Appendix: All Solutions to the Problems
969
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂ a
+ ∇ ·
⎛
⎝ ∂ F
∂∇ a ×
·
∇ b
× ∇b
⎞
⎠ = 0
∂ F
∂ b
− ∇ ·
⎛
⎝ ∂ F
∂∇ a ×
·
∇ b
× ∇a
⎞
⎠ = 0
(4)
The natural boundary conditions are
n ·
⎛
⎜
⎝
∂ F
∂∇ a ×
· ∇ b
× ∇b
⎞
⎟
⎠
S
= 0, n ·
⎛
⎜
⎝
∂ F
∂∇ a ×
· ∇ b
× ∇a
⎞
⎟
⎠
S
= 0
(5)
10.58 Derive the Euler equations and natural boundary conditions of the functional
J =
V F(a, b, ∇ a ·
× ∇ b)dV , where the double inner product of tensors of second
order conforms to the law of parallel operation.
Solution The variation of the functional is
δ J =
V
⎛
⎜
⎝
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂∇ a ·
× ∇ b
· δ∇ a ·
× ∇ b
⎞
⎟
⎠dV
(1)
Let A and B be both the tensors of order two, A = a 1 a 2 , B = b 1 b 2 , c is a vector,
then
c · ( A ·
× B) = c · (a 1 a 2 ·
× b 1 b 2 ) = c · (a 1 · b 1 )(a 2 × b 2 ) = (a 1 · b 1 )(a 2 · b 2 × c)
A : B × c = −(a 1 · b 1 )(a 2 × c · b 2 ) = −A × c : B
Putting c =
∂ F
∂∇ a ·
×
∇ b
, for the latter variation term of Eq. (1), there is
c · δ∇ a ·
× ∇ b = c · ∇δa ·
× ∇ b + c · ∇a ·
× ∇δb = ∇b × c : ∇δa − ∇a × c : ∇δb
= ∇ · (∇ b × c · δa) − ∇ · (∇ b × c) · δa − ∇ · (∇ a × c · δb)
+ ∇ · (∇ a × c) · δb
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂ a
− ∇ · (∇ b × c)
· δa +
∂ F
∂ b
+ ∇ · (∇ a × c)
· δb
dV