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Let ${\varphi }_t(x_0)$ denote the flow map generated by $\dot{x}=f(x)$.

is nonpositive on $E$.

The First Lyapunov theorem says that the existence of such a function makes the origin Lyapunov stable. To prove it, choose a sufficiently small closed ball ${\overline{B}}_{\epsilon }\subset E$. By compactness and positive definiteness,By continuity, some $\delta >0$ satisfies $V(x)<\alpha$ whenever $\Vert x\Vert <\delta$. Along a trajectory beginning in $B_{\delta }$, $V$ cannot increase. Such a trajectory therefore cannot reach the sphere $\Vert x\Vert =\epsilon$, where $V\ge \alpha$. Hence it remains in $B_{\epsilon }$ for every $t\ge 0$, which is Lyapunov stability.

The LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set on which $\dot{V}\le 0$, then it approaches the largest invariant subset of $\{x:\dot{V}(x)=0\}$.

Eliminate $R=N-S-I$ to obtain the two-dimensional autonomous differential equationIts Jacobian matrix at the endemic equilibrium isSince $I^{\ast }>0$,The trace-determinant stability criterion shows that both eigenvalues have negative real part. The Endemic equilibrium of the SIR model with waning immunity is therefore locally asymptotically stable.

Let $x_{\ast }$ be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of $x_{\ast }$ is a continuously differentiable function $V$ such thatand whose derivative along trajectories satisfies

The First Lyapunov theorem says that the existence of such a positive-definite $V$ proves Lyapunov stability of $x_{\ast }$. If $\dot{V}<0$ away from $x_{\ast }$, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot{V}\le 0$, then it approaches the largest invariant subset ofIn particular, if that largest invariant subset consists only of $x_{\ast }$, every trajectory in $\Omega$ approaches $x_{\ast }$.

At a positive equilibrium, division by $N$ givesSince $b>d$, this has the positive solution

Write the vector field as $F(N)=N(b{e}^{-aN}-d)$. Its derivative isAt $N=N^{\ast }$, the first two terms cancel and $b{e}^{-a{N}^{\ast }}=d$, soThe one-dimensional linear stability analysis therefore shows that $N^{\ast }$ is locally asymptotically stable.

At an endemic equilibrium, $I^{\ast }>0$. The equation $\dot{I}=0$ then givesSince also $S^{\ast }>0$, the equation $\dot{S}=0$ givesThis infective population is positive precisely whenwhich is exactly the assumed instability condition for the disease-free equilibrium. Thus the endemic equilibrium of the susceptible-infective model with exponential birth exists.

At this equilibrium the relations above reduce the Jacobian matrix toHenceBy the trace-determinant stability criterion, both eigenvalues have negative real part. The linearization stability theorem therefore proves that the endemic equilibrium is locally asymptotically stable.

For $\dot{x}=f(x)$, a fixed point $x^{\ast }$ satisfies $f(x^{\ast })=0$. It is Lyapunov stable when, for every $\epsilon >0$, there is $\delta >0$ such that

A Lyapunov function on a neighbourhood of $x^{\ast }$ is a continuously differentiable function $V$ such thatThe First Lyapunov theorem states that such a function makes $x^{\ast }$ Lyapunov stable. If $\dot{V}<0$ away from $x^{\ast }$, then $x^{\ast }$ is locally asymptotically stable.

To prove stability, choose a closed ball contained in the domain of $V$. On its boundary sphere $|x-x^{\ast }|=\epsilon$, compactness and positive definiteness giveContinuity at $x^{\ast }$ supplies $\delta <\epsilon$ such that $|x-x^{\ast }|<\delta$ implies $V(x)<m$. Along a trajectory, $V$ cannot increase. Such a trajectory therefore cannot first reach the boundary sphere, where its value would be at least $m$. This proves Lyapunov stability.

If $\dot{V}<0$ away from $x^{\ast }$, take a sufficiently small compact sublevel set of $V$. The LaSalle invariance principle says that every trajectory in it approaches the largest invariant subset of $\{\dot{V}=0\}$, which is just $\{x^{\ast }\}$. Hence the equilibrium is also asymptotically stable.

