A Generalized Function-Theory and the Related Dirichlet Problem: II

Annals of Mathematics

A Generalized Function-Theory and the Related Dirichlet Problem: IIAuthor(s): James SandersSource: Annals of Mathematics, Second Series, Vol. 66, No. 1 (Jul., 1957), pp. 141-154Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1970120 .

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ANNALS OF MATHEMATICS

Vol. 66, No. 1, July, 1957 Printed in U.S.A.

A GENERALIZED FUNCTION-THEORY AND THE RELATED DIRICHLET PROBLEM: II*

BY JAMES SANDERS

(Received October 1,1954) (Revised April 19, 1956)

Introduction

This paper is a continuation of Part I and uses the notation and results of that paper [5].

Our main result is given in Theorem 27, which solves the Dirichlet problem for the equation N'2Q() = 0. The method used is a generalization of the "double- layer" method of potential theory. The Dirichlet problem for the biharmonic equation was first solved in this manner by Lauricella [1] and the method was slightly extended later by Schr6der [7]. To handle the more general case almost all details of the calculation are done differently from Schrdder and Lauricella. This is especially true in the discussion of the continuity properties on the boundary of our generalized simple and double layers and in Theorem 25 giving an explicit solution of a homogeneous system of integral equations.

We have restricted ourselves to bounded domains and to the Dirichlet problem. However for A2(4) = 0, Schrdder has used the double-layer method to obtain results originally due to Muskhelisvili [3] for unbounded domains and a somewhat similar method was used by Pleijel [4] to handle other types of boundary conditions. The method therefore holds promise of solving other problems for N'2(0) = 0 and indeed for handling other higher order equations.

The Dirichlet problem for the equation N'2(O) = 0 consists of finding a function f which is C(4) in a domain D, C(1) in the closure of D and which in D satisfies N'2(k) = 0 and on the boundary C satisfies 4 = f(s), (a4)/(an) = g(s) where f(s) and g(s) are given continuous functions of the arc length s of C.

The first section will consider uniqueness theorems, that is, it will give dif- ferent conditions under which f = 0 is the only solution of the Dirichlet problem when f(s) = g(s) = 0. One result of this section we would like to emphasize is Theorem 16, which shows that the Dirichlet problem is equivalent to the problem of determining a 22-monogenic function from its values on the boundary. This result is used in some of the uniqueness theorems and is the basis of all the work on the existence of a solution of the Dirichlet problem.

In the second section, "simple layers" and "double layers" are defined and some of their properties obtained.

Finally, the third section sets up a system of integral equations whose solutions give the solution of the Dirichlet problem.

* This research was supported by the United States Air Force, through the Office of Scientific Research of the Air Research and Development Command.

141



142 JAMES SANDERS

Uniqueness theorems

We first note the following result: THEOREM 15. Suppose D is a bounded domain whose boundary C is composed

of a set of simple closed curves Cl , C2 * * * Cn+? with Cn+? D Ci (i = 1, 2, n) and suppose each C. has continuous first derivatives. Suppose also that in D, f is C04) ao(x, y) and o-2(X, y) are C), f(x, y) is C(2), and 01, a2, f are positive while ca , 02 , X, Ox I O, , fL1(O) and its conjugate are continuous in D + C. Then the only solution of the Dirichlet problem

L1[fL(G,)] = O in D, O = 0 a = on C is 0 = 0 in D + C.

PROOF. This follows from Theorem 7 (Part I) as in the classical case of the biharmonic equation. As usual, we have first applied Green's identity to a closed interior domain and then let this domain approach D. In this manner the re- strictions on 0 are lightened. Also an integration by parts device due to Musk- helisvili [3] in the biharmonic case has been used to avoid the assumption that P = fL1(O) has continuous derivatives in D + C. Instead, the conjugate Q of P defined by oi(x, y)Px = Qy, o-2(x, y)P, = Qx is assumed to be continuous. Q.E.D.

