Difference between revisions of "User:Tohline/SSC/Stability/BiPolytrope0 0"

From VistrailsWiki
Jump to navigation Jump to search
(Replaced content with '__FORCETOC__ <!-- __NOTOC__ will force TOC off --> =Radial Oscillations of a Zero-Zero Bipolytrope= This is a chapter that summarizes [[User:Tohline/SSC/Stability/BiPolytrop…')
Line 2: Line 2:
<!-- __NOTOC__ will force TOC off -->
<!-- __NOTOC__ will force TOC off -->
=Radial Oscillations of a Zero-Zero Bipolytrope=
=Radial Oscillations of a Zero-Zero Bipolytrope=
This is a chapter that summarizes [[User:Tohline/SSC/Stability/BiPolytrope0_0Details|an accompanying, detailed derivation]].
{{LSU_HBook_header}}
{{LSU_HBook_header}}


==Groundwork==
==Background==


In an [[User:Tohline/SSC/Perturbations#2ndOrderODE|accompanying discussion]], we derived the so-called,


<div align="center" id="2ndOrderODE">
<font color="#770000">'''Adiabatic Wave''' (or ''Radial Pulsation'') '''Equation'''</font><br />
{{User:Tohline/Math/EQ_RadialPulsation01}}
</div>
<!--
<div align="center" id="2ndOrderODE">
<font color="#770000">'''Adiabatic Wave Equation'''</font><br />
<math>
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x = 0 \, ,
</math>
</div>
-->
whose solution gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations.  According to our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|accompanying derivation]], if the initial, unperturbed equilibrium configuration is an <math>~(n_c, n_e) = (0,0)</math> bipolytrope, then we know that the relevant functional profiles are as follows for the core and envelope, separately.  Note that, throughout, we will preferentially adopt as the dimensionless radial coordinate, the parameter,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\xi</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{r}{r_i} \, ,</math>
  </td>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \chi_i \xi = q \biggl( \frac{G\rho_c^2 R^2}{P_c} \biggr)^{1 /2 }\xi  \, .</math>
  </td>
</tr>
</table>
</div>
The corresponding radial coordinate range is,
<div align="center">
<math>~0 \le \xi \le 1 </math>&nbsp; &nbsp; &nbsp; for the core, and<br /><br />
<math>~1 \le \xi \le \frac{1}{q} </math>&nbsp; &nbsp; &nbsp; for the envelope.
</div>
===Core===
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{P_c}{G\rho_c^2}\biggr)^{1 / 2} \chi
=
(qR) \xi
\, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\rho_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\rho_c \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{P_0}{P_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - \frac{2\pi}{3} \chi^2
=
1 - \frac{2\pi}{3} \biggl[ \frac{G\rho_c^2 R^2}{P_c} \biggr] q^2 \xi^2
=
1 - \frac{\xi^2}{g^2}
\, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~M_r</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \biggl( \frac{P_c^3}{G^3 \rho_c^4} \biggr)^{1 / 2}\chi^3
=
\frac{4\pi}{3} \biggl( \frac{P_c^3}{G^3 \rho_c^4} \biggr)^{1 / 2} \biggl( \frac{G\rho_c^2 R^2}{P_c} \biggr)^{3 /2 } (q\xi)^3
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4\pi}{3} ( \rho_c R^3 ) (q\xi)^3
=
\frac{4\pi}{3}  (q\xi)^3 \rho_c \biggl[ \biggl( \frac{P_c}{G\rho_c^2} \biggr)^{1 / 2} \biggl( \frac{3}{2\pi} \biggr)^{1 / 2} \frac{1}{qg}\biggr]^3
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4\pi}{3}  (q\xi)^3  \biggl[ \biggl( \frac{P_c^3}{G^3\rho_c^4} \biggr)^{1 / 2} \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} \frac{1}{q^3g^3}\biggr]
=
\frac{4\pi}{3}  \biggl[ \biggl(\frac{\pi}{6}\biggr)^{1 / 2} \nu g^3 M_\mathrm{tot}  \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} \frac{1}{g^3}\biggr]\xi^3
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
M_\mathrm{tot}  \nu \xi^3 \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g^2(\nu,q)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
\biggl\{ 1  + \biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( 1-q \biggr) +
\frac{\rho_e}{\rho_c} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \biggr\} \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{\rho_e}{\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3}\biggr) \, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{G(M_\mathrm{tot} \nu \xi^3)}{(qR\xi)^2} =
\biggl( \frac{GM_\mathrm{tot} }{R^2 } \biggr) \frac{\nu \xi}{q^2} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
G \biggl[\biggl( \frac{P_c^3}{G^3\rho_c^4} \biggr)^{1 / 2} \biggl(\frac{6}{\pi}\biggr)^{1 / 2} \frac{1}{\nu g^3} \biggr]
\biggl[\biggl(\frac{G\rho_c^2}{P_c} \biggr)^{ 1 / 2} \biggl(\frac{2\pi}{3} \biggr)^{1 / 2} qg \biggr]^2
\frac{\nu \xi}{q^2} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(P_c G)^{1 / 2} \biggl(\frac{2^3\pi}{3} \biggr)^{1 / 2} \frac{\xi}{g}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{\rho_0}{P_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\rho_c}{P_c} \biggl[ 1 - \frac{\xi^2}{g^2} \biggr]^{-1}
=
\frac{\rho_c}{P_c} \biggl( \frac{g^2}{g^2 - \xi^2} \biggr)
\, ;</math>
  </td>
</tr>
</table>
</div>
<span id="CoreWaveEq">and the wave equation for the core becomes,</span>
<!--
<div align="center" id="2ndOrderODE">
<font color="#770000">'''Adiabatic Wave Equation'''</font><br />
<math>
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} -
\biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0}
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2
+ (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x = 0 \, ,
</math>
</div>
-->
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(qR)^2} \cdot \frac{d^2x}{d\xi^2} + \biggl[\frac{4qR}{r_0} -
\biggl(\frac{qR g_0 \rho_0}{P_0}\biggr) \biggr] \frac{1}{(qR)^2} \cdot \frac{dx}{d\xi}
+ \biggl(\frac{\rho_0}{P_0} \biggr)\biggl[ \frac{\omega^2}{\gamma_\mathrm{g} }
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr)\frac{g_0}{r_0} \biggr]  x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(qR)^2} \biggl\{ \frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} -
q\biggl(\frac{P_c}{G\rho_c^2} \biggr)^{1 / 2}\biggl(\frac{3}{2\pi}\biggr)^{1 / 2} \frac{1}{qg}
(P_c G)^{1 / 2} \biggl(\frac{2^3\pi}{3} \biggr)^{1 / 2} \frac{\xi}{g} \frac{\rho_c}{P_c} \biggl( \frac{g^2}{g^2 - \xi^2} \biggr) \biggr]  \frac{dx}{d\xi} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{\rho_c}{P_c} \biggl( \frac{g^2}{g^2 - \xi^2} \biggr) \biggl[ \frac{\omega^2}{\gamma_\mathrm{g} }
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr)(P_c G)^{1 / 2} \biggl(\frac{2^3\pi}{3} \biggr)^{1 / 2} \frac{\xi}{g} \cdot \frac{1}{qR\xi}\biggr]  x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(qR)^2} \biggl\{ \frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} -
\biggl( \frac{2\xi}{g^2 - \xi^2} \biggr) \biggr]  \frac{dx}{d\xi} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{\rho_c}{P_c} \biggl( \frac{g^2}{g^2 - \xi^2} \biggr) \biggl[ \frac{\omega^2}{\gamma_\mathrm{g} }
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr)(P_c G)^{1 / 2} \biggl(\frac{2^3\pi}{3} \biggr)^{1 / 2} \frac{1}{qg}
\biggl(\frac{G\rho_c^2}{P_c}  \biggr)^{1 / 2} \biggl( \frac{2\pi}{3} \biggr)^{1 / 2} qg \biggr]  x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(qR)^2(g^2 - \xi^2)} \biggl\{ (g^2 - \xi^2)\frac{d^2x}{d\xi^2} + 
( 4g^2 - 6\xi^2 )  \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+ \frac{q^2 g^2 R^2 \rho_c}{P_c} \biggl[ \frac{\omega^2}{\gamma_\mathrm{g} }
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr) \frac{4\pi G\rho_c}{3}  \biggr]  x \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(qR)^2(g^2 - \xi^2)} \biggl\{ (g^2 - \xi^2)\frac{d^2x}{d\xi^2} + 
( 4g^2 - 6\xi^2 )  \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+ 2\biggl[ \frac{3\omega^2}{\gamma_\mathrm{g}4\pi G\rho_c}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr)  \biggr]  x \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
===Envelope===
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(qR) \xi
\, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\rho_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\rho_e \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{P_0}{P_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
1 - \frac{2\pi}{3}\chi_i^2  +
\frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_c}\biggr) \chi_i^2 \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{\xi} -
1\biggr) - \frac{\rho_e}{\rho_c} (\xi^2 - 1) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
1 - \frac{1}{g^2}\biggl\{ 1  -
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{\xi} -
1\biggr) - \frac{\rho_e}{\rho_c} (\xi^2 - 1) \biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{g^2 P_0}{P_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
g^2 -  1  +
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{\xi} -
1\biggr) - \frac{\rho_e}{\rho_c} (\xi^2 - 1) \biggr]  \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~M_r</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp;
</td>
  <td align="left">
<math>\frac{4\pi}{3} \biggl[ \frac{P_c^3}{G^3 \rho_c^4} \biggr]^{1/2} \chi_i^3\biggl[1 +\frac{\rho_e}{\rho_c}
\biggl( \xi^3 - 1\biggr) \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp;
</td>
  <td align="left">
<math>M_\mathrm{tot}
\frac{4\pi}{3} \biggl[\biggl( \frac{\pi}{6}\biggr)^{1 / 2}\nu g^3 \biggr] \biggl[ \biggr(\frac{3}{2\pi}\biggr)\frac{1}{g^2} \biggr]^{3 /2}
\biggl[1 +\frac{\rho_e}{\rho_c}  \biggl( \xi^3 - 1\biggr) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp;
</td>
  <td align="left">
<math>
\nu M_\mathrm{tot}
\biggl[1 +\frac{\rho_e}{\rho_c}  \biggl( \xi^3 - 1\biggr) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{G M_\mathrm{tot}\nu }{ R^2 q^2\xi^2}
\biggl[1 +\frac{\rho_e}{\rho_c}  \biggl( \xi^3 - 1\biggr) \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and, after multiplying through by <math>~(q^2 R^2 g^2P_0/P_c)</math>, the wave equation for the envelope becomes,
<!--
<div align="center" id="2ndOrderODE">
<font color="#770000">'''Adiabatic Wave Equation'''</font><br />
<math>
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0}
- \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0}
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2
+ (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x = 0 \, ,
</math>
</div>
-->
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^2 g^2 R^2 P_0}{P_c} \biggl\{
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0}
- \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} \biggr\}
+ \frac{q^2 g^2 R^2 \rho_0}{P_c} \biggl[ \frac{\omega^2 }{\gamma_\mathrm{g}}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)\frac{g_0}{r_0} \biggr]  x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{g^2 P_0}{P_c} \biggl\{
\frac{d^2x}{d\xi^2} + \biggl[4
- \biggl(\frac{qRg_0 \rho_e}{P_0}\biggr) \xi\biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} \biggr\}
+ \frac{q^2 g^2 R^2 \rho_e}{P_c} \biggl[ \frac{\omega^2 }{\gamma_\mathrm{g}}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)\frac{g_0}{r_0} \biggr]  x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{g^2 P_0}{P_c} \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- \biggl(\frac{qg^2Rg_0 \rho_e}{P_c}\biggr)  \frac{dx}{d\xi}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2\biggl(\frac{\rho_e}{\rho_c}\biggr) \frac{3}{4\pi G \rho_c} \biggl\{
\frac{\omega^2 }{\gamma_\mathrm{g}}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)\biggl(\frac{4\pi G \rho_c}{3}\biggr)
\biggl[ \frac{1}{\xi^3} + \frac{\rho_e}{\rho_c}\biggl(1-\frac{1}{\xi^3}\biggr) \biggr]
\biggr\} x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{g^2 P_0}{P_c} \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- 2 \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[1 +\frac{\rho_e}{\rho_c}  \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl\{
\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)
\biggl[ \frac{1}{\xi^3} + \frac{\rho_e}{\rho_c}\biggl(1-\frac{1}{\xi^3}\biggr) \biggr]
\biggr\} x
</math>
  </td>
</tr>
</table>
</div>
===Check1===
If <math>~\rho_e/\rho_c = 1</math>, this envelope wave equation should match seamlessly into the core wave equation.  Let's see if it does.  First,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g^2(\nu,q)|_{\rho_e=\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
1  + \biggl(\frac{1}{q^2} - 1\biggr) =\frac{1}{q^2} \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{g^2 P_0}{P_c} \biggr|_{\rho_e = \rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
g^2  - \xi^2 = \frac{1}{q^2} - \xi^2 \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, for the envelope,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{g^2 P_0}{P_c} \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- 2 \biggl[1 +  \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2 \biggl\{
\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)
\biggl[ \frac{1}{\xi^3} + \biggl(1-\frac{1}{\xi^3}\biggr) \biggr]
\biggr\} x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{1}{q^2} - \xi^2 \biggr) \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- 2\xi \cdot \frac{dx}{d\xi}
+ 2 \biggl\{
\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)
\biggr\} x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{1}{q^2} - \xi^2 \biggr) \frac{d^2x}{d\xi^2} 
+ \biggl\{ 4\biggl( \frac{1}{q^2} - \xi^2 \biggr) 
- 2\xi^2  \biggr\} \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+ 2 \biggl[
\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)
\biggr] x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{1}{q^2} - \xi^2 \biggr) \frac{d^2x}{d\xi^2} 
+ \biggl( \frac{4}{q^2} - 6\xi^2 \biggr)  \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+ 2 \biggl[
\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)
\biggr] x \, .
