Editing SSC/Stability/n1PolytropeLAWE/Pt3 (section)

==What About Bipolytropes?==

Here we will try to find an analytic expression for the radial displacement function, <math>~x</math>, for a bipolytropic ''envelope'' whose polytropic index is, <math>~n_e = 1</math>.  As in the above ''succinct'' derivation, the relevant LAWE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[4 - (n+1)Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
(n+1)\biggl[ \biggl(\frac{\sigma_c^2}{6\gamma_g } \biggr) \frac{\xi^2}{\theta} 
-\alpha 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 - 2Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
\biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi^3}{\sin\xi} 
- 2Q\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[2 + \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} 
+ \biggl[- 2 +  \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \frac{ x}{\xi^2} 
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>

</table>
</div>

===First Attempt===
Let's 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>~
A + \frac{B}{(\xi - F)^2} \biggl[1 - (\xi-D) \cot(\xi-C) \biggr] \, .
</math>
  </td>
</tr>
</table>
First, note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\xi}\biggl[\cot(\xi - C) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d}{d\xi}\biggl[ \frac{ \cos(\xi - C) }{ \sin(\xi - C)}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\xi - C)\biggr] \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2B}{(\xi-F)^3} \biggl[1 - (\xi-D) \cot(\xi-C) \biggr]
- \frac{B}{(\xi-F)^2} \biggl\{
\cot(\xi-C) - (\xi-D) [1 + \cot^2(\xi-C) ]
\biggr\}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{(\xi-F)^3} \biggl\{ 2 
- \biggl[3\xi  - (2D +F) \biggr]  \cot(\xi-C)
+ [ \xi^2 - (D+F)\xi + FD]
+ [ \xi^2 - (D+F)\xi + FD]\cot^2(\xi-C) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{(\xi-F)^3} \biggl\{ 
[ \xi^2 - (D+F)\xi + FD+2]
- \biggl[3\xi  - (2D +F) \biggr]  \cot(\xi-C)
+ [ \xi^2 - (D+F)\xi + FD]\cot^2(\xi-C) 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
And,

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

<tr>
  <td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3B}{(\xi-F)^4} \biggl\{ 
[ \xi^2 - (D+F)\xi + FD+2]
- \biggl[3\xi  - (2D +F) \biggr]  \cot(\xi-C)
+ [ \xi^2 - (D+F)\xi + FD]\cot^2(\xi-C) 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{B}{(\xi-F)^3} \biggl\{ 
[ 2\xi - (D+F)]
- 3  \cot(\xi-C)
- \biggl[3\xi  - (2D +F) \biggr] \frac{d \cot(\xi-C)}{d\xi}
+ [ 2\xi - (D+F)]\cot^2(\xi-C) 
+ \biggl[ \xi^2 - (D+F)\xi + FD \biggr] \frac{d \cot^2(\xi-C) }{d\xi}
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{(\xi-F)^4}{B}  \biggr] \frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3[ \xi^2 - (D+F)\xi + FD+2]
- 3\biggl[3\xi  - (2D +F) \biggr]  \cot(\xi-C)
+ 3[ \xi^2 - (D+F)\xi + FD]\cot^2(\xi-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- (\xi-F)[ 2\xi - (D+F)]
+ 3 (\xi-F)  \cot(\xi-C)
- (\xi-F)[ 2\xi - (D+F)]\cot^2(\xi-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\xi-F)\biggl[3\xi  - (2D +F) \biggr] \frac{d \cot(\xi-C)}{d\xi}
- (\xi-F)\biggl[ \xi^2 - (D+F)\xi + FD \biggr] \frac{d \cot^2(\xi-C) }{d\xi}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3[ \xi^2 - (D+F)\xi + FD+2]
- (\xi-F)[ 2\xi - (D+F)]
+ \biggl\{3 (\xi-F)  
- 3\biggl[3\xi  - (2D +F) \biggr] \biggr\}  \cot(\xi-C)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{ 3[ \xi^2 - (D+F)\xi + FD] 
- (\xi-F)[ 2\xi - (D+F)] \biggr\}\cot^2(\xi-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- (\xi-F)\biggl[3\xi  - (2D +F) \biggr] \biggl[ 1 + \cot^2(\xi - C)\biggr]
+ (\xi-F)\biggl[ \xi^2 - (D+F)\xi + FD \biggr] 2\cot(\xi-C)\biggl[ 1 + \cot^2(\xi - C)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3[ \xi^2 - (D+F)\xi + FD+2]
- (\xi-F)[ 2\xi - (D+F)]
- (\xi-F) [3\xi  - (2D +F) ] 
+ \biggl\{3 (\xi-F)  
- 3 [3\xi  - (2D +F) ] 
+ 2 (\xi-F) [ \xi^2 - (D+F)\xi + FD ]  
\biggr\}  \cot(\xi-C)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{ 3[ \xi^2 - (D+F)\xi + FD] 
- (\xi-F)[ 2\xi - (D+F)] 
- (\xi-F) [3\xi  - (2D +F) ] 
\biggr\}\cot^2(\xi-C) 
+ 2 (\xi-F) [ \xi^2 - (D+F)\xi + FD ] \cot^3(\xi - C) \, .
</math>
  </td>
</tr>
</table>

Let's set <math>~C = D = F</math> and see if these expressions match the ones above.

