Difference between revisions of "User:Tohline/Apps/Blaes85SlimLimit"

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(Begin chapter on Jacobi Polynomial solution to eigenvalue problem)
(Make first effort to summarize)
 
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=Stability of PP Tori in the Slim Torus Limit=
=Oscillations of PP Tori in the Slim Torus Limit=
{{LSU_HBook_header}}
{{LSU_HBook_header}}


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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
Line 32: Line 32:
</div>
</div>


where, <math>~\delta W^{(0)}</math> is the dimensionless enthalpy perturbation.
where, <math>~\delta W^{(0)}</math> is the dimensionless enthalpy perturbation. Making the substitution,
<div align="center">
<math>~\delta W^{(0)} ~\rightarrow~ V(\eta) \exp (ik\theta) \, ,</math>
</div>
this governing equation &#8212; now, a one-dimensional, 2<sup>nd</sup>-order ODE &#8212; 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>~
\eta^2 (1-\eta^2)\cdot \frac{d^2V}{d \eta^2}
- k^2(1-\eta^2) V
+ \biggl[ \eta (1-\eta^2)  -2 n \eta^3
\biggr] \cdot \frac{d V}{d \eta}
+ 2n\eta^2 \biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 V \, .
</math>
  </td>
</tr>
</table>
</div>
 
Making the additional substitution,
<div align="center">
<math>~V ~\rightarrow~ \eta^{|k|} \Upsilon(\eta) \, ,</math>
</div>
and appreciating that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{dV}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~|k|\eta^{|k|-1} \Upsilon + \eta^{|k|} \frac{d\Upsilon}{d\eta} \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{d^2V}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
|k|[|k|-1] \eta^{|k|-2}\Upsilon
+ 2|k|\eta^{|k|-1} \frac{d\Upsilon}{d\eta}
+ \eta^{|k|} \frac{d^2\Upsilon}{d\eta^2}\, ,</math>
  </td>
</tr>
</table>
</div>
 
the governing ODE becomes,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
\biggl\{k^2(1-\eta^2)  - 2n\eta^2 \biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2\biggr\} \eta^{|k|}\Upsilon
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2 (1-\eta^2)\cdot \biggl[ |k|[|k|-1] \eta^{|k|-2}\Upsilon
+ 2|k|\eta^{|k|-1} \frac{d\Upsilon}{d\eta}
+ \eta^{|k|} \frac{d^2\Upsilon}{d\eta^2}
\biggr]
+ \biggl[ \eta (1-\eta^2)  -2 n \eta^3
\biggr] \cdot \biggl[ |k|\eta^{|k|-1} \Upsilon + \eta^{|k|} \frac{d\Upsilon}{d\eta} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-\eta^2)
\biggl[ |k|[|k|-1] \eta^{|k|}\Upsilon
+ 2|k|\eta^{|k|+1} \frac{d\Upsilon}{d\eta}
+ \eta^{|k|+2} \frac{d^2\Upsilon}{d\eta^2}\biggr]
+ \biggl[ (1-\eta^2)  -2 n \eta^2
\biggr] \cdot \biggl[ |k|\eta^{|k|} \Upsilon + \eta^{|k|+1} \frac{d\Upsilon}{d\eta} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta^{|k|}(1-\eta^2)
\biggl[ k^2 \Upsilon
+ (2|k|+1)\eta \frac{d\Upsilon}{d\eta}
+ \eta^{2} \frac{d^2\Upsilon}{d\eta^2} \biggr]
- \eta^{|k|}\biggl[ 2 n \eta^2 \biggr] \cdot \biggl[ |k| \Upsilon + \eta \frac{d\Upsilon}{d\eta} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~
- 2n\eta^2 \biggl[\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 -|k|\biggr]  \Upsilon
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-\eta^2)
\biggl[ \eta^{2} \frac{d^2\Upsilon}{d\eta^2} + (2|k|+1)\eta \frac{d\Upsilon}{d\eta} \biggr]
- \biggl[ 2 n \eta^3  \frac{d\Upsilon}{d\eta} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
Finally, then, making the independent variable substitution,
<div align="center">
<math>~\eta^2 ~\rightarrow ~ y</math> &nbsp; &nbsp; &nbsp;
<math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp;
<math>~dy = 2\eta d\eta</math> &nbsp; &nbsp; &nbsp;
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d}{d\eta}</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~2y^{1/2}\frac{d}{dy}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{d^2}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~2\frac{d}{dy} + 4y\frac{d^2}{dy^2} \, .</math>
  </td>
</tr>
</table>
</div>
and,
 
