Throughout this paper, we let

. Then

is a Banach space, where

In this section, we always assume that
.

Lemma 3.1.

System (1.1) is equivalent to the following system of integral equations:

Proof.

It is easy to see that if

satisfies (3.2) then it also satisfies (3.2). So, assume that

is a solution to (1.1). Integrating both sides of the first equation of (1.1) of order

with respect to

gives us

for

,

. It follows that

for

,

. This, combined with the boundary conditions in (1.1), yields

Similarly, one can obtain

. Then it follows from (3.8) and the boundary condition

that

Therefore, for

,

This completes the proof.

The following two results give some properties of the Green functions
.

Lemma 3.2.

For
is continuous on
and
for
.

Proof.

Obviously,

is continuous on

. It remains to show that

for

. It is easy to see that

for

. We only need to show that

for

. For

, let

Note that
and
for
. It follows that
and hence
for
.

Therefore,
for
and the proof is complete.

Proof.

- (i)
Obviously,

for

. Now, for

, we have

where

is the function defined by (3.11). It follows that

for

. In summary, we have proved (i).

- (ii)
Again, one can easily see that

for

. When

, we have in this case that

which implies that
for
. To summarize, we have proved (ii) and this completes the proof.

Now, we are ready to present the main results.

Theorem 3.4.

Suppose that there exist functions

,

, 2,

, such that

then (1.1) has a unique positive solution.

Proof.

It is easy to see that

is a complete metric space. Define an operator

on

by

where

and

Because of the continuity of

and

, it follows easily from Lemma 3.2 that

maps

into itself. To finish the proof, we only need to show that

is a contraction. Indeed, for

, by (3.16) we have

This, combined with Lemma 3.3 and (3.17) and (3.18), immediately implies that
is a contraction. Therefore, the proof is complete with the help of Lemmas 3.1 and 2.5.

The following result can be proved in the same spirit as that for Theorem 3.4.

Theorem 3.5.

For

, suppose that there exist nonnegative function

and nonnegative constants

such that

and

for

,

. If

then (1.1) has a unique positive solution.

Theorem 3.6.

For

, suppose that there exist nonnegative real-valued functions

such that

for almost every

and all

. If

then (1.1) has at least one positive solution.

Proof.

Let
and
be defined by (3.19) and (3.20), respectively. We first show that
is completely continuous through the following three steps.

Step 1.

Show that

is continuous. Let

be a sequence in

such that

. Then

is bounded in

. Since

is continuous, it is uniformly continuous on any compact set. In particular, for any

, there exists a positive integer

such that

for

and

. Then, for

, we have

for

and

. Therefore,

which implies that
is continuous.

Step 2.

Show that

maps bounded sets of

into bounded sets. Let

be a bounded subset of

. Then

is bounded. Since

is continuous, there exists an

such that

It follows that, for

and

,

Immediately, we can easily see that
is a bounded subset of
.

Step 3.

Show that

maps bounded sets of

into equicontinuous sets. Let

be a bounded subset of

. Similarly as in Step 2, there exists

such that

Now the equicontituity of
on
follows easily from the fact that
is continuous and hence uniformly continuous on
.

Now we have shown that

is completely continuous. To apply Lemma 2.6, let

Fix

and define

We claim that there is no

such that

for some

. Otherwise, assume that there exist

and

such that

. Then

If

, then

Similarly, we can have
if
. To summarize,
, a contradiction to
. This proves the claim. Applying Lemma 2.6, we know that
has a fixed point in
, which is a positive solution to (1.1) by Lemma 3.1. Therefore, the proof is complete.

As a consequence of Theorem 3.6, we have the following.

Corollary 3.7.

If all
,
, are bounded, then (1.1) has at least one positive solution.

To state the last result of this section, we introduce

Theorem 3.8.

Suppose that there exist
and positive constants
with
such that

(i)
for

and

(ii)
, for
,

where
. Then (1.1) has at least a positive solution.

Proof.

Let

be defined by (3.19) and

. Obviously,

is a cone in

. From the proof of Theorem 3.6, we know that the operator

defined by (3.20) is completely continuous on

. For any

, it follows from Lemma 3.3 and condition (ii) that

On the other hand, for any

, it follows from Lemma 3.3 and condition (i) that, for

,

if

, whereas

if

. In summary,

Therefore, we have verified condition (ii) of Lemma 2.7. It follows that
has a fixed point in
, which is a positive solution to (1.1). This completes the proof.