= Solution
The <voltage> <nullcline> is $g(v,n)=I^*$ and the recovery <nullcline> is $n=n_\infty(v)$. In the relevant region $g_n>0$, so below the <voltage> curve $\dot v>0$ and above it $\dot v<0$. Below the recovery curve $\dot n>0$, while above it $\dot n<0$.
For small $\epsilon$, the outer <voltage> branches with $g_v>0$ attract fast <voltage> motion; the middle branch with $g_v<0$ repels it. Arrange the intersections so the equilibrium is on that middle branch, with recovery increasing along the high-<voltage> attracting branch and decreasing along the low-<voltage> attracting branch. The trajectory then drifts towards a fold, jumps rapidly to the other attracting branch at nearly constant $n$, and repeats. This is a <relaxation oscillation>.
The local <linear stability> calculation supports the construction. At an equilibrium,
$$
J=\begin{pmatrix}-g_v/\epsilon&-g_n/\epsilon\\n_\infty'&-1\end{pmatrix},\qquad\operatorname{tr}J=-g_v/\epsilon-1,\qquad\det J=\frac{g_v+g_nn_\infty'}{\epsilon}.
$$
A sufficiently steep recovery curve gives positive determinant, while $g_v<-\epsilon$ gives positive trace. Hence the equilibrium is a repeller. The bounded physical rectangle can be chosen forward invariant, and a small neighborhood of the repeller removed to give a trapping annulus. The <Poincaré-Bendixson theorem> then supplies a <periodic orbit>. These conditions can be realized because the recovery function and the applied current control the intersection and its slope; oscillation is possible, not forced for every choice.
For the illustration, $\gamma_K=0.1$, $H=0.8$, $v_K=0.2$, $\epsilon=0.01$, $I^*=0$, $M(v)=(1+e^{-25(v-0.2)})^{-1}$ and $n_\infty(v)=0.8/(1+e^{-20v})$. The rectangle $[-0.2,1]\times[0,0.8]$ has inward-pointing flow, and the displayed numerical trajectory realizes the folded-branch mechanism.
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