120 CHAPTER 4 The Approximate Scalar Potential: Properties and Shortcomings
4.4. Data f and g are flux loss densities, so their integrals over boxes,
or in the case of g, the part of the box's boundary that lies in S, balance
inter-box exchanges.
4.5. Apply Prop. 4.2 to b - ~t 0 m.
4.6. ~1 does not bound modulo S b but the union of
two
such surfaces
does, hence the equality of fluxes through each of them.
4.7. Since (Mq~) n has the dimensions of a flux, whereas - div(~t grad q0)
is a flux
density,
introduce the volume of the box
B n.
4.8. Call hq~ the vector of nodal values on S h and use the same block
forms as in Section 3.3.3, with q~ = {0q~,
h(~}. Observe
that fD
~
I grad
q)[2 =
fS
~ 3nq) q)h_ yS h ~
pq0h q)h where P is a certain linear operator. The aim of
the exercise is thus to work out the discrete analogue of this operator.
How does it relate with the reluctance of the region, "as seen from the
boundary"?
4.9. Work on x --~ 1/Ix I, and remember div(q~ u) = q~ div u + Vq0. u.
4.10. Show that 0q~ > 0 first, then work on the translate ~- I to show
that 0q~ _< I, by the same method.
4.11. Suppose a non-Delaunay tetrahedron is in the mesh. Show that its
circumscribed sphere must contain other nodes.
4.12. Note that the flux density through CC' is 1/Imnl. Treat separately
segments CC' and MC, M'C', of the cell boundary. Beware the obtuse
angle.
4.13. Same as Exer. 4.12, if all circumcenters are inside tetrahedra.
SOLUTIONS
4.1. Tetrahedron: ~ = 4 - 6 + 4 - 1 = 1. Cube without inner edge: ~ = 8 -
18+16-5=1. With one: ~=8-19+18-6=1.
4.2. The following pedestrian solution works convincingly. Call
F k
the
outgoing flux at edge k, with the labelling of Fig. 4.12. One has F 2 + Fll
+ F12 -- 0,
and eight other similar relations. Add them all, which results
in ~
<k<27
Fk = 0. On the other hand, relation (2) expressed at nodes i
and j implies ~8 -<
k < 17
Fk + F26 + F27 = 0
and ~ 18 -<
k < 25
Fk 4- F26 4- F27 -- 0
respectively. Add these, and subtract the previous one, hence
2
(F26
4- F27 ) -- Y~ 1_<
k_< 7
Fk, which is the flux exiting from ~1" But
F26 4- F27
is one of the "flux leaks", which has no reason to vanish. (If that accidentally
happened, a slight perturbation in the nodal positions would restore the
SOLUTIONS 121
generic situation.) The case on the right is similar: The flux exiting from
~2
is the sum of flux leaks at the perimeter of T, that is, ~
1 < k < 6 Fk°
Subtract F 1 + F 2
+ F 3,
which is 0, and what remains,
F 4 +
F 5
+ F 6 ,
again
does not vanish, generically.
$ ...................
°$ 1 ~
'
27
°
°
........................... °o°"
4.3. Figure 4.13.
e 2
FIGURE 4.12. Ad-hoc edge numberings for Exer. 4.2.
e I
2
FIGURE 4.13. How dual 2-cells can intersect. Left: Edges e I and e 2 have a
common node, and define face f. Then e 1. and e2* intersect along f*. Right:
they belong to the same tetrahedron, but without any common node. Then
el* n e2* is the barycenter of T.
4.4. Define fn = JB f and
gn = JSn OB n
g" Then box B receives (M
~)n
from adjacent boxes and loses
fn + gn"
The linear system to solve is thus
M ~ = f + g. The equations are identical if f and g are approximated by
mesh-wise affine functions.
4.7. Let us denote by
V n
the volume of the box
B n
and by ¥ the
diagonal matrix of the Yn's. Since (M~) n is the "flux loss at n", the
term
(Ml~)n/V n can be dubbed the "flux loss
density
about n". Since, on
the other hand, J
Gn
n. b
= S~n
g n. grad q~
= S Bn
div(g grad q0), this

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