= Solution
Define mean streaming <velocity> $u_i=\overline{v_i}$ and the <velocity dispersion> tensor $\sigma_{ij}^2=\overline{(v_i-u_i)(v_j-u_j)}=\overline{v_iv_j}-u_iu_j$. The zeroth <velocity> moment of the steady <Collisionless Boltzmann equation> gives the <continuity equation> $\sum_i\partial_i(nu_i)=0$. Therefore
$$
\sum_i\partial_i(nu_iu_j)=n\sum_i u_i\partial_i u_j.
$$
Substituting in <5/a> yields the standard steady <Jeans equations>:
$$
\boxed{\sum_i u_i\frac{\partial u_j}{\partial x_i}=-\frac{\partial\Phi}{\partial x_j}-\frac1n\sum_i\frac{\partial(n\sigma_{ij}^2)}{\partial x_i},\qquad j=1,2,3.}
$$
The unsteady version adds $\partial_tu_j$ on the left. Defining the stellar <pressure tensor> $P_{ij}=\rho\sigma_{ij}^2$ makes the analogy with fluid momentum balance explicit. The off-diagonal terms are correlations between random <velocity> components; they need not vanish or equal the diagonal terms.
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