Partial Differential Equations

PARTIAL DIFFERENTIAL

EQUATIONS

I

L

PARTIAL DIFFERENTIAL EQUATIONS

PARTIAL DIFFERENTIAL EQUATIONS E. T. COPSON FORMERLY REGIUS PROFESSOR OF MATifEMATICS IN TifE UNIVERSITY OF ST ANDRE\VS

CAMBRIDGE UNIVERSITY PRESS CAMBRIDGE

LONDON NEW YORK• MELBOURNE

Published by the Syndics of the Cambridge TJniversity Press The Pitt Building, Trumpington Street, Cambridge CB2 IRP Bentley House, 200 Euston Road, London YWI 2DB 32 East 57th Street, New York, NY10022, USA 296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia © Cambridge University Press 1975

First published 1975

Printed in Great Britain at the University Printing House, Cambridge (Euan Phillips, University Printer) Library of Congre38 Cataloguing in Publication Data Copson, Edward Thomas, 1901Partial differential equations.

Bibliography: p. 277 Includes index. 1. Differential equation, Partial. I. Title. QA377.C77

515'.353

74—12965

ISBN 0 521 20583 2 hard covers ISBN 0 521 09893 9 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, 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 students. 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. After 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 of the first order is of the form

Pp+Qq=R,

(1)

where p = au/ax, q = au/ay and F, 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 F, 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 f(x, y, u) = 0. If we regard (x, y, u) as rectangular Cartesian coordinates, f(x, y, u) = 0 is the equation of a surface; if f = 0 provides a solution of (1), the surface is called an integral surface. The fundamental problem is: given a regular arct I' in space, is there a unique integral surface through F? Alternatively, given a regular arc y in the xy-plane, is there a solution u(x, y) of (1) which takes given values on Let the parametric equations of F be

x = x0(t),

y = y0(t),

u=

u0(t).

On any surface, du = pdx + qdy. Hence, if there is an integral surface through F, the values p0(t), q0(t) of p and q on the integral surface at the point of parameter t of F satisfy (2)

where dots denote differentiation with respect to t. If we denote by Q0,

R

the point of F of parameter t, we have

P0p0+Q0q0=R0.

(3)

is not zero, Po and q0 are determined. Hence if by It is conventional to denote the second derivatives r, s, t; the fact that we have also used t to denote the parameter of F t The term regular arc is defined in Note 3 of the Appendix.

[1J

EQUATIONS OF THE FIRST ORDER

2]

[1.1

will not cause any confusion. If we differentiate (1) with respect to x, we get Pr+Qs = F(x,y,u,p,q),

so that, at the point of F of parameter t, P0r0+Q0s0 =

Since dp = rdx + sdy,

+

= Po and q0 are determined on F, and hence Similarly we can find all the partial derivatives so also are r0 and of u on F. Thus we get a formal solution as a Taylor series

u=

u0 +

{p0(x — x0) + q0(y — Yo)}

2s0(x—x0) (y—y0)+t0(y—y0)2}+

Under suitable conditions, it can be shown that the series converges in a neighbourhood of (x0, Yo' u0) of F, provided that — is not zero.

Now drop the suffix zero which has served its purpose. At a point of F, an integral surface satisfies

Pp+Qq=R, px+qy=ü. Hence and

similarly for p. If

q=

— —

—

vanishes at every point of F, this

equation is impossible unless the transport equation

Pü=Rx (or equivalently)

Qü =

satisfied. Hence, if — = 0 on F, there is no integral surface through F unless u satisfies the transport equation; and then there are an infinite number of integral surfaces since q can be chosen arbitrarily. An arc F which has this property is called a characteristic. There is one characteristic through each point of space at which F, Q, R are not all zero; a characteristic satisfies is

ü

PQR A characteristic is the curve of intersection of two integral surfaces. For if u = u1(x, y), u = u2(x, y) are two intersecting integral surfaces,

p1dx+q1dy—du = p2dx+q2dy—du =

0, 0,

EQUATIONS OF THE FIRST ORDER

[3

in an obvious notation, and so dx

dy

q1—q2

But

sothat

P =

q1—q2

p2q1—p1q2

P2—P1

Pp1 + Qq1 = R,

du

—

Pp2 + Qq2 = R

Q

R

P2P1

p2q1—p1q2

Therefore on the curve of intersection of two integral surfaces,

dx

dy

du

PQR —

—

The differential equations for a characteristic can be written as

x=P, y=Q, ü=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 arc, it can be shown that the unique integral surface through C is generated by the one-parameter family of characteristics which intersect C. Again, if the two-parameter family of characteristics is given by u) = çA (x, y, u) = a,

y,

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

= The projection y of a characteristic F on the plane u = 0 is called a characteristic ba8e-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 y, the data must satisfy the equation

Pü=Rx.

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 au

au

x—+y— = ax ay

flu.

We now prove the converse. The subsidiary equations of Lagrange are dx = dy = du x flu' y

From

these equations we get

x

Hence the general solution is u

—

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 = The characteristics are given by

1.

y=u—x, ü=x—y, whichgive 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 = 0,xy = 1. When u = 0,xy = 1, a2

= (x+y)2 = x2+y2+2xy

b+2.

The required integral surface is therefore

(x+y+u)2 = x2+y2+u2+2 or

1— xy

x+y

EQUATIONS OF THE FIRST ORDER

1.3] 1.3

[5

The general first order equation

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

x=x0(t),

u=u0(t)?

The method is to try to construct a Taylor series which satisfies (1) and converges in a neighbourhood of an arbitrary point of F. 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 q0 at the point of F of parameter t are given by F(x0, Yo' U0,p0, q0) = 0,

=

ü0.

We suppose that we can find a real pair (poe q0) 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, F, 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 F. Using the condition

dp = rdx+sdy, the values of the second derivatives of r and s on F satisfy

=

0,

= Hence

and similarly Since

=

(2)

= Xdx + Ydy + Udu + Pdp + Qdq =

(3) 0,

du = p dx + qdy, (2) and (3) are in fact the same equation. If Q0i:0 — is not zero, the values r0, t0 of the second derivatives are determined on F, and similarly for the derivatives of higher orders. Thus we again get a formal solution as a double Taylor where

____ EQUATIONS OF THE FIRST ORDER

6]

[1.3

series which can, under suitable conditions, be shown to converge in some neighbourhood of the chosen point of F, provided that —

not vanish there. Now drop the suffix zero. At a point of F, an integral surface satisfies

does

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

0,

+ qg,

Pr+Qs=—X—pU,

and

Ps+Qt= —Y—qU, where

+

=

+

=

Hence

and

If

—

vanishes at every point of F, there is no integral surface

through F 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 F. Thus we have now not an arc but a strip;

a sort of narrow ribbon formed by the arc F 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 —

—

PQpP+qQX+pUY+qU and also

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

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, y=Q, ü=pP+qQ, where

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

q=—Y—qU, 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

istic arc C is generated by a one-parameter family of characteristic strips. The first step is to construct an initial integral strip by

EQUATIONS OF THE FIRST ORDER

[7

ciating with each point of C a surface element whose normal is in the direction p: — 1, If the parametric equations of C are

x = x0(s), Y =

u=

Yo(S),

where s need not be the arc-length, we choose

p—p0(s),

so that and

du0

q=q0(s) dx0

Po

+ q0

eYo

F(x0,y0,u0,p0,q0) =

0.

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 (xe, Yo' u0) is a point on an integral surface of F = 0, the direction ratios Po: q0: — 1 of the normal there satisfy F(x0, Yo' U0, p0, q0) = 0. Hence the normals to all possible integral surfaces through the point generate a cone N whose equation is

F(x0,y0,u0, _xxo U—U0

U—U0/

=

0.

The tangent planes at (x0, Yo' u0) to all possible integral surfaces through this point envelope another cone T whose equation is obtained by eliminating Po and q0 from the equations u—u0 = p0(x—x0)+q0(y—y0), (x—x0)Q0—(y—y0)P0 =

.F(x0,y0,u0,p0,q0) =

where

0,

0,

and Q0 denote the values of eF/ep and eF/eq at (x0, Yo' U0,p0, q0).

EQUATIONS OF THE FIRST ORDER

8]

[1.3

The tangent plane to a particular integral surface at (x0, Yo' u0) goes through a generator of the cone T; the normal there lies on the cone N.

In the case of Lagrange's equation, the cone N degenerates into the plane

P0(x—x0)+Q0(y—y0)+R0(u—u0) = 0;

the cone T becomes the straight line X—

— Y — Yo — U — U0

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 = istic strips are given by the differential equations

0.

The character-

x=q, y=p, u=2pq, p=y, q=x and the relation pq = xy. It turns out that x = Aet+Be_t, y = Cet+De_t, u = ACe2t_BDe_2t+E, p = Cet — Det, q = Aet — Be_t, where the constants of integration are connected by

AD+BC = 0. x = s, y = 0, U =

On the initial curve

s.

On the initial integral strip, the equations

du = p dx + qdy, pq = xy

give p =

1,

when t =

0,

q = 0. Let t be measured from the initial curve. Then we have

A+B=s, C+D=0, AC—BD+E=s,

C-D=1, A-B=0 AD+BC=0.

where These give

A= B=

C=

—D

=

E=

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 = scosht, y = sinht, u = scosh2t,

p=

cosht,

q = ssinht.

EQUATIONS OF THE FIRST ORDER

1.4]

[9

Eliminating s and t from the first three equations, we obtain = 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 q(x0,y) = qY(y), (2) u(x0,y) = where q!(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 (x0, Yo) of the initial line x = x0. It is convenient to write u0 for And we q0 for make the assumption that f(x, y, u, q) is an analytic function of four independent variables, regular in a neighbourhood of (x0, Yo' U0, q0). 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 eq/ax = ap/ay. From equation (1) we have

ap

fx+fuP

eq

Hence u, p, q satisfy the equations eu

p_It, eq

—

I

ep

—

under the initial conditions u(x0,y) —

p(x0,y)

q(x0,y) =

This system of three equations is equivalent to equation (1) under the initial conditions (2). From the first and last of equations (3), eu\

ei

t See Notes 1 and 2 in the Appendix.

