Difference between revisions of "User:Tohline/SSC/Stability/Polytropes"

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(→‎n = 3 Polytrope: add subscript "xi" to all "Delta" variables)
(→‎Numerical Integration from the Center, Outward: Correct expression for x_2 using newly derived power-series expression)
 
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As can be seen in the following framed image, this is the form of the ''polytropic'' wave equation published by [http://adsabs.harvard.edu/abs/1985PASAu...6..222M J. O. Murphy &amp; R. Fiedler (1985b, Proc. Astron. Soc. Australia, 6, 222)], at the beginning of their discussion of "Radial Pulsations and Vibrational Stability of a Sequence of Two Zone Polytropic Stellar Models." (NOTE:  There appears to be a sign error in the numerator of the second term of their published expression; there also appears to be an error in the definition of the coefficient, <math>~\alpha^*</math>, as given in the text of their paper.)
[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline:  There appears to be a sign error in the numerator of the second term of the polytropic wave equation published by Murphy &amp; Fiedler; there also appears to be an error in the definition of the coefficient, &alpha;*, as given in the text of their paper.]]As can be seen in the following framed image, this is the form of the ''polytropic'' wave equation published by [http://adsabs.harvard.edu/abs/1985PASAu...6..222M J. O. Murphy &amp; R. Fiedler (1985b, Proc. Astron. Soc. Australia, 6, 222)], at the beginning of their discussion of "Radial Pulsations and Vibrational Stability of a Sequence of Two Zone Polytropic Stellar Models."  


<div align="center">
<div align="center">
Line 205: Line 205:




It is also the same as the radial pulsation equation for polytropic configurations that appears as equation (56) in the detailed discussion of "The Oscillations of Gas Spheres" published by [http://adsabs.harvard.edu/abs/1966ApJ...143..535H H M. Hurley, P. H. Roberts, &amp; K. Wright (1966, ApJ, 143, 535)]; hereafter, we will refer to this paper as HRW66.  The relevant set of equations from HRW66 has been extracted as a single digital image and reprinted, here, as a boxed-in image.
[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline:  As is shown in the subsection on "Boundary Conditions," below, it appears as though the term on the right-hand-side of HRW66's equation (58) is incorrect, as published; it should be preceded with a negative sign.]]It is also the same as the radial pulsation equation for polytropic configurations that appears as equation (56) in the detailed discussion of "The Oscillations of Gas Spheres" published by [http://adsabs.harvard.edu/abs/1966ApJ...143..535H H M. Hurley, P. H. Roberts, &amp; K. Wright (1966, ApJ, 143, 535)]; hereafter, we will refer to this paper as HRW66.  The relevant set of equations from HRW66 has been extracted as a single digital image and reprinted, here, as a boxed-in image.
 
 
<div align="center" id="HRW66excerpt">
<div align="center" id="HRW66excerpt">
<table border="2" cellpadding="10">
<table border="2" cellpadding="10">
Line 279: Line 281:
</div>
</div>


The correspondence with our derived expression is complete, assuming that,
<span id="HRW66frequency">The correspondence with our derived expression is complete, assuming that,</span>
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 456: Line 458:
With the exception of the leading negative sign on the right-hand side, this expression is identical to the outer boundary condition identified by equation (58) of HRW66 &#8212; see the [[User:Tohline/SSC/Stability/Polytropes#HRW66excerpt|excerpt reproduced above]].
With the exception of the leading negative sign on the right-hand side, this expression is identical to the outer boundary condition identified by equation (58) of HRW66 &#8212; see the [[User:Tohline/SSC/Stability/Polytropes#HRW66excerpt|excerpt reproduced above]].


==Yabushita's (1992) Analysis==
==Overview==
The eigenvector associated with radial oscillations in isolated polytropes has been determined numerically and the results have been presented in a variety of key publications:
* P. LeDoux &amp; Th. Walraven (1958, Handbuch der Physik, 51, 353) &#8212;
* [http://adsabs.harvard.edu/abs/1966ARA%26A...4..353C R. F. Christy (1966, Annual Reviews of Astronomy &amp; Astrophysics, 4, 353)] &#8212; ''Pulsation Theory''
* [http://adsabs.harvard.edu/abs/1966ApJ...143..535H M. Hurley, P. H. Roberts, &amp; K. Wright (1966, ApJ, 143, 535)] &#8212; ''The Oscillations of Gas Spheres''
* [http://adsabs.harvard.edu/abs/1974RPPh...37..563C J. P. Cox (1974, Reports on Progress in Physics, 37, 563)] &#8212; ''Pulsating Stars''


In the portion (&sect;5) of his analysis that is focused on the stability of pressure-truncated polytropic spheres, [http://adsabs.harvard.edu/abs/1992Ap%26SS.193..173Y S. Yabushita (1992)] examined the eigenvalue problem governed by the following wave equation:
==Tables==
<div align="center" id="HRW66excerpt">
<table border="1" align="center" cellpadding="5">
<table border="2" cellpadding="10">
<tr>
<tr>
   <th align="center">
   <th align="center" colspan="5">
Radial Pulsation Equation Extracted<sup>&dagger;</sup> from p. 182 of [http://adsabs.harvard.edu/abs/1992Ap%26SS.193..173Y S. Yabushita (1992)]<p></p>
Quantitative Information Regarding Eigenvectors of Oscillating Polytropes
"''Similarity Between the Structure and Stability of Isothermal and Polytropic Gas Spheres''"<p></p>
 
Astrophysics and Space Science, vol. 193, pp. 173-183 &copy; [http://www.springer.com/astronomy/astrophysics+and+astroparticles/journal/10509 Springer]
<math>~(\Gamma_1 = 5/3)</math>
   </th>
   </th>
<tr>
  <td>
[[File:Yabushita1992WaveEquation2.png|650px|center|Yabushita (1992)]]
  </td>
</tr>
</tr>
<tr><td align="left">
<sup>&dagger;</sup>Equations and text displayed here exactly as it appears in the original publication.
</td></tr>
</table>
</div>
Let's examine the overlap between this pair of governing relations and the ones employed by HRW66.  If we replace the variable <math>~X</math> with <math>~h</math>, set <math>~\gamma = (n+1)/n</math>, and set the dimensionless eigenfrequency, <math>~s</math>, to zero in the [[#HRW66excerpt|radial pulsation equation employed by HRW66]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="center">
<math>~0 </math>
{{User:Tohline/Math/MP_PolytropicIndex}}
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\frac{\rho_c}{\bar\rho}</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~
Excerpts from Table 1 of
\frac{d^2 h}{dx^2} + \biggl[\frac{4}{x} + (n+1) \frac{\theta^'}{\theta} \biggr] \frac{dh}{dx} + (n+1)\biggl[ 3 - \frac{4n}{(n+1)} \biggr] \biggl[ \frac{\theta^' h}{\theta x} \biggr]
 
</math>
[http://adsabs.harvard.edu/abs/1966ApJ...143..535H Hurley, Roberts, &amp; Wright (1966)]
 
<math>~s^2 (n+1)/(4\pi G\rho_c)</math>
   </td>
   </td>
</tr>
  <td align="center">
Excerpts from Table 3 of
 
[http://adsabs.harvard.edu/abs/1974RPPh...37..563C J. P. Cox (1974)]


<tr>
<math>~\sigma_0^2 R^3/(GM)</math>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>\frac{(n+1) *\mathrm{Cox74}}{3 *\mathrm{HRW66}} \cdot \frac{\bar\rho}{\rho_c}</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 h}{dx^2} + \biggl[\frac{4}{x} + (n+1) \frac{\theta^'}{\theta} \biggr] \frac{dh}{dx} + (3-n) \biggl[ \frac{\theta^' h}{\theta x} \biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This matches equation (5.3) of [http://adsabs.harvard.edu/abs/1992Ap%26SS.193..173Y Yabushita (1992)] &#8212; see the above boxed-in image &#8212; except the <math>~(4/x)</math> term appears as <math>~(2/x)</math> in Yabushita's article; giving the benefit of the doubt, <font color="red">this is most likely a typographical error</font> in [http://adsabs.harvard.edu/abs/1992Ap%26SS.193..173Y Yabushita (1992)].  According to HRW66, the corresponding central boundary condition is,
<div align="center">
<math>\frac{dh}{dx} = 0</math> &nbsp; &nbsp; &nbsp; &nbsp; at &nbsp; &nbsp; &nbsp; &nbsp; <math>x=0 \, .</math>
</div>
While &#8212; after changing the sign on the right-hand side of HRW66's equation (58) as argued in our [[User:Tohline/SSC/Perturbations#ChristyCox|accompanying discussion]] in order to align with the separate derivations presented by [http://adsabs.harvard.edu/abs/1966ARA%26A...4..353C Christy (1965)] and [http://adsabs.harvard.edu/abs/1967IAUS...28....3C Cox (1967)]  &#8212; the corresponding boundary condition at the surface is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="center">
<math>~\frac{dh}{dx}</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~1</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~- \frac{h}{x} \biggr[  3 - \frac{4}{\gamma}  + \cancelto{0}{\frac{x s^2}{\gamma q}} \biggr]</math>
<math>~1/3</math>
   </td>
   </td>
   <td align="left" colspan="2">
   <td align="center">
&nbsp;
<math>~1</math>
  </td>
  <td align="center">
<math>~1</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
<math>~1</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~3.30</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\frac{n-3}{n+1} \biggl(\frac{h}{x} \biggr) \, .</math>
<math>~0.38331</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; at &nbsp; &nbsp; &nbsp; &nbsp;
<math>~1.892</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~x = x_0 \, .</math>
<math>~0.997</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This surface boundary condition, which has been used by the astrophysics community in the context of ''isolated'' polytropic configurations, is different from the one displayed as equation (5.4) of [http://adsabs.harvard.edu/abs/1992Ap%26SS.193..173Y Yabushita (1992)].  The surface boundary condition chosen by Yabushita &#8212; effectively,
<div align="center">
<math>~\frac{d \ln h}{d\ln x} = -3 \, ,</math>
</div>
&#8212; does seem to be more appropriate in the context of a study of the stability of ''pressure-truncated'' polytropes because, as argued by [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)] and as reviewed in our [[User:Tohline/SSC/Perturbations#Set_the_Surface_Pressure_Fluctuation_to_Zero|accompanying discussion]], it ensures that the pressure fluctuation ''at the surface'' is zero.  It is worth noting that Yabushita's surface boundary condition matches the surface boundary condition chosen by [http://adsabs.harvard.edu/abs/1974MNRAS.168..427T Taff &amp; Van Horn (1974)] in their study of pressure-truncated ''isothermal'' spheres; in their words (see p. 428 of their article):  &nbsp; [Setting the surface logarithmic derivative to negative 3] <font color="green">expresses the condition that the pressure at the perturbed surface always remain[s] equal to the confining pressure exerted by the external medium in which the [pressure-truncated] sphere must be embedded</font>.
==Overview==
The eigenvector associated with radial oscillations in isolated polytropes has been determined numerically and the results have been presented in a variety of key publications:
* P. LeDoux &amp; Th. Walraven (1958, Handbuch der Physik, 51, 353) &#8212;
* [http://adsabs.harvard.edu/abs/1966ARA%26A...4..353C R. F. Christy (1966, Annual Reviews of Astronomy &amp; Astrophysics, 4, 353)] &#8212; ''Pulsation Theory''
* [http://adsabs.harvard.edu/abs/1966ApJ...143..535H M. Hurley, P. H. Roberts, &amp; K. Wright (1966, ApJ, 143, 535)] &#8212; ''The Oscillations of Gas Spheres''
* [http://adsabs.harvard.edu/abs/1974RPPh...37..563C J. P. Cox (1974, Reports on Progress in Physics, 37, 563)] &#8212; ''Pulsating Stars''


==Tables==
<table border="1" align="center" cellpadding="5">
<tr>
  <th align="center" colspan="5">
Quantitative Information Regarding Eigenvectors of Oscillating Polytropes
<math>~(\Gamma_1 = 5/3)</math>
  </th>
</tr>
<tr>
<tr>
   <td align="center">
   <td align="center">
{{User:Tohline/Math/MP_PolytropicIndex}}
<math>~1.5</math>
  </td>
  <td align="center">
<math>~5.99</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\frac{\rho_c}{\bar\rho}</math>
<math>~0.37640</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Excerpts from Table 1 of
<math>~2.712</math>
 
[http://adsabs.harvard.edu/abs/1966ApJ...143..535H Hurley, Roberts, &amp; Wright (1966)]
 
<math>~s^2 (n+1)/(4\pi G\rho_c)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Excerpts from Table 3 of
<math>~1.002</math>
 
[http://adsabs.harvard.edu/abs/1974RPPh...37..563C J. P. Cox (1974)]
 
<math>~\sigma_0^2 R^3/(GM)</math>
  </td>
  <td align="center">
<math>\frac{(n+1) *\mathrm{Cox74}}{3 *\mathrm{HRW66}} \cdot \frac{\bar\rho}{\rho_c}</math>
   </td>
   </td>
</tr>
</tr>
<tr>
<tr>
   <td align="center">
   <td align="center">
<math>~0</math>
<math>~2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~1</math>
<math>~11.4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~1/3</math>
<math>~0.35087</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~1</math>
<math>~4.00</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~1</math>
<math>~1.000</math>
   </td>
   </td>
</tr>
</tr>
Line 619: Line 573:
<tr>
<tr>
   <td align="center">
   <td align="center">
<math>~1</math>
<math>~3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~3.30</math>
<math>~54.2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.38331</math>
<math>~0.22774</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~1.892</math>
<math>~9.261</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.997</math>
<math>~1.000</math>
   </td>
   </td>
</tr>
</tr>
Line 637: Line 591:
<tr>
<tr>
   <td align="center">
   <td align="center">
<math>~1.5</math>
<math>~3.5</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~5.99</math>
<math>~153</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.37640</math>
<math>~0.12404</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~2.712</math>
<math>~12.69</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~1.002</math>
<math>~1.003</math>
   </td>
   </td>
</tr>
</tr>
Line 655: Line 609:
<tr>
<tr>
   <td align="center">
   <td align="center">
<math>~2</math>
<math>~4.0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~11.4</math>
<math>~632</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.35087</math>
<math>~0.04056</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~4.00</math>
<math>~15.38</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 670: Line 624:
   </td>
   </td>
</tr>
</tr>
</table>
=Numerical Integration from the Center, Outward=
Here we show how a relatively simple, finite-difference algorithm can be developed to numerically integrate the governing LAWE from the center of a polytropic configuration, outward to its surface. 
Drawing from our [[#Groundwork|above discussion]], the LAWE for any polytrope of index, <math>~n</math>, may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~3</math>
<math>~0 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~54.2</math>
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4 - (n+1)V(\xi)}{\xi} \biggr] \frac{dx}{d\xi} +
\biggl[\omega^2 \biggl(\frac{a_n^2 \rho_c }{\gamma_g P_c} \biggr) \frac{\theta_c}{\theta} -
\biggl(3-\frac{4}{\gamma_g}\biggr)  \cdot \frac{(n+1)V(x)}{\xi^2} \biggr]  x </math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~0.22774</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~9.261</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~1.000</math>
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi} \biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta} \biggl[ \frac{\sigma_c^2}{6\gamma_g}  -
\frac{\alpha}{\xi } \biggl(- \frac{d\theta}{d\xi} \biggr) \biggr]  x </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~3.5</math>
<math>~\sigma_c^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~153</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~0.12404</math>
<math>~\frac{3\omega^2}{2\pi G\rho_c} \, .</math>
  </td>
  <td align="center">
<math>~12.69</math>
  </td>
  <td align="center">
<math>~1.003</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Following a [[User:Tohline/Appendix/Ramblings/NumericallyDeterminedEigenvectors#Integrating_Outward_Through_the_Core|parallel discussion]], we begin by multiplying the LAWE through by <math>~\theta</math>, obtaining a 2<sup>nd</sup>-order ODE that is relevant at every individual coordinate location, <math>~\xi_i</math>, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~4.0</math>
<math>~\theta_i {x_i''}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~632</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~0.04056</math>
<math>~- \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \frac{x_i'}{\xi_i} 
  </td>
- (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g}  -
  <td align="center">
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr]  x_i </math>
<math>~15.38</math>
  </td>
  <td align="center">
<math>~1.000</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
Now, using the [[User:Tohline/Appendix/Ramblings/NumericallyDeterminedEigenvectors#General_Approach|general finite-difference approach described separately]], we make the substitutions,


