Let $\mathbb{D}=\{(x,y)\in \mathbb R^2: x^2+y^2\le 1\}$ be the unit disk. Let consider the Laplace operator $\Delta_0$ defined on $C_0^\infty(\mathbb D)$, the space of smooth functions $\mathbb D\to \mathbb R$ with support contained in $\mathbb D-\partial \mathbb D$. $\Delta_0$ is symmetric but no self-adjoint, so its deficiency spaces aren't trivial. Therefore I tried to calculate $$ N_\pm=\ker(\Delta_0^\star\mp i\mathbb I)\,, $$ where $\Delta_0^\star$ is the adjoint of $\Delta_0$. $\Delta_0^\star$ acts as the weak laplacian.
Here my attempt (with some inaccuracy).
In polar coordinates $(r,\theta)$, $\Delta_0^\star$ reads as $$ \Delta=\partial^2_r+r^{-1}\partial_r+r^{-2}\partial^2_\theta\,.$$ To find non trivial solution of $\Delta_0^\star\psi=i\psi$, I suppose $\psi(r,\theta)=R(r)\Theta(\theta)$ for some function $R,\Theta$ such that $\psi$ is in the domain of $\Delta_0^\star$ and $\Theta$ is $2\pi$-periodic. The separation of variables allow me to transoform the initial PDE onto two ODE $$ r^2\frac{\mathrm{d}^2R}{\mathrm{d}r^2}+r\frac{\mathrm{d}R}{\mathrm{d}r}-(ir^2+m) R=0\quad \text{ e } \quad \frac{1}{\Theta}\frac{\mathrm{d}^2\Theta}{\mathrm{d}\theta^2}+m=0 $$ The periodicity of $\Theta$ impose $m=n^2$ for some $n\in \mathbb Z$ and $$\Theta(\theta)=c_{1n} \cos n\theta+c_{2n}\sin n\theta \,.$$ The radial differential equation can be solved defining $$ \rho(r)=\frac{i-1}{\sqrt 2} \quad \text{ e } \quad R(r)=J(\rho(r))\,.$$ With this change of variable, I obtain the Bessel equation of order $n$ $$ \rho^2 \frac{\mathrm{d}^2J}{\mathrm{d}\rho^2}+\rho \frac{\mathrm{d}J}{\mathrm{d}\rho}-(\rho^2+n^2)J=0\,.$$ Let $J_n$ be the solution of this Bessel equation well defined at $r=0$. Then $$ \ker(\Delta_0^\star -i\mathbb I)=\text{span}\bigg(J_n\bigg(\frac{i-1}{\sqrt 2}r\bigg) \cos n\theta,J_n\bigg(\frac{i-1}{\sqrt 2}r\bigg) \sin n\theta: n\in \mathbb N_0\bigg)\,.$$
Does my idea make sense? How it can be improved? Could you indicate me some usefull references about this topic?
Thanks in advance !