Showing solutions to $u_t = \Delta u + a(x) u$ remain non-negative 
Let $D$ be a bounded domain in $\mathbb{R^n}$ with a smooth boundary and assume that the function $a(x)$ is continuous on the closure of $D.$  Show that solutions to
  $$ u_t = \Delta u + a(x) u
$$
  vanishing on the boundary of $D$ with $u(x,0) \geq 0$ remain nonnegative for all $t > 0.$

I feel like a strong minimum or strong maximum principle should help out in some way, but with the lower order variable coefficient term, I haven't been able to get very far.  The fact that the solution stays positive makes intuitive sense because it would be true for the heat equation without the lower order term, and the lower order term will only introduce exponential decay or growth in some way, meaning a nonnegative solution will remain nonnegative.  Anyone have suggestions or know of any solutions?

Edit:  After some looking around, I found that this is more or less exercise 7.7 in Evans.  He seems to lead the reader on to showing 
$$ u_t + cu = \Delta u \;\text{ with }\; c \geq \gamma > 0; \;\; u(x,0) \geq 0, \;\; u = 0 \text{ on } \partial U
$$
has the property that $|u(x,t)| \leq C e^{-\gamma t}.$
To prove this, I believe one can use the weak maximum principle.  Letting $w = e^{\gamma t} u,$ we get that $w_t - \Delta w + (c-\gamma)w = 0,$ and since $c(x) - \gamma \geq 0,$ we get that at any time, $|w(x,t)| \leq ||w(x,0)||_\infty.$  Then $|u(x,t)| \leq ||u(x,0)||_\infty e^{-\gamma t}.$
From here, I've returned to the problem and said that since $U$ is a bounded domain and $a(x)$ is continuous, $|a(x)| \leq M.$  Letting $u = e^{-\lambda t} v,$ we have that 
$$ v_t +\underbrace{(-a(x) - \lambda)}_{c}v = \Delta v.
$$
Then choosing $\lambda$ so that $c$ is positive, we can use the fact that $|v(x,t)| \leq Ce^{-\gamma t}$ to get the result.  

Does this argument (although very pieced together) seem to make sense?  Does anyone see any holes in this argument? Thanks!
 A: Perhaps I am missing something (please tell me if I have), but how does $\lvert v(x,t) \rvert \le Ce^{-\gamma t}$ conclude the problem?  This seems to tell you that $\lvert u(x,t) \rvert \le e^{-(\lambda+ \gamma)t}$ for all $x,t$ but does not imply that $u(x,t) \ge 0$ for all $x,t$. 
I think you were on the right track with the maximum/minimum principle. As you've said, $v = e^{\lambda t} u$ satisfies $$v_t + c_\lambda(x)v - \triangle v = 0$$ along with $v= 0$ on $\partial D$ and $v(x,0) = u(x,0) \ge 0$ for some $c_\lambda$ which we can take to be strictly positive in $D$ (by choosing $\lambda$ correctly). 
Fix $T > 0$ and assume that $v < 0$ somewhere in $D \times (0,T]$. Then the minimum of $v$ in $\overline D \times [0,T]$ is negative and this cannot happen at $t=0$ or on $\partial D$ so $v$ has a negative minimum at some $(x_0,t_0) \in D \times (0,T]$. At this minimum, we will have $$\triangle v(x_0,t_0) \ge 0, \,\,\,\,\,\,\, v_t(x_0,t_0) \le 0, \,\,\,\,\, c(x_0)v(x_0,t_0) < 0. $$ Then $$v_t + c_\lambda(x)v - \triangle v < 0 \,\,\,\, \text{ at } (x_0,t_0)$$ which is a contradiction. Hence $v$ is never negative in $D \times (0,T]$. Since $T > 0$ was arbitrary, this shows that $v(x,t) \ge 0$ for all $t > 0$ (and any $x$). Of course, $u$ and $v$ have the same sign so this completes the proof. 
