Editing SSC/Structure/OtherAnalyticRamblings (section)

=====First Guess=====
Now, if we are very lucky, we will find that,
<div align="center">
<math>~(a_0 + a_2x^2) = (1-x^2)</math> &nbsp; &nbsp;&nbsp; <math>~\Rightarrow</math>
&nbsp; &nbsp;&nbsp; <math>~a_0 = 1</math> &nbsp; &nbsp;&nbsp; and
&nbsp; &nbsp;&nbsp; <math>~a_2 = -1</math>;
</div>
and, simultaneously,
<div align="center">
<math>~(b_0 + b_2x^2) = (2-x^2)</math> &nbsp; &nbsp;&nbsp; <math>~\Rightarrow</math>
&nbsp; &nbsp;&nbsp; <math>~b_0 = 2</math> &nbsp; &nbsp;&nbsp; and
&nbsp; &nbsp;&nbsp; <math>~b_2 = -1</math>.
</div>
In this case, the fractional coefficient in the last term of the LAWE will become unity and 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>~
\biggl[ \alpha(1-x^2) (2-x^2) - 2x^2(2n  + m)  + 2x^4 (n+ m) \biggr](5-3x^2) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl[\sigma^2 - n (2-x^2) - m  (1-x^2)  - 4(2n  + m)  + 4(n + m )x^2+ 4n m x^2 \biggr](1-x^2)(2-x^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl[2n(n-1) (2-x^2)^2   + 2m(m-1) (1-x^2)^2 \biggr]x^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2\alpha - x^2(3\alpha +4n  + 2m)  + x^4 (\alpha + 2n+ 2m) \biggr](5-3x^2) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[(-\sigma^2 + 10n + 5m  )  - x^2( 5n+5m+4nm  )\biggr] (2-3x^2+x^4)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl\{ [8n(n-1) +2m(m-1) ] - [8n(n-1) + 4m(m-1)  ] x^2 + [ 2n(n-1)+ 2m(m-1) ] x^4 \biggr\} x^2 \, .
</math>
  </td>
</tr>
</table>
</div>

So, the coefficients of each even power of <math>~\chi_0^n</math> are:
<div align="center" id="FirstTable">
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="right"><math>~\chi_0^0</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~10\alpha + 2(-\sigma^2 + 10n + 5m  )</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^2</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~-6\alpha - 5(3\alpha +4n  + 2m) -2( 5n+5m+4nm  ) - 3(-\sigma^2 + 10n + 5m  ) - [8n(n-1) +2m(m-1) ]</math>
 </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^4</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~5(\alpha + 2n+ 2m)+3(3\alpha +4n  + 2m) + (-\sigma^2 + 10n + 5m  )  +3( 5n+5m+4nm  ) + [8n(n-1) + 4m(m-1)  ] </math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^6</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~-3(\alpha + 2n+ 2m) - ( 5n+5m+4nm  ) - [ 2n(n-1)+ 2m(m-1) ] </math>
  </td>
</tr>
</table>
</div>


Now let's begin simplification.  The <math>~x^0</math> coefficient implies,
<div align="center">
<math>~(-\sigma^2 + 10n + 5m  )=-5\alpha \, .</math>
</div>

Hence,
<div align="center">
<table border="1" cellpadding="8" align="center">

<tr>
  <td align="right"><math>~\chi_0^2</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~-6\alpha - 5( 4n  + 2m) -2( 5n+5m+4nm  )  - [8n(n-1) +2m(m-1) ]</math>
 </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^4</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~9\alpha + 5(2n+ 2m)+3(4n  + 2m)  +3( 5n+5m+4nm  ) + [8n(n-1) + 4m(m-1)  ] </math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^6</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~-3\alpha -3(2n+ 2m) - ( 5n+5m+4nm  ) - [ 2n(n-1)+ 2m(m-1) ] </math>
  </td>
</tr>
</table>
</div>

Using the <math>~x^6</math> coefficient to define <math>~\alpha</math>, that is, setting,
<div align="center">
<math>~3\alpha = \{-3(2n+ 2m) - ( 5n+5m+4nm  ) - [ 2n(n-1)+ 2m(m-1) ]\} \, ,</math>
</div>

means that the other two coefficient expressions are,
<div align="center">
<table border="1" cellpadding="8" align="center">

<tr>
  <td align="right"><math>~\chi_0^2</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~6(2n+ 2m) + 2 ( 5n+5m+4nm  ) + 4 [ n(n-1)+ m(m-1) ] - 5( 4n  + 2m) -2( 5n+5m+4nm  )  - [8n(n-1) +2m(m-1) ]</math>
 </td>
</tr>

<tr>
  <td align="right"><math>~\chi_0^4</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~-9(2n+ 2m) - 3( 5n+5m+4nm  ) - 6[ n(n-1)+ m(m-1) ] + 5(2n+ 2m)+3(4n  + 2m)  +3( 5n+5m+4nm  ) + [8n(n-1) + 4m(m-1)  ] </math>
  </td>
</tr>
</table>
</div>

The only question remaining is, what pair of values for <math>~(n, m)</math> result in both of these expressions going to zero?  Simplifying the first expression gives,
<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>~6(2n+ 2m) + 2 ( 5n+5m+4nm  ) + 4 [ n(n-1)+ m(m-1) ] - 5( 4n  + 2m) -2( 5n+5m+4nm  )  - [8n(n-1) +2m(m-1) ]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2m  -8n -4n(n-1) + 2m(m-1) </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2[m^2 - 2n(n+1)] \, . </math>
  </td>
</tr>
</table>
</div>

Simplifying the second expression gives,
<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>~-9(2n+ 2m) - 3( 5n+5m+4nm  ) - 6[ n(n-1)+ m(m-1) ] + 5(2n+ 2m)+3(4n  + 2m)  +3( 5n+5m+4nm  ) + [8n(n-1) + 4m(m-1)  ] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4n - 2m   + 2n(n-1) - 2m(m-1) </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2[n(n+1) -m^2] </math>
  </td>
</tr>
</table>
</div>

Okay.  Because it is not possible for both of these last two constraints to be simultaneously satisfied, I conclude that this last, specific eigenfunction guess is incorrect.  

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