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For the self-gravitating incompressible slab, let the fluid occupy $-H<z<H$, with mass density ${\rho }_0=\Sigma /(2H)$ and a vacuum exterior. Poisson equation and hydrostatic equilibrium giveA constant ambient pressure could be added without changing the result. Choose perturbations proportional to $e^{ikx-i\omega t}$, using rotational symmetry in the horizontal plane to put the wavevector along $x$.

For the fluid displacement, the linearized momentum equation inside the uniform slab is $-{\omega }^2\mathbf{\xi }=-\nabla (p^{\prime }/{\rho }_0+{\varphi }^{\prime })$. For $\omega \ne 0$ it follows that $\mathbf{\xi }=\nabla \chi$, with $p^{\prime }/{\rho }_0+{\varphi }^{\prime }={\omega }^2\chi$. Incompressibility gives $\nabla \cdot \mathbf{\xi }=0$, henceNeutral limits of these surface modes follow by continuity. Stationary vorticity perturbations form a separate zero-frequency sector; the nonzero-frequency surface modes are irrotational by the momentum equation itself, without an extra irrotational-flow assumption.

The Eulerian mass density perturbation vanishes in the bulk because the equilibrium mass density is uniform and the displacement is divergence free. If the two vertical surface displacements are ${\eta }_{+}={\xi }_z(H)$ and ${\eta }_{-}={\xi }_z(-H)$, the surface density perturbation of a displaced uniform interface givesThus Poisson equation reduces to Laplace equation for ${\varphi }^{\prime }$ inside and outside. The potential is continuous, while integrating through each surface givesThe perturbing potential must decay at vertical infinity. The kinematic conditions are ${\eta }_{\pm }={\partial }_z\chi (\pm H)$. The free surface pressure condition is the vanishing Lagrangian pressure perturbation, $p^{\prime }+{\eta }_{\pm }{p}_0^{\prime }=0$; at the top this gives $p^{\prime }(H)/{\rho }_0=g_s{\eta }_{+}$.

Put $q=kH>0$. For even scalar-potential symmetry, chooseThen ${\eta }_{-}=-{\eta }_{+}$: this is a thickness-changing mode, not a vertical bending mode. Write $\eta ={\eta }_{+}$; the top derivative jump gives $-kC(\cosh q+\sinh q)=4\pi G{\rho }_0\eta$. Since $\cosh q+\sinh q=e^q$,Substitute these values into ${\omega }^2\chi (H)=g_s\eta +{\varphi }^{\prime }(H)$. Using $4\pi G{\rho }_0=g_s/H$ gives the even-mode dispersion relation for the surface modes of a self-gravitating incompressible slab:The positive term comes from the restoring hydrostatic free-surface pressure in the background gravitational field. The negative term is the attraction due to the perturbed gravitational potential of the displaced surfaces, which tends to reinforce the disturbance. Both effects ultimately involve the slab's gravity, but enter the boundary condition through different terms.

For odd scalar-potential symmetry, instead takeNow ${\eta }_{-}={\eta }_{+}$, so the surfaces bend in the same direction. The jump again gives $C=-4\pi G{\rho }_0\eta /(k{e}^q)$, but $\chi (H)=\eta \tanh q/k$ and ${\varphi }^{\prime }(H)=-4\pi G{\rho }_0\eta /[k(1+\coth q)]$. The same pressure condition yieldsThe bottom kinematic, jump and pressure conditions follow with the indicated parity and give the same dispersion relation, so no boundary condition has been discarded.

To determine the gravitational instability of an incompressible slab, factor the two dimensionless expressions asFor the even mode the bracket is strictly increasing, with derivative $1+e^{-2q}>0$, is negative at zero and tends to positive infinity. It therefore has a unique positive zeroThe even mode is unstable for $0<kH<q_c$, neutral at $q_c$, and oscillatory for $kH>q_c$. Negative ${\omega }_e^2$ gives a growing mode with $\omega =i\sqrt{-{\omega }_e^2}$ under the stated time convention. For the odd mode, $1-e^{-2q}<2q$ for every $q>0$, so the bracket is always positive: all odd modes with positive wavenumber are stable.