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

<tr>
  <td align="right">
<math>~\frac{dx}{d\xi} \biggr|_{C=D=F}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{\xi^3} \biggl\{ 
2 + \xi^2
- 3\xi  \cot\xi
+ \xi^2 \cot^2\xi 
\biggr\} \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\xi^4}{B}  \cdot \frac{d^2x}{d\xi^2} \biggr|_{C=D=F}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3[ \xi^2 +2]
- (\xi)[ 2\xi ]
- \xi [3\xi  ] 
+ \biggl\{3 (\xi)  
- 3 [3\xi  ] 
+ 2 \xi [ \xi^2  ]  
\biggr\}  \cot(\xi)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3 \xi^2 +6- 2\xi^2 - 3\xi^2 
+ \biggl[ 3 \xi - 9\xi  + 2 \xi^3  \biggr]  \cot(\xi)
+ \biggl[ 3\xi^2  - 2\xi^2 - 3\xi^2 \biggr] \cot^2\xi + 2 \xi^3  \cot^3\xi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6- 2\xi^2 
+ [ - 6\xi  + 2 \xi^3  ]  \cot(\xi)
- 2\xi^2 \cot^2\xi + 2 \xi^3  \cot^3\xi  
</math>
  </td>
</tr>
</table>

===Second Attempt===

Up to this point we have been rather cavalier about the use of <math>~\xi</math> (and <math>~\xi_i</math>) to represent the envelope's dimensionless radius (and interface location).  Let's switch to <math>~\eta</math>,

<table border="0" align="center" cellpadding="8">

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

and, throughout the envelope we have,
<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( \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} \theta_i  \eta \biggl(- \frac{d\phi}{d\eta} \biggr)
\, .
</math>
  </td>
</tr>
</table>

Hence, 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{\rho^*}{P^*}\biggr] \biggl[ \frac{ M_r^*}{(r^*)} \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^4} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[ \frac{\rho^*}{ P^* } \biggr] \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  ~-~\frac{\alpha_e }{(r^*)^2} \biggl[ \frac{M_r^*}{r^*} \biggr] \biggr\}  x 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </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}}  
~-~\alpha_e \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i (4\pi) \eta^{-1} \biggr] \biggl(- \frac{d\phi}{d\eta} \biggr)  \biggr\}  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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} \, .
</math>
</div>

<span id="Consider">Now consider</span> the,
<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>

From our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|accompanying discussion]], we recall that the most general solution to the <math>n=1</math> Lane-Emden equation can be written in the form,
<div align="center">
<math>
\phi = A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, ,
</math>
</div>
where <math>A</math> and <math>B</math> are constants whose values can be obtained from our [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|accompanying parameter table]].  The first derivative of this function is,
<div align="center">
<math>
\frac{d\phi}{d\eta} = \frac{A}{\eta^2} \biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr] \, .
</math>
</div>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q = -\frac{d\ln\phi}{d\ln\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\eta}{\phi} \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>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-B) \biggr] \, .
</math>
  </td>
</tr>
</table>
What is this in terms of the dimensionless radius, <math>~r^*/R^*</math>?  Well,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>\frac{~r^*}{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 \biggl[\frac{\sqrt{2\pi}~\theta_i^2}{\eta_s} \biggl(\frac{\mu_e}{\mu_c}\biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{\eta}{\eta_s} = \frac{\eta}{(\pi + B)} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \eta</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi + B \biggr) \, .</math>
  </td>
</tr>
</table>
Also,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>~\eta-B</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi + B \biggr) -B = \pi \biggl( \frac{r^*}{R^*}\biggr) - B\biggl[1-\biggl( \frac{r^*}{R^*}\biggr)\biggr]</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\pi  - (\pi + B)\biggl[1-\biggl( \frac{r^*}{R^*}\biggr)\biggr] \, .</math>
  </td>
</tr>
</table>


<font color="red">'''[12 January 2019]:'''</font>  Here's what appears to work pretty well, empirically:

<table border="1" width="60%" align="center" cellpadding="8"><tr><td align="left">
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\eta^2} \biggl\{1- \eta\cot[\eta-(\pi - 0.8)] \biggr\} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi - 0.6\pi \biggr) \, .</math>
  </td>
</tr>
</table>
</td></tr></table>


