Consider the Sturm - Liouville problem:
$$y''+\lambda y=0,$$ $$ y(0)=c\pi y'(\pi)-y(\pi)=0$$
(a) Find the condition $c$ must satisfy, so that the problem has negative eigenvalues.
(b) How many are the negative eigenlaues (in case they exist)?
(c) For $c=1$, prove that $\displaystyle 1<\lambda_2-\lambda_1<\frac{9}{4}$, where $\lambda_1,~\lambda_2$ are the first two eigenvalues of the problem above.
Attempt. (a) $$\lambda \int y^2dx=-\int y''ydx=\ldots=\int(y')^2dx-c\pi (y'(\pi))^2,$$ therefore the condition $c$ must satisfy is $\int(y')^2dx<c\pi (y'(\pi))^2$ for all eigenfunctions $y$ (although I guess a condition should be independent of $\lambda,~y$).
(b) According to the theory, there are infinitely countably many (negative) eigenvalues.
(c) I was thinking of the equation of $(i)$, but didn t get me anywhere.
Thank you!
Every non-trivial solution of $y''+\lambda y=0$ with $y(0)=0$ is a scalar multiple of the solution where $y(0)=0$, $y'(0)=1$, which has unique solution $$ f_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}. $$ (The solution at $\lambda=0$ is the limit of the above as $\lambda\rightarrow 0$, which is $f_0(x)=x$.)
Therefore, the eigenvalue equation with $c\pi y'(\pi)-y(\pi)$ and $y(0)=0$ has a non-trivial solution iff $\lambda$ satisfies the power series equation $$ 0=c\pi f_{\lambda}'(\pi)-f_{\lambda}(\pi)= c\pi\cos(\sqrt{\lambda}\pi)-\frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}. $$ The equation where $\lambda=0$ is a limit of the above equation, which is $$ c\pi - \pi = 0 \implies c = 1. $$ So $\lambda=0$ is a solution iff $c=1$. There is never such a solution where $\cos(\sqrt{\lambda}\pi)=0$, even if $c=0$, because that would require $\sin(\sqrt{\lambda}\pi)=0$ and $\cos(\sqrt{\lambda}\pi)=0$. Eliminating such cases gives an equivalent equation $$ \tan(\sqrt{\lambda}\pi)=c\sqrt{\lambda}\pi $$ The only solutions $\lambda$ must be real because the Sturm-Liouville problem has only real eigenvalues $\lambda$. There are infinitely many positive real values of $\sqrt{\lambda}$ satisfying the above, which is easily seen by overlaying the graphs of the two sides and looking for intersections. Nothing is gained by looking at the negative solutions $\sqrt{\lambda}$ because only the squares of the solutions determine $\lambda$. The negative values of $\lambda$ are determined by the imaginary solutions $\sqrt{\lambda}$, and only $\sqrt{\lambda}=i\rho$ for $\rho > 0$ need be considered because the eigenvalues $\lambda$ are squares of such $i\rho$ solutions. Substituting into the equation gives $$ \tan(i\pi\rho) = c\pi i\rho \\ \tanh(\rho\pi)=c\pi\rho. $$ Looking at the graph of $\tanh(\rho\pi)$ for $\rho > 0$, there can never be more than one solution of the above, and the borderline case is where $y=\tanh(\rho\pi)$ and $y=c\pi\rho$ are tangent at the origin, which is where $$ \pi = c\pi \implies c = 1. $$ For $0 < c < 1$ there is exactly one positive solution $\rho$, which gives rise to one negative eigenvalue $\lambda=(i\rho)^2 = -\rho^2$. So there is a negative eigenvalue iff $0 < c < 1$, and there is only one in that case.