Suppose $(x,t)\in\mathbb{R}\times [0,\infty]$ and $u(x,t)$ is scalar and solves a linear parabolic partial differential equation with $u(\cdot,0)\geq 0$.
Can one apply some maximum or minimum principle to immediately get $u(\cdot,t)>0$ for all $t>0$?
I am new to this business and my feeling is that this should hold but I cannot give a proof.
I give an example of a sufficient set of conditions. Assume linear elliptic operator \begin{equation*} Lu = -\sum_{ij} a_{ij}u_{x_{i}x_{j}} + \sum_{i} b_{i}u_{x_{i}}+cu \end{equation*} with, for positive constants $A$, $B$ and $C$, \begin{equation*} 0 \le a_{ii}\le A \qquad -b_{i}x_{i} \le B\left(1+\lVert x \rVert^{2}\right) \qquad -c \le C\left(1+\lVert x \rVert^{2}\right). \end{equation*} Also smooth $u$ satisfying \begin{equation*} 0 \le u_{t}+L u\quad\text{in}\quad \Omega=\mathbb{R}^{n}\times (0,\infty). \end{equation*} Also the existence of a positive number $k$ such that \begin{equation*} 0 \le \limsup_{r\to\infty}\inf_{\substack{(x,t)\in \bar{\Omega}\\ \lVert x \rVert=r}} u\!\left(x,t\right)e^{-k\left(1+\lVert x \rVert^{2}\right)} \end{equation*} (that is the growth condition on $u$).
If we can prove for some fixed $T$ that $(x,t)\in \Omega_{T}=\mathbb{R}^{n}\times (0,T]$ implies $0\le u(x,t)$ then we are done, since we can repeat the argument as necessary in order to reach any larger value of $t$. To that end, consider \begin{eqnarray*} g\!\left(x,t\right) &=& e^{k\left(1+\lVert x \rVert^{2}\right)e^{t/T}}\\ g_{t}\!\left(x,t\right) &=& k/T\left(1+\lVert x \rVert^{2}\right)e^{t/T}g\!\left(x,t\right)\\ g_{x_{i}}\!\left(x,t\right) &=& 2kx_{i}e^{t/T}g\!\left(x,t\right)\\ g_{x_{i}x_{j}}\!\left(x,t\right) &=& 2k\left(2kx_{i}x_{j}e^{t/T}+\delta_{ij}\right) e^{t/T}g\!\left(x,t\right) \end{eqnarray*} satisfying \begin{equation*} g_{t}+Lg = \left( \frac{1+\lVert x\rVert^{2}}{T} -2\sum_{i}\left(a_{ii}-b_{i}x_{i}\right) +\frac{c}{ke^{t/T}} -4k e^{t/T}\sum_{ij}a_{ij}x_{i}x_{j} \right) k e^{t/T}g. \end{equation*} We should like $T$ such that $ 0 < g_{t}+Lg$ in $\Omega_{T}$. That is, we want \begin{equation*} 4k e^{t/T}\sum_{ij}a_{ij}x_{i}x_{j} + 2\sum_{i}\left(a_{ii}-b_{i}x_{i}\right) -\frac{c}{ke^{t/T}} < \frac{1+\lVert x\rVert^{2}}{T}. \end{equation*} Because the matrix $\left(a_{ij}\right)$ is symmetric and positive definite, \begin{equation*} \left\lvert a_{ij}x_{i}x_{j} \right\rvert\le \sqrt{a_{ii}a_{jj}} \left(1+\lVert x \rVert^{2}\right) \le A \left(1+\lVert x \rVert^{2}\right) \end{equation*} \begin{equation*} \frac{4k e^{t/T}}{1+\lVert x \rVert^{2}} \sum_{ij}a_{ij}x_{i}x_{j} \le 4 e k n^{2} A. \end{equation*} Also \begin{equation*} \frac{2}{1+\lVert x \rVert^{2}}\sum_{i}\left(a_{ii}-b_{i}x_{i}\right) \le 2n\left(A+B\right) \end{equation*} \begin{equation*} -\frac{c}{\left(1+\lVert x \rVert^{2}\right)ke^{t/T}}\le \frac{C}{k}. \end{equation*} Thus we can ensure $0 < g_{t}+Lg$ in $\Omega_{T}$ by setting \begin{equation*} T = \frac{1}{1+4ek n^{2}A+2n\left(A+B\right)+C/k}. \end{equation*} With $w=u/g$ \begin{eqnarray*} u &=& g w\\ u_{x_{i}} &=& g w_{x_{i}}+ g_{x_{i}}w\\ u_{x_{i}x_{j}} &=& g w_{x_{i}x_{j}}+ g_{x_{j}}w_{x_{i}}+g_{x_{i}}w_{x_{j}} +g_{x_{i}x_{j}}w\\ Lu &=& g \left(L-c\right) \! w + 2\sum_{ij}a_{ij}g_{x_{i}}w_{x_{j}}+ \left(L g \right)w\\ \end{eqnarray*} so dividing $0\le u_{t}+Lu$ by $g$ gives us \begin{equation*} 0 \le w_{t} + \left(L-c\right) \! w + \frac{2}{g} \sum_{ij}a_{ij}g_{x_{i}}w_{x_{j}}+ \frac{1}{g}\left(g_{t} + L g \right) w. \end{equation*} Since \begin{equation*} 0 \le \limsup_{r\to\infty}\inf_{\substack{(x,t)\in\bar{\Omega}_{T} \\ \lVert x \rVert = r}}w(x,t) \end{equation*} there exists sequence $r_{n}$ with $r_{n}\to \infty$ and \begin{equation*} -\frac{1}{n}\le \inf_{\substack{(x,t)\in\bar{\Omega}_{T}\\ \lVert x \rVert = r_{n} }} w(x,t). \end{equation*} We shall prove that for any $n$, and with $\left(x,t\right) \in \bar{\Omega}_{T}$ satisfying $\lVert x \rVert \le r_{n}$, \begin{equation*} -\frac{1}{n}\le w(x,t). \end{equation*} implying $0 \le w$ in $\Omega_{T}$, and $0 \le u$ in turn.
Suppose it false, i.e. that \begin{equation*} \inf_{\substack{(x,t)\in\bar{\Omega}_{T}\\ \lVert x \rVert \le r_{n} }} w(x,t)<-\frac{1}{n}. \end{equation*} Then $w$ must attain its minimum in $\left\{(x,t)\in \bar{\Omega}_{T} \mid \lVert x \rVert \le r_{n}\right\}$ at some point $\left(x_{0}, t_{0}\right)$ away from the parabolic boundary. We have \begin{equation*} 0 < -\frac{1}{g}\left(g_{t}+Lg\right)w \!\left(x_{0},t_{0}\right) \le w_{t}\!\left(x_{0},t_{0}\right)-\sum_{ij} a_{ij}w_{x_{i}x_{j}}\!\left(x_{0},t_{0}\right). \end{equation*} But we know that $w_{t}\!\left(x_{0},t_{0}\right)$ has to be nonpositive and that $\sum_{ij} a_{ij}w_{x_{i}x_{j}}\!\left(x_{0},t_{0}\right)$ has to be nonnegative (skipping over some linear algebra). Contradiction.