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

Authors: Shiying Song; Hongyu Li; Yumei Zou
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, the existence of extremal solutions for fractional differential equations with integral boundary conditions is obtained by using the monotone iteration technique and the method of upper and lower solutions. The main equations studied are as follows:−D0+αut=ft,ut,t∈0,1,u0=0,u1=∫01utdAt, where D0+α is the standard Riemann–Liouville fractional derivative of order α∈1,2 and At is a positive measure function. Moreover, an example is given to illustrate the main results.


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## Body

## 1. Introduction

The purpose of this paper is to study the following differential equations with integral boundary conditions:(1)−D0+αut=ft,ut,t∈0,1,u0=0,u1=∫01utdAt,where D0+α is the standard Riemann–Liouville fractional derivative and 1<α<2 and At is a positive measure function, and it satisfies ∫01GtdAt<G1.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 α∈1,2, and in [44, 45, 49, 55], the authors considered the Riemann–Liouville fractional differential equation of order α∈0,1, while in [46], the authors considered the boundary value problem with Riemann–Liouville fractional order α∈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: Section2 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 α>0 of a function u:0,∞⟶R is given by(2)I0+αut=1Γα∫0tt−sα−1usds,provided that the right-hand side is point-wise defined on 0,+∞.Definition 2.
(see [2, 5]). The Riemann–Liouville fractional derivative of order α>0 of a function u:0,+∞⟶ℝ is given by(3)D0+αut=1Γn−αddtn∫0tt−sn−α−1usds,where n=α+1, α denotes the integer part of number α, provided that the right-hand side is point-wise defined on 0,+∞.
LetE=u∈C0,1:D0+αu∈C0,1 be endowed with the norm u=maxu∞,D0+αu∞ in which u∞=maxt∈0,1ut, and then E,⋅ is a Banach space.Definition 3.
We say thatu∈E is an upper solution of (1) if it satisfies(4)−D0+αut≥ft,ut,t∈0,1,u0=0,u1≥∫01utdAt.Definition 4.
We say thatv∈E is a lower solution of (1) if it satisfies(5)−D0+αvt≤ft,vt,t∈0,1,v0=0,v1≤∫01vtdAt.
Denote(6)gt=α−2Γα−1+∑k=1+∞tkΓk+1α−2.
It is easy to check that (see [57, 58])(7)g0=α−2Γα−1<0,g′t>0,t∈0,+∞,limt⟶+∞gt=+∞.
Therefore, there exists a unique numberb∗>0 such that(8)gb∗=0.
SetGt=tα−1Eα,αbtα, where Eα,αx=∑k=0+∞xk/Γk+1α is the Mittag-Leffler function (see [2, 5]).
In this article, we list the following assumption for convenience:H1: the parameter b satisfies b∈0,b∗.H2:At is a positive measure function and 0<G1−∫01GtdAt.H3: assume that w0,v0∈E are the upper and lower solutions of problem (1), respectively, and v0t≤w0t, t∈0,1.H4:f∈C0,1×ℝ,ℝ and ft,w−ft,v≥−bw−v for v0t≤v≤w≤w0t,t∈0,1.
Next, we will consider the following auxiliary linear boundary value problem:(9)−D0+αut+but=yt,t∈0,1,u0=0,u1=∫01utdAt+c,c∈ℝ.Lemma 1.
Suppose that (H1) and (H2) hold and y∈C0,1. Then, fractional boundary value problem (9) has the following unique solution:(10)ut=∫01Kt,sysds+cGtG1−∫01GtdAt,where(11)Kt,s=K0t,s+Gths,hs=∫01K0t,sdAtG1−∫01GtdAt,K0t,s=1G1GtG1−s,0≤t≤s≤1,GtG1−s−Gt−sG1,0≤s≤t≤1.Proof.
The main idea of Lemma1 comes from [57]. By [2], we first find the solution of the fractional differential equations(12)−D0+αut+but=yt,t∈0,1,with two-point boundary condition(13)u0=0,u1=a∈ℝ,which can be expressed by(14)ut=−∫0tGt−sysds+c1Gt+c2G′t.
Sinceu0=0, we have c2=0. Then, by the condition u1=a, we calculated that(15)c1=∫01G1−sysds+aG1.
Therefore, the solution of (12) and (13) is(16)ut=−∫0tGt−sysds+∫01G1−sysds+aG1Gt=∫01K0t,sysds+u1G1Gt.
Next, we consider problem (9). Integrating equality (16) with respect to At, we have(17)∫01utdAt=∫01∫01K0t,sysdsdAt+u1∫01GtdAtG1.
Making use of the conditionu1=∫01utdAt+c in the above equality yields(18)u1−c=∫01∫01K0t,sysdsdAt+u1∫01GtdAtG1,and then we get(19)u11−∫01GtdAtG1=∫01∫01K0t,sysdsdAt+c.
Obviously,(20)u1=G1∫01∫01K0t,sysdsdAt+cG1G1−∫01GtdAt.
Therefore,(21)ut=∫01K0t,sysds+GtG1∫01∫01K0t,sysdsdAt+cG1G1G1−∫01GtdAt=∫01K0t,sysds+Gt∫01∫01K0t,sysdsdAt+cG1−∫01GtdAt=∫01K0t,sysds+∫01Gt∫01K0t,sdAtysdsG1−∫01GtdAt+cGtG1−∫01GtdAt=∫01K0t,sysds+cGtG1−∫01GtdAt.
The proof is completed.Lemma 2.
Suppose that (H1) and (H2) hold, and u∈E satisfies(22)−D0+αut+but≥0,u0=0,u1≥∫01utdAt.
Then, fort∈0,1, ut≥0.Proof.
Letyt=−D0+αut+but and c=u1−∫01utdAt. From (12), we have yt≥0, c≥0, and(23)−D0+αut+but=yt,u0=0,u1=∫01utdAt+c.
By Lemma1, we obtain that problem (23) has unique solution u, which can be expressed as follows:(24)ut=∫01Kt,sysds+cGtG1−∫01GtdAt.
From [57], it follows that K0t,s≥0 for t,s∈0,1. This together with (H1) and (H2) yields(25)Kt,s=K0t,s+Gt∫01K0t,sdAtG1−∫01GtdAt≥0,t,s∈0,1.
Hence, we conclude that(26)ut≥cGtG1−∫01GtdAt≥0,t∈0,1,which completes the proof.


