Editing Appendix/Mathematics/Hypergeometric (section)

==Parabolic Density Distribution==

===Relevant, Parabolic LAWE===

In the case of a parabolic density distribution, we have found that the equilibrium configuration is defined by the relations:
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>x</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{r_0}{R} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\rho_0}{\rho_c}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{P_0}{P_c}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2)^2 (1 - \tfrac{1}{2} x^2) \, ,</math> &nbsp; &nbsp; &nbsp; where, &nbsp; &nbsp; &nbsp; <math>P_c = \biggl(\frac{4\pi}{15}\biggr) G\rho_c^2 R^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>g_0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl(\frac{4\pi}{3}\biggr) G\rho_c R ~x(1- \tfrac{3}{5}x^2) \, ,
</math>
  </td>
</tr>
</table>
in which case,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>\mu = \frac{g_0 \rho_0 r_0}{P_0}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl(\frac{4\pi}{3}\biggr) G\rho_c R ~x(1- \tfrac{3}{5}x^2) \rho_c (1-x^2) Rx 
\biggl[ \biggl(\frac{4\pi}{15}\biggr) G\rho_c^2 R^2 (1-x^2)^2 (1 - \tfrac{1}{2} x^2)\biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
x^2(5 - 3x^2) \biggl[(1-x^2) (1 - \tfrac{1}{2} x^2) \biggr]^{-1} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr)</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl(\frac{\omega^2 R^2}{\gamma_g}\biggr) \rho_c (1-x^2)  
\biggl[ \biggl(\frac{4\pi}{15}\biggr) G\rho_c^2 R^2 (1-x^2)^2 (1 - \tfrac{1}{2} x^2)\biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl(\frac{15 \omega^2 }{4\pi G \rho_c\gamma_g}\biggr)   
\biggl[ (1-x^2) (1 - \tfrac{1}{2} x^2)\biggr]^{-1}
\, .
</math>
  </td>
</tr>
</table>

Hence, in the case of a parabolic density distribution, the [[#Kopal48Expression|Kopal (1948) LAWE]] becomes,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{d^2 f}{dx^2} + \frac{(4-\mu)}{x} \cdot \frac{d f}{dx} 
+ \biggl\{
\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr) 
- 
\frac{\alpha \mu}{x^2} 
\biggr\} f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{d^2 f}{dx^2} + \frac{1}{x}\biggl\{ 
4 - x^2(5 - 3x^2) \biggl[(1-x^2) (1 - \tfrac{1}{2} x^2) \biggr]^{-1}
\biggr\}\cdot \frac{d f}{dx} 
+ \biggl\{
\biggl(\frac{15 \omega^2 }{4\pi G \rho_c\gamma_g}\biggr)   
- 
\alpha  
(5 - 3x^2) 
\biggr\}\biggl[(1-x^2) (1 - \tfrac{1}{2} x^2) \biggr]^{-1} f \, .
</math>
  </td>
</tr>
</table>
Multiplying through by <math>[(1-x^2) (1 - \tfrac{1}{2} x^2)]</math> gives,

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2) (1 - \tfrac{1}{2} x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 
(1-x^2) (4 - 2 x^2)  - x^2(5 - 3x^2) 
\biggr] \cdot \frac{d f}{dx} 
+ \biggl[
\biggl(\frac{15 \omega^2 }{4\pi G \rho_c\gamma_g}\biggr)   
- 
\alpha  
(5 - 3x^2) 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2) (1 - \tfrac{1}{2} x^2) \cdot \frac{d^2 f}{dx^2} 
+ \frac{1}{x}\biggl[ 
4 - 11x^2  + 5 x^4  
\biggr] \cdot \frac{d f}{dx} 
+ \biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha x^2 
\biggr] \cdot f 
</math>
  </td>
</tr>
</table>

where,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>\mathfrak{F}</math>
  </td> 
  <td align="center">
<math>\equiv</math>
  </td> 
  <td align="left">
<math>
2\biggl[\biggl(\frac{3\omega^2}{4\pi \gamma_\mathrm{g} G \rho_c} \biggr) 
- \alpha \biggr] \, . 
</math>
  </td>
</tr>
</table>

===Change of Variable===

In an effort to shift this LAWE into a 2<sup>nd</sup>-order ODE that has the form of an hypergeometric equation, let's try &hellip;

