Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 32B (Dynamical Systems)

At an equilibrium, each product inmust vanish. This gives the two parameter-independent equilibriathe equilibria on $y=0$and the equilibria with both coordinates nonzerowhere matching signs are used.

The $y=0$ pair is born at $k=-9/4$, suggesting a saddle-node bifurcation. At $k=0$, one of these equilibria meets $(0,0)$, suggesting a transcritical bifurcation, while the two equilibria near $(1,2)$ are simultaneously born in a saddle-node. At $k=1$, the minus branch $(1-\sqrt{k},2-\sqrt{k})$ meets $(0,1)$, suggesting another transcritical bifurcation. At $k=4$, that branch meets the $y=0$ equilibrium $(-1,0)$, again suggesting a transcritical bifurcation. Thus the bifurcation parameter values are

The Jacobian matrix isAt $(0,0)$ its eigenvalues are $-k$ and $-1$. It is therefore a stable node for $k>0$, a saddle for $k<0$, and nonhyperbolic at $k=0$, consistent with the proposed transcritical bifurcation.

At $(0,1)$ the matrix iswith eigenvalues $1-k$ and $1$. It is an unstable node for $k<1$, a saddle for $k>1$, and nonhyperbolic at $k=1$, again consistent with a transcritical bifurcation.

There are two bifurcations in $0\le k\le 1/2$, both at $k=0$.

Near $(0,0,0)$, append $\dot{k}=0$. The plane $y=0$ is invariant and tangent to the extended center subspace, so it is an extended center manifold. The reduced equation is exactlyIts leading terms $-kx-3x^2$ have two branches, $x=0$ and $x=-k/3+O(k^2)$, which cross and exchange their center-direction stability. Hence this is a transcritical bifurcation.

For the bifurcation at $(1,2,0)$, setThe extended system becomesSolving the centre-manifold invariance equation for $v=h(X,k)$ givesSubstitution into the $X$ equation yieldsFor $k>0$ this has two nearby equilibria $X=\pm \sqrt{k}+O(k)$, while for $k<0$ it has none. It is therefore a saddle-node bifurcation. These reductions are collected in bifurcations of the 2022 Cambridge quadratic-cubic system.