Editing SSC/Stability/BiPolytropes/RedGiantToPN/Pt2 (section)

==Specific Case of (n<sub>c</sub>, n<sub>e</sub>) = (5,1)==
Drawing from our [[SSC/Structure/BiPolytropes/Analytic51/Pt2#Profile|"Table 2" profiles]], let's evaluate <math>\mathcal{H}</math> and <math>\mathcal{K}</math> for the two separate regions of bipolytrope model.

===The n<sub>c</sub> = 5 Core===
<div align="center">
<math></math>

<math>r^*= \biggl( \frac{3}{2\pi}\biggr)^{1 / 2}\xi</math>
</div>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{\rho^*}{P^*}</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl(1 + \frac{\xi^2}{3}\biggr)^{- 5 / 2}
\biggl(1 + \frac{\xi^2}{3}\biggr)^{6 / 2}
=
\biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~\mathcal{H}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4 - 2\xi^2 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 3 / 2} \biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
=
4 - 2\xi^2 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 1}
\, .
</math>
  </td>
</tr>
</table>
Also,

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~\mathcal{K}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
\biggl\{ \biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr)\frac{2\pi }{3} 
- \biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) 2\xi^2 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 3 / 2} \biggl(\frac{2\pi}{3}\biggr)\xi^{-2} \biggr\}
=
\frac{2\pi}{3}\biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
\biggl\{ \biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr) 
- 2\biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 3 / 2}  \biggr\}
</math>
  </td>
</tr>
</table>
Hence, the LAWE becomes,

<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{2\pi}{3} \biggr] \frac{d^2 x}{d\xi^2}
+ \biggl[ \mathcal{H} \biggr] \frac{2\pi}{3} \xi^{-1}\frac{dx}{d\xi} 
+ 
\frac{2\pi}{3}\biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
\biggl\{ \biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr) 
- 2\biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 3 / 2}  \biggr\}
x 
\, .
</math>
  </td>
</tr>
</table>
Multiplying  through by <math>3\xi^2/(2\pi)</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>
\xi^2 \frac{d^2 x}{d\xi^2}
+ \biggl[ 4\xi - 2\xi^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 1} \biggr]\frac{dx}{d\xi} 
+ 
\xi^2\biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
\biggl\{ \biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr) 
- 2\biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 3 / 2}  \biggr\}
x 
\, .
</math>
  </td>
</tr>
</table>

<table border="1" cellpadding="8" width="80%" align="center"><tr><td align="left">
Let's compare this with the equivalent expression [[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|presented separately]], namely,

<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>
The [[SSC/Structure/Polytropes/Analytic#n_=_5_Polytrope|primary E-type solution]] for n = 5 polytropes states that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\theta_{n=5}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 + \frac{\xi^2}{3}\biggr]^{-1 / 2}
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ Q = - \frac{\xi}{\theta}\frac{d \theta}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- ~\xi \biggl[1 + \frac{\xi^2}{3}\biggr]^{1 / 2}\frac{d}{d\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>
+~\xi \biggl[1 + \frac{\xi^2}{3}\biggr]^{1 / 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>
\frac{\xi^2}{3}\biggl[1 + \frac{\xi^2}{3}\biggr]^{-1} 
\, .
</math>
  </td>
</tr>
</table>
Hence, the LAWE may 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>
\frac{d^2x}{d\xi^2} + \biggl[4 - 6Q\biggr] \frac{1}{\xi}\cdot \frac{dx}{d\xi}
+6\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g}}\biggr) \frac{\xi^2}{\theta} - \biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) Q\biggr]\frac{x}{\xi^2}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x}{d\xi^2} + \biggl\{ 4 - 2\xi^2\biggl[1 + \frac{\xi^2}{3}\biggr]^{-1} \biggr\} \frac{1}{\xi}\cdot \frac{dx}{d\xi}
+ \biggl\{ \biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr) \biggl[1 + \frac{\xi^2}{3}\biggr]^{1/2} 
- 2\biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl[1 + \frac{\xi^2}{3}\biggr]^{-1}\biggr\} x
</math>
  </td>
</tr>
</table>
Versus above,
<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>
\xi^2\frac{d^2 x}{d\xi^2}
+ \biggl[ 4\xi - 2\xi^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 1} \biggr]\frac{dx}{d\xi} 
+ 
\xi^2\biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
\biggl\{ \biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr) 
- 2\biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 3 / 2}  \biggr\}
x 
\, .
</math>
  </td>
</tr>
</table>

