Prove perturbation theory breaks down I want to find the $\mathcal{O}(1)$ and $\mathcal{O}(\epsilon)$ terms in the pedestrian expansion $y = y_0 + \epsilon y_1 + \epsilon ^2 y_2 + \dots$, where $y$ satisfies the following second order ODE:
$\frac{d^2y}{dt^2} + y\cos(y) = 0$,
with $|{\epsilon}|\ll 1$ and subject to $y(0) = \epsilon\alpha$, $y'(0) = 0$.
I know that Regular Perturbation Theory breaks down but am struggling to prove it. I have already tried using a Taylor expansion but am unsure of how to proceed.
Any help would be much appreciated.
 A: For $ϵ=0$ you get the initial conditions $y(0)=y'(0)=0$. As the term $y\cos y$ is also zero at this point, you get a constant solution equal to zero, $y_0=0$.
Then next you get $y=ϵy_1+...$, with $y_1(0)=α$, $y_1'(0)=0$ and
$$
0=ϵy_1''+ϵy_1\cos(ϵy_1)+O(ϵ^2)\implies 0=y_1''+y_1
$$
To get the higher order terms in a more systematic manner set $u=\sin(y)$, $v=\cos(y)$ so that $\partial_ϵu=v\partial_ϵy$, $\partial_ϵv=-u\partial_ϵy$, that is
\begin{align}
u_1+2ϵu_2+3ϵ^2u_3+.. &= (1+ϵv_1+ϵ^2v_2+...)(y_1+2ϵy_2+3ϵ^2y_3+...)\\
v_1+2ϵv_2+3ϵ^2v_3+.. &= -(ϵu_1+ϵ^2u_2+...)(y_1+2ϵy_2+3ϵ^2y_3+...)\\
y_1''+ ϵy_2''+ϵ^2y_3''+..&=-(y_1+ ϵy_2+ϵ^2y_3)(1+ϵv_1+ϵ^2v_2+...)
\end{align}
so that in components
\begin{align}
u_1&=y_1,&  v_1&=0, & y_1''&=-y_1\\
2u_2&=2y_2+v_1y_1,& 2v_2 &=-u_1y_1,& y_2''&=-y_2-y_1v_1\\
3u_3&=3y_3+2y_2v_1+y_1v_2,& 3v_3&=-y_1u_2-2y_2u_1,& y_3''&= -y_3-y_2v_1-y_1v_2
\end{align}
Solving this in the direction $y\to u\to v$ results in
\begin{align}
y_1&=α\cos(t),& u_1&=α\cos(t),& v_1&=0 \\
y_2&=0,& u_2&=0,& v_2&=-\tfrac12α^2\cos^2(t)\\
y_3''+y_3&=\tfrac12α^3\cos^3(t)\\&=\tfrac18α^3(\cos(3t)+3\cos(t))\\
y_3&=-\tfrac1{64}α^3(\cos(3t)-\cos(t))+\tfrac3{16}α^3t\sin(t)
\end{align}

Remark: One could of course also shift the perturbation order one place and start again with the rescaled problem $y''+y\cos(ϵy)=0$, $y(0)=α$, $y(0)=0$ and find that the perturbation term only occurs in even orders,
$$
y''+y=\frac{ϵ^2}2y^3-\frac{ϵ^4}{24}y^5+...
$$
This will also work well for the first non-zero perturbation term, but becomes more difficult to pursue systematically for the higher-order terms.

Breakdown: The equation itself is of the conservative type $y''+f(y)=0$ with $f(0)=0$, $f'(0)>0$ so that close to the zero solution you get periodic oscillations.
The term $t\sin(t)$ in $y_3$ destroys that periodic quality of the solution. As the solution itself has amplitude $O(ϵ)$, the diverging term becomes as large as the solution if $ϵ^2t=O(1)$.

Remark: One can repair this divergent term by combining it with $y_1$ to give
$$
ϵα\cos(t)+\tfrac3{16}ϵ^3α^3t\sin(t)=ϵα\cos\left(t+\tfrac3{16}ϵ^2α^2t\right)+O(ϵ^5)
$$
introducing a frequency correction or a two-scale perturbation expansion.
