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This nondimensional two-variable reaction–diffusion system combines a fast cubic local response with slow negative feedback. With $D>0$, $0<\alpha <1$, $\beta >0$ and $\epsilon \ll 1$, its spatially uniform rest state at $I=0$ is linearly stable: the Jacobian has trace $-\alpha -\epsilon \beta$ and determinant $\epsilon (1+\alpha \beta )$. For suitable parameters a finite perturbation can nevertheless cause a large excursion followed by recovery. Frozen recovery gives a bistable cubic reaction-diffusion equation; the full coupled travelling-wave boundary-value problem determines whether front pairs form a pulse. This is a generic excitable-medium formulation of the FitzHugh-Nagumo model, without assigning any biological variable or application.

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.

At an equilibrium $(v_{\ast },k{v}_{\ast })$, set $p=f^{\prime }(v_{\ast })$. The Jacobian matrix, its trace and its determinant areThe eigenvalues and their discriminant areThus the complete local linear stability test isIn the last two cases, $\mathcal{D}>0$ gives a node, while $\mathcal{D}<0$ gives a focus; the sign of $T$ distinguishes inward and outward trajectories. At $\mathcal{D}=0$, a nonzero repeated eigenvalue gives a degenerate node. At $p=\gamma <k$ the eigenvalues are purely imaginary, so linear stability analysis alone does not decide asymptotic stability; with the required crossing and nondegeneracy this is a Hopf bifurcation. At $p=k$ there is a zero eigenvalue, requiring a nonlinear test and generically marking the saddle-node bifurcation in the preceding answer.

For a unique simple equilibrium, $p<k$, so it can be a sink equilibrium or a source equilibrium. The number of equilibria alone does not decide stability. If $k\le \gamma$, the unique simple equilibrium is always a sink equilibrium. If $\gamma <k$, it can become a source equilibrium where $\gamma <p<k$. The same conclusion follows particularly simply if $\gamma >M$: the trace is negative at every nonsaddle equilibrium.

For three simple equilibria ordered by their $v$ coordinates, the middle one has $p>k$ and is a saddle equilibrium; the two outer ones have $p<k$. When both outer equilibria are sink equilibria, each has an attracting neighborhood, giving bistability. In the usual two-basin portrait, the stable manifold of the saddle equilibrium separates their basins. Each branch of its unstable manifold initially leaves towards a different side; its eventual destination requires a global argument. If $k\le \gamma$, both outer equilibria must be sink equilibria. If $\gamma <k$, either or both outer equilibria can instead be source equilibria; the same test $p\lessgtr \gamma$ applies separately to each. The three lower panels illustrate all three possibilities up to exchanging left and right.

The arrows in the phase plane follow directly from the nullclines: below $w=f(v)+I$ the horizontal component points right, and above it left; below $w=kv$ the vertical component points up, and above it down. Near a stable focus trajectories spiral inward; near an unstable focus they spiral outward. Near a saddle equilibrium they approach along its stable manifold and leave along its unstable manifold. These directions, together with the local tests, determine the sketches.

There is also a useful global conclusion when the unique equilibrium is a source equilibrium. Withthe leading term in $v$ is $-\beta v^4$, so sufficiently large ellipses are trapping regions. A small repelling neighborhood of the hyperbolic source equilibrium can be removed to form an invariant annulus. The Poincaré-Bendixson theorem then guarantees at least one periodic orbit in that annulus. The numerical sketch illustrates the resulting sustained oscillation; this argument does not assert uniqueness of a limit cycle. A stable equilibrium can instead represent a resting membrane potential, with perturbations producing damped relaxation; near a threshold, a sufficiently large perturbation can produce a large excursion before recovery. Global basin boundaries and coexistence of periodic orbits need more information than the number and local stability of equilibria.

