Limit cycles as closed geodesics (in negatively or positively curved space)

Question

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 unique limit cycle of the Van der Pol equation does 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?:

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

2. 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

Answer 1

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"...

Answer 2

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.

Answer 3

good question !

i have found the proof of the fact that this system has at most one limit circle in the following 2 papers :

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

2. 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?:

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

2. 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 )