FitzHugh-Nagumo model

The FitzHugh-Nagumo model represents an excitable cell by a fast activation variable and a slower recovery variable. A common nondimensional form is $\dot{V}=V-V^3/3-W+I$, $\dot{W}=\epsilon (V+a-bW)$. Its cubic activation response and recovery feedback can produce an excitable resting state or sustained oscillations through bifurcations. Particular reductions and travelling-wave equations can have additional state variables; their signs and parameter conventions must be stated separately.

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.

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.

For $\dot{v}=v(\alpha -v)(v-1)-w+I$, $\dot{w}=\beta v-\gamma w$, set $k=\beta /\gamma$. Intersections of the nullclines solve $I=h(v)=v^3-(1+\alpha )v^2+(\alpha +k)v$. Three simple equilibria are possible precisely when $k<M=(1-\alpha +{\alpha }^2)/3$ and $h(u_{+})<I<h(u_{-})$, where $u_{\pm }=(1+\alpha \pm \sqrt{1-\alpha +{\alpha }^2-3k})/3$. To prove this, observe that $h^{\prime }=k-M+3(v-(1+\alpha )/3{)}^2$ has two real zeros exactly when $k<M$; monotonicity on the three resulting intervals counts the intersections. At either endpoint there are a double and a simple root, and at $k=M$, $I=h((1+\alpha )/3)$ there is one triple root.