= Solution
Write $M(v)=m^3(v)$. In the physically relevant strip $-v_K<v<1$, $0<n<H$, the zero-current <nullcline> satisfies
$$
\boxed{\frac{n^4}{H-n}=\frac{(1-v)M(v)}{\gamma_K(v+v_K)}.}
$$
The 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:
$$
g_n=4\gamma_K(v+v_K)n^3+(1-v)M(v)>0,
$$
$$
g_v=\gamma_Kn^4+(H-n)M(v)-(1-v)(H-n)M'(v),\qquad N'(v)=-g_v/g_n.
$$
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, $M'$ becomes small and $g_v>0$ again. This <folded conductance-model nullcline> has the familiar shape: the curve has \b[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.
\Image[/past-exam-of-the-mathematical-physiology-course-of-the-university-of-oxford/2011-gating-phase-plane.png]
{title=Activation-dependent voltage roots and a planar conductance-model relaxation oscillation}
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