=n = 1 Polytrope=
==Setup==
From our derived [[User:Tohline/SSC/Structure/Polytropes#n_.3D_1_Polytrope|structure of an n = 1 polytrope]], in terms of the configuration's radius <math>R</math> and mass <math>M</math>, the central pressure and density are, respectively,
<div align="center">
<div align="center">
<math>P_c = \frac{\pi G}{8}\biggl( \frac{M^2}{R^4} \biggr) </math> ,
<table border="0" cellpadding="5" align="center">
</div>
 
and
<tr>
<div align="center">
  <td align="right">
<math>\rho_c = \frac{\pi M}{4 R^3} </math> .
<math>~x_i'</math>
</div>
  </td>
Hence the characteristic time and acceleration are, respectively,
  <td align="center">
<div align="center">
<math>~\approx</math>
<math>  
  </td>
\tau_\mathrm{SSC} = \biggl[ \frac{R^2 \rho_c}{P_c} \biggr]^{1/2} =
  <td align="left">
\biggl[ \frac{2R^3 }{GM} \biggr]^{1/2} =
<math>~
\biggl[ \frac{\pi}{2 G\rho_c} \biggr]^{1/2},
\frac{x_+ - x_-}{2 \Delta_\xi}   \, ;
</math><br />
</math>
  </td>
</tr>
</table>
</div>
</div>
and,
and,
<div align="center">
<math>
g_\mathrm{SSC} = \frac{P_c}{R \rho_c} = \biggl( \frac{GM}{2R^2} \biggr) .
</math><br />
</div>


The required functions are,
* <font color="red">Density</font>:
<div align="center">
<div align="center">
<math>\frac{\rho_0(\chi_0)}{\rho_c} = \frac{\sin(\pi\chi_0)}{\pi\chi_0} </math> ;
<table border="0" cellpadding="5" align="center">
</div>


* <font color="red">Pressure</font>:
<tr>
<div align="center">
  <td align="right">
<math>\frac{P_0(\chi_0)}{P_c} = \biggl[ \frac{\sin(\pi\chi_0)}{\pi\chi_0} \biggr]^2 </math> ;
<math>~
x_i''
</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{x_+ - 2x_i + x_-}{\Delta_\xi^2} \, ,</math>
  </td>
</tr>
</table>
</div>
</div>
 
which will provide an approximate expression for <math>~x_+ \equiv x_{i+1}</math>, given the values of <math>~x_- \equiv x_{i-1}</math> and <math>~x_i</math>.  Specifically, if the center of the configuration is denoted by the grid index, <math>~i=1</math>, then for zones, <math>~i = 3 \rightarrow N</math>,
* <font color="red">Gravitational acceleration</font>:
<div align="center">
<div align="center">
<math>
<table border="0" cellpadding="5" align="center">
\frac{g_0(r_0)}{g_\mathrm{SSC}} = \frac{2}{\chi_0^2} \biggl[ \frac{M_r(\chi_0)}{M}\biggr]  =
\frac{2}{\pi \chi_0^2} \biggl[ \sin (\pi\chi_0 ) - \pi\chi_0 \cos (\pi\chi_0 ) \biggr].
</math><br />
</div>
 
So our desired Eigenvalues and Eigenvectors will be solutions to the following ODE:
 
<div align="center">
<math>
\frac{d^2x}{d\chi_0^2} + \frac{2}{\chi_0} \biggl[ 1 +  \pi\chi_0 \cot (\pi\chi_0 ) \biggr]  \frac{dx}{d\chi_0} + \frac{1}{\gamma_\mathrm{g}} \biggl\{ \frac{\pi \chi_0}{\sin(\pi\chi_0)} \biggl[ \frac{\pi \omega^2}{2G\rho_c} \biggr] + \frac{2}{\chi_0^2 } (4 - 3\gamma_\mathrm{g}) \biggl[ 1 - \pi\chi_0 \cot (\pi\chi_0 ) \biggr] \biggr\}  x = 0 ,
</math><br />
</div>
<br />
or, replacing <math>\chi_0</math> with <math>\xi \equiv \pi\chi_0</math> and dividing the entire expression by <math>\pi^2</math>, we have,


<div align="center">
<math>
\frac{d^2x}{d\xi^2} + \frac{2}{\xi} \biggl[ 1 +  \xi \cot \xi \biggr]  \frac{dx}{d\xi} + \frac{1}{\gamma_\mathrm{g}} \biggl\{ \frac{\xi}{\sin \xi} \biggl[ \frac{\omega^2}{2\pi G\rho_c} \biggr] + \frac{2}{\xi^2 } (4 - 3\gamma_\mathrm{g}) \biggl[ 1 - \xi \cot \xi \biggr] \biggr\}  x = 0 .
</math><br />
</div>
<br />
This is identical to the formulation of the wave equation that is relevant to  the (n = 1) core of the composite polytrope studied by [http://adsabs.harvard.edu/abs/1985PASAu...6..222M J. O. Murphy &amp; R. Fiedler (1985b)]; for comparison, their expression is displayed, here, in the following boxed-in image.
<div align="center">
<table border="2" cellpadding="10" id="MurphyFiedler1985b">
<tr>
<tr>
   <th align="center">
   <td align="right">
n = 1 Polytropic Formulation of Wave Equation as Presented by [http://adsabs.harvard.edu/abs/1985PASAu...6..222M Murphy &amp; Fiedler (1985b)]
<math>~\theta_i \biggl[ \frac{x_+ - 2x_i + x_-}{\Delta_\xi^2} \biggr]</math>
   </th>
  </td>
<tr>
   <td align="center">
   <td>
<math>~=</math>
[[File:MurphyFiedlerN1formulation.png|700px|center|Murphy &amp; Fiedler (1985b)]]
  </td>
   <td align="left">
<math>~- \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{x_+ - x_-}{2 \xi_i \Delta_\xi}  \biggr]
- (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g}  -
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr] x_i </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


{{LSU_WorkInProgress}}
==Attempt at Deriving an Analytic Eigenvector Solution==
Multiplying the last expression through by <math>~\xi^2\sin\xi</math> gives,
<div align="center">
<math>
(\xi^2\sin\xi ) \frac{d^2x}{d\xi^2} + 2 \biggl[ \xi \sin\xi +  \xi^2 \cos \xi \biggr]  \frac{dx}{d\xi} +
\biggl[ \sigma^2 \xi^3  - 2\alpha ( \sin\xi - \xi \cos \xi ) \biggr]  x = 0 \, ,
</math><br />
</div>
<br />
where,
<div align="center">
<table border="0" cellpadding="5">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\sigma^2</math>
<math>~\Rightarrow ~~~ \theta_i \biggl[ \frac{x_+ }{\Delta_\xi^2} \biggr] + \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{x_+ }{2 \xi_i\Delta_\xi}  \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~\frac{\omega^2}{2\pi G\rho_c \gamma_g} \, ,
-\theta_i \biggl[ \frac{- 2x_i + x_-}{\Delta_\xi^2} \biggr]
</math>
- \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{- x_-}{2 \xi_i \Delta_\xi}  \biggr]
- (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g} -
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr]  x_i </math>
   </td>
   </td>
</tr>
</tr>
Line 835: Line 777:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\alpha</math>
<math>~\Rightarrow ~~~ x_+ \biggl[2\theta_i +\frac{4\Delta_\xi \theta_i}{\xi_i} - \Delta_\xi (n+1)(- \theta^')_i\biggr] </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~3-\frac{4}{\gamma_g}
x_- \biggl[\frac{4\Delta_\xi \theta_i}{\xi_i} - \Delta_\xi (n+1)(- \theta^')_i - 2\theta_i\biggr]
\, .
+  x_i\biggl\{4\theta_i - 2\Delta_\xi^2(n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g} -
</math>
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr]  \biggr\} </math>
   </td>
   </td>
</tr>
</tr>


</table>
</div>
The first two terms can be folded together to give,
<div align="center">
<table border="0" cellpadding="5">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~ \frac{1}{\xi^2 \sin^2\xi} \cdot \frac{d}{d\xi}\biggl[ \xi^2 \sin^2\xi \frac{dx}{d\xi} \biggr]
&nbsp;
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 863: Line 798:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~\frac{1}{\xi^2 \sin\xi} \biggl[ 2\alpha ( \sin\xi - \xi \cos \xi ) - \sigma^2 \xi^3 \biggr]  x
x_- \biggl[\frac{4\Delta_\xi \theta_i}{\xi_i} - \Delta_\xi (n+1)(- \theta^')_i - 2\theta_i\biggr]
</math>
+  x_i\biggl\{4\theta_i - \frac{\Delta_\xi^2(n+1)}{3}\biggl[ \frac{\sigma_c^2}{\gamma_g}  -  
2\alpha \biggl(- \frac{3\theta^'}{\xi}\biggr)_i\biggr]  \biggr\} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
In order to kick-start the integration, we will set the displacement function value to <math>~x_1 = 1</math> at the center of the configuration <math>~(\xi_1 = 0)</math>, then we will draw on the [[User:Tohline/Appendix/Ramblings/PowerSeriesExpressions#PolytropicDisplacement|derived power-series expression]] to determine the value of the displacement function at the first radial grid line, <math>~\xi_2 = \Delta_\xi</math>, away from the center.  Specifically, we will set,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~
x_2
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 877: Line 821:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~- \biggl[ \frac{2\alpha}{\xi^2} \biggl( \frac{\xi \cos \xi}{\sin\xi} -1\biggr) + \sigma^2 \biggl( \frac{\xi}{\sin\xi}\biggr) \biggr]  x
x_1 \biggl[ 1 - \frac{(n+1) \mathfrak{F}  \Delta_\xi^2}{60} \biggr] \, ,</math>
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \biggl[ \frac{2\alpha}{\xi^2} \frac{\xi^2}{\sin\xi}  \cdot \frac{d}{d\xi} \biggl( \frac{\sin\xi}{\xi} \biggr)
+ \sigma^2 \biggl( \frac{\xi}{\sin\xi}\biggr) \biggr]  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \biggl[ \frac{2\alpha}{\xi} \frac{d}{d\xi}\biggl( \frac{\sin\xi}{\xi} \biggr) + \sigma^2 \biggr]  \biggl( \frac{\xi}{\sin\xi}\biggr) x \, ,
</math>
  </td>
</tr>
</table>
</div>
where, in order to make this next-to-last step, we have recognized that,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~ \frac{d}{d\xi} \biggl( \frac{\sin\xi}{\xi} \biggr)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\frac{\sin\xi}{\xi^2} \biggl[ \frac{\xi \cos\xi}{\sin\xi} - 1 \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
It would seem that the eigenfunction, <math>~x(\xi)</math>, should be expressible in terms of trigonometric functions and powers of <math>~\xi</math>; indeed, it appears as though the expression governing this eigenfunction would simplify considerably if <math>~x \propto \sin\xi/\xi</math>.  With this in mind, we have made some attempts to ''guess'' the exact form of the eigenfunction.  Here is one such attempt.
 
===First Guess (n1)===
Let's try,
<div align="center">
<math>~x = \frac{\sin\xi}{\xi}  \, ,</math>
</div>
which means,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~x^' \equiv \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\frac{\sin\xi}{\xi^2} \biggl[ \frac{\xi \cos\xi}{\sin\xi} - 1 \biggr]
\, .
</math>
  </td>
</tr>
 
</table>
</div>
Does this satisfy the governing expression?  Let's see.  The right-and-side (RHS) gives:
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \biggl[ \frac{2\alpha}{\xi} \frac{d}{d\xi}\biggl( \frac{\sin\xi}{\xi} \biggr) + \sigma^2 \biggr]  \biggl( \frac{\xi}{\sin\xi}\biggr) x
= - \biggl[ \frac{2\alpha x^'}{\xi} + \sigma^2 \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
At the same time, the left-hand-side (LHS) may, quite generically, be written as:
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{x^'}{\xi}
\biggl\{ \frac{\xi}{(\xi^2 \sin^2\xi)x^'} \cdot \frac{d[ (\xi^2 \sin^2\xi)x^']}{d\xi} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{x^'}{\xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
Putting the two sides together therefore gives,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~ \frac{x^'}{\xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} +2\alpha \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~-\sigma^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~ \Rightarrow ~~~~~
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']^{1/(2\alpha)}}{d\ln\xi} +1 \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \frac{\sigma^2}{2\alpha }  \biggl( \frac{\xi}{x^'} \biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~ \Rightarrow ~~~~~
\frac{d\ln[ (\xi^2 \sin^2\xi)x^']^{-1/(2\alpha)}}{d\ln\xi}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~1 + \frac{\sigma^2}{2\alpha }  \biggl( \frac{\xi}{x^'} \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>
[<font color="red">Comment from J. E. Tohline on 6 April 2015:</font> I'm not sure what else to make of this.]
 
 
===Second Guess (n1)===
Adopting the generic rewriting of the LHS, and leaving the RHS fully generic as well, we have,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~ \frac{x^'}{\xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \biggl[ \frac{2\alpha}{\xi} \frac{d}{d\xi}\biggl( \frac{\sin\xi}{\xi} \biggr) + \sigma^2 \biggr]  \biggl( \frac{\xi}{\sin\xi}\biggr) x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~~ \frac{x^'}{x}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- 2\alpha\biggl( \frac{\xi}{\sin\xi}\biggr) \frac{d}{d\xi}\biggl( \frac{\sin\xi}{\xi} \biggr)
~- \sigma^2\biggl( \frac{\xi^2}{\sin\xi}\biggr) 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~~ \frac{d\ln(x)}{d\ln \xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- 2\alpha\biggl[ \frac{d\ln(\sin\xi/\xi)}{d\ln \xi} \biggr]
~- \sigma^2\biggl( \frac{\xi^3}{\sin\xi}\biggr)  \, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>
~\Rightarrow ~~~~ - \sigma^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{\sin\xi}{\xi^3}\biggr) \biggl\{\frac{d\ln(x)}{d\ln \xi}
\biggl[\frac{d\ln[ (\xi^2 \sin^2\xi)x^']}{d\ln\xi} \biggr] +
2\alpha\biggl[ \frac{d\ln(\sin\xi/\xi)}{d\ln \xi} \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
[<font color="red">Comment from J. E. Tohline on 6 April 2015:</font> I'm not sure what else to make of this.]
 
 
===Third Guess (n1)===
Let's rewrite the polytropic (n = 1) wave equation as follows:
<div align="center">
<math>
~\sin\xi \biggl[ \xi^2 x^{''} + 2\xi x^' - 2\alpha x \biggr]
+ \cos\xi \biggl[ 2\xi^2 x^' + 2\alpha \xi x \biggr]
+\sigma^2 \xi^3 x = 0 \, .
</math>
</div>
It is difficult to determine what term in the adiabatic wave equation will cancel the term involving <math>~\sigma^2</math> because its leading coefficient is <math>~\xi^3</math> and no other term contains a power of <math>~\xi</math> that is higher than two.  After thinking through various trial eigenvector expressions, <math>~x(\xi)</math>, I have determined that a function of the following form has a ''chance'' of working because the second derivative of the function generates a leading factor of <math>~\xi^3</math> while the function itself does not introduce any additional factors of <math>~\xi</math> into the term that contains <math>~\sigma^2</math>:
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] [ A\sin\xi + B\cos\xi]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ \frac{d[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})]}{d\xi} \cdot [ A\sin\xi + B\cos\xi]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ [ A\sin\xi + B\cos\xi]
\biggl\{5a \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2})
+ b\biggl[ \frac{5}{2} \xi^{3/2}\cos^2(\xi^{5/2})  - \frac{5}{2} \xi^{3/2}\sin^2(\xi^{5/2}) \biggr]
- 5c \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2})  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ [ A\sin\xi + B\cos\xi]
\biggl\{5(a-c) \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2})
+ \frac{5b}{2} \xi^{3/2}\biggl[ 1  - 2\sin^2(\xi^{5/2}) \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x^{''}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d^2[ A\sin\xi + B\cos\xi]}{d^2\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+
\biggl\{5(a-c) \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2})
+ \frac{5b}{2} \xi^{3/2}\biggl[ 1  - 2\sin^2(\xi^{5/2}) \biggr]
\biggr\} \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ [ A\sin\xi + B\cos\xi] \biggl\{
\frac{15}{2}(a-c) \xi^{1/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2})
+\frac{25}{2}(a-c) \xi^{3}  \cos^2(\xi^{5/2})
- \frac{25}{2}(a-c) \xi^{3}  \sin^2(\xi^{5/2})
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \frac{15b}{4} \xi^{1/2}\biggl[ 1  - 2\sin^2(\xi^{5/2}) \biggr]
- 25b \xi^{3}\sin(\xi^{5/2}) \cos(\xi^{5/2})
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[a\sin^2(\xi^{5/2}) + b\sin(\xi^{5/2})\cos(\xi^{5/2}) + c\cos^2(\xi^{5/2})] \cdot \frac{d^2[ A\sin\xi + B\cos\xi]}{d^2\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+
\biggl\{5(a-c) \xi^{3/2}  \sin(\xi^{5/2}) \cos(\xi^{5/2})
+ \frac{5b}{2} \xi^{3/2}\biggl[ 1  - 2\sin^2(\xi^{5/2}) \biggr]
\biggr\} \cdot \frac{d[ A\sin\xi + B\cos\xi]}{d\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ [ A\sin\xi + B\cos\xi] \biggl\{ \frac{15}{4}\xi^{1/2} \biggl[
2(a-c) \sin(\xi^{5/2}) \cos(\xi^{5/2})
+ b\biggl( 1  - 2\sin^2(\xi^{5/2}) \biggr) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \frac{25}{2} \xi^3 \biggl[
- 2b \sin(\xi^{5/2}) \cos(\xi^{5/2})
+(a-c) \biggl( 1- 2\sin^2(\xi^{5/2}) \biggr) \biggr] \biggr\}
</math>
  </td>
</tr>
 
</table>
</div>
[<font color="red">Comment from J. E. Tohline on 9 April 2015:</font> I'm not sure what else to make of this.]
 
