- Research Article
- Open Access
Existence of Solutions for Fourth-Order Four-Point Boundary Value Problem on Time Scales
Boundary Value Problems volume 2009, Article number: 491952 (2009)
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.
Very recently, Karaca  investigated the following fourth-order four-point boundary value problem on time scales:
for and And the author made the following assumptions:
(A2) If then
The following key lemma is provided in .
Lemma 1.1 (see [1, Lemma 2.5]).
Assume that conditions () and () are satisfied. If then the boundary value problem
has a unique solution
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 , from which one can see clearly that [1, Lemma 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.
The study of dynamic equations on time scales goes back to its founder Hilger  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 [4–6]. And an increasing interest in studying the existence of solutions to dynamic equations on time scales is observed, for example, see [7–16].
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 [17–19] 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
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
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 ).
Assume that are two regressive functions, then
The following well-known fixed point theorem will play a very important role in proving our main result.
Theorem 2.4 (see ).
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:
For convenience, we denote
First, we present two lemmas about the calculus on Green functions which are crucial in our main results.
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.
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).
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:
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.
The following boundary value problem:
has a unique solution
where the Green's function of (2.33) is as follows:
After some rearrangement of (2.35), one obtains
Green function (2.37) associated with BVP (2.33) which is a special case of (2.13) is coincident with part of [21, Lemma 1].
Assume that conditions ()–() are satisfied. If then boundary value problem
has a unique solution
and is given in Lemma 2.5.
Consider the following boundary value problem:
The Green's function associated with the BVP (2.41) is . This completes the proof.
In [1, Lemma 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
Assume ()–() are satisfied. Moreover, suppose that the following condition is satisfied:
where are continuous, with
and there exists a continuous nonnegative function such that If
then BVP (1.8) has a solution .
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 ).
The right hand side tends to uniformly zero as Consequently, Steps 1–3 together with the Arzela-Ascoli theorem show that is completely continuous.
Now we consider the following boundary value problem:
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
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
Thus for any and when it follows that
In a similar way, we also conclude that for any
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.
We give an example to illustrate our result.
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, ()–() are satisfied. And
Since for each we have the following.
By simple calculation we have
On the other hand, we notice that
That is, is satisfied. Thus, Theorem 3.1 guarantees that (4.1) has at least one nontrivial solution
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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).