Mathematical Analysis and Numerical Methods for Science and Technology || Integral Equations

Chapter XI. Integral Equations

Introduction

Numerous stationary problems (see Chaps. IA and IB) are of the following type: find u satisfying

(1) Au =f

where A is a linear operator (continuous or not) in F, where F is a topological vector space and where f is given in F. The operator A can be a differential operator in an open set Q c IR" with boundary conditions at the boundary r of Q, or perhaps an integral operator(l), or even (as in the case of neutronics, see Chap. lA, §5) an integro-differential operator. This obliges us to examine, in those cases useful for applications where A is an operator from E to F, and E and F are spaces of functions or distributions, the structure of the operator A associated with our problem. The fundamental result in this sense is the Kernel theorem of L. Schwartz(2): if Q and 8 are open sets in IR" and if A is a continuous linear operator from £0'(Q) -+ £0'(8), then there exists a unique kernel distribution Ax.y = A(x, y) (on writing a distribution as a function) such that

(2) (Au)(x) = L A(x, y)u(y) dy , 'r/u E £0(Q) .

The kernel A(x, y) is a distribution on 8 x x Qy. Hence in this case, (1) will be written (symbolically), with Q = 8,

(3) L A(x, y)u(y)dy = f(x)

whenever problem (1) admits a unique solution

(4) u = Gf

(I) Some examples of integral equations (obtained from partial differential equations) have been studied earlier in Chap. II. There the search for solutions of the Dirichlet or Neumann problems (for the Laplacian in an open domain Q c IR") under the form of a simple layer or double layer potential leads to integral equations on the boundary r of Q; here we will study these examples in a different mathematical framework. (2) Which has been seen at §3.l2 in the Appendix "Distributions" in Vol. 2.

R. Dautray et al., Mathematical Analysis and Numerical Methods for Science and Technology© Springer-Verlag Berlin Heidelberg 2000

34 Chapter XI. Integral Equations

depending continuously on f, and we will write symbolically

(5) u(x) = 1 G(x, y)f(y) dy ,

where G(x, y) generally takes the name of the Green's kernel of the problem. In order to go further, it is now necessary for us to make precise the structure of this kernel and thus of A. We have seen in Chap. V that if the operator A is linear from 2C(Q) -+ 2C'(Q) and if A diminishes supports(3) then A is a differential operator. This has shown us why all local and linear phenomena in physics and mechanics are modelled with differential operators.

Remark 1. The (linear) differential operators with t(joo coefficients also diminish singular supports(3), but there is no question of this being a characteristic property: the "pseudo-differential" operators (see for example Chazarain-Piriou [1]) also diminish singular supports.

Remark 2. If the linear operator A diminishes supports, then G = A - 1 cannot diminish supports unless

A = kI (where I is the identity) .

Proof If that were the case, then we would have conservation of supports. Let us then suppose thatfis concentrated at {O};fwould be written

f = L C.D·c5(x); I_I .; k

there would exist u, equally concentrated at {O}:

u = L X.D·c5(x) 1.1 .; r

with Au = f, that is

P(D)u = f where P is a polynomial, and that is impossible when k < deg P. Hence A - 1 (if it exists) is not a differential operator except if P(D) is of order 0, that is to say, if

A = kI (I = identity) . o On many occasions we have encountered integral operators that are the inverses of differential operators. 0

Although the terminology is not very clearly fixed, we ca~ speak of the equation (1) as an integral equation, each time the kernel associated with the operator A is not

(3) An operator A diminishes supports if (see Chap. Y, §1.l):

't/ u E £iI(D) supp Au c supp u .

The singular support of a distribution W E £iI'(D) (see Chap. Y, §2.1) is the complement of the largest open set D' ,; D such that the restriction of w to D' shall be 'Coo; it is denoted by: sing supp w.

Introduction 35

that of a differential operator, that is to say, not of the form

A(x, y) = L Ca(x). D~b(x - y) (4), (6)

I_I " m

the distribution being supported by the diagonal of Q x Q. The particular name then given to the integral equation being considered depends on

- regularity properties of the kernel A(x, y) (5),

-properties of the support of A(x, y), - properties of invariance under groups.

Let us give some examples.

Example 1. We suppose

A(x, y) E rcO(Q x Q) .

Then (3) (and hence (1)) is a Fredholm integral equation(6). This situation can be generalised to kernels

(7) A(x,y) E U(Q x Q) , pE[1,oo]

for example, or more generally of the form

(8) A(x,y) = b(x - y) + A 1(x,y) , A 1(x,y)EU(Q x Q).

Equation (3) is then written

(9) u(x) + I A 1 (x, y)u(y) dy = f(x) .

Example 2. We assume that Q = [R and

(10) A (x, y) = b(x - y) + K(x, y)

where K is a regular function with support in the domain

{(x, y) E [R2; X ~ 0, y ~ x}

Equation (3) is then written

(11 ) u(x) + r K(x, y)u(y)dy = f(x)

and is called a Volterra integral equation of the first kind.

Example 3. We take Q = [R and

(12) A(x, y) = K(x, y) ,

(4) Or the same with m infinite, C. a locally finite family (see Chap. V). (5) See the Appendix "Singular Integrals" if A has a singularity.

o

o

(6) This terminology is taken here and subsequently in the large sense, i.e. that the solution of (1) is not necessarily subject to the Fredholm alternative.


K being as in Example 2, equation (3) is then written

(13) r K(x, y)u(y)dy = f(x) ,

and is called a Volterra integral equation of the second kind. o Example 4. Let us suppose that Q = IR" and that A commutes with all translations of IR", so:

(14) A (x, y) = A(x - y)

Then (3) is a convolution equation.

Example 5.

(15) { We suppose that A(x, y) admits a singularity at x = y , V x, Y E Q c IR"

for example, then (3) is a singular integral equation. Amongst the most important special cases, let us note

(16) 1

A (x, y) = 1 1"-2 x-y

and also in IR

(17) A(x, y) = pv_1_ or Fp_l_ (7)

x-y x-y

o

(Hilbert transformation) and at the same time all the cases seen in the Appendix "Singular Integrals". Remark 2 explains why, each time we have wished to solve a problem (1) modelling local phenomena, the solution has been found by inverting an integral equation (see Chaps. II and VII).

Remark 3. The method of Fourier integral operators (8) consists of writing a priori extremely general integral operators, but with a particular structure based on the Fourier transformation, then calculating the functions remaining free in the kernel in such a manner that the operator so constructed shall be "as close as possible" to the inverse of a differential operator. This is also the idea that we find in the complex variable methods developed in this Chap. XI. 0

In the modelling of physical phenomena, the integral operators appear directly whenever we wish to model non local phenomena, for example, action at a distance phenomena.

Example 6. The various phenomena of electromagnetic action by a potential responding to spatial distance: "Newtonian" potential (also called "Coulomb"

(7) Using the notations: pv = Cauchy principal value, Fp = finite part. (8) See Hormander [1].

Introduction 37

potential), potential of a simple layer or potential of a double layer, have been seen in Chap. II, §3 (for example

Au = L En(x - y)u(y)dy(y) (9)

for the potential of a simple layer). o Example 7. The phenomena of memories (action at a distance in time) have been modelled in Chap. lA, §3, with integral operators (for example

AUkh = I bijkh(t - S)Ukh(S) ds) .

o Example 8. In the equation of neutron transport seen in Chap. lA, §5, there arises the collision term

Ku(x, v) = Lf(X, v, v')u(x, v')dv' .

For all v' describing V and different from v, the collision models a transfer at a distance in the velocity space V. 0

At the same time as making precise the structure of the kernel of the operator A, it is convenient to ensure generally through the choice of functional spaces that the problem is well posed in the Hadamard sense. This generally fixes the functional framework which, whilst there are many possibilities, is chosen to be the simplest possible with regard to those properties of the model that are to be emphasised. The same holds for the methods to be used for solving or approaching the solution of the problem considered. This explains why in the study of the solution of a problem, it is sometimes not necessary in practice, despite its importance, to make explicit reference to the kernel itself.

(9) Recall: En is the elementary function of the Laplacian in [Rn, let 1/kn Ix - yln-2 for n ;;;, 3, with kn = (2 - n)un , Un being the area of the unit sphere in [Rn.

Part A. Solution Methods Using Analytic Functions and Sectionally Analytic Functions

Introduction

This part A presents a collection of different methods of solving integral equations. These methods all share a common tool, the use of analytic functions of the complex variable in the complex plane Co They enable the solution, explicit in many cases, of integral equations in a single variable (real or complex). We also obtain with the aid of the theory of the Hilbert problem, results concerning the existence and uniqueness of solutions for certain of these integral equations. In the last section we present many physical problems which can be solved by employing the methods developed here for these integral equations.

§1. The Wiener-Hopf Method

Introduction. Wiener-Hopf Equations

We consider the following problem:

Problem 1. Find f, null for t < 0, satisfying

(1.1) [+00 Jo k(t - to)f(to) dto = g(t) ,

where k is a given function defined on ~, called the kernel; g is a given function defined for t > 0.

t E ]0, ro[ ,

(1.1) is an integral equation called a Wiener-Hopfintegral equation of the first kind.

Example 1. The problem of linearly filtering a linear stationary system in the presence of an input signal leads to an equation of type (1.1). In this case, the kernel k depends on the statistical characteristics of the system studied. The input signal is f The given function g is the output signal of the system. In this problem, g is assumed to be obtained through measurements. 0

§l. The Wiener-Hopf Method 39

Another problem is:

Problem 2. Find a function f defined for all t E IR, satisfying:

(1.2) f(t) = L+oo k(t - to)f(to)dto + g(t)

where the kernel k and thefunction 9 are definedfor all t E IR. Equation (1.2) is called a Wiener-Hopf integral equation of the second kind. 0

Example 2. In neutronics the Milne problem, which we shall consider later, can be reduced to a Wiener-Hopf integral equation of the second kind.

1. The Wiener-Hopf Method

Let us note that the int~gral equation (1.1) is in fact a convolution equation, so that on making hypotheses of growth at infinity on the data and the solution, we can use the Fourier transformation. This is the transformation which is at the base of the Wiener-Hopf method and which we shall now make explicit for the solution of equation (1.1). We make the following hypotheses: the kernel k is bounded and satisfies:

(1.3) k(t) = O(exp( - b Itl)) , t -+ ± 00; b > 0 .

The given function 9 is locally bounded and such that

(1.4) g(t) = O(exp ct), t -+ + 00, 0 < c 0

g+(t) = o t < 0

f+ (t) = {fo(t) t > 0 t < 0 ;

t < 0

t > 0

we then verify that

(1.9) g_(t) = O(expbt) as t -+ - 00 .


Let us formally consider the Fourier transforms of these functions:

(1.10) F + (A) = t+: f+ (t)eiJ.t dt = 1+ 00 f(t)eiJ.t dt

and similarly for G +, G _ , K. The decreasing exponential type of behaviour of the kernel k shows that the Fourier transform K is defined for all A E 1Ft It further extends to the entire complex plane as an analytic function(lO) for all A such that

- b < 1m (A) < b .

Similarly the function G _ extends to an analytic function in the complex half-plane defined by

Im(A) a , Im(A) > c .

By formal Fourier transformation, (1.1) comes down to finding F + and G_ satisfying:

(1.11)

a relation which is valid for all A belonging to the common strip of analyticity of the functions K, F +, G _, G +, that is to say for A satisfying

(1.12) {A E C; max (a, c) < Im(A) < b} .

Then the type of equation (1.11) becomes part of the following general framework: let us consider:

Problem 3. Find F + and F _ such that

A 1-+ F + (A) is analytic for Im(A) > a, ..1,1-+ F _ (A) is analytic for Im(A) < b, a, b E IR, a < b, satisfying

(1.13)

where the given functions A, B, C, of the complex variable A are defined for

a < Im(.,1.~ < b .

(10) It will be noted that the analytic continuation K(2) with - b < 1m 2 < b is identified after a change of variable p = - i2, with the Laplace transform p f-+ K(ip) of the function k. See Chap. XVI for the principal properties of the Laplace transform. (11) The functions G _ (ip), G + (ip) and F + (ip) are also the Laplace transforms of the functions g ~, g + J+. The analytic properties of G ~, G +, F + result from the fact that the supports of the functions g ~ (resp. g + J+) are bounded on the right (resp. the left) and the properties at infinity of/and g. For the proof we refer to Chap. XVI. It will be noted that the change of the variable p to 2 = ip transforms the analytic properties in the half-planes Re p ;;. IX (IX E ~) of the Laplace transforms of the functions with support bounded on the left into analytic properties in the half-planes 1m 2 ;;. IX.


The solution of Problem 3 is then based on two ideas, namely the factorisation and the decomposition of an analytic function defined in a strip in the complex plane. More precisely if we know how to factorise AlB into the form:

(1.14)

with

(1.15)

A(2)

B(2)

H+(2)

H_(2)

H + (2) of. 0, for all 2 (with 1m (2) > a)

H _ (2) analytic in 1m (2) < b; 1

H + (2) analytic in 1m (2) > a;

H _ (2) of. 0, for all 2 (with 1m (2) < b) ,

then equation (1.13) is equivalent to

(1.16) H + (2)F + (2) + H _ (2)F _ (2) = C(2~~) (2)

C(2)H+(2) . If we then know how to decompose A(2) mto the form:

(1.17)

(1.18)

C(2)H + (2) = G (2) _ G (2) (12)

A(2) + -

{ G + (2) analytic in 1m (2) > a G _ (2) analytic in 1m (2) < b ,

then equation (1.16) can be written

The two members of equation (1.19) are two functions analytic in the strip a < 1m (2) < b, where they are equal. They are therefore both equal to the same entire function E(2). In practice, this entire function is completely determined by its behaviour at infinity in the complex plane. We finally deduce all the solutions of equation (1.13):

1 F (2) = E(2) + G + (2)

+ H+(2)'

(1.20) F (2) = E(2) + G - (2) - H _ (2)

Remark 1. The Wiener-Hopf method of solving equation (1.13) gives solutions which do not always lead to solutions of equation (1.1). The hypotheses made on the function f imply in particular that:

1m (2). F + (2) remains bounded when 1m (2) ;::: a 1 > a .

(12) The G + and G _ used here must not be confused with those in formula (1.11).


It is then possible to characterise the Fourier transforms of J+ (t) with the aid of a Paley-Wiener theorem (see Rudin [1]). 0

We shall now give information about the decomposition, and then theJactorisation, of an analytic function defined in a strip in the complex plane.

2. Decomposition of an Analytic Function Defined in a Strip in the Complex Plane

We prove

Theorem 1. Let A 1--+ G(A) be aJunction analytic in the strip IX <::: 1m (A) 0, C > 0; VA, a ~ Im(A) ~ b ,

then it admits a decomposition oj the Jorm:

(1.22)

where

(1.23) { G + is analytic Jor 1m (A) > a

G _ is analytic Jor 1m (A) < b .

Proof We consider the following contour C /' in the complex plane.

Im(),.)

Ii

- d + ib b d + ib

c~

- d + ia a d + ia

a

Fig. 1

Re

From the Cauchy formula, for all A within the interior of the domain ofthe contour C,. we have

(1.24) G(A) = ~ r G(~) d~ . 2mJc ~ - A ,.

If G satisfies (1.21), then as d -+ + 00 in (1.24), we get:

(1.25) G(A) = -. --d~ - -. --d~ 1 f+ oo + ia G(~) 1 f+OO+ib G(~)

2m -oo+ia ~ - A 2m -oo+ib ~ - A

§1. The Wiener-Hopf Method 43

Putting

(1.26) ! 1 f+oo+ib G(~) G+(A) = -. --d~

2m -oo+ib ~ - A

__ 1 f+ 00 +ia G(~) G - (A) - 2 . .;; _ 1 d~ ,

1[1 _ 00 + IQ" A

G + and G _ are analytic from (1.21) for 1m A a respectively and Theorem 1 is proved. 0

Remark 2. The decomposition obtained through Theorem 1 is not unique. In particular, we can add to G + and G _ the same entire function. 0

Remark 3. The hypothesis (1.21) implies that the function G + given by (1.26) satisfies 1 G + (A) 1 ~ C,} for A such that 1m A - a ~ 151 Rd I, t5 > 0, and tends to zero as 1m A..... + 00 for Re A fixed; whenever y > 1, we further have

1 G + (A)I ~ C, for 1m A - a ~ t: > 0 .

3. Factorisation of an Analytic Function Defined in a Strip in the Complex Plane

We now prove

Theorem 2. If H is a function analytic in the strip

r:J. < Im(A) 0, y > 0, VA, r:J. < ImA a > r:J. respectively ImA a and 1m A < b.


Proof We wish therefore to factorise H(A) into the form (1.28) with

H + analytic for 1m (A) > a,} b p . a<a< < .

H _ analytIc for 1m (A) < b,

Whenever H(A) is such that LogH(A) is well defined in the strip a < Im(A) < p and satisfies hypothesis (1.21), it suffices to apply Theorem 1 to LogH(A) in order to obtain

now under the conditions (1.27), there exists an integer p E ;£,(14) (see §3, Proposi-

- H(A) (A - ia)p tion 3 and Theorem 3) such that for H(A) = --,------:-p the function H(oo) A - I

LogH(A) is defined in a unique way and satisfies (1.21). Whence Theorem 2. 0

Remark 4. If the function H does not satisfy (1.27)i) (but does satisfy (1.27)ii)), then the method of Theorem 2 cannot be applied directly. Suppose H has zeros with finite multiplicities within the strip a < 1m A < P and denote by al , a2 , ••• ,am the zeros, distinct or not, of H. Let us introduce the function L given by

(1.30) L(A) = H(A)(A - ip)m (A - a l ) ... (A - am)

This function satisfies the hypothesis (1.27); it therefore factorises into the form

(1.31) L(A) = L+ (A) . L_ (A) ,

L+ (resp. L_) being analytic for 1m A > a (resp. 1m A < b), we then deduce from (1.30)-( 1.31):

(1.32)

o

Remark 5. Decomposition of a function is not unique. In the same way as for factorisation, one can multiply the numerator and denominator by the same entire function having no zeros in C.

. h . f l' - , H+ (A) k WIt the notatIons 0 the proof of Theorem 2, lor H(A) = -_--, we can ta e H_(A)

(14) Note the analogy between the problems treated here by Theorems 1 and 2 and the Hilbert problem which we will meet in §3. (15) This method of eliminating zeros preserves the strip Q( < 1m A. < fJ (otherwise one would of course have to modify this strip in order to eliminate the zeros).


it (A) = eG ± ().) with (1.26), and the behaviour of H+ (A) (hence of H+ (A)) at infinity can be deduced from that of G ± (A) by Remark 3. - - 0

4. Application to the Wiener-Hopf Integral Equation of the Second Kind

4.1. Generalities

Let us now consider Problem 2 which consists of having to find f satisfying (1.2) (Wiener-Hopf integral equation of the second kind). We suppose that the kernel k satisfies (1.3) and that we have this time defined g + and g _ by:

(1.33) {g(t), t > 0

g + (t) = 0 t < 0 ' {o t > 0

g _(t) = . g(t), t < 0

We suppose that g + and g _ satisfy:

(1.34) {g+(t) = O(exp(ct)), t -+ + 00, o < c 0

t ~ 0 .

We now propose to search for fbounded on all bounded sets in IR and such that

(1.36) f(t) = O(exp(at)) , t -+ + 00, 0 < a a

G + analytic in 1m (A) > c

F _ analytic in 1m (A) < b

G _ analytic in 1m (A) < b

K analytic in - b < 1m (A.) < b .

The functions F +, G +, F _, G _, K have in common the strip of analyticity max (a, c) < 1m (A) < b and satisfy there (from (1.2) by Fourier transformation):

(1.38)

an equation of type (1.13) which can be solved by factorisation and decomposition. We now study in detail an example of the Wiener-Hopf equation of the second kind.


4.2. Study of an Example

Let us consider the following integral equation:

(1.39) f(t) + 4 L+oo e -It - to~(to)dto = e- t , 0 < t < + 00 .

Putting g _ (t) = 0, equation (1.39) leads to (1.38) with

1 - K(') = A.2 + 9 G (') 1 G (') 0 I\. A. 2 + l' + I\. = 1 _ iA.' - I\. = ,

the strip of common analyticity being:

- 1 < 1m (A.) < + 1 ;

an obvious factorisation of 1 - K (A.) consists of taking, for example,

(1.40) A. + 3i A. - i

H+(A.) = T+i' H_(A.) = A. - 3i '

which leads to the following decomposition:

(141) G (A.)H (A.) _ A. - i _ 1 i. . + - - (1 - iA.)(A. - 3i) - 2(1 - iA.) + 2(A. - 3i) ,

we then have

whence

(1.42) 1 F (A.) = (A. + i)E(A.) 'i

+ A. + 3i + 2(A. + 3i)

A. - 3i i F_(A.) = - ~E(A.) + 2(A. _ i) ;

whenever E(A.) = 0, f+ (t) is given by

(1.43) 1

f(t) = 2e-3t; t > 0 .

In fact this solution is the unique solution of the integral equation (1.39) satisfying the boundary condition (1.36) with a = - 3. In effect, F + (A.) is bounded (for A. =1= - 3i) and tends to zero at infinity and we have:

E(A.) = A. + 3i F (A.) ___ -,-A. + i + 2(1 - iA.)

whence it follows that the entire function E(A.) is bounded and tends to zero at infinity. Liouville's theorem then implies that E(A.) = O. 0

§1. The Wiener-Hopf Method 47

5. Application to the Milne Problem

i) Physical Introduction

The (stationary) transport of neutrons in a medium with domain X has been introduced in Chap. lA, §S.3. It is described by equation IA (S.19)-(S.20). To conform with the notation in the present chapter, we will here denote the domain by Q. We now assume that Q = {(x, y, z) E 1R 3, X > O}, and is occupied by a homogeneous(16) medium and that the condition (S.l9)ii) on the boundary aQ (with equation x = 0) is the absence of a neutron current entering Q, namely:

g(x, v) = 0, v X E aQ and V v > 0 .

We will further assume that all velocities v have the same modulus I vi which we take equal to 1, that diffusion at the time and place of collision will be isotropic, that there is no absorption of neutrons and that the problem is invariant under rotations about an axis parallel to the Ox-axis and invariant under every parallel translation in the Oyz-plane. Taking the length 1/17(17) to be unity, characterising the direction of the velocity v by

cos (v, Ox) = Ji

and noting that f is a constant because of the hypotheses of isotropy and non absorption of neutrons, (S.19) and (S.20) can here be written (the unknown, the angular density of neutrons, being here denoted by 1/1):

(1.44) 1 al/l 1 f + 1 Ji ax (x, Ji) + I/I(x, Ji) - 2 -1 I/I(x, Ji) dJi = ° ,

x > 0, - 1 < Ji < + 1

(I.4S) 1/1(0, Ji) = 0, Ji > ° . We will denote by:

I/Io(x) = f:: I/I(x, Ji)dJi

the total neutron density at the point x. The current density j(x) in the direction of the velocity v, denoted in the general case by

j(x) = Iv vl/l(x, v)dv

(16) That is to say the given 1: and f appearing in IA (5.20) are independent of (x, y, z). (17) Which is the mean free path of the neutron since I vi = 1. .


is here written, in the direction opposite to the Ox-axis:

j(x) = - f-+ll J-lt/l (x, J-l) dJ-l, v X E IR+

(1.44) and (1.45) imply that j(x) is constant. We will put

(1.46) j(x) = 1 ,

which amounts to saying that a current, constant and equal to 1, of neutrons towards the positive infinity of the Ox-axis, flows everywhere through the medium under study. We seek to calculate the angular distribution of neutrons leaving the medium occupying the domain Q, namely:

(1.47) t/I(O, Jl) for J-l < 0 (i.e. - 1 < J-l < 0) .

ii) Utilisation of the Wiener-Hopf Methods

We now propose first to solve (1.44) with (1.45)-(1.46), and then to determine J-l H t/I(O, J-l) for J-l E ] - 1,0[. In order to do this, we will transform the problem in such a way as to reduce the solution of (1.44)-(1.45)-(1.46) to the solution of a Wiener-Hopf integral equation of the second kind. More precisely, we will prove

Proposition 1. The problem defined by (1.44), (1.45), (1.46) reduces to the solution of a Wiener-Hopf integral equation of the second kind with the kernel k having an O(e - IXI) behaviour at infinity.

Proof For fixed J-l equation (1.44) can be considered as a differential equation in x which can be integrated; we get:

J-l>0

(1.48)

J-l < 0;

on integrating with respect to J-l we deduce:

(1.49)

t/lo(x) = ::0 _uE(IX - tl)t/lo(t)dt 1 1 (+00

E(t) = r ~du, t > 0 ; Jt u

(1.49) is a Wiener-Hopf integral equation of the second kind, homogeneous (g = 0), with kernel k = E/2 having the behaviour at infinity stated in the proposition. If we know how to determine t/lo satisfying (1.49) and (1.46) then

§l. The Wiener-Hopf Method

t/I(O, Jl), Jl < ° will be given by:

(1.50) 1 1+ 00 t t/I(O, Jl) = - -2 t/lo(t) eli dt , Jl < °

Jl 0

whence Proposition 1.

We can now prove

49

o

Theorem 3. The angular distribution of neutrons leaving the region at x = 0, namely t/I(O, Jl), Jl < 0, is given by:

(1.51) t/I(O, Jl) = - 2~ 4'>0 ( - ~), Jl E ] - 1,0[

where

(1.52) j3(s + 1) 4'>o(s) = 2 r+(s) , s > °

s

(1.53) ( 1 f-/l+iooLOgr(A) ) r + (s) = exp -. dA , s > 0, f1 > °

2m -/l-ioo A - S

S2 - 1 (argtanhs) ° r(s) = --2 - 1 - , s > . s s

(1.54)

Proof We shall follow the method described in Sects. 1 and 4, noting that x E IR + = ] 0, + 00 [; it is more convenient here to use the Laplace transformation rather than the Fourier transformation. Thus let us put

(1.55)

1+00 f 4'>(s, Jl) = 0 t/I(x, Jl) e -sx dx

l SEC; Re (s) > So (so to be determined) ;

integrating with respect to Jl, we get

(1.56)

and (1.51) then follows immediately from (1.50) and (1.56). To determine 4'>o(s), we note that, from the Laplace transform of (1.48), we have:

14'>(S' Jl) = G 4'>o(s) + Jlt/l(O, Jl)) 1 : SJl

(1.57) SEC, So < Re (S), (so to be determined)

Integrating (1.57) between Jl = - 1 and Jl = + 1, we obtain:

14'>0(S)[1 - argtanhsJ = - !fO 4'>0( - 1/Jl) dJl S 2 -1 1 + SJl

SEC, So < Re (S) < 1, (so to be determined) (1.58)


Note that (1.58) is the equation we would have obtained directly from (1.44)-(1.45) by Laplace transformation. We shall now apply the Wiener-Hopf method in order to solve (1.58). To do this, we must determine the domain of analyticity for the functions which intervene and in particular the constant so' Let us consider equation (1.58); the second member is bounded for Re (s) < 1. A Laurent expansion of the first member in the neighbourhood of s = 0 shows that <1>o(s) admits a singularity at s = O. We therefore take

(1.59) So = 0 ;

hence we seek

(1.60) S 1-+ <1>o(s) analytic for Re (s) > 0, SEC.

Since

1 arg tanh s . I" s 1-+ - IS ana ytlc III - 1 < Re(s) < 1, SEC

s

1 fO <1>0( - 1 Ill) . . . S 1-+ g(s) = - -2 dll IS analytic III Re(s) < 1, SEC,

-1 1 + Sll

the three functions considered have a common strip of analyticity

(1.61 ) o < Re (s) < 1, SEC.

argtanhs . Noting that s 1-+ 1 - has a double zero at s = 0, we are led, III order to

s use the factorisation method(l8), into introducing

r(s) = (S2 - 1)(1 _ argtanhs) S2 s

The function r is analytic in a strip defined in (1.61)(19), and for f3 E ]0, 1[ there exists a constant C(f3) > 0 such that

C(f3) Ir(s) - 11 :::; W for 1 < - f3 < Re (s) < f3 < 1 .

Thus r satisfies the hypotheses of Theorem 2. From that theorem it factorises into the form

(1.62) r(s) = r - (s) r + (s)

(18) This choice essentially permits the application of Theorem 1 to the function Log ,(s) leading to a decomposition given by Remark 3, and to the subsequent obtaining of the functions, + (s) and, _ (s) having the indicated properties. (19) Or even in the strip - 1 < Re(s) < I.

