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The replicator equation says that a strategy grows in frequency when its expected payoff exceeds the population's current mean payoff. It is an ordinary differential equation for a composition, rather than for total population size. The subtraction of the mean is what makes relative success matter: even a type growing in absolute number can shrink as a fraction if other types grow faster.

Let $x_i(t)$ be the frequency of pure strategy $i$, with $n$ available pure strategies andIn a symmetric bimatrix game, let the payoff matrix $A$ have entry $a_{ij}$ equal to the payoff to strategy $i$ against $j$. The other player's payoff in that encounter is $a_{ji}$: symmetry between player roles does not require $A=A^T$. Uniform random encounters give the expected payoffs and their population meanAfter choosing the time unit so that the growth-rate scale is one, the continuous-time replicator equation isHere a payoff is measured on a scale proportional to net growth per unit time, not merely as an arbitrary preference ranking. More generally one can specify locally Lipschitz frequency-dependent payoffs $f_i(x)$ directly; the same equation with $\bar{f}={\sum }_i{x}_i{f}_i$ then defines a unique frequency trajectory. The linear pairwise model and its reproductive interpretation are given in Taylor and Jonker (1978), sections 2–3; the formulas here also follow by the following direct normalization.

Suppose the number $N_i$ of type $i$ obeys ${\dot{N}}_i=N_i[b(t)+f_i(x)]$, where the baseline $b(t)$ is identical for all types. Set $N={\sum }_i{N}_i>0$ and $x_i=N_i/N$. The quotient rule givesThus the common baseline cancels. Adding the same constant to every entry of $A$ leaves the replicator equation unchanged; multiplying $A$ by a positive constant changes only the speed along trajectories. The same payoff matrix can therefore describe frequency evolution at different growth-rate baselines and time scales.

The probability simplex is invariant in forward time. On ${\sum }_i{x}_i=1$,Normalization alone would not exclude negative frequencies. Nonnegativity follows as well: along a solution,An absent type remains absent, and a positive frequency remains positive at every finite time. For the fixed payoff matrix model, bounded rates on the compact probability simplex give global forward solutions. Frequencies can still tend to zero as $t\to \infty$. Each face is invariant, so this model has no spontaneous introduction of new pure strategies; mutation or outside introductions would require additional terms. For two present types the logarithm of their frequency ratio satisfies $\frac{d}{dt}\log (x_i/x_j)=f_i-f_j$, directly expressing the accumulated payoff advantage.

For two pure strategies, write the composition as $(x,1-x)$, where $x$ is the frequency of strategy 1, and letTheir payoff difference and the resulting scalar ordinary differential equation areIn the interior $0<x<1$, the sign of the payoff difference determines the direction of motion. The endpoints are always equilibrium points, and an additional interior equilibrium point exists precisely when the affine payoff difference has a zero strictly between them. At a simple interior zero $x_{\ast }$, $F^{\prime }(x_{\ast })=x_{\ast }(1-x_{\ast })\Delta f^{\prime }(x_{\ast })$: a negative value gives asymptotic stability, a positive value gives instability. At the endpoints $F^{\prime }(0)=b-d$ and $F^{\prime }(1)=c-a$, with stability understood relative to $[0,1]$. A zero derivative requires examining the actual flow rather than this strict sign test.

For an original coordination game, imagine two conventions that reward matching, with convention 1 giving the larger matching reward. ChooseThe crossing is $x_{\ast }=2/5$. Below it, convention 1 loses frequency; above it, convention 1 gains frequency. The endpoint equilibrium points $0$ and $1$ have one-sided asymptotic stability, while $2/5$ is unstable: the corresponding derivatives are $-2,-3$ and $6/5$, respectively. Every initial $0<x(0)<2/5$ tends to $0$, every initial $2/5<x(0)<1$ tends to $1$, and exactly $x(0)=2/5$ stays there. Even the convention with the larger matching reward can disappear if it starts below the threshold. This is bistability, not selection of a universally best convention.

