 Solution

Put the rigid lid at $z=0$, the interface at $z=-h(x,y,t)$ and ${\rho }_1={\rho }_0-\Delta \rho$. Assume hydrostatic vertical balance and a resting deep layer. Its pressure is $p_2=C-{\rho }_{0}gz$, with horizontally uniform $C$. Pressure continuity at the interface then fixes the upper-layer pressure:Hence its horizontal pressure gradient at fixed $z$ is ${\nabla }_h{p}_1=\Delta \rho g{\nabla }_{h}h$, independent of depth. Taking the horizontal momentum equation and a vertically uniform upper-layer velocity givesThe exact density denominator for upper-layer acceleration is ${\rho }_1$; the familiar ${\rho }_0$ denominator is the small-density-contrast Boussinesq approximation. The restoring pressure is weaker than for a free surface because only the density difference contributes.

The horizontal shallow-layer equations use the vertical component of planetary rotation: $f=2\Omega \sin \phi$. Thus the source's vector notation is interpreted as the projected vector $f\mathbf{k}$ after the traditional approximation. Taking the full magnitude $2\Omega$ as a midlatitude Coriolis parameter would discard the latitude dependence. Together with layer-volume conservation, the model has ${\partial }_{t}h+{\nabla }_h\cdot (h\mathbf{u})=0$.