= Solution
{c}
At $x=L/2$, the integration <interval> first moves through positive arguments, then crosses zero, and finally contains all of $[-L,L]$. The result is
$$
\boxed{y\!\left(\frac L2,t\right)=\begin{cases}
Lt/2,&0\leq ct\leq L/2,\\[4pt]
\dfrac{L^2+(ct-L/2)^2}{4c},&L/2\leq ct\leq3L/2,\\[4pt]
\dfrac{L^2}{2c},&ct\geq3L/2.
\end{cases}}
$$
The final plateau is the <integral> of the entire <initial velocity> profile divided by $2c$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v-wave-traces.png]
{title=Two traces of the D'Alembert solution for the compactly supported initial velocity}
{description=The spatial profile has two peaks at half the initial support radius. The time trace at that location eventually reaches a constant plateau. Axes use dimensionless variables; plotted curves use $L=c=1$.}
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