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\section{Steady onshore wind in a   shallow Sea}
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Let us model the effect of turbulence by a constant eddy viscosity.  Assume that convective inertia is negligible, the seabed is horizontal   and  vertical shear is important,   the governing equation in a shallow sea are 
\be \f{\p u}{\p x} + \f{\p v}{\p y} + \f{\p w}{\p z} = 0 \ee
\be  \f{\p u}{\p t} - fv = -g \f{\p \eta}{\p x} + \nu \f{\p^2 u}{\p z^2} \ee
\be  \f{\p v}{\p t} + fu = -g \f{\p \eta}{\p z} + \nu \f{\p^2 u}{\p z^2} \ee
The boundary conditons are 
\be u=v=w=0, ~~ z = -h\ee
\be \mu\f{\p u}{\p z} = \tau_x^S, ~~ \mu \f{\p v}{\p z} = \tau_y^S\ee
As in a thin boundary layer, the vertical shear dominates.

Integrating over depth and defining the  horizontal transport rates,
\be  U = \int^0_{-h} \, udz, \qquad V = \int^0_{-h} \, vdz \ee
then the 
\begin{eqnarray*}
\f{\p \eta}{\p t} & + & \f{\p U}{\p x} + \f{\p V}{\p y} = 0 \\
\f{\p U}{\p t} & - & fV = -gH \, \f{\p \eta}{\p x} + \f{\tau^S_x}{\rho} - \f{\tau^B_x}{\rho} \\
\f{\p V}{\p t} & + & fU = -gH \, \f{\p \eta}{\p y} + \f{\tau^S_y}{\rho} - \f{\tau^B_y}{\rho} .
\end{eqnarray*}
where
\be \tau_x^S = 
\rho \nu\lb\f{\p  u}{\p z }\rb_{z=0} , \quad \tau_y^S= \rho \nu\lb\f{\p v}{\p z}\rb_{z=0} \ee
\be \tau_x^B = 
\rho \nu\lb\f{\p  u}{\p z }\rb_{z=-h} , \quad \tau_y^B= \rho \nu\lb\f{\p v}{\p z}\rb_{z=-h} \ee

 As an order estimate
\[U   \sim   \f{\tau^S}{\rho f} \sim \f{u^2_*}{f} \] 
where $u_*$ is the friction velocity. 
 A vertical boundary layer (of Ekman) can exist wherein Coriolis force ${\rho f U}$ balances the viscous stress 
 \[ \tau^B   \sim   \rho \nu \, \f{\p u}{\p z} \sim \rho \nu \, \f{U}{h} \, \f{1}{\delta}\]
  Therefore, the Ekman layer thickness is \[ \delta = O\lp \sqrt{\f{\nu}{f}}\rp \]
 A typical value of eddy viscosity is $\nu=1 ~cm^2/s$ and  $ f=10^{-4} ~1/s$. Therefore the Ekman layer thicknss is $O(1)$  m. 

Our strategy is to get the horizontal  transport,  then the details of the boundary layers. 


\subsection{Wind setup due to steady onshore wind}
 Consider   an infinitely long coastline along the $x$ axis.   Assume $\tau_y^S $ to be a given constant.  Consider the steady state $\p/\p t = 0$ and ignore $\tau^B$ first.    Beginning from the equations :
\[ \f{\p U}{\p x} + \f{\p V}{\p y} = 0 . \]
\begin{eqnarray*}
-fV & = & -gH \, \f{\p \eta}{\p x} \\
+ fU & = & -gH \, \f{\p \eta}{\p y} + \f{\tau^S_y}{\rho} 
\end{eqnarray*}
with 
\be V=0, ~~y=0\ee
Because of the infinite coast, \[ U=\f{\p \eta}{\p x} = 0, \] we must have \[ \f{\p V}{\p y}  = 0,\] It follows that  $V=0$ everywhere, and 
\be   gH \, \f{\p \eta}{\p y} = \f{\tau^S_y}{\rho} . \ee
meaning that there is a sea-level   {\bf Set-up}.

\begin{eqnarray*}
\eta & = & \f{\tau^S_y}{gH} \, y + \; \mbox{constant} \\
& = & \f{\rho u^2_*}{\rho gH} \, y = \f{u^2_*}{gH} \, y . \end{eqnarray*}
If $u_* = 1$ cm/sec then $ \tau^S_y = 0.1$ Pa.  If $g = 10$ m/sec$^2$  {and} $ H = 30$ m
\[ \f{u^2_*}{gH} = \f{\lp 10^{-2} \rp^2}{10 \cdot 30} = 3 \times 10^{-7} . \]
Note that \[ 1 \, \mbox{atm} = 10^5 N/m^2, \quad    1 N/m^2 = 1 Pa = \f{1}{670} \, psi . \]
For $g = 10 m/s^2 \qquad H = 30 m$ the set up is calculated as follows. 
 \vskip 1 cm
\[ 
\begin{tabular}{|c|c|c|c|}\hline
 $\tau_y^S$ & $\f{\p \eta}{\p y}$ & $\Delta y$ & $\Delta \eta$ \\ \hline
0.1\,  Pa & $3\times 10^{-7}$ & 100 \, km & 3 \,cm \\ \hline
3 \,Pa&& 300 \, km &3 \, m \\
\hline
\end{tabular}
 \]
 
Although there is no mean flow (or flux), there is internal flow. We now look at the detailed distribution in $z$, by deviding the depth into three parts: the  geostrophic interior, the surface Ekman layer, and the bottom Ekman layer.  

