Timeline for A harmonic function
Current License: CC BY-SA 4.0
31 events
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Nov 26, 2019 at 22:31 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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Nov 26, 2019 at 15:25 | comment | added | Alexandre Eremenko | Various inequalities can be written on the whole $(0,1)$ using the power series expansion. | |
S Nov 26, 2019 at 15:24 | history | suggested | user135093 | CC BY-SA 4.0 |
Fix some typo in the previous edit
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Nov 26, 2019 at 13:14 | review | Suggested edits | |||
S Nov 26, 2019 at 15:24 | |||||
Nov 26, 2019 at 8:51 | comment | added | user135093 | Thank you, but I think this gives only a local inequality (near $0$ and $1$) and not on the whole intervals $(0,1)$. | |
Nov 25, 2019 at 14:38 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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Nov 25, 2019 at 14:18 | comment | added | Alexandre Eremenko | In my previous remark it must be $Ct^{\pi/2\alpha}$: I was confused with your picture, what do you call alpha. | |
Nov 25, 2019 at 14:15 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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Nov 25, 2019 at 10:30 | comment | added | user135093 | this is exactly what I expected. How can I prove that? | |
Nov 25, 2019 at 4:51 | comment | added | Alexandre Eremenko | It is $Ct^{\pi/\alpha}$, but $C$ has to be found numerically (or by expanding the integral into a power series. | |
Nov 24, 2019 at 18:24 | comment | added | user135093 | Okey. In fact, I want to get an estimate such $f(t)\ge Ct$ or $C t^c$ in the interval $(0, 1)$ (for some constant $C, c>0$) , or something similar, where $f$ is the underlying harmonic function. Thanks in advance. | |
Nov 24, 2019 at 17:55 | comment | added | Alexandre Eremenko | If you state exactly what sort of estimate you need, I may think how to obtain it. | |
Nov 24, 2019 at 17:53 | comment | added | Alexandre Eremenko | I do not have the book. Probably I know this principle for harmonic functions but did not know that it is "Giraud". | |
Nov 24, 2019 at 17:48 | comment | added | user135093 | It's maximum principle for elliptic equation. It's in Isakov book "Inverse problems for partial differential equations" (2017), pp 107. He used it to find lower bound of harmonic functions. See pp 57. | |
Nov 24, 2019 at 17:42 | comment | added | Alexandre Eremenko | Not by name. What is this? | |
Nov 24, 2019 at 17:23 | comment | added | user135093 | Hi @Alexandre do you know Giraud's maximum principle? | |
Nov 22, 2019 at 1:45 | comment | added | Alexandre Eremenko | This was a misprint. I corrected. | |
Nov 22, 2019 at 1:44 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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Nov 21, 2019 at 21:51 | comment | added | user135093 | you wrote $(0,\pi,\infty)\mapsto (-1,1,\infty)$ | |
Nov 21, 2019 at 14:19 | comment | added | Alexandre Eremenko | @Do Iii: function $G$ seems fine: it is the conformal map of the upper half-strip onto the upper half-plane with $(0,1,\infty)\mapsto (0,1,\infty)$. | |
Nov 21, 2019 at 12:40 | comment | added | user135093 | I think something missed in the function $G$. | |
Nov 20, 2019 at 22:40 | vote | accept | CommunityBot | ||
Nov 20, 2019 at 17:11 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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Nov 20, 2019 at 16:50 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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Nov 20, 2019 at 13:12 | comment | added | Alexandre Eremenko | If $F$ maps the straight strip onto the broken strip then the solution of Dirichlet problem is $\Re F^{-1}(z)/T$, the inverse function! | |
Nov 20, 2019 at 1:17 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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Nov 20, 2019 at 1:14 | comment | added | user135093 | Thank you. I will try again. | |
Nov 20, 2019 at 1:10 | comment | added | Alexandre Eremenko | You should map upper half of the straight strip onto the upper half-plane (by an elementary function), and then the upper half-plane to the broken strip (by that integral). The boundary correspondence should be $0\to 0,\; T\to T,\;\infty\to\infty$. By reflection this map extends to a map between the whole strips. | |
Nov 20, 2019 at 1:06 | comment | added | user135093 | Should I map straight strip -> upper half plane -> upper half of the broken strip? I thought that I can go directly. | |
Nov 19, 2019 at 19:15 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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Nov 19, 2019 at 19:07 | history | answered | Alexandre Eremenko | CC BY-SA 4.0 |