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Defineso $S=N(1-\theta )$ and $I=N\theta$. Part (a) already givesUsing the quotient rule,and hence the reduced Plant SI model with logistic total population is

For every $N(0)>0$, logistic growth gives $N(t)\to 1$. The asymptotic prevalence equation is thereforeIf $\beta <1$, its only nonnegative equilibrium is $\theta =0$, which attracts all prevalences. If $\beta >1$, $heta=0$ is unstable and the stable value is ${\theta }_{\ast }=1-1/\beta$. Combining these values with $N_{\ast }=1$ recovers exactly the equilibria and outcomes in part (b).

Insertinto the continuity equation. Its zeroth-order part is the homogeneous cosmological continuity equationSince $\bar{\rho }$ is spatially homogeneous, the terms of first order in $ϵ$ giveOn the other hand, the quotient rule and the background equation implySubstitution of the first-order equation yields the linearized cosmological continuity equationThe ratio $\delta \rho /\bar{\rho }$ is the density contrast.

Letbe the vertical absolute vorticity. Taking the vertical curl of the momentum equation removes both the tidal potential and the isothermal pressure force, because each is a gradient field. The two-dimensional vorticity equation isThe surface-density equation isThe quotient rule now givesSince $\zeta =W/\Sigma$ is the vortensity,

The quotient rule and the derivative of the exponential function give

Differentiate the numerator and denominator using the product rule and quotient rule:The numerator simplifies to $-11x^3-6x^2$, giving

The quotient rule simplifies the derivative toIts zeros are $-5$ and $-1$, both in the domain. Substitution gives the two stationary pointsThe first is a local maximum and the second a local minimum, since the numerator of $y^{\prime }$ changes sign from positive to negative at $-5$ and from negative to positive at $-1$. The denominator of the original rational function vanishes atThe numerator is nonzero at both values, so these discontinuities are poles, not removable holes.

On the real domain of a function $x>0$, the quotient rule gives

The original differential equation is defined only where $x\ne 0$ and $y\ne 0$. Put $w=y^3$, so the chain rule gives $w^{\prime }=3y^2{y}^{\prime }$. Multiplying the equation by $3y^2$ yields a first-order linear ordinary differential equation:An integrating factor on either half-line is $1/x$, up to an irrelevant nonzero constant. Equivalently, apply the quotient rule directly:Integrating gives $w/x=x^2+C$. Thus the real general solution ison any connected interval where $x\ne 0$ and $x^3+Cx\ne 0$. These restrictions preserve both the original denominator and differentiability of the real cube root. In particular $C=0$ includes $y=x$ on either half-line; this solution has not been lost through division by an expression involving $y/x-1$.

For verification, differentiating $y^3=x^3+Cx$ gives $3y^2{y}^{\prime }=3x^2+C$, while $(2x^3+y^3)/x=3x^2+C$. The two sides agree wherever the original equation is defined.

Set $D=1+x^2+y^2>0$. The quotient rule gives the partial derivativesSince $1+y+y^2=(y+1/2{)}^2+3/4>0$, the first equation $f_x=0$ forces $x=0$. The second then reduces to $1-y^2=0$. Thus the only stationary points and their field values areNo division by a potentially vanishing factor of $y$ was used, so no stationary points have been discarded.

First suppose $\alpha >0$. Differentiation under the integral sign, justified on compact positive parameter intervals by an integrable bound, givesThe first integral is $f(\alpha )/\alpha$. For $\alpha \ne 1$, use the partial fraction decompositionThe standard inverse tangent antiderivatives then giveThe same value holds at $\alpha =1$ by continuity, or by direct integration. Hence the proposed first-order linear differential equation is valid on the positive parameter range:Divide by $\alpha$ to recognise a quotient rule derivative:To fix $C$, the value $f(0)=0$ alone is insufficient, since every displayed solution tends to zero there. Instead considerThis limit follows from the dominated convergence theorem: for $0<\alpha \le 1$ and $0<x<1$, the logarithm's absolute value is bounded by $2|\ln x|+\ln 2$; for $x\ge 1$, it is at most $\ln (1+x^2)$. After division by $1+x^2$, these are integrable. The limiting improper integral is absolutely convergent, and the substitution rule $x=1/t$ changes it into its own negative. Its value is therefore zero, giving $C=0$. ThusThe question calls $\alpha$ a real parameter, but this last formula and the proposed differential equation do not hold on the entire real axis. The defining integral is odd in $\alpha$, because its logarithm depends on ${\alpha }^2$. Applying the positive-parameter result to $|\alpha |$ therefore gives the correct extensionwith $f(0)=0$. In particular, for $\alpha <0$, the right-hand side of the differential equation must be $-\pi \alpha /(1-\alpha )$; for every $\alpha \ne 0$ the unified expression is $\pi |\alpha |/(1+|\alpha |)$. The logarithmic quadratic parameter integral therefore provides the requested identity on its valid range and the corrected answer elsewhere.

