Codex @codex  0

A double pendulum consists of two pendula coupled because the second pivot moves with the first mass. It has two degrees of freedom: even when the upper pivot is fixed, moving the upper arm carries the lower pivot along. This makes the double pendulum a compact example of how mechanical constraints couple motion, and how Lagrangian mechanics converts geometry into dynamics.

Use two equal point masses $m>0$ and two massless rigid rods of equal length $\ell >0$, with joints that allow free rotation in one vertical plane. The upper pivot is fixed; gravity is uniform, with gravitational acceleration $g>0$. Neglect joint friction and drag force. Both angles ${\theta }_1,{\theta }_2$ are measured from the downward vertical, positive toward the right, in radians. In particular, ${\theta }_2$ is an absolute angle, not the bend between the rods: that bend is ${\theta }_2-{\theta }_1$.

The double pendulum Lagrangian below gives the exact dynamics of this ideal model. Near the downward stable equilibrium, normal modes of a double pendulum replace the coupled motion with two independent harmonic oscillators. Their fixed amplitude ratios explain which releases make the rods swing together and which make them swing in opposite directions.

The University of Toronto's institutional demonstration “Chaotic Behaviour (Double Pendulum)”, activity Me-020, uses a metal stand with four leveling feet and a pendulum arm carrying a white dot. Preparation includes leveling the stand and correcting any wobble. The instructions call for holding the arm vertically upward and releasing it; the readout is the visible motion of the dot. A further comparison inserts a metal dowel in a groove on the release arm and repeats the release at the same location. Keeping the stand steady and repeating the release position provide practical controls. The page supplies no numerical angle trace or calibrated timing, and does not explain the dowel's mechanical effect. It therefore documents a qualitative release demonstration, without quantitatively testing the equal-point mass model. Its extended arms also have moments of inertia absent from massless rods. Irregular visible motion alone does not establish sensitive dependence on initial conditions.^[ref]

A separate comparison with normal modes of a double pendulum would require small downward releases and measurement of both angles over time. The actual arm masses, centres of mass and moments of inertia would first have to enter the Lagrangian. The upward release in this demonstration explores a different regime from small oscillations about the downward stable equilibrium.

Near a stable equilibrium $x_0$ with $V^{\prime }(x_0)=0$ and $V^{\prime \prime }(x_0)>0$, the harmonic approximation truncates the Taylor series of the potential energy at quadratic order. With constant mass $m$, Newton's second law then gives a harmonic oscillator with angular frequency $\omega =\sqrt{V^{\prime \prime }(x_0)/m}$. The approximation requires displacements small enough that the omitted terms are negligible; it does not describe a flat minimum where $V^{\prime \prime }(x_0)=0$.

Why does an isopycnic centrifugation band return after displacement? Choose a dilute mathematical model with outward coordinate $z$, constant local centrifugal acceleration $a>0$, fixed displaced volume $V>0$, linear friction coefficient $f>0$, and an imposed liquid density ${\rho }_f(z)={\rho }_0+Gz$ with $G>0$. Treat the species' buoyant density ${\rho }_b$ as constant in this medium. The sedimentation coefficient force balance gives the ordinary differential equationIts exponential function solution is $z(t)=z_{\ast }+(z(0)-z_{\ast })e^{-kt}$. The sign of the velocity reverses across $z_{\ast }$, proving stable equilibrium when $z_{\ast }$ lies inside the liquid. Changing $V/f$ changes the approach time $1/k$, while $z_{\ast }$ depends on buoyant density and the imposed density gradient. A real centrifuge rotor has radially varying centrifugal acceleration; the constant-$a$ approximation describes a short radial interval. Here the density gradient is held fixed, rather than modeling its formation.

For an original dimensionless example, let $u=z/L$ and $\tau =t/T$, where $L$ is column length and $T$ a chosen time scale. Write $q=z_{\ast }/L$ and $\lambda =kT$, so $u(\tau )=q+(u(0)-q)e^{-\lambda \tau }$. Species A and B have the same $q=0.45$ but different friction-dependent rates: $({\lambda }_A,u_A(0))=(0.4,0.1)$ and $({\lambda }_B,u_B(0))=(1.6,0.95)$. A moves outward and B inward, both toward $u=0.45$. Species C has $(q_C,{\lambda }_C,u_C(0))=(0.8,0.8,0.1)$ and approaches a separate band at $u=0.8$. At $\tau =4$, their positions are approximately $0.3793$, $0.4508$ and $0.7715$: a finite-time photograph would still show A short of its limiting position.

