Limit cycles as closed geodesics (in negatively or positively curved space) Updated 1/25/2023  I just added  a  related post below:
Jacobi fields, Conjugate points and limit cycle theory
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"
My another motivation was the following: Almost at the same years of reading the above mentioned phrase in De Carmo book, I attended  a talk in Sharif university of Technology presented by Hessam Tehrani about variational problems. A particpant commented "I think existence of  closed geodesics is investigated by the same methods" His comment was a motivation for me  to consider a closed geodesic approach to limit cycle theory.

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

 A: I think we could gerneralize the problem: the foliation determined by Van der Pol equation are formed by maximal integral curves from a vector field in the manifold M. Geodesics are second order curves, in the sense that they are projections of vector fields defined in the tangente bundle TM. Is it possible to find equivalence between both foliations ?
This problem reminds me those trated in the Gardner's book "The Method of Equivalence and Its Applications"...
A: What I would try (using brute force again :) ) is the following. So if the Van der Pol vector field, call it 
$$Y=Y^1(x^1,x^2) \frac{\partial}{\partial x^1} + Y^2(x^1,x^2) \frac{\partial}{\partial x^2} = \Big(x^2- \big((x^1)^3 - x^1\big)\Big) \frac{\partial}{\partial x^1} - x^1 \frac{\partial}{\partial x^2},$$ defines geodesics for some metric, then since the geodesic equation is $\nabla_{\dot{\gamma}}\dot{\gamma} = 0$ and the solutions $\gamma(t)$ of Van der Pol satisfy $\dot{\gamma} = Y(\gamma)$, then we are looking for a Riemannian metric $\big(g_{ij}(x^1,x^2)\big)$ whose Levi-Civita connection $\nabla$ satisfies the equations $\nabla_Y\, Y = 0.$ In addition to that we want the tensor $g_{ij}$ to be (i) positive definite on the punctured plane and (ii) to have strictly negative Gaussian curvature on the punctured plane:
$$K = -\frac{1}{E} \left( \frac{\partial}{\partial x^1}\Gamma_{12}^2 - \frac{\partial}{\partial x^2}\Gamma_{11}^2 + \Gamma_{12}^1\Gamma_{11}^2 - \Gamma_{11}^1\Gamma_{12}^2 + \Gamma_{12}^2\Gamma_{12}^2 - \Gamma_{11}^2\Gamma_{22}^2\right) < 0.$$
Let's look at $\nabla_Y\, Y = 0,$ which written in coordinates is
$$Y^1\frac{\partial Y^1}{\partial x^1} + Y^2\frac{\partial Y^1}{\partial x^2} + \Gamma^{1}_{1 1}(Y^1)^2 + 2 \Gamma^{1}_{12} \, Y^1 Y^2 + \Gamma^{1}_{22}(Y^2)^2 = 0$$
$$Y^1\frac{\partial Y^2}{\partial x^1} + Y^2\frac{\partial Y^2}{\partial x^2} + \Gamma^{2}_{1 1}(Y^1)^2 + 2 \Gamma^{2}_{12} \, Y^1 Y^2 + \Gamma^{2}_{22}(Y^2)^2 = 0.$$ The unknowns are the Christoffel symbols, which depend on the metric and it's first partial derivatives. I guess you do have some degree of freedom. To add more degrees of freedom, one can even consider reparametrization of $Y$ by multiplying it to an unknown nonzero function $\lambda=\lambda(x^1,x^2).$ Maybe to simplify the equations above, one can consider a diagonal metric, i.e. $g_{12}(x^1,x^2) \equiv 0$. I don't know... maybe it could work, but it looks like a lot of computations.
A: good question !
i have found the proof of the fact that this system has at most one limit circle in the following 2 papers :


*

*http://arxiv.org/pdf/chao-dyn/9705006.pdf

*http://dml.cz/bitstream/handle/10338.dmlcz/107715/ArchMathRetro_036-2000-1_4.pdf
your questions are :
Is there a Riemannian metric on $\mathbb{R}^{2}-\{0\}$ with the following two properties?:


*

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

*Each leaf of the corresponding foliation of $\mathbb{R}^{2}-\{0\}$  is  a geodesic.
you can also find the solution to your 2 questions in papers 1 (question 2 is a result directly follows from question 1)
since the van der pol equation is the special case of Li´enard equation, so if you want to generalize this problem to arbitrary polynomial vector fields, it seems that you can define a metric that satisfy the Lienard equation:
$\dot{x}=y-F(x), $$ \dot{y}=-g(x)$
you can get some tips for how to construct it from paper 2 (main theorem, page 25 )
