Planar reduction of the Hodgkin-Huxley model

Replacing the fast activation gate by $m=m_{\infty }(V)$, taking ${\tau }_n(V)={\tau }_h(V)$ and assuming $n_{\infty }(V)+h_{\infty }(V)=K$ constant reduces the Hodgkin-Huxley model to equations for $V,n$ when $n(0)+h(0)=K$. Indeed $q=n+h$ satisfies $\tau (V)\dot{q}=K-q$, so $q=K$ and $h=K-n$. The FitzHugh-Nagumo model additionally simplifies the two resulting response functions to a cubic activation equation and a linear recovery equation.

For $g(v,n)={\gamma }_K(v+v_K)n^4-(1-v)(H-n)M(v)$, with rapidly increasing saturating activation $M$, the zero-current nullcline satisfies $n^4/(H-n)=(1-v)M(v)/[{\gamma }_K(v+v_K)]$ in the physical strip. The left side increases strictly, so it defines one height for each voltage. Since $g_n>0$, its slope is $-g_v/g_n$. Rapid activation can make $g_v$ negative between two positive-slope current regimes, giving two descending attracting branches and an ascending repelling branch in the voltage–recovery phase plane. Sufficient activation steepness is a parameter condition; the shape is not forced by monotonicity alone.