Mathematical Analysis and Numerical Methods for Science and Technology || Examples in Electromagnetism and Quantum Physics

Chapter IX. Examples in Electromagnetism and Quantum Physics *

Introduction

The tools developed in the preceding chapters permit the solution of diverse problems in mechanics and physics. In this Chap. XI we give a general survey of applications drawn from two domains, electromagnetism and quantum physics, making particular use of the notions of operator and spectral theory from Chaps. V, VI, and VIII, and the variational methods from Chap. VII in the framework of Sobolev spaces or Hilbert spaces constructed in an analogous fashion to Sobolev spaces. We shall give other examples in the following chapters, in particular from the domain of fluid dynamics (Tricomi's equation, Chap. X) and neutron mechanics (Chap. XII, 6).

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

Part A. Examples in Electromagnetism

1. Basic Tools: Gradient, Divergence and Curl Operators

1. Introduction. Definitions (Gradient, Divergence, Curl)

The problems of electromagnetism of which we have given examples in Chap. lA, 4 (and, moreover, many other problems in physics and mechanics such as those of hydrodynamics using the Navier-Stokes equations (see Chap. lA, 1)) make it necessary to introduce in an essential way the linear differential "vector" operators generally called gradient, divergence and curl which we shall study in this section in order to deal with the problems in electromagnetism considered in later sections. Let Q be an open set in IRn (as is usual in applications, n = 2 or 3, but for the sake of generality we assume that n can be any positive integer). For every v E .!?tl'(Q), we put grad v = (ov/ox;), i = 1 to n, which defines the linear differential operator denoted by grad 1 (for gradient) from .!?tl' (Q) into .!?tl' (Q)". 2 We then define the linear differential operator denoted by div 3 (for divergence) from .!?tl'(Q)" into .!?tl'(Q) by

div v = itl :~: for all v = (v l , .. , Vn) E .!?tl'(Q)" . In the case where n = 3, for all v = (Vl' V2 , v3 ) E .!?tl'(Q)3, we put

(1.1 )

which defines the linear differential operator, denoted by curl4 in .!?tl'(Q)3. In the case n = 2, we put

oV2 OV l 2 curlv = - - - forall v = (V l ,V2)E.!?tl'(Q) , OX l OX2

(1.1)'

1 Or alternatively V (nabla) or D. 2 Throughout this section, we use the shorter notations !?il'(Q)" (or !?il(Q)" or e(Q)") in place of(!?il'(Q))", (resp. (!?il(Q))", (e(Q))"). 3 Or again V. or !?il. 4 Or again VA, or "rot" (for rotation).

202 Chapter IX. Examples in Electromagnetism and Quantum Physics

thus defining the linear differential operator, also denoted by curl, from ~'(Q)2 into ~'(Q). We also need to introduce, in the case n = 2, the linear differential operator, denoted by Curl, from ~'(Q) into ~'(Q)2 defined by:

(1.1 )" Curlv = (::2' -::J for all v E @'(Q).5 The relations between these various differential operators and their principal properties can be proved directly from the properties of the exterior derivative of differential forms in differential geometry (see for example Choquet-Bruhat [1], Abraham-Marsden [1]). We shall not develop here this differential geometry point of view, but will content ourselves with noting in passing the properties so obtained. We simply remark that for n = 3, the diagram:

(or on replacing ~'(Q) by function spaces:

Cflk + 3(Q) ~ Cflk + 2(Q)3 ~ Cflk + 1(Q)3 ~ Cflk(Q) kEN

is such that

thus:

curl grad v = 0

divcurl v = 0

'

I. Gradient, Divergence and Curl Operators

In the case n = 2, we have diagrams:

(or again

with

thus:

curl grad v = 0

divCurlv= 0

'tIv E !!&'(Q) 'tIv E !!&'(Q? ,

{ 1m grad c ker curl 1m Curl c ker div .

203

kEN)

Here the operator curl appears as the (formal) transpose of the operator Curl, that is to say:


which is a Hilbert space for the norm:

IlvIIH(diV,Q) = (11v1l 2 + IldivvIl2)1/2. We denote by Ho(div, Q) the closure of C(Q)" in the space H(div, Q). In the case where n = 3, we also define:

(1.3) def H(curl, Q) = {v E U(Q)3, curl v E U(Q)3} which is a Hilbert space for the norm:

IlvIIH(curt,Q) = (1lvll l + IlcurlvI12)I/l, and we denote by Ho(curl, Q) the closure of C(Q)3 in H(curl, Q). In the case where n = 2, we similarly have:

(1.3)' def H(curl, Q) = {v E U(Q)2, curl VEL 2(Q)} , a Hilbert space for the norm (11v11 2 + IlcurlvI12)1/2, and we again denote by Ho(curl, Q) the closure of C(Q)2 in H (curl, Q). Otherwise, in the case where n is an arbitrary positive integer, we use the notation:

_ def C(Q)n = {cpl!] , cp E C(~n)n} .

The principal trace properties of the spaces H(div, Q) and H(curl, Q) are given by the following two theorems:

Theorem 1 (Trace theorem for H(div, Q. Let Q be an open set in ~n (n E N*), with a bounded, Lipschitz boundary r. 8 Then: i) the space C(Q)" is dense in H(div, Q); ii) the trace map, Yn: v f-+ v. n 1/ defined on C(Q)n extends by continuity to a continuous linear mapping-still denoted by Yn~from H(div, Q) onto H- 1/2(r); iii) the kernel ker Yn of this mapping Yn is the space H o(div, Q). Theorem 2 (Trace theorem for H(curl, Q. Let Q be an open set in ~n, n = 3 or 2, with a bounded, Lipschitz boundary r. 8 Then: i) the space C(Q)n is dense in H(curl, Q); ii) the trace map, Yt: v f-+ v II n I/O defined on C(Q)n extends by continuity to a continuous linear mapping-still denoted by Yt-from H(curl, Q) into H - 1/2(r)3 if n = 3 and into H- 1/ 2 (T) ifn = 2; iii) the kernel ker Yt of this mapping Yt is the space Ho(curl, Q). We first note that in the case n = 2, Theorem 2 reduces to Theorem 1 in the following way: let (X be the mapping defined on ~2 (or 1[2, or L 2 (Q)2) by: (1.4)

8 Note that Q is not necessarily therefore bounded. We assume further (and often implicitly in what follows) that Q is locally situated on one side of r. 9 Here n denotes the unit normal to r, orientated towards the exterior of Q. 10 In the case Q c 1Il 2 , one can replace v II nil by writing ['. rll' where r is a unit vector tangent to r.

1. Gradient, Divergence and Curl Operators 205

This mapping has a 2 = - I (where I is the identity on the space considered) and (av, aw) = (v, w) for all v and w

(where ( , ) denotes the scalar product on that space); hence the mapping a is unitary in /R 2 (or 1[2, or U(Q)2). Furthermore, for all v = (VI' V2) E L2(Q)2 (or similarly E .@'(Q)2),

I iJVI iJv 2 d' d' I cur av = ;;-- + ;;-- = IVV; Ivav = -cur v. uX I UX 2

Lastly, if n = (nl' n2) denotes a unit vector normal to r, an = (- n2' nl ) is a unit vector normal to n, thus tangent to r. We thus have v E H(curl, Q) if and only if av E H(div, Q), and hence Theorem 2 follows at once from Theorem 1.

Proof of Theorem 1. i) Let WE H(div, Q) be orthogonal to '@(Q)": ~ -(1.5) ((w, v)) = (w, v) + (div w, div v) = 0 "Iv E '@(Qt .

der Put Wo = div w, and denote by wand Wo the extensions of wand Wo to /R", which are 0 outside Q. Again denoting by ( , ) the scalar product in L 2(/R")", or L2(/R"), we obtain from (1.5):

(w, q + (wo, div E '@(/R"). This implies that w = grad Wo in .@'(/R") and thus Wo E H I (/R"); hence (see Chap. IV), Wo E HA(Q). Since '@(Q)is dense in H6(Q) there exists a sequence (q>k)k EN' with q>k E .@(Q) which converges to Wo in HI (Q). Then for all v E H(div, Q),

((w,v)) = lim (gradq>k' v) + (


It then follows that the mapping Yn: v 1--+ v. n Ir defined on ~(Qt equipped with the norm of H(div, Q), has values in HI/2(r) and is continuous. It thus extends by continuity to the space H(div, Q) since ~(Qt is dense in H(div, Q) (from point i)). Denoting again by Yn II the trace map so defined from H(div, Q) into H -1/2(r), we show that Yn is surjective. Let /1 E H - 1/2(r). Then (see Chap. VII, 1), there exists u E HI (Q) as a solution to the Neumann problem:

def

-Ju + u = 0 III Q

au I - /1 on r. on r

The function v = grad u is then such that v E H(div, Q) and YnV = /1, and point ii) follows. It will be noted that from the denseness of ~(Qr in H(div, Q), one can deduce from (1.8) the inequality: (1.9) IIYnvIIWl!2(r) ~ IlvIlH(div,Q)' and from (1.6) the "generalised" Green's formula: (1.10) (v,gradq + (divv,q = lr> ,

"Iv E H(div, Q), q> E HI(Q) , where the brackets denote the duality between H - 1/2(F), H 1/2(r). iii) Let W E kerYn, orthogonal to ~(Qr in H(div, Q): (1.11) ((v, w)) = (v, w) + (div v, div w) = 0, "Iv E ~(Qt .

def Putting Wo = div w, (1.11) implies

w = grad wo'

Thus Wo E HI(Q), and we can apply Green's formula (1.10) with q> = Wo and v E kerYn; then (1.11) gives:

((v, w)) = = 0, so that w = 0, whence iii) and Theorem 1. o To demonstrate Theorem 2 for n = 3, we shall have to make use of the following lemma (see Girault-Raviart [1J). Lemma 1. Let Q be an open set in 1R3 with a bounded, Lipschitz boundary r. Let u E H(curl, Q) such that: (1.12) (u, curl q - (curl u, q = 0, V q> E ~(Q)3 . Then u E Ho(curl, Q). We shall give the proof of Lemma 1 after the proof of Theorem 2.

11 We will again make use of the notation y.v = v. nlr for v E H(div, Q).

1. Gradient, Divergence and Curl Operators

Proof of Theorem 2 for n = 3. i) Let W E H(curl, .0) be orthogonal to ~(Q)3, thus:

def (1.13) w, v)) = (w, v) + (curl w, curl v) = 0, "tv E ~(Q)3 . def

Put Wo = curl w. From (1.13) we deduce:

curlwo = - w,

thus Wo E H(curl, .0), and satisfies -(curl wo, v) + (wo, curl v) = 0, "tv E ~(Q)3 .

By Lemma 1, Wo E Ho(curl, .0). Let {I/Idk EN be a sequence convergent to wo, with I/Ik E ~(.Q)3. Then

( ( w, v)) = lim (- curl 1/1 k' v) + (1/1 k' curl v) = 0, "t v E H (curl, .0) .

Whence w = 0, and point i) follows. ii) Starting from Green's formula:

(1.14) ( v, curl


iii) Let w e kerr,; then from (1.17), w is such that (w,curlIP) - (IP,curlw) = 0, VIP E ~(a)3.

