 Solution

Let $\mathbf{U}=(u,v)$, $\zeta =v_x-u_y$ and $D/Dt={\partial }_t+\mathbf{U}\cdot \nabla$. The inviscid shallow water equations areTaking the vertical curl of the momentum equation, with constant $f$, gives $D\zeta /Dt=-(f+\zeta )\nabla \cdot \mathbf{U}$. Combining this with the depth equation provesThus shallow-water potential vorticity is materially conserved: changes in absolute vorticity are exactly compensated by stretching or compression of the layer depth.

The relation to a Bernoulli function presupposes steady flow. In that case define $B=g(h+H)+\frac{1}{2}|\mathbf{U}{|}^2$. The vector identity $(\mathbf{U}\cdot \nabla )\mathbf{U}=\nabla (\frac{1}{2}|\mathbf{U}{|}^2)+\zeta \hat{\mathbf{z}}\times \mathbf{U}$ givesSteady volume conservation allows a transport stream function with $h\mathbf{U}=({\psi }_y,-{\psi }_x)$. Hence $\hat{\mathbf{z}}\times (h\mathbf{U})=\nabla \psi$, soIn particular, $\mathbf{U}\cdot \nabla B=0$: the Bernoulli function is constant along each streamline. Where the flow permits a single-valued dependence on the stream function, the displayed derivative describes how that constant changes between neighbouring streamlines. For an unsteady flow, shallow-water potential vorticity remains conserved but the stated steady Bernoulli function relation does not generally hold.