Positive Solutions for a System of Fractional Integral Boundary Value Problems Involving Hadamard-Type Fractional Derivatives

Abstract

In this paper, we use fixed-point index to study the existence of positive solutions for a system of Hadamard fractional integral boundary value problems involving nonnegative nonlinearities. By virtue of integral-type Jensen inequalities, some appropriate concave and convex functions are used to depict the coupling behaviors for our nonlinearities .

1. Introduction

In this paper, we study the existence of positive solutions for the system of Hadamard fractional integral boundary value problems:where is a real number with , , and is the Hadamard fractional derivative. The nonlinearities . Moreover, the function h on satisfies the condition: (H0) with .

In recent years, the fractional calculus and fractional differential equations are of importance in mathematics, physics, electroanalytical chemistry, capacitor theory, electrical circuits, biology, control theory, and fluid dynamics [1–20]. For example, in [1], the author considered the fractional -type conjugate boundary value problems:where , , and is the Riemann–Liouville’s fractional derivative. By means of Leray–Schauder type and Krasnosel’skii’s fixed-point theorems, the author derived an interval of parameter λ such that (2) has multiple positive solutions when any λ lies in the interval.

On the other hand, we note that coupled systems of fractional differential equations have also been investigated by many authors, see [21–32]. For example, in [21], the authors used a fixed-point theorem of increasing φ--concave operators to establish the existence and uniqueness of solutions for a system of four-point boundary value problems involving Hadamard fractional derivatives:where and are two positive parameters. In [22], the authors established positive solutions for the coupled Hadamard fractional integral boundary value problems:where the nonlinearities satisfy either of the following conditions: : there exists such that and  : there exists such that and .

Inspired by the aforementioned works, in this paper, we use the fixed-point index to consider the existence of positive solutions for system (1) of fractional integral boundary value problems involving Hadamard-type fractional derivatives. Based on integral-type Jensen inequalities, some appropriate concave and convex functions are used to depict the coupling behaviors for the nonlinearities . Moreover, our a priori estimates for positive solutions are derived by developing some appropriate nonnegative matrices when grow sublinearly at . These conditions here are different from that in and .

2. Preliminaries

In this paper, we only provide some necessary definitions and lemmas for the Hadamard fractional derivative. For more details about Hadamard fractional calculus, see the book [33].

Definition 1. The Hadamard derivative of fractional order q for a function is defined aswhere , denotes the integer part of the real number q, and .

Definition 2. The Hadamard fractional integral of order q for a function is defined as

Lemma 1. Let and . Then, the Hadamard fractional differential equation has the solutionwhere , , and .

Lemma 2. Let and . Then, we have the following formula:where and n are as in Lemma 1 and .

Lemma 3. Suppose that (H0) holds. Let . Then, the boundary value problemshas a unique solutionwhere

Proof. Using Lemma 2, we havewhere , . By , we have . Hence,Then, we know Using the condition , we haveThen, (H0) implies thatAs a result, we haveThis completes the proof.

In what follows, we study some useful inequalities for Green’s functions in (11). We first provide a result in [1].

Let , and then the Riemann–Liouville boundary-value problemhas a unique solution , where

Moreover, Green’s function H satisfies the inequalities:where .

Comparing with H, using to replace , from (11) and (19), we obtain the function satisfies the inequalities:

This, for all , implies thatand

Lemma 4. Let . Then there existsuch that

Proof. Using (20)–(22), for all , we haveThis completes the proof.

From Lemma 3, we know (1) is equivalent to the following Hammerstein-type integral equations:

Let Then becomes a real Banach space and P a cone on E. Moreover, is a Banach space with the norm , and is a cone on . Therefore, we define operators and A as follows:

Note that are nonnegative continuous functions, so the operators and are three completely continuous operators. Moreover, if is a fixed point of A, then is a positive solution for (1). Therefore, in what follows, we turn to study the existence of fixed points of the operator A.

Lemma 5. Let p be a continuous concave function. Then, if φ is an integrable function on , we have

Proof. Let , for all , and . Then, note that , for all , we haveThis completes the proof.

