#### Abstract

In this paper, we consider the existence of positive solutions for a Hadamard-type fractional differential equation with singular nonlinearity. By using the spectral construct analysis for the corresponding linear operator and calculating the fixed point index of the nonlinear operator, the criteria of the existence of positive solutions for equation considered are established. The interesting point is that the nonlinear term possesses singularity at the time and space variables.

#### 1. Introduction

In this paper, we focus on the existence of positive solutions for the following singular Hadamard-type fractional differential equation: where ; is a differential operator denoted by , that is, ; are the Hadamard fractional derivative of order ; and is a continuous function with singularity at and .

Singularity refers to a point or a domain where the given mathematical object is not defined or not “well-behaved.” Near a singular point or zone, a minor change of the variable will lead to major changes of the property of the target object. Many physical phenomena in natural sciences and engineering often exhibit some singular behaviour. For example, Fisk [1] found that in certain materials the quantum fluctuations at absolute zero may push a system into a different phase or state, as result, the process loses its continuity, and then, the singular behaviour happens near the quantum critical points. In fluid mechanics, when a fluid is subjected to a severe impact to form a fracture, singular points or singular domains also follow the fracture. Normally, at singular points and domains, the extreme behaviour such as blow-up phenomena [2, 3], impulsive influence [4–9], and chaotic system [10–13], often leads to some difficulties for people in understanding and predicting the corresponding natural problems. Hence, the study of singularity for complex systems governed by differential equations [14–27] is important and interesting in deepening the understanding of the internal laws of dynamic system.

On the other hand, since the fractional differential operator is nonlocal, some often use it to describe viscoelastic behaviour and memory phenomena in various natural science fields such as the silicone gel with the property of weak frequency dependency [28, 29] and advection dispersion in anomalous diffusion [30–34]. In most cases, some are interested in the qualitative properties of solutions for the corresponding fractional equations; for the detail, see [35–65]. In particular, in order to obtain the qualitative properties of solutions, many nonlinear analysis methods, such as fixed point theorems [66–71], iterative techniques [72–80],variational methods [81–98], and upper and lower solution methods [29, 44], have been developed and employed to study the qualitative properties and numerical results of solutions for various types of differential equations. For example, by using the fixed point index theory, Wang [69] established the existence and multiplicity of positive solutions for the following nonlocal singular fractional differential equation: where denotes the standard Riemann-Liouville fractional derivative, may be singular at , and is nonnegative. In recent years, to enrich the theory of fractional calculus, the fractional derivative and integral were extended to many different forms such as Hadamard, Erdelyi-Kober, Hilfer derivatives, and integrals. In particular, it is more difficult to obtain the qualitative properties of solutions for Hadamard-type fractional differential equations since Hadamard derivatives possess a singular logarithmic kernel [99–106].

In this paper, we focus on the existence of positive solutions for the Hadamard-type fractional differential equation (1) with singularity in space variables. Our work has some new contributions. Firstly, the equation contains a Hadamard-type fractional derivative which has a singular logarithmic kernel. Secondly, the nonlinearity can have strong singularity in time and space variables. Thirdly, a new limit condition of integral type is introduced to overcome the difficulty of singularity. The rest of this paper is organized as follows. In Section 2, we firstly introduce the concept of Hadamard fractional integral and differential operators and then give the logarithmic kernel and Green function of the boundary value problem and their properties. Our main results are summarized in Section 3.

#### 2. Preliminaries and Lemmas

Before the main results, we firstly recall the definition of the Hadamard-type fractional integrals and derivatives; for detail, see [107].

Let and be a finite or infinite interval of . The -order left Hadamard fractional integral is defined by and the left Hadamard fractional derivative is defined by

In what follows, we consider the following linear auxiliary problem:

It follows from [99] that problem (5) has a unique solution where is Green’s function of equation (5). Now, let , then the Hadamard-type fractional differential equation (1) reduces to the following convenient form:

It follows from (6) that equation (8) is equivalent to the following integral equation: where

As a result, in order to find the positive solutions of equation (1), it is sufficient to search the fixed point of the following operator:

Lemma 1 (see [99]). *Let Then, Green’s functions has the following properties:
*(i)*For all , the following inequalities hold:**Let be a cone of Banach space , for any , define
**Now, we state the following lemmas which will be used in the rest of the paper.*

Lemma 2 (see [108]). *Assume is a completely continuous operator.
*(i)*If there exists such that**then, the fixed point index *(ii)*If**then, the fixed point index *

Lemma 3 (Krein-Rutmann, see [108]). *Let be a continuous linear operator, be a total cone, and If there exist and a positive constant such that , then the spectral radius and has a positive eigenfunction corresponding to its first eigenvalue *

Lemma 4 (Gelfand’s formula, see [108]). *For a bounded linear operator and the operator norm , the spectral radius of satisfies
*

In this paper, we use the following assumption:

(B1) and for any , where .

Now, let , then is a Banach space equipped with the norm . Let and then, is a cone in the Banach space and . Let us define a nonlinear operator and a linear operator :

Thus, in order to solve equation (1), we only need to find the fixed point of operator equation . To do this, we firstly establish some lemmas.

