[proofplan] We prove Lyapunov stability by trapping sufficiently small initial conditions in sufficiently small positively invariant sublevel sets of $V$. For global attractivity, we fix an arbitrary initial point and use the proper compact sublevel containing its forward orbit to construct a nonempty omega-limit set. The strict decrease of $V$ away from $x^*$ forces every omega-[limit point](/page/Limit%20Point) to equal $x^*$, and compactness of the orbit then upgrades uniqueness of the omega-limit set to convergence of the whole trajectory. [/proofplan]

[step:Trap small initial conditions inside small invariant sublevels]Fix $\varepsilon>0$. By the local compatibility assumption, choose $c_\varepsilon>0$ such that \begin{align*} K_{c_\varepsilon}\subset B(x^*,\varepsilon). \end{align*} Since $V\in C^1(D;\mathbb{R})$, the map $V:D\to\mathbb{R}$ is continuous. Since $V(x^*)=0<c_\varepsilon$ and $D$ is open, there exists $\delta>0$ such that \begin{align*} B(x^*,\delta)\subset D \end{align*} and \begin{align*} V(x)<c_\varepsilon \end{align*} for every $x\in B(x^*,\delta)$. Hence \begin{align*} B(x^*,\delta)\subset K_{c_\varepsilon}. \end{align*} If $x_0\in D$ satisfies $|x_0-x^*|<\delta$, then $x_0\in K_{c_\varepsilon}$. Since $K_{c_\varepsilon}$ is positively invariant, $\varphi_t(x_0)\in K_{c_\varepsilon}$ for every $t\geq 0$. Therefore \begin{align*} \varphi_t(x_0)\in B(x^*,\varepsilon) \end{align*} for every $t\geq 0$. This proves Lyapunov stability of $x^*$ relative to $D$.[/step]

[guided]We want to prove stability in the usual relative-to-$D$ sense: given a target ball around $x^*$, choose a smaller ball of initial data whose forward trajectories never leave the target ball. The hypothesis on compact sublevels is designed exactly for this. Given $\varepsilon>0$, the local compatibility assumption supplies a positive level $c_\varepsilon>0$ with \begin{align*} K_{c_\varepsilon}\subset B(x^*,\varepsilon). \end{align*} So it is enough to make sure that initial data sufficiently close to $x^*$ lie in $K_{c_\varepsilon}$. Because $V\in C^1(D;\mathbb{R})$, the function $V:D\to\mathbb{R}$ is continuous. Since $V(x^*)=0$ and $c_\varepsilon>0$, continuity gives a neighbourhood of $x^*$ on which $V<c_\varepsilon$. Since $D$ is open and $x^*\in D$, we may choose this neighbourhood to be a Euclidean ball contained in $D$: there exists $\delta>0$ such that \begin{align*} B(x^*,\delta)\subset D \end{align*} and \begin{align*} V(x)<c_\varepsilon \end{align*} for every $x\in B(x^*,\delta)$. Thus every $x\in B(x^*,\delta)$ satisfies $V(x)\leq c_\varepsilon$, so \begin{align*} B(x^*,\delta)\subset K_{c_\varepsilon}. \end{align*} Now let $x_0\in D$ satisfy $|x_0-x^*|<\delta$. The previous inclusion gives $x_0\in K_{c_\varepsilon}$. The assumed positive invariance of $K_{c_\varepsilon}$ means precisely that $\varphi_t(K_{c_\varepsilon})\subset K_{c_\varepsilon}$ for every $t\geq 0$. Since $x_0\in K_{c_\varepsilon}$, this gives \begin{align*} \varphi_t(x_0)\in K_{c_\varepsilon} \end{align*} for every $t\geq 0$. Combining this with $K_{c_\varepsilon}\subset B(x^*,\varepsilon)$ yields \begin{align*} \varphi_t(x_0)\in B(x^*,\varepsilon) \end{align*} for every $t\geq 0$. This proves Lyapunov stability relative to the domain $D$.[/guided]

[step:Place each nonstationary forward orbit in a compact sublevel] Fix $x_0\in D$. If $x_0=x^*$, then $\varphi_t(x_0)=x^*$ for every $t\geq 0$ because $f(x^*)=0$, so convergence is immediate. Assume $x_0\neq x^*$. Define the trajectory map $\gamma:[0,\infty)\to D$ by $\gamma(t)=\varphi_t(x_0)$. Since $V(x_0)>0$, set \begin{align*} c_0:=V(x_0). \end{align*} Then $c_0>0$ and $x_0\in K_{c_0}$. By positive invariance, \begin{align*} \gamma(t)\in K_{c_0} \end{align*} for every $t\geq 0$. Since $K_{c_0}$ is compact in $\mathbb{R}^n$, the forward orbit $\gamma([0,\infty))$ is contained in a compact subset of $D$. [/step]