<span id="tagJanuary2019">
Let's work through the analytic derivatives again.  Keeping in mind that,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\eta - B) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\eta - B)\biggr] \, ,
</math>
  </td>
</tr>
</table>
and starting with the ''guess'',

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

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-B) \biggr] \, ,
</math>
  </td>
</tr>
</table>
we have,

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B) \, .
</math>
  </td>
</tr>
</table>
The second derivative then gives,
<table border="0" cellpadding="5" align="center">

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

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

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^4} \biggl[
6 - 3\eta^2 - 3\eta\cot(\eta-B)  -  3\eta^2\cot^2(\eta - B) 
\biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{\eta^4}{b}\cdot \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6 - 3\eta^2 - 3\eta\cot(\eta-B)  -  3\eta^2\cot^2(\eta - B) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~+~ 2\eta^2 + \eta \cot(\eta-B)  +  2\eta^2 \cot^2(\eta - B) 
-\eta^2 - \eta^2  \cot^2(\eta - B) 
-  2\eta^3\cot(\eta-B)  
-  2\eta^3\cot^3(\eta-B)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl[ 3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^2\cot^2(\eta - B) 
-  \eta^3\cot^3(\eta-B) 
\biggr]  \, .
</math>
  </td>
</tr>
</table>

<span id="Recalling">Recalling</span> that,
<div align="center">
<math>~Q = \biggl[1- \eta\cot(\eta-B) \biggr] \, ,</math>
</div>
plugging these expressions into the relevant envelope LAWE gives,

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

<tr>
  <td align="right">
LAWE
  </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} 
~-~  2 Q \cdot \frac{x}{\eta^2} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{\eta^4} \biggl\{
\frac{\eta^4}{b} \cdot \frac{d^2x}{d\eta^2} + \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b} \cdot \frac{dx}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 x}{b} 
\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\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^2\cot^2(\eta - B) 
-  \eta^3\cot^3(\eta-B) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \biggl[\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)\biggr]
~-~  \biggl[1- \eta\cot(\eta-B) \biggr] \biggl[1- \eta\cot(\eta-B) \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\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^2\cot^2(\eta - B) 
-  \eta^3\cot^3(\eta-B) 
+ \biggl[\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)\biggr]
~-~  \biggl[1- \eta\cot(\eta-B) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \eta\cot(\eta-B) \biggl[\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)\biggr]
~+~\eta\cot(\eta-B) \biggl[1- \eta\cot(\eta-B) \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\{
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^3\cot^3(\eta-B) 
~+~2\eta\cot(\eta-B) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \eta^3\cot(\eta-B) 
-2 \eta\cot(\eta-B) 
+ \eta^2\cot^2(\eta-B) 
+ \eta^3\cot^3(\eta - B)
~+~\eta\cot(\eta-B) 
~-~\eta^2\cot^2(\eta-B) 
\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\{
[- \eta \cot(\eta-B)  
- \eta\cot(\eta-B) 
~+~2\eta\cot(\eta-B) ]
+ [\eta^3\cot(\eta-B) 
- \eta^3 \cot(\eta-B)  ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ [\eta^2\cot^2(\eta-B) 
~-~\eta^2\cot^2(\eta-B) ]
+ [\eta^3\cot^3(\eta - B)
-  \eta^3\cot^3(\eta-B) ]
\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>

Okay.  Now let's determine at what value of <math>~\eta</math> the logarithmic derivative of <math>~x_P</math> goes to negative one.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\ln x_P}{d\ln \eta} = \frac{\eta}{x_P} \cdot \frac{dx_P}{d\eta}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\eta^3}{b }\biggl[1- \eta\cot(\eta-B) \biggr]^{-1}
\cdot \frac{dx_P}{d\eta}
</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]^{-1}
\biggl[ \eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)  \biggr] \, .
</math>
  </td>
</tr>
</table>
Setting this to negative one, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
-\biggl[1- \eta\cot(\eta-B) \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
\Rightarrow~~~1
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2\biggl[ 1 +  \cot^2(\eta - B) \biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~
\Rightarrow~~~1
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\eta^2}{\sin^2(\eta - B)} \, .
</math>
  </td>
</tr>
</table>
And this occurs when,
<div align="center">
<math>~\biggl(\frac{A}{\phi } \biggr)^2 = 1 \, .</math>
</div>

===Third Attempt===
====Prior to the Brute-Force Trial Fit====

Let's work through the analytic derivatives again.  Keeping in mind that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\eta - C) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\eta - C)\biggr] \, ,
</math>
  </td>
</tr>
</table>
and starting with the ''guess'',

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

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-C) \biggr] \, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right">
<math>~ \biggl( \frac{\eta^3}{b} \biggr) \frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C) \, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\eta^4}{b}\cdot \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl[ 3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-C)  
-  \eta^2\cot^2(\eta - C) 
-  \eta^3\cot^3(\eta-C) 
\biggr]  \, .
</math>
  </td>
</tr>
</table>


