Behavior of solution to the wave equation as $t\to \infty$ The equation $u_{tt}-9u_{xx}=0$ has the initial data $$u(x,0)=f(x) = \begin{cases}
 1, &\vert x\vert\leq2  \\ 0, & \vert x \vert>2 
\end{cases}$$ $$u_t(x,0)=g(x)=\begin{cases}1,&\vert x\vert\leq2 \\ 0, & \vert x\vert>2\end{cases}$$
I want to compute the limit $\lim_{t \to \infty}u(x_0,t)$ for some fixed $x_0$. Might as well take the this limit from the solution by d'Alembert $$\lim_{t\to \infty}\frac{1}{2}(f(x_0-ct)+f(x_0+ct))+\frac{1}{6}\int_{x_0-ct}^{x_0+ct}g(z)dz$$
The sum with $f(x_0)$ goes trivially to zero. For the integral term, since $t\to \infty$ I can assume that $x_0-ct<0$ and $x_0+ct>0$. With a bit of algebra I can conclude that we have $$\vert x_0-ct \vert >2,\enspace \vert x_0+ct\vert>2$$
when $t>2/3$. Therefore ( Why exactly? ) the integral is trivially zero outside of the interval $z\in[x_0-2, \enspace x_0+2]$. I conclude that $$\lim_{t\to\infty}u(x_0,t)=\frac{1}{6}\int_{x-2}^{x+2}g(z)dz=\frac{1}{6}(x+2-x+2)=\frac{2}{3}$$
Is there a qualitative description for the number $t=2/3$? Also, can someone explain the behaviour  of the integral term in the d'Alembert solution under this limit operation?
 A: Consider the wave equation $\partial_{tt}u=9\partial_{xx}u$, with boundary conditions $$\begin{cases}u(0,x)=\varphi(x)=\textbf{1}_{x\in[-2,2]}\text{ or }\chi_{[-2,2]}(x)\\\partial_tu(0,x)=\psi(x)=\textbf{1}_{x\in[-2,2]} \end{cases}.$$
The solution is given by
$$\begin{align}u(t,x)& =\frac{1}{2}[\varphi(x+3t)+\varphi(x-3t)]+\frac 1 6\int_{x-3t}^{x+3t}\psi(y)dy\\&=\frac{1}{2}[\textbf{1}_{x+3t\in[-2,2]}+\textbf{1}_{x-3t\in[-2,2]}]+\frac{1}{6}\int_{x-3t}^{x+3t}\textbf{1}_{y\in[-2,2]}dy.
\end{align}$$
Now observe that $x\pm3t\in[-2,2]\iff -2\le x\pm3t\le 2\iff -2\mp3t\le x\le 2\mp 3t$   hence $\textbf{1}_{x+3t\in[-2,2]}$ is equivalent to $\textbf{1}_{x\in[-2-3t,2-3t]}$ and $\textbf{1}_{x-3t\in[-2,2]}$is equivalent to $\textbf{1}_{x\in[-2+3t,2+3t]}$.
This shows that for a fixed $x\in\mathbb R$, both $\textbf{1}_{x\in[-2+3t,2+3t]}$, $\textbf{1}_{x\in[-2-3t,2-3t]}\longrightarrow c(x)\equiv 0$.
The result of $\int_{x-3t}^{x+3t}\textbf{1}_{y\in[-2,2]}dy$ depends on the intersection $[-2,2]\cap(x-3t,x+3t)$.
Is it  more clear now?
