Timeline for Existence of Green's functions for PDEs
Current License: CC BY-SA 3.0
7 events
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| Dec 8, 2014 at 16:28 | comment | added | Igor Khavkine | Differential operators are usually constrained to be of finite (or at least locally finite order). Otherwise, the line gets blurred, e.g., you could consider the translation $f(x) \mapsto f(x+y)$ as a differential operator of infinite order as well. | |
| Dec 7, 2014 at 9:47 | comment | added | Lars Lau Raket | Dear @PeterMichor. Thank you for your answer. Excuse my lack of knowledge, but why is the infinite series expansion of differential operators that have a Gaussian Green's function not a linear differential operator? | |
| Dec 6, 2014 at 15:44 | history | edited | Lars Lau Raket | CC BY-SA 3.0 |
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| Dec 6, 2014 at 14:28 | comment | added | Peter Michor | For example, the Gaussian $G(x,y) = \exp(-|x-y|^2/\sigma^2)$ corresponds to a differential operator of infinite order. | |
| Dec 6, 2014 at 13:07 | comment | added | Igor Khavkine | I think the simple answer is No, simply because it's not so hard to find examples of integral kernels whose inverses are not differential operators. Also, see this earlier question. | |
| Dec 6, 2014 at 12:47 | review | First posts | |||
| Dec 6, 2014 at 13:02 | |||||
| Dec 6, 2014 at 12:42 | history | asked | Lars Lau Raket | CC BY-SA 3.0 |