= Solution
{c}
Put $p=y'$. The <dynamical system> is
$$
y'=p,\qquad p'=\frac{p-e^y}{\epsilon}.
$$
The <nullclines> are $p=0$ and $p=e^y$. There are no equilibria, since they never intersect. To the right in the $(y,p)$ plane means $p>0$, and to the left means $p<0$. The arrows point upwards above $p=e^y$ and downwards below it.
Near the repelling <critical manifold> $p=e^y$, an orbit can move on the slow $x=O(1)$ scale, with $y'\simeq e^y$. More precisely the slow trajectory has $p=e^y+\epsilon e^{2y}+\cdots$. Away from this narrow region, $p$ changes rapidly on $x=O(\epsilon)$; an $O(1)$ change of $y$ in a fast layer has $p=O(\epsilon^{-1})$.
\Image[/past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2009-nonlinear-phase.png]
{title=Slow motion near the exponential nullcline and fast terminal departure}
For example, take a fixed $A<-\log2$. A trajectory starting at $y=A$ can shadow the slow curve, then depart upwards so that $y$ rapidly increases to zero at the right endpoint. This gives a terminal <outflow boundary layer> at $x=1$, of width $O(\epsilon)$. For $-\log2<A<0$, the outer trajectory reaches a positive value near one, and the terminal adjustment is downwards: the orbit turns below the nullcline, crosses $p=0$, and returns to $y=0$. The boundary layer is still at $x=1$. These directions follow from the phase portrait and agree with the matched solution below.
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