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.

  • $\begingroup$ is $\sigma$ a constant or a function? $\endgroup$ – ofer zeitouni Oct 19 '13 at 22:33
  • $\begingroup$ $\sigma$ is a constant. $\endgroup$ – Kosh Oct 19 '13 at 22:53
  • $\begingroup$ Dear @Josh: I removed the tags concerning PDEs as the question is not about PDEs. Feel free to revert my change. $\endgroup$ – Ricardo Andrade Oct 19 '13 at 23:26
  • $\begingroup$ I do not agree, cause it is something that comes out when looking for traveling waves solution of reaction-diffusion PDE. And if you see the question, the result is stated in two monographs concerning PDE. So maybe who works with parabolic PDE it is supposed to be more involved in the question. But for me it is ok, without the PDE tags :-) $\endgroup$ – Kosh Oct 20 '13 at 0:24
  • $\begingroup$ Dear @Josh: You certainly make a good point about which I was slightly ambivalent. Like I said before, please feel free to revert my changes. I apologize for the inconvenience. $\endgroup$ – Ricardo Andrade Oct 20 '13 at 0:42

@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!

  • $\begingroup$ I have to read your answer carefully, but in Volpert the theorem is used to prove that $f(w_+)=f(w_{-})=0$. While you first prove that $f(w_+)=f(w_{-})=0$ and than the statement. This suggest to me that there must be a more immediate proof (or to be more precise, an immediate consequence of some famous theorem on dynamical systems theory), that I do not know why I cannot see :). $\endgroup$ – Kosh Oct 20 '13 at 10:41
  • $\begingroup$ I see... have you tried Gronwall lemma for $|w''+\sigma w'|\leq C$? $\endgroup$ – leo monsaingeon Oct 20 '13 at 12:21
  • $\begingroup$ I think I have an elementary (and fast) proof of the following fact: in the hypotheses of the statement one has \begin{equation} \lim_{t\rightarrow\pm\infty}w''(t)=0\qquad \textbf{and}\qquad \lim_{t\rightarrow\pm\infty}w'(t)=-\frac{1}{\sigma}f(w_\pm) \end{equation} so that at least Evans is good cause he assumes that $f(w_\pm)=0$. The question is still open on if one necessarily has $f(w_\pm)=0$. PS I have to recheck my argument, but can I post it as an answer? $\endgroup$ – Kosh Oct 20 '13 at 21:27
  • $\begingroup$ I don't understand: you ASSUME now that $\lim w'=-\frac{1}{\sigma}f(w_{\pm})$, is that right? If so, you clearly have that $f(w_{\pm})=0$ (otherwise $w'\to cst\neq 0$ so $w$ blows linearly and cannot converge to $w_{\pm}$). Once you know that $\lim w'=-\frac{1}{\sigma}f(w_{\pm})=0$ you immediately get from the equation that $\lim w''=0$, so you don't even need to assume it to start with as your previous comment suggests (unless I missed something?) $\endgroup$ – leo monsaingeon Oct 20 '13 at 21:45
  • $\begingroup$ I'm not assuming, I'm able to prove the two limit that I wrote both exist and are equal to what written. So that with your comment we have finished. If you are interested I can write the proof. But if I write it as a comment I do not see the preview :-) $\endgroup$ – Kosh Oct 20 '13 at 21:51

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}

  • $\begingroup$ My observation $t\to-t\Leftrightarrow \sigma\to-\sigma$ absolutely doesn't imply that you can assume $\sigma<0$. It only means that if you can prove your statement on one side (say $t\to+\infty$) then the other side follows. Let's say you want to look at $t\to+\infty$: you still have to deal with both cases $\sigma>0$ and $\sigma<0$ so in (1) you should replace $-|\sigma|$ by just $\sigma$. I agree that your proof above works if $\sigma<0$ by letting $t_0\to\infty$ in (1). But if $\sigma>0$ sending $t_0\to\infty$ makes things diverge. Hint: if $\sigma>0$ rather let $t_0\to-\infty$!! $\endgroup$ – leo monsaingeon Oct 21 '13 at 13:26
  • $\begingroup$ If $\sigma>0$ the proof is the same. Just let $t_0\rightarrow -\infty$ and then $t\rightarrow \pm \infty$ $\endgroup$ – Kosh Oct 21 '13 at 14:41
  • $\begingroup$ Ok it is exactly what you have written :) $\endgroup$ – Kosh Oct 21 '13 at 14:43
  • $\begingroup$ hehehe ;-) but I insist that the proof is not exactly the same when $\sigma<0$ and $\sigma>0$. In the first case you use De L'Hopital's rule, while in the second case $\sigma>0$ you find an equivalent of the divergent integral $\int\limits^t_{-\infty}e^{\sigma s}f(w(s))ds\underset{t\to\infty}{\sim}f(w_+)\frac{e^{\sigma t}}{\sigma}$ (even though this is still some kind of De L'Hopital's rule...) $\endgroup$ – leo monsaingeon Oct 21 '13 at 15:29

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