Past exam of the differential equations 1 course of the University of Oxford / 2021 / 3 / a / iii

The nullclines areOn $y=\beta$, away from the equilibrium points, the vector field is vertical. Its crossing direction is downwards for $x<-1$, upwards for $-1<x<0$, downwards for $0<x<1$, and upwards for $x>1$, because $\dot{y}=x(x^2-1)$. On each of $x=-1,0,1$, the vector field is horizontal: it points right when $y>\beta$ and left when $y<\beta$.

Throughout each complementary region, $\dot{x}$ has the sign of $y-\beta$, and $\dot{y}$ has the four signs just listed. These signs orient the level sets from part (ii). In particular, the inner periodic orbits are clockwise. The saddle level is $h=1/4$, whose separatrices satisfyFor $-1<x<1$, the upper arc $y-\beta =(1-x^2)/\sqrt{2}$ goes from $(-1,\beta )$ to $(1,\beta )$; the lower arc goes back from $(1,\beta )$ to $(-1,\beta )$. These two heteroclinic orbits form a closed boundary around the family of periodic orbits. The remaining branches of the same saddle level extend outwards. The sketch shows the surrounding nonclosed trajectories as well as the nullclines and equilibrium points.