Editing SSC/Stability/n5PolytropeLAWE (section)

=Search for Analytic Solutions to the LAWE=

==Eureka Moment==

<font color="red">Note from J. E. Tohline on 3/6/2017:</font>&nbsp;  Yesterday evening, after I finished putting together the [[#n5TruncatedMovie|above animation sequence]] using an Excel workbook, I noticed that the eigenfunction of the fundamental mode for the marginally unstable model <math>~(\sigma_c^2 = 0)</math> resembles a parabola.  In an effort to see how well a parabola fits at least the central portion of this eigenfunction, I returned to my Excel spreedsheet and, in a brute-force manner, began to search for the pair of coefficients that would provide a best fit.  What I discovered was that a parabola with the following formula fits perfectly!
<div align="center" i="AnalyticSoln">
<table border="1" cellpadding="10" align="center">
<tr>
  <th align="center" colspan="1">Fundamental Mode Eigenfunction <br />
when <math>~\sigma_c^2 = 0</math> and <math>~\gamma = 6/5 ~\Rightarrow~\alpha=- 1/3</math></th>
</tr>

<tr>
  <td align="center">
<math>~x = x_0 \biggl[ 1 - \frac{\xi^2}{15} \biggr]</math>
  </td>
</tr>
</table>
</div>
For the ''specific'' normalization used in the above animation sequence, <math>~x_0 = \tfrac{5}{2}</math>.  Let's demonstrate that this eigenvector provides a solution to the LAWE for <math>~n=5</math> polytropes; for simplicity, we will set <math>~x_0 = 1</math>:
<div align="center" id="Proof">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi} = -\frac{2\xi}{15} \, ;</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left"> 
<math>~\frac{d^2x}{d\xi^2} = -\frac{2}{15} \, .</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
\biggl[ \cancelto{0}{\frac{\sigma_c^2}{3^{1/2} \gamma_g }} \cdot (3+\xi^2)^{3/2} + 2\biggr]  x 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~
-\frac{2}{15}(3+\xi^2)  -\frac{2}{15} \biggl[12 - 2\xi^2 \biggr]  + 2\biggl[1 - \frac{\xi^2}{15}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~
\biggl( - \frac{6}{15} - \frac{24}{15} + 2\biggr)
+\xi^2 \biggl( -\frac{2}{15} + \frac{4}{15} - \frac{2}{15} \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>
</div>

Q. E. D.  &nbsp;I don't think that anyone has previously appreciated that the LAWE in this case admits to an analytic eigenvector solution. 

Now, let's see how the boundary condition comes into play.  We see that the logarithmic derivative of the parabolic eigenfunction is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\ln x}{d\ln \xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{2\xi^2}{15} \biggl[ 1 - \frac{\xi^2}{15}\biggr]^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{2\xi^2}{(15-\xi^2)} \, .</math>
  </td>
</tr>
</table>
</div>
We desire a surface boundary condition that gives, <math>~d\ln x/d\ln\xi = -3</math>.  This will only happen when,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~- \frac{2\xi^2}{(15-\xi^2)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-3</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~2\xi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3(15 - \xi^2)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3 \, .</math>
  </td>
</tr>
</table>
</div>
Hence, although the parabolic eigenfunction provides an accurate solution to the <math>~n=5</math> LAWE throughout the entire configuration &#8212; that is, for all <math>~\xi</math> &#8212; the desired surface boundary condition will only be satisfied if the polytrope is truncated at <math>~\xi_\mathrm{surf} = 3</math>.  The parabolic eigenfunction is therefore only physically relevant to the model that sits at the point along the equilibrium sequence that is associated with the <math>~P_\mathrm{max}</math> turning point.

<!--  TRY MORE GENERIC CASE ...
Finally, let's see how this plays out for arbitrary values of <math>~\alpha</math>.   Guess a more general eigenfunction of the form,
<div align="center" i="AnalyticSoln">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center">
<math>~x = 1 - \frac{\xi^2}{A} </math>
  </td>
</tr>
</table>
</div>
In this more general case, we have,

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

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi} = -\frac{2\xi}{A} \, ;</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left"> 
<math>~\frac{d^2x}{d\xi^2} = -\frac{2}{A} \, .</math>
  </td>
</tr>
</table>
</div>
The LAWE then gives,

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

<tr>
  <td align="right">
<math>~
(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
\biggl[ \cancelto{0}{\frac{\sigma_c^2}{3^{1/2} \gamma_g }} \cdot (3+\xi^2)^{3/2} -6\alpha \biggr]  x 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~
-\frac{2}{A}(3+\xi^2)  -\frac{2}{A} \biggl[12 - 2\xi^2 \biggr]  -6\alpha \biggl[1 - \frac{\xi^2}{A}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~-\frac{1}{A}\biggl\{ 
2(3+\xi^2)  + 2 \biggl[12 - 2\xi^2 \biggr]  + 6\alpha \biggl[A - \xi^2\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~-
\frac{1}{A}\biggl[  \biggl(6 + 24 + 6\alpha A \biggr) + \biggl(2 -4-6\alpha \biggr)\xi^2\biggr]
</math>
  </td>
</tr>
</table>
</div>
Hence, a parabolic eigenfunction does not work for arbitrary <math>~\alpha</math>.
END MORE GENERIC CASE -->

Let's express the parabolic displacement function, <math>~x</math>, as a function of the Lagrangian mass coordinate, instead of as a function of <math>~\xi</math>.  Drawing upon our [[Appendix/Ramblings/NonlinarOscillation#Exploration|accompanying discussion]] where we have used <math>~\tilde\xi</math> to denote the truncation edge, we know that,

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

<tr>
  <td align="right">
<math>~r_\xi(\xi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi \biggl\{ 
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
and that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
r_\xi (m_\xi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde{r}_\mathrm{edge}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
</math>
  </td>
</tr>
</table>
</div>
<span id="DefineTildeC">where,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tilde{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr) \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\tilde{r}_\mathrm{edge}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
\, .
</math>
  </td>
</tr>
</table>
</div>

