)} cos 0 .jr sin n , as r""",*oo. If ¢n = un exp (-ikct), it can be shown that, for any integer n, (i) as r""",*oo, un.jr is bounded, uniformly with respect to 0, (ii) .jr{oun/or-ikun}""",*O as r""",*oo, uniformly with respect to O. These two conditions were called by Sommerfeld the finiteness condition and the radiation condition. They play an important part in the theory of the solution of the reduced wave equation in an unbounded domain. Let D be the domain which is bounded internally by a regular closed curve C and externally by a circle r of large radius a. Then if (x o, Yo) is a point of D, we obtain, as in §10.6 the expression

°

for the value of a solution u of V'2U + k2u = at the point (x o, Yo). This is obtained by combining formula (2) and formula (1) with v = Jo(kR). We may take r to be the circle R = a. Hence

u(xo, Yo )

="4i

f

r

{OU kR ) - u oR 0 H(l) } oR H(l) 0 ( 0 (kR) ds,

+£f {U~H&I)(kR) - Ou 4 c on on

H&l)

(kR)} ds

'

where n is the unit normal vector drawn out of the bounded domain whose frontier is C, this being oppositely directed to N. In the integral over r, the integrand is

-ikU) v- (OV -ikv) U ( OU oR oR ' where

v

=

H&I)(kR).

This function v satisfies Sommerfeld's finiteness and radiation conditions. There exists a constant K I such that Ivl < KI/.jR; and, for

202]

[10.7

ELLIPTIC EQUATIONS

every positive value of e, there exists a positive number a l such that ov

'k

loR-~

I

v <

.JeR '

when R > aI' Suppose that u also satisfies the finiteness and radiation conditions. Then there exists a constant K 2 such that lui < K 2 f.jR; and, 'with the same value of e, there exists a positive number a 2 such that ou 'k e oR-~ u < _,JR' when R > a 2 • Choose a greater than max (a V a 2 ). Then

I

l

I~Jr{:~H&l)(kR)-Uo~H&I)(kR)}dsl < !1T(KI +K )e, 2

so that the integral round r tends to zero as a-+oo. Therefore, at any point (xo, Yo) outside 0, the solution u of the reduced wave equation has the value u(xo,Yo) =

~Jc {u :n H&I) (kR) -

:: H&l) (kR)} ds

provided that u satisfies Sommerfeld's conditions. This formula expresses a diYergent wave function u in terms of the values taken by u and ou/on on O. But it cannot be used to solve scattering problems in acoustics because, although 8us/on is known on the rigid boundary 0, Us is unknOVi'll there. To eliminate the unknown values of Us on 0, we should have to find a Green's function. Usually this is not possible; we give a simple example when it can be done in the next section. vVe shall discuss Sommerfeld's conditions in more detail in Ch. 11 when we consider the reduced wave equation in three-dimensional space.

10.8

The half-plane problem Let u be a solution of V'2U + k2u = which is continuously differentiable in y ~ 0, has continuous second derivatives and satisfies Sommerfeld's

°

conditions at infinity. Let u = f(x) or u y = g(x) when y Yo> 0, u(xo,Yo) = -!i g(x)H&I)(kRo)dx,

= 0. Then, when

J:oo

01'

where

u(xo,Yo) =

-ti-:-Joo uYo

-00

f(x)H&IJ(kRo)dx,

10.8]

ELLIP TIC EQUAT IONS

[203 These formul ae expres s u in terms of the bound ary values of u or of ou/oN. Let D be the domain defined by

If (xo, Yo) is a point of D, so that Yo > 0,

u(xo,yo) =

where

~fiJD {:; H&lJ(kR) -u o~ H&lJ(kR)} ds, R2 = (X-X O)2+(y_YO)2;

°= ~fiJD {:;H&lJ(kR)-U O~H&lJ(kR)}dS,

but where

R2 = (x-xo)2+(Y+Yo)2

since (x o, -Yo) is in Y < 0. As in §10.7, the integra ls round the semicircle x 2+ y2 = a 2, y ~ tend to zero as a -+ 00. On y = O,ou/oN = -g(x). Also

°

..?-H(lJ(kR) = -H(l.)( kR)koR = _H(lJ( kR)k(y -yo) oyO 1 oy 1 R'

so that, on y = 0,

~H(lJ(kR) = -H(lJ( kR ) kyo oN 0 1

0

R o'

where Similarly, on y = 0,

.!.. H(lJ(kR) = oN 0

H(lJ(kR ) kyo lOR' o

Therefore, when Yo > 0, u(xo,Yo) = -£foo 4 -00 and

°

=

g(X)H&lJ(kRo)dx+~4'foo f(x)H~lJ(kRo)kyRodx,

_£fOOg(x)H&lJ(kRo)dX-~4' foo f(x)H~lJ(kRo) 4 -00

-00

Adding, we have u(xo,Yo) =

0

-00

-ii f:oo g(x) H&ll(kRo) dx.

kyRodx. 0

204]

[10.8

ELLIPTIC EQUATIONS

Subtracting, or

u(xo, Yo) = u(xo,Yo) = -

ii

f:

00

l)

f(x) Hi (kRo)

ti :, foo v:/o

-

7£0 dx,

f(x) H&l) (kR o)dx. 00

To give a rigorous proof of these results, we should have to impose suitable conditions on f(x) and g (x) and consider the limits of these integrals as Yo -+ + o.

10.9

A boundary and initial value problem

We discussed in Ch. 7 the Cauchy problem for the equation of wave motions

namely to find u when t = 0, given that u = f(x, y), u t = h(x, y) when = O. In the theory of small vibrations of a uniform stretched membrane with a fixed rim, we have to solve a boundary value and an initial value problem; we are given the values of u and u t on a plane domain D when t = 0 and also given that u vanishes on the boundary of D when t ~ O. The solution is a sum of terms, each representing a normal mode of vibration, in which u = U(x,y) cos (wt+e). The function U satisfies the reduced wave equation

t

V 2 U +w 2 U = O.

In the simplest case, D is a rectangle 0 ~ x method of separation of variables; we obtain

~

a, 0

~

y

~

b. By the

. m1TX • n1TY U = sm-asm-b-'

where m and n are positive integers because U vanishes on cD, and W = wm •n where 2 m2n 2 n2n 2 wm •n =7+""1)"2· Hence we get a formal solution u =

'" (A £..;

m,n=l

" . m1TX • n1TY m.ncoswm.nt+ B m.nsmwm,nt)sln-a smb '

where the coefficients Am,n and B m.n have to be chosen so that u = f· = h when t = 0, For a rigorous discussion we need to know the difficult theory of double Fourier series. The calculation of the coeffi·

Ut

10.9]

[205

ELLIPTIC EQUATIONS

cients is easy; the usual Fourier rule gives 4 A m,n = ab

Wm,n B m,n

If If

4 = ab

D

f( x,y ). m1TX sm • n1TY dx d y sm--;-

b

D

h( x,y ). m1TX • n1TY d x d y. sm~ sm

b

When D is the disc r < a in polar coordinates, the reduced wave equation

has solutions· In(wr) cos (nO + a),

Yn(wr) cos (nO + fJ).

Since U is one-valued on the disc n is a positive integer or zero; and the Bessel function of the second kind cannot occur since it tends to infinity as r -+ O. Since U vanishes when r = a, In(wa) = O. The equation In(z) = 0 has a multiple root at the origin and all its other roots are real and simple; they are ±jn,l' ±jn, 2' •••• The positive roots from an increasing sequence, andjn,k tends to infinity with k. Write wn,k = jn,kia. Then we get a formal solution 00

u

=

~

n=O

00

~ (An,k cos nO +Bn,ksin nO) In(wn,kr) cos (Wn,kt+sn.k)'

k=l

where the coefficients have to be chosen so that the initial conditions on r ~ a are satisfied. This involves the theory of Fourier-Bessel expansions, of which an account will be found in Ch. 18 of G. N. ·Watson's Treatise on the Theory of Bessel Functions.

Exercises 1. Deduce from the mean value theorem of § 10.6 that a solution of Y'2u+k 2u = 0 cannot have a positive minimum or a negative maximum.

2. Monochromatic sound waves of small amplitude are propagated along a cylindrical wave guide of uniform cross section D. Taking the z-axis along the guide, show that a complex velocity potential satisfying ~2

Y'2 U = _1 0 U c2

is of the form

at 2

206]

ELLIPTIC EQUATIONS

where U satisfies the reduced wave equation

[PU

02U

_ + _ + ( k2_ y 2) U = 0, ox2 oy2

with boundary condition OU IoN = 0 on the surface of the wave guide. If the cross section of the guide is the square 0 ~ x ~ a, 0 ~ y ~ a, prove that a complex velocity potential is u

= cos mrrx cos nrry ei(yz-kct) a

where

a

'

y2 = k2_(m2+n2)rr2ja2,

m and n being positive integers. Show that this mode is propagated only if k > rr '\f(m 2 + n 2 )ja. If this mode is propagated, show that the phase velocity V = kcjy is greater than c and that the group velocity d(kc)jdy is less than c.

11

EQUATIONS OF ELLIPTIC TYPE IN SPACE

11.1 Laplace's equation The simplest equation of elliptic type in three dimensions is Laplace's equation 2 _ 82u 82 u 82u V u = 8x 2 + 8y 2 + 8z 2 =

o.

The theory of the solutions of this equation resembles the theory of harmonic functions in two dimensions, with some important differences. The theory of harmonic functions in two dimensions can be made to depend on the theory of analytic functions of a complex variable x + iy. There is nothing corresponding to the theory of functions of a complex variable x+iy in three dimensions. The nearest approach is given by Whittaker's general solution 217

u(x,y,z) =

J 0

f(z+ixcosu+iysinu,u)du

of Laplace's equation. The applications of Green's theorem run, in general, parallel to those in two dimensions, with the difference that the elementary solution in three dimensions is l/R which vanishes at infinity, whereas in two dimensions it is log l/R. There are coordinate systems in which Laplace's equation in three dimensions is separable, and these lead to solutions as infinite series whose terms are the various special functions of analysis.

11.2 Polynomial solutions A solution of Laplace's equation, analytic in a neighbourhood of some fixed point which we may take as origin is expansible as a convergent series 00 2:.fn(x, y, z), o

where fn(x, y, z) is a homogeneous polynomial of degree n which satisfies Laplace's equation. Now the most general homogeneous [ 207 I

208]

ELLIPTIC EQUATIONS IN SPACE

[11.2

polynomial fn of degree n contains i(n+2) (n+ 1) terms and so has the same number of independent coefficients. Since y2fn is a homogeneous polynomial of degree n-2, it has tn(n-1) different terms. If y2fn = 0, we have tn(n - 1) independent linear relations connecting the t(n+ 2) (n+ 1) constants infn' Hence there are only t(n+2) (n+ 1) -tn(n -1) = 2n+ 1

independent constants in a homogeneous polynomial of degree n which satisfies Laplace's equation. Thus there are 2n+ 1 independent homogeneous polynomials ofdegree n which satisfy Laplace's equation in three dimensions. For example, if n = 1, the three solutions are x,y,z; if n = 2, the five solutions are x 2-z 2,y2- z2,yz,zx,xy. In contrast to this, in two dimensions, there are only two independent homogeneous polynomials of degree n which satisfy Laplace's equation, viz. r n cos nO and r n sin nO in plane polar coordinates. A particular homogeneous polynomial of degree n which satisfies Laplace's equation is ( . . . ) z + ~x cos w + ~y sm w n, where w is a parameter. This can be expanded as a trigonometric polynomial n

tao(x,y,z)+

'1:

(Gm(x,y,z)cosmw+bm(x,y,z)sinmw).

m=l

Since w is independent of x, y, z, each coefficient am, bmis a homogeneous polynomial which satisfies Laplace's equation. The highest power of z in am(x, y, z) is zn-m, and am is an even function of y. Similarly the highest power of z in bm(x, y, z) is zn-m, but bm is an odd function of y. Thus the 2n + 1 polynomials am and bm are linearly independent, and are the required 2n + 1 harmonic polynomials of degree n. They are given explicitly by Gm(x,y,z) =

~Jl7

(z+ixcosw+iysinw)ncosmwdw

!Jl7

(z+ixcosw+iysinw)nsinmwdw.

1T

bm(x,y,Z)

=

1T

-17

-17

Every harmonic polynomial of degree n is therefore of the form

J:l7 (z+ixcosw +iy sinw)nf(w) dw, where f(w) is a trigonometric polynomial. A harmonic polynomial of degree n is usually called a spherical ha1·monic.

11.2]

ELLIPTIC EQUATIONS IN SPACE

[209

Now, if V(x, y, z) is a harmonic function analytic in a neighbourhood of the origin, it is of the form 00

V(x, y, z) = 1; 1i~(x, y, Z), 1

where ~(x,y,z) is a harmonic polynomial of degree n, the series being convergent near the origin. Hence

V(x,y,z) =

~

f:"

(z+cxcosw+iysinw)nfn(w)dw,

where fn(w) is a trigonometric polynomial. From this follows Whittaker's general solution of Laplace's equation

V(x,y,z) =

f:"

F(z+ixcosw+iysinw,w)dw,

where F(I;, w) is expressible as a power series in 1;, convergent in a neighbourhood of I; = O. This solution holds under less stringent conditions. All we need is that F is such that differentiation under the sign of integration is valid. Shifting the origin, we get a solution

V(x,y,z) =

f:"

F(z-zo+ixcosw-ixocosw+iysinw-iYosinw,w)dw

valid near (xo, Yo' zo). The generalisation is rather trivial since this solution is merely

V(x,y,z) =

f:"

G(z+ixcosw+iysinw,w)dw.

There is no special merit in giving z priority; we can interchange x, y, z as we please. It should be noted that ·Whittaker's solution can represent different harmonic functions in different parts of the plane. For example 1 21T

f"_"z+ixcosw+iysinw dw 1

is equal to

1

.j(X 2+y2+ Z2)'

;: =

when z > 0, but is equal to - l/r when z < O. ·When z = 0, the integral is not convergent, but has the principal value zero. We also have

-1 21T

f"

.

dw

"

_"x+~ycosw+~zsmw

1 =+-, -r

210]

ELLIPTIC EQUATIONS IN SPACE

[11.2

where the sign is + or - according as x is positive or negative. Moreover, we have found in the region x > 0, Z > two different integral representations of l/r, and there does not appear to be any simple transformation of the one integral into the other. It should be noted that, if we put z = ict, V\Thittaker's formula gives a solution -17 H(ct+xcosw+ysinw,w)dw

°

Jl7

of the two dimensional wave equation. Further applications of Whittaker's solution will be found Chapter 18 of 'Yhittaker and Watson's Modern Analysis.

III

11.3 Spherical harmonics There are 2n + 1 linearly independent spherical harmonics of degree n which we denoted by am(x,y,z) =

~Jl7 1T

(z+ixcosw+iysinw)ncosmwdw

-17

(m

bm(x,y,z) =

~Jl7

1T

=

0,1,2, ... ,n),

(z+ixcosw+iysinw)nsinmwdw

-17

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

In spherical polar coordinates

x = r sin 8 cos rp, these are

n

am(x,y,z) = r 1T

and

bm(x,y,z) = r

Jl7

y = r sin 8 sin ¢,

z = r cos 8,

{cos8+isin8cos(w-rp)}ncosmwdw,

(1)

{cos 8+isin 8 cos (w-rp)}nsinmwdw.

(2)

-17

n

1T

Jl7 -17

Now

(cos8+isin8cost)n = Pn(cos8) +2

i:

m=l

i- m ( n! ),P,:::(cos8)cosmt, n+m. (3)

where Pn(ji,) is Laplace's polynomial and P':::(Jl) is the associated function defined when - 1 < Jl < 1 byt m Pr;:(Jl) = (_1)m (1- Jl2)i md :;::)

t

This is Hobson's definition. The factor (_l)m is sometimes omitted. Formula (3) will be found in Hobson, The Theory oj Spherical and Ellipsoidal Harmonics, (Cambridge, 1931), p. 98; 'Whittaker and Watson, in Oourse oj l1iodern Analysis, (4th edn (Cambridge, 1927), p. 392) omit the factor. For properties of the Legendre' polynomials, see also Copson, Functions oj a Oomplex T'm'iable (Oxford, University Press, 1935) and Erdelyi et al. Higher Transcendental Functions (New York, 1953), Vol. 1.

