Largest circle in basin of attraction of the origin. We're given dynamical system:
$$
\dot x = -x + y + x (x^2 + y^2)\\
\dot y = -y -2x + y (x^2 + y^2)
$$
Question is what's the largest constant $r_0$ s.t. circle $x^2+y^2 < r_0^2$ lies in the origins basin of attraction.
So far with relatively easy algebra I've got:
$$ \dot r  = \frac{r}{2}(-2-\sin(2 \phi)+2r^2) \\
   \dot \phi = -(1+\cos^2(\phi))
$$
Which immediately shows $r_0 \geq \sqrt{1/2}$. How to show that there is no better bound?

After solving the system
$$ 
\dot r  = \frac{r}{2}(-2-\sin(2 \phi)+2r^2) \\
\dot \phi = -(1+\cos^2(\phi))
$$
we have
$$
\left\{
\begin{array}{rcl}
 r & = & \frac{\sqrt{3} \sqrt{3-\cos \left(2 \sqrt{2} \left(t-2 c_1\right)\right)}}{\sqrt{3 e^{2 t} c_2-\cos \left(2 \sqrt{2} \left(t-2
   c_1\right)\right)+\sqrt{2} \sin \left(2 \sqrt{2} \left(t-2 c_1\right)\right)+9}} \\
\phi & = & -\tan ^{-1}\left(\sqrt{2} \tan \left(\sqrt{2} t-2 \sqrt{2} c_1\right)\right) \\
\end{array}
\right.
$$
Regarding the  $r$ behavior the line of maxima after considering $c_1 = c_2 = 0$ is given by
$$
r_{max} = \frac{\sqrt{3} \sqrt{3-\cos \left(2 \sqrt{2} t\right)}}{\sqrt{\sqrt{2} \sin \left(2 \sqrt{2} t\right)-\cos \left(2 \sqrt{2} t\right)+9}}
$$
with $\min r_{max} = \sqrt{1-\frac{1}{\sqrt{13}}}\approx 0.850088$
