Editing SSC/Structure/OtherAnalyticModels (section)

==Promising Avenue of Exploration==
What follows is a direct extension of what is referred to in our "Ramblings" chapter as [[SSC/Structure/OtherAnalyticRamblings#Third_Guess|the ''third guess'' under "Exploration2"]].  We pursue this line of reasoning, here, because it appears to be a particularly promising avenue of exploration.

In the case of a parabolic density distribution, the LAWE becomes,

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

<tr>
  <td align="right">
<math>~\frac{2}{(1-x^2)(2-x^2)}  \biggl[ \biggl( \alpha + \frac{x \mathcal{G}_\sigma^'}{\mathcal{G}_\sigma}\biggr)(5-3x^2) -\sigma^2 \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl(\frac{\mathcal{G}_\sigma^{' '}}{\mathcal{G}_\sigma}\biggr) +\frac{4}{x^2} \cdot \frac{x \mathcal{G}_\sigma^'}{\mathcal{G}_\sigma}  \, .
</math>
  </td>
</tr>
</table>
</div>


We have chosen to examine the suitability of an eigenfunction of the form,
<div>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="center">
<math>~\mathcal{G}_\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(a_0 + a_2x^2)^n \cdot (2 - x^2)^m \, ,</math>
  </td>
</tr>
</table>
where, for a given value of <math>~\alpha</math>, the four parameters, <math>~a_0</math>, <math>~a_2</math>, <math>~n</math> and <math>~m</math> are to be determined in concert with a value of the square of the eigenfrequency, <math>~\sigma^2</math>.  From the [[SSC/Structure/OtherAnalyticRamblings#Third_Guess|accompanying discussion]] we have determined that the following five coefficient expressions must independently be zero in order for this trial eigenfunction to satisfy the LAWE:
<div align="center" id="FirstTable">
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="right"><math>~x^0</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~ \alpha(10a_0^2) + \sigma^2(- 2a_0^2)   -20n a_0a_2 + 10ma_0^2 </math>
  </td>
</tr>

<tr>
  <td align="right"><math>~x^2</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~\alpha(- 11a_0^2 +20 a_0a_2) + \sigma^2(a_0^2 - 4 a_0a_2) + 60n a_0a_2  -20na_2^2- 25m a_0^2    
+ 20m a_0a_2 + 8n m a_0a_2   
</math><p><math>
- [ 8n(n-1)a_2^2 + 2m(m-1)a_0^2 ]
</math></p>
 </td>
</tr>

<tr>
  <td align="right"><math>~x^4</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~ 
\alpha(10a_2^2  - 22 a_0a_2 +3a_0^2) - \sigma^2(2a_2^2- 2a_0a_2 ) - 47n a_0a_2+ 60n a_2^2  - 50m a_0a_2    +11m a_0^2+ 10ma_2^2-12 n m a_0a_2    + 8n m a_2^2
</math><p>
<math>
~ + 16n(n-1)a_2^2 - 4m(m-1)a_0 a_2 + 2m(m-1)a_0^2 
</math></p>
  </td>
</tr>

<tr>
  <td align="right"><math>~x^6</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~ 
\alpha(6a_0a_2 - 11a_2^2)  + \sigma^2(a_2^2) + 11 n a_0a_2  +22 m a_0a_2  - 47n a_2^2 - 25m a_2^2  -12 n m a_2^2 + 4n m a_0a_2
</math><p>
<math>~
- 10 n(n-1)a_2^2 -2m(m-1)a_2^2  +4m(m-1)a_0 a_2 
</math></p>
  </td>
</tr>

<tr>
  <td align="right"><math>~x^8</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~ \{ 3\alpha + [ 4n m  + 11n + 11m   ] + [ 2n(n-1)   + 2m(m-1) ]\}a_2^2
</math>
  </td>
</tr>

