Positive Solutions for a System of Nonlinear Semipositone Boundary Value Problems with Riemann-Liouville Fractional Derivatives

Abstract

We study the existence of positive solutions for the system of nonlinear semipositone boundary value problems with Riemann-Liouville fractional derivatives , , , and where is a real number and is the standard Riemann-Liouville fractional derivative of order . Under some appropriate conditions for semipositone nonlinearities, we use the fixed point index to establish two existence theorems. Moreover, nonnegative concave and convex functions are used to depict the coupling behavior of our nonlinearities.

1. Introduction

In this paper, we investigate the existence of positive solutions for the system of nonlinear semipositone boundary value problems with Riemann-Liouville fractional derivativeswhere is a real number and is the standard Riemann-Liouville fractional derivative of order . The nonlinear terms , are bounded below; that is, satisfy the following.

(H1) there exists a real number , such that ,

Existence and multiplicity of solutions for fractional differential equations are widely studied in the literature; see [1–14] and the references therein. For example, in [1], the authors used the Guo-Krasnosel’skii fixed point theorem to investigate the existence of positive solutions for the singular fractional differential systemwhere satisfy

Condition (3) is used to study various types of fractional systems (see [1–12] and the references therein).

In this paper we use the fixed point index to study the existence of positive solutions for the system of nonlinear semipositone fractional boundary value problem (1). Under some appropriate conditions for , we use the fixed point index to obtain our results. Moreover, nonnegative concave and convex functions are used to depict the coupling behavior of our nonlinearities (see [13–15]), which depend on the unknown functions and their derivatives , .

2. Preliminary

Definition 1 (see [16, 17]). The Riemann-Liouville fractional derivative of order of a continuous function is given bywhere with denoting the integer part of a number , provided that the right hand side is pointwise defined on .
We first study the Green functions of problem (1). Let Then we have

Lemma 2. Let be as in (1). Then (1) is equivalent towhich takes the formLet , . Then an argument similar to that in [18, Lemma 2.7] and [19, Lemma 3] establishes the result (we omit the standard details).

Lemma 3 ([19, Lemma 4]). The functions Moreover, the following inequalities are satisfied:

Lemma 4 ([19, Lemma 5]). Let for all LetThen

Lemma 5. (i) If is a positive solution of (7), then is a positive solution of the following differential equation:where and are continuous, and(ii) If is a solution of (13) and , , then is a positive solution of (7).

Proof. If is a positive solution of (7) then (note ) we obtain and Similarly, we have that is, satisfies (13). Therefore, (i) holds. Similarly, it is easy to prove (ii). This completes the proof.

From Lemma 5, to obtain a positive solution of (7), we only need to find solutions , of (13) satisfying , , If , are solutions of (13), then , satisfy

Let , , Then is a real Banach space, and is a cone on . We denote for Now, note that , solve (1) if and only if , are fixed points of operator Therefore, if is a positive fixed for with , for , then is a positive solution for (1). Moreover, from the continuity of and , we know that , are continuous and completely continuous operators.

Lemma 6. Let , Then is a cone in and

Proof. From (9) for we have Also from (9) and the above inequality, for every , we obtain for all . Similarly, . Therefore This completes the proof.

To obtain a positive solution of (1), we seek a positive fixed point of with , (note mean that , for ). From Lemma 6, we have . For we have As a result, for if Similarly, if , we have , for .

Lemma 7 (see [20]). Let be a bounded open set and a continuous or completely continuous operator. If there exists such that for all and , then , where denotes the fixed point index on .

Lemma 8 (see [20]). Let be a bounded open set with . Suppose is a continuous or completely continuous operator. If for all and , then

3. Main Results

Let , In the sequel, we use and to stand for different positive constants. Now, we list our assumptions on .

(H2) There exist such that(i) is concave and strictly increasing on (and );(ii)there exist , , for all such that (iii) for

(H3) For all , there is a constant such that

(H4) There exist , such that(i) is convex and strictly increasing on (and );(ii)for all , (iii)there exist such that , for .

(H5) There exist , , , for all , such that where

Theorem 9. Suppose that (H1)–(H3) hold. Then (1) has at least one positive solution.

Proof. We first prove that there exists such thatwhere is a given function. Suppose there exist , with , then , for . From (i), (ii) of (H2) we haveFrom (ii) of (H2) we haveFrom (30) and (i) of (H2) we obtainThis together with (iii) of (H2) yieldsThen (32) is substituted into (29) and we obtainMultiplying by for (33) and integrating over , we haveusing the fact thatwhich can be derived from (12) in Lemma 4. From (34) we obtainNote that (note that and from Lemma 6 and ) and we haveFrom (29) we haveMultiplying by and integrating over we obtain Consequently, we have Note that we may assume for Then and . For , we have Hence, . Note , and thus there exists such that . Therefore if , with then and . Thus if we take then (28) is true. Lemma 7 impliesLet . From (H3) we have so . Similarly . Hence for Thus It follows from Lemma 8 thatFrom (42) and (45) we have Therefore the operator has at least one fixed point in and so (1) has at least a positive solution. This completes the proof.

Theorem 10. Suppose that (H1), (H4), and (H5) hold. Then (1) has at least one positive solution.

Proof. We first show that there exists such thatSuppose there exist , with , then , for From (i), (ii) of (H4), we haveFrom (ii) of (H4), we getFrom (49) and (i), (iii) of (H4) we haveThen substitute (50) into (48) and we obtainMultiplying by for (51) and integrating over , from (35), we obtainConsequently, we have Note that (note and ) and we have From (50) and Lemma 3 we have so is bounded. Note , and thus there exists such that Therefore if with then and . Thus if we take then (47) is true. Lemma 8 impliesLet and consider It follows from (H5) that and, hence, . Similarly for . Therefore for all Thus It follows from Lemma 7 thatFrom (56) and (59), we have Therefore the operator has at least one fixed point on and so (1) has at least one positive solution, which completes the proof.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The research is supported by the National Natural Science Foundation of China (Grant no. 11601048), Natural Science Foundation of Chongqing (Grant no. cstc2016jcyjA0181), the Science and Technology Research Program of Chongqing Municipal Education Commission (Grant no. KJ1703050), and Natural Science Foundation of Chongqing Normal University (Grant nos. 16XYY24, 15XLB011).