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A smooth wire has instantaneous shape $z=f(r)$ in cylindrical coordinates and rotates at constant angular velocity $\dot{\theta }=\omega$. The tangent vector at fixed time is ${\mathbf{e}}_r+f^{\prime }(r){\mathbf{e}}_z$. Its dot product with the constraint force vanishes. Projecting Newton's second law along this tangent vector gives the equation in the title.

Multiplying that equation by $\dot{r}$ proves conservation ofThis is a conserved quantity in the rotating description; the motor can exchange ordinary mechanical energy with the bead. A relative equilibrium at $a>0$ satisfies $U^{\prime }(a)=g{f}^{\prime }(a)-{\omega }^{2}a=0$. Its linearized equation is $(1+f^{\prime }(a{)}^2)\ddot{ϵ}=-U^{\prime \prime }(a)ϵ$. A strict local minimum with $U^{\prime \prime }(a)>0$ gives Lyapunov stability, while $U^{\prime \prime }(a)<0$ gives an unstable equilibrium. If $U^{\prime \prime }(a)=0$, higher orders are needed.

Ice-ocean albedo feedback is the ocean-surface case of ice-albedo feedback: loss of bright sea ice, often covered by snow, exposes darker water, lowers surface albedo and increases absorbed sunlight. The extra stored ocean heat content can promote further melting and delay refreezing. Reversing the sequence can amplify cooling. This is a positive climate feedback, meaning that it reinforces the initial change; the word positive does not mean beneficial, and it does not guarantee runaway warming.

For unchanged downward solar irradiance $S_{\downarrow }$ on a newly exposed opaque patch, the extra absorbed heat flux isFor illustrative local surface albedos $0.60$ and $0.10$, and $S_{\downarrow }=200\mathrm{W}{\mathrm{m}}^{-2}$, the patch absorbs an additional $100\mathrm{W}{\mathrm{m}}^{-2}$. This is a local sunlight calculation, not a whole-planet radiative forcing: newly exposed area, illumination, clouds and season determine its global contribution. With no incident sunlight this particular heating pathway vanishes. NASA's notes describe the contrast between reflective ice and snow and darker water. ^[ref]

To isolate how a positive climate feedback competes with Planck feedback, retain the assumptions of the zero-dimensional planetary energy balance and prescribe the smooth broadband Bond albedoThe exponential function makes a smooth transition from a bright cold state near $0.6$ to a darker warm state near $0.3$. The midpoint $240\mathrm{K}$ and width $3\mathrm{K}$ are chosen to illustrate the dynamics. They are not a seawater freezing law, an observed global relation, or a calibration of modern Earth. No evolving ice thickness or latent heat is represented.

An equilibrium satisfies $F(T_{\ast })=0$. For a small perturbation $\delta T$, linear stability follows fromSince $C>0$, $F^{\prime }<0$ gives an asymptotically stable equilibrium, while $F^{\prime }>0$ gives an unstable equilibrium. With $x=(T-240\mathrm{K})/(3\mathrm{K})$, the derivative isThus the optical term $-S{\alpha }^{\prime }/4$ is destabilizing, while the thermal radiation term $-4\sigma T^3$ is stabilizing. Their sum, not the optical term alone, decides the linear stability.

For the same $S=1360\mathrm{W}{\mathrm{m}}^{-2}$ and Stefan-Boltzmann constant used above, the bisection method on the intervals $(220,223)$, $(238,241)$ and $(253,255)$, with temperature in kelvins, gives:

Unrounded numerical zeros of a function have $|F(T_{\ast })|<10^{-10}\mathrm{W}{\mathrm{m}}^{-2}$; the table rounds them for readability. In the positive-temperature domain the sign of $F$ alternates between heating, cooling, heating and cooling across these three equilibria. The middle equilibrium separates the basins of attraction of the two stable equilibria. For example, a starting temperature of $235\mathrm{K}$ cools toward the cold state, whereas $245\mathrm{K}$ warms toward the warm state: this is bistability at identical external illumination.

The same model illustrates hysteresis. Its equilibrium curve can be written as $S(T)=4\sigma T^4/[1-\alpha (T)]$. The two turning points, where $4[1-\alpha (T)]+T{\alpha }^{\prime }(T)=0$, are saddle-node bifurcations: approximately $(T,S)=(232.637\mathrm{K},1567.801\mathrm{W}{\mathrm{m}}^{-2})$ and $(245.893\mathrm{K},1250.481\mathrm{W}{\mathrm{m}}^{-2})$. For illumination strictly between these thresholds there are two stable equilibria and one unstable equilibrium. Under sufficiently slow increase of $S$, the cold branch ends at the upper threshold; under sufficiently slow decrease, the warm branch ends at the lower threshold. The ensuing warming and cooling transitions differ, so the state depends on its history. Finite sweep rates can delay the transitions; these thresholds describe this model's static equilibrium curve, not predicted tipping points for modern Earth.