</table>
</div>

===First Constraint===
We begin by manipulating the last expression &#8212; that is, the coefficient expression for the <math>~x^8</math> term.  Rejecting the trivial option of setting <math>~a_2 = 0</math>, in order for this expression to be zero the terms inside the curly braces must sum to zero.  Rewriting this expression in terms of the ''sum'' of the exponents,

<div align="center">
<math>~s_{nm} \equiv n + m\, ,</math>
</div>

we obtain the quadratic 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>~3\alpha + [ 4n m  + 11n + 11m   ] + [ 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>~3\alpha + 4n m  + 9n + 9m + 2n^2 + 3m^2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\alpha + 9s_{nm} + 2s_{nm}^2 \, .</math>
  </td>
</tr>
</table>
</div>

This means that, once the physical parameter, <math>~\alpha = (3 - 4/\gamma_g)</math>, has been specified, the sum of the exponents must be,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~s_{nm}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4}\biggl[ -9 \pm (81 - 24\alpha)^{1/2} \biggr]</math>
  </td>
</tr>

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

===Second Constraint===
Next we examine the expression that serves as the coefficient of <math>~x^0</math>.  Setting that coefficient expression to zero while replacing <math>~m</math> in favor of <math>~s_{nm}</math> &#8212; via the relation, <math>~m = (s_{nm}-n)</math> &#8212; 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>~\alpha(10a_0^2) + \sigma^2(- 2a_0^2)   -20n a_0a_2 + 10ma_0^2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2a_0^2 \biggl[5\alpha -\sigma^2 + 5(s_{nm}-n)  - 10n \biggl(\frac{a_2}{a_0} \biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2a_0^2 \biggl[5\alpha -\sigma^2 + 5s_{nm} -5n\biggl(1  - \frac{2a_2}{a_0} \biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \frac{\sigma^2}{5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\alpha + s_{nm}) -n(1  - 2\lambda) \, ,</math>
  </td>
</tr>
</table>
</div>
where, we have set,
<div align="center">
<math>~\lambda \equiv \frac{a_2}{a_0} \, .</math>
</div>
So, once <math>~\alpha</math> is specified and <math>~s_{nm}</math> is known from the first constraint, we can use this expression to replace <math>~\sigma^2</math> in the other three coefficient expressions.  

===Intermediate Summary===
The three remaining constraints emerge from the remaining three coefficient expressions, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="right"><math>~x^2</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~\alpha(- 11a_0^2 +20 a_0a_2) + \sigma^2(a_0^2 - 4 a_0a_2) + 60n a_0a_2  -20na_2^2- 25m a_0^2    
+ 20m a_0a_2 + 8n m a_0a_2   
</math><p><math>
- [ 8n(n-1)a_2^2 + 2m(m-1)a_0^2 ]
</math></p>
 </td>
</tr>

<tr>
  <td align="right"><math>~x^4</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~ 
\alpha(10a_2^2  - 22 a_0a_2 +3a_0^2) - \sigma^2(2a_2^2- 2a_0a_2 ) - 47n a_0a_2+ 60n a_2^2  - 50m a_0a_2    +11m a_0^2+ 10ma_2^2-12 n m a_0a_2    + 8n m a_2^2
</math><p>
<math>
~ + 16n(n-1)a_2^2 - 4m(m-1)a_0 a_2 + 2m(m-1)a_0^2 
</math></p>
  </td>
</tr>

<tr>
  <td align="right"><math>~x^6</math></td>
  <td align="center">&nbsp; : &nbsp;</td>
  <td align="left">
<math>~ 
\alpha(6a_0a_2 - 11a_2^2)  + \sigma^2(a_2^2) + 11 n a_0a_2  +22 m a_0a_2  - 47n a_2^2 - 25m a_2^2  -12 n m a_2^2 + 4n m a_0a_2
</math><p>
<math>~
- 10 n(n-1)a_2^2 -2m(m-1)a_2^2  +4m(m-1)a_0 a_2 
</math></p>
  </td>
</tr>

</table>
</div>

Written in terms of the three remaining unknowns, <math>~n</math>, <math>~a_0</math>, and <math>~\lambda</math>, the three constraints are:

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

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + 60n \lambda  -20n \lambda^2- 25m    
+ 20m \lambda + 8n m \lambda  - [ 8n(n-1)\lambda^2 + 2m(m-1) ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + 5(1 - 4 \lambda)[ (\alpha + s_{nm}) -n(1  - 2\lambda)  ] + 60n \lambda  -20n \lambda^2 - 8n(n-1)\lambda^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm}-n)[- 23  + 20 \lambda + 8n \lambda]   - 2(s_{nm}-n)^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-6\alpha+ 5(1 - 4 \lambda)s_{nm} -5n(1 - 4 \lambda)(1  - 2\lambda) + 60n \lambda  -20n \lambda^2 - 8n(n-1)\lambda^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm}-n)[- 23  + 20 \lambda + 8n \lambda]   - 2(s_{nm}-n)^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x^4:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(10\lambda^2  - 22 \lambda +3) - 10 (\lambda^2- \lambda ) [ (\alpha + s_{nm}) -n(1  - 2\lambda) ] - 47n \lambda+ 60n \lambda^2  + 16n(n-1)\lambda^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm} - n)[ 9 - 46 \lambda + 10\lambda^2-12 n  \lambda    + 8n  \lambda^2]  + (2-4\lambda)(s_{nm} - n)^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 12 \lambda +3) - 10 (\lambda^2- \lambda ) s_{nm} + n10 (\lambda^2- \lambda ) (1  - 2\lambda)  - 47n \lambda+ 60n \lambda^2  + 16n(n-1)\lambda^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm} - n)[ 9 - 46 \lambda + 10\lambda^2-12 n  \lambda    + 8n  \lambda^2]  + (2-4\lambda)(s_{nm} - n)^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(6\lambda - 11\lambda^2)  + 5\lambda^2  [ (\alpha + s_{nm}) -n(1  - 2\lambda) ] + 11 n \lambda  - 47n \lambda^2 - 10 n(n-1)\lambda^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm} - n)[18  \lambda  - 23 \lambda^2  -12 n  \lambda^2 + 4n  \lambda ] + 2 \lambda (2 - \lambda ) (s_{nm} - n)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6\lambda \alpha( 1 - \lambda)  + 5\lambda^2 s_{nm} - 5n\lambda^2 (1  - 2\lambda)  + 11 n \lambda  - 47n \lambda^2 - 10 n(n-1)\lambda^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm} - n)[18  \lambda  - 23 \lambda^2  -12 n  \lambda^2 + 4n  \lambda ] + 2 \lambda (2 - \lambda ) (s_{nm} - n)^2 \, .
</math>
  </td>
</tr>
</table>
</div>

At first glance, this is ''still'' not as promising as I had hoped.  In practice there are only ''two'' unknowns &#8212; because the parameter, <math>~a_0</math>, has divided out &#8212; while there are three constraints.  So the problem remains over constrained.

===Remaining Group of Three Constraints===

Let's adopt another approach.  Let's assume that the parameter, <math>~\alpha</math>, is also initially unspecified and replace it in all three remaining constraint expressions, in favor of <math>~s_{nm}</math>, using the [[#First_Constraint|above-specified, first constraint]], namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-3\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 9s_{nm} + 2 s_{nm}^2 \, .</math>
  </td>
</tr>
</table>
</div>
 
This gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2(9s_{nm} + 2 s_{nm}^2)+ 5(1 - 4 \lambda)s_{nm} -5n(1 - 4 \lambda)(1  - 2\lambda) + 60n \lambda  -20n \lambda^2 - 8n(n-1)\lambda^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm}-n)[- 23  + 20 \lambda + 8n \lambda]   - 2(s_{nm}-n)^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x^4:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-(9s_{nm} + 2 s_{nm}^2)(1- 4 \lambda ) - 10 (\lambda^2- \lambda ) s_{nm} + n10 (\lambda^2- \lambda ) (1  - 2\lambda)  - 47n \lambda+ 60n \lambda^2  + 16n(n-1)\lambda^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm} - n)[ 9 - 46 \lambda + 10\lambda^2-12 n  \lambda    + 8n  \lambda^2]  + (2-4\lambda)(s_{nm} - n)^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\lambda (9s_{nm} + 2 s_{nm}^2)( 1 - \lambda)  + 5\lambda^2 s_{nm} - 5n\lambda^2 (1  - 2\lambda)  + 11 n \lambda  - 47n \lambda^2 - 10 n(n-1)\lambda^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm} - n)[18  \lambda  - 23 \lambda^2  -12 n  \lambda^2 + 4n  \lambda ] + 2 \lambda (2 - \lambda ) (s_{nm} - n)^2 \, .
</math>
  </td>
</tr>
</table>
</div>

The three unknowns are:  <math>~n</math>, <math>~s_{nm}</math>, and <math>~\lambda</math>.

===Prasad's Work===

====Overview====
{{ Prasad49 }} performed a semi-analytic analysis of the radial oscillations and stability of structures having a parabolic density distribution.  Let's examine his tabulated results to see if they help us understand more fully whether or not our analysis is on the right track.  For example, from his Table I, we see that <math>~\mathfrak{F} = 0</math> when <math>~\alpha = 0</math>, where, according to his equation (3),

<div align="center">
<math>~\mathfrak{F} \equiv \sigma^2 - 5\alpha \, .</math>
</div>

This means that, also, <math>~\sigma^2 = 0</math>.  Now, from our derived [[#Second_Constraint|second constraint]], we deduce that,

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

<tr>
  <td align="right">
<math>~\mathfrak{F} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~5[s_{nm} -n(1  - 2\lambda)] \, .</math>
  </td>
</tr>
</table>
</div>
Hence, since <math>~\mathfrak{F} = 0</math>, we conclude that,

<div align="center">
<math>~s_{nm} = n(1  - 2\lambda) \, .</math>
</div>

Also, since by definition <math>~s_{nm} = n + m</math>, we conclude that,

<div align="center">
<math>~\frac{m}{n} = - 2\lambda \, .</math>
</div>

Next, given that <math>~\alpha = 0</math>, we conclude from our derived [[#First_Constraint|first constraint]], that
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~s_{nm} = \frac{3^2}{2^2}\biggl[ -1 \pm 1\biggr] </math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow</math>&nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~s_{nm}^{+} =0 </math>&nbsp; &nbsp; and &nbsp; &nbsp; <math>~s_{nm}^{-} = -\frac{9}{2} \, .</math>
  </td>
</tr>
</table>
</div>

=====The Minus Root=====

Combining these two results for the "minus" solution, we furthermore conclude that, for this ''specific'' mode, the relationship between the two exponents and <math>~\lambda</math> are,


<div align="center">
<math>~n^- = - \frac{9}{2(1-2\lambda)}</math> &nbsp;  &nbsp;  &nbsp; and  &nbsp;  &nbsp;  &nbsp;  <math>~m^- = (s_{nm} - n^-) = \frac{9\lambda}{(1-2\lambda)} \, .</math>
</div>

=====The Plus Root=====
Next, let's examine the "plus" solution.  Because <math>~s_{nm}^{+} =0 </math>, this solution implies that,


<div align="center">
<math>~m^+ = -n^+</math> &nbsp; &nbsp; &nbsp;<math>~\Rightarrow</math>&nbsp; &nbsp; &nbsp;<math>~\frac{m}{n} = -1</math>.
</div>

In this case, then, we deduce that,


<div align="center">
<math>~\lambda = -\frac{1}{2}\biggl(\frac{m}{n}\biggr) = +\frac{1}{2}</math>.
</div>

So, even though these first two constraints have not revealed the ''value'' of either of the exponents, <math>~n</math> and <math>~m</math>, we see that the resulting trial eigenfunction must be,
<div>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="center">
<math>~\mathcal{G}_\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0^n(1 + \lambda x^2)^n \cdot (2 - x^2)^m </math>
  </td>
</tr>

<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0^n\biggl[\frac{(1 + \tfrac{1}{2} x^2)}{(2 - x^2)}\biggr]^n </math>
  </td>
</tr>

<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{a_0}{2}\biggr)^n\biggl[\frac{(2 + x^2)}{(2 - x^2)}\biggr]^n \, .</math>
  </td>
</tr>
</table>

Interesting!

