Editing SSC/Stability/BiPolytropes/RedGiantToPN/Pt4

__FORCETOC__
=Main Sequence to Red Giant to Planetary Nebula=
<table border="1" align="center" width="100%" colspan="8">
<tr>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/BiPolytropes/RedGiantToPN|Part I:&nbsp; Background &amp; Objective]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/BiPolytropes/RedGiantToPN/Pt2|Part II:&nbsp; ]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/BiPolytropes/RedGiantToPN/Pt3|Part III:&nbsp; ]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue"><br />[[SSC/Stability/BiPolytropes/RedGiantToPN/Pt4|Part IV:&nbsp;]]
&nbsp;
  </td>
</tr>
</table>

==Succinct==

===Generic===
<div align="center" id="2ndOrderODE">
<font color="#770000">'''Adiabatic Wave''' (or ''Radial Pulsation'') '''Equation'''</font><br />

{{Math/EQ_RadialPulsation01}}
</div>

may also be written as &hellip;

<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{d^2x}{dr*^2} + \biggl\{ 4 -\biggl(\frac{\rho^*}{P^*}\biggr)\frac{ M_r^*}{(r^*)}\biggr\}\frac{1}{r^*} \frac{dx}{dr*} 
+ \biggl(\frac{\rho^*}{ P^* } \biggr)\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  
- \biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr)\frac{ M_r^*}{(r^*)^3}\biggr\}  x \, .
</math>
  </td>
</tr>
</table>

In shorthand, we can rewrite this equation in the form,

<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>~
x'' + \frac{\mathcal{H}}{r^*} x' + \mathcal{K}x \, ,
</math>
  </td>
</tr>
</table>
where,
<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>~\frac{dx}{dr^*}</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~x''</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d^2x}{d(r^*)^2} \, ;</math>
  </td>
</tr>
</table>
and,
<div align="center">
<math>~\mathcal{K} \equiv ~\biggl(\frac{\rho^*}{ P^* } \biggr)\biggl[ \biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr)\frac{2\pi }{3} 
- \biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \frac{M_r^*}{(r^*)^3} \biggr] \, ;</math>
</div>
and,
<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>~
\biggl\{ 4 -\biggl(\frac{\rho^*}{P^*}\biggr)\frac{ M_r^*}{(r^*)}\biggr\}\, .
</math>
  </td>
</tr>
</table>

===Specific Polytropes===

In a [[SSC/Stability/Polytropes#Adiabatic_(Polytropic)_Wave_Equation|separate discussion]], we have shown that configurations with a polytropic equation of state exhibit a characteristic length scale that is given by the expression,

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

<tr>
  <td align="right">
<math>a_n</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\biggl[\frac{(n+1)K}{4\pi G} \cdot \rho_c^{(1-n)/n} \biggr]^{1/2} \, ;</math>
  </td>
</tr>
</table>
and, once the dimensionless polytropic temperature, <math>\theta(\xi)</math>, is known, the radial dependence of key physical variables is given by the expressions,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3">&nbsp;</td>
  <td align="center" bgcolor="lightgray" rowspan="8">&nbsp;&nbsp;</td>
  <td align="center" colspan="3">
if, as [[SSC/Stability/BiPolytropes/Pt3#Foundation|in a separate discussion]], <math>n=5</math> and <math>\theta = (1+\xi^2/3)^{-1 / 2}</math> &hellip;
  </td>
</tr>

<tr>
  <td align="right">
<math>r_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>a_n \xi \, ,</math>
  </td>

  <td align="right">
&nbsp; &nbsp; &nbsp;<math>r_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[\frac{K_5}{G} \cdot \rho_c^{-4/5} \biggr]^{1/2} \biggl(\frac{3}{2\pi}\biggr)^{1 / 2} \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 \theta^{n} \, ,</math>
  </td>

  <td align="right">
&nbsp; &nbsp; &nbsp;<math>\rho_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\rho_c \theta^{5} \, ,</math>
  </td>

</tr>

<tr>
  <td align="right">
<math>~P_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~K\rho_0^{(n+1)/n} = K\rho_c^{(n+1)/n} \theta^{n+1} \, ,</math>
  </td>

  <td align="right">
&nbsp; &nbsp; &nbsp;<math>P_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>K_5\rho_c^{6/5} \theta^{6} \, ,</math>
  </td>

</tr>

<tr>
  <td align="right">
<math>M(r_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>4\pi \rho_c a_n^3 \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr)
=
\rho_c^{(3-n)/(2n)} \biggl[\frac{(n+1)^3 K^3}{4\pi G^3}  \biggr]^{1/2} \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr)
\, ,</math>
  </td>

  <td align="right">
&nbsp; &nbsp; &nbsp;<math>M(r_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{ K_5^3}{ G^3}\cdot \rho_c^{-2/5}  \biggr]^{1/2}
\biggl(\frac{2\cdot 3^3 }{\pi }  \biggr)^{1/2} 
\biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr)
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~g_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{GM(r_0)}{r_0^2} = \frac{G}{r_0^2} \biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
\, ,</math>
  </td>

  <td align="right">
&nbsp; &nbsp; &nbsp;<math>g_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl\{~\biggl[\frac{K_5}{G} \cdot \rho_c^{-4/5} \biggr]^{1/2} \biggl(\frac{3}{2\pi}\biggr)^{1 / 2} \xi ~\biggr\}^{-2}
\biggl[\frac{6K_5}{4\pi G} \cdot \rho_c^{-4/5} \biggr]^{3/2}\biggl[4\pi  G \rho_c \biggr]
\biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
</math>
  </td>

</tr>

<tr>
  <td align="center" colspan="3">
&nbsp;
  </td>

  <td align="right">
&nbsp; 
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{2^3\pi^2}{3}\biggr) 
\biggl[\frac{6}{4\pi} \biggr]^{3/2}
\biggl[\frac{G}{K_5} \cdot \rho_c^{4/5} \biggr] \biggl[\frac{K_5}{G} \cdot \rho_c^{-4/5} \biggr]^{3/2}
\biggl[ G \rho_c \biggr]
\biggl(-~ \frac{d\theta}{d\xi}\biggr) \biggr]
</math>
  </td>

