Weak maximum principle for a perturbation of the Laplacian

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) $$$$D:=\frac{\partial^2}{\partial r^2}{\color{red} -}\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}.$$$$ Writing the Laplacian, $$\Delta$$ in polar coordinates we see in fact that $$$$D+\frac{2}{r}\frac{\partial}{\partial r }=\Delta.$$$$ For the Laplacian it is well known that the WMP (Weak Maximum Principle) holds (even the strong one), and since $$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 $$\mathbb{R}^2\setminus K$$ ?

• (I assume there is a typo, and you meant $\partial/\partial r$ rather than $\partial r / \partial r^2$.) If your domain does not touch the origin, you're good to go: even the strong maximum principle is satisfied, see Theorem 3.5 in Gibarg–Trudinger. On the other hand, if the domain contains $0$, then it is not immediately clear what one means by a solution. Mar 6, 2020 at 11:50
• You need a "boundary condition" at 0 since the 1d operator D^2-1/r D (the generator of a Bessel process) has 0 as an exit boundary (see for example Chapter Vi, Section 4 of the book Engel-Nagel "One parameter semigroups..."). This explains what happens on radial functions; in the non radial case expansion in spherical harmonics shows that the radial case is the worst. In case you need I can send some references where these operators have been studied in detail, in Nd. Mar 6, 2020 at 12:38
• @MateuszKwaśnicki, Typo corrected, thanks. Mar 6, 2020 at 12:50
• @GiorgioMetafune Thanks a lot for the response. If you could send me some references for such operators would be great. Mar 6, 2020 at 12:52
• Please, write to me [email protected] Mar 6, 2020 at 12:57