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In this question we are interested in the number of limit cycles which appear 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 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}

Motivated by a famous result of A. Varchenko about the finitness 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 Differantiable maps Vol II), I have the following two questions:

Questions:

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

2) If the answer to the above question is affirmative, can we control the number of zeroes of $I$, in terms of 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 the situation of this 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 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 just opposite to the concept of "Isocline". For example this is the case for the lienard 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 Poincare 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 clock wise. Consider the simple closed curve $\gamma $ consist 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 the integral of $\int_{\gamma} \kappa_{g} \neq 2\pi$ since two different points of the line $\ell$ has different slops. (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 start 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 later 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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