Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 38B / Solution

Substitution of $e^{i(kx-\omega t)}$ givesThe phase velocity is negative for all nonzero wavenumbers, so crests travel left. The group velocity is negative for $|k|<\ell$, zero at $|k|=\ell$, and positive for $|k|>\ell$. Its minimum is $-\beta /{\ell }^2$ at zero and maximum $\beta /(8{\ell }^2)$ at $|k|=\sqrt{3}\ell$; it tends to zero from above at infinity. The odd dispersion relation has extrema at $k=\pm \ell$, with $\omega (\ell )=-\beta /(2\ell )$.

The Fourier solution is $\phi (x,t)={\int }_{\mathbb{R}}A(k)e^{i[kx-\omega (k)t]}dk$. The stationary-phase statements require enough amplitude regularity and decay, for example $A$ smooth and rapidly decreasing; realness and evenness alone do not guarantee a convergent integral or a stationary-phase expansion. For $x=Vt$, set ${\Phi }_V(k)=kV-\omega (k)$ and solve ${\omega }^{\prime }(k)=V$ implicitly.