Asymptotic stability without Lyapunov

lyapunov-functionsordinary differential equationsstability-in-odes

Consider the following system of equations

$$\begin{cases}
x'=-x^3y^2 \\
y'=-2x^2y^3 \\
\end{cases}
$$
Using the Lyapunov function $V(x,y)=x^2+y^2$ we get
$$\dot{V}(x,y)=-2x^4y^2-4x^2y^4\leq 0$$
So, $(0,0)$ is a stable equilibrium of the system. But what about asymptotic stability, can we say something about it? It is easy to see that every punctured neighborhood of $(0,0)$ contains points of the form $(\epsilon,0)$ and $\dot{V}$ vanishes on them. So, we can't use $V$ to get asymptotic stability.

Best Answer

The points $(\epsilon,0)$ are themselves equilibria, so clearly $(0,0)$ is not asymptotically stable.

Related Question