[proofplan] The proof uses $V$ as a barrier. On each sphere around the equilibrium, positivity and compactness give a positive lower bound for $V$, while the chain rule and the sign condition on $\dot V$ make $V(x(t;x_0))$ nonincreasing along trajectories before they leave the Lyapunov ball. This prevents first exit and, together with the ODE continuation criterion on compact subsets of the domain, gives global forward existence. Under strict negativity of $\dot V$, a compact-annulus argument shows that any trajectory trapped near $0$ cannot return infinitely often away from $0$, because each such return would force a fixed drop in the nonnegative convergent quantity $V(x(t;x_0))$. [/proofplan]

[step:Choose a Lyapunov sublevel set inside the prescribed ball] Fix $\varepsilon \in (0,r]$. Define the sphere $S_\varepsilon \subset \mathbb{R}^n$ by \begin{align*} S_\varepsilon=\{x\in \mathbb{R}^n: |x|=\varepsilon\}. \end{align*} Since $S_\varepsilon$ is compact and $V:S_\varepsilon\to\mathbb{R}$ is continuous, the Extreme Value Theorem applies (citing a result not yet in the wiki: Extreme Value Theorem). Because $V(x)>0$ for every $x\in S_\varepsilon$, the number \begin{align*} m_\varepsilon=\min_{x\in S_\varepsilon} V(x) \end{align*} is well-defined and satisfies $m_\varepsilon>0$. Since $V$ is continuous at $0$ and $V(0)=0$, there exists $\delta_\varepsilon\in (0,\varepsilon)$ such that \begin{align*} |x_0|<\delta_\varepsilon \implies V(x_0)<m_\varepsilon. \end{align*} Set $\delta=\delta_\varepsilon$. [/step]

[step:Prevent first exit from the ball by monotonicity of $V$] Let $x_0\in B(0,\delta)$, and write $T=T_{\max}(x_0)$. Define the trajectory \begin{align*} \gamma:[0,T)\to D, \qquad \gamma(t)=x(t;x_0). \end{align*} The chain rule along $C^1$ functions and $C^1$ ODE solutions gives, for every $t\in [0,T)$ with $\gamma(t)\in B(0,r)$, \begin{align*} \frac{d}{dt}V(\gamma(t))=\nabla V(\gamma(t))\cdot \dot{\gamma}(t)=\nabla V(\gamma(t))\cdot f(\gamma(t))=\dot V(\gamma(t))\leq 0. \end{align*} Here we use the standard chain rule for differentiable curves and $C^1$ functions (citing a result not yet in the wiki: Chain rule along $C^1$ Lyapunov functions). We prove that $\gamma(t)\in B(0,\varepsilon)$ for every $t\in [0,T)$. Suppose not. Since $\gamma(0)=x_0\in B(0,\delta)\subset B(0,\varepsilon)$ and $\gamma$ is continuous, the first exit time \begin{align*} \tau=\inf\{t\in [0,T): |\gamma(t)|=\varepsilon\} \end{align*} is well-defined and satisfies $0<\tau<T$. For every $t\in[0,\tau)$, one has $|\gamma(t)|<\varepsilon\leq r$, so the monotonicity calculation applies on $[0,\tau)$. Integrating the derivative inequality on $[0,t]$ for $t<\tau$ gives \begin{align*} V(\gamma(t))\leq V(\gamma(0))=V(x_0)<m_\varepsilon. \end{align*} Letting $t\uparrow \tau$ and using continuity of $V\circ\gamma$ gives \begin{align*} V(\gamma(\tau))\leq V(x_0)<m_\varepsilon. \end{align*} But $\gamma(\tau)\in S_\varepsilon$, so $V(\gamma(\tau))\geq m_\varepsilon$, a contradiction. Therefore $\gamma(t)\in B(0,\varepsilon)$ for every $t\in[0,T)$. [/step]

