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I'm having trouble solving an integral equation. It appears to me to be a homogenous fredholm equation of the second kind. However, I'm being told that this can't be a fredholm equation, because it is non-linear. Could someone help me in trying to figure out how to classify an integral equation as linear or non-linear. Also, I'll post the equation I need to solve below, and it would be great if anyone could also give me some tips on how to try and solve it. Thank you to all who reply.

The equation is

$\phi(x) = (x^2 - x)\int\limits_0^1 \mathrm{d}y \frac{\phi(y)}{(y-x)^2}$

Also, is this by chance related to an eigenvalue problem? I know that might sound like a strange question, but I've seen some people treating these as eigenvalue equations.

By the way, I want to solve the equation for $\phi(x)$

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    $\begingroup$ Anybody care to comment on why I'm being down voted? I'm not necessarily opposed to it. But, I'd like to know why in order to learn how to use the site and to learn the right etiquette. $\endgroup$
    – adhanlon
    Commented Jun 14, 2011 at 20:07
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    $\begingroup$ I can only conjecture about the down votes: your question is interesting, but fits better with the goals of math.stackexchange.com This site is designed for research-level questions. The equation is linear in the unknown you seek. It is a Fredholm IE because the limits of integration are constant. You may want to look at the book by Kress on Linear Integral Equations. $\endgroup$
    – Nilima Nigam
    Commented Jun 14, 2011 at 21:32
  • $\begingroup$ Thank you, I was unaware of the existence of math.stackexchange.com. $\endgroup$
    – adhanlon
    Commented Jun 14, 2011 at 22:05
  • $\begingroup$ Funny that you should also mention it's designed for research-level questions though. This is a problem I'm trying to solve for a research project, but I might be in over my head... $\endgroup$
    – adhanlon
    Commented Jun 15, 2011 at 1:42
  • $\begingroup$ silmaril89: are you sure you don't mean $y-x$ instead of $(y-x)^2$ in the integral? That would make a bit more sense, because then the positive/negative parts of $y-x$ cause some cancellation; look up the Hilbert Transform. See also my comment to Michael Renardy's answer below. $\endgroup$
    – Zen Harper
    Commented Jun 15, 2011 at 2:44

1 Answer 1

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The integral diverges unless $\phi(x)=0$. Now look at what happens when you plug $\phi=0$ into the equation. By the way, the equation is linear.

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    $\begingroup$ How do you know it will diverge unless $\psi(x) = 0$? I've seen that a possible way to solve this, is by a Liouville-Neumann_series, $\psi(x) = \sum\limits_{n=0}^{\infty}\lambda^n\psi_n(x)$. If so, I could probably cut it off somewhere, but I'm not sure how to find any of the values for $\lambda^n$ or $\psi_n(x)$. $\endgroup$
    – adhanlon
    Commented Jun 14, 2011 at 19:15
  • $\begingroup$ Sorry, should have all the $\psi$'s as $\phi$'s $\endgroup$
    – adhanlon
    Commented Jun 14, 2011 at 19:16
  • $\begingroup$ Also, it's possible that I'm leaving out something important. I believe the integral is supposed to have the letter p in front of it, which apparently stands for a finite part integral. Do you know what that means? I've never heard of it. Thanks. $\endgroup$
    – adhanlon
    Commented Jun 14, 2011 at 19:41
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    $\begingroup$ – silmaril89: The $1/(y-x)^2$ term is always positive and diverges when integrated through $x$, so if $\phi$ is continuous then $\phi(x)$ must be zero for every $0<x<1$ for the integral to be finite. If you don't assume $\phi$ is continuous, you get into murkier territory involving measure theory, Lebesgue points etc. but I expect you'd still get $\phi = 0$ almost everywhere. This is why I think you've mistyped the equation; see my comment above. $\endgroup$
    – Zen Harper
    Commented Jun 15, 2011 at 2:44

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