Timeline for linear versus non-linear integral equations
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2011 at 2:44 | comment | added | Zen Harper | – 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. | |
| Jun 14, 2011 at 19:41 | comment | added | adhanlon | 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. | |
| Jun 14, 2011 at 19:16 | comment | added | adhanlon | Sorry, should have all the $\psi$'s as $\phi$'s | |
| Jun 14, 2011 at 19:15 | comment | added | adhanlon | 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 $\psi_n(x)$. | |
| Jun 14, 2011 at 18:04 | history | answered | Michael Renardy | CC BY-SA 3.0 |