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
  <td align="right">
<math>~
- 2ny \biggl[\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 -|k|\biggr]  \Upsilon
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-y) y \frac{d^2\Upsilon}{d\eta^2} + (2|k|+1)(1-y)y^{1/2} \frac{d\Upsilon}{d\eta}
- 2 n y^{3/2}  \frac{d\Upsilon}{d\eta}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4(1-y)y^2 \frac{d^2\Upsilon}{dy^2} + 2(1-y) y \frac{d\Upsilon}{dy} + 2(2|k|+1)(1-y)y \frac{d\Upsilon}{dy}
- 4 n y^{2}  \frac{d\Upsilon}{dy}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~~
- \frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 -|k|\biggr]  \Upsilon
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-y)y \frac{d^2\Upsilon}{dy^2} + \frac{1}{2}(1-y) \frac{d\Upsilon}{dy} + \frac{1}{2}(2|k|+1)(1-y)\frac{d\Upsilon}{dy}
- n y  \frac{d\Upsilon}{dy}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-y)y \frac{d^2\Upsilon}{dy^2} + (|k|+1)(1-y)\frac{d\Upsilon}{dy}
- n y  \frac{d\Upsilon}{dy}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-y)y \frac{d^2\Upsilon}{dy^2} + (|k|+1)\frac{d\Upsilon}{dy} -y (|k|+1+n)\frac{d\Upsilon}{dy} \, .
</math>
  </td>
</tr>
</table>
</div>
This matches equation (3.9) of Blaes85.  According to [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)], this equation "<font color="green">&hellip; is a standard eigenvalue problem whose only solutions are the Jacobi polynomials &hellip;</font>" 
==Solution==
My own background training and experience has not previously exposed me to the general class of Jacobi polynomials.  In my effort to understand this class of polynomials and, specifically, their relationship to the ''Singular Sturm-Liouville Equation'', I have found the following references to be useful:
* [http://lsec.cc.ac.cn/~hyu/teaching/shonm2013/STWchap3.2p.pdf PDF-formatted class notes] presented by [http://lsec.cc.ac.cn/~hyu/ Professor Haijun Yu]
* [http://people.uncw.edu/hermanr/mat463/ODEBook/Book/SL.pdf PDF-formatted class notes] presented by [http://people.uncw.edu/hermanr/ Professor Russell Herman]
* [http://people.math.sfu.ca/~cbm/aands/page_773.htm A couple of scanned pages from &sect;22 of Abramowitz &amp; Stegun (1972)] posted online by [http://people.math.sfu.ca/~cbm/ Colin Macdonald]
===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.
===Jacobi Differential Equation===
Instead, let's pursue the following lead.  Wolfram MathWorld provides a discussion of the [http://mathworld.wolfram.com/JacobiDifferentialEquation.html Jacobi Differential Equation].  Equation (1) of this reference appears to be identical to the singular Sturm-Liouville equation presented above.  However, in addition, this MathWorld chapter points out that [http://faculty.uml.edu/Dimitris_Christodoulou/Teaching/documents/handbook_DE.pdf Zwillinger (1997, p. 123)] gives a related differential equation that is referred to as "Jacobi's equation," namely,
<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>~x(1-x)y'' + [\gamma - (\alpha+1)x]y' + j(\alpha+j)y \, .</math>
  </td>
</tr>
</table>
</div>
If we rewrite our above-derived ODE &#8212; which is identical to eq. (3.9) of Blaes85 &#8212; as,
<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>~
(1-y)y \frac{d^2\Upsilon}{dy^2} +\biggl[ (|k|+1) -y (|k|+1+n)\biggr] \frac{d\Upsilon}{dy}
+ \frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 -|k|\biggr]  \Upsilon
</math>
  </td>
</tr>
</table>
</div>
we see that its form is identical to this so-called "Jacobi's equation."  The overlap is complete if we make the following parameter associations:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\gamma</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~|k|+1 \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~|k|+n \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~j(|k|+n + j)</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~\frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 -|k|\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
We can infer from this third association that, for the j<sup>th</sup> eigenfunction solution of the governing eigenvalue problem, the relevant oscillation frequency is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~|k| + \frac{2j}{n}(|k|+n + j)</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{n}\biggl[n|k| + 2j(|k|+n + j)\biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{n}\biggl[2j^2 + 2jn + 2j|k| + n|k| \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
This is identical to the eigenfrequency identified in eq. (3.12) of [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes85].  So, we must be on the right track!  Now I just need to figure out how to express the corresponding eigenfunction in terms of Jacobi polynomials.
From expression 22.5.1 on p. 777 of A&amp;S, we understand that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~G_j\biggl(\alpha+\beta + 1, \beta+1, \frac{x+1}{2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~P_j^{\alpha,\beta} (x) \biggl[ \frac{j! \Gamma(j + \alpha + \beta + 1)}{\Gamma(2j+\alpha+\beta+1)} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
In addition, we see from eq. (3.11) of Blaes85 that, to within a factor of a leading constant, the physically relevant eigenfunction is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Upsilon(y)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A_{jk} G_j(|k|+n, |k|+1, \eta^2) \, .</math>
  </td>
</tr>
</table>
</div>
By association, we conclude from this that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\beta</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~|k| \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\alpha + |k|+1</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~|k|+n \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{x+1}{2}</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~\eta^2 =y\, .</math>
  </td>
</tr>
</table>
</div>
From this third association, we conclude that the argument of the Jacobi polynomial should be,
<div align="center">
<math>x = 2y-1 \, .</math>
</div>
Following up on this clue, let's plug this coordinate transformation into Blaes85 expression (3.9) and see whether it takes the form of the Sturm-Liouville ODE.  Specifically, making the substitutions,
<div align="center">
<math>~y ~\rightarrow~\tfrac{1}{2}(x+1)</math>&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; <math>~dy = \tfrac{1}{2}dx \, ,</math>
</div>
