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Consider the ODE $$ \left\{ \begin{align*} \dot{x}=&y\\ \dot{y}=&-x^2-bx-c. \end{align*}\right. $$

Under the assumption that $b^2-4c>0$, we find the equilibria $P_1=\left(\frac{-b+\sqrt{b^2-4c}}{2},0\right)$ and $P_2=\left(\frac{-b-\sqrt{b^2-4c}}{2},0\right)$. Around $P_1$, the linearized system allow us to conclude that $P_1$ is a center and orbits around $P_1$ are periodic.

Using the first integral $F(x,y)=\frac{y^2}{2}+\frac{x^3}{3}+\frac{bx^2}{2}+cx$ around $P_1$, we can use Morse Lemma to prove that these periodic solutions also occur on the non linear system as well, and therefore, stable.

I want to prove, however, that these periodic solutions (for the non linear system) are Lyapunov stable. Looking at the system on polar coordinates gave me nothing. Can someone give me a hint (not an answer)?

As reference, this is from exercise 5.4 from Verhulst's "Nonlinear Differential Equations and Dynamical Systems".

Thanks in advance!

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    $\begingroup$ Hint: when the periods are different, clearly the periodic orbits are $\ldots$ $\endgroup$
    – John B
    Jun 13, 2021 at 18:44
  • $\begingroup$ @JohnB in the case of different period, they are NOT Lyapunov stable, right? $\endgroup$
    – Mand
    Jun 13, 2021 at 18:48
  • $\begingroup$ Precisely. We have seen too often people saying that they are. $\endgroup$
    – John B
    Jun 13, 2021 at 18:49
  • $\begingroup$ Good! I should try to look at this system in polar coordinates; didn't think it would work at first. $\endgroup$
    – Mand
    Jun 13, 2021 at 18:51
  • $\begingroup$ Many times you don't need that. For example, if some periodic orbits are close to homoclinic or heteroclinic orbits, necessarily their period approaches infinity there and so at least some periods will be different! $\endgroup$
    – John B
    Jun 13, 2021 at 18:54

2 Answers 2

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It's a bit hard to give a hint without giving away the answer, but here's an attempt:

The periodic trajectories form a family of closed curves around $P_1$ (at least if $b>0$, which perhaps is understood). Like this, for example: WA plot. So if you're circling around on one of those curves and make a small perturbation, where do you end up?

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  • $\begingroup$ On another circle? $\endgroup$
    – Mand
    Jun 13, 2021 at 18:38
  • $\begingroup$ Yes. So what can you conclude about stability? $\endgroup$ Jun 13, 2021 at 18:38
  • $\begingroup$ The circles are epsilon-close (in Lyapunov's sense)? But why does this argument does not hold, for example, on the mathematical pendulum $x''+\sin(x)=0$? $\endgroup$
    – Mand
    Jun 13, 2021 at 18:41
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    $\begingroup$ Oh, I see. We might be talking about different notions of stability. I'm referring to orbital stability: for every $\epsilon>0$ there's a $\delta>0$ such that if you perturb by at most $\delta$, you will stay for all future in an $\epsilon$-neighbourhood of the original trajectory. But in that situation, for a given time $t>0$, you might of course be at completely different points on the perturbed and the unperturbed trajectories, so it's not stable in that stronger sense. $\endgroup$ Jun 13, 2021 at 18:55
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    $\begingroup$ No, the other way around, it's more permissive, as the pendulum example shows: orbitally stable but not Lyapunov stable. Since your question (and Verhulst's book) explicitly refer to Lyapunov stability, but I thought about orbital stability nevertheless, this answer was perhaps not my best ever... But I'll leave it here anyway, since there is some info in the comments which may be of use to someone. $\endgroup$ Jun 13, 2021 at 19:01
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I don't know if I answered your question. Let $\lambda_{1,2}=\frac{-b\pm\sqrt{b^2-4c}}{2}$ and $u=x-\lambda_1,v=y$ and then the equations become $$ u'=v,v'=-u^2-\sqrt{b^2-4c}u. $$ Let $$ V=\frac13u^3+\frac12\sqrt{b^2-4c}u^2+\frac12v^2 $$ and then $\dot{V}=0$. So $V\equiv C$ is a trajectory for any $C$. For small $C>0$, $V\equiv C$ is a closed trajectory. So $P_1$ is not Liapunov stable.

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