Existence of Positive Solutions for Some Superlinear Fourth-Order Boundary Value Problems

Abstract

We are concerned with the following superlinear fourth-order equation , where are nonnegative constants such that and is a nonnegative continuous function that is required to satisfy some appropriate conditions related to a class satisfying suitable integrability condition. Our purpose is to prove the existence, uniqueness, and global behavior of a classical positive solution to the above problem by using a method based on estimates on the Green function and perturbation arguments.

1. Introduction

A natural motivation for studying higher order BVPs lies in their applications. It is well-known (see for instance [1]) that the deformation of an elastic beam in equilibrium state, whose both ends clamped, can be described by fourth-order BVP subject to the boundary conditions When the nonlinearity is nonnegative such problems have been extensively investigated by many researchers, and various forms of the equation and boundary condition have been discussed; see, for example, [1–15] and references therein.

In particular, in 2005, Ma and Tisdel [1] studied the existence of positive solutions to the sublinear BVP where and is continuous which may be singular at both ends and and satisfying some integrable conditions. Using the method of lower and upper solutions for fourth-order boundary value problems, they have given some necessary and sufficient conditions for the existence of regular positive solutions to the boundary value problem (3).

In [8], the authors used fixed-point index results, to prove the existence of a positive solution for the boundary value problem where is nondecreasing continuous function allowed to be superlinear and is continuous which may be singular at both ends and satisfying some adequate conditions.

Recently, in [4], the author considered problem (3), with and is a nonnegative continuous function on satisfying some hypotheses related to Karamata regular variation theory. Using the Schauder fixed-point theorem, he established the existence and uniqueness of a positive solution to (3).

Here, by using a method based on estimates on the Green function and perturbation arguments, we show that, for negative nonlinearity , problem (1) has a unique positive classical solution subject to some boundary conditions. More precisely, we are concerned with the following superlinear fourth order problem where are nonnegative constants such that . Our goal is to answer the questions of existence, uniqueness, and global behavior of a classical positive solution to problem (5), where the nonnegative nonlinear term is required to satisfy some appropriate conditions related to the following class .

Definition 1. A Borel measurable function in belongs to the class if satisfies the following condition:
We will often refer in this paper to the unique solution of the problem
Observe that, for , where .
Also we denote by the Green function of the operator , with the conditions , which can be explicitly given by where and (see [2]).
The outline of the paper is as follows. In Section 2, we give some sharp estimates on Green’s function , including the following 3G-inequality: for each , where .
In particular, we derive from this 3G-inequality that, for each , one has
In Section 3, our purpose is to study the superlinear fourth-order problem (5). The nonlinearity is required to satisfy a combination of the following assumptions. (H₁) is a nonnegative continuous function in . (H₂)There exists a nonnegative function with such that for each , the map is nondecreasing on . (H₃)For each , the function is nondecreasing on .
We will first exploit the 3G-inequality to prove that the inverse of fourth-order operators that are perturbed by a zero-order term are positivity preserving. That is, if is a positive measurable function and is a nonnegative function belonging to the class with , then the following problem has a positive solution. It turns out to prove that problem (13) admits a positive Green function .
Based on the construction of this Green function and by using perturbation arguments, we prove the following.

Theorem 2. Assume -, then problem (5) has a positive solution in satisfying where is a constant in .
Moreover, if hypothesis is also satisfied, then the solution to problem (5) satisfying (14) is unique.

Corollary 3. Let be a nonnegative function in such that the map is nondecreasing on . Let be a nonnegative continuous function on such that the function belongs to the class . Then for , the following problem has a unique positive solution in satisfying

Observe that in Theorem 2, we obtain a positive classical solution to problem (5) whose behavior is not affected by the perturbed term. That is, it behaves like the solution of the homogeneous problem (8). As typical example of nonlinearity satisfying , we quote , where ,    is a positive continuous function on such that and , with . Note that by using (9) and (17) the function belongs to the class .

As usual, let be the set of Borel measurable functions in and let be the set of nonnegative ones.

We define the kernel by

Let . We define the kernel by

Remark 4 (see [1]). Let such that the function is continuous and integrable on , then is the unique solution in of

2. Estimates on the Green’s Function

From the explicit expression of Green’s function (10) we derive the following.

Proposition 5 (see [4, 7]). For , one has

Using (22), we deduce the following.

Corollary 6. For , one has(i) (ii)For , the function is continuous on if and only if the integral converges.

Theorem 7 (3G inequality). For each , one has where .

Proof. To prove the inequality, we denote by and we claim that is a quasi-metric, that is, for each ,
By symmetry, we may assume that .
Using (21), we deduce that
To show (25), we separate the proof into three cases.
Case 1  . In this case, one has
Case 2  . We obtain
Case 3  . One has
This completes the proof.

In the sequel, for any , we recall that and we denote by

Proposition 8. Let be a function in ; then(i) (ii)For , one has (iii)For , one has
In particular, for , one has

Proof. Let be a function in . (i) Using (24) and (23), one has, for ,  Hence  (ii) Since, for each , one has , then we deduce by Fatou’s lemma and (12), that  which implies that, for ,  (iii) Similarly, we prove inequality (34) by observing that Inequality (35) follows from Proposition 8 (ii)-(iii) and the fact that . This completes the proof.

3. Proofs of Main Results

In this section, we aim at proving Theorem 2 and Corollary 3. So, we need the following preliminaries results.