At long wavelengths $F_e=-q+2q^2+O(q^3)$ and $F_o=q-2q^2/3+O(q^3)$: a thickness or column-density disturbance has a gravitational instability, whereas a bending disturbance is restored. At short wavelengths both approach $q-1/2$, so both surface branches are stable. The slab equilibrium is therefore linearly unstable overall, because it admits the long-wavelength even modes despite the stability of its odd sector.

Write $\mathcal{R}=k/(\mu m_H)$ for the specific gas constant, with the printed $m_H$ denoting the atomic mass unit, and $a_{\mathrm{r}}=4\sigma /c$ for the radiation constant. In the low-mass envelope the enclosed mass is approximately $M$ and the outward luminosity is approximately $L$. The stellar hydrostatic equation, stellar radiative diffusion, and ideal gas equation giveThe Kramers' opacity law becomes $\kappa =({\kappa }_0/\mathcal{R})P{T}^{-9/2}$. Dividing the two differential equations eliminates radius and mass density:Integrating from a surface matching point $(P_s,T_s)$ givesDeep compared with the photosphere, the surface terms are negligible. The radiative-zero white-dwarf envelope therefore hasThis is the leading envelope approximation used in the later parts, rather than an exact atmospheric boundary condition at $T_{\mathrm{eff}}$.

At the base, temperature, mass density and pressure match the isothermal degenerate core. Let $K_{\mathrm{NR}}=K_1{\mu }_e^{-5/3}$, where ${\mu }_e$ is the mean molecular weight per electron. Equating the gas and electron degeneracy pressure givesHence $P_b={\mu }_e^{5/2}{\mathcal{R}}^{5/2}{K}_1^{-3/2}{T}_c^{5/2}$. Substituting this into the envelope relation at $T=T_c$ yieldsThe matching of a white-dwarf envelope to a degenerate core thus gives $L/M\propto T_c^{7/2}$, the envelope relation underlying Mestel's cooling law. It depends on pressure as well as mass density continuity at the interface; otherwise the supplied core equation would not determine the matching mass density.

In the radiative-zero white-dwarf envelope, $dP/dT=(17/4)P/T=(17/4)\mathcal{R}\rho$. Substitute this derivative into the stellar hydrostatic equation:Integrating from the idealized zero-temperature surface at $R$ givesThis also gives $d\log T/d\log P=4/17$. A finite surface temperature adds $T_s$ to the integrated profile if the pure power-law pressure relation is retained locally. Near the actual photosphere, the integration constant in the pressure law also matters; the displayed expression is the radiative-zero limit implicit in the printed result and accurate in the deep envelope when $T_c\gg T_{\mathrm{eff}}$.

Let $S$ denote specific entropy, reserving $s$ from the first question for envelope thickness. The stellar adiabatic exponents are defined at fixed composition byIt is useful to introduce ${\Gamma }_3-1=(\partial \log T/\partial \log \rho {)}_S$, so $({\Gamma }_2-1)/{\Gamma }_2=({\Gamma }_3-1)/{\Gamma }_1$.

Take the stellar gas to be a fully ionized, nonrelativistic monatomic ideal gas, with fixed specific gas constant $\mathcal{R}$. Its mixture with thermal radiation hasHere $u$ is specific internal energy, and the stellar gas-pressure fraction is $\beta =\mathcal{R}\rho T/P$. The assumption of monatomic gas fixes its heat capacity; the perfect-gas pressure law alone would not determine it.

At fixed $T$ and fixed $\rho$, respectively, the pressure derivatives areFor an isentropic process, the first law of thermodynamics gives $du=Pd\rho /{\rho }^2$. Differentiate the internal energy rather than artificially holding $\beta$ fixed during the perturbation:The parenthesis is $(P/\rho T)[(3/2)\beta +12(1-\beta )]$, givingSince $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$,Using the relation between the exponents gives the adiabatic exponents of a monatomic gas-radiation mixtureThe requested values, including the adiabatic temperature gradient, areThe pure-radiation and pure-gas rows are understood as limits of the mixture.

