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.