## Recent Advance in Function Spaces and their Applications in Fractional Differential Equations

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Keyu Zhang, Jianguo Wang, Wenjie Ma, "Solutions for Integral Boundary Value Problems of Nonlinear Hadamard Fractional Differential Equations", *Journal of Function Spaces*, vol. 2018, Article ID 2193234, 10 pages, 2018. https://doi.org/10.1155/2018/2193234

# Solutions for Integral Boundary Value Problems of Nonlinear Hadamard Fractional Differential Equations

**Academic Editor:**Xinguang Zhang

#### Abstract

In this paper using fixed point methods we establish some existence theorems of positive (nontrivial) solutions for integral boundary value problems of nonlinear Hadamard fractional differential equations.

#### 1. Introduction

In this work we study the following integral boundary value problems of nonlinear Hadamard fractional differential equationswhere , , and are three positive real numbers with , , and , is the -Laplacian for , and is a continuous function on . Moreover, let with . In what follows, we offer some related definitions and lemmas for Hadamard fractional calculus.

*Definition 1 (see [1, Page 111]). *The th Hadamard fractional order derivative of a function is defined bywhere , , and denotes the largest integer which is less than or equal to . Moreover, we here also offer the th Hadamard fractional order integral of which is defined bywhere is the gamma function.

Lemma 2 (see [1, heorem 2.3]). *Let , . Thenwhere .*

In recent years, there have been some significant developments in the study of boundary value problems for nonlinear fractional differential equations; we refer to [2â€“11] and the references therein. For more related works, see also [12â€“49]. For example, by using monotone iterative methods, Wang et al. [3] investigated a class of boundary value problems of Hadamard fractional differential equations involving nonlocal multipoint discrete and Hadamard integral boundary conditions and established monotone iterative sequences, which can converge to the unique positive solution of their problems. Similar methods are also applied in [4, 5, 12â€“15].

For differential equations with the -Laplacian, see, for example, [6, 7, 15â€“20] and the references therein. In [6], Wang considered the nonlinear Hadamard fractional differential equation with integral boundary condition and -Laplacian operatorwhere grows â€“sublinearly at , and by using the Schauder fixed point theorem, a solution existence result is obtained. In [7], Li and Lin used the Guo-Krasnoselâ€™skii fixed point theorem to obtain the existence and uniqueness of positive solutions for (1) with .

However, we note that these are seldom considered Hadamard fractional differential equations with the -Laplacian in the literature; in this paper we are devoted to this direction. We first utilize the Guo-Krasnoselâ€™skii fixed point theorem to obtain two positive solutions existence theorems when grows â€“superlinearly and â€“sublinearly with the -Laplacian, and secondly by using the fixed point index, we obtain a nontrivial solution existence theorem without the -Laplacian, but the nonlinearity can allow being sign-changing and unbounded from below. This improves and generalizes some semipositone problems [21â€“31].

#### 2. Preliminaries

In this section, we first calculate Greenâ€™s functions associated with (1) and then transform the boundary value problem into its integral form. For this, we give the following lemma.

Lemma 3. *Let , , , , and , be as in (1). Then (1) can take the integral formwhereand*

*Proof. *Use to replace in (1). Let . Then from Lemma 2 we have Note that implies , and then . Therefore, we obtainNext, we calculate and :andThe condition enables us to obtainSubstituting into (10) givesNote that , and hence we obtainThen, if we let , from Lemma 2 we obtainThe condition implies that . Then we substitute into the first derivative of , and we calculate as follows:As a result, from (16) we haveThis completes the proof.

Lemma 4. *Greenâ€™s functions defined by (7) and (8) have the following properties:**(i) are continuous, nonnegative functions on ,**(ii) , for .**From [7, emma 7] and [8, emma 2.2] we easily obtain this lemma, so we omit its proof.**LetThen we obtain the following lemma.*

Lemma 5. *There exist , such that*

*Proof. *We only prove the left inequality above. From Lemma 4(ii) we haveThis completes the proof.

Let be the Banach space equipped with the norm . Then we define two sets on as follows:Consequently, are cones on . From Lemma 3 we can define an operator on as follows:The continuity of implies that is a completely continuous operator and the existence of solutions for (1) if and only if the existence of fixed points for .

Lemma 6 (see [50]). *Let be a Banach space and a bounded open set in . Suppose that is a continuous compact operator. If there exists such thatthen the topological degree .*

Lemma 7 (see [50]). *Let be a Banach space and a bounded open set in with . Suppose that is a continuous compact operator. Ifthen the topological degree .*

Lemma 8 (see [50]). *Let be a Banach space and a cone in . Assume that are open subsets of with , and let be a completely continuous operator such that either**(G1) , , and , ,**or**(G2) , , and , .**Then has a fixed point in .*

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

Let for . Now, we first list our assumptions on :

(H1) ,

(H2) there exist , such that , , uniformly on , where ,

(H3) there exists such that , , where ,

(H4) , , uniformly on , where ,

(H5) there exist , , such that , where

Lemma 9. *Suppose that (H1) holds. Then .*

*Proof. *If , from Lemma 4 we haveOn the other hand,This completes the proof.

*Remark 10. *Our aim is to find operator equation has fixed points in , and from Lemma 9, these fixed points must belong to the cone . Therefore, our work space can be chosen rather than .

In what follows, we discuss the existence of positive solutions for (1) in .

Theorem 11. *Suppose that (H1)-(H3) hold. Then (1) has at least two positive solutions.*

*Proof. *From (H3), when , we haveHence, we obtainOn the other hand, by the second limit inequality in (H2), there exists such thatNote that if , from the definition of we haveThis, together with (31), implies thatBy the first limit inequality in (H2), there exist and such thatNote that can be chosen large enough, and if , together with (32), there exists such thatCombining this and (33), we findwhere . Consequently, we haveIn summary, from (30), (33), and (37) with , Lemma 8 enables us to obtain that (1) has at least two positive solutions in and . This completes the proof.

Theorem 12. *Suppose that (H1), (H4)-(H5) hold. Then (1) has at least two positive solutions.*

*Proof. *If , we have , and . Hence, from (H5) we obtainThis indicates thatOn the other hand, by the second limit inequality in (H4), there exists such thatThis, if , implies thatThis givesBy the first limit inequality in (H4), there exist and such thatConsequently, if with large enough, we obtainwhere . Hence, we haveIn a word, from (39), (42), and (45) with , Lemma 8 enables us to obtain that (1) has at least two positive solutions in and . This completes the proof.

*Example 13. * Letwhere , , and are defined by (H3). ThenMoreover, for , we haveTherefore, (H1)-(H3) hold.

*Example 14. * Letwhere , , and are defined by (H5). ThenMoreover, for we haveTherefore, (H1), (H4)-(H5) hold.

#### 4. Nontrivial Solutions for (1)

In this section we consider the boundary value problem (1) without the -Laplacian, i.e., . In this case, (1) can be transformed into its integral form as follows:As said in Section 3, we define an operator, still denoted by , as follows:In what follows, we aim to find the existence of fixed points of . For this, we list our assumptions on :

(H6) ,

(H7) There exist nonnegative functions with and such thatMoreover,

(H8) , uniformly in ,

(H9) , uniformly in .

Theorem 15. *Suppose that (H6)-(H9) hold. Then (1) has at least one nontrivial solution.*

*Proof. *From (H9) there exist and such thatFor this , we show thatIf otherwise, there exist , such thatand hence, we obtainMultiply both sides of the above inequality by and integrate from to and together with Lemma 5 we obtainThis implies that , and for the fact that , for , which contradicts