Editing Appendix/Mathematics/Hypergeometric (section)

==Part III==

Now, suppose that <math>f = (a_0 + b_0z)</math>.  We have,

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

<tr>
  <td align="right">
LAWE
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>\biggl[
\xi^{-2}(1+3z)^2 + 2n(1+3z)z^2 \theta^{n-1} + n^2 \xi^2 \theta^{2(n-1)}z^4
\biggr]
\cdot \cancelto{0}{\frac{d^2f}{dz^2}}
+ (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1}
+ \alpha \theta^{n-1} z\biggr](a_0 + b_0z) 
+ \biggl\{
(n-1)  + (n - 3)z  
\biggr\} \theta^{n-1} z \cdot b_0  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1}
\biggr](a_0 + b_0z) 
+
\biggl\{
(n+1)\alpha (a_0 + b_0z) 
+ 
b_0(n-1)  + b_0(n - 3)z    
\biggr\} \theta^{n-1} z 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
(n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1}
\biggr](a_0 + b_0z) 
+
\biggl\{
(n+1)\alpha (a_0 ) 
+ 
b_0(n-1)  
+ 
\biggl[(n - 3)    
+ 
(n+1)\alpha \biggr]b_0z 
\biggr\} \theta^{n-1} z
\, . 
</math>
  </td>
</tr>
</table>
Now, in order for the last term to be zero, we need,

<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[(n - 3)    
+ 
(n+1)\alpha \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \alpha</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\frac{3-n}{n+1} \, .
</math>
  </td>
</tr>
</table>
This is precisely the  relation that results from the definition of <math>\alpha \equiv (3 - 4/\gamma_g)</math> if the model is evolved assuming <math>\gamma_g = (n+1)/n</math>.  We simultaneously seek the relation,

<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>
(n+1)\alpha (a_0 ) 
+ 
b_0(n-1)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ b_0</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
\biggl[ \frac{n+1}{1-n} \biggr] \alpha (a_0 ) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
a_0 \biggl[ \frac{3-n}{1-n} \biggr]  
</math>
  </td>
</tr>
</table>
It appears as though the leading coefficient, <math>a_0</math>, is arbitrary, so we will set it equal to unity. This means that the displacement function is,

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

<tr>
  <td align="right">
<math>f</math>
  </td> 
  <td align="center">
<math>=</math>
  </td> 
  <td align="left">
<math>
1 + \biggl[ \frac{3-n}{1-n} \biggr]z \, .
</math>
  </td>
</tr>
</table>
This expression for the displacement function, <math>f</math>, is identical to the [[SSC/Stability/InstabilityOnsetOverview#ExactPolytropicSolution|expression found inside the square brackets of our separately derived exact solution of the polytropic LAWE]].  Furthermore, given the notation, <math>(\sigma_c^2/\gamma_g) = (\mathfrak{F}-2\alpha)</math>, the first term on the RHS of the LAWE will go to zero when, <math>\mathfrak{F} = 2(3-n)/(n+1)</math>.