Given a simple harmonic oscillator with a quadratic perturbation
,
![\begin{displaymath}
\ddot x+{\omega_0}^2x-\alpha\epsilon x^2=0,
\end{displaymath}](s1_1309.gif) |
(1) |
find the first-order solution using a perturbation method. Write
![\begin{displaymath}
x\equiv x_0+\epsilon x_1+\ldots,
\end{displaymath}](s1_1310.gif) |
(2) |
so
![\begin{displaymath}
\ddot x= \ddot x_0+\epsilon \ddot x_1+\ldots.
\end{displaymath}](s1_1311.gif) |
(3) |
Plugging (2) and (3) back into (1) gives
![\begin{displaymath}
(\ddot x_0+\epsilon\ddot x_1)+({\omega_0}^2x_0+{\omega_0}^2\epsilon x_1)-\alpha\epsilon(x_0+2x_0x_1\epsilon+\ldots)=0.
\end{displaymath}](s1_1312.gif) |
(4) |
Keeping only terms of order
and lower and grouping, we obtain
![\begin{displaymath}
(\ddot x_0+{\omega_0}^2x_0)+(\ddot x_1+{\omega_0}^2x_1-\alpha{x_0}^2)\epsilon=0.
\end{displaymath}](s1_1314.gif) |
(5) |
Since this equation must hold for all Powers of
, we can separate it into the two differential equations
![\begin{displaymath}
\ddot x_0+{\omega_0}^2x_0=0
\end{displaymath}](s1_1315.gif) |
(6) |
![\begin{displaymath}
\ddot x_1+{\omega_0}^2x_1=\alpha {x_0}^2.
\end{displaymath}](s1_1316.gif) |
(7) |
The solution to (6) is just
![\begin{displaymath}
x_0=A\cos(\omega_0 t+\phi).
\end{displaymath}](s1_1317.gif) |
(8) |
Setting our clock so that
gives
![\begin{displaymath}
x_0=A\cos(\omega_0 t).
\end{displaymath}](s1_1319.gif) |
(9) |
Plugging this into (7) then gives
![\begin{displaymath}
\ddot x_1+{\omega_0}^2x_1=\alpha A^2\cos^2(\omega_0 t).
\end{displaymath}](s1_1320.gif) |
(10) |
The two homogeneous solutions to (10) are
The particular solution to (10) is therefore given by
![\begin{displaymath}
x_p(t)=-x_1(t)\int{x_2(t)g(t)\over W(t)}\,dt+x_2(t)\int {x_1(t)g(t)\over W(t)}\,dt,
\end{displaymath}](s1_1173.gif) |
(13) |
where
![\begin{displaymath}
g(t)=\alpha A^2\cos^2(\omega_0 t),
\end{displaymath}](s1_1323.gif) |
(14) |
and the Wronskian is
Plugging everything into (13),
Now let
Then
Plugging
and (21) into (2), we obtain the solution
![\begin{displaymath}
x(t)=A\cos({\omega_0}t)-{\alpha A^2\over 6{\omega_0}^2}\epsilon [\cos(2{\omega_0}t)-3].
\end{displaymath}](s1_1344.gif) |
(22) |
© 1996-9 Eric W. Weisstein
1999-05-26