</math>
  </td>
</tr>
</table>
</div>
Whereas, for the core,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{1}{q^2} - \xi^2 \biggr)\frac{d^2x}{d\xi^2} + 
\biggl( \frac{4}{q^2} - 6\xi^2 \biggr)  \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+ 2\biggl[ \frac{3\omega^2}{\gamma_\mathrm{g}4\pi G\rho_c}
+ \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr)  \biggr]  x \, ,
</math>
  </td>
</tr>
</table>
</div>
which matches exactly.
===Boundary Condition===
In order to [[User:Tohline/SSC/Perturbations#Ensure_Finite-Amplitude_Fluctuations|ensure finite pressure fluctuations]] at the surface of this bipolytropic configuration, we need the logarithmic derivative of <math>~x</math> to obey the following relation:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{d\ln x}{d\ln r_0}\biggr|_\mathrm{surface}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\gamma_g} \biggl( 4 - 3\gamma_g + \frac{\omega^2 R^3}{GM_\mathrm{tot}}\biggr)  \, .</math>
  </td>
</tr>
</table>
</div>
Now, according to our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#MassRadius|accompanying discussion of the equilibrium mass and radius of a zero-zero polytrope]], we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{R^3}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{P_c}{G\rho_c^2}\biggr)^{3 / 2} \biggl( \frac{3}{2\pi}\biggr)^{3 / 2} \frac{1}{(qg)^3} \biggl(\frac{G^3 \rho_c^4}{P_c^3}\bigg)^{1 / 2} \biggl(\frac{\pi}{2\cdot 3}\biggr)^{1 / 2} \nu g^3
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3}{4\pi \rho_c} \biggr) \frac{\nu}{q^3} \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, a reasonable surface boundary condition is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{d\ln x}{d\ln r_0}\biggr|_\mathrm{surface}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} \biggl( \frac{\nu}{q^3}\biggr) - \biggl( 3 - \frac{4}{\gamma_\mathrm{g}}\biggr)  \, .</math>
  </td>
</tr>
</table>
</div>
<!--
From an [[User:Tohline/SSC/Stability/Polytropes#n_.3D_5_Polytrope|accompanying, introductory discussion]], we know that the outer boundary condition is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{x}{\gamma_g \xi} \biggl[ 4 - 3\gamma_g + \omega^2 \biggl( \frac{1}{4\pi G \rho_c } \biggr) \frac{\xi}{(-\theta^')}\biggr] </math>
&nbsp; &nbsp; &nbsp; &nbsp; at &nbsp; &nbsp; &nbsp; &nbsp; <math>~\xi = \xi_1 \, .</math>
  </td>
</tr>
</table>
</div>
Given that, [[User:Tohline/SSC/Structure/Polytropes#n_.3D_0_Polytrope|for n = 0 polytropes]],
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\xi^2 \frac{d\Theta_H}{d\xi} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{1}{3}\xi^3</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\xi}{(- \theta^' )}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3 \, ,</math>
  </td>
</tr>
</table>
</div>
this boundary condition becomes,
<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>~\biggl[  \frac{3\omega^2}{4\pi G \rho_c }  -\alpha \biggr] </math>
&nbsp; &nbsp; &nbsp; &nbsp; at &nbsp; &nbsp; &nbsp; &nbsp; <math>~\xi = \frac{1}{q} \, .</math>
  </td>
</tr>
</table>
</div>
Well &hellip; the boundary condition will be ''something'' like this, but we need to check to make sure this is properly phrased in the context of the envelope's wave equation.
-->
==Attempt to Find Eigenfunction for the Envelope==
Adopting some of the notation used by [http://adsabs.harvard.edu/abs/1937MNRAS..97..582S T. E. Sterne (1937)] and enunciated in our [[User:Tohline/SSC/UniformDensity#Setup_as_Presented_by_Sterne_.281937.29|accompanying discussion of the uniform-density sphere]], we'll define,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~3 - 4/\gamma_\mathrm{g} \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathfrak{F}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3\omega^2 }{2\pi \gamma_\mathrm{g}  G \rho_c} - 2 \alpha \, ,</math>
  </td>
</tr>
</table>
</div>
in which case the wave equation for the core becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(qR)^2(g^2 - \xi^2)} \biggl\{ (g^2 - \xi^2)\frac{d^2x}{d\xi^2} + 
( 4g^2 - 6\xi^2 )  \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+ \mathfrak{F}  x \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
and the wave equation for the envelope becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{g^2 P_0}{P_c} \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- 2 \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[1 +\frac{\rho_e}{\rho_c}  \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl\{
\mathfrak{F} + 2\alpha
\biggl[1 - \frac{1}{\xi^3} - \frac{\rho_e}{\rho_c}\biggl(1-\frac{1}{\xi^3}\biggr) \biggr]
\biggr\} x \, .
</math>
  </td>
</tr>
</table>
</div>
===A Specific Choice of the Density Ratio===
Now, let's focus on the ''specific'' model for which <math>~\rho_e/\rho_c = 1/2</math>.  In this case,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g^2(\nu,q) \biggr|_{\rho_e/\rho_c=1/2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
1  + \frac{1}{2}  \biggl[ 1-q  +
\frac{1}{2} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{1}{4q^2}\biggl\{
4q^2  + \biggl[ 2q^2 - 2q^3  + 1-q^2 \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[
\frac{1+5q^2  - 2q^3 }{4q^2} \biggr] \, ;
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{g^2 P_0}{P_c}\biggr|_{\rho_e/\rho_c=1/2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
g^2 -  1  +
\frac{1}{2} \biggl[ \biggl( \frac{1}{\xi} -
1\biggr) - \frac{1}{2} \biggl(\xi^2 - 1 \biggr) \biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
g^2 -  1  -
\frac{1}{4} \biggl[  \xi^2 + 1  - \frac{2}{\xi} \biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
g^2 - \frac{\xi^2}{4} \biggl[  1 + \frac{5}{\xi^2}  - \frac{2}{\xi^3} \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>
Note that this last expression goes to zero at the surface of the bipolytrope, that is, at <math>~\xi = 1/q</math>.  For this ''specific'' case, the wave equation for the envelope becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{g^2 P_0}{P_c} \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- \biggl[1 +\frac{1}{2}  \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi}
+ \frac{1}{2}  \biggl\{
\mathfrak{F} + 2\alpha
\biggl[1 - \frac{1}{\xi^3} + \frac{1}{2}\biggl(-1 + \frac{1}{\xi^3}\biggr) \biggr]
\biggr\} x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ g^2 - \frac{\xi^2}{4} \biggl[  1 + \frac{5}{\xi^2}  - \frac{2}{\xi^3} \biggr]  \biggr\} \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- \frac{1}{2}\biggl[1 +  \xi^3 \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi}
+ \frac{1}{2}  \biggl\{
\mathfrak{F} + \alpha
\biggl[1 - \frac{1}{\xi^3}  \biggr]
\biggr\} x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4\xi^3} \biggl\{
\biggl[ 4g^2\xi^3 - \xi^5 \biggl(  1 + \frac{5}{\xi^2}  - \frac{2}{\xi^3} \biggr)  \biggr] \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- 2\xi (1 +  \xi^3 )  \frac{dx}{d\xi}
+ 2 \xi^3 \biggl[ \mathfrak{F} + \alpha \biggl(1 - \frac{1}{\xi^3}  \biggr) \biggr] x
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4\xi^3} \biggl\{
\biggl[ 4g^2\xi^3 - \xi^5 - 5\xi^3  + 2\xi^2  \biggr] \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- 2\xi (1 +  \xi^3 )  \frac{dx}{d\xi}
+ \biggl[ 2 \xi^3 (\mathfrak{F} + \alpha) - 2\alpha \biggr] x
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4\xi^3} \biggl\{
\biggl[ 2 + (4g^2 - 5)\xi - \xi^3  \biggr] \biggl[
\xi^2 \cdot \frac{d^2x}{d\xi^2} + 4\xi \cdot \frac{dx}{d\xi} \biggr]
- 2(1 +  \xi^3 ) \biggl[ \xi \cdot \frac{dx}{d\xi} \biggr]
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] x
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4\xi^3} \biggl\{
\biggl[ 2 + (4g^2 - 5)\xi - \xi^3  \biggr] \biggl[
\xi^2 \cdot \frac{d^2x}{d\xi^2} \biggr]
+\biggl[ 3 + (8g^2 - 10)\xi - 3\xi^3  \biggr] \biggl[
2\xi \cdot \frac{dx}{d\xi} \biggr]
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] x
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{x}{4\xi^3} \biggl\{
\biggl[ 2 + (4g^2 - 5)\xi - \xi^3  \biggr] \biggl[
\frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr]
+\biggl[ 6 + 4(4g^2 - 5)\xi - 6\xi^3  \biggr] \biggl[
\frac{\xi}{x} \cdot \frac{dx}{d\xi} \biggr]
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
===Idea Involving Logarithmic Derivatives===
Notice that the term involving the first derivative of <math>~x</math> can be written as a logarithmic derivative; specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d\ln x}{d\ln \xi} \, .</math>
  </td>
</tr>
</table>
</div>
Let's look at the second derivative of this quantity.
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d}{d\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\xi}{x} \cdot \frac{d^2x}{d\xi^2}
+ \frac{dx}{d\xi} \cdot \biggl[  \frac{1}{x}  -  \frac{\xi}{x^2} \cdot \frac{dx}{d\xi}\biggr]
</math>
</td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\xi}{x} \cdot \frac{d^2x}{d\xi^2}
+ \frac{1}{x} \biggl[  1 -  \frac{d\ln x}{d\ln \xi} \biggr]\cdot \frac{dx}{d\xi}
</math>
</td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]
- \biggl[  1 -  \frac{d\ln x}{d\ln \xi} \biggr]\cdot \frac{d\ln x}{d\ln \xi} \, .
</math>
</td>
</tr>
</table>
</div>
Now, if we ''assume'' that the envelope's eigenfunction is a power-law of <math>~\xi</math>, that is, ''assume'' that,
<div align="center">
<math>~x = a_0 \xi^{c_0} \, ,</math>
</div>
then the logarithmic derivative of <math>~x</math> is a constant, namely,
<div align="center">
<math>~\frac{d\ln x}{d\ln\xi} = c_0 \, ,</math>
</div>
and the two key derivative terms will be,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} = c_0 \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} = c_0(c_0-1) \, .</math>
  </td>
</tr>
</table>
</div>
Hence, in order for the wave equation for the envelope for the ''specific'' density ratio being considered here to be satisfied, we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2 + (4g^2 - 5)\xi - \xi^3  \biggr] \biggl[
c_0(c_0-1) \biggr]
+\biggl[ 6 + 4(4g^2 - 5)\xi - 6\xi^3  \biggr] \biggl[
c_0 \biggr]
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2 + (4g^2 - 5)\xi - \xi^3  \biggr] c_0(c_0-1)
+c_0\biggl[ 6 + 4(4g^2 - 5)\xi - 6\xi^3  \biggr]
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[2c_0(c_0-1) + 6c_0 - 2\alpha  \biggr]
+ \biggl[(4g^2-5)(c_0^2 - c_0 + 4c_0 ) \biggr]\xi
+ \biggl[ -c_0(c_0-1) -6c_0 + 2(\mathfrak{F}+\alpha) \biggr]\xi^3
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2\biggl[c_0^2  + 2c_0 - \alpha  \biggr]
+ \biggl[(4g^2-5)(c_0^2 + 3c_0 ) \biggr]\xi
+ \biggl[ 2(\mathfrak{F}+\alpha)  - c_0(c_0+5) \biggr]\xi^3 \, .
