= Solution
Write $u_i=\langle v_i\rangle$ and use summation over repeated Cartesian indices. A mass-normalized <galactic distribution function> obeys the <Collisionless Boltzmann equation>
$$
\partial_tf+v_i\partial_{x_i}f-(\partial_{x_i}\phi)\partial_{v_i}f=0.
$$
Assume $f$ decays in <velocity> sufficiently fast that the necessary boundary terms vanish. Its zeroth velocity moment gives the <continuity equation>
$$
\partial_t\rho+\partial_{x_i}(\rho u_i)=0.
$$
For the first velocity moment, multiply by $v_j$ and integrate. In the gravitational term, <integration by parts> gives $\int v_j\partial_{v_i}f\,d^3v=-\delta_{ij}\rho$. Therefore
$$
\partial_t(\rho u_j)+\partial_{x_i}\bigl(\rho\langle v_iv_j\rangle\bigr)=-\rho\partial_{x_j}\phi.
$$
The <velocity-dispersion tensor> is the centered second moment, so $\langle v_iv_j\rangle=u_iu_j+\sigma_{ij}^2$. Expanding the streaming terms, the factor multiplying $u_j$ cancels by the <continuity equation>:
$$
\partial_t(\rho u_j)+\partial_{x_i}(\rho u_iu_j)
=\rho\partial_tu_j+\rho u_i\partial_{x_i}u_j.
$$
Hence the <Jeans equation> is
$$
\boxed{\rho\bigl(\partial_tu_j+u_i\partial_{x_i}u_j\bigr)
=-\rho\partial_{x_j}\phi-\partial_{x_i}(\rho\sigma_{ij}^2).}
$$
Its fluid analogue is the momentum equation with pressure tensor $P_{ij}=\rho\sigma_{ij}^2$. For isotropic random motion, $P_{ij}=P\delta_{ij}$, and this becomes the <Euler equations for an inviscid fluid>:
$$
\boxed{\rho\left(\partial_t\boldsymbol u+(\boldsymbol u\cdot\nabla)\boldsymbol u\right)
=-\rho\nabla\phi-\nabla P.}
$$
The tensor form allows anisotropic stellar stresses. In a fluid an <equation of state> can provide a pressure closure; the <Jeans equation> alone does not determine the <velocity-dispersion tensor> of a <collisionless stellar system>.
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