10]

EQUATIONS OF THE FIRST ORDER 3u

sothat

is identically zero, so that q By the initial conditions, The second equation of (3) gives

af

ap

since

au/ay.

eq

a2u

a2u

ax

axay

ayax

ap ay

when u is analytic. Hence

p =f(x,y,u,q)+a2(y). Again, by the initial conditions,

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 rj defined by ax

ey'

ax

under the initial conditions

7/YYo' when x = x0. Since 7/ is independent of x, 7/ = y — Yo for all x. Then ag/ax = 1 so that = x — x0. If we put x = x0 + y = Yo +7/ in f (x, y, u, p, q) we get an analytic function u,p, q) of five variables regular in a neighbourhood of (0,0, u0,p0, q0). We now have a system of five equations au a7/ ap

ap

a

eq — ap ax — ay' — a7/

ax — ay

ax

EQUATIONS OF THE FIRST ORDER

1.5]

[11

with the initial conditions

u=

çb(y),

p =f(x0,y,çb(y),çb'(y)),

q=

when x = x0. This system is of the form = 1,2, ...,n)

with the initial conditions

(i =

=

1,

2, ..., n)

when x = x0. The coefficients are analytic functions of the dependent variables only, regular in a neighbourhood of

regular in a neighbourhood of and the data 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 au —

ax av

au

av =A—+B— ay ay

au

av

—= C—+D-ax ay ay

under the initial conditions

u=

v

=

when x = x0, where çA and are analytic functions regular in a neighbourhood of Yo' For simplicity in writing, shift the origin so that x0 and Yo are zero. The coefficients A, B, C, D are analytic functions of u V0 andy, regularinaneighbourhood of (u0, v0) ,whereu0 = Again, by a shift of origin in the uv-plane, we can take u0 and v0 to

be zero. Since çb,

and the coefficients are analytic, we can find constants

M and R so that q!(y) and are dominated by My/( 1 — y/R) in ku t0, we have as

when t > 10+ r, where r cosh s = t — t0; 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

= was given by Poisson; this is certainty valid if f" 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) ha8 continuous partial derivatives of the first and second orders, the equation of wave motion8 (1)

ha8 the solution

u =

t.fi(p; x, y, z;

(2)

t),

t> 0, where .4' denotes the mean value of p over the sphere with centre (x, y, z) and radiu3 t; this solution satisfies the initial COflditiO'n8 u = 0, = p when t = 0 and ha8 continuous derivatives of the first and

when

seccmd orders. The

proof uses Boussinesq's spherical potential — —

11

ii s

— x)2 +

—

y)2 +

— z)2}

dS

WAVE MOTIONS

94)

16.3

where S is the sphere with centre (x. y, z) and radius t. If (1, rn, n) are

the direction cosines of the outward normal to 5, T7(x, y, z, t) =

tff

p(x + it, y + mt, z

+

nt)

(3)

is the unit sphere 12+ m2 + n2 = 1. Evidently V has continuous derivatives of the first and second orders, and vanishes when where

Now

V2V(x,y,z,t) = 11'

—

dS

t

V t

t

(4)

=

where D is the =

interior of 5. This may be written

j7

ir'

as

rr

Hence V

1

=

By (4)

=

V2p(x+it,y+rnt,z+nt)dc2. +tII JJ'

tff

V2p(x+it,y+mt,z+nt)dc2 = V2V,

(5)

ci

so that V satisfies the equation of wave motions. Moreover

so that

Since V =

T'(x,y,z,O) =

471p(x,y,z).

(6)

it follows that u = tdi(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 'a = 0

WAVE MOTIONS

6.31

[95

<e, we have = p. We see later that it is unique. Moreover, if UI R1 and when 0 0. u3

=

u4

When rn = lim m

1

1,

'b(t — r cosh

?fr

u2 and u4 are the same. A second solution is then rcz)

rn—i jI o

—r

cosh

— ri_rn sinhim

— r cosh log (r sinh2 = Similarly, since u1 and u3 are the same when rn = 1, the second solution is then I

Jo t Volterra, Hobson,

Acta math. 18

+ r cosh (1894),

log (r sinh2

161.

Proc. London Math. Soc. (1)

22 (1891), 431—449.

WAVE MOTIONS

104]

Exercises 1. Use Poisson's formula to find the solution of the equation Utt—Uxx—Uy1w—Uzz = 0,

given that

when

= x2+xy+z2

u = 0,

t = 0.

2. Use Poisson's formula to solve the wave equation, given that, when = 0, U = 0, and Ut = 1 for x2+y2+z2 a2, Ut = 0 for x2+y2+z2 a2. Examine the discontinuities of the solution.

=

— 3. U iS a solution of Utt — — spherical polar coordinates. Prove that

2av

a2v

Hence

show that

0

of the form v(r, t) cos 0 in

2

a

U = cosO— ar

r

where ç5 and are arbitrary functions. Obtain the same result from the fact that, if U is a wave function, so also is

+yw)_N satisfies the equation

4. Show that (1

—+ a2U

N

axay

/ aU

aU\

I—+—-1=0. ayj

By considering its expansion in ascending powers of w,

prove

that the

equation has homogeneous polynomial solutions

=

p+q=n

(N) (N)

p.q.

(N +p —1). Prove that, if K = Nj and

where

B = max(lxj,jyl)

then

jF,,(x, y)j

(2K),,

5. Prove that Tricomi's equation of mixed type =0 = constant in the elliptic half-plane y> 0 and real characteristics x ± y)1 = constant in the hyperbolic imaginary characteristics x ±

half-plane y < 0. Prove that, if y < 0, the equation can be written as

1 aU a¼ ——— =0, —+— ar2 3r ar a2U

where

r=

t = x.

WAVE MOTIONS 6.

M au N au a¼ axay+x+y ax+x+y ay

If u satisfies

[105 — —

satisfies M and N are real constants, prove that v = (x + the same equation with M and N replaced by 1 —y and 1 —M respectively

where

satisfies the equation of Ex. 6. By

7. Show that u = considering

= r

J c(C_x)N(C+y)M

where

C is the double circuit contour of fig. 7, deduce the solution

=

f

Fig. 7 (after Whittaker & Watson, A Co'urse of Modern Analysis).

when 0 <M < 1, 0 < N < 1. Obtain also the solution U2 =

These solutions are the same when M + N =1. Obtain the second solution

when M+N= 1. 8. If u satisfies the equation of Ex. 6, show that —

x+y

N+lav — = 0. —+ M+lav —+ x+y ay x+y ax ?/zv

satisfies

axay

9. v(x, y,

z; t;

T)

is

-

the solution of the equation of wave motions a2v

a2v

=0 L(v) ——————— az2 3t2

ay2

which satisfies the initial conditions

v(x,y,z;T;T) =0, for

all r.

If

u(x, y, z; t) =

=h(x,y,z;T)

I v(x, y, Jo

z; t; T)dT

WAVE MOTIONS

1061

prove that u satisfies the equation L(v) = h(x, y, z; t) under the initial = 0. conditions 'u(x,y,z;O) = 0, value solution, show that

By using Poisson's

u(x,y,z;t) where

=

R2 = (x—

+(y

+ (z —

= h(x, 1) which satisfies the initial 10. v(x, t) is the solution of — conditions u = 0, = 0 when t = 0. By considering the integral of — over a suitable triangle, prove that 1 tt tx+r u(x,t) = — d'r

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 V¼ 0 which has continuous derivatives of the second order inside and on a simple regular closed and let P0(x0, Yo' z0) be any point inside surface The elementary

solution which plays an important part in potential theory is 1 V

= If we apply Green's transformation to 111 (uV2v — vV2u) dxdydz = 0, JJJ V where V is bounded externally by and internally by a small sphere S with centre (x0, Yo' z0), v being this elementary solution, we obtain 1'!'

au\

/

1'I'

au\

/

JJ

=

denotes differentiation along the normal drawn out of V. If we now make the radius of S tend to zero, we find that

where

u(x0,y0,z0)

lj'j'fau

av\

= which expresses u at a point P0 inside in terms of the values taken by u and au/aN on It does not solve the problem of Cauchy, since we cannot assign u and au/aN arbitrarily on given u on au/aN 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)

u

u

——-——— = 0.

We might expect that an application of Green's transformation to

111

{uL(v)—vL(u)}dxdydt =

0

1

V{(t_to)2_ (x_xo)2_ (YYo)2} [107]

1081

MARCEL RIESZ'S METHOD

[7.1

would enable us to find u in terms of the boundary values of u and 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—x0)2+(y—y0)2

so we take V to be bounded in part by the characteristic cone (x—x0)2+(y—y0)2 = (t—t0)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—x0)2+ (YYo)2

0

t

t0.

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—'

\/{(x—xO)+ (YYo)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

1°

u(x0, Yo' t)

and hence a formula for u. The values of u and bufbN on the character-

istic cone do not appear in the solution because of the properties of Volterra's function. Hadamard, in his 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

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

MARCEL RIESZ'S METHOD

7.1J

[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 a2u

(1)

at

to find u, given the values taken by u and its first derivatives on some duly-inclined n-dimensional manifold S in the (n + 1)-dimensional Euclidean space with rectangular Cartesian coordinates is

(x1,x2,

In the case of the initial value problem, S is 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 (x1, x2, ..., The equation of the characteristic cone with vertex P(x; t) is

I'

=0

(t—'r)2—

a variable point in the space-time manifold. This cone divides space-time into three domains, viz. I'> 0, 'r > t; I'> 0, T < t; where

T) is

I' < 0. The region in which I' < 0 lies outside the cone. I' = 0 is a double cone; the part for which 'r 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, 5) the domain bounded by S and the retrograde cone. In the particular case of the initial value problem, t is positive and V(P, 5) is the set of points T) for which I' 0, 0 T t. The Riesz integral of fractional order of a continuous function / is defined to be t) =

1

...

T) T(

V(P,S)

where t is chosen so that V(P, 5) is not an empty set.

The constant H(cz, n) is chosen so that Then HfrL, n) et =

I

Jy

=

et

eT

t Acta math. 81 (1949), pp.

1—223.

when S is 'r =

—00.

