Editing SSC/Stability/n3PolytropeLAWE (section)

==Play With Form of LAWE==
===Logarithmic Derivative Rewrite===

We have noticed that the LAWE that governs the eigenfunction associated with the fundamental mode of the marginally unstable model (FMMUM),
<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[ 4 - (n+1) Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}  
- (3-n)  Q \frac{x}{\xi^2} \, ,
</math>
  </td>
</tr>
</table>
</div>

may be rewritten entirely as an expression that relates the logarithmic derivatives of <math>~x, \xi,</math> and <math>~\theta</math>.  Multiplying through by <math>~\xi^2/x</math>, then drawing on a differential relation that has been [[SSC/Stability/BiPolytrope00Details#Idea_Involving_Logarithmic_Derivatives|derived in a separate context]], namely,
<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[  \frac{d\ln x}{d\ln \xi} -1 \biggr]\cdot \frac{d\ln x}{d\ln \xi} \, ,
</math>
 </td>
</tr>
</table>
</div>
this LAWE associated with the FMMUM 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}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]
+ \biggl[  \frac{d\ln x}{d\ln \xi} -1 \biggr]\cdot \frac{d\ln x}{d\ln \xi} + \biggl[ 4 - (n+1) Q \biggr] \frac{d\ln x}{d\ln\xi}  - (3-n)  Q 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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]
+ \frac{d\ln x}{d\ln \xi} \cdot \biggl[  \frac{d\ln x}{d\ln \xi} + 3 - (n+1) Q \biggr]   - (3-n)  Q 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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]
+ \frac{d\ln x}{d\ln \xi} \cdot \biggl\{  \frac{d\ln [x \xi^3 \theta^{(n+1)}] }{d\ln \xi} \biggr\}   + \frac{d\ln \theta^{(3-n) }}{d\ln\xi} \, .
</math>
  </td>
</tr>
</table>
</div>

I'm not sure if anyone else has previously appreciated that the "fundamental mode" polytropic LAWE can be written in this form.  I'm even less sure that this form sheds light on its solution.

Play a little more &hellip; &nbsp; Start by letting, <math>~A \equiv  [x \xi^3 \theta^{(n+1)}] \, ,</math> in which case 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}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]
+ \frac{d\ln x}{d\ln \xi} \cdot \biggl\{  \frac{1}{A} \cdot \frac{dA}{d\ln \xi} \biggr\}   + \frac{d\ln \theta^{(3-n) }}{d\ln\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{A} \cdot \frac{d}{d\ln\xi} \biggl[ A \cdot \frac{d\ln x}{d\ln \xi} \biggr]  + \frac{d\ln \theta^{(3-n) }}{d\ln\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{d}{d\xi} \biggl[ A \cdot \frac{d\ln x}{d\ln \xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{A}{\theta^{(3-n)}} \cdot \frac{d\theta^{(3-n) }}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{d}{d\xi} \biggl[ x \xi^3 \theta^{(n+1)} \cdot \frac{d\ln x}{d\ln \xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{x \xi^3 \theta^{(n+1)}}{\theta^{(3-n)}} \cdot \frac{d\theta^{(3-n) }}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{d}{d\xi} \biggl[ \xi^4 \theta^{(n+1)} \cdot \frac{dx}{d\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- x \xi^3 \theta^{2(n-1)} \cdot \frac{d\theta^{(3-n) }}{d\xi} \, .
</math>
  </td>
</tr>
</table>
</div>

===One Feeble Guess===

Now, what if, <math>~x \equiv [\xi^{-2}\theta^{-n}]</math> &nbsp; ?

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\xi^2 \theta^{(n+1)} \cdot \frac{d(\xi^{-2}\theta^{-n})}{d\xi} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d\theta}{d\xi} - \frac{1}{\xi^{2}\theta^{n}} \cdot \frac{d(\xi^2 \theta^{(n+1)} )}{d\xi} \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case, the LAWE becomes,


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

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

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~
-\xi^2\theta^n - \frac{d}{d\xi} \biggl[\frac{1}{\theta^{n}} \cdot \frac{d(\xi^2 \theta^{(n+1)} )}{d\xi}\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \xi  (3-n)\frac{d\theta }{d\xi} 
</math>
  </td>
</tr>
</table>
</div>

===Behavior for Known n=5 Solution===
We know that the [[SSC/Stability/n5PolytropeLAWE#Eureka_Moment|FMMUM for pressure-truncated, n = 5 polytropic configurations]] takes the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1- f_n(\xi) \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~f_5(\xi) = \frac{\xi^2}{15} \, .</math>
</div>
The governing LAWE therefore gives,

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- (3-n)x \xi^3 \theta^{n} \cdot \frac{d\theta }{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
(3-n)(1-f_n) \xi^3 \theta^{n} \cdot \frac{d\theta }{d\xi} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\xi} \biggl[ \xi^4 \theta^{(n+1)} \cdot \frac{df_n}{d\xi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^4 \theta^{(n+1)} \cdot\frac{d^2f_n}{d\xi^2} 
+ \frac{df_n}{d\xi} \cdot \frac{d}{d\xi} \biggl[ \xi^4 \theta^{(n+1)} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^4 \theta^{(n+1)} \cdot\frac{d^2f_n}{d\xi^2} 
+ \frac{df_n}{d\xi} \cdot \biggl\{ 
4\xi^3 \theta^{(n+1)} + (n+1)\xi^4 \theta^n \frac{d\theta}{d\xi} 
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{1}{\theta} \frac{d\theta }{d\xi} \biggl[ (3-n)(1-f_n) - (n+1)\xi \frac{df_n}{d\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi \cdot\frac{d^2f_n}{d\xi^2} 
+ 4 \cdot \frac{df_n}{d\xi} \, .
</math>
  </td>
</tr>
</table>
</div>
Let's check to see whether the known <math>~f_5(\xi)</math> function properly satisfies this last ODE when n = 5.

