# The Existence of Solution for ak-Dimensional  System of Multiterm Fractional Integrodifferential Equations  with Antiperiodic Boundary Value Problems

Authors: Dumitru Baleanu; Sayyedeh Zahra Nazemi; Shahram Rezapour
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

There are many published papers about fractional integrodifferential equations and system of fractional differential equations. The goal of this paper is to show that we can investigate more complicated ones by using an appropriate basic theory. In this way, we prove the existence and uniqueness of solution for ak-dimensional system of multiterm fractional integrodifferential equations with antiperiodic boundary conditions by applying some standard fixed point results. An illustrative example is also presented.


---

## Body

## 1. Introduction

Fractional differential equations have recently been studied by many researchers for a variety of problems (see, e.g., [1–33] and the references therein). Antiperiodic boundary value problems occur in the mathematical modeling of a variety of physical processes (see, e.g., [3–6, 29, 30] and the references therein). On the other hand, the study of a coupled system of fractional order is also very significant because this kind of system can often occur in applications (see, e.g., [7, 15, 24, 28, 29] and the references therein). We are going to investigate a complicated case in this work. Let T>0 and I=[0,T]. In this paper, we study the existence and uniqueness of solution for the k-dimensional system of multiterm fractional integrodifferential equations
(1)cDα1x1(t)=f1(cDβ22x2(t),…,Dcβ2kxk(t)t,x1(t),x2(t),…,xk(t),ϕ11x1(t),ϕ12x2(t),…,ϕ1kxk(t),cDμ11x1(t),Dcμ12x2(t),…,cDμ1kxk(t),Dcβ11x1(t),cDβ12x2(t),…,Dcβ1kxk(t)),cDα2x2(t)=f2(cDβ22x2(t),…,Dcβ2kxk(t)t,x1(t),x2(t),…,xk(t),ϕ21x1(t),ϕ22x2(t),…,ϕ2kxk(t),cDμ21x1(t),Dcμ22x2(t),…,cDμ2kxk(t),Dcβ21x1(t),cDβ22x2(t),…,Dcβ2kxk(t)),⋮cDαkxk(t)=fk(cDβk2x2(t),…,Dcβkkxk(t)t,x1(t),x2(t),…,xk(t),ϕk1x1(t),ϕk2x2(t),…,ϕkkxk(t),Dcμk1x1(t),cDμk2x2(t),…,Dμkkcxk(t),cDβk1x1(t),cDβk2x2(t),…,Dcβkkxk(t)),(t∈I),
with antiperiodic boundary conditions xi(0)=-xi(T), cDpixi(0)=-cDpixi(T), and cDqixi(0)=-cDqixi(T) for i=1,2,…,k, where cD denotes the Caputo fractional derivative, αi∈(2,3], pi,μij∈(0,1), qi,βij∈(1,2) for i,j=1,2,…,k, (ϕijxj)(t)=∫0t‍λij(t,s)xj(s)ds, and fj∈C(I×ℝ4k,ℝ), λij:I×I→[0,∞) are continuous functions for all i,j=1,2,…,k. Hereafter, we will use vector notations. Define the space X={u(t):u(t)∈C2(I)} endowed with the norm ∥x∥X=sup⁡t∈I|x(t)|+sup⁡t∈I|x′(t)|+sup⁡t∈I|x′′(t)|. In fact, (X,∥·∥X) and the product space (Xk=X×X×⋯×X︸k,∥·∥*) endowed with the norm ∥(x1,x2,…,xk)∥*=∥x1∥X+∥x2∥X+⋯+∥xk∥X are Banach spaces. The Riemann-Liouville fractional integral of order q is defined by Iqf(t)=(1/Γ(q))∫0t‍(f(s)/(t-s)1-q)ds (t>0 and q>0), provided the integral exists. The Caputo derivative of order q for a function f∈Cn([0,∞),ℝ) is defined by cDqf(t)=(1/Γ(n-q))∫0t‍(f(n)(s)/(t-s)q+1-n)ds=In-qf(n)(t) for t>0 and n-1<q<n [25]. Recently, Wang et al. proved the following result [30].Lemma 1.
