Editing SSC/Stability/n1PolytropeLAWE (section)

===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 [[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, [[SSC/Stability/Polytropes#Second_Guess|above, in connection with our "second guess]]."]