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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.

Local equilibrium and Hopf bifurcations do not locate the larger cycles' endpoints. In the large-$s$ case, the stable origin cycle is confined between two symmetry-related saddles. A natural global completion is a heteroclinic cycle formed by the two separatrix connections between those saddles. The stable periodic orbit expands into that connecting cycle and disappears; the two connections occur together because $(q,p)\mapsto (-q,-p)$ exchanges them. This is a global transition in the positive-$\kappa$, positive-$\lambda$ region where the Bendixson-Dulac criterion permits cycles. Beyond it, the cubic inverted potential allows trajectories to leave the local region; the local reduction does not specify their ultimate fate in the original four-dimensional system.

In the small-$s$ case, the two stable cycles born around the nonzero well equilibria can encounter the central saddle simultaneously. Their limiting orbits form a symmetric figure-eight pair of homoclinic orbits. A symmetric homoclinic gluing bifurcation can then replace the two one-well cycles by one larger cycle visiting both wells. There is also an outer unstable cycle associated with the subcritical origin Hopf branch, continued into the double-well region. A saddle-node bifurcation of periodic orbits between the large stable and unstable cycles provides a possible later termination. Thus the global curves needed for the usual completion are a symmetric heteroclinic connection in the large-$s$ case, and paired homoclinic gluing together with a fold of cycles in the small-$s$ case.

These connections, and any resulting basin changes, are not detected by the equilibrium Jacobians. Their precise parameter curves require a global return-map or separatrix calculation, which is not part of this requested qualitative description. Symmetry forces paired connections where indicated, but the local centre manifold and Hopf calculations alone should not be presented as numerical locations or a proof of the full large-amplitude diagram.

Introduce the vector $x=(t,y,y^{\prime },\dots ,y^{(k-1)})\in {\mathbb{R}}^{k+1}$. Its autonomous ordinary differential equation iswhere the coordinate indices start at zero. Appending the clock removes the explicit time dependence.

For a vector field with local existence and uniqueness, its flow ${\varphi }_t(x)$ is the solution at elapsed time $t$ starting at $x$. The uniqueness theorem for ordinary differential equations gives ${\varphi }_s({\varphi }_t(x))={\varphi }_{s+t}(x)$ whenever both sides are defined. The orbit is $O(x)=\{{\varphi }_t(x):t\text{ lies in its maximal interval}\}$. For a forward-complete solution its omega-limit set isequivalently the set of limits along sequences $t_n\to \infty$. A homoclinic orbit is a nonconstant complete orbit approaching the same equilibrium as $t\to -\infty$ and $t\to \infty$. This last definition concerns a general autonomous system; the appended clock itself cannot approach an equilibrium.

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 small $\epsilon$, the conserved quantity from part (b) instead obeys $H^{\prime }=-\epsilon (\lambda +u)v^2$. The Melnikov energy-balance method evaluates the first-order net change along the unperturbed homoclinic orbit:Using the homoclinic curve $v^2=\frac{2}{3}(s-u{)}^2(u+2s)$, these are integrals of $vdu$ on its two halves. Their weighted mean is $I_1/I_0=-5s/7$. Thus the first-order splitting vanishes at $\lambda =5s/7$. Since its derivative with respect to $\lambda$ is $-I_0\ne 0$, the implicit function theorem gives a nearby curve where the stable and unstable saddle separatrices reconnect. This is the homoclinic balance for a quadratic-force oscillator and therefore a global, rather than local, bifurcation.

Returning to the original parameters, $q={\epsilon }^2\lambda$ and $\sqrt{p}={\epsilon }^2\sqrt{\kappa }$. With fixed positive $\kappa$ and $\epsilon \to 0$, the homoclinic bifurcation curve has leading formThe energy integral locates the leading curve; the perturbed orbit itself is no longer a constant-$H$ curve.

Assume a generic saddle-focus homoclinic orbit at $\mu =0$, a smooth nondegenerate global reinjection along the selected unstable branch, and a sufficiently small neighborhood in which the linearized local flow supplies the leading passage map. Use linearizing coordinates if available; otherwise the calculation is a leading local approximation, not an exact consequence of eigenvalues alone. Choose the incoming section near $(\rho ,0,0)$ in the plane $y=0$, with coordinates $(\xi ,z)$ where $r=\rho +\xi$ and $\rho >0$. This section is transverse since $\dot{y}=\omega r\ne 0$. Choose an outgoing section $z=z_0>0$. We mark one return after a global excursion; intermediate crossings of $y=0$ during the spiral are not counted as separate global returns.

