# New Existence Results for Nonlinear Fractional Integrodifferential Equations

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

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## Abstract

This paper discusses a boundary value problem of nonlinear fractional integrodifferential equations of order1<α≤2 and 1<β≤2 and boundary conditions of the form x0=x1=Dcβx1=Dcβx0=0. Some new existence and uniqueness results are proposed by using the fixed point theory. In particular, we make use of the Banach contraction mapping principle and Krasnoselskii’s fixed point theorem under some weak conditions. Moreover, two illustrative examples are studied to support the results.


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## Body

## 1. Introduction

Fractional differential equations are relevant in many fields of science, such as chemistry, fluid systems, and electromagnetic; for more details about the theory of fractional differential equations and their applications, we invite the readers to see [1–16] and the references therein.Some physical applications of fractional differential equations include viscoelasticity, Schrodinger equation, fractional diffusion equation, and fractional relaxation equation; for more details, we refer the readers to [17].In addition, fractional integrodifferential equations are used as an important tool to gain insight into some emerging problems from several science areas, for more details, we give the following references [18–23].More recently, in [24], the existence and uniqueness of positive solutions for the fractional integrodifferential equation were proved.In [25], the authors discussed the existence and uniqueness of solutions for nonlinear integrodifferential equations of fractional order with three-point nonlocal fractional boundary conditions. The existence of solutions for nonlinear fractional integrodifferential equations has been studied in [26].Motivated by all these works and by the fact that there are no papers dealing with the new existence results for nonlinear fractional integrodifferential equations, in this work, we consider the existence and uniqueness of solutions for the following problem:(1)DcαcDβxt=ft,xt,Φxt,ψxt,t∈0,1,x0=x1=Dcβx1=cDβx0=0,where 1<α≤2,1<β≤2,cDα,cDβ are the Caputo fractional derivatives, f:0,1×ℝ3⟶ℝ is a continuous function, and
(2)Φxt=∫0tλt,sxsds,ψxt=∫0tδt,sxsds,whereλ,δ:0,1×0,1−⟶0,+∞, with ϕ∗=supt∈0,1∫0tλt,sds<∞, ψ∗=supt∈0,1∫0tδt,sds<∞.This paper is organized as follows. In Section2, we present some preliminaries and notations that will be required for the later sections. After that, in Section 3, we establish the main results by using the fixed point theory. And, in the last section, we give two examples to illustrate the results.


