SSC/Stability/n1PolytropeLAWE

Radial Oscillations of n = 1 Polytropic Spheres

Groundwork

In an accompanying discussion, we derived the so-called,

Adiabatic Wave (or Radial Pulsation) Equation

$\frac{d^{2}x}{d{r}_0^2}+[\frac{4}{r_0}-(\frac{g_0{\rho }_0}{P_0}\left)\right]\frac{dx}{d{r}_0}+\left(\frac{{\rho }_0}{{\gamma }_{\mathrm{g}}{P}_0}\right)[{\omega }^2+(4-3{\gamma }_{\mathrm{g}})\frac{g_0}{r_0}]x=0$

whose solution gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations. Because this widely used form of the radial pulsation equation is not dimensionless but, rather, has units of inverse length-squared, we have found it useful to also recast it in the following dimensionless form:

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

where,

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

In a separate discussion, we showed that specifically for isolated, polytropic configurations, this linear adiabatic wave equation (LAWE) can be rewritten as,

$0$	$=$	$\frac{d^{2}x}{d{\xi }^2}+\left[\frac{4-(n+1)V(\xi )}{\xi }\right]\frac{dx}{d\xi }+\left[\frac{{\omega }^2}{{\gamma }_g\theta }\right(\frac{n+1}{4\pi G{\rho }_c})-(3-\frac{4}{{\gamma }_g})\cdot \frac{(n+1)V(x)}{{\xi }^2}]x$
	$=$	$\frac{d^{2}x}{d{\xi }^2}+[\frac{4}{\xi }-\frac{(n+1)}{\theta }(-\frac{d\theta }{d\xi }\left)\right]\frac{dx}{d\xi }+\frac{(n+1)}{\theta }[\frac{{\sigma }_c^2}{6{\gamma }_g}-\frac{\alpha }{\xi }(-\frac{d\theta }{d\xi }\left)\right]x,$

where we have adopted the dimensionless frequency notation,

${\sigma }_c^2$

$\equiv$

$\frac{3{\omega }^2}{2\pi G{\rho }_c}.$

Here we focus on an analysis of the specific case of isolated, $n=1$ polytropic configurations, whose unperturbed equilibrium structure can be prescribed in terms of analytic functions. Our hope — as yet unfulfilled — is that we can discover an analytically prescribed eigenvector solution to the governing LAWE.

Search for Analytic Solutions to the LAWE

Setup

From our derived structure of an n = 1 polytrope, in terms of the configuration's radius $R$ and mass $M$, the central pressure and density are, respectively,

$P_c=\frac{\pi G}{8}\left(\frac{M^2}{R^4}\right)$ ,

and

${\rho }_c=\frac{\pi M}{4R^3}$ .

Hence the characteristic time and acceleration are, respectively,

${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}=[\frac{R^2{\rho }_c}{P_c}{]}^{1/2}=[\frac{2R^3}{GM}{]}^{1/2}=[\frac{\pi }{2G{\rho }_c}{]}^{1/2},$

and,

$g_{\mathrm{S}\mathrm{S}\mathrm{C}}=\frac{P_c}{R{\rho }_c}=\left(\frac{GM}{2R^2}\right).$

The required functions are,

• Density:

$\frac{{\rho }_0({\chi }_0)}{{\rho }_c}=\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}$ ;

• Pressure:

$\frac{P_0({\chi }_0)}{P_c}=[\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}{]}^2$ ;

• Gravitational acceleration:

$\frac{g_0(r_0)}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}=\frac{2}{{\chi }_0^2}\left[\frac{M_r({\chi }_0)}{M}\right]=\frac{2}{\pi {\chi }_0^2}[\sin (\pi {\chi }_0)-\pi {\chi }_0\cos (\pi {\chi }_0)].$

So our desired Eigenvalues and Eigenvectors will be solutions to the following ODE:

$\frac{d^{2}x}{d{\chi }_0^2}+\frac{2}{{\chi }_0}[1+\pi {\chi }_0\cot (\pi {\chi }_0)]\frac{dx}{d{\chi }_0}+\frac{1}{{\gamma }_{\mathrm{g}}}\left\{\frac{\pi {\chi }_0}{\sin (\pi {\chi }_0)}\right[\frac{\pi {\omega }^2}{2G{\rho }_c}]+\frac{2}{{\chi }_0^2}(4-3{\gamma }_{\mathrm{g}})[1-\pi {\chi }_0\cot (\pi {\chi }_0)\left]\right\}x=0,$

or, replacing ${\chi }_0$ with $\xi \equiv \pi {\chi }_0$ and dividing the entire expression by ${\pi }^2$, we have,

$\frac{d^{2}x}{d{\xi }^2}+\frac{2}{\xi }[1+\xi \cot \xi ]\frac{dx}{d\xi }+\frac{1}{{\gamma }_{\mathrm{g}}}\left\{\frac{\xi }{\sin \xi }\right[\frac{{\omega }^2}{2\pi G{\rho }_c}]+\frac{2}{{\xi }^2}(4-3{\gamma }_{\mathrm{g}})[1-\xi \cot \xi \left]\right\}x=0.$

This is identical to the formulation of the wave equation that is relevant to the (n = 1) core of the composite polytrope studied by J. O. Murphy & R. Fiedler (1985b); for comparison, their expression is displayed, here, in the following boxed-in image.

n = 1 Polytropic Formulation of Wave Equation as Presented by Murphy & Fiedler (1985b)
Murphy & Fiedler (1985b) Murphy & Fiedler (1985b)

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Attempt at Deriving an Analytic Eigenvector Solution

Multiplying the last expression through by ${\xi }^2\sin \xi$ gives,

$({\xi }^2\sin \xi )\frac{d^{2}x}{d{\xi }^2}+2[\xi \sin \xi +{\xi }^2\cos \xi ]\frac{dx}{d\xi }+[{\sigma }^2{\xi }^3-2\alpha (\sin \xi -\xi \cos \xi )]x=0,$

where,

${\sigma }^2$	$\equiv$	$\frac{{\omega }^2}{2\pi G{\rho }_c{\gamma }_g},$
$\alpha$	$\equiv$	$3-\frac{4}{{\gamma }_g}.$

