I have the system $$u_{tt} = u_{xx}$$ for $x \in \mathbb{R}$ and $t>0$ with initial conditions $$u(x,0) = \phi(x) = 0, \hspace{5pt} u_{t}(x,0) = \psi(x) = \left\{ \begin{array}{ll} 1 & |x|\leq 1 \\ 0 & |x|> 1 \\ \end{array} \right. $$
Using d'Alembert's formula, I got $u(x,t) = \left\{ \begin{array}{ll} t & |x|\leq 1 \\ 0 & |x|> 1 \\ \end{array} \right.$. Now I'm asked to show that $0 \leq u(x,t) \leq 2 \hspace{5pt} \forall x \in \mathbb{R}, t<0$, but as far as I know, $u(x,t)$ is unbounded. Am I missing something? Or is my initial $u(x,t)$ is incorrect? Any insights would be appreciated.
Since $\phi(s) \equiv 0$ and $\psi(s) \leq 1$ for $|s| < 1$, the d'Alembert's formula gives
$$u(x,t) = \frac{1}{2}\int^{x+t}_{x-t}\psi(s) \mathrm{d} s.$$
However, since $\psi$ is only $1$ on $[-1,1]$ and the integral domain is $[x-t,x+t]$, then this is equivalent to $$u(x,t) = \frac{1}{2}|[-1,1] \cap [x-t,x+t]|.$$
This is different from the answer you have provided, since if $x = 0.5$ and $t = 5$, we get $u(0.5,5) = 1$. I'll leave you to use this expression to prove a bound on $u(x,t)$.