966
Appendix: All Solutions to the Problems
Taking note that the strain tensor e is the symmetric tensor, the Euler equations
of the functional are
⎧
⎨
⎩
∇ · σ + ¯
F = 0
∂ A(e)
∂ e
− σ = 0
e =
1
2
(∇u + u∇)
(1)
The boundary condition is
( ¯
u − u)| S u = 0
(2)
10.56 Discuss the variational problem of the functional J =
V F(a, ∇ a · a∇)dV .
Solution The variation of the functional is
δ J =
V
∂ F
∂ a
· δa +
∂ F
∂∇ a · a∇
: δ∇ a · a∇
dV
(1)
Let A and B be both the tensor of second order, A = a 1 a 2 , B = b 1 b 2 , then
( A · B) T = (a 1 a 2 · b 1 b 2 ) T = [a 1 (a 2 · b 1 )b 2 ] T = b 2 (b 1 · a 2 )a 1 = b 2 b 1 · a 2 a 1 = B T · A T
Putting B =
∂ F
∂∇ a·a∇
, for the latter variation term of Eq. (1), there is
B : δ∇ a · a∇ = B : ∇δa · a∇ + B : ∇a · δa∇
(2)
For Eq. (2), it can be transformed into three forms, the first form is to transform
the gradient terms into the form of the left gradient, there is
B : δ∇ a · a∇ = B : ∇δa · a∇ + B : ∇a · δa∇ = B · ∇a : ∇δa + B
T
· ∇a : ∇δa
= ∇ · {[(B + B
T
) · ∇a] · δa} − {∇ · [(B + B
T
) · ∇a]} · δa
(3)
Obviously B + B
T is the symmetric tensor.
The second form is to transform the gradient terms into the form of the right
gradient, there is
B : δ∇ a · a∇ = B : ∇δa · a∇ + B : ∇a · δa∇ = a∇ · B
T
: δa∇ + a∇ · B : δa∇
= {δa · [a∇ · (B
T
+ B)]} · ∇ − {[a∇ · (B + B
T
)] · ∇} · δa
(4)
The third form is to reserve the form of a left gradient and the form of a right
gradient respectively, there is
B : δ∇ a · a∇ = B : ∇δa · a∇ + B : ∇a · δa∇ = B · ∇a : ∇δa + a∇ · B : δa∇