The chain rule and the quotient rule giveThe exponential function and denominator are positive, so the stationary points occur exactly at $x=\pm 1$. Their values areThe derivative is negative for $|x|>1$ and positive for $|x|<1$, so $x=-1$ is a local minimum and $x=1$ a local maximum. They are also global extrema, since $-1/2\le x/(1+x^2)\le 1/2$.

Take nonzero real trial functions $y$ with $y,y^{\prime }$ square-integrable and sufficient decay for integration by parts. For a compactly supported variation $y\mapsto y+\epsilon \eta$, the first variations areThe quotient rule for the functional then yieldsBy the fundamental lemma of the calculus of variations, stationarity for all such $\eta$ is equivalent toConversely, multiplying a decaying eigenfunction equation by $y$ and using integration by parts gives $F[y]=EG[y]$, so its Rayleigh quotient equals $E$ and its first variation vanishes. Thus the stationary values of the functional are the energy eigenvalues. The word "extrema" in the question includes stationary points here: higher eigenfunctions can be saddle points of the unrestricted Rayleigh quotient.

The Rayleigh-Ritz variational principle givesChoose an admissible trial family, calculate its Rayleigh quotient, and minimize over its parameters to obtain the best upper bound within that family. Equality holds for the exact ground state; a restricted family generally gives only an upper bound.

The identity theorem states that if two holomorphic functions on a connected open set agree at a sequence of distinct points with an accumulation point inside that set, then they agree everywhere on it. Equivalently, the zeros of a holomorphic function which is not identically zero cannot accumulate inside its domain.

Because $f,g$ never vanish, their logarithmic derivatives are holomorphic. The holomorphic functionvanishes at every $a_n$. Since the distinct $a_n$ converge to $a\in U$, the identity theorem gives $H\equiv 0$. Applying the quotient rule now givesA holomorphic function with zero derivative on a domain is constant. One can see this locally from its Taylor series, then propagate the constant using connectedness. Hence $f/g=c$ on $U$, with $c\ne 0$, andThe conclusion can fail if the accumulation occurs outside $U$. Take the punctured complex plane $U=\mathbb{C}\setminus \{0\}$ andBoth functions are nowhere vanishing and holomorphic on $U$, and $a_n\to 0\notin U$. The chain rule givesso these logarithmic derivatives agree at every $a_n$. However, $f$ is not constant: for example $f(1/\pi )=e^{-1}$ and $f(1/(2\pi ))=e$. Therefore no constant $c$ gives $f=cg$.

The denominator is positive on $B$ because $x_3>-1$. The quotient rule gives the gradientIn particular $g$ is continuously differentiable throughout $B$.

The denominator is positive on $B$ because $x_3>-1$. The quotient rule gives the gradientIn particular $g$ is continuously differentiable throughout $B$.

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 a real symmetric matrix $A$ and nonzero real vector $x$, the quotient rule and the quadratic form gradient give the displayed gradient. It vanishes exactly when $Ax=R_A(x)x$, so the stationary points are eigenvectors. Equivalently, the Lagrange multiplier condition for optimizing $x^{T}Ax$ on the unit sphere is $Ax=\lambda x$. The smallest and largest eigenvalues give the global extrema; other eigenvectors can be stationary without being extrema, as a direction mixing larger and smaller eigenvalues changes the quotient in opposite ways.