Editing SSC/Stability/n3PolytropeLAWE (section)

=Analytic Inquiry=

<font color="red"><b>NOTE</b> (from J. E. Tohline in April, 2017):</font> &nbsp; The following subsections present some exploratory ideas that were pursued while I was searching for analytic solutions to the polytropic LAWE.  For the most part this material has been superseded by a [[SSC/Stability/InstabilityOnsetOverview#Polytropic|separate discussion]] in which we describe the desired analytic solution, which we discovered in March, 2017.

==Fundamental-Mode, Homentropic Oscillations==

The LAWE, [[#Background|presented above]], that is relevant to polytropic spheres, may be rewritten as,
<div align="center">
{{ Math/EQ_RadialPulsation02 }}
</div>

Now, if we assume that oscillations occur adiabatically with an adiabatic index that is consistent with the chosen polytropic index &#8212; that is to say,
<div align="center">
<math>~\gamma_g = \frac{n+1}{n}</math>&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; <math>~\alpha =  \frac{3-n}{n+1} \, ,</math>
</div>
in which case the configuration remains homentropic as it oscillates &#8212; and if we look only for a (marginally unstable) configuration that has <math>~\sigma_c^2 = 0</math>, then the relevant LAWE 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^2x}{d\xi^2} + \biggl[ 4 - (n+1) Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}  
- (3-n)  Q \frac{x}{\xi^2} \, .
</math>
  </td>
</tr>
</table>
</div>

==Specific case of n = 3 Polytropes==
===Homologous Collapse===

If we examine only an <math>~n=3</math> polytropic configuration, then the last term disappears.  This means that ''in this very special case'', a perfectly valid solution to the LAWE is <math>~x = \mathrm{constant}</math>.  This is presumably the eigenfunction that Schwarzschild deduced; the fundamental-mode "oscillations" are perfectly homologous. Given that the model is marginally unstable, an ensuing dynamical collapse will presumably begin in a perfectly homologous fashion.  This is precisely the type of "free-fall" collapse that was discussed and modeled by [[Apps/GoldreichWeber80#Homologous_Solution|Goldreich &amp; Weber (1980)]].

===Another Potential Option===
We have wondered whether, in this very special case, one or more additional fundamental-mode eigenfunction(s) might satisfy the governing LAWE.  Here is a relevant line of arguments, beginning with the LAWE for the n = 3 polytropic sphere.
<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} + 4( 1 +  Q ) \frac{1}{\xi} \cdot \frac{dx}{d\xi}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dy}{d\xi} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 4( 1 +  Q ) \frac{y}{\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{4}\cdot \frac{d\ln y}{d\ln\xi} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- ( 1 +  Q ) \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~y \equiv \frac{dx}{d\xi} \, .</math>
</div>

But, by definition, the function <math>~Q(\xi)</math> is a logarithmic derivative of the Lane-Emden function.  Hence, we also can write,

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

<tr>
  <td align="right">
<math>~-1 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{4}\cdot \frac{d\ln y}{d\ln\xi} + \frac{d\ln\theta}{d\ln\xi}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{d\ln (\theta y^{1/4})}{d\ln\xi} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~- d\ln\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ d\ln (\theta y^{1/4}) \, . </math>
  </td>
</tr>
</table>
</div>
Integrating this equation once gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ln(\theta y^{1/4}) + \ln\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ln (c_0)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \xi \theta y^{1/4} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c_0</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>~\biggl( \frac{c_0}{\xi\theta}\biggr)^4 \, .</math>
  </td>
</tr>
</table>
</div>

Referring to the [[Appendix/Ramblings/PowerSeriesExpressions#PolytropicLaneEmden|power-series expansion of the polytropic Lane-Emden function]], <math>~\theta(\xi)</math>, about the configuration's center, we see that the product, <math>~\xi\theta</math>, goes to zero as the first power of <math>~\xi</math>.  This means that the right-hand side of this last differential equation blows up at the center.  This, therefore, does not appear to provide a physically viable avenue by which to identify an alternative fundamental-mode eigenfunction.

