Solutions for a category of singular nonlinear fractional differential equations subject to integral boundary conditions

We concentrate on a category of singular boundary value problems of fractional differential equations with integral boundary conditions, in which the nonlinear function f is singular at t=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$t=0$\end{document}, 1. We use Banach’s fixed-point theorem and Hölder’s inequality to verify the existence and uniqueness of a solution. Moreover, also we prove the existence of solutions by Krasnoselskii’s and Schaefer’s fixed point theorems.


Introduction
The current work concentrates on the existence and uniqueness of solutions for a category of singular nonlinear fractional differential equations (NFDEs) subject to integral boundary conditions (BCs). Specifically, we discuss the problem x(t)), 0 < t < 1, x(0) = x (0) = 0, (1.1) where c D α 0 + stands for the Caputo derivative of order α, α and γ are real numbers satisfying 2 < α ≤ 3 and 0 < γ < 1, respectively, and the function f (t, x(t)) has singular characteristics verified the existence of solutions for the BVP ⎧ ⎨ ⎩ c D q 0 + x(t) = f (t, x(t)), 0 < t < 1, 0 < q ≤ 1, where c D α 0 + stands for the Caputo derivative, and f : [0, 1] × R → R is a continuous function. Various fixed point theorems state the existence and uniqueness of solutions.
Vong [32] verified the following nonlocal BVP for a class of singular NFDEs: where n ≥ 2, α ∈ (n -1, n), μ(s) denotes a bounded-variation function, which can be singular at t = 0. Motivated by all the mentioned studies, we aim to demonstrate the existence and uniqueness of solutions to problem (1.1). We use some typical fixed point theorems and the generalized Hölder inequality to obtain fundamental results.

Preliminaries
This subsection contains the required concepts and features of the fractional calculus and some lemmas necessary to prove our essential results.
respectively, where is the gamma function.
respectively, where D stands for the derivative operator, and n = [

Lemma 2.1 ([1])
The general solution of the fractional-order equation ( c D α a + y)(x) = 0 can be obtained as In particular, for a = 0, it can be presented as i! (i = 0, 1, . . . n -1) stand for certain constants.
Proof By Lemma 2.1 we easily get for some c 0 , c 1 , c 2 ∈ R. From the BCs in (2.1) we have c 0 = c 1 = 0 and Hence Integrating both sides of (2.2) from γ to 1 yields By switching and rearranging this equation we have Substituting this equation into equation (2.2), we get The proof is finished.
The conclusions of this paper are mainly derived from the following fixed point theorems.

bounded set. Then T has a fixed point in X.
Finally, we introduce some basic knowledge of L p space and present the Hölder inequality and its generalized form [36].
Let ⊂ R n be an open set (or a measurable set), let f (x) be a real-valued measurable function on . For 1 ≤ p < ∞, since |f (x)| p is also measurable on , the integral |f (x)| p dx makes sense. Then the function space L p ( ) is defined as follows: For f ∈ L p ( ), the following norm can be defined:

Lemma 2.6 ([36] Hölder's inequality) Let ⊂ R n be an open set, let p, q be conjugate
This inequality can be generalized as follows: provided that f i (x) ∈ L p i ( ), 1 < p i < ∞, and n k=1 1 p i = 1.

Fundamental results
Let Throughout this paper, we make the following assumption on the singularity of nonlinear function f (t, x(t)) in (1.1): (H1) f (t, x(t)) has a singularity at t = 0 and t = 1, that is, Moreover, there exist constants 0 < θ 1 < 1 and 0 < θ 2 < 1 such that Based on condition (H1), we know that there is a positive constant M 0 such that Then the solutions of problem (1.1) include the FPs of A.
that is, Ax ≤ L 2 , for all x ∈ D. Thus the operator A is bounded on D. This yields the compactness of A. For every t ∈ [0, 1], we have Now the following inequality holds for t 1 , t 2 ∈ [0, 1] and t 1 < t 2 : (Ax)(t 2 ) -(Ax)(t 1 ) = Therefore A is equicontinuous on D. Thus, by the Arzelà-Ascoli theorem the operator A is completely continuous on X.
Proof For x, y ∈ X = C([0, 1]) and t ∈ [0, 1], by (H2) we have By (H3) and the Hölder inequality we have Noticing (3.3), we conclude that A is a contraction mapping. Thus by Lemma 2.3 it has a unique FP, which is also the unique solution to problem (1.1).
Proof We fix a constant Consider a ball B r = {x ∈ X = C([0, 1], R) : x ≤ r}. Define two operators A 1 and A 1 on B r as For x, y ∈ B r , by (3.1) we can check that So A 1 x + A 2 y ∈ B r . Like in the proof of Theorem 3.1, from (H2), (H3), and (3.4) we can conclude that the operator A 2 is also a contraction mapping. Lemma 3.2 and (H1) ensure the continuity of the operator A 1 . For any x ∈ B r , we have Thus A 1 is uniformly bounded on B r . For all t 1 , t 2 ∈ [0, 1] such that t 1 < t 2 , we obtain By Lemma 3.1 we have (A 1 x)(t 2 ) -(A 1 x)(t 1 ) = (α -1)M 0 (α) B(1θ 1 , αθ 2 -1)(t 2t 1 ).
This means that A 1 is equicontinuous and relatively compact on B r . Accordingly, by the Arzelà-Ascoli theorem A 1 is compact on B r . Accordingly, Lemma 2.4 ensures the existence of a solution for problem (1.1) in [0, 1].
The Schaefer fixed point theorem gives the last result.
Hence we have x ≤ L 2 .
This shows that the set V is bounded. Lemma 2.5 ensures the existence of fixed points of A. Accordingly, there is at least one solution to problem (1.1) in [0, 1].

Examples
We introduce three examples to clarify the performed work.  We have f (t, x(t)) = 5