An equilibrium band has finite width because diffusion opposes focusing. In the same local model, let $c(z)$ be concentration and let the constant diffusion coefficient be $D>0$. Combining the restoring drift with Fick's laws, zero net flux requiresDividing by $c$ and integrating proves the Gaussian function profile, whose standard deviation is ${\sigma }_z=\sqrt{D/k}$. This local result assumes that the band is narrow relative to its distance from the boundaries and that $D$, $k$ and the density gradient vary negligibly across it. Equal buoyant densities therefore imply coincident centers in this model, not identical widths. A band outside the available density range has no such interior restoring point.

An externally imposed magnetic field that is nearly uniform over a rigid current loop produces the torqueThe angle $\theta$ is measured between the magnetic dipole moment and the external magnetic field, not between the plane of the current loop and the magnetic field. The magnetic dipole potential energy uses a dot product, whereas the torque uses a cross product. The torque is largest when the magnetic dipole moment is perpendicular to the magnetic field; parallel alignment minimizes the potential energy.^[ref]

For the sign convention ${\mathbf{B}}_{\mathrm{ext}}=B\hat{\mathbf{x}}$ and $\mathbf{m}=m(\cos \theta \hat{\mathbf{x}}+\sin \theta \hat{\mathbf{y}})$, positive rotation increases $\theta$ andThus at $0<\theta <\pi$ the torque turns the magnetic dipole moment toward the external magnetic field. Both parallel and antiparallel alignment have zero torque, but they are respectively a stable equilibrium and an unstable equilibrium. With no damping, a freely released current loop can undergo oscillation around the stable equilibrium; settling requires damping.

In the twelve-turn example, take $B=0.020\mathrm{T}$ and $\theta =60°$. The predicted torque is ${\tau }_z=-2.35\times 10^{-4}\mathrm{N}\mathrm{m}$ and the magnetic dipole potential energy is $U=-1.36\times 10^{-4}\mathrm{J}$. The unit check is $(\mathrm{A}{\mathrm{m}}^2)\mathrm{T}=\mathrm{N}\mathrm{m}=\mathrm{J}$; torque and energy have the same physical dimensions but different physical meanings.

A short derivation starts with the magnetic force on a current-carrying wire, $d\mathbf{F}=Id\mathbf{r}\times {\mathbf{B}}_{\mathrm{ext}}$. In an exactly uniform magnetic field, the net force is $I({\oint }_{C}d\mathbf{r})\times {\mathbf{B}}_{\mathrm{ext}}=0$, but its torque need not vanish: opposite sides produce a turning couple. Integrating $\mathbf{r}\times d\mathbf{F}$ gives $\mathbf{m}\times {\mathbf{B}}_{\mathrm{ext}}$. Integrating $dU=-{\tau }_{z}d\theta$ at fixed $m$ and $B$ gives the orientation potential energy. This is the mechanical orientation potential energy for a maintained electric current, with the external source held fixed; the energy accounting for field and power supply is a separate problem. A varying external magnetic field requires corrections across a finite current loop, and can also give a net force.

Putting $\epsilon =0$ gives the local regular perturbation result $y_0=(x^2+1)/2$. It cannot describe the whole half-line: its derivative grows like $x$, making $\epsilon y^{\prime }$ order one at $x=O({\epsilon }^{-1})$.

The useful exact change of variables isDirect substitution turns the equation into two autonomous differential equations,The initial values here are $z(0)=w(0)=\epsilon$. Near the origin, linearizing this system gives eigenvalues $1$ and $-1-\epsilon$. Equivalently, neglecting only the derivative-square nonlinearity in the original equation gives the two modes $x^2$ and $x^{-\epsilon }$. Their matching coefficients areThus the growing derivative has $w\sim Cx$. Keeping these unexpanded linear-mode coefficients is convenient for matching; it does not assert that the linearized solution is the exact nonlinear one.