Then Lemma 1 implies that WE Ho(curl, D), whence point iii). o def

Proof of Lemma J. i) Let us again put Uo = curl u; let U and Uo be the extensions of u and Uo respectively, which are 0 outside D; (1.12) implies

Uo = curl u in ~'(1R3)3, hence u e H(curl, 1R3 ), with suppu e a. To show the existence of a sequence of elements IPk E ~(D)3 convergent to u in H(curl, D), we proceed by truncation and regularisation (see Chap. IV) as follows. ii) In the case where D is unbounded in 1R3, we take a function IP e ~(1R3), with o :::::; IP :::::; 1,

{ I if Ixl:::::; 1 IP(x) = 0 if Ixl ~ 2. Put, for all a > 0, IPa(x) = IP(x/a); then IPa' U E H(curl, 1R3 ) and lim IPaU = U in

a - 00

H(curl, 1R 3 ). Since IPaU has compact support in a, we can hence limit our consideration to U with compact support in a. iii) If D is a bounded open set which is strictly star-like, that is to say there exists y e D such that, with respect to y (which we here take to be the origin):

OaeD, VOe[O,I[, then the family of functions ue defined for all 0 e [0, 1 [ by

ue(x) = U( ~) (hence with supp ue e D) converges as 0 ~ 1_ to U in H(curl, D). It only remains to regularise the functions ue: let p e ~(1R3), p ~ 0, p(x) = 0 for Ixl ~ 1, r p(x)dx = 1 . Then the family of functions p., e > 0, defined by JR' P.(x) = p(x/e)/e3 , is such that:

P. e ~(1R3), P. ~ 0, P.(x) = 0 for Ixl ~ e, r p.(x)dx = 1 JR' lim P. = b in t9"(1R3), .-0

which implies that for all v E L2(1R3), P. * v ~ v in L2(1R3). Thus P. * U ~ U in H(curl, 1R 3 ) as e ~ 0, and for e sufficiently small supp(P. * ue) e D thus P. * ue e ~(D)3. This establishes the existence of a sequence P., * ue, in ~(D)3 convergent to U in H(curl, D). iv) In the general case, we employ a covering ofa by a family of open sets (9i' i = to N such that Di = (9i n D for i = 1 to N shall be strictly star-like with a

I. Gradient, Divergence and Curl Operators 209

Lipschitz boundary, and such that (90 c 00 c Q. We also employ a partition of unity (IX;), i = 0 to N, subordinate to this covering, that is to say:

N IXi E f0((9J, 0 ~ IXi(X) ~ 1, L IXi(X) = 1, X E Q .

Thus i = 0

N

U = L IXiU In i= 0

with IXJi E H(curl, [R3), SUPP(IXiU) c Qi'

[R3 ,

Thus we are reduced to the case iii) for i = 1 to N. For i = 0, we have SUpp(IXoU) c Q, and it suffices to regularise IXoU by convolution. Whence Lemma 1. 0 It is important to make precise the connection between the spaces H(curl, Q), H (div, Q) and the Sobolev space HI (Q)". Theorem 3. Let Q be a bounded open set in [R3, which is regular (of class ~2), with boundary r. Then the Sobolev spaces:

(1.18)

are such that:

(1.19) { H tIO(Q)3 = Ho{curl, Q) n H(div, Q) H~0(Q)3 = H(curl, Q) n Ho(div, Q), and on these spaces, the norm II v II H' (0)3 is equivalent to the norm:

(1.20) def { r }1/2 Ilvll = Ju [lcurlvl2 + Idivvl2 + Ivl2Jdx F or the proof of this theorem, we make use of the following lemma (for the proof of which see Peetre [IJ, Tartar [IJ, Lions-Magenes [IJ, p. 171). Lemma 2 (Peetre's lemma). Let Eo, EI> E2 be three Banach spaces, let Al and A2 be two continuous linear mappings, respectively from Eo to EI and from Eo to E2, with: i) A2 is a compact mapping;

ii) there exists a constant c > 0 such that: (1.21 ) Then: i) ker A 1 has finite dimension and 1m A I is closed;

ii) there exists a constant Co > 0 such that: (1.22) inf IIv + wllEo ~ colIAlvll EI .

wker AI


Proof of Theorem 3. We start with the following identity, written using the Einstein summation convention on repeated indices, and the derivative notation

oq> q>, i

ox;'

i vv dx = i v .(v . . - v .)dx + i v v dx "Iv E q}(Q)3 .12 l,) 1,1 1.1 I,) J.t 1,) J,t 'J Q Q Q By double application of Green's formula on the last term of this identity, we obtain:

(1.23) L Igradvl 2 dx = L [lcurlvl 2 + Idivvl 2 ]dx + I" "Iv E q}(Q)3 on putting

(1.24)

or again:

I, = L [(niVj - njv;)vi,j + njViVi,j - njvA,Jdr , which can also be written:

(1.25) r [ ov]-1,= J, 2(n/\v).curlv-n.vdivu+va;; dr, VVEq}(Q)3. i) Case where n /\ v I, = O. We make use of the following expression for the divergence (see Bendali [1], p. 9):

(1.26) d d' a IVV = Ivr(nv) + 2(n.v)H + on (n.v) on r,

where nv(x), for x E r, denotes the projection of v(x) onto the tangent plane to x at r; divr is the surface divergence of a tangent field to r; H(x) is the mean curvature of at X.13 Whenever v is such that n /\ v = 0 on r, (1.26) gives:

(1.27) ( ) d - 2 12 AU n . v IV V = I v H + v . on .

Then (1.25) becomes

(1.28) -L 2H(x)lv(xWdr ,

12 Here we implicitly assume that the spaces in question are complex, in order to show that the result is stilI true in that case. 13 That is to say, the mean of the eigenvalues of the curvature tensor at x E T.


def and with C = 2 max IH(x)l, (1.23) gives:

(129) In I grad Vl2 dx ~ In [Icurl Vl2 + Idiv V12] dx + C L Iv(xW dr . ii) Case where n. vir = O. Beginning directly with (1.24), we get:

Ir = r n;vA.jdr = r vj(nJ5J. jdr - r vAn;,jdr Jr Jr Jr .

211

on having extended the function x 1-+ n(x) in a neighbourhood of r into a function of class C(j 1, in such a way that n;, j makes sense. Since n. vir = 0, the operator

o . . Vj ox. IS a tangentIal operator on r, hence vj(n;ijJ. j = 0 on r. Thus:

J

(1.30)

and on putting C = max In;,jl, (123) further gives: XEr

(1.31) L Igradvl 2 dx ~ In [lcurlvl 2 + Idivvl 2 ]dx + C L Iv(xWdr. iii) It only remains now to apply Peetre's lemma with Eo = H ,10 (Q)3 (resp. Eo = H~o (Q)3, with (1.18,

E1 = L2(Q)3 x L2(Q) X L2(Q)3, E2 = L2(r)3 , and the mappings:

A 1:v E Eo 1-+ (curl v, divv, v) E E1 A 2:v E Eo 1-+ vir E L2(r)3.

We use the fact that the mapping A2 is continuous from Eo into (Hl/2(r3, (Eo being a closed subspace of H1(Q)3) and that the injection Hl/2(r) --+ L2(r) is compact (see Chap. IV). The condition (121) being satisfied as a result of inequality (1.29) (resp. (1.31, Peetre's lemma can be applied. Since ker A1 = {O}, (1.22) implies the equivalence of the norms II V II HI (Q)3 and Ilvll given by (1.20) on the spaces (1.18). 0 Remark 1. Let us put:

def H(curl, diy, Q) = H(curl, Q) n H(div, Q)

= {v E U(Q)3, curl v E L2(Q)3, divv E L2(Q)}

(1.32) { def

Hro(curl,div,Q) = {vEH(curl,div,Q),v A nlr = O} def

Hno(curl, div, Q) = {v E H(curl, div, Q), v. nlr = O} .


Theorem 3 therefore allows us to make the following identifications:

(1.33) { H (0 (Q)3 = HtO(curl, div, Q) H ~0(Q)3 = Hno(curl, div, Q) .

Furthermore, on putting:

def (1.34) Ho(curl, diy, Q) = Ho(curl, Q) n Ho(div, Q),

we can deduce from Theorems 1, 2, and 3, the identification:

(1.35) which (with (1.32)) gives

H(curl, diy, Q) c Hloc(Q)3 . We can also deduce from Theorem 3 the corollary (a priori more general):

Corollary 1. Under the hypotheses of Theorem 3 on Q, the spaces: {v E H(curl, diy, Q) = H(curl, Q) n H(div, Q) ,v. nil' E H 1/2(r)} , {vEH(curl,div,Q), v 1\ nlrEHI/2(r)3}

are identified with the space HI (Q)3, and there exist constants c I and C2 > 0 such that:

(1.36) IlvIlHl(Q)3 ~ cdllvll + IIcurlvll + Iidivvil + Ilv.nIIHli2(r)} (1.37) IIvIIH1(Q)3 ~ c2{ilvll + Ilcurlvil + Iidivvil + Ilv 1\ nil H1/1(r)3} ,

Vv E HI(Q)3 . Proof i) Let v E H(curl, diy, Q) with v.nll' E HI/2(r). For all g = v.nlr E HI/2(r), there exists a lifting


operator h E Hl/2(r)3 ~ Roh E Hl(Q)3 (thus such that Rohlr = h), then such a lifting R is given by

Rg = Ro(-g 1\ n). def def

Putting cp = Rg, and w = v - cp, we verify that W E Hto(curl, div, Q), and therefore, from (1.33), that WE Htb(Q)3; it follows immediately that v = W + cp E HI (Q)3, whence the inclusion:

{v E H(curl,div,Q) , v 1\ nlrEHl/2(r)3} c Hl(Q)3. The inverse inclusion being obvious, the identity between these two spaces IS established; inequality (1.37) follows without difficulty. 0 Remark 2. Theorem 3 remains true under each of the following conditions: i) the boundary r of Q is of class ~ 1, 1; 14 ii) the open set Q is unbounded but its boundary r is bounded and of class ~2 (the proof is analogous), or again of class ~ 1,1; iii) the open set Q is a convex bounded polyhedron (see Girault-Raviart [2J); iv) under the same conditions but with the dimension n of the space equal to 2.

Making use of conclusion i) in Peetre's lemma 15 - and Theorem 3 - we deduce: Corollary 2. The spaces:

{ {VEH 1(Q)3,V 1\ nlr = O,curiv = O,divv = A}, (1.38) {vEH 1 (Q)3,v.nl r = O,curiv = O,divv = O} are of finite dimension, and the spaces:

and

(1.39) { {h = curiv,vEH 1(Q)3,V 1\ nlr = O,divv = a}, and {h = curiv,vEH1(Q)3,v.nlr = O,divv = O} are closed in L2(Q)3. These spaces will be made explicit in the next section.

3. Kernel and Image of the Gradient, Divergence and Curl Operators. Introduction

o

o

It will be necessary for us in what follows to know the kernels and images of the operators grad, div and curl. The determination of these spaces is equally import-ant to many problems other than those of electromagnetism envisaged here,

14 For the definition of this notion, see Chap. II, 3, and Necas [1J. 15 On using, for example, Eo = {v E HI (Q)3, div v = 0, v 1\ nlr = 0 (resp. v. nlr = O)}

EI = U(Q)3, E2 = L 2(Q)3 X U(f)3, with Alv = curl v , A2v = {v, vir}. One could also take E2 = L2(f)3, A2v = vir E U(r), by using (1.29) (resp. (1.31) and Poincare's inequality (see Chap. IV).


notably in the framework of fluid mechanics - the case of hydrodynamics under the Navier-Stokes equations (see Chap. lA, 1 and Chap. XIX) - (see also, in Chap. IXA, 4.3, the problem of the thin airfoil section). One essential point will be made especially precise - in the context of square integrable functions - namely the relation of the images of the gradient and the curl with respect to the kernel of the curl and the divergence. We have already seen in the introduction the inclusions (as distributions):

1m grad c ker curl, 1m curl c ker div .

When may these inclusions be replaced by equalities? This natural question is important in order to characterise the images of the gradient and curl, or again, starting from the relation curl u = (resp. div u = 0), to affirm the existence of a "potential" cp such that u = grad cp (resp. of a "vector potential" v such that u = curl v). A partial answer to this question is given by Poincare's lemma that we will state here only in the case n = 3 (in view of physical applications), in a classical framework.

Lemma 3 (Poincare's lemma). Let Q be an open set in [R3, and let u = (Ui); = I t03 with Ui E C(jl(Q), i = 1 to 3. i) If u is such that curl u = 0, then for each open block l6 0 c Q, there exists a function cp E C(jZ(0) such that u = grad cpo U) If u is such that div u = 0, then for each open block 0 c Q, there exists a vector function w = (W;)i = I to 3 with Wi E C(j I (0) such that u = curl W.

This lemma therefore implies that if u is such that curl u = (resp. div u = 0), then locally u is of the form u = grad cp (resp. u = curl w). This lemma also has the global consequences: if Q = [R3, then u satisfying curl u = (resp. div u = 0) implies that u has the form u = grad cp, cp E C(jZ([R3) (resp. u = curl w, w E C(j 1 ([R3)); if Q is a connected and simply connected open set, then in the case where u satisfies curl u = 0, one can paste together the various functions cp = CP@ (for each open block 0 c Q) such that ul(l = grad cp, simply by the addition of constants.

