SSC/Structure/OtherAnalyticModels

Other Analytically Definable, Spherical Equilibrium Models

Linear Density Distribution

Structure

In an article titled, "A Survey with Analytic Models," 📚 R. F. Stein (1966, in Stellar Evolution, Proceedings of an International Conference held at the Goddard Space Flight Center, Greenbelt, MD, U.S.A., edited by R. F. Stein & A. G. W. Cameron, pp. 1-105) defines the "Linear Stellar Model" as a star whose density "varies linearly from the center to the surface," that is (see his equation 3.1),

$\rho (r)={\rho }_c(1-\frac{r}{R}),$

where, ${\rho }_c$ is the central density and, $R$ is the radius of the star. Both the mass distribution and the pressure distribution can be obtained analytically from this specified density distribution. Specifically, following our general solution strategy for determining the equilibrium structure of spherically symmetric, self-gravitating configurations,

$M_r(r)$	$=$	$\underset{0}{\overset{r}{\int }}4\pi r^2\rho (r)dr$
	$=$	$\frac{4\pi {\rho }_c{r}^3}{3}[1-\frac{3}{4}(\frac{r}{R}\left)\right],$

in which case we have,

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\equiv M_r(R)=\frac{\pi {\rho }_c{R}^3}{3},$

and we can write,

$g_0(r)\equiv \frac{G{M}_r(r)}{r^2}$

$=$

$\frac{4\pi G{\rho }_{c}r}{3}[1-\frac{3}{4}(\frac{r}{R}\left)\right].$

Hence, proceeding via what we have labeled as "Technique 1", and enforcing the surface boundary condition, $P(R)=0$, 📚 Stein (1966) determines that (see his equation 3.5),

$P(r)$	$=$	$-\underset{0}{\overset{r}{\int }}{g}_0(r)\rho (r)dr$
	$=$	$\frac{\pi G{\rho }_c^2{R}^2}{36}[5-24(\frac{r}{R}{)}^2+28(\frac{r}{R}{)}^3-9(\frac{r}{R}{)}^4],$

where, it can readily be deduced, as well, that the central pressure is,

$P_c=\frac{5\pi }{36}G{\rho }_c^2{R}^2.$

Stability

Lagrangian Approach

As has been derived in an accompanying discussion, the second-order ODE that defines the relevant Eigenvalue problem is,

$\left(\frac{P_0}{P_c}\right)\frac{d^{2}x}{d{\chi }_0^2}+\left[\right(\frac{P_0}{P_c})\frac{4}{{\chi }_0}-(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\left)\right]\frac{dx}{d{\chi }_0}+\left(\frac{{\rho }_0}{{\rho }_c}\right)\left(\frac{1}{{\gamma }_{\mathrm{g}}}\right)[{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2{\omega }^2+(4-3{\gamma }_{\mathrm{g}})(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\left)\frac{1}{{\chi }_0}\right]x=0.$

where the dimensionless radius,

${\chi }_0\equiv \frac{r_0}{R},$

$g_{\mathrm{S}\mathrm{S}\mathrm{C}}\equiv \frac{P_c}{R{\rho }_c}$           and           ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}\equiv (\frac{R^2{\rho }_c}{P_c}{)}^{1/2}.$

For Stein's configuration with a linear density distribution,

$g_{\mathrm{S}\mathrm{S}\mathrm{C}}=\frac{5\pi G{\rho }_{c}R}{36}$           and           ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}\equiv (\frac{36}{5\pi G{\rho }_c}{)}^{1/2}=(\frac{12}{5}\cdot \frac{R^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{1/2}.$

Hence,

$\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}$

$=$

$\frac{48}{5}\cdot {\chi }_0(1-\frac{3}{4}{\chi }_0).$

and the governing adiabatic wave equation takes the form,

$0$	$=$	$\frac{1}{5}(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)\frac{d^{2}x}{d{\chi }_0^2}+\left[\frac{1}{5}\right(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)\frac{4}{{\chi }_0}-(1-{\chi }_0)\frac{48}{5}\cdot {\chi }_0(1-\frac{3}{4}{\chi }_0\left)\right]\frac{dx}{d{\chi }_0}$
		$+(1-{\chi }_0)\left(\frac{1}{{\gamma }_{\mathrm{g}}}\right)\left[\frac{12}{5}\right(\frac{{\omega }^2{R}^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}})+(4-3{\gamma }_{\mathrm{g}})\frac{48}{5}\cdot {\chi }_0(1-\frac{3}{4}{\chi }_0\left)\frac{1}{{\chi }_0}\right]x$
$0$	$=$	$(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)\frac{d^{2}x}{d{\chi }_0^2}+\frac{4}{{\chi }_0}\left[\right(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)-12(1-{\chi }_0\left){\chi }_0^2\right(1-\frac{3}{4}{\chi }_0\left)\right]\frac{dx}{d{\chi }_0}$
		$+12(1-{\chi }_0)\left(\frac{1}{{\gamma }_{\mathrm{g}}}\right)\left[\right(\frac{{\omega }^2{R}^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}})+4(4-3{\gamma }_{\mathrm{g}})(1-\frac{3}{4}{\chi }_0\left)\right]x$
$0$	$=$	$(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)\frac{d^{2}x}{d{\chi }_0^2}+\frac{4}{{\chi }_0}\left[\right(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)-(12{\chi }_0^2-21{\chi }_0^3+9{\chi }_0^4\left)\right]\frac{dx}{d{\chi }_0}$
		$+(1-{\chi }_0)\left[\right(\frac{12}{{\gamma }_{\mathrm{g}}}\left)\right(\frac{{\omega }^2{R}^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}})+(\frac{12}{{\gamma }_{\mathrm{g}}})(4-3{\gamma }_{\mathrm{g}})(4-3{\chi }_0\left)\right]x$
$0$	$=$	$(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)\frac{d^{2}x}{d{\chi }_0^2}+\frac{4}{{\chi }_0}[5-36{\chi }_0^2+7{\chi }_0^3]\frac{dx}{d{\chi }_0}$
		$+(1-{\chi }_0)[{\mathrm{\Omega }}^2+(\frac{12}{{\gamma }_{\mathrm{g}}})(4-3{\gamma }_{\mathrm{g}})(4-3{\chi }_0\left)\right]x,$

where, following 📚 R. Stothers & J. A. Frogel (1967, ApJ, Vol. 148, pp. 305 - 309),

${\mathrm{\Omega }}^2\equiv \frac{12}{{\gamma }_{\mathrm{g}}}\left(\frac{{\omega }^2{R}^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right).$

Eulerian Approach

In his book titled, The Pulsation Theory of Variable Stars, S. Rosseland (1969) defines the relevant eigenvalue problem for adiabatic, radial pulsations in terms of the governing relation (see his equation 2.23 on p. 20, with the adiabatic condition being enforced by setting the right-hand-side equal to zero),

$\frac{\partial }{\partial r}(\gamma P_0\mathrm{\nabla }\cdot \overrightarrow{\xi })+({\omega }^2+\frac{4g_0}{r}){\rho }_0\xi$

$=$

$0,$

where,

$\overrightarrow{\xi }={\hat{\mathbf{e}}}_r\xi (r).$

Realizing that, for a spherically symmetric system,

$\mathrm{\nabla }\cdot \overrightarrow{\xi }=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\xi \right)=\frac{\partial \xi }{\partial r}+\frac{2\xi }{r},$

and remembering that,

$\frac{\partial P_0}{\partial r}=-g_0{\rho }_0,$

we can rewrite this relation in the more familiar form of a 2^nd-order ODE, namely,

