682
10 Variational Problems of Functionals with Vector, Tensor …
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⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ 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
a × n| S = 0, [(∇v × ∇w) × ∇ × a] · n| S = 0,
[∇w × (∇u × ∇ × a)] · n| S = 0, [∇w × (∇u × ∇ × a)] · n| S = 0 (6)
Example 10.5.13 The function with the rotation 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)
Putting a =
∂ F
∂∇×(∇u×∇v)
, b = δ(∇u × ∇v) = ∇δu × ∇v + ∇u × ∇δv, c = ∇ × a,
and using the parallelepipedal product formula of trivector, we have
a · ∇ × δ(∇u × ∇v) = a · ∇ × b = −∇ · (a × b) + (∇ × a) · b = −∇ · (a × b) + c · b
= −∇ · (a × b) + c · (∇δu × ∇v + ∇u × ∇δv)
= −∇ · (a × b) + (∇v × c) · ∇δu − (∇u × c) · ∇δv
− ∇ · (a × b) + ∇ · [(∇v × c)δu] − (∇ · ∇v × c)δu
− ∇ · [(∇u × c)δv] + (∇ · ∇u × c)δv
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
− ∇ · ∇v × ∇ × a
δu +
∂ F
∂v
+ ∇ · ∇u × ∇ × a
δv
dV
+
V
{−∇ · (a × b) + ∇ · [(∇v × ∇ × a)δu] − ∇ · [(∇u × ∇ × a)δv]}dV
(4)
The Euler equations of the functional are
10 Variational Problems of Functionals with Vector, Tensor …
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ 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
a × n| S = 0, [(∇v × ∇w) × ∇ × a] · n| S = 0,
[∇w × (∇u × ∇ × a)] · n| S = 0, [∇w × (∇u × ∇ × a)] · n| S = 0 (6)
Example 10.5.13 The function with the rotation 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)
Putting a =
∂ F
∂∇×(∇u×∇v)
, b = δ(∇u × ∇v) = ∇δu × ∇v + ∇u × ∇δv, c = ∇ × a,
and using the parallelepipedal product formula of trivector, we have
a · ∇ × δ(∇u × ∇v) = a · ∇ × b = −∇ · (a × b) + (∇ × a) · b = −∇ · (a × b) + c · b
= −∇ · (a × b) + c · (∇δu × ∇v + ∇u × ∇δv)
= −∇ · (a × b) + (∇v × c) · ∇δu − (∇u × c) · ∇δv
− ∇ · (a × b) + ∇ · [(∇v × c)δu] − (∇ · ∇v × c)δu
− ∇ · [(∇u × c)δv] + (∇ · ∇u × c)δv
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
− ∇ · ∇v × ∇ × a
δu +
∂ F
∂v
+ ∇ · ∇u × ∇ × a
δv
dV
+
V
{−∇ · (a × b) + ∇ · [(∇v × ∇ × a)δu] − ∇ · [(∇u × ∇ × a)δv]}dV
(4)
The Euler equations of the functional are
