Editing Appendix/Ramblings/51AnalyticStabilitySynopsis

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=More Focused Search for Analytic EigenVector of (5,1) Bipolytropes=

The ideas that are captured in this chapter have arisen after a review of [[Appendix/Ramblings/BiPolytrope51AnalyticStability|a previous hunt for the desired analytic eigenvector]] and as an extension of our [[SSC/Structure/BiPolytropes/Analytic51Renormalize|accompanying "renormalization" of the Analytic51 bipolytrope]].

==Review of Attempt 4B==

===Structure===
From a separate search that we labeled [[Appendix/Ramblings/BiPolytrope51AnalyticStability#Attempt_4B|Attempt 4B]], we draw the following information regarding the structure of the envelope.
<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_0 \biggl[ \frac{\sin(\eta - b_0)}{\eta} \biggr] \, ,</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{d\phi}{d\eta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{a_0}{\eta^2} \biggl[ \eta \cos(\eta - b_0) - \sin(\eta - b_0) \biggr] \, ,</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{d^2\phi}{d\eta^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{a_0}{\eta} \cdot \sin(\eta - b_0)
-
\frac{2a_0}{\eta^2} \cdot \cos(\eta - b_0) 
+
\frac{2a_0}{\eta^3} \cdot \sin(\eta - b_0) \, .
</math>
  </td>
</tr>
</table>
This satisfies the Lane-Emden equation for any values of the parameter pair, <math>~a_0</math> and <math>~b_0</math>.  Note that,

<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>\biggl[1 - \eta \cot(\eta - b_0) \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ \eta \cot(\eta - b_0) </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(1 - Q ) \, .</math>
  </td>
</tr>
</table>

===LAWE===

Now, guided by a [[SSC/Stability/n1PolytropeLAWE#Succinct_Demonstration|separate parallel discussion]] we also showed in [[Appendix/Ramblings/BiPolytrope51AnalyticStability#Attempt_4B|Attempt 4B]] that, in the case of a bipolytropic configuration for which <math>n_e=1</math>, the
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="4"><font color="maroon"><b>Trial Displacement Function</b></font></td>
</tr>
<tr>
  <td align="right">
<math>\sigma_c^2 = 0</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>x_P </math>
  </td>
  <td align="left">
<math>\equiv \frac{3c_0 (n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\eta \phi^{n}}\biggr) \frac{d\phi}{d\eta}\biggr] </math>
  </td>
</tr>
<tr>
  <td align="right" colspan="3">
&nbsp;
  </td>
  <td align="left">
<math>= -\biggl( \frac{3c_0}{\eta \phi}\biggr) \frac{d\phi}{d\eta} = \frac{3c_0}{\eta^2} \cdot Q \, , </math>
  </td>
</tr>
</table>

precisely satisfies the

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

<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Governing LAWE</b></font></td>
</tr>
<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x_P}{d\eta^2} + \biggl[4 - 2Q\biggr]\frac{1}{\eta}\cdot \frac{dx_P}{d\eta} - 2Q\cdot \frac{x_P}{\eta^2} \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="5" width="80%"><tr><td align="left">
Note for later use that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{d\ln x_P}{d\ln\eta} = \frac{\eta}{x_P} \cdot \frac{d}{d\eta}\biggl[ \frac{3c_0}{\eta^2} \cdot Q \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3c_0\eta \biggl[ \frac{\eta^2}{3c_0\cdot Q}  \biggr] \cdot \frac{d}{d\eta}\biggl[ \frac{Q}{\eta^2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\eta^3}{Q}  \biggr] \cdot \biggl[ \frac{1}{\eta^2} \frac{dQ}{d\eta} - \frac{2Q}{\eta^3}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{d\ln Q}{d\ln \eta} - 2\biggr] \, .
</math>
  </td>
</tr>
</table>

