Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2007 / A / 2

Define $y=k^{-1}\dot{x}+x^3/3-x$. Differentiation and the original equation give $\dot{y}=(\lambda -x)/k$, while rearrangement gives $\dot{x}=k(y-x^3/3+x)$. Write $h(x)=x^3/3-x$.

On the fast scale $\tau =kt$,Fast trajectories are horizontal. Their equilibria lie on the cubic $y=h(x)$; the transverse linearization is $-h^{\prime }(x)=1-x^2$, so the outer branches $|x|>1$ attract and the middle branch repels. On a horizontal line, arrows point right when $y>h(x)$ and left when $y<h(x)$.

The source prints the slow time as $T=k/t$. The slow variable needed here is $T=t/k$. With this correction,At leading order $y=h(x)$ andThe reciprocal printed variable would give $-T^2{y}_T=\lambda -x$, rather than the autonomous slow equations used in the stated period formula.

For $-1<\lambda <1$, the left attracting branch runs from $x=-2$ to the fold $x=-1$, where $y=2/3$. The horizontal fast jump at this height lands at $x=2$. The right branch then runs from $2$ to the fold $1$, where $y=-2/3$, followed by a fast jump to $-2$. The equilibrium at $x=\lambda$ is on the repelling middle branch. The two slow segments dominate the period:The jump times are negligible on this $O(k)$ scale. Substituting $u=-x$ in the first integral also gives the compact valueFor $\lambda =1/2$, the negative slow segment takes $k[1-\frac{3}{4}\log (5/3)]$, and the positive one takes $k[2-\frac{3}{4}\log 3]$. Thus $x(t)$ rises slowly from $-2$ to $-1$, jumps to $2$, decreases more slowly to $1$, then jumps back to $-2$; the jumps are nearly vertical in the time trace.

For $\lambda >1$ the fast horizontal field and attracting branches are unchanged. The right slow branch has arrows toward $x=\lambda$ from both directions; the left branch still moves toward $-1$ and jumps right. The equilibrium $(\lambda ,h(\lambda ))$ is stable, so the large relaxation cycle is replaced by approach to this equilibrium. If $\lambda >2$, a landing at $2$ moves right toward $\lambda$, rather than left. The sketches display the representative case $\lambda =3/2$.