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For the scalar reaction-diffusion equation $v_t=D{v}_{xx}-(v-r_1)(v-r_2)(v-r_3)$ with $D>0$ and $r_1<r_2<r_3$, an exact decreasing travelling wave isIndeed $V^{\prime }=-(V-r_1)(r_3-V)/\sqrt{2D}$, and substitution into $D{V}^{\prime \prime }+c{V}^{\prime }-(V-r_1)(V-r_2)(V-r_3)=0$ leaves the factor $(r_1+r_3)/2-r_2-c/\sqrt{2D}$, proving the speed formula. A zero speed gives the Maxwell balance for a scalar reaction-diffusion front. In a FitzHugh-Nagumo model this calculation requires treating the recovery variable as frozen externally: retaining $w_t=\beta v-\gamma w$ with exactly constant $w$ forces $v$ to be constant.

There is a consistency issue in the printed model. If $w\equiv w_0$ is imposed while retaining its evolution equation, then $0=\beta v-\gamma w_0$ forces $v\equiv \gamma w_0/\beta$. Thus the full two-equation system has no nonconstant travelling wave with exactly constant $w$. Its constant solution exists only when $I=w_0-f(\gamma w_0/\beta )$. The intended wave calculation must treat the recovery variable as externally frozen, leaving the scalar reaction-diffusion equationFor this frozen-recovery reduction, letThe cubic polynomial $f$ has a local minimum at $a_{-}$ and a local maximum at $a_{+}$. Hence the conditiongives three distinct real roots $r_1<r_2<r_3$ of $g$, withThe outer roots are stable for the scalar spatially uniform dynamics and the middle root is unstable. Seek a travelling wave $v(x,t)=V(z)$, where $z=x-ct$. The travelling-wave reduction of a reaction-diffusion system givesLet $\delta =r_3-r_1$ and chooseIt has $V(-\infty )=r_3$, $V(+\infty )=r_1$ andSubstitution leavesConsequently the frozen-recovery cubic front isReflection $x\mapsto -x$ reverses the front and its speed. A nonzero speed requires $r_2\ne (r_1+r_3)/2$; equality gives a stationary front and is the Maxwell balance for a scalar reaction-diffusion front. Since the roots sum to $1+\alpha$, this stationary value is $I=w_0-f((1+\alpha )/3)$. For the simple admissible choice $I=w_0$, the roots are $0,\alpha ,1$ andThis particular front moves when $\alpha \ne 1/2$. At $\alpha =1/2$, other currents inside the displayed open interval give moving fronts.

Use the three standard reduction assumptions: $m$ relaxes much faster than $V,n,h$ and can be replaced by $m_{\infty }(V)$; the two slower gates have the same time constant, ${\tau }_n(V)={\tau }_h(V)=\tau (V)$; and their limiting open fractions have a voltage-independent sum, $n_{\infty }(V)+h_{\infty }(V)=K$. The constant-sum assumption is essential: equality of two time constants by itself does not preserve a moving voltage-dependent sum.

For $q=n+h$, addition of the two gate ordinary differential equations givesThe specified initial condition is $q(0)=K$. Uniqueness for this first-order linear differential equation gives $q(t)=K$, hence $h=K-n$ for all time. The remaining two ordinary differential equations areThese assumptions give a planar reduction of the Hodgkin-Huxley model. The FitzHugh-Nagumo model keeps its fast activation and slow recovery variable geometry, approximating the voltage response by a cubic polynomial and the recovery response by a linear function after nondimensionalization. The polynomial equations in part (c) are a further model simplification; they are not an exact algebraic consequence of arbitrary original gate functions.

A recovery variable represents feedback that reduces activation after an excitation in a FitzHugh-Nagumo model. Its equation must be specified separately from the activation equation. Holding it externally constant gives a scalar reduction; requiring it to remain constant while keeping its original evolution equation generally imposes an additional constraint and need not permit the same solutions.