§1. The Wiener~Hopf Method

with

(1.63) { t + (s) #- 0 analytic (and without zeros) for s E 1[:, Re (s) > 0

L (s) #- 0 analytic (and without zeros) for s E 1[:, Re (s) < 1 .

Hence (1.58) can be written

(1.64) S2 c1>o(s) = (s - l)g(s) = E(s) (s + l)t + (s) L (s)

h () 1 fO c1>0( - 1/J.l)d d (). . I' f . were 9 s = - - J.l an E s IS an entIre ana ytIc unctIon. 2 ~ 1 1 + SJ.l

51

In order to determine E(s) it is necessary to examine its behaviour at infinity.

Noting that ~() (resp. _1(_) is bounded for {s E 1[:; Re (s) > O)} (resp. t + S t ~ s)

S E 1[:; Re(s) < 1) and that the same holds for the function s 1-+ sg(s) for (s E 1[:;

Re(s) < 1), we see that s 1-+ E(s) is bounded in the whole complex plane if sc1>o(s) is analytic and bounded for Re(s) > 0, s E IC. If we seek 1/10 such that

(1.65)

then the Laplace transformation s 1-+ c1>o(s) is analytic for Re (s) > 0 and sc1>o(s) is bounded for s E 1[:, Re (s) > O. Hence the entire function E(s) bounded in the whole complex plane is a constant C to be determined. We thus obtain

(1.66) Is + 1

c1>o(s) = C . --2 - r + (s)

C constant toS

be determined;

(1.66) gives all the solutions 1/10 of (1.49) having a polynomial behaviour at infinity from (1.65). In order to determine the constant C, we make use of (1.46). We first note that from (1.66) we have:

(1.67)

From (1.58), whose second member g(s) tends to j = 1 as s -+ 0, we deduce, taking into account from this that

(1.68)

arg tanh s S2 1 - --s - 3 '

3 c1>o(s) ~ 2" as s -+ 0 ,

S Res> 0 .

On comparing (1.67) and (1.68) we obtain:

(1.69) Cr + (0) = 3


It therefore remains to evaluate r + (0) in order to determine the constant C; now for all f3 E JO, 1 [, we have(20):

(1.70) r+(O) = exp(~f-/i+iOO Logr(A.) dA.) . 2m -/i-ioo A.

U sing the classical method which involves calculating the integral of (1.70) around the contour LI avoiding the origin with a semicircle of centre 0 and radius e (see Fig. 2) and noting that thanks to the sign of r, this integral reduces to the contribution from the semicircle of centre 0 and radius e, we get on letting

e ~ 0: r + (0) = Ji Hence from (1.69) we deduce:

(1.71)

whence Theorem 3. o

- I - ~ -E

Fig. 2

Remark 6. Let us note that from the asymptotic behaviour of <1>o(s) as s ~ 0 or s ~ 00, there results:

1) the total density of neutrons at x = 0 is

t/to(O) = J3 ; 2) the total density of neutrons when x ~ + 00 is equivalent to

x ~ + 00 .

o

(20) This formula is also given by Proposition 3 of §3.


6. Application to the Dock Problem


The dock problem is the one which consists in investigating the waves propagating on the surface of the sea whose depth is finite (and assumed for simplicity to be constant) and which is partially covered by a dock. We have seen in Chap. I, §1.9, the problem of the oscillatory motion of water in a canal. The problem studied here will be formulated in a completely analogous way: we will especially compare the relations (1.73), (1.74), (1.75) and (1.76) which follow, with relations (1.97) in Chap. IA; as in Chap. I, §1.9, we will consider the water to be a heavy homogeneous fluid, incompressible and without viscosity. The sea is assumed to occupy the region Q c IR 3 defined by

(1.72) Q = {(x,y,z) E 1R3; x E IR, - d ~ y ~ 0, Z E IR} .

The free surface (y = 0) is partially covered by the dock which is assumed to occupy the region

D = {(X,y,Z)EIR 3;y = O,x < O,ZEIR}

The free surface of the sea is the region

S = {(x, y, z) E 1R3; y = 0, x > 0, Z E IR}

The motion of the water at a point (x, y, z) and a time t is represented by a velocity potential cp(x, y, Z, t) whose gradient is the fluid's velocity at this point. It is modelled by the equations of motion of irrotational fluids which here reduce to the single equation

(1.73) Llcp = 0, X E Q ;

It is now necessary to express the boundary conditions

a) over the bottom of the sea, b) over the dock, c) at the free surface.

Over the bottom of the sea and over the dock the normal velocity of the fluid will be null, namely:

(1.74) 8cp 8y (x, - d, z) = 0, X E IR , Z E IR

(1. 75) 8cp 8y (x, 0, z) = 0, x < 0 , Z E IR .

Finally, at the free surface we impose the so-called linearised free surface equation:

(1.76) 82cp 8cp 8t 2 + 9 8y = 0, (x, y, z) E S .

(g is the acceleration of gravity). If we use the method of separation of variables to search for particular solutions of equation (1.73) with (1.74), (1.75), (1.76), which are periodic in time t and space Z of


the form

(1.77) { qJ(x, y, Z, t) = <P(x, y) exp i(kz - wt)

k and w > 0 ,

we are led to these equations:

(1.78) A 0, where

w2 m =-.

9

Since the problem thus posed involves unbounded regions, for which the problem must be well-posed, it is necessary to augment conditions (1.78) to (1.81) with hypotheses about the behaviour at infinity. Consequently we impose on the function <P the supplementary conditions:

{i) as x ~ + 00, <P is bounded ,

(1.82) ii) as x ~ - 00, <P(x, y) = O(e - klxl) .

Condition (1.82)ii) results from the fact that the dock forces the water velocity to zero as x ~ - 00 and one can show that it is legitimate to seek a function <P having exponential decrease of the form e - klxl. Furthermore, the unknown function <P will present a singularity at the point x = 0, y = 0 and an inspection of equation (1. 78) in polar coordinates shows that this singularity will be of the form:

<P(x, y) ~ f(r) Log r + g(r) , r = Jx 2 + y2

where f and 9 are two functions both regular at the origin.

ii) Utilisation of the Wiener-Hopf Method

We will now show that in order to solve (1.78) to (1.81) we can reduce the problem to the solution of an equation of type (1.13). To do this, we will introduce the partial Fourier transformation in the variable x of the function <P(x, y) defined by

(1.83) I/I(A, y) = I-+ 0000 <P(x, y) eih dx, A E IR .

Equations (1.78), (1.79) become

) 02 1/1 2 2 (1.84 oy2 (A, y) - (A + k ) 1/1 (A, y) = 0, - d < y < 0 ,

01/1 (1.85) oy (A, - d) = 0, A E IR ,


whence we deduce

(1.86) 1 I/!(A.,y) = /~'\~~d[e-/lY + e/l(2d+ y )]

J1. = (A.2 + k2)1/2 ;

Now (1.86) shows that in order to solve our problem completely, it suffices to calculate

I/! (A., 0) = f-+oooo tP(x,O)ei;'xdx, A. E IR .

Let us therefore introduce the following functions:

(1.87) I/! + (A., y) = L + 00 tP(x, y) ei.!x dx ( = f-+ 0000 tP + (x, y) ei.!x dX) ,

(1.88) I/! - (A., y) = f 00 tP(x, y) ei.!x dx ( = f-+: tP - (x, y) ei;'x dX) .

Equation (1.81) implies:

(1.89)

and (1.80) implies:

(1.90) iJ.p_ ay(A.,O) = 0, A. E IR .

Then making use of the relation I/! = I/! + + I/! _, we deduce from (1.89)-(1.90)

(1.91 ) vI/! + vI/! _ _ .1. _ vI/! vy + vy - m'f' + - vy .

We now calculate ~~ (A., 0) with the aid of (1.86) and finally obtain:

(1.92)

f [1 - ;coth(J1.d)J I/! + (A., 0) + I/!_(A.,O) = 0, A. E IR

l J1. = JA. 2 + k2 •

Equation (1.92) is of the type (1.13) and thefactorisation method allows us to obtain all the solutions. From the hypotheses on the bahaviour at infinity (1.82) we see that

A. 1--+ I/! + (A., 0) is analytic for 1m (A.) > ° A. 1--+ I/! _ (A., 0) is analytic for 1m (A.) < k

and the common strip of analyticity is then ° < 1m (A.) < k.

56 Chapter Xl. Integral Equations

Let us therefore examine the problem of factorising H defined by

(1.93) H(A) = 1 - J m coth(dJA2 + k2 ) • A2 + k2

This function does not satisfy hypothesis (1.27)i) of Theorem 2, because it has zeros in the strip 0 < 1m (A) < k, but it nevertheless satisfies hypothesis (1.27)ii) of that theorem because

(1.94)

J v. > 0, there ;x;sts C. > 0 with

lIH(A) - 11 < IAI; IAI.... + 00 , o < ex < 1m A < k .

We also note that this function H has a pole wherever

sinh(Jld) = 0, namely wherever Jld = 2inn, n E 7L

that is to say for A2 = - k 2 - (4n 2 n 2 /d 2 ).

These poles, situated on the imaginary axis, are all outside the strip 0 < 1m (A) < k which interests us. The zeros of the function H are the roots of the equation

(1.95) coth (Jld) = Jl/m

which admits only purely real and purely imaginary roots. The purely imaginary roots lead to values of A outside the strip which interests us. Equation (1.95) admits two real and opposite roots ± Jlo with Jlo > 0, Jlo > m. i) If Jlo ~ k, then A~ = Jl6 - k2 ~ 0, and H(A) has no zero in the strip o < ImA < k. ii) If m < Jlo < k, then A~ = Jl6 - k2 < 0, Ao has the form Ao = ± iXo with

Xo = Jk 2 - Jl6 < k and H(A) has a single zero iXo in the strip 0 < 1m A < k. Study of the function H(A) leads one to put:

(1.96) {i) H(A) = H(A)(A - ik)/A if k < Jlo

ii) H(A) = H(A)(A - ik)/(A - Ao) with Ao = iXo for k > Jlo ,

where the function H(A) so defined is such that Log H (A) satisfies Theorem 1 (with y = 1)(21); hence H(A) decomposes into:

(1.97)

with:

H + (A) (resp. H _ (A)) analytic and without zeros for 1m A > 0 (resp. < k) .

(21) Or again, following the terminology of §3, Proposition 3, the index of H()') relative to the line 1m). = fJ (with fJ fixed, 0 < fJ < k) is null.

§2. Sectionally Analytic Functions

Moreover (from Remark 3) ! I H + (A) I ~ C b for every angular sector 1m A - a ;;:: c51 ReA. I ,

a and c5 > 0 (1.98) IH_(A)I ~ Cb for every angular sector ImA - a' ~ -c5IReA.1 ,

a' < k, c5 > 0 ,

57

and H+(A) and H_(A) tend to 1 as Izl -+ + 00, with ReA. fixed. Furthermore, because of hypotheses (1.82), there exists a constant C > 0 such that:

(1.99) { It/I+(A,O)I ~ CjImA, ImA > 0 ; It/I_(A,O)I ~ C/(k - ImA) , ImA < k .

i) For k < /1-0' equation (1.92) gives, using (1.96)i) and (1.97) :

(1.100)

where E(A) is an entire function, bounded in the angular sectors of (1.98). Admitting further that E(A) is constant, we get:

(1.101) { t/I + (A, 0) = C/(AH+ (A)) , t/I_(A,O) = - C/((A - ik)H_(A)) ,

and t/I(A,O) = t/I + (A, 0) + t/I- (A, 0) gives the solutions of the problem being studied(22), by (1.86) and (1.83). ii) For /1-0 < k, equation (1.92) gives, using (1.96)ii) and (1.97):

(1.100)' (A - Ao)H+(A)t/I+(A,O) = - (A - ik)H_(A)t/I_(A,O) = E(A).

Admitting further that E(A) = C (constant), we get:

(1.101)' { t/I+(A,O) = Cj((A - Ao)H+0)) , t/I_(A, O) = - C/((A - ik)H_(A)) ,

whence we deduce the solution of the problem studied, by (1.86) and (1.83).


Introduction

We introduce in this section the notion of sectionally analytic functions which we will denote for brevity by s.analytic functions. These functions are analytic throughout the exterior of a curve L in the complex plane and discontinuous across this curve. We will associate with these functions their Cauchy integrals along the curve L. The Plemelj formulas then permit us to make precise the limits of

(22) Note that for this solution it is necessary to take ). real. ii + ().) is then obtained by - - - ~ H + ().) = lim H(z)H _ (z) and has zeros, if J1.0 ;;. k, for ).0 = ± v' J1.0 - k . z ..... A., Imz > 0


s. analytic functions from either side of the curve L. The Hilbert inversion formula permits the "inversion" of the Cauchy integrals and provides us with a powerful tool for solving certain integral equations posed on the curve L.

1. S. Analytic Functions

We will now define the s. analytic functions. To that end we recall some definitions relating to curves in the plane 1R2.

Definition 1. An arc of a differentiable curve (or differentiable arc) is the set of coordinate points (x, y)

{X = x(t) t ~ t ~ t (23)

Y = y(t) a "" "" b '

where the functions x(t) and y(t) are differentiable and such that

( dx)2 (dy)2 dt + dt # 0; V t, ta ~ t ~ tb .

The extremities, denoted by a and b, are the images ofta and tb' We assume that there is no double point, which is expressed by the property

{X(t 1) = X(t 2)

¢> t1 = t2 . y(td = y(t2 ) o

Definition 2. By a differentiable contour we shall mean a closed differentiable arc, that is to say a differentiable arc whose extremities a and b coincide as also do the tangents at these two points. We express this last property by

o Arcs and contours have a sense of direction corresponding to increasing t. We shall assume that contours are positively orientated in accordance with this sense of direction, which corresponds to the illustration below:

L

Fig. 1

(23) We assume here (for simplicity) that t. and tb are finite, so that in what follows an arc of a differentiable curve will be a compact set in [R2.

§2. Sectionally Analytic Functions 59

Definition 3. By a differentiable curve we mean a finite union of differentiable arcs or contours, with empty intersection. A piecewise differentiable curve is the finite union of differentiable arcs whose extremities can coincide. In this case, the tangents at these points are in general different. 0

Throughout the rest of this exposition, an arc, a contour or a curve will designate, in the absence of any other precision, a differentiable arc, a differentiable contour or a differentiable curve. We will denote by s the curvilinear abscissa along arcs or contours, which is defined by:

ds J(dX)2 (dy )2 dt = dt + dt ' s(a) = 0 ;

I = s(b) is the total length of the arc or contour.

Definition 4. A function ljJ of the plane 1R2 is said to be sectionally continuous (s. continuous) relative to an arc or a contour L, if it is continuous at every point z, z E IR 2 \ L, and if it is continuous up to both sides of the arc or contour L (with the possible exception of the two extremities of the arc). 0

An s. continuous function therefore admits possibly different limits on either side of the curve L. We will denote by ljJ e and ljJi these limits in the case of a contour, ljJi being the limit in the interior of the contour and ljJ e the limit in the exterior. The functions ljJi(t) and ljJe(t) are thus continuous functions of the abscissa t.

Definition 5. A function ljJ().) of the complex plane is said to be sectionally analytic (or s. analytic) relative to a differentiable curve L if:

1) it is analytic at every point not situated on the curve L, 2) it is s. continuous relative to the curve L, 3) it satisfies in the neighbourhood of every extremity a of the arcs of the curve L in

this condition:

c IljJ(A11 ::::; I). _ ala; ). rt L 0::::; a ::::; 1, where C is constant.

o In the rest of this exposition, we will say only "s. analytic function", without each time making precise that it is "relative to the curve L". Condition 3) has been introduced by N.I. Muskhelishvili.

2. Cauchy Integrals and Plemelj Formulas

Let L be a contour or an arc. We shall now examine the properties of Cauchy integrals of the form:

(2.1) ljJ().) = ~ [ f(z) dz, ). E C\L 2mlz -).

where f(z) is a function defined along the curve L.


Whenever the functionJ(z) is integrable, the function 4> (A) is defined and analytic at every point A not situated on the curve L. But it has no meaning when the point A is

on the curve L, because then the function zJ~) A is no longer integrable whenever

z runs along the curve L. Under certain hypotheses concerning the functionJwe shall define, whenever A is on the curve L, a function related to the integral (2.1) which will be called the Cauchy principal value.

Definition 6. By the Cauchy principal value, which we will denote by

1 f J(z) d h fi II . I" h . . -2' --, z, we mean teo owmg Imlt, w en It eXists: TCl r.Z-1I.

(2.2) _1 J. J(z) dz ~ lim _1 r J(z) dz A E L (24), 2niJr.z - A .~o2nil\(LC1B(i..'))z - A '

where B(A, e) is the ball(25) oj centre A and radius e in the complex plane.

Proposition 1. Let L be a contour or an arc oj class rc1,p (26). Let us suppose that the Junction J(z) be Holderian on the curve L, i.e. satisfies:

(2.3) IJ(Zl) - J(z2)1 ~ Clz1 - z21P ; VZl' Z2 E L ,where 0 < f3 ~ 1 .

Then the Cauchy principal value given by (2.2) exists at each point A oj the curve L (except at the extremities in the case oj an arc) and is a continuous Junction oj A (except at the extremities in the case oj an arc).

Proof We first consider the case where L is a contour. We have

(2.4) {

_I r J(z) dz = _1 r J(z) - J(A) dz 2nijL\(LC1B(i.,'»Z - A 2niJL\(LC1B(i.,,» z - A

J(A) 1 dz +- --. 2ni L\(L C1 B(i.,'» z - A

The hypotheses (2.3) shows that the first term in the second member takes for its limit, as e --+ 0, the following integral

_1 r J(z) - J(A) dz . 2ni JL z - A

The limit of the second term in the second member is J(A)/2 (the contour being

(24) We assume thatJis a measurable function on L, with for exampleJ E LP(L), I .;; p < + 00, which ensures the existence for almost all A. E L of the Cauchy principal value. (25) We will use here and subsequently the word "ball" instead of "disc". (26) Recall (see Chap. II, §3, Definition 3) that a regular open set Q has a boundary r of class CC I,p 0 < f3 .;; 1 if in a neighbourhood of each point z E r, there exists a normal parametric representation (R, U, (D, (X) (see Chap. II, §1, 3.a.) such that

Igrad(X(x') - grad(X(xli .;; Clx - x'IP 'V x, x' E (D .


taken with positive orientation here). We thus have

(2.5) _1 l f(z) dz = _1 r f(z) - f(A) dz + f(A) . 2ni l z - A 2nijL z - A 2

The continuity of this expression with respect to A then follows from the Lebesgue theorem. Lastly let us examine the case where L is an arc. We can find a positively orientated contour l which contains the arc L. We will then consider the function/which is the extension by zero of the function f to all the contour l. Then

_1 r f(z) dz = _1 r ](z) dz 2nijL z - A 2nijr z - A '

and we are thus reduced to the previous case. The only difference is that the function 1 is discontinuous at the extremities a and b of the arc (except if f(a) = f(b) = 0) and hence the Cauchy principal value is not in general bounded at the extremities of the arc. We have likewise the formula analogous to (2.5):

(2.6) _1 l f(z) dz = _1 r f(z) - f(A) dz + f(A) Log b - A 2ni l z - A 2ni JL z - A 2ni A - a '

AEL, Ai=a,b. o

Remark 1. Let L be an orientated arc. We can immerse it in a positively orientated contour l. The bounded domain delimited by this contour is called the interior. This permits the definition, for s. continuous functions on the arc L, of the interior limit <Pi and the exterior limit <Pe (see Fig. 2) in a unique manner.

b ___ ~ __ ~

a

Fig. 2

The Cauchy integrals defined by (2.1) are in fact s. analytic functions and the limits <Pi and <Pe are calculated with the aid of the Cauchy principal value. This is the object of the next theorem.

Theorem 1. Let L be a contour or an arc of class rt l.P; then the function

<p(A) = ~ r f(z) dz, A ¢ L 2mlz - A

is s. analytic whenever the function f is Holderian (i.e. satisfies (2.3)). In this case the non tangential limits <Pi and <Pe of this function, on both sides of the curve L, satisfy the


following Plemelj formulas :

{ (MA) - ¢e(A) = f(A) , A E L (27),

¢i(A) + ¢e(A) =~! f(z) 1 dz, A E L (27),

mJLz - /I.

(2.7)

where the Cauchy principal value is given by (2.2).

Proof Let us first examine the case where L is a contour. Consider the case where the point A is exterior to the contour. We have

(2.8) _1 r f(z) dz = _1 r f(z) - f(A) dz + f(A) r ~ . 2ni 1 z - A 2niJL z - A 2ni 1 z - A

The last integral on the right hand side vanishes because the variation of Log (z - A) as z runs round the contour is null. Let J.l be a point on the contour. Whenever A tends to J.l whilst remaining exterior to the contour L, the first integral on the right hand side of (2.8) admits a limit from the Lebesgue theorem and we therefore have

(2.9) ¢AJ.l) = ~ r f(z) - f(J.l) dz, J.l E L . 2ml z - J.l

Whenever the point A is interior we similarly obtain

~ r f(z) dz = ~ r f(z) - f(A) dz + f(A) , 2mJJ.z - A 2mJL z - A

and whenever A tends to a point J.l on the contour whilst remaining interior, we obtain

(2.10) 1 i f(z) - f(A)

¢JJ.l) = -2 . dz + f(J.l); J.l E L . m L z - J.l

The Plemelj formulas then result from (2.9), (2.10) and (2.5). We have thus proved Theorem 1 (28) in the case of a contour. I n the case of an arc the proof of the Plemelj formulas is strictly analogous at every point J.l distinct from the extremities. It only remains to show that in this case the function ¢ satisfies condition 3) of Definition 5. Let us examine the origin a. The function J.ll-+ f(J.l) admits a limitf(a) at this point. The function

_ { f(J.l) - f(a) , J.l E L g(J.l) = -

0, J.lEL,J.l¢L,

is H61derian in a neighbourhood of the point a of the contour I which contains the arc L.

(27) With, if L is an are, A. different from the extremities of L. (28) For the analyticity of tjJ in C\L in the case of either a contour or an arc L, see for example Dieudonne [1] and Hormander [1].


We have

(2.11 ) ¢(A) = ~i f(z) - f(a) dz + f(a~i ~, A If L . 2m L z - A 2m L z - A

The first integral in (2.11) is a bounded function of A in a neighbourhood of the point a (from the results obtained in the case of a contour). The second integral is

f(a) Log (A - b) 2ni A - a

which has a logarithmic singularity at the point a; this establishes condition 3) of Definition 5. 0

We now indicate what becomes of the above properties of the Cauchy integrals and the Cauchy principal values in the case where the integral is taken along an itifinite line. In this case, the Cauchy integral is defined by

1 fOO+ib f(z) (2.12) ¢(A) = -2 . --1 dz; b leal

nl - 00 + ib Z - II.

Im(A) #- b .

We have

Proposition 2. Letf(z) be Holderian on the line 1m (z) = b and satisfy the condition at itifi nit y

(2.13) f(z) = f(oo) + O(lzl- P); P > 0; Z E IC, f(oo) E IR .

Then the integral (2.12) exists(29) and is continuous at every point such that Im(A) #- b. The function ¢ thus defined is s. analytic(30). The Cauchy principal value defined by:

(2.14) 1 fOO+ib f(z) der 1 . (f.l.-' f(z) - --dz = -lim --dz 2ni -oo+ibZ - A 2ni ' .... 0 -a+ibZ - A

a"" 00

+ --dz (31) la+ib f(z) )

.1.+, Z - A

1m A = b, I Re A I < a ,

(29) In the following sense if f( (0) # ° and 1m (A) # b:

1 fa+ib f(z) CP(A) = -: lim --dz .

21t1 a ...... + 00 -Q+ib Z - A

(30) Definition 5 of an s. analytic function is here extended to the case of a line by eliminating condition 3) from that definition.

(31) Hence for b = 0, --dA = nHf(z) where Hfis the Hilbert transform of the functionf(see f+ 00 f(A)

-00 z - A Appendix "Singular Integrals").


is such that we have the Plemelj formulas:

(2.15) { (/J;(A) - 4>e(A) = f(A) ;

1 oo+ib f(z) (/J;(A) + 4>e(A) = -J --, dz ;

nl J- 00 + ib Z - /I.

ImA = b ;

ImA = b .

where 4>i(A) is the "superior" limit(32) of 4>(Jl), Jl --+ A, 1m Jl > b; and 4>e(A) is the "inferior" limit.

Proof Consider the following integral

(2.16) 1 1 fa+ib f(z) 4>a(A) = 2ni -a+ib Z _ Adz, 1m (A) #- b ;

1 fa+ib f(z) - f(oo) f(oo) fa+ib 1 = - -'---'---'----'-dz + -- --- dz .

2ni -a+ib Z - A 2ni -a+ib Z - A

From hypothesis (2.13), the first term on the right hand side of (2.16) has for its limit as a tends to infinity the integral

- dz. 1 foo+ib f(z) - f(oo)

2ni _ 00 + ib Z - A

The second term has for its limit

{ ~f(OO) if Im(A) > b

- ~f(OO) if Im(A) (A))

(2.17) 4>(A) = ~foo+ib f(z) - f(oo) dz ± ~ f(oo); (+ if Im(A) > b 2m -oo+ib Z - A 2 - if Im(A) < b) .

The portion of the line 1m (A) = b defined by I Re (A) I ~ a being an arc, we will be able, by using Theorem 1, to define the Cauchy principal value:

-. --dz, Im(A) = b, IRe(A)1 < a , 1 fa+ib f(z)

2m -a+ib Z - A

and we have ((4>a)i is the. "superior" limit, (4)a)e the "inferior" limit, taking into account the orientation of the arc):

{(4)aMA) - (4)a)e(A) = f(A)

(2.18) 1 fa+ib f(z) Im(A) = b; IRe(A)1 < a . (4)aMA) + (4)a)e(A) = ~ --, dz ;

nl -a+ib Z - /I.

(32) There should be no ambiguity with lim sup! Nor similarly with lim inf.


In order to obtain formulas (2.15), it suffices to show that (¢a)i (resp. (¢a)e) tends to ¢i (resp. ¢e) as a tends to infinity. We make use of the following equality (deduced from (2.16) and (2.17)):

- dz 1 (f-a+ib J(z) - J(oo)

2ni _ 00 + ib Z - A

(2.19) + dz f oo+ibJ(Z) - J(oo) )

a+ib Z - A

1 ( f a+ib dZ) + -.1(00) ± ni - --2m -a+ibZ-A

The second member converges uniformly to zero (for I Re (A) I < (x, (X E IR) when a tends to infinity. The result is then deduced on observing that equality (2.19) likewise permits the definition of the Cauchy principal value as the limit of

--dz: f

a + ioo J(z)

-a+iooZ - A

--dz = dz = lIm dz . foo+ib J(z) foo+ib J(z) - J(oo) . fa+ib J(z) - J(oo)

-oo+ib Z - A -oo+ib Z - A a--+oo -a+ib Z - A

Remark 2. The function ¢ defined by (2.12) is such that

{ <p(A} ~ ~f(~} when Im(A} ~ 00

¢(A) -+ - 2J (00) when Im(A)-+

and

-00

3. The Poincare-Bertrand Formula and the Hilbert Inversion Formula

o

o

We now give here two important properties of the Cauchy principal value on a contour or an arc L. The first, called the Poincare-Bertrand formula, gives a rule for inverting the order of integration in such "integrals". The second exhibits the inverse operator and is called the Hilbert inversion formula. Let L be a contour or an arc and let A, Jll--+ J(A, Jl) be a function defined on L x L. We have

Theorem 2. Let J(A, Jl) be Holderian on L x L, i.e. satisJying

(2.20) IJ(AI , Jll) - J(A2, Jl2)1 ~ C(IJlI - Jl21 + IAI - A2i)P, fJ > 0

Jor all AI' A2, JlI' Jl2 E L with L oj class rel,p.


