Triple Positive Solutions for a Type of Second-Order Singular Boundary Problems

This paper deals with the existence of triple positive solutions for a type of second-order singular boundary problems with general differential operators. By using the Leggett-Williams fixed point theorem, we establish an existence criterion for at least three positive solutions with suitable growth conditions imposed on the nonlinear term.

In this paper, we study the existence of triple positive solutions for the following second-order singular boundary value problems with general differential operators:

where , , and with

It is easy to see that and may be singular at and/or

When or and , the two kinds of singular boundary value problems have been discussed extensively in the literature; see [1–10] and the references therein. Hence, the problem that we consider is more general and is different from those in previous work.

Furthermore, we will see in the later that the presence of brings us three main difficulties:

(1)the Green's function cannot be explicitly expressed;

(2)the equivalence between BVP (1.1) and its associated integral equation has to be proved;

(3)the compactness of associated integral operator has to be verified.

We will overcome the above mentioned difficulties in Section 2. Also, although the Leggett-William fixed point theorem is used extensively in the study of triple positive differential equations, the method has not been used to study this type of second-order singular boundary value problem with general differential operators. We are concerned with solving these problems in this paper.

To state our main tool used in this paper, we give some definitions and notations.

Let be a real Banach space with a cone . A map is said to be a nonnegative continuous concave functional on if is a continuous and

for all and . Let be two numbers such that and a nonnegative continuous concave functional on . We define the following convex sets:

Theorem 1.1 (Leggett-Williams fixed point theorem).

Let be completely continuous, and let be a nonnegative continuous concave functional on such that for all . Suppose that there exist such that

(i) and , for ;

(ii), for ;

(iii), for with .

Then has at least three fixed points in satisfying and .

Remark 1.2.

We note the existence of triple positive solutions of other kind of boundary value problems; see He and Ge [11], Zhao et al. [12], Zhang and Liu [13], Graef et al. [14], and the references therein.

The rest of the paper is organized as follows. In Section 2, we overcome the above-mentioned difficulties in this work. The main results are formulated and proved in Section 3. Finally, an example is presented to demonstrate the application of the main theorems in Section 4.

Throughout this paper, we assume the following:

(H1);

(H2), and ;

(H3) is continuous and does not vanish identically on any subinterval of , and ;

(H4) is continuous.

Lemma 2.1.

Suppose that (H1) and (H2) hold. Then

(i)the initial value problem

has a unique solution and ;

(ii)the initial value problem

has a unique solution and .

Proof.

We only prove (i). (ii) can be treated in the same way.

Suppose that and is a solution of (2.1), that is,

Let

Multiplying both sides of (2.3) by , then

Since and , integrating (2.5) on , we have

Moreover, integrating (2.6) on , , we have

Let

Clearly, , and (2.7) reduces to

By using Fubini's theorem, we have

Therefore,

which implies that is a solution of integral equation (2.11).

Conversely, if is a solution of (2.11) with , by reversing the above argument we could deduce that the function is a solution of (2.1) and satisfy and . Therefore, to prove that (2.1) has a unique solution, , and is equivalent to prove that (2.11) has a unique solution .

To do this, we endow the following norm in :

Let be operator defined by

Since

then, is well defined. Set

Then, for any ,

and subsequently,

Thus,

Since , has a unique fixed point by Banach contraction principle. That is, (2.11) has a unique solution .

Remark 2.2.

Lemma 2.1 generalizes Theorem of [1], where .

Lemma 2.3.

Suppose that (H1) and (H2) hold. Then

(i) is nondecreasing in ;

(ii) is nonincreasing in .

Proof.

We only prove (i). (ii) can be treated in the same way.

Suppose on the contrary that is not nondecreasing in . Then there exists such that

This together with the equation implies that

which is a contradiction!

Remark 2.4.

From Lemmas 2.1 and 2.3, there exist positive constants , , , and such that

In fact, since

we have that and . Then, there exist constants and , such that

that is

In the following, we will show that . Suppose on the contrary, if there exist , such that

then, , which is a contradiction!

