Editing ThreeDimensionalConfigurations/Challenges (section)

====Uncluttered Setup====

Let's simply look at the vortensity expression as defined in [[#Part_II|Part II, above]], namely, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~g(\Psi)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(2\Omega_f + \zeta_z)}{\rho_c\sigma} \, ,</math>
  </td>
</tr>
</table>
and recognize that we are ultimately interested in the function, <math>~F_B(\Psi)</math>, defined such that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dF_B(\Psi)}{d\Psi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-g(\Psi) \, .</math>
  </td>
</tr>
</table>

We start with the expression for the z-component of the vorticity,
<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>

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

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

<tr>
  <td align="right">
<math>~\frac{1}{\sigma} \frac{\partial \sigma^q}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
q \sigma^{q-2} \frac{\partial \sigma}{\partial x_i}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{q}{q-1} \biggr) \frac{\partial \sigma^{q-1}}{\partial x_i} \, .
</math>
  </td>
</tr>
</table>

Hence,
<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{q\Psi_0}{\rho_c (q-1) } \biggl\{ \frac{\partial }{\partial x}\biggl[ \frac{\partial \sigma^{q-1}}{\partial x}  \biggr] 
+ \frac{\partial }{\partial y} \biggl[ \frac{\partial \sigma^{q-1}}{\partial y}  \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~ \biggl[ \frac{q\Psi_0}{\rho_c (q-1) } \biggr] \nabla^2\sigma^{q-1}
\, .
</math>
  </td>
</tr>
</table>

So, the vortensity is,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~g(\sigma) = \frac{(2\Omega_f + \zeta_z )}{\rho_c\sigma}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{2\Omega_f}{\rho_c}\biggr)\sigma^{-1} 
- 
\biggl[ \frac{q\Psi_0}{\rho_c^2 (q-1) } \biggr] \sigma^{-1} \nabla^2\sigma^{q-1} \, .
</math>
  </td>
</tr>
</table>

Let's switch to the stream-function via the ''assumed'' relation, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\Psi}{\Psi_0}\biggr)^{1/q} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ g(\Psi) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{2\Omega_f}{\rho_c}\biggr)\biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q}  
- 
\biggl[ \frac{q\Psi_0}{\rho_c^2 (q-1) } \biggr] \biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q} \nabla^2\biggl( \frac{\Psi}{\Psi_0}\biggr)^{(q-1)/q} \, .
</math>
  </td>
</tr>
</table>

This expression gives the vortensity in what appears to be the desired form &#8212; that is, expressed strictly in terms of the stream function, <math>~\Psi</math> &#8212; for a wide range of values of the exponent, <math>~q</math>.  [<font color="red">'''CAUTION:'''</font> &nbsp; the <math>~\nabla^2</math> operator is an exception.]  It is not yet (13 October 2020) clear to me how &#8212; or if &#8212; the second term on the right-hand-side of this expression can be integrated to give <math>~F(\Psi)</math>.  But the first term can be obtained from,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~F_{1^\mathrm{st}}(\Psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl(\frac{2\Omega_f}{\rho_c}\biggr) \frac{q}{(q-1)} \biggl( \frac{\Psi}{\Psi_0}\biggr)^{(q-1)/q}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dF_{1^\mathrm{st}}(\Psi)}{d\Psi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl(\frac{2\Omega_f}{\rho_c}\biggr)  \biggl( \frac{\Psi}{\Psi_0}\biggr)^{-1/q} \, .
</math>
  </td>
</tr>
</table>