Editing ThreeDimensionalConfigurations/Challenges (section)

===Hybrid Scheme===

In a separate chapter we have detailed the [[Appendix/Ramblings/HybridSchemeImplications#Hybrid_Scheme|hybrid scheme]].  For steady-state configurations, the three components of the combined Euler + Continuity equations give,

<table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left">
<div align="center">'''Hybrid Scheme Summary for ''Steady-State'' Configurations'''</div>

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right"><math>~\boldsymbol{\hat{k}:}</math></td>
  <td align="right">
<math>~
\bold\nabla \cdot (\rho v_z \bold{u})
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{\hat{k}} \cdot (\rho \bold{a}) \, ;</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\bold{\hat{e}_\varpi:}</math></td>
  <td align="right">
<math>~
\bold\nabla \cdot (\rho v_\varpi \bold{u})
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{\hat{e}}_\varpi \cdot (\rho \bold{a}) + \frac{v_\varphi^2}{\varpi} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\bold{\hat{e}_\varphi:}</math></td>
  <td align="right">
<math>~ 
\bold\nabla \cdot (\rho \varpi v_\varphi \bold{u})
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{\hat{e}}_\varphi \cdot (\rho \varpi \bold{a}) \, .</math>
  </td>
</tr>
</table>

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

In this context, the vector acceleration that drives the fluid flow is, simply,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bold{a}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\nabla(H + \Phi_\mathrm{grav} ) \, .</math>
  </td>
</tr>
</table>

Then, for the velocity flow-patterns in Riemann S-type ellipsoids, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla \cdot (\rho v_z \bold{u})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0</math>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(because <math>~v_z = 0</math>);
  </td>
</tr>

<tr>
  <td align="right">
<math>~\nabla \cdot (\rho v_\varpi \bold{u})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda^2}{\varpi^3} \biggl[\frac{a}{b} - \frac{b}{a}  \biggr] \biggl\{
y^4  \biggl(\frac{a}{b}\biggr) 
- 
x^4 \biggl(\frac{b}{a}\biggr) 
\biggr\}\rho \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\nabla \cdot (\rho \varpi v_\varphi \bold{u})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \lambda xy \Omega_f \biggl[\frac{a}{b} - \frac{b}{a} \biggr]\rho  \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\varpi v_\varphi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-
\biggl[ \lambda \biggl(\frac{b}{a}\biggr) - \Omega_f\biggr]x^2
-
\biggl[ \lambda \biggl(\frac{a}{b}\biggr) - \Omega_f\biggr]y^2
\, .
</math>
  </td>
</tr>
</table>

<font color="red">'''Vertical Component:'''</font> &nbsp;  Given that <math>~\bold{\hat{k}}\cdot (\rho \bold{a}) = 0</math>, we deduce that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~H_0  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi G \rho c^2 A_3  \, . </math>
  </td>
</tr>
</table>

<font color="red">'''Azimuthal Component:'''</font> &nbsp; Irrespective of the <math>~(x, y, z)</math> location of each fluid element, this component requires,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ 
- a b \lambda \Omega_f 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pi G \rho \biggl[ \frac{( A_1 - A_2 )a^2b^2}{b^2 - a^2}  - c^2 A_3 
\biggr] \, .
</math>
  </td>
</tr>
</table>

<font color="red">'''Radial Component:'''</font> &nbsp; After inserting the "azimuthal component" relation and marching through a fair amount of algebraic manipulation, we find that Irrespective of the <math>~(x, y, z)</math> location of each fluid element, this component requires,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\frac{2\pi G \rho }{(a^2 - b^2) }
\biggl[  A_1 a^2 - A_2 b^2  \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \lambda^2 + \Omega_f^2\biggr] \, .
</math>
  </td>
</tr>
</table>