# Ak-Dimensional System of Fractional Neutral Functional Differential Equations with Bounded Delay

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

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## Abstract

In 2010, Agarwal et al. studied the existence of a one-dimensional fractional neutral functional differential equation. In this paper, we study an initial value problem for a class ofk-dimensional systems of fractional neutral functional differential equations by using Krasnoselskii’s fixed point theorem. In fact, our main result generalizes their main result in a sense.


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## Body

## 1. Introduction

As you know, many researchers are interested in developing the theoretical analysis and numerical methods of fractional equations, because different applications of this area have been foundedin various fields of sciences and engineering (see, e.g., [1–37]). In this paper, we investigate the initial value problem of a k-dimensional system of fractional neutral functional differential equations with bounded delay:
(1)Dα1c(x1(t)-g1(t,xt))=f1(t,xt),Dα2c(x2(t)-g2(t,xt))=f2(t,xt),⋮Dαkc(xk(t)-gk(t,xt))=fk(t,xt),x1t0=ϕ1,x2t0=ϕ2,…,xkt0=ϕk,
where t0≥0, a>0, and r>0 are constants, t∈(t0,∞), 0<αi<1, for i=1,2,…,k, Dc is the standard Caputo’s fractional derivative, fi,gi:[t0,∞)×C([-r,0],ℝn)×C([-r,0],ℝn)×⋯×C([-r,0],ℝn)→ℝn are given functions (i=1,2,…,k) satisfying some assumptions that will be specified later,xt=(x1t,x2t,…,xkt), and ϕi∈C([-r,0],ℝn) for i=1,2,…,k. If x∈C([t0-r,t0+a],ℝn), then for each t∈[t0,t0+a] define xt by xt(θ)=x(t+θ) for all θ∈[-r,0]. One-dimensional version of the problem has been studied by Agarwal et al. (see [4]). We show that the problem (1) is equivalent to an integral equation and by using Krasnoselskii's fixed point theorem, we conclude that the equivalent operator has (at least) a fixed point. This implies that the problem (1) has at least one solution. One can find the following lemma in [38].Lemma 1 (Krasnoselskii’s fixed point theorem).
LetX be a Banach space and E a closed convex subset of X. Suppose that S and U are two maps of E into X such that Sx+Uy∈E for all x,y∈E. If S is a contraction and U is completely continuous, then the equation Sx+Ux=x has a solution on E.LetI be an interval in ℝ and X=C(I,ℝn) with the norm ∥x∥=sup⁡t∈I|x(t)|, where |·| denotes a suitable complete norm on ℝn. Consider the product Banach space (Xk=X×X×⋯×X︸k,∥·∥*) with the norm ∥(x1,x2,…,xk)∥*=max⁡{∥x1∥,∥x2∥,…,∥xk∥}. The fractional integral of order q with the lower limit t0 for a function f is defined by Iqf(t)=(1/Γ(q))∫t0t(f(s)/(t-s)1-q)ds for t>t0 and q>0, provided the right-hand side is pointwise defined on [t0,∞). Here, Γ is the gamma function. Also, Caputo’s derivative of order q with the lower limit t0 for a function f:[t0,∞)→ℝ is defined by
(2)Dqcf(t)=1Γ(n-q)∫t0t‍f(n)(s)(t-s)q+1-nds=In-qf(n)(t)
for t>t0 and n-1<q<n ([34]).


