Editing SSC/Dynamics/IsothermalSimilaritySolution (section)

==Mathematical Solution==

===Summary===
A similarity solution becomes possible for these equations when the single independent variable,
<div align="center">
<math>~\zeta = \frac{c_s t}{r} \, ,</math>
</div>
is used to replace both <math>~r</math> and <math>~t</math>.  Then, if <math>~M_r</math>, <math>~\rho</math>, and <math>~v_r</math> assume the following forms,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_r(r,t)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{c_s^3 t}{G}\biggr) m(\zeta) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\rho(r,t)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~v_r(r,t)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- c_s U(\zeta) \, ,</math>
  </td>
</tr>
</table>
</div>

<span id="CoupledODEs">the three coupled partial differential equations reduce to two coupled ordinary differential equations for the functions,</span> <math>~\Rho (\zeta)</math> and <math>~U(\zeta)</math>, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dU}{d\zeta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\zeta U +1) [\Rho (\zeta U +1) -2)]}{[ (\zeta U +1)^2 - \zeta^2]} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{dP}{d\zeta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\zeta \Rho [2-\Rho (\zeta U +1)]}{[ (\zeta U +1)^2 - \zeta^2]} \, ,</math>
  </td>
</tr>
</table>
</div>

and a single equation defining <math>~m(\zeta)</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~m(\zeta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Rho \biggl[ U + \frac{1}{\zeta} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

The parameters <math>~\zeta, m, \Rho</math>, and <math>~U</math>, and this summary set of equations are exactly those used by [http://adsabs.harvard.edu/abs/1977ApJ...218..834H Hunter (1977)] in his analysis of this problem.  But they differ in form from the relations used by [http://adsabs.harvard.edu/abs/1969MNRAS.145..271L Larson (1969)], [http://adsabs.harvard.edu/abs/1969MNRAS.144..425P Penston (1969)], and [http://adsabs.harvard.edu/abs/1977ApJ...214..488S Shu (1977)] primarily because these authors chose to use a similarity variable, 
<div align="center">
<math>~x = \pm \frac{1}{\zeta} \, ,</math>
</div>
instead of <math>~\zeta</math>.  Hunter's analysis is the most complete and his relations will be used here, but a transformation between his presentation and those of the other authors can be easily obtained from Table 1 of [http://adsabs.harvard.edu/abs/1977ApJ...218..834H Hunter (1977)] which, for convenience, is reproduced here.

<div align="center">
<table border="1" cellpadding="5" align="center" width="60%">
<tr>
  <th align="center" colspan="5">
Analogous to Table 1 from [http://adsabs.harvard.edu/abs/1977ApJ...218..834H Hunter (1977)]<br />
''Relations Between the Variables Used by Different Authors''
  </th>
</tr>
<tr>
  <td align="center" width="20%">Physical<br />Quantity</td>
  <td align="center" width="20%">Herein</td>
  <td align="center" width="20%">[http://adsabs.harvard.edu/abs/1969MNRAS.145..271L Larson (1969)]</td>
  <td align="center" width="20%">[http://adsabs.harvard.edu/abs/1969MNRAS.144..425P Penston (1969)]</td>
  <td align="center">[http://adsabs.harvard.edu/abs/1977ApJ...214..488S Shu (1977)]</td>
</tr>
<tr>
  <td align="center"><math>~\frac{c_s t}{r}</math></td>
  <td align="center"><math>~\zeta</math></td>
  <td align="center"><math>~- \frac{1}{x}</math></td>
  <td align="center"><math>~- \frac{1}{x}</math></td>
  <td align="center"><math>~+ \frac{1}{x}</math></td>
</tr>
<tr>
  <td align="center"><math>~- \frac{v_r}{c_s}</math></td>
  <td align="center"><math>~U</math></td>
  <td align="center"><math>~\xi</math></td>
  <td align="center"><math>~- V</math></td>
  <td align="center"><math>~-v</math></td>
</tr>
<tr>
  <td align="center"><math>~\frac{4\pi G\rho r^2}{c_s^2}</math></td>
  <td align="center"><sup>&dagger;</sup><math>~\Rho</math></td>
  <td align="center"><math>~x^2\eta</math></td>
  <td align="center"><math>~x^2 e^Q</math></td>
  <td align="center"><math>~x^2\alpha</math></td>
</tr>
<tr>
  <td align="center"><math>~\frac{GM_r}{c_s^3 t}</math></td>
  <td align="center"><math>~m</math></td>
  <td align="center">&hellip;</td>
  <td align="center"><math>~-N</math></td>
  <td align="center"><math>~m</math></td>
</tr>
<tr>
  <td align="center"><math>~\ln(4\pi G\rho t^2)</math></td>
  <td align="center"><math>~Q</math></td>
  <td align="center"><math>~\ln\eta</math></td>
  <td align="center"><math>~Q</math></td>
  <td align="center"><math>~\ln\alpha</math></td>
</tr>
<tr>
  <td align="center"><math>~\frac{r}{(- c_s t)}</math></td>
  <td align="center"><math>~y</math></td>
  <td align="center"><math>~x</math></td>
  <td align="center"><math>~x</math></td>
  <td align="center"><math>~-x</math></td>
</tr>
<tr>
  <td align="left" colspan="5">
<sup>&dagger;</sup>Adopting Hunter's notation, this dimensionless variable name, <math>~\Rho</math> (the capital Greek letter, <math>~\rho</math>), should not be confused with the variable name, <math>~P</math>, that represents herein the ideal gas pressure.
  </td>
</tr>
</table>
</div>