In cylindrical coordinates the local flow isFor $0<z\ll z_0$, the time to the outgoing section is $t_{\ell }={\lambda }_{+}^{-1}\log (z_0/z)$. Set $\delta =-{\lambda }_{-}/{\lambda }_{+}>0$ and $\Omega =\omega /{\lambda }_{+}$. The outgoing coordinates areThe global part of the Poincaré return map is smooth around $(X,Y)=(0,0)$ and takes the unstable-manifold point to $({\xi }^{\prime },z^{\prime })=(0,-\mu )$. Expand it aswhere the coefficients may depend smoothly on $\mu$ and the matrix is generically nonsingular. Absorbing the fixed $z_0^{-\delta }$ into them gives the leading two-dimensional return mapThe remainder refers to the smooth global-map Taylor approximation with bounded $\xi$; corrections to the local linear approximation are also neglected. The global map supplies the additive splitting parameter and the local map supplies the power and logarithmic rotation. For an iterated excursion on this same unstable branch, the next coordinate must again satisfy $z^{\prime }>0$. A nonpositive image leaves this local return domain; the formula must not be iterated through negative logarithms.

For general $k$, the same energy gives $\dot{E}=-k{y}^2$. For $k=0.1$ the eigenvalues at the origin are $-0.05\pm i\sqrt{3.99}/2$, so it is a stable focus. At $(1,0)$ they are $(-0.1\pm \sqrt{4.01})/2$, so it remains a saddle equilibrium. The vector field has constant divergence $-0.1$: there are no periodic orbits, by the planar divergence criterion. Nor can the former homoclinic orbit survive, since $E$ strictly decreases along every nonconstant orbit except at isolated crossings of $y=0$.

The stable manifold of the saddle separates escape to positive $x$ from convergence to the origin. Its two branches approach the saddle in forward time with tangent slope $(-0.1-\sqrt{4.01})/2$. The left unstable branch spirals into the origin, whereas the right unstable branch escapes. The shaded component on the origin's side of the stable manifold in the right panel is the basin of attraction. The boundary itself is excluded: its trajectories converge to the saddle. This boundary is the damped stable manifold, rather than the conservative energy contour. energy decrease prevents bounded recurrent motion, and every bounded orbit off the saddle's stable manifold consequently converges to the origin. The plotted manifold was obtained by integrating backwards from the saddle's stable eigendirection.

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.

The circle $\gamma :r=1$ is a periodic orbit, since $\dot{r}=0$ there and $\dot{\theta }=1$. Its period is $2\pi$. Linearizing in $(\delta r,\delta \theta )$ along it givesThe monodromy matrix is triangular with diagonal entriesThe unit Floquet multiplier comes from translating an autonomous orbit in time: its tangent vector is a periodic solution of the variational equation, independently of $g$. A sufficient condition for asymptotic stability is ${\int }_0^{2\pi }g<2\pi$, making the normal multiplier less than one. The reverse strict inequality gives instability; equality requires nonlinear analysis.

For $g=2\sin \theta$ the normal multiplier is $e^{-2\pi }$, so the circle is asymptotically stable. That does not mean it attracts every initial point. Since $r>0$ makes $\theta$ strictly increasing, divide the radial equation by $\dot{\theta }$:For $r>1$, put $v=r-1>0$. Then $v^{\prime }=(2\sin \theta -1)v-v^2\le (2\sin \theta -1)v$, whose integrated linear coefficient decreases by $2\pi$ per revolution. Thus every exterior orbit approaches $r=1$.

For $0<r<1$ put $w=(1-r{)}^{-1}>1$. It satisfies the linear equationLet $w_c$ be its solution with $w_c(\pi )=1$. It defines the separatrix $r_c=1-1/w_c$ wherever $w_c>1$. At $\theta =\pi$, $r_c=0$, $r_c^{\prime }=0$, and differentiation gives $r_c^{\prime \prime }=2$, so $r_c\sim (\theta -\pi {)}^2$. The approach to this endpoint takes infinite physical time because $dt=d\theta /r_c$.

To locate its other endpoint, use the integrating factor $e^{-A(\theta )}$ with $A(\theta )=\theta +2\cos \theta$. It givesIn particular $w_c(0)>e^{4-\pi }>1$, whereasOn $(-\pi ,0)$, crossing $w=1$ has $w^{\prime }=-2\sin \theta >0$, so there is one crossing ${\theta }_b\in (-\pi ,0)$. There cannot be another crossing before $\pi$: on $(0,\pi )$ crossings have negative derivative, while the local quadratic form at $\pi$ is positive on its left. Thus $r_c>0$ for ${\theta }_b<\theta <\pi$, and its closure is a homoclinic orbit to the origin. At ${\theta }_b$, $r_c^{\prime }=-2\sin {\theta }_b>0$, so departure from the origin likewise takes infinite time.