For eachy∈C([0,T]), the unique solution of the boundary value problem
(2)cDαx(t)=y(t),(t∈[0,T],T>0,2<α≤3)x(0)=-x(T),cDpx(0)=-cDpx(T),cDqx(0)=-cDqx(T),(0<p<1,1<q<2)
is given by x(t)=∫0T‍Gα(t,s)y(s)ds, where Gα(t,s) is Green's function defined as
(3)Gα(t,s)={(t-s)α-1-(1/2)(T-s)α-1Γ(α)+Γ(2-p)(T-2t)(T-s)α-p-12Γ(α-p)T1-p-[pT2-4Tt+2(2-p)t2]Γ(3-q)(T-s)α-q-14(2-p)Γ(α-q)T2-q,s≤t,-(T-s)α-12Γ(α)+Γ(2-p)(T-2t)(T-s)α-p-12Γ(α-p)T1-p-[pT2-4Tt+2(2-p)t2]Γ(3-q)(T-s)α-q-14(2-p)Γ(α-q)T2-q,t≤s.One can find the next result in [34].Theorem 2.
LetE be a Banach space and T:E→E a completely continuous operator. Suppose that the set V={u∈E:u=μTu,0≤μ≤1} is bounded. Then T has a fixed point in E.We will use the last two results for solving the problem (1).


## 2. Main Results

Now, we are ready to state and prove our main results. For eachi=1,2,…,k, put
(4)Mi=(32Γ(αi+1)+Γ(2-pi)2Γ(αi-pi+1)+(4-pi)Γ(3-qi)4(2-pi)Γ(αi-qi+1))Tαi+(1Γ(αi)+Γ(2-pi)Γ(αi-pi+1)+Γ(3-qi)(2-pi)Γ(αi-qi+1))Tαi-1+(1Γ(αi-1)+Γ(3-qi)Γ(αi-qi+1))Tαi-2
and M=min⁡1≤j≤k{1-∑i=1k‍Mi(bij+cijλij0+dij(T1-μij/Γ(2-μij))+eij(T2-βij/Γ(3-βij)))}, where λij0=sup⁡t∈I|∫0t‍λij(t,s)ds| for all i,j=1,2,…,k. Define the operator T:Xk→Xk by
(5)T(x)(t)=(T1(x)(t)T2(x)(t)⋮Tk(x)(t)),
where x=(x1,x2,…,xk) and
(6)Ti(x)(t)=∫0T‍Gαi(t,s)f~i(s,x(s))ds
for i=1,2,…,k, where
(7)f~i(s,x(s))=fi(cDβi2x2s,x1(s),x2(s),…,xk(s),ϕi1x1(s),ϕi2x2(s),…,ϕikxk(s),cDμi1x1(s),Dcμi2x2(s),…,Dcμikxk(s),cDβi1x1(s),Dcβi2x2(s),…,Dcβikxk(s)).
Thus, for each i=1,2,…,k, we have
(8)(Ti(x))′(t)=1Γ(αi-1)∫0t‍(t-s)αi-2f~i(s,x(s))ds-Γ(2-pi)T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1f~i(s,x(s))ds+[T-(2-pi)t]Γ(3-qi)(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1f~i(s,x(s))ds,(9)(Ti(x))′′(t)=1Γ(αi-2)∫0t‍(t-s)αi-3f~i(s,x(s))ds-Γ(3-qi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1f~i(s,x(s))ds.Theorem 3.
The operatorT:Xk→Xk is completely continuous.Proof.
First, we show that the operatorT:Xk→Xk is continuous. Let 0<μij<1 and 1<βij<2 for i,j=1,2,…,k and let {(x1n,x2n,…,xkn)} be a sequence in Xk such that (x1n,x2n,…,xkn)→(x10,x20,…,xk0). Then, we have
