A phase portrait can be misleading near a sink. Several trajectories may initially move toward an equilibrium, yet a small perturbation can later escape to another region of the state space. Mere attraction is not enough: it says that some nearby initial conditions converge, but it does not protect the system from leaving a prescribed neighbourhood before convergence begins. Mere Lyapunov stability is not enough either: it keeps nearby solutions nearby, but it may allow them to circle forever without settling. Asymptotic stability is the point where these two requirements meet. The basic question is this: when does a long-time prediction survive small errors in the initial condition? In an autonomous system, the object of interest is often an equilibrium $x^*$, but the same idea applies to invariant sets, semigroups, and PDE evolutions. We want nearby trajectories to remain controlled for all positive time and to approach the target as $t \to \infty$. [example: Stability Without Attraction] Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be given by $f(x_1,x_2)=(-x_2,x_1)$. Since $f(0,0)=(0,0)$, the origin is an equilibrium. For initial value $x_0=(a,b)$, the proposed solution is \begin{align*} \varphi_t(x_0)=(a\cos t-b\sin t,\ a\sin t+b\cos t). \end{align*} At $t=0$ this gives \begin{align*} \varphi_0(x_0)=(a\cos 0-b\sin 0,\ a\sin 0+b\cos 0)=(a,b)=x_0. \end{align*} Differentiating componentwise, \begin{align*} \frac{d}{dt}\varphi_t(x_0)=(-a\sin t-b\cos t,\ a\cos t-b\sin t). \end{align*} On the other hand, \begin{align*} f(\varphi_t(x_0))=(-(a\sin t+b\cos t),\ a\cos t-b\sin t)=(-a\sin t-b\cos t,\ a\cos t-b\sin t), \end{align*} so $\varphi_t(x_0)$ solves $\dot{x}=f(x)$ with initial value $x_0$. Now compute its distance from the origin: \begin{align*} |\varphi_t(x_0)|^2=(a\cos t-b\sin t)^2+(a\sin t+b\cos t)^2. \end{align*} Expanding the two squares gives \begin{align*} (a\cos t-b\sin t)^2=a^2\cos^2 t-2ab\sin t\cos t+b^2\sin^2 t. \end{align*} Also, \begin{align*} (a\sin t+b\cos t)^2=a^2\sin^2 t+2ab\sin t\cos t+b^2\cos^2 t. \end{align*} Adding these identities cancels the mixed terms: \begin{align*} |\varphi_t(x_0)|^2=a^2(\cos^2 t+\sin^2 t)+b^2(\sin^2 t+\cos^2 t)=a^2+b^2=|x_0|^2. \end{align*} Hence $|\varphi_t(x_0)|=|x_0|$ for every $t\ge 0$. Therefore every initial condition close to $0$ remains equally close to $0$ for all positive time, but if $x_0\ne 0$ then $|\varphi_t(x_0)|=|x_0|>0$ for all $t$, so $\varphi_t(x_0)$ cannot converge to $0$. The system is stable in the no-escape sense, but it has no damping toward the equilibrium. [/example] The rotating system has perfect control of errors but no loss of energy. The page builds the missing notion in stages: stability, attraction, their combination, the linear test, and then the Lyapunov-function method that lets us prove asymptotic stability without solving the equation. Before naming any kind of stability, we also need to say exactly what the time-evolution maps are. The same symbol $\varphi_t$ is used throughout the page, and the definition below records both its domain of allowed initial data and its codomain of states reached at time $t$. [definition: Forward Flow] Let $D \subset \mathbb{R}^n$ and let $f:D\to\mathbb{R}^n$ be a vector field. A forward flow generated by $f$ is a family of maps $(\varphi_t)_{t\ge 0}$ with \begin{align*} \varphi_t:\Omega_t &\to D, \end{align*} where $\Omega_t\subset D$ is the set of initial data whose unique solutions are defined at least up to time $t$, such that $\varphi_0=\operatorname{id}_D$ and, whenever $s,t\ge 0$ and both sides are defined, \begin{align*} \varphi_{t+s}(x_0)=\varphi_t(\varphi_s(x_0)). \end{align*} Moreover, whenever $x_0\in\Omega_t$, the curve $s\mapsto \varphi_s(x_0)$ is defined for $0\le s\le t$ and solves $\dot{x}=f(x)$ on the interval $[0,t]$ with initial value $x_0$. [/definition] This convention keeps finite-time blow-up visible rather than hidden. If $x_0\notin\Omega_t$ for some $t$, then the trajectory has not survived long enough for a long-time stability conclusion. ## Controlled Targets and Invariance To speak about stability, the target must be meaningful under the dynamics. If a solution starts in the target and instantly leaves it, then asking whether nearby solutions stay near that target confuses motion along the target with motion away from it. The first object is therefore invariance. [definition: Positively Invariant Set] Let $D \subset \mathbb{R}^n$, let $f: D \to \mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $E \subset D$. The set $E$ is positively invariant if, for every $x_0 \in E$ and every $t\ge 0$, one has $x_0\in\Omega_t$ and $\varphi_t(x_0) \in E$. [/definition] For an equilibrium, positive invariance is automatic once $f(x^*)=0$, since the constant solution stays at $x^*$. The next problem is error control: even when the target is invariant, a nearby initial condition might make a large excursion before returning. When the target is a set rather than a point, nearness is measured by the Euclidean distance to that set. [definition: Distance to a Set] Let $D\subset\mathbb{R}^n$ and let $E\subset D$ be nonempty. The distance-to-set function associated to $E$ in the ambient Euclidean metric is the function $\operatorname{dist}(\cdot,E):D\to [0,\infty)$ defined by \begin{align*} \operatorname{dist}(x,E)=\inf_{y\in E}|x-y|. \end{align*} [/definition] The distance function turns "near the target" into a numerical condition that can be checked at every future time. This is needed because finite-time continuity of the flow only controls a trajectory on a fixed time interval; stability asks for a single initial tolerance that prevents escape for all $t\ge 0$. [definition: Lyapunov Stability of an Invariant Set] Let $D \subset \mathbb{R}^n$, let $f: D \to \mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $E \subset D$ be positively invariant. The set $E$ is Lyapunov stable if, for every $\varepsilon>0$, there exists $\delta>0$ such that every $x_0\in D$ with $\operatorname{dist}(x_0,E)<\delta$ satisfies $x_0\in\Omega_t$ and \begin{align*} \operatorname{dist}(\varphi_t(x_0),E)<\varepsilon \end{align*} for every $t\ge 0$. [/definition] Lyapunov stability is a uniform-in-time statement, stronger than continuity of any single map $\varphi_t$. It still