The striking feature is that a smooth albedo law can produce distinct long-time states and abrupt transitions, although no discontinuity was inserted into the optical law. It is useful for understanding why positive climate feedback can amplify a response while stabilizing thermal radiation still limits it. Here $0.3<\alpha (T)<0.6$ bounds the incoming heating, whereas outgoing thermal radiation grows as $T^4$; high enough temperatures therefore cool rather than run away indefinitely.

Historically, M. I. Budyko's 1969 paper coupled changing albedo from expanding ice cover to a latitude-dependent temperature and heat transfer calculation, showing how that coupling amplified the effect of changed sunlight. ^[ref] William D. Sellers's 1969 paper constructed an annual energy-budget model in 10-degree latitude belts, including planetary albedo, atmospheric infrared transmission and atmospheric and oceanic exchange. ^[ref] These primary papers provide historical examples of ice-sensitive energy-budget modeling; the smooth single-temperature example above is a separate illustration, not a reproduction of either paper's equations or predictions.

The logistic differential equation is the nonlinear ordinary differential equationAs a mathematical model of population dynamics, it treats abundance $N$ as a continuous nonnegative quantity. The constant $r$ has units of inverse time, while the carrying capacity $K$ has the same units as $N$. The useful idea is to combine growth resembling exponential growth at low abundance with a slowing approach to a finite abundance. The three trajectories below already show an important distinction: starting above the carrying capacity gives a decreasing curve, whereas starting below it gives an increasing one.

For $N>0$, dividing the derivative by abundance gives the instantaneous per-capita rate:It decreases linearly with $N$: it approaches $r$ at low abundance, is $r/2$ at $N=K/2$, vanishes at the carrying capacity, and is negative above it. This per-capita rate differs from the total rate $N^{\prime }$, which also counts how many individuals contribute. It is an instantaneous rate, rather than a fixed percentage increase over a whole day. The linear per-capita assumption and the resulting qualitative picture are developed in MIT 18.03SC's logistic-model notes.

Writing $f(N)=rN(1-N/K)$, the equilibrium points satisfy $f(N)=0$, so the constant solutions of a differential equation are $N(t)=0$ and $N(t)=K$. The phase line points upward for $0<N<K$ and downward for $N>K$. Also $f^{\prime }(0)=r>0$ and $f^{\prime }(K)=-r<0$: the criterion for fixed point stability for an autonomous differential equation makes zero an unstable equilibrium and $K$ an asymptotically stable equilibrium point. Every positive initial condition approaches $K$ as $t\to \infty$; $N_0=0$ stays exactly zero. Instability of zero means that a small positive population grows away from it, not that a zero population spontaneously appears.

For every $N_0>0$, the unique solution of a differential equation for all $t\ge 0$ isThe denominator is positive on this domain, including when $N_0>K$. For $0<N_0<K$, this is a translated and scaled logistic function; its extension to all real times is a sigmoid function. For $N_0>K$ the coefficient of the exponential function is negative, and the future curve instead decreases toward $K$. No nonconstant positive solution of a differential equation crosses $K$ or reaches it at finite time. A short derivation uses the dimensionless variable $z=K/N-1$: the chain rule gives $z^{\prime }=-rz$, so exponential decay yields $z(t)=(K/N_0-1)e^{-rt}$. The substitution divides by $N$, so the zero equilibrium point must be included separately.

The total growth rate is a downward-opening quadratic function of abundance. Completing the square gives its maximum without solving the differential equation:The balance is beautiful: more individuals contribute to growth, but each contributes less, and the strongest total increase occurs halfway to the carrying capacity. This is a maximum over possible abundances. A trajectory starting above $K/2$ never passes through that abundance in its future.

The chain rule connects this maximum to the shape of the time trajectory through its second derivative:For $0<N_0<K/2$, the trajectory initially has convexity and then concavity, with its one future inflection point atFor $N_0=K/2$, this point lies at the initial time, with concavity for $t>0$. For $K/2<N_0<K$, the inflection point belongs to the past extension, so the future trajectory has concavity throughout. Above $K$, the decreasing trajectory has convexity and no inflection point. The two constant solutions of a differential equation have no change of concavity. Thus an entire logistic function has an S shape, but its observed future segment need not show both bends. See also ^[ref].

