# Positive Solutions for BVP of Fractional Differential Equation with Integral Boundary Conditions

Authors: Min Li; Jian-Ping Sun; Ya-Hong Zhao
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we consider a class of boundary value problems of nonlinear fractional differential equation with integral boundary conditions. By applying the monotone iterative method and some inequalities associated with Green’s function, we obtain the existence of minimal and maximal positive solutions and establish two iterative sequences for approximating the solutions to the above problem. It is worth mentioning that these iterative sequences start off with zero function or linear function, which is useful and feasible for computational purpose. An example is also included to illustrate the main result of this paper.


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## Body

## 1. Introduction

Fractional calculus has widespread applications in many fields of science and engineering, for example, viscoelasticity, continuum mechanics, bioengineering, rheology, electrical networks, control theory of dynamical systems, and optics and signal processing [1, 2].In the past decades, the existence of solutions or positive solutions for boundary value problems (BVPs for short) of nonlinear fractional differential equations attracted considerable attention from many authors, see [3–19] and the references therein.Recently, the monotone iterative method has been applied to study BVPs of nonlinear fractional differential equations. For example, in [20], Cui et al. discussed the BVP(1)D0+qut+ft,ut=0,t∈0,1,u0=u′0=0,u1=0,where 2<q≤3 and D0+q denotes the standard Riemann–Liouville fractional derivative of order q. The authors obtained the existence of maximal and minimal solutions and the uniqueness result for BVP (1). In 2014, Sun and Zhao [21] investigated the following BVP with integral boundary conditions:(2)D0qut+qtft,ut=0,t∈0,1,u0=u′0=0,u1=∫01gsusds,where 2<q≤3 and D0+q is the standard Riemann–Liouville fractional derivative of order q. By means of the monotone iterative method, they proved the existence of a positive solution and established an iterative sequence for approximating the solution to BVP (2). For relevant results, one can refer to [22–25].Motivated by the aforementioned works, in this paper, we consider the following BVP of nonlinear fractional differential equation with integral boundary conditions:(3)DC0+qut+ft,ut=0,t∈0,1,u″0=0,αu0−βu′0=∫01h1susds,γu1+δDC0+σu1=∫01h2susds,where DC0+q and DC0+σ denote the standard Caputo fractional derivatives of order q and order σ, respectively. Throughout this paper, we always assume that 2<q≤3,0<σ≤1 and α,β,γ,andδ are nonnegative constants satisfying 0<ρ≔α+βγ+αδ/Γ2−σ<βγ+δΓq/Γq−σ, and f:0,1×0,+∞⟶0,+∞ and hii=1,2:0,1⟶0,+∞ are continuous.The main tool used is the following theorem [26].Theorem 1.
LetK be a normal cone of a Banach space E and υ0≾ω0. Suppose thata1:T:υ0,ω0⟶E is completely continuousa2:T is monotone increasing on υ0,ω0a3:υ0 is a lower solution of T, that is, υ0≾Tυ0a4:ω0 is an upper solution of T, that is, Tω0≾ω0
Then, the iterative sequences(4)υn=Tυn−1,ωn=Tωn−1,n=1,2,3,…,satisfy(5)υ0≾υ1≾···≾υn≾···≾ωn≾···≾ω1≾ω0,and converge to, respectively, υ and ω∈υ0,ω0, which are fixed points of T.