</math>
  </td>
</tr>
</table>
</div>
This means that three algebraic relations must simultaneously be satisfied, namely:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\xi^{0}:</math>
  </td>
  <td align="left">
<math>~c_0^2  + 2c_0 - \alpha =0</math>
  </td>
  <td align="center">
<math>~\Rightarrow~</math>
  </td>
  <td align="left">
<math>~c_0 = -1 \pm (1+\alpha)^{1 / 2} \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\xi^{1}:</math>
  </td>
  <td align="left">
<math>~g^2 = \frac{5}{4}</math>
  </td>
  <td align="center">
<math>~\Rightarrow~</math>
  </td>
  <td align="left">
<math>~q=\biggl(\frac{1}{2}\biggr)^{1 / 3} </math> &nbsp; &nbsp; and, hence, <math>~\nu = \frac{2}{3} \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\xi^{3}:</math>
  </td>
  <td align="left">
<math>~2(\mathfrak{F}+\alpha)  = c_0(c_0+5)</math>
  </td>
  <td align="center">
<math>~\Rightarrow~</math>
  </td>
  <td align="left">
<math>~\frac{2}{3}\cdot \sigma^2 = (\alpha-1) \pm \sqrt{\alpha+1} \, .</math>
  </td>
</tr>
</table>
</div>
===More General Solution===
Leaving the density ratio unspecified, let's try to write the wave equation for the envelope in the same ''form'', and see if the logarithmic derivatives can be manipulated in a similar fashion.
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{g^2 P_0}{P_c} \biggl[
\frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr]
- 2 \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[1 +\frac{\rho_e}{\rho_c}  \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl\{
\mathfrak{F} + 2\alpha
\biggl[1 - \frac{1}{\xi^3} - \frac{\rho_e}{\rho_c}\biggl(1-\frac{1}{\xi^3}\biggr) \biggr]
\biggr\} x
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{\xi^3}{x} \cdot 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{g^2\xi P_0}{P_c} \biggl[ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr]
+ \biggl\{ \frac{4g^2 \xi P_0}{P_c}
- 2 \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[\biggl( 1-\frac{\rho_e}{\rho_c} \biggr) +\biggl(\frac{\rho_e}{\rho_c}  \biggr) \xi^3  \biggr] \biggr\} \frac{\xi}{x} \cdot \frac{dx}{d\xi}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\xi^3 \biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl\{
\biggl[ \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr] + 2\alpha\biggl( \frac{\rho_e}{\rho_c} -1\biggr) \cdot \frac{1}{\xi^3} 
\biggr\} 
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{g^2\xi P_0}{P_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math> \xi \biggl\{
g^2 -  1  +
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{\xi} -
1\biggr) - \frac{\rho_e}{\rho_c} (\xi^2 - 1) \biggr] 
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math> \xi \biggl\{
\biggl[ 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggr] \frac{1}{\xi} 
+\biggl[ g^2 -  1 
- 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl(1 - \frac{\rho_e}{\rho_c} \biggr)
+ \biggl(\frac{\rho_e}{\rho_c}\biggr)^2  \biggr]
- \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^2
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggr] 
+\biggl[ g^2 -  1 
- 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl(1 - \frac{\rho_e}{\rho_c} \biggr)
+ \biggl(\frac{\rho_e}{\rho_c}\biggr)^2  \biggr]\xi
- \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggr] 
+\biggl[ g^2 -  1 
- 2\biggl(\frac{\rho_e}{\rho_c}\biggr) 
+ 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2  \biggr]\xi
- \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3 \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, the wave equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \mathcal{A} + (g^2-\mathcal{B}) \xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] \biggl[ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr]
+ \biggl\{ 4\biggl[ \mathcal{A} + (g^2-\mathcal{B}) \xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] 
- \mathcal{A}
- 2\biggl(\frac{\rho_e}{\rho_c}  \biggr)^2 \xi^3 \biggr\}
\frac{\xi}{x} \cdot \frac{dx}{d\xi}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 + 2\alpha\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl( \frac{\rho_e}{\rho_c} -1\biggr) 
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \mathcal{A} + (g^2-\mathcal{B}) \xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] \biggl[ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr]
+ \biggl\{ 3\mathcal{A}  + 4(g^2-\mathcal{B}) \xi - 6\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3 \biggr\}
\frac{\xi}{x} \cdot \frac{dx}{d\xi}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha \mathcal{A} 
\biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~2\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl(1 - \frac{\rho_e}{\rho_c} \biggr)  \, ;
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~1  + 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  - 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 
\, .
</math>
  </td>
</tr>
</table>
</div>
As before, if we ''assume'' a power-law solution, the wave equation for the envelope becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \mathcal{A} + (g^2-\mathcal{B}) \xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] \biggl[ c_0(c_0-1) \biggr]
+ \biggl\{ 3\mathcal{A}  + 4(g^2-\mathcal{B}) \xi - 6\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3 \biggr\} c_0
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha \mathcal{A} 
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^0 \biggl[ \mathcal{A}c_0(c_0-1) + 3\mathcal{A}c_0 -\alpha\mathcal{A} \biggr]
+ \xi^1 \biggl[ (g^2-\mathcal{B})c_0(c_0-1) +4(g^2-\mathcal{B})c_0  \biggr]
+ \xi^3 \biggl[\biggl(\frac{\rho_e}{\rho_c}\biggr)^2c_0(1-c_0) - 6\biggl(\frac{\rho_e}{\rho_c}\biggr)^2c_0 +\biggl(\frac{\rho_e}{\rho_c}\biggr) 
\biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^0 \biggl[ c_0(c_0-1) + 3c_0 -\alpha \biggr]\mathcal{A}
+ \xi^1 \biggl[ (g^2-\mathcal{B})(c_0^2+3c_0)\biggr]
+ \xi^3 \biggl[( \mathfrak{F} +  2\alpha )  - \biggl(\frac{\rho_e}{\rho_c}\biggr)(5c_0 +c_0^2 + 2\alpha) 
\biggr]\biggl(\frac{\rho_e}{\rho_c}\biggr) \, .
</math>
  </td>
</tr>
</table>
</div>
This means that three algebraic relations must simultaneously be satisfied, namely:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\xi^{0}:</math>
  </td>
  <td align="left">
<math>~c_0^2  + 2c_0 - \alpha =0</math>
  </td>
  <td align="center">
<math>~\Rightarrow~</math>
  </td>
  <td align="left">
<math>~c_0 = -1 \pm (1+\alpha)^{1 / 2} \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\xi^{3}:</math>
  </td>
  <td align="left">
<math>~(\mathfrak{F}+2\alpha)  = \biggl(\frac{\rho_e}{\rho_c}\biggr)(5c_0 +c_0^2 + 2\alpha)</math>
  </td>
  <td align="center">
<math>~\Rightarrow~</math>
  </td>
  <td align="left">
<math>~\sigma^2 \equiv \frac{3\omega^2}{2\pi G\rho_c \gamma_\mathrm{g}}= 3\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ (\alpha-1) \pm \sqrt{\alpha+1} \biggr] \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\xi^{1}:</math>
  </td>
  <td align="left">
<math>~g^2 = 1  + 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  - 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2</math>
  </td>
  <td align="center">
<math>~\Rightarrow~</math>
  </td>
  <td align="left">
<math>~q^3 = \frac{(\rho_e/\rho_c)}{2[1-(\rho_e/\rho_c)  ]} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="left">
&nbsp;
  </td>
  <td align="center">
and, hence,
  </td>
  <td align="left">
<math>~\nu = \frac{1}{3[1-(\rho_e/\rho_c)  ]} \, .</math>
  </td>
</tr>
</table>
</div>
===Surface Boundary Condition===
Given that, with this solution, the ratio,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\nu}{q^3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3[1-(\rho_e/\rho_c)  ]} \biggl\{ \frac{2[1-(\rho_e/\rho_c)  ]}{(\rho_e/\rho_c)}  \biggr\} = \frac{2}{3(\rho_e/\rho_c) } \, ,</math>
  </td>
</tr>
</table>
</div>
we see that the [[#Boundary_Condition|desired surface boundary condition]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{d\ln x}{d\ln r_0}\biggr|_\mathrm{surface}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} \cdot \frac{2}{3(\rho_e/\rho_c) } - \biggl( 3 - \frac{4}{\gamma_\mathrm{g}}\biggr)  </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\sigma^2}{3(\rho_e/\rho_c) } - \alpha  </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c_0 \, .  </math>
  </td>
</tr>
</table>
</div>
But, for our identified solution, this is the logarithmic derivative of <math>~x</math> throughout the envelope as well as at the surface.  So the boundary condition is automatically satisfied.
==Match to a Core Eigenfunction (First Blundering)==
If we define,
<div align="center">
<math>~\eta \equiv \frac{\xi}{g} \, ,</math>
</div>
the [[#CoreWaveEq|above wave equation for the core]] becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1 - \eta^2)\frac{d^2x}{d\eta^2} + 
( 4 - 6\eta^2 )  \frac{1}{\eta} \cdot \frac{dx}{d\eta}
+ \mathfrak{F} x \, .
</math>
  </td>
</tr>
</table>
</div>
Not surprisingly, this is identical in form to the eigenvalue problem first presented by [[User:Tohline/SSC/UniformDensity#Setup_as_Presented_by_Sterne_.281937.29|Sterne (1937)]] in connection with an examination of radial oscillations in uniform-density spheres.  For the core of our zero-zero bipolytrope, we can therefore adopt any one of the [[User:Tohline/SSC/UniformDensity#Sterne.27s_General_Solution|polynomial eigenfunctions and corresponding eigenfrequencies]] derived by Sterne.  We will insist that the eigenfrequency of the envelope match the eigenfrequency of the core; and, following [http://adsabs.harvard.edu/abs/1985PASAu...6..222M J. O. Murphy &amp; R. Fiedler (1985b)] (see the top paragraph of the right-hand column on p. 223 of their article), we seek solutions for which there is continuity in both the eigenfunction and its first derivative at the interface <math>~(\xi = 1)</math>.
===Try Quadratic Core Eigenfunction===
Let's begin with Sterne's quadratic function and see if we can match it to the envelope's power-law eigenfunction. Keeping in mind that the overall normalization is arbitrary, from Sterne's presentation, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a\biggl[ 1-\frac{7}{5}\eta^2 \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a\biggl[ 1-\frac{7}{5}\biggl( \frac{\xi}{g}\biggr)^2 \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
and the associated eigenfrequency is obtained by setting,
<div align="center">
<math>~\mathfrak{F} = \sigma^2 - 2\alpha =  14 \, .</math>
</div>
In this case, then, the eigenfrequency for the envelope will match the eigenfrequency of the core if,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~14 + 2\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 3\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ (\alpha-1) \pm \sqrt{\alpha+1} \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{\rho_e}{\rho_c} \biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3}\cdot  \frac{7 + \alpha}{ (\alpha-1) \pm \sqrt{\alpha+1} }</math>
  </td>
</tr>
</table>
</div>
Now, the eigenfunction for the envelope is,
<div align="center">
<math>~x_\mathrm{env} = \xi^{c_0} \, ,</math>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~c_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
The ''value'' of this function will match the ''value'' of its core counterpart at the interface <math>~(\xi=1)</math> if,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~a\biggl[ 1-\frac{7}{5}\biggl( \frac{1}{g}\biggr)^2 \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 1-\frac{7}{5}\biggl( \frac{1}{g}\biggr)^2 \biggr]^{-1} \, .</math>
  </td>
</tr>
</table>
</div>
Finally, the slope (first derivative) of the core eigenfunction will match the slope of the envelope eigenfunction ''at the interface'' if,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~c_0 \xi^{c_0-1}\biggr|_\mathrm{\xi=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{14a}{5g^2} \cdot \xi\biggr|_\mathrm{\xi=1}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
-\frac{14}{5c_0}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{g^2}{a} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ g^2-\frac{7}{5}  </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{2}{5} + 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  - 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 - 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  + \frac{2}{5}\biggl( 1-\frac{7}{c_0}\biggr)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  0 \, .</math>
  </td>
</tr>
</table>
</div>
The solution to this quadratic equation gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \biggl(\frac{\rho_e}{\rho_c}\biggr)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{6} \biggl[
2 \pm \sqrt{4-\frac{24}{5}\biggl( 1-\frac{7}{c_0}\biggr)}
\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 \pm \sqrt{1-\frac{6}{5}\biggl( 1-\frac{7}{c_0}\biggr)}
\biggr]
</math>
  </td>
</tr>
</table>
</div>
In order for this condition to hold while also meeting the demands of the eigenfrequency, we need <math>~\alpha</math> to satisfy the relation,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{2}{3}\cdot  \frac{7 + \alpha}{ (\alpha-1) \pm \sqrt{\alpha+1} }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{3} \biggl[
1 \pm \sqrt{1-\frac{6}{5}\biggl( 1-\frac{7}{c_0}\biggr)}
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~  \frac{14 + 2\alpha}{\alpha + c_0 }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
1 \pm \biggl[ \frac{5c_0}{5c_0}-\frac{6}{5}\biggl( \frac{c_0-7}{c_0}\biggr) \biggr]^{1 / 2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~  \biggl[ \frac{14 + \alpha -c_0}{\alpha + c_0 } \biggr]^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{42-c_0}{5c_0}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~  5c_0 (14 + \alpha -c_0)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
(42-c_0)(\alpha + c_0)^2 \, ,
</math>
  </td>
</tr>
</table>
</div>
where, keep in mind,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~c_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
<font color="red"><b>RESULT:</b></font>&nbsp; After examining a range of physically reasonable values of <math>~\alpha</math>, we do not find any values for which the left-hand-side of this condition matches the right-hand-side.
