Editing SSC/Structure/BiPolytropes/Analytic51Renormalize (section)

==Attempt at Constructing Analytic Eigenfunction Expression==
===Background===

In our [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|accompanying discussion of eigenvectors associated with the radial oscillation of pressure-truncated polytropes]], we derived the following,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Exact Solution to the Polytropic LAWE</b></font></td>
</tr>
<tr>
  <td align="right">
<math>\sigma_c^2 = 0</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>x_P \equiv \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] \, .</math>
  </td>
</tr>
</table>
Drawing on the definition of <math>\theta(\xi)</math> for n = 5 polytropes, as given [[SSC/Structure/Polytropes#Primary_E-Type_Solution_2|in an accompanying chapter]], we deduce that,

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

<tr>
  <td align="right">
<math>x_P\biggr|_{n=5}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{5}\biggl[1 + \frac{1}{2}\biggl( \frac{1}{\xi \theta^{5}}\biggr) \frac{d\theta}{d\xi}\biggr]_{n=5} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{5} - \frac{3}{5\xi} \biggl( 1 + \frac{\xi^2}{3} \biggr)^{5/2} \frac{\xi}{3} \biggl( 1 + \frac{\xi^2}{3} \biggr)^{-3/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{5} - \frac{1}{5} \biggl( 1 + \frac{\xi^2}{3} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \frac{\xi^2}{15} \, .
</math>
  </td>
</tr>
</table>

And, given that [[SSC/Structure/Polytropes#Primary_E-Type_Solution_2|for n = 1 polytropes]], 
<div align="center">
<math>\theta(\xi) = \frac{\sin\xi}{\xi} \, ,</math>
</div>

we also find,

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

<tr>
  <td align="right">
<math>x_P\biggr|_{n=1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-3 \biggl[ \biggl( \frac{1}{\xi \theta}\biggr) \frac{d\theta}{d\xi}\biggr]_{n=1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{\xi} \biggl( \frac{\xi}{\sin\xi}\biggr) \biggl[\frac{\sin\xi}{\xi^2} - \frac{\cos\xi}{\xi}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{\xi^2}\biggl[ 1- \xi \cot\xi  \biggr] \, .
</math>
  </td>
</tr>
</table>

===Core===

Allowing for an overall leading scale factor, <math>\alpha</math>, a viable displacement function for the <math>(n = 5)</math> core of our bipolytropic configuration is,

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

<tr>
  <td align="right">
<math>\frac{x_\mathrm{core}}{\alpha}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 - \frac{\xi^2}{15} \biggr] \, .
</math>
  </td>
</tr>
</table>

Throughout the core, the corresponding Lagrangian radial coordinate, <math>\tilde{r}</math>, is given by the expression,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\tilde{r}_\mathrm{core}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^{-2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{4} \biggl(\frac{3}{2\pi}\biggr)^{1/2} \xi 
\, .</math>
  </td>
</tr>
</table>

For "model <b>A</b>" the range is,
<div align="center">
<math>0 \le \xi \le \xi_i = 9.0149598 \, .</math>
</div>

<table border="1" align="center" width="80%" cellpadding="5">
<tr><td align="left">

<font color="red">SLOPE:</font>&nbsp;  What is the slope of the function, <math>x_\mathrm{core}(\tilde{r})</math>, at the interface?
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{dx_\mathrm{core}}{d\xi} \biggr|_i</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>- \frac{2\alpha \xi_i}{15}
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{dx_\mathrm{core}}{d\tilde{r}} \biggr|_i</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>- \frac{2\alpha \xi_i}{15} \biggl[ 
\mathcal{m}_\mathrm{surf}^{-2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{4} \biggl(\frac{3}{2\pi}\biggr)^{1/2} 
\biggr]^{-1}
= -~707.53765~\alpha
\, ,</math>
  </td>
</tr>
</table>
where, for "model <b>A</b>," we have set <math>(\mu_e/\mu_c) = 0.31</math> and <math>\mathcal{m}_\mathrm{surf} = 1.938127063</math>.  Note as well that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{d\ln (x_\mathrm{core})}{d\ln\tilde{r}} \biggr|_i = \frac{d\ln (x_\mathrm{core})}{d\ln\xi} \biggr|_i</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>- \frac{2 \xi_i^2}{15}\biggl[1 - \frac{\xi_i^2}{15}\biggr]^{-1}
= +2.452697
\, .</math>
  </td>
</tr>
</table>

