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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