Then, for all z on the contour L or the arc and in the latter case, z distinct from the extremities of the arc, we have the Poincare-Bertrand formula:

(2.21)

_1 i ~(_1 i f(2, J1) dJ1) 2ni 1 2 - z 2ni 1 J1 - 2

Iii (1 i f(2,J1)d2 ) = 4f (z, z) + 2ni 1 dJ1 2ni J (2 - z)(J1 - 2)

Proof (in the case where L is a contour)(33). Consider the function

(2.22) X(2) = ~i f(2, J1) dJ1 . 2mJL J1 - 2

We know from earlier that X is a continuous function. We can show (see Remark 7 or Muskhelishvili [IJ) that it is fact H6lderian. We can thus consider the following Cauchy principal value:

(2.23) A(z) = ~i X(2) d2, z E L . 2m12 - z

We can likewise define

(2.24) _1 i f(2, J1) d2 = 1 [i f(2, J1) d2 _ i f(2, J1) d2J 2ni 1 (2 - Z)(J1 - 2) 2ni(J1 - z) 12 - z 12 - J1

From the result reached above, each of the two principal values of the second member is H6lderian and we can define

(2.25) B(z) = _1 i dJ1(_1 i fV, J1) d2), x E L . 2ni 1 2ni 1 (2 - Z)(J1 - 2)

The two principal values A(z) and B(z) are respectively associated with the following two s. analytic functions:

(2.26)

(2.27)

We likewise have

(2.28) ¢(z) = lim~ r ~(_1. r f(2, J1) dJ1)' z tt L . t-02mJL2 - z 2mJL\(B(i .. ,)nL) J1 - 2

We can, taking this in the sense of a Lebesgue integral, permute the integrations in J1 and 2 whence:

(33) In the case where L is an arc, we refer to Muskhelishvili [1].


(2.29) 4>(z) = lim-1 r d/1 r f(l, /1) dl, z rt L . .-0 2ni JL 2ni JL\(B(Jl, ,) n L) (l - z)(/1 - l)

This last expression (2.29) is also a definition of I/I(z) and we therefore have

4>(z) = I/I(z) , z rt L .

Now making use of the Plemelj formulas (2.7) we have

(2.30)

The function I/I(z) can be written, from (2.24):

(2.31) I/I(z) = ~ r ~~( r f(l, /1) dl _ J. f(l, /1) dl), z rt L . 2m JL /1 - z 2m JL l - z 1 l - /1

The Plemelj formulas (2.7) show us that for all Zo E L

(2.32) lim ~ r f(l, /1) dl = _! f(zo, /1) + ~J. f(l, /1) dl . Z _ Zo 2m JL l - z 2 2m 1 l - Zo

z exterior

Let t:(/1, z) be the difference between ~ r ~(l, /1) dl and its limit as z -+ zoo 2mJLII. - z We find (see Muskhelishvili [1]) that the difference is such that:

lim r t:(/1, z) d/1 = 0, for all z exterior to the contour L ; z-zo JL/1 - z

we arrive at:

1/1 Azo) = lim -21.r ~ ( - -21 f(zo, /1)

z - Zo nl JL /1 - z (2.33) zexterior

+ ~J. f(l'/1)(-1- __ 1_)dl) . 2m 1 l - Zo l - /1

The Plemelj formulas (2.7) lead to:

1 1 ~ d/1 I/IAzo) = 4- f(zo, zo) - 4------; f(zo, /1)--.m L /1 - Zo

(2.34)

67

1 ~ d/1 1 ~ (1 1) + - --- f(l /1) -- - -- dl. 2ni L /1 - Zo 2ni L ' l - Zo l - /1

The formula analogous to (2.34) giving I/Ii(ZO) is:

(2.35)

1 ~ d/1 1 ~ (1 1) + - ---- f(l /1) -- - -- dl . 2ni L /1 - Zo 2ni L ' l - Zo l - /1


The two expressions (2.34) and (2.35) associated with the expressions for A(z) given by (2.30) and (2.23) lead finally to:

(2.36) 1

A(z) = 4f (z, z) + B(z) ,

which is the Poincare-Bertrand formula. 0 Now let us give the Hilbert inversion formula in the case where L is a contour. The function g(z) being defined on the contour L, it is then a question of finding a function f(A.) such that

(2.37) 1 f f(A.) ----: -, - dA. = g(z) , z E L . 7tl LA-Z

The solution of this problem in the case where the function 9 is H6lderian is given by the next theorem.

Theorem 3. Let 9 be a Holderian function (i.e. satisfying (2.3» defined on the contour L of class rc l,p. Then the unique Holderian solution f of equation (2.37) is given by:

(2.38) f(z) = ~l g(A.) dA. '. z E L . 7tllA.-z

This is the Hilbert inversion formula.

Proof This is a consequence of the PlemeJj formulas. Let us in effect consider the following s. analytic function:

1 1 g(A.) t/J(z) = -2' -, -dA., z E C, z rt L . m LA - Z

(2.39)

The PlemeJj formulas (2.7) applied to this function give

(2.40) { If g(A.) t/Je(Z) + t/Ji(Z) = ----: -, -dA. ,

m LA - Z

t/J i(Z) - t/J e(z) = g(z) ,

ZEL;

Z E L .

Now let the s. analytic function ¢ be defined by:

¢(z) = { - t/J(Z) , if Z is exterior to the contour ; t/J(Z) , if Z is interior to the contour .

Formulas (2.40) are then transformed into:

(2.41) { ¢e(z) + ¢i(Z) = g(z) , Z E L ,

¢i(Z) - ¢e<z) = ~l ,g(A.) dA., Z E L , 7tlJL A - Z


deC LetJ(z) = cP;(Z) - cf>e(z); we then have

cf>(z) = ~ r J(A) dA, z E L , 2mlA - z

whence we deduce, from the Plemelj formulas, that

g(z) = ~~l J(A) dA, z E L mlA-z

This shows that J(z) is the solution of (2.37). Theorem 3 then follows at once. 0

Remark 3. We would have been able to prove Theorem 3 by making use of the Poincare-Bertrand formula in the case where the functionJ(A, /1) depends only on a single variable. The proof above is more direct. 0

Remark 4. If we denote by S the operator defined by

(2.42) S(f)(z) = ~l J(A) dA, z E L , mlA-z

Theorem 3 signifies that:

(2.43)

It likewise follows that the operators

{ P = ~1(1 + S)

(2.44) Q = 2(1 - S)

satisfy respectively

(2.45) { P 2 = P Q2 = Q

and are thus projection operators(34). They further satisfy the properties:

(2.46) lP - Q =S

P+Q=I PQ = QP = 0

These are therefore the complementary projection operators in the Banach space of H6lderian functions on L. If J E L 2 (L), we can show (see Pr6ssdorf [1] and this result in the appendix on singular integral operators) that S is also a bounded operator L 2(L) -+ L 2 (L); P and Q are then complementary operators in this space. 0

(34) The operator P (resp. Q) projects onto the space of traces of functions analytic in the interior (resp. exterior) of L.


Remark 5. The Hilbert formula is still true for a union of non-intersecting contours, with suitable orientations, for example as shown in Fig. 3.

Fig. 3

The Hilbert formula is false for a finite arc. We will examine that case below. 0

In the case of an infinite straight line, we have the following theorem the proof of which is analogous to that of Theorem 3.

Theorem 4. Let g be a Holderian function on the line 1m A. = b, satisfying hypothesis (2.13) of Proposition 2 and further

(2.47) g(oo) = 0 ;

then the equation

(2.48) 1 foo+ib f(z) -: --, dz = g(A.) , 1m (A.) = b , 1tl - 00 + ib Z - I\.

admits a unique solution f defined on the line with ordinate b and satisfying the same hypotheses as g; this solution f is given by:

1 foo + ib g(A.) f(z) = -: -, -dA., Im(z) = b .

1tl - 00 + ib I\. - Z (2.49)

o Remark 6. Being given a contour or an arc L, and an (H6Iderian) function f given on L, the (s.analytic) function ¢ on C\L given by the Cauchy integral (see Theorem 1):

(2.50) ¢(A.) = ~ r f(z) dz, A. ¢ L 2ml z - A.

is the solution of the transmission problem: find ¢ hoi om orphic in C\L, admitting limits on each side of L such that the jump [¢ h = ¢i - ¢ e as ¢ crosses L shall be equal to f, and tend to 0 at infinity.

With the notation J = 0_ = ! (!..- + i~) for z = x + iy, this problem in ¢ can oz 2 ox oy be written:

(2.51) { a¢ = 0 in C\L

[¢JL = ¢i - ¢e = f on L

¢(z) ~ 0 as Izl ~ 00 .


In the sense of distributions in C, (2.51) expresses that the derivative a</J of </J in .s&'(C) is the measure carried by L, with L traversed in the direct sense, defined by

1 . the differential - 2if (z) dz, that IS to say:

- 1 i <o</J, 0 = - ~ f(dz V( E .s&(C) 21 L

Whence the following presentation of problem (2.51):

1- 1

'. o</J = - 2i f(z) dz in .s&' (C)

\ </J(z) -+ 0 as Izl -+ 00 . (2.52)

On recalling (see Chap. V) that E(z) = ~ is the elementary solution of a which nz

tends to 0 at infinity: aE = tJ, we immediately obtain the solution of (2.52) by

taking the convolution product of E with - ~f(z) dz, whence (2.50)(35). For a real

function J, on taking the real and imaginary parts of </J(z), we thence obtain the usual solution of the transmission problem (or the cracking problem in the case of an arc), which will be seen again in §4.1 for potentials of simple or double layers.

o

Remark 7 (on Proposition 1, and Theorem 1). Let us suppose for the moment that L is a (differentiable) contour, L being the boundary of an open set Q locally on one side of L. Let f E ~O(L), and put

1 i f(A.) (2.53) </J(z) = -2' -1 -dA., z fL. 11:1 LII. - z

Using a curvilinear coordinate s on L, and the decomposition of the differential dA. .

element -1-- on L gIven by: II. - Z

(2.54) dA. _ dLogr d + .dlogr d with r = IA. - zl , -1-- - -d- s I-d- s , II. - Z S n

we get (writing f(s) for f(A.(s))):

(2.55) </J(z) = 21 r f(S)dd (Logr)ds + -21 . r f(S)dd (Logr)ds . nJL n 11:11 s

(35) All of this can be developed in the framework of Hilbert spaces: if the givenjsatisfiesje Hl/2(L) if L is a contour, or j e H~~(L) if L is an arc (using the notations of Chap. VII), then (this results from Djaoua [I]) rfJ belongs to the space of the u e gi)'([R2IL) such that u/lzl Log(1 + Izl) e U([R2IL), gradu e (L2([R2IL»2.


Put:

v g E rt'°(L) ,

(2.56)

Hence (see Chap. II, §3), U2 (f) is the double layer potential off, U 1 (g) is the simple layer potential of g, and thus if J E rt'l (L),

(2.57) ul(f) = - Ul (:~) .

With (2.56) and the hypothesis J E rt'°(L), (2.55) can be written:

(2.58) 4> = u2 (f) - iildf) .

IfJis a real function, (2.58) is the decomposition of the analytic function (in C\L)4> into a (harmonic) real part u2 (f) and a (conjugate harmonic) imaginary part - ill (f).

The investigation of the "continuity" on the boundary of L of the function 4> reduces to that of the functions U2 (f) and ill (f). For the interior continuity (onQ) and the exterior continuity (on C\Q) of the double layer potential it must be assumed that L is a little more than of class rt'l (for this, see Chap. II, §3). Following Proposition 11, §3 in Chap. II, if L is of class rt'l +e

(using Definition 3, §3 in Chap. 11(36)), u2(f) admits extension by continuity to Q and C \ Q by putting:

{ u~(f)(z) = w(z) + ~J(Z) (for the interior)

(2.59) 1 uHf)(z) = w(z) - "2J(z) (for the exterior)

with

(2.60) w(z) = ;n I J(s) :n Log r ds Z E L .

From the proof of Proposition 13, §3 in Chap. II, if L is of class rt'l+e, and ifJis

(36) Recall (see Chap. II, §3, Definition 3) that a regular open set has a boundary r of class ~I +. if in a neighbourhood of every point Z E r, there exists a normal parametric representation (R, U, (9, IX) (see

Chap. II, § 1, 3.a) and a continuous increasing function s: ~ + ..... ~ + with r s(r) ~ < + OCJ such that Jo r IgradIX(x) - gradIX(x')1 .:; s(lx - x'l) Vx, x' E (9 .

(This condition being evidently satisfied if r is of class ~ I.P with 0 < P < 1.)


continuous in the Dini sense (that is to say - see (3.22), Chap. II - such that

IJ(z) - J(z')1 ~ e(lz - z'l) v z, z' E r (37)

for a continuous increasing function e IR + --+ IR + with r e(r) dr < + 00), then Jo r ill (f) admits a continuous extension to 1R2, Hence under the above hypotheses: L a contour of class ce 1 +£, andJ E ce£(L) (which is realised particularly if J is H6lderian on L), we have the following results (by Proposition 11, §3 and the proof of Proposition 13, §3, in Chap. II):

1) the Cauchy principal value given by (2.2) is continuous on L, 2) and has for the values of the limits:

[4>(Z)]L = 4>i(Z) - 4>e(z) = [uZ(f)(Z)]L = J(z) (z E L)

(2.61) 1 . 1 i aLogr :2(4)'(z) + 4>e(z)) = 2n L J(s)----a;;-ds

- ~pv i J(S)dd Logrds = -21 . rh J(z) 1 dz , 2n L s m j' z - I\,

which gives Proposition 1 and Theorem 1 of this §2 in the case where L is a contour. In the case of an arc, we will have analogous results (except, of course, at the extremities of the arc) since these results are local. Subsequent to §2, we will use (particularly for Theorems 2 and 3) the fact that ifJis a function H6lderian on a contour L (f E cea(L), 0 < (J. < 1) then the Cauchy principal value is also H6lderian on L ( E cea(L)). For this to hold it is also necessary to assume that L is of class cel,a: we know from Chap. II, §3 that the double layer potential U 2 (f) is then of class cea on Q and IR z \ Q and if J E cem.a(L)

with L of class cem + 1.a, m ~ 1, then ill (f) = - u1 (:~) is of class cem.a on Q and

1R2 \ Q (this result is again true for m = 0(38), with the notation ceO.a = cea), which hence implies that the Cauchy principal value is then of class cea(L). In the case of an arc L, the above result on H6lderian regularity implies that if J E cea(L) then the Cauchy principal value will be of class cea on every strict subarc L' with L' c L, i.e. if a and b (resp. a' and b') are the extremities of L (resp. L'), we have a < a' < b' < b on the arc.

(37) One also writes f E 'C'(L). (38) All this is directly connected to the Dirichlet and Neumann problems

f ,1u = 0 in Q

1 ul r = f on r (resp, :~ Ir = f on r) with f given Hiilderian (see Chap. II, §6, Remark I and Lemma 4).


§3. The Hilbert Problem

Introduction

We study in this section a general problem concerning s. analytic functions and called in general the Hilbert problem (or alternatively the Riemann-Hilbert problem). We give its general solution, under certain hypotheses concerning the data, first in the case of a contour, then of an arc, and finally in the case of an infinite straight line. Following that, we shall consider a class of integral equations on a contour involving Cauchy principal values and called singular integral equations. We solve these equations in the setting of a Hilbert problem by the so-called Carleman method. Finally, we will consider some direct applications of this theory to particular integral equations. Let L be a contour or an arc (or a straight line). The Hilbert problem consists of the search for an s. analytic function relative to the curve L (see definitions 1 to 5, §2). We further require of it that it will have at least polynomial growth at infinity and must satisfy the equation

(3.1) (/J;(z) = A(z) cPe(z) + B(z) , z E L ,

where A (z) and B(z) are given functions on the curve L. We recall that cPi(Z) and cPe(z) denote, in the case of a contour, respectively the interior limit of cP(),,) and the exterior limit of cP(),,) as )" tends to the point z on the contour. In the case of an arc or a straight line, these limits are defined in a unique manner by taking into account the orientation. The Hilbert problem is said to be homogeneous if B == O. We will suppose throughout this section that A has no zero on the curve L. We shall now examine the solution of the Hilbert problem.

1. The Hilbert Problem in the Case Where L is a Contour

We shall assume for convenience that the origin of coordinates is interior to the contour L. Let us first examine the homogeneous problem. In the case where the function A(z) is such that Log A (z) admits a single-valued definition on the contour L, the homogeneous Hilbert equation

(3.2) cPi(Z) = A(z)cPe(z) , z E L ,

is equivalent to

(3.3) Log cPi(Z) - Log cPe(z) = Log A(z) .

Hence a solution is given in this case on making use of the Plemelj formulas (if

§3. The Hilbert Problem 75

Log A (.?c) is H6lderian, and if L is of class tt' 1.P - see Theorem 1 in §2.2) by:

(3.4) ( 1 r LogA(.?c) ) <jJ(z) = exp 2ni JL .?c _ z d.?c , z E IC, z ¢ L .

In the general case, we shall follow the same idea in order to find all the solutions of the homogeneous Hilbert problem.

Definition 1. We shall call the index of the Hilbert problem that integer p (p E Z) which is such that the argument of A (z) increases by 2np when the point z makes a circuit of the contour L, and we shall write

We then have

1 p = -2 var arg z

n ZEL

Proposition 1. Let A(z) be afunction defined and Holderian on the contour L(39) and has no zero on this contour. If p is the index given by Definition 1, then the homogeneous Hilbert problem (3.2) admits as a particular solution the following s. analytic function <jJ(z):

1 og J:P exp 2ni L .?c _ z d.?c , (

L (A (.?c)) ) if z is interior to the contour ,

(3.5) <jJ(z) =

( ( A (.?c)) ) 1 1 Log J:P

p exp -2 . r , d.?c, if z is exterior to the contour , z 7tl JL A - Z

All the other solutions of the homogeneous Hilbert problem are of the form

(3.6) e(z) = <jJ(z) lP(z) , where lP(z) is an arbitrary polynomial .

Proof The origin being interior to the contour, the function

A(z) z 1-+ Ao(z) = - zP

has a zero index and never vanishes on the contour L. Its logarithm is uniquely determined and hence a solution t/J of the homogeneous Hilbert problem corresponding to the function Ao is given by formula (3.4):

( 1 r LogAo(.?c) ) t/J(z) = exp 2ni JL .?c _ z d.?c , z E IC, z ¢ L .

(39) We further assume that L is of class «fl,p,


Consider then the function ¢ defined by

1 I/I(z) , if z is interior to the contour, ¢(z) = I/I(z)

---;;-' if z is exterior to the contour .

It is clear that ¢ is a solution of the homogeneous Hilbert problem relative to the function A. This proves the first part of the proposition. Finally, let us prove (3.6). Thus, let 8 be another solution of our problem. We have

(3.7) 8 e(z) = 8 i (z) Z E L ¢e(z) ¢i(Z) , .

B . h f ',J, • h . h 1 1 h f . 8(z) . ut SInce t e unctIOn 'I' never vams es In t e comp ex pane, t e unctIOn ¢(z) IS

analytic.

On the other hand, having polynomial growth at infinity, this function :~; is a polynomial in virtue of Liouville's theorem. Conversely, it is clear that every function of the form ¢(z) [p>(z) where [p> is a polynomial, is a solution of our problem. 0

The study of the Hilbert problem is derived from that of the homogeneous Hilbert problem in the following way:

Theorem 1. Let A, H6lderian of index p and B, Holderian, be defined on the contour L (40); the Hilbert problem of equation (3.1) admits for its only solutions the following s.analytic functions ¢:

(3.8) ¢(z) r B(A)

q>(z) = 2ni JL ¢i(A)(A _ z) dA + [P>(z)¢(z)

where [p> is an arbitrary polynomial and where ¢ is the solution of the homogeneous Hilbert problem given by (3.5), ¢i(Z) being the interior limit of ¢(z) on the contour L:

(3.9) 1 A(z) 1 Logy

¢i(Z) = exp 2 Log ---;;- + 2ni £ A _ z dA , z E L . (

A ().) )

Ifwefurther impose that the solutions must tend to zero at infinity, we have according to the value of the index p:

1) p = 0, there exists a unique solution corresponding to [p> == 0; 2) p > 0, there exists an infinity of solutions corresponding to each polynomial [p> of degree less than or equal to p - 1;

(40) Again with L of class C(j'l,p.


3) p < 0, there exists a unique solution if and only if the function B(z) satisfies the conditions:

(3.10) r Am B(A) l 41i(A) dA = 0, ° ~ m ~ - (p + 1) .

Proof Let 41(z) be the particular solution of the homogeneous Hilbert problem given by (3.5). The corresponding functions 41i(Z) and 41e(z) never vanishing on the contour L, we can replace equation (3.1) by the equation

(3.11) <Pi(Z) <Pe(z) B(z) -----=--, zEL. 41i(Z) 41e(z) 41i(Z)

Fom the Plemelj formulas (see Theorem 1, §2), a paticular solution is given by

1 r B(A) (3.12) <p(z) = 41(z) 2nijL 41i(A)(A _ z) dA, z ¢ L, z E if .

We obtain all the solutions by adjoining to this particular solution all the solutions of the corresponding homogeneous equation which are given by (3.6). Thus we have derived expression (3.8). Let us examine the behaviour at infinity of these solutions according to the index p. If p = 0, 41(z) tends to 1 as z tends to infinity. The solution therefore has at infinity the behaviour of lP'(z) plus that of a function which tends to zero like 1/lzl. Only IP' == ° leads to solutions which vanish at infinity. If p > 0, 41(z) is equivalent to z- P at infinity. Thus all polynomials of degree less than p - 1 lead to solutions which tend to zero at infinity. If p < 0, 41(z) is equivalent to z-P at infinity. Thus on the one hand only IP' == ° is allowed. On the other hand the solution

41(z) r B(A) dA 2ni l41i(A)(A - z)

tends to zero at infinity only if the conditions

r Am B(A) l 41i(A) dA = 0, ° ~ m ~ - (p + 1)

are satisfied. In that case, the solution is unique. o

2. The Hilbert Problem in the Case Where L is an Arc

Let L be an arc(41) with extremities a and b (see Definition 1, §2). The Hilbert problem consists of searching for an s. analytic function <P relative to this arc (see

(41) We again assume that L is of class C(j1.P (see footnote (26) to Proposition \, §2, above).


Definition 5, §2) and satisfying

o/i(Z) = A(z)o/e(z) + B(z) , Z E L

where A and B are given functions on the arc L. We assume that A(z) has no zero on the arc L, including the extremities. Then its logarithm is uniquely determined on the arc (which was not the case when L was 'a contour(42»). One solution of the homogeneous Hilbert problem (3.1) (i.e. with B = 0) is thus given by:

(3.13) I/I(z) = exp (2~i 1 L~g:~l) dl) (43).

But this solution is not generally s. analytic because it does not necessarily satisfy condition 3) of Definition 5, §2, which specifies that the behaviour at an extremity a shall be:

(3.14) c

I o/(l) I ~ Il _ ala' 0 ~ IX < 1, lEe.

Let us therefore consider a function (m and n are integers E Z)

(3.15) 4>(z) = (z - a)m(z - b)"exp(2~il L~g:~l)dl) We have seen (formula (2.11)) that the singularity of the function

_1 r LogA(l) dl 2niJL l - z

at the point a has the form

(3.16) _1 r LogA(l) dl = _ Log~(a)Log(a _ z) + g(z) 2ni JL l - z 2m

where g is bounded in a neighbourhood of the point a. Thus 4> given by (3.15) behaves at the point a like

I (z - a)m - 2"i LogA(a) .

To ensure that condition (3.15) is satisfied, we will thus choose the integer m such that

(3.17) o < m - Re(~LOgA(a)) + 1 ~ 1 2m

which determines the integer m uniquely.

(42) Recall that 0 is taken to be interior to the contour. (43) With the hypothesis that A is Hiilderian.


Similarly, we will guarantee condition (3.14) at the point b by choosing the integer n such that

(3.18) o < n + Re(~LOgA(b)) + 1 ~ 1 . 2m

Let 8 be another s. analytic solution of the homogeneous Hilbert problem. We have

(3.19) 8 i(z) _ 8Az) = 0, ZEL. ¢i(Z) ¢e(z)

H h f . 8(z). I' h h h .. ence t e unctIOn ¢(z) IS ana ytIc everyw ere except per aps at t e extremItIes

a and b where it has singularities satisfying (3.14). It is therefore analytic. We have proved:

Proposition 2. Let A be a Holderian function having no zero on the arc L of class <'(j1.p. Then an s. analytic solution of the homogeneous Hilbert problem:

(3.20) ¢i(Z) = A (z)¢e(z), Z E L ,

is given by:

(3.21) ¢(z) = (z - a)m (z - bt exp C~i L L~g:~A) dA)

where m and n are integers fixed by the conditions (3.17) and (3.18). All the s. analytic solutions of (3.20) have the form:

(3.22) 8(z) = lP(z) ¢(z) , where IP is an arbitrary polynomial . o

Let us examine the solution of the non homogeneous Hilbert problem (3.1). We have

Theorem 2. Let A and B be Holderianfunctions on the arc L of class <'(jl,p, with A(z) having no zero on this arc. Then every s. analytic solution of the Hilbert problem (3.1) has the form:

¢(z) r B(,{) (3.23) q>(z) = 2ni JL ¢i(A)(A _ z) dA + lP(z)¢(z) , z E C, Z ¢ L ,

where IP is an arbitrary polynomial, and ¢ is given by (3.21), namely:

¢(z) = (z - a)m(z - b)"exp(2~i L L~g:~A)dA)' z E C, Z ¢ L ,

and ¢i(Z) (for z E L) is given by:

( 1 1 f LogA(A) ) (3.24) ¢;(z) = (z - a)m(z - b)" exp - Log A(z) + -2' A dA, 2 m L - Z


the integers m and n being chosen in a unique way by (3.17) and (3.18):

o < m - Re(2~i LOgA(a)) + 1 ~ 1 ,

o < n + Re(2~iLOgA(b)) + 1 ~ 1 .

Let p be the problem's index given by:

(3.25) p = - (m + n) .

Then, according to the index p, the existence and uniqueness of solutions tending to zero at i~nity are given by:

1) p = 0, there exists a unique solution given by (3.23) with IP = 0; 2) p > 0, there exists an i~nity of solutions given by (3.23) with degree IP ~ p - 1; 3) p < 0, there exists a unique solution if and only if the following conditions are satisfied:

(3.26) r Al B(A) JL (/>;(A) dA = 0, 0 ~ I ~ - (p + 1) .

Proof Equation (3.1) can be written

(3.27) cp;(z) cP e(z) B(z) -----=-- ZEL (/>;(z) c/> e(z) c/>;(Z) ,

The second member of (3.27) admits a limit at the extremities a and b. In effect, B is Holderian and therefore has a limit at the extremities. The function c/>; has the same behaviour as c/> at the point a i.e.:

m - Re(~ Log A(a») (1) 2m exp "2 Log A(a) (a - z)

Hence it tends to infinity at this point, or is bounded and nonzero, according to

whether Re( 2~i . Log A(a)) is integral or fractional. It never vanishes on the arc L.

Consequently we can use the Plemelj formulas of Theorem 1 in 2.2, whence it follows that the s. analytic solutions of (3.26) are:

cp(z) = _1 r B(A) dA + lP(z) c/>(z) 2ni 1 c/>;(A)(A - z)

where IP is an arbitrary polynomial. The second part of Theorem 2 follows from the fact that the function c/> behaves at infinity like zP where p is the index given by (3.25). 0

Remark 1. We can, artificially, regard a contour as being an arc on choosing an origin. In this case, it is easily verified that the two definitions of the index p, for the arc and the contour, coincide. Studying carefully the apparent singularity at the


point of origin of the arc, one sees that the solutions constructed in Theorem 2 coincide in this case with those constructed in Theorem 1 (there is no singularity at this point of origin). This remark shows the usefulness of the conditions on the behaviour at the extremities (3.14). 0

Remark 2. In the case of an arc, and in the case of a contour regarded as an arc, we can construct solutions which are not s. analytic, by choosing parameters m and n in such a way that the function <jJ will have singularities of high order at the extremities. 0

3. The Hilbert Problem in the Case of a Straight Line

Let us examine the solution of the Hilbert problem (3.1) in the case of the straight line L with equation Im(z) = b. The solutions of the homogeneous problem are given by

Proposition 3. Let A be a Holderian function having no zero on the straight line Im(z) = b, and satisfying:

(3.28) A(z) = 1 + 0(_1_), Y > 0, when Izl -+ 00 Izl Y

and let the integer p, called the index, be defined by:

(3.29) p = 21 [lim arg A(z) - lim argA(Z)]' 1[. Re(z) - 00 Re(z) - - 00

with z E C and Im(z) = b. An s. analytic solution relative to the line for the homogeneous Hilbert problem is given by:

(Im(lX) > band Im({3) < b) .

