Editing SSC/Stability/n1PolytropeLAWE

__FORCETOC__
=Radial Oscillations of n = 1 Polytropic Spheres (Pt 1)=
<table border="1" align="center" width="100%" colspan="8">
<tr>
  <td align="center" rowspan="1" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/n1PolytropeLAWE|Part I: &nbsp; Search for Analytic Solutions]]<br />&nbsp;</td>
  <td align="center" rowspan="1" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/n1PolytropeLAWE/Pt2|Part II:&nbsp; New Ideas]]<br />&nbsp;</td>
  <td align="center" rowspan="1" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/n1PolytropeLAWE/Pt3|Part III:&nbsp; What About Bipolytropes?]]<br />&nbsp;</td>
  <td align="center" rowspan="1" bgcolor="lightblue"><br />[[SSC/Stability/n1PolytropeLAWE/Pt4|Part IV:&nbsp; Most General Structural Solution]]<br />&nbsp;</td>
</tr>
</table>


As far as we have been able to ascertain, the first technical examination of radial oscillation modes in <math>n=1</math> polytropes was performed &#8212; using numerical techniques &#8212; in 1951 by L. D. Chatterji; at the time, he was in the Mathematics Department of Allahabad University.  His two papers on this topic were published in, what  is now referred to as, the ''Proceedings of the Indian National Science Academy'' (PINSA).  The citations that immediately follow this opening paragraph provide inks to both of these papers by Chatterji, but the links may be insecure.  (Citations/links to articles that provide analyses of models having other polytropic indexes are provided at the [[#See_Also|bottom of this chapter]].)  Apparently [https://www.springer.com/journal/43538 Springer] is archiving recent PINSA volumes, but their holdings do not date back as early as 1951.

* {{ Chatterji51full }}, ''Radial Oscillations of a Gaseous Star of Polytropic Index I'' 
* {{ Chatterji52full }}, ''Anharmonic Pulsations of a Polytropic Model of Index Unity''

A detailed review of {{ Chatterji51hereafter }} is provided in [[SSC/Structure/BiPolytropes/Analytic51Renormalize/Pt2#Setup|an accompanying discussion]].

==Groundwork==

In an [[SSC/Perturbations#2ndOrderODE|accompanying discussion]], we derived the so-called,

<div align="center" id="2ndOrderODE">
<font color="#770000">'''Adiabatic Wave''' (or ''Radial Pulsation'') '''Equation'''</font><br />

{{Math/EQ_RadialPulsation01}}
</div>

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 [[SSC/Perturbations#Dimensionless_Expression|dimensionless form]]:
<div align="center">
<math>
\frac{d^2x}{d\chi_0^2} + \biggl[\frac{4}{\chi_0} - \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} \biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \biggr] \frac{dx}{d\chi_0} + \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} \biggl(\frac{1}{\gamma_\mathrm{g}} \biggr)\biggl[\tau_\mathrm{SSC}^2 \omega^2 + (4 - 3\gamma_\mathrm{g})\biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \frac{1}{\chi_0} \biggr]  x = 0 ,
</math><br />
</div>
where,
<div align="center">
<math>~g_\mathrm{SSC} \equiv \frac{P_c}{R\rho_c} \, ,</math> &nbsp; &nbsp; &nbsp; and  &nbsp; &nbsp; &nbsp; <math>~\tau_\mathrm{SSC} \equiv \biggl[\frac{R^2 \rho_c}{P_c}\biggr]^{1/2} \, .</math>
</div>

In a [[SSC/Stability/Polytropes#Radial_Oscillations_of_Polytropic_Spheres|separate discussion]], we showed that specifically for isolated, ''polytropic'' configurations, this linear adiabatic wave equation (LAWE) can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4 - (n+1)V(\xi)}{\xi} \biggr] \frac{dx}{d\xi} + 
\biggl[\frac{\omega^2}{\gamma_g \theta} \biggl(\frac{n+1 }{4\pi G \rho_c} \biggr) - 
\biggl(3-\frac{4}{\gamma_g}\biggr)  \cdot \frac{(n+1)V(x)}{\xi^2} \biggr]  x </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} + 
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma_g } - 
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr]  x \, ,</math>
  </td>
</tr>
</table>
</div>
where we have adopted the dimensionless frequency notation,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sigma_c^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3\omega^2}{2\pi G \rho_c}  \, .</math>
  </td>
</tr>
</table>
</div>

Here we focus on an analysis of the ''specific'' case of isolated, <math>~n=1</math> polytropic configurations, whose unperturbed equilibrium structure can be prescribed in terms of analytic functions.  Our hope &#8212; as yet unfulfilled &#8212; 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 [[SSC/Structure/Polytropes#n_.3D_1_Polytrope|structure of an n = 1 polytrope]], in terms of the configuration's radius <math>R</math> and mass <math>M</math>, the central pressure and density are, respectively,
<div align="center">
<math>P_c = \frac{\pi G}{8}\biggl( \frac{M^2}{R^4} \biggr) </math> , 
</div>
and
<div align="center">
<math>\rho_c = \frac{\pi M}{4 R^3} </math> .
</div>
Hence the characteristic time and acceleration are, respectively,
<div align="center">
<math> 
\tau_\mathrm{SSC} = \biggl[ \frac{R^2 \rho_c}{P_c} \biggr]^{1/2} = 
\biggl[ \frac{2R^3 }{GM} \biggr]^{1/2} = 
\biggl[ \frac{\pi}{2 G\rho_c} \biggr]^{1/2},
</math><br />
</div>
and,
<div align="center">
<math> 
g_\mathrm{SSC} = \frac{P_c}{R \rho_c} = \biggl( \frac{GM}{2R^2} \biggr) .
</math><br />
</div>

The required functions are,
* <font color="red">Density</font>: 
<div align="center">
<math>\frac{\rho_0(\chi_0)}{\rho_c} = \frac{\sin(\pi\chi_0)}{\pi\chi_0} </math> ;
</div>

