Editing SSC/Stability/n1PolytropeLAWE/Pt2 (section)

==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 [[SSC/Stability/BiPolytrope00Details#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.