Editing SSC/Stability/n5PolytropeLAWE (section)

===The Wave Equation===

====Starting from our Key Adiabatic Wave Equation====

The [[#Adiabatic_.28Polytropic.29_Wave_Equation|adiabatic wave equation]] therefore becomes,

<div align="center">
<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}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} 
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl\{
\biggl[ \frac{2^7\cdot 3^{7}}{\pi} \biggr]^{1 / 2}   \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \frac{1}{\xi} 
-   \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1}
\biggr\} \frac{dx}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2} 
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-3/2} 
( 3 + \xi^2 )^{1 / 2}  
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr]
+ \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2} 
\biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{15/2}  
( 3 + \xi^2 )^{-3/2}  
\biggr\}  x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl[ \frac{2^3\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2} 
( 3 + \xi^2 )^{1 / 2}   \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{-3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}
+  
( 3 + \xi^2 )^{-3/2}  
\biggr\}  x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0} 
+ \frac{6}{\gamma_g R_*^2}   \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}   \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}( 3 + \xi^2 )^{1 / 2}  
+  \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr\}  x
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>R_* \equiv R_\mathrm{norm} \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1/2} \, .</math>
</div>

Recognizing that,

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

<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_*   \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \, ,
</math>
  </td>
</tr>
</table>
</div>
we can write,

<div align="center">
<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{1}{R_*^2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}  \biggl\{
\frac{d^2x}{d\xi^2} + \biggl[
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{d\xi} 
+ \frac{6}{\gamma_g }  
\biggl[\sigma^2 ( 3 + \xi^2 )^{1 / 2}  
+  \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr]  x \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\sigma^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}   \biggl( \frac{3 + {\tilde\xi}^2}{{\tilde\xi}^2} \biggr)^{15/2} = \frac{\sigma_c^2}{2\cdot 3^{3/2}}\, .</math>
  </td>
</tr>
</table>
</div>

Finally, if &#8212; because we are specifically considering the case of <math>~n=5</math> &#8212; we set <math>~\gamma_\mathrm{g} = 1 + 1/n = 6/5</math>, we have,

<div align="center">
<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\xi^2} + \biggl[ \frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi} 
+ \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}} ( 3 + \xi^2 )^{1 / 2}  +  \frac{2}{( 3 + \xi^2 ) }\biggr]  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{( 3 + \xi^2 ) } \biggl\{ ( 3 + \xi^2 )\frac{d^2x}{d\xi^2} 
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi} 
+ \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}}  ( 3 + \xi^2 )^{3 / 2}  +  2 \biggr]  x \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
which matches exactly the [[#n5LAWE|form of the LAWE derived above]], if in ''that'' expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.


====Starting from the HRW66 Radial Pulsation Equation====
More directly, if we begin with the [[SSC/Stability/Polytropes#HRW66excerpt|HRW66 radial pulsation equation]] that is already tuned to polytropic configurations, the wave equation appropriate to <math>~n=5</math> polytropes is,
<div align="center">
<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^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi}
+ \frac{5(-\theta_5^') }{6\theta_5 \xi} \bigg[ \frac{\xi (s^')^2}{\theta^'_5} + \frac{12}{5} \biggr] X
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{(3 + \xi^2)} \biggr]\frac{d X}{d\xi}
+ \frac{1}{(3 + \xi^2)} \bigg[ -\frac{5(s^')^2(3 + \xi^2)^{3 / 2}}{2 \cdot 3^{3 / 2}} + 2 \biggr] X
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{(3+\xi^2)} \biggl\{
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
+ \bigg[ -\frac{5(s^')^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
which is identical to the brute-force derivation just presented, allowing for the mapping,
<div align="center">
<math>\sigma^2 ~~ \Leftrightarrow ~~ -\frac{(s^')^2}{2 \cdot 3^{3 / 2}} \, .</math>
</div>

Finally, remembering that the [[SSC/Stability/Polytropes#HRW66frequency|HRW66 dimensionless frequency definition]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(s^')^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\omega^2 \biggl[\frac{n+1 }{4\pi G \rho_c} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
we recognize that, specifically for the case of <math>~n=5</math>, we can make the substitution, <math>~(s^')^2 \rightarrow -\sigma_c^2</math>, in which case the LAWE becomes,

<div align="center">
<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{1}{(3+\xi^2)} \biggl\{
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
+ \bigg[ \frac{5\sigma_c^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
which matches exactly the [[#n5LAWE|form of the LAWE derived above]], if in ''that'' expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.

