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At a hydraulic control, the specific-energy function of a zero-potential-vorticity rotating channel flow is stationary with respect to wall depth at fixed discharge and width. Writing $v=u_1+fb/2$ gives $v(v+fb/2)=g{h}_1$ and $Q_c=b{v}^3/g$, hence the displayed discharge. The fully wetted positive-flow solution requires $fb<\sqrt{2g{h}_1}$; at equality the other wall dries and the first wall velocity vanishes. Beyond that limit the assumed two-wet-wall profile must be replaced.

Material conservation of shallow-water potential vorticity carries $P=0$ from the reservoir into the channel. At leading order in a long, slowly varying channel, $v$ and $v_x$ are negligible, so $\zeta \simeq -u_y$. Zero absolute vorticity therefore implies $u_y=f$. Integration from $y=0$, followed by cross-channel geostrophic balance, gives the zero-potential-vorticity rotating channel flow:In particular, $u_2=u_1+fb$ and $h_2=h_1-(fb/g)(u_1+fb/2)$. The velocity increases towards the shallower wall. Differentiating the depth-based specific energy confirmsThis cross-channel constancy is also the $P=0$ consequence of $\nabla B=-P\nabla \psi$.

For a reservoir of finite depth, exact $P=0$ requires relative vorticity $\zeta =-f$, an anticyclonic rotation cancelling the planetary vorticity. A reservoir at rest in the rotating frame has $\zeta =0$ and $P=f/h$, so it does not have exactly zero shallow-water potential vorticity. A very deep, nearly quiescent reservoir can nevertheless supply the approximation $P\simeq 0$ because $f/h$ is small. The idealization must describe vanishing absolute vorticity or this deep-reservoir limit, rather than ordinary finite-depth rest in the rotating frame.

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.