SSC/Structure/OtherAnalyticRamblings

Ramblings Regarding the Stability of Other Analytically Definable, Spherical Equilibrium Models

The material presented in this chapter was originally developed as a subsection of a chapter that discusses "Other" Analytic Equilibrium Models."

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Material that appears after this point in our presentation is under development and therefore
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Generic Setup

Dividing the above, 2^nd-order ODE through by the quantity, $[R^2(P_0/P_c)]$, gives,

$\frac{d^{2}x}{d{r}_0^2}+[\frac{4}{r_0}-\frac{1}{R}(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\left)\right(\frac{P_c}{P_0}\left)\right]\frac{dx}{d{r}_0}-\left[\frac{1}{R}\right(\frac{P_c}{P_0}\left)\right(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\left)\frac{\alpha }{r_0}\right]x=-\frac{1}{R^2}\left(\frac{P_c}{P_0}\right)\left(\frac{{\rho }_0}{{\rho }_c}\right)\left[\right(\frac{{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2{\omega }^2}{{\gamma }_g}\left)\right]x,$

which matches Prasad's (1949) equation (1), namely,

$x^{{\text{'}}^{\prime }}+[\frac{4}{r_0}-\frac{\mu (r_0)}{r_0}]x^{\text{'}}-\left[\frac{\alpha \mu (r_0)}{r_0^2}\right]x$

$=$

$-\left(\frac{P_c}{P_0}\right)\left(\frac{{\rho }_0}{{\rho }_c}\right)\left[\frac{n^2{\rho }_c}{{\gamma }_g{P}_c}\right]x,$

where, primes indicate differentiation with respect to $r_0$, and,

$\mu (r_0)\equiv \frac{r_0}{R}\left(\frac{P_c}{P_0}\right)\left(\frac{{\rho }_0}{{\rho }_c}\right)\left(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\right).$

(Note that Prasad's equation has the awkward units of inverse length-squared.) Regrouping terms in Prasad's governing equation, multiplying through by $R^2$ (to make the equation dimensionless), and now letting primes denote differentiation with respect to the dimensionless radial coordinate, ${\chi }_0$, we quite generally can write the linear adiabatic wave equation as,

$-\left(\frac{P_c}{P_0}\right)\left(\frac{{\rho }_0}{{\rho }_c}\right){\sigma }^{2}x$	$=$	$[x^{{\text{'}}^{\prime }}+\frac{4x^{\text{'}}}{{\chi }_0}]-\frac{\mu ({\chi }_0)}{{\chi }_0}[x^{\text{'}}+\frac{\alpha x}{{\chi }_0}]$
	$=$	$\frac{1}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4{x}^{\text{'}}\right)-\frac{\mu ({\chi }_0)}{{\chi }_0}\left[\frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\right({\chi }_0^{\alpha }x\left)\right].$

Defining,

$A$	$\equiv$	$\left(\frac{P_0}{P_c}\right)\left(\frac{{\rho }_c}{{\rho }_0}\right),$
$B$	$\equiv$	$\frac{A\mu ({\chi }_0)}{{\chi }_0}=\left(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\right),$

the governing equation becomes,

${\sigma }^{2}x$ $=$ $\frac{B}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left({\chi }_0^{\alpha }x\right)-\frac{A}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4{x}^{\text{'}}\right)$   $=$ $B[\frac{\alpha x}{{\chi }_0}+x^{\text{'}}]-A[\frac{4x^{\text{'}}}{{\chi }_0}+x^{{\text{'}}^{\prime }}].$

Notice that, because,

$g_0=-\frac{1}{{\rho }_0}\frac{d{P}_0}{d{r}_0},$

at every radial location throughout the configuration, it must also be true that, for any equilibrium configuration,

$B$	$=$	$-(\frac{{\rho }_0}{{\rho }_c}{)}^{-1}\frac{d(P_0/P_c)}{d{\chi }_0}$
$\Rightarrow \frac{B}{A}$	$=$	$-\frac{d}{d{\chi }_0}[\ln (\frac{P_0}{P_c}\left)\right].$

The following table shows that this relationship holds for a collection of analytically described equilibrium structures.

Table 1a: Properties of Analytically Defined Equilibrium Structures
Model	$\frac{{\rho }_0}{{\rho }_c}$	$\frac{P_0}{P_c}$	$\frac{d}{d{\chi }_0}\left(\frac{P_0}{P_c}\right)$
Uniform-density	$1$	$1-{\chi }_0^2$	$-2{\chi }_0$
Linear	$1-{\chi }_0$	$\frac{1}{5}(5-24{\chi }_0^2+28{\chi }_0^3-9{\chi }_0^4)$	$\frac{1}{5}[-48{\chi }_0+84{\chi }_0^2-36{\chi }_0^3]$
Parabolic	$1-{\chi }_0^2$	$\frac{1}{2}(2-5{\chi }_0^2+4{\chi }_0^4-{\chi }_0^6)$	$-5{\chi }_0+8{\chi }_0^3-3{\chi }_0^5$
$n=1$ Polytrope	$\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}$	$[\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}{]}^2$	$\frac{2\sin (\pi {\chi }_0)}{({\pi }^2{\chi }_0^3)}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)]$

Table 1b: Properties of Analytically Defined Equilibrium Structures
Model	$\frac{{\rho }_0}{{\rho }_c}$	$B\equiv \frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}$	$A\equiv \left(\frac{P_0}{P_c}\right)(\frac{{\rho }_0}{{\rho }_c}{)}^{-1}$	$\left(\frac{P_0}{P_c}\right)(\frac{{\rho }_0}{{\rho }_c}{)}^{-2}$
Uniform-density	$1$	$2{\chi }_0$	$1-{\chi }_0^2$	$1-{\chi }_0^2$
Linear	$1-{\chi }_0$	$\frac{48}{5}({\chi }_0-\frac{3}{4}{\chi }_0^2)$	$\frac{1}{5}(1-{\chi }_0)(5+10{\chi }_0-9{\chi }_0^2)$	$(1+2{\chi }_0-\frac{9}{5}{\chi }_0^2)$
Parabolic	$1-{\chi }_0^2$	$5{\chi }_0-3{\chi }_0^3$	$\frac{1}{2}(1-{\chi }_0^2)(2-{\chi }_0^2)$	$(1-\frac{1}{2}{\chi }_0^2)$
$n=1$ Polytrope	$\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}$	$\frac{2}{\pi {\chi }_0^2}[\sin (\pi {\chi }_0)-\pi {\chi }_0\cos (\pi {\chi }_0)]$	$\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}$	$1$

Leaning on this new expression for the ratio, $B/A$, let's play with the form of the governing equation.

$-{\sigma }^{2}x$	$=$	$A\left\{\frac{1}{{\chi }_0^4}\frac{d}{d{\chi }_0}\right({\chi }_0^4{x}^{\text{'}})+\frac{d}{d{\chi }_0}[\ln \left(\frac{P_0}{P_c}\right)]\cdot \frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}({\chi }_0^{\alpha }x\left)\right\}$
	$=$	$A\left(\frac{P_0}{P_c}{)}^{-1}\right\{\left(\frac{P_0}{P_c}\right)\frac{1}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4{x}^{\text{'}}\right)+\frac{d}{d{\chi }_0}\left(\frac{P_0}{P_c}\right)\cdot \frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left({\chi }_0^{\alpha }x\right)\}$

Polytropic Configurations

Let's compare this presentation of the LAWE to the form of the LAWE that has been derived specifically for polytropic equilibrium configurations, namely,

$\frac{d^{2}x}{d{\xi }^2}+\left[\frac{4-(n+1)V(\xi )}{\xi }\right]\frac{dx}{d\xi }+\left[{\omega }^2\right(\frac{a_n^2{\rho }_c}{{\gamma }_g{P}_c})\frac{{\theta }_c}{\theta }-(3-\frac{4}{{\gamma }_g})\cdot \frac{(n+1)V(x)}{{\xi }^2}]x$

$=$

$0,$

where,

$V(\xi )$

$\equiv$

$-\frac{\xi }{(\theta /{\theta }_c)}\frac{d(\theta /{\theta }_c)}{d\xi }=\frac{g_0}{a_n}\left(\frac{a_n^2{\rho }_0}{P_0}\right)\frac{\xi }{(n+1)}.$

[Note that ${\theta }_c=1$ and, therefore for all practical purposes, it can be dropped. This notation was introduced in our separate discussion of the polytropic LAWE in order to make it clear how our derivations have overlapped earlier published work.] Regrouping terms, we have,

$-\left[{\omega }^2\right(\frac{a_n^2{\rho }_c}{{\gamma }_g{P}_c}\left)\frac{{\theta }_c}{\theta }\right]x$	$=$	$\frac{d^{2}x}{d{\xi }^2}+\left[\frac{4-(n+1)V(\xi )}{\xi }\right]\frac{dx}{d\xi }-\left[\frac{\alpha (n+1)V(x)}{{\xi }^2}\right]x$
	$=$	$[\frac{d^{2}x}{d{\xi }^2}+(\frac{4}{\xi }\left)\frac{dx}{d\xi }\right]-\left[\frac{(n+1)V(\xi )}{\xi }\right][\frac{dx}{d\xi }+\frac{\alpha x}{\xi }]$
	$=$	$\frac{1}{{\xi }^4}\frac{d}{d\xi }\left({\xi }^4\frac{dx}{d\xi }\right)-\left[\frac{(n+1)V(\xi )}{\xi }\right]\left[\frac{1}{{\xi }^{\alpha }}\frac{d}{d\xi }\right({\xi }^{\alpha }x\left)\right]$
	$=$	$\frac{1}{{\xi }^4}\frac{d}{d\xi }\left({\xi }^4\frac{dx}{d\xi }\right)+[(n+1)\frac{d\ln (\theta /{\theta }_c)}{d\xi }]\left[\frac{1}{{\xi }^{\alpha }}\frac{d}{d\xi }\right({\xi }^{\alpha }x\left)\right].$

Next we note that, written in terms of the traditional polytropic radial coordinate, $\xi$, the fractional radius,

${\chi }_0\equiv \frac{r_0}{R}=\frac{\xi }{{\xi }_1}=\frac{a_n\xi }{R}.$

Hence, multiplying the polytropic LAWE through by the quantity, $(R/a_n{)}^2$, gives,

$\frac{1}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4\frac{dx}{d{\chi }_0}\right)+[(n+1)\frac{d\ln (\theta /{\theta }_c)}{d{\chi }_0}]\left[\frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\right({\chi }_0^{\alpha }x\left)\right]$	$=$	$-\left[{\omega }^2\right(\frac{R^2{\rho }_c}{{\gamma }_g{P}_c}\left)\frac{{\theta }_c}{\theta }\right]x$
	$=$	$-\left(\frac{{\theta }_c}{\theta }\right){\sigma }^{2}x.$

Finally, noting that, for polytropic configurations,

$\frac{\theta }{{\theta }_c}$

$=$

$\left(\frac{P_0}{P_c}\right)(\frac{{\rho }_0}{{\rho }_c}{)}^{-1}=(\frac{P_0}{P_c}{)}^{1/(n+1)},$

we can rewrite the polytropic LAWE in the form,

$\frac{1}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4\frac{dx}{d{\chi }_0}\right)+\left[\frac{d\ln (P_0/P_c)}{d{\chi }_0}\right]\left[\frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\right({\chi }_0^{\alpha }x\left)\right]$

$=$

$-\left(\frac{P_0}{P_c}{)}^{-1}\right(\frac{{\rho }_0}{{\rho }_c}){\sigma }^{2}x,$

which precisely matches the general expression for the LAWE presented at the end of our generic setup, directly above.

This seems to be a particularly insightful way to write the LAWE, as the only structural functions that appear explicitly are $P_0({\chi }_0)$ and ${\rho }_0({\chi }_0)$. It appears as though the eigenfunctions that describe adiabatic radial pulsations do not explicitly depend a priori on the radial dependence of the equilibrium gravitational acceleration.

Examine Structural Pressure-Density Relation

Derivation

One striking property exhibited by the example configurations tabulated above is the structural relationship between the chosen function, ${\rho }_0({\chi }_0)$, and the corresponding radial pressure distribution, $P_0({\chi }_0)$, that is dictated by,

Hydrostatic Balance

$\frac{1}{\rho }\frac{dP}{dr}=-\frac{d\mathrm{\Phi }}{dr}=-\frac{G{M}_r}{r^2}$ ,

As has been detailed in the last column of Table 1b, in all four cases, the ratio, $(P_0/P_c)({\rho }_0/{\rho }_c{)}^{-2}$, is an analytically prescribed polynomial expression. That is, the pressure is "evenly divisible" by the square of the density. Let's examine how broadly reliable this behavior is. [Note that, for simplicity in typing, hereafter throughout this subsection we will drop the subscript zero and, rather than ${\chi }_0$, we will use $z\equiv r/R$ to denote the dimensionless radial coordinate.]

Assume a mass-distribution given by the general quadratic function,

$\frac{\rho }{{\rho }_c}$

$=$

$1-az-b{z}^2,$

where both coefficients, $a$ and $b$, are positive.

