# New Existence of Solutions for Fractional Integro-Differential Equations with Nonseparated Boundary Conditions

Authors: Lahcen Ibnelazyz; Karim Guida; Khalid Hilal; Said Melliani
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

Results reported in this article prove the existence and uniqueness of solutions for a class of nonlinear fractional integro-differential equations supplemented by nonseparated boundary value conditions. We consider a new norm to establish the existence of solution via Krasnoselskii fixed point theorem; however, the uniqueness results are obtained by applying the contraction mapping principle. Some examples are provided to illustrate the results.


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## Body

## 1. Introduction

Fractional differential equations have been an important tool to describe many problems and processes in different fields of science. In fact, fractional models are more realistic than the classical models. Fractional differential equations appear in many fields such as physics, economics, image processing, blood flow phenomena, aerodynamics, and so on. For more details about fractional differential equations and their applications, we provide the following references [1–13].Recently, fractional integro-differential equations were investigated by many researchers in different problems, and a lot of papers were published in this matter (see, for example, [14–16]).Furthermore, many boundary conditions were considered for the fractional-order integro-differential equations; some of these conditions are the classical, periodic, anti-periodic, nonlocal, multipoint, and the integral boundary conditions (for details, see [17–22]).On the other hand, many papers have considered the nonseparated boundary conditions as they are a very important class of boundary value conditions (we refer the readers to [23–28]).Motivated by the above discussion, in this paper, we establish the existence and uniqueness of solutions for a class of fractional integro-differential equations with nonseparated boundary value conditions as follows:(1)DαcDβcxt=ft,xt,ϕxt,ψxt,t∈0,1,x0=λ1x1,x′0=λ2x′1,Dβcx0=λ3Dβcx1,Dβ+1cx0=λ4Dβ+1cx1,where 1<α≤2,1<β≤2,λ1,λ2,λ3,λ4∈ℝ−0,1, Dβc, Dαc are Caputo’s fractional derivatives, f:0;1×ℝ3⟶ℝ is a given continuous function, and(2)ϕxt=∫0tλt,sxsds,ψxt=∫0tδt,sxsds,where λ,δ:0,1×0,1⟶0,+∞, with ϕ∗=supt∈0,1∫0tλt,sds<∞, ψ∗=supt∈0,1∫0tδt,sds<∞.Our motivation comes from the fact that not many papers have considered the existence and uniqueness results of nonlinear integro-differential equations with nonseparated boundary conditions. On top of that, we show the existence results under some weak conditions. The main results in this paper can be viewed as an extension of those provided in [22].This paper is divided into five sections. In Section2, we provide some notations and basic known results. In Section 3, we study the new existence results to problem (1) under some weak conditions, and after that, we show the existence and uniqueness using Banach’s contraction principle. In Section 4, we give two examples to illustrate the results. We end the paper with a conclusion.


