164
CHAPTER 6 The "Curl Side": Complementarity
(1) roth=0 in D, (3) divb=0 in D,
(2) nxh=0 on S h (4) n.b=0 on S b,
(5) b=~th in D.
The problem is made well-posed by imposing one of the two conditions
(6) ~c ~" h = I, (7) ~c n. b = F,
2 •
as we saw m 2.4.1. These equations are the same as (2.20)-(2.26), and
little changed in the former model due to geometrical symmetry, except
for one thing: The flux F in (7) is now the flux through one-quarter of
the device, and the computed reluctance will be relative to a quarter
as well.
The layout of Eqs. (1-7) underlines a symmetry of another kind,
which will be our main concern in this chapter: the symmetry of the
magnetostatic equations
with respect to the
b-h
interchange.
This can be
made even more patent by setting the problem as follows: We look for
pairs
{F, I} for which problem (1-7) has a solution. By linearity, they all
lie on the characteristic line I = RF in the F-I plane, and the problem
thus consists in finding R. (This remark, though of moderate interest in
the present linear case, is the key to nonlinear generalization [B3].)
6.1 A SYMMETRICAL VARIATIONAL FORMULATION
We shall strive to preserve this symmetry in the search for a variational
formulation. The functional point of view continues to prevail: We look
for the solution as (first step) an element of some predefined functional
space that (second step) can be characterized as the minimizer of some
easily interpretable, energy-related quantity.
6.1.1 Spaces of admissible fields
By "the" solution, now, we mean the
pair
{h, b}. What are the eligible
fields, a priori? Both h and b will certainly belong to IL2(D), since
magnetic energy is finite. Moreover, rot h = 0 and div b = 0, and if we
had to generalize what we do to cases where a given current density j
2The relative arbitrariness in the choice of c and C is reminded (cf. Exers. 2.5, 2.6
and Fig. 4.6). In precise language, integrals (6) and (7) depend on the
homology classes
of c
and C, mod S h and mod S , respectively.
6.1 A SYMMETRICAL VARIATIONAL FORMULATION 165
exists in the cavity, rot h = j would be square-integrable, since Joule
dissipation must remain finite. This points
to
IL2rot(D) as the space in
which to look for h. Symmetrically, b will belong to IL2div(D).
We can do better, by anticipating a little on the discretization process
yet to come. Some a priori constraints on the solution are easy to enforce
at the discretized level (those are the "essential" or "Dirichlet-like"
conditions mentioned in 2.4.4), and it pays to take them into account
from the onset in the definition of admissible fields, since this reduces
the scope of the search. Boundary conditions (3) and (4) are of this kind.
So let us define (recall we are now using the
weak
grad, rot, div)
IH = {h e ILarot(D) • rot h = 0, n x h = 0
on sh},
IB = {b E
ILadiv(D) " div b = 0, n. b = 0 on sB}.
These are closed subspaces of ILa(D), after Exer. 5.3. We note that fields
of the form h = grad q0 belong to IH if q0 is a potential which assumes
constant values on both parts of the magnetic boundary S h (not necessarily
the same constant on each), and we recycle the symbol • to denote the
space of such potentials:
(I3 =
{q)
E
L2grad(W) " n x grad q0 = 0
on sh},
(It's not exactly the same as the earlier ~, beware: q0 is a constant on
sh0, but not necessarily the constant 0.) Similarly, on the side of b, let's
introduce
A = {a
e ILarot(D)"
n. rot a = 0 on S b},
and remark that fields of the form b = rot a, with a e A, belong to IB.
Moreover, if D is contractible (no loops, no holes 3 ) then, by the Poincar6
lemma (understood in its extended version of 5.1.4),
(8) IH = grad ~, IB = rot A,
instead of mere inclusions.
Conditions (6) and (7) also can be enforced a priori. Let's define
linear functionals y : IH --+ IR and 9- : IB --+ IR as follows. First, if h
and b are smooth, set
y(h) = Sc ~ . h, y-(b) = S c n. b,
3The reader who suspects that (8) may hold in spite of the existence of loops or holes
in D, for more complex geometries, is right. Only
relative
loops and holes (rood S h and
S b) are harmful.
166
CHAPTER 6 The "Curl Side"" Complementarity
Then, let us prove the following:
Proposition 6.1.
y and 1t- have extensions,
continuous
with respect to the
metric of
ILa(D),
to
IH
and lb.
Proof.
Let q01 be a smooth function assuming the values 0 on sh0 and 1
on
Shl . For b e lb, then,
fD b.
grad
q)l __
fS n
b
q)l = Jshl n. b = Y-(b), as
we saw with Exer. 2.6, and the map b-4"JD b . gract q01, which is
2
IL -continuous, is thus the announced extension. The proof for y is a bit
more involved. (Doing now Exercise 6.6 on p. 187, as preparation, may
help.) Pick a smooth vector field
a I such
that n. rot
a I =
0
on S b
and
Ic n. rot a 1 - 1. For h = grad q~ e IH, then,
fD h.
rot a 1 = - Js n x h. a 1 =
-
~ n x grad q~.
a 1 = ~s n x a I .
grad q0 = - ~s div(n
x a 1) q) = - ~s n.
rot
a 1 q) =
--
JS h n. rot d q0 = y (h) Jc n. rot a 1 = y (h). Again, the continuity of h -4
fD
h.
rot a I proves the point. From now on, we let y and I/- denote the
extended continuous functionals. ~)
Remark 6.1. Integrals such as y and 9 r stand no chance of being
ILa(D)-continuous if o~ne tries to enlarge their domains beyond IH and
lb. The conditions rot h = 0 and div b = 0 are necessary. ~?