ASIDE: Surface Location

The surface of the configuration will be defined by the radial location, $z_s$, at which the density first goes to zero. If $b=0$, then the surface will be located at $z_s=a^{-1}$; and if $a=0$, it will be located at $z_s=b^{-1/2}$. More generally, however, the roots of the quadratic equation that results from setting $\rho /{\rho }_c$ to zero are, $z_{\pm }$ $=$ $-\frac{a}{2b}[1\mp (1+\frac{4b}{a^2}{)}^{1/2}].$ Because only the $z_{+}$ solution provides positive roots, we conclude that, when both $a$ and $b$ are nonzero, the radial coordinate of the surface is, $z_s=z_{+}$ $=$ $\frac{a}{2b}\left[\right(1+\frac{4b}{a^2}{)}^{1/2}-1].$ We acknowledge, as well, that the density profile can now be written in terms of these roots; specifically, $\frac{\rho }{{\rho }_c}$ $=$ $b(z_{+}-z)(z-z_{-}).$ In the discussion, below, it may be advantageous to adopt the following notation: ${\ell }^2\equiv \frac{4b}{a^2}\Rightarrow \ell =\frac{2b^{1/2}}{a},$ in which case, $a{z}_s=\frac{2}{{\ell }^2}\left[\right(1+{\ell }^2{)}^{1/2}-1]$         and         $b^{1/2}{z}_s=\frac{1}{\ell }\left[\right(1+{\ell }^2{)}^{1/2}-1].$

This specified density profile implies a mass distribution,

$M_r$	$=$	$4\pi R^3{\rho }_c\underset{0}{\overset{z}{\int }}(1-az-b{z}^2)z^{2}dz$
	$=$	$4\pi R^3{\rho }_c(\frac{z^3}{3}-\frac{a{z}^4}{4}-\frac{b{z}^5}{5}).$

The hydrostatic balance condition therefore implies,

$\frac{1}{R{\rho }_c}(\frac{\rho }{{\rho }_c}{)}^{-1}\frac{dP}{dz}$	$=$	$-\frac{G}{R^2{z}^2}\left[4\pi R^3{\rho }_c\right(\frac{z^3}{3}-\frac{a{z}^4}{4}-\frac{b{z}^5}{5}\left)\right]$
$\Rightarrow \left[\frac{1}{4\pi G{R}^2{\rho }_c^2}\right]\frac{dP}{dz}$	$=$	$-(\frac{z}{3}-\frac{a{z}^2}{4}-\frac{b{z}^3}{5})\left(\frac{\rho }{{\rho }_c}\right)$
	$=$	$-(\frac{z}{3}-\frac{a{z}^2}{4}-\frac{b{z}^3}{5})+a(\frac{z^2}{3}-\frac{a{z}^3}{4}-\frac{b{z}^4}{5})+b(\frac{z^3}{3}-\frac{a{z}^4}{4}-\frac{b{z}^5}{5})$
	$=$	$-\left(\frac{1}{3}\right)z+z^2(\frac{a}{4}+\frac{a}{3})+z^3(\frac{b}{5}-\frac{a^2}{4}+\frac{b}{3})+z^4(-\frac{ab}{5}-\frac{ab}{4})-z^5\left(\frac{b^2}{5}\right)$
	$=$	$-\left(\frac{1}{3}\right)z+z^2\left(\frac{7a}{12}\right)+z^3(\frac{8b}{15}-\frac{a^2}{4})-z^4\left(\frac{9ab}{20}\right)-z^5\left(\frac{b^2}{5}\right)$
$\Rightarrow \left(\frac{P}{P_n}\right)$	$=$	$\int [-(\frac{1}{3})z+z^2(\frac{7a}{12})+z^3(\frac{8b}{15}-\frac{a^2}{4})-z^4(\frac{9ab}{20})-z^5(\frac{b^2}{5}\left)\right]dz$
	$=$	$[-(\frac{1}{6})z^2+z^3(\frac{7a}{36})+z^4(\frac{2b}{15}-\frac{a^2}{16})-z^5(\frac{9ab}{100})-z^6(\frac{b^2}{30}\left)\right]+C.$

where, $C$, is an integration constant, and,

$P_n\equiv 4\pi G{\rho }_c^2{R}^2.$

The integration constant — which also proves to be the normalized central pressure — is determined by ensuring that the pressure goes to zero at the surface of the configuration, $z_s$. That is,

$C=\frac{P_c}{P_n}$

$=$

$-[-(\frac{1}{6})z_s^2+z_s^3(\frac{7a}{36})+z_s^4(\frac{2b}{15}-\frac{a^2}{16})-z_s^5(\frac{9ab}{100})-z_s^6(\frac{b^2}{30}\left)\right].$

Hence, the pressure profile is,

$\frac{P}{P_n}$

$=$

$\left(\frac{1}{6}\right)(z_s^2-z^2)-\left(\frac{7a}{36}\right)(z_s^3-z^3)-(\frac{2b}{15}-\frac{a^2}{16})(z_s^4-z^4)+\left(\frac{9ab}{100}\right)(z_s^5-z^5)+\left(\frac{b^2}{30}\right)(z_s^6-z^6).$

Let's check this general expression against the specific cases described above.

Example1

First, let's set $b=0$, but leave $a$ general. As described in the above ASIDE, this means that $z_s=a^{-1}$. So the pressure distribution is,

$\frac{P}{P_n}$	$=$	$\left(\frac{1}{6}\right)(z_s^2-z^2)-\left(\frac{7a}{36}\right)(z_s^3-z^3)+\left(\frac{a^2}{16}\right)(z_s^4-z^4)$
	$=$	$\left(\frac{1}{6}\right)(a^{-2}-z^2)-\left(\frac{7a}{36}\right)(a^{-3}-z^3)+\left(\frac{a^2}{16}\right)(a^{-4}-z^4)$
	$=$	$a^{-2}\{\frac{1}{6}[1-(az{)}^2]-\frac{7}{36}[1-(az{)}^3]+\frac{1}{16}[1-(az{)}^4]\}.$

And from the expression for the integration constant, we have,

$\frac{P_c}{P_n}$	$=$	$\left[\right(\frac{1}{6})a^{-2}-a^{-3}(\frac{7a}{36})+a^{-4}(\frac{a^2}{16}\left)\right]$
	$=$	$a^{-2}[\frac{1}{6}-\frac{7}{36}+\frac{1}{16}]$
	$=$	$\frac{5}{2^4\cdot 3^2{a}^2}.$

Hence, dividing one expression by the other, we obtain,

$\frac{P}{P_c}$

$=$

$\frac{1}{5}\{24[1-(az{)}^2]-28[1-(az{)}^3]+9[1-(az{)}^4]\}.$

Let's check to see if this "general linear" pressure distribution is evenly divisible by the square of the density distribution which, in this case, is,

$\frac{\rho }{{\rho }_c}$

$=$

$1-az.$

Strategically rewriting the expression for the pressure distribution gives,

$\frac{P}{P_c}$	$=$	$\frac{1}{5}\{24[1-(az)][1+(az)]-28[1-(az)][1+(az)+(az{)}^2]+9[1-(az)][1+(az)][1+(az{)}^2]\}$
	$=$	$\frac{1}{5}\left(\frac{\rho }{{\rho }_c}\right)\{24[1+(az)]-28[1+(az)+(az{)}^2]+9[1+(az)][1+(az{)}^2]\}$
	$=$	$\frac{1}{5}\left(\frac{\rho }{{\rho }_c}\right)\{5+24(az)-28[(az)+(az{)}^2]+9[(az)+(az{)}^2+(az{)}^3]\}$
	$=$	$\frac{1}{5}\left(\frac{\rho }{{\rho }_c}\right)\{5+5(az)-19(az{)}^2+9(az{)}^3\}.$

And, as luck would have it, the expression inside the curly braces can be "divided evenly" by the quantity, $[1-(az)]$, one more time. Specifically, the expression becomes,

$\frac{P}{P_c}$

$=$

$\frac{1}{5}(\frac{\rho }{{\rho }_c}{)}^2[5+10(az)-9(az{)}^2].$

Example2

First, let's set $a=0$, but leave $b$ general. As described in the above ASIDE, this means that $z_s=b^{-1/2}$. So the pressure distribution is,

$\frac{P}{P_n}$	$=$	$\left(\frac{1}{6}\right)(z_s^2-z^2)-\left(\frac{2b}{15}\right)(z_s^4-z^4)+\left(\frac{b^2}{30}\right)(z_s^6-z^6)$
	$=$	$\left(\frac{1}{6}\right)(b^{-1}-z^2)-\left(\frac{2b}{15}\right)(b^{-2}-z^4)+\left(\frac{b^2}{30}\right)(b^{-3}-z^6)$
	$=$	$\frac{1}{b}\left\{\right(\frac{1}{6})[1-b{z}^2]-(\frac{2}{15})[1-(b{z}^2{)}^2]+\left(\frac{1}{30}\right)[1-(b{z}^2{)}^3]\}.$

And from the expression for the integration constant, we have,

$\frac{P_c}{P_n}$	$=$	$\left[\right(\frac{1}{6})z_s^2-z_s^4(\frac{2b}{15})+z_s^6(\frac{b^2}{30}\left)\right]$
	$=$	$\frac{1}{b}[\frac{1}{6}-\frac{2}{15}+\frac{1}{30}]$
	$=$	$\frac{1}{3\cdot 5b}.$

Hence, dividing one expression by the other, we obtain,

$\frac{P}{P_c}$

$=$

$\frac{1}{2}\{5[1-b{z}^2]-4[1-(b{z}^2{)}^2]+[1-(b{z}^2{)}^3]\}.$

Let's check to see if this "general parabolic" pressure distribution is evenly divisible by the square of the density distribution which, in this case, is,

$\frac{\rho }{{\rho }_c}$

$=$

$1-b{z}^2.$

Strategically rewriting the expression for the pressure distribution gives,

$\frac{P}{P_c}$	$=$	$\frac{1}{2}\{5[1-(b{z}^2)]-4[1-(b{z}^2)][1+(b{z}^2)]+[1-(b{z}^2)][1+(b{z}^2)+(b{z}^2{)}^2]\}$
	$=$	$\frac{1}{2}\left(\frac{\rho }{{\rho }_c}\right)\{5-4[1+(b{z}^2)]+[1+(b{z}^2)+(b{z}^2{)}^2]\}$
	$=$	$\frac{1}{2}\left(\frac{\rho }{{\rho }_c}\right)[2-3(b{z}^2)+(b{z}^2{)}^2]$

Again, as luck would have it, the expression inside the square brackets can be "divided evenly" by the quantity, $[1-(b{z}^2)]$, one more time. Specifically, the expression becomes,

$\frac{P}{P_c}$

$=$

$\frac{1}{2}(\frac{\rho }{{\rho }_c}{)}^2[2-(b{z}^2)].$

Example3

In the most general quadratic case, we should rewrite the pressure distribution as,

$\frac{P}{P_n}$	$=$	$\frac{1}{2^4\cdot 3^2\cdot 5^2}\{2^3\cdot 3\cdot 5^2(z_s^2-z^2)-2^2\cdot 5^2\cdot 7a(z_s^3-z^3)-[2^5\cdot 3\cdot 5b-3^2\cdot 5^2{a}^2](z_s^4-z^4)+2^2\cdot 3^{4}ab(z_s^5-z^5)+2^3\cdot 3\cdot 5b^2(z_s^6-z^6)\}$
	$=$	$\frac{z_s^2}{2^4\cdot 3^2\cdot 5^2}\{600(1-{\zeta }^2)-700(a{z}_s)(1-{\zeta }^3)-[480(b{z}_s^2)-225(a{z}_s{)}^2](1-{\zeta }^4)+324(a{z}_s)(b{z}_s^2)(1-{\zeta }^5)+120(b{z}_s^2{)}^2(1-{\zeta }^6)\},$

where, $\zeta \equiv z/z_s$. Similarly, let's rewrite the integration constant as,

$\frac{P_c}{P_n}$

$=$

$\frac{z_s^2}{2^4\cdot 3^2\cdot 5^2}\{600-700(a{z}_s)-[480(b{z}_s^2)-225(a{z}_s{)}^2]+324(a{z}_s)(b{z}_s^2)+120(b{z}_s^2{)}^2\}.$

So the pressure can be written as,

$\left[\frac{2^4\cdot 3^2\cdot 5^2}{z_s^2}\right]\left[\frac{P}{P_n}\right]$

$=$

$\left[\frac{2^4\cdot 3^2\cdot 5^2}{z_s^2}\right]\left[\frac{P_c}{P_n}\right]-600{\zeta }^2+700(a{z}_s){\zeta }^3+[480(b{z}_s^2)-225(a{z}_s{)}^2]{\zeta }^4-324(a{z}_s)(b{z}_s^2){\zeta }^5-120(b{z}_s^2{)}^2{\zeta }^6.$

The question that remains to be answered is: Is this expression for the pressure distribution "evenly divisible" by the square (or even the first power) of the normalized density distribution which, as defined above for the general quadratic case, is,

$\frac{\rho }{{\rho }_c}=1-az-b{z}^2=1-(a{z}_s)\zeta -(b{z}_s^2){\zeta }^2.$

In attempting to answer this question, it may prove advantageous to refer back to the above ASIDE discussion of the roots of this quadratic function and, in particular, that,

$a{z}_s=\frac{2}{{\ell }^2}\left[\right(1+{\ell }^2{)}^{1/2}-1]$         and         $b^{1/2}{z}_s=\frac{1}{\ell }\left[\right(1+{\ell }^2{)}^{1/2}-1],$

where, ${\ell }^2\equiv 4b/a^2$.

Uniform Density

In the case of a uniform-density configuration, the governing equation is,

${\sigma }^{2}x$

$=$

$2{\chi }_0[\frac{\alpha x}{{\chi }_0}+x^{\text{'}}]-(1-{\chi }_0^2)[\frac{4x^{\text{'}}}{{\chi }_0}+x^{{\text{'}}^{\prime }}],$

where,

${\sigma }^2\equiv \frac{{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2{\omega }^2}{{\gamma }_g}=\frac{6}{{\gamma }_g}\left[\frac{{\omega }^2}{4\pi G\overline{\rho }}\right].$

The following individual mode analyses should be compared with the results found in our discussion of Sterne's general solution.

Mode 0

Try an eigenfunction of the form,

$x=a_0,$

in which case,

$x^{\text{'}}=x^{{\text{'}}^{\prime }}=0.$

In order for this to be a solution, we must have,

${\sigma }^2{a}_0$	$=$	$2{\chi }_0\left[\frac{\alpha a_0}{{\chi }_0}\right]$
$\Rightarrow \frac{6}{{\gamma }_g}\left[\frac{{\omega }^2}{4\pi G\overline{\rho }}\right]$	$=$	$2\alpha =2(3-\frac{4}{{\gamma }_g})$
$\Rightarrow \frac{{\omega }^2}{4\pi G\overline{\rho }}$	$=$	${\gamma }_g-\frac{4}{3}.$

Mode 2

Try an eigenfunction of the form,

$x=a_0+a_2{\chi }_0^2,$

in which case,

$x^{\text{'}}=2a_2{\chi }_0$        and         $x^{{\text{'}}^{\prime }}=2a_2.$

In order for this to be a solution, we must have,

${\sigma }^2(a_0+a_2{\chi }_0^2)$	$=$	$2[\alpha (a_0+a_2{\chi }_0^2)+{\chi }_0(2a_2{\chi }_0)]-(1-{\chi }_0^2)[\frac{4(2a_2{\chi }_0)}{{\chi }_0}+2a_2]$
	$=$	$2\alpha a_0+{\chi }_0^2[2a_2(2+\alpha )]-10a_2(1-{\chi }_0^2)$
$\Rightarrow {\sigma }^2{a}_0-2\alpha a_0+10a_2$	$=$	${\chi }_0^2[-{\sigma }^2+2(2+\alpha )+10]a_2.$

Given that the coefficients on both sides of this expression must independently be zero, we have:

${\sigma }^2$	$=$	$2(2+\alpha )+10$
$\Rightarrow \frac{{\omega }^2}{4\pi G\overline{\rho }}$	$=$	$\frac{{\gamma }_g}{6}[14+2(3-\frac{4}{{\gamma }_g}\left)\right]$
	$=$	$\frac{{\gamma }_g}{6}[20-\frac{8}{{\gamma }_g}]=\frac{1}{3}(10{\gamma }_g-4),$

and

$\frac{a_2}{a_0}$	$=$	$\frac{1}{10}[2\alpha -{\sigma }^2]$
	$=$	$\frac{1}{10}\{2\alpha -[14+2\alpha )]\}=-\frac{7}{5}.$

Parabolic Density Distribution

In the case of a parabolic density distribution, the governing equation is,

${\sigma }^{2}x$

$=$

$(5{\chi }_0-3{\chi }_0^3)[\frac{\alpha x}{{\chi }_0}+x^{\text{'}}]-\frac{1}{2}(1-{\chi }_0^2)(2-{\chi }_0^2)[\frac{4x^{\text{'}}}{{\chi }_0}+x^{{\text{'}}^{\prime }}],$

where,

${\sigma }^2\equiv \frac{{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2{\omega }^2}{{\gamma }_g}=\frac{15}{{\gamma }_g}\left[\frac{{\omega }^2}{4\pi G\overline{\rho }}\right].$

