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Consider an operator of the form $$L(\phi):=\Delta \phi + \gamma \phi_{rr}$$ here the $r$ denotes derivative with respect to the radial variable (we are in $ R^N$ say where $N \ge 3$).

I am curious whether I can apply some abstract results to $L$ that require $L$ to be hypo-elliptic (I kinda know what the word means but have absolutely zero experience).

My belief is this is not hypo-elliptic since I can use separation of variables (using spherical harmonics) to come up with a non smooth solution of $L(\phi)=0$ but there is a nonzero chance that I am screwing up the computation and really the equation is $ L(\phi)=\mu$ where $\mu$ some distribution supported at the origin.
thanks Craig

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    $\begingroup$ What is $\gamma$? What does your solution look like when you convert it back to Cartesian coordinates? $\endgroup$
    – Deane Yang
    Commented Oct 15, 2018 at 17:55
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    $\begingroup$ If $\gamma<0$ then $L$ is (strictly) hyperbolic, therefore not hypo-elliptic. If $\gamma>0$ then $L$ is elliptic, in particular hypo-elliptic. Here, I assume that $\Delta$ is the traditional (negative) Laplacian. $\endgroup$
    – user80744
    Commented Oct 15, 2018 at 21:07
  • $\begingroup$ So $ \Delta$ is the usual one (not the negative one) and I forgot to add $ \gamma>0$. So it appears from the above comments its not hypo-elliptic. Regarding what my solution looks like after putting it back to Cartesian (I am not sure.... I am a little clueless on doing things like this). I have a more general question that I might start new thread on. $\endgroup$
    – Math604
    Commented Oct 16, 2018 at 17:46
  • $\begingroup$ But I think i can clarify my confusion with a radial example. If we consider the radial function $ \phi(x)= \frac{1}{|x|^\frac{N-2-\gamma}{1+\gamma}}$ for $ 0<\gamma<N-2$ this solves the pde (at least on the punctured space). Since $\phi$ is better behaved than the fundamental solution of $\Delta$ then i assume it solves the pde on the full space in sense of distributions (but i guess i should attempt to check this). So would this be sufficient to say the operator is not hypo-ellipitic? $\endgroup$
    – Math604
    Commented Oct 16, 2018 at 17:50

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