Scalar-solution comparison now gives the basin picture. Initial points inside that loop have $w<w_c$ at the same angle and approach $r=0$ before reaching angle $\pi$. Points on its boundary return to the origin along the homoclinic trajectory. Points outside the loop reach angle $\pi$ with $w>1$ or lie outside the unit circle; they approach $r=1$. To check the inner exterior region quantitatively, the return map at $\theta =\pi$ is $w\mapsto e^{2\pi }(w-I)$, where $I={\int }_0^{2\pi }{e}^{-u-2+2\cos u}du<1-e^{-2\pi }$. It sends every $w>1$ increasingly to infinity, equivalently $r\to 1$.

There are no other nonconstant periodic orbits: an orbit reaching the section with $w>1$ has strictly increasing successive $w$ values, and the remaining inner trajectories tend to the origin. The origin, defined by the continuous Cartesian extension, attracts the interior of the loop but is not Lyapunov stable, since arbitrarily small lower-sector initial points outside the loop approach the unit circle. This is the homoclinic basin boundary in a sinusoidal radial flow.

For $\lambda >0$, the equilibrium points are $x=\pm \sqrt{\lambda }$, $y=0$. The positive branch is a saddle; the negative branch has a subcritical Hopf bifurcation at $\mu =\sqrt{\lambda }$. A homoclinic orbit occurs at $\mu =(5/7)\sqrt{\lambda }+O(\lambda )$ near the double-zero point. Between this homoclinic curve and the Hopf curve, an unstable periodic orbit surrounds a stable equilibrium point. The factor $5/7$ follows from a Melnikov energy-balance method, using the integrals of $v^2$ and $u{v}^2$ along the leading conservative saddle loop.

// Destination: dynamical-systems.bigb

For $\dot{u}=u(\mu +u-v)$, $\dot{v}=\lambda u^2$ in $u>0$, the time change $d\tau /dt=u$ gives the affine linear system $u_{\tau }=\mu +u-v$, $v_{\tau }=\lambda u$. Shifting $w=v-\mu$ removes the forcing. Its eigenvalues are $(1\pm \sqrt{1-4\lambda })/2$. For $\lambda >1/4$, putting $\omega =\sqrt{4\lambda -1}/2$ givesFor fixed $\mu >0$ and initial $h,v_0$ tending to zero, the first return to small positive $u$ occurs after approximately $\pi /\omega =2\pi /\sqrt{4\lambda -1}$. This is a half-turn, not a full turn. At $\lambda =1/4$, $u=e^{\tau /2}[h+(\mu +h/2-v_0)\tau ]$ does not return and grows without bound. The loss of return diagnoses escape to infinity rather than a finite-amplitude homoclinic orbit. Physical time obeys $dt=d\tau /u$ and need not share the divergent desingularized travel time.

For $\dot{x}=\alpha x-y+y^3$, $\dot{y}=-x$, the energy $H=x^2-y^2+y^4/2$ satisfies $\dot{H}=2\alpha x^2$. At $\alpha =0$, its compact level sets give two families of closed periodic orbits around $(0,\pm 1)$ and an outer family around both; $H=0$ is a figure-eight homoclinic orbit pair through the saddle point at the origin. The two minima of $H$ are center equilibria for zero damping, stable foci for small negative $\alpha$, and unstable foci for small positive $\alpha$. For $\alpha \ne 0$, strict monotonicity along nonconstant solutions excludes periodic orbits and homoclinic orbits.

For a hyperbolic equilibrium point with one unstable eigenvalue ${\lambda }_{+}>0$ and weakest stable decay rate ${\lambda }_{-}<0$, the saddle index is $\delta =-{\lambda }_{-}/{\lambda }_{+}$. For a stable complex pair, use its real part. An incoming distance $x$ from the stable manifold takes time $T={\lambda }_{+}^{-1}\log (h/|x|)$ to reach an outgoing section at unstable coordinate magnitude $h$. A stable coordinate is multiplied by $e^{{\lambda }_{-}T}=(|x|/h{)}^{\delta }$, giving a power-law local passage. A smooth nondegenerate reinjection consequently has derivative tending to zero when $\delta >1$ and an unbounded leading derivative when $\delta <1$. This determines local contraction or expansion near a homoclinic orbit; existence of a particular attractor also depends on the global return geometry.