Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 2 / 8A / b

With $y=\dot{x}$, the system isAt equilibrium $y=0$ and $x^2=1$, so the equilibrium points are $(1,0)$ and $(-1,0)$.

The Jacobian isAt $(1,0)$ its eigenvalues are $1,-4$, so this point is a saddle. At $(-1,0)$ they are $(-3\pm i\sqrt{7})/2$, so this point is a stable spiral.

The sign of $\dot{x}$ is the sign of $y$. The $\dot{y}=0$ nullcline is$\dot{y}$ is positive below this curve and negative above it. Combining the upper and lower half-planes with the regions above and below this even curve labels the four requested sign combinations.

Draw the even $\dot{y}=0$ nullcline through $(\pm 1,0)$ and use the arrows from part (iii). Trajectories near $(-1,0)$ spiral clockwise inward because, immediately to its right on the $x$-axis, $\dot{y}<0$. At $(1,0)$ the stable and unstable eigendirections have slopes $-4$ and $1$. Their four separatrix branches divide the portrait; the only trajectory actually reaching $(1,0)$ at finite time is the constant one, while these branches approach it as $t\to \pm \infty$. All remaining curves follow the sign field, with those in its basin spiralling into $(-1,0)$.