Abstract
Combining the properties of the Green function with some point theorems, we consider the existence of nontrivial solutions for fractional equations with -Laplacian operator , , and , , where are constants and is continuous.
1. Introduction
Fractional-order models are better than integer order models to describe the real word, which appear frequently in various fields, such as electrical circuits, biology, material, control theory, and physics (see [1–5]). With the rapid development of the theory of fractional differential equations, during the last two decades, the existence of nontrivial solutions of fractional differential equations has been studied by many researchers in nonsingular case as well as singular case. See [6–19]. Usually, the proof is based on either the method of upper and lower solutions, fixed point theorems, alternative principle of Leray-Schauder, topological degree theory, or critical point theory. To our attention, based on a fixed point theorem in cones, K. Lan and W. Lin [20] obtain some new results on existence of multiple positive solutions of systems of nonlinear Caputo fractional differential equations with some of general separated boundary conditions where , is continuous on , and is the Caputo differential operator of order . are positive real numbers. The relations between the linear Caputo fractional differential equations and the corresponding linear Hammerstein integral equations are studied, which show that suitable Lipschitz type conditions are needed when one studies the nonlinear Caputo fractional differential equations.
Recently, fractional differential equations with -Laplacian operator have gained its popularity and importance due to its distinguished applications in numerous diverse fields of science and engineering, such as viscoelasticity mechanics, non-Newtonian mechanics, electrochemistry, fluid mechanics, combustion theory, and material science. There have appeared some results for the existence of solutions or positive solutions of boundary value problems for fractional differential equations with -Laplacian operator; see [15, 21–26] and the references therein. For example, under different conditions K. Hasib, W. Chen and H. Sun [21] apply some classical fixed-point theorems to study the existence of positive solution for a class of singular fractional differential equations with nonlinear -Laplacian operator in Caputo sense where is Caputo fractional derivative, is -Laplacian operator, and is continuous functions. In addition, Hyers-Ulam stability of the proposed problem is also considered.
Inspired by the references, based on some fixed point theorems in cones, under different combinations of local superlinearity and local sublinearity of the function , we will deal with the existence of nontrivial solutions for a certain -Laplacian fractional differential equation where are constants with satisfying , is continuous, and is -Laplacian operator, where and denotes inverse of -Laplacian operator. Now we give some notations as follows.
Definition 1 (see [2]). The fractional integral of order of a function is given byprovided that the right-hand side integral is pointwise defined on , where .
Definition 2 (see [2]). The Caputo fractional derivative of order of a continuous function is given by for , where is used for the integer part of , provided that the right-hand side integral is pointwise defined on
Lemma 3 (see [2, Theorem 2.22]). Let , Thenwhere ,
The paper is organized as follows. In Section 2, we give some notations and the Green function is examined whether it is increasing or decreasing and positive or negative function. In Section 3, we will give the main results, which are illustrated by some examples.
2. Preliminaries
Lemma 4. Let and . Then the solution of the fractional differential equation with -Laplacian operator can be expressed bywhere
Proof. Via some computations, from Lemma 3 it follows thatSince , we have Integrating both sides from to , we can obtain Due to , we get and According to the boundary conditions , we have and . Sinceit is clear thatSince , we have which follows that Then substituting and , we obtainwhich implies the expression of the Green function .
Lemma 5. Assume that , and The Green function has the following properties: (i), for ;(ii) is an increasing function and ;(iii)For , there exists a = ∈ such that
Proof. (i) For ,Since and , we have and
For , LetVia some computations, we get Let + + . Since = and + , we obtain , which implies that and is increasing on . Furthermore, for , we have (ii) For , since we obtain that is increasing on in , which implies that .
For , from the proof of (i) it follows that is increasing on in and .
Then from the above discussion, we can obtain the conclusion(iii) For , For , LetOn one hand, since we have . On the other hand, it is not hard to obtain that . Then is increasing on , which implies that Hence, we get . Furthermore, is increasing on and Then, we have Therefore, from the above discussion, we get the conclusion and it is clear to see that
At the end of this section, we give some notations and crucial lemmas.
