Editing ThreeDimensionalConfigurations/Challenges (section)

===Trial #2===

Still restricting our discussion to nonaxisymmetric, thin disks, let's try, <math>~\Psi = \Psi_0 (\rho/\rho_c)^2</math>, and
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\rho</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\rho_c \biggl\{1 - \biggl[ \frac{y^2}{b^2} + \frac{x^2}{a^2}\biggr]\biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{\partial^2 \rho}{\partial x^2} = -\frac{\partial}{\partial x}\biggl\{ \frac{2\rho_c x}{a^2}\biggr\} = - \frac{2\rho_c}{a^2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; and, &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{\partial^2 \rho}{\partial y^2} = - \frac{\partial}{\partial y}\biggl\{ \frac{2\rho_c y}{b^2} \biggr\} = - \frac{2\rho_c}{b^2} \, .</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="10"><tr><td align="left">
<font color="red">'''IMPORTANT NOTE''' (by Tohline on 22 September 2020):</font>&nbsp; &nbsp; As I have come to appreciate today &#8212; after studying the relevant sections of both [[Appendix/References#EFE|EFE]] and [[Appendix/References#BT87|BT87]] &#8212; this is an example of a heterogeneous density distribution whose gravitational potential has an analytic prescription.  As is discussed in a [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Inhomogeneous_Ellipsoids_Leading_to_Ferrers_Potentials| separate chapter]], the potential that it generates is sometimes referred to as a ''Ferrers'' potential, for the exponent, n = 1.

In our [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#GravFor1|accompanying discussion]] we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{ \Phi_\mathrm{grav}(\bold{x})}{(-\pi G\rho_c)} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2} I_\mathrm{BT} a_1^2 
- \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) 
~+ \biggl( A_{12} x^2y^2 + A_{13} x^2z^2 + A_{23} y^2z^2\biggr)
~+ \frac{1}{6}  \biggl(3A_{11}x^4 +  3A_{22}y^4 + 3A_{33}z^4  \biggr) 
\, ,
</math>
  </td>
</tr>
</table>

where,
<table border="0" align="center" cellpadding="10" width="80%">
<tr>
  <td align="center" width="50%">

<table border="0" cellpadding="5" align="center">
<tr><td align="center" colspan="3">for <math>~i \ne j</math></td></tr>
<tr>
  <td align="right">
<math>~A_{ij}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-\frac{A_i-A_j}{(a_i^2 - a_j^2)} </math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;21, Eq. (107)</font> ]</td></tr>
</table>

  </td>
  <td align="center" width="50%">

<table border="0" cellpadding="5" align="center">
<tr><td align="center" colspan="3">for <math>~i = j</math></td></tr>
<tr>
  <td align="right">
<math>~2A_{ii} + \sum_{\ell = 1}^3 A_{i\ell}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{a_i} </math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;21, Eq. (109)</font> ]</td></tr>
</table>

  </td>
</tr>
</table>
More specifically, in the three cases where the indices, <math>~i=j</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3A_{11}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{a_1^2} - (A_{12} + A_{13}) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~3A_{22}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{a_2^2} - (A_{21} + A_{23}) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~3A_{33}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{a_3^2} - (A_{31} + A_{32}) \, .
</math>
  </td>
</tr>
</table>

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

This means that,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{ \partial \Psi}{\partial x} = \biggl( \frac{\Psi_0}{\rho_c^2}\biggr) 2\rho \frac{ \partial \rho}{\partial x} = -2\rho \biggl( \frac{2\rho_c x}{a^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c^2}\biggr)</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{ \partial \Psi}{\partial y} = \biggl( \frac{\Psi_0}{\rho_c^2}\biggr)2\rho \frac{ \partial \rho}{\partial y} = -2\rho \biggl( \frac{2\rho_c y}{b^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c^2}\biggr)  \, .</math>
  </td>
</tr>
</table>

<span id="ConstraintTrial2">&nbsp;</span>
<table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left">
<div align="center">'''The Elliptic PDE Constraint Equation'''</div>

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

<tr>
  <td align="right">
<math>~ 
2\Omega_f 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~C_0 \rho+~\cancel{C_1 \rho \Psi} + \frac{\partial }{\partial x}\biggl[\frac{1}{\rho} \frac{\partial \Psi}{\partial x}  \biggr] + \frac{\partial }{\partial y} \biggl[\frac{1}{\rho} \frac{\partial \Psi}{\partial y}  \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~C_0 \rho
- \frac{\partial }{\partial x}\biggl[ \biggl( \frac{4  x}{a^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c}\biggr)  \biggr] 
- \frac{\partial }{\partial y} \biggl[ \biggl( \frac{4  y}{b^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c}\biggr)  \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~C_0 \rho - \frac{4\Psi_0}{\rho_c}\biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr]
</math>
  </td>
</tr>
</table>
</td></tr></table>


The momentum density vector is governed by the relation,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\rho\bold{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath} \biggl[ \frac{\partial \Psi}{\partial y} \biggr]
- \boldsymbol{\hat\jmath} \biggl[  \frac{\partial \Psi}{\partial x}\biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \boldsymbol{\hat\imath} \biggl[ 2\rho \biggl( \frac{2\rho_c y}{b^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c^2}\biggr) \biggr]
+ \boldsymbol{\hat\jmath} \biggl[  2\rho \biggl( \frac{2\rho_c x}{a^2} \biggr)\biggl( \frac{\Psi_0}{\rho_c^2}\biggr)\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\rho \biggl\{
- \boldsymbol{\hat\imath} \biggl[ \frac{4\Psi_0 y}{\rho_c b^2}  \biggr]
+ \boldsymbol{\hat\jmath} \biggl[  \frac{4\Psi_0 x}{\rho_c a^2}  \biggr] 
\biggr\} 
\, .
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<math>~\Psi</math> has units of "density &times; length<sup>2</sup> per time"]</td>
</table>

As above, let's see if the steady-state continuity equation is satisfied:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla \cdot (\rho \bold{u})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial (\rho u_x)}{\partial x} + \frac{\partial (\rho u_y)}{\partial y}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\partial }{\partial x}\biggl[ \frac{4\Psi_0 y \rho}{\rho_c b^2}  \biggr] + \frac{\partial }{\partial y}\biggl[  \frac{4\Psi_0 x \rho }{\rho_c a^2}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\Psi_0}{\rho_c} \biggl\{
- \frac{y}{b^2} \cdot \frac{\partial \rho}{\partial x} + \frac{ x }{ a^2} \cdot \frac{\partial \rho}{\partial y}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\Psi_0}{\rho_c} \biggl\{
\frac{y}{b^2} \biggl[ \frac{2\rho_c x}{a^2} \biggr] - \frac{ x }{ a^2} \biggl[ \frac{2 \rho_c y}{b^2} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>&nbsp; &nbsp; &nbsp; <font color="red">Yes!</font>
  </td>
</tr>
</table>

Next, as above, let's determine the z-component of the vorticity and the vortensity:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\zeta_z</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ 4\Psi_0}{\rho_c} \biggl\{
\frac{\partial }{\partial x}\biggl[ \frac{x}{a^2} \biggr] + \frac{\partial }{\partial y}\biggl[ \frac{y}{b^2 } \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ 4\Psi_0}{\rho_c} \biggl[ \frac{1}{a^2}  +  \frac{1}{b^2} \biggr] 
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{(\zeta_z + 2\Omega_f)}{\rho}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\rho} \biggl\{ 
\frac{ 4\Psi_0}{\rho_c} \biggl[ \frac{1}{a^2}  +  \frac{1}{b^2} \biggr] 
+ 2\Omega_f \biggr\}\, .
</math>
  </td>
</tr>
</table>
This means that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
(\vec\zeta + 2\vec\Omega) \times \bold{u} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ \frac{(\zeta_z + 2\Omega_f)}{\rho} \biggr\} \boldsymbol{\hat{k}} \times (\rho \bold{u} )</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ 4\Psi_0 (\zeta_z + 2\Omega_f)}{\rho_c} \boldsymbol{\hat{k}} \times 
\biggl\{
- \boldsymbol{\hat\imath} \biggl[ \frac{y}{b^2}  \biggr]
+ \boldsymbol{\hat\jmath} \biggl[  \frac{x}{a^2}  \biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{ 4\Psi_0 (\zeta_z + 2\Omega_f)}{\rho_c} 
\biggl\{
\boldsymbol{\hat\imath} \biggl[  \frac{x}{a^2}  \biggr] 
+ \boldsymbol{\hat\jmath} \biggl[ \frac{y}{b^2}  \biggr]
\biggr\} 
</math>
  </td>
</tr>
</table>

Now, let's examine the gradient of <math>~F_B(\Psi)</math>.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla F_B(\Psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \boldsymbol{\hat\imath} \frac{\partial}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial}{\partial y} \biggr]C_0 \Psi</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ C_0 \Psi_0}{\rho_c^2} \biggl[ \boldsymbol{\hat\imath} \frac{\partial \rho^2}{\partial x} + \boldsymbol{\hat\jmath} \frac{\partial \rho^2}{\partial y} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2C_0 \rho \Psi_0}{\rho_c^2} \biggl\{ 
-\boldsymbol{\hat\imath} \biggl[ \frac{2 \rho_c x}{a^2} \biggr] 
- \boldsymbol{\hat\jmath}  \biggl[ \frac{2 \rho_c y}{b^2} \biggr]  
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{ 4 C_0 \rho \Psi_0}{\rho_c} \biggl\{ 
\boldsymbol{\hat\imath} \biggl[ \frac{x}{a^2} \biggr] 
+ \boldsymbol{\hat\jmath}  \biggl[ \frac{y}{b^2} \biggr]  
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
which is identical to the immediately preceding expression if we set,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\zeta_z + 2\Omega_f)}{\rho} \, .
</math>
  </td>
</tr>
</table>

Continuing with the [[#ConstraintTrial2|above examination of the elliptic PDE "constraint" equation]], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2\Omega_f</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~C_0 \rho - \zeta_z
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(2\Omega_f + \zeta_z) - \zeta_z
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\Omega_f \, .</math>&nbsp; &nbsp; &nbsp; <font color="red">Hooray!</font>
  </td>
</tr>
</table>