[step:Build the omega-limit set and record its compactness] Define the omega-limit set of $x_0$ by \begin{align*} \omega(x_0):=\{y\in D:\text{there exists a sequence }(t_k)_{k=1}^{\infty}\subset[0,\infty)\text{ with }t_k\to\infty\text{ and }\gamma(t_k)\to y\}. \end{align*} Since $\gamma([0,\infty))\subset K_{c_0}$ and $K_{c_0}$ is compact, every sequence $(\gamma(s_k))_{k=1}^{\infty}$ with $s_k\to\infty$ has a convergent subsequence in $K_{c_0}$. Taking, for instance, $s_k=k$, proves that $\omega(x_0)$ is nonempty. The same compactness argument shows that $\omega(x_0)\subset K_{c_0}$. The set $\omega(x_0)$ is closed in $K_{c_0}$: if $(y_j)_{j=1}^{\infty}\subset\omega(x_0)$ and $y_j\to y\in K_{c_0}$, choose for each $j$ a time $t_j\geq j$ such that \begin{align*} |\gamma(t_j)-y_j|<\frac{1}{j}. \end{align*} Then $t_j\to\infty$ and $\gamma(t_j)\to y$, so $y\in\omega(x_0)$. Hence $\omega(x_0)$ is compact as a closed subset of the compact set $K_{c_0}$. [/step]

[step:Show that the Lyapunov function is constant on the omega-limit set] Define $a:[0,\infty)\to\mathbb{R}$ by \begin{align*} a(t):=V(\gamma(t)). \end{align*} The chain rule gives, for every $t\geq 0$, \begin{align*} a'(t)=DV_{\gamma(t)}(\gamma'(t))=DV_{\gamma(t)}(f(\gamma(t)))=\dot{V}(\gamma(t)). \end{align*} For $x\in D\setminus\{x^*\}$ the hypothesis gives $\dot{V}(x)<0$. At the equilibrium, $f(x^*)=0$, so $\dot{V}(x^*)=DV_{x^*}(0)=0$. Hence $\dot{V}(x)\leq 0$ for every $x\in D$, and the function $a$ is nonincreasing. Since $V\geq 0$ on $D$, $a$ is bounded below by $0$. Therefore the limit \begin{align*} \ell:=\lim_{t\to\infty}a(t) \end{align*} exists in $[0,\infty)$. If $y\in\omega(x_0)$, choose a sequence $(t_k)_{k=1}^{\infty}\subset[0,\infty)$ such that $t_k\to\infty$ and $\gamma(t_k)\to y$. By continuity of $V$, \begin{align*} V(y)=\lim_{k\to\infty}V(\gamma(t_k))=\lim_{k\to\infty}a(t_k)=\ell. \end{align*} Thus $V$ is constant on $\omega(x_0)$ with value $\ell$. [/step]

[step:Use invariance of omega-limit points to exclude every point except the equilibrium]We first record forward invariance of $\omega(x_0)$. Let $y\in\omega(x_0)$ and let $s\geq 0$. Choose $(t_k)_{k=1}^{\infty}$ with $t_k\to\infty$ and $\gamma(t_k)\to y$. Because the flow is forward complete on $D$, the point $y$ lies in the domain of the fixed-time map $\varphi_s:D\to D$. By the fixed-time continuity included in the definition of the forward-complete flow generated by the $C^1$ vector field $f$, this map is continuous. Hence \begin{align*} \varphi_s(y)=\lim_{k\to\infty}\varphi_s(\gamma(t_k))=\lim_{k\to\infty}\gamma(t_k+s). \end{align*} Since $t_k+s\to\infty$, this proves $\varphi_s(y)\in\omega(x_0)$. Now suppose, toward a contradiction, that $y\in\omega(x_0)$ and $y\neq x^*$. Since $\dot{V}(y)<0$ and the map $z\mapsto\dot{V}(z)$ is continuous, the function $s\mapsto V(\varphi_s(y))$ has negative derivative at $s=0$. Therefore there exists $s_0>0$ such that \begin{align*} V(\varphi_{s_0}(y))<V(y). \end{align*} By forward invariance, $\varphi_{s_0}(y)\in\omega(x_0)$. But the previous step proved that $V$ equals $\ell$ at every point of $\omega(x_0)$, so \begin{align*} V(\varphi_{s_0}(y))=\ell=V(y), \end{align*} a contradiction. Therefore every point of $\omega(x_0)$ equals $x^*$, and since $\omega(x_0)$ is nonempty, \begin{align*} \omega(x_0)=\{x^*\}. \end{align*}[/step]

