Editing ThreeDimensionalConfigurations/Challenges (section)

===Trial #3===

If the density distribution has been specified, then <math>~\Psi</math> is the "stream-function" from which all rotating-frame velocities are determined.  Specifically,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bold{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\rho} \biggl[
\boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial y} - \boldsymbol{\hat\jmath} \frac{\partial \Psi}{\partial x} 
\biggr] \, .
</math>
  </td>
</tr>
</table>
Most importantly, as has been [[#Related_Useful_Expressions|detailed above]], the term on the left-hand-side of the steady-state Euler equation becomes,
<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] \nabla\Psi \, ,</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[https://digitalcommons.lsu.edu/gradschool_disstheses/6650/ Saied W. Andalib (1998)], &sect;4.2, p. 83, Eq. (4.13)</td></tr>
</table>
where,
<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 }{\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>
</table>

Still restricting our discussion to infinitesimally thin, nonaxisymmetric disks, let's assume that,
<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{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] </math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{\partial \rho}{\partial x} = -\biggl( \frac{2\rho_c x}{a^2} \biggr)</math>
  </td>
  <td align="center">
&nbsp; &nbsp; and, &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{\partial \rho}{\partial y} = -\biggl( \frac{2\rho_c y}{b^2} \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>
And, let's assume that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Psi_0 \biggl( \frac{\rho}{\rho_c} \biggr)^q \, ,</math>
  </td>
</tr>
</table>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{\partial\Psi}{\partial x}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\rho^{q-1}\biggl(\frac{q \Psi_0}{\rho_c^q}\biggr) \frac{\partial\rho}{\partial x} 
=
- \rho^{q-1}\biggl(\frac{q \Psi_0}{\rho_c^q}\biggr) \biggl(\frac{2\rho_c x}{a^2} \biggr)
=
- \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0x}{a^2} \biggr) \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
and, similarly, &nbsp; &nbsp; &nbsp; <math>~\frac{\partial\Psi}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr) \, .
</math>
  </td>
</tr>
</table>

This means that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bold{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~
\frac{1}{\rho} \biggl\{
\boldsymbol{\hat\imath} \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr)  \biggr] 
- \boldsymbol{\hat\jmath} \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0x}{a^2} \biggr) \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\bold{u}\cdot \bold{u} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\rho^2} \biggl\{
\biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr)  \biggr]^2 
+ \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0x}{a^2} \biggr) \biggr]^2
\biggr\} \, .
</math>
  </td>
</tr>
</table>

