250 CHAPTER 9 Maxwell's Model in Harmonic Regime
9.2 THE "CONTINUOUS" PROBLEM
9.2.1 Existence
Let's first prove an auxiliary result. Let D be a
regular bounded domain in
E3,
with boundary S.
For simplicity (but this is not essential), assume D
simply connected. Let ~0 be the space of restrictions
to D of functions ~
of L2grad
(E3) for which grad
= 0 outside D. (If S is connected, they belong to
the Sobolev space H10(D), but otherwise, it's a
slightly larger space, for ~ can be a nonzero constant
on some parts of S, as shown in inset.) Call V the following closed
subspace of IL2(D):
V={veIL2(D) • (v, grad~')=0 V ~'e~°}.
Proposition 9.1. Let j be given in
IL2(E3)
, with div j= 0 and supp(j) c D.
Suppose g'_> go in D. There exists a unique A
E IE°(D)
such that
(6)
(B-1 rot A, rot A') = (Jr A') ~ A'E IE O
as well as lEA e V, and the map G = j ---> lEA is compact in V.
(6')
Before giving the proof, note that such a field A verifies
rot(~t -1 rot A) = J, div
lEA --
0 in D, n x A
----
0
on S,
but these conditions are not enough to determine it, unless S is connected.
In that case, V = {v ~ IL2(D) • div v = 0}. But otherwise, V is a strictly
smaller subspace, characterized by ~ n. v = 0 on each connected component
of S, hence as many similar conditions 2 on A, to be appended to the
"strong formulation" (6). Note also that j e V, under the hypotheses of
the statement, so G does operate from V to V.
Proof of Prop. 1. Uniqueness holds, because the kernel of rot in IE ° is
precisely grad H ~° (this is why ~0 was defined this way). The proof
will consist in showing that one passes from j to lEA by composing
continuous maps, one of which at least is compact.
Set u = (X * J), with ~ = x ~ 1/(4~ I xl ), and take its restriction DU
to D. The map j ~ DU thus defined is compact in ILa(D) (cf. [Yo]).
Therefore, the map j ---> rot u e IL2(E3) is compact, too, for if {Jn} is a
2This is one of the advantages of weak formulations: They foster thoroughness, by
reminding one of conditions which one might have overlooked in the first place.
9.2 THE "CONTINUOUS" PROBLEM 251
sequence of V such that
Jn----~ J
(weak convergence--cf. A.4.3), then
u n --~ u by continuity, and hence (dot-multiply by a test field A' and
integrate by parts) rot u ~ rot u. Moreover, J I rot
U n
I 2 -- ~D Jn "
Un*, and
DU n tends to DU, SO the norm of rot u~ converges towards that of rot u,
therefore rot u n tends to rot u. Setting H
=
rot DU, one has thus proved
the compactness of the map j ~ H.
Now let
¢~
L2grad(W) be such that
(p (H + grad ¢), grad ¢')= 0 V
¢'e
L2grad(D).
The solution of this problem is unique up to an additive constant only,
but grad ¢ is unique, and the map H ---) grad ¢ is continuous. Let us set
B
--
~I,(H 4- grad ¢). Then div B = 0 and n. B = 0, SO the prolongation by 0
of B outside D is divergence-free. If one sets A 0 = D(rot(x * B))wagain,
the restriction to Dwthen rot A 0 = B, and the mapping H ~ A 0 is
continuous.
Notice that the tangential trace of A 0 is a gradient, by the Stokes
theorem, for the flux of B through a closed circuit drawn on S vanishes,
2
D
since n. rot A 0 = n. B = 0. For this reason, the set of the • e L
grad ( )
for
which n x (A 0 + grad ~) - 0 is not empty, and there is one among them
(unique up to an additive constant) for which
(e (% + grad ~), grad ~')= 0 V ~'e ~0.
Then A = A 0 + grad • is the desired solution, and grad • continuously
depends
on A0~
with respect to the norm of ILa(D). The map j --~ A is
therefore compact, hence the compactness of the operator G = j ~ CA,
whose domain is the subspace V. 0
Let's call
singular
(or
resonating)
the nonzero values of co for which
the homogeneous problem associated with (5) has a nontrivial solution,
i.e., E ~ 0 such that
(7) (io~
E E, E') 4-
((io~p) -1 rot E, rot
E') -- 0 '~' E' fE IE 0.
