Editing SSC/Structure/OtherAnalyticRamblings (section)

===Exploration===

====Compare LAWE to Hydrostatic Balance Condition====

Returning to the [[SSC/Structure/OtherAnalyticModels#Generic|generic formulation derived earlier]], we have,

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

<tr>
  <td align="right">
<math>~- \sigma^2 \biggl(\frac{\rho_0}{\rho_c}\biggr) x </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{P_0}{P_c}\biggr)\frac{1}{\chi_0^4} \frac{d}{d\chi_0}\biggl( \chi_0^4 x^' \biggr) 
+ \frac{d}{d\chi_0}\biggl(\frac{P_0}{P_c}\biggr) \cdot \frac{1}{\chi_0^\alpha}\frac{d}{d\chi_0}\biggl(\chi_0^\alpha x\biggr) \, .
</math>
  </td>
</tr>
</table>
</div>

Dividing this entire expression through by <math>~(P_0/P_c)x</math> gives,

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

<tr>
  <td align="right">
<math>~- \sigma^2 \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{x \chi_0^4} \frac{d}{d\chi_0}\biggl( \chi_0^4 x^' \biggr) 
+ \frac{d}{d\chi_0}\biggl[ \ln \biggl(\frac{P_0}{P_c}\biggr)\biggr] \cdot \frac{1}{x\chi_0^\alpha}\frac{d}{d\chi_0}\biggl(\chi_0^\alpha x\biggr) \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{x^'}{x} \biggr) \cdot \frac{d}{d\chi_0}\biggl[\ln\biggl( \chi_0^4 x^' \biggr) \biggr]
+ \frac{d}{d\chi_0}\biggl[ \ln \biggl(\frac{P_0}{P_c}\biggr)\biggr] \cdot \frac{d}{d\chi_0}\biggl[\ln\biggl(\chi_0^\alpha x\biggr)\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Now, let's step aside from the LAWE and look directly at the differential relationship between the mass-density and the pressure, as dictated by combining the [[SSCpt2/SolutionStrategies#Spherically_Symmetric_Configurations_.28Part_II.29|two principal governing relations]], the

<div align="center">
<span id="HydrostaticBalance"><font color="#770000">'''Hydrostatic Balance'''</font></span><br />

<math>\frac{1}{\rho}\frac{dP}{dr} =- \frac{d\Phi}{dr} </math> ,<br />
</div>
and,
<div align="center">
<span id="Poisson"><font color="#770000">'''Poisson Equation'''</font></span><br />

<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d \Phi}{dr} \biggr)  = 4\pi G \rho </math> .<br />
</div>
In combination, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-4\pi G \rho_0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{r_0^2}\frac{d}{dr_0}\biggl[ \frac{r_0^2}{\rho_0} \frac{dP_0}{dr_0}\biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ -4\pi G \rho_0 \biggl(\frac{R^2\rho_c}{P_c}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\chi_0^2}\frac{d}{d\chi_0}\biggl[ \frac{\chi_0^2}{(\rho_0/\rho_c)}\cdot \frac{d}{d\chi_0}\biggl(\frac{P_0}{P_c}\biggr)\biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ -[4\pi G \rho_c \tau_\mathrm{SSC}^2] \biggl( \frac{\rho_0}{\rho_c} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\rho_0}{\rho_c} \biggr)^{-1}\frac{1}{\chi_0^2}\frac{d}{d\chi_0}\biggl[ \chi_0^2 \frac{d}{d\chi_0}\biggl(\frac{P_0}{P_c}\biggr)\biggr] 
+ \frac{d}{d\chi_0}\biggl(\frac{P_0}{P_c}\biggr) \cdot \frac{d}{d\chi_0}\biggl( \frac{\rho_0}{\rho_c}\biggr)^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ -[4\pi G \rho_c \tau_\mathrm{SSC}^2] \biggl( \frac{\rho_0}{\rho_c} \biggr)^2\biggl(\frac{P_0}{P_c}\biggr)^{-1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\chi_0^2 (P_0/P_c)}\frac{d}{d\chi_0}\biggl[ \chi_0^2 \frac{d}{d\chi_0}\biggl(\frac{P_0}{P_c}\biggr)\biggr] 
+ \frac{d}{d\chi_0}\biggl[\ln\biggl(\frac{P_0}{P_c}\biggr)\biggr] \cdot \frac{d}{d\chi_0}\biggl[\ln\biggl( \frac{\rho_0}{\rho_c}\biggr)^{-1}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{p^'}{p}\biggr)
\frac{1}{\chi_0^2 p^'}\frac{d}{d\chi_0}\biggl[ \chi_0^2 p^'\biggr] + \frac{d\ln(p)}{d\chi_0}\cdot \frac{d}{d\chi_0}\biggl[\ln\biggl( \frac{\rho_0}{\rho_c}\biggr)^{-1}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d\ln(p)}{d\chi_0}\cdot
\frac{d}{d\chi_0}\biggl[ \ln(\chi_0^2 p^')\biggr] + \frac{d\ln(p)}{d\chi_0}\cdot \frac{d}{d\chi_0}\biggl[\ln\biggl( \frac{\rho_0}{\rho_c}\biggr)^{-1}\biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~p^'</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{d}{d\chi_0}\biggl(\frac{P_0}{P_c}\biggr) \, .</math>
  </td>
</tr>
</table>
</div>

