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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.

A timelike equatorial Schwarzschild geodesic with specific energy $E=1$ and angular momentum $J=4m$ obeys $(du/d\varphi {)}^2=2u(u-1/4{)}^2$ for $u=m/r$. Substituting $q=2\sqrt{u}$ reduces it to $(dq/d\varphi {)}^2=(q^2-1{)}^2/8$, integrating to the two displayed branches. One lies outside $r=4m$ and the other inside; each approaches the same unstable circular orbit exponentially in angle and only after infinite proper time. Linearization of $u^{\prime \prime }=1/16-u+3u^2$ about $u=1/4$ gives $\delta u^{\prime \prime }=\delta u/2$, confirming instability. These are separatrix approaches or departures, not stable oscillations around the circular orbit.

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.

For an equilibrium point $p$, a homoclinic orbit is a nonconstant trajectory $x(t)$ withAt a hyperbolic equilibrium point, the trajectory leaves a neighborhood of $p$ along its unstable manifold and returns along its stable manifold. It never reaches $p$ in finite time when local solution uniqueness holds. Its closure includes the equilibrium and can form a loop-shaped separatrix.

A completely explicit example iswith Hamiltonian function $H(x,v)=v^2/2-x^2/2+x^4/4$. Its derivative is zero along every trajectory, and the origin is a saddle point with eigenvalues $\pm 1$. The curvelies on $H=0$. Differentiating gives $\dot{x}=v$ and $\ddot{x}=x-x^3$, while the hyperbolic secant tends to zero at both time infinities. Thus it is a homoclinic orbit. Changing both signs gives the second loop.

Each loop bounds periodic orbits around one potential minimum. This is an integrable Hamiltonian system, so a homoclinic loop by itself does not establish chaos. A Shilnikov bifurcation and a simple-zero Melnikov energy-balance method calculation require additional dimensional, spectral or forcing hypotheses.

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 one-dimensional conservative motion, a particle moves along one coordinate $x$ under a force $F(x)=-dU/dx$ derived from a time-independent potential energy. For fixed mass $m$, multiplying its equation of motion by $\dot{x}$ shows thatis constant. Motion is allowed only where $U(x)\le E$; the equality identifies possible turning points.

Plotting $U(x)$ turns the dynamics into a geometric picture. A local minimum can trap sufficiently low-energy trajectories and gives approximately simple harmonic motion when its curvature is positive. A local maximum separates regions of motion and may generate a separatrix. Away from turning points, the speed follows directly from $|\dot{x}|=\sqrt{2(E-U(x))/m}$, so travel times can be obtained by integration rather than solving the second-order differential equation explicitly.

Set $\ell =\sqrt{EI/F}$. The Euler elastica equation becomes ${\theta }^{\prime \prime }={\ell }^{-2}\sin \theta$. Multiplying by ${\theta }^{\prime }$ gives the first integralStraight asymptotic ends require ${\theta }^{\prime }\to 0$ and a horizontal tangent, so $C=F$. This is the separatrix of a simple pendulum with its angle shifted by $\pi$. A nonconstant solution moves from $\theta =0$ to $2\pi$, with the same physical tangent at both ends, taking infinite arc length to approach either limit. One explicit representative isUsing $\cos \theta =1-2{\mathrm{sech}}^2(s/\ell )$ and $\sin \theta =-2\mathrm{sech}(s/\ell )\tanh (s/\ell )$ and integrating $r^{\prime }=(\cos \theta ,\sin \theta )$ gives the tensile elastica loopThis solution is nontrivial, tends to horizontal straight tails, and has finite excess bending energy. It is also necessarily self-intersecting. Indeed $y$ is even and $x$ is odd. For $z>0$, let $f(z)=z-2\tanh z$. Near zero $f(z)<0$, while $f(2)>0$, so there is a root $z_{\ast }\in (0,2)$. At the two distinct parameters $s=\pm \ell z_{\ast }$, both points have $x=0$ and the same $y>0$. Numerically $z_{\ast }\simeq 1.91501$.

Any nonconstant smooth solution with the same straight asymptotic conditions has this first integral. Between consecutive horizontal orientations, separating variables gives the displayed solution up to translation, reflection, and reversal. It cannot pass through a straight equilibrium at a finite parameter, by uniqueness of the ordinary differential equation. Thus all such nontrivial branches have the same loop crossing. This is an ideal zero-thickness rod solution; a physical rod with self-contact excluded cannot realize the crossing without adding contact forces.

Let $f(r)=1-2m/r$ and work in the exterior $r>2m$, with $G=c=1$. The affinely parametrized geodesic Lagrangian isIts polar-angle equation isThe initial data $\theta =\pi /2$, $\dot{\theta }=0$ therefore remain in that plane. Rotational symmetry allows the axes to be chosen so that any nonradial initial trajectory has those data: its conserved angular-momentum direction fixes an orbital plane through the centre. Radial trajectories can also be placed in an equatorial plane.

