This question is motivated by similar considerations for the Kohn-Laplacian in the Heisenberg group, but it seems that I cannot even give an answer in the Euclidean case, so here we go.

Suppose that I consider the differential operator (in polar coordinates) \begin{equation} D:=\frac{\partial^2}{\partial r^2}{\color{red} -}\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}. \end{equation} Writing the Laplacian, $\Delta $ in polar coordinates we see in fact that \begin{equation} D+\frac{2}{r}\frac{\partial}{\partial r }=\Delta. \end{equation} For the Laplacian it is well known that the WMP (Weak Maximum Principle) holds (even the strong one), and since $D$ is a perturbation of the Laplacian by a differential operator of first order, we expect that the WMP to hold for $D$ as well, the problem being that the coefficient of $\frac{\partial}{\partial r}$ has a singularity.

So, is there something we can say about the WMP for $D$ ?

EDIT: Some clarification about the solutions of the PDE $Du=0$. One can verify that a radial solution is $u_0(r,\theta)=r^2$. Suppose now that we have a compactly supported positive Borel measure $\mu, supp(\mu)=K$. Is it true that $$ u:=u_0*\mu $$ satisfies the WMP in $\mathbb{R}^2\setminus K$ ?