Given $\nabla^2\phi + \lambda\phi = 0$ on a circular disk of radius R with boundary condition $\phi + a\nabla\phi \cdot n = 0$ where $a$ is an arbitrary constant and $n$ is the outward unit normal vector, for what values of $a$ is $\lambda \geq 0$. Also, what would be a specific value of $a$ such that $\lambda > 0$.
I know I have to use the Rayleigh quotient: $\lambda = \frac{-\int_{dR}^{} \phi\nabla\phi\cdot n \cdot ds + \int_{R}^{} |\nabla\phi|^2 dx dy}{\int_{R}^{} \phi^2 dx dy}$
but I still don't know for what values of $a$ we have $\lambda \geq 0$ and $\lambda > 0$.
If we choose $a = \phi$, $\lambda \geq 0$ assuming $\int_{dR}^{}\phi \geq 0$, but I don't think we can make that assumption...
Am I overthinking this? Is it something simple like all $a \in \mathbb{R}$?
It would be usual for the eigenfunction associated with the smallest eigenvalue to be independent of $\theta$. That is, $\phi=\phi(r)$ would satisfy $$ 0=\nabla^2\phi(r)+\lambda\phi(r)= \frac{1}{r}(r\phi'(r))'+\lambda\phi(r) \\ \phi(1)+a\phi'(1)=0. $$ The equation for $\phi(r)$ is $$ r^2\phi''(r)+r\phi'(r)+\lambda r^2\phi(r)=0. $$ Letting $\phi(r)=J(\sqrt{\lambda}r)$ gives $$ (\sqrt{\lambda}r)^2J''(\sqrt{\lambda}r)+(\sqrt{\lambda}r)J'(\sqrt{\lambda}r)+(\sqrt{\lambda}r)^2J(\sqrt{\lambda}r)=0 \\ s^2J''(s)+sJ'(s)+s^2J(s)=0. $$ The solution $J$ that is regular at $r=0$ is the zeroeth order Bessel function $J_0(s)$. The boundary condition then leads to $\phi(r)=J_0(\sqrt{\lambda}r)$ satisfying $$ J_0(\sqrt{\lambda})+a\sqrt{\lambda}J_0'(\sqrt{\lambda})=0. $$ $\sqrt{\lambda}$ can be positive or negative in the above equation because $J_0$ is a power series. So, we're looking for all complex zeros of $J_0(z)+azJ_0'(z)=0$, and $\lambda=z^2$. There are no imaginary solutions $z$ for $a > 0$, which means there are no negative $\lambda$ for $a > 0$.