[<font color="red">Additional comment from J. E. Tohline on 15 April 2015:</font> It is perhaps worth mentioning that there is a similarity between the argument of the trigonometric function being used in this "third guess" and the [[User:Tohline/SSC/Structure/Polytropes#Srivastava.27s_F-Type_Solution|Lane-Emden function derived by Srivastava for <math>~n=5</math> polytropes]]; and also a similarity between Srivastava's function and the functional form of the LHS that we constructed, [[User:Tohline/SSC/Stability/Polytropes#Second_Guess|above, in connection with our "second guess]]."]
 
===Fourth Guess (n1)===
Again, working with the polytropic (n = 1) wave equation written in the following form,
<div align="center">
<math>
~\sin\xi \biggl[ \xi^2 x^{''} + 2\xi x^' - 2\alpha x \biggr]
+ \cos\xi \biggl[ 2\xi^2 x^' + 2\alpha \xi x \biggr]
+\sigma^2 \xi^3 x = 0 \, .
</math>
</div>
Now, let's try:
<div align="center">
<math>~x = a_0 + b_1 \xi \sin\xi + c_2 \xi^2 \cos\xi  \, ,</math>
</div>
which means,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~x^' </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~b_1 \sin\xi + b_1 \xi \cos\xi + 2c_2 \xi \cos\xi - c_2\xi^2 \sin\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~(b_1 - c_2\xi^2 ) \sin\xi + (b_1  + 2c_2)\xi \cos\xi \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~x^{''} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~(- 2c_2\xi ) \sin\xi + (b_1 - c_2\xi^2 ) \cos\xi
+ (b_1  + 2c_2 )\cos\xi - (b_1  + 2c_2 )\xi \sin\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~-(2c_2+b_1 + 2c_2 ) \xi \sin\xi + (2b_1 + 2c_2 - c_2\xi^2  ) \cos\xi  \, .
</math>
  </td>
</tr>
 
</table>
</div>
The LHS of the wave equation then becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\sin\xi \biggl\{ \xi^2 \biggl[ -(2c_2+b_1 + 2c_2 ) \xi \sin\xi + (2b_1 + 2c_2 - c_2\xi^2  ) \cos\xi \biggr]
+ 2\xi \biggl[ (b_1 - c_2\xi^2 ) \sin\xi + (b_1  + 2c_2)\xi \cos\xi  \biggr]
- 2\alpha \biggl[ a_0 + (b_1 \xi) \sin\xi + (c_2 \xi^2) \cos\xi  \biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \cos\xi \biggl\{ 2\xi^2 \biggl[ (b_1 - c_2\xi^2 ) \sin\xi + (b_1  + 2c_2)\xi \cos\xi  \biggr]
+ 2\alpha \xi \biggl[ a_0 + (b_1 \xi) \sin\xi + (c_2 \xi^2) \cos\xi  \biggr] \biggr\}
+\sigma^2 \xi^3 \biggl[ a_0 + b_1 \xi \sin\xi + c_2 \xi^2 \cos\xi  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\sin\xi \biggl\{  \biggl[ -(2c_2+b_1 + 2c_2 ) \xi^3 \sin\xi + (2b_1 + 2c_2 - c_2\xi^2  ) \xi^2\cos\xi \biggr]
+ \biggl[ 2(b_1 - c_2\xi^2 )\xi \sin\xi + 2(b_1  + 2c_2)\xi^2 \cos\xi  \biggr]
- 2\alpha \biggl[ a_0 + (b_1 \xi) \sin\xi + (c_2 \xi^2) \cos\xi  \biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \cos\xi \biggl\{ \biggl[ 2(b_1 - c_2\xi^2 )\xi^2 \sin\xi + 2(b_1  + 2c_2)\xi^3 \cos\xi  \biggr]
+ \biggl[ 2a_0\alpha \xi  + 2b_1\alpha \xi^2 \sin\xi + 2c_2 \alpha \xi^3 \cos\xi  \biggr] \biggr\}
+\sigma^2 \biggl[ a_0\xi^3  + b_1 \xi^4 \sin\xi + c_2 \xi^5 \cos\xi  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\sin\xi \biggl\{-  2\alpha a_0 +  \biggl[ -(2c_2+b_1 + 2c_2 ) \xi^3  + 2(b_1 - c_2\xi^2 )\xi  - 2\alpha (b_1 \xi) \biggr]\sin\xi
+ \biggl[ (2b_1 + 2c_2 - c_2\xi^2  ) \xi^2 + 2(b_1  + 2c_2)\xi^2  - 2\alpha (c_2 \xi^2) \biggr] \cos\xi  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \cos\xi \biggl\{ + 2a_0\alpha \xi  + \biggl[ 2(b_1 - c_2\xi^2 )\xi^2 + 2b_1\alpha \xi^2 \biggr] \sin\xi 
+ \biggl[ 2(b_1  + 2c_2)\xi^3  + 2c_2 \alpha \xi^3\biggr] \cos\xi  \biggr\}
+\sigma^2 \biggl[ a_0\xi^3  + b_1 \xi^4 \sin\xi + c_2 \xi^5 \cos\xi  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\sigma^2 a_0 \xi^3 +  \biggl[ -(2c_2+b_1 + 2c_2 ) \xi^3  + 2(b_1 - c_2\xi^2 )\xi  - 2\alpha (b_1 \xi) \biggr]\sin^2\xi
+ \biggl[ 2(b_1  + 2c_2)\xi^3  + 2c_2 \alpha \xi^3\biggr] \biggl(1-\sin^2\xi \biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[ (2b_1 + 2c_2 - c_2\xi^2  ) \xi^2 + 2(b_1  + 2c_2)\xi^2  - 2\alpha (c_2 \xi^2) \biggr] \sin\xi \cos\xi 
+ \biggl[ 2(b_1 - c_2\xi^2 )\xi^2 + 2b_1\alpha \xi^2 \biggr] \sin\xi  \cos\xi 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\sigma^2 \biggl[ b_1 \xi^4 \sin\xi + c_2 \xi^5 \cos\xi  \biggr] + 2a_0\alpha \xi\cos\xi  -  2\alpha a_0 \sin\xi 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\biggl[ \sigma^2 a_0  +  2(b_1  + 2c_2)  + 2c_2 \alpha \biggr]\xi^3 +  \biggl\{+ 2(b_1 )\xi  - 2\alpha (b_1 \xi)  +[-2c_2
- 2(b_1  + 2c_2)  - 2c_2 \alpha  -(2c_2+b_1 + 2c_2 )] \xi^3  \biggr\} \sin^2\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl\{ [ (2b_1 + 2c_2 )  + 2(b_1  + 2c_2)  - 2\alpha (c_2 )  + 2(b_1 ) + 2b_1\alpha ] \xi^2 
-3c_2\xi^4 \biggr\} \sin\xi  \cos\xi 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+
\sin\xi \biggl[\sigma^2b_1 \xi^4 -  2\alpha a_0  \biggr] + \xi \cos\xi \biggl[\sigma^2 c_2 \xi^4 + 2a_0\alpha \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~[ \sigma^2 a_0  +  2(b_1  + 2c_2)  + 2c_2 \alpha ]\xi^3 + 
\biggl\{2 b_1(1-\alpha) - [2c_2(5+\alpha) + 3b_1] \xi^2 \biggr\} \xi \sin^2\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl\{ 2(3-\alpha)( b_1+c_2 )    -3c_2\xi^2 \biggr\} \xi^2 \sin\xi  \cos\xi 
+\sin\xi \biggl[\sigma^2b_1 \xi^4 -  2\alpha a_0  \biggr] + \xi \cos\xi \biggl[\sigma^2 c_2 \xi^4 + 2a_0\alpha \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
===Fifth Guess (n1)===
Along a similar line of reasoning, let's try a function of the form,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~x_s \sin\xi + x_c \cos\xi + x_1 \sin^2\xi + x_2 \cos^2\xi + x_3 \sin\xi \cos\xi\, ,
</math>
  </td>
</tr>
 
</table>
</div>
where <math>~x_s, x_c, x_1, x_2,</math> and <math>~x_3</math> are five separate, as yet, unspecified (polynomial?) functions of <math>~\xi</math>.  This also means that,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~x^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~(x_s^' - x_c)\sin\xi + (x_c^' + x_s)\cos\xi + (x_1^' - x_3)\sin^2\xi + (x_2^' + x_3)\cos^2\xi + (x_3^' + 2x_1 -2x_2)\sin\xi \cos\xi \, ;
</math>
  </td>
</tr>
 
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~x^{''}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~(x_s^{''} - 2x_c^{'} - x_s)\sin\xi + (x_c^{''} + 2x_s^' -x_c)\cos\xi
+ (x_1^{''} -2x_3^' -2x_1 + 2x_2)\sin^2\xi + (x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\cos^2\xi
+ (x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\sin\xi \cos\xi \, .
</math>
  </td>
</tr>
 
</table>
</div>
Hence the LHS of the polytropic (n = 1) wave equation becomes,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~~\sin\xi \biggl\{ \xi^2 \biggl[~(x_s^{''} - 2x_c^{'} - x_s)\sin\xi + (x_c^{''} + 2x_s^' -x_c)\cos\xi
+ (x_1^{''} -2x_3^' -2x_1 + 2x_2)\sin^2\xi + (x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\cos^2\xi
+ (x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\sin\xi \cos\xi  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ 2\xi \biggl[~(x_s^' - x_c)\sin\xi + (x_c^' + x_s)\cos\xi + (x_1^' - x_3)\sin^2\xi + (x_2^' + x_3)\cos^2\xi + (x_3^' + 2x_1 -2x_2)\sin\xi \cos\xi \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- 2\alpha \biggl[ x_s \sin\xi + x_c \cos\xi + x_1 \sin^2\xi + x_2 \cos^2\xi + x_3 \sin\xi \cos\xi\biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \cos\xi \biggl\{ 2\xi^2 \biggl[~(x_s^' - x_c)\sin\xi + (x_c^' + x_s)\cos\xi + (x_1^' - x_3)\sin^2\xi + (x_2^' + x_3)\cos^2\xi + (x_3^' + 2x_1 -2x_2)\sin\xi \cos\xi \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ 2\alpha \xi \biggl[ x_s \sin\xi + x_c \cos\xi + x_1 \sin^2\xi + x_2 \cos^2\xi + x_3 \sin\xi \cos\xi\biggr]  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\sigma^2 \xi^3 \biggl\{ x_s \sin\xi + x_c \cos\xi + x_1 \sin^2\xi + x_2 \cos^2\xi + x_3 \sin\xi \cos\xi \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\biggl[(x_s^{''} - 2x_c^{'} - x_s)\xi^2 + 2\xi (x_s^' - x_c) -2\alpha x_s + \sigma^2 \xi^3 x_1  \biggr]\sin^2\xi
+ \biggl[(x_c^{''} + 2x_s^' -x_c)\xi^2 + 2\xi (x_c^' + x_s) - 2\alpha x_c + 2\xi^2 (x_s^' - x_c) + 2\alpha \xi x_s + \sigma^2 \xi^3 x_3 \biggr] \sin\xi \cos\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\biggl[ (x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 + 2\xi (x_1^' - x_3) - 2\alpha x_1 \biggr] \sin^3\xi
+ \biggl[(x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\xi^2 + 2\xi (x_2^' + x_3) - 2\alpha x_2 + 2\xi^2 (x_3^' + 2x_1 -2x_2)+  2\alpha \xi  x_3 \biggr]\sin\xi \cos^2\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[(x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\xi^2 + 2\xi (x_3^' + 2x_1 -2x_2) - 2\alpha x_3 + 2\xi^2 (x_1^' - x_3) +  2\alpha \xi x_1  \biggr] \sin^2\xi \cos\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[2\xi^2 (x_c^' + x_s ) +  2\alpha \xi x_c + \sigma^2 \xi^3 x_2 \biggr]\cos^2\xi  + \biggl[2\xi^2 (x_2^' + x_3)+  2\alpha \xi x_2 \biggr]\cos^3\xi 
+ \sigma^2 \xi^3 x_s \sin\xi + \sigma^2 \xi^3 x_c \cos\xi 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\biggl[(x_s^{''} - 2x_c^{'} - x_s)\xi^2 + 2\xi (x_s^' - x_c) -2\alpha x_s + \sigma^2 \xi^3 x_1  \biggr]\sin^2\xi
+ \biggl[(x_c^{''} + 2x_s^' -x_c)\xi^2 + 2\xi (x_c^' + x_s) - 2\alpha x_c + 2\xi^2 (x_s^' - x_c) + 2\alpha \xi x_s + \sigma^2 \xi^3 x_3 \biggr] \sin\xi \cos\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\biggl[ (x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 + 2\xi (x_1^' - x_3) - 2\alpha x_1 + \sigma^2 \xi^3 x_s \biggr] \sin\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[(x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\xi^2 + 2\xi (x_2^' + x_3) - 2\alpha x_2 + 2\xi^2 (x_3^' + 2x_1 -2x_2)+  2\alpha \xi  x_3
- (x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 - 2\xi (x_1^' - x_3) + 2\alpha x_1 \biggr]\sin\xi \cos^2\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[(x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\xi^2 + 2\xi (x_3^' + 2x_1 -2x_2) - 2\alpha x_3 + 2\xi^2 (x_1^' - x_3) +  2\alpha \xi x_1
-2\xi^2 (x_2^' + x_3) -  2\alpha \xi x_2 \biggr] \sin^2\xi \cos\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[2\xi^2 (x_c^' + x_s ) +  2\alpha \xi x_c + \sigma^2 \xi^3 x_2 \biggr]\cos^2\xi 
+ \biggl[2\xi^2 (x_2^' + x_3)+  2\alpha \xi x_2 + \sigma^2 \xi^3 x_c \biggr]\cos\xi 
</math>
  </td>
</tr>
 
</table>
</div>
So, the five chosen (polynomial?) functions of <math>~\xi</math> must simultabeously satisfy the following, seven 2<sup>nd</sup>-order ODEs:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\sin\xi </math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~(x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 + 2\xi (x_1^' - x_3) - 2\alpha x_1 + \sigma^2 \xi^3 x_s =0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\sin^2 \xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~(x_s^{''} - 2x_c^{'} - x_s)\xi^2 + 2\xi (x_s^' - x_c) -2\alpha x_s + \sigma^2 \xi^3 x_1 =0</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\sin^2\xi \cos\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~(x_3^{''} + 4x_1^' -4x_2^' - 4x_3)\xi^2 + 2\xi (x_3^' + 2x_1 -2x_2) - 2\alpha x_3 + 2\xi^2 (x_1^' - x_3) +  2\alpha \xi x_1 
-2\xi^2 (x_2^' + x_3) -  2\alpha \xi x_2  =0</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\sin\xi \cos\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~(x_c^{''} + 2x_s^' -x_c)\xi^2 + 2\xi (x_c^' + x_s) - 2\alpha x_c + 2\xi^2 (x_s^' - x_c) + 2\alpha \xi x_s + \sigma^2 \xi^3 x_3 =0</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\sin\xi \cos^2\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~(x_2^{''} + 2x_3^'+ 2x_1 - 2x_2)\xi^2 + 2\xi (x_2^' + x_3) - 2\alpha x_2 + 2\xi^2 (x_3^' + 2x_1 -2x_2)+  2\alpha \xi  x_3
- (x_1^{''} -2x_3^' -2x_1 + 2x_2)\xi^2 - 2\xi (x_1^' - x_3) + 2\alpha x_1 =0</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\cos^2\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~2\xi^2 (x_c^' + x_s ) +  2\alpha \xi x_c + \sigma^2 \xi^3 x_2 =0</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\cos\xi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; : &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~2\xi^2 (x_2^' + x_3)+  2\alpha \xi x_2 + \sigma^2 \xi^3 x_c =0</math>
  </td>
</tr>
</table>
</div>
 
====Example 1====
Let's work on the coefficient of the <math>~\cos\xi</math> term:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~x_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^\beta (A_c)</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~x_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^\beta (C_2 \xi^2)</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~x_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^\beta (B_3 \xi)</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~</math> &nbsp; &nbsp; &nbsp; Coefficient of "<math>~\cos\xi</math>" term
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^\beta [2\xi^2 (2C_2\xi + (B_3 \xi))+  2\alpha \xi (C_2 \xi^2) + \sigma^2 \xi^3 (A_c)]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^{\beta+3} [2(2C_2 + B_3 )+  2\alpha C_2  + \sigma^2 (A_c)]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \sigma^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{2}{A_c} \biggl[B_3 +  (2+\alpha) C_2 \biggr]</math>
  </td>
</tr>
</table>
</div>
 
 
===Sixth Guess (n1)===
====Rationale====
From our [[User:Tohline/SSC/Structure/Polytropes#.3D_1_Polytrope|review of the properties of <math>~n=1</math> polytropic spheres]], we know that the equilibrium density distribution is given by the sinc function, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\rho}{\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\sin\xi}{\xi} \, ,</math>
  </td>
</tr>
 
</table>
</div>
where,
<div align="center">
<math>~\xi \equiv \pi \biggl(\frac{r_0}{R_0} \biggr) \, .</math>
</div>
The total mass is,
<div align="center">
<math>~M_\mathrm{tot} =  \frac{4}{\pi} \cdot \rho_c R_0^3 \, ,</math>
</div>
and the fractional mass enclosed within a given radius, <math>~r</math>, is,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{M_r(\xi)}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\pi} [\sin\xi - \xi \cos\xi] \, .</math>
  </td>
</tr>
 
</table>
</div>
Let's guess that, during the fundamental mode of radial oscillation, the sinc-function profile is preserved as the system's total radius varies.  In particular, we will assume that the system's time-varying radius is,
<div align="center">
<math>R = R_0 \biggl( 1 + \frac{\delta R}{R_0} \biggr) = R_0 ( 1 + \epsilon_R) \, ,</math>
</div>
and seek to determine how the displacement vector, <math>~\epsilon \equiv \delta r/r_0</math>, varies with <math>~r_0</math> in order to preserve the overall sinc-function profile.  As is usual, we will only examine small perturbations away from equilibrium, that is, we will assume that everywhere throughout the configuration, <math>~|\epsilon| \ll 1 </math>.
 