## 2. Preliminaries and Notations

In this section, we state some notations, definitions, and lemmas which are used in this paper.Definition 1 (see [5]).
The fractional integral of orderα>0 with the lower limit zero for a function f can be defined as(3)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(4)Dαcft=1Γn−α∫0tt−sn−α−1fnsds,where n∈ℕ, 0≤n−1<α<n, t>0.Theorem 1 (see [29]).
LetM be a bounded, closed, convex, and nonempty subset of a Banach space X. Let A and B be 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.Theorem 2 (see [30]) .
(Arzelá–Ascoli theorem).
A set of functions inCa,b is relatively compact if and only it is uniformly bounded and equicontinuous on a,b.Theorem 3 (see [30]).
If a set is closed and relatively compact, then it is compact.Lemma 1 (see [5]).
Letα,β≥0; then, the following relation holds:(5)Iαtβ=Γβ+1Γα+β+1tα+β.Lemma 2 (see [5]).
Letn∈ℕ and n−1<α<n. If f is a continuous function, then we have(6)IαDαcft=ft+a0+a1t+a2t2+⋯+an−1tn−1.Lemma 3.
Lety∈C0,1,ℝ. Then, a unique solution of the following boundary value problem:(7)DαcDβcxt=yt,t∈0,1,x0=λ1x1,x′0=λ2x′1,Dβcx0=λ3Dβcx1,Dβ+1cx0=λ4Dβ+1cx1,is given by(8)xt=1Γα+β∫0tt−sα+β−1ysds+A1tΓα∫011−sα−1ysds+A2tΓα−1∫011−sα−2ysds+λ11−λ1Γα+β∫011−sα+β−1×ysds+A3tΓα+β−1∫011−sα+β−2ysds,where(9)A1t=λ31−λ3tβΓβ+1+λ11−λ1Γβ+1+λ2λ11−λ21−λ1Γβ+tλ21−λ2Γβ,A2t=λ41−λ4tβλ3Γβ+11−λ3+tβ+1Γβ+2+λ2λ11−λ21−λ1Γβ+1+λ11−λ1Γβ+2+λ3λ2λ11−λ31−λ21−λ1Γβ+λ3λ11−λ31−λ1Γβ+1+λ3λ2t1−λ31−λ2Γβ+tλ21−λ2Γβ+1,A3t=λ21−λ2t+λ11−λ1.Proof.
By Lemma2, we have(10)Dβcxt=Iαyt+c0+c1t,Dβ+1cxt=Iα−1yt+c1,xt=Iα+βyt+Iβc0+Iβc1t+c2+c3t,where c0,c1,c2,c3∈ℝ. Using the condition Dβ+1cx0=λ4Dβ+1cx1, we get(11)c1=λ41−λ4Γα−1∫011−sα−2ysds.
By applyingDβcx0=λ3Dβcx1, we have(12)c0=λ31−λ3Γα∫011−sα−1ysds+λ3λ41−λ41−λ3Γα−1×∫011−sα−2ysds.
Now we usex′0=λ2x′1 to get(13)c3=λ21−λ2Γα+β−1∫011−sα+β−2ysds+λ2λ41−λ41−λ2Γβ+1Γα−1∫011−sα−2ysds+λ2λ3λ41−λ41−λ31−λ2ΓβΓα−1∫011−sα−2ysds+λ2λ31−λ31−λ2ΓαΓβ∫011−sα−1ysds.
In view ofx0=λ1x1, we have(14)c2=λ11−λ1Γα+β∫011−sα+β−1ysds+λ1λ21−λ11−λ2Γα+β−1∫011−sα+β−2ysds+λ1λ31−λ11−λ3Γβ+1+λ1λ2λ31−λ11−λ21−λ3Γβ×1Γα∫011−sα−1ysds+λ1λ3λ41−λ11−λ31−λ4Γβ+1+λ1λ41−λ11−λ4Γβ+2+λ1λ2λ41−λ11−λ21−λ4Γβ+1+λ1λ2λ3λ41−λ11−λ21−λ31−λ4Γβ1Γα−1∫011−sα−2ysds.
Substituting the value ofc0, c1, and c2, we get the desired results. Directly computing, one can prove the converse of the lemma.