It also means that,
<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 }{\partial x}\biggl[\biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 x}{\rho_c a^2} \biggr)  \biggr] 
+ 
\frac{\partial }{\partial y} \biggl[\biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 y}{\rho_c b^2} \biggr) \biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 }{\rho_c } \biggr) \biggl[ \frac{1}{a^2} + \frac{1}{b^2}\biggr] 
~+ ~
\biggl(\frac{2q \Psi_0 x}{\rho_c a^2} \biggr)  \biggl[ (q-2) \rho_c^{2-q} \rho^{q-3}  \frac{\partial \rho}{\partial x}\biggr]
~+ ~
\biggl(\frac{2q \Psi_0 y}{\rho_c b^2} \biggr) \biggl[ (q-2) \rho_c^{2-q} \rho^{q-3}  \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>~
\biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl(\frac{2q \Psi_0 }{\rho_c } \biggr) \biggl[ \frac{1}{a^2} + \frac{1}{b^2}\biggr] 
~- ~
\biggl(\frac{2q \Psi_0 x}{\rho_c^2 a^2} \biggr)  \biggl[ (q-2) \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3}  \frac{2\rho_c x}{a^2}\biggr]
~- ~
\biggl(\frac{2q \Psi_0 y}{\rho_c^2 b^2} \biggr) \biggl[ (q-2) \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3}  \frac{2\rho_c y}{b^2}\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q\Psi_0}{\rho_c} \biggl\{
2\biggl( \frac{\rho}{\rho_c}\biggr)^{q-2} \biggl[ \frac{1}{a^2} + \frac{1}{b^2}\biggr] 
~- ~
\biggl(\frac{2 x}{ a^2} \biggr)  \biggl[ (q-2) \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3}  \frac{2x}{a^2}\biggr]
~- ~
\biggl(\frac{2 y}{ b^2} \biggr) \biggl[ (q-2) \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3}  \frac{2 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{q\Psi_0}{\rho_c} \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3} \biggl[
2\biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
~- ~ 4(q-2) \biggl(\frac{x^2}{ a^4} \biggr) 
~- ~ 4(q-2) \biggl(\frac{y^2}{ b^4} \biggr)  \biggr]
</math>
  </td>
</tr>
</table>
Hence,
<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[ \boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial x} +   \boldsymbol{\hat\jmath} \frac{\partial \Psi}{\partial y} \biggr]
\biggl\{
\frac{ 2\Omega_f}{\rho_c}\biggl( \frac{\rho}{\rho_c}\biggr)^{-1} 
+
\biggl( \frac{\rho}{\rho_c}\biggr)^{-1} \frac{ \zeta_z }{\rho_c}  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~
\biggl[ \boldsymbol{\hat\imath} \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 x}{a^2} \biggr) +   \boldsymbol{\hat\jmath} \biggl( \frac{\rho}{\rho_c}\biggr)^{q-1} \biggl(\frac{2q \Psi_0 y}{b^2} \biggr) \biggr]
\biggl\{
\frac{ 2\Omega_f}{\rho_c}\biggl( \frac{\rho}{\rho_c}\biggr)^{-1} 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\frac{q\Psi_0}{\rho_c^2} \biggl(\frac{\rho}{\rho_c}\biggr)^{q-3} \biggl[
2\biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
~- ~ 4(q-2) \biggl(\frac{x^2}{ a^4} \biggr) 
~- ~ 4(q-2) \biggl(\frac{y^2}{ b^4} \biggr)  \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>

====Exponent q = 2====
Notice that, if <math>~q = 2</math>,

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

<tr>
  <td align="right">
<math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{8\Psi_0}{\rho_c}\biggl\{ \Omega_f + \frac{2\Psi_0}{\rho_c} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggr\} 
\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, if we choose a function,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~F_B(\Psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{D_1}{2} \Psi^{1 / 2} 
= \frac{D_1}{2}  \Psi_0^{1 / 2} \biggl(\frac{\rho}{\rho_c}\biggr)
= \frac{D_1}{2}  \Psi_0^{1 / 2} \biggl[1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr]
\, ,</math>
  </td>
</tr>
</table>
we obtain,
<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>~
-~\frac{D_1}{2}  \Psi_0^{1 / 2} \cdot \nabla \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~D_1  \Psi_0^{1 / 2}  
\biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) +   \boldsymbol{\hat\jmath}  \biggl(\frac{y}{b^2} \biggr) \biggr]
\, .
</math>
  </td>
</tr>
</table>
This is consistent with the elliptic PDE constraint if,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~D_1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{8\Psi_0^{1 / 2}}{\rho_c}\biggl\{ \Omega_f + \frac{2\Psi_0}{\rho_c} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggr\} \, .</math>
  </td>
</tr>
</table>

Also if <math>~q = 2</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bold{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\rho} \biggl\{
\boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial y} - \boldsymbol{\hat\jmath} \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>~
\frac{1}{\rho} \biggl\{
-\boldsymbol{\hat\imath} \biggl( \frac{\rho}{\rho_c}\biggr) \biggl(\frac{2q \Psi_0 y}{b^2} \biggr)
+ 
\boldsymbol{\hat\jmath}  \biggl( \frac{\rho}{\rho_c}\biggr) \biggl(\frac{2q \Psi_0 x}{a^2} \biggr) 
\biggr\}
</math>
  </td>
</tr>

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

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

Keep in mind that, [[#AndalibBernoulli|as discussed above]], we are trying to satisfy the scalar Bernoulli relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_B - H - \Phi_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
F_B(\Psi) + \frac{1}{2}u^2 - \frac{1}{2}\Omega_f^2 (x^2 + y^2)  
</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\{ \Omega_f + \frac{2\Psi_0}{\rho_c} \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) \biggr\}\biggl[1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] 
+ \frac{1}{2}\biggl( \frac{4 \Psi_0}{\rho_c}\biggr)^2 \biggl[ \frac{x^2}{a^4} + \frac{y^2}{b^4} \biggr]  
- \frac{1}{2}\Omega_f^2 (x^2 + y^2)  \, .
</math>
  </td>
</tr>
</table>
The right-hand-side of this expression does not appear to be rich enough to balance the gravitational potential (on the left-hand-side) which, as [[#Trial_.232|detailed above]], contains <math>~x^2 y^2</math> and <math>~x^4</math> and <math>~y^4</math> terms.


====Exponent q = 3====
Alternatively, if <math>~q = 3</math>,

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

<tr>
  <td align="right">
<math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~6\Psi_0
\biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) +   \boldsymbol{\hat\jmath}  \biggl(\frac{y}{b^2} \biggr) \biggr]
\biggl( \frac{\rho}{\rho_c}\biggr) \biggl\{
\frac{ 2\Omega_f}{\rho_c}
+
\frac{3\Psi_0}{\rho_c^2} \biggl[
2\biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
~- ~ 4 \biggl(\frac{x^2}{ a^4} \biggr) 
~- ~ 4 \biggl(\frac{y^2}{ b^4} \biggr)  \biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 6\Psi_0
\biggl\{
\frac{ 2\Omega_f}{\rho_c} \biggl( \frac{\rho}{\rho_c}\biggr) 
-
\frac{12\Psi_0}{\rho_c^2} \biggl[
\frac{x^2}{ a^4}  
+ \frac{y^2}{ b^4}  \biggr]\biggl( \frac{\rho}{\rho_c}\biggr) 
+
\frac{6\Psi_0}{\rho_c^2} 
\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
\biggl( \frac{\rho}{\rho_c}\biggr)^2 
\biggr\} \boldsymbol{\hat{f}} \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{\hat{f}}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) +   \boldsymbol{\hat\jmath}  \biggl(\frac{y}{b^2} \biggr) \biggr] \, .
</math>
  </td>
</tr>
</table>

If we choose a function,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~F_B(\Psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
D_1\Psi^{2/3} + D_2 \Psi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
D_1\Psi_0^{2/3} \biggl(\frac{\rho}{\rho_c}\biggr)^2
+ D_2 \Psi_0 \biggl(\frac{\rho}{\rho_c}\biggr)^3 \, ,
</math>
  </td>
</tr>
</table>

we obtain,
<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[ 2D_1 \Psi_0^{2/3} \biggl(\frac{\rho}{\rho_c}\biggr) + 3D_2\Psi_0 \biggl( \frac{\rho}{\rho_c}\biggr)^2\biggr] 
\cdot \nabla \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\biggl\{ 
2D_1 \Psi_0^{2/3} \biggl[1 - \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) \biggr] 
+ 3D_2\Psi_0 \biggl[1 - \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) \biggr]^2
\biggr\} 
\cdot \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{2x}{a^2} \biggr) +   \boldsymbol{\hat\jmath}  \biggl(\frac{2y}{b^2} \biggr) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\biggl\{ 
2D_1 \Psi_0^{2/3} \biggl[1 - \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) \biggr] 
+ 3D_2\Psi_0 \biggl[1 - 2\biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) + \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr)^2 \biggr]
\biggr\} 
\cdot \biggl[ \boldsymbol{\hat\imath} \biggl(\frac{2x}{a^2} \biggr) +   \boldsymbol{\hat\jmath}  \biggl(\frac{2y}{b^2} \biggr) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\biggl\{ \biggl(2D_1 \Psi_0^{2/3}  + 3D_2\Psi_0 \biggr) -
(2D_1 \Psi_0^{2/3} +6D_2\Psi_0)\biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) 
+ 3D_2\Psi_0 \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr)^2 
\biggr\} 
\boldsymbol{\hat{f}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\biggl\{ \biggl(2D_1 \Psi_0^{2/3}  + 3D_2\Psi_0 \biggr) -
(2D_1 \Psi_0^{2/3} +6D_2\Psi_0)\biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2}\biggr) 
+ 3D_2\Psi_0 
\biggl[ \frac{x^4}{a^4} + 2\biggl( \frac{x^2 y^2}{a^2 b^2} \biggr) + \frac{y^4}{b^4}\biggr] 
\biggr\} 
\boldsymbol{\hat{f}} \, .
</math>
  </td>
</tr>
</table>


----


Let's reorganize and expand the terms in both of these expressions in order to ascertain whether or not they can be matched.  First &hellip;

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

<tr>
  <td align="right">
<math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 6\Psi_0
\biggl\{
\frac{ 2\Omega_f}{\rho_c} \biggl[ 1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] 
-
\frac{12\Psi_0}{\rho_c^2} \biggl[
\frac{x^2}{ a^4}  
+ \frac{y^2}{ b^4}  \biggr] \biggl[ 1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\frac{6\Psi_0}{\rho_c^2} 
\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
\biggl[ 1 - \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) \biggr]^2 
\biggr\} \boldsymbol{\hat{f}} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 6\Psi_0
\biggl\{
\frac{ 2\Omega_f}{\rho_c}  - \frac{ 2\Omega_f}{\rho_c} \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) 
-
\frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4}  + \frac{y^2}{ b^4}  \biggr]  
+
\frac{12\Psi_0}{\rho_c^2} \biggl[ \frac{x^2}{ a^4}  + \frac{y^2}{ b^4}  \biggr] \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\frac{6\Psi_0}{\rho_c^2} 
\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
\biggl[ 1 - 2\biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) + \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr)^2 \biggr] 
\biggr\} \boldsymbol{\hat{f}} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~ 6\Psi_0
\biggl\{
 \biggl[ \frac{ 2\Omega_f}{\rho_c} +\frac{6\Psi_0}{\rho_c^2} \cdot \frac{(a^2 + b^2)}{a^2b^2} \biggr]  
- \biggl[ \frac{ 2\Omega_f}{\rho_c}  + \frac{12\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggr]\biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-
\frac{12\Psi_0}{\rho_c^2} \biggl( \frac{x^2}{ a^4}  + \frac{y^2}{ b^4}  \biggr)  
+
\frac{12\Psi_0}{\rho_c^2}  
\biggl( \frac{x^4}{ a^6} + \frac{x^2 y^2}{ a^4 b^2} +  \frac{x^2 y^2}{ a^2 b^4} + \frac{y^4}{ b^6}\biggr)
+ \frac{6\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggl( \frac{x^4}{a^4} +  \frac{2x^2 y^2}{a^2 b^2}+ \frac{y^4}{b^4} \biggr)  
\biggr\} \boldsymbol{\hat{f}} \, ,
</math>
  </td>
</tr>
</table>

In order for the zeroth-order terms to match, we need,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2D_1 \Psi_0^{2/3}  + 3D_2\Psi_0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6\Psi_0 \biggl\{ \frac{ 2\Omega_f}{\rho_c} +\frac{6\Psi_0}{\rho_c^2} \cdot \frac{(a^2 + b^2)}{a^2b^2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 2D_1 \Psi_0^{2/3}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\Psi_0 \biggl\{ \frac{ 4\Omega_f \rho_c a^2 b^2}{\rho_c^2 a^2 b^2} +\frac{12\Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} \biggr\}
- 3D_2\Psi_0 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\Psi_0 \biggl\{ \frac{ 4\Omega_f \rho_c a^2 b^2}{\rho_c^2 a^2 b^2} +\frac{12\Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} -D_2\biggr\} \, .
</math>
  </td>
</tr>
</table>
Then we also need,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ 6D_2\Psi_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~6\Psi_0 \biggl[ \frac{ 2\Omega_f}{\rho_c}  + \frac{12\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggr]
- 2D_1 \Psi_0^{2/3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~6\Psi_0 \biggl[ \frac{ 2\Omega_f}{\rho_c}  + \frac{12\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggr]
- 
3\Psi_0 \biggl\{ \frac{ 4\Omega_f \rho_c a^2 b^2}{\rho_c^2 a^2 b^2} +\frac{12\Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} -D_2\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ 3D_2\Psi_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~6\Psi_0 \biggl[ \frac{ 2\Omega_f \rho_c a^2b^2}{\rho_c^2 a^2 b^2}  + \frac{12\Psi_0}{\rho_c^2}\frac{(a^2 + b^2)}{a^2b^2} \biggr]
- 
6\Psi_0 \biggl[ \frac{ 2\Omega_f \rho_c a^2 b^2}{\rho_c^2 a^2 b^2} +\frac{6\Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~6\Psi_0 \biggl[ \frac{6\Psi_0(a^2 + b^2)}{\rho_c^2a^2b^2}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ D_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{12\Psi_0(a^2 + b^2)}{\rho_c^2a^2b^2} \, .
</math>
  </td>
</tr>
</table>