Let's compare the form of this "equilibrium" relation with the form of the LAWE just constructed:
<div align="center" id="Compare">
<table border="1" align="center" width="75%">
<tr><td align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<math>~- [4\pi G \rho_c \tau_\mathrm{SSC}^2] \biggl( \frac{\rho_0}{\rho_c} \biggr)^2\biggl(\frac{P_0}{P_c}\biggr)^{-1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{p^'}{p} \biggr) \cdot 
\frac{d}{d\chi_0}\biggl[ \ln(\chi_0^2 p^')\biggr] + \frac{d\ln(p)}{d\chi_0}\cdot \frac{d}{d\chi_0}\biggl[\ln\biggl( \frac{\rho_0}{\rho_c}\biggr)^{-1}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<font size="+1">''versus''</font>
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
<math>~- \sigma^2 \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{x^'}{x} \biggr) \cdot \frac{d}{d\chi_0}\biggl[\ln\biggl( \chi_0^4 x^' \biggr) \biggr]
+ \frac{d\ln(p)}{d\chi_0}\cdot  \frac{d}{d\chi_0}\biggl[\ln\biggl(\chi_0^\alpha x\biggr)\biggr] 
</math>
  </td>
</tr>
</table>

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

I like this layout because it unveils similarities in the way the differential operators interact with the functions that describe the radial profiles of variables &#8212; specifically, the mass-density, the pressure, and the fractional radial displacement, <math>~x</math>, during pulsations.  However, it is not yet obvious how best to translate between the two differential equations in order to aid in solving for the unknown variable, <math>~x(\chi_0)</math>.

====Dabbling with Equilibrium Condition====
In the meantime, I've found it instructive to play with the first of these two expressions to see how it might be restructured in order to most directly confirm that it is satisfied by the expressions presented in [[SSC/Structure/OtherAnalyticModels#Examples|Table 1]].  Adopting the shorthand notation,
<div align="center">
<math>~\Gamma \equiv 4\pi G\rho_c \tau_\mathrm{SSC}^2</math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
<math>~\varpi \equiv \frac{\rho_0}{\rho_c} \, ,</math>
</div>
and multiplying the "equilibrium" relation through by <math>~(-\varpi p)</math>, we have,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<math>~\Gamma \varpi^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \varpi p^'\biggl\{
\frac{1}{\chi_0^2 p^'} \frac{d}{d\chi_0} (\chi_0^2 p^') - \frac{1}{\varpi}\frac{d\varpi}{d\chi_0}
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
p^' \varpi^' - \varpi \frac{dp^'}{d\chi_0}  -\frac{2\varpi p^'}{\chi_0}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{p^' }{\chi_0}\biggl[ \chi_0 \varpi^' - 2\varpi \biggr] - \varpi \frac{dp^'}{d\chi_0}  \, ;
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\Gamma \varpi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{p^' }{\chi_0 \varpi}\biggl[ \chi_0 \varpi^' - 2\varpi \biggr] - \frac{dp^'}{d\chi_0}  \, .
</math>
  </td>
</tr>
</table>
</div>

