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Problem: Let

\begin{align} u_{tt} &= 9u_{xx} \\ u(0,x) &= 0 \\ u_{t}(0,x) &= f(x) \end{align}

where $ f(x) = \begin{cases}1, && 0<x<1 \\ 0, && \text{otherwise} \end{cases} $

Find the function $u(2,x)$

Solution. ( My Attempt ) We can directly apply d'Alembert's formula

$$ u(t,x) = \frac{f(x+at) + f(x-at)}{2} + \frac{1}{2a}\int_{x-at}^{x+at} g(s)\ ds $$

Here, $a=3$, so

\begin{align} u(t,x) &= \frac{f(x+3t) + f(x-3t)}{2} + \frac{1}{6}\int_{x-3t}^{x+3t} g(s)\ ds \\ \implies u(2,x) &= \frac{f(x+6) + f(x-6)}{2} + \frac{1}{6}\int_{x-6}^{x+6} g(s)\ ds \end{align}

How to proceed next?

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