Appendix: All Solutions to the Problems
953
Solution Putting F = ((u)
2
− 2 f u, the partial derivatives are
∂ F
∂u
= −2 f ,
∂ F
∂∂u
=
2u, the Euler equation of the functional is
(2u) − 2 f = 0 or
2 u = f
10.37 Find the Euler equation and boundary condition of the functional J =
V
1
2ρC V
(∇ · H)
2
+
H
k
·
∂ H
∂t
dV +
S θ H · ndS.
Solution Putting F =
1
2ρC V
(∇· H)
2
+
H
k
·
∂ H
∂t
, the partial derivatives are
∂ F
∂∇·H
=
∇·H
ρC V
,
∂ F
∂ H
=
1
k
∂ H
∂t
, the Euler equation of the functional is
∇∇ · H −
ρC V
k
∂ H
∂t
= 0
The boundary condition is
(∇ · H + ρC V θ)| S = 0
10.38 Find the Euler equations of the energy functional J [E, E
∗
] =
V (∇ × E
∗
· μ
−1
∇ × E − ω
2 E
∗
· ε E)dV .
Solution Let F = ∇ × E
∗
· μ
−1
∇ × E − ω
2 E
∗
· ε E, find the various partial
derivatives, we get
∂ F
∂ E
= −ω 2 ε E ∗ ,
∂ F
∂∇ × E
= μ −1 ∇ × E ∗ ,
∂ F
∂ E ∗ = −ω 2 ε E,
∂ F
∂∇ × E ∗ = μ −1 ∇ × E
The Euler equations of the functional are
∇ × (μ
−1
∇ × E
∗
) − ω
2
ε E
∗
= 0
∇ × (μ
−1
∇ × E) − ω
2
ε E = 0
10.39 In the electromagnetic field problem of the linear, static, inhomogeneous,
anisotropic medium, the Lagrange density function is
L =
H −
B
˜
μ
·
∇ × E +
∂ B
∂t
+ K
−
E −
D
˜
ε
·
∇ × H −
∂ D
∂t
− J
where, B is the magnetic induction; D the electric displacement vector; E is the
electric field intensity; H is the magnetic field intensity; J is the current density; K
is the magnetic current density; The magnetic permeability ˜
μ and the permittivity (or
dielectric constant) ˜
ε are only the functions of coordinates, and B = ˜
μH, D = ˜
ε E.
Find the corresponding Euler equations.