Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 14C / ii

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.