= Solution
Assume an axisymmetric <thin disc> rotating in the fixed potential of a dominant central mass, with $\Omega=(GM_*)^{1/2}r^{-3/2}$ independent of time and height. Neglect vertical mass loss and vertical <angular-momentum flux> at the two faces, as well as <self-gravity> and radial <pressure> corrections to the rotation law. Define the <surface density> and density-weighted <kinematic viscosity> by
$$
\Sigma=\int\rho\,dz,\qquad \bar\nu=\frac1\Sigma\int\rho\nu\,dz,
$$
and let $v_r=\Sigma^{-1}\int\rho u_r\,dz$. These assumptions give the <vertically averaged viscous disk equations>
$$
\partial_t\Sigma+\frac1r\partial_r(r\Sigma v_r)=0,\qquad
\partial_t(\Sigma l)+\frac1r\partial_r(r\Sigma v_rl-r^3\bar\nu\Sigma\Omega')=0,
$$
where $l=r^2\Omega$ is <specific angular momentum>. Subtract $l$ times <conservation of mass> from <conservation of angular momentum>. Since $l$ is fixed in time,
$$
r\Sigma v_r l'=\partial_r(r^3\bar\nu\Sigma\Omega'),\qquad
v_r=-\frac3{\Sigma r^{1/2}}\partial_r(r^{1/2}\bar\nu\Sigma).
$$
Substitution into <conservation of mass> proves the <Keplerian viscous diffusion equation>
$$
\boxed{\partial_t\Sigma=\frac3r\partial_r\left[r^{1/2}\partial_r(r^{1/2}\bar\nu\Sigma)\right].}
$$
No assumption of height-independent <kinematic viscosity> is needed; its density-weighted average is the one appearing in the integrated stress. A wind or surface magnetic stress would add terms and must not be silently discarded.
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