</tr>

<tr>
  <td align="center" colspan="3">
&nbsp;
  </td>

  <td align="right">
&nbsp; 
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
(2^3\cdot 3 \pi)^{1 / 2}
\biggl[K_5 G \cdot \rho_c^{6/5} \biggr]^{1/2}
\biggl(-~ \frac{d\theta}{d\xi}\biggr) \biggr]
\, .</math>
  </td>

</tr>
</table>

Combining variable expressions from the above right-hand column, we find that for <math>n=5</math> polytropes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{g_0 \rho_0 r_0}{P_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(2^3\cdot 3 \pi)^{1 / 2}\biggl[K_5 G \cdot \rho_c^{6/5} \biggr]^{1/2}\biggl(-~ \frac{d\theta}{d\xi}\biggr) \biggr]
\cdot \rho_c \theta^5
\cdot \biggl[\frac{K_5}{G} \cdot \rho_c^{-4/5} \biggr]^{1/2} \biggl(\frac{3}{2\pi}\biggr)^{1 / 2} \xi
~\biggl[K_5\rho_c^{6/5} \theta^{6}  \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
6~\biggl(-~\frac{\xi}{\theta} \frac{d\theta}{d\xi}\biggr) \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="5" width="80%"><tr><td align="left">
More generally, combining variable  expressions from the above left-hand column, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{g_0 \rho_0 r_0}{P_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{G}{a_n^2 \xi^2} \biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
\cdot \rho_c \theta^{n} \cdot a_n \xi \cdot \biggl[ K\rho_c^{(n+1)/n} \theta^{n+1} \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{4\pi G }{K} \biggl[ \rho_c^{1- 1/n} \biggr] \biggl(-\xi \frac{d\theta}{d\xi}\biggr) 
\cdot \theta^{-1} \cdot a_n^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(n+1)\biggl(- \frac{\xi}{\theta} \cdot \frac{d\theta}{d\xi}\biggr) 
\, ;  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\rho_0 r_0^2}{P_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\rho_c \theta^{n} \cdot \biggl(a_n \xi\biggr)^2 \cdot \biggl[ K\rho_c^{(n+1)/n} \theta^{n+1} \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
K^{-1}\rho_c^{-1/n}  \cdot a_n^2 \cdot \frac{\xi^2}{\theta}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{(n+1)}{4\pi G \rho_c} \biggr] \cdot \frac{\xi^2}{\theta}
\, .
</math>
  </td>
</tr>
</table>

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

As a result, for polytropes we can write,
<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{d^2x}{dr_0^2} + \biggl[4 - \biggl(\frac{g_0 \rho_0 r_0}{P_0}\biggr) \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
+ \biggl(\frac{\rho_0 r_0^2}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  \frac{x}{r_0^2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x}{dr_0^2} + \biggl[4 - \biggl(\frac{g_0 \rho_0 r_0}{P_0}\biggr) \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
+ \biggl[\frac{\omega^2}{\gamma_\mathrm{g}}\biggl(\frac{\rho_0 r_0^2}{ P_0} \biggr) 
- \biggl(3 - \frac{4}{\gamma_\mathrm{g}} \biggr)\biggl(\frac{g_0 \rho_0 r_0}{P_0}\biggr)  \biggr]  \frac{x}{r_0^2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x}{dr_0^2} + \biggl[4 - (n+1) Q \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
+ (n+1)\biggl[\frac{\omega^2}{\gamma_\mathrm{g}}\biggl[\frac{1}{4\pi G \rho_c} \biggr] \cdot \frac{\xi^2}{\theta} 
- \biggl(3 - \frac{4}{\gamma_\mathrm{g}} \biggr)Q \biggr]  \frac{x}{r_0^2} 
\, . 
</math>
  </td>
</tr>
</table>
Finally, multiplying through by <math>a_n^2</math> &#8212; which everywhere converts <math>r_0</math> to <math>\xi</math> &#8212; gives, what we will refer to as the,

<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>

==BiPolytrope==

Let's stick with the dimensional <math>(r_0)</math> version and set <math>\omega^2 = 0</math>, in which case the Polytropic LAWE is,
<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{d^2x}{dr_0^2} + \biggl[4 - (n+1) Q \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
- \biggl[(n+1)\alpha Q \biggr]  \frac{x}{r_0^2} 
\, . 
</math>
  </td>
</tr>
</table>

===Core (n = 5)===
For the <math>n=5</math> core, we know that <math>\theta = (1 + \xi^2/3)^{-1 / 2}</math>.  Hence,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~ Q_5 \equiv - \frac{d\ln \theta}{d\ln\xi} = - \frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\xi}{3}\cdot \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2} \biggr]\cdot \xi \biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\xi^2}{3+\xi^2}\biggr)  \, .
</math>
  </td>
</tr>
</table>
Now, given that,

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

<tr>
  <td align="right">
<math>a_5^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{3}{2\pi}\biggl[K_5 G^{-1} \rho_c^{-4/5} \biggr] \, ,</math>
  </td>
</tr>
</table>
we can everywhere make the substitution,

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

<tr>
  <td align="right">
<math>\xi^2</math>
  </td>
  <td align="center">
<math>~~\rightarrow ~~</math>
  </td>
  <td align="left">
<math>\biggl(\frac{r_0}{a_5}\biggr)^2
=
\frac{2\pi}{3}\biggl[K_5^{-1} G \rho_c^{4/5} \biggr]r_0^2 \, .</math>
  </td>
</tr>
</table>
Note, also, that throughout the core, the relevant LAWE is,
<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{d^2x}{dr_0^2} + \biggl[4 - 6 Q_5 \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
- \biggl[6\alpha Q_5 \biggr]  \frac{x}{r_0^2} 
\, . 
</math>
  </td>
</tr>
</table>