Proof of Lemma 3. i) LetuEC(jI(Q)3,curlu = 0; let XO = (x?,x~,x~)EQ;then the function cp defined for all x = (Xl' Xz, x 3) such that n [x?, x;] c Q by:

; = 1 to 3

IX

3 IX2 IXI cp(x) = u3(xl,xz,x~)dx~ + uz(xl,x2,x~)dx2 + UI(X'I,X~,x~)dx'l

~ ~ ~ is such that u = grad cp whence point i).16' ii) Let u E C(j1(Q)3, divv = 0; let XO = (x?, x~, x~) E Q; then the vector functions v = (VI' v2 , v3) and VO = (v?, v~, v~) defined for all x = (Xl' Xz, x 3 ) such that

16 I.e. a rectangular parallelepiped ("pave"). 16' The generalisation to the case of arbitrary n poses no difficulty.


f1 [x?, x;] c Q by: i = 1 to 3

with u defined by:

curlv = -3u + Uo , curl Vo = - 2uo

UO(x l , x 2, X3) = (udx?, x 2, X3), U2(XI, xg, X3), U3 (X I, X2, xm ;

thus w = - ~ ( v + ~ VO ) satisfies curl w = u, whence ii).

215

o

For the rest of this section we shall work in the context of L 2 functions; from the global view point let us say at once that - in general- the conditions u E L 2 (Q)3, curl u = 0 (resp. div u = 0) do not imply the existence of functions q> E HI (Q) (resp. v E HI (Q)3 or L2(Q)3),17 such that u = grad q> (resp. u = curl v). These results, well known in differential geometry, will be obtained when we determine the kernels and images of the operators grad, div and curl. Our results will first be stated for the case of bounded open sets Q c ~"which are assumed to be connected (and sufficiently regular) for simplicity. We will then generalise to Q which are not connected or not bounded. The kernel of the grad operator (from HI (Q) into L 2 (Q)"), being comprised of constant functions, has dimension 1 for bounded connected Q. The image of this operator is given by the following proposition.

Proposition 1 (Image of the gradient). Let Q be a connected open set in ~", with a lipschitz boundary r. Then the spaces: grad HI (Q), grad H ~ (Q) and grad Jf'l (Q), with:

def (1.40) Jf'1(Q) = {u E HI(Q), L1u = 0 in Q} = HI(Q) n Jf'(Q),18 are closed subspaces of L2(Q)". Moreover, the spaces grad HJ (Q) and grad Jf' I(Q) are orthogonal and supplementary in grad H 1 (Q). Finally, the spaces orthogonal in L 2 (Q)", to grad HI (Q) and grad H J (Q) respectively, are:

{ Ho(divO,Q) ~ {uEL2(Q)",divu = O,n.ulr = O}

(1.41) def

H(divO,Q) = {uEL2(Q)",divu = O}

17 Assuming, for example, that 0 is bounded. 18 Where '(0) denotes (as in Chap. II) the set of functions harmonic on O.


Because of this proposition, the space L 2(Q)" admits the following orthogonal decompositions:

(1.42) ii) L2(Q)" = gradHJ(Q) EB H(divO, Q) {i) L2(Q)" = gradHI(Q) EB Ho(divO,Q)

iii) L2(Q)" = gradHJ(Q) EB grad,I(Q) EB Ho(divO, Q).

The proof of Proposition 1 can take a number of different forms, all more or less equivalent.

Proof (With the aid of Peetre's lemma). i) The space grad H I (Q) (resp. grad HJ (Q), grad ' I (Q is closed in L 2 (Q)"; this results directly from Peetre's lemma applied with (using the notations from that lemma) Eo = HI(Q) (resp.HJ(Q), ,I(Q, EI = L2(Q)", E2 = L2(Q) and the mappings Al = grad cp and A2 the identity from Eo into E2 which is compact under the hypotheses imposed on Q (see Chap. IV and Netas [1] ).19 ii) Let vEL 2 (Q)" be orthogonal to the space grad H I (Q); then

(v, grad cp) = - < div v, cp> = 0, Vcp E ~(Q) hence div v = 0, V E H(div, Q), and on applying Green's formula (1.10), we obtain v. nlr = 0, hence

(grad H I (Q.L c Ho(div 0, Q) ; the inverse inclusion again follows directly from formula (1.10). Similarly, it is immediately verifiable that the spaces grad H J (Q) and grad ' I (Q) are orthogonal and supplementary in grad H I (Q). 0

Given an element U E L 2 (Q)", its decomposition with respect to each of the spaces indicated in Proposition 1 can be realised in a constructive manner in the following way (which incidentally will give another proof of this proposition). First, there exists P E HI (Q), a solution, unique to within an additive constant, to the "varia-tional" Neumann problem:

( 1.43) (grad p, grad cp) = (u, grad cp), Vcp E HI (Q) , since the bilinear form a(tf;,cp) = (grad tf;, gradcp) is coercive on HI(Q)jlR (see Chap. VII, 1); we then have

der Putting v = u - grad p, we verify that div v = 0, and with the aid of Green's formula (1.10), that v.nlr = 0, hence v E Ho(divO, Q). Elements Po E H J (Q) and q E ' I (Q) can then be found, such that p = Po + q,

19 Note that inequality (1.22) written in this framework is then Poincare's inequality with respect to the space HI(Q) (or H/,(Q) .. . ), see Chap. IV.


which are solutions of the Dirichlet problem:

and:

{A Po = div u in Q , (o.r again (grad Po, grad ep) = (u, grad ep) , wIth Polr = 0,

Aq = in Q q = p on r.

We then finally obtain the promised decomposition:

Yep E HJ(Q ,

(1.44) u = gradp + v = gradpo + gradq + v. Proposition 1 immediately implies:

217

o

Corollary 3. Let Q c IRn with the hypotheses of Proposition 1. A necessary and sufficient condition for an element u E L 2 (Qt to be of the form u = grad p with p E HI (Q) is that u shall be orthogonal to the space Ho(div 0, Q)(see (lAl, that is to say:

(u, ep) = 0, Yep E L2(Qt with divep = 0, ep.nl r = 0. An analogous result can be stated in the space H - 1 (Qt , the details of which may be found in Girault-Raviart [2]. We now move on to study the kernel of curl (in the case where the dimension n of the space is 2 or 3). We will make the following hypotheses: (lAS) i) the open set Q c IRn (n = 2 or 3) is bounded and connected, its boundary r is a manifold of class


def Determination of the space IHIl = H(curl 0, Q) n Ho(div 0, Q)

{uEL2(Q)n,curlu = 0, div u = 0, u. nlr = O} Y

We first remark that from Theorem 3 the set IHIl is contained in the space H 1 (Q)n, hence, by Corollary 2, the space IHIl has finite dimension. 22 Let u E IHI l ; then u is locally the gradient of a harmonic function q and hence (by analytic extension) there exists a function q harmonic in the simply connected open set Q such that u = grad q in Q. The hypothesis u E L 2 (Qt implies (see23 Magenes-Stampacchia [1]) or Necas [1]) that q E U(Q), hence:

def q E l(Q) = Hl n (Q)

(note that the function q is '(fro up to the edges of the cuts I').24 Further:

u.nlr = Oql = 0 on r

(which has a sense in H- l / 2 (r), see Lemma 1, Chap. VII, 1). Lastly, let us denote by I'/ and I'i- the two edges of the cut I'i' and orientate the normal n to I'i from I'/ to I'i- . For every function


(1.46)

We can thus state:

L1q = in Q,

Oql = , an r [q]Ii = constant, i = 1 to N ,

[Oq] = 0, i = 1 to N . an Ii

219

Proposition 2. Let Q be an open set (bounded and connected) in fR" (n = 2 or 3) satisfying the conditions (1.45) i), ii). The kernel of curl in L 2 (Q)", denoted by H (curl 0, Q) is the sum of two orthogonal spaces grad H 1 (Q) and 1HI1 (= 1HI1 (Q)) with 1HI1 a vector space of finite dimension N (equal to the number of cuts needed to render the open set simply connected),25 defined by:

1HI1 = {u E L2(Q)", u = grad q in the classical sense26 in Q with q E Hl(Q) a solution of problem (1.46)}

= {u E L2(Q)2, curlu = 0, divu = 0, u.nlr = O} . Furthermore, if the boundary r ofQ is of class C(1r,6 with r ~ 2,0 < f) < 1,27 then:

1HI1 C C(1r-l,6(Q)" .

Proof Let IHI ~ be the set of solutions q of (1.46). It only remains for us to prove that the space 1HI1 = {u E L 2 (Q)", u = grad q, q E IHI D, possesses the properties asserted. i) Let us first prove that the homogeneous problem: find q E H 1 (Q) satisfying:

L1q = 0 in Q, oq = 0 an on

r, (1.47)

[q]Ii = i = 1 to N , [ :! 1 = 0, i = 1 to N , admits only constant functions as solutions. To do that, let us show that if q E HI (Q) satisfies (1.47), then q E HI (Q) satisfies

25 The space ~ 1 is isomorphic to the first cohomology space, that is to say the quotient of the closed differential I-forms in Q by the exact differential I-forms in Q and N is called the first Betti number (see for example M. Berger [I]). 26 In the sense of distributions in Q. 27 See the definition in Chap. II, 3.


( 1.48)

In effect, for all


Neumann problem in Q inhomogeneous at the boundary, To avoid any difficulties over regularity at points of Xj n r, it is convenient to extend the cut Xj beyond Q,

l~t us call it I j , and to replace the function XEj by a regular (~"') extension Xj to Xj Let iJJ be the doub~ layer potential for the function Xj;28 iL is harmonic in W \ Xj (hence in Q \ Xj ) and satisfies the jump conditions (1.49)j iii) and iv). Furthermore (see Chap. II, 3, comments preceding Proposition 13) qj is of class ~ 00 up to both sides of X j which assures that the function

def _ 2 8j = oq)onl r E L (0.

def Put Pj = qj - qj; Pj must then satisfy:

Hence, from Green's formula applied to the open set Q, we verify that:

f f oq f oq. f 8j dr = _J dr = _J dr = Llqjdx = 0; r r an f an li

it then follows that the Neumann problem (1.5% admits one and only one solution, to within an additive constant (see Chap. II, 6 or VII, 2, Remark 18); the same therefore holds for problem (l.49t. iv) Finally, we prove the regularity result: !HI! C ~r-!,8(Qt if r is of class ~r,8 with r ~ 2, < 8 < 1. The function 8j = grad qj. n Ir appearing in (1.50)j is then of class ~r - 1,8 at every point x E r, x r n Xj . From the regularity results in Chap. II, 6 (see Lemma 4), the function Pj' solution of (1.50)j, is of class ~r.8 up to the boundary with the possible exception of x E Xj n r; the same therefore holds for the solution qj = Pj + qj of problem (1.49)j. Hence uj = grad qj is of class ~r-!,8 up to the boundary with the possible exception of x E Xj n r. On remarking that the vector space !HI! is in fact independent of the particular choice of cuts and that !HI! has finite dimension, we can deduce that each U E !HI! is of class ~r-!,8(Q)". 0

Remark 3. It can be verified without difficulty - by applying Green's formula-that the orthogonal complement of the space !HI! in H o(div 0, Q) is the space of elements v E H o(div 0, Q) satisfying f . v. n dX = 0, j = 1 to N.

E, (Note that these integrals are independent of the particular cuts chosen.) For later

28 Note that in Chap. II, we only considered potentials of double layers relative to surfaces L which form the boundary of an open set in ~n but the results of that Chap. II generalise without difficulty (one can also complete fj to such a surface).


use we denote this space by:

(1.51) deC { f H~(divO, Q) = V E Ho(divO, Q), Ii v.ndl' = Determination of the Image of curl

1 to N} . o

Proposition 3. Let Q be an open set (bounded and connected) in [Rn, n = 2 or 3 with the hypotheses (1.45). The image of curl, curlH!(Q)3 ifn = 3, CurlHl(Q) ifn = 2, is closed in L2(Q)", and identical to the space: (1.52)

Hr(divO, Q) ~ {u E U(Q)" , divu = 0, ti u.ndr } 29 = 0, i = to m . Its orthogonal complement in L2(Q)" is the space: (1.53) Ho(curlO, Q) = {u E U(Q)", curlu = 0, u A nir = O}

= {u = gradp, p E Hl(Q), Piri = constant, i = to m} and its orthogonal complement in the space H(div 0, Q) (kernel of the divergence) is the space:

(1.54) 1H1 2(= 1H12(Q)) = {u = gradep, epEH1(Q), Jep = 0, epiri = constant, i = to m po

(of dimension m), thus giving the orthogonal decomposition: 3! (1.55) H(divO, Q) = kerdiv = 1H12 EB curl Hl(Q)3

1H12 EB CurlHl(Q) if n = 3,

if n = 2.

