Analytic Approximation of Fredholm equation of second kind

Question

Consider an integral equation of the type \begin{equation} f(x)=g(x)+\int_{-a}^{a} K(x-x')f(x')\mathrm dx' \end{equation}

Say $a$ is very small. Is there a systematic way to approximate the solution $f(x)$ as a power series in $a$?

Answer 1

The integral on the RHS of your equation may be expanded

$$\tag{1} \int\limits_{-a}^a dx' \ K(x-x')f(x') \sim 2aK(x)f(0)+a^2\left[K(x)f'(0)-K'(x)f(0) \right] \ ,\quad a\to0 $$

I've expanded the integrand as a Taylor series around $x'=0$ then integrated term by term. You can continue to get more terms. Let

$$\tag{2} f(x)=\sum\limits_{n=0}^\infty a^nf_n(x) $$

Substitute (1) and (2) into your equation then match coefficients of $a$. The first few are

$$\tag{3} f_0(x)=g(x) \\ f_1(x)=2K(x)f_0(0)\\ f_2(x)=2K(x)f_1(0)+K(x)f_0'(0)-K'(x)f_0(0) $$

There is a pattern (which may sound more confusing than it is). Consider $\int\limits_{-a}^a dx' \ K(x-x')f(x')$, and we ask what is the coefficient of $a^k$ in this expression. There are two contributions:

1. A factor of $2 a^m/m!$ from the $m$th term in the integrated Taylor series, which sits next to the $m$th derivative of $(Kf)|_{x'=0}$. This can be simplified using the binomial theorem, if you like.

2. A factor of $a^n$ coming from the expansion $f=\sum\limits_n a^nf_n$.

The overall coefficient of $a^k$ coming from this expression is the sum of all the coefficients such that $k=m+n$.

Actually, all of this amounts to forming the expression

$$\tag{4} \sum\limits_{n=0}^\infty a^n f_n(x)=g(x)+\sum\limits_{n=0}^\infty a^n \int\limits_{-a}^a dx' K(x-x')f_n(x') $$

Then differentiating (4) $k$ times wrt $a$ and setting $a=0$, then solving for the $f_n$ recursively. To differentiate the integral bounds you use Leibniz integral rule.