Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 37B / Solution

For the local phase $S$, define $k_i={\partial }_{i}S$ and $\omega =-{\partial }_{t}S$. The dispersion relation is the Hamilton-Jacobi equation $S_t+\Omega (\nabla S,\mathbf{x},t)=0$. Differentiate it in space and follow a ray with group velocity ${\dot{x}}_i={\Omega }_{k_i}$; symmetry of the phase Hessian cancels the terms involving ${\partial }_i{k}_j$. Thus the ray-tracing equations areHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ means differentiation along the moving ray; subscripts on $\Omega$ mean partial derivatives holding its other independent arguments fixed.

The depth is stationary and independent of $y$, so $\omega ={\omega }_{\infty }$ and $l$ are constant. Choose the incident direction with positive $l$; the other sign reflects the picture. In the far deep water,Since $\Omega$ depends isotropically on the horizontal wavevector, the group-velocity ratio is $l/k$. Therefore the ray equation isNear shore, $\kappa h\to 0$, so the shallow-water dispersion relation gives $\kappa \sim {\omega }_{\infty }/(\sqrt{g\alpha }{x}^{p/2})$ and $k\sim -\kappa$. Integrating the ray slope yieldsFor $p<2$, the near-shore phase is $S\sim ly-{\omega }_{\infty }t-{\omega }_{\infty }{x}^{1-p/2}/[\sqrt{g\alpha }(1-p/2)]$. A constant-phase wave crest thus hasIts tangent becomes parallel to the shoreline, while rays approach perpendicular to it. The exponent is between zero and one, giving the curved crests sketched below for $p=1$.

These are the formal geometrical optics asymptotics. For fixed $\alpha$ and $p<2$, $\kappa x\to 0$ arbitrarily close to the shore, so the slowly varying wave approximation eventually fails there; the sketch describes the ray-theory regime before that final inner region.