Research Article | Open Access

# Positive Solutions for (*k*, *n* − *k*) Conjugate Multipoint Boundary Value Problems in Banach Spaces

**Academic Editor:**Mohamed-Aziz Taoudi

#### Abstract

By means of the fixed point index theory of strict-set contraction operator, we study the existence of positive solutions for the multipoint singular boundary value problem , , , , , , , in a real Banach space , where is the zero element of As an application, we give two examples to demonstrate our results.

#### 1. Introduction

The theory of ordinary differential equations in Banach spaces has become a new important branch (see, e.g., [1–13] and the references cited therein). In 1988, Guo and Lakshmikantham [4] discussed multiple solutions for two-point boundary value problems of second-order ordinary differential equations in Banach spaces. In [7], Guo obtained the existence of positive solutions for a boundary value problem of *n*th-order nonlinear impulsive integrodifferential equations in a Banach space by means of fixed point index theory and fixed point theory of completely continuous operators, respectively. Liu et al. in [6] obtained the existence of unbounded nonnegative solutions of a boundary value problem for *n*th-order impulsive integrodifferential equations on an infinite interval in Banach spaces by means of the M*ddoto*ch fixed point theory in a Banach space. Zhang et al. in [9] dealt with the existence, nonexistence, and multiplicity of positive solutions for a class of nonlinear three-point boundary value problems of *n*th-order differential equations in Banach spaces. Zhao and Chen in [8, 12] investigated the existence of at least triple positive solutions for nonlinear boundary value problem by upper and low solution methods.

In this paper, the author considers the existence of positive solutions of the following higher-order conjugate multipoint boundary value problems (BVPs): in a real Banach space , where is the zero element of . is continuous and allowed to be singular at and .

In scalar space, because of the widely applied background in mechanics and engineering, the nonlinear higher-order boundary value problems have received much attention (see Chyan and Henderson [Appl. Math. Letters 15 (2002) 767–774]). In [14], Eloe and Ahmad had solved successfully the existence of positive solution to the following *n*th-order boundary value problems:
Recently, the existence of solutions and positive solutions of nonlinear focal boundary value problem
and its special cases has been studied by many authors (see, e.g., [15–23]).

By using the Krasnoselskii fixed point theorem, Eloe and Henderson in [15], Agarwal and O'Regan in [16], and Kong and Wang in [20] have established the existence of solutions for following the conjugate boundary value problem:

Very recently, by using the fixed point theory in a cone for strict-set contraction operators, Jiang and Zhang in [11] have discussed the existence of positive solutions for the above boundary value problem (1.4) in a Banach space, where the nonlinear term is continuous but not allowed to have singularity at , where .

The organization of this paper is as follows. We shall introduce some lemmas and notations in the rest of this section. The preliminary lemmas are in Section 2. The main results are given in Section 3. Finally, two examples are presented to demonstrate our main results in Section 4.

Let the real Banach space with norm be partially ordered by a cone of , that is, if and only if , and denotes the dual cone of , that is, . A cone is called a solid cone if the set of interior points is not empty.

The closed balls in spaces and are denoted by and , respectively.

The basic space used in this paper is . For any , evidently, is a Banach space with norm , and for is a cone of the Banach space . A function is called a positive solution of the boundary value problem (1.1) if it satisfies (1.1) and .

Let be continuous, and the abstract generalized integral can be similarly defined as in the scalar spaces and . If exist, then we say that the abstract integral is convergent, otherwise the abstract integral is divergent.

At the end of this section, we state some definitions and lemmas which will be used in Sections 2 and 3 (for details, see [1–3]).

*Definition 1.1 (Kuratovski noncompactness measure). *Let be a real Banach space, and is a bounded set in . We denote , all the diameters of .

In the following, denotes the Kuratowski measure of noncompactness in and .

*Definition 1.2 (strict-set contraction operator). * Let be real Banach spaces, and . is a continuous and bounded operator. If there exists a constant , such that , then is called a -set contraction operator. When is called a strict-set contraction operator.

Lemma 1.3. *If is bounded and equicontinuous, then is continuous on and , where .*

Lemma 1.4. *Let be a cone in a real Banach space and let be a nonempty bounded open convex subset of . Suppose that is a strict-set contraction operator and , where denotes the closure of in . Then the fixed-point index .*

Lemma 1.5. *Let be a cone in a real Banach space and let be a bounded open subset of , and suppose that is a strict-set contraction operator. *(i)*If , and . Then .*(ii)*If there exists , such that . Then .*

#### 2. The Preliminary Lemmas

To prove the main results, we need the following lemmas.

