Editing Appendix/Mathematics/Hypergeometric (section)

=Example Density- and Pressure-Profiles=

<table border="1" cellpadding="5" align="center" width="90%">
<tr>
  <th align="center" colspan="6">Properties of Analytically Defined, Spherically Symmetric, Equilibrium Structures
Note: &nbsp;<math>x \equiv \frac{r_0}{R}</math>
  </th>
</tr>
<tr>
  <td align="center" width="10%">Model</td>
  <td align="center"><math>\frac{\rho_0(x)}{\rho_c}</math></td>
  <td align="center"><math>\frac{P_0(x)}{P_c}</math></td>
  <td align="center"><math>\frac{R}{P_c} \cdot P_0^'(x)</math></td>
  <td align="center">
<math>
\mu(x) = - \biggl[ \frac{P_0(x)}{P_c} \biggr]^{-1} \cdot \biggl[ \frac{R}{P_c} \cdot P_0^'(x) \biggr]x
</math>
  </td>
  <td align="center"><math>\frac{P_c}{\rho_c} \cdot \frac{\rho_0(x)}{P_0(x)}</math></td>
</tr>
<tr>
  <td align="center">[[SSC/Stability/UniformDensity#The_Stability_of_Uniform-Density_Spheres|Uniform-density]]</td>
  <td align="center"><math>1</math></td>
  <td align="center"><math>1 - x^2</math></td>
  <td align="center"><math>-2x</math></td>
  <td align="center"><math>\frac{2x^2}{ (1 - x^2)}</math></td>
  <td align="center"><math>\frac{1}{ (1 - x^2)}</math></td>
</tr>
<tr>
  <td align="center">[[SSC/Structure/OtherAnalyticModels#Linear_Density_Distribution|Linear]]</td>
  <td align="center"><math>1-x</math></td>
  <td align="center"><math>(1-x)^2(1 + 2x - \tfrac{9}{5}x^2)</math></td>
  <td align="center"><math>-\tfrac{12}{5}x(1-x)(4-3x)</math></td>
  <td align="center"><math>\frac{\tfrac{12}{5}x^2(4-3x)}{(1-x)(1 + 2x - \tfrac{9}{5}x^2)}</math></td>
  <td align="center"><math>\frac{1}{(1-x)(1 + 2x - \tfrac{9}{5}x^2)}</math></td>
</tr>
<tr>
  <td align="center">[[SSC/Structure/OtherAnalyticModels#Parabolic_Density_Distribution|Parabolic]]</td>
  <td align="center"><math>1-x^2</math></td>
  <td align="center"><math>(1-x^2)^2(1  - \tfrac{1}{2} x^2)</math></td>
  <td align="center"><math>-x(1-x^2)(5-3x^2)</math></td>
  <td align="center"><math>\frac{x^2(5-3x^2)}{ (1-x^2)(1  - \tfrac{1}{2} x^2) }</math></td>
  <td align="center"><math>\frac{1}{(1-x^2)(1  - \tfrac{1}{2} x^2)} </math></td>
</tr>
<tr>
  <td align="center">[[SSC/Stability/Polytropes#n_.3D_1_Polytrope|<math>~n=1</math> Polytrope]]</td>
  <td align="center"><math>\frac{\sin x }{ x}</math></td>
  <td align="center"><math>\biggl(\frac{\sin x}{x}\biggr)^2</math></td>
  <td align="center"><math>\frac{2}{x} \biggl[ \cos x - \frac{\sin x}{x} \biggr]\frac{\sin x}{x}</math></td>
  <td align="center"><math>2 (1 - x \cot x )</math>
  <td align="center"><math>\frac{x}{\sin x}</math>
</tr>
</table>

==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,

==Uniform Density==

In the case of a uniform-density, incompressible configuration, 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 - \frac{2x^2}{(1-x^2)} \biggr] \frac{d f}{dx} 
+ \biggl[\biggl(\frac{\omega^2\rho_c R^2}{\gamma_\mathrm{g} P_c} \biggr) \frac{1}{(1-x^2)}
- \biggl(\frac{2\alpha }{1-x^2}\biggr) \biggr] f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 6x^2 \biggr] \frac{d f}{dx} 
+ \biggl[\biggl(\frac{\omega^2\rho_c R^2}{\gamma_\mathrm{g} P_c} \biggr) 
- 2\alpha \biggr] f \, .
</math>
  </td>
</tr>
</table>
Given that, in the [[SSC/Structure/UniformDensity#Isolated_Uniform-Density_Sphere|equilibrium state]],
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>\frac{\rho_c R^2}{P_c}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{6}{4\pi G \rho_c}
</math>
  </td>
</tr>
</table>
we obtain the LAWE derived by {{ Sterne37full }} &#8212; see his equation (1.91) on p. 585 &#8212; namely,
<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) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 6x^2 \biggr] \frac{d f}{dx} 
+ \biggl[6\biggl(\frac{\omega^2}{4\pi \gamma_\mathrm{g} G \rho_c} \biggr) 
- 2\alpha \biggr] f 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 6x^2 \biggr] \frac{d f}{dx} 
+ \mathfrak{F} 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>
\biggl[6\biggl(\frac{\omega^2}{4\pi \gamma_\mathrm{g} G \rho_c} \biggr) 
- 2\alpha \biggr] \, . 
</math>
  </td>
</tr>
</table>

<table border="1" width="80%" cellpadding="8" align="center">
<tr>
  <th align="center">Summary for PowerPoint Slide</th>
</tr>
<tr>
  <td>
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<b>LAWE:</b> &nbsp; &nbsp;<math>0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(1-x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 6x^2 \biggr] \frac{d f}{dx} 
+ \mathfrak{F} 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>
\biggl[\biggl(\frac{\omega^2\rho_c R^2}{\gamma_\mathrm{g} P_c} \biggr) 
- 2\alpha \biggr] \, . 
</math>
  </td>
</tr>
</table>

  </td>
</tr>
</table>
This also matches, respectively, equations (8) and (9) of {{ Kopal48full }}, aside from what, we presume, is a type-setting error that appears in the numerator of the second term on the RHS of his equation (8): &nbsp;<math>(4 - x^2)</math> appears, whereas it should be <math>(4 - 6x^2)</math>.