## 2. Main Results

Consider the problem (1). Let δ and γ be positive constants, I0=[t0,t0+δ], and
(3)A(δ,γ)={(x1,x2,…,xk):xit0=ϕi,sup⁡t0≤t≤t0+δ|xi(t)-ϕi(0)|≤γ,∀i=1,2,…,k(x1,x2,…,xk):xit0=ϕi},
where xi∈C([t0-r,t0+δ],ℝn). For obtaining our results, we need the following conditions:(H1)
fi(t,φ1,φ2,…,φk) is measurable with respect to t on I0 for all i=1,2,…,k,(H2)
fi(t,φ1,φ2,…,φk) is continuous with respect to φj on C([-r,0],ℝn) for all i,j=1,2,…,k,(H3)
there existαi1∈(0,αi) and a real-valued function mi(t)∈L1/αi1(I0) such that
(4)|fi(t,xt)|≤mi(t)
for all (x1,x2,…,xk)∈A(δ,γ), t∈I0, and i=1,2,…,k,(H4)
gi(t,xt)=gi1(t,xt)+gi2(t,xt) for all (x1,x2,…,xk)∈A(δ,γ),(H5)
gi1 is continuous and
(5)|gi1(t,xt)-gi1(t,yt)|≤li∥x-y∥*
for all x=(x1,x2,…,xk),y=(y1,y2,…,yk)∈A(δ,γ), and t∈I0, where li∈(0,1) is a constant, for all i=1,2,…,k,(H6)
gi2 is completely continuous and the family {t⊢gi2(t,xt):(x1,x2,…,xk)∈Λ} is equicontinuous on C(I0,ℝn)×C(I0,ℝn)×⋯×C(I0,ℝn)︸k for all bounded set Λ in A(δ,λ) and i=1,2,…,k.Lemma 2.
Suppose that there existδ∈(0,a) and γ∈(0,∞) such that (H1)–(H3) hold. Then the problem (1) for t∈(t0,t0+δ] is equivalent to the equation
(*)xi(t)=ϕi(0)-gi(t0,ϕ1,ϕ2,…,ϕk)+gi(t,xt)+1Γ(αi)∫t0t(t-s)αi-1fi(s,xs)ds
with conditions xit0=ϕi for i=1,2,…,k and t∈I0.Proof.
It is easy to see thatfi(t,xt) is Lebesgue measurable on I0 by using conditions (H1) and (H2) for all i=1,2,…,k. Also, a direct calculation shows that (t-s)αi-1∈L1/(1-αi1)([t0,t]) for t∈I0. By using Holder's inequality and condition (H3), we get that (t-s)αi-1fi(s,xs) is Lebesgue integrable with respect to s∈[t0,t] for all t∈I0, i=1,2,…,k, and (x1,x2,…,xk)∈A(δ,γ), and
(6)∫t0t|(t-s)αi-1fi(s,xs)|ds≤∥(t-s)αi-1∥L1/(1-αi1)([t0,t])∥mi∥L1/αi1(I0).
It is easy to see that if x=(x1,x2,…,xk) is a solution of the problem (1), then x is a solution of (*). Now, suppose that x=(x1,x2,…,xk) is a solution of the equation (*) and t∈(t0,t0+δ]. Then xit0=ϕi and cDαi(xi(t)-gi(t,xt))=fi(t,xt) for all t∈(t0,t0+δ] and i=1,2,…,k. Thus, x=(x1,x2,…,xk) is a solution of the problem (1). This completes the proof.Theorem 3.
Suppose that there existδ∈(0,a) and γ∈(0,∞) such that (H1)–(H6) hold. Then the problem (1) has at least one solution on [t0,t0+η] for some positive number η.Proof.
Since condition (H4) holds, the equation (*) is equivalent to the equation
