Timeline for Fixed points which are not locally attractive can have distant basins of attraction?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 10, 2011 at 7:31 | vote | accept | Navcar | ||
| Oct 10, 2011 at 7:31 | |||||
| Oct 10, 2011 at 7:31 | vote | accept | Navcar | ||
| Oct 10, 2011 at 7:31 | |||||
| Oct 9, 2011 at 19:26 | answer | added | Aaron Golden | timeline score: 1 | |
| Jul 26, 2011 at 14:22 | history | edited | Navcar | CC BY-SA 3.0 |
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| Jul 25, 2011 at 20:12 | comment | added | Robert Israel | You might look up "semi-stable equilibrium". Another interesting example is $$ \eqalign{x' &= x^2 - y^2\cr y' &= 2 x y\cr}$$ where the trajectories off the $x$ axis are circles tangent to the $x$ axis at the origin. The basin of attraction of the origin is the complement of the positive $x$ axis. | |
| Jul 25, 2011 at 15:03 | history | edited | Navcar | CC BY-SA 3.0 |
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| Jul 25, 2011 at 11:23 | comment | added | Navcar | Do you know any reference in which this specific issue is discussed? | |
| Jul 25, 2011 at 11:11 | comment | added | Navcar | No, my question is related to your example. | |
| Jul 23, 2011 at 16:57 | answer | added | Anthony Quas | timeline score: 6 | |
| Jul 23, 2011 at 12:13 | comment | added | Martin M. W. | Under one interpretation of your terms, here's an example. The flow defined by $x' = x^2$ has a non-locally-attracting fixed point at 0, but any open set of negative numbers is attracted to it. But perhaps you mean something else? | |
| Jul 23, 2011 at 11:30 | history | asked | Navcar | CC BY-SA 3.0 |