11.3]

ELLIPTIC EQUATIONS IN SPACE

[211

It follows that

ao(x,y,z) = 2rnPn(cose), am(x,y,z)

=

n'. ),p'::(Cose)Cosmrp, 2r ni-m ( n+m.

bm(x,y,z) = 2rni-m ( n! ),p~(Cose)sinmrp. n+m. Every spherical harmonic of degree n is of the form [AoPn(COS e) + m~1 (Am cos mrp + Bmsin mrp) P:;:(cos e)] rn,

where the coefficients A k and B k are constants. The expression inside the square brackets is often called a surface harmonic. Of particular interest are solutions of Laplace's equation with symmetry about an axis, say the axis of z. Now such a solution which is analytict in a neighbourhood of the origin is expansible as a series 00

'1: un(x, y, z), o

where un is a homogeneous harmonic polynomial of degree n, independent of rp. Hence it is of the form 00

U(x,y,z)

=

'1: A n rllPnCu), o

where Jl = cose. If this solution takes on the sphere r = a the values f(Jl), we have formally 00

'1: AnanPn(Jl) = f(Jl)· o

Using the orthogonal properties of the Legendre polynomials, we get

~1 Anan = Jl f(Jl) Pn(,u) dlt. n+ -1 Thus a formal solution of the problem is V(x,y,z) =

2n+ 1 r n '1: 2 - - Pn(Jl) 00

Jl

f(t)Pn(t)dt. an -1 The problem now is to discuss whether the series on the right converges for r < a and whether its sum tends to f(Jl) as r~a. A full discussion of this is outside the scope of this book. We refer the reader to Ch. VII of Hobson's The Theory of Spherical and Ellipsoidal Harmonics. The problem of Dirichlet for a sphere is discussed later in this chapter by the use of Poisson's integral. t We shall see later that if a solution of Laplace's equation is harmonic in a neigho

bourhood of the origin, it is necessarily analytic there.

212]

11.4

ELLIPTIC EQUATIONS IN SPACE

[11.4

Green's theorem

Let D be a domain whose function is a regular closed surface S, as defined in Kote 6. S is then the union of a finite number of smooth caps; on each cap, there is a continuously varying normal direction. Let u, 'V, W be continuous in 15 and have bounded cont,inuous first derivatives in D. Then, if the triple integral exists

If

s(lu+mv+nw)dS=

Iff ( D

OU ov OW) 8x+ay+az dxdydz,

(1)

where (l, m, n) are the direction cosines of the outward drawn normal. The fact that there may be a finite number of curves on S, the rims of the caps, across which the direction of the unit normal vector N = (l, m, n) is discontinuous, does not affect the truth of this result. If, in (1), we put

u=

U~,

v = UVy,

w

=

U~,

we obtain

If we interchange U and V and subtract, we obtain

JIs (U:;-

V:~) dS =

IIID (UV 2V - VV2U)dxdydz,

(3)

provided that U, V, grad U, grad V are continuous on 15, that U and V have bounded continuous second derivatives on D, and that the triple integral exists. 'Ve sometimes need the case when 8D consists of a finite number of non-intersecting regular closed surfaces. For example, D might be the domain a < r < b in spherical polar coordinates; then aD is the pair of non-intersecting spheres r = a and r = b. In such a case, N is the unit normal vector drawn out of D.

11.5 Harmonic functions A function is said to be harmonic in a bounded domain D if it has continuous derivatives of the second order and satisfies Laplace's

11.5]

ELLIPTIC EQUATIONS IN SPACE

[213

equation there. The elementary solution l/R, where R2 = (X-XO)2+(Y_YO)2+(Z-ZO)2,

plays the same part as the logarithm does in the theory of harmonic functions in the plane. In spherical polar coordinates

where .Ie is a linear operator involving only differentiations with respect to the angle variables. If we put r = a 2 /s,

Hence It follows that, if V (x, y, z) is harmonic, so also is ~V r

(ar 2x' a2y a2z) r2 , r2 • 2

V(x, y, z) is said to be harmonic at infinity if ~V r

(a 2x a2y2 a2z)2 2

r ' r ' r

is harmonic at the origin. This enables us to extend the definition of harmonic functions to unbounded domains. If U = 1 and V is harmonic in D, equation (2) of § 11.4 becomes

IIs:;dS = o.

(1)

If V = U and V is harmonic, the equation becomes (2) fIs V:;dS = IIID (V~+ ~+ ~)dxdydz. Hence if V8V/8N vanishes on S, V. all vanish on D and so V ~,~,

is a constant. From this follow the uniqueness theoreIilll for the problems of Dirichlet and Neumann. If U1 and Uz are two functions which are harmonic in D and take the same continuous boundary values on S, V = U1 - Uz, is harmonic in D and vanishes on S. Hence V vanishes on D, and U1 = U2 there. Again, if U1 and U2 are two functions which are harmonic in D, and

214]

[11.5

ELLIPTIC EQUATIONS IN SPACE

if oUI/oN and oUz/81\T take the same continuous boundary values on S, V = UI - Uz is harmonic on D and eV/oN vanishes on S. Hence V is a constant on D, and the functions UI and Uz differ by a constant. As in the plane case, we cannot assign arbitrarily the values of the normal derivative on S of a function V harmonic in D; the boundary values must satisfy equation (1). In equation (3) of § 11.4, suppose that V is harmonic on D and that, U = 1/1'. Then

provided that the origin is not a point of D. In particular, if D is bounded by two concentric spheres r = a and r = R, then, if V is harmonic in r :::; R,

ff (~r °orV + rV) dS = rf Z

r=R

•

r=a

(~r

°OrV + rV) dS Z

.

As a-+O, the expression on the right tends to 47TV(0, 0, 0). The expression on the left is

Iff

R

oV -~

r=R Or

dS+ RzIff

T7 r=R

dS.

The first term vanishes by equation (1). Hence

V(O, 0, 0)

=

47T~Z

J

f=R

V

dS,

which is called Gauss's mean value theorem. Suppose that V is harmonic in a bounded domain D and continuous in D. If V is not constant it follows from the mean value theorem that V cannot have a relative maximum or a relative minimum at a point of D. Hence V attains its supremum and its infimum on D at points of oD. In particular if V is constant on oV, V is constant on D. This provides an alternative proof of the uniqueness of the solution of the problem of Dirichlet, if such a solution exists. If UI and Uz are two functions, harmonic in D and continuous in 15, which take "the same continuous boundary values on oD, V = UI - Uz is also harmonic in D and continu.ous in 15, and vanishes on oD. Hence V is zero on D.

11.6 Green's equivalent layer Let S be a regular closed surface bounding a domain D. If U and V have continuous first derivatives in D and have continuous second derivatives in D where they satisfy Laplace's equation, it follows

11.6]

[215

ELLIPTIC EQUATIONS IN SPACE

from (3) of § 11.4 that

IIs(U;;-V:~dS=O.

(1)

If R is the distance from a fixed point (x o,Yo, zo) to the variable point (x, y, z), we can put V = 1/R provided that (x o,Yo' zo) lies outside D. If (xo,Yo' zo) is a point of D, let So be the sphere R = £, where £ is so small that So lies in D. Then

If (

01

1

OU)

s U oNR-R olf dS+

If (

01

1

OU)

s, U oNR-R oN dS =

o.

Now on So, %N is - OjoR. Hence

IIs (U

o~ ~-~ :~) dS =

-

IIs,

(~2 U +~ :~) dS

= - IIs.( U +£ :~) dw , where dw is the element of solid angle subtended at (xo, Yo, zo) by the surface element dS of So. Since U and oU/oR are continuous in D, we obtain -47TU(Xo,yo,zo) as the limit ofthe right-hand side as £~o. Hence

U(xo,Yo,zo)

=

1

47T

If

(1 0 U

0 1)

s RoN- U oNR dS.

(2)

This result is known as the Green's equivalent layer theorem. It expresses U on D as the sum of potentials of a simple layer of density (oU /oN)/47T on S and a double layer of density - U/47T on S. Combining equations (1) and (2), we have

1 U(xo,Yo,zo) = 47T IIs {( V +~)

~~- U o~ (V +~)}dS.

If we could choose the harmonic function V so that

V+

1

R

vanished at all points of S, the derivative oU/oN would not occur in this integral. Such a function V + 1/R is simply the electrostatic potential of a unit point charge at (xo,Yo' zo) inside an earthed conductor S, and is called the Green's function. This physical argument suggests that there should be a Green's function. If we denote the Green's function by G(xo,yo,zo;x,y,z) or, more briefly, by G(po,P), a function harmonic in D has the representation

U(Po)

= -

:1TIIs U(P) o~G(Po,P)dS.

(3)

216]

ELLIPTIC EQUATIONS IN SPACE

11.6]

Thus the existence of a solution of the problem of Dirichlet is equivalent to the existence of a Green's function. By the argument used in the plane case, we can prove the symmetry property G(Po,P) = G(P, Po)' Lastly we assumed that the frontier of D was a regular closed surface. It could equally well have been a finite number of nonintersecting regular closed surfaces.

11.7 Green's function for a sphere The image of a unit point charge at a point Po inside a perfectly conducting sphere S of radius a and centre is a charge - a/OPo at the point PI which is the inverse of Po with respect to the sphere. For if P is any point of the sphere, the triangles 0PoP and OP fJr are similar, and so 1 a PoP = 0Po.fJrP'

°

Hence the Green's function for the sphere is 1 a G(po,P) = PoP- OPO.PIP'

If U is harmonic in the sphere, U(Po) = -

4~JJsU(P)O~{P:P-OPo~fJrP}dS.

It is convenient to use spherical polar coordinates. If Po is (r, 8, ¢) where r < a, then PI is (a 2/r,8,¢). If Pis (p,a,j3) POP2 = R5 = r 2+p2-2rpcosy, 4

fJrP2 =

where Since

2

a a Ri = -+p2-2-pcosy r2 r '

cosy = cos8cosa+sin8sinacos(¢-j3). o 1 rcosy-p 0 1 a2 cosy-rp op R o = R5 op R I = rRr

it follows that, when P is on the surface of the sphere,

where Hence

R2 = r 2+a2-2arcosy.

11.7]

ELLIPTIC EQUATIONS IN SPACE

[217

)JJn(a2+r2-2arcosy)t' U(a,J.,f3)dO.

2 2 U 8 ¢ = a(a -r (r, , ) 47T

or

where dO. is sinadadf3, the element of solid angle. Thus if U(r, 8, ¢) is equal to f (8, ¢) when r = a, we should expect that 2 2 U(r 8 ¢ = a(a -r )ff f(a,f3)d0. (1) , ,) 47T n(a 2+r2-2arcosy)t· This is called Poisson's integral. It is readily verified that the expression on the right is harmonic when r < a. We have to show that it tends to f(8, ¢) as r-+aunder appropriate conditions on f. If f is constant on r = a, U is constant when r :::; a. Hence

°

As we can rotate the axes in any way we please, it suffices to consider the limit when 8 = -i7T, ¢ = 0. Then 2 2 U(r, 17T , 0) - f(i 7T , 0) = a(a -r f(a, 13) - f~i7T, 0) dO.. 47T n(a2+r2-2arsmacosf3)1

)ff

We assume thatf(8, ¢) is continuous. For every positive value of e, there exists a positive number 0 such that If(a,f3)-f(i7T,O)1 < e, when - 0 < 13 < 0, i7T - 0 < a < i7T + O. Call the part of 0. on which this holds 0.0' and the rest 0.1' Firstly, we have

l

~ a(a2-r2)f'r 47T

f(a,f3) -f(i7T , 0) d0.1 2ar sin a cos 13)1

Jn. (a 2+ r2-

,::: ~ a(a2- r2)ef'r dO. ~ 47T . In.(a 2+r 2 -2arsinacosf3)! :::; ..!..a(a2-r2)efI dO.. < e. 47T n(a 2+r2-2arsmacosf3)!

Since f(a,f3) is continuous, it is bounded on 0.. Hence there exists a constant .11l such that

If(a, 13) - f(17T, 0)1 :::; M on 0., and, in particular, on 0.1' On 0.1, sin a cos 13 :::; cos 0, and so a2+r2- 2ar sin a cos 13 ~ a2+r2- 2arcos O. 8

CPO

218]

[11.7

ELLIPTIC EQUATIONS IN SPACE

Therefore

< (a2+r2-2arcoso)~'

which tends to zero as r -+ a - 0. Hence with the same value of we can choose 1'1 sufficiently near to u, so that

14~ a(a 2-r2)JJn,(a2!;~!da~{i~:'C;Sj3)tdQ I < when r 1 < r < a. Therefore, when r 1 <

l'

€,

€,

< a,

lU(r,!7T,O)-f(t 7T , 0)1 < 2€.

As tis arbitrary, U(r, t7T, 0) tends to f(!7T, 0) as r-+ a- 0. It follows that, whenf(8, ¢) is continuous, the problem of Dirichlet has a solution, necessarily unique, given by Poisson's integral. The conditions on f can be lightened. Iff is integrable, Poisson's integral represents a function harmonic in r < a, such that U(1', 8, ¢)-+ f(8, ¢)

as r -+ a - Oat every point of continuity off. If U(x, y, z) is harmonic in r > a, including the condition at infinity, and assumes the valuesf(8,¢) on r = a, then

r:: U r

z)

2 2 (a x a2y a 2 2 r ' r ' r2

is harmonic in r < a and assumes the same boundary values on r = u,. It follows that the solution of the external problem of Dirichlet for r > a is U(· 8 ¢ - 1 (2 2 f(a,j3)dQ ,1,

where

,

) -

47Ta r -a)

If

n{a2+r2-2a1'cosy)t'

cosy = cos8cosa+sin8sinacos(¢-j3).

11.8 The analytic character of harmonic functions Let U(x,y,z) be harmonic in a domain D. With any point of D as origin, there exists a closed neighhourhood r ;::;; a contained in D. U is continuous in D, and so takes continuous values f(8, ¢) on r = a

11.8]

ELLIPTIC EQUATIONS IN SPACE

[219

in polar coordinates. By Poisson's integral, if (;l::,y,z) is a point with polar coordinates (r, e, ¢) where r < a,

1 2 2 U(x,y,z) = 4 17 a (a -r ) JJn

(a2!r~a~~~~~OSy)t'

(1)

cosy = cosecosa+sinesinacos(¢-j3).

where

From the generating function

(1-2hcosy+h 2)-:i

<Xl

=

L;hnpn(cosy) o

for the Legendre polynomials, it follows that

1-h2 <Xl ')h hOlt = ~ (2n+ 1) hn.P,,(cosy). -:.. cosy+· 0 n U(x,y,z) = ~JJ !(a,j3)i;(2n+1)r Pn(cosy)dQ. 417 n 0 an (1

Hence

(2)

(3)

Now IPn(cosy)1 :::; 1. It follows that the infinite series under the sign of integration is absolutely convergent when 0:::; r :::; b < a, and that the convergence is uniform. vVe may therefore integrate termby-term to obtain 1

U(x,y,z) = 4

~ (2n+ 1) an rnJJ !(a,j3) Pn(cosy) dQ. n

(4)

17 0

But Pn(cosy) is a polynomial of degree n in cosy. Hence

rn JJn!(a,j3)Pn(cos y )dQ is a homogeneous polynomial of degree n in x, y and z which satisfies Laplace's equation; it is a spherical harmonic. vVe have thus expressed U(x,y,z) as a series of polynomial terms which converges uniformly and absolutely in r :::; b. Thus U is analytic in any sphere contained in D; this proves the analytic character of U. We can deduce from (4) the formula for the expansion in terms of the standard spherical harmonics by using the addition-theorem

Pn(cosy) = Pn(cos e) Pn(cosa) + 2

(n-m)' ( ); P;:(cos e) P;:(cos a) cosm(¢ - 13). m=l n+m . <Xl

~

It follows that

U(x,y,z) = A o+ frn{AnPn(COse) + il(A~COSm¢ +B~sinm¢)p;:(coSe)}, 8'2

220]

ELLIPTIC EQUATIONS IN SPACE

[11.8

where A"

.

2n+1J1T =' - da J21T dflsinaPn(cosa) f(a,fJ), 21Ta?'

0

0

'>n+1 (n-m)'J1T J21T ); da 27Ta n+m, 0 ° dflsinaP';(cosa)cosmfl f(a,fl),

A~ = - - -n - (

B7: =

'>n+ 1 (n-m)IJ1T J21T n ( ); da dflsinaP~'(cosa)sinmflf(a,fl)· 7Ta n+m. 0 0

=----2

If f is independent of 13, this reduces to the expansion found in § 11.3. In a similar way we can use Poisson's formula for the solution of the exterior problem of Dirichlet with data on the sphere r = a. The solution is of the form

U(x,y,z) = A a a2n+1 { < X l } +:Z: n+1 AnPn(cos8)+ ~ (A~'cosm¢+B~sinm¢)~(cos8) r 1 r m=l OCJ

_0_

with the same constants.

