Editing SSC/Stability/n5PolytropeLAWE (section)

==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.