View source for SSC/Stability/n1PolytropeLAWE

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=Radial Oscillations of n = 1 Polytropic Spheres=

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 immediately following citations contain inks to both of these papers by Chatterji, but &#8212; at least at the time (October 2021) of this posting &#8212; they appear to be insecure.  Apparently [https://www.springer.com/journal/43538 Springer] is archiving recent PINSA volumes, but their holdings do not date back as early as 1951.

* [https://insa.nic.in/writereaddata/UpLoadedFiles/PINSA/Vol17_1951_6_Art06.pdf L. D. Chatterji (1951, Vol. XVII, No. 6)], ''Radial Oscillations of a Gaseous Star of Polytropic Index I'' &hellip; insa.nic.in; PINSA = '''Proceedings of the Indian National Science Academy''', [https://www.springer.com/journal/43538 Springer]
* [https://www.researchgate.net/publication/266722347_Anharmonic_Pulsations_of_a_Polytropic_Model_of_Index_Unity L. D. Chatterji (1952, Vol. XVIII, No. 3)], ''Anharmonic Pulsations of a Polytropic Model of Index Unity''
==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, here, in the following boxed-in image.

<div align="center">
<table border="2" cellpadding="10" id="MurphyFiedler1985b">
<tr>
  <th align="center">
n = 1 Polytropic Formulation of Wave Equation as Presented by [http://adsabs.harvard.edu/abs/1985PASAu...6..222M Murphy &amp; Fiedler (1985b)]
  </th>
<tr>
  <td>
[[File:MurphyFiedlerN1formulation.png|700px|center|Murphy &amp; Fiedler (1985b)]]
  </td>
</tr>
</table>
</div>


{{ SGFworkInProgress }}


<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!

==New Idea Involving Logarithmic Derivatives==

===Simplistic Layout===
Let's begin, again, with the relevant LAWE, as [[#Attempt_at_Deriving_an_Analytic_Eigenvector_Solution|provided above]].  After dividing through by <math>~x</math>, we have,

<div align="center">
<math>
(\sin\xi )\frac{\xi^2}{x} \cdot  \frac{d^2x}{d\xi^2} + 2 \biggl[ \sin\xi +  \xi \cos \xi \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + 
\biggl[ \sigma^2 \xi^3  - 2\alpha ( \sin\xi - \xi \cos \xi ) \biggr]   = 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>

Now, in addition to recognizing that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d\ln x}{d\ln \xi} \, ,</math>
  </td>
</tr>
</table>
</div>
in a [[SSC/Stability/BiPolytrope00Details#Idea_Involving_Logarithmic_Derivatives|separate context]], we showed that, quite generally,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} 
</math>
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]
- \biggl[  1 -  \frac{d\ln x}{d\ln \xi} \biggr]\cdot \frac{d\ln x}{d\ln \xi} \, .
</math>
 </td>
</tr>
</table>
</div>

Hence, if we ''assume'' that the  eigenfunction is a power-law of <math>~\xi</math>, that is, ''assume'' that,
<div align="center">
<math>~x = a_0 \xi^{c_0} \, ,</math>
</div>
then the logarithmic derivative of <math>~x</math> is a constant, namely,
<div align="center">
<math>~\frac{d\ln x}{d\ln\xi} = c_0 \, ,</math>
</div>
and the two key derivative terms will be,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} = c_0 \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} = c_0(c_0-1) \, .</math>
  </td>
</tr>
</table>
</div>
In this case, the LAWE is no longer a differential equation but, instead, takes the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-\sigma^2 \xi^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0(c_0-1) \sin\xi  + 2c_0 [ \sin\xi +  \xi \cos \xi ] - 2\alpha ( \sin\xi - \xi \cos \xi )  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sin\xi [c_0(c_0-1) +2c_0 -2\alpha ]  + \xi \cos \xi [2(c_0+\alpha) ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sin\xi [c_0^2 + c_0  -2\alpha ]  + \xi \cos \xi [2(c_0+\alpha) ] \, .
</math>
  </td>
</tr>
</table>
</div>
Now, the cosine term will go to zero if <math>~c_0 = -\alpha</math>; and the sine term will go to zero if,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \gamma_g</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\infty \, . </math>
  </td>
</tr>
</table>
</div>
If these two &#8212; rather strange &#8212; conditions are met, then we have a marginally unstable configuration because, <math>~\sigma^2 = 0</math>.  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,
<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>~\xi^{c_0} \biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^{c_0} \frac{d}{d\xi}\biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr] + c_0\xi^{c_0-1} \biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^{c_0}  \biggl[ b_0\cos\xi - d_0 \xi\sin\xi +d_0\cos\xi\biggr] + c_0\xi^{c_0-1} \biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{d\ln x}{d\ln \xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi \biggl[ b_0\cos\xi - d_0 \xi\sin\xi +d_0\cos\xi\biggr]\biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr]^{-1} 
+ c_0 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ (b_0+d_0)\xi\cos\xi - d_0 \xi^2\sin\xi \biggr]\biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr]^{-1} 
+ c_0 </math>
  </td>
</tr>
</table>
</div>


===Another Viewpoint===

====Development====
Multiplying through the [[#Simplistic_Layout|above LAWE]] by <math>~(x \xi^{-3})</math> gives,

<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{\sin\xi }{\xi} \cdot  \frac{d^2x}{d\xi^2} + 2 \biggl[\frac{ \sin\xi +  \xi \cos \xi }{\xi^2}\biggr] \frac{dx}{d\xi} + 
\biggl[ \sigma^2   - 2\alpha \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr]x
</math>
  </td>
</tr>
</table>
</div>

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

<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} + \frac{\cos\xi}{\xi}
</math>
  </td>
</tr>

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

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

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

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

<tr>
  <td align="right">
<math>~\frac{d^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d}{d\xi} \biggl\{ 
 \biggl(\frac{\sin\xi}{\xi}\biggr)\frac{dx}{d\xi}  + x\frac{d}{d\xi} \biggl[ \biggl(\frac{\sin\xi}{\xi}\biggr) \biggr] 
\biggr\}</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\sin\xi}{\xi} \cdot \frac{d^2 x}{d\xi^2}
+ 2\frac{dx}{d\xi} \cdot \biggl[ \frac{\xi\cos\xi - \sin\xi }{\xi^2} \biggr]
+ x \cdot \biggl\{ -\frac{\sin\xi}{\xi} + 2\biggl[ \frac{\sin\xi -\xi\cos\xi}{\xi^3} \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
This means that we can rewrite the LAWE 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^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\} - 2\frac{dx}{d\xi} \cdot \biggl[ \frac{\xi\cos\xi - \sin\xi }{\xi^2} \biggr]
- x \cdot \biggl\{ -\frac{\sin\xi}{\xi} + 2\biggl[ \frac{\sin\xi -\xi\cos\xi}{\xi^3} \biggr] \biggr\}
+ 2 \biggl[\frac{ \sin\xi +  \xi \cos \xi }{\xi^2}\biggr] \frac{dx}{d\xi} + 
\biggl[ \sigma^2   - 2\alpha \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr]x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\} 
+ 4 \biggl[\frac{ \sin\xi  }{\xi^2}\biggr] \frac{dx}{d\xi} 
+ \biggl\{ \frac{\sin\xi}{\xi} 
+ \sigma^2   - 2(1+\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr\} \cdot x \, .
</math>
  </td>
</tr>
</table>
</div>

We recognize, also, that,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
4\biggl(\frac{\sin\xi}{\xi^2} \biggr)\frac{dx}{d\xi} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4}{\xi} \cdot \frac{d}{d\xi}\biggl[ \biggl(\frac{\sin\xi}{\xi} \biggr) x \biggr]
+ 4\biggl[ \frac{\sin\xi - \xi\cos\xi }{\xi^3} \biggr]x
\, .
</math>
  </td>
</tr>
</table>
</div>
So the LAWE becomes,

<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^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\} 
+ \frac{4}{\xi} \cdot \frac{d}{d\xi}\biggl[ \biggl(\frac{\sin\xi}{\xi} \biggr) x \biggr]
+ 4\biggl[ \frac{\sin\xi - \xi\cos\xi  }{\xi^3} \biggr]x
+ \biggl\{ \frac{\sin\xi}{\xi} 
+ \sigma^2   - 2(1+\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr\} \cdot x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\} 
+ \frac{4}{\xi} \cdot \frac{d}{d\xi}\biggl[ \biggl(\frac{\sin\xi}{\xi} \biggr) x \biggr]
+ \biggl\{ \frac{\sin\xi}{\xi} 
+ \sigma^2   + [4- 2(1+\alpha)] \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr\} \cdot x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 \Upsilon}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi}
+ \Upsilon 
+ \biggl[ \sigma^2   + 2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x \, ,
</math>
  </td>
</tr>
</table>
</div>
where we have introduced the new, modified eigenfunction,
<div align="center">
<math>\Upsilon \equiv \biggl( \frac{\sin\xi}{\xi} \biggr) x \, .</math>
</div>

Alternatively, the LAWE may be written 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^2 \Upsilon}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi}
+ \biggl[ \sigma^2   + 2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) + \frac{\sin\xi}{\xi} \biggr] \cdot x \, ;
</math>
  </td>
</tr>
</table>
</div>
or,