The linear equation of elliptic type If the coefficients of the second deri,'atives in a linear equation of elliptic type are constants, we can find a non-singular linear transformation of the independent variables which turns the equation into 11.9

L(u) == V'2u +2aux +2bu y +2cuz +du

=

F(x,y,z),

(1)

where the coefficients a, b, c, d and F are assumed to be analytic on a bounded domain D whose frontier aD consists of a finite number of non-intersecting regular closed surfaces. We denote by (l,m,n) the direction cosines of the normal to aD drawn out of D. The operator L* adjoint to L is defined by L*(v) == V'2 V - 2(av)x- 2(bv)y- 2(cv)z+dv. vL(u) -uL*(v) =

Then

ax aY ez

-~-+-~-+-;:;-

ox

oy

oz'

where

x

=

VU X

-uv,.+2auv,

r = VU y -uv y +2buv,

Z =

vuz-uvz+2cu~·.

Green's theorem then gives JJJ D {vL(u) -

uL*(~,)} dxdydz =

J Jan (lX + m r + nZ) dS,

11.9]

[221

ELLIPTIC EQUA nONS IN SPACE

a result which is true if u and v have bounded continuous second derivatives in D and continuous first derivatives in D. In particular, if L(u) = F, L*(v) = 0, we obtain

JJJD vFdxdydz = JJuD {v :;-u :;+ 2(la+mb+nC)uv} dS. (2)

As Hadamard showed, the elementary solution of L*(v) = singularity at (x o, Yo, zo) is of the form

°with

vo(x, y, z) V=

R

'

where and V o is analytic and takes the value 1 at (x o,Yo, zo). We can substitute this elementary function for v in (2) provided that (x o' Yo, zo) is not a point of D. If (x o,Yo' zo) is a point of D, we have u(xo,Yo' zo) =

4~ JJUD {v : ; - u :1~ +2(la+mb+nC)Uv}dS-

:17JffD vFdxdydz.

(3)

This formula will not solve the problem of Dirichlet since it involves the boundary values of u and au/oN. If we could find a solution VI of L*(v) = with no singularities on D such that

°

(4)

vanished on aD, formula (3) would not involve au/oN, and so would solve the problem of Dirichlet. The function (4) would be the Green's function. Thus, if we could solve a particular case of the problem of Dirichlet for L*(v) = 0, (3) would give a solution of the problem of Dirichlet for L(u) = F. But we should still have to show it is unique. Using the argument of § 10.4, we can show that: If d :::; 0, the equation L(u) = F has at most one solution which is continuously differentiable on D, which has bounded continuous second derivatives on D and which takes given continuous values on aD.

11.10 The equation with constant coefficients. If the coefficients a, b, c, din V 2 U + 2aUx + 2bUy + 2cUz + dU = (x, y, z)

222]

ELLIPTIC EQUATIONS IN SPACE

are constants, the change of variable U =

U

[11.10

exp ( - ax - by - cz) gives

y2 U + AU = P(x,y,z),

(1)

A = d-a 2-b 2-c 2

where

°

is a constant. If Ais positive, \,2'11 + AU = is called the reduced wave equation; it arises when we try to find 'monochromatic' wave functions in three-dimensional space. Let D be a domain bounded by one or more non-intersecting regular closed surfaces. If the problem of Dirichletfor (1) has two solutions U 1 and U2' U = u 1 - U2 satisfies y2U+ AU = and vanishes on cD. But, by Green's transformation,

°

If

8'11 dS = oD U a.l\-

Iff {8

8

8}

D ax (uu x ) + oy (uu y ) + 8z (uu z ) dxdydz

= JJJD {u~+u;+u;+uV2u}dxdydz = JfJD {u~+U~+U;-AU2}dxdydz. Hence

JJJD{U~+U~+U;}dXdYdZ= AJJJD n 2dxdydz.

If A :::; 0, both sides of this equation are equal to zero. Hence u x ' u y, U z vanish on D, and so U is constant on D. Therefore U1 = U2. Thus, if the problem of Dirichlet has a solution when A :::; 0, it is unique. If A > 0, the solution may not be unique. For example, if A = k2 and Dis r < a, there is a solution

sinkr u=-r-' which vanishes on r = a if ka = mT. Thus there are values of k for which the problem of Dirichlet on r :::; a does not have a unique solution.

The mean value theorem If U 1~S a solution of y2 U + k 2u = which has continuous second derivatives in a domain contain1:ng (x-:r o)2+(Y_Yo)2+(z-zo)2:::; a 2, the mean value of 11 over (x - x o)2 + (y - YO)2 + (z - ZO)2 = R2, where R :::; a, i8 11.11

°

sinkR

kR u(xo,Yo' zo)'

11.11]

ELLIPTIC EQUATIONS IN SPACE

[223

This reduces to Gauss's mean value theorem when k = 0. "We may take (x o, Yo' zo) to be the origin, and use spherical polar coordinates (r, e, ¢) so that x = lr,y = mr,z = nr, where

m = sin esin ¢,

1 = sinecos¢,

n = cose.

o denotes the whole solid angle at 0; dO = sineded¢.

The mean value of u over the sphere ~, X 2+y2+ Z 2 = r2, where r :::; a is I(r) = ~JJ ndS = ~Jf u(lr, mr, nr) dO. 417r 1: 417 n Hence

dI 1 ff n (lux + muy + nuz ) dO = 4m2 1 Jf E oN ou dS, dr = 417

where %N is differentiation along the outward normal. By Green's theorem = 4:r 2fJJ Y2u(;,17,S)d;d17 d S

~~

= -

4~:2 f f f

u(;, 17, s) d;d1J d s,

where integration is over ~2 + 17 2+ S2 :::; r2 • Therefore

dd

I r

2

= - k 2JT s2dsff u(ls, ms, ns) dO 417r

n

0

= - k:fT s2I(s)ds. r

Therefore or Hence

0

di r2 dI = -k 2r2I dr dr d2I dI 2r2I = 0. r 2-+2r-+k dr 2 dr I

= r~(Acoskr+Bsinkr).

But I tends to u(O, 0, 0) as r -+ 0, and so A 1

=

= 0, and

sinkr -r;:u(O, 0, 0).

The corresponding result for y2,u - k2u =

°is

sinhkr I = ~u(O,O,O).

224]

ELLIPTIC EQUATIONS IN SPACE

[11.11

From this it follows that a solution of V'2U - k2u = 0 which satisfies the usual differentiability conditions on a bounded domain D cannot have a positive maximum or a negative minimum at any point of D. This implies the uniqueness theorem; for if 1(, "anished on oD, it would be identically zero on D.

11.12 The solution of V 2 u- k 2 u = 0 in polar coordinates Suppose that we wish to solve the equation in the domain r < a in polar coordinates, given that u = f(8, ¢) when l' = a. Sincef(8, ¢) can be expanded as a series OCJ

~Sn(8,¢),

o

where Sn(8, ¢) is a surface harmonic of order n, we try to find solutions of the form RSn (8, ¢) where R depends on r alone, The differential equation in polar coordinates is 82u 28u 1 -+ - - - k2u+-Yu = 0' or 2 r or r2

where Since Y Sn

= -

n(n

+ 1) Sn' the corresponding function R

2 dR dr~ 9

If we put R =

_ {k 2 n(n+ + r~ dR dr + r2

satisfies

i)} R = O.

S/.jr, this becomes 2

d S +~ dS -{k2+ (n+t)2}s dr 2 r dr r2

=

0,

which is Bessel's equation for the functions of purely imaginary argument. It has the independent solutionst

where

_ "<Xl (tz )2m 1"(z) - (tz) m~o m! r(v+m+ 1)'

Hence we have two solutions

.

I n +! (kr) S (8 "') \/(kr) n ' Y' ,

t

See 'Watson, A Treatise on tl~f. Theory oj Bessel Functions, 2nd edn (Cambridge, 1944), p. 77.

11.12]

ELLIPTIC EQUATIONS IN SPACE

[225

When n = 0, these solutions are constant multiples of sinh kr

cosh kr

---,;;;:- , ---,;;;:Now Iy(z)/zY is an integral function of the complex variable Z and has no real zeros. If z is small, In+t(z)/.jz and Ln-t(z)/.jz behave like multiples of zn and z-n-l respectively. As the solution is to be finite at the origin, it cannot contain the function L n-!. Hence if <Xl

f(8,¢) = 1:.8n (8,¢), o

where 811, is a surface harmonic of order n, u

=

i; In+t(kr) o

.j(ka) 811,(8, ¢). .j(kr) In+t(ka)

There is no restriction on a, since In+t(ka) never vanishes. If we wish to solve the external problem of Dirichlet with data on r = a, we have to use another solution of Bessel's equation since In+t(kr) and Ln-t(kr) tend to infinity exponently as r.-+oo. But Kn+t(kr) = (- 1 )n-lj7T{ In+t(kr) - Ln-t(kr)}

,. . , (..!!--) t e2kr

kr

as r.-+oo. The required solution of the external problem is then u =

i

o

Kn+t(kr) .j(ka) 811,(8, ¢). .j(kr) Kn+t(ka)

Again, there is no restriction on the value of a since Ky(z) has no positive zeros. Lastly, if R2 = (x- X O)2 + (y - YO)2 + (z - ZO)2, the function

cosh kR/R

is the elementary solution of Y'2u - k2u

=

0.

11.13 The solution of V2u+k2u = 0 in polar coordinates If <Xl f(8, ¢) = 1:. 811,(8, ¢), o

the solution of the problem of Dirichlet of the equation Y'2u+k2u = for r < a, given that u = f(8, ¢) on r = a is u =

i o

In+t(kr) .j(ka) 811,(8, ¢). .j(kr) In+t (ka)

°

226J

ELLIPTIC EQUATIONS IN SPACE

[11.13

This can be obtained by repeating the argument of § 11.2 or, more simply, by replacing k by ik. This is satisfactory, subject to convergence considerations, provided that ka is not a zero of J,,+!(z). The positive zeros of J,,(z) form an increasing sequence; and, if the least positive zero is j,., then < jv < jp+l' Since

°

J~(z) =

J

(:z) sin z,

the least positive zero of J!(z) is 17. Therefore, if ka < 17, ka is not a zero of In+*(z) for any positive integer n. The formal solution we have found thus- holds if the radius of the sphere is sufficiently small. But if, say, In+!(ka) = 0, the equation has a solution

J,,~ (kr) S (8 "') ~(kr)

" , 'f/

,

which vanishes on r = a; and the problem of Dirichlet for r < a does not have a unique solution. The particular solutions

do not depend on 8 and

ep, and are proportional to sinkr -r-'

respectively. If

coskr l'

R2 = (X-X O)2+ (Y-YO)2+ (Z-ZO)2,

coskR

the function

Jr

is the elementary solution of V'2U + k2u = 0. Sometimes it is more convenient to use the Hankel functions H;~~ ~

(z) = J,,+! (z) - (- 1)"iJ_,,_~ (z)

H~~t (z) = J,,+! (z)

+ (-1 )"iJ_,,-t (z),

n being a positive integer or zero. These functions are of the form

11.13)

ELLIPTIC EQUATIONS IN SPACE

H~) (kr)

The solutions

H~2) (kr)

~(kr)' eikr

are multiples of

[227

~(kr) e- ikr

r'

r

11.14 Helmholtz's formula In dealing with problems of wave propagation which vary simpleharmonically with the time, it is often convenient to use a complex wave function U = uexp (-ikct), (1) where u does not depend on t, and to take real parts at the end. If

u is a complex solution of the reduced wave equation V'~L + Pu =

o.

(2)

The only wave functions in which u depends only on the distance R from a fixed point are exp (ikR - ikct) R

exp ( - ikR - ikct) . R '

the former represents expanding spherical waves, the latter contracting spherical waves. If we wish uexp (-ikct), where u is a solution of (2), to represent expanding waves, we want u to behave like exp (ikr)/r, when r is large. This is the origin of Sommerfeld's conditions, which appear later in this section. We have seen that the reduced wave equation (2) has an elementary solution cos kR/R, where R is the distance from (X o,Yo' zo) to (x,y,z). We can, if we wish, use the complex elementary solution exp (ikR)/R. We first state Helmholtz's formula, which is the analogue for the reduced wave equation of Green's equivalent layer formula in potential theory. Let D be the domain bounded by a regular closed surface B. Let u be a solution of V'~+k~ = 0 which is continl£onsly differentiable in J5, and has bOl£nded continl£ol£s second derivatives in D. Then the vall£e of

~ff {exp (ikR) 01£ _ 417 s R oN

1£

~ exp (ikR)} dB oN R '

228J

[11.14

ELLIPTIC EQUATIONS IN SPACE

R2 =

where

(X-X O)2+(Y-YO)2+(Z-ZO)2

is u(x o, Yo' :::0) or zero according as (x o' Yo' zo) is inside or outside S.

This is a straightforward consequence of Green's theorem. 'Vc note that, if n is real, the value of

~ { { {COS kR ~UT -11 r,C cos leR) dS T

417 ~ ~ s

R

cIo.

0]\

R

J

is u(xo, Yo, zo) or zero according as (xo' Yo, zo) is inside or outside S. But the value of

~

417

Jfs {Sin

kR au _ '11 ~ sin leR} dS R aN Oi.V R .

is zero everywhere, as we should expect; for sin kR/R is a solution of the reduced wave equation without any singularity. Let Li be the unbounded domain exterior to a regular closed surface S Let '11 be a solution of \12'11+ k 2u = 0 which is continuously differentiable in 3. and has bounded continuous second derivatives in Li. Then if n is the unit normal vector to S drawn into Li, the value of

~Jf (eXP(ikR) 0'11 -u!.... eXp(ikR))dS 417 s R an on R is - '11(:<:0' Yo' zo) or zero according as (x o, Yo' zo) lies outside or inside S, provided that '11 satisfies Sommerfeld's conditions at infinity.

Let L be the sphere R = a, where a is so large that L contains S. Then the value of

Jfs (

exp (ileR) au _

R

~ exp (ikR)) dS

oN u aN

R

J(

oN u aN

+::

eXP(ikR) 0'11 _

R

~ eXP(ikR)) dL

R

is 41TU(X O' Yo' zo) or zero according as (xo, Yo' zo) lies outside or inside S. On S, N is the unit normal vector drawn out of Li, and so is - n. Hence to prove the result, we have to show that the integral over L: tends to zero as a -+ co. Xow the double integral over 2: is

JJ [eXP(ikR)~U_ueXP(ikR)(ik_~)]

a2

Q

R

oR

R

R

R~a

dD,

11.14]

ELLIPTIC EQUATIONS IN SPACE

[229

where dO. is the element of solid angle. This can be written as exp(ika)

IIn

[R

IIn

:Now

G~-ikU) +U]R=a dO.. [u]R=a do.

tends to zero as a-+oo if u-+O as R-+oo uniformly with respect to the polar angles e, 9; in particular, it tends to zero if Ru is uniformly bounded as R-+oo, a condition which Sommerfeld called the finiteness condition. t The other integral tends to zero if

tends to zero as R -+00, uniformly 'with respect to the polar angles, a condition known as Sommerfeld's radiation condition. If these two conditions are satisfied, the result follows. The two conditions lead to a result similar to Liouville's theorem in the theory of functions of a complex variable. Let u be a solution of V'2u + k2u = 0 which has bounded continuous second derivatives everywhere and which satisfies Sommerfeld's conditions at infinity. Then u is identically zero. For let S be any regular closed surface. Then, if (x o' Yo' zo) is any point inside s,

If ( s

exp (ikR) ou _ ~ exp (ikR)) dS R oN uoN R '

where %N is differentiation along the normal out of the domain bounded by S, is equal to 47TU(X O' Yo' zo) by Helmholtz's result but is equal to zero by Sommerfel9-'s conditions. Hence u is zero everywhere. Sommerfeld's radiation condition implies that u exp (- ikct), or, rather, its real part, is the velocity potential of an expanding harmonic wave motion. The analogue of Liouville's theorem implies that such a wave motion must be due to sources at which the differentiability conditions do not hold.

11.15 The exterior problem of Dirichlet Let D be the unbounded domain outside a regular closed surface S. The exterior problem of Dirichlet for V'2U + cu = 0 is to find a solution which takes given values on S and which has continuous first derivat

Sommerfeld, Jahresberichte der D.ll!. V. 21 (1912),309-353. See also Magnus, ibid., 52 (1943), 177-188 and Rellich, ibid., 53 (1943), 57-65.