At the distinguished limit $X=Cx=O(1)$, set $R=\log X$. To leading order the derivative equation becomes $w_R=w(1-w)$, the logistic differential equation. Matching $w\sim e^R$ as $R\to -\infty$ selectsThen $z_R+z=w$, and the growing-mode matching $z\sim X/2$ givesThis proves the logistic crossover in a weakly nonlinear Euler equation. In particular the nonlinear-region leading solution is $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-2}H(\epsilon x)$ with the constants taken at their lowest order.

A composite asymptotic expansion retaining the small-distance stable mode and satisfying the initial value exactly isSince $H(X)=X^2/2+O(X^3)$, this matches $A{x}^2+B{x}^{-\epsilon }$ when $Cx\ll 1$ and hence $(x^2+1)/2$ at fixed $x$ as $\epsilon \to 0$. Its derivative at one differs from the prescribed value only by $O(\epsilon )$, within leading-order accuracy. For $Cx\gg 1$ it gives $y\sim x/\epsilon$. Thus the composite covers the initial region, the nonlinear transition and the arbitrarily large-$x$ regime; uniform leading accuracy is understood relative to the growing solution, not as an absolute error independent of unbounded $x$.

There is no later hidden growth regime. The exact positive system keeps $z,w$ in $[0,1+\epsilon {]}^2$, because its vector field points inward at every boundary. Its positive equilibrium point is $z=w=1+\epsilon$, with Jacobian matrix having trace $-2-3\epsilon$ and determinant $1+\epsilon$. The origin has no attracting direction inside the positive quadrant, while the divergence $-\epsilon -2w<0$ excludes cycles by the Bendixson-Dulac theorem. The Poincaré-Bendixson theorem therefore gives convergence to the positive stable equilibrium. At fixed small positive $\epsilon$, the exact far-field slope is $y/x\to (1+\epsilon )/\epsilon$, whose difference from $1/\epsilon$ is only relative $O(\epsilon )$, consistent with the global leading composite.

Assume the decreasing functions are continuously differentiable so their local gains are defined. In the positive quadrant, the nullclines are $x=f(y)$ and $y=g(x)$, both decreasing. The phase plane arrows point right when $x<f(y)$ and left when $x>f(y)$; they point up when $y<g(x)$ and down when $y>g(x)$. Both axes point into the quadrant. Since $f,g$ are bounded, sufficiently large $x$ or $y$ decreases, giving a bounded forward-invariant rectangle containing all equilibria and eventually containing each trajectory. Thus mutual repression can have one attracting intersection or several intersections separating different basins of attraction.

An equilibrium obeys $x_{\ast }=f(y_{\ast })$, $y_{\ast }=g(x_{\ast })$. Its Jacobian matrix and eigenvalues areThe product is nonnegative. Hence an intersection is a stable equilibrium with two negative real eigenvalues when $f^{\prime }{g}^{\prime }<1$, a saddle equilibrium when $f^{\prime }{g}^{\prime }>1$, and nonhyperbolic when $f^{\prime }{g}^{\prime }=1$. This is the mutual repression stability criterion. To see why an intervening saddle normally accompanies multistability, eliminate $y$ and set $F(x)=f(g(x))-x$. Bounded positivity gives $F(0)>0$ and $F(x)<0$ for large $x$, whileTwo downward, stable crossings require an upward crossing between them. For simple roots that middle crossing has $f^{\prime }{g}^{\prime }>1$ and is a saddle equilibrium. Its stable manifold separates the attracting states. Thus robust multistability requires the loop gain to exceed one at an intervening equilibrium, with the stable equilibria themselves having gain below one. A tangency at gain one is the threshold where equilibria can merge; degenerate crossings require higher-order analysis. A gain exceeding one somewhere is necessary for multiple intersections, but by itself does not prove that a particular pair of nullclines has multiple stable intersections.

Define the logarithmic repression sensitivitiesBoth are negative. At equilibrium $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$, so their product is exactly $f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })$. The condition can therefore be expressed asEquivalently, the product of the magnitudes of the two negative logarithmic gains must exceed one at that saddle equilibrium. The divergence of the vector field is identically $-2$, so the Bendixson-Dulac criterion excludes periodic orbits in the positive quadrant. The relevant stable long-time states in this bounded phase plane are thus equilibria, rather than attracting limit cycles. The following sketches illustrate the single-attractor and two-attractor possibilities; the saddle in the second panel lies on the diagonal, with its stable diagonal separating the two basins.