<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{\xi^2}{\Upsilon} \cdot \frac{d^2 \Upsilon}{d\xi^2} 
+ \frac{4\xi}{\Upsilon} \cdot \frac{d\Upsilon}{d\xi}
+ \biggl[ \sigma^2   + 2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) + \frac{\sin\xi}{\xi} \biggr] \cdot \frac{\xi^3}{\sin\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\xi^2}{\Upsilon} \cdot \frac{d^2 \Upsilon}{d\xi^2} 
+ \frac{4\xi}{\Upsilon} \cdot \frac{d\Upsilon}{d\xi}
+ \biggl[ \sigma^2 \biggl(\frac{\xi^3}{\sin\xi} \biggr)   + 2(1-\alpha) \biggl( 1 - \xi \cot \xi \biggr) + \xi^2 \biggr]   
</math>
  </td>
</tr>
</table>
</div>


Now, if we adopt the homentropic convention that arises from setting, <math>~\gamma = (n+1)/n</math>, then for our <math>~n=1</math> polytropic configuration, we should set, <math>~\gamma = 2</math> and, hence, <math>~\alpha = 1</math>.  This will mean that the lat term in this LAWE naturally goes to zero.  Hence, we have,

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

<tr>
  <td align="right">
<math>~- \sigma^2 x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 \Upsilon}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi} + \Upsilon
\, ;
</math>
  </td>
</tr>
</table>
</div>
or,

<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^2 \Upsilon}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi} + \biggl[1 + \sigma^2 \biggl(\frac{\xi}{\sin\xi}\biggr) \biggr] \Upsilon \, ;
</math>
  </td>
</tr>
</table>
</div>
or,

<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{\xi^2}{\Upsilon} \cdot \frac{d^2 \Upsilon}{d\xi^2} 
+ \frac{4\xi}{\Upsilon} \cdot \frac{d\Upsilon}{d\xi} + \biggl[\xi^2 + \sigma^2 \biggl(\frac{\xi^3}{\sin\xi}\biggr) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Does this help?

====Check for Mistakes====

Given the definition of <math>~\Upsilon</math>, its first derivative is,

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

<tr>
  <td align="right">
<math>~\frac{d\Upsilon}{d\xi} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{dx}{d\xi} +x\biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and its second derivative is,

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

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

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

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

Hence, the "upsilon" LAWE becomes,

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

<tr>
  <td align="right">
<math>~-\sigma^2 x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 \Upsilon}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi}
+ \Upsilon 
+ \biggl[  2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + 2 \biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \cdot \frac{dx}{d\xi} 
+ x \biggl[ -\frac{\sin\xi}{\xi} - \frac{2\cos\xi}{\xi^2}  + \frac{2\sin\xi}{\xi^3} \biggr] 
+ \frac{4}{\xi} \cdot \biggl\{  \biggl( \frac{\sin\xi}{\xi} \biggr) \frac{dx}{d\xi} +x\biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \biggr\} 
+ \biggl[\frac{\sin\xi}{\xi} +   2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + \biggl\{\biggl( \frac{4\sin\xi}{\xi^2} \biggr)  + 2 \biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \biggr\}\cdot \frac{dx}{d\xi} 
+ \biggl[ -\frac{\sin\xi}{\xi} - \frac{2\cos\xi}{\xi^2}  + \frac{2\sin\xi}{\xi^3} 
+  \frac{4\cos\xi}{\xi^2} - \frac{4\sin\xi}{\xi^3} 
+ \frac{\sin\xi}{\xi} +   2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + 2\biggl[ \frac{\sin\xi}{\xi^2} + \frac{\cos\xi}{\xi} \biggr] \cdot \frac{dx}{d\xi} 
+ \biggl[-2\alpha \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x \, .
</math>
  </td>
</tr>
</table>
</div>
This should be compared with the first expression, [[#Development|above]], namely,
<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{\sin\xi }{\xi} \cdot  \frac{d^2x}{d\xi^2} + 2 \biggl[\frac{ \sin\xi +  \xi \cos \xi }{\xi^2}\biggr] \frac{dx}{d\xi} + 
\biggl[ \sigma^2   - 2\alpha \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr]x \, ,
</math>
  </td>
</tr>
</table>
</div>
and it matches!  Q.E.D.

==Motivated by Yabushita's Discovery==

===Initial Exploration===
This subsection is being developed following our realization &#8212; see the [[SSC/Stability/InstabilityOnsetOverview#Overview:_Marginally_Unstable_Pressure-Truncated_Configurations|accompanying overview]] &#8212; that the eigenfunction ''is'' known analytically for marginally unstable, pressure-truncated configurations having <math>~3 \le n \le \infty</math>.  Specifically, from the work of [http://adsabs.harvard.edu/abs/1975MNRAS.172..441Y Yabushita (1975)] we have the following,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Exact Solution to the Isothermal LAWE</b></font></td>
</tr>
<tr>
  <td align="right">
<math>~\sigma_c^2 = 0</math>
  </td>
  <td align="center">
&nbsp;and &nbsp;
  </td>
  <td align="left">
<math>~x = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
  </td>
</tr>
</table>
</div>
And from our own recent work, we have discovered the following,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Precise Solution to the Polytropic LAWE</b></font></td>
</tr>
<tr>
  <td align="right">
<math>~\sigma_c^2 = 0</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~x_P \equiv \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] </math>
  </td>
</tr>
</table>
</div>
if the adiabatic exponent is assigned the value, <math>~\gamma_g = (n+1)/n</math>, in which case the parameter, <math>~\alpha = (3-n)/(n+1)</math>.  Using this polytropic displacement function as a guide, let's try for the case of <math>~n=1</math>, an expression of the form,
<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>~A - B\biggl[ \biggl( \frac{1}{\xi \theta}\biggr) \frac{d\theta}{d\xi} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A - B \biggl[ \biggl( \frac{1}{\sin\xi}\biggr) \frac{d}{d\xi} \biggl( \frac{\sin\xi}{\xi} \biggr)\biggr]</math>
  </td>
</tr>

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

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

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

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


<table border="1" cellpadding="8" align="center" width="85%"><tr><td align="left">
What if, instead, we try the more generalized form,
<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>~A + \frac{B}{(\lambda \xi)^2} \biggl[ 1-\frac{\lambda \xi \cos(\lambda \xi)}{\sin(\lambda \xi)}  \biggr] \, .</math>
  </td>
</tr>
</table>

Then we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\lambda B} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{ (\lambda \xi)^3} \biggl\{
2 - (\lambda \xi)^2 - \frac{\lambda \xi\cos(\lambda \xi)}{\sin(\lambda \xi)} 
- \frac{(\lambda \xi)^2 \cos^2(\lambda\xi)}{\sin^2(\lambda\xi)} 
\biggr\} 
\, ,
</math>
  </td>
</tr>
</table>
Probably this also means,

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

<tr>
  <td align="right">
<math>~\frac{1}{\lambda^2 B} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(\lambda\xi)^4} \biggl\{ 6 - 2 (\lambda \xi)^2 
- \frac{2 \lambda \xi\cos(\lambda\xi)}{\sin(\lambda\xi)} 
- \frac{2(\lambda\xi)^2 \cos^2(\lambda\xi)}{\sin^2(\lambda\xi)} 
- \frac{2(\lambda\xi)^3 \cos(\lambda\xi)}{\sin(\lambda\xi)} 
- \frac{2(\lambda\xi)^3 \cos^3(\lambda\xi)}{\sin^3(\lambda\xi)} \biggr\} \, .
</math>
  </td>
</tr>
</table>


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


Let's check against the [[SSC/Stability/Isothermal#Derivation_of_Polytropic_Displacement_Function|more general derivation]], which gives after recognizing that, <math>~B \leftrightarrow (3-n)/(n-1)</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3-n}{n-1}\biggr) \biggl\{ \frac{1}{\xi} 
+ \frac{n(\theta^')^2 }{\xi \theta^{n+1}} + \frac{3\theta^' }{\xi^2 \theta^{n}} \biggr\} 
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B}{\xi^3} \biggl\{ \xi^2 
+ \biggl[ \frac{\xi \cos\xi}{\sin\xi} - 2 \biggr] 
+ \biggl(\frac{\xi \cos\xi}{\sin\xi}\biggr)^2  \biggr\} 
\, .
</math>
  </td>
</tr>
</table>
</div>
This matches the preceding, direct derivation.

