Editing MathProjects/EigenvalueProblemN1 (section)

===Try a Polynomial Expression for the Eigenfunction===
Let's ''guess'' that the proper eigenfunction is a polynomial expression in <math>~x</math>.  Specifically, let's try a solution of the form,
<div align="center">
<math>\mathcal{F}_\sigma = a + bx + cx^2 + dx^3 + fx^4 + gx^5 + \cdots </math>
</div>
truncated at progressively higher- and higher-order terms.

====Lowest-order mode (Mode 0)====  
Try, 
<div align="center">
<math>\mathcal{F} = a \, ,</math>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\mathcal{F}}{dx}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, ,</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2\mathcal{F}}{dx^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>
So, the governing 2<sup>nd</sup>-order ODE reduces to,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(3\sigma^2 - 2 \alpha )  a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, ,</math>
  </td>
</tr>
</table>
</div>
which ''will'' be satisfied as long as, <math>~\sigma = (2\alpha/3)^{1/2} \, .</math>  We conclude, therefore, that the eigenvector defining the lowest-order (the simplest) solution to the governing ODE has an eigenfunction given by,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{F}_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a = \mathrm{constant} \, ,</math>
  </td>
</tr>
</table>
</div>
with a corresponding eigenfrequency whose value is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sigma_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{2\alpha}{3} \biggr)^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>

====Second Guess====  
Try, 
<div align="center">
<math>\mathcal{F} = a + bx \, ,</math>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\mathcal{F}}{dx}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~b \, ,</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2\mathcal{F}}{dx^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>
Plugging this trial eigenfunction into the governing 2<sup>nd</sup>-order ODE gives,
<div align="center">
<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{4}{x}\biggl[1 -  \frac{3}{2}x^2 \biggr] b +  \biggl[3\sigma^2 - 2 \alpha \biggr]  (a + bx)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4b}{x} + (3\sigma^2 - 2 \alpha -6) bx +  (3\sigma^2 - 2 \alpha ) a \, .</math>
  </td>
</tr>
</table>
</div>
But this expression can be satisfied for all values of <math>~x</math> only if <math>~b = 0</math>, in which case the trial eigenfunction reduces to the earlier solution, <math>~\mathcal{F}_0</math>.  We conclude, therefore, that our "second guess" does not generate a new solution to this eigenfunction problem.

====Third Guess====  
Try, 
<div align="center">
<math>\mathcal{F} = a + cx^2\, ,</math>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\mathcal{F}}{dx}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2cx \, ,</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d^2\mathcal{F}}{dx^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2c \, .</math>
  </td>
</tr>
</table>
</div>
Plugging this trial eigenfunction into the governing 2<sup>nd</sup>-order ODE therefore gives,
<div align="center">
<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>~2c(1 - x^2) + 8c(1 -  \frac{3}{2}x^2 ) +  (3\sigma^2 - 2 \alpha )  (a + cx^2)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[10c + (3\sigma^2 - 2 \alpha )a\biggr] + \biggl[ (3\sigma^2 - 2 \alpha )  -14 \biggr]cx^2 \, .</math>
  </td>
</tr>
</table>
</div>
This relation will be satisfied for all values of <math>~x</math> if both expressions inside the square brackets are simultaneously zero, that is, if, 
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~10c + (3\sigma^2 - 2 \alpha )a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, ,</math>
  </td>
</tr>
</table>
</div>
and, simultaneously,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(3\sigma^2 - 2 \alpha )  -14 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>




* <font color="purple">Mode 1</font>:
: <math>x_1 = a + b\chi_0^2</math>, in which case,
<div align="center">
<math>
\frac{dx}{d\chi_0} = 2b\chi_0; ~~~~ \frac{d^2 x}{d\chi_0^2} =  2b;
</math>
</div>

<div align="center">
<math>
\frac{1}{(1 - \chi_0^2)}  \biggl\{ 2b (1 - \chi_0^2) + 8b \biggl[1 -  \frac{3}{2}\chi_0^2 \biggr]  +  A_1 \biggl(1 + \frac{b}{a}\chi_0^2 \biggr) \biggr\} = 0 ,
</math><br />
</div>
where,

<div align="center">
<math>
A_1 \equiv \frac{a}{\gamma_\mathrm{g}}\biggl[ \biggl( \frac{3}{2\pi G\rho_c} \biggr) \omega_1^2+ 2(4 - 3\gamma_\mathrm{g}) \biggr] .
</math>
</div>
Therefore,

<div align="center">
<math>
(A_1 + 10b) + \biggl[ \biggl(\frac{b}{a}\biggr) A_1 - 14b \biggr] \chi_0^2  = 0 ,
</math>
<br />
<br />
<math>
\Rightarrow ~~~~~ A_1  = - 10b ~~~~~\mathrm{and} ~~~~~ A_1 = 14a
</math>
<br />
<br />
<math>
\Rightarrow ~~~~~ \frac{b}{a} = -\frac{7}{5}  ~~~~~\mathrm{and} ~~~~~ \frac{A_1}{a} = 14 = \frac{1}{\gamma_\mathrm{g}}\biggl[ \biggl( \frac{3}{2\pi G\rho_c} \biggr) \omega_1^2+ 2(4 - 3\gamma_\mathrm{g}) \biggr] .

</math>
</div>
Hence,
<div align="center">
<math>
\biggl( \frac{3}{2\pi G\rho_c} \biggr) \omega_1^2 = 20\gamma_\mathrm{g}  -8
</math>
<br />
<br />
<math>
\Rightarrow ~~~~~ \omega_1^2 = \frac{2}{3}\biggl( 4\pi G\rho_c \biggr) (5\gamma_\mathrm{g}  -2)
</math>
</div>
and, to within an arbitrary normalization factor,
<div align="center">
<math>
x_1 = 1 - \frac{7}{5}\chi_0^2 .
</math>
</div>