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# Existence of Solutions for Fourth-Order Four-Point Boundary Value Problem on Time Scales

*Boundary Value Problems*
**volumeÂ 2009**, ArticleÂ number:Â 491952 (2009)

## Abstract

We present an existence result for fourth-order four-point boundary value problem on time scales. Our analysis is based on a fixed point theorem due to Krasnoselskii and Zabreiko.

## 1. Introduction

Very recently, Karaca [1] investigated the following fourth-order four-point boundary value problem on time scales:

for and And the author made the following assumptions:

(A_{1}) and

(A_{2}) If then

The following key lemma is provided in [1].

Lemma 1.1 (see [1, Lemmaâ€‰â€‰2.5]).

Assume that conditions () and () are satisfied. If then the boundary value problem

has a unique solution

where

Here , and are given as follows:

Unfortunately, this lemma is wrong. Without considering the whole interval the author only considers in the Green's function Thus, the expression of (1.3) which is a solution to BVP (1.2) is incorrect. In fact, if one takes then (1.1) reduces to the following boundary value problem:

The counterexample is given by [2], from which one can see clearly that [1, Lemmaâ€‰â€‰2.5] is wrong. If one takes , here is a constant, then (1.1) reduces to the following fourth-order four-point boundary value problem on time scales:

The purpose of this paper is to establish some existence criteria of solution for BVP (1.8) which is a special case of (1.1). The paper is organized as follows. In Section 2, some basic time-scale definitions are presented and several preliminary results are given. In Section 3, by employing a fixed point theorem due to Krasnoselskii and Zabreiko, we establish existence of solutions criteria for BVP (1.8). Section 4 is devoted to an example illustrating our main result.

## 2. Preliminaries

The study of dynamic equations on time scales goes back to its founder Hilger [3] and it is a new area of still fairly theoretical exploration in mathematics. In the recent years boundary value problem on time scales has received considerable attention [4â€“6]. And an increasing interest in studying the existence of solutions to dynamic equations on time scales is observed, for example, see [7â€“16].

For convenience, we first recall some definitions and calculus on time scales, so that the paper is self-contained. For the further details concerning the time scales, please see [17â€“19] which are excellent works for the calculus of time scales.

A time scale is an arbitrary nonempty closed subset of real numbers . The operators and from to

are called the forward jump operator and the backward jump operator, respectively.

For all we assume throughout that has the topology that it inherits from the standard topology on The notations and so on, will denote time-scale intervals

where with

Definition 2.1.

Fix Let Then we define to be the number (if it exists) with the property that given there is a neighborhood of with

Then is called derivative of

Definition 2.2.

If then we define the integral by

We say that a function is regressive provided

where which is called graininess function. If is a regressive function, then the generalized exponential function is defined by

for is the cylinder transformation, which is defined by

Let be two regressive functions, then define

The generalized function has then the following properties.

Lemma 2.3 (see [18]).

Assume that are two regressive functions, then

(i) and

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

The following well-known fixed point theorem will play a very important role in proving our main result.

Theorem 2.4 (see [20]).

Let be a Banach space, and let be completely continuous. Assume that is a bounded linear operator such that is not an eigenvalue of and

Then has a fixed point in

Throughout this paper, let be endowed with the norm by

where And we make the following assumptions:

() and

() and

()

Set

For convenience, we denote

First, we present two lemmas about the calculus on Green functions which are crucial in our main results.

Lemma 2.5.

Assume that and are satisfied. If then is a solution of the following boundary value problem (BVP):

if and only if

where the Green's function of (2.13) is as follows:

where are given as (2.12), respectively.

Proof.

If is a solution of (2.13), setting

then it follows from the first equation of (2.13) that

Multiplying (2.17) by and integrating from to we get

Similarly, by (2.18), we have

Then substituting (2.18) into (2.19), we get for each that

Substituting this expression for into the boundary conditions of (2.13). By some calculations, we get

Then substituting (2.21) into (2.20), we get

By interchanging the order of integration and some rearrangement of (2.22), we obtain

Thus, we obtain (2.14) consequently.

On the other hand, if satisfies (2.14), then direct differentiation of (2.14) yields

And it is easy to know that and satisfies (2.13).

Corollary 2.6.

