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Attraction in a dynamical system means convergence of nearby forward trajectories toward a specified invariant state or set. Attraction and Lyapunov stability are separate properties; together they give the usual local asymptotic stability of an equilibrium.

A quadratic activator can have both basal production and activation inhibited by a second species. A normalized example has reaction rates $f(u,v)=1+r{u}^2/v-u$ and $g(u,v)=q(u^2-v)$, with $r,q>0$ and positive concentrations. At its spatially homogeneous equilibrium, $v=u^2$, so $u_{\ast }=1+r$ and $v_{\ast }=(1+r{)}^2$. Writing $a=(r-1)/(r+1)$, its reaction Jacobian matrix is $J=\left(\begin{array}{cc}a& -r/(1+r{)}^2\\ 2q(1+r)& -q\end{array}\right)$. Thus $\mathrm{tr}J=a-q$ and $\det J=q>0$, giving linear asymptotic stability when $q>a$ and instability when $q<a$. Basal production makes $a$ change sign at $r=1$; a positive activation feedback needed for a conventional Turing instability requires $r>1$.

Bounded-input bounded-output stability means that every bounded plant input produces a bounded plant output from zero initial state. For a causal continuous-time linear time-invariant system with a proper rational transfer function, BIBO stability is equivalent to every pole of the reduced transfer function having strictly negative real part. Cancel common factors first: an apparent denominator root may be absent from the reduced transfer function. For $K/(\tau s+1)$ with $\tau >0$, the only pole is $-1/\tau$, so the first-order low-pass filter is BIBO stable. ^[ref]

Internal asymptotic stability also requires every unforced internal state to decay. A canceled factor can hide a growing internal state. For example, consider the linear system of ordinary differential equationsIts zero-state transfer function can be written $(s-1)/((s+1)(s-1))=1/(s+1)$ and has BIBO stability. Yet from zero initial conditions and a unit-step plant input, $x_2(t)=e^t-1$ grows without bound while the plant output remains bounded. The reduced transfer function conceals that second state. Even a transfer function with no visible cancellations can be implemented with additional hidden states; its poles alone do not establish internal asymptotic stability of an arbitrary implementation.

A node equilibrium with a repeated nonzero real eigenvalue. If the planar Jacobian matrix has a one-dimensional eigenspace, its linearized solutions contain a term $t{e}^{\lambda t}$ and the node is also called improper. A negative repeated eigenvalue still gives asymptotic stability; a repeated value does not by itself indicate a bifurcation.

A feedback delay is a time interval between an event and its returning influence. Multiple biochemical steps can introduce an effective lag; a fixed delay is an idealization, not a universal description of signal transduction. To isolate timing, let $x$ be a signed displacement from an equilibrium point and use normalized dimensionless variables:The local removal is $-x$ and the returning influence is negative at every delay. At $d=0$ the ordinary differential equation reduces to the decay already illustrated under signaling feedback. For $d>0$ a delay differential equation requires a history on $[-d,0]$, not just an initial condition at one instant. Negative values of $x$ are displacements, not negative chemical concentrations.

An exponential function $x=e^{\lambda t}$ gives the characteristic equation of a delay differential equationPutting $\lambda =i\omega$ and equating real and imaginary parts gives $\cos (\omega d)=-1/2$ and $\omega =2\sin (\omega d)$. The first crossing with $\omega >0$ therefore occurs atFor this particular mathematical model, the equilibrium point is asymptotically stable for $0\le d<d_c$. No nonnegative real delayed feedback characteristic root can satisfy the equation; the first loss of asymptotic stability is the indicated pair of imaginary delayed feedback characteristic roots. Differentiating the characteristic equation of a delay differential equation along a root yieldsThus the pair moves into growth as the delay increases through $d_c$. The neutral mode at $d_c$ is a periodic function, but its amplitude depends on history: this linear mathematical model does not select an attracting limit cycle.

The figure plots $x(t)=e^{at}\cos (bt)$ using a delayed feedback characteristic root $\lambda =a+ib$ and that same formula on the history interval. At $d=0.7$ the displayed mode decays; at $d=1.5$ it grows. These are exact modes up to numerical root accuracy, not step responses with identical histories. Molecular bounds and a changed return law would be needed to interpret what happens after an unstable biological response becomes large.

