**EDIT:** Here is a related post which concern quadratic vector fields rather than Van der Pol equation. In this linked post we see that the convexity of limit cycle play a crucial role. On the other hand the unique limit cycle of Van der Pol equation is convex. So there is a Riemannian metric on $\mathbb{R}^2 \setminus C$ such that all solutions of the Van der Pol equation are geodesics. Here $C$ is the algebraic curve $yP-xQ=0$ where $P,Q$ are the components of the Van der Pol equation. Moreover the limit cycle of the Van der Pol equation do not intersect this algebraic curve $C$.

The classical Van der Pol equation is the following vector field on $\mathbb{R}^{2}$:

\begin{equation}\cases{\dot{x}=y-(x^{3}-x)\\ \dot{y}=-x}\end{equation}

This equation defines a foliation on $\mathbb{R}^{2}-\{ 0\}$. It is well known that this vector field has a unique limit cycle (isolated closed leaf) in the (punctured) plane.

I search for a geometric proof for a particular case of this fact. In fact I search for an

alternative proof of the fact that this system has **at most** one limit cycle.

Here is my question:

**Question:**

Is there a Riemannian metric on $\mathbb{R}^{2}-\{0\}$ with the following two properties?:

The Gaussian curvature is nonzero at all points of $\mathbb{R}^{2}-\{0\}$.

Each leaf of the corresponding foliation of $\mathbb{R}^{2}-\{0\}$ is a geodesic.

Obviously from the Gauss Bonnet theorem we conclude that existence of such metric implies that there are no two distinct simple closed geodesics on $\mathbb{R}^2\setminus \{0\}$, otherwise we glue two copy of the annular region surrounded by closed geodesics along the boundary then we obtain a torus with non zero curvature.(So this gives us an alternative proof for having at most one limit cycle for the Van der Pol equation)

For a related question see Conformal changes of metric and geodesics

My initial motivation for this question goes back to more than 15 years ago, when I was reading a statement in the book of De Carmo, differential geometry of curves and surface, who wrote that:

**A topological cylinder in $\mathbb{R}^{3}$ whose curvature is negative, can have at most one closed geodesic.**

After this, I asked my supervisor for a possible relation between limit cycles and Riemannian metrics. As a response to my question, he introduced me a very interesting paper by Romanovski entitled "Limit cycles and complex geometry"

Note 1:For the moment we forget "negative curvature".We just search for a metric compatible to the Van der Pol foliation. In this regard, one can see that for every metric on $\mathbb{R}^2 \setminus \{0\}$, with the property that all solutions of the Van der Pol equations are (non parametrized) geodesics, then either the metric is not complete or the punctured plane does not possess a polynomial convex function or an strictly convex function. This is a consequence of Proposition 2.1 of this paper and also the following fact.

**Note 2:** What is the answer if we replace the Van der Pol vector field by an arbitrary foliation of $\mathbb{R}^{2}\setminus \{0\}$ with a unique compact leaf?

**Remark:** The initial motivation is mentioned in page 3, item 5 of this arxiv note.

** Edit Feb 1, 2020** A reference we just found whose subject is some what similar to this post: https://arxiv.org/abs/1809.02783

Is there a negatively curved metric on the punctured plane $\mathbb{R^2}\setminus 0$ such that all solutions curves of the van der Pol equations are geodesics albeit with a possibly different parameterisation?$\endgroup$ – alvarezpaiva Mar 23 '14 at 7:54