====New Independent Variable====
Guided by our [[Appendix/Ramblings/NonlinarOscillation#Conjectures|conjecture]] regarding the proper shape of the radial eigenfunction, let's switch the dependent variable to,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~u  \equiv 1 + \frac{3}{\xi^2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~3 + \xi^2 = \frac{3u}{(u-1)}  \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\xi = 3^{1 / 2} (u-1)^{-1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
This implies that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\xi}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow ~~~</math>
  </td>
  <td align="left">
<math>~-\frac{2}{\sqrt{3}}(u-1)^{3 / 2} \frac{d}{du} \, ,</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow ~~~</math>
  </td>
  <td align="left">
<math>~\frac{4}{3}(u-1)^3 \frac{d^2}{du^2} + 2(u-1)^{2} \frac{d}{du} \, .</math>
  </td>
</tr>
</table>
</div>
Hence, the governing wave equation becomes,

<div align="center">
<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>~( 3 + \xi^2 )\frac{d^2x}{d\xi^2} 
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi} 
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2}  +  2 \biggr]  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3u}{(u-1)} \biggl[\frac{4}{3}(u-1)^3 \frac{d^2x}{du^2} + 2(u-1)^{2} \frac{dx}{du}\biggr]
+ 4(2u-3)(u-1)\frac{dx}{du} 
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2}  +  2 \biggr\}  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4u(u-1)^2 \frac{d^2x}{du^2} 
+ (14u-12)(u-1)\frac{dx}{du} 
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2}  +  2 \biggr\}  x \, .
</math>
  </td>
</tr>
</table>
</div>

{{ SGFworkInProgress }}


If we ''assume'' that <math>~\sigma^2 = 0</math>, then the governing relation is,

<div align="center">
<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>~4u(u-1)^2 \frac{d^2x}{du^2} 
+ (14u-12)(u-1)\frac{dx}{du} 
+ 2 x \, .
</math>
  </td>
</tr>
</table>
</div>
Now, again, guided by our  [[Appendix/Ramblings/NonlinarOscillation#Conjectures|conjecture]], let's guess an eigenfunction of the form:

=====First Guess (n5)=====
<div align="center">
<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^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}   \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{A^3}{2} \biggl[ (u - 1)^{-1 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{1 / 2} (A u - 1 )^{-3 / 2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{
-\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[
(Au-1) +3A (u-1)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

<!-- EXTRA 2nd DERIVATIVE
<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{A^3}{2^2} \biggl[ -(u - 1)^{-3 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} 
- A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} + 3A^2(u - 1)^{1 / 2} (A u - 1 )^{-5 / 2} \biggr]
</math>
  </td>
</tr>
END EXTRA 2nd DERIVATIVE -->

So the governing relation becomes:

<div align="center">
<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>~4u(u-1)^2 \biggl\{ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (14u-12)(u-1) \biggl\{ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \biggr\}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~u(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (7u-6)(u-1)^{1 / 2}  A^3(A-1)  (Au-1)^{-3 / 2} 
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(u-1)^{1 / 2}  \biggl\{ uA^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  
+ (7u-6) A^3(A-1)  (Au-1)^{-3 / 2} 
+ 2 A^3 (A u - 1 )^{-1 / 2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u(A-1) \biggl[(3A+1) - 4Au \biggr]  
+ (7u-6) (A-1)  (Au-1) 
+ 2 (A u - 1 )^{2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ - 4u^2 A(A-1)  + u(A-1) (3A+1) 
+ (7u-6)  [A(A-1)u +1 - A]
+ 2  (A^2u^2 - 2Au +1) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u^2 \biggl[ - 4A(A-1) +7A(A-1) +2A^2 \biggr]    
+ u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr]  + 2(3A-2)  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]    
+ u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr]  + 2(3A-2)  \biggr\} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]    
+ u\biggl[ -3A^2 -7A  +6\biggr]  + 2(3A-2)  \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

