I have $u(x,t)$ to be defined as the solution to the following partial differential equation for heat diffusion over the domain $S = (0, 1) \times (0, \infty)$.
$$ \begin{cases} u_t - u_{xx} &= 0 \\ u(0, t) &= c \\ u(1, t) &= c \\ u(x, 0) = x(1-x) \\ \end{cases} $$
Find all $\alpha, \beta > 0$ such that $u(x, t) \leq w(x,t) = \alpha x (1 - x) e^{-\beta t}$ on $S$ not $\overline{S}$ and take the limit as $t \to 0$ (yes, $0$) to show that $\lim_{t \to 0} u(x, t) = 0$ uniformly.
Additional Stipulations: You must apply the maximum principle twice: first to $u(x, t)$ and second to $w(x, t) - u(x, t)$. You cannot explicitly solve for the function $u(x, t)$.
Attempt: First off, I believe that the last half of the question meant to use $\infty$ instead of $0$ sot that it would instead read, prove that $\lim_{t \to \infty} u(x, t) = 0$ uniformly.
From a previous result, we know that $u(x, t) \in C^{2, 1} (S) \cap C(\overline{S})$, and that the maximum principle shall apply to any solution $v$ of the heat diffusion equation
$$ v_t - v_{xx} = f(x, t) \leq 0 $$
necessitates
$$ \max_{(x, t) \in \partial_P S} v(x, t) = \max_{(x, t) \in \overline{S}} v(x, t) $$
when we use the term "max" loosely without regard to topological compactness.
First we can prove that $u(x, t)$ is always positive by applying the identity for any reasonable domain $\Omega$
$$ \min_{(x, t) \in \Omega} u(x, t) = - \max_{(x, t) \in \Omega} - u(x, t) $$
so that linearity of the PDE implies $-u(x, t)$ is also a solution and that for any $(x, t) \in overline{S}$,
$$ -u(x, t) \leq \max_{\partial_P S} -u \leq 0 $$
because $x(1-x)$ is zero at $x \in \{0, 1\}$. By multiplying through by $-1$, we conclude that
$$ u(x, t) \geq 0 \in \overline{S} $$
Although no requirement for the sign of $c$ was added, I assume that $c\geq 0$ must hold, because I have defined $u(x, t) = 0$ to be absolute $0$.
Now, the final task is to somehow prove that $w(x,t) \geq u(x, t)$ by making a comment about $w(x, t) - u(x, t)$. Intuitively, I thought that showing the maximum of $u(x, t) - w(x, t)$ is negative would be the simplest means to demonstrate the desired inequality, but I will try to prove it the way delineated by the question.
In order that the maximum principle even works, I have to show that $w(x,t)$ is a solution of the heat equation to invoke linearity. This constraint imposes a fairly stringent requirement upon $\beta$. The only means to produce a solution is by setting $\beta = 2$, so I presume that this asymptotic value shall produce an upper bound on the rate of collapse.
$$w(x,t) = \alpha (x-x^2)e^{-2 t}$$
As a result, choosing $\beta \in (0, 2]$ will suffice because any other choice of $w(x,t)$ may be easily compared with the $\beta = 2$ case.
I record some preliminary observations in preparation for my attempt.
$$\max_{\partial_P S} u(x, t) = \max\{\frac{1}{4}, c\}$$ $$\max_{\partial_P S} w(x, t) - u(x, t) = \max{\frac{1}{4} (\alpha - 1), -c}$$
because $$w(x, t) = 0$$ on $$(\{0\}\cup \{1\}) \times (0, \infty)$$.
$$ \begin{align} \max_{\partial_P S} w(x, t) &= \max_{\partial_P S} (w(x, t) - u(x, t)) + u(x, t) \\ &\leq \max_{\partial_P S} w(x, t) - u(x, t) + \max_{\partial_P S} u(x, t) \end{align} $$
$$ \max_{\partial_P S} u(x,t) - \max_{\partial_P S} w(x,t) \geq \min_{\partial_P S} u(x,t) - w(x,t) $$
This final inequality is not helpful for numerous reasons. What have I overlooked? If I had to guess an answer, $\alpha > \frac{1}{4}$ and $0< \beta <2$.