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

===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>