Regularity of a nonlinear ODE [Traveling wave solutions of parabolic systems]

Question

In the book of Volpert on Traveling wave solutions of Parabolic Systems (AMS), one reads "the following assertion is readily proved and we shall not discuss it in detail". The same result is tacitely assumed in Evans book on Partial Differential Equations when dealing with traveling wave solutions of the bistable equation.

Proposition: Let $\sigma \in \mathbb{R}$. If a function $w:\mathbb{R} \rightarrow\mathbb{R}$ in $C^2(\mathbb{R})\cap C^1_b(\mathbb{R})$ satisfies the ODE \begin{equation} w''+\sigma w'+f(w)=0\qquad \text{and}\qquad\lim_{t\rightarrow \pm\infty}w(t)=w_\pm\in\mathbb{R} \end{equation} then there exist (and are zero) the two limits \begin{equation} \lim_{t\rightarrow \pm\infty}w''=\lim_{t\rightarrow \pm\infty}w'=0 . \end{equation}

PS the hypotheses on $f$ are not explicitely written, but I think that $f\in C^0_b(\mathbb{R})$ is sufficient.

Can someone give me a reference for that kind of results?

Thanks in advance, Josh.

Answer 1

@Josh: I don't have any precise reference in mind for that, maybe in the Coddington-Levinson? but as far as I remember it is mostly for linear ODE's

Note first that reversing time $t\to -t$ is equivalent to changing $\sigma\to-\sigma$, so you only need to study one side (say $t\to+\infty$). The case $\sigma=0$ is a borderline one that I am not too sure how to deal with. I assume below that $\sigma\neq 0$ (but I guess $\sigma$ is the propagation speed of the wave, so it should be OK to discard stationary waves).

The starting point is to rewrite $w''+\sigma w'+f(w)=0$ as $$ (e^{\sigma t}w')'=-e^{\sigma t}f(w).\hspace{2cm}(E) $$

Step 1 The first thing you need to show is that $w_{\pm}$ are necessarily steady-states, i-e $f(w_{\pm})=0$ (stationary equilibrium solutions of the ODE, usually one is stable while the other is unstable). In order to see this assume by contradiction that $f(w_+)\neq 0$ (again, it is enough to look at $t\to+\infty$).

• If $\sigma>0$ then by (E) we have $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ not integrable when $t\to\infty$, so $e^{\sigma t}w'\sim C\int e^{\sigma t}=Ce^{\sigma t}$ hence $w'\sim C\neq 0$. This shows that $w$ blows-up linearly and contradicts $w(\infty)=w_+$.
• If now $\sigma<0$ then $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ becomes integrable, and thus $e^{\sigma t}w'\to C$ for some limit $C\in \mathbb{R}$. If $C\neq 0$ then $w'\sim Ce^{-\sigma t}$ blows exponentially, which contradicts again $w(\infty)=w_+$. Thus $C=0$, and integrating $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ from $t$ to $\infty$ you get $e^{\sigma t}w'-0\sim C(e^{\sigma t}-0)$, hence again linear blow-up $w'\sim C\neq 0$.

Step 2 Once you know that $f(w)\to f(w_+)=0$ (here I am definitely using the continuity of $f$) the heuristic idea is quite simple: the initial ODE $w''+\sigma w+f(w)=0$ roughly becomes a 1st order linear ODE in $v=w'$ \begin{equation} v'+\sigma v=-f(w)\approx -f(w_+)= 0,\qquad v=w'.\hspace{2cm}(E') \end{equation}

This linear ODE $v'+\sigma v=0$ gives either the trivial solution $v'=v=0$ (which means precisely $w''=w'=0$), or $v'$ and $v$ proportional to $e^{-\sigma t}$. If $\sigma>0$ you see that both the trivial and exponential cases are admissible and lead to $v',v=w'',w'\to 0$ when $t\to\infty$. If now $\sigma<0$ the exponential blow-up $v=w'\sim e^{-\sigma t}$ is excluded because you assume $w(t)\to w^+=cst$, so the only possibility is again $w'',w'=u',u=0$.

Of course this step 2 is only formal, and rigorously justifying (E') from (E) requires tedious and technical computations similar to those in step 1. For that you may want to use again (E) with now $f(w)\to f(w_+)=0$ hence $$ (e^{\sigma t}w')'=o\left(e^{\sigma t}\right), $$ and distinguish again integrability or linear/exponential blowup at infinity as in step 1.

I hope this helps!

Answer 2

As @Leo Monsaingeon pointed out the equation can be rewritten in the equivalent form $( e^{\sigma t} w' ( t ) )' =-e^{\sigma t} f ( w ( t ) )$. For the time being let us suppose $\sigma<0$. Integrating the previous relation we get \begin{equation} e^{- | \sigma | t} w' ( t ) =e^{- | \sigma | t_{0}} w' ( t_{0} ) - \int_{t_{0}}^{t} e^{- | \sigma | s} f ( w ( s ) ) \text{d} s . \qquad \qquad \textbf{(1)} \end{equation} Next, we observe that since $w\in C^1_b(\mathbb{R})$ and $\lim_{t \rightarrow \pm \infty} w ( t ) =w_{\pm}\in \mathbb{R}$, by taking the limit for $t_{0} \rightarrow+ \infty$ in $\textbf{(1)}$ we get \begin{equation} w' ( t ) = \frac{1}{ e^{- | \sigma | t}} \int_{t}^{+ \infty} e^{- | \sigma | s} f ( w ( s ) ) \text{d}s. \end{equation} On the other hand, for $t \rightarrow \pm \infty$ one has [Hôpital rule]: \begin{equation} \lim_{t \rightarrow \pm \infty} w' ( t ) = \lim_{t \rightarrow \pm \infty} \frac{-e^{- | \sigma | t} f ( w ( t ) )}{- | \sigma | e^{- | \sigma | t}} = \frac{1}{| \sigma |} f ( w_{\pm} ) , \end{equation} and this also implies that $f ( w_{\pm} ) =0$ cause $w$ has been assumed to have finite limits around $\pm \infty$. Therefore also $w''(\pm \infty)=0$. For the case $\sigma>0$ just first take $t_0\rightarrow -\infty$ and then $t\rightarrow \pm \infty$.

Remark. The argument is valid in a little bit more general setting. More precisely: if $v\in C^0(\mathbb{R})\cap C^2(\mathbb{R})\cap C^1_b(\mathbb{R})$ and if \begin{equation} \lim_{t\rightarrow \pm \infty} f(w(t))=f_\pm \in \mathbb{R} \end{equation} then \begin{equation} \lim_{t\rightarrow \pm \infty} w''(t)=0\qquad \textbf{and}\qquad \lim_{t \rightarrow \pm \infty} w' ( t )=-\frac{f_\pm}{\sigma} . \end{equation}