# FitzHugh- Nagumo caricature of the Hodgkin-Huxley equations

I've been trying to solve this problem from Elements of Applied Bifurcation Theory, but even with the hints given I haven't found out how to proceed yet. I would appreciate any further hints or insights you could give me in order to make some progress.

It's exercise 2.6.(2):

The following system of partial differential equations is the FitzHugh- Nagumo caricature of the Hodgkin-Huxley equations modeling the nerve impulse propagation along an axon:

$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}-f_a(u)-v$,

$\frac{\partial v}{\partial t} = bu$,

where $u(x,t)$ represents the membrane potential, $v=v(x,t)$ is a "recovery" variable, $f_a(u)=u(u-a)(u-1)$, $1>a>0,b>0,-\infty < x < \infty$, and $t>0$.

Traveling waves are solutions to these equations of the form

$u(x,t)=U(\xi)$, $v(x,t)=V(\xi)$, $\xi = x+ct$, where $c$ is an a priori unknown wave propagation speed. The functions $U(\xi)$, $V(\xi)$ are the wave profiles.

(a) Derive a system of three ordinary differential equations for the profiles with "time" $\xi$. (Hint: introduce an extra variable: $W=\dot{U}$.

(b) Check that for all $c>0$ the system for the profiles (the wave system) has a unique equilibrium with one positive eigenvalue and two eigenvalues with negative real parts. (Hint: First, verify this assuming that eigenvalues are real. The, show that the characteristic equation cannot have roots on the imaginary axis, and finally, use the continuous dependence of the eigenvalues on the parameters.

(c) Conclude that the equilibrium can be either a saddle or a saddle- focus with a one-dimensional unstable and a two-dimensional stable in- variant manifold, and find a condition on the system parameters that de- fines a boundary between these two cases. Plot several boundaries in the $(a,c)$-plane for different values of $b$ and specify the region corresponding to saddle-foci. (Hint: at the boundary the characteristic polynomial $h(\lambda)$ has a double root $\lambda_0:h(\lambda_0)=h'(\lambda_0)=0.$).

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Why not try substituting $U(x+ct)$ and $V(x+ct)$ into the PDE and see what you get? –  Ｊ. Ｍ. Oct 6 '10 at 17:38
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## 1 Answer

Here's something to get you started: $u_t=c U'(\xi)$, $u_{xx}=U''(\xi)$, and $v_t=c V'(\xi)$. Plugging this into the system of PDEs gives you a system of two ODEs for $U(\xi)$ and $V(\xi)$. The first of these ODEs is second order in $U$, but you can turn it into a first order equation by replacing $U'(\xi)$ by an auxiliary function $W(\xi)$ and replacing $U''(\xi)$ by $W'(\xi)$, and then augmenting the system with the extra equation $U'=W$. Now you have a system of three first order ODEs (i.e., a dynamical system), and can proceed to part (b).

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