The compositions giving a symmetric Nash equilibrium in this coordination game are the two pure conventions and the mixed composition $(2/5,3/5)$. Both pure conventions are evolutionarily stable strategies because a mutant convention has strictly lower payoff against the pure resident. The mixed symmetric Nash equilibrium is not an evolutionarily stable strategy: against that resident both conventions tie, but after a pure convention-1 mutant is introduced its payoff advantage is $5(x-2/5)>0$. Equal payoffs at a symmetric Nash equilibrium need not provide resistance to invasion or asymptotic stability.

For an original coexistence model, choose payoffs making each convention advantageous when sufficiently rare:Here $x_{\ast }=3/4$ attracts every initial composition with both types present: $\dot{x}>0$ below it and $\dot{x}<0$ above it. The interior derivative is $F^{\prime }(3/4)=-3/4$, while $F^{\prime }(0)=3$ and $F^{\prime }(1)=1$ make both endpoints unstable to introduction of the other type. The endpoints nevertheless remain stationary if the other type is exactly absent; the attraction statement excludes those boundary initial states.

The sole symmetric Nash equilibrium in this coexistence model is $p=(3/4,1/4)$. It is also an evolutionarily stable strategy. For any different mixed strategy $q=(y,1-y)$, the first comparison in the two-condition criterion for evolutionary stability ties because $Ap=(3/2,3/2)$. Its second comparison is strictly positive:Thus restoring frequency feedback can protect a mixture even though all present types earn equal payoffs at the mixture. Coexistence here means different individuals using different pure strategies; it does not require every individual to randomize with a mixed strategy.

Frequency evolution need not increase the mean payoff. In the coexistence example $\bar{f}=5x-4x^2$, whose maximum occurs at $x=5/8$, whereas the attracting composition is $3/4$. For $5/8<x<3/4$, the frequency $x$ increases while $\frac{d}{dt}\bar{f}=(5-8x)\dot{x}<0$. Individual payoff advantages drive the equation; maximizing collective mean payoff is a different optimization problem.

More generally, a rest point only equalizes payoffs among present types: $x_i>0$ implies $(Ax{)}_i=x^{T}Ax$. A symmetric Nash equilibrium additionally requires $(Ax{)}_i\le x^{T}Ax$ for every absent type. All vertices are rest points even when an absent type could invade. For finite symmetric bimatrix games, a rest point with Lyapunov stability on the full probability simplex is a symmetric Nash equilibrium; an interior trajectory that converges has a symmetric Nash equilibrium as its limit. These are conditional results, not a convergence theorem for every trajectory. See Cressman and Tao (2014), Theorem 1.

The rock paper scissors payoff matrix gives an explicit obstruction to universal convergence:Since $A^T=-A$, $x^{T}Ax=0$, and in the interior $\frac{d}{dt}\log (x_1{x}_2{x}_3)={\sum }_i(Ax{)}_i=0$. Thus $x_1{x}_2{x}_3$ is conserved. Its maximum on the probability simplex occurs only at $(1/3,1/3,1/3)$, the unique symmetric Nash equilibrium, so an interior trajectory starting elsewhere cannot converge to that composition. Every other interior product level is a closed curve without a rest point, giving a periodic orbit. The existence of these cycles is also discussed in Cressman and Tao (2014), Example 1. The useful lesson is to inspect the specified payoff feedback and dynamical system, rather than infer convergence from the existence of an evolutionarily stable strategy or a Nash equilibrium.

An autonomous differential equation in three variables. Its linear planar rotation couples to a nonlinear vertical variable through $xz$. Depending on parameters, its solutions can approach an equilibrium point, a periodic orbit, or a more complicated attractor. The specialization $b=a$ is used in the 2011 Oxford combined paper.

A Smale horseshoe is a compact hyperbolic invariant set of a smooth map with expanding and contracting directions and at least two strips that stretch across a common rectangle. Iterating the strips and their inverse images produces a Cantor-set product on which the dynamics are conjugate to a full shift. It supplies infinitely many saddle periodic orbits and chaotic itineraries; hyperbolicity alone does not make it an attracting invariant set. The geometric construction explains why phase-dependent reinjection near a Shilnikov bifurcation can create chaos even when the associated symmetric amplitude equations have no such mechanism.

