 Solution

Let $L$ be the streamwise evolution scale, $\delta$ the jet width, and $U_c$ its velocity scale. A slender jet has $\epsilon =\delta /L\ll 1$; mean incompressibility gives $V=O(\epsilon U_c)$. 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 $\hat{u}\sim -l{U}_y$; a comparable transverse fluctuating speed gives the down-gradient mixing length closureThe 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 $u^{\prime 2}\sim U_c^2\delta /L=\epsilon U_c^2$. The transverse Reynolds stress gradient then has size $u^{\prime 2}/\delta \sim U_c^2/L$, whereas the streamwise normal-stress gradient is $u^{\prime 2}/L\sim \epsilon U_c^2/L$. At sufficiently high Reynolds number, molecular transverse diffusion is smaller providedStreamwise molecular diffusion is smaller still by ${\epsilon }^2$. 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 $\overline{\hat{u}\hat{v}}=\mathrm{sgn}(y)l^2(U_y{)}^2$ off the centreline, and the leading equation isAt 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.