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

Define the auxiliary variable $y=ϵ\dot{x}-x+x^3/3$. Differentiating gives $\dot{y}=ϵ\ddot{x}+(x^2-1)\dot{x}=-x$, while rearrangement gives $ϵ\dot{x}=x-x^3/3+y$. Conversely eliminating $y$ recovers the second-order equation, so this phase plane system is equivalent. At an equilibrium $\dot{y}=0$ forces $x=0$, and the first equation then forces $y=0$. Its Jacobian matrix isThus $ϵ{\lambda }^2-\lambda +1=0$. Expanding $\sqrt{1-4ϵ}=1-2ϵ-2{ϵ}^2-4{ϵ}^3+O({ϵ}^4)$ givesFor sufficiently small positive $ϵ$ both are positive: the equilibrium is an unstable node. For negative $ϵ$ the fast root is negative and the slow root positive, so it becomes a saddle equilibrium.

The nullcline is $\boxed{y=h(x)=x^3/3-x}$. Write $F=x-x^3/3+y$. Away from it at an order-one distance, $F$ is order one on the indicated finite phase-space scales. Thus $\dot{x}=O({ϵ}^{-1})$ while $\dot{y}=-x=O(1)$: the trajectory changes $x$ on the fast time in a singular perturbation $t/ϵ$, with only an $O(ϵ)$ change in $y$. Equivalently $dy/dx=-ϵx/F=O(ϵ)$, which explains the nearly horizontal motion.

For positive $ϵ$, $F_x=1-x^2$ makes the branches $|x|>1$ fast attracting and the middle branch $|x|<1$ fast repelling. The arrows point right where $y>h(x)$ and left where $y<h(x)$, carrying a generic off-nullcline point towards an attracting branch. The reduced description changes near a fold, where $F_x=0$; the rapid/slow distinction there needs a separate transition layer.

The left panel shows positive $ϵ$; the right panel shows negative $ϵ$, including the three specified starting points. Horizontal arrows indicate the leading fast motion.

On the attracting right branch, $x>1$ and $\dot{y}=-x<0$, so the motion goes from $(2,2/3)$ towards the fold $(1,-2/3)$. At essentially fixed $y=-2/3$ it jumps to the other attracting root $x=-2$. The left branch then carries it to $(-1,2/3)$, where it jumps right to $x=2$. This gives the clockwise limiting relaxation oscillation shown in the preceding figure.

Along a slow branch $y=h(x)$, differentiation gives $(x^2-1)\dot{x}=-x$. The right-branch time isReflection symmetry gives the same time on the left. The fast jumps take a time tending to zero as $ϵ\to 0^{+}$, as do the fold-transition corrections at leading order. HenceThis is a leading asymptotic period, not an exact finite-$ϵ$ equality.

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.