# On a Nonlocal Multipoint and Integral Boundary Value Problem of Nonlinear Fractional Integrodifferential Equations

Authors: Lahcen Ibnelazyz; Karim Guida; Said Melliani; Khalid Hilal
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The aim of this paper is to give the existence as well as the uniqueness results for a multipoint nonlocal integral boundary value problem of nonlinear sequential fractional integrodifferential equations. First of all, we give some preliminaries and notations that are necessary for the understanding of the manuscript; second of all, we show the existence and uniqueness of the solution by means of the fixed point theory, namely, Banach’s contraction principle and Krasnoselskii’s fixed point theorem. Last, but not least, we give two examples to illustrate the results.


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## Body

## 1. Introduction

In the last few years, fractional differential equations have gained much attention among mathematicians because of the rapid growth and for their applicability in several fields, such as physics, biology, economics, control theory, and engineering; for more details about the theory of fractional differential equations and their applications, we recommend the following articles [1–15] and the references therein.Furthermore, fractional differential equations with multipoint boundary conditions have provoked a great deal of attention by many authors; a lot of works have been published on this topic; for more details, we give the following references [16–22].In a recent paper [23], the existence of solutions for a four-point nonlocal boundary value problem of nonlinear integrodifferential equations of fractional order was proven. In [24], the authors discussed the existence of solutions for fractional differential equations with multipoint boundary conditions. The existence and uniqueness of solutions for multiterm nonlinear fractional integrodifferential equations have been studied in [25]. The existence results for sequential fractional integrodifferential equations with nonlocal multipoint and strip conditions were established in [26], and finally, in [27], the authors studied the existence of solutions for nonlinear fractional integrodifferential equations.Motivated by all these works, and by the fact that there are no papers dealing with nonlinear fractional integrodifferential equations with multipoint and integral boundary value conditions, in this work, we consider the existence and uniqueness of solutions for the following problem:(1)DcβDcαxt=ft,xt,ϕxt,φxt,t∈0,1,x0=∑i=1naixti,x′0=0,μ1x1+μ2x′1=∑j=1mbj∫cjdjxsds,where 0<α<1, 1<β≤2, with α+β−2≥0, μ1, μ2, ai, bj∈ℝ; 0<ti<cj<dj<1, i=1,2,⋯,n, j=1,2,⋯,m, Dcα, Dcβ are the Caputo fractional derivatives, and f:0,1×ℝ3→ℝ is a continuous function and ϕxt=∫0tλt,sxsds, φxt=∫0tδt,sxsds, where λ, δ:0,1×0,1→0,+∞, with λ0=supt∈0,1∫0tλt,sds<∞, δ0=supt∈0,1∫0tδt,sds<∞.This paper is organized as follows: in the second section, we give some preliminaries and notations that will be useful throughout the work; after that, in the third section, we establish the main results by using the fixed point theory; and in the last section, we give some examples to illustrate the results.


