Editing SSC/Structure/BiPolytropes/AnalyzeStepFunction (section)

====Step 6====

<font color="red"><b>STEP 6:</b></font> &nbsp; Given that when a proper solution has been obtained,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{d^2r}{dt^2}</math>
  </td>
  <td align="center">
<math>\rightarrow</math>
  </td>
  <td align="left">
<math>
- \omega^2 \cdot {\delta r}_i
\, ,
</math>
  </td>
</tr>
</table>
at each radial shell we can determine what the value of <math>\omega^2</math> would be as a result of our <math>{\delta r}_i</math> ''guess'' by rewriting the 

<div align="center">
<span id="PGE:Euler"><font color="#770000">'''Euler + Poisson Equations'''</font></span><br />
<math>\frac{d^2 r}{dt^2} = - 4\pi r^2 \frac{dP}{dM_r} - \frac{GM_r}{r^2} </math><br />
</div>
to read,
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<tr>
  <td align="right">
<math>+\biggl[ \frac{\omega_i^2}{G\rho_c} \biggr]</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
\frac{1}{{\delta r}_i \cdot r_i^2} \biggl[ 4\pi (r_i^*)^4 \cdot \frac{dP^*}{dm^*} + m^*\biggr] \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="5%" width="80%">
<tr><td align="left">

<div align="center">
<span id="PGE:Euler"><font color="#770000">'''Euler + Poisson Equations'''</font></span>
</div>

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<tr>
  <td align="right">
<math>\biggl[ \frac{K_c^{1 / 2}}{G^{1 / 2} \rho_c^{2 / 5}} \biggr] \frac{d^2 r^*}{dt^2} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- 4\pi (r^*)^2 \biggl[ \frac{K_c^{1 / 2}}{G^{1 / 2} \rho_c^{2 / 5}} \biggr]^2 
\frac{dP^*}{dm} \biggl[ K_c \rho_c^{6 / 5} \biggr] \biggl[ \frac{G^{3 / 2}\rho_c^{1 / 5}}{K_c^{3 / 2}} \biggr]
- \frac{Gm}{(r^*)^2} \biggl[ \frac{K_c^{3 / 2}}{G^{3 / 2}\rho_c^{1 / 5}} \biggr]
\biggl[ \frac{G^{1 / 2} \rho_c^{2 / 5}}{K_c^{1 / 2}} \biggr]^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ \biggl[ \frac{K_c^{1 / 2}}{G^{1 / 2} \rho_c^{2 / 5}} \biggr] \frac{d^2 r^*}{dt^2} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- 4\pi (r^*)^2 \biggl[ K_c^{1 / 2}G^{1 / 2}\rho_c^{3 / 5}\biggr] \frac{dP^*}{dm} 
- \frac{m}{(r^*)^2} \biggl[ K_c^{1 / 2}G^{1 / 2}\rho_c^{3 / 5} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ -\biggl[ \frac{1}{G \rho_c} \biggr] \frac{d^2 r^*}{dt^2} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi (r^*)^2 \frac{dP^*}{dm} 
+ \frac{m^*}{(r^*)^2} 
</math>
  </td>
</tr>
</table>

----

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<tr>
  <td align="right">
<math>
\frac{d^2r}{dt^2}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
\frac{d}{dt}\biggl[(i\omega) r_0 x~e^{i\omega t}\biggr] = - \omega^2 r_0 x~e^{i\omega t}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
r^2 \frac{dP}{dm}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
r_0^2 \biggl[1 + x~ e^{i\omega t}  \biggr]^2 \biggl\{\frac{dP_0}{dm} \biggl[1 + p~ e^{i\omega t}  \biggr]  + P_0~e^{i\omega t} \frac{dp}{dm}   \biggr\} \approx r_0^2 \frac{dP_0}{dm} \biggl[1 + (2x+p)~ e^{i\omega t}  \biggr] + P_0 r_0^2~e^{i\omega t} \frac{dp}{dm}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
\frac{Gm}{r^2}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
\frac{Gm}{ r_0^2} \biggl[1 + x~ e^{i\omega t}  \biggr]^{-2} \approx \frac{Gm}{ r_0^2} 
\biggl[1 -2 x~ e^{i\omega t}  \biggr] \, .
</math>
  </td>
</tr>
</table>

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

After introducing a perturbation, we find that &hellip;

