Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 33A (Dynamical Systems)

A transcritical bifurcation has normal formIts equilibrium branches are $x=0$ and $x=\mu$. Linearization gives derivative $\mu$ on the first and $-\mu$ on the second, so the branches cross at $(\mu ,x)=(0,0)$ and exchange stability: for $\mu <0$, $x=0$ is stable and $x=\mu$ unstable; for $\mu >0$ the labels reverse.

A small constant perturbationhas equilibria $x=(\mu \pm \sqrt{{\mu }^2+4\epsilon })/2$. For $\epsilon >0$ the two branches avoid the crossing; for $\epsilon <0$ there is a parameter interval with no equilibrium, bounded by two saddle-node points. Thus arbitrarily small perturbations change the bifurcation diagram, so the crossing is not structurally stable.

At $r=1$, the Jacobian at the origin isIts eigenvalues are $0,-2,-1$. The zero eigenvalue makes the equilibrium nonhyperbolic.

Eigenvectors for $0,-2,-1$ are respectivelyHence

With $x=v+w$ and $y=v-w$, the equations at $r=1$ becomeThe symmetry $(v,w,z)\mapsto (-v,-w,z)$ suggestsSubstitute these into the invariance equationsComparison at orders $v^2,v^3,v^4$ givesThus the center manifold of the 2024 Cambridge cubic system is

Substitution of the center-manifold series into the $v$ equation gives at cubic orderKeeping the $v^5$ terms from $3a{v}^{2}w$, $zw$, and the quartic term in $z$ gives

The transverse eigenvalues $-2$ and $-1$ are stable, so the center manifold reduction determines local asymptotic stability. If $a<1$, the leading center term has negative coefficient and $v$ flows toward zero from both sides. If $a>1$, it flows away. At $a=1$ the cubic term vanishes but the quintic coefficient isso the origin is again unstable. Therefore