#### Abstract

Using fixed point methods we establish some existence theorems of positive (nontrivial) solutions for a class of Hadamard fractional boundary value problems with sign-changing nonlinearity.

#### 1. Introduction

In this paper, using fixed point methods we study the existence of positive (nontrivial) solutions for the Hadamard fractional boundary value problems with sign-changing nonlinearity:where is a real number, is the left-sided Hadamard fractional derivative of order , , and is a sign-changing function; i.e., there exists a constant such that

(H0) for all .

As is known, fractional differential equations have been paid special attention by many researchers for the reason that they serve as an excellent tool for wide applications in various disciplines of science and engineering such as mechanics, electricity, chemistry, and control theory; for more details, we refer to books [1–3]. In recent years, there have been a large number of papers dealing with the existence of solutions of nonlinear initial (boundary) value problems of fractional differential equations by using some techniques of nonlinear analysis, such as fixed-point results [4–13], iterative methods [14–23], the topological degree [24–29], the Leray-Schauder alternative [30, 31], and stability [32].

In [4], the authors studied the following abstract evolution of the system for HIV-1 population dynamics, which takes the form in fractional sense:where and are two semipositone functions. By using the Guo-Krasnosel’skii fixed point theorem, they not only obtained the existence of positive solutions for (2) but also discussed the effect of parameters on the existence of solutions.

In [14], the authors adopted generalized -contractive map to study some fractional integro-differential equations with the Caputo-Fabrizio derivation and obtained the approximate solutions for these problems by using of some appropriate Lipschitz conditions for their nonlinearities.

In [15], Cui used the convergence of Cauchy sequences in complete spaces to obtain the unique solution for the fractional boundary value problems:where is a Lipschitz continuous function, with the Lipschitz constant associated with the first eigenvalue for the relevant operator. This method can also be applied in papers [16, 17] and references therein.

However, as a generalization of fractional calculus by Riemann and Liouville, Hadamard fractional equations have seldom been studied in the literature; we only refer to [8–10, 22, 23, 29, 32] and references therein. In [8], the authors used the Guo-Krasnosel’skii fixed point theorem on cones to establish the existence and nonexistence of positive solutions for (1) with nonnegative nonlinearity and considered solvability for the influence of the parameter intervals.

In [32], the authors used Banach and Schauder fixed point theorem to obtain the existence and Hyers-Ulam stability of solutions for Hadamard fractional impulsive Cauchy problems of the form where satisfies a Lipschitz condition.

In this paper, motivated by works aforementioned, we used fixed point methods to study the existence of solutions for (1) with sign-changing nonlinearity. We have the main results: (i) when the nonlinear term grows both superlinearly and sublinearly at , we use the fixed point index theory to obtain two existence theorems of positive solutions for (1); (ii) when satisfies an appropriate Lipschitz condition, we obtain a unique solution for (1) and establish a sequence of iterations uniformly converges to the unique solution.

#### 2. Preliminaries

*Definition 1 (see [1–3]). *The left-sided Hadamard fractional derivative of order , of a function is given bywhere is the Gamma function.

We now offer Green’s function for (1). From Lemma 2.1 of [8], (1) is equivalent to the integral equationwhere

Lemma 2 (see [10, Lemma 2.2]). *Let be defined by (7). Then the following inequalities are satisfied:where and .*

Lemma 3. *Let and . Then the following inequalities are satisfied:This is a direct result from Lemma 2, so we omit its proof. Moreover, for convenience, let**Let , , and Then becomes a real Banach space and is a cone on . Define for in the sequel.**Define and . Then is also a cone on . In what follows, we verify that when , ; we have , , where is a solution for the problemwhere is defined by (H0). From (1) and (6), takes the form as follows:where is defined by (7). Indeed, when , from (7) we have **For semipositone condition (H0), we need to construct an appropriate operator with which to study problem (1). Hence, we consider the modified problemwhere is a solution for (11). Clearly, we are easy to show that if solves (14), solves (11), and and ; then is a positive solution for (1). Consequently, we turn to study the modified problem (14). From (1) and (6), (14) is equivalent to the integral equationHence, we can define an operator as follows:It is not difficult to prove that if , then is a solution for (14). Moreover, from Lemma 2 we easily have .*

Now, we offer some basic theorems for fixed point methods used in our problem.