* <font color="red">Pressure</font>: 
<div align="center">
<math>\frac{P_0(\chi_0)}{P_c} = \biggl[ \frac{\sin(\pi\chi_0)}{\pi\chi_0} \biggr]^2 </math> ;
</div>

* <font color="red">Gravitational acceleration</font>: 
<div align="center">
<math> 
\frac{g_0(r_0)}{g_\mathrm{SSC}} = \frac{2}{\chi_0^2} \biggl[ \frac{M_r(\chi_0)}{M}\biggr]  =
\frac{2}{\pi \chi_0^2} \biggl[ \sin (\pi\chi_0 ) - \pi\chi_0 \cos (\pi\chi_0 ) \biggr].
</math><br />
</div>

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

<div align="center">
<math>
\frac{d^2x}{d\chi_0^2} + \frac{2}{\chi_0} \biggl[ 1 +  \pi\chi_0 \cot (\pi\chi_0 ) \biggr]  \frac{dx}{d\chi_0} + \frac{1}{\gamma_\mathrm{g}} \biggl\{ \frac{\pi \chi_0}{\sin(\pi\chi_0)} \biggl[ \frac{\pi \omega^2}{2G\rho_c} \biggr] + \frac{2}{\chi_0^2 } (4 - 3\gamma_\mathrm{g}) \biggl[ 1 - \pi\chi_0 \cot (\pi\chi_0 ) \biggr] \biggr\}  x = 0 ,
</math><br />
</div>
<br />
or, replacing <math>\chi_0</math> with <math>\xi \equiv \pi\chi_0</math> and dividing the entire expression by <math>\pi^2</math>, we have,

<div align="center">
<math>
\frac{d^2x}{d\xi^2} + \frac{2}{\xi} \biggl[ 1 +  \xi \cot \xi \biggr]  \frac{dx}{d\xi} + \frac{1}{\gamma_\mathrm{g}} \biggl\{ \frac{\xi}{\sin \xi} \biggl[ \frac{\omega^2}{2\pi G\rho_c} \biggr] + \frac{2}{\xi^2 } (4 - 3\gamma_\mathrm{g}) \biggl[ 1 - \xi \cot \xi \biggr] \biggr\}  x = 0 .
</math><br />
</div>
<br />

This is identical to the formulation of the wave equation that is relevant to  the (n = 1) core of the composite polytrope studied by [http://adsabs.harvard.edu/abs/1985PASAu...6..222M J. O. Murphy &amp; R. Fiedler (1985b)]; for comparison, their expression is displayed, in the following boxed-in image.

<div align="center">
<table border="2" cellpadding="10" id="MurphyFiedler1985b">
<tr>
  <td align="center">
n = 1 Polytropic Formulation of the LAWE as Presented by &hellip;

{{ MF85bfigure }}
<!--[http://adsabs.harvard.edu/abs/1985PASAu...6..222M Murphy &amp; Fiedler (1985b)] -->
  </td>
<tr>
  <td>
<!-- [[File:MurphyFiedlerN1formulation.png|700px|center|Murphy &amp; Fiedler (1985b)]] -->
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="center">
<math>
\frac{d^2\eta}{d\zeta^2} 
+ \frac{1}{\zeta}\biggl[ 4 + \frac{2(\zeta \cos\zeta - \sin\zeta)}{\sin\zeta}\biggr]\frac{d\eta}{d\zeta}
+ \biggl[ \frac{\omega_k^2 \zeta}{\sin\zeta} + \frac{2\alpha^* (\zeta \cos\zeta - \sin\zeta)}{\zeta^2\sin\zeta} \biggr]\eta 
= 0
</math>
  </td>
</tr>
</table>
  </td>
</tr>
</table>
</div>

{{ SGFworkInProgress }}

<span id="WorkInProgress">&nbsp;</span>

<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">
From an [[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|accompanying discussion]], we find the,
<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>
For an isolated n = 1 <math>(\gamma_g = 2, \alpha = 1)</math> polytrope, we know that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\theta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>\frac{\sin\xi}{\xi}</math>
  </td>
  <td align="right">
&nbsp; &nbsp; <math>\Rightarrow ~~~ Q(\xi) \equiv - \frac{d \ln \theta}{d\ln \xi}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>\biggl[1 - \xi\cot\xi\biggr] \, .</math>
  </td>
</tr>
</table>
Hence, the relevant LAWE is,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{d^2 x}{d\xi^2} 
+ 
\biggl[4 - 2(1 - \xi\cot\xi)\biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+
2 \biggl[\biggl( \frac{\sigma_c^2}{12}\biggr)\frac{\xi^3}{\sin\xi} - (1 - \xi\cot\xi)  \biggr]\frac{x}{\xi^2}
</math>
  </td>
</tr>
</table>

<div align="center">
<font color="maroon"><b>LAWE for n = 1 Polytrope</b></font><br />
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{d^2 x}{d\xi^2} 
+ 
\frac{2}{\xi} \biggl[1 +  \xi\cot\xi\biggr] \frac{dx}{d\xi}
+
\frac{1}{2} \biggl[\biggl( \frac{\sigma_c^2}{3}\biggr)\frac{\xi}{\sin\xi} - \frac{4}{\xi^2} \biggl(1 - \xi\cot\xi \biggr)  \biggr]x
</math>
  </td>
</tr>
</table>
</div>
This matches precisely the expression derived immediately above.