230J

[11.15

ELLIPTIC EQUATIONS IN SPACE

tives on fj, bounded continuous second derivatives on D. The boundary condition does not determine 11. uniquely; conditions at infinity are needed. To show that a solution, if it exists, is unique, we have to show that, if 11 is zero everywhere on S, 11. vanishes everywhere on D. \Ve write the equation in the form Li2U + k 2v. = 0, but do not require k to be real, so that the cases c > 0 and c < 0 are both covered. F. V. Atkinsont proved that, when k is real, Sommerfeld's two conditions at infinity ensure uniqueness. But he also showed that the two conditions can be replaced by one condition which suffices to ensure uniqueness even when k is complex. Atkinson proves first two lemmas, using the notation of the first paragraph of this section. Lemma 1. Let 11. be such that (i) V'2U + k 2u = 0 outside S,

(ii) (iii)

11.

is continuously differentiable on fj, has bounded continuous second derivatives on D,

11.

(iv) rexp (ikr) {(ik-n

11.-

~~)

tends to zero as r -+ 00 uniformly with respect to the polar angles Then 11. can be expressed in the form 11.

= exp (ikr)

e and ¢.

2: anr- n, 1

where the coefficients an are independent of r and are continuous functions of direction. ~lJforeover, there exists a constant a such that the series is absolutely convergent when r ~ a and uniformly convergent there with respect to e and ¢. This is not an expansion in terms of the Hankel functions of order n + t and surface harmonics of order n. A typical term exp (ikr) an r- n does not satisfy the reduced wave equation. If Po(xo, Yo' zo) lies outside S, it follows by the argument of § 11.14 that, if 11. satisfies condition (iv) of the lemma

fJ (

':' ) = ~ ( . Yo' ~o 4 IT.

11. X Ol

S

au)

~ exp (ikE) _ exp (ikR) dS v. un
where R is the distance from Po to the integration point P(x, y, z) on Sand n is the unit normal vector drawn into D. If OPo = TO' OP = r, then

t

Phil. Mag. 40 (1949), 645-651.

11.15]

where

ELLIPTIC EQUATIONS IN SPACE

[231

Vr is the angle PoOP. This gives R

= ~ ~ (1 -

2rs cos Vr + r2~?),

where S = 1lro' If we suppose, more generally, that S is a complex variable, R has a simple pole at S = and branch points at

°

s = exp ( ± iVr)lr. We then have p (ikro) _ (1 2 Y exp (ikR)jeX ,/r 2Y2)-! ----.:'-'-~ - r'<J cos 't' + r '<J R ro

x exp

[i; {(

l

1 - 2rS cos Vr + r2S2)t - 1}

If we take the branch of the square root which is equal to unity when S = 0, the expression on the right of this equation is an analytic function of S, regular when lsi < 11r. But S lies everywhere at a finite distance from 0, so that r :::; ta say, on S. The analytic function is then certainly regular when lsi < 21a, and so can be expanded as a power series, which is absolutely convergent when lsi:::; 11a, the convergence being uniform with respect to rand Vr. It follows that ex p (ikro))-lIf exp (ikR) AU dS ( ro s R On

can be expanded as a power series

where the coefficients depend only on the direction of O~ and depend on it continuously. Hence, when r o ~ a, exp (ikR) oU dS = ('k); An ,;, exp ~ r o.I..J +1' R Ifs un 0 ~ A similar argument applies to u~ exp (ikr) dS,

Ifs

on

r

and the result of the lemma follows. Note that it was not assumed that k > 0; the result holds for all real or complex values of k,

[11.15

ELLIPTIC EQUATIONS IN SPACE

232J

Lemma 2. Let u be such that (i) V'2u +k2u = outside S, (ii) u is continuously differentiable on 15, (iii) u has bounded continuous second derivatives on D. If im k ~ 0, the conditions

°

(A)

rexp (ikr) {(ik -~) u- ~~} -+ 0,

(B)

r 3 exp (-ikr)

{(ik-~) u- ~~} = 0(1),

as l' -+ 00, are equivalent. They are also equivalent to (0)

u-+O,

(D)

ru

r(iku-~~)-+oasr-+oo,

= 0(1),

andto

1'2 (ikU- ~~) = 0(1) as 1'-+00.

Write k = p+iq, where p and q are real and q (B) holds. Then

/1 ex p (i7cr) o

~ 0.

Suppose that

{(ik-~) u- ~~} I

= Ir3 exp (- ikr) {(ik -~) u- ~~} exp ;:ikr) / = 0 (ex

p

(r~ 2qr))

so that (A) holds. If (A) holds, we can apply the result of Lemma, 1, to obtain p r 3 exp (- ikr) {(ik -~) u- ~~} = _1'2 :1' (u jex ;ikr))

= _ 1'2 ~ or

:r 00

an = ~ (n - 1) an r n- 1 2' r n- 2 '

the series being absolutely and uniformly convergent when l' ~ a. Hence (B) holds. (0) implies (A) since lexp (ikr)j = exp (-qr) :::; 1. If (A) holds, then 00

u = exp (ikr) 2; 1

when

l'

~

a -.E n

r

'

a, and so u-+o as 1'-+00. Also l'

'k OU) ( t u-~ ur

nan = exp(t'k 1')2;n 00

1

r '

11.15]

ELLIPTIC EQUATIONS IN SPACE

[233

which tends to zero as r-HXJ. Thus (A) implies (0). Similarly we can show that (D) implies (A) and (A) implies (D). And thus (0) and (D) are also equivalent. Thus when im k ~ 0, Sommerfeld's two conditions can be replaced by one condition (A) . The only function which satisfies the conditions of Lemma 1 and vanishes everywhere on S is identically zero. From this follows the uniqueness theorem for the external problem of Dirichlet. We assumed that k = p + iq, where q ~ O. There are three cases to consider, viz. (i) p 4= 0, q > 0, (ii) p = 0, q > 0, (iii) P 4= 0, q = 0. (i) Suppose that p 4= 0, q > 0, and that there is a solution which vanishes on S. Let ~ be the sphere r = b, where b > a. Then S lies inside~. Let Ll be the domain bounded externally by~, internally by S. If u is the complex conjugate of u,

fff~ (u'VZU-u'V~)dxdydz = ff~ (u:;-U:;)d~, since u and u vanish on S. The expression on the left is (k

2

-k 2 ) fff~ uudxdydz =

4ipq

fft. 1u21

dxdydz.

00

By Lemma 1,

u = exp (ipr - qr)

2: anr- n , 1 00

u = exp ( - ipr - qr) 2: ii n 1,-n, 1

when r

~ a,

and so u au _ or

u au = or

0 (ex p ( - 2qr )) r2

as r-+oo. Therefore the integral over 1: tends to zero as b-+oo, and so, since pq 4= 0,

Hence u vanishes everywhere on D. (ii) Suppose, next, that p = 0, q > so that k 2 = - q2. Let u be a solution which satisfies the differentiability conditions of Lemma 1 and vanishes on S. Then

°

234]

ELLIPTIC EQUATIONS IN SPACE

[11.15

because u vanishes on 8. Since

uOU

or

as r -+ 00, the integral over

~

=0

(exp (r- 2qr )) 2

tends to zero as b -+ 00. Therefore

and so u is again zero everywhere on 15. (iii) Lastly suppose that p =!= 0, q = O. If u is a solution which vanishes on 8 and satisfies the differentia bility conditions,

JJr~ (u'V2U - u'V 2u) dxdydz = JJ2: (u :~ -- U :;T) d~. The triple integral vanishes. By Lemma 1,

If we substitute this in

b2JJn (u ou - Uau) or or

dQ

=0

r=b

and make b-+oo, we find that la 1 12 = O. Proceeding in this way, we find that every coefficient an is identically zero, and u vanishes everywhere on D. Exercises 1. (i) Prove that, if 0 < h < 1,0 < k < 1, l dfl __ 1_ l+,.j(hk) 10 _1,!(1-2hfl+h2 ),/(1-2kfl+k2 ) - ,/(hk) gl-.j(hkr

J

(ii) The Legendre pOl~'1lOmials have the generating function

Deduce from (i) that

ELLIPTIC EQUATIONS IN SPACE

2. u is harmonic in a domain containing r a, u = f{ft,) , where/" = cosO. Prove that

~ a

[235

in polar coordinates. On

"=

where

_-,--_d_u _ o z+ix cos u+iysin 1/, 2"

J

3. Prove that

27T r

where z > 0. Deduce that, if r < 1,

27T

2" _ _--:-_du_-,--,-_ o 1 - z - ix cos u - iy sin u

J

where /" = cosO. Hence show that

Pn(cosO)

1 f2" = 27T 0 (cosO+isinOcost)ndt

Deduce that 4. If

Rfi = (X-XO)2+(y_YO)2+(Z-ZO)2,

Rr = (X-XO)2+(y_YO)2+(Z+::0)2, 1 1 Ro R1

where Zo > 0, prove that

is the Green's function for the Laplace's equation in z > 0. Show that 1 1 ---R = Ro 1

where

00

2: 1

rn

--fJ. {Pn(coseX) -Pn(cosj1)}, r +

= cos 0 cos 00 + sin 0 sin 00 cos (p - ¢o), cosj1 = -cosO cos Oo+sinOsin 0ocos (p -Po), (r, 0, p) being the polar coordinates of (xo, Yo, zo) and cos eX

(ro, 0 0 , Po), (x, y, z). Show that the series is absolutely convergent in r ~ a, for any a > ro, and that the convergence is uniform with respect to 0 and p. If u is harmonic in z > 0, prove that

Zo

u(xo'yo,zo) provided that as

r~oo

= 27T

ffoo -00

u(x,y,O) dxdy

{(X-XO)2+(Y_YO)2+ zfi}!

au 2u -+-~O Or r

uniformly with respect to 0 and

p.

236)

ELLIPTIC EQUATIONS IN SPACE

5. Shm\' that, in spherical polar coordinates, Laplacc's equation is 2 ~ ~ 2 CV) + _1_ ~ (sin 17 8F) +_1_ 8 F = o. 2 2 2 r 2 or or r sin 17 8B 017 r sin 17 8¢2

(r

By the method of separation of variables, obtain the solution (Arn+Br- n) cos (m¢+c) 0';;(J.l),

where J.l = cos 17: and 0';; satisfies 2 (1_/l 2)d 0 _2J.ld0 +[n(n+1)-~]0 = dJ.l2 dJ.l 1- J.l2

o.

6. Show that, in cylindrical coordinates, Laplace's equation is 82 V 18V 1 82V 82V -+--+--+-= 0. 8p2 P 8p p2 8¢2 8z 2 By the method of separation of variables, show that a solution is (Ael:z+Be- kz ) (CJm(kp) + DYm(kp)) cos(m¢+c),

where A, B, C, D and c are constants. If this solution is one-valued and has no singularity, prove that m is an integer and that D is zero. 7. Let u(x,y,z) be a non-negative function, harmonic in a bounded domain D. Let 2 (X-X O)2+(Y_YO)2+(Z-ZO)2"; a

be a closed sphere contained in D. If (x, y, z) is a point of the sphere at a distance r < a from its centre, prove that a(a-r) a(a+r) u(xo,yo,zo) .,; u(x,y,z) .,; --)-2 u(xo,Yo,zo)· (a+r) (a-r

---2

Deduce that a non-negative function harmonic in every bounded domaill is a constant. 8. Prove that Harnack's first and second theorems on convergence. proved in §8.14 and §8.16 for plane harmonic functions, also hold fa)" sequences of harmonic functions in space of three dimensions.

Rfi =

9. If

(X-X O)2+(Y_YO)2+(Z-ZO)2,

eikRo

prove that

eikr [

-- =-

Ro

r

00

l+"a

t

rn] "r n , ...Q.

where an is a function of cosa = cos 17 cos 170 + sin 17 sin 00 cos (¢-90) alone, (r o, 170 : 90) and (r, 17, ¢) being the polar coordinates of (x o,Yo' zo) and lanl .,; 1 and that the series is absolutely convergent when r ;:, a for any a > ro, and that the convergence is uniform. (x, y, z). Prove that

ELLIPTIC EQUATIONS IN SPACE

[237

If k > 0, Zo > 0, show that

Ro

R1

Ri = (X-X O)2+(Y-YO)2+(Z+ZO)2,

where

°

is the Green's function of the equation 'V 2u + k 2u = for the half-space : > 0. Hence show that if u is continuously differentiable on Z ~ 0, has bounded continuous second derivatives on z > and satisfies 'V 2u + k2u = 0, then ikR Z e u(xo,Yo,zo) = 2~ -co R3 (l-ikR)u(x,y,O)dxdy,

°

ffco

where

R2

provided that as

r~oo,

=

(X-XO)2+(Y_YO)2+Z~,

ou +~ (2-ikr) -+ or r

uniformly with respect to () and

p.

°

12

THE EQUATION OF HEAT

12.1

The equation of conduction of heat

\Yhen heat flows along an insulated uniform straight rod of thermal conductivity K, density p and specific heat c, the temperature u at time t at a distance x from a fixed point of the rod satisfies the equation

a2u cx 2

pc 8u

Kat·

Since K: p: c are constants, this can be written in the form 82u

ox2

8u 8t

by a change of time-scale. This is the simplest linear equation of parabolic type with two independent yariables. It is called the equation of heat or the equation of diffusion. It has one family of characteristics: namel:-- the lines t = constant in the xt-plane. The simplest problem is that of the infinite rod with a given initial temperature distribution u(x, 0) = f(x). On physical grounds, it is obyious that the temperature at any subsequent instant is uniquely determined. The problem is to find conditions satisfied by f(x) so that this is true: and to find an explicit formula for u. 12.2

A formal solution of the equation of heat 82u

If, in

we put u we have

ox2

= XT

OU

= at'

(1 )

where X and T are functions of x and t respectively. X" ¥-',

where dashes and dots denote differentiation with respect to :r anel t. Hence

12.2]

EQUATION OF HEAT

[239

where a is the separation constant. Thus we have a solution 2

u

= exp ( - a 2(t - to)) cos a(x - x o),

where X o and to are constants. In the physical problems of heat conduction, u cannot increase indefinitely with t, so we assume that a is real. A more general solution, valid when t > to, is u

= f:<Xl exp(-a 2(t-to))cosa(x-xo)da

f

.J1T

= .J(t-to) exp \ -

(X-X O)2}

4(t-to) .

If x =1= X O' this solution tends to zero as t -+ to + O. Other formal solutions can be obtained by integration. For example, (2)

is a solution valid when t > O. If we put; = x + 21J.Jt we obtain U

= T1 f<Xl f(x + 21J.Jt) exp (_1J2) d1J, \/1T

-<Xl

(3)

when t > O. The limit of this as t -+ + 0 is f(x). Hence (2) is the formal solution of the initial value problem for the infinite rod. Iff(x) is zero when x < a and when x > b where a < b, the solution (2) becomes 1 U = 2.J(1Tt) a f(;) exp (-!(x - ;)2jt) df

fb

If, in addition, f(x) is positive when a < x < b, u(x, t) is positive when t > 0 for all values of x. The effect of an initial non-zero temperature distribution on a finite interval is immediately felt everywhere. This result is quite different from that for the equation of wave motions where an initial disturbance restricted to a finite interval is propagated with a finite velocity. It is convenient to write k(x,t)

=

2.J~1Tt)exP(-ix2jt),

so that the formal solution (2) becomes (4)

240)

[12.2

EQUATION OF HEAT

when t > O. This result can be justified if we assume thatf(x) possesse~ a continuous second cleri,-ative and satisfies suitable conditions at infinity to ensure the uniform com'ergence of the integrals obtained from (3) h~' differentiation under the sign of integration. But tlw result holds under very much less restricti,-e conditions. The solution (4) was obtained by integrating a multiple of lc(x-~,t-7)

along a path in the

~7-plane. Another

u(x,l)

formal solution is

= J~ ¢(7)k(:r,I-7)d7,

(5)

where t > 0, the upper limit being t since k(x, t - 7) is complex when 7 > t. This solution is an even function of x. Its value when x = 0 i" u(O,i)

1 Jt

d7

= 2,..,f7T 0 ¢(7) '1/(t- 7)'

This is Abel's integral equation for 9. If u(O, t) is continuous and vanishes 'when t = 0, the solution of the integral equation is ¢(t)

2 dJt

= -, -d .J7T t

0

d7 u(O, 7) -'(--)' ,\'

t-

7

Thus (5) is a formal solution of the equation of heat in terms of the ,-alues taken by the solution when x = O. Yet another formal solution can be obtained by differentiating the expression on the right of (5) with respect to x. It is a multiple of u(X,t) = It ¢(7)....!.-k(x,t-7)d7,

t- 7

• 0

(6)

t being positive. The expression (6) is an odd function of x. If ,,'e make the substitution t- 7 = ix 2 jv. when x> O,f, > 0, (6) gives u(x, t)

=

1 -1-

J

ro

'\ 7T ix'it

2

9 ( t -X- ) 4u

u-t e- u d7L

The limit of this as x-+ + 0 is ¢(t) when t > 0; but the limit as x....,.. is - ¢(t). since 71(X, I) is an odd function of x.

