I'm looking for a solution to an integral equation:
Given a function $f(x)$, find a function $g(x)$ such that $$\int_0^1 \frac{g(s)}{x-s} ds = f(x), \forall x \in [0,1]$$ satisfies.
Function $f(x)$ is actually a series of trigonometric functions, but let us have a simple start, say $f(x)=\cos x$. The work I'm doing now is to solve the integral numerically using Gauss-Chebyshev quadrature. Then I discretize $x$ and have a system of equations, which will give me discrete $g(s)$. I just feel the tricky part may be the discretization on $x$ and $s$ as they have the same range, and this can really affect my result.
My questions are:
- Is there an analytical solution to $g(x)$ and is it unique?
- What would be a good treatment for the discretization to solve the integral?
Thank you for your time to reading this. Any thoughts will be appreciated.
To partially answer your question, referring to Singular integral equations - Estrada, Kanwal, section 3.2. The Cauchy integral equation $$P.V.\int_0^1 \frac{g(t)}{t-s} dt = f(s)$$ Has a solution given by $$g(s) = \frac{c}{\pi \sqrt{s(1-s)}} + \frac{1}{\pi^2 \sqrt{s(1-s)}} \int_0^1 \frac{\sqrt{t(1-t)}f(t)}{s-t}dt$$ where $c$ is an arbitrary constant. And thus, you have an analytic solution.
Another source that comes to my mind right now is the Handbook of integral equations - Polyanin Manzhirov, section 12.4-3, where it shows solutions to a similar equation in terms of the boundability of the solution. The downside of this one is it doesn't show any of the theory behind it, only states the solution.
That's what I can say by now.