## 2. Preliminaries

First, we present the definitions of Riemann–Liouville fractional integral and fractional derivative and Caputo fractional derivative on a finite interval of the real line, which may be found in [1].In this section, we always assume thatℕ=1,2,3,…,μ,ν>0 and μ denotes the integer part of μ.Definition 1.
The Riemann–Liouville fractional integralI0+μu of order μ on 0,1 is defined by(6)I0+μut:=1Γμ∫0tusdst−s1−μ.Definition 2.
The Riemann–Liouville fractional derivativeD0+μu of order μ on 0,1 is defined by(7)D0+μut:=ddtnI0+n−μut=1Γn−μddtn∫0tusdst−sμ−n+1,where n=μ+1.Definition 3.
LetD0+μust≡D0+μut be the Riemann–Liouville fractional derivative of order μ. Then, the Caputo fractional derivative CD0+μu of order μ on 0,1 is defined by(8)CD0+μut≔D0+μus−∑k=0n−1uk0k!skt,where(9)n=μ+1,μ∉ℕ,μ,μ∈ℕ.Lemma 1 (see [2]).
Letν>μ. Then, the equation CD0+μI0+νut=I0+ν−μut,t∈0,1 is satisfied for u∈C0,1.Lemma 2 (see [1]).
Letn be given by (9). Then, the following relations hold:(1)
Fork∈0,1,2,…,n−1,CD0+μtk=0.(2)
Ifν>n,thenCD0+μtν−1=Γν/Γν−μtν−μ−1.
For convenience, we denote(10)Pi=1ρ∫01αs+βhisds,Qi=1ρΓ2−σ∫01γΓ2−σ1−s+δhisds,i=1,2.Lemma 3.
Let1−Q11−P2≠P1Q2. Then, for any y∈C0,1, the BVP(11)D0+qCut+yt=0,t∈0,1,u″0=0,αu0−βu′0=∫01h1susds,γu1+δD0+σCu1=∫01h2susds,has a unique solution(12)ut=∫01Ht,sysds,t∈0,1.
Here,(13)Ht,s=Gt,s+∑i=12ϕit∫01Gτ,shiτdτ,t,s∈0,1×0,1,where(14)Gt,s=αt+βργ1−sq−1Γq+δ1−sq−σ−1Γq−σ−t−sq−1Γq,0≤s≤t≤1,0,0≤t≤s≤1,ϕ1t=Γ2−σQ2αt+β+1−P2γΓ2−σ1−t+δρΓ2−σ1−Q11−P2−P1Q2,t∈0,1,ϕ2t=Γ2−σ1−Q1αt+β+P1γΓ2−σ1−t+δρΓ2−σ1−Q11−P2−P1Q2,t∈0,1.Proof.
In view of the equation in (11), Theorem 3.24 [1], and u″0=0, we have(15)ut=−I0+qyt+u0+u′0t,t∈0,1.
By (15), Lemma 1, and Lemma 2, we obtain(16)D0+σCut=−I0+q−σyt+u′0Γ2−σt1−σ,t∈0,1.
