Solutions for a Singular Hadamard-Type Fractional Differential Equation by the Spectral Construct Analysis

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.

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][5][6][7][8][9], and chaotic system [10][11][12][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][15][16][17][18][19][20][21][22][23][24][25][26][27] is important and interesting in deepening the understanding of the internal laws of dynamic system.
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.

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 α ∈ ℂ, ReðαÞ > 0, n = ½ReðαÞ and ða, bÞ be a finite or infinite interval of ℝ + . The α-order left Hadamard fractional integral is defined by and the α 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 xðtÞ = −D t β zðtÞ, then the Hadamard-type fractional differential equation (1) reduces to the following convenient form: x 1 ð Þ = σz 1 ð Þ = σx e ð Þ = 0: 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 ψ i ðtÞ = ln tð1 − ln tÞ i−2 , i = α, β, t ∈ ½ 1, e: Then, Green's functions H, G has the following properties: 2 Journal of Function Spaces (i) For all t, s ∈ ½1, e, the following inequalities hold: Let Q be a cone of Banach space E, for any 0 Now, we state the following lemmas which will be used in the rest of the paper.
then, the fixed point index iðT, Q r , QÞ = 0 then, the fixed point index iðT, Q r , QÞ = 1 Lemma 3 (Krein-Rutmann, see [108]). Let L : E → E be a continuous linear operator, P be a total cone, and LðPÞ ⊂ P: If there exist ψ ∈ E \ ð−PÞ and a positive constant c such that cLðψÞ ≥ ψ, then the spectral radius ρðLÞ ≠ 0 and has a positive eigenfunction corresponding to its first eigenvalue λ = ρðLÞ −1 : Lemma 4 (Gelfand's formula, see [108]). For a bounded linear operator L and the operator norm k·k, the spectral radius of L n satisfies In this paper, we use the following assumption: (B1) f ∈ Cðð0, 1Þ × ð0,+∞Þ × ð0,+∞Þ, ½0,+∞ÞÞ and for any 0 < r < R < +∞, where Ω n = ½1, 1 + ð1/nÞ ∪ ½e − ð1/nÞ, e. Now, let E = C½1, e, then E is a Banach space equipped with the norm kxk = max t∈½1,e | xðtÞ | . Let P = fx ∈ E : xðtÞ ≥ 0, t ∈ ½1, eg and then, Q is a cone in the Banach space E and Q R Q r ⊂ Q ⊂ P. Let us define a nonlinear operator T : Q R \ Q r → P and a linear operator L : E → E: Thus, in order to solve equation (1), we only need to find the fixed point of operator equation x = Tx. To do this, we firstly establish some lemmas. Proof. Firstly, it follows from Lemma 2 that, for any x ∈ Q, one has which imply that L : Q → Q. Since Hðt, sÞ, t ∈ ½1, e × ½1, e is uniform continuity, then the operator L : Q → Q is completely continuous.
Proof. Firstly, for any x ∈ Q R \ Q r , t ∈ ½1, e, it follows from Lemma 2 that that is Similarly, one also has which implies TðQÞ ⊂ Q and then Tð Q R \ Q r Þ ⊂ Q. On the other hand, from (B1), we know that there exists a natural number l such that Thus, for any x ∈ Q R Q r and 1 + ð1/lÞ ≤ t ≤ e − 1/l, we have Take where So, it follows from (26), (27) and (28) that which implies T is uniformly bounded for any bounded set. Secondly, we shall prove that T : Q R \ Q r → Q is continuous. To do this, let x n , x 0 ∈ Q R Q r and ∥x n − x 0 ∥→ 0 (n → ∞). For any ε > 0, it follows from (B1) that there exists a natural number m > 0 such that In other words, for the above ε > 0, there exists a positive integer N such that n > N, we have Thus, by (32) and (35), for any n > N, we have Therefore, T : Q R Q r → Q is continuous. In the end, we shall prove that T is equicontinuous. Firstly, it follows from (B1) that for any ε > 0, there exists a positive integer k such that Take where Notice that Gðt, sÞ is uniformly continuous on ½1, e × ½1, e, so for the above ε > 0 and fixed s ∈ ½1 + ð1/kÞ, e − ð1/kÞ, there exists δ > 0, when |t 1 − t 2 | <δ, t 1 , t 2 ∈ ½1, e, we have It follows from the above argument that, for |t 1 − t 2 | <δ, t 1 , t 2 ∈ ½1, e, one has which implies that T is an equicontinuous operator. According to the Arzela-Ascoli Theorem, T : Q R \ Q r → Q is completely continuous. The proof is completed.

