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For unit-mass motion in $V(x)=3x^2-2x^3$, the mechanical energy is $E=v^2/2+V(x)$. The origin is a center equilibrium, and $(1,0)$ is a saddle equilibrium. Energies $0<E<1$ have trapped periodic orbits around the origin as well as a separate escaping component to the right. The barrier energy has a homoclinic orbit returning to the saddle after a turn at $x=-1/2$, plus two saddle branches on $x>1$. Above the barrier, particles traverse the well and eventually escape to the right.

For small positive forcing-frame frequency $\omega$, use $\mu =\hat{\mu }{\omega }^2$, $C=\omega z$ and $\tau =\omega T$ in the three-to-one spatially forced amplitude equation. The leading system is $z_{\tau }+iz={\overline{z}}^2$. Writing $z=x+iy$ gives the Hamiltonian system $(x_{\tau },y_{\tau })=(H_y/2,-H_x/2)$ and the displayed first integral. The origin is a center equilibrium; three saddle equilibria at $(0,1)$ and $(\pm \sqrt{3}/2,-1/2)$ lie on $H=1/3$. The factorization $H-1/3=(1+2y)[x^2-(y-1{)}^2/3]$ reveals a triangular heteroclinic cycle, containing closed periodic orbits for every $0<H<1/3$.

The autonomous planar system $\dot{x}=y-2y^3$, $\dot{y}=-x$ preserves $K$. Its origin is a center equilibrium and $(0,\pm 1/\sqrt{2})$ are saddle equilibria. The level $K=1/4$ is $x=\pm (y^2-1/2)$: its two inner arcs are heteroclinic orbits enclosing periodic motion, while its outer branches connect saddles to escaping trajectories. Since $|\dot{y}|\sim y^2$ on escaping levels, infinity is reached in finite time. Adding $\alpha x/2$ to $\dot{x}$ gives $\dot{K}=\alpha x^2$, so positive $\alpha$ produces outward crossing of central levels and breaks the equal-energy heteroclinic connections.

A limit cycle is an isolated periodic orbit of a dynamical system. It is attracting if nearby orbits approach it in forward time. A continuous family of closed orbits around a center equilibrium does not consist of limit cycles, because its members are not isolated.

Let $X,Y$ be prey and predator abundances. With constant positive demographic and predation coefficients, the Lotka-Volterra equations areTheir positive equilibrium is $X_{\ast }=\gamma /\delta$, $Y_{\ast }=\alpha /\beta$. Its Jacobian matrix and eigenvalues areThe linearized populations oscillate with period $2\pi /\sqrt{\alpha \gamma }$, with the predator peak following the prey peak. This is a center equilibrium, not an attracting equilibrium. The conclusion is supported beyond linearization by the Logarithmic first integral of the Lotka-Volterra equations: in these units one may useDirect differentiation gives $\dot{V}=0$. Its Hessian matrix is a positive-definite matrix in the positive quadrant, and its minimum is at $(X_{\ast },Y_{\ast })$. Nearby level curves are closed periodic orbits, so there is no damping toward a preferred oscillation amplitude.

Introduce $H=x^2-y^2+y^4/2$. Along a solution,At $\alpha =0$, $H$ is a first integral. The identity $H+1/2=x^2+(y^2-1{)}^2/2$ shows its level sets are bounded: neither $x$ nor $y$ can escape to infinity on a fixed level. Consequently trajectories remain bounded as $t\to \infty$, including those starting far from the origin. This bound also prevents finite-time escape; no explicit solution is needed.

The equilibrium points are $(0,0)$ and $(0,\pm 1)$. The Jacobian matrix at $(0,y_{\ast })$ isAt the origin its eigenvalues are $(\alpha \pm \sqrt{{\alpha }^2+4})/2$, with opposite signs, so the origin is a saddle point in all three cases. An eigenvector for $\lambda$ lies on $y=-x/\lambda$; the stable direction has positive slope and the unstable direction negative slope. At $(0,\pm 1)$ the eigenvalues are $(\alpha \pm \sqrt{{\alpha }^2-8})/2$.

