2.6 Variational Problems Depending on Several Functions of One Variable
145
holds, where, λ is a undetermined constant. The equation is called the Fredholm
integral equation of the third kind.
Proof Taking the first variation to the function, we obtain
δ J [ϕ] =
[
b
a f (x)ϕ(x)dx] 2 2[
b
a
b
a K (x, y)ϕ(x)δϕ(y)dxdy +
b
a ϕ(y)δϕ(y)dy]
[
b
a f (x)ϕ(x)dx] 4
−
[
b
a
b
a K (x, y)ϕ(x)ϕ(y)dxdy +
b
a ϕ 2 (x)dx]2
b
a f (x)ϕ(x)dx
b
a f (x)δϕ(x)dx
[
b
a f (x)ϕ(x)dx] 4
= 0
or
b
a
f (x)ϕ(x)dx
b
a
b
a
K (x, y)ϕ(x)δϕ(y)dxdy +
b
a
ϕ(y)δϕ(y)dy
−
b
a
b
a
K (x, y)ϕ(x)ϕ(y)dxdy +
b
a
ϕ
2
(x)dx
b
a
f (x)δϕ(x)dx = 0
Let
λ =
[
b
a
b
a K (x, y)ϕ(x)ϕ(y)dxdy +
b
a ϕ
2
(x)dx]
b
a f (x)ϕ(x)dx
then there is
b
a
b
a
K (x, y)ϕ(x)δϕ(y)dxdy +
b
a
ϕ(y)δϕ(y)dy − λ
b
a
f (y)δϕ(y)dy
=
b
a
b
a
K (x, y)ϕ(x)dx + ϕ(y) − λ f (y)
δϕ(y)dy
= 0
Because δϕ(y) is arbitrary, so that
b
a
K (x, y)ϕ(x)dx + ϕ(y) − λ f (y) = 0
or
ϕ(y) = λ f (y) −
b
a
K (x, y)ϕ(x)dx = 0
Quod erat demonstrandum.