The following pair of images are reproductions of (left) Figure 1 and (right) Figure 3 from [http://adsabs.harvard.edu/abs/1977ApJ...218..834H Hunter (1977)].  The solid curves show how (left) the dimensionless velocity, <math>~U</math>, and (right) the dimensionless density, <math>~\Rho</math>, behave as a function of the similarity variable, <math>~\zeta</math>, for models having several different prescribed values of Hunter's parameter, <math>~Q_0</math>.  For each value of <math>~Q_0</math>, the table of numbers immediately below the pair of images provides corresponding values of several other numerical constants.

<div align="center">
<table border="1" cellpadding="5">
<tr><td align="center" colspan="2">
Figures extracted from [http://adsabs.harvard.edu/abs/1977ApJ...218..834H Hunter (1977)]<p></p>
"''The Collapse of Unstable Isothermal Spheres''"<p></p>
ApJ, vol. 218, pp. 834 - 845 &copy; American Astronomical Society
</td></tr>
<tr>
<td>
[[File:Hunter77Fig1.png|400px|center|Figure 1 from Hunter (1977, ApJ, 218, 836]]
</td>
<td>
[[File:Hunter77Fig3.png|400px|center|Figure 3 from Hunter (1977, ApJ, 218, 836]]
</td>
</tr>
<tr>
  <td align="center" colspan="2">
{| class="wikitable" style="text-align:center;"
|- 
| style="width:60px; text-align:center; border-bottom:2px solid black; "| Model 
| style="width:5px; text-align:center; "| &nbsp; 
| style="width:60px; text-align:center; border-bottom:2px solid black; "| <math>~Q_0</math> 
| style="width:5px; text-align:center; "| &nbsp; 
| style="width:60px; text-align:center; border-bottom:2px solid black; "| <math>~U_0</math> 
| style="width:5px; text-align:center; "| &nbsp; 
| style="width:60px; text-align:center; border-bottom:2px solid black; "| <math>~\Rho_0</math> 
| style="width:5px; text-align:center; "| &nbsp; 
| style="width:60px; text-align:center; border-bottom:2px solid black; ;"| <math>~m_0</math>
|- 
| LP || &nbsp;
| 0.5139  || &nbsp;
| 3.278  || &nbsp;
| 8.854  || &nbsp;
| 46.915
|- 
| H(b) || &nbsp;
| 11.236  || &nbsp;
| 0.295  || &nbsp;
| 2.378  || &nbsp;
| 2.577
|-  
| H(d) || &nbsp;
| 20.975  || &nbsp;
| 0.026 || &nbsp;
| 2.023  || &nbsp;
| 1.138
|-  
| EW || &nbsp;
| <math>~+ \infty</math>  || &nbsp;
| 0.000  || &nbsp;
| 2.000  || &nbsp;
| 0.975
|}
  </td>
</tr>
</table>
</div>

===Proof===

Plugging the similarity solution expressions for <math>~M_r</math> and <math>~\rho</math> into the first of the three governing equations gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial}{\partial r} \biggl[ \biggl(\frac{c_s^3 t}{G}\biggr) m(\zeta)  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4\pi r^2 \biggl[ \biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \biggr]  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (c_s t )  \frac{\partial}{\partial r} \biggl[ m(\zeta)  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Rho (\zeta) \, .</math>
  </td>
</tr>
</table>
</div>

