158 CHAPTER 5 Whitney Elements
5.3. Of course the kernels are closed in the stronger norm, as pre-images
of the closed set {0}, but one cannot employ this argument about the IL 2
norm, in which rot and div are not continuous, only closed. Use (2),
and its analogue for rot.
5.4. See the cotangent formula of 3.3.4 and Lemma 4.1. The latter is
especially useful (if applied with a measure of creative laziness).
5.5. In terms of the nodes-to-edges incidence matrix elements, one has
W e -- GmeW m Vw n + Gne WnVW m"
..
Develop, and use Exer. 3.10. Set g'J= Vw i . Vwj. (The analogy with the
metric coefficients gij of Riemannian geometry is not accidental.)
5.6. First show that x. Vw n - Wn(X) is a constant inside each tetrahedron
(Lemma 4.1). Then develop (Vw m x VWn) X X.
5.8. Look at Fig. 4.3 and express the numbers N', E', F' relative to the
refined mesh in terms of N, E, F.
5.9. Use Proposition 5.5, second part first, then first part.
5.11. Begin with d- 1. Then D = ]a, b[, and q0(x)
= fa x 3q)(~)
d~. Use
Cauchy-Schwarz, then sum with respect to x. For d > 1, note that, in the
arrowed notation where "X --~ Y" means "all functions from X to Y",
the functional space IR x ... [d times] ... x IR ~ IR can be identified with
IR --~ (IR x ... [d - 1 times].., x IR --~ IR).
5.12. In dimension 1 first, q0(y)- q0(x) = ~x y 3q0(~) d~, hence q0(x)- q0(y) _<
C 113q011, for all pairs of points {x, y} in [a, b]. Integrate with respect to y
to get q0(x) - q0 < C 113q011, then invoke Cauchy-Schwarz. Adapt this to d
dimensions as in the previous case.
5.13. In Cartesian coordinates, (rot rot a)i= ~j 3j(3id- 3jai). For (18),
integrate by parts.
SOLUTIONS
5.1. If {0, u} is in the closure of GRAD, i.e., is the limit of some
sequence {~n' ~rad ~n} the terms ofj, which belong to GRAD, ,then
fD ~n
div j' =
- JD
grad
~Jn"
J for all in C0~(D),
hence fD U. j -- 0
V j' ~ C0°°(D) at the limit, and hence u = 0. If {0, u} is in the closure of
ROT, i.e., the limit of
some
{an,
rot an}
of ROT,
then
fD an"
rot h' =
--
fD rot a n . h' for all h' in C0°°(D),
hence fD U. h' -- 0 V h' E
~0°°(D) at
the limit, and u = 0.
SOLUTIONS 159
5.2. On D = {x : I xl < 1}, functions of the form x -o I xl -~ foot the bill,
if cz > 0 (in order to have a singularity at 0),
~1
r-2~
dr < oo (for the
function to be square-summable)
and So I
r d-
1-~(1 + cz)dr
< oo (for its
gradient to be square-summable). This happens for 0 < cz < 1/2 and
1 + ~z < d/2, the latter constraint being redundant if d > 2. For d = 2,
look at the function x-o I xl log lxl.
5.3. After (2), "JD b. grad q0'= 0 V q0'e C0~(D) '' characterizes elements
of ker(div), and if a sequence of fields
{bn}
which all satisfy this predicate
converges to some b in IL 2 (D), this also holds for b, by continuity of
the scalar product. Same argument for fields b such that n. b = 0, since
they are characterized by
~D div
b q0'
+
~D b. grad q0' = 0 V q0'
e
L2grad(D),
and for fields such that n x h = 0, by using the similar formula in rot.
5.4. Let h be the height of node n above the plane of f (observe how
"above" makes sense if space is oriented, as well as f). Then vol(T) =
h area({k, ¢~, m})/3 = area({k, g, m})/(3 1Vw n I). There are many ways to
derive these relations, but the most illuminating is to remark that 3 x 3
matrices such as (Vw k, Vw¢, Vw m) and (nk, ng, nm) are inverses, by
Lemma 4.1, and that vol(w) = det(nk, nl, nm)/6, that area({k, g, m}) =
det(k~, km) / 2, etc.