From now on we shall consider only the special case of N'2(o) = 0 instead of Ll[fL1(O)] = 0. First, then, we relate the Dirichlet problem for N'2(0) = 0 to the theory of 12-rnonogenic functions.

THEOREM 16. Suppose the domain D and its boundary C are as described in Theorem 15. Suppose also that o-(x) and r(y) are C(3) in D, continuous in D + C and that o-r > 0 in D + C. Then, the Dirichlet problem

(2.1) N'2(4) = 0 in D, 4 = f(s) and ak/On = g(s) on C where 4 e C(4) in D and C0') in D + C while f e C-') and g is continuous on C

is equivalent to the problem of determining in D a 22-monogenic function, u - iv, such that u = f(s), v = g(s) on C and

(2.2) f af(s) dx + rg(s) dy = 0

where u and v are C(3) in D, continuous in D + C and f(s) and g(s) continuous. PROOF. Given a solution 0 of (2.1), set u = (1/o->)x, v = (1/r)4Y. Then,

ouY = TvX = Ox, and 0 = aux + rvT = N(0) and so N'(0) = 0. Also, the boundary values f(s) and g(s) of u and v are easily seen to be de-

termined as linear combinations of g(s) and (df(s)/ds). Finally, we get (2.2) since, k being single-valued,

0 Ox dx + Oy dy = J of(s) dx + rg(s) dy.

Next, given a 2-monogenic function u -iv, there exists a function p such that Ox = au and ha = rV (since (ou), = (rv)x). The other conditions in (2.1) now



A GENERALIZED FUNCTION-THEORY 143

follow easily. The condition (2.2) ensures that 0 is single-valued since, if F is any simple closed curve homologous to Cj, in D, then

f dx dx + oy dy = f dx k5dy+ f(au - rvx) dx dy

= o af dx + Tg dy + f (au - rvx) dx dy = 0. Q.E.D.

Now, using the concept of a :2-monogenic function, we obtain the same result as Theorem 15 gives in the case of N'2(0) = 0.

THEOREM 17. If u - iv is 22-monogenic in D and u = v = 0 on C, then u = v - 0 in D + C provided u, v are C) in D and continuous in D + C, 6 and its conjugate w are C0') in D and continuous in D + C, and provided the domain D and boundary C are as in Theorem 15.

PROOF. It follows from Theorem 8 (Part I) that

lID1 O' dx dy = L w(au - rvx) dx dy + f w(au dx + rv dy)

+ f 0 (2 u dy-- v dx)

and so, under our assumptions, 0 = 0. Therefore ouy + rvx = 0, O.u - Tvx = O Thus N(u) = 0 in D and u = 0 on C. Therefore by Paraf's theorem u = 0 in D + C and so its conjugate v = 0 in D + C. Q.E.D.

We shall find it necessary later to use the following theorem, relating to another type of boundary value problem.

THEOREM 18. Suppose u- iv is 12-monogenic in D and u, v are C(') in D and 0) in D + C while co is continuous in D + C. Suppose also that

fTU dx + rv dy = OIf {Y(u, v) =- on C

then

u~yf dx v = y J Yand 0 + iw = 2(y + i8)

where 5 and y are real constants. As always D and C satisfy the conditions of Theorem 15, a and r are C(3) in D and continuous in D + C.

PROOF. From Theorem 10 (Part I) we see that

il {- (UY)2 +

(vX)

+ 2

[(or)0

u - (T/av]2} dxd

- A [uX(u, v) + vY(u, v)] ds = 5 f au dx + rv dy = 0,

and so u, = vx = 0 and au3 = rvy in D. Therefore u = (dx/o-(x)), v



144 JAMES SANDERS

f(dy/r(y)) and so 0 = 2y. Thus X is a constant which must be 26 in order

that X(u, v) = -8Ban2 and Y(u, v) = 5rn1. Q.E.D. Finally, we obtain a result for unbounded domains. THE OREM 19. Suppose D is a domain exterior to a simple closed curve C with

a continuous tangent. Suppose u - iv is 22-monogenic in D, such that u, v are C() in D and C"') in D + C while w is continuous in D + C. Suppose also that W, ux, uy 2 vx 2 vy are 0 (1/r2) for r large and that u and v are bounded as r -*o .

