Consider the vander pol equation on the plane. Is there a reimannian metric on $\mathbb{R}^{2}-{0}$ such that the trajectories of the vander pol equation would be geodesics, moreover the curvature of the metric is negative? The motivation is that: existence of such metrics would give us an alternative proof for uniquness of limit cycle of the vander pol equation.
I cannot provide an answer, but a way of attacking the problem: In the geometric theory of ODEs (see e.g. the article in MR1327940 by M. Fels), there are computable invariants which allow you to decide whether the solutions of such an ODE can be realized as geodesic paths of a connection. This then defines a projective equivalence class of connections, and the problem whehter such a class contains a Levi-Civita connection is quite well studied, too, as "metrizability of projective structures".
Hi, I don't know Van der Pol's equation, but you might like to look at this paper:
Gerlich, Gerhard
Topological affine planes with affine connections. (English) Zbl 1080.51007
Adv. Geom. 5, No. 2, 265-278 (2005).
Gerlich studies the question whether there exists a connection or a Riemannian metric whose geodesics are certain preassigned curves forming a so-called stable plane (something like an open subset in an affine plane).
It seems to me that the VP equation is defined by a non zero vector field(zero only at the origin).If we denote this vector field by X, we can now find Y a vector field perpendicular to X, of unit length and so that the frame X,Y is positive(with respect to the euclidean metric and the positive orientation of the plane). Y is uniquely determined by X. Now define the Riemannian metric as the unique Riemannian metric that has X,Y as an orthonormal frame at any point of the plane. Then X is a constant (parallel vector field) with respect to g and therefore its trajectories are geodesics.
Dear Mihail Cocos
Thank you for your answer.
Assume that X=(P,Q). Are you considering Y=1/ \sqrt{P^2+Q^2) (Q, -P). For the Riemannian metric with such fram {X,Y} do you say that \Delta^X_{X}=0? If yes, could you please more explain on your computation?
Thank you.
For a "conformal " version see the following MO post
http://mathoverflow.net/questions/220456/conformal-changes-of-metric-and-geodesics
Dear Ali,
This is what I think : Let X=(P,Q) and Y=(-Q,P). Let g be the riemannian metric that is defined by g(X,X)=1,g(X,Y)=0 and g(Y,Y)=1. Let \Delta be its LeviCivita connection.Since \Delta_Xg=0, it follows that g(\Delta_X X,X)=0 and g(\Delta_X X,Y)=0 which implies \Delta_X X=0. Please correct me if I am wrong.
HI Ali,
I'll actually have to take that back. For a reason I wrongly assumed that \Delta_X Y doesn't have an X component. That is clearly wrong.
However if I can choose Y to commute with X([X,Y]=0). Then choosing g to be the metric that has X,Y as an orthonormal frame. If \Delta is now the Levi Civitta connection with respect to g, then \Delta_X X=0. Will this work?
Dear Mihail Thank you for your answer. Sorry for my delay. I ll think about your answer. I'll come back soon.
Hi Ali,
I just noticed you also wanted a negative metric, so choosing Y to be unit length will not do. I guess you'll have to play around with its magnitude while keeping X of unit length. Surely the problem is more complex than what I thought at a first glance. Good luck with it and please let me know of any outcome.
Dear Mihail Thank you for your answer. I am sorry for my delay.
If you have time, can I ask you to read the comments on this MO question, the conversation between me and Loic Teyssier, and others and also the answer by futurologist on this subjec?
http://mathoverflow.net/questions/160945/limit-cycles-as-closed-geodesicsin-negatively-curved-space
Thank you,
Best regards
Ali
A little progress is that a quadratic vector field is geodesible
Please see the following MO post
https://mathoverflow.net/questions/273635/finding-a-1-form-adapted-to-a-smooth-flow/273648#273648
It seems that the same argument works for the Van der pol equation since the limit cycles of the Van der Pol equation is a convex closed curve.
but I think that the "curvature sign" is still a non trivial question
Instead of geodesibility of flows, we may consider a weaker property as described in the following post. This weaker property is more achivable and works for our purpose
https://mathoverflow.net/questions/323126/a-concept-weaker-than-geodesibility-of-flows-which-is-possibly-usefull-in-limit
Indeed, I cannot answer this question but I remember that in two dimensions, any path space is projectively equivalent to a Finsler space. Plese see
M. Matsumoto, Every path space of dimension two is projectively related to a Finsler space, Open System. & Inform. Dynamics, 3 (1995), 291–303.
best wishes
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