$0$	$=$	$\frac{1}{\gamma }({\omega }^2+\frac{4g_0}{r}){\rho }_0\xi +\mathrm{\nabla }\cdot \overrightarrow{\xi }\left(\frac{\partial P_0}{\partial r}\right)+P_0\frac{\partial }{\partial r}(\mathrm{\nabla }\cdot \overrightarrow{\xi })$
	$=$	$\frac{\xi {\rho }_c}{\gamma }({\omega }^2+\frac{4g_0}{r})\left(\frac{{\rho }_0}{{\rho }_c}\right)-{\rho }_0{g}_0[\frac{\partial \xi }{\partial r}+\frac{2\xi }{r}]+P_0\frac{\partial }{\partial r}[\frac{\partial \xi }{\partial r}+\frac{2\xi }{r}]$
	$=$	$\frac{\xi {\rho }_c}{\gamma }({\omega }^2+\frac{4g_0}{r})\left(\frac{{\rho }_0}{{\rho }_c}\right)+[-{\rho }_0{g}_0+\frac{2P_0}{r}]\frac{\partial \xi }{\partial r}+P_0\frac{{\partial }^2\xi }{\partial r^2}+\xi [-(\frac{2{\rho }_0{g}_0}{r})-\frac{2P_0}{r^2}]$
	$=$	$P_0\frac{{\partial }^2\xi }{\partial r^2}+[\frac{2P_0}{r}-{\rho }_0{g}_0]\frac{\partial \xi }{\partial r}+\left[\right(\frac{{\omega }^2{\rho }_c}{\gamma }+\frac{4{\rho }_c{g}_0}{\gamma r}\left)\right(\frac{{\rho }_0}{{\rho }_c})-(\frac{2{\rho }_c{g}_0}{r}\left)\right(\frac{{\rho }_0}{{\rho }_c})-\frac{2P_0}{r^2}]\xi .$

Multiplying through by $(R^2/P_c)$ and, again, letting ${\chi }_0\equiv r/R$, we have,

$0$

$=$

$\left(\frac{P_0}{P_c}\right)\frac{{\partial }^2\xi }{\partial {\chi }_0^2}+\left[\frac{2}{{\chi }_0}\right(\frac{P_0}{P_c})-\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}(\frac{{\rho }_0}{{\rho }_c}\left)\right]\frac{\partial \xi }{\partial {\chi }_0}+\left\{\right[\frac{{\omega }^2{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2}{\gamma }+\frac{2}{{\chi }_0}(\frac{2}{\gamma }-1)\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\left]\right(\frac{{\rho }_0}{{\rho }_c})-\frac{2}{{\chi }_0^2}(\frac{P_0}{P_c}\left)\right\}\xi .$

Now, plugging in the functional expressions that specifically apply to the linear model gives,

$0$	$=$	$\frac{1}{5}[5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4]\frac{{\partial }^2\xi }{\partial {\chi }_0^2}$
		$+\left\{\frac{2}{5{\chi }_0}\right[5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4]-\frac{48}{5}{\chi }_0(1-\frac{3}{4}{\chi }_0\left)\right(1-{\chi }_0\left)\right\}\frac{\partial \xi }{\partial {\chi }_0}$
		$+\left\{\right[\frac{{\mathrm{\Omega }}^2}{5}+\frac{96}{5}(\frac{2}{\gamma }-1)(1-\frac{3}{4}{\chi }_0)\left]\right(1-{\chi }_0)-\frac{2}{5{\chi }_0^2}[5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4\left]\right\}\xi ,$

and, multiplying through by $(5{\chi }_0^2)$ gives,

$0$	$=$	$(5{\chi }_0^2-24{\chi }_0^4+28{\chi }_0^5-9{\chi }_0^6)\frac{{\partial }^2\xi }{\partial {\chi }_0^2}$
		$+\left[2{\chi }_0\right(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)-12{\chi }_0^3(4-7{\chi }_0+3{\chi }_0^2\left)\right]\frac{\partial \xi }{\partial {\chi }_0}$
		$+\left[{\mathrm{\Omega }}^2{\chi }_0^2\right(1-{\chi }_0)+24{\chi }_0^2(\frac{2}{\gamma }-1\left)\right(4-7{\chi }_0+3{\chi }_0^2)-2(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4\left)\right]\xi$
	$=$	$(5{\chi }_0^2-24{\chi }_0^4+28{\chi }_0^5-9{\chi }_0^6)\frac{{\partial }^2\xi }{\partial {\chi }_0^2}+(10{\chi }_0-96{\chi }_0^3+140{\chi }_0^4-54{\chi }_0^5)\frac{\partial \xi }{\partial {\chi }_0}$
		$+\left[{\mathrm{\Omega }}^2\right({\chi }_0^2-{\chi }_0^3)+(\frac{2}{\gamma }-1\left)\right(96{\chi }_0^2-168{\chi }_0^3+72{\chi }_0^4)+(-10+48{\chi }_0^2-56{\chi }_0^3+18{\chi }_0^4\left)\right]\xi$
	$=$	$(5{\chi }_0^2-24{\chi }_0^4+28{\chi }_0^5-9{\chi }_0^6)\frac{{\partial }^2\xi }{\partial {\chi }_0^2}+(10{\chi }_0-96{\chi }_0^3+140{\chi }_0^4-54{\chi }_0^5)\frac{\partial \xi }{\partial {\chi }_0}$
		$+[-10+{\chi }_0^2({\mathrm{\Omega }}^2+\frac{192}{\gamma }-48)-{\chi }_0^3({\mathrm{\Omega }}^2+\frac{336}{\gamma }-112)+{\chi }_0^4(\frac{144}{\gamma }-54\left)\right]\xi ,$

where, following 📚 Stothers & Frogel (1967) (see their equation 3 on p. 305),

${\mathrm{\Omega }}^2\equiv \frac{12}{{\gamma }_{\mathrm{g}}}\left(\frac{{\omega }^2{R}^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right).$

Parabolic Density Distribution

Equilibrium Structure

In an article titled, "Radial Oscillations of a Stellar Model," 📚 C. Prasad (1949, MNRAS, Vol 109, pp. 103 - 107) investigated the properties of an equilibrium configuration with a prescribed density distribution given by the expression,

$\rho (r)={\rho }_c[1-(\frac{r}{R}{)}^2],$

where, ${\rho }_c$ is the central density and, $R$ is the radius of the star. Both the mass distribution and the pressure distribution can be obtained analytically from this specified density distribution. Specifically, following our general solution strategy for determining the equilibrium structure of spherically symmetric, self-gravitating configurations,

$M_r(r)$	$=$	$\underset{0}{\overset{r}{\int }}4\pi r^2\rho (r)dr$
	$=$	$\frac{4\pi {\rho }_c{r}^3}{3}[1-\frac{3}{5}(\frac{r}{R}{)}^2],$

in which case we can write,

$g_0(r)\equiv \frac{G{M}_r(r)}{r^2}$

$=$

$\frac{4\pi G{\rho }_{c}r}{3}[1-\frac{3}{5}(\frac{r}{R}{)}^2].$

Note that the total mass is obtained by setting $r=R$ in the expression for $M_r(r)$, namely,

