Past exam of the mathematical physiology course of the University of Oxford / 2011 / 1 / b

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.

The voltage nullcline is $g(v,n)=I^{\ast }$ 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 $ϵ$, 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,A sufficiently steep recovery curve gives positive determinant, while $g_v<-ϵ$ 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, ${\gamma }_K=0.1$, $H=0.8$, $v_K=0.2$, $ϵ=0.01$, $I^{\ast }=0$, $M(v)=(1+e^{-25(v-0.2)}{)}^{-1}$ 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.