# Existence and Compactness Results for a System of Fractional Differential Equations

Authors: Cheikh Guendouz; Jamal Eddine Lazreg; Juan J. Nieto; Abdelghani Ouahab
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The existence and uniqueness, boundedness, and continuous dependence of solutions for fractional differential equations with Caputo fractional derivative is proven by Perov’s fixed point theorem in vector Banach spaces. We study the existence and compactness of solution sets and the u.s.c. of operator solutions.


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## Body

## 1. Introduction

In the past twenty years, the fractional differential equation has aroused great consideration not only in its application in mathematics but also in other applications in physics, engineering, finance, fluid mechanics, viscoelastic mechanics, electroanalytical chemistry, and biological and other sciences [1–7].In recent decades, the Riemann-Liouville, Caputo, and Hadamard fractional calculus are paid more attention; see the monographs [5, 8–13].Applied problems requiring definitions of fractional derivatives are those that are physically interpretable for initial conditions containingy0,y′0, etc. The same requirements are true for boundary conditions. Caputo’s fractional derivative satisfies these demands. For more details on the geometric and physical interpretation for fractional derivatives of both the Riemann-Liouville type and the Caputo type, see Podlubny [12] and Diethelm [14].The theory of fractional differential equations and inclusions has been extensively studied and developed by many authors; see [15–21] and the references therein.Perov in 1964 [22] and Perov and Kibenko [23] extended the classical Banach contraction principle for contractive maps on space endowed with a vector-valued metric. Later, they attempted to generalize the Perov fixed point theorem in several directions which has a number of applications in various fields of nonlinear analysis, semilinear differential equations, and system of ordinary differential equations.In [24], Dezideriu and Precup studied the following system of semilinear equations
(1)A1x=F1x,yA2y=F2x,ywhere A1,A2:DA⊂X⟶X are linear operators and F1,F2:J×X×X⟶X are nonlinear operators.Precup, in [25], explained the advantage of vector-valued norms and the role of matrices that are convergent to zero in the study of semilinear operator systems.Many authors studied the existence of solutions for a system of differential equations and impulsive differential equations by using the vector version fixed point theorem; their results are given in [26–30].Our goal of this paper is to treat the systems of fractional differential equations. More precisely, we will consider the following problem:(2)cDαxt=ft,xt,yt,cDβyt=gt,xt,yt,t∈J,x0=x0,y0=y0,where cDα and cDβ are the Caputo fractional derivatives, α,β∈0,1,J=0,∞,f,g:J×ℝ×ℝ⟶ℝ are given functions, and x0,y0∈ℝ.In the case whereα=β=1, the above system was used to analyze initial value problems and boundary value problems for nonlinear competitive or cooperative differential systems from mathematical biology [31] and mathematical economics [32] which can be set in the operator from ((2)).The plan of this paper is as follows: in Section2, we introduce all the background material used in this paper such as some properties of generalized Banach spaces, fixed point theory, and fractional calculus theory. In Section 3, we state and prove our main results by using Perov’s fixed point type theorem in generalized Banach spaces. By the Leray-Schauder fixed point in vector Banach space, we prove the existence and compactness of solution sets of the above problems.


