Clarifying the Existence and Uniqueness of Solutions to an Initial Value Problem Given Local Lipschitz Continuity

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.
• The soultion is $x(t)=1/(1-t)$. Commented Feb 26 at 20:06
• I still see the wrong solution $x(t)=(1-t)$. Commented Feb 28 at 9:16