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Feb 19, 2019 at 9:30 history edited mcd CC BY-SA 4.0
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Feb 19, 2019 at 9:27 comment added mcd Yes, you are right, this is not a counterexample.
Feb 18, 2019 at 15:00 comment added Bazin I have now strong doubts about your answer for the case $c=-1$ In fact, I believe that there is a self-adjoint extension in that case.
Feb 11, 2019 at 9:39 comment added Bazin I looked at the Sjöstrand reference and did not see anything on the case $c\in (-\infty, 0)$. He studies in that book the case $c\in \mathbb C\backslash(-\infty, 0[$ which indeed are all non-self-adjoint if $c$ is not positive, but the case $c\in (-\infty, 0)$ is apparently not tackled in that reference. Thanks in advance for your help.
Feb 8, 2019 at 19:57 comment added Bazin OK, I want ellipticity and the characteristic set to be a manifold, which is not the case of your example $\xi=\pm x$.
Feb 8, 2019 at 16:36 comment added mcd In my example with $a_1(x,\xi) = \xi^2 - x^2$ and $a_0 = 0$, we have that $|\nabla p_2(x,\xi)|^2 = 4(\xi^2 + x^2)$, which is elliptic.
Feb 8, 2019 at 15:50 comment added Bazin I mean that $a=a_1+a_0$ with $a_j\in \Sigma^j$ is such that $\vert \nabla a_1\vert^2$ is elliptic in $\Sigma_1$.
Feb 8, 2019 at 15:44 comment added mcd What do you mean by principal symbol? Shubin or classical sense?
Feb 8, 2019 at 15:40 comment added Bazin Thanks, I modified the question with a real principal type symbol $a$.
Feb 8, 2019 at 15:31 history answered mcd CC BY-SA 4.0