we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
- \frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 -|k|\biggr]  \Upsilon
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl[1-\frac{(x+1)}{2}\biggr] \frac{(x+1)}{2} \frac{d^2\Upsilon}{dx^2} +2\biggl[ (|k|+1) - \frac{1}{2}(x+1) (|k|+1+n)\biggr] \frac{d\Upsilon}{dx}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[2-(x+1)\biggr] (x+1) \frac{d^2\Upsilon}{dx^2} +\biggl[2 (|k|+1) - (x+1) (|k|+1+n)\biggr] \frac{d\Upsilon}{dx}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-(x^2-1) \frac{d^2\Upsilon}{dx^2} -\biggl[(n - |k|-1) + x (|k|+1+n)\biggr] \frac{d\Upsilon}{dx}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 -|k|\biggr]  \Upsilon
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x^2-1) \frac{d^2\Upsilon}{dx^2} +\biggl[(n - |k|-1) + x (|k|+1+n)\biggr] \frac{d\Upsilon}{dx} \, ,
</math>
  </td>
</tr>
</table>
</div>
which is precisely in the standard Sturm-Liouville format.  The j<sup>th</sup> eigenfunction should therefore be, <math>~\Upsilon_j = J_j^{\alpha,\beta}(2\eta^2-1)</math>, and the associated eigenfrequency should be,
<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>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\alpha-\beta</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~n - |k| -1 \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\alpha + \beta +2</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~|k|+1+n \, .</math>
  </td>
</tr>
</table>
</div>
Adding these two expressions together implies that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~n-1 \, ,</math>
  </td>
</tr>
</table>
</div>
while subtracting them implies,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\beta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~|k|\, .</math>
  </td>
</tr>
</table>
</div>
Hence, the eigenfrequency is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2 -|k|\biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~j(j+\alpha+\beta + 1)</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~j(j+n + |k|)</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ \biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2j}{n}\cdot (j+n + |k|) + |k|</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{n}\biggl[2j^2+2jn + 2j|k| + n|k|\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Again, this precisely matches equation (3.12) from Blaes85, so we appear to be on the right track.  Putting this all together, [[#Table2|Table 2]] presents the first three eigenvectors that are relevant to the stability analysis of slim, PP tori.
{{LSU_WorkInProgress}}
<div align="center">
<table align="center" cellpadding="5" border="1">
<tr>
  <th>Intermediate Detail</th>
</tr>
<tr><td align="left">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Upsilon_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{A}{8} + \biggl(\frac{B}{4} \biggr) x + \biggl(\frac{C}{8} \biggr) x^2</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{8} \biggl[A-2B+C \biggr] + \biggl[B-C\biggr] \frac{\eta^2}{2} + C\cdot \frac{\eta^4}{2} \, ,</math>
  </td>
</tr>
</table>
where, from [[#Table1|Table 1]],
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~A</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-4 - \alpha + \alpha^2-\beta + \beta^2 - 2\alpha\beta</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(-2+|k|+k^2)-n(3+2|k|) + n^2 \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~B</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~3\alpha + \alpha^2 - 3\beta-\beta^2</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-(2+3|k|+k^2) + n + n^2 \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~C</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~12+7\alpha + \alpha^2 + 7\beta+\beta^2+ 2\alpha\beta</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(6 + 5|k|+k^2) +n(5+2|k|) +n^2 \, .
</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Upsilon_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{8} \biggl\{[(-2+|k|+k^2)-n(3+2|k|) + n^2]
-2[-(2+3|k|+k^2) + n + n^2]
+[(6 + 5|k|+k^2) +n(5+2|k|) +n^2]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \biggl\{
[-(2+3|k|+k^2) + n + n^2]
-[(6 + 5|k|+k^2) +n(5+2|k|) +n^2]
\biggr\} \frac{\eta^2}{2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \biggl\{(6 + 5|k|+k^2) +n(5+2|k|) +n^2 \biggr\}\frac{\eta^4}{2} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2} \biggl\{2 + 3|k| + k^2\biggr\}
- \biggl\{4 + 4|k|+k^2 +n(2+|k|)\biggr\} \eta^2
+ \biggl\{(6 + 5|k|+k^2) +n(5+2|k|) +n^2 \biggr\}\frac{\eta^4}{2} \, .
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>
<div align="center" id="Table2">
<table align="center" border="1" cellpadding="5">
<tr>
  <th align="center" colspan="3"><font size="+1">Table 2:  Example ''Slim'' PP-Tori Eigenvectors</font></th>
</tr>
<tr>
  <td align="center"><math>~j</math></td>
  <td align="center"><math>~\Upsilon_j = J_j^{n-1,|k|}(2\eta^2-1)</math></td>
  <td align="center"><math>~\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)_j^2</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>~(n + 1 + |k|)\eta^2 - (1 + |k|)</math>
  </td>
  <td align="center">
<math>~\tfrac{1}{n}[2+2n+(2+n)|k|]</math>
  </td>
</tr>
<tr>
  <td align="center">
<math>~2</math>
  </td>
  <td align="center">
<math>~
\tfrac{1}{2}[(6 + 5|k|+k^2) +n(5+2|k|) +n^2 ]\eta^4
- [4 + 4|k|+k^2 +n(2+|k|)]\eta^2
+  \tfrac{1}{2} [2 + 3|k| + k^2]
</math>
  </td>
  <td align="center">
<math>~\tfrac{1}{n}[8+4n + (4+n)|k|] </math>
  </td>
</tr>
</table>
</div>
==Summary==
The spatial structure of the normalized enthalpy perturbation associated with each <math>~(j,k)</math> mode of oscillation in a slim PP torus is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\delta W_j^{(0)}(\eta,\theta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta^{|k|}  J_j^{n-1,|k|}(2\eta^2-1) \cdot \exp (ik\theta)  \, ,</math>
  </td>
</tr>
</table>
</div>
and the associated oscillation frequency is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl( \frac{\sigma}{\Omega_0} + m  \biggr)^2  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{n}\biggl[2j^2+2jn + 2j|k| + n|k|\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Because the oscillation frequency is always positive, all modes are dynamically stable.  Because the meridional plane amplitude is everywhere real, the azimuthal distortion associated with each nonaxisymmetric mode, <math>~m</math>, is completely uninteresting.


