Maximum of the solution of a parabolic PDE 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? 
 A: 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.
A: 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.
