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For a planar magnetic field $\mathbf{B}=({\psi }_y,-{\psi }_x,0)$, ideal induction advects the Cartesian magnetic flux function: $D_t\psi =0$. Incompressible flow also preserves the area inside each material contour. During relaxation, the arms of a magnetic separatrix can approach each other while their field-line connectivity remains fixed. A transition of width $ϵ$ with a finite change of tangential field has electric current density of order ${ϵ}^{-1}$ and can approach a current sheet. This is a possible limiting mechanism, not a claim that every planar equilibrium is singular.

An experiment of size $L$ stops the renormalization-group flow at a finite ${\ell }_{\max }$. In the leading scaled model take the original initial conditions $Y_0=0.06$ and $X_0=-0.10,-0.06,-0.03$. The invariant is respectively $0.0064,0,-0.0027$. The first approaches $X=-0.08$, hence $K=(2.08)/\pi$; the second follows the critical separatrix; the third eventually leaves the small-vortex fugacity regime. These are chosen model trajectories, with no fitted experimental parameters.

On the critical separatrix, $X=-Y$ implies $Y^{\prime }=-Y^2$, henceEven a critical finite sample therefore has a phase stiffness above its asymptotic jump value, with a slow logarithmic correction in $L$. Slightly above the transition, a sample with $L\ll \xi$ can resemble the algebraic phase before the flow reaches unbound vortices. A trap additionally varies the local density and phase stiffness; neither its whole-cloud coherence nor a finite observation window automatically measures the homogeneous thermodynamic limit.

A global bifurcation changes an invariant set through a global orbit connection or collision that cannot be identified by a single equilibrium point's linearization alone. For example, a global bifurcation through a saddle homoclinic orbit occurs when an outgoing unstable separatrix returns to the same saddle equilibrium's stable manifold. A smooth global Poincaré return map then combines with a local passage map to determine which side contains a periodic orbit. A heteroclinic orbit instead joins distinct equilibrium points; an energy-balance or Melnikov integral can locate its breaking in a nearly Hamiltonian system.

A magnetic island in a planar magnetic field is a region of closed magnetic field lines surrounding an extremum of the Cartesian magnetic flux function. A surrounding separatrix can pass through a saddle of that function and separate the island from other flux regions. In ideal incompressible flow, the flux-function value carried by each fluid particle and the area enclosed by each material contour are conserved.

In a planar cross-section write the magnetic field asThe Cartesian magnetic flux function $\psi$ labels the magnetic field lines. Ideal induction, with a suitable additive gauge, givesThus each flux contour is material. Incompressibility also preserves the area inside any material contour. Neither a magnetic island's flux distribution nor the connections of a separatrix can be freely altered during relaxation.

A configuration with two or more magnetic islands separated by an X-type saddle illustrates the mechanism. As the magnetic islands reshape to lower magnetic energy, the arms of the separatrix can be pressed together. Magnetic reconnection would change the topology, so it is excluded in a perfect conductor. The X-type structure can instead flatten into an extended interface, with oppositely directed tangential fields on its two sides. The transition shrinks and the electric current density concentrates into a current sheet. Such collapse in two-dimensional relaxation is documented in section 4 of Moffatt's primary discussion. It is a possible asymptotic outcome, not a finite-time singularity assertion or a property of every planar field.

The local equations show precisely how the interface can support equilibrium. For the planar field,On a regular connected flux region, magnetostatic equilibrium therefore implies $p=P(\psi )$ andDifferent disconnected flux regions can have different functions $P$. Their limiting fields need not join with continuous first derivatives of $\psi$ across a common separatrix; continuity of total pressure allows the tangential-field jump.

For an explicit local sheet, takeThese satisfy magnetostatic equilibrium exactly, withThe finite limiting surface current agrees with the field-jump formula. This is a local equilibrium demonstration of the sheet balance, not a claim that this particular family is the full ideal relaxation trajectory from the given initial data.