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.

The ideal operational-amplifier rules are zero input currents and, when stable negative feedback maintains unsaturated linear operation, equal input voltages. Infinite open-loop gain motivates the latter; it must not be assumed to guarantee asymptotic stability for every external network.

For the numerical sketch and usefulness discussion, use the transfer function derived in part (c). The chosen values give $K=5$ and $\tau =RC=10^{-4}\mathrm{s}$. With $x=\omega \tau$,The formal amplitude falls from 5 to zero, with high-frequency slope $-40$ dB per decade. Its unwrapped phase rises from zero through $+90^{\circ }$ at $\omega =10^4{\mathrm{s}}^{-1}$ to $+180^{\circ }$; the corresponding characteristic frequency is 1.59 kHz. At that frequency the amplitude is 2.5.

These are formal sinusoidal particular-solution curves, not a stable settled filter response. For the formal sinusoidal response of an unstable linear system, replacing $i\omega$ by the Laplace variable $s$ gives $G(s)=5/(1-s\tau {)}^2$, with a repeated pole at $s=+1/\tau$. A homogeneous output contains $(A+Bt)e^{t/\tau }$ and grows, so real perturbations lead to supply-limited nonlinear behaviour. With equal positive resistors and capacitors in the frequency-setting network, the general denominator is $1+(3-K)s\tau +s^2{\tau }^2$: the Sallen-Key equal-component stability bound is $K<3$. For example $K=3-\sqrt{2}$ gives a Butterworth filter response. Thus the Sallen-Key low-pass filter architecture is useful for active filtering with suitable gain and damping, but the stated $K=5$ component set is unsuitable for linear filtering. The component values are retained in this diagnosis.^[ref]

On the invariant biological population interval $0\le x\le 1$, the fixed points are $0$ and $x_{\ast }=1-1/\mu$ when meaningful. The zero fixed-point multiplier is $\mu$: extinction attracts for $0\le \mu <1$. At $\mu =1$, the nonzero branch crosses the origin in the full real-line family, giving a transcritical bifurcation. The nonzero point lies in the biological population interval for $\mu >1$ and has fixed-point multiplier $2-\mu$, so it is stable for $1<\mu <3$ and loses stability through $-1$ at $\mu =3$.

Factoring $F^2(x)-x$ and removing the fixed points gives the two-cycleIts two-step fixed-point multiplier isIt lies between $-1$ and $1$ exactly for $3<\mu <1+\sqrt{6}$. At the upper endpoint it crosses $-1$, creating the next period doubling. The two-cycle amplitude tends to zero like $\sqrt{\mu -3}$ at its birth, so the first flip is supercritical. At the equality endpoints linear fixed-point multipliers alone do not establish asymptotic stability.

Period-doubling cascade universality concerns the continuation of this period-doubling cascade in smooth unimodal maps with a quadratic maximum. After suitable rescalings their repeated two-step maps approach the same limiting shape; successive parameter spacings have the Feigenbaum ratio approximately $4.6692$ and the spatial rescaling has magnitude approximately $2.5029$. These are asymptotic statements involving later doublings, not numbers that can be established from the first two thresholds alone. The accumulation near $\mu \simeq 3.56995$ is outside the first interval considered here. One should not claim chaos throughout the stable two-cycle interval.

Four useful dynamical properties to distinguish are sensitivity to initial conditions, topological transitivity or mixing, dense periodic points, and aperiodic orbits with persistent complex evolution. They concern the invariant set, not an assertion that every orbit must be aperiodic. A positive Lyapunov exponent for typical initial data is an additional quantitative diagnostic.

At $\mu =4$, set $x_0={\sin }^2{\theta }_0$. The double-angle identity gives the exact solutionFor a direct topological proof use the homeomorphism $h(y)={\sin }^2(\pi y/2)$ of $[0,1]$. Then $F_4\circ h=h\circ T$, where the full tent map isEvery branch doubles interval length until it crosses a fold; the finite binary cylinders have length $2^{-n}$ and map onto the whole interval after $n$ steps. Thus every open interval eventually has an iterate covering the full interval: the map is topologically mixing and sensitive. Repeating a finite left/right branch word gives a periodic itinerary; arbitrarily long prescribed prefixes locate such periodic points in arbitrarily small cylinders, proving density. Nonperiodic binary itineraries yield uncountably many aperiodic orbits. Conjugacy transfers these properties to $F_4$, proving chaotic behaviour while retaining its dense special periodic orbits.

