Editing SSC/VariationalPrinciple/Pt2 (section)

===For All Polytropic Indexes===

====Generalized Governing Integral Relation====
Given that the derivation just completed works for the special case of n = 5, let's generalize it to all polytropic indexes
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\int_0^R \sigma^2 \rho r^4 x^2 dr</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr
- \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr
 +3\Gamma_1 P_e R^3 x_\mathrm{surface}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{R^5 \rho_c}{R^3 P_c}\int_0^R \sigma^2 \biggl( \frac{\rho}{\rho_c}\biggr) \biggl(\frac{r}{R}\biggr)^4 x^2 \frac{dr}{R}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R \biggl(\frac{r}{R}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \biggl(\frac{P}{P_c}\biggr) \biggl[\frac{dx}{d(r/R)}\biggr]^2 \frac{dr}{R}
-  \int_0^R \biggl[3\biggl(\frac{n+1}{n}\biggr)  - 4\biggr] \biggl(\frac{r}{R}\biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \frac{dr}{R}
 +3\biggl(\frac{n+1}{n}\biggr) \biggl( \frac{P_e}{P_c}\biggr)  x_\mathrm{surface}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{R^2 \rho_c}{P_c} \int_0^{\tilde\xi} \sigma^2 \theta^n \biggl(\frac{\xi}{\tilde\xi}\biggr)^4 x^2 \frac{d\xi}{\tilde\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \biggl(\frac{\xi}{{\tilde\xi}}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \theta^{n+1}  \biggl[\frac{dx}{d(\xi/\tilde\xi)}\biggr]^2 \frac{d\xi}{\tilde\xi}
~+  \int_0^{\tilde\xi} \biggl(\frac{n-3}{n}\biggr) \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 x^2 \biggl[ \frac{d\theta^{n+1}}{d(\xi/\tilde\xi)} \biggr] \frac{d\xi}{\tilde\xi}
~+~3\biggl(\frac{n+1}{n}\biggr) {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{n R^2\rho_c}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \sigma^2 \theta^n \xi^4 x^2 d\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^4 \theta^{n+1}  \biggl[\frac{dx}{d\xi}\biggr]^2 d\xi
~+  \int_0^{\tilde\xi} (n-3) \xi^3 \theta^n x^2 \biggl[ \frac{d\theta}{d\xi} \biggr] d\xi
~+~3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{n R^2 G \rho_c^2}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 d\xi
~+  \int_0^{\tilde\xi} (n-3) \xi^2 \theta^{n+1} x^2 \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] d\xi
~+~3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 
+ \int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl\{ \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 + (n-3)  \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] \biggr\} d\xi
</math>
  </td>
</tr>
</table>
</div>

For additional clarification, let's rewrite the leading coefficient on the lefthand-side of this expression.
<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{n R^2 G \rho_c^2}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n}{(n+1)} \biggr] \biggl[ \frac{G R_\mathrm{norm}^2}{P_\mathrm{norm}} \biggr] 
\biggl(\frac{R}{R_\mathrm{norm}^2}\biggr) \biggl( \frac{\rho_c}{ {\bar\rho}}\biggr)^2
\biggl[ \frac{3M}{4\pi R^3}\biggr]^2
\biggl(\frac{P_\mathrm{norm}}{P_e} \biggr) 
\biggl(\frac{P_e}{P_c} \biggr) \biggl[ \frac{1}{{\tilde\xi}^2} \biggr]
\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n}{(n+1)} \biggr] \biggl[ \frac{G M_\mathrm{tot}^2}{P_\mathrm{norm}R_\mathrm{norm}^4} \biggr] 
\biggl(\frac{R_\mathrm{norm}}{R}\biggr)^4 \biggl( \frac{\rho_c}{ {\bar\rho}}\biggr)^2
\biggl[ \biggl(\frac{3}{4\pi}\biggr)\frac{M}{M_\mathrm{tot}}\biggr]^2
\biggl(\frac{P_\mathrm{norm}}{P_e} \biggr) 
\biggl(\frac{P_e}{P_c} \biggr) \biggl[ \frac{1}{{\tilde\xi}^2} \biggr]
\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n}{(n+1)} \biggr]  
\biggl(\frac{P_\mathrm{norm}}{P_e} \biggr) 
\biggl(\frac{R_\mathrm{norm}}{R}\biggr)^4 \biggl( - \frac{\tilde\xi}{3 {\tilde\theta}^'}\biggr)^2
\biggl[ \biggl(\frac{3}{4\pi}\biggr)\frac{M}{M_\mathrm{tot}}\biggr]^2
\biggl[ \frac{{\tilde\theta}^{n+1}}{{\tilde\xi}^2} \biggr]
\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>
</table>
</div>

