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Let $\Vert\;\Vert$ be the Euclidean norm on $\Bbb{R}^n$. We consider the following O.D.E

\begin{align}(a)\qquad\qquad\begin{cases}x'(t)=f(t,x(t)) & t\geq 0,\\x(0)=x_0\in \Bbb{R}^n&\end{cases}\end{align} where \begin{align}f:\Bbb{R}^{+}\times \Bbb{R}^{n}\to \Bbb{R}^{n}\end{align}is continuous. Assuming that there exists a continuous function $\alpha:\Bbb{R}^+\to \Bbb{R}^+$ such that \begin{align}\langle f(t,y),y\rangle\leq \alpha(t)\left(1+\Vert y \Vert^2\right),\;\forall\;t\geq 0,\;y\in\Bbb{R}^n.\end{align} So, how do I prove that $(a)$ has a global solution defined on $\Bbb{R}^+?$ Thanks for your time and help!

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  • $\begingroup$ Is $f$ Lipschitz? $\endgroup$
    – copper.hat
    Aug 30 '18 at 2:05
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    $\begingroup$ I think a global solution exists by Peano. The condition should be used together with a smart application of Gronwall's lemma to prove uniqueness, I guess. $\endgroup$
    – amsmath
    Aug 30 '18 at 2:16
  • $\begingroup$ Sorry, I had the Bellman Grönwall part written up by the time you added your comment! $\endgroup$
    – copper.hat
    Aug 30 '18 at 2:21
  • $\begingroup$ @copper.hat I also had that $u(t)\le u(0)e^{\int_0^t\alpha(s)\,ds}$, where $u(t) = 1+\|x(t)\|^2$. But what does that imply concerning the uniqueness? $\endgroup$
    – amsmath
    Aug 30 '18 at 2:25
  • $\begingroup$ Uh oh. I just saw that only existence of a global solution has to be proved - not its uniqueness. Now, I am confused. $\endgroup$
    – amsmath
    Aug 30 '18 at 2:27
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Note that if $V(t) = 1+\|x(t)\|^2$ then $\dot{V}(t) = 2 \langle \dot{x}(t), x(t) \rangle \le 2\alpha(t) ( 1 + \|x(t)\|^2)= 2\alpha(t)V(t)$, so Bellman Grönwall gives $V(t) \le V(0) e^{2\int_0^t \alpha(\tau) d \tau}$.

Let $G= \{ (t,y) | t \ge 0, 1+\|y\|^2 \le V(0) e^{2\int_0^t \alpha(\tau) d \tau} \} $. Observe that $G_T=G \cap ( [0,T] \times \mathbb{R}^2 ) $ is compact for any $T \ge 0$. In particular, if $x$ is any solution of the ODE defined on an interval $[0,T]$ then we have $(t,x(t)) \in G_T$ for all $t \in [0,T]$.

It is a little easier to work with the equivalent integral formulation of the ODE. That is, if $I$ is an interval containing $t_0$ then $x$ is a solution of $\dot{x} = f(t,x), x((t_0) = x_0$ with $x\in C^1(I)$ iff $x$ is a solution of the $x(t) = x_0+\int_{t_0}^t f(s, x(s))ds$ with $x \in C(I)$. The advantage is that differentiability is 'built in'.

Peano's theorem shows the existence of a local (in time) solution, we need to demonstrate that a solution exists for all $t \ge 0$.

Suppose $x$ is a solution defined on $[0,T]$. Then Peano shows that there is some $\delta>0$ such that $x$ has an extension to $[0,T+\delta)$. Since $G_{T+\delta} $ is compact, $f$ is bounded on $G_{T+\delta} $ and hence $x(T+\delta)=\lim_{t \to T+\delta} x(t)$ exists and satisfies the integral equation (both of these are straightforward to see from the integral formulation) and hence $x$ has an extension to $[0,T+\delta]$.

Now let $T^* = \sup \{ T | x \text{ has an extension to } [0,T] \}$ and note that if $T $ is finite we can extend the solution, which is a contradiction. Hence we can extend $x$ to $[0,\infty)$.

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  • $\begingroup$ Thanks a lot! I'll go through the proof and then ask questions where necessary! $\endgroup$ Aug 31 '18 at 19:58

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