946
Appendix: All Solutions to the Problems
The Euler equation of the functional is
εεϕ + (1 − e
−x
2 /λ
2 )ρ = 0
The natural boundary conditions are ∇ϕ · n| S = 0.
10.19 The functional of the time-dependent Schrödinger equation is
J [ϕ] =
V
[(V ϕ + 2iϕ 0 − iϕ)ϕ + k∇ϕ · ∇ϕ]dV
Find its Euler equation.
Solution Putting F = (V ϕ +2iϕ 0 −iϕ)ϕ +k∇ϕ ·∇ϕ, the various partial derivatives
are
∂ F
∂ϕ
= 2(V ϕ + iϕ 0 − iϕ),
∂ F
∂∇ϕ
= 2k∇ϕ,
The Euler equation of the functional is
kϕ − V ϕ + i(ϕ − ϕ 0 ) = 0
10.20 Find the Ostrogradsky equation of the functional J [u]
=
1
m
[
˝
V (2 f u + |∇u|
2
)dV ]
n , where, n ≥ 1.
Solution The exponent n does not affect variation. Putting F = 2 f u + |∇u|
2 ,
finding the various partial derivatives, we have
∂ F
∂u
= 2 f,
∂ F
∂|∇u|
= 2|∇u|
The Euler equation of the functional namely the Ostrogradsky equation is
u =
∂
2 u
∂ x 2 +
∂
2 u
∂ y 2 +
∂
2 u
∂z 2 = f
10.21 Find the Euler equations and corresponding boundary conditions of the
functional J [u] =
1
2
˝
V
1
μ
|∇u|
2
+ iωσ u
2
− 2 ju
dV .
Solution Putting F =
1
2
1
μ
|∇u|
2
+ iωσ u
2
− 2 ju
, the various partial derivatives
are
F u = iωσ u − j,
∂ F
∂∇u
=
∇u
μ
The Euler equation of the functional is
Appendix: All Solutions to the Problems
The Euler equation of the functional is
εεϕ + (1 − e
−x
2 /λ
2 )ρ = 0
The natural boundary conditions are ∇ϕ · n| S = 0.
10.19 The functional of the time-dependent Schrödinger equation is
J [ϕ] =
V
[(V ϕ + 2iϕ 0 − iϕ)ϕ + k∇ϕ · ∇ϕ]dV
Find its Euler equation.
Solution Putting F = (V ϕ +2iϕ 0 −iϕ)ϕ +k∇ϕ ·∇ϕ, the various partial derivatives
are
∂ F
∂ϕ
= 2(V ϕ + iϕ 0 − iϕ),
∂ F
∂∇ϕ
= 2k∇ϕ,
The Euler equation of the functional is
kϕ − V ϕ + i(ϕ − ϕ 0 ) = 0
10.20 Find the Ostrogradsky equation of the functional J [u]
=
1
m
[
˝
V (2 f u + |∇u|
2
)dV ]
n , where, n ≥ 1.
Solution The exponent n does not affect variation. Putting F = 2 f u + |∇u|
2 ,
finding the various partial derivatives, we have
∂ F
∂u
= 2 f,
∂ F
∂|∇u|
= 2|∇u|
The Euler equation of the functional namely the Ostrogradsky equation is
u =
∂
2 u
∂ x 2 +
∂
2 u
∂ y 2 +
∂
2 u
∂z 2 = f
10.21 Find the Euler equations and corresponding boundary conditions of the
functional J [u] =
1
2
˝
V
1
μ
|∇u|
2
+ iωσ u
2
− 2 ju
dV .
Solution Putting F =
1
2
1
μ
|∇u|
2
+ iωσ u
2
− 2 ju
, the various partial derivatives
are
F u = iωσ u − j,
∂ F
∂∇u
=
∇u
μ
The Euler equation of the functional is
