= Solution
Take $\gamma>\gamma_c$ and let $\varepsilon\to0^+$ with the other parameters fixed. The <nullclines> are the <cubic polynomial> graph $w=f(v)$ and the line $w=\gamma v$. In the <phase plane>, $v$ increases below the <cubic polynomial> graph and decreases above it; $w$ increases below the line and decreases above it. The fast <ordinary differential equation>, in time $s=t/\varepsilon$, freezes $w$ and gives $dv/ds=f(v)-w$.
At $w=0$, the fast roots are $0,a,1$. Their <linear stability> follows from $f'(0)=-a<0$, $f'(a)=a(1-a)>0$ and $f'(1)=a-1<0$. Thus a fixed $0<v^*<a$ falls towards the resting branch, whereas a fixed $a<v^*<1$ rises towards $v=1$. If $v^*>1$, the first fast segment instead settles down towards $1$, followed by the same recovery excursion. The threshold $a$ is the leading <singular perturbation> threshold; finite-$\varepsilon$ corrections matter for initial values approaching $a$ with $\varepsilon$.
Let
$$
d=\sqrt{1-a+a^2},\qquad
v_L=\frac{1+a-d}{3},\qquad v_R=\frac{1+a+d}{3}.
$$
These are the roots of $f'(v)=0$. The outer branches $v<v_L$ and $v>v_R$ of the <critical manifold> are attracting under fast motion; the middle branch is repelling. The full excursion has four stages.
* \b[Fast activation:] nearly horizontal movement towards $(1,0)$, on the $t=O(\varepsilon)$ <time scale>, apart from the usual matching tails.
* \b[First slow phase:] motion along $w=f(v)$ from $v=1$ down to $v=v_R$. Here $w$ rises and $v$ slowly falls, because $\dot w=\gamma v-f(v)>0$ on the positive outer branch. This takes order-one time for fixed $\gamma$.
* \b[Fast repolarization:] near the right <fold point of a critical manifold>, the attracting branch ends. With $w_R=f(v_R)$ nearly constant, $v$ drops to the negative root $v_J$ of $f(v_J)=w_R$. The jump takes the fast <time scale>; a thin neighborhood of the fold has an intermediate <fold delay>.
* \b[Second slow phase:] motion along the negative attracting branch from $v_J$ towards $0$ from below, while $w$ falls towards zero. This is the refractory recovery and has order-one duration to a fixed small cutoff; exact return to the <equilibrium> takes infinite time.
For this <cubic polynomial>, the landing point is explicit. Since $f(v)-f(v_R)=-(v-v_R)^2(v-v_J)$, comparison of the quadratic <coefficients> gives
$$
\boxed{v_J=1+a-2v_R=\frac{1+a-2\sqrt{1-a+a^2}}3<0.}
$$
Thus a graph of $v(t)$ has a rapid rise, a slowly falling elevated plateau, a rapid fall through zero to a negative undershoot, and a slow approach back to zero. The figure shows both the <phase plane> and a computed time trace. The smaller initial perturbation stays near the resting branch and produces no elevated plateau.
\Image[/past-exam-of-the-mathematical-physiology-course-of-the-university-of-oxford/2023-excitable-pulse.png]
{title=Excitation threshold and the four stages of a FitzHugh-Nagumo pulse}
The four-stage description concerns the <singular limit>. Close to a simple <fold point of a critical manifold>, balancing the local quadratic activation term with the slow drift gives $\Delta v=O(\varepsilon^{1/3})$ and $\Delta t=O(\varepsilon^{2/3})$ for fixed parameters. These layers shrink compared with the slow phases and do not alter the leading time integrals below.
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