## 3. Main Results

Forv0,w0∈E with v0t≤w0t for t∈0,1, we denote an ordered interval:(27)v0,w0=u∈E:v0t≤ut≤w0t,t∈0,1.Theorem 1.
Suppose (H1)–(H4) hold, and then there exist monotone iterative sequences vn,wn⊂v0,w0 such that vn⟶v∗,wn⟶w∗ as n⟶∞ uniformly in v0,w0, and v∗,w∗ are a minimal and a maximal solution of (1) in v0,w0, respectively.Proof.
Forvn−1,wn−1∈E, n≥1, we define two sequences vn,wn⊂E satisfying the following fractional differential equation:(28)−D0+αvnt+bvnt=ft,vn−1t+bvn−1t,t∈0,1,vn0=0,vn1=∫01vntdAt,(29)−D0+αwnt+bwnt=ft,wn−1t+bwn−1t,t∈0,1,wn0=0,wn1=∫01wntdAt.
By consideration of Lemma1, for any n≥1, problems (28) and (29) have a unique solution vn+1t,wn+1t, respectively, which are well defined.
Firstly, we need to show that, for anyt∈0,1, v0t≤v1t≤w1t≤w0t. Let xt=v1t−v0t, and the definition of v1t together with H3 yields(30)−D0+αxt+bxt=−D0+αv1t−v0t+bv1t−v0t=−D0+αv1t+bv1t−−D0+αv0t+bv0t≥ft,v0t+bv0t−ft,v0t−bv0t=0,x0=v10−v00=0,x1=v11−v01≥∫01v1tdAt−∫01v0tdAt=∫01xtdAt.
In the light of Lemma2, we have xt≥0,t∈0,1, namely, v1t≥v0t. Similarly, it can be shown that w0t≥w1t, t∈0,1.
Secondly, we makeht=w1t−v1t. From H4, we get(31)−D0+αht=ft,w0t−bw1t−w0t−ft,v0t+bv1t−v0t≥−bw0t−v0t−bw1t−w0t+bv1t−v0t=−bht.
Also,h0=0 and h1=∫01htdAt. Thus, Lemma 2 implies that, for any t∈0,1, w1t≥v1t.
Thirdly, we prove thatw1,v1 are upper and lower solutions of problem (1), respectively. Note that(32)−D0+αv1t=ft,v0t+bv0t−bv1t−ft,v1t+ft,v1t=ft,v0t−ft,v1t+ft,v1t−bv1t−v0t≤bv1t−v0t+ft,v1t−bv1t−v0t=ft,v1t.
And by assumptionH4, v10=0 and v11=∫01v1tdAt. This shows that v1 is a lower solution of problem (1). Similarly, we can infer that w1 is an upper solution of (1).
Using mathematical induction, it is easy to verify that(33)v0t≤v1t≤⋯≤vnt≤⋯≤wnt≤⋯≤w1t≤w0t.
