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Appendix: All Solutions to the Problems
+ [(∇ × b) × ∇u] · δa
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
+ ∇ · (a × ∇ × b)
δu +
∂ F
∂ a
− ∇u × ∇ × b
· δa
dV
+
V
{∇ · (b × c) − ∇ · [(a × ∇ × b)δu]}dV
(3)
The Euler equations of the functional are
∂ F
∂u
+ ∇ ·
a ×
∂ F
∂∇×(∇u×a)
= 0
∂ F
∂ a
+ ∇u × ∇ ×
∂ F
∂∇×(∇u×a)
= 0
(4)
The natural boundary conditions are
n ×
∂ F
∂∇ × (∇u × a)
S
= 0, n ×
a × ∇ ×
∂ F
∂∇ × (∇u × a)
S
= 0 (5)
10.16 Derive the Euler equations and natural boundary conditions of the functional
J =
V F(a, b, 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)
+
∂ F
∂(∇ × a) × (∇ × b)
· δ(∇ × a) × (∇ × b)
dV
(1)
Putting c =
∂ F
∂(a×∇×b)
, d =
∂ F
∂(∇×a)×(∇×b)
, the integrals of the latter two terms in
Eq. (1) can be written as
c · δ(a × ∇ × b) = c · (δa × ∇ × b + a × ∇ × δb)
= [(∇ × b) × c] · δa − (a × c) · ∇ × δb
= [(∇ × b) × c] · δa + ∇ · [(a × c) × δb] − [∇ × (a × c)] · δb
(2)
d · δ(∇ × a) × (∇ × b) = d · (∇ × δa) × (∇ × b) + d · (∇ × a) × (∇ × δb)
= [(∇ × b) × d] · ∇ × δa + (d × ∇ × a) · ∇ × δb
= ∇ · [δa × (∇ × b) × d] − [∇ × (d × ∇ × b)] · δa
+ ∇ · [δb × (d × ∇ × a)] + [∇ × (d × ∇ × a)] · δb
(3)