1 ( 1 foo + ib ( (A - {3)P) 1 ) (z - {3)pexp 21[.i _ 00 + ib Log A(A) A - IX (A _ z)" dA ,

Im(z) > b ; (3.30) <jJ(z) =

1 ( 1 foo + ib ( (A - {3)P) 1 ) (z - lX)p exp 21[.i _ 00 + ib Log A(A) A - IX (A _ z)" dA ,

Im(z) band Im({3) < b .


Consider the function Ao defined on the line L:

Ao(z) = (z - f3)P A(z) , Z E L . Z - ex

(3.32)

It is H6lderian on this line, satisfies hypothesis (3.28), and its index is zero. We can thus define Log Ao(z) in a unique manner on the line L. The homogeneous Hilbert problem can be written in the following way:

(3.33) Log (4)i(Z)(Z - f3)P) - Log (4)e(z) (z - ex)P) = Log Ao(z) , Z E L .

The right hand side of (3.33) satisfies the hypotheses of Proposition 2, §2, with LogAo( (0) = 0, and hence, from that proposition, we have the formula (3.30). The proof of (3.31) is analogous to the case of the contour already considered. 0

Concerning the solution of the non homogeneous Hilbert problem (3.1), we have

Theorem 3. Let A be a Holderianfunction having no zero on the straight line L with equation 1m (z) = b, and satisfying:

(3.34) c

I A(z) - 11 ~ ~' y > 0, Z E L ,

and let B be a Holderian function on the line L and such that:

(3.35) c

I B(z) - B(oo)1 ~ ~., y > 0, Z E L .

Then the only s. analytic solutions relative to the line L of the Hilbert problem (3.1) are given by:

4>(Z) foo + ib B()') d)' (3.36) qJ(z) = -2 . . -1..( 1) ~ + IP (z) 4> (z), Z ¢ L ,

7t1 _ 00 + ,b '1', 1\ 1\ Z

where 4> is given by (3.30) and 4>i is given (for Z E L) by:

1 [1 ( (z - f3)P) 4>i(Z) = (z _ f3)pexp 2" Log A(z) Z - ex

(3.37) 1 1.00 + ib ((). f3)P ) d)' ]

+ 27ti J- 00 + ib Log ). - ex A()')). - Z '

where IP is an arbitrary polynomial, and where ex and f3 are complex numbers such that Im(ex) > band Im(f3) < b.

Proof This follows from Proposition 3 above, and trom Proposition 2, §2, on observing that equation (3.1) reduces to

(3.38) qJi(Z) qJe(z) 4>i(Z) - 4>e(Z) , Z E L ,

4>i(Z) and 4>e(z) being the interior and exterior limits of the function 4> given by (3.30). 0


Remark 3. The solutions given by (3.37) of the Hilbert problem in the case of a line L, have a behaviour at infinity, in a direction non parallel to the line L, like 1 z 1- p. This allows us to characterise, as in Theorems 1 and 2, the solutions which tend to zero at infinity in a direction non parallel to the line L according to the value of the index p defined by (3.29). 0

Remark 4. Whenever the limit at infinity A(oo) of the function A(z) is different from 1, the function Log (Ao(z)) does not have zero as its limit at infinity. Theorem 3 nevertheless remains valid on condition that one uses the definition (see formula (2.17)) of the Cauchy integral on the line L. 0

4. Some Problems Reducible to a Hilbert Problem

We present here on the one hand singular integral equations on a contour, and on the other applications of the theory of the Hilbert problem to some integral equations.

4.1. Singular Integral Equations of Cauchy Type

Let L be a contour. By a singular integral equation of Cauchy type we mean the following equation in the unknownf and where a, b, c are given Holderianfunctions on the contour L:

(3.39) 1 ~ f(A.) a(z)f(z) + b(z)---; -, -dA. = c(z) , z E L . m LA - Z

We recall that the Cauchy principal value is defined, whenever the function f is Holderian on the contour, by:

(3.40) J. f(A.) dA. = lim r f(A.) dA., z E L . JLA. - z £~oJL\(B(z,')"L) A. - z

We refer to §2 for the definition and properties of this Cauchy principal value. We will now apply the Carleman method in order to reduce our problem to a Hilbert problem on the contour L. We assume that the origin of the coordinates is interior to the contour L. Let us consider the following s. analytic function <P:

<p(z) = -21 . f ,I(A.) dA., z E C\L, A.EL. m LA - Z

(3.41 )

Using the Plemelj formulas (see Theorem 1, §2), equation (3.39) becomes

a(z)(<p;(z) - <pAz)) + b(z)(<p;(z) + <pAz)) = c(z) , z E L ,

let:

(3.42) a(z) - b(z) c(z)

<p;(z) = a(z) + b(z) <Pe(z) + a(z) + b(z) , z E L .


The solution of equation (3.42) is a Hilbert problem(44) whose unknown is the s. analytic function cp(z) defined by (3.41). We then have

Theorem 4. Let a, b, c, be three Holderian functions defined on the contour L of class q;l,p.

Suppose that:

(3.43) (a2 (z) - b2 (z)) has no zero on the contour L .

Let p be an integer E 7L, called the index of the singular integral equation (3.39), defined by:

1 (a(z) - b(Z)) p = -var arg . 2n ZEL a(z) + b(z)

Then equation (3.39) admits, according to the value of the index p, different types of solutions: 1) if p > 0, an infinity of solutions: With the notations:

(3.44) </>(z) =

( a(A) - b(A) )

( _1 1 Log AP(a(A) + b(A)) dA exp 2 . 1 ' nz L II.-Z

if z is interior to the contour

( a(A) - b(A) )

! exp (_1 r Log AP(a( A) + b( A)) dA zP 2ni l A - z '

if z is exterior to the contour

(44) Note that conversely, the Hilbert problem (3.1) reduces to (3.39) on putting:

1 + A(z) 1 - A(z) a(z) = , b(z) = , c(z) = B(z) .

2 2


the solutions of (3.42) are given by:

(346) () = ¢(z) i C(A) dA IP( )A.(z) J L • cP z 2ni L (a(A.) + b( A)) ¢i (A.)( A _ z) + z 'I' ,z 'F ,

where lP(z) is an arbitrary polynomial of degree less than or equal to p - 1, whence the solutions of (3.39):

(3.47) f(z) = CPi(Z) - CPe(z) , z E L ;

2) if p = 0, a single solution, given by the formulas (3.44) to (3.47) with IP == 0; 3) if p < 0, a single solution: this solution is given by the formulas (3.44) to (3.47) with IP == 0, if and only if the following conditions are satisfied:

(348) i AmC(A.) A ( 1) (45) • L ¢i(A)(a(A) + b(A)) d = 0, 0:::; m :::; - p + .

Proof This theorem is a direct corollary of Theorem 1. We observe in effect that the s. analytic solutions of equation (3.42) are given by formula (3.8). In our case, only those solutions which tend to zero at infinity have the form (3.41) from the Plemelj formulas. 0

Remark 5. The above theorem is often called Nother's theorem. It can be generalised in diverse directions. We can first weaken the hypotheses on H6lderian regularity on the second member c(z) by choosing it in a space of square integrable functions L 2(L) or even in a space of p-th power integrable functions U(L). The solutions will be sought in the same space. We refer to Zabreiko [IJ for references and the principal results in this sense.

We can on the other hand suppose that the symbol (:~:~ : :~:D admits zeros on

the contour. We refer to S. Pr6ssdorf [IJ for results of this type and the solution of the corresponding singular integral equations. 0

Remark 6. We will not state here the results concerning the solution of singular integral equations for an arc or an infinite straight line. But the Carleman method cal) be used in the same fashion as in the case of a contour and lead to similar results which one can easily deduce from Theorems 2 and 3 above. We refer to Muskhelishvili [1J for these results and also to the example treated below on the inversion of the Cauchy integral in the case of an arc. 0

Remark 7. Returning to the notations of Remark 4, §2, we can write equation (3.39) in the form

(3.49)

(3.50)

a(z)f(z) + b(z)(Sf)(z) = c(z)

Sf(z) = ~ l f(A) dA . mlA-z

(45) Note that Theorem 3, §2, is the special case of this Theorem 4 in which a = 0, b = 1.


Again, on putting

(3.51 )

1 p = 2(1 + S)

1 Q = 2(1 - S) ,

equation (3.39) can be written:

(3.52) (a(z) + b(z) P f(z) + (a(z) - b(z)) Qf(z) = c(z) .

If (a 2 (z) - b2 (z)) is nonzero on the contour, we can consider the operator

(3.53) 1 1

R = P a(z) + b(z) + Q a(z) - b(z)

On denoting by .xl the operator

(3.54) .xl = (a(z) + b(z)) P + (a(z) - b(z)) Q ,

the properties of P and Q show that

1 1 (3.55) .xl R = (a(z) + b(z)) P a(z) + b(z) + (a(z) - b (z)) Q a(z) _ b(z) .

It can also be shown that in the spaces of H6lderian functions (or in the spaces of p-th power summable functions), the commutators

[p, a(z) ! b(Z)] and [Q, a(z) ~ b(Z)]

are compact operators(46). Thus we obtain

(3.56) J.xIR = P + Q + [p, a(z)! b(zJ + [Q, a(z) ~ b(Z)]

1 = 1 + K

where K is a compact operator in the chosen space offunctions. Then making use of the Fredholm theory (see Chap. VIII) we can recover the results of Theorem 4. Using this technique, Pr6ssdorf [1] has obtained diverse generalisations of these results on singular integral equations. 0

4.2. The Hilbert Inversion Formula in the Case of an Arc

Let L be an arc of class C(}l,p with extremities a and b. Consider the following problem: find a function f H6Iderian(47) on L (with supplementary conditions

(46) See Prossdorf [1]. (47) Except possibly at the extremities of the arc L.

§3. The Hilbert Problem

which we will indicate later) such that:

(3.57) ~ i f(l) dl = g(z) , z E L 1tIJl-z

where g is a given function, H6lderian on the arc L.

87

The Carleman method reduces us to searching for the s. analytic function(48) cp such that:

(3.58) 1 i f(l) cp(z) = -2' -, -dl, z If. L, z E C . 1tI LII-Z

Equation (3.57) can then be written (from the Plemelj formulas) as

(3.59) cp(Z) = - CPe(z) + g(z) , z E L .

The solutions of the Hilbert problem are given by Theorem 2 in this section. The solution of the corresponding homogeneous problem is unique up to a multiplicative constant and is given by:

1 4>(z) = - , z If. L (49).

J(z - a)(z - b) (3.60)

It is uniquely determined in the complement of the arc L. The index p which corresponds to it is from (3.17) and (3.18):

(3.61) p = 1 .

We are only interested here in the solutions of equation (3.59) which tend to zero at infinity. From Theorem 2, these are all of the form

(3.62) cp(z) = 1 r g(l) dl + c 2niJ(z - b)(z - a) 14>i(l)(l - z) J(z - a)(z -:- b)

where c is an arbitrary complex constant and where the function 4>i is given by:

(3.63) 1

4>i(Z) = , Z E L ; J(z - a)(z - b)

(48) Assuming a priori that f(A) in a neighbourhood of each extremity a, b of the arc L has the form

00·) f().) = ---, 0 ~ IX < 1, c = a or b ,

I). - cl"

with O()') Holderian on L up to c, one can further show (see e.g. Muskhelishvili [1]) that the function qJ defined by (3.58) is s. analytic and satisfies the Plemelj formula. (49) Note that iff is the discontinuity of cP across the arc L, then

del 1 t f().) Sf(z) = - --d)' = cPi + cPe = 0 ,

ni ). - z

so that the operator S is here not injective (contrary to the case where L is a contour), this being due to the fact that S is here not an operator in the space of functions Holderian on all L.


the choice of the square root being the one obtained when the point z tends to the arc L whilst remaining interior (see §2 for this notion).

r-~----

We remark that the function g(A)/(MA) = g(A).j(A - a)(A - b) is H6lderian on the arc L and zero at the extremities. The s. analytic function

r g(A) dA JL (MA)(A - z)

is thus bounded at the extremities a and b and in a neighbourhood. This shows that the solutions given by (3.62) behave in a neighbourhood of the point a like

(3.64) cp(z) _ 1 . r g(A) dA + c . 2nij(a - b)(z - a) JL ¢i(A)(A - a) J(a - b)(z - a)

The following choice of the constant c:

(3.65) c = __ 1 r g(A) dA 2ni JL ¢i(A)(A - a)

leads to a solution which is bounded at the point a and is infinite at the point b like IX ~(IXEq.

yZ - b One can exchange the roles of a and b: we can chose the constant c in order to obtain a solution bounded at the point b and infinite at the point a. Whenever the constant c thus chosen is found to be the same for both the extremities a and b, we obtain a solution which is bounded at both extremities. It can be verified that this is the case whenever the function 9 satisfies the equality:

(3.66) r g(A) dA = 0 . 1 j(A - a)(A - b)

Calculation shows that the solution cp so obtained, bounded at the point a, can be written:

(3.67) cp(z) = ~ Jz - a r g(A) JA - b dA, z ¢ L , 2m z - b JL A - z A - a

and that whenever (3.66) is satisfied, this solution can also be written:

1 i g(A) dA (3.68) cp(z) = -2 .J(z - a)(z - b) j " z ¢ L .

m L (A-a)(A-b)JI.-Z

Finally, the solution of (3.57) bounded at a is given by:

(3.67)' J(z) = ~Jz - a J. g(A) JA - b dA, z E L m z-blA-z A-a

and, if (3.66) is satisfied, by:

(3.68)' J(z) = ~j(z - a)(z - b) J. g(A) dA, z E L . m lj(A - a)(A - b)A - z


We can make precise the behaviour of the solution given by (3.67) at the point a. Let us observe that we have the identity:

1 = ~ r 1 dA. tj L J(z - a)(z - b) ni lJ(A. - a)(A. _ b)A. - z' z .

We then deduce (for z tj L):

(3.69)

<p(z) = g(a) + ~Jz - a( r (g(A.) - g(a)) ~~ 2 2m z - b 1 ~ A. - z

+ r g(a) dA.) 1 J(A. - a)(A. - b) .

4.3. Wiener-Hopf Integral Equations of the Second Kind

Earlier we have studied the Wiener-Hopf integral equations of the second kind in §l. Here we take up that study again by the Hilbert problem method, which allows us to weaken the hypotheses. Let us recall that it is a question of finding a function f(t) as a solution of

(3.70) f(t) = L<Xl k(t - to)f(to) dto + g(t) , - 00 < t < 00 ,

where g(t) and k(t) are given functions defined for all real t. We assume that there exists b real, b > 0, such that:

(3.71) k(t)exp( - bt) and g(t)exp( - bt) are integrable on the real axis.

We seek a solutionf(t) such thatf(t)exp( - bt) shall be integrable on the real axis. Let US introduce the following Fourier transforms:

(3.72) F + (A.) = L<Xl f(t)eiA'dt , Im(A.) = b ;

(3.73) F - (A.) = - f <Xl f(t)e iA' dt , Im(A.) = b ;

(3.74) K(A.) = f~ <Xl k(t)eiA'dt , Im(A.) = b ;

(3.75) F(A.) = f~<Xlf(t)ei.<tdt = F + (A.) - F -(A.) Im(A.) = b ;

(3.76) G(A.) = f~ <Xl g(t)eiA'dt , Im(A.) = b ;

These Fourier transforms are defined for A. on the line 1m (A.) = b in virtue of the


hypotheses (3.71). Furthermore, F + is analytic for 1m (A) > band F _ is analytic for Im(A) < b. Recall that the hypotheses (3.71) imply that all these Fourier transforms converge to zero at infinity; in particular

K(A) -+ 0 when IRe(A)1 -+ 00, Im(A) = b .

Equation (3.70) implies

(3.77) (1 - K(A))F + (A) - F _( A) = G(A) , Im(A) = b .

Considering the s.analytic function C{J relative to the line L with equation Im(A) = b, defined by:

(3.78) { F + (A) if Im(A) > b F _ (A) if Im(A) < b ,

equation (3.77) can be considered as a Hilbert problem for this function C{J. On putting (taking account of the orientation of the line L from left to right):

(3.79) { C{Je(A) = F - (A)} A E C, Im(A) = b , C{Ji (A) = F + (A)

this problem can be written

(3.80) C{Ji(A) = 1 _ lK (A) C{Je(A) + 1 ~c;)(A)' Im(A) = b .

All the s. analytic solutions of this equation are given by Theorem 3. We are only interested in those solutions which tend to zero at infinity. We then have

Proposition 4. Let k and g be two given functions satisfying (3.71) and such that their Fourier transforms K and G are Holderian on the straight line L with equation 1m (A) = b. Assuming further that the Fourier transform K satisfies

(3.81) C

IK(A)I~D:r' y>O, AEL,

and that

(3.82) (1 - K (A)) has no zero on the line L .

Then the Fourier transform F of each solutionf(t) of equation (3.70) is given by

F()') = C{Ji(A) - C{Je(A) , A E L ,

where the s. analytic function C{J is the solution of (3.80) and is given on putting:

(3.83) 1 [. (1) ( 1 )] p = -. lIm arg - lim arg , 2m Re(i.) ~ 00 1 - K (A) Re(A) ~ _ 00 1 - K (A)

§4. Application to Some Problems in Physics 91

1 (1 foo + ib ( (l - P)P ) dl ) (z - p)pexp 2ni _ 00 + ib Log (1 - K (l»(l - a)P l - z '

Im(z) > b (3.84) tJ>(z) =

(l - P)P ) dl ) - K(l»(l - a)p l - z '

-:---_1--:-p exp (_1 . f 00 + ib Log ( (z - a) 2m _ 00 +ib (1

Im(z) (z) foo + ib G (l) dl (3.86) q>(z) = 2ni _ 00 + ib tJ>i(l)(1 - K (l»(l _ z) + l?(z)tJ>(z) , z rt L ,

where I?(z) is a polynomial such that:

1) if p > 0, then degree (I?) :r::; p - 1; 2) if p = 0, then I? == 0; 3) if p < 0, then there are no solutions except if

r lmG(l) (3.87) JLtJ>i(l)(1 _ K(l»dl = 0, O:r::; m:r::; - (p + 1),

and then the unique solution corresponds to I? == O. o

§4. Application to Some Problems in Physics

Introduction

We present here some applications of the Cauchy integral and the Hilbert problem to some problems with a physical origin: the electrostatic problem, the lifting profile problem, the elasticity problem.

1. Simple Layer and Double Layer Problems(50)

We now resume our study (see §2, Remark 7) of the connections between the Cauchy integral and the so-called simple layer and double layer potentials.

(SO) See Chap. II.


Let us consider, in the case of a contour L, the Cauchy integral

(4.1) (j)(Z) = -21 . i !().) d)', Z ¢ L, Z E IC , 7tl L/I,-Z

where J is a real function of class ~1 defined on the contour L. Let us separate the real and the imaginary parts in the expression (4.1). We denote by n the unit normal exterior to the contour L and by t the unit tangent orientated positively. We have

(4.2) ). - Z = I). - zi eiO , () = arg(). - z) ,

so that on taking the logarithmic derivative

d)' ). _ z = d(Log(l). - zl)) + id() .

We denote by r the Euclidean distance I). - al and we write the integral using the curvilinear coordinate s; after integration by parts, and taking

(4.3) d() d ds = dn(Log r) ,

into account, we get

(4.4) (j)(z) = -21 i J(s) d( ~Og r) ds + ~ i ddJ(S) Log rds, z E IC; z ¢ L . n L n 2n L s

{he real part of expression (4.4) is called the double layer potential and the imaginary part the simple layer potential. More generally (see Chap. II), if g(s) denotes a continuous real function on L, one calls the simple layer potential v defined by the charge density g and L, the expression

v(z) = 21n L g(z) Logrds , with r = Is - zl ,

and if J(z) denotes a continuous real function on L, one calls the double layer potential w of moment density J on L, the expression

w(z) = 21n LJ(S) :n (Logr)ds .

These names come from applications of these potentials to problems in electrostatics(51) which we will see later. We have

Proposition 1.(51)

i) Whenever g(s) is a continuous Junction on the contour L, with L oj class ~1 +. (52),

the simple layer potential defined by

(51) See also Chap. II. (52) See Remark 7, §2.


(4.5) v(z) = 2~ L g(s) Log 1A.(s) - zl ds (53)

is a continuous function of z throughout the complex plane C. ii) Whenever f is a continuous function, the double layer potential defined by

(4.6) w(z) = 2~Lf(S)ddn(LOglA.(S) - zl)ds, zEC\L,

93

is an s. continuous function relative to the contour L. The interior and exterior limits of w(z) are given by:

(4.7) 1 1 r d

(W(Z))i = '2f(z) + 2n If(s) dn (Log(IA.(s) - zl))ds, z E L ;

(4.8) 1 1 1 d (W(Z))e = - -2f (z) + - f(s)-(Log(IA.(s) - zl))ds, z E L . 2n L dn

Proof We refer to Chap. II, §3, Propositions 11 and 13, and in this Chap. XI to §2, Remark 7 and Theorem 1 (iff and g are Holderian). 0

We note that formulas (4.7), (4.8) are directly connected to the Plemelj formulas. It must be remarked that the double layer potential:

(4.9) ~L (IA. _ I) = n(s).(A.(s) - z) dn og z IA. _ Zl2

is a function of class 'fJ 00 of s whenever z is on the contour L, if the latter is 'fJ 00. The kernel of the simple layer potential, Log (I A. - z I), is an integrable function on the contour L whenever z is on the contour L. 0

Remark 1. The Cauchy principal value can be written:

(4.10)

1 f f(A.) lId -: -, -dA. = - f(s)-(LoglA.(s) - zl)ds m LA - Z n L dn

i 1 df + - -d (s) Log 1A.(s) - zl ds , n L s

whenever the function f is differentiable on L.

2. Determination of the Charge Density on the Surface of a Cylindrical Body at Potential V(54)

Z E L ,

o

We propose here to determine the density of charge p on the surface S of a cylindrical body (infinite in 1R3 ) raised to a known potential V (in the framework of

(53) Whenever 9 has the form g(s) = df Ids withf a real function of class ~' on L, then v(z) = 1m cp(z) with cp(z) defined by (4.1).


electrostatics(54), if the cylindrical body is a conductor in equilibrium, then V is constant on S). We will suppose that V is a continuous function and also differentiable on S. We will denote by L a right cross section of the cylinder, and the axis X3

will be chosen in the direction of the cylinder's generators. We will suppose that the potential V is independent of X3 and that the unknown charge density is also independent of X 3 ' and consequently that the potential u throughout the space is also independent of X 3 • The potential u is connected to the charge density p by the Poisson relations (see Chap. II):

(4.11) L1u = - pb(L).

It will be observed that the cylinder in question is infinite; this situation must not in general be considered to be the limiting case of a cylinder of length 1 tending to infinity (on this subject see especially Chap. II). We will further impose that the potential u tends to zero at infinity (that is to say for

r = J xi + X~ -+ (0). This implies the following consequences. 1) From Chap. II, §3, there exists one and only one solution of equation (4.11) with this condition at infinity, and this solution is the Newtonian (logarithmic) potential of - p (see Chap. II, §3(55»):

(4.12) u(t) = - 2~ L p(s)Log(ls - tl)ds, t E 1R2 •

From Proposition 1, the function u is continuous in 1R2 if p is a continuous function on L, and the limit of u on L is the given function V, so that p and V are connected by:

(4.13) - 2~ L p(s)Logls - tlds = V(t) , t,s E L .

2) From Chap. II, §3 (Proposition 3) the charge density must satisfy:

(4.14) L p(s)ds = 0

which implies that the total charge borne by the cylindrical medium is zero. 3) The energy (per unit of length along x 3 ):

(4.15)

is finite.

We will seek the density pes) in the form

(4.16)

(54) See Chap. lA, §4.S.

df pes) = -(s)

ds

(55) Or again the simple layer potential of charge density p.

o


wheref(s) is a real differentiable function on the contour L assumed bounded and sufficiently regular. We will consider the following s. analytic function:

(4.17) cp(z) = ~ r f().) d)', ZEC, z¢L. 2m JL). - Z

From (4.4) we have:

(4.18) ImCPi(z) = Imcpe(z) = - V(Z) , Z E L ,

where CPi (z) and CPe (z) are respectively the interior and exterior limits of the s. analytic function cP (z). In the case where the contour L is the circle of centre the origin and radius 1, we shall now explicitly solve equation (4.18). To the function cP we associate the s. analytic function ¢:

(4.19) { cp(z) if Izi < 1

¢(z) = (1) ip i if Izi > 1

The function ¢ is bounded at infinity by construction. Hence from (4.18) we have

(4.20) ¢i(Z) - ¢e(z) = - 2iV(z) , Z E L .

And from the Plemelj formulas (2.7), Theorem 1, §2, since V is assumed differentiable, the bounded solutions of equation (4.20) are given by:

(4.21) ¢(z) = ~ r - 2iV()') d)' + C 2m JL ). - Z

where C is an arbitrary complex constant. We likewise associate to the function cP the s. analytic function 1/1:

(4.22) { cp(z) if Izi > 1

I/I(z) = ~G) if Izl < 1

Since cp(z) tends to zero at infinity, we have

(4.23) 1/1 (0) = 0 and 1/1 -+ 0 at infinity .

From equality (4.18) we have

(4.24) I/Ii(Z) - I/Ie(z) = 2i V(z) , z E L .

And from the Plemelj formulas, the solutions of equation (4.24) which tend to zero at infinity are

(4.25) I/I(z) = ~ r 2i V()') d)' . 2ml). - z


The first condition (4.23) shows that the potential V must satisfy:

(4.26) r V(2) d2 = 0 l 2

or again (expressing (4.26) in curvilinear coordinates on the circle L)

(4.27) 1 V(s) ds = 0 .

In conclusion, we have:

(4.28)

Using (4.27) and (4.10) we can verify that f: ~2~ d2 is real, and on using (4.40), we

have thus obtained the density p:

(4.29) p(z) = d~z (~ 1 ~~ (s) Log 12 (s) - zl dS), z E L (56).

Let us summarise the results obtained in

Proposition 2. Let L be the circle of radius 1 and centre the origin. Let V(t) be a differentiable function on this circle and satisfying (4.27):

1 V(s)ds = 0 .

Then the unique (continuous) solution of the electrostatic equation (4.13) satisfying condition (4.14) is given by:

(4.30) d (2 r d V ) p(s) = dsz ; JL d;(s)Log(12(s) - zl)ds , z E L .

o Remark 2. In the case where the contour L delimits a simply connected bounded open set, there exists a conformal transformation which establishes a bijection between the open disc of radius 1 and this open set. Making use of this transformation, we can reduce equation (4.18) posed on the contour L to an analogous equation posed on the circle L and thus explicitly solve the electrostatic problem. It must nevertheless be noted that we do not in general know an explicit expression for this conformal transformation, except for very particular contours. 0

(56) With Sz denoting the curvilinear coordinate of z on L.


3. The Problem of the Thin Aerofoil Profile(57)


The problem of the thin aerofoil profile consists of seeking the velocity It = (u l , u2 ) of a fluid which flows around the wing of an aircraft which is assumed to be cylindrical and infinite, of right cross-section (f) with contour L. The complement of (f) in ~2 is denoted by Q. The fluid is assumed to be irrotational which is expressed by the equation

I -+ O· oU 2 oU I O· Q cur U = I.e. -;--- - -;--- = In . UX I UX 2

(4.31)

It is assumed to be incompressible which is expressed by the equation

(4.32) div It = oU I + oU2 = 0 In Q. oX I oX2

The boundary conditions on the wing are written

(4.33) It· it IL = 0 where it denotes the normal to the contour L (58).

The velocity of the fluid at infinity is known and we assume it to be constant and orientated along the axis x I' which is expressed by

(4.34) I It (Xl' x2 ) - u: I -+ 0 as Ixi -+ 00 ;

(4.35) -+ -+ U oo = U oo e l

The profile has the following form:

L

e: the first basis vector .

F

Fig. I

It delimits a simply connected interior and presents a sharp point F called the trailing edge which possesses two distinct tangents(59). We will see below that the physical problem is only completely determined if we impose the supplementary condition that at the trailing edge the velocity is not infinite. This condition is called the Joukowsky condition.