For the plotted example with $K=1000$, $r=0.4{\mathrm{day}}^{-1}$ and $N_0=100$, time $t$ measured in days givesThe initial total rate is $N^{\prime }(0)=0.4\cdot 100\cdot 0.9=36$ individuals per day, and the initial per-capita rate is $0.36{\mathrm{day}}^{-1}$. The fastest total increase is $rK/4=100$ individuals per day, reached at $t_{1/2}=\ln 9/0.4\approx 5.49306$ days. At that time the per-capita rate has already fallen to $0.2{\mathrm{day}}^{-1}$.

These parameters are illustrative choices, rather than an empirical fit. The equation specifies deterministic evolution under a phenomenological assumption: it summarizes how net growth depends on abundance without separately modelling every process that affects it. The carrying capacity is a parameter for a specified setting, not a universal constant for all populations. In his 1838 paper "Notice sur la loi que la population suit dans son accroissement", Correspondance mathématique et physique 10, pp. 113–121, Pierre-François Verhulst proposed $p^{\prime }=mp-n{p}^2$, equivalent to this equation with $r=m$ and $K=m/n$. He also noted that different assumed growth-retarding terms could describe the observations similarly while giving very different limiting populations. This is an early illustration of why a good-looking curve alone does not establish a unique long-term prediction.^[ref]

The logistic differential equation describes continuous time. The logistic map instead advances in discrete steps and can exhibit a period-doubling cascade and Devaney chaos; the monotone positive trajectories here cannot. Their connection through the forward Euler method is explained under Euler discretization of logistic growth, where a large step can introduce behaviour absent from the original differential equation.

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.

For Hopf thresholds in a two-pool calcium model, with the unique equilibrium parametrized by $u=\mu$, locate changes of stability by solving the two equationsIf there are two simple roots ${\mu }_{-}<{\mu }_{+}$ with positive trace between them, the equilibrium is a repeller throughout that interval, since its determinant stays positive. At a nondegenerate crossing a Hopf bifurcation is possible, with linear frequency $[-f_v/ϵ{]}^{1/2}$. A complete classification of criticality requires a nonlinear coefficient, not merely the trace calculation. At leading singular order the thresholds approach the folds determined by $f=0$ and $f_v-\gamma f_u=0$, equivalently $V^{\prime }=-1/\gamma$.

Periodic solutions are also supported globally in the positive quadrant. The axes point inward for positive input, $v^{\prime }\le (\beta -\delta v)/ϵ$ bounds $v$, and the $u$ equation then bounds $u$ by a linear restoring term plus a bounded source. A large rectangle can thus be chosen forward invariant. For an initial point distinct from the repelling unique equilibrium, its omega-limit set cannot contain that equilibrium or a boundary equilibrium. The Poincaré-Bendixson theorem supplies a periodic orbit. In the strong separation regime it follows the two attracting branches and jumps between them, a relaxation oscillation.

The interval requires the maximum of $f_v-\gamma f_u$ along the critical manifold to exceed $ϵ$; the listed scale hierarchy alone is not a substitute for that test for arbitrary remaining constants. As an explicit example, $\beta =0.1$, $\delta =10^{-5}$, $ϵ=5\times 10^{-4}$ and $\gamma =0.1$ give approximately ${\mu }_{-}=0.22618$ and ${\mu }_{+}=0.37350$. At $\mu =0.3$ the trace is positive, whereas at $0.03$ and $1$ it is negative. The figures use this example and label their singular-limit portions accordingly.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot{y}-y\dot{x})/r^2$. The polar coordinates satisfyHence the rotation is clockwise. The origin is an unstable equilibrium, while $r=1$ is a limit cycle of period $2\pi$. Writing $u=r^2$ gives the logistic differential equation $\dot{u}=2u(1-u)$. For $r(t_0)=r_0>0$,All nonzero solutions tend to the unit-circle attracting limit cycle as $t\to \infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to -\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes atThis finite-time blow-up of an ordinary differential equation occurs backward in time. Such exterior solutions have no $t\to -\infty$ behavior: their maximal time interval is $(t_{\ast },\infty )$. The zero solution exists for all time and has no defined polar angle.

At an equilibrium, the second equation gives $x=y=:s$. The first equation then givesHence the fixed-point branches in the $(a,x)$ plane areandThe three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt{a}$, so the second bifurcation value is

The Jacobian matrix at a general point isOn the branch $P_0=(a^2,a^2)$ it is triangular, with eigenvaluesThus $P_0$ is unstable for $a<0$, asymptotically stable for $0<a<1$, and unstable for $a>1$.

