Timeline for Characterization of inverse differential operators
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15 events
| when toggle format | what | by | license | comment | |
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| Jun 5, 2016 at 5:53 | answer | added | Pedro Lauridsen Ribeiro | timeline score: 0 | |
| Jul 15, 2012 at 20:39 | answer | added | Rafe Mazzeo | timeline score: 5 | |
| Apr 22, 2012 at 15:27 | answer | added | Bazin | timeline score: 1 | |
| Apr 10, 2012 at 17:20 | comment | added | Deane Yang | The simplest example of a hyperbolic operator is $\partial_n$ and when you solve $\partial_n u = f$, if $f$ has a singularity at a point, then $u$ might have a singularity not just at that point but along the integral curve of $\partial_n$ through that point. Any hyperbolic or even PDO of real principal type has the same property. This is why the kernel of the inverse will have a singular set off the diagonal. | |
| Apr 10, 2012 at 17:17 | comment | added | Deane Yang | I believe that you can recover the principal symbol of a pseudodifferential operator (and therefore also a partial differential operator) $P$ by looking at the asymptotic behavior for $\xi$ large of $e^{-ix\cdot\xi}P(e^{x\cdot\xi}\chi_\epsilon)$, where $\chi_\epsilon$ is an approximate identity near a given point. | |
| Apr 10, 2012 at 17:14 | comment | added | Deane Yang | PDO's, $\psi$DO's, and FIO's are all examples of singular integral operators (SIO's), which have a kernel that is a distribution on $M \times M$, where $M$ is the domain of the functions acted on by the operators. Roughly speaking (I don't remember any details by now), an SIO is a PDO if the kernel is supported on the diagonal and a $\psi$DO if the wavefront set of the kernel is in the conormal bundle of the diagonal. The kernel of an FIO will have its wavefront set propagating off the diagonal. | |
| Apr 10, 2012 at 15:27 | comment | added | Igor Khavkine | Piero, thanks for the suggestion. Which specific aspects of the question do you think would be covered there? Or is it just a general reference? | |
| Apr 10, 2012 at 13:26 | comment | added | Piero D'Ancona | I think you may start by reading Chapter VI of Stein's book "Harmonic Analysis" | |
| Apr 10, 2012 at 13:11 | comment | added | Igor Khavkine | Also, this is probably quite basic, but I don't think I've appreciated the difference between PDOs and FIOs to understand why the inverse of a hyperbolic differential operator is necessarily not a PDO. Is there a quick way to see that? Thanks a lot for any comments! | |
| Apr 10, 2012 at 13:08 | comment | added | Igor Khavkine | Doing some reading, I've found that there are characterizations of some classes of PDOs due to Beals and also Bony. I don't know if there is a similar characterization of FIOs; it would be interesting to learn just that. Now, supposing I knew that $G(x,y)$ is a PDO, how do I recover the principal symbol? Is it enough to write $G(x,y)=H(x,x-y)$ and the the Fourier transform of $H(x,y)$ in the second argument? How would I get the symbol if I knew that $G(x,y)$ is a FIO? After the principal symbol is known, what about lower order terms? | |
| Apr 10, 2012 at 12:36 | comment | added | Deane Yang | It's reasonable to talk about a right inverse of a differential operator without specifying any boundary conditions. As for characterizing such an operator, I don't see any easy way to do so. First, the inverse operator is not even necessarily a pseudodifferential operator. If it is, then a necessary condition is that the principal symbol is the reciprocal of a polynomial (i.e., the symbol of a differential operator). But things get more complicated with the right inverse of a hyperbolic PDE, which is a Fourier integral and not a pseudodifferential operator. | |
| Apr 10, 2012 at 12:16 | history | edited | Willie Wong |
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| Apr 3, 2012 at 9:53 | comment | added | Igor Khavkine | Certainly. However, I left the issue of boundary conditions and other restrictions on the domain of the operators (which are needed to define an inverse in the presence of such "zero modes") undiscussed because I think it might actually distract from the main thrust of the question. However, if fixing such conditions is crucial, I'm happy to restrict to the case of hyperbolic operators with retarded or advanced boundary conditions, or equally well to elliptic operators without zero modes. | |
| Apr 3, 2012 at 8:41 | comment | added | Liviu Nicolaescu | You have to be more precise when you speak about the "inverse" of a partial differential operator. Take for example the Laplacian on a Riemann manifold. It has a nontrivial kernel consisting of constant functions. When the manifold is the Euclidean spane $\mathbb{R}^n$, do you consider this operator invertible? | |
| Apr 3, 2012 at 0:38 | history | asked | Igor Khavkine | CC BY-SA 3.0 |