 
Let's begin by defining a new dimensionless coordinate,
<div align="center">
<math>~\eta \equiv \pi \biggl(\frac{r}{R} \biggr) = \pi \biggl[\frac{r_0(1+\epsilon)}{R_0(1+\epsilon_R)} \biggr]
\approx \xi (1 + \epsilon) \, ,</math>
</div>
and recognize that, in the new perturbed state, the fractional mass enclosed within a given radius, <math>~r</math>, is,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{M_r(\eta)}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\pi} [\sin\eta - \eta \cos\eta] \, .</math>
  </td>
</tr>
 
</table>
</div>
In order to associate each mass shell in the perturbed configuration with its corresponding mass shell in the unperturbed, equilibrium state, we need to set the two <math>~M_r</math> functions equal to one another, that is, demand that,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\sin\xi - \xi \cos\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sin\eta - \eta \cos\eta</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\sin[\xi(1+\epsilon)] - \xi(1+\epsilon) \cos[\xi(1+\epsilon)]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \sin\xi \cos(\xi\epsilon) + \cos\xi \sin (\xi\epsilon) \biggr]
- \xi(1+\epsilon) \biggl[ \cos\xi \cos(\xi\epsilon) - \sin\xi \sin (\xi\epsilon) \biggr]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\sin\xi \biggl[1 - \frac{1}{2}(\xi\epsilon)^2 \biggr] + (\xi\epsilon)\cos\xi
- \xi(1+\epsilon) \cos\xi \biggl[1 - \frac{1}{2}(\xi\epsilon)^2 \biggr]  + \xi^2 \epsilon(1+\epsilon) \sin\xi </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>
~\sin\xi  -\xi\cos\xi - \frac{1}{2}(\xi\epsilon)^2 \sin\xi  + (\xi\epsilon)\cos\xi
- (\xi \epsilon) \cos\xi  + \frac{1}{2} \xi^3 \epsilon^2\cos\xi
+ \xi^2 \epsilon \sin\xi + (\xi \epsilon)^2\sin\xi 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~~- \xi^2 \epsilon \sin\xi </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{(\xi \epsilon)^2}{2} \biggl[\xi \cos\xi+ \sin\xi  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~~\frac{1}{\epsilon}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-
\frac{1}{2} \biggl[\xi \cdot \frac{\cos\xi}{\sin\xi} + 1 \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~~\epsilon</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-
2\biggl[1 + \xi \cdot \frac{\cos\xi}{\sin\xi} \biggr]^{-1}
= - 2\sin\xi \biggl[\sin\xi + \xi \cos\xi \biggr]^{-1} \, .
</math>
  </td>
</tr>
</table>
</div>
 
====Resulting Polytropic Wave Equation====
So, let's try,
<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>~
2\sin\xi \biggl[\sin\xi + \xi \cos\xi \biggr]^{-1}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3} 2\sin\xi  \biggl[\sin^2\xi + 2\xi \sin\xi \cos\xi + \xi^2 \cos^2\xi \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case,
<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>~
2\cos\xi \biggl[\sin\xi + \xi \cos\xi \biggr]^{-1} -2\sin\xi \biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl[2\cos\xi  - \xi \sin\xi \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl\{
2\cos\xi \biggl[\sin\xi + \xi \cos\xi \biggr] -2\sin\xi \biggl[2\cos\xi  - \xi \sin\xi \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl[
2\cos\xi \sin\xi + 2\xi \cos^2\xi  -4\sin\xi \cos\xi  + 2\xi \sin^2\xi
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl[\xi-\sin\xi \cos\xi \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3} 2
\biggl[\xi \sin\xi + \xi^2 \cos\xi  -\sin^2\xi \cos\xi - \xi \sin\xi \cos^2\xi \biggr] \, ,
</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>~=</math>
  </td>
  <td align="left">
<math>~2
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-2}\biggl[1-\cos^2\xi + \sin^2\xi \biggr]
-
4\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3}\biggl[\xi-\sin\xi \cos\xi \biggr]\biggl[2\cos\xi  - \xi\sin\xi \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3}\biggl\{ \sin^2\xi  \biggl[\sin\xi + \xi \cos\xi \biggr]
-
\biggl[\xi-\sin\xi \cos\xi \biggr]\biggl[2\cos\xi  - \xi\sin\xi \biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4
\biggl[\sin\xi + \xi \cos\xi \biggr]^{-3}\biggl[ \sin^3\xi + \xi \sin\xi \cos\xi
+ \xi^2 \sin\xi -2\xi \cos\xi - \xi\sin^2\xi \cos\xi + 2\sin\xi \cos^2\xi \biggr]
</math>
  </td>
</tr>
</table>
</div>
 
====Graphical Reassessment====
[[Image:TrialN1Eigenfunction.png|border|300px|right]]Before plowing ahead and plugging these expressions into the polytropic wave equation, I plotted the trial eigenfunction, <math>~\epsilon(\xi/\pi)</math> (see the blue curve in the accompanying "Trial Eigenfunction" figure), and noticed that it passes through <math>~\pm \infty</math> midway through the configuration.  This is a very unphysical behavior.  On the other hand, the inverse of this function (see the red curve) exhibits a relatively desirable behavior because it increases monotonically from negative one at the center.  As plotted, however, the function has one node.  In searching for the eigenfunction of the fundamental mode of oscillation, it might be better to add "1" to the inverse of the function and thereby get rid of all nodes.  (Keep in mind, however, that the red curve might be displaying the eigenfunction associated with the first overtone.)
 
Let's therefore try,
<div align="center">
<math>~x = 1 + \frac{1}{\epsilon} = 1 - \frac{1}{2} \biggl[\xi \cdot \frac{\cos\xi}{\sin\xi} + 1 \biggr]
= \frac{1}{2} \biggl[1- \xi \cdot \frac{\cos\xi}{\sin\xi} \biggr] \, .</math>
</div>
 
In this case we have,
<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>~
\frac{1}{2}\biggl[\xi - \frac{\cos\xi}{\sin\xi} + \xi \cdot \frac{\cos^2\xi}{\sin^2\xi} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\sin^2\xi}\biggl[\xi \sin^2\xi - \sin\xi \cos\xi + \xi \cos^2\xi \biggr] \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\sin^2\xi}\biggl[\xi - \sin\xi \cos\xi \biggr] \, ,
</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>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\cos\xi}{\sin^3\xi}\biggl[\xi - \sin\xi \cos\xi \biggr] + \frac{1}{2\sin^2\xi}\biggl[1 - \cos^2\xi  + \sin^2\xi \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{\cos\xi}{\sin^3\xi}\biggl[\xi - \sin\xi \cos\xi \biggr] \, .
</math>
  </td>
</tr>
 
</table>
</div>
 
Now let's plug these expressions into the polytropic (n = 1) wave equation, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~-\sigma^2 \xi^3 x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sin\xi \biggl[ \xi^2 x^{''} + 2\xi x^' - 2\alpha x \biggr]
+ \cos\xi \biggl[ 2\xi^2 x^' + 2\alpha \xi x \biggr] \, .
</math>
  </td>
</tr>
 
</table>
</div>
 
The first term inside the square brackets on the right-hand-side gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\xi^2 x^{''} + 2\xi x^' - 2\alpha x </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^2 - \frac{\cos\xi}{\sin^3\xi}\biggl(\xi^3 - \xi^2\sin\xi \cos\xi \biggr)
+ \frac{1}{\sin^2\xi}\biggl(\xi^2 - \xi \sin\xi \cos\xi \biggr)
- \alpha \biggl(1- \xi \cdot \frac{\cos\xi}{\sin\xi} \biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\sin^3\xi} \biggl[
\xi^2 \sin^3\xi - \cos\xi (\xi^3 - \xi^2\sin\xi \cos\xi )
+ \sin\xi (\xi^2 - \xi \sin\xi \cos\xi )
- \alpha (\sin^3\xi - \xi \cos\xi \sin^2\xi )
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\sin^3\xi} \biggl[
\xi^2 \sin\xi (1-\cos^2\xi) + \xi^2\sin\xi \cos^2\xi - \xi^3 \cos\xi
+ \xi^2\sin\xi - \xi \sin^2\xi \cos\xi
- \alpha \sin^3\xi + \alpha \xi \cos\xi \sin^2\xi
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \alpha + \frac{1}{\sin^3\xi} \biggl[
2\xi^2 \sin\xi  - \xi^3 \cos\xi - \xi \sin^2\xi \cos\xi + \alpha \xi \cos\xi \sin^2\xi
\biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>
 
and the second term inside the square brackets on the right-hand-side gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~2\xi^2 x^' + 2\alpha \xi x </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\sin^2\xi}\biggl(\xi^3 - \xi^2 \sin\xi \cos\xi \biggr)
+\frac{\alpha}{\sin\xi} \biggl(\xi \sin\xi - \xi^2 \cos\xi \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>
 
Put together, then, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{1}{\sin^2\xi} \biggl[
2\xi^2 \sin\xi  - \xi^3 \cos\xi - \xi \sin^2\xi \cos\xi + \alpha \xi \cos\xi \sin^2\xi
\biggr] +
\frac{\cos\xi}{\sin^2\xi}\biggl(\xi^3 - \xi^2 \sin\xi \cos\xi \biggr) -\alpha\sin\xi
+ \alpha \cdot \frac{\cos\xi}{\sin\xi} \biggl(\xi \sin\xi - \xi^2 \cos\xi \biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{1}{\sin^2\xi} \biggl[
2\xi^2 \sin\xi  - \xi^3 \cos\xi - \xi \sin^2\xi \cos\xi + \alpha \xi \cos\xi \sin^2\xi
+ \xi^3\cos\xi - \xi^2 \sin\xi \cos^2\xi \biggr] + \frac{\alpha}{\sin\xi} \biggl[ -\sin^2\xi
+ \xi \sin\xi \cos\xi - \xi^2 \cos^2\xi \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{\xi}{\sin\xi} \biggl[
2\xi  - \sin\xi \cos\xi - \xi \cos^2\xi \biggr]
- \frac{\alpha}{\sin\xi} \biggl[ \sin^2\xi+\xi^2 \cos^2\xi \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{\xi^2}{\sin\xi} \biggl[
1+\sin^2\xi  - \biggl(\frac{\sin\xi}{\xi}\biggr) \cos\xi  - \alpha \biggl( \frac{\sin^2\xi}{\xi^2} +\cos^2\xi \biggr)\biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~-\frac{\xi \sigma^2}{2} \biggl( \frac{ \xi^2 }{\sin\xi} \biggr) \biggl[\sin\xi - \xi \cos\xi \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
If our trial eigenfunction is a proper solution to the polytropic wave equation, then the difference of these two expressions should be zero.  Let's see:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>\frac{\sin\xi}{\xi^2} \biggl( \mathrm{RHS} - \mathrm{LHS} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
1+\sin^2\xi  - \biggl(\frac{\sin\xi}{\xi}\biggr) \cos\xi  - \alpha \biggl( \frac{\sin^2\xi}{\xi^2} +\cos^2\xi \biggr) 
+ \frac{\xi \sigma^2}{2} \biggl[\sin\xi - \xi \cos\xi \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
This expression clearly is not zero, so our trial eigenfunction is not a good one.  However, the terms in the wave equation did combine somewhat to give a fairly compact &#8212; ''albeit nonzero'' &#8212; expression.  So we may be on the right track!
 
==New Idea Involving Logarithmic Derivatives==
 
===Simplistic Layout===
Let's begin, again, with the relevant LAWE, as [[#Attempt_at_Deriving_an_Analytic_Eigenvector_Solution|provided above]].  After dividing through by <math>~x</math>, we have,
 
<div align="center">
<math>
(\sin\xi )\frac{\xi^2}{x} \cdot  \frac{d^2x}{d\xi^2} + 2 \biggl[ \sin\xi +  \xi \cos \xi \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} +
\biggl[ \sigma^2 \xi^3  - 2\alpha ( \sin\xi - \xi \cos \xi ) \biggr]  = 0 \, ,
</math><br />
</div>
<br />
where,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\sigma^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
~\frac{\omega^2}{2\pi G\rho_c \gamma_g} \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
~3-\frac{4}{\gamma_g}
\, .
</math>
  </td>
</tr>
 
</table>
</div>
 
Now, in addition to recognizing that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d\ln x}{d\ln \xi} \, ,</math>
  </td>
</tr>
</table>
</div>
in a [[User:Tohline/SSC/Stability/BiPolytrope0_0Details#Idea_Involving_Logarithmic_Derivatives|separate context]], we showed that, quite generally,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]
- \biggl[  1 -  \frac{d\ln x}{d\ln \xi} \biggr]\cdot \frac{d\ln x}{d\ln \xi} \, .
</math>
</td>
</tr>
</table>
</div>
 
Hence, if we ''assume'' that the  eigenfunction is a power-law of <math>~\xi</math>, that is, ''assume'' that,
<div align="center">
<math>~x = a_0 \xi^{c_0} \, ,</math>
</div>
then the logarithmic derivative of <math>~x</math> is a constant, namely,
<div align="center">
<math>~\frac{d\ln x}{d\ln\xi} = c_0 \, ,</math>
</div>
and the two key derivative terms will be,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} = c_0 \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} = c_0(c_0-1) \, .</math>
  </td>
</tr>
</table>
</div>
In this case, the LAWE is no longer a differential equation but, instead, takes the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~-\sigma^2 \xi^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0(c_0-1) \sin\xi  + 2c_0 [ \sin\xi +  \xi \cos \xi ] - 2\alpha ( \sin\xi - \xi \cos \xi ) 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sin\xi [c_0(c_0-1) +2c_0 -2\alpha ]  + \xi \cos \xi [2(c_0+\alpha) ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sin\xi [c_0^2 + c_0  -2\alpha ]  + \xi \cos \xi [2(c_0+\alpha) ] \, .
</math>
  </td>
</tr>
</table>
</div>
Now, the cosine term will go to zero if <math>~c_0 = -\alpha</math>; and the sine term will go to zero if,
<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>~3 </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \gamma_g</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\infty \, . </math>
  </td>
</tr>
</table>
</div>
If these two &#8212; rather strange &#8212; conditions are met, then we have a marginally unstable configuration because, <math>~\sigma^2 = 0</math>.  This, in and of itself, is not very physically interesting.  However, it may give us a clue regarding how to more generally search for a physically reasonable radial eigenfunction.
 
