Ekman layer

Remarks

Example: at a bottom wall $z = 0$ with no slip and $u_g = (U, 0)$, the solution that tends to $u_g$ is $$u = U \big( 1 - e^{-z/\delta} \cos(z / \delta) \big), \qquad v = U e^{-z/\delta} \sin(z / \delta).$$ To check, set $w = (u - U) + i v$; the equation becomes $w'' = (i f / \nu) w = \big( (1 + i)/\delta \big)^2 w$, and $w = -U e^{-(1 + i) z / \delta}$ is the decaying solution with $w(0) = -U$.

At a free surface with stress $\tau$ and $u_g = 0$, integration over the layer gives the Ekman transport $M = \int u_h \, dz = \dfrac{1}{\rho f} \, \tau \times e_z$. It is at right angles to the stress, to its right in the northern hemisphere.

Non-example: in a non-rotating fluid ($f = 0$) there is no Ekman layer, since the thickness $\sqrt{2 \nu / f}$ is infinite. A layer driven by a surface stress then keeps deepening by diffusion and never reaches a steady state.