Editing ThreeDimensionalConfigurations/Challenges (section)

====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>