If then BVP (2.13) reduces to the following problem:

From Lemma 2.5, BVP (2.25) has a unique solution

where the Green's function of (2.25) is as follows:

where

Proof.

If is a solution of (2.25), take then Hence, from (2.20) we have

Substituting this expression for into the boundary conditions of (2.25). By some calculations, we obtain

where is given as (2.28). Then substituting (2.31) into (2.30), we get

where are as in (2.29), respectively. By some rearrangement of (2.32), we obtain (2.26) consequently.

From the proof of Corollary 2.6, if take we get the following result.

Corollary 2.7.

The following boundary value problem:

has a unique solution

where the Green's function of (2.33) is as follows:

where

After some rearrangement of (2.35), one obtains

Remark 2.8.

Green function (2.37) associated with BVP (2.33) which is a special case of (2.13) is coincident with part of [21, Lemmaâ€‰â€‰1].

Lemma 2.9.

Assume that conditions ()â€“() are satisfied. If then boundary value problem

has a unique solution

where

and is given in Lemma 2.5.

Proof.

Consider the following boundary value problem:

The Green's function associated with the BVP (2.41) is . This completes the proof.

Remark 2.10.

In [1, Lemmaâ€‰â€‰2.5], the solution of (1.2) is defined as

where and are given as (1.4) and (1.5), respectively. In fact, is incorrect. Thus, we give the right form of as the special case in our Lemma 2.9.

## 3. Main Results

Theorem 3.1.

Assume ()â€“() are satisfied. Moreover, suppose that the following condition is satisfied:

where are continuous, with

and there exists a continuous nonnegative function such that If

where

then BVP (1.8) has a solution .

Proof.

Define an operator by

where is given by (2.40). Then by Lemmas 2.5 and 2.9, it is clear that the fixed points of are the solutions to the boundary value problem (1.8). First of all, we claim that is a completely continuous operator, which is divided into 3 steps.

Step 1 ( is continuous).

Let be a sequence such that then we have

Since are continuous, we have which yields That is, is continuous.

Step 2 ( maps bounded sets into bounded sets in ).

Let be a bounded set. Then, for and any we have

By virtue of the continuity of and , we conclude that is bounded uniformly, and so is a bounded set.

Step 3 ( maps bounded sets into equicontinuous sets of ).

Let then

The right hand side tends to uniformly zero as Consequently, Steps 1â€“3 together with the Arzela-Ascoli theorem show that is completely continuous.

Now we consider the following boundary value problem:

Define

Obviously, is a completely continuous bounded linear operator. Moreover, the fixed point of is a solution of the BVP (3.8) and conversely.

We are now in the position to claim that is not an eigenvalue of

If and then (3.8) has no nontrivial solution.

If or suppose that the BVP (3.8) has a nontrivial solution and then we have

which yields

On the other hand, we have

From the above discussion (3.11) and (3.12), we have . This contradiction implies that boundary value problem (3.8) has no trivial solution. Hence, is not an eigenvalue of

At last, we show that

By then for any there exist a such that

Set and select such that

Denote

Thus for any and when it follows that

In a similar way, we also conclude that for any

Therefore,

On the other hand, we get

Combining (3.18) with (3.19), we have

Theorem 2.4 guarantees that boundary value problem (1.8) has a solution It is obvious that when for some In fact, if then will lead to a contradiction, which completes the proof.

## 4. Application

We give an example to illustrate our result.

Example 4.1.

Consider the fourth-order four-pint boundary value problem

Notice that To show that (4.1) has at least one nontrivial solution we apply Theorem 3.1 with and Obviously, ()â€“() are satisfied. And

Since for each we have the following.

By simple calculation we have

On the other hand, we notice that

Hence,

That is, is satisfied. Thus, Theorem 3.1 guarantees that (4.1) has at least one nontrivial solution

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## Acknowledgments

The authors would like to thank the referees for their valuable suggestions and comments. This work is supported by the National Natural Science Foundation of China (10771212) and the Natural Science Foundation of Jiangsu Education Office (06KJB110010).

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Yang, D., Li, G. & Bai, C. Existence of Solutions for Fourth-Order Four-Point Boundary Value Problem on Time Scales.
*Bound Value Probl* **2009**, 491952 (2009). https://doi.org/10.1155/2009/491952

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DOI: https://doi.org/10.1155/2009/491952