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

If we set <math>\gamma_\mathrm{g} = 6/5</math> and we set <math>\sigma_c^2 = 0</math>, this becomes,
<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>
\xi^2 \frac{d^2 x}{d\xi^2}
+ \biggl[ 4\xi - 2\xi^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 1} \biggr]\frac{dx}{d\xi} 
+ \frac{2}{3} 
\xi^2 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 1}x 
</math>
  </td>
</tr>
</table>
Next, try the solution, <math>x = (1 - \xi^2/15)~\Rightarrow ~dx/d\xi = -2\xi/15</math> and <math>d^2x/dx^2 = -2/15</math>:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{2}{15} \xi^2
- \biggl[ 4\xi - 2\xi^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 1} \biggr]\frac{2\xi}{15} 
+ \frac{2}{3} 
\xi^2 \biggl(1 + \frac{\xi^2}{3}\biggr)^{- 1}\biggl[1 - \xi^2/15\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~15\biggl(1 + \frac{\xi^2}{3}\biggr)</math> &nbsp;LAWE
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- 2\xi^2 \biggl(1 + \frac{\xi^2}{3}\biggr)
- \biggl[ 60\xi\biggl(1 + \frac{\xi^2}{3}\biggr) - 30\xi^3 \biggr]\frac{2\xi}{15} 
+ 10 \xi^2 \biggl[1 - \xi^2/15\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-2\xi^2 -\frac{2\xi^4}{3}
- \biggl[ 60\xi^2  - 10\xi^4 \biggr]\frac{2}{15} 
+ \frac{10 }{15} \biggl[15\xi^2 - \xi^4\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-2\xi^2 -\frac{2\xi^4}{3}
- \biggl[ 4\xi^2  - \frac{2}{3}\xi^4 \biggr]2 
+ \biggl[10\xi^2 - \frac{2}{3}\xi^4\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-2\xi^2 -\frac{2\xi^4}{3}
-  8\xi^2  + \frac{4}{3}\xi^4  
+ \biggl[10\xi^2 - \frac{2}{3}\xi^4\biggr] 
</math>
  </td>
</tr>

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

===The n<sub>e</sub> = 1 Envelope===

[[SSC/Stability/BiPolytropes/Pt3#Profile|Throughout the envelope]] we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r^*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\rho^*}{P^*}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)\theta_i^5 \phi \biggr]
\biggl[\theta_i^{-6} \phi^{-2}\biggr]
=
\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{M_r^*}{r^*}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\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)
\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggr]^{-1}
=
2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \frac{\theta_i}{\eta} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)
\, .
</math>
  </td>
</tr>
</table>

Hence,
<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\}
=
4 - \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi^{-1} \biggr]  
\biggl[ 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \frac{\theta_i}{\eta} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr) \biggr]
= 
4 - 2 \biggl(- \frac{d\ln\phi}{d\ln\eta} \biggr)
\, ;
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\mathcal{K}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\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]
=
\biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr)\frac{2\pi }{3}\biggl(\frac{\rho^*}{ P^* } \biggr)
- \biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \frac{\rho^*}{ P^* }\cdot \frac{M_r^*}{(r^*)} \cdot \frac{1}{(r^*)^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr)\frac{2\pi }{3}\biggl[\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi^{-1}\biggr]
- 2\biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl(-\frac{d\ln\phi}{d\ln\eta}\biggr) 
\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta\biggr]^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{\sigma_c^2}{\gamma_\mathrm{g}}\biggr)\frac{2\pi }{3}\biggl[\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \biggr]\frac{1}{\phi}
- 2\biggl(3 - \frac{4}{\gamma_\mathrm{g}}\biggr)  
\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{2} \theta^{4}_i (2\pi)\biggr] \frac{1}{\eta^2}\biggl(-\frac{d\ln\phi}{d\ln\eta}\biggr) \, .
</math>
  </td>
</tr>
</table>

