Editing SSC/Stability/n1PolytropeLAWE/Pt2 (section)

==Motivated by Yabushita's Discovery==

===Initial Exploration===
This subsection is being developed following our realization &#8212; see the [[SSC/Stability/InstabilityOnsetOverview#Overview:_Marginally_Unstable_Pressure-Truncated_Configurations|accompanying overview]] &#8212; that the eigenfunction ''is'' known analytically for marginally unstable, pressure-truncated configurations having <math>~3 \le n \le \infty</math>.  Specifically, from the work of [http://adsabs.harvard.edu/abs/1975MNRAS.172..441Y Yabushita (1975)] we have the following,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Exact Solution to the Isothermal LAWE</b></font></td>
</tr>
<tr>
  <td align="right">
<math>~\sigma_c^2 = 0</math>
  </td>
  <td align="center">
&nbsp;and &nbsp;
  </td>
  <td align="left">
<math>~x = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
  </td>
</tr>
</table>
</div>
And from our own recent work, we have discovered the following,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Precise Solution to the Polytropic LAWE</b></font></td>
</tr>
<tr>
  <td align="right">
<math>~\sigma_c^2 = 0</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~x_P \equiv \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] </math>
  </td>
</tr>
</table>
</div>
if the adiabatic exponent is assigned the value, <math>~\gamma_g = (n+1)/n</math>, in which case the parameter, <math>~\alpha = (3-n)/(n+1)</math>.  Using this polytropic displacement function as a guide, let's try for the case of <math>~n=1</math>, an expression of 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>~A - B\biggl[ \biggl( \frac{1}{\xi \theta}\biggr) \frac{d\theta}{d\xi} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A - B \biggl[ \biggl( \frac{1}{\sin\xi}\biggr) \frac{d}{d\xi} \biggl( \frac{\sin\xi}{\xi} \biggr)\biggr]</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{B}{\xi^3} \biggl\{
2 -\xi^2 - \frac{\xi\cos\xi}{\sin\xi} 
- \frac{\xi^2 \cos^2\xi}{\sin^2\xi} 
\biggr\} 
\, ,
</math>
  </td>
</tr>
</table>
</div>


<table border="1" cellpadding="8" align="center" width="85%"><tr><td align="left">
What if, instead, we try the more generalized form,
<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 + \frac{B}{(\lambda \xi)^2} \biggl[ 1-\frac{\lambda \xi \cos(\lambda \xi)}{\sin(\lambda \xi)}  \biggr] \, .</math>
  </td>
</tr>
</table>

Then we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\lambda B} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{ (\lambda \xi)^3} \biggl\{
2 - (\lambda \xi)^2 - \frac{\lambda \xi\cos(\lambda \xi)}{\sin(\lambda \xi)} 
- \frac{(\lambda \xi)^2 \cos^2(\lambda\xi)}{\sin^2(\lambda\xi)} 
\biggr\} 
\, ,
</math>
  </td>
</tr>
</table>
Probably this also means,

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

<tr>
  <td align="right">
<math>~\frac{1}{\lambda^2 B} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(\lambda\xi)^4} \biggl\{ 6 - 2 (\lambda \xi)^2 
- \frac{2 \lambda \xi\cos(\lambda\xi)}{\sin(\lambda\xi)} 
- \frac{2(\lambda\xi)^2 \cos^2(\lambda\xi)}{\sin^2(\lambda\xi)} 
- \frac{2(\lambda\xi)^3 \cos(\lambda\xi)}{\sin(\lambda\xi)} 
- \frac{2(\lambda\xi)^3 \cos^3(\lambda\xi)}{\sin^3(\lambda\xi)} \biggr\} \, .
</math>
  </td>
</tr>
</table>


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


Let's check against the [[SSC/Stability/Isothermal#Derivation_of_Polytropic_Displacement_Function|more general derivation]], which gives after recognizing that, <math>~B \leftrightarrow (3-n)/(n-1)</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B}{\xi^3} \biggl\{ \xi^2 
+ \biggl[ \frac{\xi \cos\xi}{\sin\xi} - 2 \biggr] 
+ \biggl(\frac{\xi \cos\xi}{\sin\xi}\biggr)^2  \biggr\} 
\, .
</math>
  </td>
</tr>
</table>
</div>
This matches the preceding, direct derivation.

