Editing Appendix/Ramblings/BiPolytrope51AnalyticStability (section)

====Attempt 5====

=====New Strategy=====
In the vast majority of our prior attempts to derive an analytic expression for the envelope's eigenvector, we have started with the presumption &#8212; as voiced by [http://adsabs.harvard.edu/abs/1988Ap%26SS.147..219B Beech (1988)] and repeated in [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|our accompanying derivation of the structure of the <math>~(n_c, n_e) = (5, 1)</math> bipolytrope's structure]] &#8212; that the most general solution to the n = 1 Lane-Emden equation can be written in the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\phi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, .</math>
  </td>
</tr>
</table>
But, as we have [[SSC/Structure/Polytropes#Primary_E-Type_Solution|emphasized in a separate context]], another expression that satisfies the relevant Lane-Emden equation has the form,
<table border="0" cellpadding="5" align="center">

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

<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
We fully appreciate that for appropriately chosen and ''different'' values of the parameter, <math>~B</math>, these two functions can be made equal to one another.  But, for now, let's work through our analysis pretending that they are different functions.
</td></tr></table>

This alternate "cosine" expression was not the solution of choice when we were seeking a mathematical description of the structure of an ''isolated'', n = 1 polytrope because it does not satisfy the relevant central boundary conditions.  However, it occurs to us  that this alternate expression might work in the context we are considering now <font color="red">[20 April 2019]</font>, which deals with properties of the envelope of a bipolytrope.  In this case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q \equiv - \frac{d\ln\phi}{d\ln\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A \biggl[\frac{\cos(\eta-B)}{\eta^2} + \frac{\sin(\eta-B)}{\eta}  \biggr]\cdot \frac{\eta^2}{A\cos(\eta-B)}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\cos(\eta-B) + \eta\sin(\eta-B)  \biggr]\cdot \frac{1}{\cos(\eta-B)}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \eta\tan(\eta-B)  \, ;</math>
  </td>
</tr>
</table>
and, following along the lines of our earlier '''[[#Attempt_4B|Attempt 4B]]''' discussion, a reasonable ''guess'' for the dimensionless displacement function is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1 Q}{\eta^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl[1 + \eta\tan(\eta-B)\biggr] \, .</math>
  </td>
</tr>
</table>

What are the first and second derivatives of this trial eigenfunction?

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

<tr>
  <td align="right">
<math>~\frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^2}\biggl[\tan(\eta - B) + \frac{\eta}{\cos^2(\eta-B)}\biggr]
- \frac{6c_1}{\eta^3} \biggl[1 + \eta\tan(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^3} \biggl[\frac{\eta^2}{\cos^2(\eta-B)} -2 - \eta\tan(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
3c_1 \frac{d}{d\eta}\biggl[\eta^{-3}\cos^{-2}(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3c_1}{\eta^3\cos^2(\eta-B)}\cdot \frac{d}{d\eta}\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
3c_1 \biggl[2\eta^{-3}\sin(\eta-B)\cos^{-3}(\eta-B) -3\eta^{-4}\cos^{-2}(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[2\eta + 4\sin(\eta-B)\cos(\eta-B) 
- \sin(\eta-B)\cos(\eta-B)- \eta \cos^2(\eta-B)+ \eta\sin^2(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} 
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
\cdot \biggl[2\eta\sin(\eta-B) -3\cos(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[2\eta^2\cos(\eta-B) + 4\eta \sin(\eta-B)\cos^2(\eta-B) 
- \eta \sin(\eta-B)\cos^2(\eta-B)- \eta^2 \cos^3(\eta-B)+ \eta^2\sin^2(\eta-B)\cos(\eta-B)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl\{
2\eta^3 \sin(\eta-B) - 4\eta\sin(\eta-B)\cos^2(\eta-B) - 2\eta^2 \sin^2(\eta-B)\cos(\eta-B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-3\eta^2 \cos(\eta-B) + 6\cos^3(\eta-B) +3 \eta\sin(\eta-B)\cos^2(\eta-B)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl\{6\cos^3(\eta-B) 
+ \eta^3 \biggl[2\sin(\eta-B)\biggr] 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[ 6\cos^3(\eta-B) 
+ 2\eta \sin(\eta-B)\cos^2(\eta-B)
- 2 \eta^2 \cos(\eta-B) + 2\eta^3 \sin(\eta-B)
\biggr] \, .
</math>
  </td>
</tr>
</table>

