Monotone Iterative Schemes for Positive Solutions of a Fractional -Difference Equation with Integral Boundary Conditions on the Half-Line

Abstract

In this paper, we study the boundary value problem of a fractional -difference equation with nonlocal integral boundary conditions on the half-line. Using the properties of the Green function and monotone iterative method, the extremal solutions are obtained. Finally, an example is presented to illustrate our main results.

1. Introduction

In this paper, we are concerned with the following fractional -differential equation with integral boundary value problem on the half-line:where , are the fractional -derivative of Riemann–Liouville type of order and , is the -derivative of order , , and is a given function and satisfies some conditions which will be given later.

The -difference calculus was initially developed by Jackson [1, 2] and had proved to have important applications in many subjects, such as quantum mechanics, complex analysis, and hypergeometric series. The fractional -difference calculus had its origin in the works by Al-Salam [3] and Agarwal [4]. Due to the development of the fractional differential equations, fractional -differential equations, regarded as fractional analogue of -difference equations, have been studied by several researchers, especially, about the existence of the solutions for the boundary value problems (see [5–22] and the references therein).

In [21], Yang studied the coupled integral boundary value problem for systems of nonlinear fractional -difference equations as follows:where are three parameters with and , are two real numbers and , and are the fractional -derivative of Riemann–Liouville type. By applying the nonlinear alternative of Leray–Schauder type and Krasnoselskii’s fixed point theorems, the authors obtain some sufficient conditions for the existence of positive solutions to the considered problem.

In [14], Wang et al. investigated the following nonlinear fractional -difference equation with -integral boundary condition:where , is a parameter with , , and is the fractional -derivative of Riemann–Liouville type. By applying the hybrid monotone method, the existence and uniqueness of the positive solution of the -integral boundary problem are obtained.

In [17], the authors investigate the existence of solutions for the following boundary value problem of nonlinear fractional -difference equations on the half-linewhere , . is the -derivative of Riemann–Liouville type of order . By means of Schauder fixed point theorem and Leggett–Williams fixed point theorem, some results on existence and multiplicity of solutions to the above boundary value problem are obtained.

To the best of our knowledge, there are few papers that consider the boundary value of nonlinear fractional -difference equations with nonlocal conditions on the half-line, although the study of such problems is very important. In [17], the authors only proved the existence and multiplicity of solutions, for the uniqueness of positive solutions without being given. Moreover, how to seek the solutions? It is very important and helpful for computational purpose. This thought motivates the research of iterative schemes of positive solutions for problem (1). We should mention that the paper has some new features. Firstly, we consider the infinite interval problem for higher-order nonlinear fractional -difference equation with integral boundary conditions. Secondly, the nonlinearity relies on the lower-order fractional -derivative of unknown function. Our main purpose of this paper is by constructing a suitable Banach space, defining appropriate operators, and using the monotone iterative method, which is different from the method in [17] to obtain the existence and uniqueness of positive solutions.

2. Preliminaries on -Calculus and Lemmas

Here we recall some definitions and fundamental results on fractional -integral and fractional -derivative. For more information regarding fractional -calculus, see [23].

Let and define

The -analogue of the power function with is

More generally, if , then

If , then . The -gamma function is defined byand satisfies . The -derivative of a function is defined byand -derivative of higher order by

The -integral of a function defined in the interval is given by

If and is defined in the interval , its integral from to is defined by

Similar to that for derivatives, an operator can be defined, namely,

The fundamental theorem of calculus applies to these operators and , i.e.,and if is continuous at , then

The basic properties of the two operators can be found in the book [23].

The following definition was considered first in [4].

Definition 1 (see [4]). Let and be a function defined on . The fractional -integral of the Riemann–Liouville type is and

Definition 2 (see [24]). The fractional -derivative of the Riemann–Liouville type of order is defined by andwhere is the smallest integer greater than or equal to .

Lemma 1 (see [4, 24]). Let and be a function defined on . Then, the following formulas hold:(1)(2)

Lemma 2 (see [25]). Let and n be a positive integer. Then, the following equality holds:

Lemma 3 (see [24]). For , the following is valid:

In the following, we introduce the hypotheses that will be used in subsequent proof: (H₁): satisfies and . (H2): there exist nonnegative functions and constants satisfying (H₃): there exist nonnegative functions satisfying (H₄): function is increasing with respect to the variables .