Plugging the similarity solution expressions for <math>~M_r</math>, <math>~\rho</math>, and <math>~v_r</math> into the second of the three governing equations gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial}{\partial t} \biggl[ \biggl(\frac{c_s^3 t}{G}\biggr) m(\zeta)  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 4\pi r^2 \biggl[ \biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \biggr]  \biggl[ -c_s U(\zeta)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\partial}{\partial t} \biggl[ t m(\zeta)  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Rho (\zeta) U(\zeta) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ m(\zeta) + t \biggl[ \frac{\partial m(\zeta)}{\partial t} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Rho (\zeta) U(\zeta) \, .</math>
  </td>
</tr>
</table>
</div>

And, plugging the similarity solution expressions for <math>~M_r</math>, <math>~\rho</math>, and <math>~v_r</math> into the third of the three governing equations gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial }{\partial t} \biggl[ - c_s U(\zeta) \biggr] + \biggl[ - c_s U(\zeta) \biggr] \frac{\partial }{\partial r} \biggl[ - c_s U(\zeta) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- c_s^2 \biggl[\biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta)   \biggr]^{-1} \frac{\partial }{\partial r}\biggl[  \biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \biggr] 
- \frac{G}{r^2}\biggl[ \biggl(\frac{c_s^3 t}{G}\biggr) m(\zeta) \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\partial }{\partial t} \biggl[ U(\zeta) \biggr] - c_s U(\zeta) \frac{\partial }{\partial r} \biggl[ U(\zeta) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{c_s r^2}{\Rho (\zeta)}   \biggr]\frac{\partial }{\partial r}\biggl[  \biggl(\frac{\Rho (\zeta)}{r^2}\biggr)  \biggr] 
+ \biggl[ \frac{c_s^2 t}{r^2} \biggr] m(\zeta)  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\partial U}{\partial t} - (c_s U) \frac{\partial U}{\partial r} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{c_s}{\Rho}   \biggl[ \biggl( \frac{\partial \Rho}{\partial r}\biggr) -\frac{2\Rho}{r} \biggr]  
+ \biggl[ \frac{c_s^2 t}{r^2} \biggr] m(\zeta)  \, .</math>
  </td>
</tr>
</table>
</div>

Now, from the functional dependence of <math>~m(\zeta)</math> on <math>~\Rho(\zeta)</math> and <math>~U(\zeta)</math>, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial m}{\partial r}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial r}
+ \Rho \biggl[ \frac{\partial U}{\partial r} - \frac{1}{\zeta^2} \frac{\partial \zeta}{\partial r}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial r}
+ \Rho \biggl[ \frac{\partial U}{\partial r} + \frac{1}{r \zeta} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial m}{\partial t}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial t}
+ \Rho \biggl[ \frac{\partial U}{\partial t} - \frac{1}{\zeta^2} \frac{\partial \zeta}{\partial t}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial t}
+ \Rho \biggl[ \frac{\partial U}{\partial t} - \frac{1}{t \zeta} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Hence, the first two governing equations become, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rho </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(r\zeta)  \biggl\{
\biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial r}
+ \Rho \biggl[ \frac{\partial U}{\partial r} + \frac{1}{r \zeta} \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ r\zeta U + r \biggr] \frac{\partial\Rho}{\partial r}
+ (r\zeta \Rho ) \frac{\partial U}{\partial r} + \Rho 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \zeta U + 1\biggr] \frac{\partial\Rho}{\partial r}
+ (\zeta \Rho ) \frac{\partial U}{\partial r}  \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rho (\zeta) U(\zeta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\Rho\biggl[ U + \frac{1}{\zeta}\biggr] + t \biggl\{ 
\biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial t}
+ \Rho \biggl[ \frac{\partial U}{\partial t} - \frac{1}{t \zeta} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
t \biggl\{ 
\biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial t}
+ \Rho \biggl[ \frac{\partial U}{\partial t} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \zeta U + 1 \biggr] \frac{\partial\Rho}{\partial t}
+ (\zeta \Rho) \frac{\partial U}{\partial t}  \, .
</math>
  </td>
</tr>
</table>
</div>