Remark 1. If p is a continuous convex function in Lemma 5, then (28) can be changed into the inverse inequality:

Lemma 6 (see [34]). Let E be a real Banach space and P a cone on E. Suppose that is a bounded open set and that is a continuous compact operator. If there exists a such thatthen , where i denotes the fixed-point index on P.

Lemma 7 (see [34]). Let E be a real Banach space and P a cone on E. Suppose that is a bounded open set with and that is a continuous compact operator. Ifthen .

3. Main Results

Lemma 8. Let . Then , wherewhere

Proof. From the definition of G, for all , we haveThen if , we haveNote that the arbitrariness of , we haveThis completes the proof.

Let . Then, . Now, we list our assumptions for : (H1) . (H2) There exist and such that(i) is a strictly increasing concave function on and (ii)(iii) such that  (H3) There exist and such that(i) is a strictly increasing convex function on and (ii)(iii) such that  (H4) There exist and such that(i) is a strictly increasing concave function on (ii)(iii) such that  (H5) There exist and such that

Define for . We adopt the convention in the sequel that stand for different positive constants.

Theorem 1. Suppose that (H1)–(H3) hold. Then, (1) has at least one positive solution.

Proof. Let , where are two given elements. Then, we claim that is a bounded set in . We define operators as follows:Now, if there exists , then we have and . From Lemma 8, we haveMoreover, together with (H2) (ii), we can obtain thatUsing (H2) (i) and (iii), we haveTherefore, we haveRecall that . Therefore, we multiply both sides of the above by , integrate over , and use Lemma 4 to obtainSolving this inequality, from (40), we haveOn the other hand, we estimate the norm of . Multiplying both sides of the first inequality of (41) by , integrating over , and using Lemma 4, we obtainThis implies thatWithout loss of generality, we may assume , then . Note that , we haveCombining (H2) (i) (), there exists such that .
Up to now, we have proved the boundedness of . Taking and ( is defined by (H3)), we haveThen, Lemma 6 enables us to obtainNext, we show thatIf this claim is not true, then there exist such thatCombining (H3) (ii), we obtainFrom (H3) (i) and (iii), we haveConsequently, we haveMultiplying both sides of the above by , integrating over , and using Lemma 4, we obtainNote that , we have and for . Moreover, using (54), we have for . From (H3) (i), we have for . Therefore, this contradicts to , . This also implies that (51) holds. Then, Lemma 7 enables us to obtainFrom (50) and (57), we haveTherefore, the operator A has at least one fixed point on . Equivalently, (1) has at least one positive solution. This completes the proof.

Theorem 2. Suppose that (H1) and (H4)-(H5) hold. Then, (1) has at least one positive solution.

Proof. For in (H4), we first show thatwhere are two given elements. Indeed, if this claim is false, there exist such thatThis, together with (H4) (ii), implies thatSimilar to (42), we haveFrom (H4) (iii), we haveMultiplying both sides of the above by , integrating over , and using Lemma 4, we obtainwhere . Consequently, implies that and for . Note that (65), should beFrom (H4) (i), this indicates that and for . Therefore, contradicts to and (59) holds. Then, Lemma 6 enables us to obtainLet . Then, we prove that is a bounded set in . If , then we haveFrom Lemma 8, we haveMoreover, by (H5), we haveMultiplying both sides of the above by , integrating over , and using Lemma 4, we obtainConsequently, we haveSolving this matrix inequality, we haveThis implies thatNote that , we haveTaking and ( is defined by (H4)), we haveThen, Lemma 7 enables us to obtainFrom (66) and (76), we haveTherefore, the operator A has at least one fixed point on . Equivalently, (1) has at least one positive solution. This completes the proof.
In (1), let , . Then, and (H0) holds. Moreover, we can calculate as follows:

Example 1. Let , , , for , where . Then, we have(i), for all (ii), for all (iii), for all (iv), for all (v)(vi)Therefore, (H2)-(H3) hold.

Example 2. Let , then we calculate , andLet , , and , for , where . Then for all , we haveAs a result, (H4)-(H5) hold.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (Grant no. 11601048), University Natural Science Foundation of Anhui Provincial Education Department (Nos. KJ2017A442 and KJ2018A0452), and Natural Science Foundation of Chongqing Normal University (Grant no. 16XYY24).