Editing Appendix/Ramblings/NumericallyDeterminedEigenvectors (section)

==Finite-Difference Representation==

===General Approach===

Working with the Taylor series expansion, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x(\xi)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x(a) + (\xi - a)  x_a' + \tfrac{1}{2} (\xi-a)^2 x_a'' \, ,
</math>
  </td>
</tr>
</table>
</div>
and letting <math>~\xi_\pm = a \pm \Delta </math>, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_+</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x(a) + \Delta \cdot  x_a' + \tfrac{1}{2} \Delta^2 x_a'' \, ,
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_-</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x(a) - \Delta \cdot  x_a' + \tfrac{1}{2} \Delta^2 x_a'' \, .
</math>
  </td>
</tr>
</table>
</div>
Subtracting the second of these two expressions from the first gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_+ - x_-</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2 \Delta \cdot  x_a' 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x_a'</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{x_+ - x_-}{2 \Delta}   \, ;
</math>
  </td>
</tr>
</table>
</div>
while, adding the two expressions together gives,

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

<tr>
  <td align="right">
<math>~\frac{x_+ - 2x_a + x_-}{\Delta^2}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x_a'' \, .
</math>
  </td>
</tr>
</table>
</div>

===Integrating Outward Through the Core===
From the LAWE for the core, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a (g^2 - a^2) x_a''</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  ( 4g^2 - 6a^2 ) x_a' - a \mathfrak{F}_\mathrm{core} x_a \, .
</math>
  </td>
</tr>
</table>
</div>

So, putting these last three expressions together gives an approximate relation between <math>~x_+</math> and the previous two values of the function, <math>~x_-</math> and <math>~x_a</math>, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a (g^2 - a^2) \biggl[ \frac{x_+ - 2x_a + x_-}{\Delta^2} \biggr]</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
-  ( 4g^2 - 6a^2 ) \biggl[\frac{x_+ - x_-}{2 \Delta} \biggr] - a \mathfrak{F}_\mathrm{core} x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
a (g^2 - a^2) \biggl[ \frac{x_+ }{\Delta^2} \biggr] +  ( 4g^2 - 6a^2 ) \biggl[\frac{x_+ }{2 \Delta} \biggr]
</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~a (g^2 - a^2) \biggl[ \frac{2x_a - x_-}{\Delta^2} \biggr]
+  ( 4g^2 - 6a^2 ) \biggl[\frac{x_-}{2 \Delta} \biggr] - a \mathfrak{F}_\mathrm{core} x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
\Rightarrow~~~x_+[2a (g^2 - a^2)  +   \Delta( 4g^2 - 6a^2 )  ]
</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~2a (g^2 - a^2) [ 2x_a - x_- ]
+  ( 4g^2 - 6a^2 ) [\Delta x_- ] - 2\Delta^2 a \mathfrak{F}_\mathrm{core} x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~[4a (g^2 - a^2)  - 2\Delta^2 a \mathfrak{F}_\mathrm{core} ]x_a
+ [ \Delta( 4g^2 - 6a^2 )  - 2a (g^2 - a^2)]  x_-  \, .
</math>
  </td>
</tr>
</table>
</div>

Now, at the very center of the configuration, <math>~(a = 0)</math>, we expect the function, <math>~x(\xi)</math>, to be symmetric; that is, we expect <math>~x_- = x_+</math>.  So for this case alone, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
x_+[2a (g^2 - a^2)  +   \Delta( 4g^2 - 6a^2 )   - \Delta( 4g^2 - 6a^2 )  + 2a (g^2 - a^2)]  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[4a (g^2 - a^2)  - 2\Delta^2 a \mathfrak{F}_\mathrm{core} ]x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
x_+[2(g^2 - \cancelto{0}{a^2})  + 2(g^2 - \cancelto{0}{a^2})]  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[4 (g^2 - \cancelto{0}{a^2})  - 2\Delta^2 \mathfrak{F}_\mathrm{core} ]x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
x_+  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 - \frac{\Delta^2 \mathfrak{F}_\mathrm{core}}{2g^2} \biggr]x_a  \, .
</math>
  </td>
</tr>
</table>
</div>
For all other coordinate locations, <math>~a = \xi</math>, in the range, <math>~0 < \xi < 1</math>, we will use the general expression, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\Rightarrow~~~x_+
</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{[4a (g^2 - a^2)  - 2\Delta^2 a \mathfrak{F}_\mathrm{core} ]x_a
+ [ \Delta( 4g^2 - 6a^2 )  - 2a (g^2 - a^2)]  x_-  }{[2a (g^2 - a^2)  +   \Delta( 4g^2 - 6a^2 )  ] } \, .
</math>
  </td>
</tr>
</table>
</div>