The coordinates $t,\varphi$ are cyclic in $L$, so their canonical momenta are constant:The corresponding Killing vector fields are ${\partial }_t$ and ${\partial }_{\varphi }$. With the displayed signature, $g({\partial }_t,\dot{x})=E$ and $g({\partial }_{\varphi },\dot{x})=-J$, explaining the sign in the second canonical momentum. Thus $E$ is conserved energy per unit rest mass for a massive particle parametrized by proper time, and $J$ is its axial angular momentum per unit rest mass. For a photon, affine rescaling rescales both constants; after the tangent is normalized to the physical four-momentum they represent its energy at infinity and axial angular momentum. Their ratio $b=J/E$ is the signed impact parameter.

In the equatorial plane, the photon satisfies the null geodesic constraintwhich givesFor $J\ne 0$, put $u=m/r$ and $\alpha =Em/J$. Thenand consequentlyThe $J=0$ radial case is not represented by $\varphi$ as an orbit parameter.

There is a circular photon orbit at $r=3m$, but it is unstable. The radial effective potential isA circular orbit requires $V^{\prime }(r)=0$ and $E^2=V(r)$, givingIts second derivative isIt is a maximum, rather than a minimum, so radial perturbations depart from the orbit. Equivalently the second-order orbit equation is $u^{\prime \prime }=3u^2-u$. Linearizing about $u_c=1/3$ gives $\delta u^{\prime \prime }=\delta u$, with modes $e^{\varphi }$ and $e^{-\varphi }$. The existence of a growing mode directly contradicts the printed stability claim. Exact initial data can remain circular, but that does not make the orbit stable.

The supplied implicit expression describes the critical-energy family with ${\alpha }^2=1/27$, as fixed by the circular orbit. DefineRationalizing once gives $F=(\sqrt{3}-s)/(2(\sqrt{3}+s))$ andDifferentiating $F(u)=A{e}^{\varphi }$ gives $F^{\prime }{u}^{\prime }=F$, soThis verifies the first-order equation for every allowed integration constant $A$ on its real branch. For $A=0$ the derivative argument is replaced by direct substitution: $u=1/3$ is constant and solves the critical equation. The displayed family does not solve the equation at an independently arbitrary value of $\alpha$.

For $A>0$, write $q=A{e}^{\varphi }$. On the physical exterior branch,As $\varphi \to -\infty$, $q\to 0^{+}$ andThus for $A=0$ the photon remains exactly on the circular orbit. For $A>0$ it approaches $r=3m$ from outside through indefinitely many revolutions in the negative-$\varphi$ limit. With increasing $\varphi$, this same branch peels away outward and reaches infinity when $q=1-\sqrt{3}/2$. Reversing the direction of motion gives an incoming critical ray asymptotically approaching the photon circle. These are separatrix trajectories, not evidence of stability.

The potential maximum and the outward-growing radial separation display the same instability in two complementary descriptions.

Put $f=1-2m/r$, use proper time $s$, and denote its derivatives by dots. The timelike geodesic equations, away from the Schwarzschild coordinate singularity, arewith normalization $f{\dot{t}}^2-f^{-1}{\dot{r}}^2-r^2{\dot{\theta }}^2-r^2{\sin }^2\theta {\dot{\varphi }}^2=1$. Spherical symmetry permits an equatorial plane, $\theta =\pi /2$, and the cyclic coordinates give $\boxed{E=f\dot{t},J=r^2\dot{\varphi }}$. Normalization then gives ${\dot{r}}^2=E^2-f(1+J^2/r^2)$.

For $J\ne 0$ and $u=m/r$, $du/d\varphi =-m\dot{r}/J$. ThusAt $E=1$, $J=4m$, the right side factors as $2u(u-1/4{)}^2$. Set $q=2\sqrt{u}$, so $u=q^2/4$ and $(q^{\prime }{)}^2=(q^2-1{)}^2/8$. Integrating the regions $0<q<1$ and $q>1$ gives, up to angular shifts and time reversal,The constant $u=1/4$, or $r=4m$, is the corresponding circular timelike orbit. The hyperbolic-tangent branch lies at $r>4m$ and approaches that orbit from outside as $|\varphi |\to \infty$; the hyperbolic-cotangent branch lies at $0<r<4m$ and approaches it from inside. Near the limit $u-1/4$ decays exponentially as $e^{-|\varphi |/\sqrt{2}}$. Since $\dot{\varphi }$ tends to $J/(4m{)}^2=1/(4m)$, the limiting approach also takes infinite proper time. Neither nonconstant branch reaches and joins the circular orbit at a finite angular parameter.

Differentiating the orbit equation gives $u^{\prime \prime }=m^2/J^2-u+3u^2$. A perturbation about $u=1/4$ obeys $\delta u^{\prime \prime }=\delta u/2$, showing that the circular orbit is unstable. The two explicit curves are its separatrix branches, not ordinary small stable oscillations. For $E=1$ the energy is marginally bound, so the outer critical branch is the limiting orbit between scattering and Schwarzschild marginally bound capture from infinity. Time-reversed inner and outer branches describe the corresponding approach or departure from that unstable orbit. These are the Explicit separatrices of the marginally bound Schwarzschild orbit.

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.