# Fredholm integral equations and the resolvent kernel

Consider the *inhomogeneous Fredholm integral equation*:

The unknown to be solved for is \(\psi(\mu)\) where \(a \le \mu \le b\) is the
independent variable, and the known function \(\sigma(\mu, \mu')\) is known as
the *kernel*. This is an *inhomogeneous* equation because the known function
\(S(\mu) \ne 0\).

It is a *Fredholm* equation because the limits on the integral are constants;
if they were variables then the equation would be a *Volterra* equation.

If \(H = 0\), then the equation is of the *first kind*; \(H = 1\) gives rise to a
Fredholm equation of the *second kind*, and otherwise the equation is of the
*third kind*.

For Fredholm equations of the second kind, where \(H = 1\), we can look for iterative solutions, i.e.:

When this is substituted into the original equation, we obtain:

This can be carried on until the iterative solution converges to some
desired level of accuracy. The process is called *Neumann expansion*. The
condition for the series solution to be convergent is:

where the square of the norm of the kernel is given by:

(See M. Masujima, *Applied Mathematical Methods in Theoretical Physics*, 2005.)

An alternative approach is called *Fredholm theory*. In Neumann iteration, we
repeatedly operate on \(S(\mu)\) using the kernel \(\sigma(\mu, \mu')\), to obtain
a converged answer for \(\psi(\mu)\). One could ask instead: what *single*
operation, involving the kernel, could be applied to \(S(\mu)\) to obtain the
same value of \(\psi(\mu)\)? That is, what \(R(\mu, \mu')\) for which:

The function \(R(\mu, \mu')\) is called the *resolvent kernel*, and it is easy to
find \(\psi(\mu)\) once this known. The proof of the above equation is complex
(see J. Kondo, *Integral Equations*, 1991), but the general method of obtaining
\(R(\mu, \mu')\) is quite simple.

We first define the *iterated kernels*:

(The notation \(x^{(n)}\) indicates \(x\) with \(n\) primes). So we can re-write:

and further:

can be written:

or:

Bringing the summation inside the integral:

we can write:

where:

Like Neumann iteration, this is an iterative process, but it has two significant advantages:

- Firstly, finding the resolvent kernel requires fewer iterations than Neumann expansion.
- Secondly, finding the resolvent kernel requires knowledge of only the kernel, \(\sigma(\mu, \mu')\), not the function \(S(\mu)\). This is a practical advantage if the same kernel is applied in different circumstances involving different functions \(S(\mu)\).

## Example 1

As an example we can take the simplest possible equation of this form, for which the kernel and the source term are constants:

so that:

becomes:

This is easily solvable; because the right-hand side clearly has no dependence on \(\mu\), we can see that \(\psi(\mu)\) must be a constant \(\psi_0\), hence:

Now using the resolvent kernel method to obtain the solution, we write:

where:

We have:

hence:

and:

This is equal to:

as long as \(|\lambda\sigma_0(b-a)| < 1\), hence:

which is the same solution as earlier, obtained by solving directly.

## Example 2

Masujima (2005) gives the following example, with \(\sigma(\mu, \mu') = e^{\mu - \mu'}\), \(a = 0\) and \(b = 1\):

The iterated kernels simplify to:

for all \(n\), and hence the resolvent kernel is:

which is:

as long as \(|\lambda| \lt 1\); in which case:

## Example 3

We now consider the example where \(\sigma(\mu, \mu') = \mu\mu'\) and \(a = 0\) and \(b = 1\).

The iterated kernels are:

Hence the resolvent kernel is:

The solution is therefore:

Consider three cases.

### Case 1: \(S(\mu) = S_0\)

For constant source:

For \(\lambda = 3/4\):

For \(\lambda = 1\):

### Case 2: \(S(\mu) = \mu\)

In this case:

For \(\lambda = 1\)

### Case 3: \(S(\mu) = 3\mu^2/2\)

For \(\lambda = 1\):