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,