Mathematical Analysis and Numerical Methods for Science and Technology || The Laplace Operator

Chapter II. The Laplace Operator

Introduction

This chapter studies, by methods based on classical junction spaces (continuously differentiable functions), the theory of harmonic functions and of other classical concepts attached to the Laplacian operator ~1: Newtonian potential, capacities, the Dirichlet problem; the radiation conditions associated with the operator ,1 + k2 - are similarly introduced, The Laplacian operator admits a very rich structure, and is the source of numerous problems and numerous generalizations; certain of these are indicated here in sections (marked by *) which can be passed over, at least on a first reading. "Functional" aspects, based on the calculus of variations and on the introduction of the Sobolev spaces are not used here (or scarcely touched on implicitly); they will be taken again, in a context going far beyond the Laplacian operator. in the following chapters.

§1. The Laplace Operator

1. Poisson's Equation

We give the name the Laplace operator or the Laplacian of dimension 11, to the linear differential operator with constant coefficients in [R"

From an elementary point of view, being given a point x a function u defined in the neighbourhood of x.

is defined under the condition of the existence of the partial derivatives (~2U(X)! ('xf, ... , ?2 U(X)/ ax;;: that is to say on the hypothesis of the existence of the gradient o/,u

grad 11 = (.~I , ... , ~u. \) , ,?x 1 ('x,,/

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

~ 1. The Laplace Operator 221

in the neighbourhood of x, we have

Llu(x) = div(gradu)(x)

where for a field p = (PI' ... , Pn) defined in the neighbourhood of x, the divergence of p at x is defined by

. CPI dlv p(x) =~-(x) +

oX 1

'" ::J . . " I d' . OPI UPn. with the assumption that the partla envatlves -, -(x) , . " . , -;;-"(x), eXIst.

ox 1 ex" We notice immediately the fundamental property of the in variance of the Laplacian under a Euclidean transformation. A Euclidean transformation is an isometry of the Euclidean distance in [p;"

d(x, y) = Ix - yl = ((XI - YY + . " . + (xn - Yn)2)1!2 ,

that is to say, a mapping T of a part D of [P;" into [p;n satisfying"

I T(x)- T(y)1 = Ix - yl v x, Y ED"

ft can be shown that a Euclidean transformation defined on a connected open set in [P;" is composed of a translation and of an orthogonal transformation (see Dieudonne [2J). We then have the following property, which is, moreover, characteristic of the Laplacian (see Chap. V, §2).

Proposition 1. Let x E [p;n and let T be a Euclidean trallsj()rmation defined ill the neiyhbourhood ofx and u a twice differentiablefimction defined in the neighbourhood

def of x. Theil Tu = U·cO T - 1, defined in the neiqhbourllOod of y = Tx is twice differ-

entiable in y and

( l.l ) ll(Tu)(T(x)) = L1u(x) .

This follows immediately from the fact that the Euclidean transformation T of an open set Q is of class (6 < satisfying: 7"'(x) is an orthogonal transformation, T"(x) = 0 for all x E Q, and the formula for the differentiation of composition of functions applied to the Laplacian gives

Lemma 1. For Y E rr~n, let h = (11 I, .. " , h") be defined ill the neighbourhood of y alld be twice differentiable at y; also let u be defined ill the neiqhhourhood o(x = h(y) and he (wice differentiahle at x. Theil 11 II is twice differentiable at )', and

( 1.2)

In this formula, in the classical manner the dot between grad hk(y) and grad hl(y) denotes the scalar product of these two vectors: for two vectors P = (PI' " .. , Pn) and q = (q l' ... , qn) the scalar product is defined by the equation


We now consider Q an open set in IR". For every positive integer III = 0, I, ... , the Laplacian <1 maps «(, m+ 2(Q) in «(, m(Q), in the sense in which for II E (6 m+ 2(Q), the function

<111: x E Q -> ..11I(x)

is of class «(, "'. Thus <1 maps '(, , (Q) into itself. Being given a function): on a prescribed set Q, we give the name Poisso/l's cqllatio/l to the partial differential equation

..111 = I on Q.

From an elementary point. a solution is a function 11 defined on Q and possessing l' . I d' . (~II . f partla envatlves -:,-, on Q and satls ying ex;:

..11I(X) = I(x) , \Ix E Q

In fact. to make use of the invariance under a Euclidean transformation, we shall need to suppose that II is twice differentiable on Q. This elementary point of view is sometimes interesting: meanwhile consideration of the case n = I, where Poisson's equation reduces to the ordinary differential equation II" =.1: shows clearly that this framework is not satisfying: the class of twice differentiable functions possesses very few properties. For that reason, we state the

Definition l. Being given a function I defined on Q, we say that u is a classical solutioll of Poisson's equation

.111=/ on Q

if u E «(, 2 (Q) satisfies

JII(X) = f(x) \Ix E Q.

The above definition imposes, for the existence of a solution of class ((, 2(Q), the continuity of the givenf Even in the ease n = I, the use of non-continuous data is interesting in describing the differential equation u" = f For example to remain in the mathematical domain, the cO/wcx functions on an open set of Q of IR can be characterised as the solutions of the differential equation

11" = / on Q

with/a positive Radon measure on Q. The notion of a solution must then be taken in the sense of distributions. Let us recall that a distribution on Q is a continuous linear form on the space Ct'(Q)

of ((, , -functions with compact support in Q. Following the elementary theory of distributions 1, we formulate the

Definition 2. Being given a distribution / on Q, we call a distrihution solution of Poisson's equation

Llu = I on Q,

I See Appendix "Distributions" at the end of vol. 2. for the concept of a distribution.

~ I. The Laplace Operator 223

each distribution u on Q satisfying

< u, J ( > = < f, ( > V( E £0(Q).

Remark 1. We recall that a function f E C6'0(Q) defines a distribution

(1.3) T/ ( E 2J(Q) --+ f ((x)f(x)dx

and that the mapping f --+ TJ is injective with the result that we can identify fwith the distribution that it defines. Being given u E C6'2(Q), we have

< u, J( > = f u(x)J((x)dx = f Ju(x)((x)dx = < Ju. ( >

with the result that for f E ~,o(Q). u is a classical solution iff u E C6'2(Q) and is a distribution solution. In the case n = 1 for a given f E 'fo,o(Q) every distribution solution of the differential equation u" = f is a classical solution. This situation is no longer true in dimension n ;::, 2: being given an open set Q c IR" with n ;::, 2. there exist functionsf E 'fo'O(Q) such that there do not exist classical solutions of the corresponding Poisson's equation (see §3, Remark 5). 0

From Proposition 1, being given T a Euclidean transformation of Q, if u is a classical solution of Ju = f on Q, then v defined by v = Tu = U" T- 1 is a classical solution of Jv = Tf on T(Q). This extends to the body of distribution solutions through the use of the notion of an image of a distribution: let us consider a diffeomorphism h of Q into IR"; for f E reO(Q). we have for all ( E '@(h(Q)),

< hf, ( > = ff(h- 1 (YH((Y)dY = f f(x)((h(x) IDeth'(x)ldx

Supposing that h is of class rex, ("h IDeth'l E £I7(Q) and hence2

(1.4) < hf. 0 = <f. (,' h I Det h' I > . For a distributionf on Q, this formula defines a distribution on h(Q) and serves as the definition of the distribution hf. image off by h. In the case of a Euclidean transformation T. I det T'I = 1 and

< Tf, (> = <.I; (" T> .

Since J((" T) = (JO" T, we have from Proposition 1:

Proposition 2. Let T be a Euclidean tran.~{ormation o{Q,f a distribution on Q and u a distribution solution of Ju = f on Q. Then [1 = Tu is a distribution solution of L11' = T( on T(Q).

Remark 2. For greater clarity we have considered general distributions. In fact, in practice, we shall use only the following four types of distributions: (l) the continuous functions: they are defined everywhere; two continuous functions equal in the sense of distributions. are equal everywhere;

1 In this Chap. " h' denotes the derivative of h.

224 Chapter II. The Laplace Operator

(2) the locally integrable functions (in the sense of Lebesgue): they are defined almost everywhere (except on a set negligible in the sense of Lebesgue); they define a distribution by the same formula (1.3); two locally integrable functions equal in the sense of distributions (that is to say defining the same distribution) are equal almost everywhere; (3) the Radon measures or distributions of order 0; an important example is the Dirac measure (or mass) at a point x E Q:

bx : ( E .'7(Q) -+ (x) ;

(4) distributions of order \, that is to say of the form

(E !iI(Q) -+ <f~, D + /'1;, {'l) + '" \ (XI

whereJo,h, ... ,In are Radon measures on Q.

2. Examples in Mechanics and Electrostatics

o

Characteristic of the property of invariance under a Euclidean transformation, the Laplacian occurs in very many physical problems. Here we make precise the situations in mechanics and in physics in which Poisson's equation in [R3 occurs directly. In mechanics, a certain quantity of matter distributed in the space [R3 with a (mass) density p(x), gives rise at each point x E [R3 a gravitational potential Vg(x) which satisfies

JVg = kgP

where kg is a constant depending on the chosen system of units. The gravitational force f(x) being exerted on a mass m placed at x E [R3 is given by

f = - mgrad 1'g •

The work W done by the force of gravity when we deplace the mass m along a Jordan curve I': [0, \] -+ [R3 is given by

W = f f(x)dx = II f(y(t))}"(t)dt = - II m grad l'g()'(t)))"(t)dt ;' 0 0

= - m I ~Vg(},(t))dt = m(vg()'(O)) - t'g()'(1)).

In other words, the difference Vg(x I) - Vg(X 2) of the gravitational potential is the work done by the force of gravity when a unit mass is displaced along an arbitrary curve with an initial point x I and end-point x 2'

Let us now state precisely the' dimensions with respect to the fundamental quantities: length (L), mass (M), time (T): the mass density p has the dimension M L - 3, the force of gravity f has the dimension M LT- 2 with the result that grad Vg must have the dimension LT- 2 and the gravitational potential has the dimension L 2 T- 2.

* 1. The Laplace Operator 225

The constant kg has the dimension L 3 M -I r 2.

In fact (for reasons which will appear soon), we shall put

kg = 4nG

where G, which has the same dimension L 3 A1 IT- 2 , is called the universal gravitational constant. In the C.g.S. system oI units where the fundamental units are the centimetre (cm) for L, the gram (g) for M and the second (s) for T the constant G is given by G = 6.673 x 1O- 3 cm 3 g- l s- 2 . In the S.I. system (Systeme International)3, in which the fundamental units are the metre (m) for L, the kilogram (kg) for M, and the second (5) for T, we have

G = 6.673 X 10- 5 m 3 kg I S2.

The unit of energy or of work is then the louie (11 = 1 m 2 kgs- 2 ) and the gravitational potential is expressed in lkg-I. In electrostatics, a certain quantity of electric charge distributed in [R3 with a density (of electric charge) p(x) gives rise at each point an electrostatic (or Coulomb) potential 1'c' solution of Poisson's equation

LiVe =, - kcp .

The electrostaticforcef(x) exerted on an (electric) charge q placed at x is given by

I = ..... q grad 1\ .

To the fundamental quantities of mechanics L, M, T must now be added another quantity which can be either the charge (Q) or the intensity (I), these two quantities being related by Q = J T. The charge density p has the dimension QL - 3 and the electrostatic force M LT- 3 with the result that: the electrostatic potential has the dimension ML 1 T- 2 Q-1 = ML 2 T- 3J- 1, the constant kc has the dimension ML 3 T- 2Q2 = ML 3 T- 4 /- 2.

Poisson's equation for the electrostatic potential is valid in a homogeneous, perfect dielectric4 (see Chap. lA, §4) that we suppose here to occupy the whole space, the constant kc depends on the permittivity of the medium4 ; we define the permittivity constant by /; = like (we consider also D' = 4ru;) which has the dimension M- 1L- 3 T 2 Q2 = M 1[3T4 J2. We have kc = 1/;: = 4n/8'. We denote by Do (resp. e~), the permittivity constant oI the vacuum. In the system of units u.e.s.c.g.s., we use the units of the c.g.S. system and we fix the permittivity constant of the vacuum

4n (resp. f.~ = 1).

That defines the units of charge (Q = M ~ LiT 1) and of intensity (l = M l LiT- 2 ).

J This is the official system in France. 4 See Chap. lA, ~4.


We then have for the vacuum

In the S.l. system of units. we use in addition to the mechanical units. the ampere (A) as the unit of intensity. The unit of electric charge is the coulomb (e). 1 C = 1 ampere second; the unit of electrostatic potential is the volt (V); I V = 1 joule/second. These correspond numerically5 to

1: 0 8.8542 x 10 12 m .1 kg 1 S .. A2 .6

"co 1.1294 X lOll m J kgs- 4 A-'

3. Green's Formulae: The Classical Framework

We do not propose to give a full account of differential geometry here, but only to specify clearly the background against which we shall use the formulas of Green, basic tool for the study of the Laplacian and, more generally, for the study of boundary values. We shall consider a bounded open set Q with boundary r. The point of departure is Ostrogradski's formula for a field p = (PI' ... , Pn)

f div p dx = r. p . 11 d';' S2 " I

11.5)

whose significance we shall make precise below. Considering functions u. I' and applying Ostrogradski's formula to the field p = l' grad u and making use of the identity

( 1.6) div (1' grad 11) = grad 11. grad r + t"i1u •

we obtain

( 1.7) If! 1'/1udx

\vhere we have put

Jf' grad u . grad l'dx , Q

( 1.8) ('[I

= grad lI.n . ;'/1

We refer to (1.7) as Green's f(mnula fe)J' illteyratioll by parts. In effect in the case 1/ = I, Q = ]a, he, the formula (1.7) can be written

I" 'n u"(x)l'(x)dx = u'(h)p(h)- u'(a)1'(a) - J l/fx)r'(x)dx.

a u

, By using i:o i'o C 2 ~ L I'll ~ 471.10' - and the value of c the speed of light (sec Fournet LIJI. " Or again Fm - 1 where F. the farad, is the unit of capacity. See ~5.

~ 1. The Laplace Operator 227

Finally, interchanging u and /J in the formula (1.7) and subtracting the two formulae, we obtain Green's formula

(1.9) r (uAv - vAu)dx JQ f (u c .. v - v~lj)dY . I en on

3a. Elementary Theory

In the classical theory, which we are now considering, we suppose that Q is a regular open set in the sense in which its boundary r is a hypersurface of class ({/m

with m :;:;, I, Q being locally on one side of this surface. In other words, for all a E r, we can find an orthonormal reference frame R, a neighbourhood (open set) U of a in [R", an open set (I. of [R" - J and a function ,I. of class C(jm on @, such that denoting

by (x', x~) = (x'J, ... , x~ _ J ' x~) the coordinates in the reference frame R, we would have

r n U

QnU

{(x', ex(x')); x' E (9 }

[x E U; x;, < ex(x')} .

The data of (R, U) defines (C"', ex) perfectly; we can always suppose

U = [(x', ex(x') + t); x' E (I), It I < (5}.

We shall say that (R, U) defines a normal local parametrisation of r (in the neighbourhood of a) and we shall employ, without recalling the definitions, the notations C", ex, 6, (x', x~). For all x E r, nix) denotes the unit normal vector to r, exterior to Q. In a normal local parametrisation, for x = (x', ex(x')), the vector nix) is given in the base of the reference frame R by

(1.10) (- grad ex(x'), I)

n( x) = ....... --.----.---- (1 + I grad ex(x'W )112

where here the gradient and its norm are taken in G~n ... I. In particular the normal field: x E r ...... nIx) is of class '{?m" J

We denote by d}, the surface measure or the element olarea on 1'. It is sufficient to define it locally; in a normal local parametrisation, we have for every function ( continuous on r with compact support in U.

(1.11)

We then prove classically Ostrogradski's formula (l.5) for a field p of class 'C 1 in the neighbourhood of Q: first of all we localise, by using a partition of unity, reducing in a normal local parametrisation (R, U) to the case: supp p c U (the case supp pc Q is trivial); in the reference frame R we denote p(x',ex(x') + t) = (p'(x', t),


p~(x', t)); using (1.10) and (1.11) we find that the formula (1.5) can be written

J dx' r" (div p)(x', o:(x') + I )dt =7·1 (p~(x', 0) p'(x', 0) . grad cx(x'))dx' .

We verify it immediately by using supp(p', p;,) c {' x ] (), 0] and

(~P' ?p' (divp)(x',cx(x') + I) = (div,p'(x',t)---(x',t).grado:(x') + . "(x',t)).

?t 2t

We deduce the formulae of Green (1.7) (resp. (1.9)) for functions u of class (resp. (6 2 ) in the neighbourhood of Q. This formulation, which we shall call "elementary" of the formulae of Ostrogradski and Green is often not sufficient for applications: the regularity of the neighbourhood of the boundary, the regularity even of the functions on Q or of the open set Q are constraints which we should like to avoid. In § 1.3c we give a statement within the classical framework: Q regular and functions regular in Q. We shaH extend these considerations to the case of non-regular open sets Q and to nonregular functions in Q, in §6.

3b. Normal Differentiation

We propose to define p. nand tu/(ln on r for fields p and functions u defined only on Q. We remain always with the classical framework: we suppose that Q is reyu/ar, p a cOll1inuous field on Q and u a junction of class ((;'[ on Q and we define p. nand i'u/?n as continuousfimctions on r. Being given a E r and a unit vector e leaviny Q at a, that is to say such that coste, n(a)) > 0, we can consider

( 1.12) lim p(x - le).n(x)

and prove the

x-~a,xEf

t ~ D. I> n

Lemma 2. Ifllle limit (1.12) existsfiJr at least one unit vector e /eQ1:iIlY Q at a, then it I!xistslor all such vectors. More precisely, beiny yiven a compact set K ofT such that j(if all a E K the limit (1.12) o::istsf()r at least one unit vector e leuriny Q at 11, thenfil/' aI/ I: > 0, the limit (1.12) exists uniformly jiJr a E K and e a unit vector satisFyiny cos (e, /1(a)) '? I;.

This lemma permits us to defille p. n(a)jc)r a E r hy

( 1.13) p.ll(a) =

if this limit exists and to prove

lim \" ---~ 11, '( c I 1 ~ 0.1> ()

p(x - til (a»). Il(x)

Proposition 3. Thelunction p.n is continuous on the set ofrhe points a E rjiJr which it is defined (hy (LJ3)). More precisely, heing given ii continuous 011 Q extendiny the Ilormaifield n, p.n is the trace 011 r offile extension by continuity to Q olthefilllctioll p. ii (cOlli inuolls on Q).

§ I. The Laplace Operator 229

We define eu/en by applying Proposition 3 to p defined by

grad u: at a point a E r, eu/en is

(1.14) eu -;;- (a) en

lim grad u(x - tn(a). n(x) x-a. x E r { ..• o. { > 0

or again from Proposition 3, being given ii continuous on Q extending the normal field n, eu/en is the trace on r of the extension by continuity to Q of the function n. grad u (continuous on Q), this trace being independent of the choice of the extension. By construction, if eu/en is defined on r, it is continuous on r. We can connect this definition with the elementary notion of the directional derivative and show that if eu/en is defined on r, then (1) u extends by continuity to Q.

eu I' u(a - tn(a» - u(a) h l' . b . 'f f ' (2) l(a) = 1m .-~.-.--.--, t e ImIt emg um orm or a E I. en ,-0 t

t> ° We shall use the notation {e; (Q) to denote the class offunctions u E {f/ 1 (Q) with eu/en defined on r; hence for u E ~',; (Q), u extends by continuity to Q (u E 'tjO(Q» and for each continuous extension ii of the normal field to Q, i'i. grad u extends by continuity to Q (ii. grad u E (6 o(Q».

3c. Statement of the Classical Formulae of Green

With these definitions of p. n and au/an which give a meaning to the integrals over r in the formulae of Ostrogradski and Green, we can extend the elementary domain (p of class ~d in the neighbourhood of Q, u of class '{,2 in the neighbourhood of Q) of application of these formulae, yet with the condition of giving a meaning to the integrals on Q. Within the classical framework, we state:

Proposition 4 (Formulae of Ostrogradski and Green in the classical context). Let Q be a regular bounded open set and (I) p a .field of class ((, 1 on Q with p. n defined on rand div p integrable 0/1 Q. Then

i div pdx = f. p.nd)' ; JQ [ (2) 14 E 'f;2(Q) (\ ((';(f2) and r: E ((,l(Q) (\ (6 0 (Q) with vAu and grad u.gradl' integrahle on Q. Then

i vAudx = f t' (l L/ dr, - i gradu.gradvdx; JQ I (1/1 JQ

(3) u, l' E ((,2(Q) (\ ((';(Q) with (pAu - u;Jv) integrable on Q. T1len

(1' A 1/ - u A l' ) dx = v :;- - u ~ d I' . i j~ (au ?!u) Q I en ('II

Let us show how to deduce Proposition 4 from the elementary formulation of Green's formulae (regular data in the neighbourhood of .0) with the help of the following geometrical lemma:


Lemma 3. Let Q be a given reyular open set {olclass '(, m). There exists a/unction cP or class (6 m on an open neiyhbouriJood U of 1, reyular on U (that is to say grad (P(x) =t O.f()r all x E U) such that

r QnU

[ X E U: cP(x) = 0 J

:XEU; cP(xl<O}.

Proof ol Propositiol1 4. Using the function cP of Lemma 3, let us put for 6 > 0

Q;; = (Q\U) U [XE U; cP(x) < -6}.

For () sufliciently smail, this is a regular, relatively compact open set in Q, with boundary ro = : x E U; (P(x) = - () J . Let us put

il(x) grad cP(x)

I grad cP(x) :

The field i1 is of class '(, m . 1 on U and extends the normal field IJ of r, thus, also, the normal field /1.8 of ro' For a field p of class «(, 1 on Q, we can apply Ostrogradski's elementary formula on Q 8 since Q is a neighbourhood of Qo

( 1.15) r divpdx = f. p.iidi'ii Ju. I,

where dr'li is the surface measure of r,j' In the limit when 6 -+ 0, we obtain the Ostrogradski formula for p on Q, since div p is integrable on Q and p. n is continuous on Q n U 7.

The case of Green's formulae is treated in an identical manner. 0

Remark 3. Recalling the above proof, being given p of class (6 1 on Q, we have (1.15) for all 6 > O. Suppose that p . n is defined on r, the second integral and hence also the first integral converges when 6 -+ O. That shows that the integral

,.

J div pdx Q

is convergent. Also, making use of the FatouLebesgue theorem, we are assured of the integrability of div p on Q when div p is bounded below or bounded ahove hy a function integrahle 011 Q. 0

In particular, recalling the reasoning for Green's formula of integration by parts with v = u, on applying to p = u grad u, we obtain the

For every function ~ continuous on U n 15, l ~ d'/

local parametrisations.

lim J' :; d';, . This is shown directly by b • (j r",


Proposition 5. Let u E (6 2 (Q) n 16,; (Q). Suppose that uAu is bounded below by a function integrable on Q. Then 1 grad u 12 and uAu are integrable Q and

i f au J~ Q

uA u dx = r u - d,' - Q 1 grad u 12 dx . en

4. The Laplacian in Polar Coordinates

The Laplace operator being invariant under a Euclidean transformation, is, in particular, invariant under rotations in 1R". For that reason, it will often be interesting to use the representation of 1R" in polar coordinates. Let us begin with the case 0/ the plane 1R 2 The mapping

h: (r, 0) -+ (rcosO, rsinO).

is analytic on IRl\ {OJ x lR. It is only locally bijective: for Go E 1R, h ia a diffeomorphism of JO, x[ x JOo - n, 00 + n[ onto 1R2\D8~ with DI~ = {h()., ( 0 );

;. ~ O}; we shall call the inverse diffeomorphism, the representatioll o/proper polar coordinates 0/ 1R2\DB;,. However. it will be interesting to consider r < 0: for examplehisadiffeomorphismoflR* x JOo,Oo + n[onIR2\DoowithDoo = {h()', 00 ); ;. E 1R}. Now, let us take (ro, (0) E 1R 2 with "0 i= 0 and let us consider u defined in the neighbourhood of Xo = h(ro, 00 ) and twice differentiable at Xo. An elementary calculation shows that:

(u h) ..... 7'02 (1'0' lJo )

which, with abuse of notation, can be written

(1.16)

This equation is called thefimnula for the Laplacian in polar coordinates. We note also that we have

( 1.1 7)

This formula is still true i/ u is a distribution on an open set Q of the plane not containing the origin: in the neighbourhood of a point of h - I (Q), h is a diffeomorphism with the result that we define the image distributions u c h = h· 1 U and Auah = h-1Au; we then have

( 62 1 (~ I ci 2 ) (Au) n h = ar2 +; cir + ;2 aiP. (u" h)

the derivatives and the products by l/r (r i= 0) being taken in the sense of distributions. The validity of this formula can be assured by general arguments of

232 Chapter II, The Laplace Operator

continuity and density in [f'. In order to persuade a reader with slight acquaintance with the general theory of distributions, we give a direct proof. First of all

i" - f cos (i

c",.

('

_ -rsin () cr

i' -fCOS 0 (10

(~

- r sin () ao

r

with the result that by applying (1.4), and using u = h(h-lul, we have

V(EOQ(Q).

In particular,

Applying the formula for the Laplacian in plane polar coordinates to ( and using (1.17) we obtain

or, by the definition of the differentiation of distributions,

which, account being taken of (1.17), proves the formula. o We now propose to generalize this formula in Ir~". We denote by l' the unit sphere ill [H;11, defined by

lal = ((Ji + .. , + a~ )12 = 1 l ' This is an analytic hypersurface in [H;", the boundary of the unit ball. The map (r, a) --> ra is an analytical diffeomorphism of]O,::X:;[ x L onto [H;"\{O}. We call the inverse diffeomorphism x EO [H;" \ { O} -> (I x I, x / I x I) the representation in polar coordinates in [H;". We note the ambiguity in the case n = 2: in the representation in polar coordinates (without other specifications), the argument (j = x/I x I is a point on the unit circle in Ire (which we can identify with the group of rotations of the plane, with the group of complex numbers with modulus 1, or again with the additive group [H;/2nZ:'); in the representation in suitable polar coordinates, we choose for eo a suitable parametrisation of the unit circle with the point with polar angle eo + n removed.

§ I. The Laplace Operator 233

We note also that in the case n polar coordinates is

LX { - 1, + 1 } and the representation in

X E lR * ---> (I x I, sign x) where slgnx = {+ 1 - 1

x > 0 x < 0

x --------~

Ixl Because of this case n = 1, we shall sometimes employ for x E lRn \ {O}, the notation sign x = x/lxl. We denote by dO" the surface measure ofX. For every function u integrable on lR", we have the formula

( 1.18) f udx = f'" r" -- 1 dr f u(m) dO" . WoE

In the case 11 1 this formula can be written

r udx = IX (u{r) - u(- r»dr, JiR 0

and in the case n = 2,

( 1.l9) r udx J[~2 fI f21t

= 0 I'dI' 0 u(rcose, rsinO)dO.

We note that being given a unit vector e, the map (r, 0") ---> m is an analytic diffeomorphism of lR* x X + (e) on lR"\n(e) where

X + (e) = {O" EX; 0". e > O} and nee) = {x E lR" ; x. e = O} .

In this representation, the formula (1.18) can be written

(1.20) L, udx = f~x r,,-ldr f,+(e) u(m)do-

which gives for n = 2

( 1.21) f~2 udx = f:£ rdr f: u(rcose, rsinti)d8.

We say that a function u(x) is radial if it depends only on r = I xl- It is naturally defined on an annulus J x X where J is a part [O,:x::{; it is clear that a function defined on J x L: is radial iff it is invariant under rotation. Since Ll is invariant under rotation, ttu is a radialfunclion, Llu is also radial. We have the followingformulafor the Laplacian of a radial fimction:

J n the case 11 3, we have also

(1.23)

We can easily verify these formulae by elementary calculations.


Being given a function u defined on I xL.;, it is often interesting to consider the radialised fUllctioll of U,

( 1.24) ii(x) = t u(lxlo-)do-.

We then prove

Proposition 6. The operatioll ofradialisation commutes with the Laplacian, that is to say, with the definitioll (1.24),

( 1.25)

Proof We put v(r)

(1.22),

ju(x) = ,1u(x) = t (ju)(lxlo-)do- .

t u(m) dO" with the result that u(x) v(lxl) and from

_ 1 d ( . dr) llu(x) = ~n--=l dr rn - l & (lxl)·

We can always suppose that u defined on IR" can be replaced by ~(Ixl)u(x) where ~ E Si'(l), ~ = I on a neighbourhood of the norm of the point at which we evaluate lltl. We then have

dr f -(r) = gradu(m).o-do-dr 1: r" - I

and hence by Green's formula

dr 1 I' ju dx .

Using (1.18)

thus

- (1') dr

dl' r" - 1 - (r)

dr

r n -- ,.;R(O,r)

r 5,,-1 ds t C·1u)(sO") do- ;

~_(rn-l ~c (r)) = r,,-1 II' (llu)(m) do- . dr . dr ~1:

o

We say that a function u(x) is spherical if it depends only on 0- = x/lxl. It is naturally defined on a "cone" IR + x (! where (! is an open set of 2.-; it IS

clear that a function defined on IR + x (! is spherical iff it is invariant under the homotheties H;.: x -'> i.x (/ > 0). Now for a function u defined on IR + x C. A (If ;u) = (I 2) H ;(ju). In particular, applying this result with}, = I x I, we obtain for a spherical function

(1.26) I (I X ) .;Ju(x) = -Jdu)-- - . Ixl- Ixl


We call the Laplacian on the sphere, or the Laplace-Beltrami operator the trace A" of the Laplacian on the sphere: in other words, for each function u of class '6" 2 on an open set {f of I", A"u(O"} = .1 x u(x/lxllx=". For a function u of class (6"2 on an open set Q of IR", we then have theformulafor the Laplacian in polar coordinates in IR"

( 1.27) 1 c( c) 1 (Au)(nr) = -_~ ~ .. - rn-1~ u(nr) + ·······A"u(nr). r" 1 Cr or r2

The simplest method of verifying this formula is to make use of the density in '{; 2 (I X (f) of functions with separable variables u(x) = vi I x I) w(x/I x I); the verification of (1.27) for such functions follows immediately from (1.22) and (1.26). We can also adapt the proof of Proposition 6. This formula is obviously still true for distributions. In the case n = 2, the formula (1.27) coincides with the formula (1.16). That merely expresses that the Laplace-Beltrami operator on the unit circle of [f,g 2 is given by

(';2 A"u(cos 0, sin 0) = -;;--2 u(cos 0, sin 0) .

cO

In the case n = 3, we can refer locally to points on the unit sphere by the Euler an~Jles (0, <p) defined for 0" = (x, y, z) by

x = sin 0"(0, <p) is an analytic diffeomorphism from J 00 - n, 00 + n [ x JO, n[ onto the sphere I" deprived of the semi-circle {( ~- sin <p cos °0 ,

- sin <p sin 0o, cos <p); 0 0( <p 0( n}. In this representation, the Laplace-Beltrami operator is given by the equation

( 1.28) ( I 82 I a ( a )) --;-- ~~ + -;-- -- sin(p u(O"(O, <p)) Sill 2 <p c0 2 Sill <p C<p ?i<p

called the formula for the Laplacian in spherical coordinates. This formula can be generalized for the sphere in IR" with 11 arbitrary (see Dieudonne [1 J, Vol. III). Another classical representation of the sphere is the representation in stereographic coordinates. The stereographic projection is the mapping

, "-1, (2x',lx'12-1) x = (x t , ... , x"_ d E IR --> ()(x) =-;-2--

1 + Ixl

This is a diffeomorphism from IR" - 1 onto the sphere I" with the "north pole" (0, ... , 0, 1) removed; it is in fact the trace on the hyperplane IR" - 1 X {O} of the

polar inversion, with pole the north pole and of radius fl. We shall return to the use of inversion for the study of the Laplacian in §2. Finally we draw attention to the representation of IR" = IR m x IR" - m

(2 0( In < n) in cylindrical coordinates which consists of representing x = (x', xl!), x' E IRm, xl! E IR" - m by the polar coordinates of x', r = Ix'l, () = x'llx'l and the

236 Chapter It The Laplace Operator

point x" E 1P1"- m. In this representation

(1.29) (.~uJ(r(T.x") = C'}~-l ;:(rm - 1£) + AiJ a + iJ x )U(rcJ.x,,),

where f1" is the Laplace-Beltrami operator of the unit sphere in IP1m. I n particular, in the ease n = 3, the cylindrical coordinates of (x. y. z) are (r. 0, z)

where (I'. 0) are the polar coordinates in the (x, y) plane. We have

( 1.30)

iJu(rcosO, rsinO, z) = (.~ ~~ (I' ;)' +.~ '.(~~2 + ~~2»)u(rcos(), rsin8, z). r or or r (!o oz·

--the formula for the Laplacian in cylindrical coordinates in the space 1P13. This account of the passage to polar coordinates expressed in the classical framework remains valid within the framework of distributions.

§ 2. Harmonic Functions

1. Definitions. Examples. Elementary Solutions

Definition I. Let Q be an open set of IP1n. We define a harmonicjimetion on Q to be a function u E (6 2 (Q) satisfying

L1u(x) = 0 "Ix E Q.

In other words, a harmonic function on Q is a classical solution of the equation

ljU = 0 on Q

called Laplace's equalioll 011 Q.

Remark 1. We shall prove later (see Corollary 7) that every harmonic function on Q is analytic on Q. We shall prove also (see Proposition 1 of §3 and also Corollaries 7 and 10 of this §2) that every distribution solution of Laplace's equation is a harmonic function (and hence analytic on Q). 0

Case 11 = J. The harmonic junctiolls Oil all interval of U;g are the affine junctiolls. This is immediate since d 2U / dx 2 = o. Case n = 2. Identifying the plane u;g2 with the field C of the complex numbers, we have the classical result linking a holomorphic function to a harmonic function. We denote here classically by (x, y) a point of u;g2 which we identify, topologically (but not algebraically) with the complex number z = x + iy of C.

Proposition I. (a) Let Q be an open set ofC andfa holomorphicjullctioll of the complex variable:. Then the/unction u(x, .v) = Ref(x + iy) is harmonic on Q.

~2. Harmonic Functions 237

(b) Let Q be a simply connected open set ofIR 2 and u a real-valued harmonic function on Q. Then there exists a holomorphic functionf of the complex varia hie z = x + iy, ((x, y) E Q) such that u(x, y) = Ref(x + iy)for all (x, y) E Q.

Proof Being given f(x + iy) = u(x, y) + iv(x, y) with u and P, real and differentiable in (x, y), the function f is holomorphic (that is to say differentiable with respect to the complex variable) in z = x + iy iff the real functions u and v satisfy the Cauchy-Riemann conditions

au ax

av ay'

eu ely

av ax at (x, y) .

(a) We make use of the fact that a holomorphic function on Q is of class re 2 (and even analytic) on Q (with respect to the complex variable). We deduce immediately that u and v are of class ce 2 (and even C(;' X) on Q (with respect to the real variables). Differentiating the Cauchy-Riemann conditions we obtain

(b) We have to find a function v (at least of class C(; 1), satisfying (with u given) the Cauchy-Riemann conditions, that is to say

grad v = (__al~, ~~) . oy ax

~ 1

Since Q is simply-connected and - ~u and ~u are of class C(;' I, that is equivalent to uy uX

o ( au) e (au) oy - Dy = 7J:; a:; ,

that is to say, to Llu = O. o Remark 2. Let us consider u harmonic on an open set Q of IR 2 If Q is simply connected, there exists v E (f,2(Q) such that f(x + iy) = u(x, y) + iv(x, y) is hoI om orphic. The function v is defined to within an additive constant and is also harmonic (v = Re( - il): we sometimes call it the harmonic conjugate function of u. If Q is not simply connecteq, the conjugate harmonic function is defined locally and is multiple-valued on QB.

Now, every open set is locally simply connected, hence every function, harmonic on an open set oflR 2 is locally the real part of a holomorphicfunction. Making lise, then, of the fact that a holomorphic function is analytic, we obtain a simple proof of the analyticity of a harmonic function on an open set of the plane.

8 More generally, let Q be an open set of IR:" and p a field of class ({, 1 (Q) satisfying

curl p = 0, div p = 0 .

Locally, p is the gradient of a harmonic function (defined to within a constant). If Q is simply connected, p is the gradient of a function harmonic on Q (defined to within a constant), but if Q is not simply connected, this function can be multiple-valued.

Chapter II. The Laplace Operatllf

In a general manner the theory of hoiomorphic functions will he a powerful tool fi)r the study o(the Laplacian in [R2. We note that the reason that the case n = 2 is so special, is due to the decomposition of the Laplacian in [R L

/1 = ,r 2o + ~~\ = (I ~' + i ( ) ( (_ i ,(»'

c:c rr- , C'X ;'Y \ (ix (v'

The linear differential operator (with complex coefficients)

is called the Cauchy-Riemann operator (the Cauchy-Riemann conditions can be written "in summary", a(u + iv) = 0), We can show easily that jar n ?: 3, it is impossible to write A as the product of two linear differential operators of the jirst order in C (see Chap, V). We now give some immediate properties of the set oj/unctions, harmonic in an open set Q or [R" which we shall denote hy ,ff (Q),

Proposition 2 (a)1f( Q) is a vector space: every liner comhination of harmonic limetions is ayain harmonic: (b) ,Yf( Q) is stable under differentiation: the deriratives ('1/ / cX i ofa harmoniejimction 1/ are harmonic functions (while awaiting the proof that a harmonic function is (6 ' we have to suppose that u E (f, 3(Q),

(c) ,ff( Q) is closedfor the topology ofunijc)rm topology on erery compact set (but also

fi)r the induced topology from that or distrihutions on Q): every limit of harmonic jimctiolls is harmonic (until we prove that a harmonic distribution is a harmonic function, we have to take the limit in 'IS 2 (Q), i,e, the uniform limit in every set of Q,

of u and its first and second derivatives), We shall return to this property later (see Harnack's theorems, Proposition 13 from which come the Corollaries 8 and 9), (tl) If u, v E .ff (Q), thell Ul' EW (Q) iff grad 11 ,grad r = 0, This result immediately follows from

(2.1 ) L'l(w:) = uL'll' + 2gradu,gradv + 1'/1u,

We now give some (fundamental) examples of harmonic functions:

Example 1. The affil1efimctiolls are harmonic Oil [R" (and hence on every open set of [R "). In the case 11 = I, these are the only harmonic functions (on an interval of rr«), In the case n ?: 2, there exist numerous harmonic polynomials; we note that a polynomial is harmonic on a non-empty open set of [R" iff it is so on the whole of

[R", A quadraticfcmn L aijxixj is harmonic iflits tract! L au = 0, From Proposition i.j

l, a polynomial in two variables L (lklXk)'1 is harmonic iff it is the real part of a k,1

polynomial of the complex variable x + iy, L Ck (x + iy )k. We shall return to the k

study of harmonic polynomials in ~ 7, Let us note

~2" Harmonic Functions 239

Proposition 3. The harmonic polynomials are the only tempered distribution solutions of Laplace's equation on ~".

Proof We shall make use of the Fourier transform for tempered distributions (see Vol. 2, Appendix "Distributions'} Let us consider a tempered distribution solution u of

Llu = O.

Its Fourier transform 12 is a solution of the algebraic equation

y212(y) = O.

We then deduce that 12 is with support contained in to} and hence is a distribution of finite order mEN * and a linear combination of the Dirac measure at the origin b and its derivatives (see Vol. 2, Appendix "Distributions"):

12 ==

Using the inverse Fourier transform, we obtain for u:

u == L a,x' 10 .

-xl ~ m

o

We note that this' proposition shows in particular that a tempered distribution solution of Laplace's equation is a (harmonic) analytic function. We note also the corollary, which we can obtain in the case n = 2 by means of Liouville's theorem:

Corollary 1. A harmonic function bounded on the whole oj" ~n is constant.

Proof A (locally integrable) function, bounded on ~n, is a tempered distribution (see Vol. 2, Appendix "Distributions"). From Proposition 2, a harmonic function bounded on ~il is hence a polynomial; being bounded, it is constant. 0

We shall later extend this result by another method to the case of harmonic functions on IR" bounded above or below (see Proposition 7).

Example 2. Let us now determine the radial harmonic functions. Suppose that we are given an interval I of] 0, .CXj [ and that we consider a radial function

u(x) = v(lxll

defined on the ring Q = {x E IR"; Ixl E I}. Making use of the formula of the La placian of a radial function (see (1.22)), u is harmonic on Q iff v is of class Cfi 2 on I

q We adopt here the multi-indicial notation

ex = ('X l' .. " , ~n) EN" •

10 The coefficients a, are obviously not arbitrary, the polynomials L QxX' having to be harmonic. In the case 11 1, we have necessarily In ~ 1. hut II ~ 2, there exist harmonic polynomials of arbitrary degree.

240

and

this is equivalent to

(2.2)

~.. d (r" -1 dr: ) r"- 1 dr dr

liv

dr

o

on 1 with (' an arbitrary constant. We distinguish Case 11 = 2. The integration of (2.2) gives

l'(r) = cLogI' + Co 11

Case 11 ? 3. The integration of (2.2) gives

c


rEI:

r ( r) = _.- . ,.. + co. (n - 2)r,,-2

We can therefore state

Proposition 4. Erery I"(IdiuljilllCl iOll, harmonic Oil ill) ill/llUlliS ill lR:" is of' the limn

c Log Ix , + Co if n

c

Ix I" 2 + Co if Il? 3 ,

where Co' care cOllstants.

We note that this proposition proves that a radial harmonic function is analytic; also. if a radialjimction, harmonic on {O < , x I < r 0 } is bounded, then it is constant. We remark also the preceding calculation remains valid for a radial distribution; this proves that a radial distribution solution of'Laplace's equation 011 r Ix I E I;. where I is all intermi oj' ] 0, .x; [ is a (harmonic) analytic fUllctioll. Now thc function Log Ix I in the case n = 2, and the function I xf - n in the case 1/ ? 3, is integrable in the neighbourhood of the origin; it therefore defines a distribution u on lR:". Let us now calculate the distribution Au defined by:

(!Ju,D = (11, lIS") = fu ds = }i~LI>f!/!lS"dX. To calculate this limit, let us begin with the case 11 = 2. Changing to plane polar coordinates, we have

i'

J ,, !I,1S"dx Ixl > ,.

j" r Log r dr JI'2 IT ,J sir cos 0, r sin ()) dO . I: U

I I The notation Log denotes the Naperian logarithm.


Using the formula for the Laplacian in polar coordinates we see that

J~2" _ f2IT (I a ( a.() 1 a2( ) O

A((rcos(}, rsinO)dO - -- r- + - d8 o r i'!r cr , Nl2 I

= -- r-- ((rcosO,rsinO)dO 1 d ( d f2rr ) r dr dr 0

en (12Y since Jo (O~ dO = 0, the function ((rcosO, rsinO) being periodic

Integrating by parts in the integral obtained by inserting the latter result into the integral of uA ( we obtain

r uA(dx = {x LOgr~(r.~ .. err ((rcOSO,rSinOJdO)' dr .;Ixl>f. J, dr , dr Jo

d f2IT I - [; Logt:- ((rcosO,rsinO)d8 dr 0 r ~ f.

err + Jo ((£cosO,esin8)dO.

When e -> 0, ijIe first term tends to 0 and the second to 27[((0). Therefore

(Au,O = 27[((0), that is to say, .finally

Au = 2n() ,

where c5 is the Dirac measure at the origin. We note that we should be able to simplify this calculation by remarking that Au is a radial distribution; hence it is enough to determine (Au, () for a radial function ( = (( I x I)· We shall carry out the calculation in the case n ~ 3. Here

f uA(dx = Jlxl > f.

f·x ... _1 . x r"·· 1 dr r -.. ~j (1''''' 1 dd'r'~)d() f. 1''' - 2 oJ L 1''' - 1 ell'

() fX ...... _I ~ (r"" 1 ~~) elr " , rn .. 2 dr \ dr '

where ()n = L dO' is the total surface area of the unit sphere in [R;". Integrating by

parts, we have

f u L1 (dx = (J" ( ... f, ~~ (E) - (n .. 2) (( [;) ) J Ixl > ,

and hence in the limit as e -> 0,

242

This calculation leads us to denote

(2.3)

I - Loglxl 2n

Chapter I!. The Laplace Operator

with - (11 - 2)a n .

With this notation, the (locally integrable) function E" is a distribution solution of the Poisson equation

We pose the

Definition 2. An elementary (orjimdamental) solutioll of the Laplacian in !KI" is any distribution E on fRn which satisfies the Poisson equation

With this definition, l~" is all elementary solution of' the Laplacian: since the restriction to!KI" : O} of an elementary solution is a solution of Laplace's equation, taking account of Proposition 4 (see also the remark which follows), En is aell (to

\\'ilhill all wIditire cOllstant) tile ollly elemental'r radial solution. The notion of an elementary solution plays a fundamental role in theory of linear differential operators (see Chap. V). The elementary solution En will playa constant role throughout the whole of the theory of the Laplacian (see also its physical interpretation in ~2.6). As we have noted above, tiJis elementary solution En is unal),'tic 011

[Fg": 0 l: using general theorems on linear differential operators, we deduce (see Chap. V, ~2.2.4) that eray distrihution Oil all open set Q satis{ving Laplace's equatioll 011 Q (alld hellce, ill partic!llar, erery harmonicfill1ctioll 011 Q) is a (harmol1ic) analytic .limetion 011 Q. We shall give another direct proof of this result later. Finally, we havc so far considered only thc casc 11 ~ 2. In the case Il = I it is immediate that

is an elementary solution of d 2 ; dx 2 : to within a constant, it IS the only even elementary solution.

Example 3. Every derivative of a harmonic function is again harmonic. In particular the function (lE,,/ clx n is harmonic on !KIn '\ [O}: a simple calculation shows that for all 11 ~ 2

DEn 2xn

?Xn O'n Ixln

This cylindrical function (it depends only on r = (xi + ... + x~ _ 1)1 and x n )

will be particularly useful in the study of the Laplacian for problems with cylindrical symmetry, or more generally for problems in which one of the directions

§2. Harmonic Functions 243

in space (usually taken to be the x" direction) is privileged. The Laplacian then behaves like an evolution operator

D

2. Gauss' Theorem. Formulae of the Mean. Maximum Principle

Let us consider a regular bounded open set and two functions u, v E ff( Q) f1 ~ ~ (Q), that is to say in the adopted nomenclature, harmonic on Q with normal derivatives defined on r (in the classical sense of § l.3b). The application of Green's formula (Proposition 4 of § 1.3c) gives

(2.4) f (V au - u ov ) dy = 0 , r an an

which we shall call "Green's formula for harmonic functions". Particularizing with v = I, we obtain Gauss' theorem (stated in the classical framework).

Proposition 5. Let Q be a regular bounded set and u E ff( Q) f1 ~ ~ (Q). Then

t:~ dy = o.

We shall show later (see Proposition 14) that this property when it holds for every open set with a regular boundary, characterises the harmonic functions. We shall return also to §2.6 of this chapter for the important physical interpretation of this theorem. Suppose that now we are given an open ball B = B(xo, ro) of centre Xo E [R" and radius ro > 0:

B(xo,ro) = {x E W; Ix - xol < ro}.

We denote its boundary by oB; we have

oB(xo, ro) = Xo + roL

where L always denotes the unit sphere in [R". We keep the notation dy for the surface measure on oB; we have

(2.5) rudy = r~-l f U(Xo + roa)da, J8B L

where da always denotes the surface measure on L. Now let u E ff(B). From Gauss' theorem (Proposition 5) we have, for all o < r < ro:

(2.6) (' au J -d)' = 0,

('B, an where B, B(xo, r) .


Applying now Green's formula for harmonic functions to u and 1'( x) =

En(.x - xo), where En is the elementary solution defined by (2.3), on the annulus Q = [X;I: < Ix, xol < r}withO < t: < r < ro.TheboundaryrofQismade up of? By and (IBt ; the restriction to a Br (resp. eBE ) of the normal field to r exterior (0 Q is the normal to (I Br exterior to Br (resp. the opposite of the normal field to ('B, ex terior to B,). Since r is constant on a Br and a B, we have from (2.6)

Now x

gradE,,(x) = (Tnlxl" lixi

this being true as well for II = 2 ((T 2 = 2n) as for II 0 3. Hence

f u (lV d" =.. 1_ JI' U d" - .. 1 -- Jr. U d" . l t n --- 1 I .ri --. 1 {

r (./1 (TnI' iB, (Tn L ,'B,

Using (2.5), we obtain in the limit when I' --> 1'0 and I; -> 0, thejimnula olthe mean (stated in the classical framework):

Proposition 6.

(2.7)

Remark 3. To be able to apply this formula, it is clear that it is not necessary to suppose 'that 11 is continuous on B. Denoting for i. E JO, I [

1I,,(.x) = 1I(l.xo + (I - nx) ,

the function 11 ... is harmonic in the neighbourhood of B. Applying the formula of the mean to !i;., since !i;(xo) = !i(xo). it suffices to be able to pass to the limit when ). --> O. That is true in particular from Lebesgue's theorem, if

r 1/(x) = lim u(i.xo + (1 - i.)x) exists d~'-a.e. xEi'B ~ A'· 0

Llu(i.xo +(I-i)x)lo(lIo(.X) d},-a.e.xErB with 1Io EI}("B)

We shall return later to the extension of integral formulae in a more general context (see ~6). D

We now give a variant of the formula (2.7): applying it for 0 < r < ro, we have for 11 E .f( (B),

Hence if we suppose u integrable on B, we have

In udx = f:"rn-1dr lU (X O + r(T)d(T

We can then state:

§2. Harmonic Functions

Corollary 2. Let B

(2.8)

B(xo,ro) and U E fi(B) II L!(B). Then

u(xo) = ~ ( udx . (Jnro JH

We note that (Jnln is the volume of the unit ball.

245

The formula of the mean is a powerful tool. A first example of its use is the generalisation of Corollary 1:

Proposition 7. A harmonic function on [Rn, bounded above or below, is constant.

Proof Let us suppose, for example, that u is bounded below; by the addition of a constant, we can always suppose that u ~ 0. Being given XI' X 2 E [R" and 1'1 ~ 1'2 + Ix! - X21 ~ 1'2 > 0, we have

B2 = B(X2,r2) c B(x!,r!) = B!

Applying (2.8) we obtain

u(x1 ) ="~n f udx:::; ~ f udx = ('i)" u(xJ)' (Jnl'Z H2 (Jnrz HI 1'2

Taking /'J = 1'2 + IXI - x 2 1 and letting 1'2 -> C/.), we find that

u(xz ) :::; u(x l ) .

Interchanging the roles of Xl and xz, we have u(x I ) :::; u(x1 ) and hence u(Xj) = U(X2)' 0

Let us now state the principle of the maximum

Proposition 8. Let Q be a connected set and U E $' (Q). If there exists Xo E Q such that

U(X):::; u(xo)(resp.u(x) ~ u(xo )) forall xEQ,

thell u is constallt 011 Q.

In other words: a non-constant harmonic function on a connected open set can have neither a maximum nor a minimum on that open set.

Proof First of all let us consider Xo E Q and 1'0 > ° such that

B(xo, 1'0) c Q and u(x):::; u(xol for all X E B(xo, 1'0)

and show that then u is constant on B(xo, 1'0)' We have in effect from (2.8), since (Jnl'Z/n is the volume of the ball B = B(xo, 1'0)

L (u(xo) - u(x))dx 0 .

If u(xo) - u(x) ~ 0 for X E B, then necessarily

u(xo) - u(x) = 0 \Ix E B .

Now let us consider

\Ix E Q}

246 Chapter II. The Laplace Operatm

From the hypothesis flo =f= 0; from what has gone before, Qo is open. Finally Qo is closed in Q. since

r Qo = lXoEQ; u(xo )

'I max II ( .

Q )

Hence. since Q is connected, Q o = Q. [J

Remark 4. Note that in the above proof we have used only that 1/ should satisfy the formula of the mean (2.8); we could just as well have used the formula of the mean (2.-7); in fact we shall show later that these formulae characterise the harmonic functions (see Proposition 15). Note also that. making usc of the analyticity of a harmonic function. we see that if 1/

is constant in the neighbourhood of a point. it is constant everywhere (on a connected open set): that shows that a non-constant harmonic function on a connected open set can have neither a local maximum nor a local minimum. 0

Let liS givc some useful corollaries of this maximum principle.

Corollary 3. Let Q he a bounded opel1 set (ll1d U E# (Q) n (f, 0 (rh Theil

minll ::S: lI(x) z, maxu jin' a/l SEQ.

1/; ill additiun. Q is cOl1llecled and II is IIOIl-oJ/lslanl. thell

minll < lI(x) < max!! fill' al/ x E Q. r

Proot: The function II being continuous on the compact set {';! attains its maxi-

mum at a point Xo E Q. If u( Xo ) were strictly superior to max 1I, -"0 would belong to I

Q and 11 would not be constant on the connected component of Xu in Q; this contradicts the principle of the maximum. The second assertion is an immediate corollary of the (irst assertion and of the principle of the maximum. C

This corollary IS 111 general false for an unbounded open set: for Q = ; X E ,~/; Ixi > I}. the function II(X) = E2(x) (case 11 = 2) or u(x) = l/k" - E" (x) is harmonic and positive on Q but is zero on i1Q = L. It is clear that to generalise the statement of Corollary 3 to the case of an unbounded open set. we must introduce infinity as belonging to the boundary. We can state:

Corollary 4. Lei Q he (/// unhounded open set and II E .# (Q) n ((, (] (Q). BeillY !firen AL . M, E ~. (M + > tvl ) let LIS suppose that

(i) A1 ::S: II(X) ::S: AJ f ji)!' all x E r.

(ii) AI ::S: Iiminfll(x) < Iimsupu(x) < AL I xl" , ! x I ~ ,

Theil

/'\11 ::S: 1I(x) < Al+ for all x E Q.


Proof Being given I: > 0, we deduce from hypothesis (i), that there exists R > ° such that

X E Q, Ixl ~ R=> M_ - I: ~ u(x) ~ M+ + 1:.

We apply the preceding corollary to the restriction of u to the bounded open set QR = Q Ii B(O, R); since the boundary ()QR c r u (Q Ii 13B(O, R», we have

M _ - f. ~ min u ~ max u ~ M + + I: , ?QR (lQR

so

We have also

j,\;L - f. ~ u(x) ~ M+ + f. foral! xEQ\QR = {xEQ;lxl > R}.

Therefore j'yL -- [; ~ u(x) ~ M+ + (; foral! XEQ.

In the limit when £ ~ 0, we obtain the result.

Finally we note the following useful results:

o

Corollary 5. Let Q be an open set and u 1 , Hz E ff( Q) Ii (& 0 (,(2). If Q is unbounded, we suppose in addition that

lim u1(x) Ixl ~-x;

Then on

on

lim uz(x) Ixl - CN

r = HI ~ uz r = u l u 2 .

0.

on Q

Proof: It is sufficient to apply the preceding corollaries to U = U I - U2•

Corollary 6. Let Q be an open set and (Uk) a sequence Ol./illlCtiOlls harmonic on Q and continuous on Q. If Q is unbounded, we suppose in addition that

lim udx) = ° for all k. !xl- OCJ

Then if(u'(x)) converges uniformlyfor x E r, the sequence (Uk) converges uniformly 011 Q.

Proof: From Corollaries 3 and 4, we have for all k and I

sup IUk Q

(if Q is unbounded, lim Ilk - u/ Ixl- x

II/I :( sup I Ilk -- III i r

o

12 We recall that a sequence of functions U;) defined on an (arbitrary) sct converges uniformly if

lim SUplJk - J; I = O. k. I • I


3. Poisson's Integral Formula; Regularity of Harmonic Functions; Harnack's Inequality

Let us begin by proving the

Lemma 1. Let B = B(xo, ro), x E B ;xo] and II E II (Bl n (6 0 (B). Then

(2.9) lI(X) = r lI(t) ~J E//(I - x) - ( .. _I'rJ __ )""'2 E,,(t - Y))dr,(t) .dB (11 \ Ix - xol

where

(2.10) r~

r=x+ (\ x) . 0 I·· x I" -. () . . - Xo •

Prool We can always suppose that 1I E ((, ,: (Ii): it is then enough to apply the result to Br = B(xo, r) for ° < r < ro and pass to the limit with I' --> roo We note first of all that Y ¢ B, with the result that the function v(t) = E// (t - y) is inYl (B) n (f, ~ (.ih Applying Green's formula for harmonic functions (2.4), we have

(2.11 ) y) t'U ) E,,(t - y) -::;. - (t) d·I'(t). en

Now let us consider 0 < (; < 1'0 - i x - Xo I and the open set

Q =; I E B ; It - xl > I:: : thefunetion t( t) = E" (1- x) is inYr (Q) n ((, ~ ([2). The boundary ciQ is made up of a B and a B (x, I:). Using the fact that v is constant on n B (x, /:) and Gauss' theorem for 1I, we have

Making use of the value of grad t.//, we have (see the proof of the formula of the mean):

Applying Green's formula for the harmonic functions 1I and l' on Q, and passing to the limit as r. --> 0, we obtain:

f ( elI: . ?u ) u(x)= . lI(t):;-~(t-x)-I:,,(t-x),-(t) d}'(t).

tB ('/1 I'll

1 J We recall (see ~ IAI that x; is the unit sphere of J'I" and d(J the surface measure of 2:.

~2. Harmonic Functions

Combining with (2.11) we deduce that for all Cf. E IR:,

f ( cEn CEn) d ) u(x) = u(t) (t - x) - ct~(t - y) y(t DB en un

We can show that for

f au - (En(t - x) - ctEn(t - y» ~ (t)dy(t) .

DB en

ct = ( ..... /"0_)" 2

Ix - xol

249

the second integral is zero, which will prove the lemma. This comes from the identity

(2.12) 1i=;r = I x ~o Xu I ' which we can verify by making use of (2.10). In the case n ~ 3, we therefore have

Vt E oB

En(t - x) = exEnU - y)

and the result is proved. In the case n = 2, we have ex = I and

c = 217[ LOgl~~ Xo I, with the result that the second integral reduces to

J" au c I'Bon(t)dy(t)

which is zero from Gauss' theorem. o Making explicit the right hand side of(2.9), we prove now Poisson's integralformula (for a ball):

Proposition 9.

(2.13)

Let B = B(xo, ro ) and U E .yt'(B) n 'fj0(ii). Then,flJr all x E B,

1 j' r6 ..... Ix - xol2 u(x) = ----- u(t)d}'(t) .

roan 8B It - xl n

Remark 5. We can apply Remark 3 on the non-necessity of the condition u E (1:'°(13) (as for the formula of the mean). With respect to the formula orthe mean, Poisson's integral formula is concerned with expressing the value of the harmonic function 11 at each point of B as a function of its values on (1 B; it is allied to Cauchy's integral formula for holomorphic functions, but has the inconvenience as far as this latter result is concerned of applying only for a ball. In the meantime we shall also obtain Poisson integral formulae for the exterior of a ball and for a half-space (see

25{) Chapter II. The Laplace Operator

Propositions 21 and 22), without forgetting the case II = 2 where we are able, at least in theory, to obtain integral formulae for any simply connected set whatsoever (see the end of ~2.51. 0

Proof Wc resume the notation of the lemma and its proof. We have

rE" . . (t -- xl.II(t) ~--(t - x) = gradl:."u - X).II(t) =- -. III IT" I t - X I"

Also, laking account of (2.10), (2.12) and (I

rl~(I _ .1') = (I - y).n(lI, 1_ (·I.~=- .\:01 •. ')"[ro -- _I'(~ 2(' 'o).nUll .. ' III ITrll-l'l" IT,,roll,1 Ix xul

Hence, taking account of the value of Y.

.1')

r~ ~ 1x

ro IT" It Taking this back into (2.9), wc obtain (2.13).

) .'1'0 -

xl"

o First of ail, let us apply Poisson's formula to give a direct proof of the analyticity of harmonic functions. We give a statement a little more refined than will be useful in the sequel:

Proposition 10. Let B = B(xo. 1'0) he a hall and (ud a sequence ofOfilllctiolls, harmonic on B and continuous 011 B. We suppose that (Uk) COlJl'eryes ullifrmnly 011 ?B when k ---+ CL. Then (ud conrerges ullif()rmly 011 B, its derivatives converge uiliformly Oil ('per}' compact set of B alld its limit is harmonic and analytic.

We note the corollary:

Corollary 7. l:.rcry continuous fUllctiull solution ill the scnsc of distributiol1S of Laplace's equation Oil all open set Q of IR: ", is harmonic analyt ic 011 that open set.

Proof of Proposition 10. The uniform convergence of (ud follows from the maximum principle; from Corollary 4, m,!xluk - 1111 ~ maxlu k - uII. To

B eB

simplify the notation, we suppose that Xo ~C. 0,1'0 = L which takes away nothing from the generality of the proof. Poisson's formula for the functions Uk can be written

(2.14) o I JO' 1 ..... 1.'1'12 . ) = . -- . . -.-,./i (CT)da

(In 2:(1 - T(x, IT))" 2 k

with T (x, (J) = 1 - lIT - X 12 .

For (x, IT) E 13(0,1 - r.) x L, we have

!(J - xl ? f. and hence I T(x, IT)I ~ 1- ,,2 .


Since T is a polynomial (of second degree) in (x, cr), developing ( I T 2 ) - "/2 in a power series on {I TI ~ 1 - C;2}, we obtain a development in power series of

k(x,cr) = (1 - Ix1 2 )(1 - T(X,crW n ;2

converging normally on 8(0, 1 -- t:) x E. It follows from (2.14) by integration of this development, that Uk can be developed in power series which are convergent normally on B(O, 1 - e) and that these power series converge normally when k --> cc; in particular all the derivatives of Uk converge uniformly on B(O, 1 - £) when k --> x. This clearly proves the proposition. 0

Proof 01 Corollary 7. Let us consider U EO (g (Q) with

f u,;J(dx = ° for all (EO 9(Q),

It is enough to show that the function u is harmonic on every bounded open set Q'

with {i' c Q, We fix such an open set Q' and consider 0 < ro < dist (Q', oQ). We regularise u by convolution: that is to say we prescribe a function p EO 7(IF£") with P ~ 0 and such that

fPdX = 1 and suppp c B(OJo ) '

For all k ~ I, we put Pk(X) ''''' k"p(b) with the result that Pk E 'Y(IR"), and we have:

and ( ro \ sUPP Pk c B 0, -r ) ,

We then put Uk = II * Pb that is to say

IIk(X) = fU{YlPd\: y)dy fU(X - .yjpk(y)dy

which is well defined for all x EO Q', We have Uk EO (f, 2 (Q') and [rom the hypothesis

f U(y)!1Pk(X - y)dy = ° for all x E Q' ,

that is to say Uk EO Jt (Q') ,

We then obtain the result as a corollary of Proposition 10, since Uk -> U uniformly on Q', This classical fact is immediately obvious by

!udx) - u(xll = If(U(X - y) -- lI(X))Pk(Yldyl ~ max lu(x - y) !yl ~ rt

lI(x)l·

D

Let us now apply Poisson's formula to prove H amack's incqualit y (for a hall),


Proposition 11. 1'1 = IXII. 1'2

Let B = B(xo , 1'0)' U E ,#(B) with u ? 0, XI' x2 E B. Denotin!J IX21, we have

(2,] 5)

Proot: We can always suppose U E (6 (B), it is sufficient in effect to apply the result to the ball B(xo , r) with max(r l • 1'2) < I' < ro and pass to the limit with I' -+ 1'0'

We thus have Poisson's formula

u( x j )

Now, for I E DB.

with the result that

1'0 - rj ( 1~--+-r~)n=l

We deduce that

r °

L 2,

"0 + rj r;:;;-~~y~-f .

Multiplying by U(I) ? ° and integrating, we obtain the resulL o

Considering B = B(xo.,.o)' li E ,11 (B) with 11 ? 0 and 0 < I' < I'{). we have by application of (2.15).

(2.16) for all

which can just as well be written

max II c(

III '''.r) ( r + r)" o . . mm 1I,

r 0 - I' / HI.' r) '"

By an argument of connectedness, we obtain as a generalisation the H amllck illequalityfi)1' a cOllllected open seC

Proposition 12. Lei Q be a conllected open set and K a compact suhset of (2, Thl!re exists (/ constant C such that jill' all U E .# (Q) with 11 ? O.

(2,17) maxu c( Cminu, K K

Proof Using the connectedness of Q and the compactness of K. we can find a finite sequence of balls B(XI' 21'1)"'" B(x N • 2rN ) contained in Q such that the balls B(xl.rtl".,. B(X:v,l'N) cover K and arc connected two by two: S{x;, I' j l n B(xj + I' I'j+ I) contains a point xj forj = L .... N - L


We show that the proposition is satisfied with C 3nN• That reduces to showing that being given u E .~ un with u ~ 0, for all x, Y E U B(x j , 'j)' then

j

u(x) :::; 3nN u(y). The worst situation will arise when x E B(x l' r1 ), Y E B(XN' rN); applying (2.16) we have

u(x) :::; 3"u(x'j), u(x'Jl ~ 3"u(x~), ... , u(x~ _ 1) ~ 3"u(y) . D

We note that Harnack's inequality implies the principle of the m~ximum: if u E .~(Q) and u(xo) = minu for an Xo E Q, changing u to u - u(xo), we can suppose u(xo) = 0; then Proposition 12 shows that max u ~ 0 and hence u = °

K

on every compact set containing Xo; obviously then u = u(xo ) on Q. Harnack's inequality is a much more powerful tool than the principle of the maximum, as will be shown in the proofs of the following compactness theorems called Harnack's theorems. Given an open set Q and a sequence (ud of harmonic functions on Q, it follows from Proposition 10 that ij"(ud converyes unif()rmly on every compact set of Q, then all the derivatives of Uk also converye uniformly on every compact set ofQ and the limit is harmonic on Q. We then say that Uk converyes in Yi'" (Q).

Proposition 13. Let Q be a connected open set and (Uk) a sequence of harmonic functions on Q. Suppose that: (i) there exists a function Uo harmonic on Q houndiny the functions Uk helow (1.1 0 (\") ~ udx)for all x E Q and all k), (ii) there exists Xo E Q such that (Uk (xo» is bounded abO/ceo Then there exists a suh-sequence converyent in .ff( Q).

Proof Replacing Uk by Uk 110 we can suppose Uk ~ O. Since convergence in ff (Q) is uniform convergence on every compact set, from Ascoli's theorem 14, it is enough to prove that, for all x E Q,

(a) (uk(x» is bounded, (b) the functions Uk are equicontinuous in x, i.e. that the limit lim udy) = udx) is

J~X

uniform with respect to k. From (ii) and Harnack's inequality, the sequence (Uk) is bounded on every compact K of Q: in eflect we can always suppose Xo E K; from Proposition 12 we then have

maxuk ~ Cminuk ~ Cudxo) ~ C' . K K

Let us hence prove (h); to simplify the notation, we take x B(xo , r) < Q we denote

mill Uk'

8(x(I' r)

max Uk'

8(x". rl

xo. For r > 0 with

14 For the topological ideas used here, we refer to Dieudonne [1] or Schwartz [2]. See also Lemma 4 of ~4.4.

~54 Chapter [I. The Laplace Operator

Applying (2.16) to the functions Uk mdr) and Mdr) Uk harmonic and positive on B(xo, r), we obtain the inequalities:

I r ) 3" (/Ilk (;)

\

'V[k ( 2 1II,(r) <- Iflk{r)) ~,

/

IIlk ( ; )

I

Mk (~)): AIdr) <- 3" (AIdr) '"

from which, on adding, and putting wd r) B(xo, rl, we have

iHdr) - mdr), the usc illation otUk 011

wdr) + (l)k (~) :( 3" ( wk(r) - OJ, (~)) :

which is eq uivalent to

Uh (;) :( Owdr) with Ii

By recurrence for all j = L 2 ....

(2.18)

3" - I < 1 .

3" + I

Since ,lvI, (r) is bounded. and 0 < Ii < I. we deduce that W k ( ',1'., --> 0 as j -. ~ 21 )

uniformly with respect to k, which proves (h). 0

We note the corollaries:

Corollary 8. LeT Q he a hounded Opl.'ll set and (Ll k ) (/ SI.'(jw'ncl.' o!:tilllctiolls harmollic 011 Q and cOlltillllOliS 011 Q. satis(rill?j

Theil till.'rl.' exists (/ slIh-sl.'(jllI.'IlCI.' of( lid cOlll'l.'ryilly ill ,11 (Q).

Proof From the maximum principle Vie have

IU k (x) I :(:\1 for all ,x E Q and all k

and hence in particular the conditions (i) and (ii) of Proposition 13. In the case Q is connected. the corollary is proved. In the case of an arbitrary Q. it suffices to apply Proposition 13 and use a "diagonal procedure"ls since Q has at most a denumerable number of connected components. 0

l' The diayullal f'roc~dllr(' i, the follmving topologicallefllIl1a: given a double sequencc (auk, (of a topological space) satisfying: for all fixed I and for every suh-sequcnce (a",), we can extract a convergent sequence: then there exists an increasing scqucnce (kil, such that for alii. thc sequence (a" ,I, is convergent.


Corollary 9. Let Q be a connected open set and (Uk)k a monotonic sequence (increasing or decreasing) of functions harmonic on Q. Suppose that there exists X o E Q such that (udxo» is bounded. Then the sequence (Uk) converges in £(Q).

Proof Considering for example (ud increasing, we have the conditions needed for the application of Proposition 13. But the sequence being monotonic, if a sub-sequence converges, the whole sequence converges. We note that the monotonicity of the sequence permits a direct proof: for a compact set K of Q (containing xo), applying (2.17) to Uk - u/ «Uk) increasing k ~ l)

Since the sequence (uk(xo» is increasing and bounded above, it converges; hence (Uk) converges uniformly on K. 0

4. Characterisation of Harmonic Functions. Elimination of Singularities

Given an open set Q of IRn, we have seen that a function u E £ (Q) satisfies: (a) Gauss' theorem: for every regular bounded open set w, with eli c Q,

f au dy = 0; Jew an

(b) the formula of the mean on spheres:

for each point x and r > 0 with R(x, r) c Q, u(x) = ~1 f udy; (ln r J8B(x.r)

(c) the formula of the mean on balls:

for each point x and r > 0 with R(x, r) c Q, U(x)=----,",-"f udx. (ln r JB(x.r)

We propose to show that these propositions characterise the harmonic functions. Let us begin with Gauss' theorem, which we note has a (classical) meaning when u E C(jl(Q). We prove:

Proposition 14. Let u E C(j 1 (Q) possess the Gauss property: for every regular

bounded open set w with eli c Q, f au dy = O. Then u is harmonic on Q. J8<O an

Proof We note that if u E C(j 2 (Q), the result follows immediately from the formula of Green (Ostrogradski): for Xo E Q and r > 0 with R(xo, r) c Q, we have

f Llu dx = f au dy = 0 . B(xo.r) JilB(Xo.r) an

Hence there exists Xr E B(xo, r) such that Llu(xr ) = O. In the limit when r -+ 0, Llu(xo) = O.


I n the general case considered, u is only of class ({, 1, we shall regularize u by convolution (see the proof of Corollary 7) using a radial function p. Hence we take

If. 1 r,,·tp(r)dr =',

n (J" supp p c JO, 1 J '

fI I in the neighbourhood of 0 :

for all k ;::: L we put Pk(X) = k"p(klxi) which satisfies

PkE'/(IR"j, r ilk dx = 1 , ~

I I '\ supp Pk C 8 (0, k ) .

We put Uk = U * Pk which is defined and is of class Y; f on

dist (x, r12) > ~} . k

We have for x E 12k

!judx) = f U(y)Llpk(X - y)dy = f gradu(y).gradpk(x .. yjdy

k"Tl gradu(y). -(klx - yl)· _cody f dp x - r

dr !x - v I

= kid 1 . (kr)r,,-l dr gradu(x - m).(Jd(J. 11 k dp f n dr E

But

JI' 1,1 "1" f (~!I d'I' -- 0 . :-gradu(x- mj.O"d(J = . .: r ,-B(x. rJ I'll

Hence Lluk(x) = O.

In other words, Uk is harmonic on 12k , From Proposition 10, since Uk --> II uniformly on each compact set of Q, U E .ff( 12). D

Let us now consider the case of the formulae of the mean. We have seen in the proof of Corollary 2 that the formula of the mean on spheres implies the formula of the mean on balls. We shall therefore consider only this latter case. Note that the mean on a ball B of a function u can be defined when u ELI ( B). Hence, let us consider a function u, locally intewable on 12 and for all x E Q and 0 < r < dist (x. (~Q j, define the function M by the equation

M(x, r) = /J r u(yjdy, anr tf alB(x. r)

the mean of u integrable on B(x, r)(8(x, 1') C 12). We can also write

/J f M(x, r) = 1I(x + ry)dy (In H

(2.19)

~~. Harmonic Functions

where B = B (0, I). We see therefore 1h that the function

(x, r) -> M(x, r)

~57

IS continuous on the open set U = [( x, r); x E Q, 0 < I' < dis (x, aQ)j of Q x JO,%[. In the case U E ((; 2 (Q), we have M E (f, 2 ( U) and by differentiation under the integral sign and the use of Green's formula, we obtain

J,M(x,r) = n f ...1u(x + ry)dy = ~ f grad u(x + m). O"du. Un B Un E

In the same way we have

tl'v! 11 f -l-(x,r) = - gradu(x + ry).ydy (. r u" B

(2.20)

II I'r If' = -.;;-+1 J ,"d, gradu(x + Tu).udu.

unl 0 .;E

Therefore M satisfies the partial differential equations

(2.21 ) JxM = 1 C.'.' (I''' + 1 ?!;!) on U. 1''' ?r cr

For U ollly in L loc (Q), M always satisfies the partial differential equation (2.21) provided that it is taken in the sense of distributions in U. This follows from a general argument of the density of (f, 2 (Q) in L 110e (Q), of the continuity u -> At of L loe (Q) into (6 ( U ) and of the continuity of the derivatives inJ" ( U ); the reader will also be able to verify it by duality. Now let us suppose

(2.22) M(x. r) = u(x) a.e. (x. r) E U .

In the case U E <;s2(Q). we deduce immediately from (2.21) that LlxM(x. r) = 0 for all (x. r) E U and hence u is harmonic. In the general case u E L loe (Q) this is again true by using the following reasoning. From (2.22) ? M /?r = 0 in ~l '( U ) and hence from (2.21). Ll xf M) = 0 in .J' '( U ). Consider a bounded open set w wi th w c Q,

U :::::> (J) x ] o. "0 [ with "0 = dist (w. ?Q); since M is continuous on UJ x JO, 1'0 [.

using the definition of ...1, M in './' ((!) x J O. r 0 [) and Corollary 7, we see immediately that for all r E J o. 1'0 [. the function x E (!) ---> M (x. 1') is harmonic. We have thus proved

Proposition 15. Let u E LI1oe(Q). Suppose that

(2.23)

n f u(x) = --n u(y)dy a.e. (x. 1') O""r B(x. r)

with X E Q. 0 < r < dist(x.?Q).

'" We make use of thc continuity of the translation in L '.

25K Chapter II. The Laplace Operator

Theil there exists a function u, harmonic on Q such lhal

u(x) = I~(X) a.e. x E Q .

Remark 6. We have used only very little of the information we have on the function A1. If !I E (f, 0 (Q). it follows from (2.19) that M E (f, 0 (U u Q x : O}) and

1H(x.O) = u(x) forall XEQ.

Similarly if U E (f, 1 ( Q). then A1 E 'f, 1 (U u Q x : O}) and

? 1\11 IJ JI" -:; (x.O) = -- grad u(x) .\'dy . ('I' (J'" IJ

In other words, M is a solution of the "Cauchy problem"

(2.24)

M( .. O) lIo = U • i'M

( .• 0) ?r

II JI" -- gradu.ydy (J'" IJ

on U

on Q.

Let us consider the case Q = 1M". U = 1M" x R +. We can show (see DelsarteLions [IJ) that a solution of (2.24) cannot come back to two points r 1 =1= /'2 on its initial value Uo without being constant, except for a discrete set of "spectral" values /'1 and /'2' In other words. if for /'1 =1= /'2' not belonging to a discrete set

then 11 is harmonic. o We note the Corollary:

Corollary 10. A locally inteyrahle funcrio/J 0/1 Q. solution in the sense of'distriinltions of'Laplace's equation. is equal, almost et'erywIJere to a harmonic function.

Proot: Regularizing by convolution (see proof of Corollary 7) for every bounded open set Q' with Q' c Q, there exists a sequence of harmonic functions u" on Q'

such that !I" -+ U in L 1 (Q'). Passing to the limit in the formulae of the mean applied to the harmonic functions UrI' we see that 11 satisfies (2.23) on Q'. The corollary then follows from Proposition 15. 0

We shall use this corollary to "eliminate the sinyuiarilies". The problem is the following: given an open set Q and a closed subset F such that Q' = Q \ F is dense in Q (e.g. a variety of dimension m < 11); given 1/ Ell (Q') we seek to extend by continuity II to a harmonic function on Q. If it is possible, we shall then say that we have eliminated the set of singularities F. To specify the problem, we consider Q = B( O. r o }, F = {O} and a radialfullcI ion u harmonic on the punctured ball Q\\{O}. From Proposition 4,

II(X) = cE,,(x) + Co on Q {O}.

~~. Harmonic Functions 259

and 1I extends to a harmonic function on Q iff c O. We see that a sufficient (and necessary) condition to eliminate the singularity at 0 is

if n?: 3 ,

if 11 2,

if 11 = I ,

lim Ixln - 2 u(x) = 0; I xi ---+ 0

u(x) bounded when x ~ 0,

du -- (x) converges when x ---> 0 . dx

We now consider the general case olan isolated singularity: we are given an open set Q of [R", Xo E Q and u E J( (Q \ {x o }). The case n = 1 is immediate: u is linear to the left and to the right of Xo; u and du/dx possesses limits to the left and to the right; u extends to a harmonic function iff these limits are equal. Let us now consider the case n ?: 2 and prove the

Proposition J6. Let Q be an open set in [Rn (n ?: 2), Xo E Q and U E .W( Q\ [xo }). Suppose that the function I x -- Xo I" - 2 u(x) is bounded as x --> Xo' Then there exist c, Co E U~ such that the function i'i defined on Q by

~I( x) { u(x)

Co

c :.~:::::~~1;;:::2

il x E Q\ {xo]

if x Xo

is harmonic 011 Q.

III particular, c = lim I x Xo I" - 2 u( x) exists and if n = 2 or, when n ?: 3, if x -j. Xo

c = 0 the function u extends by continuity to a function harmonic 011 Q.

Prooj: For 0 < r < ro = dist(xo• aQ), we define

1 ~

v(r) =-J u(xo + ru)du . Un L

We know (see Proposition 6 of § 1) that the radial function v(lxl) is harmonic on B(O, ro )\ {OJ. Hence (see Proposition 4 of this §2)

L'(lxl) = ('1 Enex) + Co .

We distinguish then: in the case 11 ?: 3, we take c = c 1/ kn' in the case 11 = 2. since u(x) is bounded when x ---> X o ' it is the same for r(r) as r ---> 0, with the result that ('1 = O. We take c = O.

c Replacing u(x) by u(x)- - -- ....... _ - Co' we can therefore suppose v(r) = 0

Ix - Xo I" 2

and we have to prove that u, extended by 0 for x = Xo. is harmonic on Q. Tn fact u, defined almost everywhere on Q is locally integrable on Q since I x - Xo 12 - n is locally integrable on [Rn. From Corollary 10, it is sufficient to show that u is a distribution solution of Laplace's equation on Q: in effect there will then exist u,

260 Chapter Il. The Laplace Operator

harmonic on Q such that u(x) = ii(x) for almost all x E Q and we necessarily have

ii(x) = u(x) for all x E Q : xo:

1 ~

ii(xo) = . J ii(xo + rCT)dCT = O. O"fI :.:

We therefore suppose l'(r) = 0 and show that given:; E U (Q)

f 11 Li:; dx = 0 .

Taking P E '7 (iR) with 0 ~ p < I, supp pc] -x, IJ and p

bourhood of O. For k ?: I we put

:;dx) = :;(x)p(klx xol)·

1 in the neigh-

We have :;k E '/ (Q) and :;k = :; in the neighbourhood of Xo; since u is harmonic on Q \, : xo}

We have

I d' d ') + k 2:;(,)_:. __ (rn~l_ p(klx r" I dr dr /

xol)

and

( 1 \

SUPP:;k c B,o' k) . Using the change of variables .\ --+ k(x - "0) \ve have

f' • /.)( / '

J ui1:;dx = J lI(xo + .:~ < p( .\'1).1:;1 Xo + Y) HIO. I I 1\. 1 \ k

y) k

1 o( i r i) \. - . d ) .) dr dr ' '.. J k II

From the hypothesis 11I(xo + y/k)1 ~ Ck" ~ 2 for r E B(O. 1). it follows that the first two terms in the integral tend to 0 when k --+ x; hence

fULi:;dX = .lim f II(XO + ~):;(xo + y\ _I _ d k ~ 1. HIO. 11 k, k,1 r" ~ 1 dr

x r"~l_ p(lyl)~--'=-i (I d ) dr

dr k" -

§L Harmonic Functions

which, on changing to polar coordinates, we may write

f 1'1 d ( d ) I' ( u,,1(dx = lim J - r,,-1 ~(r) drJ u Xo k~ x 0 dr dr r

Using u(r) = 0, we have

with

"k = (In sup Ix - Xo I" 2Iu(x)1 sup I ((x) - ((xo)1 . Blx".;) B(x".t1

From the hypothesis, lim "k = 0. Since

e I_~ (r" - 1 dp (r l) I __ d~' < en , Jo dr dr r" 2

we deduce that

J u,,1( dx = 0.

261

o

These techniques can be extended to the case of a variety in Q. We shall first of all consider a hypersurjclce V of Q. We suppose that V is regular and separates Q \ V into two open sets Qt and Q _ slich that

Given a function u defined on Q\ V we can consider its restrictions 11 + and 11_ to Q+ and Q_ respectively. If 11 E ((,0 (Q \ Ii'), we can consider the traces u + and 1/_ on V if they exist: for Xo EO V

lim 1/(X) , lim u(x).

The jUllction u call be extended by continuity to Q ifTu + and u ... exist and are equal on V. Similarly, if U E ((, 1 (Q \ V), we can consider the normal derivatives of u + (resp. lL) to V exterior to Q+ (resp. Q_ ): if they exist, we denote by aul (!n + and aul cn_ these Ilormal derivatires. If u EO '{,- 1 (Q), it is clear that eu/ en + = - (lui CIl_ on V; by contrast, contrarily in the case of the traces of 11 + and 11 being zero, the existence of (1111(~1l +, hl/CIl .. satisfying eulcn+ = - (lu/ cn _ does not imply u E ((, 1 (Q): in effect that says nothing about the existence of tangential derivatives on V. Finally we recall (see § 1.3b) that the existence of normal derivatives cui an +

(resp. (lu/I'IL) implies the existence of traces 1.1+ (resp. u). We now prove the

Proposition 17. Suppose that V is a regular hypersurface in Q and that U E ,if (Q \ V). Suppose also that the normal deriratives (;U / (In + and au / (~n _ exist 011


V and that

11+ = lI_ and on v.

Theil II extends hy cOlltinuity 10 a jilllctioll harmonic 0/1 Q.

Prou{ By the same hypothesis. u extends by continuity to Q. From Corollary 7, it is sufficient to show that u is a solution in the sense of distributions of Laplace's equation on Q, that is to say, for all ( E 'J' (Q)

Jf'UA(dX = r 1I+,1(dx + r lI_A(dx = O. . Ju. Ju

We apply Green's formula (~\, Proposition 4) on Q+; taking account of ( E 'J' (Q+), u + harmonic on Q +, and that (1U +! ('/1 = ('ul 0n + exists on V = (~[L n Q, we obtain

Similarly

Adding and uSlIlg the hypotheses on u and Z(/?n +

f 1/ 11 (dx = O.

?(/ ('IL we obtain

o

We note the corollary of this proposition concerning the uniqueness of" the C auelly prohlem:

Corollary II. Let Q he a regular connected open with houndary r, 20 E rand u E'Yf(Q). We suppose that ('11(2)/('11 exists and {':U(Z)/(i n = u(z) = Ofor all z ill tile neiyhhourhood of Zo ill r. Then II == 0 on Q.

Proo{ We fix r > 0 sufficiently small with the result that

(' 11 o (z) exists and ('11

I'll , (z) = lI(Z) = 0 for all Z E B(zo, r) n r = V. ell

Since Q is regular, V is a regular hypersurface of B(zo, r) which separates B(zo, r)\ V into two open sets Q~ and Qr: such that

V = N2~ n B(zo, r) = cQr: n B(20 , r) ;

we can suppose, for r sufficiently small, that Q~ = Q n B(zo, r). We define LID on B(zo.rl\Vby

Uo = u on Q~, lIO = 0 on Q~.

From the hypothesis and Proposition 17, lIo extends by continuity to a harmonic function ilo on B(zo, rl. From Corollary 7, Uo is analytic on B(zo, r); being zero on


QO , it is identically zero on B(zo, r). Hence u, which is analytic on Q, being zero on Q n 8(:::0' r), is identically zero on Q. 0

Finally, we shall consider the following result for a variety V of dimension III :s; 11 - 2:

Proposition 18. Let V be a closed regular I'ariely in Q oldimellsiol1 m :s; n - 2 and 11 E .ff (Q\ V). We suppose that

lim d(x,v),,-m-2 u (x) =0 forall XoEV17 x .--, Xo

Thcn 1I eXlends by contilluity to a fUllcTion harmollic 011 Q.

Proof The result being local, we can suppose that, possibly by changing the reference frame, that there exists an open set t of [R m and a regular map Y. of (' into [R" - m such that

v = ((x', ct(x')) ; x' E C }

Q = {(x'.x(x')) + t: X'EC. tE[Rn-m.ITI < 6o }.

Given K a compact sub-set of (i, we see. using the regularity of x, that there exists o < Ii < 60 and c > 0 such that

d(x, V) ? clTi for x = (x', ct(x') + t). x' E K.lri :s; b.

In particular the function d(x. V)2 -II + m is locally integrable on Q, since Itl 2 - tI + m

is locally integrable on [R" m Using the hypothesis the function d( x, nil - m - 2 u( x) can be extended by continuity to Q and hence a fimiori is locally bounded; we deduce that 11 defined almost everywhere on Q ( V is negligible for the Lebesgue measure), is locally integrable on Q. It suffices therefore. from Corollary 10, to prove that, for all ~ E CJ (Q).

r lii1 ~ dx = 0 . oJ

We follow the proof of Proposition 16: We take PE t.7(W- m ) with O:s:. p:S; I. suppp c 8(0.1) and p neighbourhood of 0, and we define

~dx) = ~(x)p(kl) for x (x', y.(x') + I) .

We have

SUPP~k C {(X" xix') + t): x' E K, It I :s; ~} where K is the projection of the support of ~ onto [R;"'.

Calculating /,j ~k' we see that

IA~kl :s; Ck 2

where C depends only on ~, p and x but not on k.

10 Where d(\:. n is the distance of the point x from the variety r'.

III the

2M

Then

with


I r l

ei: - j" (' k 2 dt u 11 C dx < ---"- dx' J _ ------- ---

~ k -......:.:, fl - In 2 I I rI - ttl -" 2 ~ (.K HIO.:) t

sup Ix-. >1' I -\- II

\ f' ".111

d ( x, V)" - m- 2 Ii (x) .

The latter integral is independent of k; from the hypothesis, i: k --> ° when k -+'1 .. I'

Hence J u 11 C dx = 0. 0

5. Kelvin's Transformation; Application to Harmonic Functions in an Unbounded Set; Conformal Transformation

Let us recall that an inl'ersion with pole Xo EO 1R" ul1d radius ro > ° is the transformation, denoted by I (xo' ro) of 1R"\ : ° J onto itself defined by

r~ 1(.'0,1"0): x -+ Y = Xo +, - '2 (x - xo)·

IX Xol

Inversion is involutive, that is to say that: I (xo ' r) I = I (xo, r); point by point it conserves the sphere (1 B (xo, 1'0), conserves globally the hyperplanes passing through x o , and transforms the hyperplane passing through x -# Xo and orthogonal

punctured at the pole Xo. The inversion J (xo. 1'0) is an (analytic) diffeomorphism oflR"\iO;

I(x o• I"o)(x) = Xo + r~grad Loglx - xol .

A simple calculation shows that the Jacobian matrix of I (xo' 1"0) at the point x is

(2.25) (/(~l'- \) r~ l(xo , '-o)'(x) = (,_,'_,-),- (x) = I-x- ' T(x

, - Xo I~

with

T(x)

which is an orthogonal matrix. A simple calculation shows also that

- 2(11 - 2)rG 111(xo,l"0)(x) = 4 (x - xo)·

ix - xol


Using Lemma 1 of § 1.1, we see that given a function u defined in the neighbourhood of y = J (xo , ro)(x) and twice-differentiable in y, we can show that w(x) = u(l(xo.ro )(x)) is twice differentiable in x and

r4 2(n - 2)r~ (2.26) ,jw(x) = 0 Llu(y) - grad u(y). (x - xo) .

. Ix - Xo Ix - Xo

This relation leads to

Definition 3. The Kelvin transformation of pole Xo and radius 1'0 > 0 is the name given to the map denoted H(xo, ro) which maps anyfi.mction u defined on a part Q of ~~"\{O} 10 a v = H(xo,ro)u defined on J(xo,ro)-l(Q)=J(xo,rol(Q) by

rZ- 2 u(l(xo,rolx) v(x) = ~1---1'~2--

x - Xo

It is immediate that Kelvin's transformation is involutive: if v = H(xo, ro)u, then u = H (xo, ro) l'. Thus Kelvin's transformation leaves inuariant the fimctions defined on the sphere t:B(xo , 1'0). We now prove the

Proposition 9. For a function u defined in the neighhourhood of y # Xo and twice differentiable at y, the Kelvin transformation H (xo, ro)u is twice differentiable at x = I(xo,ro)yand

(2.27)

Proof Since the function rZ- 21x with the above notation

Xo 12 - II is harmonic on ~n \ {xo }, we have

n - 2 ro LlH(xoJo)u(x) = -----_=_ Llw(x) + 2grad ---

Ix - Xo I" 2 Ix

From (2.25)

n - ! r 0 -

Xo I"

r2 gradw(x} =-°2 T(x - xo}·gradu(y),

Ix--xol

and hence since T(x - x o ) is orthogonal,

2. grad w(x) .

rZ - 2 (11 - 2) rZ grad ----, -_~ . grad w( x) = + ----- - -- grad u( y). (x- Xo 1 ,

Ix - xoi" - Ix - xol"+ 2

which, as a consequence of (2.26), prove (2.27). o We deduce immediately the illl'ariance of harmonic fimctions under Kelvin's transformation:

Corollary 12. Kelvin's transformation H (xo, ro)u of a function u, harmonic on an open set Q ofW\ {xo}' is harmonic on I(xo' ro)(Q).

Remark 7. For a distribution u on an open set Q of ~n \ {xo }, we can define its Kelvin transformation H(xo, ro)u: w = ucJ(xo, ro) is defined as the distribution image 1 (xo, ro) - 1 U and the function I x - Xo 12 -" being analytic on ~n\ {xo }, we

2h6 Chapter II. The Laplace Operator

can def1ne the product rI - 2 ro r = ---- ---- -IV

Ix Xo In - 2 .

The formula (2.21) is obviously still true for distributions. o Inversion maps the point at infinity into its pole Xo so Kelvin's transformation is a

very useful tool for reducing problems in unbounded open sets to problems in bounded open sets. We give here some applications of this method. First of all we shall apply Proposition 14 to the sinyularity at infinity of the exterior prohlem, that is to say, of the problem on an open set Q of [Rn such that [R" '\ Q is a non-empty bounded set: such an open set is then unbounded; it contains also all x E [R 11 with I x I sufficiently large: also its boundary r is bounded (and hence compact). We now prove the

Proposition 20. Let Q he all open set of [R" with II ~ 2 and [R" \ Q a IlOn-empty hounded s<'1 lIlId 1/ E -Yf(Q). Suppose u(x) is hounded as Ixl --> X_. Then there exist c and Co E [R such that

I/(x) = C + - -- - + () -- - ,,,hen Ixl -->X tX . Co (I) Ixl,,-2 1-<,,-2

Proo( Let us fix x ll E [R" {j and denote Q' = 1(xo, I)Q u(xo }: this is all open neighbourhood of Xo. From Corollary 12, the Kelvin transformation, r ::= H (xo. l)u is harmonic on Q! \, : Xo :-_ Since u(x) is bounded when Ixl -----* x I

the function 1.1' x o l"- 2 1'(J) = u(/(xo' 1).1') is bounded when y ---> Xo. Using Proposition 16, there exist c Co E [R such that

C 1'( r) = --- + C + (i( 1) when y -> x() . . ly-xol"- 2 ()

Returning to 11 = H(xo , I)l', we have

( 1 \

II( x) = c -+ Co _ ----j + () - - -- ) I x - xol" - - I x - Xo I" - 2

when Ixi -> x ,

from which the result follows, since

, 1 ) ()--eXIIlZ

when ixi --> x . o

We apply the same method to obtain Poisson's intewal.fi)rmuia Ii)r the exterior of' a hall:

Proposition21. LetQ = [R1l\.Bbeti1eexteriorojahaIlB = B(xo,ro)o{[R"alld U E ,),{,(Q) n (f,o(Q). LeI us suppose u to he hounded and write c = lim u(x)

Ixl ---)- f

1 H We recall that given two functions f and q deli ned in the neighbourhood of a point x" we denote by

/(x) = o(q(x» when x ~ Xo the situation in which lim U(x):'y(xj} = O. The limit ixl --+1

corresponds to Yo ·1_.

~2. Harnl0nic Functions 267

(which exists by Proposition 20). Then

(2.28) u( xl ( I _~A=~_ .... )c + .... Lf Ix -- xol2 - r6u(t)di:(t) . Ix - xol"- 2 ro(J" i'B it - xl"

/(Yr all x E Q.

Proof Let us use the inversion J (x 0, r 0): given t E (lB, x E Q and y we have Y E B and (see (2.12))

Hence

Ix - Xo 12 - r6 it - xl"

fO ...................... It Iy - xol

r6 - Iy - Xo 12 .----.. -- X

It - yl"

YI·

Iy - Xo In - 2 ----------------n--=-i

fO

In particular, applying Poisson's integral formula for the constant function v(y) = c (see Proposition 9), we have:

I j'IX-xoI2-r2 !y-xol"-2 1 r r6-Iy-xo I2

~-;;,: PB' It -xl" 0 cd/,(t) =, --r~- r0r:.;(lB It ._ }~In" cd;:(t)

Replacing u(x) by u(x) - c, we can therefore always suppose that c = O. The Kelvin transformation v = 11(xo, ro)u is harmonic on the punctured ball B\{xo } = l(xo,ro)(Q); but

lim Iy - x o ln - 2 v(y) = lim r~-2u(x) = O. J' --+ Xo I x I --.. 'f.

From Proposition 16, v extends to a function, harmonic on Q, which we again denote by v, On the other hand VE({j'°(ii), since B\{xol = J(xo,ro)(th We thus have the Poisson integral formula for v:

1 J~ r6 -. Iy - xol2

v(y)=-- . .. " V(tjdi'(t) ro(J" i'B II - yl

i X - xol . x - -'0 ..... ()u(t.)d'r'(t) . I ,n-2 I r I . 12 r2

.... -;:z 2-- ~o (J" ~ fB -Tt-- .~~-

Returning to u, we obtain (2.28). o Finally, using Kelvin's transformation in the proof of Poisson's illtegraif()rmuia.for a half space, we have:

Proposition 22. Let Q = {x E IRn; (x - xo ). e > O} with Xo E IR" and e a unit vector, be an open half-space, alld let u E .Jt'(Q) n (g,O(Q) he bounded on Q. Then

(2.29) u(x) =~(:~_::::-_x_ol~ f ~t)d}'(I~ .for all x E Q. (J" rlt-x+xol


Note that, by a change of the reference frame, we can always take

Q = ix = (x',x,,): x' E IR,,-l, x" > 0:,

that is to say, Xo = 0 and e is the 1l-1h of the base vectors of the reference frame. The formula (2.29) can then be written

(2.30) . 2x n II' 1I(x"x,,) = --

('ill .J

u(t) dl

It is clear that the integral has a meaning since 11 is bounded and t -+ (Ie x'! 2

+ X ~) - ".2 is integrable on IR n - 1 for all (x', XII) E Q.

Finally we notice that by the change of variable t -+ (t ~ X')i Xn we can write the formula (2.30) as

(2.31 ) ~ ('

1I(x', Xli) = -~ J O"n ;:,r:" 1

lI(X' + t-,,,)dt

Proof of Proposition 22. We can suppose Q = :(x', x,,): x' E IR" t, Xn > 0:. From formula (2.31), it is sufficient to prove (2.30) (or (2.31)) for x' = 0, Xn = I. We consider the inversion J (Yo, I) with \'0 = (0, ~ I): it transforms Q to the ball mho,}), the plane 1 =:(x',O):x' E R"-I: to the punctured sphere (":B· :.1'0) and the point (0,1) to ho. The Kelvin transformation l' = HU'o, 1)11 is harmonic on B and continuous on B\ {Yo:. We prove that although r is not continuous at Yo E tB. we have all thc same the formula of the mean (see Proposition 6 and Remark 3)

(2.32) ( 1") ~n-l f r .~) = -._.. r(s)di'(S) .

\ - / UII tH

Admitting this formula for the moment and returning to 1/, we shaH have

1'(.1'0 2) 11(0,1) = 1" 2

But carrying out the change of variables s -> t = I (.1'0' I )S, making use of (2.25)

1 and I., ~ \' I = ---.-~.-.~ . we shall have

.01 (1 + II! 2)!

I ('

11(0, I) = .. I an .)

This calculation proves the proposition with the reservation that (2.32) is satisfied: it shows also that the integral in (2.32) is well defined. To justify (2.32) it IS sufficient (sec Remark 3) to dominate the function r;( s) = r(,P Yo + (1 ~ i.)s) by a function integrable on {'B, Now. since 11 is bounded

c

)) . 0 ( \' , 1

with C sup 11 Ii

§L Harmonic Functions 269

On the other hand

/>;0 + (1 n(~o-s)/ ~ /Yf/ -~ for (YO _ s) Yo >- 0 2 22---

I~o + (1 _ )")(~o - s)1 ~ Iyo - sl for (~o-s)Yi ~ 0;

hence

Iv)Js)1 ~ C ( 2n - 2 + Iyo _ sin 2)'

From the above calculation, Iyo - sI2-", which is the Kelvin transformation of the constant function 1, is integrable on oB. We have thus dominated Vic(S). 0

Remark 8. Using the homogeneity, we have established Poisson's integral formula in a half-space, by applying Kelvin's transformation to the formula of the mean. Returning to the ball, we thus obtain another proof of Poisson's integral formula in a ball. 0

Let us now consider the particular case n = 2. Kelvin's transformation is then simply the inversion map J(xo, ro)

(do not forget that J (xo, r 0) - 1 = J (xo, r 0»' In particular, the harmonic functions are invariant under inversion. In fact more generally, we have

Proposition 23. Plane harmonic functions are invariant under conformal transformations. More precisely, given an open set Q of 1R2 and a conformal transformation H defined on Q, the function u defined on Q is harmonic iff its image Hu = u c H -1 is harmonic on H(Q).

We recall that a conformal tran.~formation of an open set Q (of IR" in general) is a diffeomorphism H of Q into IR" preserving the angles; given two (regular) hypersurfaces II and I2 of Q meeting at an angle e 19 at the point x, their images H (1'1 )

and H(I 2 ) meet at H(x) at the same angle e. A linear conformal transformation is a (linear) similitude, product of a homothetic transformation by an orthogonal transformation. It is clear that a diffeomorphism H of Q into IR n is a conformal transformation iff its derivative H' (x) is at every point x E Q, a (linear) conformal transformation, that is to say, if

(2.33) H'(x) = c(x) T(x) for all x E Q

with c(x) > 0 and T(x) an orthogonal transformation.

'9 The angle between two hypersurfaces L, and 2:2 meeting at x is the angle between the normals to L, and L2 respectively at x: the angle between two straight lines D1 and D2 being the number (} E [0, ~"J such that cos fI = Ie, .e2 1 for e, and e2 unit vectors of D, and D2 respectively.

no Chapter II. The Laplace Operator

We return now to the case II 2. If we identify topologically

[R" = :(x.y): with f: = (:: = x + ir:.

a transformation of an open set Q E [R 2 into [R 2 is identified with a (complex) function of a complex variable. If we put

HI\ + ir) = U(x.y) + iV(x,r).

the condition (2.33) is written

( ('U)2 (·rU'2 (d"2 /d/)2 ('X + (~.I') = ,/x) + ( /i.1"

, ;"U c-

i'x

that is to say

l'U ;'V I'U (' V

rv -

(X

(; - I: - with (; = ± ('X i\ I'y ('X

(oU n' + ---- 0

(y cy

1

We recognize there the Cauchy -Riemann equations: H or its conjugate li = U - i V is (locally) holomorphic. I n other words: a conformal transformatio/1 of the plane [R 2 is (to within Ii symmetry) a 11O/olnorphic trolls/ormation of (:20.

We can now give the

Proof oj Proposition 23. Let us consider 11 harmonic on Q and H a conformal transformation of Q. The problem being local. we can suppose (see Proposition I)

that 11 = Ref where f is holomorphic on Q.

Also the Laplacian being invariant by symmetry, we can suppose that H is a holomorphic transformation. Then H u = Re (f H - 1 ) is the real part of a holomorphic function: hence Hu is harmonic. 0

In a similar way to the holomorphic functions. the conformal transformations are a very useful tool for the study of the Laplacian in the plane. Principally by the use of the theorem or collfiJrlnal represelllation: every simply connected open set Q of the plane, distinct from the whole plane. is an image of the unit disk by a conformal transformation H. This theorem and Proposition 23, gives in theory an ""integral formula" for an arbitrary simply connected open set Q. a formula giving a conformal representation II mapping the unit disk onto Q. We can meanwhile refer the reader to dictionaries of conformal transformations for particular open sets (see Kober [I], Nehari [I], Laventriev-Shabat [IJ). Let us give an example of the use of Proposition 23.

Example 4. Let us consider the holomorphic transformation H: :: ---+ I + :::2'

We verify easily that it maps the semi-disk

Q = (::: Im:: > O. i:::1 < 1: onto the half-plane ::::" 1m::' > 0:. the punctured semi-circle

,n The equivalence is true locally or more generally in a simply connected open set.

*2. Harmonic Functions 271

onto the part {x; I x I ~ I} of the real axis and conserves (globally) the interval {x; - 1 < x < I} of the real axis. Given U E Jf(Q) n ~O(Q) the image by H, v = Hu is harmonic on the half-space H(Q), continuous and bounded on H(Q).We can then apply to it Poisson's integral formula for a half-space (Proposition 22).

, ., Y' f oc, v(t)dt u(x + zy) = - (t _ ')2 + ,,2 for all x' E ~,y' > O.

n -00 x }

Returning to u we shall have

u(z) = 1m H(z) foo u(H -1(t» dt n -00 It - H(z)12 '

which gives, after an elementary calculation, Poisson's integral formula for the halfcircle using the polar coordinates

x = r cos e, y = r sin e, 0 < r < 1, 0 < e < n

that is

u(x,y)

with o

6. Some Physical Interpretations (in Mechanics and Electrostatics)

6a. Elementary solutions and the laws of Newton and Coulomb

We go back to the examples in mechanics and electrostatics of § 1.2. In mechanics, a point mass mo placed at a point Xo E ~ 3 is modelled by a Dirac distribution mo(jxo' From this we have said in § 1.2, it has given rise to a gravitational potential Vg in ~ 3, solution of Poisson's equation

(2.34) L1Vg = kgmo(jxo in £t)'(~3).

A solution of (2.34) is

(2.35) 4nlx - xol .

We note from Proposition 4 that vg , given by (2.35) is, to within an additive constant, the only radial solution of (2.34); we shall see 21 also that Vg is the only solution of (2.34) tending to zero at infinity.

21 See Proposition 3 of §3; it follows from Corollaries 12 and I that t'. given by (2.35) is the only locally integrable solution of (2.34) tending to zero at infinity.


With this gravitational potentiall'g, the force exerted on a point mass III placed at a point x E IR 3 is then given by

kgmmo x - Xo --------

4n: Ix xol3 fIx) = -- 111 grad t'q(x)

We thus obtain Newton's law: a point mass /no placed at point Xo E IR J exerts on a point mass m placed at x E IR J an attractit'e f(J/-ce of intensity inversely proportional to the square of the distance I x - xo:. We note that conrersely we can "recover" the gravitational potential Vg and Poisson's equation starting from Newton's law. More precisely, let us now take as our starting point the most directly physically comprehensible affirmation of Newton's law: a point mass 1110 placed Xo E 1R3 exerts on a point mass m placed at x E IR J, [xo } an attractive force (carried by the straight line joining Xo to x)

proportional to the masses 1110 and Ill, that is to say of the form

x - Xo j(x) = - lIlomh(x) --

1\ - \01 l1(x) > O.

The work done by the forcef(x) when we displace the mass In from the point Xl to the point x 2 is independent of the path taken iff h(x) depends only on the distance ix - xol; the field E(x) = moh(x)(x - Xo x - xol is then the gradient of a potential v( x) = V( I x - Xo I). To express the of Newton's law, that is to say thatf is inversely proportional to the square of the distance, leads to the result that the expression for V( r) is of the form - kmo/( 4n:1') + c, that is to say from Proposition 4, that l' is harmonic on IR J \ {OJ, or again that r is a solution of Poisson's equation

Wc can develop similar considerations in electrostatics with the following changes: (il a point charge qo placed at a point Xo E IR 3 creates an electrostatic potential 1'(

in U~ 3 satisfying

(2.36)

(ii) We adopt as solution of (2.36),

kci/o ['((,x) = - kcqo~J(.\" - .\"0) =

4n:lx - .\"01

and the electrostatic force exerted on a point charge q placed at .\" EO IR 3 is given by

kci/qo x - Xo f(x) =- q grad vAx) = ------3

4n I x - xol

n X -o

7-1 --,J to x - xol

(iii) thus we obtain Coulomb's law which is of a similar form to Newton's law, with the.essential difference that the electrostatic force is repulsive if q and i/o are of the same sign, attractive if these charges are of opposite signs,

'.2 ;;;, ~ 471:' h,. is the permittivity constant of the vacuum (see ~ 1.2).


6b. Differential Equations (of Poisson or Laplace) and Integral Formulae (Gauss' Law)

We recall here (see Chap. IA § 4.5) that Poisson's equation for an electrostatic field ill'c = - kcp (in an open set Q)

is a consequence of Maxwell's equations

{ div E kcp

curl E = 0

with the assumption that the electric field is derivable globally from a potential:

E = - grad Ve .

This implies that the work W done by the electrostatic force when the charge q is displaced from a point Xo E Q to a point Xl E Q does not depend on the path followed (see § 1.2)23. In addition the integral form of the equation div E = kcp, that is to say

(2.37) f E. n dy = f keP dx Jew (J)

for every open set w (with w c Q) denotes that the flux of the electric field across every "regular closed" surface ow is equal to the total charge contained in the domain w "interior" to the surface Dw (with multiplicative constant kcl. In the special case P = 0 in Q, and for E = - grad Ve , Proposition 14 implies that there is an equivalence, for Ve E ~U(Q) between the integral form (2.37), i.e.

f. eVe dy = 0 Jew on

(expressing that no charge can be found in the domain w, following Gauss' law), and Laplace's equation:

The condition Vc E ~'l(Q), equivalent to E E ,&O(Q)3, expresses that the motion of a point charge in the vacuum does not have a discontinuity in the acceleration. In fact, we know that then the potential, and hence also the field, will be analytic in Q (Corollary 7), which implies that the motion of a point charge ill vacuo is then analytic (it is the same for the gravitational field, and the motion of a point mass).

6c. Principle of the Maximum (Minimum) and Equilibrium

We have seen in Proposition 8 that a harmonic potential u in a connected open set Q cannot achieve a minimum in Q (we also say, in a colourful way that we cannot have potential wells in Q). It is however possible to find points Xo in the domain Q such that grad u(xo), that is to say points at which the force is zero. Such points

23 Note that if the considered field is not derived globally from a potential. this result is false (sec particularly in hydrodynamics and magnetostatics).


are equilibrium points: a charge or a point mass finding itself at Xo at the instant 1 = 0 without initial velocity, will remain at this point for all t > O. In a general way (see Arnold [1], p. 104) such equilibrium points are unstable. Let us illustrate this through the following example. For Q c IR 2 with 0 E Q. the harmonic potential

(2.38)

is such that grad u(O) potential (2.38) are

(2.39)

k, , u = 2 (xl - X2)

O. The equations of motion of a point mass for the

For k > 0, we put (J) = ,/1<; supposing that the initial velocities are zero, the motion is given by the equations

{Xl = X ~ cos wt

x 2 = X ~ ch wt ,

which shows that the point mass moves off indefinitely when x~ f= O. (er. Fig. I.) More generally, we can show in a similar fashion that for every harmonic potential 11, a point Xo such that grad u(xo) = 0 and the Hessian (!2u(xol/cx;i1Xj f= 0 is an unstable equilibrium point. Finally, we note that these considerations can be applied to potentials other than harmonic potentials; in particular, we can obtain the same result for superharmonic potential (see 94.1 and §4.6l.

Fig. 1

~J. Newtoman Potentials 275

§3. Newtonian Potentials

1. Generalities on Newtonian Potentials of a Distribution with Compact Support

We recall (see *2.1) that we have defined E,,(x) = E,,(lxi)

- r if II = I 2

1 - Log r if 11 2IT

if 11 ~ 3 with kIf =" - (n

which is analytic harmonic on IR" \ {O}, locally integrable on IR" and an elementary solution of the Laplacian:

(in the sense of distributions).