## 3. Main Results

Denote byX the Banach space of all continuous functions from 0,1⟶ℝ endowed with the norms x=supxt:t∈0,1 and yυ=supt∈0,1yt/eυt, where υ>1+ϕ∗+φ∗/Γα+βσ, and σ will be defined later.By Lemma3, we transform problem (1) into a fixed point problem as x=Px, where P:X⟶X is given by(15)Pxt=1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxsds+A1tΓα∫011−sα−1fs,xs,ϕxs,ψxsds+A2tΓα−1∫011−sα−2fs,xs,ϕxs,ψxsds+λ11−λ1Γα+β∫011−sα+β−1fs,xs,ϕxs,ψxsds+A3tΓα+β−1∫011−sα+β−2fs,xs,ϕxs,ψxsds.For computational convenience, we set(16)A1=λ31−λ31Γβ+1+λ11−λ1Γβ+1+λ2λ11−λ21−λ1Γβ+λ21−λ2Γβ,A2=λ41−λ4λ3Γβ+11−λ3+1Γβ+2+λ2λ11−λ21−λ1Γβ+1+λ11−λ1Γβ+2+λ3λ2λ11−λ31−λ21−λ1Γβ+λ3λ11−λ31−λ1Γβ+1+λ3λ21−λ31−λ2Γβ+λ21−λ2Γβ+1,A3=λ21−λ21+λ11−λ1.Theorem 4.
Suppose thatH1 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 σ∈C0,1;0,∞.H2ft,x,y,z≤θt,∀t,x,y,z∈0,1×ℝ3 with θ∈C0,1;ℝ+.
Then, problem (1) has at least one solution.Proof.
DefineBr′=x∈X:xυ≤r′ with(17)r′≥θυeυ−1υ1Γα+β+A1Γα+A2Γα−1+λ11−λ1Γα+β+A3Γα+β−1.
We introduce the decompositionP=P1+P2, where(18)P1xt=1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxsds,P2xt=A1tΓα∫011−sα−1fs,xs,ϕxs,ψxsds+A2tΓα−1∫011−sα−2fs,xs,ϕxs,ψxsds+λ11−λ1Γα+β∫011−sα+β−1fs,xs,ϕxs,ψxsds+A3tΓα+β−1∫011−sα+β−2fs,xs,ϕxs,ψxsds.
Forx,y∈Br′, we have(19)P1xtυ≤supt∈0, 11eυt1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxsds≤supt∈0, 11eυt1Γα+β∫0tt−sα+β−1θsds≤supt∈0, 11eυt1Γα+β∫0tt−sα+β−1θseυseυsds≤supt∈0, 1θυeυtΓα+β∫0tt−sα+β−1eυsds≤supt∈0, 1θυeυtΓα+β∫0teυsds≤supt∈0, 1θυυΓα+βeυt−1eυt≤θυeυ−1υΓα+β,P2ytυ≤supt∈0, 11eυtA1tΓα∫011−sα−1fs,ys,ϕys,ψysds+A2tΓα−1∫011−sα−2fs,ys,ϕys,ψysds+λ11−λ1Γα+β∫011−sα+β−1fs,ys,ϕys,ψysds+A3tΓα+β−1∫011−sα+β−2fs,ys,ϕys,ψysds≤supt∈0, 11eυtA1Γα∫011−sα−1θseυseυsds+A2Γα−1∫011−sα−2×θseυseυsds+λ11−λ1Γα+β∫011−sα+β−1θseυseυsds+A3Γα+β−1∫011−sα+β−2θseυseυsds≤supt∈0, 1θυeυtA1Γα∫011−sα−1eυsds+A2Γα−1∫011−sα−2eυsds+λ11−λ1Γα+β∫011−sα+β−1eυsds+A3Γα+β−1∫011−sα+β−2eυsds≤supt∈0, 1θυeυtA1Γα∫01eυsds+A2Γα−1∫01eυsds+λ11−λ1Γα+β×∫01eυsds+A3Γα+β−1∫01eυsds≤supt∈0, 1eυ−1υeυtθυA1Γα+A2Γα−1+λ11−λ1Γα+β+A3Γα+β−1≤eυ−1υθυA1Γα+A2Γα−1+λ11−λ1Γα+β+A3Γα+β−1.
Thus,(20)P1x+P2yυ≤θυeυ−1υ1Γα+β+A1Γα+A2Γα−1+λ11−λ1Γα+β+A3Γα+β−1.
Hence,P1x+P2y∈Br′.
Forx,y∈Br′, consider(21)P1yt−P1xtυ≤supt∈0, 11Γα+βeυt∫0tt−sα+β−1×fs,ys,ϕys,ψys−fs,xs,ϕxs,ψxsds≤supt∈0, 11Γα+βeυt∫0tt−sα+β−1σsys−xs+ϕys−ϕxs+ψys−ψxsds≤supt∈0, 1σΓα+βeυt∫0teυsy−xυ+ϕ∗y−xυ+ψ∗y−xυds≤supt∈0, 11+ϕ∗+ψ∗συΓα+βeυt−1eυty−xυ≤1+ϕ∗+ψ∗σy−xυΓα+βυ.
By using the condition of the new norm, we have thatP1 is a contraction.
Next, we will show thatP2 is compact and continuous.
Continuity off implies that the operator P2 is continuous. Also, P2 is uniformly bounded on Br′ as(22)P2yυ≤eυ−1υθυA1Γα+A2Γα−1+λ11−λ1Γα+β+A3Γα+β−1.
