Timeline for What does the flow of the principal symbol of the differential operator tell us about the PDE?
Current License: CC BY-SA 3.0
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| Mar 24, 2018 at 20:41 | history | edited | Bombyx mori | CC BY-SA 3.0 |
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| Mar 24, 2018 at 20:28 | comment | added | Bombyx mori | @MichaelBächtold: Sorry for the messed up notation. $Q$ is really the $P$ in OP's question, $P$ is the square root of Lapacian as mcd suggested. | |
| Mar 24, 2018 at 16:55 | comment | added | Michael Bächtold | @mcd Thanks. I think I can make of the principal symbol part now. So was $P$ meant to be the one from the OP's question? There was nothing about ellipticity and self-adjointness stated there. Is it a necessary assumption? | |
| Mar 24, 2018 at 12:32 | comment | added | mcd | Yes, think of $Q$ as a localization in phase space. Then the proposition is that if you evolve $Q$ according to the classical flow you obtain a solution of the equation (modulo lower order terms, but you can iterate them away). | |
| Mar 24, 2018 at 11:15 | comment | added | Michael Bächtold | @mcd: thanks. So the statement holds for an arbitrary self adjoint elliptic $P$, unrelated to $Q$? | |
| Mar 24, 2018 at 9:51 | comment | added | mcd | The operator $P \in \Psi^1$ is a elliptic self-adjoint operator. Take for example the square root of the Laplacian $\sqrt{\Delta}$ on a compact manifold. | |
| Mar 24, 2018 at 7:54 | comment | added | Michael Bächtold | What is $P$ here? And I can't parse the sentence: "and having for each $t\in\mathbb{R}$ the principal symbol..." What is it saying? | |
| Mar 24, 2018 at 2:34 | history | answered | Bombyx mori | CC BY-SA 3.0 |