SOLUTIONS 87
3.12. Combine Exers. 3.10 and 3.11.
3.13. The same trick as in Prop. 3.1.
3.14. Grid cells must be cut in two, so if point P for instance is linked
with O by an edge, making them "neighbors" on the finite element
mesh, one expects a nonzero entry in the stiffness matrix at row O and
column P, which is
not
the case of the finite-difference scheme (24).
Explaining why this term vanishes is the key. It has to do with the right
angle, obviously.
3.15. If paving was possible, tetrahedra around a given edge would join
without leaving any gap, so the dihedral angle would have to be 2rc/n
for some integer n. Is that so?
3.17. Suppose I = 1 for simplicity. Then R1-1 = inf(~ D
~1 i
Vq) l 2" q)
E (I)l}.
Replace q0 by q02, the solution for ~t = ~t 2.
3.18. Map the problem concerning the enlarged region onto the reference
one, and see how this affects ~t.
3.19. Consider two problems corresponding to permeabilities gl and
~t 2, all other things being equal. Denote the respective solutions by
q)l
and
(4)2"
Let
[Iq)l] 1 or
IIq0112 and (q0,
q)')l or
(q0,
q)')2
stand for IIq011, and
(% q0'),, depending on the value of It. One has
(25)
fD ~i
grad
q)i"
grad q0' = 0 V q0' ~ q)0, for i = 1, 2.
Choose appropriate test functions, combine both equations, and apply
the Cauchy-Schwarz inequality.
3.20. If the deformation is a homeomorphism, the same mapping trick as
in Exer. 3.18 reduces the problem to analyzing the dependence with
respect to ~t, with a new twist, however, for ~t will become a tensor.
You will have to work out a theory to cover this case first.
SOLUTIONS
3.1. Let M(q0)
=
fD m. grad q0. Since M(q0 + )~q)') = M(q0) + k~f (q0'), the
directional derivative of M is q0' --~ lim x_, 0(M (q0 + Kq0') - M (q0))/)~, that
is, q0' --~ ~D m. grad q)', formally 21 the same as M itself. This holds for all
88 CHAPTER 3 Solving for the Scalar Magnetic Potential
linear functionals, so we shall not have to do it again. The variational
forms of (2.34) and (2.36) thus consist in minimizing the functionals
1 2
q[}---~--~IDl.I 0 Igradq}l
nt-~.10I D
m.gradcp on cI} I ,
1 2 Ci)* I
q) ~ -2 f D gO
I grad ~1 - Fy(~)
on
respectively.
3.2. On cI}, IIq}ll, = 0 implies a constant value of {p, but not c 1} = 0, so II II,
is not a norm, whereas its restriction to cI}* is one. This hardly matters,
anyway, since two potentials which differ by an additive constant have
the same physical meaning. So another possibility would be for us to
define the quotient cI}/IR of cI} by the constants, call that c~, and give it
the norm IIq} II = inf{c ~ IR : IIq} + cll,}, where {1} is a member of the class
cp ~ c~. Much trouble, I'd say, for little advantage, at least for the time
being. Later, we'll see that what happens here is a general fact, which
has to do with
gauging:
It's
equivalence classes
of potentials, not potentials
themselves, that are physically meaningful, so this passage to the quotient
I have been dodging here will have to be confronted.
3.3. Take ~ ~ cI}*, and let I =y(~). Then (Fig. 3.2) IIq}(I)ll, < I1~11,. Using
Prop. 3.2, we thus have
ly(~) I : I II : [llcp(1)ll~
]-1
Ilcp(I)ll~ < [llq}(1)ll~
]-1 iil]/]ll.t"
3.4. If m
>
n, ~lf-fl
--I[1/n, 1/m ]
zero.
dx/x/x = 2/qn- 2/q--m < 2/q-n tends to
3.5. Let f(z)= P(x, y) + i Q(x, y).
Holomorphy
of f inside D means
differentiability in the
complex
field C, that is, for all z ~ D, existence of
a complex number 3f(z) such that f(z + dz) = f(z) + 3f(z) dz + o(dz) for
all dz in {E. Cauchy conditions for holomorphy are 3xP = 3,Q and 3yP
= - 3KQ, so 3xxP = 3xy Q
= 3y KQ
= - 3yyP, hence AP = 0, and the same for Q.
In dimension 2,
conformal mappings
(those which preserve angles, but not
distances) are realized by holomorphic maps from (E to {E, and holo-
morphy is preserved by composition.
3.6. A harmonic function in the upper half-plane y > 0 which vanishes
for y = 0 is {x, y} ~ y, the function denoted Im (for imaginary part).
The
fan map
g = z --~ i
Z 3/2
sends the upper half-plane to the domain
21But not
conceptually. The argument of M is a point in an
affine
space, whereas qY, in
the expression of the directional derivative, is an element of the associated
vector
space.
SOLUTIONS 89
{{x, y} • x < 0 or y < 0}. Composition of Im and g-1 yields the desired
function (cf. Fig. 3.6), better expressed in polar coordinates:
q0(r, 0) = r2/3 sin((20 - ~)/3).
(Note that q0 cannot be extended to the whole plane. Note also that it is
not piecewise k-smooth for k > 0, in the sense we adopted in Chapter 2.)
