Suppose that I have a parabolic PDE, (to fix notation) let's say it's of the form: $$ \frac{\partial u}{\partial t}(x,t) + \mu(x,t) \frac{\partial u}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2 u}{\partial x^2}(x,t) -V(x,t) u(x,t) + f(x,t) = 0. $$
Let $\Omega$ be a fixed compact cube in $\mathbb{R}^{d+1}$, are there conditions guaranteeing that the PDE has a solution whose $f$ mapping $\Omega$ into $[0,1]$ (or if it's easier into $(0,1)$)?
I'm familiar with boundary conditions but this is a bit different...
Added: This is a backwards heat equation. It's easy to convert this into a forward equation by replacing $t$ with $T-t$ where $T$ is the terminal time. That moves the $\frac{\partial u}{\partial t}$ term to the right hand side but otherwise does not change things.
For demonstration purposes, take the case $d = 1$ (one spatial dimension). Then $\Omega = [a,b] \times [0,T]$ and one may prescribe terminal data for $u$ at $x \in [a,b], t = T$ and additional boundary conditions at $x \in \{a,b\}, 0 \le t \le T$.
Clearly the terminal data must be such that $u(x,T) \in [0,1]$. Then a general comparison principle that works also in higher space dimensions and for a broad class of boundary conditions says: If $f \ge 0$ everywhere, then also $u \ge 0$ everywhere (and in fact $u > 0$ for $t > 0$).
To obtain conditions such that $u \le 1$, you could consider $v = 1-u$ which satisfied the same equation as $u$, but with $f$ replaced by $V - f$. Then $u \le 1$ everywhere iff $v \ge 0$ everywhere, and this is true if $V-f \ge 0$ everywhere. So if $0 \le f \le V$ and $0 \le u(x,T) \le 1$, then $0 \le u \le 1$ everywhere.