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In this question, we are interested in the number of limit cycles which appears in the following perturbational system:

\begin{equation}\cases{ x'=y -x^{2}+\epsilon P(x,y) \\ y'=-x+\epsilon Q(x,y) } \end{equation} where $P$ and $Q$ are polynomials of degree $n$. The unperturbed system (i.e, $\epsilon=0$) is not a Hamiltonian vector field, so we multiply the above vector field by the integrating factor
"$e^{-2y}$". After this multiplication, the (new) unperturbed system is a Hamiltonian vector field with Hamiltonian \begin{equation} H(x,y)=e^{-2y}(y-x^{2}+1/2) \end{equation}

The unperturbed system has a unique singularity at the origin, which is of center type. The region of closed orbits surrounding the center is $\{(x,y)|0\leq H(x,y) \leq 1/2\}$. According to the theory of Abelian integrals and its relation to the second part of the Hilbert 16th problem, the number of limit cycles of the perturbed system is (equal to and ) closely related to the number of zeroes of the following integral function $I:[0,\;1/2]\to \mathbb{R}$:

\begin{equation} I(c)=\int_{H^{-1}(c)} e^{-2y}(Pdy-Qdx) \end{equation}

I have been motivated by a famous result of A. Varchenko about the finiteness of the number of the zeros of abelian integrals of polynomial perturbation of polynomial Hamiltonians (which is proved in "Arnold Gusein Zadeh Varchenko, Singularities of Differentiable maps Vol II). Hence, I have the following two questions:

Questions:

  1. Assume that $P$ and $Q$ are fixed (given) polynomials. Is it true that either $I$ is identically zero, or it has only a finite number of zeros? Is there an explicit formula for the function $I(c)$? Or at least can we compute $\lim_{c \to 0} I(c)$?
  1. If the answer to the above question is affirmative, can we control the number of zeroes of $I$, in terms of the degree of polynomials $P,Q$?

Note 1: The proof of Varchenko is essentially based on algebraic geometry. In fact, they consider the polynomial perturbation of polynomial Hamiltonian systems. (Please see page 5, section 3.5 of this paper. But in this particular question, the non-algebraic integrating factor $e^{-2y}$ destroys the algebro-geometric feature of the problem. So how can one remedy this problem?

Note 2: As a possible resolution to avoid the non-algebraic term $e^{-2y}$, we consider the following approach, which can be applied for every algebraic perturbation of algebraic vector fields, where the unperturbed system has a band of closed orbit (center). But note that it is not a Hamiltonian vector field, and we have a non-algebraic integrating factor.

Consider the polynomial vector field $$\begin{cases} x'=P+\epsilon A\\ y'=Q+\epsilon B \end{cases}$$

where the unperturbed system has a band of closed orbit. We assume that there is a straight line $\ell$ parametrized by a real parameter $h\in \mathbb{R}$ with the following property: Periodic orbits of the unperturbed system intersect $\ell$ transversally and $s'(h) \neq 0$ for all $h$ where $s(h)$ is the slope of the unperturbed vector field at point $h$. This is opposite to the concept of "isocline". For example, this is the case for the Liénard vector field \begin{equation}\cases{ x'=y -x^{2}+\epsilon P(x,y) \\ y'=-x+\epsilon Q(x,y) } \end{equation}

when $\ell$ is the $x$ axis. But it is not the case for the isocline $y$ axis.

We take a point $h \in \ell$. By $p(h)$, we mean the value of the Poincaré return map $p$ at $h$. Hence $p(h) \in \ell$ is the first return map for the orbit starting at $h \in \ell$. WLOG we may assume that the orientation of the solution curve is anti-clockwise. Consider the simple closed curve $\gamma$ consisting of the solution curve $h \to p(h)$ and the part of the straight line $p(h) \to h$. If $p(h) \neq h$, then the integral of $\int_{\gamma} \kappa_{g} \neq 2\pi$ , since two different points of the line $\ell$ has different slopes. (This is a consequence of the Gauss–Bonnet theorem). So for $\epsilon$ sufficiently small, the above curvature integral is $2 \pi$ if and only if $p(h)=h$.

