# The Exact Iterative Solution of Fractional Differential Equation with Nonlocal Boundary Value Conditions

Authors: Jinxiu Mao; Zengqin Zhao; Chenguang Wang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We deal with a singular nonlocal fractional differential equation with Riemann-Stieltjes integral conditions. The exact iterative solution is established under the iterative technique. The iterative sequences have been proved to converge uniformly to the exact solution, and estimation of the approximation error and the convergence rate have been derived. An example is also given to demonstrate the results.


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## Body

## 1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines; see [1–5]. Much attention has been paid to study fractional differential equations both with initial and boundary conditions; see, for example, [6, 7]. In [8, 9], they focused on sign-changing solution for some fractional differential equations. In [10], they get the existence of solutions for impulsive fractional differential equations. In [11–13], they get the existence and multiplicity of nontrivial solutions for a class of fractional differential equations. The mainly techniques authors need are fixed point theory, variational method, and global bifurcation techniques.Also, ordinary differential equations and partial differential equations involving nonlocal boundary conditions have been studied extensively in recent years, see [14–22], including integral boundary conditions and multipoint boundary conditions.In [23], authors obtained results on the uniqueness of positive solution for problem (1)Dpxt+ptft,xt+qt=0,t∈0,1,x0=x′0=0,x1=0,where 2<p≤3 is a real number. Under the assumption that (2)ft,u-ft,v≤kλ1u-v, where k∈[0,1), and λ is the first eigenvalue of the corresponding linear operator.Motivated by the above works, we study the following nonlocal boundary value problems:(3)Dqxt+ft,xt=0,t∈0,1,x1=x′1=0,x0=∫01xtdΛt,where Dqx denotes the left-handed Riemann-Liouville derivative of order q and 2<q≤3 is a real number. λ[x]=∫01‍x(t)dΛ(t) denotes a Stieltjes integral with a suitable function Λ of bounded variation. Different from [23] and other works, we only use the iterative methods to obtain the existence and uniqueness of positive solution. Moreover, the estimation of the approximation error and the convergence rate have also been derived.For clarity in presentation, we also list below some assumptions to be used later in the paper.( H 1 ) : f : ( 0,1 ) × ( 0 , + ∞ ) → [ 0 , + ∞ ) is continuous, and for (t,u)∈(0,1)×(0,+∞), f is increasing with respect to u and there exists a constant k∈(0,1) such that, for ∀σ∈(0,1],(4)ft,σu≥σkft,u.It is easy to see that if σ∈1,+∞, then ft,σu≤σkft,u.H 2 : 0 ≤ A = ∫ 0 1 1 - t q - 1 d Λ t < 1 ,   q ∈ 2,3,H 3 : ∫ 0 1 d Λ t ≥ 0 , ∫ 0 1 d Λ t ≠ 1 .


## 2. Preliminaries

For the convenience of the reader, we present here some necessary definitions from fractional calculus theory. These definitions and properties can be found in the recent monograph [23].Definition 1.
The Riemann-Liouville fractional integral of orderq>0 of a function x:(0,∞)→R is given by (5)Iqxt=1Γq∫0tt-sq-1xsds, provided that the right-hand side is pointwise defined on (0,∞).Definition 2.
The Riemann-Liouville fractional derivative of orderq>0 of a continuous function x:(0,∞)→R is given by (6)Dqxt=1Γn-qddtn∫0txst-sq-n+1ds, where n1≤q<n, n=[q]+1,q>0, provided that the right-hand side is pointwise defined on (0,∞). In particular, (7)Dnxt=xnt,n=1,2,3,….Lemma 3 (see [13]).
Assume that(H1)-(H3) hold. Let y∈L1((0,1),[0,+∞)). Then boundary value problem(8)Dqut+yt=0,t∈0,1,u1=u′1=0,u0=∫01utdΛt has the unique solution given by the following formula: (9)ut=∫01Gqt,sysds,where (10)Gt,s=1-tq-1sq-1,0≤s≤t≤1;1-tq-1sq-1-s-tq-1,0≤t≤s≤1,Gqt,s=Gt,s+1-tq-11-A∫01Gt,sdΛt.One can prove thatG(t,s),Gq(t,s) have the following properties.Lemma 4.
Note thatGq(t,s) is the Green function of problem (8).Lemma 5 (see [12]).
Fort,s∈[0,1], one has (11)1Γqmtks≤Gt,s≤1Γq-1ks,1Γqmtks≤Gt,s≤1Γq-1mt, where (12)ks=sq-11-s,mt=t1-tq-1.Lemma 6.
(13) g s 1 - t q - 1 ≤ G q t , s ≤ M 1 - t q - 1where M>0 is a constant and g(s)∈L1(0,1) is nonnegative for any s∈(0,1).Proof.
We have the estimation(14)Gqt,s≤1Γq-11-tq-1+1-tq-11-A∫011Γq-11-tq-1dΛt=1Γq-11+∫011-tq-1dΛt1-A1-tq-1=1Γq-111-A1-tq-1=M1-tq-1,(15)Gqt,s≥1Γq1-tq-11-A∫01t1-tq-1sq-11-sdΛt=1Γq1-tq-11-A∫01t1-tq-1dΛtsq-11-s=1Γq∫01t1-tq-1dΛt1-Asq-11-s1-tq-1=gs1-tq-1, where M=1/Γq-11/1-A and g(s)=1/Γq∫01‍t(1-t)q-1dΛ(t)/1-Asq-1(1-s). Thus, (13) holds.


