Within the planetary boundary layer, particularly within the Prandtl layer, the vertical gradients of meteorological variables are relatively large. They can only be sampled sufficiently well by a model at vertical grid-point spacings on the order of one meter. Therefore, the approaches valid in the free atmosphere for the vertical turbulent fluxes at the lower boundary of the lowest model layer are no longer applicable. Hence, in this region, the findings on the planetary boundary layer developed in Sect. 17.2.1 are used to determine the vertical turbulent fluxes.
According to Eq. (17.80), the Monin-Obukhov length reads
\[ \begin{align} L \coloneqq -\frac{T_Au_\star^3}{kg\newoverline{\left(w'\theta'\right)}}. \end{align} \]
What is problematic here is the determination of $\newoverline{\left(w'\theta'\right)}_\text{sfc}$. Substituting here, one obtains
\[ \begin{align} L = -\frac{T_Au_\star^3}{kg\frac{Q_H}{\rho c_v}} = -\frac{T_Au_\star^3}{kg\frac{T_G - T_A'}{r_H}} = -\frac{T_Au_\star^3}{kg\left(T_G - T_A'\right)}r_H. \end{align} \]
The friction velocity $u_\star$ and the flux resistance of the sensible heat $r_H$ in turn depend on $L$:
\[ \begin{align} L = -\frac{T_Au_\star^3\left(L\right)}{kg\left(T_G - T_A'\right)}r_H\left(L\right).\tag{37.3}\label{eq:mo_length_model_ansatz} \end{align} \]
Substituting here, one obtains
\[ \begin{align} L = -\frac{T_Au_\star^3}{g\left(T_G - T_A'\right)}\frac{1}{u_\star k^2}\left[\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)\right] = \frac{T_Au_\star^2}{k^2g\left(T_A' - T_G\right)}\left[\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)\right]. \end{align} \]
Inserting Eq. (17.49) further, one obtains
\[ \begin{align} L = \frac{T_AU^2}{g\left(T_A' - T_G\right)}\frac{\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2}. \end{align} \]
Here the aim is to determine $L$, with all other quantities being known. The quantity
\[ \begin{align} R_\text{i,b} \coloneqq \frac{zg\left(T_A' - T_G\right)}{T_AU^2}. \end{align} \]
is called the bulk Richardson number. With the auxiliary function
\[ \begin{align} s\left(L\right) &\coloneqq \frac{\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2} \end{align} \]
and
\[ \begin{align} a \coloneqq \frac{R_\text{i,b}}{z} \end{align} \]
one can write this in the form
\[ \begin{align} L = \frac{z}{R_\text{i,b}}s\left(L\right) = \frac{s\left(L\right)}{a} \Leftrightarrow aL - s\left(L\right) = 0\tag{37.9}\label{eq:mo_l_num_deriv_1} \end{align} \]
The sign of $L$ can already be determined in advance from Eq. (37.3). In the case $T_G < T_A'$, which corresponds to stable conditions, $L > 0.$ In this case, one has
\[ \begin{align} \psi_m\left(\frac{z}{L}\right) = \psi_h\left(\frac{z}{L}\right) = -4,7\cdot\frac{z}{L}. \end{align} \]
Now define
\[ \begin{align} h\left(L\right) \coloneqq \ln\left(\frac{z}{z_0}\right) + 4,7\cdot\frac{z}{L}, \end{align} \]
then it follows from Eq. (37.9)
\[ \begin{align} aL - s\left(L\right) &= aL - \frac{h\left(L\right) + \ln\left(7\right)}{h^2\left(L\right)} = 0\nonumber\\ \Leftrightarrow f_1\left(L\right) &\coloneqq aLh^2\left(L\right) - h\left(L\right) - \ln\left(7\right) = 0.\tag{37.12}\label{eq:mo_l_num_deriv_2} \end{align} \]
Eq. (37.12) cannot be solved analytically and must therefore be solved numerically. This corresponds to determining the root of $f_1$. For this, one uses a Newton iteration with
\[ \begin{align} f_1\left(L_i\right) + f_1'\left(L_i\right)\cdot\left(L_{i+1} - L_i\right) = 0, \end{align} \]
here $i$ is the iteration step. One has
\[ \begin{align} f_1'\left(L\right) = ah^2\left(L\right) + 2aLh\left(L\right)h'\left(L\right) - h'\left(L\right). \end{align} \]
Assuming $f_1' \not= 0$, one can reformulate this as
\[ \begin{align} \frac{f_1\left(L_i\right)}{f_1'\left(L_i\right)} + L_{i+1} - L_i &= 0\nonumber\\ \Leftrightarrow L_{i+1} &= L_i - \frac{f_1\left(L_i\right)}{f_1'\left(L_i\right)} \end{align} \]
In the case $T_G > T_A$, which corresponds to unstable conditions, $L < 0.$ In this case, one has
\[ \begin{align} f_2\left(L\right) &\coloneqq aL - s\left(L\right) = aL - \frac{\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2} = 0. \end{align} \]
From this it follows
\[ \begin{align} f_2'\left(L\right) &= a - g'\left(L\right) = a - \frac{\frac{z}{L^2}\psi'_h\left(\frac{z}{L}\right)\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2 - \left[\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)\right]2\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]\frac{z}{L^2}\psi'_m\left(\frac{z}{L}\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^4}\nonumber\\ &= a - \frac{\frac{z}{L^2}\psi'_h\left(\frac{z}{L}\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2} + \frac{2\left[\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)\right]\frac{z}{L^2}\psi'_m\left(\frac{z}{L}\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^3}. \end{align} \]