===Try Quartic Core Eigenfunction===
Let's begin with Sterne's quartic function and see if we can match it to the envelope's power-law eigenfunction. From Sterne's presentation, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a\biggl[ 1-\frac{18}{5}\biggl( \frac{\xi}{g}\biggr)^2  +\frac{99}{35} \biggl( \frac{\xi}{g}\biggr)^4 \biggr]</math>
  </td>
</tr>
</table>
</div>
and the associated eigenfrequency is obtained by setting,
<div align="center">
<math>~\mathfrak{F} = \sigma^2 - 2\alpha =  36 \, .</math>
</div>
In this case, then, the eigenfrequency for the envelope will match the eigenfrequency of the core if,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~36 + 2\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 3\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ (\alpha-1) \pm \sqrt{\alpha+1} \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{\rho_e}{\rho_c} \biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{36 + 2\alpha}{ 3[(\alpha-1) \pm \sqrt{\alpha+1}] }</math>
  </td>
</tr>
</table>
</div>
The eigenfunction for the envelope is, as before.  The ''value'' of this envelope function will match the ''value'' of its core counterpart at the interface <math>~(\xi=1)</math> if,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~a\biggl[ 1-\frac{18}{5}\biggl( \frac{1}{g}\biggr)^2  +\frac{99}{35} \biggl( \frac{1}{g}\biggr)^4 \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 1-\frac{18}{5}\biggl( \frac{1}{g}\biggr)^2  +\frac{99}{35} \biggl( \frac{1}{g}\biggr)^4 \biggr]^{-1} \, .</math>
  </td>
</tr>
</table>
</div>
Finally, the slope (first derivative) of the core eigenfunction will match the slope of the envelope eigenfunction ''at the interface'' if,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl( \frac{c_0}{a}\biggr) \xi^{c_0-1}\biggr|_\mathrm{\xi=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
-\frac{36}{5g^2} \cdot \xi + \frac{4\cdot 99}{35g^4} \cdot \xi^3 \biggr]_\mathrm{\xi=1}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~c_0 \biggl[ 1-\frac{18}{5}\biggl( \frac{1}{g}\biggr)^2  +\frac{99}{35} \biggl( \frac{1}{g}\biggr)^4 \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{36}{5g^2}  + \frac{4\cdot 99}{35g^4} 
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~c_0 \biggl[ 35g^4-7\cdot 18g^2  +99  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 7\cdot 36 g^2  + 4\cdot 99 \, .
</math>
  </td>
</tr>
</table>
</div>
==Eureka Regarding Prasad's 1948 Paper==
===Envelope Solution Outline===
[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline on 5 December 2016:  Yesterday, I stumbled on this key paper by Prasad (1948) while I was looking back through the published literature to catalog who has solved the polytropic wave equation numerically.]][http://adsabs.harvard.edu/abs/1948MNRAS.108..414P C. Prasad (1948, MNRAS, 108, 414-416)] has examined a closely related problem and, as it turns out, the mathematical approach that he used to solve that problem analytically is gratifyingly useful to me here.  If, [[#More_General_Solution|as above]], we restrict our investigation to configurations for which,
<div align="center">
<math>~g^2 = \mathcal{B} \, ,</math>
</div>
and if we multiply through by <math>~x/(\mathcal{A}\xi^2)</math>, our governing wave equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 - \frac{1}{\mathcal{A}} \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] \frac{d^2x}{d\xi^2}
+ \biggl[ 3 - \frac{6}{\mathcal{A}}\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3 \biggr]
\frac{1}{\xi} \cdot \frac{dx}{d\xi}
+ \biggl[
\frac{1}{\mathcal{A}}\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha 
\biggr]\frac{x}{\xi^2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 - \mathcal{D} \xi^3\biggr] \frac{d^2x}{d\xi^2}
+ \biggl[ 3 - 6\mathcal{D} \xi^3 \biggr]
\frac{1}{\xi} \cdot \frac{dx}{d\xi}
+ \biggl[
\mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha 
\biggr]\frac{x}{\xi^2} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>\mathcal{D} \equiv \frac{1}{\mathcal{A}} \biggl(\frac{\rho_e}{\rho_c}\biggr)^2
= \biggl(\frac{\rho_e}{\rho_c}\biggr)^2\biggl[2\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl(1 - \frac{\rho_e}{\rho_c} \biggr)\biggr]^{-1}
= \biggl[2 \biggl(\frac{\rho_c}{\rho_e}-1 \biggr)\biggr]^{-1} \, .</math>
</div>
This wave equation is very similar to equation (2) of [http://adsabs.harvard.edu/abs/1948MNRAS.108..414P Prasad (1948)].  If, following Prasad's guidance, we then assume a series solution of the form,
<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>~
\xi^{c_0} \sum_0^\infty a_k \xi^k \, ,
</math>
  </td>
</tr>
</table>
</div>
the indicial equation gives,
<div align="center">
<math>~c_0 = -1 \pm \sqrt{1+\alpha} \, .</math>
</div>
This is precisely the value of the exponent, <math>~c_0</math>, that we derived &#8212; in a more stumbling fashion &#8212; [[#More_General_Solution|above]] and, as is shown by the following framed image, it is identical to the exponent derived by [http://adsabs.harvard.edu/abs/1948MNRAS.108..414P Prasad (1948)].
<div align="center">
<table border="2" cellpadding="10" width="75%">
<tr>
  <th align="center">
Equation and accompanying text extracted<sup>&dagger;</sup> from [http://adsabs.harvard.edu/abs/1948MNRAS.108..414P C. Prasad (1948)]<p></p>
"''Radial Oscillations of a Particular Stellar Model''"<p></p>
Monthly Notices of the Royal Astronomical Society, vol. 108, pp. 414-416 &copy; Royal Astronomical Society
  </th>
<tr>
  <td>
[[File:Prasad1948Eq4.png|400px|left|Prasad (1948)]]
  </td>
</tr>
<tr><td align="left"><sup>&dagger;</sup>Displayed here exactly as presented in the original publication.</td></tr>
</table>
</div>
Using equation (7) from [http://adsabs.harvard.edu/abs/1948MNRAS.108..414P Prasad (1948)] as a guide, we hypothesize that the eigenfrequency of the ''j''<sup>th</sup> mode in the envelope is given by the relation,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\sigma_j^2 \biggr|_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(c_0 + 3j)(c_0 + 3j+5) + 2\alpha \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~
\sigma_j^2\biggr|_\mathrm{env} \equiv
\biggl(\mathfrak{F} + 2\alpha\biggr)\frac{\rho_c}{\rho_e} = \frac{3\omega^2}{2\pi \gamma_\mathrm{g} G\rho_e}
\, .
</math>
</div>
And guided by equation (6) from [http://adsabs.harvard.edu/abs/1948MNRAS.108..414P Prasad (1948)], we hypothesize that successive coefficients in the (truncated) series that defines the radial structure of each mode is governed by the recurrence relation,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{\mathcal{D}}\cdot  \frac{a_{k+3}}{a_k}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(c_0+k)(c_0+k+5) - (\sigma_j^2 - 2\alpha)}{(c_0 + k + 3)(c_0+k+5) - \alpha} \, .
</math>
  </td>
</tr>
</table>
</div>
===Example Envelope Eigenvectors===
====Mode j = 0====
Here we assume that the series defining the eigenfunction has only one term.  This should match our [[#More_General_Solution|earlier restricted solution]].  Specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_{j=0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 \xi^{c_0}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{1}{\xi} \cdot  \frac{d x_{j=0}}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 c_0 \xi^{c_0-2} \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{d^2 x_{j=0}}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 c_0(c_0-1) \xi^{c_0-2} \, .</math>
  </td>
</tr>
</table>
</div>
In this case, the wave equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 \xi^{c_0-2}\biggl\{
\biggl[1 - \mathcal{D} \xi^3\biggr]  c_0(c_0-1) 
+ \biggl[ 3 - 6\mathcal{D} \xi^3 \biggr]
c_0
+ \biggl[
\mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha 
\biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
The coefficients of the <math>~\xi^0</math> terms will sum to zero if the above-defined indicial  exponent condition is satisfied; that is, by setting,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~c_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
In order for the coefficients of the  <math>~\xi^3</math> terms to sum to zero, we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  \biggr) - 2\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c_0(c_0-1) +6c_0 </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \sigma^2_{j=0}\biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c_0(c_0+5) + 2\alpha \, .</math>
  </td>
</tr>
</table>
</div>
====Mode j = 1====
Here we assume that the series defining the eigenfunction has two terms:  <math>~k=0</math> and <math>~k=3</math>.  Specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_{j=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 \xi^{c_0} + a_3 \xi^{c_0+3}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{1}{\xi}\cdot  \frac{d x_{j=1}}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 c_0 \xi^{c_0-2} + a_3 (c_0+3) \xi^{c_0+1} \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{d^2 x_{j=1}}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 c_0(c_0-1) \xi^{c_0-2} + a_3 (c_0+3)(c_0+2) \xi^{c_0+1}  \, .</math>
  </td>
</tr>
</table>
</div>
In this case, after factoring out <math>~a_0\xi^{c_0-2}</math>, the wave equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 - \mathcal{D} \xi^3\biggr] \biggl[ c_0(c_0-1)  + \frac{a_3}{a_0} (c_0+3)(c_0+2) \xi^{3} \biggr]
+ \biggl[ 3 - 6\mathcal{D} \xi^3 \biggr] \biggl[c_0  + \frac{a_3}{a_0} (c_0+3) \xi^{3}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[
\mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha 
\biggr] \biggl[1  + \frac{a_3}{a_0} \xi^{3}  \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>
Again, the coefficients of the <math>~\xi^0</math> terms will sum to zero if the above-defined indicial  exponent condition is satisfied; that is, by setting,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~c_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
In order for the coefficients of the  <math>~\xi^3</math> terms to sum to zero, we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\mathcal{D} c_0(c_0-1)+ \frac{a_3}{a_0} (c_0+3)(c_0+2)
-6\mathcal{D}c_0 + \frac{3a_3}{a_0} (c_0+3)
+ \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)-\frac{\alpha a_3}{a_0}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{a_3}{a_0} \biggl[ (c_0+3)(c_0+2) + 3(c_0+3)-\alpha\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathcal{D}\biggl[  c_0(c_0-1)
+6c_0
-\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  \biggr) + 2\alpha \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{1}{\mathcal{D}} \cdot \frac{a_3}{a_0} \biggl[ (c_0+3)(c_0+5) -\alpha\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[  c_0(c_0+5) -\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  \biggr) + 2\alpha \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{1}{\mathcal{D}} \cdot \frac{a_3}{a_0}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ c_0(c_0+5) - (\sigma^2_{j=1} - 2\alpha) }{ (c_0+3)(c_0+5) -\alpha} \, .
</math>
  </td>
</tr>
</table>
</div>
In addition, we must also examine what condition is required for the <math>~\xi^6</math> terms to sum to zero.  We have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{D}\frac{a_3}{a_0}\biggl[
- (c_0+3)(c_0+2) -6(c_0+3)
+ \biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \sigma^2_{j=1}\biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(c_0+3)(c_0+8)  +2\alpha \, .
</math>
  </td>
</tr>
</table>
</div>
====Mode j = 2====
Here we assume that the series defining the eigenfunction has three terms:  <math>~k=0</math>, <math>~k=3</math>, and <math>~k=6</math>.  Specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_{j=2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 \xi^{c_0} + a_3 \xi^{c_0+3} + a_6 \xi^{c_0+6}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{1}{\xi}\cdot  \frac{d x_{j=2}}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\xi} \biggl[ a_0 c_0 \xi^{c_0-1} + a_3 (c_0+3) \xi^{c_0+2} + a_6(c_0+6)\xi^{c_0+5} \biggr] \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{d^2 x_{j=2}}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 c_0(c_0-1) \xi^{c_0-2} + a_3 (c_0+3)(c_0+2) \xi^{c_0+1} + a_6(c_0+6)(c_0+5) \xi^{c_0+4} \, .</math>
  </td>
</tr>
</table>
</div>
In this case, after factoring out <math>~a_0\xi^{c_0-2}</math>, the wave equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 - \mathcal{D} \xi^3\biggr] \biggl[ c_0(c_0-1)  + \frac{a_3}{a_0} (c_0+3)(c_0+2) \xi^{3} + \frac{a_6}{a_0}(c_0+6)(c_0+5) \xi^{6} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 3 - 6\mathcal{D} \xi^3 \biggr]
\biggl[ c_0  + \frac{a_3}{a_0} (c_0+3) \xi^{3} + \frac{a_6}{a_0}(c_0+6)\xi^{6} \biggr]
+ \biggl[
\mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha 
\biggr] \biggl[1 + \frac{a_3}{a_0} \xi^{3} + \frac{a_6}{a_0} \xi^{6}  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Again, the coefficients of the <math>~\xi^0</math> terms will sum to zero if,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~c_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
In order for the coefficients of the  <math>~\xi^3</math> terms to sum to zero, we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\mathcal{D}c_0(c_0-1) + \frac{a_3}{a_0} (c_0+3)(c_0+2) - 6\mathcal{D}c_0 + 3\cdot \frac{a_3}{a_0} (c_0+3)
+ \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr) - \alpha\cdot \frac{a_3}{a_0}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{1}{\mathcal{D}} \cdot \frac{a_3}{a_0} \biggl[ (c_0+3)(c_0+2)  + 3 (c_0+3)  - \alpha\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0(c_0-1)  + 6c_0 +2\alpha
- \biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{1}{\mathcal{D}} \cdot \frac{a_3}{a_0}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ c_0(c_0+5)  +2\alpha - \sigma^2_{j=2} }{ (c_0+3)(c_0+5) - \alpha }
</math>
  </td>
</tr>
</table>
</div>
In order for the coefficients of the  <math>~\xi^6</math> terms to sum to zero, we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{a_6}{a_0}(c_0+6)(c_0+5) - \mathcal{D} \cdot \frac{a_3}{a_0} (c_0+3)(c_0+2)
+3 \cdot \frac{a_6}{a_0}(c_0+6) - 6\mathcal{D}\cdot \frac{a_3}{a_0} (c_0+3)
-\alpha \cdot \frac{a_6}{a_0} + \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr) \cdot \frac{a_3}{a_0}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
\frac{a_6}{a_0} \biggl[ (c_0+6)(c_0+5)  +3 (c_0+6) -\alpha \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathcal{D} \cdot \frac{a_3}{a_0} \biggl[ (c_0+3)(c_0+2)
+6 (c_0+3) +2\alpha - \sigma^2_{j=2} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{1}{\mathcal{D}} \cdot
\frac{a_6}{a_3}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ (c_0+3)(c_0+8) - (\sigma^2_{j=2}-2\alpha) }{(c_0+6)(c_0+8)  -\alpha  } \, .
