= Solution
{c}
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 $\tau=kt$,
$$
\frac{dx}{d\tau}=y-h(x),\qquad \frac{dy}{d\tau}=\frac{\lambda-x}{k^2}\sim0.
$$
Fast trajectories are horizontal. Their <equilibria> lie on the cubic $y=h(x)$; the transverse <linearization> is $-h'(x)=1-x^2$, 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 \b[$T=t/k$]. With this correction,
$$
k^{-2}\frac{dx}{dT}=y-h(x),\qquad \frac{dy}{dT}=\lambda-x.
$$
At leading order $y=h(x)$ and
$$
(x^2-1)\frac{dx}{dT}=\lambda-x.
$$
The reciprocal printed variable would give $-T^2y_T=\lambda-x$, 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>:
$$
\boxed{\mathcal T\sim k\left[\int_{-2}^{-1}\frac{x^2-1}{\lambda-x}\,dx+
\int_1^2\frac{x^2-1}{x-\lambda}\,dx\right].}
$$
The jump times are negligible on this $O(k)$ scale. Substituting $u=-x$ in the first <integral> also gives the compact value
$$
\mathcal T\sim k\left[3+(\lambda^2-1)\log\frac{4-\lambda^2}{1-\lambda^2}\right].
$$
For $\lambda=1/2$, the negative slow segment takes $k[1-\tfrac34\log(5/3)]$, and the positive one takes $k[2-\tfrac34\log3]$. 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 $\lambda=3/2$.
\Image[/past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2007-relaxation-orbit.png]
{title=Fast horizontal flow, slow cubic branches, and the asymmetric relaxation cycle}
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