 Solution

An antiderivative of the initial velocity is $-2c{e}^{-s^2/2}$. Thus D'Alembert's formula givesThis is a positive Gaussian function moving right and a negative Gaussian function moving left. Their cancellation at $t=0$ gives the zero initial displacement, while direct differentiation gives ${\psi }_t(x,0)=2cx{e}^{-x^2/2}$.

For every $t>0$, the profile is an odd function of $x$, vanishes at $x=0$, is positive for $x>0$ and negative for $x<0$, and tends to zero at both spatial infinities. The signs follow from $(x+ct{)}^2-(x-ct{)}^2=4xct$. The positive and negative lobes move farther apart as $t$ increases; for large separation they approach the individual unit-height Gaussian functions centred at $x=\pm ct$. At finite separation the actual extrema lie slightly beyond these centres: writing $a=ct$, the positive global maximum occurs at the unique $x>a$ satisfying $(x-a)/(x+a)=e^{-2ax}$.