(7)xi(t)=ϕi(0)-gi1(t0,ϕ1,ϕ2,…,ϕk)-gi2(t0,ϕ1,ϕ2,…,ϕk)+gi1(t,xt)+gi2(t,xt)+1Γ(αi)∫t0t(t-s)αi-1fi(s,xs)ds
and xit0=ϕi for all t∈I0 and i=1,2,…,k. Let (ϕ~1,ϕ~2,…,ϕ~k)∈A(δ,γ) be defined by ϕ~it0=ϕi and ϕ~i(t0+t)=ϕi(0) for all t∈[0,δ] and i=1,2,…,k. If x=(x1,x2,…,xk) is a solution of problem (1) and xi(t0+t)=ϕ~i(t0+t)+yi(t) for t∈[-r,δ] and i=1,2,…,k, then xit0+t=ϕ~it0+t+yit for t∈[0,δ] and i=1,2,…,k. Thus,
(**)yi(t)=-gi1(t0,ϕ1,ϕ2,…,ϕk)-gi2(t0,ϕ1,ϕ2,…,ϕk)+gi1(t0+t,y1t+ϕ~1t0+t,y2t+ϕ~2t0+t,…,ykt+ϕ~kt0+t)+gi2(t0+t,y1t+ϕ~1t0+t,y2t+ϕ~2t0+t,…,ykt+ϕ~kt0+t)+1Γ(αi)∫0t(t-s)αi-1fi(t0+s,y1s+ϕ~1t0+s,y2s+ϕ~2t0+s,…,yks+ϕ~kt0+s)ds
for t∈[0,δ] and i=1,2,…,k. Since gi1, gi2 are continuous and xit is continuous in t for all i=1,2,…,k, there exists δ′>0 such that |gi1(t0+t,y1t+ϕ~1t0+t,y2t+ϕ~2t0+t,…,ykt+ϕ~kt0+t)-gi1(t0,ϕ1,ϕ2,…,ϕk)|<γ/3 and |gi2(t0+t,y1t+ϕ~1t0+t,y2t+ϕ~2t0+t,…,ykt+ϕ~kt0+t)-gi2(t0,ϕ1,ϕ2,…,ϕk)|<γ/3 for 0<t<δ′ and i=1,2,…,k. Put η=min⁡1≤i≤k{δ,δ′,(γΓ(αi)(1+βi)1-αi1/3Mi)1/(1+βi)(1-αi1)}, where βi=(αi-1)/(1-αi1)∈(-1,0) and Mi=∥mi∥L1/αi1(I0) for all i=1,2,…,k. Define
(8)E(η,γ)={(y1,y2,…,yk):yi∈C([-r,η],ℝn),yi(s)=0,∥yi∥≤γfors∈[-r,0],i=1,2,…,k}.
In fact, E(η,γ) is a closed, bounded, and convex subset of C([-r,η],ℝn)×C([-r,η],ℝn)×⋯×C([-r,η],ℝn). Define the operators S and U on E(η,γ) by
(9)S(y1,y2,…,yk)(t)=(S1(y1,y2,…,yk)(t)S2(y1,y2,…,yk)(t)⋮Sk(y1,y2,…,yk)(t)),U(y1,y2,…,yk)(t)=(U1(y1,y2,…,yk)(t)U2(y1,y2,…,yk)(t)⋮Uk(y1,y2,…,yk)(t)),
where
(10)Si(y1,y2,…,yk)(t)={0t∈[-r,0],-gi1(t0,ϕ1,ϕ2,…,ϕk)+gi1(t0+t,y1t+ϕ~1t0+t,y2tllllllllllllll+ϕ~2t0+t,…,ykt+ϕ~kt0+t)t∈[0,η],Ui(y1,y2,…,yk)(t)={0t∈[-r,0],-gi2(t0,ϕ1,ϕ2,…,ϕk)+gi2(t0+t,y1t+ϕ~1t0+t,y2t+ϕ~2t0+t,…,ykt+ϕ~kt0+t)+1Γ(αi)∫0t‍(t-s)αi-1×fi(t0+s,y1s+ϕ~1t0+s,y2sllllllllllll+ϕ~2t0+s,…,yks+ϕ~kt0+s)dst∈[0,η],
for i=1,2,…,k. It is easy to check that the operator equation y=Sy+Uy has a solution y=(y1,y2,…,yk) if and only if yi is a solution for (**) for all i=1,2,…,k. In this case, xi(t0+t)=yi(t)+ϕ~i(t0+t) will be a solution of the problem (1) on [0,η]. Thus, the existence of a solution of the problem (1) is equivalent to the existence of a fixed point for the operator S+U on E(η,γ). Hence, it is sufficient that we show that S+U has a fixed point in E(η,γ). We prove it in three steps.
Step  I. Sz+Uy∈E(η,γ) for all z=(z1,z2,…,zk),y=(y1,y2,…,yk)∈E(η,γ).
