 Solution

Vertical hydrostatic balance with atmospheric pressure at $y=h(x)$ gives $p(x,y)=p_0+\rho g(h(x)-y)$ to this order. Therefore $p_x=\rho g{h}^{\prime }$ and integrating the gradient from part (i), with $h\to 0$ on both sides, givesFor positive $U$ the upstream surface rises and the downstream surface falls. Thus the displacement scale is $\mu |U|\sqrt{a}/(\rho g{b}^{3/2})$, and the flat-geometry consistency condition isIndeed ${\max }_s|s/(1+s^2{)}^2|=9/(16\sqrt{3})$, so this scale controls the maximum, not only a typical value. Together with $b/a\ll 1$ and $\rho |\Omega |\sqrt{a}{b}^{3/2}/\mu \ll 1$, these conditions ensure a thin, weakly displaced, viscosity-dominated gap. The slope is smaller still, of order $(|h|/b)\sqrt{b/a}$.