For an open interval $I$ and an open region $ \Omega \subset \mathbb{R}^{n}$ let $F: I \times \Omega \rightarrow \mathbb{R}^{n},( t, x) \mapsto F(t, x)$, continuous and locally Lipschitz continuous with respect to $x$.
For each $\left(t_{0}, x_{0}\right) \in I \times \Omega $ then there is [...] one solution $\varphi: I \rightarrow \mathbb {R}^{n} $ of Initial value problem $\varphi^{\prime}(t)=F(t, \varphi(t)) $ with $\varphi\left(t_{0}\right)=x_{0} $.
Here [...] should be replaced with: exactly, at least, at most
I was sure it has to be "exactly", because of lipschitz, but it says it's false because: "Only uniqueness follows from local Lipschitz, but not global existence."
Now I think that it's " at least" one solution because of local lipschitz continuity and Peano theorem
An example would help a lot!
This depends on the interval $I$.
If $F$ is as in the assumptions, solutions exist and are unique on small open intervals about $t_0$. But they may not exist on the entire interval. Therefore at most is correct.
Example: Consider $x'= x^2, x(0) = x_0 = 1$. The unique solution is $x(t) = (1-t)^{-1}$. It exists on the set$(-\infty,1)$. But on the interval $(-2,2)$ this solution does not exist.