= Solution
For stationary horizontally homogeneous means, <incompressibility> gives $\bar w_z=0$ and the impermeable boundary makes $\bar w=0$ throughout. With no imposed horizontal mean pressure gradient, the balances reduce to
$$
\frac{d}{dz}\left(-\kappa\bar b_z+\overline{w'b'}\right)=0,\qquad
\frac{d}{dz}\left(-\nu\bar u_z+\overline{w'u'}\right)=0,\qquad
\bar p_z=\rho_0\bar b-\rho_0\frac{d}{dz}\overline{w'^2}.
$$
The last equation includes the vertical <Reynolds stress>; omitting it requires an additional small-normal-stress approximation. The total vertical buoyancy flux is $B_0$ and the horizontal momentum flux is a signed constant $J_M$ with $|J_M|=U_*^2$. These are fluxes per unit reference density.
The local down-gradient closure defines <eddy viscosity> and <eddy diffusivity> by
$$
\overline{w'u'}=-K_M\bar u_z,\qquad
\overline{w'b'}=-K_B\bar b_z.
$$
A parcel moving upwards typically retains the smaller momentum or scalar value from its previous height in an increasing mean profile; its fluctuation then correlates negatively with upward motion. This motivates positive coefficients, but is a closure assumption, not a universal theorem about turbulent or convective fluxes. The mean gradients obey
$$
\boxed{\bar u_z=-\frac{J_M}{\nu+K_M},\qquad
\bar b_z=-\frac{B_0}{\kappa+K_B}.}
$$
For positive applied traction on a fluid above a bottom boundary, $J_M=+U_*^2$ and the mean velocity decreases away from the forcing boundary. The usual atmospheric convention specifies the stress absorbed by the boundary and has the opposite sign of $J_M$; scalar and Richardson-number results are unchanged. At large <Reynolds number> and <Péclet number>, outside the molecular wall sublayers, $K_M\gg\nu$ and $K_B\gg\kappa$, so molecular transport can be neglected in these flux-gradient relations. Large bulk numbers do not justify doing this exactly at a smooth solid wall.
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