Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 14C (Dynamical Systems)

A steady-state bifurcation is a change in the local equilibrium branches or their stability as a parameter varies. At such a bifurcation the Jacobian matrix at the equilibrium has a zero eigenvalue, so the implicit function argument for a unique persistent local equilibrium fails.

For this system,At $(0,\mu )$ its eigenvalues are $1-{\mu }^2,-\mu$; at $(1,0)$ they are $-2,\mu -1$. Both have one simple zero eigenvalue at $\mu =1$, with the other eigenvalue negative. Interior equilibrium branches, obtained below, meet these boundary equilibria there, establishing the actual bifurcations rather than merely a necessary eigenvalue test.

Near $(1,0)$ put $u=x-1$, $v=y$, $ϵ=\mu -1$, with $\dot{ϵ}=0$. The extended system isThe extended centre manifold for a parameter has $u=h(v,ϵ)$, $h(0,0)=0$, $Dh(0,0)=0$. Its centre-manifold invariance equation is $h_v\dot{v}=\dot{u}$. The left side starts at degree three. At degree two the right side is $-2h_2-v^2$, so $h_2=-v^2/2$, with no $vϵ$ or ${ϵ}^2$ term. Substitution givesThe leading normal form of a dynamical system $\dot{v}=ϵv-v^2$ has branches $v=0$ and $v\simeq ϵ$, exchanging stability at $ϵ=0$: this is a transcritical bifurcation. In the physical quadrant the interior branch is present only for $ϵ>0$.

The equations extended across $x=0$ are equivariant under $x\mapsto -x$, so near $(0,1)$ the reduced equation is odd in $x$. To verify its nondegenerate pitchfork bifurcation, set $\eta =\mu -1$ and $z=y-\mu$. ThenThe local centre manifold has $z=-x^2+O(|\eta |x^2+x^4)$, obtained by balancing the quadratic terms of its invariance equation. Consequently $\dot{x}=-2\eta x+x^3+O({\eta }^2|x|+|\eta ||x{|}^3+|x{|}^5)$: the cubic coefficient is nonzero. This is a subcritical pitchfork, with the two symmetry-related interior branches existing for $\eta >0$ in the extended plane; only the $x>0$ branch belongs to the physical quadrant.

At an interior equilibrium point, $x^2=1-y^2=\mu -y$, soFor $1<\mu <5/4$ both values satisfy $0<y<1$, giving two physical equilibria. At such an equilibrium,The branch $y_{-}<1/2$ is asymptotically stable by the trace-determinant stability criterion; $y_{+}>1/2$ is a saddle equilibrium. In fact $J$ is symmetric, so the stable branch has two negative real eigenvalues.

At $\mu =5/4$ the equilibria merge at $(x,y)=(\sqrt{3}/2,1/2)$. Writing $y=1/2+s$, the equilibrium condition is $s^2+\mu -5/4=0$, the characteristic branch geometry of a saddle-node bifurcation. The stable node and saddle annihilate as $\mu$ increases through $5/4$.

The bifurcation diagram has boundary branches $y=0$ at $(1,0)$ and $y=\mu$ at $(0,\mu )$. The former is stable for $0<\mu <1$ and a saddle for $\mu >1$; the latter is a saddle for $0<\mu <1$ and stable for $\mu >1$. The origin is an unstable equilibrium for every $\mu >0$; its $y=0$ branch coincides in this projection with the $(1,0)$ branch, so it is shown separately in the caption.

The stable interior branch $y_{-}$ rises from $0$ at $\mu =1$ to $1/2$ at $\mu =5/4$. The unstable interior branch $y_{+}$ falls from $1$ to $1/2$ over the same interval. These meet at the saddle-node bifurcation; at $\mu =1$ they meet the transcritical and pitchfork points respectively. There is bistability between the lower interior node and $(0,\mu )$ for $1<\mu <5/4$.