MARCEL RIESZ'S METHOD

1101

[7.2

V is the region I' 0, T t. By a change of origin, we may assume that x = 0; and if we put T = t — we get where

n) =

I

dT',

,J TV

where W is defined by

I' =

0,

T'

0,

and

If we replace (ti'

by spherical polar coordinates and drop

the dashes, we have

f

H(cx, n)

dTf dr

—

f

= is the element of solid angle in n-dimensional Eucliclean

where space, and so

dTf

n) =

Since

the total solid angle

dr

—

this gives

is

n) =

—

+

n), With this choice of t) is an analytic function of the complex variable regular when > n—i. The index notation is used because

=

We prove this first when the real parts of and /3 exceed n +1, and then use analytic continuation. When we do this, all the integrands are continuous. If the real parts of and /3 exceed n + 1, n)H(/3,n)PIfif(x; t)

T')

fV(P,S) where Q is

V(Q,S)

the point

'r) and

F1 =

(T — T')2 —

—

continuous we can invert the order of integration to get n)H(/3, n)

where

g(x, t;

t)

T')

=

=

f

f

T') g(x, t; V(P.S)

T')

When / is. dT',

MARCEL RIESZ'S METHOD

7.2J

The domain W is defined by F> 0, F1> 0, -r'
0,0 < 'r < t.

By a notation of axes, we can replace x in this integral by

(r,0,0,...,0), where r

=x

Then, by a Lorentz transformation

=

cosh y + t' sinh y,

t=

sinh y + t' cosh y

we can replace r by zero, and t by s =

— r2). Hence

g(x,t; 0, 0) = g(0,s;O,0).

But g(0, s; 0,

0)

=

f

{(s — r)2 —

—

where

T

0, g( x,

/3)

F2 = (t — r')2 — I x —

where Hence

t) =

n)

T') V(P,S)

= n), H(/3, n), /3), + /3, n) do not depend on The coefficients the functionf(x; t) nor on the choice of the hypersurface S. Considering the case when f( x; t) = et and S is t = — oo, we see that

= and so

t) =

t) when the real parts of

and /3 exceed

n + 1. By analytical continuation, the result holds when the real parts of and /3 exceed n —1.

MARCEL RIESZ'S METHOD

112J

[7.2

Lastly, when the real part of is suffIciently large, we can tiate under the sign of integration in the equation

=

1

T)

fV(P,S)

t occur not only in the despite the fact that the variables x1, x2, ..., integrand but also in the equations defining the boundary of V(P, 5). FOr,

if h> 0, we have 4-2,

n)

=

f

x;

t + h) — T)

x; t)} —

h

V(P,S) T)

where

is the conical shell bounded by F = 0, and = 0 and S. The volume of the shell is 0(h) and F+ vanishes on the outer boundary of the shell. Hence the second term tends to zero as h—k Oifre > n— 1. A similar argument applies when h is negative. Hence, when re n +1, t)

=

r)

1)f V(P,S)

Similarly, when re

> n +3, t)

=

1)f

T)

T)

V(P, S)

1)

J V(P,S)

r) I

J V(P.S)

MARCEL RIESZ'S METHOD Therefore, when re n +3,

[113

7.2J

1)f

t) =

T)

V(P,S)

= or

This result holds when re > n — 1, since the expressions on both sides of the equation are analytic functions, regular when re n —1. The operators L and I do not commute. For 1 t)

f

=H

where

V(P,S)

v a2

na2

vT

1VSk

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 re > n +1, but the integral over the relevant part of S does not in general vanish. In what follows we write for 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> 0. In this ca8e, t)

=

where Vis the region on which 1' this becomes =

where

(1)

T)

1

0,0

'r

t. If we write

= (2)

r2 =

and W is the hemisphere 0 r t, 0 in the (n+ 1)-dimensional Sometimes it is space with Cartesian coordinates ..., more convenient to use polar coordinates

MARCEL RIESZ'S METHOD

114J

[7.3

1. In polar coordinates, (2) becomes

where

Pf( x; t)

drf dof

=

x + ir sin 0; t — r) 0

x

where 1 is the vector (1k, 12, ..., in) and

0,

(3)

is the element of solid angle

space. From this it follows at once that Pf(x; t) is an analytic function of regular in re > max (0, n — 1), on the assumption that f( x; t) is continuous. To extend the region of regularity, we assume that / is continuously differentiable to a suitable in

order. Since

—

=

n + 1) x;

we may write (2) in the form

+

t)

t

- r)

= 2) Integrating by parts, we obtain, when re

1

> n — 1,

I

where

is

+

= t2,

—

0

= pn_ldpdcl = Then, on The last term in (4) is =

+ 2)f

tn

4

t.

= x + ip, where p

On

't

C1

t sin 0.

0 cos

dOfdIln/( x + it sin 0; 0)

0

0.

When t> 0, this is an analytic function regular when re x> n —3. The term involving /t' which by hypothesis is continuous, is rt

Idri

Jo

x

dO

x + ir sin 0; t — r)

0

0

max (—1, n— 3).

The remaining term is J1

= x

which

is regular when re

max (0, n — 3).

0

0

MARCEL RIESZ'S METHOD

7.3J

(i) When n =

2, 2irJ3 is

[115

equal to

l)f: iosinocosof

dçbf(x+1tsin0, 0),

(x, y), I = (cos çb, sin çb) so that = dçz5. This is an analytic function of when t > 0, regular when re — 1. Hence J3 is zero when = 0. Next, 2irJ2 is equal to

where x =

of

dO sin0

F(+

dcbft(x +Ir sin 0; t — r).

This again is regular when re —1; and vanishes when Lastly, when re 0, 2irJ1 is equal to

F(+ 1)f drrfdosino cosxof

=

0,

dqSf(x+IrsinO; t—r).

Integration by parts is valid since / is continuously differentiable. Hence 2irJ1 is equal to of27T

sin0

—

dçz5/(x +It sin 0; 0)

i)f drrf dOsinOcosOf

which is regular when r27T I

JO

JO

—1. When

dqS

is zero, 2irJ1 becomes

dçb/(x+ItsinOsinO;0)

= JO dOsinO JO

Therefore, when n =

2

Riesz's integral

y; t) is regular in

= 2irf(x;t).

and /(x, y; t) is continuously differentiable, I°/(x, y; t) = /(x, y; t).

— 1, and

MARCEL RIESZ'S METHOD

1161

(ii) \\Then n =

and re

3

[7.3

0,

= where us the vector (sin cos sin sin 1ff is continuously differentiable, we

cos

and

= sin

4 dVi.

may integrate by parts with

respect to 0 and obtain =

dOf

±

cosa 0

{sin Of( x + it sin 0; 0)}. — 1 and vanishes when = 0. To deal with J1 and J2 we make the additional assumption that / has continuous derivatives of the second order. When re 0, we integrate by parts in

Hence J3 is regular when re

1

I't I

o

drl

Jo

dOI

Jci,

x

Hence

is also regular when re x> — 1 and vanishes when

respect to 8 in J1=

1

drl

0.

dOI

Jo

x f( x + ir

Jc23

sin

0; t — r)

sin2

drf dOf

±

F=

where

t't I

and obtain

=

=

0, we may again integrate by parts with

Lastly, when re

0

0

cosa OF

(f(x + ir sin 0; t — r) sin 0),

since, by hypothesis, F is continuous. In fact, F is continuously differentiable, so we may integrate by parts with respect to r, to get =

+

dOf

F(i +

dof

MARCEL RIESZ'S METHOD

7.31

Thus J1 is regular when re

1

0,

dof

dOf

=

=

—1. When

[117

r

do1

dO

=

{f( x; t) sin O}

=f(x;t). Thus when f( x; t) has continuous derivatives of the second order, Jaf( x; t) is an analytic function of z, regular in re

—1, and

I°f(x;t) =f(x;t). (iii) The argument can be repeated for larger values of n. Iaf( x; f)

is regular in re

—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 Lu

— at2

aau

aau

ax2

ay2'

Lu = F(x, y; t),

the problem is to solve

(1)

when t> 0, given that

u =f(x,y), ut = h(x,y), when t = 0. Heref, h and 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 ij, 'r, it is convenient to write A

Let P be any fixed point (x, y; t) in space—time with t> 0, and let V be the region I' 0,0 r t where

I' =

MARCEL RIESZ'S METHOD

1181

[7.4

By Green.s transformation r)

f =

/

by

J st \

ciT

1'

1

I

r)

r) —

/

au\

r)}

/ \

au\

av

arj

av

udS

ci;iJ) alJ \ where 8 is the boundary of V and (A,,u, v) are the direction cosines of

the normal to 8 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 = Then Au =

r), and Av = —1)

f =

f

Then, when re

1)

3,

—

y) + v(r— t)}

1)

— vu7}] dS. + + Since re > 3, the integral over the retrograde cone part of S vanishes; the rest of S is specified by

On

A

=p =

f

r = 0,

I'0

v=

— 1.

0,

0.

Hence

—1)

—

f

+

—1)

tu}

= Using the initial conditions, this is, in Riesz's notation,

JauJa+2F

')f

= +

+1)

LL

Now the expressions on each side of this equation are analytic func— 1; although we have proved the tions of regular when re result only for re > 3, it is true in this larger domain. In particular. taking = 0, we have u(x,y; t) = 12F(x,y; 1

MARCEL RIESZ'S METHOD or

r)

u(x, y; t) =

—

[119

r)2— (x—

(y {t2

- (x-

(2)

When F is identically zero, this reduces to the solution found in § 6.4.

Ifweput the first term in (2) becomes

= x F(x +p cos 1

277 o = —J

r) {(t —

i'2ir

drJ

x

y +p sin

o

o

F(x + p cos

y + p sin

t — r)

{r2

If we then put p = tp', r = tr', we get t1(x,y;t) = F(x+tp cosçb, y+tp sin (3) dropping the dashes. This representation as a repeated integral is valid since F is, by hypothesis, continuous. The expression on the x

right of (3) is continuous in t 0. If F(x, y; t) has continuous derivatives of the second order, so also has 11(x, y; t); and and at1/at vanish when t =0.

The second term in (2) is C12(x,y;t) =

which vanishes when t =

=

rJ'pdpf

0.

Since h is continuous and

MARCEL RIESZ'S METHOD

1201

[7.4

exists and is equal to 1

If h has

C'

pdp \/(1_p2)Jo 4 = h(x,y).

continuous derivatives of the second order, so also has

y; t). The third term in (2) is

= 1

ri I

differentiation

jo

under the sign of integration being certainly valid

are continuous; and then y; t) may have continuous partial derivatives of the second order, it suffices that / should have continuous derivatives of the third order. We have thus proved that equation (2) solves the Cauchy problem when

the partial derivatives

and

y; 0) = f(x, y). In order that

for

a2u

a2u

a2u

= F(x, y; t);

when F(x, y; t) and h(x, y) have continuous partial derivatives of the

second order, /(x, y) of the third order, the solution (2) has continuous second order derivatives. 7.5

The equation of wave motions in three dimensions

The Riesz solution for Lu

a2u

a¼

a2u

= F(x,y;z,t)

(1)

in three spatial dimensions starts like the corresponding solution in two dimensions. The initial conditions are

u =/(x,y,z), Ut = h(x,y,z), when t =

(2)

0. Here /, h and F are continuous functions, differentiable to orders which suffice to ensure the existence of a solution with

continuous second derivatives.