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}$

$=$

$\frac{4\pi {\rho }_c{R}^3}{3}\left[\frac{2}{5}\right]=\frac{8\pi {\rho }_c{R}^3}{15}$       $\Rightarrow$       $2\pi {\rho }_c=\frac{15M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4R^3}.$

---

Following "Technique 1", we appreciate that,

$\frac{d\mathrm{\Phi }}{dr}$	$=$	$g_0$
$\Rightarrow \mathrm{\Phi }$	$=$	$\frac{4\pi G{\rho }_c}{3}\int [1-\frac{3}{5}(\frac{r}{R}{)}^2]rdr$
	$=$	$\frac{4\pi G{\rho }_c}{3}\{\int rdr-(\frac{3}{5R^2})\int r^{3}dr\}$
	$=$	$\frac{4\pi G{\rho }_c}{3}\{\frac{1}{2}{r}^2-(\frac{3}{5R^2}\left)\frac{1}{4}{r}^4\right\}+C_{\mathrm{\Phi }}$
	$=$	$\frac{2\pi G{\rho }_c{R}^2}{15}\left\{5\right(\frac{r}{R}{)}^2-\frac{3}{2}\left(\frac{r}{R}{)}^4\right\}+C_{\mathrm{\Phi }}$
	$=$	$\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4R}\left\{5\right(\frac{r}{R}{)}^2-\frac{3}{2}\left(\frac{r}{R}{)}^4\right\}+C_{\mathrm{\Phi }}.$

Let's choose a constant, $C_{\mathrm{\Phi }}$, such that the potential is $-G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}/R$ at the surface $(r/R=1)$, that is,

$-\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{R}$	$=$	$\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4R}\{5-\frac{3}{2}\}+C_{\mathrm{\Phi }}$
$\Rightarrow C_{\mathrm{\Phi }}$	$=$	$-\frac{7G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{8R}-\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{R}=-\frac{15G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{8R}.$

Hence, the normalized gravitational potential is given by the expression,

${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$	$=$	$\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4R}\left\{5\right(\frac{r}{R}{)}^2-\frac{3}{2}\left(\frac{r}{R}{)}^4\right\}-\frac{15G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{8R}$
	$=$	$\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{8R}\{-15+10(\frac{r}{R}{)}^2-3\left(\frac{r}{R}{)}^4\right\}.$

Note that in terms of Cartesian coordinates, this expression becomes,

${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$	$=$	$-\frac{\pi G\rho R^2}{15}\{15-\frac{10}{R^2}(x^2+y^2+z^2)+\frac{3}{R^4}(x^2+y^2+z^2\left)\right(x^2+y^2+z^2\left)\right\}$
	$=$	$-\frac{\pi G\rho R^2}{15}\{15-\frac{10}{R^2}(x^2+y^2+z^2)+\frac{3}{R^4}(x^4+x^2{y}^2+x^2{z}^2+y^2{x}^2+y^4+y^2{z}^2+x^2{z}^2+y^2{z}^2+z^4\left)\right\}$
	$=$	$-\frac{\pi G\rho R^2}{15}\{15-\frac{10}{R^2}(x^2+y^2+z^2)+\frac{6}{R^4}(x^2{y}^2+x^2{z}^2+y^2{z}^2)+\frac{3}{R^4}(x^4+y^4+z^4\left)\right\}$
	$=$	$-\pi G\rho R^2\{1-\frac{2}{3R^2}(x^2+y^2+z^2)+\frac{2}{5R^4}(x^2{y}^2+x^2{z}^2+y^2{z}^2)+\frac{1}{5R^4}(x^4+y^4+z^4\left)\right\}.$

---

Hence, proceeding via what we have labeled as "Technique 1", and enforcing the surface boundary condition, $P(R)=0$, 📚 Prasad (1949) determines that,

$P(r)$	$=$	$-\underset{0}{\overset{r}{\int }}{g}_0(r)\rho (r)dr$
	$=$	$-\frac{4\pi G{\rho }_c^2{R}^2}{15}\underset{0}{\overset{r}{\int }}[1-(\frac{r}{R}{)}^2\left]\right[5-3\left(\frac{r}{R}{)}^2\right]\left(\frac{r}{R}\right)\frac{dr}{R}$
	$=$	$-\frac{4\pi G{\rho }_c^2{R}^2}{15}\underset{0}{\overset{r}{\int }}\left[5\right(\frac{r}{R})-8(\frac{r}{R}{)}^3+3\left(\frac{r}{R}{)}^5\right]\frac{dr}{R}$
	$=$	$\frac{2\pi G{\rho }_c^2{R}^2}{15}[2-5(\frac{r}{R}{)}^2+4(\frac{r}{R}{)}^4-(\frac{r}{R}{)}^6]$
	$=$	$\frac{4\pi G{\rho }_c^2{R}^2}{15}[1-(\frac{r}{R}{)}^2{]}^2[1-\frac{1}{2}(\frac{r}{R}{)}^2],$

where, it can readily be deduced, as well, that the central pressure is,

$P_c=\frac{4\pi }{15}G{\rho }_c^2{R}^2.$

As has been explained in the context of our discussion of supplemental relations, the enthalpy distribution $(H)$ can be obtained from the relation,

$dH$

$=$

$\frac{dP}{\rho }.$

That is,

$\frac{dH}{dr}$	$=$	$\frac{4\pi G{\rho }_c{R}^2}{15}[1-(\frac{r}{R}{)}^2{]}^{-1}\frac{d}{dr}\left\{\right[1-\left(\frac{r}{R}{)}^2{]}^2\right[1-\frac{1}{2}\left(\frac{r}{R}{)}^2\right]\}$
	$=$	$\frac{4\pi G{\rho }_c{R}^2}{15}[1-(\frac{r}{R}{)}^2{]}^{-1}\left\{\right[1-\left(\frac{r}{R}{)}^2{]}^2\right[-\left(\frac{r}{R^2}\right)]+2[1-\left(\frac{r}{R}{)}^2\right][-2(\frac{r}{R^2}\left)\right][1-\frac{1}{2}(\frac{r}{R}{)}^2\left]\right\}$
	$=$	$\frac{4\pi G{\rho }_c{R}^2}{15}\left\{\right[1-\left(\frac{r}{R}{)}^2\right][-(\frac{r}{R^2}\left)\right]+2[-2(\frac{r}{R^2}\left)\right][1-\frac{1}{2}(\frac{r}{R}{)}^2\left]\right\}$
	$=$	$-\frac{4\pi G{\rho }_c{R}^2}{15}\left\{\right[1-\left(\frac{r}{R}{)}^2\right]\left(\frac{r}{R^2}\right)+4\left(\frac{r}{R^2}\right)[1-\frac{1}{2}(\frac{r}{R}{)}^2\left]\right\}$
	$=$	$-\frac{4\pi G{\rho }_c{R}^2}{15}\left\{\right[1-\left(\frac{r}{R}{)}^2\right]+[4-2(\frac{r}{R}{)}^2\left]\right\}\left(\frac{r}{R^2}\right)$
	$=$	$-\frac{4\pi G{\rho }_c{R}^2}{15}[5-3(\frac{r}{R}{)}^2\left]\right(\frac{r}{R^2}).$

