#### Abstract

We consider a fourth-order four-point boundary value problem for dynamic equations on time scales. By the upper and lower solution method, some results on the existence of solutions of the fourth-order four-point boundary value problem on time scales are obtained. An example is also included to illustrate our results.

#### 1. Introduction

Let be a closed nonempty subset of , and let have the subspace topology inherited from the Euclidean topology on . In some of the current literature, is called a time scale (or measure chain). For notation, we shall use the convention that, for each interval of , will denote the time scales interval, that is, . Some preliminary definitions and theorems on time scales can be found in the books [1, 2], which are excellent references for calculus of time scales.

In this paper, let be a time scale and the forward jump function in . We are concerned with the following fourth-order four-point boundary value problem on time scales : for . We will assume that the following conditions are satisfied.(H1), (H2). If , then , (H3).

The upper and lower solution method has been used to deal with the boundary value problems for dynamic equations in recent years. In most of these studies, two-point boundary value problem for second-order dynamic equations is considered .

Pang and Bai  studied the following fourth-order four-point BVP on time scales: for , and and are nonnegative constants satisfying . They establish criteria for the existence of a solution by developing the upper and lower solution method and the monotone iterative technique. Our problem is more general than the problems in , and our results are even new for the differential equations as well as for dynamic equations on general time scales.

#### 2. Preliminaries

To prove the main results in this paper, we will employ several lemmas. We consider the linear boundary value problem

Denote by and , the solutions of the corresponding homogeneous equation under the initial conditions so that and satisfy the first and second boundary conditions of (2.1), respectively. Let us set Using the initial conditions (2.3), we can deduce from (2.2) for and the following equations:

Lemma 2.1. Under the conditions (H1) and (H2), the following inequalities yield.

Proof. We apply the induction principle for time scales to the statement where .
(I) The statement is true, since and .
(II) Let be right-scattered and let be true, that is, and . We need to show that and . By the definition of -derivative, we have Further, by the definition of -derivative and (2.2) for , we have From (2.9), we get , and then from (2.10), we get .
(III) Let be right-dense, be true and such that and is sufficiently close to . We need to prove that is true for .
From (2.2) with , the equations follow. To investigate the function appearing in (2.12), we consider the equation where is the desired solution. Our aim is to show that with sufficiently close to , (2.13) has a unique continuous solution satisfying inequality We solve (2.13) by the method of successive approximations, setting If the series converges uniformly with respect to , then its sum will be, obviously, a continuous solution of (2.13). To prove the uniform convergence of this series, we let Then the estimate can easily be obtained. Indeed, (2.17) evidently holds for . Let it also hold for . Then from (2.15), we get for , Therefore, by the usual mathematical induction principle, (2.17) holds for all .
Now choosing appropriately, we obtain . Then (2.13) will have a continuous solution Since , it follows that thereby proving the validity of inequality (2.14). To prove uniqueness of solution of (2.13) for , suppose it has two solutions and , and passing on to the modulus, we get Thus, Since , hence it follows that for .
From (2.12) and (2.13) in view of the uniqueness of solution, we get that . Therefore, Hence by making use of the induction hypothesis being true, we obtain for . Taking this into account, from (2.11), we also get for . Thus, is true for all .
(IV) Let and assume is left-dense and such that is true for all , that is, Passing on here to the limit as , we get by the continuity of and that and , thereby verifying the validity of .
Consequently, by the induction principle on time scales, (2.8) holds for all .
From (2.9) and (2.8) for , we also get . So the statements (2.7) for are proved.
We can prove the statements of the lemma for similarly applying the backward induction principle on time scales. The lemma is proved.

Lemma 2.2. Under the conditions (H1) and (H2), the inequality holds.

Proof. By (2.4) and (2.5), we have Since for , from (2.8), we have If , then in (2.25) the equality holds. From the condition (H2), we get . This proof is completed.

Lemma 2.3. Assume that the conditions (H1) and (H2) are satisfied. If , then the boundary value problem has a unique solution where Here are as in (2.4), (2.5), and (2.6), respectively.

Proof. Taking we have By Corollary  3.14 in , since the Wronskian of any two solutions of (2.2) is independent of , we get Hence we get So the general solution of equation has the form where and are arbitrary constants. Substituting this expression for in the boundary conditions of BVP (2.26), we can evaluate and . After some easy calculations, we can get (2.27) and (2.28).

Lemma 2.4. Under the conditions (H1) and (H2), the Green’s function of BVP (2.26) possesses the following property:

Proof. The lemma follows from (2.28), Lemmas 2.1 and 2.2 immediately.

Lemma 2.5. Assume that the conditions (H1) and (H2) are satisfied. If , then the boundary value problem has a unique solution where Here are as in (2.4), (2.5), and (2.6), respectively.

Proof. Let us consider the following BVP: The Green’s function associated with the BVP (2.40) is . This completes the proof.

Lemma 2.6. Assume that the conditions (H1)–(H3) are satisfied. If satisfies then and .

Proof. Let where .
It is easy to check that and can be given by the expression where and are as in (2.38) and (2.39), respectively. The hypothesis of the lemma implies that for , for , for , and for . Therefore, we get for and for . The proof is completed.

#### 3. Upper and Lower Solution Method

In this section, we present existence results for the BVP (1.1) by using the method of upper and lower solutions. We define the set Definition 3.1. Letting on , we say is a lower solution for the problem (1.1) if satisfies

Definition 3.2. Letting , on , we say is an upper solution for the problem (1.1) if satisfies
We assume that the function satisfies the following condition.
(H4) is continuous and satisfies where , are lower and upper solutions, respectively, for the BVP (1.1), and satisfy .

Theorem 3.3. Assume that the conditions (H1)–(H4) are satisfied. Then the problem (1.1) has a solution with for and , respectively.

Proof. Consider the BVP, where By Lemma 2.5, it is clear that the solutions of the BVP (3.6) are the fixed points of the operator where and are as in (2.38) and (2.39), respectively. It is clear that is continuous. Since the function satisfies the conditions (3.4), Thus, there exists a positive constant such that , which implies that the operator is uniformly bounded. Moreover, the operator is equicontinuous. Therefore, from the Arzela-Ascoli theorem, the operator is a compact operator. Thus, by Schauder’s fixed point theorem, there exists a solution of the BVP (3.6).
Suppose is a solution of the BVP (3.6). Since satisfies the conditions (3.4), we know that Thus, By virtue of Lemma 2.6, for and for . If is right-scattered, by using the inequality , we get the inequality . So we get on . If is right-dense, it is trivial that the inequality holds on . Similarly, one can show that for and for . This completes the proof.

Theorem 3.4. Assume that the conditions (H1)–(H4) are satisfied. Then there exist two monotone sequences and , nonincreasing and nondecreasing, respectively, with and , which converge to the extremal solutions in of the problem (1.1).

Proof. For any function which satisfies for , consider the following problem: Clearly, this problem is type (2.12). Obviously, it has a unique solution given by the expression
Step 1. on and on .
Let . Thus Using Lemma 2.6, we obtain that on and , respectively. In case that is right scattered, the assumptions and imply that . Similarly, we show that on and , respectively.
Step 2. If ; , and , then we have and .
Let and . Thus Hence we get that and on and , respectively.
Now, we define the sequences and by From the properties of , we have But then and defined by are extremal solutions of (1.1).

Example 3.5. Consider the BVP where and . It is easy to check that , are lower and upper solutions of the BVP (3.19), respectively, and that all assumptions of Theorem 3.3 are fulfilled. So the BVP (3.19) has a solution satisfying for , for .