Finally, we need,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3D_2\Psi_0 
\biggl[ \frac{x^4}{a^4} + 2\biggl( \frac{x^2 y^2}{a^2 b^2} \biggr) + \frac{y^4}{b^4}\biggr] \cdot \frac{1}{6\Psi_0}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-
\frac{12\Psi_0}{\rho_c^2} \biggl( \frac{x^2}{ a^4}  + \frac{y^2}{ b^4}  \biggr)  
+
\frac{12\Psi_0}{\rho_c^2}  
\biggl( \frac{x^4}{ a^6} + \frac{x^2 y^2}{ a^4 b^2} +  \frac{x^2 y^2}{ a^2 b^4} + \frac{y^4}{ b^6}\biggr)
+ \frac{6\Psi_0}{\rho_c^2}\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr)\biggl( \frac{x^4}{a^4} +  \frac{2x^2 y^2}{a^2 b^2}+ \frac{y^4}{b^4} \biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{(a^2 + b^2)}{a^2b^2} 
\biggl[ \frac{x^4}{a^4} + 2\biggl( \frac{x^2 y^2}{a^2 b^2} \biggr) + \frac{y^4}{b^4}\biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2 \biggl( \frac{x^2}{ a^4}  + \frac{y^2}{ b^4}  \biggr)  
+ 2  \biggl( \frac{x^4}{ a^6} + \frac{x^2 y^2}{ a^4 b^2} +  \frac{x^2 y^2}{ a^2 b^4} + \frac{y^4}{ b^6}\biggr)
+ \frac{(a^2 + b^2)}{a^2b^2} \biggl( \frac{x^4}{a^4} +  \frac{2x^2 y^2}{a^2 b^2}+ \frac{y^4}{b^4} \biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl( \frac{x^2}{ a^4}  + \frac{y^2}{ b^4}  \biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{x^4}{ a^6} + \frac{x^2 y^2}{ a^4 b^2} +  \frac{x^2 y^2}{ a^2 b^4} + \frac{y^4}{ b^6}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{x^2}{ a^4}  + \frac{y^2}{ b^4} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{x^4}{ a^6} + \frac{y^4}{ b^6} + \frac{x^2 y^2(a^2 + b^2) }{a^4 b^4}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
x^4 b^6 + y^4 a^6
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
a^2 b^2 [x^2 b^4 - (a^2+b^2)x^2 y^2 + y^2a^4]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
a^2 b^2 [x^4 b^4 - (a^2+b^2)x^2 y^2 + y^4a^4]  
+a^2b^2[x^2b^4 + y^2a^4]
-a^2b^2[x^4b^4 + y^4a^4]
</math>
  </td>
</tr>
</table>

----

Keeping in mind that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Psi_0 \biggl( \frac{\rho}{\rho_c} \biggr)^q \, ,</math>
  </td>
</tr>
</table>
and that, after setting <math>~q = 3</math>, we have chosen,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~F_B(\Psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
D_1\Psi^{2/3} + D_2 \Psi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
D_1\Psi_0^{2/3} \biggl(\frac{\rho}{\rho_c}\biggr)^2
+ D_2 \Psi_0 \biggl(\frac{\rho}{\rho_c}\biggr)^3 \, ,
</math>
  </td>
</tr>
</table>

let's try again.

