Timeline for unique continuation property for overdetermined elliptic PDE
Current License: CC BY-SA 3.0
14 events
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| Aug 4, 2013 at 3:37 | comment | added | Deane Yang | BS, thanks but I wasn't claiming that any determined elliptic PDE of higher order satisfies UCP. Just that you can find sufficient conditions for UCP for an overdetermined system using known sufficient conditions (whatever they might be) for a determined system. | |
| Aug 3, 2013 at 20:52 | comment | added | BS. | @Deane : but there are counterexamples (by Plis and then by Hormander, if I remember right), of fourth order linear elliptic operators with smooth coeffs and without the UCP. But presumably not of the form $P^*P$, so the question remains open (and badly phrased). | |
| Aug 3, 2013 at 4:17 | comment | added | Deane Yang | Have you tried googling "elliptic pde system unique continuation"? This seems to yield some papers that seem relevant. Also, if $P$ is an overdetermined elliptic PDO, then $P^*P$ is a determined elliptic PDO and it seems likely to me that $P$ satisfies the UCP if and only if $P^*P$ does. So it might suffice to consider only determined elliptic operators. | |
| Aug 3, 2013 at 4:03 | comment | added | user38064 | I cannot use Holmgren's theorem, because it requires real-analytic coefficients. It is a system of PDE for one real-valued function, but each one of those equations is alone not elliptic. | |
| Aug 3, 2013 at 4:00 | history | edited | user38064 | CC BY-SA 3.0 |
I added the definition of overdetermined ellipticity.
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| Aug 2, 2013 at 17:37 | answer | added | Bazin | timeline score: 3 | |
| Aug 2, 2013 at 13:11 | comment | added | Ben McKay | So you have two elliptic PDEs, or one elliptic and one something else, or more than two? I think that if you already have one elliptic PDE, you should be able to apply: Aronszajn, N. A unique continuation theorem for solutions of elliptic partial differential equations or inequalities of second order. | |
| Aug 2, 2013 at 12:51 | comment | added | Liviu Nicolaescu | You need to specify what you mean by overdetermined. Many but not all classes of eliptic systems satisfy the unique continuation property. | |
| Aug 2, 2013 at 12:23 | comment | added | BS. | What do you mean exactly by overdetermined ? Is $P$ vector (or complex) valued ? | |
| Aug 2, 2013 at 10:49 | comment | added | Michael Renardy | Is there any reason you cannot use Holmgren's theorem? | |
| S Aug 2, 2013 at 10:42 | history | suggested | Branimir Ćaćić | CC BY-SA 3.0 |
LaTeX and grammar cleanup
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| Aug 2, 2013 at 10:01 | review | Suggested edits | |||
| S Aug 2, 2013 at 10:42 | |||||
| Aug 2, 2013 at 8:27 | review | First posts | |||
| Aug 2, 2013 at 11:03 | |||||
| Aug 2, 2013 at 8:09 | history | asked | user38064 | CC BY-SA 3.0 |