# Existence of bounded solution to Parabolic PDE

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...

• Your question is unclear. Are you assuming that $f$ has values in $[0,1]$ or do you want conditions such that the solution $u$ has values in $[0,1]$? – Hans Engler Nov 8 '18 at 23:40
• I want conditions such that $u$ is guaranteed to take values in $[0,1]$; assuming that $f$ takes values in $[0,1]$. – AIM_BLB Nov 8 '18 at 23:57

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.