Timeline for Allocation of $\mathbb R^2$
Current License: CC BY-SA 4.0
21 events
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| Jul 14, 2023 at 20:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 14, 2023 at 19:37 | comment | added | Fawen90 | @ChristianRemling Inspired by your comments, it seems that $n=2$ is a counter-example. I'm just not sure for this case whether $\tau_{x^*}=\infty$ or not, where $x^*=:(z_1+z_2)/2$ | |
| Jun 14, 2023 at 19:35 | answer | added | Fawen90 | timeline score: 0 | |
| Jun 14, 2023 at 18:24 | comment | added | Fawen90 | @ChristianRemling I think you definitively provide a way for counter-examples, while the difficulty here is how to choose such $n$ and $z_1,\ldots, z_n$.... | |
| Jun 14, 2023 at 17:33 | comment | added | Christian Remling | I think this is mainly a matter of plotting some typical $F$'s. A tool to establish existence of periodic orbits is Poincare-Bendixson, though that's of course the opposite of what you want here. | |
| Jun 14, 2023 at 17:15 | comment | added | Fawen90 | @ChristianRemling Thanks again Christian. Do you have promising arguments to eliminate these two scenarios? Or do you have an example where there exists a non-neglieable set of points $x$ such that $Y_x(t)$ does not converges to $z_i$ | |
| Jun 14, 2023 at 16:50 | comment | added | Christian Remling | If $F(x)\not= 0$, it could still be that $Y_x\to z_j$, but there are dangers such as attracting equilibrium points (maybe not likely) or periodic orbits. | |
| Jun 14, 2023 at 16:49 | comment | added | Christian Remling | Yes, if $F(x)=0$, then the solution with this initial value $x$ is $Y_x(t)=x$, so doesn't converge to one of the $z_j$. | |
| Jun 14, 2023 at 15:12 | comment | added | Fawen90 | @ChristianRemling I've made some edits to clarify that I pursue the convergence for almost every $x\in\mathbb R^2$. Do you think this may make my claim true? | |
| Jun 14, 2023 at 15:09 | history | edited | Fawen90 | CC BY-SA 4.0 |
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| Jun 14, 2023 at 15:02 | history | edited | Fawen90 | CC BY-SA 4.0 |
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| Jun 14, 2023 at 14:55 | comment | added | Fawen90 | @ChristianRemling Thanks for the answer. If my understanding is correct, $Y(t)$ will only converge to the equilibrium points $y$, i.e. $F(y)=0$. Is this right? Does this convergence hold for all initial points $x$ (probably different to $z_i$)? | |
| Jun 14, 2023 at 14:44 | comment | added | Christian Remling | It's false for $n\ge 2$ because there are equilibrium points (for example, if $n=2$, then $F((z_1+z_2)/2)=0$). | |
| Jun 14, 2023 at 14:32 | history | edited | Fawen90 | CC BY-SA 4.0 |
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| Jun 14, 2023 at 14:17 | comment | added | Fawen90 | @ChristianRemling Absolutely right. Thanks for pointing out the typo | |
| Jun 14, 2023 at 14:17 | history | edited | Fawen90 | CC BY-SA 4.0 |
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| Jun 14, 2023 at 8:15 | history | asked | Fawen90 | CC BY-SA 4.0 |