It is even (see Proposition 4 of § 2), to within an additive constant the only radial elementary solution. In other words, E" is the only radial elementary solution satisfying

if 11 I, £,,(0) = 0

(3.1) if II 2, Ef/(x) = 0 for Ixl

if II ~ 3 , lim E,,(x) = O. 1 x I·,

From the invariance under translation, we see that for all Y E IR", the function /l(Y) = En(.Y - y) is a solution of Poisson's equation

where ()y is the Dirac mass at y. By linearity, for Y I' ... , .1'" E U;;:", Y. I , ... , Y.II E ~"t the function 1/(x) = Ly.iE,,(x - Yi) is a solution of the Poisson equation

The distribution (measure) f = L" Y.J:" ( x- Vi) corresponds to a distribution of point masses at .1'1' ... , YN of values Y. I , ... , :X.v' More generally, givenl E t' H, a distribution on IR", with compact support, we can define the COI1!'Oilitiol1 distrihution

24 Following usual practice 6' denotes the space of distributions / on iR" with compact support. i.c. I = 0 on : x: I x I > Ro} for Ro sufficiently large.


of Ell byF 5

(3.2)

This clearly has meaning since the convolution En * ~ of the locally integrable function En by a function ~ of class (6 x with compact support is a function of class rr, x :

En * ~(x) = f En(x - Y) ~(y) dy = f E,,(y) ~(, - r) dr.

Definition 1. Given f E' ;}', a distribution on [R" with compact support, the distribution Ell * f is called the Newtonian potential off

We note some immediate properties:

Remarks I (a) En is the Newtonian potential of (j, the Dirac mass at the ongm. More generally, u(x) = 2;CJ. i En (x - J';) is the Newtonian potential off = 2."CJ. i (5'i· (h) The Newtonian potential is linear: if u (, 112 are the Newtonian potentials of /;,/; E ;;;" then i.tU j + ).2112 is the Newtonian potential of ;'(/; + ).2.[; for all scalars A( and A 2 .

(e) The derivative aul?x; of the Newtonian potential 11 or/E';}' is the Newtonian •• , 0 I'll (f En. ..

potential of (~f / ex i . We note also that , ,-- * f, the convolutIOn of! E' ;}' (lXi CX i

with the locally integrable function

-- -

O",,I-,I n

These two properties are deducible from the general theorems concerning the differentiation of convolutions 25

(d) The Newtonian potentialu of IE ;}' is a solution of Poisson's equation

(3.3)

In effect by definition. I;j~ E' ;;

(f1u. D

/111 = f on [R".

fll * 11~(x) = (Ell' j~x> ~JO) = ~(x) with ~x(y) ~(X - y) .

We shall see below how to characterise this solution of Poisson's equation (proposition 3). (e) A distribution U E;}' is the Newtonian potencial of its Laplacian f = ,.JU(E 8'). This is a particular case of Proposition 3 below. but it is seen directly from the identities concerning convolutions 26:

" We consider here the convolution of the acn/imclioll En hy a distrihution: it is a simple particular CJse of the convolution of distributions (sec Vol. 2. Appendix "Distrihutions"). ,<> The formula for differentiating the convolution of distributions is deduced easily from the dc!initinn beginning with the formula for functions (sec Vol. 2. Appendix "Distrihutions").

§3. Newtonian Potentials 277

U) The point (e) above shows that every distribution is locally a Newtonian potential. More precisely,

Lemma 1. Let Q be an open set of fM",j E 9'(Q) and u a solution (in the sense of distributions) of Poisson's equatiol1 Llu = f on Q.

Thel1for every hounded open set Q I with QI c Q there existsJ; E If' such that

(3.4) J; = f on Q, u = the Newtonian potential of J; 011 Q 1 .

In effect, considering p E 9(Q) with p = 1 on Qj, the distribution pu is the Newtonian potential of J; = Ll (pu)~ see (e) above - andl;. = f on Q l' The remark U) permits us to reduce the local study of any solutiol1s whatsoever of Poisson's equation on an arbitrary open set Q, to the local study of Newtonian potentiais 27. 0

In particular, we prove

Proposition 1. Let Q he an open set of fM" f a distrihution on Q and u a solution (in the sense of distributions) of Poisson's equation

LlII = f on Q.

Then (1) iff = 0 on Q, II is harmonic on Q; (2) iffE (6 W (Q), II E (f,"(Q)28.

Proof Suppose that we are given a bounded open set QJ with QJ c Q. From Remark 1 (f), there exists J; E If' such that

.h = f,

Then we select an open set Qz with Qz c Q I and p E 9(Qd with p = 1 on Q2'

From the Remark l(b), u = U I + U 2 on Q I where "l and liz are the Newtonian potentials of PJ; and (1 - p)J; respectively. For every x E Qz, the function (pAy) = (1 - p(y))En(x - y) is of class C(;'1. on fM": we can therefore consider <h, (Px)' In fact, since the map (x, y) ...... (pAy) is of class C(f co on Q2 X fM", the map x ~-+ <.1;., ((Jx) is of class rtf' co on Qz; we verify easily that this is the restriction of Uz to Q2' In other words, Uz is C(fX on Q2' Since Lluz = 0 on Qz, the function Uz is also harmonic on Q 2 and hence analytic on Q 2

(see Corollary 7 of § 2). Let us suppose that f = 0 (resp. of class C(; CD) on Q. Then p j; = 0 on fM" (resp. P'h E 9(fMn)); we then deduce that "l = 0 (resp. of class C(fX) on fM" and hence u = II 1 + Uz on Q 1 is harmonic (resp. of class (6' ex.) on Q 2; this being true for every open set Qz with Q2 c Q I where Q j is a bounded open set with QI c Q. 0

27 All the statements made in this Remark J, can be applied without change to any elementary solution whatsoever of any linear differential operator (see Chap. V). 28 This proposition is a particular case of general results concerning linear differential operators; the point (2) will serve as the definition of the notion of a hypoelliptic operator (see Chap. V. 92). We shall find also that the same property (2) is (rue on replacing "(6 7 " by "analytic": if u is an elementary solution of Poisson's equation ill< = I on Q and f is analytic on Q, then u is analytic on Q (see Chap. V, §4).

27S Chapter 1/. The Laplace Operator

It follows in particular from Proposition I that, given f E 8', the Newtonian potential u off is harmonic analytic on the exterior of B(O, Ro) for Ro sufficiently large. We give the behaviour at infinity of u:

Proposition 2. Letf E <1' and let u he the IV ewtol1ial1 potential olI Thellj(Jr every multi-il1dex'Y. E N 29

(""U IVE (I (3.5) , (xl = (/; 1> , "(x) + a + I' 1-

(1.'(3: (IX.:1 Ixl'l 'J

III pari icuiar

if IJ ;?- 3, lim u(x) I x I .. , x °

if IJ;?- 2. lim grad u(x) I xl ., ,

We shall make use of the following lemma:

O.

Lemma 2. For erery multi-index '1 E N" with m

C (depending Oil!.)! Oil IJ and .'1) such that I'll ;?- 1. there exists a constallt

(3.6) I!" En Ie. 1_-- (x) ~ ---+--~-,. jill' all i C'X' I X I" m c

xE[R"\{O}.

Proof oj' Lemma 2. The result is trivial for 11 = 1:

IdE I 1 ,_1 (xj = .•

dx 2

for all x f= O. Let us suppose hence that 11 ;?- 2. We have then

By recurrence on m l:x I. we see that

("E P(x) _ "(x) ,. ('x' Ixln + 2(m-ll'

where P, is a polynomial of degree m. Hence the lemma is proved. 0

Proof of Proposition 2. Let us consider Ro such that supp f c: B(O, Roj.

Given x E [Rn\B(O, Ro), the function y --> E"(.\ - y) is of class C(,x. in the neighbourhood of supp f Hence

u(x) = (f(y), En(x - y)

cO For the multi-index notation sec footnote Y above. 30 Given two functionsf(x) and glx). we say thalf = 0('1) when x .. ·• Xo if there exists a constant .",1 such that I fi x)1 ,;; M I'll x) I for all x sulficiently ncar to .\0' Here we apply the idea to the case Xn = Y .

~3. Newtonian Potentials 279

and

(3.7) o'u . o'En /

~ (x) - <.I, I) '. • (X) = ( f ( y) , ex ex \

The distributionfis of finite order (see Vo!' 2, Appendix, "Distributions") i.e. there exist measures Up), f3 E Nn, IPI :( m such that

'fJr f = L C J% (in the sense of distributions)31 ex

Obviously we can suppose that supp j, c 13(0, Ro). Denoting by R.(x) the two members of the equality (3.7) we have

From Lemma 2, for I yi :( Ro < R j :( Ixl we have

la.+IJEn I C C 1

T~;-11J (x - y) :(i;=YY+'.+fJl=2':( I~'(!n~ for 1:( 1f31 :( m

IC'En a'En I I a"En .. I ~ (x - y) -~ (x) :( sup ,grad ~-. (x -- I,Y). Y ex ex 0",;.",11 ex

where C1 depends only on l:xl, Ill, Ro• R I . We deduce

, C I 2.)vlp IR.(x)1 :( Gln+~ for all x ¢ B(O, Rd,

where

MfJ = sup I f((X)dffJ(X) I I( I ", 1

is the total variation of the measure .I~. o We then have characterisation:

Proposition 3. Let f E ,g' and u he a distrihution on IW with n ~ 2. Then u is the N eutonian potential off iff u is a solution of

Au = f on iR n

31 In the examples that we shall consider, f will be a function (sect. 2). a measure (sect. 3). or a distribution of order 1 (sect. 4).


and satisfies the conditions at illfi 11 it y:

if 11 ~ 3. lim u( x) 0 . Ixl ~ ,

ifn = 2. (1 ) lim. { u(xl-- (l L) Logixl = O.

Ixl'X \ 2n:

Proof The necessary condition is a consequence of Remark I(d) and of Proposition 5. To show that the condition is sufficient, replacing u by u -- it where il is the Newtonian potential of!; we can always suppose f = 0 and we then have to show that u = O. In other words. taking Proposition 1 into account, we have a function u, harmonic on IR:" which tends to zero at infinity. Using Corollary 1 (or Corollary 5) of § 2 we deduce that u is identically zero. We note the Corollary resulting from invariance under an orthogonal transformation of the characterisation of Proposition 3:

Corollary 1. The N eMollian pOlelltial o(a radial distrihution with compact support (t hat is to say illl'ariallt under orthoyonal transformation) is radial.

Remark 2. We have excluded from the characterisation of Proposition 3, the case II = I, which can (and ought to) be treated directly. The Newtonian potentialu of a distribution f on IR: with support contained in ] - Ro, + Ro [ is defined for Ixl ~ Ro by

lI( x) / I \

\/ f(Y).~ Ix - .vi .. ) ~ /

I 2. <I: J) ixl

I x 2 <t: y) ~I

I f d2u . . . To characterise the so ution 11 0 Poisson's equation - = f whIch IS the

dx 2 .

Newtonian potentiaL it is necessary to have two independent boundary conditions since the solutions of Laplace's equation on IR: are affine functions depending on two arbitrary constants. For example. we could choose the boundary conditions

. lI(X) I.. lim -- -- =- <! 1\ J .' /

x---+ +.f.x .-

lim !I(x) + !I(-- xl= O. o

Remark 3. III the case II = 2, the Newtonian potential of a distribution f e (1'

does not tend to zero at infinity. More precisely, if u is the Newtonian potential of f e 8', the following statements are equivalent (only in the case n = 2)

(il lim !I(x) = 0:

(ii) u(x) is bounded as Ixl -+ x

(iii) U: I> = 0 .

This follows immediately from Proposition 2. o

~3. Newtonian PotentJais 281

We note also the following proposition which characterises the Newtonian potential to within an additive constant.

Proposition 4. Let fE&" and let u he a disrrihution on ~" with n ~ 2. Then, to within an additive constant, u is the Newtonian potential off iff u is a solution of Poisson's equation

Au = f and satisfies at infinity one or other of the following conditions:

(3.8) . u(x) llm- = 0

Ixi - x Ix! or

(3.9) lim grad u(x) = O. !xl~ -Yo

Proof Taking account of Proposition 2, and reasoning as in the proof of Proposition 3, we see that it is sufficient to show that a function, harmonic on ~n satisfying one or other of the conditions (3.8) or (3.9) is constant. In the case of the condition (3.9), since the derivatives eluj (Xi are also harmonic, we deduce that they are zero: hence u is constant. In the case of the condition (3.8), we can remark that it implies that u is a tempered distribution and hence, from Proposition 3 of ~ 2, a polynomial; but only the constant polynomials satisfy (3.8). 0

In Remark I (f), we have seen that every distribution is locally a Newtonian potential. Using Green's formula, we have in fact a global representation on a bounded open set 0 of a distribution on Q: we state here the result within the classical framework.

Proposition 5. Let 0 he a regular hounded open set and lei U E (6 2 (0) n (6~(Q) with Au ELI (0). Then

u = u(I + u I + U 2 on 0

where !to, 111' U l are the N e~l't()nian potentials of the distrihutions j~, j;, fl on ~" defined hy

We note that!o,j;,f2 E J':

<Fa, :; ) = J' :; Au dx Q

f(l is an integrable function on ~n with support contained in [2, II is a measure on ~n with support contained in r, I2 is a distribution of order I on !R" with support contained in r.


We say that U t is the simple layer (resp. double layer) potential defined by the function - au/en (resp. u) continuous on r. In a general way, we have the

Definition 2. Given a regular bounded open set D of llin and !.f! E (,!j0(T), we give the name simple layer (resp. double layer) potential of (p to the Newtonian potential of the distribution./: with compact support, defined by

U:D = j'r((t)(PU)di·(t) (resp. r (,l, (t)!.f!(t)dyU)) , ~ r(:11

We shall study these potentials, as well as their physical interpretation 111 the following sections.

Pro%/ Proposition 5, We fix Xo E D and denote for 0 < r, < dist(xo , T)

D, = Q\B(xo, r,) .

Applying Green's formula (see Proposition 4 of S 1) to u and ['(xl = En(x - xo) on the open set Q" and noting that v is harmonic in the neighbourhood of D" we have

f f J~ ( ?u) f i'v o = U t1l' dx = [: Llu dx + t' - 1 dy + .;; .. u di' . Q, n, dl, (·n if!, (;/1

Let us determine the limit of each of these integrals when t; -+ O.

(I) r [lLludx -+ j' En(x - -'o)Llu(x)dx = uo(.xoJ. JQ, Q

This follows from Lebesgue's theorem, since the function En('x integrable on Q

-'0) Llu(x) IS

(2) f v(- ~)dl' -+ f En(x - xo)(' - (:u(X))d{'(X) = uj(xo)· iQ, ('/1 r ('n

In effect (lQ, is made up of rand ('B(-,o, n, On (l8(xo, I:) we have t' = I'; where

2 I: if 11 = 1

1 r I i Log I: if II 2

if 11 ~ 3 .

On the other hand, since u is of class ,(, I in the neighbourhood of Xo, I ('U ?nl ~ C on (l8(xo, r.) where C is independent of I: -+ O. Therefore


In all cases VE 8"- I -> 0 when B --> 0, from which we have the limit.

This reduces to showing that

f au u dy -> - u(xol . ('B(x".') an

Dv Now ~. - ---~1 on cB(xo. e) (we note also that the normal derivative is

en (T"S"

exterior to Q, and hence interior to B(xa, s)). Therefore we have

o

Remark 4. The Newtonian potentials 111 and 112 are harmonic on Q sincej~ andfz are with support in T. Hence 11 and 110 have the same regularity on Q.

Given 11 E '6'2(Q)!1 '6~(Q) with,1u E Ll(Q), the proposition states in particular that there exists 110 the Newtonian potential of an integrable function j~ with compact support, III the simple layer potential ofa function ({Jl E y;o(Q) and tl2 the double layer potential of a function ({J2 E ~O(l) such that

We note though that this decomposition is not unique:.f~ = Llll on Q is determined on Q only if we impose on it the condition of having compact support on Q; even in this case ({Jl and ({J2 and hence III and lI2 are not determined uniquely since the sum of a single layer potential and a double layer potential can be zero on Q without each of the terms being zero. 0

2. Study of Local Regularity of Solutions of Poisson's Equation

We study the local regularity of solutions of Poisson's equation

Llu = f on Q

as a function of the regularity of the distributionfgiven on the open set Q of IR". In the case 11 = 1, the problem is simple, since Poisson's equation reduces to the dilferential equation

Iff is of class (6 m , then u is of class (6 m + 2; if{ is a bounded (measurable function),


then II is of class (6 1 with Lipschitzian J2 derivative dll/dx; iff is an integrable function, then u is of class (6 1 with absolutely continuous derivative 33 du/dx; iffis a bounded (Radon) measure, then u is Lipschitzian 33 with a derivative dll/dx of bounded variation 33 : if f is a distribution of order I, then 1/ is of bounded variation J ]: etc. The situation is entirely different in the case 11 :;?- 2. Taking account of Proposition 5, we shall content ourselves to studying the regularity of solutions of Poisson's equation when! is a distrihlll ion of at least order I. Let us note the

Proposition 6. Let Q he (Ill open set o/[K" (n > 2),1a distrihutioll at least o{order 1 OIJ Q and II a distriblllioll solution of Poisson's equation Llu = fOil Q. Theil 14 is (/ (locally inteyrahle)jilllct iOIl 011 Q. More precisely: (I) in the case ola ueneral distrihutiollf%rder 1. UELh,c(Q)f()r all p < n/(n - I): (2) iff is a measure 011 Q, then 11 E L foe ( Q )j()r all p < n / (/1 - 2); the deriratil'es (lll;'(~X; are (locally inteyrahle)jilll("{iolls 011 Q and i'u/?x;ELfoc(Q) 'ip < l1(n - 1). (3) iff is a jilllctioll heiollyiny to L!;,e(Q) with q > 111 (resp. q > 11), lilen 11 is c(}l1tinuOlls (resp. 0/ class (f, 1) 011 Q.

This proposition will follow, with the help of the Remark 1 (f) from the following lemma on the cOIl1"oilltioll of" a measure alld a fUllctioll.

Lemma 3 34• Let/he (/ 1I1eWilire 011 [K" \\"ith compact support alld if (/ Borel./imctioll heionyiny to L 1,,,. ([KIl) Irith 1 ~ p < y:. Theil (1) fin' almost all x E [K", filet/me/iol1 y --> !/(x - y) is il1tewahie with respect to the measure t: (2) the colll"oilllioll if * f is (i jilllclioll helollyiny to Lioe( IM") and defined almost el'er\'whcrc 011 [K" hr

y * f(x) f q(x - y) df(y) ;

.12 A function r on an interval 1 IS:

(il Lip,ciJir:iall if there exists C such that 11"('1 l(y)l,; Cis ····\·1 for all x. y E 1: (ii) ah,oilllelr colltillW!1IS if VI: > 0, there exists ,i > 0 such that for every sequence of intervals

Ja"h,[ c l.twobytwodisjointl:(h, - <1,1 < ,i=2.·lr(h,j -r(a,)1 <;:; (iii) o(bowuied wrialioll if there exists C that for every sequence il, < il2 < < a, of elements of I. 1:lr(I1,) - ria, ,)1'; C. From Lehesgue's ditTercntiation theorem, every function of bounded variation is a.c. ditTerentiahle and ils derivative (defined a.c.) is integrable. A function is absolutely continuous itTit is of bounded variation and the integral of its derivative; a function is Lipsehitzian iffit is locally ahsolutely continuous and has an (essentially) bounded derivative. Finally a distribution on I is defined by a Lipschitzian (resp. ahsolutely continuous, re.'p. of bounded variation) function iff its derivative (in the sense of distributions) is defined by bounded measurable function (resp. an integrable function, resp. a bounded measure) (see Schwartz [2]. Rudin [I]) . . 'J For these definitions see the preceding footnote. 34 We grant a privileged place to the notation of distributions: a measure is a distribution of order 0; a measurefis a function if it is defined by a (locally integra hie) funclion. However we make use also of the theory of integration with respect to a measure: a distribution I of order 0 defines a (unique) regular Borel measure which we shall always denote by f: we know then to say what it means that a Borel function is integrable with respect to./: what is its own integral. etc. (see Marlc [1] or Rudin [IJI.


(3) iff is a function belonging to Lq( [Rn) where q is the conjugate of p (p - 1 + q- 1

= 1), then 9 * f is continuous on [Rn.

Proof of Lemma 3. The points (1) and (2) result from Fubini's theorem. Taking

ro > ° such that B(O, ro) contains supp f, denoting by J.l the variation off, we have f = 8/1 where c: is a Borel function taking the values + 1 or - I (Lebesgue's decomposition theorem or the Radon-Nikodym theorem), Given r > 0, we have

r dx Ilg(x - y)e(y)1 d/1(y) = 1_ dJl(Y) r Ig(x)1 dx "BIO,r) J JBIO.ro) JB(Y,r)

~ (j d/1 ) (Lo,ro + r)r1g(x)ldx ) < W.

Hence (by Fubini's theorem), for almost all x E B(O, r), the function y -;. g(x - y )8(y) is integrable with respect to Jl and the function

x -;. j g(x - y)c:(y) dll(Y) is integrable on B(D, r) ,

In other words, we have the point (1) and the function defined a.e. on [Rn by

u(x) ,= j 9(X - y) df(y)

is locally integrable on [R". To obtain the proof of (2), it is enough to note that

<9 *[,0 = jdf(Yl jg(X - y)((x)dx = jU(X)((X)dX

and for supp ( c B(O, r), by Holder's inequali ty 35

1

1<9 *.[, 01 ~ ( (dll) ( r 19(XWdX)P lJ(ilLq , J "B(O",,+r)

Finally to prove the point (3), we can make use of

19 *f(xd - 9 *f(xz)1 ~ UllLq (( Ig(x + (Xl - X2» J BIO.r" + r)

1

- g(x)JP dx y' a.e. Xl' X 2 E B(D, r)

and the continuity of the translation in LFc,c( [Rn)( 1 ~ p < w). We can also use the fact that the map 9 -. 9 * f from Lioc([R;n) into LI~c([R;n) is continuous; for 9 continuous 9 * I is continuous; since the continuous functions are dense in

35 H,i'/da's il1equality. A function r, locally integrable on Q, belongs to U(Q) iff there exists C such that Ifr~dxl ~ Cjl( L'; the smallest constant Cis C = Ill'ilL"; p-! + q! = 1.

2i16 Chapter II. The Laplace Operator

Lfoc([Rn), for all g E Lfoc([R"), g 7< fwill be the uniform limit, on every compact set, of continuous functions and hence will be continuous,

Proot'olPropositioll 6, Given a bounded open set Q I with Q1 c Q and pE9(Q) with p = 1 on Q I' we put ./;) = pI andf; = 11 (p/) - k From Remark 1. (f), U = Uo + U I on 121 where un' III are the Newtonian potentials off~,f;, Since!; = 0 on Q l' II 1 is harmonic on Q 1 so II and Uo have the same regularity on 12 I; alsofo on [R" the same regularity asfon 12 1, In other words we call always suppose 12 = [Rn, f a distribution with compact support and Ii the Newtonian potential off: this we do, If/is a distribution of order 1.

at I'f f' = /' + ... ,I + ' , ' + --,-". . .10 ~ ~

('X 1 ex"

where.f~,,/;, ' , , ,.f~ are measures with compact support,

where lIo, Ii I' ' , , ,Un are the Newtonian potentials of J~, ?f,) ex l' , , , , ?f~/ rx" respectively, Applying Lemma 3 with f = j~J and q = E" which belongs to Lfoc([Rn) for all p < /J/(n - 2). we deduce that Uo is a function belonging to Lfoc(o~n) for all p < /1/(/1 - 2), If. in addition, ./~ E Lq(IR")withq > 1J/2 which is the conjugate of /1/(/1 - 2), then Llo is continuous on [R",

N f 'I E ('Ii i'E" /' A I' L 3' h./' ., 'ow. or I = .'" .11, Lli = '" 7< -.,". = ~x .Ii' PP ylng emma Wit =.Ii eX i eX i

<"Ell ' and (} = - which belongs to Lt;,c ([R") for all p < 11:' /J - 1. we deduce that

, ?Xi

Ui E Lfoc([RIl) for all p < IL(n ~. I),

To complete the proof, note that c'En

= .. - 7< j~. with the result that ('Xi

( LIo ~ ... E Lt;,JIRn) for all p < 11/(11 - I) and if in addition IE U(IRn) with q > 11 L'i

which is the conjugate of n/(n - 1). then ?u/i1x i is continuous on [R", D

I t is clear that we can obtain by this method other results concerning the regularity of the solution as a function of the given( In particular, by considering only the Lebesgue spaces U. we lose information since the elementary solution has more properties than those associated with belonging to Lfoc(Q) for all p < n/(n - 2), We can make this precise by making use of more sophisticated spaces: the spaces of Marcinkiewicz or more generally th~ spaces of Lorentz (see Vo!' 4. Appendix "Singular Integrals" and Butzer~Berens [IJ), We shall be content here to make these properties more precise in the case of a radial solution, We shall also be content to restrict ourselves to the case in which I is a function 36,

We shall prove the

.In The case of a distrihution lof order 0 or 1) will form the study of 93,

~3, Newtonian Potentials 21;7

Proposition 7. LeI f be an integrable function on the ball B(O, ro) and u a radial solution of Poisson's equation

Au = f on B(O, ro) .

Then

(I) u is of class '6 1 on B(O, ro)\{O},

. u(x) hm -- = 0 and lim !x!n-I grad u(x) = 0 ; x~o En(.x) x-o

(2) if En(x)f(x) (resp.f(x)/!x!n- t) is integrahle on B(O, ro), then u is continuous (resp. of class '6' 1) on B(O, ro); (3) the second derivatives e 2 u/ eXi ax} are locally integrable functions on B(O, 1'0) \ {O}. II f(x) Log! x! is integrable, then az u/ aXi eXj are locally integrahle on B(O,ro)'

a2 u If IE U(B(O, 1'0)) with 1 ~ p ~ CfJ, then ~,-~ E Lf.,c(B(O, ro))·

cxJ!xj

(4) If I is of class (6 m on B(O, 1'0)' then u is of class ,/&,m + 2 on B(O, ro). We shall use the

Lemma 4. Let I be a radial integrable function on ~n with compact support. The Newtonian potential u ollis (>fc/ass 'i&" on ~n\{o}, and we have

(3.10) u(x) = En(.x) f I(y) dy + r" EnLv)I(y) dy JB(O.lxl.l oil< ,B(O.lxl)

(3.11) x i grad u(x) ,= --n f(y) dy . un!x! B(O.lxll

The second derivatives are locally integrahle on ~n \ {O} and

(3.12) clzu XiX}, ( xiX j ) n i . -~-(x) = -2I(x) + ~ -, -2 ---, -n I(y)dy clxiox j !X! n Ixl UniX! B(O.lxl)

where c\j is the Kronecker symbol

()ii = {I if i=j

Oil ii'j.

Proof' of Lemma 4. From Proposition 3, it wili be enough to verify that the function u defined by (3.10) is clearly a solution of Poisson's equation satisfying the condition at infinity. It is more satisfying to "discover" these formulae. First of ail, we note from Corollary 1 that the Newtonian potential u has to be radial; the fact that this is a function (locally integrable on ~n) follows from Proposition 6. We therefore have

f(x) = gil xl), u(x) = u(!xl)


where g, t' are (measurable) functions defined a.e. on JO, x [. The function 9 has support contained in [0, roJ and fori to be integrable y satisfies

Jf'f r"-lu(r) dr < I, o

The function r" - 1 1,( r) is locally integrable on ] O,X [ and the function l' satisfies the differential equation

(3.13) r" - I

d ( dl') dr r" - I dr = y on ] O.J~. [ .

This expresses that 11 satisfies Poisson's equation (see the Laplacian of a radial function in § 1.4). We see then that r"- 1 [·'(r) is absolutely continuous on JO, x [; hence 11 is of class (6 1 on IR"\{O} and the second derivatives c' 2U/?Xi ?Xj are locally integrable on r,g" :0]. Now from Proposition 2, when r --> x,

dr -(r) = grad u(ra-).IT = <r 1> grad £1I(ra-).IT dr

+ o(~.) I'"

--;;1 If" t"-lg(tJdt + 0(1). r 0 r"

We integrate (3.13) to obtain

(r)= .. ,. t"ly(l)dt dl I f' dr I'" - 1 0

for all r > o.

This proves (3.11) and by differentiation (3.12). To obtain (3.10), it is enough to integrate (3.11). using the condition at infinity given by Proposition 2.

u(x) = E (X) J"f(\,) dr + 0(' __ 1_ ,)' when Ixi ->1_., 0 n .., I X I" I

Proof 0( Propositiol1 7, Taking up the reasoning of the beginning of the proof of Proposition 6, we can always reduce it to the case where f is integrable with compact support and u is the Newtonian potential off; if fI is radial pu will be radial and so also f The point (I) is then included in Lemma 4 or results from the formulae (3.10) and

. u(x) (3.11) by the use of Lebesgue's theorem. To prove that hm ,-- = 0, we observe

x ~ 0 t'/I(x)

that

l ~nWf'(I')1 ~ If( .)! for 0 < Ixi. ~ 11'1 ~ 1. En( x)'.! ",..). - ,

Similarly the point (2) follows from (3.10) and (3.11), the use of Lebesgue's theorem


and the inequalities

I x 1--- If(y)1

IEn{x)f(y)1 :::; IEn(y)f(y)1 and IxIJ(y) "" Iyln-I

for 0 < Iyl :::; Ixi :::; 1 .

The first part of the point (3) is stated in the lemma and we have from (3.12)

_-.Y. _ tn-1g(t)dt b .. ) n ir

n rn 0

with r = Ixi and g(r) = I f(x)l· We have trivially

II 02 u II ( b .. ) OXjOXj Leo:::; 2 - ~J Ilfllu·

Also

Ilc.:j

2;XJ!L1IBI0.IH:::; IlfllL' il dr ir + (n - bij)CTn - tn-1g(t)dt

oro

IlfllL' + (n - bjj)f If(y)1 LOg_Ill dy. BIO, I) Y

Finally for 1 < P < 00,

11 ~02~ II :::; IlfilF + CTn(n - bij)( (Rdr( I'tn-1g(t)dt)P)I!P uX j uXj UIBIO, R)) Jo r Jo

But from Holder's inequality, with q the conjugate of p

289

J:tn-Ig(t)dt:::; (J:tn-ldty,q(J:tn-lg(WdtY'P :::; (~y,qll:~I~~p,

and

(3.14) II ~II ~llfll (l+(l_ bij )(CTnRn)I-*) OXjOXj UIBIO,R)) '" . F n (p - 1)l/p .

This completes the proof of point (3). For the point (4), it is clear by (3.12) that u is of class ,£m+2 on IRln\ {O}. Now using Taylor's formula

f(x) = L ak lxI 2k + o(lxlm)37, o ~ k ~ 'f-

37 If f( x) = g( I xl) E '6 m( IR"), 9 is of class (6 m on ] O,~ [ and the odd derivatives of 9 are zero at O. In I d 2k

the expansion a, = -' -g(O). (2k)! dx 2 '


putting this back into (3.12), we obtain

from which the result follows by induction over m 3a

Let us return now to the yelJCI"a/ cuse and study the local reyularity of the second order derir:atil'es. We can generalize the point (3) of Proposition 7 for I < p < x and state:

Proposition 8. Let Q he (/11 opell .,er ill IT:);", I < p <x, fE Lfoc(Q) and u a solution of Poisson's equation

Llu = I OIJ Q.

Theil its secolJd deril 1(l1ir:es (l2U / (oX; ?Xj arefimctiolls belonging to L[~c (Q).

In the case of the statement of the Proposition: I < p <XI arbitrary, this proposition reduces to the Calderon-Z yymund theorem: there exists a constant C p

StIch that

II ~_r~s II ~ CpilLlsll v fi)r all s E C/(rr:gn) ex, eXj f.i'

(see Morrey [IJ for a "'quick" proof of this result: See also Volume 4. Appendix "singular integrals"). We give here a proof, in the ease p = 2, by using the Fourier transform.

Proof of Propositio/l!:l ill the case p = 2. As we have seen in the proof of Proposition 6, we can always suppose that Q = rr:gn, IE L2(rr:gn) with eompact support and u, the Newtonian potential of f It is clear that u is a tempered distribution (if 11 ?ec 3, u( x) tends to zero at infinity: if n = 2 or I, u( x) grows at the most like Log Ixi or Ixi at infinity. We can therefore make use of the Fourier transform (see Vo\. 2, Appendix, "Distributions"). We have

I ·F(;;-?2~~ )1 = i- Y;.\'j.F(II)! ~ 1- lyI2.F(u)1 = .F(,11ull e'\ic'\j/

I.F (nl .

,) ('-u Making use of Plancherel's theorem we have - E L 2 (rr:gn) and

(~X; ('x j

.F (f) II L' = II f II u .

Remark 5. The conclusion of Proposition 8 is not true for p = I: this is clear even in the radial case from the formula (3.12). It is no longer true either in the general case for p ="k: also in yeneralfor afunctiollf continuous on Q, the solutions

3K A function ,. E '(, ", '( Q) and of class '6 m on Q : 0: is of class '6 ",( Q) itT,' admits an expansion limited to the order In at \:0'


of the corresponding Poisson's equation do not belong to the class <e 2 (Q); we note as we have seen in Proposition 7 that in the radial case, f continuous implies that the solutions of Poisson's equation are of class ~'2. We can show the impossibility of these properties in general, in the non-radial case by a functional method; using reductio ad absurdum, let us suppose the property to be true; for every bounded (resp. continuous) function f on [Rn with compact support the functions i3 2 (En * f)/oxiox j would be locally bounded (resp. continuous) on [Rn; from the closed graph theorem 39 : for all R > 0, there would exist a constant C such that

II ;:;-~2_ (En * 0 II ~ ell (11 1/ (;XiOX j LX(8(O,R))

for every regular function ( with support in B(O, R). Now being givenfintegrable with support in B(O, R), we consider its Newtonian potential u:

which contradicts the fact that in general 02U/eXi oX j is not integrable. 0

To obtain the regularity of the second derivatives we must therefore use a stronger hypothesis than continuity: the class of Holder functions 40 is well adapted for this purpose as is shown in the

Proposition 9. Let Q be an open set in [R", 0 < Cf. < 1, m ~ 1. For every function f E <em, a( Q)40, the solutions of Poisson's equation

Llu = f on Q are of class <em + 2,a on Q.

We prove, first of all, the

Lemma 5. Let f be a H iilder function on iR" with compact support and u the Newtonian potential off. Then Ii is of class <e 2 on [Rn. For every regular open set Q

('"',2

39 See Chap. VI. Here the mapf --> - ...... (E, * f) is continuous from.5' into [/': if it maps the space PXi

X of bounded (resp. continuous) functions with support in 8(0. R) into the space Y of bounded (resp. continuousJ functions on B(O, Rl, it is continuous from X into Y. 40 We recall that a function I defined on a part K of ~n is called a Hiilder funCiioll of' order x (0 < "l < I) on K if there exists a constant C such that

II(x) - f(,1JI ~ Ix - 1'1' \Ix, l' E K

A functionl' E «(, m,,( Q) iff is of class 'is m and its derivatives of order m are Holder functions of order "l on every compact subset of Q.


containing supp f and x E Q, we have4-1

(3.15)

We note that these two integrals clearly have a meaning: it is clear for the second since x ¢: (lQ; for the first, supposingfto be a Holder function of order '1., we have

(3.16) 1,?2~n (x - y)(f(y) - f(X))1 ~ I~I _ C'ln_, exi(x} an X }

which is integrable on Q. In addition the function on the right is continuous at x E Q.

Pro()/ of Lemma 5. Given ( E 0' (Q) we have

ff (1En (1~ = - ~(x - y)f(Y)-l-(x) dxdy

(Xi (x)

lim ff(y)v,(Y)dY f. ~ 0

where

i (1 En e~ [',Lv) = - ~(x - y)-(x)dx.

Q Bly,D) (Xi ex)

By use of Ostrogradski's formula

Therefore

41 This formula is still true for a.e. x E Q ifF E U( [R") with support in Q (see Vol. 4, Appendix "singular integrals").


with

i 1 a2E Ii = dy ((x) IT (x - y)(f(y) - f(x)) dx

n Q\B(y,e) Xi Xj

I~ = r dy r (x) ~En (x - y)(f(y) - f(x»nj(x) dy(x) In J8(Q\B(Y,C)) Xi

i l aZE I~ = dy (x)f(x) IT (x - y) dx

n Q\B(y, e) Xi Xj

i i a~ I~ = dy (x)f(x) a (x - y)nj(x) dy(x) .

n 8(Q\B(y, ell Xi

We study the limit of each integral as e --+ 0:

(1) Ii --+ r (x) dx r ~a2 :n (x - y)(f(y) - f(x» dy ; In In uXiuxj

this follows from Lebesgue's theorem and the estimate (3.16);

(2) I~ --+ 0 ,

in effect using the fact that f is a Holder function of order IX

i f eea ~ dy I(x + eO')I-dO';

n I O'n

(3) I~ = r (x)f(x) dx r ~a2 :n (x - y) dy J n JQ\B(X, e) UXi uXj

i i aEn - (x)f(x) dx :l (x - y)nj(Y) dy(y)

n 8(Q\B(x, ell UXi

- r ((x)f(x) dx (f ~En (x - y)niy) dy(y) + Cij) , In iJQ UX i

where

Cij = r aEn (x - y) Xj - Yj dy(y) = ~ f YiYj dO'(y) JilB(X,e) aXi IXj - yjl O'n I

is independent of e.

(4) I~ = r dy r (x)f(x) ~En (x _ y) IYj - Xjl dy(x)

In JeB(Y,e) uXi Yj - Xj

= - r dy f ((y + eX)f(y + eX)XiXj dO'(x) . JnO'n I


Hence

l~ ---> - cij L ~(y)f(y) dy .

Grouping together these limits. we obtain (3.15). o

Remark 6. The formula (3.15) and the proof of Lemma 5 shows that a sufficient condition for the solutions of Poisson's equation

,i,l = f on Q

to be continuous at Xo is

If(x) - /(xo)1 ~ D(ix - xol) for all x in the neighbourhood of X O•12 ,

where r: IR + ---> IR + is (continuous) increasing and satisfies

j" d c:(r)r < ""j~ .

• 0 r

We can verify that this condition+ 2 is necessary in the sense: given c:: IR + ---> IT~ +

increasing and continuous with

[ dr I:(r) -- = x ,

,0 r

there exists a continuous function f with compact support satisfying

1/(x)1 ~ dlxl) \:Ix,

and such that the Newtonian potential of f is not twice differentiable at O. 0

Proof" of" Proposition 9. First of all it is sufficient to demonstrate the proposition for In = 0, since the derivatives of u are solutions of the Poisson equation corresponding to the derivatives of}: Also (see the proof of Proposition 6), we can supposefhas compact support in Q and that 11 is the Newtonian potential off Let us consider two points XI' X 2 E Q and put () = Ix( - x o!, Xo = !(x l + x 2 ). We can always suppose by translation that Xo = 0, and can always take Q = B(O, R) with R ? (). Applying the formula (3.15) we then have

where

k 1,2

42 A function satisfying this condition is said to be continuous in the sense of Dini. or to satisfy Dini's condition at 'n.

93. Newtonian Potentials 295

i aZE 13 = -ri-_n (x 1 - y)dy

B(O, R)\B(O, b) ex; ax)

i ( a2 En a2 En ]4 = -1-.- (Xl - y) - -1--- (X 2

B(O, R)\B(O, b) eX i aXj 0Xi aX) - y) ) (f(y) - /(X2)) dy

](X) = f. aEn (X - y)nj(Y) dy(y) . <'B(O, R) ax;

We then have the estimates

with x, = tX l + (I - t)x 2 .

Calculating the third derivatives of En' we find

1]41 0:::: C n(n + 5) (5 rl dt f Iy -- x 2 1a dy

'" an Jo B(O,R)\B(O,()jly-x,ln+l

~ Cn(n +

with

roc (r + 1) a

c, ~ Cnln + 5) J, (, _ ~ Y" ,. -, d,

which depends only on C, nand ()c

(4) 1](x 2 ) - ](xdl ~ e dtf I grad (:E!'(x t - Y),(Xl - x2 )ld i '(y) Jo ,'B(O, R) (Xi


Regrouping these estimates, we obtain

I r'2U ;,2 11 I 112"

, , ( X 1 )--, , ( X 2 ) I :0( C 2 ()' + J> f ( Xl) I () eX i (X j (Xi ex) ,

C 2 = 2('11 (' ~)' + 2" ~ 1 C + C t + I" 1 C 1. _

with

depending only on C 11, x. In the limit when R -> f,

for all XI' X 2 EO [R".

3. Regularity of Simple and Double Layer Potentials

We consider a regular bounded open set Q with boundary 1'. Given (jJ E (f, ll( n, we have defined (see Definition 2) the simple layer potential of (jJ as the Newtonian potential 111 of the measure (pd';' on r: this is a harmonic function on [R" defined by

('

(3.17) 1I\(X) = 1 E,,(t -- x)(p(t)d;'ur".

We have also detlned the double layer potential of (p as the Newtonian potential 112

of the distribution div ((plld';,) with support on r: it is a harmonic function on lR" r detlned by

(3.18)

where n( f) is the unit vector normal to r at t and exterior to Q.

We propose to study fhl! rl!!}u/arily of lit alld tl2 ill thl! Ill!iyhhourhood of r as (/

lilllelioll ojlhul of((J. It is clear that the datum is the compact regular hypersurface r, rather than 0: this hypersurface separates U;G" into two open sets: the regular bounded set 0 with boundary r which by an abuse of language we shall call the illterior of r and denote by Oi' the open set [R" Q which we shall call, again by abuse of language~~. thl! exterior of r and denote by Q/". Given a function 11

defined on [R". we shall denote by 1/ (resp. 1/") tiJe restricriolJ olll to Q j (resp. Qe)'

As usual the case II = 1 is immediate, A regular bounded open set of iR is a finite union of open intervals; it is clear by additivity of the Newtonian potentials, that it is enough to study the case of an open interval Q = ]a, he; we then have

r=(u,h}, d~'=()a+(\' lI(a)=-I, ll(h)=+1

",\ The n. dimension of the space ['1;" and the 1/ normal vector are obviously unconnected. H Topologically, the hypersurface I' has an empty interior and its exterior is iR;" r, "' Q i and Q c are well dclined by r: two regular hounded open sets which have the same boundary are equal.


For a function ep = (ep(a), ep(b», we have therefore for x E {a, b}

1 . ut(x) = 2(~?(a)ix al + ep(b)lx - bl)

~~(x) = i(p(a) sign (x - 0) + (p(b) sign (x - bl) dx 2

i u2 (x) = 2(ep(a) sign (x - a) - ep(b) sign (x - b)) .

We see therefore that, in the case n = 1, U I is continuous on IR, dul/dx and U2 have a limit on the left and a limit on the right at every point z of r with a jump

dU 1 ---(z+)· dx

dU I (z -) = ep(z)

dx

Uz(z +)- u2 (z -) = ..... n(z)ep(z).

In addition U il , u~ (resp. u~, u;) are C(,' x on Qi (resp. tiel. The radial case is equally simple and gives rise to the same phenomena. A radial regular bounded open set in IR" is the union of a finite number (possibly zero) of annuli B(O, 1'1 )\B(O, 1'0), (0 < 1'0 < rd, and, if it contains 0, of a ball B(O, r). We note that

c(B(O, rtJ\B(O, roll = (lB(O, I'd u DB(O, ro) ,

the exterior normal to the ring being exterior (resp. interior) on aB(O, I'd (resp. aB(O, 1'0)) to the ball B(O, rd (resp. B(O, 1'0»' Because of the additivity of the Newtonian potentials, it sLltrices to study the case of a ball Q = B(O, 1'); we sLlppose that ep is radial, i.e. constant on (18(0,1').

We then prove the

Lemma 6. The simple layer potential U I defined by the constant function <(J on rB(O, r) is continuous on IR", constant 011 B(O, rl. alld given by

B(O, r) .

Proot: We know (see Corollary 1) that U j is radial. Being harmonic on B(O. r) it is constant on B(O, r} (see Proposition 4 of §2) so

III (xl = udO) = f. (pE,,(t) d~,(t) for all x E B(O, r) . iBIO. r)

Being harmonic on IRn \ B( 0, r), it is of the form (see the same proposition)

U j (x) = aEn(x) + b for all x E IRn\B(O, 1') ,

where a and b are determined by the condition at infinity of Proposition 2

uj(x) = (JI' epdi·(t)) En(x) + 0(1 ~1) when Ixl-> x. . iBIO.r! ,xl ,

Chapter II, The Laplace Operator

Hence

a= [ rpdy(l)= r,,-I(J"rp, h O. ",'B(O. r)

It is clear then that til is continuous on [R;". o For the double layer potential, we have in a more general way the

Lemma 7. The douhle layer potelltial Uz defined hv a constant junction rp on the houlldary r o{ a reyular bounded open set Q oj"lR" is yil;ell on [R;" \ r hy

(rp if x E Q; U I = ~ .. lO If X E Q e

Proof Applying Proposition 5 to the constant function u U 2 = U = rp on Q; = Q. Now if x E Q e the function t -> E"(t on Q; by Gauss' theorem (see Proposition 5 of ~2).

rp on Q, we have - x) is harmonic

r DE" u2 (x) = rp l. (t - x) dr,(t) = O.

~ r (11 o

We see therefore that again in the radial case (and more generally for the double layer potential of a function (p constant on each connected component of n, U I is continuous on [R;", 11\, tI~ (resp. lI~, u~) are ((, C on Q; (resp. Qe), lI2 and rill) len have a jump at every point z of r, given by

u~(z) - tI~(z) rp(.::)

We propose to generalize these properties. We begin with the continuity oj" the simple layer potential:

Proposition 10. Let r he the houndary of a regular hounded open set and rp E ((,O(r). Then the simple la.ver potentia/ul oj" (p is continuous 011 [R;n.

Proof We have only to prove the continuity at a point.:: E r. Let us consider a normal parametric representation ((R, U, t. C()46 in the neighbourhood of - (x~, 'l.(x~))); we jix ro > 0 suftlciently small that

For

with

H(x;), "0) c C, Oro = [(x', xix') + t); X· E B(x~, 1'0), ITI ~ ro; c U

X E [R;"\r, let us write utfx) = Itfx) + [2(X)

.h See ~ 1.3.a. We recall that R is a reference frame. U a neighbourhood of~, C an open set of M" '. Y. a function of class '{, m on (' such that (x'. x~) EO [R:" -, x [R: being the coordinates in the reference frame R:

U(rcsp. C .'"\ I. l n 0) -~ [(x'. (X(x') + r): " E (I, ill < ,)(rcsp. I _0 (), - ,) < r < (l): .

;$3. Newtonian Potentials 299

Since En{t - x) is bounded (t, x) E (1\ Oro) X Oro/2' we have, from Lebesgue's theorem,

Ij(x) -> (. _ En(t - z)<p(t)d,,(t) when x ->;::.

Jr\ L\,

We have therefore only to consider 12 (x). For x = (x', ex(x') + r), we have (see formula (1.11) of § 1.3a)

12 (x) = f En(t' - x', ex(t') - ex(x') - r)<p(t',ex(t'))(l + Igrad ex(I'lI2)id( . B(x;), r;)

Now the term under the integral sign is majorized by

Cy(t' - x')

where

C sup 1<p(t',ex(t'))1 (1 + Igrad ex(t')12)! 8(x;,.ro)

and

g(t') = { ~-:- 2)~.!t'j' -,

2:; Log itl

if n?: 3

Since 9 is locally integrable on [R" - 1, the function

t E 1 n Oro -> En{t - ;::) <p(t)

is integrable with respect to d)' and

12 (x) -> fnc,,, En{t - z)(p(t)dl'(t) when x -->;::. o

Remark 7. In fact the function Y E Lfc,c([Rn- 1) for all p < (n - 1)/(n - 2); it follows (see Lemma 3) that the result of the proposition remains true for the simple layer potential defined by a function (P E U(n with q > n - 1. We study later the simple layer potentials defined by functions which are integrable on r or even by measures on 1 (see §6). D

We now study the "continuity" of double layer potentials. We shall have to suppose that 1 is a little more than of class {I, 1; we introduce the definition:

Definition 3. We shall say that a regular open set has boundary r of class (6 1 + £ if in the neighbourhood of every point z E 1, there exists a normal parametric representation (R, U, C, :x) and an increasing continuous function f;: [R' --> [R + with

( c(r) ~r < oc such that Jo I'

(3.19) I grad :x(x') - grad :xCX=')) ~ dlx' - .x'l) for all x'. X' E (i .

47 We can always suppose that It' - x'i < 1 with ro sufficiently small and x near to z.

·lOO Chapter II. The Laplace Operator

We note that this condition is obviously satisfied if r is of class ({,2 or more generally of class ({, 1.0 with 0 < () < 148

Proposition II. I,el r he tht:' /JoLl/utar}' o{class ({, 1 +, ora regular bounded open set and let (p EO (f,0(J'). Then:

;, 1:_." (I) If: EO I'. tilelullc/io/l I ->:;-- (t - :j(p(t) is inteyrahle Oil r (wit}Jrespectto di');

en (2) the jimctioll It' defined Oil r hy

is continuous on r. (3) the internal double layer potentialll~ (resp. external u~) defined by (p extends by continuity to Qi (resp. f2e) by putting on r

. I ( 1 ) 1I~(:::) = 11'(:::) + 2 (p(:::) resp. 1I~(:::) = IV(:::) - 2 cp(;;) .

Proof We fix ::: EO r and consider a normal parametric representation (R, U, C, -x)

in the neighbourhood of::: satisfying (3.19). With the notation of the proof of Proposition 10, for x EO [Fg". r, 112 (x) = J 1 (x)

+ 12 (x) with ~

a·;" II (x) J, - (t - X)CP(t)di'(t)

1. C r'n

J 2( x) 1 (~E" x) (p It ) d~' (t) . -,- (t -

", (!r ('11

Also I'

(~E" ld x ) --> J -;;-(t :::)qJ(tJ d;'U) when X-->:.

r ell

The existence of the integral does not pose a problem. Now, at x = (x', -x(x') + r), we have 49

-x(x')- r - grad -x(1').(t'- XC) , , , ------- ---- - - . - cp(1 •. -x(t )) dt

x'I2 --1- l-x(t') - ex(x') - rI 2 ),,2

48 r is said to be of class 'f, m.iI if in normal parametric representations (R. U. (' . ~ I. the function y is of class ((; m,l) on C .

a:." 11- ,).11(11 "" We make usc of -::----- (I ---- .\) = - and for I = I t'. ~I( II

til IT"II - .\ I"

flU I = (1 + igrad'X(nl'):

grad y ( t'). I ) (see formula (1.10) of ~ 1.3.<1.1


with

, 1 JI' oc(t') - oc(x')- gradoc(t').(t' - x') '( '))d' 13 (x, r) = -;;- .' (It' _ x'12 + loc(l') _ oc(x') __ rI2)"/2 cp(t, oc t t

n B(x o. rol

and , r r cpU', CI(t')) dt'

14(X, r)= (JII JB(x~Jo) (It;---- x'12 + IOC(t')- oc(x') _ rI2)n/2 .

Making use of the hypothesis (3.19), the term under the integral sign in 13 IS

majorized by Cg(t' - x') with

C= sup Im(t', ':1.((')1 and (t')=-~ y g It'I"-l .

H(x;,,,o)

By the hypothesis on /: given in (3.19), the function g is locally integrable on 1R" ... 1

We deduce the points (l) and (2) of the proposition; and also

f oEII 13(x',rl...... -~-(t - zlcp(t)d}'U) when x = (x',r) ...... z.

1-,[, en

To complete the proof, it is sufficient therefore to show that

With the aim of showing that this limit is uniform with respect to x~, it is enough to consider a normal limit, since w(z) and cp(z) are continuous at Z50. To simplify the calculations, we can also suppose that the reference frame is chosen with centre z and with n(z) as the last of the base vectors, with the result that

(3.20) x~ = 0, ':1.(0) = 0 and gradoc(O) = 0

and we have to show that

r f cp(t', ':I.(t')) dt' cp(O) I (r) - ...... when r ...... 0, r > 0 .

± - (J" H(O.ro)(Tf'i2+ (oc(t;)-+ r)i)"72 -2-

Changing the variables we have

I ± ( r) = ~ f ___ ({)~Tt',-~( rtJ)~t'2;;T2 . (In JB(O.ao/T) (it'l2 + (1 + OC(:t ») )

1 ':1.( Tt') I But using (3.19) and (3.20), :( 1 t' 1 /:( r 1 t' I), with the result that for r

r sufficiently small, we obtain:

on B(O, 1)

on IR" - B(O, 1) .

50 If a sequence (xm) of IR"\ r converges to Z E r. there exists a sequence (2m) of r converging and Am --+ 0, such that Xm = Zm + imn(zm)·


In the limit,

o

In the same way we study the "continuity" o(the normal deriratire o{a simple la ..... er potential.

Proposition 12. Let 1 he the howulary orelass (f, 1 + c ora regular bounded open set, and let cp EO <'(;' 0 (n. Then

(1) j()r all : EO r, the/tllletioll t --+ grad En(: (2) the function h defined on r hy

t).Il(:)cp(t) is illteyrable 011 r:

f 1 I' (: - rj.Il(:) h(:) = grad En(: - t). n(:) cp(t) d~,(t) = J _ n (p(t) d~,(t)

r an r I~ - 11

is continuous 011 1:

(3) the normal deril'atire ~ resp. ~ o{the interior (resp. exterior) simple layer ?u i (' (c U " )

(n cn potential defined by cp, is defined 011 1'51, and girell hy

?u\ 1 --:;--(:) = 11(:) - - (p(:) Cll 2

( ?u" 1 \

resp.-c-~ (:) = hI:) +) (p(:) ) . ( 11 _ /

Remark 8. It is necessary to distinguish clearly between the function h defined in this proposition and the function 11'(:) defined in Proposition 11. We note that

(3.21) >1'(:) + ill:) = Lgrad En{t - :).(n(t) - n(:))(p(t)di'(t)· o

Proof of Proposition 1]. The points (I) and (2) are obtained as corollaries of the points (I) and (2) of Proposition II and of the formula (3.21) of Remark 8: in effect making use of the hypothesis (3. I 9), we have

[11(t) - 11(:)1 ~ 1:(lt - :1)

with;;: [R1 + --+ [R11 continuous, increasing and satisfying r 1:( r) eli' < J~. Making Jo r

use of arguments from the proofs of the preceding propositions, we shall easily show that the term under the integral sign in (3.21) is integrable and IV + II thus defined is continuous. Now considering x EO [R1"\1, we have by differentiation under the integral sign

grad u1(x) = J' grad En(x - t)cp(t)d/'lt). r

'1 In the sense of ~1.3.b: ui, E '6;([2,). fI'; E '6;li2,.). We note however that n is directed along the t1ut'

external normal to Q = Q. and hence that' (~) is here in the opposite direction to that of the external I (;11

normal derivative to Q,.

~3" Newtonian Potentials

Using the same argument as In the proof of Proposition 11, we see that it is sufficient to demonstrate that

lim grad ul(z ± rn(z)).n(z) r --.;. 0 r > 0

We have

with

1dz,r)=

1 hi:::) ± - <pi -) uniformly for ::: E r . 2 -

It is clear that the problem is for 1 z - t 1 small; reasoning geometrically, we see that ::: - t is then tangent to r with the result that

:::- t ± w(.:;)! 2 ~ I::: - tI 2 +r2. We deduce that

I 1(':;' r) -> h(.:;) when r -> 0,

and using the proof of Proposition that

1 when 12 (z, r) -> 2<P(z) r -> O. 0

We propose finally to study the regularity of the deril'Gtives of interior alld exterior simple layer potentials right up to the boundary 1. In the same way as for the local regularity considered in §3.2, the continuity of if> is not sufficient to ensure that U il

(resp. u ~ ) of class 'fj I on tii (resp. tie); also <P E Cf," 1(1) is not sufficient to ensure that u~ (resp. u~) is of class (1'5 1 on tij (resp. tie)' By combining the geometrical techniques used in the proofs of Propositions 10, 11 and 12 and the "singular integrals" techniques used in the proof of Proposition 9 we can demonstrate that: i/ r is ofc/ass ({, m + 1., and <p of class ({, m" with mEN alld 0 < ex < 1, then the interior (resp, exterior) simple layer potential defined hy <p is of class (6 m + 1. 7 on tij (resp. tie)' (I!.ld the in~erior (resp. exterior) double layer potential defined by <p is of class Cf,m,~ 011

Q j (resp. Q,,).

We shall content ourselves by proving, by way of an example, the

Proposition 13. Let r be the boundary o/class (f, I + £ o/a regular bounded open set and IRi + increasing and continuous lI'itl!

(3.22) f 8(1') dr < ocsuch that I <p(:::) - <pU)1 :s.:; dl.:;- til, a r

for all t, Z E r 52 .

52 ThaI obviously implies that <f! is continuous on i; it is clear also that this condition is satisfied if <f! is a Holder function on r.


Then the interior (resp. exterior) single layer potential uit (resp. u~) defined by ({Y is or class 'f51 on Qi (resp. Qe).

Remark 9. Since u~ = u\ (Proposition 10), the tangential derivatives of u\ al1d u~ coincide on r 53. 0

Pro oj: We know already that the normal derivatives (~u'tl ('11 and cur / 011 exist (Proposition 11); it is sufficient therefore to consider the tangential derivatives; since they have to be equal on r (Remark 8) we therefore have to show that. for all :: EO r

gradtanguJx) = gradudx) ~ (gradu t (.x). lJ(z))IJ(z)

converges when x ...... z, X EO [Rn\ r. This comes to the same thing as showing that

gradlangUl (z + TIl(z)) converges when r ...... 0, r # 0

with the condition the convergence is uniform with respect to z. To simplify the notation, we can consider a normal parametric representation (R U, (i',:x) centred at z (z = (0, 0)) and whose last base vector is I1(Z}. For x = :: + TIl(::) = (0, r) and t = (t', :xU'))

x - t grad E,,(x ~ t) =.. .

a"lx ~ tin (~t', r ~ :xU')).

a"lx ~ tin

We can always suppose that supp ~Q c U n 1'. We thus have

gradtangudx) = ( 1 K(r, t')t/J(t')dt', 0), with

K(r, t') ----- -- --

a"(1t'l 2 + (r- :x(t'))2)",2

Using the hypotheses on rand ({Y, we see that t/J satisfies (3.22) on (I; also t/J is with compact support in t. The kernel K is singular in the sense in which

I IK(r t')1 ~. . .. when (I, t') -> 0,

, . anlt'I" 1

with the result that we cannot pass to the limit directly be Lebesgue's theorem (I I t'l" - 1 is not locally integrable). But

K(I, t') = grad, En{t', I ~ :xU')) + R(I, t')

.;3 In the case of a double layer potential of a function q; E 'f, 2 + '( r), 1I~ and 11~. arc of class 'f, 1 on 12,. and Q i respectively. As we have seen, u~ - u~ = If' on r. with the result that the tangential derivatives of!i~ and !i~ differ on r by the tangential derivatives of q;: on the contrary, we can show that the normal derivatives (lU~/(!11 and (!U~/(l'l coincide on r. (See Gunther [I] and Courant-Hilbert [Ij).

~ 3. Newtonian Potentials 305

with

This kernel R is not singular and even

R(r, .) -> 0 in Lfoc((ii) when r -> O.

In effect for a compact set .ff of (n we have

I iX(rt') I , ,1 1 - ~r-' Igradx(rt')1 ,

f . IR(r, t )Idt = - f·.· ( ( (r "dl (J • iX rt') )n,_ .;f n .J! Ilrl It'12 + I __ ~_

r /

But Ix(rl')l :::; "[G(r) for WI :::; 1, iX and grad x are bounded on.J(, gradx(rt') -> 0 when r -> 0, for all t'; hence by Lebesgue's theorem

L IR(r, t'lldt' -> 0 when r -> O.

It follows from this calculation that grad tang U j (x) converges with

f Kdr, t'll/l(t')d/' , and that ('

where K j (r, t') = grad,. En(t', r - x(n) .

Let us take p E £i!((!) with p = 1 in the neighbourhood ofsuppl/l. We have

with

1 Kj(r, t')I/I(t'jdt' = I Kdr, t')p(t')I/I(t')dt' = J(r) - fer)

I(r) = I grad/.(En (l', r - iX(t'))p(t'))(I~(n - 1/1(0)) dr'

f(r) = I En{t', r - a(t'))gradp(t')I/I(t')dt' .

In addition to the formula for the differentiation of a product, we have used

1 grad,. (En{t', r - a(t'))p(t'))dt' = O.

From Lebesgue's theorem, these two integrals converge when r -> O. In effect from the hypothesis (3.22)

f dr 11/1(1') - 1/1(0)1 :::; B(It'1) with 0 £(r) -;: < .X!

,l06 Chapter II. The Laplace Operator

and it follows immediately that

Igrad,(E,,(t', r ~ cx(t'))p(t'))1 ~ c( 1 + It)~ 1 )

( 1 \

where C depends only on 11 and p. Since £( I t' I) 1 + ~ 'y~l ) is locally inte-

grable in ;R" - I, we have the result for I(r). The result for J(r) is immediate since

which is locally integrable on lR: n - 1. o The regularity results for simple layer potentials allows to obtain the regularity on the surface of discontinuity of the Newtonian potential of a function f having discontinuities of the first kind. Applying Propositions 12 and 13 we obtain the

Proposition 14. Let Q he (/ regular hounded open set with houndary of class (f, I + c,

f E (f, (Q) and u the Newtonian potential offxQ 54. We suppose that (in the sense 0/

distrihutions in Q) the deri['(ltives satisf.~! all aXI E U(Q) with p > n. Then the restriction ui (resp. II'") of 1/ to Q (resp. lR:" \ Q) is of class ((,'2 on Q (resp. lR:" \ Q). III addit iOll, [(ir I = l. .... 11

where Ilk denotes the k-th component of the (ullit) normaiI.'ector exterior to Q.

Proo{ We have

/j ( (,!! \ = "(~ (,114) = J~(fXQ) \ (lx I ) eX I aXI

since from Ostrogradski's formula (see § 1.3)

On the other hand. since (see Proposition 2)

(11/ lim ~ (x) = 0,

Ixl'" f ('XI

from Proposition 3, iJu/ ()x I is the Newtonian potential of the distribution with compact support (ef/ax l ) XQ ..... fll[d". Since (('flax l ) XQ E U(lR:n) with p > 11, from Proposition (6.3), its Newtonian potential VI is of class (6' 1 on lR:".

'" Throughout this chapter we denote by III the characteristic function of the set 12: InC,) = I if x E Q.

il!IX) = 0 otherwise. Hence the Newtonian potential (Jfjl.!! is u(x) = r 1::,(x - r)((rjdr. for all .f!

'( E R'. We recall (see Proposition 6 point (3)) that Ii is of class (f, 1 on iR".


From Propositions 12 and 13, the restriction wi (resp. Wn to Q (resp. IR"\Q) of the simple layer Newtonian potential of the restriction of f to r is in ((51 (~2) (resp. (f;l(IRn\Q)) and

awl an

Dw7 -- - f on r. an .

wI', we deduce the proposition. D

The above reasoning and the results on simple layer potentials stated above allows us to prove that given a regular bounded open set Q with boundary of" class '{} and fE {f; f (ti), the restrictions to Q and IRn \Q of the Newtonian potential offxfi are of class ((; f on ~2 and IRn\Q respectively. In effect, from what we have seen above, au/eXt is the Newtonian potential of (aj! axtl xo fn t dy; since the interior and exterior Newtonian potentials offn1dy are of class C(;.~ on Q and IRn\Q respectively (see the statement preceding Proposition 13), the above affirmation is deduced by induction

= VIE <"(;X(Q), VI = 1, ... , IJ,

4. Newtonian Potential of a Distribution Without Compact Support

We have defined the Newtonian potential of a distributionfwith compact support as the distribution u = E * ./; that is to say

this has clearly a meaning: for ( E D«(IR"), we have En * ( E 6'(IR") which allows the definition of (r. Ell * (> for alif E 6" (IR") that is to say for every distributionfwith compact support. But the function En * (is not just any function of (['(IR"): it has a well defined behaviour at infinity (see Proposition 2). This will enable us to define (J En * (> for distributions f without compact support, but having a certain beha vior at infinity. We pose the

Definition 4. We say that a distributionf OIJ IR" is regular at infinity (with respect to the Laplacian) if there exists a distribution u on IRn such that

for all P E CI(IR") with p = I in the neighbourhood of 0

(u, D = lim (./; P,( En * (» for all (ESI'(IR") (3.23) , - 0

where p,(x) = p(cx) .


The distribution u (which is obviously unique) is then called the Newtonian potential of the distrihution f regular at infinity.

In other words, f is regular at infinity if the Newtonian potentials of the distributions pi; with compact support, converge in .'0'([Rn) when Ii --+ 1, Ii EO Ci"([Rn); the meaning of this convergence is specified by (3.23), The limit is the Newtonian potential off This /lotion clearly gellerali::es that of the New(()nian potential ofa distrihution with compact support: if f is a distribution with compact support, for p EO 9([R") with p = I in the neighbourhood of 0, for I: sufficiently smail, Pr. = 1 in the neighbourhood of supp f and hence Pr. f = f However this notion does not co~'er all the cases of distrihutionsf(Jr which we can consider a "natural" potential. For example, let us consider for I ~ m ~ n - 1 the distributionf = I Hm ® bw "', that is to say

U; D = J" ((x',O)dx' W'

(3.24)

where [R" = [R'" X [R"" m = {(x', x"): x' EO [RIM, x" E [R" .. In:. It will follow from Proposition 15 below that 1 W' ® ()I>" m is regular iff m < n - 2 (or m = 0 obviously); its Newtonian potential will then be the function

(3.25)

It is clear that for all 0 ~ m ~ 11 - 1, the function u can he considered as the "Newtonian potential" 1.,,'., ® bl,,·, "'. The problem encountered in the cases III = 11 - 2 and In = n - 1, which come in fact from the particular properties of the Laplacian in [R2 and [R, will be considered later (see §3.4.b).

4a. Positive Functions and Measures, Regular at Infinity

We begin by studying the (locally integrable) functions and the positive (Radon) measures, regular at infinity on [Rn and their Newtonian potentials. We have the characterisa tion:

Proposition 15. Let f he a positive Radon measure on a:g n;

(1) f is reqular at infinity iff

case IJ ~ 3,

(3.26) ease /J 2,

case n = 1,

r d/(x)

J (1 + Ixl)"

I LogO + Ixl)dj(x) < x

I Ixl df(x) < x-

(2) supposing this condition satisfied, e'E" (a) .lin" almost all x EO [R", the functions y -> En(x -- y) and y -> - (x - y) are

inteqrable with respect to the measure f ()xi .


(b) the Newtonian potential u off and its derir;atives (lU/ (~Xi are (locally integrable) functions all IR:" given by

u(X) f EnC'" - y)df(y) a.e. x E IR:"

eu I'

cE" Z)x i

(x) J -;;:(x y)df(y) a.e. x E IR:" ; CX i

(e) .If)/" et:ery boullded Borel function p with compact support and with P in a neighbourhood of 0,

lim f t~O B(O.R)

( I ~ 'I) (U . CUt Iu(x) - u,(x) + I l--:-(x) -:::(x) dx = °

(Xi (Xi

fCJI' all R > 0, where UE is the Newtonian potential of the measure with compact support pJwith Pt(x) = p(£x).

We note the

Corollary 2. Let f be a Radon measure (not necessarily positive) on IR:" satisfying:

(3.27) {

case n ~ 3, the measure (I + Ixl)-" + 2fis bounded on IR:";

case n = 2, the measure Log(l + I X I) I is bounded on rr~ 2;

case n = I, the measure xl is hounded on IR:

Then f is regular at infinity.

That follows immediately from the proposItIOn, from the fact that the set of distributions, rt;gular at infinity, is a vector space, and from the decomposition of a Radon measure into positive and negative parts (I + and I - parts respectively):

for ( ~ 0, <1+, D = sup {<.f, p); p :s; (]

.r = (- It, f == f + - f -, I I I = f + + f-

A measure g is bounded iff f dlgl(x) < C(j55.

We note also that the Newtonian potential of a distribution I regular at infinity being a linear function off; under the hypothesis of Corollary 2, we have the conclusion (2) of Proposition 15.

Remark 10. The essential difference between the cases 11 ~ 3 and n = I, 2 appears clearly in the statements above. In particular: (1) in the case n ~ 3, every hounded Radon measure on IR:" is regular at infinity. There exist unbounded measures which are also regular at infinity: every function fE U(IR:") with q < n/2 is regular at infinity; in effect the function 1/(1 + Ixl)" ···2

belongs to U(IR:") for all p > n/(11 - 2).

55 See, for example, Bourbaki [2], Schwartz [2] or Marie [IJ for the study of Radon measures. lUis a (locally integrable) function, these ideas arc c1assical:f+ = sup (O,!).


(2) in the case n 1, 2, every positive measure, regular at infinity, is necessarily bounded. A bounded measure is not however necessarily regular at infinity. 0

Before proving Proposition 15, let us give some examples:

Example 1. Let us consider a function I on [R;" and suppose that

If(xll = O(g(!xl)) when Ixl-.x

where girl is a positive (measurable) function defined for I" > 1"0' Then the regularity at infinity off is assured by

if 11 of J t: rg(r) dr < J~ ~,

(3.28) r if n 2, r Log (1 + r)g(r)dr < cr~ . r"

If when Ixl -+ x, f? 0 and y(lxl) = 0U(x)), then the condition (3.28) is necessary to ensure the regularity at infinity of.r In particular, we note that a function I on [R;" satisfying

i.f'(x)1 ~ c

for with i: > 0

is regular at infinity. o Example 2. For 1 ~ In ~ IJ - 1, the measure 1 ~m ® 6G<" m is regular at infinity if I In < 11 - 2. It is in effect a positive measure and condition (3.26) reduces to 111 < 11 - 2. The Newtonian potential u is then given by

u(x', x") = f E"(x' y', x") dy' = I r ,~'" k" JG<'" (Ix'

dy'

Example 3. Let us consider the case where we are given a regular open set Q of [R;",

but whose boundary r is unhounded 56 . For rp E ,&O(r), the measure I = rpd}' is regular at infinity under the hypothesis expressing (3.27):

< 'JJ

(3.29)

Log(! + IxIJirp(x)ld;(x) <x .

l

'h For example a half-space or a cylinder.


If <p :;:, 0, this is a necessary condition for <pd~' to be regular at infinity. The Newtonian potential of f is the simple layer potential U j , harmonic on [R"\[',

defined by

udx) = t EII(x - t)<p(t)d,'(t) for all x E [R"\r .

Let us suppose that Q is "uniformly regular", that is to say that the function 1/(1 + I x I)k is integrable with respect to d,' if k > n -- 1. In this case, the measure <pd}' is, in particular, regular at infinity if

C l(p(x)1 :(1~lji' for Ixl > ro, x E J with i: > o.

We shall now prove Proposition J 5; we give in a lemma the principal argument of its proof:

Lemma 8. Let f be a positive measure on [R", g a (Borel) function, locally integrable on [R" and w: [R + --> [R +, continuous and satisfying

(3.30) w(lx - yl) :( C(lxi)w(lyl) for all x, Y E [R" ,

with C: [R + --> [R + continuous. (I) Let us suppose

Then

Ig(x)1 = O(O)(lxl)) when Ixl --> x

f w(lxlldf(x) < x .

(a) jar almost all x E [R", thefunctioll Y --> g(x - .v) is integrable with respect tor

(h) the jimction v: x --> f g(x - y) df(y) is locally integrable 011 [R";

(c) jt)r every bounded Boreljunction p with compact support and with p 1 in the neighbourhood of 0

lim J~ Ig * (pJ) - rldx = 0 for all R > 0 ,~o B(O.R)

where, as before: p,(.') = p(l:x) .

(2) Let us suppose that when I x I --+x

y(x) :;:, 0 and w(lxl) = O(y(x)) .

Then f w(lxl)df(x) <x is a necessary condition.!!)r the existence ora distrihution

t' on [R" satisfying

{

for all p E C/'([RII) with P =v 1

<v. s> for all s E Sl([RII) .

ill the neiyhhourllOod of 0

lim <q ". p,.!: ~) , ~ 0

J 1 ~ Chapter II. The Laplace Opera lor

Proof of Propositio/l 15. We note first of all that the functions

1 (/)1(1") = ! + r, wdr) = ! + Log(! + r), (I) dr) = ..... (k ~ 0)

(! + r)k

satisfy the condition (3.30). In effect we have

wl(lx yl):;;:; + Ixl + 1.1'1:;;:; w 1 (lxl)wdlyl)

(3.31) wiLx 1'1):;;:; + Log(! + Ixl + iyl) :;;:; wdixl)wdlyl)

(I) dlx -- 1'1) :;;:; (!)dlxl)(1) ,(IYi)

the last inequality resulting from

! + Iyi :;;:; ! + ix - .1'1 + Ixl :;;:; (1 + Ixl)(1 + Ix .1'1)·

Applying the lemma with

case 11 ~ 3, Y ----- r~n' (J ) = (JJ2

case 11 = 2, if [;2 (J) = (!) l.

case II = I, il [;1 (!) = WI

we obtain the point (I) and the part of the point (2) relevant to the Newtonian potential u. Applying the lemma with?J = DEn/ax" W = WI II we obtain the part of the point (2) relative to the derivatives ?u/ ("iXi of the Newtonian potential. 0

Proof of LCnlllla 8. First of all we prove (1). The hypothesis y(x) = O(w(lxi)) can be written

(3.32)

We fix R > 0 and Po continuous with compact support with 0 :;;:; Po :;;:; ! and Po = I on B(O, R + "0)' The measure Po fbeing of compact support we can apply Lemma 3 to it; this gives the results (a) and (b) on replacing f by Po f Now (1- f)o) I has support in [P;" 13(0, R + r o ); hence. using (3.32) and (3.30), for x E B(O. R) and Y E supp(1 ..... Po)'/: wc have

Iy(x - .1)1 :;;:; Cow(lx - .1'1) :;;:; C(R)w(lyl)

with C(R) = Co sup C(r) , r -:::; R

we deduce that

r dx f Iy(x - y)i(1 olB(O. R)

PoCv))df(y) <x

and hence by Fubini's theorem, the points (a) and (h) on replacingfby (1 - Po) f Combining the two results we complete the proof of points (a) and (h); the point (c) is then proved easily by the use of Lebesgue's theorem. To prove (2), we observe that the hypothesis w(lxl) = O(glxl) can be written

g(x) ~ cw{lxl) for Ixl ~,.o with (' > O.


We fix p, 'E g&([Rn) with ° ~ P ~ 1,' ~ 0, P = 1 on B(O, ro + 1), supp' c B(O, 1). We have

<g * (pJ), 0 = 10 + Ie

with

10 = <g * pf, 0

Ie = f '(x)dx f g(x - y)(p(ey) - p(y))df(y)·

For e > ° sufficiently small, Pe ~ P and Pe = P on B(O, ro + 1). Then, since Ix - yl ~ ro for x E supp' and y E sUPP(Pe - p)

Ie ~ f '(x)dx f cw(lx - YI)(p(ey) - p(y))df(y) .

From (3.30),

w(lyi) = w(lx - (x - y)l) ~ C(lxl)w(lx - yl) ,

hence

Ie ~ C1 f '(x)dx f w(IYIHp{ey) - p(y))df(y) ,

with c

sup C(r) r ~ 1

This proves (2). o We now propose to generalize the characterisations of Propositions 3 and 4. I t is clear that the Newtonian potential of a distribution f on [Rn regular at irifinity is a solution of Poisson's equation

.du = f on [Rn.

In effect, given , E £2!([Rn) we choose p E Qt([Rn) such that p = 1 in the neighbourhood of 0; for e sufficiently small, Pe = 1 in the neighbourhood of supp C now the Newtonian potential Ue of Pe f satisfies the Poisson equation

.due = Pef on [Rn,

and hence

.dur. = f in the neighbourhood of supp C

We deduce that if u is the Newtonian potential off,

<.du,O = <u, .dO = lim <up .dO = <f, 0 .

This being true for all C, u is clearly a solution of Poisson's equation. We wish to


characterise u by a condition at infinity: the situation is not however as simple as in the case of a distribution with compact support. Let us prove the following result in the radial case:

Lemma 9. Let I he a positive radial measure 0/1 IRn, regular at i~finity and u its Newtonian potential. Then when Ixl -+ ~57

case n ): 3, u(x) -+ 0

case IJ = 2, u(x) - E2 (x) f df(y) -+ 0

case 11 ): 2, Igradu(x)l- o(-~) - Ixl'

Proof We have

u(x) En(x) f df(y) + r EnCv)df(y) B(O.lx!1 J~" BIO.lxll

x f grad u(x) = ~-n dflv) . anlxl BiO.lxll

We see easily that the formulae established in Lemma 4 for a function with compact support are still valid for a positive measure regular at infinity. In the case IJ ): 3, we have for Ixl ): R > 0

lu(x)1 :( IEn(x)1 f dI(y) + r IEn(Y)1 df(y) . B(O. RI JW\BIO. RI

Since

r IEn!Y)1 df(y) :( ~1 (1 + ~)n - 2 r ... _. df(Y)n J~"\B(O. RI Iknl R JW\ B(O. RI (1 + lyl) 2 '

this integral tends to zero when R -+ x. For E > 0, we can therefore choose R > 0 such that

t"\BiO. RI IEn(Y)ldf(y) < ~; R being fixed, we can find ro such that

Ixl > ro => l£n(x)1 f df(y) < ~ , B(O. RI ~

and hence lu(xH < E. This shows that u(x) -+ 0 when Ixl -+ x.

'7 For a function r defined a.e. on IR", we say that !'(x) .... ° when Ixl .... y_ if there exists ro and a function c(r) defined for all r > ro such that

lim I:(r) = 0, 11'(x)I';; <:(Ixl). a.e. x with Ixl > r" .


In the case n = 2,/ is a bounded measure; we have for I x I ~ I

f 1 i !vl . o ~ lI(X) - E2(x) df(y) = -2 '., Log 1"'1 dJ (y) IT [f.l2\BI0.!,11 .\

~ ~ f Log(J + Iyl)df(y)· 2IT if.l2" B(O.lxll

Finally in the case n ~ 2, for I x I > I

I f 2" - 2 f . Igradu(x)1 = ... df(v) <: ---- ---. - df(y) (T,lxl n - 1 B((I.I-II .. --..c: (In xl(1 + Ixl)" - 2 B(O.lxll .

2" - 2 f df(y) ~ O'nlxl (I+I}'I)n- 2 . D

Remark 11. The results of the lemma still remain true for a measure f on [R", dominated at infinity by a radial positive measure q, reiJular at infinity, that is to say that for an 1'0 > 0, we should have

1<./: DI ~ f ((x)diJ(x) for (~O with supp( c B(O, ro) .

In the case of functions, that means (see Example 1 above)

if(x)1 ~ iJ(lxl) for Ixl > '-0'

In effect, since the result is true for a distribution with compact support (see Proposition 2) we can always suppose that

suppl c [R"\B(O, ro).

It is clear also that f is regular at infinity and that its Newtonian potential is:

u(x) = f En(x y)df(y) .

In the case 11 ~ 3, we have immediately

lu(x)1 ~ f IE,Jx - y) diJ(Y)

where [' is the Newtonian potential of iJ. Therefore

- v(x)

u(x) ---> 0 when I x I ---> cc .

We shall similarly prove the other results. On lhe other hand, the results ofthe lemma are false in general iff is not radial: the example off = 1 J.!''' (8) ()I-!" m shows this clearly; it is a positive measure, regular at infinity if III < n - 2, whose Newtonian potential

u(x', x") = En _ m(.x")

does not tend to zero at infinity (except obviously in the directions of XU); the


regularity of the measure at finite distance has nothing to do with this property: we can easily construct a function IE 't) x ([J;£n), regular at infinity and whose Newtonian potential, in the case n ;?- 3, does not tend to zero at infinity: it will be sufficient that its support is contained in a band.

D

However we can get round this difllculty by replacing the topological limits hy limits in mean. We pose the

Definition 5. We say that a (locally integrable) function I on W tends to :ero at infinity ill mean if

kli~, L lu(kx)ldx = 0 for every compact set K of [J;£" \ [O} .

It is clear that if u(x) -> 0 when Ixl+ :c (in the topological sense), then it tends to zero in mean. But the converse is not true: a function U E LP([J;£n) with I :( p < x tends to zero at infinity in mean (whereas in general it does not tend to zero in the topological sense); in effect by Iioider's inequality:

1

J~ u(h)ldx = k "J" lu(x)ldx :( k- II (' r I( k 'I( Jk 'I(

dx ) 1 - P- Ii u If

= k -;; (L dX} -~ We now give the characterisation:

Proposition 16. LI'I fhl' a positivI' Ifll'aSlire on [J;£n, regular at infinity and let 11 he a locally integrahle fUllction 0/1 R<": (I) Case n ;?- 3: u is the Newtonian potential olf iff' u is a solution of Poisson's equatio/l

lind tends 10 :ero al ill/inity in mean; (2) Casl' n = 2: II is the Newtonian potential off iff' [/ IS a solution of Poisson's equat ion

/1u = f on

and the.!illlctioll u ~ E2 f df(y) tellds to :1'1"0 at infinity in mean.

(3) Case n ;?- 2; 11 is, 10 within an additiL'e COlJstwu, the Newtonian potemial off iff u is a solution (){ Poisson's equation

Llu = I on [J;£n

and tizejilllctio/l u/(l + Ixl) tends to zero at infinity ill mean.

Proof: First, we prove the necessary conditions. We havc already seen that the Newtonian potential was a solution of Poisson's equation; we have thus only to

~3. Ncwtlmian Potentials 317

show that in the case n ~ 3 (resp. n = 2) U (resp. U - E2 J df(y)) tend to zero at infinity in mean. In the case n ~ 3, the fundamental solution En, also u, is negative. The function U

will tend to zero at infinity in mean iff it is the same for its radialized function

u(x) = f u(:xla)dlJ. 2.

Now u is the Newtonian potential of the radialized function? of{: by commutation of the Laplacian and of the radialization (see Proposition 6 of § 1), that follows from the characterization of Newtonian potentials in the case of a distribution with compact support (see Proposition 3); since U is the limit of Newtonian potentials of p, r; choosing p radial, that is true for an arbitrary distribution regular at infinity. From Lemma 9, ii(x) -Jo 0 when I x I -Jo 'x and hence a fortiori u tends to zero at infinity in mean. In the case n = 2,

u(x)-E 2 (x) f df(y) =1 fLO/X - yl df(y) ~ J f LOg( 1 + l1)df (J). 2n Ixl 2n Ixl

From Lebesgue's theorem, since r Log (1 + I yl) df( y) <x , we have

f ( IYI). Log I +, df(y) -Jo 0 when Ixl -Jo x . Ixl

Hence U E2 f df(y) is majorised by a function tending to zero at infinity. It

follows that u - E2 r d[(y) tends to zero at infinity in mean ill' its radialized

11 - E2 f d[(y) tend: to zero at infinity in mean. We take up again the above

reasonmg. Finally we show that the conditions are sufficient. By linearity, it is enough to show that a harmonic function U on [Rn tending to zero at infinity in mean (resp. such that u/O + Ix I) tends to zero at infinity in mean) is zero (resp. constant). Admitting for the moment that u is a tempered distribution; then from Proposition 3 of §2, u is a (harmonic) polynomial; denoting its degree by m and its principle part by Urn' we have t 1/I(kx)1 dx ~ k m t lu"Jx)1 dx

(' resp. f Ju(kx)l dx ~ km - 1 f LUm(X)l dX ) when k -Jo 'x . K I + Ikxl K Ixl

Hence III = 0 and 11m = 0, that is to say 1/ = 0 (resp. III = I and Um = 0 or m = 0, that is to say u = constant). It remains therefore to show that a (locally integrable) function v tending to zero at


infinity in mean is a tempered distribution: that follows from the estimate ('

J lr(x)1 dx ~ C(I + Rn) . BIO.R)

We can establish this estimate by noting that

r Ir(x)ldx = k"j il'(kx)ldx ~ Cok". Jlk~!xl:(2k~ :1~ixl(2f

PuttingC 1 - j' Idx)ldx,wethenhaveforR> I,choosingk): I such that - • B(O.l)

2k - 1 < R ~ 2\

k

~ C 1 + Co I 2" li - 1) ~ C 1 + CoRn. 1

o The limit at infinity in mean, if it gives a simple characterization of the Newtonian potentials of a positive measure, regular at infinity, is not a very precise instrument. We can in fact evaluate the heizariour otthe N ell'lolJian potential at infinity in a much more precise fashion. We indicate two methods hcre: a method of weiyhts and a truncation method. We shall restrict ourselves to the case 11 ): 3.

Proposition 17. Letfhe a {JOsilil'e measure. reyu[ar at infinity on [Rn with 11 ): 3 and u its Newtonian pot entia!. Then we have t he est imates 5 8:

( I )

(3.33)

(2) jilr ° < '1 < 11 - 2.

f lu(x)ld~

(I + Ix1 2 )2 + 1

(3.34)

(3) for alit> 0,

(3.35) (ffIU(X)1

with

~ '1(11

1)+ dX)"- 2 ~ ~~" (f df(X))"

en = ( 11 )'" IJ 2 4" -- 1 ,.

(J~

Proof Since U is the limit of Newtonian potentials of p, r: using the Fatou lebesgue lemma we see that we can always suppose that f has compact support.

," In each estimate. we have the convention that if the integral on the right is finite. then the integral on the left is finite and we have the inequality_


In regularizing eventually, we can also suppose thatfis regular, with the result that u is a classical solution of Poisson's equation

Llu = f on IR·.

For rx > 0, we put ga = (1 + IxI 2 )-a/2 . We have

grad ga = - rxxga + 2

Llga = - rxga + 4(n + (n - 2 - rx)lxI2) .

Applying Green's formula on the ball B(O, R), we have

f gJdx = f uLlgadx + 11 - 12 JB(O.R) JB(O.R)

with

11 = f g ou dy iJB(O.R) a on

f oga 12 = U - dy.

8B(O.R) on Using the values of ga' grad ga and the estimates of u and grad u at infinity, namely

(u(x) = O(1/lxl· - 2), I grad u(x)1 = O(1/lxl· - 1)

(see Proposition 2) we have

1111 ~ CR- a,

where C depends on n, f, rx but not on R. Therefore 11 and 12 tend to ° when R --> 00.

Now u = E. * f is negative, so if ° < rx ~ n - 2

uLiga = rxlulga + 4(n + (n - 2 - rx)lxI2) ~ 0.

In the limit, we deduce

f gJdx = rx f lulga + 4(n + (n - 2 - rx)lxI 2)dx . J~n J~n

For rx = n - 2, we obtain (3.33) (with equality even). For ° < rx < n - 2, (n + (n - 2 - rx)lxI2) ~ (n - 2 - rx)(l + IxI 2), from which we have (3.34). To prove (3.35), we use quite another method. First of all, we show that for every compact set K of IR·,

f ( I K I ) 2 /•• n !. (3.36) J K I E.(x)1 dx ~ c. -;;::-. wIth c. = 2 x (n - 2)2(. - 1)

where I K I denotes the Lebesgue measure of the compact set K.

320 Chapter II. 'The Laplace Operator

In effect, for R > 0,

r IE,Jx)ldx,s; I f~Jx)i dx oJ K

(n ..... 2)u1l J '

This being true for all R > 0,

r' I (R2 I K I )' J IE (x)1 dx ,s; min -- -- + _I ---- . K II R" 0 n - 2 2 (JnRI! .. 2

Calculating this minimum, we obtain (3.36) Now we have u = Ell * t; so

r lu(x)ldx = I", dx f IEn!x - y)lf(y)dy = If(y)dy j~ EIl(x)ldx &JK ... K J ... K+r

, J' ) ( I K I ) 2 '" ,s; ( [(y)dy en .-;:

since IK + .vI = IKI· In particular, applying this inequality with K(t) = {x; lu(x)1 ? t], we have

tIK(t)1 ~-:; f KIt)

from which we derive

IK(t)I,s; ((~ fnYldyr"-2 x (J2,rtJ --- 2

II

Finally, using the fact that

d I'

dt J (lui - t) + dx

we obtain

- I K(tll

11 - 2 x-·

2 ((J," fill - 21 .

o

4b. Other Examples of Newtonian Potentials of Distributions Without Compact Support

In the preceding §4.a., we have characterised the positive measures, regular at inflnity. This gives a criterion for an arbitrary measure (see Corollary 2), but there exist measures (or/unctions), regular at infinity, without their absolute value bein{} so: this corresponds to the case of integrals or series which are semi-convergent (i.e. convergent, but not absolutely convergent).


We illustrate this by considering the case of a functionfintegrable on III We define its primitive:

F(x) = f 7C f(y)dy .

Given p E £/(IR) with p = 1 in the neighbourhood ofO, the Newtonian potential of pJ is given by

u£(x) = ~ I Ix - Ylp(ey)f(y)dy .

Integrating by parts we may this as

1 IX 1 f y, e I u,(X) =:2 _ f p(ey)F(y)dy -:2 x p(ey)F(y)dy -:2 Ix - Ylp'(ey)F(y)dy .

The convergence of u£ as t: ---> ° is assured by the integrability of F on III In effect

1£lx - Ylp'(ey)1 :( 81xlllp'IIc + Ilyp'(y)IIL'

with the result that then

r; I Ix - Ylp'(sy)F(y)dy ---> 0 uniformly for x bounded

This condition of integrability of a primitive off is different from the condition I I x II 1(x)1 dx < XJ which characterizes the regularity at infinity of the absolute

value off: for example

f sin x 2

. (x) =

is integrable on IR; its absolute value is not regular at infini ty 59, butf can be written in the form

d fIx) = fo(x) + -d F(x)

x

wherefo is of absolute value regular at infinity and F is integrable60; it is therefore regular at infinity. In a general manner, we prove in this direction the

59 flsin"'ldX = I' Isinxl dx = ± rrrsi~Xdx = + 7~ .

• x J 0 x 0 J 0 x + kIT

60 Choosing fJ E (/(IR) with fJ = 1 in the neighbourhood of 0 (to eliminate the singularity at 0) take

1 cos Xl

F(x) (p(x) - l)~

2 x"

sin x 2 3 cos x 2 1 cos x 2

J~(x) = - p(x) -~ + - (1 - p(x))-Xl 2 .\"4

.... p'(x)··· 2 . x-'


Proposition 18. Letfbe a distribution onlR" and let us suppose that it can be written

{=

where (f~) is a jamily of measures 011 IRn satish'iIlY:

for 11 ~ 2 alld 11 + I etl ~ 3, the measuref~/(1 + I x I)" + 1'1 - 2 is houmled on IR"

(3.37) for IJ = 2 and et = 0, the measllre Log( 1 + Ixl)f~ is hounded 0/1 IR n

.f{)f' II = 1, the measllres xj~ alldf~ (et ~ 1) are hOllnded on IR"

Then f is reYlIlar at infinity.

Proof We fix p, ~ E 9(1R") and consider

with

Y. = (1+ Ix I)n +-rar=2 { f~ for IJ ~ 2, II + letl ~ 3

(1 + Log(l + Ixl))j~ for

{

(1' - fJ (1 + Ixl)" + Ixl - 2_ -(E *V)

(Ix' - fJ " "

Vo.p = £2 * ( 1 + Log(1 + Ixl)

(we shall consider the case /I = I separately). Using Proposition 2, for all :J. ~ fi ~ 0

IJ = 2, et = 0

for /I ~ 2,

for IJ = 2,

v •. p(x) = O(lxI IPI ) when Ixl --> x . Now

1I+letl~3

et = 0

bl We use Leibnitz's formula: for:l = (:I" ...• :X,,), /i = (/i , .... , Ii" )./i ~ :x means /ii ~ :Xi for i = I. ))

...• 11::1 - fi = (:I, - {i" ... ,:I" - fi"): q = n -----. i /ii' ('Y. i - fli)'


afip . . Hence ........... p .. < v" p lS bounded on [R", umformly for [; > 0 and when (; -> 0, ax

c.p p, {P(O)Va,o if /3 = 0 --v -> i3x iJ "f! 0 if IPI > 0 ,

Since, by hypothesis, the measures g, are bounded we deduce that

This shows that f is regular at infinity; its Newtonian potential is defined by

where u, is the locally integrable function on [R" defined by

323

The case n 1 is treated in the same way, noting that in this case, when 1 x 1 -> Cf)

d El * ((x) = O(lx:), dx (El * () = 0(1),

dk

Ci;;,,(E I * 0 = 0 for k ~ 2, o

This proposition can be applied to the study of the regularity at infinity offunctions as we have done above in the case n = 1. It can also be used for distributions, We now give the example of the distribution of order 1 defininy a double layer potential:

Example 4. Let Q be a regular open set of [R", but with unbounded boundary r and q; E ((jo(r), For the distribution of order 1,/ = - div(q;nd/)62, the condition (3,37) can be written

(3,38) t dP~~~~1V~1 < Cf) ,

The distribution is then regular at infinity; its Newtonian potential is the double layer potential u2 , harmonic on [Rn\r and defined by

f cE" u2 (x) = -~-(t - x)q;(t)dl'(t) for all x E [R"\r , r on

If we suppose that Q is uniformly regular (see Example 3), the condition (3,37) is

62 Which defines the double layer potential of ({J (see Definition 2).


satisfied, in particular, if

c 1<p(x)1 ::;; [XT'. for Ixl > ro, x E r with I: > O. D

As we remarked at the beginning of this sect. 4 of §3, we can consider "natural" potentials for distributions, non-regular at infinity. We indicate here the method of subtraction of an "infinite constant". To present this method we consider the example of a measure f = 1 cr;!" - 1 @ fz (n ~ 3), where fz is a measure with compact support in [Rz. As we have seen (Example 2) iffz is positive, this measure is not regular at infinity. We now prove the

Proposition 19. Letj~ be a measure with compact support on [Rl, Uz the Newtonian potential offz, p a Borelfunction with compact support in [R' - z(n ~ 3) with p = 1 in the neighbourhood ofO. For all I: > 0, consider u' the Newtonian potential of the measure Pc @.I; with compact support in [R', pAx) = p(l:x). Then setting

C = - df ( U) P Y d ' 1 (i ) ( i (I: ') ) , k. iR ' Z Y iR" - 1 (1 + I y' 12) i-I Y ,

we have for all x' E [Rn - 2 and almost all XU E [RZ

UC(x', XU) - C, -> u2(x") when c -> O.

M ore precisely, for all R > 0,

lim ( ( sup lu'(x', x") - C, - uZ(X")I)dX" = O. ,- 0 jilX"1 '" RI {Ix'i '" RI

Proof We have

u'(x', x") = fw-, p(ey')dy' fcr;!l En(x' - y', x" - y")dfz(y")·

Making use of

1 170 t dt E.(x', XU) = - - (-t 2-+-1 -'I~Z-)n/=2

(In Ix"l x


u'(x', XU) = - ~ ( dfz(y") (00 tdt ( 2 p(ey') d ' (In jcr;!l jlX" _ )'''1 jcr;!" 1 (t + lx' - y'12)n/Z y ,

and letting I: -> 0, we obtain

1 r p(I:Y') d' I r dy' (In jcr;!" 1 (tZ + lx' - y'12)n/Z y -> (In jcr;!" 1 (t2 + lx' _ y'IZ)n/2

(J n - 2 f 7. r' - 3 dr = (J.t 2 0 (I + r2 r/2

*3. Newtonian Potentials 325

the convergence being uniform with respect to x' bounded in IRn - 2 and dominated with respect to t. Hence, since!2 has compact support, we find when /; -+ 0

1 f J"IX" - Y"I r p(Gy') (J" H2 df2(y") 1 tdt Jw 2 (t2 + lx' _ y;T.iY;/idy'

-+ ~ d!2(y") - -~ = Ez(x" - y")dj~(y") = U2(X") , i fix" y"1 d i 2n 0<2 1 t [R'

the convergence being uniform with respect to x' bounded in IRn - 2 and dominated with respect to XU E IRz. It is sufficient therefore to prove that

when (; -+ 0, uniformly for x' bounded in IR" - 2.

By integration over t in Ie and a change of variable, we obtain

I = (- ~ r. dj' (y,,)) ( r ~(y' + ex') -=::P(Y'2 dV') . , kn J~2 Z J~, 2 (£2 + Ili 2 )(,,/2) - 1 .

Using the hypothesis p = 1 in the neighbourhood of 0 and the continuity of the translation in L I, we obtain the limit I, -> O. D

Proposition 19 shows on the one hand that for the measure!2, with compact support in IR z the distribution! = lw 2 ®!2 in 1R" (n ? 3) is regular at infinity iff

IT{2 d!z(x") = O. On the other hand, when f~2 d!2(.x") =I 0, with the condition

that when the constant C, (tending to infinity) is subtracted, the Newtonian potentials of p, ® !2 converge when £ -> O. We shall show in the same way that forj~ a measure with compact support in IR and a bounded Borel function p with compact support in 1R" - 1 (n ? 3) with p = 1 in the neighbourhood of 0, there exists C, such that

En * (p, ®!1) + C, -. lW- 1 ® (EI * j~) when (; -+ O.

A calculation identical with that above will show that we can take

C, = (L dfl{x") )(fw I p(n')E"{X')dX') '

and hence that 1~, 1 ®!1 is regular at infinity in IR" iff t dfl (x") = O.

We can show that, in the two preceding examples, the distributions!, not regular at infinity in IRn satisfy: for all p E ~(lRn), there exists C" tending to infinity when Ie: -> 0, such that En * (p, f) + C, converges. The limit has to be independent of p (when the "infinite" constants C, depend on p); we call this limit the Newtonian potential off We emphasise however that in this method the Newtonian potential is defined to within an additive constant (we can

326 Chapter [I, The Laplace Operator

always add a "finite" constant to C, which will always be "infinite"), From a mathematical point of view, that has an important consequence: if the sum of two distributions is regular at infinity, the sum of their Newtonian potentials is not necessarily the Newtonian potential of their sum; it is only it to within an additive constant. The method of subtraction of an infinite constant does not permit the definition of a Newtonian potential for any distribution whatsoever: for example, the function f == 1 (which is not regular at infinity), does not admit a Newtonian potential by the method of subtraction of an infinite constant. That can be seen easily by remarking that for p E '.I'(iR"),

grad (p, * E,,)(x) = p, * grad E,,(x) = -.~ f grad (y + £x)E"C)') dy

diverges when I: -> O. Now, if{ E .(I"(iR") admits a Newtonian potential to within an additive constant, then

grad [(p,f) .x. En] converges when /: -> 0,

This convergence is a priori in 0/'(iRn); we can show, in fact, in the examples, that it is stronger. For example under the hypotheses of Proposition 19, we have

lim ( (sup Igrad x 1/'(x', x")1 + igrad x ,1/'(x', x") - grad 1/2(x")1 )dx" =0, l~oJ{lx"l <RI Ilx'I"'RI /

5. Some Physical Interpretations (in Mechanics and Electrostatics)

Newtonian Potentials. In mechanics, let p be the density of matter distributed in a bounded domain of iR 3 ,

Let us assume that the gravitational force f which acts then on a point mass m placed at point x is the sum of the forces obeying Newton's law due to each "element of matter"' p( y) d.\', With the aid of Newton's law (see §2,6) we thus obtain:

f(x) = - f .. .1 mkggrad E 3 (x - y)p(y)dy , 1<

This force is derived from a potential 1'" given by:

[',,(xl = kg l." E 3 (x - y)p(y)dy

which is thus (to within the coefficient kg) the Newtonian potential of the density of matter p, and which is therefore a solution of Poisson's equation

;1[:" = kqp ,

We stress the fact that we have thus chosen a particular solution of Poisson's equation, namely the one which tends to zero at infini ty63,

6.1 The possihility of this choice means to a physicist the absence of mass at infinity.


We can act similarly in electricity, for a charge density p: [R3 ---> [R with the Coulomb force. We again obtain, in a perfect homogeneous medium (or in the vacuum), a force f deriving from a potential

vc(x) = - k, 1" E3(X - y)p(y)dy .

This potential is caned the Coulomb potential. Examples of the Coulomb potential are given by the simple and double layer potentials. (i) A density of electric charge f3 distributed over a regular bounded surface F (for example the surface of a conductor) creates a simple layer (Coulomb) potential in the vacuum, this potential being given at a point x E [RJ by:

vAx) = -~ f P(t) I dy(t) = - k, f E 3 (x - t)f3(t)dy(t)64 . 4n r It - x

The electric field E being related to the Coulomb potential Vc by: E = -- grad Vc the jump in the normal derivative of Vc across F (see Proposition 1) means a discontinuity in the normal component of the electric field across F, proportional to the charge density f3:

. [CVe J [ E nJ[· = Ee n - E' n = - _.- = k f3 • • • "'I c· on

(for f3 E cgO(F) .

On the contrary, there is no jump in the components, tangential to F of the electric field (see Remark 9). (ii) A density of electric moment et. (see Jackson [1] p. 36, Jouget [1], p. 169) on a surface F, creates in the vacuum, at a point x E [R3, a pouble layer (Coulomb) potential vc(x) given by:

vc(x) = -:~ 1 et.(t) ~?SI(~=:'12n(t» dy(t)

- ke f : E3(t - x)ot(t)d]!(t) r ()n

By Proposition 11, the potential is discontinuous across r:

implying a discontinuity of the tangential components of the electric field E (for et. E cg 2 + '(F)), and then that the normal component of E is continuous across F . Examples of simple and double layer potentials can also be found in magnetostatics. and in fluid mechanics (see Brard [IJ).

Problems with Cylindrical (or Plane) Geometry. Especially in electricity, we are led to study distributions of charge which are cylindrical or plane (for example an

04 Do not confuse E3 elementary solution in [R.J of the Laplacian with the third component of the electric field.

.12K Chapter II. The Laplace Operatllf

electric wire or the plates of a plane condenser); it is important to recognise the new difficulties due to the existence of charges at infinity when we take, in Ilrst approximation, the cylinder or the plates considered to be inllnite. This is given by Proposition 19 (with 11 = 3). The direct study of the case of a distribution of electric charges p constant on a rectilinear wire AS. infinitesimally thin, of length L is interesting. The Coulomb ("Newtonian") potential created at a point x E [R;" is given by:

k 1'1 + I'lJ + 1",.(X) = 'p Log

4n rj + rlJ

with

Denoting by r the distance of the point x from the x 3-axis which carries the wire An. For I -> x, the (Coulomb) potential lAx) has an asymptotic form 1':" given by

k, I /" (x) = {dog·

as . In r k(

p log! 2n

k, ( , -- p loa r L

2n eo

which thus appears (to within the coefficient kcp) as the elementary solution t~J(x) in 1r~2 Similarly a uniform distribution of charge in a plane leads to the occurrence of the elementary solution E 1 (x) in R as a consequence of Proposition 19.

§4. Classical Theory of Dirichlet's Problem

L Generalities on Dirichlet's Problem P(Q, (p) in the case Q Bounded: Classical Solution, Examples, Outline of Perron's Method, Generalized Solutions, Regular Point of the Boundary, Barrier Function

We consider an open set Q in [R;" with boundary rand (p defIned on 1'. We pose the problem of Ilnding II satisfying

{.' Li is harmonic on Q

P(Q,<p) , 11 = (p on I

This is Dirichlet's problem/c)r Laplace's equatio/J. Within the classical framework of this ~4, we shall suppose that <p E 0,o(r): the case of a less regular given function <p

will be considered in ~6. We pose the

h.' In the third member of this equation, the two numbers I and r are here considered to he dimensionless.

~4. Classical Theory of Dirichlet's Problem 329

Definition 1. Given an open set Q of ~n with boundary rand cp E cgo(r); we call a function u a classical solution of P(Q, (p) if u is harmonic on Q and is such that

lim u(x) = cp(zl for all z E r . X --4-:: XEQ

In other words u is the classical solution of P(Q, cp) if u E £(Q) !l cgO(Q) and cp is the trace of u on r. Now let us suppose that Q is hounded. We know then (see Corollary 3 of §2) that for all u c ,Jt(Q) !l cgO(Q),

sup lui = max lui; Q r

in other words, the (linear) map which takes u to its trace cp on r is an isometry of J'1(Q) n (to(Q) with the uniform convergence norm on Q, that is to say

Ilull = sup lui Q

into ~0(T') given the uniform convergence norm on r: II cp II = max I cp I .

r

We have likewise seen (Proposition 10 of §2) that every uniform limit of harmonic functions on Q was harmonic on Q. It follows that the set of traces cp on r of the u E £(Q) !l cgO«(J) is a closed vector su b-space of cg0(J'). From the definition of a classical solution of P(Q, <p) we have the

Proposition I. Let Q he a bounded open set of ~n with boundary r. (1) The set E(Q) of the cp E '?50(T) such that the Dirichlet problem P(Q, cp) admits a classical solution is a closed vector sub-space of~'O(T); (2) for all <P E E(Q), there exists one and only one classical solution u(cp) of P(Q, cp); the map cp ~ u(cp) is a (linear) isometry of E(cp), with norm the uniform convergence norm on r, onto ,*,'(Q) !l (lIo(!'2), with norm the uniform convergence norm on Q.

From the principle of the maximum, we have for cp E E(Q)

(4.1) min ((J :S; u(cp) ~: max (P on Q [' J

and even more precisely if Qo is a connected component of Q with boundary r 0

(4.2) { cp constant on I'o = U(lp) constant on Qo

(p not constant on 1'0 = min cp < u(cp) < max cp sur Qo .

Let us now give some examples.

Example 1. Q = ] a, h [ a hounded interval of~. The boundary of Q is r = {a, h}. The functions of ,)f)(Q} are the affine functions u(x) = AX + Ii. It is immediate that the solution of P(Q, cp) is

(4.3) . cp(a)(b - x) + <p(b)(x - a)

u(x) = - b _ a ...... _ .......... . o

J30 Chapter II. The Laplace Operator

Example 2. Q = B(xo, ro) a ball in [R".

Let B = B(xo, ro) and cP E C(;;'o(('B). Poisson's integral formula 66 gives us an indication that the classical solution of P(B, cp), if it exists, is given by

(4.4) I J" rJ - I x - xo 11

1I(X} = ;'o~~ ('B--It--=~I"-- cp(t)d~,(t) .

Let us now prove the

Proposition 2. Beiny oiven B = B(xo' ro) and cp E C(, o((lB), tile function u de.fined hy (4.4) is the classical solution of' P(B, cpl.

Proof: F or all r E (lB, the function

rJ - Ix - x o l1 X ---> - --------

(JII It - xln

is harmonic on B: this in effect is the function

gradEn(t - x).(t + x - 2xo )

a product each term of which is harmonic on B, hence of the Laplacian

( (' ) (1 24 ;,-gradEn(t - x) ,;--.-(t + x - 2xo) = 2JEn(t - x) = o.

, (Xi I.X i

Now from the proof of Poisson's formula 66 , we have, for x E B,

I rJ - Ix - x ol1 ---- --- ------

(JII It - xi" ( r )" - 1

x) - 1';--=\01 En(t

where

y

We therefore see that

u(x) = u2 (x} _ ( . r o_. )" - 2 112 (.1'), Ix - .xol

where U z is the double layer potential of cp on (lB (see Definition 2 and above all Proposition 11 of ~3). Given:: E (1B, we know that

LB (1

::)cp(t) dy(t) cp(::)

lim uz(X) = ~ En(t - + x---lo= ('n 2 XEB

LB i'

- ::)cp(t) d),(t) -cp(::)

lim u2(Y) ~ - EII(t -----

( n 2 \'-----+:

YEW\B

"6 Sec Proposition 9 of *2.


Hence

( ' )" -2) f 0 !i~ u(x) = 1 (IZ ~o xul 8B on E,,(t - z)<p(t)dy(t) xER

+ (1 + ( __ ~)n -2)(~(~! = <p(z) . Iz - xol 2

o

Example 3. Q = B(xo, ro)\{xo} punctured ball 67 oflR", (n :? 2). Let B = B(xo, ro) a ball of IR" and Q = B\ {xo} whose boundary is r = oB U {xo}. Being given (p EO E(Q), the solution u(<p) EO .~(Q) and

lim u(<p, x) = <p(XO)68 x -> Xo

We know (see Proposition 16 of §2) that the function u defined on B by

{ u(<p. ' x) for x EO Q u(x) =

<p(xo) for x = Xo

is harmonic on B and hence the solution of P(B, <Po) where <Po is the restriction of <p to aB. From the formula of the mean:

(4.5) <p(xo) = anr~ -lIB (p(tjdy(t)

and from Poisson's integral formula

(4.6) for x E Q .

The converse can be deduced from Proposition 2, in the case oj' a punctured ball Q = B(xo, rol\{xo} oj'lR" with n :? 2, E(Q) = {<p E rco(oQ) satisj'ying (4.5)} and for all <p E E(<p), u((p) is given by (4.6). 0

Example 4. A Dirichlet problem in an anllulus

Q = {x; r l < Ixl < r2 } .

LetQ = {x;rl < Ix: < r 2}withO < r l < r2 < cx:aringinlR"whoseboundary is r = riB(O, r 1 ) U oB(O, r 2 ).

Also let <p be a radial function Oil r, that is to say

<p == <p(rJ on cB(O, r;) (i = 1, 2) .

We look for a radial solution of P(Q, <p): if it exists, then from uniqueness this will be the solution u(<p). A radial function u, which is harmonic on Q, is ofthe form (see Proposition 4 of §2)

u(x) = Co + C 1 En(x) on Q.

67 Punctured = deprived of a point (the point xoJ 68 We use here (and in the sequel) Ihe notation:

dcr

u(rp. x) = u(rp)(x) .


The boundary condition can be written

(p(rJ = Co + ('1 En(rJ (i 1,2) ,

which determines the constants and gives

(4.7) <p(r2 )(E,,(x), E,,(r 1 )) + (p(rtl(E,,(r 2 ) - En(x))

u(x) =---En (r 2 ) En(r!)

Note that the problem for a non-radial function will be very much more complicated. 0

Example 5. A Dirichlet problem .fiJI' (J rectangle ill We consider a rectangle Q = Ja t • (/2 [ X ]b t • 172 [of r = r I u r 2 u r 3 u r 4 where

: (x. yJ: with boundary

r .>

raj, (/2] x :h t }.

[at: x [b t , 17 2 ] and

r

r 1

4

[a 1 ,a2 J x [b 2 } ,

ra 2 } x [17 1 , b1J .

Let (P E (6°(f) be zero al the rertices (a i • hj ) Ii.j = L 2) of Q with the result that <Pi = <PXJ,' If, for i = I, 2. 3, 4, [Ii is the classical solution of P(Q, <p;) then 1.1 = u 1+ u2 + u 3 + U 4 is the classical solution of P(Q, (.p). In other words, we can, within a change of notation reduce the problem to one in which (P = 0 on r {all x Jb t , 17 2 [, To within a translation, we can suppose that Q = ] 0, a [ x J a, h [ . We are thus led to the following problem: given <P E (6([0. h]) with <prO) = <p(b) = 0, to find u harmonic on Q = JO. a[ x JO, 17[. continuous on Q and satisfying

{ ufO, y) = <pry) for Y E JO, b[

uta. y) = u(X, 0) = u(x, b) = ° for x E [0, aJ and y E [0, bJ .

We shall solve this problem by developing u in a Fourier series. We seek u(x, y) in the form

(4.8) u(X, y) = I uJx) sin nwy with (!)

tl?: 1

-

h

The boundary conditions can be written

<pry) = I untO) sinlJwy that is to say

o = I un(a) sin nwy that is to say

The condition that u is harmonic can be written

that is to say

04. Classical Theory of Dirichlet's Problem 333

from which by using the boundary conditions we derive the formula

(4.9) 2 sh nUJ(a - x) j'b .

un{X) = - ----... sm (nUJy)<p(y) dy . b sh nUJa °

We have carried out a formal calculation: however the reader familiar with the elements of Fourier series will see that by supposing <p to be a little more than continuous (e.g. satisfying a Holder condition), the series (4.8) where (un) is given by (4.9), converges uniformly on Q to the classical solution of the Dirichlet problem P(Q, tpX!(J} x [O.b])· 0

Example 3 shows that there does not always exist a classical solution of P(Q, tp): furthermore, it shows also that we can, at least in this case, consider a "natural" solution of P(Q, tp) without it being a classical solution. The use of the subharmonic fi.mctions developed in Perron's mel hod will permit us to define a generalized solution of P(Q, <p), for every bounded open Q of [R0 with boundary r and every tp E C€o(f). We introduce these notions here without proof; the development of this method will be treated in §4.6. First, we give

Definition 2. Let Q be an open set of [Rn. We say that v E C€o(Q) is sub-harmonic on Q if

,11) ~ 0 in '2iJ'(Q}

that is to say

(AI', 0 = f vt1(dx ~ 0 for all (E .~i'(Q) with (~O.

We say that H" E (Q) is superharmol1ic on Q if r = w is sub-harmonic on Q. Note that we have defined here sub-harmonicity and super-harmonicity for a continuous function: we shall come back to this notion in a more general setting69.

From Definition 2, it is immediate that a function u E (fio(Q) is harmonic iff it is at the same time subharmonic and supcrharmonic; also the set o{the 1) E (6°(Q) which are slIbharmol1ic on Q is a closed coile of{6°(Q), that is to say that

for VI, V2 E C€O(Q), ),1' )'2 ~ 0,

1'1' 1'2 subharmonic = ;'11'1 + ).21'2 sub-harmonic

for a sequence (v n ) of (fio(Q) converging to 1) E C€o(Q),

Un subharmonic "In = l' sub-harmonic.

69 An example of a sub-harmonic function is a quadratic form

V(x) = 2.:2.: aUx i "" with x = (x, .. .. , Xi" .. , Xl" .. , X o ),

with trace 2.: a ii positive. i = 1


We shall principally make use of the principle of the maximum jC)/' subharl1lonic continuous functions, which we state here in the form:

Property 1. Let Q be a bounded open set of 1R1/1 with houndary I, u E ji'(Q) n ,{;o(Q) and r E ((,o(Q) suhharmonic. Suppose that

Theil

lim sup [,(x) ~ u(z) jiJr all Z E r . x -:: 'E Q

[' ~ u ~ max u on Q. r

This property will be proved in a more general setting in *4.6 (see Propositions 20 and 21): its localization is characteristic and will serve to give a definition to the notion of a general subharmonic function (see Definition 15). This property can be reformulated in terms of Dirichlet's problem: let Q be a bounded open set in 1R1" with boundary I and <p E 'C0(T); we use the notation

det .f _ (<p) = {[' E ((j0(Q) sub-harmonic such that

lim sup v(x) ~ <p('::), V.:: E Jj (4.1O) x-+=

def .f+ (<p) =: Ii' E ,{;o(Q) super-harmonic such that

lim infw(x) ? ep(z), V.:: En. The principle of the maximum (Property 1) shows immediately:

(4.11)

and u is the classical solution of" P(Q, (p) iiI u E .1_ (ep) n f + (ep). Note that .1_ (ep)

and .f + (<p) are not empty since they contain v == min ep and w == max (p respect-/" ["

ively. The property (4.11) therefore permits us to consider for all ep E ,{;o(n and all x E Q

def def

(4.12) U (ep, x) = sup dx) and u+(P, xl = inf w(x). 1 E J (<pI \\'E.1+(<pI

The essential result of Perron's method is Wiener's theorem which we now state:

Property 2. Let Q he a hounded open set ill 1R1" with boundary r and let ep E (6°(r). (I) for all x E Q, (with the notation ol(4.12)), we have:

lL(ep, x) = U+((P, x);

70 Apply the Property I to the sub-harmonic function r - \\' and u '" o.

*4. Classical Theory of Dirichlet's Problem 335

(2) the function u(cp) defined on Q by

(4.13) u(cp)(x) = u(cp, x) = u_(cp, x) = u+(cp, x), x E Q

is harmonic on Q.

We shall prove this theorem in §4.6 (Proposition 22). Admitting it, we pose the

Definition 3. Given Q a bounded open set in ~. with boundary rand cp E rcO(r), we call the function u(cp), harmonic on Q, defined by (4.13), the generalized solution of P(Q, cp).

We observe that the generalized solution of P(Q, cp)' is defined for all cp E rcO(r). From the preceding remarks, it is clear that, if it exists, the classical solution of P(Q, cp) is the generalized solution: in other words the map which with cp E rcO(Q) associates the generalized solution u(cp), extends the map which with every cp E E(Q) associates the classical solution u(cp). We show that this extension possesses interesting properties:

Proposition 3. Given a bounded open set Q in ~. with boundary r, (1) for all cp E (eo(r)

(4.14) min cp ,,; u(cp) ,,; max cp ; r r

(2) The map cp --+ u(cp) is linear and continuous from rcO(r) into Jlt'b(Q) the space of bounded harmonic functions on Q with norm the uniform convergence norm.

Proof The point (1) follows from the definition since the functions

v = mincp and w = maxcp r r

belong to .1_(cp) and f+(cp) respectively. We now note that from (4.10), (4.12). we have

{ u+(- cp) = - u_(cp). U_(Acp) = ),u_(cp)

(4.15) u-(cpd + u_(cpz) ,,; U-(CPl + CP2)'

Since u(cp) = u_(cp) = u+(cp) it follows that the map cp--+ u(cp) IS linear. The continuity is a consequence of (4.14)

II u(cp) II = sup lu(cp)1 ,,; max Icpl = IIcpll . o Q r

To return to the classical Dirichlet problem, we introduce the

Definition 4. Given an open set Q in ~. with boundary rand z E r we say that z is a regular boundary point of Q if there exists r > 0 and v satisfying

v is continuous and sub-harmonic on Q n B(z, r)

(4.16) for all 0 < b < r, sup v < 0

Q n B(z, r)\B(z, 6)

lim v(x) = o. x~z


A function v satisfying (4.16) will be called a barrier function ofz on Q n 8(z, r). We shall specify it in §4.6 (see Proposition 23).

Property 3. Let Q be a bounded open set in IRn with boundary rand z a regular point of the boundary ofQ. Then, for all qJ E ~o (r), the generalized solution u(qJ) of P(Q, qJ) satisfies

lim u(qJ, x) = qJ(z)

We deduce the

Theorem 1. Let Q be a bounded open set in IRn with boundary r. The following assertions are equivalent: (i) for all qJ E ~O(r), P(Q, qJ) admits a classical solution:

(ii) all the points of the boundary of Q are regular.

Proof First, we prove that (i) => (ii). Given z E r, using (i) we consider the classical solution u of P(Q, qJ) with qJ(x) = - Ix - zl. Then u is a barrier function

of z on Q: in effect u E ff(Q), lim u(x) = 0, and from the principle of the

maximum we have, for all b > 0

sup u < 0 Q\B(:.6)

since u E ~O(Q) and u < 0 on r\ {z}. The implication (ii) => (i) follows from Property 3: for qJ E ~O(r) the generalized solution of P(Q, qJ) satisfies, with the use of (ii),

lim u(qJ, x) = qJ(z) for all z E r ,

and hence is the classical solution of P(Q, qJ). o We emphasize that the useful part of theorem, (ii) => (i), is also the difficult part which springs from the Properties 1, 2 and 3 which we shall prove in §4.6. The interest of this result is to reduce the existence of a classical solution of Dirichlet's problem to the study of the regularity of boundary points of Q, that is to say to the search for barrier functions. We give here some elementary examples.

Example 6. The case 11 = 1. Given Q an open set of IR with boundary rand z E r, the function v(x) = - Ix - zl is a barrier function of z on Q. We deduce from Theorem 1, that for every bounded open set Q of IR with boundary r and every qJ E ~O(r), there exists a (unique) classical solution of P(Q, qJ). This can be shown directly: an open set Q of IR is the union of intervals ]a i , bi [ two by two disjoint, their connected components; supposing that the intervals ]a i , b;[ are bounded, we have {ai' bi} c r. Given qJ E ~0(r), we can, using the formula (4.3) of Example 1, consider the function u E ff(Q) defined by

qJ(a;)(bi - x) + qJ(b;)(x - a;) b u(x) = b. _ a. for x E ]ai , i[·

I I

§4. Classical Theory of Dirichlet's Problem 337

Let us show that this function u is the classical solution of P(Q, (p). For that we consider Z E r and a sequence (xk ) of Q such that X k -> z; if z is an isolated point of r, then it belongs to the boundary in at least one and at the most two intervals ]ai , bi [ it is clear that u(xd -> <p(z); if z is a limit point of r, considering the intervals] ai., bik [ containing X k we shall have ai• -> z, bi• -> z; since (p(z), <p(b i.) -> <p(z); from which u(xd -> <p(z) since

lu(xd - <p(zll ~ max(l<p(aiJ - <p(z)l, 1<p(b;J - <p(z)I)· o Example 7. Exterior ball condition. We say that an open set Q of fRn satisfies the exterior ball condition at a point z of its boundary if there exists Xo E fR" \ {z} such that

zl) n Q = 0.

Replacing Xo by Axo + (1 - ),)z with 0 < ). < 1, it comes to the same thing to suppose that

B(xo, Ixo - zl) n t2 = {z} .

The function v(x) = En{z - x o) - En(x - xo) is then a barrier function of z on Q. We note that, in particular, a convex open set satisfies the exterior ball condition at every point of its boundary: in effect from the Hahn-Banach theorem (in a finite dimensional space), an open set Q is convex iff for every point z of its boundary there exists Xo E fRn\{z} such that

Q n {x E fRn; (x - z). (xo - z) ): O} = 0 .

Hence from Theorem 1, we have

Corollary 1. Given a convex bounded open set Q with boundary r, for all <p E ~O(r), there exists a (unique) classical solution of P(Q, (p). This criterion of the exterior ball is very rough: we now give a finer criterion, distinguishing the cases n = 2 and IJ ): 3.

Example 8. Case /J = 2: exterior segment condition. We say that an open set Q in ~~" satisfies the external segment condition at a point z of' ils boundary if there exists Xo E IR"\{:::} such that

ho + (I - ;.)zrj:Q forall }.E[O,I].

In the case n = 2, Ihis condition is sufficient to show that z is a regular point of the boundary. In effect we specify a point x of fR2 by its polar coordinates r= Ix zi, e = angle (z - X o, x - z). The condition can be written

Q n B(z, ro) c {x; 0 < r < ro, - n < () < nJ where ro = Ixo zl .

We can always suppose that 0 < ro < 1 and we then verify without difficulty that

Logr 1'(x) = ............. ·7

(Log r)2 + H-

is a barrier function of z on Q n B(z, ro).

338 Chapter IT. The Laplace Operator

Using conformal transformations (see §2.5) we can obtain other criteria 71 for the regularity of a point on the boundary of an open set of [R2. In the case n ~ 3, the exterior segment condition is no longer sufficient for the regularity of a boundary point.

Example 9. Exterior cone condition. We say that an open set ill [Rn satisfies the exterior cone condition at a point z of its boundary if there exists an open convex cone with vertex 0 and ro > 0 such that

Q n (z + C) n B(z, ro) = 0.

in other words if there exists Xo E [Rn\[zl. 0 < 00 < n. ro > 0 such that

(4.17) Q n B(z. ro) c (x; angle(z - x o, x - z) < ()0]12 .

We shall see that this condition is sufficient to show that z is a regular point of the holt/utary of Q. For that, we must suppose that n ~ 3, and we shall seek a barrier function of the form

l'(x) = - r~fW).

where r = Ix - zl, 0 = angle (z - x o, x - z) E [0, 00 [. with ), > 0 and f continuous and strictly positive on [0, 00 ], still to be determined, but such that .d t' ~ 0 in ct'(Q).

Denoting by x' the projection of x on the straight line D = z + [R(xo - ;:),

p = r sin () = Ilx' - xii, ~ = r cos 0 = x' - z •

and using the formula for the Laplacian in cylindrical coordinates (see § 1.4) we have

-p cp

III './ '(Q\D) •

from which, passing to polar coordinates (r, 0) in the ~p-plane, and taking account (~r ?t' p ?r cos 0

of -' +-' -- .. we have ?p (~r r i':() r'

7, The open set Q (cf. Fig. 2) does not satisfy the exterior segment condition (at :) but has a barrier function if the curve is analytic. 72 The open set (cf. Fig. 3) does not satisfy the external ball condition at z. It satisfies the external cone condition at z iff a > O.

--v Fig. 2 Fig. 3


;]2u n - 1 au 1 (aZp au) LIp = ~ + ~~ - + - - + (n - 2) cotg e-cr2 r or rZ ae2 ae

1 a ( n _ 1 au) 1 a ( . en _ Z at') = rn -1 or r ar + rZ sin en -Z ae Sill ae

We fix eo < e~ < n and choose

(4.18) f(!~ ( t ) n-Z I(e) = -. - dt

8 Sill t


I ~ /(00 ) > ° on [0, eo] , and

. ((n - 2)8n--3 LIt' = r,,-2

sin en - z - W. + n - 2)I(e))

(n - 2)en - 3

Since I(e) ~ I(O) and . e Z Sill n

n - 2 ~ --}-, choosing

(0

(4.19) ) -~~ I n - 2 (( 4) I/Z

. - 2 + e6.f(0)Z

on Q\D.

339

we have ), > ° and Llv ~ ° on Q\D. To complete the proof we must show that Llu ~ ° in 9'(Q). For that it is enough to show that

lim f uLlC dx ~ ° for C E 9(Q) , C ~ ° f;--+O t.\;~; < 0 < 00 :

with C depending only on rand 0. Now

dx = an_lpn-zdpd~ = an_lrn-lsinen-zdrdO.

Hence, using the formula of the Laplacian in the coordinates (r, 8) and integrating by parts, remembering that supp ( c ]0, OJ [x] 0, 00 [.

I f fOG fC1c a ( OC) ~- v LlC dx= sin on-Z dO u:1 rn- 1 ;) dr an - l [,,<0<0,,1 t 0 (·r (.r

+ rn -3 dr u - sin on -2 --- de fx fOG a ( O() o t 00 ae

1 f . = ~- I" Llvdx an - 1 '1'<0<0 I t· OJ


from which

We have therefore proved that, under the hypothesis (4.17) the fUllction 1'( x) = - r;/(O) wheref is yiren hy (4.18) and j, hy (4.19) is a h1lrrier/illlction of:: 011

Q n B(O, 1'0)'

We sum up in a Proposition the results of Examples 6, 8 and 9:

Proposition 4. Let Q he an open set ill !R" with houlldary rand:: E f. (i) Case n = 1,:: is a reyular point of f;

(ii) case Il = 2, ifQ satisfies the exterior seyment cOlldition at :: (see Example 8) then :: is a reyular point of f:

(iii) case 11 ;:?; 3, ifQ satisfies the exterior LOne condition at :: (see Example 9) thell :: is a reyular point or r.

When Q is a regular open set (of class C(;' 1) (see §1.3.a.) it is clear that it satisfies the exterior cone condition V:: E r: we have also

J[ .

for all R, for all 00 > 2, , there eXIsts '-0 such that

Q n B(z, 1'0) c(x; angle (- n(z), x z) < 00 1

for all :: Ern B(O, R) .

From Proposition 4 and Theorem I, we therefore have the

Corollary 2. Let Q be a houllded regular open set (oFelass 1(, I) of!R" ,vith howulary r. Theil for all <P E ((i,o(r), there exists a unique classical solutioll of P(Q, <p).

We shall study later the regularity of the (classical) solution of P(Q, <p) as a function of that of Q and <p (see §6). We complete this section by the

Remark 1. In the case 11 ;:?; 2, given an open set Q in !R", and Xo E Q, X ll is not a

regular point of the boundary of Qo = Q\{xo}. More precisely, assuming Q to be bounded (which it can always be made to be), for all <Po E 1(,0(3QO) the ycnerali::ed so/ulion o{P(Q(), <Pol is the restrictiollto Qo oj"tize yeneralized solution o{P(Q, (p) with 0 the function

l'oe,) = [(x) + i:(EIl(x - xo) - Ell (max I:: - xo!)) i' fl

IS 111ft (<Po); therefore

u(<Po);:?; u((p) + I:(£/I(.X - xo) - EIl(maxi: til

x o l)) ;

in the limit as I: --+ 0, u( (Po) ;:?; u( tp); applying to - (p (or repeating the argument


with IV E .J + (tp)), u(tpo) = u(tp) on Qo. This property will be extended in §5 to Q\K where K is a compact set of "zero capacity". 0

2. Generalities on the Dirichlet Problem P(fJ, rp,f) and the Green's Function of fJ, a Bounded Open Set

Let us suppose that Q is an open set in [Rn with boundary r, tp defined on r, f defined on Q. We now consider Dirichlet's problem for Poisson's equation

P(Q, tp,f) {J:: ~ :: ~. Dirichlet's problem for Laplace's equation P(Q, tp) is thus the particular case P(Q, (P, 0). We shall consider also the particular case P(Q, OJ):

PH(Q,f) { Ju = f on

u = 0 on

Q

r, which we shall call the homogeneous Dirichlet problem (for Poisson's equation). Generalizing the Definition 1, we pose the

Definition 5. Let Q be an open set in [R" with boundary r, tp E ~O(T) and f E «50(Q). We call by the name classical solution of P(Q, tp, f) every function u E «5'2(Q) n ~O(Q) such that

{ J u(x) == f(x) for all x E Q

u(z) = tp(z) for ail z E r . As in the case of Poisson's equation (see § 1.1) this (classical) notion of solution is not sufficient: it is necessary to introduce a concept of solution in the sense of distributions; however, we shall use in this classical setting only distributions on Q,

The introduction of distributions on the boundary r will be postponed until later (see §6). In a natural fashion in the classical setting, given u E .sd'(Q) and Z E r, we shall say that

lim u(x) = I

if for all c; > 0, there exists f > 0 such that

I - I: ~ U ~ I + [, in '}1'(Q n B(z, r» , that is to say

l(u,O - II ~ s for all (E .SC(Q n B(z, r» (:;:;, 0, f (dx = 1 .

If U E ~o(Q) this idea of a limit clearly coincides with the classical topological idea; more generally if u is continuous in the neighbourhood of z in Q, that is to say, if there exists fO > 0 and fi E (6'0(Q n B(z, fO» such that 11 = ii in 01'(Q n B(z, ro»,


then lim u(x) I in the sense specified above iff lim i/(x) = I in the classical sense. x---'=

We then pose the

Definition 6. Given Q an open set in IR" with boundary T, <P E (6°(T) andf E C/'(Q) a quasi-classical solutioll o{P(Q, <p,n is the name given to a distribution u satisfying

;lu = f in CI'(Q)73

lim u(x) = <p(z) for all ;; E 1'74

x---+=

Remark 2. It is clear that 1I is a classical of P(Q, (PJ) iff u E '6 2(Q) and is a quasiclassical solution of P(Q, (P, fl. In the case 11 = L for f E (6,0(Q), the notions of classical and quasi-classical solutions coincide 75; in the case 11 ?: 2Jcontinuous is no longer sufficient, but from the theorems on local regularity of the solutions of Poisson's equation (see Propositions 1 and 9 of §3) iffis locally a H Olderful1ction on

Q (and hence a fortiori iff == ° 011 Q) the flO/iollS of' classical and quasi-classical solUI ions 0/ P(Q, <p, f) coincide 7 5. 0

Remark 3. Dirichlet's problem P(Q, <PJ) is linear with respect to the data (<pJ): if Iii is a classical (resp. quasi-classical) solution of P(Q, <Pi'.t;) and ;'i E IR, then II = Xli ui is a classical (resp. quasi-classical) solution of P(Q, L J. i <Pi' l' ;.;/;l. In particular

P(Q, <PJ) ~ P(Q, <pi + PH(Q,fJ

in the sense in which /I,p is a classical solution of P(Q, <pi and ur is a classical (resp. quasi-classical) solution of PH(Q,f), 11 = u<p + ur is a classical (resp. quasiclassical) solution of P(Q, (PJ). In fact Dirichlet's prohlem liir Poisson's equation is the "sum" of Dirichlet's prohlem f{J/' Laplace's and {(n' Poisson's equatiolJ: if' 110 E (6 2 (Q) (resp. Uo E C/'(Q)), satisj'yillY

lim uo(x) = (Po(::) exists {(ir all : E T , x-::;

is a solution of Poisson's equation LJu o = fon Q, then u is a classical (resp. quasiclassical) solution of P(Q. (pJ) iff 11 - Lin is a classical solution of prO, (P - <Po).

o

We suppose noll' that Q is hounded. First we note that there is then uniqueness of a quasi-classical so/utioll of P(Q, <p, f): in effect if uland 112 are two quasi-classical solutions of P(Q, tp, fl, by linearity u = 111 - U 2 is a quasi-classical solution of P(Q, 0, 0) and hence (see Remark 2) the classical solution u == ° of P(Q, 0). Givenf E 7 '(Q), we shall denote

'.l Sec Definition 2 of ~ 1. 74 In the sense specified above.

'5 With the condition of identifying U E «(,2{Q) with the distribution which it defines.


E(QJ) = {cp E c("O(T); there exists a quasi-classical solution of P(Q, CPJ)} and for cP E E(Q, f), u( cP, f) denotes the quasi-classical solution of P(Q, cP, f) From Remark 3, if CPo E E(Q,f)

E(QJ) = CPo + E(Q) ,

and for cP E E(QJ),

where E(Q) E(Q, 0) is a closed subspace of C(,O(T) (see Proposition I) and u(cp - CPo) is the classical solution of P(Q, cP - CPo). Now for all (,0 E ~,o(T) we can consider the generalized solution u(cp - CPo); the function u(CPoJ) + u(cp (Po) is independent of the choice of CPo: indeed, if CPo is another element of E(Q, f)

u(CPoJ) + u(P - (Po) = [u(PoJ) + lI(CPo - CPo)] + u(cp - CPo)

u(ooJ) + [lI(CPo - CPo) + u(cp -- CPo)]

u(CPoJ) + lI(cp - CPo)

where we have used the linearity of the quasi-classical solution and the linearity of the generalized solution. This permits us to define a generalized solution of P(Q, cP,f) when E(Q,f) of 0, that is to say when there exists 140 E 'Y'(Q) a solution of Poisson's equation

L1uo = f, and satisfying lim uo(x) exists for all Z E Q.

We pose the

Definition 7. Let Q be a bounded open set in IR" and letf' E 9'(Q). We suppose that there exists Uo E 0' '(Q) such that

{ Lluo = f in 9'(Q)

~i~ uo(x) = CPo(z) exists for all

(4.20)

Z E r.

For all (Po E (to(T) we then define the generalized solution of' P(Q, cP,1) to be the function uo + u(cp- CPo) where u(cp (Po) is the generalized solution of P(Q, (P - (Pol (see Definition 3). As we have shown above this function depends only on Q, (P andf it is independent of the choice of Uo satisfying (4.20).

We introduce now the Green's function of' Q. Given Y E Q the distribution uo(x) = En(.x - y) satisfies (4.20) with f = ()y and CPo = CPy defined by

(4.21) (P,.(z) = En(z - y) for all Z E r . The distribution Uo - u((Py) is the generalized solution of PH(Q, .5,.). We pose the

Definition 8. Given Q a bounded open set in IR", we call the Green's jUllctioll of Q,

the function G defined on Q x Q\D, where D = [(x, Y) E Q x Q; x = y} is the


diagonal of Q x Q, by

(4.22) G(x. y) = EIl(x - .v) - u(q>,., xl

where for all J' E Q, u((Py) is the generalized solution of P(Q, (Pr)' (PI' being given by (4.21 J. In other words, as we have seen, G is the function defined on Q x Q.D by the property:jor all Y E Q, G(., y) is rhe gelleralized solution of PH(Q, (),.). Let us begin by giving two examples of Green's functions:

Example 10. Green's junctio/J 0/ a hounded inten:al ]a, b[ For y E ]a, h[, we have

(p\.(a) 1

( \' '") .

from which, applying (4.3) we have

a), (Pr(h)

1 (y lI(Pr' x) =

a) (h - x) + (h y) (x- a)

2 h - a

and hence

G(x, r) = ~ {IX which can be ,written

(x - b)(y - aJ h

if x > y - a

(4.23) G(x, y) h) (x - a)( y -

if x < y h - a

or, alternatively

G(x, y) (max(x, .1') - h)(min(x, y) a) ---------- -- --

h - a

Example II. Green's junction of (/ Ball ill [R" (n ? 2).

o

Let B B(xo,I"oJ. Using Poisson's integral formula, its Green's function is given by

G(x, y) = EIl(x - y) - 0 .. __ -;;..Jl.. EIl(t - y)di'(t) . I f ,.2 - Ix - X 12

rO{)1l ('/J It - xl

We shall obtain a more explicit formula by using the inversion operator I(xo, 1'0)'

Given y E B\{xo}. the function

rlJ /(xo, ro)y = Xo + - ------:-.12 (r - x o) ,

Ir - .\0 w(x) = E,,(x - I(X o.fo)}'). where

is harmonic in the neighbourhood of B since I(x o. 1'0))' l' B. For z E ?B we have


from the relation (2.12) of §2,

case n ~ 3, ( IY xl)n-2 w(z) = ~o 0 E.(z - y)

1 I y - Xo I w(z) = E2 (z - y) - - Log ---- .

2n '0 case n = 2,

From the uniqueness of the Dirichlet problem P(B, cp), we therefore have

u(cpy, x) = (I '0 1)·-2 w(x) in the case n ~ 3 y - Xo

1 I y - xol u(cpy, x) = w(x) + -2- Log in the case n = 2,

n ro

from which we deduce

case n ;;. 3, G(x, y) = E,(x - Y) - En(-I}-' -_x_ol (x _ xo) _ ._r_o_ (y - X O))

ro Iy-xol

(4.24) case n = 2,

I Ix - yl G(x, y) = -. Log .

211 I-I}-' -_xo_1 (x _ x o) _ _ ._ro_ (y - xo)1 ro I y - X o I

These formulae are valid for Y E B\ {xo} and x E B\ {y}. But we note that G is symmetric (as in the case n = 1 where this is immediate); in effect, replacing x - X o, Y - Yo by x, y respectively, that comes down to showing that

IUi x - '0 yl = lL1y - ~xl ro Iyl ro Ixl

that we verify immediately by squaring both sides. Hence for x E B\ {xo},

(4.25) { En(x - xo) - En(ro)

G(x, x o) = G(xo, x) = I Ix - xol - Log ----'-2n '0

if n ~ 3

if n = 2.

We now give the general properties of the Green's functions.

Proposition 5. Let Q be a bounded open set in IR· and G its Green's function (I) G is symmetric:

(4.26) G(x, y) = G(y, x) for all (x, y) E Q x Q\D;

(2) gillen x, y E Q not connected in Q,

G(x, y) = 0;

o

346 Chapter !I.The Laplace Operator

(3) gil'ell x, y E 12, x i= y. belonging to the same connected component Q o o( Q

(4.27) E,,/x -.~ .1')

wilerc ()(\. y) = min ( max i x - ::: I, l~ax y - :::1 \ : \ l.Qil f Q(I )

(4) C is allalytic Oil Q x Q.[) (//l1i fi)!' erery OpCIl set Q() with Qo c Q. the lIlap

J' -> C(., y) is ww/ylicji'om Q o in!o#h(QQo);

(5) suppose that ill addition

(4.28) lim G(x. y) o ji)/' all (:::, Y) E r x Q; .\ __ -l' ::

(lnd i" (]

lim (x.r) () Ii)/' all (;;. y) E r x Q.

Proo/ oj the points (2) to (5). We shall in fact demonstrate the properties on the function (x, y) E Q x Q -+ 1I(i/),.x). where Ip,. is defined by (4.21) and U(ip,) is the generalized solution of P(Q. (1',).

It is clear that the function y -+ (P,. is analytic from 12 into (I, °1F). Since the map !.(J> U(lp) is linear and continuous from (I,f'(n intolfh(Q) (with norm the uniform convergence norm) (sec Proposition 3). the map y _ . .,. lI(!.(Jy) is analytic from Q into Yf b(Q) and hence (x. y) -+ 11(1',.. x) is analytic on Q x Q. This proves (4). We have also

The supplementary hypothesis comes down to saying that for all Y E Q. II( is the classical solution of P(Q. (p,.). that is to say. (P, E £(Q). Since Lim is a closed subspace of (f,0(f) and (I' --> l/(p) IS an isometry of E(Q) into (f,'\f2). ?\Pr!iy' E E(Q) for ally. E r~" and Y E Q and the map.\' --> 11(,0,.) from Q into ,(,o(Q) is analytic. This proves the point (5). We consider now 120 a connected component of Q and Y E Q. The restriction lIi) of 11(11',) to Q() is clearly the generalized solution of P(QI), (P(j) where (Po(;;) f~,,(: - r)

is the restriction of (,0, to (cQ(). If r $ Q". we ha ve r $ Q Il and {lo! x) = Fnl x - r). proving the point (2). Now let us suppose y (c Uo. hom (4. J 4),

110 :S; max (Po = LII (max I Y - ;;1 '.) , ,-!!() r'fll) I

This proves the first inequality of (4.27). under the claim of the symmetry of (j which we shall prove later. Finally. we prove lIo(X) > LnC\ .. y) for all x E!to :y:. In the case 11 l. Q o is an interval and this is satisfied immediately by the explicit formula (4.23). Let us suppose then that 11 ? 2; then. firstly. Q o [y: is connected: since l/olx) E,,(x - J) is harmonic on Q o :y:, we see. from the principle of the

94. Classical Theory of Dirichlet's Problem 347

maximum, that it is enough to show that uoC,) ~ En(x ,~" y) for all x E Qo\ Lv}. By the definition of a generalized solution, it is enough to prove that

for a continuous function lV, superharmonic on Qo satisfying

(4.29) lim inf w(x) ~ En(;:- Y) for all Z E (lQO we have

w(x) ~ E,,(x ~- y) for all x E Qo\.IY] .

Since lim En( x - y) = - =c there exists 0 < r < dist (y, (lQO) such that

w(x) ~ E,,(x - y) for x E E(y, r)\[y} .

Now the function 1'(x) = En (.,

Q 1 = Qo\E(y, r) and satisfies y) w(x) is continuous and sub-harmonic on

lim sup ['(xl ~ 0 for all ;: E ?Q I = rQo u (lB( y. r) . x-;:

Hence from the principle of the maximum for sub-harmonic functions (see Property I), r(x) ~ 0 on QI' 0

To prove the point (I) of Proposition 5, we shall make use of a result on the continuous dependence of the Green's function as a function of Q:

Proposition 6. Let (Q,) be an increasing sequence of'sets ill [R" wirh hounded union Q. Denote hy Gk thefimetion defined on Q x Q\D hy

- , 'l - {GdX ; y) if (x, Y,) E Q k x Q k GdX,} - . o If (x, y) rt Q k x Q k

where Gk is t he Green's function of' Qk' Then for all (x, y) E Q x Q\D the sequence (Gk(x, .1')) ('oncerges by decreasing to G(x, .v) the Green's function of' Q.

Let us admit this proposition, for the moment, to complete the proof of Proposition 5. For that we shall make use of two other results:

Lemma 1. Let Q he all open set ill [R". There exists all increasing sequence ()f'regular bounded open sets with houndaries o( class ((,X whose union is Q.

We leave to the reader the details of the proof of this lemma: using a covering of Q

by a sequence of balls Bk with 13k c:: Q we are led to prove that. given 131, . , •• EN, closed balls contained in fl, there exists flo, a regular, bounded open set of class ((, f

containing 131 u . .. U 13.\. with Qo c Q; this is seen by induction over N.

Property 4. Let Q be a re?fll/ar hOllnded open set ill [R" with houndary of' class ((,'1+c76. Then for all Y E Q, the/ullctioll x ---> G(x,y) is ill ((,~(Q\{.r}).

7n Sec Definition J of ~3.


We shall prove this result later using the integral method (see Corollary 8). Now we give the

Proaf'oj'Painl (1) a/Proposition 5. Making use of Lemma 1 and Proposition 6, we see that it is sutricient to prove the result for an open set Q as regular as we wish. From Property 4, stated above, we call therefore suppose Q to be regular ami for all Y E Q, G(., Y) E ((;~(~2\{)}). Let us then fix .1'1' .1'2 E Q, YI "# Y2 and

1 0< r < min(dist(y 1J),dist(YzJ)'2 IYI - Y21)

with the result that Q() = Q\JB(YI' r) u B(yz, r)l is regular with boundary

r u cB(YI,r) u ?BLbr).

The functions 1'1 (x) = G(x, .I'd and 1'2(X) = G(x, Yl) are in ·Yf(Qol n (6'~ (ao). Since 1'1 = 1'2 = 0, on L Green's formula applied to 1'1 and 1'2 on Qo yields the result

(4.30) f (1"1 (~1'2 - 1'2 (~1"1 )d'/ = f (1'2 (~VI - VI ~;~3)d';' ,'R(Y1. r ) \ ('11 ell ("R(.\'1.') ell ell

in which the normal derivatives can all be taken to point inwards to Q() with the result that they point outwards from B(.h, r) and B(Yl, r). Putting !II = u(tp.\',) and applying Green's formula to U1 and 1'2 harmonic on B(YI' r) we have

(ll'

Since 1'1 = E (r) - u and-~ " I (111

of (4.30) is

(1u 1 • ,. --;;-- on eB( .\'1' rl, the lelt hand Side I I

("/1

namely II == ['l(Yt), by applying Gauss' theorem and the formula of the mean to the function 1'2 E fi(B( Y I' r)) n ((;1 (S( Y I' r)). Similarly, the right hand side of (4.30) is 1'1 (yz) from which GCh, .I'll = G(Yl • . h l. D

Proof' oj' Proposition (). From the points (2) and (3) of Proposition 5

(4.31 )

It is sutricient therefore to show that

and that for (x. y) E Q/ X Q/\P. lim Cdx. y) = G(x, y). k ~ , k > I

We can thus fix Y E Q and without loss of generality, suppose that y = 0 E Q o. We denote by tpk the trace of E" on rlQk and by Uk the generalized solution of


P(Qk , (pd· From (4.31) we have

(4.32)

For I :S k, Uk is therefore a harmonic function on Q/ satisfying the condition

By the definition of a generalized solution

Now Uk :S max <Pk :S En(5) where (5 = max Ixl. Applying Harnack's theorem (see f~ n

Corollary 9 of ~2), the function u(x) = lim uk(x) is harmonic on Q. From (4.32), k - x

U :;, En on r. Now given w, continuous and superharmonic on Q satisfying

lim inf w(x) :;, En(z) for all Z E DQ, we have w :;, En on Q; this has been proved in

the proof of Proposition 5(3) (see (4.28)). The case IJ = 1 is immediate directly, a continuous super-harmonic function on Q, being concave on each component interval of Q; hence w :;, Uk on Qk for all k and in the limit w :;, u on Q. By definition U is therefore the generalized solution of P(Q, <pl· D

The principal interest of the Green's function resides in the formula of tlJl.' intearal representation of the solutions of P(Q, <P, fl. First, let us suppose

(4.33) Q regular (ofciass ((;1) and for all x E Q, G(x,.) E (6,~(~2\{x})

or what comes to the same thing since G is symmetric,

for all x E Q, u(<Px) E (f,,~(~2) .

From Proposition 5 of ~3, for U E '(;2(Q) n (6~(Q) with Llu E J}(Q) we have for x E Q,

u(x) = r En(t - x);1u(t)dt + r ~:n (t - x)u(t)d;'(t) J!J JI ('11

r au - EnU ..... x) .:c; .... (t) d},(t) .

• 1 on

Also from Green's formula applied to u(<Px) and u, we have

1 1 i'U(Px) jl'. ('U o = !I(~?x, t)Llu(t)dt + --" (t)u(t)d,,(t) - u(P,·, tJ -::;- (t)d;·(t); n 1 ell r CIl

from which by subtracting from 14.22), since Q being regular, u(<pJ is a classical solution of P(Q, <p,J (see Corollary 2), that is to say u(<p" t) = E"U - x), t E r, we obtain

1 J' "G (4.34) u(x) = G(t,x);1u(tjdt + ~(t,x)u(t)di'(t)

n 1 CIl


which is, for the moment, valid for every function

11 E ((;2(Q) II ((',:(Q) with !1u E Ll(Q)

and under the hypothesis (4,33). Let us first extend this formula to the generalized solutions of PH(Q, f) for an arbitrary bounded open set Q.

From Proposition 5

(4.35) t~,,(x - I) - 1':;,,((») ~ C(x, r) ~ 0 on Q x f.2·.D

where () is the diameter of Q (() = max Ix Yi). x, YE!}

From Lemma 3 of ~3 and Lebesgue's theorem, we deduce that, heill9 yil'ell f a hounded measure 011 Q

(a) for a.e. x E Q, the function C(x, .) is integrable with respect to f: (h) the jilllctioll Cf defined (a.e. on Q) hy

Cf(x) = J'C(X,J')df(Y) !i

(4.36)

is inreyrahle Oil Q. It is even in U(Q) for all (I < 11.(11 2); in the casc II = I. it is continuolls on Q. In the case 11 ~ 2, iffE U(Q) with P > 'ill, Cfis continuous and bounded on Q; if in addition G satisfies (4.28), thcn

lim G/(xl = 0 for all :: E r .

All of this follows from Lemma 3 of ~3 and from (4.35) by the use of Lebesgue's theorem. The function Cj defined in this way is a solution (in the st:l1se of distributions) of Poisson's equation

(4.37) ,111 = f on Q.

Indeed by Fubini's theorem

(/1G/; ~ > = f Gf'(x) /1 s(x)dx = f d/(y) f G(x, y)/lS'{x)dx = f ~(y)df(y)· In particular, if s E C/(Q), Cs E ((; (; this allows us to defillef()r aery distrihUliollI with compact SUPP0f[ ill Q, the distrihutioll Gf 011 Q hy

(4.38)

This definition clearly extends that given by (4.36), since by the symmetry of C, for a bounded measure'/: the function Gf satisfies

f Cf(x)~('()dx = f df(y) f G(x, Y)s(x)dx = f G~( yldf(y)

for all ~ E I/'(Q) (and even for ( E U(Q) with p > ~11).


Hence more generally, Gf is defined by (4.38) for every distribution f = fo + fl wherefo is a distribution with compact support in Q andfl a bounded measure on Q. We can consider this distribution Gf as a solution in a weak sense of PH(Q,f); this point of view will be developed in § 6. We shall prove here the

Proposition 7. Let Q be a bounded open set in [Rn with Green's function G and let f = j~ + fl where fo is a distribution with compact support in Q and fl E U(Q) with p > !n (fl a bounded measure on Q in the case n = 1). Then the distribution Gf is the generalized solution of PH(Q,f). If, in addition, G satisfies (4.28), then Gf is a quasiclassical solution.

Proof For f E C'(Q), the space of distributions with compact support in Q, the Newtonian potential En * fis harmonic on [Rn\supp f; we denote by cP f its trace on r. The map f ~ cP f is linear and continuous from C'(Q) into 'C0(r); hence the map f ~ u(cp f) the generalized solution of P(Q, cP f) is linear and continuous from C'(Q) into Xb(Q). The map f ~ G{ is, by definition, even linear and continuous from C'(Q) into .Q"(Q). Forf = <5 y with Y E Q, Gf = En * f - u(CPf); since {by; y E Q} is total in 8'(Q), this equality is true for allf E t&"(Q); hence Gfis a generalized solution of PH(Q,f). In the same way for f E U(Q) with p > !n, the Newtonian potential En * 1where

(X) - . f- - {f(X) if X E Q o If X E [Rn\Q ,

is continuous on Q (see Proposition 6 of §3). We can define its trace cP f on r and the map f ~ u(CPf) is continuous from U(Q) into Yfb(Q). Then from (4.35) and Lebesgue's theorem,f ~ Gfis continuous from U(Q) into 'C~(Q). Since

G{ = En *1- u(CPf} for fE U(Q) n 6"(Q).

this is still true for all f E U(Q). If G satisfies (4.28), CPy E E(Q) for all Y E Q; E(Q) being closed in q,o(r), by continuity of f ~ cP f and density

cP f E E(Q) for all f E 8'(Q)+ U(Q) . o We now extend the representation formula (4.34) to every classical solution of P(Q, cp):

Proposition 8. Let Q be a regular bounded open set in [Rn with Green's function G satisfying G(x,.) E 'C~(Q\{x})for all x E Q and cP E rt°(r). Then the (classical) solution u(cp) is given by

1 (IG u(cp, x) = T (x, t)cp(t)d},(t) for all x E Q.

r en

Proof First of all, let us suppose that there exists Uo E 'i5'2(Q) n 'C~(Q) with Auo E U(Q), p > !n such that Uo = cP on r.


From (4.34) we have

j' I' c'C 11 0 (.\) = G(t, x) /luo(t) dt + I ~- (t. x)(p(i) d~,(t) .

!! oJr I'll

Now 110 - U(lp) is a quasi-classical solution of PH(Q, c111(J From Proposition 7,

11(1 (x) - u(P, x) = J' G(l, x) /ll1 o(t) dt SJ

and hence I' ('G

lI(p.x) = I ,(I.x)lp(l)d~·(t) . • r ( 11

In particular this formula is true for the traces (P on r of the fu nctions tin E (f, 2(.12).

From the Stone-Weierstrass theorem, these traces form a dense part of (('0(1). Being given an arbitrary 1I((P. xl

('G and since _ ( .. x) E (f,0(r).

I'll

Hence in the limit

J' ('C; l/(<p. xl = _ (1, x)(p(f)d;,(t) .

r (11

3. Generalities on Dirichlet's Problem in an Unbounded Open Set

o

We now take Q to be an llllhowu/ed open set in IF.£/1 with boundary r and take (P E (f, °(l'),fE (I'(Q). We consider Dirichlet's problem P(Q. <PJ). We note firstly that if all the connected components of Q are bounded. there is a unique quasiclassical solution of P(Q, <PJ): in effect if u is such a solution, its restriction to each connected component of Q will be determined and hence also u. It will not be the same if Q admits an unbounded connected component: we have to introduce a condition at infinity. First of all, let us give some examples:

Example 12. Case oj' an interval] a,x; [ of IF.£.

Let Q = ] a, Cf': [ with a > - x with the result that r = {a}. We take tp E IR and, for simplicity f E f'f,([a, x D. The (classical) solutions of P(Q, <p, f) are given by


(4.39) I'x

u(x) = cP + J a (x - t)f(t)dt + c(x - a) X E [a, 00 [

where c is an arbitrary constant. This constant c is given as a function of u by

(4.40) I· u(x) fX ( I C = 1m -

X--+"X) X a

..... t )fU) dt . x

In particular iff(t) ELI (]a, oo[), then

. u(x) fLf d c = hm '-.... - (t) t; x-x~ X a

and, changing the notation, for all c E !R, there exists one and only one (classical) solutioll of P(Q, cp, f) satisfying the condition at infinity

(4.41) lim u{x) = c . .'( - x~ X

This solution is gil'en by

u(x) = cp + r (x t)f(t)dt + (c - f:'f(t)dt)(.X - a),

namely

(4.42) u(x) = cp - jf' G(x, t)f(t) dt + c(x - a) • a

with

(4.43) G(x, y) = a - min (x, y) .

This function will be the Green's fUllction of the unbounded interval ]a, oc{. Note that it is the limit when b ->x of the Green's jimction of the bounded interval ]a, b[ (given by (4.23)). It is characterised by

{for all Y E ]a, x [, u(x) = G(x, Y.). is the quasi-classical solution

(4.44) _ .. . u(x) of PH (]a, XJ [, (\,) satIsfymg lIm _. = 0 .

. x- ,Y .. X

o

Example 13. Case of the exterior of a ball !R", n ): 2. Suppose we take Q = !R"\B the exterior of the ball B = B(xo, ro) whose boundary is r = ?B. Propositions 20 and 21 of *2, tell us that, given u a bounded classical solution of P(Q, f.jJ), then

exists and

(4.45) 1/(x)

c = lim u(x) Ixl ~.f.

( ( ro )"-2) 1 f Ix - xol2 - 1'2 1 - . c + - I _ .'1" 0 f.jJ(t)di'(t) . Ix - xol ro(J" tB t,


Conversely, we show as in Example 2 (Proposition 2), that the function u given by (4.45) is a classical solution of P(Q, cp) (we can make use of a Kelvin transformation as in Proposition 21 of §2 to reduce the problem to Proposition 2). But for the behaviour of the solution at infinity we have to distinguish between two cases

(i) case n ~ 3, it is immediate that u(x) --> c when Ixl -->x

(ii) case n = 2, we see easily that

u(x) --> _1 __ f cp(t)dy(t) when Ixl --> ·XJ . 2nro ,'B

Hence in the case n ~ 3, for all cp E ~o(oB) and all c E R there exists a unique classical solution of P(Q, cp) satisfying

lim u(x) c

given by (4.45). While in the case n 2, for all cp E (6 0(clB), there exists a unique classical solution of P(Q, cp) satisfying the condition at infinity:

u is bounded on Q.

This solution will again be given by (4.45) (where c is of little importance), it will have a limit at irifinity but which is imposed by the giving of cp contrarily to the case n ~ 3:

. I f hm. u(x) = ~- cp(t)d},U) . Ixl ~ , ~nro ,'B

This difference between the cases II = 2 and n ~ 3 on the conditions at infinity appears clearly also in considering the radial solutions of P(Q, 0): such a solution is necessarily of the form

case n ~ 3,

and the constant c is determined by

c = r~-2 lim u(x); Ixl ~ .,

case n Ix - xol

2, u(x) = c(Log Ix - xol - Log ro) = c Log and then ro

(4.46) c

We shall prove later in the case of almost any unbounded open set in [R12

that this condition at infinity (4.46) determines the solution of P(Q, cp) (see Corollary 3). D


Example 14. Case of a half-space of [R", n ;?: 2. We consider Q = [R" - 1 X [R + of which the boundary r = [R" 1 X {O} is identified with [R" - 1 and take qy to be a continuous function, bounded on [R" - 1.

Poisson's integral formula for a half-space (see Proposition 22 of §2) shows that a hounded classical solution of P(Q, (p) is of the form

(4.47)

Conversely the function u given by (4.47) is harmonic on Q since

2x" (It (J1I

Using f . dt 2 ,2 = (In, and then Lebesgue's theorem we see that U IS a E' I (I + It I )" 2

bounded classical solution of P(Q, <pl. Hence for euery continuous qy, hounded on [Rn- 1, there exists a unique classical solution of p([RlI-l x [R +. (p) satisf.\'iny the condition at infinity:

II is hounded Oil [Rn 1 X [R +

given by (4.47). We note that this solution does not, in general, have a limit (in the topological sense) when Ixl -->J'~; however

(4.48) lim u(x) = lim qy(x') Ixl ~ f Ix'l ~ x'

if this latter limit exists. o

We have seen appearing in these examples, various conditions at infinity assuring the existence and uniqueness of a solution of P(Q, qy). Firstly, we prove a sufficiently general result on the uniqueness of a classical or quasi-classical solution of P(Q, (P, fl, grouping together the various conditions which have appeared:

Proposition 9. Let Q he all unhounded open set ill [R" with boundary r and let qy E 'f,0(r)J E Y' (Q), u f E C/'(Q) and h E ,.j( (Q) n (6° (6) with h(x) > Of or Ixllarge. Theil there exists at the most one quasi-classical solution of P(Q, qyJ) satisfyiny the condition at infinity:

(4.49) . u(x) hm-- - lII(X)

lxi' y h(x)

CC When 1<,'11 - u, is a distribution. the limit is taken in the sense (see just before Definition 6): for all I: > 0, there exists R > 0, such that

~ c in !/'(Q\8(O, R))


Proot: Given two solutions, the difference u is a classical solution (see Remark 2) of P(Q, 0) and satisfies

lim tI(x) = ° . Ixl .~ f il(x)

From the hypothesis, we fix Ro such that h(x) > ° on Q .13(0, Ro). Given I: > 0, there exists R, ;:::: Ro such that

III I ~ ,. on Q\B(O, R,l . i h

The function l' = 1/ - !:II is harmonic on Q (1/ and hare!)

l' ~ 0 on [t\E(O, RJ (by hypothesis on R,) ,

l' ~ ° on F\E(O, RoJ (since u = 0 and h ;:::: OJ .

Hence from the maximum principle:

l' ~ max (,+ = I: max h-(h (- /z)+ = max(O, - h)) In/l(O.Ro) TnB(O.Ro)

and

max h ). Inll(O.Ro)

We can show similarly that

1:( h + max h" ) !,,, 1l(O. R"i

and hence when I: ---> 0, 11 = O.

Obviously this proposition contains the condition at infinity

lim lI(X) = (' . Ixl ~(,

D

But it also contains, in the case n = 2, the condition (4.46) with the reservation that Q =I- [R2: in effect then, (4.46) is equivalent

where -'0 EO

the \Q, il(x)

. tI(x) hm -----. - c = °

Ixl ~ , Loglx - xol

Loglx - xol satisfies the hypotheses. We therefore have

Corollary 3. Let Q he (In open set in [R2 with l2 =I- [R2 and let <Po EO (6°(nJ EO V'(Q) and 1/ f EO 1/'(£2). Then there exists at most one quasi-classical solution of P(Q, <PJ),

satis/}'iny the conditioll at infinity

(4.50) I. lI(X) 1m - Ii x (x) = 0 .

Ixl ~ •. Loglxl

III particular, there will exist at //lost one so/lltio/l satist:rill?l

(4.51 ) 1/ - u. is hounded at in/illily.


This situation will occur for erery open set Q satisj);ing

(4.52) there exists h E ,:If (Q) n 'C0(Q) such that lim h(x) = + ex; . Ixl ~ x

We state the

Corollary 4. Let Q be all unbounded open set in [Rn satisfving (4.52), cp E q;0(1), f E 0"'(Q), u x E 0"(Q). Then there exists at most one quasi-classical solution of P(Q,

cp, f) sarisj)'ing the coneiition (4.51).

Example IS. Open sets satisfyinu the condition (4.52). (a) n = 1, every open set Q i= IR: in effect if X o E [R\Q,

h(x) = Ix - xol is suitable.

(b) n 2, every open set Q with Q i= [R2: in effect Xo E [R2\Q,

h(x) = Loglx - xol is suitable.

(e) n ); 3, every open set Q contained in a half-space. In effect we can always suppose f2 c u;gn-I X IR+ = {(x', x n ); Xn > O}. For all o < 'Y. < I, the function

(4.53) f ix' + xntl' h,(x) =, --~'-C- dt H" 1 (l + W)"!2

is suitable (see Example 14, formula (4.47)). o Finally we note that in the case u x == 0, the condition at infinity (4.49) IS

independent of h.

Corollary S. Let Q be (Ill ullbounded open set ill [Rn, cp E '(,0(1)./ E SI'(Q). Then there exists ar most olle qllasi-classicul soilitioll of P(Q, (p. f) satisfvillq the condition at i Ilji 11 it y

there exists h EYf (Q) n ((,o(Q) with h(x) > 0 flJr Ixl larqe so that

(4.54) . !/(x) hm - = O.

Ixl - , h(x)

Proot: Let 11 1, 112 be two quasi-classical solutions of P(Q, (P, f) and hI' 112 E

X(Q) n ((, o(Q) such that

uj(X) hj(x) > 0 for Ixllarge, we have lim ~- = 0 i = 1,2.

Ixl - , hJx)

We have II = hI + lIz E Y1 (Q) n ((,O(Q), h(x) > 0 for Ixllarge, and since

o < I~~x) < 1 h(x)

for I x I large,

from which Corollary 5 follows.

lim uj(x) = 0 i = I, 2 Ixl - x h(x)

o

35X Chapter II. The Laplace Operator

This leads us to frame the

Definition 9. Let Q be an unbounded open set in [R:". We say that U E ;/'(Q)

satisfies the Ilull conditioll at infinity if it satisfies (4.54).

We note that a solution u of P(Q. «(!.f) satisfying the null condition at infinity does not necessarily satisfy

(4.55) lim II(X) = 0 . Ix! ----; f

For example for C E R r/zt'jililclioll 11 == c is a classical solution of P(Q, c) and it SMisfit's the null condition at infinity iff (' = 0, or Q satisfies (4.52). We note also that although this is the case for the exterior of a ball in [R:2 (see Example 13). in general a solution of P(Q, (P, fJ satisfying the null condition at infinity is not necessarily bounded at infinity. For example. for Q c [R:n-l x H> and 0 < y. < I. the unbounded function 11 = h, given by (4.53) solution of P(Q. (p). where tp is the trace of il, on r. satisfles the null condition at inflnity: indeed lim u(x)/h(x) = 0 for h = hfJ with Y. < if < 1.

IX! ----+ f

Finally. we note that tht' Ilull condition a/ infillily is equimienllO the conditio/) (4.55) iff Q satisfies

(4.55)' el't'ry millori:ed /ill1ctioll ofj( (Q) n ((; 0([2) is huunded.

Example 16. The t'xlerior Q o( (/ compact set of R" wilh n ): 3 satisrvillY (4.55)'. In effect let hE'/( (Q) n {(,o(Q) minorized. Let us take ~ E (/([R:") such that ~ = 1 in the neighbourhood of [R:",Q and put

which is harmonic on R". We know (sec Proposition 2 of ~3J. since IJ ): 3 that E" * f is bounded; since h 1 Cx) = h(x) for Ixi large, ,; is minorized and hence constant (see Proposition 7 of ~2) we deduce that hI and hence hare bounded. 0

Let us study now, the existence of P(Q, (p,f). If there exists a quasi-classical solution of P(Q. (,oJ) satisfying (4.49). ajimiori there exists !io EC/'(Q) satisfying

/1110 = I In '/(Q)

(4.56) . tlo(x)

hm --- - 11, (x) = 0 Ixl ~, h(x)

lim !inC.:) = tpo(::) exists for all:: E r . x-;;

Conversely, given 1'0 E'/'(Q) satislviny (4.56). u is a solution of' P(Q, (PJ) satisf'yinq (4.49) iffu- 110 is a classical solution of P(Q, tp - (Po) .'wtishilliJ the 1l1l11 condition at

infinity. As in the bounded case, we shall obtain 110 E '/(Q) satisfying (4.56) with the help of Newtonian potentials. We note carefully that the condition at infinity (u" h)


cannot be taken arbitrarily: the possible weights h are determined by the open set and u x depends to within an arbitrary constant on f, <p and h. We are thus led to study the existence (if a solution of P(Q, <pi satisfying the /lull condition at infinity. We can, as in the bounded case develop Perron's method to define a generalized so/ut ion. Given (f E (fjo(n, we define

f(ip) = {v E '6 0 (Q) sub-harmonic satisfying

lim sup u(x) ~ <p(z) for all Z E r and there exists x ._-':::

(4.57) h E .y(~ (Q) n (fjo (Q) with h(x) > 0 for Ixl large such

. u(x) } that hm sup -. -. - ~ 0 Ixl ~ f lI(x)

f +(<p) = - .f _( - (p) .

U sing Property I and adapting the proofs of Proposition 9 and of Corollary 5, we show without difficulty that

(4.58)

Unlike the bounded case .1 _«,.0) and f,(ip) can be empty, and it is not true in general, when they are non-empty, that sup .f(ep) = inLJ1+(ep). We pose the

Ddinition 10. Let Q be an unbounded open set in [R;" with boundary r and ip E (60(1'). We say that Dirichlet's problem P(Q, (p) with null condition at infinity admits a generalized solution if the sets f _(<pi and .1,(ep) defined by (4.57) are non-empty and if sup f _rep) = inf.J1 +(<p). We call the function u(rp)=sup f(<p) = inf.f +(<p) the iJeneralized solution of/he problem P(Q, Ip) with the Ilull condition at infinity.

It is clear that a classical solution of P(Q, (p) satisfying the null condition at infinity is a generalized solution of P(Q, ip) with the null condition at infinity. We note that in general u( ~0), if it exists, does not satisfy the null condition at infinity, and it no longer satisfies

lim u(<p, x) = <p(z) for ;:: E r .

We see easily as in the bounded case that the map Ip -> u(<p) is linear. This follows from

:J _OJP) = ;. f -(cp) for i. > 0, f(<pd + :J _(IP2) c :J - (cp] + <P2) .

We have also immediately

(PI ~ (,02 on r = u(ipd ~ U(CP2) on Q.

360 Chapter II. The Laplace Oper~lor

We deduce that when Q satisfies (4.52)

(4.59) inf qJ :( u(qJ) :( sup qJ on Q'8 r r

since in this case u(e) == c for all C ElM. In the general case. we have merely

(4.60) min (0, i~f (,0 ) :( u(p) :( max( 0, su1 09) on Q,8

which follows from l' == min (0, m;n qJ ) E J -(4)).

We shall not develop further Perron's method in the unbounded case. We shall rather take an interest in Kelvin's inversion method for which we shall need to suppose Q i= iR" with 11 :?-' 2. We fix Xo E IM"\.,Q, ro > 0 and denote hy Q' the image l(xo• ro) (Q) of Q under the inversion l(x o' 1'0); it is a bounded open set whose boundary (iQ' consists of f' = I (xo• 1'0) (n and (xo ; 79. Given 1l. L" etc., defined on Q

we denote by 11'. lC, elc .. their Kelvin transforms defined on Q'. We know already that 11 Eff(Q) iff [i' E ,1((Q'). Given l' E ((,o(Q), we have 1" E ((,o(Q') and z.: is subharmonic on Q iff r' is sub-harmonic on Q': that follows immediately from Dc!lnition 2 and from Proposition 19 of ~2. We fix (f! E ((,O(r); its Kelvin transform is a function qJ' E (('0(["). For all 2 E r. we have:

lim sup t(x) :( (,0(.:::) ilf lim sup l"(X') :( 0/(':::') x --..: '\:' --t:;'

where .:::' = l(xo, 1'0)'::: E r , We now distinguish two cases,

Case I. r is b01lndrd (exterior problem), In this case Q' = Q' u {xo} is a bounded open set with boundary r. We can consider 17' the generalized solution of P(l2', 4)')

and u the Kelvin transform of the restriction u' of 1'/' to Q', We have

lim Ix - xoln 2U(X) = r~J· 2 inx ll ) ,

Ixl - f

Now by the definition of II',

l~ = sup [1' E ((,O(Q), sub-harmonic, lim sup l'(X) :( (,0(.:::) for all .::: E r, and 1" x ---.::

extends to a continuous function. sub-harmonic on [2';: hence

u :( sup f._ (p) ,

Similarly 14 :?-' .'! +(tp) and hence 11 = u(cp), generalized solution of P(Q, 0)) with the null condition at infinity, We state already the result

-" A I'riori (,0 is not bounded on r: but the inequality i, trivial when there is ~n , -hi We refer to ~2.5 for all the ideas concerning Kelvin's transformation,


Proposition 10. Let Q be an unbounded set non-dense in IR" with n ;? 2 with bounded boundary rand rp E (to(n Then the Dirichlet problem P(Q, rp) with null condition at infinity admits a generalized solution u(rp) satisfying the condition that

(4.61) lim Ixln-Zu(P, x) exists. Ixl '" ex.

Also for X o E IR"\Q and ra > 0, u(rp) is the Kelvin transform H(xo, ro)u' of the restriction to Q' = [(xo, ro)Q of the generalized solution of P(Q' u {xo}, rp').

Case 2. r is unbounded. In this case r is dense in oQ'. Firstly, let us suppose that rp' is extended by continuity to il/ E «,fO(oQ'), that is to say that lim Iz - xol" - 2rp(Z) = r~ - 2(p' (x o) exists. We can then consider the generalized

Izl~ 0:'

solution u' of P(Q'. 41') and its Kelvin transform u. By definition of u'

u = sup { V E «,f0(Q) sub-harmonic; Iil~ :~lP v(x) ~ rp(z) for all Z E rand

Ix - xol n - 2 v(x) ~ r~-241'(xo)} ~ sup .§_(rp) } .

As above u = u(rp). Finally, we note that, since u' is bounded, Ix - xoln-Zu(x) is bounded. We consider now sup Ix - xol n - 2h(x) , Ix - xol !1

we can always suppose that h > I) on Q. We take a sequence (rpd of continuous functions with compact support in r such that

when k -+ CX] •

From the preceding case Uk = U(4?k) exists and we have

Uk = Slip .j1~ (rpd = inLj'fh+ (rpk)

where

.§h_ (rp) = {v E «,fo(Q) sub-harmonic: lim sup v(x) ~ rp(z) Vz E rand

. v(x) } hmsup'- ~ 0 Ixl ~ oc hex)

362

and J~ (ep)

We shall show that, given D > 0


(4.63) Uk - III ,s; max (epk - epl - c:h)+ + t:h on 1 1

It follows immediately that

(4.64)

and hence

that is to say uk/h converges uniformly on Q and in particular Uk converges uniformly on every bounded set of Q. Denoting its limit by 14 we show that

Given x E Q, for alii there exists!'1 E ,1h. (epl) such that uI(.x) ,s; vlx) + "I; we have r = VI - [;lh E J~(ep) since (PI - [;lh ,s; ep by the definition of "I. Applying (4.64) in the limit when k -+ x we have

U(x) ,s; v(x) + "d2h(x} + I).

This shows clearly that u,s; sup fh~ (ep); similarly, we can show that U ;.. inf J~(ep). We shall therefore have proved

Proposition 11. Let Q he an open set, 11011 del1se in [R" with n ;.. 2 and ullhounded houndary 1, ep E ((;O(I'), and h E .Y((Q) n y;o(m such that

(4.65) lim Ixln~2 h(x) = +x, Ixl ~'X

. ep(z) hm -- = o.

I=,~' h(z)

Then rhe Dirichlet prohlem P(Q, ep) with null condition at infinity admits a generalized solution satisfYing

(4.66) u(ep) ,s; max (ep - t:h)+ + eh on Q f()r all I: > ° . r

If; in addition, lim Izl"~2 ep(z) exists, then for all Xo E [Rn\Q and '-0 > 0, u(ep) Izl ~x

is the Kelvin transfiJrll1 H(xo, ro)u' of the generalized solution of P(Q', ii/) where Q' = /(xo, fo)Q and

for

. (lz-xol)n~2 11m -- - ep(z) f(Jr 1=1 '" , ro

z' = Xo .


Remark 4. With the above notation, being given q/ E C(j'O(r') satisfying

{there exists h' E £ (Q') n (€O(Q'\{x o}), with ~im h'(x') = + ex)

x --+ Xo

. (P'(z') such that hm - ..... - ,= °

z' -xo 11'(zl) ,

the Kelvin transform q; = H(xo, ro)(p' satisfies (4.62) with h = H(xo, roW. We can therefore consider the Kelvin transform u' of the generalized solution u(p). This function is quite naturally a "generalized solution" of the problem P(Q', q;') where q;' is not continuous on i'JQ'. We come back in §6, to the definition of generalized solutions of Dirichlet problems with data which are not continuous. 0

Conclusion of the Proof of Proposition 11. Proposition 11 has been demonstrated above, with the exception of (4.66), under the condition (4.63). In fact (4.63) is a particular case of (4.66) (Uk - u/ = U(Pk - (PI)) if we can demonstrate this last result for all q; such that

I· q;(z) d 1m _. = ° an

Ixl.'x h(z) u(q;) = supf"-.(q;).

For that it is sufficient to prove that, given I: > 0 and v E f"-. (q;)

v :s; max(p ..... ch)+ + sh on Q. r

From the hypothesis, there exists R such that v :s; ch on Q\B(O, R). The function z; - f;h is sub-harmonic and continuous on Q\B(O, R) and

lim sup v(x) - ,;h(x) :s; (p(z) - r.h(z))+ on D(Q n B(O, R)) . x --~ z

The proof of the Property 1 is then complete. o The method of Kelvin's transformation likewise provides a result on the existence of a classical solution: in effect with the above notation for z E r and z; E (,so(Q n B(z, r» is a barrier function 80 of z iff v' is a barrier function of z' = I(xo, ro)z, and hence:

z is a regular point oIthe boundary ofQ iflz' is a regular point oflhe boundary of Q'.

Applying Theorem 1, we deduce

Theorem 2. Let Q be an unbounded set, not dense in [R" (n :;:, 2) with boundary r. (1) The following assertions are equimlent: (i) for every continuous q; with compact support in r, there exists a classical solution of P(Q, q;) satisfying the null condition at in/inity; (ii) all the points of the houndary of Q are regular80 .

80 See Definition 4.

364 Chapter II. The Laplac,~ Operator

(2) When these assertiolls are satisfied, then: (a) Case of r hounded (exterior problem): fi)r all cP E ,(;O(I"), there exists a unique classical solution u of P(Q, (p) satisfying the null condition at infinity. I n addition

lim Ixl" -2 II(X)

exists. (b) Case of r ul1bounded:.J()r all (P E 'to(I") and h EYf(Q) n (6 0(Q) satisfying (4.65), there exists a unique classical solution u of P(Q, cp) satisfying the null condition at infinity; in addition

lim u(x) =, 0 . Ixl;> h(x)

Let us complete this section by the introduction of Greel1'sjimction olan unbounded open Si.:'t. Given Y E Q we seek to solve the Dirichlet problem PH (Q, 6,.) with the null condition at infinity. Applying the general method, to take us to a Dirichlet problem P(Q, cp) with null condition at infinity, we have to find Uo E Y'(Q) solution of

:1110 = (i, in (/(Q) ,

110 satisfies the null condition at infinity,

lim uo(x) exists for all :: E r . x ---'::

In the case 11 ?: 3, there is no problem it is enough to consider uo(x) = E)x - y). In the case 11 = 1, if Q =I R we can take 11 0 (:,) = ±ix yl !ix - xol, where Xo is a fixed point of In thc case II = 2, to carry out the same operation, we must suppose that Q =I 1R2:

k 1 Ix - vi h . f' d '. f 1Tll2 " we can ta e 110 (x) = Log - . were Xo IS a Ixe' POll1t 0 "'" \tt. 2rc I x -.... Xo I

Once Uo is chosen. we have to solve P(Q, CPo) with null condition at infinity. We know how to solve this problem, by Propositions 10 and 11 when Q =I IR"; in addition, when all the points of the boundary of Q are regular, we shall obtain by Theorem 2. a classical solution. We can equally well use a method o{passaye to the limit: we consider an increasing sequence (Qk) of bounded open sets with union Q (for example, Qk = Q n 8(0, Rk),

with (Rd an increasing sequence of positive real numbers Rk --+ 'x. as k····> (xJ). We can study the limit of the sequence 1 Gk ) where Gk is the Green's function of the open set Qk; in the case in which Q is bounded, this limit gives the Green '5 function of Q

(see Proposition 6); also in the Example 12, we have seen that the Green's functions of the intervals ]a, b[ converge when b --> :'.C. to a function which we have called the Green's function of ]a, c{.) [.

In a general way. we prove

Proposition U. Let (Qd be an increasiny sequellce o{houllded open sets ill u:g" with unioll Q bounded or not. We define. as in Proposition 6, the ntensiolls Gk by 0 ()fthe Green's functions Gk of the bounded opel1 sets Qk'


(1) The following assertions are equivalent: (i) for all (x, y) E Q x Q\D, the sequence (Gk(x, y)) converges; (ii) for every connected component Q o of Q, there exists (x, y) E Q o x Q o \D such that Gk ( x, y) is minorized; (iii) there exists u E £(Q) minorized such that

lim infu(x) - E,,(x) > - XJ Ixl ~ ctj

(2) When the assertions of (1) are satisfied, the function

G(x, y) = lim Gk(x, y) k ~ 00

satisfies

{

for all y E Q, _the ji-mction x ...... G( x, y) is the largest solution

(4.67) v E rgO(Q\{y}) II Ll1oc(Q) satisfying

v ~ 0 on Q\{y}, Llv = by in q'(Q) ,

or, which is equivalent:

(4.68) {

for all y E Q, the function x ...... E,,(x - y) -

function u, harmonic on Q, satisfying

u(x) ;:0: E,,(x - y) for all x E Q .

G(x, y) is the smallest

The existence of a function G defined on Q x Q\D satisfying (4.67) or (4.68), which is equivalent to it, implies the assertion (iii) of the point (l): in effect, for y E Q and o < r < dist(y, oQ), the function x ...... E,,(x - y) is minorized on mJ3(y, r) by En(r) and

case 11;:0: 2, lim E,,(x - y) - E,,(x) = 0 Ixl -. ex!

case n = 1, I

lim inf E,,(x - y) - E,,(x) ;:0: - -2- IY1 . Ixl -; x.

We pose the

Definition 11. Let Q be an open set of [R". The function G defined on Q x Q\D, satisfying (4.67) or (4.68) if it exists is called the Green's fimction of Q.

From Proposition 12, the Green's function of Q exists iff the sequence of Green's functions of Q II B(O, Rk ) (extended by 0) converge when Rk ...... OC', the limit being then the Green's function of Q. Also the Green's function of Q exists iff the assertion (iii) of the point (1) of the proposition is satisfied. In the case 11 ;:0: 3, every open set ()([R" admits a Green'sfunction: indeed, the function u == 0 satisfies the assertion (iii).


In the case n I, an open set Q of[R admits a Green'sjunction ifIQ 1= [R; in effect if Xo E [R\Q, u(x) = '~'Ix - xol satisfies assertion (iii); conversely a function u, harmonic on [R, is linear; it cannot therefore satisfy

I. . 1 1m mf u (x) -- I x I > ... -x; lxi' "-, 2

[n the case n = 2, Q = [R2, does not admit a Green's function: by Liouville's theorem (see Proposition 7, §2), a harmonic function u minorized on [R2 is constant

and therefore cannot satisfy lim u(x) - Log Ixl > .- 'x;. We have the same Ixl ~ OJ

conclusion for Q = [R2\ {xo}; to prove it we make use of the following argument: the Green's function of B(xo, r)\{xo} is the restriction Gr of that of B(xo, r) (this follows from Remark I). Now the explicit formula (4.24) shows that C r ( x, y) -+ - 00 when r -+ + c;c. Finally, we note that a non dense open set Q of [Rz admits a Green's function: indeed if Xo E [R2\t2, u(x) = Ez(x - xo) is

harmonic and minorized on Q and lim u(x) - E2 (x) = 0. Ixl ~ f

Proof of Proposition 12. Let us first fix y and suppose in the first instance that:

(4.69) {for every c~nnected component Q o of Q, there exists x E Q o

such that Gk(x, y) is minorized .

Putting uk(x) = E,,(x - y) - Ck(x, y), we recall from the proof of Proposition 6 that the sequence (Uk) is increasing. From the hypothesis (4.49) and Harnack's theorem (see Coroliary 9 of §2, Uk E J¥' (QI)' for all k ): I), Uk converges uniformly on every compact subset of Q; its limit u is harmonic on Q; since Ck ,,; 0, u( x) ): En(x - y) for all x E Q; finally considering LI E .ff (Q) and satisfying u(x) ): En(x - y) for all x E Q, we shall have u ): Uk on Qk and hence u ): u.

In summary, given y E Q satisfying (4.69), for all x E Q\{y}, Cdx, y) converges in decreasing and the function

u(x) = E,,(x - y) - lim Ck(x, y)

is the smallest harmonic function on Q verifying u( x) ): En(.x - y). Also as we have seen above u is suitable for (iii). To conclude it is sufficient therefore to show that (iii) implies:

Ck(x, y) is minorized for all (x, y) E Q x Q\D;

now if u satisfies (iii), m(y) = inf u( x) - En( x -- y) > eX), for y E Q fixed, the XE []

function u(x) = u(x) - m(y) is harmonic on Q and u{x) ): En(x - y) for all x E Q; hence u(x) ): En{x - y) -Gk(x - y) for all x E Qk; in short

Ck(x, y) ): mintO, E,,(x - y) + m(y) - u(x») . o Finally let us give some examples.


Example 17. Green's function of the exterior of a ball in IH", n :?: 2. Let Q = IHn\B(xo, ro). We look back to Example 11; the function G glVen by the formulae (4.24) is defined for all (x, y) E IH" x IH" with y of X o, x of y and x of /(xo, ro)y and satisfies

G = 0 on cB(xo, ro) x (IHn\{xo} )\D

case n:?: 3, lim G(x, y) = 0 Ixl ~ x

case n= 2, " I Iy-xol lim G(x, y) = - -- Log ---"'-' .

Ixl~x 2rr ro

Hence the Green's function of Q = 1H"\B(xo, ro) is given by the same formulae for all (x, y) E Q x Q\D. 0

Example 18. Green's function of a half-space IH" - 1 X IH + . The symmetry T:(y', Yn) --> (y', --Yn) maps IH"-l x IH+ onto IHn - 1 x IH. It is clear, identifying IH" - 1 with IHn - 1 X {O} that

E,,(x' - y) = Enfx' - 1)) for all x' E IHn-1 .

We deduce immediately that the Green's function on 1H,,-1 x IH+

G(x, y) = En(x -- y) En(x - Ty},

that is to say

G( (x', x n), (y', Yn)) = En(x' - y', Xn - Yn) - En(x' - y', Xn + y,,) .

We note that in the case n = I, we recover (4.43) (with a = 0). o

4. The Neumann Problem; Mixed Problem; HopPs Maximum Principle; Examples

We take Q to be an open set of IHn with boundary r and consider the problem

{ u. harmonic on Q

PN(Q,t/J) cu ~ = t/J on r. en

where t/J is given and defined on r. This is the Neumann problemfor Laplace's equation. In the classical setting, to define au/an, we must suppose Q to be regular. We pose in a natural way the

Definition 12. Let Q be a regular open set of IH" with boundary rand let t/J E (6°(T). By a classical solulion of PN(Q, t/J) we mean any solution

36~ Chapter II. The LapJace Operator

U E j{ (Q) n 'f; ~ (a) such that

ou -;;- (x) = t/J(z) for all z E r . ell

It is clear that there is no unique solution or the Neumann prohlem: if u is a classical solution of P N (Q, t/J), for every connected component QtJ of Q and all C E

11 + ex!]" is also a classical solution of PN (Q, ~/) since

grad (1/ +- C;(flo) = grad 11 on Q.

On the other hand, a solutioIJ does not existfor all t/J E (6 0 (T): from Gauss' theorem, if P N (Q, t/J) admits a classical solution, then for every bounded connected component Q o of Q,

Making use of Fredholm's integral method, we shall prove the

Theorem 3. Let Q be a bounded regular opelJ set of ~n with boundary r of class ((,11' and let t/J E C(,°U').

(l) The Neumann prohlem P N (Q, t/J) admits a solution iff t/J satisfies Gauss' conditiolls

(4.70) for every connected componellt Qo of Q, J. t/J di' = 0 . (' flo

This solution is then unique to l,vithin an additil'£, constant on each component of Q .

(2) The exterior Neumann problem PN(~n\Q, t/J) admits a solution iff" t/J satisfies Gauss' exterior cOllditiolls:

(4.71 ) tiu' eu'n' hounded cOlJnected compollent Q;) of' o.

Theil (a) in the case 11 ? 3, there exists a solution u o( PN(~"\Q, t/J) satis(yilJfj

lim u(x) = 0 : JxJ ~f'

this solutio/1 is Ullit/lIt' to with ill (Ill additil'e CO/lstunt 011 each houllded cOllnected compOllcnt ol[R" {j.

(h) in the case n ? 2, there exists a solution u of PN ([R"\fl, ~/) satisFyinIJ

lim grad u(x) = O. ixl- f

This solutiolJ is unique to within all additive constant on each n)f1nected component (hounded or not) of ~n\Q.

In the case n = I, we shall determine the solutions explicitly (see Example 19). We shall prove this theorem in the course of §4.5 (see Theorems 4 and 5 and


Proposition 17). Fredholm's integral method of solution allows us to deal with, more generally, mixed boundary value problems. Given Q an open set in [R" of boundary r we consider the problem

u harmonic on Q

u = qJ on ro

eu . .f, - + AU = 'I' on en

where r o is a part of the boundary r, qJ is defined on r o and t/t and). are defined on r \ro. We note that the Dirichlet problem P(Q, qJ) corresponds to Fo = I" and the Neumann problem PN(Q, tIt) to 1"0 = 0 and ;. == O. In the classical theory, we can always assume that r o is closed: in effect if u E <C0(Q) satisfies U = qJ on r a, qJ can be extended to a function ijJ E C60(to) and u = ijJ on to. Also to give a meaning to au/en on I" \fo we must suppose

{I" \ro is a hypersurfa.ce (of class (fc, 1) with Q locally

(4.72) on the one side of I" \1"0

We shall then always be able for U E '6' 1 (Q) and Z E I' \1'0 to define cu/cn( z) (see ~ 1.3). We have the

Definition 13. Let Q be an open set of [R", 1'0 a closed part of its boundary 1 with 1"\10 satisfying (4.72) and Jet qJ E (gO(ro), tIt, ). E v,O(F\ro)'

We give the name classical solution of PM(Q, 1'0' qJ. tIt, i) to any fUI,lction U E JI (Q) n '(,O(Q) satisfying

{

u(z) = qJ(z) for all ZEro

eu eu ~. (z) exists and -.. (z) + J.(z)u(z) = t/t(z) for all z E 1"\1"0 . cn en

This definition clearly generalizes, at the same time, the definition of a classical solution of a Dirichlet problem P(Q, qJ) and that of a classical solution of a Neumann problem PN(Q, tIt). We prove a "uniqueness' result for the mixed boundary value problem; more generally. we prove the following maximum principle:

Proposition 13. Let Q he a connected open set of[Rn, 1"0 a closed part ofits boundary r satisfying (4.72), qJ E '6°(1"0). 1/1. J. E 'f,°(r\ro) and u a classical solution oj PM(Q, 1'0. qJ, tjI, I,).

Suppose that qJ ;:;: 0 on 1"0' 1/1 ;:;: 0 and i. ;:;: 0 on I" \1'0 and in the case when Q is unbounded

lim infu(x) ;:;: O. Ixl ~ ~.

Then u is constant on Q or u > 0 on Q.


We then have the

Corollary 6. LeI Q he an open set in }l;", Fo a closed part oj' its boundary r with T\To satisfViny (4.72), (P E 0°(10), l/J,). E 0°(1'J'0) and in the case when Q is IInholinded 1I, E ((,o(Q). Lei) ?: 0 on r\ro . Then a soiutiollll ofPM(Q, 1o, (P, !II, i) satis/}!ing, in the case when Q is unbounded the condition at ir!finity

lim !I(x) - U rf.,<j = 0 Ixl- ex

is. if it exists, unique up 10 within an addilil'e constant on each bounded connected component Qo of Q such that

(4.73)

Proof of the Corollary. By linearity, we need only consider u a solution of PM(Q, To, 0, 0 • ..1.) satisfying in the case of Q unbounded lim u(x) = O. From

Ix! - ,x

Proposition 13 u is constant on each connected component Qo of Q. We have

u = 0 on aQo n 10 and [z E (Xlo ; ;.(z) > O}

and in the case of unbounded !to. lim u( x) 0; hence this constant is zero Ixl- ,

unless Qo is bounded satisfying (4.73). o Proof of' the Proposition 13. From the maximum principle (see Proposition 8 of §2) u is constant on Q on u(x) > infu for all x E Q; we can therefore suppose that

.Q

jnf < O. We take 0 < I: < -inf u and fI E ((,1(1R;) with U u

f! = Oon[D. x[ and 0 < p' oS L - loS fi < 0 on J-x, -D[.

We put l' = p(u). From the hypothesis

1'( X)e. 0 for x E Q sufficiently near 10

and also in the case when Q is unbounded

1'( x) = 0 for I x I sumciently large.

Hence there exists a bounded open set !Lo c:: Q such that

i'Qo\Q c:: r\/,o and l' == 0 on Q\Qo .

Using the hypothesis (4.72), we can suppose that Q o is regular. We then have

U E Yf (Qo) n '(;~ (t2o), r E 'f, ~ (Do), (see Proposition 5 of § 1)

grad ('.grad u p'(ullgrad ul z E Ll(QoJ


Applying Green's formula for integration by parts (see Proposition 4 of § 1),

L/'(U)lgradu2Idx = LQ"Q (vljJ - i,vu)dy ~ 0 .

Hence p'(u)lgraduI 2 = 0 on Qo. We deduce that

gradv = p'(u)gradu = 0 on Qo .

371

We can always suppose that Qo is connected; hence v is constant on Qo. By

hypothesis i~fv = P(i~fU) < 0; hence v = p(u) = c<Oandu == pl{C). 0

We note another means of approaching the principle of the maximum for boundary value problems by the use of Hopf's maximum principle; first, we state

Lemma 2. Let Q be a connected open set of [RII, zEoQ and n a unit vector of [R"

satisfying

(4.74)

there exists an open set Qo c Q and VE«?O(Qo) sub-harmonic on Qo

such that {x E oQo; v(x) > O} is relatively compact in Q,

v(z ),n) Z - An E Qofor ). > 0 small and lim inf ... ~....... . ............. > 0 .

. '-10 ),

Then,for all UE.ft(Q)n«?O(Q) non constant

. . u(z) - u(z - ),n) u(z) = mill u => lIm sup ----.--~~-~ < O.

Q ),"0 I,

Proof We can always suppose that u(z) = min u = 0; from the maximum prin-

ciple u > 0 on Q and hence in particular

We put £

We have

and

m

[}" l'

u ? m > 0 on {x E (1Qo; v(x) > O} .

if max v > 0 and £ = 1 otherwise.

t;V ~ m ~ u on {x EiWo;v(x) > O}

ell ~ 0 ~ u on {xEciQo; v(x) ~ O} .

Therefore we deduce from the maximum principle for sub-harmonic functions (Property 1) that

from which we have

. u(z) ~- u(z~ ;.n) hm sup --'----

.lIO A

. . rU(z - },n) - hm III •

. qo ) .

1· . r v(z - i,n) 0 ~ - £ 1m III < 0 .

. '10 ),


The most common example in which the hypotheses of the lemma are satisfied is the case of an interior ball condition: an open set Q of iR(" satisfies an interior ball condition at ::: E (lQ if there exists Xo E Q such that B(xo, i Xo ~ z I) c Q. Putting

::: Xo 11 = I::: ~ xol ' I' = Ixo ~ zl ,

for y.?2n/r2, the function v(x)=e-",2~e llx···xo is sub-harmonic on Qo: III

effect L1v(x) = 2y.(2y.lx ~ x o l2 ~ /l)e *'01'

We have then {x E ilQo ; v(x) > O} = i1B(XO'~) compact in Q,

. v(z ~ J.n) _ "., hm---. -~ = 2,ue > 0 . Al ° I~

We deduce from the lemma, the so-called HopI's maximum principle:

Proposition 14. Let Q be a connected open set in iR(n with boundary r, 1I E Jr(Q) (\ rgO(Q) and C E IR\. Let us suppose that (l) for al/ Z E I' with u(z) ~ c, there exists a unit vector n satisfying (4.74) such that

. u(z) ~ u(z ~ hl) hm sup' . ? 0;

)·10 I~

(2) there exists x E Q such that u(x) > (' :

(3) When Q is unbounded, lim infu(x) ? c: I x! ---+ J

then u > c on Q.

Proof From the maximum principle for harmonic functions. if the conclusion is false, u attains a minimum which is less than or equal to c at a point Z E T. which contradicts the lemma. 0

This gives the

Corollary 7. Let Q be an open set of iR(", T a closed part of its houndary T, (P E ((,0 (ro ), tj; and i. d(fined OIJ T \ro wit II ). ? 0 and ill the case of Q unhounded II f E ((,o(Q). For all ::: E ]"['0' we choose (/ IIllit L'ector n(z) satisfyill?} (4.74) (we suppose that it exists). We thell hare the conclusion of Corollary 6, fi)r a solution of PM (Q, 1'0' (P, tj;, )~) in the sense

{

UE<#(Q) (\ '6'0 (Q), U(.Z) = .:p(z)

(4.75) . u(z) ~ u(z ~ )"n(z)) , hm , + ),(z)1I(z) ;q 0 A

for all Z E r ° '

tj;(z) .lor all ::. E r,ro .

The concept of a solution of PAI(Q, ro, .:p, tj;, J.) defined by (4.75) is weaker than that of the classical solution when, T \T 0 satisfying (4.72), for every point z E T! ro


the exterior normal vector n(z) satisfies (4.74); this is in particular the case if we have the interior ball condition at each point z E r\ro. Note that mixed problems appear naturally when we apply a Kelvin transformation to a Neumann problem. Let us take an open set Q and Xo E [R", ro > 0; we denote Q' = I(xo, roHQ\{xo}) whose boundary aQ' is

1" = I(xo, ro)(aQ\{xo}) or 1" u {xo}

according as Q is bounded or not. Given a closed part roof r = aQ with r\r 0 satisfying (4.72), we find that

r~ = aQ'\I(xo, ro)(r\(ro u {xo}))

is a closed part of aQ' and aQ'\r~ satisfies (4.72). Given u E '"6'1 (Q) its Kelvin

transformS! u' = H(xo, ro)u E <;&1 (Q'); if ~~ (z) is defined for z E rvro \{xo})

au' then an' (z') is defined for z' = I(xo, ro)z and making use of the formula (2.25) of

§2, we obtain

au' , _ au, (Zl - xu). n'(z') ,~, au' (z) - H(xo,ro)an (z) + 2--~ru(,-),

Hence if u is a classical solution of PM (Q, r 0' <p, if;, ).) satisfying, in the case when Q is unbounded, a condition at infinity

lim Ixln - 2 u(x) = c, I xl ~ 00

its Kelvin transform u' = H(xo, ro)u is a classical solution of the mixed problem PM(Q', r~, (pi, if;',).') where Q' = [(xo, ro)(Q\{xo }), with boundary

r~ = aQ'\I(xo, ro)(aQ\(ro u {xo})),

and the functions q/ and l' defined by

I(xo, ro)ro ,

Xo if Q is bounded.

and ) '( ') = '(I( ) ') _ 2 (z' - xo)·n'(z') . z ). xo, r 0 z o. r~ o

If Xo E Q, then lim I x' In - 2 u'(x') = c' with (" = u(xo ); it is the same in the case Ix'i ~ 7)

Xo E ro with c' = <p(xo). Finally, in the case Xo E aQ\T'o we shall likewise have a mixed condition at infinity. Let us conclude this section with some examples:

8' See §2.S.


Example 19. Case n = l. Let us consider Q = ]a, h[ c [R;. A function u harmonic on Q is of the form u(x) =:xx + fi. The boundary T of Q is {a, b}, {a} or {b} according as Q

bounded, Q = ]a, CIJ[ or Q = ] - CIJ, be. A part 10 of 1 could be 1>, {a}, {b} or the whole of r. We give in Table I the solutions of PM(]a, bE, 1 0 , <p, tf;, ;.) with <p

defined on 1'0' I~ and i defined on l' To in the different possible cases.

Example 20. The radial case. Let us consider a radial connected open set in [R;" with n ~ 2. Excluding the cases Q = [R;n and Q = [R;"\ {a}, it is one of the following four types: Q = B(O, r) an open ball with boundary 1 = 8B(0, r),

Q = B(O, r)\{O} a punctured open ball with boundary r = 8B(0, r) u {a}, Q = {x; 1'1 < Ixl < r2 with ° < r1 < r2 <x, a ring with boundary

r = DB(0,r 1 ) u aB(0,r2 l.

Q = [R;"\B(O, r), the exterior of a ball with boundary r = ?B(O, rl. A radial part To of the boundary can be 1>, {a}, 3B(0, r) or the whole of T according to type. We take 0Q to be a radial function defined on To, tf; and I. radial functions defined on r\ r o' From uniqueness (possibly to within an additive constant), a solution of the mixed problem is necessarily radial and hence of the form:

u(x) = Co + ('1 E"(lxl) .

We shall give in Table 2 of the following page, the solutions in the different possible cases. 0

Example 21. Neumann prohlem in the halFspace. We take Q = [R;" - 1 X [R; + with boundary r identified with [R;" 1 and tf; E '6 0 ( 1). By definition u is a solution of P N (Q, tf;) iff U E .J1' (Q) and - ?u/ ex" is a solution of the Dirichlet problem P(Q, tf;). From Example 14, supposing tf; to be bounded, u is a solution of PN(Q, tf;) with eu/()x" bounded iff

tf;(x' + (X,,) dt = 2 r ~E" (x - t)tf;(t)dt , It xi" Jw 1 ex"

or, on integration

(4.76)

- En(x' - t, Xn))tf;(t)dl .

We now distinguish between the cases n ~ 3 and n = 2. Case n ~ 3. Let us suppose

there exists f;: [R;+--+ [R;+ decreasing with jf E{r)dr < T o

and such that I tf; (t) I ~ t: (I t I) . (4.77)

Tab

le I

To

=

f =

{a

, b}

1 .l.

(b)

¥---~

b -

a

:€ II 1

.l.(b

) =

---

" '-

. b

-a

" " <= ::;

0 ..0

L

....J

"'" ),(

h)

-),(

a)

'" A

(al'.

{h)

¥-r-, II

h -

a

C;

0.

-J. (

aj),

(b)

= ).(

b) +

),(a

) ¥-

O

b-a

II 0

'-.

I A

(a)

=

},(h)

=

0

r~'-

. -----------

....

. --~-."--.---.. -

----~~---

'-'

fo

=f=

{a}

8 '" r-,

._._

----

II fo

=

{a}

C;

Sol

utio

ns o

f P

M(Q

, T

o,

rp,

1jJ, .

l.)

rp(a

)(b

-x)

+ r

p(b)

(x

-a)

b -

a

rp(a

)(l +

),(b

)(b

-x)

) +

ljJ(b

)(x

-a)

I +

).(b

)(b

a)

rp(a

) +

c(x

-a)

ljJ(a

) -

ljJ(h

) +

ljJ(

a)),(

h)(h

-

x) +

ljJ(

h)A

(a)(

x -

a)

j,(a)

.l.(b

)(b

-a)

-

(}.(b

) -

),(a)

)

ljJ(a

) +

c(

x _

a +

_1

) A

(a)

A(a

)

..

ljJ(a

)(a

-xl

+ c

rp(a

) +

cia

-

x)

ljJ(a

)(a

-x)

+ e

[l +

Ma

)(x

-a

)]

Rem

arks

Dir

ichl

et's

pro

blem

exis

tenc

e an

d un

ique

ness

c ar

bitr

ary

cons

tant

ex

iste

nce

iff

<p(a

) +

(b

-a)

ljJ(b

) =

0

exis

tenc

e an

d un

ique

ness

c ar

bitr

ary

cons

tant

ex

iste

nce

iff

ljJ(b

)j A

(b)

=

ljJ(a

)/ .l.

(a) .-~.-

c ar

bitr

ary

cons

tant

ex

iste

nce

iff

ljJ(a

) +

ljJ(

b)

=

0

c ar

bitr

ary

cons

tant

no

con

diti

on f

or e

xist

ence

c ar

bitr

ary

cons

tant

no

con

diti

on f

or e

xist

ence

I

"'"

:l"- n [ n'

e:..

-I

::r '" o ~ s.. o ::;

. g: ~

",'

'"0

(3 cr" " 3 w

---l

v.

Table 2. The radical case (Example 20)

.:: ci

$1&

Solution(s) of PM(Q. 1 0 , (P.IjI. i.) = (; B (():-~- ---"--"--, --------------------------------------

i

i(r) "" 0

",(r)

ljI(r)

Remarks

Dirichlet's problem

existence and uniqueness

I

II II f-------------j----.... ----------------c: - i J.(r) = 0 t c an arbitrary

r -[' ~tl ___ , __________ ~-----------------~---c(-)n-s-t-an-t--~ existence itT ljI(r) = 0

I'o - I - ,'B(O. r) u [0: generalized solution of

P(Q. "') is classical iff ",(0) = ",(r)

~f-'--------------------~---------------------------+-----------~ 1'0 - :n: ",(0) uniqueness and

o <p(r)

C ~ Cl:l

?, s Cl:l

I

c:

1'" = r = ('B(O.'"I u ('BIO. r,l

-.:-ci

~ -II c -

(ljI(r 2 ) - i.lr,)(plr, ))(E,(x) - E,(r,)) ",Ir,) +

"'(',) + e(E,(xl

f-.j ........ ------------------------'--------j-.--'--------------- ---------------------------------- -------------

L',(', )1

+ ilr, )

+ "" 0

& II

~o

ilr 2 ) o

I' ('B(O. r)

I'll -- 0

ljI(r, I ( + (' E,(xl i.(r, )

U(X)!li

",Ir) + dE,lx) £,Ir))

£,Ir))('i(r) - I/Jlr)) + ('

existence iff i.(rl",(O) + I/J(r) = 0

Dirichlet's problem

existence and

uniqueness

c arbitrary constant

existence iff 1/J('2) = i.(r2)",(r,)

existence and uniqueness

t an arbitrary constant:

existence iff I/J(r" ljI(r,1

c an arbitrary constant;

existence iff r; 'ljI(r ,I

c arbitrary constant

no condition for existence

(' arbitrary constant

no condition for existence

(' i,(r,1 ilr,l \). ,

il! with u(x) = ",ilr,)i.lr,j{E,lr,i -. E,lr,)) + ;-", T --,-" x '::. " /

I/Jlr,) _I/Jlr,)). F,lx)1 + .

r~ - 1 r~ I


Then for all x E II{", the function t -> E,,(x - t)t/J(t) is integrable on II{"-l and

I}i~cc L" 1 En(x - t)t/J(t)dt = O.

In effect

Making xn ->x in (4.76), taking account of Corollary 6, in the case n ? 3, for t/J E (gO(II{,,-l) satisfying (4.77), we see that there exists one and only one solution u of the Neumann prohlem PN(lI{n-l x II{+, t/J) tending to zero at infinity, given hy

(4.78) u(x',xn) = (;~:~i}~:L" dt~~'t~;!n2t~1 dt.

Case n 2. Let us suppose that

(4.79) r Log(1 + (2 )1t/J(t)ldt <:0 and that f t/J(t)dt = O. JR R

Then for all (x, y)EII{2 the function t -> E2 (x - t, y)t/J(/) is integrable, since t -> E2 (x - t, y) is locally integrable and

Now for x E II{,

SInce

and

Ez(x - t,y) = O(Log(1 + (2)) when It I -> CYj.

}~~ r E2 (x - t, y)t/J(t)dt = 0,

r t/J(t)dt = 0, .,

( (X-t)2) Log 1 + -0 ,,0 when y /'x . r .

Passing to the limit in (4.76), we see then that in the case n = 2,for t/J continuous and hounded on II{ satis(ving (4.79), there exists one and only one solution u of P N (II{" - 1 X II{ + , t/J) satisfying

eu (4.80) oy is bounded on II{ >< II{+, lim u(x, y) = 0 for all x Ell{,


given by

(4.81) u(x,y) = .~ .. }'J" !/I(x + yt)Log(l + t 2 )dt. 2n H

o

Example 22. Kelvin transji)rm of the N eumanll problem ill a half space. We consider a ball B = B(xo, ro) and Zo E (1 B. The image of B by the inversion I (zo, 2ro) is the half-space

Q' = :x' E iR n ;(x' + ::0 - 2xo)(xo - zo) > O}.

Let us take ~I E «(, °U"'B,:::o) and consider its Kelvin transform !/I' = H(zo,2roJ!/I defined on oQ'. As we have explained in the general case above, if we are given u' a solution of the Neumann problem PN(Q', !/I') its Kelvin transform

u = H(zo,2ro)u' E ,ff(B) n (('~(B\{zo])

and satisfies

(4.82) au (zo~~ z)(z xo ) :i-(z) + 3 . u(z) en 2ro

!/I(z) forall zEcB\{zo]

Also if u' tends to zero at inJinity, then

(4.83) x ---+ =0

We restrict ourselves to the case 11 ? 3. The condition (4.77) on !/I' E «(,0 (c':Q'), can be expressed as:

there exists C: iR + -+ iR + increasing with If:, dr C(r)~7 < + x

r" o (4.84)

such that C(lz - zol)

I !/I ( z) I :s: --.:-~ ~ In=- 2. - -0

By Kelvin's transformation, ill the case n ? 3,for all !/I E «(,0 (a B\ {zo }J satisfjJillg (4.84), there exists a 1Illil/ue solution u olthe mixed houndary l'a/ue prohlem specified hy (4.82) and (4.83).

We notice that the function A(z) (zo - z).(z-xoJ

< ° on (1 B\ { zo} .

Example 23. A mixed prohlem in a rectangle. We take a rectangle Q = JO, a[ x JO, h[ and consider the mixed problem:

(4.85) ax (0, y) + ),u(O, y) = !/I(y) for y E JO, h[ {

~uharmonic in Q

uta, y) = u(x,O) = u(x, h) = 0 for x E [0, aJ, y E JO, h[

where if; is continuous and bounded on J 0, b [ and A. E iR.

o


We apply the same method as in Example 5, looking for a solution in the form

(4.86) n

u(x,y) = I unCx)sinnwy with w = y;' n?: 1

We must have

(4.87)

(4.88)

d 2 un 2 --2 = (nw) Un dx

ural = 0

dUn . 2fb. '/')d -(0) + AU.(O) = -b smnwy,/,(y y. dx 0

- tlW coth(nwa), (4.87) admits one and only one solution

2 shnw(x - a) fb. un(x) = -b h . h smnwyt/l(y)dy.

nw c nwa + I" S nwa 0

If A = -nwcoth(nwa), (4.87) admits a solution iff

I sinnwyt/l(y)dy = 0,

and then the solutions are

un(x) = cshnw(x--a) or c, an arbitrary constant

To sum up: for all )" E IR\ { - kw coth kwa; k = 1,2, ... }, the problem (4.85) admits one and only one solution given by (4.86) where,for all n, u.{x) is given by (4.88); irA. = -kwcothkwa,for an integer k ~ 1, then (4.85) admits a solution iff

I sinkwYI/J(y)dy = 0,

and when this condition is satisfied, the solutions of(4.85) are given by (4.86) wherej()r all n 0/- k, un(x) is given by (4.88) and

uk(x) = cshkw(x- a) with c an arbitrary constant.

As we have remarked in Example 5, the formal calculation can be justified: we verify in particular that for t/I being locally a Holder function on] 0, a [, the solution found is clearly a classical solution of the problem. 0

5. Solution by Simple and Double Layer Potentials: Fredholm's Integral Method

In this section we consider Q a regular bounded open set in IR" with n ~ 2. We shall suppose that the boundary r ofQ is ofclass82 '6 1 +, so that we can apply the results

82 See Definition 3 of §3.

3RO Chapter II. The Laplace Operator

of §3.3 on simple and double layer potentials. We seek to find solutions of the Dirichlet problem, the Neumann problem, or more generally, the mixed boundary value problem, in the form of simple layer or double layer potentials. First we recall the results of §3.3 and introduce further notation. Given 1.E'f,°(n, from Propositions 10 and 12 of §3, the interior simple layer

potential defined on Q by

(4.89) u(x) = i E,,(x ~ t)1.(t)dj·(t) x E Q

belongs to .ft (Q) n 'f,~ (12); for all Z E r. the functions

t -> En(z ~ t)1.(t) and t -> grad E,,(:; ~ t) ./l(z)1.(t)

are integrable on rand

u(z) = i En(z ~ t)1.(t)d~,(t)

i'u 1 ~ (:) = grad E,,(:; (' n r

t ) . n ( :; )1. ( t ) d j' ( t ) 1.(Z)

2

For all 1. E 'f, ° (n. we denote by L1. the function defined on r by

(4.90) L1.(:) = r En(: ~ r)1.(t)d,/(t) Jr

and we denote by J Y. the function defined on r by

(4.91 )

I'

= 2 J gradE,,(: ~ t).Il(:)y.(t)d,/(t) r

:2 j' (:; , .... I).n(:;) = - -1_ ~ 1.(t)d~·(t).

(In I 1- tl

From what is recalled above Ly. and J1. are continuous on F; also the maps 1. -> L 1. and 1. ....... J ,1.: L alld J are lillear operators 011 «(i ° (I'). Also what was recalled above can be expressed in the following manner: (I) solution of'the Dirichlet problem P (Q, (p) by a simple layer potential: being given <p E '(,O(n and 1. E '6°(Q), the interior simple layer potential defined by Y. (given by equation (4.89)) is the solution of P(Q, <p) iff

(4.92) <p(:) = Ly.(z) = i E,,(: ~ tl1.(t)d},(t) for all Z E r;

(2) solution of'the Neumann problem PN(Q.Ij;) by a simple layer potential: given Ij; E {(io (n and 'X E «(,0 (n. the interior simple layer potential defined by r:J. is the solution of PN(Q,Ij;) iff

(4.93)

hi:) ~r:J.(:) i (: ~ 1).11(:) . 'X(:;) , Ij;(:) = ..... --,.,-- = --_ ----;; 'X(t)di'(t) ~-)-- for all Z E J;

"- I (I"I_ tl ..


(3) more generally given a closed part F 0 of F,

the interior simple layer potential defined by rx is the solution of the mixed problem PA1(Q,Fo,~,~,2)iff

(4.94) {

Lrx(z) = ~(z) for all Z E Fo

,!~(z) + 2A(z~La(Z) - a(~) = ~(z) for all z E F\Fo .

The Propositions 10 and 12 of §3, indicates to us that also the exterior simple layer potential defined on IR"\,Q by

(4.95) uJx) = r E,,(x - t)rx(t)dy(t) x E IR"\,Q .1r

belongs to 't5'~ (IRn\Q) and with the notation

(4.96) { ue(z) = La(z)

GUe(z) = __ Ja(z) + rx(z) an 2

for all z E F

for all Z E F .

Also from Proposition 2 of §3

(4.97) u,,(X) = (L :xdy )E"(X) + 0CXI!-l) when Ixi --> 00

grad ue(x) = ({ :Xd}")~n + o(~) when Ixl .... ~ 00 . Jr CT"lxl Ixl

Also, given [3 E 't5'0 (r), from Proposition 11 of §3, the interior double layer potential

(4.98)

i eJE" i (1- x). nit) vex) = -;)-(t - x)/J(t)d},(t) = n [3(t)dy(t), run ra"lt-xl

aE belongs to £(Q) n 't5'0(,Q); for all Z E r, the function t --> _;:;_"(1 - z)[3(t) is

en integrable on F and

( aE" [3(z) viz) = Jra~(t - z)[3(t)d),(t) + 2 .


For all f3E~O(T), we denote by KfJ the function defined on r by

1 DEn 21 (t - z).n(t) (4.99) KfJ(z) = 2 "(t - z)fJ(t)di(t) = - fJ(t)dy(t) .

T on (In r It - zl"

The function Kf1 is continuous on r and the map If -> K f3 is linear: K is a linear operator on (6° (T). We have the solution of the Dirichlet problem P( Q, cp) by a double layer potential: given (P E cgO(T) and fJ E cgO(T), the interior double layer potential defined by fJ (given by (4.98)) is the solution of P(Q, cp) iff

{ cp(z) = K(3(z) 2+ 11(z) = r (t ~ z).n:!) fJ(t)dy(t) + ~'2(z)

(4.100) Jr (In t - z!

for all Z E r .

Proposition 11 of §3, indicates to us also that the exterior double layer potential defined on IRn\Q by

(4.101) J~ DE ve{.x) =;;n(t -- x)fJ(t)d-y(t) ,

/ en

belongs to X(IR"\Q) n ~O(lRn\Q) and

(4.102) K(3(z) - (3(z)

ve(z) = ~~2--- for all Z E r .

In addition, from Proposition 2 of §3,

(4.103)

when Ixl->x

since I'" is the Newtonian potential of the distribution I with compact support

which is such that <f, I) = o. These references back, with the notation introduced, show that the solution oj' Dirichlet, Neumann, or mixed problems, interior or exterior, by simple or double layer potentials leads to the solution of the following integral equations:

fI) = L':J. fI) = _K.:..-fJ_±---,--fJ 'r , 'r 2'

J':J. + ':J. - t/J = 2- ,etc ...

We shall therefore study these equations, that is to say the operators L, K and J First we note some properties of these operators83 . We introduce the following

83 Other properties will be developed in ~5.


notation:

(4.104)

QJ, ... , Q N the connected components of a 0'1' ... , a:", the bounded connected components of IR"\Q a~ the unbounded connected component of IRn\,Q

r l' ... , r N , r~, F'J, . , . , r N , the boundaries of OJ, ... , ON'

a~, 0'1 , .. , a~, respectively.

Note that {r l' ... , r N } and {r~, F'J, ... , r:"I'} are two partitions of the boundary r of a (which is also the boundary of W\Q).

n;,

Fig. 4

~/fllllJ I 1'0 = I'

Fig. 5

Fig. 6

N = J IV' = 1

IV = 2 N' = I


It must be emphasised that the connectedness of Q and IRn\Q are independent. With the notation of (4.1 04): Q connected iff N = 1 and IRn\Q connected iff N' = O. Figs. 4 to 6 give examples of different situations. By construction, Q being regular the set ri n ri , with i = 1, ... ,N, and j = 0, ... ,N' is open and closed in r with the result that the characteristic functions XI',,, I'} are continuous on r84. (We shall show later that the number of connected components of r is exactly N + N'). It is obviously the same for Xr and XI',' '

Proposition 15. The operators L, J, K 011 (6°(F) being defined by (4.90), (4.91) and (4.99) (1) L is symmetric, that is

f rJ.Lf3 d i' = f f3LrJ.d}' forallrJ.,f3ErcO(F); Jr JI'

(2) J and K are adjoints one with the other, that is

L rJ.Kf3dy = L f3JrJ.dy for all rJ., f3 E rco(F) ;

(3) With the notation (4.1 04), we have

(4.105) KXI', = XI',' 1: JrJ.d}' = II': rJ.d}' for i = 1, ... ,N, rJ. E (to(F)

(4.106) KXr = - Xr, f JrJ.dy = - f rJ.di' for j = 1, •.. ,N', rJ. E %,o(F) ; J J Jrj Jrj

(4) L is a bijection from %O(F) onto ImL85, subject in the case 11=2 to the following hypothesis:

(4.107) there exists xo E IRn\Q~ such that r~naB(xo, 1) = 0.

Proof The points (1) and (2) follow from the definitions of L, J, K and Fubini's theorem:

L rJ.Lf3dy = L rJ.(z)dy(z) L En(z - t)f3(t)dy(t)

= L f3(t)dy(t) L En(t - z)rJ.(z)dy(z) = L f3LrJ.d),

{I if x E A,

84 Given a part A, the characteristic function XA(X) = o if x If A,

85 Given a linear map T: X -+ Y where X, Yare vector spaces,

1m T = {TIX; IX EX} , kef T = : Y. EX: Ty. = 0:


i i 2i(t-z).n(t)

!XKfJdy = !X(z)dy(z)- I _ I" fJ(t)dy(t) r r (T" r t z

i 2 i (t - z). n(t) i = fJ(t)dy(t) - I I" !X(z)dy(z) = fJJ!Xdy.

r (T" r t - z r

Now given !X E ~O(n, the interior simple layer potential U (resp. exterior potential ue ) defined by !X is a solution of

( J!X - !X) ( (- J!X + !X)) PN Q, 2 resp.PN IR"\Q, - 2 .

Hence for i = 1, ... , N (resp. j = 1, ...• N'), the restriction of u to Q i (resp. Ue

to Qj) is a solution of PN(Q i • 1/1) (resp. PN(Qj, 1/1» where 1/1 is the restriction of j(J!X - !X) (resp. - j(J!X + !X» to r i (resp.Tj). From Gauss' theorem

r J!X - !X (f J!X + !X ) Jr 2 dy = 0 resp. r' 2 dy = 0 ,

, J

that is

f J!Xdy = f !Xdy (resp. f ,J!Xdy = - f, !XdY). r, r, rj rj

To complete the proof of point (3), we use (2): we have

r !XKXr,dy= r xrJ!Xdy=f J!Xdy= r !XX.dy Jr Jr I; Jr I,

( resp. r !XKXr,dy= r XrJ!Xdy= r !X(-XrJdY) Jr J Jr J Jr J

for all !X E ~O(n. The stated result follows immediately. Finally we prove the point (4): Let us consider !X E ~O(n such that L!X == 0 on r. The interior simple layer potential U defined by !X being a solution of the Dirichlet problem P(Q, L!X) is identically zero: we have already deduced

that is

(4.108)

au _ J!X - !X _ 0 an - --2- = on r,

J!X ==!X on r. The exterior simple layer potential U e is a solution of the Dirichlet problem P(IR"\Q, L!X): in particular for j = 1, ... , N', Ue == 0 on Qj and therefore

aU e

ane Making use of (4.108), we have

J!X + !X 2 = 0 on rj.

!X == 0 on r\r~.


In the case II ~ 3, from (4.97), lim ue(x) = 0; we have thus also lie == 0 on Q~, Ixl ~ xc'

from which as above,:X == 0 on r~. In the case n = 2 if 1 :x d}' = 0, then still from

(4.97), we shall have again lim ue(x) = 0 and hence:x == 0 on r;]. To complete I xl ~ j

the proof in the case 11 = 2, let us suppose, for example, that f:x dr' > O.