[step:Use compact containment to obtain global forward existence] The preceding step shows that $\gamma([0,T))\subset B(0,\varepsilon)\subset \overline{B}(0,r)$. Since $\overline{B}(0,r)\subset D$ and $\overline{B}(0,r)$ is compact, $\overline{B}(0,r)$ is a compact subset of the open ODE domain $D$. In particular, for every bounded interval $[0,b]\subset[0,T)$, the image $\gamma([0,b])$ is contained in the same compact subset $\overline{B}(0,r)$ of $D$. Thus the trajectory remains in a compact subset of $D$ throughout its maximal interval of existence and cannot approach the boundary of $D$ in finite time. By the ODE continuation criterion on compact subsets of the domain (citing a result not yet in the wiki: ODE continuation criterion on compact subsets of the domain), a maximal forward solution that remains in such a compact subset on every bounded subinterval cannot have finite forward lifetime. Hence $T=\infty$. Since $\varepsilon\in(0,r]$ was arbitrary and $\delta=\delta_\varepsilon$ was chosen for that $\varepsilon$, the equilibrium $0$ is Lyapunov stable. [/step]

[step:Trap small trajectories inside a ball where $\dot V$ is strictly negative away from the origin] Assume now that there exists $r_a\in(0,r]$ such that $\dot V(x)<0$ for every $x\in B(0,r_a)\setminus\{0\}$. Choose a radius $\rho\in(0,r_a)$. Applying the Lyapunov stability result already proved with $\varepsilon=\rho$, there exists $\delta_a\in(0,\rho)$ such that, whenever $x_0\in B(0,\delta_a)$, the corresponding maximal forward solution is global and satisfies \begin{align*} x(t;x_0)\in B(0,\rho) \end{align*} for every $t\geq 0$. Fix such an $x_0$. If $x_0=0$, then uniqueness of the initial value problem and $f(0)=0$ imply $x(t;0)=0$ for every $t\geq 0$, so convergence to $0$ follows. Hence assume $x_0\neq 0$. Define the global trajectory \begin{align*} \gamma:[0,\infty)\to D, \qquad \gamma(t)=x(t;x_0). \end{align*} Because $\gamma(t)\in B(0,\rho)\subset B(0,r_a)$ for every $t\geq 0$, the chain rule gives \begin{align*} \frac{d}{dt}V(\gamma(t))=\dot V(\gamma(t))\leq 0. \end{align*} Thus the function $W:[0,\infty)\to\mathbb{R}$ defined by \begin{align*} W(t)=V(\gamma(t)) \end{align*} is nonincreasing. Since $V\geq 0$ on $D$, the limit \begin{align*} \ell=\lim_{t\to\infty} W(t) \end{align*} exists in $[0,\infty)$. [/step]

