Research Article | Open Access
Sotiris K. Ntouyas, Jessada Tariboon, "Nonlocal Boundary Value Problems for -Difference Equations and Inclusions", International Journal of Differential Equations, vol. 2015, Article ID 203715, 12 pages, 2015. https://doi.org/10.1155/2015/203715
Nonlocal Boundary Value Problems for -Difference Equations and Inclusions
We study boundary value problems for -difference equations and inclusions with nonlocal and integral boundary conditions which have different quantum numbers. Some new existence and uniqueness results are obtained by using fixed point theorems. Examples are given to illustrate the results.
In this paper we introduce a new class of boundary value problems for -difference equations with nonlocal boundary conditions given by where is such that is continuous at , , , is a fixed constant, , , , and , are given constants such that .
The study of -difference equations, initiated by Jackson [1, 2], Carmichael , Mason , and Adams  in the first quarter of 20th century, has been developed over the years; for instance, see [6–8]. In recent years, the topic has attracted the attention of several researchers and a variety of new results can be found in the papers [9–21].
Nonlocal conditions were initiated by Bitsadze . As remarked by Byszewski [23, 24], the nonlocal condition can be more useful than the standard initial condition to describe some physical phenomena. For example, may be given by where , are given constants and .
In Section 3 we give some sufficient conditions for the existence and uniqueness of solutions and for the existence of at least one solution of problem (1). The first result is based on Banach’s contraction principle and the second on a fixed point theorem due to O’Regan . Concrete examples are also provided to illustrate the possible applications of the established analytical results.
In Section 4, we extend the results to cover the multivalued case, considering the following boundary value problem for -difference inclusions with nonlocal and integral boundary conditions: where is a multivalued map and is the family of all nonempty subsets of .
We give an existence result for the problem (2) in the case when the right hand side is convex valued by using the nonlinear alternative for contractive maps.
The paper is organized as follows: in Section 2 we recall some preliminary facts that we need in the sequel, in Section 3 we prove our main results for single-valued case, and in Section 4 we prove our main results for multivalued case.
Let and a function defined on a -geometric set ; that is, for all . The -difference operator is defined by provided that the limit exists and does not depend on . The higher order -derivatives are given by
The Jackson -integration  is where , provided that the series converge. Here we remark that the integral is understood as a right inverse of the -derivative.
For , is called -regular at zero if for every , . It is important to note that continuity at zero implies -regularity at zero but the converse is not true (see an example on page 7 in ).
Definition 1. Let be a function defined on a -geometric set . Then is -integrable on if and only if exists for all .
The -integration by parts rule is provided that and are -regular at zero functions.
Let be a -regular at zero function defined on a -geometric set containing zero. Then is -regular at zero, where is a fixed point in . Furthermore, exists for every and Conversely, if and are two points in , then
We denote by the Banach space of all continuous functions from which are -regular at zero.
To define the solution for the problem (1), we find the solution for its associated linear problem.
Lemma 2. Let , be a continuous function such that it is continuous at and . The solution of the -difference equation subject to the boundary conditions, is given by where
Proof. Integrating twice the given equation and changing the order of integration, we get
for some constants (for functions not necessarily continuous at zero, the constants , are -periodic functions ).
In particular, for , we get Taking -derivative for (14), for , we obtain
Taking the -integral for (14) from to , we obtain
Then we have
By applying the boundary conditions we get to the system from which we have Substituting into (14) the values of and we obtain (12).
3. Existence Results: The Single-Valued Case
In view of Lemma 2, we define an operator by
For convenience we set
Theorem 3. Let and be continuous functions. Assume that, , , ;, for all ;.Then the boundary value problem (1) has a unique solution.
Proof. For and for each , from the definition of and assumptions and , we obtain Hence As , by , is a contraction map from the Banach space into itself. Thus, the conclusion of the theorem follows by the contraction mapping principle (Banach fixed point theorem).
Example 4. Consider the following nonlocal boundary value problem of nonlinear -difference equation
Here, , , , , , , , , , and . We find that , , and .
As and , therefore, and are satisfied with and , respectively. Hence . By the conclusion of Theorem 3, the boundary value problem (27) has a unique solution on .
Next, we introduce the fixed point theorem which was established by O’Regan in . This theorem will be adopted to prove the next main result.
