Theorem (concatenation of solutions):
Assume we have two functions , satisfying
respectively. Then the function
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.
We note that the equation is supposed to solve is equivalent to
by the fundamental theorem of calculus; but this follows (by the fundamental theorem of calculus) from the equations satisfied by and for and separately.
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.
The first theorem we need is from a broader context in the sense that it has applications throughout analysis.
Let be open, and let be compact. Then there exists such that .
This can be interpreted as stating that a compact set has a minimum distance to the boundary of .
This theorem admits different proofs, dependent on the method and the characterisation of compactness that is used.
Assume otherwise. Then there exists a sequence and such that
Since is compact, the sequence is bounded and hence contains a convergent subsequence to a limit . The corresponding sequence converges to the same limit by the triangle inequality:
Now on the one hand, since is compact, it is closed, and hence since the sequence is contained within . But on the other hand, since is closed, and thus . This is a contradiction.
For each , choose an open ball (possible since is open). Now the sets
trivially form an open cover of , and due to the compactness of we may extract a finite subcover . Set
We claim that is as desired. Indeed, let . Then is contained within some , and hence for any point
by the second triangle inequality.
is Lipschitz continuous (with Lipschitz constant ), since
by the second triangle inequality. Since is compact, assumes a minimum on , and this must be greater zero, since otherwise there exists such that we could pick a sequence such that
and hence and due to the closedness of , contradiction. This minimum is the desired .
Now we have proven enough generalities to proceed to the specific theorems tailored to our claim.
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 , .
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 .
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 .
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 .