(57) See Chap. lA, §1.5. (58) In what follows, if' denotes in this instance the exterior normal to the obstacle. (59) We suppose that L is regular (at least of class ~1.P) away from F.


ii) Mathematical Formulation

Taking account of the equation

div (it - u:) = 0 ,

we now seek it u: in the form of a Curl ( := Curl) of a function r/J

(4.36)

We shall represent It in the form

(4.37) it (z) = Curl(r/J(z) + U oo x2 ), z = Xl + iX2 ,

where r/J(z) is a single-valued function throughout the exterior of the profile L. Using (4.31) and (4.33), we then deduce an equation which r/J satisfies:

(4.38)

or again

(4.39)

1 LI r/J = 0 in Q (Q := C\(!))

o r/J I = - u:.;:( on L, os L

{ LI r/J = 0 in Q

r/JIL = - U oo X2 on L.

It must further satisfy the condition

(4.40) grad r/J -+ 0 as Izl -+ 00,

and the Joukowsky condition. We now solve equation (4.39) for r/J using a conformal transformation of the exterior Q of the profile (!) onto the exterior B of the circle rc of centre 0 and radius 1. There exists such a transformation denoted by Z(60). We remark that it is singular at the trailing edge F, and this singularity is locally like z«, where IX is a number in ]0, 1[ which depends on the angle between the two tangents at the sharp point F. Let us look for r/J in the form

(4.41) r/J (z) = 1m cP (z) ,

where the function cP is s. analytic relative to the contour L(61).

We put

(4.42) <p (z) = cP 0 Z - I (z), z E B .

(60) This transformation is known by the name of Schwarz-Christoffel (see Iyanaga-Kawada [I] and Milne-Thomson [I]). (61) In fact the "contour" L being non-differentiable, it is rather more suitable to say that I/> is s. analytic relative to the "are" L \ F, (this notion is easily derived from Definition 5, §2).


Let X be the imaginary part of q>. We have

{AX(Z) = 0, Z e B

(4.43) xl~ = - u oo x2(Z -l(Z)) , zerc,

and the function q> is s. analytic relative to the circle rc(62). Equation (4.43) then reduces to:

(4.44)

A solution of this equation is given by the expression (4.25)

(4.45) () ____ 1 f2i(-UOOX2(Z-1(A)))dl B q> Z. 1 11., ze .

2m ~' z - II.

From this we deduce the expression for X and hence that for t/I

(4.46) t/I(z) = ~ 1m [L -u; ~ ~(:)1 (A)) dAJ, z e Q .

This solution satisfies (4.39) and (4.40), but does not satisfy the 10ukowsky condition because the singularity of the transformation Z induces a singularity at the sharp point F in the function t/I. A solution of the homogeneous problem in the exterior B of the circle rc is given by

(4.47) 1

cx(z) = Im(iLogz) = 2"Log(xr + xD .

Let us put

A(z) = iLogZ- 1 (z).

the function t/I + y 1m A is another solution of equation (4.38). We can then choose the constant y in such a way that the 10ukowsky condition will be satisfied by t/I + ylmA. It can then be shown that the velocity 1/ of the fluid at the sharp point F is zero. We remind ourselves that 1/ is given by (4.37). We have

Proposition 3. Let L be a contour possessing a sharp point (see Fig. 1). The problem of determining the velocity 1/ of a fluid around this profile as a solution of the equations (4.31), (4.32), (4.33) and the conditions (4.34) and

(4.48) it (F) = 0

admits a unique solution of the form

1/ = Curl(t/I + ylmA + u oo x2 )

where t/I is given by (4.46) and y being such that (4.48) is satisfied.

(62) Or rather the "arc" ~\Z-I(ZF)'


Proof The only part remaining to be proved is the uniqueness of the solution. It must be shown that the conditions

{Curl It = 0, div It = 0 in Q c [R2

It . niL = 0, It --+ 0 at infinity, and It (F) finite

imply It = o. To do this, let a be the mapping It = (u 1 , u2 ) 1-+ v = (u 2 , - u1 ) = alt. Then v is the solution of

{ Curl v = 0, div v = 0 in Q

v /\ niL = 0, It --+ 0 at infinity, and v (F) finite .

Hence there exists a function cp (a priori locally, or multi valued in Q) such that v = grad cp (hence u = Curl cp) satisfying:

{ Llcp = 0 in Q

~: IL = 0 (s tangent to L), grad cp --+ 0 at infinity, cp (F) finite.

The condition ~:IL = 0 implies CPIL = constant (because L includes only one

sharp point) and therefore cp is single-valued in Q. Propositions 3 and 4 of Chap. II, §3, can then be applied, reducing the condition at infinity:

grad cp --+ 0 at infinity ,

to the condition: there exists K (constant) such that

cp ( x) - K Log I x I --+ 0 as I x I --+ 00.

Hence, we are reduced to solving the problem:

(4.48)' {LI cp = 0 in Q cp IL = constant, cp(x) - K Log Ixl --+ 0 as Ix I --+ 00,

grad cp ( F ) finite .

The at most logarithmic behaviour of the solution at infinity implies that the latter is in the weighted Sobolev space:

w~ 1 ([R2) = {t/I E EC'([R2), t/I(1 + r2) E L2([R2),

(1 + r2)-1/2 gradt/lE(L2 ([R2)f}

on extending cp onto [R2 by cp = CPIL in [R2\Q. Now from Djaoua [lJ, pp. 93 to 97, the problem:

{ Llcp = 0 in Q and [R2\,Q ,

CPIL = CPo (constant) ,

cp(x) - KLoglxl--+ 0 as Ixl--+ 00 and gradcp(F) finite,


admits one and only one solution, which is therefore qJ = constant in [R2, whence the uniqueness of the solution of our problem, stated in the proposition. 0

It must be noted that if L has no sharp point, then there is no uniqueness of the solution of (4.48)': hence for Q = [R2 \ D where D is the disc of centre 0, and radius a, qJ = constant and qJ = K Loglxl are both solutions of (4.48)'. In the case where L has many sharp points(63), we can reduce it to qJlL = constant on all of L with the aid of cuts starting at the sharp points, and the aid of functions constant on each of the connected domains so obtained. The uniqueness indicated in Proposition 3 (for the velocity ofthe fluid) remains true, on noting that one can only impose the condition that the velocity be finite at a single sharp point of L (the fluid velocity at the other sharp points then being infinite). 0

Remark 3. As in the case of the electrostatic problem, the conformal transformation Z is explicitly known and usable only for very particular profiles and the above method is hence only constructive in these cases. 0

Remark 4. The constant}' allows us to determine the resultant of the fluid forces on the profile. The projection of this force on the Ox2-axis is called the lift; the projection on the 0 x l-axis is called the drag. 0

The Case of the Profile Reduced to a Segment. Here we study explicitly the case where the profile is reduced to a rectilinear segment ab which makes an angle rx with the Ox l-axis; the origin is chosen at the middle of the segment abo (see Fig. 2). The equations of the velocity It are the same as in the previous case of the thin profile.

a

Fig. 2

We shall seek an s. analytic function qJ relative to the segment [a, b] such that:

(4.49) t{I(Z) = Im qJ(z) , Z ¢ [a, b], (see (4.41)) .

The problem (4.39) reduces to:

(4.50) { Im(qJe(Z)) = - u oo x 2 (z) , Z E [a,b] Im(qJi(z)) = - u oo x 2 (z) , Z E [a,b]

where [a, b] denotes the closed segment abo

(63) With, again, (f) connected (and simply connected).


We deduce from it the equation (taking account of the orientation from b towards a):

(4.51)

We put

(4.52) 1 i b f(A.) cp(z) = -2. -, -dA., z ¢ [a, b] 1t1 a /I. - Z

where f is a function defined on the segment [a, b]. The Plemelj formulas then give us:

(4.53)

(4.54)

cp;(z) - CPe(z) = f(z) , z E [a, b]

1 ~b f(A.) cp;(z) + CPe(z) = -; -, -dA., z E [a, b] ma/l.- z

Hence on using (4.51), we have:

(4.55) Imf(z) = 0, Z E [a,b] .

Observe that, the segment being rectilinear, the expression:

(4.56)

is real for z on the segment abo We can then use (4.54) and (4.50) in order to deduce

1 ~b f(A.) (4.57) --; -, -dA. = - 2iu<XJ X2 (z), z E [a, b] 1tl a /l.-Z

The search for f(z) is a problem involving the Hilbert inversion formula which we have solved in §3. The solutions are given by the formula (3.62). We are interested in the solution bounded at the point b which is given by the formula (3.67) (written with a and b interchanged). We have, on recovering cp(z):

(4.58) cp(z) = ~ Jz - bib - 2iu<XJ 1m (A.) JA. - adA., z ¢ [a, b] . 2m z - a a (A. - z) A. - b

The right hand side being an odd function on the segment [a, b], the solution cP is in fact antisymmetric and the condition (3.66) is satisfied. It can therefore equally be written, using equation (3.68) as:

(4.59)

1 ib - 2iu<XJ 1m (A.) dA. cp(z) = -2 .J(z - a)(z - b) J -,-, z¢[a,b].

m a (A. - a) ( A. - b) /I. - z

An s. analytic solution of the homogeneous problem associated with (4.50) is given


by:

(4.60)

The function

(4.61)

R-b cpdz) = --.

z - a

<p(z) = cp(z) + YCP1(Z)

103

is therefore another solution of the problem, which does not tend to zero at infinity. We can then choose the constant Y in such a way that the velocity u is zero at the point b. To do this we use the formula (3.69) which shows that the function cP has a "singularity" in ~ (64) at the point b which thus gives the value of Y which annulls this singularity (and leads to a "singularity" in (z - b)3/2!). We obtain

(4.62) Y = ! [ f.b u oo 1m (A. - b)~ dA. + f.b U oo 1m (b) dA.] 1t a ~ A. - b a J(A. - a)(A. - b)

Hence we have

Proposition 4. The irrotational and incompressible fluid flow throughout the exterior of a segment [a, b], whose equations are (4.31), (4.32), (4.33) and (4.34) and which satisfy the condition:

(4.63) it (b) =0

admits one and only one solution given by the relations (4.37), (4.49), (4.59), (4.60), (4.61) and (4.62). The velocity in a neighbourhood of the point a has a singularity

(4.64) --+ c u (z) ~ c-;:..'

(v' z - a)3

4. Plane Elasticity and the Biharmonic Equation

Here we are situated within the framework of a plane deformation problem in classical elasticity (see Chap. lA, §2, Sect. 4.2.2) for which the volume forces have been neglected (hence 11/1 = 0). We further suppose that rF! = r F2 = r F and similarly ru! = r U2 = ru' The notations are modified as follows: the Airy function X becomes W. The displacement functions Va becomes Va (for rJ.. = 1,2), on Ll which replaces aou = ru n 1R1x"x,). The given stresses Fa become ga on L2 which replaces aOF (65). We are thus led to determining a scalar function W

(4.65) Ll2 W = 0 (biharmonic function in 0 c 1R2)

(64) Which leads to a singularity of u+ in (z - b)-1/2 at the point b. (65) We assume that the given (v.), (g.), Q are sufficiently regular (at least g. continuous on L 2 , v. Hiilderian on L, and r = oQ of class ee"").


connected to the stress tensor through the relations:

(4.66) { 0"13 : 0"23 = 0, ~33 = V(O"ll + 0"~2) 0"11 - »':22 0"12 - - »':12 0"22 - W,ll

which lead to the boundary conditions to be ascribed to Won L 2 • It can be verified that a function on the plane 1R2 is biharmonic in an open domain Q c 1R2 if and only if it can be expressed in the form

(4.67) W( Xl' X2) = Re(Zcp(z) + X(z)), z = Xl + iX2 (66)

where cp (z) and X (z) are hoI om orphic functions in Q. We will put

(4.68) { cp(Z) = CPI(X I ,X 2 ) + .iCP2(Xl ,X2), CPI = Recp, CP2 = Imcp X(z) = xdxl ,X2 ) + IX2(X p X2), Xl = Rex, X2 = ImX,

which leads to

(4.69)

where CPI' CP2, Xl satisfy the relations

(4.70)

and to the Cauchy conditions:

(4.71) OCPI oX I

whence we deduce

(4.72) A W = 2(OCPI + OCP2) = 4 OCPI = 4 OCP2 . oX I oX2 oX l oX2

Using the relations (4.66) and the laws of elastic behaviour, the strain tensor can be written:

(4.73)

1 + v 1 + v [ 02 W ] 8 11 = -- [0"11 - V(O"ll + 0"22)] = -- - ~ + (1 - V)A W

E E UXI

1 + v 1 + v [ 02 W ] 8 22 = -E-[0"22 - V(O"ll + 0"22)] = -E- - OX~ + (1 - v)A W

1 + v 1 + V 02 W 8 12 = -E-O"12 = - E OX I OX2 '

(66) See for example Mikhlin [1], p. 179. (67) One also has .6.X2 = O.


then with the relations (4.69) and (4.72):

(4.74)

These relations can further be expressed with the Lame coefficients A. and f.1 and the Poisson coefficient v (see Chap. lA, §2, formulas (2.27), (2.28) and (2.24)):

(4.75)

1 + v

E 3 _ 4 = A. + 3f.1

2f.1 ' v A. + f.1 '

The displacement field is then easily integrated beginning with

and the solution is given, to within a field of rigid displacements, by

(4.76)

In effect, ~ know (see §2.2 in the Appendix "Mechanics" to Chap. I), that the solutions U R of the equation

6ij (U) = 0, Vi and j = 1,2,3 ,

correspond to the displacement fields of an undeformable solid (all the rigid displacements); it is easy to verify these have the form

(4.77) v: = {a + f3x 1

y - f3x 1


Let it = (n i , n2 ) be the normal vector on the boundary L. We orientate the tangent vector on L in such a way that

Observe that if we denote by s the curvilinear abscissa on the boundary L, we have:

(4.79)

Let hi (s) and h2 (s) be functions defined on L2 to within a constant, respectively by:

(4.80) { :s (hds)) = 21Ji gds)

d 1 ds (h2 (s)) = 2Ji g2 (s) .

We can in summary write a system of equations which determine the solutions of our elasticity problem:

Problem P. Find cp(z) and X(z) holomorphic in Q and continuously differentiable on Q such that:

(4.81)

1 [(A + 3 Ji) - - ] 2Ji A + Ji cp(z) - zq/ (z) - x' (z) = Vi (z) + iV2 (z) ,

Whenever cp(z) and x(z) are known, the displacements it and the stresses (J are given respectively by the formulas (4.76) and (4.78) with (4.68). If we put

(4.82)


we have

(4.83) 1 [(A + 3Jl) - , 'J u(z) = 2Jl A + Jl IP(z) - ZIP (z) - X (z) , Z E Q .

o We shall first examine the case where the domain Q is bounded, sufficiently regular, and simply connected and where either Ll or L2 is empty. The boundary conditions are thus of only one type: displacements or stresses imposed on the whole contour L. The domain Q being simply connected, there exists a conformal transformation w(z) which is a bijection of the ball B of centre the origin and radius 1 onto the domain Q. The image of the circle ~, of centre the origin and radius 1, is the contour L. We put:

(4.84)

and for the data:

{ ¢(Z) = IP (w(z)) , tjJ(z) = x'(w(z)) ,

Z E B

z E B ,

(4.85) { V(z) = 2Jl (Vl (w(z)) + iV2 (w(z)) , z E ~

H(z) = 2Jli(hdw(z)) + ih2(w(z)) , z E ~ .

In the case where the displacements are imposed on the whole boundary L (the case where L2 = 0), the problem P becomes:

Problem Pl. Find ¢(z) and tjJ(z) holomorphic in the ball B and continuous on B(68),

such that:

(4.86) ( 2 + 3Jl) w(z) - -A + Jl ¢(z) - OJ' (z) ¢' (z) - tjJ (z) = V(z) , Z E ~ ,

with V a Holderian function given on ~.

In the case where it is the stresses that are imposed on the whole boundary L (the case where Ll = 0), the problem P becomes:

Problem P2. Find ¢(z) and tjJ(z) holomorphic in the ball B and continuous on B(68),

such that:

(4.87) ¢(z) + :,~~ cl>'(z) + ti/(z) = H(z) , z E ~ ,

with H a Holderian function given on ~. We have

Theorem 1. Let Q be a simply connected, regular, bounded domain. Then the plane elasticity problem with displacements imposed on the boundary L (4.65), (4.66)(69)

(68) With, further, t/J' continuous on B. (69) With the regularities on the data as indicated on page 103, footnote (65). For the solution indicated, it needs in fact a little more regularity (see (4.105H4.106)) A(z) having a derivative which tends to a limit at the boundary of B.


reduced to the problem (4.81) admits a unique solution analytic in Q and continuous up to the boundary L. The plane elasticity problem with stresses imposed on the boundary L admits a solution analytic in Q and continuous up to the boundary L, unique to within a displacement of the whole body (given by (4.77)) if the stresses on the boundary satisfy the conditions:

(4.88) JJ 1 Lgl(S)dS = 0, f/ 2(S)dS = 0

L(gl(S).X2(S) - g2(s)x l (s))ds = 0 .

Proof Let us first examine the uniqueness of the solutions. It suffices to examine the case where the data on the boundary are null. It follows from the equilibrium equations(70) and Stokes' formula that we have, taking into account the null boundary conditions:

(4.89) A. t (ell (it) + e22 ( it ))2 dX l dX2 + 2Jl. f}(e ll (it ))2 + (e22 ( it))2

+ 2(e12( it ))2] dX l dX2 = 0 .

It follows from this that

(4.90) e( it) = 0 ,

and hence the only solutions of the homogeneous system are the displacements of the whole body given by (4.77). In the case where the boundary conditions are those of type "imposed displacements", the homogeneous system admits only the null solution. Let us examine the problem of the existence of solutions. We shall now solve equations (4.86) and (4.87). We use a lemma.

Lemma 1. Let a(z) and fJ(z) be two functions holomorphic in the ball Band continuous on jj and satisfying the equation:

(4.91) a;(z) + /J;(z) = G(z) , z E Cfi ,

where G is a Holderian function given on the circle Cfi. Then we have:

1 i G(A.) a(z) = -2' -, - dA. + y, z E B, Y E C 1tl 'if/\' - Z

(4.92)

(70) See Chap. lA, §2. Here, those formulas can be written:

in Q.


(4.93) P(Z) = ~ f G(A.) dA. - y, Z E B . 2m ~A. - Z

Proof of Lemma 1. We will write equation (4.91) in the form:

(4.94) rli(z) + Pi(Z) = Re G(z) + i(lm G(z) + 21m Pi(Z)) , z E rtf .

We then use the results of Sect. 2 (see (4.21)) which give

1 f Re G(A.) (4.95) rl(z) + P(z) = ~ A. dA. + 15 1 , z E B ,

nl ~ - z

Similarly we can write

(4.96) rli(z) - Pi(Z) = Re G(z) - 2 Re Pi(Z) + i 1m G(z) , z E rtf .

Whence:

1 film G(A.) (4.97) rl(z) - P(z) = ~ A. dA. + 15 2 , z E B, 152 E C . nl ~ - z

Using (4.95) and (4.97) we can then deduce (4.92) and (4.93). o Completion of the Proof of Theorem 1. Using Lemma 1 we can replace equations (4.86) and (4.87) by the respective equations (for z E B)(71):

(4.98) (A. + 3Jl) </>(z) __ 1 f w(A.) 4>'(11.) dA. = _1 f V(A.) dA. + '}' . A. + Jl 2ni ~w'(A.)A. - z 2ni ~A. - z '

(4.99) </>(z) + _1 f w(A.) (f'(A.) dA. = _1 f H(A.) dA. + '}' . 2ni ~w'(A.)A. - z 2ni ~A. - z

Let us then consider the function 8, s. analytic relative to the circle rtf, defined by:

o ifzEB

(4.100) (f'(-z~) (f'(O)

w'G) - w'(O) ,

8(z) = z ¢ B .

From the Plemelj formulas (see §2.2), we have

" - _1 f (f'(A.) - _1 f (f'(O) ~ z E C rtf (4.101) 8(z) - 2ni ~ w'(A.)(A. _ z) dA. 2ni ~w'(O)A. - z ' \.

We can now, taking account of (4.100) and (4.101), replace equations (4.98) and

(71) By using here the fact that the data V (resp. H) is Holderian on ~ and by supposing a priori that

:,i~ 4i'().) is Holderian on ~.


(4.99) by the following equations:

(4.102) (A. + 3J1.)4>(Z) __ 1 f w(A.) - w(z) ¢'(A.) dA. A. + J1. 2ni 'IJ A. - z w'(A.)

4>'(0) 1 f V(A.) = w'(O)w(z) + y + 2ni 'lJA. _ zdA., z E B ,

(4.103) 4>(z) + _1 f w(A.) - w(z) ¢'(A.) dA. 2n i 'IJ A. - z w'(A.)

- _ ¢'(O) _1 f H(A.) A. - w'(O) w(z) + y + 2ni 'lJA. _ z d , Z E B .

We can replace equation (4.102) by the system of equivalent equations (with unknowns 4>'(0) and 4>(z), z E ~):

( A. + 3J1.) 4>(z) __ 1 f w(A.) - w(z) ¢'(A.) dA. A. + J1. 2ni 'IJ A. - z w'(A.)

(4.104)

= ~'(O) w(z) _1 f V(A.) d' B Y + 'I' -'(0) + 2.' 11., Z E , W m 'lJ1I. - z

( A. + 3J1.) 4>'(0) __ 1 f w(A.) - w(O) - A.W'(O) ¢'(A.) dA. A. + J1. 2ni 'IJ A. 2 w'(A.)

_ ¢'(O) w'(O) = _1 f V(A.) dA. . w' (0) 2ni 'IJ A. 2

On differentiating(72) and then taking values on the boundary, we get:

(4.105)

(4.106)

( A. + 3J1.) 4>' (z) __ 1 f ~ (w(A.) - w(Z)) ¢'(A.) A. + J1. 2ni 'lJdz A. - z w'(A.)

= ¢'(O) ~~) + Ai(z) , z E ~ ,

A. + 3J1. 4>'(0) + _1 f w(A.) - w(O) - A.W'(O) ¢'(A.) dA. A. + J1. 2ni 'IJ A.2 w'(A.)

_ ¢'(O) w'(O) = _1 f V(A.) dA. . w'(O) 2n i 'IJ A. 2

1 f V(A.) A(z) = -2. -, -dA. . nl 'lJ1I. - Z

(72) We assume here a little more regularity than previously on the function V (i.e. V is of class CC 1. P on CC) te, permit this operation.


In this system of equations, the second equation is scalar and the essential part is:

A. + 3J1. (j)'(O) _ w'(O) (j)'(O) A. + J1. 0)'(0)'

which is invertible because J1. > O. The first equation is of Fredholm type (see Chaps. II~ VII and VIII) in the space of

. f' b h' I' kId (W(A.) - W(Z») . I contmuous unctions ecause t e mtegra s erne dz A. _ Z IS regu ar.

The Fredholm alternative and the earlier proved result on uniqueness therefore show the existence of a solution such that ¢' is continuous on the circle re. This proves the theorem in the case of the imposed displacements. In the case of the imposed stresses, the system equivalent to (4.99) is:

¢'(z) + _1 f ~(W(A.) - W(Z») (j)'(A.) dA. 2ni ~dz A. - z w'(A.)

= _ ;:'(0) w'(z) + A~(z) z E re 'I' w'(O) I' ,

(4.107) A,'(O) :1:'(0) w'(O) _1 f w(A.) - w(O) - A.W' (0) (j)'(A.) dA. 'I' + 'I' 0)'(0) + 2ni ~ A. 2 O)'(A.)

= _1 f H(A.) dA. 2 · ,2 ' nl ~ A

A(Z) = ~f H(A.) dA. . 2nl ~A. - z

In this equation system, the second equation has the essential part:

(4.108) ¢/(O) + (j)/(O) ~~; ,

which is not invertible. Putting

(4.109)

that equation can also be written:

(4.110)

I = ¢'(O) w(O) ,

1+7=g,

and I is determined to within an imaginary constant from g if g is real. On coming back to the initial domain Q, and to the initial unknowns u: it is easy to see that it is defined to within a displacement of the whole and that the compatibility conditions (1m g = 0) are expressed in the form (4.88). The Fredholm alternative then applies to the first equation and this estabilishes the result (73). 0

(73) For more developments of this method, see for example Mikhlin [1], pp. 194-195.


Remark 5. For certain domains, the above method of solution of the equations of elasticity of a shell furnish an explicit solution. This is the case of the ball B with boundary the circle qj. In effect, the system (4.104) reduces in this case to:

! (). + 311) - 1 f V()') , c/>(z) = y + lz + -. -, - d)', z E B , A + 11 2nl '(; A - Z

( ). + 311) I - r = _1 f V()') d)' . ). + 11 2ni '(; ).2

(4.111)

D

The Problem of the Elastic Half-Space Submitted to the Pressure of a Plug. We examine here the particular case of the above problem of plane elasticity where the elastic domain consists of a half-plane X2 > O. The boundary conditions are supposed to be of mixed type:

(4.112)

(4.113)

where f(x l) is a given H6lderian function. The solution is defined to within a translation parallel to the Ox l-axis. The equations which correspond to equation (4.81) and expressing the boundary conditions (4.112) and (4.113) are written(74)

(4.114) ( ) 1m cp(z) + zcp'(z) + x'(z) = 0, z real ,

(4.115)

(4.116)

Re(cp(Z) + zcp'(z) + x'(Z)) = 0, x 2 = 0, IXll > a ,

2~ 1m (C). : 3:) cp(z) - (zcp'(z) + x'(Z))) = f(x l ) ,

x2 = 0, IXll < a .

In order to solve this problem, we put

(4.117) cp(z) + zcp'(z) + X'(z) = G(z) , z real,

and G(z) is a real unknown function, defined to be nonzero for X2 = 0 and IXll < a. We have, by analogy with the case of the circle (Lemma 1)(75):

(4.118) cp(z) = ~ fa G()') d)' , 2m -a). - Z

(74) For functions cp and X holomorphic in the half-plane XI > 0, continuously differentiable in XI ~ 0 (except perhaps at the points (± a, 0) and tending to zero at infinity. (75) Assuming a priori that G(z) is "Holderian" on the whole segment ( - a + e, a - e) with a behavi-

our in a neighbourhood of + a like G(A) = G(A) , cr. E [0, 1[, with G(A) Holderian on the segment - IA ± al"

( - a, + a) and in a neighbourhood of ± a, which gives a sense to formulas (4.188)-(4.121) (for this, see §3.4.2 and also Muskhelishvili [1], Chap. 4) and which will be satisfied by the solution found.


(4.119) 1 f" G(A.) I/I(z) == up'(z) + X'(z) = -2' -, -dA. . 1t1 _III\. - Z

The boundary value of qJ is

(4.120) ( ) = G(z) _1 fa G(A.) d' I I qJ z 2 + 2' , 1\., Z = Xl' Xl < a . 1t1 _III\. - Z

Equation (4.116) is then expressed by:

A. + 2p. 1 fa G(A.) (4.121) (' )2 ~dA. = - f(x l ) ,

p. I\. + P. 1t -a I\. Z

Solving (4.121) for G(A.) is a problem of inversion of the Hilbert problem on the arc [ - a, a]. We have seen (see §3, formula (3.62)) that the solutions are given by the function analytic for 1m z # 0:

(4.122) qJ(z) = 2p.(A. + p.) 1 f" f(A.)J A. 2 - a2 dA. + C .

A. + 2p. 21tJZ 2 - a2 _II A. - Z J Z2 - a2

The constant C is real and depends on the resultant offorces which act on the plug. We likewise have (since G is real):

(4.123) I/I(z) = qJ(z)

(4.124)

o

Part B. Integral Equations Associated with Elliptic Boundary Value Problems in Domains in 1R3

In this second part B we will study different integral equations posed on surfaces in the space 1R3. They all come from representations in integral form of the solutions of the elliptic partial differential equations of physics or mechanics. In contrast with the previous part A, we here study problems only in a bounded domain in [R3 or in the complement of a bounded domain in [R3. Also in contrast with the previous part A, we will not make use of analytic functions. The essential tools here will involve the use of Hilbert spaces of the Sobolev space type together with variational methods. The following notations will be used throughout this part B:

IPk will denote the space of polynomials of degree k on [R3.

Q will denote a regular, bounded open set (locally on one side of the boundary r defined below) in [R3.

r will denote the boundary of Q which is an orientated differentiable surface. n will denote the unit normal to r orientated towards the exterior of the open

set Q

Qf will denote the open set the complement of Q in [R3; Qf = [R3\,Q.

