Ordinary Differential Equations/Blow-ups and moving to boundary

Definition:

Let an ordinary differential equation
 * $$\begin{cases}

x'(t) = f(t, x(t)) & \\ x(t_0) = x_0 \end{cases}$$ be given, where $$f$$ is continuous. The maximal interval of existence around $$(t_0, x_0)$$ is the maximal (w.r.t. set inclusion) interval $$I$$ such that $$t_0 \in I$$ and there exists a solution $$x$$ defined on $$I$$ to the equation above.

Note that only the preceding theorem on concatenation of solutions ensures that the definition of a maximal interval of existence makes sense, since otherwise it might happen that there are two intervals $$I_1 = (a_1, b_1)$$ and $$I_2 = (a_2, b_2)$$ ($$a_1 < a_2 < t_0 < b_1 < b_2$$) such that $$t_0$$ is contained within both intervals and a solution is defined on both intervals, but the solutions are incompatible in the sense that none can be extended to the "large" interval $$(a_1, b_2)$$. The theorem on concatenation makes sure that this can never occur.

We now aim to prove that if we walk along the solution graph $$(t, x(t))$$ as $$t$$ approaches the endpoints of the maximal interval of existence $$I$$, then in a sense we move towards the boundary of $$D$$, where $$D$$ is required to be open and is the domain of definition of $$f$$. This shall mean that for any compact set $$K \subset D$$, if we pick $$t$$ large or small enough, $$(t, x(t))$$ is outside $$K$$. The proof is longer and needs preparation.

Corollary:

Let $$f: D \to \mathbb R^n$$ be the right hand side of a differential equation for the special case $$D = (c, d) \times \mathbb R^n$$ for an interval $$J = (c, d)$$. Let $$I = (a, b)$$ be the maximal interval of existence of a solution around $$(t_0, x_0) \in D$$. Then either $$a = c$$ or $$\|x(t)\| \to \infty$$ as $$t \to a$$. Similarly, either $$b = d$$ or $$\|x(t)\| \to \infty$$ as $$t \to b$$.

Proof:

From the preceding theorem, the solution eventually leaves every compact $$K \subset D$$ as $$t \to a$$ or $$t \to b$$. In particular, this holds for the compact sets $$\overline{D_k}$$. But to leave this implies either $$\|(t, x(t))\| \ge k$$ or $$|c - t| \le 1/k$$ or $$|d - t| \le 1/k$$, since the distance of $$(t, x(t))$$ to $$\partial D$$ is exactly the distance of $$t$$ to the nearest of the interval endpoints $$c$$, $$d$$. Hence, if not $$a = c$$, then $$\|x(t)\| \to \infty$$ as $$t \to a$$, and the analogous statement for $$b$$ and $$d$$.