<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
Note that the relevant logarithmic derivative is,

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

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

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

If we know the logarithmic slope and the value of <math>~\eta</math> at the interface, then we can solve for 
<div align="center">
<math>~y_i \equiv  \eta_i \cot(\eta_i-C) \, ,</math>
</div>
via the quadratic relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1- y_i ) \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta_i^2 -2 + y_i  +  y_i^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta_i^2 -2 + y_i  +  y_i^2
- (1- y_i ) \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 y_i^2 
+ y_i \biggl\{1 + \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i\biggr\}
+\biggl\{  \eta_i^2 -2 - \biggl[\frac{d\ln x_P}{d\ln\eta}\biggr]_i \biggr\} \, .
</math>
  </td>
</tr>
</table>
(In practice it appears as though the "plus" solution to this quadratic equation is desired if the quantity inside the last set of curly braces is positive; and the "minus" solution is desired if this quantity is negative.)  Once the value of <math>~y_i</math> is known, we can solve for the key coefficient, <math>~C</math>, via the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tan(\eta_i - C)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{\eta_i}{y_i}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~C</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta_i - \tan^{-1}\biggl(\frac{\eta_i}{y_i}\biggr)\, .</math>
  </td>
</tr>
</table>

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


Recalling that,
<div align="center">
<math>~Q = \biggl[1- \eta\cot(\eta-B) \biggr] \, ,</math>
</div>
plugging these expressions into the relevant envelope LAWE gives,

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

<tr>
  <td align="right">
LAWE
  </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} 
~-~  2 Q \cdot \frac{x}{\eta^2} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{\eta^4} \biggl\{
\frac{\eta^4}{b} \cdot \frac{d^2x}{d\eta^2} + \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b} \cdot \frac{dx}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 x}{b} 
\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\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-C)  
-  \eta^2\cot^2(\eta - C) 
-  \eta^3\cot^3(\eta-C) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \biggl[\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C)\biggr]
~-~  \biggl[1- \eta\cot(\eta-B) \biggr] \biggl[1- \eta\cot(\eta-C) \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\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-C)  
-  \eta^2\cot^2(\eta - C) 
-  \eta^3\cot^3(\eta-C) 
+ \biggl[\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C)\biggr]
~-~  \biggl[1- \eta\cot(\eta-C) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \eta\cot(\eta-B) \biggl[\eta^2 -2 + \eta\cot(\eta-C)  +  \eta^2\cot^2(\eta - C)\biggr]
~+~\eta\cot(\eta-B) \biggl[1- \eta\cot(\eta-C) \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\{
(\eta - \eta^3)\cot(\eta-C)  
-  \eta^3\cot^3(\eta-C) 
+ \eta\cot(\eta-B) \biggl[\eta^2 -1   +  \eta^2\cot^2(\eta - C)  \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\{
(\eta - \eta^3) [ \cot(\eta-C)  - \cot(\eta-B) ]
+ \eta^3 \cot^2(\eta - C) [\cot(\eta-B)-  \cot(\eta-C)] 
\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[ \cot(\eta-C)  - \cot(\eta-B) \biggr] 
\biggl[ \eta - \eta^3 - \eta^3 \cot^2(\eta - C) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^3} \biggl[ \cot(\eta-C)  - \cot(\eta-B) \biggr] 
\biggl\{ 1 - \eta^2\biggl[1 + \cot^2(\eta - C)\biggr] \biggr\}\, .
</math>
  </td>
</tr>
</table>

This will go to zero if <math>~C = (B-2m\pi), </math> where <math>~m</math> is a positive integer.  When <math>~m =1</math>, for example,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cot(\eta-C)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cot[\eta - (B-2\pi)] = \cot(\eta -B) \, . </math>
  </td>
</tr>
</table>


Okay.  Now let's determine at what value of <math>~\eta</math> the logarithmic derivative of <math>~x_P</math> goes to negative one.

====Brute-Force Trial Fit====
<table align="left" width="100%" cellpadding="0"><tr><td align="left">
<table border="0" align="right"><tr><td align="center">
[[File:BruteForceWhiteBoardsmall.png|500px|Photo of white board with steps showing development of trial eigenfunction.  This should be paired with an Excel spreadsheet.]]
</td></tr></table>
Using a couple of separate Excel spreadsheets &#8212; FaulknerBipolytrope2.xlsx/mu100Mode0 and AnalyticTrialBipolytropeA.xlsx/Sheet2, both stored in a DropBox account under the folder Wiki_edits/Bipolytrope/LinearPerturbation &#8212; we used an inelegant and inefficient trial &amp; error technique in search of an eigenfunction that had the same analytic ''form'' as the one represented above for <math>~x_P</math>, but that, when plotted, appeared to qualitatively match the numerically determined envelope eigenfunction.  Then, on a whiteboard &#8212; see the photo, here on the right &#8212; we formulated a concise expression for a trial function that seemed to work pretty well.  Our primary finding was that <math>~\alpha</math>, appearing as the argument to the <math>~\tan\alpha</math> function, needed to be shifted by something like <math>~-3\pi/4</math>.