If (1) u = v = 0 on C, then u = v = 0 in D + C. (2) X(u, v) = Y(u, v) = 0 on C, then u and v are constant in D + C. In par-

ticular if u and v approach 0 as r -* o then u = v = 0 in D + C. [a and r are C(3) in D, continuous in D + C and bounded at infinity].

PROOF. Let C be a large circle containing C. Then, under our assumptions

f[uX(u, v) + vY(u, v)] ds approaches 0 as r x-> o. Therefore once again we have

lID {4 (uy)2 + ()2 ? v [(2/_r) Ux - (r/r_)l V,]2} dx dy

= ti [uX(u, v) + vY(u, v)] ds.

Therefore when u = v = 0 in C, we conclude, as previously that u =

ly (dxlof(x)), v = -y (dy/r(y)). But, since u = v 0 on C, -y = 0 and so u =

v = OinD + C.

When X(u, v) = Y(u, v) = 0 on C, then once again u = (dx/o-(x)), v =

ly (dy/r(y)). But, ux = vy = 0 at infinity and we have assumed that a- and r

are finite at infinity. Therefore, u and v are constant in D + C. Q.E.D.

Generalized simple layers and double layers

From now on we shall make the following assumptions: HYPOTHESIS A. D is a bounded domain whose boundary C is a simple closed

curve x = x(s), y = y(s) such that x(s) and y(s) are C(') and x2 + y2 i 0. HYPOTHESIS B. o-(x) and r(y) are C(') in D and C(" in D + C. The assumption that D is simply connected is made for simplicity. The other

assumptions in Hypothesis A and B are necessitated by our proof of the existence of the fundamental matrix in Part I.

DEFINITION 15. A simple layer is a pair of functions U1(%, -), V1(, q) such that

U(, i = - A [C(s)ul + D(s)vl] ds 7r

(2.3) V1(, ) = -- [C(s)u2 + D(s)v2] ds

7r C




where C(s) and D(s) are continuous functions. U1, u2, v1 and v2 are the elements of the fundamental matrix and the variables (x, y) in u1(x, y: , n) etc. are functions of the arc length s of C.

DEFINITION 7. A double layer is a pair of functions U2(, 71), V2(t, 71) such that

U2(t, n) = -- ! [C(s)X(u1 , v1) + D(s)Y(ui, vD)] ds 7rc

(2.4) V2(t, n) = - I [C(s)X(u2 , v2) + D(s)Y(u2, v2)I ds

7rc

where C(s) and D(s) are continuous functions. The remarks made in Definition 6 about u1, v1, u2, v2, hold here.

It is clear that both a simple layer U1 - iV1 and a double layer U2 - iV2 are 12-monogenic functions of t + in1, when this point is inside or outside of C, since both functions are linear combinations of u1-iu2 and v1 - iv2 . Our purpose in this section is to obtain results analogous to the classical results of potential theory on the behaviour of simple and double layers when t + in1 approaches the boundary, C.

First, we have the elementary result. THEOREM 20. When (%, ) approaches a point (go, -qo) on C, from either the in-

side or outside of C, the simple layer { U1(t, r), V1(t, % } approaches the values obtained by substituting (t, -qo) for (%, -) in Equation (2.3).

PROOF. This follows from the results of potential theory since u1, v1, u2, v2 are all 0 (log r). Q.E.D.

The behaviour of double layers is considered next. As in potential theory, the properties of double layers are more recondite than those of single layers.

First, let s and so denote points on the curve C. Then, we make the following definition, valid for s 5 so.

DEFINITION 8.

ai(s, So) = X(u1(s, So), v1(s, So))

251(S, So) = Y(u1(s, So), v1(s, So))

(2)a(S, SO) = X(U2(S, SO), V2(S, SO))

12(S, SO) = Y(u2(s, So), V2(S, So)).

