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For $\omega \ne 0$, put $\mu =\hat{\mu }{\omega }^2$, $C=\omega z$ and $\tau =\omega T$, where $z=\hat{C}$. Then $C_T={\omega }^2{z}_{\tau }$ and division by ${\omega }^2$ givesWe may take $\omega >0$ without loss of the dynamics by conjugating the original equation if necessary; otherwise $\tau$ reverses the time orientation. The scaling is singular at $\omega =0$ and is not a transformation for that exactly zero-frequency case.

The Hamiltonian limit of three-to-one forcing drops the terms proportional to $\omega$. For $z=x+iy$ the remaining real system isThe proposed first integral isIndeed $x_{\tau }=H_y/2$ and $y_{\tau }=-H_x/2$, hence $H_{\tau }=0$. The unperturbed equation is a planar Hamiltonian system, with a center equilibrium at the origin and three saddle equilibria atAll saddle equilibria have $H=1/3$. The factorizationshows that their central separatrix consists of the three sides of an equilateral triangle. Each level $0<H<1/3$ inside this triangle is a closed periodic orbit. Indeed, inside the triangle $0<r<1$ and ${\partial }_{r}H=2r(1+r\sin 3\theta )>0$. Each ray from the origin therefore meets each such level once, giving a compact simple closed contour with no equilibrium point on it. The nonzero vector field traverses this contour periodically; the period grows without bound as the separatrix is approached. This supplies an infinite family, not a claim that every level outside the central region is closed.

Restore the small radial perturbation. Its exact effect on the first integral isConsequently the continuum of Hamiltonian system orbits generally does not persist. The origin becomes a weak attracting focus for $\hat{\mu }<0$ or a repelling focus for $\hat{\mu }>0$, and the three hyperbolic saddle equilibria persist with perturbed stable manifold and unstable manifold. For a small positive $\hat{\mu }$, outward drift on very small orbits balances cubic damping on somewhat larger ones, selecting a stable limit cycle rather than an arbitrary energy level. Near the center equilibrium $H\simeq r^2$, so its leading radius is $r\simeq \sqrt{\hat{\mu }}$ when $\hat{\mu }$ is also small.

For a more general closed unperturbed orbit ${\Gamma }_h$, the averaged area criterion for perturbed Hamiltonian cycles says that persistence requires the averaged energy drift ${\oint }_{{\Gamma }_h}{H}_{\tau }d\tau$ to vanish. Since the unperturbed speed is $|\nabla H|/2$, the planar divergence theorem converts this leading drift towhere ${\mathcal{D}}_h$ is the enclosed region. Isolated zeros select candidate periodic orbits; a drift changing from positive inside to negative outside gives an attracting limit cycle. The separatrix triangle has mean $r^2=1/4$, so its leading flux changes sign at $\hat{\mu }=1/2$. This marks the leading possible heteroclinic transition, with higher-order corrections needed to locate it precisely.

As a cycle approaches the saddle equilibria, long residence times and splitting of the heteroclinic cycle become important. Orbits can instead drift inward to the equilibrium point at the origin or leave the periodic island and approach one of the stable states with phase locking of the full canonical equation. Those upper-branch threefold phase-locked equilibria have $C=O(1)$, so they lie outside the local $C=O(\omega )$ scaling. Thus the small perturbation gives energy selection, attracting or repelling oscillations, and possible switching/locking transitions; it does not preserve a conserved $H$ or an infinite family of neutral periodic solutions. This qualitative picture does not assume all global parameter values have the same attractor.

The linearized perturbation obeys $\dot{\eta }=D\mathbf{f}(\mathbf{X}(t))\eta$, so its fundamental matrix satisfiesThe Floquet multipliers are the eigenvalues of the monodromy matrix $\Phi (T)$. Differentiating the orbit equation shows that $\dot{\mathbf{X}}(t)=\mathbf{f}(\mathbf{X}(t))$ is a solution of the variational equation. For a nonconstant periodic orbit this tangent solution is nonzero and periodic, giving one multiplier equal to $1$.

The Liouville formula for a fundamental matrix gives $(\det \Phi {)}^{\prime }=(\mathrm{tr}D\mathbf{f})\det \Phi$. Since $\det \Phi (0)=1$, the product of the two multipliers is $\exp ({\int }_0^T\nabla \cdot \mathbf{f}(\mathbf{X}(t))dt)$. The other multiplier is therefore exactly this exponential.

The Poincare-Bendixson theorem says that a nonempty compact omega-limit set of a smooth planar flow containing no equilibrium is a periodic orbit; use its allowed cylinder version here. For $F>1$, choose $p_{-}=(F-1)/(2k)$ and $p_{+}=2(F+1)/k$. At the lower boundary $\dot{p}>0$ and at the upper boundary $\dot{p}<0$, so the compact annulus $p_{-}\le p\le p_{+}$ is positively invariant. It has no equilibrium because $F-\sin \theta >0$. The theorem supplies a periodic orbit, and $\dot{\theta }=p>0$ makes it encircle the cylinder.

Every periodic orbit must lie in $p>0$: for $p\le 0$, $\dot{p}>0$, and a crossing of $p=0$ can only be upward. In $p>0$ a periodic orbit is a graph of a function satisfyingTwo distinct function graphs cannot cross by uniqueness of this scalar differential equation. If $p_1>p_2$, their difference obeysIt contracts by a factor strictly less than $1$ during one circuit, contradicting periodicity of both function graphs. The scalar return map is order preserving, so a multi-circuit periodic graph of a function also cannot have a period larger than one circuit. Therefore there is exactly one periodic orbit, and it encircles the cylinder. The standard auxiliary results used are existence/uniqueness of smooth differential equations and invariance at inward-pointing boundaries.

Because $X=\mathbb{N}$ is a countable set and $\mu (X)=1$, some $x\in X$ satisfies $\mu (\{x\})>0$. Invariance and the inclusion $\{T^{n}x\}\subseteq T^{-1}\{T^{n+1}x\}$ giveIf the forward orbit contained infinitely many distinct points, their disjoint singletons would each have measure at least $\mu (\{x\})>0$, contradicting that $\mu$ is a probability measure. The orbit therefore eventually enters a finite periodic orbit $Y$.

Now $T(Y)=Y$, so $Y\subseteq T^{-1}Y$. Since $\mu$ is invariant,Thus $Y$ belongs to the invariant sigma-algebra modulo null sets. It has positive measure, so the fact that $T$ is an ergodic transformation forcesThis is the ergodic invariant probability on a countable state space has finite cyclic support.