= 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> are
$$
V(H-h_0)=HU,\qquad VHU=HU^2+\frac g2(H^2-h_0^2).
$$
The first gives $R=H/h_0=V/(V-U)=v/(v-1)>1$. Substituting into the second gives $F^2v=(R^2-1)/2$, where $v=V/U$ and $F=U/\sqrt{gh_0}$. Hence the <piston-driven shallow-water bore> obeys
$$
\boxed{F^2v^2=\frac{v(2v-1)}{2(v-1)^2}=1+\frac{3}{2(v-1)}+\frac1{2(v-1)^2},\qquad v>1}.
$$
The positive depth increase is consistent with the energy-loss condition in <3/a/ii>, now viewed in the bore's own <shock frame>.
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