9.2 THE "CONTINUOUS" PROBLEM 255
(13) with respect to the ith coordinate, we see that 3iu and 3iv satisfy
(13-14), so the previous reasoning can be applied to derivatives of u
and v of all order: They all vanish at x, and now the analyticity
argument works, and yields the announced conclusion.
The anti-air-mode result now comes by setting f2 = D -C and E =
3C, and by choosing a system of units in which t 0 ~t 0 0~ 2 = 1. (Another
corollary is that one cannot simultaneously impose n x H and n x E on a
part of nonzero area of the cavity boundary.)
We may thus conclude that no resonant modes exist if there is a
charge in a microwave oven, and that Maxwell equations have a unique
solution in that case, assuming the tangential part of one of the fields H
or E is specified at every point of the boundary.
9.2.3 More general weak formulations
Thus satisfied that (4), or better, its weak form (5), has a unique solution,
we shall try to get an approximation of it by finite elements. But first, it's
a good idea to generalize (5) a little, as regards source terms and boundary
conditions, without bothering too much about the physical meaning of
such generalizations.
First, let's exploit the geometrical symmetry, by assuming the source
current jg is symmetrically placed with respect to plane E. Then, for
x e E, one has H(x) = - S,H(X), where s is the mirror reflection with
respect to E, and s, the induced mapping on vectors. The boundary
condition to apply is therefore, if n denotes the normal as usual,
(9) n×H=0 on Z.
Therefore, our customary splitting of the boundary into complementary
parts is in order: S = S e u S", with n x E = 0 and n x H
= 0,
respectively,
on se and sh.
Second generalization: Introduce a right-hand side K g in the second
equation (4). This term does not correspond to anything physical (it
would be a magnetic charge current, if such a thing existed), but still it
pops up in a natural way in some modellings. For instance, if the field is
decomposed as H g + ~, where H g is a known field, a term K g -- -- k0 ~t H g
appears on the right-hand side of the equation relative to the reaction
field ~.
This suggests also preserving the possibility of
non-homogeneous
boundary conditions: n x E = n x E g in (4) and n x H ---- n x H g in (9),
256 CHAPTER 9 Maxwell's Model in Harmonic Regime
where E g and H g are given fields, of which only the tangential part on S
will matter. For instance, in the case of Fig. 9.1, one might wish to limit
the computation to the oven proper, that is, the rightmost part of the
cavity. Indeed, the middle part is a waveguide, in which the
shape
of the
electric field, if not its amplitude, is known in advance. Hence the
source-term E g up to a multiplicative factor.
All this considered, we shall treat a general situation characterized
by the following elements: a regular bounded domain D limited by S,
the latter being partitioned as S
= S e k.J S h
and given fields jg
K g
(in
In2div(D)),
H g,
E g (in ILarot(D)). We denote, with the dependence on D
understood from now on, IE
=
IL2rot(D) and also IH
=
]n2rot(D), then
lEg={E~ IE: nXE=nXE g on se},
IE °={E~IE: nXE=0 on se},
IH g -- {H E
IH : n x
H "-
n
x H g on sh},
IH 0-- {He
IH: nXH=0 on sh},
and we set the following two problems:
find E ~ lEg such that
(ic0e E, E') + ((i¢op) -1 rot E, rot E') =
(10)
- (jg
E') + (riCO ~t) -1
K g
rot E') + Ss n x H g . E' V E' ~ IE °,
find
H E
IH g
such that
(i¢0
Ix H, H') + ((iO)E)-I
rot H, rot
H') =
(11) (K g H') + ((ico¢) -1Jg, rot H') -- fs n x E g . H' ~/ H' ~ IH °.
(Beware, there is no relation, a priori, between K g and
H g or
between jg
and Eg.)
Integrating by parts, one easily sees that
each
weak formulation (10)
or (11) solves the following "strong" problem (compare with (4)):
(12)
--i¢0¢E + rot
H= Jg
in D, nXE
=
nXE g
on S e,
i¢.O[LtH+rOtE=K g in D, nXH=nXH g
on S h.
One may therefore solve (12), approximately, by discretizing either (10)
or (11). But (and this is
complementarity,
again!) one obtains solutions
which slightly differ, and yield complementary information about the
exact solution.
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