Editing SSC/StabilityConjecture/Bipolytrope51 (section)

===Trial Core Eigenvector===
Borrowing from our [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|analysis of the stability of pressure-truncated n = 5 polytropes]], for the marginally unstable model <math>(\sigma_c^2 = 0)</math> let's try a radial displacement function of the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x_\mathrm{trial}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
A + B\xi^2 \, .
</math>
  </td>
</tr>
</table>
Plugging this guess into the LAWE and noting that <math>\alpha_g = (3-10/3) = -1/3</math> (for the n = 5 core), we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(2B) 
+ 
(2B)\biggl[ 4 - 2 \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] 
+ 
\biggl\{ 
\biggl(\frac{\cancelto{0}{\sigma_c^2}}{\gamma_\mathrm{g}}\biggr) \biggl[ \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2} \biggr]
+\frac{2}{3} \biggl[ \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr]
\biggr\} \biggl[A + B\xi^2 \biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{3}\biggl(1 + \frac{1}{3}\xi^2 \biggr)^{-1}\biggl\{
10B \biggl(3 + \xi^2 \biggr)
- 
12 B\xi^2 
+ 
2 
\biggl[A + B\xi^2 \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2}{3}\biggl(1 + \frac{1}{3}\xi^2 \biggr)^{-1}\biggl\{
15B + A 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
In order for this RHS to be zero, we therefore need, <math>B = -A/15</math>.  Okay.  This matches our earlier derivation.  In order to compare this trial eigenfunction with the "core" portion of the eigenfunction that we obtained empirically via the B-KB74 conjecture, we need to know how <math>x_\mathrm{trial}</math> varies with <math>M_r^*</math>.

Well we know that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>m^* \equiv \biggl(\frac{\pi}{2\cdot 3}\biggr)^{1 / 2}M_r^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl( 1 + \frac{1}{3}\xi^2 \biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\xi^2}{(m^*)^{2/3}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 3 + \xi^2 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3\xi^2}{(m^*)^{2/3}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 3(m^*)^{2/3} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\xi^2 \biggl[3 - (m^*)^{2/3}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~  \xi^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{3(m^*)^{2/3}}{3 - (m^*)^{2/3}}\biggr] \, .
</math>
  </td>
</tr>
</table>
Hence, the trial eigenfunction may be written as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x_\mathrm{trial}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
A\biggl\{
1 - \frac{1}{15}\biggl[\frac{3(m^*)^{2/3}}{3 - (m^*)^{2/3}}\biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Now, from [[#Expressions_for_q_and_.CE.BD|above]], we know that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>m^*_\mathrm{core} \equiv \biggl(\frac{\pi}{2\cdot 3}\biggr)^{1 / 2}\biggl[ \frac{G^3 \rho_0^{2/5}}{K_c^3} \biggr]^{1 / 2} M_\mathrm{core}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{\pi}{2\cdot 3}\biggr)^{1 / 2}3^2\biggl(\frac{2}{\pi} \biggr)^{1 / 2}\ell_i^3(1 + \ell_i^2)^{-3 / 2} 
=
3^{3/2} \ell_i^3(1 + \ell_i^2)^{-3 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~
[m^*_\mathrm{core}]^{2 / 3} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3 \ell_i^2(1 + \ell_i^2)^{-1} \, ,
</math>
  </td>
</tr>
</table>
where, <math>\ell_i \equiv \xi_i/\sqrt{3}</math>.  For example, if <math>\xi_i = 9.014959755</math>, then <math>\ell_i = 5.204789446</math> and <math>[m^*_\mathrm{core}]^{2 / 3} = 2.893199794</math>.