= Solution
{c}
Let $\delta\ll L$ be a representative mixing-layer thickness, $\epsilon_m=\delta/L$, and let $U_s$ be a representative streamwise <velocity>. Use the scales
$$
x\sim L,\quad y\sim\delta,\quad U\sim U_s,\quad
V\sim\epsilon_m U_s,\quad P\sim\rho U_s^2.
$$
The transverse <velocity> scale follows from the mean <continuity equation>. For turbulent transverse transport to balance streamwise inertia,
$$
\frac{U_s^2}{L}\sim\frac{\nu_TU_s}{\delta^2},\qquad
\nu_T\sim\frac{U_s\delta^2}{L}.
$$
With this ordering, the streamwise eddy-stress <divergence> is
$$
\partial_x(2\nu_TU_x)+\partial_y[\nu_T(U_y+V_x)]
=\partial_y(\nu_TU_y)+O(\epsilon_m^2U_s^2/L).
$$
Thus transverse diffusion of streamwise <linear momentum> is retained, while streamwise diffusion and the $V_x$ contribution are smaller by $\epsilon_m^2$. Molecular <kinematic viscosity> is neglected where $\nu_T\gg\nu$.
In the transverse momentum equation, inertia and the retained eddy stresses have scale $\epsilon_m U_s^2/L$, whereas an order-one transverse <pressure> variation would give $P_y/\rho\sim U_s^2/(\epsilon_m L)$. Hence the leading dimensionless transverse balance is $P_y=0$. The resulting <boundary layer> equations for the <planar turbulent mixing layer> are
$$
\boxed{\begin{aligned}
U_x+V_y&=0,\\
UU_x+VU_y&=-\rho^{-1}P_x+\partial_y(\nu_TU_y),\\
0&=-\rho^{-1}P_y.
\end{aligned}}
$$
Here $\nu_T=\mu_T/\rho$. The effective <pressure> convention explained in part (a)(iii) is understood. Since the outer streams have uniform horizontal <velocities>, their inviscid momentum equations give $P_x=0$; choosing the far-field <gauge pressure> as zero gives $P=0$ throughout this approximation.
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