Editing ThreeDimensionalConfigurations/Challenges (section)

===Part I===
Returning to the above-mentioned,

<div align="center">
<font color="#770000">'''Eulerian Representation'''</font><br />
of the Euler Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\frac{\partial \bold{u}}{\partial t} + (\bold{u}\cdot \nabla) \bold{u} = - \frac{1}{\rho} \nabla P - \nabla \Phi_\mathrm{grav} 
- {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) - 2{\vec{\Omega}}_f \times \bold{u} \, ,</math> </div>

we next note &#8212; as we have done in our [[PGE/Euler#in_terms_of_the_vorticity:|broader discussion of the Euler equation]] &#8212; that,
<div align="center">
<math> 
(\bold{u} \cdot\nabla)\bold{u} = \frac{1}{2}\nabla(\bold{u} \cdot \bold{u}) - \bold{u} \times(\nabla\times\bold{u})
= \frac{1}{2}\nabla(u^2) + \boldsymbol\zeta \times \bold{u} ,
</math>
</div>
where, as above, <math>\boldsymbol\zeta \equiv \nabla\times \bold{u}</math> is the [https://en.wikipedia.org/wiki/Vorticity vorticity].  Making this substitution, we obtain what is essentially equation (7) of [http://adsabs.harvard.edu/abs/1996ApJS..105..181K KP96], that is, 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>
Hence, in steady-state, the Euler equation becomes,
<div align="center">
<math>
\nabla F_B  + \vec{A} = 0 ,
</math>
</div>
where, the scalar "Bernoulli" function,
<div align="center">
<math>
F_B  \equiv  \frac{1}{2}u^2 + H + \Phi - \frac{1}{2}|\Omega\hat{k} \times \vec{x}|^2 ;
</math>
</div>
and,
<div align="center">
<math>
\vec{A} \equiv ({\boldsymbol\zeta}+2{\vec\Omega}_f) \times {\bold{u}} .
</math>
</div>


For later use &hellip;
<table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left">
<ol>
<li><font color="red">Curl of steady-state Euler equation:</font>  
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\nabla F_B + \bold{A}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\nabla \times \biggl[ \nabla F_B + \bold{A} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\nabla \times  \bold{A} \, .</math>
  </td>
</tr>
</table>
This last step is justified because the [https://en.wikipedia.org/wiki/Vector_calculus_identities#Curl_of_gradient_is_zero curl of any gradient is zero].
</li>
<li>
<font color="red">KP96 only deal with two-dimensional motion in the equatorial plane and, hence, there is no vertical motion:</font><br />
Hence, <math>~\bold{u}</math> lies in the equatorial plane; both <math>~\vec\zeta</math> and <math>~\vec\Omega_f</math> only have z-components; and, <math>~\bold{A}</math> is perpendicular to both <math>~\vec\Omega_f</math> and <math>~\bold{u}</math>.  Also, given that <math>~\bold{A}</math> necessarily lies in the equatorial plane, its curl will only have a z-component, that is,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\nabla \times  \bold{A} = 0</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~\Leftrightarrow</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~[\nabla \times  \bold{A}]_z = 0 \, .</math>
  </td>
</tr>
</table>
</li>
<li>
<font color="red">"Dot" <math>~\bold{u}</math> into the steady-state Euler equation:</font>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\nabla F_B + \bold{A}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{u} \cdot \biggl[ \nabla F_B + \bold{A} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{u} \cdot \nabla F_B  \, .</math>
  </td>
</tr>
</table>
This last step is justified because <math>~\bold{A}</math> is necessarily always perpendicular to <math>~\bold{u}</math>.
</li>
</ol>
</td></tr></table>


----


We will leave discussion of the Euler equation, for the moment, and instead look at the continuity equation.  As viewed from the rotating frame of reference,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial (\rho \bold{u})}{\partial t} + \nabla\cdot (\rho\bold{u})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>

If we are able to write the momentum density (in the rotating frame) in terms of a stream-function, <math>~\Psi</math>, such that,
<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>~
\nabla \times (\boldsymbol{\hat{k}} \Psi) 
=
\boldsymbol{\hat\imath} \biggl[ \frac{\partial \Psi}{\partial y} \biggr]
- \boldsymbol{\hat\jmath} \biggl[  \frac{\partial \Psi}{\partial x}\biggr] \, ,</math>
  </td>
</tr>
</table>
then satisfying the steady-state continuity equation is guaranteed because the [https://en.wikipedia.org/wiki/Vector_calculus_identities#Divergence_of_curl_is_zero divergence of a curl is always zero].  Note, as well, that when written in terms of this stream-function, the z-component of the vorticity will be,
<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{\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>

Note that the steady-state continuity equation may be rewritten in the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla\cdot \bold{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \bold{u} \cdot \nabla [ \ln \rho] \, .
</math>
  </td>
</tr>
</table>



----



It can also be shown that the condition, <math>~[\nabla \times \bold{A}]_z = 0</math> can be rewritten as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla\cdot \bold{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \bold{u} \cdot \nabla [ \ln(2\Omega_f + \zeta_z] \, .
</math>
  </td>
</tr>
</table>

By combining these last two expressions, we appreciate that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\bold{u} \cdot \nabla \ln \biggl[ \frac{(2\Omega_f + \zeta_z)}{\rho} \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
This means that, in the steady-state flow whose spatial structure we are seeking, the velocity vector, <math>~\bold{u}</math> (and also the momentum density vector, <math>~\rho \bold{u}</math>) must everywhere be tangent to contours of constant ''vortensity'', <math>~[(2\Omega_f + \zeta_z)/\rho]</math>.

We need a function <math>~g(\Psi) </math> such that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~g(\Psi) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\rho} \biggl\{
\zeta_z + 2\Omega_f
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\rho} \biggl\{
2\Omega_f
-
\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] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>

Let's try, <math>~\Psi = \rho^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>

Then,

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

<tr>
  <td align="right">
<math>~g(\Psi) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\rho} \biggl\{
2\Omega_f
-
\frac{\partial }{\partial x}\biggl[\frac{1}{\rho} \frac{\partial \rho^2}{\partial x}  \biggr] 
- \frac{\partial }{\partial y} \biggl[\frac{1}{\rho} \frac{\partial \rho^2}{\partial y}  \biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\rho} \biggl\{
2\Omega_f
-
2 \biggl[ \frac{\partial^2 \rho}{\partial x^2}  \biggr] 
- 2 \biggl[\frac{\partial^2 \rho}{\partial y^2}  \biggr] 
\biggr\} 
</math>
  </td>
</tr>

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

Hence,

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

<tr>
  <td align="right">
<math>~g(\Psi) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
2\Omega_f
+
4\rho_c \biggl[ \frac{1}{a^2} + \frac{1}{b^2}  \biggr] 
\biggr\} \Psi^{-1 / 2} \, .
</math>
  </td>
</tr>
</table>

Next, given that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dF_B}{d\Psi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- g(\Psi) \, ,</math>
  </td>
</tr>
</table>
we conclude that, to within an additive constant,

<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>~
- \int g(\Psi) d\Psi
= - g_0 \int \Psi^{-1 / 2} d\Psi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2g_0 \Psi^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2g_0 \rho \, , 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~g_0</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl\{ 2\Omega_f + 4\rho_c \biggl[ \frac{1}{a^2} + \frac{1}{b^2}  \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>

Here's what to do:

Given <math>~g(\Psi)</math>, write out the functional forms of <math>~\bold{A}</math> and <math>~F_B(\Psi)</math>.  Then see if <math>~\nabla F_B = - \bold{A}</math>.