Furthermore, if the boundary r of Q is of class Cfjr,8 with r ~ 2, < {} < 1, then: 1H12 C Cfjr-1.8(Q)" .

Proof We give the proof in the case n = 3 (the case n = 2 can be treated analogously). i) Let u E Hl(Q)3. Then (see (1.42)), there exists ep E Hl(Q) and v E Ho(divO, Q) such that u = grad ep + v. But ep is such that:

Jep = div u E L2(Q) !:Ir = u.nirEH1/2(r), 29 Note that we continue to make (improper) use of the notation f instead of the brackets r for the duality H 1/2(r), H -1/2(r), which we will often do in the sequel without comment. 30 For n = 3, the space !HI2 is isomorphic to the second cohomology space, i.e. the quotient of closed differential 2-forms in Q by the exact differential 2-forms in Q and m is called the second Betti number. For n = 2, the space !HI2 is isomorphic to !HI 1 and m = N. 31 This decomposition is also a Hodge decomposition.

\. Gradient, Divergence and Curl Operators

hence cP E H2(Q); it then follows that gradcp E HI(Q)3, v E HI(Q)3, Since curl u = curl v, we then have:

(1.56) Thus (see Corollary 2) this space is closed in L 2(Q)3. ii) We now determine the orthogonal complement of curlHl(Q)3 in L2(Q)3. Let W E (curlHI(Q)3).l; then

(w, curl cp) = 0, Vcp E ~(Q)3 , hence curl w = 0. Applying Green's formula (1.17), we have:

(w, curll/!) = (curlw, I/!) + 1. (w 1\ n).I/!dr = 0,32 VI/! E HI(Q)3, whence it follows that w 1\ nlr = 0, and we have the characterisation:

(curlH 1(Q)3).l = {w E L2(Q)3, curlw = 0, w 1\ nlr = O}.

223

But if WE (curl H 1(Q)3).l, then there exists a function p E HI(Q) such that w = grad p (see Proposition 2); the condition w 1\ nlr = implies grad p P nlr = 0, that is to say, the tangential derivative of p along r is null: therefore p is constant on each connected component ri of rand pEH1(Q), whence the characterisation (1.53) of the space (curlHI(Q)3).l. Noting that the space curl H I (Q)3 is a closed subspace of the space H(divO, Q) = kerdiv (since divcurl v = 0, Vv E H I(Q)3), we deduce from (1.53) that the orthogonal complement of curl H I (Q)3 in H(div 0, Q) is the space 1H12 given by (1.54). Hence we have:

(curl H I (Q.l = grad HJ (Q) EEl 1H12 iii) We prove (1.52); since the space curlHI(Q)3 is closed, u E curl Hl(Q)3 if and only if u is orthogonal to every v E (curlHI(Q)3).l, and hence of the form v = grad p, p E HI (Q), Plr. = constant (see (1.53. This immediately implies div v = 0, and by Green's formula:

(u, v) = r u.gradpdx = f u.npdr = ~ Plr. r u.ndr = 0, JQ r ,Jr. thus f.. u. n dr = 0, whence (1.52). iv) Finally, the regularity result: 1H12 C ~r-1.8(Q)3 if r is of class ~r.8, is an immediate consequence of the regularity results stated in Chap. II, 6. 0 From Proposition 3 we have this characterization:

32 t again denotes (improperly) the brackets


Corollary 4. Let Q be an open set (bounded and connected) in IR", n = 2 or 3, with the hypotheses (1.45). A necessary and sufficient condition for an element u E L2(Q)" to be of the form u = curl v with v E H l(Q)3 ifn = 3 (resp. u = Curl v with v E H 1 (Q) if n = 2) is that u satisfies:

div u = 0, f u. n dr = 0, i = to m . r,

(1.57)

As a consequence of Propositions 1 and 3, we can state:

Corollary 5. Let Q c IR" with the hypotheses (1.45), n = 2 or 3. Each element u E L 2 (Q)" has the unique decomposition:

(1.58) { u = grad p + curl w if n = 3 ; u = grad p + Curl w if n = 2 ;

with p E H 1 (Q) unique to within an additive constant, and with w E H 1 (Q)3 if n 3, WE Hl(Q) ifn = 2, such that

n. curl wlr = if n = 3 (resp. n. Curl wl r = 033 if n = 2) Remark 4. We have seen (from (1.56) for n = 3) that the image by the curl of the space Hl(Q)3 is the same as that of the space Hl(Q)3 n Ho(divO, Q). On decomposing the space H o(div 0, Q), following Remark 3, into

Ho(divO, Q) = !HIl EB Ht(divO, Q) and remarking that curl u = for all u E !HI l , and that !HIl c H 1 (Q)3, we obtain the following property: The operator curl is an isomorphism34 of the space HI(Q)3 n Ht(divO, Q) (with the norm HI) onto the space curIHI(Q)3 c L2(Q)3. Hence in the decomposition (1.58), the element w appearing in the decomposition of u E L2(Q)3 is unique if we further demand that w E H6(divO, Q); hence:

div w = 0, w. nlr = 0, f w. ndI' = 0, i = 1 to N . I,

An analogous result holds again in the case where n = 2.

We note that another decomposition similar to (1.58) can also be given: Corollary 5'. Let Q c IW with the hypotheses (1.45), n = 2 or 3. Each element u E L 2(Q)n has the unique decomposition:

(1.58)' { u = grad p + curl w if n = 3 ; u = grad p + Curl w if n = 2 ;

o

33 Note that this condition is equivalent to: the tangential derivative of won r is nUll, therefore wlr is constant (on each connected component of n. 34 That is to say bijective, continuous, with continuous inverse, this being due to the fact that the space curl H' (Q)3 is closed.

I. Gradient. Divergence and Curl Operators 225

der with P E H; (Q) = {p E HI (Q), pi" = constant, i = 0 to m}, (unique to within an additive constant), and with w E HI (Q)3 if n = 3, w E HI (Q) if n = 2. In this Corollary 5', the element P appearing in (1.58)' has the decomposition: P = Po + PI' where: Po E HJ(Q) and is a solution of the Dirichlet problem:

der

{ L1po = div u in Q Pol, = 0,

PI E!HI1 = {cp E HI(Q), L1cp = 0 in Q, cpl" = constant, i = 0 to m},

der hence grad PI E !HI2; PI is determined by its boundary values Yj = Pll'j for j = 0 to m in the following way: for all X E !HI L we have from applying Green's formula (see (1.1 0:

(1.59) In ugradxdx = In grad PI gradxdx = f. a:nl . XdT ; on taking X = Xi with Xi E Yf I (Q), Xi I,} = bij , and using the matrix (Cij) of influence coefficients (see Chap. II, 5), with Cij = f., (~~ dT)' we get: (1.60) r ugradXidx = f CijYj; JQ j=O PI being determined only to within an additive constant, we can always choose Yo = 0; the system (1.60) is then invertible (see Chap. II, 5) and gives the sought for values Yj. We have consequently obtained a new orthogonal decomposition of the space U(Q)":

{ U(Q)3 = gradH6(Q) EB !HI2 EB curlHI(Q)3 if n = 3; (1.61) U(Q)2 = grad Hb(Q) EB !HI2 EB CurlHI(Q) if n = 2. o

We can also prove

Proposition 4. Let Q be an open set (bounded and connected) in [R3, with the hypotheses (l.45). The image set

curiH/O(Q)3 = {curl v , v E HI(Q)3, v 1\ nl, = O} is the closed vector subspace of L2(Q)3 defined by (see (1.51)):

HJ(divO, Q) ~{u E L2(Q)3, divu = 0, u.nl, = 0, L,u.ndI'i=O' i=ltoN}.

Proof For each u E H I(Q)3 such that u 1\ nl, = 0, there exists the (unique) solution cp E HJ(Q) to the equation: L1cp = divu; since divu E L2(Q), we


have (see Chap. VII) q> E H2(Q) so that gradq> E HI(Q)3, which implies that def

V = U - grad q> E HI (Qf; further, we will have

v /\ nlr = u /\ nlr - grad q> /\ nlr = 0 , since grad q> /\ nlr is the derivative of q> tangential to r. Thus curl u = curl v, so that

curlH tIO(Q)3 = curl(Htb(Q)3 n H(divO, Q)). From Corollary 2 (see (1.39)), this space is closed in U(Q)3. Let us determine its orthogonal complement (curl Htb(Q)).l in L 2(Q)3. Let D E (curl Htb(Q)).l; then for all u E 2& (Q)3,


Proof Only the last part of the corollary needs proving. From the proof of Proposition 4 we see that we can take div w = 0. The kernel of the curl in the space H,~(Q)3 n H(div 0, Q) is then

{u E HI(Q)3, divu = 0, curlu = 0, u 1\ nlr = o} which is obviously the space 1H12 (see (1.54. Let v E HI~(Q)3 n H(divO, Q) be orthogonal to 1H12 relative to the scalar product:

def ((v, u = (curl v, curl u) + (div v, div u) + (v, u) = (v, u) Vu E 1H12 .

Then for all X E 1HI1, that is to say X E ytl(Q), Xlrj = constant = a j

In v.gradX dx = 0, which, using Green's formula, gives:

t v.n XdT = 0, hence r v . n dT = 0, i = to m. Jri The orthogonal complement of 1H12 in the space H,~(Q)3 n H(divO, Q) is then

(1.66) {v E HI(Q)3, divv = 0, v 1\ nlr = 0, t, v.ndT = 0, i = to m}, whence follows Corollary 6. o Let us make precise the element hiE 1HI1 which appears in (1.64); hi has the form: hi = grad q with q E HI (Q) satisfying (1.46), hence q can be written:

(1.67) N

q = L Yjqj' with qj a solution of (1.49)j j~ 1

(or again: the jump in the function q as it crosses the cut Ei will be [qJr, = Yi' i = 1 to N). We determine the. values Yj for given u E U(Q)3. With the help of Green's formula (in the open set Q) we have:

Let us put (as in the case of influence coefficients (see Chap. II, 5:

(1.69) def f oq Lij = r, o~ dE i .


Then Lij appears as the flux of grad qj across the cut 2:;; we also note that: Lij = (gradqj' grad q;) ,36

which implies that the matrix (Lij) is symmetric (Lij = Lj;), and invertible. 37 The quantities Lij are called the coefficients of inductance. With (1.69), (1.68) becomes:

(1.70) t u. grad q; dx = jtl LijYj , which determines the Yj (by inverting the matrix (Lij)), and hence hI' From (1.70) we deduce the expression (called the "energy"):

(1.71) If 2 If 2 1~ - Ihll dx = - I grad ql dx = - L LijYjY; 2 Q 2 Q 2 ;.j= I

which shows that the matrix (Lij) is positive definite. 38 Remark 6. The case where Q is a bounded but not connected open set. Let us assume that the open set Q is again bounded and regular (see 0.45)), but not connected, and is the union of a finite number of connected components Q;, i = 1 to n(Q). The previous results transfer without particular difficulty by passing to the connected components of Q; the decompositions of the spaces L2(Q) and HI(Q) into:

n(m n(Q) U(Q) = Et> L2(Q;) , HI(Q) = Et> HI(Q;)

;= I ;= I

imply the decompositions of the two cohomology spaces IHl dQ) and 1Hl2 (Q) (see Propositions 2 and 3) into:

n(Q) n(Q) IHlI (Q) = Et> IHlI (Q;) , 1Hl2(Q) = Et> 1Hl2(Q;) .

i= 1 i::= 1

Then, denoting by n(r) the number of connected components r; of the boundary r of Q, and by n(Q') the number of bounded connected components Q; of the complement Q' of Q, let us put:

lHli(Q) = {

l. Gradient, Divergence and Curl Operators 229

Now dim 1HI1(Q) = n(r), dim Cn = n(Q), dim 1HI1*(Q) the Alexander relation (see Chap. II, 4, Remark 6):

n(Q'), so that on using

dim 1H12(Q) = n(r) - n(Q) = n(Q') , whence it follows that

1HI1(Q) = 1HI1*(Q) EB Cn (direct sum) and the mapping 3p E HI(Q) such that u = gradp ii) u E L2(Q)3, divu = 0 => 3v E HI(Q)3 such that u = curlv. (v can further be taken 39 such that divv = 0, v.nl r = 0). The hypothese of the lemma are in particular satisfied if the open set Q (assumed connected and with a very regular, i.e. (fry oc variety, boundary r = aQ), is such that Q is "contractible to a point Xo E Q", that is to say such that: there exists a mapping H, of class ffj 00, Q x [0, 1] -. Q with

{ H(x, 1) : x, "Ix E Q , H(x, 0) - Xo .

