EXERCISES 57
HINTS
2.1. The definition doesn't say
bounded
regions. Recall that continuous
functions are bounded on closed
bounded
(hence, compact) regions of a
finite-dimensional space.
2.2. Just redo the proof of Prop. 2.1, reintroducing f at the right places.
Observe the way a minus sign appears.
2.3. Cauchy-Schwarz. Observe (this is for experts) that a certain condition
on the supports of the q0nS should be satisfied.
2.4. Same as Exer. 2.2.
2.5. Build a circuit on which to apply Stokes.
2.6. Build a volume to which Ostrogradskii-Gauss may apply, part of its
surface being Shl .
2.7. The question is, if A is an n x n matrix, and b an n-vector, why
does uniqueness of x such that Ax = b imply the existence of a
solution, whatever the right-hand side?
2.8. All that matters is div b = 0, where b = ~ (m + grad q0), and the
proof of Prop. 2.3 handles that. For (35) vs (34), apply (9) to A.
2.9. ~* being
larger
than ~0, the proof of Prop. 2.3 can be recycled in its
entirety, hence (23) and (24). So concentrate on (26), using (9).
2.10. Imitate (35), which can be understood as describing a flux injection
n. (~m) on Z.
SOLUTIONS
2.1. The constant 1 is not integrable over all
E3, so
the restriction to
bounded domains is certainly necessary. Now suppose f is smooth
over each region R i of a finite family. The way we understand "over", f
is smooth, and in particular continuous, in a bounded domain D i con-
taining the closure of R i ~ D, which is thus compact, so f is bounded
there, hence integrable. Pieces being in finite number, f is integrable
on D.
2.3. Since supp(Ip) is compact, one can build the q0nS SO that there exists
a compact K containing all the supp(q0n). Then (Exer. 2.1) b is
58 CHAPTER 2 Magnetostatics: "Scalar Potential" Approach
bounded on K, and hence, applying the Cauchy-Schwarz inequality,
[fD b. grad(qo
- q)n)] 2 _< fD ]b]2 ID
I grad(q~
- q)n)] 2
tends to zero.
2.5. Take c 1 and c 2 from S h to S h, with the same orientation, and join
the extremities by two paths lying in S h
and Shl
respectively, in order
to make a closed circuit. As rot h = 0, and by the Stokes theorem, the
circulations along c I and c 2 are equal (those along the boundary links
are 0 by (21)). One says that c I and c 2 are
homologous.
(The relation
between them is an equivalence, called
relative homology modulo
S h. We'll
have more to say about this in Chapter 5. Cf. [GH]. )
2.6. See the inset. Surface C is what is commonly called a "cut": Its
boundary is entirely in S b, and it separates D into two parts, each
containing one piece of S h. Moreover, C has
an external orientation (provided by a normal
field n), compatible with that of Shl . Now, as
n. b = 0 on S b, the fluxes through C and Shl
are equal, by Ostrogradskii, since div b = 0 in
D. All possible cuts of this kind will do,
including S h and sh0,
but
the latter must be
oriented the other way with respect to S. Again
S b
c
we have here an equivalence relation (relative
homology, but now modulo sB), and "cuts" are elements of a same class
of surfaces, of which one says they are
homologous
(mod sD). We'll return
to this in Chapter 4, and again, more formally, in 5.2.5.
2.7. Because then A is regular.
2.8.
fD m. grad q0'= ~a m. grad q0' = - ~a div m q0' + ~z n. m q0'. If div m
0, there is no special avantage to this formulation over (34), but otherwise
(35) may be easier to implement in the subsequent finite element modelling.
Be careful about the correct orientation of the normal on Z when doing
that. (You may worry about what happens when A touches S. This is a
good question, but no more a simple exercise.)
2.9. By (9), and using the information brought by the proof of Prop. 2.3
(div b = 0, n. b = 0
on sb),
plus q0'= 0 on S~ (36) reduces to
(37)
Ffl(q0')=~shn.b q0'=-Jshln.b q)' Vq0'~*,
and since the value of q)' on S h is precisely y(q0'), by definition of y, we
have F y (q0')
=
(Ish n. b) y(q0') for all q0',
hence
~S h n. b = F.
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