<table border="1" cellpadding="8" width="80%" align="center"><tr><td align="left">
Let's compare this with the equivalent expression [[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|presented separately]], namely,

<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>
The [[SSC/Structure/Polytropes/Analytic#n_=_5_Polytrope|equilibrium, off-center equilibrium solution]] for n = 1 polytropes states that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\phi_{n=1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{A}{\eta} \cdot \sin(B-\eta) 
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{d\phi}{d\eta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- A \frac{d}{d\eta}\biggl\{
\eta^{-1} \cdot \sin(B-\eta) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
A \biggl\{
\eta^{-2} \cdot \sin(B-\eta)
+ 
\eta^{-1} \cdot \cos(B-\eta) 
\biggr\}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ Q = - \frac{\eta}{\phi}\frac{d \phi}{d\eta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \eta\biggl[- \frac{A}{\eta} \cdot \sin(B-\eta) \biggr]^{-1}
A \biggl\{
\eta^{-2} \cdot \sin(B-\eta)
+ 
\eta^{-1} \cdot \cos(B-\eta) 
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr]
</math>
  </td>
</tr>
</table>
Hence, the LAWE may 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>
\frac{d^2x}{d\eta^2} + \biggl\{ 4 - 2\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr]\biggr\} \frac{1}{\eta}\cdot \frac{dx}{d\eta}
+2\biggl\{\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g}}\biggr) \frac{\eta^2}{\phi} - \biggl(3 
- \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr] \biggr\}\frac{x}{\eta^2}
</math>
  </td>
</tr>
</table>

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

===Blind Alleys===

====Reminder====
From a [[SSC/Stability/n1PolytropeLAWE/Pt3#Second_Attempt|separate discussion]], we have demonstrated that the LAWE relevant to the envelope 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}{d\eta^2} + \biggl\{ 4 - \biggl[ \frac{2  \eta}{\phi} \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^5 \phi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1}  \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}   \biggr\}  x 
~-~ \alpha_e  \biggl[ \frac{2\eta}{\phi} \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr]   \frac{x}{\eta^2} \, .
</math>
  </td>
</tr>
</table>

If we assume that, <math>~\alpha_e = (3 - 4/2) = 1</math> and <math>~\sigma_c^2 = 0</math>, then the relevant envelope 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}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  \biggl[ 2 Q \biggr]   \frac{x}{\eta^2} \, ,
</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~
Q \equiv - \frac{d \ln \phi}{ d\ln \eta} 
= \biggl[1- \eta\cot(\eta-B) \biggr] = \biggl[1 + \eta\cot(B - \eta) \biggr]\, .
</math>
</div>

Also separately, [[SSC/Stability/n1PolytropeLAWE/Pt3#Consider|we have derived]] the following,
<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>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b(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>~-b\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{b}{\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{bQ}{\eta^2} \, .</math>
  </td>
</tr>
</table>
</div>

====First Try====

<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>
\eta^{-m}~~~\Rightarrow ~~~ \frac{dx}{d\eta} = -m\eta^{-m-1}
</math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp;
<math> \frac{d^2x}{d\eta^2} = -m(-m-1)\eta^{-m-2} \, ,</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-m(-m-1)\eta^{-m-2} 
+ \biggl\{ 4 - 2\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr]\biggr\} \frac{1}{\eta}\cdot \biggl[-m\eta^{-m-1} \biggr]
+2\biggl\{\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g}}\biggr) \frac{\eta^2}{\phi} - \biggl(3 
- \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr] \biggr\}\eta^{-m-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \eta^{m+2} \times \mathrm{LAWE}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m(m+1) 
-m \biggl\{ 4 - 2\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr]\biggr\} 
+2\biggl\{\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g}}\biggr) \frac{\eta^2}{\phi} - \biggl(3 
- \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr] \biggr\}
\, .
</math>
  </td>
</tr>
</table>
Now set <math>\sigma_c^2 = 0</math> and set <math>\gamma_\mathrm{g} = 2</math>:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \eta^{m+2} \times \mathrm{LAWE}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m(m+1) 
-m \biggl\{ 4 - 2\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr]\biggr\} 
-2\biggl\{\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m(m+1) 
- 4m  
+2m \biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr] 
-2\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr] 
</math>
  </td>
</tr>
</table>
We see that the complexity of the LAWE reduces substantially if we set <math>m = +1</math>; specifically, this choice gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[\eta^{m+2} \times \mathrm{LAWE} \biggr]_{m\rightarrow 1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-2 \, .
</math>
  </td>
</tr>
</table>
<font color="red">Close, but no cigar!</font>