===More general Assumption===
Try,
<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>~\xi^{c_0} \biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr] </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^{c_0} \frac{d}{d\xi}\biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr] + c_0\xi^{c_0-1} \biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr] </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^{c_0}  \biggl[ b_0\cos\xi - d_0 \xi\sin\xi +d_0\cos\xi\biggr] + c_0\xi^{c_0-1} \biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr] </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{d\ln x}{d\ln \xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi \biggl[ b_0\cos\xi - d_0 \xi\sin\xi +d_0\cos\xi\biggr]\biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr]^{-1}
+ c_0 </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ (b_0+d_0)\xi\cos\xi - d_0 \xi^2\sin\xi \biggr]\biggl[a_0 + b_0\sin\xi + d_0 \xi\cos\xi \biggr]^{-1}
+ c_0 </math>
  </td>
</tr>
</table>
</div>
 
===Another Viewpoint===
 
====Development====
Multiplying through the [[#Simplistic_Layout|above LAWE]] by <math>~(x \xi^{-3})</math> 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>~
\frac{\sin\xi }{\xi} \cdot  \frac{d^2x}{d\xi^2} + 2 \biggl[\frac{ \sin\xi +  \xi \cos \xi }{\xi^2}\biggr] \frac{dx}{d\xi} +
\biggl[ \sigma^2  - 2\alpha \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr]x
</math>
  </td>
</tr>
</table>
</div>
 
Notice that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d}{d\xi}\biggl[\frac{\sin\xi}{\xi}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\sin\xi}{\xi^2} + \frac{\cos\xi}{\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\xi\cos\xi - \sin\xi }{\xi^2} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
And, hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d^2}{d\xi^2}\biggl[\frac{\sin\xi}{\xi}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\xi}\biggl[ \frac{\cos\xi }{\xi} -  \frac{\sin\xi }{\xi^2}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\cos\xi}{\xi^2} -\frac{\sin\xi}{\xi} + \frac{2\sin\xi}{\xi^3} - \frac{\cos\xi}{\xi^2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\sin\xi}{\xi} + 2\biggl[ \frac{\sin\xi -\xi\cos\xi}{\xi^3} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
So, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d}{d\xi} \biggl\{
\biggl(\frac{\sin\xi}{\xi}\biggr)\frac{dx}{d\xi}  + x\frac{d}{d\xi} \biggl[ \biggl(\frac{\sin\xi}{\xi}\biggr) \biggr]
\biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\sin\xi}{\xi} \cdot \frac{d^2 x}{d\xi^2}
+ 2\frac{dx}{d\xi} \cdot \biggl[\frac{d}{d\xi}\biggr(\frac{\sin\xi}{\xi}\biggr)  \biggr]
+ x \cdot \frac{d^2}{d\xi^2} \biggl(\frac{\sin\xi}{\xi}\biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\sin\xi}{\xi} \cdot \frac{d^2 x}{d\xi^2}
+ 2\frac{dx}{d\xi} \cdot \biggl[ \frac{\xi\cos\xi - \sin\xi }{\xi^2} \biggr]
+ x \cdot \biggl\{ -\frac{\sin\xi}{\xi} + 2\biggl[ \frac{\sin\xi -\xi\cos\xi}{\xi^3} \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
This means that we can rewrite the LAWE 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>~
\frac{d^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\} - 2\frac{dx}{d\xi} \cdot \biggl[ \frac{\xi\cos\xi - \sin\xi }{\xi^2} \biggr]
- x \cdot \biggl\{ -\frac{\sin\xi}{\xi} + 2\biggl[ \frac{\sin\xi -\xi\cos\xi}{\xi^3} \biggr] \biggr\}
+ 2 \biggl[\frac{ \sin\xi +  \xi \cos \xi }{\xi^2}\biggr] \frac{dx}{d\xi} +
\biggl[ \sigma^2  - 2\alpha \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr]x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\}
+ 4 \biggl[\frac{ \sin\xi  }{\xi^2}\biggr] \frac{dx}{d\xi}
+ \biggl\{ \frac{\sin\xi}{\xi}
+ \sigma^2  - 2(1+\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr\} \cdot x \, .
</math>
  </td>
</tr>
</table>
</div>
 
We recognize, also, that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{1}{\xi} \cdot \frac{d}{d\xi}\biggl[ \biggl(\frac{\sin\xi}{\xi} \biggr) x \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\xi\cos\xi - \sin\xi }{\xi^3} \biggr]x + \biggl(\frac{\sin\xi}{\xi^2} \biggr)\frac{dx}{d\xi} \, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
4\biggl(\frac{\sin\xi}{\xi^2} \biggr)\frac{dx}{d\xi}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4}{\xi} \cdot \frac{d}{d\xi}\biggl[ \biggl(\frac{\sin\xi}{\xi} \biggr) x \biggr]
+ 4\biggl[ \frac{\sin\xi - \xi\cos\xi }{\xi^3} \biggr]x
\, .
</math>
  </td>
</tr>
</table>
</div>
So the LAWE becomes,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\}
+ \frac{4}{\xi} \cdot \frac{d}{d\xi}\biggl[ \biggl(\frac{\sin\xi}{\xi} \biggr) x \biggr]
+ 4\biggl[ \frac{\sin\xi - \xi\cos\xi  }{\xi^3} \biggr]x
+ \biggl\{ \frac{\sin\xi}{\xi}
+ \sigma^2  - 2(1+\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr\} \cdot x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2}{d\xi^2} \biggl\{ \biggl( \frac{\sin\xi}{\xi}\biggr)x \biggr\}
+ \frac{4}{\xi} \cdot \frac{d}{d\xi}\biggl[ \biggl(\frac{\sin\xi}{\xi} \biggr) x \biggr]
+ \biggl\{ \frac{\sin\xi}{\xi}
+ \sigma^2  + [4- 2(1+\alpha)] \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr\} \cdot x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 \Upsilon}{d\xi^2}
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi}
+ \Upsilon
+ \biggl[ \sigma^2  + 2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x \, ,
</math>
  </td>
</tr>
</table>
</div>
where we have introduced the new, modified eigenfunction,
<div align="center">
<math>\Upsilon \equiv \biggl( \frac{\sin\xi}{\xi} \biggr) x \, .</math>
</div>
 
Alternatively, the LAWE may be written 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>~
\frac{d^2 \Upsilon}{d\xi^2}
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi}
+ \biggl[ \sigma^2  + 2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) + \frac{\sin\xi}{\xi} \biggr] \cdot x \, ;
</math>
  </td>
</tr>
</table>
</div>
or,
 
<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>~
\frac{\xi^2}{\Upsilon} \cdot \frac{d^2 \Upsilon}{d\xi^2}
+ \frac{4\xi}{\Upsilon} \cdot \frac{d\Upsilon}{d\xi}
+ \biggl[ \sigma^2  + 2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) + \frac{\sin\xi}{\xi} \biggr] \cdot \frac{\xi^3}{\sin\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\xi^2}{\Upsilon} \cdot \frac{d^2 \Upsilon}{d\xi^2}
+ \frac{4\xi}{\Upsilon} \cdot \frac{d\Upsilon}{d\xi}
+ \biggl[ \sigma^2 \biggl(\frac{\xi^3}{\sin\xi} \biggr)  + 2(1-\alpha) \biggl( 1 - \xi \cot \xi \biggr) + \xi^2 \biggr] 
</math>
  </td>
</tr>
</table>
</div>
 
 
Now, if we adopt the homentropic convention that arises from setting, <math>~\gamma = (n+1)/n</math>, then for our <math>~n=1</math> polytropic configuration, we should set, <math>~\gamma = 2</math> and, hence, <math>~\alpha = 1</math>.  This will mean that the lat term in this LAWE naturally goes to zero.  Hence, we have,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~- \sigma^2 x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 \Upsilon}{d\xi^2}
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi} + \Upsilon
\, ;
</math>
  </td>
</tr>
</table>
</div>
or,
 
<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>~
\frac{d^2 \Upsilon}{d\xi^2}
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi} + \biggl[1 + \sigma^2 \biggl(\frac{\xi}{\sin\xi}\biggr) \biggr] \Upsilon \, ;
</math>
  </td>
</tr>
</table>
</div>
or,
 
<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>~
\frac{\xi^2}{\Upsilon} \cdot \frac{d^2 \Upsilon}{d\xi^2}
+ \frac{4\xi}{\Upsilon} \cdot \frac{d\Upsilon}{d\xi} + \biggl[\xi^2 + \sigma^2 \biggl(\frac{\xi^3}{\sin\xi}\biggr) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
Does this help?
 
====Check for Mistakes====
 
Given the definition of <math>~\Upsilon</math>, its first derivative is,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d\Upsilon}{d\xi} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{dx}{d\xi} +x\biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and its second derivative is,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d^2\Upsilon}{d\xi^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d}{d\xi} \biggl\{
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{dx}{d\xi} +x\biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + 2 \biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \cdot \frac{dx}{d\xi}
+ x \cdot \frac{d}{d\xi} \biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + 2 \biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \cdot \frac{dx}{d\xi}
+ x \biggl[ -\frac{\sin\xi}{\xi} - \frac{2\cos\xi}{\xi^2}  + \frac{2\sin\xi}{\xi^3} \biggr]
</math>
  </td>
</tr>
</table>
</div>
 
Hence, the "upsilon" LAWE becomes,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~-\sigma^2 x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 \Upsilon}{d\xi^2}
+ \frac{4}{\xi} \cdot \frac{d\Upsilon}{d\xi}
+ \Upsilon
+ \biggl[  2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + 2 \biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \cdot \frac{dx}{d\xi}
+ x \biggl[ -\frac{\sin\xi}{\xi} - \frac{2\cos\xi}{\xi^2}  + \frac{2\sin\xi}{\xi^3} \biggr]
+ \frac{4}{\xi} \cdot \biggl\{  \biggl( \frac{\sin\xi}{\xi} \biggr) \frac{dx}{d\xi} +x\biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \biggr\}
+ \biggl[\frac{\sin\xi}{\xi} +  2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + \biggl\{\biggl( \frac{4\sin\xi}{\xi^2} \biggr)  + 2 \biggl[ \frac{\cos\xi}{\xi} - \frac{\sin\xi}{\xi^2} \biggr] \biggr\}\cdot \frac{dx}{d\xi}
+ \biggl[ -\frac{\sin\xi}{\xi} - \frac{2\cos\xi}{\xi^2}  + \frac{2\sin\xi}{\xi^3}
+  \frac{4\cos\xi}{\xi^2} - \frac{4\sin\xi}{\xi^3}
+ \frac{\sin\xi}{\xi} +  2(1-\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + \biggl[ \frac{2\cos\xi}{\xi} + \frac{2\sin\xi}{\xi^2} \biggr] \cdot \frac{dx}{d\xi}
+ \biggl[- 2\biggl(  \frac{\sin\xi -\xi\cos\xi}{\xi^3} \biggr)
+  (2-2\alpha) \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\sin\xi}{\xi} \biggr) \frac{d^2x}{d\xi^2} + 2\biggl[ \frac{\sin\xi}{\xi^2} + \frac{\cos\xi}{\xi} \biggr] \cdot \frac{dx}{d\xi}
+ \biggl[-2\alpha \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr] \cdot x \, .
</math>
  </td>
</tr>
</table>
</div>
This should be compared with the first expression, [[#Development|above]], 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>~
\frac{\sin\xi }{\xi} \cdot  \frac{d^2x}{d\xi^2} + 2 \biggl[\frac{ \sin\xi +  \xi \cos \xi }{\xi^2}\biggr] \frac{dx}{d\xi} +
\biggl[ \sigma^2  - 2\alpha \biggl( \frac{\sin\xi - \xi \cos \xi}{\xi^3} \biggr) \biggr]x \, ,
</math>
  </td>
</tr>
</table>
</div>
and it matches!  Q.E.D.
 
=n = 5 Polytrope=
==Setup Using Lagrangian Radial Coordinate==
 
===Individual Terms===
From our [[User:Tohline/SSC/FreeEnergy/PowerPoint#Case_M_Equilibrium_Conditions|accompanying discussion]], we have, for pressure-truncated, <math>~n=5</math> polytropic spheres
 
<div align="center">
<table border="0" cellpadding="3">
 
<tr>
  <td align="right">
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)}
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} \, ,
</math>
  </td>
</tr>
</table>
</div>
which matches the expression derived in an [[User:Tohline/SSC/Structure/Polytropes#Lane-Emden_Equation|ASIDE box found with our introduction of the Lane-Emden equation]], and
<div align="center">
<table border="0" cellpadding="3">
 
<tr>
  <td align="right">
<math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{(n+1)/(n-3)}
\tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\tilde\theta^{6}( -\tilde\xi^2 \tilde\theta' )^{6} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="3">
 
<tr>
  <td align="right">
<math>~R_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} = \biggl( \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^{2}  \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~P_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)}  = \frac{K^{10}}{G^{9} M_\mathrm{tot}^{6} }  \, ,</math>
  </td>
</tr>
</table>
</div>
 
and, from [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|our more detailed analysis]],
 
<table border="0" cellpadding="3" align="center">
<tr>
  <td align="right">
<math>
~{\tilde\theta}_5 = 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2}
</math>
  </td>
 
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
 
  <td align="right">
<math>
~\biggl(- {\tilde\xi}^2 {\tilde\theta}^'_5\biggr)  = 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \, .
</math>
  </td>
</tr>
</table>
 
Hence,
 
<div align="center">
<table border="0" cellpadding="3">
 
<tr>
  <td align="right">
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\tilde\xi \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2}  \biggr]^{-2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\tilde\xi \biggl[ 3^{-1} {\tilde\xi}^{-6} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{4\pi}{2^5\cdot 3^7}\biggr]^{1/2}
{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\biggl[  3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2} \biggr]^{6} \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2}  \biggr]^{6}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\biggl[  3^{3} \biggl( 3 + {\tilde\xi}^2\biggr)^{-3} \biggr] \biggl[ 3^{3} {\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-9}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^5}{4\pi}\biggr]^{3}
{\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-12} \, .
</math>
  </td>
</tr>
</table>
</div>
 
Now, given that the [[User:Tohline/SSC/Virial/FormFactors#Summary_.28n.3D5.29|structural form-factors for <math>~n=5</math> configurations]] are,
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathfrak{f}_M</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
( 1 + \ell^2 )^{-3/2}  = 3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathfrak{f}_W</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{5}{2^4} \cdot \ell^{-5} 
\biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathfrak{f}_A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{2^3} \ell^{-3[ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]  \, ,
</math>
  </td>
</tr>
</table>
we understand that the central density is,
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\rho_c = \frac{\bar\rho}{ {\tilde\mathfrak{f}}_M }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2}  \biggr]^{-1} \biggl[ \frac{3 M_\mathrm{tot}}{4 \pi R_\mathrm{eq}^3} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl( \frac{3}{4\pi}\biggr)
\biggl[ \frac{2^5\cdot 3^6}{4\pi}\biggr]^{ 3 / 2} (3 + {\tilde\xi}^2)^{3 / 2} M_\mathrm{tot} \biggl[ R_\mathrm{norm}
{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]^{-3}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} {\tilde\xi}^{15}  (3 + {\tilde\xi}^2)^{-15 / 2} M_\mathrm{tot} R^{-3}_\mathrm{norm}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} M_\mathrm{tot}^{-5} \biggl( \frac{G}{K} \biggr)^{-15/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \, .
</math>
  </td>
</tr>
</table>
 
<span id="r0">Now let's derive the prescription for the Lagrangian radial coordinate in the context of pressure-truncated,</span> <math>~n=5</math> polytropes. 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~r_0 \equiv a_5 \xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \rho_c^{-2/5}  \xi</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \xi \biggl\{
\biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}
\biggr\}^{-2/5}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]  \biggl( \frac{G^3M_\mathrm{tot}^2}{K^3} \biggr)
\biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</math>
  </td>
</tr>
</table>
</div>
 
<span id="m0">Also,</span>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m_0 \equiv M(r_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
\, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^2\pi \biggl\{ R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}  \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3
\biggl\{ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}  \biggr\}
\biggl\{  3^{1 / 2} \xi^3 \biggl( 3 + \xi^2\biggr)^{-3/2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 3^{1 / 2} \biggl[ 2^4 \pi^2\biggr]^{1 / 2} \biggl[ \frac{\pi^3}{2^9\cdot 3^{21}}\biggr]^{1 / 2} \biggl[ \frac{2^5\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} 
\biggl\{  \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3
\biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} R_\mathrm{norm}^3
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl\{  \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} M_\mathrm{tot}
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
 
Hence,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~g_0 = \frac{Gm_0}{r_0^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}
\biggl\{  \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} 
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
\biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]
\biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} 
\xi ( 3 + \xi^2 )^{-3/2} \, ;
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{g_0 }{r_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]
\biggl\{  \tilde\xi^{9} (3 + {\tilde\xi}^2)^{-9 / 2} \biggr\}  \biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-1}
\xi ( 3 + \xi^2 )^{-3/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2}
\biggl[  \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} 
( 3 + \xi^2 )^{-3/2}
\, ;
</math>
  </td>
</tr>
</table>
</div>
 