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  <td align="right">
RHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi \biggl[ r^* \biggr]^2 \frac{dP^*}{dm} 
+ m^*\biggl[ r^* \biggr]^{-2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi r_0^2\biggl[ 1 + \frac{\delta r}{r_0} \biggr]^2 \frac{d(P_0 + \delta P)}{dm} 
+ \frac{m^*}{r_0^2}\biggl[ 1 + \frac{\delta r}{r_0} \biggr]^{-2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
4\pi r_0^2\biggl[ 1 + 2x\biggr] \frac{d(P_0 + \delta P)}{dm} 
+ g_0\biggl[ 1 - 2x \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
4\pi r_0^2\biggl[ \frac{d(P_0 + \delta P)}{dm}\biggr]
+ g_0
+
4\pi r_0^2\biggl[ 2x\biggr] \frac{d(P_0 + \cancelto{\mathrm{small}}{\delta P})}{dm} 
-2x g_0 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
4\pi r_0^2\biggl[ \frac{d(P_0 + \delta P)}{dm}\biggr]
+ g_0
+ 2x
\biggl[ 4\pi r_0^2 \frac{dP_0}{dm} 
-g_0 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\biggl\{
4\pi r_0^2\biggl[ \frac{d(P_0 + \delta P)}{dm}\biggr]
+ g_0 \biggr\}
- 4xg_0
\, .
</math>
  </td>
</tr>
</table>

This "RHS" expression must be paired with &hellip;

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  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\omega_i^2}{G\rho_c}\biggr] \delta r_i
=
\biggl[\frac{\omega_i^2}{G\rho_c}\biggr] x r_0 \, .
</math>
  </td>
</tr>
</table>

The term inside the curly braces on the RHS is easy to evaluate inside our finite-difference scheme.  For our example parabolic displacement function, we expect this term to give,

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

<tr>
  <td align="right">
<math>\biggl\{
4\pi r_0^2\biggl[ \frac{d(P_0 + \delta P)}{dm}\biggr]
+ g_0 \biggr\}
</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
\frac{P_0}{\rho_0} \cdot \frac{dp}{dr_0} - pg_0
=
4xg_0
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\alpha_\mathrm{scale} \biggl(1 + \frac{\xi^2}{3}\biggr)^{-1 / 2}  
\biggl(\frac{8\pi}{15}\biggr)r_0 
- \biggl\{
\alpha_\mathrm{scale} \biggl[\biggl(\frac{2\pi}{9}\biggr)r_0^2  -3    \biggr] \frac{6}{5}
\biggr\}
\biggl( \frac{8\pi}{3 } \biggr)^{1/2} \biggl[ \xi \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \alpha_\mathrm{scale} \biggl[ \xi \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]r_0^2  
\biggl( \frac{2^7\pi^3 }{3^3 \cdot 5^2 } \biggr)^{1/2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\alpha_\mathrm{scale}  
\biggl( \frac{2^5 \cdot 3^3\pi}{5^2 } \biggr)^{1/2}  \xi \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2}  
+
\alpha_\mathrm{scale}\biggl(\frac{2^3\pi}{3\cdot 5}\biggr) \biggl(1 + \frac{\xi^2}{3}\biggr)^{-1 / 2}  r_0  
- 
\alpha_\mathrm{scale} r_0^2 
\biggl( \frac{2^7\pi^3}{3^3\cdot 5^2 } \biggr)^{1/2} \xi \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\alpha_\mathrm{scale} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggl\{
\biggl( \frac{2^5 \cdot 3^3\pi}{5^2 } \biggr)^{1/2}  \xi   
+
\biggl(\frac{2^5\pi}{3\cdot 5^2}\biggr)^{1 / 2} \biggl(1 + \frac{\xi^2}{3}\biggr) \xi   
- 
\biggl( \frac{2^5\pi}{3 \cdot 5^2 } \biggr)^{1/2} \xi^3 
\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\alpha_\mathrm{scale} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggl\{
\biggl[ \biggl( \frac{2^5 \cdot 3^3\pi}{5^2 } \biggr)^{1/2}   
+
\biggl(\frac{2^5\pi}{3\cdot 5^2}\biggr)^{1 / 2}\biggr] \xi   
+
\biggl[ \biggl(\frac{2^5\pi}{3^3\cdot 5^2}\biggr)^{1 / 2}    
- 
\biggl( \frac{2^5\cdot 3^2\pi}{3^3 \cdot 5^2 } \biggr)^{1/2}\biggr] \xi^3 
\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\alpha_\mathrm{scale} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggl\{
\biggl(\frac{2^7\pi}{3}\biggr)^{1 / 2} \xi   
- 2\biggl(\frac{2^5\pi}{3^3\cdot 5^2}\biggr)^{1 / 2}  \xi^3 
\biggr\}  \, .
</math>
  </td>
</tr>
</table>
For comparison, note that this expression is identical to the expression for <math>4xg_0</math>, namely,

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

<tr>
  <td align="right"><math>4xg_0 </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
4 \alpha_\mathrm{scale} \biggl[1 - \frac{\xi^2}{15}\biggr] 
\biggl( \frac{8\pi}{3 } \biggr)^{1/2} \biggl[ \xi \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\alpha_\mathrm{scale}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2}  
\biggl\{
\biggl( \frac{2^7\pi}{3 } \biggr)^{1/2}\xi - \biggl( \frac{2^7\pi}{3^3\cdot 5^2 } \biggr)^{1/2}\xi^3\biggr\} 
\, .
</math>
  </td>
</tr>
</table>