 Solution

Let the Eulerian time average commute with space and time derivatives and let the statistically steady mean have no time derivative. Define $U=\overline{u}$, $V=\overline{v}$, $\hat{u}=u-U$ and $\hat{v}=v-V$, so $\overline{\hat{u}}=\overline{\hat{v}}=0$. Averaging instantaneous incompressibility givesand subtracting gives ${\hat{u}}_x+{\hat{v}}_y=0$ for the fluctuations. Start from the instantaneous streamwise Navier-Stokes equationUse incompressibility to write $u{u}_x+v{u}_y={\partial }_x(u^2)+{\partial }_y(uv)$. The averaged products are $\overline{u^2}=U^2+\overline{{\hat{u}}^2}$ and $\overline{uv}=UV+\overline{\hat{u}\hat{v}}$. The mean conservative terms reduce back to $U{U}_x+V{U}_y$ because $U_x+V_y=0$. The Reynolds-averaged momentum equation is consequentlyThe signs use the positive covariance convention for the Reynolds stress. In particular, both quadratic averages involve the fluctuations, not products of the mean $U$ with a fluctuation; the converted TeX loses the overbar/hat structure here.