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

Evolution is deterministic when a specified initial condition and evolution rule select a unique future orbit. For a discrete dynamical system, applying the same map to the same state gives the same next state. For an ordinary differential equation, uniqueness of the initial value problem is required; merely writing a differential equation does not guarantee it.

Determinism permits both asymptotic stability and sensitive dependence on initial conditions. It does not by itself imply reliable prediction from approximate measurements, and sensitivity does not introduce a random choice into the evolution rule.

Evolutionary game theory studies how a population's composition changes when the success of a pure strategy depends on which other pure strategies it encounters. The central feedback is a frequency-dependent payoff: becoming common changes the opponents a strategy meets, which can change whether it does well. In a coordination game, common behaviour can reinforce itself; in the coexistence example below, being rare is an advantage. This makes evolutionary game theory useful for asking which behaviours can persist and which initial compositions lead to different outcomes, without assuming that each participant solves an optimization problem.

In the basic model, a large, well-mixed population has finitely many inherited pure strategies; interactions sample opponents according to their population frequencies. Assume that a strategy's expected payoff determines its per capita net reproductive growth rate, with a common baseline allowed. These are modelling assumptions: reproduction must transmit the strategy, encounters must match the proposed sampling rule, and the conversion from payoff to growth must be justified for an application. The replicator equation below describes the resulting continuous-time frequencies. Taylor and Jonker's ^[ref] explicitly derive this growth model.

For learned behaviour, reproduction can instead be replaced by an assumed copying rule: an individual samples another individual uniformly and copies a better-performing pure strategy at a rate proportional to the positive payoff difference. At frequencies $x_i,x_j$, the net flow from $j$ to $i$ is $x_i{x}_j(f_i-f_j)$, because the two directional positive parts subtract to $f_i-f_j$. Summing over $j$ gives $x_i(f_i-\bar{f})$, the replicator equation. This derivation states a particular mechanism; an arbitrary rule for learning or imitation need not give the same dynamical system.

Three questions must be distinguished. A symmetric Nash equilibrium tests whether an individual can improve its expected payoff against the resident composition. An evolutionarily stable strategy tests whether each distinct rare mutant is outperformed after it is introduced. Dynamical stability tests the response of an actual dynamical system to small changes of composition, and asymptotic stability additionally requires nearby trajectories to converge. For a finite symmetric bimatrix game under the continuous-time replicator equation, every evolutionarily stable strategy is a symmetric Nash equilibrium and a locally asymptotically stable rest point; the converse from asymptotic stability to an evolutionarily stable strategy fails in general. See Ross Cressman and Yi Tao, The replicator equation and other game dynamics (2014), Theorems 1–2 and Example 1. These statements refer to stability on the full probability simplex, including perturbations introducing currently absent pure strategies.

A concrete experiment makes the frequency feedback visible. In 2009, Gore, Youk and van Oudenaarden competed two yeast strains: a producer of the enzyme invertase, which enables sucrose hydrolysis, and a mutant lacking the corresponding SUC2 gene. Released sugars can feed neighbours, so avoiding production can be advantageous. They used shaken 5 mL cultures with 5% sucrose at $30{}^{\circ }\mathrm{C}$, diluted daily after 23 hours of growth. Different labelled proteins allowed the two strains to be counted separately by flow cytometry; selective plating periodically checked the counts. The cooperator required externally supplied histidine, so restricting that amino acid imposed an adjustable growth disadvantage. See Snowdrift game dynamics and facultative cheating in yeast, pp. 253 and 255–256, Methods Summary.

Starting with rare producers and, separately, rare nonproducers tested both invasion directions. Under the reported coexistence conditions each rare strain increased; changing histidine or added glucose shifted the eventual composition. Their explanation used nonlinear growth benefits, rather than a fixed pairwise payoff matrix. This motivates testing both rarity regimes before inferring an evolutionarily stable strategy or fitting a replicator equation; it does not identify the illustrative tables below with measured payoffs. See Gore, Youk and van Oudenaarden, Figures 1–3.