Examples of such an open set are provided by regular open sets which are "starlike with respect to a point Xo E Q", i.e. such that:

for all x E Q, the segment [xo, x] is contained in Q . We then take H(x, t) = Xo + t(x - xo). o

Remark 7. The case of an open set Q' with bounded complement. In order to treat the case of open sets with bounded complements we are led to use, in place of Sobolev spaces, the Beppo-Levi spaces. We shall briefly recall their definition40 and some principal properties; only those with dimension n = 3 will be considered here. Let Q' be an open set in [R3, with bounded complement Q.

39 Then v is unique. 40 Certain of these spaces were earlier introduced and used in Chap. II, &5.3 (see Definition 3 and the preceding note).


Let us put:

def {au } B1(Q') = U E 0'(Q') , aXi

E L 2(Q') for i = 1 to n ,

def _ Wt(Q') = closure of 0(Q') under the semi-norm ilgradcpilu(Q')n'

def WJ(Q') = closure of &(Q') under the norm II grad CPilL2(Q')n .41

When Q' is connected, grad cp = for cp E 0(Q') implies cp = 0, and II grad cp IIL 2 (Q')n is then a norm on &(Q'), so that W1 (Q') is a Hilbert space. By the Poincare inequality (see Chap. IV, 7), the elements of WI (Q') are in H l~c(Q') that is to say for each bounded open set (I)' c Q', u E Wl(Q') = Ul~i E H 1(I)'). Hence W1(Q') c B 1(Q'). Beppo-Levi spaces can equally be defined for orders of differentiation greater than 1, and notably:

def _ ( 3 1 azcp 12 )1/2 W2(Q') = closure of 0(Q') for the (semi-)norm . ~ -a a-I,J~ 1 Xi Xj U(Q')

Replacing Q' by its unbounded connected component (Q~) if need be, we shall assume henceforth that Q' is connected. We shall now give the fundamental trace theorem for the space WI (Q'), the proof of which will be left until Chap. XIB.

Theorem 4. Let Q' be an unbounded open (connected) subset in 1R 3, with bounded complement. The trace mapping: u E 0(6') ~ ulr extends by continuity to a surjec-tive continuous mapping from W 1(Q') onto H 1/2(T) whose kernel is W6(Q'). We will again denote by ulr the trace of an arbitrary element u of W 1(Q'). The results of this section remain true on replacing the Sobolev spaces HI (Q) and H 6(Q) (with regular bounded open Q) by the Beppo-Levi spaces WI (Q') and

. W6(Q') for regular unbounded open Q' with bounded complement. More precisely: 1) In the framework of Proposition 1: The spaces grad W 1(Q'), grad WJ(Q') and gradyt'I(Q') with

def yt'1(Q') = Wl(Q') n yt'(Q') are closed subspaces of U(Q')3 .

The proof is immediate on remarking that the operator grad is an isometry from W 1 (Q') into LZ(Q')3. 2) In the framework of Proposition 2 (case n = 3): The kernel of curl is H (curl 0, Q') = grad WI (Q') EEl !HI I (Q'), with

def !HII(Q') = {u E U(Q')3, curlu = 0, divu = 0, u.nlr = o} ;

41 We also often use these notations: BL(Q') for Wi (Q'); BJ(Q') or g 1 (Q') for W':(Q'). We shall see in Chap. XI a characterisation of these spaces as weighted Sobolev spaces.

I. Gradient, Divergence and Curl Operators 231

The elements of 1HI1 (Q') are also of the form u = grad q (in the classical sense in Q'), with q defined as in (1.46) (with q E WI(Q') here in place of q E Hl(Q)), 3) In the framework of Propositions 3 and 4 (case n = 3): The spaces curl Wl(Q')3 and curl Wr~(Q')3 with

Proof The proof is analogous to the case where Q is bounded (see Proposition 3): i) Let u E Wi (Q')3; we intend to decompose u into the form

u = gradp + v with v E WI(Q')3, divv = 0, v,nl r = 0,

Then P must accordingly be a solution of the Neumann problem:

(1.72)

with grad P E Wi (Q')3 . def _

Let us put f = div u, f = extension of f to [R3, equal to 0 outside Q'. Then by Fourier transformation, we can verify that there exists a function Po E W 2([R3)

a2p (thus such that -a aO E L 2 ([R3) for all i,j = 1 to 3), a solution of LJpo = jin [R3. Xi Xj

Hence since W 2([R3) c HI~c([R3), aJnlr

E HI/2(r). Put: P = Poln' + PI; the problem (1.72) then reduces to:

(1.73) {::111 = 0 in Q' apo I an r = u.nlr - a;; r = g, with 9 given,

Now this Neumann problem can be solved by a simple layer potential (see Chap. II for the classical framework, and Chap. XI for the case where the given 9 belongs to H -1/2(r). With the regularity hypotheses made on Q (see (l.45)), and thus on Q', the solution PI of (1.73) will be in HI~c(Q') (i.e. Pl10 E H 2((9) for each bounded open set (9 c Q'); further (see Chap. II, 3, Proposition 2 and Lemma 2):

apl Xi ( 1) . aX i = C IxI 3 + 0 ~ I = 1 to 3 as Ixi -> 00

- 3C XiX j + O(~4) i,j = 1 to 3 as Ixi -> 00 Ixls Ixi


(with C = constant), therefore PI E WI(Q') n W2(Q'), and because of P = Pol!]' + PI E W2(Q') ,

we have gradp E Wl(Q'). ii) We have thus proved that:

curl Wl(Q')3 = {curl v, v E Wl(Q')3, divv = 0, v.nlr = O} . Using (Peetre's) Lemma 2 with

Eo = {v E Wl(Q')3, divv = 0, v.nlr = O}, El = L 2(Q')3, E2 = L2(r)3, with A 1 V = curl v, A 2v = vir, and with inequality (1.31), we obtain, as in Corollary 2, that the space curl W 1 (Q')3 is closed in e(Q')3. It could be proved in an analogous fashion that the space curl Wtb(Q')3 is closed in L2(Qy42 0 It could again be proved that the spaces orthogonal (in e(Q')3) to curl Wi (Q'? and to curl Wtb(Q')3 are the spaces denoted by Hr(divO, Q'), and H~'(divO, Q') respectively (see (1.52) Propositions 3 and 4) with characterisations analogous to the case of a bounded domain. On defining the space 1H1 2 (Q') by (see (1.54)):

1H1 2 (Q') = {u = gradq>, q> E Wl(Q'), Liq> = 0 In Q', q>lri = constant for i = 0 to m}

we again get the orthogonal decomposition (Hodge decomposition): H(divO, Q') = kerdiv = 1H1 2(Q') EB curl Wl(Q')3 .

It can also be verified that the spaces IHII (Q') and 1H12 (Q') have finite dimension. In the case where Q' = 1R 3, we have IHId1R3) = 1H1 2(1R 3) = {O}, and we "recover" the Poincare lemma in the form:

Lemma 4' (The "global" Poincare lemma in 1R 3 ). Let u E e(1R 3 )3. i) If curl u = 0, then there exists P E Wi (1R3) (unique)

such that u = grad p. ii) If div u = 0, then there exists v E Wi (1R 3 )3

such that u = curl v; the element v is unique if one further has div v = O.

In the case where Q' c 1R3 is a regular open set with connected bounded com-plement, and such that the connected components of Q' are simply connected, then IHIdQ') = {O}, but dim 1H12(Q') = 1. Therefore, for Q and Q' connected and simply connected,

IHII(Q') = IHII(Q) = {O}, 1H12(Q) = {O}, dim 1H12(Q') = 1 . 0 Remark 8. Image of the operator div. If Q is a regular bounded open set in IRn (connected or not), then:

42 We also have curl W6(Q')3 = curl W,b(Q')3

I. Gradient, Divergence and Curl Operators

div Hl(Qt = L2(Q)

div Hb(Qt = {u E L2(Q), L, udx for every connected component Q; in Q},

and if Q' is a regular open set in fRn with bounded complement, then: div W1(Q,)n = U(Q')

div wt (Q')n = {u E L 2(Q'), r. u dx = for every connected (bounded) J Q, component Q; in Q'}.

233

The fact that each of the above images are closed in L2(Q) (or L2(Q')) again follows from (Peetre's) Lemma 2 and the inequalities (1.29), (1.31). It then suffices to determine the orthogonal complement of each of these images, which can be done by using Green's formula (1.10). From our knowledge of the images of the operator diy, we deduce the following property.43

For all 9 E H 1/2(r)n satisfying Ln, g. n dT ~ 0, for each connected component Q; in Q, there exists a lifting u = Rg E H(divO, Q) such that ul r = 9 (the mapping 9 H Rg being continuous from Hl/2(r)n into H(divO, Q)). Proof Let v E H 1 (Q)n such that vir = g; by Green's formula:

r div v dx = f g. n dT = . J n, en, But then there exists W E H A(Q)n such that div w = div v. On putting u = v - w, we see that ul r = 9 and div u = 0. The continuity of the lifting relative to 9 E H 1/2(r)n is a result of the possibility of choosing mappings 9 H V and div v H w which are themselves continuous. 0

Remark 9. Let us also note the following property (used in a more systematic fashion for the Navier-Stokes equation rather than in electromagnetism) (the proof of which will also be left till later in Chap. XIX; see also Foias-Tenman [1]): The open set Q C fRn being bounded and regular, the set 11 = {u E f0 (Q)n, div u = O} is dense in the closed subspace {v E H b(Q)n, div v = O} of H 1 (Q)n, and in the closed subspace Ho(divO, Q) = {vEL2(Q)n, divv = 0, v.nl r = O} of L2(Q)n.44 In other words, the elements of H t (Q)" have zero divergence, and those of Ho(divO, Q) can be approximated by regular functions (in f0(Q)n) with zero divergence (which also implies that the dual space of {v E Ht(Q)", divv = O} can be considered as a space of distributions with zero divergence). 0

43 This property will be used many times in the sequel- in 2 - and in Chap. XVI. 44 This result shows in particular that, in the case n = 3, the set r is not dense in the space {v E U(Q)3, curl v E U(Q)3, div v = 0, v /\ nil = O} equipped with the norm (II v 112 + II curl v 112)112 which is a closed subspace of H 1 (Q)3, a result which will be dealt with in Chap. XVI.


4. Some Results on Regularity

Throughout this section we assume for simplicity that Q is a bounded open set which is very regular, that is to say r = oQ is a variety of class re 00: thus we shall use hypothesis (l.45) with r = +00. Let us first state

Lemma 5. Let Q be a "very regular" connected bounded open set in [R". Then for all u E Hk(Q)", kEN, the decomposition (see 1.44: (\.74) u = gradp + v, with p E Hl(Q), v E e(Q)", divv = 0, v.nl l' = 0, is such that p E Hk+ 1 (Q) and v E Hk(Q)". Proof In effect, p is the solution (to within an additive constant) of the Neumann problem:

{LJP = divu E Hk-1(Q. ), oPj k-' ;;-- = u.nll'EH '(T), un r

which implies (see Chap. VII, 3, and Agmon-Douglis-Nirenberg [I]) that: p E Hk+ 1 (Q), thus grad p E Hk(Q)" and v = U - grad p E Hk(Q)" 0

With the regularity hypotheses made here on the open set Q, the cohomology spaces 1Hl1 and IHl z are contained in re


Hence:

(1.76) { ker curlk = grad Hk + 1 (Q) EEl !HIl , kerdivk = curlH k+ l (Q)3 EEl!HI z if ker divk = Curl Hk+ 1 (Q) EB!HI z if

Similarly, the space

n = 3, n = 2.

