Region of attraction of : $x'=-y-x^3,y'=x-y^3$ via Lyapunov Function **PROBLEM** : > $1)$ Show that the stationary point $O(0,0)$ is asymptotic stable > > $2)$ Find a region of attraction for the system : > > $$x'=-y-x^3$$ $$y'=x-y^3$$ given the Lyapunov's function: $$V=x^2+y^2$$ First, I differentiate my Lyapunov's function and i take: $$ \dot{V}=-2 \cdot (x^4+y^4)<0 $$ So, the stationary point $$O(0,0)$$ is asymptotic stable as $$\dot{V}<0$$ everywhere outside the origin. For the second question of my problem, I believe that the region of attraction is the circle : $$x^2+y^2=c $$ Is this right?How can I find $c$ and the boundary of the estimation of the region of attraction? If i am right with the circle can anyone help me to write it in a good mathematical way? Last days I am trying to understand how regions of attraction work so , I would really appreciate a thorough solution and explanation about how to find this region of attraction. Thanks in advance!

Your argument that the origin is asymptotically stable is correct. To find a region of attraction, identify a region where the Lyapunov function decreases along trajectories.

Since you found that $\dot{V} < 0$ everywhere (except at the origin), this means that all initial data are attracted to the origin - there is no restriction.

Of course any smaller set (e.g. any circle) also works since the Lyapunov function argument shows that trajectories always stay within any circle in which they start.