674
10 Variational Problems of Functionals with Vector, Tensor …
n ×
∂ F
∂∇ × ∇ × (a × b)
S
= 0, n × ∇ ×
∂ F
∂∇ × ∇ × (a × b)
S
= 0
(6)
Corollary Let the functional
J =
V
F(a, b, ∇
n
× (a × b))dV
(1)
The Euler equations of the functional are
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂ a
+ b × ∇
n
×
∂ F
∂∇ n × (a × b)
= 0
∂ F
∂ b
− a × ∇
n
×
∂ F
∂∇ n × (a × b)
= 0
(2)
The natural boundary conditions are
n ×
∂ F
∂∇ n × (a × b)
S
= 0, n × ∇ ×
∂ F
∂∇ n × (a × b)
S
= 0, . . . , n × ∇
n−1 ×
∂ F
∂∇ n × (a × b)
S
= 0
(3)
or
n × ∇
i−1
×
∂ F
∂∇ n × (a × b)
S
= 0 (i = 1, 2, · · · , n)
(4)
Example 10.5.6 The functional with the rotation of cross product of a vector and
the rotation of another vector. Let the functional
J =
V
F(a, b, ∇ × (b × ∇ × a))dV
(1)
Find its Euler equations and natural boundary conditions.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂∇ × (b × ∇ × a)
· δ∇ × (b × ∇ × a)
dV
(2)
Let c =
∂ F
∂∇×(b×∇×a)
, d = δ(b × ∇ × a) = δb × ∇ × a + b × ∇ × δa, there is
c · ∇ × d = −∇ · (c × d) + (∇ × c) · d = −∇ · (c × d) + (∇ × c) · (δb × ∇ × a + b × ∇ × δa)
= −∇ · (c × d) + (∇ × a) × (∇ × c) · δb − b × (∇ × c) · ∇ × δa
= −∇ · (c × d) + (∇ × a) × (∇ × c) · δb + ∇ · [b × (∇ × c) × δa] − ∇ × (b × ∇ × c) · δa
(3)
10 Variational Problems of Functionals with Vector, Tensor …
n ×
∂ F
∂∇ × ∇ × (a × b)
S
= 0, n × ∇ ×
∂ F
∂∇ × ∇ × (a × b)
S
= 0
(6)
Corollary Let the functional
J =
V
F(a, b, ∇
n
× (a × b))dV
(1)
The Euler equations of the functional are
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂ a
+ b × ∇
n
×
∂ F
∂∇ n × (a × b)
= 0
∂ F
∂ b
− a × ∇
n
×
∂ F
∂∇ n × (a × b)
= 0
(2)
The natural boundary conditions are
n ×
∂ F
∂∇ n × (a × b)
S
= 0, n × ∇ ×
∂ F
∂∇ n × (a × b)
S
= 0, . . . , n × ∇
n−1 ×
∂ F
∂∇ n × (a × b)
S
= 0
(3)
or
n × ∇
i−1
×
∂ F
∂∇ n × (a × b)
S
= 0 (i = 1, 2, · · · , n)
(4)
Example 10.5.6 The functional with the rotation of cross product of a vector and
the rotation of another vector. Let the functional
J =
V
F(a, b, ∇ × (b × ∇ × a))dV
(1)
Find its Euler equations and natural boundary conditions.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂∇ × (b × ∇ × a)
· δ∇ × (b × ∇ × a)
dV
(2)
Let c =
∂ F
∂∇×(b×∇×a)
, d = δ(b × ∇ × a) = δb × ∇ × a + b × ∇ × δa, there is
c · ∇ × d = −∇ · (c × d) + (∇ × c) · d = −∇ · (c × d) + (∇ × c) · (δb × ∇ × a + b × ∇ × δa)
= −∇ · (c × d) + (∇ × a) × (∇ × c) · δb − b × (∇ × c) · ∇ × δa
= −∇ · (c × d) + (∇ × a) × (∇ × c) · δb + ∇ · [b × (∇ × c) × δa] − ∇ × (b × ∇ × c) · δa
(3)