=====Illustration=====

<table border="1" align="center" width="710px" cellpadding="8">
<tr><td align="center">
[[File:Frame06good.png|700px|Attempt5 Trial Eigenfunction]]
</td></tr>
<tr><td align="left">
''Above &amp; Below:'' &nbsp;The solid, light-blue circular markers trace how the function, <math>~x_P(\eta)</math>, varies over the "radial" range, <math>~- \pi \le \eta \le \pi</math>, when <math>~c_1 = 1</math>, for various values of the parameter, <math>~B</math>; in the above panel as well as in each frame of the animation, the chosen value of <math>~B</math> is recorded in the upper-right corner of the image. (These values are also recorded in the table immediately below the animation.)  The red vertical dashed line segment identifies the value of <math>~\eta</math> at which the argument of the tangent function goes to <math>~\tfrac{\pi}{2}</math> and, hence, where the function <math>~x_P(\eta)</math> flips discontinuously from plus- to minus-infinity.  As explained further, below, the solid purple curve shows how the x<sub>P</sub>-''intercept'' function varies with <math>~\eta</math>; as defined, this curve is independent of the parameter, <math>~B</math>, so it is unchanging in the animation sequence.  The single larger yellow-circular marker (with a black border) shows where this "intercept" curve intersects the <math>~x_P</math> function and, therefore, where along this trial eigenfunction <math>~d\ln x_P/d\ln\eta = -1</math>; the coordinates (abscissa &amp; ordinate) of this yellow marker are recorded in the accompanying table, for each illustrative value of <math>~B</math>.
</td></tr>
<tr><td align="center">
[[File:TangentTrialAnalytic51movie.gif|700px|Animation of Attempt5 Trial Eigenfunction]]
</td></tr>
<tr><td align="center">
<table border="0" cellpadding="8">
<tr>
  <th align="center" colspan="2">&nbsp;</th>
  <th align="center" colspan="2"><math>~\frac{d\ln x_P}{d\ln\eta} = -1</math><br /><hr /></th>
</tr>
<tr>
  <th align="center">Animation<br />Frame</th>
  <th align="center"><math>~B</math></th>
  <th align="center">Abscissa<br /><math>~\eta</math></th>
  <th align="center">Ordinate<br /><math>~x_P(\eta)</math></th>
</tr>
<tr>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~1.02 \times \tfrac{\pi}{2} = 1.6022</math></td>
  <td align="center"><math>~0.014</math></td>
  <td align="center"><math>~27,165</math></td>
</tr>
<tr>
  <td align="center"><math>~2</math></td>
  <td align="center"><math>~1.2 \times \tfrac{\pi}{2} = 1.8850</math></td>
  <td align="center"><math>~0.157</math></td>
  <td align="center"><math>~242.7</math></td>
</tr>
<tr>
  <td align="center"><math>~3</math></td>
  <td align="center"><math>~1.4 \times \tfrac{\pi}{2} = 2.1991</math></td>
  <td align="center"><math>~0.311</math></td>
  <td align="center"><math>~60.31</math></td>
</tr>
<tr>
  <td align="center"><math>~4</math></td>
  <td align="center"><math>~1.6 \times \tfrac{\pi}{2} = 2.5133</math></td>
  <td align="center"><math>~0.462</math></td>
  <td align="center"><math>~26.52</math></td>
</tr>
<tr>
  <td align="center"><math>~5</math></td>
  <td align="center"><math>~1.8 \times \tfrac{\pi}{2} = 2.8274</math></td>
  <td align="center"><math>~0.606</math></td>
  <td align="center"><math>~14.68</math></td>
</tr>
<tr>
  <td align="center"><math>~6</math></td>
  <td align="center"><math>~2 \times \tfrac{\pi}{2} = 3.1416</math></td>
  <td align="center"><math>~0.739</math></td>
  <td align="center"><math>~9.187</math></td>
</tr>
<tr>
  <td align="center"><math>~7</math></td>
  <td align="center"><math>~0.2 \times \tfrac{\pi}{2} = 0.3142</math></td>
  <td align="center"><math>~0.856</math></td>
  <td align="center"><math>~6.200</math></td>
</tr>
<tr>
  <td align="center"><math>~8</math></td>
  <td align="center"><math>~0.4 \times \tfrac{\pi}{2} = 0.6283</math></td>
  <td align="center"><math>~0.949</math></td>
  <td align="center"><math>~4.381</math></td>
</tr>
<tr>
  <td align="center"><math>~9</math></td>
  <td align="center"><math>~0.5 \times \tfrac{\pi}{2} = 0.7854</math></td>
  <td align="center"><math>~0.981</math></td>
  <td align="center"><math>~3.724</math></td>
</tr>
<tr>
  <td align="center"><math>~10</math></td>
  <td align="center"><math>~0.5 \times \tfrac{\pi}{2} = 0.9425</math></td>
  <td align="center"><math>~0.998</math></td>
  <td align="center"><math>~3.178</math></td>
</tr>
<tr>
  <td align="center"><math>~11</math></td>
  <td align="center"><math>~0.8 \times \tfrac{\pi}{2} = 1.2566</math></td>
  <td align="center"><math>~0.955</math></td>
  <td align="center"><math>~2.313</math></td>
</tr>
<tr>
  <td align="center"><math>~12</math></td>
  <td align="center"><math>~0.9 \times \tfrac{\pi}{2} = 1.4137</math></td>
  <td align="center"><math>~0.840</math></td>
  <td align="center"><math>~1.944</math></td>
</tr>
<tr>
  <td align="center"><math>~13</math></td>
  <td align="center"><math>~0.98 \times \tfrac{\pi}{2} = 1.5394</math></td>
  <td align="center"><math>~0.545</math></td>
  <td align="center"><math>~1.632</math></td>
</tr>
</table>
</td></tr>
</table>

