Editing Apps/Blaes85SlimLimit (section)

===Singular Sturm-Liouville Problem===
Drawing on Theorem 3.16 from [http://lsec.cc.ac.cn/~hyu/teaching/shonm2013/STWchap3.2p.pdf Yu's class notes], we find that each one of the set of <math>j=0 \rightarrow \infty</math> ''Jacobi polynomials'', <math>~J_j^{\alpha,\beta}(x)</math>, is an eigenfunction of the singular Sturm-Liouville problem whose mathematical definition is provided by the 2<sup>nd</sup>-order ODE,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{L}_{\alpha,\beta}J_j^{\alpha,\beta}(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_j^{\alpha,\beta}J_j^{\alpha,\beta}(x) \, ,</math>
  </td>
</tr>
</table>
</div>
where the differential operator,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{L}_{\alpha,\beta}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
-(1-x)^{-\alpha}(1+x)^{-\beta} \cdot \frac{d}{dx} \biggl[ (1-x)^{\alpha+1}(1+x)^{\beta+1} \cdot \frac{d}{dx} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x^2-1)\cdot \frac{d^2}{dx^2} + [\alpha - \beta + (\alpha+\beta+2)x]\cdot \frac{d}{dx} 
\, ,</math>
  </td>
</tr>
</table>
</div>
and the corresponding j<sup>th</sup> eigenvalue is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_j^{\alpha,\beta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~j(j+\alpha+\beta + 1) \, .</math>
  </td>
</tr>
</table>
</div>
(Note that we have used "j" instead of the more traditional use of "n" to identify the specific Jacobi polynomial, because we are already using "n" to denote the polytropic index.)  According to Theorem 3.17 from [http://lsec.cc.ac.cn/~hyu/teaching/shonm2013/STWchap3.2p.pdf Yu's class notes], for each specified value of the index, <math>~j</math>, the eigenfunction solution to this eigenvalue problem &#8212; that is, the relevant Jacobi polynomial &#8212; is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~J_j^{\alpha,\beta}(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-x)^{-\alpha}(1+x)^{-\beta} \biggl\{
\frac{(-1)^j}{2^j j!} \cdot \frac{d^j}{dx^j}\biggl[ (1-x)^{j+\alpha}(1+x)^{j+\beta} \biggr] 
\biggr\} \, .</math>
  </td>
</tr>
</table>
</div>

<!-- OLD NOTE
Note:  When setting <math>~j=1</math> then performing the differentiation specified in the above definition of <math>~J_j^{\alpha,\beta}</math>, I don't actually obtain the expression provided by eq. (36) of [http://mathworld.wolfram.com/JacobiPolynomial.html Wolfram MathWorld] or (which is the same) eq. (3.110) in [http://lsec.cc.ac.cn/~hyu/teaching/shonm2013/STWchap3.2p.pdf Yu's class notes].  
-->
As an example, let's carry out the prescribed differentiation to determine the Jacobi polynomial for the case of <math>~j=2</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~J_2^{\alpha,\beta}(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-x)^{-\alpha}(1+x)^{-\beta} \biggl\{
\frac{(-1)^2}{2^2 \cdot 2!} \cdot \frac{d^2}{dx^2}\biggl[ (1-x)^{2+\alpha}(1+x)^{2+\beta} \biggr] 
\biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^3}(1-x)^{-\alpha}(1+x)^{-\beta} \frac{d}{dx}\biggl\{-(1+x)^{2+\beta} (2+\alpha)(1-x)^{1+\alpha} + (1-x)^{2+\alpha}(2+\beta)(1+x)^{1+\beta}
\biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^3}(1-x)^{-\alpha}(1+x)^{-\beta} \biggl\{
(2+\alpha)(1+\alpha)(1+x)^{2+\beta} (1-x)^{\alpha} - (2+\alpha)(2+\beta)(1+x)^{1+\beta} (1-x)^{1+\alpha}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (2+\beta)(1+\beta)(1-x)^{2+\alpha}(1+x)^{\beta} - (2+\beta)(2+\alpha)(1-x)^{1+\alpha}(1+x)^{1+\beta}
\biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^3} \biggl\{
(2+\alpha)(1+\alpha)(1+x)^{2} + 2(2+\alpha)(2+\beta)(x^2-1) + (2+\beta)(1+\beta)(1-x)^{2} 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^3} \biggl\{
(2+\alpha)(1+\alpha)(1+2x + x^2) + 2(2+\alpha)(2+\beta)(x^2-1) + (2+\beta)(1+\beta)(1-2x +x^2) 
\biggr\} 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2^3} \biggl\{
[-4 - \alpha + \alpha^2-\beta + \beta^2 - 2\alpha\beta] 
+ 2x[3\alpha + \alpha^2 - 3\beta-\beta^2)] 
+ x^2[12+7\alpha + \alpha^2 + 2\alpha\beta + 7\beta+\beta^2] 
\biggr\} 
\, .</math>
  </td>
</tr>
</table>
</div>


Table 1 presents this "j=2" eigenfunction &#8212; along with its corresponding eigenfrequency &#8212; as well as the eigenfunctions and eigenfrequencies for "j=0" and "j=1."
<div align="center" id="Table1">
<table align="center" border="1" cellpadding="5">
<tr>
  <th align="center" colspan="3"><font size="+1">Table 1:  Example Jacobi Polynomials</font></th>
</tr>
<tr>
  <td align="center"><math>~j</math></td>
  <td align="center"><math>~J_j^{\alpha,\beta}(x)</math></td>
  <td align="center"><math>~\lambda_j^{\alpha,\beta}</math></td>
</tr>
<tr>
  <td align="center">
<math>~0</math>
  </td>
  <td align="center">
<math>~1</math>
  </td>
  <td align="center">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>~1</math>
  </td>
  <td align="center">
<math>~\tfrac{1}{2}(\alpha+\beta+2)x + \tfrac{1}{2}(\alpha-\beta)</math>
  </td>
  <td align="center">
<math>~(\alpha+\beta+2)</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>~2</math>
  </td>
  <td align="center">
<math>~
\tfrac{1}{8}(12+7\alpha + \alpha^2 + 7\beta+\beta^2+ 2\alpha\beta ) x^2
+ \tfrac{1}{4}(3\alpha + \alpha^2 - 3\beta-\beta^2) x
+ \tfrac{1}{8}(-4 - \alpha + \alpha^2-\beta + \beta^2 - 2\alpha\beta) 
</math>
  </td>
  <td align="center">
<math>~2(\alpha+\beta+3)</math>
  </td>
</tr>
<tr>
  <td align="left" colspan="3">See also, eqs. (35)-(37) of [http://mathworld.wolfram.com/JacobiPolynomial.html Wolfram MathWorld]. 
</tr>
</table>
</div>

Okay, so now I understand how to identify the complete set of eigenvectors for a stability problem in which the governing ODE takes the form of the singular Sturm-Liouville equation.  But, with the recognition that the Blaes85 ODE does not have precisely the same form as the singular Sturm-Liouville equation, it is unclear how we will win by pursuing this specific line of investigation.