Proposition (concatenation of solutions to ODEs):
Assume we have a continuous function and two functions , satisfying
and
respectively. Then the function
solves
Proof: We prove differentiability at as follows: We claim that the derivative of is given by . To prove our claim, we note that
where and ; this is because
- .
In the case where both are both contained in the same of the two intervals , , the convergence is clear anyhow.
Definition (maximal interval of existence):
Let an ordinary differential equation
be given. The maximal interval of existence is defined to be the interval , where
- and .
Proposition (existence of maximal interval of existence in the continuous case):
Definition:
Let an ordinary differential equation
be given, where is continuous. The maximal interval of existence around is the maximal (w.r.t. set inclusion) interval such that and there exists a solution defined on 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 and () such that 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 . The theorem on concatenation makes sure that this can never occur.
We now aim to prove that if we walk along the solution graph as approaches the endpoints of the maximal interval of existence , then in a sense we move towards the boundary of , where is required to be open and is the domain of definition of . This shall mean that for any compact set , if we pick large or small enough, is outside . The proof is longer and needs preparation.
Proof: Since is compact, it is bounded. Therefore, for a sufficiently large . Furthermore, due to the above, has a certain minimum distance to the boundary, and we may choose such that . Choose . Then
- and .
Hence, .
Proof: Suppose otherwise. Then without loss of generality, we have a sequence such that and (an analogous supposition for the other end of the interval is led to a contradiction analogously). Since is compact, the sequence has an accumulation point . We claim that in fact
- .
Pick such that . Let be arbitrary. We may restrict ourselves to sufficiently small such that . Since is continuous, it is bounded on the compact , say by . Now pick such that . If we assume that leaves for , the intermediate value theorem applied to the function
yields the existence of an such that . But
- ,
contradiction.
Hence, . But on the other hand, by Peano's existence theorem and concatenation of solutions we may extend the solution at for every to the left by a fixed amount (namely for , where which exists due to continuity of and compactness of ), and doing so for sufficiently small yields the contradiction that is not the maximal interval of existence.
Corollary:
Let be the right hand side of a differential equation for the special case for an interval . Let be the maximal interval of existence of a solution around . Then either or as . Similarly, either or as .
Proof:
From the preceding theorem, the solution eventually leaves every compact as or . In particular, this holds for the compact sets . But to leave this implies either or or , since the distance of to is exactly the distance of to the nearest of the interval endpoints , . Hence, if not , then as , and the analogous statement for and .