§1. Study of Certain Weighted Sobolev Spaces

Introduction

In these §1 and 2 of Chap. XIB we seek the solutions u, with finite energy, of the interior (i.e. in a bounded open set Q) and exterior (i.e. in Qf, the complement of Q) problems of Dirichlet and Neumann (see (2.13), (2.14), (2.45), (2.47» making use both of the methods of Chap. II (by employing potentials of simple and double layers) and the variational methods of Chap. VII (for example the Lax-Milgram theorem). Thus the use of simple layer potentials allows us in particular to reduce the search for a finite energy solution u to the interior and exterior Dirichlet problems (with Uo E Hl/2(r) given on the boundary r of Q) to the search for the solution q E H - 112 (r) of a variational problem corresponding to a coercive bilinear form on H- 1/2(r) (see Theorem 3, §2 of this Chap. XIB); the Lax-Milgram theorem then allows us to conclude the existence and uniqueness of the solution u of the problems studied.

§1. Study of Certain Weighted Sobolev Spaces 115

- In relation to Chap. II, the framework of the Sobolev spaces Hl(Q) or HA(Q) in the interior case, and the weighted(l) Sobolev spaces Wl (Q') or Wb(Q') in the exterior case lead to notable simplifications. We can compare in particular Theorem 3, §2 of this Chap. XIB with Proposition 18, §4 of Chap. II where for all given Uo E ct'0 (F), there would not necessarily exist any solution u E ct' ~ (Q) or u E ct'~(Q') to the interior or exterior Dirichlet problem. We further solve here (see Theorem 6, §2) the interior and exterior Neumann problems by double layer potentials, again by a variational method on the boundary r of Q.

- In relation to Chap. VII, we treat the Dirichlet and Neumann problems in unbounded open domains (complements of bounded sets), which necessitates the introduction ofthe weighted Sobolev spaces Wl(Q') and wA(Q'). We further have an explicit formulation of the solution u of the problem considered (due to the use of simple or double layer potentials, see (2.24), (2.38), (2.51) for the problem (2.47), and (2.59»-with the aid of "intermediary" functions q or ({J,

which are solutions of variational problems on the boundary r (thus involving two variables in place of the three for Q c jR3, which simplifies numerical applications).

These intermediary functions q or ({J are often of further direct physical interest: thus in electrostatics, q can correspond to the surface charge density (and in certain physical problems, we propose to calculate u to find q). Finally, thanks always to the use of simple and double layer potentials, we can simultaneously solve the interior and exterior problems (of Dirichlet or Neumann). This is of special interest in electromagnetism where the problems in fact concern the whole space (jR3). The method used adapts naturally, and without particular difficulty, to the transmission problems (see (2.43), Proposition 2 of §2). The solution of the Dirichlet or Neumann problems associated with the Helmholtz equation will also benefit from similar advantages. 0

In this section we first recall the properties of the spaces Hm(Q) and H'O(Q) defined in Chap. IV where Q is a bounded open set in jR3. Then we shall study some Sobolev spaces weighted at infinity and defined in the open set Q'. Let us recan<2):

Definition 1. Let m be a positive or zero integer, then Hm(Q) is defined by:

Hm(Q) = {u E L2(Q); Da u E L2(Q), (X = ((Xl' (X2' (X3)

I(XI = (Xl + (X2 + (X3' I(XI ~ m} and H'O(Q) is the closure of .!?&(Q) in the space Hm(Q)(3),

We denote by IUlm,Q the seminorm defined by its square:

(1.1) lul~Q = f IDmul2 dx = f L IDaul2 dx , Q QI.I=m

(1) Or again the Beppo-Levi spaces. (2) See Chap. IV. (3) With Hm(Q) endowed with the topology defined by (1.2).

o


whilst the usual norm on the space Hm(Q) will be denoted by II U 11m,0:

(1.2) II U 11~0 = L f IDa uI 2 dx . lal.;; m 0

We recall that the injection of the space Hm(Q) into L 2(Q) is compact if Q is a sufficiently regular bounded open set (see Chap. IV). We have

Proposition 1. The quantity

(1.3) [u]m,o = [IUI~o + L If D.l.udx12J1/2 O.;;I.!I';;m-i 0

is a norm on Hm(Q) equivalent(4) to the norm II u 11~0. The seminorm Iulm,o is a norm on H'/j(Q).

Proof It is evident that

lulm,o ~ Gil u lim,o . Let us reason by contradiction to prove the reverse inequality. If that inequality were false then there would exists a sequence (un) in Hm(Q) such that

1 j[Un]m,o ~ -

II un lim,o = n1 . (1.4)

But the compact injection property allows us to extract a subsequence such that

Daun -+ Dau, IIXI ~ m - 1 ,

the convergence being in L 2(Q). On the other hand we have from (1.4) that

{ Dmun -+ 0 in

LDaundX -+ 0 , IIXI ~ m - 1 .

It follows that u is null which contradicts (1.4). o

We have shown in Chap. VII how to make use of the spaces Hi(Q) and Hb(Q) to construct variational solutions to boundary value problems associated with the Laplace operator A on the domain Q.

In the case of the exterior domain Q', the spaces which lead to a variational formulation for the Laplace operator A, are not H 1(Q') and HA(Q'). Here we have to define the Hilbert spaces in the domain Q'.

(4) With the hypothesis stated, namely, Q a regular bounded open set.

§l. Study of Certain Weighted Sobolev Spaces 117

Definition 2. Let r = lxi, be the Euclidean distance of x E [R3 from the coordinate origin. We define:

W I (r)l) _ { . U 2( ') au 2(£)1) 1 2 3} ~& - U, ( 2)1/2 E L D, -a EL ~& , i = " , 1 + r Xi

WMD') = closure of g)(D') in W(D').

Remark 1. The restriction of every function in WI (D') to a bounded open set (9 in [R3 is in HI ((9).

The weights (1 1 2 1/2 fix the behaviour at infinity of the function in WI (D'). + r )

o We will denote by lulu]' the seminorm given by its square:

(1.5) 3 f lau 12 luli,D' =.L -a (x) dx, ,=1 Q' x,

and by II u II I,D' the norm given by its square:

2 f lu(xW 2 (1.6) II u Ill,!!' = --2 dx + lull,Q' . D,1 + r

Let Ba be the ball of centre the origin and radius a and let B~ be the interior of its complement. We have the lemma:

Lemma 1. The expression

(1.7) lull,B; = Ctl LI::i(X)12 dx y/2

is a norm on WMB~) equivalent to the norm II u III B"

Proof. The space g)(B~) being dense in Wl(B~), it suffices to prove the corresponding inequalities for a regular function with compact support. One of the inequalities is trivial. It suffices to show that:

lIU(X)12 1

(1.8) -1--2 dx ~ Clull B • B: + r ..

We have, on passing to spherical coordinates (r, e, rp):

(1.9) l IU(XW lIU(X)12 100 f . -1--2dx < --2 -dx = dr lu(x)1 2slflrpdedrp. B: + r B: r a

For every functionf(r) zero for r sufficiently large and such thatf(a) = 0 we have:

foo roo df (1.10) a If(rW dr = - Ja 2rf(r) dr (r)dr

( roo )1/2 ( roo Idf 12 )1/2 ~ 2 Ja If(rW dr Ja dr (r) r2 dr


Using inequality (1.10) withf(r) = u(r, e, tp), we obtain

r lu(xW foo fl ou l2 r lou 12 JB; 1 + r2 dx ~ 4 a dr or sin tp r2 de dtp = 4 JB; or (x) dx.

Whence we deduce (1.8) and Lemma 1. o Theorem 1. The seminorm

(1.11 ) ( 3 f lou 12 )1/2 lulu}' =.L -;-:(x) dx ,=1 D' uX,

, X E Q' C 1R3 ,

is a norm on the spaces W l (Q' ) and W6(Q' ), equivalent to the norm II u IkD"

Proof It is immediate that we have

lull.D' ~ lIulkD' .

To prove the reverse inequality, we reason by contradiction. Thus suppose that there exists a sequence (un) in W l (Q' ) such that:

(1.12)

1 llunkD' ~ ~

II un II!,D' = 1 .

Let a be large enough for the ball Ba to contain the open set Q, and let tp and IjI be two very regular functions such that

{

tp + IjI = 1 in Q'

(1.13) tp = 0 for Ixl ~ 2a IjI = 0 for Ixl ~ a tp~O, 1jI~0.

Differentiating the products IjIv and tpv, we have, for v E Wl (Q' )

(1.14) {IIjIV kB; ~ IvkB; + Clv lo,B 2.nD' ; ItpvkD' ~ IvkB 2.nD' + Cl vlo,B2.nD' .

The injection of Hl(B2a (') QI) into L2(B2a (') QI) being compact, we can extract a subsequence such that

un -+ U in L2(B2a (') QI) .

Then it follows from inequalities (1.14) and (1.12) that the sequences (tpun) and (ljIun) are Cauchy sequences in Hl(Bza (') QI) and Wb(B~) respectively. Making use of Lemma 1 likewise, we deduce that

{tpun -+ W l in Hl(B2a (') QI)

ljIun -+ W z in Wb(B~),

and hence that un = (tp + ljI)un -+ W, W E Wl(Q' ). From (1.12) we have Dw = 0 and thus W = C, C E IR. The behaviour at infinity of the functions in Wl(Q' ) then imposes C = 0, which contradicts (1.12). 0

§2. Integral Equations Associated with the Boundary Value Problems of Electrostatics 119

Remark 2. It follows from Lemma 1 and Theorem 1 that the spaces Wi (Q') and WA(Q') are identical to the Beppo-Levi spaces, generally denoted by BL(Q') and ~ 1 (Q'), and defined by:

BL(Q') is the closure of £C(Q') (or again of H 1(Q')) for the norm (1.11) (see Theorem 1); ~l(Q') is the closure of £C(Q') (or again of HA(Q')) for this same norm.

In effect we have Hi (Q') C Wi (Q'), hence BL(Q') c Wi (Q'). Conversely, if u E W 1(Q'), then using p E £C([Rn) such that p(x) = 1 if x ~ 1, p(x) = 0 if x ~ 2, we form the sequence un(x) = u(x)p(x/n) contained in BL(Q') and which converges to u for the norm (1.11). Hence we have proved, by truncation, that BL(Q') = Wi (Q'). 0

We will now consider the "fractional" Sobolev spaces HS(r) defined on the surface r (see Chap. IV). We know that the functions in the space Hm(Q) admit traces on the surface r up to order m - 1 if the surface r is sufficiently regular. The functions in the space Wi (Q'), being locally in Hi, similarly admit a trace on the surface r. Let us recall this in the

Theorem 2. The trace mapping, which to a differentiable function u defined on Q (respectively Q') associates its trace on r, extends in a unique manner to a surjective and continuous mapping, denoted by ulr,from Hi (Q) onto H 1/2 (r) (respectively from Wi (Q') onto Hl/2(r)) the kernels of which are respectively HA(Q) and WA(Q'). 0

§2. Integral Equations Associated with the Boundary Value Problems of Electrostatics

Here we study the boundary value problems associated with the Laplace operator Lt. We have earlier examined certain of these problems in §4, part A, of this Chap. XI in the two spatial dimension case. We shall now obtain integral representations of solutions of these problems, then we will examine different integral equations whose solution allow the solution of the boundary value problems at issue. We refer to Chap. VII for a detailed study of boundary value problems for the operator Lt.

1. Integral Representations

We give here a general result concerning integral representations of functions harmonic in Q and Q'. Let us first give a result which plays the role, in the case of [R3, of Liouville's theorem in the case of [R2 and holomorphic functions.

Theorem 1. Let P be a linear differential operator on [R3 with constant coefficients, such that

(2.1)


Then every solution of the equation

(2.2) P(D)u = 0 (With D = (iO~l' iO~2' io~J) which is a tempered distribution (u E 9" (1R3)) is a polynomial.

Proof Let u be the Fourier transform of the solution u which is equally a tempered distribution. We have

(2.3)

Thus, u is a distribution with support reduced to the origin. It is therefore a finite sum of derivatives with Dirac mass at the origin. By the inverse Fourier transformation, we hence deduce that u is a polynomial. 0

Notation. Let u be an s.continuous function relative to the surface r, i.e. continuous up to the boundary in Q and in Q' respectively. We will denote by Ui the interior limit of this function and by Ue the exterior limit. We will denote by [u] the jump in this function as it crosses r: (2.4) [u] = U i - Ue (5)

We have

Theorem 2. Let u E ~OO(.Q) n ~OO(Q') n 9"(1R3 )(6) be a function which further satisfies:

(2.5) LI u = 0 in Q and in Q" ,

we then have the identities

(2.6)

"Ix rf. r

(5) The notation [.] will be made larger or smaller (examPle [:: J) according to what is enclosed, but

the meaning of (2.4) will remain the same. (6) This can be generalised in the following sense if r is of class '61 +, (see Chap. II, §3):

u E '600 (Q) n '6~(Q) n '600 (Q') n '6~(Q') n Y"(1R 3 )

au That is to say u is '6 00 at every point outside r, has tempered growth, and u and - tend towards an

. au continuous limits on both sides of r (we will similarly apply this theorem to the case where u and -an admit boundary values in HI/2(r) and H- 1/2(r), different on each side of r with r Lipschitzian, equalities (2.6) and (2.7) then being satisfied a.e. x).


2 4n r an Ix - yl (2.7) 1

Ui(X) + ue(x) = ~f [au (Y)] 1 dy(y)

- 41n 1 [u(y)] a~y (IX ~ YI) dy(y) + p(x) , "Ix E r ,

where p(x) is a polynomial harmonic in 1R3 (7).

Proof This follows from the fact that 1/4nr is the elementary solution of the operator -.1. Let B. be the ball of centre x and radius e and Ba the ball of centre o and radius a large enough to contain Q. For x E Ba n Q', x ¢ r, we apply Green's formula in Q and Q' n Ba , to the functions U and 1/4nr. By summation and passage to the limit as e -+ 0, we obtain:

(2.8) 1 u(x) = 4~f[::(Y)],X ~ Yldy(y) - 4~f [U(Y)]a~y(lx ~ YI)dY(Y)

+ 41n faB. (U(Y) a~y (IX ~ YI) - :~ (y) Ix ~ YI)dY(Y) .

Whenever x E Q'\Ba , we obtain:

(2.9) 1 0 = 4~ f [:: (y)] Ix ~ yl dy(y) - 4~ f [u(y)] a~y (IX ~ YI) dy(y)

+ 41n LB. (U(Y) a~y (IX ~ YI) - :~ (y) Ix ~ YI) dy(y) .

We put

(2.1 0) v = 41n LB. ( u(y) a~y (IX ~ YI) - :~ (y) Ix ~ YI) dy(y) .

Now let us consider the function w(x) defined by:

w(x) = {v (x), x E Ba v(x) + U(x) , X E Q'\Ba .

By construction w E ~OO(Ba) and

(2.11) .1w = 0, X E Ba .

In the open set Q', we have

(2.12) w(x) = u(x) - 41nl([:~(Y)],x ~ yl - [U(Y)]a~yCx ~ YI))dY(Y) ,

and hence we deduce that w E ~OO(Q') and is harmonic in Q'. It then follows that w is harmonic throughout 1R3 and from Theorem 1 it is a harmonic polynomial. The expression (2.7) is obtained in an analogous manner on introducing the two half-balls of centre x, B. n Q and B. n Q'. 0

(7) See Chap. II, §7.


2. Dirichlet Problems Relative to the Operator J

Let us consider the Dirichlet boundary value problem relative to the operator LI in o and 0' respectively. We shall call the problem

(2.13) { LI u(~ = 0, xeD, ulr - Uo .

the interior Dirichlet problem(S). This problem was first examined in Chap. II, then in Chap. VII where it was shown that it has a unique solution in Hi (0) if Uo e Hl/2(r). We shall call the problem

(2.14) {LlU(~ = 0 , xeD' , ulr - Uo .

the exterior Dirichlet problem(S). We have

Proposition 1. The interior and exterior Dirichlet problems (2.13) and (2.14) have a unique solution in Hl(O) and Wi (0') respectively, ifuo e Hl/2(r).

Proof The case of the interior problem having already been seen, let us examine the exterior one. From Theorem 2 §1, there exists Ruo e Wi (0') such that

(2.15) Uo = RUoir .

The problem (2.14) can be written:

(2.16) {LI(U - Ruo) = - LlRuo (u - Ruo)lr = 0 .

A variational formulation of this problem is: find v e Wo(O') such that:

, xeD'

(2.17) f t ~ ow dx = - f t o(Ruo) ow dx, Vw e wA (0') . 0' ;=1 ox; ox; 0';=1 ox; ox;

The coercivity of the bilinear form associated with this problem follows from Theorem 1, §1. The Lax-Milgram theorem(9) then shows the existence and uniqueness of the solution. 0

Let us thence consider the function u, defined in all R3, which is the solution in 0 of the interior Dirichlet problem, and the solution in Q' of the exterior Dirichlet problem. We will denote by ouJon and oue/on the normal derivatives respectively in the interior and the exterior of r. The integral representation Theorem 2 shows

(8) See also Chap. II. (9) See Chap. VI and VII.


that:

(2.18) u(x) = 4~f[~~(Y)] Ix ~ yldy(y) + p(x) , Vx E [R3 .

When x --+ 00, the first part of the second member in (2.18) behaves like 1/r and thus belongs to W1 (Q'). The belonging of u to this same space implies that:

(2.19) p = 0 .

We have

Theorem 3. Let uo(x) be given in Hl/2(r); the integral equation in q:

(2.20) 1 f q(y) 4n r Ix _ yl dy(y) = uo(x) , X E r ,

admits a unique solution in the space H -1/2(r). This problem has the variational formulation

(2.21) b(q, q') = f uo(y)q'(y)dy(y) , Vq' E H- 1/2(r) ,

where the bilinear form b(q, q') is defined on H- 1/ 2 (r) by:

b( ') = ~f f q(y)q'(x) d ( )d ( ) q, q 4 I I y x y y . nrrx-y

(2.22)

The bilinear form b(q, q') is "coercive" on the space H- 1/ 2 (r), in thefollowing sense:

(2.23) b(q, q) ~ Pllqll~-1/2(l")' P > 0 .

The solutions of the interior and exterior Dirichlet problem (2.13) and (2.14) are then given by:

(2.24) u(x) = 41n f Ix q~) yl dy(y) , Vx E [R3 ,

where q is the solution of (2.20). This representation is called the simple layer potential(lO).

Proof We put

q(x) = [:~ (X)] .

From the representation (2.18) and (1.29) we have shown (2.24). On taking the trace of u on the surface r which is the same on both sides by construction, we show that the unknown function q is connected to the given Uo by equation (2.20). Let us prove the continuity of the corresponding mappings.

(10) See also Chap. II.


Since u is harmonic in Q, we have

(2.25) f aUi f ~ au av 1 -a vdy = .f.... -a .-a .dx, VVE W (Q).

r n Q I: 1 X, X,

The left hand side of this identity is a continuous linear form on HI (Q) and thus

on H 1/2 (r), in virtue of the trace theorem. This defines ~:i as an element of the dual

space of Hl/2(r), i.e. H- 1/2(r) (from the theorem on the surjectivity of the trace). Similarly, we have

f aUi f ~ au av 1 I

(2.26) -a vdy = - .f.... -a .-a .dx, "Iv E W (Q) . r n fI' ,: 1 X, X,

which allows us to define ~:e as an element of H- 1/ 2 (r). These results show that

q = [:~J is defined in the space H- 1/2 (r) and that we have:

(2.27) f 3 f au av r qv dy = i~1 J Ie aXi aXi dx, "Iv E WI (1R3)

Observe that the function u is in the space WI (1R 3 ).

Thus if the function q is given in H - 1/2 (r), we have (2.27) which can be considered as a variational problem in u, with bilinear form

3 i au av a(u, v) = L -a -a dx.

i:l R' Xi Xi

From Theorem 2, §1, this bilinear form is coercive on W 1 (1R3) and this defines in a unique way the function u and hence its trace Uo in Hl/2(r). We have shown that the mapping Uo H q is an isomorphism of Hl/2(r) onto H- 1/2(r). We have

b(q,q) = f qUody = f .f laau.12 dx ~ .8l1qll~-l"(r) r JR'I:l X,

from the continuity of the corresponding mappings and Theorem 2, §l. 0

Remark 1. If we put (see equation (2.20) and Chap. II (4.90), but with the change of notation 2 = - L):

(2.28) 1 f q(y) 2 q = 4n r Ix _ yl dy(y) ,

we can show that 2- 1 is an isomorphism of HS(r) onto H S- 1 (r), for all real s. o

Remark 2. The results obtained here in a Hilbert space framework can be compared with those of Chap. II in the framework of functions continuous on the boundary. 0


Remark 3. As we have earlier indicated in part A, §4.1 in the case of two dimensions, the name "simple layer potential" comes from electrostatics. Equation (2.20) is in effect the equation which gives the electric charge density q on the surface of a conductor r at potential Uo (see Chap. lA, §4 and Chap. II).

We shall now examine the integral equation associated with the solution of the exterior Dirichlet problem when it is represented with the aid of a double layer potential. This problem has been considered earlier in Chap. II with continuous data on the boundary. Here, we will work in a Hilbert space framework, with the data on the boundary in HI/2(r). We will now assume that Q is connected, with connected complement. In order to solve the exterior Dirichlet problem (2.14) in WI (Q') by a double layer potential u, it is necessary (see (2.6), Theorem 2) that

(2.29) [ou] = oU;1 _ OUel = 0 . on an r an r

Hence on putting

(2.30) _ OUel _ oU;1 g-- --- an r on r '

the double layer potential u must be the solution ofthe interior Neumann problem:

{ Llu = 0 in Q,

(2.31) ::Ir = g .

This problem admits a solution (see Chap. VII, §2, and also Chap. II), unique up to a constant if g satisfies:

(2.32) fgd)' = 0 .

This condition can be equivalently written as a condition on uo. To do this, let us introduce v as a solution of

(2.33)

and let

(2.34)

{LlV(~ = 0 , vir - 1 ,

X E Q', V E WI(Q') ,

OV ho = an .

Then, condition (2.32) can equally be written, on applying Green's formula, namely:

L (uLlv - vLlu)dx = f(u:~ - V:~)d)'


as

(2.35) fUOhOdY = O.

The double layer potential u, the candidate for the solution of (2.14), can be written:

(2.36) u(x) = - 41n f [u(y)] a!y (IX ~ YI) dy(y) , X E Q u Q .

It remains to determine [u(y)]. Assuming here that [R3 \Q is connected, we have:

Theorem 4. The integral equation in cp,

cp(x) 1 f a ( 1 ) (2.37) -2- + 4n r cp(y) any Ix _ yl dy (y) = - uo(x) , x E r ,

admits a solution, unique to within a constant, in the space Hl/2(r) whenever Uo E Hl/2(r) and Uo satisfies condition (2.35). The solution u of equation (2.14) of the exterior Dirichlet problem admits a representation as a double layer potential:

(2.38) u(x) = - 4~ f cp(y) a!y (IX ~ YI) dy(y) , x E Q' ,

where cp is the solution of (2.37).

Proof Let us put

(2.39) cp(x) = [u(x)] , X E r . from (2.36) we have the representation (2.38) and from the equality (2.7) we have

uo(x) = - cp;x) - 41n f cp(y) a!y ex ~ YI) dy(y) , x E r .

This shows that a solution of equation (2.37) is given by (2.39). From the trace theorem it is in the space Hl/2(r) and is defined to within a constant because the solution of problem (2.31) is unique to within a constant. In order to construct the inverse mapping, we introduce the Hilbert spaces:

{X = Hl(Q)/1I10 x W1(Q') ,

(2.40) { [~unJ -_ o} , :f{" = U E X, Au = 0 in Q and in Q', u

where we denote by 1110 the following equivalence relation in H 1 (Q):

u == u' mod 1110 if and only if u - u' = constant .

The quotient space H 1 (Q)/1I1 0 (that is to say the space of equivalence classes of functions in H 1 (Q) for this equivalence relation) endowed with the norm

111111 = inf IluIlH'(l1), for all equivalence classes 11 E Hl(Q)/1I10 , UEii


is then identified with the subspace IPt of functions u E H 1 (0) such that

fa udx = 0 ,

endowed with its subspace topology from H 1 (0). We have seen that u E X and ,1u = 0 in 0 and Q' allows us to define ou;!on and oue/on in H- 1/2 (r). The condition [au/an] = 0 therefore cefined a closed vector subspace of X, and .Y( is a Hilbert subspace of X. Green's formula gives us:

(2.41 ) 3 (f au ov f au ov ) f ov L --dx + --dx = <p-dy, \Iv E.Y( . i=1 aoXioXi Q'oXioXi r an

It follows from Proposition 1 and Theorem 1, §1, that the bilinear form of the first member of (2.41) is coercive on the space .Y(.

This shows that the mapping <p 1--+ U is continuous from H1/2(r)/lPo into .Y(, and the trace Theorem 2, §1 shows the result sought. 0

Let us put (as in Chap. II, see (4.99))

K<p(x) = - 21nI <p(y) o~y (IX ~ YI)dY(Y) , <p E H1/2(r) ;

this allows us to write (2.37) in the form

1 2"(- <p + K<p) = Uo

We shall now make some remarks about this operator K.

Remark 4. The integral kernel of K:

(2.42) a ( 1 ) (ny. (x - y)) any Ix - yl = Ix - yI 3

has a singularity in 1/lx - yl whenever x and yare "neighbours" on the surface r. This kernel is therefore integrable in the Lebesgue sense, but [lOt in the space L2(r), when we consider it as a functions of Y for fixed x on r. The corresponding integral

I <p(y) o~y ex ~ YI) dy (y)

is defined in the Lebesgue sense when <p E H1/2(r) and is thus a continuous function on r. All this is no longer true in a neighbourhood of a singularity of the surface r, if the latter is not differentiable. 0

Remark 5. The double layer potentials behave at infinity as l/r2 (11). This explains why the condition (2.35) is necessary. 0

(II) See for example Chap. II, (4.103).


Remark 6. It can be shown that the integral operator K:

cp -. ~nl I cp(y) o~y (IX ~ YI)dY(Y)

is continuous from HS(r) into H S+ 1 (r) for all real s. We then deduce that (l - K) is an isomorphism of HS(r)/'P 0 (12) onto the subspace of W(r) of functions satisfy~g~3~ 0

Remark 7. The interior capacity operator C defined in Chap. II, §5, which makes correspond to every element Uo E D(C)(uo E <'CoW)), the boundary value au/on of the normal derivative of the solution u of the interior Dirichlet problem:

Problem PD i

{Au = 0, in Q,

ulr = Uo

can again be defined here in an analogous manner, with the following further simplification regarding to the domain of C. It follows from Theorem 3 and Theorem 4, that the capacity operator C given again (see Chap. II, §5, (5.13) with L = - 2') by:

is an isomorphism of the space H1/2(r)/'Po(l3) onto the subspace Ho 1/2(r) of

functions f E H- 1/ 2 (r) such that J/ dy = 0 and more generally, because of

Remarks 1 and 6, that C is an isomorphism of HS(r)Wo onto the subspace

H''o-1 (r) of functions f E W- 1 (r) satisfying If dy = O.

Hence the natural framework of the capacity operator C which permits one to express the energy W of the solution u of Problem PDi by:

W = ~f Igrad ul 2 dx = ~f uCudr Q r

is that of Sobolev spaces. This framework allows one to treat the study of the (interior) capacity operator with much flexibility and simplicity. We do not take up this method here. 0

Remark 8. We can likewise associate with the interior Dirichlet problem an integral representation in terms of a double layer. The integral equation with which

(12) With a definition analogous to that given above (in the proof of Theorem 4) for the space HI (Q)/IP' o' (13) Again assuming Q and Q' are connected.


this is associated is

cp(x) 1 f a ( 1 ) -2- - 4n r cp(y) any Ix _ yl dy(y) = uo(x) , X E r ,

or again 1 2"(cp + Kcp) = Uo .