[guided]The omega-limit set consists of points that the trajectory approaches along sequences of times going to infinity. To use the strict Lyapunov decrease at such a point, we need to know that if $y$ is an omega-limit point, then the forward trajectory starting from $y$ stays inside the same omega-limit set. Let $y\in\omega(x_0)$ and fix $s\geq 0$. By definition of $\omega(x_0)$, there is a sequence $(t_k)_{k=1}^{\infty}\subset[0,\infty)$ such that $t_k\to\infty$ and \begin{align*} \gamma(t_k)\to y. \end{align*} For the fixed time $s$, the point $y$ is in the domain of $\varphi_s$ because the flow is forward complete on $D$. The formalized statement includes the fixed-time continuity of the flow map $\varphi_s:D\to D$, the usual continuous-dependence conclusion for a $C^1$ vector field. Therefore \begin{align*} \varphi_s(y)=\lim_{k\to\infty}\varphi_s(\gamma(t_k)). \end{align*} The flow property gives $\varphi_s(\gamma(t_k))=\gamma(t_k+s)$. Hence \begin{align*} \varphi_s(y)=\lim_{k\to\infty}\gamma(t_k+s). \end{align*} Since $t_k+s\to\infty$, this says exactly that $\varphi_s(y)$ is also an omega-limit point. Therefore $\omega(x_0)$ is positively invariant. Now suppose that some $y\in\omega(x_0)$ satisfies $y\neq x^*$. The strict Lyapunov hypothesis gives \begin{align*} \dot{V}(y)<0. \end{align*} For the trajectory starting at $y$, the chain rule gives \begin{align*} \frac{d}{ds}V(\varphi_s(y))\bigg|_{s=0}=\dot{V}(y)<0. \end{align*} Thus for some sufficiently small $s_0>0$, \begin{align*} V(\varphi_{s_0}(y))<V(y). \end{align*} But positive invariance of the omega-limit set gives $\varphi_{s_0}(y)\in\omega(x_0)$. We recall why $V$ has the same value at every point of $\omega(x_0)$. Define $a:[0,\infty)\to\mathbb{R}$ by $a(t)=V(\gamma(t))$. The chain rule gives $a'(t)=\dot{V}(\gamma(t))\leq 0$, where the inequality follows from the strict Lyapunov hypothesis off $x^*$ and from $\dot{V}(x^*)=DV_{x^*}(0)=0$. Since $V\geq 0$ on $D$, the nonincreasing function $a$ has a limit $\ell\in[0,\infty)$. If $z\in\omega(x_0)$, then for some sequence $(r_k)_{k=1}^{\infty}\subset[0,\infty)$ with $r_k\to\infty$ we have $\gamma(r_k)\to z$, so continuity of $V$ gives \begin{align*} V(z)=\lim_{k\to\infty}V(\gamma(r_k))=\lim_{k\to\infty}a(r_k)=\ell. \end{align*} Applying this first to $z=y$ and then to $z=\varphi_{s_0}(y)$ gives \begin{align*} V(\varphi_{s_0}(y))=\ell=V(y), \end{align*} which contradicts the strict inequality. Hence no point of $\omega(x_0)$ other than $x^*$ can occur. Since $\omega(x_0)$ is nonempty, we conclude \begin{align*} \omega(x_0)=\{x^*\}. \end{align*}[/guided]

[step:Upgrade the unique omega-limit point to convergence of the whole trajectory] We prove that $\gamma(t)\to x^*$ as $t\to\infty$. Suppose not. Then there exist $\varepsilon_0>0$ and a sequence $(t_k)_{k=1}^{\infty}\subset[0,\infty)$ with $t_k\to\infty$ such that \begin{align*} |\gamma(t_k)-x^*|\geq\varepsilon_0 \end{align*} for every $k\in\mathbb{N}$. Since $\gamma(t_k)\in K_{c_0}$ and $K_{c_0}$ is compact, after passing to a subsequence there exists $y\in K_{c_0}$ such that \begin{align*} \gamma(t_k)\to y. \end{align*} Then $y\in\omega(x_0)$, so $y=x^*$. This contradicts \begin{align*} |y-x^*|\geq\varepsilon_0, \end{align*} which follows from the continuity of the Euclidean norm and the inequalities $|\gamma(t_k)-x^*|\geq\varepsilon_0$. Therefore \begin{align*} \lim_{t\to\infty}\varphi_t(x_0)=x^*. \end{align*} Since $x_0\in D$ was arbitrary, $x^*$ attracts every point of $D$. Together with Lyapunov stability proved above, this is global [asymptotic stability](/page/Asymptotic%20Stability) in $D$. [/step]