====Second Try====
Next, let's set <math>\sigma_c^2 = 0</math> but let's leave <math>\gamma_\mathrm{g}</math> unspecified:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \eta^{m+2} \times \mathrm{LAWE}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m(m+1) 
-m \biggl\{ 4 - 2\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr]\biggr\} 
-2\biggl\{\biggl(3 
- \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m(m+1) -4m
+ 2m\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr] 
-2\biggl(3 
- \frac{4}{\gamma_\mathrm{g}}\biggr) \biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m(m-3)
+\biggl\{2m - 2\biggl(3  - \frac{4}{\gamma_\mathrm{g}}\biggr)\biggr\} 
\biggl[ 1 + \eta \cdot \cot(B-\eta) \biggr]
\, .
</math>
  </td>
</tr>
</table>
The first term goes to zero if we set <math>m=3</math>; then, in order for the second term to go to zero, we need &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>
\biggl\{6 - 2\biggl(3  - \frac{4}{\gamma_\mathrm{g}}\biggr)\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \gamma_\mathrm{g} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\infty \, .</math>
  </td>
</tr>
</table>
This means that the envelope is incompressible.

====Third Try====

Note that,
<div align="center">
<math>~
Q \equiv - \frac{d \ln \phi}{ d\ln \eta} 
= \biggl[1- \eta\cot(\eta-B) \biggr] = \biggl[1 + \eta\cot(B - \eta) \biggr]\, ,
</math>
</div>
and that,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{d^2}{d\eta^2}\biggl[\cot(B-\eta)\biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
+~2\biggl[\cos(B-\eta)\biggr]^{-3}
</math>
  </td>
</tr>
</table>


Let's try &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x = x_1 + x_2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{b}{\eta^2} + \frac{c}{\eta} \cdot \cot(B-\eta)
\, .
</math>
  </td>
</tr>
</table>

If we assume that, <math>~\alpha_e = (3 - 4/2) = 1</math> and <math>~\sigma_c^2 = 0</math>, then the relevant envelope LAWE is the <i>sum</i> of the pair of sub-LAWEs,

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

<tr>
  <td align="right">
<math>\mathrm{LAWE}_1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x_1}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx_1}{d\eta} 
~-~  \biggl[ 2 Q \biggr]   \frac{x_1}{\eta^2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\mathrm{LAWE}_2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x_2}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx_2}{d\eta} 
~-~  \biggl[ 2 Q \biggr]   \frac{x_2}{\eta^2} \, .
</math>
  </td>
</tr>
</table>

One at a time:

----

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

<tr>
  <td align="right">
<math>\frac{dx_1}{d\eta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{2b}{\eta^3}\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{d^2x_1}{d\eta^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6b}{\eta^4} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~\mathrm{LAWE}_1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6b}{\eta^4}
+ \biggl\{ 4 -2Q \biggr\} \biggl[-\frac{2b}{\eta^4} \biggr]
~-~  \biggl[ 2 Q \biggr]   \frac{b}{\eta^4} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{\eta^4}\biggl\{
6b -2b\biggl[4-2Q\biggr]
~-~  \biggl[ 2b Q \biggr]    
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{2b}{\eta^4}\biggl[ Q-1  \biggr]
\, .
</math>
  </td>
</tr>
</table>

