Editing Appendix/Ramblings/51BiPolytropeStability/BetterInterface (section)

=Work With Pair of First-Order Linearized Equations=

==Equilibrium Structures Using Preferred Normalizations==
Working from our [[SSC/Structure/BiPolytropes/Analytic51Renormalize#BiPolytrope_with_(nc,_ne)_=_(5,_1)|earlier "new" normalization]] &#8212; which was done in the context of our [[SSC/Structure/BiPolytropes/Analytic51Renormalize#Model_Pairings|examination of the B-KB74 conjecture]] &#8212; that is, by setting,

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

<div align="center"><b>New Normalization</b></div>

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

<tr>
  <td align="right"><math>\tilde\rho</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>\rho \biggl[\biggl( \frac{K_c}{G} \biggr)^{3 / 2} \frac{1}{M_\mathrm{tot}} \biggr]^{-5} \, ;</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{P}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>P \biggl[K_c^{-10} G^{9} M_\mathrm{tot}^{6}  \biggr] \, ;</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{r}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>r \biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]\, ,</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{M}_r</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>\frac{M_r}{M_\mathrm{tot}} \, ;</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{H}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>H \biggl[K_c^{-5 / 2} G^{3 / 2} M_\mathrm{tot} \biggr] \, .</math></td>
</tr>
</table>
</td></tr></table>

and after adopting the notation,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\mathcal{m}_\mathrm{surf}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>\biggl(\frac{2}{\pi}\biggr)^{1 / 2} \theta_i^{-1}\biggl(-\eta^2 \frac{d\phi}{d\eta}\biggr)_s 
= \biggl(\frac{2}{\pi}\biggr)^{1 / 2} \frac{A\eta_s}{\theta_i} \, ,</math></td>
</tr>
</table>

(see [[#interfaceValues|definitions of <math>A</math>, <math>\eta_s</math>, and <math>\theta_i</math> given below]]) we have throughout the core,

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

<tr>
  <td align="right"><math>\tilde{\rho}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\mathcal{m}_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\tilde{P}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\tilde{r}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\mathcal{m}_\mathrm{surf}^{-2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{4} 
\biggl(\frac{3}{2\pi}\biggr)^{1/2} \xi \, ;</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\tilde{M}_r</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2}  \, .</math>
  </td>
</tr>
</table>

<table border="1" width="80%" align="center" cellpadding="8"><tr><td align="left">
[[#FromEarlier|For later use]], note that,

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

<tr>
  <td align="right"><math>\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ \mathcal{m}_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \biggr]
\biggl[
\mathcal{m}_\mathrm{surf}^{-6} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{12} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>\times
\biggl[ 
\mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2}  \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]
\biggl[
\mathcal{m}_\mathrm{surf}^{4} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-8} 
\biggl(\frac{3}{2\pi}\biggr)^{-1} \xi^{-2}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^2 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-4} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi~ 
\biggl(\frac{2^3\pi}{3}\biggr)^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl[\frac{\xi}{\tilde{r}} \biggr]
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
</math>
  </td>
</tr>
</table>
Note as well that,

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

<tr>
  <td align="right"><math>\frac{\tilde{r}^3}{\tilde{M}_r}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ 
\mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2}  \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]^{-1}
\biggl[
\mathcal{m}_\mathrm{surf}^{-2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{4} 
\biggl(\frac{3}{2\pi}\biggr)^{1/2} \xi \biggr]^{3}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ 
\mathcal{m}_\mathrm{surf} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} 
\biggl( \frac{4\cdot 3}{2\pi } \biggr)^{-1/2}  \xi^{-3} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \biggr]
\biggl[
\mathcal{m}_\mathrm{surf}^{-6} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{12} 
\biggl(\frac{3}{2\pi}\biggr)^{3/2} \xi^3 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ 
\mathcal{m}_\mathrm{surf}^{-5} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{10} 
\biggl( \frac{3}{4\pi } \biggr) \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \biggr]
</math>
  </td>
</tr>

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

Similarly, throughout the envelope, we find,

<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td align="right">
<math>\tilde\rho</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^5 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-9} \theta^{5}_i \phi \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tilde{P}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} \theta^{6}_i \phi^{2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tilde{r}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\mathcal{m}_\mathrm{surf}^{-2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{3} \theta^{-2}_i (2\pi)^{-1/2}\eta \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tilde{M}_r</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^{-1}~ \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr) \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\phi</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>A\biggl[ \frac{\sin(\eta - B)}{\eta}\biggr] \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>\frac{d\phi}{d\eta}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{A}{\eta^2}\biggl[ \eta \cos(\eta-B) - \sin(\eta-B)\biggr] \, ,</math></td>
</tr>
</table>