allows circular or oscillatory motion forever, so it does not answer the long-time question. The missing condition is that nearby trajectories should lose their displacement from the target as time increases. [definition: Quasi-Asymptotic Stability] Let $D \subset \mathbb{R}^n$, let $f: D \to \mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $E \subset D$ be positively invariant. The set $E$ is quasi-asymptotically stable if there exists $r>0$ such that every $x_0\in D$ with $\operatorname{dist}(x_0,E)<r$ satisfies $x_0\in\Omega_t$ for every $t\ge 0$ and \begin{align*} \lim_{t\to\infty}\operatorname{dist}(\varphi_t(x_0),E)=0. \end{align*} [/definition] ## Definition No-escape control and eventual convergence solve different problems. Lyapunov stability alone still permits persistent oscillation near the target, while attraction alone can allow large transient excursions before convergence. The useful combined notion requires both uniform control for all future time and convergence back to the invariant set. [definition: Asymptotic Stability of an Invariant Set] Let $D \subset \mathbb{R}^n$, let $f: D \to \mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $E \subset D$ be positively invariant. The set $E$ is asymptotically stable if $E$ is Lyapunov stable and quasi-asymptotically stable with respect to this flow. [/definition] This is the page's central definition. Everything that follows is a way of recognizing the two ingredients in concrete settings: first for equilibria and basins, then for linear systems, Lyapunov functions, and semigroup evolutions. ## Equilibria and Basins To use this definition in phase portraits, we need the point-target version used for equilibria. This removes the distance-to-set notation and turns convergence to an invariant set into ordinary convergence to a fixed point. [definition: Asymptotically Stable Equilibrium] Let $D \subset \mathbb{R}^n$, let $f: D \to \mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $x^*\in D$ satisfy $f(x^*)=0$. The equilibrium $x^*$ of $\dot{x}=f(x)$ is asymptotically stable if $\{x^*\}$ is an asymptotically stable invariant set with respect to this flow. [/definition] Equivalently, $x^*$ is Lyapunov stable and there exists $r>0$ such that $|x_0-x^*|<r$ implies \begin{align*} x_0\in\Omega_t \quad \text{for every } t\ge 0, \qquad \lim_{t\to\infty}\varphi_t(x_0)=x^*. \end{align*} When $x^*=0$, this is the familiar condition $\varphi_t(x_0)\to 0$ as $t\to\infty$ for all sufficiently small initial data. The radius in this statement raises a new question: which initial conditions outside that small ball still converge to the same target? [definition: Basin of Attraction] Let $D \subset \mathbb{R}^n$, let $f: D \to \mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $E\subset D$ be positively invariant. The basin of attraction of $E$ is the set \begin{align*} \mathcal{A}(E)=\{x_0\in D : x_0\in\Omega_t \text{ for every } t\ge 0 \text{ and } \operatorname{dist}(\varphi_t(x_0),E)\to 0 \text{ as } t\to\infty\}. \end{align*} [/definition] The radius $r$ in the definition of attraction says only that some neighbourhood of $E$ is contained in $\mathcal{A}(E)$. Determining the whole basin is usually a harder global problem, and even one-dimensional examples show how sharply basins can separate. [example: A Local Basin] Consider $\dot{x}=x-x^3=x(1-x^2)$ on $\mathbb{R}$. The equilibria are exactly the roots of $x(1-x^2)=0$, namely $x=-1$, $x=0$, and $x=1$. For $x_0>0$, the solution remains positive, and setting $y(t)=x(t)^2$ gives \begin{align*} y'(t)=2x(t)x'(t)=2x(t)^2(1-x(t)^2)=2y(t)(1-y(t)). \end{align*} When $y\ne 1$, separation of variables gives \begin{align*} \frac{y'}{y(1-y)}=2. \end{align*} Since $\frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y}$, integration gives \begin{align*} \log\left|\frac{y(t)}{1-y(t)}\right|=2t+\log\left|\frac{y_0}{1-y_0}\right|. \end{align*} Solving for $y(t)$ yields \begin{align*} y(t)=\frac{1}{1+\left(\frac{1-y_0}{y_0}\right)e^{-2t}}, \end{align*} where $y_0=x_0^2$. Hence, for every $x_0>0$, \begin{align*} x(t)^2\to 1. \end{align*} Because $x(t)>0$, this implies $x(t)\to 1$. The sign of $x(1-x^2)$ also gives the direction of motion: it is positive on $(0,1)$ and negative on $(1,\infty)$, so positive solutions move monotonically toward $1$. Thus nearby positive initial data stay between $x_0$ and $1$, which gives Lyapunov stability of $1$, and the basin of attraction of $1$ is $(0,\infty)$. The origin is not Lyapunov stable. If $0<x_0<1$, then $x'(t)=x(t)(1-x(t)^2)>0$ as long as $x(t)\in(0,1)$, and the computation above gives $x(t)\to 1$. Therefore arbitrarily small positive perturbations of $0$ eventually leave, for example, the interval $(-1/2,1/2)$. [/example] This example separates local and global language. Asymptotic stability is local unless a larger basin is specified. ## Linear Systems and Spectral Decay Linear systems provide the cleanest test case because every solution is given by a matrix exponential. The question becomes spectral: which eigenvalues of $A$ force $e^{tA}x_0$ to decay to zero for all small, or all, initial data? Before stating the criterion, we isolate the linear form of the definition. Since linearity scales initial data, a local convergence statement automatically extends to every $x_0\in\mathbb{R}^n$ once it holds in a ball around the origin. [definition: Asymptotic Stability for a Linear System] Let $A\in\mathbb{R}^{n\times n}$ and consider $\dot{x}=Ax$. The equilibrium $0$ is asymptotically stable if it is Lyapunov stable and there exists $r>0$ such that, for every $x_0\in\mathbb{R}^n$ with $|x_0|<r$, the solution satisfies \begin{align*} e^{tA}x_0\to 0 \end{align*} as $t\to\infty$. [/definition] For linear systems we will also use the stronger quantitative notion of exponential stability: there are constants $C\ge 1$ and $\alpha>0$ such that \begin{align*} \|e^{tA}x_0\|\le C e^{-\alpha t}\|x_0\| \end{align*} for every $x_0\in\mathbb{R}^n$ and every $t\ge 0$. This implies asymptotic stability, but it records a uniform rate of decay rather than only convergence. The definition mentions a radius, but the spectral criterion has no radius because linearity removes the distinction between local and global decay. The theorem below is the main computational