Given the following system

$\dot{x} = y$

$\dot{y} = y(9-x^2-2y^2) - x$

verify whether it has periodic solutions and if so are they attracting or repelling.

I thought:

The critical points or fixed point is (0,0) but is this correct and if the answer is yes then is that a periodic solution? And in general, how does one find out the periodic solution?

Finding fixed points is easy, you set $\dot{x}$ and $\dot{y}$ equal to zero and to verify the stability you find the derivatives of the fixed points. But I don't know how to find the periodic solutions...

Maybe rewriting the system in polar coordinates helps somehow?

  • $\begingroup$ Yes, $(0,0)$ is a stationary solution, in particular (trivially) periodic. Linearizing the system around $(0,0)$ shows that it is repelling. Writing it in polar coordinates you will get that $\dot{r}$ is positive for $0<x^2+2y^2<9$, and negative for $x^2+2y^2>9$. Since there are no other stationary points and two-dimensional systems only have limit cycles and stationary attractors, there must be at least one limit cycle. If there is just one, it has to be attracting, but I am not sure how to show this. $\endgroup$ Commented Nov 5, 2014 at 20:38
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    $\begingroup$ Maybe one should stress more clearly, as already indicated in a comment, that the accepted answer is incomplete. $\endgroup$
    – Did
    Commented Nov 6, 2014 at 20:29

1 Answer 1


In polar form we have $$r\dot{r} = x\dot{x} +y\dot{y} = r^2\sin^2\theta(9-r^2(1+\sin^2\theta)).$$ Now $$\dot{r} <0 \implies r^2 \ge\frac{9}{1+\sin^2\theta}\ge9$$ and $$\dot{r}>0 \implies r^2\le\frac{9}{1+\sin^2\theta} \le \frac{9}{2}$$ This tells us that orbits cannot leave the annulus $3/\sqrt{2}\le r\le3$.

By Poincare-Bendixson theorem, the limit set of the of orbits entering the annulus must be either a limit cycle, a fixed point, or some sort of homoclinic or heteroclinic connection. It is thus sufficient to show that there are no fixed points in this region.

Returning to the original equations; we find that the fixed point is at $(x,y)=(0,0)$. Which is not inside the annulus, hence the limit set must be a limit cycle, since the orbits can't leave the annulus, we know that the limit cycle must be stable. Hence, there is at least one stable limit cycle and it is inside the annulus.

Edit 1: Some comments

I don't have my Guckenheimer and Holmes text with me at the moment; but according to Scholarpedia, Poincare-Bendixson is sometimes stated as follows:

If a trajectory enters and does not leave a closed and bounded region of phase space which contains no equilibria, then the trajectory must approach a periodic orbit as $t\rightarrow \infty$.

So the points $\theta=n\pi$ are no problem as the trajectories remain in the closed annulus.

  • $\begingroup$ There is only one fixed point, at $(0,0)$, and you still need to rule out the case that there might be more than one limit cycle in the annulus. $\endgroup$ Commented Nov 6, 2014 at 0:38
  • $\begingroup$ Fixed; intuitively there is only one. To prove there is only one, i would recast the problem in normal form; which i may do later today. $\endgroup$
    – MrSlunk
    Commented Nov 6, 2014 at 1:00
  • $\begingroup$ I first thought one could use negative divergence to show that there is only one limit cycle, but my argument relied on a stupid algebra error. I don't see how the uniqueness of the limit cycle is intuitive, and I don't quite understand what kind of normal form you are trying to use. Numerically it looks pretty obvious that there is only one, that is true. $\endgroup$ Commented Nov 6, 2014 at 1:08
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    $\begingroup$ I think this solution gives the right general idea but is missing a couple of details which have bugged me since I first read this problem this morning. For instance, when $\theta = n\pi$ for integral $n$, $\dot r = 0$ always. So it takes more work to show that the flow heads into the annulus, and I don't know quite how to do it, and I don't know if the argument fails because $\dot r = 0$ when $\theta = n\pi$! Good start, though, +1! Cheers! $\endgroup$ Commented Nov 6, 2014 at 3:27
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    $\begingroup$ @RobertLewis I'm not quite sure, but I believe that Bony-Brezis theorem still can help here, even if we have a tangency at some points ( en.wikipedia.org/wiki/Bony-Brezis_theorem ) $\endgroup$
    – Evgeny
    Commented Nov 6, 2014 at 20:29

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