Editing SSC/Stability/Isothermal (section)

===Yabushita (1975)===
====Evolving from 1968 and 1974, to 1975====
As we have [[#Yabushita_.281968.29|discussed, above]], {{ Yabushita68 }} couched his stability analysis in terms of the search for a solution to, what we have labeled as, the [[#Yabushita68LAWE|Yabushita68 Isothermal LAWE]].  In taking this approach, he introduced (a rather unconventional) [[#gDefinition|perturbation variable]],
<div align="center">
<math>g \equiv \xi^2 \frac{d}{d\xi}\biggl( \frac{\delta\Phi}{c_s^2}\biggr) \, .</math>
</div>
He pointed out that, defined in this way, the density variation, <math>\delta\rho/\rho_c</math>, can be determined from <math>g</math> via the differential relation,
<div align="center">
<math>\frac{\delta \rho}{\rho_c} = \frac{1}{\xi^2} \frac{dg}{d\xi} \, .</math>
</div>
It was in this context that {{ Yabushita74 }} discovered that the eigenvector for the fundamental mode of radial oscillation in the configuration that has the [[SSC/Structure/BonnorEbert#Pressure-Bounded_Isothermal_Sphere|Bonnor-Ebert limiting mass]], can be specified analytically.  Specifically, the eigenfrequency is zero &#8212; that is, the configuration is marginally [dynamically] unstable &#8212; and the eigenfunction is, precisely,
<div align="center">
<math>g_Y = M - \xi^3 e^{-\psi} = \xi^2 \frac{d\psi}{d\xi} - \xi^3 e^{-\psi} \, .</math>
</div>

{{ Yabushita75full }}, on the other hand, adopted a more conventional approach and couched his discussions of radial oscillations and stability in terms of the search for solutions to, what we have referred to as, the (standard) [[#IsothermalLAWE|Isothermal LAWE]], involving the radial displacement function, <math>x \equiv \delta r/r_0</math>.  In order to facilitate a transition from his earlier work, we demonstrate, here, the relationship between the perturbation variable, <math>g</math>, and the displacement function, <math>x</math>.

[[File:ImageOfDerivations07GoodLedouxWalraven.png|250px|right|LW58 derivation]]Following the [[SSC/PerspectiveReconciliation#Approach_by_Ledoux_and_Walraven|derivation of the linearized wave equation]] presented, as by {{ LW58full }} &#8212; see, also, equation <font color="green">&#x2460;</font> in the associated derivation diagram shown here, on the right &#8212; the linearized continuity equation may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{\delta\rho}{\rho_c} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \frac{1}{r^2}\frac{d}{d r} \biggl[ r^3 \biggl(\frac{\rho}{\rho_c}\biggr) \frac{\delta r}{r_0} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \frac{1}{\xi^2}\frac{d}{d \xi} \biggl[ \xi^3 \biggl(\frac{\rho}{\rho_c}\biggr) x \biggr]\, .</math>
  </td>
</tr>
</table>
</div>
Hence, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{1}{\xi^2} \frac{dg}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \frac{1}{\xi^2}\frac{d}{d \xi} \biggl[ \xi^3 \biggl(\frac{\rho}{\rho_c}\biggr) x \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ g</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \xi^3 \biggl(\frac{\rho}{\rho_c}\biggr) x </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ x</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \frac{g}{\xi^3}~ e^{\psi} \, . </math>
  </td>
</tr>
</table>
</div>

So, in terms of the radial displacement function, the eigenfunction for the marginally unstable configuration that has the Bonnor-Ebert limiting mass is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x_Y</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \frac{g_Y}{\xi^3}~ e^{\psi} </math>
  </td>
</tr>

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

====Verification of Analytic Solution====
{{ Yabushita75 }} states that the radial eigenfunction for the marginally unstable isothermal configuration is (see his equation 5.6),
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\psi '}{\xi} - e^{-\psi}\biggr] e^{\psi} \, .</math>
  </td>
</tr>
</table>
</div>
As we have just demonstrated, this expression for the displacement function is, indeed, equivalent to the expression that was derived by {{ Yabushita74 }} for the perturbation function, <math>g_Y</math>.  (The overall sign has been reversed on the expression, but this leading factor of negative one is arbitrary.)   In order to see if this, indeed, satisfies the

<div align="center">
<font color="maroon"><b>Isothermal LAWE </b></font> with <math>\sigma_c^2=0</math><br />
<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}{d\xi^2} + \biggl[4 - \xi \psi ' \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} 
+  \biggl[
\frac{\psi ' }{\xi} \biggr]  x \, ,
</math>
  </td>
</tr>
</table>
</div>
we need to evaluate the first and second derivatives of the eigenfunction.  We also need to keep in mind that from the isothermal Lane-Emden equation, we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{1}{\xi^2} \frac{d}{d\xi}\biggr[ \xi^2 \psi ' \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>e^{-\psi}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \psi '' </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>e^{-\psi} - \frac{2 \psi '}{\xi} \, .</math>
  </td>
</tr>
</table>
</div>
So, the first derivative of the eigenfunction is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{d}{d\xi} \biggl\{ \biggl[ \frac{\psi '}{\xi} - e^{-\psi}\biggr] e^{\psi}  \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\psi '}{\xi} - e^{-\psi}\biggr] \psi ' e^{\psi}  
+ \biggl[ \frac{\psi ''}{\xi} - \frac{\psi '}{\xi^2} + \psi ' e^{-\psi}\biggr] e^{\psi}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl\{
\biggl[ \frac{\psi '}{\xi} - e^{-\psi}\biggr] \psi '  
+ \biggl[ \frac{1}{\xi}\biggl( e^{-\psi} - \frac{2 \psi '}{\xi} \biggr) - \frac{\psi '}{\xi^2} + \psi ' e^{-\psi}\biggr] \biggr\} e^{\psi}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{(\psi ')^2}{\xi}  - \frac{3\psi '}{\xi^2}\biggr]e^{\psi} + \frac{1}{\xi}  
\, .
</math>
  </td>
</tr>
</table>
</div>
And the second derivative is,

