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# A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition

*Boundary Value Problems*
**volumeÂ 2011**, ArticleÂ number:Â 569191 (2011)

## Abstract

The aim of this paper is to study a fourth-order separated boundary value problem with the right-hand side function satisfying one-sided Nagumo-type condition. By making a series of a priori estimates and applying lower and upper functions techniques and Leray-Schauder degree theory, the authors obtain the existence and location result of solutions to the problem.

## 1. Introduction

In this paper we apply the lower and upper functions method to study the fourth-order nonlinear equation

with being a continuous function.

This equation can be used to model the deformations of an elastic beam, and the type of boundary conditions considered depends on how the beam is supported at the two endpoints [1, 2]. We consider the separated boundary conditions

with , .

For the fourth-order differential equation

the authors in [3] obtained the existence of solutions with the assumption that satisfies the two-sided Nagumo-type conditions. For more related works, interested readers may refer to [1â€“14]. The one-sided Nagumo-type condition brings some difficulties in studying this kind of problem, as it can be seen in [15â€“18].

Motivated by the above works, we consider the existence of solutions when satisfies one-sided Nagumo-type conditions. This is a generalization of the above cases. We apply lower and upper functions technique and topological degree method to prove the existence of solutions by making a priori estimates for the third derivative of all solutions of problems (1.1) and (1.2). The estimates are essential for proving the existence of solutions.

The outline of this paper is as follows. In Section 2, we give the definition of lower and upper functions to problems (1.1) and (1.2) and obtain some a priori estimates. Section 3 will be devoted to the study of the existence of solutions. In Section 4, we give an example to illustrate the conclusions.

## 2. Definitions and A Priori Estimates

Upper and lower functions will be an important tool to obtain a priori bounds on , , and . For this problem we define them as follows.

Definition 2.1.

The functions verifying

define a pair of lower and upper functions of problems (1.1) and (1.2) if the following conditions are satisfied:

(i), ,

(ii), ,

(iii).

Remark 2.2.

By integration, from (iii) and (2.1), we obtain

that is, lower and upper functions, and their first derivatives are also well ordered.

To have an a priori estimate on , we need a one-sided Nagumo-type growth condition, which is defined as follows.

Definition 2.3.

Given a set , a continuous is said to satisfy the one-sided Nagumo-type condition in if there exists a real continuous function , for some , such that

with

Lemma 2.4.

Let satisfy

and consider the set

Let be a continuous function satisfying one-sided Nagumo-type condition in .

Then, for every , there exists an such that for every solution of problems (1.1) and (1.2) with

for and every , one has .

Proof.

Let be a solution of problems (1.1) and (1.2) such that (2.7) and (2.8) hold. Define

Assume that , and suppose, for contradiction, that for every . If for every , then we obtain the following contradiction:

If for every , a similar contradiction can be derived. So there is a such that . By (2.4) we can take such that

If for every , then we have trivially . If not, then we can take such that or such that . Suppose that the first case holds. By (2.7) we can consider such that

Applying a convenient change of variable, we have, by (2.3) and (2.11),

Hence, . Since can be taken arbitrarily as long as , we conclude that for every provided that .

In a similar way, it can be proved that , for every if . Therefore,

Consider now the case , and take such that

In a similar way, we may show that

Taking , we have .

Remark 2.5.

Observe that the estimation depends only on the functions , , , and and it does not depend on the boundary conditions.

## 3. Existence and Location Result

In the presence of an ordered pair of lower and upper functions, the existence and location results for problems (1.1) and (1.2) can be obtained.

Theorem 3.1.

Suppose that there exist lower and upper functions and of problems (1.1) and (1.2), respectively. Let be a continuous function satisfying the one-sided Nagumo-type conditions (2.3) and (2.4) in

If verifies

for and

where means and , then problems (1.1) and (1.2) has at least one solution satisfying

for .

Proof.

Define the auxiliary functions

For , consider the homotopic equation

with the boundary conditions

Take large enough such that, for every ,

Step 1.

Every solution of problems (3.6) and (3.7) satisfies

for , for some independent of .

Assume, for contradiction, that the above estimate does not hold for . So there exist , , and a solution of (3.6) and (3.7) such that . In the case define

If , then and . Then, by (3.2) and (3.10), for , the following contradiction is obtained:

For ,

If , then

and . If , then and so . Therefore, the above computations with replaced by 0 yield a contradiction. For , by (3.11), we get the following contradiction:

The case is analogous. Thus, for every . In a similar way, we may prove that for every .

By the boundary condition (3.7) there exists a , such that . Then by integration we obtain

Step 2.

There is an such that for every solution of problems (3.6) and (3.7)

with independent of .

Consider the set

and for the function given by

In the following we will prove that the function satisfies the one-sided Nagumo-type conditions (2.3) and (2.4) in independently of . Indeed, as verifies (2.3) in , then

So, defining in , we see that verifies (2.3) with and replaced by and , respectively. The condition (2.4) is also verified since

Therefore, satisfies the one-sided Nagumo-type condition in with replaced by , with independent of .

Moreover, for

every solution of (3.6) and (3.7) satisfies

Define

The hypotheses of Lemma 2.4 are satisfied with replaced by . So there exists an , depending on and , such that for every . As and do not depend on , we see that is maybe independent of .

Step 3.

For , the problems (3.6) and (3.7) has at least one solution .

Define the operators

by

and for , by

with

Observe that has a compact inverse. Therefore, we can consider the completely continuous operator

given by

For given by Step 2, take the set

By Steps 1 and 2, degree is well defined for every and by the invariance with respect to a homotopy

The equation is equivalent to the problem

and has only the trivial solution. Then, by the degree theory,

So the equation has at least one solution, and therefore the equivalent problem

has at least one solution in .

Step 4.

The function is a solution of the problems (1.1) and (1.2).

The proof will be finished if the above function satisfies the inequalities

Assume, for contradiction, that there is a such that , and define

If , then and . Therefore, by (3.2) and Definition 2.1, we obtain the contradiction

If , then we have

By Definition 2.1 this yields a contradiction

Then and, by similar arguments, we prove that . Thus,

Using an analogous technique, it can be deduced that for every . So we have

On the other hand, by (1.2),

that is,

Applying the same technique, we have

and then by Definition 2.1 (iii), (3.44) and (3.46), we obtain

that is,

Since, by (3.44), is nondecreasing, we have by (3.49)

and, therefore, for every . By the monotonicity of ,

and so for every .

The inequalities and for every can be proved in the same way. Then is a solution of problems (1.1) and (1.2).

## 4. An Example

The following example shows the applicability of Theorem 3.1 when satisfies only the one-sided Nagumo-type condition.

Example 4.1.

Consider now the problem

with . The nonlinear function

is continuous in . If , then the functions defined by

are, respectively, lower and upper functions of (4.1) and (4.2). Moreover, define

Then satisfies condition (3.2) and the one-sided Nagumo-type condition with , in .

Therefore, by Theorem 3.1, there is at least one solution of Problem (4.1) and (4.2) such that, for every ,

Notice that the function

does not satisfy the two-sided Nagumo condition.

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## Acknowledgments

The authors would like to thank the referees for their valuable comments on and suggestions regarding the original manuscript. This work was supported by NSFC (10771085), by Key Lab of Symbolic Computation and Knowledge Engineering of Ministry of Education, and by the 985 Program of Jilin University.

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Song, W., Gao, W. A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition.
*Bound Value Probl* **2011**, 569191 (2011). https://doi.org/10.1155/2011/569191

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DOI: https://doi.org/10.1155/2011/569191