The first two terms can be folded together to give,

$\frac{1}{{\xi }^2{\sin }^2\xi }\cdot \frac{d}{d\xi }\left[{\xi }^2{\sin }^2\xi \frac{dx}{d\xi }\right]$	$=$	$\frac{1}{{\xi }^2\sin \xi }[2\alpha (\sin \xi -\xi \cos \xi )-{\sigma }^2{\xi }^3]x$
	$=$	$-\left[\frac{2\alpha }{{\xi }^2}\right(\frac{\xi \cos \xi }{\sin \xi }-1)+{\sigma }^2(\frac{\xi }{\sin \xi }\left)\right]x$
	$=$	$-[\frac{2\alpha }{{\xi }^2}\frac{{\xi }^2}{\sin \xi }\cdot \frac{d}{d\xi }(\frac{\sin \xi }{\xi })+{\sigma }^2(\frac{\xi }{\sin \xi }\left)\right]x$
	$=$	$-\left[\frac{2\alpha }{\xi }\frac{d}{d\xi }\right(\frac{\sin \xi }{\xi })+{\sigma }^2]\left(\frac{\xi }{\sin \xi }\right)x,$

where, in order to make this next-to-last step, we have recognized that,

$\frac{d}{d\xi }\left(\frac{\sin \xi }{\xi }\right)$

$=$

$\frac{\sin \xi }{{\xi }^2}[\frac{\xi \cos \xi }{\sin \xi }-1].$

It would seem that the eigenfunction, $x(\xi )$, should be expressible in terms of trigonometric functions and powers of $\xi$; indeed, it appears as though the expression governing this eigenfunction would simplify considerably if $x\propto \sin \xi /\xi$. With this in mind, we have made some attempts to guess the exact form of the eigenfunction. Here is one such attempt.

First Guess (n1)

Let's try,

$x=\frac{\sin \xi }{\xi },$

which means,

$x^{\text{'}}\equiv \frac{dx}{d\xi }$

$=$

$\frac{\sin \xi }{{\xi }^2}[\frac{\xi \cos \xi }{\sin \xi }-1].$

Does this satisfy the governing expression? Let's see. The right-and-side (RHS) gives:

RHS

$=$

$-\left[\frac{2\alpha }{\xi }\frac{d}{d\xi }\right(\frac{\sin \xi }{\xi })+{\sigma }^2]\left(\frac{\xi }{\sin \xi }\right)x=-[\frac{2\alpha x^{\text{'}}}{\xi }+{\sigma }^2].$

At the same time, the left-hand-side (LHS) may, quite generically, be written as:

LHS	$=$	$\frac{x^{\text{'}}}{\xi }\{\frac{\xi }{({\xi }^2{\sin }^2\xi )x^{\text{'}}}\cdot \frac{d[({\xi }^2{\sin }^2\xi )x^{\text{'}}]}{d\xi }\}$
	$=$	$\frac{x^{\text{'}}}{\xi }\left[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }\right].$

Putting the two sides together therefore gives,

$\frac{x^{\text{'}}}{\xi }[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }+2\alpha ]$	$=$	$-{\sigma }^2$
$\Rightarrow [\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}{]}^{1/(2\alpha )}}{d\ln \xi }+1]$	$=$	$-\frac{{\sigma }^2}{2\alpha }\left(\frac{\xi }{x^{\text{'}}}\right)$
$\Rightarrow \frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}{]}^{-1/(2\alpha )}}{d\ln \xi }$	$=$	$1+\frac{{\sigma }^2}{2\alpha }\left(\frac{\xi }{x^{\text{'}}}\right).$

[Comment from J. E. Tohline on 6 April 2015: I'm not sure what else to make of this.]

Second Guess (n1)

Adopting the generic rewriting of the LHS, and leaving the RHS fully generic as well, we have,

$\frac{x^{\text{'}}}{\xi }\left[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }\right]$	$=$	$-\left[\frac{2\alpha }{\xi }\frac{d}{d\xi }\right(\frac{\sin \xi }{\xi })+{\sigma }^2]\left(\frac{\xi }{\sin \xi }\right)x$
$\Rightarrow \frac{x^{\text{'}}}{x}\left[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }\right]$	$=$	$-2\alpha \left(\frac{\xi }{\sin \xi }\right)\frac{d}{d\xi }\left(\frac{\sin \xi }{\xi }\right)-{\sigma }^2\left(\frac{{\xi }^2}{\sin \xi }\right)$
$\Rightarrow \frac{d\ln (x)}{d\ln \xi }\left[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }\right]$	$=$	$-2\alpha \left[\frac{d\ln (\sin \xi /\xi )}{d\ln \xi }\right]-{\sigma }^2\left(\frac{{\xi }^3}{\sin \xi }\right).$
$\Rightarrow -{\sigma }^2$	$=$	$\left(\frac{\sin \xi }{{\xi }^3}\right)\left\{\frac{d\ln (x)}{d\ln \xi }\right[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }]+2\alpha [\frac{d\ln (\sin \xi /\xi )}{d\ln \xi }\left]\right\}.$

[Comment from J. E. Tohline on 6 April 2015: I'm not sure what else to make of this.]

Third Guess (n1)

Let's rewrite the polytropic (n = 1) wave equation as follows:

$\sin \xi [{\xi }^2{x}^{{\text{'}}^{\prime }}+2\xi x^{\text{'}}-2\alpha x]+\cos \xi [2{\xi }^2{x}^{\text{'}}+2\alpha \xi x]+{\sigma }^2{\xi }^{3}x=0.$

It is difficult to determine what term in the adiabatic wave equation will cancel the term involving ${\sigma }^2$ because its leading coefficient is ${\xi }^3$ and no other term contains a power of $\xi$ that is higher than two. After thinking through various trial eigenvector expressions, $x(\xi )$, I have determined that a function of the following form has a chance of working because the second derivative of the function generates a leading factor of ${\xi }^3$ while the function itself does not introduce any additional factors of $\xi$ into the term that contains ${\sigma }^2$:

$x$	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})][A\sin \xi +B\cos \xi ]$
$\Rightarrow x^{\text{'}}$	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+\frac{d[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]}{d\xi }\cdot [A\sin \xi +B\cos \xi ]$
	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+[A\sin \xi +B\cos \xi ]\{5a{\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+b[\frac{5}{2}{\xi }^{3/2}{\cos }^{}({\xi }^{5/2})-\frac{5}{2}{\xi }^{3/2}{\sin }^{}({\xi }^{5/2})]-5c{\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})\}$
	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+[A\sin \xi +B\cos \xi ]\{5(a-c){\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+\frac{5b}{2}{\xi }^{3/2}[1-2{\sin }^{}({\xi }^{5/2})\left]\right\}$
$\Rightarrow x^{{\text{'}}^{\prime }}$	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d^2[A\sin \xi +B\cos \xi ]}{d^2\xi }$
		$+\{5(a-c){\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+\frac{5b}{2}{\xi }^{3/2}[1-2{\sin }^{}({\xi }^{5/2})\left]\right\}\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+[A\sin \xi +B\cos \xi ]\{\frac{15}{2}(a-c){\xi }^{1/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+\frac{25}{2}(a-c){\xi }^3{\cos }^{}({\xi }^{5/2})-\frac{25}{2}(a-c){\xi }^3{\sin }^{}({\xi }^{5/2})$
		$+\frac{15b}{4}{\xi }^{1/2}[1-2{\sin }^2({\xi }^{5/2})]-25b{\xi }^3\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})\}$
	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d^2[A\sin \xi +B\cos \xi ]}{d^2\xi }$
		$+\{5(a-c){\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+\frac{5b}{2}{\xi }^{3/2}[1-2{\sin }^{}({\xi }^{5/2})\left]\right\}\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+[A\sin \xi +B\cos \xi ]\left\{\frac{15}{4}{\xi }^{1/2}\right[2(a-c)\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+b(1-2{\sin }^{}({\xi }^{5/2}))]$
		$+\frac{25}{2}{\xi }^3[-2b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+(a-c)(1-2{\sin }^{}({\xi }^{5/2})\left)\right]\}$

[Comment from J. E. Tohline on 9 April 2015: I'm not sure what else to make of this.]

[Additional comment from J. E. Tohline on 15 April 2015: It is perhaps worth mentioning that there is a similarity between the argument of the trigonometric function being used in this "third guess" and the Lane-Emden function derived by Srivastava for $n=5$ polytropes; and also a similarity between Srivastava's function and the functional form of the LHS that we constructed, above, in connection with our "second guess."]

Fourth Guess (n1)

Again, working with the polytropic (n = 1) wave equation written in the following form,

$\sin \xi [{\xi }^2{x}^{{\text{'}}^{\prime }}+2\xi x^{\text{'}}-2\alpha x]+\cos \xi [2{\xi }^2{x}^{\text{'}}+2\alpha \xi x]+{\sigma }^2{\xi }^{3}x=0.$

Now, let's try:

$x=a_0+b_1\xi \sin \xi +c_2{\xi }^2\cos \xi ,$

which means,

$x^{\text{'}}$	$=$	$b_1\sin \xi +b_1\xi \cos \xi +2c_2\xi \cos \xi -c_2{\xi }^2\sin \xi$
	$=$	$(b_1-c_2{\xi }^2)\sin \xi +(b_1+2c_2)\xi \cos \xi ,$
$x^{{\text{'}}^{\prime }}$	$=$	$(-2c_2\xi )\sin \xi +(b_1-c_2{\xi }^2)\cos \xi +(b_1+2c_2)\cos \xi -(b_1+2c_2)\xi \sin \xi$
	$=$	$-(2c_2+b_1+2c_2)\xi \sin \xi +(2b_1+2c_2-c_2{\xi }^2)\cos \xi .$

The LHS of the wave equation then becomes,

LHS	$=$	$\sin \xi \left\{{\xi }^2\right[-(2c_2+b_1+2c_2)\xi \sin \xi +(2b_1+2c_2-c_2{\xi }^2)\cos \xi ]+2\xi [(b_1-c_2{\xi }^2)\sin \xi +(b_1+2c_2)\xi \cos \xi ]-2\alpha [a_0+(b_1\xi )\sin \xi +(c_2{\xi }^2)\cos \xi \left]\right\}$
		$+\cos \xi \left\{2{\xi }^2\right[(b_1-c_2{\xi }^2)\sin \xi +(b_1+2c_2)\xi \cos \xi ]+2\alpha \xi [a_0+(b_1\xi )\sin \xi +(c_2{\xi }^2)\cos \xi \left]\right\}+{\sigma }^2{\xi }^3[a_0+b_1\xi \sin \xi +c_2{\xi }^2\cos \xi ]$
	$=$	$\sin \xi \left\{\right[-(2c_2+b_1+2c_2){\xi }^3\sin \xi +(2b_1+2c_2-c_2{\xi }^2){\xi }^2\cos \xi ]+[2(b_1-c_2{\xi }^2)\xi \sin \xi +2(b_1+2c_2){\xi }^2\cos \xi ]-2\alpha [a_0+(b_1\xi )\sin \xi +(c_2{\xi }^2)\cos \xi \left]\right\}$
		$+\cos \xi \left\{\right[2(b_1-c_2{\xi }^2){\xi }^2\sin \xi +2(b_1+2c_2){\xi }^3\cos \xi ]+[2a_0\alpha \xi +2b_1\alpha {\xi }^2\sin \xi +2c_2\alpha {\xi }^3\cos \xi \left]\right\}+{\sigma }^2[a_0{\xi }^3+b_1{\xi }^4\sin \xi +c_2{\xi }^5\cos \xi ]$
	$=$	$\sin \xi \{-2\alpha a_0+[-(2c_2+b_1+2c_2){\xi }^3+2(b_1-c_2{\xi }^2)\xi -2\alpha (b_1\xi )]\sin \xi +[(2b_1+2c_2-c_2{\xi }^2){\xi }^2+2(b_1+2c_2){\xi }^2-2\alpha (c_2{\xi }^2)]\cos \xi \}$
		$+\cos \xi \{+2a_0\alpha \xi +[2(b_1-c_2{\xi }^2){\xi }^2+2b_1\alpha {\xi }^2]\sin \xi +[2(b_1+2c_2){\xi }^3+2c_2\alpha {\xi }^3]\cos \xi \}+{\sigma }^2[a_0{\xi }^3+b_1{\xi }^4\sin \xi +c_2{\xi }^5\cos \xi ]$
	$=$	${\sigma }^2{a}_0{\xi }^3+[-(2c_2+b_1+2c_2){\xi }^3+2(b_1-c_2{\xi }^2)\xi -2\alpha (b_1\xi )]{\sin }^2\xi +[2(b_1+2c_2){\xi }^3+2c_2\alpha {\xi }^3](1-{\sin }^2\xi )$
		$+[(2b_1+2c_2-c_2{\xi }^2){\xi }^2+2(b_1+2c_2){\xi }^2-2\alpha (c_2{\xi }^2)]\sin \xi \cos \xi +[2(b_1-c_2{\xi }^2){\xi }^2+2b_1\alpha {\xi }^2]\sin \xi \cos \xi$
		$+{\sigma }^2[b_1{\xi }^4\sin \xi +c_2{\xi }^5\cos \xi ]+2a_0\alpha \xi \cos \xi -2\alpha a_0\sin \xi$
	$=$	$[{\sigma }^2{a}_0+2(b_1+2c_2)+2c_2\alpha ]{\xi }^3+\{+2(b_1)\xi -2\alpha (b_1\xi )+[-2c_2-2(b_1+2c_2)-2c_2\alpha -(2c_2+b_1+2c_2)]{\xi }^3\}{\sin }^2\xi$
		$+\{[(2b_1+2c_2)+2(b_1+2c_2)-2\alpha (c_2)+2(b_1)+2b_1\alpha ]{\xi }^2-3c_2{\xi }^4\}\sin \xi \cos \xi$
		$+\sin \xi [{\sigma }^2{b}_1{\xi }^4-2\alpha a_0]+\xi \cos \xi [{\sigma }^2{c}_2{\xi }^4+2a_0\alpha ]$
	$=$	$[{\sigma }^2{a}_0+2(b_1+2c_2)+2c_2\alpha ]{\xi }^3+\{2b_1(1-\alpha )-[2c_2(5+\alpha )+3b_1]{\xi }^2\}\xi {\sin }^2\xi$
		$+\{2(3-\alpha )(b_1+c_2)-3c_2{\xi }^2\}{\xi }^2\sin \xi \cos \xi +\sin \xi [{\sigma }^2{b}_1{\xi }^4-2\alpha a_0]+\xi \cos \xi [{\sigma }^2{c}_2{\xi }^4+2a_0\alpha ].$

