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A resident trait $x$ occupies fraction $1-m/c(x)$ when its colonization rate exceeds fire extinction. With net competitive replacement given by ability difference $B(y)-B(x)$, a rare mutant has the displayed per-capita growth. The selection gradient is $g(x)=m{c}^{\prime }(x)/c(x)+[1-m/c(x)]B^{\prime }(x)$. A singular trait with negative mutant-fitness curvature is a strict local evolutionarily stable strategy. Global strict uninvadability requires that $mc(y)+[c(x)-m]B(y)$ be smaller than its resident value for every other trait. Monotonicity alone does not specify the shape or existence of an optimum.

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.

Hawk fights for a resource of value $V$, while Dove retreats from Hawk and shares with another Dove. Hawk–Hawk competition incurs damage cost $D$. The expected payoff difference between Hawk probabilities $p,r$ against population probability $q$ is $(p-r)(V-Dq)/2$. For positive $V,D$, pure Hawk is an evolutionarily stable strategy when $V>D$. At $V=D$, it remains stable because a mutant $p<1$ loses $V(1-p{)}^2/2$ against its own population. When $0<V<D$, the stable mixture has Hawk probability $p^{\ast }=V/D$: all first-condition payoffs tie, and its second-condition advantage is $D(p-p^{\ast }{)}^2/2>0$. For nonpositive resource values, pure Dove is stable if $V<0$ or if $V=0<D$. At $V=D=0$ all payoffs vanish and there is no evolutionarily stable strategy.

Under cooperative initial memories, matches involving only Tit for tat and always cooperate contain only cooperation, so all four payoffs equal $R$. The two-condition criterion for evolutionary stability has both differences zero. Thus Tit for tat is not an evolutionarily stable strategy against a strategy space containing unconditional cooperation, even if it is a Nash equilibrium. This counterexample is valid for finite horizons and discounting as well as infinite averages.

Let $a,b,e$ be fractions of patches occupied by the two competitors and empty, with $a+b+e=1$. Assume well-mixed propagule dispersal, equal independent fire extinction rate $m>0$, colonization constants $c_A>c_B>0$, and net replacement advantage $q=k_B-k_A>0$ for the poorer colonizer. The Levins metapopulation model with replacement isReplacement fluxes $k_{B}ab$ and $k_{A}ab$ cancel in the total occupancy equation. At every face the outward flux is zero or points into the patch probability simplex, so these equations preserve valid fractions. The constants include both propagule arrival and establishment probability; the model does not require unequal fire susceptibility.

Put $\Delta c=c_A-c_B$. Solving the two occupied-patch balances and their sum constraint givesThus strict coexistence is equivalent toToo little disturbance lets the better competitor exclude the colonizer; too much disturbance favors colonization or destroys occupancy. A positive equilibrium automatically has $m<c_A$. If the poorer colonizer cannot survive alone, coexistence can nevertheless be possible because it repeatedly replaces established patches of the better colonizer.

These bounds also have an invasion interpretation. The $A$-only equilibrium is $a=1-m/c_A$ when $c_A>m$; the growth rate of rare $B$ there is $q-m(q+\Delta c)/c_A$, positive at the upper-bound condition above. At a feasible $B$-only equilibrium the rare-$A$ growth is $-q+m(q+\Delta c)/c_B$, giving the lower bound. At coexistence the occupied-patch Jacobian matrix has trace $-c_A{a}_{\ast }-c_B{b}_{\ast }<0$ and determinant $a_{\ast }{b}_{\ast }q(q+\Delta c)>0$, proving local asymptotic stability. The Bendixson-Dulac criterion with multiplier $1/(ab)$ gives divergence $-c_A/b-c_B/a<0$ in the interior, excluding periodic competition cycles.

To specify selection, a monotone tradeoff alone is insufficient: one must say how two different trait values determine replacement. A natural model uses a trait $x$ with increasing competitive ability $B(x)$ and decreasing colonization $c(x)$, and net replacement $B(y)-B(x)$ by a mutant $y$ against a resident $x$. The resident metapopulation occupies $1-m/c(x)$, requiring $c(x)>m$. The rare-mutant per-capita growth, or invasion fitness, isIt is zero at $y=x$. Differentiating with respect to the mutant trait gives the selective pressureHigher colonization is valuable in empty habitat; higher competitive ability is valuable where the resident already occupies patches. This is colonization-competition tradeoff invasion fitness.