Also,

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{B}{\xi^4} \biggl\{
\biggl( \frac{3\cos\xi}{\sin^2\xi}\biggr) \biggl[ - \xi^2 \cos\xi + \xi \sin\xi \biggr]
+\biggl[ 6 - 3\xi^2 - \frac{6\xi\cos\xi}{\sin\xi} \biggr]
+ \biggl[\frac{1}{\sin\xi} + \frac{2\cos^2\xi}{\sin^3\xi}  \biggr] \biggl[ - \xi^3 \cos\xi + \xi^2 \sin\xi \biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{B}{\xi^4} \biggl\{
\biggl[ - \frac{3\xi^2 \cos^2\xi}{\sin^2\xi} + \frac{3\xi \cos\xi}{\sin\xi} \biggr]
+\biggl[ 6 - 3\xi^2 - \frac{6\xi\cos\xi}{\sin\xi} \biggr]
+ \biggl[ - \frac{\xi^3 \cos\xi}{\sin\xi} + \xi^2 \biggr]
+ \biggl[ - \frac{2\xi^3 \cos^3\xi}{\sin^3\xi} + \frac{2\xi^2 \cos^2\xi}{\sin^2\xi} \biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B}{\xi^4} \biggl\{ 6 - 2\xi^2 
- \frac{2\xi\cos\xi}{\sin\xi} 
- \frac{2\xi^2 \cos^2\xi}{\sin^2\xi} 
- \frac{2\xi^3 \cos\xi}{\sin\xi} 
- \frac{2\xi^3 \cos^3\xi}{\sin^3\xi} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
Let's also check this against the [[SSC/Stability/Isothermal#Derivation_of_Polytropic_Displacement_Function|more general derivation]], which gives after again recognizing that, <math>~B \leftrightarrow (3-n)/(n-1)</math>, 

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

<tr>
  <td align="right">
<math>\frac{d^2 x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ \frac{4}{\xi^2} + \frac{2n(\theta^')}{\xi \theta} 
+ \frac{12\theta^' }{\xi^3 \theta^{n}}+ \frac{8n(\theta^')^2}{\xi^2 \theta^{n+1}}  
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-B
\biggl\{ \frac{4}{\xi^2} + \frac{2}{\xi \theta} \biggl[ \frac{\sin\xi}{\xi^2}\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr]
+ \frac{12}{\xi^3 \theta}\biggl[ \frac{\sin\xi}{\xi^2}\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr]
+ \frac{8 }{\xi^2 \theta^{2}}  \biggl[ \frac{\sin\xi}{\xi^2}\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr]^2
+ \frac{2 }{\xi \theta^{3}} \biggl[ \frac{\sin\xi}{\xi^2}\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr]^3\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{B}{\xi^4}
\biggl\{ 4\xi^2 + 2\xi^2 \biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)
+ 12\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)
+ 8 \biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)^2
+ 2\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)^3\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{2B}{\xi^4}
\biggl\{ 2\xi^2 + \frac{\xi^3\cos\xi}{\sin\xi} - \xi^2+ \frac{6\xi\cos\xi}{\sin\xi} - 6
+ \frac{4\xi^2\cos^2\xi}{\sin^2\xi} - \frac{8\xi\cos\xi}{\sin\xi} + 4
+ \biggl(\frac{\xi^2\cos^2\xi}{\sin^2\xi} - \frac{2\xi\cos\xi}{\sin\xi} + 1\biggr) \biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr\} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B}{\xi^4}
\biggl\{6 -2 \xi^2  - \frac{2\xi\cos\xi}{\sin\xi} 
- \frac{2\xi^2\cos^2\xi}{\sin^2\xi}  
- \frac{2\xi^3\cos\xi}{\sin\xi}  
- \frac{2\xi^3\cos^3\xi}{\sin^3\xi}   
\biggr\} 
\, .
</math>
  </td>
</tr>
</table>
</div>
A cross-check with the first attempt to derive this second derivative expression initially unveiled a couple of coefficient errors.  These have now been corrected and both expressions agree.

===Succinct Demonstration===
Given that, for <math>~n=1</math>, we should set <math>~\gamma_\mathrm{g} = (n+1)/n = 2 \Rightarrow \alpha = (3-4/\gamma_\mathrm{g}) = +1</math>, and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q \equiv - \frac{d\ln\theta}{d\ln\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\xi^2}{\sin\xi} \cdot \frac{d}{d\xi}\biggl[ \frac{\sin\xi}{\xi}\biggr]
=
1 - \xi \cot\xi \, .
</math>
  </td>
</tr>
</table>
</div>
If we then employ the displacement function,
<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>~A + \frac{B}{\xi^2} \biggl[ 1 - \xi \cot\xi \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>

the LAWE becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[4 - (n+1)Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
(n+1)\biggl[ \biggl(\frac{\sigma_c^2}{6\gamma_g } \biggr) \frac{\xi^2}{\theta} 
-\alpha Q\biggr] \frac{ x}{\xi^2} </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[4 - 2Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
\biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi^3}{\sin\xi} 
- 2Q\biggr] \frac{ x}{\xi^2} </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[2 + \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} 
+ \biggl[- 2 +  \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \frac{ x}{\xi^2} 
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[- 2 +  \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \biggl[ \frac{A}{\xi^2} + \frac{B}{\xi^4} \biggl( 1-\frac{\xi \cos\xi}{\sin\xi}  \biggr)\biggr]
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{2B}{\xi^4} \biggl[ 1 -  \frac{2\xi\cos\xi}{\sin\xi} + \biggl(\frac{\xi \cos\xi}{\sin\xi}  \biggr)^2 \biggr]
+ \frac{2A}{\xi^2}\biggl[\frac{\xi\cos\xi}{\sin\xi}  -1\biggr] 
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2A}{\xi^2}\biggl[\frac{\xi\cos\xi}{\sin\xi}  -1\biggr] 
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>
</table>
</div>
Pretty amazing degree of cancelation!  So the above-hypothesized displacement function ''does'' satisfy the <math>~n=1</math>, polytropic LAWE &#8212; for any value of the coefficient, <math>~B</math> &#8212; if we set <math>~A = 0</math> and <math>~\sigma_c^2=0</math>.  If we set <math>~B = 3</math>, the function will be normalized such that it goes to unity at the center.   In summary, then, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_P\biggr|_{n=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{\xi^2} \biggl[ 1 - \xi \cot\xi \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

<!--
Let's play with this a bit more to see if we can uncover a displacement function that works for nonzero values of <math>~\omega_c^2</math>.  Leaving both <math>~A</math> and <math>~B</math> unspecified for the time being, we have,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2A}{\xi^2}\biggl[\frac{\xi\cos\xi}{\sin\xi}  -1\biggr] 
+ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggl[ A + \frac{B}{\xi^2} \biggl( 1-\frac{\xi \cos\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(\frac{A\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} 
+ \biggl[ \biggl(\frac{B\sigma_c^2}{6 } \biggr) \frac{1}{\xi\sin\xi}  
- \frac{2A}{\xi^2} \biggr] \biggl( 1 - \frac{\xi\cos\xi}{\sin\xi}  \biggr) 
</math>
  </td>
</tr>
</table>
</div>
-->


==What About Bipolytropes?==

Here we will try to find an analytic expression for the radial displacement function, <math>~x</math>, for a bipolytropic ''envelope'' whose polytropic index is, <math>~n_e = 1</math>.  As in the above ''succinct'' derivation, the relevant LAWE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[4 - (n+1)Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
(n+1)\biggl[ \biggl(\frac{\sigma_c^2}{6\gamma_g } \biggr) \frac{\xi^2}{\theta} 
-\alpha Q\biggr] \frac{ x}{\xi^2} </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[4 - 2Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
\biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi^3}{\sin\xi} 
- 2Q\biggr] \frac{ x}{\xi^2} </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[2 + \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} 
+ \biggl[- 2 +  \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \frac{ x}{\xi^2} 
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>

</table>
</div>

===First Attempt===
Let's try,
<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>~
A + \frac{B}{(\xi - F)^2} \biggl[1 - (\xi-D) \cot(\xi-C) \biggr] \, .
</math>
  </td>
</tr>
</table>
First, note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\xi}\biggl[\cot(\xi - C) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d}{d\xi}\biggl[ \frac{ \cos(\xi - C) }{ \sin(\xi - C)}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\xi - C)\biggr] \, .
</math>
  </td>
</tr>
</table>

Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2B}{(\xi-F)^3} \biggl[1 - (\xi-D) \cot(\xi-C) \biggr]
- \frac{B}{(\xi-F)^2} \biggl\{
\cot(\xi-C) - (\xi-D) [1 + \cot^2(\xi-C) ]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{(\xi-F)^3} \biggl\{\biggl[2 - 2(\xi-D) \cot(\xi-C) \biggr]
- (\xi-F) \biggl[
\cot(\xi-C) - (\xi-D) [1 + \cot^2(\xi-C) ]\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{(\xi-F)^3} \biggl\{ 2 
-  \cot(\xi-C)\biggl[ 2(\xi-D) +  (\xi-F) \biggr] 
+ (\xi-F)  (\xi-D) [1 + \cot^2(\xi-C) ]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{(\xi-F)^3} \biggl\{ 2 
- \biggl[3\xi  - (2D +F) \biggr]  \cot(\xi-C)
+ [ \xi^2 - (D+F)\xi + FD]
+ [ \xi^2 - (D+F)\xi + FD]\cot^2(\xi-C) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{(\xi-F)^3} \biggl\{ 
[ \xi^2 - (D+F)\xi + FD+2]
- \biggl[3\xi  - (2D +F) \biggr]  \cot(\xi-C)
+ [ \xi^2 - (D+F)\xi + FD]\cot^2(\xi-C) 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
And,