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

<tr>
  <td align="right">
<math>~
-\xi (3+\xi^2)^{-1}  \biggl[ -2 \biggl(1-\frac{\xi^2}{15} \biggr) - \frac{12\xi^2}{15} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2\xi}{15} + \frac{8\xi}{15} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
2 + \frac{2\xi^2}{3} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{3} (3+\xi^2) \, ,
</math>&nbsp; &nbsp; &nbsp; Yes!
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\theta_5^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{3+\xi^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\xi^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl[ \frac{1}{\theta_5^2} - 1 \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
we could ''presume'' that, when defined in terms of <math>~\theta_5</math>, the defining function,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_5(\theta_5) = \frac{1}{n}\biggl[ \frac{1}{\theta_5^2} - 1 \biggr]</math>
  </td>
  <td align="center">
&nbsp; &nbsp;<math>~\Rightarrow~</math>&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~ 
\frac{df_5}{d\xi} = - \frac{2}{n\theta_5^3} \cdot \frac{d\theta_5}{d\xi}
</math>
  </td>
  <td align="center">
&nbsp; &nbsp;<math>~\Rightarrow~</math>&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~ 
\frac{d^2f_5}{d\xi^2} = - \frac{2}{n\theta_5^3} \cdot \frac{d^2\theta_5}{d\xi^2} + \frac{6}{n\theta_5^4} \cdot \biggl( \frac{d\theta_5}{d\xi} \biggr)^2 \, .
</math>
  </td>
</tr>
</table>
</div>
In this case, the governing 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>~
\xi \cdot\frac{d^2f_n}{d\xi^2} 
- \frac{(3-n)}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ 1-f_n \biggr]
+ \biggl[ 4 + \frac{(n+1)\xi}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggr] \biggl[ \frac{df_n}{d\xi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi \biggl\{ - \frac{2}{n\theta_5^3} \cdot \frac{d^2\theta_5}{d\xi^2} + \frac{6}{n\theta_5^4} \cdot \biggl( \frac{d\theta_5}{d\xi} \biggr)^2\biggr\}
- \biggl[ 4 + \frac{(n+1)\xi}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggr] \biggl[ \frac{2}{n\theta_5^3} \cdot \frac{d\theta_5}{d\xi}\biggr]
- \frac{(3-n)}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ \frac{(n + 1)\theta_5^2 -1  }{n\theta_5^2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{n\theta_5^3}\biggl\{
\biggl[ 2\xi \cdot \frac{d^2\theta_5}{d\xi^2} - \frac{6\xi}{\theta_5} \cdot \biggl( \frac{d\theta_5}{d\xi} \biggr)^2\biggr]
+ 2\biggl[ 4 + \frac{(n+1)\xi}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggr]  \frac{d\theta_5}{d\xi}
+ (3-n) \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ (n + 1)\theta_5^2 -1 \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{n\theta_5^3}\biggl\{
2\xi \cdot \frac{d^2\theta_5}{d\xi^2} 
+ (n-2) \frac{2\xi}{\theta_5}\biggl( \frac{d\theta_5}{d\xi} \biggr)^2
+ \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ (3-n) (n + 1)\theta_5^2 + 5+ n  \biggr]   
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
Now, from the polytropic Lane-Emden equation, we also know that,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>
That is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2\xi \biggl(\frac{d\theta_5}{d\xi}\biggr) + \xi^2 \cdot \frac{d^2\theta_5}{d\xi^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \xi^2 \theta_5^n</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~  \frac{d^2\theta_5}{d\xi^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \theta_5^n - \frac{2}{\xi} \biggl(\frac{d\theta_5}{d\xi}\biggr)</math>
  </td>
</tr>
</table>
</div>
So, the LAWE becomes,

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

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(n-2) \frac{2\xi}{\theta_5}\biggl( \frac{d\theta_5}{d\xi} \biggr)^2
+ (n+1)\biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ (3-n) \theta_5^2 + 1 \biggr]   
- 2\xi \theta_5^n \, .
</math>
  </td>
</tr>
</table>
</div>
Again, let's check to see if the case of n=5 works &hellip;

<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>~
\biggl( \frac{d\theta_5}{d\xi} \biggr)^2
+ \frac{\theta_5}{\xi} \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ 1 - 2 \theta_5^2 \biggr]   
- \frac{\theta_5^{6}}{3} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3+\xi^2)^{-3}\biggl\{
3\xi^2 -3\biggl[ (3+\xi^2)  -6 \biggr]  - 3^2 \biggr\} 
</math> &nbsp; &nbsp; Yes!
  </td>
</tr>
</table>
</div>