At an interior candidate for a local evolutionarily stable strategy, $g(x_{\ast })=0$. A strict, nondegenerate local evolutionarily stable strategy follows ifsince nearby distinct mutants then have negative invasion fitness. For a strict global evolutionarily stable strategy, every feasible distinct mutant must satisfyThis is the supporting-line condition on the tradeoff curve. At a trait boundary replace the vanishing-gradient condition by the appropriate one-sided inequality. Equality against another mutant, or zero local curvature, needs the second condition of evolutionary stability or higher-order terms; the strict tests above are sufficient and are the usual nondegenerate conditions, not a claim that every tie is unstable.

For evolutionary approach to a singular point, the selection gradient must also point toward it, locally $g^{\prime }(x_{\ast })<0$. This is logically distinct from testing mutant fitness. In this particular additive-ability model with $c^{\prime }<0<B^{\prime }$, negative mutant curvature implies negative $g^{\prime }$ becauseIf replacement instead has a general kernel $\kappa (y,x)$, replace $B(y)-B(x)$ in $f$ by that kernel and differentiate it; monotonicity does not justify importing the displayed additive formula without that modeling assumption. The tradeoff shape, disturbance rate and replacement law together determine selection and local or global uninvadability.

In a two-player symmetric finite game, represent mixed strategies by probability vectors and let $e(x,y)=x^{T}Ay$ be the payoff to a player using $x$ against one using $y$. A resident strategy $x^{\ast }$ is an evolutionarily stable strategy if every distinct mutant strategy $y$ has an invasion barrier ${\epsilon }_0(y)>0$ such that, for $0<\epsilon <{\epsilon }_0(y)$,Thus when mutants are sufficiently rare, the resident has strictly larger fitness against the population mixture. The barrier may depend on the mutant; a common barrier is not part of this definition.

For fixed $y\ne x^{\ast }$ putBy bilinearity the resident-minus-mutant fitness difference isIf this is positive for all sufficiently small positive $\epsilon$, taking $\epsilon \downarrow 0$ gives $a_y\ge 0$. If $a_y=0$, strict positivity requires $b_y>0$. Conversely, if $a_y>0$, choose $0<{\epsilon }_0\le a_y/[2(a_y+|b_y|)]$; then the difference is at least $a_y-\epsilon (a_y+|b_y|)>a_y/2>0$. If $a_y=0$ and $b_y>0$, it is $\epsilon b_y>0$ for every $0<\epsilon <1$. This proves the two-condition criterion for evolutionary stability: for every $y\ne x^{\ast }$, eitherorThe weak first comparison is the symmetric Nash equilibrium condition; the second excludes neutrally competitive mutants when that comparison is an equality.

For the Hawk-Dove game, let $x^{\ast }=(1/2,1/2)$ and $y=(p,1-p)$, $0\le p\le 1$. Against the resident, both pure strategies give $1/2$, soDirect multiplication of the payoff matrix givesConsequentlyThe equality case of the two-condition criterion for evolutionary stability therefore applies to every distinct mutant. More explicitly, the fitness difference in a mixed population is $2\epsilon (p-1/2{)}^2$, positive for every $0<\epsilon <1$. Hence $\boxed{(1/2,1/2)\text{ is an evolutionarily stable strategy}}$, with the common invasion barrier one.

In the usual positive-parameter case $0<V<D$, set $p^{\ast }=V/D\in (0,1)$. Every mutant ties the resident against the resident population, because $V-D{p}^{\ast }=0$. Against a mutant population, however,The second condition of the two-condition criterion for evolutionary stability holds, soIndeed its advantage at invasion frequency $ϵ$ is exactly $ϵD(p^{\ast }-p{)}^2/2$. The interior frequency balances the two pure-strategy payoffs in the Hawk-Dove game.