An early motivation was understanding why animal contests can involve restraint even when escalation might win an individual encounter. Maynard Smith and Price's ^[ref] introduced the evolutionarily stable strategy in this setting. Taylor and Jonker's ^[ref] then studied explicit continuous and discrete dynamics. Static resistance to invasion and the paths populations follow are related questions, but they require separate definitions.

If a population $N$ has per-capita birth rate $b{e}^{-aN}$ and per-capita death rate $d$, thenWhen $b>d$, its positive equilibrium isand the derivative of the scalar vector field there is $-ad{N}^{\ast }<0$, so the equilibrium is locally asymptotically stable.

A feedback delay means a controller reacts to an earlier plant output. An accurate measurement can still be stale: when the plant output has already crossed the reference signal, a controller using the earlier value may continue pushing it away from the target. In a pure constant-delay model, $y_m(t)=y(t-\tau )$ with $\tau \ge 0$. This isolates timing from measurement noise; it does not describe every sampling or communication mechanism. ^[ref]

For the ideal scalar controlled plant $\dot{x}=u$, a zero reference signal, no control disturbance, and proportional control with constant feedback gain $k>0$, the equations areWith $\tau =0$, this is an ordinary differential equation with exponential decay $x(t)=x(0)e^{-kt}$. With $\tau >0$, it is a delay differential equation, requiring an initial history of delayed feedback. The delay stability of proportional integrator control depends on the product $k\tau$: even this simple controlled plant loses asymptotic stability at $k\tau =\pi /2$.

Substitution $s=i\omega$ into an algebraic transfer function can produce finite amplitude and phase even when the causal system has unstable poles. It then describes a sinusoidal particular solution whose growing homogeneous response is not generically absent. A physically settled frequency response requires transient asymptotic stability; for an unstable causal transfer function, its Laplace region of convergence need not include the imaginary axis.

Here is an explicitly synthetic mathematical model of a warming disturbance and its compensatory cooling. It is a teaching analogy for thermoregulation, not a measured human heat budget. Write $x=(T-T_r)/\Delta T$, where $T$ is a lumped internal temperature, $T_r$ is a fixed reference and $\Delta T>0$ is a chosen temperature scale. Let $d$ be an additional heating rate and $u$ an additional cooling rate, both scaled to the same reference rate. All three are dimensionless variables, while $\tau >0$ is a time constant in the chosen time units. The ordinary differential equation isThe term $-x$ idealizes increased passive heat transfer with increased internal temperature, around the baseline. The term $-u$ opposes warming through proportional control with feedback gain $k$. This local signed model does not treat a negative $u$ as negative sweating: it would represent decreased cooling relative to baseline. Changes of the environment and multiple body compartments are omitted.

For a constant disturbance $d=D$ and initial condition $x(0)=0$, the exponential function solution isSubstitution checks the ordinary differential equation directly. Increasing $k$ reduces the lasting offset and, in this instantaneous model, shortens the response time constant to $\tau /(1+k)$. It does not make $x_{\ast }$ zero for a nonzero disturbance and finite $k$: a persistent cooling command requires a persistent displacement under proportional control. At the equilibrium point, the extra heating is balanced by passive loss $x_{\ast }$ plus active cooling $u_{\ast }$. Zero change of temperature means balanced rates, rather than zero rates, consistent with conservation of energy.

The first diagram uses $D=1$, $\tau =4$, and a pulse from $t=2$ to $t=10$, comparing $k=0$ with $k=3$. The latter approaches $x_{\ast }=1/4$ during the pulse, compared with $1$ without active feedback control. After the pulse ends, each displacement decays as $x(10)e^{-(1+k)(t-10)/\tau }$. These are computed curves, not physiological measurements. The $k=0$ curve still has passive loss; it is not a claim that an unregulated body warms without limit.