Now we compute the integral of the curvature:

$$T(h)=\int_{\tilde{\gamma}}(\frac{(P+\epsilon A)(Q_{x}+\epsilon B_{x})+(Q+\epsilon B)(Q_{y}+\epsilon B_{y})}{P^2+Q^2})dx -(\frac{(Q+\epsilon B)(P_{y}+\epsilon A_{y})+(P+\epsilon A)(P_{x}+\epsilon A_{x})}{P^2+Q^2})dy$$

where $\tilde{\gamma}$ is the same as $\gamma$, but we remove the straight line $p(h) \to h$. (Similar to the orbit in the following picture which starts from $r_{0}$ and ends to the $p(r_{0})$.

https://www.google.com/search?q=poincare+map+planar++vector++field&source=lnms&tbm=isch&sa=X&ved=0ahUKEwis8LKS5OTTAhWQZlAKHQCQAAoQ_AUIBigB&biw=1280&bih=933#imgrc=fRd5Ci4aSvlh_M:

The above integral is equal to $2\pi +\epsilon c(h) + O(\epsilon^2)$

where $c(h)$ is the following:

$$c(\alpha)= \int_{\alpha} (\frac{(AQ_{x}+PB_{x}+BQ_{y}+QB_{y})(P^2+Q^2)+2AP+2BQ}{{(P^2+Q^2)}^2})dx-\int_{\alpha}(\frac{(BP_{y}+QA_{y}+AP_{x}+PA_{x})(P^2+Q^2)+2AP+2BQ}{{(P^2+Q^2)}^2})dy$$

where $\alpha $ is the closed orbit of the unperturbed system starting at $h$.

Now in the integral $c(h)$, we do not have any non-algebraic term.

So this would possibly imply that the zero set of $c(h)$ is equal to the zero set of $I(h)$ where $I(h) $ is the standard abelian integral $I(h)= \int_{\alpha(h)} Ady-Bdx$. Because both zero sets are corresponding points of generating limit cycles.

Is the latter statement true? Is the computation of $c(h)$ correct? Does this situation make facilities for computations of abelian integrals to count the number of generating limit cycles?

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  • $\begingroup$ (I hope you don't mind my attempt to illustrate $H(x,y)$...) $\endgroup$ – Joseph O'Rourke Nov 16 '14 at 2:20
  • $\begingroup$ @JosephO'Rourke Prof. O'Rourke thank you very much for your illustration. But I think a familly of closed curve shoud suround the origion(0,0), can I ask you to replace it with a new one? Thank you $\endgroup$ – Ali Taghavi Nov 16 '14 at 4:09
  • $\begingroup$ please "google image" "Phase portrait center". $\endgroup$ – Ali Taghavi Nov 16 '14 at 4:55
  • $\begingroup$ I removed my picture of the $xy$-region $0 \le H(x,y) \le \frac{1}{2}$. I should not meddle so far from my expertise. $\endgroup$ – Joseph O'Rourke Nov 16 '14 at 14:52
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    $\begingroup$ There's a typo before the Google link (maybe the link should be formatted too) $\endgroup$ – YCor Nov 26 '20 at 22:31
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This paper contains a conjecture and a partial result about the abelian integral under discussion :"Computer-assisted techniques for the verification of the Chebyshev property of Abelian integrals"

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    $\begingroup$ Correct me if I am wrong, but is this more of a comment and not an answer as your questions listed above are still unanswered? $\endgroup$ – Gabriel Fair Jan 27 at 0:32
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    $\begingroup$ @GabrielFair Thank you for your comment. The Chebyshev property discussed in the linked paper is a direct partial result (in low degree) to my question. On the other hand you are right it is not a complet answer to my question which asks about uniform upper bound $I(n)$ for the number of zeros of abelian integrals associated to an arbitrary polynomial perturbation of $(y-x^2)\partial_x-x\partial_y$. To be honnest I am interested in this question since 20 years ago. $\endgroup$ – Ali Taghavi Feb 1 at 19:23

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