Strictly planar fields of the form above have zero magnetic helicity: choose $\mathbf{A}=\psi {\mathbf{e}}_z$, giving $\mathbf{A}\cdot \mathbf{B}=0$. Thus this strictly planar example interprets the last part as a separate illustration of topology-constrained relaxation. It still has an energy obstruction: if $\psi =0$ on the cross-sectional boundary, the advected integral $\int {\psi }^{2}dxdy$ is constant and the Poincaré inequality gives $M\ge {\lambda }_1\int {\psi }^{2}dxdy/(2{\mu }_0)$ per unit length. If the earlier nonzero-helicity condition is retained instead, a possible extension is a field depending on only two coordinates but permitting an axial component, with periodic boundary conditions in the invariant direction. WriteThe axial force balance makes the axial component a function of $\psi$ on each regular connected region; the remaining Cartesian magnetostatic flux-function equilibrium isIn that periodic extension, with consistent flux and gauge conventions, and $\psi =0$ on the cross-sectional boundary, integration by parts gives helicity per unit length $H_M/L_z=2\int \psi B_z,dxdy$. It can be nonzero. The same separatrix-collapse mechanism still concentrates current in a sheet. Three-dimensional linkage is therefore unnecessary for sheet formation; planar frozen flux-contour topology already supplies the constraint.

Use pressure per unit constant density, $\pi$, with the gravitational potential and centrifugal potential absorbed into it. Strict geostrophic balance is the exact balancewith material acceleration omitted, for a homogeneous incompressible flow. Writing $f=2\Omega \ne 0$, its components are $fv={\pi }_x$, $fu=-{\pi }_y$ and ${\pi }_z=0$. Differentiating the first two equations with respect to $z$ gives $u_z=v_z=0$. Alternatively, the vector calculus identityshows, on taking the curl of the balance, that $(\mathbf{\Omega }\cdot \nabla )\mathbf{u}=0$. Thus $w_z=0$ too, proving the Taylor–Proudman theorem:The physical pressure still includes its background hydrostatic pressure; it is the reduced pressure $\pi$ that is independent of $z$. Between horizontal impermeable boundaries, the vertically constant $w$ is zero.

For boundaries $z=z_b(x,y)$ and $z=z_t(x,y)$, impermeability condition gives $w_b={\mathbf{u}}_H\cdot {\nabla }_H{z}_b$ and $w_t={\mathbf{u}}_H\cdot {\nabla }_H{z}_t$. A nearly columnar flow crossing contours of $H=z_t-z_b$ must therefore have $w_t-w_b\simeq {\mathbf{u}}_H\cdot {\nabla }_{H}H\ne 0$. Exact columnarity must give way to small corrections from ageostrophic flow. The leading horizontal geostrophic flow can remain nearly independent of $z$, and the leading reduced pressure has ${\pi }_z=0$, but $w_z$ is now nonzero and exact three-dimensional strict geostrophic balance no longer holds. For small slopes, the dominant vertical vorticity equation iswhere $q=v_x-u_y$ is relative vorticity. Equivalently, $D[(q+f)/H]/Dt\simeq 0$. Crossing geostrophic contours stretches or squashes columns: increasing $H$ produces cyclonic relative vorticity, decreasing $H$ produces anticyclonic relative vorticity. The full vorticity equation also contains tilting terms, neglected at this columnar order.

For the circular shallowing, fluid arriving from upstream has potential vorticity $f/h$. Its height is $h(1-ϵ)$ inside, so conservation givesThis uses the stated column-stretching description and initially open streamlines. To obtain the leading disturbance velocity field, use the conventional streamfunction definition $u=-{\Psi }_y$, $v={\Psi }_x$, and write $\Psi =-Uy+\psi (r)$. Then $q={\nabla }_H^2\psi =r^{-1}(r{\psi }_r{)}_r$. Regularity at $r=0$ gives ${\psi }_r=-fϵr/2$ inside. Outside, ${\psi }_r=B/r$; matching tangential velocity at $r=a$ fixes $B=-fϵa^2/2$. Choose the irrelevant additive constant so thatBoth $\psi$ and ${\psi }_r$ match at $a$, so there is no extra vorticity sheet on the rim. This is the circular topographic anticyclone. The radial expression specifies the disturbance; the complete streamfunction is $\Psi =-Uy+\psi$, since a radial function alone cannot represent the required uniform far-field current. Reversing the streamfunction sign convention reverses the displayed proportionality constant, but leaves the physical velocity unchanged.

