# Prove that maximal solutions are defined in $\mathbb{R}$

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 $$\dot{x}=f(t,x)$$?

• 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. – Ian Nov 10 '14 at 0:44
• @Ian You're right, I missed that part when I was copying the problem. I updated the post with new attempt to solve it. – Cure Nov 10 '14 at 1:21
• 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. – Artem Nov 10 '14 at 2:15
• @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. – Cure Nov 10 '14 at 3:02
• 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) – Artem Nov 10 '14 at 3:18

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?
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.