An upper-lower solution method for the eigenvalue problem of Hadamard-type singular fractional differential equation

. In this paper, we are concerned with the eigenvalue problem of Hadamard-type singular fractional differential equations with multi-point boundary conditions. By constructing the upper and lower solutions of the eigenvalue problem and using the properties of the Green function, the eigenvalue interval of the problem is established via Schauder’s ﬁxed point theorem. The main contribution of this work is on tackling the nonlinearity which possesses singularity on some space variables.

In recent years, fractional-order nonlinear problems have attracted the attention of many researchers from mathematics and other applied science due to its wide range of applications in applied mathematics, physics, bioscience, engineering, chemistry, etc. A large number of contributions have been made for fractional differential equations in the sense of the Riemann-Liouville fractional derivative or the Caputo fractional derivative, [1, 4-6, 8, 9, 12-20]. However, the Hadamard-type fractional integral and derivative differ from the Riemann-Liouville and the Caputo fractional derivative since the kernels of the Hadamard-type integral and derivative contain logarithmic functions of arbitrary exponent and thus are regarded as a different kind of weakly singular kernels. Thus it is more difficult to explore the existence of solutions for the Hadamard-type fractional differential equations, [2,10,11,21].
In the recent work [21], by analysing the spectral construct of a linear operator and calculating the fixed point index of the corresponding nonlinear operator, Zhang et al. considered the existence of positive solutions for the following Hadamard-type fractional differential equation: where 2 < α, β 3, σ is a differential operator denoted by t(d/dt), that is, σz(t) = t(d/dt)z(t), D t α and D t β are the Hadamard fractional derivatives of order α, β, f ∈ (1, e) × (0, +∞) × (0, +∞), [0, +∞) is a continuous function, and the criteria of the existence of positive solutions were established. Recently, based on Leray-Schauder-type continuation, El-Sayed and Gaafar [3] established the existence of positive solutions to a class of singular nonlinear Hadamard-type fractional differential equations with infinitepoint boundary conditions or integral boundary conditions. However when f possesses singularities on space variables, especially for the eigenvalue problem, few results are established on Hadamard-type fractional differential equations. Inspired by the above works, the aim of this paper is to establish the existence of positive solutions for the eigenvalue problem of the Hadamard-type fractional differential equation (1) when f possesses singularity on space variables.
The rest of this paper is organized as follows. In Section 2, we firstly recall the concepts and properties of Hadamard fractional integral and derivative and then give the logarithmic Green kernel. Our main results are summarized in Section 3. https://www.journals.vu.lt/nonlinear-analysis

Basic definitions and preliminaries
In this section, we firstly present the definition of Hadamard-type fractional integral and derivatives as given in [7]. Then we give some basic lemmas, which will be used in the rest of the paper.
Suppose α ∈ C, n = [Re α], Re α > 0, and (a, b) is 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 The relationship between fractional integration and derivative is introduced as follows.
The following formula holds: subject to the multi-point boundary conditions has a unique solution x ∈ AC [1, e] if and only if x is a solution of the integral equation The Green's functions G has the following properties: (ii) For all t, s ∈ (1, e), the following inequalities hold: We make the following assumptions throughout this paper: (H1) f : [1, e] × (0, +∞) → [0, +∞) is continuous and is nonincreasing in x > 0; (H2) For all r ∈ (0, 1), there exists a constant > 0 such that, for any (t, Remark 1. For r 1, by (H2), we have the following equivalent conclusion: for any In fact, for r 1 and any (t, Proof. By Lemma 2, the conclusion is obvious, and we thus omit the proof here. https://www.journals.vu.lt/nonlinear-analysis

Main results
Let then we state our main result as follows.
Define an operator T λ in E: It follows from Lemma 2 that the fixed point of the operator T λ is the solution of the eigenvalue problem (1).
In what follows, we prove that the operator T λ is well defined and T λ (Q) ⊂ Q. To do this, for any x * ∈ Q, it follows from the definition of Q that there exists a positive number l * x * such that x * (t) l * x * κ(ln t) for any t ∈ [1, e]. Choose l x * = min{1/2, l * x * }, then we have x * (t) l x * κ(ln t) for any t ∈ [1, e]. So by Lemma 3, (H2) and (H3), we gets Next, take B = max{2, max t∈ [1,e] x * (t)}, then it follows from Lemma 3 and (H2) that (5) and (7) indicate that T λ is well defined and T λ (Q) ⊂ Q.
https://www.journals.vu.lt/nonlinear-analysis Now we shall try to construct the upper and lower solutions of the eigenvalue problem (1). As the operator T λ is decreasing on x, let then, similar to 7, for all t ∈ [1, e], one gets i.e., where On the other hand, notice that f (t, x) is decreasing in x > 0, thus, for any λ > λ 1 , it follows from Lemma 3 and (H3) that By (H2), for any t ∈ [1, e], we have Thus it follows from Lemma 3 that Let then by Lemma 2, for any t ∈ [1, e], we have It follows from (8) and (9) that φ(t), ψ(t) ∈ Q and which implies Thus, by (10), (11), we have −λ * f t, κ(ln t) + λ * f t, κ(ln t) = 0.
Next, construct a function F : For any λ ∈ (λ 1 , λ * ), consider the following modified eigenvalue problem: We define an operator A λ in E: It follows from the assumption that F : [0, +∞) → [0, +∞) is continuous. Thus it is clear that a fixed point of the operator A λ is a solution of the modified eigenvalue problem (15). For all y ∈ E, it follows from Lemma 3, (14) and ψ(t) κ(ln t) that So A λ is bounded. It is easy to see that A λ : E → E is continuous from the continuity of F (y) and G(t, s). To do this, let w(t) = φ(t) − y(t), t ∈ [1, e]. Since φ(t) is the upper solution of the eigenvalue problem (1) and y is a fixed point of A λ , we have It follows from the definition of F , (10) and (11) that i.e., which implies that − H D α w(t) 0. It follows from Lemma 4 that w(t) 0, that is, y(t) φ(t) on [0, 1]. By the same way, we have y(t) ψ(t) on [0, 1], thus we get By (14), we have F (y(t)) = f (t, y(t)), t ∈ [1, e]. Consequently, y(t) is a positive solution of the eigenvalue problem (1). Finally, we prove the asymptotic properties of solutions. Firstly, from (19) we get On the other hand, it follows from (20) and Lemma 3 that Thus we get the asymptotic properties of solutions κ(ln t) y(t) ρ(ln t) α−1 .