Ordinary Differential Equations/Blow-ups and moving to boundary

Proposition (concatenation of solutions to ODEs):

Assume we have a continuous function and two functions , satisfying


respectively. Then the function


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):


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.

Proposition (Containing a compact set in a canonical compact set):

Let be the right hand side of a differential equation, where is open (in fact, only the set matters for this lemma). Set

for . If is any compact set, then for sufficiently large , .

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, .

Proposition (solutions of ODEs extend up to the boundary):

Let be the right hand side of a differential equation (where is open), and let be a solution to that equation, where is the interior of the maximal interval of solution. If is sufficiently close to either or , then for each compact the point lies outside .

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



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.


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 .


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 .