so I'll start with my main question, in short, giving details below:
If I'm using a theorem to prove the existence and uniqueness of solutions to $\dot{x} = f(x,t)$ which states "$J = [t_{0}, t_{1}]$ is a finite, closed interval", and I find that all other conditions apply ánd that $t_{0} < t_{1}$ can be chosen arbitrarily (large), is it correct to assume that solutions exist on $J = [t_{0}, \infty)$? No, right? My colleagues disagree and say it ís possible, but I find that it violates the assumption of a finite interval thus making the proof invalid.
Specifically, the dynamical system was given as
$$ \dot{x}(t) = t^{2} \cdot x(t) $$
and the theorem used (based on the book Nonlinear Systems by Khalil) was given as
"Lemma V.2 If $f$ is piecewise continuous in $t$ and globally Lipschitz in $x$, then the system has a unique solution on $[t_{0}, t_{1}]$ for any initial state $x_{0} \in \mathbb{R}$."
where the piecewise continuity is trivially satisfied, and Lipschitz continuity is defined as
"Throughout this section, we consider systems of the form
$$\dot{x} = f(t, x), \quad x(t_{0}) = x_{0}, \\ f : J \times D → \mathbb{R}^{n}, \quad J = [t_{0}, t_{1}], \quad D \subset \mathbb{R}^{n} \: domain.$$
Note that J is a finite, closed interval (see Section I)."
and
"Lemma IV.2. If $f$ and $[\partial f / \partial x]$ are both continuous, then $f$ is locally Lipschitz. If there furthermore exists a $B > 0$ s.t. $$ \left\Vert \left[ \frac{\partial f(t,x)}{\partial x} \right] \right\Vert \leq B, \quad \forall x \in D, t \in J,$$ then $f$ is also Lipschitz with Lipschitz constant $L = B$. [...]. It is globally Lipschitz (uniformly in $t$) if furthermore $D = \mathbb{R}^{n}$."
Now we can show that Lemma IV.2 holds with $B =t_{1}^{2}$ on the domain $D = \mathbb{R}$, making the function globally Lipschitz for any choice of $t_{1}$. Thus Lemma V.2 is satisfied with $D = \mathbb{R}$ and $J = [t_{0}, t_{1}]$. However, I believe that this is all that we can conclude. My colleagues give the argument that since $t_{1}$ can be chosen arbitrarily large, we can say $J = [t_{0}, \infty)$. Is this correct and if so/not, why? I believe it is not correct since this interval is no longer finite.
Any insight into this question would be greatly appreciated: since we are teaching this course to students I would like to have a cohesive and properly justified argumentation; thanks for reading and sorry for the long post!
EDIT:
I wanted to clarify that what I'm mainly interested in is whether the argumentation using Lemma V.2 and Lemma IV.2, and the definitions given, is correct or not: I 100% believe solutions exist on $[t_{0}, \infty)$ and that we can show the solutions don't blow up in finite time or any other theorem, but that's not really my point: is the proof given here valid, and if so, why does it not matter that we violated the condition that $J$ must be finite?
The reason I am asking is that I think is that we need to remove this condition on $J$ being finite on our formula sheet unless one of you knows why this condition is there in the first place. Hope this clarifies things.
If there is a solution on $[t_0, t_1]$ for arbitrarily large values of $t_1$, then there is a solution on $[t_0, \infty)$. Namely, $x(t)$ is equal to its value in the solution on $[t_0, t_1]$ for any $t_1 > t$. There is no ambiguity because of the uniqueness.