Find the smallest annular trapping region of the following:
$r'=r(1-2r^2+sin(2\theta)r^2)$
$\theta ' = -1$
I really do not understand how to do this. I have been trying to figure it out from things I found online but I really only found general examples, none on to actually work through a problem like this. Could someone please walk me through this?
Thanks in advance!
We can write $$\dot{r} =r[1-2r^2+\sin(2\theta)r^2]\\ =r[1-r^2-(\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta)r^2] \\ =r[1-r^2-(\cos\theta-\sin\theta)^2r^2] \\ =r\left\{1-r^2\bigg[1+2\cos^2\Big(\theta+\frac{\pi}{4}\Big)\bigg]\right\}$$ The above identity shows exactly what Did wrote in his/her comment i.e. that $\dot{r}<0$ for every $r>1$ and $\dot{r}>0$ for all $r\in(0,1/\sqrt{3})$ (the region $(1/\sqrt{3},1)$ is stable).