Given Xo E iRn\Q~, we have with (4.97)

uc(x) f lim = :xd:: > 0, lxi''" E 2 (x - xo)

and hence adapting the proof of Proposition 9,

tie ~ 0 on Q~:


and hence

ex ~ 0 on J" o .

Now, we have, since Xo E iR" ,Q~ c {j U Q'1 U ... u Q;v ,

I'

Jr :x(t)E2(XO - t)d,/(t) = O.

" Choosingxosuchthatr~ n 8B(xo , 1) = 0,on making use of(4.107), E 2 (x O - 1)

is not identically zero for t E r~ and hence keeps the same sign: we deduce :x 0 forr~. 0

Remark 5. From the proof, we see that we always have

(4.109) kerLn{:XE'6 0 (l'); l:Xd'r' = o} = [O}.

It is false in general in the case II = 2 that L is injective: for example, considering Q = B(O, 1), we have

L J == 0 on L = ('Q.

This particular feature of the case 11 = 2 arises from the normalization of the elementary solution E2 of the Laplacian in iR 2 : we can in an equivalent manner suppress the hypothesis (4.107) by replacing, in the definition of the simple layer potential, the elementary solution E 2 (x) by E 2 (x/i5) = E 2 (x) - E2 (15) where () > 0 is chosen as follows:

(»minmaxlx -:1 or ()<maxminlx - zl· Q r Q r


In other words, we can suppress the hypothesis (4.107) by changing the unit of length in the plane.

Remark 6. Alexander's Relation. An interesting application of the formulae (4.105) and (4.106) is the equality (called Alexander's relation: N + N' = number of connected components of T. Let us take, in effect, To, a connected component of T; there exists a bounded open set Qo such that To = 8Qo; from Lemma 70f§3, the interior (resp. exterior) double layer potential of Xr is constant on Qo and zero on 1R"\t20 and hence a fortiori zero on 0Q~ and constant on Q1' ... , QN' Q'j, ... , Q~.; we deduce from (4.100) and (4.102) that

(I + Khro is constant on I' j, ... , TN

(l - Khro is zero on r~ and constant on r'j' ... ,r~ .. Hence, denoting by F the space of the functions which are constant on each

N N'

connected component of T, V = LXI, IR and V' = L Xr' R we have thus i= 1 j= 1 J

shown that

from which

(J + K)F c V and (I - K)F c V';

F c V + V' ,

On the other hand from (4.105) and (4.106), we have

K <p = <p for <p E V and K <p = - <p for <p E V' .

Hence V n V' = {O}. Since obviously V and V' are contained in F, F = V EB V', from which we have N + N' = dim F = number of connected components of F. o We shall now make use of the RieszFredholm theory. Let us recall the prin~ipal results (see, for example, Lang [1 J).

Definition 14. Let X be a Banach space and T a map of X into itself. We say that T is a compact operator if for every bounded sequence (un) in X, there exists a subsequence (u".) such that the sequence Tunkconverges in X when k .....,X). Given a compact linear operator T, the operator I - T: u ....., u - Tu is a Fredholm operator: we shall now state the theorem of Riesz and Schauder.

Lemma 3 (The RieszSchauder Theorem). Let X he a Banach space and T a compact linear operator (~f X into itself, then (1) 1m (I - T) is a closed sub-space of X; (2) ker (I T) is a sub-space of X offinite dimension; (3) dim ker (f -- T) = codim 1m (I - T)86

We shall obviously apply this theory with X = ,&,o(r) provided with the uniform

86 The codimension of a sub-space F of a vector space X is the dimension of the quotient space X! F; we have codim F = N iff there exists a sub-space X 0 of dimension N such that X ,= FEB X o.


convergence norm

11:x11 supl:xl· r

The essential point is

Proposition 16. The operators L, J, K defined hV (4.90), (4.91) and (4.99) are compact operators on ~o(n.

To prove this proposition, we shall make use of Ascoli's theorem (see Dieudonne [1], Vol. 1, Schwartz [2J) which we state in the form:

Lemma 4 (Ascoli's Theorem). A map T oj'0 0 (n into itselfis compact ifffor I!1wy bounded sequence (:x k ) ojjimctions elk E ((o(r) and aery.::o E r,

(i) sup I T:xd'::o)1 <x k

Proof or Proposition 16. Let us take a bounded sequence of functions (:x k ) E (6 0 (T) and.:: 0 E r. To prove the compactness of L we go back to the proof of Proposition 10 of §J. Let us consider a normal parametric representation (R, U, (I, :x) in the neighbourhood of '::0' We can always suppose that B(O, ro) c f', '::0 = (0,0). For o < r ,;:; ro, x' E B(O, r), .:: = (x', :xix')) we put

bd.::) = (P~(.::) + I~(x') ,

with

where rr = {it', :x(t')): i'EB(O,r)}

1 k (x') r En{t' - x',:x(1') - :x(x')):xdt',:x(t'))(l + Igrad:x(t'W) 12 dt'. JBlo.rl

We have

II~(x')1 ,;:; CkJI" y(t' - x')dr'';:; ckJ". . y(t')dt' BIO.r) BIO.2rJ

with Ck = :xd sup (I + ! grad :xU' W )12

B(O.ro)

y(t')= { (~-=---2) ~lt~l-;;- 0

I I, - Log- If II 2n I t' I

if 11:? 3

2.

H7 For the details. see the proof of Proposition )0 of ~3.


Since g is locally integrable on IRn - 1

sup sup II;;(x')1 --+ 0 when r --+ O. k'" 1 x' E B(O,r)

On the other hand,

ICPk(Zo)1 ~ II:Xkll f IEn(t - zo)ldy(t) nr,

Icp:;(z) - cp;;(Zo) I ~ Ilakllf IEn(t - Z) - En(t - zo)ldy(t) nr,

with the result that for 0 < r ~ ro fixed

sup Icp:;(z) - cp;;(zo)1 --+ 0 when Z --+ Zo' k

389

We verify thus the points (i) and (ii) of Lemma 4, from which we deduce the compactness of L. The compactness of K and J will be proved in the same way as in Propositions 11 and 12. 0

We are now in a position to prove the

Theorem 4. Let 0 be a regular bounded open set in IR n with n ~ 2, with boundary r of class ~1+' and, in the case n = 2, also satisfying the hypothesis (4.107). On the other hand, let A, t/J E ~O(F) with A ~ 0 on r. Then the following properties are equivalent: (i) there exists a classical solution of the mixed problem

(4.110) { u harmonic on 0

au on + ).u = t/J on r;

(ii) there exists a E ~o (F) solution of the integral equation

(4.111) J:x - a

2 + ALa = t/J on r,

where the operators Land J are defined by (4.90), (4.91); (iii) For i = 1, ... , N

).;=0 on r;=> r t/Jdy =0, Jr, where r l' ••• , r N are defined by (4.104). In addition, the solution of(4.11O) is unique to within an additive constant on each 0;, i = 1, ... , N such that). == 0 on r;;for every solution a of(4.111) the simple layer potential u defined by IX (by (4.89» is a solution of (4.110) and the map a --+ u is a bijection from the set of solutions of (4.111) onto the set of solutions of (4.110).


Proof Given j, E ,??O (r) fixed, we denote by T the operator on (6'0 (r) defined by

Tex(z) = J:x(z) + 2)_(z)La(z) ,a E ,??O(r) , z E r ,

and B the map of £ (Q) n (6' ~ (Q) into C(i ° (r) defined by

au, (1 ~ Bu(z) =~(z) + 1_{Z)U(z) , U E.Yf Q)n'??n(Q),

on Z E r .

With the notation (4.104), we can always suppose:), == 0 on r 1, ... , rr and), == 0

on 1:+ 1"", rN with 0 :::; r :::; N. We denote

F = {i/J E (6'0(r) ; L i/J dy = 0 for i = 1, ... , r} . With this notation for i/J E ((j0(r), (4.110) is written Bu = i/J and (4.111) is written i( Trx - rx) = i/J; in other words the equivalence of (i), (ii) and (iii) is expressed by

(4.112) T - 1

Im-- = 1mB = F 2 .

From Gauss' theorem, 1m B c F. From what has been recalled above (see (4.94)) for ex E '??O (T), the interior simple layer potential u defined bya is a solution of (4.110) with i/J = (Ta - a)/2; we deduce

(4.113) T-I

1m -- c 1m B c F . 2

From Proposition 15(4), the map a -> u, the interior simple layer potential of rx, is L: for a bounded sequence (an) of '??O (T), we can extract a sub-sequence rxn• such that the sequences J ank and Lrxnk converge and hence, immediately, so does Trx nk . Hence 1m (I - T) = 1m (T - I)/2 is a closed sub-space of finite codimension equal to dim ker(I - T) (see Lemma 3). On the other hand, using for example, the projection theorem in the Hilbert space L 2 (T) on which is defined the scalar product

(cp,i/J) = r cpi/Jdy, J Slilce

F = {i/JE,??O(T);(X'i,i/J) = 0 i = I, ... ,r},

we have C(j'O(r) = FED Xr,1R ED'" ED Xr,1R

and hence codim F = r. From (4.113), the equality (4.112) is therefore equivalent to

(4.114) dimker(I - T}:::; r

From Proposition 15(4), the map rx -> U, the interior simple layer potential of rx, is injective; in effect La is the trace of U on r with the result that u == 0 on Q implies that Lex == 0 on r. Now, if a E ker(I - T), i/J = (Trx - a}/2 == 0 on r with the result that, from Corollary 6, when A :;::, 0 on r, u is constant on Ql' ... , Q 7 and


u == 0 on Q,+ I' ... , QN' In other words, when )~ ~ 0 on r, the map et. ....... II is an injection from ker(I - T) into the space Xa, ~ EEl ... EEl XQrR This proves (4.114). 0

Similarly, we have for the exterior problem in the case n ~ 3:

Theorem 5. With the same hypotheses as in Theorem 4 with n ~ 3, the following properties are equivalent: (i) there exists a classical solution of the exterior mixed problem

U e harmonic on ~n\f1

(4.115)

lim ueCx) = 0 ; I x --t 'x.:

(ii) there exists et. E <go (r) solution of the integral equation

(4.l16)

(iii) fC)I' j 1, ... , N'

Jet. +et. ALy. - ---- = 1/1 on r;

2

i. == 0 on rj = f .1/1 di' = 0, r;

where r'" ... , r;v' are defined by (4.104). In addition the solution ()[{4.115) is unique to within an additive constant on each Qj, j = 1, ... , N' such that)~ == 0 on rj;for every solution et. of (4.116) the exterior simple layer potential Ue defined by a. (by (4.95)) is solution of (4.115) and the map a. ....... ue is a bijection of the set o[soiutions 0[(4.116) onto the set o[solutions (if(4.115).

The proof is absolutely identical with that of Theorem 4, making use of what was recalled above concerning exterior simple layer potentials (see (4.96), (4.97)).

Theorems 4 and 5 admits as a particular case, corresponding to i == 0 on r, the points (I) and (2a) of Theorem 3. The conclusion of Theorem 3 follows immediately from

Proposition 17. Let Q be a regular bounded open set in ~~n with n ~ 2, with boundary r of class ((,,' +, and 1/1 E '6 0 (r). Theil the following assertions are equivalent: (i) there exists a classical solution Uc of the exterior Neumann problem PN(~n\Q, 1/1) satisfyiny

(4.117) lim grad ue{x) = 0 ; Ixl ~ £

(ii) there exists a solutioll ('j, E <'6'0 (r) of the integral equation

JY. + Y. ./, on r,' 2 = 'f'

(4.118)

392 Chapter If. The Laplace Operator

(iii) For j = I, ... , N', f t/Jd}' = 0; where 1't,· .. ,1'". are defined hy (4.104). r"'

In additiOlJ, plr el'ery classi~'al so/ution II" of PN (1R:"'.~2, t/J) satisfyill!J

(4.119) gradlle(x) = ( when

(4.120) (' = lim. (u,,(x) + (j' t/Jdi')E,,(.X» exists, I xl ~f r

and there exists a unique solution :x of (4.118) such that

(4.121) lIe(X) = C + r En(x - t):x(t)di'(t) . <iF

Proofof Propositiol1 17. Fist we take Ue to be a classical solution of P N (IR:" Q, t/J) satisfying (4.117). Given (E ((, x ( IR:") with ( == 0 in the neighbourhood of Q, == I

for Ixi ~ Ro, r = (11" E ((,' (IR:") and f = ,11' E c/(IR:").

Since r = tie for I x 1 ~ Ro it satisfies also (4.117). From Proposition 4 of 92, there exists c such that

r = E" )( f + c.

We have therefore from Proposition 2 of 93 that when Ixl ---> x

But

lI,,(x) = C + (ffdX) En(x) + 0C:~/-=I ) grad ue(.x) = (ffdX) (J"I\~ln + 0 C~I;; ) .

j"fdX = r A l' dx = J'.. c'. ~d}' = J' (~.lIedi' = - r t/J d',' , JBIO.R o) ,·BIO.R,,)CIl (BIO.Rol?n Jr

the last equality commg from Gauss' theorem applied to lie harmonic on B(O, Rol\Q. In short, every classical solution II" of P N ( IR:"Q, t/J) satisfying (4.117) satisfies (4.119) and (4.121). On the other hand, given :x solution of (4.118) and C E IR:, the function 11" given by (4.121) is a classical solution of PN(rH"\J,2, t/J) satisfying (4.117). In addition identifying (4.97) with (4.119) and (4.120), we have

(4.122) J t/Jd,' = - J :xdi'·

We are hence led to showing that the assertions (ii) and (iii) are equivalent to: (i)' there exists a classical solution of PN(rH"\Q, t/J) satisfying

(4.123)


In the case n ? 3, this is a corollary of Theorem 5. In the case n 2, it is necessary to take up again the reasoning of the proof of Theorems 4 and 5. We have

1m (I + J) C {~Ue; Ue E £ ([R"\Q) n (&'~ ([Rn\Q) satisfying (4. J 17)} ene

c{rjJE'CO(T);L.rjJdY = 0 for j = 1, ... ,N'}, j

with the result that by the Riesz-Schauder theorem, we are led to show that dim ker (I + J) ~ N'. But the map!Y. -+ ue , the exterior simple layer potential, is an injection of ker(l + J) into XQ'j[R EB ... EB Xl2,[R; indeed if!Y. E ker(I + J),

then Ue is a solution of P N ([R"\{2. 0) and satisfies lim ue(x) = 0 hence u" == 0 Ixl'" OX>

on [Rn\Q. o We have thus proved Theorem 3 which had been stated without proof in sect. 4.4 and more precisely that Neumann's problem, when it admits a solution is solved by a simple layer potential. It is not always the same for the Dirichlet problem: more precisely let us prove the

Proposition 18. Let Q be a regular bounded open set Q in [R" (n ? 2) with boundary r of class '6 1 +, and also in the case n = 2 the hypothesis (4.107). On the other hand, let (P E (to (T). Then the following assertions are equivalent:

(i) cP E 1m L; (ii) the solution of P(Q, cp) is in ~~ (Q); (iii) there exists a solution of P([R"\Q, cp) ill (g~([Rn\Q); (iv) every solution of P( [Rn\Q, cp) is ill (t~ ([Rn\Q).

We notice that cp E 1m L iff the solution u( cp) of P( Q, cp) can be written as the single layer potential of a function Y. E ((forT), which, from point (4) of Proposition 15 is unique; for the exterior problem we have also cp E 1m L iff P( [R"\,Q, cp) admits a solution which is written as an exterior simple layer potential: in the case n ? 3, it is the solution tending to zero at infinity which is no longer true in general in the case /J = 2.

ProojolPropositiolJ 18. Supposing cp E ImL, the solution u(cp) of P(Q, cp) is in ((f,; (Q) since it can be written as a simple layer potential; conversely if the solution u(cp) of P(Q, cp) is in '6~{~2), the function rjJ = Du(cp)/Dn E ~o(T) and satisfies the Gauss conditions; from Theorem 4, u( cp) solution of P N (Q, 1(1) can hence be written as a simple layer potential, that is to say that (,0 E 1m L. Hence (i) <=> (ii). We also have (i) => (iii) and (iv) => (i); for this latter point, it is enough to apply Proposition 17, noting that the constants are in 1m L which follows from the equivalence (i) <=> (ii). To complete the proof, i.e. to prove that (iii) => (iv), we are led, by linearity, to prove that

(4.124) every solution of P([R"\Q, 0) is in 'O~(lRn\Q).


Let us consider, therefore, a solution u of P ( IR" .);2,0); we fix fI EO Y (IR") such that p = I in the neighbourhood of Q: we have pu EO C(,. x (IR"\Q) and L1 (pu) EO V( IR"' Q). Let us denote by f the extension of L1 (pu) by 0 to the whole of IR". We Ila ve f EO '/ (IR") and the Newtonian potential off is harmonic in the neighbourhood of ~2; hence the trace (p of E" )0(. f is in 1m L. The exterior simple layer potential lie

ofx = L ... 1 (j) is in «(,,~ (IR" Q): the function /: = pu - f~" * I + 11,. is solution of P( IR" . Q, 0).

III the case IJ ? 3, lim r(x) = 0 and so [(x) == 0 on IR" .12: we deduce that i x i --4 f

1I = E,,* f- Ue in the neighbourhood of r in IR" Q and so u EO ({,~ (IR"\Q). III the case 11 = 2, this is again true; obviously, we always have r == 0 on Q'] , ... , Q:v: also from Proposition 2 of ~2 (see also (4.97)) when I x I -+ x

• 1 ' ['(X) = CE2(x) + 0(: .) with

Ix , c = I~ Y. d·,

• i

.r JldX.

If Q~ = .8(xo, ro ), then, from the uniqueness of the Dirichlet problem (see Proposition 9), r is radial with respect to Xo on Q~; since 1'0 T I from the hypothesis (4.107), necessarily C = 0 and r =: O. In the general case, we take R > 1 such that B(O, R) :::) £2 and put Q] = (1R2\Q) n B(O, R) and ['] defined on 1R 2 ,f2 j = Qu(1R 2 \.8(O, R)) by

/:](\)={ 0 on Q _ . r(x) on ];);lB(O,R).

It is clear that r 1 is harmonic on IR 2 \ .121 and belongs to ((, ~ ( IR 2 \ Q): since the result is established when the unbounded component is the exterior of a ball, the restriction of 1; to Q], solution of P( Q I ' Ip 1 ) where cp] is the trace of r 1 on (IQ l' is in '(,'~(QI) and hence again LlE(6~(\Q). 0

We notice the

Corollary 8, Let Q he a regular bounded open set ill IR" with houndary r of class

((,'1 +e. Then the Green'sfimction88 G of Q (resp. 1Rt/\Q) satisjies

{ {(iran J.' E f~(r.esp .. ntl \Q)"the/lI"l1ctiOIl x -t G(x,),)

IS In (('~(Q\[y))(resp.{6,~(IR"\(Q u iy})).

Proof We have H8 G(x,y) = Et/(x - y) - U((P"x) (resp.ue((p\" x)) where cPy(z) = Ell (z y). Since the function x -+ E"(x - y) is a solution of p(J«n\~2, (PyJ (resp. P(Q, (Pr )),

and is '{;'''' in the neighbourhood of r, from the proposition, u( (Pv) (resp. Ue (cpr)) is in ,{;,~(Q) (resp. '{;'~(n"\Q))89. From which we have the result. 0

88 See §§4.2 and 4.3. "" In the case /I = 2. we can always make the hypothesis (4.107).


We now give the solution of Dirichlet's problem by a double layer potential. From Proposition 18, the functions

Xl" ... , Xl N , Xl',' ... , XI:, E 1m L where r i, rj are defined by (4.104)

Taking account of point (4) of Proposition 15 we can therefore define

(4.125) { - L - 1 X . . - 1 ~T

:X; ~_ 1. [.i l, ~ , ... ~ l' I

:Xj - L Xr' ] - 1, ... , N . J

The interior (resp. exterior) simple layer potential deflned by :Xi (resp. :xj) is a solution of the Dirichlet problem P( Q, XrJ (resp. P( [R"\Q, X~.)) hence XQ, (resp. xQJ; from which we deduce J

J

(4.126) { J:Xi = :Xi for i = 1, ' .. ,N

JLtj = -':J.j for j = 1, ... ,lV'

In other words, taking into account Theorems 4, 5 and Proposition 17 in which we have shown that

dim ker (I - J) = N and dim ker (l + J) = N' ,

we see that {:xj, ... ,exN } (resp'{:X'j""':X:""}) form a base of ker(J -- J) (resp. ker(I + J)) Using point (2) of Proposition 15 we are able to prove:

Proposition 19. Let Q be a bounded open set of [R" with n ?: 2 ~vith boundary r of class (gl +£ sati.~fying the hypothesis (4.107) when n = 2. Also let <p E egO (F). (1) Thefollowing assertions are equivalent: (i) there exists fJ E egO (F) such that the solution oj P( Q, <p) be the interior double layer potential of fJ yiven by (4.98); (ii) <p satisfies

fr <p(t)exj(t)dy(t) = 0 for j = 1, ... , N'

where ex} is defined by (4.125). In addition, then fJ is defined to within an additive constanl on each rj, j = 1, ... , N'. (2) The following assertions are equivalent: (il there .exists fJ E (,Ij'0(f') such that the exterior double layer potential of 13 given by (4.101) is a solution of P( [R"\!2, <p); (ii) <p satisfies

fr <p(t)(Xi(tJdy(t) = 0 for i = 1, ... , N

where (Xi is defined by (4.125). In addition, fJ is defined then to within an additive constant on each r i ,

i = 1, ... ,N.

396 Chapter II. The l.aplace Operator

Proof Given a part M of (6° (1'), let us denote

1\1"' = : i{) E (6 0 ( r); L (Xi{) d}' = 0 for all (X EM} .

It is clear. by the projection theorem in L 2(r) that for a sub-space of finite dimension M of (6° (l),

in other words codim At c = dim M .

The operators K and J being compact (see Proposition 16), ker(1 + K) and kerf 1 + J) are spaces of finite dimension (see Lemma 3) and

codim 1m (1 ± K)

codim Im(l ± J)

dim ker(l ± K)

dim ker (/ ± J) .

Now from Proposition 15(2), K and J are adjoints the one of the other, from which we derive immediately that

We deduce that

Im(l ± K) c ker(l ± ])"

Im(l ± J) c ker(l ± K)~.

dimkcr(l ± J)=codimker(l ± J)~ ~ codimlm(I ± K)=

= dimker(l ± K)= codimker(l ± K)~ ~ codimlm(I ± ]);

from which we have the equalities

codim ker(l ± J)-'- = codim lm(l ± K) ,

and Im(l ± K) = ker(I ± J)~.

That is to say, account being taken of what was recalled above

1m (i+ K) = :y'l' .... Y;v } ~

Im(/-K)= {(Xl"" 'YN:~'

From Proposition 15 (3)

(4.127) { ker (f + K) :=J

ker(I - K) :=J

XI, [~ EB ... EB X/v iR:

But from what was proved above

dim ker (l + K)

dim ker(l K)

XI,iR: EB ... EB X/,iR:·

codim Im(l + K) N'

codim Im(l K) N'

from which we have equality in the inclusions (4.127). This, together with (4.100) and (4.102) proves the proposition. D


Remark 7. If, in addition to the hypotheses of the proposition, IR"\Q is connected, i.e. N' = OJor all qJ E ct'0 (r), the solution of P (Q, qJ) can be written in a unique way as a double layer potential. We shall see in §7, that in the case Q convex (which implies IR"\Q connected, the solution of P(Q, qJ) can be obtained by an iterative procedure with the aid of the integral operator K: this is Neumann's method. In the general case, for all qJ E ct'0(r), there exist Aj , ••• ,AN E IR unique and such that the solution u of P( Q, qJ) is of the form

u = Al u j + ... + J.N,U N • + v

where uj is the interior simple layer potential of 'X} and v is an interior double layer potential. 0

6. Sub-Harmonic Functions. Perron's Method

In this section, we develop the solution of Dirichlet's problem by the method of Perron of which we have given the broad outlines in §4.1. As we have indicated, this method is based on the properties of sub-harmonic functions. First, let us prove

Proposition 20. Let Q be an open set in IR" and v a distribution on Q. The following assertions are equivalent: (i) Llv ? 0 in ft'(Q), i.e.

< Llv, 0 = < v, LI, > ? 0 for all 'E .s1(Q), , ? 0 ;

(ii) v is a (locally integrable) function satisfying

v(x) !( ~f. v(t)dt a.e. x E Q andfor all 0 < r < dist(x, oQ); (in r B(,,r)

(iii) there exists v: Q -+ [ - oc, + ex; [ u.s.c. 90 and locally integrable satisfying

(4.128) v(x)!( ~f v(t)dt for all x E Q and 0 < r < dist(x, oQ), anr B(x,r)

such that v = fi in .@'(Q), that is

<v,O = fV(t)'(t)dt forall 'E.S0(Q).

I n addition, then, v is unique.

Proposition 21. Let Q be an open set in IR" and a map v: Q -+ [ - 00, co [ u.s.c. The following assertions are equivalent: (i) for every open set w c Q and U E Yf (w) n ct'0 (w) sati~rying

lim sup v(x) !( u(z) for all z E aw ,

90 U.S.C. is the abbreviation for upper semi-continuous: {x E Q; fIX) < ),} is open for all;. E R:, or again in an equivalent way. lim sup [,(x) 00 f(x,,). for all "0 E Q.

x ---> Xo

398

and in the case when W is unbounded

we have

limsupv(x) - u(x) ~ 0, Ixl ~x

XEW


v(x) ~ u(x) for all XEW;

(ii) for all Xo E Q, 0 < r < dist(xo , oQ)

(4.129) v(x) ~ _1_ f r2 -~~ - x o l2 v(t)d),(t)91 for all x E B(xo, r); r(In (lB(x",r) It - xl n

(iii) for all x E Q and 0 < f. ~ dist(x, DQ), there exists 0 < r < [; such that

v(x) ~ -; f v(t)dt91 .

(In r B(x.r)

These characterisations lead us to frame the definitions:

Definition 15. Let Q be an open set in IR n :

(1) a distribution l' on Q is said to be slIhharmollic if J v ;" 0 in C/" (Q) (see Proposition 20(i». (2) a map v: Q -> [ - x, x] is said to be suhharmollic if it satisfies the condition (i) of Proposition 21. A distribution (resp. map) w is said to be superharmonic if v = - II' subharmonic.

Propositions 20 and 21 show that given a distribution v, subharmonic on Q, there exists a unique map v: Q -> [ - 00, 00 [ u.s.c., locally integrable and subharmonic, such that v = v in £(I' (Q): v is the sub-harmonic u.s.c. representative of v. F or a function v E '&0 (Q), more generally for v: Q -> [ - 00, 00 [ u.s.c., locally integrable, the definitions of sub-harmonic in so far as the map or distribution coincide; in particular the Definition 15 generalizes the Definition 2 and contains the property 1 given in §4.1. In the case 11 = 1 and Q = ]a, b[, a map v: ]a, b[ -> [ - cXJ,X,)] is sub-harmonic iff the epigraph of [1, i.e. the set

{(x,Y)E]a,b[xlR;v(x) ~ Y}, is convex. In other words, 11 is sub-harmonic iff it satisfies:

(i) I {x E ]a, be; v(x) < oo} is an interval (ii) v == - 00 on I or v is finite and convex on I,

"' The function I' being u.s.c. for every compact set K of Q. t' is majorized on K and even attains its maximum on K (see, for example, Bourbaki [I], Chap. IV). It follows that for every positive Borel

measure i1 on K. we can define r l'(t)df1(t), putting r l'(t)dll(t) = -'l. when t' is not integrable on K. JK Jx


This follows immediately from the fact that the harmonic functions on an interval are the affine functions on this interval. In the case n = 1, sub-harmonic coincides with convex; this is obviously no longer true in the case n ~ 2. Also, in the case n = 1, a function v: ] a, b [ -> [ - 00, 00 [ (not taking the value + 00) subharmonic and not identically - 'f. .. is finite and continuous. This is no longer true either in the case n ~ 2.

Example 24. For n ~ 2, the map v: [R" -> [ - 00, 00 [ defined by

{ En(x) if x =f. 0

v(x) = - 00 if x = 0

is sub-harmonic. This follows in fact from the characterisation of Proposition 20 (v is u.s.c., locally integrable, and L1v = b ~ 0) as well as of the characterisation (iii) of Proposition 21).

Proof of Proposition 20. Let us note that the assertion (i) implies that Ll v is a (positive) Radon measure on Q; from Proposition 6 of §3, if v satisfies (i), it is then a (locally integrable)92 function on Q. We can therefore, without loss of generality, suppose v to be locally integrable on Q. Let us take P E .SC (If~n), P ~ 0, f p(x)dx = 1 and supp p c B(O, 1) and put for all k ~ I,

We have

Qk = {x E Q; dist (x, aQ) < k -I} ,

vk(x) = fp(t)V(X + k-1t)dt for xEQk'

Vk ....• v in LLc(Q/) when k -> 00 for all I ~ 1

We deduce immediately that v satisfies (i) iff for all k ~

Llvk{x) ;?: 0 for all x E Q k .

Also

f vk(s)ds = fp(t)dt f v(s)ds; B(x, r) Blx + kit, r)

so v satisfies (ii) iff for all k ~ 1

vk{x) ~ ~f vk{s)ds for xEQk' 0 < r < dist(x,oQk)' (Jn r Blx. r)

In other words it is enough to prove the equivalence (i) ¢> (ii) when I.' E '6'c.( (Q). Hence, let us suppose that v E ((i2(Q) satisfying ,11.' ~ 0 on Q. We put for x E Q and

n ev n 92 We have even v E Lfo,(Q) for p <-- and - E Lfo,(!2) for p <

n 2 8x i n - 1


o < r < r(x) = dist (x, rJQ)

M(x, r) =--;1 v(t)dt (Jn r Blx.r)

.~ ... 1. v(x + rt)dt . (J n BiO. I)

We have (see (2.21) in ~2)

= n JI' /1 v(x + rt)dt =~~: (rn + 1 rJ~ (x, r)) (In BIO. I) r" -] {/ , 1//

~ (r"+1 a:: (x,r)) ~ 0. cr cr

and therefore

In particular:

r" +] aM j)r..1

-- (x, r) ~ lim r" + 1 --;.- (x, r) ar ,-0 cr

nr"+ 1 1 lim --- grad v(x + rt). t dt r-O (In BIO, II

, (1,,,,/ from which we deduce - ~ 0 and hence cr

M(x, r) ~ lim M(x, r) = v(x) , r ---) 0

0,

This proves (i) = (ii). Conversely, let us consider now v E (€2 (Q) satisfying (ii), and let us put (I) = {x E Q; /1v(x) < O}. Since v satisfies (ii) and -v satisfies (i) and hence (ii) on (I), we have

v(x) = lvl (x, r) for all x E (jJ and 0 < r < dist (x, w) .

From Proposition 15 of §2, v is therefore harmonic on (I), i.e. /1 v = 0 on (I), This is contradictory unless w = 0, i,e, ,1v ~ 0 on Q, The implication (iii) = (ii) being trivial, it remains to prove (ii) = (iii) and the uniqueness of 13. Let us suppose that v satisfies (ii). We prove first the uniqueness of i), Let, first of ail, iJ satisfy (iii): we shall then have for x E Q

13(x) ~ ~ r . D(t)dt = ~ 1. v(t)dl for all 0 < r < r(x) (Jn r ",H(x,rI (Jn r Hlx.rl

smce v = iJ a.e. on Q; hence

. n f £;(x) ~ mf -;; v(ndl. 0<, < r(x) (J n r B(x.rl'

Now since v is u.s.c. for)' > 6(x), there exists 0 < I' < r(x) such that L' ~ ;. on B(x, 1') and so

n f 11 f - , ---n v(t)dt =---;, . v(t)dt ~ ) .. anr Blx,r) anr IiI x,,)


We deduce that necessarily

(4.130) n f v(x) = inf -n v(t)dt. ° < r < r(x) an r B(x, r)

Conversely, the function v defined by (4.130) satisfies (iii). In effect the function x -4 r(x) is continuous. For X o E Q, fixed and 0 < r < r(xo) fixed, the function

X -4 ~ f v(t)dt = ~ f v(x + rt)dt anr B(x.r) an B(O.I)

is defined and continuous in the neighbourhood of Xo from the continuity of translations in LIloc. Hence v the lower envelope of continuous functions is u.s.c. On the other hand, from the hypothesis (ii), for almost all x E Q

v(x) :::; ~ f v(t)dt for all 0 < r < r(x) . an r B(x. r)

Also, from Lebesgue's differentiation theorem, for almost all x E Q,

v(x) = lim ~n_ f v(t) dt . r ~ ° an rn B(x, r)

Hence v(x) = v(x) for almost all x E Q.

Finally by the definition of V, for all x E Q and 0 < r < r(x),

v(x) :::; ~f p(t)dt = ~f v(t)dt; an r B(x. r) an r B(x.r)

which concludes the proof of the proposition. o Proof of Proposition 21. To prove (i) => (ii), we note that given Xo E Q,

o < r < r(xo) and cP E ~o(oB(xo, r)), the function u defined on (J) = B(xo, r) by

1 f r2 - Ix - xol2

u(x) = - I In cp(t) dy(t) ran (lB(.<o, r) t - x

is a solution of P(w, cp) (see Example 2); hence if p satisfies (i) and v :::; cp on iJB(xo, r), we have v ~ u on B(xo, r). Now, since v is u.s.c. 93 , there exists CPk E '6'0 (iJB(xo, r)) such that

cpdz) ! v(z) when k i ''XJ for all z E oB(xo, r) .

We deduce that (i) => (ii). The proof of (ii) => (iii) is identical to that of Corollary 2 of §2 since (4.129) implies

v(xo) ~ --hI v(t)dy(t) forall 0 < r < r(xo )· anr <'B(xo.r)

Finally let us prove (iii) => (i). Let us take w c Q and u E £(w) n '6'°(w)

93 See, for example, Dieudonne [I] vol. I, Bourbaki [1], Chap. IV.


satisfying

(4.131 ) {

lim sup v(x) ~ u(z)

Ii~~~p. ['(x) u(x) ~ 0 I xl··' x

for all :: E Dw

when UJ is unbounded.

Supposing that (iii) is satisfied, it is the same for the restriction of l' to U) and, by use of the formula of the mean for U E ,!If ((j)), for l' - U on w. In other words, we can suppose u == 0 on (I): we have to show l' ~ 0 on UJ. By contradiction, suppose

sup r > 0: from (4.131) and the u.s.c. hypothesis

F = Jx E w;r(x) = supr} 1 C)

is a compact non-empty subset of w; considering X o E (~F and r;

from hypothesis (iii), there exists 0 < r < c: such that

sup r = L'(xo) ~ -!n r r(tJdt ~ sup r. () (Jn r ... BC\o,r) lJ(\o.r)

dist(xo,I'W),

Hence, [' == sup r on 8(xo, r), a contradiction with the choice of Xo E /, F. 0

Remark 8 (I) The set of sub-harmonic distributions on Q is a cone of Cr' (Q) the space of distributions on Q; similarly the set of maps (resp. u.s.c.), sub-harmonic on Q into [ X), x)] (resp. [ - x, x[) is a cone. The map which to a sub-harmonic distribution [' makes correspond its subharmonic u.s.c. representative 1:; satisfies

(2) The cone ol sub-harnwnic distribul ions is closed for convergence in the sense of distributions: if (l'k) is a sequence of sub-harmonic distributions such that

lim < l'k' ~) exists for all (E Ct(Q) , k~=

then < 1', ~ ) = lim < 1\, ~ ) is a sub-harmonic distribution. In general, however,

the sub-harmonic u.s.c. representative of l' is not the (point) limit of the subharmonic u.c.s. representatives of the l'k. (3) The cone ol sub-harmonic maps or Q into [ -- .I., x J is stable under the upper envelope: if (v;) is a family of sub-harmonic maps of Q into [ - x, Xi], then the map

r: x E Q -> sup ['i(x)

is sub-harmonic.


In general, however, the upper envelope of sub-harmonic u.s.c. maps of Q into [ - 00, 00 [ is not u.s.c. (4) The increasing convex functions act in the cone of sub-harmonic u.s.c. maps of Q into [- 00, 00 [: if CP: ~ -+ ~ is increasing and convex and v: Q -+ [ - 00, 00] is sub-harmonic u.s.c., then CP(v) is subharmonic u.s.c., with the convention CP( - 00) = infCP(~). This follows from the characterisation (iii) and Jensen's inequality

cp(Il(~) t V(t)dll(t») ~ Il(~) L CP(v(t))dll(t) .

This same inequality shows that given CP: ~ -+ ~ convex (arbitrary) and u E £(Q), CP(u) is sub-harmonic (continuous) on Q.

Let us now develop Perron's method for the solution of Dirichlet's problem. We consider a bounded open set Q in ~n with boundaty rand qJ E ~O(r). We take up again the notation (4.10):

.§ ~(qJ) = {v E ~o(Q); v subharmonic on Q and

lim sup v(x) ~ qJ(z) for all Z E r}

.§+ (qJ) = {w E <6'0(Q); w subharmonic on Q and

liminfw(x) ~ qJ(z) for all Z E r} ;

we have v ~ w on Q for all v E f ~ ( qJ ) and w E .j" + ( qJ )

and .'I ~ (qJ) and f + (qJ) are non-empty since they contain

v == min qJ and w == max qJ respectively. r r

We can thus define, with the notation (4.12), for all x E Q

(4.132) u~ (qJ, x) = sup v(x), u+ (qJ, x) = inf w(x). VE.f (ep) WE.f+(ep)

We then prove Wiener's theorem (see Property 2 of §4.1):

Proposition 22. Let Q be a bounded open set in ~n with boundary rand qJ E <6'0(r). The functions u ~ (qJ ) and u + (qJ) defined by (4.132) are harmonic and identical on Q.

First, we state some lemmas:

Lemma 5. Let Q be an open set in ~n, vE<6'°(Q) sub-harmonic on Q and a ball B = B(xo, ro ) with Ii c Q. Let us construct the function.

{ v(x) if x E Q\B

TBv(x) = 1 r6 - Ix - xol2 - r I In v(t)dy(t) if X E B . ro(Jn J,'B t - X


Then (1) v ~ TB v on Q; (2) TBv E y;o(Q) is subharmollic 011 Q; (3) TB v is harmonic Oil B.

Proof ol Lemma 5. The point (1) is a result of the characterisation (ii) of Proposition 21. The final point and the continuity of TB v follows from the fact that, from Example 12, Ta z; is a solution of P(B, v). Hence, it remains to prove that TB t: is sub-harmonic on Q: it is clearly so in the neighbourhood of each point of Q\}B. Now for x E (lB and 0 < r < dist(x, cQ), we have

TBV(X) = v(x) ~ 11_ f v(t)dt ~ n n JI' Tav(t)dt (Jn r" B(x.rl (Jn r Blx.r)

from which the result follows. D

Lemma 6. Let Q be all ope/) set ill [Rn, B B (x o , "0) a ball with jj c Q alld ,F a jClmily of suhlwrmollic jill1Cliolls l.' E ((,0([2). We suppose

IJor all v E ,F, TBL' E ,F ;

~ fi)r allt· 1., t'2 E ,F, max (VI ,v2 ) E.F ;

l [ r(xo); r E .~} is majorized in [R .

Theil, .Ii" all x E B, {v(x); r E .~} is majorized in R and the function

X E B --> sup dx) t' E.F

is harmonic on B,

Proolol Lemma 6, We put, for all x E B

U(x) = sup dx) E] - T, ex] , /' E ,F

Let us consider a sequence of points (x k ) of B, dense in B. For all k, there exists l'k E .F such that

rrdxk)+k-1~U(xd if U(Xk)~X

ll'k(Xk) ~ k if u(x k )

We then construct the sequence (i\) according to the recurrence relation

1\ = TB v1 , i\+ 1 = T~(max(i\, l'k+ 1)) .

From the hypotheses and Lemma 5, we have for all k

i\ E.F. 1\ harmonic on B •

and the seq uence (i\) is increasing. Using Corollary 9 of~2. since i\(xo) is bounded, (i\l converges uniformly on every compact set of B. We denote its limit by i'i and show that 11 = U. We have u ~ 11 by the definition of u.


Now, for all k,ii ? Vk and therefore

(4.133) { ~(Xk) + k- 1 ? u(xkl if u(xk) < 0:

u(x k ) ? k if u(xd = oc .

Let us suppose that at point x E B, ii(x) < u(x); since U is an upper envelope of continuous functions, it is u.s.c.; there exists, therefore, r > 0 and b > 0 such that U ? ii + b on B(x, r) c B. The sequence (xd being dense, there exists kl -+x;

such that x k , E B(x, r) for alii E N. Since ii is bounded on B(x, r), from (4.133) for I sufficiently large, U(Xk,) <0: and

u(.xk ,) ~ ii(x k ,} + k1- 1 < u(x,,) + 3

from which we have a contradiction. o

Proof of Proposition 22. We should notice that these two lemmas prove the first part of the proposition: the functions U _. ((p) and u + (rp) are harmonic on Q; it is sufficient to apply Lemma 6 to

.F = .f ( rp ) and .'? = { - tv; W E .f + ( (p) }

for every ball B such that B c Q.

Now we have immediately by definition of u _ (rp) and u + (rp)

(4.134)

We derive

J u - (rp) ~ u + (rp) =··11 ( ..... (p), lL U (p) = i.u - (p)

u-(rpd + u(rpz) ~ lL(rpl + rp2)'

u (rp + c) = lL (ep) + c for every constant c

l (PI ~ epz = U_(epl) ~ U-(ep2)'

for all

= u+(~) + u_(rp) + sup(ep r

~) + sup(~ - (p), J

from which we have

(4.135)

We deduce that the set

E = {rp E ~o(n; U (rp) = U+ (ep)}

is a closed vector sub-space of ~'I\n. To prove the proposition it is sufficient therefore to show that the set E contains a dense part of<;&,o(r). We observe that E contains the traces on r of the functions v E '1I,O(Q) which are sub-harmonic on Q;

in effect such a function v with trace ep is trivially in .1_ (rp) with the result that

v ~ u_(ep).


It follows that for ali Z E r

lim infu_ (<p, x) ? lim r(x) = <p(.:) , X-4= x-::

and hence, since lL (<p) is harmonic (from which a priori subharmonic) on Q,

u _ ( <p) E ,1+ ( <p) and u - ( <p) = u + ( <p ).

Given u E (PU2), we have u = 1'1 - 1'2 with

(')(x) = ilx:2, l'2(X) = ilxl2 - u(x);

let us choose I, 1 , +

:.,-sup(l1u(x))· with the result that ... 11 Q

111'1 = Vn? 0 and /J1'2 = 21.11 -l1u? 0;

the functions r l' 1'2 E ((,0 (Q) sub-harmonic on Q have their traces in E; hence 1: contains the traces of the functions u E ((,2(Q). From the Stone-Weierstrass theorem, the result is proved. 0

Wiener's theorem led us to define the function 1I(<p) = lL(<p) = 1I+(<p) as the generalized solution of P(Q, <p) (see Definition 3). We have next introduced the notion of a regular point of the boundary of an open set (see Definition 4); to conclude the proof of Theorem 1, it remains to us to prove Property 3 of §4.1.

Proposition 23. Ll!t Q bl! a hounded open set of ;\;" and::: a regular point of the houndary r of Q. Then/ilr all (,0 E ((,0 (r), the generali:::ed solution u( <p) of P (Q, (,0)

satisfies

lim lI(ifJ, xl = <p(:::).

Proof Let us take a barrier function l' E (('()(Q n B(:::, r)) (see Definition 4). For I: > 0, let us choose 0 < r, < r such that

1(,0 - <p(:::) 1 ~ /; on r n B(:::,r,)

and put III = , sup

Q c' BI:.I") BI:J, I

By hypothesis Ill, < 0; we then detlne

1'/(x)

w,(:::) =

1 k, =- sup I(p - (,0(:::)1

1m,! I

{ (p(:::) - I: + k,max(r(x), m,) on

~ry(:::) - i: + kiln, on

{ ifJ(:::_) + I,:, -- k.,.m.ax(l'(x),m r ) on

ifJ(~) + L -- k,Ill, on

By construction 1'( E .f _ (ifJ) and w, E .f + (ifJ); hence

Q n B(:::,r,)

Q(B(:::,I",)

Q n Bf:::,I",)

Q\B(:::, r/).

§S. Capacities 407

and in particular

lu(cp) - cp(z)1 ~; t: - k£v on Q n B(z, r,l,

which proves

limsuplu(cp,x) - cp(z) I :::; /: forall f. > O. o x~z

§5. Capacities*

1. Interior and Exterior Capacity Operators

We take a regular bounded open set Q of~" with boundary r. Given (P E 'fi0(T), we can consider the solution u( cp) of the Dirichlet problem P( Q, cp) (see §4); supposing u(cp) E 'fi~(Q), we can consider I/J = au(cp)jan E 'fj0(T); the map cp -> I/J defines an operator C of D( C) into 'fio (T) where

(5.1) D(C) = {cpE(6°(T);U(CP)E<t~(Q)}.

We shall call this operator the interior capacity operator ()f'fi° (T). This terminology will be jusfied by the physical interpretation which we shall make in Sect. 2. We observe that C is an unbounded linear operator (not defined everywhere) of<to(T). By definition also, given cp, I/J E (eo (T), cp E D (C) and I/J = C cp iff the solution of'the Dirichlet problem P(Q, cp) is a solution of the Neumann Problem PN(Q, I/J). Similarly, given cp E (eo (T), we can consider the solution ue ( cp) of the exterior Dirichlet problem P( ~"\Q, cp) satisfying the null condition at infinity (see §4.3); supposing that Ue(CP)E'fi~(~"\Q) we can consider I/J = (~ue(cp)/on: the map cp -> I/J defines an operator C e of D (C e) into 'fi0 (T), where

(5.2)

We call this operator, the exterior capacity operator of(f,·o(T). By definition, givell cp, I/J E "'-a (T), cp E D (C e) alld I/J = Ce cp iff the solutioll of the exterior Dirichlet problem P( ~nV2, cp) satisfyill(l (he Ilull condition at infinity is a solutiO/l of PN( ~~"\Q, I/J). We have defined the interior and exterior capacities in "'o( T) for r the boundary of a regular bounded open set in ~"; this choice of ",0 (T) corresponds to the classical theory. We can equally well consider these operators in other contexts (see §6 and Chap. XI): the capacity operators appear in fact as operators on the boundary r. We note however that they are neither differential operators nor integral operators: they are in fact pseudo-differential operators. We shall not enlarge on this remark, contenting ourselves to explaining it by an example on the unit circle.

Example 1. Capacity operators 011 the unit circle in ~2. We consider in ~ 2, referred to polar coordinates, the unit disk Q with boundary the unit circle. We can identify functions on L with periodic functions of e of period 2n. Given (p = L eneinO , the

408 Chapter n. The Laplace Operator

solution of P(Q, cp) is u(r, 0) = 2.' ('nrlnleinO whose normal derivative is

we are assured that cP EO D (C) if I' Illc" I <x, but more generally for all cp EO D( C), !j;= C cp is defined by

. 1 f27r . !j;(0) = 2.'lnlcnetnO where c" = 2; ° cp(OJe-wIJdO.

In other words, C is the restriction to (eo (E J of I he pseudo-differential operator ( _ d 2/d( 2 )L2.

Similarly for cp (0) = I' Cn einO, the solution of P( [Rn\Q) satisfying the null condition at infinity is u (r, 0) = L Cn 1'-1 n Ie inO of which the normal derivative (exterior to [R"\Q) is

!j;(0) =

In other words, ill the case of the unit circle in [R2, C = Ceo It should be emphasized that this equality C = Ce is particular to the unit circle in [R2 (see Propositions 2 and 3). 0

As usual, we consider directly the case 11 = 1.

Example 2. Case n = 1. We take a regular bounded open set of [R, that is to say a finite union of bounded open intervals, two by two disjoint

Q = ]a j ,a2 [u ... u]aN-1,aN[ with a l < il2 < a3 < ... <aN(Neven);

we have r = (0 1, (/2, ... , (fry} and (6 (I (I') is identified with [R!V Given (P E [RN the derivative of the solution u of P(Q, Ip) is

si nee u is affine on ] ak , ak + j [ (k odd). Hence

{ CPk+l - iPk

if k is odd

(5.3) (C CP)k = ak + 1 - ak

CPk CPk- 1 if k is even Ok ak ···1

Similarly, the derivative of the solution Ue of P([R\Q, cp) is

{

u'(x)

u'(x)

o

§5. Capacities

Hence

(5.4)

We now state the

o

CPk - CPk-1 ak - ak ···1

if k = 1 and k = N

if k is even, k < N

if k is odd, k > 1 .

Proposition 1. Let Q be a regular bounded open set in [R" of boundary r.

409

o

(1) For all cP E D(C), the solution u(cp) of P(Q, cp) satisfies Igradu(cp)I Z E LI(r) and

(5.5) r Igradu(cp)1 2 dx = r cpCcpdy. JQ Jr

More generally for CPt, CP2 E D(C)

(5.6) r CPl CCP2 d i = r gradu((Pl}·gradu{cpz}dx, Jr ~Q

and in particular C is positive and symmetric. (2) For all cpED(Ce ), the solution ue(cp) of the exterior problem P([R"\Q,cp) satisfying the null condition at infinity, satisfies Igradue(cpW E Ll([R"\Q} and

(5.7) l-!"\Q Igradue(cpWdx = fr (pCecpdy.

More generally for CPl' CP2 E D(C,,)

(5.8) fr CP l C"CP2 d }' = LgradUe(CPll.gradUe(cpz)dX

and in particular Ce is symmetric and positive.

Proof The point (1) is an immediate corollary of Propositions 4 and 5 of § 1 and of the definition of the interior capacity operator. For the point (2), from the same propositions, for R sufficiently large such that Q c B(O, R), we have

I grad ue ( cP W ELl (B(O, R)\!2)

and

III the case 11 ~ 3, we have, (see the proof of Proposition 17 of ~4)


when Ixl ---> :r~ with the result that

JA ~u.(~) ( . U" (Ip) d 'I' = 0 R n - 1 X

,'BIO, R) 1 IJ R"- 1 x ---' .) R n - Z

when R --->x, and

which proves (5.7). We prove (5.8) in the same way. III the case 11 = 1. we can verify it directly hy the formulae (5.4) of Example 2; we

h . h' d ave moreover 1I1 t IS case 1I)lp. x) = ° for Ixllarge. dx

III the case Il = 2, we know (see also Proposition 17 of 94) that

(' = lim lIc(~' x) exists and I grad ue(~. x)1 = O(1!lxl)


. lJr' ('lI c (qJ) Jr' (~Ue J lim ~ u.(lp)dj' - C 0 (qJ)d}' = O. R··' ~ .. <'BIO.R) ell 'BIO.R) (11

Now, applying Gauss' theorem to uc(qJ) on B(O, R).Q

j" (~Ue(qJ)d~' = _ r C,,(pd t' . • 'BIO,R) (11 Jr

The point (2) is then deduced from the following lemma.

Lemma t. Let Q he a reyular hounded open set in ~ 2 For all Ip E' D (C,,)

i c.. Ip d;' = ° . Proof We have (see Proposition 17 of ~4)

lim. (Ue(qJ,X) + 1:: z (x) r[.(CcIP)d/)\ exists Ixl- y JI

D

Since by hypothesis I }j

in.1 f U" (p. x) exists then necessarily i C" Ip d ',' = O. D

From Example 2, the result of Lemma! is true for every open set in iR; on the contrary when n ~ 3, the result of Lemma 1 is false for every open set in ~n; this appears clearly in the following Proposition. We recall the notation of ~4.5 (see (4.104) of §4):

(5,9)

Q I' ... , Q,y denote the connected components of Q .

Q'1' ...• Q'". the bounded connected components of ~n Q •

Q~ the unbounded connected component of W' Q (IJ ~ 2) .

r I ..... r N • r~). r'I ..... r'\ the boundaries of QI" ... Q". Q~ .. ..• Q:\.

§5. Capacities 411

Proposition 2. Let Q be a regular bounded open set of IR". Then

(1) kerC = XrjlR E8 ... E8 Xr,lR;

ifn?3,

ifn 2,

ker Ce

kerC"

XI'" IR E8 ... E8 Xr , IR ; 1 ,'"

XI"~ }~ E8 ... ED XI" rrtl o , .... ' If'il. •

(2) Let us suppose 1 is of class ((j 1 + I 94

1m C = { 1/1 E ,&0(1); ti !/J cl l' 0 for

ifn ? 3, ImC" {!/J E ,&0(1'):1, !/Jd;: = 0 for J

i = I, ... , N };

.i = 1, ... , N'}'

ifn = 2, ImC" = {!/JEY5·0 (1); f !/Jd"r' = 0 for .i = 0, ... ,N'}' .. 1.1

Proof The solution of P(Q, XI;) is Xn i which belongs to 16,: (Q) with normal derivative zero on r. Hence XI'; E D(C) and CXr, == O. Conversely if (P E ker C, from Proposition I, the solution u(p) of P(Q, (p) satisfies

1 grad u (p W ELl (Q)

and

L 1 grad u «p ) 12 d X = Ir <P C <p d y = 0 .

Hence grad II (<p) == 0 on Q and u (<p) is constant on each connected component of Q; therefore (p is constant on each Ii' The kernel of the exterior capacity operator is determined in the same manner by noting: (a) if n ? 3, the solution u .. ( <p) of P( IR"\Q, <p) which satisfies the null condition at infinity tends to zero at infinity; if therefore tie (<p) is constant on each connected component of IRn\Q, it is null on Q~ and hence <p == ° on 1;); (b) ifn = 2, the solution of p(1R2\Q, XI) satisfying the null condition at infinity is XQ~; hence Xr:J E kerCe ·

It is clear that 1m C is the set of the !/J E (1) such that Neumann's problem P N (Q, !/J) admits a classical solution, Similarly, 1m C" is the set of the V' E (eo en such that P N ( IR"\Q, !/J) admits a classical solution satisfying the null condition at infinity. Hence, the point (2) follows from Theorem 3 of §4. 0

Theorem 3 of §4 has been proved by the integral method with the help of the integral operators L, J, K (see §4.S). We recall (see Proposition 15 of §4) that L is a bijection 01''(,°°(1') onto 1m L95 ; we shall dellote hy L- 1, the ilwerse hijectiollfi"om D(L '" 1) = 1m L onto (('0(1)

We then state:

94 See Definition 3 of §3. 95 With the reservation that in the case n = 2 we also have the hypothesis (4.107) of §4.


Proposition 3. Let Q he a regular bounded open set of[Rn with boundary iJQ of class (6 1 +e satisfvin{J in addition ill the case 11 = 2 the hypothesis (4.107) of §4. Then (I) D(C) = D(Ce ) = 1m L and,for all q; E 1m L

1 (5.10) Cq;="2(J-1)L- ' q;;

(5.11 )

(5.12)

1 if n ~ 3, Ceq; = -"2 (J + I) L - 1 q; ,

if n~2, c.q>~ -~(J+I)L'q>+(i(L-'q»d')f."' , 0:0 dy

r with 0:0 = L - 1 1 .

(2) For all q; E 1m L, Kq; E 1m L alld

(5.13 )

(5.14) ifn ~ 3,

(5.15) ifn = 2,

1 1 Ceq;= -"2 L - (K + I)q;

1 1 ( ( 1 ) 0(0 -:i L - (K + I)q; + JJ (L- q;)dy r ( O(ody

Jr Proof The first part of the point (1) follows from Proposition 18 of §4, account being taken of the definitions of D(C) and D(Ce ). Now, for q; E ImL, the interior simple layer potential of 0: = L - 1 q; is a solution of P( Q, q; ) (see (4.92) of §4) and of PN(Q, !(Jo: - o:»)(see (4.93) of§4); hence, by the definition of the interior capacity operator

1 1 _ 1 Cq; = :i(J - I)o: = :i(J - 1)L q;.

In the case n ~ 3, the exterior simple layer potential of 0: = L - 1 q; is a solution of p([Rn\Q, q;) and PN([Rn\Q, -!(o: + Jo:» and tends to zero at infinity (see (4.96), (4.97) of §4); hence

1 1 _I Ceq; = -"2 (J +1)0: = -"2 (J + I)L q;.

In the case n = 2, the exterior simple layer potential Ue of 0( = L - 1 q; is always a solution of P([R2\Q,q;) and PN([R2\Q, -!{JO( + 0:)); but then

ue(x) = (L O:dY)E2(X) + °C.:I) when Ixi -> en.

Let us consider -'oEQ and q;o(:::) = E2 (:: - -'0). The function:

ue(x) - (1. o:dy )E2(X - xo) is a solution of p( [R2\Q, q; - (L O(dy )q;O )

§ 5. Capacities 413

and tends to zero at infinity; hence

is, by the linearity of C e'

(5.16) Cecp = _Ja; a + (fr ady )t{lo

with

But the formula (S.16), where a = L-1cpistrueforallcpEImL = D(Ce);taking q) == 1, we obtain

Cel = 0 = -~(J + 1)ao + (fraodY)t/lo. This gives (S.12), since by (S.1O), .lao = ao. For the point (2), we prove first that for cp E D(C) we have

(S.17) 1

LCcp = 2. (Kcp - (p).

We now consider the solution u( cp) of P( Q, cp). Since u( cp) E 'tj~ (Q\ from Proposition 5 of §3,

(S.18)

h . h' . . I I . I f au( cp) C d h were U 1 IS t e mtenor simp e ayer potentIa 0 - -- = - cp an U z t e an interior double layer potential of cp. Nowu1isasolutionofP(Q,L( - Ccp)) and U z is a solutionofP(Q,1(Kcp + cp)) (see (4.100) of §4), so that passing to the limit in (S.18), we have

Kcp + cp cp = -LCcp + --2- on r,

from which we have (5.17). We deduce that for (P E 1m L = D( C), we have

Kcp = cp + 2LCcp E ImL and (5.13).

Identifying (5.10) and (5.13), we deduce that

(5.19) JL -lcp = L -I Kcp for all (P E ImL.

Taking this value of JLlcp into (5.11) and (5.12), we obtain (5.14) and (5.15). From the equality (5.19) we derive immediately

(5.20) L.la = K La for all a E 'tjo (T) .

This shows in particular that the operator L exchanges the kernels and the images of


I ± J and I ± K; more precisely, we have

(5.21 )

L(ker(J - J))

L(ker(I + J))

L - 1 (1m (I - K))

= ker(1 K) = lr,iR EB ... EB It,iR

= ker(l + K) = lrl iR EB ... EB lr, IF.!(

= Im(l - J) = ker(l K)~

={I/JE(6 0 (r):f V/d~' = 0 for i =

L -1 (Im(l + K)) = Im(l + J)) = ker(I + K)-

1 I, ... , N J

={ I/JE(6 0 (l):fl/Jd}' = 0 for j = I, ... ,W}. These equalities result from ~4.5 (see the remarks preceding Proposition 19). Taking account of Proposition 2, we have also

ker(l ..... K) = kerC, Im(I - K) = fmC;

if 11 ~ 3, ker(I + K) = kerc.,. Im(l + K) = ImC

(5.22) if 11 = 2. ker C e = ker(l + K) EB h;, IF.!(

and ImC e = Im(l+K)n{I/JE(f,O(l):Jl/Jd}' = o}. We note also the immediate consequences of (5.10) and (5.15):

for all <;:) ED(C) = D(Ce ) = D(L- 1 );

if 11 ~ J, C (p + c., (p = - L ... 1 rp

(5.23) if 11 = 2, C (p + Cc (p =- L - 1 ({J

also. because of (5.19)

/ !' ) - 1:;(0 + ( 1 ( L (p) d;.' ------, r ;;(0 d}' Jr

(5.24) { for all ~ E D(C), = [)(~e), ~rp E D(C) = D(Ce )

CKrp - JC(P. LeKrp ..... J(erp.

The equality for C. in (5.24) is immediate from (5.14) in the case 11 ~ 3; in the case f1 = 2, it must be noted that in addition J;;(o = :;(0 (which follows from (5.10)) and

i(L-1rp)d;:= i(L iKrp)d)',

which follows from (5.19) and

(see Proposition 15(3) of ~4).

§S. Capacities

In §4.S we have considered the problems:

(5.25)

(5.26)

{ Llu = 0 on Q

au - -+ AU = if! on r; an

Llue = 0 on [R;"\Q

au _e -+ AU, = if! on r one

Ue satisfies the null condition at infinity

415

where ), E ~O(r) was supposed positive or zero. We have solved them (see Theorems 4 and 5) using the capacity operators J and L. We note that the problems (5.25) (resp. (5.26)) can be written in terms of the capacity operators in the following manner: u(resp. ue ) is solution of (5.25) (resp. (5.26)) iff u(resp. ue ) is the solution of P( Q, qJ) (resp. P ([R;"\Q, qJ) satisfying the null condition at infinity) where qJ E D(C) (resp D(Ce )) is a solution of

(5.27)

We have thus reduced the boundary value differential problems on Q to pseudodifferential problems on the boundary r = eQ. The study of the equations (5.27) will be greatly facilitated by the numerous properties of the capacity operators: we have already seen (Proposition 1) that C and C e are symmetric and positive. that is to say that for all (P E D( C)

tqJCqJdY ~ 0, tqJCeqJd}' ~ O.

This property will be exploited in all its force in the use of variational methods. We note here two other properties directly applicable in the classical theory: the compactness and positivity ()f the resoivants UI -+ C) -1 with ), > O. 0

First of all we give the following version of Hopf's maximum principle (see Proposition 14 of §4):

Proposition 4. Suppose that Q is a regular bounded open set in [R;" with boundary r satisfying the condition of the interior (resp. exterior) ball 96 , qJ E D(C) (resp. D(Ce)) alld Z E r i (resp. r;) such that

qJ(Z) = ~;n qJ (resp. qJ(z) = mi[~ qJ

Then

with qJ(z)::::; 0 when j = 0 ) .

either qJ == qJ(z) and CqJ == 0 (resp. CeqJ == 0) on r;(resp. rjl

96 For all = E T, there exists -'<0 E Q (resp. !l\\"\Q) such that BC<o.l-'<o - zl) c Q. (resp. Il\\" \fl) (see §4, Example 7 and commentary on Lemma 5).


or

Proot: It is sufficient to apply Lemma 2 of ~4 with the connected open set Q; (resp. Qj) ~nd u the restriction to Q; (resp. Qi) of the solution of P(Q, cp)

(resp. P( [Rn \ Q, cp) satisfying the null condition at infinity). From the classical maximum principle (see Corollary 3 of ~2 and ~4.3), cp(.::) = min 11 (resp.

fl,

(p(.::) = min u). From the condition of the interior (exterior) ball, the hypothesis Qj

(4.74) of ~4 is satisfied. so the conclusion of Lemma 2 of ~4 indicates to us that

either

or

11 == cp(.::) on Q i (resp. Qil ,

tu , (.::) < O. ('/1

from which follows the conclusion of the Proposition. We obtain then the

o

Proposition 5. Let Q he a regular hounded open set in [Rn with boundary r satisfying the hypotheses oj Propositions 3 and 4 and i E cgo(T) satisfying

{ ) :? ~ on I' and :. not i~e:icaIlY .::ero o:~ri (resp. ri) jorl - 1, .... N (resp.} - 0.1. .... j\ J.

(1) The operator if + C (resp. n + Ce ) is a bijection oID(C) (resp. D(Ce )) onto (6°(T). (2) The inverse operator ().1 + C) - 1 (resp. (n + C) - 1) is a compact operator on «;o(T). (3) For all I/; E (6°(T) with I/; :? 0011 r.

(i.1 + C) 1 I/; :? 0 (resp. (i.! + C,r 1 VI :? 0) 011 r. 1/; ill addition, I/; is not identically .::ero Oil r i (resp. I'jl. then

~i;n(if + C) 11/; > 0 (resp.I~i,n().1 + Ce )-11/; > 0). Proot: From Theorem 4 (resp. 5) of ~4. we know that for all I/; E (f,'o(T) there exists a unique solution 11 (resp. 1Ie) of the problem (5.25) (resp. (5.26)). From the equivalence of these problems with the equation (5.27). the point (1) of the Proposition appears as a corollary of this theorem. Let us prove the point (3). Given I/; E (6°(T) with I/; :? 0 on r. we put cp = (n + C)-II/; (resp.(n + C,rll/;)andconsiderforiE fl, ... ,N}(resp.jE {O, .... N' }) fixed, .:: E Fi (resp. I'jl such that cp(z) = min cp (resp. min cp).

From Proposition 4 I; rj

either (p == (p(.::) and C(P == 0 (resp. C,,4) == 0) on I'i (resp. I'i)

or C(p(z) < 0 (resp. C"cp(z) < 0) .

97 Hence (e.g. in the interior casej ifz E r,issuchthat<jJ(z) = max <jJ, then either <jJ =' <jJ(z)andC<jJ = O.

or C<jJ(z) > O. r;


Hence, since A([J + C([J = tjJ (resp. A([J + Ce([J = tjJ) on r,

either

or

([J == ([J(z) and )-([J == tjJ on Ti (resp. Ti) ,

),(z) ([J(z) > tjJ(z) .

From the hypotheses on )_, since tjJ ? ° on Ti (resp. Til, necessarily in one or the other case ([J(z) ? 0: more precisely: if ).(z) ([J(z) > tjJ(z), then (p(z) > 0; if ([J == ([J(z) and).([J == tjJ on Ti (resp. Til then ([J(z) > 0 or tjJ == ° on r i (resp. Tj). This proves the point (3). From the point (3) we deduce the continuity of the operator

In effect, putting

(Po = (n + C)-II (resp. (U + Ce) 11), for all tjJ E 'tj0(r) the function ([J defined by (p = (J.I + C)ltjJ (resp. (U + Ce)-ltjJ) satisfies

(5.28) min ( 0, m~n tjJ ) (Po ~ ([J ~ max ( 0, m:x tjJ ) ([Jo on r

and hence, in particular

II ([J II = m:x I ([J I ~ ( m:x ([Jo ) ( m~x I tjJ I) = ( m:x ([Jo ) II tjJ II .

Finally, we prove the compactness of the operator (AI + C) - I (resp. (n + Ce ) -1). From the continuity proved above that reduces to proving:

{for. every sequence (([In). of elements of D(C) (resp. D(Ce)) bounded in '5~(r), with (C([J,,) (resp. Ce([J,,), bounded in 'tj0(r), we can extract a sub-

sequence (nd such that «[JnJ converges in ,&O(r).

Given such a sequence (([In), using the compactness of the operators K and L (see Proposition 16 of §4) we can extract a subsequence (([In.l such that

From Proposition 3 (2), we deduce that the sequence

([Jnk = K([Jnk - 2LC([Jnk converges in 'b,o(r)

(resp. if n ? 3, ([Jnk = - (K([J"k + 2LC e ([J"J converges in ,&o(F) ; if n 2,

converges in (t'°(r) and hence again after possibly extracting a subsequence ([J"k converges in «jo(F). 0

4lR Chapter II. The Laplace Operator

2. Electrical Equilibrium. Coefficients of Capacitance

2a. Electrical Equilibrium: Electrostatic Problem: Capacitors

We say that a homogeneous ohmic conductor 98 K (see Chap. I, §4.2.6) is in e/a/ric equilihrium if the electric field E in the interior of K 99 is null. Hence, ij' K is ill

electric equilihrium. it can hare nOll-zero charges only on the houndary oj' K and (supposiny K to he cOllllected) the electrostatic potential is constant on K. Let us consider now a system S of homogeneous ohmic conductors placed in a domain D of the space [R3. We say that the system S is in electrical equilibrium if each of the components of S is in electrical equilibrium. We suppose that the domain D \. S is occupied by a perfect dielectric; therefore if the system S is in electrical equilibrium (i) the electrostatic potential 1\ in D is constant on each homogeneous component (assumed to be connected) of S. Denoting the components of S by K I' ... , Km we have

(5.29) 1'('=1'; on K; for i=I, ... ,/11

where 1: 1 , ... , 1'm are constants. (ii) the electric charges in D are distributed on the boundaries of the conductors K;. We denote by Pi the (surj'ace) charge density on the houndary DK; oj' K i . The total electric charge of the conductor K, is

q; = [. Pi(Z) dYi(z) ~tl\i

where dy; is the surface measure of the boundary elK;, assumed to be regular. Let us consider a problem where a priori the densities Pi and the potentials 1'; are again unknown. They are related through the equation (see § 1.2)

(5.30)

where (; is the permittivity of the medium occupying D\S. We can rewrite (5.29) and (5.30) in the form

(5.31 )

Ll1', = 0 in the open set D\(K I u ... U K lI )

r(' = t:;

IT,

?:ni

on i"K; for i = I, .. . ,m

where i'/(llli is the normal derivative exterior to K i .

n The word homogeneous employed here indicates that the material Llsed in every element K; is the same at every point of K;, and therefore that the electrical properties are the same at every point of K;. This qualification will be implied in the following pages. qq This holds at every instant t, being equivalent to the absence of current in K. The presence of a llllnzero current in an ohmic conductor is linked, through Ohm's law, 10 a dissipation of energy, from which we deduce the impossibility of a stationary regime without an external source of energy (see Fournet [I] I.

§5. Capacities 419

We see appearing in (5.31) the notion of a capacity operator introduced in Sect. 1. In order to make this more precise, we distinguish between two situations:

Case 1: The exterior problem. The domain D is supposed to be the whole space [R3.

We then impose on the electrostatic potential t'c the condition that it is zero at infinity. We denote hy Q j the (connected) interior of the conductor K i , fi = cQ, = oK j , Q

the union of the Q i whose boundary f is the union of the f i ; Qj, ... , Qm are the connected components of Q.

From (5.31), L'c is the solution of the exterior problem p([R3\Q, L't\Xr;) tending to zero at inflnity and aVe! rine = (I! E )Pi on fi' where c! Olle is the exterior normal derivative to [Rn \ Q. In other words

Pi = cCe(L'VjXr) on fi'

where C" is the exterior capacity operator of '6'0(f). Using the linearity

(5.32) on f, = ?Ki

and the total electric charge qj of the conductor Ki is

namely

(5.33)

where

(5.34)

qi = r Pi(Z) dl'(z) = I; L: Vj r CeXli dy , JG J JG

qi = L:CLvj j

Cf.j = I; r CeXljd}'. Jr,

We call Cr,j the exterior coefficient of capacitance of the conductor Ki on the conductor K j in the system S placed in the dielectric of permittivity I:: (occupying all of the space [R3\S).

Case 2: The interior problem. The domain D is supposed bounded containinfj all of the cavity of a uniForm ohmic conductor K ° in equilihrium (see Fig. 7).

Q = parts hachured

!'l'= parts hachured \\\\\\\\

\

Fig. 7


We denote by Qo the exterior of D (Q a = 1R 3 \15) whose boundary is T;) = aKo n 15; for j = 1, ... , m, Qj the (connected) interior of the conductor K),

Tj = ('Qj = aK j; Q' = Qo U Q'l U ... U Q~

and Q = 1R3 \ D' whose common boundary is r = To U r'l U ... U T~; the connected components of Q' are denoted by Q a. Q'[, ... , Q~.

The conductor Ko being supposed in equilibrium. the electrostatic potential is constant in K 0 and hence

Fe 1'0 on DKo·

In other words, from (5.31), 1'(' IS the solution of the interior problem

p(Q. f 1')11)' Hence j = 0 J

Pi = c;C ( f ['jlr) on )=0 I

where C is the interior capacity operator of (6°(r). From the linearity of C

(5.35) m

Pi = I; I rjCXf, on j=O

and the total charge of the conductor Ki is

(5.36)

where

m

qi = I Ci.jf; j=O

('

r ,

Ci . j = I: J .. C Xr di'· I ,

(5.37)

r ,

The equalities (5.35), (5.36) are also true for i = 0, Po denoting the density of the electric charges on To and qo the total electric charge on To 100.

Let us consider the system S consisting of the set of m ohmic conductors, K l' ...•

Km placed in the dielectric D filling the cavity of an ohmic conductor K(). We shall call C i. j , defined by (5.37), the interior coefficient ofcapacilance 1ol of the conductor Ki on the conductor K j (for i,j = 0, ... . 11/).

We observe that in both the exterior case and the interior case, the coefficient of capacitance of Ki on K j is the total charge o(the conductor K i I02 , when the conductor K j is at unit potential and the other conductors are al zero potential. The coetllcients of influence therefore have dimension Q V - 1, where Q is the dimension of electric charge and V the dimension of electrostatic potential.

100 Which is not a priori the total electric charge of Ko. 101 We also say. quite simply. "coellicient of capacitancc". ,02 In the case of an interior problem for i = 0, this is the total charge of r". the interior boundary of

Ko·

§S. Capacities 421

In the S.l. system of units (see § 1.2), the unit of coefficients of capacitance is the Farad (F):

IF = 1 coulomb/volt = 1 m-2kg~ls4A2.

In electricity, we call a set of several conductors, which strongly influence one another, a capacitor. We shall call a system of an arbitrary number of conductors a capacitor (in the wide sense).

2b. Properties of the Matrix of the Coefficients of Capacitances

We have defined the exterior (and interior) coefficients of capacitance by the formulae (5.34) and (5.37), where in the two cases Q denotes a bounded open set and rio r j are defined as in § 5.1 (see (5.9)) 1 03. To within the multiplicative constant [" the matrix of the interior (resp. exterior) coefficients of capacitance is the matrix

that we can consider for a regular bounded open set Q in [R0 with n ~ 2. The case n = 2 is particularly interesting for applications in electrostatics since it corresponds to cylindrical capacitors (see § 5.2). In the general case, we state the

Proposition 6. Let Q be a regular hounded open set in [R". With the notation (5.9) we put I 04

Ci.j f CXr;dy for i,j 0, .. . ,N' Jr: cf. j = f Ce '1.fj dy fiJr i,j 1 .... , N ;

1',

(1) the matrices (Ci.Ji.j=O .. .. N· and (cLkj= 1. .... N are symmetric and positive; (2) given i oF j = 0, ... , N' (resp. 1, ... , N), if'r; and rj (resp. r i and rjl are connected in Q (resp. [Rn\Q) then ci. j < ° (resp. cf.i < 0), otherwise cij = ° (resp. cL = 0);

(3) Ci • i = L j = 0 ..... N'

j *- i

if n ~ 3

if n 2

c· . I.j

and

and

for i = 0 .... , N';

L j = 1, ... , N

j:ti

L j = 1. .... N

i*i

C e . I.j

c ~~ .. t,j ,

103 We observe that the interpretation of Q is different in the exterior case from that in the interior casco 104 In the case n = 3. c,.j and cL are, to within the constant t. the coefficients of influence.


(4) the matrix (Ci.)i.j=O ..... N' is afrank N';

if n ): 3, the matrix (eL)i.j = 1. .... N is invertible,

ifn 2, the matrix (Ci'.j)i.j= 1 ....• N is of rank N - I.

Proof Given (id, Uti) E [RN' + 1 (resp. [RN), we have from Proposition I,

LL cjj jJ.1j = LL )'jJ.1j ( Xr;CXrj d)' Jr

( (L AjXr)C(L ujXr) d)' Jr 1

L grad u( ((J). grad u( I/J) dx

( resp. LL cfj·jJ.1j = ( _ grad ue(({J)·grad ue(l/J) dX), Jw Q

where ({J = L)·iXr;, I/J = LJ.1iXr; (resp. ({J = .I/iXr;. I/J = LJ.1iXr), u(({J) (resp. ue(({J)) is the solution of P(Q, ({J) (resp. p([Rn\Q, ((J) satisfying the null condition at infinity), This proves the point (I). Let us fix i i= .i = 0, ... , N' (resp. I, ... , N); given Qk (resp. Q~) a connected component of Q (resp. [Rn\,Q) whose boundary meets r; (resp. r;); we distinguish two cases:

Case /: r k n rj i= 0 (resp. r~ n rj i= 0) From the maximum principle

(5.38)

and

o < u(Xr) < 1 on Q k (resp. 0 < ue(Xr ) < 1 on Q~) J J

U(XI"') = min u(Xr,} on r; n r k J - 1

Q,

( resp. Ue(Xr) = min ue(,~r) on r i n r~) , Q;

From Hopf's principle of the maximum (see Proposition 4) 105,

a CXr = - u(Xr) < 0

1 on 1

(5.39)

( resp. CeXrj 0;= ;: UeUf) < 0 on r i n r~ ) .

105 The use of Proposition 4 demands (in the interior case) the hypothesis of the interior ball. We can in fact, construct a direct argument without making use of Hopf's maximum principle. We have in effect (without any hypothesis of regularity), CZI"' ,:; 0 on r; n r, and C/.t' not identically zero on r; n r, as

otherwise (see Corollary 6 of§4) u( "1.1') w'o:ld be zero on Q,; hence I1 C,I,.d;· < 0 which suffices to J r;nl'k ,J

prove (2).

§5. Capacities 423

Case 2: Ik n Ij = 0 (resp. I; n rj = 0), then u(Xr) = 0 on r k (resp. U.,(Xr) = 0 on r~); hence u(Xr) = 0 on Qk (resp. ue(Xr) = 0 on Qd and

(5.40) C XI"" == 0 on r k (resp. C Xr == 0 on r~). J ~ j

Since

N i ( N' i ) ci • j = I CXr; dy resp. cL = I cexr) dr' k =! r; n r k k = 0 r, n r;

the point (2) follows from (5.39) of (5.40). From Gauss's theorem, for i = 0, ... , N'

jto Ci • j = I t CXr; dy = L ~:l U(Xr) dy = 0 ; J

hence

We have also for i = 1, ... , N

f cf. j = I' CeXr, dy = r ~3 ue(Xr) dy . j=! ~r Jr cn '

In the case n = 2, from Lemma 1, the integral is zero. In the case n ?: 3, from Gauss' theorem

r : ueCXr,l dy = ° for k = I, ... ,N' , Jr; n


(5.41 ) ~ ce . = 5 U (x . ) d'J . N I rl L. ' E,j :l e J. I'

j = ! r;, en ' N

case r~ n r i = 0: ue(Xr) == 0 on Q~ and I cf. j = ° ; j = 1

case r~ n Ii # 0: since uAXr) tends to zero at infinity, from the maximum principle,

(5.42)

Let us fix R sufficiently large to ensure that B(O, R) => !R"\Q~: we have from Gauss' theorem, for all r > R

(5,43) - r"~' 1 u'(r)

where u is the radialized of u(Xr) (see §1,4)

u(r) "" L u(Xr) (ra) dO' .


Being the radialized of a harmonic function, ii can be written (see Proposition 6 of § 1 and Proposition 4 of §2)

ii( r) = a

r" '" 2 + b with a, h E G~ .

Since Uc(XI) > 0 (see (5.42)) and tends to zero at infinity, h = 0 and a > O. Hence _, (n - 2)a U (r) = - "":: {' < 0, which, from (5.41) and (5.43), shows that

r

Cf.; > L c;,) . j = I.. .,'11

jF1

This concludes the proof of the point (3). We know already from the point (3) that the matrix (c i . j ), i, j = 0, ... , N', is of rank ~ N' and also in the case n = 2 that the matrix (cf.J, i,.i = 1, ... , N of rank < N. Let us consider (I.;) E ~N' (resp. ~N); since the matrix (c i. j ),

i,j = 1, ... , N' (resp. cL i,j = 1, ... , N) is symmetric and positive,

:V' ( ,'II I ifi.) = 0 for i = 1, ... ,N' resp.I i/i.) = 0 for

j = 1 .1 1

i = I, ... , N )

iff II i·)/'i.j = 0 (resp. II A)jcf.j = 0); namely using

(P = i=1.~ .N.iixr:(resp.<p i=l~ .NAixri)'

from the proof of the point (1) above, iff

r igrad u(<pW dx = 0 (resp. r" .,igrad ue(<pW dX) , JQ JR \Q

that is to say iff u( (p) is constant on Q l' ... , Q'I (resp. Ue ( <p) is constant on Q;j, ... , Q:'1)' i.e. iff

<p E itt XI, ~ (resp., in the case 11 = 2, (P E ito Xr:~ , N'

and in the case n ? 3, <p E I Xi:~' since ue ( <p) tends to zero at infinity). i ::: 1

Using the fact that "I ."1'

I Xri~ n I Xr;~ = {O} , i = 1 i = 1

we deduce that the matrix (c i . j ), i, j = I, ... , N', is invertible and hence that the matrix (ci,J, i,j = 0, ... , N', is of rank N'; we deduce also in the case n ? 3, that the matrix (ej'.j)' ij = 1, ... , N is invertible and in the case n 2 that the matrix (cfj) is of rank N - 1. 0

Remark 1. Returning to the problem ot' electrical equilihrium presented in §5.2(/, the results of Proposition 6 arc interpreted in the following manner.

§5. Capacities 425

(1) Two conductors of a system influence each other if and only if they are connectable (in the dielectric without crossing other conductors). In other words, given a system in equilibrium, if two conductors are connectable, every change in the potential of the one implies a change in the total charge of the other; conversely if two conductors are not connectable, we can change the potential of the one without destroying the equilibrium of the other. This follows from point (2) of the Proposition. (2) The influence of two conductors of a system is symmetrical. This follows from the symmetry of the matrices of the coefficients of capacitance (point (1) of the Proposition). (3) Case of the exterior problem: given a system S of conductors K 1" .. , Km placed in a dielectric occupying all of the space lR 3 \S, for all q1" .. , qm' there exists one and only one state of equilibrium such that the total charge of each conductor Ki is qi' That is to say that there exists a unique set V1, ... , Vm (Vi E lR) such that the system is in equilibrium when Ki is at the potential Vi with the total charge density qi' This follows from the invertibility of the matrix of the exterior coefficients of capacitance (point (4) of Proposition 6). (4) Case of the interior problem. Given a system formed by conductors K 1,· .. , Km placed in a dielectric in the cavity of a conductor Ko,for al'l q1" .. , qm and vo, there exists one and only one state of equilibrium such that the total charge on each conductor Ki is qi and Ko is at the potential Vo' We note that then, the total charge

m

on the surface r~ of the interior boundary of Ko is qo = - I qi' This follows z:= 1

from points (3) and (4) of the Proposition. We observe that we should be able also to fix the potential of one of the conductors Kio and the total charge of each of the conductors Ki, i = 1, ... , m, i of. io and of the interior boundary of Ko. (5) As we have just seen, the equilibrium of a system of conductors is determined by the prescription of the total charge of each of the conductors in the exterior case and of that of the internal conductors and of the potential of external conductor in the interior case. The system being in equilibrium, the distribution of electrical charges Pi on the boundary 8Ki of the conductor Ki is then perfectly determined: it is given by the formulae (5.32) and (5.35) according to the exterior or interior case. The argument of the proof of the point (2) of the Proposition indicates that given a system of conductors in equilibrium, on the one hand the charge is zero on every part of the boundary of a conductor which is not connectable (in the dielectric) to any other conductor, while, on the other hand, that the charge is strictly negative (resp. positive) on every part of the boundary of a conductor Ki whose potential is minimum (resp. maximum) with conditions that this part is connectable to a conductor of strictly higher (resp. lower) potential. This is true in the interior case and also in the exterior case by considering at infinity a fictitious conductor at zero potential. 0

We conclude this sub-section by the explanation of the form of the matrix of the coefficients of capacitance and of the distributions of charge in some simple cases:

Example 3. System of two conductors. We have several possible situations:


(1) An exterior problem

The matrix of the coefficicnts ( C 1.1 C2 • 1

Fig. g

( 12 )h 'f . as to satJs y C2 . 2

min (C1.1' C 2 . 2 ) > - C1.2 = -C2 . 1 > O.

In a state of equilibrium, K I and K 2 being at the potentials VI. 1'2 respectively: (i) the interior boundary 1'1 of the hollow conductor K 2 will not be charged: this hollow conductor will behave like the solid conductor K 2 obtained by filling in the interior; (ii) if 1'1 ~ 0 ~ 1'2.1'1 + 1'2' the charge will be strictly negative (resp. positive) on every part of the boundary of K 1 (resp. of the exterior boundary of K 2); (iii) if 0 < VI < 1'2' the charge will be strictly positive on every part of the exterior boundary of K 2 and there will be positive charges and negative charges on the boundary of K 1: (iv) if 0 < r 1 = 1'2, the charge will be strictly positive on the boundary of K 1 and the exterior boundary of K 2'

(2) Two examples of interior problems.

Fig. 9 Fig. 10

I h t· f h fr . j' . (/C o 0 n t e two cases, tile matrIx 0 t c coe llClents 0 capacitance " C 1. 0

satisfy Cn. o = - CO. I = -C1 . 0 = C1.1 > O.

Co 1) . must C1.1

§S. Capacities 427

It only depends therefore on a single constant C, called the capacity of the

capacitor, the matrix of the coefficients of capacitance being ( _ ~ -~). In a state of equilibrium, in the case of Fig. 10, the interior boundary r'l of K 1 will not be charged: the capacity of the capacitor will depend only on the exterior boundary of the interior conductor (and also only on the interior boundary of the exterior conductor). (3) Another exterior problem.

00

Fig. 11

In this example the "capacitor" (K 1, K 2) is immersed in the dielectric occupying all of [R3 \ (K 1 u K 2)' If we denote by C the capacity of the capacitor (K l' K 2)' it is clear that the exterior coefficients of influence of K 2 on itself, and of K 2 on K 1 will be C and - C. As for the exterior coefficient of influence of K 1 on itself, it will be C~.l = C + C e, where

Ce = r oUe dyl06 Jro an

where r~ is the exterior boundary of K l' boundary of the exterior Q~ of the condenser and Ue the solution of P(Q~, 1) tending to zero at infinity. We have (see proof of the point (3) of Proposition 6)

ce > 0.

In brief, the matrix of the interior coefficients of capacitance of this system is

( c + ce -C

- C) C .

2c. Energy Balance. Examples of Capacities

o

We consider a system of conductors immersed in a dielectric, The electrostatic field E is given by (see Chap. lA, §4)

E = - grad Vc in D

106 Here the permittivity B of the exterior electrical medium is supposed to be equal to 1.


where l'e is the electrostatic potential. The electrical energy 1 07 of the system is given by

(5.44) ;; = ~LIEI2dX = iLlgradl;eI2dX.

We distinguish two cases. Case of the exterior problem. With the notation of §5.2a, l'e is the solution of P( iR" \{2, Xl'; h) tending to zero at infinity with the result that (see Proposition 1),

r .. lgradrcI2dx = j' (Lv;Xr;lClLI')Xr)di' .JOt" Q r

II vir) r Cch, di' . Jr;

Hence, taking account of the definition of the exterior coefficients of capacitance (see (5.34)), we have

(5.45)

Case of the interior problem. With the notation of §5.2.a, Vc IS the solution of

P ( Q, ;~o viXr;) with the result that as above

(5.46)

where C i .) are the coefficients of influence of the capacitor (sec (5.37)). In these formulae the (numerical values olthe) coefficients olcupacitance Cr.) and Ci.j appear as twice (the numerical values) of the energy of the system in equilibrium when the conductor K; is at unit potentia/' the others being at zero potelltial. For that reason 108, we call CL the exterior capacity of the conductor K i in the "exterior" system and C;. i the interior capacity of the conductor K i in the "interior" system. This coincides totally with the definition of Example 3.2. More generally, let So be a family of conductors of the system; we call the capacity (exterior or interior) of So in the system, twice the energy of the system when we put the conductors of the family So at unit potential and the remaining conductors at zero potential. In particular, the exterior capacity of an entire system is douhle the energy of the system when all the conductors are at unit potential; it is therefore

C" = II Cr.).

Also the interior capacity oj'the "interior" system also called the capacity of the

107 In fact, the Fee energy for perfect media; see Chap. IA. ~4 where we used the notation fV for II lOB They measure the capacity to store energy for a given potential. The capacities (of conductors in a system) are coefficients of capacitance: the unit of capacity ill the 5.1. is thell the Farad.

§5. Capacities 429

capacitor is twice the energy of the system when all the interior conductors K 1, ... , Km are at unit potential, namely

C = .. L Ci,j' 1,1 = 1, ...• m

We note that C = Co,o, capacity of the conductor Ko in the "interior" system: this

follows from the relations " C. . = - C· ° for i = ° to m. ~ '.} I, j = 1, ... , m

We observe that the capacity (exterior or interior) of a family of conductors in a system is always less than the sum of the capacities of its components. This follows from that the coefficients of influence Cf,j or Ci,j (for i #- j) are negative. This capacity can even be less than the maximum of the capacities of its components: this is clear for the system of Example 3(3) (Fig. 11), where the exterior capacity of the system (K 1, K 2 ) is

C + Ce + (- C) + (- C) + C = ce which is strictly less than the exterior capacity of K 1 in the system. The notion of capacity has very many applications going far beyond the electrostatic problem: we shall study it in Sect. 3. We shall content ourselves here with giving some examples.

Example 4. Spherical capacities. We consider a radial system of conductors, i.e. a system in which all the conductors of the system are in the form of concentric spherical layers (see Fig. 12). (1) Exterior capacity of the system. The conductors being all at unit potential, the electrostatic potential Vc is equal to 1 on B(O, r) where r is the radius of the exterior conductor: the exterior capacity of the system is the same as that of a solid

Fig. 12


conductor occupying all of the ball B(O, r); by definition it is

where Ue is the solution of P(lR: 3 \8(0, r), I) tending to zero at infinity. We have ue(x) = r/lxl on lR:\ so

and

(5.4 7)

r

Izl2 on ?B(O, 1') ,

r

(2) Capacity of the capacitor. This is the capacity of the exterior conductor in the "interior" system. The interior conductors all being at zero potential, the electrostatic potential is zero on 8( 0, r'l) where 1"1 is the radius of the exterior houndary 0/ the largest interior COllduclOr: the capacity of the capacitor is the same as that of a capacitor with two conductors where the interior conductor occupies the whole ball 8(0,1'1)' Denoting by rlJ the radius of the interior boundary of the exterior conductor, we have

where u is the solution of P({r'j < Ixi < 1'0], X,8(0-,o))' From Table 2 of Example 20, §4, we have

u(x) = (1'\ -I~T )(}t - rio)! on [1"1 < Ixl < 1'0)

from which

and

(5.48)

or agam

( 5.49)

i1u -(z) Dn

_12(_1, _ 1)"1 1'0 r 1 1'0

I

C

c 47[1'01"1 /;

ro ---- 1"1

on i-:8(0,ro)

,,)9 In this expression. I: denotes the permittivity of the exterior dielectric medium: Ixl > r:, and 11,. is directed towards the centre of the baiL 110 In this expression, E denotes the permittivity of the interior dielectric medium which in general is

different from that of the exterior dielectric medium.


(3) Capacities and coefficients of capacitance of the components of the system. Let us consider first the exterior conductor Ko: its interior capacity (in the capacitor) is given by (5.48); from Example (3.3) the exterior capacity (of Ko in the system) is C~.o = C + ce where C is given by (5.48) and ce by (5.47). Finally Ko influences only the largest interior conductor K 1 and (see Example 3(2) and (3)) C~.I = CO• I = - C. The exterior and interior coefficients of capacitance of the interior conductors are equal. The smallest conductor Km is influenced only by the smallest conductor containing it - Km I; its capacity and its coefficient of capacitance on K m -1 are the same as if it were the interior conductor of the capacitor (Km _ I, Km). Hence, from (5.48),

where for a capacitor K i , r i (resp. r;) denotes the radius of the interior (resp. exterior) boundary of K i •

Finally for 0 < i < m, the conductor Ki is influenced by Ki ____ 1 and K; + I: its capacity in the system is the sum of the capacities of the capacitors (K; _ 1, K,) and (Ki' Ki + I); its coefficient of capacitance on Ki _ 1 (resp. K; + 1) is the same as it would be if it were the interior (resp. exterior) conductor of the capacitor (Ki l' K i ) (resp. (K i , Ki + 1 )), i.e. the opposite of the capacity of that capacitor. Therefore

c· 1.1

C i. i - I = o - ------ -----" , , /'i - I -- r i

Remark 2. The above considerations on the coefficients of capacitance of the conductors Ki relatively to the capacities of the capacitors (Ki _ I' K,) are obviously valid for any system of conductors whatsoever in which they are contained the one within the other. 0

Example 5. Plane capacitor. Let us consider a system of two identical plane conductors K 1 and K 2 placed parallel to each other at a distance d apart (see Fig. 13). The system being in electrical equilibrium, each conductor K i is at the constant potential Vi' If the plates were infinite, the field between each plate would be

Fig. 13


uniform and given by

where n is the unit vector, normal to the direction of the plates, directed from K I

to K 2 .

The electrical energy in the domain D of the dielectric determined by the surface r is therefore

Ie: . , J /; 8 = 2 x i£l- x Volume of D = :2 (1'\ - V2 )2 d x surface of r.

The quantity tid is the capacity per unit area oj'sur/cu'e of the plane capacitor (K 1 , K 2 ). The quantity eSjd where S is the surface area of each of the plane conductors K l' K 2 is the capacity of the ideali:ed plane capacitor. This theoretical capacity differs from the real capacity - twice the electrical energy in the dielectric contained between the two plates when the potential difference between the plates is unity: indeed the field is not really uniform between the plates because of the "edge effects"; however, the difference is negligible in many applications (if

.JS ~d). 0

Example 6. Cylindrical capacitors. We consider the system of two coaxial cIrcular cylindrical conductors K 0 and K 1 of the same length I (see Fig. 14). Supposing nevertheless that the cylinders are infinite, the system being in equilibrium, the electric field will be orthogonal to 0: and independent of :. Hence

E = - grad 1'c where re(x, y.:) = u(x, y)

with II the solution of the Dirichlet problem P( Q, coXi" + l"1i.r 1 ) where Q is the

y

x

Fig. 14

§S. Capacities 433

annulus of [R2 contained between the circles r 0 and ["1 respectively the crosssections of the outer and the inner conductors. In the present case, where we have circular symmetry Q = {r'l < r < ro} and u is given by (see Example 20 of §4)

u(x, y) = ( )( )-1 ro r ra

VI Log - + Vo Log -;- Log -,-r r 1 r 1

where r = (x 2 + y2)t. Hence the electrical energy in the domain of the dielectric of length I is

,g' = ~ x LigradUl2dXdY x I = ~ x 2nl x (LOg;~-)-I(va -- Vj )2.

The quantity 2ne/Log (ra/r'd is the capacity per unit length of the cylindrical capacitor (K o, K I)' The quantity 2nd/Log (ro/r'd is the theoretical capacity of the cylindrical capacitor of length l. We can show, in this case also, that it differs little from the real capacity in many applications. 0

Condensers, generally consisting of two conductors separated by an insulating or dielectric medium, are used in electrical circuits (continuous, variable (alternating, ... )). In electronic circuits, their great utility is linked to the fact that they can absorb or part with large quantities of electric charge, without much changing their potential. The S.L unit of capacity is, as we have seen, the Farad (1 Farad = 1 Coulomb/volt). In practice, charges are much weaker than the Coulomb and measurements are currently made in terms ranging from the picofarad (10 - 12 F) in high frequency circuits to some microfarads (10 - 6 F) in the filters of electrical power supply. 0

3. Capacity of a Part of an Open Set in IRn

3a. Capacity of a Compact Set; Definition, Properties, Examples

In this part we introduce the mathematical idea of the capacity of a part of [R". We shall call by a regular compact set of [R" every K = 6) where (j) is a regular bounded open set of [R". Given K a regular compact set of [Rn and Q a regular bounded open set containing K, Q\ K is a regular bounded open set. We denote by uK . Q the function defined on [R" by

(5.50) { uK,Q = 1 on K

uK.Q = 0 on [R"\Q

uK,Q is continuous on [R" and harmonic on Q\K .

In other words, uK,Q is defined on Q\K as the solution of the Dirichlet problem


P(Q\ K, h). We have

Uu ! E '6~(Q\K) (see Proposition 18 of§4)

igrad uK •n l 2 E LI(Q) and (see Proposition 5 of§l)

r IgraduK.oI 2 dx= r ~luK.odi" Jo K JKnr'(12K)cn

From Gauss' theorem and account being taken of

K n e(Q\K)

(5.51)

(: and ........ U ,<: 0

ell K,O '" on eQ,

This can be interpreted in the sense of distributions: The derivatives euK,o/ ax; (in the sense of distributions) are functions of L 2 (ih\;n), JuK.o (in the sense of distributions) is a Radon measure on ih\;", supported by a(Q\K), negative on Q, positive on ih\;"\K and

total mass of

(5.52) total mass of AuK,o in ih\;n\ K

i x the total variation of JUK,Q in ih\;" ,

By definition, this quantity is the capacity of' K on Q. which we denote by caPu K Ill:

Example 7. Case n

(5.53)

L We note that we have trivially in the case n arbitrary

capo K = I capo, K;

where Q1' ... , Q m are the connected components of Q and K; = K n Q;. We CUll

therefore always suppose Q is connected, that is to say in the case n = I, Q = J a, h [. A regular compact set of ih\; is of the form

K = [al,h l ] u ... u [am, h.,] with al < hi < a2 < ... < hm ·

I I I If K is the system of ohmic conductors placed in the dielectric D occupying the domain Q

surrounded by an ohmic conductor Ko. cap" K is to within the multiplicative constant t. the capacity of the condenser thus constructed (see §5.20a and c).

§S. Capacities 435

f"; Fig. 15

It is clear that the graph of uK,Q is given by Fig. 15 and hence

1 1 (5.54) capa K = --- + ~~-

a l - a b - bm

where, we recall, Q = ]a, bL a l = min K, bm = max K. o Example 8. Radial case. From (5.53), we can suppose that Q is connected. A radial bounded connected open set is either a ball or a ring. A radial regular compact set is of the form

m

K = U {ri ~ Ixl ~ r;} with ° ~ rl < r'l < r2 < ... < r;" . i = 1

Making use of the results of Example 20 of §4, Case (a). Q = B(O, ro): uK,a is given (on Q) by

and

o

on B(O, r;")

on uK,a = O'nr~ - 1 X En(r;n) - En(ro} on aQ

and

(5.55) 1

caPB(O, '0) K = £n(ro) - En(r)

where K being radial, r = r~, = max I x I ; K

case (b). Q = {ro < Ixl < r~}

En(ro) -- En{x)

En(ro) - En(r'd on {ro < Ixl < r'd

uK,Q = 1 on {r'J ~ Ixl ~ r;"}

En(r~) -- En(x)

En(r~) - En(r;") on {r;" < Ixl < r~}

4j6 Chdpkr If The Laplace Operator

and

(5.56)

where K being radial I' = 1"1 = min I x I, r' = r~ K

We shall now prove the fundamental result:

max Ixl. D K

Proposition 7. Let K be a reyular compact set of IRn and Q a re?Jular bounded ope II sel of IR" containing K. Then

(5.57) capQK ='/~J~!)L,lgradUI2dx lI'lz!:'re Y'(K, Q) is tlz!:' set of the 11 E ((,o(IR") suclz that th!:' deriUltires ('11. (~Xi (ill liz!:'

sense of distributions) are in L 2 (IRn) and 11 ~ 1 on K, 11 ~ 0 on IR" \Q.

To simplify the writing we introduce .1 (lI) = J~. I grad u 12 dx defined on the space IR'

ill B 1 (IR") =; 11 E L~c(J~n): the del' irati res -;;-- E L 2 (IR") (ill the seilSI' o{dislrihwioIJS):.

eX i

The proposition expresses that UK Q minimises the energy functional .1 (11) on the set .'J'(K, Q) = ~u E ((,o(lRn ) n Bl(lRn): 11 ? Ion K, Ii ~ 0 on IR'" Q}.

Proof Since 11 K.ii E Y' (K, Q), it is sufficient to prove that, for tI E .'/ i K, Q),

Let us put l' = U

(5.58)

On the other hand

capli K = .I(l/KQ) ~ .I(u) .

UK.!!: we have

.1 (1') + 2 L, grad UK.'" grad t' dx

~ 2 L grad 1/K.Q' grad t' dx K

From (5.51) and (5.58) we therefore have .I(u) ? J(lIK.Q)'

This proposition leads us to frame the

D

Definition 1. Let Q he (Ill (arhitrary) OpCI1 sct ill IRn alld K il compact set ill Q. I,Ve define the capacity of K ill Q to he

inf .1(11), '/IK.121

§5. Capacities 437

where 9"(K, Q) is the set of the u E ICO(IR") n Bl(IR") with u ? 1 on K, and u ~ 0 on IR"\Q, lim sup u(x) ~ O.

Ixl- 00

This definition clearly generalizes that given in the case where Q is a regular bounded set and K is regular. In general the energy function J(u) does not attain its minimum on 9"(K, Q); this is due to the fact that the Dirichlet problem

P(Q\K, XK) does not always admit of a classical solution (i.e. continuous on Q\K); however in the case where Q is bounded, the Dirichlet problem P(Q\K, xd always has a generalized solution (see Definition 3 of §4); in the case of Q unbounded, we can also consider, if it exists, the generalized solution of P(Qj K, XK) satisfying the null condition at infinity (see Definition 10 of §4). We define the function UK,Q by

uK,Q = 1 on K

uK,Q = 0 on IR"\Q

uK •Q is equal on Q\K to the generalized solution of P(QjK,XK) (5.59)

satisfying, in the case of unbounded Q, the null condition at infinity.

We note that this function uK.Q is well defined on IR" in the case of Q bounded, and if Q is unbounded, if the Dirichlet problem P(Q\K, XK) admits a generalized solution satisfying the null condition at infinity. By definition 112 u K.Q is defined on Q \ K as the lower envelope of.~ + (Q\K, XK) the set of the continuous functions Wo superharmonic on Q\ K satisfying

lim infwo(x) ? XK(Z) for all Z E o(Q\K) ,

and in the case where Q is unbounded, min(wo, 0) satisfies the null condition at infinity (relative to Qj K). Such a function Wo is positive or zero on Q\K and

lim min (wo(x), 1) = 1 for all Z E oK .

Hence, the function defined on Q by

w = 1 on K, w = min (wo, 1) on Q\K,

is positive or zero on Q; it is also superharmonic on Q (see Remark 8(3) of §4). It follows that UK. Q is defined on Q as the lower em'elope o{the set.'i' + (K, Q) o{positive, continuous functions, superharmonic 0/1 Q, satis{j'ino w ? 1 on K. This leads us to formulate

Definition 2. Let Q be an (arbitrary) open set of IR" and K an (arbitrary) compact set in Q. We denote by uK,Q the function defined on IR" by

{ 0 IR"\Q

::::(:) = o~nf w(x) for all x E Q 'I'+(K,Q)

112 See §4.! in the case in which Q is bounded and (4.3) in the case in which Q is unbounded.

Chapter 11. The Laplace Operator

where .'/ t (K, Q) is the set of functions Ir, continuous, superharmonic and positive on Q with w ? 1 on K 113

In the case where the solution of (5.59) is well defined, as we have just seen, this solution coincides with the function IIKQ of Definition 2. These definitions 1 and 2 are justified by the

Proposition 8, LeI Q be an (arbitrary) open sel in [Rn and K an (arbitrary) compact sct ill Q. 11-'1117 Ihe definitions

(I) capf}K = inf .f(II}, where '/IK.m

Y(K, OJ = {liE St(Q); 0 z; 11 ::::; I, U= I in the neighhourhood of K)

(2) UK. f}E Bl([Rn)114 and cap!J K = .f(uK . f1):

(3) The distrihution UK. f} is suhharmonic on Q115, hannonic on QJK and

capD K = ( d( - "1u K . !J)116 JrK

In the proof of this proposition we shall make use of the following essential lemma which will be proved at the end of this sub-section 3.a.

Lemma 2. IU

10('1 f2 he (//1 opel/ set of [Rn alld U E L/oc(Q) with E L/oc(O) (in the tXj

sellse of distrihutiolls). Theil

(5.60) and , c -+-U' aX i

a.e. 011 Q117

(5.61 ) o a.c. on {xd2; U(x) = OJ.

Proof of Proposition 8.( 1) We have 9t(K, Q) c ,Y'(K, Q) with the result that

capn K::::; inf f(u).

0"(K. Q)

We now consider uoEyJ(K, Q). For £>0, we put

llc = ((I + 2<:)uo - c)+ !\ IIIB

113 Note that Y·+(K. Q) is not the cone of the positive functions of Y(K. Q).

114 The function UK." is a bounded Borel function on Ihl" and hence in the class L~,J Ihl"). 115 That is -duK .,,;" 0 in .0"(Q) (see Definition of §4). I 10 _ d UK. f! is a positive !Radon measure on Q supported by (~K. 117 u+ = max(u, 0), Xi" > 0; is the characteristic function of the set (x E Q; u(x) > 0:.

118 For two functions!: q:f t, g = min(j; q) =f- U- g) +

§5. Capacities

We have Uc E <6'°(IR"); from Lemma 2 the derivatives ~Ue E L]loc(lRn) and (Xi

a.e. on

with the result that Ue E Bl (IRn) and

§(u,) ~ (1 + 2S)2§(UO)'

Now ° ~ tI, ~ I and

u, on U, = {uo ~ f{-fi}

u, = ° on V, = {uo ~ i~2~} From the hypothesis on Uo, there are r > 0, R > ° such that

u, = on K + B(O, r)

U,= ° on (IRn\(QnB(O,R)))+B(O,r).

Let us take p E f0(B(O, r» with p ~ 0, r p dx = 1 Jw We have: p * U, E !0(K, Q) and

Therefore §(p * u,) = II p * grad Ue III, ~ ,",(Ue) .

inf §(u) ~ (1 + 21:)2§(UO)' [I(K. Q)

IR" ,

439

This being true for alII: > 0, and for all Uo E.5P(K, Q), we obtain point (1). 0

To continue the proof of Proposition 8, we shall use the property of the continuity of cap il K and UK, il with respect to K and Q. We group together different properties of cap il K and UK. il in the

Proposition 9. We have the following properties of cap fJ K and UK, f1: (l) cap [} K is increasing with respect to K and decreasing with respect to Q :

KI c K2 C Q 2 C Q 1 = capil Kl ~ cap[}2K2 ;

(2) UK, [} is increasing with respect to K and Q:

Kl C Q 1 C 5:22 , Kl C K2 C Q 2 = UK"Q, ~ UK2 . il2 '

(3) For every increasing sequence of open sets Q k with union Q and every decreasing sequence of compact sets Kk c Q k with intersection K

cap [} K = lim cap ilk Kk

and lim uKk . !h(x) = UK [}(x), for all X E IRn with the reservatioll that if Q k # Q that UK. il be the solutioll of (5.59).


(4) The capacity is strongly sub-additive with respect to K and Q

caPo[uO,K1 u Kl + capo[no,K1nK1 :::; capO[Kl + caPn,K 1 .

(5) For every family of open sets Qi' two by two disjoint, with union Q and every compact set K in Q.

capn K = L caPn,(K n Q;)

uK . n = UKnO"O, on Qiforalli,

(6) Given a diffeomorphism h of class ~l of an open set U of IW into IR" and Q a relatively compact open set of U, there exists c such that

capDK :::; c caph(f1)i1(K).

for every compact set K in Q.

Proof (~( Proposition 9, The point (1) follows from Definition 1, since

Kl c K2 C Qz c Q[ => Y'(K 2, Ql) c Y(KI' QI)

The point (2) follows from Definition 2, since under the hypothesis

K 1 c Q I C Q 2 and K I C K 2 C Q I ,

for all w E y+ (K 2' Q2) the restriction of w to Q1 is in ,'1'+ (K I' Qd. For the point (3), with its notation, we easily see that caPI1K = lim cap I1 k Kk by using point (1) of Proposition 8 (proved above) and the fact that the sequence of sets rJl(Kk, Qk) is increasing with union 9(K, Q); besides we note that capmKk converges by decreasing to capD K. The situation is not so simple for the sequence of functions Ilk = II Kk , nk'

We note first that lim sup IIk(X) :::; UK, f1(.'<) for all x, It is trivial for x E IR"\Q; for x E Q and (; > 0 from the definition there exists w E .Cf?+ (K, Q) such that w(x) :::; UK, n (x) + e; since w is continuous on Q and ~ 1 on K, {w ~ (1 + fr I} is a neighbourhood of K and hence contains Kk for k sufficiently large; in other words (1 + e)w E Y+(Kk,Q) and

udx) :::; UK., f1(X) :::; (1 + e)w(x) :::; (1 + £)(uK, f1(.x) + £)

for k sufficiently large, which proves the statement. When Qk = Q, Uk ~ UK, nand hence the sequence uk(x) converges to 11K, n (x) for all x (decreasing moreover). Now if UK, f1 is the solution of (5,59), it is also defined on Q\K as the upper envelope of the set.f _(Q\K, XK) of continuous functions v, subharmonic on Q\K satisfying the inequality

lim sup v(x) :::; h(Z) for an Z E ('(Q\K) ,

and in the case of Q unbounded, there exists h E .R'(Q\K)n~O(Q\K) with , v(x)

h(x) > 0 for Ixllarge so that hm sup -h -; :::; o. Ixl ~ C1j (XI

For such a function t' and I: > 0, we put t'£ = max(O, L' - I:h o) where ho = 1 in the case of Q bounded: in the case of Q unbounded ho = h + 1 - min h, h being the

§5. Capacities 441

function defining the null condition at infinity. We verify without difficulty that Ve is continuous, subharmonic on Q\K with compact support contained in Q and lim sup ve(x) ~ 1 for all Z E 8K, with the result that for k sufficiently large

x---+z

ve(x) ~ w on Q k \Kk for all WE Y+(Kk' Qd and hence Ve ~ Uk on Q k \K k • We deduce that v,(x) ~ lim inf Uk (x) for all x E Q \K. Letting £ -> 0 and taking the upper envelope of the v, we have

UK, Q (x) ~ lim inf Uk (x) for all x E Q\K ,

which completes the proof of the point (3). For the point (4) we consider Uj E ,c/}(Kj , Q;), i = 1,2 and put

v = max(u l , uz) = U I + (u z - Ult, W = min(u 1 , U 2 ) = U z - (U z - ult .

We have

v, W E (t&'0(~n), lim v(x) = lim w(x) = 0 , Ixl~oo Ixl~JJ

v :;:, I on K 1 U K z ,

O· h h h d f L 2 h d' . av rJW 1 n teat er an, rom emma ,t e envatlves -- ,- E L]oA~n) and aXj ax;

au

cJw ~

OX i

from which we have

~u] (I oX j

V E .'I'(KI U K 1 , Q] uQ1l, WE yJ(K] nK2' Q l nQ2) and noting that

I - X:u, > "I: = X{u, '" "I: ' we have

(5.62)

Therefore

capQluQ,Kl uK l + capQlnf2IKj nK2 ~ ,Y'(v) + ,.¢'(w) = J(u l ) + J(u 2 ).

