 Solution

Substitution of $e^{i(kx-\omega t)}$ in the Klein-Gordon equation gives $\boxed{{\omega }^2=1+k^2}$. On the positive-frequency branch, the phase velocity is $\boxed{c_p=\sqrt{1+k^2}/k}$ for $k\ne 0$, and the group velocity is $\boxed{c_g=k/\sqrt{1+k^2}}$. The group speed is less than one, although the phase speed exceeds one in magnitude.

With Fourier transform convention $\hat{f}(k)=\int f(x)e^{-ikx}dx$, the initial velocity has transform $2/(1+k^2)$. Each Fourier mode solves a harmonic-oscillator equation, givingFor $x=Vt$, symmetry makes this the imaginary part of ${\pi }^{-1}\int (1+k^2{)}^{-3/2}{e}^{it(\sqrt{1+k^2}-Vk)}dk$. When $0<V<1$, the phase has a unique stationary point $k_{\ast }=V/\sqrt{1-V^2}$, with phase value $\sqrt{1-V^2}$ and positive second derivative $(1-V^2{)}^{3/2}$. The stationary phase method givesFor $V>1$ there is no real stationary point. More precisely, finite propagation speed makes the solution at $x>t$ depend only on the initial data on $[x-t,x+t]\subset (0,\infty )$. On that half-line the initial velocity is $e^{-x}$, and $t{e}^{-x}$ exactly solves the equation with zero initial displacement. Uniqueness on the domain of dependence therefore givesThis exponential initial-velocity tail for the Klein-Gordon equation is exponentially small, rather than identically zero: the initial data already have a nonzero tail outside the origin.