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Assume that $\gamma$ is an analytic simple closed curve in $\mathbb{R}^2$ which surrounds origin.

Is there a polynomial vector field on the plane which is tangent to $\gamma$? In the other word, can an arbitrary analytic simple closed curve be realized as a closed orbit or a limit cycle of a polynomial vector field on the plane?

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I see this as very unlikely. A polynomial vector field would have a slope function that is a rational function of two variables with a finite number of coefficients.

As a consequence, if you take the field generated over $\mathbb Q$ by the coordinates $(x_k, y_k), k=1,\ldots, n$ of the points on the curve, as well as the slopes $c_k$, this field will have transcendence degree at most $2n+{\rm constant}$. I see no reason for such bound to be true for an arbitrary closed analytic curve.

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  • $\begingroup$ My apology for not giving the bounty on time. Thank you for your very interesting answer. $\endgroup$ Nov 1, 2017 at 6:43
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The answer is "no". In fact, it is still "no" for germs of curves : generically, a germ of an analytic curve $\gamma : (\mathbb R,0)\rightarrow (\mathbb R^2,0)$ is not tangent to any polynomial vector field. Otherwise, it would mean that any such germ would be solution to a polynomial differential equation, which is not true by a Baire-like argument: the space of polynomial vector fields is a countable union of closed sets with empty interior (for convenient locally convex topologies on the involved spaces). The key argument is that the flow mapping $$\mathbb R\{x,y\}^2\longrightarrow\mathbb R\{x,y,t\}^2 \\ X\longmapsto \Phi_X^t(x,y)$$ is an analytic map and the image of $\mathbb R[x,y]^2$ is "analytically meager", meaning it is included in a countable union of proper analytic subvarieties. But $\mathbb R\{t\}^2$ is "analytically Baire" : it cannot be written as the countable union of such subvarieties. You can find details in my paper "Analyticity in spaces of convergent power series and applications" (see Corollary B and section 1.2.2).

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  • $\begingroup$ Thank you very much for your interesting answer. I am sorry that I can not accept two answers simultaneously. meta.mathoverflow.net/questions/1491/… So I considered "The first appearing correct answer". :) $\endgroup$ Nov 1, 2017 at 6:40

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