# Poincaré Maps of a planar system

Consider a planar system $$\begin{array}{c} \dot{x}=x-y-x\left(x^2+y^2\right) \\ \dot{y}=x+y-y\left(x^2+y^2\right) \end{array}$$ This is the cross section: $$\Sigma=\left\{(x,y)\in \mathbb{R}^2 |x>0, y=0 \right\}$$ In polar coordinates $\rho=\sqrt{x^2+y^2}$, $\theta=\arctan(y/x)$ the system becomes $$\begin{array}{c} \dot{\rho}=\rho\left(1-\rho^2\right) \\ \dot{\theta}=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$$ and the section: $$\Sigma=\left\{(\rho,\theta)\in \mathbb{R}^+ |\rho>0, \theta=0 \right\}$$ It easy to solve the system to obtain the flow: $$\phi_t (\rho_0, \theta_0)=\left( \begin{array}{c} \left(1+\left(\frac{1}{\rho_0 ^2}-1\right)e^{-2t}\right)^{-1/2} \\ t+\theta_0 \end{array}\right)$$ My textbook (Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields of J. Guckenheimer and P. Holmes) writes that $$P(\rho_0)=\left(1+\left(\frac{1}{\rho_0 ^2}-1\right)e^{-4\pi}\right)^{-1/2}$$ 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 $$P(q)=\phi_\tau (q)$$ 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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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 $\Sigma$. –  Alex R. Dec 13 '12 at 22:54
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. –  Mark Dec 14 '12 at 8:42
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. –  Alex R. Dec 17 '12 at 16:31