</math>
  </td>
</tr>
</table>
</div>
And the frequency determined from setting to zero the sum of coefficients of the <math>~\xi^9</math> terms is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\mathcal{D} \cdot  \frac{a_6}{a_0}(c_0+6)(c_0+5) - 6\mathcal{D} \cdot \frac{a_6}{a_0}(c_0+6) + \mathcal{D} \biggl(\frac{\rho_c}{\rho_e}\biggr)  \biggl( \mathfrak{F} +  2\alpha  -2\alpha\frac{\rho_e}{\rho_c} \biggr)\cdot \frac{a_6}{a_0}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\mathcal{D} \cdot  \frac{a_6}{a_0} \biggl[ (c_0+6)(c_0+5) + 6(c_0+6) - (\sigma^2_{j=2} - 2\alpha) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (\sigma^2_{j=2} - 2\alpha) </math>
</td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(c_0+6)(c_0+11) \, .
</math>
  </td>
</tr>
</table>
</div>
==Match Prasad-like Envelope Eigenvector to the Core Eigenvector==
If we define,
<div align="center">
<math>~\eta \equiv \frac{\xi}{g} \, ,</math>
</div>
the [[#CoreWaveEq|above wave equation for the core]] becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1 - \eta^2)\frac{d^2x}{d\eta^2} + 
( 4 - 6\eta^2 )  \frac{1}{\eta} \cdot \frac{dx}{d\eta}
+ \mathfrak{F} x \, .
</math>
  </td>
</tr>
</table>
</div>
Not surprisingly, this is identical in form to the eigenvalue problem first presented by [[User:Tohline/SSC/UniformDensity#Setup_as_Presented_by_Sterne_.281937.29|Sterne (1937)]] in connection with an examination of radial oscillations in uniform-density spheres.  For the core of our zero-zero bipolytrope, we can therefore adopt any one of the [[User:Tohline/SSC/UniformDensity#Sterne.27s_General_Solution|polynomial eigenfunctions and corresponding eigenfrequencies]] derived by Sterne.  We will insist that the eigenfrequency of the envelope match the eigenfrequency of the core; and, following [http://adsabs.harvard.edu/abs/1985PASAu...6..222M J. O. Murphy &amp; R. Fiedler (1985b)] (see the top paragraph of the right-hand column on p. 223 of their article), we seek solutions for which there is continuity in both the eigenfunction and its first derivative at the interface <math>~(\xi = 1)</math>.
===Eigenfrequencies===
We must note that, heretofore, we have used the following dimensionless frequency notations:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\sigma^2|_\mathrm{core} \equiv \frac{3\omega^2_\mathrm{core}}{2\pi \gamma_g G\rho_c} \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp;
and
&nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\sigma^2|_\mathrm{env} \equiv \frac{3\omega^2_\mathrm{env}}{2\pi \gamma_g G\rho_e} \, .</math>
  </td>
</tr>
</table>
</div>
This means that, demanding that the two ''dimensional'' frequencies <math>~(\omega)</math> be the same requires that the ratio of the ''dimensionless'' frequencies <math>~(\sigma)</math> be,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{ \sigma^2|_\mathrm{core}}{\sigma^2|_\mathrm{env} }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\rho_e}{\rho_c} \, .</math>
  </td>
</tr>
</table>
</div>
Now, according to [[User:Tohline/SSC/UniformDensity#Sterne.27s_General_Solution|Sterne's derivation]], the dimensionless eigenfrequency associated with the <math>~j^\mathrm{th}</math> mode in the core is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\sigma^2_j|_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\alpha + \mathfrak{F}_j = 2\alpha + 2j(2j+5) \, .</math>
  </td>
</tr>
</table>
</div>
And, as we have just discussed, the dimensionless eigenfrequency associated with the <math>~\ell^\mathrm{th}</math> Prasad-like mode in the envelope is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\sigma^2_\ell|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\alpha + \mathfrak{F}_\ell = 2\alpha + (c_0 + 3\ell)(c_0 + 3\ell +5)  \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~c_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
Hence, in order for any specific pair of modes to have the same ''dimensional'' eigenfrequencies, we must have an envelope-to-core density ratio given by the expression,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\rho_e}{\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2\alpha + 2j(2j+5) }{ 2\alpha + (c_0 + 3\ell)(c_0 + 3\ell +5) } \, .
</math>
  </td>
</tr>
</table>
</div>
Put another way, the ratio of the ''dimensional'' eigenfrequencies is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{\omega^2_\mathrm{env}}{\omega^2_\mathrm{core}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\sigma^2_\mathrm{env}}{\sigma^2_\mathrm{core}} \biggl( \frac{\rho_e}{\rho_c} \biggr)
= \frac{ 2\alpha + (c_0 + 3\ell)(c_0 + 3\ell +5) }{ 2\alpha + 2j(2j+5) } \biggl( \frac{\rho_e}{\rho_c} \biggr)\, .
</math>
  </td>
</tr>
</table>
</div>
We also should keep in mind that, in our particular case, the envelope density must not be greater than the core density.  So, demanding that the (dimensional) eigenfrequencies be equal and, simultaneously, that <math>~\rho_e/\rho_c \le 1</math>, implies the following constraint on the ''integer''  index, <math>~j</math> for each choice of the index, <math>~\ell</math>:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~2\alpha + 2j(2j+5)</math>
  </td>
  <td align="center">
<math>~\le</math>
  </td>
  <td align="left">
<math>~2\alpha + (c_0 + 3\ell)(c_0 + 3\ell +5) </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Rightarrow~~~2j</math>
  </td>
  <td align="center">
<math>~\le</math>
  </td>
  <td align="left">
<math>~ c_0 + 3\ell </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Rightarrow~~~j</math>
  </td>
  <td align="center">
<math>~\le j_\mathrm{max} \equiv</math>
  </td>
  <td align="left">
<math>~\mathrm{INT}\biggl[  \frac{1}{2}(c_0 + 3\ell ) \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
The following table lists values of <math>~j_\mathrm{max}</math> for various values of the companion index, <math>~\ell</math>, and an assumed value of the parameter,
<div align="center">
<math>~\alpha \equiv 3 - 4/\gamma_g = \frac{3-n}{n+1} \, .</math>
</div>
<!--[The table lists assigned values of the polytropic index along with this parameter in pairs as, <math>~(\alpha,n)</math>.]-->
<div align="center">
<table border="1" cellpadding="5" align="left">
<tr>
  <td align="center" colspan="4">Limiting Index, <math>~j_\mathrm{max}</math>, assuming <br /><math>c_0 = \sqrt{1+\alpha} -1</math></td>
</tr>
<tr>
  <td align="center" rowspan="2"><math>~\ell</math></td>
  <td align="center" colspan="3"><math>~(n,\gamma_\mathrm{g}, \alpha,c_0)</math></td>
</tr>
<tr>
  <td align="center"><math>~(0,\infty,3,1)</math>
  <td align="center"><math>~(3,\tfrac{4}{3},0,0)</math>
  <td align="center"><math>~(\infty,1, -1,-1)</math>
</tr>
<tr>
  <td align="center">0</td>
  <td align="center">0</td>
  <td align="center">0</td>
  <td align="center">---</td>
</tr>
<tr>
  <td align="center">1</td>
  <td align="center">2</td>
  <td align="center">1</td>
  <td align="center">1</td>
</tr>
<tr>
  <td align="center">2</td>
  <td align="center">3</td>
  <td align="center">3</td>
  <td align="center">2</td>
</tr>
<tr>
  <td align="center">3</td>
  <td align="center">5</td>
  <td align="center">4</td>
  <td align="center">4</td>
</tr>
<tr>
  <td align="center">4</td>
  <td align="center">6</td>
  <td align="center">6</td>
  <td align="center">5</td>
</tr>
<tr>
  <td align="center">5</td>
  <td align="center">8</td>
  <td align="center">7</td>
  <td align="center">7</td>
</tr>
<tr>
  <td align="center">6</td>
  <td align="center">9</td>
  <td align="center">9</td>
  <td align="center">8</td>
</tr>
<tr>
  <td align="center">7</td>
  <td align="center">11</td>
  <td align="center">10</td>
  <td align="center">10</td>
</tr>
<tr>
  <td align="center">8</td>
  <td align="center">12</td>
  <td align="center">12</td>
  <td align="center">11</td>
</tr>
<tr>
  <td align="center">9</td>
  <td align="center">14</td>
  <td align="center">13</td>
  <td align="center">13</td>
</tr>
<tr>
  <td align="center">10</td>
  <td align="center">15</td>
  <td align="center">15</td>
  <td align="center">14</td>
</tr>
<tr>
  <td align="center" colspan="4">
[[File:C0Plus2.png|350px|C0 Plus]]
  </td>
</tr>
</table>
<table border="1" cellpadding="5">
<tr>
  <td align="center" colspan="4">Limiting Index, <math>~j_\mathrm{max}</math>, assuming <br /><math>c_0 = -\sqrt{1+\alpha} -1</math></td>
</tr>
<tr>
  <td align="center" rowspan="2"><math>~\ell</math></td>
  <td align="center" colspan="3"><math>~(n,\gamma_\mathrm{g}, \alpha,c_0)</math></td>
</tr>
<tr>
  <td align="center"><math>~(\infty,1, -1,-1)</math>
  <td align="center"><math>~(3,\tfrac{4}{3},0,-2)</math>
  <td align="center"><math>~(0,\infty,3,-3)</math>
</tr>
<tr>
  <td align="center">0</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
</tr>
<tr>
  <td align="center">1</td>
  <td align="center">1</td>
  <td align="center">0</td>
  <td align="center">0</td>
</tr>
<tr>
  <td align="center">2</td>
  <td align="center">2</td>
  <td align="center">2</td>
  <td align="center">1</td>
</tr>
<tr>
  <td align="center">3</td>
  <td align="center">4</td>
  <td align="center">3</td>
  <td align="center">3</td>
</tr>
<tr>
  <td align="center">4</td>
  <td align="center">5</td>
  <td align="center">5</td>
  <td align="center">4</td>
</tr>
<tr>
  <td align="center">5</td>
  <td align="center">7</td>
  <td align="center">6</td>
  <td align="center">6</td>
</tr>
<tr>
  <td align="center">6</td>
  <td align="center">8</td>
  <td align="center">8</td>
  <td align="center">7</td>
</tr>
<tr>
  <td align="center">7</td>
  <td align="center">10</td>
  <td align="center">9</td>
  <td align="center">9</td>
</tr>
<tr>
  <td align="center">8</td>
  <td align="center">11</td>
  <td align="center">11</td>
  <td align="center">10</td>
</tr>
<tr>
  <td align="center">9</td>
  <td align="center">13</td>
  <td align="center">12</td>
  <td align="center">12</td>
</tr>
<tr>
  <td align="center">10</td>
  <td align="center">14</td>
  <td align="center">14</td>
  <td align="center">13</td>
</tr>
<tr>
  <td align="center" colspan="4">
[[File:C0Minus2.png|350px|C0 Minus]]
  </td>
</tr>
</table>
</div>
===Implications===
Keeping in mind that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
1  + \biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( 1-q \biggr) +
\frac{\rho_e}{\rho_c} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]  \, ,
</math>
  </td>
</tr>
</table>
</div>
and that, in order for the Prasad-like modes to be relevant in the envelope, we must have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g^2 = \mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~1  + 2\biggl(\frac{\rho_e}{\rho_c}\biggr)  - 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(1  - \frac{\rho_e}{\rho_c}\biggr)\biggl[  3\biggl(\frac{\rho_e}{\rho_c}\biggr)+1\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
we recognize that once the eigenfrequency match is used to define the relevant value of the density ratio, <math>~\rho_e/\rho_c</math>, the relevant values of both <math>~q</math> and <math>~\nu</math> are set as well.  Specifically, as derived [[#More_General_Solution|above, in the context of our "more general" envelope solution]],
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~q^3 = \frac{(\rho_e/\rho_c)}{2[1-(\rho_e/\rho_c)]}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp;
and
&nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\nu = \frac{1}{3[1-(\rho_e/\rho_c)]} \, .</math>
  </td>
</tr>
</table>
</div>
This also means that the parameter,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathcal{D}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[2\biggr(\frac{\rho_c}{\rho_e} - 1\biggr)\biggr]^{-1} = q^3 \, .</math>
  </td>
</tr>
</table>
</div>
===Eigenfunctions===
The eigenfunction associated with the <math>~j^\mathrm{th}</math> Sterne-like mode of the core is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_j|_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sum_{i=0}^{j} a_{2i} \biggl( \frac{\xi^2}{g^2}\biggr)^{i} \, ,</math>
  </td>
</tr>
</table>
</div>
where, for the specified, <math>~j^\mathrm{th}</math> mode, the value of the leading coefficient, <math>~a_0</math>, is arbitrary, but for all other coefficients,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{a_{k+2}}{a_k }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{k^2 + 5k - \mathfrak{F}_j}{(k+2)(k+5)} \, .</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{k^2 + 5k - 2j(2j+5)}{(k+2)(k+5)} \, .</math>
  </td>
</tr>
</table>
</div>
The eigenfunction associated with the <math>~\ell^\mathrm{th}</math> Prasad-like mode of the envelope is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_\ell|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^{c_0} \sum_{i=0}^\ell b_{3i} \xi^{3i} \, ,
</math>
  </td>
</tr>
</table>
</div>
where, for the specified, <math>~\ell^\mathrm{th}</math> mode, the value of the leading coefficient, <math>~a_0</math>, is arbitrary, but for all other coefficients,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{q^3}\cdot  \frac{b_{k+3}}{b_k}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(c_0+k)(c_0+k+5) - \mathfrak{F}_\ell}{(c_0 + k + 3)(c_0+k+5) - \alpha}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(c_0+k)(c_0+k+5) - (c_0 + 3\ell)(c_0 + 3\ell +5)}{(c_0 + k + 3)(c_0+k+5) - \alpha} \, .