235

curl H7o+ 1 (Q)3 = {curl v, v E Hk+ 1 (Q)3, V /\ nlr = O}, for n = 3 , has !HIl as supplementary space in the (closed) space Hk(Q)3 1\ H o(div 0, Q), and is characterised by:

(1.77) curlH~+l(Q)3 = {UEH k(Q)3,diVU = O,u.nlr = 0 f u.ndI'i = 0, i = 1 to N}.

I;

Hence

(1.78) Corresponding to Propositions 2, 3 and 4, we can then deduce:

Proposition 5. Let Q be a very regular, connected bounded open set in [Rn. The sets gradHk+l(Q), and curlHk+l(Q)3 and curl H,kO+ l(Q)3, if n = 3, and Curl Hk+ 1 (Q), if n = 2, are closed subspaces in Hk(Q)". Remark 10. We note that, as in the case where k = 0, we have, for n = 3:

(1.79) { CUrlHk+ l (Q)3 = curl(Hk+l(Q)3 1\ Ho(divO, Q)) ,

curlH,kO+l(Q)3 = curl(H;O+l(Q)3 1\ H(divO, Q)). The first identity is a simple consequence of Lemma 5; the second identity is proved in an analogous way, by decomposing each element u E H~+ 1 (Q) into: (1.80) u = gradp + v with p E HJ(Q) a solution of the Dirichlet problem:

{ Lip = div u E Hk(Q) , Plr = 0,

which, because of the regularity theorem (see Chap. II, 6, Corollary 1, and Chap. VII, 3 and Agmon-Douglis-Nirenberg [1]) implies p E Hk+Z(Q), hence gradp E Hk+l(Q), and v = u - gradp E Hk+l(Q); furthermore v /\ nlr = u /\ nlr = 0, whence the assertion (1.79). 0 Proposition 6. Let Q be a very regular, connected bounded open set in [Rn. Then: ifn = 3: (1.81)

Hk+l(Q)3 {u E U(Q)3, curlu E Hk(Q)3, divu E Hk(Q), u.nlr E Hk+~(n}


and ifn = 2: (1.81 )'

Hk+ 1(Q)2 =. {u E L 2(Q)2, curl u E Hk(Q), div u E Hk(Q), u. nlr E Hk+t(r)} . Proof We take the case n = 3 (the case n = 2 can be treated in an analogous fashion). The set

{u E U(Q)3, curlu E Hk(Q)3, divu E Hk(Q), u.nlr E H k+(1/2l(r)} is obviously contained in the space Hk+ 1(Q)3; we will prove the converse. For that we shall use the decomposition (1.74); p E H1(Q) will be the solution of the Neumann problem:

{LIp = divu E Hk(Q) ,

~:Ir = u. nlr E H k+(1/2l(r) , therefore p E Hk+2(Q), gradp E Hk+1(Q)3, and v = u - gradp is such that

v E L2(Q)3, curl v = curlu E H k(Q)3, divv = 0, v.nl r = O. But then (see (1.75) and remark 10), there exists w E H k+ 1(Q)3 n Ho(div 0, Q) such that curl w = curl v.

def The element Wo = w - v then satisfies:

Wo E L2(Q)3, curl Wo = 0, div Wo = 0, Wo' nlr = 0, thus Wo E 1HI1' which gives Wo E Hk + 1 (Q)3, and

v = W - Wo E Hk+1(Q)3, u = v + gradp E Hk+1(Q)3. o It will be noted that the identification of the spaces (1.81) in the case n = 3,45 is accompanied by the equivalence of the norms:

II u II Hk+ 1(Q)3 and Illulll = (IIulli2(Q)3 + Ilcurlull~k(m' + IIdivull~k(Q) + Ilu.nI11H 1(r1/2.

In effect the space Hk+ 1(Q)3 is a Hilbert spac~ for each of these norms, and there exists a constant C > 0 such that Illulll :::;; CliuIlHk+l(Q)3. Hence the identity mapping on Hk+ 1 (Q)3 is continuous (and bijective) for 11.IIHk+ 1(Q)3 --+ 111.111, and thus bicontinuous, which implies that there exists C' > 0 such that IluIlHk+l(Q)3 :::;; C'llIulll. 0 This Proposition 6 has a number of consequences.

Corollary 7 (Regularity of the decompositions in Corollaries 5, 5' and 6). Let Q be a very regular, connected bounded open set in IRn, (n = 2 or 3). Then for all u E Hk(Q)", the decompositions (1.58), (1.58)' (with div W = 0, w. nlr = 0) and (1.64)

45 The case n = 2 can be treated similarly.


(with divw = 0, W 1\ nlr = 0) are such that P E Hk+I(Q) and WE Hk+I(Q)3 if n = 3,wEHk+I(Q)i{n = 2. Proof We give the proof for the case n = 3 (the case n = 2 can be treated analogously). i) Decomposition (1.58) (Corollary 5 and Remark 4). By Lemma 5 we have P E Hk + I (Q) and curl W E Hk(Q)3; but then by Proposition 6, W E Hk + I (Q)3. ii) Decomposition (1.58)' (Corollary 5'). In this case, P = Po + PI with PI E IHI i therefore PI E ~OO(Q) (see Corollary 5') and Po is a solution of the Dirichlet problem:

{ ,dPo = divu E Hk-I(Q) , Poll = ;

thus (see Chap. II, 6, and Chap. VII, 3) Po E Hk+ I(Q), whence P = Po + PI E Hk+ I (Q), grad P E H k(Q)3, curl W = u - grad P E Hk(Q)3, with div W = 0, w.nl l = 0, and WE H I (Q)3. By Proposition 6, WE Hk+ I (Q)3. iii) Decomposition (1.64) (Corollary 6). By Lemma 5, P E Hk+I(Q) and since hi E IHII C ~OO(Q), we have:

{ CUrlW = u - gradp - hi E Hk(Q)3, with the conditions div W = 0, W 1\ nil = .

By Proposition 5 (see (1.78 there exists Wo E Hk+ I (Q)3 such that: curl Wo = curl w, with div Wo = 0, Wo 1\ nil = .

But then the difference Wo - W = WI satisfies

curl WI = 0, div WI = 0, WI 1\ nil = , hence WI E 1H12' which implies W = Wo - WI E Hk+I(Q)3. We have seen (part iii) in Corollary 7) that the set:

{w E L2(Q)3 , curl WE Hk(Q)3, div W = 0, W 1\ nil = o} is contained in the space Hk+ I(Q)3. Now we go on to deduce

Proposition 6'. Let Q be a very regular, connected bounded open set in (Rn. Then, if n = 3:

( 1.82)

o

Hk+I(Q)3 = {u E U(Q)3, curlu E Hk(Q)3, divu E Hk(Q), u 1\ nlr E Hk+t(r)3} and ifn = 2:

(1.82)' Hk+I(Q)2 = {u E U(Q)2, curlu E Hk(Q), divu E Hk(Q), u 1\ nil E Hk+t(r)2} .

Proof (in the case n = 3).


As in Proposition 6, it suffices to prove that if u E U(Q)3, curl u E Hk(Q)3, div u E Hk(Q), u 1\ nlr E Hk+t(r)3 ,

def def then u E Hk+1(Q)3. Put g = u 1\ nlr and h = - g 1\ n E H k+(1/2)(r). From the trace theorem on Sobolev spaces (see Chap. IV, 4), for h = (h 1, h2' h3) with hi E Hk+t(r), i = 1 to 3, there exists Uo = (u?, u~, u~) with u? E Hk+1(Q) and u?lr = hi for i = 1 to 3, hence Uo E Hk+1(Q)3 with uOlr = h. But then:

UO 1\ nlr = h 1\ n = - (g 1\ n) 1\ n = g - (g.n)n = g. . def

Consequently v = u - UO satisfies the conditions:

vEL2(Q)3,curlvEHk(Q)3,divvEHk(Q),v 1\ nlr = 0. Now using the following decomposition of v (see (1.58)'): v = grad p + w, with P = Po + P1' P1 E IHI i c COO(Q) and Po E HJ (Q) such that .1p = div v E Hk(Q), and hence Po E Hk+2(Q), and P = Po + P1 E Hk+2(Q), we have grad P E Hk+ 1(Q)3 and also, since plr, is constant for i = 1 to m, gradp 1\ nlr = 0. Thus W is such that:

WE U(Q)3, curl W = curl v E Hk(Q)3, div W = 0, W 1\ nlr = . HenceWEHk+1(Q)3 andu = W + gradp + uOEHk+1(Q)3. o As we did in Proposition 6, we can verify the equivalence of the norms:

II u IIHk+ '(m3 and

From Propositions 6 and 6' we can now deduce the following "iterated" properties:

Corollary 8. Let Q be a very regular, connected bounded open set in /R 3 Then the spaces:

(1.83) X k ~ {v E L2(Q)3, divv = 0, curlPv E L2(Q)3, P = 1 to k, with v 1\ nlr = 0, (curl 2P v) 1\ nlr = 0, p = 1 to [k ; 1 T6}

(1.84) Yk ~ {v E L2(Q)3 , div v = 0, curlP v E L2(Q)3, P = 1 to k , with v.nlr = 0,(curl2p - 1V) 1\ nlr = O,p = 1 to[~T6}

are contained in Hk(Q)for each k E N*.

46 With the notation [r] = integer part of r.

2. Static Electromagnetism

Proof For k = 1, we have: Xl {vEL2(Q)3,divv = O,curlvEL2(Q)3,V 1\ nlr o}

Ho(curl, Q) n H(divO, Q) c Hl(Q)3 , Y1 {v E L2(Q)3, divv = 0, curl v E L2(Q)3, v.nlr O}

H(curl, Q) n Ho(divO, Q) c Hl(Q)3 . Moreover:

(1.85) { X k = {v E X k - l' curl v E Yk - d , Yk = {v E Yk - l' curl v E Yk - d .

With the recurrence hypothesis:

Xp c HP(Q? and Yp c HP(Q)3 for p = 1 to k - 1 ,

239

we see, if U E X k (resp. Yd, that U E Hk-l(Q)3 and curlu E Hk-l(Q)3, with divu = 0, u 1\ nlr = (resp. u.nlr = 0), hence u E Hk(Q)3 from Propositions 6 and 6' above. 0


- Magnetostatics of a surface current - Electrostatics of a surface charge

1. Magnetostatics of a Surface Current

Let us consider two perfect magnetic media47 (see Chap. lA, 4, Remark 17) with magnetic permeability 11 and 11', occupying respectively the domains48 Q and Q' with Q' = /R3\Q and Q a regular49 bounded open set in /R 3. In particular, the domain Q' will be occupied by empty space (then 11' = 110)' We assume that there is zero current density throughout each of the media considered, and hence that only surface currents - at the surface of separation r of the two media - can exist. We propose to determine the magnetic induction B in /R 3 . Because of Maxwell's equations (see Chap. lA, (4.1, B must satisfy:

(2.1) { i) div B = ii) curl B =

in III

/R3

Q and in Q' .

47 A generalisation to the case of a finite family of perfect media (occupying bounded domains) immersed in an infinite perfect medium (the vacuum, for example) can also be treated in an analogous fashion, see Duvat-Lions [1] for problems related to such situations. 48 The term "domain" is here taken in the loose sense (of a region not neccesarily connected in [R3), and not in the mathematically precise sense of a connected open set. 49 The boundary r of Q can be assumed, for example, to be of class '{j 2, or '{j I. I.


We shall concern ourselves here only with the solutions B of (2.1) of finite ("free") energy, that is to say (see Chap. lA, 4) such that:

(2.2) W = -21 [ IBI2dx + _1, [ IBI2dx < +00; J1 JQ 2J1 JQ'

since J1 and J1' are strictly positive constants, we shall therefore impose the condition B E L2(1R3)3 . a) Let us first consider the case where the surface current Iron r = oQ is known; we are then reduced to solving the following problem:

find B E L2(1R3)3 satisfying:

i) div B = 0 in 1R 3 ,

(2.3) ii) curl B = 0 In Q and in Q',

iii) [ - ; 1\ n 1 = Ir (Ir given50 with divr 1r = 051 ), where [- ~ 1\ nJ denotes the jump in the quantity - ~ 1\ n (with n normal to

J1 r J1 r = oQ orientated to the exterior of Q) across r, that is to say on denoting by BQ ,

. . [B ] (BQ.lr BQlr) and BQ the restnctlOns of B to Q' and Q, -;; 1\ n r = ---;;' - --;;- 1\ n. Put:

(2.4) { def _ _

V = H(divO, 1R 3 ) = {B E L2(1R3)3 ; div B = O} def _ _

W = {A E W 1(1R 3)3; divA = O}.

