Introduction

We consider the Nagumo reaction-diffusion equation
$$\begin{align*}\text{PDE} & \quad & \frac{\partial u}{\partial t} & = \frac{\part... ...arepsilon \sech (\beta x), & \quad &-\infty < x < \infty, \nonumber \end{align*}$$
The initial condition parameter $\varepsilon=0.01$ is chosen to give a small globalized perturbation to the unstable steady state solution u=0.For any $\beta \le \beta_0 = 2/3$, the initial condition will evolve into a monostable travelling wave solution, i.e., we have

$$\lim_{t \rightarrow \infty} \lvert u(x,t) - \overline{u}(x-ct-\phi) \vert = 0,$$

with

$$c = \beta + \frac{4}{9}\frac{1}{\beta}$$

the monostable wavespeed relationship between the constant travelling wave velocity c and the initial condition parameter $\beta$.

If we set $\beta = \beta_0$ then we will observe evolution to a wave of minimum wavespeed

$$c_0 = \min_{\beta} \biggl\{ \beta + \frac{4}{9}\frac{1}{\beta}\biggr\} = \frac{4}{3}$$

Any travelling wave solution takes the form $\overline{u}(z)=\overline{u}(x-ct).$For the Nagumo PDE it follows that $\overline{u}(z)$ must satisfy the ordinary differential equation

$$\frac{d\overline{u}}{dz^2} + c \frac{d\overline{u}}{dz} + \frac{4}{9} \overline{u}(1-\overline{u})(1+2\overline{u}) =0$$

is satisfied. For $c=c_0 = \frac{4}{3}$ we have the exact travelling wave solution

$$\overline{u}(z) = \frac{1}{1+e^{\beta_0 z}} = \frac{1}{1+e^{\frac{2}{3} z}}.$$

Now the initial condition and this minimum speed travelling wave have the same asymptotic behaviour as $x \rightarrow \infty$, i.e.,

$$u(x,0) \sim e^{-\beta x} \qquad \text{and} \qquad \overline{u}(x) \sim e^{-\beta x} \qquad \text{as } x \rightarrow \infty,$$

This is sufficient to ensure that the IC will evolve to the exact travelling wave solution under time integration. That is,

$$\lim_{t \rightarrow \infty} \biggl\lvert u(x,t) - \frac{1}{1+Ke^{\frac{2}{3}(x-4/3t)}} \biggr \vert = 0,$$

when $\beta=2/3$.