Minimizing on U j E ,Y'(K j , Q;), we establish the point (4). The point (5) is immediate and for the point (6) it is sufficient to note that for U E 9(h(K), h(Q)) the function r defined on ~n by

v(x) = u(h(x)) for x E Q ,

is of class (6 1, t' :;:, 1 on K and

v(x) = 0 for x E ~n\Q

Igrad v(x): ~ I grad u(h(x))llh'(xll

442 Chapter I!. The Laplace Operator


capnK ~ 1(1') ~ flgrad u(h(x))1 2 Ih'(x)1 2 dx ~ c.f(u)

with Ih'(xlI2 c ,,= nlax ............ __ .

II Idel h'(x)1 o

We can now give the

Pro%/PropositioIl8.( and (3). Suppose, in the first instance, that UK.n is the solution of (5.59) and take an increasing sequence of regular bounded open sets Q k

containing K with union Q, and a decreasing sequence of compact sets Kk c Q k

with intersection K 119. From the hypothesis of regularity, the functions ---

Uk = UKk . fh E Y;~(Qk K k ) and from Proposition 7

From the Proposition 9.(3)

capnk Kk-+caPflK and uk(x) -+ lI K. fl (X) for all x E [R".

We deduce l20 that liK.flE B 1([R") and

(5.63)

Also for ~ E'.I(Q), ~ ;::>- 0. since ?ud?n ~ 0 on DKk

lim -

i.c. the distribution ilK.!l is sllperharmonic on Q. For; E '.I(K,12), ~ for k sufficiently large, and hence

I, d( - Ilu K . o ) = (-- Llux. n, ~> Also, always for ~ E'.I(K,12) ,

capa K .

(- fiu Ko .D = fgradIlK.ograd~dX ~ grad ux.allullgrad (,IL2

11 Y Such sequences exist (see Lemma I of 94). 120 For ( E0'([R;"j" .

Hence grad 1IK . .fl E' L I().!"j" and II grad UK. nill.' ,-;; (cap" K)I.'1 .

§S. Capacities

and therefore

capflK = <- JUK.[l,D ~ ,,1(UK ,fl)112 f(Ol/2.

Minimizing on ( E g(K, Q), we have, from the point (1)

capflK ~ ,,1 (UK. fl)1!2(capfl K)1!2

from which we deduce that cap fl K = ,,1 (UK. n) with (5.63).

443

In the case where UK. n is not a solution of (5.59), Q is necessarily unbounded. We approach K by a decreasing sequence of compact sets Kk with non-empty interior; from Proposition 11 of §4, there exists a generalized solution of P(Q\Kb XKJ satisfying the null condition at infinity. We can therefore apply the results of the proposition to uKb fl' Since, from Proposition 9(3), UKk . ,Ax) -> UK. a(X) we can pass to the limit as in the preceding case. D

Remark 3. As the limit of subharmonic functions (or distributions) in the neighbourhood of each point of IRn\K, the distribution UK. f} is sub-harmonic on IR"\K, AUK. [lis a positive Radon measure on IRn\K carried by aQ. In the case ofQ bounded, choosing ( E Q(lRn) with ( = 1 in the neighbourhood of d, we have

f d(JUK. [1) + f d(JIIK . [l) = <JUK.!b D = Ju K . [lJ(dx = ° . ("'f) flK

Hence, we have the equality which makes (5.52) precise

(5.64)

f'

capn K = J d(JuK • a) . fa

In the case of Q ullbounded, this equality is again true when IJ = 1 or 2. In effect, considering P c:C1(B(O, 1), B(O, 2)), the function Pk(X) = p(xik) -> 1 and therefore by Lebesgue's theorem

-- lim r grad UK. [2, grad fik dx . .., Hn

By Schwarz's inequality,

I t, grad UA.!}· grad Pkdxl ~ (Jk < i,l < 2k: igrad UK. d 2dxy!2 ,,1(pd12

kn - 2 f(p) and by Lebesgue

f I grad UK. d 2 dx --> o. :k < Ixl < 2k]

We deduce (5.64) in the case n = 1 or 2 where ,,1 (pd is bounded; in the case n ~ 3, ,,1(pd -> co and the argument falls. In fact the equality (5.64) is Ilot ill general true in the case n ~ 3. Indeed, considering Q = IR",?Q = 0 and so (5.64) can be written cap~,K = 0; now, as we shall see in the example below, when n ~ 3, the capacity of a compact set in IR" is often non-zero. 0


Remark 4. As the lower envelope of continuous superharmonic functions, the fimctioll UK, a is an U.5.(', map, superharmol1ic Oil Q (see Remark 8 of §4). Now the distribution UK, a being superharmonic on Q, defines a unique I.s.c. map LI K . n, stlperharmonic on Q, given by

(5.65) UK. a(x) = lim II i UK. dt)dt . r ---. 0 (J nrll J B(x. r)

We have obviously, UK. a = UK, f} on Q\}K; but in general, the set

(5.66) K= is strictly contained in K. We note that the set K is the set of the x E K points of continuity of UK, a. In particular, in addition to the interior of K, K contains all the Z E oK, regular points of the boundary of Q\ K (in the sense of Definition 4 of §4). We note also that K is not in general is compact set; in the following sub-section, we shall define the capacity of an arbitrary part in an open set in [R1n, and, with the help of Proposition 12, we can see that cap!} K = cap!} K: the capacity of K depends only Of! the part K ol K. 0

Remark 5. Given K a compact set in an open set Q in [R1"

(5.67)

In effect if IV is continuous, superharmonic on Q ::::0 K,

\V ? min w on K therefore Y'{ (K. Q) = yf (i'K. Q) ,

,'I( UK,!} = U,'K, nand from Proposition 8, caPnK = caPa<"K. With the notation of Remark 4, we obviously have also UK, fl = Ua .. , f1 and therefore

,-~/

(5.68) r'K = K n i'K . o Finally let us give some examples.

Example 9. Capacity ora compact set ill all open set of[R1. From Proposition 9 (5), we can restrict ourselves to calculating the capacity of a compact set K in an open interval Q = ] a, h[ of R Putting (11 = min K, h1 = max K we have

cap{]K::::; cap{][a1,bd =c cap.a{a1,bd ::::; capnK,

the Ilrst and last inequalities resulting from the increase of the capacity with respect to K (Proposition 9.(1)) and the equality in the middle from Remark 5.

From Example 7, we thus have for every compact set K c:: ] C/, h[

(5.69)

CaPla,h[ K caPJd,hL[min K, max KJ +------min K - (I b - max K

§5, Capacities 445

\

QF .- X

Fig, 16

In fact, this equality has been proved for

- aJ < a < min K < max K < b < aJ

but from Proposition 9 (3) it is still true in the general case, Similarly uK,a = u[al,bll,nwhose graph is given in the <!,ifferent cases in Fig, 16, Finally, we note (see Remark 4) that UK, n = UK, nand K = K, 0

Example 10. Some compact sets of zero or non-zero capacity, (a) In the case n = l,from Example 9 above

cap n K = 0 iff Q = ~ or K = 0 .

(b) In the case n = 2, it follows from Remark 3 above that

capQ;l2 K = 0 for every compact set K of ~2 121.

This follows also because of Example 8 and of Proposition 9, on putting

r = max Ixl. with K

lim capRIO. '0) B(O, r) TO ---+ X

lim 2n

TO ---+ ex: r 0 Log

r

0,

(c) In the case n ): 3. for every point Xo of an open set Q of ~n, cap a {xo} 0. In effect, considering ro > ° with B(xo, ro) c Q,

capn {xo} ~ capR(xu.ro){xo} = lim capR(xQ,ro)B(xo, r) , - 0

= lim -----, = 0 , r - 0 En{ro) En{r)

More generally, given Va regular manifold oj'dimension n - 2 in an open set Q of ~n and K c V. then capn K = 0.

121 In IR and 1R2, we must not confuse the sets of zero capacity with the polar sets for the H' norm (for definition see later): because a compact set has zero capacity in i:£ and 1R2, even though it is not obviously polar in general.

446 Chapter [I. The Lap[ace Operator

Indeed, from the sub-additivity, we can suppose that there exists a diffeomorphism h of an open neighbourhood U of Kin Q on [Rn = [(x', x"); x' EO [Rn-2, x" EO [R2}

such that

h(U n V) = {(x',O); x' EO [Rn-2}.

Considering a bounded open set Qo containing K with Qo c U, we have from Proposition 9 (6)

capa K :( caPao K :( ccaph(fJo)h(K).

We are thus led to proving the result in the case Q = Q' x Q", K = K' x {O}, where K' is a compact set of an open set Q' of [Rn 2 and Q" is an open neighbourhood of {a} in [R2. Let us fix u' EO !:0(K', Q') and for ° < B < dist (0, oQ"), take

U" EO !l'( {a}, B(O, e)l such that r [grad u"(x"Wdx" < I; ; Jj,j2 such a u" exists since capB(O.E){O}. Putting u(x', x")=u'(x')u"(x"), we have u EO .0tJ(K, Q) and

J'(u) = (L, , [grad u'(x'WdX')( L, u"(x")2dx")

+ (r U'(X')2 dX')( r [grad U"(X"l[2 d X ")

Jj,j", JR' ~ ~

:( m:2 J [grad u'(x')[2dx' +£ J u'(x')2dx' j,j' -, w

Letting [; --> 0, we have cap a K = 0. 0

(d) On the contrary, let V he a regular hypersur/clce (manijold of dimension n I) K a compact set containing V and Q a bounded open set containing K. Then capa K > O. Transforming a piece of V by a diffeomorphism, we can always suppose that K contains the hemisphere

2:'+ = {x = (x',xn); Xn = (l - :X'(2)-1-,[X'[ < 1}

and hence that K u ( - K) contains the unit sphere 2:'. Considering ro

we have

capB(O.rO)2:' :( capQu(_QlK u (- K)

sup[X[, {J

:(capaK + caPIQ)( - K) = 2capaK,

where, in order, we have used the relation (5.55), Remark 5, Proposition 9, (1) then (4) and the obvious fact that

o (e) [n the case n ;:, 3, we have the result of (d) for an unbounded open set: for every compact set K of [R" containing a regular hypersurface, caPwK > O. In effect, with

§ 5. Capacities

the same reasoning, we are led to cap~"B(O, 1) > 0; now from (5.55)

cap~"B(xo, r) = lim capB(xQ,ro)B(xo, r) TO ---+ X

3b. Capacity of an Arbitrary Part. Applications

447

Given an open set Q of IRn, we have defined cap n K for every compact set K of Q; this set function K ~ cap n K presents some some similarities with a measure: it is increasing, continuous for decreasing sequences of compact sets, sub-additive (see Proposition 9). But note clearly that this is not a measure: it is not additive (that appears clearly in Remark 5 and in Examples 9 and to), We propose, given a part A of an open set Q of IR", to define cap n A as the infimum of the energy functional.~ (u) on a family.? (A, Q) of elements u E Bl (IR") satisfying "in a certain sense"

u :::; 0 on IR"\Q, u ~ I on A.

We obviously want this de11nition to generalize that given when A is compact; but we also want the set function A E ,:J'(Q) ~ caPnA to be an exterior measure (in the sense of measure theory), that is, satisfies the following condition:

(5,70) {for every increasing sequence of parts Ak of Q with union A,

the sequence caPnAk is increasing and converges to caPnA.

We note that this condition imposes the value caPnUfor every open set U ofQ or more generally for each part U, the denumerable union of compact sets 122 of Q: we must necessarily have

(5,71 ) cap n U = sup cap n K (K compact) , K c U

We note also that in the case n ~ 2, this condition imposes that we cannot, in

general, define cap!1 A as inf..1 (u), where .'i(A. Q)

.'/'(A, Q) = {u E (&,o(IR") n Bl (lR"); u :::; 0 on lR"\Q,

lim sup u(x) :::; 0 and u ~ 1 on A} . Ixi ~ x,

It is clear, in effect, that Y(A, Q) = Y(A, Q); given a compact set K of Q and a sequence (xk) dense in K, the sequence of compact sets Kk = {Xl' ... , xd is increasing with union A dense in K; the sequence capn Kk = 0 does not converge, in general, to

inf J(u) .'I'(A,Q)

122 We say then that U is aKa.

inf J(u)=caPnK . Y(K, Q)


We must therefore to define cap Q K, at least in the case n )'; I, go out of the space of continuous functions. The problem is then posed with the meaning: 11 ~ 0 on IR"\Q and 11 )'; 1 on A. In effect, a priori, 11 E Bl(lRn) is defined almost everywhere on U:g"; on the other hand the inequality u ~ 0 a.e. on IRn\Q or u )'; 1 a.e. Oil A is riot, in general, sufficient. We see this easily in the case n = I: (1) 11 := 0 satisfies 11 )'; 1 a.e. on (O} since the point 0 is negligible and J (u) = 0 < cap D{ O} for every open set Q # IR, 0 E Q. (2) Taking Q = ] - 2,2[\ {OJ and K = {-I, + I}, let us consider u whose graph is shown in Fig. 17. We have u ~ 0 a.e. on IR\Q since 0 is negligible and

.f(u) = 2 < capflK = caPJ-2.0l:- 1: + cap]o.2cll} = 4.

In order to specify the family .'F(A, Q) on which we shall minimize the energy functional we must introduce a definition of a limit in the space Bl(lRn). We shall say that a sequence (Uk) converges to u in Bl(IR") if

{ lim [ ". 1I1 2 dx = 0 for all R > 0 and

(5.72) .8(0 RI

lim 11 grad 11k - grad 1112 dx O. w

It is clear that Uk"' 11 m Bl(lRn) = .f(ud ---> .1(u)t23.

Therefore for every family § c B1(lRn),

(5.73) lim f(u) = lim .f(u)

where .¥ is the closure of.F in B1(lRn), that is, the set of limit points in Bl(lRn) of sequences of elements of .F. On the other hand, let us recall Lemma 2, which we shall complete by the

Lemma 3. For all u E B1(lRn), 11+ = max(u, 0) E Bl(lRn) and the map 11 ---> u+ of B 1 (IRn) into itsele 24 is continuous.

/ \: '" 2 -I 2 x

Fig. 17

12.' The converse is true with the reservation that there be quite a small measure of convergence of Uk to U. To the precise: we can show that if there exists a non-empty open set (t) such that Uk -+ U in .CI'(w) then .f(ud·' .f(/I) = Uk -+ U in Bl(R;n).

124 We note that this is not a linear map.

§5. Capacities 449

We shall prove this lemma with Lemma 2 at the end of this section. In order to express the condition u :::; 0 on !R"\Q, we now introduce the Beppo-Levi space B6(Q), as the closure of g(Q) in B1 (!Rn). In the case when Q is bounded from Poincare's inequality125, we have

II u Ilu :::; qQ) II grad u IIL2 for all u E .@(Q)

the space B6(Q) coincides with the Sobolev space II 6(Q), the closure of g(Q) in HI (!R")126; but the situation is not the same in the case in which Q is unbounded l27 .

We now give the

Definition 3. Let Q be an open set of !R", and A an arbitrary part of Q. (I ) We denote by g;- (A, Q) the set of U E B 1 (!R") satisfying:

u+ E B6(Q) and u ~ 1 a.e. in the neighbourhood of A.

(2) We define the capacity of A in Q

with the convention

inf f(u) 3"(A,Q)

if g;-(A, Q) = 0 .

In the case of a compact part K, this definitiori coincides with that of Definition 1: this is what we prove in the

Proposition 10 (1) For a compact part K of an open set Q (~f!Rn, capD K in the sense of Definition 1 coincides with that in the sense (if Definition 3. (2) For every increasing sequence (if parts (Ad of an open set Q, with union A, the sequence (cap.Q Ad is increasing and converges to cap.Q A.

Proof Proof of the point (/). We have trivially .Q(K, Q) c .~(K, Q). We take therefore uo E .~(K, Q); let U be an open neighbourhood of Kin Q such that Uo ~ 1 a.e. on

125 See Chap. IV, §7. t!u

126 H' (1M") = {u E L2(1M"); ._ ... E [2(W) for i = 1, ... ,n} = B' (IW) n L 2(1M") is a Hilbert space with ?Xi

norm II II, defined by

Ilu ~, = J' (luI 2 +lgrad ul 2 )dx (see Chap. IV). IT~n

127 In the case II ?o 3, BMW) = B'm") n L 2"""-2)(IM"): that follows from Sobolev's inequality (see J. Lions [1], p. 33)

1/11 L2,"" 2) ~ C II grad u ilL" Vu E g(IM") .

The situation is more delicate in the case n = I and n = 2.

450 Chapter ll. The Laplace Operatm

U and Uk E (/(.12) such that Uk --+ u6 in Bl([RII); we fix P E Y(K, U) and put

i'k = Ii, V P = 11k + (p - 11k )" .

We have rk E Y'(K, Q) and from Lemma 3,

['k ---> 1I~ + (p - 1/<))+= u; 111 B 1([RII).

Therefore

inf f(u):( liminf,1(I'd = f(lI~):( f(uol. D ·/If:. \I)

Pro%fthc point I :: I. It is immediate that capn A is increasing with respect to A. In the sequence capQ Ale is increasing with limit I :( capn A. Let us suppose I < capn A and fix I: > 0: since ('k = capll Ak :( I < y., there exists IIkE.F(A"Q) such Ihat f(lId:( ('k + 1:2- k : we can suppose 0:( 1I k :( 1 on rcplacing lik by min (U kf

, I). Wet hen define by recurrencc

We have 1'[ :( 1"2 :( ... :( r k :( ... "'; I. ric E .F(A k , .12), \Ik E:T(A k , Q) . Now from (5.62),

so

and by recurrence k

f Ud :( C/c + I: I 2 i i 1

We deduce I = lim I'k E Bi,(Q) and j' (1') :( I + I:. On the other hand 1" ~"' rk ~ I a.e. in the neighbourhood of Ak , for all k. Hence l' ~ 1 a.e, in the neighbourhood ~A. D

Remark 6. By definition] (A, (2) is the union of the .F (U, Q) for all the open sets U of Q containing A, hence by definition

(5.74) inf capnU (Liopen). ..t c" [ r !2

the capacity of an open sct U of Q being given by (5.71 )12H

capn U = sup capu K (K compact) . K c U

We say thai a part A is capaci/uhlc in Q if

cap fl A = sup cap!l K (K compact) , K c A

1." SinL'" from Ihe point 121 of Proposition 10. capJlis an extenor measure.

§5. Capacities 451

that is, more explicitly, if

(5.75) { for all G, > ,0, there exis,ts ,a co~pact ,set K" and ~.n o~en set U, of Q suchthatK, cAe U,andcaPnU, ..... caPnKr ~ I,.

Every open set, or more generally, every Ka 129 of Q is capacitable.

The capacitahilit)' theorem ofChoquet l30 shows that erery part ofQ yvhich is a Borel set is capacitahle in Q, 0

Remark 7. We can generalize the properties (1), (4), (5) and (6) of Proposition 9,

In particular, the capacity is stl'on{}lr slIh-additit'e: for every part Ai of an open set Qj. i = 1,2.

(5.76) capn,u{}2AtuA2 + caPn,nn,AI(lA 1 ~ capn,A I + capn,A 2 .

The proof is identical with that of Proposition 9 by noting that, from Lemma 3, for Uj E BI([RIl), i = 1.2

11/ E BMQj ) => max(lI t , tl2)+ E Bb(Ql u (2) and

min(u t , u2 )+ E B6(QI (I Q2)

On the other hand. Property (3) oj Proposition 9 does not qeneralize to a decreasing sequence of arhitrary parts of Q: for example let us consider in the case 11 = J

Q = ] - 2 , + 2[ and U = J -I -I + .~ [uJ I - ~ 1[' k ' k k ' ,

we have (using Example 7 or 9 and (5.71)), cap n Uk = 2. even though the sequence (Uk) is decreasing with empty intersection. 0

Given A a part of an open set Q of [R", by definition cap" A = 0 iff

(5.77) {for all /; > 0, there exists tlr E BA(Q) such that

tI, ~ 1 a.e. in the neighbourhood of A and flu,) ~ r:.

We note that this property is local:

(5.78) { c~p nA = ° iff for ally E Q, there exists an open neighbourhood U

of x 111 Q such that capn(U (I A) = 0,

since, Q being the denumerable union of compact sets,

capn A = 0 iff caPa(A (I K) = 0 for every compact set K in Q

and from the sub-additivity (see (5.76)). if U I, ... ,Uk cover K

k

capa A n K ~ I capn U j .

j I

129 That is to say (see the beginning of thIS section 3b) a denumerable union of compact sets of Q. 130 See Choquet [1].


When A is capacitable in Q and, in particular, thus from Choquefs theorem (see Remark 6), when A is a Borel set,

(5.79) {cap!J A = 0 iff for all x E A, there exists an open neighbourhood U

ofxinQsuchthatcap!J(UnA) = O.

In effect, then by the definition of capacitability

cap!J A = 0 iff cap!J K = 0 for all compact K c A.

This shows that the property cap!J A = 0 is independent of the bounded open set Q containing A (for A bounded). More precisely, we have

Proposition 11. Let A be a bounded part of [Rn. The j()llowing assertions are equivalent: (i) there exists a bounded open set Q containing A such that cap!J A = 0; (ii) for aery open set Q of [Rn, cap!J (A n Q) = 0; (iii) there exists a Borel part A contaillin?] A such that jor all [; > 0, there exists u E HI ([Rn) satish,ill?]

u ;: 1 a.e. ill the neighbourhood of A alld il u ~H I :::; I;.

Proof The implication (ii) => (i) is trivial. For the implication (i) => (iii) we note first, from (5.74), that if cap!J A = 0, there exists a sequence of open sets Uk ::::l A such that cap!J Uk -+ 0; A = n Uk is a Borel set containing A with cap!J A = O. Applying Poincare's inequality 131, we have (iii).

Finally we prove (iii) => (ii). Replacing A by A we can suppose A is a Borel set; from (5.79) (and (5.77)), it is sufficient therefore to prove that for every open set Q of [Rn, all Xo E Q n A and all [; > 0, there exists u E B6(Q) and r > 0 such that 11;: I a.e. on B(xo_ r) and .f(1I):::; f:. Let us fix Q, XO E Q n A, o < 1"1 < 1"0 = dist(xo ' i'Q) and C E v(B(xo' r 1 ), B(xo' ro)); given I; > 0 (using (iii)), there exists lie E Hl([Rn) and re > 0 such that lie ;: 1 a.e. on B(xo, r,)

1"1 2

and Illd HI :::;- '_ -. It is then clear that u = :;u, and I" = min(r,,!"l) are II grad (" III.'

suitable choices, since

.f(u) = flu,gradC + Cgrad 1I,1 2 dx :::; Ilgrad:;lll. IIII,II~I :::; /;. 0

This leads us to frame the

Definition 4. (I) We say that a part A of [Rn is polar 132 if cap!J(A n Q) = 0, for every open set Q of [Rn.

(2) We say that a map II: D c [Rn -+ IR is :::ero quasi-everywhere (we denote this by u=O q.e. on D) if there exists a polar part A such that II = 0 on D/A. Similarly, we define 111 = 112 q.e. on D, u1 :::; u2 q.e. on D, etc ....

13\ See Chap. IV. p. 132 It is polar with respect to HI (IR") (see 1. L. Lions [I] J.

§5. Capacities 453

Remark 8. The idea of a polar set and of equality quasi-everywhere is closely related to that of a negligible set and of equality almost everywhere. In effect, every denumerable union of polar sets is polar and also every polar set is contained in a polar Borel set. A polar set is obviously negligible and the converse is obviously false. In the case n = 1, a polar part is empty and equality quasi-everywhere is equality everywhere. In the case n ~ 2, a polar part A contained in a regular hypersurface V is negligible for the surface measure of this hypersurface (this follows from the trace theorems I33 ), but the converse is false l34; also a part contained in a regular manifold of dimension n - 2 is polar (that results from Example 10c). 0

We now give the application of these ideas to the obstacle problem.

Proposition 12. Let Q be an open set of [Rn and A a part of Q withfinite capacity;

(1) cap il A = min f (14) where ff + (A, Q) is the set of the u E B6(Q) superharmonic .>"(A. Q)

on Q and whose I.s.c. superharmonic representative ij135 satisfies il ~ 1 q.e. on A; (2) if cap Q A > 0, there exists a unique u A.!1 in ff + (A, Q) such that cap il A = .J' (u A. ill· In addition UA . il is characterized by

UA.ilE.'F+(A,Q), f. d(- LlU A . il ) 0, JQ\A

and satisfies 0,:;; u A. il':;; 1 a.e. Also il A • n is characterized as the smallest l.s.c. superharmonic map. il: Q -> [O,lJ satisfying il = I q.e. on A.

We shall not prove this proposition, referring the reader, for example, to KinderlehrerStampacchia [I r 36 for this and for other developments of this problem. Let us give now, still without proof,136 Wiener's criterion of regularity of the points on the boundary of an open set in [R".

Given a bounded open set Q in [R" with 11 ~ 3, and z a point of its boundary T, the following assertions are equivalent: (i) for all (P E (,.go((1Q), the generalized solution u(ip) of P(Q, (p) satisfies

lim u(ip,x) = ip(z)

x

(ii) I 2k (n - 2)capf,l,([R"\QnA(z,2- k )) = + oc k~l

where: A(z, r) = B(z, r)\8(z, r/2) .

133 For all U E U(G;ln), Ilu(.,OlilL'(W ') .:; Ihl(.,O)IIH',2(W') .:; IlufIH'(W) (see Chap. IV, ~4). 134 This follows from the theorems on lifting of trace; in particular, we show that the Cantor sct K. negligibly compact on G;l. is. identified with K x {a}, of strictly positive capacity in every bounded open set Q containing it. 135 Sec Definition 15 of §4. Note that this implies that il ~ 0. 136 See Wiener [1].


Finally we conclude this section with the

P if JL ? G' 1 ( . h CU 1 roo OJ emma ~. Iven u E Lice Q) Wit ex. E Lloc(Q), we have to show that for I

all ( E £((Q),

(5.80) f ~i( u+ dx = --- Jr' (~~- dx _ ( Xi {" > 0: eXi

A pplying to - u and subtracting, we obtain

thus

which will prove (5.61).

f DC ~--= udx (Xi

L ---

cu (, ~ dx 0,

O} ()X i

First, let us prove that for each bounded continuous function p: ~ ----> ~

(5.81) J' (Y r cu ~ Sj(u)dx = - p(u)~ (dx (Xi ~ ('X i

where j(r) = L pis) ds is thus a Lipschitzian function of class ({; I. The relation

(5.81) is obviously true if Ii E '6 I(Q) for then

To obtain with the hypotheses U E L/oc(Q), au/ax i E LI~c(Q), it is sufficient to regularize u; there exists a sequence of functions Uk such that Uk ----> U,

(iuk/ex; ----> (lu/ax; in LI~c(Q) and a.e. on Q; since j is Lipschitzian, j(ud ----> j(u) in L/oc(Q) and

p being continuous, p(uk ) -> p(u) a.e. on Q; p also being bounded, we have by aUk cu

Lebesgue's theorem p(uk ) -0- ----> p(u) -~- in LLc(Q) and OXj eX i

Passing to the limit in the equation (5.8\) for Uk' we obtain it for u. Equation (5.81) is still true for every simple bounded function p: ~ -----> u;g, i.e. p(r) = lim Pk(r) for all r E u;g where Pk E <;&,O(u;g), which we can always suppose is uniformly bounded. In

§S. Capacities 455

effect then, by Lebesgue's theorem,

au au Pk(U) ax; -> p(u) ax; In Ll1oc(Q)

and

f Pk(U) :;; (dx -> f p(u) :;; (dx ;

also j(r) = L p(s) ds = lim jk(r) are uniformly Lipschitzian, with the result that

jk(U) -> j(u) in Llloc(Q) and

Passing to the limit in the equalities (5.81) for Pk we obtain them for p.

Applying (5.81) to

{I if r > 0

p(r) = 0 l'f 0 r';:; ,

p 1""- -

/' I ;' I

,("P. I .' I

" I '" I

Ilk

Fig. 18

we obtain (5.80). Note that P is clearly a simple function as is shown by its graph (see Fig. 18) and that

j(r) = L p(s)ds = r+ . o

Remark 9 (1) Applying (5.81) with P = X{e} which is a simple function as is shown by its graph in Fig. 19.

"', I "

I " " ,p, " , I \

'" " ./ ....

c -- -k

c

Fig. 19

I c + -

k


d 1 ir d .. f eu v d an lence j(f) = pis) s == 0, we obtam agam o~ ~ x = O. o ;u=c;c,xi

In fact (5.81) is true for every Borel function and even for every measurable function. Applying to p = h where K is a negligihle set oJ'!?!., we obtain iOu/ax; = 0 a.e.oll [x: u(x) E K: (2) The same argument as in the proof of Lemma 2, leads to Kato's inequality: given u E Lioc(Q) with Ll U E LI1oc(Q) (in the sense of distributions), Ll (u +) is a Radon measure on Q, minorized by the function X:" :> 0: ~1u; in other words

(5.82)

Using the same function p as in Lemma 2, Kato's inequality (5.82) is a deduction from the more general inequality

(5.83) ,1j(u) ? p(u)l1u In CI'(Q) ,

true for every increasil1Y bounded function p: '!?!. -+ R In the case U E '6'l(Q)

l1j(u) = p(u)c1u + p'(u)lgrad ul 2 ? p(u)Llu on Q,

By double approximation as in the proof of Lemma 2, we obtain (5.83) in the general case. 0

Finally, we give the

Prooj oj" Lemma 3. Let Uk -+ U in Bt('!?!."). We have

lu:-u+I~luk-u: andhence limf luk+-u+1 2 dx O. JB(o. R)

Now from Lemma 2

(~

- ···~u

iiXi

from which it follows that

II grad uk+ - grad u + II L' ~ II grad Uk - grad u il [,2

+ (JI'I-~u [2 lx,,, > 0' - X'u > 0' 12 dx)i (X,! l I.. I J

But we can suppose that Uk -+ 11 a.e. with the result that

lim sup Ix: u, :> 0: - X:" :> 0: I ~ X:u - 0: a.e.

i'u Making use of point (2) of Lemma we have (~X X:" ~ 0; o a.e. and, as a

. I

consequence, by Lebesgue's theorem,

f(~U)2 Ix;", :> 0; - X:u > 0: 12 dx -+ O. LX,

o

*6. Regularity 457

§6. Regularity*

1. Regularity of the Solutions of Dirichlet and Neumann Problems

In §3.2, we have studied the local regularity of solutions of Poisson's equation

Llu = f on Q

as a function of that off In particular, we have shown (see Proposition 6 of §3)

(6.1) fE Lfoc(Q) with P > ~ (resp. p > n) => U E '6'0(Q) (resp. '6'1(Q))

and also (see Proposition 9 of §3) that for mEN and 0 < () < I,

(6.2)

These results establish the interior regularity of the solutions of the Dirichlet and Neumann problems. We propose now to study the regularity on the boundary of the problems P(Q, cp, f) and PN(Q, !/J,f) as a function of the data Q, cp, if;,f An important aspect is that,just as with interior regularity, regularity at the boundary is a local property. To clarify this assertion, we state the

Theorem 1. Let Q be an open set of ~n, ro a regular open part of class '6'co of its boundary cp E '6'CO (r 0) (resp. if; E (t x (I' 0))'/ E (too (Q u r 0)138 and

u E '6'°(QuFo) (resp. '6',~(QuI'o)139

solution

(6.3) rU~f III .@'(Q)

i'u if;) U (p ( resp. ~n on I'o ·

Then U E '1i co (Q u r o) .

We shall prove this theorem and other regularity results which we shall state precisely as they arise in the course of this section. We, first of all, change the statement of this theorem. Under the above hypotheses, we know already that U E '6' x (Q) (see Proposition 1 of § 3); it is sufficient, therefore, to show that the derivatives of u extend by continuity to each point ZEro. This result being local, we can always suppose that

(6.4) {Q = {~x',,:X(X:~ .~ :; x' E (!', 0 < t < b} ro = dx,:X(x»,x E(!,}

137 See Proposition 9 of §3 for the first use in this chapter of functions of class 'i\m.O

138 That isJ E ,/\,XC (Q) and all its derivatives extend by continuity to rD.

139 That is, U E '/\' 1 (Q) and Du/Dn(z) exist for all z (C r 0 (see § t.3b).


where (! is an open set of ~"-1, b > 0, '1. E (6 o(t') and (x',x") denotes the coordinates in an orthonormal reference frame in ~n. We can always take:: to be the origin in this frame and take n{z) to be the last of the base vectors in the reference system, with the result that

(6.5)

We put

We have

o E (!. x(O) = 0, grad x(O) = 0 .

£I(X', t) = lI(x', XIX') - 1)

II E ((,X ((! x J 0, () [) and

(6.6) n _. 1 ? (elll ?x DII) ? (" - 1 ('X (iU ') e2 i! _ I ----. -:1-- +-:- - -- I- ..... + (I + igradxl2) ~ = f i = 1 eX i (Xi eX i (It at i I (":Xi ?Xi ct

with lcx', t) = fix', xIx') - t).

In the case of the Dirichlet problem, we have U E '(,O((I X JO, be) and

u(x',O) = c,O(x') = <p(x',x(x')) for all X' E (! .

In the case of the Neumann problem, by the definition of a / DIl, denoting

we have

where

Bu E 'fj0(C x [0, be) and Bu(x',O) = fix') for all x' E C

~/(X') = _. (I + Igradx(x')!2) 12 f(x', x(x')).

Using the hypothesis (6.5) and changing the notation, we see that the theorem therefore reduces to the following lemma:

Lemma l. Let (! be an open neiyhborhood o{ 0 in ~n - 1, b > 0, ai) = aji and l all be/onyinq to ((, Y (C x [0, b C), ~? E '(,' ((!) (resp. tj; E ({, x (C)) and 1/ E ((,' (t x J 0, b [ sat isfyiny

(r x JO, 6 [

lIE((,O(t x [O,b[) and 1/(x',O) = <p(x') lorall X'EC

I'll (resp. Bli = Iain -1--E(6o(t x [O,6[) and BlI(x',O) = tj;(x')).

(Xi

We slippose that aij(O) = bij (the Kronecker delta). Then 1I is (){ class '{,. x. in the neiyhbollrhood o{ O.

For the theorem, from (6.6), we shall apply the lemma with

ror i, j = I, ... , 11 -

(6.7) for i = 1, ... , n - I

*6. Regularity 459

There are many approaches to the proof of Lemma 1. We shall commence, for Dirichlet's problem, by the use of the AgmonDouglis-Niremberg method, based on the Calderon-Zygmund estimate, which we recall (see Proposition 8 of § 3): for 1 < p <x;.

(6.8) for all u Eft (~n)

where C is a constant depending only on p and n. We shall now prove the

Lemma 2. Let n < p < XJ and aij = aji E L C()(~n+ )140 satisfying the condition that bij ~ aij is of bounded support, and

IIIIbij ~ au IlL' < (2C) 1

where C is the constant of (6.8). Then ji)r all.to, ... ,f;' E LP(~,,+-) with compact support, there exists one and only one solution u oj'the problem

and its derivatives

(6.9) in

U(X',O) o for all x' E ~n-l and lim u(x)

In addition eu

(a) the derivatives;i- E LI'(~,,+-), DX i

i3 f~ (b) If,formE N ,fa + Lo'-E Wm'P(~,,+-)and

uXk

Ixl ~ ex.

aijE wm+1.x(~n+) for i,j = 1, ... ,n,

thenuE~m+1(~n-1 x ~+)anduE wm+2,p(~,,+-)142.

O.

Let us show, first of all, that this lemma implies Lemma 1 and hence Theorem 1 for the Dirichlet problem:

Proof of Lemma J (The Dirichlet problem). We shall note first that we can always suppose that O)andlR+ = [O,oc[.

r Eu ,0" r r D( 14' That is to say I I au ~ ~ dx + /;,( dx = If. ~ dx,

"" ('Xjt'X.j .. ,., (Xk

142 w m . PI Q) is the space of the functions f E L P( Q) such that the derivatives (in the sense of distributions) rxP E U(Q) for all fJ == (fi, .. .. , fin) E: N n with IfJl = fJl + ... + fin ~ m.


In effect, the function u(x', xn) = u(x', xn) - <p(x') satisfies

L L ~ (aij au ) = f - fi: ~ (a ij ~<P ) aXi aXj i,j= 1 aX i oXj

and u(x', 0) = 0. Finally u is C(,' x. in the neighbourhood of 0 iff this is true ofu. We can thus suppose that <p == ° and this we now do. We fix n < p < (fj and ° < e < (2C)-1 where C is the constant of (6.8). Using the continuity of aij and the hypothesis aij(O) = i5 ij , Vi,j = 1, ... , n we see that there exists

° < 1'1 < dist (0, a(l,) and ° < 15 1 < 15

such that laij - i5ijl ~ e on B(O, I'd x [0,15 1 ]. Let us take ° < 1'0 < 1'1'

0< 15 0 < 15 1 andpoE(f,"'(~n-1 x ~+)withO ~ Po ~ I,

sUPPPo c B(O, I'd x [0,15 1] and Po == I on B(O,l'o) x [0,15 0 ],

We put aij = i5ij + po(aij - i5 ij ) with the result that aij E C(,' x (~n - 1 X ~ +),

supp(aij - i5ij) C sUPPPo, bounded,

Ilaij - i5u11I.' ~ Ilaij - ()ijlll.'(suPPPol ~ f..

We shall prove that U E C(, x, (B(O, 1'0) x [0,15 0 [) or, what is its equivalent, that

(6.10) {for all m,E N and all P E ({, '(~n -1 X ~+) with

suppp C B(O, 1'0) x [0,15 0 [' p!lE({,rn(~"-1 X ~+).

Given P E ({, x (~n - 1 X ~ +) with supp p C B(O, 1'0) x [0,15 0 [, the function ii = p!l satisfies

iiEC(,x(~n+)nC(,'O(~"-1 X ~+), il(x', O) = 0,

suppu C B(O,r) x [0,15 0 [' and hence, in particular, lim i/{x) = °

with

( 6.11)

(single p is involved since Po == I in the neighbourhood of supp p). From Lemma 2, u is the unique solution of (6.9) corresponding to aij and .~. Hence

ri-

the derivatives ~ E LP(~n+). This being true for all p, we have aX i

~6. Regularity 461

Using a recurrence argument and supposing that for a certain m ;;: 1, we would have U E W~,/(B(O, ro) x [0, bo[). Then for arbitrary fixed p,

(6.12) 1o+'I:~ Pf+U'I'Ia:j(aij:.:J GU up 7" __ Wm-1.p(rrlln) + - L, L, aij ~ ~ E Lm +. , ox, CX j

and hence from Lemma 2(b), i'i is of class ({j'm on !R n - 1 X !R + and U E wm + l,p(!R"t.). This being true for all p,

uEC(fW(B(O,l'o) x [O,()o[)n W~c+l.P(B(O,roJ x [O,boD.

By induction on m, we therefore have the stated result. o Going back to the reduction of Theorem 1 to Lemma 1 and the above proof, we have in fact proved the

Proposition 1. Let Q he an open set of !R", 10 a regular open part of its houndary, n < p < ex; and U E '(J'O(Q U 1 0 ) satisfying u = 0 on 10 ,

(1) Let us suppose that 10 is of class W 2, P and thaI

. cl.A Au = fo + 'I.-, - in 9'(Q)

(}X k

withfo, ... ,j~ E Lfoc(Q U 10)143. Then U E WI~'/(Q U 1'0)' (2) For mEN, let us suppose that 1'0 is of class wm + 2, wand

,;1 U = f in 9 , ( Q)

Remark 1. Proposition 1 expresses the regularity wm,p up to the houndary.f;)r the homo!Jeneous Dirichlet problem (the interior regularity has been seen in Proposition 8 of § 3. We note that the restriction n n; to generalize Proposition 1 (and Lemma 2) for I < fl ~ n, it will be necessary to interpret the condition U = 0 on 1'0 in a weak sense. The case of the inhomogeneous Dirichlet problem, with a condition U = cp on 1'0 can be deduced immediately, as we have seen in the proof of Lemma 1, jf there exists a "lifting" Uo having the desired regularity W~,/(Q U 1'0) such that Uo = cp on 1'0' The characterization of the functions cp defined on 1'0 admitting a lifting in

14,\ Given a part E of IR", Lio,( E) is the space offunctions I (defined a.e.) measurable on E such that IfIP is integrable (resp.[ bounded if p = + cx») in the neighbourhood of each poinl of E. The space Lio,(Q u ro) must not be confused with Lf,,,(Q): the first is the set of IE Lfo,(Q) such that ifl P is integrable (resp. I bounded) in the neighbourhood of each point of ro' The space W;:/(Q u To) is the space of the IE Lio,! Q u 1'0) all of whose derivatives (in the sense of distributions in Q) up to order m are in Lfo,(Q u roJ· A regular manifold To is of class W"'·P if in the local parametric representations (U, R, (!, ex) we have Y. E W::;/ur.).

144 See Chap. IV.


w~/ (0 u 1'0), involves fractional Sobolev spaces: these concern trace theorems which we shall not take up here (see Chap. IV). Proposition 1 has its equivalent for Holder regularity up to the boundary; we nole that for this regularity, the trace theorems are simple. We have the following result which we shall prove for the Neumann problem (see Lemma 4):

Given m E 1\1 and 0 < () < II E «(,' 0 (0 u 1'0). solulion of

(supposiny ro oj' class «(, m ~ 2. e). a ./imction

LlII = f in 0"'(Q).

isincr;2m+2.0(QuFo ) iff fE(f,m.fi(Ouro ) and (pE({;m+2.8(I·o)· 0

Finally, we note that when Q is bounded and To is the whole boundary of Q, these results give a global regularity on Q: we express this in:

Corollary 1. Let n < p < x; 0 a reyular hounded open set in lR n with houndary r of class w2 • P and to..r; .... .In E U(Q), There exists a (unique) quasi-classica[l45 solution u of the homogeneous Dirichlet problem PH(O..r~ + L 2j~/(~Xk)' In addition 11 E W!'P(Q) and

If for mE ... (lj~ 'm p

T is of class VV rn - 2 . f and j =./0 + '-:;E W . (0), then L.. tXk

U E W m + 2 . p (Q) and IIIIWrn.l") ~ CII.f':iwrn.['.

Proof of Corollary I. The problem PH(Q,j~ + XDA/Dx k ) has a (unique) quasi-classical solution: considering " 0 •... , Un the Newtonian potentials of j(~ X n, .... f" 1.!2' we know that Uk E (f, 1 (lR") with the result that r = Un + X (Uk / (~Xk E (f,' 0 (lR"). Considering w the solution of P (0, <p )

where <p is the trace of l' on T, U = v - w is the quasi-classical solution of PH(Q,fo + LcA/cxk )

From Proposition I, U E WI.P(O) = WI~/(O) and if r is of class wm + 2. J and f =.f~ + XaA/rXk c Wm·p(O), then U E W m+ 2.P(Q). The existence of a constant C follows from the closed graph theorem; similarly, we could make direct use of Lemma 2 with a partition of unity. 0

Let us now give the

Proof of Lemma 2 (1st stage). Let us first demonstrate the uniqueness of a solution of (6.9); taking account of the linearity, we consider a function u E ce O( lR" - 1 X lR +) with derivatives au/ax; E Lfoc(lR" .. ) satisfying u(x', 0) = 0 for all x' E lR"-l, lim u(x) = 0

Ixl ~ ,

(6.13)

and show that U == O. For that let us take F E 'f, 1 (lR) with F' :? 0 and F == 0 in the neighbourhood of O. The function w = F(u) is continuo LIS on lR n - 1 X lR + with

'"' Sec Definition 6 of ~4.

~6. Regularity 463

compact support in IR"+; we have ciw/cix; = F'(u)ou/cixi E L2(1R'~). Since aij cui aX i E L~oc (1R"c ), on regularizing wand passing to the limit, we see that we can apply (6.13) with ( w; that is to say

(6.14) O.

Now

;;, Igradul2( 1 ~ LLllaij ~ 6;jllc);;, ~lgradul2 on using the hypothesis and noting that, necessarily, the constant C of (6.8) satisfies C;;'l. Substituting in (6.14), we obtain

f F'(u)lgraduI 2 dx = 0 from which F'(u)lgraduI 2 = 0,

thengradw = OandhenceF(u) = w = OonlR n- J x 1R+.Thisbeingtruefor every function F E <g 1 (IR) with F' > 0, F == 0 in the neighbourhood of 0, we obtain u == 0 on IRn-l x IR.

Proof of Lemma 2 (2nd stage). We prove now the existence of a solution u of(6.9) satisfving point (a), au/ox i E LP(lRn,). Using the hypothesis, we consider a compact

'46 This estimate (6.8) has been stated for WE 9(W); by regularisation, it is still true for all IV E W 2 . p (G;\") with compact support; to extend it to the Newtonian potential of q E U(G;\") with compact support let us consider ~ E Q(G;\"): we have

~ ,~2W = iY ((w) _ (~((~ +o,~\V + w_02(.) cXi tXj uxJ3xj (lXi ax) ex) PXi t"lx i clXj

Applying (6.8) to (w, we have therefore

we also have with ~o == 1 on B(O, 1) and supp (0 c B(O, 2) and applying the inequalities with (,(x) = (o(x/k). Using

II L1 (( IV) II f-P';; (L1 IV llu + 1: ( 211 :' ~IV II + II IV :. 2~ II \). (XI LXi 'LP rX i U'

Let us fix (0 E ff(WJ

w(x) = or 1 ) "1.,,. (x) = o( I), \lxl"'2/ ?x, Ix!"'"

1 (x) supp(grad s,) c B(O, 2)\B(0, I), grad(,(x) = . grad '0 --k k

II " 1 II II '2v II Ct"k Clot' c ~k p l_~ we find that l~ and w-,-,- areO(l/k' )and therefore tend tozeroask ..... CXJ.Passing

(Xi eXj LP OX i ex) LP

to the limit we obtain the stated estim~le


set K of IR" -! X IR + containing a neighbourhood of the supports of uij - bij and k We denote by LP(K) the space offunctions 9 E LP(lRn) with support contained in K. Given 9 E LP( K), we consider the Newtonian potential w of g. From Propositions L 2, 6 and 8 of §3: Ii' E (fjl(IR"), Ii' is harmonic on U:g"\K, when x I -)o:f~

IV(X) = (fgdX )E,,(X) + O(I~~T) (~W , f I' \ (' E" ( I ') ~ ( x) = ( I 9 dX) ~ ( x) + 0 I 'I" LXi \ • eX i \ x /

and the derivatives t2W/?X/Xj with the estimate 14b

The functions wo(.x',x,,) = w(x',xn) - w(x',-x,,) and

(1\1' i'w IVh(X',X,,) = ~ (x',xII ) - -,-(x', - XII) {(II' k = 1, ... ,11

('X k OXk

satisfy >Vo E 'f,'!(lRn-1 x IR+), WI"'" >VII E 'f,°(IR,,-1 X U:g+),

>l'k(X',O) c.c. 0 for all x' E IR" - 1 ,

lim IVk(X) = 0 for k = 0, ... ,11. IX -+ f

II (IWk II :( 2C Ii 9 IIV' . i Dx; f.P

Finally since IV is harmonic on IR" 1 X ] -x, 0[,

Lht'o = U and

We now denote by IVk (g) the function defined in this way starting from 9 E LP ( K).

- 7" '\' ( • ' cwo (j~ ) f k 1 A f . . Now we put Jk =.fk + L 6kj - Ukj J-----::;-- or . = , ... , 11. unction U IS a j (-Xj

solution of (6.9), iff ~I = u- woUl is a solution corresponding to ]~= 0, J; , ... J;, which belong to U( K). For 0=· (gl' ... , y,,) E LP( K)", we consider

II

1'(Y) = I wk(gd and k = 1

WchaVC1'(g)E((j°(IR"1 x 1R+),v(g)(x',O) = 0 fmall X'EIR"-l.

lim r(y)(x) = O. Ixl ~ y

(1/'( (J) --:,-'- E U ( IR "+- )

('X j

§6. Regularity 465

and

We are therefore assured that l~ = v(O) is a solution of (6.9) corresponding to OI,];,···,1.,if

Oi "~' T;(o) = 1: for i = I, ... , n .

Now for all 0 E U(K)n, 1;(0) E U(K) and

( "IIOWk(QkJII ) II T;(olIILV ~ ~ 11()ij - aijllL' fa-;> LV '

from which

with c = 2CII 116ij - adL' .

From the hypothesis, c > 1; hence the map

T(O) == (Tl (0), ... , Tn(g))

is a (strict) linear contraction of LP (K t into itself. Hence I - T is an isomorphism of LP( K)" onto itself and there exists a unique g E LP( K r such that g - Tg = (];, ... ,1.,). For this g, the function u = Wo (fo) + v(g) is the solution of (6.9). Finally note that Dujoxj E LP(lRn+); we know that

q~(g)EU(lRn+) and WoCfo)E«51~1R"-1 x IR+); but Dw~(go) = 0("1 :I~:::T) uXj eXi x

when Ixi -> 00, so for p > n ~ 2

Proof' of' Lemma 2 (3rd stage). We shall prove the point (b) of the lemma for m = O. By hypothesis f = fo + IOJ;jOXk E LP(K) and aij E W1,x(IR"r). The solution u of(6.9) can be written u = ii + wo(f) where it is the solution of (6.9) corresponding tok = O,j~ = I(6kj - akj)cwo(f)/Dxj' Since

a2w (f) wo(f)E«51(lRn-1 IR+) and l();;-EU(IR"..),

(Xi CX j

we have A E WI. P (IRn+ ) and the conclusion

U E '6 1 ( IR n .. I X IR +) and

147 We obviously suppose that n ?- 2. the case n = 1 being trivial throughout this section.


holds iff it does for U. In other words, \l'e call always suppose .f(~= 0 and .f~ E WI. P( IR"+), which we are doing. With the notation of the proof of the 2nd stage, the solution u is equal to v((l - Tj- 1 U;, ... ,i,)) with the result that we have the estimate

(6.15) II' Du II

-=;--:-.. I ~ eII f~ L\ilu

for j = I, ... , 11

with e = 2e( 1 c) I, with (' = 2CII We now prove that for i = 1, ... , n - 1 and all j = 1, ... ,n we have ?2u/(~XiaXjE LP(lR n+). For that we shall use the method of trans/atiolJs: given II =c (/z', 0) E IR" 1 X [O}, let us consider ~I(X) = u(x + h) .~ .. u(x). The function ii is a solution of (6.9) corresponding tof() 0 (we have supposed thatf~ = 0) and

~(x) = f~(x -I Il) ·····((x) + I(ukdx + 17) I

Applying (6.15) to u, we have for j = 1. ... , 11

II (iU i'u Iii -(. + Il) - -_ aXj (X j U'

= 11~1~11 ~ C{IIIA(. + 11) -.h.llu ex) U'

and hence, h' being arbitrary in IR n - 1, for i = 1. ... , 11 _ 1149

and

(6.16)

" -ex" exn •

! 4H If T is a strict linear contraction of a Banach space X.

(1- T)~! = I T m and il(l- T) 1 ,; IITim = (1·~ liT )1.

m~O

149 Given 1 < p'; x. andIEU(Q) we have i"j/'-XiEL"(Q) if and only if 1)1'1 1 ,1((.+ h)

. r UIQ n IQ _ h)! is bounded for II a vector in the i-th direction.

§6. Regularity 467

From what we have proved ann o2u/ox; E LP(IR"+.). But

I 1 02U ann ~ 1 - lann - 11 ~ 1 - (2C)- ~.2' therefore ox; E U(IR"+.) .

It remains to prove that U E ~ I (IR n - I X IR + ). The function U is the solution of the homogeneous Dirichlet problem PH(IR"+.,f) tending to zero at infinity with

_ " ok "" a au P f - L. ~ + L.L. ~(bij - aij)~ E L (K). uXk uX i uXj

Therefore

Proof of Lemma 2 (Final stage). We shall prove the point (b) for arbitrary mEN by induction. By hypothesis

f=/O + I~k E Wm,p(IR"+.) and aijE wm+ 1. a: (IRn+) uXk

For I = 1, ... , n - 1, thefunction iI = ou/oxl is a solution of(6.9)(we know that U E ~ I(IR" - I x IR +) and a2U/ox i oxj E U(IR"+.» with

- of /0 =~,

uXI

Using the induction hypothesis we have

UE W m+l.P(lRn+) and hence 10 + L ~oX E W m-l. p(IR"+.). OXk

Applying the same hypothesis to ii, we have ou/oxl E W m + l,p(IR"+.). Thus we have shown that all the derivatives of U of order ~ m + 2 were in U(IR"+.), with the exception of om+ 2U/OX:+ 2. But from (6.16), 02 U/CX; E wm. P(IR"+.), from which the result follows. 0

Lemma 2 and hence Proposition 1 can be extended to the case of the Neumann problem. We refer to Agmon-Douglis-Nirenberg [1] for the general study of the regularity w m• p. Here, we shall prove Lemma 1 and Theorem 1 for the Neumann problem using Schauder's method based on the Holder estimates. Given 0 < () < 1, for a function U defined on a part Q of IRn, we denote

( 6.17) [ ] Iu(x) - u(y)1 u n.8 = sup 8

x.y E Q Ix - yl x # y

We have [U]n.8 < 00 iff u satisfies a Holder condition of order () uniformly on Q 150. We notice that [.] n. 8 is not a norm since [u] n. 8 = 0 iff u is constant on Q.

However, if Q is bounded, [']n.o is a norm on the space {u E ~8(U); U = 0 on oQ};

150 We then denote: U E <fi9(ih We also use the notation U E <fio.9(ih (see §3. Prop. 9).


this follows from the more general inequality

(6.17)' maxlu!:o( maxlul + R 8 [u]n.B Q ,'Q

where R = maxdist(x, c'Q). 'E Q

Proposition 9 of §3 shows that for every function IE y;B(IRl") with compact support, the Newtonian potential u of I satisfies

(6.18 )

where because of the homogeneity (see also the proof of Proposition'9 of § 3), C depends only on nand e. We notice also from Proposition 2 of § 3 that

lim D2ujDx j ?x j (x) = O;thefunctionD 2 u!(lx j rx j ,beingharmoniconlRl"\suppf Ixl ~>':,

attains its maximum on suppf An explicit calculation or the use of the closed graph theorem and of a homogeneity argument, shows that

(6.18)'

where C depends only on nand 0, and R is the radius of suppf

1 R = - max Ix - .vI

2 X.YESUPP! min max Ixo - xl.

xoEsuppf XEsuppI

Let us state

Lemma 3. Let ° < () < 1, R > o,t;, ... ,j~ E ,{;o(lRl" ---j x IRl +) .vith supports contained in B( 0, R) and let IjJ E '6' B (1Rl" 1) with supp IjJ c B (0, R). Then there exists one and only one quasi-classical solution u of' the Neumann problem P N (1Rl";. , 1jJ, L aj~ / rxk ) satisfying

lim u(x) + 2(J"'ljJdX')En (X) = 0. Ixl-' x

This solution u E (io 1.O(IRl" - 1 X IRl +) and we have an estimate (ill which C depends only 011 II and 0):-

(6.19)

Proof of Lemma 3. The uniqueness follows from Corollary 6 of §4 (see also the lemma below). For the existence and the estimate (6.19), we consider separately the problems

Fork 1, ... , 11 - I, we consider x (.x', XII) = A (x', Ix" ). We have.~ E ({,8(iRn)

§6. Regularity 469

with supp,K c B( 0, R). The Newtonian potential Uk ofj~ is even with respect to x,,; hence the restriction Uk of eu/ eXk is solution of PN ([R;~ , 0, ej~/ oxd. Applying (6.18) and (6.18)', we obtain (6.19) for Uk'

- x For k = n, let us consider In (x', x,,) = -~ (j~ (x', x,,) - f,; (x', 0).

Ix,,1

Again we have .Tn E rg8([R;") and ai/ax" = ~r/exn in .<:0'(W+). The Newtonian potential Un of .Tn being odd, the restriction u" of eu/ox" is the solution of PN([R;~, 0, or,./ ax,,). Finally, following Example 21 of §4, we find the function u(x', x,,) defined by

u(x', XII} = - 2 f En(x' - t, xn}Ijt(t)dt ];l" 1

is the solution of PN([R;~, Ijt, 0) satisfying

lim" u(x) + 2(fljtdX')En (X) = O. Ixl ~ y,

" J- "E ~~ (x', xn) = - 2 ~ "" (x' - /, xnjljt(t)df , eX i H' 1 DX i

We have

au, 2 r Ijt(x' + txn)dt 8~(x,x,,) = - ;';'-Jw- 1 (1 + 1(12)"/2

au (x' x ) =_ 3_ f Ijt(x' + tXn)fjdt ax; 'n 0'" R" I (1 + 1tI 2 )"/2 '

It is clear that au/ax" satisfies a Holder condition:

We also have immediately

1 ~;u(X"X,,)I,::; maxiljtl,::; R-O[IjtJw 'II' ex" lR" - ! '

To prove the same results for au/ax;, i = 1, . , . , n - 1, we make use of

f --tiiinTi dt = O. We shall not give the details of the proof which run H( o. R) (1 + It i ).

along identical (if not simpler) lines to those of Proposition 9 of §3 (see GilbargTrudinger [IJ and also Appendix "Singular integrals" Vol. 4). D

Let us take aij = a ji E '(j0([R;" - 1 X [R; +) with supp(aij - 6ij) C B(O, R). Given J;, ... , In, Ijt as in Lemma' 3, let us consider u the solution of

470 Chapter II. Th~ Laplace Operator

PN(IR'~, t/J, L (~j~/DXk)' We have:

LL'/-((/i}~~U\ = L-,,('-U~ +1:) m <:/'(};g".) eX i (X j ) (Xk

with

eu lj/Cx') = 1 (6 ill - aill(x'.O)) (x', 0) .

("U We have (akj - 6kj ) E ((, o( IR n ... 1 X IR + ) with support contained in B(O, R),

and taking account of (6.17)' we find that

Reasoning similarly with (6 i " - ai,,( .. O))('uic'x;( .. O), we see that T: U; , ... . r", t/J) H C~ .... ..i:, lj/) is a linear map of the space X = [U; . ... ,f~, t/J)) into itself; in addition, norming X by L'Ltd .. 0 + [t/J Jw '. Ii which makes it a Banach space, T is continuous with norm less than or equal to

CR U( LL[aij]H".1I + L[ain("O)]~" ,.(/). \ /

Following then the proof of Lemma 2. step by step, we shall prove the

Lemma 4. Let 0 < I} < 1. R > 0, aij = ajiE((,Ii(IR" I x [Ri~) salisIyiny

SllPP (aij - (Si)) c B (0. R) and

(6.20) (' = ('R O ( LL[aij]E",.1i + L[ai,,(·.O)JE" 1/1) < I

where C is lhe conslant 0/(6.19). Then/or allj~.f;, . ... /,: E ((,O(:R" 1 x [Ri T) with .supports continuolls in B(O. R), and II; E (f,8([Ri" ... 1), (here exists 0111' and ollly 0111'

solution o( the problem:

U E ((, I (IR"t ) ,

ill 'j' ([Ri'~ )

(6.21)

Bu(x',O) = t/J(x')

t/Jdx'- j~ j~dx ')EIl(X) c= O. - I

§6, Regularity 471

In addition (a) UE(gl,8(lRn-1 x IR+); (b) iffor mEN, aij E "em + 1,8(lRn - 1 X IR +) then

and

UE"e m + 2,O(W- 1 x IR+) iff ,10 + L~k E(gm,8(1R"1 X IR+) OXk

Proof of Lemma 1 (Neumann problem), Reasoning as for the Dirichlet problem, we have only to prove that there exists R > ° and p E !{f 00 (IR" '" I X IR) with compact support contained in (0 x [0, b[ (\ B(O, R) and p == 1 in the neighbourhood of 0 such that au = ()ij + p(aij - bij) satisfies (6.20). We fix PoE"e~(W-l x IR+) with sUPPPo c B(0,2) and Po = 1 on IR,,-l x IR+ (\ B(O, 1) and look for p of the form p(x) = Po(.xjR), We have

[pJIR",.o = R-8[po]w~,lIand IlpllL' = IIPol!c'so

[clijh",.,o::S R-O[poJw+,ollaij - oijIIL'(B(O,R)) + ilpoIIL·[ai}JJr~nB(O,R),O' From the hypothesis aij(O) = bii , for R > ° sufficiently small R8 [aijJ w+, 0 can be made as small as we please and hence satisfy (6.20). D

As we have said, the proof of Lemma 4 follows, step by step, that of Lemma 2: we leave the details to the reader (see also Gilbarg-Trudinger [1]). We shall give exactly only the proof of uniqueness:

Proof of Lemma 4 (Uniqueness). By linearity we are led to show that a solution u of (6.21) corresponding to,1o = I, = " , " = (;, = 0, !/J = 0 is identically zero. We take F E !(f 1 (IRn+) with F'(u) > ° and F = 0 in the neighbourhood of 0, Then w = F(u) E "e1(IR"cl and is of bounded support. Given (n E ({,'1(1R+) with suppf" c JO, Cf) [ the function ({x', XII) = w(x', xII)(nCxll ) is of class "e 1 on 1R"twith compact support in IRf~, therefore

which can also be written

~ 1

Use of the hypothesis (6.20), gives I2>ij~~- ~u ~ (I - c)lgraduI 2 . Choosing eX i GXj

(,,(XII) = ((xn/t;) where (E ,?/1(1R+) satisfies (' ~ 0, (= 0 on [0, IJ, ( = Ion [2, Cf) [, we have

r " F'(u)lgraduI2dx::S - ~ r ,(,(X,II \wBudx. JIR" 'x ]2",+co[ t JIR'-' x ];;,2;;[ E )

Since BUE,?/o(IR"-l x IR+) and Bu(x',O) = 0, passing to the limit as t; -+ 0,

472 Chapter IL The Laplace Operator

we have r F'(u)lgraduI 2 dx:( 0. We conclude as In the proof of Lemma 2 ..J H"_

(I st stage). D

2. Analytic Regularity and Trace on the Boundary of a Harmonic Function

Let us take an open set of Q of!R" and To a regular open part of its boundary which we shall suppose to be analytic 151 . We shall use the Cauchy-Kovalevsky theorem (see Chapter V, ~ 1.5); there exists an open neighbourhood U of To such that for all functions .f, cp, 1/1 analytic on Q u 1'0 152 , there exists one and only one solution u, analytic on (U n Q) u To, of the C aueil y prohlem

Au =f on QnU

u = (6.22)

cp on I' 0

i'u 1/1 = on To·

?n

Notice the difference between the Cauchy problem, in which 14 and aul hl are given on a part of the boundary of Q, and the Dirichlet (resp. Neumann) problem in which u (resp. ou/i1n) is given over the whole boundary of Q. We notice first the analytic regularity of the solutions of the Cauchy and Neumann problems.

Proposition 2. Let Q he all open set in !RI!. To all analytic regular opell part of its houlldary (IQ, and let cp (l'esp.l/I) he all analyticjimctioll on 1'0' fall allalyticfi.mction 011 Q u To and U E ((; o(Q u To) (resp.(6~(Q u 1'0)) solution of

Au = f ill Cf'(Q)

lI(Z) = CPt::) ( 1111

I'esp. , (::) en

Theil u is analytic on Q u 1'0'

I/I(Z)) for all Z E To .

Proot: We know already (see Theorem 1) that II is of class (6 f on Q u 1'0' Using the CaLlchy- Kovalevsky theorem we see that there exists an open neighbourhood U of 1'0 and Uo analytic on (Q n U) u I'o and solution of the Cauchy problem

L1 Llo =f on Qn U

1/0 = cp (resp. I/o - 0) on 1 0

i'lio ( resp.

?u 1/1)

- = 0, --- on ro· ?n on

151 That is to say that for every local parametric representation (R. U. (i. x) of 1'0' the function x is analytic on t. 152 Le. can be developed as a power series in the neighbourhood of each point of QuI'".

§6. Regularity 473

The function ii = u- Uo is therefore harmonic on Q n U, of class rt' x on (Q u U) u To and satisfies

( oii ) ii = ° resp. -.:;-- = ° on To. on

We are therefore led, at least for analytic regularity in the neighbourhood of the boundary, to the case I == ° and <p == ° (resp.!/J == 0), which we now assume. In the case where Q = (i) X ]0, b[ and To = (r;' x {O}, 0 being an open set of [R;"-l, the analyticity on 6 x ]0, b[ of a solution U E 'Ii l(Q u To) of

{

.du = 0 on Q

U = ° (resp. _~J~ = 0)' \ ex"

IS immediate. Indeed, the function ii(x', x") =~li(x',lx"l) (resp.u(x',x,,) Ix"1

lI(x',lx"l)) is clearly of class 'iiI on 6' x ]-b,()[; being harmonic on ((' x ] - b, b [\ ro it is so on all of 6 x ] - b, J [ (see Proposition 17 of § 2); hence 11 extends to an analytic function in the neighbourhood of roo Using a Kelvin transformation (see §2.5) we shall have the same result if To is contained in a sphere of [R;".

We now come to the general case. Reasoning as in Theorem 1, by the use of a local parametric representation, we are led to prove that, given (1. an open neighbourhood of 0 in [R;n - 1, b > 0, and Gu = Gjl analytic on (i1 x [- (5, b [ with ai/OJ = ()ii' every solution U E~' x((': x [0, c5[) of

tl I i~ ( aii (~:Jj) = ° on (I x [0, c5 [

u(x',O) = 0 (resp. I aiAx') ~u (x',O) = 0) for x' E (I \ (;Xi

is analytic in the neighbourhood of 0 in (1) x [0, b [. In fact, from the CauchyKovalevsky theorem, it is sufficient to prove that ou/iJXll(x', O) (resp. u(x', 0)) is analytic in the neighbourhood of 0 in (I.

Let us consider the Dirichlet problem; we fix '0 > 0 which we shall specify below, ,P ali u

and k = 1, ... , n - I; for all fJ E N we write lip - 0 -p . Differentiating the In aX k

equation and using Leibniz's formula up satisfies

on (I' x [0, b [


with

Using the hypothesis of analyticity of the coetflcients, for rlJ chosen sutflciently small

(6.23) ,,::1 on B(O, "0) ,

and hence supposing 1'0 "..~ b ,

(6.24) if'/il":: I ill;1 on B(O,ro) x [O.ro[· ; < /1

Given P E ((, f(~11 1 X ~~+) with suppp c B(O,l"o) x [O,ro[ we multiply the equation by p 2 11iJ and integrate by parts; we obtain

Using the hypothesis aij(O) = hi}' we can always suppose that

We deduce, using Schwau:'s inequality, that

!I p grad uti 2

from which, we have

(6.25) II p grad lip

with c depending only on the Ii ai) ii, ' (BIO.!" I)'

For r > 0, we denote Qr = B(O, r) x [0, r[. Given 0 < r < r + D < ro there exists P E (f, J(~n-l X ~~) with 0 ::; p ,,:: 1, suppp c Qr,,' P == Ion Qr and Igrad {II,,:: 21: - 1 With (6.25) with this function p, taking account of (6.24), we have for all 0 < r < r + f: < roo

(6.26)

where C docs not depend on {l, r and E.

In particular we have

§6. Regularity 475

from which we deduce by induction 153

(6.27)

Returning to (6.26) and to the definition of up, we have

I-gradu Ii afJ II I axe L ' (Q,,,,l

The direction X k in ~n - ] being arbitrary, we have in fact proved by changing the notation, that for /'0 sufficiently small, for every multi-index fi = (/31' ... , fi" - 1 )

(6.28) II (~fJ II '1 ,p grad u, ,,;; (CI /3:)1 P 1 154 .

(X L'IQ",)

Returning to the equation, we can write

where the coefficients cij ' Cj are analytic on B(O, 1'0)' Differentiating with respect to x' using Leibniz's formula, using (6.28) and the analyticity of the coefficients, we deduce that for fJ E' N" .. 1

1'1 :'122 ~p~ III ,,;; (CI/31)lfll. (x" (x L'IQ,,,I

U sing a trace theorem 155

,,;; (C fJl)lfJl

IS 3 For () < I: I < [,

Applying with [, = jil:i(fJ + 1) and using the induction hypothesis, we obtain (6.27). '54 In this formula and those which follow C denotes a constant with respect to /l.


from which we have the analyticity of aul aXn(x', 0) in the neighbourhood of 0 in IR n - 1, by using Sobolev's inclusions 156 (see Vol. 2, Chapter 4, §7). The Neumann problem can be treated in the same way: a supplementary difficulty comes from the necessity of also deriving the condition on the boundary. We refer the reader to Lions--Magenes [1] for a proof and for many more general results. 0

We propose now to use this result to define the trace on 10 of an arbitrary harmonic /imction u on Q. For simplicity, we shall now suppose that Q is bounded and that 10 = clQ. Hence we take Q to be a bounded open set with boundary 1 which we suppose to be analytic. Given u E off (Q) (\ (C ~ (2) with trace qJ on 1 (in the classical sense) from the classical Green's formula (see Proposition 4 of § I), we have.

(6.29) f ('":( i qJ -dl' = u.1( dx

r (In Q

for every function (E (6~(Q) (\ '02(Q) with .1( E Ll(Q) satisfying ( = 0 on I. Given arbitrary u E ,Yf (Q), the right side of (6.29) is de·fined for all ( E {6 2 ( Q) with .1 ( of compact support in Q, i.e. harmonic in the neighbourhood of I in Q. More generally, the right side of (6.29) can be defined for every distribution ( on Q

harmonic in the neighbourhood of 1 in Q. From Proposition 2, a distribution ( on Q harmonic in the neighbourhood of 1 in

Q and satisfying ( = 0 on r (in the classical sense: lim ((x) = 0 for all : E I) is x-=

necessarily analytic up to the boundary 1 and in particular cl(/ ('II exists and is an analytic function on I. In other words, denoting by n (I), the space offunctions analytic on I, for every distribution ( on Q, harmonic in the neighbourhood of I in Q and satisfying ( = 0 on r. we have /"':;/('11 E (((I).

We prove the

Proposition 3. Let Q he a hounded open set with analytic houndary I, alld 11 a /wrmonic jilllction on Q. Denote hy n (I) the space offunctions analytic 011 I. (I) There exists a IInique linear form qJ on n (I) slIch that

(6.30) r 11.1 (dx = / qJ, ? ) J Q \ (1/1

.fiJI' el'ery.tilllction ( E {(, '( Q) with .1 ( of compact slIpport ill Q and sati.~h'iny ( = 0 011 r. (2) For every distribution ( harmonic in the neiyhbourhood of rand satish'iny ( = 0 011 r, (6.31)

"6 If Q is an open set of [W, for m > 111 we have Hm(Q) E ,(,o(Q). Marc precisely

§6. Regularity

I n particular

(6.32) u(X) = / cp,~G(X'.)) forall xEQ \ on

where G is the Green's function of the open set Q.

(3) If u E ~ 0 uh then cp is defined by

(6.33) (cp, 1/1) = t u(t)I/I(t)dy(t) for all 1/1 E a(n .

This leads us naturally to set up the

477

Definition 1. Let Q be a bounded open set with analytic boundary r, and u an arbitrary harmonic function on Q. We call the trace of u on r, the linear form cp on the space a (n of analytic functions on r defined by (6.30).

From the point (3) of the Proposition, this concept of trace generalizes clearly the classical notion when U E ((;O(Q), with the condition of identifying the continuous function cp E (Co(n with the linear form on a(n

1/1 E a(n ...... t cp(t)I/I(t)dy(t).

From Weierstrass' theorem we can carry out this identification: in effect a( 1) being dense in cg ° (1), given CPl' (P2 E 'fj 0 (1), CPl = cpz on r iff r CPdt)l/I(t)dy(t) = r CP2(t)I/I(t)dy(t) for all 1/1 E a(1).

Proof of Proposition 3. Given 1; a distribution on Q, harmonic in the neighbourhood of r and satisfying C = 0 on r, we show first that (,1(, u) depends only on the normal derivative 1/1 = cU en on 1. In other words, by linearity, we must prove

DC ~~ = 0 on r = (A (, !l) = 0 . (n

Now from the uniqueness in the Cauchy-Kovalevsky theorem, if' is harmonic in the neighbourhood of r and satisfies ( = 0, eU en = 0 on r, necessarily it is identically zero on a neighbourhood U of r. Let us take p E q (Q) such that p == 1 on a neighbourhood of Q \ U. We have p == I in the neighbourhood of supp ,1(, from which we have

(LJ(, u) = (,1(, pu) = «(, A (pu) ;

but

A(pu) = pAu + 2gradp.gradu + uAp = ° in the neighbourhood of supp ( c Q\ G. Hence (,1(, u) = O. Now, given 1/1 E (}((1), from the Cauchy-Kovalevsky theorem, there exists (0

harmonic on a neighbourhood U of r satisfying (0 = 0, (J(o/Dn = 0 on r. Let us


take Po E U(Q) with Po = 1 in the neighbourhood of Q\ U. The function s = (I - Po) So E '(y" x (Q) and s = So in the neighbourhood of r. Hence lis has

compact support in Q,:; = 0 and ('s / Dn = til on r. The value r u.;1 s dx, from .Q

the preceding point depends only on IjI so we can denote it by < ip, til). We have thus proved the point (1) and (6.31).

The formula (6.32) can be deduced immediately from (6.31) since by definition, for x E Q,:; = G(x,.) is a solution of j s = (), on Q. The formula (6.33) reduces in fact to (6.29) and follows immediately from Green's formula if U E ((, ~ (t.h to prove it in the case where u is only continuous on Q, it is sufficient to consider Uk E (6~(Q) nll(Q) such that Uk -+ 11 in ((, o(Q) and to pass to the limit in (6.29). 0

It is immediate that the map U E .If (Q) -+ trace of u on r is linear. From (6.32),

this map is a bijection ofll (Q) on the set of linear forms ip on a (r) satisfying

! i'G \ XEQ -+ ("(P.~,-(X,.))

\ (11 is harmonic on Q

We can prot'ide ({ (n with a topoloyy such that these linear forms are exactly the continuous linear forms on O(T). We refer the reader to Lions Magenes [I] for a precise definition of this topology; here we note only that it can be characterized by the following notion of limit:

(6.34)

a sequence 15'7 (If; k) of elements of ({ (n converges to zero if and only if there exists an open neighbourhood U of r such that for all k we can solve, in a unique manner in U the Cauchy problem

Sk = 0 and

and the solution Sk converges to zero inY( (U).

C s = tilk on r

ell

From now on we shall suppose that ('( (I') is given this topology; we shall denote by (nn the dual off'( (n and we shall call the elements of (nn a/lalytic jimctional.

Proposition 4. Let Q be a bounded open set with analytic boundary r. For all u E jt(Q), the trace or u on r is all analytic jUllctional; the map u -> ip is (Ill

isomorphism from ]((Q) Ollto ('('(T)158.

Proof Let IjIk -> 0 in a (T) and U. Sk as in (6.34); considering Po E '/(Q) with Po == I in the neighbourhood of Q \U, we have

<ip,tilk) = f LIIl[(1 - Po)sddx (see proof of Proposition 3). Q

15' Possibly generalized. 158 )f(Q) is given the uniform convergence topology on every compact set and r!'(rl the strong topology.

§6. Regularity 479

Since (k -> 0 in ff(U),

,1[(1 - PO)(k] = - 2gradpo·grad(k - (k,1po -> 0

uniformly on Q with supp,1 [(1 - PO)(kJ c supp Po n supp (1 - Po) a compact

set in Q. Hence r u,1((1 - Po)(ddx -> O. This shows that qJ E au} JQ Using (6.34) and the definition of the Green's function, it is clear that the map

x -> ~~ (x, .) of Q into (r(1) is continuous; more precisely this is a harmonic map on

ac since for all t E r, x -> -o-(x, t) is harmonic. Hence, for all (P E a'(r), the function

on u defined by (6.32) is harmonic on Q. This shows that the map u -> qJ of £(Q) onto a'(1) is a bijection. The continuity of the maps u -> (P and qJ -> u can be proved without difficulty by the use of the topologies of .ff(Q) and 0'(1): we leave the details to the reader (see Lions-Magenes [IJ). 0

Proposition 4 can also be stated: for every analytic functional qJ on r, there exists one and only one solution of the Dirichlet problem P(Q, qJ), where, obviously, u = qJ on r is taken in the sense of Definition 1; also this solution u is given by (6.33) and the map qJ -> U of 0'(1) into ff{Q) is continuous It is clear that every distribution Oil r is an analytic jimctionai on r; more precisely we can identify 9'(r) with a sub-space of 0'(1) since from Weierstrass' theorem, 0(F) is dense in ('£"'{1)159. In particular for every distribution (P on r there exists one and only one solution of the Dirichlet problem P(Q, qJ). We note, however, that for an arbitrary harmonic function u on Q, its trace on r is not, in general, a distribution on r. Let us give some examples.

Example I. For Z E [hi: n\Q let us consider u=(x) = En(x - z). This is a harmonic function on Q, continuous on Q\ {z}; but if z E r, it is not continuous at z. The map z E [hI:n\Q -> qJz E £(Q) is continuous (and even harmonic); for z $ r, the trace of U z in the classical sense or in that of Definition 1 coincide in identifying the function qJAt) = En{t - z) with the analytic functional

< qJz, tjJ> = f En{1 - z)tjJ(t) di(t) . r

For all z E [hI:n\Q, the function qJz(t) = En{t - z) is integrable on r and the map z E [hI:n \ Q -> qJo ELI (1) is continuous. Hencejor all Z E r the trace ofuz on r is the analytic functional defined by the junction qJz(t) = En(t - z). We note that

< qJz, t/! > = r En(t - z)t/!(t) dy(t)

is the value at z of the simple layer potential defined by t/!. o

159 ,{/"C is provided with the topology of uniform convergence of all the derivatives.


, . . 2~ Example 2. Now let us consider u~(x) = l, (x - :). As above for: E IR"'\Q,

(Xi

U~Ej((Q) n (6°(f2\{'::0})andthemap.:: -> u~oflR"\Qinto.;;f'(Q)iscontinuous. i'E

But for.:: E T, the function ~ (t - .::) is not, in general integrable on r in the eX i

neighbourhood of I = .::. We denote by (p~ the trace of u~ on T; for t/J E (( (Tl the function:.: --> <q>~. 1/1) is continuous on u~n\Q. When Z E 1R/\Q, u~ E (6 O(Q) and

" . J~ c';[~I1'" (. \q>~. l/l)c .. ::;-:-(t - z)~/U)d},(t) = - ,,), IlvJZ)

I (Xi (Xi

where UojJ is the simple layer potential of t/J (see ~ 3),

u,,,(z) = f EnU .. z)I!I(tjd·/(t). r

Suppose for simplicity that 11 ;~ 3 and that IRn \ £2 is connected; we consider

:x E (6°(n such that f En(t - zl:x(t)d}'(t) = I for all Z E r. r

Such a function exists (see Proposition 18 of 94); its interior (resp. exterior) simple layer potential is the solution of the Dirichlet problem P(Q, I) (resp. P(IR" \ Q, I) tending to zero at infinity); in other words the function

I'

!Ja(z) = j I EIl(t z)cdt)d~,(t)

continuous on IR" satisfies u, == 1 on Q and from the maximum principle o < u, < 1. Now from Proposition 12 of §3, a(z) is the jump in the normal derivative of II, at Z E r; hence

. u,(z + tn(z)) - u,(:) :x(:) = lIm -- -- ;

1,0 t

also by Hopf's maximum principle (see Lemma 2 of §4), :x(.::) < 0 for all z E r. From Proposition 2, u, is analytic up to the boundary of IR" \, Q and hence :x E all). Finally since u, = I on r, the tangential derivatives of Ii, are zero and for all Z E r, (Cua/cxi)(z) = l1i(z)a(:) where l1 i(Z) is the i-th component of the normal n(;:). This being specified, we write for Z E T

/. I&(Z) \ '/'(z) . «p:, II,) = (m~.I'J - /' .... -:x \ + 'I' <' "

. • t', \ To /. a(;:) / :xl;:) q>=,'J. / .

From what we have just seen, we deduce that

Also since ( t/J

,I

«p~,a) = ?u

·····;:;'(z) = - I1;(Z)'J.(Z) . (Xi

t/l(Z) \ - -:x)(Z) = 0, we have

:x(z)

\ <p~, t/J I/I(Z) \

;(;) a J = f (~E'.'(t - :)[t/J(t) .-t(~'J.(t)J· d/(t) , r (Xi :x(z)

§6. Regularity

the integral being defined since when Ij; E !C 1 (T) and Ij;(z) = 0

IOEn I C -.-(t - z)lj;(t) :::; - .. ~~_~ . aXi It - zln 1

'. aEn . . . In short, the trace of U~(X) = -~-Ix - z) IS defined by

OX i

. f· [ Ij;(z) J (6.35) < cp~, Ij;) = r u~(t) Ij;(l) -ex(z) ex(t) dy(t) - Ij;(z)ni(z) ,

481

where ex is the normal derivative of the solution of the exterior Dirichlet problem p(lI;£n\Q, 1). We note that the trace cp~ is a distribution of order at most 1. o Example 3. Let us take Q = {(x, Y) E 11;£2; x2 + y2 :::; I} which we identify with the unit disk D = {z E iC; I z I < I} of the complex plane :c. A harmonic function u(x, y) on Q is the real part of a holomorphic functionJ(z) on D (see Proposition I of §2). The function can be developed in a power series

J

f(z) = I anzn, ZED. "=0

A function on T = aQ is identified with a periodic function g(O). If I I an I < Xl, the function f is continuous on jj and its trace on r is the continuous function g(8) = I ane ine . We show that in the genera I case, .the trigonometrical series I ane inO converges in (i' (r) and the trace off is the sum of this series. Let Ij; be analytic on T; we can extend it to a hoi om orphic function (p(z) on an annulus {1"1 < Izl < rl1}; this function is developable in a Laurent series (see Cartan [1]):

+,x

cp(z) = I hnzn, r 1 < izl < r11 • ,,= -:J~

Considering r 1 < r < I, since the series I bnrn converges, there exists M such that Ihnl :::; Mr- n• We then have

tI~O n~O

and we deduce

<00

cY'

and I < aneinO, Ij;) = 2n I Gnh- n . o n=O

Example 4. Let us again consider Q = {(x, y) E 11;£2; x 2 + y2 < I} and

u(x, y) = [exp Cz + (: ·~I)i) ] [cosCi:(~~l)i) ] .


This function is harmonic on D as it is the real part of the holomorphic function defined on C \{ i} by

> 1 .1(:;) = exp---:.

:;-/

This function is continuous on Q\ {(O, I)} and its trace <prO) is defined for 0 =1= tn by

( 1 ) 1 ( cos f) ) <p(O) = cos :2 eXP :2 1 _ sin f} .

The function <p, admitting an essential singularity for f} = tn cannot be extended to a distribution on r. This gives an example in which the trace oj a harmonic Junction is not a distribution, but only an analytic functional. 0

We conclude this section by noting that Definition 1 permits us likewise to give a meaning to the normal derivative (and also to the tangential derivatives) of a harmonic function: if u E .Yf (D), the derivatives ou/Dxi are also harmonic on D and we can consider their traces <Pi on r. On the other hand the components l1i(t) of the unit interior normal vector l1(t) are analytic functions of t E r; we can therefore consider the analytic functional

We prove the

Proposition 5. Let D be a bounded open set with analytic boundary r and let It E .Yf(D). Denote by <p, <P I' ... , <Pn the traces of u, ilu/(lX I' ... , (lu/oxn respectipeiy. (l) We then have the Green'sfcJrlnula

(6.36)

for every distribution C on D, harmonic ill the neighbourhood oj r in IRn; (2) II E '6.~ (Q) iff the analytic jilllctiollai I ni<Pi is continuous 011 r. Pro()f: To prove (6.36) we first note that we can suppose that C E .Yf(D): it is sufficient to replace C by Z the solution of the Dirichlet problem P(D, CI r) by applying the formula (6.31) to s - C. We shall then use a density argument: (6.36) is true for II E ({,I (Q) since it reduces then to the elementary Green's formula; to prove it for arbitrary u E .Yf (D), it is enough therefore, from the continuity of the trace, to show that ((,I(Q) n .Yf(D) is dense in .Yf(D). This last point is easily seen for particular open sets; in the general case, from the isomorphism between .)f'> (D) and ('{' (T), it suffices to prove that I(, x (T) is dense in IT (T); using the fact that (l(' (T) is reflexive (see Lions~Magenes [IJ), that leads to: ('{ (T) is dense in f)r'(T), which follows from the Weierstrass theorem. To prove the point (2), we note first that, given Do, a connected component of D, then applying (6.36) with C = XQo' we obtain Gauss' theorem

I (<Pi,ll iX,·Q.,> = 0. i

§6. Regularity 483

Supposing t/I = L ni CPi to be continuous on r we find, from Theorem 3 of §4, that there exists U E ~ ~ (Q) n ff (Q) a solution of P N (Q, t/I). Replacing u by u - U, we are therefore led to proving the result by assuming L niCPi = 0, and hence, using (6.36) again, < cP, o(/on) = 0 for every function" harmonic in the neighbourhood of Q. We can always suppose that Q is connected; given t/I E (l[(r) with

t t/I dy = 0, there exists a solution ( of the Neumann problem PN(Q, t/I),

which, from Proposition 2 is analytic up to the boundary r and hence can be extended to a function harmonic in the neighbourhood of Q. Applying (6.36) to this function (, we obtain

<cP,t/I) =0 forall t/lEa(r) with tt/ld)'=O,

hence

< cP, t/I) = < cP, 1) m- t t/I d}' for all t/I E (r(r)

where I r I is the total surface of r. In other words, cP is the constant function < cP, 1)11 nand 1I the solution of the Dirichlet problem P(Q, cp) is constant on Q. o

3. Dirichlet Problem with Ghen Measures or Discontinuous Functions. Herglotz's Theorem

We consider a positive concave function 1I on an interval ]a, b[ of R It is elementary to see the limits of 1I at the points a and b exist and that we have the representation formula 160

u(x) = ; = : (u(b) + f (b - t)dl1(t)) + ~ = : (u(a) + r (t - a)dl1(t))

where Ii = -d2uldx2 in the sense of distributions on ]a, b[, is a positive Radon measure. In particular, we have

f (b - t)dfl(t) < wand r (t - a)dfl(t) < ex for all x E ]a, b[

which we can summarize in r min(t -. a, b - t)dll(t) <Xc .

We propose to generalize this result to positive super-harmonic functions on an open set Q of !R". We take an open set Q c !R0 which, for simplicity we suppose

160 In the case Ja. b[ bounded; see Example 12 of ~4 for the unbounded case.


satisfies

(6.37)

Q is a connected reyular hounded opel! set with boundary r oj' class W 2 ., ! b 1.

It follows from Proposition 1, that G, the Green's function of Q is of class {6! on [(x, Y) E Q x Q; x i' y:. Also we show easily that Q satisfies the condition of the interior (and also exterior) ball at every point of the boundary [ (see example 7 of ~4) and hence from the maximum principle of Hopf (see Lemma 2 of §4) we deduce that:

(6.38) ?G , (x,.:) > 0 at every point x E Q and .: E r . (11

We recal! that for a function u sufflciently regular, we have Green's representation formula,

(6.39) I~ f ?G u(x) = G(x, t)Au(t)dt + "(x, t)u(t)d:·(t), • Q r c'n

We have proved this formula (sec Propositions 7 and 8 of §4) for U E ((, o(d) with fill E U(Q) wherc p > tn.

I'G We note first that for all x E Q, the function t -> -,-- (x, t) being continuous on [,

I'll

for every Radon measure v on r we can define

(6.40) II' (~G u(I', xl =-(x, t)dl'(r). x f~ Q .

oJ r ('11

The function 11(1'): x -7 u(I', x) is harmonic on Q since for all t E r the function

x -> (~ (x. t) is harmonic on Q. If \' is the measure of density ip with respect to the (.'/1

surface measure of r. where ip E (6 o(n that is to say dl' = ipd)', then the function 11(1') given by (6.40) is the classical solution of the Dirichlet problem P(Q, (pl. For

this reason we introduce the

Definition 2. Given Q an open set of ~" satisfying (6.37) and I' a Radon measure on r. we call the solurion of'the Dirichlet prohlem P(Q, \') the function 11(\') defined by (6.40).

Remark 2. When r is analytic, we have defined in the preceding section thc trace of an arbitrary harmonic function on Q; the Definition 2 is clearly compatible with this notion: for a Radon measure \' on T, which is ajiJl'tiori an analytic functionaL the solution u = u(v) of P(Q, v) in the sense of Definition 2 is the harmonic function 11 on Q admitting \' as trace on T in the sense of Definition I. 0

1(>1 That is to say. a little weaker than the class '(, ".

§6. Regularity 485

It is immediate that the map v ~ u(v) is an increasing linear map of the space 9Jl(T) of Radon measures on r into the space Yf (Q) offunctions harmonic on Q; this map

is increasing since ~~ (x, z) ~ 0 for all x E Q and Z E r. From (6.38), we have also for VI' V2 E 9Jl(r)

We state the first form of Herglotz's theorem:

Proposition 6. Given Q an open set in [Rn satisfying (6.37), the map v ~ u(v) is a bijection of the cone 9Jl+ (T) of positive Radon measures on r onto the cone Yf + (Q) of positive harmonic functions on Q.

Given that every Radon measure is the difference of two positive measures (see Dieudonne [2], we have the

Corollary 2. Given Q an open set of [Rn satisfying (6.37), the map v ~ u(v) is a linear bijection of the space 9Jl(T) of Radon measures on r onto the space :It + (Q) - Yf + (Q) of the differences of positive harmonic functions on Q; also this bijection is bi-increasing, i.e. for VI' v2 E 9Jl(r)

VI ~ V2 on r <0> u(vd ~ u(v2 ) on Q.

Given cp ELI (T), we can consider the measure v of density cp with respect to the surface measure dv = cpdy; identifying V and cp we denote by u(cp) the harmonic function on Q defined by

(6.4 t) f aG

u(cp, x) = Jr -an (x, t)cp(t)dy(t), x E Q.

For a constant function cp :; c, we have u(cp) :; c. Hence for v E WI(T)

\' ~ c on r <0> u(v) ~ c on Q.

We deduce:

Corollary 3. Let Q be an open set in [Rn satisfying (6.37) and u E Yf(Q). Thefunction u is bounded on Q if and only if there exists cp E L "'(T) such that u = u(cp) given by (6.41). Also, cp is unique (a.e. on T) and

sup u = sup ess cp, inf u = inf ess cp . Q r Q r

Before proving Herglotz's theorem we note

Lemma 5. Let Q be an open set of [Rn satisfying (6.37), u E Yf(Q) and v E gJl(r). Then u u(v) iffu E LI(Q) and

(6.42) L uJvdx = f :~dV for every function v E W 2 • ~(Q) with v ~ 0 on r.

4~6 Chapter II. The Laplace Operator

Proofo/Lemma 5. Let n < p < x'. From Corollary I, givenfE U(Q) we have that the (classical) solution vU) of the homogeneous Dirichlet problem PH (Q,f) is

'1 - d h f . . c'L'( no ' 111 ((, (Q) an t ere ore, In partIcular, -,-' - E ((, (I), and ell

(6.43 ) Il h(fJll --,.-.'- :( Cilf,luIQ, ( 11 'f, "Iii

where C depends only on p and Q.

Now using the same argument as in the proof of Proposition 4, u = u( v) iff we have (6.42) for every function \' = l{f) withl E Q'(Q). Suppose, therefore, that u = u(v); for all f E g(Q), making use of (6.42) with \' = rtf) and (6.43), we have

I r Ufdxl :( eli f IluIQ) II v jl~)I(n . .1 12

Hence uq(Q) where q is the index conjugate to p. Finally for l' E W 2 . p(Q) with l' = 0 on T, approximating Llv in U(Q) by the functionsh E 9(Q). we have from (6.43), av(f~)/an -+ evjall in (6° (n and hence (6.42) in the limit for v. D

ProololPropositioll 6. From Lemma 5, the fact that the map \. -+ u(v) is injective comes from the density in (6 o(n of

{e'. 1'. r E W 2 . '(Q); r = 0 on r}. ('11

Making use of a theorem on the lifting of traces (see Chap. IV), given t/J E WI. , (T).

there exists l' E W 2 . x (Q) such that r = 0 and Dr/en = t/J on T. From Weierstrass' theorem, WI., (I') is dense in ((; o(r); hence the result. It remains to show that the map is surjective. We shall, in the first instance, suppose that the open set Q is starshaped, that is to say satisfies

{there exists Xo E Q such that i.x + (I - ;.jxo E Q

for all x E Q and all i. E [0. 1 J .

By translation we can always suppose that Xo = O. Given u EYf ~ (Q), for all i, E JO, 1 [, the function ujx) = u(i.x) is harmonic in the neighbourhood of Q; we therefore have

(6.44)

As a result of (6.38),

and hence

u(i.x) = J' (~.-~ (x, r)u;(t)d'r,(t). x E Q. r (11

. i'G c mIn -,- (0, :) > 0

: E r ('11

c- 1 u(O) .

§6. Regularity 487

We deduce that the positive Radon measures V;. on r defined by dV A u;dy, are of total bounded masses; we can therefore find ).k -* 1 such that (in a vaguely defined way) the measures V Ak converge to a positive measure v on r 162 • Passing to the limit in (6.44), we see that u = u(v); this proves the theorem in the case of an open star-shaped set. We now consider the general case of an open set satisfying (6.37). We can find a finite covering of star-shaped open sets Q 1, ... , Q N contained in Q such that r = r 1 u ... r N where r i = cQ, () r. Given u E ff + (Q), the restrictions of u to Q i are obviously in ,Yf + (Q;); hence u ELI (Q;) and there exists Vi E 9R + (cQ;) such that

for all v E W 2 , Cf. (Qi ) with v = 0 on cQ i , and hence, denoting by Vi the restrictions of Vi to r i ,

for all v E W2.X(Q;) with v = 0 on r i and supp v c Qi Uri' Using partitions of unity we can "glue" together the measures Vi to define a positive measure V on r such that u = u(v). 0

Given cP E !6 0 (T), the solution u = u(cp) is continuous on {J and

cp(z) lim u(x) for all Z E r .

More generally, given v E 9Jl(T) and Z E r, [ = lim u(v, x) exists iff "I' is continuous x-;: XED

at z, with value [" in the sense in which there exists U a neighbourhood of Z and cp:

U () r -* IH bounded and measurable with lim (p(t) = I such that dv = (pdf

on U () r. t ...... ;;

tEU .. r

This follows easily from the local character of the trace v of u(v) obvious from Lemma 5 and from Corollary 3. A measure v E 9R(T) can very well fail to be continuous at any point Z E rand

hence for a function u E :if' + (Q), lim u(x) can very well not exist for any Z E r. X ~--J- Z

XEQ

However, we can show that the non-tangential limit exists for almost all Z E r. More precisely we have:

'62 See Dieudonne [12]. A sequence (v,) of Radon measures converges vaguely to the Radon measure v if < ",J) -; <I'J) for all continuousfwith compact support.

488 Chapter II. The Laplace Operalor

Proposition 7. Let Q be an ope/1 set of [f;£n satisfying (6.37) and v E ~)l(F). Thenf(JI· all 0 < 8 ~ 1

ip(;:;) =

exists fiJr almost all z E r.

lim y - : ,,~ Q

co~i: - :c 11(:)) ? fi

U(\', x)

Also ip E LI(n and is the density of the regular part O{V!63.

We shall not prove this result which appeals to delicate arguments of the theory of integral operators (see Halmos [IJ, Chap. VI and Helms [IJ, Chap. HI). We only note its significance in some examples.

Example 5. Given an open set Q of [f;£" satisfying (6.37), let us consider Z E rand

\' = (\, Dirac mass at ;:;. We have II(()=,

continuous (and even of class ,(, 1) on f2, I

We shall now prove that for all 0 E JO, I J,

(6.45) lim \" ---'f:: x E Q

co"(x - :. n(:)l ? ()

nG x) = 0 (x, z). The function

C/1 with u(6=) = 0 on F\

+£ .

1I(6J is

Notice, first, that we can always suppose that Q is a ball: in effect, because of the regularity (6.37) there exists r > 0 such that B = B(z - rn(z), r) c Q, the Green's function G B of the ball B satisfies from the maximum principle, G B ~ G on B x B; hence for x E B

ICG B . G B(X,;:; --- 1/1(;:;)) rG -0- (x, ;:;) = 11m -------- -- ~-' (x, z) . ell Ii 0 t en

By translation and homothety, we can therefore suppose that Q = B(O, 1). From Poisson's integral formula (see Proposition 9 of §2) we then have

2cos(;:;- x, z) - Ix - z!

(I"

In this form, we easily verify (6.45).

Example 6. Let us consider in the plane [f;£2 = ((x, y)] the unit disk

Q = {(x, y); x 2 + )"2 < I]

and for (p the characteristic function of the semi-circle

I" + = {(x. y); x 2 +- y2 = I, y > 0 I .

o

Using the semi-polar coordinates (r. 0) where - I < r < + I, 0 :'( 0 < 7[. the

163 From the Lebesgue decomposition, l' can be written in a unique manner as the sum of a measure "/ with density with respect to the surface measure and of a singular measure '\ (see Dicudollne [1 J).

§6. Regularity 489

function u = u(cp) is given by

u(r,8) = 1 2-n r2 f-"e- 0 1 dr 2 = ~ Arctg(~ tg ~)I" -e - 2rcosr + r n 1 - r 2 -8

The function u is continuous on

f2\ {( ± I, o)} = {(r, 0); - 1 ~ r ~ 1, 0 < 0 < 1t}

with u(l, 8) = 1,u(-1,8) = O,i.e.u = cp on L'\{(±1,0)}. Now let us consider the limit of u at the point z = (1,0) of the boundary. We have u(r,O) = 1: the normal limit at the boundary is 1 which is moreover half the sum of the tangential limits. Given ( E Q, we denote by IX the angle (z - (, z) (see Fig. 20) and show that for fixed IX

lim ~ -+ ::

(z - (. zj = "

v 1 IX u(~) = - + -- .

2 1t

Because of the symmetry we have u(x, y) + u(x, - y) = 1, with the result that we can restrict IX to ] 0, ~ n [.

Fig. 20

Making use of polar coordinates, we then have

r sin 8 tgIX = ---

I - r cos e from which we derive

1 + r

I - r

sin(O + IX) + sinIX

sin(O + IX) - sin a: .

W'h v hOI + r . 1 + r 2 tg a en I, -> z, we ave -> 0 and --- --> + 00 wlth ~~ - --- . . I-r l-r 0

490 Chapter I I. The Laplace Operator

Hence,

1 1 + I' u(r, 8)' = Arctg --- x

7[ 1 I'

1

o tg 2

1 1+1' () + Arctg- - x tg-J[ 1-1' 2

1 1 ->- Arctg (+ x) + - Arctg (tg:x)

:x - + -. 2 J[ 7[ J[

D

Example 7. We can obviously consider the Dirichlet problem P(Q. \') for an unbounded open set Q and a Radon measure \' on the boundary r of Q.

with a condition of growth at infinity. Let us consider the case of the half-space Q = IRn' 1 X IR + = {(x', Xn) } with boundary r = IR" - 1 = {x'}. Poisson's integral formula (see Proposition 22 of §2) tells us that for <p continuous and bounded on [Fgn-l, the bounded solution of P(Q, <p) is given by

(6.46) I/(x'. XII) = ~~n J. -_. _<p(~)_9t -~,.- . (In L1'" (It - x'l- + X~)",2

This formula again has a meaning for <p ELf (IR"- 1). More generally we can define the solution of P(Q, v) by

for every Radon measure \' on [Fg"- 1 satisfying

f Let us now return to the case of a function (P ELf ([Fg" .. 1 ) and show that for every Lebesgue point of <p164 and every (' > 0, the solution u given by the formula (6.46) satisfies

lim u(x'. xn) = <p(z) . x" ~ 0

lx' - ::1 !.Sc.\11

Since almost every point z E iR"- 1 is a Lebesgue point (see Sadosky [IJ) this will prove Proposition 7 in this particular case. We can always suppose that z = 0 and <p(z) = o. By a change of variables. we have

and hence for fixed R > 0

104 That is to say lim ~ r I(p(t) - rp(2)ldt = 0 r -+ 0 r OJ B(;:;,r)

96, Regularity 491

Since, if we suppose I x' I ~ ex,,,

J~ l<p(x' + xnt)ldt = x;n+1 f 1<p(t)ldt ~ x"-n+l f 1<p(t)ldt, 8(0, R) B(x'. RXn) B(O. (R + c)xn)

from the hypothesis: 0 is a Lebesgue-point of 7:;, we obtain the required result. o Let us consider now the homogeneous Dirichlet problem PH(Q, J.1) for a Radon measure J.1 on Q. For all x E Q, the function t -> G(x, t) is u.s.c. negative on Q

(taking the value - ex; for t = x), hence, for every positive Radon measure f1. on Q,

we can define the function

(6.4 7) r(J.1, xl = f G(x, t) dJ.1(t) , Q

on putting v(J.1, x) = - 'X! if the function - G(x, .) is not integrable with respect to J.1; the function v(J.1) is u.s.c. negative on Q. Let us prove

Lemma 6. Let Q be an open set o{[Rn satisfying (6.37), fl a positive Radon measure on Q. Thefimction V(fl) defined by (6.47) is locally integrable on Q (with respect to the Lebesgue measure) ~ff fl satisfies

f dist(t, cQ)dJ.1(t) < x; . Q

Then v{J;) is sub-harmonic (u.s.c. negative) on Q, and Llv(J.1) = pin ,Q'(Q).

Proof Given € > 0, we put Q, = {x EO Q; dist(x, cQ) > r:} and define the function

X EO Q.

We have, for all x EO Q,

vc(X) 1 1'(p, xl when t: 1 0 .

Using the inequality (see (4.27))

EnClC - t) - En(t5) ~ G(x, t)

where t5 is the diameter of Q, we see that

vf.(x) ): w,(x) - En(t5) f dfl(t) Q,

where We is the Newtonian potential of the measure with compact support Pc = PXa, on [R". From Lemma 3 of §3, v, is integrable on Q: it is also in U(Q)

for all p < nl(t! - 2). Also. making LIse of Fubini's theorem, we flnd that for all


f l', ,~~dx = ( dli(t) fG(X, t)[j~(x)dx = JI' ~(t)d!I(t). J~ ~

Hence Al', = fif in v' (Q) and in particular v, is sub-harmonic on Q (see Definition 15 of §4). (Jiven a compact set K in Q, for f: > 0 sufficiently small, K c Qf and hence G is of class If, I on K x Q\QI' Using (6.38), we see there exists a constant c > Osuch that, for ,: > ° sufficiently small

Cdist(t, (~Q) ;( IG(x, 1)1 ;( C I distIl, i'Q) ,

We deduce that for I: > 0 sufficiently small

X E K

rUt, x) ;( C I r dist(t, (lQ) dfi (t), x E K JQ [2,

from which the result follows. o

Remark 3. From the above proof, for a positive measure p, either 1'(fi) E LP(Q)

+ LJ~c(Q) for all p < nj(n - 2), Of V(fi, x) = -X!, a.e. x E Q. 0

Let us now consider a Radon measure fi on Q (not necessarily positive) satisfying

r dist(t, (~QJdl!dt) < x ..,Q

(6.48)

where Ipi is the total variation of the measure Ji. From the lemma and Fubini's theorem, for almost all x E Q. the function r, G(x, t) is ,u-integrable, the function

I'

1'(jI,X) = JQ G(x, t)dflU) a.c. XEQ

is locally intcgrablc on Q and

/Jt(,u) ~ •.• Ji III (/(Q).

Wc can now state the complete form of Herglotz's theorem.

Theorem 2. Let Q he an open set in iR;" salish'illY (6.37) and u: 1:2 --> [ - :1_. +x]. (I) The followiny assertions are equil'a!ent: (i) there exiSI WI' W z: Q --> [O,X!] I.s.c. locally inleyrable superharmollic such that U = WI - w2 a.e. on Q; (ii) there exists a Radon measlire f1 on Q satisfyiny (6.48) and v a Radoll measure 011 r such that i j" ?G

u(x) = G(x, t)df1(t) + "(x. tjdl'(t) a,c. Q I vn

(2) when the properties (i) and (ii) are satisfied, (a) the measures f1 and v of(ii) are unique and Au f1 in ::t' (Q);

~6. Regularity 493

(b) the/unctions IV+ and IV: Q ~ [O,:X;] defined hy l65

IV±{X) = - t G(x, t)d,H+(t) + t ~~ (x, t)dl't(t), X E Q

are l.s.c., locally integrable and superharmonic and u = It' + - IV ... a.e. on Q; (e) given WI' 1V2 satisfying (i)

IV + (x) ~ lI'dx) alld IL (x) ~ 1\'2 (x) for all x E Q

Llw, ? bV 1 and Llw ? /lw2 ill Y'(Q).

Proof Let us suppose that (ii) is satisfied. From Lemma 6, the functions v(tl -+-) and v(tl'" ) are u.s. c., negative, locally integrable and subharmonic; the functions u( v±), solutions in the sense of Definition 2 of the Dirichlet problems P(Q, I'±) are positive and harmonic. Hence the functions IV ± = - v{tl:;:) + u(1' ±) are I.s.c., positive, locally integrable and superharmonic; by definition u = IV + - IV _ a.e. on Q . We ha ve thus proved the implication (ii) => (i) of point (1) as well as (b) of point (2). Since Llv(tl±) = J~± in 0"'(Q) and u(I") E ]((Q), we have Au = tl+ - tl- == tl

in 9'(Q); hence tl is completely determined by u. From Proposition 6 (see Coroliary 2), \' is therefore also completely determined; this proves (a) of the point (2).

We consider now 11': Q ---+ [O,:x~ J u.s.c., locally integrable and superharmonic. From Propositions 20 and 21 of §4, /1 = -Llw (in the sense of distributions) is a positive Radon measure on Q. For I: > 0, the function

IV,(X) = - L,. G(x, t)dJl(t)

is a quasi-classical solution of P(Q, - lilli, ). Since

Llw f = - Pili, ? - II = ,;lw In C/'(Q) and

W,(:) = 0 ~ lim inf w(x) for ali : E r ,

we have w, ~ It' a.e, on Q. Passing to the limit

- l'Ui) = lim IV, ~ IV a.e. on Q. , ~ 0

In particular v(Jl) is iocally integrable and so from Lemma 6, tl satisfies (6.48). Also, we have Av(p) = tl = -Llwin9'(Q),solt' + v(tl)isapositiveharmonicdistribution on Q; from Proposition 6, there exists v a positive Radon measure on r such that It' + I'(p) = Il(!') a.e. on Q. But the functions IV and -1'(tl) + u(l') being l.s.c., super-harmonic and locally integrable, the equality almost everywhere implies that they are identically equal (see Proposition 20 of §4).

16' I' +, JI- (resp ,,+ , \' -) denote the positive and negative variations of I' on Q (resp. \' on J") (th3t is, for the measure 1', ,,+ = 1(11'1 + II), p = 1<1111 Id. where 1111 is the total variation of 1').


Applying the result to WI and W2 we have thus proved the implication (i) = (ii) of the point (1).

Finally let, for i = I, 2,

Wi = - 1'(/1i) + u(v;) .

From uniqueness, we have II = Ju = /12 PI and v = \'1 - 1'2; from which, by definition,/1+ :;;;: Pl,/I- :;;;: /11 onQandl'+ :;;;: l'I'V :;;;: vzonT.Wededuce(c)of the point (2). 0

For 0 :;;;: IJ :;;;: I, we denote by 9Jlg(Q) the space of Radon measures P on Q satisfying

(6.49)

Theorem 2 shows that the map (I', r) --> - d/I) + u(v) is a (linear) bijection of 9)11 (Q) x 9.l1(r) onto the space of the differences of positive locally integrable functions (defined a.e. on Q) and sub-harmonic; this map is bi-increasiIJY: for

(Pt, 1'1)' (Pl' 1'2) e 9.l1 1(Q) x 9.ll(r). /11 ~ fll on Q and \'1 ~ \'2 on T itT

- ['(PI) + u(rtl ~ - ['(fl2) + U(I'2) on Q

and similarly

(fIt, \'1) =I (/1 2 , \'2)' PI ~ fl2 on Q and \'t ~ "2 on r

¢> - CULt! + lI(vtl > - l'(pz) + u(vzl on Q.

At the beginning of this section we studied the harmonic function u( v) for \' e ~))i(r) which we defined as the solution of the Dirichlet problem P(Q, rJ. Now we study the function 1'( pl, JI e ~l)1 I (Q), considered as the solution of the homogeneous Dirichlet problem PH(Q, /l). We know that if d/I = fdt withf e U(12), where p > In, then l'Ud e ((,o(Q) and is a quasi-classical solution of PH (Q, f); for JI e 9.11 tfQ) this is no longer obviously true: even if dp = jdt with Ie {(, f (Q) (in which case [:(p) e ((, '(Q)), the

r

hypothesis Ii e 9)/1 (.12) (which then be written I dist (t, /'12) I fU) I dt <x) is not o'li

sumcient to ensure that ['(/;) = 0 on r, in the classical sense. To specify in what sense d/d = 0 on T, we shall use the Sobolcv spaces W{P(.Q). We recall 166 that for I :;;;: p :;;;: ;£, WJ.P(Q) = {u e U(Q); the derivatives (lu/Dx i (in the sense of distributions) are functions of U(Q)}; it is a Banach space, which for I :;;;: p <% has the norm

IIUilw'r = (L (11111' + L I(~,r)d.\y" By definition Wci·p(Q) is the closure of C/(.Q) in W1'''(Q).

1 "', See Chap. IV.

~6. Regularity 495

In fact, in the case considered where Q is of class W 2 ,J, W6,P(Q) can also be identif1ed with the space of the u E w1,p([Rn) such that u = 0 a,e, on [R"\Q or again u = 0 in 9'([Rn\Q) (see Chap, IV). Let us now prove

Proposition 8. Let Q be an open set in [R" satisfying (6.37) and Jl a Radon measure on Q.

