Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 32A (Dynamical Systems)

A $C^1$ function $V$ near an equilibrium at the origin is a Lyapunov function if $V(0)=0$, $V(x)>0$ for $x\ne 0$, andThe First Lyapunov theorem says these conditions imply stability. The second says that if the last inequality is strict away from the origin, then the origin is asymptotically stable.

To prove the first, fix a sufficiently small ball of radius $\epsilon$. Positive definiteness and compactness giveContinuity at zero gives $\delta >0$ such that $|x_0|<\delta$ implies $V(x_0)<m$. Since $V$ cannot increase along a trajectory, that trajectory cannot meet the sphere $|x|=\epsilon$, where $V\ge m$. It therefore remains in the $\epsilon$-ball for all forward time, which is stability.

The Jacobian at the origin is $-I$, whose two eigenvalues are $-1$. The linearization theorem for a hyperbolic equilibrium therefore implies that the origin is asymptotically stable.

SetDirect substitution of the system gives the exact factorizationCauchy--Schwarz in the ellipsoidal norm givesThus if $0<Q<1$, then $x+y<1/2$ and $\dot{Q}<0$. Every trajectory beginning in $Q<1$ remains in a compact smaller sublevel set and tends to the only point where $Q=0$, namely the origin. Hence the basin contains the stated region.

Yes. On $Q\le 1$ the same inequality gives $\dot{Q}\le 0$, so the closed ellipse is positively invariant. Equality in $\dot{Q}=0$ occurs only at the origin and at the boundary pointAt that point the vector field isso $\{p\}$ is not invariant. The largest invariant subset of $\{\dot{Q}=0\}$ is therefore the origin. LaSalle's invariance principle shows that every trajectory in the closed ellipse converges to the origin. The inequality may consequently be extended to $Q\le 1$.