----


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

<tr>
  <td align="right">
<math>\frac{dx_2}{d\eta} = \frac{d}{d\eta}\biggl[\frac{c}{\eta}\cdot \cot(B-\eta)\biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{c}{\eta^2}\cdot \cot(B-\eta)
+
\frac{c}{\eta}\cdot \biggl[\sin(B-\eta)\biggr]^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{c}{\eta^2}\biggl\{ -\frac{\cos(B-\eta)}{\sin(B-\eta)}
+
\frac{\eta}{\sin^2(B-\eta)}\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{c}{\eta^2}\biggl\{\eta ~-~\sin(B-\eta)\cos(B-\eta)
\biggr\}\biggl[\sin(B-\eta)\biggr]^{-2}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{d^2x_2}{d\eta^2} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{d\eta}\biggl\{
-\frac{c}{\eta^2}\cdot \cot(B-\eta)
\biggr\}
+
\frac{d}{d\eta}\biggl\{
\frac{c}{\eta}\cdot \biggl[\sin(B-\eta)\biggr]^{-2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{
\frac{2c}{\eta^3}\cdot \cot(B-\eta)
-\frac{c}{\eta^2}\cdot \biggl[\sin(B-\eta)\biggr]^{-2}
\biggr\}
+
\biggl\{ 
-\frac{c}{\eta^2}\cdot \biggl[\sin(B-\eta)\biggr]^{-2}
+ \frac{2c}{\eta}\cdot \biggl[\sin(B-\eta)\biggr]^{-3}\cos(B-\eta)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{2c}{\eta^3}\biggl\{
\cot(B-\eta)
-\eta\cdot \biggl[\sin(B-\eta)\biggr]^{-2}
+
\eta^2\cdot \biggl[\sin(B-\eta)\biggr]^{-3}\cos(B-\eta)
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~\mathrm{LAWE}_2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x_2}{d\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx_2}{d\eta} 
~-~  \biggl[ 2 Q \biggr]   \frac{x_2}{\eta^2} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl\{ 4 -2Q \biggr\}\cdot \frac{c}{\eta^3}\biggl\{\eta ~-~\sin(B-\eta)\cos(B-\eta)
\biggr\}\biggl[\sin(B-\eta)\biggr]^{-2} 
~-~  \biggl[ 2 Q \biggr]   \frac{c}{\eta^3} \cdot \cot(B-\eta)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \frac{2c}{\eta^3}\biggl[\sin(B-\eta)\biggr]^{-3} \biggl\{
(2 -Q )\cdot \biggl[ \eta ~-~\sin(B-\eta)\cos(B-\eta)
\biggr] \biggl[\sin(B-\eta)\biggr] 
~-~  Q \biggl[\sin(B-\eta)\biggr]^{3}  \cot(B-\eta)
\biggr\}
</math>
  </td>
</tr>

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

----

Hence,

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

<tr>
  <td align="right">
<math>\mathrm{LAWE}_1 + \mathrm{LAWE}_2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{2b}{\eta^4}\biggl[ Q-1  \biggr]
+
\frac{2c}{\eta^3}\biggl[\sin(B-\eta)\biggr]^{-3}\biggl\{
\eta^2\cdot \cos(B-\eta)
- \biggl[\sin^2(B-\eta)\cos(B-\eta)\biggr]  
+ 
(1 - Q )\cdot \biggl[\eta \cdot \sin(B-\eta) \biggr]  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{2(b-c)}{\eta^3}\biggl[\cot(B-\eta)  \biggr]
</math>
  </td>
</tr>
</table>

====Fourth Try====
Try adding an additional term that was discussed above under "First Try", namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x_3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{\eta}
\, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>
\mathrm{LAWE}_{3}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{2d}{\eta^3} \, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\mathrm{LAWE}_{1} + \mathrm{LAWE}_{2} + \mathrm{LAWE}_{3}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{2}{\eta^3}\biggl[(b-c)\cot(B-\eta) - d \biggr]
\, .
</math>
  </td>
</tr>
</table>