# About interval of definition for solution to an ODE

Let $$f:\mathbb{R}\times\mathbb{R}^{n}$$ continuous such that $$x'=f(t,x)$$ has uniqueness of solution, and $$|f(t,x)|\leq 10$$. I wnat to prove that every solution for the ODE is defined for all $$t\in\mathbb{R}$$.

Seems like the hypothesis $$|f(x,t)|<10$$ is just to say that $$f$$ is bounded.

A question: $$f$$ having uniqueness of solution implies $$f$$ is Lipschitz?

I'm asking that because, if that is true, it contradicts what the Picard Theorem says about the interval of solution.

How can I solve that?

• The answer to the question is: NO. Indeed, in a sense a "generic" RHS has the property that we have local uniqueness of the IVP. How can that contradict the Picard theorem? Apr 20, 2019 at 18:25
• Just some thoughts, I'm wrong. How can I solve that? Apr 21, 2019 at 1:46

Assume that $$f \colon \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n$$ is continuous and that there exists $$M > 0$$ such that $$\lvert f(t, x) \rvert \le M$$ for all $$(t, x) \in \mathbb{R} \times \mathbb{R}^n$$.
Suppose to the contrary that some nonextedible (to the right, for the definiteness' sake) solution $$\varphi$$ of $$x' = f(t, x)$$ is defined on an open interval whose right endpoint, $$\beta$$, is $$< \infty$$. Observe that for any two $$t_1, t_2$$ belonging to that interval there holds $$\tag{*} \lvert \varphi(t_1) - \varphi(t_2) \lvert = \Bigl\lvert \int\limits_{t_1}^{t_2} \varphi'(s) \, \mathrm{d}s \Bigr\rvert \le \int\limits_{t_1}^{t_2} \lvert \varphi'(s) \rvert \, \mathrm{d}s \le M \lvert t_2 - t_1 \rvert.$$ Now, let $$(t_k)$$, $$t_k < \beta$$, be a sequence convergent to $$\beta$$ as $$k \to \infty$$. $$(t_k)$$ is a Cauchy sequence, so, by $$(*)$$, $$(\varphi(t_k))$$ is a Cauchy sequence, too. By the completeness of $$\mathbb{R}$$, $$(\varphi(t_k))$$ has a (finite) limit, $$\tilde{\varphi}$$. Indeed, this limit is the same for all sequences converging to $$\beta$$ from the left (if there were sequences $$\check{t}_k$$, $$\hat{t}_k$$ for which the limits are different, then the limit for the sequence $$(\check{t}_1, \hat{t}_1, \check{t}_2, \hat{t}_2, \dots)$$ would not exist), and we can legitimately denote it as $$\varphi(\beta)$$. We can easily prove that the extended function $$\varphi$$ has left-hand derivative, $$\varphi'_{-}(\beta)$$, at $$\beta$$, equal to $$f(\beta, \varphi(\beta))$$.
We have thus extended $$\varphi$$ to a solution of $$x' = f(t, x)$$ on a right-closed interval with right endpoint $$\beta$$. Apply the (local) existence theorem (for instance, the Peano theorem) to the IVP $$\begin{cases} x' = f(t, x) \\ x(\beta) = \varphi(\beta) \end{cases}$$ to extend the solution $$\varphi$$ to the right of $$\beta$$. But this contradicts our assumption that $$\varphi$$ cannot be extended to the right. So the right endpoint of the interval of existence of $$\varphi$$ is $$\infty$$.
In a similar way we prove that the left endpoint of the interval of existence of $$\varphi$$ must be $$-\infty$$. Notice that we nowhere assume the uniqueness property.