 Solution

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.