(10)sup⁡t∈I|Dμijcxjn(t)-Dμijcxj0(t)|=sup⁡t∈I|1Γ(1-μij)∫0t‍(t-s)-μij(xjn)′(s)ds-1Γ(1-μij)∫0t‍(t-s)-μij(xj0)′(s)ds|=sup⁡t∈I|1Γ(1-μij)×∫0t‍(t-s)-μij[(xjn)′(s)-(xj0)′(s)]ds1Γ(1-μij)|≤T1-μijΓ(2-μij)sup⁡t∈I|(xjn)′(t)-(xj0)′(t)|≤T1-μijΓ(2-μij)∥xjn-xj0∥,sup⁡t∈I|Dcβijxjn(t)-Dcβijxj0(t)|=sup⁡t∈I|1Γ(2-βij)∫0t‍(t-s)1-βij(xjn)′′(s)ds-1Γ(2-βij)∫0t‍(t-s)1-βij(xj0)′′(s)ds|=sup⁡t∈I|1Γ(2-βij)×∫0t‍(t-s)1-βij[(xjn)′′(s)-(xj0)′′(s)]ds1Γ(2-βij)|≤T2-βijΓ(3-βij)sup⁡t∈I|(xjn)′′(t)-(xj0)′′(t)|≤T2-βijΓ(3-βij)∥xjn-xj0∥,sup⁡t∈I|ϕijxjn(t)-ϕijxj0(t)|=sup⁡t∈I|∫0t‍λij(t,s)xjn(s)ds-∫0t‍λij(t,s)xj0(s)ds|≤sup⁡t∈I|∫0t‍λij(t,s)ds|sup⁡t∈I|xjn(t)-xj0(t)|≤λij0∥xjn-xj0∥
for i,j=1,2,…,k. Since ∥xjn-xj0∥X→0 for j=1,2,…,k, the sequences {Dμijcxjn(t)}, {Dcβijxjn(t)}, and {ϕijxjn(t)} converge uniformly on [0,T] and also lim⁡n→∞Dμijcxjn(t)=Dμijcxj0(t), lim⁡n→∞Dcβijxjn(t)=Dcβijxj0(t), and lim⁡n→∞ϕijxjn(t)=ϕijxj0(t) converge uniformly on [0,T] for i,j=1,2,…,k. Since
(11)∥T(x1n,x2n,…,xkn)-T(x10,x20,…,xk0)∥*=sup⁡t∈I|(x10,x20,…,xk0)T1(x1n,x2n,…,xkn)(t)-T1(x10,x20,…,xk0)(t)|+sup⁡t∈I|(x10,x20,…,xk0)T2(x1n,x2n,…,xkn)(t)-T2(x10,x20,…,xk0)(t)|+⋯+sup⁡t∈I|(x10,x20,…,xk0)Tk(x1n,x2n,…,xkn)(t)-Tk(x10,x20,…,xk0)(t)|+sup⁡t∈I|(T1(x10,x20,…,xk0))′(T1(x1n,x2n,…,xkn))′(t)-(T1(x10,x20,…,xk0))′(t)|+sup⁡t∈I|(T2(x10,x20,…,xk0))′(T2(x1n,x2n,…,xkn))′(t)-(T2(x10,x20,…,xk0))′(t)|+⋯+sup⁡t∈I|(Tk(x10,x20,…,xk0))′(Tk(x1n,x2n,…,xkn))′(t)-(Tk(x10,x20,…,xk0))′(t)|+sup⁡t∈I|(T1(x10,x20,…,xk0))′′(T1(x1n,x2n,…,xkn))′′(t)-(T1(x10,x20,…,xk0))′′(t)|+sup⁡t∈I|(T2(x10,x20,…,xk0))′′(T2(x1n,x2n,…,xkn))′′(t)-(T2(x10,x20,…,xk0))′′(t)|+⋯+sup⁡t∈I|(Tk(x10,x20,…,xk0))′′(Tk(x1n,x2n,…,xkn))′′(t)-(Tk(x10,x20,…,xk0))′′(t)|,
by using the above inequalities and the continuity of fi (i=1,2,…,k), we get
(12)∥T(x1n,x2n,…,xkn)-T(x10,x20,…,xk0)∥*⟶0.
Thus, T is continuous in Xk. Let Ω be a bounded subset of Xk. Choose positive constants l1>0,…,lk>0 such that |f~i(s,x(s))|≤li for all x=(x1,x2,…,xk)∈Ω and i=1,2,…,k. Thus, for each x=(x1,x2,…,xk)∈Ω we have
(13)|Ti(x)(t)|≤1Γ(αi)∫0t‍(t-s)αi-1|f~i(s,x(s))|ds+12Γ(αi)∫0T‍(T-s)αi-1|f~i(s,x(s))|ds+Γ(2-pi)|T-2t|2T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1|f~i(s,x(s))|ds+|piT2-4Tt+2(2-pi)t2|Γ(3-qi)4(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))|ds≤liΓ(αi)∫0t‍(t-s)αi-1ds+li2Γ(αi)∫0T‍(T-s)αi-1ds+liΓ(2-pi)|T-2t|2T1-piΓ(αi-pi)∫0T‍(T-s)αi-pi-1ds+li|piT2-4Tt+2(2-pi)t2|Γ(3-qi)4(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1ds≤(32Γ(αi+1)+Γ(2-pi)2Γ(αi-pi+1)+(4-pi)Γ(3-qi)4(2-pi)Γ(αi-qi+1))Tαili,|(Ti(x))′(t)|≤1Γ(αi-1)∫0t‍(t-s)αi-2|f~i(s,x(s))|ds+Γ(2-pi)T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1|f~i(s,x(s))|ds+|T-(2-pi)t|Γ(3-qi)(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))|ds≤(1Γ(αi)+Γ(2-pi)Γ(αi-pi+1)+Γ(3-qi)(2-pi)Γ(αi-qi+1))Tαi-1li,|(Ti(x))′′(t)|≤1Γ(αi-2)∫0t‍(t-s)αi-3|f~i(s,x(s))|ds+Γ(3-qi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))|ds≤(1Γ(αi-1)+Γ(3-qi)Γ(αi-qi+1))Tαi-2li