[[SSC/Stability/Polytropes#Boundary_Conditions|Surface boundary condition]]:
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>- \frac{d\ln x}{d\ln \xi} \biggr|_\mathrm{surf}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl( \frac{3-n}{n+1}\biggr) + \frac{n\sigma_c^2}{6(n+1)}
\biggl[ \frac{\xi}{\theta'}\biggr]_\mathrm{surf}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ - \frac{d\ln x}{d\ln \xi} \biggr|_\mathrm{surf}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
1 + \frac{\sigma_c^2}{12}
\biggl[ \frac{\xi^3}{(\xi \cos\xi - \sin\xi)}\biggr]_{\xi=\pi}
=
1 - \frac{\pi^2 \sigma_c^2}{12}
</math>
  </td>
</tr>
</table>

</td></tr></table>

==Attempt at Deriving an Analytic Eigenvector Solution==
Multiplying the last expression through by <math>~\xi^2\sin\xi</math> gives,

<div align="center">
<math>
(\xi^2\sin\xi ) \frac{d^2x}{d\xi^2} + 2 \biggl[ \xi \sin\xi +  \xi^2 \cos \xi \biggr]  \frac{dx}{d\xi} + 
\biggl[ \sigma^2 \xi^3  - 2\alpha ( \sin\xi - \xi \cos \xi ) \biggr]  x = 0 \, ,
</math><br />
</div>
<br />
where,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\sigma^2</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{\omega^2}{2\pi G\rho_c \gamma_g} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\alpha</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
3-\frac{4}{\gamma_g}
\, .
</math>
  </td>
</tr>

</table>
</div>

The first two terms can be folded together to give,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\frac{1}{\xi^2 \sin^2\xi} \cdot \frac{d}{d\xi}\biggl[ \xi^2 \sin^2\xi \frac{dx}{d\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{\xi^2 \sin\xi} \biggl[ 2\alpha ( \sin\xi - \xi \cos \xi ) - \sigma^2 \xi^3 \biggr]  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{2\alpha}{\xi^2} \biggl( \frac{\xi \cos \xi}{\sin\xi} -1\biggr) + \sigma^2 \biggl( \frac{\xi}{\sin\xi}\biggr) \biggr]  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{2\alpha}{\xi^2} \frac{\xi^2}{\sin\xi}  \cdot \frac{d}{d\xi} \biggl( \frac{\sin\xi}{\xi} \biggr) 
+ \sigma^2 \biggl( \frac{\xi}{\sin\xi}\biggr) \biggr]  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{2\alpha}{\xi} \frac{d}{d\xi}\biggl( \frac{\sin\xi}{\xi} \biggr) + \sigma^2 \biggr]  \biggl( \frac{\xi}{\sin\xi}\biggr) x \, ,
</math>
  </td>
</tr>
</table>
</div>
where, in order to make this next-to-last step, we have recognized that,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\frac{d}{d\xi} \biggl( \frac{\sin\xi}{\xi} \biggr)
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\sin\xi}{\xi^2} \biggl[ \frac{\xi \cos\xi}{\sin\xi} - 1 \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

It would seem that the eigenfunction, <math>x(\xi)</math>, should be expressible in terms of trigonometric functions and powers of <math>~\xi</math>; indeed, it appears as though the expression governing this eigenfunction would simplify considerably if <math>x \propto \sin\xi/\xi</math>.  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,
<div align="center">
<math>x = \frac{\sin\xi}{\xi}  \, ,</math>
</div>
which means,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>x^' \equiv \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\sin\xi}{\xi^2} \biggl[ \frac{\xi \cos\xi}{\sin\xi} - 1 \biggr]
\, .
</math>
  </td>
</tr>

</table>
</div>
Does this satisfy the governing expression?   Let's see.  The right-and-side (RHS) gives:
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{2\alpha}{\xi} \frac{d}{d\xi}\biggl( \frac{\sin\xi}{\xi} \biggr) + \sigma^2 \biggr]  \biggl( \frac{\xi}{\sin\xi}\biggr) x
= - \biggl[ \frac{2\alpha x^'}{\xi} + \sigma^2 \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

At the same time, the left-hand-side (LHS) may, quite generically, be written as:
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{x^'}{\xi}
\biggl\{ \frac{\xi}{(\xi^2 \sin^2\xi)x^'} \cdot \frac{d[ (\xi^2 \sin^2\xi)x^']}{d\xi} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{x^'}{\xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Putting the two sides together therefore gives,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\frac{x^'}{\xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} +2\alpha \biggr] 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\sigma^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']^{1/(2\alpha)}}{d\ln\xi} +1 \biggr] 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{\sigma^2}{2\alpha }  \biggl( \frac{\xi}{x^'} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
\frac{d\ln[ (\xi^2 \sin^2\xi)x^']^{-1/(2\alpha)}}{d\ln\xi} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \frac{\sigma^2}{2\alpha }  \biggl( \frac{\xi}{x^'} \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>
[<font color="red">Comment from J. E. Tohline on 6 April 2015:</font> 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,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~ \frac{x^'}{\xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \biggl[ \frac{2\alpha}{\xi} \frac{d}{d\xi}\biggl( \frac{\sin\xi}{\xi} \biggr) + \sigma^2 \biggr]  \biggl( \frac{\xi}{\sin\xi}\biggr) x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~~ \frac{x^'}{x}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- 2\alpha\biggl( \frac{\xi}{\sin\xi}\biggr) \frac{d}{d\xi}\biggl( \frac{\sin\xi}{\xi} \biggr) 
~- \sigma^2\biggl( \frac{\xi^2}{\sin\xi}\biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~~ \frac{d\ln(x)}{d\ln \xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- 2\alpha\biggl[ \frac{d\ln(\sin\xi/\xi)}{d\ln \xi} \biggr]
~- \sigma^2\biggl( \frac{\xi^3}{\sin\xi}\biggr)  \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~\Rightarrow ~~~~ - \sigma^2  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{\sin\xi}{\xi^3}\biggr) \biggl\{\frac{d\ln(x)}{d\ln \xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr] +
2\alpha\biggl[ \frac{d\ln(\sin\xi/\xi)}{d\ln \xi} \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
[<font color="red">Comment from J. E. Tohline on 6 April 2015:</font> I'm not sure what else to make of this.]


