Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 31E / b / iii

The factor $x^2(2-x^2)$ is positive for $0<|x|<\sqrt{2}$. The product $y^3\sin 2y$ is positive for $0<|y|<\pi /2$. Therefore $\dot{V}<0$ away from the origin throughoutNeither side length can be enlarged while retaining a strict Lyapunov function on the whole rectangle: at $(x,0)$ with $|x|>\sqrt{2}$, the first term makes $\dot{V}>0$, while at $(0,y)$ immediately beyond $|y|=\pi /2$, the second term makes $\dot{V}>0$.

Strict decrease on this rectangle does not by itself put the whole rectangle in the domain of stability, because the rectangle has not been shown to be positively invariant; a trajectory may leave it before convergence.