As a metric check, its invariant measure has density $1/[\pi \sqrt{x(1-x)}]$ and its typical Lyapunov exponent is $\log 2>0$. This is a typical-orbit assertion, not a claim about the derivative along every exceptional critical or preperiodic orbit. The explicit trigonometric solution and tent-map conjugacy explain why deterministic iteration of a map can nevertheless amplify uncertainty.

A fixed point $x_{\ast }$ satisfies $f(x_{\ast },\mu )=x_{\ast }$. A perturbation obeys $u_{n+1}=m{u}_n+O(u_n^2)$, where the periodic-orbit multiplier is $m=f_x(x_{\ast },\mu )$. It is linearly attracting for $|m|<1$ and repelling for $|m|>1$; at $m=\pm 1$ nonlinear terms decide Lyapunov stability. A crossing through $+1$ can produce a saddle-node, transcritical or pitchfork bifurcation, depending on the mathematical model parameter dependence and symmetry. A crossing through $-1$ generically gives a period-doubling bifurcation. A real one-dimensional function cannot have the complex conjugate pair needed for a Neimark–Sacker bifurcation.

For a genuine period-two orbit, study the return map $f\circ f$ at either point. By the chain rule its periodic-orbit multiplier isIts modulus determines linear stability by the same criterion, now per two iterates.

For the real quadratic iteration map, the fixed-point equation givesThe derivative is $-2x$. The lower branch $x_{-}$ is repelling whenever $\mu >-1/4$. The upper branch satisfies $|-2x_{+}|<1$ exactly forAt $\mu =-1/4$ the two fixed points merge in a saddle-node bifurcation. With $u=x+1/2$ at that mathematical model parameter, $u^{\prime }=u-u^2$, so the point attracts from the right and is repelling from the left. At $\mu =3/4$, $x_{+}=1/2$ has periodic-orbit multiplier $-1$ and undergoes a period-doubling bifurcation. At the threshold itself $u^{\prime }=-u-u^2$ and the two-step return is $u^{\prime \prime }=u-2u^3-u^4$, proving weak asymptotic stability there despite the inconclusive linear test.

Factor the second-iterate equation:Removing the fixed-point factor yields the periodic orbit pointsThey are distinct and exchanged by $f$ exactly when $\mu >3/4$; at equality they coincide with the fixed point. Their product is $1-\mu$, hence their periodic-orbit multiplier is $4(1-\mu )$. ThusIt is repelling for $\mu >5/4$ and reaches its next period-doubling bifurcation threshold at $\mu =5/4$. At that exact endpoint it is weakly attracting: if $f^2$ is expanded about either periodic orbit point as $v^{\prime }=-v+A{v}^2+B{v}^3+O(v^4)$, then $A=2\mu -6x_i^2$, $B=-4x_i$, and $A^2+B=20\pm 10\sqrt{2}>0$. Its second return is $v^{\prime \prime }=v-2(A^2+B)v^3+O(v^4)$.

At $\mu =2$ put $h(y)=4{\sin }^2(\pi y)-2=-2\cos (2\pi y)$. The double-angle identity givesThis verifies the printed relation to the Bernoulli shift. Precisely, it is a semiconjugacy: $h(y)=h(1-y)$, so it is not invertible on the circle. To obtain a genuine interval topological conjugacy, use the homeomorphism $H(u)=-2\cos (\pi u)$ from $[0,1]$ onto $[-2,2]$, and the full tent mapThe same identity proves $f(H(u),2)=H(T(u))$.

