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A family of path laws satisfies compact containment when for every finite horizon and every error tolerance there is a compact set which contains the whole path up to that horizon with at least the complementary probability, uniformly over the family. In Euclidean state space this is the displayed condition. It prevents escape to distant regions where local coefficient approximations give no control. A nonnegative Lyapunov function with uniformly bounded generator growth often proves it by the optional stopping theorem.

A nonnegative function that is a uniformly continuous function on the interval $[0,\infty )$ and has a finite integral tends to zero. Otherwise uniform continuity gives disjoint intervals of a fixed width on which the function exceeds a fixed positive value, contradicting integrability. This supplies a useful decay argument for Lyapunov functions with merely continuous vector fields.

To make the fixed width explicit, suppose some $\eta >0$ occurs as a lower bound $f(t_n)\ge \eta$ along a sequence with $t_n\to \infty$. Choose $\delta >0$ from uniform continuity for output tolerance $\eta /2$. On every interval $(t_n-\delta /2,t_n+\delta /2)$ lying in $[0,\infty )$, the function is greater than $\eta /2$. Extract a subsequence whose centers are more than $\delta$ apart. Each disjoint interval contributes at least $\eta \delta /2$ to the integral, so the first $N$ contribute at least $N\eta \delta /2$. This contradicts finiteness of the integral. The same input width at every center is the essential step; continuity alone would allow progressively narrower peaks.

A positive-definite Lyapunov function with nonpositive orbital derivative proves Lyapunov stability of the equilibrium. Indeed, fix a sufficiently small ball $B_{\epsilon }$ in the function's domain and putContinuity and $V(0)=0$ give a ball $B_{\delta }$ on which $V<\alpha$. Since $V$ cannot increase along a trajectory, one starting in $B_{\delta }$ cannot cross the sphere $\Vert x\Vert =\epsilon$, on which $V\ge \alpha$. It therefore remains in $B_{\epsilon }$ for all forward time.

This Lotka-Volterra predator-prey model includes self-limitation of the prey through the term $-b{x}^2$, with positive constants $a,b,c,d,e$. A positive coexistence equilibrium point exists when $a>be/d$, at $x_{\ast }=e/d$, $y_{\ast }=(a-b{x}_{\ast })/c$. The Lyapunov functionsatisfies $\dot{V}=-bd(x-x_{\ast }{)}^2$: its predator-prey cross terms cancel. Its sublevel sets are compact in the positive quadrant. The only invariant subset of $\{\dot{V}=0\}$ is the coexistence equilibrium point, because staying on $x=x_{\ast }$ requires $y=y_{\ast }$. The LaSalle invariance principle therefore proves convergence to coexistence for every positive initial state.

For a unit-jump population chain with one-step probabilities $\alpha k/n$ and $\beta k^2/n^2$, its equilibrium density is $u_0=\alpha /\beta$. Speed time by $n$ and center and scale the population as $(k-n{u}_0)/\sqrt{n}$. The accelerated conditional drift is $-\alpha x-\beta x^2/\sqrt{n}$, and the second moment is $2{\alpha }^2/\beta +3\alpha x/\sqrt{n}+\beta x^2/n$. Their compact-set limits give the displayed Ornstein-Uhlenbeck process. If $2{\alpha }^2<\beta$, the invalid-probability threshold stays a fixed positive density distance above equilibrium. Before that threshold, the generator of $x^2$ is at most one because its drift contribution is nonpositive. The stopped Lyapunov function bound proves compact containment and makes the probability of hitting that threshold on a fixed accelerated horizon tend to zero. This supplies the global hypotheses of the diffusion approximation theorem.

Since the exponential factor is positive, equilibria must have $y=0$ and $x^4=1$. Thus $\boxed{(x,y)=(-1,0),(1,0)}$. At either point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -4x^3& 0\end{array}\right)$. At $(-1,0)$ its eigenvalues are $\pm 2$, so this is a saddle point. At $(1,0)$ they are $\pm 2i$: the linearization has a centre equilibrium.