Keep in mind that, when we move across the interface at <math>~a = 1</math>, we want both the value of the function, <math>~x_q</math>, and its first derivative, <math>~x_q'</math>, to be the same as viewed from ''both'' the envelope and the core.  In a numerical integration algorithm, it will be very straightforward to set the ''value'' of the eigenfunction at the interface.  In order to properly handle the first derivative, I propose that we extend the core solution and evaluate the eigenfunction at one zone beyond the interface, and identify the values of the eigenfunction that straddles the interface as,
<div align="center">
<math>~(x_-)_q</math> &nbsp;  &nbsp; &nbsp; and  &nbsp; &nbsp; &nbsp; <math>~(x_+)_q</math>.
</div>
Then define the slope of the eigenfunction at the interface by the expression,
<div align="center" id="InterfaceSlope">
<table border="1" cellpadding="8">
<tr><td align="center">Slope at the Interface</td></tr>
<tr><td align="center">

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

<tr>
  <td align="right">
<math>~x_q'</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(x_+)_q - (x_-)_q}{2\Delta} \, .</math>
  </td>
</tr>
</table>

</td></tr>
</table>
</div>


===Integrating Outward Through the Envelope===
From the LAWE for the envelope, we have,

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

<tr>
  <td align="right">
<math>~a^2( 1  - q^3 a^3 ) x_a'' </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- ( 3  - 6q^3 a^3 ) a x_a'
- 
[ q^3  \mathfrak{F}_\mathrm{env} a^3 -\alpha_e  ]x_a \, .
</math>
  </td>
</tr>
</table>
</div>
Inserting the same finite-difference expressions for the first and second derivatives, we therefore have,

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

<tr>
  <td align="right">
<math>~a^2( 1  - q^3 a^3 ) \biggl[ \frac{x_+ - 2x_a + x_-}{\Delta^2}  \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- ( 3  - 6q^3 a^3 ) a \biggl[ \frac{x_+ - x_-}{2 \Delta}   \biggr]
- 
[ q^3  \mathfrak{F}_\mathrm{env} a^3 -\alpha_e  ]x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
a^2( 1  - q^3 a^3 ) \biggl[ \frac{x_+ }{\Delta^2}  \biggr] + ( 3  - 6q^3 a^3 ) a \biggl[ \frac{x_+ }{2 \Delta}   \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
( 3  - 6q^3 a^3 ) a \biggl[ \frac{x_-}{2 \Delta}   \biggr]
- a^2( 1  - q^3 a^3 ) \biggl[ \frac{x_- - 2x_a }{\Delta^2}  \biggr] 
- [ q^3  \mathfrak{F}_\mathrm{env} a^3 -\alpha_e  ]x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
x_+ [2a^2( 1  - q^3 a^3 )  + \Delta ( 3  - 6q^3 a^3 ) a ] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ \Delta ( 3  - 6q^3 a^3 ) a 
- 2a^2( 1  - q^3 a^3 ) ] x_- + [4a^2( 1  - q^3 a^3 ) 
- 2\Delta^2 ( q^3  \mathfrak{F}_\mathrm{env} a^3 -\alpha_e  ) ] x_a 
</math>
  </td>
</tr>
</table>
</div>

Now, at the interface (only), we need to relate <math>~x_-</math> to <math>~x_+</math> in such a way that the slope gives the proper value at the interface.  Specifically, we need to set,
<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>~x_+ - 2\Delta (x_q') \, ,</math>
  </td>
</tr>
</table>
</div>
where, <math>~x_q'</math> takes the [[#InterfaceSlope|value that was determined for the core]].  Hence, ''at'' the interface <math>~(a = 1)</math>, the first step into the envelope is special and demands that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
x_+ [2a^2( 1  - q^3 a^3 )  + \Delta ( 3  - 6q^3 a^3 ) a ] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ \Delta ( 3  - 6q^3 a^3 ) a 
- 2a^2( 1  - q^3 a^3 ) ] [x_+ - 2\Delta (x_q')]  + [4a^2( 1  - q^3 a^3 ) 
- 2\Delta^2 ( q^3  \mathfrak{F}_\mathrm{env} a^3 -\alpha_e  ) ] x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
x_+ [4a^2( 1  - q^3 a^3 )   ] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ \Delta ( 3  - 6q^3 a^3 ) a - 2a^2( 1  - q^3 a^3 ) ] [- 2\Delta (x_q')]  
+ [4a^2( 1  - q^3 a^3 ) - 2\Delta^2 ( q^3  \mathfrak{F}_\mathrm{env} a^3 -\alpha_e  ) ] x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2\Delta [ \Delta ( 3  - 6q^3 a^3 ) a - 2a^2( 1  - q^3 a^3 ) ] x_q'
+ [4a^2( 1  - q^3 a^3 ) - 2\Delta^2 ( q^3  \mathfrak{F}_\mathrm{env} a^3 -\alpha_e  ) ] x_a 
</math>
  </td>
</tr>

<tr>
  <td align="right">
and, setting, &nbsp; &nbsp; &nbsp;  <math>~a = 1 ~~~~\Rightarrow ~~~
x_+ 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2\Delta [2( 1  - q^3  ) - \Delta ( 3  - 6q^3  )  ] x_q'
+ [4( 1  - q^3  ) - 2\Delta^2 ( q^3  \mathfrak{F}_\mathrm{env}  -\alpha_e  ) ] x_a }{ 4( 1  - q^3  )  } \, .
</math>
  </td>
</tr>
</table>
</div>