First Trial

Try an eigenfunction of the form,

$x=(2-{\chi }_0^2{)}^{-1}(a+b{\chi }_0^2+c{\chi }_0^4),$

in which case,

$x^{\text{'}}$	$=$	$(2-{\chi }_0^2{)}^{-1}(2b{\chi }_0+4c{\chi }_0^3)+2{\chi }_0(2-{\chi }_0^2{)}^{-2}(a+b{\chi }_0^2+c{\chi }_0^4)$
	$=$	$(2-{\chi }_0^2{)}^{-2}[(2-{\chi }_0^2)(2b{\chi }_0+4c{\chi }_0^3)+2{\chi }_0(a+b{\chi }_0^2+c{\chi }_0^4)]$
	$=$	$2{\chi }_0(2-{\chi }_0^2{)}^{-2}[(2-{\chi }_0^2)(b+2c{\chi }_0^2)+(a+b{\chi }_0^2+c{\chi }_0^4)]$
	$=$	$2{\chi }_0(2-{\chi }_0^2{)}^{-2}[(2b+4c{\chi }_0^2-b{\chi }_0^2-2c{\chi }_0^4)+(a+b{\chi }_0^2+c{\chi }_0^4)]$
	$=$	$2{\chi }_0(2-{\chi }_0^2{)}^{-2}[(a+2b)+4c{\chi }_0^2-c{\chi }_0^4];$

and,

$x^{{\text{'}}^{\prime }}$	$=$	$(2-{\chi }_0^2{)}^{-1}(2b+12c{\chi }_0^2)+2{\chi }_0(2-{\chi }_0^2{)}^{-2}(2b{\chi }_0+4c{\chi }_0^3)+2(2-{\chi }_0^2{)}^{-2}(a+b{\chi }_0^2+c{\chi }_0^4)$
		$+2{\chi }_0(a+b{\chi }_0^2+c{\chi }_0^4)[-2(2-{\chi }_0^2{)}^{-3}(-2{\chi }_0)]+2{\chi }_0(2-{\chi }_0^2{)}^{-2}(2b{\chi }_0+4c{\chi }_0^3)$
	$=$	$(2-{\chi }_0^2{)}^{-3}[2(4-4{\chi }_0^2+{\chi }_0^4)(b+6c{\chi }_0^2)+4{\chi }_0^2(2-{\chi }_0^2)(b+2c{\chi }_0^2)+2(2-{\chi }_0^2)(a+b{\chi }_0^2+c{\chi }_0^4)$
		$+8{\chi }_0^2(a+b{\chi }_0^2+c{\chi }_0^4)+4{\chi }_0^2(2-{\chi }_0^2)(b+2c{\chi }_0^2)]$
	$=$	$2(2-{\chi }_0^2{)}^{-3}\{(4b+24c{\chi }_0^2-4b{\chi }_0^2-24c{\chi }_0^4+b{\chi }_0^4+6c{\chi }_0^6)+(8b{\chi }_0^2+16c{\chi }_0^4-4b{\chi }_0^4-8c{\chi }_0^6)$
		$+(2a+2b{\chi }_0^2+2c{\chi }_0^4-a{\chi }_0^2-b{\chi }_0^4-c{\chi }_0^6)+(4a{\chi }_0^2+4b{\chi }_0^4+4c{\chi }_0^6)\}$
	$=$	$2(2-{\chi }_0^2{)}^{-3}\{(4b+2a)+{\chi }_0^2(24c-4b+8b+2b-a+4a)+{\chi }_0^4(-24c+b+16c-4b+2c-b+4b)+{\chi }_0^6(6c-8c-c+4c)\}$
	$=$	$2(2-{\chi }_0^2{)}^{-3}\{(2a+4b)+{\chi }_0^2(3a+6b+24c)+{\chi }_0^4(-6c)+{\chi }_0^6(c)\}$

In order for this to be a solution, we must have,

${\sigma }^{2}x$	$=$	$(5-3{\chi }_0^2)[\alpha x+{\chi }_0{x}^{\text{'}}]-(1-{\chi }_0^2)(2-{\chi }_0^2)[\frac{2x^{\text{'}}}{{\chi }_0}+\frac{x^{{\text{'}}^{\prime }}}{2}]$
$\Rightarrow {\sigma }^2(2-{\chi }_0^2{)}^{-1}(a+b{\chi }_0^2+c{\chi }_0^4)$	$=$	$(5-3{\chi }_0^2)[\alpha (2-{\chi }_0^2{)}^{-1}(a+b{\chi }_0^2+c{\chi }_0^4)+2{\chi }_0^2(2-{\chi }_0^2{)}^{-2}[(a+2b)+4c{\chi }_0^2-c{\chi }_0^4]]$
		$-(1-{\chi }_0^2)[4(2-{\chi }_0^2{)}^{-1}[(a+2b)+4c{\chi }_0^2-c{\chi }_0^4]+(2-{\chi }_0^2{)}^{-2}\{(2a+4b)+{\chi }_0^2(3a+6b+24c)-6c{\chi }_0^4+c{\chi }_0^6\}].$

Multiplying through by $(2-{\chi }_0^2{)}^2$ gives,

${\sigma }^2(2-{\chi }_0^2)(a+b{\chi }_0^2+c{\chi }_0^4)$	$=$	$(5-3{\chi }_0^2)\{\alpha (2-{\chi }_0^2)(a+b{\chi }_0^2+c{\chi }_0^4)+2{\chi }_0^2[(a+2b)+4c{\chi }_0^2-c{\chi }_0^4]\}$
		$-(1-{\chi }_0^2)\{4(2-{\chi }_0^2)[(a+2b)+4c{\chi }_0^2-c{\chi }_0^4]+[(2a+4b)+{\chi }_0^2(3a+6b+24c)-6c{\chi }_0^4+c{\chi }_0^6]\}$
$\Rightarrow {\sigma }^2[2a+{\chi }_0^2(2b-a)+{\chi }_0^4(2c-b)-c{\chi }_0^6]$	$=$	$(5-3{\chi }_0^2)\{\alpha [2a+{\chi }_0^2(2b-a)+{\chi }_0^4(2c-b)-c{\chi }_0^6]\}$
		$+(5-3{\chi }_0^2)\{(2a+4b){\chi }_0^2+8c{\chi }_0^4-2c{\chi }_0^6\}$
		$-(1-{\chi }_0^2)\{[8(a+2b)+32c{\chi }_0^2-8c{\chi }_0^4]-[4{\chi }_0^2(a+2b)+16c{\chi }_0^4-4c{\chi }_0^6]\}$
		$-(1-{\chi }_0^2)\{(2a+4b)+{\chi }_0^2(3a+6b+24c)-6c{\chi }_0^4+c{\chi }_0^6\}$
	$=$	$(5-3{\chi }_0^2)\{2a\alpha +{\chi }_0^2[(2b-a)\alpha +(2a+4b)]+{\chi }_0^4[(2c-b)\alpha +8c]-(2+\alpha )c{\chi }_0^6\}$
		$-(1-{\chi }_0^2)\{(10a+20b)+{\chi }_0^2[56c-a-2b]+{\chi }_0^4[-30c]+5c{\chi }_0^6\}$

So, the coefficients of each even power of ${\chi }_0^n$ are:

${\chi }_0^0$	:	$2a{\sigma }^2-10a\alpha +10a+20b=a(2{\sigma }^2-10\alpha +10)+20b$
${\chi }_0^2$	:	$(2b-a){\sigma }^2-5[(2b-a)\alpha +(2a+4b)]+6a\alpha +[56c-a-2b]-(10a+20b)$ $=a[-{\sigma }^2-5(-\alpha +2)+6\alpha -1-10]+b[2{\sigma }^2-10\alpha -20-2-20]+c[56]$ $=a[-{\sigma }^2+11\alpha -21]+b[2{\sigma }^2-10\alpha -42]+56c$
${\chi }_0^4$	:	$(2c-b){\sigma }^2-5[(2c-b)\alpha +8c]-30c+3[(2b-a)\alpha +(2a+4b)]-[56c-a-2b]$ $=a[-3\alpha +7]+b[-{\sigma }^2+5\alpha +6\alpha +12+2]+c[2{\sigma }^2-10\alpha -40-30-56]$ $=a[7-3\alpha ]+b[11\alpha -{\sigma }^2+14]+c[2{\sigma }^2-10\alpha -126]$
${\chi }_0^6$	:	$-c{\sigma }^2+(10+5\alpha )c+3[(2c-b)\alpha +8c]+5c+30c$ $=b[-3\alpha ]+c[-{\sigma }^2+10+5\alpha +6\alpha +24+35]$ $=b[-3\alpha ]+c[11\alpha -{\sigma }^2+69]$
${\chi }_0^8$	:	$-3(2+\alpha )c-5c=c[-11-3\alpha ]$

Independent Investigation of Parabolic Distribution

In the specific case of a parabolic density distribution, the leading factor on the LHS is,

$\frac{1}{A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}\equiv \left(\frac{P_c}{P_0}\right)\left(\frac{{\rho }_0}{{\rho }_c}\right)$

$=$

$\frac{(1-{\chi }_0^2)}{(1-{\chi }_0^2{)}^2(1-\frac{1}{2}{\chi }_0^2)}=\frac{1}{(1-{\chi }_0^2)(1-\frac{1}{2}{\chi }_0^2)}=\frac{2}{2-3{\chi }_0^2+{\chi }_0^4},$

and the function appearing on the RHS is,

$\mu ({\chi }_0)$

$=$

$\frac{{\chi }_0^2(1-{\chi }_0^2)(5-3{\chi }_0^2)}{(1-{\chi }_0^2{)}^2(1-\frac{1}{2}{\chi }_0^2)}=\frac{{\chi }_0^2(5-3{\chi }_0^2)}{A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}.$

Multiplying the linear adiabatic wave equation through by $A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}$, gives,

$-{\sigma }^{2}x$

$=$

$\frac{A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4{x}^{\text{'}}\right)-\frac{B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left({\chi }_0^{\alpha }x\right),$

where,

$B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}\equiv {\chi }_0(5-3{\chi }_0^2).$

Now we note that,

$\frac{d}{d{\chi }_0}\left[A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}\right]$	$=$	$\frac{A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left[{\chi }_0^4{x}^{\text{'}}\right]+{\chi }_0^4{x}^{\text{'}}\frac{d}{d{\chi }_0}\left[\frac{A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^4}\right]$
$\Rightarrow \frac{A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left[{\chi }_0^4{x}^{\text{'}}\right]$	$=$	$\frac{d}{d{\chi }_0}\left[A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}\right]-{\chi }_0^4{x}^{\text{'}}\frac{d}{d{\chi }_0}\left[\frac{A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^4}\right]$
	$=$	$\frac{d}{d{\chi }_0}\left[A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}\right]-\frac{{\chi }_0^4{x}^{\text{'}}}{2}\frac{d}{d{\chi }_0}[1-\frac{3}{{\chi }_0^2}+\frac{2}{{\chi }_0^4}]$
	$=$	$\frac{d}{d{\chi }_0}\left[A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}\right]-{\chi }_0^4{x}^{\text{'}}[\frac{3}{{\chi }_0^3}-\frac{4}{{\chi }_0^5}]$
	$=$	$\frac{d}{d{\chi }_0}\left[A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}\right]+\frac{x^{\text{'}}}{{\chi }_0}(4-3{\chi }_0^2).$

Similarly we note that,

$\frac{d}{d{\chi }_0}\left[B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right]$	$=$	$\frac{B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left[{\chi }_0^{\alpha }x\right]+{\chi }_0^{\alpha }x\frac{d}{d{\chi }_0}\left[\frac{B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^{\alpha }}\right]$
$\Rightarrow \frac{B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left[{\chi }_0^{\alpha }x\right]$	$=$	$\frac{d}{d{\chi }_0}\left[B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right]-{\chi }_0^{\alpha }x\frac{d}{d{\chi }_0}\left[\frac{B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}}{{\chi }_0^{\alpha }}\right]$
	$=$	$\frac{d}{d{\chi }_0}\left[B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right]-{\chi }_0^{\alpha }x\frac{d}{d{\chi }_0}[5{\chi }_0^{1-\alpha }-3{\chi }_0^{3-\alpha }]$
	$=$	$\frac{d}{d{\chi }_0}\left[B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right]-{\chi }_0^{\alpha }x[5(1-\alpha ){\chi }_0^{-\alpha }-3(3-\alpha ){\chi }_0^{2-\alpha }]$
	$=$	$\frac{d}{d{\chi }_0}\left[B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right]-x[5(1-\alpha )-3(3-\alpha ){\chi }_0^2].$

Hence, the LAWE can be rewritten as,

$-{\sigma }^{2}x$

$=$

$\frac{d}{d{\chi }_0}\left[A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}\right]+\frac{x^{\text{'}}}{{\chi }_0}(4-3{\chi }_0^2)-\frac{d}{d{\chi }_0}\left[B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right]+x[5(1-\alpha )-3(3-\alpha ){\chi }_0^2];$

then multiplying through by ${\chi }_0$, and rearranging terms gives,

$-x\{[5(1-\alpha )+{\sigma }^2]{\chi }_0-3(3-\alpha ){\chi }_0^3\}$

$=$

${\chi }_0\frac{d}{d{\chi }_0}[A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}-B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x]+x^{\text{'}}(4-3{\chi }_0^2).$

Next, we note that,

$A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}$	$=$	$\frac{d}{d{\chi }_0}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)-x\left[\frac{d}{d{\chi }_0}\right(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}\left)\right]$
	$=$	$\frac{d}{d{\chi }_0}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)-\frac{x}{2}\left[\frac{d}{d{\chi }_0}\right(2-3{\chi }_0^2+{\chi }_0^4\left)\right]$
	$=$	$\frac{d}{d{\chi }_0}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)+x(3{\chi }_0-2{\chi }_0^3)$
$\Rightarrow A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}{x}^{\text{'}}-B_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x$	$=$	$\frac{d}{d{\chi }_0}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)+[(3{\chi }_0-2{\chi }_0^3)-(5{\chi }_0-3{\chi }_0^3)]x$
	$=$	$\frac{d}{d{\chi }_0}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)-(2{\chi }_0-{\chi }_0^3)x.$

So, the LAWE becomes,

$-x\{[5(1-\alpha )+{\sigma }^2]{\chi }_0-3(3-\alpha ){\chi }_0^3\}$	$=$	${\chi }_0\frac{d}{d{\chi }_0}\left[\frac{d}{d{\chi }_0}\right(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x)-(2{\chi }_0-{\chi }_0^3)x]+x^{\text{'}}(4-3{\chi }_0^2)$
	$=$	${\chi }_0\frac{d^2}{d{\chi }_0^2}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)-{\chi }_0\frac{d}{d{\chi }_0}[(2{\chi }_0-{\chi }_0^3)x]+x^{\text{'}}(4-3{\chi }_0^2)$
	$=$	${\chi }_0\frac{d^2}{d{\chi }_0^2}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)-\left\{\frac{d}{d{\chi }_0}\right[(2{\chi }_0^2-{\chi }_0^4)x]-(2{\chi }_0-{\chi }_0^3)x\}+\left\{\frac{d}{d{\chi }_0}\right[(4-3{\chi }_0^2)x]+6{\chi }_{0}x\}$
	$=$	${\chi }_0\frac{d^2}{d{\chi }_0^2}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)+\frac{d}{d{\chi }_0}[(4-3{\chi }_0^2)x-(2{\chi }_0^2-{\chi }_0^4)x]+(2{\chi }_0-{\chi }_0^3)x+6{\chi }_{0}x$
	$=$	${\chi }_0\frac{d^2}{d{\chi }_0^2}\left(A_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{b}}x\right)+\frac{d}{d{\chi }_0}[(4-5{\chi }_0^2+{\chi }_0^4)x]+(8{\chi }_0-{\chi }_0^3)x.$