====Third Constraint====

=====The Minus Root=====

Let's insert all of these relations into the algebraic expression that we have derived from the <math>~x^2</math> coefficient:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2s_{nm}(9 + 2 s_{nm})+ 5(1 - 4 \lambda)s_{nm} -5n(1 - 4 \lambda)(1  - 2\lambda) + 60n \lambda  -20n \lambda^2 - 8n(n-1)\lambda^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm}-n)[- 23  + 20 \lambda + 8n \lambda]   - 2(s_{nm}-n)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0
-\frac{45}{2}\biggl(1 - 4 \lambda\biggr) +\frac{45}{2}\biggl(1 - 4 \lambda\biggr) + n\lambda \biggl\{ 60  -20 \lambda + 8\biggl[\frac{11-4\lambda}{2(1-2\lambda)}\biggr]\lambda \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{9\lambda}{(1-2\lambda)} \biggl\{- 23  + 20 \lambda - \biggl[ \frac{36\lambda}{(1-2\lambda)} \biggr] \lambda \biggr\}   - 2\biggl[ \frac{9\lambda}{(1-2\lambda)}\biggr]^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)} \biggl\{ -30  +10 \lambda - 2\biggl[\frac{11-4\lambda}{(1-2\lambda)}\biggr]\lambda 
- 23  + 20 \lambda - \biggl[ \frac{36\lambda}{(1-2\lambda)} \biggr] \lambda  - \biggl[ \frac{18\lambda}{(1-2\lambda)}\biggr]\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)} \biggl\{ -53  +30 \lambda - \biggl[\frac{58-8\lambda}{(1-2\lambda)}\biggr]\lambda 
- \biggl[ \frac{18\lambda}{(1-2\lambda)}\biggr]\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)^2} \biggl\{( -53  +30 \lambda)(1-2\lambda) - (58-8\lambda)\lambda - 18\lambda  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)^2} \biggl[ -53  +30 \lambda  + 106\lambda -60\lambda^2 - 58\lambda + 8\lambda^2 - 18\lambda  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)^2} \biggl[ -53   + 60\lambda -52\lambda^2   \biggr]
</math>
  </td>
</tr>

</table>
</div>

Let's repeat this step, but start from an earlier expression for the <math>~x^2</math> coefficient, namely,

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

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + 60n \lambda  -20n \lambda^2- 25m    
+ 20m \lambda + 8n m \lambda  - [ 8n(n-1)\lambda^2 + 2m(m-1) ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + n\biggl[60 \lambda  -20 \lambda^2 + \frac{m}{n}\biggl(- 25    
+ 20\lambda \biggr)\biggr] + 8n m \lambda  - [ 8n(n-1)\lambda^2 + 2m(m-1) ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + n\biggl[60 \lambda  -12 \lambda^2 + \frac{m}{n}\biggl(- 23    
+ 20\lambda \biggr)\biggr] + 2n^2\biggl[- 4\lambda^2 +  4 \biggl(\frac{m}{n}\biggr) \lambda  -  \biggl(\frac{m}{n}\biggr) ^2  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

The first two terms on the RHS immediately go to zero because, for this ''specific'' eigenfunction, both <math>~\alpha</math> and <math>~\sigma^2</math> are zero.  Plugging in our determined expressions for <math>~n^-</math> and <math>~(m^-/n^-)</math> gives,