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\rho_0}{P_0} = \frac{\rho_0}{K\rho_0^{1+1/n}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[K^5 \rho_c \theta^5 \biggr]^{-1/5}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \theta^{-1}
\biggl\{ K^5 \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}\biggr\}^{-1/5}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
\biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[{\tilde\xi}^{-3}  (3 + {\tilde\xi}^2)^{3 / 2} \biggr]^{-3} } \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1 / 2}
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[ {\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} }
( 3 + \xi^2 )^{1 / 2}
\, ;
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
\biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \biggl[{\tilde\xi}^{-3}  (3 + {\tilde\xi}^2)^{3 / 2} \biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2}
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2}  ( 3 + \xi^2 )^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2}
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2}  ( 3 + \xi^2 )^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~  \times ~
\biggl( \frac{G^2M_\mathrm{tot}^2}{R_\mathrm{norm}^4} \biggr)^{1 / 2}\biggl[ \frac{2^6\cdot 3^{14}}{\pi^2}\biggr]^{1 / 2}
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{-9 / 2} 
\xi ( 3 + \xi^2 )^{-3/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl( \frac{G^5 M_\mathrm{tot}^4}{K^5} \biggr)^{1 / 2}  R_\mathrm{norm}^{-2}
\biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl( \frac{K^5}{G^5 M_\mathrm{tot}^4} \biggr)^{1 / 2} 
\biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2}
\xi ( 3 + \xi^2 )^{-1}
\, .
</math>
  </td>
</tr>
</table>
</div>
 
===The Wave Equation===
 
====Starting from our Key Adiabatic Wave Equation====
 
The [[#Adiabatic_.28Polytropic.29_Wave_Equation|adiabatic wave equation]] therefore 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>~
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0}
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl\{
\biggl[ \frac{2^7\cdot 3^{7}}{\pi} \biggr]^{1 / 2}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \frac{1}{\xi}
-  \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1}
\biggr\} \frac{dx}{dr_0}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2}
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-3/2}
( 3 + \xi^2 )^{1 / 2} 
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr]
+ \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2}
\biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{15/2} 
( 3 + \xi^2 )^{-3/2} 
\biggr\}  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl[ \frac{2^3\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
\frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2}
( 3 + \xi^2 )^{1 / 2}  \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{-3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}
( 3 + \xi^2 )^{-3/2} 
\biggr\}  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
\frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0}
+ \frac{6}{\gamma_g R_*^2}  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}( 3 + \xi^2 )^{1 / 2} 
+  \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr\}  x
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>R_* \equiv R_\mathrm{norm} \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1/2} \, .</math>
</div>
 
Recognizing that,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \, ,
</math>
  </td>
</tr>
</table>
</div>
we can write,
 
<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>~\frac{1}{R_*^2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}  \biggl\{
\frac{d^2x}{d\xi^2} + \biggl[
\frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{d\xi}
+ \frac{6}{\gamma_g } 
\biggl[\sigma^2 ( 3 + \xi^2 )^{1 / 2} 
+  \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr]  x \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
 
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\sigma^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}  \biggl( \frac{3 + {\tilde\xi}^2}{{\tilde\xi}^2} \biggr)^{15/2} \, .</math>
  </td>
</tr>
</table>
</div>
 
Finally, if &#8212; because we are specifically considering the case of <math>~n=5</math> &#8212; we set <math>~\gamma_\mathrm{g} = 1 + 1/n = 6/5</math>, we have,
 
<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>~
\frac{d^2x}{d\xi^2} + \biggl[ \frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi}
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{1 / 2}  +  \frac{2}{( 3 + \xi^2 ) }\biggr]  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{( 3 + \xi^2 ) } \biggl\{ ( 3 + \xi^2 )\frac{d^2x}{d\xi^2}
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi}
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2}  +  2 \biggr]  x \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
 
====Starting from the HRW66 Radial Pulsation Equation====
More directly, if we begin with the [[#HRW66excerpt| HRW66 radial pulsation equation]] that is already tuned to polytropic configurations, the wave equation appropriate to <math>~n=5</math> polytropes is,
<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>~
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi}
+ \frac{5(-\theta_5^') }{6\theta_5 \xi} \bigg[ \frac{\xi (s^')^2}{\theta^'_5} + \frac{12}{5} \biggr] X
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{(3 + \xi^2)} \biggr]\frac{d X}{d\xi}
+ \frac{1}{(3 + \xi^2)} \bigg[ -\frac{5(s^')^2(3 + \xi^2)^{3 / 2}}{2 \cdot 3^{3 / 2}} + 2 \biggr] X
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{(3+\xi^2)} \biggl\{
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
+ \bigg[ -\frac{5(s^')^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
which is identical to the brute-force derivation just presented, allowing for the mapping,
<div align="center">
<math>\sigma^2 ~~ \Leftrightarrow ~~ -\frac{(s^')^2}{2 \cdot 3^{3 / 2}} \, .</math>
</div>
 
====New Independent Variable====
Guided by our [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#Conjectures|conjecture]] regarding the proper shape of the radial eigenfunction, let's switch the dependent variable to,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~u  \equiv 1 + \frac{3}{\xi^2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~3 + \xi^2 = \frac{3u}{(u-1)}  \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\xi = 3^{1 / 2} (u-1)^{-1 / 2} \, .</math>
  </td>
</tr>
</table>
</div>
This implies that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d}{d\xi}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow ~~~</math>
  </td>
  <td align="left">
<math>~-\frac{2}{\sqrt{3}}(u-1)^{3 / 2} \frac{d}{du} \, ,</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d^2}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow ~~~</math>
  </td>
  <td align="left">
<math>~\frac{4}{3}(u-1)^3 \frac{d^2}{du^2} + 2(u-1)^{2} \frac{d}{du} \, .</math>
  </td>
</tr>
</table>
</div>
Hence, the governing wave equation 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>~( 3 + \xi^2 )\frac{d^2x}{d\xi^2}
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi}
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2}  +  2 \biggr]  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3u}{(u-1)} \biggl[\frac{4}{3}(u-1)^3 \frac{d^2x}{du^2} + 2(u-1)^{2} \frac{dx}{du}\biggr]
+ 4(2u-3)(u-1)\frac{dx}{du}
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2}  +  2 \biggr\}  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4u(u-1)^2 \frac{d^2x}{du^2}
+ (14u-12)(u-1)\frac{dx}{du}
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2}  +  2 \biggr\}  x \, .
</math>
  </td>
</tr>
</table>
</div>
If we ''assume'' that <math>~\sigma^2 = 0</math>, then the governing relation is,
 
<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>~4u(u-1)^2 \frac{d^2x}{du^2}
+ (14u-12)(u-1)\frac{dx}{du}
+ 2 x \, .
</math>
  </td>
</tr>
</table>
</div>
Now, again, guided by our  [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#Conjectures|conjecture]], let's guess an eigenfunction of the form:
 
=====First Guess (n5)=====
<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>~
A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}  \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{A^3}{2} \biggl[ (u - 1)^{-1 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{1 / 2} (A u - 1 )^{-3 / 2} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \, ;
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{
-\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 2}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[
(Au-1) +3A (u-1)\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
<!-- EXTRA 2nd DERIVATIVE
<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{A^3}{2^2} \biggl[ -(u - 1)^{-3 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2}
- A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} + 3A^2(u - 1)^{1 / 2} (A u - 1 )^{-5 / 2} \biggr]
</math>
  </td>
</tr>
END EXTRA 2nd DERIVATIVE -->
 
So the governing relation 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>~4u(u-1)^2 \biggl\{ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (14u-12)(u-1) \biggl\{ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \biggr\}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~u(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (7u-6)(u-1)^{1 / 2}  A^3(A-1)  (Au-1)^{-3 / 2}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(u-1)^{1 / 2}  \biggl\{ uA^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] 
+ (7u-6) A^3(A-1)  (Au-1)^{-3 / 2}
+ 2 A^3 (A u - 1 )^{-1 / 2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u(A-1) \biggl[(3A+1) - 4Au \biggr] 
+ (7u-6) (A-1)  (Au-1)
+ 2 (A u - 1 )^{2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ - 4u^2 A(A-1)  + u(A-1) (3A+1)
+ (7u-6)  [A(A-1)u +1 - A]
+ 2  (A^2u^2 - 2Au +1) \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u^2 \biggl[ - 4A(A-1) +7A(A-1) +2A^2 \biggr]   
+ u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr]  + 2(3A-2)  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
+ u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr]  + 2(3A-2)  \biggr\} \, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
+ u\biggl[ -3A^2 -7A  +6\biggr]  + 2(3A-2)  \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
 
=====Second Guess (n5)=====
<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>~
(u - 1)^{b / 2} (A u - 1 )^{-a / 2}  \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{2}(u-1)^{b/2-1} (A u - 1 )^{-a / 2}
- \frac{aA}{2}(u - 1)^{b / 2} (A u - 1 )^{-a / 2-1}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \biggl[
\frac{b}{2}(u-1)^{-1}
- \frac{aA}{2}  (A u - 1 )^{-1} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{(u-1)}{x} \frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (A u - 1 )^{-1} \biggl[
\frac{b}{2} (A u - 1 )
- \frac{aA}{2} (u-1) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{2(A u - 1 )} \biggl[
b (A u - 1 ) - aA (u-1) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1 }{2(A u - 1 )} \biggl[
(aA - b) + A(b    - a)u \biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]\frac{dx}{du}
+ x \frac{d}{du}\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]^2
+ x  \biggl[ -\frac{b}{2}(u-1)^{-2} + \frac{aA^2}{2}  (A u - 1 )^{-2}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{x}{4(u-1)^2 (Au-1)^2} \biggl\{
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
Hence, the governing wave equation 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>~2u \biggl\{ \frac{(u-1)^2}{x} \frac{d^2x}{du^2} \biggr\}
+ (7u-6)\biggl\{ \frac{(u-1)}{x} \frac{dx}{du} \biggl\} + 1
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2u}{4 (Au-1)^2} \biggl\{
\biggl[ (aA - b) + A(b    - a)u \biggr]^2
+  \biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{(7u-6) }{2(A u - 1 )} \biggl[
(aA - b) + A(b    - a)u \biggr]
+ 1
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+  2u\biggl[  2aA^2 (u^2 - 2u + 1)  -2b (A^2 u^2 - 2Au + 1 ) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2(A u - 1 )(7u-6) \biggl[
(aA - b) + A(b    - a)u \biggr]
+ 4 (Au-1)^2 \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+  2u\biggl[ 2A^2(a-b)u^2 + 4A(b - aA) u + 2(aA^2  -b)  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[
(aA - b) + A(b    - a)u \biggr]
+  (4A^2u^2-8Au + 4) \biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
If <math>~b=a</math>,
 
<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>~
2u\biggl[ (aA - b)^2  \biggr]
+  2u\biggl[ 4A(b - aA) u + 2(aA^2  -b)  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[
(aA - b) \biggr]  +  (4A^2u^2-8Au + 4)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2a^2u (A - 1)^2 
+  2au [ 4A(1 - A) u + 2(A^2  -1) ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2a(A - 1) \biggl[7Au^2 - (6A+7)u  +6 \biggr] +  (4A^2u^2-8Au + 4)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [4a (1 - A) +  7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2  -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>
</table>
</div>
This should then match the [[#First_Guess_.28n5.29|"first guess"]] algebraic condition if we set <math>~a=1</math>.  Let's see.
<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>~
2Au^2 [4 (1 - A) +  7(A - 1) + 2A] + 2u [ (A - 1)^2 +  2(A^2  -1) - (A - 1) (6A+7) - 4A] + 4[ 3(A-1) + 1]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [4  - 4A +  7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) +  2A^2  -2 + (1-A ) (6A+7) -4A] + 4[ 3A-2]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [5A - 3] + 2u [  - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, .
</math>
  </td>
</tr>
</table>
</div>
And we see that this expression ''does'' match the one derived earlier.
 
Going back a bit, before setting <math>~a=1</math>, we have the 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>~
2Au^2 [4a (1 - A) +  7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2  -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [  3aA  -3a + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2  -1) - a(6A^2+A-7) - 4A] + 4[ 3a(A-1) + 1]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2Au^2 [  3a(A  - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A] + 4[ 3a(A-1) + 1] \, .
</math>
  </td>
</tr>
</table>
</div>
Now, in order for all three expressions inside the square-bracket pairs to be zero, we need, first,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~3a(A  - 1) + 2A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2A}{3(1-A)} \, ;</math>
  </td>
</tr>
</table>
</div>
and, third, by simple visual comparison with the first expression,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~3a(A-1) + 1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3a(A-1) + 2A</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3} \, ;</math>
  </td>
</tr>
</table>
</div>
which forces the second expression to the value,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{2}{3}\biggr)^2 \biggl(-\frac{1}{2} \biggr)^2 +  \frac{2}{3}\biggl[  -1-\frac{1}{2} +5 \biggr] - 2</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{9}  +  \frac{7}{3}  - 2</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4}{9}  \, ,</math>
  </td>
</tr>
</table>
</div>
which is ''not'' zero.  Hence our pair of unknown parameters &#8212; <math>~a </math> and <math>~A</math> &#8212; do not simultaneously satisfy all three conditions.  (Not really a surprise.)
 
==Setup Using Lagrangian Mass Coordinate==
 
===Alternative Terms===
Let's change the independent coordinate from <math>~r_0</math> to <math>~m_0</math>.  In particular, the derivative operation will change as follows:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d}{dr_0}</math>
  </td>
  <td align="center">
<math>~~\rightarrow~~</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{dm_0}{dr_0} \biggr)\frac{d}{dm_0}
= \biggl( \frac{dm_0}{d\xi} \cdot \frac{d\xi}{dr_0}  \biggr)\frac{d}{dm_0}
\, ,</math>
  </td>
</tr>
</table>
</div>
so what is the expression for the leading coefficient?  From [[#r0|above]], we have,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} r_0 \, .
</math>
  </td>
</tr>
</table>
</div>
 
Also, from [[#m0|above]], we know that,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dm_0}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
\biggl\{ 3\xi^2 ( 3 + \xi^2 )^{-3/2}
- 3 \xi^4 ( 3 + \xi^2 )^{-5/2}\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3\xi^2 (3 + \xi^2)^{-5/2}
\biggl\{ ( 3 + \xi^2 )
- \xi^2  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dm_0}{dr_0}</math>
</td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2}
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
</td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{M_\mathrm{tot} }{R_*}
\biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} \, .
</math>
  </td>
</tr>
</table>
</div>
 
To simplify expressions, let's [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#DefineTildeC|borrow from an accompanying derivation]] and define,
<div align="center">
<math>\tilde{C} \equiv \frac{3^2}{{\tilde\xi}^2} \biggl( 1 + \frac{ {\tilde\xi}^2}{3} \biggr) = 3 \biggl[ \frac{( 3 + {\tilde\xi}^2 )}{ {\tilde\xi}^2} \biggr] \, .</math>
</div>
Then we have,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{m_0}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{\tilde{C}}{ 3}\biggr]^{3 / 2}
\biggl[ \frac{\xi^2}{ ( 3 + \xi^2 )} \biggr]^{3/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{\xi^2}{ ( 3 + \xi^2 )}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~( 3 + \xi^2 )\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\xi^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~3 m_*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \xi^2 (1-m_*)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\xi^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{3m_*}{(1-m_*)} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~m_* \equiv \biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3} \, .</math>
</div>
 
In summary:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ 
\frac{\xi^2}{ ( 3 + \xi^2 )} = m_* \, ;
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; while, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~ 
\frac{ {\tilde\xi}^2}{ ( 3 + {\tilde\xi}^2 )} = \frac{3}{\tilde{C}} \, ;
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
= R_*  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggr[ \frac{3m_*}{ (1-m_*) }\biggr]^{1 / 2} \, ;
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{6}{R_*} \biggl[  \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2) }\biggr]^{9}  \frac{\xi}{ ( 3 + \xi^2 )}
=
\frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  \frac{m_*}{ \xi }
=
\frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2} \, ;
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{g_0 }{r_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{GM_\mathrm{tot}}{R_*^3}
\biggl[  \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2)}\biggr]^{15/2}  \frac{1}{\xi^3}
\biggl[ \frac{ \xi^2 }{ ( 3 + \xi^2 ) }\biggr]^{3/2}
=
\frac{GM_\mathrm{tot}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} \, ;
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{\rho_0}{\gamma_g P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
</div>
So, the wave equation may be written 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>~
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0}
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2}
+ \biggl\{ \frac{4}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2} 
- \frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2}  \biggr\} \frac{dx}{dr_0}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{ \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{GM_\mathrm{tot}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} \biggr\}  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2}
+ \frac{1}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggl\{ 4 
- 6\biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{6}  m_* \biggr\} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{\sigma^2  +  (1-m_*)^{3 / 2} \biggr\}  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl\{
R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}
R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \frac{dx}{dr_0}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
 
where,
<div align="center">
<math>~\sigma^2 \equiv (4 - 3\gamma_\mathrm{g})^{-1} \frac{R_*^3}{GM_\mathrm{tot}} \biggl[  \frac{ \tilde{C} }{3 } \biggr]^{15/2} \omega^2 \, .</math>
</div>
 
Now, let's look at the differential operators, after defining.
<div align="center">
<math>~c_0 \equiv  3^{1 / 2} R_*  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} ~~~~\Rightarrow ~~~~R_*  = c_0 3^{-1 / 2}  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} \, .</math>
</div>
 
