Editing SSC/Stability/n5PolytropeLAWE (section)

==Eureka Moment==

<font color="red">Note from J. E. Tohline on 3/6/2017:</font>&nbsp;  Yesterday evening, after I finished putting together the [[#n5TruncatedMovie|above animation sequence]] using an Excel workbook, I noticed that the eigenfunction of the fundamental mode for the marginally unstable model <math>~(\sigma_c^2 = 0)</math> resembles a parabola.  In an effort to see how well a parabola fits at least the central portion of this eigenfunction, I returned to my Excel spreedsheet and, in a brute-force manner, began to search for the pair of coefficients that would provide a best fit.  What I discovered was that a parabola with the following formula fits perfectly!
<div align="center" i="AnalyticSoln">
<table border="1" cellpadding="10" align="center">
<tr>
  <th align="center" colspan="1">Fundamental Mode Eigenfunction <br />
when <math>~\sigma_c^2 = 0</math> and <math>~\gamma = 6/5 ~\Rightarrow~\alpha=- 1/3</math></th>
</tr>

<tr>
  <td align="center">
<math>~x = x_0 \biggl[ 1 - \frac{\xi^2}{15} \biggr]</math>
  </td>
</tr>
</table>
</div>
For the ''specific'' normalization used in the above animation sequence, <math>~x_0 = \tfrac{5}{2}</math>.  Let's demonstrate that this eigenvector provides a solution to the LAWE for <math>~n=5</math> polytropes; for simplicity, we will set <math>~x_0 = 1</math>:
<div align="center" id="Proof">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi} = -\frac{2\xi}{15} \, ;</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left"> 
<math>~\frac{d^2x}{d\xi^2} = -\frac{2}{15} \, .</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
\biggl[ \cancelto{0}{\frac{\sigma_c^2}{3^{1/2} \gamma_g }} \cdot (3+\xi^2)^{3/2} + 2\biggr]  x 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~
-\frac{2}{15}(3+\xi^2)  -\frac{2}{15} \biggl[12 - 2\xi^2 \biggr]  + 2\biggl[1 - \frac{\xi^2}{15}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~
\biggl( - \frac{6}{15} - \frac{24}{15} + 2\biggr)
+\xi^2 \biggl( -\frac{2}{15} + \frac{4}{15} - \frac{2}{15} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~
0 \, .
</math>
  </td>
</tr>
</table>
</div>

Q. E. D.  &nbsp;I don't think that anyone has previously appreciated that the LAWE in this case admits to an analytic eigenvector solution. 

Now, let's see how the boundary condition comes into play.  We see that the logarithmic derivative of the parabolic eigenfunction is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\ln x}{d\ln \xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{2\xi^2}{15} \biggl[ 1 - \frac{\xi^2}{15}\biggr]^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{2\xi^2}{(15-\xi^2)} \, .</math>
  </td>
</tr>
</table>
</div>
We desire a surface boundary condition that gives, <math>~d\ln x/d\ln\xi = -3</math>.  This will only happen when,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~- \frac{2\xi^2}{(15-\xi^2)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-3</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~2\xi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3(15 - \xi^2)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3 \, .</math>
  </td>
</tr>
</table>
</div>
Hence, although the parabolic eigenfunction provides an accurate solution to the <math>~n=5</math> LAWE throughout the entire configuration &#8212; that is, for all <math>~\xi</math> &#8212; the desired surface boundary condition will only be satisfied if the polytrope is truncated at <math>~\xi_\mathrm{surf} = 3</math>.  The parabolic eigenfunction is therefore only physically relevant to the model that sits at the point along the equilibrium sequence that is associated with the <math>~P_\mathrm{max}</math> turning point.

<!--  TRY MORE GENERIC CASE ...
Finally, let's see how this plays out for arbitrary values of <math>~\alpha</math>.   Guess a more general eigenfunction of the form,
<div align="center" i="AnalyticSoln">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center">
<math>~x = 1 - \frac{\xi^2}{A} </math>
  </td>
</tr>
</table>
</div>
In this more general case, we have,

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

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi} = -\frac{2\xi}{A} \, ;</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left"> 
<math>~\frac{d^2x}{d\xi^2} = -\frac{2}{A} \, .</math>
  </td>
</tr>
</table>
</div>
The LAWE then gives,

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

<tr>
  <td align="right">
<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[ \cancelto{0}{\frac{\sigma_c^2}{3^{1/2} \gamma_g }} \cdot (3+\xi^2)^{3/2} -6\alpha \biggr]  x 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~
-\frac{2}{A}(3+\xi^2)  -\frac{2}{A} \biggl[12 - 2\xi^2 \biggr]  -6\alpha \biggl[1 - \frac{\xi^2}{A}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~-\frac{1}{A}\biggl\{ 
2(3+\xi^2)  + 2 \biggl[12 - 2\xi^2 \biggr]  + 6\alpha \biggl[A - \xi^2\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left"> 
<math>~-
\frac{1}{A}\biggl[  \biggl(6 + 24 + 6\alpha A \biggr) + \biggl(2 -4-6\alpha \biggr)\xi^2\biggr]
</math>
  </td>
</tr>
</table>
</div>
Hence, a parabolic eigenfunction does not work for arbitrary <math>~\alpha</math>.
END MORE GENERIC CASE -->

Let's express the parabolic displacement function, <math>~x</math>, as a function of the Lagrangian mass coordinate, instead of as a function of <math>~\xi</math>.  Drawing upon our [[Appendix/Ramblings/NonlinarOscillation#Exploration|accompanying discussion]] where we have used <math>~\tilde\xi</math> to denote the truncation edge, we know that,

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

<tr>
  <td align="right">
<math>~r_\xi(\xi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi \biggl\{ 
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
and that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
r_\xi (m_\xi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde{r}_\mathrm{edge}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
</math>
  </td>
</tr>
</table>
</div>
<span id="DefineTildeC">where,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tilde{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr) \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\tilde{r}_\mathrm{edge}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
\, .
</math>
  </td>
</tr>
</table>
</div>

By equating <math>~r_\xi(\xi)</math> with <math>~r_\xi(m_\xi)</math>, we find,

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

<tr>
  <td align="right">
<math>~\xi \biggl\{ 
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \xi 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
This means that,
<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>~
x_0 \biggl\{1 - 
\frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr] 
\biggr\} \, ;
</math>
  </td>
</tr>
</table>
</div>
and, specifically for the critical case of <math>~\tilde\xi = 3</math>, in which case, <math>~\tilde{C} = 4</math>,
<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>~
x_0 \biggl\{1 - 
\frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{4 - 3 m_\xi^{2/3}}\biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>


{{ SGFworkInProgress }}