We saw that the displacement of the plucked string is, on the one hand,
$$
u(x, t)=\frac{4 L}{\pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2 n-1)^{2}}
\cos \frac{(2 n-1) \pi a t}{L} \sin \frac{(2 n-1) \pi x}{L}, 0 \leq x \leq L,
t \geq 0
$$
and, on the other hand,
$$
u(x, \tau)=\left\\{\begin{aligned}
x, & 0 \leq x \leq \frac{L}{2}-a \tau, \\
\frac{L}{2}-a \tau, & \frac{L}{2}-a \tau \leq x \leq \frac{L}{2}+a \tau, \\
L-x, & \frac{L}{2}-a \tau \leq x \leq L.
\end{aligned}\right.
$$
if \(0 \leq \tau \leq L / 2 a\). The first objective of this exercise is to show
that (B) can be used to compute \(u(x, t)\) for \(0 \leq x \leq L\) and all \(t>0\)
(a) Show that if \(t>0,\) there's a nonnegative integer \(m\) such that either
(i) \(t=\frac{m L}{a}+\tau\) or
(ii) \(t=\frac{(m+1) L}{a}-\tau\),
where \(0 \leq \tau \leq L / 2 a\)
(b) Use (A) to show that \(u(x, t)=(-1)^{m} u(x, \tau)\) if (i) holds, while
\(u(x, t)=(-1)^{m+1} u(x, \tau)\) if (ii) holds.
(c) Perform the following experiment for specific values of \(L\) and \(a\) and
various values of \(m\) and \(k:\) Let
$$
t_{j}=\frac{L j}{2 k a}, \quad j=0,1, \ldots k
$$
thus, \(t_{0}, t_{1}, \ldots, t_{k}\) are equally spaced points in \([0, L / 2 a]
.\) For each \(j=0,1,2, \ldots, k,\) graph the \(m\) th partial sum of (A) and
\(u\left(x, t_{j}\right)\) computed from (B) on the same axis. Create an
animation, as described in the remarks on using technology at the end of the
section.