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

<tr>
  <td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3B}{(\xi-F)^4} \biggl\{ 
[ \xi^2 - (D+F)\xi + FD+2]
- \biggl[3\xi  - (2D +F) \biggr]  \cot(\xi-C)
+ [ \xi^2 - (D+F)\xi + FD]\cot^2(\xi-C) 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{B}{(\xi-F)^3} \biggl\{ 
[ 2\xi - (D+F)]
- 3  \cot(\xi-C)
- \biggl[3\xi  - (2D +F) \biggr] \frac{d \cot(\xi-C)}{d\xi}
+ [ 2\xi - (D+F)]\cot^2(\xi-C) 
+ \biggl[ \xi^2 - (D+F)\xi + FD \biggr] \frac{d \cot^2(\xi-C) }{d\xi}
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{(\xi-F)^4}{B}  \biggr] \frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3[ \xi^2 - (D+F)\xi + FD+2]
- 3\biggl[3\xi  - (2D +F) \biggr]  \cot(\xi-C)
+ 3[ \xi^2 - (D+F)\xi + FD]\cot^2(\xi-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- (\xi-F)[ 2\xi - (D+F)]
+ 3 (\xi-F)  \cot(\xi-C)
- (\xi-F)[ 2\xi - (D+F)]\cot^2(\xi-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\xi-F)\biggl[3\xi  - (2D +F) \biggr] \frac{d \cot(\xi-C)}{d\xi}
- (\xi-F)\biggl[ \xi^2 - (D+F)\xi + FD \biggr] \frac{d \cot^2(\xi-C) }{d\xi}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3[ \xi^2 - (D+F)\xi + FD+2]
- (\xi-F)[ 2\xi - (D+F)]
+ \biggl\{3 (\xi-F)  
- 3\biggl[3\xi  - (2D +F) \biggr] \biggr\}  \cot(\xi-C)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{ 3[ \xi^2 - (D+F)\xi + FD] 
- (\xi-F)[ 2\xi - (D+F)] \biggr\}\cot^2(\xi-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- (\xi-F)\biggl[3\xi  - (2D +F) \biggr] \biggl[ 1 + \cot^2(\xi - C)\biggr]
+ (\xi-F)\biggl[ \xi^2 - (D+F)\xi + FD \biggr] 2\cot(\xi-C)\biggl[ 1 + \cot^2(\xi - C)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3[ \xi^2 - (D+F)\xi + FD+2]
- (\xi-F)[ 2\xi - (D+F)]
- (\xi-F) [3\xi  - (2D +F) ] 
+ \biggl\{3 (\xi-F)  
- 3 [3\xi  - (2D +F) ] 
+ 2 (\xi-F) [ \xi^2 - (D+F)\xi + FD ]  
\biggr\}  \cot(\xi-C)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{ 3[ \xi^2 - (D+F)\xi + FD] 
- (\xi-F)[ 2\xi - (D+F)] 
- (\xi-F) [3\xi  - (2D +F) ] 
\biggr\}\cot^2(\xi-C) 
+ 2 (\xi-F) [ \xi^2 - (D+F)\xi + FD ] \cot^3(\xi - C) \, .
</math>
  </td>
</tr>
</table>

Let's set <math>~C = D = F</math> and see if these expressions match the ones above.

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

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi} \biggr|_{C=D=F}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{\xi^3} \biggl\{ 
2 + \xi^2
- 3\xi  \cot\xi
+ \xi^2 \cot^2\xi 
\biggr\} \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\xi^4}{B}  \cdot \frac{d^2x}{d\xi^2} \biggr|_{C=D=F}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3[ \xi^2 +2]
- (\xi)[ 2\xi ]
- \xi [3\xi  ] 
+ \biggl\{3 (\xi)  
- 3 [3\xi  ] 
+ 2 \xi [ \xi^2  ]  
\biggr\}  \cot(\xi)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3 \xi^2 +6- 2\xi^2 - 3\xi^2 
+ \biggl[ 3 \xi - 9\xi  + 2 \xi^3  \biggr]  \cot(\xi)
+ \biggl[ 3\xi^2  - 2\xi^2 - 3\xi^2 \biggr] \cot^2\xi + 2 \xi^3  \cot^3\xi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6- 2\xi^2 
+ [ - 6\xi  + 2 \xi^3  ]  \cot(\xi)
- 2\xi^2 \cot^2\xi + 2 \xi^3  \cot^3\xi  
</math>
  </td>
</tr>
</table>

===Second Attempt===

Up to this point we have been rather cavalier about the use of <math>~\xi</math> (and <math>~\xi_i</math>) to represent the envelope's dimensionless radius (and interface location).  Let's switch to <math>~\eta</math>,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>~r^*</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta</math>
  </td>
</tr>
</table>

<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\eta^2} + \biggl\{ 4 -\biggl(\frac{\rho^*}{P^*}\biggr)\frac{ M_r^*}{(r^*)}\biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^4} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl(\frac{\rho^*}{ P^* } \biggr)\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  ~-~\frac{\alpha_\mathrm{g} M_r^*}{(r^*)^3}\biggr\}  x \, .
</math>
  </td>
</tr>
</table>

and, throughout the envelope we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\rho^*}{P^*}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{M_r^*}{r^*}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)
\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggr]^{-1}
=
2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i  \eta \biggl(- \frac{d\phi}{d\eta} \biggr)
\, .
</math>
  </td>
</tr>
</table>

Hence, the LAWE relevant to the envelope is,

<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\eta^2} + \biggl\{ 4 -\biggl[ \frac{\rho^*}{P^*}\biggr] \biggl[ \frac{ M_r^*}{(r^*)} \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^4} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[ \frac{\rho^*}{ P^* } \biggr] \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  ~-~\frac{\alpha_e }{(r^*)^2} \biggl[ \frac{M_r^*}{r^*} \biggr] \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\eta^2} + \biggl\{ 4 -\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1} \biggr] \biggl[ 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i  \eta \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^4} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1} \biggr] \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  
~-~\alpha_e \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggr]^{-2} \biggl[2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i  \eta \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \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\eta^2} + \biggl\{ 4 - \biggl[ \frac{2  \eta}{\phi} \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^5 \phi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1}  \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  
~-~\alpha_e \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i (4\pi) \eta^{-1} \biggr] \biggl(- \frac{d\phi}{d\eta} \biggr)  \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\eta^2} + \biggl\{ 4 - \biggl[ \frac{2  \eta}{\phi} \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^5 \phi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1}  \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}   \biggr\}  x 
~-~ \alpha_e  \biggl[ \frac{2\eta}{\phi} \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr]   \frac{x}{\eta^2} \, .
</math>
  </td>
</tr>
</table>

If we assume that, <math>~\alpha_e = (3 - 4/2) = 1</math> and <math>~\sigma_c^2 = 0</math>, then the relevant envelope LAWE is,

<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\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  \biggl[ 2 Q \biggr]   \frac{x}{\eta^2} \, ,
</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~
Q \equiv - \frac{d \ln \phi}{ d\ln \eta} \, .
</math>
</div>

Now consider the,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Precise Solution to the Polytropic LAWE</b></font></td>
</tr>

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\eta \phi^{n}}\biggr) \frac{d\phi}{d\eta}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-b\biggl[ \biggl( \frac{1}{\eta \phi}\biggr) \frac{d\phi}{d\eta}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{\eta^2}\biggl[ -\frac{d\ln \phi}{d\ln \eta}\biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{bQ}{\eta^2} \, .</math>
  </td>
</tr>
</table>
</div>

From our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|accompanying discussion]], we recall that the most general solution to the <math>n=1</math> Lane-Emden equation can be written in the form,
<div align="center">
<math>
\phi = A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, ,
</math>
</div>
where <math>A</math> and <math>B</math> are constants whose values can be obtained from our [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|accompanying parameter table]].  The first derivative of this function is,
<div align="center">
<math>
\frac{d\phi}{d\eta} = \frac{A}{\eta^2} \biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr] \, .
</math>
</div>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q = -\frac{d\ln\phi}{d\ln\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\eta}{\phi} \cdot \frac{A}{\eta^2} \biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1- \eta\cot(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-B) \biggr] \, .
</math>
  </td>
</tr>
</table>
What is this in terms of the dimensionless radius, <math>~r^*/R^*</math>?  Well,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>\frac{~r^*}{R^*}</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggl[\frac{\sqrt{2\pi}~\theta_i^2}{\eta_s} \biggl(\frac{\mu_e}{\mu_c}\biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{\eta}{\eta_s} = \frac{\eta}{(\pi + B)} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \eta</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi + B \biggr) \, .</math>
  </td>
</tr>
</table>
Also,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>~\eta-B</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi + B \biggr) -B = \pi \biggl( \frac{r^*}{R^*}\biggr) - B\biggl[1-\biggl( \frac{r^*}{R^*}\biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\pi + \pi \biggl[ \biggl( \frac{r^*}{R^*}\biggr)-1\biggr] - B\biggl[1-\biggl( \frac{r^*}{R^*}\biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\pi  - (\pi + B)\biggl[1-\biggl( \frac{r^*}{R^*}\biggr)\biggr] \, .</math>
  </td>
</tr>
</table>


