# Why the maximal interval of existence isn't always $\mathbb{R}$

We know from the Picard-Lindelof Theorem that given an IVP with $$f(x)\in C(E)$$ where $$E$$ is an open subset of $$\mathbb{R}^n$$ then we have a local solution define in some interval $$]t_0-\alpha,t_0+\alpha[$$. Now suppose we want the maximal interval of existence , I am interest in figuring out what are the factors that stop us from this maximal interval being $$\mathbb{R}$$. One thing I think could happen is that when we try to apply the Picard Lindelof again to a point in $$]t_o-\alpha,t_0+\alpha[$$ we obtain a new interval but the point is that this sum of lenghts of the intervals could be converging, and so we could not get to the whole space of $$\mathbb{R}$$ ?

Is there anything more that can stop us from getting a solution defined in the whole line ? We need to have control over the values of $$x(t)$$ but I think the picard-Lindelof theorem gives us that $$x(t)$$ is in $$E$$ for all $$t$$ in the interval given by picard lindelof. I guess a problem could be that if we try to use picard-lindelof in $$t_0-\alpha$$ we need to check that $$x(t_0-\alpha)$$ is defined and so this could become a problem too.

Also I have seen both version where the interval where the solution is defined is closed and sometimes I have seen it open so I guess I am confused about that too, I can see why they could equivalent but I don't understand why we jut don't fix one.

Is there anthing I am missing or missunderstanding? Thanks in advance.

Try $$x(0) = 0,$$ $$\dot{x} = 1 + x^2$$
• Yeah we get $tan(x)$ and we will have the problem that it's going to $\infty$ as we aprroach $\frac{\pi}{2}$ and so our solution won't be defined. I still don't quite understand if what I talked about makes sense or not :) Oct 13 '20 at 21:54
If a solution $$y(t)$$ exists in an open interval $$(a,b)$$ where $$b$$ is finite, but not in any larger open interval, then either $$\lim_{t \to b-} y(t)$$ does not exist at all or it is a point outside $$E$$. Either of these is possible.