We note that
is a solution of (1.1) if and only if
Let
be endowed with the norm
and define the cone of
by
Clearly,
is a Banach space with the norm
. For each
, extend
to
with
for
.
Define
as
We seek a fixed point,
, of
in the cone
. Define
Then
denotes a positive solution of BVP (1.1).
It follows from (2.3) that the following lemma holds.
Lemma 2.1.
Let
be defined by (2.3). If
, then
-
(i)
-
(ii)
is completely continuous.
The proof of Lemma 2.1 can be found in [15].
We need to define further subsets of
with respect to the delay
. Set
Throughout this paper, we assume
and 
Lemma 2.2.
Suppose that (H1)–(H5) hold. Then there exists a
such that the operator
has a fixed point
at
, where
is the zero element of the Banach space
.
Proof.
Set
We know that
Let
where
From above, we have
Let
and
Then
By the Lebesgue dominated convergence theorem [16] together with (H3), it follows that
decreases to a fixed point
of the operator
The proof is complete.
Lemma 2.3.
Suppose that (H1)–(H6) hold and that
for some
. Then there exists a constant
such that for all
and all possible fixed points
of
at
, one has 
Proof.
Set
We need to prove that there exists a constant
such that
for all
If the number of elements of
is finite, then the result is obvious. If not, without loss of generality, we assume that there exists a sequence
such that
, where
is the fixed point of the operator
defined by (2.3) at 
Then
We choose
such that
such that
In view of (H6) there exists an
sufficiently large such that
For
we have
which is a contradiction. The proof is complete.
Lemma 2.4.
Suppose that (H1)–(H5) hold and that the operator
has a positive fixed point
in
at
. Then for every
the operator
has a fixed point
at
, and 
Proof.
Let
be the fixed point of the operator
at
. Then
where
Set
and
Then
where
is also defined by (2.6), which implies that
decreases to a fixed point
of the operator
, and
The proof is complete.
Lemma 2.5.
Suppose that (H1)–(H6) hold. Let
have at least one fixed point at
in
. Then
is bounded above.
Proof.
Suppose to the contrary that there exists a fixed point sequence
of
at
such that
Then we need to consider two cases:
-
(i)
there exists a constant
such that 
-
(ii)
there exists a subsequence
such that
which is impossible by Lemma 2.3.
Only (i) is considered. We can choose
such that
, and further
. For
, we have
Now we consider (2.18). Assume that the case (i) holds. Then
leads to
which is a contradiction. The proof is complete.
Lemma 2.6.
Let
Then
where
is defined just as in Lemma 2.5.
Proof.
In view of Lemma 2.4, it follows that
We only need to prove
In fact, by the definition of
, we may choose a distinct nondecreasing sequence
such that
Let
be the positive fixed point of
at
By Lemma 2.3,
is uniformly bounded, so it has a subsequence denoted by
converging to
Note that
Taking the limitation
to both sides of (2.21), and using the Lebesgue dominated convergence theorem [16], we have
which shows that
has a positive fixed point
at
The proof is complete.
Theorem 2.7.
Suppose that (H1)–(H6) hold. Then there exists a
such that (1.1) has at least two, one, and no positive solutions for
and
respectively.
Proof.
Assume that (H1)–(H5) hold. Then there exists a
such that
has a fixed point
at
In view of Lemma 2.4,
also has a fixed point
and
Note that
is continuous on
. For
there exists a
such that
Hence,
From above, we have
Set
for
and
. We have
for
By Lemma 2.1,
In view of (H6), we can choose
such that
Set
Similar to Lemma 2.3, it is easy to obtain that
In view of Lemma 2.1,
By the additivity of fixed point index,
So,
has at least two fixed points in
. The proof is complete.