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10 Variational Problems of Functionals with Vector, Tensor …
+
V
{∇ · ϕa − ∇ · [ψ(∇v × ∇w)δu] − ∇ · [ψ(∇w × ∇u)δv] − ∇ · [ψ(∇u × ∇v)δw]}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
+ ∇ ·
(∇u × ∇v)∇ ·
∂ F
∂∇[∇u · (∇v × ∇w)]
= 0
(5)
The natural boundary conditions are
ϕ| S = 0, [(∇v × ∇w)∇ · a] · n| S = 0, [(∇w × ∇u)∇ · a] · n| S = 0, [(∇u × ∇v)∇ · a] · n| S = 0
(6)
Example 10.3.12 The functional with the gradient of divergence of a sine function
multiplied by a vector. Let the functional
J =
V
F(u, a, ∇∇ · sin kua)dV
(1)
where, k is a constant, Derive the Euler equations and natural boundary conditions
of the functional.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂∇∇ · sin kua
· δ∇∇ · sin kua
dV
(2)
where, the variation of sin kua can be written as δ sin kua = k cos kuaδu +sin kuδa.
Putting b =
∂ F
∂∇∇·sin kua
, the integral of the last term in Eq. (2) can be written as
b · δ∇∇ · sin kua = b · ∇δ∇ · sin kua = ∇ · (bδ∇ · sin kua) − (∇ · b)δ∇ · sin kua
= ∇ · (bδ∇ · sin kua) − ∇ · [(∇ · b)δ sin kua] + ∇∇ · b · (k cos kuaδu + sin kuδa)
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
+ k cos kua · ∇∇ · b
δu +
∂ F
∂ a
+ sin ku∇∇ · b
· δa
dV
+
V
{∇ · (bδ∇ · sin kua) − ∇ · [(∇ · b)δ sin kua]}dV
(4)