Let $f : (0,T) \rightarrow \mathbb{R}$ be differentiable and $u_0 \in \mathbb{R}$. Consider the differential problem $(DP)$ given by $$ u^{\prime} (t) = f(u(t)), $$ $$ u (0) = u_0 .$$
What is the precise meaning of the following sentence:
the function $v : [0,T] \rightarrow \mathbb{R}$ solves the differential problem (DP) on the compact interval $[0,T]$.
It seems crystal clear, that this must imply the following points:
- $v(0) = u_0$.
- $v \in \mathcal{C}^{1}((0,T))$.
- It must hold, that $v^{\prime}(t) = f(v(t))$ for all $t \in (0,T)$
But what else is required?
In particular, what is supposed to happen at the boundaries $v(0)$ and $v(T)$ ?
Does $v$ have to be continuous on the whole compact interval $[0,T]$? It seems rather natural to at least require $v \in \mathcal{C}([0,T))$.
Is this sufficient, or do we need/want $v \in \mathcal{C}([0,T])$ as well? If so, why?