Moving the last term on the RHS of this expression to the LHS, and factoring the polynomial coefficients of the terms inside of the first and second derivatives gives,

$x\{(5\alpha -13-{\sigma }^2){\chi }_0+(10-3\alpha ){\chi }_0^3\}$

$=$

$\frac{{\chi }_0}{2}\frac{d^2}{d{\chi }_0^2}[(1-{\chi }_0^2)(2-{\chi }_0^2)x]+\frac{d}{d{\chi }_0}[(1-{\chi }_0^2)(4-{\chi }_0^2)x].$

First Trial (Same as Above)

Try an eigenfunction of the form,

$x=(2-{\chi }_0^2{)}^{-1}(a+b{\chi }_0^2+c{\chi }_0^4),$

in which case,

LHS	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{(a+b{\chi }_0^2+c{\chi }_0^4)(2-{\chi }_0^2)[(5\alpha -13-{\sigma }^2)+(10-3\alpha ){\chi }_0^2]\}$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}(a+b{\chi }_0^2+c{\chi }_0^4)\{(10\alpha -26-2{\sigma }^2)+({\sigma }^2-11\alpha +33){\chi }_0^2+(3\alpha -10){\chi }_0^4\}$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{a(10\alpha -26-2{\sigma }^2)+{\chi }_0^2[a({\sigma }^2-11\alpha +33)+b(10\alpha -26-2{\sigma }^2)]$
		$+{\chi }_0^4[a(3\alpha -10)+b({\sigma }^2-11\alpha +33)+c(10\alpha -26-2{\sigma }^2)]+{\chi }_0^6[b(3\alpha -10)+c({\sigma }^2-11\alpha +33)]+c(3\alpha -10){\chi }_0^8\}$

RHS (1st term)	$=$	$\frac{{\chi }_0}{2}\frac{d^2}{d{\chi }_0^2}[(1-{\chi }_0^2)(a+b{\chi }_0^2+c{\chi }_0^4)]$
	$=$	$\frac{{\chi }_0}{2}\frac{d^2}{d{\chi }_0^2}[a+(b-a){\chi }_0^2+(c-b){\chi }_0^4-c{\chi }_0^6]$
	$=$	$\frac{{\chi }_0}{2}\frac{d}{d{\chi }_0}[2(b-a){\chi }_0+4(c-b){\chi }_0^3-6c{\chi }_0^5]$
	$=$	$\frac{{\chi }_0}{2}[2(b-a)+12(c-b){\chi }_0^2-30c{\chi }_0^4]$
	$=$	$(b-a){\chi }_0+6(c-b){\chi }_0^3-15c{\chi }_0^5$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{(4-4{\chi }_0^2+{\chi }_0^4)[(b-a)+6(c-b){\chi }_0^2-15c{\chi }_0^4]\}$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{[(4b-4a)+(24c-24b){\chi }_0^2-60c{\chi }_0^4]+[(-4b+4a){\chi }_0^2+(-24c+24b){\chi }_0^4+60c{\chi }_0^6]+[(b-a){\chi }_0^4+6(c-b){\chi }_0^6-15c{\chi }_0^8]\}$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{(4b-4a)+[(24c-24b)+(-4b+4a)]{\chi }_0^2+[-60c+(-24c+24b)+(b-a)]{\chi }_0^4+[60c+6(c-b)]{\chi }_0^6-15c{\chi }_0^8\}$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{(4b-4a)+[24c-28b+4a]{\chi }_0^2+[-84c+25b-a]{\chi }_0^4+[66c-6b]{\chi }_0^6-15c{\chi }_0^8\}$

RHS (2nd term)	$=$	$\frac{d}{d{\chi }_0}[(1-{\chi }_0^2)(4-{\chi }_0^2)(2-{\chi }_0^2{)}^{-1}(a+b{\chi }_0^2+c{\chi }_0^4)]$
	$=$	$(2-{\chi }_0^2{)}^{-1}\frac{d}{d{\chi }_0}[(4-5{\chi }_0^2+{\chi }_0^4)(a+b{\chi }_0^2+c{\chi }_0^4)]$
		$+(4-5{\chi }_0^2+{\chi }_0^4)(a+b{\chi }_0^2+c{\chi }_0^4)\frac{d}{d{\chi }_0}[(2-{\chi }_0^2{)}^{-1}]$
	$=$	$(2-{\chi }_0^2{)}^{-1}\frac{d}{d{\chi }_0}[4a+{\chi }_0^2(4b-5a)+{\chi }_0^4(4c-5b+a)+{\chi }_0^6(b-5c)+c{\chi }_0^8]$
		$+[4a+{\chi }_0^2(4b-5a)+{\chi }_0^4(4c-5b+a)+{\chi }_0^6(b-5c)+c{\chi }_0^8]\frac{d}{d{\chi }_0}[(2-{\chi }_0^2{)}^{-1}]$
	$=$	$(2-{\chi }_0^2{)}^{-2}\{(2-{\chi }_0^2)[{\chi }_0(8b-10a)+{\chi }_0^3(16c-20b+4a)+{\chi }_0^5(6b-30c)+8c{\chi }_0^7]$
		$+2{\chi }_0[4a+{\chi }_0^2(4b-5a)+{\chi }_0^4(4c-5b+a)+{\chi }_0^6(b-5c)+c{\chi }_0^8]\}$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{[(16b-20a)+{\chi }_0^2(32c-40b+8a)+{\chi }_0^4(12b-60c)+16c{\chi }_0^6]$
		$-[{\chi }_0^2(8b-10a)+{\chi }_0^4(16c-20b+4a)+{\chi }_0^6(6b-30c)+8c{\chi }_0^8]$
		$+[8a+{\chi }_0^2(8b-10a)+{\chi }_0^4(8c-10b+2a)+{\chi }_0^6(2b-10c)+2c{\chi }_0^8]\}$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{[(16b-20a)+8a]+{\chi }_0^2[(32c-40b+8a)+(10a-8b)+(8b-10a)]$
		$+{\chi }_0^4[(12b-60c)+(20b-4a-16c)+(8c-10b+2a)]+{\chi }_0^6[16c+(30c-6b)+(2b-10c)]-6c{\chi }_0^8\}$
	$=$	${\chi }_0(2-{\chi }_0^2{)}^{-2}\{[16b-12a]+{\chi }_0^2[(32c-40b+8a)]+{\chi }_0^4[-2a+22b-68c]+{\chi }_0^6[36c-4b]-6c{\chi }_0^8\}$

So, the coefficients of each even power of ${\chi }_0^n$ are:

${\chi }_0^0$	:	$a(10\alpha -26-2{\sigma }^2)-(4b-4a)-[16b-12a]=a(10\alpha -10-2{\sigma }^2)-20b$
${\chi }_0^2$	:	$[a({\sigma }^2-11\alpha +33)+b(10\alpha -26-2{\sigma }^2)]-[24c-28b+4a]-[(32c-40b+8a)]$ $=[a({\sigma }^2-11\alpha +21)+b(10\alpha +42-2{\sigma }^2)]-56c$
${\chi }_0^4$	:	$[a(3\alpha -10)+b({\sigma }^2-11\alpha +33)+c(10\alpha -26-2{\sigma }^2)]-[-84c+25b-a]-[-2a+22b-68c]$ $=[a(3\alpha -7)+b({\sigma }^2-11\alpha -14)+c(10\alpha +126-2{\sigma }^2)]$
${\chi }_0^6$	:	$[b(3\alpha -10)+c({\sigma }^2-11\alpha +33)]-[66c-6b]-[36c-4b]$ $=[b(3\alpha )+c({\sigma }^2-11\alpha -69)]$
${\chi }_0^8$	:	$c(3\alpha -10)+15c+6c=c[3\alpha +11]$

The expressions for the coefficients presented in this table exactly match the entire set of expressions derived earlier, except for the adopted sign convention — every term in this second derivation has the opposite sign to the corresponding term in the earlier derivation.

Conjecture

Returning to the generic formulation derived earlier, we have,

$-{\sigma }^2\left(\frac{{\rho }_0}{{\rho }_c}\right)x$

$=$

$\left(\frac{P_0}{P_c}\right)\frac{1}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4{x}^{\text{'}}\right)+\frac{d}{d{\chi }_0}\left(\frac{P_0}{P_c}\right)\cdot \frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left({\chi }_0^{\alpha }x\right).$

Now, suppose that the expression on the RHS is of the form,

$\mathrm{R}\mathrm{H}\mathrm{S}$

$=$

$udv+vdu,$

where $u\equiv (P_0/P_c)$? Then the function,

$v\equiv \frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left({\chi }_0^{\alpha }x\right),$

and the eigenvector, $x$, must satisfy both of the relations:

Relation I	$:$	$-{\sigma }^2\left(\frac{{\rho }_0}{{\rho }_c}\right)x=\frac{d}{d{\chi }_0}\left[\right(\frac{P_0}{P_c}\left)\frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\right({\chi }_0^{\alpha }x\left)\right]$
Relation II	$:$	$\frac{d}{d{\chi }_0}\left[\frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\right({\chi }_0^{\alpha }x\left)\right]=\frac{1}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4{x}^{\text{'}}\right)$

The validity (or not) of this conjecture can be tested against both configurations whose --- ABANDON!

Exploration

Compare LAWE to Hydrostatic Balance Condition

Returning to the generic formulation derived earlier, we have,

$-{\sigma }^2\left(\frac{{\rho }_0}{{\rho }_c}\right)x$

$=$

$\left(\frac{P_0}{P_c}\right)\frac{1}{{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4{x}^{\text{'}}\right)+\frac{d}{d{\chi }_0}\left(\frac{P_0}{P_c}\right)\cdot \frac{1}{{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left({\chi }_0^{\alpha }x\right).$

Dividing this entire expression through by $(P_0/P_c)x$ gives,

$-{\sigma }^2\left(\frac{{\rho }_0}{{\rho }_c}\right)(\frac{P_0}{P_c}{)}^{-1}$	$=$	$\frac{1}{x{\chi }_0^4}\frac{d}{d{\chi }_0}\left({\chi }_0^4{x}^{\text{'}}\right)+\frac{d}{d{\chi }_0}[\ln (\frac{P_0}{P_c}\left)\right]\cdot \frac{1}{x{\chi }_0^{\alpha }}\frac{d}{d{\chi }_0}\left({\chi }_0^{\alpha }x\right).$
	$=$	$\left(\frac{x^{\text{'}}}{x}\right)\cdot \frac{d}{d{\chi }_0}[\ln ({\chi }_0^4{x}^{\text{'}}\left)\right]+\frac{d}{d{\chi }_0}[\ln (\frac{P_0}{P_c}\left)\right]\cdot \frac{d}{d{\chi }_0}[\ln ({\chi }_0^{\alpha }x\left)\right].$

Now, let's step aside from the LAWE and look directly at the differential relationship between the mass-density and the pressure, as dictated by combining the two principal governing relations, the

Hydrostatic Balance

$\frac{1}{\rho }\frac{dP}{dr}=-\frac{d\mathrm{\Phi }}{dr}$ ,

and,

Poisson Equation

$\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\mathrm{\Phi }}{dr}\right)=4\pi G\rho$ .

In combination, we have,

$-4\pi G{\rho }_0$	$=$	$\frac{1}{r_0^2}\frac{d}{d{r}_0}\left[\frac{r_0^2}{{\rho }_0}\frac{d{P}_0}{d{r}_0}\right]$
$\Rightarrow -4\pi G{\rho }_0\left(\frac{R^2{\rho }_c}{P_c}\right)$	$=$	$\frac{1}{{\chi }_0^2}\frac{d}{d{\chi }_0}[\frac{{\chi }_0^2}{({\rho }_0/{\rho }_c)}\cdot \frac{d}{d{\chi }_0}(\frac{P_0}{P_c}\left)\right]$
$\Rightarrow -[4\pi G{\rho }_c{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2]\left(\frac{{\rho }_0}{{\rho }_c}\right)$	$=$	$\left(\frac{{\rho }_0}{{\rho }_c}{)}^{-1}\frac{1}{{\chi }_0^2}\frac{d}{d{\chi }_0}\right[{\chi }_0^2\frac{d}{d{\chi }_0}\left(\frac{P_0}{P_c}\right)]+\frac{d}{d{\chi }_0}(\frac{P_0}{P_c})\cdot \frac{d}{d{\chi }_0}(\frac{{\rho }_0}{{\rho }_c}{)}^{-1}$
$\Rightarrow -[4\pi G{\rho }_c{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2]\left(\frac{{\rho }_0}{{\rho }_c}{)}^2\right(\frac{P_0}{P_c}{)}^{-1}$	$=$	$\frac{1}{{\chi }_0^2(P_0/P_c)}\frac{d}{d{\chi }_0}\left[{\chi }_0^2\frac{d}{d{\chi }_0}\right(\frac{P_0}{P_c}\left)\right]+\frac{d}{d{\chi }_0}[\ln (\frac{P_0}{P_c}\left)\right]\cdot \frac{d}{d{\chi }_0}[\ln (\frac{{\rho }_0}{{\rho }_c}{)}^{-1}]$
	$=$	$\left(\frac{p^{\text{'}}}{p}\right)\frac{1}{{\chi }_0^2{p}^{\text{'}}}\frac{d}{d{\chi }_0}\left[{\chi }_0^2{p}^{\text{'}}\right]+\frac{d\ln (p)}{d{\chi }_0}\cdot \frac{d}{d{\chi }_0}[\ln (\frac{{\rho }_0}{{\rho }_c}{)}^{-1}]$
	$=$	$\frac{d\ln (p)}{d{\chi }_0}\cdot \frac{d}{d{\chi }_0}[\ln ({\chi }_0^2{p}^{\text{'}})]+\frac{d\ln (p)}{d{\chi }_0}\cdot \frac{d}{d{\chi }_0}[\ln (\frac{{\rho }_0}{{\rho }_c}{)}^{-1}],$

where,

$p^{\text{'}}$

$\equiv$

$\frac{d}{d{\chi }_0}\left(\frac{P_0}{P_c}\right).$

Let's compare the form of this "equilibrium" relation with the form of the LAWE just constructed:

$-[4\pi G{\rho }_c{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2]\left(\frac{{\rho }_0}{{\rho }_c}{)}^2\right(\frac{P_0}{P_c}{)}^{-1}$ $=$ $\left(\frac{p^{\text{'}}}{p}\right)\cdot \frac{d}{d{\chi }_0}[\ln ({\chi }_0^2{p}^{\text{'}})]+\frac{d\ln (p)}{d{\chi }_0}\cdot \frac{d}{d{\chi }_0}[\ln (\frac{{\rho }_0}{{\rho }_c}{)}^{-1}]$   versus   $-{\sigma }^2\left(\frac{{\rho }_0}{{\rho }_c}\right)(\frac{P_0}{P_c}{)}^{-1}$ $=$ $\left(\frac{x^{\text{'}}}{x}\right)\cdot \frac{d}{d{\chi }_0}[\ln ({\chi }_0^4{x}^{\text{'}}\left)\right]+\frac{d\ln (p)}{d{\chi }_0}\cdot \frac{d}{d{\chi }_0}[\ln ({\chi }_0^{\alpha }x\left)\right]$

I like this layout because it unveils similarities in the way the differential operators interact with the functions that describe the radial profiles of variables — specifically, the mass-density, the pressure, and the fractional radial displacement, $x$, during pulsations. However, it is not yet obvious how best to translate between the two differential equations in order to aid in solving for the unknown variable, $x({\chi }_0)$.

