10.5 The Euler Equations of Rotation Type Functionals
683
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂u
− ∇ · ∇v × ∇ ×
∂ F
∂∇ × (∇u × ∇v)
= 0
∂ F
∂v
+ ∇ · ∇u × ∇ ×
∂ F
∂∇ × (∇u × ∇v)
= 0
(5)
The natural boundary conditions are
a × n| S = 0, (∇v × ∇ × a) · n| S = 0, (∇u × ∇ × a) · n| S = 0
(6)
Example 10.5.14 The functional with the double rotation of product of a sine function and a vector. Derive the Euler equations and natural boundary conditions of the
functional
J =
V
F(u, a, ∇ × ∇ × sin kua)dV
(1)
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂∇ × ∇ × sin kua
· δ∇ × ∇ × sin kua
dV (2)
Let b =
∂ F
∂∇×∇×sin kua
, c = δ sin kua = k cos kuaδu + sin kuδa, there is
b · δ∇ × ∇ × sin kua = b · ∇ × δ∇ × sin kua
= −∇ · (b × δ∇ × sin kua) + (∇ × b) · δ∇ × sin kua
= −∇ · (b × δ∇ × sin kua) − ∇ · [(∇ × b)
× δ sin kua] + ∇ × ∇ × b · (k cos kuaδu + sin kuδa)
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
+ k cos ku(∇ × ∇ × b) · a
δu +
∂ F
∂ a
+ sin ku∇ × ∇ × b
· δa
dV
−
V
{∇ · (b × δ∇ × sin kua) + ∇ · [(∇ × b) × δ sin kua]}dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂u
+ k cos kua · ∇ × ∇ ×
∂ F
∂∇ × ∇ × sin kua
= 0
∂ F
∂ a
+ sin ku∇ × ∇ ×
∂ F
∂∇ × ∇ × sin kua
= 0
(5)
The natural boundary conditions are