Positive Solutions of Singular Initial-Boundary Value Problems to Second-Order Functional Differential Equations

Positive solutions to the singular initial-boundary value problems are obtained by applying the Schauder fixed-point theorem, where on and may be singular at and . As an application, an example is given to demonstrate our result.

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 above-mentioned references depend on .

The more difficult case is that the term depends on for second-order 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 second-order initial-boundary value problems:

where . By Leray-Schauder fixed-point 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.

(H₁) is continuous on .

(H₂) 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].

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 initial-boundary 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.

Example 3.1.

Consider the singular IBVP(3.1):

where .

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.

The research was supported by NNSF of China (10571111) and the fund of Shandong Education Committee (J07WH08).

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