<div align="center">
<font size="+3">
<p>&nbsp;</p><p>THIS SPACE</p><p> INTENTIONALLY</p><p> LEFT BLANK</p>
</font>
</div>
</td></tr></table>

====Following Up on the Brute-Force Trial Fit====
In an [[SSC/Stability/BiPolytropes#Is_There_an_Analytic_Expression_for_the_Eigenfunction.3F|accompanying discussion]] &#8212; see especially [[SSC/Stability/BiPolytropes#Attempt_2|Attempt #2]] &#8212; we have determined by visual inspection that a decent fit to the envelope's eigenfunction is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_\mathrm{trial}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \biggl[ \frac{\tan(\eta_i - \Lambda - 3\pi/4) + f_\alpha}{1 - f_\alpha \cdot \tan(\eta_i - \Lambda - 3\pi/4)} \biggr] \biggr\} - a_0 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \cot(\Lambda - E)\biggr\} - a_0 \, ,
</math>
  </td>
</tr>
</table>

<table border="1" cellpadding="5" align="right">
<tr>
  <th align="center" colspan="4">Limiting Parameter Values</th>
</tr>
<tr>
  <td align="center">&nbsp;</td>
  <td align="center">min</td>
  <td align="center">max</td>
  <td align="center"><math>~\alpha = \alpha_s</math>
</tr>
<tr>
  <td align="center"><math>~\eta_\mathrm{F}</math></td>
  <td align="center"><math>~\eta_i</math></td>
  <td align="center"><math>~\eta_s</math></td>
  <td align="center"><math>~\frac{8}{\pi} ( \eta_s - \eta_i )^2 + 2\eta_s - \eta_i</math></td>
</tr>
<tr>
  <td align="center"><math>~\alpha</math></td>
  <td align="center"><math>~-\frac{\pi}{2}</math></td>
  <td align="center"><math>~-\frac{5\pi}{8}</math></td>
  <td align="center"><math>~\eta_i - \eta_s - \frac{3\pi}{4}</math></td>
</tr>
<tr>
  <td align="center"><math>~\Lambda</math></td>
  <td align="center"><math>~\eta_i - \frac{\pi}{4}</math></td>
  <td align="center"><math>~\eta_i - \frac{\pi}{8}</math></td>
  <td align="center"><math>~\eta_s</math></td>
</tr>
</table>
where, over the range, <math>~\eta_i \le \eta \le \eta_s \, ,</math>

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

<tr>
  <td align="right">
<math>~E</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\eta_i - \frac{5\pi}{4} + \tan^{-1} f_\alpha \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Lambda(\eta)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\eta_i + g_\mathrm{F}  \biggl[ \eta_i - 2\eta_s + \eta \biggr] = \Lambda_0 + g_\mathrm{F}\eta 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{1}{f_\alpha} = \tan(\alpha_s)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\tan[ - (\eta_s - \eta_i + \tfrac{3\pi}{4}) ]
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~g_\mathrm{F}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{\pi}{8(\eta_s - \eta_i)}
\, .</math>
  </td>
</tr>
</table>


----

Here, we reference a [[SSC/Stability/BiPolytropes#Attempt_1|separate discussion of the bipolytrope's underlying equilibrium structure]]

<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" width="50%"><math>~B = \eta_i - \frac{\pi}{2} + \tan^{-1}f</math></td>
  <td align="center"><math>~E = \eta_i - \frac{5\pi}{4} + \tan^{-1}f_\alpha</math></td>
</tr>
<tr>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow~~~\cot(\eta_i - B)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\eta_i - B)]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\tfrac{\pi}{2} - \tan^{-1}f)]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~f</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~f</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan(B + \tfrac{\pi}{2} - \eta_i )</math>
  </td>
</tr>
</table>
  </td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow~~~\cot(\eta_i - E)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\eta_i - E)]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\tfrac{5\pi}{4} - \tan^{-1}f_\alpha)]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan( \tan^{-1}f_\alpha - \tfrac{3\pi}{4} )</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\tan( \tfrac{3\pi}{4} - \tan^{-1}f_\alpha )</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\cot( \tan^{-1}f_\alpha )</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{1}{f_\alpha}</math>
  </td>
</tr>
</table>
  </td>
</tr>
<tr>
  <td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
Hence &hellip; &nbsp; &nbsp; <math>~\cot(\eta - B)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\eta - B)]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - \eta + \eta_i - \tfrac{\pi}{2} + \tan^{-1}f]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\eta_i - \eta + \tan^{-1}f]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ \tan(\eta_i-\eta) + f }{ 1 - f \cdot \tan(\eta_i - \eta)}</math>
  </td>
</tr>
</table>
  </td>
  <td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
Hence &hellip; &nbsp; &nbsp; <math>~\cot(\Lambda - E)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\tfrac{\pi}{2} - (\Lambda - E)]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan[\eta_i - \Lambda - \tfrac{3\pi}{4} + \tan^{-1}f_\alpha]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\tan(\eta_i - \Lambda - \tfrac{3\pi}{4}) + f_\alpha  }{1 - f_\alpha \cdot \tan(\eta_i - \Lambda - \tfrac{3\pi}{4})  }</math>
  </td>
</tr>
</table>
  </td>
 </tr>
<tr>
  <td align="center">Also &hellip; &nbsp; &nbsp;<math>~B = \eta_s - \pi</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ f = \cot(\eta_i - B)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cot(\eta_i - \eta_s + \pi)</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{f}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan(\eta_i - \eta_s + \pi)</math>
  </td>
</tr>
</table>
  </td>
  <td align="center"><math>~</math></td>
</tr>
</table> 