Fifth Guess (n1)

Along a similar line of reasoning, let's try a function of the form,

$x$

$=$

$x_s\sin \xi +x_c\cos \xi +x_1{\sin }^{}\xi +x_2{\cos }^{}\xi +x_3\sin \xi \cos \xi ,$

where $x_s,x_c,x_1,x_2,$ and $x_3$ are five separate, as yet, unspecified (polynomial?) functions of $\xi$. This also means that,

$x^{\text{'}}$

$=$

$(x_s^{\text{'}}-x_c)\sin \xi +(x_c^{\text{'}}+x_s)\cos \xi +(x_1^{\text{'}}-x_3){\sin }^{}\xi +(x_2^{\text{'}}+x_3){\cos }^{}\xi +(x_3^{\text{'}}+2x_1-2x_2)\sin \xi \cos \xi ;$

and,

$x^{{\text{'}}^{\prime }}$

$=$

$(x_s^{{\text{'}}^{\prime }}-2x_c^{\text{'}}-x_s)\sin \xi +(x_c^{{\text{'}}^{\prime }}+2x_s^{\text{'}}-x_c)\cos \xi +(x_1^{{\text{'}}^{\prime }}-2x_3^{\text{'}}-2x_1+2x_2){\sin }^{}\xi +(x_2^{{\text{'}}^{\prime }}+2x_3^{\text{'}}+2x_1-2x_2){\cos }^{}\xi +(x_3^{{\text{'}}^{\prime }}+4x_1^{\text{'}}-4x_2^{\text{'}}-4x_3)\sin \xi \cos \xi .$