</math>
  </td>
</tr>
</table>
</div>
====Example11====
Let's try <math>~(\ell,j) = (1,1) \, .</math>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_1 |_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl[1 + \frac{a_2}{a_0} \biggl(\frac{\xi}{g}\biggr)^2 \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl\{1 + \biggl[ \frac{k^2 + 5k - 2j(2j+5)}{(k+2)(k+5)} \biggr]_{k=0} \biggl(\frac{\xi}{g}\biggr)^2 \biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl[1 -  \frac{7}{5}  \biggl(\frac{\xi}{g}\biggr)^2 \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dx_1}{d\xi}\biggr|_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-  a_0 \cdot \frac{14}{5}  \biggl(\frac{\xi}{g^2}\biggr) </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~x_1|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0}\biggl[1 +  \frac{b_{3}}{b_0} \xi^{3} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0}\biggl\{ 1 +  \biggl[ \frac{(c_0+k)(c_0+k+5) - (c_0 + 3\ell)(c_0 + 3\ell +5)}{(c_0 + k + 3)(c_0+k+5) - \alpha}\biggr]_{k=0} \xi^{3} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0}\biggl\{ 1 +  \biggl[ \frac{c_0(c_0+5) - (c_0 + 3)(c_0 + 8)}{(c_0 + 3)(c_0+8) - \alpha}\biggr] \xi^{3} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dx_1}{d\xi}\biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0-1}\biggl\{ c_0 +  c_0\biggl[ \frac{c_0(c_0+5) - (c_0 + 3)(c_0 + 8)}{(c_0 + 3)(c_0+8) - \alpha}\biggr] \xi^{3}
+\biggl[ \frac{3c_0(c_0+5) - 3(c_0 + 3)(c_0 + 8)}{(c_0 + 3)(c_0+8) - \alpha}\biggr] \xi^{3}  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0-1}\biggl\{ c_0 +  \biggl[ \frac{c_0^2(c_0+5) - c_0(c_0 + 3)(c_0 + 8)+ 3c_0(c_0+5) - 3(c_0 + 3)(c_0 + 8)}{(c_0 + 3)(c_0+8) - \alpha}
\biggr] \xi^{3}  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0-1}\biggl\{ c_0 +  (c_0+3)\biggl[ \frac{c_0(c_0+5) - (c_0 + 3)(c_0 + 8)}{(c_0 + 3)(c_0+8) - \alpha}
\biggr] \xi^{3}  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0-1}\biggl\{ c_0 -  \biggl[ \frac{6(c_0+3)(c_0+4)}{(c_0 + 3)(c_0+8) - \alpha}
\biggr] \xi^{3}  \biggr\}
</math>
  </td>
</tr>
</table>
</div>
Let's define both <math>~a_0</math> and <math>~b_0</math> such that the ''values'' of both eigenfunctions is unity at the interface <math>~(\xi=1)</math>.  This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~a_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1 -  \frac{7}{5g^2}  \biggr]^{-1} =\biggl[\frac{5g^2}{5g^2-7}  \biggr] \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~b_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ 1 +  \biggl[ \frac{c_0(c_0+5) - (c_0 + 3)(c_0 + 8)}{(c_0 + 3)(c_0+8) - \alpha}\biggr]  \biggr\}^{-1} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(c_0 + 3)(c_0+8) - \alpha}{c_0(c_0+5)- \alpha }\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Hence, in order for the first derivative of both eigenfunctions to be equal ''at the interface'', we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~-  a_0 \cdot \frac{14}{5}  \biggl(\frac{1}{g^2}\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \biggl\{ c_0 -  \biggl[ \frac{6(c_0+3)(c_0+4)}{(c_0 + 3)(c_0+8) - \alpha}
\biggr]  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
-  \biggl[\frac{5g^2}{5g^2-7}  \biggr] \cdot \frac{14}{5g^2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \frac{c_0[(c_0 + 3)(c_0+8) - \alpha ]-6(c_0+3)(c_0+4)}{(c_0 + 3)(c_0+8) - \alpha}
\biggr\}\biggl[ \frac{(c_0 + 3)(c_0+8) - \alpha}{c_0(c_0+5)- \alpha }\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
\frac{14}{7-5g^2}   
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c_0[(c_0 + 3)(c_0+8) - \alpha ]-6(c_0+3)(c_0+4)}{c_0(c_0+5)- \alpha }
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
\frac{7-5g^2}{14}   
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c_0(c_0+5)- \alpha }{c_0[(c_0 + 3)(c_0+8) - \alpha ]-6(c_0+3)(c_0+4)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
g^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{7}{5}+
\frac{14}{5} \biggl[ \frac{\alpha -c_0(c_0+5)}{c_0[(c_0 + 3)(c_0+8) - \alpha ]-6(c_0+3)(c_0+4)}\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
====Example12====
Let's try <math>~(\ell,j) = (1,2) \, .</math>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_1 |_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 + a_2\biggl(\frac{\xi}{g}\biggr)^2 + a_4\biggl(\frac{\xi}{g}\biggr)^4</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl[ 1 + \frac{a_2}{a_0}\biggl(\frac{\xi}{g}\biggr)^2 + \frac{a_4}{a_2} \cdot \frac{a_2}{a_0} \biggl(\frac{\xi}{g}\biggr)^4 \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl\{1
+ \biggl[ \frac{k^2 + 5k - 2j(2j+5)}{(k+2)(k+5)} \biggr]_{k=0} \biggl(\frac{\xi}{g}\biggr)^2
+ \biggl[ \frac{k^2 + 5k - 2j(2j+5)}{(k+2)(k+5)} \biggr]_{k=2}  \biggl[ \frac{k^2 + 5k - 2j(2j+5)}{(k+2)(k+5)} \biggr]_{k=0} \biggl(\frac{\xi}{g}\biggr)^4
\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl[ 1
- \frac{18}{5} \biggl(\frac{\xi}{g}\biggr)^2
+  \frac{99}{35} \biggl(\frac{\xi}{g}\biggr)^4
\biggr] \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dx_1}{d\xi}\biggr|_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0 \biggl[ - \frac{36}{5}  \biggl(\frac{\xi}{g^2} \biggr) +  \frac{4\cdot 99}{35} \biggl(\frac{\xi^3}{g^4}\biggr) \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a_0}{35g^4} \biggl[ - 7\cdot 36 g^2 \xi  + 4\cdot 99 \xi^3 \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
In order for the core's <math>~x_2</math> eigenfunction to have the value of unity at <math>~\xi=1</math>, we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl[ 1
- \frac{18}{5} \biggl(\frac{1}{g}\biggr)^2
+  \frac{99}{35} \biggl(\frac{1}{g}\biggr)^4
\biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a_0}{35g^4} \biggl[ 35 g^4
- 7\cdot 18g^2 + 99
\biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
a_0
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~35g^4 \biggl[ 35 g^4 - 7\cdot 18g^2 + 99 \biggr]^{-1} </math>
  </td>
</tr>
</table>
</div>
Hence, in order for the first derivative of both eigenfunctions to be equal ''at the interface'', we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\frac{ a_0}{35g^4}\biggl[ - 7\cdot 36 g^2    + 4\cdot 99  \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \biggl\{ c_0 -  \biggl[ \frac{6(c_0+3)(c_0+4)}{(c_0 + 3)(c_0+8) - \alpha}
\biggr]  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~35g^4 \biggl[ 35 g^4 - 7\cdot 18g^2 + 99 \biggr]^{-1}
\frac{ 1}{35g^4}\biggl[ - 7\cdot 36 g^2    + 4\cdot 99  \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{(c_0 + 3)(c_0+8) - \alpha}{c_0(c_0+5)- \alpha }\biggr] \biggl\{ c_0 -  \biggl[ \frac{6(c_0+3)(c_0+4)}{(c_0 + 3)(c_0+8) - \alpha}
\biggr]  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ 36\biggl[ 35 g^4 - 7\cdot 18g^2 + 99 \biggr]^{-1}
\biggl[ 11 - 7g^2  \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{c_0[ (c_0 + 3)(c_0+8) - \alpha] - 6(c_0+3)(c_0+4)}{c_0(c_0+5)- \alpha} \biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\biggl[ 11 - 7g^2  \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\mathcal{H} } \biggl[ 35 g^4 - 7\cdot 18g^2 + 99 \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
35 g^4 - 7\cdot 18g^2 + 99  - \mathcal{H}\biggl( 11 - 7g^2  \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
35 g^4 - 7( 18-\mathcal{H} )g^2 + 11 (9-\mathcal{H}) \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathcal{H}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~36\biggl[ \frac{c_0(c_0+5)- \alpha}{c_0[ (c_0 + 3)(c_0+8) - \alpha] - 6(c_0+3)(c_0+4)} \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>
The solution to this quadratic equation gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{70} \biggl\{7( 18-\mathcal{H} ) \pm \sqrt{7^2 ( 18-\mathcal{H} )^2 - 44 \cdot 35(9-\mathcal{H}) }
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{7( 18-\mathcal{H} )}{70} \biggl\{1 \pm \sqrt{1 - \frac{44 \cdot 35(9-\mathcal{H})}{7^2 ( 18-\mathcal{H} )^2} }
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
====Example21====
Let's try <math>~(\ell,j) = (2,1) \, .</math>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_1 |_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl[1 + \frac{a_2}{a_0} \biggl(\frac{\xi}{g}\biggr)^2 \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl\{1 + \biggl[ \frac{k^2 + 5k - 2j(2j+5)}{(k+2)(k+5)} \biggr]_{k=0} \biggl(\frac{\xi}{g}\biggr)^2 \biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl[1 -  \frac{7}{5}  \biggl(\frac{\xi}{g}\biggr)^2 \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dx_1}{d\xi}\biggr|_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-  a_0 \cdot \frac{14}{5}  \biggl(\frac{\xi}{g^2}\biggr) </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~x_2|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0}\biggl[1 +  \frac{b_{3}}{b_0} \xi^{3} +  \frac{b_{3}}{b_0}\cdot \frac{b_{6}}{b_3}  \xi^{6} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0}\biggl\{ 1 +  \mathcal{D} \biggl[ \frac{(c_0+k)(c_0+k+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + k + 3)(c_0+k+5) - \alpha}\biggr]_{k=0} \xi^{3}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  \mathcal{D}^2 \biggl[ \frac{(c_0+k)(c_0+k+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + k + 3)(c_0+k+5) - \alpha}\biggr]_{k=0}
\biggl[ \frac{(c_0+k)(c_0+k+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + k + 3)(c_0+k+5) - \alpha}\biggr]_{k=3}\xi^{6} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0}\biggl\{ 1 +  \mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr] \xi^{3}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  \mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr]\xi^{6} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dx_2}{d\xi}\biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \xi^{c_0-1}\biggl\{ c_0 + (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr] \xi^{3}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr]\xi^{6} \biggr\}
</math>
  </td>
</tr>
</table>
</div>
Let's define both <math>~a_0</math> and <math>~b_0</math> such that the ''values'' of both eigenfunctions is unity at the interface <math>~(\xi=1)</math>.  This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~a_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1 -  \frac{7}{5g^2}  \biggr]^{-1} =\biggl[\frac{5g^2}{5g^2-7}  \biggr] \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~b_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ 1 +  \mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
+  \mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}^{-1}
</math>
  </td>
</tr>
</table>
</div>
Hence, in order for the first derivative of both eigenfunctions to be equal ''at the interface'', we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~-  \frac{1}{b_0} \cdot \frac{14}{5}  \biggl(\frac{1}{g^2}\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{a_0} \biggl\{ c_0 + (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ 0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 -  \frac{7}{5g^2}  \biggr]\biggl\{ c_0 + (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \frac{14}{5}  \biggl(\frac{1}{g^2}\biggr)
\biggl\{ 1 +  \mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  \mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
But, we can show that,
<div align="center">
<math>\frac{1}{g} = (1+2\mathcal{D}) \, .</math>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ 0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 -  \frac{7}{5}\biggl(1+\mathcal{D}\biggr)^2  \biggr]\biggl\{ c_0 + (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \frac{7}{5}  \biggl(1+\mathcal{D}\biggr)^2
\biggl\{ 2 +  2\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  2\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ c_0 + (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \frac{7}{5}  \biggl(1+\mathcal{D}\biggr)^2
\biggl\{ (2 -c_0 ) +  2\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
- (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  2\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ c_0 + (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \frac{7}{5}  \biggl(1+\mathcal{D}\biggr)^2
\biggl\{ (2 -c_0 ) - (c_0+1)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-  (c_0 +4)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha}\biggr] \biggr\}
</math>
  </td>
</tr>
</table>
</div>
==Allow Different Adiabatic Exponents==
Let's allow the core to have a different adiabatic exponent, <math>~\gamma_c</math>, from the envelope, <math>~\gamma_e</math>; this also means that <math>~\alpha</math> will be correspondingly different in the two regions.  First, note that the ratio of the eigenfrequencies is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{3\omega_e^2}{2\pi G \gamma_e \rho_e} \biggl[\frac{2\pi G \gamma_c \rho_c}{3\omega_c^2}\biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2\alpha_e + (c_0 + 3\ell)(c_0 + 3\ell +5) }{ 2\alpha_c + 2j(2j+5) } \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~c_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm\sqrt{1 + \alpha_e} - 1 \, .</math>
  </td>
</tr>
</table>
</div>
So, if we want the ''dimensional'' eigenfrequencies to be identical, this means we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{\gamma_c \rho_c}{\gamma_e \rho_e} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2\alpha_e + (c_0 + 3\ell)(c_0 + 3\ell +5) }{ 2\alpha_c + 2j(2j+5) } \, .