[step:Rule out infinitely many returns to a compact annulus]We prove that $\gamma(t)\to 0$ as $t\to\infty$. Suppose, for contradiction, that $\gamma(t)$ does not converge to $0$. Then there exist $\eta\in(0,\rho)$ and an unbounded sequence $(s_k)_{k=1}^\infty$ in $[0,\infty)$ such that \begin{align*} |\gamma(s_k)|\geq \eta \end{align*} for every $k\in\mathbb{N}$. Define the compact annulus $A_\eta\subset\mathbb{R}^n$ by \begin{align*} A_\eta=\{x\in\mathbb{R}^n: \eta/2\leq |x|\leq \rho\}. \end{align*} Since $\rho<r_a$, one has $A_\eta\subset B(0,r_a)\setminus\{0\}$. The function $\dot V:A_\eta\to\mathbb{R}$ is continuous, so the Extreme Value Theorem applies (citing a result not yet in the wiki: Extreme Value Theorem). Since $\dot V(x)<0$ on $A_\eta$, the number \begin{align*} a=-\max_{x\in A_\eta}\dot V(x) \end{align*} satisfies $a>0$, and hence \begin{align*} \dot V(x)\leq -a \end{align*} for every $x\in A_\eta$. Define the compact ball $K_\rho\subset\mathbb{R}^n$ by \begin{align*} K_\rho=\{x\in\mathbb{R}^n: |x|\leq \rho\}. \end{align*} The function $f:K_\rho\to\mathbb{R}^n$ is continuous, so it is bounded on $K_\rho$ by the Extreme Value Theorem. Define \begin{align*} M=\max_{x\in K_\rho}|f(x)|. \end{align*} Because $A_\eta$ is nonempty and $\dot V<0$ on $A_\eta$, $f$ is not identically zero on $K_\rho$, so $M>0$. Set \begin{align*} \tau=\frac{\eta}{2M}. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. For each $k$ and each $t\in[s_k,s_k+\tau]$, the trajectory remains in $K_\rho$ by trapping. Therefore \begin{align*} |\gamma(t)-\gamma(s_k)|\leq \int_{s_k}^{t} |\dot{\gamma}(\sigma)|\,d\mathcal{L}^1(\sigma)=\int_{s_k}^{t}|f(\gamma(\sigma))|\,d\mathcal{L}^1(\sigma)\leq M(t-s_k)\leq \eta/2. \end{align*} Hence $|\gamma(t)|\geq \eta/2$ for every $t\in[s_k,s_k+\tau]$. Since also $|\gamma(t)|<\rho$, we have $\gamma(t)\in A_\eta$ for every $t\in[s_k,s_k+\tau]$, and so \begin{align*} \frac{d}{dt}W(t)=\dot V(\gamma(t))\leq -a \end{align*} on that whole interval. Because the sequence $(s_k)$ is unbounded, pass to a subsequence, still denoted $(s_k)$, such that the intervals $[s_k,s_k+\tau]$ are pairwise disjoint and ordered increasingly. For every $N\in\mathbb{N}$, integrating the previous differential inequality over these $N$ intervals gives \begin{align*} W(s_N+\tau)\leq W(0)-Na\tau. \end{align*} For $N$ large enough, the right-hand side is negative, contradicting $W(t)=V(\gamma(t))\geq 0$. Therefore $\gamma(t)\to 0$ as $t\to\infty$.[/step]