Lemma 4 (see [33]). *Let be a real Banach space and a cone on . Suppose that is a bounded open set and that is a continuous compact operator. If there exists such that then , where denotes the fixed point index on .*

Lemma 5 (see [33]). *Let be a real Banach space and a cone on . Suppose that is a bounded open set with and that is a continuous compact operator. If then .*

#### 3. Positive Solutions for (1)

Let , and . Then we give some assumptions for nonlinear term .

(H1) uniformly on ,

(H2) there exist such that where ,

(H3) uniformly on ,

(H4) there exist , such that where ,

(H5) , and there exists such that for ,

(H6) for .

We now state our main results and offer their proofs.

Theorem 6. *Suppose that (H0)-(H2) hold. Then (1) has at least a positive solution.*

*Proof. *We first prove that there exist large enough such thatwhere is a given element. If false, there exits such that . Note that ; then for , and thus . This also implies that for .

Note that ; then , and for . From (H1) we haveuniformly on . As a result, there exist and such that This implies that Note that (12) is multiplied by on both sides of the above and integrated over and use Lemma 3 to obtain Consequently, we have Noting that , we have Therefore, if we choose , then (21) holds true. From Lemma 4 we haveOn the other hand, we prove thatIf false, there exist such that ; this implies , , and . However, from (H2) we have This indicates that for . This has a contradiction, and thus (29) holds true. From Lemma 5 we haveFrom (28) and (31), we obtain Therefore the operator has at least one fixed point in with , and then is a positive solution for (1). This completes the proof.

Theorem 7. *Suppose that (H0), (H3), and (H4) hold. Then (1) has at least a positive solution.*

*Proof. *We first prove that there exist large enough such thatIf false, there exist such that , and for the fact that when . This also implies .

Note that ; then , and for . From (H3) we haveuniformly on . As a result, there exist and such that This implies that Note that (12) is multiplied by on both sides of the above and integrated over and use Lemma 3 to obtain Consequently, we have Noting that , we obtain Taking , then (33) is satisfied. From Lemma 5 we haveOn the other hand, we prove thatwhere is a given element. If false, there exist such that ; this implies , and thus . However, from (H4) we havefor . This has a contradiction, and thus (41) holds true. From Lemma 4 we haveFrom (40) and (43), we obtain Therefore the operator has at least one fixed point in with , and then is a positive solution for (1). This completes the proof.

From (6), we define an operator as follows: Then is a completely continuous operator, and is a solution for (1) if and only if is a fixed point of .

Theorem 8. *Suppose that (H5)-(H6) hold. Then (1) has only a nontrivial solution, denoted by , and for all , the sequence uniformly converges to .*

*Proof. *(H6) ensures that is not a solution for (1). Then if (1) has a solution, this solution is nontrivial. For all , from Lemma 3 we haveOn the other hand, letting and in (H5), we have Noting that , we obtain Therefore, for all , we have Consequently, for all , we have This implies is a Cauchy sequence, and from ’s completeness, there exists such that . Taking the limits for sequence and we have ; i.e., is a nontrivial solution for (1).

Next we prove that (1) has only a solution. If are solutions for (1) and , then and for all . By (H5) we obtainThis implies that Noting that , then there exists , when , , and thus a contradiction for the above inequality. This obtains the uniqueness of solutions for (1). This completes the proof.

In what follows, we offer some examples for our main results. Let , . Then , , and .

*Example 9. *Let for all , , and . Then uniformly on , and when , . Moreover, . Therefore, (H1) and (H2) hold.

*Example 10. *Let for all , , and . Then uniformly on , and when , . Moreover, . Therefore, (H3) and (H4) hold.

*Example 11. *Let , where and with for . Therefore, (H5) and (H6) hold.

#### Data Availability

No data were used to support this study

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Acknowledgments

This work is supported by the Natural Science Foundation of Shandong Province (ZR2018MA009, ZR2015AM014).