Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 31E (Dynamical Systems)

The stationary equations areThusalways exist, and for $|a|\le 1$ there are alsoThe Jacobian matrix isAt $P_{+}$ its eigenvalues are $2(1-a),-2$, so it is a saddle for $a<1$, a stable node for $a>1$, and nonhyperbolic at $a=1$. At $P_{-}$ they are $-2(1+a),2$, so it is a saddle for $a>-1$, an unstable node for $a<-1$, and nonhyperbolic at $a=-1$.

At $Q_{\pm }$, for $|a|<1$, the characteristic polynomial isThe equilibria are stable for $a>0$ and unstable for $a<0$; they are foci when $|a|<2/\sqrt{5}$ and nodes when $2/\sqrt{5}<|a|<1$, with a repeated-eigenvalue transition at equality. For $a=0$ the eigenvalues are purely imaginary. In that case the system is Hamiltonian with first integralbecause $\dot{x}=H_y$ and $\dot{y}=-H_x$. The Hessian of $H$ is positive definite at $(1,0)$ and negative definite at $(-1,0)$, so nearby regular level sets are closed curves. Hence both nonhyperbolic equilibria are genuine nonlinear centers.

Finally,For $a\ne 0$ this has a strict constant sign throughout the simply connected plane, so the Bendixson-Dulac criterion proves that there are no periodic orbits. These results comprise the phase portrait of x dot equals two x times y minus a.

The line $x=0$ is invariant for every $a$; on it, $\dot{y}=1-y^2$, so the segment from $P_{-}$ to $P_{+}$ is a heteroclinic orbit. For $a=0$, the zero-energy set isThese separatrices divide the two families of closed orbits around the centers from the exterior trajectories. For $a=1/2$, both axial equilibria are saddles and $Q_{\pm }=(\pm \sqrt{3}/2,1/2)$ are stable foci. For $a=3/2$, the off-axis equilibria have disappeared, $P_{+}$ is a stable node, and $P_{-}$ remains a saddle.

The arrows, fixed-point types, invariant axis, and the exact $a=0$ separatrix are shown in the three phase portraits.

The off-axis equilibria exist for $|a|\le 1$ and coalesce with an axial equilibrium at each endpoint. Thus the stationary bifurcations occur atFor the positive bifurcation, $a_0=y_0=1$. Putso thatWrite the extended center manifold as $Y=h(x,\mu )$. Its invariance equation isUnder the scaling $\mu =O(x^2)$, comparison of the terms of order $x^2$ givesSubstitution into the $x$ equation givesWith the reversed parameter $\nu =-2\mu =2(1-a)$, this is the pitchfork bifurcation normal form $\dot{x}=\nu x-x^3$. Therefore the positive stationary bifurcation is a supercritical pitchfork in the parameter $\nu =2(1-a)$, or equivalently the stable daughter equilibria exist on the $a<1$ side.