______

MARCEL RIESZ'S METHOD

[121

With any fixed point (x, y, z; t) in space—time, let V be the region in 0, where

r? 0 bounded by I' = 0 and r = I, =

If u and v have continuous second derivatives, Green's transformation gives

r)

f

r)

r) —

('1 1ev

=I

\

aT

r)}

eu\

1ev

eu

arj

\ /

au\

au\)

I

where S is the boundary of V and (A, 4u, w) are the direction cosines of the outward normal. As before, Au = u be the desired solution of our problem, and let v=

Av =

Then

We assume in the first instance that re x> 4, and then use analytical

continuation. We have

2)

Js

—

x)

— y)

z) + w(r — t)}

+ eu

I

euljdS.

eu

If we divide through by = we obtain Jau(x, y, z; t) _Ia+2F(x, y, z; t) —

I

I

°

aT

+ (z— 2)

where and

(z— I'0 = t2— (x — (y is the region of the plane -r = 0 on which 0

0))

t2.

The reason for this is that when re x> 4, the integrand vanishes on the part of S belonging to the retrograde cone. On the rest of 5, 5

CPD

1221

we

have r =

0,

[7.5 MARCEL RIESZ'S METHOD A = = v = 0 and w = — 1. Using the initial condi-

tions, y, z; t)

z; t)

= ij,

(3)

The expression on the left is an analytic function of regular when re — 1 when u has continuous derivatives of the second order; its value when = 0 is u(x, y, z; t) — 12F(x, y, z; t).

If we introduce spherical polar coordinates

?)=y+mr, = sin 0 sin the first term on the right of (3) is where

1 = sin 0 cos

2)5:

=

r2drf

where

n

m

cos 0,

h(x + ir, y + mr, z +

—

dfI3 = sin

and

Integration by parts gives 1

1'

rt

d

= =

dr (t2 — r2)a12 {h(x

+

+r which

+lr, y + mr, z + nr)

h(x + ir, y

+ mr, z + nr)},

is valid if h is continuously differentiable. Now = =

Hence

= = =

47r.

MARCEL RIESZ'S METHOD

where Ji(h; x,

z;

t) is the mean value of

over the sphere

iii,

=

[123

t2.

A similar expression follows for the second term on the right of (3). Lastly, by §7.3 (2), Jz+2F(x,y, z; t) =

where

ij,

t

= r2+r2

B2 =

with a slight change of notation. Here W is 0

B

t, -r

integrate by parts with respect to r, r varies from 0 to then have :i '

'

r

1

=

r.t_R\1v(P--r2) / B Jo

I

I

+

/

0. When we — r2). We I

vL

— B) —

where V is We

+ (y —

(x —

+

(z —

t2.

have transformed Ia+2F(x, y, z; t) into a function regular when

rez> —1.Hence 12F(x,y,z;t)

=

B

47T

J'V(P—r') a

It follows that u(x,y,z; t) =

=

if

t—B) t—

t)

dr}

r)

x,y,z; t) (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 f(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

MARCEL RIESZ'S METHOD

1241

[7.5

The solution of (1) which satisfies the conditions u =

t= —ais U(X,y,z;t)

where

is now

=

If

+

we

=

0

on

r

— x)2 + (ij — y)2 +

If we make a

0,

(t + a)2.

z)2

get the well-known retarded potential t

U(X, y, z; t)

= 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 a2u

a2U

a2u

(1)

The theory of this equation is well known, but rather difficult. We 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

vanish when t =

0,

J2Lu = LI2U =

U.

Hence

U + k2I2U = 12F.

This implies that

=

+

Omitting convergence considerations, U = (12—k214+k416— ...)F.

But y, z; t)

where

=

I' =

(t — r)2 — (x —

and V is the region on which I'

r

r)

(r— — (i'

0,0

-r

—

— (z —

t. This formula holds when

2. Hence U(X,y,Z;t) = 12F(x,y,z;t)

+

1)!

r)

MARCEL RIESZ'S METHOD

7.61

[125

Jfwewrite B2 = I' = (t—r)2—r2,

this becomes

u(x,y,z; t) =

1f —

where W is the sphere 0 r t, and J1(lcR) is the Bessel function of order 1. The form of the solution used by Bhabha differs slightly from this;

his result can be obtained by assuming that u and Ut vanish on t=

—a

and making a

+ cc.

7.7 A mixed boundary and initial value problem The problem is to find the solution of a2u

(1)

in y> 0,t> 0, given that U =f(x,y),Ut = h(x,y) on t = O,y ? 0 and t) on y = 0, t 0. The data are assumed to be continuous, U= and, in particular,f(x, 0) =

0), h(x, 0) =

0).

t) We may assume that 0. For, if not, the substitution t) leads to the same problem with different F, / and h U= v+

and with v zero on y = 0, t> 0. We wish to find U(X, y; t) when y> 0, t> 0. The retrograde teristic cone — — (t —

r)2 — (x

—

(y

=0

cuts the plane r = 0 in the circle = t2. If t y, the disc with this circle as boundary lies in ij 0 on which u and Ut are given, so that the solution is that given in § 7.4. But if t> y, the circle crosses ij = 0 and bounds a disc on part of which U and Ut are unknown. From now on, assume that t> y > 0. As in § 7.4, — x)2 +

— 1)

(ij

— y)2

—

1)

—y)+

—t)}

+ F(a-l)12{AUg +/LU,. —

MARCEL RIESZ'S METHOD

1261

[7.7

where V is the region bounded by S on which I' = 0,7/ = 0 and r = 0, v) are the direction cosines of the normal to S drawn out of and V. Since t > y> 0, S consists of three parts. The part of S belonging to the retrograde cone contributes nothing when re > 3. Next there of a disc in r = 0, defined by is a segment — x)2

Lastly there is a region

+ in

t2,

—

= 0 defined by y2,

= 0, v = — 1; on , A = , A = on r = = ditions u r ? 0, we have

0

r

t—y.

= 0, ,u =

On

l)fv

0.

7/

— 1.

Using the con0 on = 0,

0, and u =

0,7/

7/;

r)

where

I'0

=

(2)

I'i =

t) is the unknown value of at (x, 0; t). We get over the difficulty that is unknown on y = of reflection. Let — I' = and

0

by a method

I' = 0 is the characteristic cone with vertex (x, — y; t). The retrograde part of T = 0 cuts off from V a region V1 whose boundary Then

is part of the retrograde cone, the same area ment defined by t2,

On

u

rex> 3,

= f, —1)

=

/t.

L

0.

Repeating the argument, we find that, when r) -4-1)

is equal to

y

on y = 0, and a seg-

dr r)

dr

MARCEL RIESZ'S METHOD 27TF(cx + 1)

{J'

ij)

[127 ij)

+

r)

—J

(3)

r0 = = = If we subtract (3) from (2) the terms involving cancel. The expressions which appear in (2) and (3) are analytic functions where

of

regular when rear.> —1. LetV2andV3bethepartsofVonwhicht—y

Then

1

r)

—1) J'v2

is Riesz's integral of order

-r

tandO

-r

t—y.

dr

corresponding to the initial plane -r = t—y;

its limit as cx-.O is u(x,y;t). But the limit of F(

is zero since (x, y; t) is outside V3. Similarly the limit of

r)

1)5v1

is zero.

It follows from the limiting forms of (2) and (3) that

u(x,y;t) =

+ d

d

This is precisely the result we should have got if we had extended the range of definition of F,f and /t to y 0.

So V2U is not positive anywhere on D; similarly it is not negative. Hence

V2U =

All we have proved is that, if there is a function U belonging to for which 1(u) attains the infimum m, then U is harmonic. The assump-

tion that there is a function which minimises the integral 1(u) was called by Riemann Dirichlet's 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 (x0, Yo' 0) of D has electrostatic potential

where

1?2

= (x —x0)2 + (y

POTENTIAL THEORY

146]

[8.7

If the conducting cylinder is earthed, a surface charge of density ois 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 cylinder has potential U

u +2 e log

The

total field inside the

1

The electric force is grad U. On first sight, we should expect on physical grounds, that U would have continuous first derivatives on D. But since U is zero on the conductor, the electric force is normal to the conductor and has magnitude As we go round 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 either tends to zero or to infinity as we approach the corner. Since the charge density o- is equal to (1/47T) the total charge per unit length is 1 U

may be infinite at a and this is equal to — e. Thus, although finite number of points of t9D, it is integrable. The potential U, with e = is called the Green's function, and is denoted usually by G(x, y; x0, 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. consists of a The same physical argument can be applied when 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 If u is continuously differentiable in and if u has continuous second derivatives in D, then u(x, y)

—

R) ds +

?J)log R

ID when (x, y) is a point of D, provided that the double integral exists: here B2 =

POTENTIAL THEORY

8.81

[147

If v is harmonic in D and continuously differentiable in D, we also have

1

Subtracting,

u(x,y) =

r

au\

/ av

1

j'j'

ir ía

au —

ds — log R}

lID

Now suppose that D has a Green's function = which vanishes when u(x,

=

ii) is a point of

With this value of v,

ds —

—

G is continuously differentiable with respect to in D, or, more generally, if is integrable round

If u(x, y) satisfies Poisson's equation V¼ =

— 2iro-(x,

and

y), this be-

comes u(x,

+ff

=

x,

Presumably this would give the solution of Poisson's equation in D It is necessary to prove the existence of when u is known on Green's function and to show that u(x,

ds

x, y)

= tends to f(x0, Yo) as (x, y) tends, on D, to any point (x0, Yo) of The corresponding solution for Laplace's equation is u(x,y) =

(2)

(3)

Properties of Green's function Let D be a bounded domain whose frontier consists of a finite number 8.9

of non-interesting regular closed curves and which possesses a Green's function G(x, y; x0, Yo) = v(x, y; X0, Yo) —logR,

where

R2 = (x—x0)2+(y—y0)2,

POTENTIAL THEORY

148]

[8.9

(x0, 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 Li and continuously differentiable in round is — The integral of This follows from equation (1) of § 8.8 if we put u 1. The physical has corners. argument suggests that it holds even if For every pair of points (x, y) and

(x0,

Yo) of D, G(x, y; x0, Yo) > 0.