Lemma 5. Let be an open set in a closed, convex set of a Banach space . Assume . Also assume that is bounded and that is given by , in which is continuous and completely continuous and is a nonlinear contraction (i.e., there exists a continuous nondecreasing function satisfying for , such that for all ). Then, either has a fixed point orthere exist a point and with , where and , respectively, represent the closure and boundary of on .
Theorem 6. Let be a continuous function. Suppose that holds. In addition we assume that;there exists a nonnegative function and a nondecreasing function such that , where and are defined in (23) and (24), respectively.Then the boundary value problem (1) has at least one solution on .
Proof. Consider the operator as that defined in (30). We decompose into a sum of two operators
From there exists a number such that We will prove that the operators and satisfy all the conditions in Lemma 5.
Step 1. The operator is continuous and completely continuous. We first show that is bounded. For any we have This proves that is uniformly bounded.
In addition for any , , we have which is independent of and tends to zero as . Thus, is equicontinuous. Hence, by the Arzelá-Ascoli Theorem, is a relatively compact set. Now, let with . Then the limit uniformly valid on . From the uniform continuity of on the compact set it follows that is uniformly valid on . Hence as which proves the continuity of . Hence Step 1 is completely proved.
Step 2. The operator is contractive. This is a consequence of . Indeed, we have or
Step 3. The set is bounded. By and we imply that for any . This, with the boundedness of the set , implies that the set is bounded.
Step 4. Finally, it is to show that the case () in Lemma 5 does not occur. To this end, we suppose that () holds. Then, we have that there exist and such that . So, we have and With hypotheses , and follows the computations of Step 1, we have which implies Thus, which contradicts to (32). Consequently, we have proved that the operators and satisfy all the conditions in Lemma 5. Hence, the operator has at least one fixed point , which is the solution of the boundary value problem (1). The proof is completed.
Example 7. Consider the following nonlocal boundary value problem of nonlinear -difference equation
Here, , , , , , , , , , and . We find that , , and .
As with and ; therefore, and are satisfied, respectively. Since , we choose and . We can show that Therefore, by Theorem 6, the boundary value problem (42) has at least one solution on .
4. Existence Results: The Multivalued Case
For a normed space , let , , , and . A multivalued map is convex (closed) valued if is convex (closed) for all . The map is bounded on bounded sets if is bounded in for all (i.e., . is called upper semicontinuous (u.s.c.) on if for each , the set is a nonempty closed subset of , and if, for each open set of containing , there exists an open neighborhood of such that . is said to be completely continuous if is relatively compact for every . If the multivalued map is completely continuous with nonempty compact values, then is u.s.c. if and only if has a closed graph; that is, , imply . has a fixed point if there is such that . The fixed point set of the multivalued operator will be denoted by . A multivalued map is said to be measurable if, for every , the function, is measurable.
Let denote the space of all functions defined on such that .
Definition 8. A function is a solution of the problem (2) if , , and there exists a function such that it is continuous at , and on and
Definition 9. A multivalued map is said to be Carathéodory (in the sense of -calculus) if is upper semicontinuous on . Further a Carathéodory function is called —Carathéodory if there exists such that for all on for each .
For each , define the set of selections of by
The following lemma will be used in the sequel.
Lemma 10 (see ). Let be a Banach space. Let be an —Carathéodory multivalued map and let be a linear continuous mapping from to . Then the operator is a closed graph operator in .
To prove our main result in this section we will use the following form of the nonlinear alternative for contractive maps [31, Corollary 3.8].
Theorem 11. Let be a Banach space and a bounded neighborhood of . Let and two multivalued operators satisfying the following:(a) is contraction,(b) is u.s.c and compact. Then, if , either(i) has a fixed point in or(ii)there is a point and with .
Theorem 12. Assume that holds. In addition we suppose that is such that is u.s.c. on ;there exists a continuous nondecreasing function and a function such that
there exists a number such that
where and are defined in (23) and (24), respectively.
Then the boundary value problem (2) has at least one solution on .
Proof. Transform the problem (2) into a fixed point problem. Consider the operator defined by
Now, we define two operators as follows. by and the multivalued operator by
Then . We will show that the operators and satisfy all the conditions of Theorem 11 on . For better readability, we break the proof into a sequence of steps and claims.
Step 1. We show that is a contraction on . The proof is similar to the one for the operator in Step 2 of Theorem 6.
Step 2. We will show that the operator is compact and convex valued and it is completely continuous. This will be given in several claims.
Claim I. maps bounded sets into bounded sets in . To see this, let be a bounded set in . Then, for each , , there exists such that
Then for we have