For uniform composition, the Schwarzschild criterion for stellar convective instability isTo test whether a radiative configuration becomes unstable, use its required stellar radiative temperature gradient ${\nabla }_{\mathrm{rad}}$ for $\nabla$. Since $dP/dr<0$, the equivalent radial condition isAn outward-displaced parcel then cools less than its new surroundings, remains less dense at the same pressure, and is further accelerated outward.

The requested Eddington-model convective-core mass fraction follows from the usual global constant-$\beta$ closure. At the outer radiative surface, put $m=M$, $L_r=L$, and $P_{\mathrm{rad}}=(1-\beta )P$. Dividing the radiation-pressure gradient by the stellar hydrostatic equation givesThere is no energy generation outside the convective core, so use $L_r=L$ there. The local radiative-gradient formula isThe same representative $\beta$ has been used in the global luminosity closure and at the boundary. Setting this gradient equal to ${\nabla }_{\mathrm{ad}}$ at $m=M_c$ producesThis is $1$, $77/80$, and $5/8$ for $\beta =0,1/2,1$, respectively, with $\beta =0$ a marginal radiation-dominated limit.

There is a consistency qualification to this last model estimate. It cannot be an exact stellar solution if all the printed assumptions are enforced pointwise. An exactly uniform nonzero radiation fraction would give $d{P}_{\mathrm{rad}}/dP=1-\beta$ at every radiative point. The exact equations would then requireIf $L_r$ is constant throughout an envelope with positive mass density, $m$ increases with radius, so this equality cannot hold there. Equivalently uniform $\beta$ fixes the actual gradient to $1/4$, whereas ${\nabla }_{\mathrm{ad}}>1/4$ for $\beta >0$. The boxed core fraction is the intended Eddington closure and boundary estimate, which relaxes exact constancy of $\beta$ in the detailed envelope. The constant-beta radiative-envelope obstruction identifies the missing approximation; it is not legitimate to silently assert exact compatibility.

Use the fully developed flow branch of Hartmann flow. Translation invariance along the walls makes the velocity and induced magnetic field functions of $z$ alone. Incompressible flow and zero normal velocity at the walls give $u_z^{\prime }=0$ and hence $u_z=0$. The zero divergence of the magnetic field makes $B_z$ constant, equal to the imposed $B_0$. There is no forcing in $y$; the homogeneous $y$-components obey the same coupled viscous-resistive equations as the $x$-components, with zero boundary data. Multiplying them by $u_y$ and $b_y/({\mu }_0\rho )$, integrating by parts and adding givesThus $u_y=b_y=0$, establishing the asserted forms on this fully developed flow branch. This is a symmetry reduction of the steady channel model, rather than a claim that every possible flow in a channel is translation invariant.

The current density and Lorentz force density areThe advective acceleration vanishes because $\mathbf{u}\cdot \nabla =u(z){\partial }_x$ acts on fields independent of $x$. The $x$-component of the magnetohydrodynamic momentum equation therefore givesThe $z$-component must also balance: it requires ${\partial }_z(p+b^2/(2{\mu }_0))=0$. Hence a compatible pressure is $p(x,z)=p_{\ast }-\rho Gx-b(z{)}^2/(2{\mu }_0)$; the specified streamwise pressure gradient does not require the ordinary pressure to be uniform in $z$. This accounts for the magnetic pressure of the induced field.

The resistive induction equation with constant magnetic diffusivity gives ${\partial }_t\mathbf{B}=\nabla \times (\mathbf{u}\times \mathbf{B})+\eta {\nabla }^2\mathbf{B}$. Since $\mathbf{u}\times \mathbf{B}=(0,-B_{0}u,0)$, its curl has $x$-component $B_0{u}^{\prime }$. Steadiness consequently givesFinally the no-slip boundary condition gives $u(\pm L)=0$, and the normal magnetic field boundary condition gives $b(\pm L)=0$. The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.

