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CHAPTER 6 The "Curl Side": Complementarity
Remark 6.3. As Fig. 6.3 suggests, the radius r of the hypercircle is
given by [E(b, h)]1/2/2. This information allows one to get bilateral
bounds on other quantities than the reluctance. Suppose some quantity
of interest is a linear continuous functional of (say) h, L(h). There is a
Riesz vector h L for this functional (cf. A.4.3), such that L(h)
= fD ~ hL"
h.
What is output is L(h). But one has I L(h) - L(h) l = I J" D ~t h L . (h - hm)]
< Ilhrll IIh- h~l, < r IIhLIl,, hence the bounds. There is a way to express
the value of the potential at a point x as such a functional [Gr]. Hence
the possibility of
pointwise
bilateral estimates for the magnetic potentials.
This was known long ago (see bibliographical comments at the end), but
seems rarely applied nowadays, and some revival of the subject would
perhaps lead to interesting applications. 0
6.1.4 Constrained linear systems
With such incentives, it becomes almost mandatory to implement the
vector potential method. All it takes is some Galerkin space AF , and
since the unknown a is
vector-valued,
whereas q0 was
scalar-valued,
let's pretend we don't know about edge elements and try this: Assign a
vector-valued
DoF a_ n to each node, and look for the vector field a as a
linear combination
a = ~n~N a--nWn"
(We shall have to refer to the space spanned by such fields later, so let us
name it Iplm, on the model of the p1 of Chapter 3, the "blackboard" style
reminding us that each DoF is a vector.) Now (21) is a quadratic
optimization problem in terms of the Cartesian components of the vector
DoFs. The difficulty is, these degrees of freedom are not
independent,
because a must belong to AF. As such, it should first satisfy n. rot a =
0 on S b that is, on each face f of the mesh that belongs to S b. Assume
again flat faces, for simplicity, and let nf be the normal to f. Remembering
that N(f) denotes the subset of nodes that belong to f, we have, on
face f,
n. rot a = Ev
~ N(f)
nf. rot(av w) = ~v, N(f)nf" (V w v x a~)
= E ~ x(f) (n~ x Vw~). a_~,
since n x Vw~ vanishes for all other nodes v (Fig. 6.4), hence the linear
constraints to be verified by the DoFs:
Z~ N(f) (nf x Vw~).
a_~ = 0