Hence the LHS of the polytropic (n = 1) wave equation becomes,

LHS	$=$	$\sin \xi \left\{{\xi }^2\right[(x_s^{{\text{'}}^{\prime }}-2x_c^{\text{'}}-x_s)\sin \xi +(x_c^{{\text{'}}^{\prime }}+2x_s^{\text{'}}-x_c)\cos \xi +(x_1^{{\text{'}}^{\prime }}-2x_3^{\text{'}}-2x_1+2x_2){\sin }^{}\xi +(x_2^{{\text{'}}^{\prime }}+2x_3^{\text{'}}+2x_1-2x_2){\cos }^{}\xi +(x_3^{{\text{'}}^{\prime }}+4x_1^{\text{'}}-4x_2^{\text{'}}-4x_3)\sin \xi \cos \xi ]$
		$+2\xi [(x_s^{\text{'}}-x_c)\sin \xi +(x_c^{\text{'}}+x_s)\cos \xi +(x_1^{\text{'}}-x_3){\sin }^{}\xi +(x_2^{\text{'}}+x_3){\cos }^{}\xi +(x_3^{\text{'}}+2x_1-2x_2)\sin \xi \cos \xi ]$
		$-2\alpha [x_s\sin \xi +x_c\cos \xi +x_1{\sin }^{}\xi +x_2{\cos }^{}\xi +x_3\sin \xi \cos \xi ]\}$
		$+\cos \xi \left\{2{\xi }^2\right[(x_s^{\text{'}}-x_c)\sin \xi +(x_c^{\text{'}}+x_s)\cos \xi +(x_1^{\text{'}}-x_3){\sin }^{}\xi +(x_2^{\text{'}}+x_3){\cos }^{}\xi +(x_3^{\text{'}}+2x_1-2x_2)\sin \xi \cos \xi ]$
		$+2\alpha \xi [x_s\sin \xi +x_c\cos \xi +x_1{\sin }^{}\xi +x_2{\cos }^{}\xi +x_3\sin \xi \cos \xi ]\}$
		$+{\sigma }^2{\xi }^3\{x_s\sin \xi +x_c\cos \xi +x_1{\sin }^{}\xi +x_2{\cos }^{}\xi +x_3\sin \xi \cos \xi \}$
	$=$	$[(x_s^{{\text{'}}^{\prime }}-2x_c^{\text{'}}-x_s){\xi }^2+2\xi (x_s^{\text{'}}-x_c)-2\alpha x_s+{\sigma }^2{\xi }^3{x}_1]{\sin }^2\xi +[(x_c^{{\text{'}}^{\prime }}+2x_s^{\text{'}}-x_c){\xi }^2+2\xi (x_c^{\text{'}}+x_s)-2\alpha x_c+2{\xi }^2(x_s^{\text{'}}-x_c)+2\alpha \xi x_s+{\sigma }^2{\xi }^3{x}_3]\sin \xi \cos \xi$
		$+[(x_1^{{\text{'}}^{\prime }}-2x_3^{\text{'}}-2x_1+2x_2){\xi }^2+2\xi (x_1^{\text{'}}-x_3)-2\alpha x_1]{\sin }^3\xi +[(x_2^{{\text{'}}^{\prime }}+2x_3^{\text{'}}+2x_1-2x_2){\xi }^2+2\xi (x_2^{\text{'}}+x_3)-2\alpha x_2+2{\xi }^2(x_3^{\text{'}}+2x_1-2x_2)+2\alpha \xi x_3]\sin \xi {\cos }^{}\xi$
		$+[(x_3^{{\text{'}}^{\prime }}+4x_1^{\text{'}}-4x_2^{\text{'}}-4x_3){\xi }^2+2\xi (x_3^{\text{'}}+2x_1-2x_2)-2\alpha x_3+2{\xi }^2(x_1^{\text{'}}-x_3)+2\alpha \xi x_1]{\sin }^2\xi \cos \xi$
		$+[2{\xi }^2(x_c^{\text{'}}+x_s)+2\alpha \xi x_c+{\sigma }^2{\xi }^3{x}_2]{\cos }^2\xi +[2{\xi }^2(x_2^{\text{'}}+x_3)+2\alpha \xi x_2]{\cos }^3\xi +{\sigma }^2{\xi }^3{x}_s\sin \xi +{\sigma }^2{\xi }^3{x}_c\cos \xi$
	$=$	$[(x_s^{{\text{'}}^{\prime }}-2x_c^{\text{'}}-x_s){\xi }^2+2\xi (x_s^{\text{'}}-x_c)-2\alpha x_s+{\sigma }^2{\xi }^3{x}_1]{\sin }^2\xi +[(x_c^{{\text{'}}^{\prime }}+2x_s^{\text{'}}-x_c){\xi }^2+2\xi (x_c^{\text{'}}+x_s)-2\alpha x_c+2{\xi }^2(x_s^{\text{'}}-x_c)+2\alpha \xi x_s+{\sigma }^2{\xi }^3{x}_3]\sin \xi \cos \xi$
		$+[(x_1^{{\text{'}}^{\prime }}-2x_3^{\text{'}}-2x_1+2x_2){\xi }^2+2\xi (x_1^{\text{'}}-x_3)-2\alpha x_1+{\sigma }^2{\xi }^3{x}_s]\sin \xi$
		$+[(x_2^{{\text{'}}^{\prime }}+2x_3^{\text{'}}+2x_1-2x_2){\xi }^2+2\xi (x_2^{\text{'}}+x_3)-2\alpha x_2+2{\xi }^2(x_3^{\text{'}}+2x_1-2x_2)+2\alpha \xi x_3-(x_1^{{\text{'}}^{\prime }}-2x_3^{\text{'}}-2x_1+2x_2){\xi }^2-2\xi (x_1^{\text{'}}-x_3)+2\alpha x_1]\sin \xi {\cos }^{}\xi$
		$+[(x_3^{{\text{'}}^{\prime }}+4x_1^{\text{'}}-4x_2^{\text{'}}-4x_3){\xi }^2+2\xi (x_3^{\text{'}}+2x_1-2x_2)-2\alpha x_3+2{\xi }^2(x_1^{\text{'}}-x_3)+2\alpha \xi x_1-2{\xi }^2(x_2^{\text{'}}+x_3)-2\alpha \xi x_2]{\sin }^2\xi \cos \xi$
		$+[2{\xi }^2(x_c^{\text{'}}+x_s)+2\alpha \xi x_c+{\sigma }^2{\xi }^3{x}_2]{\cos }^2\xi +[2{\xi }^2(x_2^{\text{'}}+x_3)+2\alpha \xi x_2+{\sigma }^2{\xi }^3{x}_c]\cos \xi$

So, the five chosen (polynomial?) functions of $\xi$ must simultabeously satisfy the following, seven 2^nd-order ODEs:

$\sin \xi$	:	$(x_1^{{\text{'}}^{\prime }}-2x_3^{\text{'}}-2x_1+2x_2){\xi }^2+2\xi (x_1^{\text{'}}-x_3)-2\alpha x_1+{\sigma }^2{\xi }^3{x}_s=0$
${\sin }^2\xi$	:	$(x_s^{{\text{'}}^{\prime }}-2x_c^{\text{'}}-x_s){\xi }^2+2\xi (x_s^{\text{'}}-x_c)-2\alpha x_s+{\sigma }^2{\xi }^3{x}_1=0$
${\sin }^2\xi \cos \xi$	:	$(x_3^{{\text{'}}^{\prime }}+4x_1^{\text{'}}-4x_2^{\text{'}}-4x_3){\xi }^2+2\xi (x_3^{\text{'}}+2x_1-2x_2)-2\alpha x_3+2{\xi }^2(x_1^{\text{'}}-x_3)+2\alpha \xi x_1-2{\xi }^2(x_2^{\text{'}}+x_3)-2\alpha \xi x_2=0$
$\sin \xi \cos \xi$	:	$(x_c^{{\text{'}}^{\prime }}+2x_s^{\text{'}}-x_c){\xi }^2+2\xi (x_c^{\text{'}}+x_s)-2\alpha x_c+2{\xi }^2(x_s^{\text{'}}-x_c)+2\alpha \xi x_s+{\sigma }^2{\xi }^3{x}_3=0$
$\sin \xi {\cos }^{}\xi$	:	$(x_2^{{\text{'}}^{\prime }}+2x_3^{\text{'}}+2x_1-2x_2){\xi }^2+2\xi (x_2^{\text{'}}+x_3)-2\alpha x_2+2{\xi }^2(x_3^{\text{'}}+2x_1-2x_2)+2\alpha \xi x_3-(x_1^{{\text{'}}^{\prime }}-2x_3^{\text{'}}-2x_1+2x_2){\xi }^2-2\xi (x_1^{\text{'}}-x_3)+2\alpha x_1=0$
${\cos }^2\xi$	:	$2{\xi }^2(x_c^{\text{'}}+x_s)+2\alpha \xi x_c+{\sigma }^2{\xi }^3{x}_2=0$
$\cos \xi$	:	$2{\xi }^2(x_2^{\text{'}}+x_3)+2\alpha \xi x_2+{\sigma }^2{\xi }^3{x}_c=0$