Its gradient is infinite at the origin, where its modulus behaves like r -1/3.
Since ~0 R
(r-1/3) 2 r
dr = J0 R
r 1/3
dr converges, this is a potential with finite
associated magnetic (co)energy.
y . I
FIGURE
3.6. The function f(r, 0)=
2/3
sin((20 -~)/3) of plane polar coordinates,
for ~ /2 < 0 < 2~. Left: level lines. Right: perspective view of the graph.
Now imagine one of the level surfaces of q0 (a cylinder along 0z) is
lined up by some perfectly permeable material. This potential is then the
solution of a two-dimensional analogue 22 of our model problem, in which
the system would be infinite in the z-direction.
3.7. Two large diagonals would cut at the center, so there can't be more
than one. At least three faces must hinge on it, since dihedral angles are
less than ~, and at most six, corresponding to the six possible vertices.
So, see how to leave out one, two, or three of these. Fig. 3.7 gives the
result. Alternatively, one may consider whether opposite faces are cut
by parallel or anti-parallel small diagonals. (This is meaningful when
22A
genuinely three-dimensional example would be more demonstrative. See [Gr] for
the (more difficult) techniques by which such examples can be constructed.
90 CHAPTER 3 Solving for the Scalar Magnetic Potential
one thinks of stacking cubes in order to make a tetrahedral mesh.)
Whichever way, it's pretty difficult to prove the enumeration complete!
6 5 4
0 3 4
FIGURE 3.7. All ways to mesh a cube, depending on the number of inner faces
that have a large diagonal as one of their edges.
3.8. ~n
an ~n(Xm) --~ am,
by construction of the nodal functions, so if
~n
an £n= 0, then all a nS vanish.
3.9. One may invoke the general result about "change of variables" in
integration,
~u(D) f
Ju =
fD u'f,
where u*f is the pull-back x --+ f(u(x)) and
Ju the Jacobian of the mapping u, for there is an affine map from w
to
itself that swaps n and m,
~n
and )m which is volume preserving
(Ju = 1). A much more elementary but safer alternative is, in Cartesian
coordinates: Place the basis of tetrahedron T in plane x-y, and let n be
the off-plane node, at height h. Then )~n(x, y, Z) = z/h. If A is the area
of the basis,
then ~T ~ --
A f0 h dz (1 - z/h)az/h = hA/12,
hence fw ~n .
VOI(T)/4. Same thing for a triangle: basis on x axis, height h, etc. You
may prefer a proof by recurrence on the dimension. Anyway, once in
possession of these basic symmetry results, further computations (cf. Exer.
3.10) simplify considerably.
3.10. First compute
Inn = ST (~n)2 = a/h 2 ~h
dz (h
- z) 2 z2/h 2 =
vol(T)/10,
and similarly, obtain the equality Inn n = ~W ()n)3 = VOI(T)/20. Then Inn =
JW ~n(1 -- ~ m,n ~m) = VOI(T)/4 --
3
Inm ,
hence
I m = VOI(T)/20.
And so
on.
The general formula,
fT (~L)i (~m)j (~n)k __
6 volCT) i! j! k! /(i+j+k+3)!
(cf. [Sf]), may save you time someday. (Thanks to (13) and Remark 3.3,
SOLUTIONS 91
this is enough to sum any polynomial
over T.)
The analogue on faces
[SF] is ~f ()m)i
(~n)j ~_ 2
area(f)i! j!/(i +j + 2)!.
3.11. On [0, 1], the function w = x ~ 2x 2- x behaves as requested, that
is, w(0)= w(1/2)= 0 and w(1)= 1. Therefore,
Wnm=4
~nKm and Wnn =
~n(2~n -- 1).
3.12. Weights 1 / 5 at mid-edges and -1/20
at nodes. For the triangle, amusingly, 1 /3
at mid-edges and 0 at nodes (inset).
3.13. For all ~ 0 < (M({p + 9~), {p + L~) =
2)~ (Mq), ~) + ~2 (M~,
~), so (M(p, ~)=
0
for all ~, which implies M(p = O. <)
-1/20
0
~1/3
1/3
3.14. Use the cotangent formula (23). There are two right angles in front
of edge OP, hence the nullity of the coefficient Aop. Formula (24) comes
immediately in the case of node pairs like O-N, O-E, etc., by using (23)
and the relation tan 0 = k/h, where 0 is
the angle shown in the inset. Note how
(23) gives the same value for matrix entries
corresponding to such pairs whatever the
diagonal along which one has cut the
rectangular cell. (Further study: Consider
orthogonal, but not uniformly spaced,
grids. Generalize to dimension 3.)
0
3.15. No, the regular tetrahedron is not a "space-filling" solid. Its dihedral
angle, easily computed, is about 70o32 ', hence a mismatch. See [Ka] or
[Si] on such issues.
a
FIGURE
3.8. Left: The tetrahedron (in thick lines). Middle: Assembly of four
copies of it into an octahedra, by rotation around the middle vertical pole. Right:
Sticking a fifth copy to the upper right flank. A sixth copy will be attached to
the lower left flank in the same way, hence a paving parallellepiped.
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