Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 2 / 2B / ii / Solution

The constant solutions are fixed points of $G(u)=u^2(1+u)/2$. Solving $G(u)=u$ gives $u(u+2)(u-1)=0$, hence $u_{\ast }=0,-2,1$. For a discrete dynamical system, a small perturbation satisfies ${\eta }_{n+1}=G^{\prime }(u_{\ast }){\eta }_n+O({\eta }_n^2)$, whereA multiplier of magnitude less than one contracts perturbations; a multiplier greater than one expands them. Therefore $u=0$ is asymptotically stable, while $u=-2$ and $u=1$ are unstable. In particular, if $|u|\le 1/2$, then $|G(u)|\le (3/4)|u{|}^2\le (3/8)|u|$, proving convergence to zero without relying solely on the vanishing linear multiplier.