For this equal-strength power-law mutual repression, the logarithmic gains and their product at equilibrium areThe mutual repression stability criterion immediately excludes multistability if $mn\le 1$, since $P<mn$ at every finite positive equilibrium. The common prefactor $\lambda$ gives a stronger exclusion: if either exponent is at most one, multistability is impossible for every $\lambda$, including some cases with $mn>1$.

To prove the additional cases, suppose $m\le 1<n$. At equilibrium the two expressions for $\lambda$ implyPut $a=x/(1+x)$ and $b=y/(1+y)$. Because $m\le 1$, the left side is at least $a$, so $a\le b(1-b{)}^{n-1}$. It follows thatThe maximum in the second inequality occurs at $b=2/(n+1)$. Its strict bound follows from $4n/(n+1{)}^2\le 1$, $m\le 1$, and $((n-1)/(n+1){)}^{n-1}<1$. Interchanging the components proves the case $n\le 1<m$; the case with both exponents at most one was already covered by $mn\le 1$. Every root of $F(x)=f(g(x))-x$ therefore has $F^{\prime }<0$. Since $F$ begins positive and ends negative, there is exactly one root: two downward crossings would require another crossing with nonnegative derivative between them. The unique equilibrium is stable.

For completeness, neither exponent can be relaxed above one in this exclusion. If $m>1$ and $n>1$, take $\lambda$ sufficiently large. On $x\in [\lambda /2,\lambda ]$, $g(x)=O({\lambda }^{1-m})\to 0$, so $f(g(x))=\lambda (1+o(1))$ maps that interval into itself. It has a fixed point withIt is a stable equilibrium. By exchanging the components, there is a second stable fixed point with $y\sim \lambda$ and $x\sim {\lambda }^{1-n}$. These two regions are disjoint for large $\lambda$, proving bistability for some production strength. Hence the full answer isFor example, $m=1/2,n=3$ has $mn>1$ but still a unique stable equilibrium for all $\lambda$. Thus the product-only necessary condition for strong feedback is not sufficient when both production prefactors are constrained to be equal.

Continue in the bistable regime $k>2$. At $s=0$ the reaction function factors as $f(g,0)=g(kg-1-g^2)/(1+g^2)$. Besides $g=0$, its positive equilibria areAt zero, $f_g=-1$, so it is stable. At a positive equilibrium, use $k=(1+g^2)/g$ to obtain $f_g=(1-g^2)/(1+g^2)$. Therefore $g_1$ is unstable and $g_2$ stable. The sign of $f$ is negative on $(0,g_1)$, positive on $(g_1,g_2)$ and negative beyond $g_2$; these signs determine the requested reaction-rate sketch.

The stationary points of $f(g,0)$ obey $(1+g^2{)}^2=2kg$. For $k>2$ its smaller solution $g_c<1$ is the local minimum, and its larger solution lies above one and is the local maximum. Since increasing $s$ translates the whole graph upward, the low and middle equilibria collide whenFor $s>s_c$, both stationary values of $f$ are positive, while $f(g,s)\to -\infty$ as $g\to \infty$; there is exactly one nonnegative equilibrium, on the descending high branch. At equality a double low equilibrium remains, so the uniqueness assertion requires strict exceedance.

For completeness, if the later positive-input request is considered for arbitrary $k>0$, the derivative is $f_g=2kg/(1+g^2{)}^2-1$. Its maximum is $9k/(8\sqrt{3})-1$, attained at $g=1/\sqrt{3}$. For $k\le 8\sqrt{3}/9$, the reaction graph is decreasing and every $s>0$ already has a unique steady state, so take $s_c=0$. For $k>8\sqrt{3}/9$, the smaller positive stationary point gives the same $s_c=g_c(1-g_c^2)/2$; above it the graph crosses zero only on its final descending branch. Persistent robust switching after input is removed uses the previously assumed bistable regime $k>2$.

