Research Article | Open Access

Yanping Guo, Wenying Wei, Yuerong Chen, "Existence of Three Positive Solutions for -Point Discrete Boundary Value Problems with -Laplacian", *Discrete Dynamics in Nature and Society*, vol. 2009, Article ID 538431, 15 pages, 2009. https://doi.org/10.1155/2009/538431

# Existence of Three Positive Solutions for -Point Discrete Boundary Value Problems with -Laplacian

**Academic Editor:**Leonid Shaikhet

#### Abstract

We consider the multi-point discrete boundary value problem with one-dimensional -Laplacian operator , subject to the boundary conditions: , , where with and . Using a new fixed point theorem due to Avery and Peterson, we study the existence of at least three positive solutions to the above boundary value problem.

#### 1. Introduction

The second-order differential and difference boundary value problems arise in many branches of both applied and basic mathematics and have been extensively studied in literature. We refer the reader to some recent results for second-order nonlinear two-point [1â€“6] and multipoint [7â€“9] boundary value problems. The main tools used in the above works are fixed point theorems.

Recently, Feng and Ge in [9] considered the following multipoint BVPs:

The authors obtained sufficient conditions that guarantee the existence of at least three positive solutions by using fixed point theorems due to Avery-Peterson.

In this work, we study the existence of multiple positive solutions to the discrete boundary value problem for the one-dimensional -Laplacian:

where for , , for , and , , with .

In order to study the existence of at least three positive solutions to the above boundary value problem, we assume that , satisfy the following.

() satisfy .() is continuous.() for .We will depend on an application of a fixed point theorems due to Avery and Peterson, which deals with fixed points of a cone-preserving operator defined on an ordered Banach space to obtain our main results.

#### 2. Preliminaries

For the convenience of readers, we provide some background material from the theory of cones in Banach spaces. In this section, we also state Avery-Peterson's fixed point theorem.

*Definition 2.1. * Let be a real Banach space over . A nonempty convex closed set is said to be a cone of if it satisfies the following conditions:(i) for all and all , (ii) implies .Every cone induces an ordering in given by if and only if .

*Definition 2.2. *An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

*Definition 2.3. *The map is said to be a nonnegative continuous concave functional on a cone of a real Banach space provided that is continuous and
for all and . Similarly, we say that the map is a nonnegative continuous convex functional on a cone of a real Banach space provided that is continuous and
for all and .

Let and be nonnegative continuous convex functionals on a cone , let be a nonnegative continuous concave functional on a cone , and let be a nonnegative continuous functional on a cone . Then for positive real numbers and , we define the following convex sets:

and a closed set

To prove our results, we need the following fixed point theorem due to Avery and Peterson in [1].

Theorem 2.4. *Let be a cone in a real Banach space . Let and be nonnegative continuous convex functionals on , let be a nonnegative continuous concave functional on , and let be a nonnegative continuous functional on satisfying for , such that for some positive numbers and ,
**
for all . Suppose that
**
is completely continuous and there exist positive numbers and with such that** and for ;** for with ;** and for with .**Then has at least three fixed points , such that** for **, with ,*

#### 3. Related Lemmas

Let the Banach space be endowed with the ordering if for all , and the maximum norm

Then, we define the cone in by

Let be a natural number, such that

Let the nonnegative continuous concave functional , the nonnegative continuous convex functionals , , and the nonnegative continuous functional be defined on the cone by

for .

In order to prove our main results, we need the following lemma.

Lemma 3.1. *If , then
*

*Proof. *Suppose that the maximum of occurs at by the definition of the cone , we know , and then,
So, we have
The proof is complete.

By Lemma 3.1 and the definitions, the functionals defined above satisfy

for all . Therefore, condition (2.5) is satisfied.

Now, we show that . Here, we also suppose , and by the definitions of and the cone , we can distinguish two cases.

(i) , then we certainly have , and that is, .

(ii) , then , and that is, . So, we have .

Lemma 3.2. *Assume that hold. Then, for any ,
**
has a unique solution given by
**
or
**
where satisfies
*

*Proof. *For any suppose that is a solution of the BVPs (3.7) and (3.11). According to the property of the difference operator, it follows that
then
Let ,
and by
then
By
so that
Using the boundary condition (3.11), we can easily obtain
or
where satisfies (3.14).

On the other hand, it is easy to verify that if is the solution of (3.12) or (3.13), then is a solution of (3.10) and (3.11).

Lemma 3.3. *For any , , there exists a unique satisfying (3.14). Moreover, there is a unique , such that
*

*Proof. *Let
so that
By the continuity of , we know that there exists at least one
satisfying (3.14).

On the other hand,
Then is strictly increasing on .

So, there exists a unique satisfying (3.14).

Moreover, there exists a , such that

Lemma 3.4. *Assume that hold. If , then the unique solution of the BVPs (3.10) and (3.11) has the following properties:*(i)*, and ;*(ii)*there exists a unique , such that , which is given in Lemma 3.3.*

*Proof. *Suppose that is the solution of (3.10) and (3.11). Then we have the following.(i)By Lemma 3.2, it is easy to see that . Without loss of generality, we assume that . By , we know that . So we get that is, . Hence . So, from the concavity of , we know that .(ii)From
and by Lemma 3.3, we have , so that .

Also
and one arrives at . So, .

If there exist such that , then
which is a contradiction.

Lemma 3.5. *For any define the operator
**
Then is completely continuous.*

*Proof. *Using the continuity of and the definition of , it is easy to show that is continuous. Next, we prove that is completely continuous.

Suppose that the sequence is bounded, then there exists , such that , for any . By the continuity of , and are bounded, and we know that there exists , such that , for and . In view of the bounded sequence , there exists , such that . For the bounded sequence , there exists , such that . By repetition in this way, we have that there exists for , such that . Let ; by the definition of the norm on , there exists , such that .

Hence, is completely continuous.

#### 4. Existence of Triple Positive Solutions to (1.2)

We are now ready to apply Avery-Peterson's fixed point theorem to the operator to give sufficient conditions for the existence of at least three positive solutions to the BVPs (1.2).

Now for convenience we introduce the following notations. Let

Theorem 4.1. *Assume that conditions hold. Let and suppose that satisfies the following conditions:**, for ;** for ;**, for .**Then BVPs (1.2) have at least three positive solutions , and such that
*

*Proof. *The BVPs (1.2) have a solution if and only if solves the operator equation . Thus we set out to verify that the operator satisfies Avery-Peterson's fixed point theorem which will prove the existence of three fixed points of which satisfy the conclusion of the theorem. Now the proof is divided into some steps.

(1) We will show that implies that .

In fact, for , there is . With Lemma 3.1, there is , and then condition implies . On the other hand, for , there is , then is concave on , and and so
Thus, holds.

(2)We show that condition in Theorem 2.4 holds.

We take , for , and . It is easy to see that and . Hence . Thus for , there is , . Hence by condition of this theorem, one has for . By Lemma 3.4 and combining the conditions on and , we have
Therefore we have
Consequently, condition in Theorem 2.4 is satisfied.

(3)We now prove that in Theorem 2.4 holds.

With (3.7), we have
for with . Hence, condition