====First Try====
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>z = x^2 - 1</math>
  </td> 
  <td align="center">
<math>\Rightarrow</math>
  </td> 
  <td align="left">
<math>
x = (1+z)^{1 / 2} \, ;
</math>
&nbsp; &nbsp; &nbsp; also, &nbsp; &nbsp; &nbsp; <math>1 - \tfrac{1}{2} x^2 = \tfrac{1}{2}(1-z) \, ;</math> 
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{dz}{dx}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
2x = 2(1+z)^{1 / 2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{df}{dx} = \frac{dz}{dx} \cdot \frac{df}{dz}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
2(1+z)^{1 / 2} \cdot \frac{df}{dz} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{d^2f}{dx^2} = \frac{dz}{dx} 
\cdot \frac{d}{dz} \biggl[ 2(1+z)^{1 / 2} \cdot \frac{df}{dz} \biggr]
</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
2(1+z)^{1 / 2}  
\biggl[ (1+z)^{-1 / 2} \cdot \frac{df}{dz}  + 2(1+z)^{1 / 2} \cdot \frac{d^2 f}{dz^2}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ 2 \cdot \frac{df}{dz}  + 4(1+z) \cdot \frac{d^2 f}{dz^2}\biggr] \, .
</math>
  </td>
</tr>
</table>
Hence, the ''parabolic'' LAWE takes the form,

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2) (1 - \tfrac{1}{2} x^2) \cdot \frac{d^2 f}{dx^2} 
+ \frac{1}{x}\biggl[ 
4 - 11x^2  + 5 x^4  
\biggr] \cdot \frac{d f}{dx} 
+ \biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha x^2 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(-z) (1 - z) \cdot \biggl[ \frac{df}{dz}  + 2(1+z) \cdot \frac{d^2 f}{dz^2}\biggr] 
+ (1+z)^{-1 / 2}\biggl[ 
4 - 11(1+z)  + 5 (1+z)^2  
\biggr] 2(1+z)^{1 / 2} \cdot \frac{df}{dz} 
+ \biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha (1+z) 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
-z(1 - z) \cdot \biggl[ 2(1+z) \cdot \frac{d^2 f}{dz^2}\biggr] 
-z(1 - z) \cdot \frac{df}{dz} 
+ \biggl[ 
8 - 22(1+z)  + 10 (1+z)^2  
\biggr] \cdot \frac{df}{dz} 
+ \biggl[
\biggl( \frac{5}{2} \cdot \mathfrak{F} + 3\alpha\biggr) + 3\alpha z 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
-2z(1 - z^2) \cdot \frac{d^2 f}{dz^2} 
+ \biggl[ 
8 - 22 - 22z  + 10 + 20z + 10z^2 + z^2-z 
\biggr] \cdot \frac{df}{dz} 
+ \biggl[
\biggl( \frac{5}{2} \cdot \mathfrak{F} + 3\alpha\biggr) + 3\alpha z 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
-2z(1 - z^2) \cdot \frac{d^2 f}{dz^2} 
+ \biggl[ 
-4 - 3z + 11z^2 
\biggr] \cdot \frac{df}{dz} 
+ \biggl[
\biggl( \frac{5}{2} \cdot \mathfrak{F} + 3\alpha\biggr) + 3\alpha z 
\biggr] \cdot f 
</math>
  </td>
</tr>
</table>

====Second Try====
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>z = Fx^B - G</math>
  </td> 
  <td align="center">
<math>\Rightarrow</math>
  </td> 
  <td align="left">
<math>
x = \biggl( \frac{G + z}{F} \biggr)^{1 / B} \, ;
</math>
&nbsp; &nbsp; &nbsp; also, &nbsp; &nbsp; &nbsp; <math>1 - \tfrac{1}{2} x^2 = 1 - \frac{1}{2}\biggl( \frac{G + z}{F} \biggr)^{2 / B}
\, ;</math> 
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{dz}{dx}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
BFx^{B-1} = BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{df}{dx} = \frac{dz}{dx} \cdot \frac{df}{dz}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\frac{df}{dz}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{d^2f}{dx^2} = \frac{dz}{dx} \cdot \frac{d}{dz}\biggl[ 
BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\frac{df}{dz}
\biggr]</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\biggl\{
BF \cdot F^{-1 + 1/B} \biggl[ \frac{B-1}{B}\biggr] \biggl( G + z \biggr)^{-1 / B} \cdot \frac{df}{dz}
+
BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\frac{d^2f}{dz^2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
BF^{1/B}\biggl( G+z \biggr)^{1 - 1/B}\biggl\{
F^{1/B} (B-1) \biggl( G + z \biggr)^{-1 / B} \cdot \frac{df}{dz}
+
BF^{1/B}\biggl( G+z \biggr)^{1 - 1/B}\frac{d^2f}{dz^2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
BF^{2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[
(B-1) \cdot \frac{df}{dz}
+
B( G + z )  \cdot \frac{d^2f}{dz^2}
\biggr]
\, .
</math>
  </td>
</tr>
</table>

