10.7 Variational Problems of Functionals with Series-Type …
721
∂ F
∂ A∇
n + 1
· δ A∇ = ∇ ·
∂ F
∂ A∇
n
· δ A
− ∇ ·
∂ F
∂ A∇
n
· δ A
(10.7.21)
∂ F
∂∇ · A
n − 1
· ∇ · δ A = ∇ ·
δ A n − 1
·
∂ F
∂∇ · A
−
∂ F
∂∇ · A
∇ n
· δ A
(10.7.22)
∂ F
∂ A · ∇
n − 1
· δ A · ∇ = ∇ ·
∂ F
∂ A · ∇
n − 1
· δ A
− ∇
∂ F
∂ A · ∇
n
· δ A
(10.7.23)
∂ F
∂∇ × A
n
· ∇ × δ A = ∇ ·
δ A ×
(n − 1)·
∂ F
∂∇ × A
−
∂ F
∂∇ × A
× ∇ n
· δ A
(10.7.24)
∂ F
∂ A × ∇
n
· δ A × ∇ = ∇ ·
∂ F
∂ A × ∇
×
(n − 1)· δ A
− ∇ ×
∂ F
∂ A × ∇
n
· δ A
(10.7.25)
Making use of Eqs. (10.7.20)–(10.7.25), the Euler equation and corresponding
natural boundary conditions of the functional with series-type inner product tensors
and the Hamiltonian operators-namely with the gradient, divergence and rotation can
be conveniently obtained.
Theorem 10.7.1 Let A be a tensor of order n, the functional with the Hamiltonian
operators is
J =
V
F( A, ∇ A, ∇ · A, ∇ × A, A∇, A · ∇, A × ∇)dV
(10.7.26)
then the Euler equation of the functional is
∂ F
∂ A
−
∂ F
∂∇ A
· ∇ −
∂ F
∂∇ · A
∇ −
∂ F
∂∇ × A
× ∇ − ∇ ·
∂ F
∂ A∇
− ∇
∂ F
∂ A · ∇
− ∇ ×
∂ F
∂ A × ∇
= 0
(10.7.27)
The natural boundary conditions are
∂ F
∂∇ A
· n
S
= 0,
∂ F
∂∇· A
n
S
= 0,
∂ F
∂∇× A
× n
S
= 0
n ·
∂ F
∂ A∇
S
= 0, n
∂ F
∂ A·∇
S
= 0, n ×
∂ F
∂ A×∇
S
= 0
(10.7.28)
where, for
∂ F
∂∇· A
n
S
= 0 and n
∂ F
∂ A·∇
S
= 0, if they do not appear together with the
other boundary conditions, the vector n in the expressions can be removed.
Proof Taking the variation to Eq. (10.7.26) and making use of Eqs. (10.7.20)–
(10.7.25), we have
δ J =
V
∂ F
∂ A
n
·
δ A +
∂ F
∂∇ A
n + 1
·
δ∇ A +
∂ F
∂∇ · A
n − 1
·
δ∇ · A +
∂ F
∂∇ × A
n
·
δ∇ × A
dV