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Let X be the following vector field on the plane:

{x=yy=xx3(X)

The vector field (X) has a non isochronous center at the origin.The proof is given in Remark 2 below. The punctured plane is filled with periodic orbits of X. The vector field is geodesible on the punctured plane, in the sense that there is a Riemannian metric such that solutions of X are geodesics of the metric. The reason of geodesibility of X is explained for some similar system in this post. In fact the 1_ form ψ=dθ works for this vector field X.That is dψ=0 and ψ(X)>0. This is equivalent to geodesibility in dimension 2.

Question: What is an explicit flat Riemannian metric such that the trajectories of X are unparametrized geodesics?

Remark 1 Note that because of the following interesting comment of Prof. Goodwillie we should not expect that X has a constant length, since the center is not isochronous

Extension of a vector field to an orthonormal frame for a flat metric

Remark 2: The period is not a constant for periodic orbits surrounding the origin. A proof can be given as follows:

If we compute the integral of signed curvature along a closed orbit of period T, we obtain 2π=T+T0y2(t)x2(t)y2(t)+(x(t)+x3(t))2dt, using Gauss Bonnete theorem. Since the integrant of the latter integral is positive and tends zero when closed orbits tend origin , then period T should depend on choosing the periodic orbits, namely the period T can not be a constant. So the center is not isochronous.

1 Answers1

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Since the metric doesn't have to extend to the origin, take the flat metric g=(d(x1+x2/2))2+dy2x2+x4/2+y2. The level curves x2+x4/2+y2=r2 are geodesics for the metric, and these are the integral curves of the vector field X.

Robert Bryant
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