Next, we prove THEOREM 21. When (Q, 'q) approaches so on the boundary C, the double layer

{U2(%, a V2(%, )} approaches

U2(sO) = i C(so) - - [C(s)al(s, so) + D(s)Oi(s, so)] ds 7rc

(2.6) 11

V2(SO) = i D(so) - - f [C(s)a2(s, So) + D(s)f2(s, so)] ds. 7r C



146 JAMES SANDERS

The + (-) sign holds when (t, -q) -> so from the interior (exterior) of C. PROOF. {U = 1, v = 0, 0 + iw = 0 } is a 12-monogenic function. Substituting

these functions into "Cauchy's theorem" (Theorem 14, Part I) we obtain

1/2r X(ui, vl)ds = -1 and 1/2rf X(u2, v2)ds = 0 when (%, -) is interior to C.

Similarly, from the fact that {u = 0, v = 1, 0 + iw = O} is 12-monogenic, we

find that 1/2rf Y(ui, vl)ds = 0, 1/27r Y(u2, v2)ds -1 when (%, -) is in-

terior to C. When (%, n) is exterior to C we apply the fundamental identity (Theorem 11,

Part I) instead of "Cauchy's theorem", and so obtain

X ,(uv 1l) ds = Y(Ui , vl) ds = -IfX(U2, V2) ds = Y(u1 ,lv) ds = 0.

It is now fairly easy to show that when (Q, -) is on the boundary C,

1 Ix(ui vi) ds = 2 Y(u2, V2) ds = -2, and

14 X(u2 v2) ds = 2 f Y(u1 ,v) ds = O.

For, if (%, n) is on C, construct a small circle r, with centre (t, 7). This circle is divided into two parts Ti and F2 by the boundary C. F1 is exterior and F2 interior to C. The curve C is thus divided by F into two parts C2 and C1 where C1 D (Q, -). Now applying the above results on the value of the integrals to the con- tours C2 + F1, and C2 + F2 and noting that by the same considerations used to obtain our "Cauchy's theorem", the integrals over F1, approach the corresponding integrals over F2, we obtain the quoted results.

Next, we assert that

I X(u1 , vl) I ds, I Y(ui ,vi) I ds, IX(U2, V2) I ds, Y(U2, V2) I ds

are uniformly bounded in D + C. This follows from the fact that the singular

terms I (a log r/ln) I ds and / log r I ds are uniformly bounded, results used

in potential theory. The theorem now follows in the same manner as the corresponding result in

potential theory. In the case of U2(t, 1), (Q, -) interior to C, for instance, we write 1 !1~~~ U2(, n) = - C(so) X(ui ,l) ds- D(so) Y(ui, vi) ds

- { [C(s) - C(so)]X(ul , Vi) + [D(s) - D(so)] Y(ui , Vi) } ds

! { [C(s) - C(so)]X(u1 , Vi) + [D(s) - D(so)] Y(ui , Vi) } ds r 2

- + 2C(so) - - 1< The result now follows in the usual manner. Q.E.D.




We now prove a result analogous to the potential-theoretical theorem on the normal derivative of a simple layer.

THEOREM 22. When (t, r7) approaches so on the boundary C,

XWU1%t X7), V1%t -)) =F C(so)

- {C(S)XZ,(U1, U2) + D(s)Yz,,(vi ,V2) } ds (2.7)

27)Y(U1(% X7), V1(%, X)) :jF D(so)

-! f {C(S)Yt,(Ul, U2) + D(s)Yt,,,(v, V2)} ds. 7r

The - (+) sign holds when (%, v) approaches so from the interior (exterior) of C. PROOF. When (Q, X) is not on C,

X(U1, V1) = f {C(s)Xt.,,(u1, u2) + D(s)Xt,,(vi, v2)} ds

Y(Ul, 17)- -= f C(S)Yt,(Ul, U2) + D(s)Ye,,(vi, V2)} ds.