By equating <math>~r_\xi(\xi)</math> with <math>~r_\xi(m_\xi)</math>, we find,

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

<tr>
  <td align="right">
<math>~\xi \biggl\{ 
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \xi 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
This means that,
<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>~
x_0 \biggl\{1 - 
\frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr] 
\biggr\} \, ;
</math>
  </td>
</tr>
</table>
</div>
and, specifically for the critical case of <math>~\tilde\xi = 3</math>, in which case, <math>~\tilde{C} = 4</math>,
<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>~
x_0 \biggl\{1 - 
\frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{4 - 3 m_\xi^{2/3}}\biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>


{{ SGFworkInProgress }}

==Setup Using Lagrangian Radial Coordinate==

===Individual Terms===
From our [[SSC/FreeEnergy/Powerpoint#Case_M_Equilibrium_Conditions|accompanying discussion]], we have, for pressure-truncated, <math>~n=5</math> polytropic spheres

<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} 
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)}
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} \, ,
</math>
  </td>
</tr>
</table>
</div>
which matches the expression derived in an [[SSC/Structure/Polytropes#Lane-Emden_Equation|ASIDE box found with our introduction of the Lane-Emden equation]], and
<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}} 
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{(n+1)/(n-3)}
\tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\tilde\theta^{6}( -\tilde\xi^2 \tilde\theta' )^{6} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>~R_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} = \biggl( \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^{2}  \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~P_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)}  = \frac{K^{10}}{G^{9} M_\mathrm{tot}^{6} }  \, ,</math>
  </td>
</tr>
</table>
</div>

and, from [[SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|our more detailed analysis]],

<table border="0" cellpadding="3" align="center">
<tr>
  <td align="right">
<math>
~{\tilde\theta}_5 = 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2}
</math>
  </td>

  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>

  <td align="right">
<math>
~\biggl(- {\tilde\xi}^2 {\tilde\theta}^'_5\biggr)  = 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \, .
</math>
  </td>
</tr>
</table>

Hence,

<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} 
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\tilde\xi \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2}  \biggr]^{-2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\tilde\xi \biggl[ 3^{-1} {\tilde\xi}^{-6} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{4\pi}{2^5\cdot 3^7}\biggr]^{1/2}
 {\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}} 
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\biggl[  3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2} \biggr]^{6} \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2}  \biggr]^{6} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\biggl[  3^{3} \biggl( 3 + {\tilde\xi}^2\biggr)^{-3} \biggr] \biggl[ 3^{3} {\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-9}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^5}{4\pi}\biggr]^{3}
{\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-12} \, .
</math>
  </td>
</tr>
</table>
</div>

Now, given that the [[SSCpt1/Virial/FormFactors#Summary_.28n.3D5.29|structural form-factors for <math>~n=5</math> configurations]] are,

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

<tr>
  <td align="right">
<math>~\mathfrak{f}_M</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
( 1 + \ell^2 )^{-3/2}  = 3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathfrak{f}_W</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{5}{2^4} \cdot \ell^{-5}  
\biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathfrak{f}_A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{2^3} \ell^{-3}  [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]  \, ,
</math>
  </td>
</tr>
</table>
we understand that the central density is,

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

<tr>
  <td align="right">
<math>~\rho_c = \frac{\bar\rho}{ {\tilde\mathfrak{f}}_M }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl[3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2}  \biggr]^{-1} \biggl[ \frac{3 M_\mathrm{tot}}{4 \pi R_\mathrm{eq}^3} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl( \frac{3}{4\pi}\biggr)
\biggl[ \frac{2^5\cdot 3^6}{4\pi}\biggr]^{ 3 / 2} (3 + {\tilde\xi}^2)^{3 / 2} M_\mathrm{tot} \biggl[ R_\mathrm{norm} 
 {\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]^{-3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} {\tilde\xi}^{15}  (3 + {\tilde\xi}^2)^{-15 / 2} M_\mathrm{tot} R^{-3}_\mathrm{norm} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} M_\mathrm{tot}^{-5} \biggl( \frac{G}{K} \biggr)^{-15/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \, .
</math>
  </td>
</tr>
</table>

<span id="r0">Now let's derive the prescription for the Lagrangian radial coordinate in the context of pressure-truncated,</span> <math>~n=5</math> polytropes.  
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r_0 \equiv a_5 \xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \rho_c^{-2/5}  \xi</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \xi \biggl\{
\biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}
\biggr\}^{-2/5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]  \biggl( \frac{G^3M_\mathrm{tot}^2}{K^3} \biggr) 
\biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}   \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</math>
  </td>
</tr>
</table>
</div>

<span id="m0">Also,</span>

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

<tr>
  <td align="right">
<math>~m_0 \equiv M(r_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^2\pi \biggl\{ R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}   \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3 
\biggl\{ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}  \biggr\} 
\biggl\{  3^{1 / 2} \xi^3 \biggl( 3 + \xi^2\biggr)^{-3/2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 3^{1 / 2} \biggl[ 2^4 \pi^2\biggr]^{1 / 2} \biggl[ \frac{\pi^3}{2^9\cdot 3^{21}}\biggr]^{1 / 2} \biggl[ \frac{2^5\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2}  
\biggl\{  \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3 
\biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} R_\mathrm{norm}^3 
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl\{  \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} M_\mathrm{tot} 
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

Hence,

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

<tr>
  <td align="right">
<math>~g_0 = \frac{Gm_0}{r_0^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}
\biggl\{  \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\}  
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} 
\biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}   \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] 
\biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9}  
\xi ( 3 + \xi^2 )^{-3/2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{g_0 }{r_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] 
\biggl\{  \tilde\xi^{9} (3 + {\tilde\xi}^2)^{-9 / 2} \biggr\}  \biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}   \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-1}
\xi ( 3 + \xi^2 )^{-3/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2} 
\biggl[  \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15}  
( 3 + \xi^2 )^{-3/2} 
\, ;
</math>
  </td>
</tr>
</table>
</div>