Such an E verifies e E e V (take E' e grad wo) as well as rot(p -x rot E) =
CO 2 e E (integrate by parts). In other words, e E = 032 G e E. Thus, e E is an
eigenvector of G, corresponding to the eigenvalue co -2. (One says that
the pair {E, H}, where H = - (rot E)/io~p, is an "eigenmode" of the cavity,
for the angular frequency co.) By Fredholm's theory, there exists a denu-
merable infinity of eigenvalues for G, each with finite multiplicity, and
at the origin in the complex plane. The
not clustering anywhere except . . . 3
3Owing to uniqueness in (6), 0 is not an eigenvalue of G.
252
CHAPTER 9 Maxwell's Model in Harmonic Regime
singular values are thus the square roots of the inverses of the eigenvalues
of G. (A priori, eigenvalues are complex, unless both c and ~t are real.)
Theorem 9.1. For each non-singular value of co, problem (5) is well posed, i.e.,
has a unique solution E, and the map jg ~ E is continuous from ILa(D)
into IE(D).
Proof. Since co is not singular, uniqueness holds. Let's look for a
solution of the form E = - ico A- grad ,v, with A
E IE 0, C
A
E V,
and
~v ~ W0. Set E'= grad ,v' in (5), with ~v' ~ W0. This yields
(ico
c
(A 4- grad ,v), grad ~v')= (jg, grad ~v') V ,v'~ W0,
and hence, since
CA
is orthogonal to all grad ,v',
(8) (ico c grad ,v, grad ~v') = (jg, grad ,v') V ,v'~ ~0,
a well-posed problem in W0, hence the continuity of
Jg -'--)
grad • in
IL2(D).
This leaves A
to
be determined. After (5), one must have
(~t -1 rot A, rot A') = 0 g + icoC E, A') V A' ~ IE °
-- 0 g --
ico c grad ,v,
A') 4- co2 (GO, A') V A' E
IE °.
But this is the Fredholm equation of the second kind,
(1
-
c02G) c A
=
G 0 g - icoc grad ~v),
hence A by the Fredholm alternative, if co is not a singular value, and
provided jg - icoc grad ,v ~ Vmwhich is what (8) asserts. 0
9.2.2 Uniqueness
Hence the question: Are there singular values? For an empty cavity (It
= ~t 0 and c = c0), or with lossless materials ($t and c real and positive),
yes, since all eigenvalues of G are then real and positive. If co ¢ 0 is one
of them and E = e R a nonzero associated real solution of (7) (there is a
real one), then H
=
i hi, with real
h I .
The existence of such a solution
means that a time-periodic electromagnetic field, of the form e(x, t) =
Re[E(X) exp(icot)] - ee(X ) COS cot and h(x, t) =- hi(x ) sin cot can exist
forever in the cavity, without any power expense, and also of course
without loss.
To verify this point, let's start from the equations icoCE - rot H
=
0
and ico~t H 4- rot E = 0, dot-multiply by E and H, add, and integrate over
9.2 THE "CONTINUOUS" PROBLEM
253
D" by the curl integration-by-parts formula, and because of n x E = 0,
this gives
;D E
E 2
+
;D ~
Ha -- 0,
that is, since E = e R and H
=
i h I,
~D E [eR[2--;D ~t ]hi 12.
But the energy contained in D at time t, which is (cf. Chapter 1)
1 JD (e le(t) l 2 2)
W(t) = -~ + p I h(t) I
1 2 1 2
-- 2 ;D E
l e R I
COS 2
o~t +
-~ ;D ~
]hi[ sin2
o~t
__ 1 ;D E
leRl2
1 2
-- 2
--= 2-
~D~
Ihi[
'
is indeed constant, and is the sum of two periodic terms of identical
amplitudes, "electric energy" and "magnetic energy" in the cavity, the
former vanishing at each half-period and the latter a quarter-period later.