Now, from an [[StabilityVariationalPrinciple#Test_Virial_Equilibrium_Condition|accompanying discussion]], we know that, in equilibrium,
<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} 
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ 
\biggl[(n+1)^{-n} ( 4\pi )\biggr]^{1/(n-3)} \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(n-1)/(n-3)}
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~\frac{P_e}{P_\mathrm{norm}} 
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl[(n+1)^{3} ( 4\pi )^{-1} \biggr]^{(n+1)/(n-3)}\biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2(n+1)/(n-3)}
\tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)} 
\, ,
</math>
  </td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ 
\biggl\{ \biggl[(n+1)^{3} ( 4\pi )^{-1} \biggr]^{(n+1)}\biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2(n+1)}
\tilde\theta_n^{(n+1)(n-3)}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)} \biggr\}^{1/(n-3)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\times
\biggl\{\biggl[(n+1)^{-n} ( 4\pi )\biggr] \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(n-1)}
\tilde\xi^{(n-3)} ( -\tilde\xi^2 \tilde\theta' )^{(1-n)} \biggr\}^{4/(n-3)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \tilde\xi^{4} \tilde\theta_n^{(n+1)}
\biggl\{ (n+1)^{3(n+1)} ( 4\pi )^{(-n-1)}  \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2n-2}
( -\tilde\xi^2 \tilde\theta' )^{2n+2} (n+1)^{-4n} ( 4\pi )^4 \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(4n-4)}
( -\tilde\xi^2 \tilde\theta' )^{(4-4n)} 
\biggr\}^{1/(n-3)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \tilde\xi^{4} \tilde\theta_n^{(n+1)}
\biggl\{ (n+1)^{(3-n)} ( 4\pi )^{(3-n)}  \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{2(n-3)}
( -\tilde\xi^2 \tilde\theta' )^{2(3-n)} 
\biggr\}^{1/(n-3)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ 
(n+1)^{-1} ( 4\pi )^{(-1)}  \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{2} \tilde\xi^{4} \tilde\theta_n^{(n+1)}( -\tilde\xi^2 \tilde\theta' )^{-2} \, .
</math>
  </td>
</tr>
</table>
</div>
This means that, in equilibrium,
<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>~\biggl[ \frac{n}{(n+1)} \biggr]  
\biggl\{ (n+1) ( 4\pi )  \tilde\xi^{-4} \tilde\theta_n^{-(n+1)}( -\tilde\xi^2 \tilde\theta' )^{2} \biggr\}
\biggl( - \frac{\tilde\xi}{3 {\tilde\theta}^'}\biggr)^2
\biggl(\frac{3}{4\pi}\biggr)^2
\biggl[ \frac{{\tilde\theta}^{n+1}}{{\tilde\xi}^2} \biggr]
\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\int_0^{\tilde\xi} \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi \, .</math>
  </td>
</tr>
</table>
</div>

In summary, then, we have,

<div align="center" id="PolytropeRelation">
<table border="1" align="center" cellpadding="8">
<tr><td align="center">

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\int_0^{\tilde\xi} \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 
+ \int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl\{ \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 + (n-3)  \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] \biggr\} d\xi \, .
</math>
  </td>
</tr>
</table>

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

Perhaps this looks better if the terms are rearranged to give,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^2\theta^{n+1} x^2 \biggl\{ \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \frac{\xi^2}{\theta} 
- \biggl[ \biggl( \frac{d\ln x}{d\ln \xi}\biggr)^2 + (n-3)  \biggl( \frac{d\ln\theta}{d\ln\xi} \biggr) \biggr] \biggr\} d\xi \, .
</math>
  </td>
</tr>
</table>
</div>

====Plug in Known Marginally Unstable Solution====

As has been summarized in an [[SSC/Stability/InstabilityOnsetOverview#Marginally_Unstable_Pressure-Truncated_Gas_Clouds|accompanying discussion]], we have found that, for marginally unstable pressure-truncated polytropic configurations, the eigenvector associated with the fundamental mode of radial oscillation is prescribed analytically by the following eigenfrequency-eigenfunction pair: 
<div align="center">
<table border="0" cellpadding="5" align="center">
<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 = \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>
This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3-n}{n-1}\biggr) \biggl[
\frac{1}{\xi } 
+ \frac{3\theta^'}{\xi^2 \theta^{n}}
+ \frac{n (\theta^')^2}{\xi \theta^{n+1}}
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, also,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[1  
+ \frac{3\theta^'}{\xi \theta^{n}}
+ \frac{n (\theta^')^2}{\theta^{n+1}}
\biggr] \biggl[\biggl(\frac{n-1}{n-3}\biggr)  + \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^{-1} \, .
</math>
  </td>
</tr>
</table>
</div>

Rather, let's try:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\xi^2 x^2 \biggl[ \biggl( \frac{d\ln x}{d\ln \xi}\biggr)^2 + (n-3)  \biggl( \frac{d\ln\theta}{d\ln\xi} \biggr) \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2 \xi^2 \biggl( \frac{\xi}{x}\cdot \frac{dx}{d\xi}\biggr)^2 + (n-3) x^2 \xi^2 \biggl( \frac{\xi}{\theta} \cdot \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>~
\xi^4 \biggl\{ \frac{dx}{d\xi}\biggr\}^2 + (n-3) \biggl[ \frac{\xi^3 \theta^'}{\theta} \biggr] x^2
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^2 (n-3) \biggl[ \frac{3}{2n} \biggr]^2\biggl\{
(n-3) \biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 
+\xi \biggl( \frac{ \theta^'}{\theta} \biggr) 
\biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2
\biggr\}
</math>
  </td>
</tr>
</table>
</div>
Hence, after setting <math>~\sigma^2 = 0</math>, the [[#PolytropeRelation|above rearranged integral relation]] becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
- \frac{2^2 n^2}{3(n-3)} \biggl[ {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^2 \theta^{n+1} \biggl\{
(n-3) \biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 
+\xi \biggl( \frac{ \theta^'}{\theta} \biggr) 
\biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2
\biggr\} d\xi 
</math>
  </td>
</tr>
</table>
</div>


