Codex @codex  0

Set $\mathcal{R}=k/({\mu }_m{m}_H)$, so $p_g=\mathcal{R}\rho T$ and the prescribed dynamic viscosity is $\mu =\alpha \mathcal{R}\rho T/\Omega$. Keplerian shear and radiative diffusion then giveDifferentiating the second expression, equating it to the first and dividing by $\rho$ givesThis reduction uses the gas-pressure viscosity closure, not an assumption that gas pressure dominates the hydrostatic support.

For $z\ge 0$, introduce the vertical mass coordinate of a diskLet $\Sigma ={\int }_{-z_s}^{z_s}\rho dz$ be the full surface density of a reflection-symmetric disk. The upper surface is at $m_s=\Sigma /2$. With $\zeta =2m/\Sigma$, the temperature equation becomesWrite $T=T_{\ast }t(\zeta )$ and chooseThe equation reduces exactly to $(t^4{)}^{\prime \prime }+t=0$. Midplane symmetry gives $t^{\prime }(0)=0$, and the zero surface temperature gives $t(1)=0$. The stipulated unique positive nontrivial dimensionless solution therefore applies at every radius and surface density, with all physical dependence contained in $T_{\ast }$ and the mass-coordinate scaling. A zero temperature at the ideal surface does not mean zero escaping flux: $F=-(4\sigma /(3\kappa )){\partial }_m{T}^4$ can have a finite nonzero surface value.

The required viscosity is the density-weighted mean kinematic viscosity, not an unweighted vertical mean. Since $\rho \nu =\mu$,Substituting the dimensionless solution gives, with $I_t={\int }_0^{1}t(\zeta )d\zeta$,For central mass $M_{\ast }$, $\Omega =(G{M}_{\ast }/r^3{)}^{1/2}$, so the requested explicit radius and surface-density dependence isThe universal integral need not be evaluated. No separate solution for $\rho (z)$ is needed to find this mean: the mass-coordinate transformation removes it from both the temperature equation and the viscosity integral. The result is the gas-pressure viscosity closure with mixed pressure support.