Next, try the solution, 
<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>
\biggl[1 - \frac{r_0^2}{15a_5^2} \biggr]
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2r_0}{15a_5^2}
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{d^2x}{dr_0^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2}{15a_5^2}
\, ,
</math>
  </td>
</tr>
</table>

in which case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\mathrm{LAWE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x}{dr_0^2} + \biggl[4 - 6 Q_5 \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
- \biggl[6\alpha Q_5 \biggr]  \frac{x}{r_0^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2}{15a_5^2}
+ \biggl[4 - 6 Q_5 \biggr] \biggl[-~\frac{2}{15a_5^2} \biggr] 
- \biggl[6\alpha Q_5 \biggr]  \frac{1}{r_0^2} \cdot \biggl[1 - \frac{r_0^2}{15a_5^2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2}{15a_5^2}
+ \biggl[4 - 6 Q_5 \biggr] \biggl[-~\frac{2}{15a_5^2} \biggr] 
- \biggl[6\alpha Q_5 \biggr]  \frac{1}{r_0^2 } \cdot \biggl[\frac{15a_5^2 - r_0^2}{15a_5^2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2}{15a_5^2}
+ \biggl[4 - 6 Q_5 \biggr] \biggl[-~\frac{2}{15a_5^2} \biggr] 
- \biggl[6\alpha Q_5 \biggr]  \frac{1}{15a_5^2 } \cdot \biggl[\frac{15 - \xi^2 }{\xi^2 } \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 15a_5^2 ~\times ~\mathrm{LAWE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-10 + 12 Q_5 - \biggl[6\alpha Q_5 \biggr]  \cdot \biggl[\frac{15 - \xi^2 }{\xi^2 } \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-10 + 12 \biggl( \frac{\xi^2}{3+\xi^2}\biggr) - \biggl[6\alpha \biggl( \frac{\xi^2}{3+\xi^2}\biggr) \biggr]  
\cdot \biggl[\frac{15 - \xi^2 }{\xi^2 } \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-10(3+\xi^2) + 12 \xi^2 - \biggl[6\alpha \biggl( \frac{\xi^2}{3+\xi^2}\biggr) \biggr]  
\cdot \biggl[\frac{15 - \xi^2 }{\xi^2 } \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 15a_5^2(3+\xi^2) ~\times ~\mathrm{LAWE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-30+ 2 \xi^2 
- 6\alpha (15 - \xi^2) \, .
</math>
  </td>
</tr>
</table>
Setting <math>\alpha = -1/3</math> gives the desired result, namely,

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

<tr>
  <td align="right">
<math>\mathrm{LAWE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
0 \, .
</math>
  </td>
</tr>
</table>

===Envelope (n = 1)===
From the variable expressions in the right-hand column of [[SSC/Structure/BiPolytropes/Analytic51#Step_8:_Throughout_the_envelope_(ηi_≤_η_≤_ηs)|Step 8 of the construction chapter]],
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{g_0 \rho_0 r_0}{P_0} = \frac{GM_r}{r_0^2} \cdot \frac{\rho_0 r_0}{P_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
G~\biggl\{ \biggl[ \frac{K_5^3}{G^3 \rho_c^{2/5}} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)\biggr\}\cdot
\biggl\{ \biggl[ \frac{K_5}{G \rho_c^{4/5}} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggr\}^{-1}
\cdot \biggl\{\rho_c \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i \phi\biggr\}
\cdot \biggl\{ K_5 \rho_c^{6/5}  \theta^{6}_i \phi^{2} \biggr\}^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)\biggr\}\cdot
\biggl\{ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{2}_i (2\pi)^{1/2}\eta^{-1} \biggr\}
\cdot \biggl\{\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i \phi\biggr\}
\cdot \biggl\{ \theta^{-6}_i \phi^{-2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2~ \biggl(-\frac{\eta}{\phi} \cdot \frac{d\phi}{d\eta} \biggr) \, .
</math>
  </td>
</tr>
</table>

For the <math>n=1</math> envelope, we know [[SSC/Structure/BiPolytropes/Analytic51#Step_6:_Envelope_Solution|from separate work]] that,
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~ Q_1 \equiv - \frac{d\ln \phi}{d\ln\eta} = - \frac{\eta}{\phi}\cdot \frac{d\phi}{d\eta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- ~\frac{\eta}{A}\biggl[ \frac{\eta}{\sin(\eta - B)} \biggr] \cdot\frac{A}{\eta^2} \biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 -  \eta\cot(\eta-B) \biggr] \, .
</math>
  </td>
</tr>
</table>
<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{d^2x}{dr_0^2} + \biggl[4 - (n+1) Q_1 \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
- \biggl[(n+1)\alpha Q_1 \biggr]  \frac{x}{r_0^2} 
\, . 
</math>
  </td>
</tr>
</table>

==Numerical Integration Through Envelope==

===Finite-Difference Expressions===

The discussion  in this subsection is guided by our [[SSC/Stability/Polytropes/Pt3#Numerical_Integration_from_the_Center,_Outward|previous attempt at numerical integration]].

Here, we focus on the LAWE that is relevant to the envelope, namely,
<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{d^2x}{dr_0^2} + \biggl[4 - (n+1) Q_1 \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
- \biggl[(n+1)\alpha Q_1 \biggr]  \frac{x}{r_0^2} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x}{dr_0^2} + \biggl[4 - 2 Q_1 \biggr] \frac{1}{r_0} \cdot\frac{dx}{dr_0} 
- \biggl[2 Q_1 \biggr]  \frac{x}{r_0^2} 
\, ,
</math>
  </td>
</tr>
</table>
<span id="FD">where we have plugged</span> in the values, <math>(n,\alpha) = (1, 1)</math>.  Using the [[Appendix/Ramblings/NumericallyDeterminedEigenvectors#General_Approach|general finite-difference approach described separately]], we make the substitutions,

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

<tr>
  <td align="right">
<math>\biggl[\frac{dx}{dr_0}\biggr]_i</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
\frac{x_+ - x_-}{2 \Delta_r}   \, ;
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and,&nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>\biggl[\frac{d^2x}{dr_0^2}\biggr]_i</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\frac{x_+ - 2x_i + x_-}{\Delta_r^2} \, ;</math>
  </td>
</tr>
</table>
which will provide an approximate expression for <math>x_+ \equiv x_{i+1}</math>, given the values of <math>x_- \equiv x_{i-1}</math> and <math>x_i</math>.

<font color="orange"><b>A:</b></font> &nbsp; &nbsp; Pick <math>\xi_\mathrm{int}</math>; this will give analytic expressions for <math>\eta_\mathrm{int}</math>, <math>B</math>, and for <math>\eta_\mathrm{surf}</math>, as well as analytic expressions for <math>(r_0)_\mathrm{int}</math> and <math>(r_0)_\mathrm{surf}</math>.

<font color="orange"><b>B:</b></font> &nbsp; &nbsp; Divide the radial coordinate grid into 99 spherical shells <math>\Rightarrow~ \Delta_r = [(r_0)_\mathrm{surf} - (r_0)_\mathrm{int}]/99.</math>  Then tabulate 100 values of <math>(r_0)_i, \eta_i, (Q_1)_i = [1 -  \eta\cot(\eta-B) ]_i</math>.