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

<tr>
  <td align="right">
<math>~\frac{\rho_0}{P_0} = \frac{\rho_0}{K\rho_0^{1+1/n}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl[K^5 \rho_c \theta^5 \biggr]^{-1/5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \theta^{-1}
\biggl\{ K^5 \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}\biggr\}^{-1/5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
\biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[{\tilde\xi}^{-3}  (3 + {\tilde\xi}^2)^{3 / 2} \biggr]^{-3} } \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1 / 2}
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[ {\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} }
( 3 + \xi^2 )^{1 / 2} 
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
\biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \biggl[{\tilde\xi}^{-3}  (3 + {\tilde\xi}^2)^{3 / 2} \biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} 
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2}  ( 3 + \xi^2 )^{1/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} 
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2}  ( 3 + \xi^2 )^{1/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~  \times ~
\biggl( \frac{G^2M_\mathrm{tot}^2}{R_\mathrm{norm}^4} \biggr)^{1 / 2}\biggl[ \frac{2^6\cdot 3^{14}}{\pi^2}\biggr]^{1 / 2} 
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{-9 / 2}  
\xi ( 3 + \xi^2 )^{-3/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl( \frac{G^5 M_\mathrm{tot}^4}{K^5} \biggr)^{1 / 2}  R_\mathrm{norm}^{-2}
\biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl( \frac{K^5}{G^5 M_\mathrm{tot}^4} \biggr)^{1 / 2}  
\biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} 
\xi ( 3 + \xi^2 )^{-1}
\, .
</math>
  </td>
</tr>
</table>
</div>

===The Wave Equation===

====Starting from our Key Adiabatic Wave Equation====

The [[#Adiabatic_.28Polytropic.29_Wave_Equation|adiabatic wave equation]] therefore 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^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} 
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl\{
\biggl[ \frac{2^7\cdot 3^{7}}{\pi} \biggr]^{1 / 2}   \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \frac{1}{\xi} 
-   \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1}
\biggr\} \frac{dx}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2} 
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-3/2} 
( 3 + \xi^2 )^{1 / 2}  
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr]
+ \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2} 
\biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{15/2}  
( 3 + \xi^2 )^{-3/2}  
\biggr\}  x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl[ \frac{2^3\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2} 
( 3 + \xi^2 )^{1 / 2}   \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{-3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}
+  
( 3 + \xi^2 )^{-3/2}  
\biggr\}  x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0} 
+ \frac{6}{\gamma_g R_*^2}   \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}   \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}( 3 + \xi^2 )^{1 / 2}  
+  \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr\}  x
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>R_* \equiv R_\mathrm{norm} \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1/2} \, .</math>
</div>

Recognizing that,

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

<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_*   \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \, ,
</math>
  </td>
</tr>
</table>
</div>
we can write,

<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{1}{R_*^2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}  \biggl\{
\frac{d^2x}{d\xi^2} + \biggl[
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{d\xi} 
+ \frac{6}{\gamma_g }  
\biggl[\sigma^2 ( 3 + \xi^2 )^{1 / 2}  
+  \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr]  x \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\sigma^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}   \biggl( \frac{3 + {\tilde\xi}^2}{{\tilde\xi}^2} \biggr)^{15/2} = \frac{\sigma_c^2}{2\cdot 3^{3/2}}\, .</math>
  </td>
</tr>
</table>
</div>

Finally, if &#8212; because we are specifically considering the case of <math>~n=5</math> &#8212; we set <math>~\gamma_\mathrm{g} = 1 + 1/n = 6/5</math>, we have,

<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}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi} 
+ \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}} ( 3 + \xi^2 )^{1 / 2}  +  \frac{2}{( 3 + \xi^2 ) }\biggr]  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{( 3 + \xi^2 ) } \biggl\{ ( 3 + \xi^2 )\frac{d^2x}{d\xi^2} 
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi} 
+ \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}}  ( 3 + \xi^2 )^{3 / 2}  +  2 \biggr]  x \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
which matches exactly the [[#n5LAWE|form of the LAWE derived above]], if in ''that'' expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.


====Starting from the HRW66 Radial Pulsation Equation====
More directly, if we begin with the [[SSC/Stability/Polytropes#HRW66excerpt|HRW66 radial pulsation equation]] that is already tuned to polytropic configurations, the wave equation appropriate to <math>~n=5</math> polytropes is,
<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 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi}
+ \frac{5(-\theta_5^') }{6\theta_5 \xi} \bigg[ \frac{\xi (s^')^2}{\theta^'_5} + \frac{12}{5} \biggr] X
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{(3 + \xi^2)} \biggr]\frac{d X}{d\xi}
+ \frac{1}{(3 + \xi^2)} \bigg[ -\frac{5(s^')^2(3 + \xi^2)^{3 / 2}}{2 \cdot 3^{3 / 2}} + 2 \biggr] X
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{(3+\xi^2)} \biggl\{
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
+ \bigg[ -\frac{5(s^')^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
which is identical to the brute-force derivation just presented, allowing for the mapping,
<div align="center">
<math>\sigma^2 ~~ \Leftrightarrow ~~ -\frac{(s^')^2}{2 \cdot 3^{3 / 2}} \, .</math>
</div>

Finally, remembering that the [[SSC/Stability/Polytropes#HRW66frequency|HRW66 dimensionless frequency definition]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(s^')^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\omega^2 \biggl[\frac{n+1 }{4\pi G \rho_c} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
we recognize that, specifically for the case of <math>~n=5</math>, we can make the substitution, <math>~(s^')^2 \rightarrow -\sigma_c^2</math>, in which case 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{1}{(3+\xi^2)} \biggl\{
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
+ \bigg[ \frac{5\sigma_c^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
which matches exactly the [[#n5LAWE|form of the LAWE derived above]], if in ''that'' expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.

