Consider the following Sturm-Liouville boundary value problem:
$\text{Given parameters }c > 0 \text{ and } \beta > 0, \text{ let } y=y(x) \text{ for } 0 \leq x \leq c. \text{ We have }$
$$y''+ \lambda y=0$$
$\text{ with boundary conditions }$
$$y'(0)=\beta y(0),$$
$$y'(c)=\beta y(c),$$
By supposing $y(x)=e^{kx}$, one can show that there is only one negative eigenvalue: $-\beta^2$
Is this the only eigenvalue? That is, there are no positive eigenvalues? I have heard that there is some sort of uniqueness theorem that guarantees this to be the case. What is this theorem? If it is the case that there are positive eigenvalues, what are they and how can one obtain them?
Your problem in standard Sturm-Liouville form is: $$ y''+\lambda y = 0,\;\;\; 0 \le x \le c, \;\; c > 0, \\ y'(0)-\beta y(0) = 0, \\ y'(c)-\beta y(c) = 0. $$ Start by solving $$ y''+\lambda y=0,\;\; 0 \le x \le c, \;\; c> 0, \\ y(0)=1,\; y'(0)=\beta. $$ This has the unique solution $$ y_{\lambda}(x)=\cos(\sqrt{\lambda}x)+\beta\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}. $$ And, this solution works for $\lambda=0$ as well if you let $\lambda\rightarrow 0$: $$ y_0(x)=1+\beta x. $$
Therefore, there is a non-trivial solution of the original problem for some $\lambda$ iff $y_{\lambda}'(c)-\beta y_{\lambda}(c)=0$, which gives the eigenvalue equation $$ 0=-\sqrt{\lambda}\sin(\sqrt{\lambda}c)+\beta\cos(\sqrt{\lambda}c) \\ -\beta\cos(\sqrt{\lambda}c)-\beta^2\frac{\sin(\sqrt\lambda c)}{\sqrt{\lambda}} \\ = -\sqrt{\lambda}\sin(\sqrt{\lambda}c) -\beta^2\frac{\sin(\sqrt\lambda c)}{\sqrt{\lambda}}. $$ The limiting case where $\lambda\rightarrow 0$ has no non-trivial solution unless $\beta=0$. So $\lambda=0$ may be excluded if $\beta\ne 0$, which gives the general eigenvalue equation in $\lambda$: $$ (\lambda+\beta^2)\sin(\sqrt{\lambda}c)=0,\;\;\lambda\ne 0. $$ The eigenvalues are $$ \lambda=-\beta^2,\frac{n^2\pi^2}{c^2},\;\; n=1,2,3,\cdots. $$ So, there is only one negative eigenvalue, but infinitely many positive eigenvalues.