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

===Envelope===
As we have demonstrated [[SSC/Structure/BiPolytropes/Analytic51#Step_6:_Envelope_Solution|in a separate ''structure'' discussion]], the radial profile of the <math>(n = 1)</math> envelope of our bipolytropic configuration is governed by the ''modified'' sinc-function,

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

<tr>
  <td align="right">
<math>\phi(\eta)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
A \biggl[ \frac{\sin(\eta-B)}{\eta}\biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ -\frac{d\phi}{d\eta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{A}{\eta^2} \biggl[\sin(\eta-B) - \eta\cos(\eta - B) \biggr] \, .
</math>
  </td>
</tr>
</table>
where, for "model <b>A</b>," <math>A = 0.200812422</math> and <math>B = -0.859270052</math>.

Again allowing for an overall leading scale factor, <math>\beta</math>, a viable displacement function for the <math>(n = 1)</math> envelope of our bipolytropic configuration is,

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

<tr>
  <td align="right">
<math>\frac{x_\mathrm{env}}{3\beta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{\eta \phi} \biggl( - \frac{d\phi}{d\eta} \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{A}{\eta^3 } \biggl[\sin(\eta-B) - \eta\cos(\eta - B) \biggr] \cdot  \biggl[\frac{\eta}{A\sin(\eta - B)}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{\eta^2 } \biggl[1 - \eta\cot(\eta - B) \biggr]\, . 
</math>
  </td>
</tr>
</table>

Throughout the envelope, the corresponding Lagrangian radial coordinate is,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\tilde{r}_\mathrm{env}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\mathcal{m}_\mathrm{surf}^{-2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{3} \theta^{-2}_i (2\pi)^{-1/2}\eta \, .</math>
  </td>
</tr>
</table>

For "model <b>A</b>" the range is,
<div align="center">
<math>(\eta_i = 0.1723205) \le \eta \le (\eta_s = 2.282322601) \, .</math>
</div>

<table border="1" align="center" width="80%" cellpadding="5">
<tr><td align="left">

<font color="red">SLOPE:</font>&nbsp;  As we have [[Appendix/Ramblings/BiPolytrope51AnalyticStability#Eigenfunction_Details|detailed elsewhere]], the slope of the function, <math>x_\mathrm{env}(\tilde{r})</math>, is related to the slope of <math>x_\mathrm{core}(\tilde{r})</math> at the interface via the expression,

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

<tr>
  <td align="right">
<math>~
\biggl\{ \frac{d\ln x}{d\ln r}\biggr|_i \biggr\}_\mathrm{env}
=
\biggl\{ \frac{d\ln x}{d\ln \eta}\biggr|_i \biggr\}_\mathrm{env}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr) + \frac{\gamma_c}{\gamma_e} \biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_i \biggr\}_\mathrm{core} 
\, .</math>
  </td>
</tr>
</table>
In our case, <math>\gamma_c = 6/5</math> and <math>\gamma_e = 2 ~~\Rightarrow \gamma_c/\gamma_e = 3/5</math>.  Hence, from the point of view of the envelope displacement function, at the interface,