For values of e which are not too large, the closed disc B e lies in D. Let C0 be the circle B e, D0 the domain obtained by deleting

the disc from D. Regarded as a function of (x, y), G is harmonic in D0, continuous in On G is zero; on C0, G is positive since G-÷ + oo as R-÷ 0. By the maximum principle, G attains its supremum and infimum in on the boundary of Hence G is strictly positive when (x, y) is any point of D0. Since e can be as small

as we please, this proves the result. The Green's function possesses the symmetry property

G(x,y;x0,y0)

G(x0,y0;x,y)

for every pair of points (x, y) and (x0, y0) of D.

Write G(x,y;x0,y0)

G0(x,y),

G(x,y;x1,y1)

G1(x,y).

Let R0 and B1 be the distances of (x, y) from (x0, y0) and (x1, y1) respectively. Let C0 be the circle B0 = 8, C1 the circle B1 = e, where 8

and e are such that C0,C1 and do not intersect; then the discs B0 < 8, B1 < e lie in D. Since G0 and G1 are harmonic in the domain bounded by C'0, C1 and

=

0.

The third integra' vanishes since G0 and G1 vanish when (x, y) is on Near (x0, y0), G1(x, y) is continuous, but G0 is of the form v(x, y; x0, y0) —log B0,

where v is harmonic. It follows that the first integral tends to —

2irG1(x0, y0)

as 8-÷ 0; similarly the second integral tends to 2irG0(x1, y1) as e —÷ 0. Hence G1(x0,y0) = G0(x1,y1).

POTENTIAL THEORY

8.101

[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 which takes on the boundary the same values as f(x, y), in 8.10

then

U = u—f(x,y) and satisfies Poisson's equation

vanishes on

V2u = —2iro(x,y),

where o = V2f/4ir is also a polynomial. By (2) of § 8.8, we have U(x,

x, y)

(1)

= Hence, instead of (3) of § 8.8, we have a different formula

u(x,y)

(2)

for the harmonic function. i;i; x, y) vanishes when (x, y) is a point of by the symSince metry 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 for, if this were so, u(x, y) would tend to f(x, y) as (x, y) moves up to the boundary. The Green's function = we write as

x, y) =

ii;x,y) —loga}

where a is a constant greater than the diameter of D. On

v—

log R

is zero, and so v—log a is negative. By the maximum principle, v — log a

is negative on D. Therefore 0

of for all points Let (x0, Yo) be any point of

be the U(x,

LID.

D

of D. Then x,

ff

x,

POTENTIAL THEORY

1501

There exists a constant K

Since o is a polynomial, it is bounded on such that JoJ Kthere, and so U(x, y)J

Kff

[8.10

+ Kff

x, y)

x, y)

Suppose that (x, y) is a point of D0. The integral over D1 tends to zero as (x, y) moves to (x0, y0) on D. Hence

limsup

J

U(x, y)J

lim sup Kff

(x,v)—+(x0,v0)

D0

But, if D2 is the disc R < 2e, D2 contains D0 and so

fJ log

log 2e).

As e can be as small as we please, U(x, y) tends to zero as (x, y) moves By (2), u(x,y) tends tof(x0,y0). to (x0,y0) on We can write (1) in the form U(x,

=

+ff

ffD

x,

The first term is a logarithmic potential of the type considered in § 8.2. It is continuous and continuously differentiable everywhere. Moreover, if o 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 x, y) is a harmonic function of (x, y) in D. Regarded as a function of ii) it is harmonic in D and continuous differentiable in save possibly at corners of So the second term is harmonic in D. Hence V2U = — 2iw-. The restrictions onf(x, y) could be lightened. For example, it would suffice if f(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 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, =

POTENTIAL THEORY

8.111

[151

where V is a harmonic function. Guided by known results in electrostatics, we expect that G(P;P0)

where P1 is the point inverse to P0 with respect to the circle C.

Now if Q is any point on the circle C, the triangles OP0 Q and OQP1 are similar, and so P0Q/P1Q = 0P0/OQ.

Therefore

=

G(Q;

+ constant,

which is zero if the constant is log 0P0/OQ. Hence

P1P.oPo

G(P;P0) = log

Let P, have polar coordinates (p, ordinates (a2/r, 0). It follows that

(r, 0); then P1 has polar co-

a4—2a2prcos(0—Ø)+p2r2

G(P'P) =

It can be readily checked that this function has all the required properties. On C,

is the limit of aG/ap as p

a.

Hence at the point Q(a,

a2—r2

=

at

The function, harmonic on the disc, which takes the value the point (a cos asin of C, is therefore 1

r21T______________

u(rcos0,rsin0) =—--(a2—r2)I 2ir

Jo

2'

This is known as Poisson's integral. It reduces to Gauss's mean value formula when r = 0. Now suppose that P0 lies in the semicircle 0 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) = under suitable conditions. The discussion must be deferred until after

we have considered functions harmonic on an unbounded domain. function for the quadrant Similarly, the 0,11> 0 with singularity (x, y) is 2

again by the method of images. G is a harmonic function of ?j) on any domain which does not contain any of the points (± x, ± y), is continuous on the and so is continuously differentiable. 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 0 and 2ir. It can be shown, by using Sommerfield's multiple-valued

___________________

POTENTIAL THEORY

[153

potential,t that the Green's function for the cut plane is =

p + r —2

cos

+

= 0 and when = It has the correct logarithmic singularity when the point of polar coordinates (p, This vanishes when

moves to (r, 0) since

+r

cos

= It also has a logarithmic singularity when p = r, cos + = 1. Now when = 1 = 4n7r cos —0 where n is an integer or zero. As + are restricted to lie between 0 and 2ir, this is impossible; the 0 and Green's function has but one singularity in the cut plane. The origin is a 'corner' of the boundary of angle 2ir. When p is are both of the order O(p4) and so tend to small, aG/ap and p' infinity as the point of polar coordinates (p, moves up to the corner. 8.12

Poisson's integral

Let u(x, y) be harmonic in a domain D. If K is an open disc with centre (x0, Yo) and radius a, which is contained in D, then if (x, y) is any point of K, u(x,y) where

and

1 I

u(x0+acosq5,y0+asinq5)P(r,0—q5)dq5, o

x = x0+rcos0, y = y0+rsin0 (r < a) a2 — r2 —. P(r0)= a2— — 2arcos0+r2

(1)

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 We now ask any point of the disc K and the known values of u on what properties r2ff 1

LiT j

f(çb)P(r,0—q5)dçb

(2)

o

has when/is an arbitrary function. By a shift of origin, we may suppose x0 and Yo to be zero.

The simplest case is when

is a continuous periodic function of

t A brief account of Sommerfeld's multiple-valued potentials is given in Jeans, The Mathematical Theory of

and

5th edn (Cambridge, 1925),

pp. 279—283. 6

CPD

POTENTIAL THEORY

1541

[8.12

period 2ir. We need the following properties of the Poisson kernel P(r,O): (i) for every value of q5, P(r, 0—

is harmonic in r < a, P(r, 0)> 0 when r < a,

(ii)

1'21T

(iii)

I

P(r,0—çz5)dçz5

JO

=

2ir.

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 0, the solution may not be unique. For example, if A = Ic2 and D is r < a, there is a solution u

sin Icr

r a if ka = Thus there are values of k for which the problem of Dirichlet on r a does not have a unique

which vanishes on r = solution.

The mean value theorem If u is a solution of V2u + k2v = 0 which has 11.11

continuous

second deriva-

tives in a domain containing (x — x0)2 + (y — y0)2 + (z — z0)2 mean value of v over (x —

+ (y — yo)2 + (z — z0)2 = sin H?

u(x0y0,z0).

R2,

where R

a2,

t/i

a,

ELLIPTIC EQUATIONS IN SPACE

11.111

[223

This reduces to Gauss's mean value theorem when Ic = 0. We may take (x0, Yo' z0) to be the origin, and use spherical polar coordinates (r, 0, ç6) so that x = lr, y = mr, z = nr, where 1 = sin 0 cos

m

=

n

sin 0 sin

denotes the whole solid angle at 0;

= sin OdOdçb.

The mean value of u over the sphere 1

I(r)= Hence

udS =

dl

where theorem

x2 +

y2 + z2 =

r2,

where

1

+ muy +

=

cos 0.

dS,

=

is differentiation along the outward normal. By Green's

dl

=

where integration is over

dl — = dr

+

+

rr

k2

4rrrjo s2ds II

————i

k2

I

r2. Therefore u(is,ms,n.s)dcA

s21(s)ds.

=

dl

Therefore

— r2 — = — k2r21

dr

dr

d21

dl

dr

dr

or

r2 —i + 2r — + k2r21 = 0.

Hence

I=

(A cos kr + B sin kr).

But Itends to u(0,0,0) as r-÷0, and so A =

0,

I = sin kr The corresponding result for V2u —

=

0 is

1= sinhkr kr u(0,0,0).

and

ELLIPTIC EQUATIONS IN SPACE

224]

[11.11

From this it follows that a solution of

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 u vanished on bD, it would be identically zero on D. 11.12

The solution of V2u—k2u = 0 in polar coordinates

Suppose that we wish to solve the equation in the domain r < a in polar coordinates, given that u = f(O, when r a. Sincef(O, can be expanded as a series çf), 0

where (0, is a surface harmonic of order n, we try to find solutions of the form where R depends on r alone. The differential equation in polar coordinates is Cr2

2'u = b2u + cot 0 bu +

where Since

1

2'u = rör — k2u + — r2

+—

oO

=

— n(n +1)

dr2

If we put R =

0,

b¼

1

sin2 0

Sn, the corresponding function R satisfies

rdr

r2

R—0 —

S/\/r, this becomes dr2

rdr

r2

5—0 —

which is Bessel's equation for the functions of purely imaginary

argument. It has the independent solutionst

I (z) =

m=om!F(v+rn+1Y

Hence we have two solutions

t See Watson, A 1944), p. 77.

on thc Theory of Bessel Function.s, 2nd edn (Cambridge,

When n = 0,

ELLIPTIC EQUATIONS IN SPACE these solutions are constant multiples of

sirihkr

kr'

[225

coshkr kr

Now 4(z)/zL' is an integral function of the complex variable z and has no real zeros. If z is small, and behave like multiples of and respectively. As the solution is to be finite at the origin, it cannot contain the function Hence if = where

is a surface harmonic of order n, V(ka)

— o

V(kr)

S (0

There is no restriction on a, since 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 tend to infinity exponently as and But =

(— 1

—

\2kr) as

The required solution of the external problem is then K "ica' '

u=

4j(kr)

o

2

Again, there is no restriction on the value of a since

has no

positive zeros.

