Stability of this system

Question

I have to study the stability of the system $\dot{x} = 0, \ \dot{y} = -(y-x)^3 $. I tried with the Lyapunov method with a quadratic function but it doesn't work. Someone has any idea?

Answer 1

This system has a line of equilibrium points, which can be described as $$ E= \{(x,y)\in\mathbb R^2:\; x=y\}. $$ It is possible to study their stability by finding a general solution to the system, but it is more interesting and simpler to use the Lyapunov functions.

Consider some equilibrium point $(x_0,y_0)\in E$. The function $$ V(x,y)= (x-x_0)^2+(y-x)^2 $$ is a Lyapunov function for this point. Indeed, it is continuously differentiable, positive definite, $V(x_0,y_0)=0$; its derivative along the trajectories of the system is equal to $$ \dot V= 2(x-x_0)\cdot 0+2(y-x)\cdot(-(y-x)^3-0)=-2(y-x)^4. $$ Since $\forall (x,y)\in\mathbb R^2\;\; \dot V(x,y)\le 0$, the equilibrium point $(x_0,y_0)$ is stable.

Notice that any equilibrium point $(x_0,y_0)$ from $E$ is stable, but not asymptotically stable because the solution that starts from an arbitrary close to $(x_0,y_0)$ equilibrium point stays in place and does not tend to $(x_0,y_0)$.