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

<tr>
  <td align="right">
<math>
\biggl[ \frac{\tilde{r}}{x_\mathrm{env}} \cdot \frac{d x_\mathrm{env}}{d \tilde{r} } \biggr]_i
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{3}{5} \biggl\{ \biggl[ \frac{d\ln (x_\mathrm{core})}{d\ln \xi}\biggr]_i  - 2\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{3}{5} \biggl\{ \biggl[ 2.452697  - 2\biggr\} = 0.271618 \, .
</math>
  </td>
</tr>
</table>
Now, at the interface of any bipolytrope, the ratio <math>\tilde{r}/x</math> should have the same numerical value whether it is viewed from the point of view of the core or the envelope.  Given that, for our particular "model <b>A</b>",
<div align="center">
<math>\biggl[ \frac{\tilde{r}}{x} \biggr]_i = \frac{0.015315}{0.00485976} = 3.15139 \, ,</math>
</div>
we should expect the slope of the envelope's displacement function at the interface to be,

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

<tr>
  <td align="right">
<math>
\frac{d x_\mathrm{env}}{d \tilde{r} } \biggr|_i
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0.08619 \, .
</math>
  </td>
</tr>
</table>


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

===Trial Displacement Function===

The blue curve in the following figure results from plotting <math>x_\mathrm{core}</math> versus <math>\tilde{r}_\mathrm{core}</math> after setting the leading coefficient, <math>\alpha = - 0.0011</math>.  The red-dotted curve results from plotting <math>(x_\mathrm{env} + x_\mathrm{shift})</math> versus <math>\tilde{r}_\mathrm{env}</math> after setting the leading coefficient, <math>\beta = - 0.000062</math>, and <math>x_\mathrm{shift} = + 0.0105</math>.

<table border="1" align="center" cellpadding="2">
<tr>
  <td align="center">[[File:TrialAnalyticEigenfunction01.png|700px|Trial Analytic Eigenfunction]]</td>
</tr>
</table>

ASSESSMENT:
<ul>
  <li>Our analytically specified displacement function, <math>x_\mathrm{core}</math>, appears to be an excellent match to the displacement function obtained throughout the core by implementing the [[Appendix/Ramblings/NonlinarOscillation#Radial_Oscillations_in_Pressure-Truncated_n_=_5_Polytropes|B-KB74 conjecture]].</li>
  <li>At first glance, the plot of <math>(x_\mathrm{env} + x_\mathrm{shift})</math> appears to provide a reasonably good fit to the ''approximate'' displacement function that we have obtained throughout the envelope by implementing the B-KB74 conjecture.   But, in reality, there are two fatal flaws:
<ol type="1">
<li>We have presented the behavior of our analytically specified envelope displacement function only up to the radial coordinate, <math>\eta = 2.19707 ~~ (\tilde{r}_\mathrm{env} = 0.19526)</math>. Between this point and the surface, <math>\eta_s = 2.2823226 ~~ (\tilde{r}_\mathrm{env} = 0.2028415)</math> &#8212; where the argument of the cotangent, <math>(\eta_s - B) \rightarrow \pi</math> &#8212; the analytic function dives steeply to negative infinity.  This violently departs from the behavior derived via the [[Appendix/Ramblings/NonlinarOscillation#Radial_Oscillations_in_Pressure-Truncated_n_=_5_Polytropes|B-KB74 conjecture]].</li>
<li>While our analytically specified displacement function, <math>x_\mathrm{env}</math>, satisfies the "n = 1" polytropic LAWE, this satisfaction is destroyed by adding <math>x_\mathrm{shift}</math> to the displacement function.</li>
</ol>
  </li>
</ul>

Let's examine the slope of the displacement function at the interface.  From the perspective of the core, our analytic prescription for the displacement function matches the K-BK74-derived displacement function very well.  An analytic evaluation of the slope at the inferface &#8212; as derived [[#Core|above]] &#8212; gives,
<div align="center">
<math>\frac{dx_\mathrm{core}}{d\tilde{r}}\biggr|_i = -707.53765\alpha = +0.77829</math>.
</div>
The black-dashed line segment that appears in the following figure has this slope and goes through the point of intersection; it appears to be tangent to the analytic displacement function, as expected.  Alternatively, the orange-dashed line segment that appears in this same figure, also goes through the point of intersection, but it has a slope that matches our ''expectation'' for the envelope's displacement function; that is, it has a slope [[#Envelope|as derived]] of,
<div align="center">
<math>\frac{dx_\mathrm{env}}{d\tilde{r}}\biggr|_i = +0.08619</math>.
</div>
This orange-dashed line segment does ''not'' appear to lie tangent to the K-BK74-derived displacement function for the envelope.