Suppose that0≤t1<t2≤1. We have(23)P2yt2−P2yt1≤A1t2−A1t1Γα∫011−sα−1fs,ys,ϕys,ψys×ds+A1t2−A1t1Γα−1∫011−sα−2×fs,ys,ϕys,ψysds+A3t2−A3t1Γα+β−1×∫011−sα+β−2fs,ys,ϕys,ψysds.
Thus,(24)P2yt2−P2yt1⟶0,ast1⟶t2independently fromy∈Br′.
This proves that the operatorP2 is relatively compact on Br′. Then, by the Arzelá–Ascoli theorem, we have that P2 is compact on Br′.
Therefore, problem (1) has at least one solution on Br′.Theorem 5.
Assume thatf:0,1×ℝ3⟶ℝ is continuous function satisfyingH1 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 Σ∈L10,1;0,∞ and Σ∗=∫01Σsds.
Then, there exists a unique solution for the boundary value problem (1) provided that(25)r1=1+ϕ∗+ψ∗Σ∗1Γα+β+A1Γα+A2Γα−1+λ11−λ1Γα+β+A3Γα+β−1<1.Proof.
Setsup0≤t≤1ft,0,0,0=M.
LetBr=x∈X:x≤r be the ball with radius r, where r≥r2/1−r1, with(26)r2=MΓα+β+A1MΓα+A2MΓα−1+λ1M1−λ1Γα+β+A3MΓα+β−1.
Then,Br is a closed, convex, and nonempty subset of the Banach space X.
Our objective is to show that the operatorP has a unique fixed point on Br.
We prove thatPBr⊆Br.
Forx∈Br, t∈0,1, we have(27)Pxt≤1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxsds+A1tΓα∫011−sα−1fs,xs,ϕxs,ψxsds+A2tΓα−1∫011−sα−2fs,xs,ϕxs,ψxsds+λ11−λ1Γα+β∫011−sα+β−1fs,xs,ϕxs,ψxsds+A3tΓα+β−1∫011−sα+β−2fs,xs,ϕxs,ψxsds≤1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxs−fs,0,0,0+fs,0,0,0ds+A1tΓα∫011−sα−1fs,xs,ϕxs,ψxs−fs,0,0,0+fs,0,0,0ds+A2tΓα−1∫011−sα−2×fs,xs,ϕxs,ψxs−fs,0,0,0+fs,0,0,0ds+λ11−λ1Γα+β∫011−sα+β−1fs,xs,ϕxs,ψxs−fs,0,0,0+fs,0,0,0ds+A3tΓα+β−1×∫011−sα+β−2fs,xs,ϕxs,ψxs−fs,0,0,0+fs,0,0,0ds≤1Γα+β∫0tt−sα+β−1Σsxs+ϕxs+ψxs+Mds+A1Γα∫011−sα−1Σsxs+ϕxs+ψxs+Mds+A2Γα−1∫011−sα−2Σsxs+ϕxs+ψxs+Mds+λ11−λ1Γα+β∫011−sα+β−1Σsxs+ϕxs+ψxs+Mds+A3Γα+β−1∫011−sα+β−2Σs×xs+ϕxs+ψxs+Mds≤1+ϕ∗+ψ∗xΓα+β∫01Σsds+MΓα+β∫0tt−sα+β−1ds+1+ϕ∗+ψ∗A1xΓα∫01Σsds+A1MΓα∫0tt−sα−1ds+1+ϕ∗+ψ∗A2xΓα−1∫01Σsds+A2MΓα−1∫0tt−sα−2ds+λ11+ϕ∗+ψ∗x1−λ1Γα+β∫01Σsds+λ1M1−λ1Γα+β∫011−sα+β−1ds+A31+ϕ∗+ψ∗xΓα+β−1∫01Σsds+A3MΓα+β−1∫011−sα+β−2ds,which implies that(28)Px≤1+ϕ∗+ψ∗Σ∗xΓα+β+MΓα+β+1+ϕ∗+ψ∗Σ∗A1xΓα+A1MΓα+1+ϕ∗+ψ∗Σ∗A2xΓα−1+A2MΓα−1+λ11+ϕ∗+ψ∗Σ∗x1−λ1Γα+β+λ1M1−λ1Γα+β+A31+ϕ∗+ψ∗Σ∗xΓα+β−1+A3MΓα+β−1≤1+ϕ∗+ψ∗Σ∗Γα+β+1+ϕ∗+ψ∗Σ∗A1Γα+1+ϕ∗+ψ∗Σ∗A2Γα−1+λ11+ϕ∗+ψ∗Σ∗1−λ1Γα+β+A31+ϕ∗+ψ∗Σ∗Γα+β−1x+MΓα+β+A1MΓα+A2MΓα−1+λ1M1−λ1Γα+β+A3MΓα+β−1≤r1r+r2≤r.
Now, forx,y∈Br and for t∈0,1,(29)Pxt−Pyt≤1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxs−fs,ys,ϕys,ψysds+A1tΓα∫011−sα−1×fs,xs,ϕxs,ψxs−fs,ys,ϕys,ψysds+A2tΓα−1∫011−sα−2fs,xs,ϕxs,ψxs−fs,ys,ϕys,ψysds+λ11−λ1Γα+β×∫011−sα+β−1fs,xs,ϕxs,ψxs−fs,ys,ϕys,ψysds+A3tΓα+β−1×∫011−sα+β−2fs,xs,ϕxs,ψxs−fs,ys,ϕys,ψysds,and then(30)Px−Py≤r1x−y.
Sincer1<1, then the operator P is a contraction mapping. Thus, problem (1) has a unique solution.


