I have the following problem: In the equation $\ddot{x}+\Omega^2x+\epsilon f(x) = \Gamma \cos t$, $\Omega$ is not close to an odd integer, and $f(x)$ is an odd function of x, with expansion, $$f(a\cos t) = -a_1(a)\cos t - a_3(a)\cos3t-\cdots.$$ Derive a perturbation solution of period $2\pi$, to order $\epsilon$. I have a solution so my question is on the solution of the problem.
Solution:
In the equation $\ddot{x}+\Omega^2x+\epsilon f(x) = \Gamma \cos t$, let $$x = x_0+\epsilon x_1+\cdots$$. The two leading terms satisfy $$\ddot{x_0}+\Omega^2 x_0 = \Gamma \cos t$$ $$\ddot{x_1}+\Omega x_1 = -f(x_0+\epsilon x_1+\cdots)\approx -f(x_0).$$ The $2\pi$ periodic solution of the first equation is $$x_0 = \dfrac{\Gamma}{\Omega^2-1}\cos t$$ provided $\Omega$ is not close to $1$ or any odd integer for higher order terms. The next term $x_1$ in the perturbation satisfies
\begin{align}\ddot{x_1}+\Omega x_1 &= -f\bigg(\dfrac{\Gamma}{\Omega^2-1}\cos t\bigg) \\ &= a_1(k)\cos t+a_3(k)\cos 3t. (1) \end{align} where $k = \dfrac{\Gamma}{\Omega^2-1}$. The $2\pi$ periodic solution is $$x_1 = \dfrac{a_1(k)}{\Omega^2-1}\cos t + \dfrac{a_1(k)}{\Omega^2-9}\cos 3t.$$
My question is at step $(1)$. I understand that they are putting the solution for $x_0$ in the function and expanding it out but why only up to $a_3(k)\cos 3t$?
Bennett Gardiner correctly called out a typo in $x_1$: it should read $$x_1 = \dfrac{a_1(k)}{\Omega^2-1}\cos t + \dfrac{a_3(k)}{\Omega^2-9}\cos 3t$$ since this is what we get by solving (1).
The reason for dropping higher frequency terms is that their coefficients are implicitly assumed decaying, probably as $a_{2j+1} = O(\epsilon^j)$. This is something that the authors of the problem failed to specify after writing $$f(a\cos t) = -a_1(a)\cos t - a_3(a)\cos3t-\cdots$$ For example, if $a_{3}$ is zero but $a_5$ is not, then we should have $$x_1 = \dfrac{a_1(k)}{\Omega^2-1}\cos t + \dfrac{a_5(k)}{\Omega^2-25}\cos 5t +\dots$$ instead. And if each $a_j$ is $10^6$ times the preceding one, then all these asymptotics are junk.