# Monotone Iterative Method for Fractional Differential Equations with Integral Boundary Conditions.

1. Introduction

The purpose of this paper is to study the following differential equations with integral boundary conditions:

[mathematical expression not reproducible.], (1)

where [D.sup.[alpha].sub.0+] is the standard Riemann-Liouville fractional derivative and 1 < [alpha] < 2 and A(t) is a positive measure function, and it satisfies [[integral].sup.1.sub.0] G(t)dA(t) < G(1).

Fractional differential equations have been widely used in physics, chemistry, aerodynamics, electrodynamics of complex media, and rheology of polymers [1-7]. As a result, various nonlinear functional analysis methods have been used to study the existence of solutions for differential equations [8-38]. However, it is difficult to obtain the exact solutions of fractional differential equations, so the monotone iteration method and the upper and lower solutions are generally used to obtain the approximate solutions of fractional differential problems. This method is also applicable to both initial value problems and boundary value problems for integer-order differential equations and differential systems [39-43]. In recent years, many scholars have used this method to study various fractional differential equation problems [44-55]. For example, in [47, 48, 52], the authors paid attention to the Riemann-Liouville fractional differential equations of order [alpha] [member of] (1,2), and in [44, 45, 49, 55], the authors considered the Riemann-Liouville fractional differential equation of order [alpha] [member of] (0,1), while in [46], the authors considered the boundary value problem with Riemann-Liouville fractional order [alpha] [member of] (2,3). Of course, some papers also use the monotone iterative method to deal with nonlinear Caputo fractional differential equations [56].

Based on the upper and lower solutions, this paper presents a method to prove the existence of solutions of Riemann-Liouville fractional differential equation (1). By using the monotone iteration technique coupled with the upper and lower solution method, a new comparison principle is established and the existence of the extremal solution of integral boundary value problems (1) is proved.

This paper is mainly divided into the following two parts: Section 2 mainly introduces the preparation of this article, and then in Section 3, the monotone sequence of solutions is constructed, and the main result of integral boundary value problems (1) is given.

2. Preliminaries

In this section, we will briefly introduce some of the necessary definitions and results that will be used in the main results.

Definition 1 (see [2, 5]). The fractional integral of order [alpha] > 0 of a function u: (0, [infinity]) [right arrow] R is given by

[I.sup.[alpha].sub.0+](t) = 1/[GAMMA]([alpha]) [[integral].sup.t.sub.0] [(t - s).sup.[alpha]-1]u(s)ds, (2)

provided that the right-hand side is point-wise defined on (0, +[infinity]).

Definition 2 (see [2, 5]). The Riemann-Liouville fractional derivative of order [alpha] > 0 of a function u: (0, +[infinity]) [right arrow] R is given by

[mathematical expression not reproducible.], (3)

where n = [[alpha]] + 1, [[alpha]] denotes the integer part of number [alpha], provided that the right-hand side is point-wise defined on (0, +[infinity]).

Let E = {u [member of] C[0,1]: [D.sup.[alpha].sub.0+] u [member of] C[0,1]} be endowed with the norm [mathematical expression not reproducible.] in which [[parallel]u[parallel].sub.[infinity]] = [max.sub.t[member of][01]] [absolute value of (u (t))], and then (E, [parallel]*[parallel]) is a Banach space

Definition 3. We say that u [member of] E is an upper solution of (1) if it satisfies

[mathematical expression not reproducible.]. (4)

Definition 4. We say that v [member of] E is a lower solution of (1) if it satisfies

[mathematical expression not reproducible.]. (5)

Denote

g(t) = [alpha] - 2/[GAMMA]([alpha] - 1) + [+[infinity].summation over (k=1)] [t.sup.k]/[GAMMA]((k + 1)[alpha] - 2). (6)

It is easy to check that (see [57, 58])

[mathematical expression not reproducible.]. (7)

Therefore, there exists a unique number [b.sup.*] > 0 such that

g([b.sup.*]) = 0. (8)

Set [mathematical expression not reproducible.] is the Mittag-Leffler function (see [2, 5]).

(H1): the parameter b satisfies b [member of] (0, [b.sup.*]].

(H2): A(t) is a positive measure function and 0 < G(1) - [[integral].sup.1.sub.0]G(t)dA(t).

(H3): assume that [w.sub.0], [v.sub.0] [member of] E are the upper and lower solutions of problem (1), respectively, and [v.sub.0](t) [less than or equal to] [w.sub.0](t), t [member of] [0,1].

(H4): f [member of] C([0,1] x R, R) and f(t, w) - f(t, v) [greater than or equal to] -b(w - v) for [v.sub.0] (t) [less than or equal to] v [less than or equal to] w [less than or equal to] [w.sub.0] (t), t [member of] [0,1].