for all i=1,2,…,k. Hence,
(14)∥Ti(x)∥≤(32Γ(αi+1)+Γ(2-pi)2Γ(αi-pi+1)+(4-pi)Γ(3-qi)4(2-pi)Γ(αi-qi+1))Tαili+(1Γ(αi)+Γ(2-pi)Γ(αi-pi+1)+Γ(3-qi)(2-pi)Γ(αi-qi+1))Tαi-1li+(1Γ(αi-1)+Γ(3-qi)Γ(αi-qi+1))Tαi-2li=Mili
for all i=1,2,…,k and so ∥T(x)∥*≤∑i=1k‍Mili=M′. This implies that the operator T is uniformly bounded. Now, we show that TΩ is an equicontinuous set. Let 0≤t1<t2≤T. Then, we have
(15)|Ti(x)(t2)-Ti(x)(t1)|=|1Γ(αi)∫0t2‍(t2-s)αi-1f~i(s,x(s))ds-1Γ(αi)∫0t1‍(t1-s)αi-1f~i(s,x(s))ds-Γ(2-pi)(t2-t1)T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1f~i(s,x(s))ds+[2T(t2-t1)-(2-pi)(t22-t12)]Γ(3-qi)2(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1f~i(s,x(s))ds|≤liΓ(αi)∫0t1‍[(t2-s)αi-1-(t1-s)αi-1]ds+liΓ(αi)∫t1t2‍(t2-s)αi-1ds+[Γ(2-pi)Tαi-1liΓ(αi-pi+1)+Γ(3-qi)(3-pi)Tαi-1li(2-pi)Γ(αi-qi+1)]×(t2-t1)=liΓ(αi+1)(t2αi-t1αi)+[Γ(2-pi)Tαi-1liΓ(αi-pi+1)+Γ(3-qi)(3-pi)Tαi-1li(2-pi)Γ(αi-qi+1)]×(t2-t1),|(Ti(x))′(t2)-(Ti(x))′(t1)|≤|1Γ(αi-1)∫0t2‍(t2-s)αi-2f~i(s,x(s))ds-1Γ(αi-1)∫0t1‍(t1-s)αi-2f~i(s,x(s))ds-Γ(3-qi)(t2-t1)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1f~i(s,x(s))ds|≤liΓ(αi-1)∫0t1‍[(t2-s)αi-2-(t1-s)αi-2]ds+liΓ(αi-1)∫t1t2‍(t2-s)αi-2ds+Γ(3-qi)Tαi-2liΓ(αi-qi+1)(t2-t1)=liΓ(αi)(t2αi-1-t1αi-1)+Γ(3-qi)Tαi-2liΓ(αi-qi+1)(t2-t1),|(Ti(x))′′(t2)-(Ti(x))′′(t1)|≤1Γ(αi-2)|∫0t2‍(t2-s)αi-3f~i(s,x(s))ds-∫0t1‍(t1-s)αi-3f~i(s,x(s))ds|≤liΓ(αi-2)∫0t1‍[(t1-s)αi-3-(t2-s)αi-3]ds+liΓ(αi-2)∫t1t2‍(t2-s)αi-3ds=liΓ(αi-1)[2(t2-t1)αi-2+t1αi-2-t2αi-2]
for all i=1,…,k. As t2→t1, the right-hand side of the above inequalities tends to zero. Thus, by using the Arzela-Ascoli theorem one can conclude that the operator T:Xk→Xk is completely continuous. This completes the proof.Theorem 4.
Assume that there exist positive constantsai>0, bij≥0, cij≥0, dij≥0, and eij≥0 (i,j=1,2,…,k) such that
(16)|fi(t,x1,x2,…,xk,y1,y2,…,yk,z1,z2,…,zk,w1,w2,…,wk)|≤ai+∑j=1k‍bij|xj|+∑j=1k‍cij|yj|+∑j=1k‍dij|zj|+∑j=1k‍eij|wj|
and ∑i=1k‍Mi(bij+cijλij0+dij(T1-μij/Γ(2-μij))+eij(T2-βij/Γ(3-βij)))<1 for all xi,yi,zi,wi∈ℝ, t∈I, and i,j=1,2,…,k. Then problem (1) has at least one solution.Proof.
