# A cubic system with two nested limit cycles with opposite orientations

What is an example of polynomial vector field $$\begin{cases} x'=P(x,y)\\ y'=Q(x,y) \end{cases}$$ such that two 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 $C_2$.

The motivation for this question is the following counterexample of a nongeodesible flow on the torus and the following post.

Note that this situation can not be occurred when the degree of $P,Q$ is at most $2$. See theorem $4$ of the following paper..So we search for a cubic (or higher degree ) system.

It is not hard to concoct such an example in sufficiently high degree. For an example of degree $5$, take \begin{align} x' &= x\,(1-x^2-y^2)(x^2+y^2-3) - y\,(2-x^2-y^2)\\ y' &= y\,(1-x^2-y^2)(x^2+y^2-3) + x\,(2-x^2-y^2). \end{align} The circles $x^2+y^2=1$ and $x^2+y^2=3$ are integral curves of this vector field with opposite orientations. The only zero of the vector field is at the origin $(x,y) = (0,0)$. In the annulus $1 < x^2+y^2 < 3$, the flow always increases $x^2+y^2$ and it moves in the counterclockwise direction when $1 < x^2+y^2 < 2$ but in the clockwise direction when $2 < x^2+y^2 < 3$.
• Thank you again for your answer to this question. What about higher dimensional foliation: For example is it trivial to ask "Is the Reeb foliation, geodesible" in the sense that there is a metric on $S^3$ such that all leaves are totally geodesics? – Ali Taghavi Aug 11 '17 at 19:01
• @AliTaghavi: There is no metric $g$ on $S^3$ for which the all the leaves of some codimension 1 foliation $\mathcal{F}$ of $S^3$ are even minimal, let alone totally geodesic. The reason is that, for such a pair $(g,\mathcal{F})$, the leaves can be oriented by a unit normal vector field $N$. Then the $2$-form $\Omega = \iota(N)\bigl(\mathrm{vol}(g)\bigr)$ would be closed (since the leaves are minimal) and hence exact. But by Novikov, $\mathcal{F}$ must have a compact leaf $L$, and the integral of $\Omega$ over $L$ would have to be positive. But by Stokes' Theorem, this integral must be zero. – Robert Bryant Aug 12 '17 at 11:00