A direct elimination of pressure gives the same result as taking a double curl of the momentum equation. Incompressibility gives ${\nabla }^{2}p=\mathrm{Ra}\mathrm{Pr}{\partial }_z\theta$. Apply ${\nabla }^2$ to the vertical momentum equation and subtract this pressure derivative to obtainHere ${\nabla }_h^2={\partial }_x^2+{\partial }_y^2$ is the horizontal Laplacian. At $z=0,1$, impermeability gives $w=0$ and fixed temperature gives $\theta =0$. The stress-free boundary condition gives ${\partial }_{z}u={\partial }_{z}v=0$, since horizontal derivatives of the boundary value $w=0$ vanish. Differentiating incompressibility once in $z$ therefore givesFor a horizontal Fourier mode $X=e^{i(k_{x}x+k_{y}y)}$, ${\nabla }_h^{2}X=-{\lambda }^{2}X$ with real $\lambda =\sqrt{k_x^2+k_y^2}$. More generally the nonnegative spectrum of $-{\nabla }_h^2$ follows by integration by parts for periodic or square-integrable horizontal disturbances. Substituting the separated normal mode gives, with $D=d/dz$,The boundary conditions admit a complete Fourier sine series in $z$. For a nonzero horizontal wavenumber, take $W=W_n\sin (n\pi z)$, $\Theta ={\Theta }_n\sin (n\pi z)$, and set $a_n^2={\lambda }^2+n^2{\pi }^2$. The two amplitude equations areThus the stress-free convection growth-rate polynomial isIts discriminant is $(\mathrm{Pr}-1{)}^2{a}_n^4+4\mathrm{Ra}\mathrm{Pr}{\lambda }^2/a_n^2$, so both growth rates are real for $\mathrm{Ra}>0$. For $\mathrm{Ra}<0$, its two coefficients after the leading term are positive: their sum gives ${\sigma }_1+{\sigma }_2=-(1+\mathrm{Pr})a_n^2<0$, and their product is $\mathrm{Pr}(a_n^4-\mathrm{Ra}{\lambda }^2/a_n^2)>0$. Real roots are both negative; nonreal roots have real part $-(1+\mathrm{Pr})a_n^2/2<0$. Hence stable heating from above damps every mode. Horizontally uniform thermal modes and uncoupled horizontal viscous modes are also purely damped; impermeability and incompressibility exclude a horizontally uniform nonzero $w$.

Because the unstable-side roots are real, the onset is stationary, an exchange of stabilities in stress-free convection. Setting $\sigma =0$ gives the free-slip convection neutral curveWith $x={\lambda }^2$, its derivative has the sign of $2x-n^2{\pi }^2$. Its minimum is at $x=n^2{\pi }^2/2$, with value $27n^4{\pi }^4/4$. Minimizing also over $n\ge 1$ gives the critical values

Use the Boussinesq approximation with a constant inertial reference mass density ${\rho }_0$, keeping the background mass density gradient in buoyancy. Write $V(z)=\overline{U}(z)-c$ and $P=\hat{p}/{\rho }_0$. For $k\ne 0$, the linearized normal mode equations areThe reference mass density is chosen close to the background values, whose relative variation must be small. Hydrostatic background pressure satisfies ${\overline{p}}^{\prime }=-g\overline{\rho }$.

From incompressibility, $\hat{u}=i{\hat{w}}^{\prime }/k$. The horizontal momentum equation and the density equation then giveSubstitute these into the vertical momentum equation and multiply by $ik$:For $V\ne 0$, rearrangement gives the Taylor–Goldstein equationThe buoyancy frequency $N$ is real for stable stratification, where ${\overline{\rho }}^{\prime }<0$. Points where $\overline{U}=c$ are critical levels of an internal gravity wave; the derivation there is understood through a limiting or piecewise formulation. The printed density gradient is ${\overline{\rho }}^{\prime }$, not the pressure gradient accidentally substituted in the local TeX.

Choose the positive $x$ direction along the background flow, so $U>0$; reverse $x$ if necessary. Let $z$ measure depth below the impermeable roof, and set $\Delta \rho ={\rho }_w-{\rho }_c>0$. The deep ambient water has $p_w=p_0(x)+{\rho }_{w}gz$. Continuity of pressure at $z=h$ and hydrostatic pressure inside the buoyant CO2 giveHere $k$ is permeability of a porous medium and $\mu$ is dynamic viscosity. Since $h/H\ll 1$, the ambient return-flow correction to the imposed pressure gradient is higher order. The small aspect ratio supplies the hydrostatic pressure approximation; capillary trapping, dissolution, compressibility and dispersion are omitted in this sharp-interface model.

