Abstract and Applied Analysis

Volume 2011, Article ID 543035, 13 pages

http://dx.doi.org/10.1155/2011/543035

## Uniqueness of Positive Solutions for a Class of Fourth-Order Boundary Value Problems

Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain

Received 31 December 2010; Accepted 23 March 2011

Academic Editor: Yuri V. Rogovchenko

Copyright © 2011 J. Caballero et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The purpose of this paper is to investigate the existence and uniqueness of positive solutions for the following fourth-order boundary value problem: , , . Moreover, under certain assumptions, we will prove that the above boundary value problem has a unique symmetric positive solution. Finally, we present some examples and we compare our results with the ones obtained in recent papers. Our analysis relies on a fixed point theorem in partially ordered metric spaces.

#### 1. Introduction

The purpose of this paper is to consider the existence and uniqueness of positive solutions for the following fourth-order two-point boundary value problem: which describes the bending of an elastic beam clamped at both endpoints.

There have been extensive studies on fourth-order boundary value problems with diverse boundary conditions. Some of the main tools of nonlinear analysis devoted to the study of this type of problems are, among others, lower and upper solutions [1–4], monotone iterative technique [5–7], Krasnoselskii fixed point theorem [8], fixed point index [9–11], Leray-Schauder degree [12, 13], and bifurcation theory [14–16].

#### 2. Background

In this section, we present some basic facts which are necessary for our results.

In our study, we will use a fixed point theorem in partially ordered metric spaces which appears in [17].

Let denote the class of those functions satisfying the condition

Now, we recall the above mentioned fixed point theorem.

Theorem 2.1 (see 1, Theorem 2.1). *Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be a nondecreasing mapping such that there exists an element with . Suppose that there exists such that
**
Assume that either is continuous or is such that
**
Besides, suppose that
**
Then has a unique fixed point.*

In our considerations, we will work with a subset of the classical Banach space . This space will be considered with the standard metric This space can be equipped with a partial order given by In [18], it is proved that with the above mentioned metric satisfies condition (2.3) of Theorem 2.1. Moreover, for , as the function , satisfies condition (2.4).

On the other hand, the boundary value problem (1.1) can be rewritten as the integral equation (see, e.g., [19]) where is the Green's function given by Note that satisfies the following properties: (i) is a continuous function on , (ii), for , (iii), for .

#### 3. Main Results

Our starting point in this section is to present the class of functions which we use later. By we denote the class of functions satisfying the following conditions: (i) is nondecreasing, (ii)for any , , (iii).

Examples of functions in are with , and . In the sequel, we formulate our main result.

Theorem 3.1. *Consider problem (1.1) assuming the following hypotheses: *(a)* is continuous, *(b)* is nondecreasing with respect to the second variable, for each , *(c)* suppose that there exists , such that, for with , **
Then, problem (1.1) has a unique nonnegative solution.*

*Proof. *Consider the cone
Obviously, with is a complete metric space satisfying condition (2.3) and condition (2.4) of Theorem 2.1.

Consider the operator defined by
where is the Green's function defined in Section 2.

It is clear that applies the cone into itself since and are nonnegative continuous functions.

Now, we check that assumptions in Theorems 2.1 are satisfied.

Firstly, the operator is nondecreasing.

Indeed, since is nondecreasing with respect to the second variable, for , and , we have
On the other hand, a straightforward calculation gives us
Taking into account this fact and our hypotheses, for and , we can obtain the following estimate:
This gives us, for and ,
where .

Obviously, the last inequality is satisfied for .

Therefore, the contractive condition appearing in Theorem 2.1 is satisfied for . Besides, as and are nonnegative functions,

Finally, Theorem 2.1 tells us that has a unique fixed point in , and this means that problem (1.1) has a unique nonnegative solution.

This finishes the proof.

Now, we present a sufficient condition for the existence and uniqueness of positive solutions for our problem (1.1) (positive solution means , for ). The proof of the following theorem is similar to the proof of Theorem 3.6 of [8]. We present a proof for completeness.

Theorem 3.2. *Under assumptions of Theorem 3.1 and suppose that for certain , problem (1.1) has a unique positive solution.*

*Proof. *
Consider the nonnegative solution given by Theorem 3.1 of problem (1.1).

Notice that this solution satisfies

Now, we will prove that is a positive solution.

In contrary case, suppose that there exists such that and, consequently,
Since , is nondecreasing with respect to the second variable and , we have
and this gives us
This fact and the nonnegative character of and imply
As a·e (*s*), because is given by a polynomial, we obtain
On the other hand, as for certain and , we have that .

The continuity of gives us the existence of a set with and , where is the Lebesgue measure, satisfying that for any . This contradicts (3.14).

Therefore, for .

This finishes the proof.

Now, we present an example which illustrates our results.

*Example 3.3. *Consider the nonlinear fourth-order two-point boundary value problem

In this case, . It is easily seen that satisfies (a) and (b) of Theorem 3.1.

In order to prove that satisfies (c) of Theorem 3.1, previously, we will prove that the function , defined by , satisfies

In fact, put and (notice that, as and is nondecreasing, ). Then, from
as , then , we obtain
Applying to this inequality and taking into account the nondecreasing character of , we have
or, equivalently,
This proves our claim.

In the sequel, we prove that satisfies assumption (c) of Theorem 3.1.