Differentiation gives the full relative velocity field in the rotating frame:away from the narrow transition across the change of depth. The disturbance turns clockwise for $fϵ>0$ and decays as $1/r$. The horizontal field is the leading small-$ϵ$ columnar inversion: the exact variable-depth transport across a sharp rim needs higher-order corrections, rather than an exactly two-dimensional velocity continuous across a literal vertical step.

Take $U>0$ and $ϵ>0$. The greatest opposing zonal disturbance speed is $Ca$, attained at $(0,-a)$. If $Ca<U$, then $u>0$ everywhere and every particle advances in $x$, precluding closed streamlines. If $Ca>U$, the interior stagnation point $(0,-U/C)$ is a center, since the interior contours are circles about that point. The exterior stagnation point $(0,-C{a}^2/U)$ is a saddle: linearization there has real eigenvalues of opposite sign. Closed contours around the center are bounded by a separatrix through the saddle. The two points meet at the rim at onset, giving the closed-streamline threshold for a circular topographic anticyclone:At this threshold $U/(fa)=ϵ/2\ll 1$, consistent with small Rossby number. Equality is the onset, not a finite-area closed cell. Beyond onset, upstream data alone no longer determine the potential vorticity of trapped fluid; the threshold follows from the continuation of the supplied open-flow model.

The relevant saddle equilibrium is the origin, with eigenvaluesFor $\lambda <0$, they have opposite signs. Its saddle index isnear the homoclinic curve, because that curve has $\kappa >0$. Along its leading asymptote, $\delta =1-\frac{4}{5}\sqrt{-\lambda }+o(\sqrt{-\lambda })$.

A local passage near the saddle equilibrium contributes a power $s^{\delta }$ to the return map. Because $\delta <1$, its derivative becomes large as the orbit approaches the separatrix. The periodic orbits meeting the loops are therefore unstable, and their periods diverge logarithmically. On the side between the homoclinic and subcritical Hopf curves there are two unstable single-well periodic orbits, surrounding the two sink equilibria. On crossing the double-loop curve downward, they join into one unstable outer periodic orbit surrounding both wells. The attracting outer periodic orbit is not the one that collides with the saddle equilibrium: it persists across this global event.

The unstable outer periodic orbit must subsequently meet that attracting outer periodic orbit in a fold of periodic orbits. This extra curve is required to obtain the six-region diagram. One can see its position from the same energy-balance calculation. After normalizing $a_0=1$, an outer Hamiltonian contour with $H>0$ selectsAt $H=0$, $B=4/5$. The derivative of the denominator diverges positively there, while that of the numerator stays finite, so $B$ initially decreases. Rescaling at large $H$ gives $B(H)\propto \sqrt{H}$, so it eventually increases. At its minimum the stable and unstable selected periodic orbits meet. Thus the outer-cycle fold in a weakly perturbed double-well oscillator has ${\kappa }_{\mathrm{fold}}=c_f(-\lambda )+o(|\lambda |)$ with $0<c_f<4/5$; quadrature gives $c_f\simeq 0.75226$. The decreasing portion of $B(H)$ selects the unstable periodic orbit, and the increasing portion selects the stable one, since the energy drift has sign $\beta -B(H)$.

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.

After one iteration $y_n\in \{1,-1\}$, so work on that invariant sign sheet. For $x_n\ne 0$,since $z_n=-x_n{y}_n$ and $|y_n|=1$. This is the signed gluing-map reduction. If an arbitrary initial value of $y$ is admitted, the formula applies after that first iterate, not before it. The point $z=0$ represents a separatrix at which the original flow return is singular, even though the scalar map can be extended continuously there.

Let a scalar fixed point have magnitude $r>0$. If $z=-r<0$, then $x$ and $y$ have the same sign; the two lifts are the fixed points $(r,1)$ and $(-r,-1)$. If $z=r>0$, their signs are opposite, and the lift is the period-two orbit $(r,-1)\leftrightarrow (-r,1)$. The scalar fixed-point multiplier isThe two-return fixed-point multiplier of the period-two lift is $m^2$, so in either case dynamical stability requires $m<1$. The negative fixed point represents one of two mirror-related single-lobe periodic orbits in the flow, whereas the positive fixed point represents a double-lobe periodic orbit.

