Neumann series

23.5 NEUMANN SERIES
23.5 Neumann series
As mentioned above, most integral equations met in practice will not be of the
simple forms discussed in the last section and so, in general, it is not possible to
find closed-form solutions. In such cases, we might try to obtain a solution in the
form of an infinite series, as we did for differential equations (see chapter 16).
Let us consider the equation
b
K(x, z)y(z) dz,
(23.34)
y(x) = f(x) + λ
a
where either both integration limits are constants (for a Fredholm equation) or
the upper limit is variable (for a Volterra equation). Clearly, if λ were small then
a crude (but reasonable) approximation to the solution would be
y(x) ≈ y0 (x) = f(x),
where y0 (x) stands for our ‘zeroth-order’ approximation to the solution (and is
not to be confused with an eigenfunction).
Substituting this crude guess under the integral sign in the original equation,
we obtain what should be a better approximation:
b
b
K(x, z)y0 (z) dz = f(x) + λ
K(x, z)f(z) dz,
y1 (x) = f(x) + λ
a
a
which is first order in λ. Repeating the procedure once more results in the
second-order approximation
b
K(x, z)y1 (z) dz
y2 (x) = f(x) + λ
a
b
a
b
K(x, z1 )f(z1 ) dz1 + λ2
= f(x) + λ
b
dz1
a
K(x, z1 )K(z1 , z2 )f(z2 ) dz2 .
a
It is clear that we may continue this process to obtain progressively higher-order
approximations to the solution. Introducing the functions
K1 (x, z) = K(x, z),
b
K2 (x, z) =
K(x, z1 )K(z1 , z) dz1 ,
a
b
K3 (x, z) =
b
dz1
K(x, z1 )K(z1 , z2 )K(z2 , z) dz2 ,
a
a
and so on, which obey the recurrence relation
b
K(x, z1 )Kn−1 (z1 , z) dz1 ,
Kn (x, z) =
a
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INTEGRAL EQUATIONS
we may write the nth-order approximation as
b
n
yn (x) = f(x) +
λm
Km (x, z)f(z) dz.
(23.35)
a
m=1
The solution to the original integral equation is then given by y(x) =
limn→∞ yn (x), provided the infinite series converges. Using (23.35), this solution
may be written as
b
R(x, z; λ)f(z) dz,
(23.36)
y(x) = f(x) + λ
a
where the resolvent kernel R(x, z; λ) is given by
R(x, z; λ) =
∞
λm Km+1 (x, z).
(23.37)
m=0
Clearly, the resolvent kernel, and hence the series solution, will converge
provided λ is sufficiently small. In fact, it may be shown that the series converges
in some domain of |λ| provided the original kernel K(x, z) is bounded in such a
way that
b b
dx
|K(x, z)|2 dz < 1.
(23.38)
|λ|2
a
a
Use the Neumann series method to solve the integral equation
1
y(x) = x + λ
xzy(z) dz.
(23.39)
0
Following the method outlined above, we begin with the crude approximation y(x) ≈
y0 (x) = x. Substituting this under the integral sign in (23.39), we obtain the next approximation
1
1
λx
xzy0 (z) dz = x + λ
xz 2 dz = x +
y1 (x) = x + λ
,
3
0
0
Repeating the procedure once more, we obtain
1
xzy1 (z) dz
y2 (x) = x + λ
0
1 λ λ2
λz
dz = x +
x.
xz z +
= x+λ
+
3
3
9
0
For this simple example, it is easy to see that by continuing this process the solution to
(23.39) is obtained as
2 3
λ
λ
λ
y(x) = x +
+
+ · · · x.
+
3
3
3
Clearly the expression in brackets is an infinite geometric series with first term λ/3 and
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