Editing Appendix/Ramblings/BiPolytrope51AnalyticStability (section)

=====New Strategy=====
In the vast majority of our prior attempts to derive an analytic expression for the envelope's eigenvector, we have started with the presumption &#8212; as voiced by [http://adsabs.harvard.edu/abs/1988Ap%26SS.147..219B Beech (1988)] and repeated in [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|our accompanying derivation of the structure of the <math>~(n_c, n_e) = (5, 1)</math> bipolytrope's structure]] &#8212; that the most general solution to the n = 1 Lane-Emden equation can be written in the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\phi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, .</math>
  </td>
</tr>
</table>
But, as we have [[SSC/Structure/Polytropes#Primary_E-Type_Solution|emphasized in a separate context]], another expression that satisfies the relevant Lane-Emden equation has the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\phi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- A \biggl[ \frac{\cos(\eta - B)}{\eta} \biggr] \, .</math>
  </td>
</tr>
</table>

<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
We fully appreciate that for appropriately chosen and ''different'' values of the parameter, <math>~B</math>, these two functions can be made equal to one another.  But, for now, let's work through our analysis pretending that they are different functions.
</td></tr></table>

This alternate "cosine" expression was not the solution of choice when we were seeking a mathematical description of the structure of an ''isolated'', n = 1 polytrope because it does not satisfy the relevant central boundary conditions.  However, it occurs to us  that this alternate expression might work in the context we are considering now <font color="red">[20 April 2019]</font>, which deals with properties of the envelope of a bipolytrope.  In this case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q \equiv - \frac{d\ln\phi}{d\ln\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A \biggl[\frac{\cos(\eta-B)}{\eta^2} + \frac{\sin(\eta-B)}{\eta}  \biggr]\cdot \frac{\eta^2}{A\cos(\eta-B)}</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \eta\tan(\eta-B)  \, ;</math>
  </td>
</tr>
</table>
and, following along the lines of our earlier '''[[#Attempt_4B|Attempt 4B]]''' discussion, a reasonable ''guess'' for the dimensionless displacement function is,
<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{3c_1 Q}{\eta^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl[1 + \eta\tan(\eta-B)\biggr] \, .</math>
  </td>
</tr>
</table>

What are the first and second derivatives of this trial eigenfunction?

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

<tr>
  <td align="right">
<math>~\frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^2}\biggl[\tan(\eta - B) + \frac{\eta}{\cos^2(\eta-B)}\biggr]
- \frac{6c_1}{\eta^3} \biggl[1 + \eta\tan(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^3} \biggl[\frac{\eta^2}{\cos^2(\eta-B)} -2 - \eta\tan(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
3c_1 \frac{d}{d\eta}\biggl[\eta^{-3}\cos^{-2}(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3c_1}{\eta^3\cos^2(\eta-B)}\cdot \frac{d}{d\eta}\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
3c_1 \biggl[2\eta^{-3}\sin(\eta-B)\cos^{-3}(\eta-B) -3\eta^{-4}\cos^{-2}(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[2\eta + 4\sin(\eta-B)\cos(\eta-B) 
- \sin(\eta-B)\cos(\eta-B)- \eta \cos^2(\eta-B)+ \eta\sin^2(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} 
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
\cdot \biggl[2\eta\sin(\eta-B) -3\cos(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[2\eta^2\cos(\eta-B) + 4\eta \sin(\eta-B)\cos^2(\eta-B) 
- \eta \sin(\eta-B)\cos^2(\eta-B)- \eta^2 \cos^3(\eta-B)+ \eta^2\sin^2(\eta-B)\cos(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl\{
2\eta^3 \sin(\eta-B) - 4\eta\sin(\eta-B)\cos^2(\eta-B) - 2\eta^2 \sin^2(\eta-B)\cos(\eta-B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-3\eta^2 \cos(\eta-B) + 6\cos^3(\eta-B) +3 \eta\sin(\eta-B)\cos^2(\eta-B)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl\{6\cos^3(\eta-B) 
+ \eta^3 \biggl[2\sin(\eta-B)\biggr] 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[ 6\cos^3(\eta-B) 
+ 2\eta \sin(\eta-B)\cos^2(\eta-B)
- 2 \eta^2 \cos(\eta-B) + 2\eta^3 \sin(\eta-B)
\biggr] \, .
</math>
  </td>
</tr>
</table>