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

<tr>
  <td align="right">
<math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=3} - \nabla F_B(\Psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 6\Psi_0
\biggl\{
\frac{ 2\Omega_f}{\rho_c} \biggl( \frac{\rho}{\rho_c}\biggr) 
-
\frac{12\Psi_0}{\rho_c^2} \biggl[
\frac{x^2}{ a^4}  
+ \frac{y^2}{ b^4}  \biggr]\biggl( \frac{\rho}{\rho_c}\biggr) 
+
\frac{6\Psi_0}{\rho_c^2} 
\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
\biggl( \frac{\rho}{\rho_c}\biggr)^2 
\biggr\} \cdot \boldsymbol{\hat{f}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~\biggl[ 2D_1 \Psi_0^{2/3} \biggl(\frac{\rho}{\rho_c}\biggr) + 3D_2\Psi_0 \biggl( \frac{\rho}{\rho_c}\biggr)^2\biggr] 
\cdot \nabla \biggl( \frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 3\Psi_0
\biggl\{
\frac{ 2\Omega_f}{\rho_c}  
-
\frac{12\Psi_0}{\rho_c^2} \biggl[
\frac{x^2}{ a^4}  
+ \frac{y^2}{ b^4}  \biggr] 
+
\frac{6\Psi_0}{\rho_c^2} 
\biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
\biggl( \frac{\rho}{\rho_c}\biggr) 
\biggr\} \cdot 2\biggl(\frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~\biggl[ 2D_1 \Psi_0^{2/3}  + 3D_2\Psi_0 \biggl( \frac{\rho}{\rho_c}\biggr)\biggr] 
\cdot 2\biggl(\frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} \, .
</math>
  </td>
</tr>
</table>
Now, set &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2D_1 \Psi_0^{2/3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{6\Psi_0 \Omega_f}{\rho_c}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ D_1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3\Psi_0^{1 / 3} \Omega_f}{\rho_c} \, ;</math>
  </td>
</tr>
</table>
and, set &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3D_2 \Psi_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{18 \Psi_0^2(a^2 + b^2)}{\rho_c^2 a^2 b^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ D_2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{6 \Psi_0(a^2 + b^2)}{\rho_c^2 a^2 b^2} \, .
</math>
  </td>
</tr>
</table>
We then have,

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

<tr>
  <td align="right">
<math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=3} - \nabla F_B(\Psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 3\Psi_0
\biggl\{ 
-
\frac{12\Psi_0}{\rho_c^2} \biggl[
\frac{x^2}{ a^4}  
+ \frac{y^2}{ b^4}  \biggr] 
\biggr\} \cdot 2\biggl(\frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{72\Psi_0^2}{\rho_c^2} \biggl[
\frac{x^2}{ a^4}  
+ \frac{y^2}{ b^4}  \biggr] 
\cdot \biggl(\frac{\rho}{\rho_c}\biggr)\boldsymbol{\hat{f}} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\bold{u}\cdot \bold{u}) \nabla \biggl(\frac{\rho}{\rho_c}\biggr) \, .
</math>
  </td>
</tr>
</table>
<font color="red">'''VERY INTERESTING!''' (29 September 2020)</font>

====Exponent q = 4====

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

<tr>
  <td align="right">
<math>~\biggl[ (\vec\zeta + 2\vec\Omega) \times \bold{u} \biggr]_{q=4}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~8\Psi_0
\biggl[ \boldsymbol{\hat\imath} \biggl(\frac{x}{a^2} \biggr) +   \boldsymbol{\hat\jmath}  \biggl(\frac{y}{b^2} \biggr) \biggr]
\biggl( \frac{\rho}{\rho_c}\biggr)^{2} \biggl\{
\frac{ 2\Omega_f}{\rho_c}
+
\frac{8\Psi_0}{\rho_c^2} \biggl(\frac{\rho}{\rho_c}\biggr) \biggl[
\biggl( \frac{\rho}{\rho_c}\biggr) \biggl( \frac{1}{a^2} + \frac{1}{b^2}\biggr) 
~- ~ 4 \biggl(\frac{x^2}{ a^4} \biggr) 
~- ~ 4 \biggl(\frac{y^2}{ b^4} \biggr)  \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>