A periodic Sturm-Liouville problem: Eigenvalues of the Laplacian in One Dimension I'm following Pinchover and Rubinstein's "Introduction to Partial Differential Equations" and am trying to make sense of their analysis of the following periodic Sturm-Liouville problem.
\begin{align*}
\frac{d^{2}v}{dx^{2}}+\lambda v & =  0, & & &x\in\left(0,L\right),\\
v\left(0\right)&=v\left(L\right),&  v'\left(0\right)&=v'\left(L\right).
\end{align*}
The authors note that here we have periodic boundary conditions, which suggests
that we can extend our eigenfunctions to an $L$-periodic function
on $\mathbb{R}$, which is twice differentiable except possibly at
the points $kL,k\in\mathbb{Z}$. Assuming $\lambda\in\mathbb{R}$,
we consider three cases: $\lambda <0, \lambda = 0$ and $\lambda > 0$.
I'm trying to make sense out of the first of these. I understand that for $\lambda < 0$, we have $v\left(x\right)=\alpha\cosh\left(\sqrt{-\lambda}x\right)+\beta\sinh\left(\sqrt{-\lambda}x\right)$. The authors note that,
"In this case any non-trivial solution of the corresponding ODE is an unbounded function on $\mathbb R$. In particular, there is no periodic nontrivial solution for this equation."
Why should this necessarily be?
 A: Sometimes it is simply just better to do simple math rather than use some additional "physical" arguments. Here, assume that $\lambda<0$ then you get your solution
$$
A\cosh \mu x+ B\sinh \mu x,
$$
where $\mu=\sqrt{-\lambda}$. Use the boundary conditions:
$$
A=A\cosh \mu L+B\sinh \mu L,\\
\mu B=\mu A\sinh\mu L+\mu B\cosh \mu L,
$$
or
$$
1=\cosh \mu L+\sinh \mu L,
$$
which is actually
$$
1=\exp(\mu L)\implies \mu=0,
$$
which contradicts the assumption that $\mu>0$. This means that there are no nontrivial solutions to the Sturm-Liuoville problem with periodic boundary conditions for negative $\lambda$s.
To be fully rigorous, you should also consider case $\lambda\in\mathbb C$, which is also, however, does not yield any nontrivial solutions.
A: Another way to write your solution is $v(x)=Ce^{\sqrt{-\lambda}x}+De^{-\sqrt{-\lambda}x}$. If either $C$ or $D$ is non zero then $|v(x)|$ blows up as $x\to +\infty$ or $x\to -\infty$, respectively. The fact that periodic solutions don't become unbounded follows from the fact that we're assuming that it is continuous on its compact period $[0,L]$.
