Existence and Global Behavior of Positive Solutions for Some Fourth-Order Boundary Value Problems

Abstract

We establish the existence and uniqueness of a positive solution to the following fourth-order value problem: , with the boundary conditions , where and is a nonnegative continuous function on (0, 1) that may be singular at or . We also give the global behavior of such a solution.

1. Introduction

The purpose of this paper is to study the existence and uniqueness with a precise global behavior of a positive solution for the following fourth-order two-point boundary value problem: where and is a nonnegative continuous function on that may be singular at or and satisfies some hypotheses related to the class of Karamata regularly varying functions.

There have been extensive studies on fourth-order boundary value problems with diverse boundary conditions via many methods; see, for example, [1–9] and the references therein.

A naturel motivation for studying higher order boundary value problems lies in their applications. For example, it is well known that the deformation of an elastic beam in equilibrium state, whose both ends clamped, can be described by fourth-order boundary value problem

Our aim in this paper is to give a contribution to the study of these problems by exploiting the properties of the Karamata class of functions.

To state our result, we need some notations. We denote by the set of all continuous functions on , and we will use to denote the set of Karamata functions defined on by for some , where and such that . It is clear that a function is in if and only if is a positive function in such that

For two nonnegative functions and defined on a set , the notation , , means that there exists such that , for all . We denote by , , , for ,   , and the set of all measurable functions on .

Throughout this paper, we assume that is nonnegative on and satisfies the following condition: such that for where , , satisfying

In the sequel, we introduce the function defined on by where

Our main result is the following.

Theorem 1. Let and assume that satisfies . Then, problem (1) has a unique positive solution satisfying for

This paper is organized as follows. Some preliminary lemmas are stated and proved in the next section, involving some already known results on Karamata functions. In Section 3, we give the proof of Theorem 1.

2. Technical Lemmas

To let the paper be self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following is due to [10, 11].

Lemma 2. The following assertions hold.(i)Let and ; then, one has (ii)Let and let . Then, one has , , and .

Example 3. Let be a positive integer. Let , let , and let be a sufficiently large positive real number such that the function is defined and positive on , for some , where ( times). Then, .

Applying Karamata’s theorem (see [10, 11]), we get the following.

Lemma 4. Let and let be a function in defined on . One has the following: (i)if , then diverges and ;(ii)if , then converges and .

Lemma 5 (see [12] or [13]). Let be defined on . Then, one has
If further converges, then one has

Remark 6. Let be defined on ; then, using (4) and (12), we deduce that
If further converges, we have by (12) that

Lemma 7. Given that , then the unique continuous solution of is given by where is Green’s function for the boundary value problem (16).

Remark 8. For , we have .

In the following, we give some estimates on the Green function that will be used later.

Proposition 9. On , one has the following: (i);(ii).

Proof. (i) It follows from the fact that for we have
(ii) Since for we have , the result follows from (i).
As a consequence of the assertion (ii) of Proposition 9, we obtain the following.

Corollary 10. Let and put , for .
Then,

Proposition 11. Let be a measurable function such that the function is continuous and integrable on . Then, is the unique solution in of the problem

Proof. From Corollary 10, the function is defined on and, by Proposition 9, we have
Now, since is integrable near 0 and is integrable near 1, then, for , we have This gives
Moreover, we have .
Finally, we prove the uniqueness. Let be two solutions of (21) and put . Then, and . Hence, it follows that . Using the fact that , we conclude that and so .
In the sequel, we assume that and and we put where satisfy
So, we aim to give some estimates on the potential function .
We define the Karamata functions , by
Then, we have the following.

Proposition 12. For ,

Proof. Using Proposition 9, we have
For , we have . So, using Lemma 4 and hypothesis (26), we deduce that
Now, we have
This implies by Lemma 4 that
Hence, it follows by Lemma 5 and hypothesis (26) that, for , we get
That is, for , Now, since , we use similar arguments as above applied to instead of to obtain
This together with (34) implies that, for , we have

3. Proof of Theorem 1

In order to prove Theorem 1, we need the following lemma.

Lemma 13. Assume that the function satisfies and put for . Then, one has, for ,

Proof. Put and . Then, for , we have
Let , , , and . Then, using Proposition 12, we obtain by a simple computation that

Proof of Theorem 1. From Lemma 13, there exists such that for each where .
Put and let
In order to use a fixed point theorem, we define the operator on by
For this choice of , we can easily prove that, for , we have and .
Now, since the function is continuous on and, by Proposition 9, Corollary 10, and Lemma 13, the function is integrable on , we deduce that the operator is compact from to itself. It follows by the Schauder fixed point theorem that there exists such that . Then, and satisfies the equation
Since the function is continuous and integrable on , then by Proposition 11, the function is a positive solution in of problem (1).
Finally, let us prove that is the unique positive continuous solution satisfying (9). To this aim, we assume that (1) has two positive solutions satisfying (9) and consider the nonempty set and put . Then, and we have . It follows that and consequently
Since the function is continuous and integrable on , it follows by Proposition 11 that . By symmetry, we obtain also that . Hence, and . Since , then and consequently .

Example 14. Let and let be a positive continuous function on such that where and . Then, using Theorem 1, problem (1) has a unique positive continuous solution satisfying the following estimates: where

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.