 Solution

An eddy at height $z$ cannot extend arbitrarily far through the ground, so its largest local vertical scale is proportional to $z$. Choose $\ell \sim z$ and absorb its constant of proportionality into the coefficient. With the prescribed constant turbulent velocity scale, writeThe mixing-length closure is $-\overline{u^{\prime }{w}^{\prime }}={\nu }_T{U}_z$. Thus the full kinematic stress isFor $z\gg \delta$, neglect the molecular term. With no internal momentum source, $T$ is constant, soIntegrating between two heights givesEquivalently define an extrapolated roughness length $z_0$ and write $U=(T/C)\log (z/z_0)$. If $C=\kappa u_{\ast }$, the familiar law of the wall is $U/u_{\ast }=(1/\kappa )\log (z/z_0)$. The logarithmic profile is valid outside the inner viscous region and below scales where the constant-stress or wall-distance assumptions fail. It cannot be continued to $z=0$ to impose the smooth-wall no-slip condition.

The part b notation $\tau ={\nu }_T{U}_z$ must therefore mean $\tau =T$, a stress divided by density. Using the physical stress from part a on that right-hand side would omit a factor of $\rho$.