Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 40A / b / Solution

Both sides of the dispersion relation are positive on an interval beginning at zero phase, so a zeroth branch exists for every $k>0$. Higher branches first appear when $c=c_s$, for which the right side vanishes and $mh=n\pi$. ThusandThe $n=1$ dispersion curve begins on the line $\omega =c_{s}k$ at $(k_c^{(1)},{\omega }_c^{(1)})$, bends below it, and approaches the layer shear-wave line $\omega ={\bar{c}}_{s}k$ from above as $k\to \infty$.