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I'm trying to solve with Mathematica an integral equation. I found this excellent answer (How to solve a non-linear integral equation?) solving with a collocation method a problem which can be restated as:

$$ \int_a^b f \left( \phi \left(x \right) \right) \mathrm{d} x= 1 $$

for some function $f$, $\phi(x)$ being the unknown. My problem is slightly more complicated:

$$ \phi(x) = \int_a^b K \left( x, y \right) f \left( \phi \left( y \right) \right) \mathrm{d} y $$

and the kernel $K(x,y)$ is singular for $x=y$, it behaves as $\sim \frac{1}{\left| x - y \right|}$. I would like to know if it is possible to extend the collocation method to this case, or, alternatively, which other methods can be used to numerically solve my integral equation.

Thanks in advance!

Edit: some more details:

$$ f(\phi(x)) = \frac{\phi(x)}{\sqrt{\phi^2(x) + C^2}} $$ $$ K(x,y) = y \frac{\mathrm{e}^{-\left| x - y \right|}}{\left| x - y \right|} $$

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I do not have an answer offhand (would require, at minimum, a specific K,f pair), but here are some links that might contain helpful approaches. Check Moe, Larry, and Curly (I really need some pointers on naming my links). –  Daniel Lichtblau Feb 28 at 16:32
    
I edited the question to include some more details! Thank you for your links! –  zakk Feb 28 at 16:47
    
At a glance I'm having a hard time seeing how that integral could possibly converge for a<x<b. Assuming i'm wrong and it does converge your challenge is to develop a quadrature rule that is accurate at the singular point. Have fun..! (I think this really is a math.stackexchange question in any case) (BTW The right most "=1" is a typo right..?) –  george2079 Feb 28 at 19:53
    
@george2079: it should converge, at least it is eq. 5 in this paper: journals.aps.org/prl/pdf/10.1103/PhysRevLett.74.1633 The authors refer to an unpublished paper for the numerical part, commenting as follows: letting $x=tan(\beta)$, we set up a Gaussian-quadrature grid for $\beta$ and convert the above equations into a matrix form which can be solved iteratively. The logarithmic singularities are treated separately. –  zakk Mar 1 at 1:54

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