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An averaged amplitude equation describes the slow evolution of a periodic orbit's amplitude under a weak perturbation. If unperturbed energy is $H(r)$ with $H^{\prime }(r)\ne 0$, divide the mean energy drift over one fast period by $H^{\prime }(r)$. Simple zeros identify attracting or repelling persistent periodic orbits according to the derivative of this drift; this approximation is local to an appropriate perturbative amplitude range.

Set $H=({\dot{x}}^2+x^2)/2$. ThenOn an unperturbed circle write $x=r\cos t$, $\dot{x}=-r\sin t$. The supplied integrals giveEquivalently, the leading averaged amplitude equation isWith $s=r^2$, its nonzero stationary amplitudes solve $\beta s^2-2\alpha s+8=0$. If ${\alpha }^2>8\beta$, both roots are positive and simple:The averaged drift is positive below $r_{-}$, negative between $r_{-}$ and $r_{+}$, and positive above $r_{+}$. Thus the inner orbit is attracting and the outer orbit repelling, with leading displacement amplitudes $r_{-}$ and $r_{+}$ and period $2\pi +O(ϵ)$. The simple zeros of the Poincaré return map persist for sufficiently small $ϵ$.

If ${\alpha }^2<8\beta$, the leading quadratic is strictly positive for all $s\ge 0$; the energy increases on every nonzero unperturbed orbit, so the energy balance method yields no periodic orbits in any fixed compact range of nonzero amplitudes for sufficiently small $ϵ$. This perturbative calculation does not itself establish a uniform global exclusion at amplitudes growing as $ϵ$ tends to zero. At equality, the leading drift has a double zero at $r^2=\alpha /\beta$. It is positive on both sides, with a semistable candidate at that amplitude. First-order averaging alone cannot decide whether the exact system has zero, one, or two nearby periodic orbits at equality: higher-order drift determines the unfolding of this degenerate zero. The original PDF has ${\alpha }^2=8\beta$; the extra exponent on $\beta$ in the converted TeX is an OCR error.