Past exam of the differential equations and applications course of the University of Oxford / 2009 / A / 1 / d / Solution

For negative $ϵ$ the fast arrows reverse: the outer branches become repelling and the middle branch attracting. On the middle branch the slow equation is $\dot{x}=x/(1-x^2)$, so the slow motion goes away from the saddle equilibrium at zero towards a fold.

At $(3,1)$, $F=3-9+1=-5$, so $\dot{x}>0$: the trajectory immediately escapes to the right. At $(1,-2)$, $F=1-1/3-2=-4/3$, so it also moves rapidly right and escapes. In each case the large-$x$ balance is $\dot{x}\sim x^3/(3|ϵ|)$, with $y$ decreasing; it is not a bounded relaxation cycle.

At $(-1,-1/3)$, $F=-1$, so the initial fast motion is rightward. At nearly fixed $y=-1/3$ it approaches the attracting middle root of $x-x^3/3=1/3$, approximately $x=0.347$. It then moves slowly down/right along the middle branch towards $(1,-2/3)$, leaves that fold to the right, and eventually escapes. These are the small-negative-$ϵ$ limiting descriptions; the fold locations and fast landing point receive small corrections. The right panel of the figure distinguishes this slow excursion from the two immediate escapes.