Abstract
In this paper, we use the fixed-point index and nonnegative matrices to study the existence of positive solutions for a system of Hadamard-type fractional differential equations with semipositone nonlinearities.
1. Introduction
In this paper, we study the existence of positive solutions for the following system of Hadamard-type fractional boundary value problems:where , is the Hadamard fractional derivative of order α, (i.e., if u is ξ or η, then ; etc.), and the nonlinearities satisfy the semipositone condition:
(H0) , and there exists such thatwhere
Fractional problems arise in many applications in aerodynamics, signal and image processing, biophysics, blood flow phenomena, etc. (see, for example, [1–37] and the references therein). In [1], the authors used the Krasnoselskii–Zabreiko fixed-point theorem to study the existence of positive solutions for Riemann–Liouville-type fractional boundary value problems:where f has a semipositone nonlinearity, i.e., it is bounded below and can be sign-changing, and in [2], the authors used the Guo–Krasnoselskii’s fixed-point theorem to investigate the existence of positive solutions for fractional multipoint boundary value problems:where the nonlinear f may have singularities on the time and the space variables.
Coupled systems of fractional-order differential equations have also been investigated by many authors (see [7–22, 25–36] and the references therein). In [7], the authors used coincidence degree theory to establish an existence result for a coupled system of nonlinear fractional differential equations:with the multipoint boundary conditions:
There are a few results in the literature on Hadamard-type fractional differential equations (see [23–36]). In [23], the authors used fixed-point methods to obtain some existence theorems for Hadamard-type fractional boundary value problems:where f grows superlinearly and sublinearly at infinity and can be sign-changing. The authors in [24] studied positive solutions for the p-Laplacian Hadamard fractional differential equations with integral boundary value problems:where f has a -superlinear and -sublinear nonlinearity. In [25], the authors studied fractional impulsive Cauchy problems:where denotes the left-sided Hadamard fractional derivative and the nonlinearity f satisfies a Lipschitz condition, and in [26], the authors studied positive solutions for the system of Hadamard fractional differential equations involving coupled integral boundary conditions:where the nonlinearities grow superlinearly and sublinearly.
Motivated by the above, in this paper, we study the existence of positive solutions for the system of Hadamard-type fractional differential equation (1). The nonlinearities can be sign-changing and can depend on the unknown functions and their derivatives. We use some appropriate nonnegative matrices to characterize the coupling behavior for the nonlinearities , and the nonlinearities can grow superlinearly and sublinearly.
2. Preliminaries
For details about Hadamard fractional calculus, see the book [37].
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
Now, we establish Green’s functions associated with the system of (1). In order to do this, we first consider the following auxiliary problem:where are as in (1), and satisfies the following semipositone condition:
(H0)′ There exists such that , for .
Let for . Then, from Lemma 2.3 of [36], we havewhere . This, together with the boundary conditions , implies thatwhereLetThen, we haveTherefore, substituting into problem (12), we haveFrom (13) and (14), we see that problem (18) is equivalent to the following Hammerstein-type integral equation:
Lemma 1. Let , . Then, the functions have the following properties:(i), ,(ii)For all , the following inequalities hold:
Lemma 1 (ii) is a direct result of [1, Lemma 4], so we omit the proof.
Lemma 2. Let , , , and . Then,
These inequalities can be deduced from Lemma 1 (ii), so we omit the proof.
Let , where M is defined in (H0)′. Then, satisfies the following boundary value problem:
In what follows, we consider the following boundary value problems:where
Then, , and we have the following lemma.
Lemma 3. (i) If is a solution of (24) with , then is a positive solution of (18)(ii)If is a solution of (18), then is a solution of (24)
Proof. Let be a solution of (24) with . Then, we substitute into problem (18) and obtainConsidering (23), we haveThis, together with implies that (26) holds, and is a positive solution of (18).
Substitute into (24), and we getConsidering (23), we haveThis, together with implies is a solution of (24). This completes the proof.