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

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{9}{2(1-2\lambda)}\biggl[60 \lambda  -12 \lambda^2 -2\lambda\biggl(- 23    
+ 20\lambda \biggr)\biggr] + 2\biggl[- \frac{9}{2(1-2\lambda)}\biggr]^2\biggl[- 4\lambda^2 +  4 \biggl(-2\lambda\biggr) \lambda  -  \biggl(-2\lambda\biggr) ^2  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{9\lambda}{(1-2\lambda)}\biggl[53   -26 \lambda\biggr] 
- \frac{8\cdot 81 \lambda^2}{(1-2\lambda)^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{9\lambda}{(1-2\lambda)^2}\biggl[ (1-2\lambda)(53   -26 \lambda) + 72 \lambda \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{9\lambda}{(1-2\lambda)^2}\biggl[53 - 60 \lambda + 52\lambda^2 \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
which exactly matches the previous, but messier, derivation.  Now, the two roots of the quadratic expression inside the square brackets are,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^3\cdot 13} \biggl[ 2^2\cdot 3\cdot 5 \pm \sqrt{ -2^8\cdot 29 }\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2\cdot 13} \biggl[ 3\cdot 5 \pm \sqrt{ -2^4\cdot 29 }\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Both roots are imaginary numbers and therefore not of interest in the context of this astrophysical problem.

=====The Plus Root=====
Next, in addition to setting <math>~\alpha = \sigma^2 = 0</math>, we'll plug <math>~\lambda = \tfrac{1}{2}</math> and <math>~m^+/n^+ = -1</math> into the third constraint expression as follows:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + n\biggl[ 60 \lambda  -20 \lambda^2- 25\biggl(\frac{m}{n}\biggr)    
+ 20\biggl(\frac{m}{n}\biggr) \lambda + 8\lambda^2 + 2\biggl(\frac{m}{n}\biggr)  \biggr] + n^2\biggl[ 8\biggl(\frac{m}{n}\biggr) \lambda  - 8\lambda^2 - 2\biggl(\frac{m}{n}\biggr)^2  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n\biggl[ 60 \lambda  -20 \lambda^2+ 25    
- 20 \lambda + 8\lambda^2 - 2  \biggr] + n^2\biggl[ -8 \lambda  - 8\lambda^2 - 2  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n[ 40 \lambda  - 12 \lambda^2+ 23 ] -2 n^2[ 4 \lambda  + 4\lambda^2 +1 ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n[ 20  - 3+ 23 ] -2 n^2[ 2  + 1 +1 ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8n(5-n) \, .
</math>
  </td>
</tr>

</table>
</div>

So the nontrivial solution is <math>~n^+ = 5</math> &#8212; and, hence, <math>~m^+ = -5</math> &#8212; in which case the trial eigenfunction is,
<div>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="center">
<math>~\mathcal{G}_\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{a_0}{2}\biggr)^5\biggl[\frac{(2 + x^2)}{(2 - x^2)}\biggr]^5 \, .</math>
  </td>
</tr>
</table>



====Fifth Constraint====
=====The Minus Root=====
In a similar vein, let's insert all of the deduced relations into the algebraic expression that we have derived from the <math>~x^6</math> coefficient:

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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(6\lambda - 11\lambda^2)  + \sigma^2\lambda^2 + 11 n \lambda  +22 m \lambda  - 47n \lambda^2 - 25m \lambda^2  -12 n m \lambda^2 + 4n m \lambda
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(6\lambda - 11\lambda^2)  + \sigma^2\lambda^2 + n\biggl[ 11 \lambda - 47\lambda^2  
+22 \biggl(\frac{m}{n}\biggr) \lambda  - 25\biggl(\frac{m}{n}\biggr) \lambda^2
+ 10 \lambda^2 + 2\biggl(\frac{m}{n}\biggr)\lambda^2  - 4\biggl(\frac{m}{n}\biggr)\lambda\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+n^2\biggl[ 
-12 \biggl(\frac{m}{n}\biggr) \lambda^2 + 4\biggl(\frac{m}{n}\biggr) \lambda - 10 \lambda^2 -2\biggl(\frac{m}{n}\biggr)^2\lambda^2  +4\biggl(\frac{m}{n}\biggr)^2\lambda \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(6\lambda - 11\lambda^2)  + \sigma^2\lambda^2 + n\biggl[ 11 \lambda - 37\lambda^2  
+ \lambda\biggl(\frac{m}{n}\biggr)(18   - 23\lambda)
\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+n^2\biggl[ - 10 \lambda^2 
+ 4\lambda\biggl(\frac{m}{n}\biggr)(1-3 \lambda^2 ) + 2\lambda\biggl(\frac{m}{n}\biggr)^2 ( 2 -\lambda ) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