(l) Girel1 0 < 8 < 1, let us suppose that f.l E 9Jle(Q) (i.e, satisfies (6.49)), Then the solution 11 of PH(Q, JI} satisfies u = v(Jl} E W6'P(Q) with p = (11 - 8)/(n - 1), (2) Suppose {I E 9J'o(Q) (or 0 ~: 8 < 1. Theil the /imction u = VUI) is characterised by

V E W6' 1 (Q) and ,dv = Jl in 9'(Q),

Proof Given (1' ',., (" E 9(Q) let us consider the classical solution (

of PH( Q, I ~~:). We know from Corollary 1 and the hypothesis (6.37) that

( E C(j 1 (Q) and for 11 < q < CfJ

II grad (II Lq ~ C L Ii (Ii Lq

where C depends only on q, 11 and Q. Using the Sobolev inclusion l67 W 1 . q c::: ((jB

with 8 = q - n, since ( = 0 on aQ we have. for 0 < {] < q.- I

I((x)i ~ C dist (x, oQ)8 L II (i il "" 1.1 rJ

with a constant C (possibly different from the preceding one), depending only on e, nand Q,

Let us suppose, first of all, that Jl has compact support in Q. Then v = v(l1) is the quasi-classical solution of P(Q, PI; from the theorem of local regularity (see Proposition 6 of §3), we know that v E WI~/(Q) for all p < 11/(11 - I); from Proposition 1 (applied to n \supp P and 1'0 = an) we know that v E ((,1 (f}/ K); also v = 0 on (lQ with the result that by Green's formula, we have

i cv i o( i I-~-(idx = -- uI ~dx = - u J( dx Q ex; Q eX i Q

= - r (dp ~ CI IIU "" r dist(x, oQ)8dipl(x). JQ [If! JQ

This being true for the functions (1' ... , (/1 E 01(Q) arbitrary

(6.50) Ilgradv(I1)llu ~ C t dist(x, (lQ)odlpl(x)

n-8 n-8 where p is the index conjugate to q = ~' that is p = II _ I

167 See Adams [1].


Now let us take Il E ~Jl8(Q) with 0 < tJ < 1. We apply (6.50) to Il, = 1%2,

Ilgradu(llrlllu ~ c L dist(x, DQ)Odlpel(x) ~ c L dist(x, a,Q)8dllll(x).

Now from the growth of the map II -+ - v(pl,

Idp')l ~ IV{flll a,e. on Q.

From the definition of VUl), therefore

u(p,) -+ z;{p) in Lfoc(Q) and a.e. on Q.

In the limit we thus have again (6.50) for v(p) and, using the reflexivity of the Banach space W!'P(Q), v(pJ converges weakly to V(Il) in W1,P(Q). Since v(p,) E W~'P(Q), the same is true for v(p). It remains to show that v = Vitl) is characterized by the property

v E W6· 1 (Q) and Llv = Il in Y'(Q).

From point (1) and the linearity, we are led to prove the nullity of a harmonic function v E W6' 1 (Q). Let, therefore, v E W~·l(Q) be harmonic on Q; given IE 0?(Q), from Corollary I, the function u(n is in (6'j(Q). Since v E W6,1(Q) on approximating by the functions Vk E 0:'(Q) converging to v in Wl,1 (Q), we have

f vldx = J!' vLlv(f)dx = - f gradv. grad v(fldx . Q Q Q

Now let us approximate v(I) by the functions Uk E .@(Q) satisfying: grad Uk IS

bounded in L "'(Q) and grad Uk -+ grad v(f) a.e. in Q.

Then we shall have by dominated convergence

L gradv.gradv(ndx = lim Lgradv.gradUkdx.

But since v is harmonic and Uk E ;2(Q)

Therefore

L ufdx = - Lgrad u. grad v(f)dx = 0

This being true for all f E .@(Q), u is identically zero. 0

Remark 4. As appears in the proof, for Il E fmo(Q) with 0 < fJ < 1 and v E Lloc(Q), the following assertions are equivalent:

(i) u = v(p) a.e. on Q;

(ii) v E W6,1(Q) and Llv = JI in fW(Q);

§6. Regularity

n - e (iii) v E WJ·P(Q) with p = --1 and

n -f grad v . grad ( dx

JQ every function ( E WJ·q(Q) with q

n - e168

1 - e

4. Neumann Problem with Given Measures

497

+ L (dp 0, for

o

We recall the classical Green's formula: given Q, a regular bounded open set of ~n, u and v E yg',;UJ) n <g2(Q) with uJv - vJu E Ll(Q),

f (uJv - vJu)dx = f (u av - v ~~~)dY, JQ r an an / We also have the formula of integration by parts: for u E (t';(.(l) n <g2(Q) and v E ygO(Q) n <gl(Q) with

vJu + gradv. grad u E Ll(Q),

i f CU (vJu + gradv.gradu)dx = v:1 diJ •

Q r en

In Sects. 2 and 3, we have used these formulae to define the trace on the boundary of a function, harmonic in Q: from the Definition 2 and Lemma 5, for example, when Q satisfies (6.37), given a function u E .1f(Q), a Radon measure v on r is the trace of u on r iff

i f av u Jvdx =~dv

Q r en

for every function v E (t',; (.d) with .1v E Et:(Q) and v = ° on r. In other words, the trace of u E yt(Q) on r, is defined as the measure such that Green's formula is true for ali r varying in a space ()( "lest functions". Such a defillition will be called variational. The variational method will be developed in Chap. VII in an abstract context. Here we propose, for Neumann problems, to relate a variational notion of solution of these problems to more classical ideas; notice also that the functions considered here will vary in classical spaces of measures or continuous functions which, not being reflexive, would take badly to an abstract theory. First of all we note:~

Proposition 9. Let Q be an open set of ~n satisfying (6.37), !1 a bounded Radon measure Oil Q alld u the solution or the homogeneous Dirichlet prohlem 169

U E WJ,l(Q) , Ju = p in 9'(Q).

168 We recall that W!)·q(Q) c '(,°([2) for q = (n - 0)/(1 - 0) which shows that In ~ dp is effectively

defined for:; EO W,\·q(Q) and I' EO 9JI" (Q). 169 See Proposition 8.

49~ Chapter II. The Laplace Operator

There exists a unique Ij; E L J (l) such thai

(6.51 )

filr a/ll' E ((,'(~2). Also

(6.52) (i)

(6.52) (ii)

j' 1'1/1 d',' = r l'dp + f grad 1'. grad u dx I ~Q Q

Jl t 0, II;?: 0 on Q => inf ess Ij; > 0 . r

Proof We noticc first that the uniqueness of If ELI (l) satisfying (6.51) is immediate since, taking account of the regularity of Q,

I'

I ['Ij;d;' = 0 for all l' E ((,I(Q) => Ij; = 0 a.c. on r. oJ£!

Let us suppose first that the measure II has compact support K c Q. From Proposition 8, u is the quasi-classical solution of PH(Q, p) and hence, from Proposition 1, II E ((, 1 (Q\K). We prove that Ij; = (lU/an satisfies the Proposition. The relation (6.51) is true for every function t' E ((,1(,(2) with compact support contained in Q: it can then in effect be written

f l' dll = - f grad l' . grad u dx

which is the deHnition Of.Jll = P inC/'(Q). The relation (6.51) is also true for every function [' E ((, J (Q) with support disjoint from K: the field p = l' grad 11 is continuous on !2, of class ((, 1 on Q and

div p = grad r. grad u ELI (Q)

with the result that (6.51) amounts to the classical Ostrogradski formula (see Proposition 4 of § 1). The relation (6.51) is thus true for every function [' E ((,1 (Q) which can always be written as the sum of a function with compact support in Q and of a function with support disjoint from K. When II ;?: 0 on Q we have !i ~ 0 on Q and hence Ij; = cU/?1l ;?: 0 on r; more precisely, if litO on Q. from Hopf's maximum principle (see Lemma 2 of §4)

If; > 0 on r and since VI is continuous, min Ij; > O. Considering II + and p-, the [

positive and negative parts of the measure II. we denote by U + and u ... the (quasiclassical) solutions of PH(Q, II') and PH(Q, II·· ). We have 11 = 11" .- 11_ and hence for Ij; = ("lI/i':/1

r I If; I d',' ~ r (1(:~llt + 1(~11l I)d';' = 1 (~ll1td~' + 1 (~I,;d:' f di/ll·

Q

96. Regularity 499

The second-last equality is a consequence of Gauss' theorem

f eu I' -d'c = d,u

r en I J Q

which is merely (6.51) with v == 1 on Q. Let us now consider the general case of an arbitrary bounded Radon measure ,u on Q. We take an increasing sequence of compact sets Kk of union Q and denote Jlk = ,uXKk and Uk the (quasi-classical) solution of PIl(Q, ,ud. Using Lebesgue's theorem, we have

r dlJlk - ,Ill = r dlpl -> 0 when k ----+ x; . JQ Jl2K,

From Proposition 8 we deduce that Uk -> U in WI. 1 (Q)- and even WI. P(Q) for all 1 ~ p < 1'1/(1'1 - 1). Now, applying (6.52) (i) to the measure Pk - II (with compact support in Q), we have for k ?: 1

CUll 1 r' --:1----- di' ~ d Illk- III I = . d II i •

en Q • K, .K,

We deduce that (aUk/an) is a Cauchy sequence in L 1 (F); we denote its limit by 1/1. Passing to the limit in (6.51) and (6.52) (i) applied to I/Ik = eUk/CIl, we obtain (6.51) and (6.52) (i) for 1/1. Finally, in the case J.I ?: 0 on Q, the sequence (ud is increasing with the result that so is the sequence ((';Uk/?Il); hence

inf ess 1/1 r

for all k;

Cli if. in addition, P -t 0, then there exists k such that Ilk -t 0 and hence ~in I'llk > O.

o

Remark 5. Proposition 9 states 1 70 that the normal derivative of a function U E W6· 1 (Q) whose Laplacian is a bounded measure on Q is an integrable function on r. We notice that ill general, this normal deriL'atil'e is not more than inteyrahle 011

r. In effect, let us suppose that there exists p > I such that for every bounded measure 1-1 on Q, the normal derivative 1/1 of the solution Ii of PIl(Q, J.I) is in If(f'); from the closed graph theorem, there will exist C slIch that

Ii 1/1 11[p(/1 ~ cJ~ dlill Q

and hence, in particular, applying with Ii = (\' Dirac mass at x E Q,

II (:G (x, ·)11 :;;:; C for all x E Q , ( 11 U'

(6.53)

1 '0 See Definition 3 below.


where G is the Green's function of Q. We can easily verify in the case of a ball that

thc relation (6.53) is impossible for p > 1, by using the explicit value of r"~ (x. :) <'Il

given by Poisson's formula. Another argument showing the impossibility of (6.53) consists of noting that if (6.53) is true, then for all <p E cgo(n the solution u«p) of P(Q, (j)) will satisfy

lu(p, x)1 = I r <:G (x, Z)(P(:)d}'(Z)1 ~ C Ii (j) ilL. x E Q , ~ I (n

where q is the index conjuge to p, and therefore

max l<pl I

sup lu«p)1 ~ Cil(j)ilul Q

which is a contradiction with q < 'X!, that is, with p > 1.

Proposition 9, and Proposition 10, lead us to formulate the

o

Definition 3. Given Q a bounded open set in [R" and u E W t. 1 (Q) whose Laplacian is a bounded measure p on Q, we call normal derivative (in the sense of distributions) of 11 0/1 f the distribution \' 1 defined on [R" by

(6.54) <I'l' D = r ~dp + JI' grad~.gradudx, ~ E U([RII) . .Ill Q

Note that the distribution \'1 defined by (6.54) is of compact support In the boundary r of Q: if supp ~ n r = 0, then

~X.1i E '/(Q) and grad(~XQ) = (grad~)Xl!

with the result that by thc definition of the Laplacian in the sense of distributions

If} ~ df! ,= < ,1u, ~XQ> = - I grad (~XQ . grad 11 dx = I grad ~ . grad u dx . !]

Proposition 9 can be stated with this definition: for every bounded Radon measure f1 on Q, the solution 11 of PH(Q, p), which is in Wb'P(Q) for all p E [1, n/(n - 1)[ (see Proposition 8) has for normal derivative on r a measure with density with respect to the surface measure of f: dl' l = Ij.td)' where Ij.t E Ll(f) is defined by (6.51). We arc interested now in the case where Vl is a Radon measure (carried by r); let us prove:

Proposition 10. Let Q be an open set of [R" satisfying (6.37), jJ. a bounded Radoll measure on Q and 1'1 a Radon measure Oil f. Let us now cOllsider the N eumarln problem PN(Q, p, I'l)

(6.55) {u E rf'd.l(Q), Au = fl ill 9'(Q)

I'l is the Ilormal derivative on r of u

(1) There exists a solution of P N (Q, p, \'l) iff

(6.56) f dl'j = f dp. r f2

§6. Regularity 501

(2) When (6.56) is satisfied, the solution of (6.55) is defined to within an additive constant l71 and u E Wl.P(Q)for alii ~ p < nj{n - I).

Proof The Gauss condition (6.56) is obviously necessary: more generally, for U E Wi. I (Q) with Laplacian fJ. a bounded measure on Q, the normal derivative VI

(with compact support) satisfies

(VI' 1) = In dfJ.

(apply (6.54) with ( == 1 in the neighbourhood of Q). We prove now the solution of PN(Q, fJ., vd, if it exists, is unique to within an additive constant; by linearity, this comes down to showing that a solution of

U E Wi. I (Q), I grad(. grad udx = ° for all 'E '@(lRn) !l

(6.57)

is constant on Q. Given f E .~(Q) with I/dX = 0, we have the existence of a

solution of the classical homogeneous Neumann problem

- av v E (gl(Q) n «i2(Q) , Llv = I on Q, -;:;- = 0 on r

on

(this follows from Theorem 2 of § 4 and from the regularity results of § 6.1). If u is the solution of (6.57), we have

(6.58) I n1u dx = - I n grad v. grad u dx = 0 .

The first equality in (6.58) is obtained by approximating u by the functions Uk E «il (Q) such that Uk -> U in WI. I (Q); the second equality is obtained from (6.57) by now approximating 1) by functions (k E 01(lRn) such that (k -> v in «i1(Q). These approximations are possible due to the regularity hypothesis on Q. By the density of 9(Q) in L I (Q), we therefore have from (6.58)

InfudX = 0 forall fEC'(Q) with I fdx = 0 Q

from which we deduce that u is constant on Q.

Finally we prove the existence of a solution u of PN(Q, fJ., vtJ in WI, P(Q) for all 1 ~ p < n/(n - 1). From Propositions 8 and 9, we can reduce to the case fJ. == ° on Q: in effect considering ii the solution of PH (Q, /1), u is the solution of PN(Q, fJ., VI) iff u - ii is solution of PN(Q, 0, VI - vd where Vj is the normal derivative of ii, a Radon measure with density, from Proposition 9, Suppose, first that VI is a measure with density t/J E 'b,o(r) satisfying the Gauss condition

171 For simplicity, we supposed in (6.37) that Q was connected; in the case where Q has several connected components, we have the same result with the Gauss condition (6.56) and to within an additive constant on each component.


("

J r t/J dr' = O. From Theorem 2 of § 4, there exists a classical solution u of the

Neumann problem

('U U E ((, ~ (!2) n .11(0), -;;- = t/J on r.

en

We know (see Proposition 5 of § 1) that I grad uI2 ELI (0) and hence a forTiori U E rt71 • 1 (0). Given ~ l' ... , ~Il E 0"(0), then as we have just seen above there exists a solution t· of

on 0,

and, by integration by parts, we have

Dn ~.- = 0 on en r.

f I~il'.~~U dx = - j' u,.1l'dx = r gradu.gradvdx = f (~Uvdl" !2 .(Xi !2 vf2 r(!n

From the analogue for the homogeneous Neumann problem of Proposition 1, for all n < q < 'Y,

II rill' ~ CI ULq where C depends only on 0 and q; hence

I' j' I C~lI dx I ~ f2 ( Xi


Ilgradull LP ~ cll(:ull en Ll(n

where p is the index conjugate to q.

The function u being defined to within an additive constant, we can always impose

on it f u dx = 0; then we have from Poincare's inequality172 a

We conclude by approximating an arbitrary measure VI on r by measures with continuous density and passing to the limit. D

5. Dependence of Solutions of Dirichlet Problems as a Function of the Open Set: Hadamard's Formula

In § 4 we proved that, given an increasing sequence of open sets (Od with union Q,

the Green's function G of 0 is the decreasing limit of the Green's functions Gk of Q k

171 See Chap. IV. (Vol. 2. 126, 379).

§6. Regularity 503

(see Propositions 6 and 12 of § 4). Here, we propose to study more completely the dependence with respect to a parameter}. of the Green's function G;. of an open set Q A depending regularly on ;. and, more generally, to study solutions of Dirichlet problems P(Q}., q>;.,f;.).

We take a parameter i. varying in a domain A which, to fix ideas, we take to be real. We take a family Q;;. of open sets in IR" with boundary fA = cQA and we shall suppose that the open sets satisfy (6.37) uniformly with respect to ). E A which we shall express by the existence of U, a bounded open set in IR" containing all the boundaries fA and 1>: (I., xl E A x U -> IR of class W 2 • x. with respect to x E U uniformly for I. E ,1 such that

{ grad, 1>T().~ x) 'l 0 . for all ()., x) E ,1 x U Q) II /j - {x E U, 1>()., x) < O} for all ). E 11

with the result that for all ). E A

r A = r.y E U; 1>(1., x) = O} .

We denote by GA, the Green's function of the open set Q;. extended to zero outside of QJ. x Qic' From Proposition 8, for x E Q;., the function vi .. Ay) = GJx, y) is characterized by

Vi" x E WI. 1 (IR") , Vi" x = 0 in 0"(IR//\Q;.),

111:A• x = 6x in q'(Q,,J.

In addition, for all p E [I, n/(Il 1)[, vA.X E fV1. P(IR") and, account being taken of the uniformity with respect to i. of the regularity of Q;,

II t'.:t. x Ii K;P :( C, independent of i. and x .

We deduce that the map (i., x) -+ t'A.x = G;,(x,.) is continuous in {(I., x);), E A, x E Q;,} in W 1. P(U:g") with the weak topology, subject to the continuity of the function 1> with respect to). which ensures that for all ;.0 E A and ( E 9(Q;,o) (resp. 0'(IR"\fJAo )'

we have ( E 9(Q;.) (resp. Q(lRn\Q,J for)' sufficiently close to 1.0 , This allows us to prove

Proposition II. With the abore data in supposing 1> continuous with respect to )., leI us consider q>: A x U -+ IR continuous and denote for all J. E.1, (P;.: Z E r A -+ cp(l., z) and u). the (classical) solution ofP(Q;" cp.d. Theil thefunction (i, x) ~ u,(x) = u()., x) is continuous 011 the set {()., x); ;. E .1, x E QJ.

Proof Given ).k -> ;·0 and x k .. 4 Xo with X k E Q'k' we wish to prove that U;.J"'k) -+ u;'o(xo)· From the principle of the maximum, denoting q>k: Z E r;,. -+ q>()·o, z) and Uk the solution of P(Q;'k' cpd, we have

From the continuity of q> on A x U it is sufficient therefore to prove that Uk(Xk) --> U;,.,(xo); in other words, we can suppose q>()., x) = (p(x) independent of X


Using the same maximum principle and the density of!(l'L in !(10, we can suppose that cP E !(IW. Finally since only the values of cP on a compact set of U are involved, we can suppose that cpU-, x) = cp(x) on A x U with cp E ceCO([R"). We then have

uA(x) = cp(x) - f G;Jx, y) dcp(y)dy,

the result IS then deduced from the continuous dependence of G Jx, .) proved above. D

Returning to the Green's function G A' we have

GA(x, y) = E,,(x - y) - uA(cpxl where cpAy)

we then deduce the

r . A '

Corollary 4. With the above data and supposing <P continuous with respect to )" the function (), x, y) -+ G .lx, y) is continuous on {()., x, y); J. E 11, x of- y} .

In the same way we shall have results on continuous dependence of the solution of P(QI<' CfJ;.,{,J; we leave it to the reader to supply the details. We shall now study difTerentiability with respect to A. We state Hadamard's formula:

Proposition 12. With the above data, let us suppose <P is olclass (6 1 with respect to

X Then the function (}, x, y) -+ G lx, y) is of class !(lIon

and

(6.59)

where

[V,x,y); ;.EI1,X,YEQ;, x of- y}

D ;)" G;.(x,y) (i),

f eGA aGIo ' . . ~." (x, z) --;;-.. (y, Z)p(A, z)dYAlz)

I en en ).

a<p ~V,x) (')

.p(i" xl = --'-----;-- - . Igradx<P(A, x)1

We note that from Corollary 4 and Proposition 1. account being taken of the uniformity with respect to )" of the regularity of the open sets Q;., the function G A(x, y) is of class '6 1 with respect to (x, y) on

l (i" x, y); i" E 11, x, y E Q;., x # y} .

This shows in particular that the term on the right in Hadamard's formula (6.59) is continuous in (A, x, y). It is sufficient therefore for ;, E A, (x, y) E Q). with x # y to show that

1 lim } .... (G; + h(X. Y) - GiJx. y)) h-' 0 1

exists and is given by (6.59); similarly we content ourselves with proving it for

96. Regularity 505

lim or lim since a function differentiable on the right and with continuous h~O h~O

">0 "<0 right-derivative is continuously differentiable. We shall be content here to prove Proposition 12 under the hypothesis that the map), ~ Q;. is monotonic (increasing or decreasing); we refer the reader to Garabedian-Schiffer [lJ for a proof in the general case.

Proof under the hypothesis of monotonicity of ), ~ Qk We can suppose that A -+ Q) is monotonic increasing. For ;, E A, x, Y E Q)" X =1= y let us prove that

lim h! (C;. + h(X, y) - C;,(x, y)) h~ 0 h> 0

exists and is given by (6.59). We fix A E A, and y E Q),; for simplicity of writing we put:

Q = Q)" r = T;." C = C)" p(z) = p(}., z), u(x) = C(x, y) .

For all h > 0, smail, we put uh(x) = CHh(X,y). Since Q Hh ~ Q;._ = Q,

the function Vh = h - I(U h - u) is harmonic on Q and continuous on Q with vh(z) = h lUh(Z) for all Z E r.

. . . gradx cJ>p" z) We fix Z E T. The extenor normal n(z) IS gIven by n(z) =-. From the

Igradx cJ>(A, z) I implicit function theorem, for h small there exists a unique t{h) in a neighbourhood of 0, such that z + r(h)n(z) ETA + h' that is to say

(6.60) cJ>(;. + /1, z + t(h)n(z) = 0 ;

The function t(h) is of class (tf; 1 and by the development of (6.60)

acJ> -1~ (i, z)

dt c/, ......... (0) = - -- .. dh gradx cJ>(i, z) .11(z)

_. p{z) .

Hence by the mean value theorem

o = UI/(z + t(hln(z») = uh(.::) + t(h) grad uh(z + 8(h)ll(z)). n(z)

with 0 < f!J~) < I . t(h)

From which on regrouping, we have

with

R (h, z)

au 1Ih(z) = hp(z) -;;- (z) + R(h, z)

en

. au t(h)(grad 1I(Z) - grad uh(z + O(h)n(z))). n(z) - . (z)(t(h) - ht'(O) . en

From what we have seen before the beginning of this proof, the function (h, x) -+ grad uh(x) is continuous with the result that R(h, z) = o(h) uniformly with


respect to Z E r. Using the principle of the maximum we see that Vh -+ I' in (6°(.6) I 1) . . ell

where v IS the solutIOn of P ( Q, p -. . (~n ,

o

The Hadamard formula (6.59) allows us to obtain the derivative with respect to ;, of the solution u; of

P(Q A, t.p;,f~) where (P;(:) = t.p(i., z) and f~(x) = fk x)

with t.p andffunctions sufficiently regular on 11 x [R". Considering, for)' E ;1 fixed, VI! the solution of P(Q; + h' (PhJh) with

t.p'I(Z) = t.p(J., z), f~(x) = f()" x) .

solution of p(Q. t.p(i. + h,.J - t.p(i.,.j j'(~+ h,.j -/(;.,-1). At h, h . h .

f . . 11; + h - I'll I" rom PropOSitIOn 11 --- ---- -+ IV so utlOn of . h

We can therefore reduce the problem to the case in which t.p andfare independent of X. Let us state

Corollary 5. Under the hypotheses of Proposition 12 let

(P E W~)/([Rn) and fE Lfoc([R") fi)f' P > II .

For all i. E /1, dellote by 11.1 [he solurion of P(Qk , t.pXr ,fXu ). The funetioll (i., xl -+ 1I}Jx) is of class (6 1 Oil {()., x); ), E /\,\.: E Q;: alld

(' !' ('G· ?l' ,. u."(xl =, l A (x, z) 1" (zlp(i.. z) d~';(zl CA ~ 1', (I! ( I!

where Pi, is the solution of' PH(f2 i,. (f - ,.1t.p)XQ;) .

Proof We have 11;. = (p + l'A: replacing f by f - /1t.p we can therefore suppose that t.p == O. We then have

lI,J,X) = f G;Jx. y)f(y)dy, ,X E Q;: Q;

from which for ;, E 11 and x E Q;, fixed, we deduce

llA + ,,(xl - U.l(x)

h :1 f GA + h(X, ylf(yldy

!l;., Q;

J~ Gi.+,,(X,y) - GA(x,}'). + ----- --- - - f (l'ld)' . Q, h . . .

U sing Corollary 4 and G A + ,,(x, y) = 0 for Y E r i + II the first integral tends to zero.

§7. Other Methods of Solution of Dirichlet's Problem 507

From Proposition 12, the second integral tends to

o

§7. Other Methods of Solution of the Dirichlet Problem*

1. Case of a Convex Open Set: Neumann's Integral Method

We take up again the solution of the problems of Dirichlet and Neumann by the integral method when the open set Q is convex: we shall obtain the solution by an iterative method due to Neumann. 173 First we recall the notation of the integral method (see §4.l5). Let Q be an open set in [Rn which, in the first instance, we suppose to be regular bounded with boundary of class !C1 + '; for a E !C0(r) we define (see (4.99))

(7.1) 2 f (t - z). n(t)

Ka(z) = _. n a(t) dy(t), z E r . (In r It - zl

The function Ka E !C0(r), and the interior (resp. exterior) double layer potential

(7.2) 1 f (t - x). n(t) -

u(x) = - I In a(t) dy(t), x E Q (resp. [Rn\Q) , (In r t - X

is the solution of the Dirichlet problem P(Q, t(Ka + a)) (respectively p([Rn\Q, t(Ka - a)) with the null condition at infinity)l73. In other words, given cP E !C0(r), to solve the Dirichlet problem P(Q, cp) (resp. p([Rn\Q, cp) with the null condition at infinity) by a double layer potential we are led to solve the integral equation

(7.3) Ka + a ( Ka - a ) 2 = cp resp. 2 = cp .

Now let us suppose that the open set Q is convex, which can be expressed analytically as

(7.4) (t - x). n(t) ~ 0 for all t E r and x E Q.

173 See § 4.5 and Proposition II of § 3.


The operator K is then increasing:

(7.5)

Using the equality (see Proposition 15 of §4)

(7.6) KXr == I on r, we deduce that

(7.7) mm 'Y. ~ K 'Y. ~ max 'Y. on r r r

and hence, in particular, that

(7.8) max IKe<1 ~ max le<I . r r

We note, furthermore, that for an open set Q, the properties (7.5), (7.7) and (7.8) are all equivalent to the convexity of Q. Since Q is convex, the open sets Q and IRn\Q are both connected; we know by Fredholm's method (see Proposition 19 of § 4) that for all qJ E (6 o(r) (resp.

satisfying f r 'Y.qJ di' = 0 174 ) the integral equation (7.3) has a unique solution (resp.

to within an additive constant). We shall develop an iterative method to determine these solutions: besides, this method will prove the existence of these without any appeal being made to Fredholm theory. Given e< E '6 0 (r) we put

(' = , (7.9)

We have

and

(7.10) maxlil max'Y.

We prove:

max'Y. + mm'Y. i

'Y.

2 'Y. - (', .

.. J i for all (' E IR + C

max'Y. min'Y. min i = ------

2 min max I'Y. + c I .

r

Proposition 1. Let Q he a convex reyu!ar hounded open set of'lR" with hOlllulary r of' class 0 1 + e and qJ E 0°(r). We define recursirely the sequence of' functions 'Y. k E (f,0(r) hy

~~

(7.11) ''Y.o = rjJ, 'Y. k = K e<k - 1

where the map 'Y. --> i is defined hy (7.9) and the operator K hy (7.1).

I 7. ~ = L - I i' I is the normal derivative of the solution of the exterior Dirichlet problem P( il" Q. i'rI satisfying the null condition at infinity.

§7. Other Methods of Solution of Dirichlet's Problem

(1) L max I IJ(k I < 00 and hence, in particular, we can define k I'

(7.12) lJ(i = 2 L (-- 1 )klJ(k k

the series being uniformly convergent on r.

509

(2) There exists a unique c i E IR such that the interior double layer potential defined by (7.2) with IJ( = lJ(i + ci is the solution of the Dirichlet problem P(Q, cp). (3) There exists a unique ce E IR such that the exterior double layer potential defined by (7.2) with IJ( = lJ(e is the solution of the Dirichlet problem p(lRn\Q, cp + ce) tending to zero

The proof rests on

Lemma 1. Let Q be a regular bounded convex open set in IRn with boundary r of class ~I + '. There exists 15 E ]0, 1 [ such that

max IKexl ~ 15 max lal for all IJ( E ct'0(r) . I' I'

Proof of Proposition 1. Using Lemma 1, we have

maxlexkl = maxlKlJ(k-ll ~ 15maXlak - l l = 15maxllJ(k-ll

and hence . maxllJ( I L maxllJ(kl ~ maxllJ(ol L (}k = °

k 1 - 15

which proves the point (1).

Since the operator K maps «j0(r) continuously into itself, we deduce from (7.8) that

But

also

Hence k k

max cp + minq: 2

L (- I)/(KIJ(, + ex,) = cp + c<p - L (- I)'c/ + (- I)k+ IlJ(k+ I' ,= 0 ,= °

We deduce that D - l)kCk is convergent and k


Similarly the series LCk is convergent and

KIJ." - IJ."

2 = (P + c" with D

Proof of Lemma I. Given IJ. E (6°(F), we have from (7.6) that KIJ.

(7.10) that Ky. and from

maxIK~!

We put

1 (max Ka- min Ka) . 2

T+(a)

and obtain

[t E T; itt) > OJ. r (i) l t E T; i (t) < O} ;

K N(Z) 2 f (t - z).Il(t)_ d () - +(-) -'A.:::;- -I ------:-1" ex(t) i'l :::;(j(T ex)maxex (1" /'I'd t ~

where for every measurable part To of T

2 r (t - :).11(1) MTo) = max d;'(t).

I (1 n • I It - z In Similarly, we have

Ki(z) ~ 6(r-(a))mini ,

with the result that, using (7.10), we obtain

(7.13) _ (5(T+(i)) + b(r-(i)) _

maxlKrxi :::;--2-"- ·maxl:xl·

We note now that all of the preceding can be extended to a function ex ELf (Tl: we can define

Kex(z) = --- - - :x(t)d,,(t) for all 2 J" (t - z). n(t)

Un I It· .::1" .:: E r.

From the compactness of the operator K of (6° (T) into itself (see Proposition 16 of ~4). Krx E ({, o(T) and the map:x -> Kex is continuous on the bounded sets of L' (n provided with the weak-star topology. into ({, 0(1') with the uniform convergence topology. We can define Y. replacing max and min by sup ess and inf ess

I r r I

respectively. hom the compactness of the unit ball of L f. (r) for the weak-star topology, there exists :Xo E L' (I') with 11:(0 II L' :::; 1 such that

From (7.13) it is sufficient, therefore, to prove the lemma by showing that given :x E L" (r), IJ. =i= 0, we must have

b(F C (ex)) + ()(/'- (:x))

2 < I.

~7. Other Methods of Solution of Dirichlet's Problem 511

where b(T ± (rxl 1 is well defined, since r ± (rxl is defined to within a negligible set;

moreover b(T± (rx)) = max KXr+IxI' 1

Given a measurable part To of r, KXro = 1 - KXr\rO' with the result that 15(T 0) ~ 1 and there is equality iff there exists Zo E r such that KXr fo (zol = 0, that is to say account being taken of the convexity of Q,

T\To c Ho = {t EO 1R";(t - zol.n(zo) = O} (to within a negligible set).

Given two measurable parts T + and r _ of T, we therefore have

and there is equality iff there exist hyperplanes H + and.H _ such that

(7.14) T\T ± c H ± (to within a negligible set)

In the case T ± = r±(rx) with rx '# 0, we have r + n r _ = 0 and F t , f_ not both negligible, with the result that (7.14) is impossible. 0

Remark 1. In the proof of Lemma 1, we have used the extension of the integral operator K to L'Y:'(r). Given (P E LX(r), we can define as in Proposition 1 the sequence (rxk) by (7.11), the sums (Xi, rx C by (7.12), as well as the constants c i and ceo By a continuity argument, we show that the interior (resp. exterior) double layer potential of rx = (Xi + ci (resp. rx = rx C ) is the solution of P(Q, rp) (resp. P(IR"\Q, rp + eC ) tending to zero at infinity) in the sense of Definition 2 of§6. 0

In § 4.5, we have likewise considered the operator J defined by

2 f (::: - t). n(:::) h(:::) = ........ --~---n- rx(t) d~,(t), ::: E T .

(J" r I~ - tl

For (X EO ({,o(r), we have hE (6°1r) and the interior (resp. exterior) single layer potential

u(x) = L E,,(x - t)(x(t) di'(t) , x E Q (resp. IR"\Q)

is the solution of the Neumann problem PN(Q, 1(h - e<)) (respectively PN(IR"\Q,

-!(h + e<)) with u(x) = (j'(Xd/) En(x) + o( ---::-~:::I) when Ixi --+ ee). \.1 "xl

When Q is convex, the operator J is increasing:

(Xl ;:?:. :X 2 on r = J(Xl ;:?: Je<2 on T.

Let us now prove

Lemma 2. Let Q be a regular bounded convex open set in IR" with boundary r of class 'iS'1 +, and Ii a Radon measure 0/1 r.

. (z-l).n(z) .. (I) For d}' - almost all z EO T the jimctlOn t --+_._- IS I J11-l1ltegrable on T; and

Iz - tl"


the function 2 r (.:: ~ t).n(.::)

J .u(.::) = ....... - _ ~-,,- dll(t) d/ (J".!lltl

a.e. .:: E r

is d~'-inteyrabl('. (2) The interior (resp. exterior) simple layer potential

u(x) = LE,,(X ~ t)dJl(t) x E Q (resp. IR" \Q)

is solutiol1 of the Neumann problem PN(Q. 1(JJl fi)) (resp. PN(IR"Q.

t(- Jil + II)) with u(t) = (fId Ji ) En(x) + O(lxlr~-':T) when ix! -> xl ill the

sense ()f Definition 3 of ~ 6.

(3) When Ldll 0=.0 0, we hare LJ 11('::) d~'(,::) = 0 and

liJpIILl(J) ~ 6 fdiP' where () is the constant of Lemma I.

P f· S· n . h f' (.:: ~ t). 11('::) . .. F F b' . roo. mce 'l IS convex, t e unctIOn t -> - ---- IS posItIve. rom u 1I11S . I.: - [in

theorem

j~ d'I'('::) " 2(.:: .. ~ t) 11('::) dllll(rl r ~ I (Jill.: ..... t IF! . .

= J" K X ,(t) dill! (t) = I' d II-I I ( Il r wI

giving point (I). More generally, we have for all 'J. EO (f, °Ul

,:x(.::) JIl(.::)dj'('::) = r K:x(t)d/dtl. ~ , ~ , (7.15)

I n particular I,' .lId.::) d-,-(:::) = j' KXr(tJ dJl(l) = I" dJl(t). ~I r wI

When r dp = O. we have j' }fldi' = 0 and for all :x EO 'f, 0(1'1, "I" r

(0 (' r ,'" f' I 'J.Jlldj· = I K'J.dp = J .k;dJI ~ ()maxliIJ dip ~ ()maxl'J. I dlpl .)1 .,J r f I [.;f"

from which point (3) follows. Finally for point (2), we consider the interior (resp. exterior) simple layer potential

1/(x) = " E,,(x .. - t) dJd!) x E Q (resp. IR"Q) ~ r


When J.l is the measure of density :x E Cfo°(I'), then u is the classical solution of PN(Q, -!-(J:x -:x)) (resp. PN(IR"V2, - -!(J:x + a))); then, for all (E ~(lRn)

(7.16) f /~;~ (dy = 1..! grad (. grad u dx

( f h+a 1 ) resp. (d}' + _ grad (. grad u dx = 0 . r 2 W\Q

n From Proposition 10 of § 6, for 1 ~ P < ................ _ .... we have

n - l'

(7.17) IlullwlPIQ) ~ cpll la ; all ~ Cpll:xIIL'Ir) VIr)

(resp. given R such that Q c 8(0,· R)

1111 II w' P(B(O.R).Q) ~ Cp , R 11:x IlvI[")) .

Approximating an arbitrary measure fl by measures of densities ak E ceO(r) such that L (a k dy --> L (dfl for all (E ceO(r)

(and hence Ilak Ilv(r) bounded), the corresponding simple layer potentials converge in .~(lRn\r) to 11; from (7.16), Iludw1P(Q)(resp. Iluk ll w ,p(B(o.R).2)) is bounded. Finally, taking account of (7.15) and (7.16), for ( E .9(IR"), we have

r v J fl dy - dfl f K ( - ( ~ = ---dll

~ r 2 r 2

. f v lak:Xk = hm ~ di'

r 2

= lim f!J grad ( . grad Uk dx = f!J grad ( . grad u dx

( resp. similarly f (!J1~t_+~fl + f _ grad ( . grad u dx = 0) . r 2 R"\Q

o

Remark 2. Points (1) and (2) of Lemma 2 remain valid in the non-convex case: indeed, the operator K being a compact map of Cfoo (r) into itself, its transpose I K is a compact map of ml(r) into itseIr (see Dieudonne [1 J). 0

Now let us prove

Proposition 2. Let Q he a regular bounded convex open set oflR" with boundary r of class '6 1 + E and VI a Radon measure on r with zero integral. We define recursively the sequence offunctions 'Yk E L 1(r)

where 1 is the extension considered in Lemma 2.1 of the classical operator


(1) L liC(kllL'(r) < !YJ and hence-in particular we can define k ~ 1

r

C(i = - 2 I CX k , k=l

the series converging in mean on r.

~.

':Xc = 2 I (- 1)k+ 1 ':Xk '

k = 1

(2) The interior (resp. exterior) simple layer potential defined by the measure fli = r) - 2\'1 (resp. fl" = ':X e - 2v 1) is the solution of the Neumann problem PN(Q, VI} (resp. PN(~n\Q, vd tending to zero at infinity).

Proof From the point (3) of Lemma 2, we have for all k ): 1,

t. C(kdl' = t. dV I = ° and IlakllLv) ~ 15 k fr dlvll·

Point (1) of the Proposition follows. By the continuity of 1 in L 1 (1), we have

{ lC(i =

1 ':X"

Therefore

- 2 I ak+ 1 = 2C( 1 + a i

k ~ t

2 I (- 1 t + 1 ':Xk + 1 = 2a 1 - a" . k ~ 1

= VI and lfle + fl"

2 = VI'

Account being taken of Lemma 2 (2), this proves the Proposition. o

Remark 3. We can generalize the preceding results to the case of an arbitrary convex open set Q of ~n. We note first that the boundary r of Q satisfies locally a Lipschitz condition (and Q is situated on one side of its boundary); this allows us to define the surface measure y of its boundary and the exterior normal to dl' a.e. z E r; besides, at every point: E r, we can define

N(z) = {I1E~n;lnl = 1,(: - x).n): ° forall xEQ}

the set of exterior unit "normals" to Q at :. The "multi-map" : -+ N(z) is "continuous" in the sense

{\:/:Er, \:/8>0, :16>0 such that t E r (\ B(z, 6) ,11 E N(t) = dist(n, N(:)) < /;

and dy. a.e. : E r, N(z) is reduced to a point n(z). Using this (very weak) continuity and the (very strong) property (: - x). tI(:) ): 0, we can take up again the theory of simple and double layer potentials (see § 3.3) and generalize Propositions 1 and 2. 0

We note, however, that the operators K and J are no longer compact when the open set Q admits a corner (irregular point); these operators are then singular


Fig. 21

integrals of the type: . f (t - z). n(t) Ko:(z) = hm I In o:(t) dy(t)

• ~ 0 nB(Z.8) t - z

(see Y. Meyer [1] for a study of these operators in the L 2(r) spaces).

2. Alternating Procedure of Schwarz

We take two open sets Ql' Q 2 of IRln with boundaries r 1, r 2 respectively and consider Q = Q 1 U Q 2 whose boundary is denoted by r. The alternating procedure of Schwarz gives an approximation method of solving the Dirichlet problem P(Q, cp) from the solutions of the Dirichlet problems P(Q 1 , CPl) and P(Q2 , C(2)' We denote


r;

Fig. 22

516 Chapter Il. The Laplace Operator

Given (P E '(j0(1), it is clear that there exists a classical solution of P(Q, ip) iff there exist ipi E 't}°(r;) and Ui classical solution of P(Qi, ip;) such that

on r; for = 1, 2

on Qt n Qz .

Such a solution u of P(Q, ip) is then defined by

II = lI i on Q i for = 1,2.

We restrict ourselves to the case (~fQ hounded and we}ix ip E '(j0(1). For i 1,2 we denote

on 1" l i J

and we suppose that for all ipi E Ei(ip), the Dirichlet prohlem P(Q j , ipi) admits a classical solution denoted by u;(ipY 75.

Given ip lEE I (ip), we can associate with it the function T2 ip I defined on r 2 by

(7.18) { ip(z) on r~ Tzipdz) = ( ~)

. U t ip I , ~ on r 2 n Q j

r 2 = r'2 U (1'2 n ad and r~ n 1'2 n Q 1 = 1'1 n r 2 .

Similarly with I.f! 2 E E 2 ((p) we associate the function Tl ipz E Ed (p) defined by

(7.19)

Now for ipj E Ei(p),

iffl76

on

on

r 1

lIZ(ipZ) on Q j n Q 2

u2(l.f!z) on (1'1 n Q 2l u (1'2 ('! Ql)

that is, with the above notation, iff

Tzl.f!l = (pz and T1(pz = (Pl'

In other words. wilh lhe abore notation, the Dirichlet problem P(Q, ip) admits a classical solution iff the map T

((PI' (,02) E EI(rp) x E 2(1.f!) -> T((Pl' I.f!z) = (TJ (,02, T 2 1.f!!)

admits afixed point I7 !. From the uniqueness of the Dirichlet problem P(Q, (,0), if it exists, this fixed point (ipl' ipz) will be unique and the solution of P(Q, ip) will he defined by: II = lIj(ipj) on Q j for i = 1,2.

],' In particular this is true if all the boundary points of the opcn sets Q i arc. regular. (see Theorem I. ~4). ] 't, This follows from the maximum principle and from the fact that the boundary of Q] n Q, is contained in (I] n Q,) u (r, n Qd u (I] n r,). ]" Given a map T of a set E into itself, a fixed point of T is an clement of E mapped onto itself

(invariant under T).


We note that (CPI' cpz) is a fixed point of Tiff CPI (resp. 0/2) is a fixed point of the map SI = TI T2 (resp. Sz = TzTI ) and cpz = TzCPI (resp. CPI = TI cpz). There exist many ways of obtaining fixed points; the simplest method is certainly the method of successive approximations: starting from cp? EEl (cp) (resp. cP~ E E2 (cp)), we construct by recurrence

cP~ = TI Tzcp~ - I (resp. cP~ = Tz TI cP~ - I)

that is to say

<p~ E EI((p) gIven, <pg = T2<P~, <pi = TI<Pg, <pi = Tz<pi, etc .. .

(resp. <pg E Ez(<p) given, <p~ ,= TI <pg, cp1 = T2 cP~, (Pi = Tl <pL etc ... )

hence the name "alternating procedure": starting from a given value on the boundary of one of the open sets having the value <P on the part common with r, we construct, by solving a Dirichlet problem in this open set, a function on the boundary of the other open set with the value <P on the part in common with l' and we recommence working alternately with each open set. Given <PI' <PI E EI(<p), the function VI UI(<PI) - UI(<PI)' harmonic on QI' con-tinuous on Q 1 and satisfies

VI = 0 on ["'I' VI = <PI - <PIon r l (\ Q2 •

From the maximum principle, supposing CPI "¥= <PI' we have

I V I I < max I (P I - r 2\Ql

IT2 <PI - Tz<PII < maxl<pI - <PIlon 1'2· I,

In short, using the uniform norm in ("~O(1';)1 78,

II Tz<pj - 1'z<pj II < II <pj - <PI:I for all <pj, <pj EEl (<p), <PI"¥= <PI

and similarly

(7.20) II TI <pz - 1'1 <Pzll < II <pz - <Pzll

for all (pz, <Pz E Ez(<p), <Pz"¥= <Pz .

Although strict, these inequalities are not, in general, sufficient to ensure the convergence of the successive approximations (<pD. For that reason we shall introduce the quantity

(7.21)

where v~ is the solution of the Dirichlet problem P(QI' Xr; n n)

17" II'Pili = maxl'P'!' r;


The function Xl', n 1'2 not being continuous on 11, v? must be considered in a weak sense: when QI is sufficiently regular, we can put (see Definition 2 of §6)

° f cG I t'l (x) = -~- (x, z) dy(z), x E Q T, n n, en

(7.22)

where GI is the Green's function of QI; in the general case, it is sufficient to take

(7.23) v?(x) = sup{ vI(x); ['I E .ff(Qd (\ C(j O(Qd, VI = 0 on r'1' 0 ~ VI ~ I} .

This clearly generalizes that given by (7.22) in the case in which Q I is regular, since then, for all VI E £(Qd (\ '6°(Qtl

['dx) = f ('::1 (x, Z)VI (z)dl'(z) . I' (n

By linearity, we have for all VI E .if (Qd (\ (6 o(Qd with VI == 0 on r'1 ,

It' II ~ u? max I r I I on Q I Tl

and therefore

It'll ~ b(QI,QZ)max(utl on QI (\ l z Tl


II TZCfJI - Tz(PIII ~ b(QI,QzlliCfJI - cPlli forall CfJI,cPIEEI(CfJ)·

Similarly, we have

II TICfJz - T1cPzII ~ b(Qz, QtlllCfJz - (Pzli for all CfJz, (Pz E Ez(CfJ)

and hence

ilSiCfJi - SicPill ~ b(QI,Q2)b(Q2,QtlIICfJi - cPill

for all CfJi, cPi E Ei(CfJ) ,

where, we recall, SI = TI Tz, Sz = TzTI . Supposing that b = b(QI' QZ)b(Q2' Qd < I, the successive approximations CfJ~ converge uniformly on Ii; their limit CfJi is the fixed point of Si and we have the estimate

The Dirichlet problem P(Q, cp) admits a classical solution u which, by the maximum principle, satisfies

(7.24)

We have thus proved:

Proposition 3. Let QI, Qz be two bounded open sets in [R;n with boundaries II' 1 2 ,

Q = Q I U Qz with boundary 1 alld CfJ E (Ii 0(1). We put

1; = Ii n 1 and Ei(CfJ) = {CfJi E (6 0 (li ): (Pi = CfJ Oil 1;).


We suppose that for all CPi E Ei(cp), the Dirichlet problem P(Qi, CPJ admits a classical solution denoted by ui(cpJ Finally, we suppose that 0 = 0(Q1, Q:z}0(Q2, Ql) < 1, where O(Ql' Q2) is defined by (7.2\). Then the Dirichlet problem P(Q, cp) admits a classical solution u which can be approximated by Ui(cpn uniformly on Q; with the estimate (7.24), where the sequences (cpD are defined recursively by

cp?EEi(CP) given, cP~ = T2CP~,cp~+1 = TICP~ forall k ~ ° with T J, T2 given by (7.\8) and (7.\9).

We have always o(Q 1, Q2) ::;; 1, with the result that 0(Q1, Q2)0(Q2 , Ql) < 1 iff O(Ql' Q2) < 1 or 0(Q2' Qd < 1. We shall now give geometrical hypotheses sufficient for the condition 0(Q1, Q2) < 1 to be satisfied. We shall establish the results in the case n = 2:

Proposition 4. Let Q 1 and Q 2 be two bounded open sets in ~2 with boundaries r 1 and r 2 respective/yo We suppose that for all z E r 1 n (T2 n Ql)' (1) Q1 and Q2 satisfies the condition of the exterior segment at z, that is to say that there exists Xo E ~2 such that

ho + (1 - ).)z ¢ Q 1 U Q 2 for all A E [0, 1] ,

(2) Q 1 n Q2 satisfies the condition of the interior cone at z, that is to say that there exist Yl and Y2 in ~2, not aligned with z, such that

)'IYl + )'2Y2 + (\ - )'1 - )'Z)ZEQ I n QzforalU1,Az > ° with Al + )'2 < 1. Then odQ1, Q2) < 1.

Proof Considering v? the solution of P(Q 1, XF. nIl) we have to show that I 2

01 = sup v? < 1. By the maximum principle, v? < 1 on Q1 n r 2 with the {}t nr2

result that it is sufficient to show that:

for all

lim sup v?(x) < 1 . X --t z

Hence we fix Z E r1 n (Q 1 n r 2); the hypothesis is represented by Fig. 23. For r > ° sufficiently small, we denote by Q'l the disk sector

B(z,r)\{z + }'o(xo - z) + A1(YI - z);}'O')'1 ~ o}

and by V'I the solution of P(Q~, 1 - Xs) where S is the segment

rho + (1 - }.)z;). E [0, 1]} .

The boundary of Q 1 n QI1 is made up of

r 1 n B(z, r) which for r sufficiently small is disjoint from Q2 with the result that v? = ° ~ V'I on this part of the boundary;

iJQ~ n Q 1 which is disjoint from S with the result that v? ::;; 1 = V'I on this part of the boundary.


Fig. 23

By the maximum principle, v? ~ Vii on DI n D~. Now, for r sufficiently small, r 2 n Din B(z, r) is disjoint from the cone

C =:z + i.le}') - z) + idyz - z); ).1' ;'2 > O}

and therefore

lim sup t:?(.x) ~ lim sup 1"1 (x) . x--+ :: X ---t;:

XE 12rl fl, XEn; c

In other words, we are led to prove that

lim sup V'I (xl < I . x-z

XE f);',C

To within a similitude, this is a result of the following lemma. o Lemma 3. For 0 < 'Y. < 2n, we denote by Dx the sector oj'the ullit disk in [Rz formed by the points oj'polar angle B E JO, 'Y.[; let t', be the solutioll oj'P(D" 1 - Xs) where S is the segment [0, 1] of the axis oj'the abscissae. Theil, f()r all ° < fJ < 'Y.,

. [J hm sup v,(x) ~ x~o'Y.

XEO'J

Proof We shall identify U:g 2 with the complex plane C Given ° < 'Y., # < 2n, the conformal transformation z -> zP!X transforms Do to DfJ; using the invariance of harmonic functions under a conformal transformation (see Proposition 23 of § 2), we have va{z) t'/l(zlJ;,) for all z E Da' In particular

1'a(z) = l'n(ZIT;") for all z E Da .

An explicit calculation of Vn gives (see Sect. 5, (7.99))

(7.25) . 1 [ (1 - p cos <p ) v,,(pe'<P) = - n - Arctg - ---;---n p 5111 <p

( p - cos <p) l + Arctg --.---SlI1(P


a formula by means of which we verify that for 0 < CPo < n

. . 1 [n ] lim sup v,,(pe"P) ~ lim v,,(pe''I'O) = - -2 - Arctg cotg CPo p~o p~o n

0<'1' < '1'0

Returning to Va' we find for 0 < f3 < ex

lim sup va(z) = z~o

ZEQp

CPo n

o

In the case n ~ 3, the situation is more delicate. We can show that c5(Q 1, Q2) < 1 if for all Z E r I n r 2, r I and r 2 are regular in the neighbourhood of Z with exterior unit normal vectors n1 (z) and n2 (z) satisfying

I n l (z). niz) I < 1 .

More generally we can moreover suppose that Ql' Q2 are, in the neighbourhood of z, the union of regular open sets Q}, ... , Q;, Q~, ... , Q~ with boundaries passing through z and with unit exterior normal vectors

n~(z), . .. ,n;(z) , nHz), ... , n~ (z)

linearly independent (which implies that r + s ~ n). In the case n = 3, admitting as intuitive the notions of vertex and edge, these conditions express that (1) r l (resp. r 2 ) contains no vertex of r 2 (resp. rd, (2) an edge of r 1 (resp. r 2) which meets r 2 (resp. r 1) cuts it in a regular point. We shall not prove these results (see Courant-Hilbert [1]). The alternating procedure of Schwarz can be applied to problems other than that of Dirichlet. For example, let us consider the mixed problem:

Llu = 0 on Q = Q 1 U Q2

(7.26) U = cP on r'1 = r l n r

au = !/J on F\r'1 . an

For CPl E E1(cp), we shall consider the solution of the mixed problem on Q2:

Llu2 = 0 on Q2

(7.27) U2 = U1 (cpd on r 2 n Q1

aU2 = !/J on r\r'1 = r 2 \Q1 an

where U 1(CPI) is the solution of P(Q1 , CPl). We shall suppose that (7.27) admits a classical solution for all CPl E El (cp); we recall that it is then unique (see Corollary 6

522

of ~4). We shall then define SI.{JI by

(7.28) r' 1


Given I.{J l' iP 1 E Ed I.{J), u 2' fi2 the corresponding solutions of (7.27), the function 1'2 = u2 il2 is the solution of the mixed problem

,11'2 0 on Q2

112 l' 1 011 r 2 II Q 1

a1'2 0 l'2\Q j = on all

where l'l = U 1 (P tl - u tf (P 1 ).

From the principle of the maximum for mixed problems (see Proposition 13 of ~ 4) we have

11'21 ~ max 11'11 ~ ()(QI,Q2)ilcPl - iPd· i,

Therefore

In other words we have proved:

Proposition 5. Let Q l' Q 2 he two hounded open sets o( [R" \vitl! boundaries r l' r 2'

Q = Q I U Q 2 with houndary r, 1"1 = r II r l' I.{J E (,gO(1"d, I/J E ~O(r\1"tl. Suppose thatIor aery (PI E (6°(1-1) with I.{JI = (P on 1"1' rhe Dirichlet problem P(Q j , I.{Jtl

admits a classical solution u 1 (I.{J d, the mixed problem (7.27) admits a classical solution 1I 2(CP1) alld (5 = ()(QI,Q2) < 1. Then the mixed problem (7.26) admits a classical solution II which can be approximated uniformly on Q j and Q 2 by the solutions u 1 (I.{J;) and U2(1.{J~) \\,llh the estimates

where (I.{J;) is the sequence constructed recursil'l'iy by: IP7 EEl (Ip) qil'en, (P~' 1 = S(P~ where S is given hy (7.28).

Remark 4. The alternating procedure of Schwarz gives a method of approximate calculation for fairly general open sets arising from the knowledge of the solutions for elementary open sets (ball, cube, etc ... ), since every open set is the union of an increasing sequence of finite unions of elementary open sets. For complementary results and a probabilistic interpretation, see P.L. Lions [I} This method can be exploited at the level of numerical calculation within the context of the so-called methods of "sub-domains" (see Frailong-Paklesa [1], GlowinskiPeriauxDinh [IJ and Miellou [1]179).

I C9 In this last article, it is a question of work directed towards parallel calculation relative to the use of mUltiprocessors.


3. Method of Separation of Variables. Harmonic Polynomials. Spherical Harmonic Function

Let us first of all consider an open set Q of IRn of the form Q = Q l X Qz where Ql,

Qz are open sets of IRn" IRn2 respectively with 1 ~ nz ~ nl < n l + nz = n. Denoting by Xl' x2 the variables oflRn " IRn2 respectively, afunction U(Xl' Xz) on Q is said to have separable variables if it can be written in the form

Such a function with separable variables is harmonic on Q iff

that is to say iff there exists a constant A such that

{ L1U l (X l ) = AU1(X l ) for all

L1u2 (X Z ) + AUz(Xz ) = 0 for all

In other words, the set of functions with separable variables, harmonic on Q = Q l X Q 2 is obtained by solving, for every constant A, the equations

(7.29)

The equations (7.29) are a little more complicated than Laplace's equation (corresponding to A = 0); by contrast the open sets Qi are of dimension ni strictly less than n, which a priori makes the search for harmonic functions on Q with variables separable much easier. For example, in the case n = 2, we have nl = n2 = 1 and we are led to consider the pair of ordinary differential equations

In the case n = 3, we have n l = 2, n2 = 1 and we are led to consider

(i) an ordinary differential equation d2 u22 + AU2 = 0,

dX2

(ii) a partial differential equation L1u l = AU on an open set Q l of 1R2, for which we can use the special techniques of theory of functions of a complex variable.

The interest in the search for harmonic functions with separable variables on Q rests in Weierstrass' theorem: every function continuous on Q = Q l X Q 2 is the limit (uniformly on every compact set of Q) of a sequence of polynomials and hence of a sequence of the sum of functions with variables separable. Indeed, we shall see, at least in a certain number of examples, that every harmonicfunction u on Q can be expressed with the aid of harmonic functions, with variables separable on Q and even often can be written

(i) as the sum of a series L Uk' k


or

(ii) as an integral f U; di.,

where

are harmonic functions with separable variables. To illustrate this statement, we shall consider examples in the case 112 = 1. We put Q = Q' x I where I is all open interval IR and Q' an open set in IR" ... 1. A function with variables separable u(x'. t) = u'(x')v(t) is harmonic on Q iff there exists a constant I. such that

ju' = i.u' on Q',

The second (ordinary differential) equation can be solved classically:

tEl.

where (J) is a square root (in q of ;. and 'i., {1 are arbitrary constants. In other words, a harmonic function with variables separable 011 Q = Q' x I can be writtell in the form

(7.30) u(x', t) = u'(x')(xe i('/ + {ie-i(")

where w, (x, f3 E IC and u' is a solution or (7.31 )

We now show in examples how a harmonic function on Q can be developed with the aid of harmonic functions with variables separable.

Example L HarmonicfunCTion 011 Q = Q' x JO, l[ (with 0 < I <-:c) contilltums 011 fl' x [0, I J and ::.ero 011 Q' X r 0, I;. Let us consider first a function u with variables separable given by (7.30); it is continuous on Q' x [0, I]. u is not identically zero on Q' x J o. I [ and zero on Q' x i 0, I: iff

1/' is not identically zero on Q'. (!) # 0 .

l'i.l + I fiI # 0, 'J. + fl = 0 and xei<Jl + {Jc - i()[ 0 ,

namely ifl'

u' is not identically zero on Q', {i 'l. # 0

OJ # 0 and sin wi = 0 .

Now sin wi = 0 iff w = krc/l with k E JI.. Hence, the harmonic functions with separable l.'ariahles on Q' x J 0, I [ and zero on Q' x [0, I}

(7.32) 14(.\', t) krcl

u~(x')sin I

p. Other Methods of Solution of Dirichlet's Problem 525

with k = 1, 2, . .. and u~ solution of

(7.33) (kn)2 Llu~ = T u~ on Q'.

Now, let us consider an arbitrary function u harmonic on {2' x JO, I[ continuous on Q' x [0, lJ and zero on {2' x fO, I}. We can extend it (in a unique manner) to a function u on Q' x IR odd and periodic with period 2{ with respect to t. The function u is harmonic on Q' x IR as it is so on Q' x (1R\1:n/[) and at each point (x', knl):

u ( x', ~F -+- ° ) = ii ( x', kin - ° ) = ° , = ~ii(x·,kn _ 0).

ot I

Cfi( ,kn 0) , x, 1 + ot

For all x' E Q', we develop the periodic function of period 21, iix·(t) Fourier series; since it is odd we have

fi(x', t) as a

x . krrt iiAt) = k~l u~(x')slll-l-

with

u~(x') I r + I - . krrt - fi(x', t) Sill ~ dt I oJ I I

2 fl , . knt I 0 lI(x, t)smT dt;

we have

1 f+1 knt 1 f+1 i]2ii knt - Llx' fi(x', t) sin - dt = - -~i (x', t) sin -- dt 2 _ I I 2 _{ ot 1

(k{nr ~f~: fi(x"t)Sin~yrdt = (~Fr lI~(x'). Hence every harmonicfunction on Q' x ]0, I[ continuous 011 [2' x [0, I] and zero on [2' x {O, I} can be written

(7.34) x. knt

u(x', t) == I u~(x') sin ~ , k ~ 1 I

with u~ a solution of (7.33).

W h h . ~ '( ') . knt 'f I e note t at t e senes k f-l Uk X Sill -[-- converges um orm y on K' x [0, IJ for

every compact set K' c Q' or, more precisely, from the classical theory of Fourier series on D' x [0, IJ for each part D' of [2' such that the modulus of uniform continuity

c;(r) = sup sup lu(x', t) - u(x', s)1 x' Ell' I. SE [0. II

II - "I "r

satisfies e(r) - < ex . fl dr

o r o


Example 2. Function harmonic on Q variables separable given by (7.30),

Q' x IR. We consider a function with

u(x', t) = u'(x')(oceiwt + {3e' iwr ) with a, /3, OJ E IC .

Two cases arise: (i) if OJ E IR, then u is bounded on D' x IR for every part D' c Q' on which u' is bounded; (ii) if w rf.: IR, then for all x' E Q' the function uAr) = u(x', t) is neither identically zero nor unbounded on IR and even in this case there will exist c > 0 such that eclrlux·(t) is unbounded and hence for all mEN, ux.(t)!(l + t 2 )m will not be bounded. In other words: A fimctioll with separahle l'ariables, harmonic 011 Q' x IR satisfying (7.35) .If)r all

, 'h h I lu(x', t)1 f h . X E Q, t ere exists mEN sue t wt sup ---'---2- < 'l~, is 0 t e form

tEM (l + 1 )m '.

u(x', t) = u'(x')(ae i "" + {3e- iwt )

with wEIR and u' a solution of ju' = w 2u' 011 Q'.

Let us now consider an arbitrary function (not with variables separable), harmonic on Q' x IR. We suppose, in the first place that U E ,(,o(Q'; L t (IR)) with the result that for all x' E Q', we can define the classical Fourier transform

(7.36) u(x', w) = f e- i ()[ u(x', t)dt, OJEIR.

Supposing that we also have U E (6 2 (Q'; L 1 (IR)), we can apply the classical Fourier inversion formula

I ~ , u(x',t) = --J u(x',w)e''''tdOJ.

2n

Now, assuming u E (6 0 (Q', L 1 (IR)) and using (7.36), we have

f ~ , 32u jx'u(x',w) = e- iwt .1 x ,u(x',t)dt = - Je-wH-::;,-(X',t)dl

('(2

(1)2 f e - iOl[ U (x', 1) dt = (1)2 U (x', w) .

In short, if we denote El(lR) = {vEL1(1R); (;EL1(1R)} every function u E ~O(Q'; El(IR)) (\ ~2(Q'; Ll(lR)) harmonic 011 Q' x IR can be written

( ') f' ( J) iw[ d U x , t = Um x e (J)

withfor all WEIR, jU;n = (J)2 U;,j' The important restriction on u(x', t) for It I ->x.: can be lifted by use of the generalized Fourier transform. Let us consider II harmonic on Q' x IR satisfying only (7.35). Then for all x' E Q' we can define the generalized Fourier transform of uAt) = u(x', t):

~7. Other Methods of Solution of Dirichlet's Problem

UX' is the distribution defined by

(7.37) ( E £i1(IR) .

The right hand side clearly has a meaning: in effect, on the one hand, since

from (7.35),

(7.38)

f lu(x', t)1 d lu(x', t)1 t ~ n sup ----=-

(1+t2)m+l tE~(1+t2r

for all x' E Q' there exists mEN such that f J~(.\",t)1 dt < 00 . , (1 + t 2 )m

On the other hand

I f e - its (( t) dt I = I (l + S2) -m f (( t) (I - :t22 ) m e - its dt I

527

= (1 + S2)-mlfe-itS(I - :t22 )m((t)dtl ~ (1 + s2)-mll(I - :t22 )m(IIU(I1)

Now we have the Fourier inversion formula: Ux' = ~!lx" namely 2n

(7.39) ~E~(IR).

This formula dearly has a meaning, for, as we have seen in justifying (7.37) < fix" ( >

is defined for every function (E ,&2m(lR) with (I - :t22 ) m ( E e(lR) where

m is an integer corresponding to x' in (7.38). Now for ~ E £i1(IR), the function

1 f . ((t) = 2; e"S~(s)ds is '/,j00 and for all mEN,

This calculation shows that the two members of (7.39) are perfectly defined and

equal for every function ~ E '/,j2(1R1 with (I - t;i)(1 + s2)m( E Ll(lR) where m

is an integer corresponding to x' in (7.38).

52)'\ Chapter II. The Laplace Operator

Now, let us fix p EO q(lR) with prO) = l. The Fourier transform p then satisfies

(1 -- ;;2)'(1 + S2)'" {I EO Ll(lR)

and fpCS)dS = 2n. We deduce

for all r, m EO N

(7.40) u(x', t) = lim uk(x', t) for all (x', t) EO Q' x IR k- ,x

with

, lr(' s), udx,t)=2n.Ju x,t+ k p(s)ds.

I n effect

( s) u x', t + k Pis) -> u(x', t)t3(s) for all s EO IR ,

and we can apply Lebesgue's theorem since

But, by a change of variables

1 I'

uk(x', t) = - J u(x', s)kt3(k(s - t))ds 2n

and hence applying (7.39) with ¢(s) = k{J(k(s - t)), noting that

1 1"" eic>t J~ .w '( w ) 2~ J elf"' kt3(k(s - t))ds =i~- e ',,' {J(s)ds = e"M P k,

we obtain

" _ (I I, iwt ( W ) \ udx,/) - \-ux.(wj,e p - ,I.

2n k /

·1 It is clear that the map u': x' -> .... l1x' is continuous (and even analytic) from Q into

2n Ct'( IR) and we have

(7.41) Llu'(x') = w 2 u'(x') in (./'(U:g) for all x' EO Q' .

In brief: every fimction u, harmonic on Q' x IR satisfying (7.35) can be written in the form

(7.42) u(x', t) = (u'(x')(w), eio")


with the (analytic) map u' of Q' into SC' (IR) satisfving (7.41), the equality (7.42) being taken in the sense

( . (w)\ u(x', t) = lim. U'(X')(W), e,wt p -k ) k~ eYCo • I

for each function p E ~(IR) with p(O) = l. The examples given above can be generalized for the study of harmonic functions on Q = Q 1 X Q 2 satisfying linear conditions independent of x I E Q I on the boundary Q I x CQ2- In effect the method of separation of variables leads to the study of the constants A and the functions U z on Q 2 satisfying

(7.43) { LlU2 + ),u2 = 0 on Q2

boundary conditions on i)Q 2 _

In other words, considering the (linear) space X 2 of the functions U 2 on Qz,

satisfying the boundary conditions on aQz, we are led to seeking the eigenvalues I, and the eigenfunctions U z of the operatar- Ll on X 2- The cases considered in Examples 1, 2 will be found in the general context which will be studied in § 8. By means of theorems of spectral resolution, every harmonic function on Q I x Q2

satisfying the conditions on the boundary Q 1 x aQ2 can be written as a sum (possibly in a generalized sense, as in Example 2) of harmonic functions with separable variables u1 ex 1 )u~ Cx 2 ), where u~ is a solution of (7.43) and u1 is a solution of

,dul = ),.u~ on £1 1 ,

We note clearly that we have not imposed any condition on the boundary a£1 1 and that u~ is not, strictly speaking, an eigenfunction of the Laplacian, We shall now generalize the method of separation of variables which will lead us to study harmonic polynomials. Let us consider an open set £1 of IRn of the form Q = h (M 1 X M 2) where IH l' M 2. are differentiable man[folds of dimension n I, 112 respectively with 1 :( n2 :( n l < n l + n2 = nand h a diffeomorphism of MIx A12 on £1: (YI' Y2) E lvl) x M 2 is a system of curdlinear coordinates of x = h(Yl, Yz)· The function u defined on Q is with separable variables in this system of curvilinear coordil1ates if

(7.44)

where u I' Uz are functions defined on All' M 2 respectively. In the general case, the change of variables will give a very complicated form to (Llu)(lJ(YI' yz)); let us suppose that for every Junction u E (6 2 (£1)

(7.45) (,:).u )(h(Yl, Y2)) = e(y I' yz) { J't (YI' ~)) + Pz (Y2' a~J } (u "h)(Yl' Y2)

where p), Pz are differential operators (of order 2) on M I and M)80, respectively

180 See Chap. V. ~ 1.3 for the definition of a ditTerential operator on a manifold.


and c is a non-zero function defined on MIx M 2. Then a function with variables separable u satisfying (7.44) is harmonic iff there exists a constant A. such that

on Ml

on M 2 •

The hypothesis (7.45) depends on the parametrization h of Q. It generalizes the case in which Q = Q l x Q2: when c == 1, PI = Ll y1 , P2 = Ll yz • An important case where (7.45) is satisfied is the parametrization in polar coordinates

((J,r)EM x I-->raEQ,

where I is an open interval of ]0, CN [, M an open set of the unit sphere 1: of [Rn,

Q = {ra; rEI, (J E M} is a sector of an annulus in [Rn. We know then that for all u E 1€2(Q) (see (1.27) of § 1)

( (12 (11 - 1) (1 I )

(Llu)(ra) = (lr 2 + --r-- (1r + r2 LI" u(ra)

where LI" is the Laplace-Beltrami operator on the unit sphere 1:. A function with variables separable in the parametrization in polar coordinates is of the form

u(x) = c(r)v((J) , r = lxi,

Such a function will be harmonic iff there exists ), such that

(7.46) { ~:~,:~~l~»~o on J

LI"l - ).V on M.

The first equation of (7.46) is solved classically: we put

(7.47) 11

To = I - -2·

x

r

(I) If ;. #- T6, then c(r) = rtO(art + pr- t) where T is a square root of T6 I..

(2) If). = T6, then c(r) = rtO(a + pLogr). Hence the fUllctions with separable variables (in the parametrization in polar coordinates) harmonic on Q = {x; r = Ixl E I, (J = xlr EM} are of the form

(7.48) rtO(a + pLogr)v((J) with Lluv = T6V on M

or for TEe with T #- 0,

(7.49)

where TO is given by (7.47) and a, pEe are arbitrary constants.


We notice the in variance under Kelvin's transformation: if u(x) = c(r)v(a) is a function with variables separable (in polar coordinates) on M x I, its Kelvin transform 181 H(O, l)u is u'(x') = Ix'12-" u(I(O, l)x') where 1(0,1) is the inversion leaving the unit sphere 1: invariant; that is to say

u'(x') = c'(r')v(a) with c'(r') = r'2 - "c(r' - 1) .

In particular, if c( r) = rto + t (resp. rto - r, rto Log r), then

c'(r') = r'to-t (resp. r'to+r, - rft°Log r').

We note also that the radial functions u(x) = c(r) are obviously with separable variables: we recover the characterisation of the harmonic radial functions (see Proposition 4 of § 2) by taking v(a) == 1 in (7.48) and (7.49); (a) in (7.48), v(a) == 1 is the solution of J"v = '5v iff '5 = 0, that is to say n = 2 and the harmonic radial functions are of the form

~ + f3 Logr

(b) in (7.49), v(a) == 1 is the solution of J"v = ('5 - ,2)V iff ,2 = '5, namely, since , =1= 0, in the case n # 2 and the harmonic radial functions are of the form

~ ~r2to + f3 = ~- + f3 . r" - 2

We note finally that the harmonic functions on the ball B(O, ro) are identified with the bounded harmonic functions on the punctured ball

B(O, ro)\{O} = {ra; ° < r < ro, a E 1:}

(see Proposition 16 of § 2). Hence from (7.48) and (7.49) the functions with separable variables, harmonic on B(O, ro) are of the form rto+tv(a) with, E C, Re , ;;:, in - 1 and v a solution of J.,.l' = ('5 - ,2)V on 1:. In fact we shall see later (see Proposition 6) that the harmonic functions with separable variables on B(O, ro) are the harmonic homogeneous polynomials. We can develop the same argument as in Examples 1 and 2 by imposing a condition on the boundary M x 01; we leave the details of such a study to the reader, preferring to develop the discussion of the second equation of the pair (7.46). To introduce the fundamental ideas, let us commence with the case n = 2.

Example 3. Harmonic function on a sector of an annulus of [R2.

In the case n 2, the unit circle is parametrized by the polar angle 8;

L = {( cos (), sin (); () E [R}

and

(LI"v)(cos(), sin() = dd;2 v (cos 8, sin().

In other words, a function v defined on M is a solution of the second equation of the

181 See §2.5.


pair (7.46) iff w(8) = v(cos 8, sin 8) is a solution of

(7.50) d2 w d82 = }.w.

We ought, however, to distinguish two cases: case of a complete annulus: Q = {(rcosO, r sinO); r 1 < r < r 2 , 8 E ~},

case of a strict sector of an annulus:

Q = {(rcosO, rsinO); r 1 < r < r2 ,81 < fJ < fJ 2 } with fJ 2 - 61 1 ~ 2n.

In the second case the parametric representation

is bijective, whereas in the first case there exists no parametric representation bijective from the unit circle onto an open interval of ~. We identify the plane ~2 = {(x, y)} with the complex plane IC = {z = x + iy}, the unit circle 1; with few; 8 E ~}. (1) Case of a strict sector of an annulus. Since To = 1 - 1 n = 0, the solutions of fJ.,. = T6v (resp. iI"v = (T6 - T2)V with T i= 0) on AI = few; 81 < 8 < 02 }

where O2 - 0 1 ~ 2n, are v(eiO) = Ct.' + {J'O (resp. v(eiO) = Ct.'e itO + {J'e- itO ). Hence from (7.48) and (7.49), the functions with variables separable, harmonic on the sector {reW; r 1 < r < r2 ,81 < 0 < Oz} with Oz - 01 ~ 2n are of the form

(7.51) {(Ct. + {J Logr)(Ct.' + {J'fJ) (Ct.r t + {Jr-r)(Ct.'eir8 + {J'e- irO ) for T E IC, T i= O.

We can impose a condition on the boundary FlU F z where

j = 1,2.

To change from Example 1, let us suppose that fJ 2 - 01 < 2n and consider the mixed boundary condition:

(7.52) u(reie, ) = 0, :fJ u(reie2 ) = 0 ,

A function with separable variables c(r)v(O), not identically zero, satisfies this condition iff

(7.53)

A function v(eiB ) = Ct.' + {J'O cannot satisfy (7.53) without being identically zero; a function v(eiO) = Ct.'e ire + {J'e - ite satisfies (7.53) iff

(7.54) {,.Ct.'e ir81 + {J.'e"it81 = 0 Ct.' eir82 - {J' e - irO, = 0

a system which admits a non-zero solution (Ct.', {J') iff

cos T(02 - Dj ) = 0


namely iff T = ( k + ~) O2

proportional to

n with k E 'l'. Then the solutions of (7.54) are - °l 1 . "

a' = - e- lt02

2 f3' 1 " ____ ettv2

2 to which corresponds

v(e iO ) = cos T(02 0) .

In summary, therefore, account taken of (7.51), the functions with variables separable, harmonic on the sector Q: {reW ; r J < I' < rb ° 1 < 0 < 82 } with O2 - Ol < 2n and satisfying the boundary conditions (7.52) are of the form

(aI'" + f3r-'k)coSTk{02 - 0)

with Tk = (k + ~)~O for k = 0, 1, .... 2 82 1

Using a development in Fourier series, we see, as in Example 1, that every harmonic function u on Q satisfying (7.52) can be written

a'

I (akTk + f3k T- k)COS Lk( 8z - 0) . k=Q

(2) Case of a complete annulus. The functions v defined on the complete unit circle are identified with periodic functions of period 2n on R The differential equation (7.50) admits a periodic solution of period 2n, not identically zero iff }. = - k2 with k E 'l'; the solutions are then of the form w(8) = a'e ikO + f3'e ikO•

Hence from (7.48) and (7.49), the functions with separable variables, harmonic on the annulus {reW, 1'1 < r < 1'2,0 E IR} are of the form

{a + f3 Log I' (radial function)

(7.55) or (ark + f3rk)(c/e ikO + f3'e iW), k = 1,2, ....

In other words, putting aside the radial functions a + f3 Log I' the functions with variables separable, which are harmonic on the annulus {z E iC; r 1 < I z I < 1'2} are of the form

k = 1,2, ....

There it is a question of functions with complex values (a, a', Ii, 11' E iC); the functions with real values are of the form

rk(acoskO + f3sinkO) + r-k(a'coskO --t- f3'sinkO)

which can also be written in the classical manner

with C(, a', (P, q/ E R Every function harmonic on the annulus {z E iC; r 1 < I z I < r 2} can be written in the form

+ :c'

f3o LogJzl --t- I (akzk + a~zk) k=

534 Chapter [I. The Laplace Operator

or again in the case of real-valued functions

x

:Xo + fJoLogr + L (:xkrkCOsk(O - rpd + Ci.~r-kcosk(e - rp~)). k·o j

This can be seen either as a development in Fourier series, or as a Laurent series of a holomorphic function on an annulus in the complex plane. D

We propose now to extend to the case of the arbitrary dimension n the results we have just obtained for harmonic functions on an annulus in [R2. We first prove:

Lemma 4. Every polynomial homogeneous oldegree k in n variables Xl' ... , Xn can he written in the form

(7.56)

where hk _ 2p is a homogeneous harmonic polynomial of degree k - 2p (p an integer).

Proof This comes down to showing the every homogeneous polynomial f of degree k can be written in the form

(7.57) f(x 1, ... ,xn) = h(x j , ••• ,XII) -+ (xi + ... + X~)g(Xl"" ,XII)

where h is a homogeneous harmonic polynomial of degree k and g a homogeneous polynomial of degree k - 2. Using Euler's identity

. I n Dl j (x l' ... , Xn) = -k L Xj -~ - (x l' ... , Xn) ,

,= 1 eXi

and reasoning by recurrence on k, we are led to proving (7.57) for a polynomial f(x 1 , ..• ,xlI)=xjhj(x j ••..• XII) where h j is a homogeneous harmonic polynomial of degree k - 1.

2 2 ah j • h(x1,···,xn ) = xihi(xj •... ,Xn ) - /"(X 1 + .,. + Xn)~O (X1,···,Xn )

Xi

ah we have, since h j and~' are harmonic,

eX i

But since ah/ax i is homogeneous of degree k - 2, we have, from Euler's identity

therefore

II ;Yh oh L Xj ~ ~' = (k - 2);:;', j= 1 (jX;fjX j (IX i

ah L1h = (2 - ..t(2n + 4(k - 2))) ~ ;

eXi

choosing J. = l/(n + 2(k - 2)), we see that h is harmonic. D

A homogeneous function is a function with separable variables (in the parametrization in polar coordinates); from (7.48) and (7.49), a homogeneous harmonic


polynomial h of degree k can therefore be written

hex) = rk Y(a)

with

namely

(7.58)

Let us now prove

Proposition 6

,1" Y = - k(k + n - 2) Y on E

535

(1) For all k, (k = 0, 1, ... ), the set qlf~ of the functions Y solutions 0{(7.58) where E is the unit sphere in [Rn is a finite-dimensional vector space; we hare

dim (;7!Z = 1 , dim qlf1 = n and

in the case II = 2, dim ~l/f 2 for all k ~ 1

in the case n = 3, dim ~l/~ 2k + 1 for all k ~ ° in the case n = 4, dim'fY: = (k + 1)2 for all k ~ 0.

In general k

(7.59) dim Zl/~ = I dim qlf?-l [=0

(2) The spaces{l/k = !l/Z are two-hy-two orthogonal in L2(L:), and L2(E) is the Hilhert sum o{the suhspaces ~l/k181a. (3) Every harmonic junction on a ball B(O, '0) (resp. all anllulus B(O, '2)\8(0,'1)) can be written in a unique manner in the form

ktolYk (a) (resp. hrkYda) + (5n.2{3oLOg,)

with YkE'!lfk(resp. YkE'!lfkifk ~ 0, Yk = 0if2 - n < k < Oand Yk E(;7!z-ll-k if k ~ 2 - n), the series being uniformly convergent on every compact set of B(O, ro) (resp. B(O, r 2 )\B(O, r1 )).

Proof Identifying Y E ~l/Z with the function h (x) = rk Y( a), we see from (7.49), that qy~ is identified with the space of all homogeneous functions (of degree k) harmonic on [R"; we know that every tempered distribution, harmonic on [R", is a harmonic polynomial (see Proposition 3 of § 2); hence qYk is identified with the space of homogeneous harmonic polynomials of degree k. Now let Yk E!Yk , Y[ E{Y[ with k * I.

18" Every element r of L2(J:) can be written r = L Proj'~k r where Proj,q, is the orthogonal k~O

projection on the sub-space :1/ k'


By homogeneity (Yk is identified with the harmonic polynomial rk Yk(O")) we have, from Euler's identity

where a/en is differentiation external to B(O, I). Since Yk and Y1 are harmonic, we have, from Green's formula

Therefore L. YdO") Y/(O"jdO" = O.

From Lemma 4, for every homogeneous polynomial rkv(O") we have:

rEI ";lf k 2p' From the Stone-Weierstrass theorem I '!Yk is dense in ~O(1:) O~p~~2 k

and hence also in L 2 (1'); hence L 2 (1:) is clearly the Hilbert sum of the sub-spaces (If k' Also the space g:>~ of homogeneous polynomials of degree k can be written as the direct sum of the'!YZ-2p for 0 ~ p ~ tk with the result that

dim ;{P~ dim{lf~ - 2p .

Since o ~ p ~ k.i2

k

dim·0"Z = I dim9'?-l , 1=0

we deduce (7.59) and the values dim~lf~ for n = 2, 3,4 beginning with dimW~ = 0 for k ? 2. Given a function U E Lroc({r 1 < Ixl < rz }), it can be decomposed in the unique manner

of

(7.60) u(x) = I udx) k=O

where for all kEN and r 1 < r < r 2 , the function uk(rO") is the orthogonal projection of U(rCT) on i{lfk ; if (YUi=1. .... 1K is an orthonormal base of q]jk'

Ik = dim '!Yk'

(7.61) ik I c~(r) Yi(O") ,

i = 1 r lxi,

x

r

with

(7.62)


Supposing u to be harmonic, we shall have

( d2 n - 1 d) . fl. -d 2 + -- -d c~(r) = - -2 Llau(ro-) n(O")dO"

r r r E r

1 r . = -"2 u(ro-)Ll" n(O")dO" r ~ E

k(k + n - 2)f . r2 E u(ra) YUa)da

where '0 1 - (nI2); going back to the proof of (7.48) and (7.49), we deduce

cUr) = :xil + ptr-(k+n-2)

unless n 2, k = 0 (dim @'o = 1) in which case Co = ()(o + Po log r. Hence

lK h uk(x) = rk E rx~ Y£(a) + r-(n-2+k) E Pi Yi((J)

i = 1 i= 1

unless n = 2, k = 0 when uo(x) = rxo + Po Logr. If u is harmonic on B (0, r 0), then the functions c~ (r) given by (7.62) on JO, r 0 [, are bounded as I' --+ 0 and hence

c1(r) = rxirk and uk(x) == rk Ec4 Yi((J).

Finally the uniform convergence follows from Harnack's theorems (see §2.3) since the functions Uk are harmonic. 0

In the case n = 2, the sub-spaces i{tlf are of dimension 2 for all k ~ 1. As we have seen in Example 3, in the complex plane

~lj~ = {rxe ikO + pe- ike ; C(, pEe}

while in real terms q7jt = {:xcosk8 + psink8; rx, P E IR}

which we can rewrite as:

?lj~ = {rxcosk(8 - <p); C( E IR, <p E IR (modulo2nl}.

In other words the functions of /ilif are to within a multiplicative constant the circular/unctions Y(8) = cosk(8 - <p). In a general way we pose the

182 We have used the formula of the Laplacian in polar coordinates (see (1.27))

( ,12 n - I (1 1 ')

(L1u)(ra) =:,,- + -- -:::-" + -,L1", /lira) (T2 r cr r2

and the fact deduced from it that L1" is a symmetric operator

t vd(J)..1a v2(a)dO" = t v2(0")LlITvda)dO" .


Definition l. The elements of the space qI/'k are called the spherical harmonic functions of order k in IR".

Let us determine the spherical harmonics in the space 1R3 = {x, y, z}. The unit sphere E of 1R3 is represented by

(7.63) {

X = sin () cos cp

y = sin () sin cp

z = cos ()

where () E [0,70] is the angle of a(x, y, z) with the polar axis Oz and cp the polar angle of (x, y) in the plane 1R2 defined modulo 270 (and even undefined at the poles () = 0 and () = 7o. We prove

Proposition 7. The space!lJ ~ of spherical harmonic functions of order k in the space 1R3 a4mitsfor orthonormal base the functions (Y;:')-k';; m H defined in the parametric representation (7.63) by

(7.64)

where for 0 ~ m ~ k, P;:'(z) is the solution of

(7.65) (1_z2)d 2 P _ 2z dP + (k(k + 1) - -1 m2 z)p = 0, dz2 dz - z

satisfying the normalization

(7.66) 270

Such a function is given by m

(7.67) m m (-ll(1 - z2)'+ zzk-m-21

Pk(z) = (Xk O"'I~'T"-411!(l + m)!(k- m - 2/)!

where the constant (x;:' is chosen so that (7.66) is satisfied.

Proof We note first of all that the functions Y;:' defined by (7.64) where P;:' satisfies (7.66) form an orthonormal system on L 2 (E). We have in effect

t y;:'t(a)Y;:'2(a)da = {"ei(mt-m2)'PdCP IPlmtl(cos()Plm21(cos()sin()d()

= 27O() f+l plmtl(z)plm21(zldz ml,m2 k k .

-1

Now we know that, expressed in the parametric representation 183 (7.63) the

18~·See·(1.28) of§l.


operator ,1", becomes

1 0 (. 0 ) 1 02

,1", = sinO oe smO 00 + sin2 0 Ocp2 .

Hence a function. Y(u) = P(cos8)eimq> is in qy~ iff

m2 . L1",Y = (P"(cosO)sin2 0 - 2P'(cosO)cosO - --:-yn P(cosO»e'mq>

sm u

= - k(k + I)P(cosO)eimq>

namely iff P is a solution of (7.65).

539

To determine the explicit formula (7.67), we use a method other than verification. We know that the spherical harmonic functions of order k are the trace on L of the homogeneous harmonic polynomials of order k. A homogeneous polynomial of degree k can always be written

(7.68) h(x, y, z) = L ap,q(x + iy)P(x - iy)qzk- P- q . p,qE'IJ

p+q';;'k

We have

L1h(x,y,z) = L ap,q[4pq(x + iy)P-1(X - iy)q-1 zk- P- q p+q"'k

+ (k - p - q)(k - p - q - l)(x + iy)P(x - iy)qzk- p- q-2] .

Hence h given by (7.68) is harmonic iff

(7.69) 4pqap,q + (k + 2 - p - q)(k + 1 - p - q)ap-l,q-1 = O.

. k - Iml Puttmg, for - k ~ m ~ k and 0 ~ I ~ -~2~ ,

m {a,+m" m ~ 0 a, = a','-m m ~ 0

we have ap,q = a{;,i;,ip,q) and (7.69) can be written

41(l + Iml)a;" + (k + 2 - Iml - 21)(k + 1 - Iml - 2I)a;"-1 = 0

that is

with

(7.70)

m _ (-1)' m n' (k + 2 - Iml - 2j)(k + 1 - Iml - 2j) a, - ao

j= 1 4j(j + Iml)

= (-I)'a~(k -lml)!lml!c;"

1 - = 4' l!(l + Iml)!(k - Iml - 21)! c;"

To sum up, passing to the trace on L specified by (7.63) we find that the functions of


~Ij~ are of the form

I '1meim<p I (- 1)IC;"(sinB)21+lml(COSO)k Iml21 k~m~k O~I~k-t'~L

from which (7.67) follows. D

We note that P~(z) = '1(1 - Z2 ti2 with

1 f+ 1 --2 = (I - z2 tdz 2ncx_ 1

22k + 1 (k!)2 184 -------------------------

(2k + I)!

Therefore

(7.71 )

22k(k!)2 ------.------

k(2k + I)!

Hence

(7.72)

Propositions 6 and 7 show that every function u(x, y, z) harmonic on B(O, ro) can be written in a unique manner.

(7.73)

with

(7.74)

u(x, y, z) , I I c~rkP~(c6sG)ei/ll(p k~O -k"'rn~;k

('~ = r- k f:" e im<p dip t P~(cos U) u(rsin B cos <p, rsin 0 sin <p, rcos 0) sin 0 dO .

We note that although I' appears in the expression giving c~, this is in fact independent of 1'; in the case where k = m = 0, this independence is the formula of the mean

u (0, 0, 0) =.-(.~- =~ f2" dip fIT u( r sin 0 cos <p, r sin 0 sin <p, r cos 0) sin () d () . 2y n 4n 0 0

In this case, we can consider (7.74) as generalizing the formula of the mean.

1 H4 I, = f' (l .... ~2)k d~ = ~. J' 1 (1 _ ~2 )k d~ satisfies the recurrence relation (' I-+-l ') Ie = I, l'

" - - 1 2k,

namely. with 10 , 21

1. I," n _.-:- . 1= 121 + 1


We have considered an orthonormal base of ~7!~ in the complex domain; in the real domain, an orthonormal base of ~7J f is formed of the (2k + 1) functions

P~(cos8), J2PI:'(cos8)cosnup, J2PI:'(cos8)sinmcp, m = 1, ... , k.

We note that every real harmonic fimction on B(O, ro) can be written

'iC k

(7.75) u(x,y,z) = L L al:'rkPI:'(cos8)cosm(cp - CPI:'), k=O m=O

where a~ c~ (real) and for m = 1, ... , k

al:' = 21cl:'1 ,

cl:' being given by (7.74). It is enough to use (7.73) noting that c,;m = cr'. A particular case concerns the harmonic solutions u on B(O, ro) with cylindrical symmetry, that is to say invariant under rotations about the polar axis Oz. We then will have cl:' = 0 for all m # O. Hence every harmonic fimction u with cylindrical symmetry can be written

Cfj

u(x, y, z) = I C~(X2 + y2 + Z2)k!2 P~(z) k=O

To within a multiplicative constant, P~ is the Legendre polynomial of degree k, it is also the Gegenbauer polynomial c,t. More generally the Gegenbauer polynomials

(-- 1)1(2t)k-21 f(p + k - l)

k!(k - 21)!r(p) q(t) = (7.76)

solutions of the differential equation

d 2 P dP (l - t 2 )_. - (2p + l)t dt + k(k + 2p)P = 0

dt 2

allow us to express the spherical harmonic functions of IR" with cylindrical symmetry: we can show (see Vilenkin [1]) that the spherical harmonic functions of u;gn = {(x', x,,); x' EO IR" - 1, Xn EO IR} of order k with cylindrical symmetry (that is to say invariant under rotations about the axis Ox,,) are proportional to C til -1 (Xn). Finally we note that to within a multiplicative constant PI:'(z) is the polynomial

d"' (1 - Z2 )m/2 -d P ~ (z). This is easily verified by change of the unknown function zm P = (1 - Z2 ytm u in (7.65) and use of Leibniz's formula in (7.65) taken for m = O. The functions PI:' (z) are also called the associated Legendre functions of the first kind (see Robin [1]). We shall not develop further the theory of spherical harmonic functions: we note that it is intimately linked with the theory of the group of orthogonal transformations

542 Chapter II The Laplace Operator

of [R"; in effect it is clear that {Ij~ is invariant under the group of orthogonal transformations, for if T is a linear transformation of [R" and h a homogeneous polynomial of [R", the image Th = h T is a homogeneous polynomial of the same degree; now if h is harmonic and T orthogonal, then Th is harmonic (see Proposition 2 of § 1), from which the above statement follows. We refer the reader to Vilenkin [IJ for a systematic study of symmetric harmonic functions based on the theory of groups.

4. Dirichlet's Method

Let us first consider Dirichlet's problem P (Q, ({1) where Q is a set of [Rn with boundary rand ({1 E ~o (1). We suppose first of all that Q is regular and bounded and that the solution u of P(Q, ({1)185 is in (t~Uh Then we know 186 that IgraduI 2 EL 1 (Q) and that for every function VE<'gl(Q)n<'gO(Q) with Igradvl 2 E U(Q), we have

r gradu.gradvd.x = f v ~u di'· v12 r en

In particular

(7.77) L grad u . grad v dx = 0 for every function t' E '6 1 (Q) Ii ({,O U2 )

satisfying Igradvl 1 EL1(Q) and ['=0 on r.

This property (7.77) can be expressed in an equivalent manner by

I r Igradul 2 dx:( JI' 1 grad It' 2dx for every function 11'E«,1(Q)n«,O(Q) (7.78) ~.(2 12

l satisfying Igrad\\fELl(Q) and 1I'=({1 on r.

In effect applying (7.77) with r = IV - u and using

1 grad u 12 =- 1 grad \\' 12 - 2 grad u . grad r

we obtain (7.78). Conversely applying (7.78) with It' = u + iv, (i E [R), we obtain

JI' graduj 2 dx:( f (lgradul 2 + 2i.gradu.gradr + 1.2IgradvI2)dx • Q Q

from which we deduce that if ). > 0 (resp. i. < 0)

Jf' grad u. grad r dx ;0: (resp. :() - ~ f I grad t' 12 dx . Q 2 12

Letting} ---> 0, we obtain (7.77).

185 Which exists from Theorem 1 of ~4. 18h See Proposition 5 of ~ I.


In other words, ifQ is a regular bounded open set ofW and if the solution of P(Q, qJ)

is in «?~ (Q), then that solution minimises the energy functional

,;1(u) = r Igradul 2 dx JQ

on {u E (€l(Q) n «?o(.{2); Igradul 2 E Ll(Q) and u = qJ on r}. We note that we have already applied this result in the study of the capacity of a regular compact set K contained in a regular bounded open set (see Proposition 7 of § 5). Dirichlet's method consists of using this property of the minimization of the energy functional to define a solution of P(Q, qJ). Before developing the method in a general framework, let us recall that given Q, an arbitrary open set I.!flR", we denote

Bl (Q) = {UELtoc(!"l); the derivatives ~u EL2(Q) (in the sense of distributionS)}, OX j

and by Bb(Q) the closure of g(Q) in Bl (Q) which is given naturally the topology defined by the convergence

1 {Uk -> U in L~oc(Q) Uk -> U in B (Q) iff

grad Uk -> grad u in L 2 (Q) .

In general, Bl(Q) contains strictly the Sobolev space187

the denvatlves ............ E L (Q) = B (Q) n L (Q). .. au 2 } 1 2

aX j

If Q is bounded (or, more generally, of finite measure), then Bb(Q) coincides with Hb(Q), the closure of ~(Q) in HI (Q)188. The energy functional

,;1(u) = t Igradul 2 dx is definedfor all u E 8 1 (Q).

We now prove

Proposition 8. Let Q be an open set oflR" and qJ E Bl (Q). We suppose only Q =f. IR if n = 1 and Q =f. IR 2 if n = 2. (1) There exists a unique function u minimizing the energy functional on {qJ + v; V E Bb(Q)}; (2) u is the unique solution of the problem:

(7.79) { u is harmonic on Q ,

U - qJ E Bb(Q) .

187 Even in the case Q bounded: for example Q = U ] ~I_, ~[ an open set of Q;£ k~1 k+l k

u = I)X],;,I[EB'(Q)\H 1 (Q).

'88 With the norm Ilulllf'lQ) = (J, (IU I2 + IgradUl2jdXY·


Proof Consider first u minimizing Jon qJ + B~(Q). Since u + B~(Q) = qJ + B~(Q), we have

J(u) ~ J(u + A.v) for all v E B~(Q) and A E IR

from which we deduce, as above, that

(7.80) L gradu.gradvdx = 0 for all v E B~(Q).

Conversely if U E qJ + B~(Q) satisfies (7.80), then it minimizes J on qJ + BA(Q). We note also that by definition of BA(Q), for U E Bl (Q), (7.80) is equivalent to

<L1u,O = - fgradU. grad ,dx = 0 for all 'E E0(Q) ,

that is to say u is harmonic on Q. To prove existence, we consider a sequence (vd of BA(Q) such that

m = inf J = lim J (qJ + vk );

we have <p + B6(Q) k~ 00

1 ( Vk + VI) 2J (vk - VI) = J(qJ + Vk) + ,/"(qJ + VI) - 2J qJ + 2

~ J (qJ + vd + J (qJ + VI) - 2m ,

so that lim J(Vk - VI) = O. To conclude the proof of existence, it must be k,l -+ 00

shown that (vk ) (or a sub-sequence extracted from it) converges in Lfoc (Q); then (vk )

will converge in Bl (Q) and, considering its limit v, the function qJ + V will minimize J on qJ + B6(Q). In the case of Q bounded (or of finite measure) we can make use of Poincare's inequality (see Chap. IV, § 7 - Vol. 2, p. 126; see also Adams [1 J, Gilbarg-Trudinger [lJ) to obtain

II Vk - v,III2 ~ C(Q)J(uk - vzl

which proves that (vk ) converges in L2(Q).

In the case n ~ 3, Q being arbitrary, we can use Sobolev's inequality (see Gilbarg-Trudinger [lJ, Adams [lJ)

Ilvk - v/lli"~"2 ~ CJ(Vk - VI)

which proves that (vk ) converges in L 2n/(n-2)(Q).

In the case n = 1, if Q i= IR we have

I(vk - vl)(xW ~ dist(x, OQ)J(Vk - vzl

and hence (vd converges uniformly on every bounded set of Q.


In the case n '# 2, if Q = 1R2 we can always suppose 0 ¢ Q. For v E BA(Q) let us consider

w(y) = VC:12) (image of v under the inversion 1(0, 1» ;

using the fact that 1(0, 1)'(y) = lyl-2SR(0,28), where S = G _~) is the

symmetry with respect to the OY1-axis, R(O, 28) is the rotation about 0, with angle 28 (see §2.25), we have

I grad w(y)1 = ,y\2IgradvC:,2)1

and

r Igradv(x)1 2dx = r Igradw(yWdy with Q' = 1(0, l)Q. JQ JQ'

Since Q' is bounded, using Poincare's inequality we obtain

t, Iw(y)1 2dy = t 1~~~12 dx ~ C(Q')J(v);

we deduce that the sequence (vk I xl- 2) converges in L 2(Q) and hence (vk) converges in Lroc(Q). From the different cases considered, we see that, generally, for v E BA(Q),

J(v) = 0 => v = 0 ;

this proves the uniqueness of a function minimizing .~ on qJ + BA(Q). 0

Remark 5. In the case Q ='!R or Q = 1R2, BA(Q) = Bl(Q). We show first that BA(Q) contains the constants: given ( E ~(!R) (resp. ~(1R2», the sequence of functions (dx) = (x/k) converges in effect in Bl(lR) (resp. weakly in Bl(1R2)) to the constant function v(x) == (0); because clearly

and

J«(d = :2 f('(~ Y dx = ~ f ('(x)2dx ---. 0

(resp. J«(k) = :2 f flgrad(X/kWdX = f flgrad(X)12dX is bounded).

We consider now qJ E Bl (Q) and consider a sequence (Vk ) of B6(Q) such that J (qJ + vk ) ---. inf J; we know that lim .~ (Vk - v,) = 0 (see the proof of

Proposition 8). '" + B6(fJ) k.1 ~ 00

In the case Q = IR, we have qJ E <cO (IR) and Vk E <cO (IR); replacing Vk by Vk - vdO) - qJ(O) E BA(IR), we can suppose that qJ(O) + vk(O) = 0; we then have by the Schwarz inequality


and hence (t\l converges uniformly on every compact set of 1M; the limit v E Bb(lM) and u = l' + <fJ minimizes I on <fJ + B6 (1M) and satisfies u(O) = 0; as we have seen u satisfies (7.79), and in particular, is harmonic on necessarily u == ° and therefore <fJ E Bb(IM).

In the case Q = 1M 2, we can suppose f l'k(x)dx BID. i)

Poincare's inequality l89, since

0,

we have for all R > 1 I'

- J" ,<fJ(xldx; from B(O. II

J Irk(x) - l"1(x)1 2 dx:( C(R),f(vk - rtl B(O. R)

and hence (vk ) converges in Lfoc(1M2); the limit r E Bb(iJ;£ 2 l and u = r + <fJ IS

harmonic on a:g 2 ; since f. u(xldx = 0, necessarily u == 0 19°. 0 B(O.i)

Proposition 8 and this Remark 5 lead us to frame:

Definition 2. Given an open set Q in 1M" and (P E BI (Q) we call a generalized solution (in the sense of Dirichlet) of P (Q, <fJ) every solution u of (7.79), or what is equivalent every function u minimizing the energy functional

I(u) = f Igradul1dx on ~ry + B6(Q)· Q

Unlike the classical Dirichlet problem we assume that we are given (P on Q and not on the boundary r of Q. We note however that given <fJI, <fJ2 E Bl (Q) with <fJ I - <fJ2 E B6 (Q), u is a generalized solution of P (Q, (P d iff it is a solution of P (Q, <fJ 2); ill other words the notion of a generalized solutioll (ill the Dirichlet seilsI')

'"" Poincare's inequality can bc stated: given Q a connected regular bounded open set and \\ f: 1.'(Q).

\\. # O. there exists C(Q. w) such that. for U E f{' (m ( = H\ (Q))

! tl "L' ,-:; C( Q. \\') I grad II i L ..

where

(see Lemma 5 below). '90 H' (~") n .W" (~") is reduced to constant function: in efl'ect the derivatives

(....,u 'i = - E L 2 (n") nil ('F!")

( .... x i

are since their Fourier transforms t, E L 2 (IR") and satisfy

I~i't,(;) 0 a.e. ~ E J,!".

*7. Other Methods of Solution of Dirichlet's Problem 547

of P (Q, <p) depends only on "trace" of <p on r and in the case of Q unbounded, at infinity. Here, the notion of trace is the equivalence class in B'I (Q) modulo B6(Q); we note that Proposition 8 can be stated: the map u 1--* class of u in Bl (Q) modulo B~(!;2) is a bijection of £(Q) n B1 (Q) = {u harmonic on Q with energy

§(u) = t Igradul2dxfinite} onto the quotient space B'(Q)/B6(Q)·

When Q is a regular, bounded open set,191 B1 (Q) = HI (Q) and the trace theorem (see Chap. IV) states that we can extend the map:

u E HI (Q) n rca U2) --+ u E HI (Q) and the trace of u on r = 0 (in Ht(r» .

It follows that, in the case Q regular, bounded, BI(Q)/B6 .:::,: Hi(r) and more precisely that the map u --+ trace of u on r is a bijection of £(Q) n Bl (Q) onto H~ (r). We shall not develop this aspect further, but content ourselves with making the link with the classical theory presented in §4. First, we prove:

Proposition 9. Let Q be an open set of[Rn, with only Q i= [R ;fn = I, and Q i= [R2

if n = 2, <p E B 1 (Q) and u the generalized solution (in the Dirichlet sense) ofP (Q, (p). (1) If <p ~ 0 a.e. on Q, then u ~ 0 on Q; more generally

infess<p ,,::; u(x) ,,::; supess<p for all x E Q. Q Q

(2) Let (Qkl be an increasing sequence of open sets with union Q, and Uk the generalized solution of P(Qb <p IQkLfor all k = 1, ... Then Uk --+ u in BI (Q).

Proof Considering m ~ supess, the function u /\ m = min(u, m), as a result of Q

Lemmas 2 and 3 of §5, satisfies

(7.81) u /\ mE BI (Q), §(u /\ m) ,,::; §(u) and u /\ m - <p E B6(Q) ,

from which u /\ m = u implies u ,,::; m. For the last point of (7.81), note that if a sequence (vd of '@(Q) is such that

V k --+ u - <p in BI(Q),

then[(v. + cp) /\ m] - cp is a function of Bl(Q) with support contained m SUPPVk'

191 [t is sufficient to suppose Q to have a boundary satisfying a Lipschitz condition. 192 We can characterize Hl(r) in many ways; for example

{ J"f IIp(x) - 4'(y) 12 l Hl(T) = 4' E L2(T); - d},(x)d,(y) < .XcJ .

FxF Ix-yl


hence a function of Bb(Q) and

[(Vk + qJ) /\ mJ - qJ --+ U /\ m - qJ III Bl(Q).

This proves point (1). For point (2), by definition of Uk

r I grad Uk 12 dx :::; r I grad qJ 12 dx :::; r I grad qJ 12 dx Jfh JQk JQ

and there exists V k E f?j) (Q k ) such that

1 :1 grad(uk - qJ Vk) II [.l(Q,) :::; k

and for every compact set K of Q k there exists C K such that

CK II Uk -- (P - Vk II[lIK) :::;k

To prove that Uk --+ U in Bl (Q), it is sufficient to prove that Vk --+ U - qJ in Bl (Q). We have

(7.82) lim sup II grad (Vk + qJ H L '(Q) :::; II grad U II L'(Q) .

In eifect since U qJ E B6(Q), there exists a sequence W k E !ZJ(Q) such that

W k --+ U - qJ in BI(Q);

we can always suppose that Wk E .@(QkJ and hence that

II grad uk!1 e(Q.) :::; II grad (qJ + wd L'Ul,) :::; II grad (qJ + wd II L'(Q) .

We then have

from which we obtain (7.82) in the limit.

II grad(vk + qJ) 11[2(Qd

1 il grad Uk IIL'(Q,) + k

It follows from (7.82) that grad L\ is bounded in L 2 (Q); from the proof of Proposition 8 we see that Vk is bounded in L?oc(Q). Given a sub-sequence (k[) such that

1\, ~ v in Bl(Q),

we have that the function qJ + v is the generalized solution of P (Q, qJ): in effect FE B6(Q) and for ( E 9(Q) ,

JI' grad (qJ + v). grad (dx = lim r grad (qJ + t: k ,). grad (dx = 0 Q [~ex. JQ

f7. Other Methods of Solution of Dirichlet's Problem

since for I sufficiently large, supp' c Q kl and hence

f gradukl·grad(dx = 0. Q"

In other words, Vk --"'- U - (p in 8 1 (Q); using (7.82) again, we have

grad(vk + cp) -+ gradu III e(Q), namely

gradvk ----? grad(u - cp) III L2(Q); we deduce that

Vk-+U-CP III 8 1 (Q)

549

which concludes (he proof of point (2). 0

We can then prove

Proposition 10. Let Q be a bounded open set (ifal n and cP E 8 1 (Q) n <eO (Q). Then the generalized solution (in the Dirichlet sense) of P(Q, cp) coincides with the generalized solution of P(Q, cP Ir) (in the sense of Definition 3 of §4). In particular the generalized solution (in the Dirichlet sense) u of P (Q, cp) satisfies

lim u(x) = cp(z) X-l'Z

for every regular point z of the boundary (in the sellse of Definitioll 4 of §4).

We observe that Dirichlet's method allows us to obtain a solution of P(Q, cp) for all cP E 8 1 (Q), whereas the classical method would give a solution for all cP E (6'0(1').

Proposition 10 shows that the two notions coincide for cP E 8 1 (Q) n <eO (Q). When Q is regular, we can identify 8 1 (Q)/ Bb(Q) with H1(F), and the Dirichlet method and the classical method give the same solution for cP E <e0(r) n H-l:(F).

Proof Let us consider u the generalized solution (in the Dirichlet sense) of P(Q, cp). From the definition of the generalized solution of P(Q, <p Id, we have to prove that given w E (6'0 (Q) superharmonic satisfying

lim infl-v(x) ~ cp(z) for all z E r

we have

w(x) ~ u(x) for all x E Q .

Let us consider for k = I, ...

Uk = {x E Q; d(x. r) < ~,W(X) > cp(x) - ~} . The set Q\Uk is a compact set of Q; we can always find a regular open set Q k =:J Q\Uk with boundary Tk of class 0, we put

qJ,(x) = r qJ(x + £t)p(t)dt .

" The function (P, is 'fix on {x E Q; dist(x, f) > c} . ForO < i: < dist(Qk,f),~etusconsidertheclassicalsolutionuk.,ofP(Qk,qJ,Ir/ we know that Uk., is Cf, CD (Q) (see Theorem I of §6).

1 The function w(x) + k + max i qJ - qJ, I is superharmonic on Qk' continuous on _ [,

Qk and greater than or equal to qJ, = Uk., on r k. Hence

I w(x) + k + max I(p - qJ,i ~ uk.,(x) for all x E Qk .

r,

Since Q k and Uk.! are regular, Uk., is the generalized solution (in the Dirichlet sense) of P(Qk' qJ,IQ ); denoting by Uk the generalized solution (in the Dirichlet sense) of P(Qk' qJ IQ ), trom point (1) of Proposition 9, ,

Since qJ, -> qJ uniformly on every compact set of Q,

1 w(x) + - ~ udxJ for all x E Q k ;

k

using point (2) of Proposition 9, in the limit as k ---> ,X),

w(x) ~ u(x) for all x E Q . o Proposition 10 extends to the case of an unbounded open set Q; with reservation 12 of. lRifn = I andQ of. 1R 2 ifn = 2,foreverYqJ E BI(Q) (\ (f,0(Q) satisfying the null condition at infinity (see Definition 9 of §4), the generalized solution (in the sense of Dirichlet) of P (Q, qJ) coincides with the generalized solution satisfying the null condition at infinity of P(Q, (P Ir) (see Definition 10 of §4). We leave to the reader the proof of this assertion. Dirichlet's method is linked with the notion of capacity and of the capacity operator. First, let us take two open sets Q I' Q2 at" IR" such that the set (Ql,Q2) u (Q 2 \Q I ) is a polar part (for the norm HI(IR"), see Definition 4 of §5). Then Bl (Q 1) = B 1 (Q2); this equation clearly has a sense, since every polar part being negligible (see Remark 8 of §5), a function defined a.e. on Q 1

is defined a.e. on Q2; we prove that the equation BI(Q 1 ) = Bl(Q2) is true. For that,itsuffices.givenqJ E Bl(Q I (\ Q2)andqJ, = (~qJ/Dx,inQ'(Ql (\ Q2),toshow that (P E L~)c(Ql) (resp. L~c(Q2)) and

f r 0~ qJj(dx = - qJOl-=-dx for all (E 0"(Q 1 ) (resp . . 0'(Q 2 ));

~ GXi (7.83 )

we note, in effect, that qJj E £2(Q)) = £ 2(Q2) = L 2(Q 1 (\ Q2)' We suppose first that qJ E Lloc(Q j ); given (E 0"(Q 1 ), K = supp(\Q2 is a polar compact set and hence of zero capacity in every open set containing it (see Definition 4 of §5). From


the characterization of the capacity of a compact set in an open set (see Proposition 8 of §5), given e > 0, there exists p, E g(Q1 ) such that 0 ~ Pe ~ 1, p, = 1 in the neighbourhood of K,

sUPPPe c Ke = {x;d(x,K) ~ e} and Ilgradpellu ~ 8;

the function 'e = (1 - Pe)( belongs to g(Q J n (2) and

II( - (,IIH' ~ IIp,(ilu + IIp,gradO L2 + II(gradp,llu

~ [(IIULT + IlgradULr)IKIt+~ + 11(11L"J£193.

