# proof of unique solution given that $\|F(x,t)\| \leq C$

Let $C>0$ be constant and let $F: \mathbb{R} \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuously differentiable, where $\|F(x,t)\| \leq C$ for all $(x,t)\in \mathbb{R} \times \mathbb{R}^n$. Then there exists a unique solution $x:\mathbb{R} \rightarrow \mathbb{R}^n$ of the initial value problem $x'(t)=F(x,t), x(t_0)=x_0$ for every $(t_0, x_0) \in \mathbb{R} \rightarrow \mathbb{R}^n$.

Can I deduce from the condition $\|F(x,t)\| \leq C$ that $F(x,t)$ satisfies the Lipschitz condition($|F(t,x_1)-F(t,x_2)|\leq L |x_1-x_2|$) and also $\|\frac{\partial F}{\partial y}\| \leq k$ is bounded for some constant $k$ ? Or is there another useful property by $\|F(x,t)\| \leq C$?

• The derivative may not be bounded – Eduardo Longa Dec 15 '17 at 22:26

$F$ is locally Lipschitz, just from the fact that it is continuously differentiable, but is not necessarily globally Lipschitz. Locally Lipschitz is enough for the Existence and Uniqueness Theorem. Global, rather than just local, existence of solutions follows from the bound, because $$\|x(t) - x(t_0)\| = \left\| \int_{t_0}^t x'(s)\; ds \right\| \le C |t - t_0|$$