The full tent map has dense periodic points, is topologically mixing and has sensitive dependence on initial conditions. To see these properties directly, its $n$th iterate has $2^n$ monotone branches on dyadic intervals of length $2^{-n}$, each mapping onto $[0,1]$. Every nonempty open interval contains such a branch domain, and a later iterate covers all of $[0,1]$, proving topological mixing. On any branch, the function graph of the iterate has a fixed point by the intermediate value theorem, so periodic points occur in every open interval. Given any initial point and any neighborhood, some iterate of that neighborhood covers the interval; one can choose a nearby point whose image differs from the initial point's image by at least a fixed positive distance. This proves sensitive dependence on initial conditions. Uniform continuity of the inverse homeomorphism transfers these properties to $f$ on $[-2,2]$.

Consequently the real quadratic iteration map is chaotic on its invariant interval in the Devaney sense: topological transitivity, dense periodic points and sensitive dependence on initial conditions, with all these orbits bounded. There are also aperiodic orbits: periodic points form a countable union of finite sets of roots of an equation of the equations $f^n(x)=x$, whereas the interval is uncountable. The semiconjugacy alone would not justify treating $h$ as invertible; the tent-map argument supplies the missing rigorous transfer.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the two differential equations:Substituting $x=r\cos \theta$, $y=r\sin \theta$ gives the required equations in polar coordinates. The origin is handled in Cartesian coordinates, where it is an equilibrium point; an angular coordinate is not defined there.

A Lyapunov function near an equilibrium point is a continuously differentiable positive-definite scalar function $V$ whose derivative along solutions is nonpositive. This implies local Lyapunov stability under the usual comparison bounds; if its derivative is negative definite, the Second Lyapunov theorem gives local asymptotic stability. Here takeIf $a<f(0,0)$, continuity gives a small ball and $\delta >0$ such that $f(x,y)-a\ge \delta$ inside it. Consequently $\dot{V}\le -2\delta V$. A sufficiently small initial sublevel set stays inside the ball, and $V(t)\le V(0)e^{-2\delta t}$. The origin is asymptotically stable, in fact locally exponentially stable. If $a>f(0,0)$, continuity instead gives $\dot{V}\ge 2\delta V$ in a small ball. Every nonzero initial point there exits that ball in forward time, proving instability of the origin. These inequalities hold along all local solutions; continuity of $f$ alone need not supply uniqueness for a general vector field.

If $f=g(r)$, a nonzero constant radius $r_0$ satisfying $a=g(r_0)$ gives a periodic orbit of period $2\pi$, since $\dot{\theta }=1$. Linearizing the scalar radial equation there givesThus the periodic orbit is orbitally asymptotically stable for $g^{\prime }(r_0)>0$ and unstable for $g^{\prime }(r_0)<0$. The direction along the periodic orbit is neutral, as expected for an autonomous periodic orbit; orbital asymptotic stability means attraction in a dynamical system to the orbit with a possible phase shift.

For the specified radial response,The origin is therefore stable for $0<a<3/2$ and unstable for $a>3/2$. The nonzero-radius equation has roots of an equationretaining only positive radii. For $1<a<3/2$, the square root exceeds one, so only $r_{+}>1$ exists and $g^{\prime }(r_{+})<0$: the periodic orbit is unstable. For $3/2<a<2$, both radii are positive; $0<r_{-}<1$ gives $g^{\prime }(r_{-})>0$ and a stable periodic orbit, while $r_{+}>1$ remains unstable.

As $a$ increases through $3/2$, the stable small-radius periodic orbit emerges from the origin while the origin loses Lyapunov stability. This is the supercritical Hopf-type transition with a nonsmooth radial response requested. It follows directly from the explicit radial dynamics. There is a regularity qualification: $g(r)=3/2+r/2+O(r^2)$ near zero, so the vector field in Cartesian coordinates contains $x\sqrt{x^2+y^2}$ and $y\sqrt{x^2+y^2}$ terms. It is continuously differentiable but not generally twice differentiable at the origin, and the periodic orbit radius grows as $r_{-}\sim 2(a-3/2)$ rather than the square-root law of a generic smooth Hopf normal form. One should therefore use the explicit computation here rather than invoke a smooth Hopf bifurcation theorem with unmet hypotheses. At $a=3/2$ itself the origin is still weakly attracting, because $a-g(r)<0$ for small positive $r$.