Now, we can use these two relations to replace derivatives of <math>~\Rho</math> with derivatives of <math>~U</math> &#8212; or ''visa versa'' &#8212; in the third governing relation.  In the first case, we obtain,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \frac{\Rho}{c_s} \biggl[\frac{\partial U}{\partial t} - (c_s U) \frac{\partial U}{\partial r} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\partial \Rho}{\partial r}\biggr) -\frac{2\Rho}{r}   
+ \frac{\Rho^2}{r} \biggl[\zeta U + 1\biggr]   </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~   
\frac{\Rho^2(\zeta U + 1)}{r}  -\frac{2\Rho}{r}  - \biggl( \frac{\partial U}{\partial r}\biggr) \biggl[ \frac{\zeta \Rho}{(\zeta U + 1)} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~  \Rightarrow ~~~  \frac{1}{r} \biggl[
\Rho^2(\zeta U + 1)  - 2\Rho \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{\Rho}{c_s} \biggl[\frac{\partial U}{\partial t}\biggr]  - (\Rho U) \frac{\partial U}{\partial r} 
+ \biggl( \frac{\partial U}{\partial r}\biggr) \biggl[ \frac{\zeta \Rho}{(\zeta U + 1)} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~  \Rightarrow ~~~  \biggl[
\Rho(\zeta U + 1)  - 2 \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{r}{c_s} \biggl[\frac{\partial U}{\partial t}\biggr]  
+ \biggl( \frac{\partial U}{\partial r}\biggr) \biggl[ \frac{r \zeta }{(\zeta U + 1)} - (rU) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{t}{\zeta} \biggl[\frac{\partial U}{\partial t}\biggr]  
+ r \biggl( \frac{\partial U}{\partial r}\biggr) \biggl[ \frac{\zeta - U (\zeta U + 1)}{(\zeta U + 1)} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\frac{\partial U}{\partial t}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{dU}{d\zeta} \biggr)
\frac{\partial \zeta}{\partial t}
=
\biggl( \frac{dU}{d\zeta} \biggr)\frac{c_s}{r} = \biggl( \frac{dU}{d\zeta} \biggr)\frac{\zeta}{t} \, ;</math>
  </td>
</tr>
<tr><td colspan="3" align="center">and</td></tr>
<tr>
  <td align="right">
<math>~\frac{\partial U}{\partial r}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{dU}{d \zeta} \biggr) \frac{\partial \zeta}{\partial r}
=
- \frac{c_s t}{r^2} \biggl( \frac{dU}{d \zeta} \biggr) = -\frac{\zeta^2}{c_st} \biggl( \frac{dU}{d \zeta} \biggr) 
\, ,</math>
  </td>
</tr>
</table>
</div>

we can rewrite this as an ODE of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \biggl[
\Rho(\zeta U + 1)  - 2 \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl(\frac{d U}{d\zeta}\biggr)  
-\zeta \biggl( \frac{d U}{d\zeta }\biggr) \biggl[ \frac{\zeta - U (\zeta U + 1)}{(\zeta U + 1)} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~  [
\Rho(\zeta U + 1)  - 2 ](\zeta U + 1) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl(\frac{d U}{d\zeta}\biggr)  \biggl\{(\zeta U + 1)
-\zeta \biggl[ \zeta - U (\zeta U + 1) \biggr]\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl(\frac{d U}{d\zeta}\biggr)  \biggl[ \zeta^2U^2 + 2\zeta U + 1 - \zeta^2
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \Rightarrow ~~~ 
\frac{d U}{d\zeta}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ [\Rho(\zeta U + 1)  - 2 ](\zeta U + 1)}{  [ (\zeta U + 1)^2 - \zeta^2 ] } \, .</math>
  </td>
</tr>
</table>
</div>

In the second case, we obtain,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{c_s}{\Rho} \biggl( \frac{\partial \Rho}{\partial r}\biggr)   
+ \frac{\zeta}{t}  \biggl[ \Rho (\zeta U + 1 ) - 2\biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{\partial U}{\partial t} - (c_s U) \frac{\partial U}{\partial r} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-\biggl[ \frac{\zeta U +1}{\zeta \Rho} \biggr] \frac{\partial \Rho}{\partial t}
+ (c_s U) \biggl[ \frac{\zeta U + 1}{\zeta\Rho} \biggr] \frac{\partial \Rho}{\partial r}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
\frac{\zeta}{t}  \biggl[ 2- \Rho (\zeta U + 1 ) \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl\{ \frac{c_s}{\Rho}   
- (c_s U) \biggl[ \frac{\zeta U + 1}{\zeta\Rho} \biggr] \biggr\}\frac{\partial \Rho}{\partial r}
+ \biggl[ \frac{\zeta U +1}{\zeta \Rho} \biggr] \frac{\partial \Rho}{\partial t}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
\frac{\zeta^2 \Rho}{c_s t}  \biggl[ 2- \Rho (\zeta U + 1 ) \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \zeta  
- U (\zeta U + 1 ) \biggr] \frac{\partial \Rho}{\partial r}
+ \frac{1}{c_s}\biggl[ \zeta U +1 \biggr] \frac{\partial \Rho}{\partial t} \, .
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\frac{\partial \Rho}{\partial t}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{d\Rho}{d\zeta} \biggr)
\frac{\partial \zeta}{\partial t}
=
\biggl( \frac{d\Rho}{d\zeta} \biggr)\frac{c_s}{r} = \biggl( \frac{d\Rho}{d\zeta} \biggr)\frac{\zeta}{t} \, ;</math>
  </td>
</tr>
<tr><td colspan="3" align="center">and</td></tr>
<tr>
  <td align="right">
<math>~\frac{\partial \Rho}{\partial r}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{d\Rho}{d \zeta} \biggr) \frac{\partial \zeta}{\partial r}
=
- \frac{c_s t}{r^2} \biggl( \frac{d\Rho}{d \zeta} \biggr) = -\frac{\zeta^2}{c_st} \biggl( \frac{d\Rho}{d \zeta} \biggr) 
\, ,</math>
  </td>
</tr>
</table>
</div>

