 Solution

Write $M(v)=m^3(v)$. In the physically relevant strip $-v_K<v<1$, $0<n<H$, the zero-current nullcline satisfiesThe left side increases strictly from zero to infinity, so there is exactly one nullcline height $n=N(v)$ for each interior voltage. If $M$ is positive at the endpoints, its curve joins $(-v_K,H)$ to $(1,0)$; when low-voltage activation is idealized as exactly zero, the limiting left branch can lie on $v=-v_K$.

The signs of the derivatives explain its folds:At low voltage, potassium's approximately linear contribution dominates and $g_v>0$. During rapid sodium activation the last negative term can dominate, giving $g_v<0$. Once activation saturates, $M^{\prime }$ becomes small and $g_v>0$ again. This folded conductance-model nullcline has the familiar shape: the curve has descending outer branches and an ascending middle branch in the $(v,n)$ plane. Equivalently, an intermediate fixed $n$ can give three zeros of $g$ as voltage rises, with positive, negative and positive $g_v$ at those crossings.

The rapid-activation description permits this folded shape; exact fold positions depend on $M,{\gamma }_K,H,v_K$. The figure supplies an explicit admissible sigmoid example. Its right panel shows $g=0$ as well as the recovery nullcline and an oscillating trajectory.