[guided]We now prove the attraction part carefully. The trapping step already gives a global trajectory \begin{align*} \gamma:[0,\infty)\to D, \qquad \gamma(t)=x(t;x_0), \end{align*} with $\gamma(t)\in B(0,\rho)$ for every $t\geq 0$, where $\rho<r_a$. Define \begin{align*} W:[0,\infty)\to\mathbb{R}, \qquad W(t)=V(\gamma(t)). \end{align*} The chain rule gives \begin{align*} W'(t)=\nabla V(\gamma(t))\cdot \dot{\gamma}(t)=\nabla V(\gamma(t))\cdot f(\gamma(t))=\dot V(\gamma(t)). \end{align*} Since $\gamma(t)\in B(0,r_a)$, this derivative is nonpositive, and it is strictly negative whenever $\gamma(t)\neq 0$. Thus $W$ is nonincreasing and bounded below by $0$, so $W(t)$ has a finite limit as $t\to\infty$. The remaining question is: could the trajectory keep returning away from $0$ while the values of $V$ merely drift down to a positive limit? We show that this is impossible because strict negativity of $\dot V$ is uniform on every compact annulus away from $0$. Assume, for contradiction, that $\gamma(t)$ does not converge to $0$. Then there is some distance $\eta>0$ from the origin that the trajectory reaches at arbitrarily large times. More precisely, because the trajectory is trapped in $B(0,\rho)$, we may choose $\eta\in(0,\rho)$ and an unbounded sequence $(s_k)_{k=1}^\infty$ such that \begin{align*} |\gamma(s_k)|\geq \eta \end{align*} for all $k$. Define the compact annulus \begin{align*} A_\eta=\{x\in\mathbb{R}^n: \eta/2\leq |x|\leq \rho\}. \end{align*} The strict inequality $\rho<r_a$ is important here: it ensures that $A_\eta$ lies entirely inside the region where $\dot V$ is strictly negative away from $0$. Since $A_\eta$ is compact and $\dot V$ is continuous, the Extreme Value Theorem applies (citing a result not yet in the wiki: Extreme Value Theorem). The maximum of $\dot V$ on $A_\eta$ is still negative, so defining \begin{align*} a=-\max_{x\in A_\eta}\dot V(x) \end{align*} gives $a>0$ and \begin{align*} \dot V(x)\leq -a \end{align*} for every $x\in A_\eta$. This uniform negativity alone is not enough: we also need to know that after a visit to $|\gamma|\geq \eta$, the trajectory remains in the annulus for a uniform amount of time. This is where boundedness of the vector field on the trapped compact ball enters. Define \begin{align*} K_\rho=\{x\in\mathbb{R}^n: |x|\leq \rho\}. \end{align*} The map $f:K_\rho\to\mathbb{R}^n$ is continuous, so the Extreme Value Theorem gives a finite bound \begin{align*} M=\max_{x\in K_\rho}|f(x)|. \end{align*} Since $\dot V<0$ somewhere on the nonempty annulus $A_\eta$, the vector field cannot vanish identically on $K_\rho$, and hence $M>0$. Set \begin{align*} \tau=\frac{\eta}{2M}. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Fix $k$. For $t\in[s_k,s_k+\tau]$, the trapping property gives $\gamma(t)\in K_\rho$. Therefore \begin{align*} |\gamma(t)-\gamma(s_k)|\leq \int_{s_k}^{t}|\dot{\gamma}(\sigma)|\,d\mathcal{L}^1(\sigma). \end{align*} Since $\dot{\gamma}(\sigma)=f(\gamma(\sigma))$ and $\gamma(\sigma)\in K_\rho$, the definition of $M$ gives \begin{align*} \int_{s_k}^{t}|\dot{\gamma}(\sigma)|\,d\mathcal{L}^1(\sigma)=\int_{s_k}^{t}|f(\gamma(\sigma))|\,d\mathcal{L}^1(\sigma)\leq M(t-s_k)\leq \eta/2. \end{align*} Thus \begin{align*} |\gamma(t)|\geq |\gamma(s_k)|-|\gamma(t)-\gamma(s_k)|\geq \eta-\eta/2=\eta/2. \end{align*} The upper bound $|\gamma(t)|<\rho$ still holds by trapping. Hence $\gamma(t)\in A_\eta$ throughout the interval $[s_k,s_k+\tau]$. On each such interval, \begin{align*} W'(t)=\dot V(\gamma(t))\leq -a. \end{align*} Since the times $s_k$ are unbounded, we may choose a subsequence so that the intervals $[s_k,s_k+\tau]$ are pairwise disjoint and occur in increasing order. Integrating $W'(t)\leq -a$ over the first $N$ of these disjoint intervals gives \begin{align*} W(s_N+\tau)\leq W(0)-Na\tau. \end{align*} For sufficiently large $N$, this right-hand side is negative. That contradicts $W(t)=V(\gamma(t))\geq 0$ for all $t\geq 0$. Therefore the assumption of infinitely many returns away from $0$ is false, and so \begin{align*} \lim_{t\to\infty}\gamma(t)=0. \end{align*}[/guided]

[step:Conclude local asymptotic stability] We have shown that the equilibrium $0$ is Lyapunov stable. Under the additional strict negativity assumption, we chose $\delta_a>0$ such that every solution with $x_0\in B(0,\delta_a)$ is global, remains in $B(0,\rho)\subset B(0,r_a)$, and satisfies \begin{align*} \lim_{t\to\infty}x(t;x_0)=0. \end{align*} This is exactly local asymptotic stability of the equilibrium $0$. [/step]