First, we show thatV={x=(x1,x2,…,xk)∈Xk:x=μT(x)  for  some  μ∈[0,1]} is bounded. Let x=(x1,x2,…,xk)∈V. Then, for each t∈I we have
(17)xi(t)=1Γ(αi)∫0t‍μ(t-s)αi-1f~i(s,x(s))ds-12Γ(αi)∫0T‍μ(T-s)αi-1f~i(s,x(s))ds+Γ(2-pi)(T-2t)2T1-piΓ(αi-pi)×∫0T‍μ(T-s)αi-pi-1f~i(s,x(s))ds-[piT2-4Tt+2(2-pi)t2]Γ(3-qi)4(2-pi)Γ(αi-qi)T2-qi×∫0T‍μ(T-s)αi-qi-1f~i(s,x(s))ds.
Hence,
(18)xi′(t)=1Γ(αi-1)∫0t‍μ(t-s)αi-2f~i(s,x(s))ds-Γ(2-pi)T1-piΓ(αi-pi)∫0T‍μ(T-s)αi-pi-1f~i(s,x(s))ds+[T-(2-pi)t]Γ(3-qi)(2-pi)Γ(αi-qi)T2-qi×∫0T‍μ(T-s)αi-qi-1f~i(s,x(s))ds,(19)xi′′(t)=1Γ(αi-2)∫0t‍μ(t-s)αi-3f~i(s,x(s))ds-Γ(3-qi)Γ(αi-qi)T2-qi×∫0T‍μ(T-s)αi-qi-1f~i(s,x(s))ds
for all i=1,2,…,k. Thus, we get
(20)|xi(t)|=μ|Ti(x)(t)|≤[ai+∑j=1k‍bij∥xj∥+∑j=1k‍cijλij0∥xj∥+∑j=1k‍dijT1-μijΓ(2-μij)∥xj∥+∑j=1k‍eijT2-βijΓ(3-βij)∥xj∥]×(32Γ(αi+1)+Γ(2-pi)2Γ(αi-pi+1)+(4-pi)Γ(3-qi)4(2-pi)Γ(αi-qi+1))Tαi,|xi′(t)|=μ|(Ti(x))′(t)|≤[ai+∑j=1k‍bij∥xj∥+∑j=1k‍cijλij0∥xj∥+∑j=1k‍dijT1-μijΓ(2-μij)∥xj∥+∑j=1k‍eijT2-βijΓ(3-βij)∥xj∥]×(1Γ(αi)+Γ(2-pi)Γ(αi-pi+1)+Γ(3-qi)(2-pi)Γ(αi-qi+1))Tαi-1,|xi′′(t)|=μ|(Ti(x))′′(t)|≤[ai+∑j=1k‍bij∥xj∥+∑j=1k‍cijλij0∥xj∥+∑j=1k‍dijT1-μijΓ(2-μij)∥xj∥+∑j=1k‍eijT2-βijΓ(3-βij)∥xj∥]×(1Γ(αi-1)+Γ(3-qi)Γ(αi-qi+1))Tαi-2.
Hence, ∥xi∥≤Mi[ai+∑j=1k‍bij∥xj∥+∑j=1k‍cijλij0∥xj∥+∑j=1k‍dij(T1-μij/Γ(2-μij))∥xj∥+∑j=1k‍eij(T2-βij/Γ(3-βij))∥xj∥] for i=1,2,…,k. This implies that
(21)∑i=1k‍∥xi∥≤∑j=1k‍[∑i=1k‍Mi(bij+cijλij0+dijT1-μijΓ(2-μij)+eijT2-βijΓ(3-βij))]∥xj∥+∑i=1k‍Miai
and so ∥x=(x1,x2,…,xk)∥*≤(∑i=1k‍Miai)/M. Therefore, the set V is bounded. Now by using Theorem 2, the operator T has at least one fixed point. This implies that the problem (1) has at least one solution.Theorem 5.
Suppose that there exist nonnegative constantsηij≥0, θij≥0, νij≥0, and ξij≥0 for i,j=1,2,…,k such that
(22)|(z1′,z2′,…,zk′,w1′,w2′,…,wk′)fi(t,x1,x2,…,xk,y1,y2,…,yk,z1,z2,…,zk,w1,w2,…,wk)-fi(t,x1′,x2′,…,xk′,y1′,y2′,…,yk′,z1′,z2′,…,zk′,w1′,w2′,…,wk′)|≤∑j=1k‍ηij|xj-xj′|+∑j=1k‍θij|yj-yj′|+∑j=1k‍νij|zj-zj′|+∑j=1k‍ξij|wj-wj′|,(23)∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij)≤12kMi
for all t∈I, xi,yi,zi,wi,xi′,yi′,zi′,wi′∈ℝ, and i=1,2,…,k. Then the problem (1) has a unique solution.Proof.
