Timeline for When is a limit cycle generated by a Hamiltonian oval stable?
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Jun 28, 2016 at 0:37 | comment | added | Futurologist | In terms of literature, I am not too sure where exactly one can see the proof, although I have seen $I(t)$ used in all articles related to the topic of infinitesimal 16th Hilbert problem. But maybe in the books written by some of the experts in the field? | |
Jun 28, 2016 at 0:32 | comment | added | Futurologist | Yes, it is exactly as you say: for small enough $\varepsilon >0$ the limit cycle is stable if $I'(t_0)>0$ and unstable if $I'(t_0)<0$. The function $I(t)$ conveniently encodes a lot of information about the qualitative behavior of the dynamical system in the relevant neighborhood: existence of periodic orbits as well as stability of the latter. Otherwise, to hope for semi-stability, assuming the dynamical system is at least real analytic, or even polynomial, one probably needs to have $I(t_0)=0, \, I'(t_0) = 0$ but $I''(t_0) \neq 0$. BTW, I've already fixed the $\ker$ typo. Thanks. | |
Jun 28, 2016 at 0:00 | history | edited | Futurologist | CC BY-SA 3.0 |
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Jun 24, 2016 at 13:44 | comment | added | hsp99 | Thank you for a very nice answer. If I understand correctly, the punchline is: Assume $\epsilon >0$. Then the limit cycle is stable if $I'(t_0)>0$ and unstable if $I'(t_0)<0$? This seems remarkably simple, but I'm left wondering if there is any notion of semi-stability? Moreover, do you have a good reference to confirm your proof of Pontryagin's criterion? I am a little unsure how much rigour is contained in the one you have provided. (and by $ker(a)=0$ do you just mean $ker(a)$?) | |
Jun 24, 2016 at 4:48 | history | answered | Futurologist | CC BY-SA 3.0 |