\subsection{Geostrophic core}
 
Outside the boundary layers,   we have
\be -fv_g= -g\f{\p \eta}{\p x}  =0,\ee
\be 
 fu_g= -g\f{\p \eta}{\p y}=-\f{u_*^2}{H}\ee
In this geostropic  balance, there is a  longshore current  in the core.
  
 \subsection{Surface Ekman layer}
Now viscosity is important, so that
\begin{eqnarray*}
-fv & = & -g \, \f{\p \eta}{\p x} + \nu \, \f{\p^2 u}{\p z^2} \\
fu & = & -g \, \f{\p \eta}{\p y} + \nu \, \f{\p^2 v}{\p z^2} . 
\end{eqnarray*}
For this example
\[ \f{\p \eta}{\p x} = 0 , ~~~  \f{\p \eta}{\p y} = \f{\tau^S_y}{\rho g H}= \f{\rho  u_*^2}{\rho g H}=\f{u^2_*}{gH} \]
hence 
 \be  -fv   = \nu \, \f{\p^2 u}{\p z^2} \ee
\be fu  = - \f{u^2_*}{H} + \nu \, \f{\p^2 v}{\p z^2} . \ee  
 As long as $\delta/H \ll 1$, bounday-layer approximation can be made.
 Let us make some estimates based on empirical data, cited from 
Csanady  : 
\[\delta  = 0.1 \, \f{u_*}{f},  \quad \nu = \f{u^2_*}{200 f},  \quad Re_*   = \f{u_* \delta}{\nu} = 20 \]
 Pedlosky :  \[ \nu = 1 \sim 10^3 cm^2/sec \]
\[ \delta = \sqrt{\f{1 \sim 10^3}{10^{-4}}} cm = 10^2 cm \sim 3 \times 10^3 cm. \]

Let the total  velocity in the surface boundary layer be  \be u=u_g+u_E= -\f{{u^*}^2}{fH}+u_E, ~~v=v_g+v_E=v_E. \ee
so that $(u_E,v_E)$ are  the boundary layer corections. Then for $z<0$, we have
\be -f v_E = \nu \f{\p^2 u_E}{\p z^2 } \ee
\be  f u_E   = \nu \, \f{\p^2 v_E}{\p z^2} \ee
The boundary conditions are
\[ \nu \, \f{\p u_E}{\p z} = 0, \qquad \nu \, \f{\p v_E}{\p z} = u^2_* \qquad \mbox{on} \; z = 0 \]
\[ u_E, \; v_E \to 0 \qquad z \to -\infty . \]
This is the {Ekman boundary-layer problem}. 
The solution is best obtained by introducing the complex velocity,
\[ q_E = u_E + iv_E  \]   then 
\[   q_E = \nu \, \f{\p^2  q_E}{\p z^2} \, \]
\[\hbox{or}\quad  \f{d^2q_E}{dz^2} - \f{if}{\nu} \, q_E = 0 \] 
Let the solution be of the form,
\[  q_E \propto e^{Dz} \]
then
\[ D^2 - \f{if}{\nu} = 0 \] 
Since \[  (i)^{1/2} = \pm e^{i\pi/4} = \pm \f{1 + i}{\sqrt{2}} . \]
We get
\be q_E = A \, \exp\lp {\f{1+i}{\sqrt{2}} \, \f{z}{\sqrt{\nu/f}}}\rp =A \, e^{(1+i)^{z/\delta}} . 
\ee
Let
\be
\delta = \sqrt{\f{2\nu}{f}} \ee
denote the boundary layer thickness. Apply the boundary condition on the sea surface, \[ \nu \, \left. \f{\p q_E}{\p z} \right|_0 = i \, u^2_* \]
hence
\[ A = \f{i \, u^2_* \, \delta}{(1+i)\nu} . \]
The solution is
\be q_E = u_E + i \, v_E = \f{i\delta \, u^2_*}{(1+i)\nu} \, e^{(1+i){z/\delta}} = \f{\delta \, u^{*^2}}{2\nu} (1+i) \, e^{z/\delta} \, \lp \cos \, \f{z}{\delta} + i \, \sin \, \f{z}{\delta} \rp \ee 
Separating real and imaginary parts, we get the velocity components,
\be u_E =\f{\delta}{2} \, \f{u^2_*}{\nu} \, e^{z/\delta} \, \lb \cos \, \f{z}{\delta} - \sin \, \f{z}{\delta} \rb \ee
\be v_E = \f{\delta}{2} \, \f{u^2_*}{\nu} \, e^{z/\delta} \, \lb \cos \, \f{z}{\delta} + \sin \, \f{z}{\delta} \rb .
\ee