For $\delta >1$, the nearby fixed point is $z=\mu +o(\mu )$ and its fixed-point multiplier tends to zero. Thus two stable single-lobe limit cycles for $\mu <0$ glue into one stable double-lobe limit cycle for $\mu >0$. Their periods diverge as the separatrix is approached; $\mu =0,z=0$ is not a finite-period limit cycle.

For $0<\delta <1$, the small branch instead satisfies $z\sim -\mathrm{sgn}(\mu )(|\mu |/A{)}^{1/\delta }$ and its fixed-point multiplier tends to infinity. Hence one unstable double-lobe limit cycle on $\mu <0$ splits into two unstable single-lobe limit cycles on $\mu >0$. Additional finite-size stable branches depend on the surrounding map; near the resonance $\delta =1$ with $A<1$ they are resolved by the saddle-node bifurcations in part (c).

Write $v=\dot{x}$. The simple pendulum becomes $\dot{x}=v$, $\dot{v}=-\sin x$. Its equilibria are $(k\pi ,0)$. The linearization is $\left(\begin{array}{cc}0& 1\\ -\cos (k\pi )& 0\end{array}\right)$. For even $k$ its eigenvalues are $\pm i$, and for odd $k$ they are $\pm 1$. The conserved energyhas a strict local minimum at each even multiple of $\pi$, proving nonlinear Lyapunov stability there. The odd multiples are saddles and are unstable. ThereforeThe stable centres are not asymptotically stable because energy is conserved.

The saddle energy is $H=2$, so the separatrix is determined byEnergy levels $0<H<2$ form closed libration curves around the centres; levels $H>2$ have velocity of constant sign and describe rotations. The separatrices join neighboring saddles in the unwrapped $(x,v)$ plane. On the angle cylinder, where $x$ is identified modulo $2\pi$, they are homoclinic to the single saddle.

At $x=0$ the positive separatrix has $\dot{x}=2$. Before reaching the neighboring saddle at $x=\pi$, the velocity is $2\cos (x/2)>0$, so the time to angle $a<\pi$ isAs $a\uparrow \pi$, $\cos (x/2)\sim (\pi -x)/2$. The integrand is therefore asymptotic to $1/(\pi -x)$, whose improper integral diverges logarithmically. Thus the pendulum separatrix approach takes infinite time:Starting with velocity $-2$ gives the reflected approach to $x=-\pi$ and the same infinite time. The velocity tends to zero but never reaches zero at a finite time.

The rescaling resolves the small, nearby nonzero saddles while retaining the weak non-Hamiltonian terms. It allows the global stable manifold/unstable manifold connection to be studied as a perturbation of the explicit heteroclinic cycle, rather than searched for only in the original variables.

Differentiate the first integral using the first-order perturbation from part (c):On the positive unperturbed connecting branch, $u=(2s-v^2)/(2\sqrt{2})$, soIntegrating the energy change from one saddle equilibrium to the other therefore givesThis is the heteroclinic Melnikov integral for a quartic Hamiltonian. It measures the first-order energy mismatch, equivalently the transverse splitting between the outgoing unstable manifold and the incoming stable manifold. A nonzero value means that the heteroclinic orbit has broken. A simple zero supplies the leading connection condition; smooth perturbation then shifts the curve by higher-order terms. The central inversion symmetry makes the two connections have the same condition, so both can be restored on one parameter curve.

For fixed $s>0$, $\partial M/\partial \mu ={\int }_{-\sqrt{2s}}^{\sqrt{2s}}(s+v^2/2)dv>0$. The coefficient is affine in $\mu$, has a unique simple zero, and changes sign across it. No explicit evaluation is necessary. Energy drift on the nearby closed Hamiltonian levels can similarly determine a periodic orbit and its stability; the cycle born at the supercritical Hopf bifurcation grows towards the separatrix and can terminate at this global bifurcation through a heteroclinic orbit. Translating the resulting balance curve back with ${\mu }_{\mathrm{old}}={\epsilon }^2\mu$, $\sigma -1={\epsilon }^{2}s$ places it near the original double-zero point.