Such a behavior seems unlikely in the case of a loaded cavity, since
the energy of the field decreases in the process of yielding heat to the
region C = supp(e") w supp(p"). How can this physical intuition be
translated into a proof? It's easy to see that E and H vanish in C:
Setting E'= E* in (7), one gets
(D;DE"
[EI2+if.OYD E' [EI2+~D(~ 1~t12)-1~ '' IrotE[ 2
+JD(iCoIpl2) -lp' [rotE[2=0,
hence, taking the real part, E = 0 or rot E = 0 on C, hence H = 0, too,
and therefore E = 0 after (2). But what opposes the existence of a
nonzero mode E that would be supported in the complementary region
D - C, what one may call an "air mode" ?
If such an air mode existed, both n x E and n x H would vanish on
3C, which is impossible if E and H are to satisfy Maxwell's equations in
region D- C. We shall prove this by way of a mathematical argument,
here encapsulated as a context-independent statement, cast in non-
dimensional form:
Proposition 9.2.
Let f2 be a regular domain of
E 3.
Let u and v satisfy
(13) i rot u = v, - i rot v = u in f2,
(14) nxu=0, nxv=0 on E,
where E is a part of F with a smooth boundary and a non-empty interior
(relative to
E).
Then u and v vanish in all f2.
254
CHAPTER 9 Maxwell's Model in Harmonic Regime
Proof.
(Though reduced to the bare bones by many oversimplifications,
the proof will be long.) Both u and v satisfy div u = 0 and -Au = u
in ~ after (13). It is known (cf., e.g., [Yo]), that every u which satisfies
(A + 1)u = 0 in some open set is
analytic
there. This is akin to the Weyl
lemma discussed in 2.2.1, and as mentioned there, this result of "analytic
ellipticity" is valid also for div(a grad) + b, where a and b are smooth.
If we could prove that all derivatives of all components of u and v
vanish at some point, u and v would then have to be 0 in all f2, by
analyticity.
The problem is, we can prove this
fact, but only for boundary points such
as x (inset), which is in the relative
interior of Z, not inside f~. So we
need to expand f2 in the vicinity of x,
as suggested by the inset, and to make
some continuation of the equations to
this expanded domain f2' in a way
which preserves analytic ellipticity. To
do this, first straighten out Z around
x by an appropriate diffeomorphism,
then consider the mirror symmetry s
r
f~
y *-- ~-* sy
~;qx
with respect to the plane where Z now lies. Pulling back this operation
to f~ gives a kind of warped reflexion with respect to Z, still denoted by
s. Let us define continuations of u and v to the enlarged domain f2'
by setting u(sy) =- s,u(y) and v(sy) = s,v(y), where s, is the mapping
induced by s on vectors (cf. A.3.4, p. 301). Now u and v satisfy in f~'
a system similar to (13-14), with some smooth coefficients added, the
solutions of which are similarly analytic.
The point about vanishing derivatives remains to be proven. This is
.... 1 2 y~~
done by working In an appropnate coordinate system y ~ {y, y,
1 2
with the above point x at the origin, ly and y charting 2 Z, and
along the normal. In this system, u = {u, u, u }, and uJ(x, x, 0} = 0, for
j = 1, 2, and the same for v. Derivatives 3.u j vanish for all j and i = 1 or
• . 1 3 • 2 1
2 at point x by hypothesis. One has also 3~u = - 13~(3~v -32 v ) = 0 on
• 3 3 3 •
Z, and in the same way, &v = O, 3.u = O, &u = 0 Since rot u =
3 2 1 3 2 z 1 1. . z " 2 1 •
{32U --33U, 03U -- 31U , 01U --32U }, which is
equal to {-
33U , 33U , 0}, lS
also 0 at x, we have 33uJ(x) = 0 for j = 1 and 2. The last derivative to
consider
is 33 u3 ~
div u- (3 1 2
•
lU + ~)2 u ) = 0, since div u = div u = 0 at
point x. This disposes of first-order derivatives.
What has just been proved for a generic point of Z implies that all
first-order derivatives of u and v vanish on Z. Thus, differentiating
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