Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 65 / 3 / f / Solution

The eddy viscosity in the prescribed mixing length model isThus at a fixed similarity coordinate $\eta$, every nonzero value grows as $x^{1/2}$. Its maximum occurs at a fixed coordinate maximizing ${\eta }^2|f^{\prime \prime }|$, so the maximum itself has the same downstream scaling.

The formal argument intended by the edge assumption is also clear. If one independently stipulates $|f^{\prime \prime }(\eta )|\le |f^{\prime \prime }({\eta }_w)|$ for $|\eta |\le {\eta }_w$, then ${\eta }^2|f^{\prime \prime }(\eta )|\le {\eta }_w^2|f^{\prime \prime }({\eta }_w)|$. This would put the maximum eddy viscosity at the edge and giveindependent of $x$, provided $f^{\prime \prime }({\eta }_w)\ne 0$. That is a conditional scaling calculation, not an admissible solution of the full momentum problem.

The printed edge claim fails for the reason established in part (e): $f^{\prime }({\eta }_w)=0$ forces $f^{\prime \prime }({\eta }_w)=0$, and thereforeA nontrivial profile has positive eddy viscosity in the interior; hence its maximum cannot occur at that finite free edge. Moreover $U(x,0)y_w/{\nu }_T(x,\pm y_w)$ has a zero denominator, so it is not a finite effective Reynolds number under the literal assumptions.

The intended downstream cancellation is valid if one uses a consistent interior reference point ${\eta }_{\ast }\in (0,{\eta }_w)$ with $f^{\prime \prime }({\eta }_{\ast })\ne 0$, for example the maximum of ${\nu }_T$. Since $U(x,0)=M_0^{1/2}{x}^{-1/2}{f}^{\prime }(0)$ and $y_w={\eta }_{w}x$,which is independent of $x$. This is a qualified correction of the reference point, not a proof of the false edge maximum.