An equilibrium is asymptotically stable when it is Lyapunov stable and attracts every initial state in some neighbourhood. Lyapunov stability means that sufficiently small initial displacement stays small for all future times; attraction additionally requires convergence to the equilibrium. A Lyapunov function is a differentiable positive-definite scalar function near the equilibrium whose derivative along solutions is nonpositive. Strict negativity, or an appropriate invariant-set argument for a non-strict derivative, gives asymptotic stability.

For $\nu =-1$, equilibria satisfy $y=0$ and $x+x^3=0$, so only $(0,0)$ is real. Its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with eigenvalues $\pm i$ for every $\kappa$. It is a linear centre equilibrium, so linearization alone cannot decide nonlinear attraction.

Use the positive proper energyFor $\kappa <0$, energy sublevel sets are compact and invariant. The zero-derivative set is the union of the axes, but its only invariant subset is the origin: on $x=0,y\ne 0$ one immediately leaves that axis, and on $y=0,x\ne 0$ one has $\dot{y}=-x-x^3\ne 0$. LaSalle's invariance principle therefore proves global asymptotic stability. At $\kappa =0$ the conserved energy levels are closed periodic orbits, giving a nonlinear centre equilibrium. For $\kappa >0$, time reversal together with $y\mapsto -y$ converts the system to the attracting negative-$\kappa$ case, so the origin is a nonlinear repeller and is unstable. Nonzero forward trajectories cannot remain bounded; this does not assert existence for all time if finite-time escape occurs.

A fixed point $z_{\ast }$ of a map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ satisfies $F(z_{\ast })=z_{\ast }$. Its linear perturbations evolve by the Jacobian matrix $DF(z_{\ast })$: both eigenvalues must lie strictly inside the unit disk for linear asymptotic stability. One inside and one outside gives a saddle; both outside give a repeller. Unit-modulus eigenvalues require nonlinear analysis, unlike the continuous-time test using signs of real parts.

Here a fixed point has $x=y=p$ and $2p=a(p-p^3/3)$. ThereforeAt zero the characteristic equation is ${\lambda }^2+\lambda -a=0$. Its negative root is less than $-1$ for every $a>0$, while its positive root crosses $+1$ at $a=2$. Thus zero is a saddle for $0<a<2$ and a real repeller for $a>2$, with a pitchfork of fixed-point branches at $a=2$.

At either nonzero point the characteristic equation is ${\lambda }^2+\lambda +2a-6=0$, whose discriminant is $25-8a$. For $2<a<3$ the fixed points are saddles. At $a=3$ their periodic-orbit multipliers are $0,-1$, giving a flip threshold. For $3<a<25/8$ they are attracting real nodes; at $25/8$ repeated-eigenvalue degenerate nodes; and for $25/8<a<7/2$ attracting foci. The Jury stability criterion conditions reduce to $a>3$ and $a<7/2$, confirming the full stable interval.

At $a=7/2$ the periodic-orbit multipliers are $e^{\pm 2\pi i/3}$. For larger $a$ they form a repelling focus equilibrium. This is a unit-circle crossing at a strong resonance of a planar map with ratio $1:3$; linearization alone does not justify asserting a generic nonresonant Neimark-Sacker bifurcation. The period-two branch at the flip threshold is derived next, including an additional family absent from the source's proposed quadratic.

Proportional-integral control is the proportional and integral part of a PID controller, with the derivative term omitted. Proportional-integral control combines proportional control with integral action. For the ideal controlled plant, a constant reference signal, exact measurement and a constant control disturbance, the ordinary differential equations giveThe characteristic polynomial is ${\lambda }^2+k_p\lambda +k_i$. Positive $k_p$ and $k_i$ imply asymptotic stability of the equilibrium point: real roots of a polynomial have negative sum and positive product, hence are both negative; a nonreal complex conjugate pair has real part $-k_p/2<0$. This is a short stability check for this ideal controlled plant, rather than a rule for every dynamical system.