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

<tr>
  <td align="right">
<math>\frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{d\xi}\biggl\{\biggl[\frac{(\psi ')^2}{\xi}  - \frac{3\psi '}{\xi^2}\biggr]e^{\psi} + \frac{1}{\xi}  \biggr\}
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e^{\psi}\biggl\{ 
\frac{(\psi ')^3}{\xi}  - \frac{8 (\psi ')^2}{\xi^2}   + \frac{12\psi '}{\xi^3}  
\biggr\}+ \frac{2\psi '}{\xi}  - \frac{4}{\xi^2}   \, . 
</math>
  </td>
</tr>
</table>
</div>

Plugging these expressions into the isothermal LAWE, we have,

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

<tr>
  <td align="right">
<math>
\frac{d^2x}{d\xi^2} + \biggl[4 - \xi \psi ' \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} +  \biggl[\frac{\psi ' }{\xi} \biggr]  x 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e^{\psi}\biggl\{ 
\frac{(\psi ')^3}{\xi}  - \frac{8 (\psi ')^2}{\xi^2}   + \frac{12\psi '}{\xi^3}  
\biggr\}+ \frac{2\psi '}{\xi}  - \frac{4}{\xi^2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[\frac{4}{\xi} - \psi ' \biggr] \biggl\{ \biggl[ \frac{(\psi ')^2}{\xi}  - \frac{3\psi '}{\xi^2}\biggr]e^{\psi} + \frac{1}{\xi}\biggr\}
+  \frac{\psi ' }{\xi}  \biggl\{ \biggl[ \frac{\psi '}{\xi} - e^{-\psi}\biggr] e^{\psi}  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e^{\psi}\biggl\{
\frac{(\psi ')^3}{\xi}  - \frac{8 (\psi ')^2}{\xi^2}   + \frac{12\psi '}{\xi^3} + \biggl(\frac{\psi '}{\xi} \biggr)^2   
+ \biggl[\frac{4}{\xi} - \psi ' \biggr] \cdot\biggl[ \frac{(\psi ')^2}{\xi}  - \frac{3\psi '}{\xi^2}\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \frac{1}{\xi}\biggl[\frac{4}{\xi} - \psi ' \biggr]  
- \frac{\psi ' }{\xi}  
+ \frac{2\psi '}{\xi}  - \frac{4}{\xi^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e^{\psi}\biggl\{
\frac{(\psi ')^3}{\xi}  - \frac{8 (\psi ')^2}{\xi^2}   + \frac{12\psi '}{\xi^3} + \biggl(\frac{\psi '}{\xi} \biggr)^2   
+ \frac{4(\psi ')^2}{\xi^2}  - \frac{12\psi '}{\xi^3} 
- \frac{(\psi ')^3}{\xi}  + \frac{3(\psi ')^2}{\xi^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>
</div>
Amazing!!

Actually, we should have started with the simpler version of the expression for the displacement function, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 - \frac{\psi ' e^\psi}{\xi} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{dx}{d\xi} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \biggl\{ \frac{\psi '' e^\psi}{\xi} + \frac{(\psi ')^2 e^\psi}{\xi}  - \frac{\psi ' e^\psi}{\xi^2} \biggr\}</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~ - \frac{d\ln x}{d\ln\xi} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{x}\biggl\{ 1  + (\psi ')^2 e^\psi  - \frac{3\psi ' e^\psi}{\xi} \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{x}\biggl[ (\psi ')^2 e^\psi  + 3x-2\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Hence, if at the surface we impose a boundary condition, <math>d\ln x/d\ln\xi = -3</math>, then this also means that, at the surface,

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

<tr>
  <td align="right">
<math>3x_\mathrm{surf}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ (\psi ')^2 e^\psi  + 3x-2\biggr]_\mathrm{surf} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ \biggl[ (\psi ')^2 e^\psi  \biggr]_\mathrm{surf} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>2 \, .</math>
  </td>
</tr>
</table>
</div>
And this is precisely the condition that Bonnor recognized was associated with the critical turning point along the equilibrium sequence.  Hence, we can now also precisely associate the marginally (dynamically) unstable configuration as being the configuration at the turning point.

<span id="CentralValue">For the record,</span> it would also be good to know what the limiting value of the Yabushita displacement function is at the center of the isothermal configuration.  Drawing on some of our already-developed [[Appendix/Ramblings/PowerSeriesExpressions#Approximate_Power-Series_Expressions|power-series expressions]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 - \frac{\psi ' e^\psi}{\xi} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 - \frac{1}{\xi} 
\biggl[1 + \psi + \frac{\psi^2}{2} \cdots  \biggr]
\frac{d}{d\xi}\biggl[ \psi \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>1 - \frac{1}{\xi} 
\biggl[1 + \biggl(\frac{\xi^2}{6} - \frac{\xi^4}{120}\biggr) + \frac{1}{2} \biggl(\frac{\xi^2}{6} - \frac{\xi^4}{120}\biggr)^2  \biggr]
\frac{d}{d\xi}\biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>1 -  
\biggl[1 + \frac{\xi^2}{6}  \biggr]
\biggl[ \frac{1}{3} - \frac{\xi^2}{30} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>1 -  
\biggl[ \frac{1}{3} - \frac{\xi^2}{30} \biggr]
+\frac{\xi^2}{18} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\frac{2}{3} + \frac{2^2\xi^2}{3^2 \cdot 5} \, .
</math>
  </td>
</tr>
</table>
</div>