= Solution
The <nullclines> are the cubic $w=f(v)$ and the straight line $w=\gamma v$. At an <equilibrium> either $v=0$, or $\gamma=(a-v)(v-1)$. The latter quadratic has maximum $(1-a)^2/4$, so \b[$\gamma>(1-a)^2/4$ leaves only the resting state $(0,0)$]. Its <Jacobian matrix> is
$$
J=\begin{pmatrix}-a/\epsilon&-1/\epsilon\\\gamma&-1\end{pmatrix},\qquad\operatorname{tr}J=-a/\epsilon-1<0,\qquad\det J=(a+\gamma)/\epsilon>0.
$$
Both <eigenvalues> have negative real parts; the rest state is <linearly stable> and <asymptotically stable>.
On the fast time $t/\epsilon$, the <recovery variable> $w$ is approximately fixed. At $w\simeq0$, the <scalar> fast equation has stable roots $v=0,1$ and an unstable root $v=a$. A small kick below $a$ decays towards rest; a kick above it moves rapidly towards the right attracting branch. For finite $\epsilon$, the threshold is displaced from the frozen root by slow recovery, but the leading threshold mechanism is this unstable middle branch.
After the fast excitation, $w_t=\gamma v-w>0$ raises $w$ while $v$ follows the right attracting branch of the cubic. At its upper fold the branch ends and $v$ makes a fast downward jump to the left branch. Recovery then lowers $w$ slowly, with a negative undershoot in $v$, before return towards $(0,0)$. Thus the <action potential> time trace has rapid depolarization, a slower high-voltage excursion, rapid repolarization, and slow recovery. On the fast segments $w$ scarcely changes; on slow segments $w\simeq f(v)$ and $f'(v)<0$. The figure shows a numerically integrated superthreshold excursion and its time trace, rather than just arrows on the <nullclines>.
\Image[/past-exam-of-the-mathematics-and-the-environment-and-mathematical-physiology-course-of-the-university-of-oxford/2010-fhn-excitability.png]
{title=FitzHugh-Nagumo resting nullclines, threshold excursion and action-potential time trace}
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