Increasing property of solution of heat equation Let $$u_t - \Delta u = f$$
with $u(t,0) = u(t,1) = 0$ and $u(0,x) = u_0(x)$ given, and $f(t,x)$ is also given. This is the heat equation on the interval $(0,1)$.

How can I choose $f$ and $u_0$ to ensure that $u$ is increasing with respect to time? I.e. if $t \geq s$ then $u(t,x) \geq u(s,x)$ for a.e. $x$?

If $f \equiv 0$, then $u$ is decreasing with respect to $t$. But I don't know how to ensure that it becomes increasing. Does anyone know?
For the Allen-Cahn equation, $u_t = \epsilon^2 u_{xx} + u - u^3$, it's possible to have $u(t, x) \geq u(s, x), (t \geq s)$. The AC equation can be viewed as the $L^2$ gradient of the Ginzburg-Landau free energy $E = \int_\Omega 1/2 \epsilon^2 |u_x|^2 + 1/4(1 - u^2)^2 \mathrm{d} x$. The second term is called a double-well potential. If $u_0 = c \in (0, 1)$ in $[a, b]$, then the solution $u$ will evolve to $1$ finally.
For the Allen-Cahn equation, $u_t = \epsilon^2 u_{xx} + u - u^3$, it's possible to have $u(t, x) \geq u(s, x), (t \geq s)$. The AC equation can be viewed as the $L^2$ gradient of the Ginzburg-Landau free energy $E = \int_\Omega 1/2 \epsilon^2 |u_x|^2 + 1/4(1 - u^2)^2 \mathrm{d} x$. The second term is called a double-well potential. If $u_0 = c \in (0, 1)$ in $[a, b]$, then the solution $u$ will evolve to $1$ finally.