We find,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~dr_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0 ~d[ m_*^{1 / 2} (1-m_*)^{-1 / 2} ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0 ~\biggl[\frac{1}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-1 / 2} + \frac{1}{2} ~m_*^{1 / 2} (1 - m_*)^{-3 / 2}
\biggr] dm_*
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c_0}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-3 / 2}~ dm_*
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{d}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{c_0} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ R_*\frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \, .
</math>
  </td>
</tr>
</table>
</div>
 
Also,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{d^2}{dr_0^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggl[ m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2} 
+\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2} \biggl[ \frac{1}{2} m_*^{-1 / 2}( 1 - m_*)^{3 / 2} - \frac{3}{2} m_*^{1 / 2}( 1 - m_*)^{1 / 2}~
\biggr] ~ \frac{d}{dm_*}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2}  +\frac{1}{2} \biggl( \frac{2}{c_0} \biggr)^{2}~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{d}{dm_*}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] 
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
</math>
  </td>
</tr>
</table>
</div>
 
So, the wave equation 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>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] 
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*}  \biggr]
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}
\biggl[\frac{2^2}{3}  \biggr] 
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*}  \biggr]
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} m_*^{1 / 2} \biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} 
\biggl\{    2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +  ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} 
+ \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} 
\biggl\{  2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 5 - 4m_*  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} 
+ \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  \biggl\{  2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} 
+ (5 - \mathcal{A}  m_*) (1-m_*)^2 \frac{dx}{dm_*} 
+ \mathcal{B} \biggl[ \frac{\sigma^2}{(1-m_*)^{1 / 2}}  +  (1-m_*) \biggr]  x  \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~4 + 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3^{5/2}(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3}  \, .</math>
  </td>
</tr>
</table>
</div>
 
==Try Again==
 
This time, let's adopt the notation used in a [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#n_.3D_5_Mass-Radius_Relation|related chapter in our ''Ramblings'' appendix]].  Specifically, the parametric relationship between <math>~m_\xi</math> and <math>~r_\xi</math> in pressure-truncated, <math>~n=5</math>  polytropes is,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m_\xi \equiv \frac{m_0}{ M_\mathrm{tot} } = \frac{M_r(\xi)}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(3 + \xi^2 \biggr)^{-3/2}
\biggl(3 +  {\tilde\xi}^2 \biggr)^{3/2} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3 / 2}\biggl[ \frac{( 3+\xi^2)}{ {\xi}^2} \biggr]^{- 3 / 2}
\, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~r_\xi \equiv \frac{r_0}{R_\mathrm{norm}} = \biggl(\frac{\xi}{\tilde\xi} \biggr) \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi \biggl\{
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1/2}
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3} \xi \, .
</math>
  </td>
</tr>
</table>
</div>
 
And we are in the fortunate situation of being able to eliminate <math>~\xi</math> to obtain the direct relation,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
r_\xi (m_\xi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde{r}_\mathrm{edge}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\tilde{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr)
= 3 \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]
\, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\tilde{r}_\mathrm{edge}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
= \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1 / 2} \biggl[ \frac{\tilde{C}}{ 3} \biggr]^{3}
\, .
</math>
  </td>
</tr>
</table>
</div>
 
If we furthermore define,
<div align="center">
<math>m_* \equiv \frac{3}{\tilde{C}} \cdot m_\xi^{2 / 3} \, ,</math>
</div>
then,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
r_\xi (m_*)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2}
\, .
</math>
  </td>
</tr>
</table>
</div>
 
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
\frac{dr_0}{R_\mathrm{norm}} = dr_\xi
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl\{
\frac{1}{2} (1-m_*)^{- 1 / 2} m_*^{-1 / 2} + \frac{1}{2}m_*^{1 / 2}(1-m_*)^{-3 / 2}
\biggr\} dm_*
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{3^{1 / 2}}{2} \biggr) \tilde{r}_\mathrm{edge} m_*^{-1 / 2} (1-m_*)^{-3 / 2} dm_*
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>\Rightarrow ~~~ R_\mathrm{norm} \cdot \frac{d}{dr_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*} \, .
</math>
  </td>
</tr>
</table>
</div>
We therefore also have,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ R^2_\mathrm{norm} \cdot \frac{d^2}{dr_0^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2}
\biggl\{
\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d^2}{dm_*^2}\biggr]
+ \biggl[ \frac{1}{2} m_*^{-1 / 2} (1-m_*)^{3 / 2} + \frac{3}{2}m_*^{1 / 2} (1-m_*)^{1 / 2}\biggr]  \frac{d}{dm_*}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
\biggl[ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}\biggr]
+ \biggl[  (1-m_*)^{3 } + 3m_* (1-m_*)^{2}\biggr]  \frac{d}{dm_*}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{d}{dm_*} \biggr\}
\, .
</math>
  </td>
</tr>
</table>
</div>
 
So the wave equation may be written,
<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>~
R_\mathrm{norm}^2 \cdot \frac{d^2x}{dr_0^2} + \biggl[\frac{4R_\mathrm{norm}}{r_0} - \biggl(\frac{g_0 \rho_0 R_\mathrm{norm}}{P_0}\biggr) \biggr] R_\mathrm{norm} \cdot \frac{dx}{dr_0}
+ \biggl(\frac{\rho_0 R_\mathrm{norm}}{\gamma_\mathrm{g} P_0} \biggr)\biggl[R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0 R_\mathrm{norm}}{r_0} \biggr]  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr)  \biggl\{ \frac{4}{r_\xi}
- \biggl[\frac{6R_\mathrm{norm}}{R_*} \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{9}  m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2} \biggr] \biggr\}
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6R_* R_\mathrm{norm}}{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{
R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g}) \frac{GM_\mathrm{tot} R_\mathrm{norm}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} 
\biggr\}  x \, .
</math>
  </td>
</tr>
</table>
</div>
Keeping in mind that,
<div align="center">
<math>~\frac{R_*}{R_\mathrm{norm}} = \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1 / 2} = {\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \, ,</math>
</div>
 
we therefore have,
<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>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr)  \biggl\{ 4 \biggl[3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2}  \biggr]^{-1}
- 6 \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{9} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-1} m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2}  \biggr\}
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 6 \biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{
\biggl[ \frac{R_*^3}{\gamma_g GM_\mathrm{tot} } \biggr] \omega^2
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}  \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} 
\biggr\}  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}^2} \biggl( \frac{2^3}{3} \biggr) 
\biggl[ 1  - \frac{3}{2} \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_*  \biggr]
(1-m_*)^{2} \frac{dx}{dm_*}
+ \frac{6}{ {\tilde{r}}_\mathrm{edge}^2 } 
\biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2}  \biggr]  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ \biggl[  5  - 6 \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* + 2m_* \biggr]
(1-m_*)^{2} \frac{dx}{dm_*}
+ 3^{5 / 2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} 
\frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}
\biggl[ \frac{\sigma^2 }{(1-m_*)^{1 / 2}} +  (1-m_*)  \biggr]  x \biggr\}
\, ,
</math>
  </td>
</tr>
</table>
</div>
where, as before,
<div align="center">
<math>\sigma^2 \equiv \biggl(  \frac{ \tilde{C} }{3 } \biggr)^{15/2} \biggl[ \frac{R_*^3}{(4 - 3\gamma_g) GM_\mathrm{tot} } \biggr] \omega^2 \, .</math>
</div>
 
=n = 3 Polytrope=
Here we perform a numerical integration of the governing LAWE for <math>~n=3</math> polytropes.  We can directly compare our results with [http://adsabs.harvard.edu/abs/1941ApJ....94..245S Schwarzschild's (1941)] published work on "Overtone Pulsations for the Standard [Stellar] Model." 
 
Drawing from our [[#Groundwork|above discussion]], the LAWE for any polytrope of index, <math>~n</math>, may be written 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>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4 - (n+1)V(\xi)}{\xi} \biggr] \frac{dx}{d\xi} +
\biggl[\omega^2 \biggl(\frac{a_n^2 \rho_c }{\gamma_g P_c} \biggr) \frac{\theta_c}{\theta} -
\biggl(3-\frac{4}{\gamma_g}\biggr)  \cdot \frac{(n+1)V(x)}{\xi^2} \biggr]  x </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi} \biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta} \biggl[ \frac{\sigma_c^2}{6\gamma_g}  -
\frac{\alpha}{\xi } \biggl(- \frac{d\theta}{d\xi} \biggr) \biggr]  x </math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\sigma_c^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3\omega^2}{2\pi G\rho_c} \, .</math>
  </td>
</tr>
</table>
</div>
 
It can be shown straightforwardly that this matches the LAWE used by [[User:Tohline/SSC/Perturbations#Schwarzschild_.281941.29|Schwarzschild (1941)]], if <math>~n</math> is set to 3.  But let's postpone making this substitution until we formulate a general approach to integrating this equation from the center of the configuration, outward.  Following a [[User:Tohline/Appendix/Ramblings/NumericallyDeterminedEigenvectors#Integrating_Outward_Through_the_Core|parallel discussion]], we begin by multiplying the LAWE through by <math>~(\xi\theta)</math>, obtaining a 2<sup>nd</sup>-order ODE that is relevant at every individual coordinate location, <math>~\xi_i</math>, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\theta_i {x_i''}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \frac{x_i'}{\xi_i} 
- (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g}  -
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr]  x_i </math>
  </td>
</tr>
</table>
</div>
 
Now, using the [[User:Tohline/Appendix/Ramblings/NumericallyDeterminedEigenvectors#General_Approach|general finite-difference approach described separately]], we make the substitutions,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~x_i'</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{x_+ - x_-}{2 \Delta_\xi}  \, ;
</math>
  </td>
</tr>
</table>
</div>
and,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
x_i''
</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{x_+ - 2x_i + x_-}{\Delta_\xi^2} \, ,</math>
  </td>
</tr>
</table>
</div>
which will provide an approximate expression for <math>~x_+ \equiv x_{i+1}</math>, given the values of <math>~x_- \equiv x_{i-1}</math> and <math>~x_i</math>.  Specifically, if the center of the configuration is denoted by the grid index, <math>~i=1</math>, then for zones, <math>~i = 3 \rightarrow N</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\theta_i \biggl[ \frac{x_+ - 2x_i + x_-}{\Delta_\xi^2} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{x_+ - x_-}{2 \xi_i \Delta_\xi}  \biggr]
- (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g}  -
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr]  x_i </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \theta_i \biggl[ \frac{x_+ }{\Delta_\xi^2} \biggr] + \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{x_+ }{2 \xi_i\Delta_\xi}  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\theta_i \biggl[ \frac{- 2x_i + x_-}{\Delta_\xi^2} \biggr]
- \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{- x_-}{2 \xi_i \Delta_\xi}  \biggr]
- (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g}  -
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr]  x_i </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x_+ \biggl[2\theta_i +\frac{4\Delta_\xi \theta_i}{\xi_i} - \Delta_\xi (n+1)(- \theta^')_i\biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x_- \biggl[\frac{4\Delta_\xi \theta_i}{\xi_i} - \Delta_\xi (n+1)(- \theta^')_i - 2\theta_i\biggr]
+  x_i\biggl\{4\theta_i - 2\Delta_\xi^2(n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g}  -
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr]  \biggr\} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
where,
In order to kick-start the integration, at the center of the configuration <math>~(\xi_1 = 0)</math>, we will set the eigenfunction value to <math>~x_1 = 1</math>; and we will use the assumed symmetry condition, <math>~x_1' = 0 ~~\Rightarrow ~~ x_- = x_+</math>, in order to determine the value of the eigenfunction at the first grid location off center <math>~(\xi_2 = \Delta)</math>.  That is, for <math>~i = 1</math>, the discretized LAWE gives,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 7,064: Line 834:
   <td align="right">
   <td align="right">
<math>~
<math>~
x_+
\mathfrak{F}
</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
x_i \biggl\{ 1
\biggl[ \frac{\sigma_c^2}{\gamma_g}  - 2\alpha\biggr]\, .</math>
-\frac{\Delta_\xi^2 (n+1)}{2\theta_i}\biggl[ \frac{\sigma_c^2}{6\gamma_g}  -  
\frac{\alpha}{\xi_i } (- \theta^')_i\biggr] \biggr\} \, .</math>
   </td>
   </td>
</tr>
</tr>
Line 7,081: Line 849:


=Related Discussions=
=Related Discussions=
* Radial Oscillations of [[User:Tohline/SSC/UniformDensity#The_Stability_of_Uniform-Density_Spheres|Uniform-density sphere]]
* Radial Oscillations of Isolated Polytropes
** [[User:Tohline/SSC/Stability/Polytropes#Radial_Oscillations_of_Polytropic_Spheres|Setup]]
** n = 1:&nbsp; [[User:Tohline/SSC/Stability/n1PolytropeLAWE|Attempt at Formulating an Analytic Solution]]
** n = 3:&nbsp; [[User:Tohline/SSC/Stability/n3PolytropeLAWE|Numerical Solution]] to compare with [http://adsabs.harvard.edu/abs/1941ApJ....94..245S M. Schwarzschild (1941)]
** n = 5:&nbsp; [[User:Tohline/SSC/Stability/n5PolytropeLAWE|Attempt at Formulating an Analytic Solution]]


* In an accompanying [[User:Tohline/Appendix/Ramblings/SphericalWaveEquation#Playing_With_Spherical_Wave_Equation|Chapter within our "Ramblings" Appendix]], we have played with the adiabatic wave equation for polytropes, examining its form when the primary perturbation variable is an enthalpy-like quantity, rather than the radial displacement of a spherical mass shell.  This was done in an effort to mimic the approach that has been taken in studies of the [[User:Tohline/Apps/ImamuraHadleyCollaboration#Papaloizou-Pringle_Tori|stability of Papaloizou-Pringle tori]].
* In an accompanying [[User:Tohline/Appendix/Ramblings/SphericalWaveEquation#Playing_With_Spherical_Wave_Equation|Chapter within our "Ramblings" Appendix]], we have played with the adiabatic wave equation for polytropes, examining its form when the primary perturbation variable is an enthalpy-like quantity, rather than the radial displacement of a spherical mass shell.  This was done in an effort to mimic the approach that has been taken in studies of the [[User:Tohline/Apps/ImamuraHadleyCollaboration#Papaloizou-Pringle_Tori|stability of Papaloizou-Pringle tori]].

Latest revision as of 20:50, 28 February 2017

Radial Oscillations of Polytropic Spheres

Whitworth's (1981) Isothermal Free-Energy Surface
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Groundwork

Adiabatic (Polytropic) Wave Equation

In an accompanying discussion, we derived the so-called,

Adiabatic Wave (or Radial Pulsation) Equation

LSU Key.png

<math>~ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x = 0 </math>


whose solution gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations. If the initial, unperturbed equilibrium configuration is a polytropic sphere whose internal structure is defined by the function, <math>~\theta(\xi)</math>, then

<math>~r_0</math>

<math>~=</math>

<math>~a_n \xi \, ,</math>

<math>~\rho_0</math>

<math>~=</math>

<math>~\rho_c \theta^{n} \, ,</math>

<math>~P_0</math>

<math>~=</math>

<math>~K\rho_0^{(n+1)/n} = K\rho_c^{(n+1)/n} \theta^{n+1} \, ,</math>

<math>~g_0</math>

<math>~=</math>

<math>~\frac{GM(r_0)}{r_0^2} = \frac{G}{r_0^2} \biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr] \, ,</math>

where,

<math>~a_n</math>

<math>~=</math>

<math>~\biggl[\frac{(n+1)K}{4\pi G} \cdot \rho_c^{(1-n)/n} \biggr]^{1/2} \, .</math>

Hence, after multiplying through by <math>~a_n^2</math>, the above adiabatic wave equation can be rewritten in the form,

<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{g_0}{a_n}\biggl(\frac{a_n^2 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{d\xi} + \biggl(\frac{a_n^2\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{a_n\xi} \biggr] x </math>

<math>~=</math>

<math>~0 \, .</math>

In addition, given that,

<math>~\frac{g_0}{a_n}</math>

<math>~=</math>

<math>~4\pi G \rho_c \biggl(-\frac{d \theta}{d\xi} \biggr) \, ,</math>

and,

<math>~\frac{a_n^2 \rho_0}{P_0}</math>

<math>~=</math>

<math>~\frac{(n+1)}{(4\pi G\rho_c)\theta} = \frac{a_n^2 \rho_c}{P_c} \cdot \frac{\theta_c}{\theta}\, ,</math>

we can write,

<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4 - (n+1)V(\xi)}{\xi} \biggr] \frac{dx}{d\xi} + \biggl[\omega^2 \biggl(\frac{a_n^2 \rho_c }{\gamma_g P_c} \biggr) \frac{\theta_c}{\theta} - \biggl(3-\frac{4}{\gamma_g}\biggr) \cdot \frac{(n+1)V(x)}{\xi^2} \biggr] x </math>

<math>~=</math>

<math>0 \, ,</math>

where we have adopted the function notation,

<math>~V(\xi)</math>

<math>~\equiv</math>

<math>~- \frac{\xi}{\theta} \frac{d \theta}{d\xi} \, .</math>

Comment by J. E. Tohline: There appears to be a sign error in the numerator of the second term of the polytropic wave equation published by Murphy & Fiedler; there also appears to be an error in the definition of the coefficient, α*, as given in the text of their paper.