Hence,

$H$	$=$	$-\frac{4\pi G{\rho }_c}{15}\int [5r-\frac{3}{R^2}(r^3\left)\right]dr$
	$=$	$-\frac{4\pi G{\rho }_c{R}^2}{15}\left[\frac{5}{2}\right(\frac{r}{R}{)}^2-\frac{3}{4}\left(\frac{r}{R}{)}^4\right]+C$
	$=$	$-\frac{1}{15}(2\pi G{\rho }_c{R}^2)\left[5\right(\frac{r}{R}{)}^2-\frac{3}{2}\left(\frac{r}{R}{)}^4\right]+C$
	$=$	$-\left(\frac{GM}{10R}\right)\left[5\right(\frac{r}{R}{)}^2-\frac{3}{2}\left(\frac{r}{R}{)}^4\right]+C.$

Let's normalize by setting $H=0$ when $r=R$; that is,

$C$

$=$

$\frac{7GM}{20R}.$

So, the enthalpy is given by the expression,

$H$	$=$	$-\left(\frac{GM}{10R}\right)\left[5\right(\frac{r}{R}{)}^2-\frac{3}{2}\left(\frac{r}{R}{)}^4\right]+\frac{7GM}{20R}$
	$=$	$\frac{GM}{20R}[7-10(\frac{r}{R}{)}^2+3\left(\frac{r}{R}{)}^4\right].$

Note that the quantity, $({\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+H)$, is constant throughout the configuration. Specifically, ${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+H$ $=$ $-\frac{GM}{20R}\{27-10(\frac{r}{R}{)}^2+3\left(\frac{r}{R}{)}^4\right\}+\frac{GM}{20R}[7-10(\frac{r}{R}{)}^2+3\left(\frac{r}{R}{)}^4\right]$   $=$ $-\frac{GM}{R}.$

Specific Entropy Distribution

From an accompanying discussion of specific entropy distributions, $s$, we realize that to within an additive constant $(s_0)$,

$s-s_0$

$=$

$c_P\ln \left(\frac{P^{1/{\gamma }_g}}{\rho }\right).$

[LL75], §80, Eq. (80.12)

Or (see a related discussion), given that

$c_P$

$=$

$\frac{{\gamma }_g}{({\gamma }_g-1)}\left(\frac{\mathrm{\Re }}{\overline{\mu }}\right),$

we can also write,

$\frac{(s-s_0)}{\mathrm{\Re }/\overline{\mu }}$	$=$	$\frac{{\gamma }_g}{({\gamma }_g-1)}\{\frac{1}{{\gamma }_g}\ln P-\ln \rho \}$
	$=$	$\frac{1}{({\gamma }_g-1)}\{\ln P-{\gamma }_g\ln \rho \}$
	$=$	$\frac{1}{({\gamma }_g-1)}\{\ln P-{\gamma }_g\ln \rho -\ln ({\gamma }_g-1)\}+\frac{\ln ({\gamma }_g-1)}{({\gamma }_g-1)}$
	$=$	$\frac{1}{({\gamma }_g-1)}\ln \left[\frac{P}{({\gamma }_g-1){\rho }^{{\gamma }_g}}\right]+\frac{\ln ({\gamma }_g-1)}{({\gamma }_g-1)}.$

Notice that if we set the constant,

$\frac{s_0}{\mathrm{\Re }/\overline{\mu }}$

$=$

$-\frac{\ln ({\gamma }_g-1)}{({\gamma }_g-1)},$

we obtain the same expression for the entropy distribution as we used in our discussions with Patrick Motl, namely,

$\frac{s}{\mathrm{\Re }/\overline{\mu }}$

$=$

$\frac{1}{({\gamma }_g-1)}\ln \left[\frac{P}{({\gamma }_g-1){\rho }^{{\gamma }_g}}\right].$

Shifting this expression for the specific entropy by another constant (not written out explicitly here) leads us to the more "normalized" expression,

$\frac{s}{\mathrm{\Re }/\overline{\mu }}$	$=$	$\frac{1}{({\gamma }_g-1)}\ln \left[\frac{P/P_c}{({\gamma }_g-1)(\rho /{\rho }_c{)}^{{\gamma }_g}}\right]$
$\Rightarrow ({\gamma }_g-1)\exp \left[\frac{({\gamma }_g-1)s}{\mathrm{\Re }/\overline{\mu }}\right]$	$=$	$\left[\frac{P/P_c}{(\rho /{\rho }_c{)}^{{\gamma }_g}}\right]$

Plugging in the expressions for the pressure and density distributions that are relevant to the 📚 Prasad (1949) model having a "parabolic" density distribution, then gives,

$({\gamma }_g-1)\exp \left[\frac{({\gamma }_g-1)s}{\mathrm{\Re }/\overline{\mu }}\right]$	$=$	$\left\{\right[1-\left(\frac{r}{R}{)}^2{]}^2\right[1-\frac{1}{2}\left(\frac{r}{R}{)}^2\right]\left\}\right[1-(\frac{r}{R}{)}^2{]}^{-{\gamma }_g}$
	$=$	$[1-(\frac{r}{R}{)}^2{]}^{(2-{\gamma }_g)}[1-\frac{1}{2}(\frac{r}{R}{)}^2].$

Stabililty

As has been derived in an accompanying discussion, the second-order ODE that defines the relevant Eigenvalue problem is,

$\left(\frac{P_0}{P_c}\right)\frac{d^{2}x}{d{\chi }_0^2}+\left[\right(\frac{P_0}{P_c})\frac{4}{{\chi }_0}-(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\left)\right]\frac{dx}{d{\chi }_0}+\left(\frac{{\rho }_0}{{\rho }_c}\right)\left(\frac{1}{{\gamma }_{\mathrm{g}}}\right)[{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2{\omega }^2+(4-3{\gamma }_{\mathrm{g}})(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\left)\frac{1}{{\chi }_0}\right]x=0,$

where the dimensionless radius,

${\chi }_0\equiv \frac{r_0}{R},$

$g_{\mathrm{S}\mathrm{S}\mathrm{C}}\equiv \frac{P_c}{R{\rho }_c}$           and           ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}\equiv (\frac{R^2{\rho }_c}{P_c}{)}^{1/2}.$

For Prasad's configuration with a parabolic density distribution,

$g_{\mathrm{S}\mathrm{S}\mathrm{C}}=\frac{4\pi G{\rho }_{c}R}{15}$           and           ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}\equiv (\frac{15}{4\pi G{\rho }_c}{)}^{1/2}=(\frac{2R^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{1/2}=(\frac{3}{2\pi G\overline{\rho }}{)}^{1/2}.$

Hence,

$\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}$

$=$

$(5-3{\chi }_0^2){\chi }_0,$

and the governing adiabatic wave equation takes the form,

$(1-{\chi }_0^2)(1-\frac{1}{2}{\chi }_0^2)\frac{d^{2}x}{d{\chi }_0^2}+\frac{1}{{\chi }_0}[4(1-{\chi }_0^2)(1-\frac{1}{2}{\chi }_0^2)-(5-3{\chi }_0^2){\chi }_0^2]\frac{dx}{d{\chi }_0}+[\frac{{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2{\omega }^2}{{\gamma }_{\mathrm{g}}}-\alpha (5-3{\chi }_0^2)]x=0,$

where,

$\alpha \equiv 3-\frac{4}{{\gamma }_{\mathrm{g}}}.$

In keeping with Prasad's presentation — see, specifically, his equations (2) & (3) — this wave equation can also be written as,

