Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 2A (Differential Equations)

An equilibrium of an autonomous differential equation is a constant $y_{\ast }$ with $f(y_{\ast })=0$. Put $y=y_{\ast }+\eta$. For differentiable $f$, the linear stability analysis givesThus the linear perturbation is proportional to $e^{f^{\prime }(y_{\ast })x}$. More directly, continuity of $f^{\prime }$ makes $f(y)$ point toward $y_{\ast }$ on both sides when $f^{\prime }(y_{\ast })<0$, and away from it when $f^{\prime }(y_{\ast })>0$. HenceThe sign refers to evolution toward increasing $x$. If $f^{\prime }(y_{\ast })=0$, linear stability analysis is inconclusive: the leading nonzero nonlinear term or the signs of $f$ on either side are needed. For example ${\eta }^{\prime }=-{\eta }^3$ is attracting whereas ${\eta }^{\prime }={\eta }^3$ is repelling, despite the same zero linear coefficient.

Factor the right-hand side as $f(y)=y(y-2)(y+1)$. Its equilibrium points are $-1,0,2$, and $f^{\prime }(y)=3y^2-2y-2$ givesIndeed $f^{\prime }(-1)=3$, $f^{\prime }(0)=-2$ and $f^{\prime }(2)=6$. The phase line has downward motion for $y<-1$, upward motion for $-1<y<0$, downward motion for $0<y<2$, and upward motion for $y>2$. The uniqueness of a smooth autonomous differential equation prevents trajectories from crossing an equilibrium point. Consequently trajectories starting between $-1$ and $2$ approach zero as $x\to \infty$, while those outside that interval move away from the three equilibrium points. The constant trajectories and representative nonconstant trajectories are shown below.