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I have the following problem to solve:

$\int_{0}^{1}K(x,y)\phi(y)dy$

where:

$K(x,y)=x(1-y), 0\leq x\leq y\leq 1$ and $K(x,y)=y(1-x), 0\leq y\leq x\leq 1$

already tried using the methods suggested in another question I found here (Find the eigenvalues and eigenvectors of an integral operator), but I do not understand how to get to the nonzero eigenvalues ​​and eigenfunctions when karnel have two different, for two different constraints. I've tried to expand the functions and stuff as I have suggested, I sent the link but I am not able to understand very well.

thanks for any help

  • What's happened with your previous problem? – Mhenni Benghorbal May 30 '13 at 14:09
  • my problem it's solved... already decided enough exercises on this subject, but never any of this kind, ie with separate karnel.Also I'm trying to solve the two exercises from the following link http://math.stackexchange.com/questions/404889/urgent-help-please-integral-eigenvectors-and-eigenvalues , now expand the functions and everything, but the most likely is have hurt expansion. – Olinda Fernandes May 30 '13 at 15:11
  • @Willie, I see you've edited the tags to include integral-equations. But the way the problem is stated it looks like OP is just stuck doing a definite integral. I don't see what eigenvalues and eigenfunctions have to do with the question, nor functional analysis, nor integral equations. But maybe I don't understand what OP was actually asking. – Gerry Myerson Jun 12 '13 at 09:35
  • @GerryMyerson I am telegraphing based on the previous question linked to by the OP. I think the OP is considering the mapping $\phi(x) \mapsto \int K(x,y)\phi(y)\mathrm{d}y$ and trying to find its eigenvalues and eigenvectors. (You probably also noticed that I retagged a bunch of OP's other questions to remove the functional-analysis tag when I am sure it doesn't apply.) In this case here, while the question is not very clearly stated, I am inclined to be charitable... (And thanks for checking on me and keeping me honest!) – Willie Wong Jun 12 '13 at 11:35

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