==Play With Form of LAWE==
===Logarithmic Derivative Rewrite===

We have noticed that the LAWE that governs the eigenfunction associated with the fundamental mode of the marginally unstable model (FMMUM),
<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[ 4 - (n+1) Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}  
- (3-n)  Q \frac{x}{\xi^2} \, ,
</math>
  </td>
</tr>
</table>
</div>

may be rewritten entirely as an expression that relates the logarithmic derivatives of <math>~x, \xi,</math> and <math>~\theta</math>.  Multiplying through by <math>~\xi^2/x</math>, then drawing on a differential relation that has been [[SSC/Stability/BiPolytrope00Details#Idea_Involving_Logarithmic_Derivatives|derived in a separate context]], namely,
<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[  \frac{d\ln x}{d\ln \xi} -1 \biggr]\cdot \frac{d\ln x}{d\ln \xi} \, ,
</math>
 </td>
</tr>
</table>
</div>
this LAWE associated with the FMMUM 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}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]
+ \biggl[  \frac{d\ln x}{d\ln \xi} -1 \biggr]\cdot \frac{d\ln x}{d\ln \xi} + \biggl[ 4 - (n+1) Q \biggr] \frac{d\ln x}{d\ln\xi}  - (3-n)  Q 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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]
+ \frac{d\ln x}{d\ln \xi} \cdot \biggl[  \frac{d\ln x}{d\ln \xi} + 3 - (n+1) Q \biggr]   - (3-n)  Q 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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]
+ \frac{d\ln x}{d\ln \xi} \cdot \biggl\{  \frac{d\ln [x \xi^3 \theta^{(n+1)}] }{d\ln \xi} \biggr\}   + \frac{d\ln \theta^{(3-n) }}{d\ln\xi} \, .
</math>
  </td>
</tr>
</table>
</div>

I'm not sure if anyone else has previously appreciated that the "fundamental mode" polytropic LAWE can be written in this form.  I'm even less sure that this form sheds light on its solution.

Play a little more &hellip; &nbsp; Start by letting, <math>~A \equiv  [x \xi^3 \theta^{(n+1)}] \, ,</math> in which case 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}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]
+ \frac{d\ln x}{d\ln \xi} \cdot \biggl\{  \frac{1}{A} \cdot \frac{dA}{d\ln \xi} \biggr\}   + \frac{d\ln \theta^{(3-n) }}{d\ln\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{A} \cdot \frac{d}{d\ln\xi} \biggl[ A \cdot \frac{d\ln x}{d\ln \xi} \biggr]  + \frac{d\ln \theta^{(3-n) }}{d\ln\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{d}{d\xi} \biggl[ A \cdot \frac{d\ln x}{d\ln \xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{A}{\theta^{(3-n)}} \cdot \frac{d\theta^{(3-n) }}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{d}{d\xi} \biggl[ x \xi^3 \theta^{(n+1)} \cdot \frac{d\ln x}{d\ln \xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{x \xi^3 \theta^{(n+1)}}{\theta^{(3-n)}} \cdot \frac{d\theta^{(3-n) }}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{d}{d\xi} \biggl[ \xi^4 \theta^{(n+1)} \cdot \frac{dx}{d\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- x \xi^3 \theta^{2(n-1)} \cdot \frac{d\theta^{(3-n) }}{d\xi} \, .
</math>
  </td>
</tr>
</table>
</div>

===One Feeble Guess===

Now, what if, <math>~x \equiv [\xi^{-2}\theta^{-n}]</math> &nbsp; ?

