942
Appendix: All Solutions to the Problems
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂u
−
a
v
· ∇
∂ F
∂∇·(ua/v)
= 0
∂ F
∂v
+
ua
v 2 · ∇
∂ F
∂∇·(ua/v)
= 0
∂ F
∂ a
−
u
v
∇
∂ F
∂∇·(ua/v)
= 0
(4)
The natural boundary conditions are
∂ F
∂∇ · (ua/v)
S
= 0
( 5 )
10.13 Derive the Euler equations and natural boundary conditions of the functional
J =
V F(a, b, ∇ · (a × ∇ × b))dV .
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂∇ · (a × ∇ × b)
δ∇ · (a × ∇ × b)
dV (1)
Putting ϕ =
∂ F
∂∇·(a×∇×b)
, c = δ(a ×∇ × b), and using the parallelepipedal product
formula of trivector and the divergence formula of cross product of two vectors, there
is
ϕδ∇ · (a × ∇ × b) = ϕ∇ · δ(a × ∇ × b) = ϕ∇ · c = ∇ · ϕc − c · ∇ϕ
= ∇ · ϕc − ∇ϕ · (δa × ∇ × b + a × ∇ × δb)
= ∇ · ϕc − (∇ × b) × ∇ϕ · δa − (∇ϕ × a) · ∇ × δb
= ∇ · ϕc − (∇ × b) × ∇ϕ · δa + ∇ · (∇ϕ × a × δb)
− [∇ × (∇ϕ × a)] · δb
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂ a
+ ∇ϕ × ∇ × b
· δa +
∂ F
∂ b
− ∇ × (∇ϕ × a)
· δb
+∇ · ϕc + ∇ · (∇ϕ × a × δb)}dV
(3)
The Euler equations of the functional are
∂ F
∂ a
+ ∇
∂ F
∂∇·(a×∇×b)
× ∇ × b = 0
∂ F
∂ b
+ ∇ ×
a × ∇
∂ F
∂∇·(a×∇×b)
= 0
(4)
The natural boundary conditions are
∂ F
∂∇ · (a × ∇ × b)
S
= 0, n ×
∇
∂ F
∂∇ · (a × ∇ × b)
× a
S
= 0
(5)