10.5 The Euler Equations of Rotation Type Functionals
681
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⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂ a
+ (b × c) × ∇
n
×
∂ F
∂∇ n × [a × (b × c)]
= 0
∂ F
∂ b
− c ×
a × ∇
n
×
∂ F
∂∇ n × [a × (b × c)]
= 0
∂ F
∂ c
+ b ×
a × ∇
n
×
∂ F
∂∇ n × [a × (b × c)]
= 0
(4)
The natural boundary conditions are
∇
i−1
×
∂ F
∂∇ × ∇ × [a × (b × c)]
× n
S
= 0 (i = 1, 2, · · · , n)|
(6)
Example 10.5.12 The functional with the rotation of cross product of trivector,
where the three vectors are all the respective gradients of scalars. Find the Euler
equations and natural boundary conditions of the functional
J =
V
F(u, v, w, ∇ × [∇u × (∇v × ∇w)])dV
(1)
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂v
δv +
∂ F
∂w
δw +
∂ F
∂∇ × [∇u × (∇v × ∇w)]
· δ∇ × [∇u × (∇v × ∇w)]
dV
(2)
Putting a =
∂ F
∂∇×[∇u×(∇v×∇w)]
, b = δ[∇u × (∇v × ∇w)] = ∇δu × (∇v × ∇w) +
∇u × (∇δv × ∇w) + ∇u × (∇v × ∇δw), c = ∇ × a, and using the parallelepipedal
product formula of trivector, there is
a · ∇ × δ[∇u × (∇v × ∇w)] = a · ∇ × b = −∇ · (a × b) + (∇ × a) · b = −∇ · (a × b) + c · b
= −∇ · (a × b) + c · [∇δu × (∇v × ∇w) + ∇u × (∇δv × ∇w) + ∇u × (∇v × ∇δw)]
= −∇ · (a × b) + [(∇v × ∇w) × c] · ∇δu − (∇u × c) · (∇δv × ∇w) − (∇uc) · (∇v × ∇δw)]
= −∇ · (a × b) + [(∇v × ∇w) × c] · ∇δu − [∇w × (∇u×)] · ∇δv + [∇v × (∇u×)] · ∇δw
= −∇ · (a × b) + ∇ · [(∇v × ∇w) × cδu] − [∇ · (∇v × ∇w) × c]δu − ∇ · [∇w × (∇u × c)δv]
+ {∇ · [∇w × (∇u × c)]}δv + ∇ · [∇v × (∇u × c)δw] − ∇ · [∇v × (∇u × c)]δw
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
+ ∇ · (∇v × ∇w) × c
δu +
∂ F
∂v
+ ∇ · ∇w × (∇u × c)
δv
+
∂ F
∂w
− ∇ · ∇v × (∇u × c)
δw
dV +
V
[∇ · a × b + ∇ · (∇v × ∇w) × cδu
− ∇ · ∇w × (∇u × c)δv + ∇ · ∇v × (∇u × c)δw]dV
(4)
The Euler equations of the functional are