Also,

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\biggl[ -2\xi - \frac{2\cos\xi}{\sin\xi} + \frac{2\xi\sin\xi}{\sin\xi} + \frac{2\xi\cos^2\xi}{\sin^2\xi}\biggr]
\biggr\} 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{B}{\xi^4} \biggl\{
\biggl[ - \frac{3\xi^2 \cos^2\xi}{\sin^2\xi} + \frac{3\xi \cos\xi}{\sin\xi} \biggr]
+\biggl[ 6 - 3\xi^2 - \frac{6\xi\cos\xi}{\sin\xi} \biggr]
+ \biggl[ - \frac{\xi^3 \cos\xi}{\sin\xi} + \xi^2 \biggr]
+ \biggl[ - \frac{2\xi^3 \cos^3\xi}{\sin^3\xi} + \frac{2\xi^2 \cos^2\xi}{\sin^2\xi} \biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B}{\xi^4} \biggl\{ 6 - 2\xi^2 
- \frac{2\xi\cos\xi}{\sin\xi} 
- \frac{2\xi^2 \cos^2\xi}{\sin^2\xi} 
- \frac{2\xi^3 \cos\xi}{\sin\xi} 
- \frac{2\xi^3 \cos^3\xi}{\sin^3\xi} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
Let's also check this against the [[SSC/Stability/Isothermal#Derivation_of_Polytropic_Displacement_Function|more general derivation]], which gives after again recognizing that, <math>~B \leftrightarrow (3-n)/(n-1)</math>, 

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

<tr>
  <td align="right">
<math>\frac{d^2 x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ \frac{4}{\xi^2} + \frac{2n(\theta^')}{\xi \theta} 
+ \frac{12\theta^' }{\xi^3 \theta^{n}}+ \frac{8n(\theta^')^2}{\xi^2 \theta^{n+1}}  
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-B
\biggl\{ \frac{4}{\xi^2} + \frac{2}{\xi \theta} \biggl[ \frac{\sin\xi}{\xi^2}\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr]
+ \frac{12}{\xi^3 \theta}\biggl[ \frac{\sin\xi}{\xi^2}\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr]
+ \frac{8 }{\xi^2 \theta^{2}}  \biggl[ \frac{\sin\xi}{\xi^2}\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr]^2
+ \frac{2 }{\xi \theta^{3}} \biggl[ \frac{\sin\xi}{\xi^2}\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr]^3\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{B}{\xi^4}
\biggl\{ 4\xi^2 + 2\xi^2 \biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)
+ 12\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)
+ 8 \biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)^2
+ 2\biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)^3\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{2B}{\xi^4}
\biggl\{ 2\xi^2 + \frac{\xi^3\cos\xi}{\sin\xi} - \xi^2+ \frac{6\xi\cos\xi}{\sin\xi} - 6
+ \frac{4\xi^2\cos^2\xi}{\sin^2\xi} - \frac{8\xi\cos\xi}{\sin\xi} + 4
+ \biggl(\frac{\xi^2\cos^2\xi}{\sin^2\xi} - \frac{2\xi\cos\xi}{\sin\xi} + 1\biggr) \biggl(\frac{\xi\cos\xi}{\sin\xi} - 1\biggr)\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{2B}{\xi^4}
\biggl\{-2 + \xi^2 - \frac{2\xi\cos\xi}{\sin\xi} + \frac{\xi^3\cos\xi}{\sin\xi}  
+ \frac{4\xi^2\cos^2\xi}{\sin^2\xi}  
- \biggl(\frac{\xi^2\cos^2\xi}{\sin^2\xi} - \frac{2\xi\cos\xi}{\sin\xi} + 1\biggr)
+ \frac{\xi^3\cos^3\xi}{\sin^3\xi} - \frac{2\xi^2\cos^2\xi}{\sin^2\xi} + \frac{\xi\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{2B}{\xi^4}
\biggl\{-3 + \xi^2  + \frac{\xi\cos\xi}{\sin\xi} + \frac{\xi^3\cos\xi}{\sin\xi}  
+ \frac{\xi^2\cos^2\xi}{\sin^2\xi}  
+ \frac{\xi^3\cos^3\xi}{\sin^3\xi}   
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B}{\xi^4}
\biggl\{6 -2 \xi^2  - \frac{2\xi\cos\xi}{\sin\xi} 
- \frac{2\xi^2\cos^2\xi}{\sin^2\xi}  
- \frac{2\xi^3\cos\xi}{\sin\xi}  
- \frac{2\xi^3\cos^3\xi}{\sin^3\xi}   
\biggr\} 
\, .
</math>
  </td>
</tr>
</table>
</div>
A cross-check with the first attempt to derive this second derivative expression initially unveiled a couple of coefficient errors.  These have now been corrected and both expressions agree.