## 2. Preliminaries and Notations

Throughout this section, we present some notations, definitions, and lemmas which will be used for the rest of the paper.Definition 1 (see [5]).
The fractional integral of orderα>0 with the lower limit zero for a function f can be defined as
(2)Iαft=1Γα∫0tt−sα−1fsds.Definition 2 (see [5]).
The Caputo derivative of orderα>0 with the lower limit zero for a function f can be defined as
(3)Dcαf=1Γn−α∫0tt−sn−α−1fnsds.
wheren∈ℕ, 0≤n−1<α<n, t>0.Theorem 3 (see [28]).
LetM be a bounded, closed, convex, and nonempty subset of a Banach space X. Let A and B be two operators such that
(i)
Ax+By∈M whenever x,y∈M(ii)
A is compact and continuous(iii)
B is a contraction mapping
Then, there existsz∈M such that z=Az+Bz.Lemma 4 (see [5]).
Letα,β≥0, then the following relation holds:
(4)Iαtβ=Γβ+1Γα+β+1tα+β.Lemma 5 (see [5]).
Letn∈ℕ and n−1<α<n. If f is a continuous function, then we have
(5)IαDcαft=ft+a0+a1t+a2t2+⋯+an−1tn−1.Lemma 6.
Leth∈C0,1,ℝ. Then, the unique solution of the boundary value problem
(6)DcβDcαxt=ht,t∈0,1,x0=∑i=1naixti,x′0=0,μ1x1+μ2x′1=∑j=1mbj∫cjdjxsds,is given by
(7)xt=1Γα+β∫0tt−sα+β−1hsds−B1tΓα+β∑i=1nai∫0titi−sα+β−1×hsds+B2t∑j=1mbjΓα+β∫cjdj∫0ss−uα+β−1hududs−μ1Γα+β∫011−sα+β−1hsds−μ2Γα+β−1×∫0t1−sα+β−2hsds,where
(8)Δ=L1K2−L2K1≠0,L1=−∑i=1naitiα+1Γα+2,L2=1−∑j=1nai,K1=μ1Γα+2+μ2Γα+1−∑j=1mbjdjα+2−cjα+2Γα+3,K2=μ1−∑j=1mbjdj−cj,B1t=1ΔK1−K2tα+1Γα+2,B2t=1ΔL1−L2tα+1Γα+2.Proof.
By applying Lemma5, we obtain
(9)Dcαxt=Iβht+θ0+θ1t,xt=Iα+βht+Iαθ0+Iαθ1t+θ2,
whereθ0,θ1,θ2∈ℝ.
This means that(10)xt=1Γα+β∫0tt−sα+β−1hsds+tαΓα+1θ0+tα+1Γα+2θ1+θ2,x′t=1Γα+β−1∫0tt−sα+β−2hsds+tα−1Γαθ0+tαΓα+1θ1,
and by using the conditionx′0=0, we get θ0=0.
As a result of the conditionx0=∑i=1naixti, we find that L2θ2+L1θ1=H1, where
(11)H1=∑i=1naiΓα+β∫0titi−sα+β−1hsds.
Now, we use the conditionμ1x1+μ2x′1=∑j=1mbj∫cjdjxsds, to obtain K2θ2+K1θ1=H2, where
(12)H2=−μ1Γα+β∫011−sα+β−1hsds−μ2Γα+β−1∫011−sα+β−2×hsds+∑j=1mbjΓα+β∫cjdi∫0ss−uα+β−1hududs.
Finally, we have(13)θ1=K2H1−L2H2Δ,θ2=L1H2−K1H1Δ.
By substituting the value ofθ0, θ1, and θ2, we get the following:
(14)xt=1Γα+β∫0tt−sα+β−1hsds−B1tΓα+β∑i=1nai∫0titi−sα+β−1×hsds+B2t∑j=1mbjΓα+β∫cjdj∫0ss−uα+β−1hududs−μ1Γα+β∫011−sα+β−1hsds−μ2Γα+β−1×∫011−sα+β−2hsds.
Conversely, by direct computations, we obtain the desired result.