The depth-integrated Darcy flux is $q=Uh-K_{b}h{h}_x$. The actual stored fluid volume is $\varphi h$ per unit horizontal area, where $\varphi$ is the aquifer porosity. Thus the porous gravity current with background flow satisfieswith $u=U/\varphi$, $a=K_b/\varphi$ and $J=Q/\varphi$. Both $Q$ and the later $V$ denote actual injected volume per unit transverse width. Away from injection, the equation is an advected porous medium equation. At each finite front impose $h=0$ and vanishing outgoing volume flux per unit width. At the injection point $h$ is continuous and $q(0^{+})-q(0^{-})=Q$. For an initially empty aquifer, $\varphi \int hdx=Qt$. The advective speed is $U/\varphi$, not $U$.

For a constant-flux porous gravity current, write characteristic depth $H_c$ and extent $L_c$ temporarily. At early times volume conservation gives $H_c{L}_c\sim Jt$; gravity spreading gives $L_c^2/t\sim a{H}_c$. HenceBackground advection becomes comparable when $L_c\sim ut$, so the transition scales areFor $t\ll t_c$, the leading similarity solution is symmetric:For $0<\eta <{\eta }_N$, its model ordinary differential equation and boundary conditions areChoose the nonnegative finite-front branch, reflect evenly in $x$, and require $F{F}^{\prime }\to 0$ at the front. The factor $1/2$ divides the injection equally between the two halves. Integration of the ordinary differential equation gives $2{\int }_0^{{\eta }_N}Fd\eta =1$; the limiting front slope is $F^{\prime }({\eta }_N^{-})=-2{\eta }_N/3$. This boundary-value problem determines the initially symmetric shape without assuming an unjustified elementary formula. Independent numerical integration gives ${\eta }_N\simeq 1.17619$ in this normalization.

For $t\gg t_c$, upstream storage approaches a finite steady value. With no upstream leakage, $q=0$ for $x<0$, giving $h_x=U/K_b$ wherever $h>0$. Downstream the uniform interior carries $q=Q$, soThe upstream volume flux vanishes even though local Darcy velocity contributions from background flow and buoyancy cancel. The upstream stored volume is $\varphi h_{\mathrm{int}}|x_{-}|/2$, a finite constant. The downstream plateau therefore extends a distance $ut+O({\ell }_c)$, and its front advances at $u$ to leading order.

The diffusive nose of an advected porous gravity current has widthIn a frame translating at $u$, balance the time derivative against the nonlinear gravity diffusion. With $\eta =(x-x_c(t))/\sqrt{a{h}_{\mathrm{int}}t}$ and $x_c(t)=ut+O({\ell }_c)$, the leading transition is $h=h_{\mathrm{int}}F(\eta )$, withIts finite front has $F\sim ({\eta }_N/2)({\eta }_N-\eta )$ as $\eta \uparrow {\eta }_N$. Independent numerical integration of this boundary-value problem gives ${\eta }_N\simeq 1.23849$, so a more accurate leading nose position is $x_N\simeq x_c+1.23849\sqrt{a{h}_{\mathrm{int}}t}$ and its local shape is $F\simeq 0.619247(1.23849-\eta )$. The entire transition has no elementary closed form. A useful explicit, approximate shape is the volume-preserving linear rampThis gives $x_N\simeq x_c+\sqrt{a{h}_{\mathrm{int}}t}$. Its ramp width is $2\sqrt{a{h}_{\mathrm{int}}t}$, and its edge obeys the exact kinematic boundary condition ${\dot{x}}_N=u-a{h}_x(x_N^{-},t)$ to the displayed order. It also has the same area as a sharp step at $x_c$. This is an approximate ramp closure, not an exact solution of the nonlinear nose ordinary differential equation. A constant phase shift $x_c=ut-|x_{-}|/2$ balances the finite upstream storage within this ramp approximation; the full solution's subleading matching may alter constant offsets.

