8. Show that the solution of the boundary-value problem
$u_t = c^2 u_{xx}$;
$\begin{cases}
u(x,0) = f(x), \quad u_t(x,0) = g(x), \quad -l < x < l \\
u(0,t) = u(l,t) = 0
\end{cases}$
is
$u(x,t) = \frac{1}{2}[F(x-ct) + F(x+ct)] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)ds$
where $F$ is the odd periodic extension of $f$.