test for constant-coefficient systems: it translates long-time behaviour of all trajectories into the position of finitely many complex numbers, and in fact identifies when the stronger exponential estimate holds. [quotetheorem:6367] The estimate is stronger than convergence: it gives a quantitative decay rate. The constant $C$ accounts for non-normal transient growth, while $\alpha$ is controlled by how far the spectrum lies inside the left half-plane. Concrete two-dimensional systems show how this estimate appears in phase portraits. [example: Spiral Sink] Let $A\in\mathbb{R}^{2\times 2}$ be defined by $A(x_1,x_2)=(-x_1-2x_2,2x_1-x_2)$. Its characteristic polynomial is \begin{align*} \det(\lambda I-A)=(\lambda+1)^2+4=\lambda^2+2\lambda+5. \end{align*} Solving $\lambda^2+2\lambda+5=0$ gives \begin{align*} \lambda=\frac{-2\pm\sqrt{4-20}}{2}=-1\pm 2i. \end{align*} Both eigenvalues have real part $-1<0$, so *Spectral Criterion for Linear Asymptotic Stability* applies and the origin is asymptotically stable. For $x_0=(a,b)$, define \begin{align*} x(t)=e^{-t}(a\cos(2t)-b\sin(2t),a\sin(2t)+b\cos(2t)). \end{align*} At $t=0$ this gives $x(0)=(a,b)$. Differentiating the first component gives \begin{align*} x_1'(t)=-e^{-t}(a\cos(2t)-b\sin(2t))+e^{-t}(-2a\sin(2t)-2b\cos(2t)). \end{align*} Differentiating the second component gives \begin{align*} x_2'(t)=-e^{-t}(a\sin(2t)+b\cos(2t))+e^{-t}(2a\cos(2t)-2b\sin(2t)). \end{align*} Since $x_1(t)=e^{-t}(a\cos(2t)-b\sin(2t))$ and $x_2(t)=e^{-t}(a\sin(2t)+b\cos(2t))$, these identities are exactly \begin{align*} x_1'(t)=-x_1(t)-2x_2(t). \end{align*} and \begin{align*} x_2'(t)=2x_1(t)-x_2(t). \end{align*} Thus $x'(t)=Ax(t)$, so $x(t)=e^{tA}x_0$. Finally, \begin{align*} |e^{tA}x_0|^2=e^{-2t}\bigl((a\cos(2t)-b\sin(2t))^2+(a\sin(2t)+b\cos(2t))^2\bigr). \end{align*} Expanding the two squares gives \begin{align*} (a\cos(2t)-b\sin(2t))^2=a^2\cos^2(2t)-2ab\sin(2t)\cos(2t)+b^2\sin^2(2t). \end{align*} and \begin{align*} (a\sin(2t)+b\cos(2t))^2=a^2\sin^2(2t)+2ab\sin(2t)\cos(2t)+b^2\cos^2(2t). \end{align*} Adding cancels the mixed terms and uses $\sin^2(2t)+\cos^2(2t)=1$: \begin{align*} |e^{tA}x_0|^2=e^{-2t}(a^2+b^2)=e^{-2t}|x_0|^2. \end{align*} Therefore $|e^{tA}x_0|=e^{-t}|x_0|$ for every $t\ge 0$, so trajectories rotate while their distance from the origin decays exponentially. [/example] The failure cases are just as instructive. An eigenvalue on the imaginary axis may preserve distance, and an eigenvalue with positive real part creates exponential escape along some direction. [example: Imaginary Spectrum Blocks Attraction] Let $A\in\mathbb{R}^{2\times 2}$ be defined by $A(x_1,x_2)=(-x_2,x_1)$. In the standard basis, its entries are $A_{11}=0$, $A_{12}=-1$, $A_{21}=1$, and $A_{22}=0$. Hence \begin{align*} \det(\lambda I-A)=\lambda^2+1. \end{align*} Thus $\lambda^2+1=0$, so the eigenvalues are $\lambda=i$ and $\lambda=-i$. For $x_0=(a,b)$, define \begin{align*} x(t)=(a\cos t-b\sin t,\ a\sin t+b\cos t). \end{align*} At $t=0$, \begin{align*} x(0)=(a\cos 0-b\sin 0,\ a\sin 0+b\cos 0)=(a,b)=x_0. \end{align*} Differentiating componentwise gives \begin{align*} x'(t)=(-a\sin t-b\cos t,\ a\cos t-b\sin t). \end{align*} Since \begin{align*} Ax(t)=(-(a\sin t+b\cos t),\ a\cos t-b\sin t)=(-a\sin t-b\cos t,\ a\cos t-b\sin t), \end{align*} we have $x'(t)=Ax(t)$ and $x(0)=x_0$, so $x(t)=e^{tA}x_0$. Now compute the distance from the origin: \begin{align*} |e^{tA}x_0|^2=(a\cos t-b\sin t)^2+(a\sin t+b\cos t)^2. \end{align*} The first square is \begin{align*} (a\cos t-b\sin t)^2=a^2\cos^2 t-2ab\sin t\cos t+b^2\sin^2 t. \end{align*} The second square is \begin{align*} (a\sin t+b\cos t)^2=a^2\sin^2 t+2ab\sin t\cos t+b^2\cos^2 t. \end{align*} Adding these two identities cancels the mixed terms and uses $\sin^2 t+\cos^2 t=1$: \begin{align*} |e^{tA}x_0|^2=a^2+b^2=|x_0|^2. \end{align*} Therefore $|e^{tA}x_0|=|x_0|$ for every $t\ge 0$. Given $\varepsilon>0$, choosing $\delta=\varepsilon$ gives $|x_0|<\delta \Rightarrow |e^{tA}x_0|<\varepsilon$ for every $t\ge 0$, so the origin is Lyapunov stable. But if $x_0\ne 0$, then $|e^{tA}x_0|=|x_0|>0$ for every $t\ge 0$, so $e^{tA}x_0$ cannot converge to $0$. Thus the imaginary eigenvalues produce bounded rotation, not attraction, and the origin is not asymptotically stable. [/example] Linear stability also explains why nonlinear analysis starts with the Jacobian. Near a hyperbolic equilibrium, the linear part gives the first-order model and the nonlinear remainder is smaller than the displacement. ## Linearisation and Nonlinear Equilibria A nonlinear vector field can behave like its linearisation near an equilibrium, but only when the linearisation has no spectrum on the imaginary axis. If the linear part decays exponentially, the nonlinear terms are perturbative in a small ball. If the linear part has a neutral direction, higher-order terms decide the outcome. To state the principle, we need the standard derivative notation at an equilibrium. The total derivative is a [linear map](/page/Linear%20Map), while its matrix representation is the Jacobian matrix, whose eigenvalues control the first-order phase portrait. [definition: Hyperbolic Equilibrium] Let $D\subset\mathbb{R}^n$ be open, let $f\in C^1(D;\mathbb{R}^n)$, and let $x^*\in D$ satisfy $f(x^*)=0$. Let $Df_{x^*}:\mathbb{R}^n\to\mathbb{R}^n$ be the total derivative at $x^*$, and let $Jf_{x^*}\in\mathbb{R}^{n\times n}$ be its Jacobian matrix in the standard basis. The equilibrium $x^*$ is hyperbolic if no eigenvalue $\lambda\in\mathbb{C}$ of $Jf_{x^*}$ satisfies \begin{align*} \operatorname{Re}(\lambda)=0. \end{align*} [/definition] Hyperbolicity is a nondegeneracy condition. It excludes centres and other borderline cases where the linear approximation does not settle the stability question. When all hyperbolic directions point inward, the nonlinear flow inherits the sink behaviour of its linear part. [quotetheorem:6852] This theorem is often the first practical test. It turns a local nonlinear problem into a finite-dimensional spectral computation, which is why the Jacobian is the first object computed in many models. [example: Damped Pendulum Near the Downward Equilibrium] For the damped pendulum \begin{align*} \dot{\theta}=\omega,\qquad \dot{\omega}=-\sin\theta-c\omega \end{align*} with $c>0$, write $f(\theta,\omega)=(\omega,-\sin\theta-c\omega)$. Since \begin{align*} f(0,0)=(0,-\sin 0-c\cdot 0)=(0,0), \end{align*} the downward position $(0,0)$ is an equilibrium. The partial derivatives are \begin{align*} \frac{\partial f_1}{\partial \theta}=0,\qquad \frac{\partial f_1}{\partial \omega}=1,\qquad \frac{\partial f_2}{\partial \theta}=-\cos\theta,\qquad \frac{\partial f_2}{\partial \omega}=-c. \end{align*} At $(0,0)$ these give $Jf_{(0,0),11}=0$, $Jf_{(0,0),12}=1$, $Jf_{(0,0),21}=-1$, and $Jf_{(0,0),22}=-c$. Therefore the entries of $\lambda I-Jf_{(0,0)}$ are $\lambda$, $-1$, $1$, and $\lambda+c$, so the characteristic polynomial is \begin{align*} \det(\lambda I-Jf_{(0,0)})=\lambda(\lambda+c)-(-1)(1)=\lambda^2+c\lambda+1. \end{align*} The eigenvalues solve \begin{align*} \lambda^2+c\lambda+1=0. \end{align*} By the [quadratic formula](/theorems/1301), \begin{align*} \lambda=\frac{-c\pm\sqrt{c^2-4}}{2}. \end{align*} If $0<c<2$, then $\sqrt{c^2-4}=i\sqrt{4-c^2}$, so both eigenvalues have real part $-c/2<0$. If $c=2$, both eigenvalues are equal to $-1$. If $c>2$, then $0<\sqrt{c^2-4}<c$, hence \begin{align*} \frac{-c+\sqrt{c^2-4}}{2}<0 \end{align*} and \begin{align*} \frac{-c-\sqrt{c^2-4}}{2}<0. \end{align*} Thus every eigenvalue of $Jf_{(0,0)}$ has negative real part for every $c>0$. By *Linearisation Criterion for Asymptotic Stability*, the equilibrium $(0,0)$ is asymptotically stable, so small perturbations of the downward pendulum remain controlled and decay back to rest. [/example] The next example shows why the hypothesis cannot be replaced by mere Lyapunov stability of the linearisation. [example: Linearisation With a Zero Eigenvalue] Consider $\dot{x}=-x^3$ on $\mathbb{R}$ with initial value $x(0)=x_0$. If $x_0=0$, then the solution is the constant solution $x(t)=0$. Now assume $x_0\ne 0$. Separating variables gives \begin{align*} x^{-3}\frac{dx}{dt}=-1. \end{align*} Since $\frac{d}{dt}x(t)^{-2}=-2x(t)^{-3}x'(t)$, the differential equation implies \begin{align*} \frac{d}{dt}x(t)^{-2}=2. \end{align*} Integrating from $0$ to $t$ gives \begin{align*} x(t)^{-2}-x_0^{-2}=2t. \end{align*} Thus \begin{align*} x(t)^2=\frac{1}{x_0^{-2}+2t}=\frac{x_0^2}{1+2t x_0^2}. \end{align*} Because the differential equation has a unique solution and the formula has the same sign as $x_0$, we obtain \begin{align*} x(t)=\frac{x_0}{\sqrt{1+2t x_0^2}}. \end{align*} This formula gives \begin{align*} |x(t)|=\frac{|x_0|}{\sqrt{1+2t x_0^2}}\le |x_0| \end{align*} for every $t\ge 0$. Hence, given $\varepsilon>0$, choosing $\delta=\varepsilon$ ensures $|x_0|<\delta \Rightarrow |x(t)|<\varepsilon$ for every $t\ge 0$, so the origin is Lyapunov stable. Also, if $x_0\ne 0$, then $1+2t x_0^2\to\infty$, and therefore \begin{align*} \frac{x_0}{\sqrt{1+2t x_0^2}}\to 0. \end{align*} Thus every initial condition converges to $0$, and the origin is asymptotically stable. For $f(x)=-x^3$, the derivative is \begin{align*} f'(x)=-3x^2. \end{align*} At the origin this gives $f'(0)=0$, so the linearised equation is $\dot{y}=0$, whose solutions are constant. The nonlinear equation still decays to $0$, so a zero eigenvalue makes the linearisation test inconclusive rather than false. [/example] When the Jacobian has an eigenvalue with positive real part, the linear model predicts that nearby points move away from the equilibrium in an expanding direction. A nonlinear remainder cannot remove that expansion in a sufficiently small neighbourhood, so the natural companion to the stability theorem is an instability test. [quotetheorem:698] Together, the two linearisation criteria handle hyperbolic equilibria. The remaining cases demand a different method, and that method is to find a quantity that decreases along trajectories. ## Lyapunov Functions Solving a nonlinear ODE explicitly is rare. A Lyapunov function replaces the solution formula with an energy-like scalar quantity. If the energy is positive away from the equilibrium and decreases along every nearby trajectory, the dynamics is forced toward lower energy levels. The relevant derivative is taken along the vector field, not independently in time. It measures the instantaneous change of $V$ along the solution curve, so it turns a scalar function on state space into a diagnostic for the whole flow. [definition: Orbital Derivative of a Lyapunov Function] Let $D\subset\mathbb{R}^n$ be open, let $f\in C^1(D;\mathbb{R}^n)$, and let $V\in C^1(D;\mathbb{R})$. The orbital derivative of $V$ along $\dot{x}=f(x)$ is the function $\dot{V}:D\to\mathbb{R}$ defined by \begin{align*} \dot{V}(x)=\nabla V(x)\cdot f(x). \end{align*} [/definition] Along a solution $x(t)$, the chain rule gives $\frac{d}{dt}V(x(t))=\dot{V}(x(t))$. To use this derivative for stability, the scalar quantity must play two roles at once: it must vanish only at the equilibrium, so small energy means small displacement, and it must not increase along trajectories, so small-energy neighbourhoods become barriers. [definition: Lyapunov Function] Let $D\subset\mathbb{R}^n$ be open, let $f\in C^1(D;\mathbb{R}^n)$, and let $x^*\in D$ satisfy $f(x^*)=0$. A function $V\in C^1(D;\mathbb{R})$ is a Lyapunov function for $x^*$ on a neighbourhood $U\subset D$ if $x^*\in U$, $V(x^*)=0$, $V(x)>0$ for every $x\in U\setminus\{x^*\}$, and $\dot{V}(x)\le 0$ for every $x\in U$. [/definition] This definition proves stability but not necessarily convergence. To force convergence to the equilibrium itself, the orbital derivative needs to be strictly negative away from the equilibrium, so that every non-equilibrium state experiences energy loss. [definition: Strict Lyapunov Function] Let $D\subset\mathbb{R}^n$ be open, let $f\in C^1(D;\mathbb{R}^n)$, and let $x^*\in D$ satisfy $f(x^*)=0$. A strict Lyapunov function for $x^*$ on a neighbourhood $U\subset D$ is a function $V\in C^1(D;\mathbb{R})$ such that $V(x^*)=0$, $V(x)>0$ for every $x\in U\setminus\{x^*\}$, and $\dot{V}(x)<0$ for every $x\in U\setminus\{x^*\}$. [/definition] The sign conditions above are designed to make sublevel sets into barriers. If $V$ cannot increase along trajectories, a solution that starts inside a small sublevel set remains inside it. This is the mechanism behind Lyapunov stability. [quotetheorem:7603] A barrier argument only keeps trajectories in small sublevel