## 4. Examples

Example 1.
Consider the following fractional problem:(31)D17/11cD15/11cxt=t4800xte−t1+xt+∫0tt+s4xse−2s+e−3s8001+xsds,t∈0,1,x0=1200x1,x′0=1200x′1,D15/11cx0=1200D15/11cx1,D26/11cx0=1200D26/11cx1.
We choose(32)β=1511,α=1711,ft,x,y,z=t4800xte−t1+xt+yte−2t1+yt+zte−3t1+zt,λt,s=δt,s=t+s4800.
We have(33)ft,x1,x2,x3−ft,y1,y2,y3≤t4800x1−y1+x2−y2+x3−y3,for allt∈0,1andx1,x2,x3,y1,y2,y3∈ℝ,ft,x1,x2,x3≤3t4800.
Clearly,ϕ∗=ψ∗=31/4000, σ=1/800.
Hence, by Theorem4, problem (31) has a least one solution.Example 2.
Consider the following boundary value problem:(34)D10/6cD8/6cxt=t3300011+xt+11000∫0tt5s4xsds,t∈0,1,x0=1201x1,x′0=1201x′1,D8/6cx0=1201D8/6cx1,D14/6cx0=1201D14/6cx1.
Hereβ=8/6, α=10/6, ft,x,y,z=t3/30001/1+xt+yt+zt, λt,s=δt,s=t5s4/2000.
We get(35)ft,x1,x2,x3−ft,y1,y2,y3≤t33000x1−y1+x2−y2+x3−y3,for allt∈0,1andx1,x2,x3,y1,y2,y3,∈ℝ.
Clearly,ϕ∗=ψ∗=1/10000, Σ∗=1/12000.
Hence, we haver1≈0.000043<1.
Then, problem (34) has a unique solution.


## 5. Conclusion

In this paper, new existence and uniqueness results have been studied for nonlinear fractional integro-differential equations equipped with nonseparated conditions. By using a new norm, we have established the new existence of solution for (1) under some weak conditions. The uniqueness of solutions has been given by using the sup norm and applying the Banach fixed point theorem. We have ended the article with some examples to illustrate the results.



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