12.3

(I

Use of integral transforms

Problems of heat conduction in an infinite or semi-infinite rod can often be soh-ed by the use of integral transforms. As it is difficult to give this method a rigorous form, ,w content ourselves with illustrative examples.

12.3]

[241

EQUATION OF HEAT

To solve the initial value problem u(x, 0) = f(x) for an infinite rod, we may use the complex Fourier transform 1 foo . U(ex, t) = ..)(21T) _oou(x, t) euxdx,

whose inverse is u(x, t)

1

= ..) (21T)

foo

.

-00

U (ex, t) e-,udex.

Assuming that differentiation under the sign of integration is valid, we have 8U = _1_foo 8u(x,t) eia.xdx = _1_foo 82u(x,t) eia.xdx 8t ..)(21T) -00 8t ..)(21T) -00 &2

= -~foo u(x t)eia.xdx = -ex 2 U(a t) ..)(21T)

-00

'

,

,

provided that u(x, t) behaves at x = ± 00 so that the integrated terms vanish. We then have U(ex, t) = U(ex, 0) exp ( - ex 2t) =

..J(~1T) exp ( -

ex 2t) f:oof(;) expiex;d;.

Using the formula for inverting the Fourier transform, u(x, t) =

2~ f:oo exp ( -

ex 2t - iax) f:oo f(;) exp (iex;) d;dex

=

2~f:oof(;) f:oo exp(-ex2t-iex(x-;))dexd;

=

2..)~1Tt) f:oo f(;) exp ( -!(x -

;)2jt) d;,

the result we obtained in § 12.2. To solve the problem for the semi-infinite rod x > 0 with data u(x,O) = f(x), u(O, t) = ¢(t) when t > 0, it is more convenient to use the Fourier sine transform

~(ex, t) whose inverse is

=

u(x, t) =

J(~) f:

J(2) foo

IT

0

u(x, t) sinaxdx, ~(ex, t)

sin axdex.

Then

8~(ex,t)_J(~)fOOOu(X,t).
242)

[12.3

EQU A nON OF HEAT

Int.egrating h~- parts and assuming that u(x, t) behans suitably at x = + CfJ, "'C haye

This gin:U.,(x,t)

= C,(x,0)exp(-a 2i)+ J(~) a J:ex p (-a 2(t-7))¢(7)dT,

where [:(x, 0) is the Fourier sine transform -F,.(a) of u(x, 0) (which is equal tof(x).) Using the formula for inverting the Fourier sine transform, we have u(x,t)

= ~ I~'" sin ax exp (-a 2i)Jro f(~)sina~d~ 7T.

(I

(I

+-2Jro asinax 7T

=

Jt exp(-a (t-7))¢(7)d7 2

(I

(I

1Jro f(t) Jro exp (- a t) {cosa(x- cosa(x = '2 ;a -- Jt ¢(7) Jt exp(-a (t-7))cosaxdadT ox ~)-

2

0

(I

1/

+~)}dad~

2

0

7T

•

=

0

( Cf'f(~){k(x-rt)-k(x+t,t)}dt-'2:ox Ji0 ¢(7)lc(x,t-7)dT (I

To

J

.f(~){k(x-t,t)-k(x+f"i)}d~+

(I

Ji ¢(7)-'-k(x,t-7)d7. X 0

t- 7

The Laplace transform of an integrable function ¢(t) is <1>(s) =

(ro ¢(t) e-stdt. a' (I

If 1¢(t)1 < Ee rd , <1>(s) is an analytic function of the complex variabJc. s, regular in re 8 > a. The inverse of this Laplace transform is ¢(t) =

~ _7T1

f

C7roi

c- ",i


where c > a. The theor.,· ofthe Laplace transform lacks the symmetry of that. ofthe Fourier transform. V~'e illustrate the theory by considering the problem of heat. conduction in a semi-infinite rod :1: > 0 under the conditions 11(:t',

U)

= () (.l· >

0),

1/(0, t)

=

¢(t) (t > 0).

12.3)

EQUATION OF HEAT

[243

If the Laplace transform of the solution u(x, t) is

= fo<Xl u(x, t) e-stdt,

U(x, s)

we have

= [u(x, t) e-·~t]~ +s

f:

u(x,t)estdt.

82 U

~-sU = 0

Hence

uX~

provided that the real part of s is so large that u(x, t) e- st tends to zero as t -+ + co. Write s = 0-2 where the branch of 0- is taken which is positive when s > 0. Then Since U (0, s) is the Laplace transform of u(O, t), which is given to be ¢(t), we have A +B = <1>(s). We need another condition in order to determine A and B. If we require u(x, t) to be such that U(x, s) is a bounded function of x, A is zero, and U(x, s)

= <1>(s) e-
which expresses U(x, s) as the product of the Laplace transforms of ¢(t),

x

t

k(x, t).

That e-(s) and 'Y(s) are the Laplace transforms of ¢(t) and Jjf(t), then <1>(s) 'Y(s) is the Laplace transform of

f:

¢(u) Jjf(t -1t) duo

Hence the sol ution of the heat conduction problem is u(x, t)

=

f

t

o

x

¢(u) _.- k(x, t-u)du. t-u

244]

[12.3

EQU A TION OF HEAT

It is impossible to give in a short space an adequate account of integral transform methods: the reader should consult the books cited below.t

12.4

Use of Cauchy-Kowalewsky theorem

The equation

82u

OU

8x 2

= of

(1)

has, by the Cauchy-Kowalewsky theorem, a unique analytic solution, regular in a neighbourhood of (x o, to), satisfying the conditions

= ¢(t), ux(xo, t) = lj;'(t),

u(xo, t)

(2)

provided that ¢(t) and lj;'(t) are regular in a neighbourhood of to' If these conditions are satisfied at every point (x o, to) of a finite inten'al y of the initial line x = x o' the problem has a unique solution regular near y. Bya change of origin, we may take X o and to to be zero. By a repeated differentiation of the differential equation, we can calculate all the derivatives of U with respect to x, and obtain the Taylor series u(x,t)

x

x3

x2

= ¢(f)+lj;'(t)1i +¢l(t)21 +Jjfl (t)3!+'" X 2n +l

x2n

+¢n(t)(2n)!+lj;'n(t)(2n+1)!+'"

where ¢" and

~'Jn

are the nth derivatives of ¢ and lj;'. ro

Let

(3)

¢(t)

= 12 an tn, o

ro

lj;'(t)

= 12 bn tn, 0

the series having radii of convergence R l and E 2 , say. Let E < Min (Ell E 2 )·

Since ¢(t) and ~'J(t) are analytic functions of the complex variable t regular in It I ~ E, we have 111 111. lanl

~ R~'

Ibn I ~ R~'

where 1111 and 1112 are the maxima of the moduli of /¢(t)/ respecti,-elyon It I = E. Therefore 111 111 lanl ~ En' Ibnl ~­ '" Rn'

t

E. C. Titchmarsh, Introduction to tlw Theory oj Fourier Integrals 2nd edn (Oxford, 1948), pp. 281-8. D ..Y. 'Widder, The Laplace T1'arl~form (Princeton, 1941). H. S. Carslaw and J. C. Jaeger, Ope1'Ot10nai .Methods 111 AppZ.ied :!JIathematics, ~nd edn (Oxford, 1948). C. J. Tranter, Integral Tmnsjorms in 111athematiwl Physics (London, 1971).

12.4]

EQUATION OF HEAT

[245

where it! = Max (Ml>M2 ). It follows that, when It I < R, the function (t) = iYJR R-t

is a majorant for ¢(t) and 1f(t), and so MRn! (R-t)n+l

is a majorant for ¢n(t) and for 1fn(t). If we substitute in (3) the Taylor series for ¢(t) and 1f(t) and their derivatives, we obtain a formal double series S. The corresponding double series S' obtained by replacing ¢(t) and 1f(t) by (t) is a majorant of S. Typical terms in S' are x

2n

n!

(t)k

111 (2n)! Rn n+ICk R and

The double series S' is therefore absolutely convergent for all values of Ixl and for all values of It I less than R. Hence the double series S is absolutely convergent for all values of Ixl and for all It I < R, and is uniformly convergent on any bounded closed subset. We can therefore differentiate S, and hence also the series (3), term-by-term. It follows that (3) satisfies the equation of heat under the given conditions when It I < R. This verifies the result of the CauchyKowalewsky theorem when u(O, t) = ¢(t),

ux(O, t) = 1f(t),

where ¢(t) and 1f(t) are analytic functions of t, regular when It I < R. This has an interesting consequence. When t = 0, u(x, t) is equal to F(x), where x x2 x3 x4 x5 F(x) = ao + bo11 + 1! a 1 2! + 1! b1 3! + 2! a 2 4! + 2! b2 5! +" .. The absolute value of the coefficient of x 2n is

I

nl n! ll-f (2n)l an ~ (2n)! Rn

I and of X 2n +1is I

n! (2n+ 1)! bn

I ~ (2n+n! 1)! Rn" M

246]

EQUATION OF HEAT

[12.4

Hence F(:r), regarded as a function of a complex variable x, is an integral function. Evidently "" 11!n1 x 2n F(x) <{ 111(1 +x) I: -(2 )1 R----"j· o . n. "n:

A

If

_ n!n! n -

we have

A n +1 An

--=

1 (2n)! (KR)U

(n + 1)2 1 1 -<-(2u+l)(2n+2)KR 4KR"

If we choose K so that 4KR > 1, {An} is a decreasing sequence and so An < 1. Therefore "" (Kx 2 )n F(x) <{ 1If(1 +x) ~ - - 1 o n.

In particular, when

since

= 111(1 +x) exp (Kx 2 ).

Ixl is large, IF(x)1 < M exp (2Kx 2 ), 11 +::rl ~ 1 +Ixl < exp (Kx 2 ).

Lastly, if we rearrange the double series S as a power-series in t, we obtain

where Fn(x) is the nth derivative of F(x). This is a solution of the equation of heat under the initial condition u(x, 0) = F(x).

12.5

An example due to Tikhonov

Tikhonov gave an example which showed that the initial value problem for the equation of heat 02U

ox

2

8u =

ot

does not have a unique solution unless v satisfies a condition lu(x, t)1

Under suitable conditions, . u(x, t)

< 1If exp (ax 2 ). ""

x 2k

= ~ ¢k(t) (27:)!'

(1) \ .

where 9/;(t) is the lcth derintive of ¢(t). is a solution. In Tikhonov's example," ¢(t) = exp (-1/t 2 ).

12.5]

[247

EQUATION OF HEAT

f

By Cauchy's Theorem,

k! ¢(z) ¢k(t) = -2' 7T~ r (Z - t)k' 1 dz, T

where r is a simple closed contour in the z-plane; the point t is inside r, the origin is outside. Take r to be the circle z = t + it eiO • Then on r 1

4

re-~Z2 S1t 2

and so l¢k(t)1

~ ;~ exp ( -

S:t 2)

f:" (~r de 2;:! =

exp ( - S:t 2)'

Hence, if 1£ is given by (1), we have

( 4)

00 2k k ! X 2k 11£(x,t)1 ~ exp - S1t2 ~ (2k)t 7' when t > O. Therefore

11£(x, t)1

~ exp ( -

S:t 2)

~

:! :~:k

= exp

(~2 - S:t 2)'

Therefore this solution tends to zero as t -J>- + 0 uniformly on any finite interval of the x-axis. If we have found a solution of the initial value problem, another solution will be obtained if we add a multiple of Tikhonov's solution. But the resulting solution does not satisfy an inequality 11£(x, t)1 ~ .M exp (ax 2) in t ~ O.

12.6 The case of continuous initial data In § 12.1, we obtained the formal solution

foo

1 l£(x,t) = 2.j(7Tt) _oof(f,) exp (_t(X-f,)2jt)

df"

(1)

which we expect to satisfy the initial condition l£(x, 0) = f(x) for all x. We now consider the simplest case when f(x) is continuous and satisfies the inequality (2) If(x) 1 ~ M eax' for all x, whereais a non-negative constant, and showthatl£(x, t) tends to f(x o) as (x, t) tends to (xo, + 0) in any manner. If (x, t) lies in the rectangle - a ~ x ~ a, 0 < jJ ~ t ~ y, we have If(f,) exp (-t(x-f,)2jt)1 •

~

M exp {af,2- 4~ (X-;)2}

~ 11'1 exp {- (4~ -a) f,2+ 2~ alf,I}.

248]

[12.6

EQUATION OF HEAT

Hence, if i' < 1j4a, the integral (1) converges uniformly on the rectangle, and so u(x, t) is continuous. Since ct may be as large as "\\'e please, u(x, t) is continuous on the strip 0 < t < 1j4a. A similar argument showi' that we may calculate the derivatives of u(x, t) by differentiation under the sign of integration, and that the derivati-ns of all orders are continuous in the same strip. Since

satisfies the equation of heat, (1) is a solution in the strip 0 < t < 1j4a. And if (2) holds for every non-negative value of a, (1) is a solution in t> O. We next show that u(x,t) tends tof(x) as t-?- +0, uniformly with respect to x on any closed interval - ct ~ X ~ ct. 'Ve have 1 JOO u(x,t)-f(x) = 2~('1Tt) _oo{f(f,)-f(x)}ex p (-i(x- fY jt)df,

(3)

Since f(x) is uniformly continuous, corresponding to any positive value of e, there exists a positive number 0 such that, if x and x + 11 belong to [ - ct, ct], then

If(x+1/)-f(;/:)1 < e, ",henever 11/1 <

o. Hence

12~~7Tt) J~8 {f(x +1/) - f(x)} exp ( < for all positive valnes of t. Next

2

i1/ jt) d1/

I

2~~7Tt) J~8 exp (-i1/ 2jt) d1/ < e

12.6]

[249

EQUATION OF HEAT

The first term does not exceed

-1- .iJ;J eaaz 2-J(7Tt)

foo exp ( - l.1J2jt) d1J 8

4

= 2 1/7T .2J;1 exp (aa 2 ) foo exp ( - H;'2) ds < 'V

if oj-Jt is sufficiently large, that is, if 0 < t < 02T1 where on a and e. The second term does not exceed M 2-J(7Tt)

e,

8/v'l

foo exp {a(x+1J) 8

2

T1

depends

1J2} d1J - 4t

~ 2J~t)faOO ex p {a(1J+ a )2- ~:}d1J =

.1.11 eaazfoo -2 / exp {- g2(1- 4at) + 2aaS-Jt} ds· 'V 7T

8/vl

Hence, if 0 < t < lj(8a), the second term does not exceed .2J;J eaazfoo -2 /7T exp{ -i{;'2+ ta S-J(2a)}ds 'V

8/v l

and this again is less than e if oj-Jt is sufficiently large, say if 0< t < 02 T2 .

Similarly for the integral from - 00 to - o. We have thus proved that, for every positive value of e, there exists a positive number to such that

lu(x, t) - f(x)1 < 5e, whenever 0 < t < to, for every x belonging to [-a,a]. Hence u(x,t) tends tof(x) as t-'?- +0, uniformly with respect to x. Lastly, suppose that X obelongs to [-a,a]. Then, if Ix-xol < 0, o < t < to'

lu(x,t)-f(xo)1 ~ lu(x,t)-f(x)I+lf(x)-f(xo)1 < 6e, so that u(x, t) tends to f(x o) as (x, t) tends to (x o,

+ 0) in any manner.

12.7 The existence and uniqueness theorem A. N. Tikhonovt proved the existence and uniqueness theorem for the initial value problem of the equation of heat under quite general . t Mat. sbornik

42 (1935), 199-216. This paper is in Russian. The present account is based on D. V. Widder's paper in Trans. Amer. ll'Iath. Soc. 55 (1944), 85-95. 9

CPO

250]

EQUA TION OF HEAT

[12.7

conditions. The formal solution found in § 12.2 is 11(X,t)

= J~~Jml:(~C-~,t)d;,

k(x, t) =

where

ui exp ( -

tx2It)/\/(7Tt).

