I'm seeking assistance in figuring out the values of 'a' that would ensure the eigenvalues of a specific regular Sturm-Liouville problem are nonnegative. The problem is presented as follows:
Given the regular Sturm-Liouville problem:
$$ \frac{d}{d x}\left(e^{-x} \frac{d \phi}{d x}\right)-x e^{-x} \phi(x)+\lambda e^{-x} \phi(x)=0 $$ $$ \phi(0)+a \frac{\partial \phi}{\partial x}(0)=0 $$ $$ (1+a) \phi(L)+\frac{\partial \phi}{\partial x}(L)=0 $$
I've attempted to tackle the problem without explicitly solving the differential equation and enforcing the boundary conditions, but I found it hard to provide a direct answer, which has been somewhat challenging.
My primary approach has been to consider one of the known general theorems for the Regular Sturm-Liouville problem, specifically, the Rayleigh quotient. In this context, it's given by:
$$ \lambda=\frac{-\left.e^{-x} \phi \frac{\partial \phi}{\partial x}\right|_0 ^L+\int_0^L\left(e^{-x}\left(\phi^{\prime}\right)^2-x e^{-x} \phi^2\right) d x}{\int_0^L e^{-x} \phi^2 d x} $$
For the eigenvalues to be nonnegative, the numerator of the Rayleigh quotient needs to be nonnegative as well. Upon evaluating each term and combining them, the numerator of the Rayleigh quotient transposes to:
$$ -e^{-L}(1+a)\phi(L) - \frac{\partial \phi}{\partial x}\bigg|_L - \phi(0)\frac{\partial \phi}{\partial x}\bigg|_0 + \frac{1}{a^2}\int_0^L e^{-x} \phi^2(0) dx + \frac{\phi^2(0)}{2} - \frac{\phi^2(L)}{2} + \frac{1}{2}\int_0^L \phi^2 e^{-x} dx $$
For the nonnegativity of the eigenvalues to be ensured, this expression should be nonnegative.
Finally, if $a=-1,0$ it surely must mean something but i cant wrap my head around it well enough to know if it will have any impact on the eigenvalues.
I'd greatly appreciate any help or insights into this problem. I hope my query is well-constructed, and thank you in advance for your time and consideration.
*Note: The presented Regular Sturm-Liouville Problem is an outcome of applying the method of separation of variables for the ensuing heat equation:
$$ \begin{aligned} & \frac{\partial u}{\partial t}=e^x \frac{\partial}{\partial x}\left(e^{-x} \frac{\partial u}{\partial x}\right)-x u, \quad 0<x<L, \quad t>0 \\ & u(0, t)+a \frac{\partial u}{\partial x}(0, t)=0 \\ & (1+a) u(L, t)+\frac{\partial u}{\partial x}(L, t)=0 \\ & u(x, 0)=g(x) \\ & \frac{\partial u}{\partial t}(x, 0)=0 \end{aligned} $$
Here is an incomplete answer. First, we notice that there is a sign error in the numerator of the Rayleigh quotient; the correct expression is \begin{align} N&=-e^{-x} \phi\, \frac{\partial \phi}{\partial x}\bigg|_0 ^L +\int_0^Le^{-x}\,[\left(\phi^{\prime}\right)^2\,{\color{red}+}\,x \phi^2]\,dx \\ &=-e^{-L}\phi(L)\phi'(L)+\phi(0)\phi'(0) +\int_0^Le^{-x}\,[\left(\phi^{\prime}\right)^2+x \phi^2]\,dx. \tag{1} \end{align} Now, using the boundary conditions, we rewrite $(1)$ as $$ N=e^{-L}(1+a)[\phi(L)]^2-a[\phi'(0)]^2 +\int_0^Le^{-x}\,[\left(\phi^{\prime}\right)^2+x \phi^2]\,dx. \tag{2} $$ The first term on the RHS of $(2)$ is nonnegative if $a\geq -1$; the second, if $a\leq 0$, and the third (the integral), for any value of $a$. We conclude that $\lambda$ cannot be negative if $a\in[-1,0]$. (This range almost certainly is too conservative, but a full solution probably requires the explicit solution of the Sturm-Liouville problem.)