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Edit 1: For a related discussion see this MSE post

I apologize in advance, if this question is obvious:

1)What is an example of a polynomial vector field on $\mathbb{R}^{2}$ with at least two limit cycles $\gamma_{1},\; \gamma_{2}$ such that they lie on the same leaf of the corresponding singular holomorphic foliation of $\mathbb{C}^{2}$?

2)I have the same question by replacing "they lie on the same leaf" with "they lie on different leaves"

An indirect (or may be direct) motivation for this question is the following lemma in a celebrated paper of Petrovski and Landis "On the number of limit cycles of the equation ${dy\over dx}={P(x, y)\over Q(x, y)}$, where $P$ and $Q$ are polynomials of 2nd degree" :

"Lemma: If two distinct limit cycles of a real polynomial vector fields lie on the same complex leaf and they are homolog to each other, then the leaf is an algebraic curve"

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    $\begingroup$ Meta post: meta.mathoverflow.net/questions/2349/down-vote-without-comment $\endgroup$
    – Todd Trimble
    Commented Jul 10, 2015 at 12:40
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    $\begingroup$ Dear Ali, have you tried to write down an example? Can you mention what you have tried so far? I particularly think about the article by Llibre and Rodriguez «Configurations of limit cycles and planar polynomial vector fields» where you encounter explicit examples of planar polynomial vector fields with prescribed topological configuration of limit cycles. I'm sure you can answer one of the two questions you raise here by following the work in the article. But you must mention the effort you provided with respect to your problem! $\endgroup$
    – Loïc Teyssier
    Commented Jul 10, 2015 at 15:12
  • $\begingroup$ @LoïcTeyssier Thanks for the comment. Could you please send a PDF link of that paper. As I wrote in the question my main motivation for this question was the Lemma of petrovski Landis which used Chow theorem. I was interested in such type of questions since many years ago, as I wrote at the last page of this paper arxiv.org/abs/math/0409594. $\endgroup$
    – Ali Taghavi
    Commented Jul 10, 2015 at 15:35
  • $\begingroup$ But I was not aware of any method to approach such questions.This is my main reason of asking such question. $\endgroup$
    – Ali Taghavi
    Commented Jul 10, 2015 at 15:36
  • $\begingroup$ @LoïcTeyssier Dear Loic Thanks again for your helpful comment. Is there a real polynomial $H(x,y)$ with a regular connected complex level set $L$ which intersection with real plane has at least 2 disjoint closed curve? $\endgroup$
    – Ali Taghavi
    Commented Jul 10, 2015 at 16:37

1 Answer 1

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Possibly a very "brute force" approach could be the following. Take an non-singular algebraic curve $H(x,y)=0$ given by a polynomial $H(x,y)$ with real coefficients that has at least two ovals in the affine real plane, and generate the one-form $dH + H \omega$, where $\omega = P(x,y)dx + Q(x,y)dy$ is a real polynomial one-form. The foliation generated by the kernel of the form $dH + H \omega = 0$ gives you the orbits of the vector field $$\dot{x}= \frac{\partial H}{\partial y}(x,y) + H(x,y) Q(x,y)$$ $$\dot{y}= - \frac{\partial H}{\partial x}(x,y) - H(x,y) P(x,y).$$ The ovals $H(x,y)=0$ are cycles of both the vector field and the form (whichever you prefer, although in the complex setting the form I think is the right object to work with). For instance, the first thing that comes to my mind is $$H(x,y) = (x^2+y^2 - 1)(x^2+2y^2-4) - 1.$$ If I am not mistaken, it is a regular curve even at infinity (it is one of these "ultra-Morse" polynomials?!). It has two nested real ovals and the rest of it is in the complex plane. I think it is a non-singular curve. If we take $\omega=y \, dx$ wouldn't that work? I haven't check whether the cycles are limit, but chances are that they are, because the form is not closed. If not, one can try a different $\omega$...

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  • $\begingroup$ Thank you for the answer, Is there a particular reason that you did not consider for example $(x^{2}+y^{2}-1)(x^{2}+y^{2}-4)-1$? Moreover what are two ovals precisely? $\endgroup$
    – Ali Taghavi
    Commented Jul 11, 2015 at 21:40
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    $\begingroup$ Yes, I think $(x^2+y^2−1)(x^2+y^2−4)−1$ has a singularity at infinity, so as a compactified Riemann surface it is singular. First take $y \mapsto iy$ to rotate the points at infinity in the real projective plane and then perform the projective transformation $x \mapsto 1/x$ and $y \mapsto y/x$. Then you see two ovals touching at two points. In my example this does not occur because it is ultra-Morse: its leading highest order term is of the form $(x-a_1y)...(x-a_ny) + \text{lower order terms}$ which makes it non-singular at infinity for different $a_j$. $\endgroup$
    – Futurologist
    Commented Jul 11, 2015 at 21:57

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