=See Also=
=See Also=


* [https://en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory Wikipedia overview of Sturm-Liouville Theory]


{{LSU_HBook_footer}}
{{LSU_HBook_footer}}

Latest revision as of 01:02, 6 May 2016


Oscillations of PP Tori in the Slim Torus Limit

Whitworth's (1981) Isothermal Free-Energy Surface
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Statement of the Eigenvalue Problem

Here, we build on our discussion in an accompanying chapter in which five published analyses of nonaxisymmetric instabilities in Papaloizou-Pringle tori were reviewed: The discovery paper, PP84, and papers by four separate groups that were published within a couple of years of the discovery paper — Papaloizou & Pringle (1985), Blaes (1985), Kojima (1986), and Goldreich, Goodman & Narayan (1986). Following the lead of Blaes (1985; hereafter Blaes85), in particular, we have shown that the relevant eigenvalue problem is defined by the following 2nd-order PDE,

<math>~0</math>

<math>~=</math>

<math>~ \eta^2 (1-\eta^2)\cdot \frac{\partial^2(\delta W)^{(0)}}{\partial \eta^2} + (1-\eta^2) \cdot \frac{\partial^2(\delta W)^{(0)}}{\partial\theta^2} + \biggl[ \eta (1-\eta^2) -2 n \eta^3 \biggr] \cdot \frac{\partial (\delta W)^{(0)}}{\partial \eta} + 2n\eta^2 \biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 (\delta W)^{(0)} \, , </math>

where, <math>~\delta W^{(0)}</math> is the dimensionless enthalpy perturbation. Making the substitution,

<math>~\delta W^{(0)} ~\rightarrow~ V(\eta) \exp (ik\theta) \, ,</math>

this governing equation — now, a one-dimensional, 2nd-order ODE — becomes,

<math>~0</math>

<math>~=</math>

<math>~ \eta^2 (1-\eta^2)\cdot \frac{d^2V}{d \eta^2} - k^2(1-\eta^2) V + \biggl[ \eta (1-\eta^2) -2 n \eta^3 \biggr] \cdot \frac{d V}{d \eta} + 2n\eta^2 \biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 V \, . </math>

