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10 Variational Problems of Functionals with Vector, Tensor …
In order to better master the properties of the Hamiltonian operator strings, below
we give two examples.
Example 10.6.13 Let A be a tensor of order n, here n = 0, 1, 2, · · · , prove that
∇ × ∇ A = 0.
Proof Expanding the operator sting ∇ × ∇, there is
∇ × ∇ =
∂
∂ x
i +
∂
∂ y
j +
∂
∂z
k
×
∂
∂ x
i +
∂
∂ y
j +
∂
∂z
k
=
∂ 2
∂ x∂ x
i × i +
∂ 2
∂ x∂ y
i × j +
∂ 2
∂ x∂z
i × k +
∂ 2
∂ y∂ x
j × i +
∂ 2
∂ y∂ y
j × j
+
∂ 2
∂ y∂z
j × k +
∂ 2
∂z∂ x
k × i +
∂ 2
∂z∂ y
k × j +
∂ 2
∂z∂z
k × k
=
∂ 2
∂ x∂ y
(i × j + j × i) +
∂ 2
∂ x∂z
(i × k + k × i) +
∂ 2
∂ y∂z
( j × k + k × j ) = 0
That is ∇ × ∇ A = 0. Quod erat demonstrandum.
Example 10.6.14 Let A be a tensor of order n, here n = 1, 2, · · · , prove that ∇ ·
∇ × A = 0.
Proof Let A = A k 1 k 2 ···k n e k 1 e k 2 · · · e k n , and putting e k 1 = i, then there is
∇ · ∇ × A =
∂
∂ x
i +
∂
∂ y
j +
∂
∂z
k
·
∂
∂ x
i +
∂
∂ y
j +
∂
∂z
k
× A k 1 k 2 ···kn e k 1 e k 2 · · · e kn
=
∂
∂ x
i +
∂
∂ y
j +
∂
∂z
k
·
∂
∂ x
i × e k 1 +
∂
∂ y
j × e k 1 +
∂
∂z
k × e k 1
A k 1 k 2 ···kn e k 2 · · · e kn
=
∂
∂ x
i +
∂
∂ y
j +
∂
∂z
k
·
∂
∂ x
i × i +
∂
∂ y
j × i +
∂
∂z
k × i
A k 1 k 2 ···kn e k 2 · · · e kn
=
∂
∂ x
i +
∂
∂ y
j +
∂
∂z
k
·
∂
∂ x
0 −
∂
∂ y
k +
∂
∂z
j
A k 1 k 2 ···kn e k 2 · · · e kn
=
−
∂
∂ x∂ y
i · k +
∂
∂ x∂z
i · j −
∂
∂ y∂ y
j · k +
∂
∂ y∂z
j · j −
∂
∂z∂ y
k · k +
∂
∂z∂z
k · j
A k 1 k 2 ···kn e k 2 · · · e kn
=
∂
∂ y∂z
−
∂
∂z∂ y
A k 1 k 2 ···kn e k 2 · · · e kn = 0
Similarly, putting e k 1 = j or e k 1 = k, the above result can also be obtained. Quod
erat demonstrandum.
Because the rotation of gradient and the divergence of rotation are both zero,
when one of the two kinds of operations appears in a Hamiltonian operator strain,
the result must be zero.
Example 10.6.15 Let A and B be both the tensors of order n, construct the functional
J =
V
( A n
· B + ∇ A n + 1
· ∇∇ × B + ∇ × ∇ · ∇ · ∇ A n − 1
· ∇ · B)dV (1)