Editing AxisymmetricConfigurations/SolutionStrategies (section)

==Double Check Vector Identities==

Let's plug a few different [[AxisymmetricConfigurations/SolutionStrategies#Simple_Rotation_Profile_and_Centrifugal_Potential|simple rotation profiles]] into the Euler equation, using a cylindrical-coordinate base to demonstrate that the three expressions are identical, namely, that
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(\vec{v} \cdot \nabla) \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\nabla \Psi \, .</math>
  </td>
</tr>
</table>

===Uniform Rotation===
In the case of uniform rotation, we have,
<div align="center">
<math>~\vec{v} = \hat{e}_\varphi (v_\varphi) = \hat{e}_\varphi (\varpi \omega_0) ~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \frac{(\varpi v_\varphi)^2}{\varpi^3} = \frac{(\varpi^2\omega_0)^2}{\varpi^3} = \varpi \omega_0^2\, ,</math>
</div>
where, <math>~\omega_0</math> is independent of radial position.  This also means that,
<div align="center">
<math>
\Psi \equiv - \int \frac{j^2(\varpi)}{\varpi^3} d\varpi = - \frac{1}{2} \varpi^2 \omega_0^2~;
</math>
</div>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec\zeta = \nabla \times \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_\varpi \biggl[ -\cancel{ \frac{\partial v_\varphi}{\partial z} }\biggr] + \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi v_\varphi)}{\partial \varpi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi^2 \omega_0 )}{\partial \varpi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z ( 2\omega_0 )
</math>
  </td>
</tr>
</table>

[A] &nbsp; Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(\vec{v} \cdot \nabla) \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi \biggl[ - \frac{(\varpi \omega_0)\cdot (\varpi \omega_0)}{\varpi} \biggr] = - \hat{e}_\varpi (\varpi \omega_0^2) \, .</math>
  </td>
</tr>
</table>

[B] &nbsp; Alternatively,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_z ( 2\omega_0 ) \times \hat{e}_\varphi (\varpi \omega_0) + \hat{e}_\varpi \frac{1}{2} \biggl[ \frac{\partial}{\partial\varpi} (\varpi^2 \omega_0^2) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \hat{e}_\varpi  \biggl\{ -( 2\omega_0 ) (\varpi \omega_0) + (\varpi \omega_0^2)  \biggr\} = - \hat{e}_\varpi (\varpi \omega_0^2) \, .</math>
  </td>
</tr>
</table>

[C] &nbsp; Or,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla \Psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi \biggl[- \frac{1}{2} \frac{\partial}{\partial\varpi} (\varpi^2 \omega_0^2) \biggr] = - \hat{e}_\varpi (\varpi \omega_0^2) \, .</math>
  </td>
</tr>
</table>
This demonstrates that, in the case of uniform angular velocity, all three expressions are identical.

===Power Law===
In the case of a power-law expression, we have,
<div align="center">
<math>~\vec{v} = \hat{e}_\varphi (v_\varphi) = \hat{e}_\varphi \biggl[ \frac{j_0}{\varpi_0^2} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(q-1)} \biggr] 
~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \biggl[ \frac{j_0^2}{\varpi_0^3} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(2q-3)} \biggr] \, ,</math>
</div>
where, <math>~j_0</math> and <math>~\varpi_0</math> are both independent of radial position.  This also means that,
<div align="center">
<math>
\Psi \equiv - \int \frac{j^2(\varpi)}{\varpi^3} d\varpi = - \frac{1}{2(q-1)} \biggl[ \frac{j_0^2}{\varpi_0^2} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{2(q-1)} \biggr]~;
</math>
</div>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec\zeta = \nabla \times \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_\varpi \biggl[ -\cancel{ \frac{\partial v_\varphi}{\partial z} }\biggr] + \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi v_\varphi)}{\partial \varpi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z~ \frac{1}{\varpi} \frac{\partial }{\partial \varpi} \biggl[ \frac{j_0}{\varpi_0} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{q} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z~ \frac{q}{\varpi} \biggl[ \frac{j_0}{\varpi_0^{q+1}} ( \varpi)^{q-1} \biggr] 
=
\hat{e}_z~ q \biggl[ \frac{j_0}{\varpi_0^{3}} \biggl( \frac{\varpi}{\varpi_0} \biggr)^{q-2} \biggr]\, .
</math>
  </td>
</tr>
</table>

[D] &nbsp; Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(\vec{v} \cdot \nabla) \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \hat{e}_\varpi \frac{1}{\varpi} \biggl[ \frac{j_0^2}{\varpi_0^4} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{2(q-1)} \biggr]  
= - \hat{e}_\varpi \biggl[ \frac{j_0^2}{\varpi_0^5} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(2q-3)} \biggr]\, .</math>
  </td>
</tr>
</table>

[E] &nbsp; Alternatively,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_z~ q \biggl[ \frac{j_0}{\varpi_0^{3}} \biggl( \frac{\varpi}{\varpi_0} \biggr)^{q-2} \biggr] \times 
\hat{e}_\varphi \biggl[ \frac{j_0}{\varpi_0^2} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(q-1)} \biggr] 
+ \hat{e}_\varpi \frac{1}{2}  \frac{\partial}{\partial\varpi} \biggl[ \frac{j_0^2}{\varpi_0^4} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(2q-2)} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\hat{e}_\varpi~ q \biggl[ \frac{j_0^2}{\varpi_0^{5}} \biggl( \frac{\varpi}{\varpi_0} \biggr)^{2q-3} \biggr]  
+ \hat{e}_\varpi(q-1)  \biggl[ \frac{j_0^2}{\varpi_0^5} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(2q-3)} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\hat{e}_\varpi~ \biggl[ \frac{j_0^2}{\varpi_0^{5}} \biggl( \frac{\varpi}{\varpi_0} \biggr)^{2q-3} \biggr]  \, .
</math>
  </td>
</tr>
</table>

[F] &nbsp; Or,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla \Psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi\frac{\partial}{\partial\varpi} \biggl\{- \frac{1}{2(q-1)} \biggl[ \frac{j_0^2}{\varpi_0^2} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{2(q-1)} \biggr] \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- ~\hat{e}_\varpi\frac{\partial}{\partial\varpi} \biggl[ \frac{j_0^2}{\varpi_0^3} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{2q-3} \biggr] </math>
  </td>
</tr>
</table>
This demonstrates that, in the case of power-law angular velocity profile, all three expressions are identical.

===Uniform v<sub>&#x03C6;</sub>===
In the case of a uniform <math>~v_\varphi</math> (i.e., a flat rotation curve), we have,
<div align="center">
<math>~\vec{v} = \hat{e}_\varphi (v_\varphi) = \hat{e}_\varphi v_0 
~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} =  \frac{v_0^2}{\varpi} \, ,</math>
</div>
where, <math>~v_0</math> is independent of radial position.  This also means that,
<div align="center">
<math>
\Psi \equiv - \int \frac{j^2(\varpi)}{\varpi^3} d\varpi = - v_0^2 \ln \biggl( \frac{\varpi}{\varpi_0} \biggr)~;
</math>
</div>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec\zeta = \nabla \times \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_\varpi \biggl[ -\cancel{ \frac{\partial v_\varphi}{\partial z} }\biggr] + \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi v_\varphi)}{\partial \varpi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z \biggl( \frac{v_0}{\varpi} \biggr) \, .
</math>
  </td>
</tr>
</table>

[G] &nbsp; Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(\vec{v} \cdot \nabla) \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] = -~\hat{e}_\varpi \biggl[ \frac{v_0^2}{\varpi} \biggr] \, .</math>
  </td>
</tr>
</table>

[H] &nbsp; Alternatively,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z \biggl( \frac{v_0}{\varpi} \biggr) \times \hat{e}_\varphi v_0
+ \hat{e}_\varpi~ \frac{1}{2} \frac{\partial}{\partial \varpi} (v_0^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\hat{e}_\varpi \biggl( \frac{v_0^2}{\varpi} \biggr) \, .
</math>
  </td>
</tr>
</table>

[I] &nbsp; Or,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla \Psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi\frac{\partial}{\partial\varpi} \biggl\{- v_0^2 \ln \biggl( \frac{\varpi}{\varpi_0} \biggr)\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ \hat{e}_\varpi v_0^2  \biggl(\frac{\varpi}{\varpi_0} \biggr)^{-1} \frac{1}{\varpi_0}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ \hat{e}_\varpi \biggl( \frac{v_0^2}{\varpi} \biggr)  \, .</math>
  </td>
</tr>
</table>
This demonstrates that, in the case of a constant <math>~v_\varphi</math> profile, all three expressions are identical.

===j-Constant Rotation===
In the case of so-called j-constant rotation, we have,
<div align="center">
<math>~\vec{v} = \hat{e}_\varphi (v_\varphi) = \hat{e}_\varphi ~\omega_c \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr] 
~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \frac{(\varpi v_\varphi)^2}{\varpi^3} = 
\frac{\omega_c^2}{\varpi} \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr]^2 =
\biggl[ \frac{\omega_c^2 A^4\varpi}{(A^2 + \varpi^2)^2}\biggr]
\, ,
</math>
</div>

where, <math>~\omega_c</math>, and the characteristic length, <math>~A</math>, are both independent of radial position.  This also means that,

<div align="center">
<math>
\Psi \equiv - \int \frac{j^2(\varpi)}{\varpi^3} d\varpi =  +\frac{1}{2}\biggl[ \frac{\omega_c^2 A^4}{(A^2 + \varpi^2)}\biggr]~;
</math>
</div>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec\zeta = \nabla \times \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_\varpi \biggl[ -\cancel{ \frac{\partial v_\varphi}{\partial z} }\biggr] + \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi v_\varphi)}{\partial \varpi} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z \biggl\{ \frac{\omega_c}{\varpi} \frac{\partial }{\partial \varpi}  \biggl[ \frac{A^2\varpi^2}{A^2 + \varpi^2}\biggr]\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z~ \frac{\omega_c}{\varpi}  \biggl\{ \biggl[ 2A^2\varpi(A^2 + \varpi^2)^{-1} \biggr] - \biggl[ 2A^2\varpi^3(A^2 + \varpi^2)^{-2} \biggr]\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z~  \biggl[2\omega_c A^4 (A^2 + \varpi^2)^{-2} \biggr] \, .
</math>
  </td>
</tr>
</table>

[J] &nbsp; Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(\vec{v} \cdot \nabla) \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\hat{e}_\varpi \frac{\omega_c^2}{\varpi} \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr]^2 
=
-~\hat{e}_\varpi \biggl[ \frac{\omega_c^2A^4 \varpi}{(A^2 + \varpi^2)^2} \biggr] \, .
</math>
  </td>
</tr>
</table>

[K] &nbsp; Alternatively,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_z~  \biggl[2\omega_c A^4 (A^2 + \varpi^2)^{-2} \biggr] \times \hat{e}_\varphi ~\omega_c \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr] + \frac{1}{2} \hat{e}_\varpi \frac{\partial}{\partial \varpi}\biggl[ \omega_c^2 A^4\varpi^2 (A^2 + \varpi^2)^{-2}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \hat{e}_\varpi ~  \biggl[ \frac{2\omega_c^2 A^6 \varpi }{(A^2 + \varpi^2)^{3}} \biggr] + \hat{e}_\varpi \biggl[  \omega_c^2 A^4\varpi (A^2 + \varpi^2)^{-2} - 2 \omega_c^2 A^4\varpi^3 (A^2 + \varpi^2)^{-3}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 \hat{e}_\varpi \biggl[ \frac{ \omega_c^2 A^4\varpi }{ (A^2 + \varpi^2)^{2}} - \frac{2 \omega_c^2 A^4\varpi^3}{ (A^2 + \varpi^2)^{3} }  
- \frac{2\omega_c^2 A^6 \varpi }{(A^2 + \varpi^2)^{3}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 \hat{e}_\varpi \biggl[  (A^2 + \varpi^2) - 2 \varpi^2  - 2A^2 \biggr] \frac{ \omega_c^2 A^4\varpi }{ (A^2 + \varpi^2)^{3}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~ \hat{e}_\varpi \biggl[ \frac{ \omega_c^2 A^4\varpi }{ (A^2 + \varpi^2)^{2}} \biggr] \, .
</math>
  </td>
</tr>
</table>

[L] &nbsp; Or,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla \Psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi ~ \frac{1}{2} \frac{\partial}{\partial \varpi}\biggl[ \omega_c^2 A^4 (A^2 + \varpi^2)^{-1} \biggr] 
= - \hat{e}_\varpi  \biggl[ \frac{ \omega_c^2 A^4 \varpi }{ (A^2 + \varpi^2)^{2}} \biggr] \, .</math>
  </td>
</tr>
</table>
This demonstrates that, in the case of a j-constant rotation profile, all three expressions are identical.