<span id="interfaceValues">and,</span>
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\theta_i</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(1+\frac{1}{3}\xi_i^2\biggr)^{-1 / 2} \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>\eta_i</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{\mu_e}{\mu_c}\biggr)~ \sqrt{3} \theta_i^2 \xi_i \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>\Lambda_i</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{\xi_i}{\sqrt{3}} 
\biggl[ \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\frac{1}{\theta_i^2 \xi_i^2} - 1\biggr] \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>A</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\eta_i\biggl(1 + \Lambda_i^2\biggr)^{1 / 2} \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>B</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\eta_i - \frac{\pi}{2} + \tan^{-1}(\Lambda_i) \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>\eta_s = \pi + B</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\frac{\pi}{2} + \eta_i + \tan^{-1}(\Lambda_i) \, .</math></td>
</tr>
</table>
Keep in mind, as well, that,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^2 \sqrt{3} ~ \biggl[ \frac{\xi_i^3 \theta_i^4}{A\eta_s}\biggr] \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>q \equiv \frac{r_\mathrm{core}}{R}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{\mu_e}{\mu_c}\biggr) \sqrt{3} ~ \biggl[ \frac{\xi_i \theta_i^2}{\eta_s}\biggr] \, .</math></td>
</tr>
</table>

==Linearized Equations With Preferred Normalizations==

===Review and Elaborate===
<div align="center">
<table border="1" cellpadding="10">
<tr><td align="center">
<font color="#770000">'''Linearized'''</font><br />
<span id="Continuity"><font color="#770000">'''Equation of Continuity'''</font></span><br />
<math>
r_0 \frac{dx}{dr_0} = - 3 x - d ,
</math><br />

<font color="#770000">'''Linearized'''</font><br />
<span id="PGE:Euler"><font color="#770000">'''Euler + Poisson Equations'''</font></span><br />
<math>
\frac{P_0}{\rho_0} \frac{dp}{dr_0}  = (4x + p)g_0 + \omega^2 r_0 x ,
</math><br />

<font color="#770000">'''Linearized'''</font><br />
<span id="PGE:AdiabaticFirstLaw">Adiabatic Form of the<br />
<font color="#770000">'''First Law of Thermodynamics'''</font></span><br />
<math>
p = \gamma_\mathrm{g} d  \, .
</math>
</td></tr>
</table>
</div>

The LHS of the "linearized Euler + Poisson" equation is rewritten as,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\mathrm{LHS} = \frac{P_0}{\rho_0} \frac{dp}{dr_0}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[K_c^{-10} G^{9} M_\mathrm{tot}^{6}  \biggr]^{-1}\tilde{P}
\biggl[\biggl( \frac{K_c}{G} \biggr)^{3 / 2} \frac{1}{M_\mathrm{tot}} \biggr]^{-5} \tilde{\rho}^{-1}
\biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr] \frac{dp}{d\tilde{r}}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{P}}{\tilde{\rho}} \frac{dp}{d\tilde{r}}
\biggl[K_c^{10} G^{-9} M_\mathrm{tot}^{-6}  \biggr]
\biggl[ G^{15/2} K_c^{-15/2}  M_\mathrm{tot}^5 \biggr] 
\biggl[K_c^{5/2} G^{-5/2} M_\mathrm{tot}^{-2}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{P}}{\tilde{\rho}} \frac{dp}{d\tilde{r}}
\biggl[K_c^{5} G^{-4} M_\mathrm{tot}^{-3}  \biggr]
\, ;</math>
  </td>
</tr>
</table>
and,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>g_0 = \frac{GM_r}{r_0^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
GM_\mathrm{tot} \tilde{M}_r
\biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]^2 \tilde{r}^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]^2 
GM_\mathrm{tot}
</math>
  </td>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[K_c^5 G^{-4}M_\mathrm{tot}^{-3}  \biggr] 
\, .
</math>
  </td>
</tr>
</table>
Therefore, multiplying the full equation through by <math>[K_c^{-5} G^4 M_\mathrm{tot}^3]</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{\tilde{P}}{\tilde{\rho}} \frac{dp}{d \tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(4x + p)\frac{\tilde{M}_r}{\tilde{r}^2}
+
\biggl[K_c^{-5} G^4 M_\mathrm{tot}^3 \biggr] \biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]^{-1} \omega^2 \tilde{r} x
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(4x + p)\frac{\tilde{M}_r}{\tilde{r}^2}
+
\tau_c^2 \omega^2 \tilde{r} x \, ,
</math>
  </td>
</tr>
</table>
where the square of the characteristic timescale, 

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

<tr>
  <td align="right"><math> \tau_c^2</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\biggl[ K_c^{-15/2} G^{13/2} M_\mathrm{tot}^{5}  \biggr] \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="8" width="75%"><tr><td align="left">
<font color="red">ASIDE:</font>&nbsp;  Building on [[SSC/Stability/Polytropes#Numerical_Integration_from_the_Center,_Outward|an associated discussion]], the square of the dimensionless frequency also can be represented by the expression,

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

<tr>
  <td align="right"><math>\sigma_c^2 \equiv \frac{3\omega^2}{2\pi G\rho_c}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3\omega^2}{2\pi \tilde{\rho}_c} \biggl[\biggl( \frac{G}{K_c} \biggr)^{15 / 2} M_\mathrm{tot}^5 \biggr]G^{-1} 
=
\frac{3\tau_c^2 \omega^2}{2\pi \tilde{\rho}_c} \, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right"><math>\tilde{\rho}_c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
m_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} \, .
</math>
  </td>
</tr>
</table>
</td></tr></table>

<span id="FromEarlier">Hence we can write,</span>
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{dp}{d \tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[(4x + p)
+
\tau_c^2 \omega^2 \biggl(\frac{\tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr] \, .
</math>
  </td>
</tr>
</table>

