Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 37C / Solution

The initial Fourier transform is $\hat{\zeta }(k,0)=2b{\zeta }_0/(1+b^2{k}^2)$. Initial rest gives zero time derivative, so each wave mode evolves by $\cos (\sqrt{g|k|}t)$. Fourier inversion therefore givesOn $x=Vt$, the product of cosines produces phases ${\phi }_{\pm }(k)=Vk\pm \sqrt{gk}$ with prefactor $b{\zeta }_0/\pi$. Only the minus phase has a stationary point:Expanding quadratically there and evaluating the Gaussian oscillatory integral gives the stationary-phase factor $\sqrt{2\pi /(t{\phi }_{-}^{\prime \prime }(k_{\ast }))}$ and phase shift $+\pi /4$. The other phase has no stationary point. Using $q=\sqrt{k}$ makes both endpoint phases smooth with nonzero endpoint derivative and amplitude vanishing at zero, so endpoint contributions are smaller. ThusThe selected wavenumber has group velocity $V$, explaining why it determines the long-time signal on this ray.