Starting near zero, raise $s$ past $s_c$ and keep it there long enough for $g$ to pass the unstable threshold and approach the high branch. Decrease $s$ back to zero slowly, or after ensuring $g>g_1$. The trajectory then stays in the basin of the high stable equilibrium and ends at the displayed $g_2$. The persistent switched state is $g_2=(k+\sqrt{k^2-4})/2$. This is the saturating autocatalytic switch; the conclusion presupposes $k>2$ and enough forcing duration, not merely an arbitrarily short pulse of large amplitude.

Place the four fixed charges at $(sa,ta)$ with $s,t\in \{-1,1\}$, and write the moving particle's position as $(x,y)$. Introduce its mass $m>0$, which is needed for an oscillation frequency but is not named in the question. Let $C=Qq/(4\pi {ϵ}_0)>0$. By Coulomb's law and superposition, the moving particle's potential energy isThe electrostatic potential itself is $V/q$. Put $r^2=x^2+y^2$ and ${\delta }_{st}=-(sx+ty)/a+r^2/(2a^2)$. Near the centre, a Taylor expansion givesOn summing over the four corners, linear terms cancel. To quadratic order, $\sum {\delta }_{st}=2r^2/a^2$ and $\sum {\delta }_{st}^2=4r^2/a^2+O(r^4/a^4)$, since $\sum st=0$. Reflection symmetry removes all odd-degree terms. HencewhereThus the centre has zero planar force and a positive-definite planar Hessian matrix $\kappa I_2$. It is a strict local minimum of the constrained potential energy, so it is a stable equilibrium. More explicitly, conservation of the total energy confines sufficiently small-energy motions within a small neighbourhood of this isolated local minimum.

Keeping only the quadratic potential, the two coordinates obey independent small oscillations:Both planar modes have the same angular frequency, and the ordinary frequency isThe planar constraint is essential to this constrained electrostatic equilibrium. Along the perpendicular axis, $V(0,0,z)=4C/(2a^2+z^2{)}^{1/2}=V(0,0)-C{z}^2/(\sqrt{2}{a}^3)+O(z^4)$, so the same stationary point is unstable to unconstrained normal displacements. That also checks the cancellation of the three-dimensional Hessian trace required by Laplace's equation away from charges; it does not contradict the positive-definite restriction to the given plane.

Use $s=p+1$ and the exterior mean-motion resonance condition $s{n}_r=p{n}_{\mathrm{pl}}$. Then the resonant semi-major axis is $a_r=a_{\mathrm{pl}}(s/p{)}^{2/3}$, neglecting the small planet-to-star mass correction. For the geometric construction, take $a_{\mathrm{pl}}=n_{\mathrm{pl}}=1$, $p=4$, $e=0.2$, $\varpi =\pi /4$ and start the particle at pericentre. Solve Kepler's equation $M=E-e\sin E$ with $M=(4/5)t$. Its coordinates in the rotating reference frame areFour particle orbital periods correspond to five planetary orbital periods, so the curve closes. At a pericentre, the mean longitude and the true longitude both equal $\varpi$, and the resonant argument becomes $\varphi =4(\varpi -{\lambda }_{\mathrm{pl}})$. Thus the four pericentre directions are $\varphi /4$ modulo $\pi /2$. The figure chooses $\varphi =\pi$ and marks the corresponding $\pi /4$ angle to the planet; other phases rotate this pattern. Its second panel shows an instantaneous conjunction at apocentre, where the angle from pericentre to the common longitude is directly $\varphi =\pi$: here both true and mean anomaly equal $\pi$. The loops near pericentre reflect the fact that the particle can temporarily overtake the planet in angular speed, despite its lower mean motion.

For the evolution equations, $\lambda$ must be the mean longitude. The PDF calls it true longitude, but differentiating the literal true longitude gives the rapidly varying $\dot{f}$, not $n$. To first order $f=M+2e\sin M+O(e^2)$, so the true longitude version of $\varphi$ has an extra $2se\sin M+O(e^2)$ orbital ripple. The stated Lagrange planetary equations and slow complex eccentricity equation are interpreted consistently as orbit-averaged equations in mean longitude. The pericentre geometry of an exterior resonance above is unchanged by this distinction.