Lastly, if the function

R2 = (x—x0)2+(y—y0)2+(z—z0)2,

cosh kR/R

is the elementary solution of V¼ — k¼ = 11.13

0.

The solution of V2u+ k2u = 0 in polar coordinates

If

=

the solution of the problem of Dirichlet of the equation V¼ + k¼ = 0 for r 0 and c < 0 are both covered. F. V. Atkinsont proved that, when Ic 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 u be such that (i) V2u + k2u = 0 outside S, (ii) u is continuously differentiable on D, (iii) u has bounded continuou$ second derivatives on D, 1\ I.

(iv) rexp(ilcr)

tends to zero as r —+ co uniformly with respect to the polar angles 0 and

Then u can be expressed in the form

u = exp (ikr) where the coefficients are independent of r and are continuou$ functions of directicrn. Jlloreover, there exists a constant a such that the series is absolutely convergent when r a and uniformly convergent there with

respect to 0 and 0. This is not an expansion in terms of the Hankel functions of order and surface harmonics of order n. A typical term exp (ilcr)

n+

does not satisfy the reduced wave equation. If y0, z0) lies outside S, it follows by the argument of § 11.14 that, if u satisfies condition (iv) of the lemma u(x0,

z0)

=

1

jj

/

b

exp(i/cR)

—

exp(ikR) 9u

dS,

—

where R is the distance from to the integration point P(x, y, z) on S and n is the unit normal vector drawn into D. If OP0 = r0, OP = r, then R2 =

Phil. Mag. 40

(1949), 645—6M.

ELLIPTIC EQUATIONS IN SPACE

[231

where Vi is the angle P0 OP. This gives

R=

c°s Vi +

—

where = 1/r0. If we suppose, more generally, that is a complex variable, R has a simple pole at = 0 and branch points at

= exp( ±iVi)/r.

We then have

exp(ikR)/exp(ikr0)

=

(1 —

x exp

{( 1—

cos

+

If we take the branch of the square root which is equal to unity when = 0, the expression on the right of this equation is an analytic function of regular when < 1/r. But S lies everywhere at a finite distance from 0, so that r say, on S. The analytic function is then certainly regular when < 2/a, and so can be expanded as a power series, which is absolutely convergent when 1/a, the convergence being uniform with respect to r and It follows that

(exp (ikr0)\'

exp (iJcR)

/ us

R

an

dS

can be expanded as a power series

where the coefficients depend only on the direction of OP0 and depend on it continuously. Hence, when r0 a, =

A similar argument applies to 11

r JJs and the result of the lemma follows. Note that it was not assumed that Ic> 0; the result holds for all real or complex values of Ic.

ELLIPTIC EQUATIONS IN SPACE

232]

[11.15

Lemma 2. Letube such that (i) V2u+k2u =

0

outside S,

—

(ii) u is continuowsly differentiable on D,

(iii) u has bounded continuou$ second derivative,s on D. If im k 0, the conditions (A)

r

are equivalent. They are also equivalent to

u —÷0,

r

I.

—÷ 0

as r

and to

—+

—

(D)

ru=O(1),

=O(1)asr-÷cc.

p + iq, where p and q are real and q (B) holds. Then

0. Suppose that

Write Ic =

rexp (i/er) =

jf.

1\

=

r3exp(_ikr){(iIc_!)

so that (A) holds. If (A) holds, we can apply the result of Lemma, 1, to obtain If. 1\ I Iexp (ikr) r

a —

the

—'p' —

series being absolutely and uniformly convergent when r

Hence (B) holds. (C) implies (A)

since

exp (ikr) J = exp

(

qr)

u= when r

a, and so v—÷O as r—÷co. Also 1.

.

—f,

a.

1. If (A) holds, then

[233 ELLIPTIC EQUATIONS IN SPACE which tends to zero as r —÷ co Thus (A) implies (C). Similarly we can show that (D) implies (A) and (A) implies (D). And thus (C) and (D) are also equivalent. Thus when im Ic ? 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 uiiiqueness theorem for the external problem of Dirichlet. We assumed that Ic = p + iq, where q 0. There are three cases to consider, viz. 11.151

(i)p + 0,q >0, (ii)p = 0,q>0, (iii)p + 0,q =

0.

(i) Suppose that p + 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 be the domain bounded externally by internally by 5. If 1Z is the complex conjugate of u,

(u

fff 1Z

—

vanish on S. The expression on the left is

= 4ipqffflu2ldxdydz.

u = exp (ipr — qr)

By Lemma 1,

r

= exp ( — ipr — qr)

when r

a, and so —

—

r —* x. Therefore the integral over

(

— 2qr)

tends to zero as b -* x, and

so,sincepq+0,

Ill u vanishes everywhere on D.

(ii) Suppose, next, that p = 0, q> 0 so that k2 = — q2. Let u be a solution which satisfies the differentiability conditions of Lemma 1 and vanishes on S. Then

fff

+

u

[11.15

ELLIPTIC EQUATIONS IN SPACE

234]

because u vanishes on S. Since

0(exp(—2qr) tends to zero as b —÷ co. Therefore

the integral over

as r —+

= 0,

and so u is again zero everywhere on D.

(iii) Lastly suppose that p 4 0, q =

0.

If u is a solution which

vanishes on S and satisfies the differentiability conditions,

(u

JJJ

--

The triple integral vanishes. By Lemma 1,

r

r

1

If we substitute this in b21( and make b —÷ co, we find that Ja1J2 = 0. Proceeding in this way, we

find that every coefficient where on

is identically zero, and u vanishes every-

Exercises 1.

(i) Prove that, if 0 < h < 1,0 < k < 1, —

J —1 \/(1

— 2h,u +

(ii) The

1

h2) \/(1 — 2kp ± k2) —

1

og

1I(hL)

polynomials have the generating function 1

Deduce from (i) that

5

5

= YhflP(/i) dji

= 2n+1

4 = 0 (rn +

[235 ELLIPTIC EQUATIONS IN SPACE 2. u is harmonic in a domain containing r a in polar coordinates. On = cosO. Prove that = a, u

u=

a

o

=

where

3.

Provethat

2n+1 I" —2

—.

I

z+

J

J du cos u +

2rr

.

.

sin u

=— r

where z > 0. Deduce that, if r < 1, du 1 —z-—ixcosu—iysinu = Jo = cos 0. Hence show that

where

2rr

1

= —I 27rj Deduce that

1.

4. If

= (x—x0)2+(y-—y0)2+(z-—z0)2, = (x—x0)2+(y-—y0)2+(z+z0)2,

where z0 > 0, prove that is

—

the Green's function for the Laplace's equation in z > 0. Show that ----i-- =

where

cosfi = —cosOcosO0+sinOsinO0cos(çb---çb0), çb0), (r, 0, çb) being the polar coordinates of (x0, Yo' z0) and (x, y, z). Show that the series is absolutely convergent in r a, for any a > r0, and that the convergence is uniform with respect to 0 and çb. If u is harmonic in z > 0, prove that

(r0,

u(x,y,0)dxdy

z0

u(x0,y0,z0) = — I I 2irjj

provided that

as r

2u

i— + — -÷0 c'r

r

oo uniformly with respect to 0 and çb.

ELLIPTIC EQUATIONS IN SPACE

236]

5. Show that, in spherical polar coordinates, Laplace's equation is 1

/

/

1

1

.

=0.

— (r2 + I + —i----- — I sin 0 cOj r2sm2O r2 Cr \ Cr) r2sinO \ —

By the method of separation of variables, obtain the solution cos

where p =

satisfies

cos 0, and

dO

d2®

6.

m21

F

0=

0.

Show that, in cylindrical coordinates, Laplace's equation is 1

1

—0. —+------+——+— p

By the method of separation of variables, show that a solution is (CJm(kp)+DYm(kP))

+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

(x — x0)2 + (y — Yo)2 + (z —

a closed sphere contained in D. If (x, y, z) distance r < a from its centre, prove that

be

a(a—r)

(a+r)2

u(x0,y0,z0)

a2 is

a point of the sphere at a

a(a+r)

u(x,y,z)

(a—r)2

u(x0,y0,z0).

Deduce that a non-negative function harmonic in every bounded domain is

a constant. 8.

Prove that Harnack's first and second theorems on convergence. in § 8.14 and § 8.16 for plane harmonic functions, also hold for

proved

sequences of harmonic functions in space of three dimensions. 9•

If

= (x — x0)2 +

(Y — Yo)2+ (z — z0)2,

r

— =— I 1+ R0 r

prove that

L

where

,

=

cos 0 cos

and (r, 0,

00+ sin 0

sin 00 cos

—

being the polar coordinates of (x0, Yo' z0) and 1 and that the series is absolutely convergent a for any a> r0, and that the convergence is uniform. 00,

(x, y, z). Prove

when r

rj

a function of cos

alone, (r0,

'

that

ELLIPTIC EQUATIONS IN SPACE If k> 0, z0> 0, show that R0

where

[237

R1

= (x_xo)Z+(y_yo)2+(z+zo)2,

is the Green's function of the equation V2u + k2u = 0 for the half-space > 0. Hence show that if u is continuously differentiable on z 0, has bounded continuous second derivatives on z> 0 and satisfies V2u + k2u = 0, then i/CR (1— ikR) u(x, y, 0) dxdy, u(x0, Yo' z0)

=

where

providedthat

B2 = U

as r -* oc, uniformly with respect to 0 and ç5.

-±0

12

THE EQUATION OF HEAT 12.1

The equation of conduction of heat

When 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

Since K,

azu

—

Pc

ax2

—

K

at

p, c are constants, this can be written in the form a2u

au

ax2 =

at

by a change of time-scale. This is the simplest linear equation of

parabolic type with two independent variables. It is called the equation of heat or the equation of diffusion. It has one family of character• istics, namely the lines t = constant in the xt-plane. The simplest problem is that of the infinite rod with a given

temperature distribution

v(x,O)

=f(x).

