718
10 Variational Problems of Functionals with Vector, Tensor …
∂ F
∂∇ A
n + 1
· ∇δ A = ∇ ·
δ A n
·
∂ F
∂∇ A
−
∂ F
∂∇ A
· ∇ n
· δ A
(10.7.5)
In Eq. (10.7.4), change A into δ A, let D =
∂ F
∂ A·∇
, then the variational formula of
the right divergence of A is obtained
∂ F
∂ A · ∇
n − 1
· δ A · ∇ = ∇ ·
∂ F
∂ A · ∇
n − 1
· δ A
− ∇
∂ F
∂ A · ∇
n
· δ A
(10.7.6)
Equations (10.7.5) and (10.7.6) are the precisely needed results.
Let A and B be both the tensors of order n, A = A j 1 j 2 ··· j n e j 1 e j 2 · · · e j n , B =
B k 1 k 2 ···k n e k 1 e k 2 · · · e k n , do one time cross product and n − 1 time inner products,
then taking the divergence, making use of the parallelepipedal product formulae of
trivector a · (b × c) = b · (c × a) = c · (a × b), we obtain
∇ · [A ×
(n − 1)· B] = −∇ · [B (n − 1)·
×
A] = ∂ i e i · [A j 1 j 2 ··· jn e j 1 e j 2 · · · e jn
×
(n − 1)· B k 1 k 2 ···kn e k 1 e k 2 · · · e kn ]
= (A j 1 j 2 ··· jn ,i B k 1 k 2 ···kn + A j 1 j 2 ··· jn B k 1 k 2 ···kn ,i )e i · (e j 1 × e kn )(e j 2 · e k n−1 ) · · · (e jn · · · e k 1 )
= A j 1 j 2 ··· jn ,i B k 1 k 2 ···kn e kn · (e i × e j 1 )(e j 2 · e k n−1 ) · · · (e jn · · · e k 1 )
+ A j 1 j 2 ··· jn B k 1 k 2 ···kn ,i e j 1 · (e kn × e i )(e j 2 · e k n−1 ) · · · (e jn · e k 1 )
= ∂ i e i × A j 1 j 2 ··· jn e j 1 e j 2 · · · e jn
n
· B k 1 k 2 ···kn e k 1 e k 2 · · · e kn
+ A j 1 j 2 ··· jn e j 1 e j 2 · · · e jn
n
· B k 1 k 2 ···kn e k 1 e k 2 · · · e kn × ∂ i e i
= ∇ × A n
· B + A n
· B × ∇ = B n
· ∇ × A + B × ∇ n
· A
(10.7.7)
where, B × ∇ is the right rotation of B.
Transposing A and B in Eq. (3.7), we have
∇ · [B ×
(n − 1)· A] = −∇ · [A (n − 1)·
×
B] = ∇ × B n
· A + B n
· A × ∇ = A n
· ∇ × B + A × ∇ n
· B
(10.7.8)
In Eq. (10.7.7), change A into δ A, let B =
∂ F
∂∇× A
, then the variational formula of
the left rotation of A can be obtained
∂ F
∂∇ × A
n
· ∇ × δ A = ∇ ·
δ A ×
(n − 1)·
∂ F
∂∇ × A
−
∂ F
∂∇ × A
× ∇ n
· δ A (10.7.9)
In Eq. (10.7.8), change A into δ A, let B =
∂ F
∂ A×∇
, then the variational formula of
the right rotation of A can be obtained
∂ F
∂ A × ∇
n
· δ A × ∇ = ∇ ·
∂ F
∂ A × ∇
×
(n − 1)· δ A
− ∇ ×
∂ F
∂ A × ∇
n
· δ A
(10.7.10)
10 Variational Problems of Functionals with Vector, Tensor …
∂ F
∂∇ A
n + 1
· ∇δ A = ∇ ·
δ A n
·
∂ F
∂∇ A
−
∂ F
∂∇ A
· ∇ n
· δ A
(10.7.5)
In Eq. (10.7.4), change A into δ A, let D =
∂ F
∂ A·∇
, then the variational formula of
the right divergence of A is obtained
∂ F
∂ A · ∇
n − 1
· δ A · ∇ = ∇ ·
∂ F
∂ A · ∇
n − 1
· δ A
− ∇
∂ F
∂ A · ∇
n
· δ A
(10.7.6)
Equations (10.7.5) and (10.7.6) are the precisely needed results.
Let A and B be both the tensors of order n, A = A j 1 j 2 ··· j n e j 1 e j 2 · · · e j n , B =
B k 1 k 2 ···k n e k 1 e k 2 · · · e k n , do one time cross product and n − 1 time inner products,
then taking the divergence, making use of the parallelepipedal product formulae of
trivector a · (b × c) = b · (c × a) = c · (a × b), we obtain
∇ · [A ×
(n − 1)· B] = −∇ · [B (n − 1)·
×
A] = ∂ i e i · [A j 1 j 2 ··· jn e j 1 e j 2 · · · e jn
×
(n − 1)· B k 1 k 2 ···kn e k 1 e k 2 · · · e kn ]
= (A j 1 j 2 ··· jn ,i B k 1 k 2 ···kn + A j 1 j 2 ··· jn B k 1 k 2 ···kn ,i )e i · (e j 1 × e kn )(e j 2 · e k n−1 ) · · · (e jn · · · e k 1 )
= A j 1 j 2 ··· jn ,i B k 1 k 2 ···kn e kn · (e i × e j 1 )(e j 2 · e k n−1 ) · · · (e jn · · · e k 1 )
+ A j 1 j 2 ··· jn B k 1 k 2 ···kn ,i e j 1 · (e kn × e i )(e j 2 · e k n−1 ) · · · (e jn · e k 1 )
= ∂ i e i × A j 1 j 2 ··· jn e j 1 e j 2 · · · e jn
n
· B k 1 k 2 ···kn e k 1 e k 2 · · · e kn
+ A j 1 j 2 ··· jn e j 1 e j 2 · · · e jn
n
· B k 1 k 2 ···kn e k 1 e k 2 · · · e kn × ∂ i e i
= ∇ × A n
· B + A n
· B × ∇ = B n
· ∇ × A + B × ∇ n
· A
(10.7.7)
where, B × ∇ is the right rotation of B.
Transposing A and B in Eq. (3.7), we have
∇ · [B ×
(n − 1)· A] = −∇ · [A (n − 1)·
×
B] = ∇ × B n
· A + B n
· A × ∇ = A n
· ∇ × B + A × ∇ n
· B
(10.7.8)
In Eq. (10.7.7), change A into δ A, let B =
∂ F
∂∇× A
, then the variational formula of
the left rotation of A can be obtained
∂ F
∂∇ × A
n
· ∇ × δ A = ∇ ·
δ A ×
(n − 1)·
∂ F
∂∇ × A
−
∂ F
∂∇ × A
× ∇ n
· δ A (10.7.9)
In Eq. (10.7.8), change A into δ A, let B =
∂ F
∂ A×∇
, then the variational formula of
the right rotation of A can be obtained
∂ F
∂ A × ∇
n
· δ A × ∇ = ∇ ·
∂ F
∂ A × ∇
×
(n − 1)· δ A
− ∇ ×
∂ F
∂ A × ∇
n
· δ A
(10.7.10)