As above, the first two terms on the RHS immediately go to zero because, for this ''specific'' eigenfunction, both <math>~\alpha</math> and <math>~\sigma^2</math> are zero.  Plugging in our determined expressions for <math>~n^-</math> and <math>~(m^-/n^-)</math> gives,


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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{9}{2(1-2\lambda)}
\biggl[ 11 \lambda - 37\lambda^2  -2 \lambda^2(18   - 23\lambda)\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+2\biggl[  -\frac{9}{2(1-2\lambda)} \biggr]^2\biggl[ - 5 \lambda^2 -4\lambda^2(1-3 \lambda^2 ) + 4\lambda^3 ( 2 -\lambda ) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{9\lambda}{2(1-2\lambda)}
\biggl[ 11 - 73\lambda  +46 \lambda^2\biggr]  
+\biggl[ \frac{9^2\lambda^2}{2(1-2\lambda)^2} \biggr]\biggl[ - 9 + 8\lambda  +8\lambda^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{9\lambda}{2(1-2\lambda)^2} \biggl\{(1-2\lambda)
[ 11 - 73\lambda  +46 \lambda^2]  
-9\lambda [ - 9 + 8\lambda  +8\lambda^2]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{9\lambda}{2(1-2\lambda)^2} \biggl[11 - 14\lambda  +120 \lambda^2 -164 \lambda^3
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

=====The Plus Root=====
Next, in addition to setting <math>~\alpha = \sigma^2 = 0</math>, we'll plug <math>~\lambda = \tfrac{1}{2}</math> and <math>~m^+/n^+ = -1</math> into the fifth constraint expression as follows:

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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
\alpha(6\lambda - 11\lambda^2)  + \sigma^2\lambda^2 + n\biggl[ 11 \lambda - 37\lambda^2  
+ \lambda\biggl(\frac{m}{n}\biggr)(18   - 23\lambda)
\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+n^2\biggl[ - 10 \lambda^2 
+ 4\lambda\biggl(\frac{m}{n}\biggr)(1-3 \lambda^2 ) + 2\lambda\biggl(\frac{m}{n}\biggr)^2 ( 2 -\lambda ) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
n\lambda [ 11 - 37\lambda  - (18   - 23\lambda) ]  
+n^2\lambda [ - 10 \lambda - 4 (1-3 \lambda^2 ) + 2 ( 2 -\lambda ) ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
n\lambda [ -7 - 14\lambda ]  +n^2\lambda [ - 10 \lambda -4 + 12 \lambda^2  + 4 - 2\lambda  ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
- 7n  - \biggl( \frac{3}{2} \biggr) n^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
- 7n\biggl[1  + \biggl( \frac{3}{14} \biggr) n \biggr] \, .
</math>
  </td>
</tr>

</table>
</div>
From this constraint, it appears that the nontrivial result is, <math>~n = -14/3</math>.