## 3. Main Results

LetX be the Banach space of all continuous functions from 0,1→ℝ endowed with norm x=sup∣xt∣:t∈0,1.Theorem 7.
Suppose thatf:0,1×ℝ3→ℝ is a continuous function satisfying
(H1) for all t∈0,1 and x1,x2,x3,y1,y2,y3∈ℝ, we have ft,x1,x2,x3−ft,y1,y2,y3≤σtx1−y1+x2−y2+x3−y3 with σt∈L10,1;0,∞.Then, there exists a unique solution for problem (1) under the following condition: r1<1, where
(15)r1=1+λ0+δ0σ∗1+B1∑i=1nai+B2∑j=1mbjdj−cj+B2μ1Γα+β+B2μ2Γα+β−1,with B1=1/ΔK1+K2/Γα+β,B2=1/ΔL1+L2/Γα+2, and σ∗=∫01σtdt.Proof.
DefineP:X→X by
(16)Pxt=1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,φxsds−B1tΓα+β×∑i=1nai∫0titi−sα+β−1fs,xs,ϕxs,φxsds+B2t×∑j=1mbjΓα+β∫cjdj∫0ss−uα+β−1fu,xu,ϕxu,φxu×duds−μ1Γα+β∫011−sα+β−1fs,xs,ϕxs,φxsds−μ2Γα+β−1∫011−sα+β−2fs,xs,ϕxs,φxsds.
Settingsup0≤t≤1ft,0,0,0=F.
We consider the following setBr=x∈X:x≤r, where r≥r2/1−r1, with
(17)r2=FΓα+β+B1Γα+β∑i=1naiF+B2∑j=1mbjFdj−cjΓα+β+B2μ1FΓα+β+B2μ2FΓα+β−1.
For eacht∈0,1 and x∈Br, we have
(18)Pxt≤1Γα+β∫0tt−sα+β‐1fs,xs,ϕxs,φxsds+B1tΓα+β×∑i=1nai∫0titi−sα+β‐1fs,xs,ϕxs,φxsds+B2t∑j=1mbjΓα+β∫cjdj∫0ss−uα+β‐1×fu,xu,ϕxu,φxududs+μ1Γα+β∫011−sα+β‐1×fs,xs,ϕxs,φxsds+μ2Γα+β−1∫011−sα+β‐2×fs,xs,ϕxs,φxsds≤1Γα+β∫0tt−sα+β‐1fs,xs,ϕxs,φxs−fs,0,0,0+fs,0,0,0ds+B1tΓα+β∑i=1nai∫0titi−sα+β‐1×fs,xs,ϕxs,φxs−fs,0,0,0+fs,0,0,0ds+B2t∑j=1mbjΓα+β∫cjdj∫0ss−uα+β−1×fu,xu,ϕxu,φxu−fu,0,0,0+fu,0,0,0duds+μ1Γα+β∫011−sα+β−1fs,xs,ϕxs,φxs−fs,0,0,0+fs,0,0,0ds+μ2Γα+β−1∫011−sα+β−2×fs,xs,ϕxs,φxs−fs,0,0,0+fs,0,0,0ds≤1Γα+β∫0tt−sα+β−1σsxs+ϕxs+φxs+Fds+B1Γα+β∑i=1nai∫0titi−sα+β−1σsxs+ϕxs+φxs+Fds+B2∑j=1mbjΓα+β∫cjdj×∫0ss−uα+β−1σuxu+ϕxu+φxu+Fduds+μ1Γα+β∫011−sα+β−1σsxs+ϕxs+φxs+Fds+μ2Γα+β−1∫011−sα+β−2σsxs+ϕxs+φxs+Fds≤1+λ0+δ0xΓα+β∫01σsds+FΓα+β∫0tt−sα+β−1ds+B1Γα+β∑i=1nai1+λ0+δ0x∫01σsds+B1Γα+β∑i=1naiF×∫0titi−sα+β−1ds+B2∑j=1mbjΓα+β∫cjdj1+λ0+δ0x×∫01σudu+B2∑j=1mbjFdj−cjΓα+β+B2μ1Γα+β1+λ0+δ0x×∫01σsds+B2μ1FΓα+β∫011−sα+β−1ds+B2μ2Γα+β−11+λ0+δ0x∫01σsds+B2μ2FΓα+β−1∫011−sα+β−2ds≤1+λ0+δ0xΓα+βσ∗+FΓα+β+B1Γα+β∑i=1nai1+λ0+δ0xσ∗+B1Γα+β∑i=1naiF+B2∑j=1mbjdj−cjΓα+β1+λ0+δ0xσ∗+B2∑j=1mbjFdj−cjΓα+β+B2μ1Γα+β1+λ0+δ0xσ∗+B2μ1FΓα+β+B2μ2Γα+β−11+λ0+δ0σ∗x+B2μ2FΓα+β−1≤1+λ0+δ0σ∗Γα+β+B1Γα+β∑i=1nai1+λ0+δ0σ∗+B2×∑j=1mbjdj−cjΓα+β1+λ0+δ0σ∗+B2μ1Γα+β1+λ0+δ0σ∗+B2μ2Γα+β−11+λ0+δ0σ∗x+FΓα+β+B1Γα+β×∑i=1naiF+B2∑j=1mbjFdj−cjΓα+β+B2μ1FΓα+β+B2μ2FΓα+β−1.
This means thatPx≤r.
Therefore,PBr⊆Br.
Next, we prove thatP is a contraction mapping.
Forx,y∈Br, we have
