Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 7E (Dynamical Systems)

The fixed points satisfy $x[rx(1-x)-1]=0$. There is always $x=0$, and the other branches areThey coincide at $(r,x)=(4,1/2)$. For $0<x<1$, $F(x)>0$ and its maximum occurs at $x=2/3$, with value $4r/27$. Consequently $\boxed{F((0,1))\subset (0,1)⟺0<r<27/4}$; equality at the upper endpoint fails because the maximum is attained inside the interval.

The fixed-point multiplier at zero is zero, so zero is attracting for every $r>0$. On either nonzero branch,The lower branch has multiplier greater than one for $r>4$ and is repelling. The upper branch is attracting for $4<r<16/3$, reaches multiplier $-1$ at $x=3/4$, and is repelling for $r>16/3$. At $r=4$ the multiplier is $+1$, and the two branches are created in a saddle-node bifurcation. The fixed point there is attracting from one side and repelling from the other: $F(x)-x=-4x(x-1/2{)}^2$.

The second change is a supercritical period-doubling bifurcation at $\boxed{(r,x)=(16/3,3/4)}$. To check its type, expand at that fixed point. The local map has quadratic coefficient $a=-20/3$ and cubic coefficient $b=-16/3$ when the multiplier is $-1$. Its second iterate has cubic coefficient $-2(a^2+b)=-704/9<0$. The multiplier decreases through $-1$ as $r$ increases, so a small attracting two-cycle is born on the $r>16/3$ side. These exhaust the fixed-point bifurcations, since their multipliers have no other crossings of $\pm 1$.