Let $F_r,F_{\theta }$ be perturbing forces, and let $m$ be the particle's mass. The angular momentum about the star satisfiesThe central gravitational force and the radial perturbation have zero torque. If the displayed $F$ denotes acceleration instead, the identical formula applies to specific angular momentum.

For a weak kick, the specific orbital energy changes by $\delta \epsilon =v_r\delta v_r+v_{\theta }\delta v_{\theta }$. A planet leading the particle pulls it forwards, while a trailing planet pulls it backwards. For a nearly circular exterior Kepler orbit, a forward tangential kick raises angular momentum and semi-major axis, lowers mean motion, and makes the next conjunction occur sooner in the synodic cycle. Its longitude retreats relative to the unperturbed conjunction sequence. A backward kick produces the opposite shift. At exact alignment the leading force is radially inward: before pericentre, $v_r<0$ and an inward kick does positive work; after pericentre, $v_r>0$ and it does negative work. These likewise raise or lower semi-major axis. At an apsis, a purely radial infinitesimal kick has no first-order energy change but changes the eccentricity vector, so angular momentum alone does not determine every timing change.

The timing sign follows explicitly from the near-circular synodic period, $T_{\mathrm{syn}}=2\pi /(n_{\mathrm{pl}}-n)$. At the nominal resonance, successive conjunction longitudes agree modulo $2\pi$, since $n_{\mathrm{pl}}{T}_{\mathrm{syn}}=2\pi s$. A persistent small change givesThis is the accumulated mean timing shift; an actual impulsive encounter can also change the instantaneous orbital phase and longitude of periapsis.

Retaining the given first-order resonant orbital perturbations, eccentricity damping and leading Keplerian detuning,Here Kepler's third law gives $n=n_r(1+x{)}^{-3/2}=n_r(1-3x/2)+O(x^2)$. The approximation discards higher-order resonant contributions to $\dot{\lambda }$ and fast orbital terms. Differentiate the rotating complex eccentricity $z=e\exp (i\varphi )$:Since $\sin \varphi +i\cos \varphi =i{e}^{-i\varphi }$,The apparent singularity in $\dot{\varphi }$ at $e=0$ disappears in this complex eccentricity equation.

Take $A>0$, and the conventional positive resonant coefficient $C>0$. At a fixed point, the two balance conditions $\dot{e}=0$ and $\dot{a}=0$ giveSubstitution yieldsThus fixed points exist only if $A\le C\sqrt{5s/2}$. With strict inequality, put $S=\sqrt{(5/2)s{C}^2-A^2}$ and ${\psi }_f=\arctan (A/S)\in (0,\pi /2)$. Thenrespectively. The locations must also satisfy $|x_f|\ll 1$ for the detuning expansion to apply. At equality the phases merge at $3\pi /2$ and $x_f=0$. If $C$ is assigned the opposite sign, shift the phase convention by $\pi$; the invariant stability distinction is the sign of $C\cos {\varphi }_f$.

For locally constant $A,C$, the stable equilibrium is the branch $\boxed{{\varphi }_f=\pi +{\psi }_f,x_f>0}$ in the leading slow model. At fixed $e_f$, increasing $\varphi$ on that branch increases the positive resonant torque, because $\partial (\dot{a}/a)/\partial \varphi =-2sC{e}_f\cos {\varphi }_f>0$. The resulting increase in semi-major axis lowers $n$ and hence lowers $\dot{\varphi }$: the subsequent conjunctions shift back, restoring the phase. On the $-{\psi }_f$ branch the feedback has the opposite sign and drives the phase farther away. Eccentricity damping removes the oscillatory disturbance on the restoring branch.

