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Let $\mathcal{R}=k/({\mu }_m{m}_H)$, take constant opacity $\kappa$, and prescribe the dynamic viscosity $\mu =\alpha p_g/\Omega =\alpha \mathcal{R}\rho T/\Omega$ in a Keplerian disk. The pressure supporting the disk may contain both gas pressure and radiation pressure. Combining viscous dissipation and radiative diffusion gives ${\partial }_m^2{T}^4+A\Omega T=0$ in the vertical mass coordinate of a disk, where $A=27\alpha \kappa \mathcal{R}/(16\sigma )$. Set $\zeta =2m/\Sigma$, $T=T_{\ast }t(\zeta )$ and $T_{\ast }^3=A\Omega {\Sigma }^2/4$. With a radiative-zero disk surface, $t$ satisfies $(t^4{)}^{\prime \prime }+t=0$, $t^{\prime }(0)=t(1)=0$. Writing $I_t={\int }_0^{1}td\zeta$, the density-weighted mean kinematic viscosity is $\bar{\nu }=\alpha \mathcal{R}{T}_{\ast }{I}_t/\Omega$, giving the displayed formula. This calculation does not assume gas-pressure domination: the closure depends on gas pressure while hydrostatic support can be mixed.

Work at one radius and write $\mathcal{R}=k/({\mu }_m{m}_p)$. Use the surface density to fix the mass density normalization and introduce the height scaleChoose the dimensionless variablesThe pressure and temperature scales make the hydrostatic and ideal gas equations parameter-free. The flux scale fixes the heating coefficient at $9/4$, and the choice of $h$ makes the diffusion coefficient unity. Direct substitution gives the universal ordinary differential equationsFor example, the dimensionless coefficient in the temperature equation is $3\alpha \kappa {\Sigma }^2{\mathcal{R}}^4/(16\sigma {\Omega }^5{h}^6)=1$.

Let the upper free surface be at $\zeta ={\zeta }_s$, so its physical semithickness is $H=h{\zeta }_s$. Midplane symmetry and the radiative-zero disk surface conditions giveThe last condition is precisely $\Sigma ={\int }_{-H}^H\rho dz$. The unknown surface position and the two positive midplane values $P(0),\Theta (0)$ are determined as part of the normalized boundary value problem. Symmetry makes $P,D,\Theta$ even and $Q$ odd; $P^{\prime }(0)={\Theta }^{\prime }(0)=0$ already follow from the equations and are not additional independent boundary conditions.

The zero conditions do not mean zero surface flux. Integrating the heating equation gives $Q({\zeta }_s)=(9/4){\int }_0^{{\zeta }_s}Pd\zeta >0$. Near a regular surface with finite outgoing flux, dividing the hydrostatic equation by the temperature equation yields $dP/d\Theta \simeq {\zeta }_s{\Theta }^3/Q_s$. Thus $P\propto {\Theta }^4$, $D\propto {\Theta }^3$ and $\Theta \propto {\zeta }_s-\zeta$, so mass density also vanishes. The apparent singular quotient in the diffusion equation has a finite regular limit. All boundary conditions and equations are independent of the physical parameters: this is homologous vertical structure of a constant-opacity alpha disk.