## 3. The Main Results

Throughout this paper, we will work in the spaceE=C[0,1], which is a Banach space if it is endowed with the norm ∥u∥=maxt∈[0,1]|u(t)| for any u∈E.Define the setP in E as follows:P = { u ∈ E ∣ there exists positive constants 0<lu<1<Lu such that lu(1-t)q-1≤u(t)≤Lu(1-t)q-1,t∈[0,1]}.And define the operatorT:E→E.(16)Tut=∫01Gqt,sfs,usds.Evidently (1-t)q-1∈P. Therefore, P is not empty.Theorem 7.
Assume that(H1)-(H3) hold. And (17)0<∫01ft,1-tq-1dt<∞. Then BVP (3) has at least one positive solution u(t), and there exist constants 0<lu<1<Lu satisfying(18)lu1-tq-1≤ut≤Lu1-tq-1,t∈0,1.Proof.
It is clear thatu is a solution of (3) if and only if u is a fixed point of T.
Claim 1.The operator T:P→P is nondecreasing.
In fact, foru∈E, it is obvious that u∈E, Tu(1)=Tu′(1)=0,Tu(0)=∫01‍Tu(t)dΛ(t), and Tu(t)>0 for t∈(0,1). For any u∈P, we have that, for t∈[0,1], (19)Tut=∫01Gqt,sfs,usds.≤M1-tq-1fs,Lu1-sq-1ds≤∫01M1-tq-1Lukfs,1-sq-1≤LTu1-tq-1 and (20)Tut=∫01Gqt,sfs,usds.≥gs1-tq-1fs,lu1-sq-1ds≥1-tq-1luk∫01gsfs,1-sq-1=lTu1-tq-1, where LTu and lTu are positive constants satisfying (21)LTu>max⁡1,∫01MLukfs,1-sq-1,0<lTu<min⁡1,luk∫01gsfs,1-sq-1. Thus, it follows that there are constants 0<lTu<1<LTu such that, for t∈[0,1], (22)lTu1-tq-1≤Tut≤LTu1-tq-1. Therefore, for any u(t)∈P, Tu(t)∈P, i.e., T is the operator P→P. From (16), it is easy to see that T is nondecreasing for u. Hence, Claim 1 holds.
Claim 2.We take e(t)=(1-t)q-1. Let δ and γ be fixed numbers satisfying(23)0<δ≤lTe1/1-k,γ≥LTe1/1-k,and assume that(24)u0=δet,v0=γet,(25)un=Tun-1,vn=Tvn-1,n=1,2,3,….Then(26)u0≤u1≤…≤un≤…vn≤…≤v1≤v0,and there exists u∗∈P such that(27)unt→u∗t,vnt→u∗t,uniformly on [0,1].
In fact,0<lTe<1<LTe since Te∈P. Therefore, 0<δ<1<γ. From (24), we have u0,v0∈P and u0≤v0.
On the other hand,(28)u1=Tu0t=∫01Gqt,sfs,δesds≥δk∫01Gqt,sfs,esds=δkTe≥δklTeet≥δkδ1-ket=u0,v1=Tv0t=∫01Gqt,sfs,γesds≤γk∫01Gqt,sfs,esds=γkTe≤γklTeet≤γkγ1-ket=v0,and since u0≤v0 and T is nondecreasing, by induction, (26) holds.
Letc0=δ/γ, and then 0<c0<1. It follows from (4) that (29)Tc0u≥c0kTu.
And for any natural numbern, (30)un=Tun-1=Tnu0=Tnδet=Tnc0γet≥c0knTnγet=c0knvn.
Thus, for any natural numbern and p∗, we have(31)0≤un+p∗-un≤vn-un≤1-c0knvn≤1-c0knγet,which implies that there exists u∗∈P such that (27) holds and Claim 2 holds.
Lettingn→∞ in un=Tun-1 and noting the fact that T is continuous, we obtain u∗(t)=Tu∗(t), which is a positive solution of BVP (3). The proof of Theorem 7 is now complete.Theorem 8.
Assume that(H1)-(H3) hold. Then
(i) BVP (3) has unique positive solution u∗(t), and there exist constants l,L with 0<l<1<L such that (32)l1-tq-1≤u∗t≤L1-tq-1,t∈0,1.
(ii) For any initial valuex0∈P, there exists a sequence xn(t) that uniformly converges to the unique positive solution u∗(t), and one has the error estimation(33)max⁡xnt-u∗t=∘1-λkn,where λ is a constant with 0<λ<1 and determined by x0.Proof.
Letu0,v0,un,vn be defined in (24) and (25).
(i) It follows from Theorem7 that BVP (3) has a positive solution u∗(t)∈P, which implies that there exist constants l and L with 0<l<L<1 such that u∗(t) satisfies (18). Let v∗(t) be another positive solution of BVP (3); then from Theorem 7 we have that there exist constants c1 and c2 with 0<c1<1<c2 such that (34)c11-tq-1≤v∗t≤c21-tq-1,t∈0,1. Let δ defined in (23) be small enough such that δ<c1 and γ defined in (23) be large enough such that γ>c2. Then (35)u0t≤v∗t≤v0t,t∈0,1. Note that Tv∗=v∗ and T is nondecreasing; we have(36)unt≤v∗t≤vnt,t∈0,1.Letting n→∞ in (36), we obtain that v∗=u∗. Hence, the positive solution of BVP (3) is unique.
(ii) From (i), we know that the positive solutionu∗ to BVP (3) is unique. For any x0∈P, there exist constants l0 and L0 with 0<l0<1<L0 such that (37)l01-tq-1≤x0t≤L01-tq-1,t∈0,1. Similar to (i), we can let δ and γ defined by (23) satisfy δ<l0 and γ>L0. Then (38)u0t≤x0t≤v0t,t∈0,1. Let xn=Txn-1,n=1,2,…. Note that T is nondecreasing; we have(39)unt≤xnt≤vnt,t∈0,1.Letting n→∞ in (39), it follows that xn uniformly converges to the unique positive solution u∗ for BVP (3), where (40)xn=∫01Gqt,sfs,xn-1sds,n=1,2,…. At the same time, (33) follows from (31). Thus, the proof of the theorem is complete.