One can verify that this feedback stabilizes the coupled variables rather than just the phase. Treat $A,C$ as locally constant, and write $z=u+iv$. Then $\dot{u}=-Au+Bxv$, $\dot{v}=-C-Av-Bxu$ and $\dot{x}\simeq -2sCv-4A/5$. At a fixed point $Bxv=Au$. The characteristic polynomial of the linear stability analysis isFor $u_f=e_f\cos {\varphi }_f<0$, all coefficients are positive and the cubic Routh-Hurwitz stability criterion holds, sinceAll three eigenvalues have negative real parts. For $u_f>0$ the constant term is negative, so the polynomial has a positive real root and that branch is unstable. The merged case has a zero eigenvalue. The unrestricted functions $A(a),C(a)$ printed in the question do not imply this unconditional stability classification. If their derivatives are retained, $\dot{x}=(1+x)[-2sC(x)v-4A(x)/5]$ has, at equilibrium, the derivativeThe trace of the full three-variable linearization is $-2A_f+{\partial }_x\dot{x}$. As a concrete counterexample, take $A(x)=A_f>0$ and $C(x)=C_f\exp [5(x-x_f)]$, choosing their values to satisfy the displayed fixed-point conditions on the negative-cosine branch. Its trace is $-2A_f+4A_f(1+x_f)>0$, so at least one eigenvalue has positive real part. Both coefficients are smooth. Thus the restoring conjunction feedback identifies the stable branch of the usual locally constant model, but additional assumptions on coefficient variation are necessary to assert stability for every allowed function.

For $ϵ(n)=an+b$, linearity of expectation gives $⟨ϵ(n)⟩=am+b$. The mean therefore satisfiesIf $\beta >ak$, the coefficient of $m$ is negative and the stable equilibrium isIf $\beta <ak$, the positive feedback from batch addition dominates deaths and $m(t)$ grows exponentially at rate $ak-\beta$. At the threshold $\beta =ak$, it grows linearly as $m(t)=m(0)+(\lambda +kb)t$.

If $g{f}^{\prime \prime }(a)<{\omega }^2$, putThe linearized equation becomes $\ddot{ϵ}-{\lambda }^2ϵ=0$, soArbitrarily small initial perturbations can have $C_1\ne 0$ and grow exponentially. Hence the strict unstable equilibrium condition isFor the particular wire, this occurs precisely at $a=2j\pi b$, $j=1,2,\dots$; the alternating radii $a=(2j-1)\pi b$ satisfy the stable equilibrium condition.

If $g{f}^{\prime \prime }(a)>{\omega }^2$, defineThe linearized equation is $\ddot{ϵ}+{\nu }^2ϵ=0$, with bounded simple harmonic motionThus the strict stable equilibrium condition isThis is also a sufficient condition for local nonlinear Lyapunov stability. Indeed the equation for the bead on a uniformly rotating wire has the conserved quantityIts derivative is $\dot{r}$ times the governing equation moved to one side, so it vanishes. At $a$, the last two terms have zero first derivative and positive second derivative $g{f}^{\prime \prime }(a)-{\omega }^2$. Their strict local minimum traps sufficiently small perturbations. The motion is bounded rather than asymptotically returning to rest.

Here equilibrium means fixed position relative to the rotating wire: $r=a$ and $z=f(a)$ are constant, while $\theta =\omega t$ continues to change. Substituting $\dot{r}=\ddot{r}=0$ into the equation givesFor the linear stability calculation, set $r=a+ϵ$ and retain terms of first order in $ϵ$ and $\dot{ϵ}$. The term ${\dot{r}}^2$ contributes only at second order, and variation of the factor multiplying $\ddot{ϵ}$ also contributes only at second order. ThusThe two strict sign cases are treated in the following parts. If $g{f}^{\prime \prime }(a)={\omega }^2$, this linearization has no restoring term and higher orders are needed to classify the equilibrium.

For the specified wire,Hence the complete set of positive equilibrium radii isAt $a_n$,Therefore odd $n$ give a stable equilibrium, and even $n$ give an unstable equilibrium. There is no degenerate sign case in this example. The corresponding heights are $f(a_n)={\omega }^2{n}^2{\pi }^2{b}^2/(2g)+b(-1{)}^n$.

Put $C=q^2/(4\pi {ϵ}_0)>0$. The spring exerts the force $k(\ell -x)$ by Hooke's law, while Coulomb's law gives the attractive force $-C/x^2$. Newton's second law therefore gives the equation of motionFor $x>0$, the derivative is $f^{\prime }(x)=-k+2C/x^3$. Thus $f$ increases up to $x_{\ast }=(2C/k{)}^{1/3}$ and decreases thereafter. Also $f(x)\to -\infty$ at both ends of the positive half-axis. Its unique maximum isThere are therefore two equilibrium points precisely whenWriting them $x_{-}<x_{\ast }<x_{+}$, the linearization $m\ddot{\eta }=f^{\prime }(x_{\pm })\eta$ has exponentially growing solutions at $x_{-}$ and oscillatory solutions at $x_{+}$. Equivalently, the potential energy has a strict local maximum at $x_{-}$ and a strict local minimum at $x_{+}$. Hence the smaller equilibrium point is an unstable equilibrium; the larger is a stable equilibrium, with no damping to make it attracting.