===Third Guess (n1)===
Let's rewrite the polytropic (n = 1) wave equation as follows:
<div align="center">
<math>
~\sin\xi \biggl[ \xi^2 x^{''} + 2\xi x^' - 2\alpha x \biggr]
+ \cos\xi \biggl[ 2\xi^2 x^' + 2\alpha \xi x \biggr]
+\sigma^2 \xi^3 x = 0 \, .
</math>
</div>
It is difficult to determine what term in the adiabatic wave equation will cancel the term involving <math>~\sigma^2</math> because its leading coefficient is <math>~\xi^3</math> and no other term contains a power of <math>~\xi</math> that is higher than two.  After thinking through various trial eigenvector expressions, <math>~x(\xi)</math>, 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 <math>~\xi^3</math> while the function itself does not introduce any additional factors of <math>~\xi</math> into the term that contains <math>~\sigma^2</math>:
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] [ A\sin\xi + B\cos\xi]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ \frac{d[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})]}{d\xi} \cdot [ A\sin\xi + B\cos\xi] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ [ A\sin\xi + B\cos\xi] 
\biggl\{5a \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2}) 
+ b\biggl[ \frac{5}{2} \xi^{3/2}\cos^2(\xi^{5/2})  - \frac{5}{2} \xi^{3/2}\sin^2(\xi^{5/2}) \biggr]
- 5c \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2})  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ [ A\sin\xi + B\cos\xi] 
\biggl\{5(a-c) \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2}) 
+ \frac{5b}{2} \xi^{3/2}\biggl[ 1  - 2\sin^2(\xi^{5/2}) \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x^{''}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d^2[ A\sin\xi + B\cos\xi]}{d^2\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+
\biggl\{5(a-c) \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2}) 
+ \frac{5b}{2} \xi^{3/2}\biggl[ 1  - 2\sin^2(\xi^{5/2}) \biggr]
\biggr\} \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ [ A\sin\xi + B\cos\xi] \biggl\{
\frac{15}{2}(a-c) \xi^{1/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2}) 
+\frac{25}{2}(a-c) \xi^{3}  \cos^2(\xi^{5/2}) 
- \frac{25}{2}(a-c) \xi^{3}  \sin^2(\xi^{5/2}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \frac{15b}{4} \xi^{1/2}\biggl[ 1  - 2\sin^2(\xi^{5/2}) \biggr]
- 25b \xi^{3}\sin(\xi^{5/2}) \cos(\xi^{5/2}) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d^2[ A\sin\xi + B\cos\xi]}{d^2\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+
\biggl\{5(a-c) \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2}) 
+ \frac{5b}{2} \xi^{3/2}\biggl[ 1  - 2\sin^2(\xi^{5/2}) \biggr]
\biggr\} \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ [ A\sin\xi + B\cos\xi] \biggl\{ \frac{15}{4}\xi^{1/2} \biggl[
2(a-c) \sin(\xi^{5/2}) \cos(\xi^{5/2}) 
+ b\biggl( 1  - 2\sin^2(\xi^{5/2}) \biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \frac{25}{2} \xi^3 \biggl[
- 2b \sin(\xi^{5/2}) \cos(\xi^{5/2}) 
+(a-c) \biggl( 1- 2\sin^2(\xi^{5/2}) \biggr) \biggr] \biggr\}
</math>
  </td>
</tr>

</table>
</div>
[<font color="red">Comment from J. E. Tohline on 9 April 2015:</font> I'm not sure what else to make of this.]

[<font color="red">Additional comment from J. E. Tohline on 15 April 2015:</font> 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 [[SSC/Structure/Polytropes#Srivastava.27s_F-Type_Solution|Lane-Emden function derived by Srivastava for <math>~n=5</math> polytropes]]; and also a similarity between Srivastava's function and the functional form of the LHS that we constructed, [[SSC/Stability/Polytropes#Second_Guess|above, in connection with our "second guess]]."]

===Fourth Guess (n1)===
Again, working with the polytropic (n = 1) wave equation written in the following form,
<div align="center">
<math>
\sin\xi \biggl[ \xi^2 x^{''} + 2\xi x^' - 2\alpha x \biggr]
+ \cos\xi \biggl[ 2\xi^2 x^' + 2\alpha \xi x \biggr]
+\sigma^2 \xi^3 x = 0 \, .
</math>
</div>
Now, let's try:
<div align="center">
<math>x = a_0 + b_1 \xi \sin\xi + c_2 \xi^2 \cos\xi  \, ,</math>
</div>
which means,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>x^' </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
b_1 \sin\xi + b_1 \xi \cos\xi + 2c_2 \xi \cos\xi - c_2\xi^2 \sin\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(b_1 - c_2\xi^2 ) \sin\xi + (b_1  + 2c_2)\xi \cos\xi \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>x^{''} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(- 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 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-(2c_2+b_1 + 2c_2 ) \xi \sin\xi + (2b_1 + 2c_2 - c_2\xi^2  ) \cos\xi  \, .
</math>
  </td>
</tr>

</table>
</div>
The LHS of the wave equation then becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\sin\xi \biggl\{ \xi^2 \biggl[ -(2c_2+b_1 + 2c_2 ) \xi \sin\xi + (2b_1 + 2c_2 - c_2\xi^2  ) \cos\xi \biggr] 
+ 2\xi \biggl[ (b_1 - c_2\xi^2 ) \sin\xi + (b_1  + 2c_2)\xi \cos\xi  \biggr] 
- 2\alpha \biggl[ a_0 + (b_1 \xi) \sin\xi + (c_2 \xi^2) \cos\xi  \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \cos\xi \biggl\{ 2\xi^2 \biggl[ (b_1 - c_2\xi^2 ) \sin\xi + (b_1  + 2c_2)\xi \cos\xi  \biggr] 
+ 2\alpha \xi \biggl[ a_0 + (b_1 \xi) \sin\xi + (c_2 \xi^2) \cos\xi  \biggr] \biggr\}
+\sigma^2 \xi^3 \biggl[ a_0 + b_1 \xi \sin\xi + c_2 \xi^2 \cos\xi  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\sin\xi \biggl\{  \biggl[ -(2c_2+b_1 + 2c_2 ) \xi^3 \sin\xi + (2b_1 + 2c_2 - c_2\xi^2  ) \xi^2\cos\xi \biggr] 
+ \biggl[ 2(b_1 - c_2\xi^2 )\xi \sin\xi + 2(b_1  + 2c_2)\xi^2 \cos\xi  \biggr] 
- 2\alpha \biggl[ a_0 + (b_1 \xi) \sin\xi + (c_2 \xi^2) \cos\xi  \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \cos\xi \biggl\{ \biggl[ 2(b_1 - c_2\xi^2 )\xi^2 \sin\xi + 2(b_1  + 2c_2)\xi^3 \cos\xi  \biggr] 
+ \biggl[ 2a_0\alpha \xi  + 2b_1\alpha \xi^2 \sin\xi + 2c_2 \alpha \xi^3 \cos\xi  \biggr] \biggr\}
+\sigma^2 \biggl[ a_0\xi^3  + b_1 \xi^4 \sin\xi + c_2 \xi^5 \cos\xi  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\sin\xi \biggl\{-  2\alpha a_0 +  \biggl[ -(2c_2+b_1 + 2c_2 ) \xi^3  + 2(b_1 - c_2\xi^2 )\xi   - 2\alpha (b_1 \xi) \biggr]\sin\xi 
+ \biggl[ (2b_1 + 2c_2 - c_2\xi^2  ) \xi^2 + 2(b_1  + 2c_2)\xi^2  - 2\alpha (c_2 \xi^2) \biggr] \cos\xi  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \cos\xi \biggl\{ + 2a_0\alpha \xi  + \biggl[ 2(b_1 - c_2\xi^2 )\xi^2 + 2b_1\alpha \xi^2 \biggr] \sin\xi   
+ \biggl[ 2(b_1  + 2c_2)\xi^3   + 2c_2 \alpha \xi^3\biggr] \cos\xi  \biggr\}
+\sigma^2 \biggl[ a_0\xi^3  + b_1 \xi^4 \sin\xi + c_2 \xi^5 \cos\xi  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\sigma^2 a_0 \xi^3 +  \biggl[ -(2c_2+b_1 + 2c_2 ) \xi^3  + 2(b_1 - c_2\xi^2 )\xi   - 2\alpha (b_1 \xi) \biggr]\sin^2\xi 
+ \biggl[ 2(b_1  + 2c_2)\xi^3   + 2c_2 \alpha \xi^3\biggr] \biggl(1-\sin^2\xi \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[ (2b_1 + 2c_2 - c_2\xi^2  ) \xi^2 + 2(b_1  + 2c_2)\xi^2  - 2\alpha (c_2 \xi^2) \biggr] \sin\xi \cos\xi  
+ \biggl[ 2(b_1 - c_2\xi^2 )\xi^2 + 2b_1\alpha \xi^2 \biggr] \sin\xi  \cos\xi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\sigma^2 \biggl[ b_1 \xi^4 \sin\xi + c_2 \xi^5 \cos\xi  \biggr] + 2a_0\alpha \xi\cos\xi   -  2\alpha a_0 \sin\xi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \sigma^2 a_0  +  2(b_1  + 2c_2)   + 2c_2 \alpha \biggr]\xi^3 +  \biggl\{+ 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  \biggr\} \sin^2\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl\{ [ (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 \biggr\} \sin\xi  \cos\xi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>+
\sin\xi \biggl[\sigma^2b_1 \xi^4 -  2\alpha a_0  \biggr] + \xi \cos\xi \biggl[\sigma^2 c_2 \xi^4 + 2a_0\alpha \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
[ \sigma^2 a_0  +  2(b_1  + 2c_2)   + 2c_2 \alpha ]\xi^3 +  
\biggl\{2 b_1(1-\alpha) - [2c_2(5+\alpha) + 3b_1] \xi^2 \biggr\} \xi \sin^2\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl\{ 2(3-\alpha)( b_1+c_2 )     -3c_2\xi^2 \biggr\} \xi^2 \sin\xi  \cos\xi  
+\sin\xi \biggl[\sigma^2b_1 \xi^4 -  2\alpha a_0  \biggr] + \xi \cos\xi \biggl[\sigma^2 c_2 \xi^4 + 2a_0\alpha \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

===Fifth Guess (n1)===
Along a similar line of reasoning, let's try a function of the form,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x_s \sin\xi + x_c \cos\xi + x_1 \sin^2\xi + x_2 \cos^2\xi + x_3 \sin\xi \cos\xi\, ,
</math>
  </td>
</tr>

</table>
</div>
where <math>x_s, x_c, x_1, x_2,</math> and <math>x_3</math> are five separate, as yet, unspecified (polynomial?) functions of <math>\xi</math>.  This also means that,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>x^'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(x_s^' - x_c)\sin\xi + (x_c^' + x_s)\cos\xi + (x_1^' - x_3)\sin^2\xi + (x_2^' + x_3)\cos^2\xi + (x_3^' + 2x_1 -2x_2)\sin\xi \cos\xi \, ;
</math>
  </td>
</tr>

</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>x^{''}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(x_s^{''} - 2x_c^{'} - x_s)\sin\xi + (x_c^{''} + 2x_s^' -x_c)\cos\xi 
+ (x_1^{''} -2x_3^' -2x_1 + 2x_2)\sin^2\xi + (x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\cos^2\xi 
+ (x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\sin\xi \cos\xi \, .
</math>
  </td>
</tr>