LetNi=sup⁡t∈I|fi(t,0,0,…,0)|<∞ for i=1,2,…,k, r≥2kmax⁡1≤i≤k{MiNi}, and
(24)Br={x=(x1,x2,…,xk)∈Xk∣∥x∥*≤r}.
We show that TBr⊆Br. Let x=(x1,x2,…,xk)∈Br. Then
(25)|Ti(x)(t)|≤1Γ(αi)∫0t‍(t-s)αi-1|f~i(s,x(s))|ds+12Γ(αi)∫0T‍(T-s)αi-1|f~i(s,x(s))|ds+Γ(2-pi)|T-2t|2T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1|f~i(s,x(s))|ds+|piT2-4Tt+2(2-pi)t2|Γ(3-qi)4(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))|ds≤1Γ(αi)∫0t‍(t-s)αi-1×[|f~i(s,x(s))-fi(s,0,0,…,0)|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds+12Γ(αi)∫0T‍(T-s)αi-1×[|f~i(s,x(s))-fi(s,0,0,…,0)f~i(s,x(s))|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds+Γ(2-pi)|T-2t|2T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1×[|f~i(s,x(s))-fi(s,0,0,…,0)|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds+|piT2-4Tt+2(2-pi)t2|Γ(3-qi)4(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1×[|f~i(s,x(s))-fi(s,0,0,…,0)|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds≤([∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij)]r+Ni)×(32Γ(αi+1)+Γ(2-pi)2Γ(αi-pi+1)+(4-pi)Γ(3-qi)4(2-pi)Γ(αi-qi+1))Tαi,|(Ti(x))′(t)|≤1Γ(αi-1)×∫0t‍(t-s)αi-2|f~i(s,x(s))|ds+Γ(2-pi)T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1|f~i(s,x(s))|ds+|T-(2-pi)t|Γ(3-qi)(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))|ds≤1Γ(αi-1)×∫0t‍(t-s)αi-2×[|f~i(s,x(s))-fi(s,0,0,…,0)|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds+Γ(2-pi)T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1×[|f~i(s,x(s))-fi(s,0,0,…,0)|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds+|T-(2-pi)t|Γ(3-qi)(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1×[|f~i(s,x(s))-fi(s,0,0,…,0)|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds≤([∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij)]r+Ni)×(1Γ(αi)+Γ(2-pi)Γ(αi-pi+1)+Γ(3-qi)(2-pi)Γ(αi-qi+1))Tαi-1,|(Ti(x))′′(t)|≤1Γ(αi-2)×∫0t‍(t-s)αi-3|f~i(s,x(s))|ds+Γ(3-qi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))|ds≤1Γ(αi-2)×∫0t‍(t-s)αi-3×[|f~i(s,x(s))-fi(s,0,0,…,0)|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds+Γ(3-qi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1×[|f~i(s,x(s))-fi(s,0,0,…,0)|+|fi(s,0,0,…,0)||f~i(s,x(s))-fi(s,0,0…,0)|]ds≤([∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij)]r+Ni)×(1Γ(αi-1)+Γ(3-qi)Γ(αi-qi+1))Tαi-2
for i=1,2,…,k. Hence,
(26)∥Ti(x)∥≤([∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij)]r+Ni)Mi≤rk
for i=1,2,…,k. Hence, ∥T(x)∥*≤r. Now, for each x=(x1,x2,…,xk),  y=(y1,y2,…,yk)∈Xk, and t∈I we get
(27)|Ti(x)(t)-Ti(y)(t)|≤1Γ(αi)∫0t‍(t-s)αi-1|f~i(s,x(s))-f~i(s,y(s))|ds+12Γ(αi)×∫0T‍(T-s)αi-1|f~i(s,x(s))-f~i(s,y(s))|ds+Γ(2-pi)|T-2t|2T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1|f~i(s,x(s))-f~i(s,y(s))|ds+|piT2-4Tt+2(2-pi)t2|Γ(3-qi)4(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))-f~i(s,y(s))|ds≤(∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij))×(32Γ(αi+1)+Γ(2-pi)2Γ(αi-pi+1)+(4-pi)Γ(3-qi)4(2-pi)Γ(αi-qi+1))Tαi×(∑j=1k‍∥xj-yj∥),|(Ti(x))′(t)-(Ti(y))′(t)|≤1Γ(αi-1)×∫0t‍(t-s)αi-2|f~i(s,x(s))-f~i(s,y(s))|ds+Γ(2-pi)T1-piΓ(αi-pi)×∫0T‍(T-s)αi-pi-1|f~i(s,x(s))-f~i(s,y(s))|ds+|T-(2-pi)t|Γ(3-qi)(2-pi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))-f~i(s,y(s))|ds≤(∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij))×(1Γ(αi)+Γ(2-pi)Γ(αi-pi+1)+Γ(3-qi)(2-pi)Γ(αi-qi+1))Tαi-1×(∑j=1k‍∥xj-yj∥),|(Ti(x))′′(t)-(Ti(y))′′(t)|≤1Γ(αi-2)×∫0t‍(t-s)αi-3|f~i(s,x(s))-f~i(s,y(s))|ds+Γ(3-qi)Γ(αi-qi)T2-qi×∫0T‍(T-s)αi-qi-1|f~i(s,x(s))-f~i(s,y(s))|ds≤(∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij))×(1Γ(αi-1)+Γ(3-qi)Γ(αi-qi+1))Tαi-2×(∑j=1k‍∥xj-yj∥)
for i=1,2,…,k. This implies that
(28)∥Ti(x)-Ti(y)∥≤Mi(∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij))×(∑j=1k‍∥xj-yj∥)
for i=1,2,…,k and so
(29)∥T(x)-T(y)∥*≤[∑i=1k‍Mi(∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νijT1-μijΓ(2-μij)+∑j=1k‍ξijT2-βijΓ(3-βij))]×(∑j=1k‍∥xj-yj∥).
