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10 Variational Problems of Functionals with Vector, Tensor …
Subtracting Eqs. (24) and (23), we obtain
H · ∇ × E − E · ∇ × H = −H ·
∂ B
∂t
− E · J − E ·
∂ D
∂t
(25)
or
∇ · (E × H) + E ·
∂ D
∂t
+ H ·
∂ B
∂t
+ E · J = 0
(26)
Putting S = E × H, then S is called the Poynting(’s) vector. It expresses energy
flow density in the electromagnetic field, it is the electromagnetic power per unit
area that the electromagnetic wave is perpendicular to the direction of propagation.
Thus Eq. (26) can be written as
∇ · S + E ·
∂ D
∂t
+ H ·
∂ B
∂t
+ E · J = 0
(27)
Equation (27) is called the Poynting theorem or Poynting equation. The
Poynting theorem was proposed in 1884, it describes the laws of energy conservation and transformation of electromagnetic field, reveals the materiality of
electromagnetic field in theory. It can also be expressed as
∇ · S +
∂
∂t
1
2
ε E
2
+
1
2
μH
2
+ E · J = 0
(28)
Example10.2.3 Find the Euler equation of the acoustic field functional J [η] =
1
2
t 1
t 0
V [ρη
2
t + 2 p∇ · η − kp(∇ · η)
2
]dV dt, where, η is the displace vector that the
gas particle in the acoustic field deviates from the normal position; ρ is the equilibrium
mass density of the gas; p is the equilibrium pressure of the gas, k is the adiabatic
index of the gas, and k is the constant.
Solution The integrand is F = ρη
2
t + 2 p∇ · η − kp(∇ · η)
2 , calculating the various
partial derivatives
∂ F
∂∇ · η
= 2 p(1 − k∇ · η),
∂ F
∂η t
= 2ρη t
(1)
The Euler equation of the functional is
−∇
∂ F
∂∇ · η
−
∂
∂t
∂ F
∂η t
= 2kp∇(∇ · η) − 2ρη tt = 0
( 2 )
or
a
2
∇(∇ · η) − η tt = 0
(3)