As can be seen in the following framed image, this is the form of the polytropic wave equation published by J. O. Murphy & R. Fiedler (1985b, Proc. Astron. Soc. Australia, 6, 222), at the beginning of their discussion of "Radial Pulsations and Vibrational Stability of a Sequence of Two Zone Polytropic Stellar Models."

Polytropic Wave Equation extracted from J. O. Murphy & R. Fiedler (1985b)

"Radial Pulsations and Vibrational Stability of a Sequence of Two Zone Polytropic Stellar Models"

Proceeding of the Astronomical Society of Australia, vol. 6, pp. 222 - 226 © Astronomical Society of Australia

Murphy & Fiedler (1985b)
Equations displayed here, as a single digital image, with layout modified from the original publication.


Comment by J. E. Tohline: As is shown in the subsection on "Boundary Conditions," below, it appears as though the term on the right-hand-side of HRW66's equation (58) is incorrect, as published; it should be preceded with a negative sign.

It is also the same as the radial pulsation equation for polytropic configurations that appears as equation (56) in the detailed discussion of "The Oscillations of Gas Spheres" published by H M. Hurley, P. H. Roberts, & K. Wright (1966, ApJ, 143, 535); hereafter, we will refer to this paper as HRW66. The relevant set of equations from HRW66 has been extracted as a single digital image and reprinted, here, as a boxed-in image.


Radial Pulsation Equation as Presented by M. Hurley, P. H. Roberts, & K. Wright (1966)

"The Oscillations of Gas Spheres"

The Astrophysical Journal, vol. 143, pp. 535 - 551 © American Astronomical Society

Hurley, Roberts & Wright (1966)

Set of equations and accompanying text displayed here, as a single digital image, exactly as they appear in the original publication.

In order to make clearer the correspondence between our derived expression and the one published by HRW66, we will rewrite the HRW66 radial pulsation equation: (1) Gathering all terms on the same side of the equation; (2) making the substitution,

<math>\theta^' \rightarrow -\frac{\theta V}{x} \, ;</math>

and (3) reattaching a "prime" to the quantity, <math>~s</math>, to emphasize that it is a dimensionless frequency.

ASIDE: In their equation (46), HRW66 convert the eigenfrequency, <math>~s</math> — which has units of inverse time — to a dimensionless eigenfrequency, <math>~s^'</math>, via the relation,

<math>~s = \biggl( \frac{4\pi G \rho_c}{1+n} \biggr)^{1/2} s^' ~~~~~~~\cdots\cdots~~~~~~~(46)</math>

Then, immediately following equation (46), they state that they will "omit the prime on <math>~s</math> henceforward." As a result, the dimensionless eigenfrequency that appears in their equations (56) and (58) is unprimed. This is unfortunate as it somewhat muddies our efforts, here, to demonstrate the correspondence between the HRW66 polytropic radial pulsation equation and ours. In our subsequent manipulation of equation (56) from HRW66 we reattach a prime to the quantity, <math>~s</math>, to emphasize that it is a dimensionless frequency. But this prime on <math>~s</math> should not be confused with the prime on <math>~\theta</math> (HRW66 equation 56) or with the prime on <math>~X</math> (HRW66 equation 57), both of which denote differentiation with respect to the radial coordinate.

With these modifications, the HRW66 radial pulsation equation becomes,

<math>~0</math>

<math>~=</math>

<math> ~\frac{d^2 X}{dx^2} + \biggl[\frac{4 - (n+1)V }{x}\biggr]\frac{dX}{dx} - \frac{V}{\gamma x^2}\biggl[\frac{x^2 (s^')^2}{\theta V} + (3\gamma -4)(n+1) \biggr]X </math>

 

<math>~=</math>

<math> ~\frac{d^2 X}{dx^2} + \biggl[\frac{4 - (n+1)V }{x}\biggr]\frac{dX}{dx} + \biggl[-\frac{(s^')^2 }{\gamma \theta } - \biggl(3 -\frac{4}{\gamma}\biggr)\frac{(n+1)V}{x^2} \biggr]X \, . </math>

The correspondence with our derived expression is complete, assuming that,

<math>~(s^')^2</math>

<math>~=</math>

<math>~-\omega^2 \biggl(\frac{a_n^2 \rho_c }{P_c} \biggr) \theta_c</math>

 

<math>~=</math>

<math>~-\omega^2 \biggl[\frac{n+1 }{4\pi G \rho_c} \biggr] \, .</math>

As has been explained in the above "ASIDE," this is exactly the factor that HRW66 use to normalize their eigenfrequency, <math>~s</math>, and make it dimensionless <math>~(s^')</math>. It is clear, as well, that HRW66 have adopted a sign convention for the square of their eigenfrequency that is the opposite of the sign convention that we have adopted for <math>~\omega^2</math>. That is, it is clear that,

<math>~s^2 ~~\leftrightarrow~~ - \omega^2 \, .</math>

Boundary Conditions

As we have pointed out in the context of a general discussion of boundary conditions associated with the adiabatic wave equation, the eigenfunction, <math>~x</math>, will be suitably well behaved at the center of the configuration if,

<math>~\frac{dx}{dr_0} = 0</math>        at         <math>~r_0 = 0 \, ,</math>

which, in the context of our present discussion of polytropic configurations, leads to the inner boundary condition,

<math>~\frac{dx}{d\xi} = 0</math>        at         <math>~\xi = 0 \, .</math>

This is precisely the inner boundary condition specified by HRW66 — see their equation (57), which has been reproduced in the above excerpt from HWR66.


As we have also shown in the context of this separate, general discussion of boundary conditions associated with the adiabatic wave equation, the pressure fluctuation will be finite at the surface — even if the equilibrium pressure and/or the pressure scale height go to zero at the surface — if the radial eigenfunction, <math>~x</math>, obeys the relation,

<math>~r_0 \frac{dx}{dr_0}</math>

<math>~=</math>

<math>~\biggl( 4 - 3\gamma_g + \frac{\omega^2 R^3}{GM_\mathrm{tot}}\biggr) \frac{x}{\gamma_g}</math>        at         <math>~r_0 = R \, .</math>

Or, given that, in polytropic configurations, <math>~r_0 = a_n\xi</math>,

<math>~\xi \frac{dx}{d\xi}</math>

<math>~=</math>

<math>~\frac{x}{\gamma_g} \biggl[ 4 - 3\gamma_g + \frac{\omega^2 (a_n \xi_1)^3}{GM_\mathrm{tot}}\biggr] </math>        at         <math>~\xi = \xi_1 \, ,</math>

where, the subscript "1" denotes equilibrium, surface values. As can be deduced from our above summary of the properties of polytropic configurations,

<math>~GM_\mathrm{tot}</math>

<math>~=</math>

<math>~4\pi G a_n^3 \rho_c (-\xi_1^2 \theta_1^') \, .</math>

Hence, for spherically symmetric polytropic configurations, the surface boundary condition becomes,

<math>~\frac{dx}{d\xi}</math>

<math>~=</math>

<math>~\frac{x}{\gamma_g \xi} \biggl[ 4 - 3\gamma_g + \omega^2 \biggl( \frac{1}{4\pi G \rho_c } \biggr) \frac{\xi}{(-\theta^')}\biggr] </math>         at         <math>~\xi = \xi_1 \, ,</math>

<math>~\Rightarrow ~~~~~(n+1)\frac{dx}{d\xi}</math>

<math>~=</math>

<math>~\frac{x}{\gamma_g \xi} \biggl[ (n+1)(4 - 3\gamma_g) + \omega^2 \biggl( \frac{1+n}{4\pi G \rho_c } \biggr) \frac{\xi}{(-\theta^')}\biggr] </math>

 

<math>~=</math>

<math>~-\frac{x}{\gamma_g \xi} \biggl[ (n+1)(3\gamma_g-4) - \omega^2 \biggl( \frac{1+n}{4\pi G \rho_c } \biggr) \frac{\xi}{(-\theta^')}\biggr] </math>         at         <math>~\xi = \xi_1 \, .</math>

Adopting notation used by HRW66, specifically, as demonstrated above,

<math>~-\omega^2 \biggl( \frac{1+n}{4\pi G \rho_c } \biggr) \rightarrow (s^')^2 \, , </math>

and, from equation (50) of HRW66,

<math>~-\theta^' \rightarrow q </math>         at         <math>~\xi = \xi_1 \, ,</math>

this outer boundary condition becomes,

<math>~(n+1)\frac{dx}{d\xi}</math>

<math>~=</math>

<math>~-\frac{x}{\gamma_g \xi} \biggl[ (n+1)(3\gamma_g-4) + \frac{\xi (s^')^2}{q}\biggr] </math>         at         <math>~\xi = \xi_1 \, .</math>

With the exception of the leading negative sign on the right-hand side, this expression is identical to the outer boundary condition identified by equation (58) of HRW66 — see the excerpt reproduced above.

Overview

The eigenvector associated with radial oscillations in isolated polytropes has been determined numerically and the results have been presented in a variety of key publications:

Tables

Quantitative Information Regarding Eigenvectors of Oscillating Polytropes

<math>~(\Gamma_1 = 5/3)</math>

<math>~n</math>

<math>~\frac{\rho_c}{\bar\rho}</math>

Excerpts from Table 1 of

Hurley, Roberts, & Wright (1966)

<math>~s^2 (n+1)/(4\pi G\rho_c)</math>

Excerpts from Table 3 of

J. P. Cox (1974)

<math>~\sigma_0^2 R^3/(GM)</math>

<math>\frac{(n+1) *\mathrm{Cox74}}{3 *\mathrm{HRW66}} \cdot \frac{\bar\rho}{\rho_c}</math>

<math>~0</math>

<math>~1</math>

<math>~1/3</math>

<math>~1</math>

<math>~1</math>

<math>~1</math>

<math>~3.30</math>

<math>~0.38331</math>

<math>~1.892</math>

<math>~0.997</math>

<math>~1.5</math>

<math>~5.99</math>

<math>~0.37640</math>

<math>~2.712</math>

<math>~1.002</math>

<math>~2</math>

<math>~11.4</math>

<math>~0.35087</math>

<math>~4.00</math>

<math>~1.000</math>

<math>~3</math>

<math>~54.2</math>

<math>~0.22774</math>

<math>~9.261</math>

<math>~1.000</math>

<math>~3.5</math>

<math>~153</math>

<math>~0.12404</math>

<math>~12.69</math>

<math>~1.003</math>

<math>~4.0</math>

<math>~632</math>

<math>~0.04056</math>

<math>~15.38</math>

<math>~1.000</math>


Numerical Integration from the Center, Outward

Here we show how a relatively simple, finite-difference algorithm can be developed to numerically integrate the governing LAWE from the center of a polytropic configuration, outward to its surface.

Drawing from our above discussion, the LAWE for any polytrope of index, <math>~n</math>, may be written as,

<math>~0 </math>

<math>~=</math>

<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4 - (n+1)V(\xi)}{\xi} \biggr] \frac{dx}{d\xi} + \biggl[\omega^2 \biggl(\frac{a_n^2 \rho_c }{\gamma_g P_c} \biggr) \frac{\theta_c}{\theta} - \biggl(3-\frac{4}{\gamma_g}\biggr) \cdot \frac{(n+1)V(x)}{\xi^2} \biggr] x </math>

 

<math>~=</math>

<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi} \biggr)\biggr] \frac{dx}{d\xi} + \frac{(n+1)}{\theta} \biggl[ \frac{\sigma_c^2}{6\gamma_g} - \frac{\alpha}{\xi } \biggl(- \frac{d\theta}{d\xi} \biggr) \biggr] x </math>

where,

<math>~\sigma_c^2</math>

<math>~\equiv</math>

<math>~\frac{3\omega^2}{2\pi G\rho_c} \, .</math>

Following a parallel discussion, we begin by multiplying the LAWE through by <math>~\theta</math>, obtaining a 2nd-order ODE that is relevant at every individual coordinate location, <math>~\xi_i</math>, namely,

<math>~\theta_i {x_i}</math>

<math>~=</math>

<math>~- \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \frac{x_i'}{\xi_i} - (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g} - \frac{\alpha}{\xi_i } (- \theta^')_i\biggr] x_i </math>

Now, using the general finite-difference approach described separately, we make the substitutions,

<math>~x_i'</math>

<math>~\approx</math>

<math>~ \frac{x_+ - x_-}{2 \Delta_\xi} \, ; </math>

and,

<math>~ x_i </math>

<math>~\approx</math>

<math>~\frac{x_+ - 2x_i + x_-}{\Delta_\xi^2} \, ,</math>

which will provide an approximate expression for <math>~x_+ \equiv x_{i+1}</math>, given the values of <math>~x_- \equiv x_{i-1}</math> and <math>~x_i</math>. Specifically, if the center of the configuration is denoted by the grid index, <math>~i=1</math>, then for zones, <math>~i = 3 \rightarrow N</math>,

<math>~\theta_i \biggl[ \frac{x_+ - 2x_i + x_-}{\Delta_\xi^2} \biggr]</math>

<math>~=</math>

<math>~- \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{x_+ - x_-}{2 \xi_i \Delta_\xi} \biggr] - (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g} - \frac{\alpha}{\xi_i } (- \theta^')_i\biggr] x_i </math>

<math>~\Rightarrow ~~~ \theta_i \biggl[ \frac{x_+ }{\Delta_\xi^2} \biggr] + \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{x_+ }{2 \xi_i\Delta_\xi} \biggr]</math>

<math>~=</math>

<math>~ -\theta_i \biggl[ \frac{- 2x_i + x_-}{\Delta_\xi^2} \biggr] - \biggl[4\theta_i - (n+1)\xi_i (- \theta^')_i\biggr] \biggl[ \frac{- x_-}{2 \xi_i \Delta_\xi} \biggr] - (n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g} - \frac{\alpha}{\xi_i } (- \theta^')_i\biggr] x_i </math>

<math>~\Rightarrow ~~~ x_+ \biggl[2\theta_i +\frac{4\Delta_\xi \theta_i}{\xi_i} - \Delta_\xi (n+1)(- \theta^')_i\biggr] </math>

<math>~=</math>

<math>~ x_- \biggl[\frac{4\Delta_\xi \theta_i}{\xi_i} - \Delta_\xi (n+1)(- \theta^')_i - 2\theta_i\biggr] + x_i\biggl\{4\theta_i - 2\Delta_\xi^2(n+1)\biggl[ \frac{\sigma_c^2}{6\gamma_g} - \frac{\alpha}{\xi_i } (- \theta^')_i\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ x_- \biggl[\frac{4\Delta_\xi \theta_i}{\xi_i} - \Delta_\xi (n+1)(- \theta^')_i - 2\theta_i\biggr] + x_i\biggl\{4\theta_i - \frac{\Delta_\xi^2(n+1)}{3}\biggl[ \frac{\sigma_c^2}{\gamma_g} - 2\alpha \biggl(- \frac{3\theta^'}{\xi}\biggr)_i\biggr] \biggr\} \, .</math>

In order to kick-start the integration, we will set the displacement function value to <math>~x_1 = 1</math> at the center of the configuration <math>~(\xi_1 = 0)</math>, then we will draw on the derived power-series expression to determine the value of the displacement function at the first radial grid line, <math>~\xi_2 = \Delta_\xi</math>, away from the center. Specifically, we will set,

<math>~ x_2 </math>

<math>~=</math>

<math>~ x_1 \biggl[ 1 - \frac{(n+1) \mathfrak{F} \Delta_\xi^2}{60} \biggr] \, ,</math>

where,

<math>~ \mathfrak{F} </math>

<math>~\equiv</math>

<math>~ \biggl[ \frac{\sigma_c^2}{\gamma_g} - 2\alpha\biggr]\, .</math>

Related Discussions

  • In an accompanying Chapter within our "Ramblings" Appendix, we have played with the adiabatic wave equation for polytropes, examining its form when the primary perturbation variable is an enthalpy-like quantity, rather than the radial displacement of a spherical mass shell. This was done in an effort to mimic the approach that has been taken in studies of the stability of Papaloizou-Pringle tori.
  • <math>~n=3</math> … M. Schwarzschild (1941, ApJ, 94, 245), Overtone Pulsations of the Standard Model: This work is referenced in §38.3 of [KW94]. It contains an analysis of the radial modes of oscillation of <math>~n=3</math> polytropes, assuming various values of the adiabatic exponent.
  • <math>~n=\tfrac{3}{2}</math> … D. Lucas (1953, Bul. Soc. Roy. Sci. Liege, 25, 585) … Citation obtained from the Prasad & Gurm (1961) article.
  • <math>~n=1</math> … L. D. Chatterji (1951, Proc. Nat. Inst. Sci. [India], 17, 467) … Citation obtained from the Prasad & Gurm (1961) article.


Whitworth's (1981) Isothermal Free-Energy Surface

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Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation