Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 6B / Solution

The Ricker population map has equilibria $N=0$ and $N=K$. Since $f^{\prime }(N)=e^{r(1-N/K)}(1-rN/K)$, their linearization multipliers areThus extinction is unstable and the positive equilibrium is locally asymptotically stable for $0<r<2$, becoming unstable for $r>2$.

To verify an actual period-doubling bifurcation, rather than only a multiplier crossing, put $N/K=1+x$ and $r=2+\epsilon$. The centered map isIts second iterate minus $x$ is $2\epsilon x-\frac{4}{3}{x}^3$ to leading order. After rescaling $x=\sqrt{\epsilon }y$, the limiting nonzero roots are simple, so the implicit function theorem continues them. Besides $x=0$, two nearby points therefore appear for $\epsilon >0$, with $x_{\pm }=\pm \sqrt{3\epsilon /2}+O(\epsilon )$. The map exchanges these points, and the derivative of its second iterate there is $1-4\epsilon +O({\epsilon }^{3/2})$, so the emerging two-cycle is attracting. The bifurcation at $r=2$ is supercritical period doubling.

For nonnegative populations, $f$ increases up to $N=K/r$ and decreases thereafter to zero. Hence every update after the initial time satisfiesThe bound is attained by choosing $N_0=K/r$; it is not a bound on an arbitrary initial population.