Editing ThreeDimensionalConfigurations/Challenges (section)

===Trial #4===
We begin with the,

<div align="center">
Euler Equation<br />
written <font color="#770000">'''in terms of the Vorticity'''</font> and<br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\frac{\partial \bold{u}}{\partial t} + (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi + \frac{1}{2}u^2 - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]</math> .
</div>

Next, we rewrite this expression to incorporate the following three realizations:
<ul>
<li>For a barotropic fluid, the term involving the pressure gradient can be replaced with a term involving the enthalpy via the relation, <math>~\nabla H = \nabla P/\rho</math>.</li>
<li>The expression for the centrifugal potential can be rewritten as, <math>~\tfrac{1}{2}|\vec\Omega_f \times \vec{x}|^2 = \tfrac{1}{2}\Omega_f^2 (x^2 + y^2)</math>.</li>
<li>In steady state, <math>~\partial \bold{u}/\partial t = 0</math>.</li>
</ul>
This means that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>- \nabla \biggl[H + \Phi_\mathrm{grav} + \frac{1}{2}u^2 - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] \, .</math>
  </td>
</tr>
</table>

If the term on the left-hand-side of this equation can be expressed in terms of the gradient of a scalar function, then it can be readily grouped with all the other terms on the right-hand-side, which already are in the gradient form.

Building on the insight that we have gained from the [[#Exponent_q_.3D_3|above examination of systems for which the exponent, q = 3]], let's change the <math>~\tfrac{1}{2}\nabla u^2</math> term on the RHS to  <math>~\tfrac{1}{2}\nabla (\rho u)^2</math> then reexamine the LHS.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ [ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}} ]_{q=3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>- \nabla \biggl[H + \Phi_\mathrm{grav}  - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] - \nabla \biggl[\frac{1}{2}u^2\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- \nabla \biggl[H + \Phi_\mathrm{grav}  - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] 
- \biggl(\frac{\rho}{\rho_c}\biggr)^{-2} \biggl\{ \biggl(\frac{\rho}{\rho_c}\biggr)^{2} \nabla \biggl[\frac{1}{2}u^2\biggr]\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- \nabla \biggl[H + \Phi_\mathrm{grav}  - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] 
- \biggl(\frac{\rho}{\rho_c}\biggr)^{-2} \biggl\{ 
\nabla \biggl[\frac{1}{2}\biggl(\frac{\rho}{\rho_c}\biggr)^{2} u^2\biggr]
- \frac{1}{2}u^2\nabla \biggl[\biggl(\frac{\rho}{\rho_c}\biggr)^{2}\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ [ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}} ]_{q=3} - \nabla F_B</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- \nabla \biggl[H + \Phi_\mathrm{grav} + F_B  - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] 
- \biggl(\frac{\rho}{\rho_c}\biggr)^{-2} \biggl\{ 
\nabla \biggl[\frac{1}{2}\biggl(\frac{\rho}{\rho_c}\biggr)^{2} u^2\biggr]
- \frac{1}{2}u^2\nabla \biggl[\biggl(\frac{\rho}{\rho_c}\biggr)^{2}\biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Now, rewriting the LHS gives,

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

<tr>
  <td align="right">
<math>~[ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}} ]_{q=3} - \nabla F_B</math>
  </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>

<tr>
  <td align="right">
&nbsp;
  </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{36\Psi_0^2}{\rho_c^2} \biggl[
\frac{x^2}{ a^4}  
+ \frac{y^2}{ b^4}  \biggr] 
\cdot \nabla \biggl(\frac{\rho}{\rho_c}\biggr)^2 \, ,
</math>
  </td>
</tr>
</table>
where we have set,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~D_1 = \frac{3\Psi_0^{1 / 3} \Omega_f}{\rho_c}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~D_2 = \frac{6\Psi_0 (a^2 + b^2)}{\rho_c^2 a^2 b^2} \, .</math>
  </td>
</tr>
</table>

Notice that when the exponent, <math>~q=3</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[ \bold{u}\cdot \bold{u} ]_{q=3} </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\}_{q=3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\rho^2} \biggl\{
\biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{2} \biggl(\frac{6 \Psi_0 y}{b^2} \biggr)  \biggr]^2 
+ \biggl[ \biggl( \frac{\rho}{\rho_c}\biggr)^{2} \biggl(\frac{6 \Psi_0x}{a^2} \biggr) \biggr]^2
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{36\Psi_0^2}{\rho_c^2}
\biggl( \frac{\rho}{\rho_c}\biggr)^2 
\biggl[ \frac{x^2}{a^4}  + \frac{ y^2}{b^4} \biggr] \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~[ (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}} ]_{q=3} - \nabla F_B</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~
\biggl( \frac{\rho}{\rho_c}\biggr)^{-2}  u^2
\cdot \nabla \biggl(\frac{\rho}{\rho_c}\biggr)^2 \, .
</math>
  </td>
</tr>
</table>