This should reduce to the "First Try" example by setting:  &nbsp;<math>(B, F, G) = (2, 1, 1)</math>.  Let's see &hellip;
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>\frac{dz}{dx}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B} = 2(1+z)^{1 / 2}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{df}{dx} </math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\frac{df}{dz}
=
2(1+z)^{1 / 2} \cdot \frac{df}{dz}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{d^2f}{dx^2} 
</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
BF^{2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[
(B-1) \cdot \frac{df}{dz}
+
B( G + z )  \cdot \frac{d^2f}{dz^2}
\biggr]
=
2(1+z)^{0}\biggl[\frac{df}{dz} + 2(1+z) \frac{d^2f}{dz^2}  \biggr]
\, .
</math>
  </td>
</tr>
</table>

<table border="1" width="80%" cellpadding="10" align="center"><tr><td align="left">
<div align="center"><b>Functional Expressions from "First Try"<b></div>
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>z = x^2 - 1</math>
  </td> 
  <td align="center">
<math>\Rightarrow</math>
  </td> 
  <td align="left">
<math>
x = (1+z)^{1 / 2} \, ;
</math>
&nbsp; &nbsp; &nbsp; also, &nbsp; &nbsp; &nbsp; <math>1 - \tfrac{1}{2} x^2 = \tfrac{1}{2}(1-z) \, ;</math> 
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{dz}{dx}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
2(1+z)^{1 / 2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{df}{dx}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
2(1+z)^{1 / 2} \cdot \frac{df}{dz} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{d^2f}{dx^2} 
</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ 2 \cdot \frac{df}{dz}  + 4(1+z) \cdot \frac{d^2 f}{dz^2}\biggr] \, .
</math>
  </td>
</tr>
</table>

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

Hence, the ''parabolic'' LAWE takes the form,

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2) (1 - \tfrac{1}{2} x^2) \cdot \frac{d^2 f}{dx^2} 
+ \frac{1}{x}\biggl[ 
4 - 11x^2  + 5 x^4  
\biggr] \cdot \frac{d f}{dx} 
+ \biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha x^2 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ 1- \biggl( \frac{G+z}{F} \biggr)^{2/B} \biggr] \biggl[ 1 - \frac{1}{2} \biggl( \frac{G+z}{F} \biggr)^{2/B}\biggr] 
BF^{2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[
(B-1) \cdot \frac{df}{dz}
+
B( G + z )  \cdot \frac{d^2f}{dz^2}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ \biggl( \frac{G+z}{F} \biggr)^{-1/B}\biggl[ 
4 - 11x^2  + 5 x^4  
\biggr] \cdot BF\biggl( \frac{G + z}{F} \biggr)^{1 - 1/B}\frac{df}{dz} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ \biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( \frac{G+z}{F} \biggr)^{2/B} 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ F^{2/B} - \biggl( G+z \biggr)^{2/B} \biggr] \biggl[ F^{2/B} - \frac{1}{2} \biggl( G+z \biggr)^{2/B}\biggr] 
BF^{-2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[
B( G + z )  \cdot \frac{d^2f}{dz^2}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+\biggl[ F^{2/B} - \biggl( G+z \biggr)^{2/B} \biggr] \biggl[ F^{2/B} - \frac{1}{2} \biggl( G+z \biggr)^{2/B}\biggr] 
BF^{-2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[
(B-1) \cdot \frac{df}{dz}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ \biggl[ 
4F^{4/B} - 11 F^{2/B}\biggl( G+z \biggr)^{2/B}  + 5 \biggl( G+z \biggr)^{4/B}  
\biggr] \cdot BF^{-2/B}\biggl( G + z \biggr)^{1 - 2/B} \cdot \frac{df}{dz} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ \biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha F^{-2/B} \biggl( G+z \biggr)^{2/B} 
\biggr] \cdot f 
</math>
  </td>
</tr>
</table>

