Page 140 - Vector Analysis

136 CHAPTER 4. Vector Calculus

Therefore, making change of variable x = ϑm(y) in each Wm we find that

ż Kż

v ¨ N dS = ÿ ζm(v ¨ N) dS

B Ω m=1 B ΩXWm

K nż (ζm ˝ ϑm)(vi ˝ ϑm)(Ni ˝ ϑm)?gm dy 1
= ÿÿ

m=1 i=1 Vmˆtyn=0u

K nż

= ÿÿ (ζm ˝ ϑm)(vi ˝ ϑm)Jm(Am)in dy 1

m=1 i=1 Vmˆtyn=0u

= K nż B [ ˝ ϑm)Jm(Am)ni (vi ˝ ] dy .
B yn (ζm ϑm)
ÿÿ

m=1 i=1 Vmˆ(´εm,0)

On the other hand, for α P t1, ¨ ¨ ¨ , n ´ 1u and i P t1, ¨ ¨ ¨ , nu,

ż B [ ˝ ϑm)Jm(Am)iα(vi ˝ ] dy = 0 ;
B yα (ζm ϑm)
Vmˆ(´εm,0)

thus the Piola identity (2.6) implies that

ż = K nż B [ ˝ ϑm)Jm(Am)ji (vi ]
B yj (ζm ˝ ϑm) dy
v ¨ N dS ÿÿ

BΩ m=1 i,j=1 Vmˆ(´εm,0)

K nż

=ÿÿ Jm(Am)ij(ζm ˝ ϑm),j (vi ˝ ϑm) dy

m=1 i,j=1 Vmˆ(´εm,0)

K nż

+ÿ ÿ (ζm ˝ ϑm)Jm(Am)ji (vi ˝ ϑm),j dy .

m=1 i,j=1 Vmˆ(´εm,0)

Making change of variable y = ϑm´1(x) in each Vm ˆ (´εm, 0) again, by the fact that

n and ż(
div ζ0v) dx = 0 ,
ÿ (Am)ij(vi ˝ θm),j = (divv) ˝ θm

i,j=1 W0

we conclude that

ż ż ( Kż Kż
v ¨ N dS = div ζ0v) dx + (v ¨ ∇x)ζm dx + ÿ
ÿ ζmdivv dx

B Ω W0 m=1 Wm m=1 Wm

Kż Kż

=ÿ (v ¨ ∇x)ζm dx + ÿ ζmdivv dx

m=0 Wm m=0 Wm

ż żż

= (v ¨ ∇x)1 dx + divv dx = divv dx . ˝

Ω ΩΩ

Letting v = (0, ¨ ¨ ¨ , 0, f, 0, ¨ ¨ ¨ , 0) = f ei, we obtain the following