If $ u_t=u_{xx} $ and $u(x,0)=4x(1-x),~x\in [0,1],~~u(0,t)=u(1,t)=0,~t\geqslant 0$, prove $0<u<1$ for $x\in (0,1),~t>0$

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Let $u$ be the solution of the heat equation initial and boundary value problem: $$\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2},~~x\in (0,1),~t>0,$$ and: $$u(x,0)=4x(1-x),~x\in [0,1]~~ \textrm{and} ~~u(0,t)=u(1,t)=0,~~t\geqslant 0.$$ Prove that $0<u(x,t)<1$ for all $x\in (0,1)$ and $t>0$.

Attempt. Instead of solving the equation, we shall work with the maximum-minimum principle. Since $u=0$ for $x=0,\,1$ and $0\leqslant u\leqslant 1$ for $t=0$, we get the estimate $0\leqslant u\leqslant 1$ on the boundary of $\varOmega=(0,1)\times [0,+\infty).$ By the maximum-minimum principle, we get $0\leqslant u\leqslant 1$ for all $x\in (0,1)$ and $t>0$.

How one can derive the strict inequalities?

Thanks in advance for the help.

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The strong maximum principle states that if the solution attains its maximum $M$ in some $0 < \bar{x} <1$ and $\bar{t} > 0$, then the solution $u(t, x) \equiv M$ for $0 \le x \le 1$ and $0 \le t \le \bar{t}$. Since the initial condition is not constant, you have a contradiction. (Similarly for the minimum)