Timeline for Weak maximum principle for a perturbation of the Laplacian
Current License: CC BY-SA 4.0
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Mar 6, 2020 at 13:15 | history | edited | an_ordinary_mathematician | CC BY-SA 4.0 |
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Mar 6, 2020 at 13:08 | history | edited | an_ordinary_mathematician | CC BY-SA 4.0 |
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Mar 6, 2020 at 12:57 | comment | added | Giorgio Metafune | Please, write to me [email protected] | |
Mar 6, 2020 at 12:52 | comment | added | an_ordinary_mathematician | @GiorgioMetafune Thanks a lot for the response. If you could send me some references for such operators would be great. | |
Mar 6, 2020 at 12:50 | comment | added | an_ordinary_mathematician | @MateuszKwaśnicki, Typo corrected, thanks. | |
Mar 6, 2020 at 12:49 | history | edited | an_ordinary_mathematician | CC BY-SA 4.0 |
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Mar 6, 2020 at 12:38 | comment | added | Giorgio Metafune | You need a "boundary condition" at 0 since the 1d operator D^2-1/r D (the generator of a Bessel process) has 0 as an exit boundary (see for example Chapter Vi, Section 4 of the book Engel-Nagel "One parameter semigroups..."). This explains what happens on radial functions; in the non radial case expansion in spherical harmonics shows that the radial case is the worst. In case you need I can send some references where these operators have been studied in detail, in Nd. | |
Mar 6, 2020 at 11:50 | comment | added | Mateusz Kwaśnicki | (I assume there is a typo, and you meant $\partial/\partial r$ rather than $\partial r / \partial r^2$.) If your domain does not touch the origin, you're good to go: even the strong maximum principle is satisfied, see Theorem 3.5 in Gibarg–Trudinger. On the other hand, if the domain contains $0$, then it is not immediately clear what one means by a solution. | |
Mar 6, 2020 at 11:25 | review | First posts | |||
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Mar 6, 2020 at 11:24 | history | asked | an_ordinary_mathematician | CC BY-SA 4.0 |