8.2 INFINITE DOMAINS: "TRIFOU"
225
The electric field has thus disappeared from this formulation. One
easily retrieves it in C, where E = r~ -1 rot H. One may find it outside C
by solving a static problem 2, formally similar to magnetostatics in region
O, but this is rarely called for. (And anyway, this outside field would be
fictitious, as already pointed out.)
To simplify, and to better emphasize the basic ideas, we first consider,
in Section 8.2 below, the case when the passive conductor is contractible
(i.e., simply connected with a connected boundary, cf. A.2.3), as in Fig.
8.4. It's obviously too strong a hypothesis (it doesn't hold in the above
case), but the purpose is to focus on the treatement of the outer region,
by the same method as for "open space" magnetostatics in Chapter 7. In
Section 8.3, we'll reintroduce loops, but forfeit infinite domains, thus
separately treating the two main difficulties in eddy-current computation.
8.2 INFINITE DOMAINS: "TRIFOU"
The key idea of the method to be presented now, already largely unveiled
by the treatment of open-space magnetostatics of Chapter 7, is to reduce
the computational domain to the conductor, in order not to discretize the
air region 3 around. The method, implemented under the code name
"Trifou", was promoted by J.C. V6rit6 and the author from 1980 onwards
[B1, BV, BV'], and provided at the time the first solution of general
applicability to the three-dimensional eddy-currents problem.
8.2.1 Reduction to a problem on C
We tackle problem (13), assuming C contractible (no current loops, no
non-conductive hole inside C). In that case, the outside region O also is
contractible.
2Namely the following problem: rotE=- i0~pH, D=¢0E, divD=Q in O, with
n x E known on the boundary S, where Q is the density of electric charge outside C. The
difficulty is that the latter is not known in region I, for lack of information on the fine
structure of the inductor. One may assume Q = 0 with acceptable accuracy if the objective
is to obtain E near C (hence in particular the surface charge on S, which is ¢0 n. E). Such
information may be of interest in order to appraise the magnitude of capacitive effects.
3It's not always advisable thus to reduce the computational domain D to the passive
conductor C. It's done here for the sake of maximum simplicity. But "leaving some air"
around C may be a good idea, for instance, in the presence of small air gaps, conductors of
complex geometry, and so forth. Methods for such cases will be examined in Section 8.3.
226 CHAPTER 8 Eddy-current Problems
s
I /
/
/
t c
FIGURE
8.5. Model problem for the study of the hybrid approach in "Trifou",
finite elements in the conductor C, and integral method over its boundary S to
take the far field into account. The support of the given current density
Jg
is the
inductor I. Contrary to Fig. 8.2, C here is loop-free, and we restrict consideration
to this case to separate the difficulties. Section 8.3 will address loops (but shun the
far-field effect).
Let's keep the notations of Chapter 7: • is the space of magnetic
potentials (the Beppo Levi space of 7.2.1), ~o is composed of the restrictions
to O of elements of ~, and the set of elements of ~o that have in
common the trace ~s is denoted ~o(%)" Let ~00 stand 4 for the
subspace {~ ~: ~=0 on D}. Set
IK g--
{H E
IH g" SE3H. grad ~'= 0 V ~'~ ~00}
(the support of the integrand reduces to O, in fact), and
IK 0--
{H E IH °" ~E3 H. grad ~'= 0 V ~' ~ O°°}.
We note that
(14) IH ° = IK ° @ grad ~00,
by construction, and that IK g
= H g + IK °, for
H g
is orthogonal
to grad ~00, since div
H g--
0.
(Cf. the inset drawing.)
By their very definition, the
elements of
IK g
and of the
parallel subspace IK ° satisfy
div H
--
0 in O. This property
y HF;
IK °
00
grad
is shared by the required solution, since div H
=
~10 -1 div B = 0 in O. One
may therefore expect to find this solution in IK g. Which is indeed the
case"
4The notation ~0 is reserved for an analogous, but slightly larger space (see below).
8.2 INFINITE DOMAINS: "TRIFOU" 227
Proposition 8.2.
The solution H of Problem
(13)
lies in
IK g.
Proof.
By letting H'= grad ~' in (13), where ~' roams in cI) °°, one gets
0 =
ico SE 3 ~l
H. grad ~'= ico~t 0
rE3 H.
grad ~' V ~' ~ cI) °°,
and hence H E IK g. ~)
To find the point of stationarity of
Z(H)
in
IH g,
it is thus enough to
look for it in IK g and to check that no other, spurious critical point is in
the way. Indeed,
Corollary of Prop. 8.2.
Problem
(13)
is equivalent to
find H
E IK g such
that
(15) if,0 SE 3 ~1, H. H' q- S C I~ -1 rot H. rot H' = 0 V H' E IK 0,
since this is the Euler equation for the search of critical points of
Z(H)
in
the affine subspace IK g and it has at most one solution.
Our effort, now, will concentrate on showing that Problem (15) is in
fact "posed on C", meaning that a field in IK g (or in IK °) is entirely
determined by its restriction to C. I expect this to be obvious "on
physical grounds", but this doesn't make the proofs any shorter. We are
embarked on a long journey, till the end of 8.2.3. The operator P of
Chapter 7 will play a prominent part in these developments.
Remark 8.2. Set cI) °={~ ~" grad~=0 on C}. By introducing as
before the orthogonal subspaces IK g° and IK °°, one would have IH ° -
IK °° @ grad ~0, and one could proceed with the same kind of reduction,
with IK g° strictly contained in IK g. This may look like an advantage, but
in practice, it makes little difference. 0
Exercise 8.1. Show that Ss n. H-
0,
and prove the analogue of Prop. 8.2
in the context suggested by Remark 8.2.