When $\alpha =0$, these latter eigenvalues are $\pm i\sqrt{2}$. The nonlinear classification follows from the strict minima of $H$ at the two points: nearby level sets are closed curves without other equilibria, and the nonvanishing vector field makes them periodic orbits. Thus the two points are center equilibria. For $-1/2<H<0$, there are two separate closed ovals, one around each center equilibrium. The level $H=0$ consists of the saddle point and two homoclinic orbits forming a figure eight, explicitlyFor $H>0$, a single outer closed orbit encloses all three equilibrium points. Motion is clockwise, as the vector points downwards wherever $x>0$. The upper homoclinic orbit leaves the origin into $x<0,y>0$ and returns through $x>0,y>0$; the lower one has the reflected orientation.

When $\alpha =0.1$, the two nonzero equilibrium points have eigenvalues $0.05\pm i\sqrt{7.99}/2$ and are unstable foci. Nearby paths spiral outward clockwise. The first integral becomes a strictly increasing energy along every nonconstant trajectory: although $\dot{H}$ can momentarily vanish when $x=0$, it cannot vanish on an interval unless the solution is an equilibrium. Thus there are no periodic orbits or homoclinic orbits. The two stable separatrices of the saddle point, traced backwards, spiral towards the corresponding unstable foci. Its unstable separatrices move onto positive energy and escape outwards. All other nonequilibrium trajectories outside the stable manifold are unbounded forwards: a bounded forward limit set would have to lie in $x=0$, whose only invariant points are the three equilibria; the two unstable foci cannot attract a nonconstant trajectory, and convergence to the saddle occurs only on its stable manifold. Escape is not finite-time blow-up, since $\dot{H}\le 2\alpha (H+1/2)$ bounds the energy on every finite interval.

When $\alpha =-0.1$, the eigenvalues at the nonzero points are $-0.05\pm i\sqrt{7.99}/2$, so they are stable foci. The energy decreases, keeping all forward trajectories in compact sublevel sets. A bounded limit set must lie in the invariant part of $x=0$, so trajectories converge to equilibria: generically to one of the two stable foci, exceptionally to the saddle along its stable manifold. The saddle's unstable separatrices spiral into the two stable foci; its stable manifold separates their attraction basins. Again, energy monotonicity excludes periodic orbits and homoclinic orbits. These three cases give the quartic double-well phase portrait with linear damping below.

Assume well-mixed populations, exponential prey growth without density regulation, predator mortality without food, and bilinear encounters. The Lotka-Volterra predator-prey model isIts coexistence equilibrium is $X_{\ast }=d/c$, $Y_{\ast }=a/b$. The Jacobian matrix there has eigenvalues $\pm i\sqrt{ad}$, so it is a center equilibrium rather than an attracting equilibrium. More globally,is a first integral: substituting the two equations into $\dot{H}$ gives zero. It has a strict minimum at coexistence, diverges at the boundary and at infinity, and its other positive-quadrant level sets are closed periodic orbits. Small oscillations have frequency $\omega =\sqrt{ad}$. In scaled variables $U=\sqrt{d}(X-X_{\ast })/X_{\ast }$ and $V=\sqrt{a}(Y-Y_{\ast })/Y_{\ast }$, the linearization is $\dot{U}=-\omega V$, $\dot{V}=\omega U$.

Model an annual fixed quota as a pulse removing $h>0$ prey, $X^{+}=X^{-}-h$, $Y^{+}=Y^{-}$, with $h<X^{-}$. Between pulses the original equations apply. In logarithmic coordinates the flow has zero divergence, but the pulse has determinant $X^{-}/(X^{-}-h)>1$. Therefore a periodic annually harvested trajectory has return map determinant greater than one, and cannot have both multipliers inside the unit circle. It is unstable, even when a leading rotation-plus-translation approximation looks neutral. Near a nonresonant small-quota forced orbit this expansion destabilizes the original center equilibrium; resonant pulse timing can add secular effects. Quotas larger than the remaining prey are not biologically realizable and require stopping the model. This is fixed-quota harvesting of Lotka-Volterra populations.

