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Use the fluid displacement $\mathbf{\xi }$, with velocity perturbation $\delta \mathbf{u}={\partial }_t\mathbf{\xi }$. Denote Eulerian perturbations by $\delta$ and the corresponding Lagrangian pressure perturbation by ${\Delta }_{L}p=\delta p+\mathbf{\xi }\cdot \nabla p$. For adiabatic perturbations, the linearized mass conservation, momentum, adiabatic equation of state and Poisson equation areThese self-gravitating adiabatic displacement equations retain the perturbation of the star's own Newtonian gravitational potential. In the uniform-density interior, $\delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, $\nabla \Phi ={\omega }_d^2\mathbf{r}$, and $\nabla p=-\rho {\omega }_d^2\mathbf{r}$. The dynamical frequency of a uniform-density star ${\omega }_d=(GM/R^3{)}^{1/2}$ sets the natural timescale. For a normal mode with time factor $e^{-i\omega t}$, replace ${\partial }_t^2$ by $-{\omega }^2$. The requested interior analysis needs no surface or exterior matching conditions.

The spatial factor $r^l{Y}_l^m$ is a regular solid harmonic. Besides the Laplace equation ${\nabla }^{2}F=0$, homogeneity gives $\mathbf{r}\cdot \nabla F=lF$. The divergence of the fluid displacement isDefine the scalar dilation amplitude $D_r=r{U}^{\prime }+(l+3)U+(l/r)V^{\prime }$; this is the quantity denoted $\Delta$ in the question, not the notation ${\Delta }_L$ for a Lagrangian pressure perturbation. Then mass conservation gives $\hat{\rho }=-\rho D_r$.

The equilibrium pressure gradient and homogeneity identity giveConsequently the adiabatic equation of state yieldsFor the force equations, the product rule givesEquating the coefficients of $F{\mathbf{e}}_r$ and $\nabla F$ in the self-gravitating adiabatic displacement equations givesFinally, applying the Laplacian to the gravitational perturbation givesso the Poisson equation becomesThese are the required interior equations for uniform-density stellar oscillation.

There is a radial degeneracy at $l=0$: $F$ is spatially constant and $\nabla F=0$, so the displacement is independent of $V$. The second force equation then cannot be inferred by equating independent vectors. For nonzero $\omega$ it may be imposed as an auxiliary definition of $V$, but it is not an additional physical radial equation. At zero frequency the radial equations should be used directly. This distinction matters for interpreting the zero factor in (c).

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

The zero-temperature internal energy isUsing part (i),Consequently the adiabatic equation of state isThe pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys $pV=N{k}_{B}T$, so its corresponding volume exponent is one and its pressure vanishes as $T\to 0$.