Generally speaking, after multiplying through by <math>r_0^2</math>, the finite-difference  representation of the envelope's LAWE takes the form,
<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>
r_0^2\biggl[\frac{x_+ - 2x_i + x_-}{\Delta_r^2}\biggr] 
+ \biggl[4 - 2 Q_1 \biggr] r_0 \biggl[\frac{x_+ - x_-}{2 \Delta_r}\biggr] 
- \biggl[2 Q_1 \biggr]  x_i 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x_+ \biggl\{ \frac{r_0^2}{\Delta_r^2} + (4-2Q_1)\frac{r_0}{2 \Delta_r} \biggr\}
+
x_i \biggl\{- \frac{2r_0^2}{\Delta_r^2} - 2Q_1  \biggr\}
+
x_- \biggl\{ \frac{r_0^2}{\Delta_r^2} - (4-2Q_1) \frac{r_0}{2 \Delta_r} \biggr\}
</math>
  </td>
</tr>
</table>
Multiplying through by <math>(\Delta_r^2/r_0^2)</math> and solving for <math>x_+</math> gives,

<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>
x_+ \biggl\{ 1 + (4-2Q_1)\frac{\Delta_r}{2r_0} \biggr\}
-
2x_i \biggl\{1 + Q_1\biggl( \frac{\Delta_r^2}{r_0^2} \biggr)  \biggr\}
+
x_- \biggl\{ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ x_+ \biggl\{ 1 + (4-2Q_1)\frac{\Delta_r}{2r_0} \biggr\}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2x_i \biggl\{1 + Q_1\biggl( \frac{\Delta_r^2}{r_0^2} \biggr)  \biggr\}
-
x_- \biggl\{ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ x_+ </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{~
2x_i \biggl[1 + Q_1\biggl( \frac{\Delta_r^2}{r_0^2} \biggr)  \biggr]
-
x_- \biggl[ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr]
~\biggr\}~\biggl[ 1 + (4-2Q_1)\frac{\Delta_r}{2r_0} \biggr]^{-1} \, .
</math>
  </td>
</tr>
</table>
Now, at the interface &#8212; as viewed from the perspective of both the core and the envelope &#8212; we know the value of <math>x_i =x_\mathrm{int}</math>, but we don't know the value of <math>x_-</math> as viewed from the envelope.  However &#8212; [[#STEPS|see <font color="maroon">STEP #4</font> below]] &#8212; we know analytically the value of the first derivative at the interface as viewed from the perspective of the envelope, namely, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[ \frac{dx}{dr_0}\biggr]_\mathrm{int}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{x_\mathrm{int}}{r_0} \cdot \biggl\{ \frac{d\ln x}{d\ln r_0}\biggr|_\mathrm{int} \biggr\}_\mathrm{env}</math>
  </td>
</tr>
</table>
Therefore, from the [[#FD|above-specified finite-difference representation]] of the first derivative, we deduce that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x_-</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x_+ - 2\Delta_r \cdot \biggl[ \frac{dx}{dr_0}\biggr]_\mathrm{int}
</math>
  </td>
</tr>
</table>
Hence, at the interface &#8212; and only ''at'' the interface &#8212; the finite-difference representation of the envelope's LAWE can be written as,

<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>
x_+ \biggl[ 1 + (4-2Q_1)\frac{\Delta_r}{2r_0} \biggr]
-
2x_i \biggl[1 + Q_1\biggl( \frac{\Delta_r^2}{r_0^2} \biggr)  \biggr]
+
\biggl\{x_+ - 2\Delta_r \cdot \biggl[ \frac{dx}{dr_0}\biggr]_\mathrm{int} \biggr\}\cdot \biggl[ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x_+ \biggl[ 1 + (4-2Q_1)\frac{\Delta_r}{2r_0} \biggr]
-
2x_i \biggl[1 + Q_1\biggl( \frac{\Delta_r^2}{r_0^2} \biggr)  \biggr]
+
x_+  \biggl[ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr]
- 2\Delta_r \cdot \biggl[ \frac{dx}{dr_0}\biggr]_\mathrm{int} \cdot \biggl[ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
x_+ \biggl[ 1 + (4-2Q_1)\frac{\Delta_r}{2r_0} \biggr]
+
x_+  \biggl[ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr]
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2x_i \biggl[1 + Q_1\biggl( \frac{\Delta_r^2}{r_0^2} \biggr)  \biggr]
+ 2\Delta_r \cdot \biggl[ \frac{dx}{dr_0}\biggr]_\mathrm{int} \cdot \biggl[ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
x_+ 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x_i \biggl[1 + Q_1\biggl( \frac{\Delta_r^2}{r_0^2} \biggr)  \biggr]
+ \Delta_r \cdot \biggl[ \frac{dx}{dr_0}\biggr]_\mathrm{int} \cdot \biggl[ 1 - (4-2Q_1) \frac{\Delta_r}{2r_0} \biggr]
\, .
</math>
  </td>
</tr>
</table>

===Steps===
<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">
<font color="maroon">STEP 1:</font>&nbsp;&nbsp;Specify the interface location from the perspective of the core; that is, specify <math>\xi_\mathrm{int}</math>, in which case,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>(r_0)_\mathrm{int} = a_5\cdot \xi_\mathrm{int}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl[ K_5 G^{-1}\rho_c^{-4/5} \biggr]^{1 / 2}\biggl(\frac{3}{2\pi}\biggr)^{1 / 2} \xi_\mathrm{int} \, .
</math>
  </td>
</tr>
</table>

<font color="maroon">STEP 2:</font>&nbsp; &nbsp;Adopting the normalization <math>\phi_\mathrm{int} = 1</math>, determine numerous additional [[SSC/Structure/BiPolytropes/Analytic51#Step_6:_Envelope_Solution|equilibrium properties]] at the interface, such as &hellip;
<table border="0" align="center" cellpadding="8" width="80%">
<tr><td align="center" colspan="4">
<table border="1" align="center" cellpadding="8"><tr><td align="center"><font color="darkgreen">Example numerical values inside parentheses assume <math>(\mu_e/\mu_c) = 1</math> and <math>\xi_\mathrm{int} = 1.668646016</math><br /><math>\Rightarrow~~~(r_0)_\mathrm{int}[ K_5^{-1} G\rho_c^{4/5} ]^{1 / 2} = 1.153014872 \, .</math></td></tr></table>
  </td>
</tr>