It follows from (15) and (16) and the boundary conditions in (11) that(17)u0=1ρβγI0+qy1+βδI0+q−σy1+γΓ2−σ+δΓ2−σ∫01h1susds+β∫01h2susds,u′0=1ραγI0+qy1+αδI0+q−σy1−γ∫01h1susds+α∫01h2susds,which together with (15) shows that(18)ut=∫0tαt+βργ1−sq−1Γq+δ1−sq−σ−1Γq−σ−t−sq−1Γqysds+∫t1αt+βργ1−sq−1Γq+δ1−sq−σ−1Γq−σysds+γΓ2−σ1−t+δρΓ2−σ∫01h1susds+αt+βρ∫01h2susds=∫01Gt,sysds+γΓ2−σ1−t+δρΓ2−σ∫01h1susds+αt+βρ∫01h2susds,t∈0,1.
From (18), we get(19)1−Q1∫01h1susds−P1∫01h2susds=∫01h1s∫01Gs,τyτdτds,−Q2∫01h1susds+1−P2∫01h2susds=∫01h2s∫01Gs,τyτdτds,and so,(20)∫01h1susds=1−P2∫01h1s∫01Gs,τyτdτds+P1∫01h2s∫01Gs,τyτdτds1−Q11−P2−P1Q2,∫01h2susds=Q2∫01h1s∫01Gs,τyτdτds+1−Q1∫01h2s∫01Gs,τyτdτds1−Q11−P2−P1Q2,which together with (18) implies that(21)ut=∫01Gt,sysds+∑i=12ϕit∫01his∫01Gs,τyτdτds=∫01Gt,sysds+∑i=12ϕit∫01hiτ∫01Gτ,sysdsdτ=∫01Gt,sysds+∑i=12ϕit∫01ys∫01Gτ,shiτdτds=∫01Gt,s+∑i=12ϕit∫01Gτ,shiτdτysds=∫01Ht,sysds,t∈0,1.
In what follows, we let(22)gs=α+βργ1−sq−1Γq+δ1−sq−σ−1Γq−σ,s∈0,1,ηs=βδΓq−Γq−σρ−βγα+βγΓq−σ+δΓq1−sσ,s∈0,1.Lemma 4.
G
t
,
s satisfies the following properties:(1)
Gt,s≤gs,t,s∈0,1×0,1.(2)
Gt,s≥ηsgs,t,s∈0,1×0,1.Proof.
Since (1) is obvious, we only need to prove that (2) holds.
First, it is clear thatGt,1≥η1g1 for t∈0,1.
Now, we verify thatGt,s≥ηsgs for t,s∈0,1×0,1. In fact, if s≤t, then(23)Gt,sgs=αt+βγΓq−σ1−sq−1+δΓq1−sq−σ−1−ρΓq−σt−sq−1α+βγΓq−σ1−sq−1+δΓq1−sq−σ−1≥βγΓq−σ1−sσ+βδΓq−ρΓq−σ1−sσα+βγΓq−σ1−sσ+δΓq≥βδΓq−Γq−σρ−βγα+βγΓq−σ+δΓq1−sσ=ηs,and if t≤s, then(24)Gt,sgs=αt+βγΓq−σ1−sq−1+δΓq1−sq−σ−1α+βγΓq−σ1−sq−1+δΓq1−sq−σ−1≥βγΓq−σ1−sσ+βδΓqα+βγΓq−σ1−sσ+δΓq≥βγΓq−σ1−sσ+βδΓq−ρΓq−σ1−sσα+βγΓq−σ1−sσ+δΓq≥βδΓq−Γq−σρ−βγα+βγΓq−σ+δΓq1−sσ=ηs.
By the definition ofη and the condition 0<ρ<βγ+δΓq/Γq−σ, we may obtain the following remark.Remark 1.
η
s
≥
0 and ηs≢0 for s∈0,1.
In the remainder of this paper, we always assume that the following condition is fulfilled:(25)C1Q1<1,P2<1,1−Q11−P2>P1Q2.
Now, we define(26)xt=1+∑i=12∫01hiτdτϕit,t∈0,1.Lemma 5.
H
t
,
s has the following property:(27)xtηsgs≤Ht,s≤xtgs,t,s∈0,1×0,1.Proof.
In view of the definition ofHt,s and Lemma 4, it is obvious.


