= Solution
{c}
The <product rule> and fluctuation <incompressibility> give the pointwise identity
$$
(u_i'u_j')_{,j}=u_j'u_{i,j}'+u_i'u_{j,j}'=u_j'u_{i,j}'.
$$
Averaging therefore gives
$$
\boxed{\overline{u_j'u_{i,j}'}=\overline{(u_i'u_j')_{,j}}
=\partial_j\overline{u_i'u_j'}.}
$$
The fluctuation term is consequently the <divergence> of a <Reynolds stress>. The <eddy viscosity> model represents this momentum transport through a coefficient multiplying the mean <strain-rate tensor>:
$$
-\rho\overline{u_i'u_j'}\simeq\mu_T(U_{i,j}+U_{j,i})
$$
for its shear or deviatoric part. For example, $-\rho\overline{u'v'}\simeq\mu_TU_y$ in a thin parallel <shear flow>. The <dynamic viscosity> $\mu_T$ is an effective flow-dependent transport coefficient, rather than a molecular material property. Large eddies carry <linear momentum> between regions of different mean <velocity>.
There is a necessary qualification to the printed tensor equation (4). Taking its <trace> literally gives $0=-\rho\overline{u_i'u_i'}$, which would exclude nonzero <turbulence>. A consistent <deviatoric eddy-viscosity closure> in $d$ fluctuating dimensions is
$$
-\rho\overline{u_i'u_j'}=\mu_T(U_{i,j}+U_{j,i})
-\frac{2\rho e_t}{d}\delta_{ij},\qquad
e_t=\frac12\overline{u_k'u_k'}.
$$
The isotropic term is absorbed into the effective <pressure> $P_{\mathrm{eff}}=P+2\rho e_t/d$. Here $d=2$ for a strictly two-dimensional fluctuation model. The subsequent equations use $P$ for this effective <pressure>, or equivalently use (4) as a shear-stress closure. This preserves the intended thin-layer model without asserting an impossible full covariance identity.
For turbulent motions of scale $l$ and <velocity> scale $u_l$, the estimate $\mu_T\sim\rho u_ll$ gives $\mu_T/\mu\sim u_ll/\nu$. It is much greater than one on energetic large scales and in the <inertial range> well above the <Kolmogorov length scale>. At the molecular dissipation scales it is no longer large, and molecular transport also dominates very near a smooth solid wall.
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