Dabbling with Equilibrium Condition

In the meantime, I've found it instructive to play with the first of these two expressions to see how it might be restructured in order to most directly confirm that it is satisfied by the expressions presented in Table 1. Adopting the shorthand notation,

$\mathrm{\Gamma }\equiv 4\pi G{\rho }_c{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2$         and         $\varpi \equiv \frac{{\rho }_0}{{\rho }_c},$

and multiplying the "equilibrium" relation through by $(-\varpi p)$, we have,

$\mathrm{\Gamma }{\varpi }^3$	$=$	$-\varpi p^{\text{'}}\{\frac{1}{{\chi }_0^2{p}^{\text{'}}}\frac{d}{d{\chi }_0}({\chi }_0^2{p}^{\text{'}})-\frac{1}{\varpi }\frac{d\varpi }{d{\chi }_0}\}$
	$=$	$p^{\text{'}}{\varpi }^{\text{'}}-\varpi \frac{d{p}^{\text{'}}}{d{\chi }_0}-\frac{2\varpi p^{\text{'}}}{{\chi }_0}$
	$=$	$\frac{p^{\text{'}}}{{\chi }_0}[{\chi }_0{\varpi }^{\text{'}}-2\varpi ]-\varpi \frac{d{p}^{\text{'}}}{d{\chi }_0};$

or,

$\mathrm{\Gamma }{\varpi }^2$

$=$

$\frac{p^{\text{'}}}{{\chi }_0\varpi }[{\chi }_0{\varpi }^{\text{'}}-2\varpi ]-\frac{d{p}^{\text{'}}}{d{\chi }_0}.$

Specific Cases

Case 1 (Parabolic):

$\varpi =1-{\chi }_0^2$	$\Rightarrow$	${\varpi }^{\text{'}}=-2{\chi }_0$
$p^{\text{'}}=-5{\chi }_0+8{\chi }_0^3-3{\chi }_0^5={\chi }_0(1-{\chi }_0^2)(-5+3{\chi }_0^2)$	$\Rightarrow$	$\frac{p^{\text{'}}}{{\chi }_0\varpi }=-5+3{\chi }_0^2.$
	Also, note:	$\frac{d(p^{\text{'}})}{d{\chi }_0}=-5+24{\chi }_0^2-15{\chi }_0^4.$

For the parabolic case, therefore, the RHS of the "equilibrium" expression is,

RHS	$=$	$(-5+3{\chi }_0^2)[-2{\chi }_0^2-2(1-{\chi }_0^2)]-(-5+24{\chi }_0^2-15{\chi }_0^4)$
	$=$	$(10-6{\chi }_0^2)+(5-24{\chi }_0^2+15{\chi }_0^4)$
	$=$	$15(1-2{\chi }_0^2+{\chi }_0^4),$

which, indeed, matches the LHS of the "equilibrium" relation, if,

$\mathrm{\Gamma }=15$        $\Rightarrow$        ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2=\frac{15}{4\pi G{\rho }_c}.$

This has all worked satisfactorily because, as presented above, this is the correct value of ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2$ in the case of the parabolic density distribution.

Case 2 (Linear):

$\varpi =1-{\chi }_0$	$\Rightarrow$	${\varpi }^{\text{'}}=-1$
$p^{\text{'}}=\frac{12}{5}[-4{\chi }_0+7{\chi }_0^2-3{\chi }_0^3]$	$\Rightarrow$	$\frac{p^{\text{'}}}{{\chi }_0\varpi }=\frac{12}{5}(-4+3{\chi }_0).$
	Also, note:	$\frac{d(p^{\text{'}})}{d{\chi }_0}=\frac{12}{5}[-4+14{\chi }_0-9{\chi }_0^2].$

For the linear case, therefore, the RHS of the "equilibrium" expression is,

RHS	$=$	$\frac{12}{5}(-4+3{\chi }_0)[-{\chi }_0-2(1-{\chi }_0)]-\frac{12}{5}(-4+14{\chi }_0-9{\chi }_0^2)$
	$=$	$\frac{12}{5}[(4-3{\chi }_0)(2-{\chi }_0)+(4-14{\chi }_0+9{\chi }_0^2)]$
	$=$	$\frac{12}{5}(12-24{\chi }_0+12{\chi }_0^2)$
	$=$	$\frac{2^4\cdot 3^2}{5}(1-2{\chi }_0+{\chi }_0^2),$

which, indeed, matches the LHS of the "equilibrium" relation, if,

$\mathrm{\Gamma }=\frac{2^4\cdot 3^2}{5}$        $\Rightarrow$        ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2=\frac{2^2\cdot 3^2}{5\pi G{\rho }_c}.$

This has all worked satisfactorily because, as presented above, this is the correct value of ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2$ in the case of the linear density distribution.

Case 3 (n = 1 polytrope):

$\varpi =\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}$	$\Rightarrow$	${\varpi }^{\text{'}}=\frac{\cos (\pi {\chi }_0)}{{\chi }_0}-\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0^2}$
$p^{\text{'}}=\frac{2\sin (\pi {\chi }_0)}{({\pi }^2{\chi }_0^3)}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)]$	$\Rightarrow$	$\frac{p^{\text{'}}}{{\chi }_0\varpi }=\frac{2}{(\pi {\chi }_0^3)}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)].$
Also, note:      $\frac{d(p^{\text{'}})}{d{\chi }_0}$ $=$ $[\frac{2\pi \cdot \cos (\pi {\chi }_0)}{({\pi }^2{\chi }_0^3)}-\frac{6\sin (\pi {\chi }_0)}{({\pi }^2{\chi }_0^4)}][\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)]+\frac{2\sin (\pi {\chi }_0)}{({\pi }^2{\chi }_0^3)}[\pi \cos (\pi {\chi }_0)-{\pi }^2{\chi }_0\sin (\pi {\chi }_0)-\pi \cos (\pi {\chi }_0)]$   $=$ $\frac{1}{{\pi }^2{\chi }_0^4}\left\{\right[2\pi {\chi }_0\cdot \cos (\pi {\chi }_0)-6\sin (\pi {\chi }_0)\left]\right[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)]+2{\chi }_0\sin (\pi {\chi }_0)[-{\pi }^2{\chi }_0\sin (\pi {\chi }_0)\left]\right\}$   $=$ $\frac{1}{{\pi }^2{\chi }_0^4}\{2{\pi }^2{\chi }_0^2{\cos }^2(\pi {\chi }_0)-8\pi {\chi }_0\sin (\pi {\chi }_0)\cos (\pi {\chi }_0)+6{\sin }^{}(\pi {\chi }_0)-2{\pi }^2{\chi }_0^2{\sin }^{}(\pi {\chi }_0)\}$   $=$ $\frac{2}{{\pi }^2{\chi }_0^4}\{3{\sin }^2(\pi {\chi }_0)-4\pi {\chi }_0\sin (\pi {\chi }_0)\cos (\pi {\chi }_0)+{\pi }^2{\chi }_0^2[1-2{\sin }^{}(\pi {\chi }_0)\left]\right\}.$

For the case of an n = 1 polytropic configuration, therefore, the equilibrium requirement is,

$\mathrm{\Gamma }{\varpi }^2$	$=$	$\frac{p^{\text{'}}}{{\chi }_0\varpi }[{\chi }_0{\varpi }^{\text{'}}-2\varpi ]-\frac{d{p}^{\text{'}}}{d{\chi }_0}$
	$=$	$\frac{2}{(\pi {\chi }_0^3)}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)][\cos (\pi {\chi }_0)-\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}-\frac{2\sin (\pi {\chi }_0)}{\pi {\chi }_0}]$
		$-\frac{2}{{\pi }^2{\chi }_0^4}\{3{\sin }^2(\pi {\chi }_0)-4\pi {\chi }_0\sin (\pi {\chi }_0)\cos (\pi {\chi }_0)+{\pi }^2{\chi }_0^2[1-2{\sin }^{}(\pi {\chi }_0)\left]\right\}$
	$=$	$\frac{2}{({\pi }^2{\chi }_0^4)}\left\{\right[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)\left]\right[\pi {\chi }_0\cos (\pi {\chi }_0)-3\sin (\pi {\chi }_0)]$
		$-3{\sin }^2(\pi {\chi }_0)+4\pi {\chi }_0\sin (\pi {\chi }_0)\cos (\pi {\chi }_0)-{\pi }^2{\chi }_0^2[1-2{\sin }^{}(\pi {\chi }_0)]\}$
	$=$	$\frac{2}{({\pi }^2{\chi }_0^4)}\{3{\sin }^2(\pi {\chi }_0)-4\pi {\chi }_0\sin (\pi {\chi }_0)\cos (\pi {\chi }_0)+(\pi {\chi }_0{)}^2[1-{\sin }^{}(\pi {\chi }_0)]$
		$-3{\sin }^2(\pi {\chi }_0)+4\pi {\chi }_0\sin (\pi {\chi }_0)\cos (\pi {\chi }_0)-{\pi }^2{\chi }_0^2[1-2{\sin }^{}(\pi {\chi }_0)]\}$
	$=$	$\frac{2}{({\pi }^2{\chi }_0^4)}\{(\pi {\chi }_0{)}^2{\sin }^2(\pi {\chi }_0)\}$
	$=$	$2{\pi }^2[\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}{]}^2$

So, the equilibrium condition is satisfied if,

$\mathrm{\Gamma }=2{\pi }^2$        $\Rightarrow$        ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2=\frac{2{\pi }^2}{4\pi G{\rho }_c}=\frac{\pi }{2G{\rho }_c}.$

This has all worked satisfactorily because, as presented in a separate chapter discussion, this is the correct value of ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2$ in the case of an n = 1 polytropic configuration.

Dabbling with LAWE

Now, let's experiment with the LAWE as presented above, that is,

$-{\sigma }^2\left(\frac{{\rho }_0}{{\rho }_c}\right)(\frac{P_0}{P_c}{)}^{-1}$

$=$

$\left(\frac{x^{\text{'}}}{x}\right)\cdot \frac{d}{d{\chi }_0}[\ln ({\chi }_0^4{x}^{\text{'}}\left)\right]+\frac{d\ln (p)}{d{\chi }_0}\cdot \frac{d}{d{\chi }_0}[\ln ({\chi }_0^{\alpha }x\left)\right].$

After multiplying though by $(-p)$, this expression may be written as,

$-{\sigma }^2\varpi$	$=$	$\frac{p}{x}[\frac{d{x}^{\text{'}}}{d{\chi }_0}+\frac{4x^{\text{'}}}{{\chi }_0}]+\frac{\alpha p^{\text{'}}}{{\chi }_0}[1+\frac{{\chi }_0{x}^{\text{'}}}{\alpha x}]$
	$=$	$\left(\frac{p}{x}\right)x^{{\text{'}}^{\prime }}+p\left[\frac{4}{{\chi }_0}\right]\frac{x^{\text{'}}}{x}+p^{\text{'}}\left[\frac{x^{\text{'}}}{x}\right]+\frac{\alpha p^{\text{'}}}{{\chi }_0}$
$\Rightarrow -[{\sigma }^2\varpi +\frac{\alpha p^{\text{'}}}{{\chi }_0}]x$	$=$	$p{x}^{{\text{'}}^{\prime }}+[\frac{4p}{{\chi }_0}+p^{\text{'}}]x^{\text{'}}$
$\Rightarrow 0$	$=$	$p{x}^{{\text{'}}^{\prime }}+[4p+{\chi }_0{p}^{\text{'}}]\frac{x^{\text{'}}}{{\chi }_0}+[{\sigma }^2\varpi +\frac{\alpha p^{\text{'}}}{{\chi }_0}]x.$

(We could have, perhaps, obtained this expression in a more direct fashion had we started directly from the form of the LAWE derived earlier.)

Uniform Density

Case 0 (Uniform density):

$\varpi$	$=$	$1;$
$p$	$=$	$1-{\chi }_0^2;$
$\frac{\alpha p^{\text{'}}}{{\chi }_0}$	$=$	$-2\alpha .$

For the uniform-density case, therefore, the the LAWE becomes,

$-{\sigma }^2$	$=$	$\frac{(1-{\chi }_0^2)}{x}[\frac{d{x}^{\text{'}}}{d{\chi }_0}+\frac{4x^{\text{'}}}{{\chi }_0}]-2\alpha [1+\frac{{\chi }_0{x}^{\text{'}}}{\alpha x}]$
$\Rightarrow (2\alpha -{\sigma }^2)x$	$=$	$(1-{\chi }_0^2)[\frac{d{x}^{\text{'}}}{d{\chi }_0}+\frac{4x^{\text{'}}}{{\chi }_0}]-2{\chi }_0{x}^{\text{'}}$
	$=$	$(1-{\chi }_0^2)\frac{d{x}^{\text{'}}}{d{\chi }_0}+(1-{\chi }_0^2)\left[\frac{4x^{\text{'}}}{{\chi }_0}\right]-2{\chi }_0{x}^{\text{'}}$
	$=$	$(1-{\chi }_0^2)\frac{d{x}^{\text{'}}}{d{\chi }_0}+\frac{2x^{\text{'}}}{{\chi }_0}(2-3{\chi }_0^2),$

where, as defined above,

$\alpha \equiv 3-\frac{4}{{\gamma }_g}.$

Mode 3

Try an eigenfunction of the form,

$x=a+b{\chi }_0^2+c{\chi }_0^4,$

in which case,

$\frac{2x^{\text{'}}}{{\chi }_0}=\frac{2}{{\chi }_0}(2b{\chi }_0+4c{\chi }_0^3)=4b+8c{\chi }_0^2$         and         $x^{{\text{'}}^{\prime }}=2b+12c{\chi }_0^2.$

In order for this to be a solution, we must have,

$(2\alpha -{\sigma }^2)(a+b{\chi }_0^2+c{\chi }_0^4)$	$=$	$(1-{\chi }_0^2)(2b+12c{\chi }_0^2)+(2-3{\chi }_0^2)(4b+8c{\chi }_0^2)$
	$=$	$(2b+12c{\chi }_0^2)-{\chi }_0^2(2b+12c{\chi }_0^2)+2(4b+8c{\chi }_0^2)-3{\chi }_0^2(4b+8c{\chi }_0^2)$
	$=$	$10b+{\chi }_0^2(12c-2b+16c-12b)-{\chi }_0^4(12c+24c)$
	$=$	$10b+{\chi }_0^2(28c-14b)-{\chi }_0^4(36c).$

So, the coefficients of each even power of ${\chi }_0^n$ are:

${\chi }_0^0$	:	$a\mathfrak{𝔉}+10b$
${\chi }_0^2$	:	$b\mathfrak{𝔉}-14b+28c$
${\chi }_0^4$	:	$c[\mathfrak{𝔉}-36]$

where, following Sterne's (1937) presentation,

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

In order for all three of the coefficients to be zero, we must have:

First:     $\mathfrak{𝔉}=36;$

Second:     $22b=-28c\Rightarrow c=-(11/14)b;$

Third:     $36a=-10b\Rightarrow b=-(18/5)a.$

Hence, choosing $a=1$ implies: $b=-18/5$        and         $c=(11/7)(9/5)=+99/35.$ This precisely matches the "j = 2" mode identified by Sterne.

