Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 37C / Solution

Let the rapidly varying phase be $\Theta$, so $\mathbf{k}=\nabla \Theta$ and $\omega =-{\Theta }_t$. The dispersion law is the Hamilton-Jacobi equation ${\Theta }_t+\Omega (\nabla \Theta ,\mathbf{x},t)=0$. Differentiating it and following a path with group velocity gives the ray tracing equationsThe final relation is obtained by the chain rule along that path; its two spatial terms cancel.

The medium is stationary and independent of $y$, so $\omega ={\omega }_{\infty }$ and $k_2=K$ are conserved. For deep water far from shore, ${\kappa }_{\infty }={\omega }_{\infty }^2/g$. Choose the incident branch with $k_1<0$ and $K=({\omega }_{\infty }^2/g)\sin {\theta }_{\infty }>0$; choosing the opposite alongshore direction changes its sign. Since the dispersion relation depends on direction only through $\kappa$, the group-velocity component ratio is $k_2/k_1$. Hence the required implicit ray equation isNear shore, $\kappa h\to 0$ and $\kappa \sim {\omega }_{\infty }/\sqrt{g\alpha }{x}^{-p/2}$. Thus $y^{\prime }\sim -K\sqrt{g\alpha }{x}^{p/2}/{\omega }_{\infty }$, whose integral givesThe rays approach the shoreline normally. A crest is a level set of $\int k_{1}dx+Ky-\omega t$. Its slope is $dy/dx=-k_1/K\sim {\omega }_{\infty }{x}^{-p/2}/(K\sqrt{g\alpha })$, so crests become parallel to the shoreline. Their separation in the shore-normal direction is $2\pi /|k_1|\sim 2\pi \sqrt{g\alpha }{x}^{p/2}/{\omega }_{\infty }$ and decreases to zero. These are formal ray asymptotics; for $p<2$, the slow-variation assumption eventually fails arbitrarily close to the shoreline. The sketch shows a representative shallow-water case $p=1$; it illustrates these limiting directions rather than assuming that value in the derivation.