= Solution
Zero initial velocity selects the cosine solution of the transformed equation:
$$
\widetilde u(k,t)=\widetilde f(k)\cos(ckt)=\frac{2(1-\cos k)}{k^2}\cos(ckt),
$$
with the removable value one at $k=0$. Since $\cos(ckt)=(e^{ickt}+e^{-ickt})/2$, the <Fourier shift theorem> gives
$$
\boxed{u(x,t)=\frac12f(x-ct)+\frac12f(x+ct)=\frac12(1-|x-ct|)_++\frac12(1-|x+ct|)_+,}
$$
where $r_+=\max(r,0)$. This is <triangular pulse splitting under the wave equation>, a special case of the <D'Alembert formula>.
The two half-height triangles move in opposite directions at speed $|c|$. Their sum initially reproduces the full triangle; once $|c|t>1$ their supports are disjoint. The field vanishes outside $|x|\leq1+|c|t$, consistent with <finite propagation speed>. For $c=0$ it simply remains $f(x)$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2-2013-ib-triangle-pulses.png]
{title=A triangular displacement released from rest splits into two translating half-amplitude pulses; their overlap adds}
Because the triangular profile has slope discontinuities, this is a <weak solution> rather than a globally twice differentiable solution. The initial velocity is zero in the distributional or square-integrable sense, and pointwise away from the initial corners. A classical pointwise time derivative need not exist at those corners; for example $u(0,t)=1-|ct|$ near $t=0$. The initial conditions must be interpreted consistently with this nonsmooth displacement.
Back to article page
