Let $ E $ be a Banach space (for our purposes $ \mathbb R^n $ will suffice). Let $ U\subset E $ be open, and let $ X\colon U\to E $ be a continuous and locally Lipschitz function (in the following, the "vector field").
It's well known that for every $ x_0\in U $ there exists a neighborhood $ V\subset U $ of $ x_0 $ and an $ \alpha > 0 $ such that for every $ x\in V $ there exists a unique integral curve $ x\colon \left]-\alpha,\alpha\right[\to U $ of $ X $ such that $ x(0) = x $ and $ x(t)\in V $ for all $ t\in \left]-\alpha,\alpha\right[ $.
We can thus define a map $$ \Phi\colon \left]-\alpha,\alpha\right[\times V\to U $$ in the obvious way, requiring that $ \Phi_t(x) $ is the image of the integral curve of $ X $ passing through $ x\in V $ at the time $ t\in \left]-\alpha,\alpha\right[ $.
My question is: can we define $ \Phi $ on the whole $ U $? Under what conditions?