In fact, for and , we can obtain

Now, we will prove that belongs to . In fact, obviously takes into itself and, as , is nondecreasing. Besides, as the derivative of is for , is strictly increasing, and, consequently, for (notice that ). Notice that if and , then is a bounded sequence because, in contrary case, and, thus, . Suppose that . Then, we can find such that, for each , there exists with . The bounded character of gives us the existence of a subsequence of with convergent. Suppose that . From , we obtain and, as the unique solution of is , we obtain . Thus, , and this contradicts the fact that for any . Therefore, . This proves that satisfies assumption (c) of Theorem 3.1. Finally, as , Problem (3.15) has a unique positive solution for by Theorems 3.1 and 3.2.

*Remark 3.4. *In Theorem 3.2, the condition for certain seems to be a strong condition in order to obtain a positive solution for Problem (1.1), but when the solution is unique, we will see that this condition is very adjusted one. More precisely, under assumption that Problem (1.1) has a unique nonnegative solution , then

In fact, if for , then it is easily seen that the zero function satisfies Problem (1.1) and the uniqueness of solution gives us .

The other implication is obvious since if the zero function is solution of Problem (1.1), then for any .

*Remark 3.5. *Notice that assumptions in Theorem 3.1 are invariant by continuous perturbations. More precisely, if for any and satisfies (a), (b), and (c) of Theorem 3.1, then , with continuous and , satisfies assumptions of Theorem 3.2, and this means that the following boundary value problem
has a unique positive solution.

#### 4. Some Remarks

In this section, we compare our results with the ones obtained in recent papers. Recently, in [19], the authors present as main result the following theorem.

Theorem 4.1 (Theorem 3.1 of [19]). *Suppose that ** is continuous, ** is nondecreasing in , for each , ** for each . ** Moreover, suppose that there exist positive numbers such that
**
where
**
with being the Green's function defined in Section 2.Then, Problem (1.1) has at least one symmetric positive solution such that and, moreover, in the uniform norm, where is the operator defined by
**
and is the function given by , for , with , for (symmetric solution means a solution satisfying , for ).*

In what follows, we present a parallel result to Theorem 3.2 where we obtain uniqueness of a symmetric positive solution of Problem (1.1).

Theorem 4.2. *Adding assumption of Theorem 4.1 to the hypotheses of Theorem 3.2, one obtains a unique symmetric positive solution of Problem (1.1).*

*Proof. *As in the proof of Theorem 3.1, instead of , we consider the following set* K*

It is easily seen that is a closed subset of . Thus, , where is the induced metric given by
is a complete metric space.

Moreover, with the induced order by satisfies condition (2.3) of Theorem 2.1, and it is easily proved that the function , for and, consequently, , satisfies condition (2.4) of Theorem 2.1.

Now, as in Theorem 2.1, we consider the operator defined by
In the sequel, we prove that, under our assumptions, applies into itself.

In fact, suppose that is symmetric, then for , we have
Making the change of variables , we obtain
Now, it is easily seen that for and taking into account assumption of Theorem 4.1 and the symmetric character of , we have
The rest of the proof follows the lines of Theorems 3.1 and 3.2.

This finishes the proof.

Now, we present an example which illustrates Theorem 4.2.

*Example 4.3. *Consider the following problem

In this case, . It is easily checked that satisfies (a) and (b) of Theorem 3.1 and , for .

On the other hand, taking into account Example 3.3, we can obtain, for and ,

Finally, as it is proved in Example 3.3, belongs to . Therefore, Theorem 4.2 tells us that Problem (4.10) has a unique symmetric positive solution for . In what follows, we prove that Problem (4.10) can be treated using Theorem 4.1. In fact, in this case, . Moreover, (see proof of Theorem 3.1); it can be proved that . As we have seen in Example 4.3, satisfies assumptions (*H1*), (*H2*), and (*H3*) of Theorem 4.1. Moreover,Consider the function , with and . Obviously, and, as , we can find such that . This means that
On the other hand, we consider the function , with and .

Then, as and is a continuous function, we can find such that
Therefore, Problem (4.10) can be treated using Theorem 4.1, and we obtain the existence of a symmetric positive solution.

Our main contribution is the uniqueness of the solution.

In what follows, we present the following example which can be treated by Theorem 4.2 and Theorem 4.1 cannot be used.

*Example 4.4. *Consider the following problem which is a variant of Example 4.3:
where is a symmetric positive function satisfying , for example,

In this case, . Taking into account Example 4.3, it is easily proved that satisfies assumptions of Theorem 4.2, and, consequently, Problem (4.15) has a unique symmetric positive solution for .

Now, we prove that does not satisfy assumptions of Theorem 4.1 and, consequently, Problem (4.15) cannot be treated using this theorem. In fact, in this case (notice that ),
and we cannot find a positive number such that
This proves that Problem (4.15) cannot be treated by Theorem 4.1.

Now, we compare our results with the ones obtained in [14]. In [14], the author studies positive solutions of the problem

using theory of bifurcation.

His main result works with functions satisfying (i), for , (ii), (iii), (iv), for ,

and the author proves that there exists a critical such that Problem (4.19) has exactly two, exactly one, or no symmetric positive solution depending on whether or .

Our Example 3.3 cannot be treated by the results of [14], because, in this case, and does not satisfy assumptions (ii) and (iv) above mentioned.

#### Acknoledgment

This paper is dedicated to Professor Antonio Martin_on on the occasion of his 60th birth-day. This research was partially supported by “Ministerio de Educaci_on y Ciencia”, Project MTM 2007=65706

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