Real physiological effectors take time to respond. To isolate that issue, set $\tau =1$ and replace the instantaneous command by another ordinary differential equation:Here the disturbance remains $1$ from $t=0$ onward and the effector time constant is also $1$. The equilibrium point is still $(x_{\ast },u_{\ast })=(1/5,4/5)$, but the characteristic polynomial is $(\lambda +1{)}^2+4$. Its eigenvalues $-1\pm 2i$ give decaying oscillations and asymptotic stability. The exact response isIn particular, $\dot{x}=e^{-t}\cos (2t)$, so the first maximum occurs at $t=\pi /4$, with $x=1/5+(2/5)e^{-\pi /4}\simeq 0.3824$. The response exceeds its eventual displacement while remaining above its baseline. The return has the same negative sign as before; the extra state changes the timing. A finite effector time constant is distinct from the fixed time lag in the delayed negative feedback model.

For a rigid body with three distinct positive principal moments of inertia $0<I_1<I_2<I_3$ and zero external torque about its centre of mass, steady rotation about the first or third principal axis is Lyapunov stable in angular velocity, while rotation about the intermediate principal axis is unstable. Here steady rotation means $\mathbf{\omega }=\Omega {\hat{e}}_i$ in the principal body frame, with constant $\Omega \ne 0$. A small sideways perturbation remains small about the extreme principal moments of inertia, but generically grows about the intermediate one. Stable does not mean that the wobble decays: without damping, there is no asymptotic stability. Coincident principal moments of inertia and a stationary body are excluded.^[ref]

No external torque is needed for a body to tumble. Conservation of angular momentum fixes $\mathbf{L}$ in the fixed space frame, but the body can turn relative to $\mathbf{L}$, so its angular velocity components in the principal body frame change. The torque-free rigid-body invariants constrain this motion. A supported gyroscope with gravitational torque has different governing assumptions.

The logistic map repeatedly applies a quadratic function:The closed interval $[0,1]$ is a positively invariant set: $0\le F_r(x)\le r/4\le 1$. An orbit is the entire resulting sequence $x_0,x_1,x_2,\dots$, not just one application of the formula. For example, with $r=2$ and $x_0=0.2$, the first states are $0.2,0.32,0.4352,0.49160192,\dots$, approaching the fixed point $1/2$.

As a simple model of population dynamics, $x_n$ is a scaled abundance in generation $n$. When abundance is small, $x_{n+1}/x_n=r(1-x_n)$ is close to $r$; increasing abundance reduces this per-capita multiplier. The falling side of the quadratic function represents sufficiently severe crowding reducing the next generation. The scale $x=1$ is not generally the carrying capacity: the positive fixed point is $1-1/r$. The logistic map is a discrete-generation model; the continuous-time logistic growth equation has a different evolution rule and does not produce its period-doubling cascade.

The practical question is how abundance behaves after the initial adjustment: does it settle, alternate, or continue varying? The same formula permits all these possibilities. The following figure shows three orbits, their cobweb plots, and a bifurcation diagram. In a cobweb plot, moving vertically to $y=F_r(x)$ calculates the next state, and moving horizontally to $y=x$ makes that value the input for the next step.

The fixed points solve $F_r(x)=x$:The second fixed point is in $[0,1]$ for $r\ge 1$ and coincides with zero at $r=1$. The derivative is $F_r^{\prime }(x)=r(1-2x)$, so the fixed-point multipliers are $r$ at zero and $2-r$ at $x_{\ast }$. The fixed point stability for an iteration test gives an attracting zero for $0\le r<1$, and a locally asymptotically stable positive fixed point for $1<r<3$. A multiplier of modulus one needs a nonlinear argument; the strict derivative test alone settles neither $r=1$ nor $r=3$.

This derivative test has a direct interpretation. A small error $e_n=x_n-x_{\ast }$ obeysAt $r=2.8$, the leading multiplier is $-0.8$: the error changes sign, explaining the alternating approach, while its magnitude eventually shrinks by approximately a factor $0.8$ per step. At $r=3.2$, the multiplier is $-1.2$, and the positive fixed point repels nearby states. However, a different outcome has asymptotic stability: the logistic map two-cycle attracts nearby orbits.

For $3<r<1+\sqrt{6}$, the attracting period-two orbit has the two valuesAt $r=3.2$ they are approximately $0.513045$ and $0.799455$, with $F_r(x_{-})=x_{+}$ and $F_r(x_{+})=x_{-}$. Their periodic-orbit multiplier is $4+2r-r^2=0.16$, so a small displacement shrinks over two steps even though the fixed point is unstable. The brief algebraic justification is to factor $F_r^2(x)-x$, remove the fixed point factors, and multiply the two derivatives using the chain rule.

