Fredholm integral equations and the resolvent kernel

By 0x7df, Sun 12 February 2017, modified Sun 12 February 2017, in category Maths

mathematics

Consider the inhomogeneous Fredholm integral equation:

$$ H \psi(\mu) = \lambda \int_a^b \sigma(\mu, \mu') \psi(\mu') d\mu' + S(\mu) $$

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.:

$$ \psi(\mu) = \psi_0(\mu) + \lambda\psi_1(\mu) + \lambda^2\psi_2(\mu) + ... + \lambda^n\psi_n(\mu) + ... $$

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

$$ \psi_0(\mu) = S(\mu) $$
$$ \psi_1(\mu) = \int_a^b \sigma(\mu, \mu') \psi_0(\mu') d\mu' $$
$$ \psi_2(\mu) = \int_a^b \sigma(\mu, \mu') \psi_1(\mu') d\mu' $$
$$ ... $$
$$ \psi_n(\mu) = \int_a^b \sigma(\mu, \mu') \psi_{n-1}(\mu') d\mu' $$

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:

$$ |\lambda| \le \frac{1}{||\sigma||} $$

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

$$ ||\sigma||^2 = \int_a^b \int_a^b |\sigma(\mu, \mu')|^2 d\mu' d\mu $$

(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:

$$ \psi(\mu) = \int_a^b R(\mu, \mu') S(\mu') d\mu' + S(\mu) $$

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:

$$ \sigma_1(\mu, \mu') = \sigma(\mu, \mu') $$
$$ \sigma_2(\mu, \mu') = \int_a^b \sigma(\mu, \mu'')\sigma(\mu'', \mu') d\mu'' $$
$$ \sigma_3(\mu, \mu') = \int_a^b \int_a^b \sigma(\mu, \mu''') \sigma(\mu''', \mu'')\sigma(\mu'', \mu') d\mu'' d\mu''' $$
$$ \sigma_n(\mu, \mu') = \int_a^b d\mu^{(n)} \int_a^b d\mu^{(n-1)} ... \int_a^b d\mu'' \sigma(\mu, \mu^{(n)}) \sigma(\mu^{(n)}, \mu^{(n-1)}) ... \sigma(\mu'', \mu') $$

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

$$ \psi_0(\mu) = S(\mu) $$
$$ \psi_1(\mu) = \int_a^b \sigma_1(\mu, \mu') S(\mu') d\mu' $$
$$ \psi_2(\mu) = \int_a^b \sigma_2(\mu, \mu') S(\mu') d\mu' $$
$$ ... $$
$$ \psi_n(\mu) = \int_a^b \sigma_n(\mu, \mu') S(\mu') d\mu' $$

and further:

$$ \psi(\mu) = \psi_0(\mu) + \lambda\psi_1(\mu) + \lambda^2\psi_2(\mu) + ... + \lambda^n\psi_n(\mu) + ... $$

can be written:

$$ \psi(\mu) = S(\mu) + \lambda \int_a^b \sigma_1(\mu, \mu') S(\mu') d\mu' + \lambda^2 \int_a^b \sigma_2(\mu, \mu') S(\mu') d\mu' + ... + \lambda^n \int_a^b \sigma_n(\mu, \mu') S(\mu') d\mu' + ... $$

or:

$$ \psi(\mu) = S(\mu) + \sum_{n=1}^\infty \lambda^n \int_a^b \sigma_n(\mu, \mu') S(\mu') d\mu' $$

Bringing the summation inside the integral:

$$ \psi(\mu) = S(\mu) + \int_a^b \sum_{n=1}^\infty \lambda^n \sigma_n(\mu, \mu') S(\mu') d\mu' $$

we can write:

$$ \psi(\mu) = S(\mu) + \int_a^b R(\mu, \mu'; \lambda) S(\mu') d\mu' $$

where:

$$ R(\mu, \mu'; \lambda) = \sum_{n=1}^{\infty} \lambda^n \sigma_n(\mu, \mu') $$

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

  1. Firstly, finding the resolvent kernel requires fewer iterations than Neumann expansion.
  2. 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:

$$ \sigma(\mu, \mu') = \sigma_0 $$
$$ S(\mu) = S_0 $$

so that:

$$ \psi(\mu) = S(\mu) + \lambda\int_a^b \sigma(\mu, \mu') \psi(\mu') d\mu' $$

becomes:

$$ \psi(\mu) = S_0 + \lambda\sigma_0\int_a^b \psi(\mu')d\mu' $$

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:

$$ \psi_0 = \frac{S_0}{1 - \lambda\sigma_0(b-a)} $$

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

$$ \psi(\mu) = S(\mu) + \int_a^b R(\mu, \mu'; \lambda) S(\mu') d\mu' $$

where:

$$ R(\mu, \mu'; \lambda) = \sum_{n=1}^{\infty} \lambda^n \sigma_n(\mu, \mu') $$

We have:

$$\sigma_1(\mu, \mu') = \sigma_0 $$
$$\sigma_2(\mu, \mu') = \sigma_0^2 (b - a) $$
$$\sigma_3(\mu, \mu') = \sigma_0^3 (b - a)^2 $$
$$ ... $$
$$\sigma_n(\mu, \mu') = \sigma_0^n (b - a)^{n-1} $$

hence:

$$ R(\mu, \mu'; \lambda) = \sum_{n=1}^{\infty} \lambda^n \sigma_0^n (b - a)^{n-1} $$

and:

$$ \psi(\mu) = S_0 + \int_a^b \sum_{n=1}^{\infty} \lambda^n \sigma_0^n (b - a)^{n-1} S(\mu') d\mu' $$
$$ \psi_0 = S_0 \left(1 + \sum_{n=1}^{\infty} \left[ \lambda \sigma_0 (b - a)\right]^n \right) $$

This is equal to:

$$ \psi_0 = S_0 \left(1 - \frac{\lambda\sigma_0(b - a)}{\lambda\sigma_0(b - a) - 1} \right) $$

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

$$ \psi_0 = \frac{S_0}{1 - \lambda\sigma_0(b - a)} $$

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 \(b = 1\):

$$ \psi(\mu) = S(\mu) + \lambda\int_0^1 e^{\mu-\mu'}\psi(\mu')d\mu' $$

The iterated kernels simplify to:

$$ \sigma_n(\mu, \mu') = e^{\mu - \mu'} $$

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

$$ R(\mu, \mu'; \lambda) = \sum_{n=1}^{\infty} \lambda^n e^{\mu-\mu'} $$

which is:

$$ R(\mu, \mu'; \lambda) = \frac{\lambda}{1 - \lambda} e^{\mu - \mu'} $$

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

$$ \psi(\mu) = S(\mu) + \frac{\lambda}{1 - \lambda} \int_0^1 e^{\mu - \mu'} S(\mu') d\mu' $$

Example 3

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

$$ \psi(\mu) = S(\mu) + \lambda \int_0^1 \mu \mu' \psi(\mu') d\mu' $$

The iterated kernels are:

$$ \sigma_1(\mu, \mu') = \mu\mu' $$
$$ \sigma_2(\mu, \mu') = \int_0^1 \mu\mu''\mu''\mu' d\mu'' = \mu\mu'/3 $$
$$ \sigma_3(\mu, \mu') = \int_0^1 \int_0^1 \mu\mu'''\mu'''\mu''\mu''\mu' d\mu''d\mu''' = \mu \int_0^1 \mu'''^2 \int_0^1 \mu''^2 d\mu'' d\mu''' = \mu\mu'/9 $$
$$ \sigma_n(\mu,\mu') = \mu\mu'/3^{n-1} $$

Hence the resolvent kernel is:

$$ R(\mu, \mu'; \lambda) = \sum_{n=1}^\infty \lambda^n \mu\mu' / 3^{n-1} $$
$$ R(\mu, \mu'; \lambda) = \frac{\mu\mu'\lambda}{1 - \lambda/3} $$

The solution is therefore:

$$ \psi(\mu) = S(\mu) + \frac{\lambda}{1 - \lambda/3} \mu \int_0^1 \mu' S(\mu') d\mu' $$

Consider three cases.

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

For constant source:

$$ \psi(\mu) = S_0 + \frac{\lambda}{1 - \lambda/3} \frac{\mu S_0}{2} $$
$$ \psi(\mu) = S_0 \left( 1 + \frac{\lambda\mu}{2[1 - \lambda/3]}\right) $$

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

$$\psi(\mu) = S_0 \left(1 + \frac{\mu}{2}\right) $$

For \(\lambda = 1\):

$$ \psi(\mu) = S_0 \left(1 + \frac{3\mu}{4}\right) $$

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

In this case:

$$ \psi(\mu) = \mu + \frac{\lambda}{1 - \lambda/3} \mu \int_0^1 \mu' \mu' d\mu' $$
$$ \psi(\mu) = \mu + \frac{\lambda}{1 - \lambda/3} \mu \frac{1}{3} $$
$$ \psi(\mu) = \frac{3\mu}{3 - \lambda} $$

For \(\lambda = 1\)

$$ \psi(\mu) = \frac{3\mu}{2} $$

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

$$ \psi(\mu) = \frac{3\mu^2}{2} + \frac{3\lambda}{2[1 - \lambda/3]} \mu \int_0^1 \mu'^3 d\mu' $$
$$ \psi(\mu) = \frac{3\mu^2}{2} + \frac{3\lambda\mu}{8[1 - \lambda/3]} $$
$$ \psi(\mu) = \frac{3}{2} \left( \mu^2 + \frac{\lambda\mu}{4[1 - \lambda/3]} \right) $$

For \(\lambda = 1\):

$$ \psi(\mu) = \frac{3}{2}\left(\mu^2 + \frac{3\mu}{8} \right) $$

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