</math>
  </td>
</tr>
</table>
</div>
===No Physical Solution11===
Let's try <math>~(\ell,j) = (1,1) \, .</math>  Skimming through the earlier [[#Example11|Example11 discussion, above]], it looks like the only thing we need to do in order to allow the core and envelope to have different adiabatic exponents is to explicitly add the "envelope" subscript to <math>~\alpha</math> on the right-hand side of the last expression for <math>~g^2</math>.  That is, in order to match both the value and the first derivative of the two eigenfunctions at the core/envelope interface, we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
g^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{7}{5}+
\frac{14}{5} \biggl[ \frac{\alpha_e -c_0(c_0+5)}{c_0[(c_0 + 3)(c_0+8) - \alpha_e ]-6(c_0+3)(c_0+4)}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
g^2 -1
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5}\biggl\{1 +
7\biggl[ \frac{\alpha_e -c_0(c_0+5)}{c_0[(c_0 + 3)(c_0+8) - \alpha_e ]-6(c_0+3)(c_0+4)}\biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
But, [[#Implications|as before]], the requirement that <math>~g^2 = \mathcal{B}</math> means that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 3\biggl( \frac{\rho_e}{\rho_c}\biggr)^2 - 2\biggl( \frac{\rho_e}{\rho_c}\biggr) + (g^2-1)</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{\rho_e}{\rho_c}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{3} \biggl[1 \pm \sqrt{1-3(g^2-1) }  \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Notice that the right-hand side of this expression only depends on the choice of the adiabatic exponent in the ''envelope''.  When combined with the condition imposed by setting the ''dimensional'' frequency ratio to unity, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{\gamma_c }{\gamma_e } </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{ 2\alpha_e + (c_0 + 3\ell)(c_0 + 3\ell +5) }{ 2\alpha_c + 2j(2j+5) }\biggr]\biggl( \frac{\rho_e}{\rho_c}\biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2\alpha_e + (c_0 + 3\ell)(c_0 + 3\ell +5) }{ 3[2\alpha_c + 2j(2j+5) ]}
\biggl[1 \pm \sqrt{1-3(g^2-1) }  \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~\gamma_c  3[2\alpha_c + 2j(2j+5) ]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\gamma_e [2\alpha_e + (c_0 + 3\ell)(c_0 + 3\ell +5) ]
[1 \pm \sqrt{1-3(g^2-1) } ] \, .
</math>
  </td>
</tr>
</table>
</div>
In this last expression, the left-hand side only depends on the adiabatic exponent in the core, while the right-hand side only depends on the adiabatic exponent in the envelope.
Evaluation (Tohline's Excel spreadsheet ''AdExp'' inside workbook ''AnalyticEigenvector.xlsx''):  &nbsp; Over the range, <math>~-1 \le \alpha \le 3</math>, we plotted <math>~\mathrm{LHS}(\alpha_e)</math> and <math>~\mathrm{RHS}(\alpha_c)</math> for the example case of <math>~(\ell, j) = (1,1)</math>.  There was a fairly wide range of pairings, <math>~(\alpha_c,\alpha_e)</math> for which the LHS = RHS; for example, at <math>~(-1, + 1.05)</math>, both sides give <math>~\approx 36</math>, and at <math>~(1.95, 1.95)</math>, both sides give <math>~\approx 200</math>.  But in all cases, the inferred density ratio was greater than unity.  Hence, this example index pairing does not seem to result in physically relevant core-envelope eigenvectors.
===Solution21===
====Setup====
Let's try <math>~(\ell, j) = (2,1)</math>.  Examining the [[#Example21|above, Example21 discussion]], we deduce that, in order for the first derivative of both eigenfunctions to be equal ''at the interface'', we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~-  a_0 \cdot \frac{14}{5}  \biggl(\frac{1}{g^2}\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b_0 \biggl\{ c_0 + (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha_e}\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ -  \biggl[\frac{14}{5g^2-7}  \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ 1 +  \mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr]
+  \mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha_e}\biggr] \biggr\}^{-1}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\times  \biggl\{ c_0 + (c_0 +3)\mathcal{D} \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  (c_0+6)\mathcal{D}^2 \biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr]
\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha_e}\biggr] \biggr\}
</math>
  </td>
</tr>
</table>
</div>
Notice that the core's adiabatic exponent does not explicitly enter into this condition.  Given that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathcal{D} = q^3 \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp;
and
&nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~g^2 = (1+2\mathcal{D})^{-2} = (1+2q^3)^{-2} \, ,</math>
  </td>
</tr>
</table>
</div>
we can view this expression as a conditional relationship between <math>~q</math> and <math>~\gamma_e</math>.  Once a <math>~(q,\gamma_e)</math> pair has been found that satisfies this condition, we can use the matching-frequency condition to give the corresponding, required value of <math>~\gamma_c</math>; specifically, for <math>~(\ell,j) = (2,1)</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \gamma_c[2\alpha_c + 2j(2j+5) ]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\gamma_e[2\alpha_e + (c_0 + 3\ell)(c_0 + 3\ell +5) ]\biggl(\frac{\rho_e}{\rho_c}\biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\gamma_e[2\alpha_e + (c_0 + 3\ell)(c_0 + 3\ell +5) ] \biggl[\frac{2q^3}{1+2q^3}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\gamma_c[2(3-4/\gamma_c) + 14 ]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\gamma_e [2(3 - 4/\gamma_e) + (c_0 + 6)(c_0 + 11) ] \biggl[\frac{2q^3}{1+2q^3}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
20\gamma_c-8
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\gamma_e [6  + (c_0 + 6)(c_0 + 11) ] \biggl[\frac{2q^3}{1+2q^3}\biggr] - 8\biggl[\frac{2q^3}{1+2q^3}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\gamma_c
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{20}\biggl\{
\gamma_e [6  + (c_0 + 6)(c_0 + 11) ] \biggl[\frac{2q^3}{1+2q^3}\biggr] + 8\biggl[\frac{1}{1+2q^3}\biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
====First Few Numerically Determined Model Parameters====
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="12">Example Solutions <math>~(\ell,j) = (2,1)</math></th>
</tr>
<tr>
  <th align="center" colspan="5">Envelope</th>
  <th align="center" colspan="3">Interface</th>
  <th align="center" colspan="4">Core</th>
</tr>
<tr>
  <td align="center"><math>~n_e</math></td>
  <td align="center"><math>~\gamma_e</math></td>
  <td align="center"><math>~\alpha_e</math></td>
  <td align="center"><math>~c_0</math> (plus)</td>
  <td align="center"><math>~\sigma^2_{\ell=2}|_\mathrm{env}</math></td>
  <td align="center"><math>~q^3</math></td>
  <td align="center"><math>~\frac{\rho_e}{\rho_c}</math> </td>
  <td align="center"><math>~\nu</math></td>
  <td align="center"><math>~n_c</math></td>
  <td align="center"><math>~\gamma_c</math></td>
  <td align="center"><math>~\alpha_c</math></td>
  <td align="center"><math>~\sigma^2_{j=1}|_\mathrm{core}</math></td>
</tr>
<tr>
  <td align="center"><math>~20</math></td>
  <td align="center"><math>~1.05</math></td>
  <td align="center"><math>~- 0.80956</math></td>
  <td align="center"><math>~- 0.56356</math></td>
  <td align="center">55.118</td>
  <td align="center"><math>~0.4564055</math></td>
  <td align="center"><math>~0.4772092</math></td>
  <td align="center"><math>~0.6376037</math></td>
  <td align="center"><math>~1.2805786</math></td>
  <td align="center"><math>~1.780897</math></td>
  <td align="center"><math>~+0.7539409</math></td>
  <td align="center">15.508</td>
</tr>
<tr>
  <td align="center"><math>~10</math></td>
  <td align="center"><math>~1.1</math></td>
  <td align="center"><math>~- 0.6363636</math></td>
  <td align="center"><math>~- 0.3969773</math></td>
  <td align="center">58.136</td>
  <td align="center"><math>~0.47763915</math></td>
  <td align="center"><math>~0.4885639</math></td>
  <td align="center"><math>~0.6517594</math></td>
  <td align="center"><math>~1.0393067</math></td>
  <td align="center"><math>~1.9621799</math></td>
  <td align="center"><math>~+0.9614509</math></td>
  <td align="center">15.923</td>
</tr>
<tr>
  <td align="center"><math>~5</math></td>
  <td align="center"><math>~1.2</math></td>
  <td align="center"><math>~- 1/3</math></td>
  <td align="center"><math>~- 0.1835034</math></td>
  <td align="center">62.247</td>
  <td align="center"><math>~0.50243157</math></td>
  <td align="center"><math>~0.50121284</math></td>
  <td align="center"><math>~0.6682877</math></td>
  <td align="center"><math>~0.7861924</math></td>
  <td align="center"><math>~2.2719532</math></td>
  <td align="center"><math>~+1.2394004</math></td>
  <td align="center">16.479</td>
</tr>
<tr>
  <td align="center"><math>~3</math></td>
  <td align="center"><math>~4/3</math></td>
  <td align="center"><math>~0</math></td>
  <td align="center"><math>~0</math></td>
  <td align="center">66</td>
  <td align="center"><math>~0.52185923</math></td>
  <td align="center"><math>~0.51069581</math></td>
  <td align="center"><math>~0.68123949</math></td>
  <td align="center"><math>~0.6071418</math></td>
  <td align="center"><math>~2.6470616</math></td>
  <td align="center"><math>~+1.4888905</math></td>
  <td align="center">16.978</td>
</tr>
</table>
</div>
Note that, in all cases, we find,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\sigma^2|_\mathrm{env}}{\sigma^2_\mathrm{core}} \biggl(\frac{\rho_e}{\rho_c} \biggr) \frac{\gamma_e}{\gamma_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 \, ,</math>
  </td>
</tr>
</table>
</div>
thereby demonstrating that the ratio of the ''dimensional'' frequencies is unity. 