Lemma 5. * is a completely continuous operator with the spectral radius . Moreover, has a positive eigenfunction corresponding to the first eigenvalue .*

*Proof. *Firstly, it follows from Lemma 2 that, for any , one has
which imply that . Since is uniform continuity, then the operator is completely continuous.

On the other hand, by (10), we know that there exists a such that Thus, from the continuity of , there exists a closed interval such that and for all Now, we take such that and for all . Then, for all one has
Thus, there exists such that for From Krein-Rutmann’s theorem, we know that the spectral radius and have a positive eigenfunction that satisfies where is its first eigenvalue. The proof is completed.

Lemma 6. *Suppose that (B1) holds, then the operator is completely continuous.*

*Proof. *Firstly, for any , it follows from Lemma 2 that
that is
Similarly, one also has
which implies and then .

On the other hand, from (B1), we know that there exists a natural number such that
Thus, for any and , we have
Take
where
So, it follows from (26), (27) and (28) that
which implies is uniformly bounded for any bounded set.

Secondly, we shall prove that is continuous. To do this, let and (). For any , it follows from (B1) that there exists a natural number such that
On the other hand, it follows form the fact that is uniformly continuous on (21) that
holds uniformly on . Thus, the Lebesgue control convergence theorem ensures
In other words, for the above , there exists a positive integer such that , we have
Thus, by (32) and (35), for any , we have
Therefore, is continuous.

In the end, we shall prove that is equicontinuous. Firstly, it follows from (B1) that for any , there exists a positive integer such that
Take
where
Notice that is uniformly continuous on , so for the above and fixed , there exists , when , we have
It follows from the above argument that, for , one has
which implies that is an equicontinuous operator. According to the Arzela-Ascoli Theorem, is completely continuous. The proof is completed.

#### 3. Main Results

We state the main results of this paper as follows.

Theorem 7. *Let be the first eigenvalue of , assume (B1) holds and
uniformly holds on . Then, equation (1) has at least one positive solution.*

*Proof. *Firstly, by Lemma 6, we know that is completely continuous. Thus, it follows from the extension theorem of a completely continuous operator (see Theorem 8.3 on page 56 of [108]) that, for any , there exists a completely continuous operator . Thus, without loss of the generality, we shall still write it as .

Next, it follows from (42) that there exists such that
Since, for any , we have
thus, for , from (43) and (44), one gets
Let be the positive eigenfunction corresponding to , i.e., . In what follows, we shall use the contradiction method to show
Firstly, we can suppose that has no fixed points on (otherwise, the proof is finished). In this case, let us suppose there exist and such that . This implies that and then
Let , then , . Thus, it follows from (45) that
which contradicts with the definition of . So (46) holds and from Lemma 2, we have
On the other hand, it follows from (42) that there exists and such that
Let Obviously, is also a bounded linear operator and . Since is the first eigenvalue of operator and , we have
From Gelfand’s formula, we have
Now, choose , it follows from (52) that there exists an enough large integer such that when , one has
Let be the identity operator and define
then is still the norm of . Let
then from (31), we have . Take
where . Notice that , so we can choose a large enough such that implies .

Next, we shall show
Otherwise, there exist and such that . Let and
Since , , there exists such that . So for any , we get and , which implies , i.e., . Thus by Lemma 2, we have
So it follows from and (59), we have
Since is a normal cone with normality constant 1 and (60), one has
which leads to
According to the selection of , we have Thus, it follows from the fact implies and (53), (54), and (62) that
Notice that , we have , and then , which is a contradiction with . So (57) is indeed valid and it follows from Lemma 2 that
Thus, (49) and (64) lead to
which implies has at least one fixed point on . Consequently, equation (1) has at least one positive solution.

Theorem 8. *Let be the first eigenvalue of , suppose (B1) holds, and
uniformly on where is any eigenvalues of . Then, equation (1) has at least one positive solution.*

To prove Theorem 8, we need some preliminaries and lemma. For any enough small , define

By Lemma 5, we know is still a completely continuous linear operator with the spectral radius , and has a positive eigenfunction corresponding to its first eigenvalue .

Lemma 9. *There exists an eigenvalue of such that
*

*Proof. *Let satisfying . Thus, for and , one has
and then
where It follows from the fact that is a normal cone with normality constant 1 that
By Gelfand’s formula, we have where is the first eigenvalue of , which implies is monotonous with lower boundedness . Let
we assert is an eigenvalue of .

In fact, let be positive eigenfunctions of corresponding to with . So we have
Since
is uniformly bounded.

Moreover, for any and , one has
Notice that is uniformly continuous on ; then, (75) implies is equicontinuous. Thus, from the Arzela-Ascoli theorem and , we know as , which yields . Take the limit for two sides of (73), we obtain
that is, and is an eigenvalue of .

*Proof of Theorem 8. *Firstly, it follows from (66) that for any , there exists such that
Thus, for any , since
we have
In the following, we prove that
Firstly, we may suppose that has no fixed point on (otherwise, the proof is completed). If (80) is not true, i.e., there exist and satisfying . Thus , it follows from (79) that
From induction, we have
which implies that
Thus, by Gelfand’s formula, we have
which contradicts with . Therefore, (