</table>
</div>
Hence the LHS of the polytropic (n = 1) wave equation becomes,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\sin\xi \biggl\{ \xi^2 \biggl[~(x_s^{''} - 2x_c^{'} - x_s)\sin\xi + (x_c^{''} + 2x_s^' -x_c)\cos\xi 
+ (x_1^{''} -2x_3^' -2x_1 + 2x_2)\sin^2\xi + (x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\cos^2\xi 
+ (x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\sin\xi \cos\xi  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ 2\xi \biggl[~(x_s^' - x_c)\sin\xi + (x_c^' + x_s)\cos\xi + (x_1^' - x_3)\sin^2\xi + (x_2^' + x_3)\cos^2\xi + (x_3^' + 2x_1 -2x_2)\sin\xi \cos\xi \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- 2\alpha \biggl[ x_s \sin\xi + x_c \cos\xi + x_1 \sin^2\xi + x_2 \cos^2\xi + x_3 \sin\xi \cos\xi\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \cos\xi \biggl\{ 2\xi^2 \biggl[~(x_s^' - x_c)\sin\xi + (x_c^' + x_s)\cos\xi + (x_1^' - x_3)\sin^2\xi + (x_2^' + x_3)\cos^2\xi + (x_3^' + 2x_1 -2x_2)\sin\xi \cos\xi \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ 2\alpha \xi \biggl[ x_s \sin\xi + x_c \cos\xi + x_1 \sin^2\xi + x_2 \cos^2\xi + x_3 \sin\xi \cos\xi\biggr]  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\sigma^2 \xi^3 \biggl\{ x_s \sin\xi + x_c \cos\xi + x_1 \sin^2\xi + x_2 \cos^2\xi + x_3 \sin\xi \cos\xi \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[(x_s^{''} - 2x_c^{'} - x_s)\xi^2 + 2\xi (x_s^' - x_c) -2\alpha x_s + \sigma^2 \xi^3 x_1  \biggr]\sin^2\xi 
+ \biggl[(x_c^{''} + 2x_s^' -x_c)\xi^2 + 2\xi (x_c^' + x_s) - 2\alpha x_c + 2\xi^2 (x_s^' - x_c) + 2\alpha \xi x_s + \sigma^2 \xi^3 x_3 \biggr] \sin\xi \cos\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\biggl[ (x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 + 2\xi (x_1^' - x_3) - 2\alpha x_1 \biggr] \sin^3\xi 
+ \biggl[(x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\xi^2 + 2\xi (x_2^' + x_3) - 2\alpha x_2 + 2\xi^2 (x_3^' + 2x_1 -2x_2)+  2\alpha \xi  x_3 \biggr]\sin\xi \cos^2\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[(x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\xi^2 + 2\xi (x_3^' + 2x_1 -2x_2) - 2\alpha x_3 + 2\xi^2 (x_1^' - x_3) +  2\alpha \xi x_1  \biggr] \sin^2\xi \cos\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[2\xi^2 (x_c^' + x_s ) +  2\alpha \xi x_c + \sigma^2 \xi^3 x_2 \biggr]\cos^2\xi  + \biggl[2\xi^2 (x_2^' + x_3)+  2\alpha \xi x_2 \biggr]\cos^3\xi   
+ \sigma^2 \xi^3 x_s \sin\xi + \sigma^2 \xi^3 x_c \cos\xi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[(x_s^{''} - 2x_c^{'} - x_s)\xi^2 + 2\xi (x_s^' - x_c) -2\alpha x_s + \sigma^2 \xi^3 x_1  \biggr]\sin^2\xi 
+ \biggl[(x_c^{''} + 2x_s^' -x_c)\xi^2 + 2\xi (x_c^' + x_s) - 2\alpha x_c + 2\xi^2 (x_s^' - x_c) + 2\alpha \xi x_s + \sigma^2 \xi^3 x_3 \biggr] \sin\xi \cos\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\biggl[ (x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 + 2\xi (x_1^' - x_3) - 2\alpha x_1 + \sigma^2 \xi^3 x_s \biggr] \sin\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[(x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\xi^2 + 2\xi (x_2^' + x_3) - 2\alpha x_2 + 2\xi^2 (x_3^' + 2x_1 -2x_2)+  2\alpha \xi  x_3 
- (x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 - 2\xi (x_1^' - x_3) + 2\alpha x_1 \biggr]\sin\xi \cos^2\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[(x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\xi^2 + 2\xi (x_3^' + 2x_1 -2x_2) - 2\alpha x_3 + 2\xi^2 (x_1^' - x_3) +  2\alpha \xi x_1  
-2\xi^2 (x_2^' + x_3) -  2\alpha \xi x_2 \biggr] \sin^2\xi \cos\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[2\xi^2 (x_c^' + x_s ) +  2\alpha \xi x_c + \sigma^2 \xi^3 x_2 \biggr]\cos^2\xi  
+ \biggl[2\xi^2 (x_2^' + x_3)+  2\alpha \xi x_2 + \sigma^2 \xi^3 x_c \biggr]\cos\xi  
</math>
  </td>
</tr>

</table>
</div>
So, the five chosen (polynomial?) functions of <math>~\xi</math> must simultabeously satisfy the following, seven 2<sup>nd</sup>-order ODEs:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\sin\xi </math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~(x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 + 2\xi (x_1^' - x_3) - 2\alpha x_1 + \sigma^2 \xi^3 x_s =0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\sin^2 \xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>(x_s^{''} - 2x_c^{'} - x_s)\xi^2 + 2\xi (x_s^' - x_c) -2\alpha x_s + \sigma^2 \xi^3 x_1 =0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\sin^2\xi \cos\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>(x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\xi^2 + 2\xi (x_3^' + 2x_1 -2x_2) - 2\alpha x_3 + 2\xi^2 (x_1^' - x_3) +  2\alpha \xi x_1  
-2\xi^2 (x_2^' + x_3) -  2\alpha \xi x_2  =0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\sin\xi \cos\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>(x_c^{''} + 2x_s^' -x_c)\xi^2 + 2\xi (x_c^' + x_s) - 2\alpha x_c + 2\xi^2 (x_s^' - x_c) + 2\alpha \xi x_s + \sigma^2 \xi^3 x_3 =0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\sin\xi \cos^2\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>(x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\xi^2 + 2\xi (x_2^' + x_3) - 2\alpha x_2 + 2\xi^2 (x_3^' + 2x_1 -2x_2)+  2\alpha \xi  x_3 
- (x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 - 2\xi (x_1^' - x_3) + 2\alpha x_1 =0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\cos^2\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>2\xi^2 (x_c^' + x_s ) +  2\alpha \xi x_c + \sigma^2 \xi^3 x_2 =0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\cos\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>2\xi^2 (x_2^' + x_3)+  2\alpha \xi x_2 + \sigma^2 \xi^3 x_c =0</math>
  </td>
</tr>
</table>
</div>