If Q' is connected, it admits a unique solution in Hl/2(r) if Uo E Hl/2(r). The proof is similar to that for Theorem 4. 0

The following proposition essentially follows from (an extension of) Theorem 2:

Proposition 2. The following transmission problem: determine the solution u E Hl(Q) x W1(Q') of:

1 i) Au = 0 in Q and Q',

ii) [u] = uilr - uelr = f E Hl/2(r) given,

... ) [au] aui/ aue/ H- 1/2(r). 11l - = - - - = g E gIVen, an an r an r

(2.43)

admits one and only one solution, and this solution is given by (see (2.6))

(2.44) u(x) = 4~ L g(y) Ix ~ yl dy(y) - 4~ L f(y) a~y (IX ~ YI) dy (y) .

Proof The uniqueness follows at once from the fact that iff = g = 0, then u is the solution of Au = 0 in 1R3, and u E Wl(1R3), thus u = O. The existence of u follows immediately from (2.44) (or from Theorems 3 and 4). As in Chap. VII, a variational method could be used. 0

To wish to solve the interior (or exterior) Dirichlet (or Neumann) problem by a simple layer potential (or a double layer potential) comes down to reducing the problem under consideration to the above transmission problem, on imposing [u] = 0 (or [au/an] = 0). Let us take for example the exterior Dirichlet problem (2.14), which we seek to solve by a double layer potential. Hence we seek to determine the jump f in u as it crosses r whenever uelr = Uo is given, with the condition [au/an] = g = O. This is only possible under the condition (2.35) on Uo if Q and Q' are connected. The jump fin u is determined by the integral equation (2.37) (with cp == f) in Hl/2(r). We will later give other examples (in the case of the Neumann problem) where we reduce it to the transmission problem (2.43). It will be noted that it is in general arbitrary to wish to solve the problem that we have studied in the open set Q #- IR" by a simple layer potential or a double layer potential, that is to say such that [u] = 0 or [au/an] = O. On the physical level, we might wish for a solution thrqughout the whole space 1R3 of a certain type, that is to say with or without discontinuity: thus in the electrostatic framework, if we consider a charge density covering the surface r of a conducting medium, the potential throughout the space 1R3 will in general be continuous as it crosses r.


3. Neumann Problems Relative to the Operator J

We shall consider in this section the Neumann problems associated with the operator Ll in the open sets Q and Q' respectively. We shall call the interior Neumann problem(14) the problem:

(2.45) { Llu(x) = 0, X E Q ,

:~Ir = g, on r .

We have seen that if Q is connected, this problem admits a unique solution in H1(Q)/IPO' whenever g E H- 1/2 (r) and satisfies the condition

(2.46) tgdY = 0 .

Furthermore, this solution can be represented in the form of a simple layer potential. We shall call the exterior Neumann problem(l4), in W1(Q'), the problem:

(2.47) { Llu(x) = 0, X E Q' ,

oul on r = g, on r .

A variational formulation of this problem is (see Chap. VII):

3 f ou ov f ' .L -;--:-;-:dx = - gvdy, VVE W1(Q). ,=1 aUx,ux, r

(2.48)

The bilinear form on the left hand side of (2.48) is coercive on W 1 (Q') from Theorem 2, §1, and thus this problem admits a unique solution. We have proved the

Proposition 3. The exterior Neumann problem (2.47) admits a unique solution in W1(Q') whenever the given g is in the space H- 1/2(r) (lfQ' is connected). The interior Neumann problem (2.45) admits a unique solution to within a constant in H1(Q), lfthe given g is in H- 1/2(r) and further satisfies (jor Q connected):

tgdY = 0 . o

We shall first give a representation as a simple layer potential of the exterior Neumann problem(15). Let us consider the function u defined in all 1R3 as the solution of the exterior Neumann problem (2.47) for x E Q', and as the solution of the interior Dirichlet

(14) See also Chap. II. (15) See also Chap. II.


problem which has the same trace on r for x E Q. We have by construction [u] = 0, and Theorem 2(16) shows that

(2.49) u(x) = 41n fJ~~ (y)] Ix ~ yl dy(y) , x E 1R3 .

It is necessary to obtain the expression for the exterior normal derivative of a simple layer potential, in order to find the integral equation which connects the given g to the unknown

(2.50) q(x) = [~~(X)], X E r . It is given by:

Lemma 1. Let u be given by(17):

(2.51) u(x) = 4~ I Ix q~) yl dy(y), x E 1R3 ,

then if q E H- 1/2 (r), we have

{ ~:i (x) = q;X) + 4~ I q(~) O~x Cx ~ YI) dy(y), x E r . (2.52) OUe q(x) 1 f 0 ( 1 )

a;:r(x) = - 2 + 4n r q(y) onx Ix _ yl dy(y), x E r . o

The proof is very closely analogous to that of Theorem 2 and we need not reproduce it here (see for example Chap. II, §3, Proposition 12 for the classical framework or again Kellog [1], p. 164). Let us suppose Q' is connected. From Lemma 1 we deduce

Theorem 5. The integral equation in q:

(2.53) q(x) 1 f 0 ( 1 ) - 2 + 4n r q(y) onx Ix _ yl dy(y) = g(x) , XEr,

admits a unique solution in H - 1/2 (r) if the given g E H - 1/2 (r). The solution of the exterior Neumann problem (2.47) is then given by the representation (2.51). On putting, as in Chap. II, see (4.91) (but with a change of sign of J):

J q(x) = 42n I q(y) O~x Cx ~ YI) dy(y) ,

equation (2.53) can again be written in the form (used in Chap. II):

1 2( - I + J)q = g

with, here, g E H- 1/ 2 (r) given, and q E H- 1/ 2 (r).

(16) In fact its extension. (17) See also Chap. II, in the case where q E 'b'(F).


Remark 9. One can show that ( - / + J) is an isomorphism of W(r) onto HS(r) ~~~~ 0

Remark 10. The interior Neumann problem also admits a solution representable as a simple layer potential. The associated integral equation is:

(2.54) q(x) 1 f a ( 1 ) -2- + 4n q(y) onx

Ix _ yl dy(y) = g(x) , X E r , r

or again:

1 2(q + Jq) = g .

The operator (/ + J) is an isomorphism of the subspace of functions in H - 1/2 (r) satisfying condition (2.46) onto the quotient subspace of H- 1/ 2 (r) by the functions proportional to the function ho(x) given by (2.34), when Q is connected. 0

Remark 11. Exterior capacity operator Ceo (Analogous to Remark 7): We note that the exterior capacity operator Ce , defined in Chap. II, §5, (see (5.14)), in the framework of continuous functions, can be defined, here in the framework of Sobolev spaces, by: the operator Ce makes correspond, to every element Uo E Hl/2(r), the boundary value of the normal derivative of the solution u to the exterior Dirichlet problem PDe:

{Au = 0 in W 1 (Q') (18)

ul r = Uo

It follows from Theorems 3 and 5 that the operator Ce , given again here by CeUO = (I - J)2- 1 uo/2, is an isomorphism of Hl/2(r) onto H- 1/2(r), (and more generally, because of Remarks 1 and 9, of HS(r) onto HS- 1 (r) if Q' is connected). Hence the natural framework of the capacity operator Ce which allows the energy W of the solution u of problem PDe to be expressed by:

W = -21 f Igrad ul 2 dx = !f uoCeuodr , Q' 2 r

is that of Sobolev spaces. o Let us now examine the representation of the solutions of the Neumann problems as double layers. Whenever we formally differentiate a double layer poential given by (2.38) in order to calculate the normal derivative at the boundary, there appears an integral

(18) It will be observed that because of Theorem 3, the solution of this exterior Dirichlet problem can be written in the form of a simple layer potential (see (2.24)), hence (see Chap. II, §3, Proposition 2) it tends to zero at infinity; and consequently the definition of the operator Ce given here agrees with that in Chap. II, §5.


operator whose kernel is

(2.55) 02 ( I' ) nx·ny _ 3(nx.(x - y))(ny.(x - y))

onXony Ix - yl = Ix - yl3 Ix _ yl5

This kernel is non-integrable because it admits a singularity in l/lx - Y13. It therefore does not define an integral operator (see Appendix "Singular Integrals"), but instead a principal value type of operator. Moreover, it follows from the representation Theorem 2 that a double layer potential likewise has a normal derivative on both sides of the surface. It is thus difficult to write down directly an integral equation associated with the representation in terms of a double layer of the interior and exterior Neumann problems. To get round this difficulty, we will now establish some variational formulations of this integral equation problem. Let u be the function defined in all 1R 3, which is the solution of the interior Neumann problem (2.45) in Q (defined up to a constant) and the solution of the exterior Neumann problem (2.47) in Qf. On the other hand, let 9 be given in the space H- 1/2 (r) and satisfying (2.46):

f gdy = 0 . r

By construction, we have u E :f( with (see (2.40)):

:f( = {v E Hl(Q)/IPO x Wl(Qf), Jv = 0 in Q and Qf, [ov/on] = O}

Let the unknown be cp(x) = [u(x)], X E r. Then the integral equation sought is the one connecting the unknown function cp to the given g. We then have

Theorem 6, Let 9 be a given function satisfying 9 E H- 1/2 (r) and (2.46). The variational equation

(2.56) b(cp, t/!) = I gt/!dy, 'tit/! E Hl/2(r)/IPo ,

where the bilinear form b(cp, t/!) is given by

(2.57) b(cp, t/!) = 8~ I I (cp(x) - cp(y))(t/!(x) - t/!(y))

02 ( 1 ) x onx any Ix _ yl dy(x)dy(y)

admits a unique solution in Hl/2(r)/IPO' Further, we have

(2.58) b(cp, cp) > PllcplI~l!2(r)!I'O' P > 0 .

The solutions of the interior and exterior Neumann problems (2.45) and (2.47) are then


given by:

(2.59) u(x) = - 41n I cp(y) o~y C x ~ YI) dy(y) , x if. r, x E [R3 ,

where cp is the cited solution of (2.56).

Proof The function g being given in H- 1/ 2 (r), we have constructed the function u in the space f and hence the function cp in Hl/2(r)/IPO. Conversely we have:

(2.60)

This variational equation defines u in f in a unique manner whenever cp is given in H 1/2 (r)/P o. The elements of f possess a unique normal derivative on r defined in H- 1/ 2 (r) and satisfying (2.46). n., the element cp in H 1/2 (r)/P 0 we therefore associate the function u in the space f and, similarly, to the function t/J we associate the function v E f. We then define the bilinear form b(cp, t/J) by:

(2.61) f ov f ou

b(cp, t/J) = r cp on dy = r t/J on dy

The inequality (2.58) then results from the continuity inequality of the isomorphism cp f--+ ou/on, whence:

(2.62) Jl (t (::,Y dx + L (::,Y dX) ~ Pllcpll~112(r)/Po .

It now remains for us to prove the equality (2.57). We use the following so-called solid angle relations

(2.63)

We have

(2.64)

Ifo( 1) 11 ifxEQ -4 -;- I _ dy(y) = 1/2 if x E r

n r uny x yl O·f Q' I X E .

b(cp, t/J) = r cp(y) o~y { - 41n r t/J(x) o~x ex ~ YI)dY(X)} dy(x) }dY(Y)

- 81nr(cp(Y)o~'{fr t/J(X)o~xCx ~ YI)dY(X)}dY(Y)

+ t/J(y) o~,{ r cp(X) o~x (IX ~ YI) dy(x) } dY(Y)) .


Let us consider the function w defined for sufficiently regular ({J and", by:

w(y) = t ({J(X) "'(x) o~x (IX ~ YI)dY(X) .

By construction wE:fl and thus t ~: dy = 0, which becomes

(2.65) 8~ t O~y {t ({J (x) t/J (x) o~x (IX ~ yl )dY(X)} dy(y) = 0 . On using relation (2.63) we have

(2.66) Of o( 1 ) - - dy(x)=O ony r onx Ix - yl ' or equally

(2.67) 81n L ({J(Y)"'(Y) o~J t o~x (IX ~ YI)dY(X) }dY(Y) = 0 .

Then adjoining the identities (2.64), (2.65) and (2.67), we obtain

b«({J, t/J) = 8~ Ir O~y {t «({J(z) - ((J(x))(t/J(z) - "'(x)) (2.68)

x o~x (IX ~ YI)dY(X) }dY(Y) ,

an expression in which we take z = Y after differentiating with resepct to y. In this expression we can move the differentiation under the integral sign and hence obtain (2.57) for the differentiable functions ({J and t/J. The final result is obtained by a density argument. 0

We now give another expression for the bilinear form b«({J, "') defined by (2.57). Let us point out that the interest in variational formulations for these problems is to lead to stable numerical approximations of the solutions ({J of the corresponding problems. These then lead to the numerical calculation of solutions u of the Neumann problems in 1R3. In this sense, the expression of the bilinear form b«({J, t/J) which we have constructed is much simpler to employ in numerical work. This expression uses differential operators on the surface r which we introduce here. Let the open set ra be a neighbourhood of the surface r in 1R3. It consists of points whose distance from the surface r is less than J. (see Fig. 1).

r

Fig. 1


For a judiciously chosen value of J every point x of r~ projects in a unique way onto the surface r in a point denoted by &>(x). To every function cP defined on r, we associate the function ip defined on r~ by:

(2.69) ip(x) = cp(&>(x)) , x E r~ . Definition 1. We associate to every differentiable function cp(x) on the surface r, the field of vectors tangent to the surface r: (2.70) curl~ cp(x) = n(x) /\ grad ip(x) (19),

where n(x) is the normal at the point x to the surface. r (grad is the gradient operator of a function defined on .an open set in 1R3 ). 0

Let X(x) be a field of vectors tangent to the surface r and differentiable. We associate with it the field of vectors X(x) defined on an open set r~ by:

(2.71) X(x) = X(&'(x)) .

Definition 2. We associate with every field of differentiable vectors tangent to r the scalar function curlr X defined by:

(2.72) curlrX(x) = n(x).curl' X(x) (19),

where curl' is the usual curl vector of a field of vectors in 1R3. 0

Remark 12. We denote by curl~ cp the curl of a function, which is a tangent vector, in order to distinguish it from curlr X, the curl of a tangent vector, which is, itself, a scalar function. It can be verified that the operator curlr has for its adjoint the operator - curl~ with respect to the scalar product in L 2 (r), i.e.:

(2.73) L curl~ cp(y). X(y) dy(y) = - L cp(y)curlr X(y)dy(y) ,

for all X, cp such that

X E L2(r)3 (20) curlrX E L2(r) ,

cp E L2(r) , curl~cp E L2(r)3 .

We now assume that r is twice continuously differentiable.

Definition 3. The Laplace-Beltrami operator on the surface r is defined for every twice differentiable function cp by:

----c> (2.74) Llrcp(x) = curlrcurlrcp(x) .

o Notice that on a plane portion of the surface r, the Laplace-Beltrami operator coincides with the Laplace operator Ll in the orthonormal axes(21).

(19) These definitions are in fact independent of the chosen extensions (2.69) and (2.71). (20, Where X is a field of vectors tangent to T. (21) Note, however, that the operator curlr then becomes - curl.


Theorem 7. Let q>(x) and t/J(x) be two differentiable functions on the closed and regular surface r. Then the bilinear form b(q>, t/J) defined in Theorem 6 by (2.57) is also equal to:

(2.75) Iff 1 ~ ~ b(q>, t/J) = -4 curlrq>(y). curlrt/J(x)dy(x)dy(y) . n r rlx - yl

The normal derivative on the surface r of a double layer potential

(2.76) u(x) = - 4~ r q>(y) O~y (IX ~ YI)dY(y), x E 1R3\r ,

is given by

ou 1 i 1 -;-(x) = - -4 I ILl rq>(y)dy(y) un n r X - Y

(2.77) + 41n f r curl;. q>(y) A grady (IX ~ YI). (n(y)

- n(x))dy(y)), x E r , ~

where the operators curlr and Llr are defined by (2.70 and 2.74). o Proof We first establish a preliminary lemma. We assume the functions q>(x) and t/J(x) to be twice differentiable. Let us consider the gradient of the function u defined by (2.76):

(2.78) grad u(x) = - 41n r q>(y) grad x (o~y (IX ~ YI) ) dy(y) ,

X E 1R3 \r .

We then have

Lemma 2.

(2.79) grad u(x) = - 41 f curl~ q>(y) A grady ( 1 ) dy(y) , n r Ix - yl

x E 1R 3\r .

Proof of Lemma 2 . We have

(2.80) ~ = - ~f q>(y)~(~( 1 ))dY(Y) x E 1R3 \r . ox! 4n r ox! ony Ix - yl '

But we also have

(2.81)


We easily verify that

(2.82)

whence we derive the identity

Let e1 be the first basis vector. Then on using the fact that 1 is a harmonic Ix - yl

function, we verify that

whence finally for x E /R3\r

(2.85) 1::1 = 41nfr cp(y)ny.cUrl;,y(e t /\ gradyCx ~ YI))dY(Y) ,

::1 = 4~L CP(Y)Curlr,y(e1 /\ gradyCx ~ YI))dY(Y).

Then using the identity (2.73), we deduce (2.79). o Let us now prove the identity (2.77). We look to see whether the derivative grad u(x). nxo has a limit when the point x tends to the point Xo along the normal nxo' We have

(2.86)

using (2.79) we have:

(2.87) gradu(x).nxo = - 41nfr CUrl;cp(y).(gradY(lx ~ YI) /\ nxo)dY(Y) ,

or again:

(2.88)

1 grad u(x).nxo = - 41nL CUrl;cp(y).(gradyCx ~ YI) /\ ny)dY(Y)

- 41nL curl; cp(y), (gradyCx ~ YI) /\ (nxo - ny))dY(Y) .


On using afresh the identity (2.73) we deduce from (2.88) that

(2.89)

In expression (2.89) the two integrands have singularities uniformly bounded by C/ly - xol when p --+ 0 (see (2.86)), and are therefore integrable. The Lebesgue dominated convergence theorem then proves that in the limit we have the identity (2.77). Let us prove the identity (2.75). By definition, we have

(2.90) r ou b(cp, t/J) = J, on (x). t/J(x) dy(x) .

Let Be(x) be the ball with centre x and radius E. The expression in the second member of (2.77) being integrable in the Lebesgue sense, we have:

Next we use the Stokes' formula on the portion of the surface r\(Be n n with boundary (on the curve n oBe(x) n r. We denote by v the normal to this boundary which is in the tangent plane to the surface r. We have:

r 1 LI,cp(y)dy(y) J I\(B,(x) n f) Ix - yl

(2.92) = r curl~ cp(y). (grady ( 1 ) /\ ny) dy(y) J I\(B,(x) n f) Ix - yl

I 1-----c-> + Icurl,LI,cp(y).vds(y), XEr.

oB,(x) n ,Ix - y

The integral on the curve oB.(x) n r tends to zero when E --+ 0 by reason of the symmetry of the ball Be and the integrand. We then deduce, for all x on the surface r, on using (2.92) and (2.91):

ou . 1 I (-----c-> ( 1 )) (2.93) a(x) = hm - 4 curl, cp(y) /\ grady I I' nxdy(y) , n e _ 0 n I\(B,(x) n n x - Y


whence from (2.90):

(2.94) 0 ! b(<p, t/J) = !~ - 41n t (t\(Bo(X) n r) curl~ <p(y)

x(gradyCx ~ YI) /\ nx)t/J(X)dY(Y))dY(X).

Now exchanging the integrations in x and y, we get:

(2.95) ! b(<p, t/J) = lim - ~ r ( r (nx /\ £--+0 4nJr In(B,(X)nr)

xcurl~<p(y)t/J(x)dy(x) )dY(Y) .

In the second member of (2.95) we have

(2.96) curl~ (IX ~ YI) = nx /\ gradx (IX ~ YI) ,

and if we denote by

(2.97)

(where Y'x is the projection operator onto the tangent plane to r at the point x), then by differentiation, we have:

~ ---:-+> ------:-+ ------:-+ (2.98) curlr X (x) = curlr t/J(x). curlr <p(y) + t/J(x)curldY'xcurlr <p(y)) .

We verify that:

(2.99)

and hence

(2.100) ---0-> ---0->

curlr X (x) = curlr <p(y). curlr t/J(x) .

Using Stokes' formula (2.73) on the surface r\{B£(y) n n, we obtain:

(2.101 )

! . 1 iii ---0-> ---0-> b(<p, t/J) = lim - 4 curlr<p(y)·curlrt/J(x)dy(x)dy(y) , £--+0 n r n(B,(y) n r) Ix - yl

+ r (f 1 t/J (x)Y'x curl; <p(y). V(X)dS(X))dY(Y) . J r cB,(y) n r Ix - yl

When t: ~ 0, the last term in the second member of (2.1 0 1) tends to zero by reason of symmetry. The first term then has, from the dominated convergence theorem, the limit sought. 0

Remark 13. It is of interest for certain physical applications to calculate the tangential derivatives on the surface r of the potential u (this is the case, for example, of the lifting profile which we have studied in §4 of part A). It can be

§3. Integral Equations Associated with the Helmholtz Equation 141

shown that

(2.102) ! curl; u,(xl ~ 1---..,+ - 2: Curlr <P(x)

- 41nt (CUrl~<P(Y) /\ gradyCx ~ YI)) /\ nxdy(y) , XEr,

where Ue is the exterior value of the potential u on the surface r. o

§3. Integral Equations Associated with the Helmholtz Equation

The search for monochromatic soiutions(22) in a bounded domain Q in 1R3 or its complement Qf, to the evolution problems relating to wave (source) equations or the equations of electromagnetism(23) often lead to a stationary equation:

(3.1) L1u(x) + k2u(x) = 0, X E Q or x E Qf, k real =f. 0 .

The desired solutions u often take complex values. This equation is called the Helmholtz equation (see Chap. II, §8). We associate with it boundary conditions on the surface r which are of the Dirichlet or Neumann type (and which express the physical nature of the body Q, see for example Chap. I, §4). The interior problems of Dirichlet and of Neumann

(3.2)

(3.3)

{L1U(X) + k2u(x) = 0 ,

ulr = Uo ,

1 L1u(x) + k2u(x) = 0 ,

oul = g , an r

XEQ,

XEQ,

have been studied in Chap. VII where it has been shown that they admit a unique solution in Hl(Q) for Uo E Hl/2(r) (resp. g E H- 1/2(r)) except for certain values of k2 called eigenvalues. The eigenvalues form a countable sequence of real numbers which diverge to infinity and the eigenvalues of the Dirichlet problem (3.2) are in general distinct from those of the Neumann problem (3.3). To each eigenvalue is associated a vector space of eigenfunctions which are solutions of the associated homogeneous problem. This space has finite dimension. Whenever k2 is an eigenvalue of the Dirichlet (resp. Neumann) problem, the

(22) See Chap. lB. (23) See Chap. lA, §4.


problem (3.2) (resp, (3.3)) admits a solution in Hl (Q) (defined to within an eigenfunction) if and only if Uo satisfies the conditions(24)

(3.4) t Uo ~~ dy = 0, for every eigenfunction v of the Dirichlet problem

(respectiVelY, if g satisfies

(3.5) t gv dy = 0, for every eigenfunction v of the Neumann problem).

Let us now consider the exterior problems

(3.6) { LlU(X) + k2 u(x) = 0 , ulr = Uo ,

X E Q' ,

and

(3.7) I Llu(x) + k2 u(x) = 0, X E Q' ,

::Ir = g .

First we examine the homogeneous problem in allIR 3

(3.8)

It is easy to show that the particular solutions with complex values are:

(3.9) I/J(x) = elk. x --+

where the constant vector k has modulus k. It can be shown, formally, by using the Fourier transformation, that "all" solutions of equation (3.8) are "combinations" of the particular solutions given by (3.9). In order to get the uniqueness of the solutions of equations (3.6) and (3.7) or (3.8), it is necessary to adjoin a condition on the behaviour at infinity which eliminates the plane waves and their "combinations". We shall seek solutions which tend to zero at infinity as l/r where r is the Euclidean distance from the origin. But this does not

I·· hi' sin kr f b e lmmate t e so utlOn -- 0 pro lem (3.8). r

It has been shown in Chap. II that it is further necessary to include the so-called radiation or Sommerfeld condition. For example, this Sommerfeld condition can be (strongly) outgoing(25):

(3.10) r2 (:: + ikU) is bounded when r = Ixl tends to infinity

(24) To verify, apply Green's formula:

r (uLlv - vLlu)dx = f (u ov _ ou V)dY JQ r on on (25) See Chap. II, §8, 7.d. The terminology ik-outgoing wave may also be used. See also the chapter on wave equations in the forthcoming volume on "Waves" for the physical interpretation of this condition.


G~ denotes the derivative normal to the ball of centre the origin and radius r)

We can also use the (strongly) incoming Sommerfeld condition(26):

(3.11) r2G~ - ikU) is bounded when r = Ixl --+ 00 .

The elementary solutions of the operator (LI + k2 I) are (see Chap. 11):

(3.12) e ikr e -ikr

E).(x) = A. 4nr + (1 - A.) 4nr ' VA. E C, r = Ixl .

Only the elementary solution

(3.13)

satisfies the outgoing Sommerfeld condition (3.10) and behaves at infinity like l/r. We can now state

Proposition 1. The exterior Dirichlet problem (3.6) and the exterior Neumann problem (3.7) each admit a unique solution in the space denoted by Xs of complex valued functions u such that:

(3.14) ii) ru is bounded when Ixl --+ 00 and ! i) u E HI~c(Q') (27)

r2 (:~ + ikU) is bounded when Ixl --+ 00

if the given Uo and g are such that Uo E HI/2(r) and g E H- 1/2(r). Proposition 1, stated for the outgoing Sommerfeld condition (3.10) is equally true with the incoming Sommerfeld condition (3.11) in the space:

X E = {u E HI~c(Q'); ru and r2 (:~ - ikU) bounded as Ixl --+ oo} .

We have

Theorem 1. Let u be a function from 1R3 into ctOO(Q) n ct oo (Q')(28) satisfying the equations

(3.15) Llu(x) + k 2 u(x) = 0, X E Q and x E Q' ,

(26) See Chap. II, §8, 7.d, and formula (8.148). Also called an ik-incoming wave. (27) That is to say for all bounded measurable sets B c !I', we have:

L [lul 2 + Igrad u1 2 ] dx < + OCJ •

(28) As for Theorem 2 in §2, it is necessary to extend this theorem to: u E 'If'''(Q) n 'If'''(Q'), has tempered growth, and u and au/an have boundary values in HI/2(r) and H-I/2(r), different on each side of T, the relations (3.16) and (3.17) then holding for a.e. x.


and satisfying the conditions (3.14); denote by Ui and ue the interior and exterior limits of the function u on rand

[ au] aUi aUe [u] = Ui - ue ' an = an - a; ;

then we have the representation:

(3.16)

1 f [au ] e -ikix - yi 1 f a (e -ikix - Yi) u(x) = 4n r an (y) Ix _ yl dy(y) - 4n r [u(y)] any Ix _ yl dy(y)

for all x E Q u Q', and

! Ui(X) + uAx) _ ~ f [au ] e - ikix - yi

2 - 4 a (y) I I dy(y) (3.17) n r n x - y

- 41n L [u(y)] a~y G: i: ~~) dy(y) , X E r .

o The proof is similar to that for Theorem 2, §2, established for the operator A. We use Proposition 1 to complete the proof instead of Theorem 1 in §2. This proposition plays in the case considered here the role of Liouville's theorem. We will again call, through a natural generalisation of the definition given for k = 0 (see the end of Theorem 3, §2 and Chap. II), a simple layer potential (outgoing or ik-outgoing) a representation of the form

1 f e - ikix - yi u(x) = -4 q(y) I I dy(y) , X E [R3 .

n r X - Y (3.18)

As in the case where k = 0 (see 2.28» we will put:

1 f e - ikix - yi

5l'kq(X) = -4 q(y) I I dy(y) , X E [R3 . n r X - Y

Whenever we seek a solution of the exterior Dirichlet problem (3.6), we are reduced to solving the integral equation obtained from (3.18) when x is on the surface r. We then have the analogue to Theorem 3, §2:

Theorem 2. The integral equation

1 f e - ikix - yi -4 q(y) I I dy(y) = uo(x) ,

n r X - Y (3.19) XEr,

with unknown q and given Uo E HS(r) (s real and arbitrary) admits one and only one solution in HS+ l(r)-and because 5l'k- 1 is an isomorphism of HS(r) onto H S + 1 (r) - whenever P is not an eigenvalue of the interior Dirichlet problem for the operator - A. The solu.ions of the interior Dirichlet problems (3.2) in Hl(Q)for Uo E Hl/2(r), and the exterior Dirichlet problems (3.6) with u E X e have then one and only one solution (see Proposition 1) and this solution is given by (3.18) with q given by (3.19).