The requested first-order approximation follows by putting $x=1+u$, $y=v$: $\dot{u}=v+O(\Vert (u,v){\Vert }^2)$ and $\dot{v}=-4u+O(\Vert (u,v){\Vert }^2)$. The linearized system conserves $4u^2+v^2$, giving the approximate ellipsesHowever, the printed nonlinear system has a weak repelling focus at $(1,0)$, rather than a centre. A centre in the linearization need not be a nonlinear centre. Here that distinction can be proved explicitly, without relying on the numerical drift in the figure.

Put $X=2(x-1)$, $Y=y$, $\tau =2t$ and $r^2=X^2+Y^2$. The weak focus of an exponential polynomial planar system has the local positive functionExpanding the vector field through cubic order and differentiating givesThe cubic correction cancels the cubic terms in the derivative of $H$, and the quartic correction leaves the positive radial term $3r^4/16$. Since $H=r^2/2+O(r^3)$ is positive and the angular velocity stays near $-1$, small nonzero trajectories spiral outward. Equivalently the reversed system has a strict local Lyapunov function and an asymptotically stable equilibrium. In particular, a periodic orbit sufficiently near the equilibrium is impossible, because $H$ would have to increase strictly and return to its starting value. This proves the failure of the printed centre assertion while retaining its correct linearized ellipse calculation.

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.

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.

The term $p$ represents prey reproduction and $-p^2/a$ limits growth by available food, giving carrying capacity $a$ in the absence of hunters. The loss $-ph/a$ represents encounters between prey and hunters. For hunters, $hp/(8b)$ is prey-supported growth and $-h/8$ is mortality; their per-capita growth changes sign at prey population $b$. The coordinate axes are invariant, and positive initial populations remain positive.

For $a=1$, the physical equilibrium points are the origin, $(1,0)$, and $(b,1-b)$ when $b<1$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1$ and $-1/8$, so the origin is always a saddle equilibrium, attracting along the hunter-only axis and repelling along the prey-only axis. At $(1,0)$ the eigenvalues areIt is a saddle equilibrium for $0<b<1/2$ and an asymptotically stable node for $b>1$.

For $0<b<1/2$, the coexistence Jacobian matrix isIts trace is negative and determinant positive. Its discriminant $b^2-(1-b)/2=(2b-1)(b+1)/2$ is negative, so $(b,1-b)$ is an asymptotically stable focus. Interior trajectories approach it in damped, counterclockwise oscillations in the $(p,h)$ plane. Both populations ultimately coexist.

To justify the global interior fate for this logistic predator-prey model, let $h_{\ast }=1-b$ and use the Lyapunov functionIt is nonnegative and its sublevel sets are compact inside the positive quadrant. Differentiation gives the explicit cancellationThe only invariant subset on which this vanishes is $p=b,h=h_{\ast }$: remaining at $p=b$ requires $\dot{p}=b(1-b-h)=0$. LaSalle's invariance principle therefore proves that every strictly positive initial state converges to this coexistence point.

For $b>1$, the formal coexistence point has negative $h$ and is outside the physical state space. The only physical equilibria are the saddle equilibrium at the origin and the stable prey-only node. Since $\dot{p}\le p(1-p)$, positive prey has $\mathrm{limsup}p\le 1$. Eventually $p<b$ with a fixed margin, so hunter population decays exponentially. When $h$ is sufficiently small, comparison with $\dot{p}=p(1-\epsilon -p)$ gives $\mathrm{liminf}p\ge 1-\epsilon$ for every $\epsilon >0$. Thus $h\to 0$ and $p\to 1$ for positive initial prey: hunters die out and prey reaches carrying capacity.

On the boundary, hunter-only initial data approach the origin; prey-only positive data approach $(1,0)$ in either parameter range. The sketches show representative values in the two requested ranges. Dashed and dotted lines are the prey and hunter nullclines; filled dots are attractors and open dots are saddle equilibria.

