In this section, we present our boundary value problem analog. This is done under the assumption that . However, with the additional assumption from Theorem 3, the same result is established for and the proof remains the same. Without this additional assumption, the derivative at would only be a right-hand derivative but the result still holds.
Proof. We will only prove part (a) as the proof of part (b) is similar. Part (c) is immediate consequence from parts (a)and (b) when coupoled with hypothesis (v).
Let , and consider In the interests of conserving space and lessening the tedious notation, we denote by as is the boundary value of interest.
Let
be as in Theorem 2,
be given, and define
Note that for every
,
Also, for every
with
Now that we have established the boundary conditions for
, we show that
solves the variational equation. To that end, for
, let
and
By Theorem 2, for
as
Using the notation of Theorem 1 for solutions of initial value problems for (
1), viewing
as the solution of an initial value problem, and denoting the solution
, we have
Then, by utilizing telescoping sums, we have
By Theorem 1 and the Mean Value Theorem, we obtain
where, for
is the solution of the variational equation (
3) along
satisfying
Furthermore, for is between and
Thus, to show exists, it suffices to show, for exists.
Now, from the construction of
we have
Hence, for
with
, we have a system of
linear equations with
unknowns:
In the system of equations above, we notice that
is not always the same. Therefore, we consider the coefficient matrix
M based on
We claim
. Suppose to the contrary that
. Then, there exist
for
not all zero such that
Then,
is a nontrivial solution of the variational equation (
3). However,
and, for
with
,
By hypothesis (v),
. Thus,
which is a contradiction to the choice of the
Hence,
Thus, as a result of continuous dependence, for and sufficiently small, implying has an inverse where is the appropriately defined matrix from the system of equations. Therefore, for each , we are able to find using Cramer’s rule.
Note as
and so for
as
where
is the
matrix found by replacing the appropriate column of the matrix defining
M by
Now, let
and by construction of
,
Furthermore,
which is a solution of the variational equation (
3) along
. In addition, for
,
This completes the argument for □