Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 2 / 2B (Differential Equations)

Put $F(y)=(y^2-4)\arctan y$. Its zeros, and hence the equilibrium points, are $-2,0,2$. For a perturbation $y=y_{\ast }+\eta$, linear stability analysis gives $\dot{\eta }=F^{\prime }(y_{\ast })\eta +O({\eta }^2)$, withAt zero, $F^{\prime }(0)=-4$, so sufficiently small perturbations decay and $y=0$ is asymptotically stable. At either outer equilibrium point, $F^{\prime }(\pm 2)=4\arctan 2>0$, so $y=-2$ and $y=2$ are unstable. The phase line confirms the result: $F$ has signs $-,+,-,+$ on the intervals cut out by $-2,0,2$, respectively. Flow points toward zero on both sides and away from both outer equilibria.

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.