Behavior of solutions to the heat equation at infinity I have read that for the solution $u$ of the heat equation
$$u_t = u_{xx},$$
with $u(x,0)= a \exp(-bx^2)$ for some $a,b >0$, 
it holds
$$\lim_{x \to \infty} u_x(x,t) = 0 = \lim_{x \to \infty} u(x,t)$$
and also
$$ \lim_{x \to -\infty} u_x(x,t) = 0 = \lim_{x \to -\infty} u(x,t) .$$
Why is that? Does anyone know how to proof this? I can't find it in my pde books.
 A: With such simple initial data, the most helpful way is to find the solution using the Poisson formula
\begin{align}
u(x,t)=\frac{1}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}ae^{-by^2}e^{-\frac{(x-y)^2}{4t}}dy=\frac{ae^{-\frac{bx^2}{1+4bt}}}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}e^{-(b+\frac{1}{4t})\bigl(y-\frac{x}{1+4bt}\bigr)^2}dy\\
=\frac{ae^{-\frac{bx^2}{1+4bt}}}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}e^{-(b+\frac{1}{4t})y^2}dy=\frac{ae^{-\frac{bx^2}{1+4bt}}}{\sqrt{\pi(1+4bt)}}\int\limits_{-\infty}^{\infty}e^{-z^2}dz=\frac{ae^{-\frac{bx^2}{1+4bt}}}{\sqrt{1+4bt}}
\end{align}
which readily implies that 
$$
\lim_{|x| \to \infty} u_x(x,t) = 0 = \lim_{|x| \to \infty} u(x,t).\tag{$\ast$}
$$
But the fundamental fact is that $(\ast)$ holds with any bounded initial data $u_0(x)\overset{\rm def}{=}u(x,0)$ vanishing at infinity, i.e.,
$$
\sup_{\mathbb{R}}|u_0|=M<\infty,\qquad\lim_{|x| \to \infty} u_0(x) = 0. 
$$
Indeed, $\forall\,\varepsilon>0\;\exists\,R_{\varepsilon}>0\,\colon\; |u_0(x)|<\varepsilon/2\;\forall\,x\in\mathbb{R}$  with modulus $ |x|>R_{\varepsilon}\,.$ Hence
\begin{align}
|u(x,t)|\leqslant\frac{1}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}|u_0(y)|e^{-\frac{(x-y)^2}{4t}}dy=\frac{1}{\sqrt{4\pi t}}\biggl(\int\limits_{|y|<R_{\varepsilon}}+\int\limits_{|y|>R_{\varepsilon}}\biggr)\\
\leqslant\frac{M}{\sqrt{4\pi t}}\int\limits_{-R_{\varepsilon}}^{R_{\varepsilon}}e^{-\frac{(x-y)^2}{4t}}dy+\frac{\varepsilon/2}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}e^{-\frac{(x-y)^2}{4t}}dy\leqslant \frac{M}{\sqrt{4\pi\delta}}\int\limits_{-R_{\varepsilon}}^{R_{\varepsilon}}e^{-\frac{(x-y)^2}{4T}}dy+\varepsilon/2\tag{$\ast\ast$}
\end{align}
for all $x\in\mathbb{R}$ and $t\in [\delta,T]$, with arbitrary fixed positive $\delta<T$.
Since the last integral in $(\ast\ast)$ is vanishing as $|x|\to\infty$, there is some $d>0$ such that
$$
\frac{M}{\sqrt{4\pi\delta}}\int\limits_{-R_{\varepsilon}}^{R_{\varepsilon}}e^{-\frac{(x-y)^2}{4T}}dy<\varepsilon/2\quad\forall\,x\colon\;|x|>d
$$
which implies that $|u(x,t)|<\varepsilon/2+\varepsilon/2=\varepsilon$ when $|x|>d$ and $t\in [\delta,T]$. Thus it has been established that 
$$
\lim_{|x| \to \infty} u(x,t)=0\quad\forall\, t\in (0,\infty).
$$
The rest can be done in a similar manner, since
$$
u_x(x,t)=-\frac{1}{2t\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}(x-y)u_0(y)e^{-\frac{(x-y)^2}{4t}}dy,
$$ 
while
$$
\frac{1}{2t\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}|x-y|e^{-\frac{(x-y)^2}{4t}}dy=
\frac{1}{2t\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}|y|e^{-\frac{y^2}{4t}}dy=
\frac{1}{t\sqrt{4\pi t}}\int\limits_{0}^{\infty}ye^{-\frac{y^2}{4t}}dy=
\frac{1}{\sqrt{\pi t}}.
$$ 