Letz,y∈E(η,γ) be given. Then, Siz+Uiy∈C([-r,η],ℝn) for all i=1,2,…,k. It is easy to check that (Sz+Uy)(t)=0 for all t∈[-r,0]. Also, we have
(11)|Siz(t)+Uiy(t)|≤|(t0+t,z1t+ϕ~1t0+t,z2t+ϕ~2t0+t,…,zkt+ϕ~kt0+t)-gi1(t0,ϕ1,ϕ2,…,ϕk)+gi1(t0+t,z1t+ϕ~1t0+t,z2t+ϕ~2t0+t,…,zkt+ϕ~kt0+t)|+|(t0+t,y1t+ϕ~1t0+t,y2t+ϕ~2t0+t,…,ykt+ϕ~kt0+t)-gi2(t0,ϕ1,ϕ2,…,ϕk)+gi2(t0+t,y1t+ϕ~1t0+t,y2t+ϕ~2t0+t,…,ykt+ϕ~kt0+t)|+1Γ(αi)∫0t|(t-s)αi-1fi(t0+s,y1s+ϕ~1t0+s,y2sllllllllllllllllllllllllllllllllllllllllllll+ϕ~2t0+s,…,yks+ϕ~kt0+s)|ds≤2γ3+1Γ(αi)(∫0t(t-s)(αi-1)/(1-αi1)ds)1-αi1×(∫t0t0+t‍(mi(s))1/αi1ds)αi1≤2γ3+Miη(1+βi)(1-αi1)Γ(αi)(1+βi)1-αi1≤γ
for all t∈[0,η] and i=1,2,…,k. Thus, ∥Siz+Uiy∥=sup⁡t∈[0,η]|(Siz)(t)-(Uiy)(t)|≤γ for all i=1,2,…,k. Hence, Sz+Uy∈E(η,γ) for all z,y∈E(η,γ).
Step  II. S is a contraction on E(η,γ).
Lety′=(y1′,y2′,…,yk′),y′′=(y1′′,y2′′,…,yk′′)∈E(η,γ). Then,
(12)(y1t′+ϕ~1t0+t,y2t′+ϕ~2t0+t,…,ykt′+ϕ~kt0+t),(y1t′′+ϕ~1t0+t,y2t′′+ϕ~2t0+t,…,ykt′′+ϕ~kt0+t)∈A(δ,γ)
and so
(13)|Siy′(t)-Siy′′(t)|=|gi1(t0+t,y1t′+ϕ~1t0+t,y2t′+ϕ~2t0+t,…,ykt′+ϕ~kt0+t)-gi1(t0+t,y1t′′+ϕ~1t0+t,y2t′′lllllllllllllllllll+ϕ~2t0+t,…,ykt′′+ϕ~kt0+t)|≤li∥y′-y′′∥*
for all i=1,2,…,k. This implies that ∥Sy′-Sy′′∥*≤l∥y′-y′′∥*, where l=max⁡{l1,l2,…,lk}. Since 0<l<1, S is a contraction on E(η,γ).
Step  III. U is a completely continuous operator.
Suppose that(14)Ui1(y1,y2,…,yk)(t)={0t∈[-r.0],-gi2(t0,ϕ1,ϕ2,…,ϕk)+gi2(t0+t,y1t+ϕ~1t0+t,y2tlllllllllllll+ϕ~2t0+t,…,ykt+ϕ~kt0+t)t∈[0,η],Ui2(y1,y2,…,yk)(t)={0t∈[-r.0],1Γ(αi)∫0t(t-s)αi-1×fi(t0+s,y1s+ϕ~1t0+s,y2s+ϕ~2t0+s,…,yks+ϕ~kt0+s)dst∈[0,η],
for i=1,2,…,k. It is clear that
(15)U=(U11+U12U21+U22⋮Uk1+Uk2).
Sincegi2 is completely continuous for all i=1,2,…,k, Ui1 is continuous and also {Ui1(y):y∈E(η,γ)} is uniformly bounded. By using condition (H6), it is easy to check that {Ui1(y):y∈E(η,γ)} is equicontinuous. On the other hand,
(16)|Ui2y(t)|≤1Γ(αi)∫0t(t-s)αi-1|fi(t0+s,y1s+ϕ~1t0+s,y2sllllllllllllllllllllllllllllllllllllll+ϕ~2t0+s,…,yks+ϕ~kt0+s)|ds≤1Γ(αi)(∫0t(t-s)(αi-1)/(1-αi1)ds)1-αi1×(∫t0t0+t(mi(s))1/αi1ds)αi1≤Miη(1+βi)(1-αi1)Γ(αi)(1+βi)1-αi1
for all t∈[0,η] and i=1,2,…,k. This implies that {Ui2y:y∈E(η,γ)} is uniformly bounded. Now, we prove that {Ui2y:y∈E(η,γ)} is equicontinuous. Let 0≤t1<t2≤η and y∈E(η,γ) be given. Then, we have