Write $c=(1-a-b)/a$. The admissible equilibrium points are $O=(0,0)$, $E=(1-a,0)$ when $a\le 1$, and $P=(b,c)$ when $a+b\le 1$. The Jacobian matrix isAt $O$ the eigenvalues are $(1-a)/a$ and $-1$; at $E$ they are $-(1-a)/a$ and $(1-a-b)/b$. At a positive coexistence equilibrium point,It is therefore asymptotically stable. It is a node for $b^2>4a(1-a-b)$, a focus for the reverse inequality, and has repeated real eigenvalues at equality. The node–focus change does not alter the attracting equilibrium point's topology.

There are three open physical parameter regions:

The axes are invariant. On $x=0$, $y$ decays exponentially; on $y=0$, prey obey the logistic differential equation. Positivity follows because each right-hand side has its population as a factor. Also $x$ is bounded by logistic comparison, while $W=x+by$ satisfies $\dot{W}=(1-a)x/a-x^2/a-by$, so all forward trajectories are bounded. The Dulac function $1/(xy)$ has divergence $-1/(ay)<0$ in the interior, excluding periodic orbits by the Bendixson-Dulac criterion.

In the extinction region, $\dot{x}\le -(a-1)x/a$, and then $\dot{y}/y=x/b-1$ is eventually negative. In the prey-only region, the upper bound $\mathrm{limsup}x\le 1-a<b$ implies predator decay and convergence of interior prey to $1-a$. In the coexistence region the Lyapunov functionhas $\dot{\mathcal{V}}=-(x-b{)}^2/a\le 0$. Its positive-quadrant sublevel sets are compact, and the only invariant subset with $x=b$ also has $y=c$. The LaSalle invariance principle proves convergence of every interior trajectory to $P$. These facts justify the complete qualitative phase portraits, not just their local arrows.

At $a=1$, $O$ and $E$ meet in a transcritical bifurcation; positive prey decay algebraically. At $a+b=1$ with $a<1$, $E$ and $P$ exchange stability. On its centre manifold, $x=b-ay+O(y^2)$ gives $\dot{y}=-a{y}^2/b+O(y^3)$, so the physically positive side is still attracting. There is no interior Hopf bifurcation since the coexistence trace never vanishes.

A fixed point $x_0$ is Lyapunov stable if for every $\epsilon >0$ there is $\delta >0$ such that solutions starting within $\delta$ of $x_0$ stay within $\epsilon$ for all forward times. It has quasi-asymptotic stability if some neighborhood of $x_0$ consists of initial points whose solutions converge to $x_0$. Asymptotic stability requires both properties.

Choose the Lyapunov functionIt vanishes only at the origin, is positive elsewhere, and has compact sublevel sets. The vector field is polynomial, hence locally Lipschitz. Along a solution,A small sublevel set is forward invariant and can be fitted inside any prescribed neighborhood, proving Lyapunov stability; boundedness also prevents finite-time escape.

For attraction, a bounded forward solution has a nonempty compact limit set, and $V$ decreases to a limit. Every point in that limit set has that same $V$ value. Continuous dependence of solutions makes the limit set forward invariant, so $\dot{V}=0$ along every trajectory in it. Such a trajectory must have $x=0$ at all times. But then $\dot{x}=-y$, forcing $y=0$. Thus the only possible limit point is the origin, and every solution converges there. This verifies the invariant-set condition behind the LaSalle invariance principle, rather than incorrectly treating $\dot{V}$ as strictly negative at every nonzero point. The origin is therefore asymptotically stable, indeed globally attracting.

For the symmetric zero-diagonal weights and asynchronous rule just specified, update only unit $k$. In the energy difference, terms involving neither $k$ nor its symmetric partner cancel. Thus asynchronous Hopfield energy descent givesIf the unit was already aligned, it does not change and $\Delta E=0$. If it flips into alignment, $x_k^{\prime }-x_k=2\mathrm{sgn}(h_k)$, soAt $h_k=0$ our tie rule retains the old activity. The network has only $2^M$ configurations. Strict descent on every actual change prevents revisiting a configuration and permits only finitely many changes. With a fair schedule every misaligned unit is eventually updated, so the terminal configuration is a fixed point. At that point each possible single-bit flip has energy cost $2x_k{h}_k\ge 0$. The network reaches a minimum against single-unit changes, not necessarily the global minimum. Its energy is a Lyapunov function for these asynchronous dynamics.