The common continuous approximation to an annual quota gives $\dot{X}=X(a-bY)-h$. Its positive equilibrium is $X_{\ast }=d/c$, $Y_h=(a-h/X_{\ast })/b$, requiring $h<a{X}_{\ast }$. Its Jacobian matrix has trace $h/X_{\ast }>0$ and determinant $bc{X}_{\ast }{Y}_h>0$, so this approximation also destabilizes coexistence. At small $h$ it is an unstable focus. A quota is a larger per-capita mortality when prey are scarce, which is the adverse feedback responsible for the result.

For abundance-triggered pulses, retain the same fixed prey removal and specify the crossing direction. The exact event-triggered harvesting by conserved energy calculation isIf prey abundance triggers the event at $X=X_c$, this increment is the same at every event. A sufficiently high prey threshold with a modest quota gives $\Delta H<0$: the orbit's oscillation energy decreases until it no longer crosses the threshold, after which harvesting stops. If $X_c\le X_{\ast }$, then $\Delta H>0$ and harvesting increases oscillations. At the balanced threshold satisfying $ch=d\log [X_c/(X_c-h)]$, energy is unchanged and the ideal orbit is neutral. Thus the useful rule is harvesting at high prey abundance, not simply using any fixed threshold; the equality case supplies no asymptotic restoration.

A predator threshold $Y=Y_c$ does not hold prey abundance fixed. On a rising-predator crossing, $X>X_{\ast }$, so a sufficiently small harvest initially lowers $H$; on a falling-predator crossing $X<X_{\ast }$, it raises $H$. Nevertheless the rising-predator rule is not a robust stabilizer in this conservative model. Its balanced return orbit has preharvest prey abundanceobtained by equating $cX-d\log X$ before and after the pulse. On the return section the derivative isIts magnitude exceeds one because $e^z-e^{-z}>2z$ for $z>0$. The balanced predator-triggered cycle is consequently unstable. At leading small-oscillation order its multiplier is $-1$, explaining a marginal alternation rather than stable damping. If the trigger is crossed in both directions, or a variable rather than fixed quota is used, a different return rule results and must be analyzed separately.

For conservation, the ideal model shows why fixed removals can be dangerous and why prey-based high-abundance feedback can be safer than a calendar quota or an unspecified predator threshold. These conclusions depend on the chosen pulse rule and on the absence of density dependence, refuges and stochasticity. They do not justify a universal numerical safe quota. A harvest policy needs a stopping threshold and the correct phase-dependent feedback, not merely a small nominal annual catch.

Differentiate the proposed first integral along the vector field:At $\alpha =0$, $K$ is conserved and the system is an inverted quartic oscillator phase portrait. Its equilibria are $O=(0,0)$ and $S_{\pm }=(0,\pm 1/\sqrt{2})$. The Jacobian matrix isAt the origin its eigenvalues are $\pm i$. Moreover $K$ has a strict local minimum there, and its nearby positive contours are closed, so the origin is a genuine center equilibrium, not merely an inconclusive linear centre. The arrows circulate clockwise: on the positive $x$-axis, $\dot{y}<0$. At either $S_{\pm }$, the eigenvalues are $\pm \sqrt{2}$, giving saddle equilibria. The stable eigendirection is $\delta x=\sqrt{2}\delta y$, and the unstable one is $\delta x=-\sqrt{2}\delta y$.

The special energy level factors asBetween the two saddles, its left arc has $x<0$ and travels upward from $S_{-}$ to $S_{+}$; its right arc has $x>0$ and travels downward from $S_{+}$ to $S_{-}$. These two heteroclinic orbits bound the central eye. Trajectories inside the eye are closed periodic orbits and do not tend to the centre as $t\to \infty$.

More explicitly, a level $K=C$ hasFor $0<C<1/4$, its bounded central component is periodic, while its exterior components turn at $|y|=\sqrt{(1+\sqrt{1-4C})/2}$ and then escape. For $C>1/4$, $x$ never vanishes and the two branches run from one vertical infinity to the other. Levels $C\le 0$ have only exterior nonconstant components, apart from the origin at $C=0$.

On the upper outer separatrix, the $x>0$ branch arrives from positive infinity and approaches $S_{+}$ in forward infinite time; the $x<0$ branch leaves $S_{+}$ towards positive infinity. On the lower outer separatrix, the $x<0$ branch arrives from negative infinity and approaches $S_{-}$, while the $x>0$ branch leaves towards negative infinity. Approaching a saddle along its stable manifold is exponential at rate $\sqrt{2}$. A trajectory close to but off a separatrix can spend a long time near the saddle before completing a periodic orbit or escaping.