we can rewrite this as an ODE of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\frac{\zeta^2 \Rho}{c_s t}  \biggl[ 2- \Rho (\zeta U + 1 ) \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
- \frac{\zeta^2}{c_s t} \biggl[ \zeta  
- U (\zeta U + 1 ) \biggr] \frac{d\Rho}{d \zeta}
+ \frac{\zeta}{c_s t}\biggl[ \zeta U +1 \biggr] \frac{d\Rho}{d\zeta} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\zeta\Rho [ 2- \Rho (\zeta U + 1 ) ] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
- \zeta\biggl[ \zeta  
- U (\zeta U + 1 ) \biggr] \frac{d\Rho}{d \zeta}
+ \biggl[ \zeta U +1 \biggr] \frac{d\Rho}{d \zeta} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl\{( \zeta U +1 )  - \zeta [ \zeta  - U (\zeta U + 1 ) ] \biggr\} \frac{d\Rho}{d \zeta}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
[(\zeta U + 1)^2 - \zeta^2]\frac{d\Rho}{d \zeta}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \Rightarrow~~~
\frac{d\Rho}{d \zeta}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ \zeta\Rho [ 2- \Rho (\zeta U + 1 ) ] }{[(\zeta U + 1)^2 - \zeta^2] } \, .
</math>
  </td>
</tr>
</table>
</div>

Thus, we are able to understand the origin of the [[#CoupledODEs|pair of 1<sup>st</sup>-order ODEs, given above]], that describe the connected relationship between the two quantities, <math>~\Rho</math> and <math>~U</math>.

<!--
===Proof2===

First, note that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \zeta}{\partial t}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{c_s}{r} = \frac{\zeta}{t} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \zeta}{\partial r}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{c_s t}{r^2} = -\frac{\zeta^2}{c_st} \, .</math>
  </td>
</tr>
</table>
</div>

Next, let's take partial derivatives, with respect to both <math>~r</math> and <math>~t</math>, of the three primary physical variables, <math>~M_r</math>, <math>~\rho</math>, and <math>~v_r</math>.   
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial M_r}{\partial r}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{c_s^3 t}{G} \biggr)\frac{dm}{d\zeta} \biggl( \frac{\partial \zeta}{\partial r} \biggr)
=
\biggl( \frac{c_s^3 t}{G} \biggr)\frac{dm}{d\zeta} \biggl(- \frac{\zeta^2}{c_s t} \biggr)
=
-\biggl( \frac{c_s^2 \zeta^2}{G} \biggr)\frac{dm}{d\zeta}  \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial M_r}{\partial t}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{c_s^3 t}{G} \biggr)\frac{dm}{d\zeta} \biggl( \frac{\partial \zeta}{\partial t} \biggr) + \biggl( \frac{c_s^3}{G}\biggr) m
=
\biggl( \frac{c_s^3 t}{G} \biggr)\frac{dm}{d\zeta} \biggl( \frac{\zeta}{t} \biggr) + \biggl( \frac{c_s^3}{G}\biggr) m
=
\biggl( \frac{c_s^3 }{G} \biggr) \biggl[ \zeta ~\frac{dm}{d\zeta}  + m \biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial v_r}{\partial r}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-c_s \frac{dU}{d\zeta} \biggl( \frac{\partial \zeta}{\partial r} \biggr)
=
-c_s \frac{dU}{d\zeta} \biggl( - \frac{\zeta^2}{c_s t} \biggr)
=
\frac{\zeta^2}{t} \frac{dU}{d\zeta} \, ;
</math>
  </td>
</tr>
</table>
</div>
-->