Making the additional substitution,

<math>~V ~\rightarrow~ \eta^{|k|} \Upsilon(\eta) \, ,</math>

and appreciating that,

<math>~\frac{dV}{d\eta}</math>

<math>~=</math>

<math>~|k|\eta^{|k|-1} \Upsilon + \eta^{|k|} \frac{d\Upsilon}{d\eta} \, ,</math>

<math>~\frac{d^2V}{d\eta^2}</math>

<math>~=</math>

<math>~ |k|[|k|-1] \eta^{|k|-2}\Upsilon + 2|k|\eta^{|k|-1} \frac{d\Upsilon}{d\eta} + \eta^{|k|} \frac{d^2\Upsilon}{d\eta^2}\, ,</math>

the governing ODE becomes,

<math>~ \biggl\{k^2(1-\eta^2) - 2n\eta^2 \biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2\biggr\} \eta^{|k|}\Upsilon </math>

<math>~=</math>

<math>~ \eta^2 (1-\eta^2)\cdot \biggl[ |k|[|k|-1] \eta^{|k|-2}\Upsilon + 2|k|\eta^{|k|-1} \frac{d\Upsilon}{d\eta} + \eta^{|k|} \frac{d^2\Upsilon}{d\eta^2}

\biggr]

+ \biggl[ \eta (1-\eta^2) -2 n \eta^3 \biggr] \cdot \biggl[ |k|\eta^{|k|-1} \Upsilon + \eta^{|k|} \frac{d\Upsilon}{d\eta} \biggr] </math>

 

<math>~=</math>

<math>~(1-\eta^2) \biggl[ |k|[|k|-1] \eta^{|k|}\Upsilon + 2|k|\eta^{|k|+1} \frac{d\Upsilon}{d\eta} + \eta^{|k|+2} \frac{d^2\Upsilon}{d\eta^2}\biggr] + \biggl[ (1-\eta^2) -2 n \eta^2 \biggr] \cdot \biggl[ |k|\eta^{|k|} \Upsilon + \eta^{|k|+1} \frac{d\Upsilon}{d\eta} \biggr] </math>

 

<math>~=</math>

<math>~\eta^{|k|}(1-\eta^2) \biggl[ k^2 \Upsilon + (2|k|+1)\eta \frac{d\Upsilon}{d\eta} + \eta^{2} \frac{d^2\Upsilon}{d\eta^2} \biggr] - \eta^{|k|}\biggl[ 2 n \eta^2 \biggr] \cdot \biggl[ |k| \Upsilon + \eta \frac{d\Upsilon}{d\eta} \biggr] </math>

<math>~\Rightarrow~~~ - 2n\eta^2 \biggl[\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 -|k|\biggr] \Upsilon </math>

<math>~=</math>

<math>~(1-\eta^2) \biggl[ \eta^{2} \frac{d^2\Upsilon}{d\eta^2} + (2|k|+1)\eta \frac{d\Upsilon}{d\eta} \biggr] - \biggl[ 2 n \eta^3 \frac{d\Upsilon}{d\eta} \biggr] \, . </math>

Finally, then, making the independent variable substitution,

<math>~\eta^2 ~\rightarrow ~ y</math>       <math>~\Rightarrow</math>       <math>~dy = 2\eta d\eta</math>      

in which case,

<math>~\frac{d}{d\eta}</math>

<math>~\rightarrow</math>

<math>~2y^{1/2}\frac{d}{dy}</math>

<math>~\frac{d^2}{d\eta^2}</math>

<math>~\rightarrow</math>

<math>~2\frac{d}{dy} + 4y\frac{d^2}{dy^2} \, .</math>

and,

<math>~ - 2ny \biggl[\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 -|k|\biggr] \Upsilon </math>

<math>~=</math>

<math>~ (1-y) y \frac{d^2\Upsilon}{d\eta^2} + (2|k|+1)(1-y)y^{1/2} \frac{d\Upsilon}{d\eta} - 2 n y^{3/2} \frac{d\Upsilon}{d\eta} </math>

 

<math>~=</math>

<math>~ 4(1-y)y^2 \frac{d^2\Upsilon}{dy^2} + 2(1-y) y \frac{d\Upsilon}{dy} + 2(2|k|+1)(1-y)y \frac{d\Upsilon}{dy} - 4 n y^{2} \frac{d\Upsilon}{dy} </math>

<math>~\Rightarrow~~~~ - \frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 -|k|\biggr] \Upsilon </math>

<math>~=</math>

<math>~ (1-y)y \frac{d^2\Upsilon}{dy^2} + \frac{1}{2}(1-y) \frac{d\Upsilon}{dy} + \frac{1}{2}(2|k|+1)(1-y)\frac{d\Upsilon}{dy} - n y \frac{d\Upsilon}{dy} </math>