====Example 1====
Let's work on the coefficient of the <math>~\cos\xi</math> term:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x_c</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\xi^\beta (A_c)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>x_2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\xi^\beta (C_2 \xi^2)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>x_3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\xi^\beta (B_3 \xi)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~</math> &nbsp; &nbsp; &nbsp; Coefficient of "<math>~\cos\xi</math>" term
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\xi^\beta [2\xi^2 (2C_2\xi + (B_3 \xi))+  2\alpha \xi (C_2 \xi^2) + \sigma^2 \xi^3 (A_c)]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\xi^{\beta+3} [2(2C_2 + B_3 )+  2\alpha C_2  + \sigma^2 (A_c)]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \sigma^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \frac{2}{A_c} \biggl[B_3 +  (2+\alpha) C_2 \biggr]</math>
  </td>
</tr>
</table>
</div>

===Sixth Guess (n1)===
====Rationale====
From our [[SSC/Structure/Polytropes#.3D_1_Polytrope|review of the properties of <math>~n=1</math> polytropic spheres]], we know that the equilibrium density distribution is given by the sinc function, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\rho}{\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\sin\xi}{\xi} \, ,</math>
  </td>
</tr>

</table>
</div>
where,
<div align="center">
<math>~\xi \equiv \pi \biggl(\frac{r_0}{R_0} \biggr) \, .</math>
</div>
The total mass is,
<div align="center">
<math>~M_\mathrm{tot} =  \frac{4}{\pi} \cdot \rho_c R_0^3 \, ,</math>
</div>
and the fractional mass enclosed within a given radius, <math>~r</math>, is,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{M_r(\xi)}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\pi} [\sin\xi - \xi \cos\xi] \, .</math>
  </td>
</tr>

</table>
</div>
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,
<div align="center">
<math>R = R_0 \biggl( 1 + \frac{\delta R}{R_0} \biggr) = R_0 ( 1 + \epsilon_R) \, ,</math>
</div>
and seek to determine how the displacement vector, <math>~\epsilon \equiv \delta r/r_0</math>, varies with <math>~r_0</math> 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, <math>~|\epsilon| \ll 1 </math>.


Let's begin by defining a new dimensionless coordinate,
<div align="center">
<math>~\eta \equiv \pi \biggl(\frac{r}{R} \biggr) = \pi \biggl[\frac{r_0(1+\epsilon)}{R_0(1+\epsilon_R)} \biggr] 
\approx \xi (1 + \epsilon) \, ,</math>
</div>
and recognize that, in the new perturbed state, the fractional mass enclosed within a given radius, <math>~r</math>, is,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{M_r(\eta)}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\pi} [\sin\eta - \eta \cos\eta] \, .</math>
  </td>
</tr>

</table>
</div>
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 <math>~M_r</math> functions equal to one another, that is, demand that,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sin\xi - \xi \cos\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sin\eta - \eta \cos\eta</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\sin[\xi(1+\epsilon)] - \xi(1+\epsilon) \cos[\xi(1+\epsilon)]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \sin\xi \cos(\xi\epsilon) + \cos\xi \sin (\xi\epsilon) \biggr] 
- \xi(1+\epsilon) \biggl[ \cos\xi \cos(\xi\epsilon) - \sin\xi \sin (\xi\epsilon) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\sin\xi \biggl[1 - \frac{1}{2}(\xi\epsilon)^2 \biggr] + (\xi\epsilon)\cos\xi 
- \xi(1+\epsilon) \cos\xi \biggl[1 - \frac{1}{2}(\xi\epsilon)^2 \biggr]  + \xi^2 \epsilon(1+\epsilon) \sin\xi </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>
~\sin\xi  -\xi\cos\xi - \frac{1}{2}(\xi\epsilon)^2 \sin\xi  + (\xi\epsilon)\cos\xi 
- (\xi \epsilon) \cos\xi  + \frac{1}{2} \xi^3 \epsilon^2\cos\xi
+ \xi^2 \epsilon \sin\xi + (\xi \epsilon)^2\sin\xi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~- \xi^2 \epsilon \sin\xi </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{(\xi \epsilon)^2}{2} \biggl[\xi \cos\xi+ \sin\xi  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~\frac{1}{\epsilon}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-
\frac{1}{2} \biggl[\xi \cdot \frac{\cos\xi}{\sin\xi} + 1 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~\epsilon</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-
2\biggl[1 + \xi \cdot \frac{\cos\xi}{\sin\xi} \biggr]^{-1}
= - 2\sin\xi \biggl[\sin\xi + \xi \cos\xi \biggr]^{-1} \, .
</math>
  </td>
</tr>
</table>
</div>

====Resulting Polytropic Wave Equation====
So, let's try,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\sin\xi \biggl[\sin\xi + \xi \cos\xi \biggr]^{-1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3} 2\sin\xi  \biggl[\sin^2\xi + 2\xi \sin\xi \cos\xi + \xi^2 \cos^2\xi \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\cos\xi \biggl[\sin\xi + \xi \cos\xi \biggr]^{-1} -2\sin\xi \biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl[2\cos\xi  - \xi \sin\xi \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl\{
2\cos\xi \biggl[\sin\xi + \xi \cos\xi \biggr] -2\sin\xi \biggl[2\cos\xi  - \xi \sin\xi \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl[
2\cos\xi \sin\xi + 2\xi \cos^2\xi  -4\sin\xi \cos\xi  + 2\xi \sin^2\xi 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl[\xi-\sin\xi \cos\xi \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3} 2
\biggl[\xi \sin\xi + \xi^2 \cos\xi  -\sin^2\xi \cos\xi - \xi \sin\xi \cos^2\xi \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^{''}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl[1-\cos^2\xi + \sin^2\xi \biggr] 
- 
4\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3}\biggl[\xi-\sin\xi \cos\xi \biggr]\biggl[2\cos\xi  - \xi\sin\xi \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3}\biggl\{ \sin^2\xi  \biggl[\sin\xi + \xi \cos\xi \biggr]
- 
\biggl[\xi-\sin\xi \cos\xi \biggr]\biggl[2\cos\xi  - \xi\sin\xi \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3}\biggl[ \sin^3\xi + \xi \sin\xi \cos\xi 
+ \xi^2 \sin\xi -2\xi \cos\xi - \xi\sin^2\xi \cos\xi + 2\sin\xi \cos^2\xi \biggr]
</math>
  </td>
</tr>
</table>
</div>