<font color="red">'''[12 January 2019]:'''</font>  Here's what appears to work pretty well, empirically:

<table border="1" width="60%" align="center" cellpadding="8"><tr><td align="left">
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\eta^2} \biggl\{1- \eta\cot[\eta-(\pi - 0.8)] \biggr\} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi - 0.6\pi \biggr) \, .</math>
  </td>
</tr>
</table>
</td></tr></table>


<span id="tagJanuary2019">
Let's work through the analytic derivatives again.  Keeping in mind that,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\eta - B) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\eta - B)\biggr] \, ,
</math>
  </td>
</tr>
</table>
and starting with the ''guess'',

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

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-B) \biggr] \, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right">
<math>~ \frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{2b}{\eta^3} \biggl[1- \eta\cot(\eta-B) \biggr] 
- \frac{b}{\eta^2} \biggl\{ 
\cot(\eta-B) - \eta \biggl[ 1 + \cot^2(\eta - B)\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \Rightarrow ~~~ \biggl( \frac{\eta^3}{b} \biggr) \frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \biggl[2- 2\eta\cot(\eta-B) \biggr] 
- \biggl\{ 
\eta \cot(\eta-B) - \eta^2 \biggl[ 1 + \cot^2(\eta - B)\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B) \, .
</math>
  </td>
</tr>
</table>
The second derivative then gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\eta}\biggl\{
\frac{b}{\eta^3} \biggl[
\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B) 
\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{3b}{\eta^4} \biggl[
\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{b}{\eta^3} \biggl\{
2\eta + \cot(\eta-B)  +  2\eta \cot^2(\eta - B) 
+ \eta\frac{d}{d\eta}\biggl[\cot(\eta-B)\biggr]  +  2\eta^2\cot(\eta-B)  \frac{d}{d\eta} \biggl[ \cot(\eta - B) \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^4} \biggl[
6 - 3\eta^2 - 3\eta\cot(\eta-B)  -  3\eta^2\cot^2(\eta - B) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{b}{\eta^4} \biggl\{
2\eta^2 + \eta \cot(\eta-B)  +  2\eta^2 \cot^2(\eta - B) 
- \eta^2\biggl[ 1 + \cot^2(\eta - B)\biggr]  -  2\eta^3\cot(\eta-B)  \biggl[ 1 + \cot^2(\eta - B)\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{\eta^4}{b}\cdot \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6 - 3\eta^2 - 3\eta\cot(\eta-B)  -  3\eta^2\cot^2(\eta - B) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~+~ 2\eta^2 + \eta \cot(\eta-B)  +  2\eta^2 \cot^2(\eta - B) 
-\eta^2 - \eta^2  \cot^2(\eta - B) 
-  2\eta^3\cot(\eta-B)  
-  2\eta^3\cot^3(\eta-B)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl[ 3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^2\cot^2(\eta - B) 
-  \eta^3\cot^3(\eta-B) 
\biggr]  \, .
</math>
  </td>
</tr>
</table>

Recalling that,
<div align="center">
<math>~Q = \biggl[1- \eta\cot(\eta-B) \biggr] \, ,</math>
</div>
plugging these expressions into the relevant envelope LAWE gives,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  2 Q \cdot \frac{x}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 -2 \biggl[1- \eta\cot(\eta-B) \biggr]\biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]   \frac{2x}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{\eta^4} \biggl\{
\frac{\eta^4}{b} \cdot \frac{d^2x}{d\eta^2} + \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b} \cdot \frac{dx}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 x}{b} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^2\cot^2(\eta - B) 
-  \eta^3\cot^3(\eta-B) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \biggl[\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)\biggr]
~-~  \biggl[1- \eta\cot(\eta-B) \biggr] \biggl[1- \eta\cot(\eta-B) \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^2\cot^2(\eta - B) 
-  \eta^3\cot^3(\eta-B) 
+ \biggl[\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)\biggr]
~-~  \biggl[1- \eta\cot(\eta-B) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \eta\cot(\eta-B) \biggl[\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)\biggr]
~+~\eta\cot(\eta-B) \biggl[1- \eta\cot(\eta-B) \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^3\cot^3(\eta-B) 
~+~2\eta\cot(\eta-B) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \eta^3\cot(\eta-B) 
-2 \eta\cot(\eta-B) 
+ \eta^2\cot^2(\eta-B) 
+ \eta^3\cot^3(\eta - B)
~+~\eta\cot(\eta-B) 
~-~\eta^2\cot^2(\eta-B) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
[- \eta \cot(\eta-B)  
- \eta\cot(\eta-B) 
~+~2\eta\cot(\eta-B) ]
+ [\eta^3\cot(\eta-B) 
- \eta^3 \cot(\eta-B)  ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ [\eta^2\cot^2(\eta-B) 
~-~\eta^2\cot^2(\eta-B) ]
+ [\eta^3\cot^3(\eta - B)
-  \eta^3\cot^3(\eta-B) ]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>

Okay.  Now let's determine at what value of <math>~\eta</math> the logarithmic derivative of <math>~x_P</math> goes to negative one.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\ln x_P}{d\ln \eta} = \frac{\eta}{x_P} \cdot \frac{dx_P}{d\eta}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\eta^3}{b }\biggl[1- \eta\cot(\eta-B) \biggr]^{-1}
\cdot \frac{dx_P}{d\eta}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1- \eta\cot(\eta-B) \biggr]^{-1}
\biggl[ \eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)  \biggr] \, .
</math>
  </td>
</tr>
</table>
Setting this to negative one, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
-\biggl[1- \eta\cot(\eta-B) \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
\Rightarrow~~~1
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2\biggl[ 1 +  \cot^2(\eta - B) \biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~
\Rightarrow~~~1
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\eta^2}{\sin^2(\eta - B)} \, .
</math>
  </td>
</tr>
</table>
And this occurs when,
<div align="center">
<math>~\biggl(\frac{A}{\phi } \biggr)^2 = 1 \, .</math>
</div>


===Third Attempt===
====Prior to the Brute-Force Trial Fit====

Let's work through the analytic derivatives again.  Keeping in mind that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\eta - C) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\eta - C)\biggr] \, ,
</math>
  </td>
</tr>
</table>
and starting with the ''guess'',

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

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-C) \biggr] \, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right">
<math>~ \biggl( \frac{\eta^3}{b} \biggr) \frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C) \, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\eta^4}{b}\cdot \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl[ 3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-C)  
-  \eta^2\cot^2(\eta - C) 
-  \eta^3\cot^3(\eta-C) 
\biggr]  \, .
</math>
  </td>
</tr>
</table>


<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
Note that the relevant logarithmic derivative is,

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

<tr>
  <td align="right">
<math>~ \frac{d\ln x_P}{d\ln\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 \biggl( \frac{b}{\eta^2} \biggr)\biggl[ \eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C) \biggr]x_P^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C) \biggr]\biggl[1- \eta\cot(\eta-C) \biggr]^{-1}
</math>
  </td>
</tr>
</table>

If we know the logarithmic slope and the value of <math>~\eta</math> at the interface, then we can solve for 
<div align="center">
<math>~y_i \equiv  \eta_i \cot(\eta_i-C) \, ,</math>
</div>
via the quadratic relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1- y_i ) \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta_i^2 -2 + y_i  +  y_i^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta_i^2 -2 + y_i  +  y_i^2
- (1- y_i ) \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 y_i^2 
+ y_i \biggl\{1 + \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i\biggr\}
+\biggl\{  \eta_i^2 -2 - \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i \biggr\} \, .
</math>
  </td>
</tr>
</table>
(In practice it appears as though the "plus" solution to this quadratic equation is desired if the quantity inside the last set of curly braces is positive; and the "minus" solution is desired if this quantity is negative.)  Once the value of <math>~y_i</math> is known, we can solve for the key coefficient, <math>~C</math>, via the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tan(\eta_i - C)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{\eta_i}{y_i}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~C</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta_i - \tan^{-1}\biggl(\frac{\eta_i}{y_i}\biggr)\, .</math>
  </td>
</tr>
</table>

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


Recalling that,
<div align="center">
<math>~Q = \biggl[1- \eta\cot(\eta-B) \biggr] \, ,</math>
</div>
plugging these expressions into the relevant envelope LAWE gives,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  2 Q \cdot \frac{x}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 -2 \biggl[1- \eta\cot(\eta-B) \biggr]\biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]   \frac{2x}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{\eta^4} \biggl\{
\frac{\eta^4}{b} \cdot \frac{d^2x}{d\eta^2} + \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b} \cdot \frac{dx}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 x}{b} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-C)  
-  \eta^2\cot^2(\eta - C) 
-  \eta^3\cot^3(\eta-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \biggl[\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C)\biggr]
~-~  \biggl[1- \eta\cot(\eta-B) \biggr] \biggl[1- \eta\cot(\eta-C) \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-C)  
-  \eta^2\cot^2(\eta - C) 
-  \eta^3\cot^3(\eta-C) 
+ \biggl[\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C)\biggr]
~-~  \biggl[1- \eta\cot(\eta-C) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \eta\cot(\eta-B) \biggl[\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C)\biggr]
~+~\eta\cot(\eta-B) \biggl[1- \eta\cot(\eta-C) \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
(\eta - \eta^3)\cot(\eta-C)  
-  \eta^3\cot^3(\eta-C) 
+ \eta\cot(\eta-B) \biggl[\eta^2 -1   +  \eta^2\cot^2(\eta - C)  \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
(\eta - \eta^3) [ \cot(\eta-C)  - \cot(\eta-B) ]
+ \eta^3 \cot^2(\eta - C) [\cot(\eta-B)-  \cot(\eta-C)] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl[ \cot(\eta-C)  - \cot(\eta-B) \biggr] 
\biggl[ \eta - \eta^3 - \eta^3 \cot^2(\eta - C) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^3} \biggl[ \cot(\eta-C)  - \cot(\eta-B) \biggr] 
\biggl\{ 1 - \eta^2\biggl[1 + \cot^2(\eta - C)\biggr] \biggr\}\, .
</math>
  </td>
</tr>
</table>