 The hodograph is shown in Figure \ref{fig:Ekman}, 
  \begin{figure} [h]\begin{center}
\includegraphics[scale=1]{Ekman.eps}\end{center}
\caption{ Ekman Boundary layer.} 
 \label{fig:Ekman}
\end{figure} 
\begin{enumerate}
\item Physical Remark \#1:\\
\no Maximum velocity occurs on $z = 0$:
\[ \f{u_E(0)}{u_*} = \f{v_E(0)}{u_*} = \f{u_*}{f\delta} \lp \gg \f{U}{u_*h} = \f{u_*}{fh} \rp . \]
and is 45 degrees to the right of wind.
\item Physical  Remark \#2 :\\
The total flux in Ekman layer is
\begin{eqnarray*}
U^E_x + iV^E_y & = & \int^0_{-\infty} \, dz \, \lp u_E + i \, v_E \rp = \int^0_{-\infty} \, dz \, q_E \\
& = & \f{\delta \, u^2_*}{2\nu} \, (1+i) \, \int^0_{-\infty} \, e^{(1+i){z/\delta}} dz \\
& = & \f{\delta \, u^2_*}{2\nu} \, (1+i) \cdot \f{\delta}{1+i} \cdot \left. e^{(1+i)^{z/\delta}} \right|^0_{-\infty} = \f{\delta^2 \, u^2_*}{2\nu} = \f{u^2_*}{f} . 
\end{eqnarray*}
where\[ \delta^2 = \f{2\nu}{f} . \]
Therefore the  total mass flux in Ekman layer is 90 degrees inclined with respect to wind.
\item Physical remark \# 3:\\
 Note that the flux in the surface Ekman layer is counter balanced by the geostrophic  return flow beneath.   This implies the assumption that the bottom Ekman layer is very weak and contributes little to the flux.  Let us check.
\end{enumerate}

\subsection{Bottom Ekman  layer}
The total flow is governed by 
\[ -fv = \nu \, \f{\p^2 u}{\p z^2} \qquad fu = fu_g + \nu \, \f{\p^2 v}{\p z^2} . \]
Let
\be u_E =u - u_g \qquad v_E = v \ee
so that
\be -fv_E = \nu \, \f{\p^2 u_E}{\p z^2} , \qquad fu_E = \nu \, \f{\p^2 v_E}{\p z^2} 
\ee
Let us shift to new coordinates with the origin on the sea bed so that   the boundary conditions are 
\be 
z \to \infty, ~~   u_E, v_E \to 0 \ee
and 
\be z = 0 , ~~  u_E = -u_g, v_E = 0 \ee 

Let
\be q_E = u_E + iv_E ~~~ q_E = A \, e^{-(1+i){z/\delta}} 
\ee 
Since there is no slip at $z = 0$
\[ q_E = -u_g . \]
We conclude,
\[ A = -u_g . \]
and 
\begin{eqnarray*}
q_E & = & \lp u_E + iv_E \rp = -u_g \, e^{-(1+i) {z/\delta}} \\
& = & -u_G \, e^{-z/\delta} \, \lp \cos \, \f{z}{\delta} - i \, \sin \, \f{z}{\delta} \rp . 
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
u_E & = & -u_g \, e^{-z/\delta} \, \cos \, \f{z}{\delta} \\
v_E & = & u_g \, e^{-z/\delta} \, \sin \, \f{z}{\delta} . \end{eqnarray*}


 The bottom shear stress is 
\[ \nu \, \f{\p q_E}{\p z} = \f{1+i}{\delta} \, u_g \, e^{-(1+i) \, z/\delta} \]
at $z = 0$
\[ \nu \, \left. \f{\p q_E}{\p z} \right|_0 = \f{1+i}{\delta} \, u_g = - \f{1+i}{\delta} \, \f{u^2_*}{fH} . \]


The total flux in bottom Ekman layer is 
\begin{eqnarray*}
U_E + iV_E & = & \int^\infty_0 \, \lp u_E + iv_E \rp dz \\
& = & -u_g \, \int^\infty_0 \, e^{- (1+i/\delta)z} dz \\
& = & -u_g \, \f{1}{-(1+i)/\delta} \, \left. e^{-(1+i/\delta)z} \right|^\infty_0 \\
& = & -u_g \, \f{\delta}{1+i} = \f{u^2_*}{fH} \, \f{\delta}{1+i} = \f{\delta}{2} (1-i) \f{u^2_*}{fH}
\end{eqnarray*}
It  is of order $\delta$ and is directed at 135 degrees to the right of wind.  

\subsection{Summary}: 
\begin{itemize}
\item Total Ekman layer flux on top is $u^2_*/f$

\item Total geostropic flux is  $- u^2_*/f$

\item Total bottom Ekman layer flux is very small  $O \lb \f{u^2_*}{f} \lp \f{\delta}{H} \rp \rb$.
\end{itemize}




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