Lemma 4. Assume that satisfies condition (H₁), then fractional -differential equationhas a unique integral representationwhere

Proof. In view of Definition 2 and Lemma 1, we see thatBy Lemma 2, we obtainwhere are some constants. Since , we getwhich implies thatFrom (29) and Lemma 1, we haveHence,The condition together with (31) implies thatSubmitting (32) to (29), we getEquation (33) is multiplied by and -integrating from 0 to , and we obtainFrom (34), we getCombining (33) and (35), we haveThe proof is completed.

Lemma 5. From Lemma 4, by direct calculation, we havewhere

Now, we shall prove a lemma which plays a key role in the next theorems.

Lemma 6. Function defined in (24) satisfies the following conditions: and    for all .

Proof. From (24), it is obvious that, for all ,The above two inequalities imply thatThe proof is completed.

Lemma 7. Function defined in Lemma 5 satisfies the following conditions:for all .

Proof. For and , by Remark 2 and Lemma 5, it is easy to getThis gives forThe proof is completed.
In the following, we will construct a suitable Banach space. Letwith the normwhere .
Similar to the proof of Lemma 2.2 in [26], we can prove is a Banach space. Note that the Arzela–Ascoli theorem fails to work in . In order to proceed, we need the following compactness criterion.

Lemma 8. Let be a bounded set. Then, is relatively compact in if the following conditions hold:(i) are equicontinuous on for any (ii)For such that for any , the following inequalities hold:

Proof. The method is similar to the proof of Lemma 2.3 in [26], so we omit the details.

Lemma 9. If condition holds, then for , we have

Proof. For , by condition , we have

Lemma 10. If condition holds, then for , we have

Proof. For , by condition , we haveDefine the cone by and define the operator as follows:We also defineIt is easy to see problem (1) has a solution, if and only if the operator has a fixed point.

Lemma 11. If conditions (H₁) and (H₂) hold, then the operator is completely continuous.

Proof. It is easy to see . Since and , we know , for . In the following, we divide the proof into four steps: Step 1: we show that is uniformly bounded. Let be any bounded subset of , i.e., there exists such that for each . We only need to show that is bounded in . For , by Lemmas 5 and 8, we have By Lemmas 5, 7, and 9, we have From (54)–(56), we obtain which means that is uniformly bounded. Step 2: we show that the operator is equicontinuous for all on any compact interval of . For any given and , we have By Lemmas 5 and 6, we have On the other hand, for all , , we have Similar to (60), for all , we have Together (59)–(61), which implies that is equicontinuous on any compact intervals of . Note that, from (53) and Lemma 5, we have where which means does not rely on . Hence, is equicontinuous on any compact intervals of . In the following, we show that , , is equicontinuous on any compact intervals of . In fact, for any given and , we have Similar to (59)–(61), we obtain Inequalities (59)–(64) imply that the operator is equicontinuous on any compact intervals of . Step 3: we show that is equiconvergent at . We need to prove that, for , such that, for any , the following inequalities hold: Note that, since the function does not rely on , it is easy to infer that i.e., is equiconvergent at . By (H₁) and Lemma 9, for any , there exists such that On the other hand, since There exists sufficiently large such that for any , we have Also because , there exists sufficiently large such that, for any and , we have Choose , for any , and let . For convenience, we denote , and we obtain By (68) and (72), we obtain Similar to (72) and (73), we obtain From (67), (73), and (74), we obtain is equiconvergent at . The above three steps and Lemma 8 imply that the operator is relatively compact. Step 4: we show that the operator is continuous. For , such that as , and we need to show that as . By Lemma 6, the continuity of function , and the Lebesgue dominated convergence theorem, we obtain Similar to (75), by Lemmas 7 and 9, we get for all . From (75)–(77), we have which means that the operator is continuous.In view of all the above arguments, we deduce that the operator is completely continuous. This completes the proof.