From Poincare's Lemma 4' (in 1 above), global in 1R 3 , for each BE V, there exists A E W, unique, such that B = curl A; and using Fourier transformation we can verify that the mapping A E W H B = curl A E V is an isometry of the Beppo-Levi space W onto the space V. Now let us define the bilinear forms:

(2.5) { a(B, B) = ~ [ B.Bdx + ~ [ B.Bdx, VB, BE V ,

2/J JQ 2/J JQ' ao(A, A) = a(curlA, curl A) , VA, A E W.

These bilinear forms are continuous respectively on V and W, and moreover are

50 See Chap. lA, 4, (4.25) or (4.44). 51 This condition comes from the "continuity" relation (4.2) of Chap. lA, 4, ensuring the conservation of surface current.

2. Static Electromagnetism 241

coercive on V and W; in effect, on putting:

we have:'

(2.6) -21 [ IBI 2 dx ~ a(B, B) ~ -21 [ IB1 2 dx, VB E V. flM J~3 flm J~3

We further note that each element A E Wadmits a trace Air E H i/2(r)3 on r. We can then state:

Proposition 1. For each given surface current 1 r such that: (2.7) lr E H- i/2(r)3, lr.n = 0, a.e. on r, divrlr = 0, the problem (2.3) is equivalent to the variational problem: find B E V (resp. A E W, B = curl A)5 2 satisfying: (2.8)

a(B, B) = ao(A, A) = + - lr.Alr dr ,53 - - 1 i -2 r

VB = curl A E V, or V A E W

-and admits one and only one solution.

Proof i) Let us show that if B E V satisfies (2.8), then (2.3) ii) and iii) are satisfied: We choose successively A in (2.8) such that A E ~(Q)3 (with div A = 0), since A E ~(Q')3 (with div A = 0); (2.8) then gives curl B = 0 in Q and Q', hence (2.3) ii). Applying Green's formula (1.17) in Q (and its analogue in QI) for A E ~(1R3)3, with div A = 0, or again for A E W, (2.8) gives, with the notation Bg = Big and Bg, = Big':

a(B, B) = ao(A, l) = + ~f (Bg A n).Adr - ~ f (Bg,. A n).Adr, 2fl r 2fl r

hence

(2.9) - - 1f[ B J-a(B, B) = ao(A, A) = - - - A n .Alrdr. 2 r fl r

Now ~he set_ {g E Hi/2(r)3, t g.ndr = o} is identical to the set of traces g = A Ir for A E W (this follows from Remark 8 in 1); then the comparison of (2.9) with (2.8) gives the condition (2.3) iii). ii) Further, the mapping L J : A E W i (1R 3 )3 ~ ~ f lr.Alrdr is a continuous 2 r 52 With Vand W given by (2.4). To simplify the proof we assume that Q and Q' are connected. 53 We use here the (improper) notation f for the brackets [" of the duality HI12(T), H ~ 112(T),


linear form since Jr E H-l/2(r)3. Moreover, for all


It can be verified that the space WI (/R 3)' can be identified with the space denoted by W- I(/R3) and defined with the help of the Fourier transformation by: W- I (/R 3) is the closure of 9' (/R 3) under the norm

u E 9'(/R 3) f-> (t3 la~)12 dy ) 1/2 Making use of the notations (2.4), (2.5), and proceeding in a way analogous to that in proposition 1, we can prove:

Proposition 1'. For each given current JEW-I (/R3)3 with div J = 0, the problem (2.10) is equivalent to the variational problem: find B E V (or A E W, B = curl A) satisfying:

- - 1 - - - -a(B, B) = ao(A, A) = 2: (1, A>, VB = curl A E V (or A E W)

(where (, > denotes the duality WI (/R 3)3, W - I (/R 3)3), and admits one and only one solution. 0 Remark 3. In the case where the magnetic media considered are of equal magnetic permeability J.L and J.L' (one can then take J.L = ti = 1 by changing B to J.LB if need be55 ), we get the following simple result: The surface current J r is equal to the jump [ - oAlon In the negative of the normal derivative to A (the solution of(2.8 across r, and A is the simple layer potential of the "charge density" J r, i.e. 56 A is given by:

(2.11 ) A(x) = ~ f J r(Y) dr(y) , a.e. x E /R3 . 4n r Ix - yl

Proof i) Using the formula: curl curl A = - LlA + grad div A ,

and (2.3) ii), we see, since div A = 0, that A is harmonic in Q and Q'; with the condition A E WI(/R 3)3, we then have (see Chap. IV and Chap. VII) that AlrEHI/2(r)3 and oAlonlr EH- 1/2(r)3, with [AJr=O, but (a priori), [oAlon]r #- 0. ii) From formula (1.26), with v replaced by A (it is necessary in fact to use a sequence Vn regularising A), we obtain:

on each side (r + and r _) of r. Taking differences, we then have: (2.12) divr[nAJr + 2H[n. AJr + [:n (n. A) 1 = 0, in H -1/2(r) . 55 Such a change is not without consequence for the units employed; in order to ensure the homogen-eity of the formulas for these units it is advisable that the coefficient J1 should still appear. 56 See Chap. II, 3 and Chap. XIB. 2.


Now [AJ, = 0, [A. nJ, = 0, [nAJ, = 0, hence (2.12) gives:

(2.13) [ :n (n. A) 1 = . We shall subsequently show57 that the condition (2.3) i) implies:

[B.nJ,=O, thus [curlA.nJ,=O; hence the jump in the tangential derivatives of A across r is null. 58 Denoting by (e 1 , e2 , e3 ) an orthonormal basis for [R3, we have:

[n. :~ 1 = [O~l (A.n) 1 =[n 1 :n (A. n)l + [n(e 1 )grad(A.n)J, = 0. Now the first component of J, is, using (2.3) iii):

Jll = - ([curlAJ, A n)l = - t nj [O;)A1 J +.t nj[~AjJ j=l uXj ' J=l ux 1 ,

thus

J ll = - [O~11 Working similarly with the other components of J" we obtain the result asserted: J, = [-oA/onJ" and (2.11), which concludes the proof. 0 The trace A I, of the simple layer potential A on r being obtained from J, by the mapping ff'59 (see Chap. II, 4 and Chap. XIB, 2, (2.28)):

(2.14) AI, = ff'J, A(x) = - dr(y) a.e. x E r , ( If J ,(y) ) 4n ,Ix - yl the (free) energy W of the solution B = curl A of (2.3) or (2.8) is then given (using (2.2), (2.5), (2.8) and (2.14)) by:

(2.15) W = ~ t J,ff'lrdr (1, E H-1/2(r)3) , which concludes Remark 3. o b) We now propose to find all solutions of(2.1) satisfying (2.2). As a first result we have:

Lemma 1. The set of solutions B of(2.1) satisfying (2.2) is identical to the set of pairs {Bg, Bg.} E U(Q)3 X L 2 (Q')3 (with Bg = Big, Bg. = Big.) satisfying:

57 See Lemma I below. 58 Which is also directly due to: [AJr = O. 59 We shaH see in Chap. XIB, 2 that ! is an isomorphism from H -11~(T) onto H ' 12(T).


(2.16) { div Bo = 0 , curl Bo = 0 div Bo' = 0, curlBo' = 0 in Q'

in Q

(2.17) n,Bolr = n.Bo,lr on r. Proof i) Being given any B satisfying (2.1) and (2.2), then Bo = Blo and Bo' = Blo' satisfy (2.16); further, for all ({J E E0(1R3), Green's formula (1.10) applied to Q and Q' gives:

(B, grad ({J) = r B grad ({J dx + r B grad ({J dx = i [B. n]r({J dr , Jo Ja r

whence (2.17). ii) Conversely, given {Bo, Bo'} E L2(Q)3 X L 2(Q')3, with (2.16), (2.1 7), the func-tion BE L2(1R3)3 such that Blo = Bo, Blo' = Bo' satisfy (2.1): in effect, for all ({J E E0(1R3), by Green's formula (1.10):

< div B, ({J> = - (B, grad ({J) = - r Bo grad ({J dx - r Bo' grad ({J dx Jo Jo'

= r (div Bo)({Jdx + r (div Bo')({Jdx Jo Jo'

o

We can now use the results of 1.3 to decompose Bo and Bo" By Proposition 2, 1 and Remark 7, 1, Bo and Bo' have the form:

{ Bo = grad o + Bo with o E yt l(Q) , (2.18) -Bo' = grad o' + Bo' with o' E yt 1 (Q') ,

U sing condition (2.1 7), we further obtain:

(2.19) oo I = oo' I on r on r'

Thefunctions o and o' are called (scalar) magnetic potentials61 (see Fournet [1] p. 214 to 220). It may be noted that o and o' must satisfy somewhat stronger conditions than o E yt1(Q) and o' E yt1(Q/). In effect from condition (2.19),

. oo I oo' I f and puttIng g = -- = -- ,cPo and cPo' must be such that g dr = 0 & r & r ~ and flO; g dr = 0 for each (bounded) connected component of Q and Q/, thus such that ti g dr = 0 for each connected component of r. 60 For all notations used here, consult I and the Table at the end of this chapter. 61 To within a coefficient -1/J1.


Hence, putting:

(2.20) yt' II (Q) = { qJ E yt' I (Q), t, ~: dr = 0 ,Ito n(T) } , yt'11(Q') = {qJ E yt'1(Q'), t, ~: dr = 0, i = 1 to n(T)} ,

we must have


We now express the surface current 1r with the aid of (2.18). Under condition (2.3) iii), with


and the energy W = WB corresponding to B (see (2.2)) is then such that:

(2.30) If 1 1 If 22 d WB = WBI + WBl = - J r A Ir dr + - J r A Ir r. 2 r 2 r

c) Some physical interpretations. (Free) energy; magnetic induction flux; current intensity. We have seen (in (2.29)) in the preceding section b) that each solution B of (2.1) with (2.2) is the sum of two different types of solution Bland B 2 of (2.1) with (2.2) (see (2.27)). Here we shall now give some physical interpretations related to each type of solution, and more particularly we shall give expressions for the energy corres-ponding to these (types of) solutions. Let us emphasise here that the surface currents J i and J i corresponding to these solutions are not assumed to be known (they will be derived from Bland B2 via (2.25)). We will successively study, first in c.l) the case where B = B2 and hence with B 1 = 0 (making the further assump-tion that the restriction of B 2 to Q' is null), then in c.2) the case where B = B 1 with B2 = O. c.1) Consider a magnetic induction B such that: (2.31 ) BIl!' = Bl!' = 0 ; B is thus the solution of (2.1) with (2.2) (because of Lemma 1), without magnetic potential (see (2.18)). Since the space !HIl (Q) has finite dimension, B is therefore of the form:

N (2.32) B = " y.B. 66 ~ J J' j= 1 with Bj Il! = grad qj, qj E H 1 (Q), where qj is a solution of (1.49)j (see (1.67)). The (free) energy corresponding to B is then given (see (2.2), (1.71) with (1.69)) by

(2.33) 1 i 2 1 ~ 67 W = - IBI dx = - ~ LijYjYi' 2J1 l! 2J1 i.j= 1

The flux cfJ i of magnetic induction B across the cut Li of the open set Q is then

(2.34)

Hence the energy W corresponding to B is:

(2.35) N 1 W = L - YicfJi' i = 1 2J1

Thus Yi appears as the dual variable to the flux in the energy expression, and must

65 For the definition of this space see Proposition 1, 1. 66 With N = b,(Q) = dim U1J,(Q) here. 67 Do not forget that for the formulas (2.33) and (2.34) we have [qj]r, = bij (see (1.49U, which will falsify any dimensional analysis involving the units.