Example 1

Let's work on the coefficient of the $\cos \xi$ term:

$x_c$	$=$	${\xi }^{\beta }(A_c)$
$x_2$	$=$	${\xi }^{\beta }(C_2{\xi }^2)$
$x_3$	$=$	${\xi }^{\beta }(B_3\xi )$
$\Rightarrow$       Coefficient of "$\cos \xi$" term	$=$	${\xi }^{\beta }[2{\xi }^2(2C_2\xi +(B_3\xi ))+2\alpha \xi (C_2{\xi }^2)+{\sigma }^2{\xi }^3(A_c)]$
	$=$	${\xi }^{\beta +3}[2(2C_2+B_3)+2\alpha C_2+{\sigma }^2(A_c)]$
$\Rightarrow {\sigma }^2$	$=$	$-\frac{2}{A_c}[B_3+(2+\alpha )C_2]$

Sixth Guess (n1)

Rationale

From our review of the properties of $n=1$ polytropic spheres, we know that the equilibrium density distribution is given by the sinc function, namely,

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

$=$

$\frac{\sin \xi }{\xi },$

where,

$\xi \equiv \pi \left(\frac{r_0}{R_0}\right).$

The total mass is,

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}=\frac{4}{\pi }\cdot {\rho }_c{R}_0^3,$

and the fractional mass enclosed within a given radius, $r$, is,

$\frac{M_r(\xi )}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$

$=$

$\frac{1}{\pi }[\sin \xi -\xi \cos \xi ].$

Let's guess that, during the fundamental mode of radial oscillation, the sinc-function profile is preserved as the system's total radius varies. In particular, we will assume that the system's time-varying radius is,

$R=R_0(1+\frac{\delta R}{R_0})=R_0(1+{ϵ}_R),$

and seek to determine how the displacement vector, $ϵ\equiv \delta r/r_0$, varies with $r_0$ in order to preserve the overall sinc-function profile. As is usual, we will only examine small perturbations away from equilibrium, that is, we will assume that everywhere throughout the configuration, $|ϵ|\ll 1$.

Let's begin by defining a new dimensionless coordinate,

$\eta \equiv \pi \left(\frac{r}{R}\right)=\pi \left[\frac{r_0(1+ϵ)}{R_0(1+{ϵ}_R)}\right]\approx \xi (1+ϵ),$

and recognize that, in the new perturbed state, the fractional mass enclosed within a given radius, $r$, is,

$\frac{M_r(\eta )}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$

$=$

$\frac{1}{\pi }[\sin \eta -\eta \cos \eta ].$

In order to associate each mass shell in the perturbed configuration with its corresponding mass shell in the unperturbed, equilibrium state, we need to set the two $M_r$ functions equal to one another, that is, demand that,

$\sin \xi -\xi \cos \xi$	$=$	$\sin \eta -\eta \cos \eta$
	$\approx$	$\sin [\xi (1+ϵ)]-\xi (1+ϵ)\cos [\xi (1+ϵ)]$
	$=$	$[\sin \xi \cos (\xi ϵ)+\cos \xi \sin (\xi ϵ)]-\xi (1+ϵ)[\cos \xi \cos (\xi ϵ)-\sin \xi \sin (\xi ϵ)]$
	$\approx$	$\sin \xi [1-\frac{1}{2}(\xi ϵ{)}^2]+(\xi ϵ)\cos \xi -\xi (1+ϵ)\cos \xi [1-\frac{1}{2}(\xi ϵ{)}^2]+{\xi }^2ϵ(1+ϵ)\sin \xi$
	$\approx$	$\sin \xi -\xi \cos \xi -\frac{1}{2}(\xi ϵ{)}^2\sin \xi +(\xi ϵ)\cos \xi -(\xi ϵ)\cos \xi +\frac{1}{2}{\xi }^3{ϵ}^2\cos \xi +{\xi }^2ϵ\sin \xi +(\xi ϵ{)}^2\sin \xi$
$\Rightarrow -{\xi }^2ϵ\sin \xi$	$\approx$	$\frac{(\xi ϵ{)}^2}{2}[\xi \cos \xi +\sin \xi ]$
$\Rightarrow \frac{1}{ϵ}$	$\approx$	$-\frac{1}{2}[\xi \cdot \frac{\cos \xi }{\sin \xi }+1]$
$\Rightarrow ϵ$	$\approx$	$-2[1+\xi \cdot \frac{\cos \xi }{\sin \xi }{]}^{-1}=-2\sin \xi [\sin \xi +\xi \cos \xi {]}^{-1}.$

Resulting Polytropic Wave Equation

So, let's try,

$x$	$=$	$2\sin \xi [\sin \xi +\xi \cos \xi {]}^{-1}$
	$=$	$[\sin \xi +\xi \cos \xi {]}^{-3}2\sin \xi [{\sin }^{}\xi +2\xi \sin \xi \cos \xi +{\xi }^2{\cos }^{}\xi ],$

in which case,

$x^{\text{'}}$	$=$	$2\cos \xi [\sin \xi +\xi \cos \xi {]}^{-1}-2\sin \xi [\sin \xi +\xi \cos \xi {]}^{-2}[2\cos \xi -\xi \sin \xi ]$
	$=$	$[\sin \xi +\xi \cos \xi {]}^{-2}\{2\cos \xi [\sin \xi +\xi \cos \xi ]-2\sin \xi [2\cos \xi -\xi \sin \xi ]\}$
	$=$	$[\sin \xi +\xi \cos \xi {]}^{-2}[2\cos \xi \sin \xi +2\xi {\cos }^{}\xi -4\sin \xi \cos \xi +2\xi {\sin }^{}\xi ]$
	$=$	$2[\sin \xi +\xi \cos \xi {]}^{-2}[\xi -\sin \xi \cos \xi ]$
	$=$	$[\sin \xi +\xi \cos \xi {]}^{-3}2[\xi \sin \xi +{\xi }^2\cos \xi -{\sin }^{}\xi \cos \xi -\xi \sin \xi {\cos }^{}\xi ],$

