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Let $\Omega$ denote a an exterior domain in $R^N$ with smooth boundary.
I am interested in Liouville Theorems related to smooth solutions of

$$\Delta \phi(x) + \gamma \sum_{i,j=1}^N \frac{x_i x_j}{|x|^2} \phi_{x_i x_j}(x) =0$$ in $\Omega$ with $ \phi=0$ on $ \partial \Omega$ and under the assumption that $ \sup_\Omega |x|^{\sigma+1} | \nabla \phi(x)| \le C$. The domain does not contain the origin in its closure.

In the case of $\Omega$ the compliment of the unit ball centred at the origin I can get exact (and optimal) conditions on $\gamma$ and $\sigma$ that give me a Liouville theorem (one writes $ \phi$ with spherical harmonics and then examines the ODE's).

In the case of $\gamma=0$ but a general domain I work on $\Omega_R:=\{x \in \Omega: |x|<R\}$ and multiply $-\Delta \phi$ by $ \phi$ and integrate by parts and then use the bound on $\phi$ to control the boundary terms on $B_R$. This gives an optimal result.

In the case of $ \gamma>0$ and general domain I can try and copy the same approach but it appears either I won't get anything, or at least I won't get optimal results.

QUESTION. Is there a general approach to trying to prove these things? If I let $$ \psi(r) = \int_{|\theta|=1} \phi(r \theta) d \theta$$ for $ r>0$ big enough so $ \partial B_r \subset \Omega$ I see (under the expected assumptions on $\gamma$ and $ \sigma$) that $ \psi(r)$ is constant. Of course since the domain isn't radial its hard to use this for much.

Is there any hope of using a continuity argument to connect the general case to the radial case?

thanks for all comments.

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  • $\begingroup$ The operator is symmetric with respcet to the measure $|x|^r\, dx$ with $r=\frac{N-1}{\gamma+1}-N+1$. Variational computations could be easier here, but I do not know if the new measure fits well with your assumptions. $\endgroup$ Commented Apr 9, 2022 at 22:37
  • $\begingroup$ Thank you very much for the comment. I had done something similar (but with a change of dimension) but then I got confused with the Laplace-Beltrami operator. Maybe your approach will allow me to copy the Laplacian approach; I will try. Thank you very much. $\endgroup$
    – Math604
    Commented Apr 11, 2022 at 16:28
  • $\begingroup$ @GiorgioMetafune. I checked it and it appears this gives me exactly what I want..i think. Thank you very much. If you cut and paste your comment as an answer I can accept it. $\endgroup$
    – Math604
    Commented Apr 11, 2022 at 17:10
  • $\begingroup$ Good if it works. It is not really an answer but only a suggestion how to go on, having done similar computations. $\endgroup$ Commented Apr 11, 2022 at 17:12

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