678
10 Variational Problems of Functionals with Vector, Tensor …
+
∂ F
∂ b
+ (a · ∇ × c)a − (a · a)∇ × c
· δb − ∇ · c × d
dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂ a
+
a · ∇ ×
∂ F
∂∇ × [a × (a × b)]
b + (a · b)∇ ×
∂ F
∂∇ × [a × (a × b)]
−2
b · ∇ ×
∂ F
∂∇ × [a × (a × b]
a = 0
∂ F
∂ b
+
a · ∇ ×
∂ F
∂∇ × [a × (a × b)]
a − (a · a)∇ ×
∂ F
∂∇ × [a × (a × b)]
= 0
(5)
The natural boundary condition is
∂ F
∂∇ × [a × (a × b)]
× n
S
= 0
(6)
Example 10.5.10 The functional with the rotation of cross product of trivector.
Derive the Euler equations and natural boundary condition of the functional
J =
V
F(a, b, c, ∇ × [a × (b × c)])dV
(1)
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂ c
· δc +
∂ F
∂∇ × [a × (b × c)]
· δ∇ × [a × (b × c)]
dV
(2)
Putting d =
∂ F
∂∇×[a×(b×c)]
, e = δ[a × (b × c)] = δa × (b × c) + a × (δb × c) +
a × (b × δc), there is
d · δ∇ × [a × (b × c)] = d · ∇ × δ[a × (b × c)] = d · ∇ × e = −∇ · (d × e) + (∇ × d) · e
= −∇ · (d × e) + (∇ × d) · [δa × (b × c) + a × (δb × c) + a × (b × δc)]
= −∇ · (d × e) + [(b × c) × ∇ × d] · δa + [(∇ × d) × a] · (δb × c) + [(∇ × d) × a] · (b × δc)
= −∇ · (d × e) + [(b × c) × ∇ × d] · δa + c × [(∇ × d) × a] · δb − b × [(∇ × d) × a] · δc
= −∇ · (d × e) + [(b × c) × ∇ × d] · δa − c × (a × ∇ × d) · δb + b × (a × ∇ × d) · δc
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂ a
+ (b × c) × ∇ × d
· δa +
∂ F
∂ b
− c × (a × ∇ × d)
· δb
10 Variational Problems of Functionals with Vector, Tensor …
+
∂ F
∂ b
+ (a · ∇ × c)a − (a · a)∇ × c
· δb − ∇ · c × d
dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂ a
+
a · ∇ ×
∂ F
∂∇ × [a × (a × b)]
b + (a · b)∇ ×
∂ F
∂∇ × [a × (a × b)]
−2
b · ∇ ×
∂ F
∂∇ × [a × (a × b]
a = 0
∂ F
∂ b
+
a · ∇ ×
∂ F
∂∇ × [a × (a × b)]
a − (a · a)∇ ×
∂ F
∂∇ × [a × (a × b)]
= 0
(5)
The natural boundary condition is
∂ F
∂∇ × [a × (a × b)]
× n
S
= 0
(6)
Example 10.5.10 The functional with the rotation of cross product of trivector.
Derive the Euler equations and natural boundary condition of the functional
J =
V
F(a, b, c, ∇ × [a × (b × c)])dV
(1)
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂ c
· δc +
∂ F
∂∇ × [a × (b × c)]
· δ∇ × [a × (b × c)]
dV
(2)
Putting d =
∂ F
∂∇×[a×(b×c)]
, e = δ[a × (b × c)] = δa × (b × c) + a × (δb × c) +
a × (b × δc), there is
d · δ∇ × [a × (b × c)] = d · ∇ × δ[a × (b × c)] = d · ∇ × e = −∇ · (d × e) + (∇ × d) · e
= −∇ · (d × e) + (∇ × d) · [δa × (b × c) + a × (δb × c) + a × (b × δc)]
= −∇ · (d × e) + [(b × c) × ∇ × d] · δa + [(∇ × d) × a] · (δb × c) + [(∇ × d) × a] · (b × δc)
= −∇ · (d × e) + [(b × c) × ∇ × d] · δa + c × [(∇ × d) × a] · δb − b × [(∇ × d) × a] · δc
= −∇ · (d × e) + [(b × c) × ∇ × d] · δa − c × (a × ∇ × d) · δb + b × (a × ∇ × d) · δc
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂ a
+ (b × c) × ∇ × d
· δa +
∂ F
∂ b
− c × (a × ∇ × d)
· δb
