Let $u$ solve $u_t-\Delta u = f(u)$ on $[0,T] \times \Omega$ where $\Omega$ is a bounded domain in $\mathbb{R}^n$ and $u=0$ on $\partial{\Omega}$. Also, let $\partial{\Omega}$ be sufficiently smooth to apply the maximum principle.
Also suppose $f(0)=0$ and there exist $m \leq 0 \leq M$ such that interval $[m,M]$ is invariant w.r.t the ODE $\frac{da}{dt}=f(a(t))$; this means that $m \leq a(0) \leq M$ implies $m \leq a(t) \leq M$ for all $t>0$.
Then, I have to show that $m \leq u(x,0) \leq M$ implies $m \leq u(x,t) \leq M$ for all $0 \leq t \leq T$.
I am just stuck at it....I cannot deal with the term $\Delta u$ at all....Could anyone please help me?
The fact that $[m,M]$ is invariant under the autonomou ODE simply means that $f(m)\geq 0$ and $f(M)\leq 0$ (draw a picture! otherwise a particle starting at $a(0)=m$ would immediately cross $m$ with a negative slope, and likewise a particle starting at $a(0)=M$ would cross $M$ with a positive slope).
As a consequence the smooth, constant functions $\underline{u}(t,x):=m$ and $\overline{u}(t,x):=M$ are sub and super solutions, respectively $$ \partial_t\underline u-\Delta \underline u-f(\underline u)=0-f(\underline u)=-f(m)\leq 0 $$ and $$ \partial_t\overline u-\Delta \overline u-f(\overline u)=0-f(\overline u)=-f(M)\geq 0 $$ Since in addition $f(0)=0$ and $m\leq 0\leq M$, and the solution $u$ satisfies homogeneous Dirichlet boundary conditions, the ordering is correct on the lateral boundary $$ \underline{u}(t,x)\leq u(t,x)\leq \overline{u}(t,x) \qquad \forall (t, x)\in[0,T]\times\partial \Omega. $$ The classical comparison principle then guarantees that the initial ordering $m\leq u_0(x)\leq M$ is preserved along the evolution.