$(1-{\chi }_0^2)(1-\frac{1}{2}{\chi }_0^2)\frac{d^{2}x}{d{\chi }_0^2}+\frac{1}{{\chi }_0}[4-11{\chi }_0^2+5{\chi }_0^4]\frac{dx}{d{\chi }_0}+[\mathfrak{𝔍}+3\alpha {\chi }_0^2]x=0,$

where,

$\mathfrak{𝔍}\equiv \frac{3{\omega }^2}{2\pi G{\gamma }_{\mathrm{g}}\overline{\rho }}-5\alpha .$

For what it's worth, we have also deduced that this expression can be written as,

$(1-{\chi }_0^2)(1-\frac{1}{2}{\chi }_0^2){\chi }_0^{-4}\frac{d}{d{\chi }_0}\left[{\chi }_0^4\frac{dx}{d{\chi }_0}\right]-(5-3{\chi }_0^2){\chi }_0^{1+\alpha }\frac{d}{d{\chi }_0}\left[{\chi }_0^{-\alpha }x\right]+\left(\frac{{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2{\omega }^2}{{\gamma }_{\mathrm{g}}}\right)x=0,$

Ramblings

The material originally contained in this "Ramblings" subsection has been moved to generate a separate chapter that stands on its own.

Promising Avenue of Exploration

What follows is a direct extension of what is referred to in our "Ramblings" chapter as the third guess under "Exploration2". We pursue this line of reasoning, here, because it appears to be a particularly promising avenue of exploration.

In the case of a parabolic density distribution, the LAWE becomes,

$\frac{2}{(1-x^2)(2-x^2)}\left[\right(\alpha +\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }})(5-3x^2)-{\sigma }^2]$

$=$

$\left(\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}\right)+\frac{4}{x^2}\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}.$

We have chosen to examine the suitability of an eigenfunction of the form,

${{𝒢}}_{\sigma }$

$=$

$(a_0+a_2{x}^2{)}^n\cdot (2-x^2{)}^m,$

where, for a given value of $\alpha$, the four parameters, $a_0$, $a_2$, $n$ and $m$ are to be determined in concert with a value of the square of the eigenfrequency, ${\sigma }^2$. From the accompanying discussion we have determined that the following five coefficient expressions must independently be zero in order for this trial eigenfunction to satisfy the LAWE:

$x^0$	:	$\alpha (10a_0^2)+{\sigma }^2(-2a_0^2)-20n{a}_0{a}_2+10m{a}_0^2$
$x^2$	:	$\alpha (-11a_0^2+20a_0{a}_2)+{\sigma }^2(a_0^2-4a_0{a}_2)+60n{a}_0{a}_2-20n{a}_2^2-25m{a}_0^2+20m{a}_0{a}_2+8nm{a}_0{a}_2$ $-[8n(n-1)a_2^2+2m(m-1)a_0^2]$
$x^4$	:	$\alpha (10a_2^2-22a_0{a}_2+3a_0^2)-{\sigma }^2(2a_2^2-2a_0{a}_2)-47n{a}_0{a}_2+60n{a}_2^2-50m{a}_0{a}_2+11m{a}_0^2+10m{a}_2^2-12nm{a}_0{a}_2+8nm{a}_2^2$ $+16n(n-1)a_2^2-4m(m-1)a_0{a}_2+2m(m-1)a_0^2$
$x^6$	:	$\alpha (6a_0{a}_2-11a_2^2)+{\sigma }^2(a_2^2)+11n{a}_0{a}_2+22m{a}_0{a}_2-47n{a}_2^2-25m{a}_2^2-12nm{a}_2^2+4nm{a}_0{a}_2$ $-10n(n-1)a_2^2-2m(m-1)a_2^2+4m(m-1)a_0{a}_2$
$x^8$	:	$\{3\alpha +[4nm+11n+11m]+[2n(n-1)+2m(m-1)]\}{a}_2^2$

First Constraint

We begin by manipulating the last expression — that is, the coefficient expression for the $x^8$ term. Rejecting the trivial option of setting $a_2=0$, in order for this expression to be zero the terms inside the curly braces must sum to zero. Rewriting this expression in terms of the sum of the exponents,

$s_{nm}\equiv n+m,$

we obtain the quadratic expression,

$0$	$=$	$3\alpha +[4nm+11n+11m]+[2n(n-1)+2m(m-1)]$
	$=$	$3\alpha +4nm+9n+9m+2n^2+3m^2$
	$=$	$3\alpha +9s_{nm}+2s_{nm}^2.$

This means that, once the physical parameter, $\alpha =(3-4/{\gamma }_g)$, has been specified, the sum of the exponents must be,

$s_{nm}$	$=$	$\frac{1}{4}[-9\pm (81-24\alpha {)}^{1/2}]$
	$=$	$\frac{3^2}{2^2}[-1\pm (1-\frac{2^3\alpha }{3^3}{)}^{1/2}].$

Second Constraint

Next we examine the expression that serves as the coefficient of $x^0$. Setting that coefficient expression to zero while replacing $m$ in favor of $s_{nm}$ — via the relation, $m=(s_{nm}-n)$ — gives,

$0$	$=$	$\alpha (10a_0^2)+{\sigma }^2(-2a_0^2)-20n{a}_0{a}_2+10m{a}_0^2$
	$=$	$2a_0^2[5\alpha -{\sigma }^2+5(s_{nm}-n)-10n(\frac{a_2}{a_0}\left)\right]$
	$=$	$2a_0^2[5\alpha -{\sigma }^2+5s_{nm}-5n(1-\frac{2a_2}{a_0}\left)\right]$
$\Rightarrow \frac{{\sigma }^2}{5}$	$=$	$(\alpha +s_{nm})-n(1-2\lambda ),$

where, we have set,

$\lambda \equiv \frac{a_2}{a_0}.$

So, once $\alpha$ is specified and $s_{nm}$ is known from the first constraint, we can use this expression to replace ${\sigma }^2$ in the other three coefficient expressions.

Intermediate Summary

The three remaining constraints emerge from the remaining three coefficient expressions, namely,

$x^2$	:	$\alpha (-11a_0^2+20a_0{a}_2)+{\sigma }^2(a_0^2-4a_0{a}_2)+60n{a}_0{a}_2-20n{a}_2^2-25m{a}_0^2+20m{a}_0{a}_2+8nm{a}_0{a}_2$ $-[8n(n-1)a_2^2+2m(m-1)a_0^2]$
$x^4$	:	$\alpha (10a_2^2-22a_0{a}_2+3a_0^2)-{\sigma }^2(2a_2^2-2a_0{a}_2)-47n{a}_0{a}_2+60n{a}_2^2-50m{a}_0{a}_2+11m{a}_0^2+10m{a}_2^2-12nm{a}_0{a}_2+8nm{a}_2^2$ $+16n(n-1)a_2^2-4m(m-1)a_0{a}_2+2m(m-1)a_0^2$
$x^6$	:	$\alpha (6a_0{a}_2-11a_2^2)+{\sigma }^2(a_2^2)+11n{a}_0{a}_2+22m{a}_0{a}_2-47n{a}_2^2-25m{a}_2^2-12nm{a}_2^2+4nm{a}_0{a}_2$ $-10n(n-1)a_2^2-2m(m-1)a_2^2+4m(m-1)a_0{a}_2$

Written in terms of the three remaining unknowns, $n$, $a_0$, and $\lambda$, the three constraints are:

$x^2:$	$0$	$=$	$\alpha (-11+20\lambda )+{\sigma }^2(1-4\lambda )+60n\lambda -20n{\lambda }^2-25m+20m\lambda +8nm\lambda -[8n(n-1){\lambda }^2+2m(m-1)]$
		$=$	$\alpha (-11+20\lambda )+5(1-4\lambda )[(\alpha +s_{nm})-n(1-2\lambda )]+60n\lambda -20n{\lambda }^2-8n(n-1){\lambda }^2$
			$+(s_{nm}-n)[-23+20\lambda +8n\lambda ]-2(s_{nm}-n{)}^2$
		$=$	$-6\alpha +5(1-4\lambda )s_{nm}-5n(1-4\lambda )(1-2\lambda )+60n\lambda -20n{\lambda }^2-8n(n-1){\lambda }^2$
			$+(s_{nm}-n)[-23+20\lambda +8n\lambda ]-2(s_{nm}-n{)}^2;$
$x^4:$	$0$	$=$	$\alpha (10{\lambda }^2-22\lambda +3)-10({\lambda }^2-\lambda )[(\alpha +s_{nm})-n(1-2\lambda )]-47n\lambda +60n{\lambda }^2+16n(n-1){\lambda }^2$
			$+(s_{nm}-n)[9-46\lambda +10{\lambda }^2-12n\lambda +8n{\lambda }^2]+(2-4\lambda )(s_{nm}-n{)}^2$
		$=$	$\alpha (-12\lambda +3)-10({\lambda }^2-\lambda )s_{nm}+n10({\lambda }^2-\lambda )(1-2\lambda )-47n\lambda +60n{\lambda }^2+16n(n-1){\lambda }^2$
			$+(s_{nm}-n)[9-46\lambda +10{\lambda }^2-12n\lambda +8n{\lambda }^2]+(2-4\lambda )(s_{nm}-n{)}^2;$
$x^6:$	$0$	$=$	$\alpha (6\lambda -11{\lambda }^2)+5{\lambda }^2[(\alpha +s_{nm})-n(1-2\lambda )]+11n\lambda -47n{\lambda }^2-10n(n-1){\lambda }^2$
			$+(s_{nm}-n)[18\lambda -23{\lambda }^2-12n{\lambda }^2+4n\lambda ]+2\lambda (2-\lambda )(s_{nm}-n{)}^2$
		$=$	$6\lambda \alpha (1-\lambda )+5{\lambda }^2{s}_{nm}-5n{\lambda }^2(1-2\lambda )+11n\lambda -47n{\lambda }^2-10n(n-1){\lambda }^2$
			$+(s_{nm}-n)[18\lambda -23{\lambda }^2-12n{\lambda }^2+4n\lambda ]+2\lambda (2-\lambda )(s_{nm}-n{)}^2.$

At first glance, this is still not as promising as I had hoped. In practice there are only two unknowns — because the parameter, $a_0$, has divided out — while there are three constraints. So the problem remains over constrained.

Remaining Group of Three Constraints

Let's adopt another approach. Let's assume that the parameter, $\alpha$, is also initially unspecified and replace it in all three remaining constraint expressions, in favor of $s_{nm}$, using the above-specified, first constraint, namely,

$-3\alpha$

$=$

$9s_{nm}+2s_{nm}^2.$

This gives,

$x^2:$	$0$	$=$	$2(9s_{nm}+2s_{nm}^2)+5(1-4\lambda )s_{nm}-5n(1-4\lambda )(1-2\lambda )+60n\lambda -20n{\lambda }^2-8n(n-1){\lambda }^2$
			$+(s_{nm}-n)[-23+20\lambda +8n\lambda ]-2(s_{nm}-n{)}^2;$
$x^4:$	$0$	$=$	$-(9s_{nm}+2s_{nm}^2)(1-4\lambda )-10({\lambda }^2-\lambda )s_{nm}+n10({\lambda }^2-\lambda )(1-2\lambda )-47n\lambda +60n{\lambda }^2+16n(n-1){\lambda }^2$
			$+(s_{nm}-n)[9-46\lambda +10{\lambda }^2-12n\lambda +8n{\lambda }^2]+(2-4\lambda )(s_{nm}-n{)}^2;$
$x^6:$	$0$	$=$	$2\lambda (9s_{nm}+2s_{nm}^2)(1-\lambda )+5{\lambda }^2{s}_{nm}-5n{\lambda }^2(1-2\lambda )+11n\lambda -47n{\lambda }^2-10n(n-1){\lambda }^2$
			$+(s_{nm}-n)[18\lambda -23{\lambda }^2-12n{\lambda }^2+4n\lambda ]+2\lambda (2-\lambda )(s_{nm}-n{)}^2.$

The three unknowns are: $n$, $s_{nm}$, and $\lambda$.

Prasad's Work

Overview

📚 Prasad (1949) performed a semi-analytic analysis of the radial oscillations and stability of structures having a parabolic density distribution. Let's examine his tabulated results to see if they help us understand more fully whether or not our analysis is on the right track. For example, from his Table I, we see that $\mathfrak{𝔉}=0$ when $\alpha =0$, where, according to his equation (3),

$\mathfrak{𝔉}\equiv {\sigma }^2-5\alpha .$

This means that, also, ${\sigma }^2=0$. Now, from our derived second constraint, we deduce that,

$\mathfrak{𝔉}$

$=$

$5[s_{nm}-n(1-2\lambda )].$

Hence, since $\mathfrak{𝔉}=0$, we conclude that,

$s_{nm}=n(1-2\lambda ).$

Also, since by definition $s_{nm}=n+m$, we conclude that,

$\frac{m}{n}=-2\lambda .$

Next, given that $\alpha =0$, we conclude from our derived first constraint, that

$s_{nm}=\frac{3^2}{2^2}[-1\pm 1]$

$\Rightarrow$

$s_{nm}^{+}=0$    and     $s_{nm}^{-}=-\frac{9}{2}.$

The Minus Root

Combining these two results for the "minus" solution, we furthermore conclude that, for this specific mode, the relationship between the two exponents and $\lambda$ are,

$n^{-}=-\frac{9}{2(1-2\lambda )}$       and       $m^{-}=(s_{nm}-n^{-})=\frac{9\lambda }{(1-2\lambda )}.$

The Plus Root

Next, let's examine the "plus" solution. Because $s_{nm}^{+}=0$, this solution implies that,

$m^{+}=-n^{+}$      $\Rightarrow$     $\frac{m}{n}=-1$.

In this case, then, we deduce that,

$\lambda =-\frac{1}{2}\left(\frac{m}{n}\right)=+\frac{1}{2}$.

So, even though these first two constraints have not revealed the value of either of the exponents, $n$ and $m$, we see that the resulting trial eigenfunction must be,

${{𝒢}}_{\sigma }$	$=$	$a_0^n(1+\lambda x^2{)}^n\cdot (2-x^2{)}^m$
	$=$	$a_0^n[\frac{(1+\frac{1}{2}{x}^2)}{(2-x^2)}{]}^n$
	$=$	$\left(\frac{a_0}{2}{)}^n\right[\frac{(2+x^2)}{(2-x^2)}{]}^n.$

Interesting!

Third Constraint

The Minus Root

Let's insert all of these relations into the algebraic expression that we have derived from the $x^2$ coefficient:

$x^2:$	RHS	$=$	$2s_{nm}(9+2s_{nm})+5(1-4\lambda )s_{nm}-5n(1-4\lambda )(1-2\lambda )+60n\lambda -20n{\lambda }^2-8n(n-1){\lambda }^2$
			$+(s_{nm}-n)[-23+20\lambda +8n\lambda ]-2(s_{nm}-n{)}^2$
		$=$	$0-\frac{45}{2}(1-4\lambda )+\frac{45}{2}(1-4\lambda )+n\lambda \{60-20\lambda +8[\frac{11-4\lambda }{2(1-2\lambda )}\left]\lambda \right\}$
			$+\frac{9\lambda }{(1-2\lambda )}\{-23+20\lambda -[\frac{36\lambda }{(1-2\lambda )}\left]\lambda \right\}-2[\frac{9\lambda }{(1-2\lambda )}{]}^2$
		$=$	$\frac{9\lambda }{(1-2\lambda )}\{-30+10\lambda -2[\frac{11-4\lambda }{(1-2\lambda )}]\lambda -23+20\lambda -[\frac{36\lambda }{(1-2\lambda )}]\lambda -[\frac{18\lambda }{(1-2\lambda )}\left]\right\}$
		$=$	$\frac{9\lambda }{(1-2\lambda )}\{-53+30\lambda -[\frac{58-8\lambda }{(1-2\lambda )}]\lambda -[\frac{18\lambda }{(1-2\lambda )}\left]\right\}$
		$=$	$\frac{9\lambda }{(1-2\lambda {)}^2}\{(-53+30\lambda )(1-2\lambda )-(58-8\lambda )\lambda -18\lambda \}$
		$=$	$\frac{9\lambda }{(1-2\lambda {)}^2}[-53+30\lambda +106\lambda -60{\lambda }^2-58\lambda +8{\lambda }^2-18\lambda ]$
		$=$	$\frac{9\lambda }{(1-2\lambda {)}^2}[-53+60\lambda -52{\lambda }^2]$

Let's repeat this step, but start from an earlier expression for the $x^2$ coefficient, namely,

$x^2:$	$0$	$=$	$\alpha (-11+20\lambda )+{\sigma }^2(1-4\lambda )+60n\lambda -20n{\lambda }^2-25m+20m\lambda +8nm\lambda -[8n(n-1){\lambda }^2+2m(m-1)]$
		$=$	$\alpha (-11+20\lambda )+{\sigma }^2(1-4\lambda )+n[60\lambda -20{\lambda }^2+\frac{m}{n}(-25+20\lambda \left)\right]+8nm\lambda -[8n(n-1){\lambda }^2+2m(m-1)]$
		$=$	$\alpha (-11+20\lambda )+{\sigma }^2(1-4\lambda )+n[60\lambda -12{\lambda }^2+\frac{m}{n}(-23+20\lambda \left)\right]+2n^2[-4{\lambda }^2+4(\frac{m}{n})\lambda -(\frac{m}{n}{)}^2].$

The first two terms on the RHS immediately go to zero because, for this specific eigenfunction, both $\alpha$ and ${\sigma }^2$ are zero. Plugging in our determined expressions for $n^{-}$ and $(m^{-}/n^{-})$ gives,

$x^2:$	RHS	$=$	$-\frac{9}{2(1-2\lambda )}[60\lambda -12{\lambda }^2-2\lambda (-23+20\lambda \left)\right]+2[-\frac{9}{2(1-2\lambda )}{]}^2[-4{\lambda }^2+4(-2\lambda )\lambda -(-2\lambda {)}^2]$
		$=$	$-\frac{9\lambda }{(1-2\lambda )}[53-26\lambda ]-\frac{8\cdot 81{\lambda }^2}{(1-2\lambda {)}^2}$
		$=$	$-\frac{9\lambda }{(1-2\lambda {)}^2}[(1-2\lambda )(53-26\lambda )+72\lambda ]$
		$=$	$-\frac{9\lambda }{(1-2\lambda {)}^2}[53-60\lambda +52{\lambda }^2],$

which exactly matches the previous, but messier, derivation. Now, the two roots of the quadratic expression inside the square brackets are,

$\lambda$	$=$	$\frac{1}{2^3\cdot 13}[2^2\cdot 3\cdot 5\pm \sqrt{-2^8\cdot 29}]$
	$=$	$\frac{1}{2\cdot 13}[3\cdot 5\pm \sqrt{-2^4\cdot 29}].$

Both roots are imaginary numbers and therefore not of interest in the context of this astrophysical problem.

The Plus Root

Next, in addition to setting $\alpha ={\sigma }^2=0$, we'll plug $\lambda =\frac{1}{2}$ and $m^{+}/n^{+}=-1$ into the third constraint expression as follows:

$x^2:$	RHS	$=$	$\alpha (-11+20\lambda )+{\sigma }^2(1-4\lambda )+n[60\lambda -20{\lambda }^2-25(\frac{m}{n})+20(\frac{m}{n})\lambda +8{\lambda }^2+2(\frac{m}{n}\left)\right]+n^2\left[8\right(\frac{m}{n})\lambda -8{\lambda }^2-2(\frac{m}{n}{)}^2]$
		$=$	$n[60\lambda -20{\lambda }^2+25-20\lambda +8{\lambda }^2-2]+n^2[-8\lambda -8{\lambda }^2-2]$
		$=$	$n[40\lambda -12{\lambda }^2+23]-2n^2[4\lambda +4{\lambda }^2+1]$
		$=$	$n[20-3+23]-2n^2[2+1+1]$
		$=$	$8n(5-n).$

So the nontrivial solution is $n^{+}=5$ — and, hence, $m^{+}=-5$ — in which case the trial eigenfunction is,

${{𝒢}}_{\sigma }$

$=$

$\left(\frac{a_0}{2}{)}^5\right[\frac{(2+x^2)}{(2-x^2)}{]}^5.$

Fifth Constraint

The Minus Root

In a similar vein, let's insert all of the deduced relations into the algebraic expression that we have derived from the $x^6$ coefficient:

$x^6:$	RHS	$=$	$\alpha (6\lambda -11{\lambda }^2)+{\sigma }^2{\lambda }^2+11n\lambda +22m\lambda -47n{\lambda }^2-25m{\lambda }^2-12nm{\lambda }^2+4nm\lambda$
			$-10n(n-1){\lambda }^2-2m(m-1){\lambda }^2+4m(m-1)\lambda$
		$=$	$\alpha (6\lambda -11{\lambda }^2)+{\sigma }^2{\lambda }^2+n[11\lambda -47{\lambda }^2+22(\frac{m}{n})\lambda -25(\frac{m}{n}){\lambda }^2+10{\lambda }^2+2(\frac{m}{n}){\lambda }^2-4(\frac{m}{n}\left)\lambda \right]$
			$+n^2[-12(\frac{m}{n}){\lambda }^2+4(\frac{m}{n})\lambda -10{\lambda }^2-2(\frac{m}{n}{)}^2{\lambda }^2+4\left(\frac{m}{n}{)}^2\lambda \right]$
		$=$	$\alpha (6\lambda -11{\lambda }^2)+{\sigma }^2{\lambda }^2+n[11\lambda -37{\lambda }^2+\lambda (\frac{m}{n})(18-23\lambda )]$
			$+n^2[-10{\lambda }^2+4\lambda (\frac{m}{n})(1-3{\lambda }^2)+2\lambda (\frac{m}{n}{)}^2(2-\lambda )].$

As above, the first two terms on the RHS immediately go to zero because, for this specific eigenfunction, both $\alpha$ and ${\sigma }^2$ are zero. Plugging in our determined expressions for $n^{-}$ and $(m^{-}/n^{-})$ gives,

$x^6:$	RHS	$=$	$-\frac{9}{2(1-2\lambda )}[11\lambda -37{\lambda }^2-2{\lambda }^2(18-23\lambda )]$
			$+2[-\frac{9}{2(1-2\lambda )}{]}^2[-5{\lambda }^2-4{\lambda }^2(1-3{\lambda }^2)+4{\lambda }^3(2-\lambda )]$
		$=$	$-\frac{9\lambda }{2(1-2\lambda )}[11-73\lambda +46{\lambda }^2]+\left[\frac{9^2{\lambda }^2}{2(1-2\lambda {)}^2}\right][-9+8\lambda +8{\lambda }^2]$
		$=$	$-\frac{9\lambda }{2(1-2\lambda {)}^2}\{(1-2\lambda )[11-73\lambda +46{\lambda }^2]-9\lambda [-9+8\lambda +8{\lambda }^2]\}$
		$=$	$-\frac{9\lambda }{2(1-2\lambda {)}^2}[11-14\lambda +120{\lambda }^2-164{\lambda }^3].$

The Plus Root

Next, in addition to setting $\alpha ={\sigma }^2=0$, we'll plug $\lambda =\frac{1}{2}$ and $m^{+}/n^{+}=-1$ into the fifth constraint expression as follows:

$x^6:$	RHS	$=$	$\alpha (6\lambda -11{\lambda }^2)+{\sigma }^2{\lambda }^2+n[11\lambda -37{\lambda }^2+\lambda (\frac{m}{n})(18-23\lambda )]$
			$+n^2[-10{\lambda }^2+4\lambda (\frac{m}{n})(1-3{\lambda }^2)+2\lambda (\frac{m}{n}{)}^2(2-\lambda )]$
		$=$	$n\lambda [11-37\lambda -(18-23\lambda )]+n^2\lambda [-10\lambda -4(1-3{\lambda }^2)+2(2-\lambda )]$
		$=$	$n\lambda [-7-14\lambda ]+n^2\lambda [-10\lambda -4+12{\lambda }^2+4-2\lambda ]$
		$=$	$-7n-\left(\frac{3}{2}\right)n^2$
		$=$	$-7n[1+(\frac{3}{14}\left)n\right].$

From this constraint, it appears that the nontrivial result is, $n=-14/3$.

Fourth Constraint

$x^6:$	RHS	$=$	$\alpha (10{\lambda }^2-22\lambda +3)-{\sigma }^2(2{\lambda }^2-2\lambda )-47n\lambda +60n{\lambda }^2-50m\lambda +11m+10m{\lambda }^2-12nm\lambda +8nm{\lambda }^2$
			$+16n(n-1){\lambda }^2-4m(m-1)\lambda +2m(m-1)$
		$=$	$\alpha (10{\lambda }^2-22\lambda +3)-{\sigma }^2(2{\lambda }^2-2\lambda )+n[-47\lambda +60{\lambda }^2-50(\frac{m}{n})\lambda +11(\frac{m}{n})+10(\frac{m}{n}){\lambda }^2-16{\lambda }^2+4(\frac{m}{n})\lambda -2(\frac{m}{n}\left)\right]$
			$+n^2[-12(\frac{m}{n})\lambda +8(\frac{m}{n}){\lambda }^2+16{\lambda }^2-4(\frac{m}{n}{)}^2\lambda +2\left(\frac{m}{n}{)}^2\right]$
		$=$	$\alpha (10{\lambda }^2-22\lambda +3)-{\sigma }^2(2{\lambda }^2-2\lambda )+n[-47\lambda +44{\lambda }^2-37(\frac{m}{n})\lambda +10(\frac{m}{n}\left){\lambda }^2\right]$
			$+n^2[+16{\lambda }^2-12(\frac{m}{n})\lambda +8(\frac{m}{n}){\lambda }^2-4(\frac{m}{n}{)}^2\lambda +2\left(\frac{m}{n}{)}^2\right]$

The Minus Root

In addition to setting $\alpha ={\sigma }^2=0$, here we plug $n^{-}=-9/[2(1-2\lambda )]$ and $m^{+}/n^{+}=-2\lambda$ into the fourth constraint expression as follows:

$x^6:$	RHS	$=$	$n\lambda [-47+118\lambda -20{\lambda }^2]+n^2{\lambda }^2[16+24-16\lambda -16\lambda +8]$
		$=$	$-\left[\frac{9}{2(1-2\lambda )}\right]\lambda [-47+118\lambda -20{\lambda }^2]+[\frac{9}{2(1-2\lambda )}{]}^2{\lambda }^2[48-32\lambda ]$
		$=$	$\left[\frac{9\lambda }{2(1-2\lambda {)}^2}\right]\{9\lambda [24-16\lambda ]-(1-2\lambda )[-47+118\lambda -20{\lambda }^2]\}$
		$=$	$\left[\frac{9\lambda }{2(1-2\lambda {)}^2}\right]\{216\lambda -144{\lambda }^2+[47-118\lambda +20{\lambda }^2]+[-94\lambda +236{\lambda }^2-40{\lambda }^3]\}$
		$=$	$\left[\frac{9\lambda }{2(1-2\lambda {)}^2}\right]\{47-4\lambda +112{\lambda }^2-40{\lambda }^3\}$

The Plus Root

Next, in addition to setting $\alpha ={\sigma }^2=0$, we'll plug $\lambda =\frac{1}{2}$ and $m^{+}/n^{+}=-1$ into the fourth constraint expression as follows:

$x^6:$	RHS	$=$	$n[-47\lambda +44{\lambda }^2+37\lambda -10{\lambda }^2]+n^2[+16{\lambda }^2+12\lambda -8{\lambda }^2-4\lambda +2]$
		$=$	$\frac{7n}{2}[1+n(\frac{16}{7}\left)\right].$

From this constraint, it appears that the nontrivial result is, $n^{+}=-7/16$.

More General Approach

The specific trial eigenfunction that we have just examined does not appear to simultaneously satisfy all constraints prescribed by the LAWE. So, in a separate chapter, we will examine an even more general trial eigenfunction. It is the one that also has previously been introduced in our "Ramblings" chapter under the subheading, "Consider Parabolic Case", having the form,

${{𝒢}}_{\sigma }$

$=$

$(a_0+a_2{x}^2{)}^n\cdot (b_0+b_2{x}^2{)}^m.$

In this accompanying chapter, we will be examining whether or not it satisfies the (same) version of the LAWE that describes stability in structures having a parabolic density profile, namely,

$\frac{2}{(1-x^2)(2-x^2)}\left[\right(\alpha +\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }})(5-3x^2)-{\sigma }^2]$

$=$

$\left(\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}\right)+\frac{4}{x^2}\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}.$

See Also