The qualifications are essential to the printed claim. Synchronous updates can cycle: for two units with $w_{12}=w_{21}=1$, the state $(1,-1)$ alternates with $(-1,1)$, keeping energy $+1$. Symmetry alone therefore does not prove convergence under simultaneous updates.

Nor does the valid descent proof guarantee a global minimum. Consider four units with $w_{12}=w_{34}=3$ and all four cross-group weights equal to $1$, symmetrically and with zero diagonal. The state $(1,1,-1,-1)$ has aligned fields $(1,1,-1,-1)$ and energy $-2$, so it is a strict single-flip local minimum. The all-positive state has energy $-10$. This explicit counterexample shows why local minima may trap retrieval even though the energy always descends under the stipulated rule.

Choose $A=B=\sigma /r$ in the proposed Lyapunov function. The cubic terms cancel and direct differentiation givesFor $0<r<1$ this is negative definite, while $V$ is positive definite and radially unbounded. Finite-dimensional equivalence of these positive quadratic forms gives $\dot{V}\le -cV$ for some $c>0$. Thus $V(t)\le V(0)e^{-ct}$, every solution remains bounded and exists for all forward times, and every solution tends to zero. Therefore the origin is globally asymptotically stable, indeed exponentially stable.

A random access network shares one communication channel among an unbounded collection of possible transmitters. Each packet occupies one slot; exactly one attempt succeeds, an idle slot carries nothing, and a collision delivers none of its attempted packets. Assume perfect slot synchronization, immediate idle/success/collision feedback, no channel errors, and independent random choices by packets. The important quantities are throughput, the size of the retransmission backlog, and delay. A good one-slot success probability alone does not establish stability of the backlog.

For a precise infinite-population slotted ALOHA model, let $A_t$ be independent Poisson random variables of mean $\eta >0$, representing new packets which attempt immediately in slot $t$. Let $B_t$ be the backlog before those arrivals. Each old packet independently retries with a constant probability $p\in (0,1)$, so conditional on $B_t=b$, its retry count is $K_t\sim \mathrm{Binomial}(b,p)$, independent of $A_t$. If $D_t=1_{\{A_t+K_t=1\}}$ records a successful slot, thenThis includes a new packet in the backlog precisely when it has not succeeded; an old packet is removed precisely when it succeeds. The total backlog is a Markov chain in this constant-probability model. Its idle and success probabilities arewith the second term in $S(0)$ interpreted as zero. A success may be either one new packet and no old attempt, or one old attempt and no new packet.

The fluid approximation of slotted ALOHA replaces the binomial retry count by a Poisson count when there are many packets with small retry probability. If $g=\eta +bp$ is total attempt intensity, the approximate success probability is $g{e}^{-g}$. Differentiation givesThis is the limiting ALOHA throughput bound, not an exact upper bound for every finite packet population: one isolated packet which certainly transmits succeeds with probability one. In a fluid model, when $0<\eta <e^{-1}$ the equation $\eta =g{e}^{-g}$ has two roots. The root below one is locally stable for constant $p$, while the root above one is unstable, because the drift derivative is $-p(1-g)e^{-g}$. These approximations explain a low-backlog operating region and a congestion threshold; they do not imply a stationary law for the infinite-population chain.

Indeed, for fixed $p>0$, the exact success probability $S(b)$ tends to zero as $b\to \infty$, so the exact drift tends to $\eta >0$. One can prove instability, rather than relying only on this positive drift. Fix $\theta >0$ and use $V(b)=e^{-\theta b}$. Since $D_t$ is zero or one,The first term is strictly less than one and the second tends to zero. Hence $V(B_{t\wedge \tau })$ is a nonnegative supermartingale outside a sufficiently large finite set, where $\tau$ is the first entrance into that set, say $\{0,\dots ,b_0\}$. The optional sampling theorem for a supermartingale, first at bounded stopping times, givesThis is an exponential-supermartingale escape bound. The chain is irreducible: a no-arrival slot with exactly one old retry can reduce any positive state by one, and Poisson arrivals can reach arbitrarily high states. Therefore the escape probability proves transience and excludes a positive recurrent Markov chain, for every $\eta >0$ and fixed $0<p<1$. A locally stable fluid operating point can consequently coexist with eventual stochastic congestion. At $p=1$, once at least two packets are backlogged, they collide forever, so constant unit retry probability also fails.

