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Let $u:\mathbb{R}\times [0, \infty) \rightarrow \mathbb{R}$ be defined by

$u_{xx} + u_x - u_t = u(u - 2)(u - 1)$

with $u \rightarrow 0$ as $|x| \rightarrow \infty$ and $u(x,0) = 3e^{-x^2}$. Now, let $g(t) = \max_x u(x, t)$.

Is there anything that can be said about $g(t)$? In particular, are $g'(t)$ and $g''(t)$ continuous?

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    $\begingroup$ Can you say something about the context of this question? $\endgroup$
    – Stefan Kohl ♦
    Commented Nov 20, 2013 at 14:54
  • $\begingroup$ @StefanKohl My objective is to find t* s.t. g(t*) = 1. I would like to explore the properties of g(t) first to rule out some ideas. $\endgroup$
    – rodms
    Commented Nov 20, 2013 at 15:59
  • $\begingroup$ @rodms $t^*=0$ works ;) $\endgroup$
    – username
    Commented Nov 20, 2013 at 16:01

2 Answers 2

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There is certainly a simpler way, but you can probably show that $g$ is decreasing from $3$ to $0$ as $t\to\infty$.

Looking in a moving frame with speed $-1$, i.e. at $v(t,x) = u(t,x-t)$, you see that $v$ satisfies a bistable reaction-diffusion equation. Moreover, your nonlinearity is balanced, i.e. $\int_{[0,2]} u(u-2)(u-1) = 0$. In this case it is known that under various assumptions on the initial datum, the solution of the Cauchy problem tends loc. unif. to $0$.

In your case, starting with initial datum that is not stacked between $0$ and $2$ might complicate things, you'll probably have to show that in some finite time, the solution is $\leq 2$.

Edit : actually neither this nor the fact that the nonlinearity is balanced are needed ; I think that $3e^{-x^2}$ has small enough mass for the solution of your equation to decay uniformly to zero. See my comment below.

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  • $\begingroup$ You can also bound $u(1-u)(u-2)$ on $\mathbb R_+$ by above by its maximal slope $au$ and compare $v$ with a solution of the equation with $f$ replaced by this slope. Then you can bound $v(t,x)$ by $e^{at} \frac{1}{\sqrt{4\pi t}} \int u_0$. Then you can optimize in $t$ ($t = 1/(2a)$ I guess) and see if this is less than $1$ for some value of $t$. If that's the case, then starting from this time, the equation will have a non-positive right hand side and basically you're done. In your case $a = \frac{1}{\sqrt{3}} - \frac{1}{3}$ and I think that this is true. $\endgroup$
    – Laurent
    Commented Nov 20, 2013 at 21:11
  • $\begingroup$ In terms of the original question, to establish that $g'$ and $g''$ exist and are continuous, since your $v$ solves an a priori bounded bistable reaction-diffusion equation, with symmetric initial data, a global classical solution exists, which is also symmetric. Additionally, since the lap number of $v$ doesn't increase, and is initially equal to 1, it follows that $g(t)=v(t,0)$. In addition, the nonlinear term is $C^\infty (\mathbb{R})$ and initial data is $C^\infty(\mathbb{R} )$, so $v$ is smooth. Finally it follows that both $g'$ and $g''$ exist and are continuous on [0,\infty )$. $\endgroup$
    – JCM
    Commented Jul 27, 2016 at 14:21
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As the initial data is perfectly smooth, I think that it is easy to apply the standard energy method to obtain a local in time existence in $H^s$ for any $s>0$. Take $s=3$. Then $g(t)=u(x^*,t)$ is a Lipschitz function. Thus, using Rademacher theorem, it has a derivative a.e. Moreover, the equation is $$ g'=[-u(u-2)(u-1)+u_x+u_{xx}]|_{x=x^*}. $$ So, $$ g'=-g(g-2)(g-1)+[u_{xx}]|_{x=x^*}\leq -g(g-2)(g-1). $$ And you can obtain $g\rightarrow1$, right? Hopefully this helps.

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