Let D be the set of points in the xt-plane defined by - R < x < R, o < t :( c. The liHes x = ± R, (I :( t :( c and t = 0, - R :( :1.: :( R belong to the frontier of D, but do not belong to D; we denote them by B. Then 15, the closure of D, is the union of D and B. A function u(x, t) with continuous first and second derivatives in a given region, which satisfies there the equation of heat, is said for brevity to belong to the set Y'f' in that region. Theorem 1. If u(x, t) belongs to Y'f' in D and if

lim inf u(x, t)

~

0

(x, I) -+ (X o, to)

fa?' every point (x o, to) of B, then u(x, t) ~ 0 in D. It is understood that (x, t) approaches (x o, to) along a path in D. If (x o, to) is any point of B, then, for every positive value of e, there exists a positive number 00 such that u(x, t) > - e whenever the point (x, t) of D lies in the open disc with centre (x o, to) and radius 00' The family of such open discs, one for each point of B is an infinite open covering of the bounded closed set B. By the Heine-Borel theorem, we can choose from this family a finite number of discs which also cover B. Hence there exists a positive number 0, the smallest radius of this finite family of discs, such that u(x, t) > - e whcnever the point (x, t) is at a distance less than 0 from B. If the theorem is false, there exists a point (Xl' t l ) of D at which u is ncgative, so that U(Xl,t l ) = -1 < O. Let where

v(x,t) = u(X,t)+K.(t-t l ) K i~

a constant. Since u satisfies the equation of heat, we have 82 v OV 8x2 = ot -K.

Choose

K

so that

(I

<

K

< llt l , as we may since t l > O. Let

o<

t

< l-Kt l ,

and let 0 be the corresponding positive number found in the preceding paragraph. Then, for all points (x, t) at a distance less than 0 from B,

v(x,t) > -e+K(t-t l ) > -l+Kt > -1.

12.7]

[251

EQUATION OF HEAT

Since v(x v tl ) = -l, the minimum of v(x, t) cannot exceed -l, and so is attained at a point (x 2 , t2 ) of D. This is impossible. For at a minimum ovlot = 0 (or possibly ovlot < 0 ift 2 = c), and o2vlox2 ~ 0, yet 02V

oV

ox2 = ot

-K,

where K > O. Hence the assumption that u(x, t) is negative at a point of D is false, and the theorem is proved.

Theorem 2. Ifu(x, t) belongs to Ye in the strip 0 < t

~

c, if

u(x, t) = 0

lim (X,t)->(x,,+O)

sup u(x, t) = O(eaXZ )

for all xu' and if

O
as x -J>- ± 00, for some positive value of a, then u(x, t) is identically zero in the strip. Note that we assume that u(x, t) tends to zero as (x, t) tends to (xo, + 0) in any manner in t > O. Let F(x) = sup lu(x,t)l. O
Then there exists a constant J.YJ such that

o~

F(x)

~

1.11 eaxz

for all x. The function

U(x,t) = F( -R)k(x+R,t) +F(R)k(x-R,t) belongs to Ye in t > 0, and, in particular, in the region D defined by -R < x < R,O < t ~ c.

U(R t) >- F(R) k(O ) = F(R) ,,,.. ,t 2,J(7Tt)"

Now Hence, if 0 < t

~

c,

lu(R,t)1 or

~ F(R) ~ F(R)~~ ~ 2,J(7TC) U(R, t)

2U(R,t),J(7TC)

± u(R, t)

~ O.

A similar result follows with R replaced by - R. Let w1 (x, t) = 2,J(7TC) U(x, t) + u(x, t).

This function belongs to Ye in 0 < t ~ C and is non-negative when x = ±R,O < t ~ c. If (x,t)-J>-(x o, +0) where -R < X o < R, then 9"Z

252}

[12.7

EQUA TION OF HEAT

+ 0), then F( - R) k(:r + R, t) tends to zero and F(Ri lc(x- R, t) is non-negative. Hence

v' l (x, t) tends to zero. If (x, t) -J>- (R,

U(x, t)

lim inf (x,I)~(R,

B." hypothesis.

~

O.

+0)

liminf 1i(X,t) = limu(:r,f) = O. (x,I)-:-(R, +0)

liminf w 1 (x, t)

Hence

~

0,

(x,I)-:-(R,+O)

with a similar result as (x, t) -J>- (-R, + 0). By Theorem 1, w 1 (x, t) in D. In the same way, we can prove that

u'2(x,f) = 2\/(7TC) U(x,t)-u(x,f) in D. Therefore

~

~

0

0

lu(x,t)1 ~ 2\/(7TC) U(x,t)

onD. The function u(x, t) does not depend on R. If we can show that U(x, t) tends to zero as R-""" 00, we shall have u(x, t) = O. Kow 7 R lIf {R" (R-X)2} F (R) tC(x,t) ~ 2\/(7Tt)ex p a -4t

so that F(R) k(x - R, t) tends to zero as R -J>- 00, provided that

o< t

< 1/(4a),

with a similar result for F( - R)k(x + R, f). Hence if c ~ 1/(4a), u(x, t) is zero on the strip 0 < t ~ c. If C > 1/(4a), we can repeat the argument with u(x, t + 1/80), and so forth as often as is necessary: u(x, t) vanishes on 0 < t ~ c for any given value of c. From this follows Tikhonov's uniqueness theorem: If u 1 (x, t) and u 2(x, t) belong to £ in 0 < t ~ c, if both tend to f(x o) as (x, t) -J>- (x o' + 0) for all values of X o, and zf sup u 1 (x, f) = O(c'';'''), O
sup u 2 (x, f) = Orear')

O
as X-J>- ± oc for some pos£tive value of a, then 1'1(::r, t) and u 2(x, t) Me ident1:cal in the sfrip. For U 1 - u 2 satisfies the conditi.ons of the~)I'em 2.

Theorem 3. Iff(x)exp(-ax 2 ) is integrable in Lebesgue's sense, over (- ce, oc) for some positive value of a, then, on the strip 0 < t < 1/(40), the /1111 ction 1

[C/O

U(.T.t) = - I() 2 f(;)exp(--i(:r-fY/t)d~ \' 7Tf • _ C/O

12.7]

[253

EQUATION OF HEAT

is continuous, has continuous derivatives of all orders and satisfies the equation of heat. If f(x) exp (-ax 2) is integrable for every positive value of a, the conclusions hold in the half-plane t > o.

Since k(x, t)

=

2~~1Tt) exp ( - t X2 /t)

satisfies the equation of heat, it suffices to prove that u and its derivatives are continuous and that differentiation under the sign of integration is valid. Writef(x) = g(x)eax', so that g(x) is integrable over (-00,00). On the rectangle R defined by -a ~ x ~ a,O < to ~ t ~ t1 < lj(4a), we have If(;)k(x-;,t)1 =

Ig(;)I 2~~1Tt) exp {a;2 - ~t (x - ;)2}

~ Ig@1 2~ (~o) exp {a;2 - ~tl (x - ;r~}

~ Ig(;)12~:1Tto) exp {a;2_ 4~1 (lxl-IW2} Ig(;)/ 2~:1Tto) exp {- (4~1 - a) ;2 + 2t1 Ixll;l- :t1 Ix12} 1

=

~ Ig(;)1 2~:1Tto) exp { - (4~1 -a) ;2 + 2~1 I;I} If 0 < t1 < 1/4a,

2~1 1;1- (4~1 -a) 1;1 2~ 4t (1a~ 4at 1

Therefore

If(;)k(x-;,t)1 ~

1

)'

Glg(;)I,

where G is a constant. Since g(;) is integrable, the integral defining u(x, t) is uniformly convergent on R. Hence u(x, t) is continuous on R for all a, to and t1 ; therefore u(x, t) is continuous on 0 < t < 1/4a. Next, if (x,t) and (x+h,t) belong to R, u(x+h, tl-U(X, t) =

f:oo g(;) exp (a;2) k(x+h-;, tl- k(x-;, t) d;.

By the mean-value theorem, u(x+h, tl-U(X, t) =

f:oo g(;) exp (a;2) kx(x+fJh-;,t) df"

254]

[12.7

EQUATION OF HEAT

where 0 < e < 1. B:: an argument similar to that just used, there exists a constant C1 such that lexp (a~2) 7:",(;1; + eh -~, t)1

< c1.

By Lebesgue'f' dominated conyergence theorem, the last integral tends to

af' 11 -7 O. Hence Cli/OX exists, and is given by

ou(x,t) _

-----;C-x c

Joo

fl:· . I: dl: (~) kx(:L -~, t) '"

-00

from which the continuity of eu/ox follows. Similar methods apply to derivatives of all orders. ,Ve can therefore differentiate under the sign of integration; and as k(x-~,t) satisfies the equation of heat, so also does u(:r, t).

Theorem 4. If, in theorem 3, the limits f(x o + 0) and f(x o - 0) exist, limsup lu(x,t)l (x,I)~(x,,-i-(l)

Iff (x) is continuous at

< max{lf(xo+O)I, If(xo-O)/}.

,1: 0 ,

lim

U(.'f, t) = f(::c o)'

(x,I)-(:>:,,-(l)

Let .JJf be the greater of !f(x o ± 0)1. Then, for eyer,\' positive value of e, there exists a positive num bel' 0 such that U(x)! < 111 + c\Yhenever /:C' - xol < o. \I'e divide up the range of integration and "\\Tite u(x,t)

=

J:,:b +.r::~: + JX~-i-J(~)lc(x-~,t)d~ = 11+12+13,

and we shift the origin so that X o is zero. Then Ixl < The term i~ casv to handle., since . 1'

o.

0

Xext, if 0 <

401

ex])

< 1.

(0~2) l.~(:r -

E,. t) = _1_ ex]) (aE,2 - '.11_\1,1' - E,)2/t) 2\/(7Tt) -

is a decreasing function of ~ when ~ > :r/(l - 4at). Hence if Ix! < p < 0, the function i~ decreasing ,,·hen.;' ;:? G, and so

12.7]

EQUATION OF HEAT

[255

Therefore

1131

:>; exp(a8 2)k(8-p,t)

r~oo exp( -aS2) If(;)1 df

Thus 13 tends to zero as t-+ +0. The integral 11 does also. Hence limsup

lu(x,t)l:>; 1l'] +e.

(x.t)-(x•• +0)

As e is arbitrary, limsup

lu(x,t)l:>;

iV] =

max{lf(xo+OI, If(xo-O)I}.

(x.t)-(x•• +0)

Iff(x) is continuous at x o' let u 1 (x, t) = u(x, t) - f(x o).

Then

u 1(x,t) = I:oo {f(;)-f(xo)}k(x-;,t)df

By the first part of the theorem, lim sup lu 1 (x, t)1 :>;

o.

(x.t)_(x•• +0)

Hence, as (x, t) tends to (x o' + 0), u 1 (x, t) tends to zero and so u(x, t) tends to f(x o)' A similar result, due to G. H. Hardy, t is that, for fixed x, u(x, t) tends to f(x) as t -+ + 0 for almost all values of x and that it tends to -Hf(x+ 0) +f(x- O)} whenever this expression has a meaning.

12.8

The equation of heat in two and three dimensions 2 The function k ( ' t) 1 (X + y2) x,y, = 41Ttexp -~ satisfies the equation

82u 82u 8u 8x 2 + 8y 2 =

at·

From this we can construct the formal solution u(x,y,t) = If we put

4~tII:oof(;'1J)exp{ ; = x+2X.jt,

(X-;)2:e(y-1J)2}d;d1J.

1J = y+2Y .jt,

we obtain 1Ioo u(x, y, t) = iT _oof(x +2X .jt, y +2 Y.jt) exp (-X2 - Y2)dXdY.

As t-+ +0, this solution tends tof(x,y) under suitable conditions. t Mess. Math., 46

(1916), 43-48.

256]

EQUATION OF HEAT

SimjlaJ'l~',

[12.8

a formal solution of the initial value problem for

a2u

(32 u

(32 u

au

+-+-=ox 2 ay 2 0::: 2 at IS

where

U(X,y,z,t)

=

_1_2

8(7Tt)~

JJJr»

-r»

f(~,1/,i;)exp(-iR2Jt)d;d1Jdl;,

R2 = (X- ;)2 + (y _1J)2 + (z - 1;)2.

12.9 Boundary conditions In the problem of heat conduction in a finite rod, there are, in addition to the initial condition, boundary conditions at the end points of the rod. Similar problems arise in the theory of heat conduction in the plane or in space. Suppose that we have a conducting solid bounded by a closed surface 8. The temperature u satisfies a2u 02U a2u au + + = 2 ax oy2 OZ2 at and is given initially everywhere inside 8. There are three possible types of boundary condition on 8. (i) The temperature may be prescribed on 8 for all time. (ii) There may be no flow of heat across 8 so that au/eN vanishes on8. (iii) If the flux of heat across 8 is proportional to the difference between the temperature at the surface and the temperature U o of the surrounding medium, it is equal to H (u o- u) where H is a positive constant. The boundary condition is then

au

K aN = H(U o-11), where a/aN is differentiation along the outward normal, and K is a positive constant. YVe write this as

au

aN+ hu =

huo,

where h is a positive constant. If the solid is bounded externally by a closed surface 81> internally by a closed surface 8 2 , we could have different types of boundary condition on 8 1 and 8 2 ,

12.10]

EQU A TION OF HEA T

[257

12.10 The finite rod In the case of an insulated uniform rod of finite length a, we have to show that

has a unique solution which satisfies the initial condition u = f(x), when t = 0, < x < a, and also satisfies conditions at the ends. There are the four types of condition at the ends. (i) The temperature is given at the ends by

°

u(O, t) = !/o(t),

u(a, t) = !/l(t),

when t > 0. We do not require u(x, t) to tend to a limit as (x, t) tends to (0, +0) in any manner or as (x,t) tends to (a, +0). For example, we might ha.ve the problem of a finite rod at a unique temperature whose end points are suddenly cooled to zero. The conditions are then u(x,O)

=1

u(O, t) = 0,

(0 < x < a),

u(a, t) =

°

(t > 0).

This is the idealisation of a real problem in which the cooling of the ends of the rod takes place in a very short time. (ii) :: - hou

= !/o(t) (x = 0, t

> 0),

where ho and hI are non-negative constants. (iii) u = !/o(t) (x = O,t > 0),

:: +h1u = !/l(t) (x = a,t > 0);

(iv) :: -hou = !/o(t) (x = O,t > 0),

u = !/l(t) (x = a,t > 0).

To prove uniqueness, we have to show that, iff, !/o and!/l are all zero, u(x, t) is identically zero for all t > 0.

Let D be the rectangle dary. In the relation

°< x < a, 0< t < b, and let r be its boun-

where (1, m) are the direction cosines of the outward normal, put ¢J = U2 , where u satisfies the equation of heat. Then

t

IfD u;dxdt

=

fr

(luu,;-t mu2 )ds.

258]

[12.10

EQUATION OF HEAT

If f, go and :11 arc all zero, this equation becomes

JJn u~dxdt+tJ: {u(x,b)}2dx

(i)

Jr 'u~d:(;dt + t r

= 0,

a

(ii)

• 1J

•

{n(x, b)}2dx

(I

+hoJ: {u(O,t)}2dt + hIJ: {u(a,t))2dt = 0, (iii) JJD

u~dxdt+tJ: {u(x,b)}2dX+h I J: {u(a,t)}2dt =

0,

(J (b {u(O,t))2dt = in the four cases. Since k and hI are not negative, we have 'u(x, b) = ° when °< x < a, for every positive value of b. Hence u(x, vanishes when °< x < a, t > which proves the uniqueness theorem. (iv)

U~dxdt+tJa {u(x,b)}2dx+h o • n o .

0,

0

o

t)

0,

12.11 The semi-infinite rod In the problem of the semi-infinite rod, there is an initial condition, an end condition and an order condition at infinity. The initial condition we take to be 'u(x, 0) = f(x) (x > 0), "'heref(x)e- ax2 is integrable over (0, co) for every positive value of a. There are various end conditions. ';Ve start by considering the cases when either (i) u = ¢(t) (x = 0, t > 0), or

(ii)

Ux

= 1fr(t)

(x = 0, t > 0),

where ¢(t) and y'r(t) are integrable over any finite interval. The first problem can be split into two simpler problems (a) when ¢(t) is iclt'ntically zero, (b) when f(x) is identically zero. The required solution is the snm of the solutions of the two simpler problems. The solution of problem (a) is an odd function of x. To solve it, we extend the initial data to x < by introducing a new function

°

F(.T) = f(x) (x > 0),

F(x)

= - f( - x) (x

< 0).

