Let $v\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R}^n)$ $(n\geq 2)$ a vector field, such that the set $E=\{v=0\}$ is a manifold of dimension $n-2$. Assume that for every $x\in\mathbb{R}^n-E$, the trajectory $(x_t)_{t\geq 0}$ such that $x_0=x$ and $\dot{x}_t=v(x_t)$ for $t\geq 0$, is periodic, with period $T(x)$, such that $T\in\mathcal{C}^0(\mathbb{R}^n-E,\mathbb{R}_+)$. Assume that for every $x\in E$ there exist an antisymmetric matrix $A(x)$ of rank 2 such that $A\in\mathcal{C}^0\big(E, \mathcal{A}_n(\mathbb{R})\big)$ (for the natural topologies of these spaces) and that for a $x'\in\mathbb{R}^n-E$ neighbor of $x$


Question: Does $T$ have a continuous extension defined over $\mathbb{R}^n$?

  • $\begingroup$ What do you mean? Typically (for instance, if $A$ has some eigenvalue with modulus >1) the period must go to infinity as the point approaches $E$, do you want the function $1/T$ rather than $T$? Still, for the linear vector.field in the plane $x'=-y$ and $y'=x$ this does not hold. $\endgroup$
    – rpotrie
    Jun 15, 2021 at 1:29
  • $\begingroup$ For $\omega>0$, the planar vector field $x'=-\omega y$ and $y'=\omega x$ has for solution $(x,y)(t)=\mathcal{R}_{\omega t}(x_0,y_0)$ where $\mathcal{R}_{\theta}$ is the rotation matrix with angle $\theta$. So if $(x_0,y_0)\neq (0,0)$, then $T(x_0,y_0)=2\pi/\omega$, so $T$ has a continuous extension such that $T(0,0)=2\pi/\omega$. This does not depend on the modulus of the eigenvalues of $A_{(0,0)}$ (which is $\omega$). $\endgroup$
    – G. Panel
    Jun 15, 2021 at 13:40


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