Dividing through by, <math>BF^{-2/B}( G+z )^{1 - 2/B}</math>, we have,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl\{  
\biggl[ F^{2/B} - \biggl( G+z \biggr)^{2/B} \biggr] \biggl[ F^{2/B} - \frac{1}{2} \biggl( G+z \biggr)^{2/B}\biggr] 
B( G + z )
\biggr\}  \cdot \frac{d^2f}{dz^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ 
\biggl\{ \biggl[ F^{2/B} - \biggl( G+z \biggr)^{2/B} \biggr] \biggl[ F^{2/B} - \frac{1}{2} \biggl( G+z \biggr)^{2/B}\biggr] 
(B-1) 
+ \biggl[ 
4F^{4/B} - 11 F^{2/B}\biggl( G+z \biggr)^{2/B}  + 5 \biggl( G+z \biggr)^{4/B}  \biggr] 
\biggr\} \cdot \frac{df}{dz} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ B^{-1}( G+z )^{-1 + 2/B}\biggl[
F^{2/B}\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr)^{2/B} 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{1}{2}\biggl[ 2F^{4/B} - 3 F^{2/B} \biggl( G+z \biggr)^{2/B} + \biggl( G+z \biggr)^{4/B}\biggr] 
B( G + z )
\cdot \frac{d^2f}{dz^2}
+ B^{-1}( G+z )^{-1 + 2/B}\biggl[
F^{2/B}\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr)^{2/B} 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ \frac{1}{2}
\biggl\{ \biggl[ 2F^{4/B} - 3 F^{2/B} \biggl( G+z \biggr)^{2/B} + \biggl( G+z \biggr)^{4/B}\biggr] 
(B-1) 
+ \biggl[ 
8F^{4/B} - 22 F^{2/B}\biggl( G+z \biggr)^{2/B}  + 10 \biggl( G+z \biggr)^{4/B}  \biggr] 
\biggr\} \cdot \frac{df}{dz} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{1}{2}\biggl[ 2F^{4/B} - 3 F^{2/B} \biggl( G+z \biggr)^{2/B} + \biggl( G+z \biggr)^{4/B}\biggr] 
B( G + z )
\cdot \frac{d^2f}{dz^2}
+ B^{-1}( G+z )^{-1 + 2/B}\biggl[
F^{2/B}\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr)^{2/B} 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ \frac{1}{2}
\biggl[ (2B+6)F^{4/B} - ( 3B + 19) F^{2/B} \biggl( G+z \biggr)^{2/B} + (B+9)\biggl( G+z \biggr)^{4/B}\biggr] 
\cdot \frac{df}{dz} 
</math>
  </td>
</tr>
</table>

Now, this should reduce to the "First Try" example by setting:  &nbsp;<math>(B, F, G) = (2, 1, 1)</math>.  Let's see &hellip;

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{1}{2}\biggl[ 2 - 3  \biggl( 1+z \biggr) + \biggl( 1+z \biggr)^{2}\biggr] 
2( 1 + z )
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ 10 - 25 \biggl( 1+z \biggr) + 11\biggl( 1+z \biggr)^{2}\biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( 1+z \biggr) 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ 2 - 3 -3z  + \biggl( 1+2z +z^2\biggr)\biggr] 
( 1 + z )
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ 10 - 25 - 25z  + 11\biggl( 1+ 2z + z^2 \biggr)\biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( 1+z \biggr) 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
z(z-1) ( 1 + z )\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ -4 - 3z  + 11 z^2 \biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( 1+z \biggr) 
\biggr] \cdot f \, .
</math>
  </td>
</tr>
</table>

