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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 Riemanian 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 vander 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?:

  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"

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 Vander 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.

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  • $\begingroup$ You haven't specified how the metric is related to the Van der Pol equation, so how can this provide an alternative proof? $\endgroup$ – Jaap Eldering Mar 21 '14 at 14:12
  • $\begingroup$ @JaapEldering The second point says "each leaf of the corresponding foliation of $\mathbb{R}^{2}-\{0\}$ is a geodesic. so there is a relation between the foliation arise from vander pol vec. field and the metric under my question. I am interested in this question, since many years ago. I approched to chapter "geodesiable flow" in the book"Geometry of foliation" By Tondeur. But I can find an answer to this question $\endgroup$ – Ali Taghavi Mar 21 '14 at 17:54
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    $\begingroup$ Am I right in thinking that your question is simply: 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
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    $\begingroup$ @AliTaghavi: The answer to my query is then "yes". $\endgroup$ – alvarezpaiva Mar 23 '14 at 15:07
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    $\begingroup$ @AliTaghavi: although your approach is interesting in that case, I don't understand how you hope to relate it to Hilbert 16th's problem when there is more than 1 limit cycle…? $\endgroup$ – Loïc Teyssier Apr 11 '14 at 7:15
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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"...

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  • $\begingroup$ thank you for your reference to Gardner book's I will look at it. Could you please more explain about your first paragraph? Are you comparing two foliations on $M$? your second foliation is the projection of a foliation of $TM \setminus M$ with geodesic flow? what metric you are considering for such geodesic flow? $\endgroup$ – Ali Taghavi May 10 '14 at 19:18
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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.

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  • $\begingroup$ thank you for your answer. I wonder why this argument should not be applicable for higher degree Lienard equation (when we replace $x^{3}-x$ by a higher degree polynomial. Note that higher defrees polynomials can have more thatn "one " limit cycle, i.e geodesic. Are there some special facilities for the particular third degree?Are there some obstructions for higher degrees? $\endgroup$ – Ali Taghavi Dec 6 '15 at 16:41
  • $\begingroup$ I mean that: to what extent your guess on "degrees of Freedom" is based on the particular case $x3−x$? As I said in the previos comment, we should not expect this degree of freedom for higher degree for example for $x5−x3+\epsilon x$ because in this case we have 2 limit cycles, so the global negative curvature is not possible. $\endgroup$ – Ali Taghavi Dec 8 '15 at 17:18
  • $\begingroup$ I can only guess, as I haven't done any attempts to bring this idea any further, but one might be able to solve the system of two equations for the Christoffel symbols in more general situations, like higher degree Lienard equation for instance, since everything is in the ring of polynomials. However, do not forget that this system provides just a connection. After that one needs to show that the metric corresponding to this connection is in fact a metric (positive definite) and then check the inequality $K < 0$ which for general Lienard systems might not be satisfied. $\endgroup$ – Futurologist Jan 7 '16 at 20:27
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    $\begingroup$ What I am trying to say is that one may get a metric in a very general context of polynomial vector fields, but this metric is rarely negatively curved. $\endgroup$ – Futurologist Jan 7 '16 at 20:28
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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 )

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    $\begingroup$ thanks for the links. I did not find, in these 2 papers, some thing related to the curvature. could you please more explain on your answer? $\endgroup$ – Ali Taghavi May 13 '14 at 16:19

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