From (13) and (14), we also find that problem (24) is equivalent to the following Hammerstein-type integral equation:Let Then, is a real Banach space and P a cone on E. Then, is also a Banach space, equipped with the norm: , where are norms in E. From (30), we define an operator as follows:Note that our functions are continuous, so the operator A is a completely continuous operator. Moreover, if there is a , a fixed point of A, and , then we have that is a positive solution of (18). From (12) and (18), we have is a positive solution of (12). Therefore, in what follows, we study the existence of fixed points of the operator A, which are greater than .
Lemma 4. Let Then, .
This is a direct result from (20) in Lemma 1, so we omit its proof.
Note that our aim is to find a fixed point of A, which is greater than . If there is a such that , and , then if , Lemma 4 enables us to obtain
Consequently, if A has fixed point with , then , and is a positive solution of (12).
Lemma 5 (see [38]). 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 6 (see [38]). 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. Positive Solutions for (1)
From Section 2, let , and then, we should consider the following system of Hadamard-type fractional boundary value problems:where
We define some operators as follows:
Then, if there is a such that with , i.e., (35) has a solution with , and thus, is a positive solution for (1) and .
Now, we list our assumptions for the functions as follows: (H1) There exist and such that and the coefficients satisfy (H2) There exists in such that (H3) There exist and such that and the coefficients satisfy (H4) There exists in and such that
Let for . Then, we have and .
Theorem 1. Suppose that (H0) to (H2) hold. Then, (1) has at least one positive solution.
Proof. For convenience, let . We next show that there is a such thatwhere is a fixed element. Suppose not, i.e., suppose there exist such thatCombining this and Lemma 4, we haveUsing (45), we obtainFrom (H1), we haveMultiplying by and integrating from 1 to e, Lemma 2 enables us to obtainThis implies thatSolving this matrix inequality, we haveConsequently, there exist such thatNote (46) and we haveChoose . Then, (44) holds. Lemma 5 givesNext, we prove thatSuppose not. Then, there exist such thatThis implies thatHowever, from (H2), we haveNote that from (H2), we have . Hence, we obtain . Similarly, . This is a contradiction. As a result, (55) holds. From Lemma 6, we haveFrom (54) and (59), we haveTherefore, the operator B has at least one fixed point on . Thus, (1) has at least one positive solution. This completes the proof.
Theorem 2. Suppose that (H0), (H3) and (H4) hold. Then, (1) has at least one positive solution.
Proof. We first show that there is a such thatIf this claim is false, there exist such thatLemma 4 implies thatMoreover, from (H3), we haveMultiply by on both sides of the above and integrate over and use Lemma 2 to obtainThis implies thatSolving this matrix inequality, we obtainNote that , there exist such thatChoose . Then, (61) holds. From Lemma 6, we haveNext, we prove thatwhere is a fixed element. Suppose not. Then, there exist such thatThis implies thatHowever, from (H4), we haveNote that from (H2), we have . Hence, we obtain . Similarly, . This has a contradiction, and thus, (70) holds. From Lemma 5, we obtainFrom (69) and (74), we haveTherefore, the operator B has at least one fixed point on . Thus, (1) has at least one positive solution. This completes the proof.
Let . Then, we have , , and .
Example 1. Let , , , andfor , where . Then, we haveMoreover, note that when , we have , . Consequently, if we choose , then .
Note that we have
As a result, (H1) and (H2) hold.
Example 2. Let , , , , andfor . Then, we have , andMoreover, when , we obtain , and if we choose and , , we also have .
Note that we have
As a result, (H3) and (H4) 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 article.
Acknowledgments
This work was supported by the China Postdoctoral Science Foundation (Grant no. 2019M652348), Technology Research Foundation of Chongqing Educational Committee (Grant no. KJQN201900539), Natural Science Foundation of Chongqing Normal University (Grant no. 16XYY24), and Doctoral Scientific Research Foundation of Qufu Normal University and Youth Foundation of Qufu Normal University (BSQD20130140).