<table border="1" align="center" cellpadding="2">
<tr>
  <td align="center">[[File:TrialEigenfunctionSlopes01.png|700px|Trial Analytic Eigenfunction with Intersection Slopes]]</td>
</tr>
</table>

===2<sup>nd</sup> Trial===

The relevant LAWE for the envelope is,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right"><math>\biggl[ \frac{A\sin(\eta-B)}{\eta}\biggr]\frac{d^2x}{d\eta^2} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{2A}{\eta}\biggl\{ \sin(\eta - B) + \eta\cos(\eta - B) \biggr\} \frac{1}{\eta} \cdot \frac{dx}{d\eta}  
+ \frac{2A}{\eta} \biggl\{ \sin(\eta - B) - \eta\cos(\eta - B) \biggr\}  \frac{x}{\eta^2}
- 2 \biggl( \frac{\sigma_c^2}{12} \biggr) x 
</math>
</td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{1}{3\beta} \cdot \frac{d^2x}{d\eta^2} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{2}{\eta}\biggl\{ 1 + \eta\cot(\eta - B) \biggr\} \biggl[ \frac{1}{3\beta } \cdot \frac{dx}{d\eta} \biggr] 
+ 2 \biggl\{ 1 - \eta\cot(\eta - B) \biggr\}  \frac{x}{3\beta\eta^2}
- 2 \biggl( \frac{\sigma_c^2}{12} \biggr) \biggl[ \frac{\eta}{3 \beta A\sin(\eta-B)}\biggr] x
\, . 
</math>
</td>
</tr>
</table>
Here, we will ''guess'' a displacement function, <math>x_\mathrm{env}</math>, of the form,

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

<tr>
  <td align="right">
<math>\frac{x_\mathrm{env}}{3\beta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{\eta^2 } \biggl[1 - \eta\cot(\eta - B_x) \biggr]
=
\frac{1}{\eta^2 }  - \frac{\cos(\eta - B_x)}{\eta\sin(\eta - B_x)} 
\, , 
</math>
  </td>
</tr>
</table>
where we will assume, quite generally, that <math>B_x \ne B</math>.  The first and second derivatives of <math>x_\mathrm{env}</math> are,