therefore be interpreted as characterising a current intensity. We shall accordingly put Yi = Ii to encapture this interpretation. Such an interpretation obliges us to make precise the units being used in (2.32). If we give to Yi the dimension of an intensity: Yi = Ii (in Ampere), then in (2.32), the magnetic induction B being expressed in tesla (i.e. in weber per m2 : Wb/m 2 ),68 Bj must be expressed in tesla per Ampere (hence in Wb/m2A); because of the relation Bj = gradqj, qj must be in Wb/mA, hence in henry per metre (Him) which is the unit of permeability. Thus we can replace qj by ii j = J.Lqj (or again in (1.49)j, [qj]Ii = bij by [iijJr. = f.1bij) where f.1 is the permeability of the medium con-sidered, which gives Bj = grad ii j The coefficients of inductance (which should be expressed in henry must now be defined (using (1.69 by:

(1.69)' - def 1 I Lij = - grad iij grad iii dx = f.1 Q

so that formula (2.33) is then written:

112 1,,-(2.33)' W = ~ B dx = - L., LII (in joule) , 2f.1 Q 2 i,j I} I } and (2.34) and (2.35) become:

(2.34)' &i = f B. n dJ: i = L Ij f ~iij dJ: i = L ii)j (in weber) , Ii j Ii un j

(2.35)'

Denoting by (iij)i,j~ lioN the matrix inverse of the matrix iij' we determine Ii from the given B by inverting (2.34)':

(2.36)

If we are given the surface current (in fact the surface current density) 1 r which creates the magnetic induction B (see (2.3, and if we denote respectively by l i , Ai the surface current and the vector potential associated with Bi that is to say:

1 li = ~ Bi /\ nlr, Bi = curl Ai with div Ai = 0, we can then write:

(2,37) N

lr = I IJj, j~l

N A = " I.A .. 69 ~ J J'

j~ 1

and since the energy W associated with B is still given (see (2.2) and (2.8 by:

(2.38) 1 f N 1 f W = -2 l r Alr dr = . ~ - IJj li A j lrdr, r I,)~ 1 2 r

68 Hence (2.32) can be written B = L IjBj . j

69 The terms J i, Ai' Bi (which are vectors) are not to be confused with the vector components J, A, B (which are scalars).


we obtain by comparison with (2.33)' (and (1.69)'):

- I faiL Lij = J;Ajdr = -;-- d1'; , r Ii un

(2.39)

and:

(2.40) t J rAk dr = jtl I j t JjAk dr = jtl ijkIj ; whence finally the current intensity Ii with the aid of the surface current J r by:

(2.41 )

Example 1. In the particular case where the domain Q is a torus, one cut 1'1 suffices to render Q simply connected, thus N = 1, the space IHlI (Q) has one dimension, and the formulas (2.34)', (2.35)" (2.33)' have the simple expressions (on using an arbitrary current intensity II with Jr = IIJ I , see (2.37)):

{ cPI = i 11 11 = flux of B across 1'i'

1 - 2 1 - 1 1 -2 W = -2 Lilli = -2 11


or again, with (2. t 9), and with the usual notations:

(2.45)

(2.46) WB = - ~ f [t] o


{ i) curiE - 0 III [R3 ,74 (2.50) ii) div E = 0 in Q and Q' ,

iii) [eE. nJr = Pr gIven.

Let us put

(2.51 ) der V = H(curl 0, [R3)

From the global Poincare lemma in [R3, for all E E V, there exists one and only one function 4> E W I ([R3) such that: (2.52) E = - grad~7S

(~ is called the potential or electric potential of the electric field E). Let us define on V the bilinear form:

(2.53) - der e f - e' f -a(E, E) = -2 E.Edx + - E.Edx, Q 2 Q'

"IE, E E V

_ der _ (or again on WI ([R3), ao(,


Q') for E ~([R3) or Wl([R3), (2.54) gives, with the notations Eo = Elo and Eo' = Elo':

thus:

(2.55) - - 1 f -a(E, E) = ao(


find E E V (or f/J E Wl(~3), E = - grad f/J) satisfying: (2.57)

a(E, E) = ao(f/J, ) = ~ (p, ), VE = - grad E V or E Wl(~3), (where


ii) Conversely, given {En' En'} E L 2(Q)3 X L 2(Q')3 with (2.60) and (2.61), the function E E L 2([R3)3 such that Eln = En, Eln' = En' satisfies (2.48): in effect for all q> E tJ([R3)3, from Green's formula (1.17):

dx + f En,curlq>dx n n'

= f (curl En)q> dx + f (curl En,)q> dx n n'

+ teEn 1\ n - En' 1\ n).q>dr = o. 0

We can now use the results of 1.3 to decompose En and En', on noting that E is of the form E = - grad, E Wl([R3); thus: (2.62) We write in a manner analogous to (2.18):

{ - - 79 En = -grad n + En with n E II(Qf9 En E 1H12(Q) , (2.63) -

En' = -gradn' + En' with n,EII(Q'), En' E lHI 2(Q') Using condition (2.61) we further obtain: (2.64) n 1\ grad nlr = n 1\ grad n' Ir ; we again write with the notation (2.23):

----+ ----+

(2.65) curlr n = curlr n' ,

which expresses that the tangential derivatives of n and n' on r are equal on r, and thus that the traces nlr and n,lr differ only by a constant function on each connected component of r. Noting that the functions n and n' are not uniquely determined from En and En' by (2.63), one can demand that it be possible to choose the functions n and n' in II(Q) and II(Q') in such a way that their traces nlr and n,lr on r, should be equal. We now go to show that this is indeed the case, but that the choice is then unique (starting from the given En and En')' To do this it will be necessary for us to recall and to give some definitions. We have already defined, in Chap. II, 5, the interior and exterior capacity operators C and Ce respectively, in a classical framework. Anticipating Chap. XIB, we give here the definitions of these operators in the framework of the Sobolev spaces HI/2(T), H- I/2(T). For all hE HI/2(T), let u E I(Q) and u' E I(Q') be such that ulr = hand

78 This is also a consequence of the relation (2.61). 79 See (2.20) and the notations in 1 (and the table at the end of this chapter).

256

u'lr

(2.66)

Chapter IX. Examples in Electromagnetism and Quantum Physics

h;80 we put:

Ch=auj EH~I/2(T), an r

It can be verified that C and C. are continuous mappings from H 1/2(T) into H ~ 1/2(T), whose images and kernels arc then:

(2.67)

ImC = {g E H~I/2(T), Lli gdT = 0, i = I to n(Q)}, ker C = {h on T, h loni = constant, i = I to n(Q)} ,

ImC. = {g E H~I/2(T), Jmi gdT = 0, i = I to n(Q')} ker C. = {h on T, hlon; = constant, i = I to n(fl') hloD" O} .

Denoting by Cn and Cg' the following spaces of constant functions:

(2.68) { Cg = {ep on Q, ep Ini = constant, i = I to n(Q)} , Cn = {ep on Q', epln; = constant, i = I to n(Q'), eplg" = O} ,

we see that ker C and ker C. are the spaces of the traces of functions in Cn and Cg .. We can now define these spaces: (2.69)

H;;I/2(T) ~ {gEH~ 1/2(T), t gdT = 0, i = I to n(T)} = 1m C n ImC., H;/2(T) ~ {hE H 1/2(T), Li ChdT = Li C.hdT = 0, i = 1 to n(T)81 }

= {h E H 1/2(T), Ch E H *~ 1/2(T), Ceh E H;; 1/2(T)} , def

;I(Q) = {ep E 1(Q), eplr E H;/2(T)} , def

;I(Q') = {ep E 1(Q'), eplr E H;/2(T)} . On recalling that the space 11(Q) is defined by (2.20) and, on again putting (see (2.20)'): (2.70) we can now state

80 U and u' are then the solutions respectively of the interior and exterior Dirichlet problems:

{ JU = 0 in Q, U E H!(Q) {JU' = 0 in Q', u' E W!(Q') ulr = h given, h E H !/2(T) , u'lr = h given, h E H 1/2(T) .

81 Note that because of (2.67) there are only n(T) independent conditions and not 2n(T). 82 We recall that

L?(Q)' = gradJf'"(Q) = (curlH!(Q)3) n (gradH!(Q)) L;.(Q')' = grad.tf"(Q') = (curl WI(Q')') n (grad W!(Q')).


Proposition 3. With the definition (2.20), (2.68) to (2.70), and (2.67), we have: i) the spaces H 1/2(r), Yl' 11 (Q), Yl' 11 (Q') have direct sum decompositions:

(2,71 ) { Hl/2(r) = H;/2(r) EB kerC EB kerCe ,83 Yl'11(Q) = Yl';I(Q) EB CQ , Yl'11(Q') = Yl';I(Q') EB CO' ;

257

ii) the gradient is an isomorphism84 of Yl'; 1 (Q) onto L;g(Q) and of Yl'; 1 (Q') onto L 2 (Q'). rg , iii) the traces: Yo (You = ulr) and y 1 (y 1 U = au/an Ir) are such that: Yo (resp, y d is an isomorphism of Yl';I(Q) (resp. Yl';I(Q')) onto H;/2(r) (resp, H;1/2(r)): iv) the capacity operators C and Ce are isomorphisms of H;/2(r) onto H;I/2(r). The proof is not difficult, beginning with the general results on traces (Chap. IV) and capacity operators (Chap. XI B, 2) (that is to say, C and Ce are continuous mappings of H 1/2(r) into H -1 /2(r)), and is left to the reader. Essential use is made of the fact that the matrices of influence coefficients Cij and C iJ (see Chap, II, 4) with i andj = 1 to n(Q') for Cij' and with i andj = 1 to n(Q) for C:} are invertible. This proposition implies that in (2.63) (for gjven En and En') one can choose


Hence the set of solutions E of (2.48) with (2.49) is also identifiable with the set of triples {, 'I'n, 'I'n'} with , 'I'n and 'I'n' E W 1([R3), electric potentials such that

= {In, In'} E Yf,!l(Q) x Yf,!l(Q'), 'I'n = {'I'nln, 'I'nln'} E lHIi*(Q) x Cn" 'I' n' = {'I' n' In, 'I' n' In'} E Cn x IHI i*(Q') ,

with:

(2.73) E = - grad - grad 'I' n - grad 'I' n' = - grad ( + 'I' n + 'I' n') . We note that the boundary condition on Q (or Q'):

E A nlr = 0 used as the boundary condition for the perfect conductor implies the vanishing of the potential ; whenever Q is further occupied by a perfect conducting medium, then one must also impose En = 0, and the set of such electric fields 86 becomes identifiable with the space 1H12(Q') (of dimension n(Q)). Let us now express the surface charge density Pr with the aid of (2.50) iii) and (2.63). With {n, n'}, E = {En' En'} we get:

(2,74) [ a] -P=- ea;.; r+[eE.n]r,

Putting

(2,74)' p} ~ - [ea] , Pi- ~ [eE.n]r; an r

we have:

(2.75) P = p} + Pi- Note that the two current densities pf. and Pi- are not a priori orthogonal in L 2 (r): in effect p} E H;l/Z(r), thus pf. is orthogonal to the functions constant on (each connected component of) r. But Pi- is not necessarily constant on each connected component of r. 0

Now let us put in (2.63) (see also (2.73)): def

{ El -gradl with l = = {n, n'} E W 1 ([R3) , (2.76) - -EZ = {En, En'} = -gradZ with z = 'I'n + 'I'n,E W 1([R3), l E Yf,!l(Q) x Yf,!l(Q'), EZ E IHI z(Q) x IHIz(Q') ,

Then (2.63) can be written: E = E 1 + EZ, and the energy W = WE (see (2.49)) associated with E is given by:

(2.77)

86 Also known as the set of electrical equilibrium states.


Using (2.54), and since WE = a(E, E), we also get:

(2.78) 1 f 1 f lId 1 f 2 A,2 d WE = - Prlr dT = - Pr Ir T + -2 Pr'l' Ir T. 2 r 2 r r

o

c) Some physical interpretations We have seen in the preceding section b) that each solution E of (2.48) with (2.49) is the sum of two different types of solutions Eland E2 of (2.48) with (2.49) (see (2.76)). Here we shall now give some physical interpretations related to each type of solution, and more particularly we shall give expressions for the energy corres-ponding to these (types of) solutions. Let us emphasise here that the surface charge densities Pf and P;' corresponding to these solutions are not assumed to be known (they will be derived from Eland E 2 via (2.47)'). We w

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