Editing SSC/Stability/n1PolytropeLAWE/Pt3 (section)

====Prior to the Brute-Force Trial Fit====

Let's work through the analytic derivatives again.  Keeping in mind that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\eta - C) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\eta - C)\biggr] \, ,
</math>
  </td>
</tr>
</table>
and starting with the ''guess'',

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

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-C) \biggr] \, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right">
<math>~ \biggl( \frac{\eta^3}{b} \biggr) \frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C) \, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\eta^4}{b}\cdot \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl[ 3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-C)  
-  \eta^2\cot^2(\eta - C) 
-  \eta^3\cot^3(\eta-C) 
\biggr]  \, .
</math>
  </td>
</tr>
</table>


<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
Note that the relevant logarithmic derivative is,

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

<tr>
  <td align="right">
<math>~ \frac{d\ln x_P}{d\ln\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 \biggl( \frac{b}{\eta^2} \biggr)\biggl[ \eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C) \biggr]x_P^{-1}
</math>
  </td>
</tr>

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

If we know the logarithmic slope and the value of <math>~\eta</math> at the interface, then we can solve for 
<div align="center">
<math>~y_i \equiv  \eta_i \cot(\eta_i-C) \, ,</math>
</div>
via the quadratic relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1- y_i ) \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta_i^2 -2 + y_i  +  y_i^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta_i^2 -2 + y_i  +  y_i^2
- (1- y_i ) \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 y_i^2 
+ y_i \biggl\{1 + \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i\biggr\}
+\biggl\{  \eta_i^2 -2 - \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i \biggr\} \, .
</math>
  </td>
</tr>
</table>
(In practice it appears as though the "plus" solution to this quadratic equation is desired if the quantity inside the last set of curly braces is positive; and the "minus" solution is desired if this quantity is negative.)  Once the value of <math>~y_i</math> is known, we can solve for the key coefficient, <math>~C</math>, via the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tan(\eta_i - C)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{\eta_i}{y_i}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~C</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta_i - \tan^{-1}\biggl(\frac{\eta_i}{y_i}\biggr)\, .</math>
  </td>
</tr>
</table>

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


Recalling that,
<div align="center">
<math>~Q = \biggl[1- \eta\cot(\eta-B) \biggr] \, ,</math>
</div>
plugging these expressions into the relevant envelope LAWE gives,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  2 Q \cdot \frac{x}{\eta^2} 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-C)  
-  \eta^2\cot^2(\eta - C) 
-  \eta^3\cot^3(\eta-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \biggl[\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C)\biggr]
~-~  \biggl[1- \eta\cot(\eta-B) \biggr] \biggl[1- \eta\cot(\eta-C) \biggr] 
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \eta\cot(\eta-B) \biggl[\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C)\biggr]
~+~\eta\cot(\eta-B) \biggl[1- \eta\cot(\eta-C) \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
(\eta - \eta^3)\cot(\eta-C)  
-  \eta^3\cot^3(\eta-C) 
+ \eta\cot(\eta-B) \biggl[\eta^2 -1   +  \eta^2\cot^2(\eta - C)  \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
(\eta - \eta^3) [ \cot(\eta-C)  - \cot(\eta-B) ]
+ \eta^3 \cot^2(\eta - C) [\cot(\eta-B)-  \cot(\eta-C)] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl[ \cot(\eta-C)  - \cot(\eta-B) \biggr] 
\biggl[ \eta - \eta^3 - \eta^3 \cot^2(\eta - C) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^3} \biggl[ \cot(\eta-C)  - \cot(\eta-B) \biggr] 
\biggl\{ 1 - \eta^2\biggl[1 + \cot^2(\eta - C)\biggr] \biggr\}\, .
</math>
  </td>
</tr>
</table>

This will go to zero if <math>~C = (B-2m\pi), </math> where <math>~m</math> is a positive integer.  When <math>~m =1</math>, for example,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cot(\eta-C)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cot[\eta - (B-2\pi)] = \cot(\eta -B) \, . </math>
  </td>
</tr>
</table>


Okay.  Now let's determine at what value of <math>~\eta</math> the logarithmic derivative of <math>~x_P</math> goes to negative one.