10.5 The Euler Equations of Rotation Type Functionals
679
+
∂ F
∂ c
+ b × (a × ∇ × d)
· δc − ∇ · (d × e)
dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂ a
+ (b × c) × ∇ ×
∂ F
∂∇ × [a × (b × c)]
= 0
∂ F
∂ b
− c ×
a × ∇ ×
∂ F
∂∇ × [a × (b × c)]
= 0
∂ F
∂ c
+ b ×
a × ∇ ×
∂ F
∂∇ × [a × (b × c)]
= 0
(5)
The natural boundary condition is
∂ F
∂∇ × [a × (b × c)]
× n
S
= 0
(6)
Example 10.5.11 The functional with the double rotation of cross product of
trivector. Let the functional
J =
V
F(a, b, c, ∇ × ∇ × [a × (b × c)])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
∂ c
· δc +
∂ F
∂∇ × ∇ × [a × (b × c)]
· δ∇ × ∇ × [a × (b × c)]
dV
(2)
Putting d =
∂ F
∂∇×∇×[a×(b×c)]
, e = δ[a × (b × c)] = δa × (b × c) + a × (δb ×
c) + a × (b × δc), there is
d · δ∇ × ∇ × [a × (b × c)] = d · ∇ × ∇ × δ[a × (b × c)]
= d · ∇ × ∇ × e = −∇ · (d × ∇ × e) + (∇ × d) · ∇ × e
= −∇ · (d × ∇ × e) − ∇ · [(∇ × d) × e]
+ (∇ × ∇ × d) · [δa × (b × c) + a × (δb × c) + a × (b × δc)]
= −∇ · (d × ∇ × e) − ∇ · [(∇ × d) × e] + [(b × c) × ∇ × ∇ × d] · δa
+ [(∇ × ∇ × d) × a] · (δb × c) + [(∇ × ∇ × d) × a] · (b × δc)
= −∇ · (d × ∇ × e) − ∇ · [(∇ × d) × e] + [(b × c) × ∇ × ∇ × d] · δa
+ c × [(∇ × ∇ × d) × a] · δb − b × [(∇ × ∇ × d) × a] · δc
= −∇ · (d × ∇ × e) − ∇ · [(∇ × d) × e] + [(b × c) × ∇ × ∇ × d] · δa
− c × (a × ∇ × ∇ × d) · δb + b × (a × ∇ × ∇ × d) · δc
(3)