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- #1
How can I reduce it?You can reduce the order of your perturbation expansion to something manageable i.e. order 2 (or just use maple).
Otherwise, looks like you'll have to get your hands dirty.
Here is how I got to this point: Is there a mistake? The professor had the $R_T = \frac{1}{16}(6R^3 - 5\beta r^2 - 8r)$.I ended up with this:
\begin{alignat*}{3}
R' & = & -\frac{1}{16}(5\beta R^5 + 8R)\\
\theta' & = & 0
\end{alignat*}
When beta is 9/40 it is supposed to be special. I don't see anything different in the plots or limit cycles.
$x(t,\varepsilon) = x_0(t,T)+\varepsilon x_1(t,T)+\cdots$ where the slow time is $T=\varepsilon t$. Let $f(x,x') = \frac{dx}{dt}\left[1 - \left(\frac{dx}{dt}\right)^2 + \beta\left(\frac{dx}{dt}\right)^4\right]$.$$
\frac{d^2x}{dt^2} + x + \epsilon\frac{dx}{dt}\left[1 - \left(\frac{dx}{dt}\right)^2 + \beta\left(\frac{dx}{dt}\right)^4\right] = 0,\quad\quad\epsilon\ll 1,
$$
Is there a smart way to do this problem? It will take forever to do directly.