716
10 Variational Problems of Functionals with Vector, Tensor …
According to the calculation formulas of double vector product and parallelepipedal product of trivector, there is
[(a × b) × c] · n = [c × (b × a)] · n = [(a · c)b − (b · c)a] · n = (b · n)(c · a) − (b · c)(n · a)
= [(b · n)c − (b · c)n] · a = [b × (c × n)] · a = [(n × c) × b] · a
= (a · c)(n · b) − (a · n)(c · b) = [(a · c)n − (a · n)c] · b = [a × (n × c)] · b
= (n · b)(a · c) − (n · a)(b · c) = [(n · b)a − (n · a)b] · c = [n × (a × b)] · c
(9)
The natural boundary conditions can also be written as
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
n ·
∂ F
∂∇(a× A)
×
(n − 1)· A
S
= 0, a ×
n ·
∂ F
∂∇(a× A)
S
= 0, n ∂ F
∂∇·(a× A)
×
(n − 1)· A
S
= 0
a ×
n ∂ F
∂∇·(a× A)
S
= 0, n ×
∂ F
∂∇×(a× A)
×
(n − 1)· A
S
= 0, a ×
n ×
∂ F
∂∇×(a× A)
S
= 0
(10)
The various functionals of Examples 10.6.17–10.6.22 are all incorporated into
the functional (10.6.30) or functional (10.6.43), the Euler equations and the
corresponding natural boundary conditions with the more contents can be obtained.
10.7 Variational Problems of Functionals with Series-Type
Inner Product Tensors and Hamiltonian Operators
The variational problems of the functional with the series type inner product tensors
and the Hamiltonian operators namely with the gradient, divergence and rotation
are discussed in this section which contain gradient, divergence and rotation are
discussed in this section, where the method used is the same as the method used in
the previous section, only the operation results are different to some extent. At the
same time it can also be seen that the series type operation of tensors can sometimes
be more complex than the parallel type operation, for some operations, some tensors
have not be represent by the whole, they must be broken down into two parts, which
reflects the singularity of mathematical beauty.
10.7.1 Variational Formula Derivations of Gradients,
Divergences and Rotations of Series-Type Inner
Product Tensors
Let A be a tensor of order n, A = A j 1 j 2 ··· j n e j 1 e j 2 · · · e j n , D is a tensor of order n − 1,
D = D k 1 k 2 ···k n−1 e k 1 e k 2 · · · e k n−1 , take ∂ i =
∂
∂ x i
, ∇ = ∂ i e i =
∂
∂ x i
e i , do n − 1 time
inner products of A and D, then take the divergence, noticing that A n − 1
· D is a
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