I was doing some preparation for my exam, and I found this interesting exercise (text is in italian).

Let's consider the planar system \begin{cases} \dot{x} &= P(x,y) \\ \dot{y} &=Q(x,y) \end{cases} where $P,Q \in C^1(\mathbb{R}^2,\mathbb{R})$. Assume that every circle centered at the origin with integer radius is invariant.

Let's denote with $\Lambda^+(x,y)$ the limit set of the point $(x,y)$. I have to answer the following questions:

1) Prove that $\Lambda^+(x,y) \neq \emptyset$ for every $(x,y) \in \mathbb{R}^2$.

it's trivially true. If the point $(x,y)$ lies in one of the circle, it's entire orbit it's contained in it, and by compactness every sequences ($t_n \to \infty$) of points of the orbit admits a convergent subsequence. A similar argument works for the annuli between the circles. I'm confident with this answer

2) Determine for which points the set $\Lambda^+(x,y)$ can be computed explicitly.

If the point is the origin, the circumferencewith radius $0$ is invariant, so the point is fixed, so $\Lambda^+(0,0)=(0,0)$. I can't say anything for the other points, because the other cycles can be a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting the points. According to me, I can't say anything else

3) Give sufficient conditions s.t. the system doesn't have limit cycles.

If the the system is Hamiltonian, then it can't have any limit cycles. If the system is a gradient system, I can't have any periodic orbit,in particular any limit cycle. if the system satisfy the hypothesis of the Bendixson-Dulac theorem then it can't have any periodic solution. I don't know any other condition

and then I added (as a challenge:) ) this question

4) Study the properties of the orbits in the annuli. It's permitted to ad some hypothesis on the system.

If the only critical points lies in the cycles, and the system is Hamiltonian, then using the Poincaré-Bendixson thm I can conclude that every point belongs to a periodic solution. I don't know how to deal with the case that fixed point could exists in the annuli

So my questions are:

Are my thoughts right?

Is there a way to deal with fixed point inside the annuli? (I doubt that there can be fixed points, because of the Continue dependence from initial value. (I can't formalize very well this fact)

A few words about this last question: if we consider the system (non hamiltonian) in polar coordinates \begin{cases} \dot{r} &= r \sin(r) \\ \dot{\theta} &= -\cos( r) \end{cases} it has a phase portrait of this kind (modulo a reparametrization to have integer radii)


In which the orbits don't have any cycles inside the annuli. So if I relax the hypothesis on being Hamiltonian, it is not true that I have cycles inside the annuli. Anyway I can't find an example with fixed point inside.

NB Highlighted texts are my answers at the question.

  • $\begingroup$ This is probably just my misunderstanding of invariant in context, but does the first bolded phrase mean $P(x,y)=Q(x,y)=0$ when $x^2+y^2=k^2$, $k$ an integer, or just that any point on an integral radius will stay on that integral radius (i.e. $\dot{r}=0$, $r$ an integer)? $\endgroup$ – Mark Hurd Jun 11 '14 at 6:36
  • $\begingroup$ @MarkHurd the second one. Any solution in the circle, will remain in the circle both for negative and positive times. $\endgroup$ – Riccardo Jun 11 '14 at 7:31

I agree with 1 and 2. For 3, observe that we don't know anything about the system in the annulus $1<r<2$, except that the annulus is a completely invariant domain. Pretty much anything can happen within this domain, and the only way to rule out limit cycles is to invoke a general condition that rules them out. The condition about invariance of integer-radius circles is of no help.

That said, observe that it's difficult for such a system to be a gradient system: the integer-radius circles would have to consist entirely of stationary points. (Could happen, of course).

To 4, same comment as in 3 applies: anything could happen within an annulus, subject to its complete invariance. You can have any number of fixed points, periodic orbits and limit cycles inside. I offer a drawing in MS Paint:


  • $\begingroup$ thanks! yeah, my problem is that I'm lacking of example, so sometimes I don't have any idea of the possibilities! beautiful sketch! clearer than a thousand words! $\endgroup$ – Riccardo Jun 9 '14 at 19:41

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