----

Let's examine the first and second derivatives of this trial eigenfunction, recognizing that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx_\mathrm{trial}}{d\eta} = \frac{d\Lambda}{d\eta} \cdot \frac{dx_\mathrm{trial}}{d\Lambda}= g_\mathrm{F} \cdot  \frac{dx_\mathrm{trial}}{d\Lambda}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{d^2x_\mathrm{trial}}{d\eta^2} 
= \frac{d\Lambda}{d\eta} \cdot \frac{d}{d\Lambda} \biggl[ g_\mathrm{F}\cdot \frac{dx_\mathrm{trial}}{d\Lambda} \biggr]
= g_\mathrm{F}^2 \cdot \frac{d^2x_\mathrm{trial}}{d\Lambda^2} \, .
</math>
  </td>
</tr>
</table>
and drawing from the [[#Prior_to_the_Brute-Force_Trial_Fit|derivative expressions already derived, above]].  For the first derivative, we have,

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

<tr>
  <td align="right">
<math>~\frac{dx_\mathrm{trial}}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
g_\mathrm{F} \biggl( \frac{b_0}{\Lambda^3} \biggr) 
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] \, .
</math>
  </td>
</tr>
</table>
And the second derivative gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2x_\mathrm{trial}}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
g_\mathrm{F}^2 \biggl(\frac{2b_0}{\Lambda^4} \biggr) \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr]  \, .
</math>
  </td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x_\mathrm{trial}}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx_\mathrm{trial}}{d\eta} 
~-~  2 Q \cdot \frac{x_\mathrm{trial}}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x_\mathrm{trial}}{d\eta^2} + \biggl\{ 4 -2 \biggl[1- \eta\cot(\eta-B) \biggr]\biggr\}\frac{1}{\eta} \cdot \frac{dx_\mathrm{trial}}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]   \frac{2x_\mathrm{trial}}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b_0}{\eta^4} \biggl\{
\frac{\eta^4}{b_0} \cdot \frac{d^2x_\mathrm{trial}}{d\eta^2} + \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b_0} \cdot \frac{dx_\mathrm{trial}}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 x_\mathrm{trial}}{b_0} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b_0}{\eta^4} \biggl\{
\frac{\eta^4}{b_0} \cdot g_\mathrm{F}^2 \biggl(\frac{2b_0}{\Lambda^4} \biggr) \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b_0} \cdot 
g_\mathrm{F} \biggl( \frac{b_0}{\Lambda^3} \biggr) 
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 }{b_0} \cdot 
\biggl[\frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \cot(\Lambda - E)\biggr\} - a_0\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b_0}{\eta^4} \biggl\{
g_\mathrm{F}^2 \biggl(\frac{2\eta^4}{\Lambda^4} \biggr) \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \cdot 
g_\mathrm{F} \biggl( \frac{2\eta^3}{\Lambda^3} \biggr) 
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  
\biggl[\frac{2\eta^2}{\Lambda^2} [ 1 - \Lambda \cot(\Lambda - E) ] - \frac{2\eta^2 a_0}{b_0} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b_0}{\Lambda^4\eta^2} \biggl\{
g_\mathrm{F}^2 \eta^2 \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\cot(\eta-B) \biggr] \cdot 
g_\mathrm{F} \Lambda \eta 
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  
\biggl[\Lambda^2 [ 1 - \Lambda \cot(\Lambda - E) ] - \frac{a_0\Lambda^4}{b_0} \biggr]
\biggr\}
</math>
  </td>
</tr>
</table>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow~~~\biggl(\frac{\Lambda^4}{2b_0}\biggr) \cdot</math> &nbsp;LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
g_\mathrm{F}^2 \biggl[ 3 - \Lambda^2 
- (\Lambda + \Lambda^3)\cot(\Lambda-E)  
-  \Lambda^2\cot^2(\Lambda - E) 
-  \Lambda^3\cot^3(\Lambda-E) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  
\frac{g_\mathrm{F} \Lambda}{ \eta }
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
~-~    \biggl(\frac{\Lambda}{\eta}\biggr)^2\biggl[  1 - \Lambda \cot(\Lambda - E)  - \frac{a_0\Lambda^2}{b_0} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ \eta\cot(\eta-B) \biggr] \biggl\{ 
\frac{g_\mathrm{F} \Lambda }{\eta }
\biggl[ \Lambda ^2 -2 + \Lambda\cot(\Lambda-E)  +  \Lambda^2\cot^2(\Lambda - E) \biggr] 
~+~ \biggl( \frac{\Lambda}{\eta}\biggr)^2 \biggl[ 1 - \Lambda \cot(\Lambda - E)  - \frac{a_0\Lambda^2}{b_0 }\biggr]  \biggr\}
</math>
  </td>
</tr>
</table>