Next, we will consider the following auxiliary linear boundary value problem:

[mathematical expression not reproducible.]. (9)

Lemma 1. Suppose that (H1) and (H2) hold and y [member of] C[0,1]. Then, fractional boundary value problem (9) has the following unique solution:

[mathematical expression not reproducible.], (10)

where

[mathematical expression not reproducible.]. (11)

Proof. The main idea of Lemma 1 comes from [57]. By [2], we first find the solution of the fractional differential equations

-[D.sup.[alpha].sub.0+]u(t) + bu(t) = y(t), t [member of] [0,1], (12)

with two-point boundary condition

u(0) = 0, u(1) = [alpha] [member of] R, (13)

which can be expressed by

u(t) = -[[integral].sup.t.sub.0] G(t - s)y(s)ds + [c.sub.1]G(t) + [c.sub.2]G'(t). (14)

Since u(0) = 0, we have [c.sub.2] = 0. Then, by the condition u (1) = a, we calculated that

[c.sub.1] = [[integral].sup.1.sub.0]G(1 - s)y(s)ds + a/G(1) (15)

Therefore, the solution of (12) and (13) is

[mathematical expression not reproducible.]. (16)

Next, we consider problem (9). Integrating equality (16) with respect to A(t), we have

[mathematical expression not reproducible.]. (17)

Making use of the condition u(1) = [[integral].sup.1.sub.0] u(t)dA(t) + c in the above equality yields

[mathematical expression not reproducible.], (18)

and then we get

[mathematical expression not reproducible.]. (19)

Obviously,

[mathematical expression not reproducible.]. (20)

Therefore,

[mathematical expression not reproducible.]. (21)

The proof is completed.

Lemma 2. Suppose that (H1) and (H2) hold, and u [member of] E satisfies

[mathematical expression not reproducible.]. (22)

Then, for t [member of] [0,1], u (t) [greater than or equal to] 0.

Proof. Let y(t) = -[D.sup.[alpha].sub.0+]u(t) + bu(t) and c = u(1) - [[integral].sup.1.sub.0] u(t)dA (t). From (12), we have y(t) [greater than or equal to] 0, c [greater than or equal to] 0, and

[mathematical expression not reproducible.]. (23)

By Lemma 1, we obtain that problem (23) has unique solution u, which can be expressed as follows:

[mathematical expression not reproducible.]. (24)

From [57], it follows that [K.sub.0](t, s) [greater than or equal to] 0 for t, s [member of] [0,1]. This together with (H1) and (H2) yields

[mathematical expression not reproducible.]. (25)

Hence, we conclude that

[mathematical expression not reproducible.], (26)

which completes the proof.

3. Main Results

For [v.sub.0], [w.sub.0] [member of] E with [v.sub.0] (t) [less than or equal to] [w.sub.0] (t) for t [member of] [0,1], we denote an ordered interval:

[[v.sub.0], [w.sub.0]] = {u [member of] E: [v.sub.0] (t) [less than or equal to] u(t) [less than or equal to] [w.sub.0](t), t [member of] [0,1]}. (27)

Theorem 1. Suppose (H1)-(H4) hold, and then there exist monotone iterative sequences {[v.sub.n]}, {[w.sub.n]} [subset] [[v.sub.0], [w.sub.0]] such that [v.sub.n] [right arrow] [v.sup.*], [w.sub.n] [right arrow] [w.sup.*] as n [right arrow] [infinity] uniformly in [[v.sub.0], [w.sub.0]], and [v.sup.*], [w.sup.*] are a minimal and a maximal solution of (1) in [[v.sub.0], [w.sub.0]], respectively.

Proof. For [v.sub.n-1], [w.sub.n-1] [member of] E, n [greater than or equal to] 1, we define two sequences {[v.sub.n]}, {[w.sub.n]} [subset] E satisfying the following fractional differential equation:

[mathematical expression not reproducible.], (28)

[mathematical expression not reproducible.]. (29)

By consideration of Lemma 1, for any n [greater than or equal to] 1, problems (28) and (29) have a unique solution [v.sub.n+1](t), [w.sub.n+1](t), respectively, which are well defined.

Firstly, we need to show that, for any t [member of] [0,1], [v.sub.0] (t) [less than or equal to] [v.sub.1] (t) [less than or equal to] [w.sub.1] (t) [less than or equal to] [w.sub.0] (t). Let x (t) = [v.sub.1] (t) - [v.sub.0] (t), and the definition of [v.sub.1] (t) together with (H3) yields

[mathematical expression not reproducible.]. (30)

In the light of Lemma 2, we have x(t) [greater than or equal to] 0, t [member of] [0,1], namely, [v.sub.1] (t) [greater than or equal to] [v.sub.0] (t). Similarly, it can be shown that [w.sub.0] (t) [greater than or equal to] [w.sub.1] (t), t [member of] [0,1].

Secondly, we make h (t) = [w.sub.1] (t) - [v.sub.1] (t). From (H4), we get

[mathematical expression not reproducible.]. (31)

Also, h(0) = 0 and h (1) = [[integral].sup.1.sub.0] h(t)dA (t). Thus, Lemma 2 implies that, for any t [member of] [0,1], [w.sub.1] (t) [greater than or equal to] [v.sub.1] (t).

