[proofplan] Assume, toward a contradiction, that a maximal trajectory has a finite right endpoint. Local essential boundedness of the control and the linear growth estimate give an integral inequality for the Euclidean norm of the state. Gronwall's inequality gives a uniform bound for the trajectory up to the endpoint; the same growth estimate then gives a uniform bound on its metric speed, so the trajectory has a finite left limit at the endpoint. The controlled ODE local well-posedness and continuation theorem applied from this limiting point extends the solution past the endpoint, contradicting maximality. [/proofplan] [step:Assume a maximal trajectory stops at a finite time] Fix $t_0 \in \mathbb{R}$, $x_0 \in \mathbb{R}^n$, and an admissible control $u:[t_0,\infty) \to U$ with $u \in \mathcal{U}$. Let $x:[t_0,T_{\max}) \to \mathbb{R}^n$ be the corresponding maximal trajectory, so $x(t_0)=x_0$, the map $x$ is locally absolutely continuous, and \begin{align*} \dot{x}(t)=f(x(t),u(t)) \end{align*} for $\mathcal{L}^1$-almost every $t \in [t_0,T_{\max})$. Suppose, for contradiction, that $T_{\max}<\infty$. Since $u$ is locally essentially bounded on $[t_0,\infty)$, there exists a constant $M \geq 0$ such that \begin{align*} |u(t)| \leq M \end{align*} for $\mathcal{L}^1$-almost every $t \in [t_0,T_{\max}]$. [/step] [step:Derive the Gronwall inequality for the trajectory norm] For every $t \in [t_0,T_{\max})$, absolute continuity of $x$ and the differential equation give \begin{align*} x(t)=x_0+\int_{t_0}^{t} f(x(s),u(s))\,d\mathcal{L}^1(s). \end{align*} Taking Euclidean norms, applying the triangle inequality for the Bochner integral in $\mathbb{R}^n$, and using the growth hypothesis together with $|u(s)| \leq M$ for almost every $s$, we obtain \begin{align*} |x(t)| \leq |x_0|+\int_{t_0}^{t} |f(x(s),u(s))|\,d\mathcal{L}^1(s). \end{align*} Thus \begin{align*} |x(t)| \leq |x_0|+\int_{t_0}^{t} \bigl(a+b|x(s)|+cM\bigr)\,d\mathcal{L}^1(s). \end{align*} Equivalently, \begin{align*} |x(t)| \leq |x_0|+(a+cM)(t-t_0)+b\int_{t_0}^{t}|x(s)|\,d\mathcal{L}^1(s). \end{align*} [guided] The goal is to convert the differential equation into an inequality involving only the scalar function $t \mapsto |x(t)|$. Since $x$ is locally absolutely continuous and satisfies the controlled ODE almost everywhere, the integral form of the equation is valid: \begin{align*} x(t)=x_0+\int_{t_0}^{t} f(x(s),u(s))\,d\mathcal{L}^1(s). \end{align*} The integral is an ordinary vector-valued [Lebesgue integral](/page/Lebesgue%20Integral) over the finite interval $[t_0,t]$. Taking Euclidean norms and using the triangle inequality for integrals gives \begin{align*} |x(t)| \leq |x_0|+\int_{t_0}^{t} |f(x(s),u(s))|\,d\mathcal{L}^1(s). \end{align*} Now the linear growth hypothesis applies pointwise to the pair $(x(s),u(s))$ for almost every $s$, and the essential bound on the control gives $|u(s)| \leq M$ for almost every $s$. Therefore \begin{align*} |f(x(s),u(s))| \leq a+b|x(s)|+cM \end{align*} for $\mathcal{L}^1$-almost every $s \in [t_0,t]$. Substituting this estimate into the integral yields \begin{align*} |x(t)| \leq |x_0|+\int_{t_0}^{t} \bigl(a+b|x(s)|+cM\bigr)\,d\mathcal{L}^1(s). \end{align*} Separating the constant part from the term involving $|x(s)|$ gives \begin{align*} |x(t)| \leq |x_0|+(a+cM)(t-t_0)+b\int_{t_0}^{t}|x(s)|\,d\mathcal{L}^1(s). \end{align*} This