Page 156 - Vector Analysis

152 CHAPTER 5. Additional Topics

which further implies that

ż dSx = ż F JAT N |JATN|dSy . (5.2)
(f n)(x) |JAT N|
B Ω1
B Ω2

Let ψ˚(dSx) denote the pull-back of the surface element dSx having the property that for
any function h defined on B Ω2 = ψ(B Ω1),

żż

h(x) dSx = (h ˝ ψ)(y)ψ˚(dSx) ;

ψ(B Ω1) B Ω1

in other words, ψ˚(dSx) = ag(y) dSy for some “Jacobian” ?g of the map ψ : B Ω1 Ñ B Ω2.
Therefore, (5.2) suggests that

ż f ndS = ż [ ˝ ] = ż ˝ ψ) JAT N |JATN|dSy .
(f n) ψ (y)ψ˚(dSx) |JAT N|
B Ω2 B Ω1 (f

B Ω1

Since f can be chosen arbitrarily, the equality above suggests that

n ˝ ψ = JATN = ATN (5.3)
|JATN| |ATN|

and (5.4)
ψ˚(dSx) = |JATN|dSy .

We finish this section by the following

Theorem 5.1 (Reynolds’ transport theorem). Let Ω Ď Rn be a smooth domain, ψ : Ω ˆ
[0, T ] Ñ Rn be a diffeomorphism, Ω(t) = ψ(Ω, t) and f (x, t) be a function defined on Ω(t).
Then

dż ż ż (σf )(x, t) dSx , (5.5)
f (x, t) dx = ft(x, t) dx +
dt B Ω(t)
Ω(t) Ω(t)

where σ is the speed of the boundary in the direction of outward pointing normal of B Ω(t);
that is, with n denoting the outward-pointing unit normal of Ω(t),

σ = (ψt ˝ ψ´1) ¨ n .

Proof. By the change of variable formula,

żż
f (x, t) dx = f (ψ(y, t), t) det(∇ψ)(y, t) dy.

Ω(t) Ω