juniorychichoa70

2022-04-22

Poincare Map for Polar ODE

I am currently trying to obtain a Poincare Map for the ODE system originally given by

$\{\begin{array}{l}\dot{x}=(1-{x}^{2}-{y}^{2})x-y\\ \dot{y}=x+(1-{x}^{2}-{y}^{2})y\end{array}$

on the region $x\in (\frac{1}{2},\frac{3}{2})$ and $y=0$. Since $x}^{2}+{y}^{2}={r}^{2$ and $\mathrm{tan}\left(\theta \right)=\frac{y}{x}$, we obtain that

$\{\begin{array}{l}\dot{r}=\frac{x\dot{x}+y\dot{y}}{r}\\ {\mathrm{sec}}^{2}\left(\theta \right)\dot{\theta}=\frac{x\dot{y}-y\dot{x}}{{x}^{2}}\end{array}\phantom{\rule{0ex}{0ex}}\u27f9\phantom{\rule{0ex}{0ex}}\{\begin{array}{l}\dot{r}=r-{r}^{3}\\ \dot{\theta}=1\end{array}$

However, I am stuck here with trying to identify the Poincare Map for the given system. Are there any recommendations for how to proceed? Moreover, how can I linearize this system at the point $(x,y)=(1,0)$ (or in polar coordinates $(r,\theta )=(1,0)$?

elvis0217t2x

Beginner2022-04-23Added 13 answers

Explanation:

Obviously you return to$y=0$ with a positive x after a full rotation, $t=\theta =2\pi$ . Now solve the Bernoulli equation for the radius

${({r}^{-2}-1)}^{\prime}=-2({r}^{-2}-1)\Rightarrow (r{\left(2\pi \right)}^{-2}-1)={e}^{-4\pi}(r{\left(0\right)}^{-2}-1)$ .

Obviously you return to

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