 Solution

Write the cosine in i as two complex exponentials. On a ray $x=vt$, the oscillatory phases are $kv\pm \omega (k)$. The stationary phase method requires $v=\mp {\omega }^{\prime }(k)$. By ii, the largest absolute horizontal group velocity for the occupied $n=1$ mode isInside this region there are stationary wavenumbers supporting appreciable wave packet contributions. Outside it no stationary wavenumber exists; oscillatory cancellation suppresses the large-time response. This explains "mainly confined": the initial Lorentzian profile already has nonzero algebraic tails, so the displayed region is not the exact support of the solution. Near its boundary the stationary points meet $k=0$ and the usual nondegenerate stationary-phase formula needs a transition approximation.