At $r=3$, the positive fixed-point multiplier passes through $-1$ and a period-doubling bifurcation creates the period-two orbit. At $r=1+\sqrt{6}\approx 3.44949$, its periodic-orbit multiplier passes through $-1$ and an attracting period-four periodic orbit emerges. At $r=3.5$, the period-four values are approximately $0.382820,0.826941,0.500884,0.874997$ in that order. Further period-doubling bifurcations accumulate near $r\approx 3.56995$. Beyond that value there are also parameter windows with attracting periodic orbits, including a period-three window near $r=3.83$. Neither a crowded numerical plot nor a parameter beyond the accumulation value proves sensitive dependence on initial conditions on the attracting set.

The endpoint $r=4$ supplies an analytically transparent example. The homeomorphism $h(u)={\sin }^2(\pi u/2)$ satisfieswhere $T$ is the full tent map. Thus these maps have topological conjugacy; stretching and folding in the tent map explain sensitive dependence on initial conditions and Devaney chaos for $F_4$ on $[0,1]$. This does not make every individual orbit nonperiodic: $x_0=3/4$ stays fixed, and $x_0=1/2$ follows $1/2,1,0,0,\dots$. Sensitivity means that these exceptional orbits also have arbitrarily close initial states that eventually depart substantially.

Robert M. May's "Simple mathematical models with very complicated dynamics" (1976), Nature 261, pp. 459–467, reviewed how simple deterministic recurrence relations can pass from stable abundances through period-doubling cascades to irregular behaviour. May presented the logistic map as an especially accessible example and discussed the relevance of such models to population dynamics. The paper explicitly describes itself as a review of results found by several researchers, bringing their findings together for a broad scientific audience.^[ref]

Negative feedback closes a causal loop whose net effect opposes an initiating change. In signal transduction, activation followed by a downstream inhibition of the earlier activating stage is one example of negative feedback. Loop sign alone does not establish asymptotic stability or complete signaling adaptation; see the worked signaling feedback example.

A useful biological example is chemotaxis feedback regulation: an Escherichia coli biological cell adjusts its response to a sustained attractant instead of remaining at its immediate response. Increased CheA activity promotes CheB phosphorylation; CheB reverses receptor methylation, opposing their activation. After attractant addition, reduced activity instead favors receptor methylation by CheR, restoring activity. This is a return from receptor activity to the machinery regulating that activity. The response need not return exactly at every dose. ^[ref]

A motivating measurement is the chemotaxis FRET assay. CheY and CheZ carrying yellow and cyan fluorescent tags were observed in immobilized Escherichia coli biological cells on a coverslip in a flow chamber. Adding and removing attractant changed the yellow-to-cyan emission ratio, reporting the association of phosphorylated CheY with CheZ. This measures a molecular readout upstream of motor behavior. Comparisons with biological cells lacking CheR, CheB, or both tested the dependence on receptor modification; these perturbations also change starting activity and sensitivity, so they are not pure changes of one feedback gain. ^[ref]

The later chemotaxis feedback regulation experiment used only the Tar receptor and nonmetabolizable methylaspartate, supplied and removed by flow. Long recordings separated slow recovery from incomplete signaling adaptation; correction for bleaching addressed drift in the optical readout. Activity recovered incompletely after sufficiently large additions, with detectable imprecision already at $10\mu \mathrm{M}$ in that preparation. Restricting the receptor type and preventing consumption of the stimulus help interpret the perturbation, but limit transfer to other receptors and conditions. The readout infers CheA activity through CheY-CheZ association, rather than directly measuring every molecular step. ^[ref]

Negative feedback can also overshoot or sustain oscillations if its return arrives too late. The worked signaling feedback example gives the contrasting instantaneous case; the delayed negative feedback model below isolates the effect of timing. Loop sign describes causation, while asymptotic stability describes what happens to displacements over time.