## 3. Main Results

For convenience, we let(28)A=1∫01gsds,a=maxt∈0,1xt.Theorem 2.
Assume thatft,0≡0 for t∈0,1 and the following condition is satisfied:(29)C2ft,u1≤ft,u2≤A,0≤t≤1,0≤u1≤u2≤a.
Then, BVP (3) possesses a minimal positive solution υ and a maximal positive solution ω.Proof.
LetE=C0,1 be equipped with the norm(30)u=maxt∈0,1ut,K=u∈E:ut≥0,t∈0,1.
Then,K is a normal cone in Banach space E. Note that this induces an order relation “≾” in E by defining u≾v if and only if v−u∈K.
Now, we defineυ0t≡0andω0t=xtfort∈0,1 and T:υ0,ω0⟶K by(31)Tut=∫01Ht,sfs,usds,u∈υ0,ω0,t∈0,1.
Then, it is easy to know thatT:υ0,ω0⟶K is completely continuous, and fixed points of T are nonnegative solutions of BVP (3).Step 1.
We assert thatT is monotone increasing on υ0,ω0.
Letu,v∈υ0,ω0 and u≾v. Then, 0≤ut≤vt≤a for t∈0,1, which together with C2 implies that(32)Tut=∫01Ht,sfs,usds≤∫01Ht,sfs,vsds=Tvt,t∈0,1.
This shows thatTu≾Tv.Step 2.
We prove thatυ0 is a lower solution of T.
For anyt∈0,1, we have(33)Tυ0t=∫01Ht,sfs,0ds≥0=υ0t,which indicates that υ0≾Tυ0.Step 3.
We show thatω0 is an upper solution of T.
In view of Lemma5 and C2, we get(34)Tω0t=∫01Ht,sfs,ω0sds≤Axt∫01gsds=ω0t,t∈0,1,which implies that Tω0≾ω0.Step 4.
We claim that BVP (3) possesses a minimal positive solution υ and a maximal positive solution ω.
In fact, if we construct sequencesυnn=0∞ and ωnn=0∞ as in the following,(35)υn=Tυn−1,ωn=Tωn−1,n=1,2,3,…,then it follows from Theorem 1 that(36)υ0≾υ1≾···≾υn≾···≾ωn≾···≾ω1≾ω0,and υnn=0∞ and ωnn=0∞ converge to, respectively, υ and ω∈υ0,ω0, which are nonnegative solutions of BVP (3).
In view of Lemma5, Remark 1, the definitions of xt and gs, and the assumption ft,0≡0 for t∈0,1, we get(37)Tυ0t=∫01Ht,sfs,0ds≥xt∫01ηsgsfs,0ds>0,t∈0,1,and so,(38)0<Tυ0t≤Tυt=υt≤ωt,t∈0,1,which shows that υ and ω are positive solutions of the BVP (3).
Moreover, ifu∈υ0,ω0 is a positive solution of the BVP (3), then it follows from the fact T is monotone increasing on υ0,ω0 that υ≾u≾ω. This indicates that υ and ω are minimal and maximal positive solutions of the BVP (3), respectively.Example 1.
Consider the following BVP:(39)D0+5/2Cut+π3u2t1474113456sinπt=0,t∈0,1,u″0=0,u0−3u′0=∫01susds,2u1+u′1=12∫01susds.
Sinceq=5/2,σ=α=δ=1,β=3,andγ=2, a simple calculation shows that(40)0<ρ=α+βγ+αδΓ2−σ=9<βγ+δΓqΓq−σ=212,A=15π16.
Moreover, in view ofh1s=s and h2s=s/2 for s∈0,1, we get(41)P1=1154,P2=11108,Q1=554,Q2=5108,which indicates that C1 is fulfilled. At the same time, we also obtain that(42)a=maxt∈0,1xt=3829.
Now, if we letft,u=π/3u2+14741/13456sinπt,t,u∈0,1×0,+∞, then it is easy to know that f:0,1×0,+∞⟶0,+∞ is continuous, ft,0≢0 for t∈0,1, and (C2) is fulfilled. Therefore, it follows from Theorem 2 that BVP (39) possesses a minimal positive solution υ and a maximal positive solution ω. In addition, the two iterative schemes are as follows:(43)υ0t≡0,t∈0,1,υn+1t=∫0132t−963045−s7/2+t−s5/2s+16t2−48t609t−s5/2+48t+3207831−s3/2+12t+802611−s1/2υn2s+1474113456sinπsds−49∫0tt−s3/2υn2s+1474113456sinπsds,t∈0,1,n=0,1,2,…;ω0t=−3t+3829,t∈0,1,ωn+1t=∫0132t−963045−s7/2+t−s5/2s+16t2−48t609t−s5/2+48t+3207831−s3/2+12t+802611−s1/2ωn2s+1474113456sinπsds−49∫0tt−s3/2ωn2s+1474113456sinπsds,t∈0,1,n=0,1,2,….



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