Timeline for Persistence of homoclinic points in the non-compact case
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 18, 2021 at 23:21 | comment | added | rpotrie | What I say is that all you need to control does not escape a compact set, so, the same proof applies. Say that $y$ is a transverse homoclinic point of a periodic point $p$. So, the forward orbit of $y$ belongs to a compact submanifold $S$ of the stable manifold of the orbit of $p$ (which is forward invariant) and the backward orbit to a compact submanifold $U$ of the unstable manifold of the orbit of $p$ (backward invariant). There is a neighborhood $V$ of $f$ in $C^1$ topology such that if $g \in V$ then $g$ and its derivative are close to $f$ in a nbhdd of $U \cup S$. | |
| Jun 18, 2021 at 10:55 | comment | added | Leon Staresinic | To be more specific, it seems implicit in your response that results for compact manifolds continue to work in the non-compact setting, as long as the property we are interested in (such as the continuity of a homoclinic point) "lives" in a compact set. Could you elaborate more on why this is so? | |
| Jun 18, 2021 at 9:40 | comment | added | Leon Staresinic | Could you please elaborate a bit on how the argument can be adapted to this setting? The above arguments seem to crucially depend on working with diffeomorphisms of compact manifolds. | |
| Jun 17, 2021 at 0:23 | comment | added | rpotrie | The periodic orbit and the orbit of some transverse homoclinic intersection belong to a compact set, so same argument works. | |
| Jun 9, 2021 at 9:42 | review | First posts | |||
| Jun 9, 2021 at 9:56 | |||||
| Jun 9, 2021 at 9:39 | history | asked | Leon Staresinic | CC BY-SA 4.0 |