In a subcritical Hopf bifurcation, the small periodic orbit repels within the critical centre manifold and lies on the side where the equilibrium point attracts within that manifold. With the critical eigenvalue real part increasing through zero, the positive cubic sign gives the exact radial Hopf normal formFor $\mu <0$, the nonzero radius is $r_{\ast }=\sqrt{-\mu }$. Radii below $r_{\ast }$ shrink to zero, whereas radii above it grow. The radial derivative at the cycle is $\mu +3r_{\ast }^2=-2\mu >0$, proving that the periodic orbit repels. For example, $\mu =-0.04$ gives an unstable circle of radius $0.2$: a radius of $0.1$ decays while one of $0.3$ grows. The unstable circle shrinks into zero as $\mu \uparrow 0$; at $\mu =0$, $\dot{r}=r^3$ already drives every small positive radius outward.

For $\mu >0$ there is no nonzero constant radius in this cubic model, and zero repels. In a general dynamical system, this says only that the nearby orbits leave the local neighbourhood. It does not guarantee a distant attracting limit cycle, a jump to a particular amplitude, or hysteresis. Additional nonlinear or global information is needed; the quintic radial Hopf equation illustrates one possible stabilizing extension. Reversing the parameter orientation reverses which numerical side contains the branch, so the defining distinction between supercritical Hopf bifurcation and subcritical Hopf bifurcation is their reduced stability and their relation to the equilibrium point, rather than the bare sign of an arbitrarily named parameter.

In a supercritical Hopf bifurcation, a small periodic orbit attracts within the critical centre manifold on the side where the equilibrium point repels within that manifold. Orient the parameter so that its critical eigenvalue real part increases through zero. For the exact Hopf normal form $\dot{z}=(\mu +i\omega )z-|z{|}^{2}z$, with $\omega >0$, the nonzero constant radius solvesFor $0<r<r_{\ast }$ the radius grows; for $r>r_{\ast }$ it shrinks. A radial disturbance $\delta r$ therefore has leading equation $\dot{\delta r}=-2\mu \delta r$, since ${\partial }_r(\mu r-r^3){|}_{r_{\ast }}=-2\mu <0$. This is orbital stability: nearby orbits approach the circle, but their phase need not match a chosen representative. The exact angular frequency is $\omega$, so the period is $2\pi /\omega$ and the frequency is $\omega /(2\pi )$.

The Jacobian matrix at zero is $\left(\begin{array}{cc}\mu & -\omega \\ \omega & \mu \end{array}\right)$, with eigenvalues $\mu \pm i\omega$. Zero attracts for $\mu <0$ and repels for $\mu >0$. Exactly at $\mu =0$, $\dot{r}=-r^3$ givesThus zero still has asymptotic stability in this model, with algebraic decay, and there is no nonzero limit cycle at the threshold itself. As $\mu \downarrow 0$ on the cycle branch, both its radius $\sqrt{\mu }$ and its radial attraction rate $2\mu$ vanish. Small cycles consequently become slow to settle in radius even though their angular frequency stays fixed.

// Target: dynamical-systems.bigb

A periodic orbit of a differentiable interval map is superstable when its periodic-orbit multiplier vanishes. A cycle passing through a critical point has zero multiplier because the derivative product includes $f^{\prime }(c)=0$. For the nonconstant quadratic family $f_{\mu }(x)=1-\mu x^2$, a superstable $2^n$ cycle satisfies $f_{\mu }^{2^n}(0)=0$ and does not have a smaller critical period.

// Target: dynamical-systems.bigb

A fixed point of the $n$th iteration of a map of a subshift of finite type is determined uniquely by its length-$n$ block with a permitted closing transition. Summing the counts of length-$n$ closed walks over their starting symbols gives the displayed trace. This counts points whose least period divides $n$. If $P_n$ counts points of least period exactly $n$, then $\mathrm{tr}(A^n)={\sum }_{d\mid n}{P}_d$, and Möbius inversion gives $P_n={\sum }_{d\mid n}{\mu }_{\mathrm{M}}(n/d)\mathrm{tr}(A^d)$. The number of distinct least-period-$n$ periodic orbits is $P_n/n$.