[proofplan] The linear growth hypothesis gives an a priori bound for a solution on every bounded time interval. We prove this bound directly by integrating the differential equation and applying the differential form of Gronwall's argument, avoiding any finite-time escape of the Euclidean norm. The [continuation criterion for ordinary differential equations](/theorems/8314) then rules out a finite right endpoint, and the same argument applied to the time-reversed equation rules out a finite left endpoint. [/proofplan] [step:Bound the solution near a finite right endpoint] Let $x:J\to\mathbb{R}^n$ be a maximal solution, where $J\subset\mathbb{R}$ is its interval of definition. Suppose, toward a contradiction, that $b:=\sup J<\infty$. Choose $c\in J$ with $c<b$, and set $K:=[c,b]$. By hypothesis, there exists $A_K>0$ such that \begin{align*} |F(t,z)|\le A_K(1+|z|) \end{align*} for all $t\in K$ and all $z\in\mathbb{R}^n$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Define \begin{align*} u:J\cap[c,b)&\to[0,\infty) \end{align*} by $u(t):=1+|x(t)|$. For $t\in J\cap[c,b)$ with $t\ge c$, the integral form of the differential equation, as in [citetheorem:8311], gives \begin{align*} x(t)=x(c)+\int_c^t F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Taking Euclidean norms, using the triangle inequality for the Bochner integral in $\mathbb{R}^n$, and applying the linear growth bound on $K$, we obtain \begin{align*} |x(t)|\le |x(c)|+A_K\int_c^t (1+|x(s)|)\,d\mathcal{L}^1(s). \end{align*} Equivalently, \begin{align*} u(t)\le u(c)+A_K\int_c^t u(s)\,d\mathcal{L}^1(s). \end{align*} Define \begin{align*} H:J\cap[c,b)&\to(0,\infty) \end{align*} by \begin{align*} H(t):=u(c)+A_K\int_c^t u(s)\,d\mathcal{L}^1(s). \end{align*} Then $u(t)\le H(t)$ and, by the [fundamental theorem of calculus](/theorems/632) for continuous functions, \begin{align*} H'(t)=A_Ku(t)\le A_KH(t) \end{align*} for every interior point $t$ of $J\cap[c,b)$. Hence \begin{align*} \frac{d}{dt}\left(e^{-A_K(t-c)}H(t)\right)=e^{-A_K(t-c)}(H'(t)-A_KH(t))\le 0. \end{align*} Therefore $e^{-A_K(t-c)}H(t)\le H(c)=u(c)$, and so \begin{align*} u(t)\le H(t)\le u(c)e^{A_K(t-c)}\le u(c)e^{A_K(b-c)} \end{align*} for all $t\in J\cap[c,b)$ with $t\ge c$. Thus \begin{align*} |x(t)|\le u(c)e^{A_K(b-c)}-1 \end{align*} for all $t\in J\cap[c,b)$ with $t\ge c$. [guided] We want to show that the solution cannot run off to infinity while the time variable remains in a compact interval. Suppose that the right endpoint $b:=\sup J$ is finite, and choose a time $c\in J$ with $c<b$. The compact time interval relevant to the endpoint is \begin{align*} K:=[c,b]. \end{align*} The hypothesis gives a constant $A_K>0$ satisfying \begin{align*} |F(t,z)|\le A_K(1+|z|) \end{align*} for every $t\in K$ and every $z\in\mathbb{R}^n$. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Define the scalar control function \begin{align*} u:J\cap[c,b)&\to[0,\infty) \end{align*} by \begin{align*} u(t):=1+|x(t)|. \end{align*} The extra $1$ is useful because the growth bound has the form $1+|x|$, so the right-hand side becomes proportional to $u$. For $t\in J\cap[c,b)$ with $t\ge c$, the interval property of $J$ gives $[c,t]\subset J$. Since $x$ solves $\dot{x}=F(t,x)$, the integral formulation from [citetheorem:8311] gives \begin{align*} x(t)=x(c)+\int_c^t F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Taking Euclidean norms and using the triangle inequality for the integral of a continuous $\mathbb{R}^n$-valued function gives \begin{align*} |x(t)|\le |x(c)|+\int_c^t |F(s,x(s))|\,d\mathcal{L}^1(s). \end{align*} Because $s\in[c,t]\subset K$, the linear growth bound applies to the integrand: \begin{align*} |x(t)|\le |x(c)|+A_K\int_c^t (1+|x(s)|)\,d\mathcal{L}^1(s). \end{align*} In terms of $u$, this becomes \begin{align*} u(t)\le u(c)+A_K\int_c^t u(s)\,d\mathcal{L}^1(s). \end{align*} Now define \begin{align*} H:J\cap[c,b)&\to(0,\infty) \end{align*} by \begin{align*} H(t):=u(c)+A_K\int_c^t u(s)\,d\mathcal{L}^1(s). \end{align*} The inequality above says $u(t)\le H(t)$. Since $u$ is continuous, the fundamental theorem of calculus gives \begin{align*} H'(t)=A_Ku(t) \end{align*} at every interior point of $J\cap[c,b)$. Combining this identity