====Special Case of 4/3 Envelope====
It is worth examining in more detail the specific case of <math>~\gamma_e = 4/3</math> because, in this case, <math>~\alpha_e</math> and <math>~c_0</math> (plus) are both zero, so the constraint equations become simpler.  For this specific case, in order for the two eigenfunctions and their first derivatives to match at the interface, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~-  \biggl[\frac{14(1+2q^3)^2 }{5-7(1+2q^3)^2}  \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ 1 +  q^3 \biggl[ \frac{-6\cdot 11}{3\cdot 5}\biggr]
+  q^6 \biggl[ \frac{- 6\cdot 11}{3\cdot 5}\biggr]
\biggl[ \frac{3\cdot 8 - 6\cdot 11}{6\cdot 8} \biggr] \biggr\}^{-1}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\times  \biggl\{ 3q^3 \biggl[ \frac{-6\cdot 11}{3\cdot 5}\biggr]
+  6q^6 \biggl[ \frac{-6\cdot 11}{3\cdot 5}\biggr]
\biggl[ \frac{3\cdot 8 - 6\cdot 11}{6\cdot 8} \biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
q^3\biggl\{ -66
+  q^3
\biggl[ \frac{6\cdot 77}{4} \biggr] \biggr\}
\times \biggl\{ 5 -22q^3
+  q^6
\biggl[ \frac{77}{4} \biggr] \biggr\}^{-1}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
66 q^3\biggl[ \frac{-4+  7q^3}{20 -88q^3 +  77q^6 } \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ -  [ 14(1+2q^3)^2] [20 -88q^3 +  77q^6]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
66 q^3[ -4+  7q^3 ] [5-7(1+2q^3)^2] \, .
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 165q^3(4-7q^3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
14(1+2q^3)^2(10+22q^3 - 77q^6) \, .
</math>
  </td>
</tr>
</table>
</div>
In the physically relevant range of the parameter, <math>~0 < q < 1</math>, the parameter value that satisfies this constraint is,
<div align="center">
<math>~q^3 = 0.52185923 \, .</math>
</div>
From the frequency-ratio constraint, therefore, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\gamma_c
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{ 2}{5}\biggl[\frac{1+24 q^3 }{1+2q^3 }\biggr] = 2.6470616 \, .</math>
  </td>
</tr>
</table>
</div>
====Roots of Quartic Equation====
TO BE DONE: &nbsp;
* In the ''specific'' case of <math>~\gamma_e = 4/3</math>, the expression governing the relevant value of <math>~q</math> can be readily viewed as a quartic equation in <math>~\Chi \equiv q^3</math>.  One of the real roots of this quartic equation should provide an analytic expression for this value of <math>~q</math>.  Indeed, it seems likely that the more ''general'' expression for <math>~q(\gamma_e)</math> can also be viewed as a quartic equation in <math>~\Chi</math> whose root is expressible analytically.
* Need to also generate a table of solutions for the case of <math>~c_0</math> (minus).
Let,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="4">&nbsp;</td>
  <td align="center">for <math>~\gamma_e = \frac{4}{3}</math>
</tr>
<tr>
  <td align="right">
<math>~A_{21}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{c_0(c_0+5) - (c_0 + 6)(c_0 + 11)}{(c_0 + 3)(c_0+5) - \alpha_e}\biggr]</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~- \frac{22}{5}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~B_{21}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(c_0+3)(c_0+8) - (c_0 + 6)(c_0 + 11)}{(c_0 + 6)(c_0+8) - \alpha_e}\biggr] </math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~\frac{-7}{8}</math>
  </td>
</tr>
</table>
</div>
Then the interface constraint equation takes the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{14(1+2\mathcal{D})^2}{7(1+2\mathcal{D})^2-5} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ c_0 + (c_0 +3)\mathcal{D} A_{21}
+  (c_0+6)\mathcal{D}^2 A_{21}
B_{21} }{  1 +  \mathcal{D} A_{21}
+  \mathcal{D}^2 A_{21}\cdot B_{21} }
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
7(1+4\mathcal{D} + 4\mathcal{D}^2) \cdot [2 +  2 A_{21}\mathcal{D} +  2A_{21}\cdot B_{21}\mathcal{D}^2 ]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[c_0 + (c_0 +3) A_{21}\mathcal{D} +  (c_0+6)A_{21}B_{21} \mathcal{D}^2 ]\cdot [7(1+4\mathcal{D} + 4\mathcal{D}^2)-5]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
-5 [c_0 + (c_0 +3) A_{21}\mathcal{D} +  (c_0+6)A_{21}B_{21} \mathcal{D}^2 ]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
7(1+4\mathcal{D} + 4\mathcal{D}^2) \biggl\{  [2 +  2 A_{21}\mathcal{D} +  2A_{21}\cdot B_{21}\mathcal{D}^2 ] - [c_0 + (c_0 +3) A_{21}\mathcal{D} +  (c_0+6)A_{21}B_{21} \mathcal{D}^2 ]\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
7(1+4\mathcal{D} + 4\mathcal{D}^2) [  (2-c_0) - (1+c_0) A_{21}\mathcal{D}  - (4+c_0)A_{21}\cdot B_{21}\mathcal{D}^2 ]
</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~5 [c_0 + (c_0 +3) A_{21}\mathcal{D} +  (c_0+6)A_{21}B_{21} \mathcal{D}^2 ] +
7(1+4\mathcal{D} + 4\mathcal{D}^2) [  (2-c_0) - (1+c_0) A_{21}\mathcal{D}  - (4+c_0)A_{21}\cdot B_{21}\mathcal{D}^2 ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[5c_0 + 5(c_0 +3) A_{21}\mathcal{D} +  5(c_0+6)A_{21}B_{21} \mathcal{D}^2 ] +
[  7(2-c_0) - 7(1+c_0) A_{21}\mathcal{D}  - 7(4+c_0)A_{21}\cdot B_{21}\mathcal{D}^2 ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ [  28(2-c_0)\mathcal{D} - 28(1+c_0) A_{21}\mathcal{D}^2  - 28(4+c_0)A_{21}\cdot B_{21}\mathcal{D}^3 ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ [  28(2-c_0) \mathcal{D}^2 - 28(1+c_0) A_{21}\mathcal{D}^3  - 28 (4+c_0)A_{21}\cdot B_{21}\mathcal{D}^4 ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathcal{D}^0[5c_0 + 7(2-c_0)  ] +
\mathcal{D}^1[5(c_0 +3) A_{21} - 7(1+c_0) A_{21} + 28(2-c_0)]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+
\mathcal{D}^2[5(c_0+6)A_{21}B_{21} - 7(4+c_0)A_{21}\cdot B_{21} - 28(1+c_0) A_{21} + 28(2-c_0) ] +
\mathcal{D}^3[- 28(4+c_0)A_{21}\cdot B_{21} - 28(1+c_0) A_{21} ] +
\mathcal{D}^4[- 28 (4+c_0)A_{21}\cdot B_{21}  ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathcal{D}^0[ 14- 2c_0  ] +
\mathcal{D}^1[28(2-c_0) + (8-2c_0) A_{21} ] +
\mathcal{D}^2[28(2-c_0) - 28(1+c_0) A_{21} + 2(1 - c_0)A_{21}B_{21}  ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+
\mathcal{D}^3[(1+c_0) + (4+c_0)B_{21} ] (-28A_{21}) +
\mathcal{D}^4[- 28 (4+c_0)A_{21}\cdot B_{21}  ] \, .
</math>
  </td>
</tr>
</table>
</div>
To solve this equation analytically, we follow the [http://en.wikipedia.org/wiki/Quartic_function#Summary_of_Ferrari.27s_method Summary of Ferrari's method] that is presented in Wikipedia's discussion of the Quartic Function to identify the roots of an arbitrary quartic equation. 
<div align="center">
<table border="1" cellpadding="8" width="85%">
<tr><td align="left">
First, we adopt the shorthand notation:
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a\Chi^4 + b\Chi^3 + c\Chi^2 +d\Chi +e \, ,</math>
  </td>
</tr>
</table>
where, in our particular case,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="4">&nbsp;</td>
  <td align="center">for <math>~\gamma_e = \frac{4}{3}</math>
</tr>
<tr>
  <td align="right">
<math>~e</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ 14- 2c_0  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~14</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~d</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~28(2-c_0) + (8-2c_0) A_{21} \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~\frac{104}{5}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~c</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~28(2-c_0) - 28(1+c_0) A_{21} + 2(1 - c_0)A_{21}B_{21}  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~\frac{1869}{10} </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-28A_{21}[(1+c_0) + (4+c_0)B_{21} ]  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~- 308</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~a</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~- 28 (4+c_0)A_{21}\cdot B_{21}  \, .</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~- \frac{2156}{5}</math>
  </td>
</tr>
</table>
Now, define,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="4">&nbsp;</td>
  <td align="center">for <math>~\gamma_e = \frac{4}{3}</math>
</tr>
<tr>
  <td align="right">
<math>~\Delta_0</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ c^2 - 3bd + 12ae  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~-1.829079\times 10^{4}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Delta_1</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ 2c^3 - 9bcd + 27b^2e + 27ad^2 - 72ace  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~1.358913\times 10^{8}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~p</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \frac{8ac - 3b^2}{8a^2}  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~-0.6247681</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~q</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \frac{b^3 - 4abc + 8a^2 d}{8a^3}  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0.1521170</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~Q</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2^{1 / 3}} \biggl[\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}  \biggr]^{1 / 3} \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~5.141760\times 10^{2}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~S</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2} \biggl[ - \frac{2p}{3} + \frac{1}{3a}\biggl(Q + \frac{\Delta_0}{Q} \biggr)  \biggr]^{1 / 2} \, .</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0.1078593</math>
  </td>
</tr>
</table>
Then the four roots of the quartic equation are,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="4">&nbsp;</td>
  <td align="center">for <math>~\gamma_e = \frac{4}{3}</math>
</tr>
<tr>
  <td align="right">
<math>~\Chi_{1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{b}{4a} - S + \frac{1}{2}\biggl[ -4S^2 - 2p + \frac{q}{S} \biggr]^{1 / 2}  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0.5218592</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Chi_{2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{b}{4a} - S - \frac{1}{2}\biggl[ -4S^2 - 2p + \frac{q}{S} \biggr]^{1 / 2}  \, ,</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>~-1.094721</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Chi_{3,4}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{b}{4a} + S \pm \frac{1}{2}\biggl[ -4S^2 - 2p - \frac{q}{S} \biggr]^{1 / 2}  \, .</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; &hellip; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
imaginary
  </td>
</tr>
</table>
The <math>~\Chi_1</math> root is the physically relevant one, and it matches the interface value of <math>~\mathcal{D} = q^3</math> associated with <math>~\gamma_e = 4/3</math> in our [[#First_Few_Numerically_Determined_Model_Parameters|above table of example solutions]].  Excellent!
</td></tr>
</table>
</div>
In summary then, for any choice of the envelope's adiabatic exponent, <math>~\gamma_e</math>, the physically relevant root of the quartic equation gives us the interface location of the model for which our analytically specified eigenvector applies; specifically,
<div align="center">
<math>~q = \Chi_1^{1/3} \, .</math>
</div>
From this value, we also know that,
<div align="center">
<math>~
g = \frac{1}{(1+2q^3)} \, ;
</math>
&nbsp; &nbsp; &nbsp;
<math>~
\frac{\rho_e}{\rho_c} = \frac{2q^3}{(1+2q^3)} \, ;
</math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
<math>~
\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}} = \biggl[3\biggr(1 - \frac{\rho_e}{\rho_c}\biggr)\biggr]^{-1} \, .
</math>
&nbsp; &nbsp; &nbsp;
</div>
===Combined Eigenfunction===
[[#Example21|From above]], we know that the eigenfunction for the core <math>~(0 \le \xi \le 1)</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_{j=1} |_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0\biggl[1 -  \frac{7}{5}  \biggl(\frac{\xi}{g}\biggr)^2 \biggr] =
\frac{5 -  7 (1+2q^3)^2 \xi^2}{5-7(1+2q^3)^2} \, .</math>
  </td>
</tr>
</table>
</div>
And the matching eigenfunction for the envelope <math>~(1 \le \xi \le q^{-1})</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x_{\ell=2} |_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^{c_0}\biggl[ \frac{ 1 +  q^3 A_{21} \xi^{3} +  q^6 A_{21}B_{21}\xi^{6} }{ 1 +  q^3 A_{21}  +  q^6 A_{21}B_{21}}\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>


=Related Discussions=
=Related Discussions=

Revision as of 01:36, 15 December 2016

Radial Oscillations of a Zero-Zero Bipolytrope

This is a chapter that summarizes an accompanying, detailed derivation.

Whitworth's (1981) Isothermal Free-Energy Surface
|   Tiled Menu   |   Tables of Content   |  Banner Video   |  Tohline Home Page   |

Background

Related Discussions

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
|   H_Book Home   |   YouTube   |
Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS |
Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation

Retrieved from "https://www.vistrails.org//index.php?title=User:Tohline/SSC/Stability/BiPolytrope0_0&oldid=12877"