Appendix: All Solutions to the Problems
753
v
2 u = uv + ∇ · [v∇((u)] − ∇ · ((u∇v)
Quod erat demonstrandum.
1.6 Verify v( pu) = u( pv) + +∇ · [pv∇u − u∇( pv)] − ∇ · [pu∇v −
v∇( pu)].
Proof Let w = pu, do the derivation similar to problem 1.5, there is
∇ · [∇(v pu)] = ∇ · [∇(vw)] = ∇ · [v∇w + w∇v] = vw + 2∇ · (w∇v) − wv
= v( pu) + 2∇ · ( pu∇v) − puv
Let the rear of the second equal sign and the fear of the final equal sign be equal,
after arrangement, to obtain
puv = v( pu) + ∇ · [pu∇v − v∇( pu)]
Transposing u and v, there is
pvu = u( pv) + ∇ · [pv∇u − u∇( pv)]
The latter two expressions are equal. Quod erat demonstrandum.
1.7 Verify:
(1) (gu)v = ∇ · [v∇(gu)] − ∇v · ∇(gu);
(2) (gu)v = ∇ · [v∇(gu) − gu∇v] + guv.
Proof (1) Let w = gu, a = ∇w, using the divergence formula ∇ · (ϕa) =
ϕ∇ · a + a · ∇ϕ, there is
∇ · [v∇(gu)] = ∇ · (va) = ∇ · (v∇w) = v∇ · ∇w + ∇v · ∇w
= vw + ∇v · ∇w = v(gu) + ∇v · ∇(gu)
(1)
The second term on the right side of the last equal sign is moved to the left side
on the first equal sign, to get
(gu)v = ∇ · [v∇(gu)] − ∇v · ∇(gu)
(2)
This is the equation that needs to be proved. Quod erat demonstrandum.
Proof (2) Let w = gu, b = ∇v, using the divergence formula again, there is
∇ · (gu∇v) = ∇ · (wb) = w∇ · b + b · ∇w = w∇ · ∇v + ∇v · ∇w
= guv + ∇v · ∇(gu)
(3)