Parabolic

Case 1 (Parabolic):

$\varpi =1-{\chi }_0^2$	$\Rightarrow$	${\varpi }^{\text{'}}=-2{\chi }_0$
$p=\frac{1}{2}(1-{\chi }_0^2{)}^2(2-{\chi }_0^2)$
$p^{\text{'}}=-5{\chi }_0+8{\chi }_0^3-3{\chi }_0^5={\chi }_0(1-{\chi }_0^2)(-5+3{\chi }_0^2)$	$\Rightarrow$	$\frac{\alpha p^{\text{'}}}{{\chi }_0}=\alpha (1-{\chi }_0^2)(-5+3{\chi }_0^2).$

For the parabolic case, therefore, the the LAWE becomes,

$0$	$=$	$p{x}^{{\text{'}}^{\prime }}+[4p+{\chi }_0{p}^{\text{'}}]\frac{x^{\text{'}}}{{\chi }_0}+[{\sigma }^2\varpi +\frac{\alpha p^{\text{'}}}{{\chi }_0}]x$
	$=$	$\frac{1}{2}{\varpi }^2(2-{\chi }_0^2)x^{{\text{'}}^{\prime }}+[2{\varpi }^2(2-{\chi }_0^2)+{\chi }_0^2\varpi (-5+3{\chi }_0^2)]\frac{x^{\text{'}}}{{\chi }_0}+[{\sigma }^2\varpi +\alpha \varpi (-5+3{\chi }_0^2)]x$
	$=$	$\frac{\varpi }{2}\{\varpi (2-{\chi }_0^2)x^{{\text{'}}^{\prime }}+[4\varpi (2-{\chi }_0^2)+{\chi }_0^2(-10+6{\chi }_0^2)]\frac{x^{\text{'}}}{{\chi }_0}+[\mathfrak{𝔎}+6\alpha {\chi }_0^2\left]x\right\}$
	$=$	$\frac{\varpi }{2}\{(2-3{\chi }_0^2+{\chi }_0^4)x^{{\text{'}}^{\prime }}+[8-22{\chi }_0^2+10{\chi }_0^4]\frac{x^{\text{'}}}{{\chi }_0}+[\mathfrak{𝔎}+6\alpha {\chi }_0^2\left]x\right\}$

where,

$\mathfrak{𝔎}\equiv 2({\sigma }^2-5\alpha ).$

Mode Inverse

Try an eigenfunction of the form,

$x=(1+a{\chi }_0^2{)}^{-\beta }=(1+a{\chi }_0^2{)}^{-(\beta +2)}(1+2a{\chi }_0^2+a^2{\chi }_0^4),$

in which case,

$x^{\text{'}}$	$=$	$-2a\beta {\chi }_0(1+a{\chi }_0^2{)}^{-\beta -1};$
	$=$	$-2a\beta ({\chi }_0+a{\chi }_0^3)(1+a{\chi }_0^2{)}^{-(\beta +2)};$
$x^{{\text{'}}^{\prime }}$	$=$	$-2a\beta (1+a{\chi }_0^2{)}^{-\beta -1}-2a\beta {\chi }_0[-2a(\beta +1){\chi }_0(1+a{\chi }_0^2{)}^{-\beta -2}]$
	$=$	$-2a\beta (1+a{\chi }_0^2{)}^{-(\beta +2)}[(1+a{\chi }_0^2)-2a(\beta +1){\chi }_0^2]$
	$=$	$-2a\beta (1+a{\chi }_0^2{)}^{-(\beta +2)}[1-a(2\beta +1){\chi }_0^2]$

In order for this to be a solution, we must have,

$0$	$=$	$(2-3{\chi }_0^2+{\chi }_0^4)x^{{\text{'}}^{\prime }}+[8-22{\chi }_0^2+10{\chi }_0^4]\frac{x^{\text{'}}}{{\chi }_0}+[\mathfrak{𝔎}+6\alpha {\chi }_0^2]x$
	$=$	$-2a\beta [1-a(2\beta +1){\chi }_0^2](2-3{\chi }_0^2+{\chi }_0^4)-2a\beta (1+a{\chi }_0^2)[8-22{\chi }_0^2+10{\chi }_0^4]+[\mathfrak{𝔎}+6\alpha {\chi }_0^2](1+2a{\chi }_0^2+a^2{\chi }_0^4)$
	$=$	$2a\beta (-2+3{\chi }_0^2-{\chi }_0^4)+2a^2\beta (2\beta +1)(2{\chi }_0^2-3{\chi }_0^4+{\chi }_0^6)$
		$+2a\beta [-8+22{\chi }_0^2-10{\chi }_0^4]+2a^2\beta [-8{\chi }_0^2+22{\chi }_0^4-10{\chi }_0^6]$
		$+\mathfrak{𝔎}(1+2a{\chi }_0^2+a^2{\chi }_0^4)+6\alpha ({\chi }_0^2+2a{\chi }_0^4+a^2{\chi }_0^6)$

So, the coefficients of each even power of ${\chi }_0^n$ are:

${\chi }_0^0$	:	$\mathfrak{𝔎}-20a\beta$
${\chi }_0^2$	:	$50a\beta +4a^2\beta (2\beta -3)+6\alpha +2a\mathfrak{𝔎}$
${\chi }_0^4$	:	$-22a\beta +2a^2\beta (19-6\beta )+12a\alpha +a^2\mathfrak{𝔎}$
${\chi }_0^6$	:	$2a^2\beta (2\beta -9)+6a^2\alpha$

Mode 3P

Try an eigenfunction of the form,

$x=a+b{\chi }_0^2+c{\chi }_0^4,$

in which case,

$\frac{x^{\text{'}}}{{\chi }_0}=\frac{1}{{\chi }_0}(2b{\chi }_0+4c{\chi }_0^3)=2b+4c{\chi }_0^2$         and         $x^{{\text{'}}^{\prime }}=2b+12c{\chi }_0^2.$

In order for this to be a solution, we must have,

$0$	$=$	$\frac{\varpi }{2}\{(2-3{\chi }_0^2+{\chi }_0^4)(2b+12c{\chi }_0^2)+(8-22{\chi }_0^2+10{\chi }_0^4)(2b+4c{\chi }_0^2)+(\mathfrak{𝔎}+6\alpha {\chi }_0^2)(a+b{\chi }_0^2+c{\chi }_0^4)\}$
	$=$	$\frac{\varpi }{2}\{(4b+24c{\chi }_0^2-6b{\chi }_0^2-36c{\chi }_0^4+2b{\chi }_0^4+12c{\chi }_0^6)+(16b+32c{\chi }_0^2-44b{\chi }_0^2-88c{\chi }_0^4+20b{\chi }_0^4+40c{\chi }_0^6)$
		$+(a\mathfrak{𝔎}+b\mathfrak{𝔎}{\chi }_0^2+c\mathfrak{𝔎}{\chi }_0^4+6a\alpha {\chi }_0^2+6b\alpha {\chi }_0^4+6c\alpha {\chi }_0^6)\}$
	$=$	$\frac{\varpi }{2}[(20b+a\mathfrak{𝔎}){\chi }_0^0+(24c-6b+32c-44b+b\mathfrak{𝔎}+6a\alpha ){\chi }_0^2+(-36c+2b-88c+20b+c\mathfrak{𝔎}+6b\alpha ){\chi }_0^4+(12c+40c+6c\alpha ){\chi }_0^6]$

So, the coefficients of each even power of ${\chi }_0^n$ are:

${\chi }_0^0$	:	$20b+a\mathfrak{𝔎}$
${\chi }_0^2$	:	$56c+(\mathfrak{𝔎}-50)b+6a\alpha$
${\chi }_0^4$	:	$b(22+6\alpha )+c(\mathfrak{𝔎}-124)$
${\chi }_0^6$	:	$c(52+6\alpha )$

This is disappointing, as it does not result in nonzero coefficient values.

Polytrope

Case 3 (n = 1 polytrope):

$\varpi =\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}$	$\Rightarrow$	${\varpi }^{\text{'}}=\frac{\cos (\pi {\chi }_0)}{{\chi }_0}-\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0^2}$
$p=[\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}{]}^2={\varpi }^2$
$p^{\text{'}}=\frac{2\varpi }{(\pi {\chi }_0^2)}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)]$	$\Rightarrow$	$\frac{\alpha p^{\text{'}}}{{\chi }_0}=\frac{2\alpha \varpi }{(\pi {\chi }_0^3)}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)].$

For the n = 1 polytropic case, therefore, the the LAWE becomes,

$0$	$=$	$p{x}^{{\text{'}}^{\prime }}+[4p+{\chi }_0{p}^{\text{'}}]\frac{x^{\text{'}}}{{\chi }_0}+[{\sigma }^2\varpi +\frac{\alpha p^{\text{'}}}{{\chi }_0}]x$
	$=$	$\left[\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}\right]\varpi x^{{\text{'}}^{\prime }}+\left\{4\right[\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}]+\frac{2}{(\pi {\chi }_0)}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)\left]\right\}\frac{\varpi x^{\text{'}}}{{\chi }_0}$
		$+\{{\sigma }^2+\frac{2\alpha }{(\pi {\chi }_0^3)}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)\left]\right\}\varpi x$
	$=$	$\frac{\varpi }{\pi {\chi }_0^2}\{[\sin (\pi {\chi }_0)]{\chi }_0{x}^{{\text{'}}^{\prime }}+[2\sin (\pi {\chi }_0)+2\pi {\chi }_0\cos (\pi {\chi }_0)]x^{\text{'}}+\{(\pi {\chi }_0^2){\sigma }^2+2\alpha {\chi }_0^{-1}[\pi {\chi }_0\cos (\pi {\chi }_0)-\sin (\pi {\chi }_0)]\}x\}$

Exploration2

Let's begin with the LAWE written in the following form (see, for example, the related context discussion):

$-{\sigma }^2{{𝒢}}_{\sigma }$

$=$

$\left(\frac{P}{\rho }\right)\frac{1}{x^4}\frac{d}{dx}\left(x^4{{𝒢}}_{\sigma }^{\text{'}}\right)+\left(\frac{P^{\text{'}}}{\rho }\right)\frac{1}{x^{\alpha }}\frac{d}{dx}\left(x^{\alpha }{{𝒢}}_{\sigma }\right).$

One advantage of beginning with this construction is that — as the following table shows — it might be reasonable to expect in general that both pre-factors — $(P/\rho )$ and $(P^{\text{'}}/\rho )$ — will have a relatively simple mathematical form. In addition, however, it appears as though both terms on the RHS want to be logarithmic derivatives.

Properties of Analytically Defined Astrophysical Structures
Model	$\rho (x)$	$\left[\frac{P(x)}{\rho (x)}\right]$	$\left[\frac{P^{\text{'}}(x)}{\rho (x)}\right]$
Uniform-density	$1$	$1-x^2$	$-2x$
Linear	$1-x$	$(1-x)(1+2x-\frac{9}{5}{x}^2)$	$-\frac{12}{5}x(4-3x)$
Parabolic	$1-x^2$	$(1-x^2)(1-\frac{1}{2}{x}^2)$	$-x(5-3x^2)$
$n=1$ Polytrope	$\frac{\sin }{x}$	$\frac{\sin x}{x}$	$\frac{2}{x}[\cos x-\frac{\sin x}{x}]$

New Form of LAWE

Multiplying the LAWE through by $[\rho /(P{{𝒢}}_{\sigma })]$ gives,

$-{\sigma }^2\left(\frac{\rho }{P}\right)$	$=$	$\left(\frac{{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}\right)\frac{1}{(x^4{{𝒢}}_{\sigma }^{\text{'}})}\frac{d}{dx}\left(x^4{{𝒢}}_{\sigma }^{\text{'}}\right)+\left(\frac{P^{\text{'}}}{P}\right)\frac{1}{(x^{\alpha }{{𝒢}}_{\sigma })}\frac{d}{dx}\left(x^{\alpha }{{𝒢}}_{\sigma }\right)$
	$=$	$\frac{d\ln {{𝒢}}_{\sigma }}{dx}\cdot \frac{d\ln (x^4{{𝒢}}_{\sigma }^{\text{'}})}{dx}+\frac{d\ln P}{dx}\cdot \frac{d\ln (x^{\alpha }{{𝒢}}_{\sigma })}{dx}$
	$=$	$\frac{d\ln {{𝒢}}_{\sigma }}{dx}[\frac{d\ln (x^4)}{dx}+\frac{d\ln ({{𝒢}}_{\sigma }^{\text{'}})}{dx}]+\frac{d\ln P}{dx}[\frac{d\ln (x^{\alpha })}{dx}+\frac{d\ln ({{𝒢}}_{\sigma })}{dx}]$
	$=$	$\frac{d\ln {{𝒢}}_{\sigma }}{dx}[\frac{d\ln (x^4)}{dx}+\frac{d\ln ({{𝒢}}_{\sigma }^{\text{'}})}{dx}+\frac{d\ln P}{dx}]+\frac{d\ln P}{dx}\left[\frac{d\ln (x^{\alpha })}{dx}\right]$
$\Rightarrow -\left[{\sigma }^2\right(\frac{\rho }{P})+\frac{d\ln P}{dx}\cdot \frac{d\ln (x^{\alpha })}{dx}]$	$=$	$\frac{d\ln {{𝒢}}_{\sigma }}{dx}\left[\frac{d\ln (x^{4}P{{𝒢}}_{\sigma }^{\text{'}})}{dx}\right]$
$\Rightarrow -\frac{{{𝒢}}_{\sigma }}{P}[{\sigma }^2\rho +\frac{\alpha P^{\text{'}}}{x}]$	$=$	$\frac{1}{x^{4}P}\left[\frac{d(x^{4}P{{𝒢}}_{\sigma }^{\text{'}})}{dx}\right]$
$\Rightarrow \frac{d(x^{4}P{{𝒢}}_{\sigma }^{\text{'}})}{dx}$	$=$	$-{{𝒢}}_{\sigma }[{\sigma }^2{x}^4\rho +\alpha x^3{P}^{\text{'}}]$
	$=$	$-x^4\rho {{𝒢}}_{\sigma }[{\sigma }^2+\frac{\alpha P^{\text{'}}}{x\rho }].$

