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Take the positive constant ratio $\lambda =p_r/p_g$. The equation of state impliesThis is a polytropic equation of state of index three. Differentiating the pressure and using vertical hydrostatic equilibrium gives $4(1+\lambda )\mathcal{R}\rho T_z=-\rho {\Omega }^{2}z$. HenceDefine the semi-thickness $H$ by $T(H)=0$ and put $s=1-z^2/H^2$. Then the constant gas-radiation ratio vertical disc structure iswithThe surface density is the integral across both faces. The supplied polynomial integral gives ${\int }_0^1(1-x^2{)}^{3}dx=16/35$, so $\Sigma =(32/35){\rho }_{0}H$. Substitution yieldsAll coefficients here are local in radial position; the derivation fixes the vertical profile rather than a global accretion rate. The zero-temperature surface is a formal boundary of this bulk model; optically thick diffusion ultimately fails in a thin photospheric layer.

The first relation is vertical hydrostatic equilibrium. In a thin disc, the central object's gravity near the midplane is $g_z=-{\Omega }^{2}z$; 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 $rd\Omega /dr=-3\Omega /2$, so its heating rate is $\mu (r{\Omega }^{\prime }{)}^2=9\mu {\Omega }^2/4$. 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. The four relations express vertical force balance, gas–radiation thermodynamics, local heating, and diffusive cooling.

Let primes denote radial derivatives of the equilibrium and put $\Delta u=u^{\prime }+2u/r$. The linearized continuity, adiabatic pressure and Poisson equation areTake the time derivative of the Poisson equation and substitute continuity. Integration in radius givesCentre regularity excludes a changing point mass at the origin, so $C(t)=0$ andThe perturbed self-gravity must be retained; this calculation does not make the Cowling approximation.

The radial linear momentum equation is $\rho {\partial }_{t}u=-{\partial }_r\delta p-\delta \rho {\Phi }^{\prime }-\rho {\partial }_r\delta \Phi$. Differentiate it in time and use the preceding equations:The equilibrium has $p^{\prime }=-\rho {\Phi }^{\prime }$ and ${\Phi }^{\prime \prime }+2{\Phi }^{\prime }/r=4\pi G\rho$. The remaining terms simplify asConsequently the radial stellar oscillation equation isThe derivative acts on the full variable coefficient $\gamma (r)p(r)$; discarding ${\gamma }^{\prime }$ would change the result.

For $u=r\xi (r)e^{-i\omega t}$, $\Delta u=(3\xi +r{\xi }^{\prime })e^{-i\omega t}$. Substitute and multiply by $-r^3$. Combining the two terms proportional to ${\xi }^{\prime }$ into a total derivative gives the radial stellar pulsation equation,

Set $P=\gamma p{r}^4$, $W=\rho r^4$ and $Q=r^3[(4-3\gamma )p{]}^{\prime }$. The Sturm-Liouville operator isFor regular physical perturbations, $\xi$ is finite at the centre and $P{\xi }^{\prime }\to 0$ at the free surface. For a nonzero-frequency mode the fluid displacement is $u/(-i\omega )$, so vanishing Lagrangian pressure perturbation is equivalent to $\gamma p(3\xi +r{\xi }^{\prime })\to 0$. With $p(R_s)=0$ and finite $\xi$, this gives $P{\xi }^{\prime }\to 0$. Zero-frequency modes use the same displacement boundary condition directly. Assume bounded $\gamma$, positive $\rho ,p,\gamma$ in the interior and the usual finite-energy endpoint domain. Integration by parts givesThus this physical self-adjoint differential operator has real eigenvalues ${\omega }^2$. The endpoint domain is essential: the fact that $p$ vanishes does not by itself allow arbitrary singular trial functions. The regular free-surface realization of the Sturm-Liouville problem is the one used here.

The Rayleigh-Ritz variational principle gives the fundamental squared frequency as the infimum of the weighted stellar pulsation Rayleigh quotient,The infimum is over admissible finite-energy functions satisfying the physical endpoint conditions. If this quotient is nonnegative for every such function, all radial frequencies are real and there is no exponentially growing radial mode. A negative value for even one trial function proves a negative eigenvalue and an exponentially growing solution, because ${\omega }^2<0$. A zero lowest value is marginal and needs separate treatment of neutral motion.

Choose the homologous trial function $\xi =1$, which corresponds to radial velocity proportional to $r$. Its gradient contribution is zero, andThe denominator $\int \rho r^{4}dr$ is positive. HenceThis pressure-weighted radial instability criterion is sufficient, not necessary: another trial function can detect instability even if this one does not. For a constant stellar adiabatic exponent it recovers instability below $4/3$. At constant $\gamma =4/3$, the homologous mode is neutral. For constant $\gamma >4/3$, $Q=(4-3\gamma )r^3{p}^{\prime }>0$ in a normally stratified hydrostatic equilibrium, so the quotient is positive and the star is radially stable.

A nonspherical configuration requires the vector form of hydrostatic equilibrium and the Poisson equation, rather than the spherical mass-coordinate equations:For the same index-one polytrope, hydrostatic balance inside the positive-density region gives $\Phi =C-2K\rho$. Thus its interior mass density must obey the Helmholtz equationA positive separated solution with the required boundary values isIts maximum is ${\rho }_c$ at the centre of the cube and it vanishes on every face. Its Laplacian is $-3{\pi }^2\rho /L^2$, so it satisfies the interior equations whenTogether with $P=K{\rho }^2$ and $\Phi =C-2K\rho$, this explicitly constructs the cubic polytropic interior.

The mass integral separates into three elementary sine integrals:ThereforeThis construction satisfies the equations inside the prescribed cube. It does not by itself establish the existence of an isolated self-gravitating star: the interior potential must also match the exterior vacuum field generated by that very mass density. That additional physical requirement is addressed in part (c).