## 4. An Example

(41) D 5 / 2 x t + sin ⁡ t x 1 / 2 t + x 1 / 3 t = 0 ,  t ∈ 0,1 , x 1 = x ′ 1 = 0 , x 0 = ∫ 0 1 x t d Λ t ,where (42)Λt=0,t∈0,12;2,t∈12,34;1,t∈34,1.Analysis 1. Let (43)q=52,ft,x=sin⁡tx1/2t+x1/3t, and then for any σ∈(0,1), we take k=1/2 and have (44)ft,σu≥σkft,u. Then (H1) holds.In addition, we have(45)0<A=∫011-tq-1dΛt=22-18<1.0<∫01dΛt=34<1. Then (H2) and (H3) hold.And(46)0<∫01ft,1-t5/2-1dt<∞.Hence all conditions of Theorem7 are satisfied, and consequently we have the following corollary.Corollary 9.
Problem (41) has unique positive solution x∗(t). For any initial value x0∈P, the successive iterative sequence xn(t) generated by (47)xnt=∫01Gqt,ssin⁡sxn-11/2s+xn-11/3sds,n=1,2,…, uniformly converges to the unique positive solution x∗(t) on [0,1]. One has the error estimation (48)max⁡xnt-x∗t=∘1-λ1/2n, where λ is a constant with 0<λ<1 and determined by the initial value x0. And there are constants l,L with 0<l<1<L such that (49)l1-t3/2≤x∗t≤L1-t3/2,t∈0,1.



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