A state-dependent retry rule removes this fixed-$p$ congestion mechanism. As an ideal benchmark with exact backlog information and known arrival intensity $0<\eta <e^{-1}$, useThis is backlog-aware ALOHA with arrival-adjusted attempt probabilities. Its total attempt intensity approaches one: the mean old-attempt count is $1-\eta$ and the new-attempt count has mean $\eta$. Substitution in the exact success formula yieldsThus the drift tends to $\eta -e^{-1}<0$. For the Lyapunov function $L(b)=b$, the drift is bounded above by a negative constant outside a finite set, and the expected one-step backlog is finite at every state. Foster theorem therefore gives positive recurrence. This proves a genuine stochastic stability result for this specified, informed controller. Real transmitters generally observe feedback rather than the exact backlog; estimating the backlog and using a probability of order its reciprocal motivates adaptive ALOHA, but an estimate's dynamics must be included in the state and analyzed before claiming the same stability result.

A different nonconstant rule uses a packet's own collision history. In geometric-attempt exponential backoff, a packet at stage $k$ attempts with probability $p_k=p_0{2}^{-k}$, where $0<p_0<1$. A collision advances every attempted packet by one stage, a success removes its packet, and new arrivals enter stage zero. For definiteness, in this version arrivals join at the end of a slot and first become eligible in the next slot. The full Markov chain state is the sequence $(N_0,N_1,\dots )$ of stage populations; the total backlog alone is insufficient. Conditional on that state,These formulas follow from independent attempts and selecting which one packet succeeds. They reduce approximately to $S\simeq G{e}^{-G}$ for many small probabilities with $G={\sum }_k{N}_k{p}_k$. The retry probabilities decrease after failure, preventing repeated aggressive attempts by the same packets. Binary exponential backoff instead chooses uniform random counters from expanding finite windows; that counter model requires residual counters as well as collision stages in its state and is not identical to geometric waiting.

Backoff also exposes a distinction between eventual delivery and acceptable delay. In the simplifying model where each attempt independently fails with a fixed probability $q<1$, the probability of reaching stage $k$ is $q^k$ and its mean waiting time is $1/p_k=2^k/p_0$. The mean delay under independent exponential-backoff failures is thereforeAlthough a packet succeeds almost surely whenever $q<1$, its mean delay can be infinite. In an actual random access network, failures are correlated with the other stage populations, so this fixed-$q$ calculation is a diagnostic approximation, not a proof of network stability. These models show why random-access analysis must track the retry rule, feedback information and full state, and assess throughput, stability and delay separately.

A Lyapunov function near an equilibrium is a continuously differentiable function with $V(0)=0$, $V(x)>0$ for nonzero nearby $x$, and $\dot{V}=\nabla V\cdot f\le 0$ along trajectories there. Strict negativity away from the equilibrium gives local asymptotic stability. Lyapunov stability itself means that every prescribed small neighborhood contains a smaller neighborhood whose forward trajectories never leave the prescribed one.

For the proposed quadratic form, direct differentiation givesWhen $\beta >1$, the quadratic part is negative definite, so it dominates the cubic remainder in a sufficiently small ball; $V$ is then a strict Lyapunov function. If $0<\beta <1$, take $x=y$ to make the leading derivative positive. At $\beta =1$, that same line gives $\dot{V}=-4x^3$, positive on its negative half. Values $\beta \le 0$ fail positive definiteness of $V$. ThereforeFor $\beta =2$ the full derivative factors, with no small-amplitude truncation:Every closed sublevel set $V\le c<1$ lies in $x>-1$, is forward invariant, and has strictly decreasing $V$ away from the origin. Compactness and the Lyapunov stability argument therefore show that the ellipse $x^2+2y^2<1$ belongs to the origin's basin of attraction. Boundary points with $V=1$ other than $(-1,0)$ also have negative derivative and enter the ellipse. The exception $(-1,0)$ is another fixed point, so the largest such sublevel ellipse cannot be included wholesale in the basin. This is a basin estimate from a Lyapunov sublevel set, not a claim that no additional points outside the ellipse converge to the origin.

