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I have the following system: $$\begin{cases} x'&=&2(z-1)y,\\y'&=&-(z-1)x,\\z'&=&-z^3. \end{cases}$$ and want to study the stability of it at the fixed point $(0,0,0)$.

Doing the linearization of the system, it predicts a linear center. Then, we cannot guarantee that it is a non-linear center.

Then, I can think about the following options:

  • Find a Lyapunov's function (if it is asymptotically stable).
  • Check if the system s reversible by doing $t\mapsto -t$, $(x,y,z)\mapsto R(x,y,z)$ where $R^2=\text{Id}$, then we would have a non-linear center.
  • Find a conserved quantity, i.e.m a real-valued continuous function $E(x,y,z)$ such that it is constant on trajectories $\left(\dfrac{dE}{dt}=0\right)$.
  • If we can write $(x',y',z')=-\nabla V$ ($V$ potential function), where it would be impossible to find closed orbits.
  • Dulac's Criterion for ruling out closed orbits.
  • Poincaré-Bendixon to see if closed orbits exist.

I unsuccessfuly tried the first three items above. Then I proceeded to draw the system on Maple and seems like there will be centers, but I need to solve this analytically. Any hint to continue this problem?

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    $\begingroup$ $x^2+2y^2$ is a conserved quantity. $\endgroup$
    – Gonçalo
    Feb 4 at 0:00
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    $\begingroup$ Another way to see this is to notice that $z\to0$ then study the asymptotic linear system in $x$ and $y$ which is periodic. $\endgroup$
    – whpowell96
    Feb 4 at 1:16
  • $\begingroup$ Thank you very much @Gonçalo. Just checked it and it worked. However, if you don't mind, how did you come up with that conserved quantity? The functions one can try are similar to Lyapunov's or any "trick" to be close? $\endgroup$
    – Fabrizio G
    Feb 4 at 18:49
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    $\begingroup$ Dividing the first equation by the second, one obtains the ODE $\frac{dx}{dy}=-\frac{2y}{x}$. Its solution is $x^2+2y^2=C$. $\endgroup$
    – Gonçalo
    Feb 4 at 19:07
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    $\begingroup$ "opposite sign of $z$" that was the key! Thank you very much. $\endgroup$
    – Fabrizio G
    Feb 4 at 20:06

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