=====Second Guess (n5)=====
<div align="center">
<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>~
(u - 1)^{b / 2} (A u - 1 )^{-a / 2}   \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{2}(u-1)^{b/2-1} (A u - 1 )^{-a / 2}
- \frac{aA}{2}(u - 1)^{b / 2} (A u - 1 )^{-a / 2-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \biggl[ 
\frac{b}{2}(u-1)^{-1} 
- \frac{aA}{2}  (A u - 1 )^{-1} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{(u-1)}{x} \frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (A u - 1 )^{-1} \biggl[ 
\frac{b}{2} (A u - 1 )
- \frac{aA}{2} (u-1) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{2(A u - 1 )} \biggl[ 
b (A u - 1 ) - aA (u-1) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1 }{2(A u - 1 )} \biggl[ 
(aA - b) + A(b    - a)u \biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]\frac{dx}{du} 
+ x \frac{d}{du}\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]^2
+ x  \biggl[ -\frac{b}{2}(u-1)^{-2} + \frac{aA^2}{2}  (A u - 1 )^{-2}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{x}{4(u-1)^2 (Au-1)^2} \biggl\{
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{4 (Au-1)^2} \biggl\{
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>
</table>
</div>

Hence, the governing wave equation becomes,

<div align="center">
<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>~2u \biggl\{ \frac{(u-1)^2}{x} \frac{d^2x}{du^2} \biggr\}
+ (7u-6)\biggl\{ \frac{(u-1)}{x} \frac{dx}{du} \biggl\} + 1 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2u}{4 (Au-1)^2} \biggl\{
\biggl[ (aA - b) + A(b    - a)u \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{(7u-6) }{2(A u - 1 )} \biggl[ 
(aA - b) + A(b    - a)u \biggr] 
 + 1 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+  2u\biggl[  2aA^2 (u^2 - 2u + 1)  -2b (A^2 u^2 - 2Au + 1 ) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2(A u - 1 )(7u-6) \biggl[ 
(aA - b) + A(b    - a)u \biggr] 
 + 4 (Au-1)^2 \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+  2u\biggl[ 2A^2(a-b)u^2 + 4A(b - aA) u + 2(aA^2   -b)  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[ 
(aA - b) + A(b    - a)u \biggr] 
 +  (4A^2u^2-8Au + 4) \biggr\}
</math>
  </td>
</tr>
</table>
</div>

If <math>~b=a</math>,

<div align="center">
<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>~
2u\biggl[ (aA - b)^2   \biggr]
+  2u\biggl[ 4A(b - aA) u + 2(aA^2   -b)  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[ 
(aA - b) \biggr]  +  (4A^2u^2-8Au + 4) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2a^2u (A - 1)^2  
+  2au [ 4A(1 - A) u + 2(A^2   -1) ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2a(A - 1) \biggl[7Au^2 - (6A+7)u  +6 \biggr] +  (4A^2u^2-8Au + 4) 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [4a (1 - A) +  7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2   -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>
</table>
</div>
This should then match the [[#First_Guess_.28n5.29|"first guess"]] algebraic condition if we set <math>~a=1</math>.  Let's see.
 
<div align="center">
<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>~
2Au^2 [4 (1 - A) +  7(A - 1) + 2A] + 2u [ (A - 1)^2 +  2(A^2   -1) - (A - 1) (6A+7) - 4A] + 4[ 3(A-1) + 1]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [4  - 4A +  7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) +  2A^2   -2 + (1-A ) (6A+7) -4A] + 4[ 3A-2]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [5A - 3] + 2u [  - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, .
</math>
  </td>
</tr>
</table>
</div>
And we see that this expression ''does'' match the one derived earlier.

Going back a bit, before setting <math>~a=1</math>, we have the expression:
 

<div align="center">
<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>~
2Au^2 [4a (1 - A) +  7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2   -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [  3aA  -3a + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2   -1) - a(6A^2+A-7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [  3a(A  - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A] + 4[ 3a(A-1) + 1] \, .
</math>
  </td>
</tr>
</table>
</div>
Now, in order for all three expressions inside the square-bracket pairs to be zero, we need, first,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3a(A  - 1) + 2A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2A}{3(1-A)} \, ;</math>
  </td>
</tr>
</table>
</div>
and, third, by simple visual comparison with the first expression,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3a(A-1) + 1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3a(A-1) + 2A</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3} \, ;</math>
  </td>
</tr>
</table>
</div>
which forces the second expression to the value,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{2}{3}\biggr)^2 \biggl(-\frac{1}{2} \biggr)^2 +  \frac{2}{3}\biggl[   -1-\frac{1}{2} +5 \biggr] - 2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{9}  +  \frac{7}{3}  - 2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4}{9}  \, ,</math>
  </td>
</tr>
</table>
</div>
which is ''not'' zero.  Hence our pair of unknown parameters &#8212; <math>~a </math> and <math>~A</math> &#8212; do not simultaneously satisfy all three conditions.  (Not really a surprise.)