For completeness, if $V<D$ but $V<0$, pure Dove is an evolutionarily stable strategy by the strict inequality $e(0,0)-e(p,0)=-pV/2>0$ for every mutant $p>0$. If $V=0<D$, that difference is zero, but $e(0,p)-e(p,p)=D{p}^2/2>0$, so Dove again satisfies the second condition. Thus for the remaining allowed values $\boxed{x^{\ast }=(0,1)\text{ when }V\le 0<D\text{ or }V<D\le 0}$; equivalently, Dove works whenever $V<D$ and $V\le 0$.

For $V=D>0$, choose pure Hawk. Every mutant $p<1$ ties it against a resident Hawk population, so the second condition must be checked. The Hawk-Dove game formula givesConsequentlyThe actual invasion advantage is $ϵV(1-p{)}^2/2>0$, despite the tie at invasion frequency zero.

If the equal parameters are negative, pure Dove is instead an evolutionarily stable strategy: for a mutant $p>0$, $e(0,0)-e(p,0)=-pV/2>0$. If $V=D=0$, every payoff is zero and no resident has a strict advantage over any mutant. Thus the full equal-parameter answer is Hawk for $V=D>0$, Dove for $V=D<0$, and no ESS for $V=D=0$.

The payoff $e(x,z)$ is the expected fitness of an individual using strategy $x$ against a population using strategy $z$. The resident $x^{\ast }$ is initially common; the mutant $y$ is introduced at small frequency $ϵ$. Convexity of the strategy set makes their population mixture admissible. If residents have higher expected fitness in every sufficiently small invasion by a fixed mutant, selection works against that mutant: this is an evolutionarily stable strategy. The allowed upper bound on $ϵ$ may depend on $y$.

For expected payoffs of mixed strategies, $e(x,z)$ is affine in the population strategy $z$; in a finite matrix game it is $x^{T}Az$. This property is the basis of the two-condition criterion for evolutionary stability. For fixed $y\ne x^{\ast }$, putAffineness gives the exact invasion differenceIf this is positive for all sufficiently small $ϵ>0$, taking $ϵ\downarrow 0$ gives $A_y\ge 0$. If $A_y=0$, the difference is $ϵB_y$, so necessarily $B_y>0$. Conversely, if $A_y>0$, it remains positive for sufficiently small $ϵ$: for example take $ϵ<A_y/(A_y+|B_y|)$. If $A_y=0$ and $B_y>0$, it is positive for every $ϵ>0$. Applying this argument separately to every $y$ provesThus an evolutionarily stable strategy is a symmetric Nash equilibrium with an additional condition against tied mutants. For a general nonlinear interaction function on a convex set, affineness would be an additional necessary hypothesis for this particular equivalence; here it follows from the mixed-strategy expected-payoff interpretation. For example, on $X=[0,1]$ take $e(x,z)=xz(x/2-z)$ and resident $x^{\ast }=0$. For every $y>0$, $A_y=0$ and $B_y=y^3/2>0$, but at population $ϵy$ the mutant earns $ϵy^3(1/2-ϵ)>0$ while the resident earns zero. Thus the two conditions alone would not suffice for an arbitrary nonlinear $e$; the expected-payoff assumption is used explicitly.

For the Hawk-Dove game, let $p$ be an individual's Hawk probability and $q$ the population's Hawk probability. The given matrix yieldsWe use this formula in each case. In the usual interpretation, resource value $V$ and damage cost $D$ are positive. The parts below also specify the boundary or negative-value alternatives because positivity is not written explicitly in the PDF.

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.

For mixed-strategy expected payoffs put $A_y=e(x^{\ast },x^{\ast })-e(y,x^{\ast })$ and $B_y=e(x^{\ast },y)-e(y,y)$. The invasion payoff difference is exactly $(1-ϵ)A_y+ϵB_y$. Positivity for all sufficiently small positive $ϵ$ forces $A_y\ge 0$ by taking the limit. When $A_y=0$, it forces $B_y>0$. Conversely $A_y>0$ makes the difference positive for sufficiently small frequency, and $A_y=0,B_y>0$ makes it positive for every positive frequency. Apply this separately to every distinct mutant. This proves the equivalence with an evolutionarily stable strategy; affineness in population composition is essential to the formula.