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10 Variational Problems of Functionals with Vector, Tensor …
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
+ ∇ · (∇v × ∇w) × ∇ϕ
δu +
∂ F
∂v
− ∇ · (∇u × ∇ϕ) × ∇w
δv
+
∂ F
∂w
+ ∇ · (∇u × ∇ϕ) × ∇v
δw
dV +
V
{∇ · ϕa − ∇ · [(∇v × ∇w) × ∇ϕδu]
+ ∇ · [(∇u × ∇ϕ) × ∇wδv] − ∇ · [(∇u × ∇ϕ) × ∇vδw]}dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂u
+ ∇ · (∇v × ∇w) × ∇
∂ F
∂∇ · ∇u × (∇v × ∇w)
= 0
∂ F
∂v
+ ∇ · ∇w ×
∇u × ∇
∂ F
∂∇ · ∇u × (∇v × ∇w)
= 0
∂ F
∂w
− ∇ · ∇v ×
∇u × ∇
∂ F
∂∇ · ∇u × (∇v × ∇w)
= 0
(5)
The natural boundary conditions are
ϕ| S = 0, [(∇v × ∇w) × ∇ϕ] · n| S = 0, [(∇u × ∇ϕ) × ∇v] · n| S = 0, [(∇u × ∇ϕ) × ∇v] · n| S = 0
(6)
Example 10.4.16 The functional with the divergence of cross product of respective
gradients of two scalars. Let the functional
J =
V
F(u, v, ∇ · ∇u × ∇v)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
∂∇ · ∇u × ∇v
δ∇ · ∇u × ∇v
dV
(2)
Let ϕ =
∂ F
∂∇·∇u×∇v
, a = δ(∇u × ∇v) = ∇δu × ∇v + ∇u × ∇δv, and using the
parallelepipedal product formula of trivector, there is
ϕ∇ · δ(∇u × ∇v) = ϕ∇ · a = ∇ · ϕa − ∇ϕ · a = ∇ · ϕa − ∇ϕ · (∇δu × ∇v + ∇u × ∇δv)
= ∇ · ϕa − (∇v × ∇ϕ) · ∇δu + (∇u × ∇ϕ) · ∇δv
= ∇ · ϕa − ∇ · [(∇v × ∇ϕ)δu] + [∇ · (∇v × ∇ϕ)]δu
+ ∇ · [(∇u × ∇ϕ)δv] − [∇ · (∇u × ∇ϕ)]δv
(3)
Substituting Eq. (3) into Eq. (2), we get