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Let $f:\mathbb{R}\times\mathbb{R}^n\to \mathbb{R}^n$ be a continuously differentiable function. Suppose that exists $v:\mathbb{R}\to[0,\infty)$ continuous such that $||f(t,x)||\leq v(t)||x||\; \forall (t,x)\in \mathbb{R}\times\mathbb{R}^n$. Prove that all maximal solutions for $\dot{x}=f(t,x)$ are defined in $\mathbb{R}$; what happens if $f$ is bounded?

I believe that the Picard–Lindelöf theorem is the way to go, but I don't know how to prove that $f$ satisfies the Lipschitz condition. In order to use it, if I understand correctly, I should prove that $v$ is bounded for every $t$ in a neighborhood of $(t_0,x_0)$ being the last a solution for the differential equation.

If $\dot{x}=f(t,x)$ then $x(t)=\displaystyle\int f(t,x)dt$. Considering the hypothesis, $$\displaystyle{||x||\leq\int||f(t,x)||dt\leq\int v(t) ||x||dt}\\1\leq\int||f(t,x)||dt\leq\int v(t)dt$$

which doesn't seem useful at all....



First update

Alright, let's try it this way: The goal is to show that we always have a is defined in $\mathbb{R}$ for any IVP. Suppose $(t_0,x_0)$ is a maximal solution for $\dot{x}=f(t,x)$ with $x(t_0)=a$, then I should prove that exists an interval $[t_0-c,t_0+c]$ in which the solution exists.

$v$ is continuos, then for any $t_0$ we can pick $c$ and $[t_0-c,t_0+c]$ will be be bounded. Define $K=\operatorname{max}\{v(t):t\in [t_0-c,t_0+c]\}$. Then $f$ is Lipschitz for every $(t,x)\in [t_0-c,t_0+c]\times \mathbb{R}^n$, this is, $||f(t,x)||\leq v(t)||x||\leq K ||x||\;\forall (t,x)\in [t_0-c,t_0+c]\times\mathbb{R}^n$

By the Picard-Lindelöf theorem, the IVP has an unique maximal solution. And $t_0,c$ were arbitrary, can I conclude that all maximal solutions are defined?


Second update

Looking about extensibility I found these notes about differential equation. The extensibility theorem and the corollary seem to be enough to prove that the solutions exists for all $\mathbb{R}$.

I'll quote both here:

  • Suppose that $f$ is $C^1$ on $\mathbb{R}^n$. Denote the unique solution by $x(t)$ and suppose $J:=(a,b)$ is the maximal interval of existence.

    Theorem 1.20 (Extensibility Theorem): For each compact set $K\subset \mathbb{R}^n$ there is a $t\in J$ such that $x(t)\notin K$; thus, $$\lim_{t\to b^-}|x(t)|=\lim_{t\to a^+}|x(t)|=+\infty .$$

    Corollary 1.21: Without loss of generality, if $f:\mathbb{R}^n\to \mathbb{R}^n$ is continuous, then the solutions to $\dot{x}=f(x)$, $x(0)=x_0$ exist $\forall t\in \mathbb{R}$.

With the corollary and the continuity of $f:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}^n$, can I affirm that all maximal solutions can be defined in $\mathbb{R}$ for the equation $\dot{x}=f(t,x)$?

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    $\begingroup$ You need some mild assumption on $v$, for example that it is continuous. If it is continuous, then it is bounded on compact intervals, which is exactly the local Lipschitz property that you need for Picard-Lindelof. $\endgroup$ – Ian Nov 10 '14 at 0:44
  • $\begingroup$ @Ian You're right, I missed that part when I was copying the problem. I updated the post with new attempt to solve it. $\endgroup$ – Cure Nov 10 '14 at 1:21
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    $\begingroup$ You are asked to prove that solutions are defined globally, for all $\mathbb R$. The Picard theorem is essentially local, it guarantees the uniqueness and existence of solution on a some small interval. To work through your problem you need several facts about extensibility of solutions. $\endgroup$ – Artem Nov 10 '14 at 2:15
  • $\begingroup$ @Artem Thanks. Could you be a bit more explicity about the $\text{several}$ facts about extensibility? I looked for it and I found the theorem I posted in the second update of the post, but I'm unaware of what other facts are needed to answer the question. $\endgroup$ – Cure Nov 10 '14 at 3:02
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    $\begingroup$ Not really. Again, the estimate $|f(t,x)|\leq T|x|$ is essential, without it any proof doomed to be wrong (see my example above) $\endgroup$ – Artem Nov 10 '14 at 3:18
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By the Picard-Lindelöf theorem, the IVP has an unique maximal solution. And $t_0,\,c$ were arbitrary, can I conclude that all maximal solutions are defined?

No, you cannot conclude this, because, as I mentioned, Picard's theorem is local.

To finish your prove you need to have $$ x(t)=x_0+\int_{t_0}^t f(\tau,x)d\tau, $$ which implies that $$ |x(t)|\leq |x_0|+K|x|(t-t_0), $$ using your notations. After this you will need a Grownwall's inequality, to find that $$ |x(t)|\leq\ldots, $$ which, together with my previous remarks, finished the proof. I think you should be able to fill in the necessary details.

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