<table border="1" width="80%" cellpadding="10" align="center"><tr><td align="left">
<div align="center"><b>Compare with LAWE from "First Try"<b></div>
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
-2z(1 - z^2) \cdot \frac{d^2 f}{dz^2} 
+ \biggl[ 
-4 - 3z + 11z^2 
\biggr] \cdot \frac{df}{dz} 
+ \biggl[
\biggl( \frac{5}{2} \cdot \mathfrak{F} + 3\alpha\biggr) + 3\alpha z 
\biggr] \cdot f 
</math>
  </td>
</tr>
</table>

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

Next, let's try setting <math>B = 2</math> while leaving <math>F</math> and <math>G</math> arbitrary.  The LAWE becomes,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{1}{2}\biggl[ 2F^{2} - 3 F \biggl( G+z \biggr) + \biggl( G+z \biggr)^{2}\biggr] 2( G + z )
\cdot \frac{d^2f}{dz^2}
+ 2^{-1}( G+z )^{0}\biggl[
F\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr) 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
&nbsp;
  </td> 
  <td align="left">
<math>
+ \frac{1}{2}
\biggl[ 10F^{2} - 25 F \biggl( G+z \biggr) + 11\biggl( G+z \biggr)^{2}\biggr] 
\cdot \frac{df}{dz} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ 2F^{2} - 3 F \biggl( G+z \biggr) + \biggl( G+z \biggr)^{2}\biggr] ( G + z )
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ 10F^{2} - 25 F \biggl( G+z \biggr) + 11\biggl( G+z \biggr)^{2}\biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
F\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr) 
\biggr] \cdot f \, .
</math>
  </td>
</tr>
</table>

----


Now, let's set <math>G = -1</math> to obtain,

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ 2F^{2} - 3 F \biggl( z - 1 \biggr) + \biggl( z - 1 \biggr)^{2}\biggr] ( z - 1 )
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ 10F^{2} - 25 F \biggl( z - 1  \biggr) + 11\biggl( z - 1  \biggr)^{2}\biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
F\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( z - 1  \biggr) 
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ 2F^{2} - 3 F z + 3F +  z^2 - 2z + 1 \biggr] ( z - 1 )
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ 10F^{2} + 25F - 25 F z + 11 z^2 - 22z + 11 \biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
F\biggl( \frac{5}{2} \biggr) \mathfrak{F} -3\alpha + 3\alpha z
\biggr] \cdot f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ (2F^{2} + 3F + 1 ) - (3 F +2) z +  z^2 \biggr] ( z - 1 )
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ ( 10F^{2} + 25F + 11)  - (25 F + 22) z + 11 z^2 \biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
F\biggl( \frac{5}{2} \biggr) \mathfrak{F} -3\alpha + 3\alpha z
\biggr] \cdot f \, .
</math>
  </td>
</tr>
</table>
Notice that if <math>F = -\tfrac{2}{3}</math>, we have,

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ z^2 - \frac{1}{9} \biggr] ( z - 1 )
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ \frac{40}{9} + \frac{50}{3} + 11  - \biggl(\frac{50}{3} + 22 \biggr) z + 11 z^2 \biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
F\biggl( \frac{5}{2} \biggr) \mathfrak{F} -3\alpha + 3\alpha z
\biggr] \cdot f
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{1}{9}\biggl[ 9z^2 - 1 \biggr] ( z - 1 )
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{18}
\biggl[ 201  - 348 z + 99 z^2 \biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
F\biggl( \frac{5}{2} \biggr) \mathfrak{F} -3\alpha + 3\alpha z
\biggr] \cdot f
</math>
  </td>
</tr>
</table>

----


Alternatively, let's set, 

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
z^2 + G\biggl[ 2F^{2} - 3 F \biggl( G+z \biggr) + \biggl( G+z \biggr)^{2}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
2GF^{2} - 3 GF \biggl( G+z \biggr) + G\biggl( G+z \biggr)^{2} + z^2 
</math>
  </td>
</tr>

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

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

in which case,

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
z^2\biggl( \frac{z}{G} - 1 \biggr)
\cdot \frac{d^2f}{dz^2}
+ \frac{1}{2}
\biggl[ 10F^{2} - 25 F \biggl( G+z \biggr) + 11\biggl( G+z \biggr)^{2}\biggr] 
\cdot \frac{df}{dz} 
+ \frac{1}{2}\biggl[
F\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr) 
\biggr] \cdot f \, .
</math>
  </td>
</tr>
</table>
The coefficient of the first-derivative term also may be rewritten ad,