Page 145 - Vector Analysis

§4.6 The Divergence Theorem 141

Since

Kż (u ¨ ∇Σζm) ˝ ψm?gm dy K ż żK
(u ¨ ∇Σζm) dS = (u ¨ ∇ ÿ ζm) dS = 0 ,
ÿ = ÿ

m=1 ψm´1(UmXΣ) m=1 UmXΣ Σ m=1

we conclude that

ż Kż [ 2 nα?gmgmαβ B ψm ]
ÿ B yβ
Σ divΣu ψm)(u ÿ
dS = (ζm ˝ ˝ ψm) ¨ dy1 .
α,β=1
m=J+1 BmXty2=0u

On the other hand,

ż Kż ζmu ¨ (T ˆ N) ds
u ¨ (T ˆ N) ds = ÿ

B Σ m=J+1 B ΣXUm

K ż [ ˇ B ψm ˇ]
(T ψmˇˇ B y1
= ÿ (ζm ˝ ψm)(u ˝ ψm) ¨ ˆ N) ˝ ˇ dy1 .
ˇ
m=J +1 BmXty2=0u

Therefore, the theorem can be concluded as long as we can show that

2 nα?gm gmαβ B ψm (T N) ˇ B ψm ˇ on Bm X ty2 = 0u . (4.19)
B yβ ˇ B y1 ˇ
ÿ ˇ ˇ

= ˆ ˝ ψm

α,β=1

Let τm = 2 nα?gmgmαβ B ψm on Bm Xty2 = 0u. Since nα = ´δ2α, we find that τ m ¨ B ψm =
B yβ B y1
ř

α,β=1

0 on Bm X ty2 = 0u; thus

τ m ¨ (T ˝ ψm) = 0 on Bm X ty2 = 0u .

Moreover, noting that τm is a linear combination of tangent vectors B ψm , we must have

B yβ

τ m ¨ (N ˝ ψm) = 0 on Bm X ty2 = 0u .

As a consequence,

τ m (T ˆ N) ˝ ψm on Bm X ty2 = 0u .

B ψm ˇ
B y2
Since (T ˆ N) points away from Σ, while ˝ ψm´1ˇˇ points toward Σ, by the fact that
B
Σ

τm ¨ B ψm = 2 nα?gmgmαβ B ψm ¨ B ψm = ´?gmgm22 ă 0,
B y2 B yβ B y2
ÿ

α,β=1