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\xi^2 \theta^{(n+1)} \cdot \frac{d(\xi^{-2}\theta^{-n})}{d\xi} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d\theta}{d\xi} - \frac{1}{\xi^{2}\theta^{n}} \cdot \frac{d(\xi^2 \theta^{(n+1)} )}{d\xi} \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case, the LAWE becomes,


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

<tr>
  <td align="right">
<math>~
\frac{d}{d\xi} \biggl[ \xi^2 \frac{d\theta}{d\xi} - \frac{1}{\theta^{n}} \cdot \frac{d(\xi^2 \theta^{(n+1)} )}{d\xi}\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- [\xi^{-2}\theta^{-n}] \xi^3 \theta^{2(n-1)} \cdot \frac{d\theta^{(3-n) }}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{d}{d\xi} \biggl[ \xi^2 \frac{d\theta}{d\xi} \biggr] - \frac{d}{d\xi} \biggl[\frac{1}{\theta^{n}} \cdot \frac{d(\xi^2 \theta^{(n+1)} )}{d\xi}\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \xi \theta^{n-2} \cdot \frac{d\theta^{(3-n) }}{d\xi}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
-\xi^2\theta^n - \frac{d}{d\xi} \biggl[\frac{1}{\theta^{n}} \cdot \frac{d(\xi^2 \theta^{(n+1)} )}{d\xi}\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \xi  (3-n)\frac{d\theta }{d\xi} 
</math>
  </td>
</tr>
</table>
</div>

===Behavior for Known n=5 Solution===
We know that the [[SSC/Stability/n5PolytropeLAWE#Eureka_Moment|FMMUM for pressure-truncated, n = 5 polytropic configurations]] takes the form,
<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>~1- f_n(\xi) \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~f_5(\xi) = \frac{\xi^2}{15} \, .</math>
</div>
The governing LAWE therefore gives,

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

<tr>
  <td align="right">
<math>~
\frac{d}{d\xi} \biggl[ \xi^4 \theta^{(n+1)} \cdot \frac{dx}{d\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- x \xi^3 \theta^{2(n-1)} \cdot \frac{d\theta^{(3-n) }}{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- (3-n)x \xi^3 \theta^{n} \cdot \frac{d\theta }{d\xi} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
(3-n)(1-f_n) \xi^3 \theta^{n} \cdot \frac{d\theta }{d\xi} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\xi} \biggl[ \xi^4 \theta^{(n+1)} \cdot \frac{df_n}{d\xi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^4 \theta^{(n+1)} \cdot\frac{d^2f_n}{d\xi^2} 
+ \frac{df_n}{d\xi} \cdot \frac{d}{d\xi} \biggl[ \xi^4 \theta^{(n+1)} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^4 \theta^{(n+1)} \cdot\frac{d^2f_n}{d\xi^2} 
+ \frac{df_n}{d\xi} \cdot \biggl\{ 
4\xi^3 \theta^{(n+1)} + (n+1)\xi^4 \theta^n \frac{d\theta}{d\xi} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
(3-n)(1-f_n) \frac{d\theta }{d\xi} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi \theta \cdot\frac{d^2f_n}{d\xi^2} 
+ \frac{df_n}{d\xi} \cdot \biggl\{ 
4 \theta + (n+1)\xi \frac{d\theta}{d\xi} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{1}{\theta} \frac{d\theta }{d\xi} \biggl[ (3-n)(1-f_n) - (n+1)\xi \frac{df_n}{d\xi} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi \cdot\frac{d^2f_n}{d\xi^2} 
+ 4 \cdot \frac{df_n}{d\xi} \, .
</math>
  </td>
</tr>
</table>
</div>
Let's check to see whether the known <math>~f_5(\xi)</math> function properly satisfies this last ODE when n = 5.