Trial Logarithmic Eigenfunction

Defining,

${ℱ}(x)\equiv [{\sigma }^2+\frac{\alpha P^{\text{'}}}{x\rho }],$

the LAWE becomes,

$\frac{d(x^{4}P{{𝒢}}_{\sigma }^{\text{'}})}{dx}$

$=$

$-x^4\rho {ℱ}{{𝒢}}_{\sigma }.$

First Try

Let's try an eigenvector of the form,

${{𝒢}}_{\sigma }$

$=$

$A(x)\ln P+B(x),$

in which case,

${{𝒢}}_{\sigma }^{\text{'}}$	$=$	$A^{\text{'}}\ln P+\frac{A\cdot P^{\text{'}}}{P}+B^{\text{'}}$
$\Rightarrow x^{4}P{{𝒢}}_{\sigma }^{\text{'}}$	$=$	$x^4[P{A}^{\text{'}}\ln P+A\cdot P^{\text{'}}+P{B}^{\text{'}}]$
$\Rightarrow \mathrm{L}\mathrm{H}\mathrm{S}\equiv \frac{d}{dx}\left[x^{4}P{{𝒢}}_{\sigma }^{\text{'}}\right]$	$=$	$x^4[P^{\prime }{A}^{\text{'}}\ln P+P{A}^{{\text{'}}^{\prime }}\ln P+P{A}^{\text{'}}(\frac{P^{\text{'}}}{P})+A^{\prime }\cdot P^{\text{'}}+P{B}^{{\text{'}}^{\prime }}+A\cdot P^{{\text{'}}^{\prime }}+P^{\prime }{B}^{\text{'}}]$
		$+4x^3[P{A}^{\text{'}}\ln P+A\cdot P^{\text{'}}+P{B}^{\text{'}}]$
	$=$	$x^4\{\ln P[P^{\prime }{A}^{\text{'}}+P{A}^{{\text{'}}^{\prime }}+\frac{4}{x}\cdot P{A}^{\text{'}}]+[A^{\text{'}}{P}^{\text{'}}+A^{\prime }\cdot P^{\text{'}}+P{B}^{{\text{'}}^{\prime }}+A\cdot P^{{\text{'}}^{\prime }}+P^{\prime }{B}^{\text{'}}+\frac{4}{x}(A\cdot P^{\text{'}}+P{B}^{\text{'}})\left]\right\}.$

Now, in order for this expression to match the RHS of the LAWE, we must have, first of all,

${ℱ}A$

$=$

$-\frac{1}{\rho }[P^{\prime }{A}^{\text{'}}+P{A}^{{\text{'}}^{\prime }}+\frac{4}{x}\cdot P{A}^{\text{'}}];$

and, second,

${ℱ}B$

$=$

$-\frac{1}{\rho }[A^{\text{'}}{P}^{\text{'}}+A^{\prime }\cdot P^{\text{'}}+P{B}^{{\text{'}}^{\prime }}+A\cdot P^{{\text{'}}^{\prime }}+P^{\prime }{B}^{\text{'}}+\frac{4}{x}(A\cdot P^{\text{'}}+P{B}^{\text{'}}\left)\right].$

Case 1 (Parabolic):

${ℱ}(x)\equiv [{\sigma }^2-\alpha (5-3x^2)]=f_0+f_2{x}^2,$

where,

$f_0\equiv {\sigma }^2-5\alpha$         and         $f_2\equiv 3\alpha .$

Also, the first condition is,

$-2(f_0+f_2{x}^2)A$

$=$

$2(-5+3x^2)x{A}^{\text{'}}+(2-3x^2+x^4)[A^{{\text{'}}^{\prime }}+\frac{4A^{\text{'}}}{x}].$

So, if we adopt a polynomial expression for the function, $A(x)$, of the form,

$A(x)=a_0+a_2{x}^2+a_4{x}^4,$

$\Rightarrow A^{\text{'}}=2a_{2}x+4a_4{x}^3$         and         $A^{{\text{'}}^{\prime }}=2a_2+12a_4{x}^2,$

the condition becomes,

$-2(f_0+f_2{x}^2)(a_0+a_2{x}^2+a_4{x}^4)$	$=$	$(-10+6x^2)(2a_2{x}^2+4a_4{x}^4)+(2-3x^2+x^4)[(2a_2+12a_4{x}^2)+4(2a_2+4a_4{x}^2)]$
$\Rightarrow (f_0+f_2{x}^2)(a_0+a_2{x}^2+a_4{x}^4)$	$=$	$(10-6x^2)(a_2{x}^2+2a_4{x}^4)-(2-3x^2+x^4)(5a_2+14a_4{x}^2)$

So, the coefficients of each even power of ${\chi }_0^n$ are:

${\chi }_0^0$	:	$f_0{a}_0+10a_2$
${\chi }_0^2$	:	$f_0{a}_2+f_2{a}_0-10a_2+28a^4-15a_2$
${\chi }_0^4$	:	$f_0{a}_4+f_2{a}_2-20a_4+6a_2-42a_4+5a_2$
${\chi }_0^6$	:	$f_2{a}_4+12a_4+14a_4$

This does not seem to work.

Another Trial

Start with a form of the LAWE found midway through the above derivation, namely,

$-\left[{\sigma }^2\right(\frac{\rho }{P})+\frac{d\ln P}{dx}\cdot \frac{d\ln (x^{\alpha })}{dx}]$

$=$

$\frac{d\ln {{𝒢}}_{\sigma }}{dx}\left[\frac{d\ln (x^{4}P{{𝒢}}_{\sigma }^{\text{'}})}{dx}\right];$

and multiplying through by $x^2$ gives,

$-\left[{\sigma }^2\right(\frac{x^2\rho }{P})+\alpha \cdot \frac{d\ln P}{d\ln x}]$	$=$	$\frac{d\ln {{𝒢}}_{\sigma }}{d\ln x}\left[\frac{d\ln (x^{4}P{{𝒢}}_{\sigma }^{\text{'}})}{d\ln x}\right]$
$\Rightarrow -\frac{d\ln P}{d\ln x}\left[{\sigma }^2\right(\frac{P^{\text{'}}}{x\rho }{)}^{-1}+\alpha ]$	$=$	$\frac{d\ln {{𝒢}}_{\sigma }}{d\ln x}[\frac{d\ln ({{𝒢}}_{\sigma }^{\text{'}})}{d\ln x}+\frac{d\ln (P)}{d\ln x}+\frac{d\ln (x^4)}{d\ln x}]$
$\Rightarrow -\frac{d\ln P}{d\ln x}[\alpha +{\sigma }^2(\frac{P^{\text{'}}}{x\rho }{)}^{-1}+\frac{d\ln {{𝒢}}_{\sigma }}{d\ln x}]$	$=$	$\frac{d\ln {{𝒢}}_{\sigma }}{d\ln x}[\frac{d\ln ({{𝒢}}_{\sigma }^{\text{'}})}{d\ln x}+4].$

Note that,

$\frac{d\ln {{𝒢}}_{\sigma }}{d\ln x}$	$=$	$\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }};$
$\frac{d\ln {{𝒢}}_{\sigma }}{d\ln x}\cdot \frac{d\ln ({{𝒢}}_{\sigma }^{\text{'}})}{d\ln x}$	$=$	$\frac{x^2{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}.$

Consider Parabolic Case

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

$\frac{2x^2(5-3x^2)}{(1-x^2)(2-x^2)}[\alpha -{\sigma }^2(5-3x^2{)}^{-1}+\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}]$	$=$	$\left(\frac{x^2{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}\right)+4\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}$
$\Rightarrow \frac{2}{(1-x^2)(2-x^2)}\left[\right(\alpha +\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }})(5-3x^2)-{\sigma }^2]$	$=$	$\left(\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}\right)+\frac{4}{x^2}\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}$

Let's try,

${{𝒢}}_{\sigma }$

$=$

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

which implies,

${{𝒢}}_{\sigma }^{\text{'}}$	$=$	$n(a_0+a_2{x}^2{)}^{n-1}(2a_{2}x)\cdot (b_0+b_2{x}^2{)}^m+m(a_0+a_2{x}^2{)}^n\cdot (b_0+b_2{x}^2{)}^{m-1}(2b_{2}x)$
$\Rightarrow \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}$	$=$	$n(a_0+a_2{x}^2{)}^{-1}(2a_2{x}^2)+m(b_0+b_2{x}^2{)}^{-1}(2b_2{x}^2)$
	$=$	$\frac{2x^2}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}[n{a}_2(b_0+b_2{x}^2)+m{b}_2(a_0+a_2{x}^2)]$
	$=$	$\frac{2x^2}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}[(n{a}_2{b}_0+m{b}_2{a}_0)+(n{a}_2{b}_2+m{b}_2{a}_2)x^2],$

and,

${{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}$	$=$	$nm(a_0+a_2{x}^2{)}^{n-1}(2a_{2}x)\cdot (b_0+b_2{x}^2{)}^{m-1}(2b_{2}x)+n(a_0+a_2{x}^2{)}^{n-1}(2a_2)\cdot (b_0+b_2{x}^2{)}^m+n(n-1)(a_0+a_2{x}^2{)}^{n-2}(2a_{2}x{)}^2\cdot (b_0+b_2{x}^2{)}^m$
		$+mn(a_0+a_2{x}^2{)}^{n-1}(2a_{2}x)\cdot (b_0+b_2{x}^2{)}^{m-1}(2b_{2}x)+m(a_0+a_2{x}^2{)}^n\cdot (b_0+b_2{x}^2{)}^{m-1}(2b_2)+m(m-1)(a_0+a_2{x}^2{)}^n\cdot (b_0+b_2{x}^2{)}^{m-2}(2b_{2}x{)}^2$
$\Rightarrow \frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}$	$=$	$8nm{a}_2{b}_2{x}^2(a_0+a_2{x}^2{)}^{-1}\cdot (b_0+b_2{x}^2{)}^{-1}+n2a_2(a_0+a_2{x}^2{)}^{-1}+n(n-1)4a_2^2{x}^2(a_0+a_2{x}^2{)}^{-2}+m2b_2(b_0+b_2{x}^2{)}^{-1}+m(m-1)4b_2^2{x}^2(b_0+b_2{x}^2{)}^{-2}$
	$=$	$\frac{2n{a}_2(b_0+b_2{x}^2)+2m{b}_2(a_0+a_2{x}^2)}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}+[\frac{4n(n-1)a_2^2}{(a_0+a_2{x}^2{)}^2}+\frac{8nm{a}_2{b}_2}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}+\frac{4m(m-1)b_2^2}{(b_0+b_2{x}^2{)}^2}]x^2$

So, we have for the LAWE:

LHS	$=$	$\frac{2}{(1-x^2)(2-x^2)}\left[\right(\alpha +\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }})(5-3x^2)-{\sigma }^2]$
	$=$	$\frac{2}{(1-x^2)(2-x^2)(a_0+a_2{x}^2)(b_0+b_2{x}^2)}\left\{\right[\alpha (a_0+a_2{x}^2)(b_0+b_2{x}^2)+2x^2(n{a}_2{b}_0+m{b}_2{a}_0)+2x^4(n{a}_2{b}_2+m{b}_2{a}_2)](5-3x^2)$
		$-{\sigma }^2(a_0+a_2{x}^2)(b_0+b_2{x}^2)\};$
RHS	$=$	$\left(\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}\right)+\frac{4}{x^2}\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}$
	$=$	$\frac{2n{a}_2(b_0+b_2{x}^2)+2m{b}_2(a_0+a_2{x}^2)}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}+[\frac{4n(n-1)a_2^2}{(a_0+a_2{x}^2{)}^2}+\frac{8nm{a}_2{b}_2}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}+\frac{4m(m-1)b_2^2}{(b_0+b_2{x}^2{)}^2}]x^2$
		$+\frac{8}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}[(n{a}_2{b}_0+m{b}_2{a}_0)+(n{a}_2{b}_2+m{b}_2{a}_2)x^2]$
	$=$	$\frac{1}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}\{2n{a}_2(b_0+b_2{x}^2)+2m{b}_2(a_0+a_2{x}^2)+8(n{a}_2{b}_0+m{b}_2{a}_0)+8(n{a}_2{b}_2+m{b}_2{a}_2)x^2$
		$+[8nm{a}_2{b}_2+\frac{4n(n-1)a_2^2(b_0+b_2{x}^2)}{(a_0+a_2{x}^2)}+\frac{4m(m-1)b_2^2(a_0+a_2{x}^2)}{(b_0+b_2{x}^2)}]x^2\}.$

Putting these together gives,

$0$	$=$	$[\alpha (a_0+a_2{x}^2)(b_0+b_2{x}^2)+2x^2(n{a}_2{b}_0+m{b}_2{a}_0)+2x^4(n{a}_2{b}_2+m{b}_2{a}_2)](5-3x^2)-{\sigma }^2(a_0+a_2{x}^2)(b_0+b_2{x}^2)$
		$-[n{a}_2(b_0+b_2{x}^2)+m{b}_2(a_0+a_2{x}^2)+4(n{a}_2{b}_0+m{b}_2{a}_0)+4(n{a}_2{b}_2+m{b}_2{a}_2)x^2+4nm{a}_2{b}_2{x}^2](1-x^2)(2-x^2)$
		$-\frac{(1-x^2)(2-x^2)}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}[2n(n-1)a_2^2(b_0+b_2{x}^2{)}^2+2m(m-1)b_2^2(a_0+a_2{x}^2{)}^2]x^2.$

---

First Guess

Now, if we are very lucky, we will find that,

$(a_0+a_2{x}^2)=(1-x^2)$      $\Rightarrow$      $a_0=1$      and      $a_2=-1$;

and, simultaneously,

$(b_0+b_2{x}^2)=(2-x^2)$      $\Rightarrow$      $b_0=2$      and      $b_2=-1$.