YVe can assign to F(O) any value we please. The problem with this initial function has a unique solution

12.11]

[259

EQUATION OF HEAT

°

This is an odd function of x, continuous when t > 0, and so u 1 (0, t) = there. At every point of continuity of F(x), u 1 (x,t)-+F(x o) as (x, t) -+(xo, +0) in any manner. The solution of problem (b) has already been found; it is

x

t

I

u 2 (x,t) =

o

¢(r)-k(x,t-r)dr. t- r

Hence the solution of problem (i) is U(X,t) =

I

OO

o

f(~){k(x-~,t)-k(x+~,t)}d~+

It 0

x ¢(r)-k(x,t-r)dr. t-r

The solution of problem (ii) can also be split into two simpler problems, (c) when 1fr(t) is identically zero, (d) whenf(x) is identically zero. To solve problem (c), we introduce a new function G(x)

The solution is

= f(x) (x u 1 (x,t)

> 0),

=

G(x)

= f( -

x) (x < 0).

I:oo G(~)k(x-~,t)d;.

At any point of continuity X o of G(x), this tends to G(xo) as t-+ + 0. In t > 0, u 1 (x, t) is continuous and continuously differentiable. Hence

o~(x,t) = ox

-Ioo

G(~)x-~k(x-~,t)d~.

-00

2t

When x = 0, t > 0, this is equal to

which vanishes since G(~) is an even function. The solution of problem (d) is

This vanishes when t = 0. Also OU 2 (x, t) ox

=

It 0

1fr( r)...!:.- k(x, t - r) dr, t- r

and this tends to 1fr(to) as (x, t) tends to ( + 0, to) in any manner when to> 0. The solution of problem (ii) is u(x, t)

=

I: f(~){k(x -~,

t) + k(x +~, t)} d~ - 2

I:

1fr( r) k(x, t - r) dr.

260]

[12.11

EQUATION OF HEAT

YVhen the end condition for the semi -infinite rod was 1i = 0 or U x = 0 when x = 0, we started with the solution

= Jooof(~)k(X-f,t)df+ Jooo g(~)k(x+~,t)df

u(x,t)

of the equation of heat and chose the function g(f) appropriately. The same method can be applied when the end condition is U x - hu = 0 where h > O. V?hen x = 0, Ux

= Jooo f(~) ~t k(~, t) d~ - Jooo g(~) ~t k(~, t) d~ =-

oo

J 0

()

{j(f) - g(~)} ()~ k(~, t) d~.

This we can integrate by parts if f(~) and g(~) are absolutely continuous. For then the derivatives l' and g' exist almost everywhere. YVe then have

ux(O, t)= -

[{j(~) - g(~)} k(~, t)]O" + Jooo {j'(S) -

g'(m k(s, t) dft

= J~ {j'(O-g'(mk(~,t)dft provided that the terms at the limits vanish. 'We can ensure this by choosing g(~) so that g(O) = f(O) and lim {j(~)- g(m k(~, t) = O.

;_00

The end condition is then

Jooo {1'(~)-hf(~)-g'(ft)-hg(~)}k(~,t)d~ = 0, for t > O. This is satisfied if g(~) satisfies the differential equation

g' + hg The solution is

g(~)elt;-g(O) =

= l' - hf.

J: {j'(1J)-hf(1J)}e hn d1J

= f(~) e"; - f(O) - 2hJ>(1J) ehn d1J. Since g(O)

= f(O),

f(f,)-g(~) = 2he-lt;J;!(17)ell1ld17. o

If

If(~)1

< M ea;2

.

12.11]

[261

EQUATION OF HEAT

for every positive value of a,

Mh p (-;2/ 4t - h;) If(;)-g(;)lk(;,t) < .j(17t)ex

I; 0

exp(a1J2+k'l)d1J

Mh < .j(17t); exp ( - ;2/4t + a£2), which tends to zero as f...:;..co when t < 1/4a. As a can be as small as we please, If(;)-g(;)1 k(;,t) tends to zero as ;.-+co for any positive value oft. The solution of the equation of heat we are trying to obtain is thus

u(x,t) = f:f(;){k(X-;,t)+k(x+;,t)}d; - 2h I: k(x +;, t) e-h;I: eh~f(1J) d1Jdf Inverting the order of integration

u(x, t) = I: f(;){k(x-;, t) + k(x +;, t)} d; - 2hI: f(1J) K(x +1J, t) d1J, K(x,t) = Ioook(x+;,t)ChSd;.

where

K(x, t) can be expressed in terms of known functions. For

x Y = 2.jt +h.jt.

where Hence

K(x, t)

=

)17 exp {hx + h 2t} Erfc (2~t +h.jt) ,

where Erfcx is the Error Function.

262]

12.12

[12.12

EQUA TION OF HEAT

The finite rod again

In the problem of heat conduction in a finite rod we have to solve the equation of heat in < x < a, t > 0, given certain end conditions and an initial condition 'lI(X, 0) = f(x), where f(x) is bounded and integrable. Suppose that the temperature is assigned at the ends, so that

°

u(O, t)

= ¢(t), 'It(a, t) = JjJ(t),

when t > 0, where ¢ and JjJ are bounded and integrable over any finite interval. This problem can be split up into three problems: in problem (a), ¢ and JjJ are zero; in problem (b)fand ¥ are zero; in problem (c), f and ¢ are zero. VIce solve problem (a) by extending the initial data, just as we did in the case of the semi-infinite rod. Since u(O, t) = 0, let u(x, t) be an odd function of x; since u(a, t) = 0, let u(a + x, t) be an odd function of x. 'Ve consider then the infinite rod with F(x)

= f(x)

(0 < x < a),

F(-x)= -F(x) F(a+x)

=

-F(a-x).

The second and third equations give F(a+x) = F(x-a)

and so

F(x) = F(x+ 2a).

The extended initial data satisfy F(x) =f(x)(O < x < a),

F(-x)

=

-F(x),

F(x+2a)

= F(x).

Since f(x) is bounded, F(x) certainly satisfies the condition IF(x)1 <

],f ccx

2

for every positive value of c, on which the theory of the infinite rod depended. The solution of problem (a) is then 'lt1(x,t) =

J~<XJ k(x-;,t)F(;)d;.

°

This function is continuous in t > for all values of x, has there continuous derivatives of all orders and satisfies the equation of heat. As (x, t) tends to (x o, + 0) in any manner, u1(x, t) tends to F(x o) at every point of continuity, and so does so alm~st everywhere. In particular it tends to f(x o) almost everywhere on x < a.

°::

12.12]

[263

EQUATION OF HEAT

Whent > 0, U1

(0,t)

= f:oo k(;,t)F(;)d; =

°

since F is an odd function. Also

u 1 (a,t)

= f:oo k(a-;,t)F(;)d; =

f:oo k(1J,t)F(a-1J)d1J =

°

since F(a-1J) is an odd function. This solution can be written as u 1 (x, t)

00

=

~

f(2n+l la

00

= ~

F(;) k(x -;, t) d;

(2n-lla

-00

fa-a F(;)k(x-;-2na,t)d;

= ~ fa0 f(;){k(x -; - 2na, t) - k(x +;- 2n'Z, t)} d; 00

= f:f(;)K(x,g,t)d;,

where

K(x, ;,t)

=2

1 00 I( ) ~ (exp{ -i(x-;-2na)2jt}-exp{ -i(x+;-2na)2jt}).

" 1Tt _ 00

Inversion of the order of integration and summation is readily justified. The function K can be expressed in terms of Jacobi's theta functiont 00 ita(zl r) = ~ exp (1Tirn 2- 2niz) -00

where im r > 0. In this notation,

_~ exp ( - (x-4~na)2)

= exp{ -i x2 jt} ita (i~li~2)

.j ~t) ita

and this is equal to

G: Ii::)

by Jacobi's imaginary transformation. Hence

K( x, \), C ) t

t

Ia

Ia

=~ {ita (1T(X2- ;) i1Tt)2 __Va<1 (1T(X2+;) i1Tt))} 2 2· a

a

a

See Whittaker and Watson, Modern Analysi8, pp. 464 a.nd 474.

264]

[12.12

EQUATION OF HEAT

The last formula gives ~r c .ti.(x,s,t)

:::. ( -n 27T 2t/ a 2) sm-Sln-. . n7TX . n7T~ = -a2 2.Jexp 1 a a

Hence we have the Fourier solution n7TX

co

uI(x, t)

= ~ bn exp (- n 27T 2t/a 2 ) sin a' bn =

where

~Jaf(~) sin n7T; d;,

a 0 a which we discuss later. The second problem is that in which u vanishes when 0 < x < a, t = 0 and when x = a, t > 0, but u = ¢(t) when x = 0, t > O. YVe have seen that i x uo(x, t) = ¢( 7) k(x, t - 7) dT

J

o t-

7

is a solution ofthe equation of heat which is continuously differentiable as often as we please in x > 0, t > 0 and in x < 0, t > o. It vanishes when t = 0 and is an odd function of x. It is discontinuous across x = 0, t > 0; for it tends to ¢(t o) at every point of continuity of ¢(t) as (x, t) tends to ( + 0, to), but it tends to - ¢(to) as (x, t) --+ (- 0, to)' Consider the function co

u 2 (x, t)

= 2: uo(x + 2na, t). - co

Since ¢(t) is bounded, say 1¢(t)1 < M, we have, when

o~ x

~

a,

0 < to

~

luo(x+2na,t)1 ~ (2n+ l)Ma

t i

Jo

~

tI,

n > 0,

d7 k(2na,7)7

( 2 2/ )d7 = (2n+2 l)MaJi I exp - n a 7 % " 7T

0

7-

i

<

( (2n+ l)Ma"ltJ 2 2/ )d7 2 I exp - n a 7 ""2 " 7T 0 7

_(2n+l)M"lt 22/) 2 0 I exp - nat n"a", 7T

-

r

(

. (2n + 1) M "It 1 ( 2 2/ ) ~ 2 0 I exp - n a to' n-a'V 7T

and similarly for luo(x- 2na, t)l. Hence the series defining u 2 (x, t)

i~uniformlyandabsolutelyconvergentwhenO ~ x ~ a,O < to ~ t ~ tl'

12.12]

[265

EQUATION OF HEAT

The series obtained by term-by-term differentiation can be discussed in the same way. It follows that u 2 (x, t) satisfies the equation of heat in < x ~ a, t > 0. 00 Write u 2(x, t) = uo(x, t) + 2;' uo(x + ~na, t),

°

-00

°

where the term n = is omitted in L'. Hence at every point of continuity of ¢(t) and so almost everywhere, we have 00

u 2( + 0, t) = ¢(t) + ~' u o(2na, t) = ¢(t), -

00

since uo(x, t) is an odd function of x. Also 00

u 2(a,t) = 2;u o((2n+l)a,t) = 0. -00

Thus u 2 (x, t) does satisfy the prescribed initial and end conditions. From this form of the solution, we can deduce the solution by Fourier series. For U 2(x,t)

It o l: It o~ 00

= -2 T

uX

= -~

00

UX_ oo

=

-iI ox

o = -~

t

¢(r)k(x+2na,t-r)dr

0

-00

0

dr ¢(t)exp (-!(x+2na)2/(t-r)).j.j( ) 1T t-r

¢(t){}

0

3

(:!.:2a Ii1T(tr)) dr a2 a

It [ +

uxo

¢(r) 1

00 n1TX] dr, 2 2; exp (-n2rr 2(t- r)/a 2) cos1 a a

and so u 2(x, t)

')1T

00

n1TX

a

1

a

= ; 2; nsin -

exp ( - n2rr2t/a2)

It ¢(

r) exp (n2rr 2r/a 2) dr.

0

The solution of the problem when the initial and end conditions are u(x,O) = 0, u(O, t) = 0, u(a, t) = 1/f(t) can be deduced by replacing x by a-x and

rp by 1/f.

12.13 The use of Fourier series The problem of the finite rod when the temperature is given initially and the ends of the rod are kept at the same constant temperature can also be solved by using Fourier series. We have to find the solution of

266]

[12.13

EQUATION OF HEAT

given that

U(X,

+ 0) =

f(X)

(0 < x < a),

'u( + 0, tt = u(a - 0, t) =

° (t >

0).

The particular solution

. n7TX exp ( - n 2 7T 2t/a 2 ) sm-a vanishes when x for f(x) is

=

°and x = a. If the Fourier half-range sine series

where

bn

=

. n7TX a2faof(x)sm-;;:dx,

we should expect co b • n7Tx U(X,t) = Z; nexp(-n 27T 2t / a-Q) sm-

a

1

(1 )

would be the required solution. IVe assume that f(x) is integrable in Lebesgue's sense. Then {b n } is a null-sequence, and so is bounded; there exists a constant K such that Ibnl < K for all values ofn. Hence, ifO ~ x ~ a, < to ~ t ~ t1 , we have

°

The series (1) therefore converges uniformly and absolutely in the rectangle, and so its sum is continuous there. In particular u(O, t) and u(a, t) vanish when t ~ to > 0. The series obtained from (1) by differentiation under the sign of summation are also uniformly and absolutely convergent. Therefore (1) defines a solution of the equation of heat which satisfies the end conditions. co

The series 2: b" sin n7Tx/a is not necessarily convergent, but it is 1

summable (C, 1)t almost everywhere; and its Cesaro sum is J(x) at every point of continuity, again almost everywhere. To complete the co

proof we use a theorem due to Bromwich,t that if 1: a k is summable o

(C, 1) with sum s, then co

lim 2: an exp (- eA.,,) = s /1-+0 0

t The term" "ummable ~

(C. 1) and Cesaro sum are defined in note S of the Appendi>:. Bromwich, An Introduction to Infinite Series, 2nd edn (1926), p. 429.

12.13]

[267

EQUATION OF HEAT

if {An} is an increasing sequence of postitive integers such that, for all n,

where K is a constant. In the present case () = 1T 2tfa 2, and An = n 2, so that Bromwich's conditions are satisfied. Hence the series in (1) tends to f(x) as t -+ + whenever f(x) is continuous. The problem ofsolving the equation of heat in x::::; a,t > O,when u(x,O) = 0, u(O, t) = ¢(t), u(a, t) = can be solved by a modification of the Fourier method. Assume that

°

°:: :;

°

00 n1TX u(x, t) = L; bn(t) sin-, 1 a

. n1TX bn(t) = -2Ia u(x, t) sm dx. a a 0

where

Since u(x, t) vanishes when t = 0, bn(O) = 0. We have

dbn(t) = ~Ia 8u(x, t) sin n1TX dx ot a dt a 0 2Ia o2u (x,t). n1Tx ~2 S l n - dx a a 0 uX

=-

. n1TX n1T n1TX] a 2n2rr2Ia . n1TX = -2 [OU - s m - - - u c o s - - -3u(x,t)sm-dx a ox

a

2n1T

n2rr2

a

a

0

a

0

a

= -a2¢ ( t ) -a-2bn(t). From we have Thus

2n1Tft 2 ¢(r)exp(-n2rr2(t-r)fa 2)dr. bn = a 0

21T 00 n1TX I t U(X,t) = 2 L;nsin- ¢(r)exp(-n2rr 2(t-r)fa 2)dr. a 1 a 0

This formal argument gives the answer found in § 12.12, but it would be difficult to give a rigorous proof. Note that u(O, t) = 0. The reason for this is that u(x, t) is an odd function of x. u(x, t) tends to ¢(t) as x -+ + 0, but it tends to - ¢(t) as x -+ - 0. The sum of the series when x = is t{u( + 0, t) + u( - 0, t)}, which is zero.

°

268]

[12.13

EQUATION OF HEAT

If we try to so1\'e the equation of heat for the finite rod when the conditions are u(x, 0) = f(x) (0 < x < a), u.x(O, t) - hu(O, t)

= 0,

ux(a, t) + Hu(a, t)

=

°

(t > 0),

where 11 and H are positive constants, we might start with the solution u = exp (- k 2t) (A cos kx + B sin kx).

In order to satisfy the end conditions, we must have (P - hH) sin ka = k(h + H) cos ka.

