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.
Sturm-Liouville problem: eigenvalues
1 Answers
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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0Yes, 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 – 2012-11-26