====Fourth Constraint====

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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(10 \lambda^2  - 22 \lambda +3) - \sigma^2(2\lambda^2- 2\lambda ) - 47n \lambda+ 60n \lambda^2  - 50m \lambda    +11m + 10m \lambda^2-12 n m \lambda    + 8n m \lambda^2
</math>
  </td>
</tr>

<tr>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
  <td align="left">
<math>
~ + 16n(n-1)\lambda^2 - 4m(m-1)\lambda + 2m(m-1) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(10 \lambda^2  - 22 \lambda +3) - \sigma^2(2\lambda^2- 2\lambda ) + n\biggl[- 47\lambda+ 60 \lambda^2  - 50\biggl(\frac{m}{n}\biggr) \lambda    
+11\biggl(\frac{m}{n}\biggr) + 10\biggl(\frac{m}{n}\biggr) \lambda^2 - 16\lambda^2 + 4\biggl(\frac{m}{n}\biggr) \lambda - 2\biggl(\frac{m}{n}\biggr)\biggr] 
</math>
  </td>
</tr>

<tr>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
  <td align="left">
<math>
~+n^2 \biggl[ -12 \biggl(\frac{m}{n}\biggr) \lambda  + 8\biggl(\frac{m}{n}\biggr) \lambda^2 + 16\lambda^2 - 4\biggl(\frac{m}{n}\biggr)^2\lambda + 2\biggl(\frac{m}{n}\biggr)^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(10 \lambda^2  - 22 \lambda +3) - \sigma^2(2\lambda^2- 2\lambda ) + n\biggl[- 47\lambda+ 44 \lambda^2  - 37\biggl(\frac{m}{n}\biggr) \lambda    
+ 10\biggl(\frac{m}{n}\biggr) \lambda^2 \biggr] 
</math>
  </td>
</tr>

<tr>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
  <td align="left">
<math>
~+n^2 \biggl[ + 16\lambda^2 -12 \biggl(\frac{m}{n}\biggr) \lambda  + 8\biggl(\frac{m}{n}\biggr) \lambda^2 - 4\biggl(\frac{m}{n}\biggr)^2\lambda + 2\biggl(\frac{m}{n}\biggr)^2 \biggr]
</math>
  </td>
</tr>
</table>
</div>

=====The Minus Root=====
In addition to setting <math>~\alpha = \sigma^2 = 0</math>, here we plug <math>~n^- = -9/[2(1-2\lambda)]</math> and <math>~m^+/n^+ = -2\lambda</math> into the fourth constraint expression as follows:


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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
n\lambda [- 47+ 118 \lambda    -20 \lambda^2 ] +n^2 \lambda^2[ 16 +24   -16 \lambda - 16 \lambda + 8 ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\biggl[\frac{9}{2(1-2\lambda)}\biggr]
\lambda [- 47+ 118 \lambda    -20 \lambda^2 ] +\biggl[\frac{9}{2(1-2\lambda)}\biggr]^2 \lambda^2[ 48   -32 \lambda]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{9\lambda}{2(1-2\lambda)^2}\biggr]\biggl\{9 \lambda[ 24   -16 \lambda]  -
(1-2\lambda) [- 47+ 118 \lambda    -20 \lambda^2 ] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{9\lambda}{2(1-2\lambda)^2}\biggr]\biggl\{216\lambda - 144\lambda^2  
+ [47- 118 \lambda    +20 \lambda^2 ] + [- 94\lambda + 236 \lambda^2    -40 \lambda^3 ]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{9\lambda}{2(1-2\lambda)^2}\biggr]\biggl\{47 -4\lambda + 112\lambda^2   -40 \lambda^3 
\biggr\}
</math>
  </td>
</tr>
</table>
</div>


=====The Plus Root=====
Next, in addition to setting <math>~\alpha = \sigma^2 = 0</math>, we'll plug <math>~\lambda = \tfrac{1}{2}</math> and <math>~m^+/n^+ = -1</math> into the fourth constraint expression as follows:

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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
n\biggl[- 47\lambda+ 44 \lambda^2  + 37 \lambda    - 10 \lambda^2 \biggr] +n^2 \biggl[ + 16\lambda^2 + 12 \lambda  - 8 \lambda^2 - 4 \lambda + 2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{7n}{2}\biggl[1+ n\biggl(\frac{16}{7}\biggr)  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

From this constraint, it appears that the nontrivial result is, <math>~n^+ = -7/16</math>.