 Solution

For a Fourier mode $e^{i(kx-\omega t)}$, the wave equation givesTaking the positive-frequency branch,For $k>0$, the phase velocity and group velocity areThe frequency rises from zero and approaches $1$; $c_p$ reaches its maximum at $k=1$, whileshows that the front velocity isThus the wave at the front hasMoreover $c_g/c_p=2/(1+k^2)$, so the packet outruns its crests for $0<k<1$, the speeds agree at $k=1$, and the crests outrun the packet for $k>1$.

The initial conditions give the exact superpositionThe condition $A^{\ast }(-k)=A(k)$ makes this solution real. For $V=x/t$ with $0<V<V_m$, the equationhas two positive roots $0<{\alpha }_1<1/\sqrt{3}<{\alpha }_2$. The second exponential has the corresponding stationary points $-{\alpha }_1,-{\alpha }_2$. Sincethe stationary phase method and pairing of complex-conjugate contributions giveFor $-V_m<V<0$, the same picture is reflected and the stationary points have negative wavenumber. For $|V|>V_m$ there are no real stationary points, so a smooth localized spectrum gives a much smaller nonstationary tail outside the wave front.