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 $\mathbb{R}^2\setminus K$ ?

  • 3
    $\begingroup$ (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. $\endgroup$ Mar 6, 2020 at 11:50
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 6, 2020 at 12:38
  • $\begingroup$ @MateuszKwaśnicki, Typo corrected, thanks. $\endgroup$ Mar 6, 2020 at 12:50
  • $\begingroup$ @GiorgioMetafune Thanks a lot for the response. If you could send me some references for such operators would be great. $\endgroup$ Mar 6, 2020 at 12:52
  • $\begingroup$ Please, write to me [email protected] $\endgroup$ Mar 6, 2020 at 12:57


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.