Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 32A / b

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 $Q\le 1$.