Proposition (concatenation of solutions to ODEs):
Assume we have a continuous function
and two functions
,
satisfying
![{\displaystyle {\begin{cases}x_{1}'(t)=f(t,x_{1}(t))&t\in [t_{0}-\gamma ,t_{0}]\\x_{1}(t_{0})=x_{0}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee536e288f3ac24ea9a4f876e527d6c5a3366339)
and
![{\displaystyle {\begin{cases}x_{2}'(t)=f(t,x_{2}(t))&t\in [t_{0},t_{0}+\delta ]\\x_{2}(t_{0})=x_{0}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2759d651b2996d969f6da427b6a3691347506874)
respectively. Then the function
![{\displaystyle x:[t_{0}-\gamma ,t_{0}+\delta ]\to \mathbb {R} ^{n},x(t):={\begin{cases}x_{1}(t)&t\in [t_{0}-\gamma ,t_{0}]\\x_{2}(t)&t\in [t_{0},t_{0}+\delta ]\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a60db6759cc0b4dd3a9485e09fa6dc5f8022a6d4)
solves
![{\displaystyle {\begin{cases}x'(t)=f(t,x(t))&t\in [t_{0}-\gamma ,t_{0}+\delta ]\\x(t_{0})=x_{0}.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a7164d279da0662ab2f10c523a140c1d1a4f3b8)
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
.