In order to see if this differential equation is of the same form as the ''hypergeometric expression'', we'll make the substitution,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>z</math>
  </td> 
  <td align="center">
<math>\equiv</math>
  </td> 
  <td align="left">
<math>
x^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ dz</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
2x dx
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{d^2f}{dx^2}</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
2z^{1 / 2} \cdot \frac{d}{dz}\biggl[ 2z^{1 / 2} \cdot\frac{df}{dz} \biggr]
=
2z^{1 / 2} \biggl[ z^{-1 / 2} \cdot\frac{df}{dz} + 2z^{1 / 2} \cdot\frac{d^2f}{dz^2}\biggr]
=
\biggl[ 2\cdot\frac{df}{dz} + 4z \cdot\frac{d^2f}{dz^2}\biggr] \, ,
</math>
  </td>
</tr>
</table>
in which case the {{ Sterne37 }} LAWE may be rewritten as,
<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-z) \biggl[ 2\cdot\frac{df}{dz} + 4z \cdot\frac{d^2f}{dz^2}\biggr] + \frac{1}{z^{1 / 2}}\biggl[ 4 - 6z \biggr]2 z^{1 /2} \frac{d f}{dz} 
+ \mathfrak{F} f
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
4z(1-z) \cdot\frac{d^2f}{dz^2} 
+ 2\biggl[ 5 - 7z \biggr] \frac{d f}{dz} 
+ \mathfrak{F} f \, .
</math>
  </td>
</tr>
</table>
This is, indeed, of the hypergeometric form if we set <math>(\alpha, \beta; \gamma ; z)</math>
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>\gamma</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{5}{2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>(\alpha + \beta +1)</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{7}{2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\alpha\beta</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
-\frac{\mathfrak{F}}{4} \, .
</math>
  </td>
</tr>
</table>
Combining this last pair of expressions 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>
-\frac{\mathfrak{F}}{4} - \alpha\biggl[\frac{5}{2}-\alpha \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\alpha^2 - \biggl(\frac{5}{2}\biggr)\alpha -\frac{\mathfrak{F}}{4} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \alpha</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{1}{2}\biggl\{
\frac{5}{2} \pm \biggl[\biggl(\frac{5}{2}\biggr)^2 - \mathfrak{F}  \biggr]^{1 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{5}{4}\biggl\{
1 \pm \biggl[1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} \biggr]^{1 / 2}
\biggr\} \, ;
</math>
  </td>
</tr>
</table>
and,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>\beta</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{5}{2} -
\frac{5}{4}\biggl\{
1 \pm \biggl[1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} \biggr]^{1 / 2}
\biggr\} \, .
</math>
  </td>
</tr>
</table>

===Example &alpha; = -1===
If we set <math>\alpha = -1</math>, then the eigenvector is,
 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>u_{1} = F\biggl (-1, \frac{7}{2}; \frac{5}{2}; x^2 \biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \biggl[\frac{ \beta}{\gamma} \biggr]x^2
=
1 - \biggl(\frac{ 7}{5} \biggr)x^2 \, ;
</math>
  </td>
</tr>
</table>
and the corresponding eigenfrequency is obtained from the expression,
 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>-1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{5}{4}\biggl\{
1 \pm \biggl[1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} \biggr]^{1 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~-\frac{9}{5}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm \biggl[1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~\frac{3^4}{5^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~\mathfrak{F} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{5^2}{4}\biggr)\biggl[ 1 -  \frac{3^4}{5^2} \biggr]
=
\frac{1}{4}\biggl[ 5^2 -  3^4 \biggr]
=
14 \, .
</math>
  </td>
</tr>
</table>
As we have reviewed in a [[SSC/Stability/UniformDensity#Sterne's_Presentation|separate discussion]], this is identical to the eigenvector identified by {{ Sterne37 }} as mode "<math>j=1</math>".

===More Generally===
More generally, in agreement with {{ Sterne37 }}, for any (positive integer) mode number, <math>0 \le j \le \infty</math>, we find,
 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\alpha_j</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- j \, ;</math>
&nbsp; &nbsp; &nbsp; 
<math>\beta_j = \frac{5}{2} + j \, ;</math>
&nbsp; &nbsp; &nbsp; 
<math>\gamma = \frac{5}{2} \, ;</math> 
&nbsp; &nbsp; &nbsp; 
<math>\mathfrak{F} = 2j(2j+5) \, .
</math>
  </td>
</tr>
</table>
And, in terms of the hypergeometric function ''series'', the corresponding eigenfunction is,
 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>u_{j} = </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
F\biggl (\alpha_j, \beta_j; \frac{5}{2}; x^2 \biggr) \, .
</math>
  </td>
</tr>
</table>