Put $k=\beta /\gamma$. The nullclines of the space-clamped FitzHugh-Nagumo model areThe first nullcline is a downward-leading cubic polynomial, shifted vertically by $I$; the second nullcline is a line through the origin with slope $k>0$. Their intersections satisfyWriting $M=(1-\alpha +{\alpha }^2)/3$, differentiation givesIf $k>M$, $h$ increases strictly, so there is exactly one equilibrium. If $k=M$, $h$ is still strictly increasing, although its derivative vanishes at one point, so the equilibrium is again unique. If $0<k<M$, defineHere $h$ has a local maximum at $u_{-}$ and a local minimum at $u_{+}$; thereforeAt either endpoint there are two distinct equilibria, one a double root. These are the saddle-node bifurcation boundaries omitted by the generic one-or-three wording. At $k=M$ and $I=h((1+\alpha )/3)$ there is one triple root. This is the Cubic FitzHugh-Nagumo nullcline criterion.

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.

The linearization of this reduction of the FitzHugh-Nagumo model at its trivial equilibrium point isA double zero eigenvalue requires both the constant and linear coefficients to vanish. Hence the codimension-two point isThere the polynomial is $m^2(m+3/2)$. The kernel of a linear map of $L$ is one-dimensional, so its zero eigenvalue has a nontrivial size-two block in Jordan normal form. This is a reflection-symmetric double-zero point of Bogdanov–Takens bifurcation type, with one stable direction; the nonlinear coefficients are computed below rather than assumed.

TODO: biological interpretations and organism-specific propagation are user-deferred. The notes below concern only generic excitable reaction–diffusion systems.

A front solution joins different asymptotic states in space. In a reaction–diffusion system, local nonlinear dynamics select possible states while diffusion couples neighboring points. A travelling wave with profile $u(x,t)=U(\xi )$, $\xi =x-ct$, converts a partial differential equation into a boundary value problem for the shape and speed. A pulse instead returns to the same resting state on both sides and can contain leading and trailing fronts.

A useful nondimensional cubic fast-slow reaction-diffusion model, in a convention of the FitzHugh-Nagumo model, iswhere $D>0$, $\epsilon \ll 1$, and $\beta >0$. Here $u$ is a fast activation variable, $w$ a slow negative-feedback variable and $I$ a generic drive; no physiological variable is assigned. Without spatial coupling, nullclines are $w=f(u)+I$ and $w=u/\beta$. At $I=0$, the resting state $(0,0)$ has Jacobian matrixIt has linear stability. With appropriate parameter choices a sufficiently large perturbation makes a fast excursion to the other outer branch of the cubic, followed by slow feedback and return. Thus a stable rest state can be excitable without being linearly unstable. Other parameters or applied drive can yield multiple equilibria or oscillations; these behaviours should not be asserted for every parameter set.

The fast-subsystem front can be obtained explicitly by freezing $w=I=0$. For the bistable cubic reaction-diffusion equation, $u=0,1$ are stable and $u=\alpha$ is unstable. The decreasing logistic front of a bistable cubic equation isTo check it, $U^{\prime }=-U(1-U)/\sqrt{2D}$ and substitution into $D{U}^{\prime \prime }+c{U}^{\prime }+f(U)=0$ cancels every term. For $\alpha <1/2$, the high state advances to the right; for $\alpha >1/2$, the direction reverses. At $\alpha =1/2$ the front is stationary. More generally, multiplying the travelling-wave equation by $U^{\prime }$ and integrating yieldsThis links propagation direction to the bias between the two stable states and recovers Maxwell balance for a scalar reaction-diffusion front at zero speed. It is a bistable selected-speed problem, rather than a universal Fisher-type minimum-speed formula.

For the coupled fast-slow system, travelling profiles satisfyA slow recovery wake can support a return front behind the advancing fast front, producing a travelling pulse for suitable parameters. Its width reflects the slow recovery time and propagation speed, while each sharp edge has the faster diffusion–reaction scale. The frozen-recovery speed above is a controlled subsystem result, not an exact speed for every coupled pulse. A definite pulse speed requires parameters and connecting boundary condition; none are specified by this brief-note question.

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.

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.