3. Main Results

In this section, we give the main results of this paper. For convenience, we denote

Let

We have the following two theorems.

Theorem 1. Assume that (H₁), (H₂), and (H₄) hold. There exists a positive such that problem (1) has two positive solutions and satisfying and . Moreover, and , and can be obtained by the following monotone iterative schemes:In addition, we have

Proof. First, Lemma 11 implies that for any and . Next, let , . For the case , the proof is similar to that of . Define , where satisfiesFor any , similar to (54)–(56), we obtainThus,That is .
Denote that , then , and , i.e.,On the other hand, (H₄) implies that the operator is increasing, and then we haveBy the induction, we have . Then, the sequence and satisfies the following relation:By Lemmas 5 and 7, we haveBy the induction, we haveIn view of the complete continuity of the operator and , is relatively compact. That is, has a convergent subsequence and there exists a such that as . This together with (90) implies . Since is continuous and , then we have , that is, is a fixed point of operator .
For the scheme , we use a similar discussion. For , by Lemma 5 and , we haveApplying the monotonicity of function , we haveBy the induction, we haveBy Remark 2 and Lemma 6, we getBy the induction, we haveUsing the complete continuity of the operator and the iterative scheme, it is easy to getFinally, we show that are the maximal and minimal positive solutions of problem (1). Let be any positive solution of (1), thenNoting that is increasing, then we haveSimilarly, for and , we can getSince and , it follows that, for and , we haveThis implies that and are the minimal and maximal solutions of problem (1), and (53) and (54) hold.
On the other hand, we infer that and are two positive solutions of (1). In fact, by the condition (H₄), we get 0 function is not the solution of problem (1), which means , , and they can be constructed by means of two monotone iterative schemes in (81) and (82).

Theorem 2. Suppose conditions (H₁) and (H₃) are satisfied. Ifthen problem (1) has a unique positive solution in . Moreover, there is an iterative scheme such that as uniformly on any finite interval of , wherehold. is defined in (80). In addition, there is an error estimate for the approximation scheme

Proof. Choose , where is defined in (H₃). Let . For any , by Lemmas 5–7 and 10, we haveThus, we havei.e., .
In the following, we show that is a contraction. For any , by condition (H₃) and Lemma 10, we obtainIn the same way, we haveFrom (108)–(110), we haveSince , then is a contraction. The Banach fixed point theorem ensures that has a unique fixed point . Hence, problem (1) has a unique positive solution .
Moreover, for any as , where . By (111), we haveLet on both sides of (113), we obtain

Remark 1. The function in problem (1) relies on lower-order fractional -derivative of unknown function which is different from [17].

Remark 2. The results in Theorems 1 and 2 extend the fractional -derivative from the low-order to the high-order fractional derivative.

Remark 3. The two iterative sequences in Theorem 1 begin with some simple functions which are helpful for computational purpose.

4. Example

Consider the following equation:where , and function

By computation, we deduce thatwhich means that hypothesis (H₂) is satisfied.

On the other hand, , and . So condition (H₁) is satisfied. Moreover, from the expression of the function , it is obvious that is increasing with respect to the variables . Hence, condition (H₄) is satisfied. By Theorem 1, it follows that problem (1) has two positive solutions, which can be given by the limits of two explicit monotone iterative schemes in (81) and (82).

5. Conclusions

In this paper, we study the boundary value problem of a fractional -difference equation with nonlocal integral boundary conditions on the half-line. We obtain some new results as follows: (1) the existence and uniqueness of positive solutions for a higher-order nonlinear fractional -difference equation with integral boundary conditions on the half-line are obtained, and (2) the monotone iterative schemes of positive solutions for the problem which considered in our paper are obtained.

Data Availability

All the data included in this study are available upon request by contact with the corresponding author.

Conflicts of Interest

The authors declare no conflicts of interest.

Acknowledgments

This research was supported by the Science and Technology Foundation of Guizhou Province (Grant nos. [2016]7075 and [2019]1162), the Project for Young Talents Growth of Guizhou Provincial Department of Education under (Grant no. Ky[2017]133), and the Project of Guizhou Minzu University under (Grant no. 16yjrcxm002).