<tr>
  <td align="right">
<math>\theta_\mathrm{int}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl[1 + \frac{\xi^2_\mathrm{int}}{3}\biggr]^{-1 / 2} \, ;
</math>
  </td>
  <td align="right">(0.720165375)</td>
</tr>

<tr>
  <td align="right">
<math>\biggl( \frac{d\theta}{d\xi} \biggr)_\mathrm{int}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
-~\frac{\xi_\mathrm{int}}{3}\biggl[1 + \frac{\xi^2_\mathrm{int}}{3}\biggr]^{-3 / 2} \, ;
</math>
  </td>
  <td align="right">(- 0.207749350)</td>
</tr>

<tr>
  <td align="right">
<math>\eta_\mathrm{int}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
3^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2_\mathrm{int}\cdot \xi_\mathrm{int} \, ;
</math>
  </td>
  <td align="right">(1.498957494)</td>
</tr>

<tr>
  <td align="right">
<math>\biggl( \frac{d\phi}{d\eta} \biggr)_\mathrm{int}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
3^{1 / 2} \theta^{-3}_\mathrm{int}\cdot \biggl( \frac{d\theta}{d\xi} \biggr)_\mathrm{int} \, ;
</math>
  </td>
  <td align="right">(- 0.963393227)</td>
</tr>

<tr>
  <td align="right">
<math>\Lambda_\mathrm{int}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{1}{\eta_\mathrm{int}} + \biggl( \frac{d\phi}{d\eta} \biggr)_\mathrm{int} \, ;
</math>
  </td>
  <td align="right">(- 0.296262902)</td>
</tr>

<tr>
  <td align="right">
<math>A</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\eta_\mathrm{int}(1 + \Lambda^2_\mathrm{int})^{1 / 2} \, ;
</math>
  </td>
  <td align="right">(1.563357124)</td>
</tr>

<tr>
  <td align="right">
<math>B</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\eta_\mathrm{int} - \frac{\pi}{2} + \tan^{-1}(\Lambda_\mathrm{int}) \, .
</math>
  </td>
  <td align="right">(- 0.359863580)</td>
</tr>

<tr>
  <td align="right">
<math>\eta_\mathrm{surf}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
B + \pi \, .
</math>
  </td>
  <td align="right">(2.781729074)</td>
</tr>
</table>

<font color="maroon">STEP 3:</font>&nbsp;&nbsp; Throughout the core &#8212; that is, at all radial positions, <math>0 \le r_0 \le (r_0)_\mathrm{int}</math> &#8212; the displacement amplitude and its first and second derivatives, respectively, are given by the expressions,
<table border="0" cellpadding="5" align="center" width="80%">
<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 - \frac{r_0^2}{15a_5^2} \biggr] = \biggl[1 - \frac{\xi^2}{15} \biggr]
\, ;
</math>
  </td>
  <td align="right">(0.814374698)</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~r_0\cdot \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2r_0^2}{15a_5^2} = -~\frac{2\xi^2}{15}
\, ;
</math>
  </td>
  <td align="right">(- 0.371250604)</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ r_0^2 \cdot \frac{d^2x}{dr_0^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2r_0^2}{15a_5^2} = -~\frac{2\xi^2}{15}
\, ;
</math>
  </td>
  <td align="right">(- 0.371250604)</td>
</tr>

<tr>
  <td align="right">
also &hellip; &nbsp; &nbsp; 
<math>
\biggl\{ \frac{d\ln x}{d\ln \xi} \biggr\}_\mathrm{core}
= 
\biggl\{ \frac{d\ln x}{d\ln r_0} \biggr\}_\mathrm{core} 
= \frac{r_0}{x} \cdot \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{15}{15 - \xi^2} \biggr] \cdot \biggl[-~\frac{2\xi^2}{15 }\biggr]
= 
\biggl[\frac{2\xi^2}{\xi^2 - 15} \biggr]
\, .
</math>
  </td>
  <td align="right">(-0.455871977)<sup>&dagger;</sup></td>
</tr>
</table>

<font color="maroon">STEP #4:</font>&nbsp; &nbsp; From the determination of the logarithmic slope of the displacement function at the edge of the core &#8212; <i>i.e.,</i> at the core-envelope interface &#8212; determine the slope as viewed from the perspective of the envelope.
<table border="0" cellpadding="5" align="center" width="80%">

<tr>
  <td align="right">
<math>~
\biggl\{ \frac{d\ln x}{d\ln r_0}\biggr|_\mathrm{int} \biggr\}_\mathrm{env}
=
\biggl\{ \frac{d\ln x}{d\ln \eta}\biggr|_\mathrm{int} \biggr\}_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr) + \frac{\gamma_c}{\gamma_e}
\biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_\mathrm{int} \biggr\}_\mathrm{core} \, .</math>
  </td>
  <td align="right">(-1.473523186)<sup>&dagger;</sup></td>
</tr>
</table>

----

<sup>&dagger;</sup>This analytically determined value matches the [[SSC/Stability/BiPolytropes/Pt3#Eigenfunction_Details|previous determination]] that was obtained via numerical integration of the LAWE.
</td></tr></table>

[[SSC/Structure/BiPolytropes/Analytic51#Step_8:_Throughout_the_envelope_(ηi_≤_η_≤_ηs)|Throughout the envelope]] &#8212; that is, over the range, <math>(\eta_\mathrm{int} \le \eta \le \eta_\mathrm{surf})</math> &#8212; the radial coordinate, <math>r_0</math>, is a linear function of <math>\eta</math> and takes on values given by the expression,
<table border="0" cellpadding="5" align="center" width="80%">

<tr>
  <td align="right">
<math>
r_0 [K_5^{-1} G \rho_c^{4/5}]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1} \theta^{-2}_\mathrm{int} (2\pi)^{-1 / 2} \biggr]\cdot \eta
</math>
  </td>
  <td align="right">(0.769211186 &times; &eta;)</td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~1.153014872
</math>
  </td>
  <td align="center">
<math>\leq r_0 \leq</math>
  </td>
  <td align="left">
<math>2.139737121
\, .
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>
</table>



[[SSC/Stability/BiPolytropes/Pt3#Eigenfunction_Details|From our earlier discussions]],


