Positive solutions to the singular initialboundary value problems are obtained by applying the Schauder fixedpoint theorem, where on and may be singular at and . As an application, an example is given to demonstrate our result.
1. Introduction
Recently, in [1–4], Erbe, Kong, Jiang, Wang, and Weng considered the following singular functional differential equations:
where and the existence of positive solutions to (1.1) is obtained. When in (1.1), Agarwal and O'Regan in [5], Lin and Xu in [6] discussed the existence of positive solutions to (1.1) also. We notice that the nonlinearities in all the abovementioned references depend on .
The more difficult case is that the term depends on for secondorder functional differential equations with delay. When has no singularity at and , there are many results on the following (1.2) (see [7–9] and references therein). Up to now, to our knowledge, there are fewer results on (1.2) when the term is allowed to possess singularity for the term at and , which is of more actual significance.
In this paper, motivated by above results, we consider the secondorder initialboundary value problems:
where . By LeraySchauder fixedpoint theorem, the existence of positive solutions to (1.2) is obtained when is singular at and .
For and , let and . Then, and are Banach spaces. Let and . Obviously, and are cones in and respectively. Now, we give a new definition.
Deffinition.
is said to be singular at for when satisfies for and is said to be singular at for when satisfies for .
And one defines some functions which one has to use in this paper.
Let
where is a Green's function. It is clear that for and on
We now introduce the definition of a solution to IBVP(1.2).
Deffinition.
A function is said to be a solution to IBVP(1.2) if it satisfies the following conditions:
(1) is continuous and nonnegative on ;
(2);
(3) and exist on ;
(4) is Lebesgue integrable on ;
(5) for .
Furthermore, a solution is said to be positive if on .
Let be a solution to IBVP(1.2). Then, it can be represented as
It is clear that
for all solutions, , to IBVP(1.2), where . For , let on throughout this paper. Obviously, and for all .
Throughout this paper, we assume the following hypotheses hold.
(H_{1}) is continuous on .
(H_{2}) There exists , such that
Lemma 1.3.
Assume that (H _{1} )(H _{2} ) hold, then there exists a , such that
for all solutions, , to (1.2).
Proof.
Suppose that the claim is false. (1.5) guarantees that there exists a sequence of solutions to IBVP(1.2) such that
Without loss of generality, we may assume that
From and (1.5), it follows that
which contradicts the assumption that and hence the claim is true provided is suitably small.
Remark 1.4.
The following inequality
holds provided that is sufficiently small, where is in Lemma 1.3.
There exist a nonnegative continuous function defined on (0,1) and two nonnegative continuous functions defined on, respectively, , such that
where and ) satisfy
Furthermore, is nonincreasing and is nondecreasing, that is,
Lemma 1.5 (see [7]).
Let be the Banach space and let X be any nonempty, convex, closed, and bounded subset of . If is a continuous mapping of into itself and is relatively compact, then the mapping has at least one fixed point (i.e., there exists an with ) .
Using Lemma 1.5, we present the existence of at least one positive solution to (1.2) when is singular at and (notice the new Definition 1.1). To some extent, our paper complements and generalizes these in [1–6, 8–10].
2. Main Results
Theorem 2.1.
Assume that (H _{1} )–(H _{3} ) hold. Then, the IBVP( 1.2 ) has at least one positive solution.
Proof.
Since , we can choose an such that
where the positive number satisfies
Let
For each , we define by
It is obvious that satisfies the hypotheses and .
We now consider the modified initialboundary value problem:
We claim that for all solutions, , to IBVP(2.5),
Suppose that the claim is false. Then there exists such that
Since on , there are the following three cases.
Case 1.
for all .
The solution of IBVP(2.5) can be represented as (notice Remark 1.4)
which contradicts (2.7).
Case 2.
There exists a such that and .
In this case, we have
which contradicts (2.7).
Case 3.
There exists a such that and .
From (1.5), we get
which contradicts (2.7).
So we have
To prove the existence of positive solutions to IBVP(2.5), we seek to transform (2.5) into an integral equation via the use of Green's function and then find a positive solution by using Lemma 1.5.
Define a nonempty convex and closed subset of by
Then, we define an operator by
From and the definition of , we have, for every
Together with the definition of , we get .
Also,
is continuous in (0,1), and
From and (2.15), we can get
which implies that is integrable on .
Now, we claim that is equicontinuous on . We will prove the claim. For any , we have
Since is continuous on and , then for any , there is a such that
By (2.6), we have, for
where is a constant number.
Put , then for
Set . Then for
Since on , the above inequality holds for .
Thus, is a relative compact subset of . That is, is a compact operator.
We are now going to prove that the mapping is continuous on .
Let be arbitrarily chosen and let converge to uniformly on as . Now, we claim that converge to uniformly as . From the definition of , we get
Thus,
that is, the claim is true.
Since is continuous with respect to for , we have
for each fixed . From the definition of and , we know that
and hence
where is a Lebesgue integrable function defined on because of . Consequently, we apply the dominated convergence theorem to get
which shows that the mapping is continuous on .
Then from Lemma 1.5, we get that there exists at least one positive solution, , to IBVP(2.5) in . The solution can be represented by (1.4), where is replaced with . So, (2.6) holds. Furthermore, from the definition of , we can get
Thus, the solution of IBVP(2.5) is also the one of (1.2). The proof is complete.
3. Application
Example 3.1.
Consider the singular IBVP(3.1):
where .
4. Conclusion
Equation (3.1) has at least one positive solution.
Now, we will check that hold in (3.1).
In IBVP(3.1), . It is clear that is continuous and singular at and . For we choose
when ; by simple computation, we can get
It is obvious that is nonincreasing and is nondecreasing.
Now, we check . For any , (notice the definition of ), we have
We define
Now, we will prove that there exists such that is decreasing on .
Obviously,
Put , then
From the continuity of , we can find such that on . Then, on . That is, is decreasing on .
Furthermore, we have
Thus,
which implies that holds.
So, from Theorem 2.1, IBVP(3.1) has at least one positive solution.
Acknowledgments
The research was supported by NNSF of China (10571111) and the fund of Shandong Education Committee (J07WH08).
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