Let $h(x)$ be a known well-behaved function, I have to solve for $\sigma(t)$: $$ \phi(x) = \int_a^b\log\left[\left(x-t\right)^2 + \left(h(x) - h(t)\right)^2\right]\sigma(t)dt $$

Where, $b>a>0$, and $\phi(x)$ is known. This is a Fredholm equation of first kind with the kernel: $$ K(x, t) = \log\left[\left(x-t\right)^2 + \left(h(x) - h(t)\right)^2\right] $$

How to do it? Any ideas? Any hints? Anything?

  • Given the kernel is symmetric, $K(x, t) = K(t, x)$, there exists an orthogonal set of eigenfunctions, in which the solution can be easily found from these eigenfunctions. But, I couldn't find the eigenfunctions itself. Is there a general method to find eigenfunctions for symmetric kernels?

  • If the Kernel is a displacement kernel, that is, $K(x, t) = K(x-t)$, then this can be easily solved (with laplace/fourier transform, and convolution). But I was unable to transform $K$ into a displacement, by variable substitution, by adding stuff, or multiplying and dividing by things.

  • I've tried to transform the eigenfunction integral equation into a differential equation, to see what kind of ODE the eigenfunctions solves. But I couldn't find an ODE. By the way, if you can do this, please answer.

  • I've even tried to transform the whole equation into an ODE, but I couldn't as well. If you can do that too, please answer.

  • I've tried expanding the logarithm into a series, but wasn't useful.

  • $\begingroup$ I think you mistake that eigenfunctions can always be "easily" found. I am not convinced then one can find a closed form expression. Is the integral of $K$ even possible in closed form? $\endgroup$ – Gregory Jun 22 '18 at 19:37
  • $\begingroup$ @Gregory Well, $K$ depends on $h$, which can be any.. so... probably no integral for $K$. And, I never said eigenfunctions can be found easily :D. I was trying to find an ODE for the eigenfunctions.. $\endgroup$ – Physicist137 Jun 22 '18 at 19:49
  • $\begingroup$ haha. Well, I don't see any obvious trick that turns this into an ODE and that also makes me wonder whether the ODE would actually be anymore friendly. Short of guessing the form of the eigenfunction... $\endgroup$ – Gregory Jun 22 '18 at 20:17
  • $\begingroup$ It's easy to just approximate the solution. Take a number of linearly independent functions, including $\phi(x)$, apply Gram–Schmidt orthogonalization and then write down the matrix elements of the integral operator in that basis, and solve for the solution in that basis. $\endgroup$ – Count Iblis Jun 23 '18 at 20:56

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