Orbital stability asks that a disturbance remain close to an invariant orbit or symmetry family, rather than to one fixed representative. In a complex Landau amplitude equation with $\beta =0$, $\mu >0$, $\lambda >0$, the circle $|A|=\sqrt{\mu /\lambda }$ attracts radial disturbances while every phase on it remains stationary. The family is attracting as a set, but its individual points do not have individual asymptotic stability because their phase is neutral.

A critical point of a dynamical system is a point where both components of the vector field vanish. For a continuously differentiable vector field, strictly negative real parts of both eigenvalues of its Jacobian matrix imply asymptotic stability of the equilibrium. If an isolated critical point is a center of a planar dynamical system and the eigenvalues are distinct, they must be a nonzero purely imaginary conjugate pair. Purely imaginary eigenvalues alone do not in general prove nonlinear stability; the nonlinear ordinary differential equation system must still be examined.

Here the critical points are $(0,0)$ and $(\pm 1,0)$, with Jacobian $\left(\begin{array}{cc}0& 1\\ 1-3x^2& 0\end{array}\right)$. At zero its eigenvalues are $1,-1$, giving an unstable saddle equilibrium. The unstable direction is $(1,1)$ and the stable direction is $(1,-1)$. Thus trajectories leaving the origin are tangent to $y=x$, and those returning are tangent to $y=-x$, with both opposite rays included.

An exact first integral isThe two homoclinic loops belong to $H=0$:They confirm the incoming and outgoing tangent directions. For example the right loop is $x=\sqrt{2}\mathrm{sech}(t-t_0)$, $y=-\sqrt{2}\mathrm{sech}(t-t_0)\tanh (t-t_0)$; changing both signs gives the left loop.

At $(1,0)$ put $x=1+X$, $y=Y$. The exact equations are $X^{\prime }=Y$, $Y^{\prime }=-2X-3X^2-X^3$, so the linearization of a dynamical system gives $X^{\prime \prime }+2X=0$. Its nonconstant solutions haveThe same linear classification holds at $(-1,0)$. For the nonlinear ordinary differential equation system, $H$ has strict local minima $-1/4$ at both points. Levels $-1/4<H<0$ are two families of smooth closed ovals, proving that both are stable centers, rather than asymptotically stable equilibria. These give actual periodic orbits of the nonlinear ordinary differential equation system, with small-amplitude periods tending to $\sqrt{2}\pi$.

The Bendixson-Dulac theorem states that if $(P,Q)$ is a continuously differentiable planar vector field on a simply connected open region $D$, and a continuously differentiable Dulac function $B$ makeshave one weak sign throughout $D$, without vanishing identically on any nonempty open subset, then there is no nonconstant periodic orbit contained in $D$. To see why, a hypothetical simple periodic orbit bounds a region in $D$. The divergence theorem gives the integral of this expression as the outward flux of $B(P,Q)$ across the orbit. The flux is zero because the vector field is tangent to its orbit, whereas the one-sign area integral is nonzero, a contradiction. The usual stricter version assumes a strictly positive or strictly negative expression everywhere.

For the printed polynomial system, the origin is an equilibrium andThus its linearization is a centre equilibrium. For the full system take $B=1$. The divergence isOn the open disk $x^2+y^2<3$ this is nonpositive, and is negative wherever $x\ne 0$. Its zero set inside the disk is a line, so it cannot vanish on any open subset. The Bendixson-Dulac theorem therefore provesThis also exposes a terminology problem in the source: a nonlinear centre equilibrium has surrounding closed orbits, so the printed claim of a centre cannot hold in that sense. The imaginary eigenvalues establish only a centre of the linearized system. In fact the full equilibrium is an asymptotically stable weak focus, as the following direct check shows.

A quartic Lyapunov correction for a weak focus makes the nonlinear damping visible. Put $r^2=x^2+y^2$ and consider the Lyapunov functionIt is positive definite sufficiently close to zero. Differentiating along the polynomial vector field givesFor clarity, the quartic terms come from $-x^4$ in the derivative of $r^2/2$, and from applying $y{\partial }_x-x{\partial }_y$ to the quartic correction. Their sum is $-3x^4/8-3x^2{y}^2/4-3y^4/8$; all other terms have degree at least six. Therefore $\dot{V}<0$ in a sufficiently small punctured neighbourhood, proving asymptotic stability by Second Lyapunov theorem. Finallyso nonconstant nearby orbits turn clockwise as they converge. The equilibrium is a weak focus despite the centre in its linearization. This interpretation resolves both the local dynamics and the absence of closed orbits without changing the printed coefficients.

