10.2 The Euler Equations of Functionals …
625
(double inner product series, p = 1, 2, · · · , r )
(10.2.40)
where, D is a partial differential operator, i(i ≥ 1) is a degree or order of derivative
of F with respect to the independent variable combination.
Of course, the functional (10.2.37) and its corresponding Euler equations namely
Eqs. (10.2.38) and (10.2.39) or Eq. (10.2.40) may be written in more complicated
forms, it follows that the Euler equations contain the Euler equations of the complete
functional and the Euler equations of the functional variational problems in the
literature [81].
It is necessary to point out that although the aforementioned formulae of the functional variational problems are listed for three-dimensional Hamiltonian operators,
they also apply for two-dimensional Hamiltonian operators, at the moment it need
only to change the triple integral domain V into its double integral domain D, and
change the closed surface S that surrounds V into the closed curve Γ that surrounds D.
If the functional has also other additional boundary conditions, then the natural
boundary conditions of the above functional need to be combined with these
additional boundary conditions, they all together constitute the total boundary
conditions.
Example 10.2.1 Find the Euler equation of the functional J [u] =
V [∇u · ∇u + 2u f (x, y, z)]dxdydz, where V is the integral domain, f (x, y, z) is
a given function, and it is continuous in V.
Solution Let F = ∇u ·∇u + 2u f (x, y, z), there are
∂ F
∂u
= 2 f (x, y, z),
∂ F
∂∇u
= 2∇u,
thus the Euler equation of the functional is
2 f (x, y, z) − ∇ · (2∇u) = 2[ f (x, y, z) − u] = 0
( 1 )
or
u =
∂
2 u
∂ x 2 +
∂
2 u
∂ y 2 +
∂
2 u
∂z 2 = f (x, y, z)
(2)
This is the Poisson(’s) equation, it is the exact same as the result of Example
2.8.6.
Example10.2.2 The electromagnetic field problem. In an electromagnetic field, E
is the electric field intensity, B is the magnetic induction, A is the magnetic vector
potential namely vector potential, ϕ is the scalar potential, ρ is the volume density
of free charge, J c is the conduction current density, J s is the source current density,
W e = ρϕ −
1
2
ε E
2 is the electric field energy density, W m = J · A −
1
2μ
B
2 is the
magnetic field energy density, there J = J c + J s is the current density; The Lagrange
density function L = W m − W e =
1
2
ε E
2
−
1
μ
B
2
− ρϕ + J · A is introduced, thus
the constructed functional is
625
(double inner product series, p = 1, 2, · · · , r )
(10.2.40)
where, D is a partial differential operator, i(i ≥ 1) is a degree or order of derivative
of F with respect to the independent variable combination.
Of course, the functional (10.2.37) and its corresponding Euler equations namely
Eqs. (10.2.38) and (10.2.39) or Eq. (10.2.40) may be written in more complicated
forms, it follows that the Euler equations contain the Euler equations of the complete
functional and the Euler equations of the functional variational problems in the
literature [81].
It is necessary to point out that although the aforementioned formulae of the functional variational problems are listed for three-dimensional Hamiltonian operators,
they also apply for two-dimensional Hamiltonian operators, at the moment it need
only to change the triple integral domain V into its double integral domain D, and
change the closed surface S that surrounds V into the closed curve Γ that surrounds D.
If the functional has also other additional boundary conditions, then the natural
boundary conditions of the above functional need to be combined with these
additional boundary conditions, they all together constitute the total boundary
conditions.
Example 10.2.1 Find the Euler equation of the functional J [u] =
V [∇u · ∇u + 2u f (x, y, z)]dxdydz, where V is the integral domain, f (x, y, z) is
a given function, and it is continuous in V.
Solution Let F = ∇u ·∇u + 2u f (x, y, z), there are
∂ F
∂u
= 2 f (x, y, z),
∂ F
∂∇u
= 2∇u,
thus the Euler equation of the functional is
2 f (x, y, z) − ∇ · (2∇u) = 2[ f (x, y, z) − u] = 0
( 1 )
or
u =
∂
2 u
∂ x 2 +
∂
2 u
∂ y 2 +
∂
2 u
∂z 2 = f (x, y, z)
(2)
This is the Poisson(’s) equation, it is the exact same as the result of Example
2.8.6.
Example10.2.2 The electromagnetic field problem. In an electromagnetic field, E
is the electric field intensity, B is the magnetic induction, A is the magnetic vector
potential namely vector potential, ϕ is the scalar potential, ρ is the volume density
of free charge, J c is the conduction current density, J s is the source current density,
W e = ρϕ −
1
2
ε E
2 is the electric field energy density, W m = J · A −
1
2μ
B
2 is the
magnetic field energy density, there J = J c + J s is the current density; The Lagrange
density function L = W m − W e =
1
2
ε E
2
−
1
μ
B
2
− ρϕ + J · A is introduced, thus
the constructed functional is