As a corollary, subspaces
IH~= {h e IH" J c z. h = I}, lbF._ {b e IB " Ic n. b = F},
are closed. By (8), we have IHI= grad
(I)I
and lbF=
rot a F,
where
(I)I --
{q~ e O: Jc "¢. ~rad
q0
=F I} and AF= {a e A • j c n. rot a = F}, the
pre-images of IH and lb.
Note these are not vector subspaces, but
affine
subspaces of IH, lb,
etc., unless I = 0 or F = 0. We consistently denote by IH °, IB °, O °, A °
the subspaces that would be obtained in the latter case. IH ~ is
parallel,
the
usual way, to IH °, and so forth. The following lemma will be important:
Lemma 6.1.
Spaces
IH °
and
IB
are
orthogonal
in
ILa(D),
i.e., ~
D h .
b = 0
if h ~ IH ° and
b ~ lb.
Similarly, IH and lbo are orthogonal.
Proof.
Let he IH and b e IB. Then h= Vq~. Let q0 and q~l be the
values of q0 on both parts of
S h.
One
has
Y D h. °b
=
Y D b. Vq0 =
-- f D q)
div b + J s q0 n. b = ~s h q0 n. b after (4), and this is equal to
(q) 1 -- q) 0 ) fS h n. b,
that is, to the product (Jc z. h)(~ c n. b). Now if h ~ IH °,
or b e IB °, one of the factors vanishes.
Remark 6.2. There is more, actually: With the simple topology we have
here, both pairs are
ortho-complements
in ILa(D), which amounts to saying
that any square-integrable field u can be written as u = h + b, with
he IH ° and be lb, orwith he IH and be IB °, i.e., as the sum of a
curl-free field and a solenoidal field. These are
Helmholtz decompositions.
6.1 A SYMMETRICAL VARIATIONAL FORMULATION
167
We won't need a thorough treatment of them, but the paradigm is
important, and will recur.
The present state of affairs is summarized by Fig. 6.2, in which one
may recognize a part of the front of Maxwell's building of Fig. 5.1. Note
how all "vertical" relations have been taken care of, in advance, by the
very choice of functional spaces. Only the "horizontal" condition (5),
which expresses the constitutive laws of materials inside D, remains to
be dealt with.
I 0
!
grad
]HI9 h b=gh
i
rot !
V
A
i div
I
F
beIB
i
! rot
F
ae A
FIGURE 6.2. Structure of the magnetostatics problem. (This is a "Tonti diagram"
[To]. Similar graphic conventions have been independently proposed by many
researchers, Roth in particular [Rt]. See [Bw] for some history.)
6.1.2 Variational characterization of the solution
The following result will make the horizontal connection.
Proposition
6.2. The problem which consists in finding h ~ IH I and b ~ IB F
such that
(9)
fD~.L -1 ]b-~thl2_<fDlLt -11b'-~th'l 2 Vh'e IH I, V b'e IB F
has a unique solution, which is the solution of (1-7) when the latter exists.
Proof. The proof of Lemma 6.1 shows that f
D
h'. b' = I F. Therefore,
(10)
fD~[ -11b'-~th'12=fDg -1
[b']2+fDg ]h'] 2- 2 IF
for all pairs {h', b'} e IH ~ x IB F. This means that problem (9) splits in two
independent minimization problems"
(11) find he IH ~ such that fDg Ih12 isminimum
168 CHAPTER 6 The "Curl Side": Complementarity
and
(12)
find b
~ IB F
such that
~D
[1-1
i bl2
is minimum.
Both problems have a unique solution by the Hilbertian projection theorem
(because IH I
and
IB F are
closed convex sets). The minima are necessarily
of the form S I 2 and R F 2, where R and S are positive constants. Then
(10) shows that
IDa1-1 Ib -~hl 2=RF 2 + SI2-2 IF _>0
whatever F and I. If (1-7) has a solution for a nontrivial pair {I, F}, the
left-hand side vanishes, which implies RS = 1 (look at the discriminant),
I = RF, and b = ~h. We call R the
reluctance
of D under the prevailing
boundary conditions.
We note that, for a given nonzero F, there is always a value of I
such that the left-hand side of (9) vanishes, namely I = RF, so this proves
the existence in (1-7) for the right value of I/F. However, the point of
Prop. 6.2 is not to prove again the existence of a solution to the
magnetostatics problem, but to introduce a new variational characteriza-
tion of it. The quantity at the right-hand side of (9) is an "error in
constitutive law" as regards the pair {h', b'}. Thus, compliance with such
a law amounts to looking for a couple of fields that minimize the discrep-
ancy, among those that satisfy all other required conditions. This old
and esthetically attractive idea, which generalizes to
monotone
nonlinear
constitutive laws, thanks to the theory of convex functions in duality [Fe,
Ro], seems to date back to Moreau [Mo, Ny], and has been increasingly
popular ever since (and rediscovered), in the "computational magnetics"
community [R§] and others [OR, LL]. It works for
time-dependent
problems
just as well [B2, A&].
There are several equivalent ways to formulate problems (11) and
(12). One consists of writing the associated Euler equations, or weak
formulations:
(13)
(14)
find
h E IH I
such that
JD ~1 h. h'= 0 V h'~ IH °,
find b ~ IB F such that
JD t 1-1b" b'= 0 V b' ~ IB °.
Another consists of using
potentials,
thanks to (8):
(15)
find {p ~ {I} i minimizing
Y
D [1
I grad q}
]2,
(16)
find
a
~ a F
minimizing
IDl1-1
Irotal 2.
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