Finding the spectrum of a Sturm-Liouville problem

Question

I have the following Sturm-Liouville problem for $0 \le x\le \pi$

$$y'' + \lambda y=0, \qquad y(0) = 0, \qquad y(\pi)+y'(\pi) = 0 $$

How do I find the spectrum of the problem? And how will the spectrum change if instead of $0 \le x \le \pi$ and $y(0)=0$, we have $0<x<\pi$ and $y(0)-y'(0)=0$ instead?

edit: for the first problem i got that the spectrum is $\lambda>0$ since for $=0$ or $<0$ i get the trivial solution (don't know if i'm correct though). As for the second problem, i've got $\lambda=1,-1$ (and also the trivial solution). are there any issues with those solutions?

Answer 1

Solving for

$$ y'' + \lambda^2 y=0, \qquad y(0) = 0 $$

we have

$$ y=c_2\sin(\lambda x) $$

and then

$$ y(\pi)+y'(\pi)=0\Rightarrow c_2\left(\lambda\cos(\lambda\pi)+\sin(\lambda\pi)\right)=0 $$

this follows for all $\lambda_k$ such that $\lambda_k\cos(\lambda_k\pi)+\sin(\lambda_k\pi)=0$ as can be depicted in the following plot

Answer 2

Every solution of $-y''=\lambda y$, $y(0)=0$ can be normalized by adding a normalization condition such as $y'(0)=1$: $$ -y''=\lambda y,\;\; y(0)=0,\; y'(0)=1. $$ Any solution of the above without the condition $y'(0)=1$ is a scalar multiple of the solution with $y'(0)=1$. The solution of this equation is unique, and is given by $$ y_{\lambda}(x)=\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}},\;\; \lambda\ne 0. $$ This also works in the limiting case as $\lambda\rightarrow 0$ to give $y_0$: $$ y_0(x)=x $$ Every solution of $-y''=\lambda y, y(0)=0$, is a scalar mutliple of $y_{\lambda}$. The eigenvalues of your problem are the values of $\lambda$ for which $$ y_{\lambda}(\pi)+y_{\lambda}'(\pi)=0 \\ \frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}+\cos(\sqrt{\lambda}\pi)=0. $$ A common way to write this eigenvalue equation is: $$ \tan(\sqrt{\lambda}\pi)=-\sqrt{\lambda}. $$ The eigenvalues of your problem are the solutions $\lambda$ of the eigenvalue equation. It is possible in such problems for there to be negative solutions $\lambda$.