 Solution

Zero initial velocity selects the cosine solution of the transformed equation:with the removable value one at $k=0$. Since $\cos (ckt)=(e^{ickt}+e^{-ickt})/2$, the Fourier shift theorem giveswhere $r_{+}=\max (r,0)$. 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 $|x|\le 1+|c|t$, consistent with finite propagation speed. For $c=0$ it simply remains $f(x)$.

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.