sets; it does not by itself rule out motion forever inside such a set. The strict version of the Lyapunov condition is therefore the part of the criterion that upgrades stability to attraction: away from the equilibrium the energy must keep decreasing, so persistent nonzero limiting behaviour is incompatible with the sign condition. A quadratic Lyapunov function often certifies linear decay and then extends to nonlinear perturbations. The next example is the algebraic model behind many energy estimates. [example: Quadratic Energy for a Linear Sink] Let $A\in\mathbb{R}^{n\times n}$ have all eigenvalues in the open left half-plane, and consider $\dot{x}=Ax$. Choose a symmetric positive definite matrix $P$ satisfying \begin{align*} A^\top P+PA=-I. \end{align*} Define $V:\mathbb{R}^n\to\mathbb{R}$ by $V(x)=x^\top Px$. Since $P$ is positive definite, $V(0)=0$ and $V(x)>0$ for every $x\ne 0$. Let $x(t)$ be a solution of $\dot{x}=Ax$. Differentiating $V(x(t))=x(t)^\top P x(t)$ by the product rule gives \begin{align*} \frac{d}{dt}V(x(t))=x'(t)^\top P x(t)+x(t)^\top P x'(t). \end{align*} Using $x'(t)=Ax(t)$, this becomes \begin{align*} \frac{d}{dt}V(x(t))=(Ax(t))^\top P x(t)+x(t)^\top P A x(t). \end{align*} Since $(Ax)^\top=x^\top A^\top$, the orbital derivative is \begin{align*} \dot{V}(x)=x^\top A^\top P x+x^\top PAx. \end{align*} Factoring out the common vector $x$ gives \begin{align*} \dot{V}(x)=x^\top(A^\top P+PA)x. \end{align*} Substituting $A^\top P+PA=-I$ gives \begin{align*} \dot{V}(x)=x^\top(-I)x. \end{align*} Therefore \begin{align*} \dot{V}(x)=-x^\top x=-|x|^2. \end{align*} Thus $\dot{V}(x)<0$ for every $x\ne 0$, so $V$ is a strict Lyapunov function for the origin. Finally, because $P$ is symmetric positive definite, the number $m=\min_{|y|=1}y^\top Py$ is positive, and hence \begin{align*} V(x)\ge m|x|^2. \end{align*} So each sublevel set $\{x:V(x)\le c\}$ is closed and bounded, hence compact in $\mathbb{R}^n$. By *Strict Lyapunov Criterion for Asymptotic Stability*, the origin is asymptotically stable. [/example] Strict negativity can be too demanding. Many mechanical systems dissipate only some variables directly, and convergence follows because the only trajectory that can remain in the zero-dissipation set is the equilibrium. To state that idea, we need a way to discuss invariant pieces of a specified set. [definition: Invariant Subset of a Set] Let $D\subset\mathbb{R}^n$, let $f:D\to\mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $M\subset D$. A subset $N\subset M$ is invariant for the flow if, for every $x_0\in N$ and every $t\ge 0$, one has $x_0\in\Omega_t$ and $\varphi_t(x_0)\in N$. [/definition] When $\dot{V}\le 0$ is not strict, an energy argument can trap a trajectory without forcing the energy to drop at every non-equilibrium point. The obstruction is the zero-dissipation set, where the Lyapunov function gives no instantaneous decrease. Convergence is then controlled by the invariant trajectories that can remain inside that set forever. [quotetheorem:7104] LaSalle's principle is the bridge from weak dissipation to convergence. It is especially useful when $\dot{V}$ sees damping but not restoring forces directly. [example: Damped Pendulum Energy] For the damped pendulum with $c>0$, \begin{align*} \dot{\theta}=\omega,\qquad \dot{\omega}=-\sin\theta-c\omega, \end{align*} define \begin{align*} V(\theta,\omega)=1-\cos\theta+\frac{1}{2}\omega^2. \end{align*} Near $(0,0)$ this is positive away from the origin: if $|\theta|<\pi$ and $(\theta,\omega)\ne(0,0)$, then either $\omega\ne 0$, giving $\frac{1}{2}\omega^2>0$, or $\omega=0$ and $\theta\ne 0$, giving $1-\cos\theta>0$. The partial derivatives of $V$ are \begin{align*} \frac{\partial V}{\partial \theta}(\theta,\omega)=\sin\theta,\qquad \frac{\partial V}{\partial \omega}(\theta,\omega)=\omega. \end{align*} Therefore the orbital derivative along the vector field is \begin{align*} \dot{V}(\theta,\omega)=\sin\theta\cdot \omega+\omega\cdot(-\sin\theta-c\omega). \end{align*} Expanding the second term gives \begin{align*} \dot{V}(\theta,\omega)=\sin\theta\,\omega-\omega\sin\theta-c\omega^2. \end{align*} The first two terms cancel, so \begin{align*} \dot{V}(\theta,\omega)=-c\omega^2\le 0. \end{align*} The zero-dissipation set is therefore \begin{align*} M=\{(\theta,\omega):\dot{V}(\theta,\omega)=0\}=\{(\theta,\omega):\omega=0\}, \end{align*} because $c>0$ and $\omega^2=0$ exactly when $\omega=0$. If a trajectory remains in $M$, then $\omega(t)=0$ for all $t$, hence $\dot{\theta}(t)=\omega(t)=0$. It must also satisfy \begin{align*} 0=\dot{\omega}(t)=-\sin\theta(t)-c\cdot 0=-\sin\theta(t). \end{align*} Thus $\sin\theta(t)=0$. In a neighbourhood with $|\theta|<\pi$, the only solution is $\theta(t)=0$, so the largest invariant subset of $M$ in that neighbourhood is $\{(0,0)\}$. Small sublevel sets of $V$ are compact and stay inside such a neighbourhood, $V$ is nonincreasing along trajectories, and the only invariant zero-dissipation point nearby is $(0,0)$. By *LaSalle Invariance Principle*, nearby damped-pendulum trajectories converge to $(0,0)$; the energy also gives Lyapunov stability, so the downward equilibrium is asymptotically stable. [/example] ## Attraction, Basins, and Global Stability Local asymptotic stability tells us what happens after sufficiently small perturbations. Applications often ask a stronger question: which initial states converge, and can the permitted set be all of the state space? The global version removes the local radius from attraction and stability. It is a strong statement and usually requires proper energy functions or special order structures. [definition: Global Asymptotic Stability] Let $D\subset\mathbb{R}^n$, let $f:D\to\mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $x^*\in D$ be an equilibrium. The equilibrium $x^*$ is globally asymptotically stable in $D$ if it is Lyapunov stable, every $x_0\in D$ satisfies $x_0\in\Omega_t$ for every $t\ge 0$, and \begin{align*} \lim_{t\to\infty}\varphi_t(x_0)=x^* \end{align*} for every $x_0\in D$. [/definition] The definition of global asymptotic stability separates convergence from the compactness needed to prove it. To make a Lyapunov function useful on all of $D$, its bounded sublevel sets must confine trajectories to regions where limiting behaviour