Here we examine some of the properties of the fundamental-mode eigenfunctions that we have found are associated with marginally unstable, <math>~(n_c, n_e) = (5,1)</math> bipolytropes.
<table border="0" align="right" width="40%">
<tr>
  <th align="center">Figure 5</th>
</tr>
<tr><td align="center">
[[File:Mod0MuRatio100.png|450px|Example eigenvector]]
</td></tr>
</table>
Consider the model on the <math>~\mu_e/\mu_c = 1</math> sequence for which <math>~\sigma_c^2=0~</math>; key properties of this specific equilibrium model are enumerated in the first row of numbers provided in [[#Equilibrium_Properties_of_Marginally_Unstable_Models|Table 2, above]].  Figure 5 shows how our numerically derived, fundamental-mode eigenfunction, <math>~x = \delta r/r_0</math>, varies with the fractional radius over the entire range, <math>~0 \le r/R \le 1</math>.  By prescription, the eigenfunction has a value of unity and a slope of zero at the center  <math>~(r/R = 0)</math>.  Integrating the LAWE outward from the center, through the model's core (blue curve segment), <math>~x</math> drops smoothly to the value <math>~x_i = 0.81437</math> at the interface <math>~(\xi_i = 1.6686460157 ~\Rightarrow~ q = r_\mathrm{core}/R_\mathrm{surf} = 0.53885819)</math>.  Our numerical integration of the LAWE showed that, at the interface, the logarithmic slope of the ''core'' (blue) segment of the eigenfunction is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\biggl\{ \frac{d\ln x}{d\ln r}\biggr|_i \biggr\}_\mathrm{core}
=
\biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_i \biggr\}_\mathrm{core}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 0.455872 \, .</math>
  </td>
</tr>
</table>
Next, following the [[#Interface|above discussion of matching conditions at the interface]], we determined that, from the perspective of the envelope, the slope of the eigenfunction at the interface must therefore be,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\biggl\{ \frac{d\ln x}{d\ln r}\biggr|_i \biggr\}_\mathrm{env}
=
\biggl\{ \frac{d\ln x}{d\ln \eta}\biggr|_i \biggr\}_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr) + \frac{\gamma_c}{\gamma_e} \biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_i \biggr\}_\mathrm{core} = -1.47352 \, .</math>
  </td>
</tr>
</table>

Adopting this "env" slope along with the amplitude, <math>~x_i = 0.81437</math>, as the appropriate ''interface'' boundary conditions, we integrated the LAWE from the interface to the surface, obtaining the green-colored segment of the eigenfunction that is shown in Figure 5.  The amplitude continued to steadily decrease, reaching a value of <math>~x_s = 0.38203</math>, at the model's surface <math>~(r/R = 1)</math>.  At the surface, this ''envelope'' (green) segment of the eigenfunction exhibits a logarithmic slope that matches to eight significant digits the value that is [[#SurfaceCondition|expected from astrophysical arguments]] for this marginally unstable <math>~(\sigma_c^2=0)</math> model, namely,
<div align="center">
<math>~
\frac{d\ln x}{d\ln \eta}\biggr|_s = \biggl[ \biggl( \frac{\rho_c}{\bar\rho} \biggr)\frac{\cancelto{0}{\sigma_c^2}}{2\gamma_e} - \biggl(3 - \frac{4}{\gamma_e}\biggr)\biggr] = -1 \, .
</math>
</div>

==Numerically Determined Marginally Unstable Models==

The following table should be compared with [[SSC/Stability/BiPolytropes/Pt3#Equilibrium_Properties_of_Marginally_Unstable_Models|Table 2 of an earlier attempt]] at identifying marginally unstable models.
<table border="1" cellpadding="8" align="center">
<tr>
  <th align="center" colspan="7">
Properties of Marginally Unstable Bipolytropes Having<br /><br /><math>~(n_c, n_e) = (5, 1)</math> and <math>~(\gamma_c, \gamma_e) = (\tfrac{6}{5}, 2)</math><br /><br />Determined from Integration of the Envelope's LAWE
  </th>
</tr>
<tr>
  <td align="center"><math>~\frac{\mu_e}{\mu_c}</math></td>
  <td align="center"><math>~\xi_i</math></td>
  <td align="center"><math>~q \equiv \frac{r_\mathrm{core}}{R_\mathrm{surf}}</math></td>
  <td align="center"><math>~\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}}</math></td>
  <td align="center">temporary 1</td>
  <td align="center">temporary 2</td>
  <td align="center">temporary 3</td>
</tr>

<tr>
  <td align="center">1.00</td>
  <td align="right">1.66869</td>
  <td align="right">0.53886</td>
  <td align="right">0.49776</td>
  <td align="center">&#8212;</td>
  <td align="center">&#8212;</td>
  <td align="center">&#8212;</td>
</tr>

<tr>
  <td align="center">0.50</td>
  <td align="right">2.27928</td>
  <td align="right">0.30602</td>
  <td align="right">0.40178</td>
  <td align="center">&#8212;</td>
  <td align="center">&#8212;</td>
  <td align="center">&#8212;</td>
</tr>

<tr>
  <td align="center"><math>\tfrac{1}{3}</math></td>
  <td align="right">2.58201</td>
  <td align="right">0.17629</td>
  <td align="right">0.218242</td>
  <td align="center">&#8212;</td>
  <td align="center">&#8212;</td>
  <td align="center">&#8212;</td>
</tr>
</table>

==Power-Series Expression for x<sub>P</sub>==

As a [[SSC/Stability/BiPolytropes/RedGiantToPN/Pt2#Reminder|reminder]], the analytic expression for <math>x_P</math> throughout the envelope is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Precise Solution to the Polytropic LAWE</b></font></td>
</tr>

<tr>
  <td align="right">
<math>~\frac{x_P}{b}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\eta \phi^{n}}\biggr) \frac{d\phi}{d\eta}\biggr]</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{Q_1}{\eta^2} \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~
Q_1 \equiv - \frac{d \ln \phi}{ d\ln \eta} 
= \biggl[1- \eta\cot(\eta-B) \biggr] = \biggl[1 + \eta\cot(B - \eta) \biggr]\, .
</math>
</div>

