= Solution
{c}
Assume $U_c=(U_a+U_b)/2>0$ and set $\Delta U=U_a-U_b>0$. With $\nu_T=C\Delta U\delta(x)$ independent of $y$, balancing streamwise transport with transverse mixing gives a layer width proportional to $x$. Write its constant slope as
$$
\boxed{\delta'=\frac{S\Delta U}{U_c},\qquad
\delta(x)=\frac{S\Delta U}{U_c}x.}
$$
Here $S>0$ is a spreading constant associated with the chosen definition of thickness; it is unrelated to the capillary $S$ in Question 1. The origin $x=0$ selects zero virtual-origin offset.
Define $a=\sqrt{S/C}$, $\xi=ay/\delta$, and introduce the <stream function>
$$
\psi=U_c\left[y+\frac\delta a F(\xi)\right],\qquad
U=\psi_y,\quad V=-\psi_x.
$$
This automatically satisfies <incompressibility> and yields
$$
\boxed{U=U_c[1+F'(\xi)],\qquad
V=\Delta U\sqrt{CS}\,[\xi F'(\xi)-F(\xi)].}
$$
For the momentum equation, use $\xi_x=-\xi\delta'/\delta$ and $\xi_y=a/\delta$:
$$
\begin{aligned}
U_x&=-U_c\frac{\delta'}\delta\xi F'',&
U_y&=U_c\frac a\delta F'',\\
UU_x+VU_y&=-\frac{U_c^2\delta'}\delta(\xi+F)F'',&
\nu_TU_{yy}&=\frac{S\Delta U\,U_c}\delta F'''.
\end{aligned}
$$
Because $U_c^2\delta'=S\Delta U\,U_c$, the common coefficient cancels, giving the <turbulent mixing-layer similarity equation>
$$
\boxed{F'''+(\xi+F)F''=0.}
$$
The <boundary layer> approximation also requires $S\Delta U/U_c\ll1$, consistently with the assumed slow spreading.
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