Finding a Lyapunov function for a given system

Question

I need to find a Lyapunov function for $(0,0)$ in the system:

\begin{cases} x' = -2x^4 + y \\ y' = -2x - 2y^6 \end{cases}

Graph built using this tool showed that there should be stability but not asymptotic stability. I'd like to prove that fact by means of Lyapunov function but cannot find the appropriate one among the most common ones such as $Ax^{2n}+By^{2m}$, $Ax^2+By^2+Cxy$, etc.

Please, give me some hint about how to proceed with the search of the suitable function or even the desired function itself it you know the easy way of finding it in this case.

Thanks in advance

Answer 1

As explained in the comments, even though simulated phase diagrams seem to exhibit cycling trajectories near the origin, they are not conclusive enough to decide whether the origin is stable or not, that is, whether trajectories cycle or spiral outwardly or spiral inwardly. Caution is advised about approximation errors in simulations.

streamplot{{-2x^4+y,-2x-2y^6},{x,-1,1},{y,-1,1}}

Answer 2

Dynamical system

$$ % \begin{align} % \dot{x} &= -2x^{4} + y \\ % \dot{y} &= -2x - 2y^{6} \\ % \end{align} % $$

There are two critical points: the origin $(0,0)$ and $$\left\{-\frac{1}{2^{6/23}},2 \left(\frac{1}{2^{6/23}}\right)^4\right\} =\left\{ -\frac{1}{2^{6/23}}, \frac{1}{2^{1/23}} \right\} \approx \{-0.834585,0.970313\}$$

Phase portrait

The phase portrait below displays the nullcline $\dot{x}=0$ with a red, dashed line and $\dot{x}=0$ in purple.

Stability

The polar transformations $$ x = r \cos \theta, \qquad y = r \sin \theta $$ lead to $$ r^{2} = x^{2} + y^{2} $$ which implies $$ \dot{r} = \frac{x\dot{x}+y\dot{y}}{\sqrt{x^2+y^2}} = \frac{-2 x^5-x y-2 y^7}{\sqrt{x^2+y^2}}, $$ plotted below.

The problem is that the sign of $\dot{r}$ changes twice in every neighborhood of the origin.