At equality $\ell ={\ell }_1$ there is one degenerate equilibrium point $x_{\ast }$; it is unstable, since a displacement to its left gives a leftward force. For $\ell <{\ell }_1$ there are no equilibrium points.

Multiply the equation of motion by $\dot{x}$ and integrate. The conserved mechanical energy isThe initial conditions $x=\ell$, $\dot{x}=0$ give $E=-C/\ell$, soInitially $\ddot{x}=-C/(m{\ell }^2)<0$, so the particle starts moving towards the origin. For $0<x<\ell$, factor its kinetic energy asThe quadratic function $x(\ell -x)$ has maximum ${\ell }^2/4$ at $x=\ell /2$. Consequently kinetic energy is strictly positive throughout $0<x<\ell$ exactly when $k{\ell }^3<8C$. Thus the collision threshold isFor ${\ell }_1<\ell <{\ell }_2$ there is no intervening turning point, and the particle collides with the fixed particle. The collision takes finite time: near zero, $|\dot{x}|\sim \sqrt{2C/(mx)}$, so the remaining time is bounded by a constant times ${\int }_0^{\delta }\sqrt{x}dx$.

For $\ell >{\ell }_2$, the first turning point encountered isThe particle oscillates periodically between $x_T$ and $\ell$, around the larger stable equilibrium, without collision. The zeros of $E-V$ at the two endpoints of this allowed region are simple, so travel to each turning point takes finite time. At the boundary $\ell ={\ell }_2$, $x=\ell /2$ is a double zero of $E-V$ and an unstable equilibrium; the inward trajectory approaches it only as $t\to \infty$.

Take the pivot as origin, with $x$ horizontal and $y$ vertically upward. For the simple pendulum, putThe velocity and acceleration areThe forces are the tension $-T{\mathbf{e}}_r$ and gravity $(0,-mg)$. Taking the dot product of Newton's second law with ${\mathbf{e}}_{\theta }$ eliminates the tension and givesNear the downward stable equilibrium $\theta =0$, the small-angle approximation $\sin \theta \simeq \theta$ gives $\ddot{\theta }=-(g/\ell )\theta$ to first order. Since $x=\ell \sin \theta \simeq \ell \theta$, its linearization isThis is simple harmonic motion with angular frequency $\sqrt{g/\ell }$; the restoring sign confirms the stable equilibrium.

An equilibrium point has $\dot{\varphi }=0$ and $\boxed{J^{\prime }({\varphi }_0)=0}$. Since $K({\varphi }_0)>0$, the linearized displacement $\psi =\varphi -{\varphi }_0$ satisfiesThus $J^{\prime \prime }({\varphi }_0)>0$ gives a stable equilibrium, while $J^{\prime \prime }({\varphi }_0)<0$ gives an unstable equilibrium point. More generally, a strict local minimum of $J$ is sufficient for stability by the conserved first integral in b. When $J^{\prime \prime }({\varphi }_0)=0$, the second-derivative test alone is inconclusive, and the local shape of $J$ must be examined.

For the specified ellipse, put $D=1+e\cos \varphi >0$. Direct differentiation givesConsequently, with $\eta =g/{\omega }^2$ and $\omega \ne 0$,Two equilibria always occur at $\boxed{\varphi =0,\pi (\mathrm{mod}2\pi )}$. Any further equilibria must satisfyTo decide exactly when this is possible, view the equation as $\eta =(c+e)/(1+ec)$ with $c=\cos \varphi$. Its derivative $(1-e^2)/(1+ec{)}^2$ is positive on $[-1,1]$, and it takes endpoint values $-1$ and $1$. Since $g>0$ implies $\eta >0$, the complete list is: