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Vertical boundary conditions

The new boundary conditions for the vertical velocity, $\omega$, in equation (36) become
\begin{displaymath}
\omega (0) = \omega (-1) = 0.
\end{displaymath} (45)

The new conditions at the surface ($\sigma = 0)$ becomes
$\displaystyle \frac{K_M}{D} \left( \frac{\partial U}{\partial \sigma},
\frac{\partial V}{\partial \sigma} \right)$ $\textstyle =$ $\displaystyle \frac{1}{\rho_0} (\tau_{0x},\tau_{0y}),$ (46)


$\displaystyle \frac{K_H}{D} \left( \frac{\partial T}{\partial \sigma},
\frac{\partial S}{\partial \sigma} \right)$ $\textstyle =$ $\displaystyle (\dot{T_0},\dot{S_0}) ,$ (47)

and at the bottom ($\sigma = -1$) the boundary conditions become
$\displaystyle \frac{K_M}{D} \left( \frac{\partial U}{\partial \sigma},
\frac{\partial V}{\partial \sigma} \right)$ $\textstyle =$ $\displaystyle \frac{1}{\rho_0} (\tau_{bx},\tau_{by}),$ (48)


$\displaystyle \frac{K_H}{D} \left( \frac{\partial T}{\partial \sigma},
\frac{\partial S}{\partial \sigma} \right)$ $\textstyle =$ $\displaystyle 0.$ (49)



Helge Avlesen 2004-05-10