Computational Electromagnetism

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