For a nonnegative function in such that , we define the function on , by where and Next, we establish some inequalities on . In particular, we deduce that is well defined.

Lemma 9. Let be a nonnegative function in such that ; then, for each and , one has (i).  In particular, is well defined in . (ii)  where and .(iii) and .(iv).

Proof. (i) The assertion is clear for .
Assume that inequality in (i) holds for some ; then by using (42) and (12), we obtain
Now, since , it follows that is well defined in .
(ii) Using (22) and (42), we obtain (43) by induction.
(iii) Since , then equalities in (iii) are clear for .
Assume that for a given integer and , one has
Using (42) and Fubini-Tonelli theorem, we obtain
On the other hand, by (42) and (45), we deduce that
(iv) Let and . By Lemma 9 (i) and (22), one has
Hence the series converges.
So we deduce by the dominated convergence theorem and Lemma 9 (iii) that

Proposition 10. Let be a nonnegative function in such that . Then the function is continuous on .

Proof. We claim that for , the function is continuous on .
The assertion is clear for .
Now assume that for a given integer , the function is continuous on . By (42), one has
Since the function is continuous on and by Lemma 9 (i) and (22)
we deduce by the dominated convergence theorem that is continuous on .
This proves our claim.
Now by Lemma 9 (i) and (22), one has, for each ,
This implies that the series is uniformly convergent on and therefore the function is continuous on .
The proof is completed.

Lemma 11. Let be a nonnegative function in such that . Then for in , one has

Proof. Since , we deduce from Lemma 9 (i) that
On the other hand, from the expression of , one has
Since the series is convergent, we deduce by (55) and (42) that
That is,
Now from (54) and Lemma 9 (i) (with ), we obtain
This implies that
So it follows that and by (57) and Lemma 9 (i) (with ), one has

In the sequel, for a given nonnegative function such that , we define the operator by

Using (53) and (22), we obtain the following.

Corollary 12. Let be a nonnegative function in such that and ; then the following statements are equivalent.(i)The function is continuous on .(ii)The integral converges.

Next, we will prove that the kernel satisfies the following resolvent equation.

Lemma 13. Let be a nonnegative function in such that and . Then satisfies the following resolvent equation:
In particular, if , one has

Proof. Let , then by (57) one has which implies by Fubini-Tonelli theorem that, for ,
On the other hand, by Lemma 9 (iii) and Fubini-Tonelli theorem, we obtain for and that is,
So we obtain
This completes the proof.

Proposition 14. Let be a nonnegative function in such that and let such that is continuous and integrable on . Then is the unique nonnegative solution in of the perturbed fourth-order equation (13) satisfying

Proof. It is clear by Corollary 12 that the function is continuous on . Using (62) and (22), there exists a nonnegative constant such that
So we deduce that
Hence by using Remark 4, the function satisfies the equation and by integration inequalities (53), we obtain (69).
It remains to prove the uniqueness. Assume that is another nonnegative solution in of problem (13) satisfying (69).
Since the function is continuous on and by (69) and (70), the function is integrable on ; then it follows by Remark 4 that the function satisfies
From the uniqueness in Remark 4, we deduce that
Hence
Now since by (69), (70), and (35), one has then by (63), we deduce that . This completes the proof.

Proof of Theorem 2. Let and with and recall that
Since satisfies , then there exists a positive function in such that and for each , the map is nondecreasing on .
Let and define the operator on by
By (62) and Proposition 8, one has and by , we obtain
So we claim that is invariant under . Indeed, using (81) and (80), one has for
Next, we will prove that the operator is nondecreasing on . Indeed, let be such that . Since the map is nondecreasing on , for , we obtain
Now, we consider the sequence defined by and , for . Since is invariant under , one has and by the monotonicity of , we deduce that
Hence by dominated convergence theorem and hypotheses -, we conclude that the sequence converges to a function satisfying
That is,
On the other hand, since by (80), one has , then by applying the operator on both sides of the above equality and using (62) and (63), we conclude that satisfies
Next we aim at proving that is a solution of problem (5). To this end, we remark by (81) and (9) that
This implies by Corollary 6 (ii) that the function is continuous on and so by (87), is continuous on .
Now, since by and (88), the function is continuous and integrable on , we conclude by Remark 4 that is the required solution.
It remains to prove that under condition , is the unique solution to problem (5) satisfying (14). Assume that is another nonnegative solution in to problem (5) satisfying (14). Since , we deduce by (88) that
So the function is continuous and integrable on and by Remark 4, we conclude that the function satisfies
From the uniqueness in problem (8), we deduce that
Now let be the function defined on by
Then by ,   and by (87) and (91), one has
On the other hand, by , we remark that and by (80) we deduce that
Hence by (63), we conclude that . This completes the proof.

Proof of Corollary 3. Let ,  , and . It is clear that hypotheses and are satisfied. Since the function belongs to the class , one has for . Moreover, by a simple computation, for and , we obtain
This implies that the function satisfies hypothesis . So the result follows by Theorem 2.

Example 15. Let and with . Let , and be a positive continuous function on such that
Since the function belongs to the class , then, for , the problem has a unique positive solution in satisfying

Example 16. Let and with . Let ,, and be a positive continuous function on such that
Let . Since the function , then, for , the problem has a unique positive solution in satisfying

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors thank the referees for their careful reading of the paper. This project was supported by King Saud University, Deanship of Scientific Research, College of Science Research Center.