Consider the Harmonic Oscillator as a Hamiltonian System on phase space with Hamiltonian $H = x^2/2 + \rho^2/2$. Now modify the system by adding a perturbation.
$H(\epsilon) = x^2/2 + \rho^2/2 + \epsilon x^3$ for some $\epsilon > 0$
Calculate the first nonzero order correction to the amplitude of the $H=1$ trajectory.
I'm very confused because I can't find anything in my class notes to help me out with this other than that (I think) we need something of the form $A(\epsilon) = A_{0} + \epsilon A_{1} + \epsilon^2 A_{2} …$ where we must determine the $A_i$ (at least up to the first nonzero one after $A_{0}$. I believe $A_0 = \sqrt{2}$ (although I'm not sure why). But how do I go about calculating the other coefficients?
The amplitude $A(\epsilon)$ is the largest value $x$ takes, and occurs (for $\epsilon\ge 0)$ when $p=0$. So $$1 = H(\epsilon)\vert_{p=0} = \frac{1}{2}A(\epsilon)^2 + \epsilon A(\epsilon)^3.$$ Writing $A(\epsilon) = A_0 + \epsilon A_1 + \ldots$, this gives to first order in $\epsilon$ $$ 1 = \frac{1}{2}(A_0^2 + 2\epsilon A_0A_1) + \epsilon A_0^3.$$ $\epsilon = 0$ implies that $A_0 = \sqrt{2}$, and then the above equation implies $A_1 = -A_0^2 = -2$. So the first order correction should be $$A(\epsilon) = \sqrt{2}-2\epsilon.$$