Appendix: All Solutions to the Problems
971
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂ a
− [(∇ b) × c] · ∇
· δa +
∂ F
∂ b
− (c × ∇a) · ∇
· δb
dV
−
V
{∇ · [δa · (∇ b) × c] − ∇ · [δb · (c × ∇a)]}dV
(3)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂ a
−
⎡
⎣ (∇ b) ×
∂ F
∂∇ a ×
·
∇ b
⎤
⎦ · ∇ = 0
∂ F
∂ b
−
⎛
⎝ ∂ F
∂∇ a ×
·
∇ b
× ∇a
⎞
⎠ · ∇ = 0
(4)
The natural boundary conditions are
⎡
⎢
⎣(∇ b) ×
∂ F
∂∇ a ×
· ∇ b
⎤
⎥
⎦ · n
S
= 0,
⎛
⎜
⎝
∂ F
∂∇ a ×
· ∇ b
× ∇a
⎞
⎟
⎠ · n
S
= 0
(5)
10.60 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 series 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 both A and B be the second order tensors, 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 2 )(a 2 × b 1 ) = (a 1 · b 2 )(a 2 · b 1 × c)
= −(a 1 · b 2 )(a 2 · c × b 1 ) = −A : c × B = −(a 1 · b 2 )(a 2 × c · b 1 ) = −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 = −c × ∇b : ∇δa − ∇a × c : ∇δb