In other words for (E g(Q 1 ) and b > 0, there exists (b E g(Q I n (2) such that Ii ( - (blIH, ~ b; we hence prove easily (7.83), since by definition

f 41i(b dx = - f 41 :~: dx .

It remains to prove that 41 E Lfoc (Q J); for that, let us consider for k > 0, the truncation 41k = max(min«p, - k), - k). We have (Pk E L~c(QI) and

J . 041k 041 41 E B (Q n Q ) wlth-· = ---x (see Lemma 2 of §'i) From the pre-k J -2 oX i ox i [I4'1 < k} . - .

ceding 41k E B 1 (QtJ and 041k = ~C!!.X!I I k} in g'(Q 1 ), from which in particular oX i oX i l 4' <

we have II grad 41k IIL2(Q ) ~ II grad (P IIL2(Q n Q,); it follows that (41k) is bounded in 2 ( ) 194 h 1 2 l_

L loe Q 1 t us that 41 E Lloc(Q d. The above argument proves similarly, which besides is simpler, that B6(Q J ) = B6(Q2 )195. By definition of the generalized solution in the Dirichlet sense, we can state:

Proposition 11. Let QJ and Q 2 be two open sets of [R" such that (Q J \(2) U (Q 2 \Qd is a polar part.. For all 41 E Bl (QJ) = Bl (Q 2 ), the generalized solution (in the Dirichlet sense) of P(QJ, 41) and P(Q 2 , 41) coincide on Q 1 n Q2 and are thus the extensions by continuity to QJ and Q2 respectively of the generalized solution (in the Dirichlet sense) of P(Q J n Q2' 41)·

Corollary 1. Let Q be an open set of [R", F a closed polar part o]D and u E Jf'(Q\F).

Then u extends to a harmonic function on Q iff f I grad u 12 dx < CfJ for every K F

compact set K of Q.

Corollary 2. Let Q be a bounded open set of [R" with boundary T, F a closed polar part of Q and 41 E rt'°(T u F). The generalized solution (in the sense of Definition 4

, , 193 We have used II p, III.' ,;: I KJ2+;; II grad p, II L' which is deduced from Sobolev's inequality. ,94 Otherwise, there would exist k, -> 'x and a ball 8(xo , fO) c Q 1 such that I, ~, II 'Pk, !I L'iBlx ,'0 II -> 'X,

and we should thus have for a sequence of functions r , = (jJk,/f, the contradiction grad r:, -> 0 in L 2(8(xo, foll, r , -> 0 a.c. on 8( xo, ro) and II r , II I 'I Blx"-,,, II = 1 (using the compact ness of H' (8(xo , roll in [2(8(xo, '0))' 195 AlsothatH ' U2,) = H ' (Q2)andHAU:2,) = H6(Q,).


of§4) ofP(Q\F, qJ) extends by continuity to Q the generalized solution of P(Q, qJl d. 1 n particular, the generalized solution of P(Q \ F, qJ) does not depend on the values of qJ

on r.

Proof of Corollary 1. The condition is obviously necessary; let us show that it is sufficient. The problem being local, we can always suppose that

r. Igradul 2 dx < 00 ~ Q·.F

and therefore u E Bl (Q\F) = Bl (Q); it is therefore the generalized solution of P(Q\F, u) which from Proposition 11 extends by continuity the generalized solution of P(Q, u) harmonic on Q. 0

Proof of Corollary 2. From Proposition 10, Corollary 2 reduces to Proposition 11 if there exists ;p E Bl (Q) n ,&0 ([2) such that ;p == qJ on r u F. Now from Weierstrass' theorem, there exists a seq uence ( qJn) of polynomials converging to qJ

on r u F; the corollary is then deduced from the continuity of the generalized solution with respect to the data on the boundary (see Proposition 3 of §4). 0

Let us now consider a Neumann problem, or more generally a mixed problem:

{ "hannon;, on Q

(7.84) u = qJ on r o

("u

l Dn !/J on 1'1

where r 0 is a closed part of the boundary r of Q and r 1 = r'J o· If Q is bounded and regular and u E '?5 ~ (Q) is the solution of (7.84), we have from Green's formula

(7.85) { r grad u . grad v dx = f !/J v d i' JQ r

for every function v E B 1 ('Q) n '6'0 (Q) with t' = 0 on

Using the same reasoning as in the proof of Proposition 8, we see that (7.85) is equivalent to

{

- u minimises the fu.n. ctional

~ r Igradul 2 dx - f !/JudI' JQ r,

on {u E B 1 (Q)n(60(Q); u = qJ on ro}.

(7.86)

We give a formulation of this problem in the case of an arbitrary open set Q. We denote by V( Q, r 0) the closure in B I (Q) of {v E B 1 (Q); t' == 0 in the neighbourhood of ro in Q}.

Heuristically, V( Q, r 0) is the space of the v E B 1 (Q) such that IJ = 0 on r 0; in the case where Q is regular, V( Q, r 0) is exactly the set {IJ E Bl (Q); trace of IJ on r is

§ 7. Other Methods of Solution of Dirichlet's Problem 553

zero on ro}; in the case with Q bounded and ro r, V(Q, r) = Bb(Q). In the case Q unbounded, we do not make a hypothesis at infinity on v E V(Q, ro); we can say that, in the case of Q unbounded, the space V(Q, r 0) corresponds to a Neumann condition at infinity. When Q is bounded and regular; for 1/1 E <C0(r), the map

(7.87) U E Bl(Q) (l (&,o{.Q) ---+ f I/Iudy 1,

extends to a continuous linear form on Bl (Q): in effect as we have recalled above (see just after Definition 2), the map u E BI (Q) (l <cO 0'2) ---+ u Ir extends to a continuous linear map of BI(Q) onto H1:(r) c L2(F)196. In the general case we can hence replace the given 1/1 on r by a continuous linear form on BI (Q) and consider the following generalization of the mixed problem (7.84): given qy E BI(Q) a 1/1 a continuous linear form on Bl(Q),

{

u minimises the functional

~LlgradUl2dx - <I/I,u>

on {qy + v;v E V(Q,Fo}}.

(7.88)

We shall interpret this general problem later; first of all, we prove

Proposition 12. Let Q be an arbitrary open set of [Rn, r 0 a closed part of 1 the boundary of Q, qy E B I (Q) and 1/1 a continuous linear form on Bl (Q). The problem (7.88) admits a solution iff

(7.89) {for every connected component Qo of Q ,

XQ" E V(Q, 10} => <1/1, XQ o> = o. Then the problem (7.88) admits a solution depending on as many arbitrary constants as there are connected components Qo ofQ such that XQo E V(Q, 10)'

The condition (7.89) is the generalized form of Gauss' theorem (see §2.2). We shall make the property XQ o E V(Qo, ro) clear later. For the proof of Proposition 12, we shall make use of

Lemma 5. Let Qo be a connected open set, 1/10 a continuous linear form on Bl (Qo) with <l/Io,XQo > i= O. Then, for every compact set Ko, there exists a constant eKo

such that

2n 2 196 From the Sobolev inclusions. HI (1) c L"( n for all q < Cfj if n = 2 and all q ,;; -- ifn ~ 3 (in

n ~ 2

particular q = 4 if n = 3); hence if.p E U(Ii) with p > 1 if n = 2 and p ~ 2 (~:-~) if n ~ 3, the map

defined by (7.87) extends to a continuous linear form on B 1 (Q).


We shall make use in the proof of this lemma, which is a generalization of Poincare's inequality, of the structure of the continuous linear forms on B 1 (Q); the following lemma will help us similarly to interpet the problem (7.88):

Lemma 6. Let Q be an arbitrary open set ollR". The continuous linear forms on BI(Q) are ol the form

f ( i":u Du ) (tf;,u) = j~u+j;~-~+···+.{,,·-~- dx

Q \ eXt eXn

wherefo ,J;, ... ,j~ E L 2 (Q) and fo is with compact support.

I ((lU au ) " . Proof of Lemma 6. The map 11 E B (Q) ---> U, ;-, •... :, ........ IS an IsomorphIsm eX I uXn

of Bl(Q) onto a sub-space of L~oc(Q) x L2(Q)"; from the Hahn-Banach theorem, a continuous bilinear form on Bl(Q) is therefore of the form

. . / au \ ! au \ ( tf;, u) = (Lo · 11) + (. L I .~--- ) + . . . + \ L I , -::;--;

\ (X I / \ ex"

where Lo is a continuous linear form on L~,c (!2) and L j, ... , Ln continuous linear forms on L 2 (Q). Hence the lemma, account being taken of the representation of continuous linear forms on L 2 (Q). 0

Proof of Lemma 5. By contradiction, given K () a compact set of Qo, let us suppose that for all kEN, there exists Uk E B I (Qo ) such that

II ( If; (J, 11k ) ~II

Uk - ;J~-;---; i > k grad Uk IIL'IQ,,) . "'/'0' Xli,,) IL 2 IK,,)

From Lemma 6, there existf() .... , j~ E L 2 ([20) with j;) having compact support such that

Let us consider an open set QI' connected, regular, bounded and such that

Ko u suppj~ C Q j C Q 1 C Q o and put

we have (tf;o,v k ) = 0, :11\11[1([2) = 1 and Ii gradvk liullio) < l/k.

From the compact injection of HI(Qtl into L 2 (Qtl 197 , there exists a subsequence (k l ) such that

l'k, -. [:0 in L 2 (QI)'

Since grad lOkI -+ 0 in L 2 (Q 1 ), grad Vo = 0 and hence, Q I' being connected, V() c,

19' See Chap. IV; we recall that Q I is a regular bounded sct.


on Q1' Since

II vde(Qd = 1, then I c I (mes Q 1 )1-

and, in particular, c ¥ O. On the other hand, since suppfo c Q1'

Lj~ vk , dx -> C Lfo dx ,

and since II grad Uk II L'(Q) -> 0,

f (r GUk GUk ) )1-ri - + ... + fn- dx -> ° .

Q (,Xl rJXn

555

Hence <1/10' vk,) = ° -> c f fodx = c<l/Io, Xilo>' a contradiction smce c ¥ 0, J!l < 1/1 0, Xila> ¥ 0. 0

Proof of Proposition 12. We denote

1 f 2 .f.p ( u) = 2 I grad u I d x - < 1/1, u > . Q

Suppose first that (7.88) admits a solution and hence, a fortiori,

rn = inf J .p(u) > - CI) •

<(J + V(Q, To)

Given Qo a connected component of Q with XQ E V(Q, To), we have, for all), E IR o

namely rn ~ .9vAcp + ),XQ) = .9.p(CP) - ;,<I/I,Xn) '

and hence < 1/1, XQ > = 0. Conversely, let us() suppose that the condition (7.89) is satisfied. Let us denote by (QJiEI the family of connected components of Q198 and

10 = {i EO I; XQ, E V(Q, To)} ;

from (7.89), for every family ()';)iElo of real numbers

U E cp + V(Q, To) -= U L }.iXQ E cp + V(Q, To) 10 I

and

and hence, in particular,

u is the solution of (7.88) -= U + L Ai XQ is the solution of (7.88) . 10 '

198 If Q is regular, bounded, the family (Q;)iEl is finite; in the general case it is at most denumerable.


For all i E f 0' we fix B j a compact set of Q j of measure I B j I > 0; applying Lemma 5 to the linear form

u -> J f u dx, for every compact K j of Qj, IBil Bi

there exists C Ki such that for all U E B 1 (Q)

For i E 1\10' since XQ ¢ V(Q, 10)' we have, from the Hahn-Banach theorem, that there exists a continuous linear form t/J j on B 1 (Q i ) ~uch that

(t/Ji, l'XQ) = 0 for all r E V(Q,ro)

(t/Ji,XQ) = 1.

Applying Lemma 5, we see that for every compact set K, of Qj, there exists C Ki such that for all v E V(Q,lol

ilvil,.'(K i ) ~ CKillgradd/'(Q;)'

Regrouping the two results, we find that for every compact K of Q, there exists C K

such that

(7.90) IlvIIL2(K) ~ CdgradvIIL2(Q)

for all v E V(Q, 1'0) with 1. v dx = 0 for all i E fo. It is sufficient to take B,

C K = max { CK,; i E f, K i = K n Q j of 0} in noting that there is only a finite number of i E I such that K n Q j of 0. After these preliminaries, we consider a sequence (vd in V( Q, 1'0) such that

J",(rp + vd -> It! = inf .f", (u) when k ->XJ . (jJ t V(Q, In)

We can always suppose r Vk dx = 0 for all i E fo: it is sufl1cient to replace L\ by Wk JBi

where

since

We note first that m > - oc. Using Lemma 6, with its notation

.f",(u) = If Igradul 2 dx -- f (f~u +fl;~ + ... +fn ~U)dX 2 Q Q LX 1 ex"

~ ~LlgradUl2dx - LU~u +ff + -.- +f;)dx.


On the other hand, let us suppose Y", (cp + vd -+ - OCJ when k -+ + OCJ; we can always suppose

Y",(cp + Vk) + L (focp + Ii + ... + I;)dx ~ 0,


~L,grad(cp + vkWdx ~ LfoVk dX ~ II.foIIL2(Q)CKIIgradvdL2(Q)

where we have applied (7.90) with K = suppfo. We deduce

LfoVk dX ~ CKlifoIIL2(Q)(llgradCPfl[2(Q) + 2(Lfo Vk dX Y), namely

and hence Y", (cp + vk ) is minorized - a contradiction. We now show that (vk ) converges in BI(Q). We have

~ L I grad (Vk - vdl 2 dx ~ Y",(cp + vd + Y",(cp + VI) - 2Y",( cP +~k-i-~l) ~ .1",(cp + vd + .1",(cp + VI) - 2m.

Therefore IIgrad(vk - v1)IIL2(Q) -> 0 as k,l-+ 00; taking account of (7.90), the sequence (vk ) converges in Bl (Q). Its limit v belongs to V(Q, r 0) and by continuity of Y"" u = cP + v minimises YIjJ on cp + V(Q, ro). To conclude the proof we have to prove uniqueness, to within arbitrary constants on the Q; for i E 10 , Given

V1 ,V2 E V(Q,ro) such that YIjJ(CP + VI) = YIjJ(CP + v2) = m,

we have

( VI + V2) - 2YIjJ cP + 2 ~ 0,

so Vl - V2 is constant on each Q;, i E l. But

(VI - V2 )xQ,E V(Q,ro) foraB iEI

and so Vl - V 2 = 0 on Q; for all i E 1\10; in other words

o

558 Chapler II. The Laplace Operator

Let us conclude this section by explaining the significance of the problem (7.88). From Lemma 6,

j' (. . . au . cu ) <I/I,U> = . loU +11;)-- + ... +1";;--- dx ~ (Xl eXn

where Jo,ft, ... ,In E L2(Ql,1~ with compact support in Q. Note carefully that Jo J, , ... , In are not unique. As we have seen in Proposition 8 for the Dirichlet problem, U is a solution of (7.88) iffuE(p + V(Q,rolandforallvE V(Q,rol

(7.91) L grad u. grad v dx = < 1/1, v> ,

that is to say

rQ grad u. grad to dx = J~ (.fa v + I; .~~.~- + ... + In av) dx . J Q eX I ax"

In particular, applying with v E .01(Q), we see that

,1u = ~li + ... +~::In - Jo III q'(Q). eX I ex"

Going back, first of all to the case qJ = 0 and

h = at; + ... + % m g'(Q) . . 0 (1XI ax"

It is clear that V( Q, n is contained in V( Q, r 0); in particular if V( Q, r 0) satisfies (7.89), it is the same for V( Q, T) and hence there exists U o a solution of

(7.92) Uo minimises .§ I}! on (P + V( Q, T) .

Putting ii = U -- uo, it is clear for example by using the characterization (7.91) that U is the solution of (7.88) iff

(7.93) ii minimises f;;; on V( Q, r 0)

where f is defined by

(7.94) < f, I' > = < 1/1, v> - r grad Uo . grad v dx . JQ In other words, U the solution of (7.88) can be written as the sum of Uo solution of (7.92) and ii solution of (7.93) where f is given by (7.94). In the case in which Q is bounded, V{Q, n coincides with B~(Q): the solution U o of (7.92) is characterized by U o E qJ + BMQ) and

jt' f ( . ov . av ) grad U o . grad v dx = Jo l' + II -1 - + . . . + In l dx Q Q LXI LX"

for alit: E g(Q). In other words, in the case Q hounded, Uo is the solution of(7.92) iff


Uo - <p E B6(Q) and

Oh a!. Au = ~ + ... + - - fo in f&'(Q);

oX 1 ax. the solution of(7.92) is the generalized solution (in the Dirichlet sense) of the Dirichlet

( oh a,) . problem P Q, <p, O~l + ... + o~~ - fo . We note that there IS uniqueness of this

solution: if u1 , u2 are two solutions, U 1 - U 2 E B6(Q) n ,Jf(Q) and so from Proposition 8, U1 = u2 •

In the case ofQ unbounded, V(Q, r) no longer coincides, in general, with B6(Q): a solution Uo of (7.92) is then a (generalized) solution of the Dirichlet problem

p(Q oh oJ. r) . f . ~T d' . '.r;' Th , <p, ~ + ... + -;;-- - JO satls ymg a l~eumann con ltton at IIIj<mty. ere ox! ux"

is not necessarily uniqueness: for example, if Q = [R", V( Q, r) = B 1 ([R") and the solutions of (7.92) are defined to within an additive constant; however, there is uniqueness if the unbounded connected components of Q have a boundary which is not a polar part of [R". Let us now consider the problem (7.93). First of all, we note that

- f ( - cv - cv ) < tf;, v) = j~ v + h -;:;- + ... +.' -:1- dx Q uX I u~

with.l = J: - ~uo and from the construction of uo (IX;

<If, v) = 0 for all v E V(Q, r)

that is to say that If is a continuous linear form on the "trace space SO (Q)j V( Q, r)". In the case in which Q is a regular bounded open set and if the functions J;, ... ,1" are regular, we have from Ostrogradski's formula (see § 1.3), and denoting by Jthe field (J;, ... ,1,,)

f (-cv - ov ) h -;:;- + ... + J. -~ - dx Q uX l ox.

thus, since

Therefore u is a solution of (7.93) iff u E V(Qo, r 0) and

Lgradu.gradVdX = tuJ.ndY foral! VE V(Q,ro)'

If u E ~O(2) n ~~(Q\ro), this expresses that u is the classical solution of the


mixed problem

Au = 0 on Q

11 = (7.95)

0 on To

IU - - = f.11 on r\rO (' 11

In the general case, we shall say that u is a yeneralized solution (ill the Dirichlet sense) of the mixed problem (7.95). We notice that the passage from the notion of generalized solution to that of classical solution requires a regularity of the data Q and/and of the solution U. The Dirichlet method will be generalized in the rest of this book: it is in fact a particular case of the variational method; we shall not develop further this method in the present chapter.

5. Symmetry Methods and Method of Images

We recall that the Laplace operator is invariant under displacements of the space: given V an open set of [Rn and T a displacement leaving V invariant .. I.e. T[V] = V-fi)r all g E CI'(V), r is a solution of Poisson's equation

jl'=q In ',/'(V)

iff w Tv is a solution of

jw = Tg In Cf'(U)

where for a function L" the function Tr is l' T- 1 and for a distribution y, the distribution Tg is defined by

<Ty,D=<q,( T)

(where T is a displacement). We now consider V, an open set of [R", symmetric with respect to a hyperplane H and put Q = V n H + where H + is one of the open half-spaces defined by H (cf.

H'

I I \ \ ,

¥x I I I I I C ----------~----r_-------f{ I I

:hx .... _------

r o

Fig. 24

/ /

I I I I

I /

n


Fig. 24). The boundary T of Q is made up of the plane part To

Tl = nro = Tn DU

Un H and of

where au is the boundary of U. Let us show to reduce a problem

LJu = f on Q

u = 0 ( au = 0) To resp. ~ on (7.96) en

au TI aou + a 1 -;;- cp on

en

to a problem

(7.97)

by using the symmetry T with respect to the hyperplane Ii. We shall work within the classical framework although this method of symmetry can also be developed in a variational framework. Let us state:

Proposition 13. Let Q = U n H+, where U is an open set of fR" symmetric with respect to a hyperplane Hand H + one or the open halj~spaces determined by H; the boundary T of Q is made up of To = U n Hand T 1 = au n T. Let f be a distribution on Q written in the form f = fa + J; where j~ E U(Q) with p > 1n (resp. p > n) and j~ is a distribution with compact support in Q; we denote by g the distribution on U extending f by 0 on U \ Q199. Finally let ao, ai' cP E (to (1' d with

ao ? 0, a 1 ? 0, ao + a 1 > 0 on T I' cP = 0 on Tin H ;

we denote by bo' b l' IjJ the continuous extensions of ao, ai' cP to a u by

ho(Z) = ao( Tz), hdz) = ad Tz), ljJ(z) = 0 for z E ou\r/oo.

We suppose that the open part of the boundary au, {z E i"U; bdz) > O} is regular (of class 'fad) and that the problem (7.97) admits a quasi-classical solution v. Then the restriction u to Q of v - Tv (resp. v + Tv) is a quasi-classical solution of the problem (7.96)201.

'4" This is possible because of the form off:

<g,S) = f fo(x)((x)dx + <f;.~xQ>' forall (E9(U). Q

200 We note that ho• h" IjJ E 'f,°(I"U), ho ~ O. h, ~ O. ho + b, > 0 on ?U. 20' This is true again if Q is unbounded: we could then introduce a condition at infinity: for example if I' is a solution of(7.97) satisfying the null condition at infinity (see §4.3), it is the same for Tv and hence the restriction of r - Tt, (resp. t' + Tr) to Q satisfies the null condition at infinity.


We note that account taken of the hypothesis that {z E a V; b , (z) > O} is a regular part of the boundary 3 V, the notion of a quasi-classical solution of (7.97) is clear (see Definition 13 of §4): the (mixed) condition on () V can again be written

{~ ho t/J :b l > OJ +- r

hi on

hI

t/J {hI = 0] (contained in {bo > O}) . r =

ho 011

For the problem (7.96), the condition on the boundary is well defined in a classical fashion on r 0 (regular open part of 1), on {z E r I; a I (z) = O} (the condition is of Dirichlet type, written u = (p/ao) and on {z E r I n H+; adz) > O} (from the hypothesis, this is a regular open part). Now at a point z E r I n H with a 1 (z) > 0, it is necessary to specify in a classical sense the condition on u, since from all the evidence r is not of class (6 1 in the neighbourhood of the points of r I n H: we shall take up this discussion again below; first let us give the

Prooj" oj" Proposition 13. The function II' = l' ~ Tv (resp. l' + Tv) is the solution of Poisson's equation

ilw = U ~ Ty (resp.y + Tyj in i:/'(V).

But the restriction ofy ± TytoV o = V\(supp}; u T(supp};)isin U(Vo) with p > 111 (resp. p > n); hence (see Proposition 6 of §3), w is continuous (resp. of class ((5 I ) on V o. The restriction u of II' to Q is therefore continuous (resp. of class Cfij I ) on Q n Vo = (Q\suppt;) u r 0 and, since Til' = ~ II' (resp. Til' = 11') we obtain

( ('U \

U = 0 resp.-;;-. = 0) ell

Now the restriction of y ± Ty to Q is .1; so

Au = f inCl'(Q).

Finally w satisfies classically the boundary condition

('II'

how+h,;;-=t/J~Tt/J (resp.t/J+Tt/J) on av. (11

Since the restriction of t/J ± Tt/J to r 1

boundary condition (l V n r is cp, u satisfies classically the

cp on r I . o We note:

Corollary 3. Let Q = U n Ht where U is a bounded open set or [R" with all its boundary points reguiar 202 , symmetric with respect to a hyperplane Hand H + one oj"

102 See Definition 4 of ~4: this hypothesis is introduced to ensure the existence of a classical Green's function C,,: the formula (7.98) is in fact true for an arbitrary symmetric open set U.


the half-spaces determined by H. Then the Green's function Gn of Q is given by

(7.98) GQ{x, y) = Gu(x, y) - Gd Tx, y), x, Y E Q, x"* y

where Gu is the Green's function of U and T the symmetry with respect to H.

Proof For Y E Q, it is sufficient to apply Proposition 13 with f = l)y, ao = 1, a l = q; = 0; in effect then g = by, bo = 1, b j = IjJ = 0 and vex) = Gu(x, y) is a classical solution of (7.97). 0

We show by some examples the use of the method of symmetry.

Example 4. Case of a half~ball. We consider a half-ball Q = B(xo, ro) n if + of [Rn, where H is a hyperplane passing through Xo. From (7.98),

Gg(x, y) = GB(., r I(x, y) - GB(, r)· (Tx, y) ; 0' () , 0, (]

thus, by use of the value of the Green's function of a bali (see Example 11 of §4),

GQ(x,y) = EII (x - y) - En(Tx - y) - CT;~.~~I))"-2 [En(l(xO,ro)x .... y) - E"(l(xoJo)Tx - y)J

where I (xo ' ro) is the inversion with pole Xo and power ro. We can observe, for example by applying (7.98) with U = [Rn (see also Example 18 of §4), that En (x - y) - En (Tx -- y) is the Green's function GH + (x, y) of the half-space H + .

'Since the symmetry T commutes with the inversion I (xo, r 0),

( I' )"- 2 GH - B( I(x, y) = Gw (x, y) - _---.Cl__ GJI+ (l(XOo ro)x, y) .

n xo,ro Ix ~ xol

We can derive this formula by the method of images (see below). In particular in the case 11 = 2, identifying [R2 with the complex plane, the Green's function of the unit semi-disk

IS

where

sll1ce

Q = {z;lzl < 1,lmz > O} = {reiB;O < I' < 1,0 < () < n},

1(z) = z Iz

Tz = z.

We use this value of the Green's function to solve the Dirichlet problem P( Q, Xs) where S is the segment [0, IJ of the real axis. The solution is given by

j'l ~G _ C Q u(z) - -- --;;-~ (z, x 2 ) dxz

o 0Y2

where (xz , yz) are the coordinates of Z2 in [R2.


Now

so

(7.99)

u(re ili ) = ~ [rrSinOj -+ --(r--~coti(J)i - L'in81-+~!tcotg&)2 J ='(Arctg(_I_- - cotgo) - Arctg(_l_' - cotgo)')

J[ r Sin () Sin 0

which is therefore the solution of P( Q, /,).

The symmetry method (Proposition 13) allows us to treat other problems. For example, let us consider the mixed problem

ju = 0 on 8(xo. 1'0) (\ H~

(' 11 (7.100) - = ~J on B(xo.l'o) (\ H

i'l1

11 = (P on (08(xo• 1'0) (\ H +

where lj; E «('0(ll) and ip E C(, o("B(xo' ro) are given. To simplify the notation. we can always suppose that

Xo = O. H - = [R" - 1 X IR-

We begin by reducing to the case lj; = 0 by considering the solution r of P IV (H +. tf;): we know (see Example 21 of ~4) that r is given by

, 2x" f 1// (x' + x,JJ [·(x. XII) = (/1 _ 2)()~W-l (l-;-tI2);~-1 dt if /1;? 3

(7.101)

['(x.r) = -2\~J tf;(x+yI)Log(1 +t2 )dt if II = 2 these formulae being justified if tf; satisfies (4.77) if 11 ;? 3 and (4.79) if 11 = 2: in fact only the values of tf; on 8(xo. /'0) interest us with the result that we can always

I'

suppose that lj; is of compact support satisfying. in addition'J 1/1 (t) dt = 0 and gn !

hence that (4.77) and (4.79) are satisfied. Putting II' = 1I - l' we see that II satisfies (7.100) iff w is a solution of

(7.102) = 0 ?n

on B(xo.ro)nH-

To solve (7.102), we use the symmetry method: let us consider 9 the function defined

p. Other Methods of Solution of Dirichlet's Problem 565

on 8B(xo, ro) by

g(x', x n ) = g(x', - x n ) for all (x', x n ) E aB(xo, ro)

g = cp - v on aB(xo,ra) n H+ ;

it is continuous and the solution of the Dirichlet problem P(B(xo, raj, g) is even with respect to Xn with the result that its restriction to B(xo, ro) n H + is the classical solution of (7.102); from Poisson's integral formula (see §2.3) we have

1 f r~ - 1 x - Xo 12 \V(X) = - n~g(z)d}'(z)

fOO"n ?B(xo.roJ 1 z - xl which, from the definition of g, can be written

(7.103) 11'( x) = _r~=----_i_x_-_x_o_12 f ( 1 z - x 1 ~ n

ro(Jn ,'B(x".ro)nH-

+ Iz - Txrnl(cp(z) - v(z))d::(z)

where T is the symmetry with respect to H. In brief, the mixed system (7.100) admits a classical so/ution 203 u = u + IV where u is given by (7.101) and 11' by (7.103). Besides the possibility of obtaining explicit solutions as in the above example, one of the interests of the symmetry method is to yield regularity results for Dirichlet, Neumann or mixed problems in non-regular open sets. We illustrate this in:

Example 5. Case of a cube. Let us take a cube of [R" or more generally n

Q = n] ai' bi [where ]ai, bi [ are bounded intervals of IR. We take! E ~a: (lJ) and i = 1

consider the problem

(7.104) u = 0 on ro au

{

11u = f on Q

a;~ = 0 on r 1

where r 0 is a union of (open) faces of Q and r I is the union of the other faces 204

with the result that 10 n r 1 = 0 and r\(r a urI) are the edges of Q. Using, for example. the Dirichlet method (see §7.4) we shall show that if fo #- 0 (resp.

10 = 0 and LfdX = 0), there exists a unique (resp. to within an additive

constant) generalized solution (in the Dirichlet sense) of (7.104). In fact using the results oflocal regularity (see §3.2) as well as the regularity at the regular boundary (see §6.1). we verify that this solution is classical and is in ,!}X (Q n f 0 urI)' The

lOJ Unique (see Corollary 6. ~4). n

204 Q admits in faces such that n ]ai • hi I: x [an}. ;-=1


real problem is the regularity in the neighbourhood of the points of the boundary of thefaces. Using the symmetry method, we prove that the solution u E '6'1 (Q) and evenUE Wl.P(Q) for all 1 < p < a:;205;ingeneralhoweverur!:'6'2(Q),although this is the case if f is zero on F\(ro uri); more generally iff is with support contained in Q u Fo U F I , then u E YiX(Q). So as to throw light on these properties, we shall restrict ourselves to the case 11 = 2. We can suppose that Q = ]0, I [ x JO, I [ and study the regularity of u in the neighbourhood of (0,0). We can always suppose

F ° => {o} x JO, I [ u JO, I [ x r O} .

rlu i'u (~2 u even if we replace u by (;x' ~ly or ex a y according as F I contains r O} x JO, 1 [,

]0, I [ x {O} or both. We consider then Q = JO, I[ x J- I, le,.idefined on Q by

1 = f on Q and ICx, - y) = -/(x, y) for all (x, y) E Q

and Fo = {OJ x J-I, l[joined possibly to {I} x J-I, I[ and ]0, I[ x [-1,1) according as fn contains {I} x JO, I [ and ]0, 1 [ x {I}. Given 1 < p < x, under the hypothesisf E U (Q), the function lEU (Q) with the result that the solution u of the corresponding problem (7.104) is in Wro·t(Q uFo) (see §3.2). Now

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

with the result that u is the restriction of u to Q and hence u is W 1 . P in the neighbourhood of (0, 0). Even within the hypothesis/ E Cf,'h (Q), in general I is not even continuous on Q: we have this property, however, if

f(x,O) = 0 for all x E JO, I [ ;

then I satisfies a Lipschitz condition on Q and hence ij E Wfo"J'(Q u Fo ) and u is W 3 . P in the neighbourhood of (0, 0) for all I < p <'JJ.

In fact the solution u is of class (6 2 ill the neighbourhood of (0, 0) ifff(O, 0) = 0. Let us prove this property. First of all, since u(x, 0) = ufO, y) = 0, then D2 u a2 u

-2··· (x, 0) = --2(0, y) = 0 for all (x, y) E Q and hence u of class (6 2 in the ax cy neighbourhood of (0, 0) implies f(0, 0) = 0. To prove the converse, we consider r defined on Q by

_ eu v = 1 on 0,

()x v(x, - y) = - u(x, y) for all (x, y) E Q .

Since

uE<6 1(QU[O, I[ x {O}u{O} x [0, I[) and

we have r E (f,O(!2 u {OJ x ] -1, + l[).

2()' Under the sole hypothesis! E U( Q) (sec Proof below).

au -- = ° on [0, 1 [ x {O}, ex

p. Other Methods of Solution of Dirichlet's Problem

We have

and also

_ of -Llv = - on Q ox

~_/ ~

ov 32 u 32u_ a-; = a-;2 = f - ayz = f on {o} x ] - 1, + ] [)

567

since u ° on {o} x [0, 1 [. Given ° < IX < 1, under the hypothesis f E ~a(.Q), since f(O, 0) = 0, the function j(O,.) E ~"([ -I, + I]); from the regularity theorems (see §6.1) we have therefore

V E ~lo:a(Q u {O} x ]- I, + I[)

and hence ~u of class 9Jll h in the neighbourbood of (0,0); using the equation ex

D2 u 32 u -;;----. 2 = f - -3 2' u is therefore of class ~2 +a in the neighbourhood of(O, 0). To sum cy x up, we have proved the property and even that iff E ~a(Q) withf(O, 0) = 0, then u is (~2+a in the neighbourhood of (0, 0). This example, not only puts in evidence the phenomena of the problems of regularity in the neighbourhood of a non-regular point of the boundary, but is fundamental for the study of the problems of regularity in the corners. We shall not deal here with the subject referring the reader to Grisvard [I J, Kondratiev [1]. We conclude this section by showing that the symmetry method is, in a certain sense a particular case of what is called the method of images. Let us consider an open set Q in IR" with boundary T. Given x E fl, we give the name (electrostatic) image of the point x with respect to T to a distribution T(x) E g' (IR"\Q) such that the Newtonian potential of T(x) coincides on T with that of bx , namely

E" * T(x)(z) = E"(x - z) for all z E T .

In the case of a half-space H +, it is clear that for all x E H +, bT(x) is the image of x with respect to H where T is the symmetry with respect to H. Knowledge of the image of the points of Q permit the determination of the Green's function of fl: in effect if T(x) is the image of x with respect to T,

GQ(x, y) = E,,(x - y) - (E" * T(x»)(y)

since the function E" * T(x) is harmonic on Q and by definition takes the value E,,(x- .) on T. We have effectively used the method of images to determine the Green's function of a half-space H + :

GJI+ Cx, y) = E,,(x - y) - E"( T(x) - y)

where T is the symmetry with respect to H. Similarly we have used the method of images to determine the Green's function of a ball 8(xo, ro ): considering /(xo, ro ) inversion with pole Xo and power ro , we have seen (see Example II of §4) that


CUSI!Il ~ 3. For all x E B(xoJo)\{xo},we have

EIl(l(xoJo)X ~ ... :::) = ex ~OXOI)"-2 En(x

hence the distribution (_~_'_·o_ .. )" ~ 2 b1(xo.rol x is the image of x with respect to ,Ix - xol

?B(xo , 1'0) and as we have seen

(' r )" - 2

G E(.. I(X. y) = E,,(x - y) - .... _0_ En(l(xO' "o)x - y). '".r" Ix ~.~ xol

We note that the Newtonian potential of a uniform distribution f1 on a sphere ('B(x o, r) is invariant under rotation about Xo and its value on the sphere a B (xo , 1'0) is flU n 1''' - 1 E" (1'0)' Hence, for r > r 0' the uniform distribution of density 1 !rJnr,,-1 on c'B(xo, r) is an image of the point Xo with respect to (1B(x o , 1'0)'

Case: n = 2. For all x E B(xo• ro)\{xo ~

I "0 E,(x - ;::) + -- Log---- for all ::: E ?8(xo• 1'0) . - 2n Ix xol

As we have seen above the constant potential c on the circle (IB(xo, 1'0) is created ('

by a uniform distribution of densitY!l = .~ .. - on the circle i' B(xo• r). Hence an /' Log 1'0

image of x with respect to i'8(xo , 1'0) is

(j. -L(·.l - LO~IX~-:._~~J)(), with r > 1'0 arbitrary. f(\".r"lx 2nr Logro 18('".rl

§ 8. Elliptic Equations of the Second Order*

In the preceding sections of this chapter. we have studied Laplace's eq uation and Poisson's equation and the Dirichlet, Neumann or mixed problems for these equations. In Chap. V (vol. 2), we shall study, and more particularly, we shall classify general linear partial differential equations. In this section we propose to show whether or not the properties of Laplace's equation generalize to more general elliptic equations of the second order. Given an open set Q in [R" we consider a partial differential equation of the second order

(8.1 ) n ('2 U 11 ?u

i.~ 1 U ij 8X;i3Xj + i~1 b i r'';X i + CU = f on Q

in which the coefficients aij , hi' c andfare real functions defined on Q. A function

~8. Elliptic Equations of the Second Order 569

U EO ct'2(Q) satisfying (8.1) for all x EO Q is called a classical solution of equation (8.1). ? a2u a1u

Given that for u E ct'-(Q), ---- = ~--~-., even if we replace aij by 1(aij + aji)' exJ3xj OXjCXi

we can always suppose that aij aji . We say that the equation (8.1) is elliptic on Q206 ijF)r all x E Q

(8.2) A(X)

n

mm L ai/x)~i~j > 0; CEIR" i.j= 1 i¢i = I

we shall say that it is strictly elliptic on Q if inf ),(x) > O. We observe that under the XED

assumption aij = aji, ).(x) is the smallest (real) eigenvalue of the symmetric matrix (aij(x»i,j, while

def n

A (xl max L aij(xK~j (ErR" i.j= 1

121 ~ 1

is the largest eigenvalue. We say that (8.1) is uniformly elliptic on Q if it is elliptic and Jc{x)

inf -- > O. XEl.! A (x) When the coefficients aij' hi' c are constants we can always reduce the equation207

by change of the coordinate axes and of the function f to the equation

(8.3) -- Ll u + cll = f where c E R The case c 0 is Poisson's equation, the case c < 0 is Helmholtz's equation which we rather write

(8.4)

where k is real constant that we can always choose to be positive. We shall study this equation in greater depth in Sect. 7 on account of its importance as much mathematical as for modelling numerous physical problems. The case c > 0, which we shall rather write

(8.5)

with k a real posItIve constant, will present practically all the properties of Poisson's equation as we shall see below: we shall sometimes give it the name of the equation of neutronics, where it occurs as a model (see Chap. lA, §5); the model closest to reality is rather the equation:

(8.6) -div(agradu)+k 2 u=f on Q

where a and k are positive functions defined on Q.

206 See Chap. V for a more general definition of ellipticity. 107 ln the elliptic case (see Chap. V).


1. The Divergence Form, Green's Formula

We can put every equation (8.1) in the divergence form

the relation between the coefficients (b i, c) 0[(8.1) and (c i, di, d) of(8.7) being given by

(8.8)

We notice that the divergence .form (8.7) is not unique. Setting aside the fact that physical problems are habitually modelled in divergence form, this provides many advantages from the mathematical point of view. First of all it allows us to generalize simply Green's formula; for that we introduce some new concepts.

n ? (I n t" ) n ? Definition 1. Let L = - L -. - L a·- + c + L d ~.- + d be a dif-

. i=lr:Xi )=1 tJC:Xj 'l i=l taXi

ferential operator of the second order in divergence form on an open set Q of [R".

We denote by VL the differential field associated with L, whose components are

(g.9) n ?

VIi = ') a· --- + c _ t....-A I J ~, ~ I

j=l (.x j

i = 1, ... , 11 .

By the co-normal deriwlive associated with L we mean the trace operator on the boundary r of Q

(8.1 0)

where n is the unit vector normal to r exterior to Q. We give the name (formal) adjoint operator of L 20s , to the operator

(8.11 ) L* = n (l

+ 'c·.-~+d ~ t'" • j = 1 eX i

Within the classical framework, the coefficients aij' Cj , d j , d will be supposed continuous on Q; we shall denote by (&f(Q) the set offunctions u E (6'1 (Q) such p = VLu E ~l(Q)" and ~~ (Q) the set of/unctions u E ~l(Q) n '{50(Q) such that

L au - (z) = p. n(z) exists for all z Erin the classical sense considered in § 1.3 209 . al1L

208 Or transposed. 209 It is clear that if U i j. Ci E (6 ' (Q) then (f, 2 (Q) c (6 t (Q); we have even 'f, t (Q) = '6 2 (Q ) if L is elliptic

or, more generally, if the matrix (uij(x)),.j is invertible for all x E Q. Also if aij , CiE«,O(Q), then (,f' (Q) c «,~JQ); we obviously do not have equality except in dimension n = !.

~g. Elliptic Equations of the Second Order 571

Given U E ~UQ) and v E ~I (Q), we have

We introduce the hilinear differential operator associated with L

II au ev n (ev eu \ (8.12) L(u, v) = .2: aij -~- ~- +2: CiU -~- + di -1') + duv,

1.]= I cX j eX i I = I eX i

with the result that we have

div(v VLu) = L(u, v) - vLu ,

from which by application of Ostrogradski's formula (see Proposition 4, § 1) with the reservation that u E (6 1 (6), v E (6°(6) and L(u, v) - I'Lu ELI (Q)

til.

(8.13) fa (L(u, I') - vLu)dx

which we shall rather write in the form

(8.14 ) f f au vLudx + v ~- di'

Q Q ellL

r I L(U,l,)dx "Q

under the condition that vLu E Ll(Q). We notice now that by the definition of the adjoint operator

L (u, I') = L * (v, u) .

Hence under adequate hypothesis

r L(U,I')dx = r L*(v,u)dx = rQUL *l'dX + J" lI;~-d}' . • Q J Q J I ellL'

from which with (8.14), we deduce Green's formula

(8.15) r (uL*l'-- rLu)dx = J" (v ~1U - II J-':'-)d" JQ r ('ilL 1'111: I'

To sum up, we have

Proposition 1. Gil'en L = i :112. (a ij /' + C i ) + i di .{: + d a differ-i.j= 1 (Xi LXj i= 1 (Xi

ential operator of second order ill dit'eryellcejc)rrn on a reyular hounded open set Q of [R" with houndary 1. With the Ilotatioll of Definition I above, jc)r all

II E '(,t(Q) n ((1 (6) and l' EO ((,1 (Q) n '(,°(6) ilL

(resp. l' EO (6i.(Q) n '6;, .(Q)) with L(u, l'), l'Lu EO LI(Q) (resp. and uL*t, EO LI(Q)), we have Green'srormul~ (8.14)for intewation by parts (resp. Green's{ormula (8.15)).


We notice as a corollary of Proposition 1, Gauss' theorem, corresponding to the particular case r: == I.

Corollary 1. With the data of Proposition 1. fin' all 11 E (6 i,(Q) Ii (6,1" (Ql with LIJ U = - di V VI. U ELI ( 0), we hWl(!

Following Proposition 5 of ~ I, for

t' E C(,1(0) Ii (6°(~2) and U E (61(0) Ii (6 1 (Q). J nL

the integrability of L (u. r) - r Lu is assured since this function is minorized by an integrable function. Let us suppose that the operator L is elliptic and denote by ), (x) the smallest eigenvalue of the symmetric matrix ({I;j(x )). For all u E ({j'1 (0) we have

L(u,u)? i.lgradui 1 - (I(c; + d;)2)1Iullgradul + du 2 •

and for all t : 0 ---+ [0. I]

(8.16 )

We deduce

Corollary 2. With the data of Proposition I, yiv(!n a measurahle function t: Q ---+ JO, 1 J, we suppose L is elliptic and

(8.17)

where}. (x) is the smallest (!igenraiue of the symmetric matrix (a;j(x)). Thenj(n' every junction u E '('i'I(O) Ii '{,I (Q) with Lu E e (Q) we have

Ill,

and

(8.18)

(1 - t)i,lgraduI 2 E L 1 (Q). (-I.-I(c; + 1i;)2 - d)U 2 E L 1 (Q) \ 4).(

Supposing always that L is elliptic, we denote by A(x) the largest eigenvalue of the symmetric matrix (a; j (x)); it is classical that

I ~ai/X)~irljl ~ A(x)I¢1111l for all ¢, '1 E [Rn . !,J

98. Elliptic Equations of the Second Order 573

Hence for u, v E ~1 (Q), we have

1 L (u, v) 1 ~ A 1 grad u i I grad v I + (L c? yt 1 u II grad v I

+ (Ld1)+llillgradul + Idllullvl

~ (~ + 1) (~ 1 grad u 12 + ~ I grad Ii 12 )

+ (r.,c? + Id l) u1 + (r.~f + Id l) v2

.II. 2 J, 2

Applying Corollary 2 with f = t, we deduce:

Corollary 3. With the data of Proposition 1, let L be elliptic and

~ E LUJ(Q)' , c2 d2 , --;-,~,dEL1(Q),

J, J, ),

where i(x), A(x) are respectively the smallest and the largest eigenvalue of the symmetric matr:Jx (a jj (x)21o. Then for every U E ~UQ) Ii ~,; (Q) and v E ~i.(Q) Ii '6~L.(Q) with Lu, L*v E L1 (Q), we have L(u, Ii) E [1 (Q) and the Green's formulae

I f eu I f Ct' I L(u,v)dx= v-~-d}'+ vLudx= U-;1-' di+ uL*vdx. Q f' enL Q r e n£* Q

We recall that the divergence form Lu = f on Q of equation (8.1) is not unique with o 0

the result that the conormal derivatives -a-'l- are not associated with the ·nL enL'

differential operator

p n

L i.j= 1

a2 a" --

'J OXiOX j

defining (8.1), but with its particular divergence form L. We notice however that the ~ '1 n a

principal parts of -~ and ~ are the same, this is L aijni~' and so they are ('·n L on I: i.j= 1 eXj

associated with P. Now let us give another statement of Green's formula

Proposition 2. Let Q be a regular bounded open set of [Rn of boundary r of class ~m + 2 where m ;;:, 0,

n p I

i.J= 1

32 a··--- +

fJ 2xJJXj

a differential operator of order 2 on Q with

n 3 I b-

i= 1 I i)X i

210 That is to say that we suppose that L is uniformly elliptic on Q.

+ ('


n D and B = 130 + jf:

1 f3j oXj a differential operator of order 1 on r with 130 E <'6 m ( r),

{31 , ... , f3n E "6'm + 1 (r) satisfying

(8.l9) /I

L {Jj(z)nj(z) #- 0 for all Z E r . j= 1

Thenfor all u, v E 'ti 2(Q) n 'til (Q) with Pu, P* l' E e (Q)

(8.20) r (vPu - uP*v)dx = f (uB*v - (XvBu)d,)', JQ r

where

(8.21) P*

with

(8.21 )' b~ = b. 2"oaij I 1+ L...:1 ' j uX j

and

(8.22)

n 0 (8.23) B* = (XB + T with T = to + L tj-

1=1 oX j

a differential operator of order 1 on r with coefficients to E <6'm(r) and t 1, t 2' ... ,

t n E <6'm + 1 (r), satisfying

(8.24) /I

L tj(z).nj(z)=O forall ZEro j = I

In other words, for P a differential operator of order 2 on Q and B a differential operator 1 on r, /lon-tangential (that is satisfying (8.19», there exists afunction (X on r and T a differential operator of order 1 on r, tangential (that is satisfying (8.24» such that we have Green'sformula (8.20) with P* the "adjoint differential operator" of P on Q (given by 8.21) and B* = (XB + T. We notice that if p* depends only on P, (X and

'f ,,0. T depends on P, Band r; we can further speCI y T = to + L... tj ~ wIth uX j

(8.25) ( oa .. ) 1 To = ~ bi + ~ 0;; nj - 2divr(TI, ... ,Tn)

(8.26) tj = 2( ~ai/Jj - (X{3j) , j = 1, ... ,n ,

where div d t \, ... , Tn) is defined by the formula for integration by parts on r. Lemma I. Let Q be a regular bounded open set with boundary r o( class 'h'm+2 and T\, ... ,TnE'h,m+I(r) satiifying (8.24). There exists a function

§8. Elliptic Equations of the Second Order 575

divr(r" ... , rn) E (Cm(n such that

f ( au) f' I~a Ij dy = UdlVr(r 1 ,···,rn )dy r .I x) r

(8.27)

for all u E '6'1 (Q).

We leave to the reader the verification of this lemma by the use of local maps or the application of general formulae of differential geometry (see Choquet-Bruhat [IJ). Let us now give the

Proof of Proposition 2. a ( 0) a Denote L = - I -- a . -- + I d· -- + c with i.j oX, I] aXj I aX i

aa . d=h+I-.'..!.

f - I j axj ·

For u E '(i'2(Q) (\ r,1 (Q) we have u E '6'i(Q) (\ ~'I (Q), Lu = Pu on Q and n" au ..

;;-- = )'. grad u on T where for ail Z E T, y(z) IS the vector wIth components un I.

n

rj(z) = I aij(z)ni(z) . i = 1

With ::i given by (8.22) and r I, ... , rn given by (8.26) we have

a 1 (') .~~ = ::i(B - flo) + .. I r·~, . cn L 2 j J oXj

On the other hand L* = - I-.i3-(a!.... + d.) + c i.j (1X i lJ (lXj l ,

(1 a 1 a .... = + 'd,n· = ::i(B - {:1 ) + 'd.n. +- 'r.' '-". ~ 1 L.lt" 0 ~lt ,L../"'I

onL' rnL i i ~ OXj

For v E ~'2(Q)(\{61(,{;h we have v E '6~ ,(Q) (\ 'Ii.~ ,(2), L*v = P*v and L L

Noting that

1 (au au ) ( av ) 1 a -~ri u-;")- - u-,- = u ~rj-~- - -~tj~(uv), 2 1 eXj ex; ) eXj 2 J

we therefore have from (8.27)

f ( au cu ) f { UP u ;:;-- - v;;-- dy = a(uBu - vBu) + u~ tj -~-r 0nL' enL r j eXj

+(~d,n, - ~divdrl, ... 'tn))UV}dY. Green's formula (8.20) is deduced from Proposition 1. o


2. Different Concepts of Solutions, Boundary Value Problems, Transmission Conditions

We have defined what we mean naturally by a classical solution of the equation (8.1). As we have seen in the case of Poisson's equation, this notion is not sufficient; in the case of a general equation of the form (8.1) or of the divergence form (8.7), we shall be led to consider several concepts of solution weakening that of a classical solution. First of all, we take a differential operator

with coefficients aij' bi , C E (6 '(Q). We have defined by the formula (8.21), the adjoint operator

P* + c*

with

Given u E Y; 2 (Q), we have from the construction of P*

f ;Pudx = f uP*~dx for all ~ E C/(Q) . !! Q

This, quite naturally, leads us to pose the

(,2 ? Definition 2. Given P = - I a . - -- + I h - + C a differential operator

j,j ~) (::xi(lX j i I tlXi

of order 2 on an open set Q of [R" with coefficients aij' hi. C E ((, Y (Q) andf E '/' (Q);

by a solution in the sense of distributions of the equation

Pu=j on Q

we mean every u on Q satisfying

(u,p*o = (t:O forall ~EC!(Q)

where P * is the adjoint operator defined by (8.21).

We notice that this definition has a meaning since P* ~ E 9(Q) for all ~ E 9(Q). It generalizes clearly that given of a solution in the sense of distributions of Poisson's equation .1u = f on Q since .1 is formally self-adjoint. The hypothesis of '6 f_

coefficients can be weakened by imposing a minimum regularity on u: in particular, if we restrict ourselves to considering solutions U E Lfoc(Q) with 1 ~ p ~ eXJ, it is sufficient to suppose that the coefficients of P* are in Lfoc(Q) with q = p/(p - 1) the conjugate index of p, with the result that uP* ~ ELI (Q) for all ( E 9( Q). Although this aspect is interesting in certain applications, we shall rather develop the notion


of a weak solution, for which we restrict u E Wi!o'c! (Q )211 in using the divergence form

ourselves to consider functions

L = - ~ ric (2:: aij -l-- + Ci) ,exi J ox)

If the coefficients of P and L are regular on Q and linked by the relations (8.8), we have, for all U E «,2(Q),

(8.28) r (Pudx = r L(u, ()dx for all ~ E !£(Q) JQ JQ where L( ., . ) is the bilinear differential operator defined by (8.12). The regularity of the coefficients is not necessary for considering L (u, 0 for (E.9 (Q) and U E WJ1o·tO(Q) n Lfo'c(Q); as we see below, it is interesting to consider operators with discontinuous coefficients. Although we can consider coefl1cients which are only measurable (see, for example, Trudinger [1], we suppose here, for simplicity that the coefl1cients are in LJ~c (Q). We therefore pose the

Definition 3. Let L = - I..,D (2:: aij ..,(i + Cj ) + 2:: di ~c + d be a differ-,DXj J (,Xj ,(;X j

ential operator of order 2 in divergence form on an open set Q of [R", whose coefficients are in LJ~c (Q) and f E 9' (Q). By a weak solution of the equation

Lu = f on Q

we mean any function u E Wilde! (Q) satisfying

(8.29) fa L(u, Odx = ({,O for all (E ost:(Q) .

If the coefficients of L are of class ({,' 7.', it is clear that every weak solution of Lu = f on Q is a solution in the sense of distributions of Pu = fan Q, where P is the differential operator of the second order whose coefficients are given by (8.8).

We note also that under the hypotheses of Definition 3, if there exists a weak solution of Lu = f on Q, the distribution f is of the form .hl + l' cfjDx j with .fo,f", ... ,/~ E Lioe(Q); the notion of a weak solution does not permit the solution of equations with arbitrary data and distribution f212. Finally, from the relation (8.28) iFthe coefficients are regular on Q, then/iJ/'fE (6oo (Q) u is the classical solution of Pu = f on Q iff u E (6 2 (Q) and is a weak solution of Lu = fOil Q. In fact, the

(lU

211 That is to say uELlo,(Q) and its derivatives ~- (in the sense of distributions) arc in L,lo,(Q). ('X i

2'2 We are limited by f E WI~'l. 1 (Q), which, all the same allows Radon measures 011 Q; when the coefficients are continuous on Q, we can consider L( u. ~) for every function U E B V"" ( Q) (that is to say 11 E L,~, (Q) and its derivatives (lU/(lX; are Radon measures on Q) and define for an arbitrary distribution of order 1. a weak solution of Lu = f

57H Chapter II. The Laplace Operator

equation (8.28) can be verified under much more general hypotheses: without considering the most general case, let us suppose that the coefficients of P are in LI~c(Q), with the result that Pu E Lloc(!2) for allu E W~'l (Q). Now to suppose (8,8) to be true in the sense of distributions comes down to supposing that (8,28) is true for U E (f, x (Q). Using the density of (f,' (Q) in I--Vfo'c l (Q) and. under the hypothesis that the coefficients of P and L are in LI~c (Q) the continuity of the maps U -+ L (u, (), li --> (P u of TV~u} ( Q), into L 1: lC (Q) we deduce that (8.28) is true for all u E wfo'cl (Q). This leads us to pose the

<,2 (~

Definition 4. Let P = - '\' a,,' - -,'-' + '\' h -- + c be a differentiable oper-~ q"" '"' L... r "" i,j eX i ('x j i eX i

ator of order 2 on an open set Q of [R", whose coefficients are in LI~c(Q), and f E Lioc(Q); by a strong soiution of the equation Pu = f on Q, we call every function u E W~o} (Q) such that

(8.30) Pu(X) = fix) a.e. x E Q .

We have shown above that, the coefficients of P and L being in LI~c(Q) and (8.8) being satisfied in the sense of distributions, II is a strollY solution of Pu = fOil Q iff u E W~d}(Q) and is a weak solution of Lu = fOil Q.

The different concepts of solutions that we have introduced for equation (8.1) or its divergence form (8.7) will have their correspondence for the boundary value problem

(8.31) { Pu = f (resp. Lu ={) on Q

BII =!l on r where P (resp. L) is a differential operator of order 2 (resp. with divergence form) on Q and B a differential operator of order less than or equal to 1.

Definition 5. Given Q an open set in [R" with boundary 1.

a differential operator of order 2 with coefficients aij , hi' C E (6°(Q) (resp. n (I

((, x (Q)), B = flo + L flj~- a differential operator of order less than or equal to I j= I oX j

with coefficients fJjE~·O(Q) (resp. (~;x(Q))213, fE((,O(Q) (resp.Q'(Q)) and Y E reo (f), we call a classical (resp. quasi-classical) solution of the boundary value problem (8.31) every function u E '(52 (Q) (resp. distribution U E 0" (Q)) such that (1) U is a classical (resp. in the sense of distributions) of the equation Pu = fon Q;

213 Taking account of the definition below. we find that it is sufficient to define fio,} in the neighbourhood of r in Q,


(2) lim Bu(x) = g(z) for all Z Erin the classical sei1se (resp. in the sense specified X --t z

in Definition 6 of§4)214.

This definition clearly generalizes those given in §4 for the concepts of classical and quasi-classical solutions of a Dirichlet, Neumann, or even mixed problem for Poisson's equation. We note that a Dirichlet (resp. Neumann) condition

u = g (resp. :: = g) on r

n a corresponds to Bu = u (resp. BL ni ;;-, where n = (nl' ... , nn) is a con-

i = 1 uX i

tinuous extension to Q of the field of the unit normal vectors to r, exterior to Q)21s. We take as given the boundary condition in the form of a differential operator

B = /30 + f fJi{- with coefficients in "6'0(.Q): it is obvious from the definition j = 1 (,xj

that the condition depends only on the values of the /3j in the neighbourhood of the boundary r, or more precisely of the "trace of the operator B onI" defined as an equivalence class modulo

{ Po + L f3 j ~a; /3j E "6'0 (Q), li j = 0 in the neighbourhood of r }216 ox j

There exist many ways of defining a notion of a weak solution of the boundary value problem (8.31), the weakening being able to bear moreover as much on the equation Pu = f (resp. Lu = f) on Q as on the condition on the boundary. We shall be content here to discuss the idea of a weak solution for a mixed boundary value in the case of an equation in divergence form:

Lu =f on Q

(8.32) u == (P on ro cu

='" r\ro Gn L

on

214 That is to say. for all [ > 0, there exists r> ° such that

I(BII,O,,(z)I":£ forall ~E.c;(QnB(z,r)).(",o,f(dX = I,

where

21, Such an extension always exists if we suppose Q to be regular (with boundary of class Cf,' 1) and we

have seen (see §1.3) that for U E (6; (ih lim grad u(x). nix) was independent of the choice of this

extension. 216 See §7.4 for another example of trace as an equivalence class.


where L is a differential operator of the second order in divergence form on the open set Q, lois a closed part of the boundary 1 of Q.

Supposing Q to be bounded, the (open) part of the boundary 1',10 regular, the coefficients of L as well as that of f continuous on Q and <p EO ('£°(1'0)'

t}I EO' ({,,0(1\ fo) we can define, in a natural way, what we call a classical solution of (8.32): with the notation of Definition I, this is a function

liE ((;f(Q) n ((;O(Q u 1'0) n ({,,;/(f2\ro)

satisfying the equalities of (8.32) at every point of Q, 1'0- rro respectively. Now. from Proposition L for such a classical solution II, we have

f L(u, r)dx = f It1 dx + r . t}ll'd}' Q Q.;I I"

for every function l' EO ((;I(Q) n (tOU'l) with SUpPI: n 1'0 0, .Ii' and L(u, z;) ELI (Q)21 7. More precisely, given

Ii EO (ti(Q) n ((j0(Q u 1'0) n (t~I(Q 1 0 ) with u = <p on 1 0 ,

the above equalities characterize the property "/I is a classical solution of (8.32)"

subject to the density in (to(r' ["0) of the set of traces on ["\.1'0 of the functions I EO (tl(Q) n ((jO(Q) with SUpPl' n 1'0 = 0,fi1 and L(u, v) EO LI(Q).

To place ourselves in the generality of the Definition 3, given Q an arbitrary open set in IR", 10 a closed part of its boundary, 1 ~ p < X., we denote by VP(Q, 1 0 ) the closure in W)lo"!,(Q)2IR of the set of the restrictions to Q of the v E 9(lRn \ 1'0)'

Definition 6. Let Q be an arbitrary open set in IR", 1'0 a closed part of its boundary and

a differential operator of order 2 in divergence form with coeflicients in L)~c(r;2\ f o): llnally, let 1 ~ p < rf~, (P EO' W)~~1(r;2\ro),fa distribution on Q of the form/~ + II with!;) a Radon measure on Q, bounded on Q n K for all compact K of Q, ro and

.I; EO' 6'(Q u ro), t}I EO (J'(IR"\ro) with support in r roo By a weak solution relative to Wic;!,(Q) o{ the mixed problem (8.32) we mean every function U E <p + VP(Q, ro)

21 C We apply Proposition I to a regular open set Q, conIained in Q such that f2, r I" ~ 0 and suppr c Q,.

'IH We recall that by an abuse of notation, for an open set Q and If) a dosed part of its boundary

L["I!l £'0) ~ : II E L,:,c\Q) ; lIX. E UiQ) for every compact sct K of £'2. ro: '

Lf:,Jl2 lli)j·

Ohviously W,;,/(f2) corresponds to 1'" = 0; if a IS bounded. /,[,,,f2) l."(Q), Wi,,/IQ) c. w'·rli.!) and VI'(a, n = W,\"(Q)<


satisfying

(S.33) f L(u, Odx = f fo~dx + <f~, 0 + <1/1, O. for all ~ E ~(iR"\ro) Q Q

where L(. , .) is the bilinear differential operator defined by (S.12). We note that from the hypotheses. (8.33) has a meaning for all, E .p(iR"\r oj. As we have done it in the preceding definitions, the Dirichlet condition on r 0 is given by a function cP E WI~~l(Q\ro); it is clear that the condition U E cP + vP(Q, ro) depends only on the trace of cp on roo defined here as the equivalence class modulo vP(Q, r oj. We notice also that in the case when Q is unbounded, this definition of a weak solution and, moreover, that of a classical or a quasi-classical solution (Definition 5) do not involve a condition at infinity: that condition can be introduced in different ways as we have seen for Laplace's equation or Poisson's equation (see §4.3). The notion of a weak solution like that we have defined above makes evident the duality between the problem (S.32) and the dual mixed boundary value problem

L *u* = f* on Q

u* -.. cp* on I' (S.34) 0

i'u* -- = 1/1* on I'\ro· all l .'

Restricting our attention to the case in which Q is bounded but not necessarily regular, we can state in a precise fashion:

Proposition 3. Let Q be a bounded set of u~n, ro a closed part of its boundary, 1 < p, p* < ox.; with p - 1 + p* - 1 = 1

a linear differential operator of order 2 in divergencef()rm with coefficients ill LX (Q)

LInd f;)o ... • f~ E U(Q) (resp. ft, ... .I,,* E U*(Q))

with ". ?{ ( n i'f*) .I -.".' resp·I -;;-.~. EJ'(Qul'o), J ~ I ( Xi , ~ 1 LXi

qo, ...• fin E U(Q) (resp. q6,' ..• Y~ E U*(Q))

with _ n (lYi ( * _ II ag1' \ .

Yo - . I"'1- resp'. {fa - . I -"1- ) 111 '~l(Xi \ '~l(Xi/

Q'(Q) .

u (resp. u*) a weak solution relatil'e to Wl.P(Q) (resp. rtTl.P*(Q)) of(8.32)

II rr ( n i1j'*) (resp. (8.34)) with f = fa - . I;:;'..' resp. f* = f6 - . I ~- ill , ~ I (Xi , 1 (Xi

2'(Q)and


in .Q'(iR"\fo), we hare

(8.35)

f (L(!p,u*)-L(u,(p*))dx= r {(f~+go)(u* - !p*)-(f(j+g(j)(u - !pJ}dx Q ~Q

Proof By definition of a weak solution of (S.32)

(8.36) J' L(u, S")dx = f (.f~ + 90)(dx + I f u; + ?};) ~S"-dx Q Q i~} Q OXi

for all S" E U(iR"\J'o). Since u* E !p* + VP*(Q, ro) the relation (8.36) will again be true for ( = 1/* - !p*: in effect using the definition of vr'(Q, ro) there exists a sequence ((II)' (II Ec/(iR"\ ["0)

(II --+ u* -!p* In W 1. P'(Q) .

From the hypotheses on the data, we can therefore pass to the limit in (8.36), Similarly, we shall have

(8.37) J' L *(u*, (*)dx = II" Uti + 96)(* dx +I r (f/ + 9n ~:dx Ii .1ft 1~1 JQ eXi

for (* = u - !po Using L *(u*, (*) = L((*, u*), and subtracting the equalities (S.36) and (S.37), we obtain (8.35). 0

We note that (S.35) is only the expression of Green's formula (S.15) for weak solutions of the dual mixed problems (8.32), (S.34). FinaJly, in this sub-section, we show how operators with discontinuous coefficients can be involved, and how to interpret the notions of weak alld stroll?} solutions. We take a domain Q separated into two parts by a regular hypersurface SeQ (see Fig. 25); we denote by Q}, Q2 the two domains thus determined. Given lI}EL/"c(Q} uS), U2EL/oc(Q2US) we can define u = Xli,U[ + Zo,U 2 ELloc (Q) that

5

Fig. 25


is to say U is defined a.e. on Q by

Even if U I' U z are very regular on Q b Q z right up to the interface S, the regularity of u on Q depends on the coincidence of U I , U z and of their derivatives on S. In a precise fashion, supposing within the classical framework that Ui E c&' I (QJ n ~O(Qi u S) with grad Ui E Ll~c(Qi U S) (i = 1, 2) we have for ( E £:0(Q)

t U grad ( dx = f ulgrad(dx + f uzgrad(dx Q, Q,

= - f (graduJdx- f (gradu 2 dx+ Is ((u1n1+uZnz)dy Q, Q ,

where nl , nz are the unit normal vectors to S exterior to Ql' Qz respectively. In other words, the gradient of u in the sense of distributions on Q, is the sum of

XQ, grad u 1 + /(Q, grad U z E LI~c(Q)

and of the measure carried by S, -(uln t + u1nz)dy; the hypothesis u; E c&' t (Q;) n ~O(Qi u S) is in fact superfluous and it is sufficient to suppose U; E Wl~~ 1 (Q; uS), which allows us to define the trace of Ui on S; we have thus shown that given Ui E Wl~~l(Qi U S) (i = 1,2), the function u = Xl2,U j + Xl2,Uz

belongs to WI~~l(Q) iffu t = U2 on S, and then gradu = Xa,gradu 1 + XQ,gradu z . Let us consider now, on each open set, Ql' Qz the equations

(8.38) {PIU l (resp. Llud = j~

Pzuz (resp. Lzuz ) = f2

on Q I

on Q2

where PI' P 2 (resp. L l, L z) are the differential operators of order 2 (resp. III

di vergence form) on Q I, Q Z •

Let us denote by P (resp. L) the differential operator on Q,

XQ,P I +XQ,PZ

(resp. - div (XQ, VL , + XQ, VL ,) + XQ, (L I + div Ill.,) + XQ,(L z + div ilL,))

whose coefficients a;j' bi , c (resp. a;j' c;, db d) are defined a.e. on Q by the values of the corresponding coefficients of PI' P z (resp. L I , L z ) on Q l' Qz respectively. Even if the coefficients of PuP 2 (resp. L t , L 2 ) are very regular (or even constants) on Q l' Q z, the coefficients of P (resp. L) are, in general discontinuous on Q and we can consider only the strong (resp. weak) solutions of the equation

(8.39) Pu (resp. Lu) = f on Q,

wheref = XQJl + XQ,fz·

It is obvious that if U is a strong (resp. weak) solution of (8.39), its restrictions U I , U z to QI, Q 2 are strong (resp. weak) solutions of (8.38) and u = XQ, U I + XQ,u 2 •


III rhe case oj' a strong solution, LI E Wl~~l (Q); hence Lli E Wl~/ (Q i u S) and from what we have seen above

(S.40) (k = J, ... , n) on S

In fact if LlI U 2 on S, the tangential derivatives of U I' U z coincide on S, with the result that (8.40) reduces to

(S.411 <'U I

+ (~U2

0 = on anI ?n2

S

Conversely, if Lli E WI~~I (Qi uS), i = 1,2 are strong solutions of (8.38) and satisfy the conditions (8.41), it is clear that LI = Xo, U I + Xo, Ll2 is a strong solution of (8.39). III the case ota weak solution, U E Wl;)~l(Q); hence U i E WI~~l(Qi U S) and U I = U 2

on S. We must have

J" L(u, ~)dx = r Il(dx + f .f~::'dx for all ~ E 0'(Q). Q olD, D2

Now

f L(u,Odx Q

and

('U i where is the conormal derivative taken in the weak sense of DefInition 6.

i"iIl L ,

Hence U is a weak solution of (8.39) iff its restrictions Ui to Q i are weak solutions of (8.38) in Wl~~ 1 (Q uS) and satisfy the conditions

(8.42) (lUI

+ rU 2

0 S 111 = 11 2 , - on ('n tllL2 L,

We sum up this discussion in

Proposition 4. With the Ilotation and hypothesis of Definition 4 (resp. Definition 3), we consider S a regular hypersurface separating Q into two open sets Qt, Q2' and denote by PI' P 2 (resp. L I , L 2 ) linear differential operators of order 2 (resp. in divergence j<mn) on Q I' Q2 whose coefficients are the restrictions to Q t, Q2 of the coefficients ofP (resp. L). Afimction u is a strong (resp. weak) solution 01(8.39) iffits restrictions 11 1 ,11 2 to £1 1 , Q2 are strono (resp. weak) solutions o{(S.3S) satisfying the conditions (S.41) (resp. (8.42)).

Conditions (8.41), (8.42) are called transmission conditions: the introduction of an operator with discontinuous coefficients enables us therefore to interpret a problem


made up of a problem of a partial differential equation on each open set Ql' Qz and of transmission across their common boundary S as a differential problem on Q. We note the difference between the transmission conditions (8.41) corresponding to the strong solutions on Q and the transmission conditions (8.42) corresponding to the weak solutions.

3. General Results on the Regularity of Elliptic Problems of the Second Order

First of all we give some results on local regularity and regularity up to the boundary in the case of operators with regular coefficients.

az a Proposition 5. Let Q be an open set of [R", P = - 2: au ;;--~ + ~ bi -~ - + c,

I,) vXiUXj , eX i

a differential operator of order 2 with <{/x_ (resp. analytic) coefficients on Q and f E <{/ ""(m (resp. analytic on Q). (1) Let us suppose P is elliptic on Q, that is to say

I:aij(X)(i(j > 0 for all x E Q, ( E [R" with (# O. i..i

Then every solution in the sense of distributions of'the equation

Pu = f on Q

is cg x. (resp. analytic on Q )219.

(2) Let lobe a regular open part of class cg en (resp. analytic) of the boundary of Q; we suppose that the coefficients of P andf are <{/ oc (resp. analytic on 10 ) on Q u Fo and P strictly elliptic on Q. On the other hand, let g E (1&":0(10 ) (resp. analytic on 1 0 )

n a and B = fJo + j~l fJj oX j a differential operator of' order less than or equal to 1 with

coefficients belonging to <(/3:J(Q u 1 0 ) (resp. analytic on Q u lol. We suppose

or

fJ 1 = ... = fJn == 0 on 10 (Dirichlet problem)

n

I fJjn j # 0 on 10 (Neumann problem)220. j=l

Then every classical solution of the problem

{ PU = f Bu = 9

on

on

is (6' (resp. analytic) on Q u r o.

219 And hence a classical solution of Pu = f on Q.

220 n} is the j-th component of the unit normal to ro exterior to Q. This condition corresponds to a Neumann problem with possibly an oblique derivative (see Proposition 2).

5X6 Chapter I!. The Laplace Operator

This proposition has been proved to within the details stated in §6 except what concerns the Neumann problem where by the given proof we are confined to the

case of an operator B = I 0'. -~ - where ". = I a .. n221 j 11 ? x j I J j lJ l •

We note that further we have shown results of regularity wm + 2, P or (f;m + 2.7 when the data have the corresponding regularity. Finally, we note that the point (I) is proved for an elliptic operator in Chap. V. We refer to AgmonDouglisNirenberg [1 J for a complete proof of the Proposition in the wm , P framework and to GilbargTrudinger [I J for the (f,m" context. In all of these results, the continuity of the coefficients aij is essential since we use principally that in the neighbourhood of Xo, the operator I (lij(x) 1'2 /?xJ1X j "is not very different" from the operator with constant coeftkients I (lijCxo)(i2/E'x;EiXj which is, to within a change of coordinates, the Laplace operator. In the case of operators with discontinuous coefikients, we cannot hope to have the regularity (f,2 (resp. Cf, 1) of strong (resp. weak) solution: to convince ourselves of this, it is enough to consider in 1 dimension. the problem:

(8.43) a d21~ (resp. d ((l dll)) = f on J - 1, I [ dx- , dx dx /

where

{ a + a(x) =

a_ on JO,I[

on J-I,O[,

at, a being distinct negative constants; iff is continuous on J - 1, + 1 [ (resp. f E Ll~c(] -I, + 1 C)) a strong (resp. weak) solution of (8.43) cannot be of class (fo2

1 ' d211 f( dll 1 f' . d ). d' , (resp. Cf,' ) smce -d 2 = '-- resp. 'd~- = - j (x) X IS Iscontmuous at O.

x a x a The De Georgi--Nash theorem shows that we have, however, even in the case of operators with discontinuous coefficients, a Holder continuity of the weak solutions; in precise form, we state222 :

Proposition 6. Lct Q be all open scI ill [Rn. L = - I --;,,(~ (I .. aij,? + Ci \ I (Xi ! ( .Xj )

t + I d,. -~ + d a linear differential operaTor or order 2 with coefficients ill Lf (Q) ,. (,Xi

and strictly clliptic, that is to say satisFying, with a constant Co > 0

Iaij(x)~i~j?:col~12 a.e. xEQ, jiJrall sEIR:"; i, j

jurthermore, let p > 11 andj;), . , , ,f~ E U(Q).

221 Or. more generally an operator B whose tangential terms TJ given by (K26) arc "small". 212 See Trudinger [IJ for the results under more general hypotheses.

98. Elliptic Equations of the Second Order 587

(1) Every weak solution of

(8.44) Lu = r _" of; on Q .JO 1.... '" ox;

satisfies a Holder condition locally on Q.

(2) Suppose Q is bounded with Lipschitz

" ogj go, ... , gn E U(Q) with go = L- oXj

boundary and let q; E Wl,P(Q) (resp.

on !Z"(Q) and suppose also that

L iU; E 6''' (Q)). ex;

Then every weak solution relative to W1.2(Q) of the Dirichlet (resp. Neumann) prohlem

J ef;

Q Lu = fo - L;J- on eX j

(8.45) u = q; on r

I (resp. ~ju = go - L~~j on r) l onL eX j

satisfies a Holder condition on Q.

In the case of the Poisson equation, - Llu = f~ - L afd ax; on Q, we know that forf~), ... ,f~ E Lfoc(Q) with p > n. the solution satisfies locally a Holder condition of order (lIn) - (lIp) on Q223; this is still true for an elliptic operator with regular coefficients. In the case of De Georgi-Nash theorem, we affirm only that a solution of Lu = f~ - I/fdaxj on Q satisfies a Holder condition locally on Q: its order on a compact set K of Q depends on K, norms in L CX(Q), of the coefficients of (l/co)L and on p > n. We refer the reader to Trudinger [I], Stampacchia [IJ or Ladyzenskaya-Ural'ceva [I] for a complete proof of the De Giorgi-Nash theorem as well as different developments on elliptic problems of the second order with discontinuous coefficients. We shall give here the essential elements of a proof of this theorem so as to show the constituents used which are very different from those used in the case of continuous coefficients: an essential technique is the truncation method introduced by De Giorgi. In this section we shall use it to obtain global estimates of a weak solution relative to W1.2(Q) of the mixed problem (8.45). We shall show in Sect. 5 below how to localize this method to obtain a Harnack inequality (see Propositions 15 and 16) from which we can deduce a proof of Proposition 6. First of all we prove:

Lemma 2. LeI Q he an open sel in [R", L a differential operator oj" order 2 in divergence form with coefficients in L 'X'(Q) and strictly elliptic, ro a part

f· h b d' . f' " (li; () Ie oun ary oj Q p -> 2, jo,"" . n E U(Q) with L--~- E 6"(Q u raj, cXi

223 In effect 11 E WJ~/(Q) (see ~3) and WJ~/(Q) c (f, 10,- :(QJ (see Brezis [IJ).

5XX Chapter II,The Laplace Operator

i:q, Yo"", gn E U(Q) with Yo = L -:;'-'- inQ'(Q), cP E Wl.P(Q) and U a weak solutio/l

(IXi

'f reiatil'e to WI~~2(Q) of the mixed problem (8.32) with f = t;) - L ~.~ ill '/"(Q) eX i

alld ~I = go .... L ~fli in U' (IR" \ roJ. For every fUfIction e, increasing 011 IR with oX i

rO(r) ~ 0 such that, puttillY W = 11 - (P, we have

W l = wIO(w)l l2 E L2(Q) and \\'2 = IIJ(wW 2 E UriP 2(Q),

then the jimctioll v = rw 1f!(rW2 dr belonys to V2(Q, ro) and

~o

where C depends only 011 the L f -norms of the coefficients of(l/col L where ('0 is the COllstant of strict ellipticity of L.

Proof: By definition IV E V2(Q, ro) and

(8.47) (L(IV, s)dx == j' ('j~)S + L.f; ~'s )dX for all s ECi'(IR"\ro ) JQ Q (Xi

where

, (I, i'(P ) f = f + y - L (/ --.' + c, (P i = I" . , , 11 . , I . I ~ j , IJ (l X j I

In the limit, (8.47) is again true for all ~ E V2(Q, ro). Supposing first (} to be

continuous and bounded on R we can therefore apply (8.4 7) with ( = J: I O(r) I dr;

using the growth of 0, we have isl ~ wO(w) and the application of (8.47) gives

Using the ellipticity and majorizing the second member by Young's inequality we obtain

cof' (I'W)2 , f 10(w)1 "., -- 10(w)IL ,: dx'S . ---- L fi- dx 2 Q (Xi Q (() i " ()


from which (8.46) follows noting that 18(W)I(::Y = (:~J2 and v2 ~ w218(w)1 = wi.

The explanation of the constant C offers nothing of interest, but we see clearly that it depends only on the LX' norms of the coefficients of (l/colL. To prove the lemma in the case of a general function 8, it is sufficient to note that there exists a sequence (8k ) of increasing bounded continuous functions such that

18k (r)! ~ 10(1') i for all k and all I' E IR

Ok (I') -> 0(1') when k -> 00, a.e. I' E IR .

Applying (8.46) for Ok' we obtain a majorization of

Vk = f: IOk(rlI 1/ 2 dr in /tTl.2(Q)

by the term on the right of (8.46) (corresponding to 0); since Vk -> v a.e. on Q, we deduce that v E V2(Q, 1'0) with the estimate (8.46). 0

When Q is bounded with a Lipschitz boundary in IR n with n ~ 3 (resp. n have the Sobolev inclusion W1.2(Q) C L 2m/(m2) with m '? n (resp. m > allows us to write (8.46):

(8.48)

2), we 2) which

Ilvllumm 2 ~ C{l + 11<pllwl.p -I- ~~LII[; + gillu}(IIW]IIL2 + Ilw2 11!'p(Pli)

where we have included the constant of the Sobolev inclusion in C which thus now depends on m and Q224.

Such an inequality (8.48) allows us to obtain estimates on u.

Lemma 3. Let m, p ~ 2 and WE L2(Q)22S. We suppose that there exists a constant Co such that for every increasing function e on IR with re(r) ~ 0

w] = wIO(w)1 1/2 E e(Q) and W z = IO(w)1 1/2 E e p /(P-2)(Q) ,

the function v IW IO(r)II/2dr E L 2m!(m-2)(Q) and satisfies

(8.49) II V 11,2mml ~ CaUl wl ll zl + II w2I1t2PIP-21) .

If p < m (resp. p > m), then WE um/(m-p)(Q) (resp. WE L OC{Q» .

We deduce:

224 We have l' EO V2(Q. To); in the case in which To = T. V2(Q. To) = WJ·2(Q) and the constant C 1 I

depends only on m and IQI" m where IQI is the measure of Q. In particular in this case, the hypothesis that the boundary satisfies a Lipschitz condition is not necessary, also if m = f1 ,"" 3, the assumption that Q is bounded is superfluous. 225 This is a lemma from the theory of integration: Q can denote an arbitrary measure space.


Proposition 7. With the data of Lemma 2, let us suppose Q is bounded with Lipschitz boundary. If p < n (resp. p > n), then u E If"/(n - p)(Q) (resp. u E L CX'(Q)) and

(8.50) II U ill.""'" 'p' (resp.llull r ,) ~ C{II<Pllw1P +L f Iii; + Yillu} (Oi~O

where C depends only on [he norms in L' of the coefficients ol(l/colL, on p and on the open set Q. This Proposition is clearly a consequence of Lemmas 2 and 3, since, Q being bounded with a boundary which satisfies a Lipschitz condition, if p < n (resp. p > /1) we have l-tl1.P(Q) c un!(n-p)(Q) (resp. W1.P(Q) c L'(Q» with the result that U E u"/{n-p)(Q) (resp. L'(Q» iff it is thus of the form w = U - <po The estimate (8.50) is obtained from (8.48) and from Lemma 3 by a homogeneity argument.

Prool ol Lemma 3. Given k ~ 0, we denote by

Q k = {x E Q; Iw(x)j > k} and If>(k) = II(lwl - k)" IlL! .

The function If> is convex, decreasing on \R+ with derivative q/(k) = -IQkl, the measure of the set Qk' Given x ~ 0, we apply (8.49) with the function 226

O(r) = (x + ])2 (signrHlrl _ k(2"

We have then

and t' = (sign wHlwl - k)+'~) ;

so if WE U,(Q) where Pa = max (2(X + 1), 2XP ) then WE Lq,(Q) p - 2 '

In where q, = 2(a + 1)--- - and for all k ~ 0,

m-2

!I(lwi - k)+ II~q; I ~ Co(x + I)Cliw(lwl - k)+'IIL' + II(iwl- k)+'ll r 2PIP 21)'

By Holder's inequality

liw(lwl - k)+'li/2 ~ 1!(lwl _. k)+'''!i{2 + k!i(lll') - k)"llr2

~ 1!(lw! and subject to p. ~ (}.,

1!(lwl - k)+'11 1 2P1P 21 ~ II(iwl- k)+ il~.q,iQkl(1 +': 'F'I} + I

HenceonconditionthatwEU"(Q),p, ~ q,andCo(a + 1)IQkll!m ~ 1, we shall have

22h We denote here (irj -- kif 2, for ((Irl _ kif )2,.

§8, Elliptic Equations of the Second Order 591

We note that k IQk 11/2 ~ II w II L' with the result that the condition Co(C( + 1)I Qkl l/m ~ t is in particular satisfied if

k ~ ka = (2Co(ot + 1»m/2I1wIIL2.

When p < m, [ m(p - 2)J {C( ~ 0, p, ~ q,} = 0, 2(m _ p) and

m(p - 2) for C( =------

2(m - p)'

p, = qa = ~; therefore, by continuity of the norm WE Lmp/(m - PI(Q) and m-p

(8.51) gives an estimate of II w II Lmp/(m p) noting that for q ~ 2

2

IIwllL' ~ 2kl-qllwll~2q + 1I(lwl - k)+ ilL';

The reader will verify that if we use k = kn we have

( m - 2)1 + m{t - ;) m - 2

IIwllL'~3 CoP--- IIwlle+ CoP--· m-p m-p

When p > m, to prove that WE LCO(Q) we shaH use (8.51) with C( = O. We then have for all k ~ ko = (2Cor/2 11 w II L2

1 1

1I(lwl - k)+ Ilem(ml ~ C1k1Qk1 Z - p

with C1 = 2Co (~o + IQkoI 1 / P ) .

Then by Holder's inequality we have

1 I

cp(k) = 1I(lwl - k)+ 11[1 ~ C\klQkl l +m-p;

(' + 'ml _pl.,)-1 which by putting f3 = 1 - 1- - > 0, and

IQki = - cp'(k) we can rewrite as

ddk(C}-licp(k)1i + kfl) ~ 0 for k ~ ko,

from which we deduce that cp(k) = 0 for

kfJ ~ C} - Ii cp(ko)P + kg namely that again w E L C0 (Q) and

IlwllL' ~ (kg + C} -flcp(ko)fl)l/fl.

recalling that

o We have detailed the proof of the estimates of Proposition 7 to convince the reader of the power of the truncation method: using this method we shall obtain the De Georgi-Nash theorem; as we have already stated we shall take up this local technique in Sect. 3 to prove Harnack's inequality. The above proof also makes obvious the estimates of Proposition 7; we shall state without proof the corresponding local result:


Proposition 8. Let Q be an open set offR", L a linear d~fJerential operator of order 2 in divergence form with coefficients in L'l) (Q) and strictly elliptic, p;::, 2, fo, ... ,f"EU(Q)anduaweaksoiutionin WI~~2(Q)ofLu =fonQ.lfp < n(resp. p > n), then u E Lf:!(Il- p) (resp. u E LI~'c(Q)) and for every compact set K in Q

n

Ilull u" PIKI(resp. iiulll.'(ld ~ C L 11.f;llulilj i = 0

\vhere C depends only on the L'-norms of the coefficients of L, on the constant of uniform ellipticity and on the distance from K to the boundary of Q.

As we have shown in the preceding section, operators with discontinuous coefficients allow us, in particular, to express problems on two different media Q j' Q 2

with transmission conditions on the common interface S to Q j and Q2' As we have seen, the global regularity of a solution in the passage to S is limited; however, supposing the operators on Q j and Q 2 to have coefficients regular right up to the interface S, we can show that the solution will likewise be regular on Q j and Q 2 up to the interface S. We state the result in the case of an operator in divergence form:

Proposition 9. Let Q be an open set ()f~n separated into two open sets Q j and Q 2 by a /zypcrswface S o{'class {{j x, L a linear differential operator of order 2 in divergence form. strictly elliptic Oil Q alldfafullctioll on Q. We suppose the restrictiolls off and of the coefficients of L to Q I and Q 2 are '6 ex: (Q IUS) and '6 x (Q 2 U S) respectively. Then eloery weak solution ill WI~/(Q) of

Lu = I 011 Q

has its restrictions u l , U z to QI' Q 2 ill '6 X (Q 1 uS), '(,5 ""(Q 2 u S) respectively and there/eire in particular classical solutions of the transmission problem

Llu l =f 011 Q I

L z Uz f 011 Q2

III Uz OIl S

(lU 1

+ ?U 2

0 S = Oil ell i'n L, I.,

where L I' L2 are the restrictions of'L to Q I' Q 2 respectil·ely.

Proof The problem being local, we can always suppose that, after a change of variables, S = {(x', 0); x' E ~n I} n Q. From Proposition 6 we know already that Ii is continuous on Q with the result that U I (resp. uz ) is a quasi-classical solution of

{ PL.' = f l' = U

on

on

where P is the differential operator of order 2 on Q\S strictly elliptic with ~>coefficients defined by (8.8) in terms of those of L. From the proof of Theorem 1 of


§6, if the derivatives with respect to x', ~~~ belong to Lroc(Q) for all IIX! :( m, then ox

the restrictions UI (resp. U2) of u to QI (resp. Q2) are in Hl~c(QI u S) (resp. Hl~c(Q2 uS)). Since n Ht;;c(Q i u S) = cg"'(Q i U S) (see Chap. IV), it is

M",O

aau enough to prove that - E LFoc(Q) for all IX.

ax'a We shall make use of the method of translations. Given hEIR, h oF 0 k = 1, ... , n - 1, for a function v defined on Q we shall use the notation

fh;. 1 ~;)~(x), ... , xn) = -(vex), ... ,Xk + h, ... ,xnl - v(x), . .. ,Xn» CXk h

defined on Qt = {(Xl' ... , xn) E Q; (X)' ... , X k + h, ... , x.) E Q} . It is clear

that if v E Ll(lRn), then -(x)dx = 0; if v E W1,2(Q), then ~ E W 1,2(Qt); f ahv . i)hv

i)~ o~ or; i)hv

finally for 1 0 and , i)xk "i)Xk . k ,

that then

i)h L . , . ... ah where - IS the operator whose coefhclents are the hillte differences ~~ - of the

~ ~ coefficients of L. Since u is a weak solution of Lu =.f on Q, we have

and hence r L(chu, ,)dX = r {~-hf _ ~~(u, O}dX. JQ i)xk JQ OXk CXk

h d ahu . k . f In ot er wor s Uh =~- IS a wea solutlOn 0

cXk

with

and

We have


where C depends only on the norms in L J:. (Q \5) of the derivatives Dfl i)x k of/and of the coefficients of L227. Given Q o relatively compact in Q, we have, from the strict ellipticity

II grad I'h II L 2 (Q,,) ,,:; C( I + 1111li JjI2(Q)l .

<iU In the limit, when h -> 0, we shall have grad:i E L2(QO}' Also I'

(X k

weak solution of

" 1.9; Lr = go + L ;; ~ eX i

where Ui are the limits of the y~. By induction, we obtain the result. o

4. Results on Existence and Uniqueness of Solutions of Strictly Elliptic Boundary Value Problems of the Second Order on a Bounded Open Set

The quickest method of stating general results on existence and uniqueness is certainly the variational method joined with the Fredholm theory: this method will be developed within an abstract framework in Chap. VII. We content ourselves here with applying it to the study of a mixed boundary value problem for a partial differential equation in divergence form.

Proposition to. Let Q be a hounded open set with a Lipschitz boundarJ' of [R;". L a dUferentiai operator of order 2 in diveryence fcmn with coefficients in L'(Q) and .\Iric( iy elliptic, loa closed part of the houndary r of Q. (I) The set N(L, 1'0) of the u, weak solutions relative to wL2(Q) of the mixed houl1darv value problem

Lu = ° on Q

u .- 0 on Q

I'll = 0 on Q

?n L

is a suh-space ofJil1ite dimension of V2(Q, 1'0)' 111 addition

dimN(L, 1 0 ) = dimN(L*, ro)

where L * is the operator in divergence form, adjoint of L.

(2) Git'en j~, ... ,.t;l E L 2(Q) with IC;~ E I:'(Q u r o), Yo' ... , Yn E L 2(Q) with ex;

tog Yo = I -~~~ in (j' (Q) and ((i E W L2 (Q), the boundary value problem

eX i

22' We can always suppose that the derivatives on [2\ S ofr and the coefficients of L are bounded.


oj; Lu = J~- I-I. on Q eX j

(8.52) u = rp on 1'0

~~u = go - I ~gj on r\ro cnr. cXj

admits a weak solution relative to wl, 2(Q) iff

(8.53) fa L(rp, u*)dx = fa {(fo + go)u* +. ~(j; + g;)~:~}dX for all u* E N(L*, 1'0) .

Proof The map (u, u*) E Vl(Q, ro) x V2(Q, ro) -> r L(u, u*)dx is a continu-JQ

ous bilinear form on the Banach space V2(Q, ro). We have

1 r [ au eu eu L(u u)dx = ') a.·---. - + " (c + d.)u -..... ,~ i.....d l) ""\:1 L! l::J

Q "Q .. i.) CXjCXj j eXj + du 2 JdX

1 ( ( ~)2 ) ~ [ 1 / ) J Co ell.) 2' 2 2 ~ -- I -. - + u" dx + J d·· -2 ~- { Co + ~ (c j + d;) u dx; 2 Q I ex; Q Co \ '.

hence considering io = II [2~0 (d + ~ (e j + d;)2 )- d J + t" we see that the

bilinear form (u, u*)·-+ r (L(u, u*) + ;'ouu*)dx is coercive on V2(Q, ro)2. JQ

From the Lax-Milgram theorem (see Chaps. VI and VII) for every continuous linear form Ion V2(Q, 1'0), there exists a unique 11 E V2(Q, ro)

r (L(u, n + )'oundx = I(n for all (E V2(Q, ro) . JQ

In particular, givenfE L2(Q), there exists a unique u E V2(Q, ro) such that

{ (L(u,O + ;'ouOdx = {f(dx for all (E V2(Q, ro). JQ JQ

We denote by Go the mapf --> 11; it is linear, continuous from L2(Q) into V2(Q, ro) and therefore a compact operator on L 2(Q)228.

By definition we have 11 E N(L, 1'0) iffu = ;.oG01l. In other words N(L, 1'0) is the kernel of the operator I - loGo on L 2 (Q). From the RieszSchauder theorem (see §4), we know that N(L, ['0) is of finite dimension, that ImU - ioGo) is closed with finite codimension and

228 Taking account of the hypothesis that Q is bounded with a Lipschitz boundary. we see that the canonical injection of W1.2(Q) into L2(Q) is compact (see Chap. IV).


Now Im(l loGo} is the orthogonal in U(Q) of the kernell - ;'oG(j where G(j is the adjoint operator of Go in U(Q}: as we see immediately G(j is the operator associated with L * as Go is that associated with L. Hence Im(l 1'0 Go) is the orthogonal of N(L*, ro) in elm. We deduce, in particular, that

dimN(L*, ro) = codimlm(1 - )'oGo)

which completes the proof of point (I). For the point (2), we put

j~ = .t~ ( ?~ d~ ) + Yo - .. Ld; .., + LXi

1; = j; ( t~ Ci~ ). + ?Ii - I au -;;-- + ex) /

By definition u is a weak solution relative to W1.2 (m of (8.52) iff II WE V2(Q, ro) and

~ + IV with

fa L(w, s)dx = r ((f~)s + It; ~~s )dX for ali ; E Ci(lR"\rol . J .;Q , I (X;

There exists a (unique) l' E V 2 (Q, ro) such that

f (L(l" n + I,()l's)dx = r (]~; + L];~'; ,)dX for all s E V2(Q, ro) . JQ Jo i LX;

Hence u is a weak solution relative to W1.2(Q) of (8.52) iff U = ~ + II' with It' E V2(Q, r 0) and \\' = l' + ;'0 Go It'. There exists therefore a weak solution relative to W1.2(Q) of (8.52) iff v E Im(I - ;.oGo}, Since Im(l- l.oG o) is the orthogonal of N(L*, r o), we have 1; E Im(l - l.oGo) iff

I'

JQ l'u*dx = 0 for all u* E N(L*, ro)·

But, for u* E N(L*, 1'0)' f L(1" lI*)dx = 0; if ;'0 f:. 0, r E Im(l -- l.oG o) if and JQ only if, for allu* E N(L*, ro)

I' Ja (L(I', 1I*) + l.olll1*)dx = ° which is clearly the condition (8.53); if ).0 = 0, the result is triviaL o Combined with the results on regularity of the preceding section, this result has a vast field of applications although it is limited as we shall see below. It shows that in a very general context the solution of the boundary value problem (8,52) leads to the study of the dual homogeneous problem. Let us give some examples.

Example 1. C as!! of dimension 1. Let us give a complete discussion of an elliptic boundary value problem on a bounded interval of R We can always reduce the


interval to ]0, IC 29 . Let a, b, c, d,f, qJ, IjJ be functions defined on [ = ]0,1[, ro a part of I' = {O, I}; we consider the boundary value problem:

- i_ ( a du + bU) + c du + du = f on [ dx dx dx

(8.54) U = qJ on ro du

a + bu = IjJ on I' \ I' 0 ; dx

The homogeneous adjoint problem can be written:

- ~ ( a dv + cv) + b ddxV + dv = 0 on [ dx dx

(8.55) v = 0 on ro dv

a - + cu = 0 on I' \ I' 0 . dx

Supposing the coefficients to be sufficiently regular and a > ° on r, we change the dependent variable from v to w, where

rx r(s) ds -0

e u(x) = /_~ w(x) with r

ya(x)

b - c

2a

The problem (8.55) can then be written

(8.56)

with

(8.57)

dZw = kw

dx2

w =0

Wx = AW

12 + ad - be k=;:+1L

on I

on r 0

on r\r o

a' - (b + c)

2a

The set N(r 0) of the solutions of (8.56) is either reduced to the null function, or to a

space of dimension 1. In effect considering WI' W2 solutions of ddz~ = kw satisfying x

the initial conditions

WZ(O) = 1 , w~(O) = 0,

we have the cases shown in Table 3.

229 We could suppose that I = I, but here we prefer to retain the length of the interval as a parameter.


Table 3

Condition for N(ro) '" {O}

r \I'd/) = 0

[I)

o i(O)\\;(l) + \V~(/) = A(/)(A(O)W,(l) + 1>'2(1))

The application of Proposition 10 then gives:

Particular solution ofN(fo)

\1:'1

A(O)W, + "'2

case N (ra) = {O}; for all functions!; I.{J, IjJ there exists a unique solution of (8,54); case N(ra) * {O}, Then N(rol = [J;l;wo, Wo being a particular (non null) solution of (8.56), Given f, I.{J, IjJ there exists a solution of (8.5) iff according to the cases

quoted J: uo(.x)f(x)dx is equal to

fa I' a(O) v~ (0) I.{J(O) a(l)v~(l)(p(l)

I'o {O} u(O)v~(O)I.{J(O) + t:a(l)IjJ(l)

1"0 {I} ~ vo(O)IjJ(O) u(l)v~(l)I.{J(I)

To 0 ~ v()(O)IjJ(O) + vo(l)IjJ(/)

j(~(.~) e r' [(sj ds b ~ c

where voCx) = " () .\'0 (x) with T=----.

2a

Moreover, then, the solution of (8.45) is unique to within the addition of a constant times the function

J'" , - [(s)ds

uo(x) = Ja(x)e () wo(x).

The last point comes from the fact that k and i., defined by (8.57), are symmetric in b and c with the result that we shall reduce the homogeneous problem of (8.54) to (8.56). The explanation of the condition of existence and uniqueness, that is to say N(rol = {O}, or of a particular solution Wo in the case N(ro) * {O}, obviously

d2w depends on the knowledge of the particular solutions WI' W 2 of (1;'2 = kw which

we cannot integrate by quadratures in the general case. If k is constant, that is simple:

shwx chww

(J)

§S. Elliptic Equations of the Second Order

Table 4. Condition for N(To) # 0

r o

r

l I I '0\

k = w 2 > 0

N(T) = :0)

;.(1) th wi - w Wo = shwx

k = 0

N(F) = {OJ

W)I-l \t'o = x

I {I) i.(O)thwl + UJ = 0 I + /J.(O) = 0 I'

I w" = sh w(l - x) 11'" = I - x

k -- (1)2 < 0

mn OJ - -----, \Va - sinwx

I

A(I)tgwl = UJ

Wo ---- SIn wx

J.(O)tgw/ -+ (!J - 0

11'0 = sin w(l - x)

• I !

I 0 [I w[i.(O) + wthw/] 'i.(I) - ).(0) = i.(O)iJ/J/! w[i.(O) - wtgwl]

I I 110 = shw(x + <pi Wo = 1 + 1[(0)\ i Wo - sinw(x -+ <pi

IiI (I) I 1 w I I <p = Arg th I <p = ... ,' Arc tg ......... .

599

- }(/)[w + i.(O)thw/] J I - i.(/)[w + A(Oltgwl]

L ' 1') ).(0) i (') )(0) _. __ L_____ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _. __ 1. ______ . ____ . __ . __ . __ ._._._. ___ . ___ . __ ._

k O. WI = X, Wz

k _ w 2 < 0, W Sill WX

I W z = coswx. W

Table 3 expresses itself then in the following Table 4 giving the condition for N (r 0) # {O} and in this case a particular solution W o' Read upside down, this table exhibits the eigenvalues of the operator d2/dx 2 having as domain

W = 0 on r\ro }.

In particular we can make clear the largest eigenvalue k (ro, l) according to the values of r o and A; this value k(ro, ;.) is specially interesting as we know 230 that, the operator d 2/dx 2 being symmetric on D(ro, A), the operator k(ro, I,) - d 2/dx2

is positive on D(r 0, ;.): we deduce that for every function k such that

(8.58)

the homogeneous problem (8.56) admits only the null solution; in effect if W is a solution of (8.56) we shall have

I (k - k(ro, ;,))w2 dx

and hence from (8.58), W = o.

230 See Chap. VI.


Returning to the problem (8.54), we see therefore that if the coefficients a, b, c, d

satisfy

(8.59) di. -- + dx

. 2 ad -- be I. + --2--;:;' kilo, Il) where

a

d --I. = --- Log /a

dx Y

b + c

2a

and kilo, i.) is the laryest eigenvalue of the operator d 2 jdx 2 having as domain

D(lo,ll) = fWE W2.2(]0,1[);w = 0 on lo,dw = i.IV 011 lllO}' L dx

then/or alljunctions(!p, Ij;, the boundary value problem (8.54) admits one and only

olle solution. We conclude this example by explaining k(lo, i.) in Table 4: case 1 = lo (pure Dirichlet problem): k(lo) = - (njl)2;

case J = (O} (resp. 1 = {I}), putting Ii = U(l) (resp. -litO)),

(a) if p > I, k(ro, A) = (Ojl)2 where 8 is the unique positive solution of pth IJ = 8; (b) if p = I, kilo, J.) = 0; (c) if Ii < I, kilo, i.) = (8//)2 where 0 is the unique solution in JO, n[ of fltg8 = f);

case r = 0, putting Po = 1).(1), fl) = = - 0.(0)

(a) if flo + fll > flofll' k(ro, X) = (0/1)2 where 0 is the unique positive solution of

the = 0'2_+_~J!. . 02 + fLoll ) ,

(b) if fio + III = flofll' kilo, j,) = 0; (c) if flo + Jil < flofll, kilo, Il) = (0/1)2 where 0 is the unique solution in ]0, n[ of tg e = (flo + fldO/W - flofl!)'

Example 2. A quasi-classical Dirichlet problem. We consider the Dirichlet problem

-I ~~-(I ai) ~l~) (Xi CX j

+ ku = f 011 Q

U = Y on 1

which we propose to solve in the "quasi-classical" sense as follows:

U E WI.2(Q) n 'fjO(Q),

(8.60) -li~(Iaij~~) + ku =f in -q'(Q)

u = 9 on r (in the classical sense)

We state

Proposition II. Let Q he a hounded open set in 1R1" with n ;:;, 2, with a Lipschil:: boundary,au = ajiEC'-(Q),withIaij~i~j;:;' col~!2f()rall~ElR1nwhereco > 0,

" k E Li+'(Q).


(1) The set N(a ij , k) of solutions of the homogeneous problem (8.60) (corresponding to f = 0 on Q, g = 0 on r) is a finite-dimensional space.

n

(2) ForfE Lz+'(Q), and g = ep on r where ep E W1.n+f.(Qf31 (£ > 0) there exists a solution of (8.60) iff for all ( E N(a ii , k)

( (l: aij ooep 00' + kep() dx = ( f( dx . JQ i,j ,Xi Xj JQ . II grad, II f2 . .

(3) With the notation c(Q) = mfi' N(b i !, - c(Q)) IS a space of d!men-(E.Si:il2) Il(IIu .

sion 1, namely iRu(Q). Now supposing that

k ~ - codQ)

we have N (au, k) = {O} except if , ou(Q)

k == - coc(Q). and l: aij ---:;;- == j eXj

Du(Q) . Co-~-, I

eX i

in which case N(a ij , k) = N(bij' '- c(Q» = iRu(Q) .

1, ... , n ,

Proof We note first that if w E w1.2(Q) n 't°(fl), then w E H J(Q) iff w = 0 on r 232 ;hence, given g = (p on r where ep E Wl.2(Q), we have u a solution of(8.60) iff u E 'to(Q) and is a weak solution relative to w1.2(Q) of

(8.61 ) Lu = f - ku on Q, u = g on r

with

n We note now the hypothesisfE v+e(Q) that ensures the existence offt, ... ,j~ in Ln+f.(Q) such thatf = l: DfdDxi in !?t'(Q)233. Using the De Georgi-Nash theorem

n

(Proposition 6), when k E Lex (Q),/ E U + '(Q) and ep E Wl.n+e(Q): a weak solution relative to Wl.2(Q) of(8.61) is continuous on Q; hence, thanks to this theorem, u is a solution of (8.60) iff u is a weak solution relative to W1.2(Q) of (8.61). In other words, when k E L 00 the points (1) and (2) of Proposition II are a particular case of Proposition 10.

To treat the general hypothesis k E L~ + '(Q) we use the following argument. First we consider N(a ij , 0) = {O}: this is an obvious particular case of point (3) which we

231 We recall that w··n+'(Q) c (6°(ii), The condition y = rp on r with rp E W'·n+'(Q) can be expressed in an equivalent manner by: g E ((j'U(r) and the (classical) solution of the Dirichlet problem J 0).

232 See Chap. IV. av

233 It is sufficient to take j, = _.- on Q where v is the Newtonian potential offx". ax,


shall prove below. We fix g q> on r with q> E wl.n+E(Q} and for aliI E LI+'(Q)

we denote by Go f the solution of the problem (8.60) corresponding to the case k == O. It is clear that u is a solution of (8.60) iff

II = Go( f - ku) .

Now, from the De Georgi ~Nash theorem, we know that Go f is not only continu~ ous on Q but even satisl1es a Holder condition. As we have noted above, the

modulus of continuity of Go/depends only on Q. liaijl:/," Co. !1(pIIJj"" and

f' I~"" and therefore Go will be a compact operator on LifE(Q) into (f, o(th Since

k E L; +I(Q), the map 11 --> f - k1l is continuous on (f, °(0) into LI+1(Q) and therefore 11 --> GnU - k1l) is a compact map of ((,0(0) into itself. It suffices then to

apply the Riesz~ Schauder theorem to conclude this part of the proof. Finally, let us prove Point (3), We refer to the Appendix to Chap. VIII (vol. 3) for the proof that N(i5ij' - c(Q)) is of dimension I, although one of the essential arguments is used in what follows, Hence. we take U E N (a ii , k); we have by definition of e(Q)

o = Jf' (I (Ii ~u (21 + kll Z )dX 3 r ('> (Ii _~'u ~~~ - ('0 I (I ~~)2\dX !i i,) J eX i (IX) JQ H J (Xi ax) ex i )

+ r (k + C O c(Q))u 2 dx . JQ

Hence if k 3 - coc(Q), we have

k == - coc(Q) a.e. on {x E Q; u(x) * 0] and

a.e. on Q

But Co minorizes the smallest eigenvalue of the symmetric matrix (ai)(x)); hence the second relation implies

all (~lI Io,-- = ('0-- a,e.on Q for i = 1" .. ,11, j IJ i"x j r'x i

Supposing 11 ~ 0, we always suppose that

Q+ = {XEQ;U(X) > OJ is non-empty; we then have U E W 1 . 2 (Q+) n (6 o(Q+), II = 0 on cQ+ and

= - c(Q)1I III

In particular, we have

II grad uI12,2Ifi.) = c(Q)IIu 11721Q I

and hence c(Q + ) c(Q) (we always have c(Q) ~ c(Q+). We deduce that Q \Q+


has zero capacity and u E WJ·2(Q) and Llu within a multiplicative constant, u = u(Q).

- c(Q)u in 9(Q). In other words, to

o

The proof shows that, for the points (l) and (2), we can replace

L = - 2: ;'Iii, (2: aij~c -) by a more general operator of the form ,ex i J ex j

o( C ) C. L = - 2: -;:;-- l>ij -~- + c, + I di -0- wIth Ci• di E L L(Q) ; 1 LXi J eXj eXi

we could even refine the result by assuming only that Ci , d i E C+r.(Q) (see Ladyzenskaya-U ral'ceva [1]); in the existence condition of point (2) we obviously ought to make use of N (L *, k).

D a When L contains the "transport terms" - I -;;, Ci + I di:i,' the uniqueness

uX i (,Xi

condition k + coc(Q) ? 0 of point (3) is no longer valid as we see in the case of dimension I (see Table 4). We must replace it by the condition

k + coc(O) ? ,-/J (c i + di ) in 9'(Q), '-' (,Xi 2

noting that if u E N(L, k) we have formally

jl' ( GU au au) o = I aij ;) -;-- + I (e i + d;)u -,- + ku 2 dx Q eXi (X j eX i

( k - I~~_((i!~'))U2)dX. eX i 2

Other conditions can be developed as is shown by the following example:

Example 3. A classical Neumann problem. We consider the Neumann problem

- '\' a,,-- + '\' b- + cu = f on Q Ltl.fl:"l ~I~ • {

a2u au (8.62) i.j , CXiUXj . i. OX i

OU I a· -- n + AU = q on r i,j lJ DXi j •

which we shall solve in the classical sense supposing that

on Q, I. J

Q being a regular bounded open set with boundary r of class C(;YJ with I., g E (CX(T). We state

Proposition 12. Under the above hypotheses (1) The set N(a ij • bi , c, )) of the solutions u E 'ti 'X'(Q) of the homogeneous problem (8.62) (with f = 0 on Q, g = 0 on n is a finite dimensional space.

604 Chapter 11, The Laplace Operator

(2) The problem (8.62) admits a solul iOI1 ill C6 f (Q) ifffor all ( E IV (a ii , bf, c*, t. *),

rf(dX + r !J(d~' = 0 .[2 .1

/'. + ',. ('J' -((~ ... (.~ .. i.Jl'· + hi \; 11" . where hf, c* are yir-en hy (8.21)' and i* = L. L. " )

(3) Whell i ~O 011 r alld c > 0 Oil ti we hare IV(a i), hi,c, ;.) = IV(a ji , hf ,c*,;*) = [O}.