====Graphical Reassessment====
[[Image:TrialN1Eigenfunction.png|border|300px|right]]Before plowing ahead and plugging these expressions into the polytropic wave equation, I plotted the trial eigenfunction, <math>\epsilon(\xi/\pi)</math> (see the blue curve in the accompanying "Trial Eigenfunction" figure), and noticed that it passes through <math>\pm \infty</math> 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,
<div align="center">
<math>x = 1 + \frac{1}{\epsilon} = 1 - \frac{1}{2} \biggl[\xi \cdot \frac{\cos\xi}{\sin\xi} + 1 \biggr]
= \frac{1}{2} \biggl[1- \xi \cdot \frac{\cos\xi}{\sin\xi} \biggr] \, .</math>
</div>

In this case we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x^'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2}\biggl[\xi - \frac{\cos\xi}{\sin\xi} + \xi \cdot \frac{\cos^2\xi}{\sin^2\xi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\sin^2\xi}\biggl[\xi \sin^2\xi - \sin\xi \cos\xi + \xi \cos^2\xi \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\sin^2\xi}\biggl[\xi - \sin\xi \cos\xi \biggr] \, ,
</math>
  </td>
</tr>

</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x^{''}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>~
- \frac{\cos\xi}{\sin^3\xi}\biggl[\xi - \sin\xi \cos\xi \biggr] + \frac{1}{2\sin^2\xi}\biggl[1 - \cos^2\xi  + \sin^2\xi \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \frac{\cos\xi}{\sin^3\xi}\biggl[\xi - \sin\xi \cos\xi \biggr] \, .
</math>
  </td>
</tr>

</table>
</div>

Now let's plug these expressions into the polytropic (n = 1) wave equation, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>-\sigma^2 \xi^3 x</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\sin\xi \biggl[ \xi^2 x^{''} + 2\xi x^' - 2\alpha x \biggr]
+ \cos\xi \biggl[ 2\xi^2 x^' + 2\alpha \xi x \biggr] \, .
</math>
  </td>
</tr>

</table>
</div>

The first term inside the square brackets on the right-hand-side gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\xi^2 x^{''} + 2\xi x^' - 2\alpha x </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\xi^2 - \frac{\cos\xi}{\sin^3\xi}\biggl(\xi^3 - \xi^2\sin\xi \cos\xi \biggr) 
+ \frac{1}{\sin^2\xi}\biggl(\xi^2 - \xi \sin\xi \cos\xi \biggr) 
- \alpha \biggl(1- \xi \cdot \frac{\cos\xi}{\sin\xi} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{\sin^3\xi} \biggl[
\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^3\xi - \xi \cos\xi \sin^2\xi )
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{\sin^3\xi} \biggl[
\xi^2 \sin\xi (1-\cos^2\xi) + \xi^2\sin\xi \cos^2\xi - \xi^3 \cos\xi 
+ \xi^2\sin\xi - \xi \sin^2\xi \cos\xi 
- \alpha \sin^3\xi + \alpha \xi \cos\xi \sin^2\xi 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \alpha + \frac{1}{\sin^3\xi} \biggl[
2\xi^2 \sin\xi  - \xi^3 \cos\xi - \xi \sin^2\xi \cos\xi + \alpha \xi \cos\xi \sin^2\xi 
\biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>

and the second term inside the square brackets on the right-hand-side gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>2\xi^2 x^' + 2\alpha \xi x </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\sin^2\xi}\biggl(\xi^3 - \xi^2 \sin\xi \cos\xi \biggr) 
+\frac{\alpha}{\sin\xi} \biggl(\xi \sin\xi - \xi^2 \cos\xi \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>

Put together, then, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{\sin^2\xi} \biggl[
2\xi^2 \sin\xi  - \xi^3 \cos\xi - \xi \sin^2\xi \cos\xi + \alpha \xi \cos\xi \sin^2\xi 
\biggr] +
\frac{\cos\xi}{\sin^2\xi}\biggl(\xi^3 - \xi^2 \sin\xi \cos\xi \biggr) -\alpha\sin\xi
+ \alpha \cdot \frac{\cos\xi}{\sin\xi} \biggl(\xi \sin\xi - \xi^2 \cos\xi \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{\sin^2\xi} \biggl[
2\xi^2 \sin\xi  - \xi^3 \cos\xi - \xi \sin^2\xi \cos\xi + \alpha \xi \cos\xi \sin^2\xi 
+ \xi^3\cos\xi - \xi^2 \sin\xi \cos^2\xi \biggr] + \frac{\alpha}{\sin\xi} \biggl[ -\sin^2\xi
+ \xi \sin\xi \cos\xi - \xi^2 \cos^2\xi \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\xi}{\sin\xi} \biggl[
2\xi  - \sin\xi \cos\xi - \xi \cos^2\xi \biggr] 
- \frac{\alpha}{\sin\xi} \biggl[ \sin^2\xi+\xi^2 \cos^2\xi \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\xi^2}{\sin\xi} \biggl[ 
1+\sin^2\xi  - \biggl(\frac{\sin\xi}{\xi}\biggr) \cos\xi  - \alpha \biggl( \frac{\sin^2\xi}{\xi^2} +\cos^2\xi \biggr)\biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{\xi \sigma^2}{2} \biggl( \frac{ \xi^2 }{\sin\xi} \biggr) \biggl[\sin\xi - \xi \cos\xi \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
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:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{\sin\xi}{\xi^2} \biggl( \mathrm{RHS} - \mathrm{LHS} \biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1+\sin^2\xi  - \biggl(\frac{\sin\xi}{\xi}\biggr) \cos\xi  - \alpha \biggl( \frac{\sin^2\xi}{\xi^2} +\cos^2\xi \biggr)  
+ \frac{\xi \sigma^2}{2} \biggl[\sin\xi - \xi \cos\xi \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
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 &#8212; ''albeit nonzero'' &#8212; expression.  So we may be on the right track!

=See Also=

{{ SGFfooter }}