The other inequality can be treated in the same manner.

Lemma 2.5.

Suppose that (H1), (H2), and (H3) hold. Then

Proof.

We only prove the first equality; the other can be treated in the same way. From Remark 2.4 and (H3), we have

Lemma of [2] together with the facts that and (H3) implies that

Combining (2.27) and (2.28), we have

Lemma 2.6.

Suppose that (H1), (H2), and (H3) hold. Then the problem

has a unique solution

where

Moreover, on .

Proof.

By Lemma 2.3 and (2.32), we have

This together with Remark 2.4 implies that the right side of (2.31) is well defined.

Now we check that the function

satisfies (2.30). In fact,

Therefore,

Equation (2.34) and Lemma 2.5 imply that

Since for , then

Let with the norm

and let be a cone in defined by

Lemma 2.7.

Suppose that (H1)–(H3) hold and is a positive solution of (2.30). Then

where

Furthermore, for any , there exists corresponding such that

Proof.

In fact, if , then

and if , then

Combining this and , we have

Take

Then Lemma 2.3 guarantees that , and Lemma 2.7 guarantees that (2.43) holds.

Remark 2.8.

From Lemma 2.7 and Remark 2.4, we have

Now, for any , we can define the operator by

Lemma 2.9.

Let (H1)–(H4) hold. Then is a completely continuous operator.

Proof.

From (H3) and (H4) and Lemma 2.6, it is easy to see that , and is continuous by the Lebesgues dominated convergence theorem.

Let be any bounded set. Then (H3) and (H4) imply that is a bounded set in .

Since

then this together with the similar proof of Lemma of [2] yields

From this fact, it is easy to verify that is equicontinuous. Therefore, by the Arzela-Ascoli theorem, is a completely continuous operator.

Let be nonnegative continuous concave functional defined by

We notice that, for each , , and also that by Lemma 2.6, is a solution of (1.1) if and only if is a fixed point of the operator .

For convenience we introduce the following notations. Let

Theorem 3.1.

Assume that (H1)–(H4) hold. Let , and suppose that satisfies the following conditions:

(S1), for ;

(S2), for ;

(S3), for .

Then the boundary value problem (1.1) has at least three positive solutions in satisfying , and .

Proof.

From Lemma 2.9, is a completely continuous operator. If , then , and assumption (S3) implies that . Therefore

Hence, . In the same way, if , then . Therefore, condition (ii) of Leggett-williams fixed-point theorem holds.

To check condition (i) of Leggett-Williams fixed-point theorem, choose . It is easy to see that and . so,

Hence, if , then . From assumption (S2) and Remark 2.8, we have

Finally, we assert that if and , then . To see this, suppose that and , then

To sum up, all the conditions of Leggett-williams fixed-point theorem are satisfied. Therefore, has at least three fixed points, that is, problem (1.1) has at least three positive solutions in satisfying and .

Theorem 3.2.

Assume that (H1)–(H4) hold. Let , and suppose that satisfies the following conditions:

(A1), for ;

(A2), for .

Then the boundary value problem (1.1) has at least positive solutions.

Proof.

When , it follows from condition (A1) that , which means that has at least one fixed point by the Schauder fixed-point Theorem. When , it is clear that Theorem 3.1 holds (with ). Then we can obtain at least three positive solutions , and satisfying and with . Following this way, we finish the proof by the induction method.

Consider the following boundary value problem:

where

Then, by computation, we have

Furthermore, for ,

In fact, let , then , and . It is easy to compute that

Then, , that is

The other inequalities in (4.4) can be proved by the same method.

Thus, we can choose that , and . By computation, we have

Let , and . Then, we can compute

Consequently,

Therefore, all the conditions of Theorem 3.1 are satisfied, then problem (4.1) has at least three positive solutions , and satisfying

The first author was partially supported by NNSF of China (10901075).

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