(17)|Ui2y(t2)-Ui2y(t1)|=|1Γ(αi)∫0t1[(t2-s)αi-1-(t1-s)αi-1]×fi(t0+s,y1s+ϕ~1t0+s,y2slllllllllllllllllllllllllllll+ϕ~2t0+s,…,yks+ϕ~kt0+s)ds+1Γ(αi)∫t1t2(t2-s)αi-1fi(t0+s,y1s+ϕ~1t0+s,y2sllllllllllllllllllllllllllllllllllllllllllllll+ϕ~2t0+s,…,yks+ϕ~kt0+s)ds1Γ(αi)∫0t1[(t2-s)αi-1-(t1-s)αi-1]|≤1Γ(αi)∫0t1[(t1-s)αi-1-(t2-s)αi-1]llllllllllllllll×|fi(t0+s,y1s+ϕ~1t0+s,y2slllllllllllllllllllllllllll+ϕ~2t0+s,…,yks+ϕ~kt0+s)|ds+1Γ(αi)∫t1t2(t2-s)αi-1|fi(t0+s,y1s+ϕ~1t0+s,y2sllllllllllllllllllllllllllllllllllllllllllllllllllll+ϕ~2t0+s,…,yks+ϕ~kt0+s)|ds≤MiΓ(αi)(∫0t1[(t1-s)αi-1-(t2-s)αi-1]1/(1-αi1)ds)1-αi1+MiΓ(αi)(∫t1t2[(t2-s)αi-1]1/(1-αi1)ds)1-αi1≤MiΓ(αi)(∫0t1[(t1-s)βi-(t2-s)βi]ds)1-αi1+MiΓ(αi)(∫t1t2(t2-s)βids)1-αi1≤MiΓ(αi)(1+βi)1-αi1(t11+βi-t21+βi+(t2-t1)1+βi)1-αi1+MiΓ(αi)(1+βi)1-αi1(t2-t1)(1+βi)(1-αi1)≤2MiΓ(αi)(1+βi)1-αi1(t2-t1)(1+βi)(1-αi1)
for all i=1,2,…,k. Thus, {Ui2y:y∈E(η,γ)} is equicontinuous. Moreover, it is clear that Ui2 is continuous for all i=1,2,…,k. This implies that U is a completely continuous operator. Now, by using Krasnoselskii's fixed point theorem we get that S+U has a fixed point on E(η,γ) and so the problem (1) has a solution x=(x1,…,xk), where xi(t)=ϕi(0)+yi(t-t0) for all t∈[t0,t0+η] and i=1,2,…,k.If we putgi1=0 for all i=1,2,…,k, then we obtain next result.Corollary 4.
Suppose that there existδ∈(0,a) and γ∈(0,∞) such that conditions (H1)–(H3) hold, gi is continuous for all i=1,2,…,k, and |gi(t,xt)-gi(t,yt)|≤li∥x-y∥* for all x=(x1,x2,…,xk), y=(y1,y2,…,yk)∈A(δ,γ), and t∈I0, where li∈(0,1) is a constant for all i=1,2,…,k. Then the problem (1) has at least one solution on [t0,t0+η] for some positive number η.If we putgi2=0 for all i=1,2,…,k, then we obtain next result.Corollary 5.
Suppose that there existδ∈(0,a) and γ∈(0,∞) such that conditions (H1)–(H3) hold, gi is completely continuous for all i=1,2,…,k, and the family {t⊢gi(t,xt):(x1,x2,…,xk)∈Λ} is equicontinuous on C(I0,ℝn)×C(I0,ℝn)×⋯×C(I0,ℝn) for all bounded set Λ in A(δ,λ). Then the problem (1) has at least one solution on [t0,t0+η] for some positive number η.


## 3. Conclusions

In this work, we study an initial value problem for a class ofk-dimensional systems of fractional neutral functional differential equations by using Krasnoselskii’s fixed point theorem. Our result generalizes some old related results in a sense.



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