(19)Pxt−Pyt≤1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,φxs−fs,ys,ϕys,φysds+B1∑i=1naiΓα+β×∫0titi−sα+β−1fs,xs,ϕxs,φxs−fs,ys,ϕys,φysds+B2∑j=1mbjΓα+β×∫cjdj∫0ss−uα+β−1fs,xu,ϕxu,φxu−fs,yu,ϕyu,φyududs+μ1Γα+β×∫011−sα+β−1fs,xs,ϕxs,φxs−fs,ys,ϕys,φysds+μ2Γα+β−1×∫011−sα+β−2fs,xs,ϕxs,φxs−fs,ys,ϕys,φysds≤1Γα+β∫0tt−sα+β−1σsxs−ys+ϕxs−ϕys+φxs−φysds+B1Γα+β∑i=1nai∫0titi−sα+β−1σs×xs−ys+ϕxs−ϕys+φxs−φysds+B2∑j=1mbjΓα+β∫cjdj∫0ss−uα+β−1σu×xu−yu+ϕxu−ϕyu+φxu−φyududs+μ1Γα+β∫011−sα+β−1σsxs−ys+ϕxs−ϕys+φxs−φysds+μ2Γα+β−1∫011−sα+β−2σsxs−ys+ϕxs−ϕys+φxs−φysds≤1+λ0+δ0x−yΓα+β∫01σsds+B1Γα+β∑i=1nai1+λ0+δ0x−y∫01σsds+B2∑j=1mbjΓα+β∫cjdj1+λ0+δ0x−y∫01σududs+B2μ1Γα+β1+λ0+δ0x−y∫01σsds+B2μ2Γα+β−11+λ0+δ0x−y∫01σsds≤1+λ0+δ0x−yΓα+βσ∗+B1Γα+β∑i=1nai1+λ0+δ0x−yσ∗+B2∑j=1mbjdj−cjΓα+β1+λ0+δ0σ∗x−y+B2μ1Γα+β1+λ0+δ0σ∗x−y+B2μ2Γα+β−11+λ0+δ0σ∗x−y≤1+λ0+δ0σ∗1+B1∑i=1nai+B2∑j=1mbjdj−cj+B2μ1Γα+β+B2μ2Γα+β−1x−y.
Sincer1<1, then P is a contraction. Therefore, system (1) has a unique solution.Theorem 8.
Assume that (H1) holds and f:0;1×ℝ3→ℝ is a continuous function. Furthermore, we suppose
(20)H3ft,x,y,z≤θt,∀t,x,y,z∈0,1×ℝ3withθ∈L10,1;ℝ+.Then, problem (1) has at least one solution on [0,1] if R<1, where
(21)R=1+λ0+δ0σ∗B1∑i=1nai+B2∑j=1mbjdj−cj+B2μ1Γα+β+B2μ2Γα+β−1.Proof.
We now consider the closed ballBr′=x∈X:x≤r′ with fixed radius r′:
(22)r′≥θL1Γα+β+1+B1Γα+β+1∑i=1naitiα+βθL1+B2∑j=1mbjθL1dj−cjΓα+β+1+μ1θL1Γα+β+1+μ2θL1Γα+β.
We define the operatorsP1 and P2 on Br′ as
(23)P1xt=1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,φxsds,P2xt=−B1t∑i=1naiΓα+β∫0titi−sα+β−1fs,xs,ϕxs,φxsds+B2t×∑j=1mbjΓα+β∫cjdj∫0ss−uα+β−1fu,xu,ϕxu,φxu×duds−μ1Γα+β∫011−sα+β−1fs,xs,ϕxs,φxsds−μ2Γα+β−1∫011−sα+β−2fs,xs,ϕxs,φxsds.
Forx,y∈Br′, we have
(24)P1xt≤θL1Γα+β+1,P2yt≤B1Γα+β+1∑i=1naitiα+βθL1+B2∑j=1mbjθL1dj−cjΓα+β+1+μ1θL1Γα+β+1+μ2θL1Γα+β+1.
Consequently,(25)P1x+P2y≤θL1Γα+β+1+B1Γα+β+1∑i=1naitiα+βθL1+B2×∑j=1mbjθL1dj−cjΓα+β+1+μ1θL1Γα+β+1+μ2θL1Γα+β.
Then,(26)P1x+P2y∈Br′.
Next, we show thatP2 is a contraction. For x,y∈Br′, we have
(27)P2xt−P2yt≤B1tΓα+β∑i=1nai∫i=1titi−sα+β−1×fs,xs,ϕxs,φxs−fs,ys,ϕys,φysds+B2t∑j=1mbjΓα+β∫cjdj∫0ss−uα+β−1×fs,xu,ϕxu,φxu−fs,yu,ϕyu,φyududs+μ1Γα+β∫011−sα+β−1fs,xs,ϕxs,φxs−fs,ys,ϕys,φysds+μ2Γα+β−1∫011−sα+β−2fs,xs,ϕxs,φxs−fs,ys,ϕys,φysds≤B1Γα+β∑i=1nai∫0tit−sα+β−1σsxs−ys+ϕxs−ϕys+φxs−φysds+B2∑j=1mbjΓα+β×∫cjdj∫0ss−uα+β−1σuxu−yu+ϕxu−ϕyu+φxu−φyududs+μ1Γα+β∫011−sα+β−1σsxs−ys+ϕxs−ϕys+φxs−φysds+μ2Γα+β−1∫011−sα+β−2σsxs−ys+ϕxs−ϕys+φxs−φysds≤1+λ0+δ0σ∗B1∑i=1nai+B2∑j=1mbjdj−cj+B2μ1Γα+β+B2μ2Γα+β−1x−y.since R<1, we conclude that P2 is a contraction. Now, we show that P1 is compact and continuous.
Continuity off implies that the operator P1 is continuous. Also, P1 is uniformly bounded on Br′ as
(28)P1x≤θL1Γα+β+1.
Suppose that0≤t1<t2≤1. We have
(29)P1xt2−P1xt1≤1Γα+β∫0t1t2−sα+β−1−t1−sα+β−1×fs,xs,ϕxs,φxsds+∫t1t2t2−sα+β−1fs,xs,ϕxs,φxsds≤θL1Γα+β+12t2−t1α+β+t1α+β−t2α+β.
Hence,P1xt2−P1xt1→0, as t1→t2 independently from x∈Br′.
This shows that the operatorP1 is relatively compact on Br′. Hence, by the Arzela-Ascoli theorem, P1 is compact on Br′.
Then, by Krasnoselskii’s fixed point theorem, problem (1) has at least one solution on Br′.


