TITLE: Properties of heat equation
QUESTION [6 upvotes]: ** I simplified the question: **
On bounded domains, the maximum principle implies that the solution to the heat equation is (strictly) positive, if the initial and boundary data is positive. 
I would like to know whether this result is also true if applied on an unbounded domain:
Let $a>0$ and $g$ a continuous function with $g(a,t)>0$ for all $t \in [0,T]$ and $u(x,0) = u_0(x) >0$ for all $x \in (a,\infty)$ with $u_0 \in C^{\infty} \cap L^{\infty}.$
Let $u(x,t)$ be the solution to the heat equation $\left(\partial_t - \partial_{x}^2 \right)u=0$  with the above boundary data.
Does there exist then a version of the maximum principle saying that $u(x,t)>0$ for $(x,t) \in [a,\infty) \times [0,T]$?

REPLY [4 votes]: As a companion to Alexandre Eremenko's answer, the answer to your question is "yes" if we assume in addition that $u$ satisfies the growth condition $|u| < Ae^{M|x|^2}$ for some $A,\,M > 0$.
To see this assume for simplicity that $a = 0$ and $T = 1$. Let $h_R = -t^{-1/2}e^{-\frac{(x-R)^2}{4t}}$ and let $b_R = Ae^{MR^2}h_R$. By the comparison principle on bounded domains we have $u \geq b_R$ on $[0,\,R] \times [0,\,1]$. A short computation shows that $b_R \geq -8A\sqrt{M}e^{-3MR^2}$ on $[0,\,R/2] \times [0,\, (64M)^{-1}]$. Taking $R \rightarrow \infty$ we conclude that $u \geq 0$ for $t \in [0,\,(64M)^{-1}]$, and repeating the argument finitely many times we get $u \geq 0$ on $[0,\,T]$. That $u > 0$ follows from the strong maximum principle.