Write $k=\sqrt{\mu }$. The equilibrium points are a stable node $(-k,0)$ and a saddle equilibrium $(k,0)$, whose unstable and stable eigenvalues are $2k$ and $-\lambda$. The saddle equilibrium's local stable manifold is $x=k$. When the right-going unstable separatrix returns at $(\nu ,h)$ with $\nu =k$, it lands exactly on that stable manifold, producing a homoclinic orbit of the saddle. Crossing this condition changes whether the returning trajectory is captured by the node or lies on the outgoing side of the saddle equilibrium.

Use an incoming section $y=h$ and outgoing section $x=h$. For entry $(x,h)$ with $k<x<h$, integrate the local flow:so the outgoing height isThe global excursion is smooth and transverse. Its return coordinate has expansion $x_{\mathrm{next}}=\nu +cY+O(Y^2)$, with $c>0$ in these section orientations. Thus the local passage map near a dissipative saddle-node givesFor small $\mu >0$, $\lambda >2k$, hence $P(k^{+})=\nu$ and $P^{\prime }(k^{+})=0$. If $\nu -k>0$ is small, the graph starts above the diagonal and, with slope less than one nearby, crosses it close to $x=\nu$. More explicitly, $P(x)-x=(\nu -k)-(x-k)+O((x-k{)}^{\beta })$, giving a unique small fixed point with Floquet multiplier $0<P^{\prime }<1$. It corresponds to an attracting periodic orbit. For $\nu <k$ no such local crossing exists on $x>k$; the unstable branch returns to the node's side instead. The Poincaré return map panel shows this geometric fixed-point test.

For $\mu >0$, the left equilibrium point $(-\sqrt{\mu },0)$ is a stable node, the right equilibrium point $(\sqrt{\mu },0)$ a saddle equilibrium, and the saddle equilibrium's left unstable branch flows directly towards the node. The right branch takes the prescribed global excursion. Its landing relative to the two equilibrium points gives the six local portraits in the figure.

If $\mu >{\nu }^2$, then $-\sqrt{\mu }<\nu <\sqrt{\mu }$ for either sign of $\nu$. The global branch lands between the equilibrium points and flows left into the node. There is no nearby Poincaré return map cycle, and the landing approaches the node from its right.

If $0<\mu <{\nu }^2$ and $\nu >0$, the branch lands to the right of the saddle equilibrium. The contracting Poincaré return map has an attracting fixed point, giving a stable periodic orbit; the node also remains as a separate attractor with its own basin. If instead $\nu <0$, the branch lands to the left of the node and approaches it from the left. There is no nearby periodic orbit in that case.

If $\mu <0$, the two equilibrium points have vanished and a stable returning cycle exists for either sign of $\nu$. For negative $\nu$ the cycle traverses the full bottleneck and has the divergent period in part (c). For fixed positive $\nu$ it enters to the right of the bottleneck; its local period approaches $1/\nu -1/h$, so it need not diverge when $\mu \to 0^{-}$. The outer arcs in the figure encode only the given return; the local equations do not specify the full vector field away from the origin.

The boundaries explain the codimension-two organization. On $\nu =\sqrt{\mu }>0$ the saddle equilibrium has a homoclinic orbit. On $\mu =0$, $\nu <0$, the return passes through the semistable equilibrium point at a saddle-node bifurcation, producing a saddle-node bifurcation on an invariant circle. On $\mu =0$, $\nu >0$, the node and saddle equilibrium instead collide away from the persistent cycle. At $\mu =\nu =0$, the returning separatrix hits the equilibrium point's strong stable manifold at the saddle-node bifurcation: this is the saddle-node separatrix-loop bifurcation. The curve $\nu =-\sqrt{\mu }$ changes which side of the node receives the returning orbit; unlike the positive branch, it is not a global bifurcation through a saddle homoclinic orbit.

Solving the two factors in each equation gives $(0,0),(3,0),(0,3),(1,1)$. The Jacobian isAt the three boundary equilibria its eigenvalues are $3,-3$. These equilibria are hyperbolic fixed points; the Hartman-Grobman theorem gives local topological conjugacy to their saddle linearizations. At $(1,1)$ the matrix is $\left(\begin{array}{cc}1& 2\\ -2& -1\end{array}\right)$, with eigenvalues $\pm i\sqrt{3}$. The linearization is a centre, but the hyperbolic theorem does not apply there.