====New Independent Variable====
Guided by our [[Appendix/Ramblings/NonlinarOscillation#Conjectures|conjecture]] regarding the proper shape of the radial eigenfunction, let's switch the dependent variable to,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~u  \equiv 1 + \frac{3}{\xi^2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~3 + \xi^2 = \frac{3u}{(u-1)}  \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\xi = 3^{1 / 2} (u-1)^{-1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
This implies that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\xi}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow ~~~</math>
  </td>
  <td align="left">
<math>~-\frac{2}{\sqrt{3}}(u-1)^{3 / 2} \frac{d}{du} \, ,</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow ~~~</math>
  </td>
  <td align="left">
<math>~\frac{4}{3}(u-1)^3 \frac{d^2}{du^2} + 2(u-1)^{2} \frac{d}{du} \, .</math>
  </td>
</tr>
</table>
</div>
Hence, the governing wave equation 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>~( 3 + \xi^2 )\frac{d^2x}{d\xi^2} 
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi} 
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2}  +  2 \biggr]  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3u}{(u-1)} \biggl[\frac{4}{3}(u-1)^3 \frac{d^2x}{du^2} + 2(u-1)^{2} \frac{dx}{du}\biggr]
+ 4(2u-3)(u-1)\frac{dx}{du} 
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2}  +  2 \biggr\}  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4u(u-1)^2 \frac{d^2x}{du^2} 
+ (14u-12)(u-1)\frac{dx}{du} 
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2}  +  2 \biggr\}  x \, .
</math>
  </td>
</tr>
</table>
</div>

{{ SGFworkInProgress }}


If we ''assume'' that <math>~\sigma^2 = 0</math>, then the governing relation is,

<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>~4u(u-1)^2 \frac{d^2x}{du^2} 
+ (14u-12)(u-1)\frac{dx}{du} 
+ 2 x \, .
</math>
  </td>
</tr>
</table>
</div>
Now, again, guided by our  [[Appendix/Ramblings/NonlinarOscillation#Conjectures|conjecture]], let's guess an eigenfunction of the form:

=====First Guess (n5)=====
<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^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}   \, ,
</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}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{A^3}{2} \biggl[ (u - 1)^{-1 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{1 / 2} (A u - 1 )^{-3 / 2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{
-\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[
(Au-1) +3A (u-1)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

<!-- EXTRA 2nd DERIVATIVE
<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{A^3}{2^2} \biggl[ -(u - 1)^{-3 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} 
- A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} + 3A^2(u - 1)^{1 / 2} (A u - 1 )^{-5 / 2} \biggr]
</math>
  </td>
</tr>
END EXTRA 2nd DERIVATIVE -->

So the governing relation 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>~4u(u-1)^2 \biggl\{ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (14u-12)(u-1) \biggl\{ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \biggr\}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~u(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (7u-6)(u-1)^{1 / 2}  A^3(A-1)  (Au-1)^{-3 / 2} 
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(u-1)^{1 / 2}  \biggl\{ uA^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  
+ (7u-6) A^3(A-1)  (Au-1)^{-3 / 2} 
+ 2 A^3 (A u - 1 )^{-1 / 2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u(A-1) \biggl[(3A+1) - 4Au \biggr]  
+ (7u-6) (A-1)  (Au-1) 
+ 2 (A u - 1 )^{2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ - 4u^2 A(A-1)  + u(A-1) (3A+1) 
+ (7u-6)  [A(A-1)u +1 - A]
+ 2  (A^2u^2 - 2Au +1) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u^2 \biggl[ - 4A(A-1) +7A(A-1) +2A^2 \biggr]    
+ u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr]  + 2(3A-2)  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]    
+ u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr]  + 2(3A-2)  \biggr\} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]    
+ u\biggl[ -3A^2 -7A  +6\biggr]  + 2(3A-2)  \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

=====Second Guess (n5)=====
<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>~
(u - 1)^{b / 2} (A u - 1 )^{-a / 2}   \, ,
</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}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{2}(u-1)^{b/2-1} (A u - 1 )^{-a / 2}
- \frac{aA}{2}(u - 1)^{b / 2} (A u - 1 )^{-a / 2-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \biggl[ 
\frac{b}{2}(u-1)^{-1} 
- \frac{aA}{2}  (A u - 1 )^{-1} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{(u-1)}{x} \frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (A u - 1 )^{-1} \biggl[ 
\frac{b}{2} (A u - 1 )
- \frac{aA}{2} (u-1) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{2(A u - 1 )} \biggl[ 
b (A u - 1 ) - aA (u-1) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1 }{2(A u - 1 )} \biggl[ 
(aA - b) + A(b    - a)u \biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]\frac{dx}{du} 
+ x \frac{d}{du}\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]^2
+ x  \biggl[ -\frac{b}{2}(u-1)^{-2} + \frac{aA^2}{2}  (A u - 1 )^{-2}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{x}{4(u-1)^2 (Au-1)^2} \biggl\{
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{4 (Au-1)^2} \biggl\{
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>
</table>
</div>

Hence, the governing wave equation 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>~2u \biggl\{ \frac{(u-1)^2}{x} \frac{d^2x}{du^2} \biggr\}
+ (7u-6)\biggl\{ \frac{(u-1)}{x} \frac{dx}{du} \biggl\} + 1 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2u}{4 (Au-1)^2} \biggl\{
\biggl[ (aA - b) + A(b    - a)u \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{(7u-6) }{2(A u - 1 )} \biggl[ 
(aA - b) + A(b    - a)u \biggr] 
 + 1 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+  2u\biggl[  2aA^2 (u^2 - 2u + 1)  -2b (A^2 u^2 - 2Au + 1 ) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2(A u - 1 )(7u-6) \biggl[ 
(aA - b) + A(b    - a)u \biggr] 
 + 4 (Au-1)^2 \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+  2u\biggl[ 2A^2(a-b)u^2 + 4A(b - aA) u + 2(aA^2   -b)  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[ 
(aA - b) + A(b    - a)u \biggr] 
 +  (4A^2u^2-8Au + 4) \biggr\}
</math>
  </td>
</tr>
</table>
</div>

