796
Appendix: All Solutions to the Problems
δ J [ϕ] =
1
2
b
a
b
a
K (x, y)[ϕ(y)δϕ(x) + ϕ(x)δϕ(y)]dxdy −
b
a
f (x)δϕ(x)dx
=
b
a
a
a
K (x, t)ϕ(x)δϕ(y)dxdy −
b
a
f (y)δϕ(y)dy
=
b
a
[
b
a
K (x, y)ϕ(x)dx]δϕ(y)dy −
b
a
f (y)δϕ(y)dy
=
b
a
[
b
a
K (x, y)ϕ(x)dx − f (y)]δϕ(y)dy
If the functional takes extremum, there must be δ J = 0, since δϕ(y) is arbitrary,
there is
b
a
K (x, y)ϕ(x)dx − f (y) = 0
or
f (x) =
b
a
K (x, y)ϕ(y)dy
Quod erat demonstrandum.
2.79 Write the Ostrogradsky equation of the functional J [u]
=
˝
V [u
2
x + u
2
y + u
2
z + 2u f (x, y, z)]dxdydz.
Solution Let the integrand F = u
2
x + u
2
y + u
2
z + 2u f (x, y, z), the various partial
derivatives are
F u = 2 f (x, y, z),
∂
∂ x
F u x = 2
∂
2 u
∂ x 2 ,
∂
∂ y
F u y = 2
∂
2 u
∂ y 2 ,
∂
∂z
F u z = 2
∂
2 u
∂z 2
Substituting them into the Ostrogradsky equation, we obtain u = f (x, y, z).
2.80 Let the electric potential in a static electric field Ω be V = V (x, y, z), the
energy E of the electric field is
E = J [V ] =
1
8π
˚
Ω
(V
2
x + V
2
y + V
2
z )dxdydz
According to the electrical knowledge, when E is the minimum, the electrostatic field
is in an equilibrium state. Find the differential equation which the potential function
should satisfy.
Solution Let the integrand F = V
2
x + V
2
y + V
2
z , the Euler equation of the functional
is