1
$\begingroup$

Using Dulac's criterion one can prove that a system has no closed orbits. Does it have any applicability for the global stability of a system?

Please help me to understand the relation between a system having no closed orbits versus the global stability of the system. Which one implies the other or are they equivalent? I am a beginner of this topic. Thanks in advance.

$\endgroup$
1
$\begingroup$

It is easy to give examples of equations without periodic trajectories for which all solutions are stable or for which some solutions are not stable.

The only thing that can be added is that the existence of at least one periodic trajectory precludes that all trajectories are stable (we are always talking about equations on the plane).

$\endgroup$
  • $\begingroup$ 'It is easy to give examples of equations without periodic trajectories for which all solutions are stable or for which some solutions are not stable.' Can you provide any one such example? $\endgroup$ – Germain Sep 18 '17 at 18:02
  • $\begingroup$ A globally stable system has no closed orbits, mentioned in Strogatz.So for the other way i,e, non existence of a closed orbit need not imply a system to be globally stable. If you can provide some easy example for this other way, it will be a great help for me to understand the theory. $\endgroup$ – Germain Sep 18 '17 at 18:05
  • $\begingroup$ Let me note that I wrote: "the existence of at least one periodic trajectory precludes that all trajectories are stable". My impression is that your notion of "globally stable system", refers only to the origin. Am I right? If so, then my statement is stronger. Otherwise, better adding the definition to your question. $\endgroup$ – John B Sep 18 '17 at 19:11
  • $\begingroup$ By globally stable what I understand is as follows: Any trajectory starting from different initial phase points will terminate at a certain point of equilibrium(not necessarily origin) of the system. $\endgroup$ – Germain Sep 19 '17 at 3:33
  • 1
    $\begingroup$ Counterexample: $x'=x$. $\endgroup$ – John B Sep 19 '17 at 7:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.