Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 31E (Dynamical Systems)

Let ${\varphi }_t(x)$ denote the flow through $x$. Its alpha-limit set and omega-limit set areEquivalently, they contain the limits of ${\varphi }_{t_j}(x)$ along sequences $t_j\to -\infty$ and $t_j\to +\infty$, respectively.

The vector field is unchanged by reflection in the $x$-axis, $(x,y)\mapsto (x,-y)$; more generally, linearity of the second equation gives the scaling symmetry $(x,y)\mapsto (x,cy)$. It is also reversible under reflection in the $y$-axis:When $k=0$, $y$ is constant, so arbitrary translations $y\mapsto y+c$ are additional symmetries.

At an equilibrium point, $x^2-1=0$ and $kxy=0$. If $k\ne 0$, the only equilibria areTheir Jacobian matrix isAt $(1,0)$ its eigenvalues are $2,k$, and at $(-1,0)$ they are $-2,-k$. Hence for $k>0$, $(1,0)$ is an unstable node and $(-1,0)$ is a stable node; for $k<0$, both points are saddles.

If $k=0$, every point on each line $x=1$ and $x=-1$ is an equilibrium. The line $x=1$ is transversely repelling because the nonzero eigenvalue is $2$, while $x=-1$ is transversely attracting because the nonzero eigenvalue is $-2$; the tangential eigenvalue is zero on both nonisolated equilibrium lines.

Away from $x=\pm 1$, the sign of $\dot{x}=x^2-1$ is fixed within each of the three invariant strips, so $x(t)$ is strictly monotone. On $x=\pm 1$, the equation is $\dot{y}=\pm ky$, whose nonconstant solutions are exponential and hence not periodic. Therefore the monotone coordinate excludes periodic orbits, and there are

For $x\ne \pm 1$, eliminating time gives the separable differential equationand henceDefine the possible limit at $x=s\in \{-1,1\}$ for a trajectory starting off the boundary lines byThe $k<0$ set is empty because $|y|\to \infty$, so there is no finite accumulation point. Since $\dot{x}<0$ on $(-1,1)$ and $\dot{x}>0$ outside $[-1,1]$, the complete classification off the boundary lines isFor $x_0>1$ the solution has forward finite-time blow-up, and for $x_0<-1$ it has backward finite-time blow-up, explaining the other empty entries under the usual infinite-time definitions.

It remains to treat $x_0=s=\pm 1$. Here $x(t)=s$ and $y(t)=y_0{e}^{kst}$. If $y_0=0$, both limit sets equal $\{(s,0)\}$. If $y_0\ne 0$, thenThis includes all boundary cases in both $k$ and the initial point.

The initial point has $x_0=0$, so its orbit remains in $-1<x<1$. Integrating the orbit equation from part (iv) and using $y(0)=y_0$ givesThis formula also covers $y_0=0$ and $k=0$ within the open strip. The endpoints are approached only as infinite time limits and have the behavior classified in part (iv).