If <math>~b=a</math>,

<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>~
2u\biggl[ (aA - b)^2   \biggr]
+  2u\biggl[ 4A(b - aA) u + 2(aA^2   -b)  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[ 
(aA - b) \biggr]  +  (4A^2u^2-8Au + 4) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2a^2u (A - 1)^2  
+  2au [ 4A(1 - A) u + 2(A^2   -1) ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2a(A - 1) \biggl[7Au^2 - (6A+7)u  +6 \biggr] +  (4A^2u^2-8Au + 4) 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [4a (1 - A) +  7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2   -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>
</table>
</div>
This should then match the [[#First_Guess_.28n5.29|"first guess"]] algebraic condition if we set <math>~a=1</math>.  Let's see.
 
<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>~
2Au^2 [4 (1 - A) +  7(A - 1) + 2A] + 2u [ (A - 1)^2 +  2(A^2   -1) - (A - 1) (6A+7) - 4A] + 4[ 3(A-1) + 1]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [4  - 4A +  7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) +  2A^2   -2 + (1-A ) (6A+7) -4A] + 4[ 3A-2]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [5A - 3] + 2u [  - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, .
</math>
  </td>
</tr>
</table>
</div>
And we see that this expression ''does'' match the one derived earlier.

Going back a bit, before setting <math>~a=1</math>, we have the expression:
 

<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>~
2Au^2 [4a (1 - A) +  7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2   -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [  3aA  -3a + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2   -1) - a(6A^2+A-7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [  3a(A  - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A] + 4[ 3a(A-1) + 1] \, .
</math>
  </td>
</tr>
</table>
</div>
Now, in order for all three expressions inside the square-bracket pairs to be zero, we need, first,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3a(A  - 1) + 2A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2A}{3(1-A)} \, ;</math>
  </td>
</tr>
</table>
</div>
and, third, by simple visual comparison with the first expression,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3a(A-1) + 1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3a(A-1) + 2A</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3} \, ;</math>
  </td>
</tr>
</table>
</div>
which forces the second expression to the value,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{2}{3}\biggr)^2 \biggl(-\frac{1}{2} \biggr)^2 +  \frac{2}{3}\biggl[   -1-\frac{1}{2} +5 \biggr] - 2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{9}  +  \frac{7}{3}  - 2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4}{9}  \, ,</math>
  </td>
</tr>
</table>
</div>
which is ''not'' zero.  Hence our pair of unknown parameters &#8212; <math>~a </math> and <math>~A</math> &#8212; do not simultaneously satisfy all three conditions.  (Not really a surprise.)

==Setup Using Lagrangian Mass Coordinate==

===Alternative Terms===
Let's change the independent coordinate from <math>~r_0</math> to <math>~m_0</math>.  In particular, the derivative operation will change as follows:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{dr_0}</math>
  </td>
  <td align="center">
<math>~~\rightarrow~~</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{dm_0}{dr_0} \biggr)\frac{d}{dm_0} 
= \biggl( \frac{dm_0}{d\xi} \cdot \frac{d\xi}{dr_0}  \biggr)\frac{d}{dm_0}
\, ,</math>
  </td>
</tr>
</table>
</div>
so what is the expression for the leading coefficient?  From [[#r0|above]], we have,

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

<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} r_0 \, .
</math>
  </td>
</tr>
</table>
</div>

Also, from [[#m0|above]], we know that,

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

<tr>
  <td align="right">
<math>~m_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dm_0}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
\biggl\{ 3\xi^2 ( 3 + \xi^2 )^{-3/2} 
- 3 \xi^4 ( 3 + \xi^2 )^{-5/2}\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3\xi^2 (3 + \xi^2)^{-5/2} 
\biggl\{ ( 3 + \xi^2 ) 
- \xi^2  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dm_0}{dr_0}</math>
 </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} 
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
 </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{M_\mathrm{tot} }{R_*} 
\biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} \, .
</math>
  </td>
</tr>
</table>
</div>

To simplify expressions, let's [[Appendix/Ramblings/NonlinarOscillation#DefineTildeC|borrow from an accompanying derivation]] and define,
<div align="center">
<math>\tilde{C} \equiv \frac{3^2}{{\tilde\xi}^2} \biggl( 1 + \frac{ {\tilde\xi}^2}{3} \biggr) = 3 \biggl[ \frac{( 3 + {\tilde\xi}^2 )}{ {\tilde\xi}^2} \biggr] \, .</math>
</div>
Then we have,

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

<tr>
  <td align="right">
<math>~\frac{m_0}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\biggl[ \frac{\tilde{C}}{ 3}\biggr]^{3 / 2}
\biggl[ \frac{\xi^2}{ ( 3 + \xi^2 )} \biggr]^{3/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{\xi^2}{ ( 3 + \xi^2 )} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~( 3 + \xi^2 )\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\xi^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~3 m_*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \xi^2 (1-m_*)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\xi^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~   \frac{3m_*}{(1-m_*)} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~m_* \equiv \biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3} \, .</math>
</div>