## 2. Preliminaries and Notations

In this section, we give some notations, definitions, and lemmas which will be required for the rest of the paper.Definition 1 [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 [5].
The Caputo derivative of orderα>0 with the lower limit zero for a function f can be defined as
(4)Dcαft=1Γn−α∫0tt−sn−α−1fnsds,
wheren∈ℕ, 0≤n−1<α<n, t>0.Theorem 3 [27].
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 [5].
Letα,β≥0; then, the following relation hold:
(5)Iαtβ=Γβ+1Γα+β+1tα+β.Lemma 5 [5].
Letn∈ℕ and n−1<α<n. If f is a continuous function, then we have
(6)IαDcαft=ft+a0+a1t+a2t2+⋯+an−1tn−1.Lemma 6.
Leth∈C0,1,ℝ. Then, the unique solution of the problem
(7)DcαDcβxt=ht,t∈0,1,x0=x1=Dcβx1=Dcβx0=0,is given by
(8)xt=1Γα+β∫0tt−sα+β−1hsds−tβ+1+tΓαΓβ+2∫011−sα−1×hsds−tΓα+β∫011−sα+β−1hsds.Proof.
By applying Lemma5, we have
(9)Dcβxt=Iαht+a0+a1t,xt=Iα+βht+Iβa0+Iβa1t+a2+a3t,where a0,a1,a2,a3∈ℝ. So
(10)xt=1Γα+β∫0tt−sα+β−1hsds+tβΓβ+1a0+tβ+1Γβ+2a1+a2+a3t.And by using cDβx0=x0=0, we obtain a0=a2=0. As a result of cDβx1=0, we have that
(11)a1=−1Γα∫011−sα−1hsds.Now, we use x1=0 to get
(12)a3=−1Γα+β∫011−sα+β−1hsds+1ΓαΓβ+2∫011−sα−1hsds.By substituting the value of a0,a1,a2,a3, we obtain the following
(13)xt=1Γα+β∫0tt−sα+β−1hsds−tβ+1−tΓαΓβ+2∫011−sα−1×hsds−tΓα+β∫011−sα+β−1hsds.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 the norms y=supyt:t∈0,1 and yυ=supt∈0,1yt/eυt, where υ>1+ϕ∗+φ∗/Γα+βσ, and σ will be defined later.Theorem 7.
Assume that
(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 σ∈C0,1;0,∞
(H2) ft,x,y,z≤θt, ∀t,x,y,z∈0,1×ℝ3 with θ∈C0,1;ℝ+. Then, the problem (1) has at least one solution.Proof.
We consider the ballBr=y∈X:yυ≤r with r≥θ/υ2eυ−1/ΓαΓβ+2+eυ/Γα+β.We define the operators F=F1+F2 on Br, where.
(14)F1yt=1Γα+β∫0tt−sα+β−1fs,ys,ϕys,ψysds,(15)F2yt=−tβ+1−tΓαΓβ+2∫011−sα−1fs,ys,ϕys,ψysds−1Γα+β∫011−sα+β−1fs,ys,ϕys,ψysds. For x,y∈Br, we have
(16)F1xtυ≤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≤θυυΓα+β,F2ytυ≤supt∈0,11eυt−tβ+1−tΓαΓβ+2∫011−sα−1fs,ys,ϕys,ψysds−tΓα+β∫011−sα+β−1fs,ys,ϕys,ψysds≤supt∈0,11eυt2ΓαΓβ+2∫011−sα−1θseυseυsds+1Γα+β×∫011−sα+β−1θseυseυsds≤supt∈0,1θυeυt2ΓαΓβ+2∫011−sα−1eυsds+1Γα+β×∫011−sα+β−1eυsds≤supt∈0,1θυeυt2ΓαΓβ+2∫01eυsds+1Γα+β∫01eυsds≤supt∈0,1θυυeυt2eυ−1ΓαΓβ+2+eυ−1Γα+β≤θυυ2eυ−1ΓαΓβ+2+eυ−1Γα+β.Therefore,
(17)F1x+F2yυ≤θυ2eυ−1ΓαΓβ+2+eυΓα+β.Then,
(18)F1x+F2y∈Br.Now, we prove that F1 is a contraction. For x,y∈Br, we have
(19)F1yt−F1xtυ≤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 conclude that F1 is a contraction.
Next, we will prove thatF2 is compact and continuous.
Continuity off implies that the operator F2 is continuous. Also, F2 is uniformly bounded on Br as
(20)F2y≤θυeυ−1υ2ΓαΓβ+2+1Γα+β.Suppose that 0≤t1<t2≤1. We have
(21)F2yt2−F2yt1≤t2β+1−t1β+1+t2−t1ΓαΓβ+2∫011−sα−1×fs,xs,ϕxs,ψxsds+t2−t1Γα+β∫011−sα+β−1fs.xs,ϕxs,ψxsds.Then, F2yt2−F2yt1⟶0, as t1⟶t2 independently from y∈Br.
This shows that the operatorF2 is relatively compact on Br. Thus, by the Arzela Ascoli theorem, we obtain that F2 is compact on Br.
By the Krasnoselskii fixed point theorem, the problem (1) has at least one solution on Br.Theorem 8.
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 the problem (1) under the following condition: r1<1, where r1=21+ϕ∗+ψ∗σ∗1/Γα+β+1/ΓαΓβ+2,
withσ∗=∫01σtdt.Proof.
DefineF:X⟶X by
(22)Fxt=1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxs,ψxsds−tβ+1−tΓαΓβ+2∫011−sα−1fs,xs,ϕxs,ψxsds−tΓα+β∫011−sα+β−1fs,xs,ϕxs,ψxsds.Setting sup0≤t≤1ft,0,0,0=P.
We consider the setBr=x∈X:x≤r, where r≥r2/1−r1, with
(23)r2=2P1Γα+β+1ΓαΓβ+2.For each t∈0,1 and x∈Br, we have
(24)Fxt≤1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxsds+tβ+1+tΓαΓβ+2∫011−sα−1fs,xs,ϕxs,ψxsds+tΓα+β∫011−sα+β−1fs,xs,ϕxs,ψxs≤1Γα+β∫01t−sα+β−1fs,xs,ϕxs,ψxs−fs,0,0,0+fs,0,0,0ds+tβ+1+tΓαΓβ+2∫011−sα−1×fs,xs,ϕxs,ψxs−fs,0,0,0+fs,0,0,0ds+1Γα+β∫01t−sα+β−1fs,xs,ϕxs,ψxs−fs,0,0,0+fs,0,0,0ds≤1Γα+β∫01t−sα+β−1σsxs+ϕxs+ψxs+Pds+2ΓαΓβ+2∫011−sα−1σsxs+ϕxs+ψxs+Pds+2Γα+β∫011−sα+β−1σsxs+ϕxs+ψxs+Pds≤1+ϕ∗+ψ∗xΓα+β∫01σsds+PΓα+β∫0tt−sα+β−1ds+21+ϕ∗+ψ∗xΓαΓβ+2∫01σsds+2PΓαΓβ+2+1+ϕ∗+ψ∗xΓα+β∫01σsds+PΓα+β≤21+ϕ∗+ψ∗xΓα+βσ∗+2PΓα+β+2PΓαΓβ+2+21+ϕ∗+ψ∗xΓαΓβ+2σ∗≤21+ϕ∗+ψ∗σ∗1Γα+β+1ΓαΓβ+2x+2P1Γα+β+1ΓαΓβ+2.Then, Fx≤r.
Therefore,FBr⊆Br.
Next, we show thatF is a contraction mapping.
Forx,y∈Br, we have
(25)Fxt−Fyt≤1Γα+β∫0tt−sα+β−1fs,xs,ϕxs,ψxs−fs,ys,ϕys,ψysds+tβ+1+tΓαΓβ+2∫0t1−sα−1×fs,xs,ϕxs,ψxs−fs,ys,ϕys,ψysds+1Γα+β∫0t1−sα+β−1fs,xs,ϕxs,ψxs−fs,ys,ϕys,ψysds≤1Γα+β∫0tt−sα+β−1σsxs−ys+ϕxs−ϕys+ψxs−ψysds+tβ+1+tΓαΓβ+2∫0t1−sα−1σs×xs−ys+ϕxs−ϕys+ψxs−ψysds+tΓα+β∫011−sα+β−1σsxs−ys+ϕxs−ϕys+ψxs−ψysds≤1+ϕ∗+ψ∗x−yΓα+β∫01σsds+21+ϕ∗+ψ∗x−yΓαΓβ+2×∫01σsds+1+ϕ∗+ψ∗x−yΓα+β∫01σsds+1Γα+β∫011−sα+β−1σsxs−ys+ϕxs−ϕys+ψxs−ψysds≤1+ϕ∗+ψ∗x−yΓα+β∫01σsds+21+ϕ∗+ψ∗x−yΓαΓβ+2×∫01σsds+1+ϕ∗+ψ∗x−yΓα+β∫01σsds≤21+ϕ∗+ψ∗x−yΓα+βσ∗+21+ϕ∗+ψ∗x−yΓαΓβ+2σ∗≤21+ϕ∗+ψ∗σ∗1Γα+β+1ΓαΓβ+2x−y. Since r1<1, then F is a contraction. Therefore, the system (1) has a unique solution.