is the precise form needed for Gronwall's inequality: the unknown scalar function is bounded by a known finite quantity plus a constant multiple of its own time integral. [/guided] [/step] [step:Apply Gronwall to obtain a uniform bound before the endpoint] Define the constant \begin{align*} A:=|x_0|+(a+cM)(T_{\max}-t_0). \end{align*} Then $A<\infty$, and for every $t \in [t_0,T_{\max})$, \begin{align*} |x(t)| \leq A+b\int_{t_0}^{t}|x(s)|\,d\mathcal{L}^1(s). \end{align*} By Gronwall's inequality applied to the nonnegative locally integrable function $y:[t_0,T_{\max}) \to [0,\infty)$ defined by $y(t)=|x(t)|$, we obtain \begin{align*} |x(t)| \leq A e^{b(t-t_0)} \end{align*} for every $t \in [t_0,T_{\max})$. Hence \begin{align*} |x(t)| \leq R \end{align*} for every $t \in [t_0,T_{\max})$, where \begin{align*} R:=A e^{b(T_{\max}-t_0)}. \end{align*} Here $\overline{B}(0,R):=\{x \in \mathbb{R}^n : |x| \leq R\}$ denotes the closed ball of radius $R$ centered at the origin. The trajectory therefore remains in the compact closed ball $\overline{B}(0,R) \subset \mathbb{R}^n$. [/step] [step:Show the trajectory has a finite endpoint limit] The growth estimate and the bounds $|x(t)| \leq R$ and $|u(t)| \leq M$ imply \begin{align*} |\dot{x}(t)|=|f(x(t),u(t))| \leq a+bR+cM \end{align*} for $\mathcal{L}^1$-almost every $t \in [t_0,T_{\max})$. Define \begin{align*} L:=a+bR+cM. \end{align*} For any $s,t \in [t_0,T_{\max})$ with $s<t$, absolute continuity gives \begin{align*} |x(t)-x(s)| \leq \int_s^t |\dot{x}(r)|\,d\mathcal{L}^1(r) \leq L(t-s). \end{align*} Thus $x(t)$ is Cauchy as $t \uparrow T_{\max}$. Since $\mathbb{R}^n$ is complete, there exists a point $x_* \in \mathbb{R}^n$ such that \begin{align*} \lim_{t \uparrow T_{\max}} x(t)=x_*. \end{align*} Moreover $|x_*| \leq R$ because $\overline{B}(0,R)$ is closed. [/step] [step:Extend the solution beyond the finite endpoint and contradict maximality] Apply the controlled ODE local well-posedness and continuation theorem at the initial time $T_{\max}$ with initial state $x_*$ and the same admissible control $u$ restricted to $[T_{\max},\infty)$. The hypotheses of that theorem are available by assumption on compact time intervals, and the restricted control is locally essentially bounded because $u$ is locally essentially bounded on $[t_0,\infty)$. Hence there exist $\varepsilon>0$ and a trajectory \begin{align*} z:[T_{\max},T_{\max}+\varepsilon) \to \mathbb{R}^n \end{align*} such that $z(T_{\max})=x_*$ and \begin{align*} \dot{z}(t)=f(z(t),u(t)) \end{align*} for $\mathcal{L}^1$-almost every $t \in [T_{\max},T_{\max}+\varepsilon)$. Define the concatenated map \begin{align*} \tilde{x}:[t_0,T_{\max}+\varepsilon) \to \mathbb{R}^n \end{align*} by setting $\tilde{x}(t)=x(t)$ for $t \in [t_0,T_{\max})$ and $\tilde{x}(t)=z(t)$ for $t \in [T_{\max},T_{\max}+\varepsilon)$. Since $x(t) \to x_*=z(T_{\max})$ as $t \uparrow T_{\max}$, the map $\tilde{x}$ is continuous at $T_{\max}$; since both pieces are absolutely continuous on compact subintervals and satisfy the same controlled ODE almost everywhere, $\tilde{x}$ is a trajectory extending $x$ beyond $T_{\max}$. This contradicts maximality of $x$ on $[t_0,T_{\max})$. Therefore the assumption $T_{\max}<\infty$ is false, and every maximal trajectory is defined for all $t \geq t_0$. [/step]