The escape is finite-time blowup, not an asymptote at $t=\infty$. On any escaping level, $|x|=\sqrt{y^4-y^2+C}\sim y^2$, so the remaining travel time is bounded by a convergent integral proportional to ${\int }^{\infty }dy/y^2$. Thus an outward upper branch has $y\sim (T-t{)}^{-1}$, $x\sim -(T-t{)}^{-2}$; an outward lower branch has $y\sim -(T-t{)}^{-1}$, $x\sim (T-t{)}^{-2}$. The statement about approaching from infinity likewise refers to a finite past endpoint. Only equilibria, periodic trajectories and trajectories tending to saddles have a future defined for all time in this conservative portrait.

For small positive $\alpha$, the equilibria remain in place, but the origin's eigenvalues become $\alpha /4\pm i\sqrt{1-{\alpha }^2/16}$, so it becomes a repelling focus. The saddles retain one eigenvalue of each sign, $\alpha /4\pm \sqrt{2+{\alpha }^2/16}$. Since $\dot{K}\ge 0$, and its time integral is strictly positive on every nonconstant trajectory, trajectories cross $K$-contours towards larger values. Near the centre this means outward spiralling. No nonconstant periodic orbit can remain, and the former saddle-to-saddle connections at equal $K=1/4$ are broken. Stable saddle trajectories approach from $K<1/4$; unstable ones depart into $K>1/4$. Generic central trajectories eventually escape, while the exceptional stable saddle manifolds still approach their saddles.

For both parameters the autonomous planar system is $\dot{x}=v$, $\dot{v}=-x(x^2+\lambda x+1)$. It has the conserved energyIndeed $\dot{H}=v\dot{v}+V_{\lambda }^{\prime }(x)\dot{x}=0$. At any equilibrium point $(x_{\ast },0)$ the eigenvalues satisfy

For $\lambda =1$, the quadratic $x^2+x+1$ is strictly positive, so the only equilibrium point is $(0,0)$, with eigenvalues $\pm i$. Since $V_1^{\prime }(x)$ has the sign of $x$, $V_1$ has a unique strict minimum at zero and tends to infinity in both directions. All positive energy levels are closed curves surrounding this stable center equilibrium. They are clockwise: rightward when $v>0$ and leftward when $v<0$.

For $\lambda =5/2$, factor $x^2+(5/2)x+1=(x+2)(x+1/2)$. The three equilibrium points and their classifications areThe two center equilibria are stable but not asymptotically stable; the saddle equilibrium is unstable. At the saddle equilibrium the local unstable and stable directions are $v=(\sqrt{3}/2)(x+1/2)$ and $v=-(\sqrt{3}/2)(x+1/2)$.

The potential values are $V(-2)=-2/3$, $V(0)=0$, and $V(-1/2)=7/192$. Consequently there are closed loops around the left center for $-2/3<H<0$, closed loops in both wells for $0<H<7/192$, and a figure-eight pair of homoclinic orbits at $H=7/192$. At $H=0$ the right center is a stationary point while the left well still has a closed orbit. For $H>7/192$, each closed curve encircles both centers and the saddle.

The two homoclinic orbits satisfy $v=\pm \sqrt{2(7/192-V(x))}$. Their outer turning points can be located exactly fromEvery nonstationary energy curve is oriented to the right in the upper half-plane and to the left in the lower half-plane. The sketches show the closed curves, the separating orbits, the stable and unstable saddle directions and the direction of flow.

Introduce the velocity variable $v=\dot{x}$. The second-order differential equation becomes the autonomous planar systemA critical point of a dynamical system is a point where both components of the vector field vanish: here $v=0$ and $f(x,0)=0$. It is a stationary solution in the phase plane. The Jacobian matrix at each such point gives the local linear stability of a planar equilibrium. For the equations below, a conserved energy also determines the nonlinear phase portrait; this is important because purely imaginary eigenvalues alone do not generally prove that an equilibrium is a center equilibrium.