On physical grounds, it is obvious that the temperature at any sub

sequent instant is uniquely determined. The problem is to find condi

tions satisfied by f(x) so that this is true, and to find an formula for U.

12.2 A formal solution of the equation of heat azuaU If,in (1 ax2 at' we put v = XT where X and T are functions of x and t respectively we have

X=T' where dashes and dots denote differentiation with respect to x and Hence = 2x, —a

D 238]

—

a2T,

EQUATION OF HEAT

12.2]

[239

where a2 is the separation constant. Thus we have a solution

u= where x0 and t0 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 > t0, is u

=Jexp(_a2(t_to))cosa(x_xo)da

-

I

—

4(t-t0)

If x x0, this solution tends to zero as t —÷ t0 +0. Other formal solutions can be obtained by integration. For example, exp

=

is a solution valid when t > 0. If we put

(2)

=x+

we

obtain

1

(3)

U=

when t > 0. 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. If f(x) is zero when x < a and when x > b where a 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 Ic(x,t) = so

that the formal solution (2) becomes u(x, t) =

k(x - t)

(4)

EQUATION OF HEAT

240]

L12.2

when t > 0. This result can be justified if we assume thatf(x) possesses

a continuous second derivative and satisfies suitable conditions at infinity to ensure the uniform convergence of the integrals obtained from (3) by differentiation under the sign of integration. But the result holds under very much less restrictive conditions. The solution (4) was obtained by integrating a multiple of lc(x—E,t—r)

along a path in the er-plane. Another formal solution is (5)

where t > 0, the upper limit being t since k(x, t — r) is complex when r > t. This solution is an even function of x. Its value when x = 0 is u(0;t) =

1

r'

dr

If u(0, t) is continuous and the solution of the integral equation is

This is Abel's integral equation for

vanishes when t =

0,

2

u(0. r)

dr

r)' = Thus (5) is a formal solution of the equation of heat in terms of the values taken by the solution when x = 0. 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)

(6)

being positive. The expression (6) is an odd function of x. If we make the substitution t — r = u(x, t) =

when x > 0,1> 0, (6) gives 2

1

— I

—

\ 4uj The limit of this as x-÷ ± 0 is when t > 0; but the limit as is — ç'(t), since v(x, t) is an odd function of x. 12.3

— U

Use of integral transforms

Problems of heat conduction in an infinite or semi-infinite rod can often be solved by the use of integral transforms. As it is difficult to give this method a rigorous form we content ourselves with illustra tive examples.

EQUATION OF HEAT

12.3]

[241

To solve the initial value problem u(x, 0) = f(x) for an infinite rod,

we may use the complex Fourier transform t)

u(x, t)

J

=

whose inverse is

u(x,t) —

Assuming that differentiation under the sign of integration is valid, we have 1 I

at

=

=

u(x, t)

—

-

— —

at

—

O)exp(—

I

t),

—

provided that u(x, t) behaves at x = ± vanish. We then have t) =

1

so that the integrated terms

exp (

= Using the formula for inverting the Fourier transform, u(x, t)

1

exp (

=

—

rJ 1

=

exp

—

exp( exp ( - 1(x

= 2V(irt)

-

the result we obtained in § 12.2.

To solve the problem for the semi-infinite rod x > 0 with data u(x, 0) = f(x), u(0, t) = çb(t) when t> 0, it is more convenient to use the Fourier sine transform = whose inverse is

2

u(x,t)

Then

/12\

t) 3at

=

at

sin cLxdx =

/12\

a¼(x,t)

sin cLxdx.

____ EQUATION OF HEAT

242]

[12.3

Integrating by parts and assuming that u(x. t) behaves suitably at

x= //2\

= This gives t)

=

0) exp

r))

+

dr,

of v(x, 0) (whieh is 0) is the Fourier sine transform equal tof(x).) Using the formula for inverting the Fourier sine transform, we have where

u(x.t) =

7T0

(

.10

('I

exp(

r))

; =

t - r) dr.

{k(x -5, t) - k(x +

=

The Laplace transform of an integrable function =

If




a.

I

is

estdt.

is an analytic function of the complex variabk The inverse of this Laplace transform is =

_7T?

j

c > a. The theory of the Laplace transform lacks the symmetry of that of the Fourier transform. We illustrate the theory by consi(lering the problem of heat. conduction in a semi-infinite rod x > 0 unde the conditions v(:L, (i) = U

(x> 0),

v(O. 1)

=

(1> 0).

EQUATION OF HEAT

12.3]

[243

If the Laplace transform of the solution u(x, t) is

U(x,s) =

—

we have

t)

=I

=

Jo

= [u(x, t)

Hence

I

—

Jo

+ s J u(x, t) estdt.

—T—sU=O

provided that the real part of s is so large that u(x, t)

tends to

zero as t—*

Write s = s > 0. Then

cr2

where the branch of o is taken which is positive when

U(x,s) =

Since U(O, s) is the Laplace transform of u(0, t), which is given to be we have

A +B =

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) = which expresses U(x, s) as the product of the Laplace transforms of

That is the Laplace transform of the second function can be found from tables of integral transforms. It can be shown that, if '1'(s) and 'F(s) are the Laplace transforms of

and

then 'D(s) 'F(s) is the Laplace transform of

J

—

u)

Hence the solution of the heat conduction problem is u(x,t) = It I

jo

x

t—u

EQUATION OF HEAT

244]

[12.3

It' is impossible to give in a short space an adequate account of integral transform methods: the reader should consult the books cited 12.4

Use of Cauchy—Kowalewsky theorem

The equation

b2u

—

bu

(1

has, by the Cauchy—Kowa]ewsky theorem, a unique analytic regular in a neighbourhood of (x0. t0), satisfying the conditions t) = (2) u(x0, t) = are regular in a neighbourhood of t0. If provided that and these conditions are satisfied at every point (x0, t0) of a finite interval y of the initial line x = x0, the problem has a unique solution regular near y.

By a change of origin, we may take x0 and t0 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

+

+ p1(t)

x2

x3

+

± (2n+

where

and

Let

are the nth derivatives of =

=

+ ...

(3)

and tb-. b,1 ti',

the series having radii of convergence R1 and R2, say. Let R <Min(R1,R2). Since cXt) and are analytic functions of the complex variable t regular in ti we have ILl2

where M1 and M2 are the maxima of the moduli of

respectively on

t

R. Therefore M

EL C. Titchmarsh, Introduction to

and

M

Theory of Fourier Integrals 2nd edn (Oxford,

1948), pp. 281—8. D. .V. 'SVidder, The Laplace (Princeton, 1941). H. S. Carsiaw and J. C. Jaeger, Operational Methods m Appl'ted

edn (Oxford, 1948). C. J. Tranter, Integral Transforms in

(London, 1971),

2nd

EQUATION OF HEAT

12.41

[245

where ill = Max (M1, M2). It follows that, when ti < 1.

[12.4

EQUATION OF HEAT

246]

1


0.

We next show that u(x, t) tends to f(x) as t + 0, uniform'y with respect to x on any c'osed interva' — We have x u(x, t) -f(x)

-f(x)} exp (-

=

J

-f(x)} exp (-

{f(x

(3)

Since f(x) is continuous, corresponding to any positive va'ue of e, there exists a positive number 8 such that, if x and x ± then

be'ong to [—

whenever

< 8. Hence

J

-f(x)} exp

— e whenever the point (x, t) of D lies in the open disc with centre (x0, t0) and radius The family of such open discs, one for each point of B is an infinite

for every point (x0, t0) of B, then. u(x, t)

open covering of the bounded c'osed set B. By the theorem, we can choose from this family a finite number of discs which a'so cover B. Hence there exists a positive number 8, the smallest radius of this finite family of discs, such that u(x, t) > — c whcnever the point (x, t) is at a distance tess than 8 from B.

if the theorem is fa'se, there exists a point (x1, t1) of D at which u is

so that u(x1,t1)

Let where K is

= —l < 0.

v(x, t) = u(x, t) + K(t — t1) a

constant. Since u satisfies the equation of heat, we have a2v

Fly

— = — —K. ax2

Choose K 50 that 0 < K < i/t1, as we may

since t1 >

0. Let

0 —6+K(t—t1) > —i+Kt > —1.

EQUATION OF HEAT

12.71

[251

—1, the minimum of v(x, t) cannot exceed —1, and so is attained at a point (x2, t9) of D. This is impossible. For at a minimum = 0 (or possibly av/at < 0 if t2 = c), and a2v/ax2 ? 0, yet

Since v(x1, t1) =

where K > 0. Hence the assumption that u(x, t) is negative at a point

of D is false, and the theorem is proved. Theorem 2. If u(x, t) belongs to Em

in the strip 0 < t

u(x,t) =

c, if

0

(x,t)-+(x0,+O)

for all x0, and if

sup u(x, t) =

as x ± oo, 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 (x0, +0) in any manner in t> 0. Let F(x) = sup Iu(x,t)I.

Then there exists a constant hi such that 0 ( F(X) for all x. The function U(x, t) = F( — R) k(X + R, t) + F(R) k(x — R, t)

belongs to

in t >

—R < x < R,0 0. We do not require u(x, C) to tend to a limit as (x, C) tends

to (0, + 0) in any manner or as (x, C) tends to (a, + 0). For example, we might have the problem of a finite rod at a unique temperature whose end points are suddenly cooled to zero. The conditions are then

u(x,0)=1 (0<x 0).

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)

—

h0u

=

g0(C)

(x =

0,

C> 0),

+

=

g1(C)

(x = a, C> 0),

where h0 and h1 are non-negative constants.

(iii)u=g0(C)(x=0,C>0), (iv)

=

g0(C)(x

= 0,C> 0),

u = g1(C)(x = a,t>

To prove uniqueness, we have to show that, if f, g0 and u(x, C) is identically zero for all C> 0.

0).

are all zero,

Let D be the rectangle 0 < x < a, 0 < C 0. When x =

0,

= =

t)

t)

-5

-

-5

t)

and This we can integrate by parts if are absolutely continuous. For then the derivatives f' and g' exist almost everywhere.

We then have

t) =

-

t)

= 50 provided that the terms at the limits vanish. We can ensure this by choosing so that g(0) = f(O) and

=

lim

0.