The nonlinear classification follows from the actual HamiltonianAlong solutions $\dot{H}=H_x{H}_y-H_y{H}_x=0$. At $(1,1)$, $H=-1$ and its Hessian is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, which is positive definite. Nearby regular level curves are closed ovals around this strict minimum. Their vector field is nonzero and tangent, so each is a periodic orbit. The nonlinear equilibrium is a centre.

The separatrix level $H=0$ consists of $x=0$, $y=0$, and $x+y=3$. Within the triangle they connect the saddles as $(3,0)\to (0,0)\to (0,3)\to (3,0)$. The interior levels $-1<H<0$ are nested clockwise periodic orbits; outside the triangle the remaining level branches are unbounded. On $y=0$, flow is left for $0<x<3$ and right for $x<0$ or $x>3$; on $x=0$, it is up for $0<y<3$ and down otherwise. These directions and the invariant levels determine the full phase portrait.

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.

The simultaneous zeros are the origin and the four corners $(\pm 1,\pm 1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, with eigenvectors $(1,1),(1,-1)$, so the origin is a saddle equilibrium of Poincaré index $-1$. At a corner $J=-2xyI$. Thus $(1,1)$ and $(-1,-1)$ are stable nodes; $(1,-1)$ and $(-1,1)$ are unstable nodes, all of index $+1$. Although the node eigenvalues coincide, the equilibria are hyperbolic and their stability follows from their strictly signed real eigenvalues.

The lines $x=\pm 1$ and $y=\pm 1$ are invariant and divide the plane into nine convex regions. A nonconstant periodic trajectory cannot cross any of these lines by uniqueness, and cannot be contained in one of them because one-dimensional autonomous flows have no nonconstant periodic solutions. It must therefore lie inside one region. The central region contains only the origin, whose index is $-1$; all other regions contain no equilibria. But a simple periodic orbit has index $+1$, equal to the sum of the enclosed equilibrium indices. Its interior stays in the same convex region, giving a contradiction in either case. There are no periodic orbits.

Away from the invariant lines, trajectory curves have the first integralsince both logarithmic derivatives equal $-2xy$. The diagonal $y=x$ is the unstable separatrix of the origin, leading toward the two sinks; $y=-x$ is its stable separatrix, coming from the two sources. The diagram shows these separatrices, the invariant lines and the direction field.

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.

Measure a signed angle $\theta$ from the upward radius $A\to O$ in the direction of falling. With $I_A=3m{a}^2/2$, gravitational torque givesThere is an idealization in the release condition: exactly upright and exactly at rest is an unstable equilibrium, so the exact ideal solution stays there. Uniqueness of this smooth equation with $\theta (0)=\dot{\theta }(0)=0$ prevents spontaneous departure. The usual falling calculation means an infinitesimal disturbance, with its energy tending to that of the upright rest state. For that limiting separatrix motion, conservation of mechanical energy givesOn the branch with increasing angle, $\dot{\theta }=\sqrt{8g/(3a)}\sin (\theta /2)$ for $0<\theta <2\pi$; the opposite branch has the opposite angular velocity. The nontrivial limiting motion approaches the upright position only as $t\to -\infty$, because $dt/d\theta$ has a logarithmically divergent integral near zero.

Let $e_r$ point from $A$ toward $O$, and $e_{\theta }$ in the direction of increasing angle, so $e_r=\sin \theta e_x+\cos \theta e_z$ and $e_{\theta }=\cos \theta e_x-\sin \theta e_z$. The centre of mass acceleration on the limiting falling branch isGravity has components $-mg\cos \theta e_r+mg\sin \theta e_{\theta }$. Applying Newton's second law to the whole disc givesThe signed component parallel to the radius, positive from $A$ toward $O$, is the required $mg(7\cos \theta -4)/3$. This same radial formula is consistent at the exact stationary upright state, where $\theta =0$, $a_O=0$, and $R=mg{e}_z$; nonzero falling angles require the limiting-release interpretation just stated.