and,

$x^{{\text{'}}^{\prime }}$	$=$	$2[\sin \xi +\xi \cos \xi {]}^{-2}[1-{\cos }^{}\xi +{\sin }^{}\xi ]-4[\sin \xi +\xi \cos \xi {]}^{-3}[\xi -\sin \xi \cos \xi ][2\cos \xi -\xi \sin \xi ]$
	$=$	$4[\sin \xi +\xi \cos \xi {]}^{-3}\{{\sin }^{}\xi [\sin \xi +\xi \cos \xi ]-[\xi -\sin \xi \cos \xi ][2\cos \xi -\xi \sin \xi ]\}$
	$=$	$4[\sin \xi +\xi \cos \xi {]}^{-3}[{\sin }^{}\xi +\xi \sin \xi \cos \xi +{\xi }^2\sin \xi -2\xi \cos \xi -\xi {\sin }^{}\xi \cos \xi +2\sin \xi {\cos }^{}\xi ]$

Graphical Reassessment

Before plowing ahead and plugging these expressions into the polytropic wave equation, I plotted the trial eigenfunction,

$ϵ(\xi /\pi )$

(see the blue curve in the accompanying "Trial Eigenfunction" figure), and noticed that it passes through

$\pm \mathrm{\infty }$

midway through the configuration. This is a very unphysical behavior. On the other hand, the inverse of this function (see the red curve) exhibits a relatively desirable behavior because it increases monotonically from negative one at the center. As plotted, however, the function has one node. In searching for the eigenfunction of the fundamental mode of oscillation, it might be better to add "1" to the inverse of the function and thereby get rid of all nodes. (Keep in mind, however, that the red curve might be displaying the eigenfunction associated with the first overtone.)

Let's therefore try,

$x=1+\frac{1}{ϵ}=1-\frac{1}{2}[\xi \cdot \frac{\cos \xi }{\sin \xi }+1]=\frac{1}{2}[1-\xi \cdot \frac{\cos \xi }{\sin \xi }].$

In this case we have,

$x^{\text{'}}$	$=$	$\frac{1}{2}[\xi -\frac{\cos \xi }{\sin \xi }+\xi \cdot \frac{{\cos }^2\xi }{{\sin }^2\xi }]$
	$=$	$\frac{1}{2{\sin }^2\xi }[\xi {\sin }^2\xi -\sin \xi \cos \xi +\xi {\cos }^{}\xi ],$
	$=$	$\frac{1}{2{\sin }^2\xi }[\xi -\sin \xi \cos \xi ],$

and,

$x^{{\text{'}}^{\prime }}$	$=$	$-\frac{\cos \xi }{{\sin }^3\xi }[\xi -\sin \xi \cos \xi ]+\frac{1}{2{\sin }^2\xi }[1-{\cos }^{}\xi +{\sin }^{}\xi ]$
	$=$	$1-\frac{\cos \xi }{{\sin }^3\xi }[\xi -\sin \xi \cos \xi ].$

Now let's plug these expressions into the polytropic (n = 1) wave equation, namely,

$-{\sigma }^2{\xi }^{3}x$

$=$

$\sin \xi [{\xi }^2{x}^{{\text{'}}^{\prime }}+2\xi x^{\text{'}}-2\alpha x]+\cos \xi [2{\xi }^2{x}^{\text{'}}+2\alpha \xi x].$

The first term inside the square brackets on the right-hand-side gives,

${\xi }^2{x}^{{\text{'}}^{\prime }}+2\xi x^{\text{'}}-2\alpha x$	$=$	${\xi }^2-\frac{\cos \xi }{{\sin }^3\xi }({\xi }^3-{\xi }^2\sin \xi \cos \xi )+\frac{1}{{\sin }^2\xi }({\xi }^2-\xi \sin \xi \cos \xi )-\alpha (1-\xi \cdot \frac{\cos \xi }{\sin \xi })$
	$=$	$\frac{1}{{\sin }^3\xi }[{\xi }^2{\sin }^3\xi -\cos \xi ({\xi }^3-{\xi }^2\sin \xi \cos \xi )+\sin \xi ({\xi }^2-\xi \sin \xi \cos \xi )-\alpha ({\sin }^{}\xi -\xi \cos \xi {\sin }^{}\xi )]$
	$=$	$\frac{1}{{\sin }^3\xi }[{\xi }^2\sin \xi (1-{\cos }^{}\xi )+{\xi }^2\sin \xi {\cos }^{}\xi -{\xi }^3\cos \xi +{\xi }^2\sin \xi -\xi {\sin }^{}\xi \cos \xi -\alpha {\sin }^{}\xi +\alpha \xi \cos \xi {\sin }^{}\xi ]$
	$=$	$-\alpha +\frac{1}{{\sin }^3\xi }[2{\xi }^2\sin \xi -{\xi }^3\cos \xi -\xi {\sin }^{}\xi \cos \xi +\alpha \xi \cos \xi {\sin }^{}\xi ];$

and the second term inside the square brackets on the right-hand-side gives,

$2{\xi }^2{x}^{\text{'}}+2\alpha \xi x$

$=$

$\frac{1}{{\sin }^2\xi }({\xi }^3-{\xi }^2\sin \xi \cos \xi )+\frac{\alpha }{\sin \xi }(\xi \sin \xi -{\xi }^2\cos \xi ).$