At the equilibrium point, $e_{\ast }=0$, $x_{\ast }=r$, and $z_{\ast }=-w/k_i$. For $w=0.2$ and $k_i=1$, the stored value $z_{\ast }=-0.2$ supplies $u_{\ast }=-0.2$ even though the tracking error is zero. This separates the eventual accuracy from the transient response: integral action cancels the lasting offset but can make the plant output pass beyond the goal. The authors discuss practical limits on integral action in Åström and Murray, Feedback Systems, Chapter 11.

For $k_p=2$, $k_i=1$, $x(0)=z(0)=0$ and $w=0$, the characteristic polynomial is $(\lambda +1{)}^2$. The initial conditions are $e(0)=1$ and $\dot{e}(0)=-2$, givingThe tracking error changes sign at $t=1$. Since $\dot{x}=(2-t)e^{-t}$, the largest plant output occurs at $t=2$ and is $1+e^{-2}\simeq 1.135335$, about $13.53\%$ above the reference signal.

For a separate comparison, start the control disturbance $w=0.2$ at $t=4$. For $t\ge 4$, write $s=t-4$. The exact responses areThe proportional control response tends to $1.1$, while the proportional-integral control response returns to $1$. At $t=8$, the values are approximately $1.099966$ and $1.017001$. The added term in the proportional-integral control response eventually vanishes, exposing how memory removes a constant control disturbance's lasting effect.

A resistive velocity term proportional to the speed squared removes energy at rate proportional to $|\omega {|}^3$. Its derivative at zero velocity vanishes, so it does not appear in the linearization of a dynamical system at rest. Nonlinear energy arguments can still prove asymptotic stability.

Consider a smooth planar system $\dot{x}=y+P_3(x,y)+O(r^5)$, $\dot{y}=-x+Q_3(x,y)+O(r^5)$, where $P_3,Q_3$ are homogeneous cubic polynomials and $r^2=x^2+y^2$. The leading derivative of $r^2/2$ is the quartic polynomial $H_4=x{P}_3+y{Q}_3$. Let $\kappa$ be the angular average of $H_4$ on the unit circle. A homogeneous quartic correction $V_4$ solvinggives $\dot{V}=\kappa r^4+O(r^6)$ for $V=r^2/2+V_4$. The right side has zero angular average. Since $y{\partial }_x-x{\partial }_y=-{\partial }_{\theta }$, its nonconstant angular Fourier modes can be integrated individually to obtain a periodic quartic correction; only modes of orders two and four can occur after removing the constant average.

The correction is $O(r^4)$, so $V$ remains positive definite close to zero. If $\kappa <0$, its orbital derivative is strictly negative there, and the Second Lyapunov theorem proves asymptotic stability. If $\kappa >0$, apply the same argument in reversed time to obtain repulsion. If $\kappa =0$, this quartic calculation alone does not decide stability. This method converts the averaged damping of a weak focus into a pointwise sign for a Lyapunov function.

An equilibrium point is quasi-asymptotically stable when some neighborhood of initial conditions has forward solutions converging to it. This attraction condition alone does not include Lyapunov stability. Asymptotic stability requires both attraction and Lyapunov stability. A positive Lyapunov function with nonpositive derivative and no nonstationary invariant trajectory in its zero-derivative set proves both when its sufficiently small sublevel sets are compact and forward invariant.

For $\dot{r}=r(a-g(r))$, $\dot{\theta }=1$, a positive root $a=g(r_0)$ gives a circular periodic orbit of period $2\pi$. The radial exponent is $-r_{0}g\prime (r_0)$, so $g\prime (r_0)>0$ gives orbital asymptotic stability and $g\prime (r_0)<0$ gives instability. The phase direction is neutral; a periodic orbit is not an asymptotically stable individual point of the flow.

The family $x_{n+1}=\mu -x_n^2$ has fixed points $(-1\pm \sqrt{1+4\mu })/2$. Its upper branch attracts linearly for $-1/4<\mu <3/4$ and loses stability in a period-doubling bifurcation. The resulting two-cycle has points $(1\pm \sqrt{4\mu -3})/2$ and multiplier $4(1-\mu )$, attracting for $3/4<\mu <5/4$. At $\mu =2$, the map is conjugate on $[-2,2]$ to the full tent map.

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.