= Solution
Let $L$ be the streamwise evolution scale, $\delta$ the jet width, and $U_c$ its velocity scale. A slender jet has $\varepsilon=\delta/L\ll1$; mean <incompressibility> gives $V=O(\varepsilon U_c)$. Both mean advection terms are then $O(U_c^2/L)$. For turbulent parcels crossing a transverse distance $l$, retention of their original streamwise momentum gives a fluctuation $\hat u\sim-lU_y$; a comparable transverse fluctuating speed gives the down-gradient <mixing length> closure
$$
\overline{\hat u\hat v}=-l^2|U_y|U_y=-\nu_TU_y,
\qquad\nu_T=l^2|U_y|.
$$
The <mixing length> is an effective distance of momentum retention, with statistical coefficients absorbed into $l$; it is not the net transverse displacement of a mean streamline. This closure models the <Reynolds stress> rather than deriving it exactly.
In a slender equilibrium jet, the fluctuation magnitude inferred from momentum balance is $u'^2\sim U_c^2\delta/L=\varepsilon U_c^2$. The transverse <Reynolds stress> gradient then has size $u'^2/\delta\sim U_c^2/L$, whereas the streamwise normal-stress gradient is $u'^2/L\sim\varepsilon U_c^2/L$. At sufficiently high <Reynolds number>, molecular transverse diffusion is smaller provided
$$
\frac{\nu U_c/\delta^2}{U_c^2/L}=\frac1{\mathrm{Re}_L\varepsilon^2}\ll1,
\qquad\mathrm{Re}_L=\frac{U_cL}\nu.
$$
Streamwise molecular diffusion is smaller still by $\varepsilon^2$. A free jet has ambient mean pressure constant to leading order; transverse momentum corrections imply only a higher-order mean streamwise pressure force. Thus high speed must be accompanied by the slender-jet ordering, not used alone to discard every other term.
For a symmetric jet decreasing away from the centre, $U_y<0$ on $y>0$ and $U_y>0$ on $y<0$. Accordingly $\overline{\hat u\hat v}=\operatorname{sgn}(y)l^2(U_y)^2$ off the centreline, and the leading equation is
$$
\boxed{UU_x+VU_y=-\partial_y\overline{\hat u\hat v}
=-\operatorname{sgn}(y)\,\partial_y\{l^2(U_y)^2\}.}
$$
At the centreline the signed form requires the stress to tend continuously to zero; otherwise differentiation would add a distributional term. The undifferentiated <Reynolds stress> form is the appropriate global equation.
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