The second part of Hilbert's 16th problem not only concerns "The number of limit cycles of a polynomial vector field", but also the position and configuration of of those limit cycles with respect to each other. However the primary and the usual interpretation of the concept "Relative Position" is that "Whether two given limit cycles are nested closed curves or not" but the following post is a motivation to consider other types of questions about configuration of limit cycles, for example "Can two nested limit cycles have opposite orientation?"

A cubic system with two nested limit cycles with opposite orientations

In this regard we focus on cubic polynomial system and ask the following question:

What is an example of Cubic polynomial vector field $$\begin{cases} x'=P(x,y)\\ y'=Q(x,y) \end{cases}$$ such that two nested closed orbits $C_1$, $C_2$ of the system surrounds an annular region $R$ such that $R$ does not contain any singular point and the flow orientation of $C_1$ is opposite to the flow-orientation of $C_2$?

There is a very interesting example of degree $5$ here but we search for degree $3$

  • $\begingroup$ Is not in the example of degree 5 that you link to the origin a singular point? $\endgroup$ – მამუკა ჯიბლაძე Jul 30 '17 at 10:04
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    $\begingroup$ @მამუკაჯიბლაძე Yes but it does not lie in the annular region between $C_1$ and $C_2$. And this is what I search for. Right? $\endgroup$ – Ali Taghavi Jul 30 '17 at 10:07
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    $\begingroup$ Sorry I did misread the condition, thanks. $\endgroup$ – მამუკა ჯიბლაძე Jul 30 '17 at 10:32
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    $\begingroup$ If $C_1$ and $C_2$ are concentric circles, $\dot x x+\dot y y=P(x,y)x+Q(x,y)y$ must vanish identically on $C_1$ and $C_2$, and on the origin. Doesn't this tell us that either P or Q has degree larger than 3? $\endgroup$ – Pietro Majer Jan 6 '18 at 22:42
  • $\begingroup$ @PietroMajer Thanks for your comment. Yes that is true. But in my question we do not assume that our limit cycles are algebraic curves(eg. circles). In fact a generic polynomial vector field does not possess alfebraic solution. $\endgroup$ – Ali Taghavi Jan 6 '18 at 23:31

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