===Succinct Demonstration===
Given that, for <math>~n=1</math>, we should set <math>~\gamma_\mathrm{g} = (n+1)/n = 2 \Rightarrow \alpha = (3-4/\gamma_\mathrm{g}) = +1</math>, and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q \equiv - \frac{d\ln\theta}{d\ln\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\xi^2}{\sin\xi} \cdot \frac{d}{d\xi}\biggl[ \frac{\sin\xi}{\xi}\biggr]
=
1 - \xi \cot\xi \, .
</math>
  </td>
</tr>
</table>
</div>
If we then employ the displacement function,
<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 + \frac{B}{\xi^2} \biggl[ 1 - \xi \cot\xi \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>

the LAWE becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LAWE
  </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} + 
(n+1)\biggl[ \biggl(\frac{\sigma_c^2}{6\gamma_g } \biggr) \frac{\xi^2}{\theta} 
-\alpha Q\biggr] \frac{ x}{\xi^2} </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[4 - 2Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 
\biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi^3}{\sin\xi} 
- 2Q\biggr] \frac{ x}{\xi^2} </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[2 + \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} 
+ \biggl[- 2 +  \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \frac{ x}{\xi^2} 
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2B}{\xi^4}
\biggl\{3 - \xi^2  - \frac{\xi\cos\xi}{\sin\xi} 
- \biggl(\frac{\xi\cos\xi}{\sin\xi} \biggr)^2 
- \frac{\xi^3\cos\xi}{\sin\xi}  
- \biggl( \frac{\xi\cos\xi}{\sin\xi} \biggr)^3  
\biggr\} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[- 2 +  \frac{2\xi\cos\xi}{\sin\xi}  \biggr] \biggl[ \frac{A}{\xi^2} + \frac{B}{\xi^4} \biggl( 1-\frac{\xi \cos\xi}{\sin\xi}  \biggr)\biggr]
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2B}{\xi^4}
\biggl\{3 - \xi^2  - \frac{\xi\cos\xi}{\sin\xi} 
- \biggl(\frac{\xi\cos\xi}{\sin\xi} \biggr)^2 
- \frac{\xi^3\cos\xi}{\sin\xi}  
- \biggl( \frac{\xi\cos\xi}{\sin\xi} \biggr)^3  
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{2B}{\xi^4} \biggl[ 1 -  \frac{2\xi\cos\xi}{\sin\xi} + \biggl(\frac{\xi \cos\xi}{\sin\xi}  \biggr)^2 \biggr]
+ \frac{2A}{\xi^2}\biggl[\frac{\xi\cos\xi}{\sin\xi}  -1\biggr] 
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2A}{\xi^2}\biggl[\frac{\xi\cos\xi}{\sin\xi}  -1\biggr] 
+ \biggl[ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggr] x 
</math>
  </td>
</tr>
</table>
</div>
Pretty amazing degree of cancelation!  So the above-hypothesized displacement function ''does'' satisfy the <math>~n=1</math>, polytropic LAWE &#8212; for any value of the coefficient, <math>~B</math> &#8212; if we set <math>~A = 0</math> and <math>~\sigma_c^2=0</math>.  If we set <math>~B = 3</math>, the function will be normalized such that it goes to unity at the center.   In summary, then, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_P\biggr|_{n=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{\xi^2} \biggl[ 1 - \xi \cot\xi \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

<!--
Let's play with this a bit more to see if we can uncover a displacement function that works for nonzero values of <math>~\omega_c^2</math>.  Leaving both <math>~A</math> and <math>~B</math> unspecified for the time being, we have,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2A}{\xi^2}\biggl[\frac{\xi\cos\xi}{\sin\xi}  -1\biggr] 
+ \biggl(\frac{\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} \biggl[ A + \frac{B}{\xi^2} \biggl( 1-\frac{\xi \cos\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(\frac{A\sigma_c^2}{6 } \biggr) \frac{\xi}{\sin\xi} 
+ \biggl[ \biggl(\frac{B\sigma_c^2}{6 } \biggr) \frac{1}{\xi\sin\xi}  
- \frac{2A}{\xi^2} \biggr] \biggl( 1 - \frac{\xi\cos\xi}{\sin\xi}  \biggr) 
</math>
  </td>
</tr>
</table>
</div>
-->