8.3 BOUNDED DOMAINS: TREES, H-~ 231
Remark
8.4. When jg is sinusoidal, using this scheme is a viable alternative
to solving (18) directly. One should use an informed guess of the solution
as initial condition, and monitor the time average over a period,
(~)2
-10~ft_ 2x/c0t
exp(i~ s) k(s) ds, duly approximated by a sum of the kind
1
N- ~._ 1 N
exp(i0~(m + 1/2 -j)St)k
m+ 1/2-j,
where N~t = 2~/0), the
period.-This will converge, relatively rapidly (in no more than three or
four periods, in practice) towards the solution K of (18). Such a "time
domain" approach to the harmonic problem can thus be conceived as
another iterative scheme to solve (18). Fast Fourier Transform techniques
make the calculation of time averages quite efficient. 0
8.3 BOUNDED DOMAINS: TREES, H-~
Now we change tack. Let D be a simply connected domain containing
the region of interest (here the conductor C and its immediate
neighborhood, Fig. 8.6), the inductor I, magnetic parts where ~t ¢ ~t 0, if
such exist, and suppose either that D is big enough so that one can
assume a zero field beyond, or that n x h = 0 on 3D for physical
reasons (as in the case of a cavity with ferromagnetic walls, used as a
shield to confine the field inside). We thus forget about the far field and
concentrate on difficulties linked with the
degeneracy
of the eddy-current
equations in regions where c~ = 0.
D
FIGURE 8.6. Computational domain D, containing the region of interest, and
large enough for the boundary condition n x h = 0 on 3D to be acceptable. (In
practice, the size of the elements would be graded in such a case, the farther from
C the bigger, the same as with Fig. 7.2.)
232
CHAPTER 8 Eddy-current Problems
8.3.1 A constrained linear system
Let m be a simplicial mesh of D. Functional spaces, a bit different now,
are IH = {H
E InLrot(W)"
n x H = 0 on 3D} and
(21) IH g = {H ~ IH : rot H
= Jg
in D - C},
(22) IH ° - {H ~ IH : rot H
-- 0
in D - C}.
Now we know the paradigm well, and we can state the problem to solve
without further ado:
find
H E IH g
such that
(23)
~D i
co ~t H. H' + ~c rs-1 rot H. rot H' = 0 V H' ~ IH °.
As we intend to enforce null boundary conditions on the boundary
of D, let us remove from N, E, 9" the boundary simplices, as we did
earlier in 7.3.1, and for convenience, still call N, E, J-, T the simplicial
sets of this "peeled out" mesh. Apart from this modification, the notation
concerning the spaces W p and the incidence matrices is the same as
before. In particular, W 1 is the spanmwith
complex
coefficients, this
m
timemof Whitney edge elements w e, for all e in E. As this amounts to
have null circulations along the boundary edges, a field in W 1 can be
m
prolongated by 0 to all space, the result being tangentially continuous
and therefore an element
of
In2rot(E3 ). So we can identify Wlm with a
subspace of
In 2
(E3)
Let us set IH = W 1 denote by IH the isomorphic
rot " m m*
finite-dimensional space C z, composed of all vectors I~ = {E e : e ~ E },
and call E = #E the number of inner edges. For v and ~' both in IH,
we set
(U, U') = ~ e~ z Ue . U'e
------E ee E
(Re[ue] + i Im[ue])" (Re[u'e] + i Im[u'e]).
(Again, beware: This is not the Hermitian scalar product.) This way, an
integral of the form fD Ct U. U', where ¢z is a function on D (such as kt,
for instance), possibly complex-valued, is equal to (Ml(Ct)u, c') when u
U'
-- Ee~E UeW e
and
U'--Ee~E eWe .
We know from experience the eventual form of the discretized
problem: It will
be find H
E IH g
such that
m
(24)
YD i
co gt H. H' + Yc rs-1 rot H. rot H' = 0 V H' ~ IH°m,
where IH g
and IH ° are parallel subspaces of IH .
m m m

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