 Solution

For a real propagating positive-frequency branch, require ${\omega }^2>a^2$ and $c\ne 0$. If $a$ is the usual nonnegative cutoff, the source condition $\omega >a$ gives this branch. With an arbitrary real negative $a$, that inequality alone is insufficient: $a=-2$ and $\omega =1$ satisfy it but make the displayed square root imaginary.

On the positive-frequency real branch, the phase velocity $v_P=\omega /k$ givesDifferentiate the dispersion relation to obtain the group velocity:ThereforeFor $c>0$ and a nonzero cutoff, $v_P>c$ while $v_G<c$; when $a=0$, both equal $c$. The constant $c$ is the velocity scale in this assumed medium, and need not independently be identified with the vacuum speed of light. A group velocity describes a narrow-band wave packet on the chosen differentiable branch, rather than automatically fixing a signal front velocity in every dispersive model.