Timeline for Heating a long cylinder: steady states
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2022 at 14:54 | comment | added | Leo Moos | Thanks for the clarifications, they were very helpful! | |
| Oct 2, 2022 at 19:44 | comment | added | Mateusz Kwaśnicki | [...] (3) I was thinking about general boundary data, but of course periodic are fine, too, save for uniqueness. (4) The normal derivative is not really a derivative, but I agree this step looks suspicious at first glance. I added some details. | |
| Oct 2, 2022 at 19:43 | comment | added | Mateusz Kwaśnicki | I expanded the answer a bit to address your comments. Roughtly: (1) Yes, the Green function is always the time integral of the heat kernel in dimensions three and above; and the same is true in dimensions one and two, as long as the complement of your domain is non-polar. In fact, this is true for much more general operators than the Laplace operator. (2) The heat kernel is defined on $\Omega \times \Omega$, too! (In fact: on $(0 \infty) \times \Omega \times \Omega$ if you include time as a variable.) [...] | |
| Oct 2, 2022 at 19:39 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
added 3148 characters in body
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| Oct 2, 2022 at 17:18 | vote | accept | Leo Moos | ||
| Oct 2, 2022 at 17:18 | comment | added | Leo Moos | [...] Two other, smaller points. When you speak of 'the' solution of the Poisson problem in $D \times \mathbf{R}$, you mean the unique periodic solution, right? Because there should be other, non-periodic solutions, no? The last point: when you use the bounds (above and below) for the Green's function to get an estimate for the Poisson kernel, are you skipping over some details? It looks a bit like differentiating an inequality. | |
| Oct 2, 2022 at 17:14 | comment | added | Leo Moos | Sorry it took me so long to get back to you. There are a couple of things I was hoping you could clarify. The main one is the expression of the Green's function as an integral of the heat kernel. Is this always true, maybe provided the integral is convergent? Perhaps on every regular domain $\Omega$ - here the cylinder? Is there a heuristic explaining it? It seemed a bit weird to me because I thought the Green's function ought to be defined on $\Omega \times \Omega$, but the heat kernel on $\Omega \times \partial \Omega$. [...] | |
| Sep 27, 2022 at 6:40 | comment | added | Mateusz Kwaśnicki | Sure, of course! | |
| Sep 26, 2022 at 23:19 | comment | added | Leo Moos | Thanks for taking the time to write up your answer! I may have some questions once I've managed to read through it - I hope that's OK. | |
| Sep 26, 2022 at 22:59 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |