The following is a problem in a text (in Portuguese) on Critical Point Theory that I am reading:
Find the eigenvalues and eigenfunctions of the problem $$ (P) \quad \begin{cases} - y'' = \lambda y \quad \text{ in } (0, \pi) \\ y(0) = y(\pi) = 0 \end{cases} $$
Here is what I thought:
Note that $y(x) = \sin(\sqrt{\lambda}x + k \pi$ with $k \in \Bbb{N}$ are functions satisfying the boundary conditions and, for all $x \in (0, \pi)$ $$ y'(x) = \sqrt{\lambda} \cos(\sqrt{\lambda}x + k \pi), $$ and therefore $$ - y''(x) = \lambda^2 \sin(\sqrt{\lambda}x + k \pi) = y(x), $$ and note that the above holds for every $\lambda > 0$.
Is the above corret? Is that all? I am studying the spectrum of the Laplacian, so it seems strange that all $\lambda > 0$ are eigenvalues. Am I missing something?
Thanks in advance and kind regards.
The functions $y(x)$ do not necessarily satisfy the boundary conditions. That is where you ran adrift.
We need to begin by identifying all possible solutions to $-y'' =\lambda y$ for a given $\lambda$. Straightforward analysis of second order linear equations tells us there are two families of solutions, depending on the sign of $\lambda$. If we set $\omega = \sqrt{\lvert \lambda \rvert}$, we obtain, $$ y(x) = \left\{ \array{ A \cos (\omega x) + B \sin (\omega x), & \text{when } \lambda =\omega^2 > 0 \\ C \cosh(\omega x) + D \sinh(\omega x), & \text{when } \lambda =-\omega^2 < 0 } \right. $$ Now apply the boundary conditions to each of these possibilities. Taking the first, when $x = 0 $, $y(x) = 0$, so we conclude $A = 0$. When $x=\pi$ we then find $B\sin \omega \pi = 0$. We are not interested in zero solutions, so exclude $B \neq 0$ and $\omega = 0$. That leaves $\omega = n$ for $n=1,2, \cdots$ (recall by definition $\omega \geq 0$). The eigenvalues are $\lambda = 1, 2^2, 3^2,4^2, \cdots$ with corresponding eigenfunctions being any non-zero multiple of the functions $\sin (\sqrt\lambda x)$.
Taking the same approach with the second set of equations, we find easily there are no additional non-zero solutions, and no eigenvalues can arise in this case.
In conclusion the complete set of eigenvalues and eigenfunctions are, $$ y_n(x) = B_n \sin n x, \lambda_n = n^2, \quad\text{where } B_n \neq 0 \text{ and } n =1,2,3,\cdots$$
Often, for brevity, the $B_n$ are given the value 1.
I hope this helps.