This equation in k has no complex roots. Its real roots occur in pairs ± kv ± k2 , .•. where {k n } is a strictly increasing sequence of positive numbers. IVhen n is large, k = n1T + a(h+H) +0 (~). n a n1T n2

Since ak.J1T is not an integer, the Fourier series method is not applicable. IVe have to use instead an expansion as a series of SturmLiouville functions. The theory is outside the scope of this book.

Exercises 1. If U(a, t) is the Fourier transform of a solution u(x, t) of

02U OU ox2 = ot +xu,

prove that and hence that

U(a, t) = F(a+ it) eJl..-P (- ii( 3).

Deduce that, if u(x, 0) = f(x) for all values of x .

u(x,t) =

.

2,\1~1Tt) eJl..-P(it3 -xt) J:oof(;)eJl..-p(-t(X-;-t 2)2/t )d;.

2. u(x,t) satisfies the equation of heat in x > O,t > u(x,O)

= f(x),

u",(O, t) = Jjr(t).

The Fourier cosine transform of u(x, t) is

°under the conditions

[269

EQUATION OF HEAT

with inverse

U(x,t)

Prove that

= J(~)I:", Uc(a,t)cos:xxda. °Zc = J

(~) ljr(t) -

Deduce that Ue(x, t) = Fc(a) e-"St_ J

(~)

a 2 Uc'

J:

ljr(r) e-"S(t-T)dr,

where Fc(x) is the Fourier cosine transform of!(x), and hence that u(x,t)

= I: !(;') {k(x-;,t) +k(x+;,t)} d;-:2 I: ljr(r) k(x,t-r) dr.

3. Prove that, if a and b are positive, I:", k(x-;,a)k(;,b)d; = k(x,a+b).

Hence show that, if the operator

t9'; is defined by

~[f] = I:",!(;)k(x-s,t)d;

(t> 0),

then t9':1 [t9':. [f]] = t9':1 + t. [f], if t1 and t2 are positive. 4. Find the solution of in

°< x < 1T, t > °such that U(x, 0) =0 (O<X<1T),

U(O,t) = l-e- t (t>O),

U(1T,t) =0 (t>O).

Prove that this solution tends to (1T-X)/1T as t-+ +00. Why would you expect this result?

5. u(x, t) is the solution of the equation of heat which satisfies the conditions U= (t = O,X ~ 0), ure-hu = ¢J(t) (x = O,t ~ 0),

° where h is a positive constant and ¢J(t) is continuous. Prove that, when x> °and t> ° e:u -hu It ¢J(r)'-:"" k(x,t-r)dr. =

0

vX

t-r

Hence show that u(x,t)

=-

'" e-h ; Itx+; --¢J(t-T)k(x+;,r)d;dT. 0 r I o

6. u(x,t) satisfies the equation of heat in x > O,t > 0, under the condi·

tions

u(x,O) =

° (x > 0),

u(O, t) = ¢J(t) (t > 0).

270]

EQUATION OF HEAT

J'" if; (

Prove that v -_ " 'here.Ll.

2 X U(X, t) = .j7T

2

X ) e-s' d;, 4;2

t-

1 / I :!x 'I,f.

°

7. Prove that the solution of the equation of heat in t ~ which satisfies the conditions u(x,O) = 1 (x> 0), u(x, 0) = -1 (:1: < 0)

x! ]• T2 [ 1-Erfc-

u(x,t) =

IS

2 'Ii t

'I 7T

8. Prove that the solution of the equation of heat in x satisfies the conditions u(x,O)

=

O(x > 0), u(x, t)

IS

u(O,t)

=

1 (0 < t < T),

2

x

'1m

'I'

t

~7T [Erfc 2

'I

= -,- Erfc -2I =

'I

(0 <

x / -Erfc t

t ~

~

0, f

~

°

which

u(O,t) = O(t > T)

T).

I x T] (f> T).

2 'I (t-

)

Verify that the solution is continuous and continuously differentiable. 9. u,(x, t) satisfies the equation of heat in

ditions u(x,O) = f(x) (0 < x < a),

°

~ x ~

0, t

u",(O, t) = u",(a, t) =

~

0, under the con·

°

(t > 0),

wheref(x) is integrable. Prove that u(x,t) =

'" n7TX tao + L;ane-n''''t/a'cos-,

a

1

where

n7Tx tao + :E'"1 an COSa

is the Fourier half.range cosine series for f(x). 10. Show that the solution of the equation of heat in x > 0, t > Osuch that u(x,O) =f(x) (x> 0),

is where

u",(O,t)-hu(O,t)

=

°

(t > 0)

u(x,t) = Jo'" {f(;)k(:/:-f"t)+g(;)k(x+;,t)}d;, g(x) = f(x)- 2he- hx

J:

f(;) eltsdf,.

APPENDIX Note 1. Analytic functions The function !(x,y) of two real variables is said to be an analytic function regular in a neighbourhood of (xo, Yo) if it can be expanded as a double series 00 Z; amn(x-xo)m (Y_Yo)n m,n=O

absolutely convergent on a disc (x - X O)2 + (y - YO)2 < R~. If (xO+R1,Yo+R1) is a point ofthis disc, {amnRf+n} is a bounded double sequence. Hence if R < R 1 the double series is uniformly and absolutely convergent on the square Ix-xol ~ R, IY-Yol ~ R. The series can therefore be differential term-by-term on that closed square as often as we please. The definition can be extended in the obvious way to any number of variables.

Note 2. Dominant functions Suppose that, by some formal process, we have obtained a double series 00 Z; amn(x-xo)m(Y-Yo)n m,n=O

and that there exists a double sequence of positive numbers {A mn } such that the double series 00

Z; Amn(x- xo)m (Y- Yo)n

m,n=O

is convergent absolutely and uniformly when Ix - xol ~ R, Iy - Yol ~ R and that lamnl < A mn for all m and n. Then, by the comparison test, 00

Z; amn(x-xo)m (Y-Yo)n

m,n=O

is also uniformly and absolutely convergent when Ix-xol ~ R, IY-Yol ~ R. If we denote the sums of the two series by! and F, we say that F is a dominant (or majorant) function of!, and we write ! ~ F. If, on the same square,! ~ F, g ~ G, then !+g~F+G, !g~FG. [ 271 ]

272]

Also

APPENDIX

of of ox ~ ox'

of of oy ~ oy;

with similar expressions for derivatives of all orders. The definition can obviously be extended to any number of variables. If f(x, y) is an analytic function, regular in a neighbourhood of some point, the origin say, it is expansible as a double series ~amnxmyn

uniformly and absolutely convergent on a square Ixl ~ R, Iyl ~ R. The double sequence {amnRm+n} is therefore bounded, Iamn Rm+n I < M. It follows that

f~ ~M~'Z: = Also

M/(l-i) (1-~).

f~!:nM(:7n~)! ~: =

M /(1-~-~).

Note 3. Re~ular arcs A set of points in the plane defined by parametric equations x =f(t), y = g(t) (a ~ t ~ b), where f(t) and g(t) are continuous functions, is called an arc. If no point of the arc corresponds to two different values of t, the arc is said to be simple. If, in addition,f(t) and g(t) are continuously differentiable, the derivatives being one-sided at the end points, we shall call such a simple arc a regular arc. A regular arc has an arc length defined in the usual way. The definition also holds with the obvious changes for arcs in space. Note 4. Re~ular closed curves A set of points in the plane defined by parametric equations x=f(t),

y=g(t)

(a~t:(b),

when f(t) and g(t) are continuous, is called a simple closed curve if no point corresponds to two different values of t except that f(a) =f(b),

g(a) = g(b).

If the simple closed curve is a chain of a finite number of regular arcs we shall call it a regular closed curve. A regular closed curve may have a finite number of corners; it has a piece-wise continuously-turning tangent. A regular closed curve in the plane divides the plane into two domains, a bounded interior domain and an unbounded exterior domain.

[273

APPENDIX

Note 5. Green's theorem in the plane Let D be the domain bounded by a regular closed curve 0. If (i) u and v are continuous in D, the closure of D, (ii) U x and vy exist and are bounded in D, (iii) the double integrals of tt x and VII over D exist, then

° °

°

where (l,m) are the direction cosines of the normal to drawn out of D. The fact that the normal to may suddenly change direction at a finite number of points of does not affect the truth of the result. And the result is evidently true if U x and vy are continuous in D. There is an alternative form of the result corresponding to Stokes's theorem in space, namely that Ic Pdx+Qdy = IID (Qx- P1J)dxdy,

°

where integration over is in the positive sense. . If D is bounded externally by a regular closed curve 01 and internally by a regular closed curve 02'

If

(ux+Vy)dxdy = I

D

c.

(lu+mv)d8+I (lu+mv)ds, c.

where (1, m) are the direction cosines of the normal to 01 or 02' again drawn out of D. The Stokes form is, in this case,

If

(Qx-Py)dxdy = I

D

c.

PdX+Qdy-I Pdx+Qdy, c.

where integration over 01 and over O2 are both in the positive sense.

Note 6. Surfaces A simple closed surface in three-dimensional Euclidean space is one which is topologically equivalent to a sphere. Two surfaces are topologically equivalent if each can be transformed into the other by continuous deformation. A simple closed surface is bounded and divides the whole space into a bounded interior domain and an unbounded exterior domain. A simple closed curve drawn on a simple closed surface divides the surface into two portions, which we call caps. A cap has parametric equations r = r(i\,,u) in vector notation, where each component is a

274]

APPENDIX

continuous function of A and fl. If r(A, j1) is differentiable, the vectors rIc and r ll are tangent to the cap, and N = (r,\ x r/J/lr,\ x rill

is a unit vector normal to the cap at the point of parameters A and fl. If rIc and r ll are continuous, the direction of N varies continuously as (A,fL) moves on the cap. "\Ye then call the cap a regular cap. If a simple closed surface is formed of a finite number of regular caps, we call it a regular closed surface.

Note 7. Green's theorem in space Let D be a domain bounded by a regular closed surface 8. If (i) U,V and ware continuous in 15, the closure of D, (ii) u x ' v y and 1Cz exist and are bounded in D, (iii) the integrals of u x ' v y and W z over D exist, then

where (l, m, n) are the direction cosines of the normal to 8 drawn out ofD. More generally, if D is bounded internally by a regular closed surface 8 0 and externally by a regular closed surface 8 1 , the triple integral is then equal to

(lu+mv+nw)d8+If (lu+mv+nw)d8, If s , . s, where (l, m, n) are still the direction cosines of the normal drawn out ofD.

Note 8. Summability a:J

The infinite series

La"

is said to be convergent with sum

8

if the

1

sequence {8,,} defined by 8n

= aO +al +a 2 + '" +a,,_1

tends to the limit 8 as n-7-ct:.:. If the series is not convergent, it may be summable in some other sense. If

[275

APPENDIX

it may happen that the sequence

{CTn }

tends to the limit

CT

as n-yoo.

00

If this is the case, the series

~

o

an is said to be summabIe by Cesaro's 00

mean, or summable (0,1), with Cesaro sum with sum 8, it is summable (0,1), and (0,1) is not necessarily convergent. 00

If the series

~

o

CT

=

If ~ an is convergent o But a series summable

CT.

8.

anxn has radius of convergence unity and has sum 00

f(x) when Ixl < 1, the series

~

an is said to be summable in Abel's

1

sense, or summabIe (A), if f(x) tends to a finite limit S as

X-y

1- 0;

00

S is called the Abel sum. If ~ an is convergent with sum o

sum is also

8,

8,

the Abel

by Abel's theorem on the continuity of power series.

00

If ~ an is summable (0,1), with sum o

CT,

its Abel sum is also

Note 9. Fourier series If f(O) is integrable in Lebesgue's sense over 0 Fourier series 00 tao + ~ (an cos nO + bnsin nO),

~

0

~ 211,

CT.

it has a

1

where

an =

~f2" f(O)cosnOdO,

11

0

bn =

~f2" f(O)sinnOdO.

11

0

The sequences {an} and {b n} are null-sequences, but this does not imply that the Fourier series is convergent. But the Fourier series is summable (0, 1) to the sum

t{f(O + 0) +f(O - OJ} for every value of 0 for which this has a meaning. If we assume that f(O) is periodic of period 211, the sum (0,1) when 0 = 0 is Mf( + 0) +f(211 - OJ}. In particular, the series is summable (0,1) to the sumf(O) at every point wheref(O) is continuous; it is summable (0, 1) tof(O) for almost all O. Since {an} and {b n} are null-sequences, the power series 00

tao + ~ (an cos nO + bn sin nO) x n 1

276]

APPENDIX

has radius of convergence unity. Since the series with x = 1 is summabIe (0,1) with sum f(O) at every point where f is continuous, it follows that . lim + (an cosnO+bnsin nO)Xn] =}(O) x-+l-O

[tao

I; 1

at every point where f is continuous. The Abel sum of the Fourier series isf(O) for almost all O.

BOOKS FOR FURTHER READING

Bateman, H. Partial Differential Equations of Mathematical Physics (Cambridge, 1932). Courant, R. and Hilbert, D. Methods of Mathematical Physics (New York, 1 (1953), 2 (1962)). This is the revised translation of Methoden der Mathematischen Physik (Berlin, 1 (1924),2 (1937)). Duff, G. F. D. Partial Differential Equations (Toronto, 1956). Epstein, B. Partial Differential Equations (New York, 1962). Garabedian, P. R. Partial Differential Equations (New York, 1964). Goursat, E. Course in Mathematical Analysis, 3 parts 1 and 2 (New York, 1964). This is the translation of tome 3 of Cours d'Analyse MatMmatique (Paris, 1923). Hadamard, J. Lectures on Cauchy's Problem in Linear Partial Differential Equations (Yale, 1923; reprinted New York, 1952). Hellwig, G. Partial Differential Equations, an Introduction (Waltham, Mass., 1964). Jeffreys, H. and B. S. Methods of Mathematical Physics, 3rd edn (Cambridge, 1956). Kellogg, O. D. Foundations of Potential Theory (Berlin, 1923; reprinted, New York, 1953). Petrovsky, 1. G. Lectures on Partial Differential Equations (New York and London, 1955). Sauer, R. Anfangswertprobleme bei Partiellen Differentialgleichungen (Berlin, 1952). Sneddon, 1. N. Mixed Boundary Value Problems in Potential Theory (Amsterdam, 1966). Steinberg, W. J. and Smith, T. L. The Theory of Potential and Spherical Harmonics (Toronto, 1944).

[ 277 ]

INDEX

adjoint linear operator, 77 analytic functions, 271 barriers, 180 boundary value problems, 45, 175, 193 Cauchy, problem of, 24, 44, 58 Cauchy and Kowalewsky, theorem of, 26,244 characteristics, 2, 28, 47 characteristic base curve, 25 characteristic strips, 6, 30 classification of second order equations, 34,39 complex variable, use of, 165

harmonic functions in the plane, 139 in space, 212 harmonics, spherical, 210 Harnack, theorems of, 158, 159, 161 heat, equation of, 50, 238 Helmholtz, formula of, 227 hyperbolic type, equations of, 16, 19, 34,54 Lagrange, linear first order equation of, 1 Lagrange and Charpit, method of, 7 Laplace, equation of, 131, 207 Laplace transform, use of, 242

descent, Hadamard's method of, 95 Dirichlet boundary value problem of, 45, 175, 193 principle of, 144 divergent waves, 200 dominant functions, 271

maximum principle for harmonic functions, 143 mean value theorem, 141,214,222

elementary solution of Hadamard, 189, 198 elliptic type, equations of, 16, 19, 34, 186, 207 Euler, Poisson and Darboux, equation of,98 existence theorem for the equation of heat, 249 for Laplace's equation, 175

parabolic type, equations of, 16, 34, 50, 238 Perron and Remak's solution of Dirichlet's problem, 175 Picard's method, 59 Poisson's integral, 153 potential theory, 131, 207

Neumann, problem of, 156, 167, 193 normal form of a second order equation, 33-42

first order equations, 1 systems of, 11, 14 Fourier series summabilityof,275 use of, 71, 155, 165 Fourier transforms, use of, 240 Gauss, mean value theorem of, 141, 214, 222 Green's equivalent layer, 133, 214 Green's function, 146, 150, 169, 216 Green's theorem, 273,274 Hadamard's method of descent, 95

radiation condition of Sommerfeld, 201, 229 reduced wave equation, 222, 225 regular are, 272 regular closed curve, 272 regular closed surface, 273 Riemann, method of, 77 Riemann-Green function, 79, 81 Riesz, method of Marcel, 107 second order equations, 24 classification of, 34, 39 normal form of, 33-42 self-adjoint linear operator, 78 spherical harmonics, 210 subharmonic functions, 175

[279 ]