Note as well that,

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

<tr>
  <td align="right">
<math>Q </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[1 - \frac{\eta \cdot \cos(\eta - b_0)}{\sin(\eta - b_0)} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{dQ}{d\eta} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\biggl[\frac{\cos(\eta - b_0)}{\sin(\eta - b_0)} \biggr] 
+
\biggl[\frac{\eta \cdot \sin(\eta - b_0)}{\sin(\eta - b_0)} \biggr] 
+
\biggl[\frac{\eta \cdot \cos^2(\eta - b_0)}{\sin^2(\eta - b_0)} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\eta 
+
\eta \cot^2(\eta-b_0) 
- \cot(\eta-b_0) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{d\ln Q}{d\ln \eta} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
Q^{-1} \biggl[\eta^2 + \eta^2 \cot^2(\eta-b_0) - \eta\cot(\eta-b_0) \biggr]
</math>
  </td>
</tr>

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

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

While it is rather amazing that we have been able to identify this analytic solution to the LAWE, the solution seems troubling because it blows up at the surface, where, <math>\eta_s - b_0 = \pi</math>.  We will ignore this undesired behavior for the time being.

===Transition at Interface===

<table border="1" align="center" width="60%" cellpadding="5"><tr><td align="left">
Here, as a numerical example, we will adopt the parameters that are relevant to [[SSC/Structure/BiPolytropes/AnalyzeStepFunction#Model_Amodel2|Amodel2 from an associated discussion]].  For example, <math>(\mu_e/\mu_c) = 0.31</math> and <math>\xi_i = 9.0149598</math>.
</td></tr></table>

Under [[Appendix/Ramblings/BiPolytrope51AnalyticStability#Attempt_1|"Attempt 1" of our accompanying discussion]], we have shown that, at the core/envelope interface (note the following mappings: &nbsp; <math>b \rightarrow 3c_0</math> and <math>B \rightarrow b_0</math>),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\eta_i \cot(\eta_i - b_0)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left" colspan="3">
<math>1 - \biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{3\xi_i^2 }{3 + \xi_i^2}\biggr]  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ Q_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left" colspan="3">
<math>\biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{3\xi_i^2 }{3 + \xi_i^2}\biggr] = 0.8968919  \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\eta_i </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left" colspan="3">
<math>3^{1 / 2} \biggl(\frac{\mu_e}{\mu_c}\biggr) \xi_i \biggl[1 + \frac{\xi^2}{3} \biggr]^{-1} = 0.1723205</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left" colspan="3">
<math>
3^{1 / 2} \biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{3\xi_i}{3 + \xi^2} \biggr] 
=
\frac{3^{1 / 2}Q_i}{\xi_i} 
\, ;
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>3c_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x_i \eta_i^2 \biggl[1 - \eta_i \cot(\eta_i - b_0) \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{5}\biggl(\frac{\mu_e}{\mu_c}\biggr)  \biggl[\frac{15-\xi_i^2}{3+\xi_i^2}\biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>b_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\eta_i - \frac{\pi}{2} + \tan^{-1}\biggl[\frac{1}{\eta_i} - \frac{\xi_i}{\sqrt{3}}  \biggr] = -0.8592701 \, .
</math>
  </td>
</tr>
</table>

As viewed from the perspective of the envelope, then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[ \frac{d\ln x_P}{d\ln\eta} \biggr]_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{d\ln Q}{d\ln \eta}\biggr]_i - 2 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-0.0700000 - 2 = -2.0700000 \, . 
</math>
  </td>
</tr>
</table>

As viewed from the perspective of the core, we have instead,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[ \frac{d\ln x_P}{d\ln\eta} \biggr]_i 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3 \biggl(\frac{\gamma_c}{\gamma_e} - 1\biggr)
+ \frac{\gamma_c}{\gamma_e}\biggl[ \frac{d\ln x}{d\ln\xi} \biggr]_i
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp; 
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3 \biggl(\frac{3}{5} - 1\biggr)
- \frac{3}{5}\biggl[ \frac{2\xi_i^2}{15-\xi_i^2} \biggr]_i
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp; 
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{5}\biggl\{\biggl[ \frac{2\xi_i^2}{\xi_i^2-15} \biggr]_i - 2 \biggr\} = + 0.2716182 \, .
</math>
  </td>
</tr>
</table>

===Playing Around===

Evidently, for our chosen example "Amodel2", <math>d\ln Q/d\ln\eta = - 7/100</math> exactly.  How can this be?

=See Also=

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