The potential factors as $V=x^2(3-2x)$, so its zeros are $0$ and $3/2$, with a double zero at the origin. Its derivatives are $V^{\prime }=6x(1-x)$ and $V^{\prime \prime }=6-12x$. Thus there is a local minimum $(0,0)$, a local maximum $(1,1)$ and an inflection at $(1/2,1/2)$. The potential tends to $+\infty$ on the far left and $-\infty$ on the far right. These features determine the left-hand sketch.

Put $v=\dot{x}$. Newton's second law gives the phase plane system $\dot{x}=v$, $\dot{v}=-V^{\prime }=6x(x-1)$. Its conserved mechanical energy and trajectories areThe linearization at $(0,0)$ has eigenvalues $\pm i\sqrt{6}$, and the surrounding energy curves are closed, so it is a center equilibrium. At $(1,0)$ the eigenvalues are $\pm \sqrt{6}$, giving a saddle equilibrium. The figure shows the cubic potential barrier phase portrait; arrows point right when $v>0$ and left when $v<0$.

For $0<E<1$, there are three turning-point roots of $V(x)=E$. The two leftmost enclose a closed orbit around the origin: the particle oscillates between them. There is also a separate open component to the right of the third root, where an incoming particle turns and escapes to $+\infty$. The closed orbits run clockwise in the $(x,v)$ plane.

For $E=1$, the factorization $1-V(x)=(x-1{)}^2(2x+1)$ gives the separatrixThe branch with $-1/2\le x<1$ is a homoclinic orbit: it leaves the saddle asymptotically, turns at $x=-1/2$ and returns asymptotically. On $x>1$, one branch approaches the saddle and the other departs towards infinity. Approaching the saddle takes infinite time. The saddle itself is also an equilibrium trajectory.

For $E>1$, the particle can cross the barrier. There is one left turning point below $-1/2$; an incoming particle passes through the well, turns on the left and escapes to the right. For $E<0$, only a right-hand open component exists, with its turning point beyond $3/2$. At $E=0$, the stationary centre coexists with a right-hand open trajectory turning at $3/2$. Thus negative energy does not imply trapping here: the cubic potential is unbounded below on the right.

With the stated time rescaling, $\dot{x}={\epsilon }^3{u}^{\prime }$ and $\ddot{x}={\epsilon }^4{u}^{\prime \prime }$, where primes refer to $\tilde{t}$. Dividing by ${\epsilon }^4$ givesAt $\epsilon =0$, put $v=u^{\prime }$ and define the HamiltonianIts orbital derivative is $H^{\prime }=v(u^2-\kappa )+(\kappa -u^2)v=0$, so this is a conserved quantity. The Hamiltonian system has $u^{\prime }=H_v=v$, $v^{\prime }=-H_u=u^2-\kappa$.

For $\kappa >0$, set $s=\sqrt{\kappa }$. The potential $U(u)=\kappa u-u^3/3$ has a minimum at $-s$ and a maximum at $s$. The corresponding center equilibrium and saddle equilibrium have energies $H_c=-2s^3/3$ and $H_s=2s^3/3$. Levels $H_c<H<H_s$ contain closed ovals around $(-s,0)$, as well as separate unbounded components on the right. At $H=H_s$, the center's ovals limit to a homoclinic orbit to the saddle, turning at $u=-2s$. Its equation isThe same energy also has external unbounded saddle separatrices for $u>s$. Levels above $H_s$ pass over the potential barrier and are open. Along the upper halves $u^{\prime }=v>0$, so the closed loops are traversed clockwise. If $\kappa =0$, the two critical points coalesce into a degenerate equilibrium; if $\kappa <0$, the potential is strictly decreasing and there is no center or homoclinic loop.

For $k=0$, the first integralhas $\dot{E}=0$. The equilibrium points are $(0,0)$ and $(1,0)$. The linearization has eigenvalues $\pm i$ at the origin and $\pm 1$ at $(1,0)$. The origin is a center equilibrium: the sufficiently small positive level sets of $E$ are closed curves, since its Hessian matrix there is positive-definite. The other equilibrium is a saddle equilibrium.