Now, from Theorem 13, Part I, it follows that the highest order terms in Xz,j(ui, u2), XZ,j(v1, v2), Yt,,(ul, u2) and Yt,j(v1, v2) are merely the negative of those in Xx.,(ul , Vl), Xz,,(u2, v2), Y ,y(ui, vl) and Yx,y(u22, v2) respectively.

The lower order terms are 0 (log r) and so by Theorem 20, give a continuous contribution. Thus, the theorem follows from Theorem 21. Q.E.D.

Finally, we shall consider the behaviour of X(U2, V2) and Y(U2, V2) on the boundary. The method used is a variation of a method used by E. Schmidt [6] in considering the behaviour of the simple and double layers of potential theory. There is a difficulty in proceeding directly in this manner, namely that while the normal derivative of any function can be formed, X(u, v) and Y(u, v) are defined only when u - iv is 22-monogenic: The main devise used to circumvent this difficulty is a form of Green's identity [Theorem 9, Part I].

THEOREM 23. If the densities C(s) and D(s) are continuously differentiable, then

[X(U2, V2)]+ - [X(U2, V2)]- = 0 (2.8)

[Y(U2, V2)]+ - [Y(U2, V2)7- = 0

where + (-) indicates the limit as (%, 7o) approaches a point so on the boundary from the interior (exterior). Equation (2.8) means that if one of limits exists, the other exists and they are equal.

PROOF. Let C be a small part of C, containing the point so . Then, the integrals over C - C obviously are continuous at so and so we only have to consider the contributions due to C.

Now, define a function 4 in a small domain D1 interior to C, with C as part of its boundary, by setting 42 = iC(s), k, = rD(s) on C. Then, assuming C(s) and



148 JAMES SANDERS

D(s) are C), we can make ' 6 C(4) in D. Thus, setting u = (l/v}+z, V = (1/Tr)4,0 we have that u and v are C) in D + C, and Ox = (aux + rvV), and OX are C(0) in D + C. Also, o-uY = rVx in D, while the boundary values of u and v on C are C(s) and D(s) respectively.

Similarly, C(s) and D(s) can be extended into functions u and v, satisfying the above conditions in a domain D2 exterior to D with C as part of its boundary.

Theorem 9 (Part I) can now be applied to D1 and D2 . In fact, proceeding in the same manner as in the proof of "Cauchy's" theorem, we obtain

U2( Q) = - I {uX(u, v1) + vY(u, vi)} ds Tr c

- ?Xu(t,~ 7Z)- k f ul A (u, v) + vi B(u, v)} ds 7rc

+ it, Ox + r O6X dx dy (2.9)

V2(% X)= 1 f {uX(U2 V2) + VY(U2 V2)} ds

- Xv(t, rl) - J {u2A(u, v) + v2B(u, v)} ds

+ J 2 fL + u + V20y} dx dy where X = 2 if (%, n) is in D, 1 if (%, q) is on C and 0 if (Q, q) is outside D + C.

In Equation (2.9), terms that are continuous when (%, -) crosses C have obviously been left out.

We shall now consider various terms in (2.9) separately. First, the terms

U3 = f u1A(u, v) + v1 B(u, v)} ds

- I

V3 = 1 f {U2A (u, v) + v2B(u, v) } ds

form a simple layer and so by Theorem 22,

[X(U3, V3)]+ - [X(U3, V3)]- = -2A(u, v). Next,

[A(Xu, Xv)]t - [A(Xu, Xv)]- = +2A(u, v). Then,

[A (I ff (1I uiO + I Vl v) dx dy, 2 ff (! U2Oz + I V20y)dX dy)]

- [A I21 Ox + I + 1 O,, dx dy,

I I U2 O + -V20 dx dy 0




since A acting on the above integrals is merely a linear combination of first derivatives of area logarithmic-potentials whose densities are C().

