 Solution

Put $\Delta =fb$ and $v=u_1+\Delta /2$, the mean of the two wall velocities. The zero-potential-vorticity rotating channel flow has $h_2=h_1-\Delta v/g$. Substitution into the volume flux formula givesThe resulting quadratic equation has two roots, $v=g{h}_1/\Delta \pm \sqrt{g^2{h}_1^2/{\Delta }^2-2gQ/(b^{2}f)}$. The positive-depth condition $h_2>0$ selects the minus sign. ThusThe discarded sign would make $h_2<0$. On the retained branch $h_2=\sqrt{h_1^2-2fQ/g}$, which also shows the admissibility condition $h_1^2\ge 2fQ/g$.

The depth-based specific energy at the first wall isBecause ${\partial }_y(h+u^2/(2g))=0$, this same specific energy holds everywhere across the channel at fixed $x$. Its sum with the bed elevation is the Bernoulli function divided by $g$. Zero shallow-water potential vorticity makes that Bernoulli function constant throughout the connected steady channel flow, soConsequently the depth-based specific energy decreases as the bed rises and increases as it falls. It is constant along a level-bed channel, even if the width varies; that width variation instead changes the wall depths and velocities compatible with the conserved volume flux and Bernoulli function.