Let $\Omega \subset \mathbb{R}^3$ a smooth bounded domain with $0\in \Omega$ and $u_\varepsilon(x)$ the solution to $$ \Delta u_\varepsilon + (\lambda-\varepsilon) u_\varepsilon = \frac{1}{|x|}\quad \text{in } \Omega, \quad u_\varepsilon(x)=0 \quad \text{on }\partial \Omega.$$ Here $\lambda_{k}<\lambda-\varepsilon<\lambda<\lambda_{k+1}$ for some $\lambda_k \in \sigma(-\Delta)$. I am interested to find sufficient conditions on the location of $0\in \Omega$ to have $u_\varepsilon(0)\neq 0$ for almost every $\varepsilon \in [0,\varepsilon_1)$.
For example, I would like to prove that locating the origin near the boundary is sufficient: if $0<d(0,\partial \Omega)<\varepsilon_0$ for some $\varepsilon_0$, then $u_\varepsilon(0)\neq 0$ for almost every $\varepsilon \in [0,\varepsilon_1)$ for some $\varepsilon_1=\varepsilon_1(\varepsilon_0)$.
The following are some observations.
To work with a non-singular source I am studying the problem for $v_\varepsilon\in C^{2,1}(\Omega)$ where $u_\epsilon(x)=\frac{1}{2}|x|+v_\varepsilon(x)$. The problem becomes $$ \Delta v_\varepsilon + (\lambda - \varepsilon)v_\varepsilon =-(\lambda-\varepsilon)\frac{|x|}{2}\quad \text{in } \Omega,\quad v_\varepsilon(x)=-\frac{|x|}{2}\quad \text{on } \partial \Omega $$
If we decompose $v_\varepsilon=v_{1,\varepsilon}+c|x|^{3}$ where $c$ is such that $\Delta \left(c|x|^3\right)=-\frac{\lambda}{2} |x|$ and we continue the process to infinity (letting $v_{k-1,\varepsilon}=v_{k,\varepsilon}+c_k|x|^{2k+1}$ and send $k\to \infty$) we reduce to an homogeneus problem for $v_{\infty,\varepsilon}(x)$ where $u_\varepsilon(0)=v_{\infty,\varepsilon}(0)$ and $$\Delta v_{\infty,\varepsilon}+(\lambda-\varepsilon)v_{\infty,\varepsilon}=0\quad \text{in }\Omega,\quad v_{\infty,\varepsilon}=\frac{1}{\lambda-\varepsilon} \frac{\cos(\sqrt{\lambda-\varepsilon}|x|)-1}{|x|} \quad \text{on }\partial \Omega.$$
For $d(0,\partial \Omega)=|x|=\delta(\varepsilon)$ small I thought it could be useful to expand $-\frac{1}{2}\delta=\sum_{k=0}^\infty D_x^k v_\varepsilon(0)\cdot x^k$ and using the strong continuation property to get a contradiction, but I don't see how to get it.
Maximum principle does not apply.
In case of $\Omega=B_1(0)$, the solution gives $u(0)=-\tan(\frac{\sqrt{\lambda-\varepsilon}}{2})$ and hence $u(0)\neq 0 \quad \Leftrightarrow \quad \varepsilon \neq \lambda-4k^2 \pi^2$ for some $k\in \mathbb{N}$. This suggests me that the condition is generic and that in general we do not need to get close to the boundary to obtain the desired property; but being close to the boundary it could helps to prove it in the general case.
I could not find $u(0)$ for $\Omega=B_1(z)$ with (for example) $z=(0,0,R)$ (with $R\in (0,1)$)). Translating the problem to $B_1(0)$ the problem becomes $$ \Delta u_\varepsilon + (\lambda-\epsilon)u_{\varepsilon}=\frac{1}{|{z-y}|} $$ with 0 BC. Writing the problem in spherical coordinates we see that $u(r,\theta,\phi)=u(r,\theta)$ is independent of the $\phi$ (having picked $z$ on the $x_3$-axis) but, of course, not radial.
Also proving the weaker statement
there is some $\varepsilon$ small such that $u_\varepsilon(0)\neq 0$
it would be enough for my purposes. Any reference or suggestion related to this problem would be helpful.
