Maximum principle for a nonlinear heat equation Suppose that $\Omega\in\mathbb{R}^n$ is bound and open with smooth boundary $\partial\Omega$. Consider the following initial-boundary value problem for a nonlinear heat equation: 
$$
\begin{cases}
%\vspace{3mm} 
u_t=\Delta u+u(1-u), \ \ &x\in\Omega,\ \ t>0,\\
%\vspace{3mm} 
u(x,0)=u_0(x),\ \ &x\in\Omega,\\
u=0, \ \ &x\in\partial\Omega,\ \ t>0,
\end{cases}
$$
where $0\le u_0\le1$. Show that $0\le u(x,t)\le1$, $x\in\Omega$, $t>0$.
We prove $u(x,t)\le1$ by contradiction. Assume for some $(x_1,t_1)$ with $x_1\in\Omega$ and $T>t_1>0$, we have $u(x_1,t_1)>1$ and
$$
\displaystyle\max_{x\in\overline{\Omega}, T\ge t\ge0} u(x,t)=u(x_1,t_1).
$$
Then $u_t(x_1,t_1)=0\ge\Delta u(x_1,t_1)$, which leads to
$$
u_t-\Delta u-u(1-u)\Big|_{(x,t)=(x_1,t_1)}>0.
$$
Therefore $u(x,t)\le1$. On the other hand, however, I have no idea how to prove $u(x,t)\ge0$. A similar argument for proving $u(x,t)\le1$ seems not work for the proof of $u(x,t)\ge0$. Any suggestion for ideas or proofs is welcome, thanks!
Another related question: When the homogeneous Dirichlet boundary condition $u=0$ is replaced by the homogeneous Neumann boundary condition $\frac{\partial u}{\partial \nu}=0$, the same conclusion remains true?

Consider $\displaystyle v\left(x,t\right)=u\left(x,t\right)-\frac{t}{4}$. Then you can get  \begin{equation}v_t-\Delta v=u_t-\Delta u+u-u^2-\frac{1}{4}\leq0, \end{equation}which implies $v$ is subsolution of the heat equation. Hence we can apply the maximum principle to $v$ to obtain\begin{equation}\max_{\Omega}v\leq\max_{\Omega}u_0\left(x\right)\leq1.\end{equation} Thus, $u\leq1$.
On the other hand, we can set $\displaystyle w\left(x,t\right)=1-u\left(x,t\right)+\frac{t}{4}$ and $w_0\left(x\right)=1-u_0\left(x\right)$. Then we can get\begin{equation}w_t-\Delta w=w\left(w-1\right)+\frac{1}{4}\geq0,\end{equation}which means $w$ is the supersolution of the heat equation. Similarly, by using the minimum principle, we have shown\begin{equation}\min_{\Omega}w\leq\min_{\Omega}w_0\leq1.\end{equation}Thus, $u\geq0$.
