970
Appendix: All Solutions to the Problems
+
V
[∇ · (∇ b × c · δa) − ∇ · (∇ a × c · δb)]dV
(3)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂ a
− ∇ ·
⎡
⎢
⎣(∇ b) ×
∂ F
∂ a ·
× ∇ b
⎤
⎥
⎦ = 0
∂ F
∂ b
+ ∇ ·
⎡
⎢
⎣(∇ a) ×
∂ F
∂ a ·
× ∇ b
⎤
⎥
⎦ = 0
(4)
The natural boundary conditions are
n ·
⎡
⎢
⎣(∇ b) ×
∂ F
∂ a ·
× ∇ b
⎤
⎥
⎦
S
= 0, n ·
⎡
⎢
⎣(∇ a) ×
∂ F
∂ a ·
× ∇ b
⎤
⎥
⎦
S
= 0
(5)
10.59 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 A and B be both 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 · (c × b 2 )(a 2 · b 1 )
= a 1 · (b 2 × c)(a 2 · b 1 ) = A : B × c = (c × a 1 ) · b 2 (a 2 · b 1 ) = c × A : 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 + c × ∇a : ∇δb
= ∇ · [δa · (∇ b) × c] − [(∇ b) × c] · ∇ · δa + ∇ · [δb · (c × ∇a)] − (c × ∇a) · ∇ · δb
(2)