Editing SSC/Perturbations (section)

==Complementary Approach==
Looking back at the set of three, coupled, linearized equations &#8212; but using <math>~\Delta</math> to denote the density fluctuation, rather than <math>~d</math>, in order to avoid confusion with the differentiation operators,

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<font color="#770000">'''Linearized'''</font><br />
<span id="Continuity"><font color="#770000">'''Equation of Continuity'''</font></span><br />
<math>
r_0 \frac{dx}{dr_0} = - 3 x - \Delta ,
</math><br />

<font color="#770000">'''Linearized'''</font><br />
<span id="PGE:Euler"><font color="#770000">'''Euler + Poisson Equations'''</font></span><br />
<math>
\frac{P_0}{\rho_0} \frac{dp}{dr_0}  = (4x + p)g_0 + \omega^2 r_0 x ,
</math><br />

<font color="#770000">'''Linearized'''</font><br />
<span id="PGE:AdiabaticFirstLaw">Adiabatic Form of the<br />
<font color="#770000">'''First Law of Thermodynamics'''</font></span><br />
<math>
p = \gamma_\mathrm{g} \Delta  \, ,
</math>
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let's combine them into a 2<sup>nd</sup>-order ODE that governs the eigenfunction of the density perturbation, <math>~\Delta</math>, rather than of the radial displacement, <math>~x</math>.  We begin by using the second equation to obtain an expression for <math>~x</math>:
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<math>~ x</math>
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<math>~=</math>
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<math>~(4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] \, .</math>
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From this, we determine that,
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<math>~ (4g_0 + \omega^2 r_0)\frac{dx}{dr_0}</math>
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<math>~=</math>
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<math>~\frac{d}{dr_0} \biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] 
+ (4g_0 + \omega^2 r_0)\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr]\frac{d}{dr_0}(4g_0 + \omega^2 r_0)^{-1} 
</math>
  </td>
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&nbsp;
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<math>~=</math>
  </td>
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<math>~\frac{P_0}{\rho_0} \frac{d^2p}{dr_0^2} + \frac{dp}{dr_0} \cdot \frac{d}{dr_0} \biggl( \frac{P_0}{\rho_0} \biggr) 
-p \frac{dg_0}{dr_0}  - g_0\frac{dp}{dr_0}  
- (4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] \biggl[ \omega^2 + 4\frac{dg_0}{dr_0}\biggr] 
</math>
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&nbsp;
  </td>
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<math>~=</math>
  </td>
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<math>~\frac{P_0}{\rho_0} \frac{d^2p}{dr_0^2} + \frac{dp}{dr_0} \biggl\{ -g_0 + \frac{d}{dr_0} \biggl( \frac{P_0}{\rho_0} \biggr) 
- (4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \biggr] \biggl[ \omega^2 + 4\frac{dg_0}{dr_0}\biggr] \biggr\}
+ p \biggl\{ (4g_0 + \omega^2 r_0)^{-1}\biggl[ g_0 \biggr] \biggl[ \omega^2 + 4\frac{dg_0}{dr_0}\biggr] -\frac{dg_0}{dr_0} \biggr\}
</math>
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Next, we substitute this expression for <math>~x</math> into the first equation &#8212; gradually replacing <math>~p</math> with <math>~\gamma_g \Delta</math>, as well &#8212; and carry out the differentiation:
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<math>~ -\Delta - 3\biggl\{(4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - \gamma_g \Delta g_0 \biggr]
\biggr\}
</math>
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<math>~=</math>
  </td>
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<math>~r_0 \frac{d}{dr_0} \biggl\{(4g_0 + \omega^2 r_0)^{-1}\biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr]
\biggr\}
</math>
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<math>~ \Rightarrow~~- \Delta \biggl[ \frac{g_0}{r_0} (4 - 3\gamma_g)+ \omega^2 \biggr] - \frac{3P_0}{r_0 \rho_0} \frac{dp}{dr_0}
</math>
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<math>~=</math>
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<math>~ \frac{d}{dr_0} \biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr] -
(4g_0 + \omega^2 r_0)^{-1} \biggl[ \frac{P_0}{\rho_0} \frac{dp}{dr_0} - pg_0 \biggr]\frac{d}{dr_0} \biggl[(4g_0 + \omega^2 r_0)
\biggr]
</math>
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</table>
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