----

Focusing on the core &hellip;

As demonstrated earlier, the leading term on the RHS of this expression can be rewritten to give,

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

<tr>
  <td align="right"><math>\frac{dp}{d \tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl[\frac{\xi}{\tilde{r}} \biggr]
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{dp}{d \xi} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr]
\, .
</math>
  </td>
</tr>
</table>
Noting as well that,

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

<tr>
  <td align="right"><math>\tilde{\rho}_c \biggl( \frac{\tilde{r}^3}{\tilde{M}_r}\biggr)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
m_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} 
\biggl[
\mathcal{m}_\mathrm{surf}^{-5} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{10} 
\biggl( \frac{3}{4\pi } \biggr) \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2}
\biggr] 
=
\frac{3}{4\pi } \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \, ,
</math>
  </td>
</tr>

</table>
we have,

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

<tr>
  <td align="right"><math>\frac{dp}{d \xi} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ (4x + p)
+
\frac{\sigma_c^2}{2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} x \biggr] \, .
</math>
  </td>
</tr>
</table>

===At the Center===
====All &sigma;<sup>2</sup>====
According to our [[Appendix/Ramblings/PowerSeriesExpressions#Displacement_Function_for_Polytropic_LAWE|discussion in an appendix chapter]], starting from the center of the equilibrium configuration, the displacement function can be represented by a power-series expression of the form,

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

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 1 - \biggl[\frac{(n+1)\mathfrak{F}}{60}\biggr]\xi^2 
\, ,</math>
  </td>
</tr>
</table>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math>, and (see, for example, [[SSC/Structure/BiPolytropes/Analytic51Renormalize#Core|here]]),

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

<tr>
  <td align="right">
<math>\xi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
m_\mathrm{surf}^2 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-4} \biggl(\frac{2\pi}{3}\biggr)^{1 / 2} \tilde{r}
\, .</math>
  </td>
</tr>
</table>
Note that, at the center of our <math>(n_c, n_e) = (5, 1)</math> bipolytrope, <math>\gamma_g = 6/5</math>, so <math>\alpha = -1/3</math>.  Hence, for this particular investigation, the central boundary condition is,

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

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 1 -  \biggl[ \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr]\xi^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 1 -  \biggl[ \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr]
\biggl[ m_\mathrm{surf}^4 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-8} \biggl(\frac{2\pi}{3}\biggr) \biggr] \tilde{r}^2
\, . 
</math>
  </td>
</tr>
</table>
Also, the derivative of this displacement function is,

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

<tr>
  <td align="right">
<math>\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> -  2\biggl[ \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr]\xi \, .
</math>
  </td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
<math>p = -\gamma_c\biggl[3x + \xi \cdot \frac{dx}{d\xi}\biggr]</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>-\frac{6}{5} \biggl\{ 3 \biggl[ 1 -  \biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr] 
-  2\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{18}{5} \biggl[ 1 -  \frac{5}{3}\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr] \, .
</math>
  </td>
</tr>
</table>
Furthermore, we find,

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

<tr>
  <td align="right">
<math>(4x + p)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\biggl[ 1 - \biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2  \biggr]
-\frac{18}{5} \biggl[ 1 -  \frac{5}{3}\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr] \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{20}{5} - 4\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2  \biggr]
+ \biggl[ - \frac{18}{5} + 6\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{2}{5} + 2\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \, .
</math>
  </td>
</tr>
</table>
Hence we have,

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

<tr>
  <td align="right"><math>\frac{dp}{d \xi} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl\{ 
\frac{2}{5} + 2\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 
+
\frac{\sigma_c^2}{2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} 
\biggl[
1 -  \biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl\{ 
24 + 2\biggl( 5\sigma_c^2 + 4 \biggr)\xi^2 
+
\frac{\sigma_c^2}{2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} 
\biggl[
60 -  \biggl( 5\sigma_c^2 + 4 \biggr)\xi^2
\biggr] 
\biggr\}\frac{1}{60}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{30}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl\{ 
24 + 8\xi^2 +  10\sigma_c^2 \xi^2 
+
30\sigma_c^2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2}  
-
\frac{\sigma_c^2}{2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \biggl( 4 + 5\sigma_c^2 \biggr)\xi^2 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{30}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl\{ 
\biggl[ 24 + 8\xi^2 +  10\sigma_c^2 \xi^2 \biggr]
+
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2}\biggl[
30\sigma_c^2  
-
2\sigma_c^2  \xi^2 
-
\frac{5}{2} \sigma_c^4 \xi^2 \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{15}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ 12 + 4\xi^2 +  5\sigma_c^2 \xi^2 \biggr]

+
\frac{\sigma_c^2 \xi}{30} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1/2}\biggl[
30  
-
2  \xi^2 
-
\frac{5}{2} \sigma_c^2 \xi^2 \biggr]
</math>
  </td>
</tr>
</table>

====Just &sigma;<sup>2</sup> = 0====
If we set <math>\sigma_c^2 = 0</math>, we have,

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

<tr>
  <td align="right">
<math>p</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{18}{5} +  \frac{2}{5} \xi^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{dp}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{4}{5} \xi \, .
</math>
  </td>
</tr>
</table>

Alternatively, from immediately above,

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

<tr>
  <td align="right"><math>\frac{dp}{d \xi}\biggr|_{\sigma_c^2=0} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{15}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ 12 + 4\xi^2 +  \cancelto{0}{5\sigma_c^2 \xi^2} \biggr]

+
\cancelto{0}{\frac{\sigma_c^2 \xi}{30} }
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1/2}\biggl[
30  
-
2  \xi^2 
-
\frac{5}{2} \sigma_c^2 \xi^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{12}{15}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ 1 + \frac{1}{3}\xi^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4}{5}\xi \, .
</math>
  </td>
</tr>
</table>

Yes!  It  matches!

====Summary====

Moving from the center, outward thorough the core &#8212; that is, interior to the interface &#8212; we can assign values of <math>x(\xi)</math> and <math>p(\xi)</math> using the following approximate (''exact'' if <math>\sigma_c^2 = 0</math>) relations:

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

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 1 -  \biggl[ \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr]\xi^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>p </math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>-\frac{18}{5} \biggl[ 1 -  \frac{5}{3}\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr] \, .
</math>
  </td>
</tr>
</table>
For all radial shells throughout the entire bipolytropic configuration, the pair of first derivatives can be evaluated using the following relations:

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

<tr>
  <td align="right">
<math>\frac{dx}{d\tilde{r}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> -\frac{1}{\tilde{r}}\biggl[3x + \frac{p}{\gamma_g}\biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\frac{dp}{d \tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr] \, .
</math>
  </td>
</tr>
</table>

Near the center, this pressure-derivative expression can be checked against the relation,

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

<tr>
  <td align="right"><math>\frac{dp}{d \xi} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{15}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ 12 + 4\xi^2 +  5\sigma_c^2 \xi^2 \biggr]

+
\frac{\sigma_c^2 \xi}{30} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1/2}\biggl[
30  
-
2  \xi^2 
-
\frac{5}{2} \sigma_c^2 \xi^2 \biggr] \, ;
</math>
  </td>
</tr>
</table>
notice that, in order to make this comparison, you need to multiply this last expression through by the ratio,

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

<tr>
  <td align="right"><math>\frac{\xi}{\tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^{2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-4} 
\biggl(\frac{2\pi}{3}\biggr)^{1 / 2} \, .</math>
  </td>
</tr>
</table>

The comparison should be especially accurate in the case of <math>\sigma_c^2 = 0</math>.

===At the Interface===

See [[#Interface|below]].

===At the Surface===

Drawing from a [[SSC/Stability/Polytropes#Boundary_Conditions|separate discussion]], the surface boundary condition is,

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

<tr>
  <td align="right">
<math>~r_0 \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 4 - 3\gamma_g + \frac{\omega^2 R^3}{GM_\mathrm{tot}}\biggr) \frac{x}{\gamma_g}</math>&nbsp; &nbsp; &nbsp; &nbsp; at &nbsp; &nbsp; &nbsp; &nbsp; <math>~r_0 = R \, ,</math>
  </td>
</tr>
</table>
that is,

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

<tr>
  <td align="right">
<math>\frac{d\ln x}{d\ln \tilde{r}}\biggr|_s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \alpha  + \frac{\omega^2 R^3}{\gamma_g GM_\mathrm{tot}} 
\, ,</math>
  </td>
</tr>
</table>
where (see also, [[SSC/Stability/BiPolytropes#Review_of_the_Analysis_by_Murphy_&_Fiedler_(1985b)|here]]),

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

<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>
Note that since,

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

<tr>
  <td align="right">
<math>R^3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\tilde{r}_s^3 \biggl[ G^{15/2} K_c^{-15/2} M_\mathrm{tot}^6\biggr]
\, ,</math>
  </td>
</tr>
</table>
in terms of our adopted normalizations, the frequency-squared term should be rewritten as,

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

<tr>
  <td align="right">
<math>\frac{\omega^2 R^3}{\gamma_g G M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\tilde{r}_s^3 \biggl[ \frac{\omega^2 \tau_c^2}{\gamma_g}\biggr]
\, .</math>
  </td>
</tr>
</table>
Note as well that, at the surface of our <math>(n_c, n_e) = (5, 1)</math> bipolytrope, <math>\gamma_g = 2</math>, so <math>\alpha = +1</math>.  Hence, for this particular investigation, the surface boundary condition is,

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

<tr>
  <td align="right">
<math>\frac{d\ln x}{d\ln \tilde{r}}\biggr|_s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \frac{1}{2}\biggl[ \tilde{r}_s^3\omega^2 \tau_c^2\biggr] -1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \frac{\tilde{r}_s^3}{6}\biggl[ 2\pi \tilde{\rho}_c\sigma_c^2\biggr] -1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \frac{\tilde{r}_s^3}{6}\biggl[ 2\pi \tilde{\rho}_c\sigma_c^2\biggr]\frac{ 3 }{4\pi \tilde{r}_s^3 \tilde{\rho}_\mathrm{mean}} -1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \frac{1}{4}\biggl( \frac{ \tilde{\rho}_c }{\tilde{\rho}_\mathrm{mean}}\biggr)\sigma_c^2  -1 \, .
</math>
  </td>
</tr>
</table>

This result should be compared with our [[SSC/Stability/BiPolytropes#Eigenfunction_Details|separate discussion of ''eigenfunction details'']].

==Discretize for Numerical Integration==

===General Discretization===

====First Approximation====
Now, let's set up a grid associated with a uniformly spaced spherical radius, where the subscript <math>J</math> denotes the grid zone at which all terms in the finite-difference representation of the governing relations will be evaluated.  More specifically,

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

<tr>
  <td align="right"><math>\tilde{r}_{J-1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\tilde{r}_J - \Delta\tilde{r}
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>\tilde{r}_{J+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\tilde{r}_J + \Delta\tilde{r} \, ;
</math>
  </td>
</tr>
</table>
also,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\biggl(\frac{dx}{d\tilde{r}}\biggr)_{J}</math></td>
  <td align="center"><math>\approx</math></td>
  <td align="left">
<math>
\frac{(x_{J+1} - x_{J-1})}{2\Delta\tilde{r}}
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>\biggl(\frac{dp}{d\tilde{r}}\biggr)_{J}</math></td>
  <td align="center"><math>\approx</math></td>
  <td align="left">
<math>
\frac{(p_{J+1} - p_{J-1})}{2\Delta\tilde{r}} \, .
</math>
  </td>
</tr>
</table>
And at each grid location, the governing relations establish the local evaluation of the derivatives, that is,

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

<tr>
  <td align="right"><math>\biggl(\frac{dx}{d \tilde{r}}\biggr)_J </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{1}{\tilde{r}_J}\biggl[ 
3x + \frac{p}{\gamma_g}\biggr]_J \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>\biggl(\frac{dp}{d \tilde{r}}\biggr)_J </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{\rho}_J}{\tilde{P}_J}\biggl[ 
(4x + p)\frac{\tilde{M}_r}{\tilde{r}^2}
+
\tau_c^2 \omega^2 \tilde{r} x \biggr]_J \, .
</math>
  </td>
</tr>
</table>
<span id="1stapprox">So, integrating</span> step-by-step from the center of the configuration, outward, once all the variable values are known at grid locations <math>J</math> and <math>(J-1)</math>, the values of <math>x</math> and <math>p</math> at <math>(J+1)</math> are given by the expressions,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>x_{J+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_{J-1} - 2\Delta\tilde{r} \biggl\{
\frac{1}{\tilde{r}}\biggl[ 
3x + \frac{p}{\gamma_g}\biggr]
\biggr\}_J \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>p_{J+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
p_{J-1} + 2\Delta\tilde{r}
\biggl\{
\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr] 
\biggr\}_J\, .
</math>
  </td>
</tr>
</table>

Then we will obtain the "<math>x_J</math>" and "<math>p_J</math>" values via the ''average'' expressions,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>x_{J}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2}(x_{J-1} + x_{J+1})
\, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>p_{J}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2}(p_{J-1} + p_{J+1})
\, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left">
Consider implementing a more ''implicit'' finite-difference analysis.  Wherever a "<math>J</math>" index appears in the source term, replace it with the ''average expressions.''  The general form of the source term expressions is,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>x_{J+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_{J-1} + 2\Delta\tilde{r} \biggl\{
\mathcal{A}x_J + \mathcal{B}p_J
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_{J-1} + \Delta\tilde{r} \biggl\{
\mathcal{A}\biggl[x_{J+1}+x_{J-1}\biggr] + \mathcal{B}\biggl[p_{J+1}+p_{J-1}\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ x_{J+1}\biggl[1 - \Delta\tilde{r} \mathcal{A}  \biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_{J-1}\biggl[1 + \Delta\tilde{r}\mathcal{A}\biggr] + \Delta\tilde{r} 
\mathcal{B}\biggl[p_{J+1}+p_{J-1}\biggr]
\, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\mathcal{A}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
- \frac{3}{\tilde{r}} \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>\mathcal{B}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
- \frac{1}{\gamma_g \tilde{r}} \, ;
</math>
  </td>
</tr>
</table>
and,

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

<tr>
  <td align="right"><math>p_{J+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
p_{J-1} + 2\Delta\tilde{r} \biggl\{
\mathcal{C}x_J + \mathcal{D}p_J
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
p_{J-1} + \Delta\tilde{r} \biggl\{
\mathcal{C}\biggl[x_{J+1}+x_{J-1}\biggr] + \mathcal{D}\biggl[p_{J+1}+p_{J-1}\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ p_{J+1}\biggl[1 - \Delta\tilde{r} \mathcal{D}  \biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
p_{J-1}\biggl[1 + \Delta\tilde{r}\mathcal{D}\biggr] + \Delta\tilde{r} 
\mathcal{C}\biggl[x_{J+1}+x_{J-1}\biggr]
\, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right"><math>\mathcal{C}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\biggl\{
\mathcal{D}
\biggl[ 4
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr)  \biggr] 
\biggr\}_J\, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>\mathcal{D}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\biggl\{
\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}
\biggr\}_J\, .
</math>
  </td>
</tr>
</table>

In both cases, the two unknowns are <math>x_{J+1}</math> and <math>p_{J+1}</math>.  Combining this pair of equations gives,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>x_{J+1}\biggl[1 - \Delta\tilde{r} \mathcal{A}  \biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_{J-1}\biggl[1 + \Delta\tilde{r}\mathcal{A}\biggr] 
+ \Delta\tilde{r} \mathcal{B}\biggl[p_{J-1}\biggr]
+ \Delta\tilde{r} \mathcal{B}\biggl[1 - \Delta\tilde{r} \mathcal{D}  \biggr]^{-1}\biggl\{ 
p_{J-1}\biggl[1 + \Delta\tilde{r}\mathcal{D}\biggr] + \Delta\tilde{r} 
\mathcal{C}\biggl[x_{J+1}+x_{J-1}\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ x_{J+1}\biggl\{ 1 - \Delta\tilde{r} \mathcal{A}  
- \Delta\tilde{r} \mathcal{B}\biggl[1 - \Delta\tilde{r} \mathcal{D}  \biggr]^{-1}\biggl[ \Delta\tilde{r} \mathcal{C}\biggr]
\biggr\}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_{J-1}\biggl[1 + \Delta\tilde{r}\mathcal{A}\biggr] 
+ \Delta\tilde{r} \mathcal{B}\biggl[p_{J-1}\biggr]
+ \Delta\tilde{r} \mathcal{B}\biggl[1 - \Delta\tilde{r} \mathcal{D}  \biggr]^{-1}\biggl\{ 
p_{J-1}\biggl[1 + \Delta\tilde{r}\mathcal{D}\biggr] + \Delta\tilde{r} 
\mathcal{C}\biggl[x_{J-1}\biggr]
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
which determines <math>x_{J+1}</math>, which then allows the straightforward determination of <math>p_{J+1}</math>.  Via the ''average'' expressions, we can also then determine &#8212; and record &#8212; the self-consistent values of <math>x_J</math> and <math>p_J</math>.

----

Dropping terms <math>\mathcal{O}[(\Delta\tilde{r})^2]</math> and higher gives,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>x_{J+1}\biggl[ 1 - \Delta\tilde{r} \mathcal{A}  
\biggr]_J</math></td>
  <td align="center"><math>\approx</math></td>
  <td align="left">
<math>
x_{J-1}\biggl[1 + \Delta\tilde{r}\mathcal{A}\biggr]_J 
+ \Delta\tilde{r} \mathcal{B}_J\biggl[p_{J-1}\biggr]
\, ,
</math>
  </td>
</tr>
</table>
and, in turn,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>p_{J+1}\biggl[1 - \Delta\tilde{r} \mathcal{D}  \biggr]_J</math></td>
  <td align="center"><math>\approx</math></td>
  <td align="left">
<math>
p_{J-1}\biggl[1 + \Delta\tilde{r}\mathcal{D}\biggr]_J 
+ \Delta\tilde{r} \mathcal{C}_J\biggl[x_{J-1}\biggr]
+ \Delta\tilde{r} \mathcal{C}_J\biggl[
x_{J+1}
\biggr] \, .
</math>
  </td>
</tr>
</table>

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

====Second Approximation====

Let's assume that we know the three quantities, <math>x_{J-1}, x_J</math>, and <math>(x_J)^' \equiv (dx/d\tilde{r})_J</math> and want to project forward to determine, <math>x_{J+1}</math>.  We should assume that, locally, the displacement function <math>x</math> is quadratic in <math>\tilde{r}</math>, that is,

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

<tr>
  <td align="right"><math>x</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a + b\tilde{r} + c\tilde{r}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{dx}{d\tilde{r}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
b + 2c\tilde{r} \, ,
</math>
  </td>
</tr>
</table>
where we have three unknowns, <math>a, b, c</math>.  These can be determined by appropriately combining the three relations,

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

<tr>
  <td align="right"><math>(x_J)^'</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
b + 2c\tilde{r}_J \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>x_J</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a + b\tilde{r}_J + c\tilde{r}_J^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>x_{J-1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a + b(\tilde{r}_{J}-\Delta\tilde{r}) + c(\tilde{r}_{J}-\Delta\tilde{r})^2 \, .
</math>
  </td>
</tr>
</table>
We have,

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

<tr>
  <td align="right"><math>b</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(x_J)^' - 2c\tilde{r}_J \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>- a</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-x_J + [(x_J)^' - 2c\tilde{r}_J]\tilde{r}_J + c\tilde{r}_J^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>- a</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-x_{J-1} + [(x_J)^' - 2c\tilde{r}_J](\tilde{r}_{J}-\Delta\tilde{r}) + c(\tilde{r}_{J}-\Delta\tilde{r})^2 \, .
</math>
  </td>
</tr>
</table>
Combining the last two expressions gives,

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

<tr>
  <td align="right"><math>-x_J + [(x_J)^' - 2c\tilde{r}_J]\tilde{r}_J + c\tilde{r}_J^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-x_{J-1} + [(x_J)^' - 2c\tilde{r}_J](\tilde{r}_{J}-\Delta\tilde{r}) + c(\tilde{r}_{J}-\Delta\tilde{r})^2
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ -x_J + (x_J)^'\tilde{r}_J - c\tilde{r}_J^2 </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-x_{J-1} 
+ (x_J)^'(\tilde{r}_{J}-\Delta\tilde{r}) 
- 2c\tilde{r}_J(\tilde{r}_{J}-\Delta\tilde{r}) 
+ c[ \tilde{r}_{J}^2 - 2r_J \Delta\tilde{r} + (\Delta\tilde{r})^2 ]
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ -x_J  - c\tilde{r}_J^2 </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-x_{J-1} 
- (x_J)^'\Delta\tilde{r} 
+ c[ \tilde{r}_{J}^2 - 2r_J \Delta\tilde{r} + (\Delta\tilde{r})^2 
- 2\tilde{r}_J^2 + 2\tilde{r}_J(\Delta\tilde{r})]
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ c(\Delta\tilde{r})^2
</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J -x_{J-1} - (x_J)^'\Delta\tilde{r}
\, .
</math>
  </td>
</tr>
</table>
Therefore, also,

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

<tr>
  <td align="right"><math>b</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(x_J)^' - \biggl[ x_J -x_{J-1} - (x_J)^'\Delta\tilde{r} \biggr]\frac{2\tilde{r}_J}{(\Delta\tilde{r})^2}
 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>a</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - (x_J)^' \tilde{r}_J 
+ \biggl[ x_J -x_{J-1} - (x_J)^'\Delta\tilde{r} \biggr]\frac{\tilde{r}_J^2}{(\Delta\tilde{r})^2}
 </math>
  </td>
</tr>
</table>

Hence,

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

<tr>
  <td align="right"><math>x_{J+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a + b(\tilde{r}_J + \Delta\tilde{r}) + c(\tilde{r}_J + \Delta\tilde{r})^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - (x_J)^' \tilde{r}_J 
+ \biggl[ x_J -x_{J-1} - (x_J)^'\Delta\tilde{r} \biggr]\frac{\tilde{r}_J^2}{(\Delta\tilde{r})^2}
+ 
(\tilde{r}_J + \Delta\tilde{r})\biggl\{
(x_J)^' - \biggl[ x_J -x_{J-1} - (x_J)^'\Delta\tilde{r} \biggr]\frac{2\tilde{r}_J}{(\Delta\tilde{r})^2}
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ 
(\tilde{r}_J + \Delta\tilde{r})^2 
\biggl[ x_J -x_{J-1} - (x_J)^'\Delta\tilde{r} \biggr]\frac{1}{(\Delta\tilde{r})^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - (x_J)^' \tilde{r}_J 
+ 
\biggl\{
\tilde{r}_J^2 
+ 2\tilde{r}_J(\tilde{r}_J + \Delta\tilde{r})
+ (\tilde{r}_J + \Delta\tilde{r})^2
\biggr\} 
\biggl[ x_J -x_{J-1} - (x_J)^'\Delta\tilde{r} \biggr]\frac{1}{(\Delta\tilde{r})^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - (x_J)^' \tilde{r}_J 
+ 
\biggl[ 2\tilde{r}_J + \Delta\tilde{r}\biggr]^2
\biggl[ x_J -x_{J-1} - (x_J)^'\Delta\tilde{r} \biggr]\frac{1}{(\Delta\tilde{r})^2}
\, .
</math>
  </td>
</tr>
</table>
<font color="red"><b>WRONG!!</b></font>  Try again &hellip;

====Third Approximation====

Let's assume that we know the three quantities, <math>x_{J-1}, x_J</math>, and <math>(x_J)^' \equiv (dx/d\tilde{r})_J</math> and want to project forward to determine, <math>x_{J+1}</math>.  We should assume that, locally, the displacement function <math>x</math> is quadratic in <math>\tilde{r}</math>, that is,

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

<tr>
  <td align="right"><math>x</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a + b\tilde{r} + c\tilde{r}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{dx}{d\tilde{r}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
b + 2c\tilde{r} \, ,
</math>
  </td>
</tr>
</table>
where we have three unknowns, <math>a, b, c</math>.  These can be determined by appropriately combining the three relations,

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

<tr>
  <td align="right"><math>(x_J)^'</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
b + 2c\tilde{r}_J \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>x_J</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a + b\tilde{r}_J + c\tilde{r}_J^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>x_{J-1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a + b(\tilde{r}_{J}-\Delta\tilde{r}) + c(\tilde{r}_{J}-\Delta\tilde{r})^2 \, .
</math>
  </td>
</tr>
</table>
The difference between the last two expressions gives,

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

<tr>
  <td align="right"><math>x_J - x_{J-1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ b\tilde{r}_J + c\tilde{r}_J^2 \biggr]
-
\biggl[ b(\tilde{r}_{J}-\Delta\tilde{r}) + c(\tilde{r}_{J}-\Delta\tilde{r})^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
b\Delta\tilde{r} 
+ 
c(2\tilde{r}_J \Delta\tilde{r} - \Delta\tilde{r}^2) \, .
</math>
  </td>
</tr>
</table>
Combining this with the first of the three expressions then gives,

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

<tr>
  <td align="right">
<math>
\Delta\tilde{r} \biggl[(x_J)^' - 2c\tilde{r}_J  \biggr]
+ 
c(2\tilde{r}_J \Delta\tilde{r} - \Delta\tilde{r}^2) 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - x_{J-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
c(2\tilde{r}_J \Delta\tilde{r} - \Delta\tilde{r}^2) 
-c\biggl[ 2\tilde{r}_J \Delta\tilde{r}  \biggr]
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
-c \Delta\tilde{r}^2 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
c  
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{\Delta\tilde{r}^2}\biggl[
-x_J + x_{J-1} + (x_J)^'\Delta\tilde{r}
\biggr] 
\, .
</math>
  </td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
<math>
b
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(x_J)^' - 2c\tilde{r}_J 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(x_J)^' + \frac{2\tilde{r}_J}{\Delta \tilde{r}^2}\biggl[
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
\biggr] 
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
a
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - b\tilde{r}_J -c\tilde{r}_J^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J
-\biggl\{ (x_J)^' + \frac{2\tilde{r}_J}{\Delta \tilde{r}^2}\biggl[
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
\biggr] \biggr\}\tilde{r}_J
- \biggl\{
\frac{1}{\Delta\tilde{r}^2}\biggl[
-x_J + x_{J-1} + (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}\tilde{r}_J^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - (x_J)^'\tilde{r}_J
-\biggl\{  \frac{2\tilde{r}_J^2}{\Delta \tilde{r}^2}\biggl[
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
\biggr] \biggr\}
+ \biggl\{
\frac{\tilde{r}_J^2}{\Delta\tilde{r}^2}\biggl[
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - (x_J)^'\tilde{r}_J
-\frac{\tilde{r}_J^2}{\Delta \tilde{r}^2}\biggl[
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
\biggr] 
\, .
</math>
  </td>
</tr>
</table>
As a result,

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

<tr>
  <td align="right"><math>x_{J+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{
a 
\biggr\}
+ (\tilde{r}_J +\Delta\tilde{r}) \biggl\{
b
\biggr\}
+ (\tilde{r}_J+\Delta\tilde{r})^2 \biggl\{
c
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{
x_J - (x_J)^'\tilde{r}_J
-\frac{\tilde{r}_J^2}{\Delta \tilde{r}^2}\biggl[
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}
+ (\tilde{r}_J +\Delta\tilde{r}) \biggl\{
(x_J)^' + \frac{2\tilde{r}_J}{\Delta \tilde{r}^2}\biggl[
x_J - x_{J-1} - (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ (\tilde{r}_J^2 + 2\tilde{r}_J \Delta\tilde{r} +\Delta\tilde{r}^2) \biggl\{
\frac{1}{\Delta\tilde{r}^2}\biggl[
-x_J + x_{J-1} + (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J - (x_J)^'\tilde{r}_J
+ \tilde{r}_J^2\biggl\{\frac{1}{\Delta \tilde{r}^2}\biggl[
-x_J + x_{J-1} + (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}
+ (x_J)^'(\tilde{r}_J +\Delta\tilde{r}) 

- 2\tilde{r}_J(\tilde{r}_J +\Delta\tilde{r}) \biggl\{
\frac{1}{\Delta \tilde{r}^2}\biggl[
-x_J + x_{J-1} + (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ (\tilde{r}_J^2 + 2\tilde{r}_J \Delta\tilde{r} +\Delta\tilde{r}^2) \biggl\{
\frac{1}{\Delta\tilde{r}^2}\biggl[
-x_J + x_{J-1} + (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J + (x_J)^'\Delta\tilde{r} 

+ \biggl[ \tilde{r}_J^2 - 2\tilde{r}_J(\tilde{r}_J +\Delta\tilde{r})+ (\tilde{r}_J^2 + 2\tilde{r}_J \Delta\tilde{r} +\Delta\tilde{r}^2)\biggr]
\biggl\{\frac{1}{\Delta \tilde{r}^2}\biggl[
-x_J + x_{J-1} + (x_J)^'\Delta\tilde{r}
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_J + (x_J)^'\Delta\tilde{r} 
+ 
\biggl[
-x_J + x_{J-1} + (x_J)^'\Delta\tilde{r}
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_{J-1} + 2(x_J)^'\Delta\tilde{r} \, .
</math>
  </td>
</tr>
</table>
<font color="red"><b>GOOD!</b></font>  This is the same as our [[#1stapprox|first approximation expression]] stated above.


<table border=1 align="center" cellpadding="10" width="80%"><tr><td bgcolor="lightgreen" align="left">
This is test ...
<table border="1" align="center" cellpadding="5">

  <tr>
<td align="center" bgcolor="white"><math>\Delta\tilde{r}</math></td>
<td align="center" bgcolor="white"><math>x_J</math></td>
<td align="center" bgcolor="white"><math>x_{J-1}</math></td>
<td align="center" bgcolor="white"><math>(x_J)^'</math></td>
<td align="center" bgcolor="white"><math>(x_{J-1})^'</math></td>
  </tr>

  <tr>
<td align="center" bgcolor="white">0.001936393</td>
<td align="center" bgcolor="white">-4.695376</td>
<td align="center" bgcolor="white">-4.547832</td>
<td align="center" bgcolor="white">-116.0119</td>
<td align="center" bgcolor="white">-76.19513</td>
  </tr>
</table>
<tr><td bgcolor="white" align="center">

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

<tr>
  <td align="right"><math>x_{J+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
x_{J-1} + 2(x_J)^'\Delta\tilde{r}
=
-4.997121
\, .
</math>
  </td>
</tr>
</table>
</td></tr>
</td></tr></table>