We will keep the notation in Section 1 and at the same time introduce the following new notation:

In this section, we always assume the following:
,
,
,
,
,
,
,
, and
,
.

be Green's function for the boundary value problems [22, 23].

The following estimates are easily obtained:

where
(
) are positive and increasing for
and
as
.

To show the main results of this section, the following lemmas are needed. The first lemma is Lemma 3.1 on page 15 in [24].

Lemma 2.1.

Let

,

, and

be real, continuous functions on

with

. If

Lemma 2.2.

Let

be a continuous function on

. If

,

, and

are nondecreasing and nonnegative on

and

Proof.

Replacing

in the two integrals of (2.6) by the expression on the right hand side in (2.6), changing the integral order of the resulting inequality and making use of the monotonicity of

,

and

, one gets

Apply Lemma 2.1 to get the conclusion.

Lemma 2.3.

Let

,

, and

. Then the problem

has a unique solution

, and

is in

and satisfies

where
.

One sees that
depends on
only and is bounded near zero.

Proof.

The existence and uniqueness of the solution comes from Theorem 5.3 on page 320 in [25].

As in [

22,

23], the solution

can be written as

By Lemma 2.1, one gets the desired inequality.

Now we show that

. As in the proofs of Lemmas 2.1 and 2.3 in [

15], one gets

. For

,

with

,

where

is a constant and

where
is a constant. Since
and
are slowly oscillating, the right-hand sides of the inequality above approaches zero as
. This means that
. The proof is complete.

Consider the following problem.

Problem 1.

*Find functions*
*and*
*such that*
It follows from (2.24) that

Let
, and let
. We have the following two additional problems for
and
, respectively.

Problem 2.

*Find functions*
*and*
*such that*
Problem 3.

*Find functions*
*and*
*such that*
Lemma 2.4.

Problems 1, 2, and 3 are equivalent to each other.

Proof.

The existence and uniqueness of the solution

of Problem 2 can be easily obtained from that of the solution

of Problem 1. Conversely, let (

) be the solution of Problem 2. We show that Problem 1 has a unique solution (

). The uniqueness comes from the uniqueness of (2.19)–(2.21). For the existence, let

Obviously,

and satisfies (2.22). Also

satisfies (2.21) because

. By (2.23) and (2.27), one sees that (2.20) is true. Finally, we show that

satisfies (2.19) and therefore, along with

, constitutes a solution of Problem 1. In fact,

Thus, we have shown the equivalence of Problems 1 and 2. Replacing (2.34) by the function

the equivalence of Problems 2 and 3 can be proved similarly. The proof is complete.

By Lemma 2.4, to solve Problem 1, we only need to solve Problem 3. By (2.30)–(2.32), we have the integral equation about

:

where
is determined by (2.37).

One can directly test that Problem 3 is equivalent to (2.37)-(2.38).

Note that for a given

, Lemma 2.3 shows that (2.30)–(2.32) (or equivalently, (2.37)) have a unique solution

. Thus, (2.38) does define an operator

. Therefore, we only need to show that the integral (2.38) has a unique solution

and

. That is,

has a fixed point in

. Let

Set

, where

. If

, then, by Lemma 2.3,

is in

, and so, by (2.38),

is in

with

Equation (2.37) gives the estimate

Choose

such that when

, one has

. It follows that

Choose

such that when

, one has

and therefore,
.

Let

,

. By (2.38),

. Note that the function

is the solution of the problem

So, by Lemma 2.3, one has

Choose
such that for
,
. Now, set
. Then
is a contraction from
into itself, and therefore, has a unique fixed point. Thus, we have shown.

Theorem 2.5.

Let functions
,
,
, and
be as above. Then, for small
, Problem 3 has a unique solution (
) in
with
and
.

Let
be the solutions of Problem 3 in
for the functions
,
,
, and
. Set
,
,
, and
. For the stability of the solution, we have the following.

Theorem 2.6.

For

, one has

where
depends on
,
,
,
,
,
,
, and
.

Proof.

Note that the function

is the solution of the problem

Using a formula similar to (2.37) and Lemma 2.2 for the function

, one gets

Applying Lemma 2.2 and (2.48), one gets the desired conclusion with

and

is majorant of

. One can specially assume that

The proof is complete.

Corollary 2.7.

Under the conditions in Theorem 2.6, the solution of Problem 3 is unique.