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

<tr>
  <td align="right">
<math>~
-\xi (3+\xi^2)^{-1}  \biggl[ -2 \biggl(1-\frac{\xi^2}{15} \biggr) - \frac{12\xi^2}{15} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2\xi}{15} + \frac{8\xi}{15} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
2 + \frac{2\xi^2}{3} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{3} (3+\xi^2) \, ,
</math>&nbsp; &nbsp; &nbsp; Yes!
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\theta_5^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{3+\xi^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\xi^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl[ \frac{1}{\theta_5^2} - 1 \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
we could ''presume'' that, when defined in terms of <math>~\theta_5</math>, the defining function,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_5(\theta_5) = \frac{1}{n}\biggl[ \frac{1}{\theta_5^2} - 1 \biggr]</math>
  </td>
  <td align="center">
&nbsp; &nbsp;<math>~\Rightarrow~</math>&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~ 
\frac{df_5}{d\xi} = - \frac{2}{n\theta_5^3} \cdot \frac{d\theta_5}{d\xi}
</math>
  </td>
  <td align="center">
&nbsp; &nbsp;<math>~\Rightarrow~</math>&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~ 
\frac{d^2f_5}{d\xi^2} = - \frac{2}{n\theta_5^3} \cdot \frac{d^2\theta_5}{d\xi^2} + \frac{6}{n\theta_5^4} \cdot \biggl( \frac{d\theta_5}{d\xi} \biggr)^2 \, .
</math>
  </td>
</tr>
</table>
</div>
In this case, the governing 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>~
\xi \cdot\frac{d^2f_n}{d\xi^2} 
- \frac{(3-n)}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ 1-f_n \biggr]
+ \biggl[ 4 + \frac{(n+1)\xi}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggr] \biggl[ \frac{df_n}{d\xi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi \biggl\{ - \frac{2}{n\theta_5^3} \cdot \frac{d^2\theta_5}{d\xi^2} + \frac{6}{n\theta_5^4} \cdot \biggl( \frac{d\theta_5}{d\xi} \biggr)^2\biggr\}
- \biggl[ 4 + \frac{(n+1)\xi}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggr] \biggl[ \frac{2}{n\theta_5^3} \cdot \frac{d\theta_5}{d\xi}\biggr]
- \frac{(3-n)}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ \frac{(n + 1)\theta_5^2 -1  }{n\theta_5^2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{n\theta_5^3}\biggl\{
\biggl[ 2\xi \cdot \frac{d^2\theta_5}{d\xi^2} - \frac{6\xi}{\theta_5} \cdot \biggl( \frac{d\theta_5}{d\xi} \biggr)^2\biggr]
+ 2\biggl[ 4 + \frac{(n+1)\xi}{\theta} \biggl( \frac{d\theta }{d\xi} \biggr) \biggr]  \frac{d\theta_5}{d\xi}
+ (3-n) \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ (n + 1)\theta_5^2 -1 \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{n\theta_5^3}\biggl\{
2\xi \cdot \frac{d^2\theta_5}{d\xi^2} 
+ (n-2) \frac{2\xi}{\theta_5}\biggl( \frac{d\theta_5}{d\xi} \biggr)^2
+ \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ (3-n) (n + 1)\theta_5^2 + 5+ n  \biggr]   
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
Now, from the polytropic Lane-Emden equation, we also know that,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>
That is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2\xi \biggl(\frac{d\theta_5}{d\xi}\biggr) + \xi^2 \cdot \frac{d^2\theta_5}{d\xi^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \xi^2 \theta_5^n</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~  \frac{d^2\theta_5}{d\xi^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \theta_5^n - \frac{2}{\xi} \biggl(\frac{d\theta_5}{d\xi}\biggr)</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>~
(n-2) \frac{2\xi}{\theta_5}\biggl( \frac{d\theta_5}{d\xi} \biggr)^2
+ (n+1)\biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ (3-n) \theta_5^2 + 1 \biggr]   
- 2\xi \theta_5^n \, .
</math>
  </td>
</tr>
</table>
</div>
Again, let's check to see if the case of n=5 works &hellip;

<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>~
\biggl( \frac{d\theta_5}{d\xi} \biggr)^2
+ \frac{\theta_5}{\xi} \biggl( \frac{d\theta }{d\xi} \biggr) \biggl[ 1 - 2 \theta_5^2 \biggr]   
- \frac{\theta_5^{6}}{3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\xi^2(3+\xi^2)^{-3}
-3(3+\xi^2)^{-2}\biggl[ 1 - 6(3+\xi^2)^{-1} \biggr]   
- 3^2(3+\xi^2)^{-3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3+\xi^2)^{-3}\biggl\{
3\xi^2 -3\biggl[ (3+\xi^2)  -6 \biggr]  - 3^2 \biggr\} 
</math> &nbsp; &nbsp; Yes!
  </td>
</tr>
</table>
</div>