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

<tr>
  <td align="right">
<math>\frac{1}{3\beta} \biggl[ \frac{dx_\mathrm{env}}{d\eta} \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{2}{\eta^3 }  
+ 
\frac{\cos(\eta - B_x)}{\eta^2\sin(\eta - B_x)} 
+ 
\frac{\sin(\eta - B_x)}{\eta\sin(\eta - B_x)} 
+ 
\frac{\cos^2(\eta - B_x)}{\eta\sin^2(\eta - B_x)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{2}{\eta^3 }  
+ 
\frac{1}{\eta} 
+ 
\frac{\cos(\eta - B_x)}{\eta^2\sin(\eta - B_x)} 
+ 
\frac{\cos^2(\eta - B_x)}{\eta\sin^2(\eta - B_x)}
\, ; 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{1}{3\beta} \biggl[ \frac{d^2x_\mathrm{env}}{d\eta^2} \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{\eta^4} - \frac{1}{\eta^2}
- \frac{2\cos(\eta-B_x)}{\eta^3\sin(\eta-B_x)}
+ \frac{1}{\eta^2}\biggl[
- \frac{\sin(\eta - B_x)}{\sin(\eta - B_x)}
-
\frac{\cos^2(\eta - B_x)}{\sin^2(\eta - B_x)}
\biggr]
- 
\frac{\cos^2(\eta - B_x)}{\eta^2\sin^2(\eta - B_x)}
+ \frac{1}{\eta} \biggl[
-
\frac{2\cos(\eta - B_x)}{\sin(\eta - B_x)}
-
\frac{2\cos^3(\eta - B_x)}{\sin^3(\eta - B_x)}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{\eta^4} - \frac{1}{\eta^2}
- \frac{2\cos(\eta-B_x)}{\eta^3\sin(\eta-B_x)}
- \frac{1}{\eta^2}\biggl[
1
+
\frac{\cos^2(\eta - B_x)}{\sin^2(\eta - B_x)}
\biggr]
- 
\frac{\cos^2(\eta - B_x)}{\eta^2\sin^2(\eta - B_x)}
- \frac{1}{\eta}\biggl[
\frac{2\cos(\eta - B_x)}{\sin(\eta - B_x)}
+
\frac{2\cos^3(\eta - B_x)}{\sin^3(\eta - B_x)}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{\eta^4} - \frac{2}{\eta^2}
- \frac{2\cos(\eta-B_x)}{\eta^3\sin(\eta-B_x)}
- \frac{2}{\eta^2}\biggl[
\frac{\cos^2(\eta - B_x)}{\sin^2(\eta - B_x)}
\biggr]
- \frac{2}{\eta}\biggl[
\frac{\cos(\eta - B_x)}{\sin(\eta - B_x)}
+
\frac{\cos^3(\eta - B_x)}{\sin^3(\eta - B_x)}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{\eta^4} - \frac{2}{\eta^2}\biggl[1 + \cot^2(\eta - B_x) \biggr]
- \frac{2\cot(\eta-B_x)}{\eta^3}
- \frac{2}{\eta}\biggl[
\cot(\eta - B_x)
+
\cot^3(\eta - B_x)
\biggr]\, .
</math>
  </td>
</tr>
</table>
These match the expressions for <math>dx_P/d\eta</math> and <math>d^2x_P/d\eta^2</math> that we separately derived in a [[Appendix/Ramblings/BiPolytrope51AnalyticStability#Attempt_4B|subsection labeled ''Attempt_4B'' of an accompanying discussion labeled]].  Plugging these three relations into the LAWE, then multiplying through by <math>\eta^4</math>, gives,
<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>
\frac{6}{\eta^4} - \frac{2}{\eta^2}\biggl[1 + \cot^2(\eta - B_x) \biggr]
- \frac{2\cot(\eta-B_x)}{\eta^3}
- \frac{2}{\eta}\biggl[
\cot(\eta - B_x)
+
\cot^3(\eta - B_x)
\biggr]
</math>
</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{2}{\eta}\biggl\{ 1 + \eta\cot(\eta - B) \biggr\} \biggl[ -\frac{2}{\eta^3 }  
+ 
\frac{1}{\eta} 
+ 
\frac{\cot(\eta - B_x)}{\eta^2} 
+ 
\frac{\cot^2(\eta - B_x)}{\eta}
 \biggr] 
</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ \biggl\{
2 \biggl[ 1 - \eta\cot(\eta - B) \biggr]  \frac{1}{\eta^2}
- 2 \biggl( \frac{\sigma_c^2}{12} \biggr) \biggl[ \frac{\eta}{ A\sin(\eta-B)}\biggr]
\biggr\} \biggl[ \frac{1}{\eta^2 }  - \frac{\cot(\eta - B_x)}{\eta}\biggr]
</math>
</td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~
6 - 2\eta^2 \biggl[1 + \cot^2(\eta - B_x) \biggr]
- 2\eta \cot(\eta-B_x)
- 2\eta^3 \biggl[
\cot(\eta - B_x)
+
\cot^3(\eta - B_x)
\biggr]
</math>
</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- 2\biggl\{ 1 + \eta\cot(\eta - B) \biggr\} \biggl[ -2  
+ 
\eta^2 
+ 
\eta\cot(\eta - B_x) 
+ 
\eta^2 \cot^2(\eta - B_x)
 \biggr] 
</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ \biggl\{
2 \biggl[ 1 - \eta\cot(\eta - B) \biggr] 
- 2 \biggl( \frac{\sigma_c^2}{12} \biggr) \biggl[ \frac{\eta^3}{ A\sin(\eta-B)}\biggr]
\biggr\} \biggl[ 1  - \eta \cot(\eta - B_x)\biggr]
</math>
</td>
</tr>
</table>