===Fourth Attempt===

====XXXX====
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,

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

<tr>
  <td align="right">
<math>~Q \equiv - \frac{d \ln \phi}{ d\ln \eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1- \eta\cot(\eta-B_0) \biggr] \, .
</math>
  </td>
</tr>
</table>


Let's work through the analytic derivatives again.  Keeping in mind that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\eta - B) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\eta - B)\biggr] \, ;
</math>
  </td>
</tr>
</table>
and that the,

<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>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-B_0) \biggr] \, .
</math>
  </td>
</tr>
</table>

As we have [[#First_Attempt|already tried once, above]], let's try a more general form of this expression, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_Q</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
A + \frac{C}{(\eta - F)^2} \biggl[1 - (\eta-D) \cot(\eta-B) \biggr] \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{dx_Q}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 - (\eta-D) \cot(\eta-B) \biggr] \frac{d}{d\eta}\biggl[ \frac{C}{(\eta - F)^2} \biggr]
- \frac{C}{(\eta - F)^2} \frac{d}{d\eta} \biggl[ (\eta-D) \cot(\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-D) \cot(\eta-B) \biggr]\biggl[ \frac{-2C}{(\eta - F)^3} \biggr]
- \frac{C}{(\eta - F)^2} \biggl[ \cot(\eta-B) \biggr]
+ \frac{C(\eta - D)}{(\eta - F)^2}\biggl[1 + \cot^2(\eta - B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^3} \biggl\{
- 2 + 2(\eta-D) \cot(\eta-B) 
- (\eta - F) \biggl[ \cot(\eta-B) \biggr]
+ (\eta - D)(\eta - F) \biggl[1 + \cot^2(\eta - B)\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^3} \biggl\{
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\} \, .
</math>
  </td>
</tr>
</table>

And,

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+\frac{C}{(\eta-F)^3}  \biggl\{
[2\eta - (D+F) ] + \cot(\eta-B)
- (\eta - 2D + F) \biggl[1 + \cot^2(\eta-B) \biggr]
+ [2\eta -(D+F) ] \cot^2(\eta - B)
- 2[\eta^2 -\eta(D+F) + DF]\cot(\eta - B)\biggl[1 +  \cot^2(\eta - B)\biggr]
\biggr\} 
</math>
  </td>
</tr>

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

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

====YYYY====
And,

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

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
-3(\eta - 2D + F) \cot(\eta-B) 
-3(\eta - D)(\eta - F) \cot^2(\eta - B)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
-3(\eta - 2D + F) \cot(\eta-B) 
-3(\eta - D)(\eta - F) \cot^2(\eta - B)
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
-3(\eta - 2D + F) \cot(\eta-B) 
-3(\eta - D)(\eta - F) \cot^2(\eta - B)
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{C}{(\eta-F)^4} \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
- (\eta-F) (\eta - 2D + F) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ (\eta-F)
-3(\eta - 2D + F) 
-2 (\eta-F) (\eta - D)(\eta - F)\biggr] \cot(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ (\eta-F) [  2\eta - (D+F) ]
-3(\eta - D)(\eta - F) 
-2 (\eta-F) (\eta - D)(\eta - F) \cot(\eta - B)
- (\eta-F) (\eta - 2D + F)\biggr] \cot^2(\eta - B)
\biggr\} 
</math>
  </td>
</tr>
</table>