can be extracted. For a global Lyapunov argument, bounded energy must also keep the state in a bounded, closed region of phase space. Otherwise a decreasing energy estimate can coexist with escape along an unbounded sublevel set, so the energy no longer gives compactness or accumulation information. This motivates the following compactness condition on the energy function itself. [definition: Proper Function] Let $D\subset\mathbb{R}^n$ and let $V:D\to\mathbb{R}$. The function $V$ is proper on $D$ if, for every $c\in\mathbb{R}$, the set $\{x\in D:V(x)\le c\}$ is compact as a subset of $\mathbb{R}^n$. [/definition] Properness is the condition that bounded energy prevents escape to spatial infinity or to the boundary of the chosen domain. It does not by itself guarantee that solutions exist for all positive time; that is a separate dynamical hypothesis in the theorem below. Combined with global existence and strict decay, properness upgrades a local barrier argument into a global one because every trajectory remains in a compact sublevel set where accumulation can be analysed. [quotetheorem:7892] The finite-time existence hypothesis is not cosmetic. A decreasing local energy does not prevent blow-up outside the region where the energy controls the state. [example: Attraction Is Not Global Stability] Consider the scalar equation $\dot{x}=-x+x^2=x(x-1)$ on $\mathbb{R}$. The equilibria are $x=0$ and $x=1$, since $x(x-1)=0$ exactly for those two values. For an initial value $x(0)=x_0$ with $x_0\ne 0,1$, separation of variables gives \begin{align*} \frac{x'(t)}{x(t)(x(t)-1)}=1. \end{align*} The partial fraction identity \begin{align*} \frac{1}{x(x-1)}=-\frac{1}{x}+\frac{1}{x-1} \end{align*} turns this into \begin{align*} -\frac{x'(t)}{x(t)}+\frac{x'(t)}{x(t)-1}=1. \end{align*} Integrating from $0$ to $t$ gives \begin{align*} -\log|x(t)|+\log|x(t)-1|=-\log|x_0|+\log|x_0-1|+t. \end{align*} Equivalently, \begin{align*} \log\left|\frac{x(t)-1}{x(t)}\right|=\log\left|\frac{x_0-1}{x_0}\right|+t. \end{align*} Solving the resulting identity with the initial sign fixed by $x(0)=x_0$ gives \begin{align*} \frac{x(t)-1}{x(t)}=\frac{x_0-1}{x_0}e^t. \end{align*} Thus \begin{align*} x(t)=\frac{x_0}{x_0+(1-x_0)e^t}. \end{align*} If $x_0<1$, then the denominator $x_0+(1-x_0)e^t$ is never zero for $t\ge 0$: for $0<x_0<1$ both terms are positive, and for $x_0<0$ the equation $x_0+(1-x_0)e^t=0$ would require $e^t=-x_0/(1-x_0)<1$, which cannot happen for $t\ge 0$. Hence every solution with $x_0<1$ exists for all positive time. Since $1-x_0>0$, \begin{align*} \lim_{t\to\infty}\frac{x_0}{x_0+(1-x_0)e^t}=0. \end{align*} Together with the constant solution from $x_0=0$, this shows that every initial condition in $(-\infty,1)$ converges to the origin. If $x_0>1$, then $1-x_0<0$, and the denominator vanishes when \begin{align*} x_0+(1-x_0)e^t=0. \end{align*} Solving for $t$ gives \begin{align*} e^t=\frac{x_0}{x_0-1}. \end{align*} Since $x_0/(x_0-1)>1$, the blow-up time \begin{align*} T=\log\left(\frac{x_0}{x_0-1}\right) \end{align*} is positive. As $t\uparrow T$, the denominator tends to $0$ through positive values, so $x(t)\to+\infty$. Therefore solutions starting at $x_0>1$ do not exist for all $t\ge 0$, and the origin is not globally asymptotically stable on $\mathbb{R}$. Its basin of attraction is $(-\infty,1)$, not the whole line. [/example] This example also shows why basins matter. Stability is local near the target, while global conclusions depend on the entire phase space. ## Semigroups and Infinite-Dimensional Evolutions Many evolution equations are not finite-dimensional ODEs. Heat flow, damped waves, and abstract Cauchy problems are naturally written as semigroups on Banach or Hilbert spaces. The same stability logic survives, but norms replace Euclidean distance. The semigroup definition records time evolution without requiring a finite-dimensional phase portrait. At this level it only encodes the algebra of time evolution; when the maps come from PDE, additional analytic hypotheses such as strong continuity or linear boundedness are often added later to prove estimates. [definition: Semigroup Equilibrium] Let $X$ be a [normed vector space](/page/Normed%20Vector%20Space) and let $(S(t))_{t\ge 0}$ be a family of maps $S(t):X\to X$ satisfying $S(0)=\operatorname{id}_X$ and $S(t+s)=S(t)S(s)$ for all $s,t\ge 0$. A point $0\in X$ is an equilibrium of the semigroup if $S(t)0=0$ for every $t\ge 0$. [/definition] For linear PDE, $S(t)$ is often a strongly continuous linear semigroup of bounded operators. The stability definition itself only needs the norm and the forward maps, because the central issue is still uniform control of all future states that start near the equilibrium. [definition: Stability for a Semigroup] Let $X$ be a normed [vector space](/page/Vector%20Space) and let $(S(t))_{t\ge 0}$ be a semigroup on $X$ with equilibrium $0$. The equilibrium $0$ is stable if, for every $\varepsilon>0$, there exists $\delta>0$ such that \begin{align*} \|x\|_X<\delta \implies \|S(t)x\|_X<\varepsilon \quad \text{for every } t\ge 0. \end{align*} [/definition] Semigroup stability controls all future iterates that start near zero, but it still allows persistent bounded motion in the chosen norm. To express decay, the same norm must also measure whether the orbit actually tends to zero; this choice is mathematically important because convergence in $L^2$ may hold even when convergence in a stronger Sobolev norm does not. [definition: Asymptotic Stability for a Semigroup] Let $X$ be a normed vector space and let $(S(t))_{t\ge 0}$ be a semigroup on $X$ with equilibrium $0$. The equilibrium $0$ is asymptotically stable if it is stable and there exists $r>0$ such that \begin{align*} \|x\|_X<r \implies \lim_{t\to\infty}\|S(t)x\|_X=0. \end{align*} [/definition] ### Exponential decay estimates For linear semigroups, exponential decay is a powerful sufficient condition. It packages stability and attraction into one norm estimate, making it possible to compare infinite-dimensional evolution with the finite-dimensional estimate $|e^{tA}x_0|\le C e^{-\alpha t}|x_0|$. [definition: Exponential Stability of a Semigroup] Let $X$ be a normed vector space and let $(S(t))_{t\ge 0}$ be a semigroup such that $S(t):X\to X$ is a [bounded linear