5.5. Up to obvious sign changes, there are only three cases:
(a) e = e'= {m, n}"
(b)
e = {m, n}, e'= {m, ~}"
(c)
e = {k, g}, e'= {m, n}"
(12 vol(T)/5!)(g nn + gmm gnm},
(6 vol(T)/5! )(2 gr¢_ gm¢_
gnm 4- gram),
(6 vol(w) / 5! )(g~n _ gkn
_ ~m 4-
gkm).
5.6. (Vw m x Vwn) X x = x. Vw m Vw n - x. Vw n Vw m -- w m VW n - w n Vw m 4-
b, where b is some vector, hence W{m,n } -- (Vw m X VWn) X X 4- b. Now,
observe (cf. 3.3.4 and Lemma 4.1) that
Vw m X VW n -- (kn x k~) x
VWn/6
VOI(T)
= kg
/ 6
VOI(T).
Using this, one has the following alternative form for the face element:
wf = 2(we kg + wmkm + w n kn)/6 vol(w),
hence the desired result (place the origin at node k).
5.7. If e = {m, n} and R fee: 0, then f = {¢, m, n} or {¢, n, m}, for some ~,.
In both cases,
RfeW f
=
Wf =
2(w¢
Vw m X Vw n 4- WmVW n X Vw~ 4- wnVw ~ x Vwm).
Therefore, summing over all faces,
160
CHAPTER 5 Whitney Elements
~fs y R feW f =
2 ~ e ~ N (we
VWm X Vw n q-... ) --
2
Vw m x Vw n.
For the divergence, just notice that div wf = 2(Vwe. V w m x Vw n + ... ) =
6 det(Vw e, Vw m, Vw n) = w w if T
= {k, ~r m, n}. Two
compensating
changes of sign occur if T--
{~, k, m, n},
the other orientation.
5.8. F'=4F, E'=2E +3F, N'=N+E, hence N'-E' +F'=N+E
- (2E + 3F) + 4F - N - E + F.
5.9. Since ker(div ; W 2) = rot W 1, its dimension is the dimension of W 1,
which is E, minus the dimension of ker(rot; W 1) = grad W °. The latter
is the dimension of W °, i.e., N, minus the dimension of ker(grad; W°),
which is 1. Project: Practice with this in the general case, to see how the
Betti numbers come to slightly modify these dimensions (but not their
asymptotic behavior when the mesh is refined).
5.11. Since D is bounded, it is contained in a set of the form P- ]a, b[ x
IR x... x IR. Extending by 0, outside D, the functions of C0~(D), one
identifies the latter space with C0~(P), which is isomorphic to
C0~([a, b] ; C0~(IR d- 1)). Then, if x - {xl,..., x d} e P, one has
q)(X) Ia 1 .. d)
d~,
__ x ()lq)(~' x2~' ., X
where
()lq)
is the partial derivative with respect to the variable x ~. By
the Cauchy-Schwarz inequality,
I q)(X)I 2~ (X-- a) 1/2 IaXl I alq)(~ .... )[2
d~_<
(b- a)l/2Iab [ VIp(~ .... )[2
d{,
and hence, by Fubini,
Ip ] q)(x) ] 2 dx ~ (b- a)l/a~p dxI.., dx d ja b ] Vq)(~, ... )]2
d{
=(b
- a)l/2fab dxl~p ] Vqo(~ .... )]2 d~ dx2.., dx d
hence c(D) < (b- a) 3/2. Of course, this is an upper bound, not the "best"
value of c(D), which can be obtained, but by very different methods.
5.13. First,
(rot rot a - grad div
a) i = £j [~j(ai aj- aja i) - ai(ajaJ)]
= ~j [~j (ai aj- aja i) - aj(aiaJ)] = -~j a jj a i.
Then I (div a) 2 + I I rot
a ]2__ £i I-Aa
i a i=
£i ~
I grad
ai] 2.
(Further
study: What of a domain D with surface S? Try to cast the surface
terms that then appear in coordinate-free form, by using adequate
curvature operators.)
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