= Solution
{c}
Since the initial <velocity> vanishes, <D'Alembert's formula> reduces to
$$
u(x,t)=\frac12\bigl[f(x-t)+f(x+t)\bigr].
$$
Each translated copy has half the original <amplitude>. To identify each requested cross-section, track the translated support intervals and add the values where they overlap.
At $t=\pi$, $f(x-\pi)$ is supported on $[-\pi,0]\cup[2\pi,3\pi]$ and $f(x+\pi)$ on $[-3\pi,-2\pi]\cup[0,\pi]$. On these intervals $\sin(x\pm\pi)=-\sin x$, so
$$
\boxed{u(x,\pi)=
\begin{cases}
-\tfrac12\sin x,&x\in[-3\pi,-2\pi]\cup[-\pi,\pi]\cup[2\pi,3\pi],\\
0,&\text{otherwise}.
\end{cases}}
$$
The pieces meet continuously at zero, where both contributions vanish.
At $t=3\pi/2$, the support of $f(x-3\pi/2)$ is $[-\pi/2,\pi/2]\cup[5\pi/2,7\pi/2]$; that of $f(x+3\pi/2)$ is $[-7\pi/2,-5\pi/2]\cup[-\pi/2,\pi/2]$. In the central overlap,
$$
\sin(x-3\pi/2)=\cos x,\qquad\sin(x+3\pi/2)=-\cos x,
$$
so the two contributions cancel. The remaining <wave> profile is
$$
\boxed{u(x,3\pi/2)=
\begin{cases}
-\tfrac12\cos x,&-7\pi/2\leq x\leq-5\pi/2,\\
\tfrac12\cos x,&5\pi/2\leq x\leq7\pi/2,\\
0,&\text{otherwise}.
\end{cases}}
$$
For the time cross-section at the origin, the initial profile is an <odd function>, so
$$
\boxed{u(0,t)=\tfrac12[f(-t)+f(t)]=0\qquad(t\geq0).}
$$
For the time cross-section at $x=\pi/2$, the left-travelling contribution is nonzero for $\pi/2\leq t\leq3\pi/2$ and equals $\cos t/2$. The right-travelling contribution is nonzero for $3\pi/2\leq t\leq5\pi/2$ and also equals $\cos t/2$. At the common endpoint both vanish. Hence
$$
\boxed{u(\pi/2,t)=
\begin{cases}
\tfrac12\cos t,&\pi/2\leq t\leq5\pi/2,\\
0,&0\leq t<\pi/2\ \text{or }t>5\pi/2.
\end{cases}}
$$
The following graphs show all four cross-sections. In particular, the last trace reaches $-1/2$ at $t=\pi$ and $1/2$ at $t=2\pi$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-v-wave-sections.png]
{title=The four requested cross-sections of the <wave equation> solution}
{description=The upper panels show the spatial profiles at $t=\pi$ and $t=3\pi/2$. The lower panels show the time traces at $x=0$ and $x=\pi/2$ for $t\geq0$. Horizontal axes are labelled in multiples of $\pi$.}
Because $u(0,t)=0$, restriction to the half-line solves the following <initial-boundary value problem> with a homogeneous <Dirichlet boundary condition>:
$$
\boxed{
\begin{aligned}
u_{tt}&=u_{xx},&&x>0,\ t>0,\\
u(0,t)&=0,&&t>0,\\
u(x,0)&=
\begin{cases}\sin x,&\pi\leq x\leq2\pi,\\0,&\text{otherwise},\end{cases}&&x\geq0,\\
u_t(x,0)&=0,&&x\geq0.
\end{aligned}}
$$
This is <wave reflection by odd extension>. If a condition at infinity is desired, take the solution with finite <wave energy>; here the profile has bounded support at every finite time.
The initial profile is <continuous> and piecewise continuously differentiable, but its first <derivative> jumps at the support endpoints. Thus the formula gives a <weak solution> globally and a piecewise <classical solution> away from the <characteristic lines for the one-dimensional wave equation> issuing from those endpoints. The initial <velocity> is zero in the <square-integrable function> sense and pointwise wherever the initial profile is <differentiable>; at the kink points a pointwise right time <derivative> need not equal zero. This regularity qualification is needed for the given data.
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