10.5 The Euler Equations of Rotation Type Functionals
673
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
∂ F
∂ a
+ b ×
∂ F
∂(a × b)
+ ∇ ×
∂ F
∂∇ × (a × b)
= 0
∂ F
∂ b
− a ×
∂ F
∂(a × b)
+ ∇ ×
∂ F
∂∇ × (a × b)
= 0
(6)
The natural boundary condition is
∂ F
∂∇ × (a × b)
× n
S
= 0
(7)
Example 10.5.5 The functional with the secondary rotation of cross product of two
vectors. Let the functional
J =
V
F(a, b, ∇ × ∇ × (a × b))dV
(1)
Derive its Euler equations and natural boundary conditions.
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
(2)
Putting c =
∂ F
∂∇×∇×(a×b)
, d = δ(a × b), we get
c · ∇ × ∇ × d = −∇ · (c × ∇ × d) + ∇ × c · ∇ × d
= −∇ · (c × ∇ × d) − ∇ · (∇ × c × d) + (∇ × ∇ × c) · (a × δb − b × δa)
= −∇ · (c × ∇ × d) − ∇ · (∇ × c × d) + (b × ∇ × ∇ × c) · δa − (a × ∇ × ∇ × c) · δb
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂ a
+ b × ∇ × ∇ × c
· δa +
∂ F
∂ b
− a × ∇ × ∇ × c
· δb − ∇ · (c × ∇ × d) − ∇ · (∇ × c × d)
dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂ a
+ b × ∇ × ∇ ×
∂ F
∂∇ × ∇ × (a × b)
= 0
∂ F
∂ b
− a × ∇ × ∇ ×
∂ F
∂∇ × ∇ × (a × b)
= 0
(5)
The natural boundary conditions are