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I have a Sturm-Liouville problem $$ y'' + \lambda^2 y = 0, \\ y'(0) + \alpha_1 y(0) = 0, \\ y'(L) + \alpha_2 y(L) = 0, $$ where $\alpha_1 \alpha_2 \neq 0$. I found that eigenvalues are positive and satisfy $$ \lambda L = \pi k + \arctan \left( \frac{\alpha_2 - \alpha_1}{\lambda^2 + \alpha_1 \alpha_2}\lambda \right), \; k \in \mathbb{Z} $$ Is it possible to say something more about eigenvalues in this case? Is it possible to find them as a series or as an integral? Does this equation have the unique solution if solution exists for some fixed $k$? All the questions are of the practical interest, I have to find eigenvalues numerically.

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Rewrite the equation of the eigenvalues as $$ \tan(L\,\lambda)=\frac{\alpha_2-\alpha_1}{\lambda^2+\alpha_1 \alpha_2}\lambda. $$ Obviously $\lambda=0$ is an eigenvalue, and if $\alpha_1=\alpha_2$, then the eigenvalues are $\lambda=k\,\pi/L$, $k\in\mathbb{Z}$, $k\ge0$.

Assume from now on $\alpha_1\ne\alpha_2$. Looking at the graphs of $\tan(L\,\lambda)$ and $\dfrac{\alpha_2 - \alpha_1}{\lambda^2 + \alpha_1 \alpha_2}\lambda$ as functions of $\lambda$, you see that there is eigenvalue in each interval $$ \Bigl(\frac{(2\,k-1)\pi}{2\,L},\frac{(2\,k+1)\pi}{2\,L}\Bigr),\quad k\ge1, $$ and depending on the values of the parameters, maybe another one on the interval $(0,\pi/(2\,L))$. You can use Newton's method to find a numerical approximation, taking for instance as a first approximation the middle point of each interval, $k\,\pi/L$.

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Your answer is very good. But question of uniqueness of solution on each interval in case $\alpha_2 < \alpha_1$ is not obvious for me (function on the left and function on the right are increasing and concave on the majority of intervals). Is it possible to say something in this case in general? –  Nimza Nov 25 '12 at 22:50
    
The main parameters are the sign of $\alpha_2-\alpha_1$ and $\alpha_2\cdot\alpha_1$, and the derivative of the right hand side at $\lambda=0$ (if it is $>1$ there will be two solutions on $(0,\pi/(2\,L))$.) Since you look for numerical results, you must have specific values of the $\alpha_i$. Draw a graph and you will "see" the eigenvalues. –  Julián Aguirre Nov 26 '12 at 12:54
    
Yes, I've already plotted it. For negative sign of $\alpha_2 - \alpha_1$ it looks like it has only one solution on each interval since some fixed interval. But it seems nontrivial to me to show this strictly because both $\tan$ and function on the RHS are increasing and concave on such intervals –  Nimza Nov 26 '12 at 12:56

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