Whenever k2 is an eigenvalue of the interior Dirichlet problem (3.2), equation (3.19) admits a solution if and only if the function Uo satisfies conditions (3.4). In this case, the normal derivatives ov/on of the eige,yunctions of the interior Dirichlet problem (3.2) are solutions of equation (3.19) with the right hand side null.

Proof Let us consider the integral kernel associated with equation (3.19). It can be written

(3.20) 1 e - iklx - yl 1 e - iklx - yl - 1

- = +-----4n Ix - yl 4nlx - yl 4nlx - yl

the kernel

(3.21) e - iklx - yl - 1

E2 (x - y) = 4 I I n x - y

is continuous and has derivatives integrable on the surface r. This implies that the corresponding operator Pk :

(3.22) def f Pkq(x) = r E 2 (x - y)q(y)dl'(Y) , x E r ,

is continuous from HS(r) into Hs+ 2 (r) (we will accept this result). It follows from

h . 1 T eorem 3, §2 (see Remark 1, §2) that the operator !f' whose kernel IS 4 I I IS

n x - y an isomorphism of HS(r) onto H s+1 (r). The operator Pk is thus a compact perturbation of the operator !f'. The Fredholm alternative applies to the operator !f' + Pk == !f' k'

From this we deduce Theorem 2 on making use of Proposition 1 and Theorem 1 which shows that the kernal of the operator !f' + It = !f'k is reduced to zero except when k2 is an eigenvalue of the interior Dirichlet problem. In the latter case the normal derivatives of the eigenfunctions of the interior Dirichlet problem are solutions of the equation

(3.23) ov

(!f' + Pd on = 0 . o

Whenever we seek a solution to the exterior Neumann problem (3.7) by using a simple layer potential (3.18), we are led to the following integral equation which connects the unknown q with the given g:

- q(x) 1 f 0 (e -iklx - YI) (3.24) 2 + 4n r q(y) on

x Ix _ yl dl'(Y) = g(x) , x E r .

We will put, by analogy with the case where k = 0,

2 f 0 (e -iklx - YI) Jkq(x) = 4n r q(y) on

x Ix _ yl dy(y);

then (3.24) can be written:


Making use of the representation Theorem 1 and the Fredholm alternative, we can now prove the following analogue of Theorem 5, §2:

Theorem 3. The integral equation (3.24) admits one and only one solution in H- 1/ 2 (r)for all g E H- 1/2(r), and - I + Jk is an isomorphism of HS(r) onto itself for all real s if P is not an eigenvalue of the interior Dirichlet problem (3.2). The solution of the exterior Neumann problem (3.7) with u E X s is given by (3.18) with q a solution of (3.24). Whenever k2 is an eigenvalue of the interior Dirichlet problem (3.2), the integral equation (3.24) admits a solution if and only if the given g satisfies:

1 r gvdy = 0 for all sol,utions v of the adjoint equation (3.25)

v(x) 1 r a (e -.klx - YI) 2 - 4n J r v(y) any Ix _ yl dy(y) = 0, X E r ,

and then the solutions of the integral equation (3.24) with its right hand side null are the normal derivatives of the eigenfunctions of the interior Dirichlet problem (3.2).

o Remark 1. We can equally make use of representations as "double layer" potentials (outgoing or ik-outgoing) that is to say of the form:

(3.26) 1 i a (e -iklx - YI)

u(x) = - -4 <p(y);:;- dy(y) , X E 1R3 \ r . n r uny Ix - yl

The integral equations obtained in this case are analogues of those obtained for the Laplace operator in §2.3. 0

Remark 2. The variational formulation of Theorem 3, §2, is likewise suitable for the numerical solution of the integral equation (3.18). It does not lead to a positive operator as in the case of the operator (- L1), but we can nevertheless show that it leads to stable numerical approximations whenever k2 is not an eigenvalue of the interior Dirichlet problem. 0

The Transmission Problem for the Helmholtz Equation (see (3.1» with the Sommerfeld Condition - Simple Layer Potential; Double Layer Potential. The Dirichlet Problem; The Neumann Problem

The following proposition results essentially from the (extension of) Theorem 1:

Proposition 2. The transmission problem: determine u E Hl(Q) x Xs (resp. H 1 (Q) x X E )(2 9) satisfying:

Ii) L1u + k2 u = 0 ,

ii) [u] = udr - uel r = f E Hl/2(r) given ,

iii) [:~ ] = ~:t -~:t = g E H- 1/ 2 (r) given,

(3.27)

(29) See the notations in Proposition I (or again: u is a function from ~3 into H' (Q) n X s. resp. H'(Q) n Xd.


admits one and only one solution, which is given (see (3.16)) by:

1 f -iklx-YI 1 f a ( -ikIX-YI) (3.28) u(x) = -4 g(y) ~ I dy(y) - -4 !(y)-;:- ~ I dy(y)

n r X - Y n r uny x - y

(resp.

1 f e iklx - yl 1 f a (e'ikIX - YI) ) (3.29) u(x) = 4n r g(y) Ix _ yl· dy(y) - 4n /(y) any Ix _ yl dy(y)

Proof Uniqueness. We reduce at once to the case! = 9 = 0; and we then seek to solve the problem: find u E Hl~c(1R3) satisfying the outgoing (resp. incoming) Sommerfeld condition (3.14), as a solution of the Helmholtz equation in 1R3:

L1u + k2 u = 0 .

Now we know (see Chap. II) that the only solution of this problem is u = 0 (this is also a direct consequence of the Rellich theorem-see Lax-Phillips [1], p. 148, and the chapter on the wave equation in the forthcoming volume on "Waves"). Existence. This results directly from (3.28) or (3.29) (see Lax-Phillips [1], p. 127, and the chapter on the wave equation in the forthcoming volume on "Waves").

o As in the case of the Laplace equations, to wish to solve the interior (3.2) or exterior (3.6) Dirichlet problem with (3.14), or the interior (3.3) or exterior (3.7) Neumann problem with (3.14) by an ("outgoing") simple layer potential (3.18), comes down to reducing the problem under consideration to the previous transmission problem in the space Hl(Q) x X s, on imposing [u] = 0(30). Theorems 2 and 3 show that there is no difficulty in passing from the interior or exterior Dirichlet problems and from the exterior(31) Neumann problem to the corresponding transmission problem with [u] = 0 whenever k2 is not an eigenvalue of the interior Dirichlet problem. We finally note that we can again define in the case of the Helmholtz equations an interior capacity operator (dependent on k) in a fashion analogous to that in Remark 7, and exterior capacity operators (outgoing or incoming) in a fashion analogous to that in Remark 11, with similar properties to those in these remarks (and in Chap. II). On the physical level, we have seen in Chap. II, §8 that the energy flux across the surface r (orientated towards the exterior) of a solution u of the wave equation is given by

(30) The asserted analogue for the solution of problems with an ("outgoing") double layer potential (3.26). There again the choice of a solution in the form of a simple or double layer potential is arbitrary. (31) It will be the same for the interior Neumann problem.


with u' = ~~ ; in the case where u is a monochromatic solution ofthe wave equation,

of the form: u(x, t) = eik1uo(x) with k E Ill, and Uo a solution of the Helmholtz equation (3.1), this energy flux becomes:

Ir = Re(ik L~:Udr) . Besides the explanation of the notions of the outgoing or incoming wave derived from the relations (3.10) and (3.11), this would again lead to an "energy" interpretation of the capacity operator (as in the case of Chap. II); see remarks 7 and 11 in §2.

§4. Integral Equations Associated with Problems of Linear Elasticity

We shall examine here a system of integral equations associated with the solution of the Dirichlet problem for the system of equations of linear elasticity homogeneous and isotropic in 1Il3 (32).

Let us recall these elasticity equations in 1Il3 in the case where the principal unknowns are the displacements u = (u1 , u2 , U3) of the elastic body (whose domain is represented by the bounded open set 0 or by the open set 0,(33)

according to case). We introduce the strain tensor Il:

(4.1) 1 (au. au.) llij(U) = 2 ax~ + ax~ , i,j = 1,2,3 .

The stress tensor is defined by 3

(4.2) Gij(U) = .Mij L 1l,,(U) + 2Jlllij(u) (32).

1= 1

The interior Dirichlet problem considered here is then, in the absence of volume body forces (and in the case of a quasi-static problem) to find u such that:

(4.3) 1 t aa Gij(U)(X) = 0, X EO, j= 1 Xj

ulr = Uo , X E r . The exterior Dirichlet problem considered here is to find U such that:

(4.4) 1 t aa GiAu)(x) = 0, X E 0' , j= 1 Xj

ulr = Uo , X E r . o

(32) See Chap. lA, §2. (33) Where, as we recall, Q' is the complement of a bounded set n in iii 3.

§4. Integral Equations Associated with Problems of Linear Elasticity 149

We shall now solve these problems by giving them a variational formulation in the Hilbert spaces (H'(Q))3 and (WI (Q'))3. The trace theorem (Theorem 2 in § 1 of this part B) allows us to find for every Uo E (HI/2(r)3 a function (a lifting) Ruo E (W I ([R3))3 whose trace on the surface r is uo. On putting

(4.5) v = u - Ruo ,

the interior (4.3) and exterior (4.4) Dirichlet problem respectively admit the following variational formulation: seek v such that:

1 a(v, w) = - a(Ruo, w), Vw E (HJ(Q»3, v E (HJ(Q»3 ,

(4.6) where we have defined the bilinear form a(v, w) by:

a(v, w) = A. L Ctl ekk(v») Ctl el/(W) )dX + 2/1 i'~ I L eij(V)eij(w)dx

1 a(v, w) = - a(Ruo, w), Vw E (WJ(Q,))3, V E (WJ(Q,»3 ,

(4.7) where we have defined the bilinear form a(v, w) by:

a(v, w) = A. L, C tl ekk(v») (,tl el/(W») dx + 2/1 i, ~ 1 L eij(V)eij(w)dx

Proposition 1. Let Uo be given satisfying Uo E (Hl/2(Q))3, then the interior (4.3) and exterior (4.4) Dirichlet problem admit a unique solution respectively in (HI (Q))3 and (WI (Q'))3.

Proof We apply the Lax-Milgram theorem(34). It suffices to show the coercivity of the bilinear forms defined in (4,6) and (4.7) respectively in (HJ(QW and (Wb(Q'))3. In the case of the interior problem, the corresponding inequality is called the Korn inequality and we refer for its proof to Chap. VII, §2. In the case of the exterior problem, this inequality is proved using Theorem 1, §l. We refer to Giroire [1J for this proof. 0

In order to associate this problems above to the integral equations which permit their solution, we first of all establish a result about the integral representation of the solutions of the elasticity equations. The latter are obtained by making use on the one hand of the elementary solution of the elasticity operator, and on the other Green's formula for this operator. We will denote by A the operator given by:

3 0 (4.8) (Au)i = - L - (Jij(u) , 1 ~ i ~ 3 ,

j= I OX j

(34) See Chap. VII.


and associated with the bilinear form a defined by (4.6) on (HJ(Q))3 or (4.7) on (WJ(Q'))3. Using Green's formula, we can establish the equalities:

(

a(v,w) = fuJ1 (Av)iwi dx + Lit1 (t1 (J'ij(v)ni)widY,

(4.9) for all v, WE (H1(Q))3, a(v, w) defined by (4.6), n being the

normal to Q with components ni , j = 1 to 3 ;

(4.10)

1 a(v, w) = r,.f (Av)iwi dx - f .f (.f (J'ij(v)ni)widY , Ja 1=1 T,=l J=l

a(v, w) being defined by (4.7)

An elementary solution of the operator A is by definition a 3 x 3 matrix denoted by E such that

(4.11) AE = M, where b is the Dirac measure and I the identity matrix.

It can be verified(35) that an elementary solution of the operator A is given by:

(4.12) 1 a a

Eij(X) = 81tJl(A. + 2Jl)lxl ((A. + 3Jl)bij + (A. + Jl) ax/ lxl). ax/ lxl )) ,

~ i,j ~ 3

(bij is the Kronecker symbol). This matrix, an elementary solution of the operator A, is symmetric. We then have

Theorem 1. The following system of integral equations, with unknowns q = (q1' q2' q3) and given Uo = (U01' U02, U03)

(4.13) it1 L Eij(x - y)qiy)dy(y) = UOi(X) , x E r, 1 ~ i ~ 3 ,

where the elementary matrix E is given by formulas (4.12), admits a unique solution in (H-1/2(r))3 if Uo E (H1/2(r))3. To it there is associated the variational formulation:

(4.14) b(q, q') = it1 L uOi(y)qi(y)dy(y) , Vq' E (H-1/2(r))3 ,

where the bilinear form b(q, q') is defined by:

(4.15) b(q, q') = iti it1 L L EiAx - y)qi(X)qj(y) dy(x) dy(y) ,

(35) See the proof in Remark 3, §5 below.

§4. Integral Equations Associated with Problems of Linear Elasticity 151

and satisfies the inequality:

(4.16) b(q, q) ~ P IlqIU~rlt2(n)" P > 0 .

The solution u of the interior (4.3) Dirichlet problem and the exterior (4.4) Dirichlet problem is then given by:

(4.17) ui(x) = jti L Eij(x - y)qiy)dy(y) , x E 1R3 ,

with (qj) a solution of(4.13).

Proof We shall only indicate the main ideas of the proof. First we establish a result about the representation of the solution u of the interior (4.3) and the exterior (4.4) Dirichlet problem. To do this we shall use Green's formulas (4.9) and (4.10) with on the one hand the solution u and on the other the different columns of the matrix E, the elementary solution defined by (4.12). From this we deduce formula (4.17) and the integral equation (4.13), and on then putting

(4.18) 3

qi(X) = L [lTijnix)] , x E r , j= 1

([<p] denotes the jump in the function <p across r, [<p] = <Pi - !Pe).lt follows from Proposition 1 and Green's formulas (4.9) and (4.10) that the mapping Uo 1-+ q is continuous from (Hi/2(r))3 into (H-i/2(r))3. In order to define the inverse mapping we obtain by adding formulas (4.9) and (4.10):

3 f 3 a(u, v) = i~i r j~i [lTij(u)nj(y)]vi(y)dy(y)

Ji L qividy(y) , "Iv E (Wi (1R 3 ))3

(4.19) where the bilinear form a is defined by:

a(u, v) = A t,Cti eu(u))(Ji eU(V))dX

+ 2J.l i.~ 1 t, eij(u)eij(v)dx .

It follows from the Korn inequality in 1R3 (see Chap. VII, §2) that the following problem, where q is given in (H-i/2(r))3:

(4.20) a(u, v) = Ji L qividy(y) , "Iv E (Wi (1R 3 ))3 ,

admits a unique solution in (Wi (1R3))3. This shows us that the mapping q 1-+ Uo is continuous from (H-i/2(r))3 onto (Hi/2(r))3. The inequ!:llity (4.16) then results from the equality (4.20) and the continuity of the mappings in question. 0


Remark 1. We have restricted ourselves here to the simplest case, that of the Dirichlet problem and its representation through a "simple layer potential". If we consider other types of boundary conditions for the elasticity system (in particular

3

with L (Jijnj imposed) and other types of integral representations, we would be led j=l

to other integral equations. Contrary to what happens in the case of the Laplace operator (see §2.2), all these integral equations behave as Cauchy principal value integrals. Nevertheless by using the variational formulations we can avoid the difficulties connected with these principal values and particularly the difficulties of numerical approximation. 0

§5. Integral Equations Associated with the Stokes System

We have seen in Chap. lA, §1 the linearised and stationary equations in 1R3 of incompressible viscous fluids with low velocity. The velocity U = (u 1 , u2 , u3 ) of the fluid at a point x E Q c: 1R3 (or Q') and the pressure p of the fluid are the unknowns. If we denote by va positive quantity(36) connected to the viscosity and the density of the fluid, these equations can be written:

1- vLlu. + op = 0

! oXi

t oU i = div U = 0, X E Q or x E Q' c: 1R3 . i= 1 OXi

XEQ or Q' E 1R3 1, 2, 3 (5.1)

To these must be added boundary conditions on the boundary denoted oQ or r of the fluid. The Dirichlet problem will be that where the velocity Uo of the fluid is known on the surface r. 0

We shall now treat here the integral equations through the "simple layer potential" associated with equation (5.1). We shall call the interior Dirichlet problem the problem: find u and p satisfying:

1-VLlUi(X) + ~P (x) = 0, X E Q, 1 ~ i ~ 3 , 3 uX i

" OU i d· L... -;-(x) = IV U = 0, X E Q , i= 1 uX i

Ui(X) = UOi(X) , x E r, 1 ~ i ~ 3 .

(5.2)

(36) In fact v = /1/ Po is the kinematic viscosity of the fluid and throughout this section we put Po = 1.

§5. Integral Equations Associated with the Stokes System

We shall call the exterior Dirichlet problem the problem: find U and p satisfying:

! -vLlui(x) + ~p.(X)=O, XEQ', 1 ~ i ~ 3 (37) 3 uXI

( 3) ~ ~Ui . 5. i:--l ox/x) = dlVU = 0, X E Q' ,

ui(x) = UOi(X) , x E r, 1 ~ i ~ 3

153

In order to give the variational formulations of problems (5.2) and (5.3), we introduce the following tensors G and 0":

(5.4) I(OU. Ou.) Gij(u) = - _I + _1 , 1 ~ i,j ~ 3 , 2 oXj oXi

(5.5)

(jij is the Kronecker symbol. With the condition div U = 0, Green's formulas associated with Stokes' equation are then:

(5.6)

and

(5.7)

where n = (n l' n2 , n3 ) is the normal exterior to the surface r. We have need of the following proposition which can be found in Ladyzenskaya [1] or Cattabriga [1] and which ensures for us the existence of a lifting.

Proposition 1. Let Uo = (UOl' U02, U03) E (H I /2(r))3. If Q, a bounded open set (Q c [R3) is simply connected, and if Uo satisfies the identity:

(5.8) L uo·ndy = 0 ,

then there exists a lifting Ruo E (HI (Q))3 such that

{RUO Ir = Uo , X E r ,

(5.9) div (Ruo) = 0, X E Q .

(37) With, let us recall, Q a bounded open set in [R3, and Q' the complement in [R3 of Q.


If Q', the complement of Q, (Q' c [R3) is simply connected, then there exists Ruo E (WI (Q'))3 such that

(5.10) {Ruo1r = uo , X E r , div(Ruo) = 0, X E Q' = [R3\,Q .

o Using this proposItIon, we transform the interior (5.2) and the exterior (5.3) Dirichlet problems to give them the following variational formulations: having taken as the new unknown v, defined by:

(5.11). v = u - Ruo , Uo E (HI/2(r))3 ,

find v satisfying (with the Einstein summation conventions):

(5.12) J Q oXj oXj J Q oXj oXj ! v r oVi OWidx = _ v r O(RUO)i OWidx ;

Vw E (WA(Q))3 satisfying divw = 0, V E (HJ(Q))3, divv = 0 .

and: find v satisfying (with the same conventions):

! i ov. ow. i 0 ow. v -' -'dx = - v -(RUO)i-'dx,

(5.13) Q,oxj oXj Q,oxj oXj

Vw E (Wb(Q'))3 satisfying divw = 0, V E (WJ(Q'))3, divv = 0 .

We have

Proposition 2. Whenever the open sets Q and Q' are simply connected, the interior (5.2) and the exterior (5.3) Dirichlet problems each admit a unique solution u in (HI(Q))3 and (WI (Q'))3 respectively, and such that:

(5.14) divu = 0

if the given Uo is in (H I /2(n)3 and satisfies condition (5.8), i.e.:

L uo·ndy = 0 (38).

Proof We apply the Lax-Milgram theorem to the problems (5.12) and (5.13) . respectively. The coercivity of the bilinear form of the first member of (5.12) on (H 6(Q))3 results from the Poincare inequality (see Chap. IV, §7). The coercivity of the bilinear form in the second member of (5.13) on the space (WJ(Q'))3 results from Theorem 1, §l. With regard to the pressure p, we have:

(38) Recall that this hypothesis is a priori not necessary for the existence of the solution u of the exterior Dirichlet problem (5.3).

§5. Integral Equations Associated with the Stokes System 155

Proposition 3. The pressure p associated with the solution of the interior Dirichlet problem (S.2) exists and is unique to within a constant in the space L 2(Q); the pressure p associated with the exterior Dirichlet problem (S.3) exists and is unique in the space L 2 (Q').

Proof We have in the two cases in question

(S.lS) gradp = vLlu, P E ~'(Q) n ~'(Q') .

and thus in the case of the interior problem

grad p E (H -1 (Q))3

Now it follows from Proposition 1 (see Cattabriga [1]) that

(S.16) IIgradpIIW'(ll»' ~ IX IlpIIL'(ll)/Po .

The result follows from the inequality (S.16). In the case of the exterior problem, we have

gradp E «WA(Q'))3), .

We now show that for p E L 2 (Q'), we have

(S.17) Ilgradpll«W~(ll'»')' ~ IX IlpIIL'(ll')' IX > 0 .

The result sought for will then flow from inequality (S.17). We now solve the problem:

(S.18) J ;:I~:~. xeD',

1 an, From for example Giroire [1], it admits a solution (not unique) whenever pEL 2 (Q'), which is such that:

Thus

(S.19) Ilgradcpll(w'(ll'»' ~ CllpIIL'(ll') ,

(S.20) v = grad cpl, E (H1/2(r))3 and satisfies condition (S.8).

We then use Proposition 1 in order to choose

(S.21) w = grad cp - Rv ,

so that we have obtained

(S.22) {diVW = p wi, = 0 ,

(S.23) Ilwll(w~(ll'»' ~ CllpIlL'(ll') .


On using (S.22) and (S.23), we have:

(S.24) t, pdiv wdx 1

IIgradpll(WMU'))')' ~ II II ~ -C IIpIIL'(u') . W (W~(lJ'))'

o In order to obtain the integral representations of solutions of the Dirichlet problems (S.2) and (S.3), we shall use Green's formulas (S.6) and (S.7) and the elementary matrix solution. We recall that an elementary matrix solution E of an operator A is defined by:

(S.2S) AE = M; where c5 is the Dirac measure and I the identity matrix .

It can be verified (see the proof in Remark 3 below) that an elementary solution of the Stokes' operator(39) is given by the following 4 x 4 matrix:

(S.26) E .. (x) = U.(x) = _ -...!:!. + _'_J I (c5.. x.x.) 'J 'J 8nv Ixi Ixl 3

= - --.!!. - --(Ixi) I (2c5.. 02 )

8nv Ixl OXiOX j , ~ i,j ~ 3

I 0 ( I ) I Xi (S.27) Pi(x) = E4i(X) = Ei4(X) = - 4n oXi ~ = 4n N3 '

(S.28) E44(X) = vc5 .

We have

Theorem 1. Let Vij and Pi be given by formulas (S.26) and (S.27). The system of integral equations with unknowns t = (t l' t 2, t 3)

(S.29) k t! f/k(y) Vdx - y)dy(y) = UOi(X) , X E r, 1 ~ i ~ 3 ,

admits a unique solution, to within a vector proportional to the normal n to the surface r, in the space (H -!/2(F»3 if Uo E (H!/2(F»3 and satisfies condition (S.8):

r uo·ndy = 0 .

The system of equations (S.29) admits the variational formulation: find t such that:

b(t, t') = it! r uOitidy, 'tit' E (H-!/2(F»3 ,

(39) The operator A is then given by:

a Aij = - VA<>ij' i,j = 1 to 3, Ai4 = A4i = -a ' i = 1 to 3, A44 = 0 .

Xi


where the bilinear form b(t, t') is defined by:

b(t, t') = i'~ ILL ti(X)tj(y)UiAx - y)dy(x)dy(y) , Vt, t' E T ,

T = (H-I/2(r))3/~ , (5.30)

where [jf is the equivalence relation

t ~ t' iff t - t' = An, A E IR (40).

The bilinear form b(t, t') satisfies:

(5.31) b(t,t) > Plltll?, P > 0, WE T.

Then the solutions (u, p) of the interior (5.2) and the exterior (5.3) Dirichlet problems are represented in the form of the F.K. Odqvist hydrodynamic potentials:

(5.32) ui(x) = ± I Uki(X - y)tk(y)dy(y), x E 1R3 , i = 1 to 3 k= I Jr

(5.33) p(x) = ktl Ir Pk(x - y)tk(y)dy(y), x E 1R 3 \r .

Proof Using Green's formulas (5.6) and (5.7) with the function u and the different columns of the elementary matrix, we obtain the representations (5.32) and (5.33) on putting

(5.34)

ti(x) = [jtl O'ij(u, p)nj(x) J, x E r, 1:::; i :::; 3, O'ij defined by (5.5) .

Green's formulas and definition (5.5) likewise show that t = (tl' t 2 , t 3 ) is in the space (H-I/2(r))3 and is defined to within An where n is the normal to r and A E IR (because the interior pressure is defined to within a constant). This shows that the mapping Uo E (HI/2(r))3 1--+ t E (H- I/2(r))3 is continuous. To show the continuity of the inverse mapping we solve the problem:

l t3i'~I;::;::dX = LJI tividy, VVE V (5.35)

V = {v E (WI (1R 3))3; divv = O} .

The coercivity of the bilinear form on the left hand side of (5.35) results from Theorem 1, §2. This allows the continuity of the mapping t ..... Uo to be shown. Finally the inequality (5.31) follows from the coercivity of the bilinear form on the left hand side of (5.35) and from the continuity established earlier. 0

Remark 1. As in the case of the elastic system (§2.3), by considering other types of boundary conditions we would obtain other integral representations and other

(40) In fact this relation can be written:

t - t' iff t(x) - t' (x) = In(x) , x E r, l E IR .


integral equations. These equations involve non integrable kernels and hence lead to the introduction of principal values. 0

Remark 2. The property (5.31) allows the introduction of the stable numerical approximations of Dirichlet problems into the Stokes equation. 0

Remark 3. It can be verified, by Fourier transformation, that the operators A of elasticity (§4) and Stokes (§5) admit as elementary solutions the expressions E given by (4.12) and (5.26) to (5.28).

a) In the elasticity case, (4.11) is equivalent to:

(5.36)

with

- (A. + Ji)-{!-(div Ek) - Ji,dEJ = ~jk~(X) uXj

Ek =(EJ)j=lto3, E~=Ejk'

By Fourier transformation, this can be written

(5.37)

Multiplying by Yj and summing over j, we get:

(A. + 2Ji)y2(y.Ek) = Yk ,

hence

E~k 1 Yk Y - -. - A. + 2Ji y2

Substituting this into (5.37), we get:

(5.38) £.k = E~ = bjk _ (A. + Ji) YjYk J J Jiy2 Ji(A. + 2Ji) y4 .

Because of the formulas:

(5.39) a x· o2(lxl) ~jk XjXk

oXj (lxl) = I~' oxkoxj = ~ - IxI 3 '

,d(lxl) = I~I' ,d ( - 4n11xl) = b ,

we have:

(5.40)

whence, by inverse Fourier transformation:

bjk (A. + Ji) o2(lxl) EOk = -- - ----

J 4nJilxI 8nJi(A. + 2Ji) oxjoxk . (5.41)

We deduce without difficulty, using (5.39), the expression (4.12).


b) In the case of fluid mechanics (Stokes' system), the search for an elementary solution (see (5.25)) comes down to solving:

1 Opk

- vAEJ + oXj = bjkb(X)

div Ek = 0 (5.42)

on putting

Ek = (EJ)j=lto3, EJ = Ejk' pk = E4k = Ek4 = Pk .

By Fourier transformation, (5.42) gives:

(5.43) { .~) vy2 ~J _+ iyjpk = bjk 11) y.E - 0 .

Multiplying (5.43)i) by Yj' summing over j, and using (5.43)ii), we get:

iy3pk = Yk

whence

(5.44) 1 pk = - ;Yk y2

E~ = bjk _ YjYk J vy2 vy4

By inverse Fourier transformations, with (5.40), we obtain:

lk 0(1) p = - oXk 4n Ixi

(5.45) E~_~ __ 1_~

J - 4nvixi 8nv oxjoxk '

whence the asserted formulas (5.26) to (5.28). o Remark 4. In the framework of the problems (5.2) and (5.3), the pressure p must appear (for the proposed modelling of the system) as a positive quantity. In the case of the interior problem where the function p is determined to within an additive constant, this is possible if this function p is bounded below (minorised), but this condition is not always satisfied. On this subject we refer to Ladyzenskaya [1], p. 47. 0

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