Find the limit cycles of $\dot r =r(1-r^2)$, $\dot \theta=1$

Question

Apparently $r(t)=1$ is a limit cycle for the above system. Can anyone please explain why?

Thank you

Answer 1

I assume that $r(t)\geq0$ for all $t$. Since \begin{align} \dot{r}=r\left(1-r^2\right) \end{align} we have for some constant $t_0\in\mathbb{R}$ : \begin{align} t+t_0&=\int\frac{\mathrm{d}r}{r\left(1-r^2\right)}\\ &=\int\frac{\mathrm{d}r}{r\left(1+r\right)\left(1-r\right)}\\ &=\int\frac{\mathrm{d}r}{r}-\frac{1}{2}\int\frac{\mathrm{d}r}{1+r}+\frac{1}{2}\int\frac{\mathrm{d}r}{1-r}\\ &=\ln r-\frac{1}{2}\ln\left(1+r\right)-\frac{1}{2}\ln\left|1-r\right|\\ &=\ln\left(\frac{r}{\sqrt{\left|1-r^2\right|}}\right). \end{align} So we get \begin{align} \frac{r}{\sqrt{\left|1-r^2\right|}}=\mathrm{e}^{t+t_0} \end{align} that is \begin{align} \frac{r^2}{\left|1-r^2\right|}=\mathrm{e}^{2\left(t+t_0\right)}. \end{align}

• If $0\leq r\leq1$, then we have \begin{align} r^2=\mathrm{e}^{2\left(t+t_0\right)}\left(1-r^2\right) \end{align} and solving for $r^2$ and taking the square root yields the solution \begin{align} r(t)=\frac{\mathrm{e}^{\left(t+t_0\right)}}{\sqrt{{1+\mathrm{e}^{2\left(t+t_0\right)}}}} \end{align} so $\lim r(t)=1$ as $t\to+\infty$.
• If $1\leq r$, then \begin{align} r(t)=\frac{\mathrm{e}^{\left(t+t_0\right)}}{\sqrt{{\mathrm{e}^{2\left(t+t_0\right)}-1}}} \end{align} so we get once again $\lim r(t)=1$ as $t\to+\infty$.

Now, for any initial data $\left(r_0,\theta_0\right)\in\mathbb{R}_+\times[0,2\pi)$, we find that \begin{align} \begin{cases} \displaystyle r(t)=\frac{r_0\mathrm{e}^{t}}{\sqrt{1+r_0^2\left(\mathrm{e}^{2t}-1\right)}}\\ \theta(t)=t+\theta_0 \end{cases}. \end{align}