Existence and uniqueness of solutions for a class of fractional nonlinear boundary value problems under mild assumptions

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                </mml:math></jats:alternatives></jats:disp-formula> Under mild assumptions, we prove the existence of a unique continuous solution <jats:italic>v</jats:italic> to this problem satisfying <jats:disp-formula><jats:alternatives><jats:tex-math>$$ \bigl\vert v(x) \bigr\vert \leq cx^{\alpha -1}(1-x)\quad\text{for all }x \in [ 0,1]\text{ and some }c>0. $$</jats:tex-math><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML">
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                </mml:math></jats:alternatives></jats:disp-formula> Our results improve those obtained by Zou and He (Appl. Math. Lett. 74:68–73, 2017).</jats:p>


Introduction
Fractional differential equations have attracted great attention due to their ability to model various phenomena in applied sciences. The so-called fractional differential equations are specified by generalizing the standard integer-order derivative to arbitrary order. For more interesting theoretical results and scientific applications of fractional differential equations, we refer to the monographs of Diethelm [2] and Kilbas et al. [3] and references therein.
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Zou and He [1] investigated the problem
where D α denotes the standard Riemann-Liouville fractional derivative, and f satisfies the following conditions: (H1) f ∈ C((0, 1) × R, R) and Let L > 0 be the minimum positive constant such that where G α (x, y) is the Green's function (given later in this paper) associated with problem (1.1). By using Banach's contraction principle on some convenient Banach space they have obtained the following result. Motivated by this reault, we prove that the conclusion of Theorem 1.1 remains true under the following weaker assumptions: (A1) f ∈ C((0, 1) × R, R) and In this paper, for α ∈ [2, 3), we use the following notations: • G α (x, y) denotes the Green's function of the operator v → -D α v with boundary condi- (1). (1.5) We will prove that M is a positive constant satisfying the following range estimation: • For a ∈ R, a + := max(a, 0).
In the next remark, we list some properties of elements of C h ([0, 1]).
Our main result is the following: Our paper is organized as follows. In Sect. 2, we improve the estimates on Green's function G α obtained in [1,Lemma 2.2]. This allows us to obtain the range estimation (1.6). Our main result is proved in Sect. 3. Some examples and approximations are given at the end.

Preliminaries
(i) The Riemann-Liouville fractional integral of order γ > 0 for f is defined as where is the Euler gamma function. (ii) The Riemann-Liouville fractional derivative of order γ > 0 for f is defined as where n = [γ ] + 1, and [γ ] is the integer part of γ .

Lemma 2.2
The Green's function G α (x, y) has the following properties: Proof It is obvious that (i) holds. Now we prove (ii). From (2.1), for all x, y ∈ (0, 1), we have Since for λ > 0 and t ∈ [0, 1], we deduce that Using this fact and (2.4), we obtain Hence estimates (2.2) follow from we get Combining this fact with (2.2), we immediately obtain inequalities (i) and (ii). Therefore estimates (2.2) improve those stated in [1, Lemma 2.2].
Proof Let E = a > 0 : On the other hand, using again (2.2) and that min(x, y)(1 -max(x, y)) ≥ xy(1x)(1y) for x, y ∈ [0, 1], we deduce that for any a ∈ E, Hence for each a ∈ E, Remark 2.5 From Lemma 2.4 it is obvious that if M q,α < 1, then M := inf E < 1. Note that the inequality M q,α < 1 can be verified for a large class of functions q, including the singular cases. For example, let B(a, b) := 1 0 t a-1 (1t) b-1 dt for a > 0 and b > 0. Then by using MATLAB we obtain (i) If q ∈ C((0, 1)) with q > 0 and q ∞ ≤ 1, then

Existence and uniqueness
We need the following useful lemma.
To this end, define the operator T by We claim that T is a contraction operator from (C h ([0, 1]), · h ) into itself.
On the other hand, since v → cos v is a Lipschitz function, we obtain By Lemma 2.4 and Remark 2.5(i) we have Hence by Theorem 1.4 problem (3.5) has a unique solution v ∈ C h ([0, 1]).
On the other hand, we clearly have converges to v (see Fig. 2).