This will go to zero if <math>~C = (B-2m\pi), </math> where <math>~m</math> is a positive integer.  When <math>~m =1</math>, for example,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cot(\eta-C)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cot[\eta - (B-2\pi)] = \cot(\eta -B) \, . </math>
  </td>
</tr>
</table>


Okay.  Now let's determine at what value of <math>~\eta</math> the logarithmic derivative of <math>~x_P</math> goes to negative one.

====Brute-Force Trial Fit====
<table align="left" width="100%" cellpadding="0"><tr><td align="left">
<table border="0" align="right"><tr><td align="center">
[[File:BruteForceWhiteBoardsmall.png|500px|Photo of white board with steps showing development of trial eigenfunction.  This should be paired with an Excel spreadsheet.]]
</td></tr></table>
Using a couple of separate Excel spreadsheets &#8212; FaulknerBipolytrope2.xlsx/mu100Mode0 and AnalyticTrialBipolytropeA.xlsx/Sheet2, both stored in a DropBox account under the folder Wiki_edits/Bipolytrope/LinearPerturbation &#8212; we used an inelegant and inefficient trial &amp; error technique in search of an eigenfunction that had the same analytic ''form'' as the one represented above for <math>~x_P</math>, but that, when plotted, appeared to qualitatively match the numerically determined envelope eigenfunction.  Then, on a whiteboard &#8212; see the photo, here on the right &#8212; we formulated a concise expression for a trial function that seemed to work pretty well.  Our primary finding was that <math>~\alpha</math>, appearing as the argument to the <math>~\tan\alpha</math> function, needed to be shifted by something like <math>~-3\pi/4</math>.

<div align="center">
<font size="+3">
<p>&nbsp;</p><p>THIS SPACE</p><p> INTENTIONALLY</p><p> LEFT BLANK</p>
</font>
</div>
</td></tr></table>

====Following Up on the Brute-Force Trial Fit====
In an [[SSC/Stability/BiPolytropes#Is_There_an_Analytic_Expression_for_the_Eigenfunction.3F|accompanying discussion]] &#8212; see especially [[SSC/Stability/BiPolytropes#Attempt_2|Attempt #2]] &#8212; we have determined by visual inspection that a decent fit to the envelope's eigenfunction is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_\mathrm{trial}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \biggl[ \frac{\tan(\eta_i - \Lambda - 3\pi/4) + f_\alpha}{1 - f_\alpha \cdot \tan(\eta_i - \Lambda - 3\pi/4)} \biggr] \biggr\} - a_0 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \cot(\Lambda - E)\biggr\} - a_0 \, ,
</math>
  </td>
</tr>
</table>

<table border="1" cellpadding="5" align="right">
<tr>
  <th align="center" colspan="4">Limiting Parameter Values</th>
</tr>
<tr>
  <td align="center">&nbsp;</td>
  <td align="center">min</td>
  <td align="center">max</td>
  <td align="center"><math>~\alpha = \alpha_s</math>
</tr>
<tr>
  <td align="center"><math>~\eta_\mathrm{F}</math></td>
  <td align="center"><math>~\eta_i</math></td>
  <td align="center"><math>~\eta_s</math></td>
  <td align="center"><math>~\frac{8}{\pi} ( \eta_s - \eta_i )^2 + 2\eta_s - \eta_i</math></td>
</tr>
<tr>
  <td align="center"><math>~\alpha</math></td>
  <td align="center"><math>~-\frac{\pi}{2}</math></td>
  <td align="center"><math>~-\frac{5\pi}{8}</math></td>
  <td align="center"><math>~\eta_i - \eta_s - \frac{3\pi}{4}</math></td>
</tr>
<tr>
  <td align="center"><math>~\Lambda</math></td>
  <td align="center"><math>~\eta_i - \frac{\pi}{4}</math></td>
  <td align="center"><math>~\eta_i - \frac{\pi}{8}</math></td>
  <td align="center"><math>~\eta_s</math></td>
</tr>
</table>
where, over the range, <math>~\eta_i \le \eta \le \eta_s \, ,</math>

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

<tr>
  <td align="right">
<math>~E</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\eta_i - \frac{5\pi}{4} + \tan^{-1} f_\alpha \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Lambda(\eta)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\eta_i + g_\mathrm{F}  \biggl[ \eta_i - 2\eta_s + \eta \biggr] = \Lambda_0 + g_\mathrm{F}\eta 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{1}{f_\alpha} = \tan(\alpha_s)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\tan[ - (\eta_s - \eta_i + \tfrac{3\pi}{4}) ]
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~g_\mathrm{F}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{\pi}{8(\eta_s - \eta_i)}
\, .</math>
  </td>
</tr>
</table>


----

Here, we reference a [[SSC/Stability/BiPolytropes#Attempt_1|separate discussion of the bipolytrope's underlying equilibrium structure]]

<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" width="50%"><math>~B = \eta_i - \frac{\pi}{2} + \tan^{-1}f</math></td>
  <td align="center"><math>~E = \eta_i - \frac{5\pi}{4} + \tan^{-1}f_\alpha</math></td>
</tr>
<tr>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow~~~\cot(\eta_i - B)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\eta_i - B)]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\tfrac{\pi}{2} - \tan^{-1}f)]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~f</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~f</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan(B + \tfrac{\pi}{2} - \eta_i )</math>
  </td>
</tr>
</table>
  </td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow~~~\cot(\eta_i - E)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\eta_i - E)]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\tfrac{5\pi}{4} - \tan^{-1}f_\alpha)]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan( \tan^{-1}f_\alpha - \tfrac{3\pi}{4} )</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\tan( \tfrac{3\pi}{4} - \tan^{-1}f_\alpha )</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\cot( \tan^{-1}f_\alpha )</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{1}{f_\alpha}</math>
  </td>
</tr>
</table>
  </td>
</tr>
<tr>
  <td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
Hence &hellip; &nbsp; &nbsp; <math>~\cot(\eta - B)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\eta - B)]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - \eta + \eta_i - \tfrac{\pi}{2} + \tan^{-1}f]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\eta_i - \eta + \tan^{-1}f]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ \tan(\eta_i-\eta) + f }{ 1 - f \cdot \tan(\eta_i - \eta)}</math>
  </td>
</tr>
</table>
  </td>
  <td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
Hence &hellip; &nbsp; &nbsp; <math>~\cot(\Lambda - E)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\Lambda - E)]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\eta_i - \Lambda - \tfrac{3\pi}{4} + \tan^{-1}f_\alpha]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\tan(\eta_i - \Lambda - \tfrac{3\pi}{4}) + f_\alpha  }{1 - f_\alpha \cdot \tan(\eta_i - \Lambda - \tfrac{3\pi}{4})  }</math>
  </td>
</tr>
</table>
  </td>
 </tr>
<tr>
  <td align="center">Also &hellip; &nbsp; &nbsp;<math>~B = \eta_s - \pi</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ f = \cot(\eta_i - B)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cot(\eta_i - \eta_s + \pi)</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{f}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan(\eta_i - \eta_s + \pi)</math>
  </td>
</tr>
</table>
  </td>
  <td align="center"><math>~</math></td>
</tr>
</table> 


----

Let's examine the first and second derivatives of this trial eigenfunction, recognizing that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx_\mathrm{trial}}{d\eta} = \frac{d\Lambda}{d\eta} \cdot \frac{dx_\mathrm{trial}}{d\Lambda}= g_\mathrm{F} \cdot  \frac{dx_\mathrm{trial}}{d\Lambda}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{d^2x_\mathrm{trial}}{d\eta^2} 
= \frac{d\Lambda}{d\eta} \cdot \frac{d}{d\Lambda} \biggl[ g_\mathrm{F}\cdot \frac{dx_\mathrm{trial}}{d\Lambda} \biggr]
= g_\mathrm{F}^2 \cdot \frac{d^2x_\mathrm{trial}}{d\Lambda^2} \, .
</math>
  </td>
</tr>
</table>
and drawing from the [[#Prior_to_the_Brute-Force_Trial_Fit|derivative expressions already derived, above]].  For the first derivative, we have,