operator](/page/Bounded%20Linear%20Operator) for every $t\ge 0$. The semigroup is exponentially stable if there exist constants $M\ge 1$ and $\omega>0$ such that \begin{align*} \|S(t)x\|_X \le M e^{-\omega t}\|x\|_X \end{align*} for every $x\in X$ and every $t\ge 0$. [/definition] ### From exponential decay to asymptotic stability The definition gives a checkable estimate, but in nonlinear evolution problems the estimate is usually available only for the linearized semigroup. The remaining question is whether the nonlinear terms are small enough near the equilibrium that this linear decay still controls the full flow. In infinite-dimensional evolution problems, direct trajectory formulas are often unavailable, so stability is usually certified by combining a uniform semigroup estimate with a perturbation argument. Exponential decay supplies the integrable factor that controls accumulated nonlinear errors, while the smallness assumptions on the nonlinear part prevent those errors from overwhelming the decay. This motivates the precise implication needed here: exponential stability of the linearized semigroup can imply local asymptotic stability for a semilinear equilibrium, but only under hypotheses that keep the nonlinear remainder subordinate to the linear decay. [quotetheorem:7077] The theorem is a local result, not a blanket statement about every nonlinear perturbation. Its hypotheses are doing real work: the exponential bound controls the linearized motion, and the smallness or regularity condition on the nonlinear term keeps nearby trajectories in the region where the perturbative estimate is valid. In applications below, we first verify exponential decay for a concrete linear semigroup, then use this decay estimate as the input that a semilinear stability theorem requires. ### Heat semigroup example The [heat equation](/page/Heat%20Equation) gives the model example. Diffusion damps spatial variation, and boundary conditions remove constant modes. [example: Heat Semigroup on a Bounded Domain] Let $U\subset\mathbb{R}^n$ be a bounded connected Lipschitz domain, and consider the heat equation \begin{align*} \partial_t u-\Delta u=0 \end{align*} with zero Dirichlet boundary condition $u|_{\partial U}=0$. On $X=L^2(U)$, write $S(t)u_0=u(t)$ for the Dirichlet heat semigroup, so $0$ is an equilibrium because the zero initial datum gives the zero solution. First take $u_0$ in the operator domain of the Dirichlet Laplacian. Then $u(t)$ has zero trace on $\partial U$, and \begin{align*} \frac{d}{dt}\|u(t)\|_{L^2(U)}^2=\frac{d}{dt}\int_U |u(t,x)|^2\,dx. \end{align*} Differentiating under the integral gives \begin{align*} \frac{d}{dt}\|u(t)\|_{L^2(U)}^2=2\int_U u(t,x)\partial_t u(t,x)\,dx. \end{align*} Using $\partial_t u=\Delta u$, this becomes \begin{align*} \frac{d}{dt}\|u(t)\|_{L^2(U)}^2=2\int_U u(t,x)\Delta u(t,x)\,dx. \end{align*} Green's identity and the boundary condition $u|_{\partial U}=0$ give \begin{align*} \int_U u(t,x)\Delta u(t,x)\,dx=-\int_U |\nabla u(t,x)|^2\,dx. \end{align*} Therefore \begin{align*} \frac{d}{dt}\|u(t)\|_{L^2(U)}^2=-2\|\nabla u(t)\|_{L^2(U)}^2. \end{align*} By the *Poincare inequality* for $H_0^1(U)$, there is a constant $C_U>0$ such that \begin{align*} \|v\|_{L^2(U)}\le C_U\|\nabla v\|_{L^2(U)} \end{align*} for every $v\in H_0^1(U)$. Applying this to $v=u(t)$ gives \begin{align*} \|\nabla u(t)\|_{L^2(U)}^2\ge C_U^{-2}\|u(t)\|_{L^2(U)}^2. \end{align*} Hence, with $E(t)=\|u(t)\|_{L^2(U)}^2$, \begin{align*} E'(t)\le -2C_U^{-2}E(t). \end{align*} Now \begin{align*} \frac{d}{dt}\left(e^{2C_U^{-2}t}E(t)\right)=e^{2C_U^{-2}t}\left(E'(t)+2C_U^{-2}E(t)\right)\le 0. \end{align*} Thus $e^{2C_U^{-2}t}E(t)\le E(0)$, so \begin{align*} \|u(t)\|_{L^2(U)}^2\le e^{-2C_U^{-2}t}\|u_0\|_{L^2(U)}^2. \end{align*} Taking square roots gives \begin{align*} \|S(t)u_0\|_{L^2(U)}\le e^{-C_U^{-2}t}\|u_0\|_{L^2(U)}. \end{align*} The operator domain of the Dirichlet Laplacian is dense in $L^2(U)$, and the heat semigroup is strongly continuous on $L^2(U)$, so the same estimate passes to every $u_0\in L^2(U)$ by approximating $u_0$ in $L^2(U)$. Therefore $S(t)$ is exponentially stable with constants $M=1$ and $\omega=C_U^{-2}$, and the zero solution is exponentially stable in $L^2(U)$. [/example] This example also warns that stability depends on the space and boundary conditions. On a periodic domain, constants do not decay, so the zero equilibrium does not attract initial data with nonzero spatial average. ## Rates, Robustness, and Failure Modes Asymptotic stability is qualitative: convergence occurs, but the definition does not specify a rate. In applications, rate information controls settling time and sensitivity to perturbations. The most common rate is exponential decay. It is stronger than asymptotic stability and behaves well under many perturbations, so it deserves a separate name in finite-dimensional dynamics. [definition: Exponentially Stable Equilibrium] Let $D\subset\mathbb{R}^n$, let $f:D\to\mathbb{R}^n$ generate a forward flow $(\varphi_t:\Omega_t\to D)_{t\ge 0}$, and let $x^*\in D$ be an equilibrium. The equilibrium $x^*$ is exponentially stable if there exist constants $C\ge 1$, $\alpha>0$, and $r>0$ such that every $x_0\in D$ with $|x_0-x^*|<r$ satisfies $x_0\in\Omega_t$ and \begin{align*} |\varphi_t(x_0)-x^*|\le C e^{-\alpha t}|x_0-x^*| \end{align*} for every $t\ge 0$. [/definition] Exponential stability gives both Lyapunov stability and attraction by a single estimate. Many asymptotically stable nonlinear equilibria are not exponentially stable because the vector field is too flat at the equilibrium. [example: Asymptotic but Not Exponential] Consider $\dot{x}=-x^3$ with $x(0)=x_0$. If $x_0=0$, then $x(t)=0$ for all $t\ge 0$. Now let $x_0\ne 0$. Since $x(t)$ cannot cross $0$ before reaching it, we separate variables on any interval where $x(t)\ne 0$: \begin{align*} x(t)^{-3}x'(t)=-1. \end{align*} Because $\frac{d}{dt}x(t)^{-2}=-2x(t)^{-3}x'(t)$, the differential equation gives \begin{align*} \frac{d}{dt}x(t)^{-2}=2. \end{align*} Integrating from $0$ to $t$ gives \begin{align*} x(t)^{-2}=x_0^{-2}+2t. \end{align*} Taking reciprocals gives \begin{align*} x(t)^2=\frac{1}{x_0^{-2}+2t}=\frac{x_0^2}{1+2t x_0^2}. \end{align*} The sign of the solution is the sign of $x_0$, so \begin{align*} x(t)=\frac{x_0}{\sqrt{1+2t x_0^2}}. \end{align*} This formula gives \begin{align*} |x(t)|=\frac{|x_0|}{\sqrt{1+2t x_0^2}}\le |x_0| \end{align*} for every $t\ge 0$. Hence, for every $\varepsilon>0$, choosing $\delta=\varepsilon$ ensures $|x_0|<\delta \Rightarrow |x(t)|<\varepsilon$ for all $t\ge 0$, so the origin is Lyapunov stable. Also, since $1+2t x_0^2\to\infty$ when $x_0\ne 0$, \begin{align*} \lim_{t\to\infty}\frac{x_0}{\sqrt{1+2t x_0^2}}=0. \end{align*} Thus every sufficiently small initial condition converges to $0$, and the origin is asymptotically stable. It is not exponentially stable. Suppose, toward a contradiction, that there are constants $C\ge 1$, $\alpha>0$, and $r>0$ such that \begin{align*} |x(t)|\le C e^{-\alpha t}|x_0| \end{align*} for every $t\ge 0$ whenever $|x_0|<r$. Choose a fixed $x_0$ with $0<|x_0|<r$. Substituting the explicit solution and dividing by $|x_0|$ gives \begin{align*} \frac{1}{\sqrt{1+2t x_0^2}}\le C e^{-\alpha t}. \end{align*} Multiplying by $e^{\alpha t}$ gives \begin{align*} \frac{e^{\alpha t}}{\sqrt{1+2t x_0^2}}\le C. \end{align*} But \begin{align*} \log\left(\frac{e^{\alpha t}}{\sqrt{1+2t x_0^2}}\right)=\alpha t-\frac{1}{2}\log(1+2t x_0^2), \end{align*} and $\log(1+2t x_0^2)/t\to 0$, so this logarithm tends to $+\infty$. Therefore the left-hand side becomes larger than any fixed $C$, a contradiction. The decay is algebraic rather than exponential: for fixed $x_0\ne 0$, \begin{align*} \sqrt{t}\,|x(t)|=\frac{\sqrt{t}\,|x_0|}{\sqrt{1+2t x_0^2}}=\frac{|x_0|}{\sqrt{t^{-1}+2x_0^2}}\to \frac{1}{\sqrt{2}}. \end{align*} [/example] A second failure mode is instability hidden behind attraction along a thin set. Convergence of some trajectories does not imply stability of the equilibrium. [example: Stable Manifold Is Not Stability] Consider the saddle system \begin{align*} \dot{x}_1=x_1,\qquad \dot{x}_2=-x_2 \end{align*} with initial condition $x(0)=(a,b)$. Define \begin{align*} x(t)=(e^t a,e^{-t}b). \end{align*} At $t=0$ this gives \begin{align*} x(0)=(e^0a,e^0b)=(a,b). \end{align*} Differentiating componentwise gives \begin{align*} x'(t)=(e^t a,-e^{-t}b). \end{align*} Since $x_1(t)=e^t a$ and $x_2(t)=e^{-t}b$, this is exactly \begin{align*} x'(t)=(x_1(t),-x_2(t)). \end{align*} Thus $x(t)$ solves the saddle system with initial value $(a,b)$. If the initial condition lies on the $x_2$-axis, then $a=0$, so \begin{align*} x(t)=(0,e^{-t}b). \end{align*} Its distance from the origin is \begin{align*} |x(t)|=\sqrt{0^2+(e^{-t}b)^2}=e^{-t}|b|. \end{align*} Because $e^{-t}\to 0$ as $t\to\infty$, every trajectory starting on the $x_2$-axis converges to the origin. This convergence along one line does not give Lyapunov stability. Let $\varepsilon=1$. For any $\delta>0$, choose \begin{align*} x_0=\left(\frac{\delta}{2},0\right). \end{align*} Then \begin{align*} |x_0|=\frac{\delta}{2}<\delta. \end{align*} The corresponding solution is \begin{align*} x(t)=\left(e^t\frac{\delta}{2},0\right), \end{align*} so \begin{align*} |x(t)|=e^t\frac{\delta}{2}. \end{align*} Choose $t>\log(2/\delta)$. Then $e^t>2/\delta$, hence \begin{align*} |x(t)|=e^t\frac{\delta}{2}>1=\varepsilon. \end{align*} Thus no matter how small the initial tolerance $\delta$ is, there is an initial condition within distance $\delta$ of the origin whose trajectory later leaves the unit ball. The origin is not Lyapunov stable, and therefore it is not asymptotically stable. [/example] The saddle example is not robustly attracting because one direction expands. Hyperbolic sinks are different: when all linear directions decay, there is no neutral direction for a small modelling error to exploit. The next theorem records why these sinks are the stable objects that persist in numerical phase portraits and perturbed models. [quotetheorem:7893] The theorem is one reason hyperbolic sinks are the stable objects seen in numerical phase portraits: their stability does not rely on a finely tuned cancellation. ## Beyond and Connected Topics Asymptotic stability sits at the intersection of ODE, topology, spectral theory, and PDE. In finite-dimensional dynamics, the natural continuation is local phase portrait theory: stable and unstable manifolds explain the geometry near saddles, while bifurcation theory studies how stability changes as a parameter varies. The page [Cambridge II Dynamical Systems](/page/Cambridge%20II%20Dynamical%20Systems) is the natural course-level continuation for equilibria, linearisation, phase portraits, and Lyapunov methods. It supplies the broader dynamical vocabulary surrounding sinks, sources, saddles, periodic orbits, and bifurcations. The page [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) provides the metric-space background behind neighbourhoods, convergence, compactness, and continuity. These ideas enter whenever stability is phrased in terms of balls, distances to invariant sets, and compact sublevel sets. The page [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes) gives the foundational language of limits, continuity, differentiability, and norm estimates. Those tools support the local estimates used in linearisation and Lyapunov arguments. There is also a complex-analysis viewpoint through spectra. For finite-dimensional linear systems, the location of eigenvalues in the complex plane determines decay. The page [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis) is useful background for understanding complex eigenvalues, exponentials, and the geometry of the left half-plane. In infinite-dimensional analysis, asymptotic stability leads to [semigroup theory](/page/Semigroup%20Theory), spectral bounds, compactness methods, and dissipative PDE. A key difference is that spectral location alone may not imply uniform decay without additional semigroup hypotheses, so the finite-dimensional intuition must be refined. ## References Androma, [Cambridge II Dynamical Systems](/page/Cambridge%20II%20Dynamical%20Systems). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Hirsch, Smale, and Devaney, *Differential Equations, Dynamical Systems, and an Introduction to Chaos* (2013). Khalil, *Nonlinear Systems* (2002). Perko, *Differential Equations and Dynamical Systems* (2001). Pazy, *Semigroups of Linear Operators and Applications to Partial Differential Equations* (1983).