For a rapid injection of total volume $V$, remove the source for $t>0$ and introduce $\xi =x-ut$. The advected constant-volume porous gravity current obeys $h_t=a(h{h}_{\xi }{)}_{\xi }$ and $\int hd\xi =M=V/\varphi$. A similarity solution has depth proportional to $t^{-1/3}$ and radius proportional to $t^{1/3}$. Direct substitution gives the compactly supported Barenblatt solutionIndeed ${\int }_{-R}^R(R^2-{\xi }^2)/(6at)d\xi =2R^3/(9at)=V/\varphi$. Its centre translates at the pore speed $u$, its two edges are $ut\pm R(t)$, and its maximum depth decays as $t^{-1/3}$. The expression is exact for a point release within this reduced model; finite-width initial data approach this similarity solution at late times. If $U=0$, the constant-rate current remains in the symmetric spreading regime and the constant-volume solution has a stationary centre.

For constant-density incompressible inviscid flow with conservative body force, Euler equations for an inviscid fluid give ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla (p/\rho +\Phi )$. Using $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$, take the curl to obtain ${\partial }_t\mathbf{\omega }=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. Since both divergences vanish, the vector identity expands this toWithout incompressibility or the constant-density/barotropic hypothesis, extra compression or baroclinic terms are present; the printed equation uses these flow assumptions.

Let $\kappa (t)=\dot{h}/h$, with $h>0$. Axisymmetric mass conservation is $r^{-1}{\partial }_r(r{u}_r)+{\partial }_z{u}_z=0$. If $u_r$ is independent of $z$, then $u_z$ is affine in $z$. The impermeable moving-plate conditions $u_z(r,\pm h,t)=\pm \dot{h}$ force $u_z=\kappa z$, and integrating the radial equation gives $u_r=-\kappa r/2+C(t)/r$. Regularity at the axis excludes $C(t)/r$. The meridional part of the axisymmetric inviscid flow between moving parallel plates is thereforeFor no swirl, $u_{\theta }=0$ completes the field. Continuity alone does not determine azimuthal motion; this must be specified separately. More generally, prescribe a smooth regular initial profile $U(r)$ independent of $z$. The azimuthal part of Euler equations for an inviscid fluid isso conservation of $r{u}_{\theta }$ along radial particle paths gives $u_{\theta }(r,t)=\lambda U(\lambda r)$, where $\lambda =\sqrt{h/h_0}$. Together with the meridional components above, this gives a family of complete fields, with a suitable pressure determined by the radial and vertical equations. The field is therefore not unique without rotational data.

With the stated solid-body rotation, $u_{\theta }=\Omega (t)r$ and $\mathbf{\omega }=2\Omega {\mathbf{e}}_z$. The vorticity equation gives $2\dot{\Omega }=2\Omega {\partial }_z{u}_z=2\Omega \kappa$, henceFor a fluid particle, $\dot{r}=-\kappa r/2$, so $r^{2}h$ is constant and $r^2\Omega$ is constant. This is conservation of angular momentum about the axis: radial spreading when the plates approach decreases the angular velocity. The meridional and rotating fields satisfy the inviscid equations with an appropriate pressure; no tangential no-slip condition is imposed on inviscid plates.

The prescribed cosmic string network scaling gives $\dot{\rho }=-2H\rho$. Comparing with the cosmological perfect-fluid continuity equation givesThe averaged pressure is therefore $P=-\rho /3$, provided the network is treated as the stipulated isotropic homogeneous component. The Friedmann acceleration equation gives $\ddot{a}=0$. On an expanding branch let $\rho =C/a^2$ and $B^2=8\pi GC/3-K>0$; thenThis coasting cosmic-string universe has no growing flatness instability, but it does not drive an initially nonflat universe toward flatness. The comoving Hubble radius remains constant, so there is also no shrinking-radius mechanism of the usual cosmic inflation type.

The global causal issue is more precise than saying that a constant comoving Hubble radius solves every horizon problem. If this ideal coasting solution extends to its zero-size time $t_B$, then ${\int }_{t_B}^{t}d{t}^{\prime }/[B(t^{\prime }-t_B)]$ diverges logarithmically, so it has no finite particle horizon. If string domination starts at a finite earlier time after another cosmological phase, its causal history must instead be evaluated from that earlier phase. The string era alone does not establish that previously disconnected regions had time to equilibrate. A static branch with $B=0$ is a separate possibility; its Hubble-normalized $\Omega$ is undefined and is not covered by the expanding formulas.

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