=====Specific Cases=====

<font color="red">'''Case 1''' (Parabolic)</font>:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi = 1 -\chi_0^2</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\varpi^' = -2\chi_0 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~p^' = -5\chi_0 + 8\chi_0^3 - 3\chi_0^5 = \chi_0(1-\chi_0^2)(-5+3\chi_0^2)</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{p^'}{\chi_0 \varpi} = -5 +3\chi_0^2  \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
Also, note: 
  </td>
  <td align="left">
<math>~\frac{d(p^')}{d\chi_0} = -5 +24\chi_0^2 -15\chi_0^4 \, .</math>
  </td>
</tr>
</table>
</div>

For the parabolic case, therefore, the RHS of the "equilibrium" expression is,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<font size="+1">RHS</font>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(-5+3\chi_0^2)\biggl[ -2\chi_0^2 - 2(1-\chi_0^2) \biggr] - (-5 +24\chi_0^2 -15\chi_0^4)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(10 - 6\chi_0^2) + (5 -24\chi_0^2 +15\chi_0^4)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~15(1-2\chi_0^2+\chi_0^4) \, ,
</math>
  </td>
</tr>
</table>
</div>
which, indeed, matches the LHS of the "equilibrium" relation, if,
<div align="center">
<math>~\Gamma = 15</math> 
&nbsp;  &nbsp; &nbsp; &nbsp;<math>~\Rightarrow</math>&nbsp;  &nbsp; &nbsp; &nbsp;
<math>~\tau_\mathrm{SSC}^2 = \frac{15}{4\pi G \rho_c} \, .</math>
</div>
This has all worked satisfactorily because, [[SSC/Structure/OtherAnalyticModels#Stabililty_2|as presented above]], this is the correct value of <math>~\tau_\mathrm{SSC}^2</math> in the case of the parabolic density distribution.


<font color="red">'''Case 2''' (Linear)</font>:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi = 1 -\chi_0</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\varpi^' = -1 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~p^' = \tfrac{12}{5}[- 4\chi_0 + 7\chi_0^2 - 3\chi_0^3]</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{p^'}{\chi_0 \varpi} = \tfrac{12}{5}(-4 +3\chi_0)  \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
Also, note: 
  </td>
  <td align="left">
<math>~\frac{d(p^')}{d\chi_0} = \tfrac{12}{5}[- 4 + 14\chi_0 - 9\chi_0^2] \, .</math>
  </td>
</tr>
</table>
</div>

For the linear case, therefore, the RHS of the "equilibrium" expression is,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<font size="+1">RHS</font>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\tfrac{12}{5}(-4 +3\chi_0)\biggl[ -\chi_0  - 2(1-\chi_0) \biggr] - \tfrac{12}{5}(- 4 + 14\chi_0 - 9\chi_0^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\tfrac{12}{5}\biggl[ (4 -3\chi_0)( 2-\chi_0 ) + (4 - 14\chi_0 + 9\chi_0^2) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\tfrac{12}{5}(12-24\chi_0 + 12\chi_0^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\tfrac{2^4\cdot 3^2}{5}(1-2\chi_0 + \chi_0^2) \, ,
</math>
  </td>
</tr>
</table>
</div>
which, indeed, matches the LHS of the "equilibrium" relation, if,
<div align="center">
<math>~\Gamma = \frac{2^4\cdot 3^2}{5}</math> 
&nbsp;  &nbsp; &nbsp; &nbsp;<math>~\Rightarrow</math>&nbsp;  &nbsp; &nbsp; &nbsp;
<math>~\tau_\mathrm{SSC}^2 = \frac{2^2\cdot 3^2}{5\pi G \rho_c} \, .</math>
</div>
This has all worked satisfactorily because, [[SSC/Structure/OtherAnalyticModels#Lagrangian_Approach|as presented above]], this is the correct value of <math>~\tau_\mathrm{SSC}^2</math> in the case of the linear density distribution.



<font color="red">'''Case 3''' (n = 1 polytrope)</font>:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi = \frac{\sin(\pi\chi_0)}{\pi\chi_0}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\varpi^' = \frac{\cos(\pi\chi_0)}{\chi_0} -  \frac{\sin(\pi\chi_0)}{\pi\chi_0^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~p^' = \frac{2\sin(\pi\chi_0)}{(\pi^2\chi_0^3)} \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr]</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{p^'}{\chi_0 \varpi} = \frac{2}{(\pi\chi_0^3)} \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr]  \, .</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="3">

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

<tr>
  <td align="right">
Also, note: &nbsp; &nbsp; &nbsp;<math>~\frac{d(p^')}{d\chi_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2\pi \cdot \cos(\pi\chi_0)}{(\pi^2\chi_0^3)} - \frac{6\sin(\pi\chi_0)}{(\pi^2\chi_0^4)}  \biggr] \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr] 
+\frac{2\sin(\pi\chi_0)}{(\pi^2\chi_0^3)} \biggl[ \pi\cos(\pi\chi_0) - \pi^2\chi_0 \sin(\pi\chi_0) - \pi \cos(\pi\chi_0) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\pi^2 \chi_0^4} \biggl\{
\biggl[ 2\pi \chi_0\cdot \cos(\pi\chi_0) - 6\sin(\pi\chi_0)  \biggr] \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr] 
+2 \chi_0 \sin(\pi\chi_0) \biggl[ - \pi^2\chi_0 \sin(\pi\chi_0) \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\pi^2 \chi_0^4} \biggl\{
2\pi^2\chi_0^2\cos^2(\pi\chi_0) -8\pi\chi_0 \sin(\pi\chi_0) \cos(\pi\chi_0)+6\sin^2(\pi\chi_0)
- 2 \pi^2 \chi_0^2 \sin^2(\pi\chi_0)  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{2}{\pi^2 \chi_0^4} \biggl\{3\sin^2(\pi\chi_0)
-4\pi\chi_0 \sin(\pi\chi_0) \cos(\pi\chi_0)+\pi^2\chi_0^2\biggl[1  - 2\sin^2(\pi\chi_0)  \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

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

For the case of an n = 1 polytropic configuration, therefore, the equilibrium requirement is,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<math>~\Gamma \varpi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{p^' }{\chi_0 \varpi}\biggl[ \chi_0 \varpi^' - 2\varpi \biggr] - \frac{dp^'}{d\chi_0}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{(\pi\chi_0^3)} \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr]\biggl[ \cos(\pi\chi_0) -  \frac{\sin(\pi\chi_0)}{\pi\chi_0} - \frac{2\sin(\pi\chi_0)}{\pi\chi_0} \biggr]   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{2}{\pi^2 \chi_0^4} \biggl\{3\sin^2(\pi\chi_0)
-4\pi\chi_0 \sin(\pi\chi_0) \cos(\pi\chi_0)+\pi^2\chi_0^2\biggl[1  - 2\sin^2(\pi\chi_0)  \biggr]
\biggr\}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{(\pi^2\chi_0^4)} \biggl\{ \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr]\biggl[\pi\chi_0 \cos(\pi\chi_0) -  3\sin(\pi\chi_0) \biggr]   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-3\sin^2(\pi\chi_0) + 4\pi\chi_0 \sin(\pi\chi_0) \cos(\pi\chi_0) - \pi^2\chi_0^2\biggl[1  - 2\sin^2(\pi\chi_0)  \biggr]
\biggr\}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{(\pi^2\chi_0^4)} \biggl\{3\sin^2(\pi\chi_0) - 4\pi\chi_0\sin(\pi\chi_0)\cos(\pi\chi_0) + (\pi\chi_0)^2 \biggl[1-\sin^2(\pi\chi_0) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-3\sin^2(\pi\chi_0) + 4\pi\chi_0 \sin(\pi\chi_0) \cos(\pi\chi_0) - \pi^2\chi_0^2\biggl[1  - 2\sin^2(\pi\chi_0)  \biggr]
\biggr\}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{(\pi^2\chi_0^4)} \biggl\{ (\pi\chi_0)^2 \sin^2(\pi\chi_0)  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi^2 \biggl[\frac{\sin(\pi\chi_0)}{\pi\chi_0}  \biggr]^2
</math>
  </td>
</tr>
</table>
</div>
So, the equilibrium condition is satisfied if,
<div align="center">
<math>~\Gamma = 2\pi^2</math> 
&nbsp;  &nbsp; &nbsp; &nbsp;<math>~\Rightarrow</math>&nbsp;  &nbsp; &nbsp; &nbsp;
<math>~\tau_\mathrm{SSC}^2 = \frac{2\pi^2}{4\pi G \rho_c} = \frac{\pi}{2G \rho_c}  \, .</math>
</div>
This has all worked satisfactorily because, [[SSC/Stability/Polytropes#Setup|as presented in a separate chapter discussion]], this is the correct value of <math>~\tau_\mathrm{SSC}^2</math> in the case of an n = 1 polytropic configuration.

