Timeline for Regularity of a nonlinear ODE [Traveling wave solutions of parabolic systems]
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 21, 2013 at 10:13 | vote | accept | Kosh | ||
| Oct 20, 2013 at 22:01 | comment | added | leo monsaingeon | Sure, no problem. If any answer to one of your questions actually helps, you should mark it as "accepted" so that other overflowers know it's a reliable one. Of course if you don't like the answer you can also vote it down to let OF-fellows know that they shouldn't trust it. Cheers | |
| Oct 20, 2013 at 21:54 | comment | added | Kosh | I will read carefully your answer, I'm sure that I will learn something new :-) So, for the time being you have been very precious. The key was to use your $(E)$ form of the equation. | |
| Oct 20, 2013 at 21:51 | comment | added | leo monsaingeon | Also: yes, it must be true in cany case that $f(w_{\pm})=0$ (and for that you only need continuity of $f$ and $\lim w=w_{\pm}$, you don't have to assume anything for $w'$ or $w''$). Double-check step 1 in my first answer, but that's really what I show. | |
| Oct 20, 2013 at 21:51 | comment | added | Kosh | 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 :-) | |
| Oct 20, 2013 at 21:45 | comment | added | leo monsaingeon | 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?) | |
| Oct 20, 2013 at 21:27 | comment | added | Kosh | 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? | |
| Oct 20, 2013 at 12:21 | comment | added | leo monsaingeon | I see... have you tried Gronwall lemma for $|w''+\sigma w'|\leq C$? | |
| Oct 20, 2013 at 10:41 | comment | added | Kosh | 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 :). | |
| Oct 20, 2013 at 1:24 | history | answered | leo monsaingeon | CC BY-SA 3.0 |