Use $\hslash =c=1$ in the field-theory calculations. Work with signature $(+,-)$ and $a\ne 0$. Since $x=(x_{+}+x_{-})$ and $t=(x_{+}-x_{-})$, ${\partial }_{+}={\partial }_x+{\partial }_t$, ${\partial }_{-}={\partial }_x-{\partial }_t$, and the Sine-Gordon equation is ${\varphi }_{+-}=\sin \varphi$. Introduce $w=(\rho +\kappa )/2$ and $d=(\rho -\kappa )/2$. The specified Sine-Gordon Bäcklund transformation becomes $d_{+}=a\sin w$, $w_{-}=a^{-1}\sin d$. Its compatibility givesAdding and subtracting these equations establishesThus a compatible transformed field satisfies the same Sine-Gordon equation. Smoothness is needed to interchange mixed derivatives. The statement cannot literally include $a=0$, since one defining equation contains $1/a$.

Generating the kink. With zero seed, the two first-order equations give ${\rho }_{+}=2a\sin (\rho /2)$ and ${\rho }_{-}=2a^{-1}\sin (\rho /2)$. On a nonconstant branch put $p=\tan (\rho /4)$; then $p_{+}=ap$, $p_{-}=a^{-1}p$. Integration givesFor $a>0,C>0$, this is the Sine-Gordon kink, after absorbing $C$ in the center position. Its speed and Lorentz factor obeyChanging the sign of $a$ or of the exponential coefficient selects the corresponding antikink orientation; $|v|<1$ whenever $a$ is finite and nonzero. Constant vacuum branches, omitted by division by $\sin (\rho /2)$, can be added separately.

The continuously parametrized Bäcklund transformation does more than produce this one solution. Successive compatible transformations and Bianchi permutability for sine-Gordon Bäcklund transformations construct multisoliton solutions; expansion of the generating relations gives the local conserved-charge hierarchy of sine-Gordon theory. Together with its Lax pair and inverse scattering transform, this is the structure behind classical integrability, elastic scattering and the absence of generic radiative energy loss in soliton collisions. Existence of a transformation alone is not a proof that every generated conserved charge is independent and in involution; those are additional properties of the integrable system.

Transforming the static kink at $a=1$. The seed obeys ${\kappa }_x=2\mathrm{sech}x$, $\sin (\kappa /2)=\mathrm{sech}x$, $\cos (\kappa /2)=-\tanh x$. Adding and subtracting the two Sine-Gordon Bäcklund transformation equations yieldsAgain set $p=\tan (\rho /4)$. These simplify to $p_x=-p\tanh x$, $p_t=\mathrm{sech}x$. The first gives $p=f(t)/\cosh x$, and the second gives $f^{\prime }=1$. ThereforeContinuous inverse-tangent branches may be chosen without creating artificial jumps. At spatial infinity the field approaches the same vacuum, so its net topological charge is zero. For large $|t-t_0|$ there are two well-separated transitions near $x_{\pm }=\pm \mathrm{arcosh}|t-t_0|$, with speeds tending to zero. The two transitions exchange their kink/antikink orientations as they pass through the collision, while the field remains smooth.

This is the Sine-Gordon threshold kink-antikink solution, a separatrix between finite-speed scattering and the bounded Sine-Gordon breather motion, rather than a finite-period breather. In fact it is the $v\to 0$ limit of the scattering field in question 3. At $t=t_0$, $\rho =0$ and ${\rho }_t=4\mathrm{sech}x$, so the reduced energy is $\int {\rho }_t^{2}dx/2=16$, exactly twice the reduced single-kink energy $8$.

For mass comparisons below, the original PDF really prints $m^2/\beta$. Restoring physical coordinates $X=x/m$, $T=t/m$ gives the angular-field action ${\beta }^{-1}\int dtdx[\frac{1}{2}(\partial \varphi {)}^2+\cos \varphi -1]$ and literal kink mass $8m/\beta$, for $\beta >0$. The usual Sine-Gordon theory instead has prefactor $m^2/{\beta }^2$ and mass $8m/{\beta }^2$. The equation of motion and every field above are unchanged by this overall normalization; the quantum scattering comparison in question 3 is not.