Why are maximal existence intervals of IVP open?

Question

In my current ODE lecture we recently introduced the proposition

Let $D \subseteq \mathbb{R} \times \mathbb{R}^{n+1}$ be open, $f: D \rightarrow \mathbb{R}^n$ be continuous and admit a local Lipschitz-condition in it's second component. Then there exists for every initial value $(t_0, y_0) \in D$ an open interval $I$ with $t_0 \in I$ and a solution $y$ on $I$ to the IVP

$$\begin{cases} y' = f(t, y) & (t, y) \in D \\ y(t_0) = y_0 \end{cases}$$

such that for every interval $J$ with $t_0 \in J$ and solution $z$ on $J$ to the IVP we already have

• $J \subseteq I$
• $z = y|_J$

The interval $I$ is called maximal interval of existence for the IVP.

Proof: (Sketch)

1. We define

   $$t^* = \sup \{t : t \geq t_0 \text{ and there exists a solution on } [t_0, t]\}$$ $$t_* = \inf \{t : t \leq t_0 \text{ and there exists a solution on } [t, t_0]\}$$

   and prove (by the local Picard-Lindelöf) that this already gives a solution $y$ on $(t_*, t^*)$ (notice that we chose the open interval here).

2. Now exactly:

   Let $z: J \rightarrow \mathbb{R}^n$ be a solution to the IVP on $J \subseteq I$. In particular $z$ is a solution to the IVP on $[\min(t_0, t), \max(t_0, t)]$ for every $t \in J$. From local Picard-Lindelöf we already have $z = y|_J$.

Question: Now I'm curious about the part "... on $J \subseteq I$." How do we only look at intervals with $J \subseteq I$. In particular why isn't it possible that the solution in step $1$ is defined on $[t_*, t^*]$ (closed). Am I missing something or does this require extra work?

EDIT: The first page of this document has the theorem in it (no proof) and the text above describes the process as well.

Answer 1

By definition, a function that is differentiable at $t$ must be defined in some open interval around $t$, because the definition of derivative requires a (two-sided) limit. In order for $y$ to be a solution of the differential equation $y' = f(t,y)$, that derivative $y'$ must exist. So solutions of differential equations quite naturally "live" on open intervals.

That said, it is quite possible that as $t$ approaches an endpoint of the maximal interval of existence, the (one-sided) limit of a solution exists. It just won't give you a point in $D$.