Consider a planar system \begin{equation} \begin{array}{c} \dot{x}=x-y-x\left(x^2+y^2\right) \\ \dot{y}=x+y-y\left(x^2+y^2\right) \end{array} \end{equation} This is the cross section: \begin{equation} \Sigma=\left\{(x,y)\in \mathbb{R}^2 |x>0, y=0 \right\} \end{equation} In polar coordinates $\rho=\sqrt{x^2+y^2}$, $\theta=\arctan(y/x)$ the system becomes \begin{equation} \begin{array}{c} \dot{\rho}=\rho\left(1-\rho^2\right) \\ \dot{\theta}=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} \end{equation} and the section: \begin{equation} \Sigma=\left\{(\rho,\theta)\in \mathbb{R}^+ |\rho>0, \theta=0 \right\} \end{equation} It easy to solve the system to obtain the flow: \begin{equation} \phi_t (\rho_0, \theta_0)=\left( \begin{array}{c} \left(1+\left(\frac{1}{\rho_0 ^2}-1\right)e^{-2t}\right)^{-1/2} \\ t+\theta_0 \end{array}\right) \end{equation} My textbook (Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields of J. Guckenheimer and P. Holmes) writes that \begin{equation} P(\rho_0)=\left(1+\left(\frac{1}{\rho_0 ^2}-1\right)e^{-4\pi}\right)^{-1/2} \end{equation} is the Poincaré Map. Why? The textbook is written that the Poincaré Map $P:U\rightarrow\Sigma$ is defined for a point $q\in U$ by \begin{equation} P(q)=\phi_\tau (q) \end{equation} where $\tau=\tau(q)$ is the time taken for the orbit $\phi_\tau (q)$ based at q to first return to $\Sigma$.

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    $\begingroup$ So you just need $t+\theta_0=0\pmod{2\pi}$, so $t=2\pi-\theta_0$, and perhaps the book assumes $\theta_0=0$ as that would corresponding to starting in $\Sigma$. $\endgroup$ – Alex R. Dec 13 '12 at 22:54
  • $\begingroup$ Thanks for the comment, but I still have some doubts (perhaps because of my English is incorrect, or my weak theoretical basis).Let us assume that $\theta_0 =0$. Why I just need $t+\theta_0=0\pmod{2\pi}$? Perhaps because this system has a unique periodic solution $\phi_t=(\rho\cos \theta,\rho \sin \theta)$? Thanks. $\endgroup$ – Mark Dec 14 '12 at 8:42
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    $\begingroup$ I think it's because the angular part of your solution is $t+\theta_0$ and that the surface $\Sigma$ exactly corresponds to having zero angular coordinate. $\endgroup$ – Alex R. Dec 17 '12 at 16:31

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