<table border="1" align="center" width="80%" cellpadding="5">
<tr><td align="left">
Let's check to see whether the terms in this last expression balance out when we plug in the functions that are appropriate for the marginally unstable, n = 5 configuration, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="left">
<math>~\theta_5 = \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2}</math>,
  </td>
  <td align="center">
&nbsp; &nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;
  </td>
  <td align="left">
<math>~\frac{d\theta_5}{d\xi} = - \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}</math>.
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
RHS Term 1
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(n-3) \int_0^{\tilde\xi} \xi^2 \theta^{n+1} 
\biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \int_0^{\tilde\xi} \xi^2 \biggl[ \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2} \biggr]^{6} \biggl\{ 
1 - \biggl[ \xi \biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2} \biggr]\frac{1}{\xi }\biggl(\frac{3+\xi^2}{3}\biggr)^{5 / 2} 
+5  \biggl[ \frac{\xi^2}{3^2}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3} \biggr] \biggl[ \biggl(\frac{3+\xi^2}{3}\biggr)^{3} \biggr]
\biggr\}^2 d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \int_0^{\tilde\xi} \frac{3^3 \xi^2}{(3+\xi^2)^3}  \biggl\{ 
1 -  \biggl(\frac{3+\xi^2}{3}\biggr) 
+ \frac{5\xi^2}{3^2}  
\biggr\}^2 d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^3}{3} \int_0^{\tilde\xi} \frac{\xi^6 ~d\xi}{(3+\xi^2)^3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^3}{3}  \biggl[ 
\frac{27\xi}{8(3+\xi^2)} - \frac{9\xi}{4(3+\xi^2)^2} + \xi - \biggl(\frac{3^{3/2}\cdot 5}{2^3} \biggr)\tan^{-1}\biggl(\frac{\xi}{3^{1 / 2}}\biggr)
\biggr]_0^{3} = \frac{2^3}{3}  \biggl[ \frac{3^5}{2^6} - \frac{3^{1 / 2}\cdot 5\pi}{2^3} \biggr] \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
RHS Term 2
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^3 \theta^{n} \theta^'
\biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2 d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \int_0^{\tilde\xi} \xi^3 \biggl(\frac{3+\xi^2}{3}\biggr)^{-5 / 2} \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}
\biggl\{ 4 - 2\frac{1}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}\biggl(\frac{3+\xi^2}{3}\biggr)^{5/2}
\biggr\}^2 d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{3}\int_0^{\tilde\xi} \biggl(\frac{3\xi}{3+\xi^2}\biggr)^{4} 
\biggl\{ 4 - \frac{2}{3}\biggl(\frac{3+\xi^2}{3}\biggr)
\biggr\}^2 d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^3}{3}\int_0^{\tilde\xi} \frac{1}{2} \biggl(\frac{\xi}{3+\xi^2}\biggr)^{4} 
\biggl\{ 15-\xi^2
\biggr\}^2 d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{2^3}{3}  \biggl[ 
\frac{123\xi}{8(3+\xi^2)} - \frac{243\xi}{4(3+\xi^2)^2} + \frac{162\xi}{2(3+\xi^2)^3} + \frac{\xi}{2} - \biggl(\frac{3^{3/2}\cdot 5}{2^3} \biggr)\tan^{-1}\biggl(\frac{\xi}{3^{1 / 2}}\biggr)
\biggr]_0^{3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^3}{3} \cdot \frac{5}{2^5}  \biggl[ 2^2\cdot 3^{1 / 2} \pi - 3^3\biggr] = -\frac{2^3}{3} \biggl[ \frac{3^3\cdot 5}{2^5} - \frac{3^{1 / 2} \cdot 5\pi}{2^3} \bigg] \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~</math> RHS Total
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^3}{3}  \biggl[ \frac{3^5}{2^6} - \frac{3^3\cdot 5}{2^5}  \biggr]
= \frac{3^2}{2^3}  \biggl[ 3^2 - 2\cdot 5  \biggr] = - \frac{3^2}{2^3}  \, .
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^2 5^2}{2\cdot 3} \biggl[ 3^3 \biggl(\frac{3}{3+3^2}\biggr)^{3} \frac{2^2}{5^2} \biggr]
=
- \frac{2^3 }{3} \biggl[\biggl(\frac{3}{2^2}\biggr)^{3} \biggr]
=
- \frac{3^2 }{2^3} \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, the LHS = RHS. &nbsp; <font size="+1" color="red"><b>Hooray!</b></font>
</td></tr>
</table>