= Solution
The <nullclines> are $w=g(v)$ and $v=0$. The two outer pieces of $g$ have slope $-1$; the middle piece has positive slope
$$
\mu=\frac{1-\beta+\alpha}{\beta-\alpha}>0.
$$
The curve is continuous, with folds $(\alpha,-\alpha)$ and $(\beta,1-\beta)$. The unique equilibrium is $(0,-\alpha/(\beta-\alpha))$. Its <Jacobian matrix> has trace $\mu/\epsilon>0$ and determinant $1/\epsilon>0$, so it is unstable. The outer branches attract the fast <voltage> motion.
Let $w_L=-\alpha$ and $w_R=1-\beta$. The hypothesis $\beta-\alpha<1$ gives $0<w_L<w_R<1$. The singular <relaxation oscillation> follows this route:
* On the right branch $v=1-w>0$, $w$ slowly increases from $w_L$ to $w_R$.
* At $w=w_R$, the fast <voltage> jumps from $v=\beta$ to $v=\beta-1<\alpha$.
* On the left branch $v=-w<0$, $w$ slowly decreases from $w_R$ to $w_L$.
* At $w=w_L$, <voltage> jumps from $v=\alpha$ to $v=1+\alpha>\beta$.
The jumps are almost horizontal in the <phase plane>, because $w$ changes negligibly on the fast $O(\epsilon)$ time. Below $w=g(v)$, <voltage> increases; above it, <voltage> decreases; $w$ increases on the right of $v=0$ and decreases on the left. These signs provide a closed trapping region around the indicated route for sufficiently small positive $\epsilon$, with the unstable equilibrium excluded, giving a <periodic orbit> by the planar return-map or <Poincaré-Bendixson theorem> argument. The continuous piecewise linear field is locally Lipschitz, so solution uniqueness is available at the corners.
\Image[/past-exam-of-the-mathematical-physiology-course-of-the-university-of-oxford/2011-piecewise-relaxation.png]
{title=Nullclines, fast jumps, and slow motion of the piecewise linear relaxation oscillator}
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