What is the expression for the logarithmic derivative of this eigenfunction guess?

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

<tr>
  <td align="right">
<math>~\frac{d\ln x_P}{d\ln\eta} = \frac{\eta}{x_P} \cdot \frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\eta^3}{3c_1} \biggl[1 + \eta\tan(\eta-B)\biggr]^{-1} \biggl\{
\frac{3c_1}{\eta^2}\biggl[\tan(\eta - B) + \frac{\eta}{\cos^2(\eta-B)}\biggr]
- \frac{6c_1}{\eta^3} \biggl[1 + \eta\tan(\eta-B)\biggr]
\biggr\}
</math>
  </td>
</tr>

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

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

At the surface of the bipolytropic configuration &#8212; that is, presumably when <math>~\eta = \eta_s</math> &#8212; we must find that this logarithmic derivative is negative one.  So, for a given value of the parameter, <math>~B</math>, what is the value of <math>~\eta_s</math>?  Well &hellip;

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

<tr>
  <td align="right">
<math>~\frac{d\ln x_P}{d\ln\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ -\cos^2(\eta-B) - \eta\sin(\eta-B) \cos(\eta-B)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta^2 - \eta\sin(\eta - B)\cos(\eta-B) - 2\cos^2(\eta-B)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta^2  - \cos^2(\eta-B)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm \cos(\eta-B)\, .</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\tan(\eta-B) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \, ,</math>
  </td>
</tr>
</table>
we see that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_P\biggr|_\mathrm{intercept}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\eta^2}}{\eta} \biggr] \biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \sqrt{1-\eta^2}\biggr\} \, .</math>
  </td>
</tr>
</table>

The solid-purple, x<sub>P</sub>-''intercept'' curve that appears in the above figure/animation is defined by this function.  Notice that this function never exceeds unity.  This presumably means that if the tangent-based <math>~x_P</math> eigenfunction is the correct solution to the envelope's LAWE, then the dimensionless radius, <math>~\eta_s</math>, of the bipolytrope must never exceed unity.

The following diagram &#8212; the original of which appears in our [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|accompanying discussion of the equilibrium properties of bipolytropic configurations having]] <math>~(n_c, n_e) = (5, 1)</math> &#8212; shows how <math>~\eta_i</math> (purple curve) and <math>~\eta_s</math> (green curve) vary with the interface location <math>~\xi_i</math> (ordinate).  The solid yellow circular markers (with black edges) identify where the logarithmic derivative of the dimensionless displacement function, <math>~d\ln x_P/d\ln\eta</math>, equals negative one.  If <math>~x_P</math> is the correct eigenfunction for the marginally unstable bipolytropic configuration, one of these yellow circular markers should coincide with the green curve, that is, it should be associated with the configuration's surface.  Since the curve identified by the yellow circular markers does not appear to intersect the green curve, we conclude that we have not yet identified the correct eigenfunction.