Let's define <math>\epsilon \equiv (\eta_\mathrm{surf} - \eta) =  (B - \eta +  \pi )</math>, which will go to zero as <math>\eta</math> approaches the surface.  Recognizing as well that <math>\cot(\epsilon - \pi) = \cot(\epsilon)</math>, we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{x_P}{b}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{(B + \pi - \epsilon)^2} \biggl\{1+ (B + \pi - \epsilon)\cot(\epsilon - \pi) \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{(B + \pi - \epsilon)^2} \biggl\{1+ (B + \pi - \epsilon)\cot(\epsilon ) \biggr\}</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
(B+\pi)^{-2}\biggl[ 1 - \lambda\biggr]^{-2} 
+
(B+\pi)^{-1}\biggl[ 1 - \lambda\biggr]^{-1}\cot(\epsilon ) 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(B+\pi)^{2}\cdot \frac{x_P}{b}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 - \lambda\biggr]^{-2} 
+
(B+\pi)\biggl[ 1 - \lambda\biggr]^{-1}\cot(\epsilon ) 
\, ,
</math>
  </td>
</tr>
</table>
where, <math>\lambda \equiv \epsilon/(B+\pi)</math>.  Drawing from the binomial series,

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

<tr>
  <td align="right">
<math>~(B+\pi)^{2}\cdot \frac{x_P}{b}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 + 2 \lambda + 3\lambda^2 + 4\lambda^3 + 5\lambda^4 + O(\lambda^5)\biggr] 
+
(B+\pi)\biggl[ 1 + \lambda + \lambda^2 + \lambda^3 + \lambda^4 + \lambda^5 + \lambda^6 + O(\lambda^7)\biggr]\cdot
\biggl[\frac{1}{\epsilon} - \frac{\epsilon}{3} - \frac{\epsilon^3}{45} + O(\epsilon^5)  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 + 2 \lambda + 3\lambda^2 + 4\lambda^3 + 5\lambda^4 + O(\lambda^5)\biggr] 
+
\frac{(B+\pi)}{\epsilon}\biggl[ 1 + \lambda + \lambda^2 + \lambda^3 + \lambda^4 + \lambda^5 + \lambda^6 + O(\lambda^7)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+~
\frac{(B+\pi)}{\epsilon}\biggl[ 1 + \lambda + \lambda^2 + \lambda^3 + \lambda^4 + \lambda^5 + \lambda^6 + O(\lambda^7)\biggr]\cdot
\biggl[- \frac{\epsilon^2}{3} \biggr] 
+~
\frac{(B+\pi)}{\epsilon}\biggl[ 1 + \lambda + \lambda^2 + \lambda^3 + \lambda^4 + \lambda^5 + \lambda^6 + O(\lambda^7)\biggr]\cdot
\biggl[- \frac{\epsilon^4}{45} \biggr] 
+ O(\epsilon^5)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 + 2 \lambda + 3\lambda^2 + 4\lambda^3 + 5\lambda^4 + O(\lambda^5)\biggr] 
+
\frac{1}{\lambda}\biggl[ 1 + \lambda + \lambda^2 + \lambda^3 + \lambda^4 + \lambda^5 + \lambda^6 + O(\lambda^7)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-~
\frac{(B+\pi)^2\lambda}{3}  
\biggl[ 1 + \lambda + \lambda^2 + \lambda^3 + \lambda^4 + \lambda^5 + \lambda^6 + O(\lambda^7)\biggr]
-~\frac{(B+\pi)^4\lambda^3}{45}  
\biggl[ 1 + \lambda + \lambda^2 + \lambda^3 + \lambda^4 + \lambda^5 + \lambda^6 + O(\lambda^7)\biggr]
+ O(\epsilon^5)
</math>
  </td>
</tr>
</table>

==Guessing Game==

===Methodical Thinking===
<ol type="A"><li>First Try
<ol>
  <li>
Pick a value of <math>r_\mathrm{norm} = r_0/r_\mathrm{surf}</math> and read off the normalized amplitude at that radial location.  For example, <math>x_\mathrm{norm} = 2.333</math> at <math>r_\mathrm{norm} = 0.500</math>.
  </li>
  <li>
The corresponding value of <math>\eta = r_\mathrm{norm}\times \eta_\mathrm{surf} = 0.500 \times 2.6243 = 1.31215</math>.
  </li>
  <li>
Notice as well that the logarithmic slope at this chosen location is (pull this from column "N" in excel "Sheet03333") - 1.52363.
  </li>
  <li>
The corresponding value of <math>\xi = (2\pi/3)^{1 / 2}\times r_\mathrm{0} = </math>
  </li>
</ol>
</li>

<li>Second Try

<ol>
  <li>Pick a value of <math>(\mu_e/\mu_c)</math>, and a value of the interface location, <math>\xi</math>; the corresponding value of <math>\eta</math> is,
<div align="center">
<math>\eta = \biggl[3^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2  \biggr]\xi \, .</math>
</div>

  </li>
  <li>We can immediately deduce that,
<div align="center">
<math>r_0 = \biggl( \frac{3}{2\pi}\biggr)^{1 / 2} \xi = \biggl[(2\pi)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]^{-1} \eta \, ;</math>
</div>
and from [[SSC/Stability/BiPolytropes/RedGiantToPN/Pt4#Steps|an accompanying series of analytic expressions]]
<div  align="center">
<table border="1" width="80%" cellpadding="8"><tr><td align="left">
&hellip; note, in particular, that
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>\Lambda</math></td>
  <td  align="center"><math>=</math></td>
  <td align="left">
<math>\frac{1}{\eta} + \biggl(\frac{d\phi}{d\eta}\biggr)</math>
  </td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td align="left">
<math>\biggl[3^{-1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\theta^{-2}  \biggr]\cdot \frac{1}{\xi} 
+ 3^{1 / 2}\theta^{-3} \biggl(\frac{d\theta}{d\xi}\biggr)</math>
  </td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ 1+\frac{\xi^2}{3} \biggr]  \cdot \frac{1}{3^{1 / 2}\xi} 
+ 3^{1 / 2}\biggl[ 1+\frac{\xi^2}{3} \biggr]^{3/2} \biggl[- \frac{\xi}{3} \biggl(1 + \frac{\xi^2}{3}  \biggr)^{-3/2}\biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ 1+\frac{\xi^2}{3} \biggr]  \cdot \frac{1}{3^{1 / 2}\xi}  - \frac{\xi}{3^{1 / 2}} \, .</math>
  </td>
</tr>
</table>

