46
CHAPTER 2 Magnetostatics: "Scalar Potential" Approach
Definition 2.2.
A piecewise smooth field h which satisfies
(13)
will be said to
be
curl-free,
or irrotational,
in the weak sense.
We can prove, in quite the same way as Prop. 2.1, what follows:
Proposition 2.2.
For a piecewise smooth h,
(13)
is equivalent to the statement
"rot h = 0
inside regularity regions and
In x h] = 0
at their interfaces".
Proof.
This should be an exercise. If rot h = 0 in the regularity regions
D 1 and D 2, and [n x h] =0 at the interface ~, then
~D h.
rot a = JD1 h. rot a
+ ~D 2 h.
rot a
-
J'D1 a. rot h - J'z nl x h. a + J'D2 a. rot h - J'z
n2 X
h. a
=-Jz [nxh]. a- 0
for all test fields in C01(D), which is (13). Conversely, assuming (13), we
first obtain rot h = 0 in D 1 and D 2 separately by the same maneuvers
as above, then, backtracking,
0 --
fD h.
rot a - JD~ h. rot a + iDa h. rot a =
=-Jzn~x h. a-Jz n2 xh. a=-Jz [nxh]. a
for all a ~ C01(D). Surface values of a are not constrained on ~, so the
only way this equality can hold is by having [n x h]z = 0. 0
One may generalize there too:
(14)
~D h.
rot a
= ~D
J" a V a ~ G0 I(D),
with j given, piecewise smooth. This means (Exercise 2.4: Make sure
you understand this) "rot h = j in the weak sense". Here also, C01(D)
can be replaced by C0~(D).
This was only a first brush with weak formulations, and the full
potential of the idea has not been exploited yet. Instead of (10) or (10'),
we could have characterized divergence-free vector fields by the weak
formulation
(15)
JD b. grad q0 = ~s n. b q0 V q0 ~ C~(E)),
for instance, which suggests (cf. Remark 2.4) that not only right-hand
sides, as in (11), but also some boundary conditions maybe accommodated.
The symmetrical formula on the curl side is
(16)
fD h.
rot a = Js n x h. a V a ~ C~(b).
2.3 WEAK FORMULATIONS
47
With experience, this flexibility turns out to be the most compelling reason
to use weak formulations.
2.3.3 The uniqueness issue
So we got rid of the ambiguities hidden in the "strong" formulations
rot h = j and div b = 0. A different kind of problem arises about the
uniqueness
of the solution, assuming there is one. Take j = 0 in (1), and
bt = ~0 in all space. The physical solution is then h = 0 and b = 0. But
this is
not
implied by Eqs. (1-3): Take h = grad % where qo is a
harmonic function in all space (for instance, to exhibit only one among
an infinity of them, q~(x, y, z) = xy, in x-y-z Cartesian coordinates).
Then rot h = 0, and div(bt0h) = bt 0 Aq~ = 0. So we have here an example of
a nonzero static field that satisfies the equations, although there is no
source to create it.
All fields of this kind, however, have in common the property of
carrying infinite energy, which is the criterion by which we shall exclude
them: We want 12 fields
with finite
energy. From Chapter 1, the expression
of the energy of the magnetic field is
= 1
fE3~ ihl2
1 yF_3 -1 2
(17)
Wmag ~ =
~ I bl .
Since bt -> bt 0 all over, the first integral is bounded from below by
bt 0 ~ I h I
2/2;
hence the eligible h's are
square-integrable:
II h II <
where II II denotes the quadratic norm, thus defined"
Ilhll
= [rE3
I h(x) 12 dx] 1/2.
If there is also an upper bound bt 1 to bt, which we assume, the same
reasoning with the other integral shows that b should be square-integrable
as well.
To be consistent with this requirement of finite energy, we shall also
assume that j, besides being piecewise smooth, has compact support;
this excludes cases such as, for instance, that of a uniform current density
in all space, which would generate a field of infinite energy.
All that is required of ~, then (last item in our critical review of
(1-3)), is not to spoil these arrangements. We want the integrals in (17)
to make sense for all eligible fields h and b, that is, square-integrable
fields, and this is the case if bt is piecewise smooth (a reasonable require-
ment, as regards a material property) and if there exist two positive
12Note
this is a
modelling choice,
justified in the present situation, not a dogma.
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