Set $u_0=\alpha /\beta$ and use the accelerated grid time $j/n$. At population $k=n{u}_0+\sqrt{n}x$, the fluctuation increment is $h_n=n^{-1/2}$, $-h_n$, or zero. Its first and second conditional moments, multiplied by $n$, areThus $b_n\to b(x)=-\alpha x$ and $a_n\to a=2{\alpha }^2/\beta$, uniformly on every compact set. The centered conditional variance has the same limit, since the squared conditional mean contributes only $b_n(x{)}^2/n$.

Here is the diffusion approximation theorem being applied. For Markov chains sampled with time step $1/n$, suppose the first and second conditional increment moments multiplied by $n$ converge uniformly on compact sets to continuous coefficients $b,a$; the large-jump conditional second moments satisfy the local Lindeberg condition; the initial laws converge; compact containment holds on finite time intervals; and the limiting martingale problem is well-posed and nonexplosive. Then the interpolated chains converge weakly, uniformly on finite time intervals, to that problem's continuous solution. The step versions converge in the Skorokhod J1 topology. Local moment bounds supply the stopped-process tightness, and the limiting test-function identities identify every subsequential limit by the martingale problem.

We verify the global condition rather than assume it. Before the stopping rule is triggered, $p_k+q_k\le 1$, and $k\ge 0$ implies $x\ge -u_0\sqrt{n}$. ThereforeFor the Lyapunov function $V(x)=x^2$, the exact one-step generator is consequentlyAfter stopping, the process is constant. Applying the elementary stopped supermartingale inequality for $V(X_{j/n}^n)-j/n$, and stopping also on first reaching $|X^n|\ge L$, givesThis is compact containment. At zero population we use the natural absorbing convention $p_0=q_0=0$; the same calculation is valid there.

The invalid-probability threshold is at the positive rootBecause $2{\alpha }^2<\beta$, we have $u_{\ast }>u_0$. The corresponding fluctuation level $\sqrt{n}(u_{\ast }-u_0)$ tends to infinity, so the stopping rule changes none of the compact-set moment limits. The initial fluctuation satisfies $|X_0^n|\le n^{-1/2}$ and hence tends to zero. All jumps have magnitude at most $n^{-1/2}$, so the Lindeberg condition is immediate. The limiting equation has globally Lipschitz coefficients and is well-posed and nonexplosive. The theorem therefore givesThis is the Ornstein-Uhlenbeck fluctuation limit at a stable population equilibrium. The limiting Ornstein-Uhlenbeck process has explicit solutionThe PDF samples the linearly interpolated population at $\lfloor nt\rfloor$, so its displayed process is actually the step version. Using $nt$ instead gives the continuous interpolation; their uniform distance is at most $n^{-1/2}$. Both therefore have the asserted continuous diffusion limit, with the step version interpreted in Skorokhod J1 topology.

Finally, if ${\tau }_n\le 1$, the stopped fluctuation reaches a value greater than $\sqrt{n}(u_{\ast }-u_0)$. The preceding Lyapunov function bound gives directlyThis also shows explicitly why the artificial stopping rule is asymptotically irrelevant.

Neither monotonic decrease alone nor equality of the initial spectra guarantees convergence. A nonnegative Lyapunov function can approach a positive limiting value, and the set where $\dot{V}=0$ can contain trajectories other than the target. Different initial spectra already obstruct exact tracking: unitary time evolution preserves the eigenvalues of each density operator, while two matrices approaching one another in norm must have the same limiting eigenvalues.

