[proofplan] We first choose a smaller ball $B(0,\rho)$ inside the ball where the Lyapunov estimates are valid. The quadratic bounds make small Euclidean balls into strict sublevel sets of $V$, and the derivative inequality prevents a trajectory starting in such a sublevel set from crossing the sphere $|x|=\rho$. Once this trapping is established, the derivative estimate can be rewritten as a closed scalar differential inequality for $V(x(t))$. Integrating that inequality and converting back through the quadratic bounds gives the exponential decay estimate. [/proofplan] [step:Choose a strict sublevel set contained in the ball of validity] Fix $\rho \in (0,r)$. Define the positive radius \begin{align*} \delta := \rho \sqrt{\frac{c_1}{2c_2}}. \end{align*} Let $x_0 \in D$ satisfy $|x_0|<\delta$, and let $x:[0,T_{\max}) \to D$ denote the maximal solution of \begin{align*} \dot{x}(t)=f(x(t)), \qquad x(0)=x_0, \end{align*} where $T_{\max} \in (0,\infty]$. By the Picard-Lindelof local existence and uniqueness theorem, local existence and uniqueness hold because $D$ is open, $x_0\in D$, and $f$ is locally Lipschitz on $D$. Since $|x_0|<\delta<\rho<r$, the upper quadratic bound gives \begin{align*} V(x_0) \leq c_2 |x_0|^2 < c_2 \delta^2. \end{align*} By the definition of $\delta$, \begin{align*} c_2 \delta^2 = \frac{c_1\rho^2}{2}. \end{align*} Therefore \begin{align*} V(x_0) < c_1\rho^2. \end{align*} [/step] [step:Trap the trajectory inside $B(0,\rho)$] We prove that $|x(t)|<\rho$ for every $t \in [0,T_{\max})$. Suppose instead that the trajectory first reaches the sphere of radius $\rho$. Define the exit time \begin{align*} \tau := \inf\{t \in [0,T_{\max}) : |x(t)|=\rho\}. \end{align*} Since $x$ is continuous and $|x(0)|<\rho$, if the set is nonempty then $\tau>0$, $|x(t)|<\rho$ for $0\leq t<\tau$, and $|x(\tau)|=\rho$. For $0\leq t<\tau$, we have $x(t)\in B(0,\rho)\subset B(0,r)$. The chain rule gives \begin{align*} \frac{d}{dt}V(x(t)) = \nabla V(x(t))\cdot f(x(t)). \end{align*} Using the derivative hypothesis, \begin{align*} \frac{d}{dt}V(x(t)) \leq -c_3 |x(t)|^2 \leq 0. \end{align*} Thus $t\mapsto V(x(t))$ is nonincreasing on $[0,\tau)$. By continuity of $V\circ x$, this also gives \begin{align*} V(x(\tau)) \leq V(x_0) < c_1\rho^2. \end{align*} But $|x(\tau)|=\rho$ and $\rho<r$, so the lower quadratic bound applies at $x(\tau)$ and yields \begin{align*} V(x(\tau)) \geq c_1 |x(\tau)|^2 = c_1\rho^2. \end{align*} This contradiction proves $|x(t)|<\rho$ for every $t \in [0,T_{\max})$. [guided] The point of this step is to avoid using the estimates at the boundary of $B(0,r)$. We work with a smaller radius $\rho<r$, so the whole closed sphere $\{x\in \mathbb{R}^n: |x|=\rho\}$ lies inside the open ball $B(0,r)$ where the hypotheses are valid. Assume, for contradiction, that the solution leaves $B(0,\rho)$. Because $x:[0,T_{\max})\to D$ is continuous and starts at $|x_0|<\rho$, there is a first time at which it reaches the sphere $|x|=\rho$. Define \begin{align*} \tau := \inf\{t \in [0,T_{\max}) : |x(t)|=\rho\}. \end{align*} If this set is nonempty, then $\tau>0$, $|x(t)|<\rho$ for every $0\leq t<\tau$, and $|x(\tau)|=\rho$. For each $t\in[0,\tau)$, the point $x(t)$ belongs to $B(0,\rho)$, hence also to $B(0,r)$. Therefore the Lyapunov derivative estimate applies along the trajectory. Since $V\in C^1(D;\mathbb{R})$ and $x$ is differentiable, the chain rule gives \begin{align*} \frac{d}{dt}V(x(t)) = \nabla V(x(t))\cdot \dot{x}(t). \end{align*} Using the differential equation $\dot{x}(t)=f(x(t))$, this becomes \begin{align*} \frac{d}{dt}V(x(t)) = \nabla V(x(t))\cdot f(x(t)). \end{align*} The hypothesis on the Lie derivative of $V$ along $f$ gives \begin{align*} \frac{d}{dt}V(x(t)) \leq -c_3 |x(t)|^2. \end{align*} Since $c_3>0$ and $|x(t)|^2\geq 0$, we obtain \begin{align*} \frac{d}{dt}V(x(t)) \leq 0. \end{align*} Thus $V(x(t))$ cannot increase before the first exit time. Passing to $t=\tau$ by continuity of $V\circ x$ gives \begin{align*} V(x(\tau)) \leq V(x_0). \end{align*} From the choice of $\delta$ and the upper quadratic bound at $x_0$, we already know \begin{align*} V(x_0) < c_1\rho^2. \end{align*} Therefore \begin{align*} V(x(\tau)) < c_1\rho^2. \end{align*} On the other hand, at the first exit time we have $|x(\tau)|=\rho$. Since $\rho<r$, the lower quadratic bound is valid at $x(\tau)$, and it gives \begin{align*} V(x(\tau)) \geq c_1 |x(\tau)|^2 = c_1\rho^2. \end{align*} The two inequalities for $V(x(\tau))$ are incompatible. Hence the assumed exit time cannot exist, and the trajectory remains in $B(0,\rho)$ for its entire interval of existence. [/guided] [/step] [step:Extend the trapped solution for all positive time] We next show that $T_{\max}=\infty$. From the previous step, \begin{align*} x([0,T_{\max})) \subset B(0,\rho) \subset \overline{B}(0,\rho). \end{align*} Because $\rho<r$ and $\overline{B}(0,r)\subset D$, the compact set $\overline{B}(0,\rho)$ is contained in $D$. Since $f$ is locally Lipschitz, it is continuous, and therefore there exists a finite constant \begin{align*} M_\rho := \sup\{|f(y)| : y\in \overline{B}(0,\rho)\}. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Thus for $0\leq s<t<T_{\max}$, \begin{align*} |x(t)-x(s)| \leq \int_s^t |f(x(a))|\, d\mathcal{L}^1(a) \leq M_\rho(t-s). \end{align*} If $T_{\max}<\infty$, this Lipschitz estimate implies that $x(t)$ has a limit $x_* \in \overline{B}(0,\rho)$ as $t\uparrow T_{\max}$. Since $x_* \in D$ and $f$ is locally Lipschitz near $x_*$, the standard ODE continuation theorem for locally Lipschitz vector fields gives a solution starting from $x_*$, which extends $x$ beyond $T_{\max}$. This contradicts maximality. Hence $T_{\max}=\infty$. [/step] [step:Convert the Lyapunov derivative estimate into a scalar decay inequality] For every $t\geq 0$, the trapping result gives $x(t)\in B(0,\rho)\subset B(0,r)$. Hence both quadratic bounds and the derivative inequality apply at $x(t)$. The upper bound $V(x(t))\leq c_2|x(t)|^2$ implies \begin{align*} |x(t)|^2 \geq \frac{1}{c_2}V(x(t)). \end{align*} Combining this with the derivative inequality along the solution gives \begin{align*} \frac{d}{dt}V(x(t)) \leq -c_3 |x(t)|^2 \leq -\frac{c_3}{c_2}V(x(t)). \end{align*} Define the scalar function \begin{align*} W:[0,\infty) &\to \mathbb{R} \end{align*} \begin{align*} t &\mapsto V(x(t)). \end{align*} Then $W$ is differentiable and satisfies \begin{align*} W'(t) \leq -\frac{c_3}{c_2}W(t) \end{align*} for every $t\geq 0$. [/step] [step:Integrate the scalar inequality and recover exponential decay of $|x(t)|$] Define the constant \begin{align*} \alpha := \frac{c_3}{c_2}. \end{align*} For the scalar function $W$ from the previous step, define \begin{align*} Y:[0,\infty) &\to \mathbb{R} \end{align*} \begin{align*} t &\mapsto e^{\alpha t}W(t). \end{align*} By the product rule, \begin{align*} Y'(t)=e^{\alpha t}(W'(t)+\alpha W(t)). \end{align*} Since $W'(t)\leq -\alpha W(t)$, we get \begin{align*} Y'(t)\leq 0. \end{align*} Therefore $Y(t)\leq Y(0)$ for every $t\geq 0$, which is \begin{align*} V(x(t)) \leq e^{-\alpha t}V(x_0). \end{align*} Using the lower quadratic bound at $x(t)$ and the upper quadratic bound at $x_0$, \begin{align*} c_1 |x(t)|^2 \leq V(x(t)) \leq e^{-\alpha t}V(x_0) \leq c_2 e^{-\alpha t}|x_0|^2. \end{align*} Hence \begin{align*} |x(t)|^2 \leq \frac{c_2}{c_1}e^{-\frac{c_3}{c_2}t}|x_0|^2. \end{align*} Taking square roots gives \begin{align*} |x(t)| \leq \sqrt{\frac{c_2}{c_1}}\, e^{-\frac{c_3}{2c_2}t}|x_0|. \end{align*} This estimate holds for every $t\geq 0$ whenever $|x_0|<\delta$. Since the right-hand side tends to $0$ exponentially and the solution remains in the neighbourhood where it is defined, the equilibrium $0$ is exponentially stable. [/step]