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

<tr>
  <td align="right">
<math>~  
\frac{\xi^2}{ ( 3 + \xi^2 )} = m_* \, ;
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; while, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~  
\frac{ {\tilde\xi}^2}{ ( 3 + {\tilde\xi}^2 )} = \frac{3}{\tilde{C}} \, ;
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
= R_*  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggr[ \frac{3m_*}{ (1-m_*) }\biggr]^{1 / 2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{6}{R_*} \biggl[  \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2) }\biggr]^{9}  \frac{\xi}{ ( 3 + \xi^2 )}
=
\frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  \frac{m_*}{ \xi }
=
\frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{g_0 }{r_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{GM_\mathrm{tot}}{R_*^3} 
\biggl[  \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2)}\biggr]^{15/2}  \frac{1}{\xi^3}
\biggl[ \frac{ \xi^2 }{ ( 3 + \xi^2 ) }\biggr]^{3/2} 
= 
\frac{GM_\mathrm{tot}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\rho_0}{\gamma_g P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
</div>
So, the wave equation 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^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} 
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} 
+ \biggl\{ \frac{4}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}  
- \frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2}  \biggr\} \frac{dx}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} 
\biggl\{ \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{GM_\mathrm{tot}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 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}{dr_0^2} 
+ \frac{1}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggl\{ 4  
- 6\biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{6}  m_* \biggr\} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} 
\biggl\{\sigma^2  +  (1-m_*)^{3 / 2} \biggr\}  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl\{
R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2} 
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} 
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} 
R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2} 
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \frac{dx}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>

where,
<div align="center">
<math>~\sigma^2 \equiv (4 - 3\gamma_\mathrm{g})^{-1} \frac{R_*^3}{GM_\mathrm{tot}} \biggl[  \frac{ \tilde{C} }{3 } \biggr]^{15/2} \omega^2 \, .</math>
</div>

Now, let's look at the differential operators, after defining.
<div align="center">
<math>~c_0 \equiv  3^{1 / 2} R_*  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} ~~~~\Rightarrow ~~~~R_*   = c_0 3^{-1 / 2}  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} \, .</math>
</div>

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

<tr>
  <td align="right">
<math>~dr_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0 ~d[ m_*^{1 / 2} (1-m_*)^{-1 / 2} ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0 ~\biggl[\frac{1}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-1 / 2} + \frac{1}{2} ~m_*^{1 / 2} (1 - m_*)^{-3 / 2}
\biggr] dm_*
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c_0}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-3 / 2}~ dm_*
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{d}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{c_0} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ R_*\frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \, .
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\frac{d^2}{dr_0^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggl[ m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2}  
+\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2} \biggl[ \frac{1}{2} m_*^{-1 / 2}( 1 - m_*)^{3 / 2} - \frac{3}{2} m_*^{1 / 2}( 1 - m_*)^{1 / 2}~
\biggr] ~ \frac{d}{dm_*}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2}  +\frac{1}{2} \biggl( \frac{2}{c_0} \biggr)^{2}~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{d}{dm_*} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr]   
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
</math>
  </td>
</tr>
</table>
</div>

So, the wave equation 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{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} 
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr]   
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*}  \biggr] 
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} 
\biggl[\frac{2^2}{3}  \biggr]   
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*}  \biggr] 
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} m_*^{1 / 2} \biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  
\biggl\{    2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +  ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*}  
+ \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  
\biggl\{  2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 5 - 4m_*  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*}  
+ \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  \biggl\{  2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  
+ (5 - \mathcal{A}  m_*) (1-m_*)^2 \frac{dx}{dm_*}  
+ \mathcal{B} \biggl[ \frac{\sigma^2}{(1-m_*)^{1 / 2}}  +  (1-m_*) \biggr]  x  \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~4 + 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3^{5/2}(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3}  \, .</math>
  </td>
</tr>
</table>
</div>

==Try Again==

This time, let's adopt the notation used in a [[Appendix/Ramblings/NonlinarOscillation#n_.3D_5_Mass-Radius_Relation|related chapter in our ''Ramblings'' appendix]].  Specifically, the parametric relationship between <math>~m_\xi</math> and <math>~r_\xi</math> in pressure-truncated, <math>~n=5</math>  polytropes is,

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

<tr>
  <td align="right">
<math>~m_\xi \equiv \frac{m_0}{ M_\mathrm{tot} } = \frac{M_r(\xi)}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(3 + \xi^2 \biggr)^{-3/2} 
\biggl(3 +  {\tilde\xi}^2 \biggr)^{3/2} </math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~r_\xi \equiv \frac{r_0}{R_\mathrm{norm}} = \biggl(\frac{\xi}{\tilde\xi} \biggr) \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi \biggl\{ 
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1/2} 
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3} \xi \, .
</math>
  </td>
</tr>
</table>
</div>

And we are in the fortunate situation of being able to eliminate <math>~\xi</math> to obtain the direct relation,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
r_\xi (m_\xi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde{r}_\mathrm{edge}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tilde{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr) 
= 3 \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\tilde{r}_\mathrm{edge}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
= \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1 / 2} \biggl[ \frac{\tilde{C}}{ 3} \biggr]^{3} 
\, .
</math>
  </td>
</tr>
</table>
</div>

If we furthermore define,
<div align="center">
<math>m_* \equiv \frac{3}{\tilde{C}} \cdot m_\xi^{2 / 3} \, ,</math>
</div>
then,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
r_\xi (m_*)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2} 
\, .
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~
\frac{dr_0}{R_\mathrm{norm}} = dr_\xi
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl\{
\frac{1}{2} (1-m_*)^{- 1 / 2} m_*^{-1 / 2} + \frac{1}{2}m_*^{1 / 2}(1-m_*)^{-3 / 2}
\biggr\} dm_*
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{3^{1 / 2}}{2} \biggr) \tilde{r}_\mathrm{edge} m_*^{-1 / 2} (1-m_*)^{-3 / 2} dm_*
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ R_\mathrm{norm} \cdot \frac{d}{dr_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*} \, .
</math>
  </td>
</tr>
</table>
</div>
We therefore also have,