## 4. Examples

In this section, we give two examples to show the applicability of our results.Example 1.
Consider the following problem:(26)cD1711cD1611xt=t3400xte−t1+xt+∫0tt+s3xscoss+sins4001+xsx0=x1=cD1611x0=cD1611x1=0.t∈0,1,Here,
(27)β=1611,α=1711,ft,x,y,z=t3400xte−t1+xt+ytcost1+yt+ztsint1+zt,λt,s=δt,s=t+s3400,σt=t3400,θt=3t3400.
It follows that(28)ϕ∗=ψ∗=151600,σ∗=11600.Then, by Theorem 7, we obtain that the problem (26) has at least one solution.Example 2.
Consider the following system:(29)cD107cD117xt=t220011+xt+1100∫0tt4s3xsds,t∈0,1,x0=x1=cD117x0=cD117x1=0.
Here,(30)B=117,α=107,ft,x,y,z=t220011+xt+yt+zt,λt,s=δt,s=t4s3200,σt=t2200.It is clear that
(31)ϕ∗=ψ∗=1800,σ∗=1600,r1≈0,0036.By Theorem 8, we conclude that the problem (29) has a unique solution.


## 5. Conclusion

In this paper, we proved the existence and uniqueness of solutions for nonlinear fractional integrodifferential equations of order1<α≤2 and 1<β≤2 using the Banach contraction mapping principle and Krasnoselskii’s fixed point theorem under some weak conditions. Furthermore, we provided two examples to illustrate the main results.



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