= Solution
The first relation is vertical <hydrostatic equilibrium>. In a <thin disc>, the central object's gravity near the midplane is $g_z=-\Omega^2z$; the <pressure> <gradient> balances it. The second is an <equation of state> combining <ideal gas> <pressure> with isotropic <radiation pressure>. Write $\mathcal R=k/(\mu_m m_p)$, so $p_g=\mathcal R\rho T$ and $p_r=4\sigma T^4/(3c)$; $\sigma$ is the <Stefan-Boltzmann constant> and $\mu_m$ the <mean molecular weight> in proton-mass units.
The third is local energy balance: the increase of upward <radiative flux> with height equals viscous dissipation. Here $\mu$ is <dynamic viscosity>, not mean molecular weight. A <Keplerian disc> has shear $r\,d\Omega/dr=-3\Omega/2$, so its heating rate is $\mu(r\Omega')^2=9\mu\Omega^2/4$. The fourth is optically thick <radiative diffusion>, with <opacity> per unit mass $\kappa$ and diffusion directed down the <temperature> <gradient>. Midplane symmetry gives $F(0)=0$ and oppositely directed fluxes from the two faces. \b[The four relations express vertical force balance, gas–radiation thermodynamics, local heating, and diffusive cooling.]
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