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

<tr>
  <td align="right">
<math>~ R^2_\mathrm{norm} \cdot \frac{d^2}{dr_0^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} 
\biggl\{ 
\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d^2}{dm_*^2}\biggr] 
+ \biggl[ \frac{1}{2} m_*^{-1 / 2} (1-m_*)^{3 / 2} + \frac{3}{2}m_*^{1 / 2} (1-m_*)^{1 / 2}\biggr]  \frac{d}{dm_*}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{  
\biggl[ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}\biggr] 
+ \biggl[  (1-m_*)^{3 } + 3m_* (1-m_*)^{2}\biggr]  \frac{d}{dm_*}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{  
2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{d}{dm_*} \biggr\} 
\, .
</math>
  </td>
</tr>
</table>
</div>

So the wave equation may be written,
<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>~
R_\mathrm{norm}^2 \cdot \frac{d^2x}{dr_0^2} + \biggl[\frac{4R_\mathrm{norm}}{r_0} - \biggl(\frac{g_0 \rho_0 R_\mathrm{norm}}{P_0}\biggr) \biggr] R_\mathrm{norm} \cdot \frac{dx}{dr_0} 
+ \biggl(\frac{\rho_0 R_\mathrm{norm}}{\gamma_\mathrm{g} P_0} \biggr)\biggl[R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0 R_\mathrm{norm}}{r_0} \biggr]  x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{  
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr)  \biggl\{ \frac{4}{r_\xi} 
- \biggl[\frac{6R_\mathrm{norm}}{R_*} \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{9}  m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2} \biggr] \biggr\} 
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6R_* R_\mathrm{norm}}{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{
R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g}) \frac{GM_\mathrm{tot} R_\mathrm{norm}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2}  
\biggr\}  x \, .
</math>
  </td>
</tr>
</table>
</div>
Keeping in mind that,
<div align="center">
<math>~\frac{R_*}{R_\mathrm{norm}} = \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1 / 2} = {\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \, ,</math>
</div>

we therefore have,
<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{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{  
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr)  \biggl\{ 4 \biggl[3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2}  \biggr]^{-1}
- 6 \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{9} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-1} m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2}  \biggr\} 
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 6 \biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{
\biggl[ \frac{R_*^3}{\gamma_g GM_\mathrm{tot} } \biggr] \omega^2 
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}  \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2}  
\biggr\}  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{  
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}^2} \biggl( \frac{2^3}{3} \biggr)  
\biggl[ 1  - \frac{3}{2} \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_*  \biggr] 
(1-m_*)^{2} \frac{dx}{dm_*} 
+ \frac{6}{ {\tilde{r}}_\mathrm{edge}^2 }  
\biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} 
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2}  \biggr]  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{  
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ \biggl[  5  - 6 \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* + 2m_* \biggr] 
(1-m_*)^{2} \frac{dx}{dm_*} 
+ 3^{5 / 2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  
\frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} 
\biggl[ \frac{\sigma^2 }{(1-m_*)^{1 / 2}} +  (1-m_*)  \biggr]  x \biggr\}
\, ,
</math>
  </td>
</tr>
</table>
</div>
where, as before,
<div align="center">
<math>\sigma^2 \equiv \biggl(  \frac{ \tilde{C} }{3 } \biggr)^{15/2} \biggl[ \frac{R_*^3}{(4 - 3\gamma_g) GM_\mathrm{tot} } \biggr] \omega^2 \, .</math>
</div>

==Take Another Approach Using Logarithmic Derivatives==

===Change Independent Variable===

Returning to the [[#n5LAWE|LAWE for n = 3 polytropes, as given, above]], and repeated here,

<table border="1" cellpadding="8" align="center">
<tr><th align="center">LAWE for <math>~n=5</math> Polytropes</th></tr>
<tr><td 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>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
\biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{3/2} - 
6\alpha \biggr]  x </math>
  </td>
</tr>
</table>
</td></tr></table>

let's make the substitution,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~u \equiv (3 + \xi^2)^{1/2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\xi^2 = u^2-3 \, .</math>
  </td>
</tr>
</table>
</div>
We must therefore also make the operator substitution,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{du}{d\xi} \cdot \frac{d}{du}</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{1}{\xi} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{u}\cdot  \frac{dx}{du} \, ;</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2}   \frac{d}{du} \biggl\{  \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2}  \frac{d}{du} \biggr\}</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{3}{u^3} \frac{dx}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr]  \frac{d^2x}{du^2} \, .</math>
  </td>
</tr>
</table>
</div>

The rewritten LAWE is therefore,
<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>~u^2 \biggl\{ \frac{3}{u^3} \frac{dx}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr]  \frac{d^2x}{du^2}  \biggr\} 
+ 2\biggl[9 - u^2 \biggr] \frac{1}{u} \cdot \frac{dx}{du} + 
\biggl[\Omega^2 u^3 - 
6\alpha \biggr]  x </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~(u^2-3)  \frac{d^2x}{du^2}  
+ (21 - 2u^2 ) \frac{1}{u} \cdot \frac{dx}{du} + 
(\Omega^2  u^3 - 6\alpha )  x \, ,</math>
  </td>
</tr>
</table>
</div>
where we have adopted the shorthand notation,
<div align="center">
<math>~\Omega^2 \equiv \frac{\sigma_c^2}{3^{1/2} \gamma_g } \, .</math>
</div>