At the saddle, $E=1/6$, and the separatrix satisfiesIts two branches with $-1/2\le x\le 1$ form a homoclinic orbit through the saddle; its remaining branches extend to $x>1$. Inside the loop, the closed trajectories run clockwise: above the horizontal axis $\dot{x}=y>0$. The saddle directions have slopes $+1$ and $-1$. Outside the loop there are unbounded trajectories. The left panel shows the closed orbits and all saddle branches.

Validity and the complete angular equation. The Lagrange planetary equations here describe a planar, circular-planet, small-orbital eccentricity, weakly perturbed exterior mean-motion resonance. Require $M_{\mathrm{pl}}/M_{\star }\ll 1$, $e\ll 1$, one isolated slowly varying resonant argument, perturbation times long compared with an orbital period, and no close planetary encounters. Other resonant harmonics and short-period terms must be negligible; coefficients may be frozen only across a narrow range of semi-major axis. Take $p>0$, $q=1$ or $2$ in a reduced integer ratio. Exactly $e=0$ makes the longitude of periapsis undefined, so the displayed angular variables must then be replaced by nonsingular eccentricity vector components.

Neglecting the perturbation to the accumulated mean longitude gives $\dot{\lambda }\simeq n(a)$; this is an accumulated-phase approximation, not differentiation of a fictitious expression $n(t)t$ while discarding $t\dot{n}$. With $n^2{a}^3=G{M}_{\star }$,Differentiating, using $\dot{n}=-3n\dot{a}/(2a)$ and the supplied orbital element equations, gives the full result within this constant-coefficient model:An equivalent formula replaces the last two terms in brackets by $q^2{C}_r{e}^{q-2}[(p+q)n-p{n}_{\mathrm{pl}}-2q{C}_s]-2q^3{C}_r^2{e}^{2q-4}\cos \varphi$. No derivatives of $C_r,C_s$ are included because they were specified as constants.

A full fixed point requires $\sin \varphi =0$ and $\dot{\varphi }=0$. Hence ${\varphi }_{\ast }=0$ or $\pi$, with its corresponding detuned semi-major axis fixed byAt either point $\dot{a}=\dot{e}=0$. The longitude of periapsis still precesses, so the fixed point refers to the reduced resonant dynamics.

Linear stability analysis and encounter geometry. Linearizing the full equation about a fixed point givesFor $q=2$, $\pi$ is a center equilibrium and zero is a saddle equilibrium. For $q=1$, $\pi$ is always a center equilibrium, while zero is a saddle equilibrium ifAt sufficiently small orbital eccentricity the full truncated equations instead admit a center at zero as well; equality is a degenerate case requiring higher-order analysis. Thus an unconditional instability claim at zero does not follow from the full equation. In the usual fixed-eccentricity weak-resonance regime the inequality holds and the stable libration center is $\pi$ for either order.

The physical explanation is resonance protection. At equal mean longitudes, the conjunction direction relative to periapsis obeys $q(\Lambda -\varpi )=\varphi$ modulo $2\pi$. For $q=1$, $\varphi =0$ places conjunction near periapsis, where the exterior particle comes closest to the planet; $\varphi =\pi$ places it near apoapsis. For $q=2$, zero includes both apsidal conjunctions, one of them near periapsis, whereas $\pi$ places the two conjunction branches near quadrature. The protected arrangements give restoring kicks in the positive-$C_r$ leading-harmonic model. This geometric explanation is conditional on its non-crossing, small-orbital eccentricity approximation; it does not override the low-$e$ term retained in the full stability calculation.

Small resonant-argument librations. Define the nominal resonant semi-major axis and the displaced center byThis follows by expanding $n=n_r[1-3(a-a_r)/(2a_r)]$ and taking $\cos (\pi +\delta )=-1+O({\delta }^2)$. It gives the requested initial semi-major axis to first order in the precession-induced detuning; $e_0=e(0)$ and $\delta =\Delta \varphi$.