 

<math>~=</math>

<math>~ (1-y)y \frac{d^2\Upsilon}{dy^2} + (|k|+1)(1-y)\frac{d\Upsilon}{dy} - n y \frac{d\Upsilon}{dy} </math>

 

<math>~=</math>

<math>~ (1-y)y \frac{d^2\Upsilon}{dy^2} + (|k|+1)\frac{d\Upsilon}{dy} -y (|k|+1+n)\frac{d\Upsilon}{dy} \, . </math>

This matches equation (3.9) of Blaes85. According to Blaes (1985), this equation "… is a standard eigenvalue problem whose only solutions are the Jacobi polynomials …"

Solution

My own background training and experience has not previously exposed me to the general class of Jacobi polynomials. In my effort to understand this class of polynomials and, specifically, their relationship to the Singular Sturm-Liouville Equation, I have found the following references to be useful:

Singular Sturm-Liouville Problem

Drawing on Theorem 3.16 from 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 2nd-order ODE,

<math>~\mathcal{L}_{\alpha,\beta}J_j^{\alpha,\beta}(x)</math>

<math>~=</math>

<math>~\lambda_j^{\alpha,\beta}J_j^{\alpha,\beta}(x) \, ,</math>

where the differential operator,

<math>~\mathcal{L}_{\alpha,\beta}</math>

<math>~\equiv</math>

<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>

 

<math>~=</math>

<math>~ (x^2-1)\cdot \frac{d^2}{dx^2} + [\alpha - \beta + (\alpha+\beta+2)x]\cdot \frac{d}{dx} \, ,</math>

and the corresponding jth eigenvalue is,

<math>~\lambda_j^{\alpha,\beta}</math>

<math>~=</math>

<math>~j(j+\alpha+\beta + 1) \, .</math>

(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 Yu's class notes, for each specified value of the index, <math>~j</math>, the eigenfunction solution to this eigenvalue problem — that is, the relevant Jacobi polynomial — is,

<math>~J_j^{\alpha,\beta}(x)</math>

<math>~=</math>

<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>

As an example, let's carry out the prescribed differentiation to determine the Jacobi polynomial for the case of <math>~j=2</math>.

<math>~J_2^{\alpha,\beta}(x)</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

 

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>


Table 1 presents this "j=2" eigenfunction — along with its corresponding eigenfrequency — as well as the eigenfunctions and eigenfrequencies for "j=0" and "j=1."

Table 1: Example Jacobi Polynomials
<math>~j</math> <math>~J_j^{\alpha,\beta}(x)</math> <math>~\lambda_j^{\alpha,\beta}</math>

<math>~0</math>

<math>~1</math>

<math>~0</math>

<math>~1</math>

<math>~\tfrac{1}{2}(\alpha+\beta+2)x + \tfrac{1}{2}(\alpha-\beta)</math>

<math>~(\alpha+\beta+2)</math>

<math>~2</math>

<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>

<math>~2(\alpha+\beta+3)</math>

See also, eqs. (35)-(37) of Wolfram MathWorld.

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.

Jacobi Differential Equation

Instead, let's pursue the following lead. Wolfram MathWorld provides a discussion of the Jacobi Differential Equation. Equation (1) of this reference appears to be identical to the singular Sturm-Liouville equation presented above. However, in addition, this MathWorld chapter points out that Zwillinger (1997, p. 123) gives a related differential equation that is referred to as "Jacobi's equation," namely,

<math>~0</math>

<math>~=</math>

<math>~x(1-x)y + [\gamma - (\alpha+1)x]y' + j(\alpha+j)y \, .</math>

If we rewrite our above-derived ODE — which is identical to eq. (3.9) of Blaes85 — as,

<math>~ 0 </math>

<math>~=</math>

<math>~ (1-y)y \frac{d^2\Upsilon}{dy^2} +\biggl[ (|k|+1) -y (|k|+1+n)\biggr] \frac{d\Upsilon}{dy} + \frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 -|k|\biggr] \Upsilon </math>

we see that its form is identical to this so-called "Jacobi's equation." The overlap is complete if we make the following parameter associations:

<math>~\gamma</math>

<math>~\leftrightarrow</math>

<math>~|k|+1 \, ,</math>

<math>~\alpha</math>

<math>~\leftrightarrow</math>

<math>~|k|+n \, ,</math>

<math>~j(|k|+n + j)</math>

<math>~\leftrightarrow</math>

<math>~\frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 -|k|\biggr] \, .</math>

We can infer from this third association that, for the jth eigenfunction solution of the governing eigenvalue problem, the relevant oscillation frequency is,

<math>~\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2</math>

<math>~=</math>

<math>~|k| + \frac{2j}{n}(|k|+n + j)</math>

 

<math>~=</math>

<math>~\frac{1}{n}\biggl[n|k| + 2j(|k|+n + j)\biggr]</math>

 

<math>~=</math>

<math>~\frac{1}{n}\biggl[2j^2 + 2jn + 2j|k| + n|k| \biggr] \, .</math>

This is identical to the eigenfrequency identified in eq. (3.12) of Blaes85. So, we must be on the right track! Now I just need to figure out how to express the corresponding eigenfunction in terms of Jacobi polynomials.

