Editing SSC/Stability/n1PolytropeLAWE (section)

===Sixth Guess (n1)===
====Rationale====
From our [[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!