Editing ThreeDimensionalConfigurations/Stability/RiemannEllipsoids (section)

===Lagrangian Displacement and Linearization===

Suppose that, at time <math>t = 0</math>, the function set <math>[\mathbf{u}_0(\mathbf{x}), \rho_0(\mathbf{x}), p_0(\mathbf{x})]</math> properly describes the properties of a &#8212; as yet unspecified &#8212; geometrically extended, fluid configuration. <!-- Now suppose that the entire fluid configuration is "perturbed." -->  According to the Euler equation and, in particular, as dictated by the flow-field,
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right"><math>\mathbf{u}_0(\mathbf{x})</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[\boldsymbol{\hat\imath} u_x(\mathbf{x}) + \boldsymbol{\hat\jmath} u_y(\mathbf{x}) + \mathbf{\hat{k}} u_z(\mathbf{x}) \biggr]_0
\ ,
</math>
  </td>
</tr>
</table>

after an interval of time, <math>t</math>, each "Lagrangian" fluid element will move from its initial location, <math>\mathbf{x}</math>, to a new position, <math>\mathbf{x} + \boldsymbol\xi</math>.  In general each Lagrangian fluid element will discover that, at its new coordinate location, the "environment" is different.  For example,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>p_0(\mathbf{x})</math></td>
  <td align="center"><math>~~\rightarrow~~</math></td>
  <td align="left"><math>p(\mathbf{x} + \boldsymbol\xi,t) \, ,</math></td>
</tr>
<tr>
  <td align="right"><math>\rho_0(\mathbf{x})</math></td>
  <td align="center"><math>~~\rightarrow~~</math></td>
  <td align="left"><math>\rho(\mathbf{x} + \boldsymbol\xi,t) \, ,</math></td>
</tr>
<tr>
  <td align="right"><math>[u_i(\mathbf{x})]_0</math></td>
  <td align="center"><math>~~\rightarrow~~</math></td>
  <td align="left"><math>u_i(\mathbf{x} + \boldsymbol\xi,t) \, .</math></td>
</tr>
</table>
With this in mind, {{ Lebovitz89ahereafter }} <font color="red">introduces a ''Lagrangian-change operator''</font>, <math>\Delta</math>, in order to mathematically indicate that this evolutionary step is being executed for any physical variable, <math>F</math>.  Specifically, 
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right"><math>\Delta F</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>F(\mathbf{x} + \boldsymbol\xi,t) - F_0(\mathbf{x}) \, .</math></td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, §2, p. 223, Eq. (3)<br />
{{ LBO67hereafter }}, p. 293, Eq. (1)
  </td>
</tr>
</table>

Following {{ Lebovitz89ahereafter }} and applying the operator, <math>\Delta</math>, to each side of the Euler equation, we can write,
<table border="0" cellpadding="3" align="center">
<tr>
  <td align="right">
<math>\Delta \biggl\{ \frac{D\mathbf{u}}{Dt} \biggr\} 
+ \Delta\biggl\{ 2\boldsymbol\omega \boldsymbol\times \mathbf{u} \biggr\}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \Delta\biggl\{ \rho^{-1} \nabla p \biggr\} 
+ \Delta\biggl\{\mathbf\nabla \biggl[ \Phi_\mathrm{L89} + \tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]\biggr\}
\, .
</math>
  </td>
</tr>
</table>

====LHS====
With the assurance provided by {{ Lebovitz89ahereafter }} that <math>\Delta</math> commutes with the Lagrangian time-derivative, <math>D/Dt</math> &#8212; see also the paragraph immediately preceding Eq. (4) in {{ LBO67hereafter }} &#8212; and that
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right"><math>\Delta \mathbf{u}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{D\boldsymbol\xi}{Dt} \, ,</math></td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, §2, p. 223, Eq. (4)
  </td>
</tr>
</table>

we can immediately appreciate that,
<table border="0" cellpadding="3" align="center">
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="right">
<math>\frac{D}{Dt} \biggl[\Delta\mathbf{u} \biggr] 
+ 2\boldsymbol\omega \boldsymbol\times \biggl[ \Delta \mathbf{u}\biggr] </math>
=
<math>\frac{D}{Dt} \biggl[\frac{D\boldsymbol{\xi}}{Dt} \biggr] 
+ 2\boldsymbol\omega \boldsymbol\times \biggl[\frac{D\boldsymbol{\xi}}{Dt} \biggr] \, .</math>
  </td>
</tr>
</table>
Hence, we obtain the (still, exact nonlinear),
<table border="0" cellpadding="3" align="center">
<tr>
  <td align="center" colspan="3">
<font color="#770000">'''(Lagrangian) Perturbed Euler Equation'''</font>
  </td>
</tr>
<tr>
  <td align="right">
<math>\frac{D^2\boldsymbol{\xi}}{Dt^2}  
+ 2\boldsymbol\omega \boldsymbol\times \biggl[\frac{D\boldsymbol{\xi}}{Dt} \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \Delta\biggl\{ \rho^{-1} \nabla p \biggr\} 
+ \Delta\biggl\{\mathbf\nabla \biggl[ \Phi_\mathrm{L89} + \tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]\biggr\}
\, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, §2, p. 223, Eq. (5)
  </td>
</tr>
</table>

For later reference, notice that the LHS may further be rewritten as,
<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{D}{Dt} \biggl[\boldsymbol{\xi}_t + (\mathbf{u}\cdot \nabla)\boldsymbol\xi \biggr] 
+ 2\boldsymbol\omega \boldsymbol\times \biggl[\boldsymbol\xi_t 
+ (\mathbf{u}\cdot \nabla)\boldsymbol\xi  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\partial }{\partial t} + (\mathbf{u}\cdot \nabla)\biggr]
\biggl[\boldsymbol{\xi}_t + (\mathbf{u}\cdot \nabla)\boldsymbol\xi \biggr] 
+ 2\boldsymbol\omega \boldsymbol\times \biggl[\boldsymbol\xi_t 
+ (\mathbf{u}\cdot \nabla)\boldsymbol\xi  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\xi}_{tt} + \frac{\partial }{\partial t} \biggl[(\mathbf{u}\cdot \nabla)\boldsymbol\xi \biggr] 
+
(\mathbf{u}\cdot \nabla)
\biggl[\boldsymbol{\xi}_t + (\mathbf{u}\cdot \nabla)\boldsymbol\xi \biggr] 
+ 2\boldsymbol\omega \boldsymbol\times \biggl[\boldsymbol\xi_t 
+ (\mathbf{u}\cdot \nabla)\boldsymbol\xi  \biggr] 
</math>
  </td>
</tr>
</table>

where we have adopted {{ Lebovitz89ahereafter }}'s shorthand notation,
<table border="0" cellpadding="3" align="center">
<tr>
  <td align="right">
<math>
\boldsymbol\xi_{t} \equiv \frac{\partial \boldsymbol\xi}{\partial t}  \, ,
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; and, &nbsp; &nbsp;
  </td>
  <td align="right">
<math>
\boldsymbol\xi_{tt} \equiv \frac{\partial^2 \boldsymbol\xi}{\partial t^2}  \, .
</math>
  </td>
</tr>
</table>
Finally, if <font color="green">&hellip; the unperturbed solution &hellip; is steady</font> &#8212; as is the case in the context of our study of the stability of Riemann S-Type ellipsoids (see more, below) &#8212; then <math>(\mathbf{u}\cdot \nabla)</math> commutes with the Eulerian time-derivative, that is,
<div align="center"><math>\frac{\partial}{\partial t}\biggl[ (\mathbf{u}\cdot \nabla) \boldsymbol\xi \biggr] ~\rightarrow ~  (\mathbf{u}\cdot \nabla) \boldsymbol\xi_t \, ,</math></div>
which means we may write,
<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\xi}_{tt} + 2\{ (\mathbf{u}\cdot \nabla)\boldsymbol\xi_t 
+ \boldsymbol\omega \boldsymbol\times \boldsymbol\xi_t \}
+ \{ (\mathbf{u}\cdot \nabla)^2\boldsymbol\xi 
+ 2\boldsymbol\omega \boldsymbol\times [(\mathbf{u}\cdot \nabla)\boldsymbol\xi ] \} \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, §2, p. 224, immediately preceding Eq. (10)
  </td>
</tr>
</table>

====RHS====

Next, {{ Lebovitz89ahereafter }} <font color="red">introduces the ''Eulerian-change operator''</font>, <math>\delta</math> (which commutes with <math>\nabla</math>),
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right"><math>\delta F</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>F(\mathbf{x},t) - F_0(\mathbf{x}, t) \, .</math></td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, §2, p. 224, Eq. (6)<br />
{{ LBO67hereafter }}, p. 293, Eq. (2)
  </td>
</tr>
</table>

<table border="1" width="80%" cellpadding="8" align="center">
<tr><td align="left">
Without immediate proof, {{ Lebovitz89ahereafter }} states that the relationship between the ''Lagranian-change operator'' and the ''Eulerian-change operator'' is, to lowest order (linear),
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right"><math>\Delta F</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\delta F + \boldsymbol\xi \cdot \nabla F \, .</math></td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, §2, p. 224, Eq. (7)<br />
{{ LBO67hereafter }}, p. 294, Eq. (3)
  </td>
</tr>
</table>

</td></tr>
</table>

Introducing this mapping into the right-hand side of the perturbed Euler equation gives:
<table border="0" cellpadding="3" align="center" width="80%">

<tr>
  <td align="right">
1<sup>st</sup> term on RHS <math>= - \Delta\biggl\{ \rho^{-1} \nabla p \biggr\}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\rho^{-2}\Delta\rho \biggl[\nabla p \biggr] - \rho^{-1} \nabla \biggl[\Delta p \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\rho^{-1}\biggl[\frac{\Delta\rho}{\rho} \biggr] \nabla p - \rho^{-1} \nabla \biggl[\delta p + (\boldsymbol\xi \cdot \nabla) p \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \rho^{-1}\biggl[\nabla \cdot \boldsymbol\xi \biggr] \nabla p 
- \rho^{-1} \biggl[\nabla \delta p + (\boldsymbol\xi \cdot \nabla)\nabla p \biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, Appendix B, p. 239, Eq. (B.2)
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \rho^{-1} \biggl\{ \nabla \delta p + (\nabla \cdot \boldsymbol\xi ) \nabla p 
+ (\boldsymbol\xi \cdot \nabla)\nabla p  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \rho^{-1} \biggl\{ \nabla \biggl[ \underbrace{\Delta p}_{\mathrm{fixed}~\mathrm{typo}} - \boldsymbol\xi \cdot \nabla p \biggr] + (\nabla \cdot \boldsymbol\xi ) \nabla p 
+ (\boldsymbol\xi \cdot \nabla)\nabla p  
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, Appendix B, p. 240, Eq. (B.3)
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \rho^{-1} 
\nabla ( \Delta p )
+ \rho^{-1} \biggl\{  
\nabla ( \boldsymbol\xi \cdot \nabla p )
- (\nabla \cdot \boldsymbol\xi ) \nabla p 
- (\boldsymbol\xi \cdot \nabla)\nabla p  
\biggr\} \, .
</math>
  </td>
</tr>
<tr>
  <td align="left" colspan="3"><font color="green">
Comments: 
<ol type="1"><li>
In order to move from the 2<sup>nd</sup> to the 3<sup>rd</sup> line of this derivation, it seems that {{ Lebovitz89ahereafter }} employs the relation: &nbsp; <math>[\Delta\rho/\rho] = - \nabla\cdot \boldsymbol\xi \, .</math>  This relation strongly resembles the continuity equation which, in Lagrangian form, is <math>[D\rho/Dt] = -\rho \nabla\cdot \mathbf{u} \, .</math></li>
<li>In order to move from the 2<sup>nd</sup> to the 3<sup>rd</sup> line of this derivation, {{ Lebovitz89ahereafter }} seems to be acknowledging that, <math>\nabla</math> commutes with <math>(\boldsymbol\xi \cdot \nabla) \, .</math> </li>
<li>A typographical error appears in Eq. (B.3) of {{ Lebovitz89ahereafter }}; <math>\Delta\rho</math> appears in the publication whereas, as noted here in the fifth line of this derivation, the term should be <math>\Delta p</math>.</li>
</ol>
  </font></td>
</tr>
</table>

<table border="0" cellpadding="3" align="center" width="80%">

<tr>
  <td align="right">
2<sup>nd</sup> term on RHS <math> =
+ \Delta\biggl\{\mathbf\nabla \biggl[ \Phi_\mathrm{L89} + \tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]\biggr\}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\mathbf\nabla  \biggl\{\Delta \Phi_\mathrm{L89} + \Delta\biggl[\tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\mathbf\nabla  \biggl\{[\delta \Phi_\mathrm{L89} + \boldsymbol\xi\cdot \nabla\Phi_\mathrm{L89}] 
+ \cancelto{0}{\delta\biggl[\tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr] }
+ \boldsymbol\xi \cdot \nabla \biggl[\tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]

\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\nabla \delta \Phi_\mathrm{L89}
+ \mathbf\nabla  \biggl\{ (\boldsymbol\xi\cdot \nabla) \Phi_\mathrm{L89} 
+ \boldsymbol\xi \cdot \nabla \biggl[\tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\nabla \delta \Phi_\mathrm{L89}
+ \boldsymbol\xi \cdot \mathbf\nabla  \biggl\{ \nabla \biggl[ \Phi_\mathrm{L89} 
+ \tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
<tr>
  <td align="left" colspan="3"><font color="green">
Comment: 
<ol type="1" start="4"><li>
A term in the 2<sup>nd</sup> row of this derivation goes to zero because there is no ''Eulerian'' variation in either of the vectors, <math>\boldsymbol\omega</math> or <math>\mathbf{x}</math>.
</li>
</ol>
  </font></td>
</tr>
</table>
Adding these two "RHS" terms together gives,

<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \rho^{-1} 
\nabla ( \Delta p )
+ \rho^{-1} \biggl\{  
\nabla ( \boldsymbol\xi \cdot \nabla p )
- (\nabla \cdot \boldsymbol\xi ) \nabla p 
- (\boldsymbol\xi \cdot \nabla)\nabla p  
\biggr\}
+
\nabla \delta \Phi_\mathrm{L89}
+ \boldsymbol\xi \cdot \mathbf\nabla  \biggl\{ \nabla \biggl[ \Phi_\mathrm{L89} 
+ \tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
That is to say,

<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \rho^{-1} 
\nabla ( \Delta p )
+ L\boldsymbol\xi \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, &sect;2, p. 224, Eq. (8)
  </td>
</tr>
</table>
where the operator, <math>L</math>, is defined such that,

<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>L\boldsymbol\xi</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math> 
\rho^{-1} \biggl\{  
\nabla ( \boldsymbol\xi \cdot \nabla p )
- (\nabla \cdot \boldsymbol\xi ) \nabla p 
- (\boldsymbol\xi \cdot \nabla)\nabla p  
\biggr\}
+
\nabla \delta \Phi_\mathrm{L89}
+ \boldsymbol\xi \cdot \mathbf\nabla  \biggl\{ \nabla \biggl[ \Phi_\mathrm{L89} 
+ \tfrac{1}{2} |\boldsymbol\omega \boldsymbol\times \mathbf{x}|^2 \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Lebovitz89ahereafter }}, Appendix B, p. 240, Eq. (B.4)
  </td>
</tr>
</table>

<font color="red">Are our four comments correct?</font>