Lemma 2.1 (see [20]). * Let be the Green function for the conjugate BVP (1.4). Then
**
Obviously, is continuous on and has the following properties. **(G _{1}) There exist nonnegative functions such that
*

*where*

*(G*

_{2}) For any satisfies*where*

Setting It is obvious that , and by the properties of the Euler integral, we have

In order to abbreviate our discussion, we give the following assumptions.(C_{0}).
() and , where is continuous and bounded and is continuous and satisfies .() For any and , is uniformly continuous on .() There exists constant such that for any and the bounded set
where

Lemma 2.2. * Let and be satisfied, then the problem
**
has a unique solution
**
where
*

*Proof. * According to the definitions of generalized integral in abstract space, the proof of this lemma is similar to the proof in scalar spaces, so we omit it.

For , we define an operator by

Lemma 2.3. * Suppose that – hold. Then is a strict-set contraction operator.*

* Proof. * For , it follows from that is well defined and bounded operator. If is a bounded subset of , then is bounded.

Next we prove that is continuous on . Let , and . Hence is a bounded subset of . Thus, there exists such that and .

According to continuity of , for all , there exists such that
for .

Then,
Therefore, for all , for any and , we get
This implies is continuous on . By the properties of continuity of , it is easy to see that is equicontinuous on .

For any , it is easy to get that functions are uniformly bounded. By Lemma 1.3, we get
where is fixed}.

Write
By , for any , we have
where .

It follows from and that the Hausdorff metric ,. Thus

We next shall estimate . For any , by , then
where .

On the other hand, using a similar method as in the proof of Lemma 2 in [11] we can get that
Therefore, it follows from (2.19), (2.25) that
Thus, we have
Combining with (2.24), we get
Therefore, we have
Notice that by (2.9) we claim that is a strict-set contraction. The proof is complete.

Further, we construct a cone by where , and where is defined in . It is easy to see that is a closed convex cone of and .

Lemma 2.4. * Suppose that ()–() hold. Then, .*

* Proof. * From , (2.6), (2.15), and (2.30), for any , we obtain
Therefore, , that is, .

#### 3. The Main Results

Let

Theorem 3.1. *Let ()–() hold. In addition, assume that the following conditions are satisfied: **(C _{4}) , where
*

*(C*

_{5}) is a solid cone, and there exist and such that , and*for all .*

*Then problem (1.1) has at least two positive solutions , and , for .*

* Proof. *We first show that there exists , such that
If (3.4) is not true, then there exist the sequences satisfying , and . Thus , which is contradiction with .

Let
Then
From and (3.6), there exist two positive constants with
such that
Therefore, for any , we have
where .

Choose
In the following, let
It is to see from (3.7) that is nonempty for , which implies . Obviously, are nonempty, convex, open sets and . So

For any , form (3.8) and (3.9), we have
which implies
From (2.15) and (3.7), we get

For any , it follows from (3.16) that . According to and , we can obtain
which implies
Combining (3.16)–(3.19) with Lemma 1.4, we have . Furthermore, using the fixed point index theory, we obtain successively

Then has at least two fixed points and which satisfy and . Then Theorem 3.1 is proved.

Theorem 3.2. * Assume that ()–(), hold. Suppose further that **() is a cone of the real Banach space , and there exist and such that , for for all .** Then problem (1.1) has at least one positive solutions with , where is defined in (3.3).*

* Proof . *As in the proof of Theorem 3.1, we need only to show that has one positive fixed point with .

Choose satisfying and let
Obviously, is a bounded closed convex set in . , for . Let . As the proof of (3.19), we have , where is given by (2.15). Thus, it follows from Lemmas 2.3 and 1.4 that has a fixed point . The proof is complete.

Theorem 3.3. *Assume that ()–() and the following conditions hold.** There exists such that , where .** There exists such that for any , and
**where , and
** There exists such that for any , and
**
where , and is defined in (3.23). Then problem (1.1) has at least two positive solutions.*

* Proof. * If holds, then there exists satisfying such that . For any , it follows from and (2.15) that
which implies , and . From Lemma 1.5 (i), we have

According to , there exist and , for any , such that
Let . We need only to show that , for any , and with . If it is false, then there exists and , such that
which implies
Hence, for any , we have

It is easy to see that . In fact, if , since , then we have and consequently , which contradicts with . By (3.29) and (3.30), we get , which is a contradiction. It follows from Lemma 1.5 (ii) that

If holds, then there exist and with , such that
Let . As in the proof of (3.31), for any , we get
Notice that . Thus, it follows from (3.26), (3.31), and (3.33) that

Then has at least two fixed points and which satisfy and . Then Theorem 3.3 is proved.

Similarly to the proofs of Theorem 3.3, we can easily get the following corollaries.

Corollary 3.4. *Assume that ()–(), and hold. Then problem (1.1) has at least one positive solution.*

Corollary 3.5. *Assume that ()–(), and hold. Then problem (1.1) has at least one positive solution.*

#### 4. Examples

Now we present two examples to illustrate our main results.

*Example 4.1. *Consider the boundary value problems in (*N*-dimensional Euclidean space and )
where , and
where is as defined in (3.10). Then the problem (4.1) can be regarded as a BVP of the form (1.1) in . In this situation, and , in which