8.2.2 The space H~, isomorphic to IK g
Let now HcI) (treated as a single symbol) stand for the vector space of
pairs {H, %}, where H is a field supported on C and ~s a surface
potential "associated with" H, in the precise following sense 5"
(16) HcI) = {{I-I~ ~s} ~ IL2rot (C) x H1/2(8) • H S -- grad s ~s},
where grad s denotes the surface gradient. Note that the projection of
HcI) on the first factor of the Cartesian product
In2rot(C) X
H1/2(8) is not
5Refer to Fig. 2.5 for H S, the tangential part of H.
228
CHAPTER 8 Eddy-current Problems
In2rot(C) in its entirety, for there are constraints that H must satisfy, in
particular n. rot H-- 0 on S. On the other hand, ~s may be any
function in H1/2(S). One provides HcI) with its natural Hilbertian norm,
induced by the norm of the encompassing Cartesian product. Then,
Proposition
8.3. HcI)
is isomorphic to
IK g
and
IK °.
Proof.
Since C is simply connected, and rot H g= 0 in C, there exists
cI) g E n2grad(C) such that H g =
grad
cI) g
in C. Let us still denote by
cI) g
the
harmonic continuation of this function outside C. Now, take H
E IK g.
One has rot H
= Jg = rot(H g --
grad ~g) in O. Since O is simply
connected, there exists a unique • in BL(O) such that the equality H
--
Hg+
grad(~-~ g) hold in O. By restricting H and • to C and S, a map
from H
to
the pair {H, ~s} of H~ is therefore defined. Conversely, such
a pair {H(C), a)s} being given, let • be the exterior harmonic continuation
of %. Set H equal to H(C) in C and to
Hg+
grad(~- ~g) outside C.
This enforces H
E
IL2rot(E3 ), because both tangential traces H(C) s and H g
+ grads( % - ~g) = grads% coincide, after (16). In O, rot H
= Jg
and div H
= 0 by construction, whence H ~ IK g. Moreover, the one-to-one
correspondence thus established (in a way that would apply as well to
IK °, just consider the special case jg = 0) is an isometry. Hence the
announced isomorphisms: For if H is the difference between two elements
of IK g then H= grad • outside C, and
SE 3 ]H [2 _ff
SC I rot
H ]2 = SC I H ]2 q_
SC I rot
H ]2 q_ SO ]
grad
• ]2
--Sc ]HI2+ S C
]rotH]2+SsP~s~s ,
which is indeed the square of the norm of the pair {H, %} in the product
Inarot(C) x H1/2(S). 0
Remark
8.3. The isomorphism depends of course on ~g which in turn
depends on jg, up to an additive constant. 0
8.2.3 Reformulation of the problem in H~
So, let
~gs
be the function on S associated with
Jg
that specifies the
above isomorphism. One has
Z(H) =
i c0 YE3 ~t
H 2 q- S C (~-1
(rot
H) 2
= ic0 ~c ~t I~ + Sc •-1 (rot
H) 2 q- i(.l)~l, 0 S O
(H g q- grad(~-
cI)g)) 2,
and we are looking for the critical point of this function in HcI). Since,
thanks to the properties of P,
8.2 INFINITE DOMAINS: "TRIFOU"
229
IO (H g -}-
grad((I)
- (I)g)) 2 -- Io (H g -
grad
(I)g) 2 q- 2 I S n. H g (I) S
+ ~s P~s ~)s- 2 ~s P(I)gs ~s'
Problem (15), equivalent to the initial problem (13) under our assumptions,
/v
amounts to finding the critical point of the function z (equal to z up to
a constant) thus defined:
Z({H, (I)s}) -- i(l) [I C ~t H a q- ,0IS
P% %] + ~c (~-1 (rot
H) 2
+ 2 i0),0 ~s (n. H g - P~)g)%,
whence, by taking the Euler equation, the following result (index S is
understood in ~s, ~)'s and ~)gs):
Proposition
8.4. When C is contractible, (13) is equivalent to find {H, ~)} in
H(I) such that
(17)
i~ [Jc ~ H. H' + go Is P~) ~)'] + Ic (~-1 rot H. rot
H'
= i0)~t0 Js ( P(I)g -n.
H g) (I)' V {H',
(I)'} E U(I).
This is the final weak formulation, on which "Trifou" was based.
The pending issue is how to discretize it. Clearly, H and H' in (17) will
be represented by edge elements, and • and ~)' by surface nodal
elements (these are compatible representations, thanks to the structural
properties of the Whitney complex). The discretization of terms fs P~) ~)'
and ~s P~g (I)' by Galerkin's method will make use of the matrix P of
Chapter 7 (Subsection 7.4.5, Eq. (48)).
8.2.4 Final discrete formulation
Let K
-- {H e "
e ~ E°(C); ~n " n ~ N (S)} be the vector of all degrees of
freedom (complex valued): one DoF ti e for each edge inside C (that is,
not contained in S) and one DoF
(I) n
for each surface node. The expression
of H in C is thus
H -- Z e~ '£0(C)He We -ff ~n ~ N (S) ~n
grad
w n.
(This is an element
of
wire(C), thanks to the inclusion grad W ° c W1.)
Then (17) becomes
(18) ico (M(B) + BoP)K + N((~) K = icoB0(P~ g
- Lg),
with obvious notations, except for L g, defined by Lgm = ~s n. H g W m for all
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