So the envelope LAWE becomes,

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

<tr>
  <td align="right">
<math>~\frac{(\eta-F)^4}{C} \cdot \mathrm{LAWE}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\eta-F)^4}{C} \cdot \frac{d^2x_Q}{d\eta^2} + \frac{(\eta-F)^4}{C} \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \cdot \frac{dx_Q}{d\eta} 
~-~ \frac{(\eta-F)^4}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2x_Q}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{
-3[(\eta - D)(\eta - F) - 2] 
- (\eta-F) (\eta - 2D + F) 
+ \biggl[ (\eta-F)
-3(\eta - 2D + F) 
-2 (\eta-F) (\eta - D)(\eta - F)\biggr] \cot(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ (\eta-F) [  2\eta - (D+F) ]
-3(\eta - D)(\eta - F) 
-2 (\eta-F) (\eta - D)(\eta - F) \cot(\eta - B)
- (\eta-F) (\eta - 2D + F)\biggr] \cot^2(\eta - B)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta-F) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl\{ 
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
+ (\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~ \frac{(\eta-F)^4}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2}{\eta^2} \biggl\{
A + \frac{C}{(\eta - F)^2} \biggl[1 - (\eta-D) \cot(\eta-B) \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-3[(\eta - D)(\eta - F) - 2] 
- (\eta-F) (\eta - 2D + F) 
+ \biggl[ (\eta-F)
-3(\eta - 2D + F) 
-2 (\eta-F) (\eta - D)(\eta - F)\biggr] \cot(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta-F) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl\{ 
[(\eta - D)(\eta - F) - 2] 
+ (\eta - 2D + F) \cot(\eta-B) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ (\eta-F) [  2\eta - (D+F) ]
-3(\eta - D)(\eta - F) 
-2 (\eta-F) (\eta - D)(\eta - F) \cot(\eta - B)
- (\eta-F) (\eta - 2D + F)\biggr] \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta-F) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl[ 
(\eta - D)(\eta - F) \cot^2(\eta - B)
\biggr]
~-~ (\eta-F)^2  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2}{\eta^2} \biggl[1 - (\eta-D) \cot(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~ \frac{(\eta-F)^4}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2A}{\eta^2} \, .
</math>
  </td>
</tr>
</table>

What does this reduce to if <math>~A = D = F = 0</math>.

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl\{ 
[(\eta )(\eta ) - 2] 
+ (\eta ) \cot(\eta-B) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ (\eta) [  2\eta ]
-3(\eta )(\eta ) 
-2 (\eta) (\eta )(\eta) \cot(\eta - B)
- (\eta-) (\eta )\biggr] \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (\eta) \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \frac{2}{\eta} \biggl[ 
(\eta )(\eta ) \cot^2(\eta - B)
\biggr]
~-~ (\eta)^2  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2}{\eta^2} \biggl[1 - (\eta) \cot(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~ \frac{(\eta)^4}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]   \frac{2A}{\eta^2} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
6 - 4\eta^2
-2 (\eta + \eta^3 ) \cot(\eta - B)
+ 2 \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr] \biggl[
\eta^2 - 2 + \eta  \cot(\eta-B) 
\biggr]
- 2\biggl[ \eta^2 + \eta^3 \cot(\eta - B) \biggr] \cot^2(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2 \biggl[ 1 + \eta\cot(\eta-B_0)  \biggr]\biggl[ 
\eta^2 \cot^2(\eta - B)
\biggr]
~-~2  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr] \biggl[1 - \eta \cot(\eta-B) \biggr]
~-~ \frac{2A\eta^2}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
6 - 4\eta^2
-2 (\eta + \eta^3 ) \cot(\eta - B)
+ 2\eta^2 - 4 + 2\eta  \cot(\eta-B) 
+ 2 \eta^3 \cot(\eta-B_0)
~-~4 \eta\cot(\eta-B_0) 
+ 2 \eta^2\cot(\eta-B_0) \cot(\eta-B) 
- 2\eta^2  \cot^2(\eta - B)
- 2 \eta^3  \cot^3(\eta - B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2 \eta^2 \cot^2(\eta - B)
+ 2 \eta^3 \cot(\eta-B_0)  \cot^2(\eta - B)
-2 + 4 \eta\cot(\eta-B_0) - 2\eta^2\cot^2(\eta-B_0)  
~-~ \frac{2A\eta^2}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2\eta^2
-2 \eta^3 \biggl[ \cot(\eta - B) + \cot^3(\eta - B) \biggr]
+ \cot(\eta-B_0) \biggl[ 2 \eta^3 
+ 2 \eta^2 \cot(\eta-B) 
+ 2 \eta^3  \cot^2(\eta - B)
\biggr] - 2\eta^2\cot^2(\eta-B_0)  
~-~ \frac{2A\eta^2}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]    
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2\eta^2
+ 2 \eta^3\biggl[ \cot(\eta-B_0) - \cot(\eta-B)\biggr] \biggl[ 1 +  \cot^2(\eta - B)\biggr] 
+ 2 \eta^2  \cot(\eta-B_0)\biggl[ \cot(\eta-B)  - \cot(\eta-B_0)  \biggr]
~-~ \frac{2A\eta^2}{C}  \biggl[ 1- \eta\cot(\eta-B_0)  \biggr]  \, .  
</math>
  </td>
</tr>
</table>