Since ∑i=1k‍Mi(∑j=1k‍ηij+∑j=1k‍θijλij0+∑j=1k‍νij(T1-μij/Γ(2-μij))+∑j=1k‍ξij(T2-βij/Γ(3-βij)))<1, T is a contraction and so, by using the Banach contraction principle, T has a unique fixed point. Now, one can easily get that the problem (1) has a unique solution.


## 3. Example

Here, we provide an example to illustrate one of our results. It is considerable that there are some examples which provide nonuniqueness of solutions for some fractional differential equations (see, e.g., [32]).Example 1.
Consider the following 3-dimensional system:(30)Dα1cx1(t)=12e-cos⁡2t+e-πt(x1(t)+x2(t)+x3(t))×((|+Dμ13cx3(t)|)9121π×(+cDμ13x3(t)|)1+|Dμ11cx1(t)+Dμ12cx2(t)+Dμ13cx3(t)cDμ11x1(t)|))-1+5+cos⁡πt8124(t+4)2×(∑k=13‍e-|xk(t)|sin2(t3+4)2kk!ln⁡(1+|ϕ1kxk(t)|))+8e-12sin2(x2(t))cos⁡3t2450(2t+387e3)×(∑k=13‍e-ktcos⁡kπ4kk!Dμ1kcxk(t))+12130(t+6)3×(∑k=13‍e-(2+cos⁡t)sin(Dμ1kcxk(t))Dcβ1kxk(t)ek+1(2+|xk(t)+Dμ1kcxk(t)|)),cDα2x2(t)=5t2+cos⁡3t6421π(2+|x1(t)+Dcβ21x1(t)|)×(∑k=13‍(t-1/2)4xk(t)3+sinxk(t))+(3-t)1/25180×(∑k=13‍(Dcβ2kxk(t))31+|Dcβ2kxk(t)|3ϕ2kxk(t))+e-5tsin(Dcβ22x2(t))1145(1+et)×(∑k=13‍e-πt(1+|xk(t)+ϕ2kxk(t)|)Dμ2kcxk(t)17+sin⁡(Dμ2kcxk(t))(1+|Dμ2kcxk(t)|))+e(1-cos⁡2x1(t))22918(1+|x2(t)|)×(∑k=13‍ln⁡⁡(1+|Dcβ2kxk(t)|)ln⁡⁡(e+cos⁡2xk(t)+sin2(Dcβ2kxk(t)))),cDα3x3(t)=15cos⁡2t+6+cos⁡t360(t+7)5×(∑k=13‍xk(t)+Dcβ3kxk(t)2+|xk(t)|+cos⁡⁡(Dμ3kcxk(t)))+e-cos⁡4t1200π(17+|x3(t)+ϕ33x3(t)|)×(∑k=13‍ln⁡(1+|ϕ3kxk(t)+Dμ3kcxk(t)|)12+sin⁡(Dμ3kcxk(t))+|Dcβ3kxk(t)|),xi(0)=-xi(1),cDpixi(0)=-cDpixi(1),cDqixi(0)=-cDqixi(1),i=1,2,3,
where p1=1/9, p2=3/5, p3=1/7, q1=3/2, q2=11/10, q3=7/6, α1=17/8, α2=7/3, α3=12/5, μ11=1/3, μ12=1/15, μ13=1/20, μ21=1/19, μ22=1/17, μ23=1/12, μ31=1/14, μ32=2/7, μ33=4/13, β11=8/7, β12=13/12, β13=15/13, β21=6/5, β22=15/8, β23=5/4, β31=10/9, β32=7/5, β33=12/11, ϕ11x1(t)=∫0t‍((t-s)e(t-s)2/919)x1(s)ds,   ϕ21x1(t)=∫0t‍((t-s)3e(t-s)4/727)x1(s)ds, ϕ31x1(t)=∫0t‍((t-s)2e(t-s)3/500)x1(s)ds, ϕ12x2(t)=∫0t‍(4s3s4+7/831)x2(s)ds,  ϕ22x2(t)=∫0t‍((3s+6)/700(s2+5s+4))x2(s)ds, ϕ32x2(t)=∫0t‍(6s/901(4+s2))x2(s)ds, ϕ13x3(t)=∫0t‍(e(t-s)/2/925)x3(s)ds, ϕ23x3(t)=∫0t‍(e(t-s)/3/1875)x3(s)ds, and ϕ33x3(t)=∫0t‍(e(t-s)/4/3200)x3(s)ds, with λ110=(e-1)/1838, λ210=(e-1)/2908, λ310=(e-1)/1500,  λ120=2(88-77)/2493, λ220=(ln⁡2+2ln⁡(5/4))/700, λ320=3ln⁡(5/4)/901, λ130=2(e-1)/925, λ230=(e3-1)/625, and λ330=(e4-1)/800. Then, we have
