Throughout this paper, we assume the following:

(H1)
;

(H2)
, and
;

(H3)
is continuous and does not vanish identically on any subinterval of
, and
;

(H4)
is continuous.

Lemma 2.1.

Suppose that (H1) and (H2) hold. Then

(i) the initial value problem

has a unique solution
and
;

(ii) the initial value problem

has a unique solution
and
.

Proof.

We only prove (i). (ii) can be treated in the same way.

Suppose that

and

is a solution of (2.1), that is,

Multiplying both sides of (2.3) by

, then

Since

and

, integrating (2.5) on

, we have

Moreover, integrating (2.6) on

,

, we have

Clearly,

, and (2.7) reduces to

By using Fubini's theorem, we have

which implies that
is a solution of integral equation (2.11).

Conversely, if
is a solution of (2.11) with
, by reversing the above argument we could deduce that the function
is a solution of (2.1) and satisfy
and
. Therefore, to prove that (2.1) has a unique solution,
, and
is equivalent to prove that (2.11) has a unique solution
.

To do this, we endow the following norm in

:

Let

be operator defined by

then,

is well defined. Set

Then, for any

,

Since
,
has a unique fixed point
by Banach contraction principle. That is, (2.11) has a unique solution
.

Remark 2.2.

Lemma 2.1 generalizes Theorem
of [1], where
.

Lemma 2.3.

Suppose that (H1) and (H2) hold. Then

(i)
is nondecreasing in
;

(ii)
is nonincreasing in
.

Proof.

We only prove (i). (ii) can be treated in the same way.

Suppose on the contrary that

is not nondecreasing in

. Then there exists

such that

This together with the equation

implies that

which is a contradiction!

Remark 2.4.

From Lemmas 2.1 and 2.3, there exist positive constants

,

,

, and

such that

we have that

and

. Then, there exist constants

and

, such that

In the following, we will show that

. Suppose on the contrary, if there exist

, such that

then,
, which is a contradiction!

The other inequality can be treated in the same manner.

Lemma 2.5.

Suppose that (H1), (H2), and (H3) hold. Then

Proof.

We only prove the first equality; the other can be treated in the same way. From Remark 2.4 and (H3), we have

Lemma

of [

2] together with the facts that

and (H3) implies that

Combining (2.27) and (2.28), we have

Lemma 2.6.

Suppose that (H1), (H2), and (H3) hold. Then the problem

Moreover,
on
.

Proof.

By Lemma 2.3 and (2.32), we have

This together with Remark 2.4 implies that the right side of (2.31) is well defined.

Now we check that the function

satisfies (2.30). In fact,

Equation (2.34) and Lemma 2.5 imply that

Since

for

, then

Let

with the norm

and let

be a cone in

defined by

Lemma 2.7.

Suppose that (H1)–(H3) hold and

is a positive solution of (2.30). Then

Furthermore, for any

, there exists corresponding

such that

Proof.

In fact, if

, then

and if

, then

Combining this and

, we have

Then Lemma 2.3 guarantees that
, and Lemma 2.7 guarantees that (2.43) holds.

Remark 2.8.

From Lemma 2.7 and Remark 2.4, we have

Now, for any

, we can define the operator

by

Lemma 2.9.

Let (H1)–(H4) hold. Then
is a completely continuous operator.

Proof.

From (H3) and (H4) and Lemma 2.6, it is easy to see that
, and
is continuous by the Lebesgue
s dominated convergence theorem.

Let
be any bounded set. Then (H3) and (H4) imply that
is a bounded set in
.

then this together with the similar proof of Lemma

of [

2] yields

From this fact, it is easy to verify that
is equicontinuous. Therefore, by the Arzela-Ascoli theorem,
is a completely continuous operator.