Velocity diagram of traveling waves for discrete reaction–diffusion equations

Abstract

We consider a discrete version of reaction-diffusion equations. A typical example is the fully overdamped Frenkel–Kontorova model, where the velocity is proportional to the force. We also introduce an additional exterior force denoted by $\sigma$. For general discrete and fully nonlinear dynamics, we study traveling waves of velocity $c=c(\sigma )$ depending on the parameter $\sigma$. Under certain assumptions, we show properties of the velocity diagram $c(\sigma )$ for $\sigma \in [\sigma ^-,\sigma ^+]$. We show that the velocity c is nondecreasing in $\sigma \in (\sigma ^-,\sigma ^+)$ in the bistable regime, with vertical branches $c\ge c^+$ for $\sigma =\sigma ^+$ and $c\le c^-$ for $\sigma =\sigma ^-$ in the monostable regime.

Appendix: Example of discontinuous viscosity solutions

We give in this section an example of a discontinuous viscosity solution.

Proposition 6.1

(Discontinuous viscosity solution for the classical Frenkel–Kontorova model) Consider $\beta >0,$ $\sigma \in {\mathbb R}$ and let $(c,\phi )$ be a solution of

$$\begin{aligned} \left\{ \begin{aligned}&c\phi '(z)=\phi (z+1)-2\phi (z)+\phi (z-1)+f(\phi (z))+\sigma \quad \text{ on } \ {\mathbb R},\quad f(x):=-\beta \cos (2\pi x) \\&\phi \ \text{ is } \text{ non-decreasing }\\&\phi (+\infty )-\phi (-\infty )=1. \end{aligned} \right. \end{aligned}$$

(6.1)

(i) (Sign of the critical velocities)

Then $\sigma ^{\pm }=\pm \beta$ and $c_-<0<c_+$.

(ii) (Discontinuous solution for large $\beta$)

Moreover, if $\beta >1$ and $|\sigma |<\beta -1,$ then $\phi \notin C^{0}$ and $c=0.$

Proof of Proposition 6.1

Step 1: proof of i)

Clearly, we have $\sigma ^{\pm }=\pm \beta$. Let $\sigma =\sigma ^{+}$ and let us show that $c^{+}>0.$ In this case, we can moreover assume that a solution $\phi$ of (6.1) satisfies

$$\begin{aligned} \phi (-\infty )=0,\quad \phi (+\infty )=1 \end{aligned}$$

Integrating over the real line the equation

$$\begin{aligned} c^{+}\phi '(z)=\phi (z+1)+\phi (z-1)-2\phi (z)+g(\phi (z)),\quad g:= f+ \sigma ^+\ge 0 \end{aligned}$$

we get that

$$\begin{aligned} c^{+}=\int _{{\mathbb R}}g(\phi (z))dz\ge 0. \end{aligned}$$

Since $g>0$ on (0, 1),  if $c^{+}=0,$ then

$$\begin{aligned} \phi (z)=0 \text{ or } 1\ \text{ almost } \text{ everywhere }. \end{aligned}$$

Then the equation itself implies that (because $g(\phi )=0$ a.e.)

$$\begin{aligned} \Delta _{1}\phi (z):=\phi (z+1)+\phi (z-1)-2\phi (z)=0\ \text{ almost } \text{ everywhere. } \end{aligned}$$

(6.2)

Because $\phi$ is monotone non-decreasing, up to translation, we have

$$\begin{aligned} \phi (z)=\left\{ \begin{array}{ll} 0&{}\quad \quad \text{ if }\quad z<0\\ 1&{}\quad \quad \text{ if }\quad z>0\\ \end{array}\right. \end{aligned}$$

This leads to a contradiction with (6.2), and shows that $c_+>0$. Similarly, we get $c_-<0$.

Step 2: proof of ii)

Let $|\sigma |<\beta -1$ and let us show that $\phi \notin C^{0}({\mathbb R}).$ For the convenience of the reader we give the proof of this result (which is basically contained in Theorem 1.2 in Carpio et al. [6]).

Assume to the contrary that $\phi \in C^{0}({\mathbb R})$. Notice that because $\phi$ is non-decreasing and $\phi (+\infty )-\phi (-\infty )=1,$ we deduce that

$$\begin{aligned} \phi (z+1)-2\phi (z)+\phi (z-1)\in [-1,1]. \end{aligned}$$

Define now

$$\begin{aligned} \psi (z)=\phi (z+1)-2\phi (z)+\phi (z-1)+f(\phi (z))+\sigma \quad \text{ with }\quad f(\phi ):=-\beta \cos (2\pi \phi ) \end{aligned}$$

Assume by contradiction that $\phi \in C^{0}$. Then we deduce that

$$\begin{aligned} \left\{ \begin{aligned}&\sup _{{\mathbb R}}\psi \ge \beta +\sigma -1>0\\&\inf _{{\mathbb R}}\psi \le -\beta +\sigma +1<0, \end{aligned} \right. \end{aligned}$$

where the strict inequalities follow from $|\sigma |<\beta -1.$ But $c\phi '=\psi$ which implies that $c\phi '$ changes sign. Contradiction. Therefore $\phi \notin C^{0}({\mathbb R})$, which also implies that $c=0.$

This ends the proof of the proposition.