Past exam of the waves and compressible flow course of the University of Oxford / 2023 / 3 / b / i / Solution

Let $H$ be the depth behind the bore. The wall's kinematic boundary condition makes the velocity there $U$; ahead of the bore the depth and velocity are $h_0$ and zero. Across a rightward hydraulic bore moving at $V$, the Rankine-Hugoniot conditions areThe first gives $R=H/h_0=V/(V-U)=v/(v-1)>1$. Substituting into the second gives $F^{2}v=(R^2-1)/2$, where $v=V/U$ and $F=U/\sqrt{g{h}_0}$. Hence the piston-driven shallow-water bore obeysThe positive depth increase is consistent with the energy-loss condition in ii, now viewed in the bore's own shock frame.