(31)|cDβ13x3(t)cDβ22x2(t),cDβ23x3(t))f1(cDβ13x3(t)t,x1(t),x2(t),x3(t),ϕ11x1(t),cDβ13x3(t)ϕ12x2(t),ϕ13x3(t),Dcμ11x1(t),cDμ12x2(t),Dcμ13x3(t),cDβ11x1(t),Dcβ12x2(t),Dcβ13x3(t))|≤12+∑j=13‍19121π|xj(t)|+∑j=13‍38124×8×2j×j!|ϕ1jxj(t)|+∑j=13‍82450×387e3×4j×j!|Dcμ1jxj(t)|+∑j=13‍14260×63×ej+1|Dcβ1jxj(t)|,|cDβ22x2(t),cDβ23x3(t))f2(cDβ23x3(t)t,x1(t),x2(t),x3(t),ϕ21x1(t),ϕ22x2(t),ϕ23x3(t),Dμ21cx1(t),cDμ22x2(t),Dμ23cx3(t),Dβ21cx1(t),cDβ22x2(t),Dβ23cx3(t))|≤5+∑j=13‍16421×64×π|xj(t)|+∑j=13‍35180|ϕ2jxj(t)|+∑j=13‍11145×4|Dμ2jcxj(t)|+∑j=13‍e2918|Dcβ2jxj(t)|,|cDβ32x2(t)cDβ22x2(t),cDβ23x3(t))f3(cDβ33x3(t)t,x1(t),x2(t),x3(t),ϕ31x1(t),cDβ33x3(t)cDβ33x3(t)ϕ32x2(t),ϕ33x3(t),Dμ31cx1(t),cDμ32x2(t),Dμ33cx3(t),Dcβ31x1(t),Dcβ32x2(t),Dcβ33x3(t))|≤15+∑j=13‍174×360|xj(t)|+∑j=13‍11200π×17×11|ϕ3jxj(t)|+∑j=13‍11200π×17×11|Dμ3jcxj(t)|+∑j=13‍174×360|Dcβ3jxj(t)|.
Here,M1≅5.401859055, M2≅5.327781471, M3≅4.713515099,
(32)∑i=13‍Mi(bi1+ci1λi10+di1Γ(2-μi1)+ei1Γ(3-βi1))≅0.015425405<1,∑i=13‍Mi(bi2+ci2λi20+di2Γ(2-μi2)+ei2Γ(3-βi2))≅0.036796259<1,∑i=13‍Mi(bi3+ci3λi30+di3Γ(2-μi3)+ei3Γ(3-βi3))≅0.01958673<1.
Thus, by using Theorem 4, 3-dimensional system (30) has at least one solution.


## 4. Conclusions

Fractional integrodifferential equations, system of fractional differential equations, and their applications represent a topic of high interest in the area of fractional calculus and its applications in various fields of science and engineering. Antiperiodic boundary value problems occur in the mathematical modeling of a variety of physical processes. The goal of this paper is to investigate a complicated case by using an appropriate basic theory. In this way, we prove the existence and uniqueness of solution for a newk-dimensional system of multiterm fractional integrodifferential equations with antiperiodic boundary conditions.



---

*Source: 896871-2014-04-24.xml*