with $u(t)\le H(t)$ yields \begin{align*} H'(t)\le A_KH(t). \end{align*} This is the differential form of the Gronwall estimate. Multiplying by the integrating factor $e^{-A_K(t-c)}$, we compute \begin{align*} \frac{d}{dt}\left(e^{-A_K(t-c)}H(t)\right)=e^{-A_K(t-c)}(H'(t)-A_KH(t))\le 0. \end{align*} Thus the map $t\mapsto e^{-A_K(t-c)}H(t)$ is nonincreasing on $J\cap[c,b)$, and hence \begin{align*} e^{-A_K(t-c)}H(t)\le H(c)=u(c). \end{align*} Therefore \begin{align*} u(t)\le H(t)\le u(c)e^{A_K(t-c)}\le u(c)e^{A_K(b-c)}. \end{align*} Since $u(t)=1+|x(t)|$, this proves the uniform endpoint bound \begin{align*} |x(t)|\le u(c)e^{A_K(b-c)}-1 \end{align*} for all $t\in J\cap[c,b)$ with $t\ge c$. [/guided] [/step] [step:Use the continuation criterion to exclude a finite right endpoint] Let \begin{align*} R:=u(c)e^{A_K(b-c)}. \end{align*} Let $\overline{B}(0,R):=\{z\in\mathbb{R}^n:|z|\le R\}$ be the closed Euclidean ball of radius $R$ centred at $0$. Then $x(t)\in\overline{B}(0,R)$ for all $t\in J\cap[c,b)$ with $t\ge c$. The set $\overline{B}(0,R)\subset\mathbb{R}^n$ is compact by the [Heine-Borel theorem](/theorems/309). For the continuation criterion [citetheorem:8314], take the ambient time interval $I:=\mathbb{R}$ and the state domain $U:=\mathbb{R}^n$. The set $U$ is open, and the hypotheses of the theorem give that $F:I\times U\to\mathbb{R}^n$ is continuous and locally Lipschitz in the state variable. Since $b<\infty$, the number $b$ is not a right endpoint of $I=\mathbb{R}$. Thus [citetheorem:8314] says that, if $b=\sup J<\infty$ for a maximal solution, then for every compact set $C\subset U$ and every $a<b$ there must exist $t\in J\cap(a,b)$ such that $x(t)\notin C$. Taking $C=\overline{B}(0,R)$ and $a=c$ contradicts the bound from the previous step. Hence $\sup J$ cannot be finite. Therefore \begin{align*} \sup J=+\infty. \end{align*} [/step] [step:Apply the same argument to the time-reversed equation] It remains to exclude a finite left endpoint. Suppose, toward a contradiction, that $a:=\inf J>-\infty$. Define the time-reversed interval \begin{align*} \widetilde{J}:=\{-t:t\in J\} \end{align*} and the time-reversed curve \begin{align*} \widetilde{x}:\widetilde{J}\to\mathbb{R}^n \end{align*} by $\widetilde{x}(r):=x(-r)$. Define also the time-reversed vector field \begin{align*} \widetilde{F}:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}^n \end{align*} by \begin{align*} \widetilde{F}(r,z):=-F(-r,z). \end{align*} Then $\widetilde{F}$ is continuous. If $\widetilde{K}\subset\mathbb{R}$ is compact and $\widetilde{C}\subset\mathbb{R}^n$ is compact, then $K:=-\widetilde{K}$ is compact, and the locally uniform Lipschitz estimate for $F$ on $K\times\widetilde{C}$ gives the corresponding estimate for $\widetilde{F}$ on $\widetilde{K}\times\widetilde{C}$. Also, if $\widetilde{K}\subset\mathbb{R}$ is compact, then $K:=-\widetilde{K}$ is compact and \begin{align*} |\widetilde{F}(r,z)|=|F(-r,z)|\le A_K(1+|z|) \end{align*} for all $r\in\widetilde{K}$ and all $z\in\mathbb{R}^n$. Thus $\widetilde{F}$ satisfies the same hypotheses as $F$. For every interior point $r\in\widetilde{J}$, \begin{align*} \frac{d\widetilde{x}}{dr}(r)=-\dot{x}(-r)=-F(-r,x(-r))=\widetilde{F}(r,\widetilde{x}(r)). \end{align*} The endpoint convention for one-sided derivatives is preserved under the change of variables $r=-t$, so $\widetilde{x}$ is a solution of the time-reversed equation on $\widetilde{J}$. Maximality is also preserved: any proper extension of $\widetilde{x}$ for $\widetilde{F}$ would reverse back to a proper extension of $x$ for $F$. Now \begin{align*} \sup \widetilde{J}=-\inf J=-a<\infty. \end{align*} Applying the already proved right-endpoint conclusion to the maximal solution $\widetilde{x}$ gives a contradiction. Therefore $\inf J$ cannot be finite, and hence \begin{align*} \inf J=-\infty. \end{align*} [/step] [step:Conclude that the maximal interval is all of $\mathbb{R}$] The interval $J\subset\mathbb{R}$ satisfies \begin{align*} \inf J=-\infty \end{align*} and \begin{align*} \sup J=+\infty. \end{align*} Since $J$ is an interval, these two endpoint identities imply $J=\mathbb{R}$. Thus every maximal solution of $\dot{x}=F(t,x)$ is defined for all real times. [/step]