<table border=0 cellpadding=2 align="center">

<tr>
  <td align="right">
<math>\Rightarrow ~~~
\biggl( \frac{\sigma_c^2}{6} \biggr) \biggl[ \frac{\eta^3}{ A\sin(\eta-B)}\biggr]
\biggl[ 1  - \eta \cot(\eta - B_x)\biggr]
</math>
</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ 1 + \eta\cot(\eta - B) \biggr] \biggl[ 4  
- 
2\eta^2 
- 
2\eta\cot(\eta - B_x) 
- 
2\eta^2 \cot^2(\eta - B_x)
 \biggr] 
</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ 
2 \biggl[ 1 - \eta\cot(\eta - B) \biggr] 
\biggl[ 1  - \eta \cot(\eta - B_x)\biggr]

-6 + 2\eta^2 \biggl[1 + \cot^2(\eta - B_x) \biggr]
+ 2\eta \cot(\eta-B_x)
+ 2\eta^3 \biggl[
\cot(\eta - B_x)
+
\cot^3(\eta - B_x)
\biggr]
</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ 4 - 2\eta^2 - 2\eta\cot(\eta - B_x) - 2\eta^2 \cot^2(\eta - B_x)\biggr] 
+
\eta\cot(\eta - B) 
\biggl[ 4 - 2\eta^2 - 2\eta\cot(\eta - B_x) - 2\eta^2 \cot^2(\eta - B_x)\biggr] 

</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ 
\biggl[ 2 - 2\eta\cot(\eta - B) - 2\eta \cot(\eta - B_x) + 2\eta^2 \cot(\eta - B)\cot(\eta - B_x)\biggr]

-6 + 2\eta^2 + 2\eta^2 \cot^2(\eta - B_x) 
+ 2\eta \cot(\eta-B_x)
+ 2\eta^3 \cot(\eta - B_x) 
+ 2\eta^3 \cot^3(\eta - B_x)
</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- 2\eta\cot(\eta - B_x) - 2\eta^2 \cot^2(\eta - B_x) 
+
4\eta\cot(\eta - B) - 2\eta^3\cot(\eta - B) - 2\eta^2 \cot(\eta - B)\cot(\eta - B_x) - 2\eta^3\cot(\eta - B) \cot^2(\eta - B_x)

</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
- 2\eta\cot(\eta - B) - 2\eta \cot(\eta - B_x) + 2\eta^2 \cot(\eta - B)\cot(\eta - B_x)
+ 2\eta^2 \cot^2(\eta - B_x) 
+ 2\eta \cot(\eta-B_x)
+ 2\eta^3 \cot(\eta - B_x) 
+ 2\eta^3 \cot^3(\eta - B_x)
</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- 2\eta^3 \biggl[\cot(\eta - B)  +\cot(\eta - B) \cot^2(\eta - B_x)
- \cot(\eta - B_x) - \cot^3(\eta - B_x) \biggr]
+ 2\eta \biggl[\cot(\eta - B) - \cot(\eta - B_x) \biggr]
</math>
</td>
</tr>
</table>

Notice that if we set <math>B_x = B</math>, the RHS of this LAWE expression goes to zero.  This [thankfully] is as expected.