= Solution
Use an upward vertical coordinate $z$, constant <mass density> $\rho$, and a positive <Coriolis parameter> $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_x\,dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho\,dz$ and including the <Coriolis acceleration> gives
$$
u_t-fv=-\frac{p_x}{\rho}+\frac{X_z}{\rho},\qquad v_t+fu=-\frac{p_y}{\rho}+\frac{Y_z}{\rho}.
$$
Here $X,Y$ are the vertical <shear stress> components of the <viscous stress tensor>, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an <Ekman layer> correction. On an <f-plane> these satisfy
$$
u_t^P-fv^P=-p_x/\rho,\qquad v_t^P+fu^P=-p_y/\rho,
$$
and
$$
u_t^E-fv^E=X_z/\rho,\qquad v_t^E+fu^E=Y_z/\rho.
$$
The exterior pressure response is in <geostrophic balance> when steady. For a <surface Ekman layer> with negligible stress at its base, its <Ekman transport> is
$$
M_x^E=\int u^E\,dz=\frac{Y^s}{\rho f},\qquad M_y^E=\int v^E\,dz=-\frac{X^s}{\rho f}.
$$
Integrating the <continuity equation> and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E=\partial_xM_x^E+\partial_yM_y^E$. \b[Thus the base velocity, positive upward, is]
$$
\boxed{w_b^E=\frac{Y_x^s-X_y^s}{\rho f}.}
$$
The PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.
For constant <kinematic viscosity>, write $W_E=u^E+iv^E$ and $\delta=\sqrt{2\nu/f}$. The steady <Ekman layer> equation is $\nu W_E''=ifW_E$. Taking the surface at $z=0$ and the ocean below it, the bounded <surface Ekman layer> solution is
$$
\boxed{W_E(z)=\frac{X^s+iY^s}{\rho\nu\lambda}\,e^{\lambda z},\qquad \lambda=\frac{1+i}{\delta},\quad z\le0.}
$$
The total velocity is $W_P+W_E$. Prescribed <wind stress> determines the coefficient of this <Ekman layer> correction; it does not determine a relation between the wind and the independent pressure-driven velocity. \b[An additional boundary condition is necessary for a <laminar Ekman boundary stress> law in terms of $u^P,v^P$.]
The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard <Bottom Ekman layer> relations. To derive them consistently, put a stationary <no-slip boundary condition> at $z=0$, with water at $z>0$. Then
$$
W_E=-W_Pe^{-(1+i)z/\delta},\qquad X_b+iY_b=\rho\nu W_E'(0)=\rho\sqrt{\frac{f\nu}{2}}(1+i)W_P.
$$
Consequently
$$
\boxed{X_b=\rho\sqrt{\frac{f\nu}{2}}(u^P-v^P),\qquad Y_b=\rho\sqrt{\frac{f\nu}{2}}(u^P+v^P).}
$$
These are <shear stress> values; the actual bottom traction on the fluid has the opposite sign. Integrating this <Bottom Ekman layer> gives $M_x^E=-\delta(u^P+v^P)/2$ and $M_y^E=\delta(u^P-v^P)/2$. Since the exterior <geostrophic flow> is horizontally nondivergent, the upward velocity above the bottom is $-\nabla_h\cdot\mathbf M_E=\delta\zeta_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yields
$$
\boxed{w_b=\frac{1}{\rho f}\sqrt{\frac{\nu}{2f}}\,\nabla_h^2p.}
$$
This recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.
For <Ekman spin-down in a shallow-water layer>, let $\alpha=\sqrt{f\nu/2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior <linearized shallow water equations> through $\eta_t+H\nabla_h\cdot\mathbf u=w_b$. Taking the curl of their momentum equations gives $\zeta_t+f\nabla_h\cdot\mathbf u=0$, hence
$$
\left(\zeta-\frac{f\eta}{H}\right)_t=-\frac{f}{H}w_b.
$$
Slow <geostrophic balance> gives $\zeta=(g/f)\nabla_h^2\eta$ and $w_b=(g\alpha/f^2)\nabla_h^2\eta$. Therefore the intended damping equation is
$$
\boxed{(\nabla_h^2\eta-R_D^{-2}\eta)_t=-\frac{\alpha}{H}\nabla_h^2\eta.}
$$
\b[The slow-adjustment assumption is $f\tau\gg1$, not the printed $f\tau\ll1$.] The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.
For a <Fourier mode> $\eta=\widehat\eta(t)e^{i\mathbf k\cdot\mathbf x}$ with $|\mathbf k|=\kappa>0$, substitution gives
$$
\widehat\eta_t=-\frac{\alpha}{H}\frac{\kappa^2}{\kappa^2+R_D^{-2}}\widehat\eta,\qquad \boxed{\tau=\frac{H}{\alpha}\left(1+\frac{1}{\kappa^2R_D^2}\right).}
$$
For scales small compared with the <Rossby deformation radius>, $\kappa R_D\gg1$ and $\tau\simeq H/\alpha$, independent of <wavenumber>. For scales large compared with the <Rossby deformation radius>, the equation becomes $\eta_t=K^E\nabla_h^2\eta$, where
$$
\boxed{K^E=\frac{\alpha R_D^2}{H}=\frac{g}{f^2}\sqrt{\frac{f\nu}{2}}.}
$$
The zero <wavenumber> mode does not decay. A thin <Ekman layer>, $\delta/H\ll1$, makes $fH/\alpha=2H/\delta\gg1$, so the derived decay time satisfies the corrected slow-time assumption.
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