Proo{ We can write (8.62) in divergence form

_I,i' (Ia, ~'u + CiU)\ + Idi~U + du =f on Q ( X, J ex) eX i

Lu

(8.63)

= Y on r

by choosing c;, Iii' d satisfying (8.8) and

I C;fli = i. on r The best way is to choose p E C6 f. (til such that

J'~ pdx = ,( i.d~' Q .; r

and to solve the 'Neumann problem

on - -

('/1 I. on 1'.

It admits a solution IV E ({, f (ti)234; we put Ci = ('..,t'/i'x i with the result that

ICill i = ;.onl'andIfcji:xi = ,uonQ:puttingthendi = hi + Ci + Ii"aij/i'x j , j

d = c + p, we obtain equivalence between the problems (8.62) and (8.63). From the theorems on regularity, a weak solution, relative to W1.2(Q), of (8.63) will be in C6'" (ti). Points (1) and (2) result from Proposition 10. To prove point (3) we make use of a truncation method. Given u E IV (a ij , hi, c, i.), we have for () E (6 1 (IR) with (j' ~ 0

o =

W ({Ja ij )2'J - -:2 I I .. --~-------- + hi uO(u)dx, 2(0' I (X j __ _

where w is a function such that 0(U)2 .:( wuO(v)O'(u). Choosing 0(1') rlrl" we can

234 We can always suppose Q to be connected.


take w = IIG( and hence for G( sufficiently large, in fact

II 1 (caij ) 211 w ( caij

G( > I ~ I -~- + bi . ' we have c > -:2 I I;:;-i 2c oe j ex) LX 2(0 i j eXj

+ bi ) 2 and hence

u8(u) = /u/ 3 + 2 = O. o Proposition 10 gives the existence of weak solutions of (8.52) for the given fo, ... ,f", go, ... ,gn E L 2(Q), qJ E W1.2(Q): these conditions are optimal since we seek a weak solution in WI. 2(Q). In particular, this Proposition does not permit the solution of the equations Lu = 6), where ()y is the Dirac mass at y: in effect, such a distribution cannot be written in the formfo - L: iJfjeXi withfo, ... ,f~ E L2(Q) (except in dimension 1). The following result provides an answer to this problem: for simplicity, we shall restrict our attention to the case of a homogeneous problem.

Proposition 13. Let Q be a regular bounded open sel of iH" with W 2 , x'-boundary L a differential operator of order 2 in divergence form with coefficients ail = aji E W1.X(Q), Ci' di , d E C(Q) and strictly elliptic. (I) For fE L"+'(Q), every weak solution relatil'e 10 wl.!(Q) of the homogeneous Dirichlet (resp. Neumann) problem

(8.64) Lu = f on Q, u = 0 (resp. ~:u = 0) 011 r , ( n L

is of class (t 1 on Q. In particular, with the notation of Proposition 10, N (L, F) (resp. N(L, 0)) is contained in ((,1 (Q).

(2) Given'/: a bounded Radon measure on Q, the Dirichlet (resp. N f/umann) problem (8.64) admits a weak solution relatively to W1.1(Q) iff

J~ (fdx = 0 for all (E N(L*, n (resp. N(L*, 0))235 . l2

I n addition, when it exists, this is unique, to within the addition of an element of N(L, n (resp. N(L, 0)) and is ill W!.P(Q)jrJr all p < 11/(11- I).

Proof For the point (1), we note first that a weak solution relative to W1.2(Q) is of class ((,' I on Q. From Proposition 6, we know already that it is continuous; the regularity of class (f,' 1 will be proved by the method of translations (see proof of Proposition 9), the hypotheses r E W 2", aij E WI. '(D) securing a recurrence method. We leave the details to the reader 236 . By Fredholm's method, used as in Propositions 11 or 12, we can always reduce the problem to the case where L is coercive. Given go' ... , gn E 9(D) let us consider l' the solution of

r:g L*u = go - I -;;'--'- on

(Xi D, r = 0 (resp. ~v = 0) on

(:11 r

235 We recall that we use J~ :;fdx to denote the integral of:; with respect to the measureI Q

236 See also the proof of Theorem I of *6. We can, if we wish, suppose r,fand the coel1icients of L to be of class (f, Y: the important point of this Proposition is the possibility of using weak solutions relative to Wl.l(Q) (rather than W1.2(Q)).


We know that r E (6 1 (6) and also from Proposition 7, for q > n n

IlrlIL' ,,:;; C L YiilLq i 0

where C depends only on the coefficients, Q and q. Let us take I to be a bounded measure on Q. If u is a weak solution relative to Wl.l(Q) of (8.64) we have

f · f f (' i1?li \ f ( eu ) Judx= L(u, z:)dx= u 90-L 1 - )dx= UYO+LYi,. dx. Q Q Q (Xi I Q ( Xi

Therefore

This being true for all Yo, ... , YII EO 9(Q), u EO WI. P(Q) and

Iltlllw" ,,:;; CJlfl dX

where p is the index, conjugate to q.

This relation shows, first of all the uniqueness of a weak solution relative to

WI. 1 (Q) (takef = 0). It also proves existence: approximatefby regular functionsf~ such that

J~ U~ dx -+ f sIdx Q Q

for all ( E 9(Q) (resp. U([RII). Considering the corresponding solutions Uk' they are bounded in WI. P(Q); we can therefore suppose, possibly after the extraction of a sub-sequence, that Uk converges weakly to u in WI.P(D); it is clear that in the limit that u is a weak solution, relative to WI. P(D), of (8.64). 0

We note the

Proposition 14. Under the hypotheses of Proposition 13, let us take (el' ... , er ),

(ei, ... , en as orthollormal bases ill N (L, fl, N (L *, fl (resp. N (L, 0), N (L *, 0)) respectively. There exists a unique function G: 6 x 6 --> IR of' class (6 1 on Q x Q\ {(x, y); x = y} and such that lor all y E D, thefimctiol1 u(x) = G(x, y) satisfies (\) u is a ,veak solution relative to Wl.l(Q) olthe Dirichlet (resp. Neumann) prohlem

n

(8.64) withf' = by - L env)et k ~ 1

(2) So u(x)ek(x)dx = 0 jC)/' k = 1, ... ,r.

This function G is the Green's function of the Dirichlet (resp. Neumann) problem for the operator L on Q; it is independent of the choice of the orthonormal bases


(e j , ••• , er ), (ef, ... , en. We shall come back to the idea of a Green's function in Sect. 6. 0

5. Harnack's Inequality and the Maximum Principle

There exist several approaches to the generalization of the principle of the maximum for sub harmonic functions which we have proved in §3, based on the formula of the mean. We shall, in the first instance, approach that based on a generalization of Harnack's inequality. We introduce the

Definition 7. Given Q an arbitrary open set, L a differential operator of order 2 in divergence form with coefficients in LX (Q) and f E 9'(Q). A subsolution (resp. supersolution) of the equation Lu = f on Q is any function u E WI~~l (Q) such that

fa L(u, n dx :( (resp. ;~) <./; ,> for all 'E .Q'(Q), , ;? 0 .

When L = - ,1, a subsolution (resp. supersolution) of - ,1u = 0 on Q is a subharmonic (resp. superharmonic) distribution on Q (see Definition 15 of §4). First of all, let us state, the generalized Harnack inequality for subsolutiolls.

Proposition 15. Let L be a differential operator olorder 2 in divergence form with coefficients in L x'(Q) and strictly elliptic. a subsolution u of Lu = 0 on Q which we suppose to be in WI~/(Q). Then u + E LI~c(Q) and for every compact set K of Q,

o < r < dist(K, cQ) and p > I,

[ ( 2) p.2 ]"IP (8.65) u:( C 1 + ~ p---'::~ II u + II V'(K + 8(0. rll a.e. on K

where C depends on/yon n and the LX-norms ol the coefficients oj (1/('0)L.

The proof rests on the

Lemma 4. Under the hypotheses of Proposition 15,for all 'Y > 1

(8.66) 1114+ Ilc'"l(K) :( C('Y, r)llu+ 111'IK + B(O.r))

with C('Y, r) [ C (1 + ~ ) ~;I ] 1/, '""here C depends only on the LX-norms

ol(1/co)L.

Proof of Lemma 4. We put Q, = K + B(O, 1') and dist(x, Q\Q,)

I1(X) = d' ( Q\ Q ) d' . The function 11 satisfies a Lipschitz condition 1St x, \ r + 1St (x, K)

on Q with index III', with compact support Dr and satisfying 0 :( 11 :( I, f/ == I on K. On the other hand, we take a bounded Borel set e on [R; satisfying 8 == 0

on J - 00, O[ and (} > 0 on JO, x [ and we put ( = 112 r T 8(r)dr; we have

(E w1.2(Q), supp' c Dr with the result that in the limit in the definition of a


subsolution t L(u, ~)dx ::;; O. Developing L(u, (), using the strict ellipticity and

Young's inequality, we obtain

:2'()f 1]20(u)lgraduI2dx::;; f _1_, I[WI(lJd i + 2~1J ~aij) + WZCi IJ j2 dx Q Q 2(0 i (Xi I

where we have put

WI = O(U)-l,Z f" x O(r)dr, W 2 UO(U)L2, 1V3 = u T r O(r)dr.

We put l' = f:< 11(1')1/2 dr, Ii' = IVI + H'2; noting that r2 + 1\'3 ::;; w 2 and re

calling that IJ ::;; 1, Igradl71 ::;; 1/1', we obtain

II grad (IJv) 11 [2 ::;; C (I + I~) II H' II UIQ,) .

Making use now of the Sobolev inclusion in the Gagliardo Nirenberg form 237 and noting that Ilrllu(Kl ::;; 111J1'111-2 ::;; IlwlluiJill we have

(8.67) [ ( I' j 12

Ild,2"'" "IK)::;; _ CIt r) wIL,w,)

where C depends always on the L x -norms of (1 )L.

r n the limit we ean apply this relation with

+" 22 we then have v = u - and w = .-

2(0: I) II +' 2 from which (8.66) folIows 238 . o

Pro(){ o{ Proposition 15. We shall usc Moser's iterative method. We fix K a compactsub-setofQ,O < r < dist(K,DQ),p > I; we put

') -k r. , 2k = p( _~_)k n - I

For k fixed, we define recursively the compact sets K, by Kk K and K i - 1 = Ki + 13(0, 8.j with the result that Ko = K + 13(0, rd. Applying (8.66) to the compact set Ki with 0: = !Y.i- 1 and r = (5i' we have

Ilu+ IIL"(K,J ::;; Cillu+ 11'""(K i 1)

'I' We have II ( i L"'" " !( II grad ( III.' and hence

II ~ Ii i ,".," " = II ~ 1 :: I"'" " !( 211 ( grad; ill' ,,: 2 KI I.' II grad ( I, L'

238 We recall that u+" denotes (U t )"/2


with

C 1+- ,-1 [ .( 1) (J} J1 ia i -,

bi ai -1 - 1

and hence

(8.68) II u + II L"(K) ~ (fIl Ci ) II u + II l'°(K + B(O. r,)) .

UsingC; ~ c(p,~irr n If' X (;-~l.};l(n~l)' 1 weobtain

( n-l)'') (2 )n(n-I)( . ('n'-I)" (n-I)"I) ... ) n \. 1 + (k - I) - k n X _____ .____ {J \ n n

n 1

Passing to the limit as k -> CfJ in (8,68), we therefore have

. II C r n II + Ii ( )n( 2 )n(n- 1)

Ilu~ I.'(K) ~ P'l -;;---=-1 p i U IIU(K+ B(O.r))

that is to say (8.65). o We now prove the generalized Harnack inequality for positive supersoiutions.

Proposition 16. Let L be a differential operator of order 2 in divergence form with coefficients in L'x (Q) and strictly elliptic, a supersolution u of Lu = 0 on Q which we suppose to be in WI~~2(Q). We suppose Q to be connected, u ): 0 on Q and u ¥= 0 on Q.

Then I/u E L;;;c(Q) andfor all Xo E Q, 0 < r ~ ro < dist(xo, i'JQ) and p < n/(n - I),

(8.69) II u II !fIBtxo,!')) ~ Cr"iPu a.e. on B ( xo, ;)

where C depends only on the Lf-norms ()/'the coefficients (Jf(l/co)L, on 11, p, ro and dist (xo, i!Q) - roo

ProoF Given a compact set K of Q and 0 < r < dist (K, i'!Q) we choose IJ as in the proof of Lemma 4 and fix I: > O. Applying the definition of a supersolution

1J2 with the test function, - -- -.-:-- ---- with 'X > ·····1; expanding L(u, 0 we

('X + 1) I u + 1:)" + I

obtain

f'o f ~J~.il.~llfdx f 1 [ 1 ( (';'7 ') 2 JQ (u + 1:)"+2 . ~ JQ 2eo ~ (et. +1)(~+-0a72 2~ au + dilJ

- CIJ __ .!.I ___ ]2dX + f ____ '1u ______ (dlJ + 2I. Ci-~)dX. 1 (U+C)~+I • JQ(a+I)(u+I:j"+l I oX i

In the case 'X = O. we obtain

(8.70)


In the case :x > 0, we obtain by the same reasoning as in the proof of Lemma 4,

(lUI) 11(11 + <':) 1111'"" 'I/<..I ~ (C(I + :.):Xy"II(U + <,.)-1111'1/<.. tBIO.rll·

Finally, in the case - I < Y. < 0, we shall ha ve by putting fi = - Y.

l ( 1 ) /3 -1 1 (J

(R.72) Ilu+i:III.I''''''II>:)~ C 1+~: 1-::-~-7iJ 111I+1:11I"ll\tBIO.rll·

where in all of these inequalities, C depends only on the L '(D) norms of the coefficients of (1 ) L. Applying Moser's iteration method to (8.71) as in the proof of Proposition 15. we deduce for all Y. > O.

(8.73) II(u + 1:) Illf 'I/<..I ~ ( C(1 + :)y.) 1 x II(u + 1:) 1 II "!I>: t BIIl,rl)'

Using (8.70), we see that Logu E Wl~/(D). In effect, thing a eompact set Ko of nonzero measure, we put

I (' k, = I Log(u + 8)dx and 1',

IKol.JKo

we have 1:, E Wl~/(Q), f lt dx = 0 and from (8.70), gradL' f is bounded in L2(K) Ko

for every compact set in Q; from Poincare's inequality we deduce that [I, is relatively

compact in Lfoc(Q); now kr ? Log u

since 11 =i' 0 on D. We deduce that

lim z:, a.e. on Q and hence kE is bounded ,: _ .... 0

Log II = lim P, + lim k, E Wl;)~ 2 (0) .

GivenaballB(xn,ro ) c Q.weputr = Logll - kwhere

k = I . f Log 11 dx . IB(xo• fo)! R(xo,r" 1

We have r E W1.2(B(x o' fo)). J~ .. rdx B(xo. Ynl

o and from (8.70), for every ball B of

};I"

1 1

IlgradvIIL2IBnBlx.,.r,,11 ~ C(ro, dist(xo, aD))IBIZ-n.

Using Lemma 5 below

where Po y.oC(r(), dist(xo, DD)) and Co depending only on n. Taking account of

I II ... x"I"1 d I ( (' )2/XO 11 U /.""IBI'". r"l) ~ -;--=-1---- e x

1/11 II 1-'''(8(x". r,,)) ,J Hlxo.r,,)


and using (8.73) and (8.72) which we iterate a sufficient number of times, we deduce that for all p < n/(n - 1) and 0 < 1'1 < 1'0

l!u-lllcIB(xo,r,»llullu(B(Xo.ro» ~ C

where C depends only on the LX(Q)-norms of the coefficients of (l/calL, of n, 1'0' dist(xo, cQ) - ro and 1'0 .- 1'1' By a dilatation argument, we deduce the Proposition. 0

We state Lemma 5, which we have used above, referring the reader to John and Nirenberg [1]:

Lemma 5. Let B (xo, 1'0) be a ball of [Rn and v E W1.l (B(xo, 1'0» satisfying

f vdx = o and IlgradvIIL'IBnB(xo.ro» ~ CIBI1-!./iH every ball B. B(xo.ro)

Then

(8.74) f ePoc'it'idx ~ CoIB(xo, 1'0)1

B(xo.ro)

where Po, Co depend only on n.

We note now, as an immediate corollary of Propositions 15 and 16, Harnack's inequality for positive solutions.

Corollary 4. Let L be a differential operator of order 2 in divergence form with coefficients in L X(Q) and strictly elliptic, and a positiue weak solution of Lu = 0 on Q which we suppose is in WI~~2(Q). (I) For all xa E Q, 0 < r ~ ro < dist(xo, ZiQ)

(8.75) sup u ~ C inf u B(xo. r) H(xo,r)

where C depends only on the L'(Q)-norms of the coefficients (l/co)L, and on II, 1'0,

dist(xo, am - 1'0'

(2) Suppose Q is connected; for every compact set K in Q

(8.76) max u ~ C min u K K

where C depends on (l/co}L, K and dist(K, clQ).

The inequality (8.75) is immediate from (8.65) and (8.69); as for (8.76), we deduce it from (8.75) as in the case of Harnack's inequality for harmonic functions. Furthermore, we note that by the same reasoning we have the analogue for a compact set of the inequality (8.69) for the positive supersolutions. In the inequalities (8.75) and (8.76), we have used sup, max, etc., in place of sup ess, etc.; in effect from the De Giorgi-Nash, we know that the weak solution u is continuous on Q. In the case

L = -)' -- a--(1 ( ?) ........ I};) (hi (Xj

612 Chapter II. The Laplace Opera tor

this continuity results from Harnack's inequality as we have seen in the proof of Proposition 13 of §22J9. In the general case, we refer the reader to LadyzenskayaUrarceva [1] as well as for complementary results such as, for example, the semicontinuity of subsolutions and supersolutions.

Another immediate corollary of Harnack's inequalities is the principle of the strolliJ maximum.

Proposition 17. LetL = -L~'-(La.-.?- + c.) + Ld~ + dheadifler-DXi !J tXj r I ('?Xi .,

entia/ operator of order 2 with coefficients in L' (Q) and strictly elliptic. u a subsolution of Lu = 0 on Q which we shall suppose to be in WI~~2(Q). We suppose Q to he connected and that we are in one of the followilliJ two cases:

cle· case 1: d - ") -;;-'- ill Ct'(Q)

~{'Xi

('c-ease 2: d :;? L?~' ill C/'(Q) sup ess u :;? 0

. , Q

239 To apply the method of the proof of Proposition 13 of ~2, we must sharpen the dependence of the constant C in (8.75): in particular, if 0 < 4ro < dist (xo, N2) C depends only on the L f -norms of the coefficients of L, IJ and '0 (see Theorem 8.20 in Gilbarg-Trudinger [1]). Then. for 0 < r ~ ro < dist(xo, ('Q) with the notation

l11(rl = inf ess u. AJ(r) = supcssu, (.')(r) = ,\1(rl - Ill(r) : H(;>':(l,rj

applying this inequality to the fUllctions u - m(r) and !'vf(r) - 11. we ohtain

:\1 (r/4) - Ill' ~ C(Ill(ri4) - m(r)), AI(r) - lI1(r'4) ~ CIMlr) - M(rA))

where C docs not depend on r. Thus

ffJ(r i 4) + <o(rl ~ C(w(r) - w(r;4))

namely

c-'" (r/4) ~ ('Jlr)

c +

and by induction ( r )\ (i C ~~~~ I)'

"J - ~ . -- - I'Jlr) . 4' C + 1,

ro ro Hence for --~ ~ ,. < -, we have

4' + 1 4'

with

Therefore, this proves that" satisfies a Hiildcr conditiun.


Then either u is constant on Q, or for every compact set K of Q

sup ess u < sup ess u . K Q

Proof Let us consider m = sup ess u. If m = + 00, the result is trivial; let us n

suppose m < 00 and put v = m -. u. It is clear that v E WI~~2(Q), v ;?: 0 on Q and for all ( E ft(Q) with ( ;?: 0

f L(v,()dx;?: f L(m,Odx = m f (d( + ICi~)dX. JQ In In 0Xi

As much as in the case 1 as in the case 2, v is thus a supersolution of Lv = 0 on Q. From Proposition 16, if v is not identically zero on Q, for every compact set K of Q, inf ess v > O. 0

K

The principle of the strong maximum gives immediately as in the case of the Laplacian, the uniqueness of a quasi-classical solution of the Dirichlet problem

(8.77) { LU = f u = 9

on

on

Q

r. More precisely, as a corollary of Proposition 17, given

L = - r. ~-(I aij~~l + Ci ) + r. di~a + d a differential operator of order 2 I (Xi J oXj I LX;

in divergence form on an arbitrary bounded 240 open set Q with coefficients in LI~'c(Q) and strictly elliptic on every compact set of Q, if d ;?: I cc;/Dx; in .sr'{Q), then L satisfies the principle of the weak positive maximum on Q namely:

(8.78) { ev.~ry suhsoiution U E W,lo}(Q) ot'Lu = 0 on Q satis/ving

lim u(x) ::s; 0 for all : E r satisfies u ::s; 0 on Q. x-z

It is clear that the principle of the weak positive maximum on Q implies the uniqueness of a quasi-classical solution of (8.77): more precisely this solution, if it exists, will be obtained by Perron's method as the outer (resp. inner) envelope of the set of subsolutions [' (resp. supersolutions w) of Lu = f on Q satisfying

lim v(x) ::s; y(:) (resp. li!n_dx) ::s; y(:» for all : E r.

Let us content ourselves with proving

24() We restrict our attention to the case in which. for simplicity, Q is bounded: the case of Q unbounded can be considered in the same way by taking a condition at infinity.


Proposition 18. Let Q be a bounded open set with Lipschitz boundary and L a differential operator of order 2 in divergence form with coefficients in LX (Q) and strictly elliptic. We suppose that L satisfies the principle of the weak positive maximum (8.78); then for all 9 E '{j'0(T), there exists a unique solution II E WJ~~2(Q) n 'to(Q) of the problem

(8.79) o on Q (in the weak sense)

{J on r (in the classical sense)

In addition, we have

(8.80) e(x)min(g 1\ 0):( u(x):( e(x)max(g V 0) foral! XEQ r r

where e is the so/lit ion of(8.79) corresponding to!J == 1 on r. (1('.

Before proving this Proposition, we note that e > 0 on Q; also if I :"l I ~ d

(' '\' Dc; ) ex; resp. L. .::;--.. :( d in 'I"(Q), then e :( I (resp. e ~ 1) on Q; moreover e 1 on Q

ox;

iff I ~Ci d and then we have for the solution II of (8.79) Dx;

(8.81 ) mm 9 :( II(X) :( max 9 for all x E Q I r

as in the case L = - Ll. We note finally that if Q is connected, and if g is not identically zero on r (resp. if 9 is not constant on T) we then have strict inequalities in (8.80) (resp. (8.81 )). All of these remarks arise immediately from the principle of the weak positive maximum (8.78) and from Harnack's inequality (Corollary 4). We can besides sum up these remarks by noting that from Harnack's inequality, if L is a differential operator of order 2 in divergence form on a connected bounded open set Q with coefficients in LI;;c(Q), strictly elliptic on every compact set in Q and satisfying the principle of the weak positive maximllm (8.78) then

(8.82)

every subsolution II E WI~/(Q) of LII = 0 on Q sati4'ying

lim u(x) :( 0 for all z E r satisfies 14

for every compact set K of Q .

o on Q or supessll < 0 K

Proof of Proposition 18. The uniqueness follows immediately from (8.78). In particular, with the notation of Proposition 10, N(L, n = {O}; from this Proposition and the De Giorgi-·Nash theorem (Proposition 6), there exists, therefore, a solution of (8.79) when 9 = (p on r where qJ E W1. nl '(Q). Thus we obtain the

existence of e corresponding to 9 == 1 on r. Applying (8.78) to 14 - (max 9 V Ole r

and (min 9 1\ Ole - u, we find the relation (8.80) is true for every solution of (8.79). r


Let us now consider g E (CO(r) arbitrary and approximated uniformly on r by a sequence of functions gk = <fJk on r where <fJk E wl.n+E(Q). There exists a unique solution Uk corresponding to gk and we have from (8.80),

maxluk - u/! ~ (maxlgk - gll)(maxe). Q r Q

Also from the local estimates for every compact set K of Q we have

II grad (Uk - u/) 11I-2(K) ~ Cd Uk - U11!L2(Q)

(see the proof of Proposition 15 for example). We deduce that the sequence (Uk) converges uniformly on Q and in Wl~~2(Q). The limit is the solution of (8.79). 0

An important tool for the study of subsolutions and of the principle of the weak positive maximum is Kato's inequality.

. " c (" a )" a Proposition 19. Let L = - L..,:- L,. aij - + Ci + L.. d;;;-- + d be a differ-,('X i J aXj uX i

ential operator of order 2 in divergence form with coefficients in L{;:;c(Q) and weakly

elliptic, that is to say satisfying I aij~j~j ?: ° a.e. on Qfor all ~ E \R". Letf E Lloc(Q) i.j

and let U E Wl~~2(Q) be a subsolution of Lu = f Then the function v = u+ E Wl~~2(Q) is a subsolution of

Lv = (Xlu>O} + JiX{u=o})f

on Q for all Ji E [0, 1].

Proof Let us fix Ji E [0, 1] and take (J E ~ 1 (\R) with (J' ?: 0, (J(O) = Ji, (J == 0 on ]-w,-l[,(J == lon[l,+co[.Wetake'1E.@(Q)with'1?: OandfofE > Oletus use the test function (. = (J(u/c) '1. We have

au ( u) a'1 (u ) a'1 au ( u ) L(u, (.) = r.. aij ;;-- (J - -:;-- + r. Cju(J - -;-:- + r. dj ;;-- e -:- '1 ',J uXj C OX i I C uX i I (~Xi C

( u ) '1 ( u ) au au u ( u ) au + due .... '1 + ()' .... I aij-;:;- + '1--: ()' - I Cj -;,-- . c c C '.j aXj (X j I; [; , ox,

The second last term is positive or zero, from the hypothesis of (weak) ellipticity. When [; -+ 0

and

with

e( ~) ~ Xlu>O} + JiX{u=O} a.e. on Q

~()' (~') -+ 0 a.e. on Q [; c J


Now

tXi a.e. on Q

?X j

o a.e. on OJ .

Hence in the limit

which can be considered as the weak form of Kato's inequality (vve compare it with (5.82) in the case in which L = - J). 0

As a particular case of Proposition 19, we have that, if 11 is a subsolution of Lu 0 on Q, then it is the same for the function 1I +. Because of that tile principle of t he weak positive maximum (8.78) can be expressed in an eq1liraiellt way for an operator L sati~fJ)ing the hypotheses of Proposition 19:

(8.83) { erer}" subsolulion U E WI~/ (Q) of LlI = 0 Oil Q suwfyinlJ

u ::::: 0 a.e. on Q and lim u(x) = 0 for all :: e r , .x -;:

is identically ::ero OIL Q.

In particular if L satisfies the prillciple of the weak positil'e maximum the same is true of L + ), ji)r all ;, ? O. This permits us to obtain the following characterization:

Proposition 20. Let Q he a hounded open set with Lipschit:: boundary rand L a differential operator of order 2 ill diL1eryence jimn with L x -coefficients and strictly eiliptic. Then L satisjies the principle of the weak positive maximum if and only if IV (L + X, n = {O] pJr all ;, ? 0241 . IF also L is .fimnally selFadjoint, thell these properties are equiralellt to the coercivity of L ill Wrj·2(Q), that is to say

(8.84) I"

there is c > 0 such that J L(u, 1/) dx ? ell u Ilrt" II

Proof If L satisfies (8.78), then for all;' ? 0, as wc shall sec, L + I. satisfies (8.78)

and hence N(L + ;" n = (O}. For the converse we shall show that:

(a) the coercivity (8.84) implies (8.78) always; (h) if L + ;, satisfies (8.78) for all),"> 0 and N(L, n = :OJ, then L satisfies (8.78); (e) if L satisfies (8.78), then there exists Ii < 0 such that L + i. satisfies (8.78) for all

), > p.

241 We recall that N(L. J') = :11 E Wl~,2(Q): Lu = 0 on Ql (sec Proposition 10).


Considering then /1 = {i. E IR; L + i:. satisfies (8.78)}; from (a), A =I- 0 (for ).Iarge, L + J. is coercive; from (e), /1 = Fo, cc[ and from (b) N(L + ),0' n =I- {OJ which clearly proves that )'0 < 0 if N(L + )., n = {OJ for all ). ); O. Let us prove (a) first (){ all. Let u E WI~~2(Q) subsolution of Lu = 0 on Q

and satisfying lim u(x) :;:; 0 for all Z E 1. For I: > 0, let us use the test function x ........ z

, = (u - c)+ which is in WI.2(Q) with compact support in Q.

Expanding L(u, (u - 1:) + ) we obtain

" + +. (, a(u - 1:)+ d( )+) L(H, ~) = L((u - 1:) ,(H - 1:) ) + I: L.. Ci ;:;- + u - 1:. , i OX i

from which, using (8.84) and fa L(u, 0 dx :;:; 0, we have

In the limit when I: -+ 0, u+ = O. Let HS now consider (b). First, let H E W1.2(Q) be a subsolution of LlI = 0 on Q with compact support in Q. Given J. > 0, we can consider the solution Ui- E W6· 2 (Q) of Lu;. + ).H A = LH on Q; since supp LH C supp H, from the De Giorgi~Nash theorem Hi. E (t(Q\SUppH) and from the hypothesis, since Lu :;:; 0 on Q, H;. :;:; 0 on Q. Now 1.1 - 1.1" E W6· 2 (Q) and L(1.I - 1.IJ = ).1.1;, on Q; since N(L, 1) = {O}, L is an isomorphism of WJ·2(Q) onto W- 1• 2 (Q) and hence a fortiori

we deduce that 1.1;. -+ H in W1.2(Q), from which 1.1 :;:; O. In the general case u E WI~~2(Q) subsolution of Lu = 0 on Q, satisfying

lim u(x) :;:; 0 for all Z E 1, we consider Ut = (u - I:e)+ where e is the solution of

Le = 0 on Q, e == 1 on 1. From Kato's inequality, He is also a subsolution of Lu = 0 on Q and Ue E W 1• 2 (Q) with compact support. Applying the above result, Ue = 0 on Q and hence U :;:; 8e; in the limit when I; --> 0, 11 :;:; 0 on Q.

To prove (e), supposing that L satisfies (8.78), let us consider the solution Uo of Lu = I on Q, u = 0 on 1. We have Uo E (6,o(Q) and Ho > 0 on Q; let us put,u = -1/ II 110 II L" We consider 0 > ). > ,u and 11 E WI~~2(Q) a subsolution of

Lu + i.u = 0 on Q satisfying lim u(x) :;:; 0 for all Z E 1. From Proposition 15,

u + E LX (Q). The function v = u + )" II u + Ii L' Uo is then in WI~~ 2 (Q), subsolution of

Lv = A( II u + II L' - u) :;:; 0 and satisfies lim v(x) :;:; 0 for all Z E r. Therefore x ~ z

u :;:; -;~llu+ IlL' on Q from which we have Ilu" 11 / ,= 0 from the choice of ; .. To complete the proof of the Proposition, we must in the symmetric case, show that N(L + )~, 1') = {OJ for ali ;. ); 0 implies the coercivity (8.84). We note that the


coercivity in W6· 1 (Q) is equivalent to the coercivity in L2(Q):

(8.85) there exists c > 0 such that IQ L(u, u)dx ? ell u IIi2

for all u E W6· 2 (Q);

we have in effect for 6 > 0,

(8.86) L(u, u)? (1 - f,)L(u, u) + E[COlgradUl2 + (d - _1- I (c i + dY )U2J: 2 2co i

therefore, assuming (8.85) and choosing

E = C (c + (~() + II~- 4 (Cj + dY _ dll ) -1 2 2co I L Y

I ECO 1

we shall have L(u, u)dx ? Til U Il w·1.2 •

Let us consider now 11 = inf r L(u, u) dx . u E W~·2IQ) JQ

!Iud u = 1

Using (8.86) with f, = 1, it is clear that 11 > - CXJ; also using the compact injection of W6,2(Q) into U(Q) we see that there exists u E W6,2 with II u Ilu = 1 such that

In L(u, u)dx = 11· Using

In L(u + "u + ()dx ? In L(u, u)dx for all (E W6,2(Q),

we see that u is a weak solution of Lu - IlU = 0 on Q. Hence if N(L + I., 1) = {O} for all ). ? 0, necessarily, we have It > O. 0

In the case where L is not formally self-adjoint, the principle of the weak positive maximum does not necessarily imply the coercivity of L, as is shown by the following example in dimension n = 1: taking

d2 d L = - - + b-- + d where dE P(]O, 1[), b E W1.1(]0, 1[);

dx2 dx

L is coercive iff d + 1[2 ? ~ and d + 1[2 of=. ~; L will satisfy the principle of the

weak positive maximum iff tb 2 + d + 1[2 ? tb' and tb 2 + d + 1[2 == tb' (see Example 1). From Proposition 20, if L satisfies the principle of the weak positive maximum, it is the samefor its adjoint L*. In particular, with the usual notation, if d ? I ad;/(lx j

in E0'(Q), then L satisfies the principle of the weak positive maximum since it is so for its adjoint from Proposition 17. In fact in this case L satisfies a principle of the maximum in L 1, dual of the classical maximum principle. Let us state this precisely


Proposition 21. Let Q be an arbitrary bounded open set of IR"

i3( i3 ) i3 L = - L -0 . L aij -0 . + Ci + L di -0 . + d x, Xl .x,

a differential operator of order 2 in divergence form with coefficients in L 00 and strictly elliptic,f ELI (Q) and u a subsolution of Lu = f on Q which we shall suppose to be in

1 2 d >.: '\' i3d£ I' ( 0 f II T" Wlo~ (Q). We suppose ?' L.., ~ in 9'(Q) 1m u x) ~ or a Z E r. ,len eX i x--+-z

r .f dx ;0 O. Jru> O}

Formally, denoting Q+ = {u > O}, we have

h au ~ . Now if u ~ 0 on r, we have u = 0 on oQ +; ence '" ~ 0 on cQ + and sillce

unL

d ;0 I ~di in 9' (Q), r (L di ~~. + dU) dx ;0 O. To be justified, this reasoning eXi Ja, ex,

would demand that the coefficients of Land u be sufficiently regular, which we might possibly suppose, but also that Q + be regular which is not, in general true. Let us give a general proof based on the same argument as that used in establishing Kato's inequality.

Proal Let us take (j E (6'1(1R) with (j' ;0 0, (j == 0 on J-oc, 1J and () == I on

[2, oc [ and use the test function (E = () ( ~) which from the hypothesis is III

W1.2(Q) with compact support in Q. Using the strict ellipticity, we have

(8.87)

where WE = f~ o(~.)drE W',2(Q) with compact support. In the term on the

right of (8,87), the first integral is obviously positive or zero; the second term also . '\' adi . thanks to the hypothesIs d ;0 L.., _. III 9'(Q), Now when £ -> 0

aXe

y u ,(u) ~£->X(u>O}, -8 - ->0, [; I:

WE -> u+ a,e, on Q


with

u (u) o ~ - 0' - ~ 110' II L' ' f, [;

Therefore in the limit in (8.87), l/l:lu> o} dx :;, O. o

Remark 1. When the coefficients au E W l. C (D), all the results of Propositions 15 to 21 remain valid for the subsolutions (or supersolutions) U E WI~~l(D) without the restriction U E WI~~2(D) which we have imposed. In effect, given u E WI~~l(D) subsolution of Lu = 0 on D, for every bounded open set Do with i20 c D, there exists a sequence of subsolutions (Uk) in !(5' 1 (Qol of Lu = 0 on Do such that Uk ~ U

in Wl.l (Do); that follows from Proposition 14. If u is a subsolution of Lu = 0 on D, the distribution Lu is a negative Radon measure on Q with the result that the tracefo of Lu on Qo is a (negative) bounded measure. Supposing Qo with Lipschitz boundary, u E Wl. I (Qo) admits a trace go ELI (eQo) on the boundary of Qo (see Netas [IJ). When aQo is a W 2 . .< boundary, we know (see Proposition 14) that N(L*, DQo) c 1f,jI(Qol and we shall have

j' U~dx = f Yo ;:;(~ -d)' for all s E N(L*, eQol; Q n ('fio cn/*

we can always approximate (.f~I' Yo) by a sequence of regular functions (fk' ud such that fk ~ 0 on Qo and

Jt' sj~dx = j', uki)~s di' forall SE N(L*,iJQol Q() (·Qo ' r*

with the result that we can solve the problem LUk =.f~ on Qo, Uk = Uk on aQo by imposing on the solution the condition of verifying

r uksdx = r usdx for all s E N(L, I:QO); JQ() JQo

from Proposition 14, Uk E y;l(QO) and Uk ~ u in WI. I(Do).

Let us consider now a differential operator of order 2,

o

Supposiny the coefficients au = a ji E Wi. x (Q), hi' eEL' (Q), we can put it in divergence form

(8.88) L = - '" - '" (/ -- + '" h· + '" --- - - + c i! ( a ) I ('au) i' i cX i j cXj i ( j ex j ('Xi

L"" L,t},-" L l ~'"'I

with coefficients in L I (Q). We can then apply all the preceding results; in particular ifP is lmijimnly elliptic and c = 0 (resp. c :;, 0) a.c. on Q supposed connected, aery function u E W~o'c' (Q) satislying Pu ~ 0 a.e. on Q (resp. and sup ess u > 0) satisfies

n


the principle of the maximum:

u == sup ess u or sup ess u < sup ess u for every compact set K of O. fJ K fJ

In effect if u E W~o'/ (0) satisfies Pu ~ 0 a.e. on 0, then u is a subsolution (in the sense of Definition 7) of Lu = 0 on 0 where L is given by (8.88); from which, taking account of Remark 1, we have the result from Proposition 17. We can obtain the same property without the hypothesis au E WI, <Xl (0). We state the following maximum property due to Bony:

Proposition 22. Let P = - I aij ':;-D: -: + I bi ;-,c + c be a differential operat-ij oxiOX j ox;

or of order 2 with coefficients in L % (0) and strictly elliptic and u E W~o't(O) with p > n such that Pu ~ 0 a.e. on 0; we suppose that 0 is connected and c = 0 a.e. 011 0

(resp. c ): 0 a.e. on 0 and sup u .): 0). Then if u is not constant 011 0, Q

u(x) n. Then the image under F of a negligible set (.for the Lebesgue measure) of 0 is negligible in IR".

We refer the reader to J. M. Bony [1] for the proof of this lemma based on the

Sobolev inclusion Wl~/(O) c 'til -~(O). Pro()f of Proposition 22. We shall first give a proof supposing c ): () > 0 a.e. on 0 that P satisfies the principle of the weak positive maximum:

(8.89) {for every bounded open set 0 0 with 0 0 cO,

14 ~ 0 on cOo => u ~ 0 on 0 0 ,

Let us suppose u ~ 0 on cOo and sup u > 0; then there exists Xo E 0 0 such that flo

u(xo) = sup 14 > 0; since c ): (j > 0, b E LCD and grad u(xo) = 0, there exists ao

ro > 0 and (50 > 0 such that

Ib .. ~.1! .. + cu ): (jo a.e. on B(xo, ro) , . I (lx i

namely, since Pu ~ 0,

Let us put v(x) IY.

u(x) - 21x - xol2 whereIY. > OsatisfiesIY.Ia ii < (jo with the


result that

To obtain a contradiction, we show that the set M of the points x E B(xo, '0) such that

l'(y) ~ v(x) + grad v(x). (y - x) for all y E B(xo, ro)

is non-negligible. This will conclude the proof since at a point x of M where v is

twice differentiable, the symmetric matrix ( ..... (i2L.' (Xl) is negative and therefore (l X J;X j

a2 v l>ij(.x) ;;--:1 -(x) ~ O.

UX i uXj

To prove that M is non-negligible, we note from Lemma 6 that it is enough to show that {grad v(x); x EM} is non-negligible. Now, given I; E fR", let us consider l(() = max v(x) - (.(x - x o) and x(O E 8(xo, '0) such that

B(.x ll , fo)

).(() = v(x(()) - (.(x(l;) - xo);

we have

v(y) ~ v(x(()) + (. (y - x(()) for all y E B(xo,ro)


x(() E B(xo, ro) => grad v(x(m = ~ and x(() EM.

and

To sum up: {grad l'(x); x EM} contains {I; E fR", I (I < ,=;o} which completes this

part of the proof. Now (8.89) is true with only c ~ 0 a.c. on Q, but for the open sets Q o of sufficiently small diameter. Let us suppose that Qo c B(O, r) and put for a > 0, li = ealxl2u.

We have

where C-: = ealxI2P(e-alxI2.) = - 4o.2 '\a··x·x. + 2a'\a. 2a"h x + c' J. L, 'J', J L, ,,- . 4- i· i ,


from which, using aii ~ Co the constant of strict ellipticity, we deduce

c ~ 20: [nc "" - r(II(Ibf)I/2t.x + 20:rIllauIILx)] on Qo .

If r < nco II( I b1)! /2 II i), for 0: > 0 sufficiently small we therefore have

infess c > O. Since Qv = ealxl ' Pu :::; 0 on Qo and v :::; ° on oQo, from (8.89) no

applied to Q and v, we have v :::; ° on Q o and so u :::; 0 on Q o. Now let us show under the hypothesis c ~ ° a.e. on Q,

for every ball B(xo, r) with B(xo, r) c Q and r sufficiently small,

and z E 3B(xo, r)

(8.90) u(z) > u(x) for all x E 8(xo, r) and u(z) ~ 0

lim u(z) - u(txo + (1 - t)z) =>~ > 0'

Ij 0 t '

We can always suppose that Xo = 0; applying (8.89) to the function v = u - u(z) + s(e'a lx j2 - e- ar2 ) on the open set Qo = B(O, r)\B(O, tr); we have from the above calculation, using c ~ 0 and the strict ellipticity

Pv :::; eP(e-alxI2) .~ ce-' lxI2 [ -0:2('or2

+ 2adl:Iaiillp + rI!lb;IIL') + Ilcil L"] on flo·

We can therefore choose 0: sufficiently large, independently of [; > 0, such that PI' :::; ° on flo. We now put m = max u; from the hypothesis u(x) < u(z) for all

118(O,r/2)

X E B(O, r), we have m < u(z); let us now choose c = (u(z) __ m)(e- ar2 / 4

- e -ar2)-1 > 0, with the result that v :::; ° on 3B(0, 1r); we have v = ° on oB(O, r) and hence by application of (8.89), v :::; ° on Qo. In particular, for t > ° sufficiently small,

and in the limit

I. u(z) - u(txo + (I - t)z) , 1m . ~ 2w.rc - w > ° .

t

We finally complete the proof of the proposition. We suppose that

F = {x E Q; !I(x) = S~F 14 }

is non-empty and distinct from Q. Since Q is connected, there exists XI E 3F n Q; let us choose X o E Q\F such that r = dist(xo, F) is strictly less than dist (xo, aQ) with the result that 8(xo, r) c Q, and on the other hand, sufficiently small for us to apply (8.90); there exists Z E aB(xo, r) n F and !I(x) < u(z) for all x E B(xo, r) c Q\F. Since u(z) = max u, grad u(z) = 0; but from (8.90), grad u(z). (xo - z) > 0 which is a contradiction. 0


We notice that in the above proof we have passed to the end through the medium of lJopI's maximum principle which we now state in the general case.

/",2 ? Proposition 23. Let P = - 2:aij -::;- --,- + I hi :,- + c he a differential oper-

eX i (Xij ('Xi

Mor of' order 2 with coefficients in L' and strictly elliptic. u EO Wfo'!(Q) with p > 11

slIch that Pu :S; 0 a.e. on Q alld ::. EO (~Q. ~Ve suppose that there exists a hall

B(.\o. 1'0) c Q such that:: E ?B(xo. 1'0), that u(z) = lim u(x) e.'(ists and is equal to X -0.::;

sup II. alld c = 0 a.e. Oil Q (resp. c ~ 0 a.e. 011 Q (llld sup LI ~ 0). a a

Then 11 is constalll on the connected component contaillillY xo. orfor all 0 E [0, n12[

. u(z) u(x) hm -------- > 0 x "0 I:: - xl x E Co

where Co = {x EO Q; (x - z)(xo - z) ~ Ix - ::llxo - ::Icos O}.

Proof: We can suppose Q to be connected; if 14 is not constant, from Proposition 22, u(x) < u(z) = sup lIo for all x EO Q. It is sufficient then to take up again the proof of (8,90), noting that

e -- >1\ lim

Ix - zl ~ 2:xre - ,,2 cos () . o

As in the case of the Laplacian (see §4.4), Hopf's maximum principle provides the proof of the uniqueness in a Neumann problem:

where B

{PU = I on Q

Bu = y on r ? I fJi-;;;-- is a homogeneous operator of order 1 non-tangential on r.

(Xi

6. Green's Functions

First of all, let us take Q a bounded open set of [M" with Lipschitz boundary I'

~ ( ~ ) ; 6 ~ 6 (.. L = - I 1 - I aij~- + c; + I d;-;,-- + Ii a differentIal operator of order 2

(X j L'(j ()X i

in divergence form with coefficients in L"(Q) and strictly elliptic on Q. To a closed part of T. Let us consider the mixed problem

r =f on Q

" ~ ~ on r (8.91) 0

011 ('n

=tjJ on I \ro · . L


Following Proposition 10, we denote by N(L, r 0) the set of weak solutions (relative to Wi. 2 (Q)) of (8.91) corresponding to f = 0, qJ = O,!j; = 0; we know that this is a finite dimensional space of the same dimension as N (L *, r 0) where L * is the formal adjoint of L. Supposing that qJ = O,!j; = ° we know from Proposition 10 that, given f E [}(Q) (8.91) admits a solution iff

I'

I Ivdx = ° for all v E N(L *, ro) "n

that is to say iff I belongs to N(L *, rol", the orthogonal subspace in L 2(Q) of N(L *, r o); finally, when this is the case a solution of (8.91) is defined to within the addition of an element of N (L, r 0): we can thus consider that, if it exists, the unique solution of (8.91) is in N(L, ro)~. Finally, we recall (see Proposition 6), that if f E U(Q) with p > 1n, a solution of (8.91), with (p = 0, !j; = 0 is continuous and bounded in Q. In particular the elements of N(L, 1 0 ) and N(L *, 1 0 ) are continuous and bounded on Q.

We now prove

Proposition 24. Under the hypotheses and notation introduced above, there exists a unique function G defined to be continuous on {(x, y) E Q x Q; x =f. y} with GEL 1 (Q x Q) satisfying

(ilfor allfEN(L*,ro)~nLY(Q), the unique solution u in N(L,lo)~ of (8.91) corresponding to (p = 0, 11 = ° is gir:en by

(8.92) u(x) = In G(x, y)f(y)dy a.e. x E Q

(ii) for all v E N(L*, 1 0 ),

In G(x, y)r(y)dy = ° a.e. x E Q

In addition, for all y E Q, the functio/1 G(., y) is a weak solutio/1 relative to wl.r(Q)

. n f for all r < - - 0 . 11- 1

(8.93) u = 0 on ro au

{

Lu = /; on Q

'~ .. ' .. = 0 0/1 rro cnL

where .I;. is the ul1ique bounded measure on Q such that

(8.94) {.I;, - by E N(L*, ro)

ff}V(X)df,(X) = v(y) foral! 1: E N(L*, 10)'

Chapter [I. The Laplace Operator

Definition 8. Under the hypotheses and with the notation introduced above, the function G considered in Proposition 24 is called the Green's function o( the mixed prohlem relatiue to (D, 1'0' L).

Prooju( Proposlioll 24. There is uniqueness of the function G for if Gland G 2 are two solutions of the problem, the difference G = G I - G 2 defined to be continuous on i (x, Y) E Q x Q; x * .l'} and integrable on Q x Q will satisfy

r G(s, y)f(yJdy = 0 for all f ELf (Q), a.e. x E Q ~a

Therefore G(x, y) = 0 for a.e. y E Q and a.e. x E Q and by Fubini's theorem and continuity G(x, y) = 0 for all (x, y) E Q x Q with x * y. To prove existence we first of all denote by G the map which with IE L2(Q)

associates the unique solution U E N(L, ro)J. of (8.91) corresponding to qJ = 0,

I/J = 0 and the projection (orthogonal in e(Q) of/on N(L*, roll. We say that G is the Green"s operator ill L 2(Q) of'the mixed prohlem relatire to (Q, 1'0' L). As we have seen in the proof of Proposition 10, G is a compact operator in e(Q) and its adjoint G* is the Green's operator in U(Q) of the mixed problem relative to (Q, r o , L *). Also, as we have recalled above, for p > !Il, G and G* map U(Q) into the space '6~(QJ of bounded continuous functions on Q. Given Y E Q, the map y E U(Q) -> (G*y)(y) is a continuous linear form on U(Q)

for p > !Il; thus, there exists Gr E Lq(Q) for q < n/(11 - 2) such that

r J G,,(x)~J(x)dx = (G*iJ)(Y) for all iJ E U(m .

f2

We shall show that the function Gy is continuous on Q {y: or more precisely that given K a compact set in Q, Q o a neighbourhood of K in D, the function G" is continuous on Q ... 0 0 uniformly for Y E K. We can therefore consider the function G defined on {(x, Y) E Q x Q; x * y} by (f(x . .v) = Gr(x); since from the definition of Gr" the map y -> G" is weakly continuous from Q into Lq(Q), the function G is continuous, Also G is bounded in Lq(Q) uniformly for Y E Q and therefore a/ortiori G is integrable on Q x Q, Given fE L'(Q), we have for all fJ E Lf(Q)

r y(x)dx f G(x, y)f(y)dy = J 11 f2 J!" f(y)dy r Gr,(x)y(x)dx

f2 ~!2

f f(y)(G*y)(y)dy !2

(' I (Gf)(x)y(x) dx .if!

and hence

f G(x, y)f(y)dy = Gj(x) a.e. x E Q , f2

By the definition of Green's operator G, we sec therefore that the function G satisfies the properties (i) and (ii),

~R. Elliptic Equations of the Second Order 627

To complete the proof let us take p E 9(lRn) with p ~ 0, f p(x)dx = 1,

supp p c B(O, 1) and consider for y E Q, the sequence of functions Uk = Gfk where A(z) = k"p(k(z - y)). We have Uk ~ Gy in Lq(Q) when k -> CfJ, since for all g E U(Q),

f uk(x)g(x)dx = f j~(z)(G*q)(z)dz -> (G*q)(y) . Q !l

Let us consider the projection X ofj~ on N (L *, 10)"-; the function Uk is the solution in N(L, 10)"- of ru

' ~J. on Q

Uk = 0 on 10

c1uk 0 r,ro ' on

i'n t

The sequence of the functions X is bounded in L 1 (Q) and converges vaguely to the measure.~. defined by (8.94). By duality, as in the proof of Proposition 13, we show that the sequence (Uk) is bounded in W1.r(Q) for all r < n/(Il - 1) and so its limit Gy is a weak solution relative to ft'l.r(Q) of(8.93). Finally, considering K a compact set in Q and Qo a neigbbourhood of K in Q, making use of the fact that N (L *, ro) is of finite dimension,j~ is bounded on Q\Qo uniformly for kEN, Y E K and with the limit G" bounded on Q\Qo uniformly. From the De Giorgi Nash theorem, since t; is bounded and continuous on Q\Qo uniformly for y E K, G" satisfies a Holder condition on Q\Qo uniformly for Y E K. 0

Remark 2. As we have seen in Proposition 14 when the coefficients aij E WI. 'X (Q) and in the pure Dirichlet case (10 = 1) or in the pure Neumann case (ro = 0), there exists a unique weak solution relative to WI. '(Q) of (8.93); this problem then defines completely the Green's function. In particular, Definition 8 generalizes that given in §4: the Green's function of a bounded open set Q (with boundary satisfying a Lipschitz condition) is the Green's function relative to the Dirichlet problem (ro = 1) in Qforthe operator L = -/1. Let us now give some properties of the Green's function generalizing those proved in §4 in the case of the Dirichlet problem for the Laplacian.

Proposition 25. Under the precedinq hypotheses and notation, lei us consider G the Green's iimetion of the mixed prohlem relative co (Q, 1 0 , L). (l) The Green's function G* of the mixed problem relative to (Q, r o, L *) is given by

G*(x, y) = G(y, x) Fir all (x, y) E Q x Q, x =f )' .

In particular, if L is formally self-adjoint (L * = L), then G is symmetric:

G(x, y) = G(y, x) for all (x, y) E Q x Q, x =f y.

(2) Let us suppose that L is coercive in V2(Q, ro) (that is to say that there exists

c > ° such [hat IQ L(u, u)dx ~ ell 11 II ~l 2 for all u E V2(Q, ro) thell G(x, y) > 0


fiJr all (x, y) E Q x Q, x and y two distinct points 0/ the same connected component o/Q.

Proof of Proposition 25. Point (1) follows immediately from the construction of G given in the proof of Proposition 24. For point (2), we notice first of all the hypothesis of coercivity implies that N(L, 1 0 ) = N(L*, 1 0 ) = {O} since, given

I'

u E N(L, 1 0 ), we have J L(u, u)dx = 0 by definition of N(L, 10)' From Pro-f}

position 20 and (8.82), it is sut11cient to prove that G(x, y) ;:: 0; if account is taken of the proof of Proposition 24, it is sufficient to prove that the Green's operator G of the mixed problem relative to (Q, 10' L) is positive in the sense

f ;:: 0 a.e. on Q => Gl ;:: 0 a.e. on Q

Let us consider f ;:: 0 a.e. on Q and let u = G!; given /; > 0, we consider r = (- u) + which is V2(Q, 10): we have L(u, r) = - L(v, v) a.e. on Q, so

C L'I!fj12 ~ r L(r,v)dx = - f L(u,v)dx =_. ffVdX ~ 0 JQ f} Q

and t' = 0, that is to say u ;:: 0 a.e. on Q. o

As we shall see in Example 4 below, in general the Green's function is not positive; in the case of a Dirichlet problem, the positivity of the Green's function is equivalent to the principle of the weak positive maximum (8.78). The Green's function has been introduced as the kernel in the solution of the homogeneous mixed problem ((p = 0, i/J = 0). In the case of a general mixed problem, we have Green's representation Jormula of the solution in N (L, 1 o)~ of (8.91) (if it exists):

(8.95)

r I' 'G G(x, y)f(yldy - J :--- (x, z)ip(z)di'(Z)

.n 1,,1'111.'

+ r .... G(x, z)i/J(z)d[,(z) . J I .I n

Formally this is deduced from Green's formula (8.15); considering x E Q, we put /:xCV) = G(x, y) so we have from (8.93) and point (I) of Proposition 25

where 9x is defined by

f w(y)dYxCrl = 0 for all Ii' E N(L, 1'0) !]

and r'in T *

o on 1\10 ,

~8. Elliptic Equations of the Second Order

Applying (S.15) formally, we obtain for the solution u in N(L, To).1 of (S.91):

u(x) = f u(y)d6x (Y) = f u(y)dgAy) = f uL *vx dx u a a

629

I' r i "'C J G(x, y)f(y)dy + G(x, z)ljJ(z)d{'(z) - ~ " (x, z}qJ(z)dy(z) a .r-..r" ro(;nL

To justify this calculation, it is sufficient, following the notation of the proof of Proposition 24, to approximate l'x by

Vk = G*gk where gk(y) = k"p(k(x - y)) .

The passages to the limit

fi] uL *vk dx = fa ugkdx ---> u(x)

f vkLudy ---> Jf' G(x, y)f(y)dy a [}

do not pose problems; for the passages to the limit

Lro vdt d)' --> L 1'" G(x, z)ljJ(z) di'(z)

i eVk d f cG ([J }' --> (x, z)qJ(z)dy(z)

r cnL ' . r anL ' o "

we shall need some justification of regularity up to the boundary. Although this is not the most general such result, we state the following proposition:

Proposition 26. With the preceding hypotheses and notation, let us suppose au E Wl,OC(Q), Q with W 2 .y. andl'\T~ = T\T0242. The Green'sfunction G of the problem relative to (Q, To, L) is in '6 1 ( {(x. y) E Q x Q; x Ie y}) and fhe solution of (8.91) in N(L, ro).1 is, if it exists, yiven by (8.95).

Proof It suffices to make use of the 'ti 1-regularity of the solutions of (S.91) for (P = 0, IjJ = ° when f E U(Q) with p > n (see Proposition 13). 0

Let us now give examples of Green's functions.

Example 4. The case of a symmetric operator with constant coefficients in dimension 1. Let us consider Q = JO, I[ a bounded open interval of iR1

d 2

and L = - J-;2 + k. The study of Table 4 of Example 1243 leads to

the determination of N(L. ro) = N(L*, ro) according to the values of k and

242 In particular, r 0 = r (Dirichlet problem) and ro = 0 (Neumann problem). 243 With I. == O.


fo c r = {O,IJ. We recall that N(L, fo) is either red uced to {O}, or is of dimension 1 generated by an eigenfunction woo When N(L, fo) = {OJ, given y E JO, I [, the function u(x) = G(x, y) is a solution of

-II" + ku = c)y III y'(]O,I[)

which we can rewrite as

(8.96)

Whenk by

{ u" = ku on J 0, y[ u J y, l[

u(y -) = u(y +), u'(y -) = u'(y +) + I

{J)2 > ° (resp. k = 0, k = (1)2 < 0), the solutions of (8.96) are given

sh (J)(x y) + u(x) = ae"'X + he wx - ---------- -

OJ

(' + _ _ sinw(x resp. ax + h - (x - y) ,ae'IOX + he-1()X -

, w

Incorporating the boundary conditions we obtain the Green's functions (see Table 5).

TableS. Grecn'sfunetionassociatcdwith(JO,I[,fo. _ d' + k)WhenN( _ d' + k.to)\ = {OJ dx 2 \ dx 2

k ~ 0

k =- - 0/

k = ()

k = (1/

/.; -

, ur

(10 ~ /') 1

Clx. \) ~--- [sh o)x sh 0)(/ - 1)- sh OJ! sh wlx - y) -- ] - oj sh (1)/

1 C(.\:, YI = Lli Ilx-- Ilx r)+ J

/

1

1111[

wioI

CI\:. r) 00 ---- --- [sin OJ'( sin O){/ .vI w sin wi

(/0 =:0: I 1

sino)lsinw(x - r)' J

(lix. yl = ----- [sh (!)x eh (u(l - vi - eh OJ/ sh w(x -- y) + J o)el! (!)I -

C(:>.:. y)= \: - (x - r)'

(lJ # ~ (m + :) . I -_ ~

Ci(\:. y) = -[sinwxcos(')(i - y) - cosw[sinw(x -- y)" ] OJ cos (1)[

1 G(x. r) = ----leI! oJX el! (1)(/ - \'l - sl! OJ/ sh OJ(-' - y) + J

• (!) sh 0)1 -

1

mIT ('J #--

I

G(\. II ~- [cOS(lJXCOSi')(/ l'I sinwlsinwlx - yl'-J - OJ sin wi

, _________________________________ ---.1


We now consider the case in which N(L, To) is of dimension 1 generated by the eigenfunction Wo which we can always assume to be real and normalized

(L IWo l2 dx = 1). Given Y E ]0,1[, the measure j~ defined by (8.94) is

by - wo(y)wo with the result that u C(x, y) is the solution of

- un + ku

u = ° on To, u' = ° on T\ro r uwodx = ().

Given a particular solution Uo of the equation

(8.97) - u~ + kuo = Wo on ]0, l[

the function ii = u + woCV)uo is the solution of

ii n + kii = by in fC'(]O,I[)

(8.98) ii = wo(y)uo on To, ii' = woCv)u~ on T\To r tlwodx = wo(Y) r uowodx.

The case N(L, To) =1= {a} occurs if

k = 0 and To = 0 .

or: k _ (n~~y and card To =1= 1 ,

or: k = - (( m + DiY and card To 1 .

When k = ° and To = 0, we can take Wo I/J! and Uo

solution ii of (8.98) is then

it = (' - (x - y)+ ,

where the constant c is determined by the last condition

(I - y)2 [2 cl = ------ - --

2 6

I y2 _ x 2

To sum up: u = :3 - y + --2-[-- - (x - y)+.

The other calculations carried out in the same way give the results shown In

Table 6.


Table 6. Green's function associated with ( JO. 1[. ro. d' + k)' when rv( -~,'~ + k. ro') oF 0 dx' dx' /

Prohlem

x +),2 !;-. () and I" = 0lNeumann; CiiX . .l1 ~ - r + - - il-- - Ix - rl' I

f---------~---.,--------------~-.---- ----+-------------~------j

k = - ,'J' with

mn sin (1).\ [' ( \' \) ..L Si,_n, '_'J.\' J-Cilx . .1') = ----- I -~, cos (OJ' ,

UJ \ /. "J/ 1'" = r (Dirichlet) and ('J =- - -. fIl - l..

x 1

I - sin<'irCOSU)x sin (!)(x - y)' I'd UJ

I I~I'~ :0: Im"cdl ,," "~~ _1 ___ _ ----------- -- ----- ------ ------ --------------j

, k OJ' (J = . III = L, I (il'. I) = ~ [~cOS(.)\,SinWX (I) I

';')sinW\'Sil1uJ'\: - sinw(x -\')' J J and r l , 0 (Neumann) (I L~~~~~~~~ ___

Example 5. Green'slime/ion of the Neumanll problem/cn' the Laplacian. Given Q.

a bounded set. we consider the Neumann problem (ro = 0) for the Laplacian (L = - /1). Let us suppose that Q is connected with the result that N(L fo) = N(L *. rn) =:constant functions on .12\ (see ~4.4).

I Given r E Q. the measure fl', defined by (8.94) is ()\,- where IQI is the Lebesgue . ',., IQI measure of the open set Q. Therefore. the function [I(.\') = G(.\'. y) is the solution of the problem:

I (\ 1111 on Q

:QI ('II

(8.99) i:/1

0 on I

J tid.\' = o. f}

Let us put 11 in the form

11(.\') = 211 QI

- E,,(x - y) + c + dx)

where Ell is the elementary solution of the Laplacian (see ~2) . .\'0 is an arbitrary


point and c = L (E"(~ of the problem

) 1~ ..... xoI2)dYTh f .. hi' - y - -i~Tfl-l- C;. e unctIOn v IS t en a so utlOI1

(8.100)

Llu = 0 on Q

ev (z - Y 7 - X ) on (z) = ;~~-=~T'; - -n i Q{ .11{Z) for z EO r

j' v(x)dx = 0 . 12

We shall obtain the function v and hence the Green's function G in the case of a ball B of Ri". We can always suppose that B = B(O, 1) and put y = ra with o ~ r < I and (f EO E; we denote by v(r, x) the solution of(8.100). We can always take Xo = 0 with the result that v(r, xl satisfies

r Llv(r, xl = 0 on x EO Q

1 ~~ (" z) ~ ~~[ ~: ~";:I~ -L I v{r, xl dx - 0 .

I J for Z E I (8.101)

We put w(r, x) = E"(rx - (f), so that we have

(8.102)

r Llw(r, x) = 0 for x E Q

I cw r r - (f. z l (1/1 (r, z) = (f" X Irz -=~T;; for Z EO I .

Noting that z, (J E I we have

irz - (JI = Iz - r(fl and r(r- (J. z) = Iz - raI2 ... ~ (1 - m. z) ,

we deduce that the function h(r, x) = vir, x) + w(r, x) is a solution of

1 Llh(r, x) = 0 for x E Q

?h (r,z) = ~[ ___ I_, - IJ for z EO I. en (f" 1m - zl" -

Differentiating with respect to r EO ]0, 1 [ we see that k(r, x) = ~h (r, x) is a solution Dr

of

f Llk(r, xl = 0 for x E Q

ok n - 2 (J. z r l-;;-- (r,z) = -- x ---en (J" Ir(J - ziti for z E 1.: .

In other words from (8.102),

(8.103) ah _- (r, x) = k(r, x) cr

(n - 2)\-\.'(r, xl - -....... ---- + err) . r


Now for r = 0, we have trivially that frO, x) == 0: but by the construction of l'

and for r E JO, 1 [

I' II" 12 v(r, 0) = G(O, ra) + E,,(r) + I ( l, .- En(~

",B \ 2a n

Using the symmetry of G (see Proposition 25) we have

G(O, ra} = G(rO", 0) =;,2. - En(r) - J~ (i~12 - E"(~))d; -() n B 2a"

and hence that

r~

(8.104) , i 1'(1',0) = + B (EIl(~) - En(~ - ra))d~ . 2rJn

Since w(/" 0) = EIl ( 1).

(/1 dr) =

2)w(r,0) I'll (11 - 2) E Il ( I) nr ... - + ,- (/'.0) = -------- + - (r,O) r tr r (lr

from which from (8.103). integrating k(l'. xl with respect to r. we find that

dr. xl = 1'(1',0) + E,,(l) - EIl(,.x - a)

~r d +(11 - 2)j (EI/(I) - EIl(px - a)) ... f..:

o p

(8.105)

the formulae (8.104) and (8.105) therefore give us an explicit expression for 1'(1', x).so that, regrouping, we obtain

I\f + I)f - (. y ) G(x, y) = - ..... -- + l: (x - r) - E I vlx - . 20-n n v tl. I y I

(8.106) I'r ( I

+ (11- 2) I En( 1) - t.) fiX ~o \

Y \)) dp ).vl p

i(/" i~)2) ,. + E,,(C;) - j- de;. B __ (J,,/

We note that in the case II = 2 the first integral vanishes and that for n ?c 3, this integral converges for all (x, y) E B x B with yolO, since

.. . ( \' ) I [1 1 '.,(1) - ,., !'x - iy j ~ ~~ 2)~ F-~~l'" -I ~ 0(1') "' I' - 00

This example has shown us the part which we can extract from the method of images for more general problems than the Dirichlet problem for the Laplacian. Let us know how to use this method for the operator - Ll + k 2 :


Example 6. Green'sfunctionfor the operator -.1 + k 2. The operator -.1 + k 2 ,

where k > 0 is coercive on WI, 2(Q)for every open set Q of [R" for any dimension n whatsoever. This allows to solve in a unique fashion every problem of the form

for any open set Q of [Rn with the datal, (P. !/J not having too great a rate of growth at infinity. This allows us also to define a Green's function of the mixed problem relative to (Q, To, - .1 + k 2 ) for every open set Q.

Without developing a complete theory for the operator - A + k2 , we shall be content with some particular results.

Proposition 27. Let k > O. (1) For every tempered distribution 244 f on [Rn, there exists a unique tempered distribution u on [Rn such thaI

(8.107)

In particular, there exists a unique tempered distribution E on [R" satisfying

- (JE + k 2 E = (j in 01'([R")

where (j is the distribution at the origin. This distribution E is analytic on radial; it is the elementary solution of - A + k 2 given by

(8.108)

or again

(8.109)

k"'-2 IX "--3 E(x) = ----, . e-klxlt(t2 - 1)-2-dt

an(n - 2). I (n ~ 2)

E(x) = l (_"--_ ')i- I K!,. (klxl) rr 2rrlxl, 2 I

(if n ~ 3)

{O} and

where K denotes a Bessel function of the second kind (in standard notation); in the particular case n 3 we hal'e the particularly simple form

(8.110) 1

E(x) = --,-- ek!xl 4rrr x i

(2) Given P E [I, (j:-], iff E U([Rn), the solution u of (8.107) is in U([R") and

I II u i u :0:; k 2 I!f Ii If .

If I 0 on [Rn (unless f == 0 on [Rn) ill the

244 See Appendix "Distributions" Vol. 2. p. 506.

636

sellse that


inf ess u > 0 for all R > o. BIO. RI

(4) Iffis a distribution with compact support, then u has exponential decay of order k at infinity, that is to say

u(x) = o(e-k!x l ) when Ixl --> .J~ .

Proof Considering u and f to be tempered distributions we have (8.107) ifT their Fourier transforms a and / satisfy

~2U + k2 a = I that is ill

f + -k2

Given that if f is a tempered distribution, the distribution //(~2 + k2 ) is also a tempered distribution, there exists a unique tempered distribution u, solution of (8.107), which is the inverse Fourier transform of//((/ + k2 ). In particular iff = (),

the solution E is the inverse Fourier transform of (~2 + e) - 1. The formulae (8.108) and (8.109) are then deduced classically (see Schwartz [1]). Supposing thatI EO 9(lRn) we find that the solution u E (fjX (IR"), and that U and all of its derivatives are in the space L 1 (IR") II LX (IR"). If we multiply equation (8.107) by O(u), where {} E 'e l (IR), and integrate by parts, we obtain

fO'(UJlgrad uI 2 dx + k2 fU8 (U)dX = fI8 (U)dX

from which we deduce that if (J' ? 0,

(8.111 ) k 2 fO(U)UdX ~ JrO(U)dX .

Applying this result with Orr) = Irl p - 1 r (approximating with a sequence of increasing functions of class '(;1, if I ~ p < 2), we have

k2 flUjPdX ~ JlIUIP-1UdX ~ lifllullullr;l

from which we have the estimate of point (2). The result is then deduced by a density argument. Now applying (8.111) with Orr) = min(r,O) = - u-, we obtain

/.:.2 ru - )ldx ~ - ffU - dx ,

from which iff ? 0, J (u - )2dx = 0 and 11 ? 0. Point (3) is deduced by density and

use of Harnack's formula for subsolutions (see Proposition 16)245.

Col' In the present we can prove Harnack's formula directly as in the case of the Laplace operator.

*8. Elliptic Equations of the Second Order 637

For point (4), we notice, first of all, that u E 'ficx)(~"\suppf) and is given by

u(x) = <f(y), E(x - y) for x E ~n\suppf

We can develop the same method as that used in the proof of Proposition 2 of § 3 to obtain the estimate at infinity. Another method consists of reducing to the radial case: considering Ro such that

suppI c B(O, Ro) , M = max iu(x)l. Ixl = Ro

we have u(x)1 ~ u(lxl) for Ixl > Ro where u is the solution of

__ 1_ ~(r"-l ~.) + k2u = 0 r,,-1 dr, dr

il(RoJ = M, u( + xl = O.

Multiplying the eq uation by 2r 2 (11 - 1) ~~ , we obtain

d l / du)2 J (j; - (rn - 1 (j~ + (kr"- 1 u)2 = 2(n

From this, having noted that u ;" 0 and dii/dr ~ 0, we deduce that

du - + hi ~ 0 and ii(r) ~ M ek(R,,- r) .

dr o

Remark 3. The functions u E (6"2 (Q) satisfying - Ll u + k2 U = 0 are called metaharmonic functions; we can develop for these functions a theory similar to that for harmonic functions; for example, we can prove a formula of the mean (see Garnir [I]):

(S.112)

where

(S.113) 'In. '. k) ~ - 2r -- r(:;)l' r' ch(k"111 vn r(~ ~_) Jo

\ "-

In particular in the case n = 3

(S.114) k f u(xo) =----- u dy 4nr sh(kr) ,cB(x",I")

which also gives by integration

(S.115) u(xo) = 47[[kr~h(l~~--=- sh(kr)] txo,r) udx.

63R Chapter II. The Laplace Operator

Finally, let us give the Green's function oj" a half-space ~n - 1 X ~ + for the operator - Lf + k2 . We note that given f a tempered distribution on ~n with support in ~"-I x ~+, there exists a unique quasi-classical tempered solution u of the Dirichlet (resp. Neumann) problem

-lfu+k21/=f on ~n-I X ~+

u( .,0) = 0 (resp. ~'1/ (.,0) = 0\) on ~" - 1 .

(J.\n ,

In effect 1/ is a solution iff its odd (resp. even) in Xn extension LI to ~n by symmetry is a solution of

- Ifu + k2 ii = J on ~"

whereJis the extension off It suffices to apply point (1) of the Proposition 27. [n particular the Green's function of the Dirichlet (resp. Neumann) problem in the half-space ~n - I X ~ +

for the operator - Lf + k 2 is

G(x, y) = E(x - y) - E(x -.\') (resp. E(x - y) + E(x - .v))

where for y = (y', Yn), .0 = (y', - y,,). 0

Example 7. Green's fUl1ction oj" a Rohin prohlem jiJr the Laplacian. We consider the problem

(8.116) I ~. U = f on (U o +:XU = 9 (11

Q

on 1'.

This problem can be put into the form of a Neumann problem

L1I = - I on Q

(;n L !J on r

by the introduction of the operator

I ,11. (,r +:Xlli) + I :Xliii;] + d with LXi ,LX, (Xi

L

where 11; are the components of a vector field, defined on Q by the unit exterior normal on r. If G denotes the Green's function of this problem, the solution of (8.116) will be given by246

u(x) = J" G(x, ::)g(z)di'(Z) - f G(x, y)I(y) dy . r !}

)46 If it exists. this will be the unique solution orthogonal to the set of solutions of J" ~ 0 on (U

Q. - + 'Ow = () on I.


We shall specify the Green's function G when Q = iR" - I X iR + and IX constant. For IX = 0, the Green's function Go is given by the method of images (see §7.5):

Ga(.x, y) = E,(x - y) + E,(x - ji)

where 5' = (y', - y") if y = (y', y,,). We denote by G(IX, X, y) the solution of

(8.117) 1-" LlG(IX,' x, y) = 6(x - Yi, ,x E Q

3G en (ex,z,y) + exG(ex,z,y) = 0, ZEro

The open set Q being unbounded, we must specify a condition at infinity. For ex ~ 0 the null condition at infinity allows us to ensure the existence and uniqueness of the solution of this problem; this solution depends regularly on ex, as can be shown a priori but which will appear a posteriori in the explicit expression for GCct, x, y). Let us suppose that we can expand

G(IX, x, y) = I exmGm(x, y) . m~O

Then substituting this form in (8.117) and equating powers of ex, we have, for m ~

1 ~ .dGJx,y) = 0,

(JG m --(z,v) + Gm - 1 (z,y) = 0,

(in "'

xEQ

Z E r.

We note that for Z E r

so that

. DG I Sillce Ll -_-(x, y) = ° = Ll(2E,,(x - ji» for x E Q, we have

ex"

For m ~ 2, we have

?G Ll ~(x, yJ = 0 = .dGm _ 1 (x, y) for x E Q

ex" and

aGm cGm --(z,y) = --_-(z,y) = Gm-1(z,y) for ZEr; ex" en

(iG hence --"'- = G m - 1 .

ax"


Regrouping we have

I'G ~ - (y, x, r) ( .\1/ .

clG o -:;-- (x. y) + 2yE,,(x - n eX tl

m :?: 2

namely. after an integration in x"

G(CJ:, x. Y) = Gdx, y) - 2 JI" e,lx,,- t) EI/(x' - y', t + J'I/)dt x"

which we can rewrite

G(y, x. y) E,J,(' - y', x" - .1'1/) + EI/(x' - r', XI/ + .1'1/)

n(18) "-t- f

2 J . e'lx" + r" - 11 tl/(X' - r', t) dt ; .\:/1 + Yn

the reader should verify that this formula defines for y > 0 the solution of (8.117).

7. Helmholtz's Equation

Helmho/t::'s equation is the name given to the equation

(8.119)

or, more generally, to the equation

(8.120)

where k is a positive real constant (k > 0). This eq uation occurs as a model of numerous physical problems, notably in vector form (both u,f are vector fields) in the theory of the radiation from electromagnetic sources (see Chap. lA, ~4 and Roubine [IJ). The link with the ware equatioll

(8.121)

is clear: let us suppose that g(X, t) ek'f(x) on Q x IR, then u(x) is a solution oj' Helmholtz's equation (8.120) iffw(x, t) = eik'u(x) is a solutioll of the waFe equation (8.121): such a solution is a stationary ware. In the case of the equation of free waves

(8.122)

the stationary waves are the solutions with variables separable w(x, t) p(t)u(x)


of (8.122) satisfying

(8.123) inf sup Iw(x, tll < (X 247 .

x E Q t ,0 G;l

641

We notice that Helmholtz's equation is similarly involved in the method of the separation of variables in the study of" Laplace's equation (see § 7.3). In the case n = I, Helmholtz' equation

u" + k2 u = 0

is the equation of the vibrating string whose general solution is

u(x) = }.e ikx + Jie ikx.

As we have already seen above, the solutions that we shall consider will be complexvalued. We shall use, as for the case of Laplace's or Poisson's equation (see Definitions 1 and 2 of § 1) the terminology of classical solutions and of solutions in the sense of distributions as well as in the case of boundary value problems, those of quasi-classical and weak solutions (see Definitions 5 and 6f48. The Helmholtz operator is elliptic (with constant coefficients) with the result that every solution of (8.119) or of (8.120) withf analytic (resp. '60'.) is analytic (resp. <gX) (see Proposition 5 as well as Chap. V). Similarly, we shall have regularity results up to the boundary of solutions of boundary value problems with regular data. On the other hand, the example of dimension I shows immediately that the real solutions of" Helmholtz's equation (8.119) do not, ill general, satisfy the maximum principle.

7a. Radial Solutions. Elementary Solutions

Just as with the Laplace operator, we find that the Helmholtz operator is invariant under a Euclidean transformation. For that reason, it is especially interesting to study the radial solutions and the spherical solutions. Given l' defined on an interval I of IR1 +, the radial function u(x) = d I x!) is a solution of (8.119) on the annulus Q = {x E 1R1"; I x I E I} iff

(8.124)

Putting H'

(8.125)

1

d 2 1' ndr2 + r

dv + k 2 v = 0 on I.

dr

rIt! - 11', we transform equation (8.124) to Bessel's equation

d2w I dw ( ,2 (11 - 2)2) , ") + - - - + k ---- ~t = 0 . dr- r dr 2r ,

We shall use as the independent particular solutions of (8.125) the Hankel func-. H(l) k) d· . (2) k) tIOn .~ - 1 ( r an Its conjugate H 1- 1 ( r.

247 In effect IV(X, t) = p(t)u(x) is a solution of(8.122) iff p"(t)u(x) "0 p(t),Ju(x) namely iff p" = cp on!R and Ju = cu on Q for a certain constant c; the condition (8.123) expresses that p is bounded on If,! and this, in turn, imposes the condition c < o. 24" Equation (8.120) can be written in divergence form

Lu = .... f with I" = I,1':,(6;j (~(:) ..... k' .


We recall that

(8.126)

where .11, is the Bessel function of the first kind

(8.127) r' fIT . . f (r) = .. '- -- e" cos () sinZ,() dO

a /1\ ( 1) , 2>r(2)rx +:2 0

and IV, is the Bessel function of the second kind or Weber's function:

(8.128) f,(r)cos Xl[ - f ___ ,(r)

IV ,(r) = --- --- -.. -- --sm XT[

extended by continuity for all integersx 249

With this notation, the radial solutions of Helmholtz's equation (8.119) are therefore the functions

/I

(8.129) ( k )2 1 u(x) = ~f OBf 1 (klxlJ + pHf 1 (klxl))

where J., p are scalar constants. Let us now seek the elementary radial solutions of Helmholtz's equation. An elementary solution E is by definition a solution (in the sense of distributions) of the equation

Ll £ + k2 £ = 0 on 0:£",

or again, account being taken of Green's formula and the regularity of the solutions of (8.119), a function E E ~"X(Q;g"\{O}) n Lioc(Q;gt/) satisfying

(8.130) lIE + k2 E = 0 on Q;gt/\ {OJ

and for a regular bounded open set Q containing a

J. (:E di' + k 2 J" £ dx = I . <'[1 I'll f2

(8.131 )

This relation will then be true for every regular bounded open set Q containing O. In the case of a radial function E(lx!), the relation (8.131) can be written for Q = B(O, r)

r" 1 £'(r) + k 2 L p" - 1 £(p)dp

and since this relation is true for all r > O.

(8.132) lim r" - 1 E'(r) r ---10 0

24'1 Sec G. Petiau [I] whose notation we have adopted: iV, is sometimes denoted Y,.


Using the relation d/dr(r- a lJ~l)(r)) = - r-2lJ~J21 (r)250, we see that

(8.133) r" -1 :r [ 0 f- 1 Hr 1 (kr) J = _. (kr)'iB~l~ 1 (kr)

therefore

d [(k)!! -1 J lim r" - 1 ~d - 2 H}1~ 1 (kr) r~O r r 2

- lim rT Hbl~ I (r) r~O 2

i(n + O! 251

In r( ~~_1_)2T - 1

Using (8.129) and (8.132), we see therefore that the radial elementary solutions of Helmholtz's equation are the functions

(8.134)

where ). is an arbitrary constant and

(8.134)'

- ('n+l) n Jnr ~2 - (2k)T- I (1)

L~ )(r) = -i;;:(~-~ 1)T- .-;:. Hi ... 1 (kr) .

Let us explain

(8.135) case n 2,

(8.136) case n = 3, 4nr

where we have used Hi1(r) = - i J3~ eir. 2 nr

We can further find the case n = 3 directly by the method oldescent: in effect if v is a

I · f 8 I If' 1 dv . . f so utlOn 0 ( . 24), t le unctIOn w = - ~ IS a solutIOn 0 . r dr

250 See G. Petiau [I], page 8l. 251 Using, for example

dw 2 d~+kw=O.


Since we know the solutions ),eiir + lie ikr of (8.124) for 11 = 1, the solutions for

IJ = 3 are - (ie,kr + fle- ,k,); since lim /"2___ = - I, the radial elementary I... d (e ikr )

r r ~ 0 dr r lk solutions in the case IJ = 3 are clearly the functions- (i,e' r + (I - i.)e,kr).

4nr We note also that in the case 11 = 1 the radial elementary solutions are always given by (8.134) with

(8.137)

7b. Solutions with Spherical Symmetry

We shall now look for solutions of Helmholtz's equation (8.119) with spherical

.. (x) symmetry, that IS to say, solutIOns of the form u(x) = h(lxilJ\!~xl .

Making use of the expression for the Laplacian in spherical polar coordinates, (see ~ 1.4). we see that II is a solution of (8.119) iff we have

( ~2'o1(r) + 11 - I ddl,~ (r))J(G) + h(r) /l"Y(G) + k2!J(r)y(G) = 0 dr- r

where Aa is the Laplace- Beltrami operator of the unit sphere I. This equation splits into

(8.138) c) - -,- h = 0 r"

where (' is a constant. We know the first equation of the pair (8.138) admits non-constant solutions only if c = 1(1 + 11 - 2) with I EC N and then !/fp the space of solutions is of finite dimension (sec Proposition 6 of ~ 7). Transforming the second equation of the pair (8.138) by changing the dependent variable from h(r) to 1;(1') = r1 n --- 1 il(r}. we therefore have for c = 1(1 + 11 - 2) that 17 is a solution of Bessel's equation

d217 I dl7 ( (7.)2) elr:?: + r ell' + k 2 r 17 = 0

with

(8.139)

To sum up: the solutiolls with spherical symmetry oj"HelmllOltz's equatiol1 (8.119) are of the form

(S.140)


where for lEN *, Y, is a spherical harmonic function of order I in [R" and H; , 11; are the Hankel functions of order :x given by (/;.139).

Let us explain the cases n = 2 and n = 3. Case n = 2. Then qy? is of dimension 2 admitting for I =F 0 the base {eilO, e ilO }

where fJ is the polar angle fJ. In (8.139), :x = I. Hence the solutions with circular symmetry of Helmholtz's equation (8.139) are for I E N* the functions

(8.141) u(rcose, rsine) = VHj!)(kr) + /lH~2)(kr))(Ct.eiW + (Je- itO )

with A, /l, Ct., {J arbitrary constants. Case IJ = 3. Then :&1 is of dimension 21 + 1 admitting the base { Y7'; - I ,::; m ,::; I} defined by (7.64). In (8.139), 0: = I +1. Hence the spherically symmetric solutions of Helmholtz's equation (8.119) are the functions

(8.142) - ~ (1) (2) (I m(. X )) Ixl '().11 I + l (klxl) + f.1 H '+-l(klxl)) m~"'ICt.mYI 1;'1 .

We observe that for Ct. > 0, N,{r) has a singularity r- a and 1'-'.1,11') is bounded as r -+ O. It follows that

-~i-~()"H~l)(lxl) + /lH~2)(lxl)) = (A + /l) ..f,~lxl) + irA - /l) Na~lxi) Ixl2 - 1 Ixl 2 - 1 Ixl2' - 1

is in Lloc ([Rn) iff J. = /l or Ct. < tn + I; in particular, for if. given by (8.139), this is not true except for I = 1.

7c. Outgoing and Incoming Flux of Energy

Unlike the case of the Laplace operator where when n ;;?: 3 there existed one and only one elementary solution, null at infinity, in the case of the Helmholtz equation we have an infinity of such elementary solutions; it will be necessary to fid criteria to distinguish among them. We introduce now a first criterion enabling us to classify them into two categories by using the stationary wave which defines them. Let us consider in a general manner a wave, that is to say a solution w of the wave equation

a2 l-V --- - Jw + y = 0 on Q x [R. Bt2 '

We suppose that Q is a regular bounded set and w sufficiently regular to justify the

following calculations. We multiply the equation by w' a~v . ~ and 1l1tegrate over Q; ot

using Green's formula and the identities

: .. ~." ~ la~wl2 ct 2 ot

ai' Re grad w. grad -at

we obtain

i ~ f (lc:w/ 2

dt 2 f} of + 1 grad Wl2 dx - Re -;;- -:;-dl' + Re y-;-dw ) f aw a,t, f r'w

,'{] en ot {] of o.


By definition the quantity

EfAIt'· I) = ~ L (I(~i'(x. If + grad w(x, tl1 2 ) dx

is the energy of the ware in the domain Q at the instant r,

is the flux of energy of the outgoing It'at:e across cQ at the instant t, and finally the

f aIt, term - Re g(x, t)--:;- (x, t) dx is the energy applied to the domain Q at the

!~ uf

instant t. The energy balance can be written

f (1\\' Re g(x. I):; (x, t)dx

!} ('t

that is to say: the variation of the energy plus the outgoing flux of energy across aQ is equal to the ellergy applied at each illstant. We consider now a stationary wave w(x, 1) = ekr u(x) with g(x, t) = eikrf(x) and u is a solution of the Helmholtz equation (8.120). The energies are then constant (from which we have the name stationary wave):

1 ,. EQ(w, t) = - J (Igrad ul 2 + k2 !uI 2 )dx

2 [}

and the energy balance can be written

Re ik J. g~ ii d" = - Re ik f Iii dx . F!2 an ' Q

We can state:

Proposition 28. Let Q be a regular boullded open set and u E ~2 (Q) II ~~ (Q) a classical solution or Helmholtz's equation (8.120) with IE L 1 (Q). The elleryy ffux going (Jut across ?Q

(8.143)

of the stationary wave eikr U(X)252 is equal to ---- Re ik r fudx. III particular, iff == 0 JQ

Oil Q, then I c!} = O. o

252 [<,{,(u) is the energy flux of the wave C ikt u(x): some authors associate with 11 the wave e - ikt 1/(x); in this case the energy flux will have the opposite sign and the terms outgoing/incoming will be reversed.


The precision given by the regularity in the classical theory253 justifies the calculations involved in the proof of the proposition. We have the corollary:

Corollary 5. Let Q be an arbitrary open set in !R", K a compact set in Q and u a solution of Helmholtz's equation (8.119) on Q' = Q\K. Then for every regular bounded open set Q o with K c Q o c 0 0 c Q, the energy flux IOilo(defined by (8.143)) is independent ofQo.

Prooj: We notice first of all that u E Y§ x (Q') and /Wo c Q' with the result that Iwo(u) is defined. Let QI, Q 2 be two regular bounded open sets satisfying

K c Q; c Q; c Q; there exists a regular bounded open set Q o such that

Q I u Q 2 C Q o c Q o c Q. Now applying the proposition with the open set Qo \0;, which is relatively compact in Q'; we have I"(Qn\fl,)(u) = 0; but

c(Qo\Q;) = (lQouaQ; and II'(Q" Q)u) = Iwo(u) - lrw,(u) ,

from which we deduce that

1, 2 .

This leads us to pose

o

Definition 9. Let K be a compact set in [R", Q an open set containing K and u a solution of Helmholtz's equation (8.119) on Q' = Q\ K. We shall call 254 by the name energy flu;>: of U to the quantity

i au flu) = Re ik ~-- lid,'

,'no en

for a regular bounded open set Q o with K c Q o c Q o c Q (J(u) = fu!lo(u) is independent of Qol. We shall say that the flux energy of u is outgoing (resp. incoming) if I(u) ? 0 (resp. I(u) ~ 0); we shall say also that u defines an outgoin(J (resp. incoming) wave2 55,

In an obvious way I (iil = - J (u): the energy flux of u is outgoing iff that of ii is incoming. In particular, if u is real, I(u) = O. In the terminology and notation of Definition 9, we have not involved the given sets K and Q; in effect it is clear that J(u) depends only on the values of u in the neighbourhood of K in [R"\K; also it does not depend on K in the measure where, given another compact set, K j c Q, if we can extend u to a solution U I

of the homogeneous Helmholtz equation on Q'j = Q\(K j n K), I (u I) = I (u). The reason for this independence is the analyticity of the solutions of the homogeneous Helmholtz equation: on every connected component of Q'j meeting Q', the extension U I of u is unique; also for a connected component Q~ of Q'J not meeting Q', we have Q~ n KI = 0 and hence if Q j is a regular bounded open set with

253 We can give a formulation within a framework of a weak solution in WI. 2(l'2).

254 By abuse of language. 255 As we have said, this terminology is linked to the choice of the stationary wave associated with u, here eik'u(x); it will be reversed for the choice e - "'u(x).

648 Chapter ll. The Laplace Operator

K 1 C Q 1 C Q 1 C Q, U l is a solution of the homogeneous Helmholtz equation on Q 1 and i"f2)ud = o. To clarify these notions we consider the radial case. Let u(x) = u(lxl) be a solution of the homogeneous Helmholtz equation on an annulus

Q' = B(O,rtl· B(O,l"o) (0 ~ ro < r 1 ~ +

The energy nux is given by

flu) Re(ika/' - 1/,'(r)[(r)) .

We know (see (8.129)) that

n

(k)'i 1(r) = ~,

1

UHhl~ 1 (kr) + t<H'1~ 1 (kr)) 2 2

from which, due to (8.133), we have

r n - 11"(r) n, (l, (2)

- (kr)2 (fH i (kr) + pH I (kr)) and

Now

N ,(r).9", t 1 (r)

dN, d.1, .1>(r)d~ (r) - N,(r) dr- (r) =

2

nr

so

(8.144)

To sum up: the energy flux of the radial solution given by (8.129) is given by (8.144);

hence the energy nux is outgoing (resp. incoming) iff i pi? I i.1 (resp. I p I ~ I i.I). In

particular with the notation (8.134), the srationary wal'e ek' qn)(x) (resp. ekl LLnl(x))

is incoming (resp. outgoing) and a superposition eik'(i.LLn)(xj + pLLn)(x)) is ()utgoiny

iff the intensity I p I or the outgoing wave pe ikl Lt"\x) exceeds the intensity I i·1 of r he incoming wave ie ikl L~I)(x).

elkl'l cik[l'-Ix l ) In the case 11 = 3, LLnl{x) = - - - and C'k! Lk')(X) - --, is the 111-

4nlxl 4nlxl eik[l - 1-' II

coming wave while eikl L}l) (x) - - is the outgoing wave. The former 4nlxl

wave is also said to be convergent and the latter to be dil'er!Jent. This terminology can be employed instead of incoming and outgoing respectively.

7d. Sommerfeld's Radiation Conditions

The notion of an outgoing or an incoming wave allowed us to classify the radial elementary solutions of the Helmholtz operator but we did not characterise


the solutions L~") and L~"), although their importance was made apparent. This characterisation will be effected because of conditions of uniqueness, formulated by Sommerfeld, called radiation conditions.

eiklxl In the case n = 3, the elementary solution Ll3 )(x) = - -- immediately

4nlxl satisfies

(8.145) eLk3) eiklxl -~ -(x) - ikLPI(X) = ........... -.

or 4nlxl2

Use of the asymptotic expansion of Hankel functions 256

(8.146) when r -> 00

and the differentiation formula (8.133) show that for arbitrary n

:r G)" H~1)(kr)- ikG), H~l)(kr) = - kG X [H~l! 1 (kr)

+ iH~II(kr)] = o( 1 ) r' + ~

and hence using the definition of the Lin),

(8.147) eL(n) ( 1 ) :....~. k_ (x) - ikLLn)(x) = 0 ----

(jr I x 1"1-'-when Ixl -> 00.

GivenIa distribution with compact support on [Rl", the distribution u = LZ * fis a solution (in the sense of distributions) on [Rl" of the Helmholtz equation (8.120) and hence u E 'gx'(Q') with Q' = [Rl"\suppI and solution on Q' of the homogeneous Helmholtz equation (8.l19). We point out the

Lemma 7. Let I he a distrihution with compact support 011 [Rl". Theil U = L~n) * f has the asymptotic hehaviour

(8.148) au ( I ) -. (x) - iku(xl = 0----iJr 'I x I!!. __ ~ ___ l when Ixl -> 00 .

We shall prove this lemma later, this condition permitting the characterisation of the solution L[") * f of the Helmholtz equation (8.120). Moreover we shall consider two other weaker conditions,

(8.149) Du ( 1 ) .~ (x) -. iku(x) = 0 - -.-ur Ixl'!i.l

when Ixl .+ 'x;

(8.1 SO) limf Icu_ikUI2dY=0. r - 'Xi f~8(O, r) cr

"" SceG. Petiau[1]'p.133.


We say that (8.149) (resp. (8.148), (8.ISO)) is the incoming classical (resp. strong, weak) radiation condition of Sommerfeld. It is clear that the classical condition (8.149) implies the weak condition. since denoting

f;(r) = sup I(~U (x) - ikU(X)IIXI"~i-~ , I x I 3 r cr

we have

f Itu 12 '~. - iku d~' ~ c;(r) . ('B(O.r) U

We can similarly define the (Jut?}oing radiation conditions of Sommerfeld " strong, classical and weak - by

(8.148b) , (x) + iku(x) = 0 ., --i"u (I I ') CI' Ixl"+1/

when Ixl -->x .

(8.149b) iku(x) = 0(" ,"I when Ixl --> ex . / I ) \Ixl/l.}-

(8.ISOb) lim f li'l! + ikul2 di' = O. r ----> J (~B(O, r) i';r

The terminology incoming and outgoing corresponds closely to the notions defined previously thanks to the

Proposition 29. Let K be a compact set and u a solution of the homogeneous Helmholtz equatio/l 011 Q' = K. Let us suppose that u satisfies the incoming (resp. outgoing) weak (8.ISO) (resp. 8. I 50b) radiation condition of Sommerfeld. Then the energy flux of u is incoming (resp. out?Joil1fJ); more precisely

(8.ISI) -J(u) (resp. J(u)) = k 2 lim r lul 2 dy. r ---l' .1_, JtB(O. r)

Proo{ We have for all r sufficiently large

J(u) = Reikf (:lAudl' = Reikf (Ill! - ikU)lld)' - k2 f luI2d). i1B(O. r) (r fB(O. r) \ I'r ,'B(O. r)

Now by Schwarz's inequality

I f ('au ) I ('" 111U 12 )t(f ')1 Re ik -::;- - i!w udi' ~ k J 1 - iku d}' IUI 2 d}' ('B(O.r) cr , <'B(O.r) cr ,'B(O.r)

first of all, we deduce from these two relations

(~1.IS2) k (LBIOr) luI 2 d}Y ~ (LBIO.J:~~ - ikul2 diy + ,/i I(u):

then (8.ISl). o We shall now state the result of ciJe ullil.juenes", 0/ exterior Dirichlet or IV eumallll problems with a radiation condition at infinity.


Proposition 30. Let fl' be a connected open set containing the exterior of a compact set C?f[Rn and whose boundary r, ifit is non-empty257, is regular ofdass (fa'I+'. Then the function u == 0 on Q' is the unique classical solution of the problem

(8.153)

f Au + k 2u = 0 on Q

l u = 0 (resp. ~~ = 0) on r

satisfying an (incoming or outgoing) radiation condition at infinity.

Proof We take u to be solution of the problem and we have to prove that u == O. We note that, because of the hypothesis of regularity, u E (fa'; (Q') and the energy fl ux of u can be calculated on r; the hypothesis of a homogeneous Dirichlet or Neumann problem hence implies that /(u) = O. From Proposition 29, we deduce

(8.154) }i~x IB(o,r) lul 2 dy = O.

Let us fix 1'0 > 0 such that B(O, 1'0) ::::J K = [R"\fl'. For I" ~ 1"0; Green's formula (or Gauss' theorem) assures us that for every solution L of the homogeneous Helmholtz equation

(8.155) i (' iJL cu) C(L) =. u ~-:;; L d}' is constant for I' > 1'0'

fB(O. r) cr cr

Let us suppose that u were to satisfy an incoming condition. We can write

IC(L)I = I ( u(~L - ikL)d i' - f L(~~ -- ikU)dl'l J"B(O. r) cr ,'B(O. r) ,( I'

:::; max I(:L - ikLI flUid)' + max ILl f leu - ikuldi'. cB(O. r) cr c8(0. r) ,'B(O. r) ?B(O. r) (ir

From Schwarz's inequality we have that, for every function Ii'

(8.156)

From the radiation condition and (8.154), we can conclude that C(L) 0, if we choose a function L satisfying

(8.157) L(x) = 0C~'(ln~2-i} ~:~ (x) = O(I~~I;~~I) when Ixl -> x .

First of all let us choose L = Lkn ,. It is clear that (8.157) is satisfied and hence C(L) = 0 which can be written

o for r > 1'0

"., We include the case Q' = ~n; then r 0 as well as the boundary condition in (8.153),


where

vo(r) = .;,1:-1 f. ud~' = f u(m)da. I ,oB(O,,) 1:

We have therefore ['0(1') = cL~n)(r). But from (8.156) at (8.154), we have

11 - 1

lim ri 1'0(1') = 0 .

1l.- 1

Taking account 0[(8.146) and (8.134)' we·find that lim r- 2 - Lln)(r) # 0 and hencc

that c = O. Thus we havc proved that

tU(ra)da = 0 for all I' > 1'0'

We now carry out the same reasoning with

n

L(x) = (II. )2 . 1 ll~l)(kiXI)Yl(~)' Ixl Ixl

putting

1/(1') = t u(m) Y/(a) da

where Y/ is a spherical harmonic function of order I and'Y. is given by (8.139). We know (see (8.140)) that L is a solution of the homogeneous Helmholtz equation; it is

dear from (8.146) that L(x) = O(---~-=I.··) . Ixl2

tL (1 \ . Also - (x) = 0 ." ' ' ) since ar Ixl!l~

~ ('., l_lJ~ll(r)) = dr n_

1'2 J

n

Therefore /,/(r) = c( ~)2 " J H~J I(kr) and we conclude that c = 0 as above.

The set of spherical harmonic functions being complete (see Proposition 6 of ~ 7) we conclude that

u(m) == 0 for a!1 a E 1.' and all r > ro .

The connexity of Q' and the analyticity of u leads to the conclusion. o Remark 4. The above proof shows in fact that a solutioll U oj'the homogeneous Helmholtz equation on a connected set with complementary compact set, whose energy flux is zero and which satisfies a radiation condition is identically zero. The regularity of the boundary of the open set is involved only to ensure that the null condition on the boundary implies the vanishing of the energy flux. Going back to


the proof of Proposition 28 we see this is true, without a regularity assumption, if u

is a weak solution relative to Wlo'c2 (Q') of (8, 153); we then have the same uniqueness result. In the case of the Dirichlet problem with r a Lipschitz boundary, this is called Meixner's uniqueness condition. 0

7e. Retarded Potentials

Given a distributionfwith compact support on [R", we call the distribution Lin) *f the retarded potential off As a corollary of Sommerfeld's uniqueness theorem we have

Proposition 31. Let f be a distribution with compact support in [R". The retarded potential off is the unique solution of the Helmholtz equation satis/ring the outgoing weak radiation condition. 0

The uniqueness is Proposition 30 with Q' = [Rn. We have already stated in Lemma 7 that Lin) *f satisfies the incoming radiation condition and hence

Lin) * f = Lin) * J satisfies the outgoing condition. Now we give the

Proof of Lemma 7. We can always write f = I ~a!~ for a family (fa) of 1'1 ·oS m ex .

bounded continuous functions with compact support contained in .8(0, ro), For Ixl > ro we have

u(x) = <flY}, Lkn)(X- y)

Therefore

where

and

with

au -;;- (x) + iku(x) cr

[ i oaF -J I ~ (x - Y).f~(y)dy + R,(x) lal oS til w ex _

F

x· wlx) = _J .

Ixl

Now when Ixl ---+ OC, we haVe;;wj(x) = o(-+rn)' . uX Ixl


e- ik I x I In the case 11

_ e-iklxl -------, F(x)

4nlxl from which when

4nlxl2

?X ( 1 ) r'x' F(x) = 0 [;(2

In the case 11 arbitrary, we shall verify, with the help of the properties of the Hankel functions, that

?'F (I) (1x' (x) = 0 -I-~-I'I+ [ .

\.\ 2

The lemma is then easily deduced as in Proposition 2 of §3.

We take note of the

o

Corollary 6. Let u be a solution or the hOlnoyelJeous Helmholtz equation 011

Q' = ~n\ K wit h compact K. Then the outgoiny (resp. incominy) strong (8.148b) (resp. (8.148)), classical (8.149b) (resp. (8. i 49)) and weak (8. i SOb) (resp. (8.150)) radiation conditions are equivalent.

Proot: It is enough to prove that the weak condition implies the strong condition. Let p EO 9(~n) with p = 1 in the neighbourhood of ~n\Q'. We put r = (I - p)u,f = ,11' + k2 v. Since [(x) = u(x) for Ixllarge, [C satisfies the weak radiation condition and hence is the retarded potential ofr: From Lemma 7, l'

therefore satisfies the strong condition and it is the same for u. 0

Corollary 7. Let Q be a regular bounded opell set and u a solution or the

homogeneous Helmholt: equation 011 Q' = ~"\Q. Suppose U E (('~(Q'), thell there exists a unique decompositiolJ U = U I + u 2 011 Q' where III is a solutiolJ of the homoyeneous Helmholtz equatio/l on ~n,

112 is a solution of the homoyeneous Helm/wit: equation 011 Q' alld satisjies the oUlyoil19 radial ion conditioll.

Proof Let us first prove the umqueness; considering two decompositions, u = !/1 + L/ 2 = 1'[ + 1'2' we shall have 1'[ - u[ = U2 - i'2' Since L/2' [12 satisfy the radiation condition, it is the same for 1'1 - U I which satisfying the homogeneous Helmholtz equation on the whole space IR;n is therefore the null solution. Hence ['[ = HI and i'2 = til'

For existence, let us put

f lL/(t) : Lkn-) (t -- x) - (~(tlL~n)(t - Xl] d~,(t) r ell ('/1

that is to say that u 2 is the retarded potential of the distribution with support in r tiu il

f = .:;- di' +~- (ud y') (see Prop. 5 of§3, or Def. 2, §3) ell en

§s. Elliptic Equations of the Second Order 655

Now from Green's formula

U 1 (x) = U{X) - U2(X) = f. [u(t) .. ~ ... LT") (t - x) - ~u (t) LI/li (t - X)] dy(t) JCB(O. r) an on

for all r such that B(O, r) :::) Q and x E B(O, r)\!2. o

In the neighbourhood of 0, L~n)(x) ~ En{x) is an elementary solution of the Laplacian. It follows that all the results on the local regularity of Newtonian potentials (see §3.2) are valid for the retarded potentials. In particular, if f is a Radon measure with compact support, its retarded potential u E Lfoc([Rn) for all

n p < + and is given by

(n - 2)

u(x) = f. L~n) (x - y)df(y) a.c. x E [Rn . Jw In the case n 3, u E Lfoc([R") for all p < 3 and

I -iklx - yl

u(x) = - : 1 _ 'I df(y) ; H' n x }

in particular

lu{x)1 ~ f. dlfl(y) a.e. x E [R3 ; Ju<, 4n lx - yl

in other words, in the case n = 3, the retarded potential of a Radon measure with compact support is dominated in modulus by the opposite of the Newtonian potential of the variation of this measure (that is to say the modulus of the density whenf is with density). We can also define simple and double layer retarded potentials given in the case n = 3 by

f e-iklz-xl

u dx) = - -4 I ~---=----'-I (p(z) dy(z) r n ~ x

f . -·kl"- I(z - x).n(z) u2 (x) = (lklz - xl - l)e I - X 7 • cp(z)dy(z)

r 4nl_ - x

where r is the boundary of a regular bounded open set Q and cp E (eO(r). We shall obtain the same regularity for these retarded potentials as we did in the case of the Newtonian potentials (see §3.3). In effect denoting by u7, ug the simple and double layer Newtonian potentials

u 1 (x) = u7 (x) + IR 1 (z - x)cp(z) d)'(z)

u2 (x) = ug(x) + L R 2 (z, x)cp(z)d}'(z)

656

with


1 _ . iklx!

4n xl

R 2(:, x) = [Ukl: - xl - 1)c- ikl= xl _ 1] ~z _-_ ~)'-'1~ - 4nlz - xl

The kernels R1(x), Ixl grad Rtfx), Ix- :IR 2 (z, xl are bounded. From Lebesgue's theorem we deduce that tI I - u? EO "6'1 (IR") and tl2 - u~ E (6'°{IR"). We can develop an integral method of solution of Dirichlet and Neumann problems (interior and exterior) for the Helmholtz equation as in the case of Laplace's equation (see Chap. XIB §3 and Kleinmann-Roach [1]). Observe, however, that unlike the case of Laplace's equation, there is, in general, neither uniqueness nor existence ()j" the solution or the interior Dirichlet problem: this obviously complicates the discussion of Fredholm's method. 0

Remark 5. In the usual way in physics, we call the retarded potential of a given charge density pix, t) in IR~ x IR,

(8.158) u(x,t) =_1 r, ____ ~ :;p('x"t--I:'-~X'I)dX' 4nJRlix xl (

(where c is the speed of light) which translates the fact that the (density of) charge at (x', 1') influences the point (x, t) only if Ix - x'i = crt - n. In the sequel we take c = 1. We denote by ~I(X. k) (resp. rJ(x, k)) the Fourier transform in t of u(x, t) (resp. pix. t)). supposing p is such this has a meaning. We have

u(x. k) - r u(x. t)e- ikl dt = J~ JrR] HJ X

e-- ik(r +!x -- x'!)

- --- - - -, ;--_. pix', r) dx' dr 4njx - xi

f e ik!x-- x'

=----:- --=----; pix', r)e - ikr dx' dr : 1<' x J.l 4n I x x I

from which, with (8.136):

18.159) u(x. k) l~fl) * ( - /i( .• k)) ,

which corresponds to the definitio/J given at the beginning or this sectio/1 of the retarded potential for fix) = - p(x, k). We verify that since u is a solution of Helmholtz's equation

(A + k2)U(X, k) = - pix, k) ,

u is a solution of the wave equation

Au = P .

Denoting by £3 the elementary solution of the wave equation with support the


future cone (see Chap. XIV, §3 formulae (3.45), (3.46)) which we write:

- I ,-2 2 £3(X,t) = ·········}(t)o(t - x)

2n

the retarded potential (8.158) can be written

(8.160) u = £3 * p. x. ,

1 b(t - Ixl);

4nlxl

657

A particular case of interest is that of the retarded potential due to a charged point particle, of charge e whose motion is given by a function xa(t) with Xo E <e I (IR, 1R 3 )

IdXO I and dt (I) < c = 1. The charge density pix, t) is then given by:

pix, t) = eb(x - xoU)) .

Let us calculate the retarded potential of such a charge density. For all rp E g(1R4) we have

(8.161) {

(£3 * {J, 4?) = (£3(X', r') ® pix, r), rp(x' + X, r' + r)

- ~ r _1_" rp(x' -I- xo(r), Ix'i + r)dx'dr. 2nJH421xI

We now make the change of variable 0: (x', r) -> (x, t) defined by

{ \" ~ < + xo(r) 1 - Ix I + r.

The Jacobian J(O) of this transformation is

O ['o(r).x' J( ) = 1 - 171' where

We verify that this transformation is invertible, with inverse:

fr ': {x' = x - Xo(T x .,)

r = t - Ix - xo(rx.,ll

where T", IS the solution of the equation t - Tu = !x - xo (rx .,lI258. Hence

(8.162) J(O)~ , ~otr",Ux - X.Q(rx • .'.n)~ 1

Ix~ xo(Tx.tll

VO(Tx.,);(~~x .. ;()(T~.t))) ~ I

258 Thus (xo('x.,), 'x.,) are the coordinates of the trajectory of the partide considered with the past cone with apex (x, t). that is to say the set {(x'. t') EO [R4; t - t' = Ix - x'i}.


and (8.161) may be written in terms of the new variable 11:

- e JM 1 (E 3 *p,<p) = -- '. ------ <p(x,t)J(U) ldxdt, 4n H' ,x - xo(Tx.,ll

which shows that the retarded potential of the charge density eCi(x - xo(tj) is given, with (8.162), by

(8.163)

u(x, t) = E3 * P x. t

I! ----- ---J(Or 1

4nlx- xo(Tx.tH

1 o

x 254

4n

Review of Chapter II

As we indicated at the beginning, the primary objective of the chapter is to present the classical theory of thl! harmonic operator, of harmonic .timctions and of the Newtonian potential. This has been attained in §§1, 2 and 3. We have applied that to the "classical" Dirichlet problem, i.e. the Dirichlet problem in a "good" open set in spaces of "regular enough" functions. This was carried out in §4, then followed, in this same section by the solution of "sharper" problems: unbounded or irregular open sets, other boundary conditions .... To proceed much further we have to introduce the notion of capacity; this is done in §5 (which may be omitted on a first reading). Various extensions of rl!iJularity can also be obtained; this is the object of §6 (which can similarly be omitted on a first reading). All of that is based (more or less directly) on the PrincipiI! of the M aximwn (a systematic exposition of which will be found in Vol. 2, Chap. V, §5). Without using the methods of functional analysis (introduced in Vo!' 2), other methods are indicated in § 7 (certain of which are important for numerical applications). Finally, §8 gives various extensions of these ideas to more general elliptic operators of the second order. These two sections §§ 7 and 8 can be omitted initially by the reader. Methods based on the theory of holomorphic functions (specific to two dimensional problems) will be seen later in the context of numerical techniques.

20" This retarded potential is called the Licnard Wiechert potential.

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