## 4. Example

In this section, we give two examples to prove the applicability of our main results.Example 9.
Let us consider the following system:(30)Dc11/7Dc5/7xt=t220011+xt+1100∫0tt4s3xsds,t∈0,1,x0=1500x119+x419+x819+x1119,x′0=0,1300x1+1300x′1=1400∫2/193/19xsds+∫5/197/19xsds+∫9/1910/19xsds.Here,m=3, n=4, β=11/7, α=5/7, a1=a2=a3=a4=1/500, μ1=μ2=1/300, b1=b2=b3=1/400, c1=2/19, c2=5/19, c3=9/19, d1=3/19, d2=7/19, d3=10/19, ft,x,y,z=t2/2001/1+xt+yt+zt, λt,s=δt,s=t4s3/200, and σt=t2/200.It follows that(31)λ0=δ0=1800,σ∗=1600,B1≈1,3367,B2≈117,95142,r1≈0,0027.By Theorem7, we obtain that problem (30) has a unique solution.Example 10.
Consider the following problem:(32)Dc18/11Dc6/11xt=t3400xte−t1+xt+∫0tt+s3xscoss+sins4001+xst∈0,1,x0=1800x1241+x1541+x1841+x2141,x′0=0,1600x1+1600x′1=1700∫13/4114/41xsds+∫16/4117/41xsds+∫19/4120/41xsds.Here,m=3, n=4, β=18/11, α=6/11, a1=a2=a3=a4=1/800, μ1=μ2=1/600, b1=b2=b3=1/700, c1=13/41, c2=16/41, c3=19/41, d1=14/41, d2=17/41, d3=20/41, ft,x,y,z=t3/400xte−t/1+xt+ytcost/1+yt+ztsint/1+zt, λt,s=δt,s=t+s3/400, σt=t3/400, and θt=3t3/400.It is clear that(33)λ0=δ0=151600,σ∗=11600,B1≈1,3641,B2≈229,3039,R≈0,000566.Then, problem (32) has at least one solution.



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