</td></tr></table>
</div>
we also deduce that,
</div>
<div align="center">
<math>r_\mathrm{surf} = \biggl[(2\pi)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]^{-1} 
\biggl[ \eta + \frac{\pi}{2} + \tan^{-1}(\Lambda)\biggr] </math>
&nbsp; &nbsp; &nbsp; <math>\Rightarrow</math> &nbsp; &nbsp; &nbsp;
<math>\frac{r_0}{r_\mathrm{surf}} =  \eta \cdot \biggl[ \eta + \frac{\pi}{2} + \tan^{-1}(\Lambda)\biggr]^{-1} \, . </math>
</div>

  </li>
</ol>
</li></ol>

===Envelope Displacement Function===
<div align="center">
<math>\xi^2 = \biggl(\frac{2\pi}{3}\biggr) r_0^2 \, .</math>
</div>

<font color="maroon">STEP 3:</font>&nbsp;&nbsp; Throughout the core &#8212; that is, at all radial positions, <math>0 \le r_0 \le (r_0)_\mathrm{int}</math> &#8212; the displacement amplitude and its first and second derivatives, respectively, are given by the expressions,
<table border="0" cellpadding="5" align="center" width="80%">
<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 - \frac{r_0^2}{15a_5^2} \biggr] = \biggl[1 - \frac{\xi^2}{15} \biggr]
\, ;
</math>
  </td>
  <td align="right">(0.814374698)</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~r_0\cdot \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2r_0^2}{15a_5^2} = -~\frac{2\xi^2}{15}
\, ;
</math>
  </td>
  <td align="right">(- 0.371250604)</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ r_0^2 \cdot \frac{d^2x}{dr_0^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{2r_0^2}{15a_5^2} = -~\frac{2\xi^2}{15}
\, ;
</math>
  </td>
  <td align="right">(- 0.371250604)</td>
</tr>

<tr>
  <td align="right">
also &hellip; &nbsp; &nbsp; 
<math>
\biggl\{ \frac{d\ln x}{d\ln \xi} \biggr\}_\mathrm{core}
= 
\biggl\{ \frac{d\ln x}{d\ln r_0} \biggr\}_\mathrm{core} 
= \frac{r_0}{x} \cdot \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{15}{15 - \xi^2} \biggr] \cdot \biggl[-~\frac{2\xi^2}{15 }\biggr]
= 
\biggl[\frac{2\xi^2}{\xi^2 - 15} \biggr]
</math>
  </td>
  <td align="right">(-0.455871977)<sup>&dagger;</sup></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{4\pi r_0^2/3}{2\pi r_0^2/3 - 15} \biggr]
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{2r_0^2}{r_0^2 - 45/(2\pi)} \biggr]
\, .
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>
</table>

<font color="maroon">STEP #4:</font>&nbsp; &nbsp; From the determination of the logarithmic slope of the displacement function at the edge of the core &#8212; <i>i.e.,</i> at the core-envelope interface &#8212; determine the slope as viewed from the perspective of the envelope.
<table border="0" cellpadding="5" align="center" width="80%">

<tr>
  <td align="right">
<math>~
\biggl\{ \frac{d\ln x}{d\ln r_0}\biggr|_\mathrm{int} \biggr\}_\mathrm{env}
=
\biggl\{ \frac{d\ln x}{d\ln \eta}\biggr|_\mathrm{int} \biggr\}_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr) + \frac{\gamma_c}{\gamma_e}
\biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_\mathrm{int} \biggr\}_\mathrm{core}</math>
  </td>
  <td align="right">(-1.473523186)<sup>&dagger;</sup></td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
\biggl\{ \frac{d\ln x}{d\ln r_0}\biggr\}_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr) + \frac{\gamma_c}{\gamma_e}
\biggl[\frac{2r_0^2}{r_0^2 - 45/(2\pi)} \biggr] </math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
\biggl\{ \frac{dx}{x}\biggr\}_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr)\frac{dr_0}{r_0} + \frac{\gamma_c}{\gamma_e}
\biggl[\frac{2r_0^2}{r_0^2 - 45/(2\pi)} \biggr]\frac{dr_0}{r_0} </math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
d\ln x \biggr|_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr)d\ln r_0 + \frac{\gamma_c}{\gamma_e}
\biggl[\frac{2r_0}{r_0^2 - 45/(2\pi)} \biggr]dr_0 </math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
\ln x \biggr|_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\ln r_0^{3(\gamma_c/\gamma_e - 1) } 
- \frac{2\gamma_c}{\gamma_e}
\int\frac{r_0\cdot dr_0}{45/(2\pi)-r_0^2}  
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\ln \biggl[r_0\biggr]^{3(\gamma_c/\gamma_e - 1) } 
+ \frac{\gamma_c}{\gamma_e}\cdot \ln\biggl[45/(2\pi)-r_0^2 \biggr] 
+ \ln \Gamma
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\ln \biggl[r_0\biggr]^{3(\gamma_c/\gamma_e - 1) } 
+ \ln\biggl[45/(2\pi)-r_0^2 \biggr]^{\gamma_c/\gamma_e} 
+ \ln \Gamma
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\ln \biggl\{~\biggl[r_0\biggr]^{3(\gamma_c/\gamma_e - 1) } 
\cdot \biggl[45/(2\pi)-r_0^2 \biggr]^{\gamma_c/\gamma_e} 
\cdot \Gamma~\biggr\}
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
x \biggr|_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl[r_0\biggr]^{3(\gamma_c/\gamma_e - 1) } 
\cdot \biggl[45/(2\pi)-r_0^2 \biggr]^{\gamma_c/\gamma_e} 
\cdot \Gamma
\, .
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>
</table>

=Related Discussions=
<ul>
  <li>[[SSC/Stability/InstabilityOnsetOverview#Fig1|Instability Onset Overview]]</li>
  <li>Analytic <math>(n_c, n_e) = (5, 1)</math>
  <ul>
    <li>[[SSC/Structure/BiPolytropes/Analytic51#BiPolytrope_with_nc_=_5_and_ne_=_1_(Pt_1)|Part 1]]</li>
    <li>[[SSC/Structure/BiPolytropes/Analytic51/Pt2#BiPolytrope_with_nc_=_5_and_ne_=_1_(Pt_2)|Part 2]]</li>
  </ul>
  </li>
</ul>

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