inordinatum

Physics and Mathematics of Disordered Systems

An interesting integral equation

Recently, I discussed an interesting problem with a friend of mine. Basically, the objective was to find an approximation for the solution of a linear $N\times N$ equation system in the limit of $N\to\infty$.

It turned out that in this limit the solution can be obtained by solving the following integral equation

$\displaystyle h(x)=\int_{-1}^1 \frac{f(x)-f(y)}{x-y}\,\mathrm{d}y$,

where $h(x)$ is known and $f(x)$ is to be found. Now, such integral equations are notoriously difficult to solve (in particular, much more difficult than differential equations). An excellent resource (which contains most known exact solutions) is the EqWorld page on integral equations (and the “Handbook of Integral Equations” by EqWorld’s authors, which you can find by googling). Unfortunately, it does not help in this case (the closest thing one finds there are the eigenfunctions for the integral operator $\int_{-1}^1 \frac{f(x)-f(y)}{|x-y|}\,\mathrm{d}y$, which, however, turn to to be competely unrelated to those of $\int_{-1}^1 \frac{f(x)-f(y)}{x-y}\,\mathrm{d}y$).

One angle of attack to our equation is the observation that plugging in a polynomial $f(x)$ gives a polynomial $h(x)$ (and vice versa). The mapping of coefficients is linear, and by inverting its matrix one can explicitely write the solution $f(x)$ for any given polynomial $h(x)$. Looking at the resulting coefficients polynomials of small degree one can guess the following expression for the solution (which is then easily verified to hold for general holomorphic $h(x)$):

$\displaystyle f(x) = \frac{1}{2\pi i} \oint_\gamma \frac{h(z)}{2 \text{arctanh}\frac{1}{z}}\frac{\mathrm{d}z}{x-z}$

Here, the contour $\gamma$ has to be chosen in such a way that all singularities of the integrand are inside $\gamma$. For example, one can choose a circle with radius $\max 1,|2x|$. For non-holomorphic $h(x)$, this formula still works but the choice of integration contour is more complicated.

In the problem which motivated that equation, we had $h(x)$ given for $x\in [-1;1]$ and wanted to obtain $f(x)$ for $x\in [-1;1]$. In this case, the contour integral formula can be rewritten as

$\displaystyle f(x)=\frac{h(x)\log \frac{1+x}{1-x}}{\left(\log \frac{1+x}{1-x}\right)^2 + \pi^2} + \int_{-1}^1 \frac{h(y)}{\left(\log \frac{1+x}{1-x}\right)^2 + \pi^2}\frac{\mathrm{d}y}{x-y}$

This integral is assumped to be regularized by taking the Cauchy principal value. Computing this (e.g. numerically) now only requires $h(x)$ for $x\in [-1;1]$, so it is no longer necessary to assume that $h(x)$ can be extended analytically to $\mathbb{C}$. A similar expression can also be found for $f(x)$ where $|x|>1$.

If somebody finds another use or application for this pretty result, please drop me a comment below 😉

Written by inordinatum

October 1, 2011 at 8:03 pm