 Solution

For a plane wave with amplitude $\alpha$, wavenumber $k>0$ and angular frequency $\omega >0$, the dispersion relation $\omega =\omega (k)$ specifies which frequencies the governing equations permit. A point of fixed wave phase obeys $kx-\omega t=\text{constant}$, so the phase speed isIt is the speed of individual crests. For a narrow wave packet, two nearby plane waves with differences $\Delta k$ and $\Delta \omega$ have a slowly varying wave envelope depending on $\Delta kx-\Delta \omega t$. Taking the limit gives the group velocityThis is the envelope's propagation speed and, in the present linear lossless wave model, the energy transport speed for a narrow packet.

A dispersive wave has phase speed dependent on wavenumber, so different components can change their relative phases and alter the packet's shape. A nondispersive wave on the chosen propagation branch has constant phase speed over the relevant band:Indeed $d{c}_p/dk=(c_g-c_p)/k$, so equality throughout a band is equivalent to constant phase speed there. Equality at an isolated wavenumber does not make the entire wave family nondispersive.