Positive initial data in a bounded domain forces a solution of a nonlinear heat equation to be positive as well Let $U\subset \mathbb{R}^n$ be open and bounded with smooth boundary, and let $f: \mathbb{R} \to \mathbb{R}$ have bounded derivative and satisfy $f(0)=0$. If $u$ solves
\begin{align}
u_t - \Delta u &= f(u) \text{ in } U \times (0,\infty)\\
u(x,t) &= 0 \text{  on } \partial U\times (0,\infty) \\
u(x,0) &= u_0(x)\text{ on } U\times \{t=0\}
\end{align}
where $u_0(x)\ge 0$, then $u\ge 0$ for all $(x,t) \in U\times (0,\infty)$.
I know the result is true (via strong maximum principle) if $f=0$ on $U_T = U\times (0,T)$ for all $T>0$, so I'm wondering if the maximum principle might also help with this problem. If not, them I'm not sure how to proceed.

Yes, you can use the maximum principle, though I'm not sure if the argument is standard. It would go something like this: Suppose first that $f(0)>0$ and $u_0(x) > 0$ for $x \in U$. Then define
$$T = \sup\big\{\tau>0 \, : \, u(x,t)>0 \text{ for all } (x,t) \in U\times [0,\tau)\big\}.$$
By our assumptions $T>0$. We just need to show that $T=\infty$. Assume to the contrary that $T<\infty$. Then $u$ attains its minimum over $U\times [0,T]$ at a point $(x,T)$ where $u(x,T)=0$. Therefore
$$f(0)=u_t(x,T) - \Delta u(x,T)\leq 0.$$
This is a contradiction if $f(0)>0$. 
The rest of the proof boils down to reducing the problem to the case where $f(0)>0$ and $u_0>0$. You should try introducing a small parameter $\varepsilon>0$ and making a small perturbation of $u$. For example, $u_\varepsilon(x) = u(x) + \alpha\varepsilon - \varepsilon \exp(\beta x_1)$ should work for appropriately chosen constants $\alpha$ and $\beta$.