Let be the Banach space and it endowed with the norm Define a subcone as For any given , let
Lemma 6 (see [27]). Let be a Banach space, a closed, convex subset of , an open subset of , and . Suppose that is completely continuous. Then either (i) has a fixed point in , or(ii)there are an (the boundary of in ) and with .
Lemma 7 (see [27]). Let be a Banach space and be a cone in . Assume are open subsets of with , , and let be a completely continuous operator such that either (i), and , ; or(ii), and ,
Then has a fixed point in .
Lemma 8 (see [21]). Let be a Laplacian operator. Then(i)If and , then(ii)If , and , then
3. Existence Results
For convenience, the following assumptions hold throughout this paper: is continuous; is continuous;there exists positive constants and such that satisfiesthere exists a positive constant such that for all ,
In addition, let
Theorem 9. Suppose that , , and hold and . Then problem (3) has a unique solution if
Proof. By Lemma 4, (3) is equivalent to the following integral equation: Define an operator byFrom and , the operator is well defined.
Let with . Then we have Let , and then we have which implies that Therefore, we proved that
For any , by Lemma 8, we have Since , from Banach’s contraction mapping principle it follows that there exists a unique fixed point for the operator , which corresponds to the unique solution for problem (3).
Lemma 10. Assume that and hold. Then the operator is completely continuous.
Proof. For any , according to Lemma 5, we can get and So we have This implies
Given , now we show that is completely continuous on .
Firstly, we will show that is continuous, and we only need to prove that for any as It is clear that From the continuity of , , we haveas Therefore, , as , and is continuous.
Next, we show that the operator is uniformly bounded. By , we get which implies that the operator is uniformly bounded on .
Finally, we show that the operator is equicontinuous. Before that, we will proceed with being bounded.
For , Since the function , and are bounded, there exists a positive constant such that . In a similar way, for , there exists a positive constant such that Choosing , we can obtain
Finally, for any with , there exists a such that Thus, the operator is equicontinuous. According to Arzela-Ascoli theorem, is compact.
Theorem 11. Suppose that and hold. In addition,(C1)there exists a continuous function such that(C2)there exists a constant such that Then problem (3) has at least one solution.
Proof. Now we show the (ii) of Lemma 6 does not hold. If is a solution of (3), then, for , we have Let . From the above inequality and , it yields a contradiction. Therefore, the operator has a fixed point in .
For , define the following functions:
Theorem 12. Suppose that and hold. In addition, there exist such that one of the following conditions satisfied:
and
or
and
Then problem (3) has a positive solution such that
Proof. We only verify the case . On one hand, for any , we have and Thus, , for any .
On the other hand, for any , we have , and Then, , for any .
Therefore, By Lemma 7, the operator has a fixed point with .
Theorem 13. Suppose that holds. In addition ;there exists a constant such that ;there exists and such that whereThen problem (3) has at least two solutions.
Proof. Since , there exist and such that , for , where satisfies For , we have Choosing . For , we have For any , choosing , we have Furthermore, we have By Lemma 7, problem (3) has at least two positive solution and .
At the end of this section, we give some examples to illustrate our main results.
Example 1. Let us consider the problem It is clear to see that , for , . In addition, there exists a sufficiently large , we have . Therefore, problem (69) has at least one solution by Theorem 11.
Example 2. Let us consider the problem Since , we have = as , and hold. By some calculations, we get = , and it is clear to see that , for , and , for Let , and then for any , we have = ≥ ≥ , and holds. Therefore, problem (70) has at least two positive solutions by Theorem 13.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Authors’ Contributions
All the authors contributed equally and significantly to writing this article. All the authors read and approved the final manuscript.
Acknowledgments
Fanglei WANG was supported by NNSF of China (no. 11501165) and the Fundamental Research Funds for the Central Universities (2018B58614).