From expression 22.5.1 on p. 777 of A&S, we understand that,

<math>~G_j\biggl(\alpha+\beta + 1, \beta+1, \frac{x+1}{2}\biggr)</math>

<math>~=</math>

<math>~P_j^{\alpha,\beta} (x) \biggl[ \frac{j! \Gamma(j + \alpha + \beta + 1)}{\Gamma(2j+\alpha+\beta+1)} \biggr] \, .</math>

In addition, we see from eq. (3.11) of Blaes85 that, to within a factor of a leading constant, the physically relevant eigenfunction is,

<math>~\Upsilon(y)</math>

<math>~=</math>

<math>~A_{jk} G_j(|k|+n, |k|+1, \eta^2) \, .</math>

By association, we conclude from this that,

<math>~\beta</math>

<math>~\leftrightarrow</math>

<math>~|k| \, ,</math>

<math>~\alpha + |k|+1</math>

<math>~\leftrightarrow</math>

<math>~|k|+n \, ,</math>

<math>~\frac{x+1}{2}</math>

<math>~\leftrightarrow</math>

<math>~\eta^2 =y\, .</math>

From this third association, we conclude that the argument of the Jacobi polynomial should be,

<math>x = 2y-1 \, .</math>

Following up on this clue, let's plug this coordinate transformation into Blaes85 expression (3.9) and see whether it takes the form of the Sturm-Liouville ODE. Specifically, making the substitutions,

<math>~y ~\rightarrow~\tfrac{1}{2}(x+1)</math>      and       <math>~dy = \tfrac{1}{2}dx \, ,</math>

we have,

<math>~ - \frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 -|k|\biggr] \Upsilon </math>

<math>~=</math>

<math>~ 4\biggl[1-\frac{(x+1)}{2}\biggr] \frac{(x+1)}{2} \frac{d^2\Upsilon}{dx^2} +2\biggl[ (|k|+1) - \frac{1}{2}(x+1) (|k|+1+n)\biggr] \frac{d\Upsilon}{dx} </math>

 

<math>~=</math>

<math>~ \biggl[2-(x+1)\biggr] (x+1) \frac{d^2\Upsilon}{dx^2} +\biggl[2 (|k|+1) - (x+1) (|k|+1+n)\biggr] \frac{d\Upsilon}{dx} </math>

 

<math>~=</math>

<math>~ -(x^2-1) \frac{d^2\Upsilon}{dx^2} -\biggl[(n - |k|-1) + x (|k|+1+n)\biggr] \frac{d\Upsilon}{dx} </math>

<math>~\Rightarrow~~~ \frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 -|k|\biggr] \Upsilon </math>

<math>~=</math>

<math>~ (x^2-1) \frac{d^2\Upsilon}{dx^2} +\biggl[(n - |k|-1) + x (|k|+1+n)\biggr] \frac{d\Upsilon}{dx} \, , </math>

which is precisely in the standard Sturm-Liouville format. The jth eigenfunction should therefore be, <math>~\Upsilon_j = J_j^{\alpha,\beta}(2\eta^2-1)</math>, and the associated eigenfrequency should be,

<math>~\lambda_j^{\alpha,\beta}</math>

<math>~=</math>

<math>~j(j+\alpha+\beta + 1) \, ,</math>

where,

<math>~\alpha-\beta</math>

<math>~\leftrightarrow</math>

<math>~n - |k| -1 \, ,</math>

<math>~\alpha + \beta +2</math>

<math>~\leftrightarrow</math>

<math>~|k|+1+n \, .</math>

Adding these two expressions together implies that,

<math>~\alpha</math>

<math>~=</math>

<math>~n-1 \, ,</math>

while subtracting them implies,

<math>~\beta</math>

<math>~=</math>

<math>~|k|\, .</math>

Hence, the eigenfrequency is,

<math>~\frac{n}{2}\biggl[\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 -|k|\biggr] </math>

<math>~=</math>

<math>~j(j+\alpha+\beta + 1)</math>

 

<math>~=</math>

<math>~j(j+n + |k|)</math>

<math>~\Rightarrow~~~ \biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 </math>

<math>~=</math>

<math>~\frac{2j}{n}\cdot (j+n + |k|) + |k|</math>

 

<math>~=</math>

<math>~\frac{1}{n}\biggl[2j^2+2jn + 2j|k| + n|k|\biggr] \, .</math>

Again, this precisely matches equation (3.12) from Blaes85, so we appear to be on the right track. Putting this all together, Table 2 presents the first three eigenvectors that are relevant to the stability analysis of slim, PP tori.



