<div class="gmail_quote">On Fri, Aug 12, 2011 at 12:26, Paul Anton Letnes <span dir="ltr"><<a href="mailto:paul.anton.letnes@gmail.com">paul.anton.letnes@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
I'm not 100% sure what you mean by "second kind integral operator", but it is a Fredholm equation of the first kind, as far as I understand (my background is physics rather than mathematics).</blockquote></div>
<br><div>Is the thing you're trying to solve with actually of the first kind, not of the second kind?</div><div><br></div><div><a href="http://en.wikipedia.org/wiki/Fredholm_integral_equation">http://en.wikipedia.org/wiki/Fredholm_integral_equation</a></div>
<div><br></div><div>The distinction is whether there is essentially in whether there is a local part to the equation or not. The issue, as I understand it, is that solving a first-kind integral equation is generally not a stable process because the eigenvalues of the integral operator decay to zero, implying that it is essentially low rank, thus not invertible. Maybe you use some regularization to get a system that is not essentially singular?</div>