Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to find an example of an autonomous differential equation on $\mathbb{R}^3$, which has an isolated limit cycle but no rest point. I come up with the following example: \begin{eqnarray*} \dot{x} & = & y\\ \dot{y} & = & -x\\ \dot{z} & = & 1-x^{2}-y^{2} \end{eqnarray*} Am I right? Is the phase portrait a sphere?

share|cite|improve this question
up vote 1 down vote accepted

It looks like at every point of the cylinder $x^2+y^2+1$ ($z$ arbitratry), the value of $\dot z$ is zero, which will cause there to be a cycle at every level of the cylinder. That is, starting at any point actually on the cylinder $x^2+y^2=1$ the vector field will cause the point to move around the circle at that level. So this is not an example where there is a single "isolated limit cycle".

But I think you're on the right track; the question is what to do about points on the cylinder to make the field go toward the origin there, say by using an extra factor pointing down or up the cylinder to force the turning cycles to all level out along the single circle $x^2+y^2=1,z=0$ [a single circle in the plane $z=0$].

Maybe if you changed to $\dot z = z^2(1-x^2-y^2)$...

Maybe better would be $\dot z=z^2+(1-x^2-y^2)^2$. This would only be zero if both $z=0$ and $x^2+y^2=1$, i.e. only on the desired single cycle of the unit circle on the plane $z=0$. At all other points it would cause the curves to move in the $z$ direction, like spirals.

share|cite|improve this answer
Sorry for the late acceptance. – user38335 Nov 20 '12 at 3:02
I think my last stab at adjusting $\dot z$ to $z^2+(1-x^2-y^2)^2$ might actually work, and give the single cycle at $z=0, x^2+y^2=1$. I'd be curious what this vector field looked like, but I don't have a good version of maple to do the vector field plot. – coffeemath Nov 20 '12 at 3:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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