Timeline for A question of uniqueness
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2020 at 14:20 | vote | accept | Paul | ||
| Aug 4, 2020 at 13:56 | comment | added | Willie Wong | @cerise: I had taken your convergence $\lim_{y\to\infty} u(x,y) = 0$ to mean convergence with respect to some reasonable function norm (say $L^\infty$). If you only wished to have pointwise convergence and no more, then as Eremenko said there are counterexamples. | |
| Aug 4, 2020 at 6:09 | history | edited | Paul | CC BY-SA 4.0 |
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| Aug 4, 2020 at 4:57 | history | edited | Paul | CC BY-SA 4.0 |
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| Aug 4, 2020 at 4:51 | history | edited | Paul | CC BY-SA 4.0 |
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| Aug 4, 2020 at 4:19 | comment | added | Paul | @Willie Wong I tried to apply the maximum principle with your indication with $ \Omega_t $ but without success. Can you give your answer because I do not see why it is easy and especially that A.Ermenko says that there is a counter example. | |
| Aug 4, 2020 at 3:13 | answer | added | Alexandre Eremenko | timeline score: 3 | |
| Aug 3, 2020 at 18:02 | comment | added | Willie Wong | Let $\Omega_t = \Omega \cap \{ y < t\}$. Apply the maximum principle for $u$ on $\Omega_t$. Take $t\to \infty$. | |
| Aug 3, 2020 at 17:25 | comment | added | Paul | @Willie Wong the maximum principale is available if $\Omega$ a bounded domain theoreme 2.17 math.ucdavis.edu/~hunter/pdes/ch2.pdf | |
| Aug 3, 2020 at 15:25 | review | Close votes | |||
| Aug 10, 2020 at 3:03 | |||||
| Aug 3, 2020 at 15:10 | comment | added | Willie Wong | Anyway, for the method you choose: $u(x,R) \frac{du}{dy}(x,R) = \frac12 \frac{d}{dy} |u(x,y)|^2 \Big]_{y = R}$. If the lim of $u(x,y)$ tends to zero, there exists some $R$ for which the integral of this derivative is negative. | |
| Aug 3, 2020 at 15:07 | comment | added | Willie Wong | Isn't it easier to just use maximum principle? | |
| Aug 3, 2020 at 14:42 | history | asked | Paul | CC BY-SA 4.0 |