Since, then, X(U2, V2) = A(U2, V2) - 20r2n2 the above results show that the only possible discontinuity in X( U2, V2) is due to the X - function of U4 iV4, where

U4 = XU + 2- ff (! u0,x + I v 0v) dx dy

V4 = XV + ?U2fx + - V20) dx dy.

We now show that this co-function is continuous across C by showing that 04 = c(U4)z + r(V4), has a continuous normal derivative across C. Since

W+ f = 1 re4ia _ ln ds

this will prove our result. Now,

04 = XO ( 0101 + a 0y02) dx dy

X6 + 1 ox (e log r + ay a log r) dx dy

+ (terms, which together with their first derivatives are continuous across C). But,

FdX) _ 8(xo)1 = 00 d Lo~xo) I+ E(M l 2 = L an Lan T n

while Schmidt's results show that the double integral gives a jump in the normal derivative of - 27r/7r(0xnl + 0yn2) = - 200/On.

We have thus obtained the first result of (2.8) when C(s) and D(s) are C). Obviously, the second result follows in the same manner.

Finally, however, we can obtain the results when C(s) and D(s) are Cal) by approximating these functions by C(3) functions. For, since the expressions X(U2, 1V2), Y(U2, 1V2) are combinations of simple layers, double layers and their derivatives, the results of potential theory show that the desired limits must exist when C(s) and D(s) are only 01). By approximating, therefore, we see that the limits must be 0, as asserted. Q.E.D.

The Dirichlet problem

The Dirichlet problem we shall solve is stated in (2.1). Actually, we shall use Theorem 16 and so find a Z2-monogenic function u - iv, with prescribed continuous values f(s) - ig(s) on the boundary C, subject to the condition

f of(s)dx + rg(s)dy = 0. We shall assume that D is a bounded simply-con-



150 JAMES SANDERS

nected domain and that the boundary C is C(3. Under these conditions the highest order terms in aR, f30, a2 , f32 defined in (2.5) are continuous functions of s and so, as well as having continuous first derivatives with respect to so.

To solve the problem we first adopt an heuristic procedure and suppose that U and V are given as double layers

U(, Ai) = f {C(s)X(ul, v1) + D(s)Y(u, vD)} ds 7rc (2.10)

V(, ?7) = -- f { C(S)X(U2 , V2) + D(s) Y(U2, V2) Ids.

Then, from Theorem 21, the densities C(s) and D(s) must satisfy the integral equations

f(sO) = C(sO) - I {C(s)ai(s, so) + D(s)X,(s, so)} ds 7rc

(2.11) g(so) = D(so) - L {IC(s)a2(S, So) + D(s)f32(s, so) } ds. c

The homogeneous system corresponding to (2.11) is

C(sO) - {C(s)ai(s, so) + D(s)t31(s, so)} ds = 0

(2.12) D(so) - f {C(s)a2(s, so) + D(s)f32(s, so) ds = 0

while the adjoint homogeneous system is

c(so) - f {O(s)ai(so, s) + D(s)a2(so, s)} ds = 0 7r

(2.13) D(so) - 1f {O(s)l(so, s) + D(s)32(sO, s)} ds = 0.

7rc

We now consider the properties of the solutions of the integral equations (2.11), (2.12) and (2.13). We first note the following consequences of the Fredholm theory:

THEOREM 24. (1) If (2.11) has solutions, they are continuous. (2) If (2.12) and (2.13) have non-zero solutions, these are also continuous. (3) If (13) has m linearly independent solutions, then (2.13) also has m linearly

independent solutions, and vice versa. We next show that the adjoint system (2.13) does have a non-zero solution. THEOREM 25. The adjoint homogeneous system (2.13) has a solution

(2.14) C(s) = n2(s)o(x(s)), D(s) = -ni(s)r(y(s)).




PROOF. Consider the integral

Ii(t, 7) = I |X(u1 ; x, y), vK%, -7; x, y))o(x)n2(x, Y)

-X(u2(%, _; x, y), v2(%, _; x, y))r(y)nl(x, y) I dsx

| , [ tQ) aui% 7 ); x, y) __ avv(, 7; x, Y)

- r(y)ni(x, y,) [oft) du2( x an ds2

[Ut) IOU2(% 77; x, Y) y) - f'Xr, x) ? 7f~l2(x, y)f {~xn( ~ ~ - ~ ,Y] dsx .

- Tyn2(X 77 Lr(xn)(,Y Y~(1 7 ,Y

Now, u1(t, _q; x, y) - iu2(, 7; x, y), vI(, 77; x, y) -iv2(, 7; n, y), w1% 77; x, y)-

iW2(Q, -q; x, y) are 22-monogenic functions of (x + iy) and so the above integrals are independent of the path. Thus, the integrals can be evaluated by finding their value when the contour is a small circle with (%, 7) as centre. We have done such calculations a number of times and so in the familiar manner, we find that

I, %i) = 2n2(%, 1i)C(Q) if (%, 71) is interior to D

= 0 if (Q, -q) is exterior to D + C.

Next, then, we show in the same manner as in Theorem 21, that

I, %-) = n2(t, r)a(t) if (t, -) is on C.

This shows, then, that C(s) = n2(s)u(x(s)) and D(s) = -nvr(y(s)) satisfy the first of Equations (2.13).

A similar method now shows that these functions also satisfy the second equation. Q.E.D.

An immediate corollary of Theorems 24 and 25 is that there is at least one non- zero solutions of the homogeneous system (2.12). Actually, this is obvious, since C(s) and D(s) equal to any constants gives a solution. What we shall show, however, is that there is only one linearly independent solution of (2.12).

THEOREM 26. There exists one and only one linearly independent, non-zero solution of (2.12) and so of (2.13).

PROOF. Let C1(s), Di(s) be the solution of (2.12) that corresponds to the solution (2.14) of (2.1.3). C1(s) and Di(s) are 0), as follows by differentiating (2.25).

Now, set

U1(, 71) = -- { C1(s)X(u1, v1) + Di(s) Y(u1, v1) } ds 7r

(2.15) V1(, 77 = - { C(s)X(u2, v2) + Di(s)Y(u2, v2)} ds. 7rc



152 JAMES SANDERS

Then, as (Q, q) -* so from the interior of C,

lim U1(Q,q) = C1(so) - If {Ci(s)a,(sso) + Di(s)31 (s, so)} ds = 0 7rc

and similarly lim V1(, -q) = 0. Thus by our uniqueness theorem (Theorem 17), U1(t, 7,) = V1(t, ) =0 for

(%, -) in D + C. Therefore 01 = a(0)(U1) + r(v)(V1), = 0 in D + C and so the conjugate functions of 01, % = a,, where a1 is a constant. Thus,

X(U1, VI) = - 4lai()n2( 7) 1 (

Y(U17 VI + 'ir(-~niQ -q)for (t, 77) in C + D. Y( U1, V1) = + 2 alr(1)n1(t, 71) j

Now, U1 -iV1 as defined by (2.15) is a double layer, and so by Theorem 23, when (Q, -) s> o from the exterior of C,

lim X(Ui, V1) = - la1oa()n2( , 1)

and lim U(Y1, V1) = + 1air(-)n(%, t7).

These last limits are zero, if a1 = 0. We shall show that this can not happen. For, our work done in obtaining the fundamental matrix, shows that X(u1, v1), Y(u1, v1), X(u2, v2), Y(u2, v2) are O(1/r) for r large and so U1 and V1 are also O(1/r). Thus, if a1 = 0, Theorem 19 shows that U1 = V1 = 0 outside D.

But when (, -) approaches the boundary C from outside,

lim U1 = -2C1(s), lim v1 = -2D(s).