In this case, the fractional coefficient in the last term of the LAWE will become unity and the LAWE becomes,

$0$	$=$	$[\alpha (1-x^2)(2-x^2)-2x^2(2n+m)+2x^4(n+m)](5-3x^2)$
		$-[{\sigma }^2-n(2-x^2)-m(1-x^2)-4(2n+m)+4(n+m)x^2+4nm{x}^2](1-x^2)(2-x^2)$
		$-[2n(n-1)(2-x^2{)}^2+2m(m-1)(1-x^2{)}^2]x^2$
	$=$	$[2\alpha -x^2(3\alpha +4n+2m)+x^4(\alpha +2n+2m)](5-3x^2)$
		$+[(-{\sigma }^2+10n+5m)-x^2(5n+5m+4nm)](2-3x^2+x^4)$
		$-\{[8n(n-1)+2m(m-1)]-[8n(n-1)+4m(m-1)]x^2+[2n(n-1)+2m(m-1)]x^4\}{x}^2.$

So, the coefficients of each even power of ${\chi }_0^n$ are:

${\chi }_0^0$	:	$10\alpha +2(-{\sigma }^2+10n+5m)$
${\chi }_0^2$	:	$-6\alpha -5(3\alpha +4n+2m)-2(5n+5m+4nm)-3(-{\sigma }^2+10n+5m)-[8n(n-1)+2m(m-1)]$
${\chi }_0^4$	:	$5(\alpha +2n+2m)+3(3\alpha +4n+2m)+(-{\sigma }^2+10n+5m)+3(5n+5m+4nm)+[8n(n-1)+4m(m-1)]$
${\chi }_0^6$	:	$-3(\alpha +2n+2m)-(5n+5m+4nm)-[2n(n-1)+2m(m-1)]$

Now let's begin simplification. The $x^0$ coefficient implies,

$(-{\sigma }^2+10n+5m)=-5\alpha .$

Hence,

${\chi }_0^2$	:	$-6\alpha -5(4n+2m)-2(5n+5m+4nm)-[8n(n-1)+2m(m-1)]$
${\chi }_0^4$	:	$9\alpha +5(2n+2m)+3(4n+2m)+3(5n+5m+4nm)+[8n(n-1)+4m(m-1)]$
${\chi }_0^6$	:	$-3\alpha -3(2n+2m)-(5n+5m+4nm)-[2n(n-1)+2m(m-1)]$

Using the $x^6$ coefficient to define $\alpha$, that is, setting,

$3\alpha =\{-3(2n+2m)-(5n+5m+4nm)-[2n(n-1)+2m(m-1)]\},$

means that the other two coefficient expressions are,

${\chi }_0^2$	:	$6(2n+2m)+2(5n+5m+4nm)+4[n(n-1)+m(m-1)]-5(4n+2m)-2(5n+5m+4nm)-[8n(n-1)+2m(m-1)]$
${\chi }_0^4$	:	$-9(2n+2m)-3(5n+5m+4nm)-6[n(n-1)+m(m-1)]+5(2n+2m)+3(4n+2m)+3(5n+5m+4nm)+[8n(n-1)+4m(m-1)]$

The only question remaining is, what pair of values for $(n,m)$ result in both of these expressions going to zero? Simplifying the first expression gives,

$0$	$=$	$6(2n+2m)+2(5n+5m+4nm)+4[n(n-1)+m(m-1)]-5(4n+2m)-2(5n+5m+4nm)-[8n(n-1)+2m(m-1)]$
	$=$	$2m-8n-4n(n-1)+2m(m-1)$
	$=$	$2[m^2-2n(n+1)].$

Simplifying the second expression gives,

$0$	$=$	$-9(2n+2m)-3(5n+5m+4nm)-6[n(n-1)+m(m-1)]+5(2n+2m)+3(4n+2m)+3(5n+5m+4nm)+[8n(n-1)+4m(m-1)]$
	$=$	$4n-2m+2n(n-1)-2m(m-1)$
	$=$	$2[n(n+1)-m^2]$

Okay. Because it is not possible for both of these last two constraints to be simultaneously satisfied, I conclude that this last, specific eigenfunction guess is incorrect.

---

Second Guess

Let's try again, keeping the same values of the $b_0$ and $b_2$ — that is,

$(b_0+b_2{x}^2)=(2-x^2)$      $\Rightarrow$      $b_0=2$      and      $b_2=-1$

— but leaving the values of $a_0$ and $a_2$ unspecified. In this case, the LAWE becomes,

$0$	$=$	$[\alpha (a_0+a_2{x}^2)(2-x^2)+2x^2(2n{a}_2-m{a}_0)-2x^4(n{a}_2+m{a}_2)](5-3x^2)(a_0+a_2{x}^2)-{\sigma }^2(a_0+a_2{x}^2{)}^2(2-x^2)$
		$-[n{a}_2(2-x^2)-m(a_0+a_2{x}^2)+4(2n{a}_2-m{a}_0)-4(n{a}_2+m{a}_2)x^2-4nm{a}_2{x}^2](1-x^2)(2-x^2)(a_0+a_2{x}^2)$
		$-[2n(n-1)a_2^2(2-x^2{)}^2+2m(m-1)(a_0+a_2{x}^2{)}^2]x^2(1-x^2)$
	$=$	$\{\alpha [2a_0]+x^2[(4n{a}_2-2m{a}_0)+\alpha (2a_2-a_0)]-x^4[(2n{a}_2+2m{a}_2)+a_2\alpha ]\}[5a_0+(5a_2-3a_0)x^2-3a_2{x}^4]-{\sigma }^2[2a_0+(2a_2-a_0)x^2-a_2{x}^4](a_0+a_2{x}^2)$
		$+[(5m{a}_0-10n{a}_2)+(4nm{a}_2+5n{a}_2+5m{a}_2)x^2](1-x^2)[2a_0+(2a_2-a_0)x^2-a_2{x}^4]$
		$-\{[8n(n-1)a_2^2+2m(m-1)a_0^2]+[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]x^2+[2n(n-1)a_2^2+2m(m-1)a_2^2]x^4\}{x}^2(1-x^2)$
	$=$	$\{\alpha [2a_0]+x^2[(4n{a}_2-2m{a}_0)+\alpha (2a_2-a_0)]-x^4[(2n{a}_2+2m{a}_2)+a_2\alpha ]\}[5a_0+(5a_2-3a_0)x^2-3a_2{x}^4]$
		$-{\sigma }^2\{2a_0^2+[2a_0{a}_2+a_0(2a_2-a_0)]x^2+[a_2(2a_2-a_0)-a_0{a}_2]x^4-a_2^2{x}^6\}$
		$+\{[(5m{a}_0-10n{a}_2)]+[(4nm{a}_2+5n{a}_2+5m{a}_2)-(5m{a}_0-10n{a}_2)]x^2-[4nm{a}_2+5n{a}_2+5m{a}_2]x^4\}[2a_0+(2a_2-a_0)x^2-a_2{x}^4]$
		$-\{[8n(n-1)a_2^2+2m(m-1)a_0^2]x^2+[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]x^4+[2n(n-1)a_2^2+2m(m-1)a_2^2]x^6\}(1-x^2).$

So, the coefficients of each even power of $x^n$ are:

$x^0$	:	$10a_0^2\alpha -2a_0^2{\sigma }^2+2a_0[(5m{a}_0-10n{a}_2)]$
$x^2$	:	$5a_0[(4n{a}_2-2m{a}_0)+\alpha (2a_2-a_0)]+\alpha [2a_0](5a_2-3a_0)-{\sigma }^2[2a_0{a}_2+a_0(2a_2-a_0)]$ $+2a_0[(4nm{a}_2+5n{a}_2+5m{a}_2)-(5m{a}_0-10n{a}_2)]+(2a_2-a_0)[(5m{a}_0-10n{a}_2)]-[8n(n-1)a_2^2+2m(m-1)a_0^2]$
$x^4$	:	$-5a_0[(2n{a}_2+2m{a}_2)+a_2\alpha ]+(5a_2-3a_0)[(4n{a}_2-2m{a}_0)+\alpha (2a_2-a_0)]-6a_0{a}_2\alpha -{\sigma }^2[a_2(2a_2-a_0)-a_0{a}_2]$ $-2a_0[4nm{a}_2+5n{a}_2+5m{a}_2]+(2a_2-a_0)[(4nm{a}_2+5n{a}_2+5m{a}_2)-(5m{a}_0-10n{a}_2)]-a_2(5m{a}_0-10n{a}_2)$ $-[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]+[8n(n-1)a_2^2+2m(m-1)a_0^2]$
$x^6$	:	$-3a_2[(4n{a}_2-2m{a}_0)+\alpha (2a_2-a_0)]-(5a_2-3a_0)[(2n{a}_2+2m{a}_2)+a_2\alpha ]+{\sigma }^2{a}_2^2$ $-(2a_2-a_0)[4nm{a}_2+5n{a}_2+5m{a}_2]-a_2[(4nm{a}_2+5n{a}_2+5m{a}_2)-(5m{a}_0-10n{a}_2)]$ $-[2n(n-1)a_2^2+2m(m-1)a_2^2]+[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]$
$x^8$	:	$3a_2{a}_2\alpha +a_2[4nm{a}_2+11n{a}_2+11m{a}_2]+[2n(n-1)a_2^2+2m(m-1)a_2^2]$

After simplification:

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

---

Third Guess

Let's try again, keeping the same values of the $b_0$ and $b_2$ — that is,

$(b_0+b_2{x}^2)=(2-x^2)$      $\Rightarrow$      $b_0=2$      and      $b_2=-1$

— but leaving the values of $a_0$ and $a_2$ unspecified. In this case, the LAWE becomes,

$0$	$=$	$[\alpha (a_0+a_2{x}^2)(2-x^2)+2x^2(2n{a}_2-m{a}_0)-2a_2{x}^4(n+m)](5-3x^2)-{\sigma }^2(a_0+a_2{x}^2)(2-x^2)$
		$+[-n{a}_2(2-x^2)+m(a_0+a_2{x}^2)-4(n{a}_{2}2-m{a}_0)+4(n{a}_2+m{a}_2)x^2+4nm{a}_2{x}^2](1-x^2)(2-x^2)$
		$-\frac{1}{(a_0+a_2{x}^2)}[2n(n-1)a_2^2(2-x^2{)}^2+2m(m-1)(a_0+a_2{x}^2{)}^2](1-x^2)x^2$
	$=$	$\{[2a_0\alpha ]+[\alpha (-a_0+2a_2)+(4n{a}_2-2m{a}_0)]x^2+[-a_2\alpha -2a_2(n+m)]x^4\}(5-3x^2)-{\sigma }^2[2a_0+(-a_0+2a_2)x^2-a_2{x}^4]$
		$+[(-10n{a}_2+5m{a}_0)+(5m{a}_2+5n{a}_2+4nm{a}_2)x^2](2-3x^2+x^4)$
		$-\frac{1}{(a_0+a_2{x}^2)}\{[8n(n-1)a_2^2+2m(m-1)a_0^2]+[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]x^2+[2n(n-1)a_2^2+2m(m-1)a_2^2]x^4\}(x^2-x^4)$
	$=$	$\{[10a_0\alpha -2a_0{\sigma }^2]+[5\alpha (-a_0+2a_2)+5(4n{a}_2-2m{a}_0)-6a_0\alpha +(a_0-2a_2){\sigma }^2]x^2$
		$+[-5a_2\alpha -10a_2(n+m)-3\alpha (-a_0+2a_2)-3(4n{a}_2-2m{a}_0)+a_2{\sigma }^2]x^4+[3a_2\alpha +6a_2(n+m)]x^6\}(a_0+a_2{x}^2)$
		$+\{[2(-10n{a}_2+5m{a}_0)]+[2(5m{a}_2+5n{a}_2+4nm{a}_2)-3(-10n{a}_2+5m{a}_0)]x^2$
		$+[-3(5m{a}_2+5n{a}_2+4nm{a}_2)+(-10n{a}_2+5m{a}_0)]x^4+(5m{a}_2+5n{a}_2+4nm{a}_2)x^6\}(a_0+a_2{x}^2)$
		$-\{[8n(n-1)a_2^2+2m(m-1)a_0^2]+[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]x^2+[2n(n-1)a_2^2+2m(m-1)a_2^2]x^4\}(x^2-x^4)$
	$=$	$\{[10a_0\alpha -2a_0{\sigma }^2]+[\alpha (-11a_0+10a_2)+(20n{a}_2-10m{a}_0)+(a_0-2a_2){\sigma }^2]x^2$
		$+[\alpha (3a_0-11a_2)+(-22n{a}_2+6m{a}_0-10a_{2}m)+a_2{\sigma }^2]x^4+[3a_2\alpha +6a_2(n+m)]x^6\}(a_0+a_2{x}^2)$
		$+\{[-20n{a}_2+10m{a}_0]+[10m{a}_2+40n{a}_2+8nm{a}_2-15m{a}_0]x^2$
		$+[-15m{a}_2-25n{a}_2-12nm{a}_2+5m{a}_0]x^4+[5m{a}_2+5n{a}_2+4nm{a}_2]x^6\}(a_0+a_2{x}^2)$
		$-\{[8n(n-1)a_2^2+2m(m-1)a_0^2]x^2+[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]x^4+[2n(n-1)a_2^2+2m(m-1)a_2^2]x^6\}(1-x^2)$

So, the coefficients of each even power of $x^n$ are:

$x^0$	:	$a_0[10a_0\alpha -2a_0{\sigma }^2]+a_0[-20n{a}_2+10m{a}_0]$
$x^2$	:	$a_0[\alpha (-11a_0+10a_2)+(20n{a}_2-10m{a}_0)+(a_0-2a_2){\sigma }^2]+a_2[10a_0\alpha -2a_0{\sigma }^2]+a_0[10m{a}_2+40n{a}_2+8nm{a}_2-15m{a}_0]$ $+a_2[-20n{a}_2+10m{a}_0]-[8n(n-1)a_2^2+2m(m-1)a_0^2]$
$x^4$	:	$a_0[\alpha (3a_0-11a_2)+(-22n{a}_2+6m{a}_0-10a_{2}m)+a_2{\sigma }^2]+a_2[\alpha (-11a_0+10a_2)+(20n{a}_2-10m{a}_0)+(a_0-2a_2){\sigma }^2]$ $+a_0[-15m{a}_2-25n{a}_2-12nm{a}_2+5m{a}_0]+a_2[10m{a}_2+40n{a}_2+8nm{a}_2-15m{a}_0]$ $-[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]+[8n(n-1)a_2^2+2m(m-1)a_0^2]$
$x^6$	:	$a_0[3a_2\alpha +6a_2(n+m)]+a_2[\alpha (3a_0-11a_2)+(-22n{a}_2+6m{a}_0-10a_{2}m)+a_2{\sigma }^2]$ $a_0[5m{a}_2+5n{a}_2+4nm{a}_2]+a_2[-15m{a}_2-25n{a}_2-12nm{a}_2+5m{a}_0]$ $-[2n(n-1)a_2^2+2m(m-1)a_2^2]+[-8n(n-1)a_2^2+4m(m-1)a_0{a}_2]$
$x^8$	:	$a_2[3a_2\alpha +6a_2(n+m)]+a_2[5m{a}_2+5n{a}_2+4nm{a}_2]+[2n(n-1)a_2^2+2m(m-1)a_2^2]$

After simplification:

$x^0$	:	$\alpha (10a_0^2)+{\sigma }^2(-2a_0^2)-20n{a}_0{a}_2+10m{a}_0^2$
$x^2$	:	$\alpha (-11a_0^2+20a_0{a}_2)+{\sigma }^2(a_0^2-4a_0{a}_2)+60n{a}_0{a}_2-20n{a}_2^2-25m{a}_0^2+20m{a}_0{a}_2+8nm{a}_0{a}_2$ $-[8n(n-1)a_2^2+2m(m-1)a_0^2]$
$x^4$	:	$\alpha (10a_2^2-22a_0{a}_2+3a_0^2)-{\sigma }^2(2a_2^2-2a_0{a}_2)-47n{a}_0{a}_2+60n{a}_2^2-50m{a}_0{a}_2+11m{a}_0^2+10m{a}_2^2-12nm{a}_0{a}_2+8nm{a}_2^2$ $+16n(n-1)a_2^2-4m(m-1)a_0{a}_2+2m(m-1)a_0^2$
$x^6$	:	$\alpha (6a_0{a}_2-11a_2^2)+{\sigma }^2(a_2^2)+11n{a}_0{a}_2+22m{a}_0{a}_2-47n{a}_2^2-25m{a}_2^2-12nm{a}_2^2+4nm{a}_0{a}_2$ $-10n(n-1)a_2^2-2m(m-1)a_2^2+4m(m-1)a_0{a}_2$
$x^8$	:	$a_2[3a_2\alpha +6a_2(n+m)]+a_2[5m{a}_2+5n{a}_2+4nm{a}_2]+[2n(n-1)a_2^2+2m(m-1)a_2^2]$

See Also