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Consider the Fredholm equation of the 2nd kind
$$ f(s) = \lambda \int_{-\infty}^{\infty} f(s') \Big(\sum_{n=1}^{N} g_n(s) h_n(s') \Big) ds' , $$ with $f(s)$ an unknown function, $\lambda$ a constant, {$h_1(s') , . . . , h_N(s')$} and {$g_1(s) , . . . , g_N(s)$} two sets of known functions.

Please note that this integral must be interpreted in the Principal Value sense. If the sets {$g_n(s)$}, {$h_n(s')$} contain only functions which decay sufficiently fast and contain no poles on the interval $(-\infty,\infty)$, then methods to solve this equation are covered in most introductory discussions of integral equations. However, what can be made of the case where at least one element of {$g_n(s)$} has at least one simple pole on the real axis?

Questions:

(1) How can this equation be classified? Is it a singular integral equation even though it does not involve poles 'along the diagonal'?
(2) Is there an accepted reference for integral equations of this type?

Note: If it helps, I am interested, to start, in a specific example: $$ f(s) = \frac{1}{a \beta} \int_{-\infty}^{\infty} f(s') \frac{s \sinh(\frac{\pi s}{2})}{s \sinh(\frac{\pi s}{2}) - 1} \frac{1}{\cosh(\frac{\pi (s-s')}{2})} ds' , $$ which has simple poles at $ s = \pm s_0 \approx \pm .7202 $. I know, from the solution of a transformed version of this equation that in $\lim_{a->\pm \infty} f(s)$ is proportional to a sum of Dirac deltas. So, clearly there is something pathological going on because the solution is not a function at all. It is not even clear to me that we may legitimately expand the kernel so that it follows the form that I have listed above.

I may need to solve several equations of this form, so I am interested in a general solution method.

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  • $\begingroup$ You have ${\large s}$ and ${\large s'}$ in the integrand. $\endgroup$ – Felix Marin Aug 30 '13 at 6:07
  • $\begingroup$ @FelixMarin Yes, we could move those terms that depend only on $s$ outside the integral. I chose to put them inside just so it was clear that the Kernel of this equation is unusual. Many people mention this when they first see the equation, can you tell me why? $\endgroup$ – Kevin Driscoll Aug 30 '13 at 15:50
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Have you seen/considered homotopy perturbation techniques given here - http://www.sciencedirect.com/science/article/pii/S0960077905010362?

They may be applicable to your problem.

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  • 2
    $\begingroup$ There actually is an integral equation very close to the one I would like to solve that has a known analytical solution (they differ by just 1 term). I have heard of this method before, but would never have remembered it for this problem. Thanks for your helpful suggestion. I will investigate this method and if it turns out to work, I will happily accept your suggestion as an answer. $\endgroup$ – Kevin Driscoll Sep 6 '13 at 15:39

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