10.4 The Euler Equations of Divergence Type Functionals
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∂ F
∂u
− ∇ · b ×
a × ∇
∂ F
∂∇ · a × (∇u × b)
= 0
∂ F
∂ a
+ (b × ∇u) × ∇
∂ F
∂∇ · a × (∇u × b)
= 0
∂ F
∂ b
− ∇u ×
a × ∇
∂ F
∂∇ · a × (∇u × b)
= 0
(5)
The natural boundary conditions are
∂ F
∂∇ · a × (∇u × b)
S
= 0,
b ×
a × ∇
∂ F
∂∇ · a × (∇u × b)
· n
S
= 0 (6)
Example 10.4.15 The functional with the divergence of cross product of trivector,
where the three vectors are the gradients of three scalars respectively. Let the
functional
J =
V
F(u, v, w, ∇ · ∇u × (∇v × ∇w))dV
(1)
Find its Euler equations and natural boundary conditions.
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 ϕ =
∂ F
∂∇·∇u×(∇v×∇w)
, a = δ∇u × (∇v × ∇w) = ∇δu × (∇v × ∇w) +
∇u × (∇δv × ∇w) + ∇u × (∇v × ∇δw), and using the parallelepipedal product
formula of trivector, there is
ϕ∇ · δ[∇u × (∇v × ∇w)] = ϕ∇ · a = ∇ · ϕa − ∇ϕ · a
= ∇ · ϕa − ∇ϕ · [∇δu × (∇v × ∇w) + ∇u × (∇δv × ∇w)
+ ∇u × (∇v × ∇δw)]
= ∇ · ϕa − [(∇v × ∇w) × ∇ϕ] · ∇δu
+ (∇u × ∇ϕ) · (∇w × ∇δv) − (∇u × ∇ϕ) · (∇v × ∇δw)
= ∇ · ϕa − ∇ · [(∇v × ∇w) × ∇ϕδu]
+ [∇ · (∇v × ∇w) × ∇ϕ]δu
+ [(∇u × ∇ϕ) × ∇w] · ∇δv − [(∇u × ∇ϕ) × ∇v] · ∇δw
= ∇ · ϕa − ∇ · [(∇v × ∇w) × ∇ϕδu]
+ [∇ · (∇v × ∇w) × ∇ϕ]δu + ∇ · [(∇u × ∇ϕ) × ∇wδv]
− [∇ · (∇u × ∇ϕ) × ∇w]δv − ∇ · [(∇u × ∇ϕ) × ∇vδw]
+ [∇ · (∇u × ∇ϕ) × ∇v]δw
(3)