Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 30A / b / Solution

For $k\ne 0$, put $t=k^2>0$ and use the positive-definite Lyapunov function $V=x^2+t{y}^2$. Its derivative isFor $V>0$ set $u=x^2/V\in [0,1]$. The brace divided by $V^2$ is $A{u}^2+2Bu(1-u)+C(1-u{)}^2$, where $A=1$, $B=\beta (1+t)/(2t)$ and $C=1/t$. Since $\beta >2$, $B>A,C$. Completing the square gives its maximumHence $\dot{V}\le -2V+2M{V}^2$. If $V(0)<1/M$, this sublevel region is positively invariant: $V$ decreases, and $\dot{V}\le -2(1-MV(0))V$. The exponential bound makes $V\to 0$. The trajectory is bounded in a compact ellipse, so smoothness of the vector field also guarantees existence for every positive time. ThereforeFor $k=0$ the displayed strict inequality has no solutions; negative $k$ gives the same ellipse as $-k$. Taking the union of the nonempty ellipses is valid because each point needs only one Lyapunov certificate.

At $k=1$ the certified disk has squared radius $2/(\beta +1)$. Along $y=0$, the limiting threshold as $k^2\to \infty$ is $4(\beta -1)/{\beta }^2$, which is strictly larger, since their difference is $2({\beta }^2-2)/({\beta }^2(\beta +1))>0$ for $\beta >2$. Choose $x^2$ strictly between these thresholds and then a sufficiently large finite $k$. This point is certified by that ellipse but lies outside the $k=1$ disk. Thus the union gives a strictly larger certified region. It need not equal the whole basin of attraction.