Thirdly, we prove that [w.sub.1], [v.sub.1] are upper and lower solutions of problem (1), respectively. Note that

[mathematical expression not reproducible.]. (32)

And by assumption (H4), [v.sub.1] (0) = 0 and [v.sub.1](1) = [[integral].sup.1.sub.0] [v.sub.1] (t)dA (t). This shows that [v.sub.1] is a lower solution of problem (1). Similarly, we can infer that [w.sub.1] is an upper solution of (1).

Using mathematical induction, it is easy to verify that

[mathematical expression not reproducible.]. (33)

Clearly, it is easy to conclude that {[v.sub.n]} and {[w.sub.n]} are uniformly bounded in C[0,1]. Moreover, by Lemma 1, problems (28) and (29) are equivalent to the following integral equation:

[mathematical expression not reproducible.], (34)

respectively. Therefore, the continuity of the functions K(t, s) allows us to conclude that {[v.sub.n]} and {[w.sub.n]} are equi-continuous in C[0,1]. Using (28) and (29) again, we know that {[D.sup.[alpha].sub.0+] [v.sub.n]} and {[D.sup.[alpha].sub.0+][w.sub.n]} are uniformly bounded and equicontinuous in C[0, 1]. So, {[v.sub.n]} and {[w.sub.n]} are uniformly bounded and equicontinuous in E. Using the standard arguments, we have {[v.sup.n]} and {[w.sub.n] } converging, say, to [v.sup.*] and [w.sup.*], uniformly on [0,1], respectively. That is,

[mathematical expression not reproducible.]. (35)

Furthermore, [v.sup.*](t) and [w.sup.*](t) are the solutions of problem (1), and [v.sub.0] [less than or equal to] [v.sup.*] [less than or equal to] [w.sup.*] [less than or equal to] [w.sub.0] on [0, 1].

Next, we need to prove that [v.sup.*] and [w.sup.*] are extremal solutions of (1) in [[v.sub.0], [w.sub.0]]. Let u [member of] [[v.sub.0], [w.sub.0]] be any solution of problem (1). We assume that [v.sub.m](t) [less than or equal to] u(t) [less than or equal to] [w.sub.m] (t), t [member of] [0,1] for some m. Take p(t) = u(t) - [v.sub.m+1](t), q(t) = [w.sub.m+1](t) - u(t). Then, by assumption (H4), we obtain

[mathematical expression not reproducible.]. (36)

By Lemma 2, we have

[v.sub.m+1](t) [less than or equal to] u(t) [less than or equal to] [w.sub.m+1](t), t [member of] [0,1]. (37)

Applying mathematical induction, one has [v.sub.n] (t) [less than or equal to] u(t) [less than or equal to] [w.sub.n] (t) on [0,1] for any n. Taking the limit, we conclude [v.sup.*] (t) [less than or equal to] u(t) [less than or equal to] [w.sup.*] (t), t [member of] [0,1]. The proof is complete.

Example 1. Consider the following problem:

[mathematical expression not reproducible.], (38)

where [alpha] = 4/3, A(t) = t, and f(t,u) = -([u.sup.3]/10)+ (sint/10).

Taking [v.sub.0] (t) = -[t.sup.1/3] and [w.sub.0] (t) = [t.sup.1/3], we have

[mathematical expression not reproducible.], (39)

which shows that [v.sub.0] (t) and [w.sub.0] (t) are a lower and an upper solution of (38), respectively, and [v.sub.0](t) [less than or equal to] [w.sub.0](t). So, (H3) holds.

Using the strictly monotone increasing property of gamma function [GAMMA](*) on [2, +[infinity]), we have

[mathematical expression not reproducible.]. (40)

By MATLAB, we obtain g(3/10) < -0,0568. Therefore, [b.sup.*] > b = (3/10). In addition, we have

f(t, w) - f(t, v) = -[[w.sup.3]/10 - [v.sup.3]/10] [greater than or equal to] -3/10 (w - v), (41)

for [v.sub.0](t) [less than or equal to] v [less than or equal to] w [less than or equal to] [w.sub.0](t), t [member of] [0,1]. Hence, (H1) and (H4) hold.

Note that G(t) = [[SIGMA].sup.+[infinity].sub.k=0][(3/10).sup.k] ([t.sup.(4k+1)/3]/([GAMMA]((4k + 4)/3))).

Then,

[mathematical expression not reproducible.]. (42)

It shows that (H2) holds. Thus, Theorem 1 ensures that problem (38) has extremal solutions in [[v.sub.0], [w.sub.0]].

https://doi.org/10.1155/2020/7319098

Data Availability

The data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This project was supported by the National Natural Science

Foundation of China (51774197 and 11801322), the Shandong Natural Science Foundation (ZR2018MA011), and SDUST graduate innovation project (SDKDYC190238).

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Shiying Song, (1) Hongyu Li [ID], (1) and Yumei Zou [ID], (2)

(1) Department of Applied Mathematics, Shandong University of Science and Technology, Qingdao 266590, China

(2) Department of Statistics and Finance, Shandong University of Science and Technology, Qingdao 266590, China

Correspondence should be addressed to Yumei Zou; sdzouym@126.com

Received 4 February 2020; Accepted 4 March 2020; Published 30 March 2020

Guest Editor: Chuanjun Chen