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Intermediate Detail

<math>~\Upsilon_2</math>

<math>~=</math>

<math>~\frac{A}{8} + \biggl(\frac{B}{4} \biggr) x + \biggl(\frac{C}{8} \biggr) x^2</math>

 

<math>~=</math>

<math>~\frac{1}{8} \biggl[A-2B+C \biggr] + \biggl[B-C\biggr] \frac{\eta^2}{2} + C\cdot \frac{\eta^4}{2} \, ,</math>

where, from Table 1,

<math>~A</math>

<math>~\equiv</math>

<math>~-4 - \alpha + \alpha^2-\beta + \beta^2 - 2\alpha\beta</math>

 

<math>~=</math>

<math>~ (-2+|k|+k^2)-n(3+2|k|) + n^2 \, , </math>

<math>~B</math>

<math>~\equiv</math>

<math>~3\alpha + \alpha^2 - 3\beta-\beta^2</math>

 

<math>~=</math>

<math>~ -(2+3|k|+k^2) + n + n^2 \, , </math>

<math>~C</math>

<math>~\equiv</math>

<math>~12+7\alpha + \alpha^2 + 7\beta+\beta^2+ 2\alpha\beta</math>

 

<math>~=</math>

<math>~ (6 + 5|k|+k^2) +n(5+2|k|) +n^2 \, . </math>

Hence,

<math>~\Upsilon_2</math>

<math>~=</math>

<math>~ \frac{1}{8} \biggl\{[(-2+|k|+k^2)-n(3+2|k|) + n^2] -2[-(2+3|k|+k^2) + n + n^2] +[(6 + 5|k|+k^2) +n(5+2|k|) +n^2] \biggr\} </math>

 

 

<math>~+ \biggl\{ [-(2+3|k|+k^2) + n + n^2] -[(6 + 5|k|+k^2) +n(5+2|k|) +n^2] \biggr\} \frac{\eta^2}{2} </math>

 

 

<math>~+ \biggl\{(6 + 5|k|+k^2) +n(5+2|k|) +n^2 \biggr\}\frac{\eta^4}{2} </math>

 

<math>~=</math>

<math>~ \frac{1}{2} \biggl\{2 + 3|k| + k^2\biggr\} - \biggl\{4 + 4|k|+k^2 +n(2+|k|)\biggr\} \eta^2 + \biggl\{(6 + 5|k|+k^2) +n(5+2|k|) +n^2 \biggr\}\frac{\eta^4}{2} \, . </math>


Table 2: Example Slim PP-Tori Eigenvectors
<math>~j</math> <math>~\Upsilon_j = J_j^{n-1,|k|}(2\eta^2-1)</math> <math>~\biggl( \frac{\sigma}{\Omega_0} + m \biggr)_j^2</math>

<math>~0</math>

<math>~1</math>

<math>~0</math>

<math>~1</math>

<math>~(n + 1 + |k|)\eta^2 - (1 + |k|)</math>

<math>~\tfrac{1}{n}[2+2n+(2+n)|k|]</math>

<math>~2</math>

<math>~ \tfrac{1}{2}[(6 + 5|k|+k^2) +n(5+2|k|) +n^2 ]\eta^4 - [4 + 4|k|+k^2 +n(2+|k|)]\eta^2 + \tfrac{1}{2} [2 + 3|k| + k^2] </math>

<math>~\tfrac{1}{n}[8+4n + (4+n)|k|] </math>

Summary

The spatial structure of the normalized enthalpy perturbation associated with each <math>~(j,k)</math> mode of oscillation in a slim PP torus is,

<math>~\delta W_j^{(0)}(\eta,\theta)</math>

<math>~=</math>

<math>~\eta^{|k|} J_j^{n-1,|k|}(2\eta^2-1) \cdot \exp (ik\theta) \, ,</math>

and the associated oscillation frequency is,

<math>~\biggl( \frac{\sigma}{\Omega_0} + m \biggr)^2 </math>

<math>~=</math>

<math>~\frac{1}{n}\biggl[2j^2+2jn + 2j|k| + n|k|\biggr] \, .</math>

Because the oscillation frequency is always positive, all modes are dynamically stable. Because the meridional plane amplitude is everywhere real, the azimuthal distortion associated with each nonaxisymmetric mode, <math>~m</math>, is completely uninteresting.

See Also

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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