Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 7C (Dynamical Systems)

For the iteration of a map $f(x)=\lambda x(1-x^2)$, a fixed point satisfies $x[1-\lambda +\lambda x^2]=0$. If $0<\lambda <1$, the bracket is strictly positive, so the sole fixed point is $0$. Its multiplier of a periodic orbit of an iteration is $f^{\prime }(0)=\lambda$, whose modulus is below one. Thus $\boxed{x=0\text{ is locally asymptotically stable}}$ by fixed point stability for an iteration.

For $\lambda >1$, the multiplier at the origin exceeds one, so it is unstable. The other fixed points areUsing fixed point stability for an iteration, $|3-2\lambda |<1$ exactly when $1<\lambda <2$, and $|3-2\lambda |>1$ when $\lambda >2$. HenceThe endpoint $\lambda =2$ has multiplier $-1$ and requires nonlinear analysis, which is not part of the asserted open intervals.

On $(0,1)$ the map is positive. Its maximum occurs at $x=1/\sqrt{3}$ and equals $2\lambda /(3\sqrt{3})$. Thus if $\lambda <3\sqrt{3}/2$, then $f((0,1))\subset (0,1)$. Since $f$ is odd, also $f((-1,0))\subset (-1,0)$. Induction givesThe case $x_0=0$ is constant. These intervals are positively invariant sets for the discrete dynamical system.

Put $a=2/\sqrt{3}$. On $[-a,a]$, the extrema of $|x(1-x^2)|$ occur at $\pm 1/\sqrt{3}$ or the endpoints. The corresponding values are $2/(3\sqrt{3})$, including the endpoint value $a(a^2-1)$. ConsequentlyThus $[-a,a]$ is a trapping region andIt contains a neighbourhood of the origin. Sign changes are possible because $f(x)$ has opposite sign to $x$ for $1<|x|<a$.