For the pendulum approximation of a mean-motion resonance, also require fractional changes in $e$ small across a libration, and precession terms small compared with $n$. In particular $C_r/(n_r{e}_0^3)\ll 1$ for $q=1$ makes the extra full-equation curvature negligible; for $q=2$, small fractional eccentricity changes require $C_r/(n_r{e}_0^2)\ll 1$. Freeze $e=e_0$ in the leading restoring coefficient and setThen $u=\varphi -\pi$ obeys simple harmonic motion. Using $u(0)=\delta$, $\dot{u}(0)=0$ and integrating the remaining Lagrange planetary equations givesThe omitted apsidal modulation comes from higher perturbative orders and $u^2$. All expressions are leading resonant approximations, valid while the omitted fractional changes remain small. The small-libration phase portrait iswith period $2\pi /{\omega }_0$. The semi-major axis half-width is $2a_r{\omega }_0|\delta |/(3p{n}_{\mathrm{pl}})$ and the orbital eccentricity half-width is $q{C}_r{e}_0^{q-1}|\delta |/{\omega }_0$.

Dividing the original $\dot{e}$ and $\dot{a}$ equations gives the exact invariant of their retained terms,Thus the $e$–$a$ plot is a segment of this increasing curve traversed back and forth, not a closed ellipse. Locally its slope is $q/[2(p+q)e_0{a}_c]$.

Finite amplitude and the separatrix. Keep the same weak-resonance approximation but allow $\delta$ to be finite. The leading angular equation is $\ddot{\varphi }={\omega }_0^2\sin \varphi$, or $\ddot{u}=-{\omega }_0^2\sin u$. Multiplication by $\dot{u}$ givesThis resonant pendulum energy is constant in the pendulum approximation of a mean-motion resonance. It is not an exact integral of the earlier full equation when $e$ and its restoring coefficient vary: the $\dot{\varphi }\sin \varphi$ term alone prevents that conclusion in general.

For $|\delta |<\pi$ the phase portrait has closed libration curves around $\pi$, withwhere $K$ is the complete elliptic integral of the first kind. The largest speed is $2{\omega }_0\sin (|\delta |/2)$ and the leading semi-major axis half-width is $4a_r{\omega }_0\sin (|\delta |/2)/(3p{n}_{\mathrm{pl}})$. As $|\delta |\to \pi$, the energy approaches $2{\omega }_0^2$, the separatrix through the saddles at zero and $2\pi$; the particle spends increasingly long intervals near those saddles. With $ϵ=\pi -|\delta |$,At exactly $\delta =\pi$, $\dot{\varphi }=0$, the particle stays at the unstable fixed point in the ideal model. Nonstationary separatrix trajectories approach the saddle only in infinite time; above separatrix energy, the resonant argument circulates and resonance protection is lost. These distinctions matter when describing the limiting motion.

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 equilibrium equations give $y=0$ and $x-x^3=0$, so the fixed points areThe Jacobian isAt the origin its determinant is $-1$, so the origin is a saddle equilibrium. At either of the other points the determinant is two and the trace is $\epsilon (1-\alpha )$. For small positive $\epsilon$, they are unstable foci when $\alpha <1$ and stable foci when $\alpha >1$; at $\alpha =1$ the linearization is nonhyperbolic. When $\epsilon =0$, each is a center equilibrium.

For $\epsilon =0$, the system is Hamiltonian withbecause $\dot{x}=H_y$ and $\dot{y}=-H_x$. Its conservative planar phase portrait has two wells centered at $(\pm 1,0)$, a saddle at the origin, periodic level curves around each center for $-1/4<H<0$, two homoclinic loops on $H=0$, and outer periodic curves surrounding both wells for $H>0$.

For positive small $\epsilon$,On the unperturbed level $H=H_0$,The upper and lower halves contribute equally because $dt=dx/y$. Hence the energy balance for the weakly perturbed double-well oscillator isso one may take

For the right homoclinic loop, $H_0=0$, $x_1=0$, and $x_2=\sqrt{2}$. Its persistence requiresThe substitution $u=1-x^2/2$ givesThus the homoclinic balance for the weakly perturbed double-well oscillator occurs atThe left loop gives the same condition by symmetry.

For an orbit surrounding one center, the corresponding balance ratio tends to $\alpha =1$ as the orbit shrinks to the center and to $\alpha =5/4$ as it approaches the homoclinic loop. The single-well periodic orbit range for the weakly perturbed double-well oscillator is therefore