6
$\begingroup$

In this post we would like to give a possible new approach to the second part of the Hilbert 16th problem

First we give a short introduction:

A quadratic system is a polynomial vector field on the plane in the form $$(X_{\alpha})\;\;\;\;\;\;\;\;\;\;\begin{cases}\dot x=P_{\alpha}(x,y)\\ \dot y=Q_{\alpha}(x,y) \end{cases}$$

where $P_{\alpha},Q_{\alpha}$ are polynomials of degree $2$ which are parametrized by a $10$ tuple parameter $\alpha$ of coefficients.

A center is a singularity of this vector field which is surrounded by a band of closed orbits.

All quadratic vector fields with center are classified as follows:

They correspond to a finite number of algebraic conditions in $\alpha$, see "Integrability of plane quadratic vector fields" Expos. Math(1990)3-25..

We denote these algebraic conditions by $Cent(\alpha)=0$

Question:

Are there a family of (polynomial) Riemannian metrics $g_{\alpha}$ on the punctured plane(after removing singularities) with Gaussian curvature $\kappa_{\alpha}$ with the following properties:

The solutions of $X_{\alpha}$, after a possible reparametrization, are geodesics of $g_{\alpha}$.Moreover $\kappa_{\alpha}$, is not zero except at a finite number of algebraic curves transverse to $X_{\alpha}$. Moreover $\kappa_{\alpha}$ is identically zero if $Cent(\alpha)=0$? The question, in particular its last part is well behaved and consistent since a quadratic system can not have a center and a limit cycle, simultaneously.

If the answer would be yes, then $H(2)$, the maximum number of limit cycles of a quadratic system, would be finite.

This question is already discussed at the comment-conversation of the following post:

Limit cycles as closed geodesics(in negatively curved space)

$\endgroup$

0

You must log in to answer this question.