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

<tr>
  <td align="right">
<math>~\frac{dx_\mathrm{trial}}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
g_\mathrm{F} \biggl( \frac{b_0}{\Lambda^3} \biggr) 
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] \, .
</math>
  </td>
</tr>
</table>
And the second derivative gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2x_\mathrm{trial}}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
g_\mathrm{F}^2 \biggl(\frac{2b_0}{\Lambda^4} \biggr) \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr]  \, .
</math>
  </td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x_\mathrm{trial}}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx_\mathrm{trial}}{d\eta} 
~-~  2 Q \cdot \frac{x_\mathrm{trial}}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x_\mathrm{trial}}{d\eta^2} + \biggl\{ 4 -2 \biggl[1- \eta\cot(\eta-B) \biggr]\biggr\}\frac{1}{\eta} \cdot \frac{dx_\mathrm{trial}}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]   \frac{2x_\mathrm{trial}}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b_0}{\eta^4} \biggl\{
\frac{\eta^4}{b_0} \cdot \frac{d^2x_\mathrm{trial}}{d\eta^2} + \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b_0} \cdot \frac{dx_\mathrm{trial}}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 x_\mathrm{trial}}{b_0} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b_0}{\eta^4} \biggl\{
\frac{\eta^4}{b_0} \cdot g_\mathrm{F}^2 \biggl(\frac{2b_0}{\Lambda^4} \biggr) \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b_0} \cdot 
g_\mathrm{F} \biggl( \frac{b_0}{\Lambda^3} \biggr) 
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 }{b_0} \cdot 
\biggl[\frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \cot(\Lambda - E)\biggr\} - a_0\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b_0}{\eta^4} \biggl\{
g_\mathrm{F}^2 \biggl(\frac{2\eta^4}{\Lambda^4} \biggr) \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \cdot 
g_\mathrm{F} \biggl( \frac{2\eta^3}{\Lambda^3} \biggr) 
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  
\biggl[\frac{2\eta^2}{\Lambda^2} [ 1 - \Lambda \cot(\Lambda - E) ] - \frac{2\eta^2 a_0}{b_0} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b_0}{\Lambda^4\eta^2} \biggl\{
g_\mathrm{F}^2 \eta^2 \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \cdot 
g_\mathrm{F} \Lambda \eta 
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  
\biggl[\Lambda^2 [ 1 - \Lambda \cot(\Lambda - E) ] - \frac{a_0\Lambda^4}{b_0} \biggr]
\biggr\}
</math>
  </td>
</tr>
</table>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow~~~\biggl(\frac{\Lambda^4}{2b_0}\biggr) \cdot</math> &nbsp;LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
g_\mathrm{F}^2 \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  
\frac{g_\mathrm{F} \Lambda}{ \eta }
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
~-~    \biggl(\frac{\Lambda}{\eta}\biggr)^2\biggl[  1 - \Lambda \cot(\Lambda - E)  - \frac{a_0\Lambda^2}{b_0} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ \eta\cot(\eta-B) \biggr] \biggl\{ 
\frac{g_\mathrm{F} \Lambda }{\eta }
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
~+~ \biggl( \frac{\Lambda}{\eta}\biggr)^2 \biggl[ 1 - \Lambda \cot(\Lambda - E)  - \frac{a_0\Lambda^2}{b_0 }\biggr]  \biggr\}
</math>
  </td>
</tr>
</table>


===Fourth Attempt===

====XXXX====
If we assume that, <math>~\alpha_e = (3 - 4/2) = 1</math> and <math>~\sigma_c^2 = 0</math>, then the relevant envelope LAWE is,

<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\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  \biggl[ 2 Q \biggr]   \frac{x}{\eta^2} \, ,
</math>
  </td>
</tr>
</table>
where,

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

<tr>
  <td align="right">
<math>~Q \equiv - \frac{d \ln \phi}{ d\ln \eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1- \eta\cot(\eta-B_0) \biggr] \, .
</math>
  </td>
</tr>
</table>


Let's work through the analytic derivatives again.  Keeping in mind that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\eta - B) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\eta - B)\biggr] \, ;
</math>
  </td>
</tr>
</table>
and that the,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Precise Solution to the Polytropic LAWE</b></font></td>
</tr>

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\eta \phi^{n}}\biggr) \frac{d\phi}{d\eta}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-b\biggl[ \biggl( \frac{1}{\eta \phi}\biggr) \frac{d\phi}{d\eta}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{\eta^2}\biggl[ -\frac{d\ln \phi}{d\ln \eta}\biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{bQ}{\eta^2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-B_0) \biggr] \, .
</math>
  </td>
</tr>
</table>

As we have [[#First_Attempt|already tried once, above]], let's try a more general form of this expression, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_Q</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
A + \frac{C}{(\eta - F)^2} \biggl[1 - (\eta-D) \cot(\eta-B) \biggr] \, .
</math>
  </td>
</tr>
</table>

Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx_Q}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 - (\eta-D) \cot(\eta-B) \biggr] \frac{d}{d\eta}\biggl[ \frac{C}{(\eta - F)^2} \biggr]
- \frac{C}{(\eta - F)^2} \frac{d}{d\eta} \biggl[ (\eta-D) \cot(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 - (\eta-D) \cot(\eta-B) \biggr]\biggl[ \frac{-2C}{(\eta - F)^3} \biggr]
- \frac{C}{(\eta - F)^2} \biggl[ \cot(\eta-B) \biggr]
+ \frac{C(\eta - D)}{(\eta - F)^2}\biggl[1 + \cot^2(\eta - B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^3} \biggl\{
- 2 + 2(\eta-D) \cot(\eta-B) 
- (\eta - F) \biggl[ \cot(\eta-B) \biggr]
+ (\eta - D)(\eta - F) \biggl[1 + \cot^2(\eta - B)\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^3} \biggl\{
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\} \, .
</math>
  </td>
</tr>
</table>

And,

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

<tr>
  <td align="right">
<math>~\frac{d^2x_Q}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\} \frac{d}{d\eta}\biggl[\frac{C}{(\eta-F)^3} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+\frac{C}{(\eta-F)^3} \cdot \frac{d}{d\eta} \biggl\{
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\} \biggl[\frac{-3C}{(\eta-F)^4} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+\frac{C}{(\eta-F)^3}  \biggl\{
[2\eta - (D+F) ] + \cot(\eta-B)
- (\eta - 2D + F) \biggl[1 + \cot^2(\eta-B) \biggr]
+ [2\eta -(D+F) ] \cot^2(\eta - B)
- 2[\eta^2 -\eta(D+F) + DF]\cot(\eta - B)\biggl[1 +  \cot^2(\eta - B)\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
- 3(\eta - 2D + F) \cot(\eta-B) 
- 3 (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+\frac{C}{(\eta-F)^3}  \biggl\{
[2\eta - (D+F) ] 
- (\eta - 2D + F) 
+ \cot(\eta-B)
- (\eta - 2D + F) \cot^2(\eta-B)
+ [2\eta -(D+F) ] \cot^2(\eta - B)
- 2[\eta^2 -\eta(D+F) + DF]\cot(\eta - B)
- 2[\eta^2 -\eta(D+F) + DF]\cot^3(\eta - B)
\biggr\} 
</math>
  </td>
</tr>
</table>

====YYYY====
And,

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

<tr>
  <td align="right">
<math>~\frac{d^2 x_Q}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\} \frac{d}{d\eta}\biggl[\frac{C}{(\eta-F)^3} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{C}{(\eta-F)^3} \cdot \frac{d}{d\eta}\biggl\{
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\}\biggl[\frac{-3C}{(\eta-F)^4} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{C}{(\eta-F)^3} \cdot \biggl\{
\frac{d}{d\eta}\biggl[ (\eta - 2D + F) \cot(\eta-B) \biggr]
+\frac{d}{d\eta}\biggl[  (\eta - D)(\eta - F) \cot^2(\eta - B)\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
-3(\eta - 2D + F) \cot(\eta-B) 
-3(\eta - D)(\eta - F) \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
(\eta-F)\frac{d}{d\eta}\biggl[ (\eta - 2D + F) \cot(\eta-B) \biggr]
+(\eta-F)\frac{d}{d\eta}\biggl[  (\eta - D)(\eta - F) \cot^2(\eta - B)\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
-3(\eta - 2D + F) \cot(\eta-B) 
-3(\eta - D)(\eta - F) \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta-F) \cot(\eta-B)  \frac{d}{d\eta}\biggl[ (\eta - 2D + F) \biggr]
+ (\eta-F) (\eta - 2D + F)  \frac{d}{d\eta}\biggl[ \cot(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+(\eta-F) \cot^2(\eta - B) \frac{d}{d\eta}\biggl[  \eta^2 -\eta(D+F) + DF \biggr]
+(\eta-F) (\eta - D)(\eta - F) \frac{d}{d\eta}\biggl[ \cot^2(\eta - B)\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
-3(\eta - 2D + F) \cot(\eta-B) 
-3(\eta - D)(\eta - F) \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta-F) \cot(\eta-B) 
- (\eta-F) (\eta - 2D + F)  \biggl[ 1 + \cot^2(\eta - B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+(\eta-F) \cot^2(\eta - B) \biggl[  2\eta - (D+F) \biggr]
-2 (\eta-F) (\eta - D)(\eta - F) \cot(\eta - B)\biggl[ 1 + \cot^2(\eta - B)\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
- (\eta-F) (\eta - 2D + F) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ (\eta-F)
-3(\eta - 2D + F) 
-2 (\eta-F) (\eta - D)(\eta - F)\biggr] \cot(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ (\eta-F) [  2\eta - (D+F) ]
-3(\eta - D)(\eta - F) 
-2 (\eta-F) (\eta - D)(\eta - F) \cot(\eta - B)
- (\eta-F) (\eta - 2D + F)\biggr] \cot^2(\eta - B)
\biggr\} 
</math>
  </td>
</tr>
</table>