Main Results
We state the main results of this paper as follows.
uniformly holds on t ∈ ½1, e. Then, equation (1) has at least one positive solution.
Proof. Firstly, by Lemma 6, we know that T : Q R \ Q r → Q 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 R > 0, there exists a completely continuous operator T * : Q R → Q. Thus, without loss of the generality, we shall still write it as T.
Next, it follows from (42) that there exists r > 0 such that Since, for any x ∈ ∂Q r , we have thus, for x ∈ ∂Q r , from (43) and (44), one gets Let ω * be the positive eigenfunction corresponding to μ 1 , 5 Journal of Function Spaces i.e., ω * = μ 1 Lω * . In what follows, we shall use the contradiction method to show Firstly, we can suppose that T has no fixed points on x ∈ ∂Q r (otherwise, the proof is finished). In this case, let us suppose there exist x 0 ∈ ∂Q r and μ 0 ≥ 0 such that x 0 − Tx 0 = μ 0 ω * . This implies that μ 0 > 0 and then Let Thus, it follows from (45) that which contradicts with the definition of μ. So (46) holds and from Lemma 2, we have On the other hand, it follows from (42) that there exists R 1 > r and 0 < κ < 1 such that LetLx = κμ 1 Lx: Obviously,L : E → E is also a bounded linear operator andLðQÞ ⊂ Q. Since μ 1 is the first eigenvalue of operator L and 0 < κ < 1, we have From Gelfand's formula, we have Now, choose ε 0 = 1/2ð1 − κÞ, it follows from (52) that there exists an enough large integer N > 0 such that when n ≥ N, one has LetL 0 = I be the identity operator and define then k·k * is still the norm of E. Let then from (31), we have M < +∞. Take where M * = kMk * . Notice that kxk * > ½κ + ε 0 N−1 kxk, so we can choose a large enough R > R 2 such that kxk ≥ R implies kxk * > R 2 . Next, we shall show Otherwise, there exist x 1 ∈ ∂Q R and μ * ≥ 1 such that T x 1 = μ * x 1 . LetxðtÞ = min fx 1 ðtÞ, R 1 g and Since x 1 ∈ C½1, e, x 1 ðtÞ ≤ kx 1 k = R, there exists 1 < t 0 ≤ e such that x 1 ðt 0 Þ = R. So for any t ∈ ½1, e, we getxðtÞ = min fx 1 ðtÞ, R 1 g ≤ min fR, R 1 g = R 1 andxðt 0 Þ = min fx 1 ðt 0 Þ, R 1 g = min fR, R 1 g = R 1 , which implies kxðtÞk = R 1 , i.e.,x ∈ ∂ Q R 1 . Thus by Lemma 2, we have So it follows fromLðQÞ ⊂ Q and (59), we have Since Q is a normal cone with normality constant 1 and (60), one has which leads to According to the selection of R 2 , we have M * < ðε 0 /2ÞR 2 : Thus, it follows from the fact kxk ≥ R implies ∥x∥ * > R 2 and 6 Journal of Function Spaces (53), (54), and (62) that Notice that μ * ≥ 1, we have ð1/4Þκ + 3/4 ≥ 1, and then κ ≥ 1, which is a contradiction with 0 < κ < 1. So (57) is indeed valid and it follows from Lemma 2 that Thus, (49) and (64) lead to which implies T has at least one fixed point on Q R \ Q r . Consequently, equation (1) has at least one positive solution.
Theorem 8. Let μ 1 be the first eigenvalue of L, suppose (B1) holds, and uniformly on t ∈ ½1, e, whereμ 1 is any eigenvalues of L. Then, equation (1) has at least one positive solution.
To prove Theorem 8, we need some preliminaries and lemma. For any enough small 0 < ε < 1, define By Lemma 5, we know L ε : Q → Q is still a completely continuous linear operator with the spectral radius ρðL ε Þ ≠ 0, and L ε has a positive eigenfunction ω ε corresponding to its first eigenvalue μ ε = ðρðL ε ÞÞ −1 .