Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 65 / 1 / a / i

Differentiating the velocity potential gives $u=a\omega e^{kz}\cos (kx-\omega t)$ and $w=a\omega e^{kz}\sin (kx-\omega t)$. Let $A=a{e}^{kz}$ and $\vartheta =kx-\omega t$. To first order in the wave steepness $ka$, evaluate these velocities at the initial parcel position and integrate from time zero:These expressions satisfy both initial conditions and distinguish initial and mean parcel labels in a surface wave. Their actual short-time limits areThe printed expressions omit the integration constants and therefore cannot be the small-time displacement from the prescribed initial point. For example, at $x=0$, they give $Z(0)-z=a{e}^{kz}\ne 0$. This is a genuine inconsistency, rather than a missing step in the calculation.

The intended oscillatory orbit is recovered by using mean parcel coordinates instead. Set $x_c=x+A\sin (kx)$ and $z_c=z-A\cos (kx)$ to this order. ThenThus a deep-water gravity wave produces approximately circular parcel orbits with radius $a{e}^{k{z}_c}$, decaying exponentially with depth. The approximation is one of small amplitude over a wave cycle, not an assertion that these oscillatory coordinates vanish at time zero.