The end condition is then =

for t > 0. This is satisfied if The solution is

satisfies the differential equation

g'+hg =f'—hf. =

I

JO

Sinceg(0)=f(0), = 0

If

0,

< 111

EQUATION OF HEAT

12.111

[261

for every positive value of a, Mh C) I

exp (-

-

5 exp

+

Mh

which tends to zero as we please, —

C)

when C < 1/4a. As a can be as small as for any positive

tends to zero as

value of C.

The solution of the equation of heat we are trying to obtain is thus {k(x -

u(x, C) =

C)

+ k(x +

C)}

_2hf 0

0

Inverting the order of integration u(x,C)

where

K(x,C) =

I

Jo K(x, C) can be expressed in terms of known functions. For K(x, C)

= =

=

e:P { —

(x —

exp {

— —

exp {hx + h2C}J exp

= -r- exp{hx + h2C} I exp { — r2} where

Hence

K(x, C) =

exp{hx + h2C} Eric

where Erfc x is the Error Function.

dii

12.12

[12.12

EQUATION OF HEAT

262]

The finite rod again

In the problem of heat conduction in a finite rod we have to solve the equation of heat in 0 < x < a, I > 0, given certain end conditions and an initial condition zt(x, 0) = f(x), where f(x) is bounded and integrable. Suppose that the temperature is assigned at the ends, so that are bounded and integrable over any finite t > 0, where and interval. This problem can be split up into three problems: in problem are zero; in problem (b)f and ifr are zero; in problem (c), (a), and and are zero. We 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. We consider then the infinite rod with

f

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) F(x) = F(x+ 2a).

and so

The extended initial data satisfy

F(x) =f(x)(0 < x

F(—x) = —F(x),

< a),

F(x±2a) = F(x).

Sincef(x) is bounded, F(x) certainly satisfies the condition IF(x)I
0 for all values of x, has there continuous derivatives of all orders and satisfies the equation of heat As t) tends to (x0, + 0) in any manner, u1(x, t) tends to F(x0) at

every point of continuity, and so does so almost everywhere. In particular it tends to f(x0) almost everywhere on 0 < x < a.

EQUATION OF HEAT

12.121

When C > 0,

F

u1(0, C)

[263

C)

= is an odd function. Also C)

= lID k(a

-

C)

k(ii,C)F(a—ii) di1 =

=5

0

since F(a — is an odd function. This solution can be written as C) =

k(x —

C)

—x,J (2n—1)a

=

k(x —

1a

=

{k(x

K(x,

=

where

C)

- 2na, C) - k(x +

C)}

C)

- - 2na)2/C} - exp { -

(exp { -

=

-

— 2na,

+

2na)2/C}).

Inversion of the order of integration and summation is readily justi-

fled. The function K can be expressed in terms of Jacobi's theta functiont (in rn2 — 2niz)

&3(zI r) =

where im r> 0. In this notation, exp I

(x — 2na)2\ —

4C

)

= exp { —

&3

liax ia2

—i-

and this is equal to

by Jacobi's imaginary transformation. Hence

K'

—

1

'Ta

fin(x —

2a

fn(x +

a2)

2a

iint) a2

f See Whittaker and Watson, Modern Anal ysia, pp. 464 and 474.

264]

EQUATION OF HEAT

[12.12

The last formula gives

nrrx.

=—

Hence we have the Fourier solution

.nTrx

exp (— n2rr2t/a2) sin

u1(x, t) =

=

where

—,

.

—

ajo

a which we discuss later. The second problem is that in which u vanishes when 0 <x < a, t = 0 when x 0, t> 0. We have seen and when x = a, 1> 0, but u

that u0(x,t)

I

T

is a solution of the equation of heat which is continuously differentiable.

as often as we please in x> 0,t > 0 and in x < 0,t> 0. It vanishes when t = 0 and is an odd function of x. It is discontinuous across x 0, t > 0; for it tends to at every point of continuity of as (x, t) tends to (+ 0. t0), but it tends to — Consider the function u2(x, t) =

Since

is bounded, say

as

(x, t) -÷ (— 0, C0).

+ 2na, t).
0, (2n+

(2n+1)Mar'

dT

JexP(—n2a2/r)---r

(2n

1)

exp (— n2a2/t0),

and similarly for 1u0(x— 2na. ')I Hence the series defining u2(x, t) is uniformly and absolutely convergent. when 0 x a, 0 < t t1.

EQUATION OF HEAT

12.12J

[265

The series obtained by term-by-term differentiation can be discussed

in the same way. It follows that u2(x, C) satisfies the equation of heat in 0 < x a,t> 0. Write u2(x,t) =

is omitted in E'. Hence at every point of con-

0

tinuity of q5(t) and so almost everywhere, we have u2( + 0, C) =

q5(t)

+

u0(2na, C) =

since u0(x, C) is an odd function of x. Also

=

u2(a,t)

0.

Thus u2(x, C) does satisfy the prescribed initial and end conditions.

From this form of the solution, we can deduce the solution by Fourier series. For a

u2(x, C) = —2

= — —

=

and so u2(x, C) =

r) dr

cx,J 0

_a

q5(r) k(x + 2na, C —

I

vX

q5(C)exp

Aft)

axJ0"

+2na)2/(C—

(

a

q5(r) [1

nsin

fl7TX

dr

(lTx iir(C —

3\2a

(—n2ir2(C—r)/a2)

exp (— n27r2C/a2)f ç5(r) exp (n2ir2r/a2) dr.

The solution of the problem when the initial and end conditions are

u(x,0) = 0,

u(0,C) = 0,

u(a,C) =

can be deduced by replacing x by a — x and 12.13

by

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 a2u

au

ax2

at

[12.13

EQUATION OF HEAT

266]

u(x. ±0) = f(x)

given that

(0 < x 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.

The series

is not necessarily convergent, but it is

sin

summable (C. 1)t almost everywhere; and its Cesàro sum is f(x) at every point of continuity, again almost everywhere. To complete the proof we use a theorem due to Bromwich4 that if

a1. is summable

(C. 1) with sum s, then lim

exp (— 0A7) = 0

t The terms summable (C. 1) and Cesàro sum are defined in note S of the Appendix. An Introduction to Infinite &ries, 2nd edn (1926), p. 429.

EQUATION OF HEAT

12.131

if

is

[267

an increasing sequence of postitive integers such that, for

aim,

> 0,

+

— —

where K is a constant. In the present case U = 1T2t/a2, and

=

n2,

so

that Bromwich's conditions are satisfied. Hence the series in (1) tends to f(x) as C —k +0 whenever f(x) is continuous. The problem of solving the equation of heat in 0 x a, C > 0, when u(a, C) = 0 can be solved by a modification u(x, 0) = 0, u(0, C) =

of the Fourier method. Assume that u(x,C) = 21's

where Since

=—

ajo

.

u(x, C) sin

=

u(x, C) vanishes when C = 0, 2

dC

— a

0.

dx.

We have

.

=—J —-------sm—dx

aj0

a

2 ('a a2u(x, C)

=—J

aj0

nirx

.

a

ax2

nirxl a ra 2 Iau . nirx fir . nirx u(x,C)sin—dx = —i—sm———ucos—i — I a a aj0 a3 Jo a a[ax

=

prom we

9n7r

have

Thus

+

dC

= —ia u(x, C)

=

n2ir2 2

2nlT —

2

q5(r) exp (— n2ir2(C — r)/a2) dr.

sin

q5(r) exp ( — n2ir2(C — r)/a2) dr.

This formal argument gives the answer found in § 12.12, but it would be difficult to give a rigorous proof. Note that u(0, C) = 0. The reason for this is that u(x, C) is an odd function of x. u(x, C) tends to q5(C) as x + 0, but it tends to — q5(C) as x

—0.

The sum of the series when x =

which is zero.

0

is

+ 0, C) + u( —0, C)),

[12.13

EQUATION OF HEAT

268]

If we try to solve the equation of heat for the ffnite rod when the conditions are 0) = f(x) (0 < x < a), v(x,

= where

ii

and H are positive

=

0,

(t > 0),

constants, we might start with the solution

exp (— k2t) (A cos kx + B sin In

0

kx).

order to satisfy the end conditions, we must have (k2 —

hH) sin ka = k(h + H) cos ka.

This equation in k has no complex roots. Its real roots occur in pairs is a strictly increasing sequence of positive ± /c1, ± /c2,.. where numbers. When n is large, nir

a(h+H)

a

nir

/1

is not an integer, the Fourier series method is not applicable. We have to use instead an expansion as a series of Sturm— Liouville functions. The theory is outside the scope of this book. Since

Exercises 1. If

t) is the Fourier transform of a solution u(x, t) of a2u — =

ax2

prove

au

that

.au

at

Deduce that, if u(x,

2.

t)

at

—+ a — =—

and hence that

u(x,

au

=

u(x, t) satisfies

it)exp

0)

= f(x)

for all values of x

(*t3_xt)J

(—

the equation of heat in x> 0, t

u(x,0) =f(x), The Fourier cosine transform of u(x, t) is

=

=

> 0 under the conditions

EQUATION OF HEAT

[269

j/9\ ('00

with inverse

u(x, t)

= U

t)

=

—

is the Fourier cosine

where

transform

of f(x), and hence that

u(x,t) =

?fr(r)k(x,t—r)dr.

JO

Jo

3. Prove that, if a and b

are

positive,

=

fT00 Hence show that, if the operator

is defined by (t> 0),

=5 =

then

if t1 and t2 are positive. au — =— ax2 at a2u

4. Find the solution of mO
0 and t> 0 hu

Hence show that u(x, t)

6. u(x, t) tions

= —s:

=5

is continuous. Prove that, when k(x, t — r) dr.

k(x

r)

the equation of heat in x> 0, t > 0, under the condiu(x,0) = 0 (x> 0), u(0,t) = çb(t) (t> 0).

satisfies

EQUATION OF HEAT

270]

Prove that

v(x, t)

where X

7. Prove that the solution of the equation of heat in t 0 which satisfies the conditions v(x,0) = 1 (x> 0), u(x,0) = —1 (x 0). 9

x

u(x,t)=

is

1(0

21

— Erfc



(x>0), u(x, t)

=5 g(x) = f(x) — 2h

0, t > 0 such

(t>O) t)

APPENDIX Note 1. Analytic functions The function f(x, y) of two real variables is said to be an analytic function regular in a neighbourhood of (x0, Yo) if it can be expanded as a double series amn(x_xo)m(y_yo)n m, n 0

absolutely convergent on a disc (x — x0)2 + (y — lfo)2