On a square-root branch write $s=\pm \sqrt{a}$. Thenwhose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt{a}$ is therefore a saddle equilibrium for every $a>0$. The positive branch is a saddle for $0<a<1$ and asymptotically stable for $a>1$; close to $a=1$ it is a stable node.

To resolve the nonhyperbolic point at $a=0$, make the prescribed substitutionand append $\dot{a}=0$. The equations becomeThe centre subspace is $Y=X$. Seek the extended centre manifold for a parameter asThe centre-manifold invariance equation isTo second order, $\dot{X}=-aX+O(3)$, so comparison of coefficients givesand hence $A=C=0$, $B=1$. ThereforeSubstitution into the $X$ equation gives the reduced flowIts central branch $X=0$ is unstable for $a<0$ and stable for $a>0$, while the two nonzero branches $X\sim \pm \sqrt{a}$ for $a>0$ are unstable. Thus the bifurcation at zero is a subcritical pitchfork bifurcation with reversed normal-form parameter $\mu =-a$, exactly as recorded by the extended centre manifold of the 2023 Cambridge quadratic-product system.

The complete bifurcation diagram is therefore as follows. For $a<0$, only $P_0$ exists and is unstable. For $0<a<1$, $P_0$ is stable while both $P_{+}$ and $P_{-}$ are saddles. For $a>1$, $P_{+}$ is stable while $P_0$ and $P_{-}$ are saddles. At $a=1$, $P_0$ and $P_{+}$ cross and exchange stability, so the bifurcation is transcritical. This gives the bifurcation diagram of the 2023 Cambridge quadratic-product system.

Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll 1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt{a},\sqrt{a})$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt{a},\sqrt{a})$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional stable and unstable manifolds; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.

Newton's equation and its energy integral areThusLet $a=-V^{\prime \prime }(x_{+}^{\ast })>0$. Near the maximum,and the nearby turning point has distance $ϵ\sim \sqrt{2E^{\ast }\delta /a}$ from $x_{+}^{\ast }$. The singular part of the period is an $\mathrm{arcosh}$ integral:HencePhysically, the limiting orbit approaches the unstable equilibrium at the barrier top with vanishing speed and spends arbitrarily long there.

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.

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 $t\to \infty$.

Put $\lambda ={\omega }^2-g/a>0$. The zero value of the first integral in b givesAt the endpoint $r=a$, this gives ${\dot{r}}^2=\lambda a^2/2$. The absolute speed includes radial, azimuthal and vertical components. Since $\dot{z}=\dot{r}$ there,Hence the limiting exit speed isThere is a small initial condition qualification in the wording. The exact initial conditions $r=\dot{r}=0$ give the stationary solution forever, by Picard-Lindelöf theorem. The displayed zero-first integral moving branch instead approaches that unstable equilibrium as $t\to -\infty$.

For a precise small disturbance, release the bead with zero relative velocity at $r=ϵ>0$. Its first integral is $J=-\lambda {ϵ}^2/2$, givingThe bead initially accelerates outward and reaches the end. Letting $ϵ\to 0^{+}$ recovers exactly the stated exit speed. The conserved $J$ is a Jacobi integral associated with the imposed rotation; ordinary mechanical energy need not be conserved because the apparatus driving the wire can do work.

First take $a\ne 0$, put $A=|a|$, and suppose $k>0$. The force is the negative derivative of the potential energy:The potential energy is an odd function, passes through $U(0)=0$, and approaches zero from above as $x\to -\infty$ and from below as $x\to +\infty$. Its stationary points are $x=\pm A$:Thus the sole stable equilibrium is the local minimum at $x=A$; the local maximum at $x=-A$ is an unstable equilibrium. The requested values at the stable equilibrium areThe force is an even function, positive for $|x|<A$, negative for $|x|>A$, and zero at $x=\pm A$. Its central maximum is $F(0)=k/A^2$. Sinceits two minima occur at $x=\pm \sqrt{3}A$, with $F=-k/(8A^2)$, and $F\to 0^{-}$ at both ends. These values fix the shapes of both requested sketches.

In one dimension, attraction towards the origin means $xF(x)<0$, while repulsion from the origin means $xF(x)>0$. With that explicitly stated convention,A positive force alone means rightward, not necessarily outward. At the origin $F(0)\ne 0$, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls $a,k$ merely constants. For $k<0$ both curves reverse sign, the stable equilibrium is $x=-A$, its potential energy is $-|k|/(2A)$, and the attraction/repulsion intervals interchange. For $k=0$ the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If $a=0$ and $k\ne 0$, then $U=-k/x$ for $x\ne 0$ and $F=-k/x^2$, with no finite equilibrium.