Ordinary Differential Equations/Blow-ups and moving to boundary

Definition:

Let an ordinary differential equation

$\{\begin{array}{cc}{x}^{\prime }(t)=f(t,x(t))& \\ x(t_0)=x_0\end{array}$

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 $f:D\to {ℝ}^n$ be the right hand side of a differential equation for the special case $D=(c,d)\times {ℝ}^n$ 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 $\Vert x(t)\Vert \to \mathrm{\infty }$ as $t\to a$. Similarly, either $b=d$ or $\Vert x(t)\Vert \to \mathrm{\infty }$ 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 $\Vert (t,x(t))\Vert \ge k$ 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 $\Vert x(t)\Vert \to \mathrm{\infty }$ as $t\to a$, and the analogous statement for $b$ and $d$.$◻$