Clearly, it is easy to conclude thatvn and wn are uniformly bounded in C0,1. Moreover, by Lemma 1, problems (28) and (29) are equivalent to the following integral equation:(34)vnt=∫01Kt,sfs,vn−1s+bvn−1sds,wnt=∫01Kt,sfs,wn−1s+bwn−1sds,respectively. Therefore, the continuity of the functions Kt,s allows us to conclude that vn and wn are equicontinuous in C0,1. Using (28) and (29) again, we know that D0+αvn and D0+αwn are uniformly bounded and equicontinuous in C0,1. So, vn and wn are uniformly bounded and equicontinuous in E. Using the standard arguments, we have vn and wn converging, say, to v∗ and w∗, uniformly on 0,1, respectively. That is,(35)limn⟶∞vnt=v∗t,limn⟶∞wnt=w∗t,t∈0,1.
Furthermore,v∗t and w∗t are the solutions of problem (1), and v0≤v∗≤w∗≤w0 on [0, 1].
Next, we need to prove thatv∗andw∗ are extremal solutions of (1) in v0,w0. Let u∈v0,w0 be any solution of problem (1). We assume that vmt≤ut≤wmt,t∈0,1 for some m. Take pt=ut−vm+1t,qt=wm+1t−ut. Then, by assumption H4, we obtain(36)−D0+αpt≥−bpt,t∈0,1,p0=0,p1=∫01ptdAt,−D0+αqt≥−bqt,t∈0,1,q0=0,q1=∫01qtdAt.
By Lemma2, we have(37)vm+1t≤ut≤wm+1t,t∈0,1.
Applying mathematical induction, one hasvnt≤ut≤wnt on 0,1 for any n. Taking the limit, we conclude v∗t≤ut≤w∗t,t∈0,1. The proof is complete.Example 1.
Consider the following problem:(38)−D0+4/3ut=−u3t10+sint10,t∈0,1,u0=0,u1=∫01utdt,where α=4/3, At=t, and ft,u=−u3/10+sint/10.
Takingv0t=−t1/3 and w0t=t1/3, we have(39)−D0+4/3v0t=0≤t10+sint10=ft,v0t,t∈0,1,v00=0,v01=−1<−34=∫01v0tdt,−D0+4/3w0t=0≥−t10+sint10=ft,w0t,t∈0,1,w00=0,w01=1>34=∫01w0tdt,which shows that v0t and w0t are a lower and an upper solution of (38), respectively, and v0t≤w0t. So, (H3) holds.
Using the strictly monotone increasing property of gamma functionΓ⋅ on 2,+∞, we have(40)gt=−23Γ1/3+∑k=1∞tkΓ4k/3−2/3=−23Γ1/3+tΓ2/3+∑k=2∞tkΓ4k−2/3+2≤−23Γ1/3+tΓ2/3+∑k=2∞tkΓk=−23Γ1/3+tΓ2/3+tet−1,t∈0,+∞.
By MATLAB, we obtaing3/10<−0.0568. Therefore, b∗>b≔3/10. In addition, we have(41)ft,w−ft,v=−w310−v310≥−310w−v,for v0t≤v≤w≤w0t,t∈0,1. Hence, (H1) and (H4) hold.
Note thatGt=∑k=0+∞3/10kt4k+1/3/Γ4k+4/3. Then,(42)∫01Gtdt=∑k=0+∞310k∫01t4k+1/3dtΓ4k+4/3=∑k=0+∞310k14k+4/3Γ4k+4/3<∑k=0+∞310k1Γ4k+4/3=G1.
It shows that (H2) holds. Thus, Theorem 1 ensures that problem (38) has extremal solutions in v0,w0.



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