Therefore, if a1 = 0, C1(s) = D1(s) = 0, which is a contradiction: Thus, we can assume that a1 # 0. Suppose, now, that (2.13) had a second non-zero solution C2(s), D2(s). As be-

fore, we form a double layer U2(%, 7), V2(%, 1) using C2 and D2 as densities. As be- fore

X(U2, V2) = - 1a2oUQ)n2( , q)

Y'(U2, V2) = + 1a2r(?7)ni(%, -)

where a2 is a constant. Since a1 # 0, we can find a constant b such that a3 = a2- ba1 = 0. Set

C3(s) = C2(s) - bCi(s)

D3(s) = D2(s) - bDi(s).

Again, form a double layer with C3 and D3 as densities. However, now X(U3, V3) = Y(U3, 1V3) = 0. Thus by the same argument used to show that a1 # 0, we find C3(s) = D3(s) = 0.




Theref ore

C2(s) = bC,(s)

D2(s) = bC2(s). Q.E.D.

An immediate consequence of Theorems 25 and 26 is that f o(x)f(s)dx +

r(y)g(s)dy = 0 is a necessary and sufficient condition for the existence of a solution of (2.12). Since we have assumed that this condition holds, we find a solution C(s), D(s) of (2.12). With this pair of functions we can form the double layer (2.10) and (2.11) then shows that U = f(s), V g(s) on C.

Finally, then, we have our fundamental result. THEOREM 27. The Dirichlet problem

N '2(+) 0 in D, 0 = f(s), (a3)/(an) = g(s) on C

f E C, g continuous

(2.16) a(x), T(y) E C(5) in D, and C) in D + C,

a > O T > 0, in D + C,

the boundary C e C)

has one and only one solution. The solution can be expressed in terms of a Green's matrix

/G i(x, Y:, 77) Hi(x, Y:t, 77)\

\G2 (X) : 77 ) H2(Xy Yg: i, a)/

This is defined as a fundamental matrix with G1 = H1 =G2 = H2 = 0 for (x, y) on C. This matrix can be formed by adding regular parts to ul, v1, u2 and v2. The determination of these regular parts leads, of course, to a Dirichlet problem.

Now, since ui and vi are logarithmically infinite, f auidx + rv,,dy = 0 and so

the condition for the solvability of the Dirichlet problem is satisfied. Thus the Green's matrix exists.

Applying the fundamental identity (Theorem 11) to the Green's matrix and our solution of the Dirichlet problem (2.16), we find, in the usual manner

(U%771)= - j {f(s)X(Gi, H1) + g(s)Y(Gi, H1) ds

VQ1~r = -2 A {f(s)X(G2, H2) + g(s) Y(G2, H2)} ds

where U = (/I a)O, V = (1/r)4). Also, we can easily show, using the fundamental identity, that the Green's

matrix is Hermitean symmetric.

UNIVERSITY COLLEGE OF THE WEST INDIES, JAMAICA



154 JAMES SANDERS

REFERENCES

1. G. LAURICELLA, Sur l'intigration de l'Nquation relative a l'Nquilibre des plaques 61astiques encastrees, Acta Math., vol. 32 (1909), pp. 201-256.

2. A. E. H. LOVE, The Mathematical Theory of Elasticity, Cambridge University Press, 1927. 3. N. MUSKHELISVILI, Rdcherches sur les proble'mes aux limites relatifs a l'Nquation bihar-

monique et aux equations de l'elasticitg a deux dimensions, Math. Ann., vol. 107 (1933), pp. 282-312.

4. A. PLEIJEL, On Green's functions for elastic plates with clamped, supported and free edges, Proc. of Symposium on Spectral Theory and Differential Problems, Stillwater, Oklahoma, 1951.

5. JAMES SANDERS, A generalized function-theory and the related Dirichlet problem, Ann. of Math., vol. 64 (1956), pp. 523-543.

6. E. SCHMIDT, Bemerkung zur Potential theorie, Mathematische Abhandlungen, H. A. Schwarz, Berlin, 1914.

7. K. SCHR6DER, Zur theorie der Randwertaufgaben der Differentialgleichung AAu = 0, Math. Zeit., vol. 48 (1942-43), pp. 553-675.



A Generalized Function-Theory and the Related Dirichlet Problem: II

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