So the envelope LAWE becomes,

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

<tr>
  <td align="right">
<math>~\frac{(\eta-F)^4}{C} \cdot \mathrm{LAWE}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\eta-F)^4}{C} \cdot \frac{d^2x_Q}{d\eta^2} + \frac{(\eta-F)^4}{C} \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \cdot \frac{dx_Q}{d\eta} 
~-~ \frac{(\eta-F)^4}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2x_Q}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
- (\eta-F) (\eta - 2D + F) 
+ \biggl[ (\eta-F)
-3(\eta - 2D + F) 
-2 (\eta-F) (\eta - D)(\eta - F)\biggr] \cot(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ (\eta-F) [  2\eta - (D+F) ]
-3(\eta - D)(\eta - F) 
-2 (\eta-F) (\eta - D)(\eta - F) \cot(\eta - B)
- (\eta-F) (\eta - 2D + F)\biggr] \cot^2(\eta - B)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta-F) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl\{ 
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~ \frac{(\eta-F)^4}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2}{\eta^2} \biggl\{
A + \frac{C}{(\eta - F)^2} \biggl[1 - (\eta-D) \cot(\eta-B) \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-3[(\eta - D)(\eta - F) - 2] 
- (\eta-F) (\eta - 2D + F) 
+ \biggl[ (\eta-F)
-3(\eta - 2D + F) 
-2 (\eta-F) (\eta - D)(\eta - F)\biggr] \cot(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta-F) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl\{ 
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ (\eta-F) [  2\eta - (D+F) ]
-3(\eta - D)(\eta - F) 
-2 (\eta-F) (\eta - D)(\eta - F) \cot(\eta - B)
- (\eta-F) (\eta - 2D + F)\biggr] \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta-F) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl[ 
(\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr]
~-~ (\eta-F)^2  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2}{\eta^2} \biggl[1 - (\eta-D) \cot(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~ \frac{(\eta-F)^4}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2A}{\eta^2} \, .
</math>
  </td>
</tr>
</table>

What does this reduce to if <math>~A = D = F = 0</math>.

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

<tr>
  <td align="right">
<math>~\frac{\eta^4}{C} \cdot \mathrm{LAWE}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-3[(\eta )(\eta ) - 2] 
- (\eta) (\eta ) 
+ \biggl[ (\eta)
-3(\eta ) 
-2 (\eta) (\eta )(\eta )\biggr] \cot(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl\{ 
[(\eta )(\eta ) - 2] 
+ (\eta ) \cot(\eta-B) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ (\eta) [  2\eta ]
-3(\eta )(\eta ) 
-2 (\eta) (\eta )(\eta) \cot(\eta - B)
- (\eta-) (\eta )\biggr] \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl[ 
(\eta )(\eta ) \cot^2(\eta - B)
\biggr]
~-~ (\eta)^2  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2}{\eta^2} \biggl[1 - (\eta) \cot(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~ \frac{(\eta)^4}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2A}{\eta^2} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
6 - 4\eta^2
-2 (\eta + \eta^3 ) \cot(\eta - B)
+ 2 \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \biggl[
\eta^2 - 2 + \eta  \cot(\eta-B) 
\biggr]
- 2\biggl[ \eta^2 + \eta^3 \cot(\eta - B) \biggr] \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2 \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr]\biggl[ 
\eta^2 \cot^2(\eta - B)
\biggr]
~-~2  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr] \biggl[1 - \eta \cot(\eta-B) \biggr]
~-~ \frac{2A\eta^2}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
6 - 4\eta^2
-2 (\eta + \eta^3 ) \cot(\eta - B)
+ 2\eta^2 - 4 + 2\eta  \cot(\eta-B) 
+ 2 \eta^3 \cot(\eta-B_0)
~-~4 \eta\cot(\eta-B_0) 
+ 2 \eta^2\cot(\eta-B_0) \cot(\eta-B) 
- 2\eta^2  \cot^2(\eta - B)
- 2 \eta^3  \cot^3(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2 \eta^2 \cot^2(\eta - B)
+ 2 \eta^3 \cot(\eta-B_0)  \cot^2(\eta - B)
-2 + 4 \eta\cot(\eta-B_0) - 2\eta^2\cot^2(\eta-B_0)  
~-~ \frac{2A\eta^2}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2\eta^2
-2 \eta^3 \biggl[ \cot(\eta - B) + \cot^3(\eta - B) \biggr]
+ \cot(\eta-B_0) \biggl[ 2 \eta^3 
+ 2 \eta^2 \cot(\eta-B) 
+ 2 \eta^3  \cot^2(\eta - B)
\biggr] - 2\eta^2\cot^2(\eta-B_0)  
~-~ \frac{2A\eta^2}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2\eta^2
+ 2 \eta^3\biggl[ \cot(\eta-B_0) - \cot(\eta-B)\biggr] \biggl[ 1 +  \cot^2(\eta - B)\biggr] 
+ 2 \eta^2  \cot(\eta-B_0)\biggl[ \cot(\eta-B)  - \cot(\eta-B_0)  \biggr]
~-~ \frac{2A\eta^2}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]  \, .  
</math>
  </td>
</tr>
</table>

=See Also=
* Radial Oscillations of [[SSC/Stability/UniformDensity#The_Stability_of_Uniform-Density_Spheres|Uniform-density sphere]]
* Radial Oscillations of Isolated Polytropes
** [[SSC/Stability/Polytropes#Radial_Oscillations_of_Polytropic_Spheres|Setup]]
** n = 1:&nbsp; [[SSC/Stability/n1PolytropeLAWE|Attempt at Formulating an Analytic Solution]]
** n = 3:&nbsp; [[SSC/Stability/n3PolytropeLAWE|Numerical Solution]] to compare with [http://adsabs.harvard.edu/abs/1941ApJ....94..245S M. Schwarzschild (1941)]
** n = 5:&nbsp; [[SSC/Stability/n5PolytropeLAWE|Attempt at Formulating an Analytic Solution]]

* In an accompanying [[Appendix/Ramblings/SphericalWaveEquation#Playing_With_Spherical_Wave_Equation|Chapter within our "Ramblings" Appendix]], we have played with the adiabatic wave equation for polytropes, examining its form when the primary perturbation variable is an enthalpy-like quantity, rather than the radial displacement of a spherical mass shell.  This was done in an effort to mimic the approach that has been taken in studies of the [[Apps/ImamuraHadleyCollaboration#Papaloizou-Pringle_Tori|stability of Papaloizou-Pringle tori]].

* <math>~n=3</math> &hellip; [http://adsabs.harvard.edu/abs/1941ApJ....94..245S M. Schwarzschild (1941, ApJ, 94, 245)], ''Overtone Pulsations of the Standard Model'':  This work is referenced in &sect;38.3 of [<b>[[Appendix/References#KW94|<font color="red">KW94</font>]]</b>].  It contains an analysis of the radial modes of oscillation of <math>~n=3</math> polytropes, assuming various values of the adiabatic exponent.

* <math>~n=2</math> &hellip; [http://adsabs.harvard.edu/abs/1961MNRAS.122..409P C. Prasad &amp; H. S. Gurm (1961, MNRAS, 122, 409)], ''Radial Pulsations of the Polytrope, n = 2''

* <math>~n=\tfrac{3}{2}</math> &hellip;  D. Lucas (1953, Bul. Soc. Roy. Sci. Liege, 25, 585) &hellip; Citation obtained from the Prasad &amp; Gurm (1961) article.

* <math>~n=1</math> &hellip;  L. D. Chatterji (1951, Proc. Nat. Inst. Sci. [India], 17, 467) &hellip; Citation obtained from the Prasad &amp; Gurm (1961) article.

* Composite Polytropes &hellip; [http://adsabs.harvard.edu/abs/1968MNRAS.140..235S M. Singh (1968, MNRAS, 140, 235-240)], ''Effect of Central Condensation on the Pulsation Characteristics''

* Summary of Known Analytic Solutions &hellip; [http://adsabs.harvard.edu/abs/1981MNRAS.197..351S R. Stothers (1981, MNRAS, 197, 351-361)], ''Analytic Solutions of the Radial Pulsation Equation for Rotating and Magnetic Star Models''

* Interesting Composite! &hellip; [http://adsabs.harvard.edu/abs/1948MNRAS.108..414P C. Prasad (1948, MNRAS, 108, 414-416)], ''Radial Oscillations of a Particular Stellar Model''


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