====Dabbling with LAWE====
Now, let's experiment with the [[SSC/Structure/OtherAnalyticModels#Compare|LAWE as presented above]], that is,

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

<tr>
  <td align="right">
<math>~- \sigma^2 \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{x^'}{x} \biggr) \cdot \frac{d}{d\chi_0}\biggl[\ln\biggl( \chi_0^4 x^' \biggr) \biggr]
+ \frac{d\ln(p)}{d\chi_0}\cdot  \frac{d}{d\chi_0}\biggl[\ln\biggl(\chi_0^\alpha x\biggr)\biggr] \, .
</math>
  </td>
</tr>
</table>
</div> 

After multiplying though by <math>~(-p)</math>, this expression may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-\sigma^2 \varpi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{p}{x}\biggl[ \frac{dx^'}{d\chi_0} + \frac{4x^'}{\chi_0}\biggr]
+ \frac{\alpha p^'}{\chi_0}\biggl[1 + \frac{\chi_0 x^'}{\alpha x}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{p}{x}\biggr) x^{' '} + p\biggl[ \frac{4}{\chi_0}\biggr]\frac{x^'}{x}
+ p^'\biggr[\frac{x^'}{x}\biggr] + \frac{\alpha p^'}{\chi_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ -\biggl[\sigma^2 \varpi + \frac{\alpha p^'}{\chi_0}  \biggr]x</math>
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
px^{' '} + \biggl[ \frac{4p}{\chi_0} + p^'\biggr]x^' 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ 0</math>
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
px^{' '} + \biggl[ 4p + \chi_0 p^'\biggr]\frac{x^'}{\chi_0} + \biggl[\sigma^2 \varpi + \frac{\alpha p^'}{\chi_0}  \biggr]x \, .
</math>
  </td>
</tr>
</table>
</div>

(We could have, perhaps, obtained this expression in a more direct fashion had we started directly from the form of the [[SSC/Structure/OtherAnalyticModels#LAWE|LAWE derived earlier]].)

=====Specific Case Attempts=====
======Uniform Density======
<font color="red">'''Case 0''' (Uniform density)</font>:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi</math>
  </td>
  <td align="center">
<math>~=</math> 
  </td>
  <td align="left">
<math>~1 \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~p</math>
  </td>
  <td align="center">
<math>~=</math> 
  </td>
  <td align="left">
<math>~1-\chi_0^2 \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\alpha p^'}{\chi_0}</math>
  </td>
  <td align="center">
<math>~=</math> 
  </td>
  <td align="left">
<math>~- 2\alpha \, .</math>
  </td>
</tr>
</table>
</div>

For the uniform-density case, therefore, the the LAWE becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-\sigma^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(1-\chi_0^2)}{x}\biggl[ \frac{dx^'}{d\chi_0} + \frac{4x^'}{\chi_0}\biggr]
-2\alpha \biggl[1 + \frac{\chi_0 x^'}{\alpha x}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~ (2\alpha -\sigma^2)x </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-\chi_0^2)\biggl[ \frac{dx^'}{d\chi_0} + \frac{4x^'}{\chi_0}\biggr] - 2\chi_0 x^' 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-\chi_0^2)\frac{dx^'}{d\chi_0}  + (1-\chi_0^2)\biggl[ \frac{4x^'}{\chi_0}\biggr] - 2\chi_0 x^' 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-\chi_0^2)\frac{dx^'}{d\chi_0}  + \frac{2x^'}{\chi_0}\biggl( 2 - 3\chi_0^2 \biggr) \, ,
</math>
  </td>
</tr>
</table>
</div>
where, [[SSC/Structure/OtherAnalyticModels#Stabililty_2|as defined above]],
<div align="center">
<math>~\alpha \equiv 3 - \frac{4}{\gamma_g} \, .</math>
</div>

'''Mode 3'''

Try an eigenfunction of the form,
<div align="center">
<math>x = a + b\chi_0^2 + c\chi_0^4 \, ,</math>
</div>
in which case,
<div align="center">
<math>~\frac{2x^'}{\chi_0} = \frac{2}{\chi_0}(2 b\chi_0 +4c\chi_0^3) = 4b+8c\chi_0^2</math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
<math>~x^{' '} = 2 b + 12c\chi_0^2 \, . </math>
</div>
In order for this to be a solution, we must have,

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

<tr>
  <td align="right">
<math>~(2\alpha -\sigma^2)( a + b\chi_0^2 + c\chi_0^4) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-\chi_0^2)(2 b + 12c\chi_0^2 )  + ( 2 - 3\chi_0^2 )(4b+8c\chi_0^2 ) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2 b + 12c\chi_0^2)  - \chi_0^2(2 b + 12c\chi_0^2 )  + 2(4b+8c\chi_0^2 ) - 3\chi_0^2 (4b+8c\chi_0^2 )  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~10b + \chi_0^2(12c-2b+16c-12b) - \chi_0^4(12c + 24c)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~10b + \chi_0^2(28c-14b) - \chi_0^4(36c) \, .
</math>
  </td>
</tr>
</table>
</div>


So, the coefficients of each even power of <math>~\chi_0^n</math> are:
<div align="center" id="FirstTable">
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="right"><math>~\chi_0^0</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~a\mathfrak{F} +10b</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^2</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~b\mathfrak{F} -14b + 28c</math>
 </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^4</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~c[\mathfrak{F} -36]</math>
  </td>
</tr>
</table>
</div>
where, following [[SSC/Stability/UniformDensity#Setup_as_Presented_by_Sterne_.281937.29|Sterne's (1937) presentation]],
<div align="center">
<math>~\mathfrak{F}  \equiv \sigma^2 - 2 \alpha \, .</math>
</div>

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

First: &nbsp;&nbsp;&nbsp; <math>~\mathfrak{F} = 36 \, ;</math>

Second: &nbsp;&nbsp;&nbsp; <math>~22b = -28c ~~~~~\Rightarrow ~~~~~ c = - (11/14)b \, ;</math>

Third: &nbsp;&nbsp;&nbsp; <math>~36a = -10b ~~~~~\Rightarrow ~~~~~ b = -(18/5)a \, .</math>

Hence, choosing <math>~a = 1</math> implies:  <math>~b = -18/5</math> &nbsp; &nbsp; &nbsp; &nbsp;and  &nbsp; &nbsp; &nbsp; &nbsp; 
<math>~ c = (11/7)(9/5) = +99/35 \, .</math>  This precisely matches the "j = 2" mode identified by Sterne.


======Parabolic======

<font color="red">'''Case 1''' (Parabolic)</font>:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi = 1 -\chi_0^2</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\varpi^' = -2\chi_0 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~p = \tfrac{1}{2}(1 -\chi_0^2)^2 (2-\chi_0^2)</math>
  </td>
  <td align="center">
&nbsp;  
  </td>
  <td align="left">
&nbsp;  
  </td>
</tr>

<tr>
  <td align="right">
<math>~p^' = -5\chi_0 + 8\chi_0^3 - 3\chi_0^5 = \chi_0(1-\chi_0^2)(-5+3\chi_0^2)</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{\alpha p^'}{\chi_0} = \alpha (1-\chi_0^2)(-5+3\chi_0^2)  \, .</math>
  </td>
</tr>
</table>
</div>


For the parabolic case, therefore, the 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>~
px^{' '} + \biggl[ 4p + \chi_0 p^'\biggr]\frac{x^'}{\chi_0} + \biggl[\sigma^2 \varpi + \frac{\alpha p^'}{\chi_0}  \biggr]x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\tfrac{1}{2}\varpi^2 (2-\chi_0^2)x^{' '} 
+ \biggl[ 2\varpi^2 (2-\chi_0^2) + \chi_0^2\varpi(-5+3\chi_0^2)\biggr]\frac{x^'}{\chi_0} 
+ \biggl[\sigma^2 \varpi + \alpha \varpi(-5+3\chi_0^2)  \biggr]x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\varpi}{2}\biggl\{
\varpi (2-\chi_0^2)x^{' '} 
+ \biggl[ 4\varpi (2-\chi_0^2) + \chi_0^2(-10+6\chi_0^2)\biggr]\frac{x^'}{\chi_0} 
+ \biggl[\mathfrak{K}+6\alpha\chi_0^2  \biggr]x 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\varpi}{2}\biggl\{
(2-3\chi_0^2 + \chi_0^4)x^{' '} 
+ \biggl[ 8-22\chi_0^2 + 10\chi_0^4\biggr]\frac{x^'}{\chi_0} 
+ \biggl[\mathfrak{K}+6\alpha\chi_0^2  \biggr]x 
\biggr\}
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{K} \equiv 2(\sigma^2 - 5\alpha) \, .</math>
</div>

'''Mode Inverse'''

Try an eigenfunction of the form,
<div align="center">
<math>x = (1 + a\chi_0^2)^{-\beta} = (1 + a\chi_0^2)^{-(\beta+2)} (1 + 2a\chi_0^2 + a^2\chi_0^4)\, ,</math>
</div>
in which case,

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

<tr>
  <td align="right">
<math>~x^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2a\beta \chi_0 (1 + a\chi_0^2)^{-\beta-1} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2a\beta (\chi_0 + a\chi_0^3)(1 + a\chi_0^2)^{-(\beta+2)} \, ;
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~x^{' '}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2a\beta (1 + a\chi_0^2)^{-\beta-1} - 2a\beta \chi_0\biggl[ -2a(\beta+1) \chi_0(1 + a\chi_0^2)^{-\beta-2}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2a\beta (1 + a\chi_0^2)^{-(\beta+2)}\biggl[(1 + a\chi_0^2)  -2a(\beta+1) \chi_0^2\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2a\beta (1 + a\chi_0^2)^{-(\beta+2)}\biggl[1 - a(2\beta+1)\chi_0^2 \biggr]
</math>
  </td>
</tr>
</table>
</div>

In order for this to be a solution, we must 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>~
(2-3\chi_0^2 + \chi_0^4)x^{' '}
+ \biggl[ 8-22\chi_0^2 + 10\chi_0^4\biggr]\frac{x^'}{\chi_0} 
+ \biggl[\mathfrak{K}+6\alpha\chi_0^2  \biggr]x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2a\beta \biggl[1 - a(2\beta+1)\chi_0^2 \biggr](2-3\chi_0^2 + \chi_0^4)
- 2a\beta (1 + a\chi_0^2) \biggl[ 8-22\chi_0^2 + 10\chi_0^4\biggr] 
+ \biggl[\mathfrak{K}+6\alpha\chi_0^2  \biggr](1 + 2a\chi_0^2 + a^2\chi_0^4) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2a\beta (-2+3\chi_0^2 - \chi_0^4)
+ 2a^2\beta (2\beta+1) (2\chi_0^2-3\chi_0^4 + \chi_0^6)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2a\beta \biggl[ -8+22\chi_0^2 - 10\chi_0^4\biggr] 
+ 2a^2\beta  \biggl[ -8\chi_0^2+22\chi_0^4 - 10\chi_0^6\biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \mathfrak{K} (1 + 2a\chi_0^2 + a^2\chi_0^4) 
+ 6\alpha (\chi_0^2 + 2a\chi_0^4 + a^2\chi_0^6) 
</math>
  </td>
</tr>
</table>
</div>



So, the coefficients of each even power of <math>~\chi_0^n</math> are:
<div align="center" id="FirstTable">
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="right"><math>~\chi_0^0</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~\mathfrak{K} - 20a\beta </math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^2</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~50a\beta + 4a^2\beta (2\beta-3) + 6\alpha + 2a\mathfrak{K}</math>
 </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^4</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~- 22a\beta + 2a^2\beta (19-6\beta)  + 12a\alpha + a^2\mathfrak{K}</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^6</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~2a^2\beta (2\beta-9) + 6a^2\alpha</math>
  </td>
</tr>
</table>
</div>



'''Mode 3P'''

Try an eigenfunction of the form,
<div align="center">
<math>x = a + b\chi_0^2 + c\chi_0^4 \, ,</math>
</div>
in which case,
<div align="center">
<math>~\frac{x^'}{\chi_0} = \frac{1}{\chi_0}(2 b\chi_0 +4c\chi_0^3) = 2b+4c\chi_0^2</math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
<math>~x^{' '} = 2 b + 12c\chi_0^2 \, . </math>
</div>
In order for this to be a solution, we must 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{\varpi}{2}\biggl\{
(2-3\chi_0^2 + \chi_0^4)(2 b + 12c\chi_0^2 ) 
+ (8-22\chi_0^2 + 10\chi_0^4 )(2b+4c\chi_0^2 ) 
+ (\mathfrak{K}+6\alpha\chi_0^2 )( a + b\chi_0^2 + c\chi_0^4 ) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\varpi}{2}\biggl\{
(4b+24c\chi_0^2 - 6b\chi_0^2 -36c\chi_0^4 + 2b\chi_0^4 + 12c\chi_0^6) 
+ (16b + 32c\chi_0^2 - 44b\chi_0^2 - 88c\chi_0^4 + 20b\chi_0^4 + 40c\chi_0^6 ) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ (a\mathfrak{K} + b\mathfrak{K}\chi_0^2 + c\mathfrak{K}\chi_0^4 + 6a\alpha\chi_0^2 + 6b\alpha\chi_0^4 + 6c\alpha\chi_0^6 )
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\varpi}{2}\biggl[ (20b + a\mathfrak{K}) \chi_0^0 +(24c - 6b + 32c - 44b + b\mathfrak{K} + 6a\alpha)\chi_0^2
+ (-36c+2b -88c+20b +c\mathfrak{K} + 6b\alpha) \chi_0^4 + (12c + 40c + 6c\alpha )\chi_0^6 \biggr]
</math>
  </td>
</tr>
</table>
</div>


So, the coefficients of each even power of <math>~\chi_0^n</math> are:
<div align="center" id="FirstTable">
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="right"><math>~\chi_0^0</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~20b + a\mathfrak{K}</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^2</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~56c + (\mathfrak{K}- 50)b + 6a\alpha</math>
 </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^4</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~b(22+6\alpha) + c(\mathfrak{K}-124)</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^6</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~c(52 + 6\alpha)</math>
  </td>
</tr>
</table>
</div>


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

======Polytrope======

<font color="red">'''Case 3''' (n = 1 polytrope)</font>:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi = \frac{\sin(\pi\chi_0)}{\pi\chi_0}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\varpi^' = \frac{\cos(\pi\chi_0)}{\chi_0} -  \frac{\sin(\pi\chi_0)}{\pi\chi_0^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~p = \biggl[\frac{\sin(\pi\chi_0)}{\pi\chi_0}\biggr]^2 = \varpi^2</math>
  </td>
  <td align="center">
&nbsp;  
  </td>
  <td align="left">
&nbsp;  
  </td>
</tr>
<tr>
  <td align="right">
<math>~p^' = \frac{2\varpi}{(\pi \chi_0^2)} \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr]</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{\alpha p^'}{\chi_0 } = \frac{2\alpha \varpi}{(\pi \chi_0^3)} \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr] \, .</math>
  </td>
</tr>
</table>
</div>



For the n = 1 polytropic case, therefore, the 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>~
px^{' '} + \biggl[ 4p + \chi_0 p^'\biggr]\frac{x^'}{\chi_0} + \biggl[\sigma^2 \varpi + \frac{\alpha p^'}{\chi_0}  \biggr]x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{\sin(\pi\chi_0)}{\pi\chi_0}\biggr] \varpi x^{' '} 
+ \biggl\{ 4\biggl[\frac{\sin(\pi\chi_0)}{\pi\chi_0}\biggr] + \frac{2}{(\pi \chi_0)} \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr]\biggr\}\frac{\varpi x^'}{\chi_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{\sigma^2  + \frac{2\alpha }{(\pi \chi_0^3)} \biggl[ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) \biggr] \biggr\}\varpi x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
   </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{\varpi}{\pi \chi_0^2} \biggl\{
[\sin(\pi\chi_0)] \chi_0 x^{' '} + [ 2\sin(\pi\chi_0) + 2\pi\chi_0 \cos(\pi\chi_0) ] x^' 
+ \{(\pi \chi_0^2)\sigma^2  + 2\alpha \chi_0^{-1} [ \pi\chi_0 \cos(\pi\chi_0) - \sin(\pi\chi_0) ] \} x
\biggr\}
</math>
  </td>
</tr>
</table>
</div>