Where $y\ne \beta$, taking the quotient of the two ordinary differential equations givesChoose the first integralThe chain rule checks conservation even at points excluded in taking that quotient:Thus each trajectory is contained in a connected component of a level set of $h$. At $(0,\beta )$ the gradient vanishes and the Hessian matrix is the identity, so $h$ has a strict local minimum there. To make the closed-trajectory argument explicit, put $z=y-\beta$ and $0<E<1/4$. The bounded component of $h=E$ isThe two arcs join at $(\pm r_E,0)$ to form a simple closed curve that is a regular level set. The gradient cannot vanish on it: the only critical point of a dynamical system in $|x|<1$ is the centre, whose level is $0$.

The vector field is tangent and nowhere zero on this compact set, so its speed has a positive minimum. It therefore traverses the curve in finite time and returns to its initial point, giving a periodic orbit. These ovals shrink to the centre as $E\downarrow 0$. This is the planar first integral criterion for a centre and proves the required nonlinear assertion; it also gives Lyapunov stability without asymptotic stability.

At an equilibrium point, $x=-y^2$ and $x^2=a^{2}y$, so $y(y^3-a^2)=0$. With $d=|a{|}^{2/3}>0$, the two equilibrium points and their Jacobian matrix areAt $(-d^2,d)$, $\det J=-3a^2<0$. The trace-determinant stability criterion therefore gives an unstable saddle equilibrium. At the origin, the eigenvalues are $\lambda =\pm i|a|$: its linearization is a center equilibrium. The linearization stability theorem does not classify this nonhyperbolic equilibrium of the nonlinear system.

Nonlinear classification. In fact the answer depends on $|a|$:Here is an explicit derivation, rather than an inference from the imaginary eigenvalues. When $a^2=1$, set $L=1-x+y$. Near the origin $L>0$, and the first integralhasIts Hessian matrix at the origin is the identity. Small level curves are closed ovals on which the vector field is nonzero, proving the center equilibrium classification and Lyapunov stability.

For $a^2\ne 1$, a higher-order Lyapunov function calculation settles the issue. Put $k=|a|$, $U=x$, $V=ky$, $\tau =kt$, $p=k^{-1}$ and $q=k^{-2}$. ThenWrite $R^2=U^2+V^2$ and introduce the differential operatorsSuccessively cancel the terms of degrees three, four and five in the derivative of $R^2+H_3+H_4+H_5$, using these homogeneous polynomials:Direct differentiation verifies $\mathcal{L}{H}_3=-\mathcal{N}{R}^2$, $\mathcal{L}{H}_4=-\mathcal{N}{H}_3$ and $\mathcal{L}{H}_5=-\mathcal{N}{H}_4$. The remaining degree-six term has circular averageFor example, the only nonzero averages of degree-six monomials needed here are $⟨U^6⟩=⟨V^6⟩=5/16$ and $⟨U^4{V}^2⟩=⟨U^2{V}^4⟩=1/16$. Since $\mathcal{L}={\partial }_{\theta }$ in polar coordinates, the zero-average part can be cancelled by a degree-six homogeneous polynomial $H_6$ satisfyingIndeed its restriction to the unit circle is obtained by integrating a zero-average trigonometric polynomial; its Fourier modes have orders $2,4,6$, so multiplication by $R^6$ gives a polynomial. Consequently the positive Lyapunov function $W=R^2+H_3+H_4+H_5+H_6$ satisfiesFor $k>1$ this is strictly negative near the origin, giving asymptotic stability. For $k<1$ it is strictly positive, and $W_{\tau }\ge C{W}^3$ in a sufficiently small neighbourhood; every nonzero starting point must eventually leave that neighbourhood. Finally ${\theta }_{\tau }=1+O(R)>0$, so the nearby trajectories rotate: these are weak foci, with the stability stated above. Merely calling the nonlinear origin a center for every $a\ne 0$ would miss this distinction.