Editing SSC/Stability/n1PolytropeLAWE/Pt2 (section)

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

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

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

</table>
</div>

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

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

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

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

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

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

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

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \gamma_g</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\infty \, . </math>
  </td>
</tr>
</table>
</div>
If these two &#8212; rather strange &#8212; conditions are met, then we have a marginally unstable configuration because, <math>~\sigma^2 = 0</math>.  This, in and of itself, is not very physically interesting.  However, it may give us a clue regarding how to more generally search for a physically reasonable radial eigenfunction.