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Find the eigenvalue and eigenfunctions of the integral operator $Ku=\int_0^\pi k(x,y)u(y)dy$. $k( x,y) = \sum\limits_{n=1}^\infty \frac{1}{n^2} \sin\big((n+1)x\big)\sin(ny)$.

This is how I approached the problem:

To find the eigenfunctions, we solve $$Ku=\lambda u$$ Since $u\in L_2[0,\pi]$, then we can expand $u(x)$ as $u(x)=\sum_{m=0}^\infty (a_m \sin(2mx)+b_m \cos(2mx))$

Now, we have

\begin{gather}\int_0^\pi \sum_{n=1}^\infty \frac{1}{n^2}\sin\big((n+1)x\big)\sin(ny)\sum_{m=1}^\infty (a_m \sin(2my)+b_m \cos(2my))\,dy=\\=\lambda \sum_{m=1}^\infty (a_m \sin(2mx)+b_m \cos(2mx))\end{gather}

$$ \sum_{n=1}^\infty \sum_{m=1}^\infty\int_0^\pi\frac{1}{n^2}\sin((n+1)x)\sin(ny) (a_m \sin(2my)+b_m \cos(2my))\,dy=\lambda \sum_{m=1}^\infty (a_m \sin(2mx)+b_m \cos(2mx))$$

I did the trivial steps, but don't know how to move next.

user3382078
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    Some thoughts: (i) A Fourier basis is not a bad idea. However are you sure you've implement it correctly? Namely, are the basis elements correct: is the constant function in there, and is your scaling of the interval OK? (ii) Can you exchange the order of summation and integration when dealing with infinite series? If not in general, when can you? (iii) Some notation: $\pi$ uses the MathJax code "\pi", and $\sin$ and $\cos$ are "\sin" and "\cos", all within the equation environment. – Stromael May 25 '15 at 08:21
  • I fixed the constant issue. How should I scale the interval ? @Stromael. – user3382078 May 25 '15 at 09:32
  • Thanks for pointing out the limit and summation issue. I have to check if conditions described here (http://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged) is verified. – user3382078 May 25 '15 at 09:36
  • My integrand is not monotonic, hence, I can't justify the exchange between the integration and summation. Am I correct? – user3382078 May 25 '15 at 09:42
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    The interval is traditionally scaled into the arguments of the Fourier basis functions in such a way that the $m$-th terms in the sum have $m$ complete oscillations in the interval. So for $m=1$, the $\sin$ term starts at $0$ at the left end, moves through 1 complete oscillation and finishes at $0$ at the right ($0$, up, down, $0$, down, up, $0$).

    For the exchange of integration and limit, you could consider Lebesgue's monotone and dominated convergence theorems.

     – Stromael May 25 '15 at 20:29
  • I'm happy to explain this in more technical detail, but I thought you might enjoy working through it yourself. I find actual Fourier calculations the most fruitful method of understand the basics of the topic. – Stromael May 25 '15 at 20:41
  • So I should use $\sin(2my)$. Actually, I find this style of feedback very beneficial. @Stormael – user3382078 May 26 '15 at 02:10
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    Great! To further the calculation, what happens when you integrate $\sin(ny)\sin(2my)$ over $(0,\pi)$? (Can you recall any tricks for evaluating this integral?) – Stromael May 26 '15 at 08:33
  • yea, I proceed with that trick and found that the sum will collapse and I will be left with only the term $b_2$ because the integral will only have a value of pi/2 when n=2. In order to justify the exchange between the integral and infinite sum I have to find a function g that is lebesgue integrable and bounds my integrand. Any hints how to find because b_j can be anything. this is my integrand :$(\sin((n+1)x)\sin(ny) )(a_m \sin(2my)+b_m \cos(2my)$ @Stromael – user3382078 May 26 '15 at 11:18
  • ok i think i figured it out. my integrand is less than g which is $g=\sum (a_j cos(jx) + b_jcos(jx) ) \in H=L_2$. So its square integrable. Am i correct – user3382078 May 26 '15 at 11:57
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    I'm not convinced the sum collapses in the way you describe. I would double-check that, as there really ought to be an infinite series left on the LHS.

    Once you have that, try comparing coefficients, and find the values of $\lambda$ for which you can get non-trivial (i.e., not identically zero) solutions. Hey-presto, these are the eigen-values, and substituting these back into your expression for the solution will give the eigen-functions.

    Dominated convergence sounds fine.

     – Stromael May 27 '15 at 18:02

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