<div align="center">
[[File:Bipolytrope51Boundaries03b.png|500px|Bipolytrope Properties]]
</div>

=====Is This Compatible With LAWE=====

In an effort to track the two <math>~Q(\eta)</math> functions separately, we will add a subscript zero to the one that applies to the ''structural'' properties of the underlying equilibrium configuration.  Again, we will be focused on finding a solution in the case where <math>~\sigma_c^2 = 0</math> and <math>~n = 1</math>, that is &#8212; see also our [[#Envelope:|above discussion]] &#8212; the relevant envelope LAWE is,
<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>~
\frac{d^2x_P}{d\eta^2} + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{dx_P}{d\eta} -2\alpha_g Q_0 \cdot \frac{x_P}{\eta^2} \, ,
</math>
  </td>
</tr>
</table>
where, drawing from our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|discussion of the n = 1 envelope's equilibrium structure]],
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ - \frac{d\ln\phi}{d\ln\eta} \biggr]_\mathrm{n=1} = - \frac{\eta}{\phi}\biggl[ \frac{d\phi}{d\eta} \biggr]_\mathrm{n=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{1}{\sin(\eta - B_0)} \biggr]\biggl[ \eta\cos(\eta-B_0) - \sin(\eta-B_0) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1 -  \eta\cot(\eta-B_0) \biggr] \, .</math>
  </td>
</tr>
</table>

Hence, after recognizing that for this specific case, <math>~\alpha_g = +1</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x_P}{d\eta^2} + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{dx_P}{d\eta} -2 Q_0 \cdot \frac{x_P}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~</math>&nbsp; LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[ 6\cos^3(\eta-B) 
+ 2\eta \sin(\eta-B)\cos^2(\eta-B)
- 2 \eta^2 \cos(\eta-B) + 2\eta^3 \sin(\eta-B)
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] 
- \frac{2 Q_0}{\eta^2} \cdot \frac{3c_1 }{\eta^2}\biggl[1 + \eta\tan(\eta-B)\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>&nbsp; LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 3\cos^3(\eta-B) 
+ \eta \sin(\eta-B)\cos^2(\eta-B)
-  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 2 - Q_0 \biggr]\cos(\eta - B)\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] 
- Q_0 \cos^3(\eta - B) \biggl[1 + \eta\tan(\eta-B)\biggr] 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- Q_0\biggl\{ \biggl[\eta^2\cos(\eta - B) -2\cos^3(\eta-B) - \eta\sin(\eta-B)\cos^2(\eta-B)\biggr] 
+ \biggl[\cos^3(\eta - B) + \eta\sin(\eta-B)\cos^2(\eta - B)\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\cos^3(\eta-B) 
- \eta \sin(\eta-B)\cos^2(\eta-B)
+  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
+ Q_0\biggl[ \cos^3(\eta-B) - \eta^2\cos(\eta - B) \biggr] 
</math>
  </td>
</tr>
</table>

<table border="1" width="85%" cellpadding="10" align="center"><tr><td align="left">
<font color="red">'''Quick Check:'''</font> &nbsp; Now, if we set <math>~Q_0 = 1 + \eta\tan(\eta-B)</math>, these RHS terms should sum to zero.  Let's check.

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

<tr>
  <td align="right">
<math>~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>&nbsp; LAWE
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~
-\cos^3(\eta-B) 
- \eta \sin(\eta-B)\cos^2(\eta-B)
+  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .
</math>
  </td>
</tr>
</table>

Excellent!
</td></tr></table>


Now, let's plug in the expression for the ''structural'' <math>~Q_0</math>.  Specifically, we want,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_0</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 - \eta\cot(\eta-B_0) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 + \eta\tan(\eta-B_0 - \tfrac{\pi}{2} \pm m\pi) \biggr] \, ,</math>
  </td>
</tr>
</table>
in which case we have,

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

<tr>
  <td align="right">
<math>~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>&nbsp; LAWE
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~
-\cos^3(\eta-B) 
- \eta \sin(\eta-B)\cos^2(\eta-B)
+  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\tan(\eta-B_0 - \tfrac{\pi}{2} \pm m\pi) \biggr]  \cdot \biggl[ \cos^3(\eta-B) - \eta^2\cos(\eta-B) \biggr] \, .
</math>
  </td>
</tr>
</table>
As we have just shown, above, in the context of a "<font color="red">'''Quick Check'''</font>", the expression on the RHS will go to zero if we adopt the transformation, <math>~[B_0 + \tfrac{\pi}{2} \mp m\pi] \rightarrow B</math>.  Does this help shift the coordinate, <math>~\eta</math>?