===Look at Logarithmic Derivative===

Multiplying through by <math>~(u^2/x)</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>~(u^2-3) \frac{u^2}{x} \cdot \frac{d^2x}{du^2}  
+ (21 - 2u^2 ) \frac{d\ln x}{d\ln u} + 
(\Omega^2  u^5 - 6\alpha u^2 )  \, .</math>
  </td>
</tr>
</table>
</div>
Now, in the context of a [[SSC/Stability/BiPolytrope00Details#Idea_Involving_Logarithmic_Derivatives|separate derivation]], we showed that, quite generally we can make the substitution,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{u^2}{x} \cdot \frac{d^2x}{du^2} 
</math>
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln u} \biggl[ \frac{d\ln x}{d\ln u} \biggr]
+ \biggl[  \frac{d\ln x}{d\ln u}-1 \biggr]\cdot \frac{d\ln x}{d\ln u} \, .
</math>
 </td>
</tr>
</table>
</div>
Hence, if we assume that the displacement function can be expressed as a power-law in <math>~u</math>, such that,
<div align="center">
<math>\frac{d\ln x}{d\ln u} = c_0 \, ,</math>
</div>
then the LAWE for <math>~n=5</math> polytropes simplifies as follows,
<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>~(u^2-3) c_0(c_0-1) + (21 - 2u^2 ) c_0 + (\Omega^2  u^5 - 6\alpha u^2 )  \, .</math>
  </td>
</tr>
</table>
</div>
This polynomial equation will be satisfied if, simultaneously, we set:
<ul>
<li><math>\Omega^2 = 0 \, ;</math></li>
<li><math>c_0^2 -3c_0 -6\alpha = 0 </math>&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math>&nbsp; &nbsp; &nbsp; <math>c_0 = \frac{3}{2}\biggl[1 \pm \biggl(1+\frac{8\alpha}{3} \biggr)^{1/2} \biggl]\, ;</math></li>
<li><math>~\alpha = 20/3 \, .</math>
</ul>

This gives us some hope that a more general solution of the following form will work:
<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>~u^{c_0} \biggl[ a + bu + cu^2 + du^3 + \cdots\biggr] \, .</math>
  </td>
</tr>
</table>
</div>

This means that, for example,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0 u^{c_0-1} \biggl[ a + bu + cu^2 + du^3 \biggr] 
+ u^{c_0} \biggl[ b + 2cu + 3du^2 \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{d\ln x}{d\ln u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c_0(a + bu + cu^2 + du^3) + bu + 2cu^2 + 3du^3}{a + bu + cu^2 + du^3} 
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0(c_0-1) u^{c_0-2} \biggl[ a + bu + cu^2 + du^3 \biggr] 
+ 2c_0 u^{c_0-1} \biggl[ b + 2cu + 3du^2 \biggr] 
+ u^{c_0} \biggl[ 2c + 6du \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{u^2}{x} \cdot \frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c_0(c_0-1) ( a + bu + cu^2 + du^3 ) + 2c_0 ( bu + 2cu^2 + 3du^3 ) + ( 2cu^2 + 6du^3 ) }{ a + bu + cu^2 + du^3}
</math>
  </td>
</tr>
</table>
</div>
So the LAWE becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~- (\Omega^2  u^5 - 6\alpha u^2 ) (a + bu + cu^2 + du^3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~(u^2-3)  [c_0(c_0-1) ( a + bu + cu^2 + du^3 ) + 2c_0 ( bu + 2cu^2 + 3du^3 ) + ( 2cu^2 + 6du^3 )]  
+ (21 - 2u^2 ) [c_0(a + bu + cu^2 + du^3) + bu + 2cu^2 + 3du^3] \,.
</math>
  </td>
</tr>
</table>
</div>
This is cute, but I don't see any way that this approach will provide an avenue to cancel the <math>~\Omega^2 u^5</math> term.


==Yet Another Guess==
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>~e^{a + b\ln\xi + c(\ln\xi)^2} \, ,</math>
  </td>
</tr>
</table>
</div>
and examine the specific case of <math>~\sigma_c^2 = 0</math>, and, <math>~\gamma = (n+1)/n = 6/5 ~~\Rightarrow~~ \alpha = (3-20/6) = -1/3</math>.  Under these conditions, the LAWE for <math>~n=5</math> polytropes becomes,

<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>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 2x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~(3+\xi^2) \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + 2\xi^2 \, .
</math>
  </td>
</tr>
</table>

And the derivatives give,
<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>~x \frac{d}{d\xi}\biggl[ a + b\ln\xi + c(\ln\xi)^2 \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~b + 2c\ln\xi \, ;</math>
  </td>
</tr>
</table>
</div>
and,
<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>~ \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr] \frac{dx}{d\xi} + x \frac{d}{d\xi}\biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \xi \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + \xi^2 \frac{d}{d\xi}\biggl[ \frac{b+ 2c\ln\xi}{\xi} \biggr]</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (b + 2c\ln\xi )^2 + 2c- (b+ 2c\ln\xi) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[ b^2 + 2c - b] + [4bc  - 2c] \ln\xi+4c^2 (\ln\xi)^2 \, .
</math>
  </td>
</tr>
</table>
</div>

Hence the "fundamental mode" 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>~(3+\xi^2) \biggl[ ( b^2 + 2c - b ) + (4bc  - 2c) \ln\xi+4c^2 (\ln\xi)^2 \biggr] 
+ (12 - 2\xi^2 ) \biggl[ b + 2c\ln\xi  \biggr] \, .
+ 2\xi^2 
</math>
  </td>
</tr>
</table>
</div>

Now, this expression cannot be satisfied for arbitrary <math>~\xi</math>.  But, here we seek a solution only at the surface for the ''specific'' model, <math>~\xi = 3</math>.  Plugging this value into the expression 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>~12 \biggl[ ( b^2 + 2c - b ) + (4bc  - 2c) \ln 3+4c^2 (\ln 3)^2 \biggr] 
+ (12 - 18 ) \biggl[ b + 2c\ln 3 \biggr] + 18
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~2 \biggl[ ( b^2 + 2c - b ) + (4bc  - 2c) \ln 3+4c^2 (\ln 3)^2 \biggr] 
-\biggl[ b + 2c\ln 3 \biggr] + 3 \, .
</math>
  </td>
</tr>
</table>
</div>
It appears as though one perfectly satisfactory solution is, <math>~c = 0</math>, in which case, we need,
<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>~2 b^2    - 3b + 3 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~b</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~
\frac{3}{4}\biggl[1 \pm \sqrt{1-\frac{8}{3} }  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Thus, <math>~b</math> is an complex number.