Even the spectral obstruction can be absent while the Lyapunov quantum control stalls. Take a qubit with $H_0=Z/2$, $H_1=X$, ${\rho }_d(0)=|0⟩⟨0|$ and $\rho (0)=|1⟩⟨1|$. Both density operators commute with $H_0$, and the commutator of $X$ with $\rho$ has zero diagonal, so $f=\mathrm{Tr}({\rho }_d[-iX,\rho ])=0$. Both states remain constant. They have the identical spectrum $(1,0)$, but $\boxed{\Vert \rho (t)-{\rho }_d(t){\Vert }_{\mathrm{HS}}=\sqrt{2}\text{ for all }t}$. Moreover $iZ,iX$ generate the full special unitary Lie algebra $\mathfrak{su}(2)$, so the failure here is a limitation of this particular feedback law, not a lack of available state-transfer controls.

The only equilibrium point is $(0,0)$. Its Jacobian matrix is $\left(\begin{array}{cc}-\mu & 1\\ -1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2+\mu \lambda +1$. For $\mu >0$ both eigenvalues have negative real parts, so the origin is locally asymptotically stable.

Use the Lyapunov function $V=(x^2+y^2)/2$. Along a solution,A Lyapunov function is positive away from the equilibrium and nonincreasing along trajectories; its sufficiently small sublevel sets prove stability. Asymptotic stability additionally means nearby trajectories converge to the equilibrium. The basin of attraction is the set of initial states whose forward solutions exist for all positive time and converge to it. Every closed disc of radius at most $\sqrt{3}$ is positively invariant, since $\dot{V}\le 0$ there. The LaSalle invariance principle says that a solution in a compact positively invariant set where $\dot{V}\le 0$ approaches the largest invariant subset of $\{\dot{V}=0\}$. Here that subset is only the origin: on $x=0$, remaining on the line would require $y=0$; the isolated points $x=\pm \sqrt{3},y=0$ on the outer disc do not remain there because $\dot{y}=-x\ne 0$. Hence the closed disc $x^2+y^2\le 3$ lies in the basin of attraction.

This estimate does not assert that the basin is exactly that disc, or the whole plane. For example, choose $R$ so large that $R^2\ge 6(\mu +1)/\mu$ and $R^2>3(\mu +2)/\mu$. The region $x\ge R,y\ge -x$ is positively invariant: on $x=R$ one has $\dot{x}>0$, and on $y=-x$ one has $(x+y{)}^{\cdot }=\mu x^3/3-(\mu +2)x>0$. Within it $\dot{x}\ge \mu x^3/6$, which forces finite-time escape. Thus the attraction is genuinely local. The Lyapunov function gives a rigorous sufficient domain for attraction without identifying its entire boundary.

The First Lyapunov theorem states that for a locally Lipschitz autonomous system $\dot{x}=f(x)$ with equilibrium zero, if a $C^1$ function $V$ is positive definite near zero and $\dot{V}=\nabla V\cdot f\le 0$, then the equilibrium is Lyapunov stable. To prove it, choose a small sphere $|x|=\epsilon$ inside that neighborhood. Compactness and positivity give $m={\min }_{|x|=\epsilon }V(x)>0$. Continuity at zero gives a smaller ball $|x|<\delta$ where $V<m$. Since $V$ cannot increase along a trajectory, a trajectory starting there cannot first cross the sphere where $V\ge m$. It stays within the sphere for all future time, with continuation justified by confinement in a compact subset of the vector field's domain. This proves stability for every sufficiently small $\epsilon$.

The LaSalle invariance principle states that on a compact positively invariant set on which $\dot{V}\le 0$, every trajectory approaches the largest invariant subset of $\{\dot{V}=0\}$. If that subset is just the equilibrium, the trapping and stability argument proves asymptotic stability.

For example take $\dot{x}=y$, $\dot{y}=-x-y$ and $V=(x^2+y^2)/2$. Then $\dot{V}=-y^2$, which is not strictly negative on the punctured plane. Nevertheless its sublevel discs are compact and invariant, and a trajectory confined to $y=0$ must have $\dot{y}=-x=0$, hence $x=0$. LaSalle gives asymptotic stability despite the non-strict orbital derivative of this particular energy function.

The printed phrase about nonexistence of a strict Lyapunov function must be understood as a limitation of the chosen candidate. In this example a strict function actually exists:More generally the converse Lyapunov theorem supplies local strict functions for asymptotically stable sufficiently smooth systems. LaSalle avoids having to find such a function; it does not show that none exists.

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.