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Simple Harmonic Motion Quadratic Perturbation

Given a simple harmonic oscillator with a quadratic perturbation $\epsilon x^2$,

\begin{displaymath}
\ddot x+{\omega_0}^2x-\alpha\epsilon x^2=0,
\end{displaymath} (1)

find the first-order solution using a perturbation method. Write
\begin{displaymath}
x\equiv x_0+\epsilon x_1+\ldots,
\end{displaymath} (2)

so
\begin{displaymath}
\ddot x= \ddot x_0+\epsilon \ddot x_1+\ldots.
\end{displaymath} (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} (4)

Keeping only terms of order $\epsilon$ 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} (5)

Since this equation must hold for all Powers of $\epsilon$, we can separate it into the two differential equations
\begin{displaymath}
\ddot x_0+{\omega_0}^2x_0=0
\end{displaymath} (6)


\begin{displaymath}
\ddot x_1+{\omega_0}^2x_1=\alpha {x_0}^2.
\end{displaymath} (7)

The solution to (6) is just
\begin{displaymath}
x_0=A\cos(\omega_0 t+\phi).
\end{displaymath} (8)

Setting our clock so that $\phi=0$ gives
\begin{displaymath}
x_0=A\cos(\omega_0 t).
\end{displaymath} (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} (10)

The two homogeneous solutions to (10) are
$\displaystyle x_1$ $\textstyle =$ $\displaystyle \cos(\omega_0 t)$ (11)
$\displaystyle x_2$ $\textstyle =$ $\displaystyle \sin(\omega_0 t).$ (12)

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} (13)

where
\begin{displaymath}
g(t)=\alpha A^2\cos^2(\omega_0 t),
\end{displaymath} (14)

and the Wronskian is
$\displaystyle W$ $\textstyle \equiv$ $\displaystyle x_1\dot x_2-\dot x_1 x_2$  
  $\textstyle =$ $\displaystyle \cos ({\omega_0}t){\omega_0}\cos({\omega_0}t)-[-{\omega_0}\sin({\omega_0}t)]\sin({\omega_0}t)$  
  $\textstyle =$ $\displaystyle {\omega_0}.$ (15)

Plugging everything into (13),


$\displaystyle x_p$ $\textstyle =$ $\displaystyle \alpha A^2\left[{-\cos({\omega_0}t)\int{\sin({\omega_0}t)\cos^2({...
...a_0}}\,dt+\sin({\omega_0}t)\int{\cos^3({\omega_0}t)\over{\omega_0}}\,dt}\right]$  
  $\textstyle =$ $\displaystyle {\alpha A^2\over {\omega_0}} \left\{{\sin({\omega_0}t)\int [1-\sin^2({\omega_0}t)]\cos({\omega_0}t)\, dt}\right.$  
  $\textstyle \phantom{=}$ $\displaystyle \left.{\mathop{-}\cos({\omega_0}t)\int\sin({\omega_0}t)\cos^2({\omega_0}t)\, dt}\right\}.$ (16)

Now let
$\displaystyle u$ $\textstyle \equiv$ $\displaystyle \sin({\omega_0}t)$ (17)
$\displaystyle du$ $\textstyle =$ $\displaystyle {\omega_0}\cos({\omega_0}t)\,dt$ (18)
$\displaystyle v$ $\textstyle \equiv$ $\displaystyle \cos(\omega_0 t)$ (19)
$\displaystyle dv$ $\textstyle =$ $\displaystyle -{\omega_0}\sin({\omega_0}t)\,dt.$ (20)

Then


$\displaystyle x_p$ $\textstyle =$ $\displaystyle {\alpha A^2\over {\omega_0}^2} \left[{\sin({\omega_0}t)\int (1-u^2)\,du+\cos({\omega_0}t) \int v^2\,dv}\right]$  
  $\textstyle =$ $\displaystyle {\alpha A^2\over {\omega_0}^2} \left[{\sin({\omega_0}t)(1-{\textstyle{1\over 3}} u^3)+\cos({\omega_0}t) {\textstyle{1\over 3}} v^3}\right]$  
  $\textstyle =$ $\displaystyle {\alpha A^2\over {\omega_0}^2}\{\sin({\omega_0}t)[1-{\textstyle{1...
...in^3({\omega_0}t)]+{\textstyle{1\over 3}}\cos({\omega_0}t)\cos^3({\omega_0}t)\}$  
  $\textstyle =$ $\displaystyle {\alpha A^2\over {\omega_0}^2}\left\{{{\textstyle{1\over 3}}[\cos^4({\omega_0}t)- \sin^4({\omega_0}t)]+\sin^2({\omega_0}t)}\right\}$  
  $\textstyle =$ $\displaystyle {\alpha A^2\over {\omega_0}^2}\left\{{{\textstyle{1\over 3}}[\cos^2({\omega_0}t)-\sin^2({\omega_0}t)]+\sin^2({\omega_0}t)}\right\}$  
  $\textstyle =$ $\displaystyle {\alpha A^2\over {\omega_0}^2}{\textstyle{1\over 3}} [\cos^2({\omega_0}t)+2\sin^2({\omega_0}t)]$  
  $\textstyle =$ $\displaystyle {\alpha A^2\over 3{\omega_0}^2}[2-\cos^2({\omega_0}t)] = {\alpha A^2\over 3{\omega_0}^2} \{2-{\textstyle{1\over 2}}[1+\cos(2{\omega_0}t)]\}$  
  $\textstyle =$ $\displaystyle {\alpha A^2\over 6{\omega_0}^2}[3-\cos(2{\omega_0}t)].$ (21)

Plugging $x_0(t)$ 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} (22)



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© 1996-9 Eric W. Weisstein
1999-05-26