10.5 The Euler Equations of Rotation Type Functionals
675
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂ a
− ∇ × (b × ∇ × c)
· δa +
∂ F
∂ a
+ (∇ × a) × (∇ × c)
· δb
dV
+
V
{∇ · [b × (∇ × c) × δa] − ∇ · (c × d)}dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂ a
− ∇ ×
b × ∇ ×
∂ F
∂∇ × (b × ∇ × a)
= 0
∂ F
∂ b
+ (∇ × a) × ∇ ×
∂ F
∂∇ × (b × ∇ × a)
= 0
(5)
The natural boundary conditions are
n ×
∂ F
∂∇ × (b × ∇ × a)
S
= 0,
b × ∇ ×
∂ F
∂∇ × (b × ∇ × a)
× n
S
= 0 (6)
Example 10.5.7 The functional with the rotation of product of the scalar and the
rotation of vector. Let the functional
J =
V
F(u, a, ∇ × u∇ × a)dV
(1)
Derive its Euler equations and natural boundary conditions.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂∇ × u∇ × a
· δ∇ × u∇ × a
dV
(2)
Let b =
∂ F
∂∇×u∇×a
, c = δ(u∇ × a) = ∇ × aδu + u∇ × δa, there is
b · ∇ × c = −∇ · (b × c) + (∇ × b) · c = −∇ · (b × c) + (∇ × b) · (∇ × aδu + uδ∇ × a)
= −∇ · (b × c) + (∇ × a · ∇ × b)δu + u∇ × b · ∇ × δa
= −∇ · (b × c) + (∇ × a · ∇ × b)δu
− ∇ · (u∇ × b × δa) + (∇ × u∇ × b) · δa
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
+ ∇ × a · ∇ × b
δu +
∂ F
∂ a
+ ∇ × u∇ × b
· δa − ∇ · b × δ(u∇ × a) − ∇ · (u∇ × b × δa)
dV
(4)
The Euler equations of the functional are