Simple Harmonic Motion Quadratic Perturbation

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

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

find the first-order solution using a perturbation method. Write

$$x\equiv x_0+\epsilon x_1+\ldots,$$ (2)

so

$$\ddot x= \ddot x_0+\epsilon \ddot x_1+\ldots.$$ (3)

Plugging (2) and (3) back into (1) gives

$$(\ddot x_0+\epsilon\ddot x_1)+({\omega_0}^2x_0+{\omega_0}^2\epsilon x_1)-\alpha\epsilon(x_0+2x_0x_1\epsilon+\ldots)=0.$$ (4)

Keeping only terms of order $\epsilon$ and lower and grouping, we obtain

$$(\ddot x_0+{\omega_0}^2x_0)+(\ddot x_1+{\omega_0}^2x_1-\alpha{x_0}^2)\epsilon=0.$$ (5)

Since this equation must hold for all Powers of $\epsilon$, we can separate it into the two differential equations

$$\ddot x_0+{\omega_0}^2x_0=0$$ (6)

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

The solution to (6) is just

$$x_0=A\cos(\omega_0 t+\phi).$$ (8)

Setting our clock so that $\phi=0$ gives

$$x_0=A\cos(\omega_0 t).$$ (9)

Plugging this into (7) then gives

$$\ddot x_1+{\omega_0}^2x_1=\alpha A^2\cos^2(\omega_0 t).$$ (10)

The two homogeneous solutions to (10) are

$$x_1 = \cos(\omega_0 t)$$ (11)

$$x_2 = \sin(\omega_0 t).$$ (12)

The particular solution to (10) is therefore given by

$$x_p(t)=-x_1(t)\int{x_2(t)g(t)\over W(t)}\,dt+x_2(t)\int {x_1(t)g(t)\over W(t)}\,dt,$$ (13)

where

$$g(t)=\alpha A^2\cos^2(\omega_0 t),$$ (14)

and the Wronskian is

$$W \equiv x_1\dot x_2-\dot x_1 x_2$$

$$= \cos ({\omega_0}t){\omega_0}\cos({\omega_0}t)-[-{\omega_0}\sin({\omega_0}t)]\sin({\omega_0}t)$$

$$= {\omega_0}.$$ (15)

Plugging everything into (13),

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

$$= {\alpha A^2\over {\omega_0}} \left\{{\sin({\omega_0}t)\int [1-\sin^2({\omega_0}t)]\cos({\omega_0}t)\, dt}\right.$$

$$\phantom{=} \left.{\mathop{-}\cos({\omega_0}t)\int\sin({\omega_0}t)\cos^2({\omega_0}t)\, dt}\right\}.$$ (16)

Now let

$$u \equiv \sin({\omega_0}t)$$ (17)

$$du = {\omega_0}\cos({\omega_0}t)\,dt$$ (18)

$$v \equiv \cos(\omega_0 t)$$ (19)

$$dv = -{\omega_0}\sin({\omega_0}t)\,dt.$$ (20)

Then

$$x_p = {\alpha A^2\over {\omega_0}^2} \left[{\sin({\omega_0}t)\int (1-u^2)\,du+\cos({\omega_0}t) \int v^2\,dv}\right]$$

$$= {\alpha A^2\over {\omega_0}^2} \left[{\sin({\omega_0}t)(1-{\textstyle{1\over 3}} u^3)+\cos({\omega_0}t) {\textstyle{1\over 3}} v^3}\right]$$

$$= {\alpha A^2\over {\omega_0}^2}\{\sin({\omega_0}t)[1-{\textstyle{1... ...in^3({\omega_0}t)]+{\textstyle{1\over 3}}\cos({\omega_0}t)\cos^3({\omega_0}t)\}$$

$$= {\alpha A^2\over {\omega_0}^2}\left\{{{\textstyle{1\over 3}}[\cos^4({\omega_0}t)- \sin^4({\omega_0}t)]+\sin^2({\omega_0}t)}\right\}$$

$$= {\alpha A^2\over {\omega_0}^2}\left\{{{\textstyle{1\over 3}}[\cos^2({\omega_0}t)-\sin^2({\omega_0}t)]+\sin^2({\omega_0}t)}\right\}$$

$$= {\alpha A^2\over {\omega_0}^2}{\textstyle{1\over 3}} [\cos^2({\omega_0}t)+2\sin^2({\omega_0}t)]$$

$$= {\alpha A^2\over 3{\omega_0}^2}[2-\cos^2({\omega_0}t)] = {\alpha A^2\over 3{\omega_0}^2} \{2-{\textstyle{1\over 2}}[1+\cos(2{\omega_0}t)]\}$$

$$= {\alpha A^2\over 6{\omega_0}^2}[3-\cos(2{\omega_0}t)].$$ (21)

Plugging $x_0(t)$ and (21) into (2), we obtain the solution

$$x(t)=A\cos({\omega_0}t)-{\alpha A^2\over 6{\omega_0}^2}\epsilon [\cos(2{\omega_0}t)-3].$$ (22)