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I have this equation:

$$\lambda \sin(2 \alpha)+ \sin(2 \alpha \lambda)=0$$

where "$\alpha$" is a known parameter and my desire is to calculate eigenvalues, "$\lambda$". I've tried some newton-raphson and muller codes in Matlab, but it didn't work since I know eigenvalues for some alphas. for example, for $\alpha=\pi/3$, first eigenvalue becomes $0.5122$ and it becomes complex for second one and after. I hope I've explained it clearly.

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Neither Newton-Raphson nor Muller work well if you don't have a good initial guess for the eigenvalues you seek... –  J. M. May 4 '12 at 11:19
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There must be a mistake somewhere; there are no non-trivial real solutions to this equation for $\alpha=\pi/3$, as you can see from this plot. –  joriki May 4 '12 at 11:34
    
@joriki: indeed there is no non-trivial real solution for $0 < \alpha < 1.276782905$ approximately. Perhaps he meant $\alpha = 5\pi/6$, for which there is a solution $\lambda \approx .5122213612$ –  Robert Israel Jul 16 '12 at 19:15

2 Answers 2

For convenience write $\alpha = x \pi/2$, so the equation becomes $\lambda \sin( \pi x)+\sin(\pi \lambda x) = 0$. Here is an implicit plot:

enter image description here

If $x$ is a nonzero integer, the equation becomes $\sin(\pi \lambda x) = 0$ so the solutions are $\lambda = k/x$ for integers $k$. If $x$ is close enough to a nonzero integer $n$, there will be a real solution near $k/n$. If $n+k$ is odd, that solution is approximately $$\dfrac{k}{n}+{\dfrac {k{\pi }^{2} ( {k}^{2}-{n}^{2} ) }{6{n}^{4}}}{(x- n)}^{3}+{\dfrac {k{\pi }^{4} ( {k}^{2}-{n}^{2} ) ( 9 k^2 -{n}^{2} ) }{120 {n}^{6}}}{(x-n)}^{5} $$

If $n+k$ is even, it is approximately $$ \dfrac{k}{n}-2{\frac { \left( x-n \right) k}{n^2}}+4\,{\frac {k \left( x- n \right) ^{2}}{{n}^{3}}}-{\frac {k \left( 48+{\pi }^{2}{k}^{2}-{ n}^{2}{\pi }^{2} \right) \left( x-n \right) ^{3}}{6{n}^{4}}} $$

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Why do you call this equation a "eigenvalue problem"? I can't see an eigenvector or eigenfunction anywhere. Your equation is transcedental, meaning you cannot globally invert it to solve for $\lambda$. So your problem is simply a root-finding problem.

If you can guess one solution or calculate it via newton iterations you could use a numeric continuation software to trace the "guessed" solution for one alpha for all lambdas.

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Maybe because it arises from an eigenvalue problem for a differential operator with certain boundary conditions. –  timur Aug 15 '12 at 23:31

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