10.7 Variational Problems of Functionals with Series-Type …
725
∂ f
∂ψ ∗ = (α − aθ)ψ + (β − bθ)ψ
2
ψ
∗
−
e
∗ A
2m ∗ (−i∇ψ − e
∗ Aψ),
∂ f
∂∇ψ ∗ =
i
2m ∗ (−i∇ψ − e
∗ Aψ)
∂ f
∂ A
=
e
∗2
m ∗2 A|ψ|
2
+
ie
∗
2m ∗2 (ψ
∗
∇ψ − ψ∇ψ
∗
),
∂ f
∂∇ × A
=
1
μ 0
∇ × A
For a vector a, since ∇ · a = a · ∇, a × ∇ = −∇ × a, the Euler equations are
(α − aθ)ψ + (β − bθ)ψ 2 ψ ∗ − e ∗ A
2m ∗ (−i∇ψ − e ∗ Aψ) − i
2m ∗ (−i∇ · ∇ψ − e ∗ ∇ · Aψ) = 0
1
μ 0
∇ × ∇ × A + e ∗2
m ∗2 A|ψ| 2 + ie ∗
2m ∗2 (ψ ∗ ∇ψ − ψ∇ψ ∗ ) = 0
(2)
According to the formula ∇ × ∇ × A = ∇∇ · A − ∇ · ∇ A = ∇∇ · A − ∇
2 A =
∇∇ · A − A, taking note that the restricted conditions of the vector and scalar
potentials are in the Coulomb gauge, ∇ · A = 0, that is ∇∇ · A = 0, substituting it
into Eq. (2), and after arrangement, the modified Ginzburg-Landau functions can
be obtained
1
2m ∗ (−i∇ − e
∗ A)
2
ψ + (α − aθ)ψ + (β − bθ)|ψ|ψ = 0
∇
2 A =
μ 0 ie
∗
2m ∗2 (ψ
∗
∇ψ − ψ∇ψ
∗
) +
μ 0 e
∗2
m ∗2 A|ψ|
2
(3)
The natural boundary conditions of the functional are
(−i∇ψ − e ∗ Aψ) · n
S = 0, ∇ × A × n| S = 0
(4)
Example 10.7.4 The Feynman expression for electrostatic energy as a first-order
variational principle in electrostatic potential ϕ is the following functional
J =
V
ε
2
(∇ϕ)
2
+ ρϕ
dV
(1)
where, ε is the dielectric constant of matter, and ρ is the electric charge density and
the integrals are over space. Find its Euler equation.
Solution Let the integrand F =
ε
2
(∇ϕ)
2
+ ρϕ, the two partial derivatives are
∂ F
∂∇ϕ
=
ε∇ϕ,
∂ F
∂ϕ
= ρ, then the Euler equation of the functional is
ρ − (ε∇ϕ) · ∇ = 0
( 2 )
or
ϕ =
ρ
ε
(3)
This is a Poisson(’s) equation.