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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?
I saw an answer to a similar question on MSE some time ago, but as I cannot find it I write the following.
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.