Put together, then, we have,

RHS	$=$	$\frac{1}{{\sin }^2\xi }[2{\xi }^2\sin \xi -{\xi }^3\cos \xi -\xi {\sin }^{}\xi \cos \xi +\alpha \xi \cos \xi {\sin }^{}\xi ]+\frac{\cos \xi }{{\sin }^2\xi }({\xi }^3-{\xi }^2\sin \xi \cos \xi )-\alpha \sin \xi +\alpha \cdot \frac{\cos \xi }{\sin \xi }(\xi \sin \xi -{\xi }^2\cos \xi )$
	$=$	$\frac{1}{{\sin }^2\xi }[2{\xi }^2\sin \xi -{\xi }^3\cos \xi -\xi {\sin }^{}\xi \cos \xi +\alpha \xi \cos \xi {\sin }^{}\xi +{\xi }^3\cos \xi -{\xi }^2\sin \xi {\cos }^{}\xi ]+\frac{\alpha }{\sin \xi }[-{\sin }^{}\xi +\xi \sin \xi \cos \xi -{\xi }^2{\cos }^{}\xi ]$
	$=$	$\frac{\xi }{\sin \xi }[2\xi -\sin \xi \cos \xi -\xi {\cos }^{}\xi ]-\frac{\alpha }{\sin \xi }[{\sin }^{}\xi +{\xi }^2{\cos }^{}\xi ]$
	$=$	$\frac{{\xi }^2}{\sin \xi }[1+{\sin }^2\xi -(\frac{\sin \xi }{\xi })\cos \xi -\alpha (\frac{{\sin }^2\xi }{{\xi }^2}+{\cos }^{}\xi \left)\right],$

and,

LHS

$=$

$-\frac{\xi {\sigma }^2}{2}\left(\frac{{\xi }^2}{\sin \xi }\right)[\sin \xi -\xi \cos \xi ].$

If our trial eigenfunction is a proper solution to the polytropic wave equation, then the difference of these two expressions should be zero. Let's see:

$\frac{\sin \xi }{{\xi }^2}(\mathrm{R}\mathrm{H}\mathrm{S}-\mathrm{L}\mathrm{H}\mathrm{S})$

$=$

$1+{\sin }^2\xi -\left(\frac{\sin \xi }{\xi }\right)\cos \xi -\alpha (\frac{{\sin }^2\xi }{{\xi }^2}+{\cos }^{}\xi )+\frac{\xi {\sigma }^2}{2}[\sin \xi -\xi \cos \xi ].$

This expression clearly is not zero, so our trial eigenfunction is not a good one. However, the terms in the wave equation did combine somewhat to give a fairly compact — albeit nonzero — expression. So we may be on the right track!

New Idea Involving Logarithmic Derivatives

Simplistic Layout

Let's begin, again, with the relevant LAWE, as provided above. After dividing through by $x$, we have,

$(\sin \xi )\frac{{\xi }^2}{x}\cdot \frac{d^{2}x}{d{\xi }^2}+2[\sin \xi +\xi \cos \xi ]\frac{\xi }{x}\cdot \frac{dx}{d\xi }+[{\sigma }^2{\xi }^3-2\alpha (\sin \xi -\xi \cos \xi )]=0,$

where,

${\sigma }^2$	$\equiv$	$\frac{{\omega }^2}{2\pi G{\rho }_c{\gamma }_g},$
$\alpha$	$\equiv$	$3-\frac{4}{{\gamma }_g}.$

Now, in addition to recognizing that,

$\frac{\xi }{x}\cdot \frac{dx}{d\xi }$

$=$

$\frac{d\ln x}{d\ln \xi },$

in a separate context, we showed that, quite generally,

$\frac{{\xi }^2}{x}\cdot \frac{d^{2}x}{d{\xi }^2}$

$=$

$\frac{d}{d\ln \xi }\left[\frac{d\ln x}{d\ln \xi }\right]-[1-\frac{d\ln x}{d\ln \xi }]\cdot \frac{d\ln x}{d\ln \xi }.$

Hence, if we assume that the eigenfunction is a power-law of $\xi$, that is, assume that,

$x=a_0{\xi }^{c_0},$

then the logarithmic derivative of $x$ is a constant, namely,

$\frac{d\ln x}{d\ln \xi }=c_0,$

and the two key derivative terms will be,

$\frac{\xi }{x}\cdot \frac{dx}{d\xi }=c_0,$

and

$\frac{{\xi }^2}{x}\cdot \frac{d^{2}x}{d{\xi }^2}=c_0(c_0-1).$

In this case, the LAWE is no longer a differential equation but, instead, takes the form,

$-{\sigma }^2{\xi }^3$	$=$	$c_0(c_0-1)\sin \xi +2c_0[\sin \xi +\xi \cos \xi ]-2\alpha (\sin \xi -\xi \cos \xi )$
	$=$	$\sin \xi [c_0(c_0-1)+2c_0-2\alpha ]+\xi \cos \xi [2(c_0+\alpha )]$
	$=$	$\sin \xi [c_0^2+c_0-2\alpha ]+\xi \cos \xi [2(c_0+\alpha )].$

Now, the cosine term will go to zero if $c_0=-\alpha$; and the sine term will go to zero if,

$\alpha$	$=$	$3$
$\Rightarrow {\gamma }_g$	$=$	$\mathrm{\infty }.$

If these two — rather strange — conditions are met, then we have a marginally unstable configuration because, ${\sigma }^2=0$. This, in and of itself, is not very physically interesting. However, it may give us a clue regarding how to more generally search for a physically reasonable radial eigenfunction.

More general Assumption

Try,

$x$	$=$	${\xi }^{c_0}[a_0+b_0\sin \xi +d_0\xi \cos \xi ]$
$\Rightarrow \frac{dx}{d\xi }$	$=$	${\xi }^{c_0}\frac{d}{d\xi }[a_0+b_0\sin \xi +d_0\xi \cos \xi ]+c_0{\xi }^{c_0-1}[a_0+b_0\sin \xi +d_0\xi \cos \xi ]$
	$=$	${\xi }^{c_0}[b_0\cos \xi -d_0\xi \sin \xi +d_0\cos \xi ]+c_0{\xi }^{c_0-1}[a_0+b_0\sin \xi +d_0\xi \cos \xi ]$
$\Rightarrow \frac{d\ln x}{d\ln \xi }$	$=$	$\xi [b_0\cos \xi -d_0\xi \sin \xi +d_0\cos \xi ][a_0+b_0\sin \xi +d_0\xi \cos \xi {]}^{-1}+c_0$
	$=$	$[(b_0+d_0)\xi \cos \xi -d_0{\xi }^2\sin \xi ][a_0+b_0\sin \xi +d_0\xi \cos \xi {]}^{-1}+c_0$

See Also