## 2. Preliminaries

In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper.Definition 1 [22].
LetX be a nonempty set. The mapping d:X×X⟶ℝ+m which satisfies all the usual axioms of the metric is called a generalized metric in Perov’s sense and X,d is called a generalized metric space.In a generalized metric space in Perov’s sense, the concepts of Cauchy sequence, convergent sequence, completeness, and open and closed subsets are similarly defined as those for usual metric space.Ifv,r∈ℝm,v≔v1,v2,⋯,vm and r≔r1,r2,⋯,rm, then by v≤r, we mean vi≤ri for each i∈1,⋯,m, and by v<r, we mean vi<ri for each i∈1,⋯,m. Also ∣v∣≔∣v1∣,⋯,∣vm∣ and maxu,v≔maxu1,v1,⋯,maxum,vm. If c∈ℝ, then v≤c means vi≤c for each i∈1,⋯,m. Denote by
(3)Bx0,r=x∈X:dx0,x<r,the open ball centered in x0 with radius r, and
(4)Bx0,r¯=x∈X:dx0,x≤r,the closed ball centered in x0 with radius r.Definition 2.
A square matrixM of real numbers is said to be convergent to zero if and only if Mn⟶0 as n⟶∞.Lemma 3 [33].
LetM∈Mm,mℝ+. The following statements are equivalent:
(i)
M is a matrix convergent to zero(ii)
The eigenvalues ofM are in open disc,  i.e., ∣μ∣<1 , for every μ∈ℂ with detM−μI=0(iii)
The matrixI−M is nonsingular and I−M−1=I+M+⋯+Mn+⋯(iv)
The matrixI−M is nonsingular and I−M−1 has nonnegative elements(v)
Mnq⟶0 and tqMn⟶0 as n⟶∞ , for any q∈ℝmExample 4.
Some examples of matrix convergent to zero are(5)1M=abab,where a,b∈ℝ+ and a+b<1,
(6)2M=aabb,where a,b∈ℝ+ and a+b<1, and
(7)3M=ab0c,where a,b,c∈ℝ+ and maxa,c<1.Definition 5 [34].
LetX,d be a generalized metric space. An operator N:X⟶X is said to be contractive if there exists a convergent to zero matrix M such that
(8)dNx,Ny≤Mdx,yforallx,y∈X.Notice now that the Banach fixed point theorem can be extended to generalized metric spaces in the sense of Perov.Theorem 6 [22, 28].
LetX,d be a complete generalized metric space and N:X⟶X be a contractive operator with Lipschitz matrix M. Then, N has a unique fixed point x∗, and for each x0∈X, we have
(9)dNkx0,x∗≤MkI−M−1dx0,Nx0forallk∈ℕ.We recall now the following Leary-Schauder type theorem.Theorem 7 [28, 35].
LetX be a generalized Banach space and let N:X⟶X be a completely continuous operator. Then, either
(i)
the equationNx=x has at least one solution, or(ii)
the setM=x∈X∣μNx=x,μ∈0,1 is unboundedWe will use the following notations. LetX,d and Y,ρ be two metric spaces and G:X⟶PY.
(10)PX=Y⊂X:Y=∅,PclX=Y∈PX:Yisclosed,PbX=Y∈PX:Ybounded.Definition 8 [28, 36].
A multivalued mapφ:X⟶PY is called upper semicontinuous (u.s.c.) at a point x0∈X provided that for every open subset V⊂Y with V⊇φx0, there exists U∈Vx0 such that
(11)∀x∈U,φx⊂V.φ is called upper semicontinuous if it is u.s.c. at every point x∈X.The mappingG is said to be completely continuous if it is u.s.c., and for every bounded subset C⊆X, GC is relatively compact, i.e., there exists a relatively compact set K=K⊂X such that
(12)GC=∪Gx:x∈C⊂K.Also,G is compact if GX is relatively compact, and it is called locally compact if for each x∈X, there exists an open set W containing x, such that GW is relatively compact.Theorem 9 [36].
LetG:X⟶PcpY be a closed locally compact multifunction. Then, G is u.s.c.Now, we recall some notations and definitions of fractional calculus theory.Definition 10 [5].
The Riemann-Liouville fractional integral of the functionh∈L10,T,ℝ+ of order α∈ℝ+ is defined by
(13)Iaht=1Γa∫0tt−sa−1hsds,where Γ is the Euler gamma function defined by Γa=∫0∞tα−1e−tdt,α>0.Definition 11 [5].
For a functionh∈ACnJ,ℝ, the Caputo fractional-order derivative of order α of h is defined by
(14)cD0αht=1Γn−α∫0tt−sn−α−1hnsds,where n=α+1.We recall Gronwall’s lemma for singular kernels, whose proof can be found in Lemma 7.1.1 of [37].Lemma 12.
Letv:0,b⟶0,∞ be a real function; w· is a nonnegative, locally integrable function on 0,bsomeb≤+∞; at is a nonnegative, nondecreasing continuous function defined on 0≤t<b; at≤M (constant); and suppose vt is nonnegative and locally integrable on 0≤t<b. Assume γ>0 such that
(15)vt≤wt+at∫0tvst−s1−γds,then
(16)vt≤wt+∫0t∑n=1∞atΓγnΓnγt−snγ−1wsds,for every t∈0,b.


## 3. Existence, Uniqueness, and Bounded Solutions

In order to define a solution for problem (2), consider the following functional spaces. Let J=0,∞ and CJ,ℝ be the space of all continuous functions from J into ℝ.
(17)Cb=y∈CJ,ℝ:yisbounded.C
b is a Banach with norm
(18)yCb=sup∣yt∣:t∈J.We need the following auxiliary result.Lemma 13 [14].
Concerning the problem,(19)cDαxt=ft,x,0<α<1,x0=x0∈ℝ,where the function f:ℝ+×ℝ⟶ℝ is continuous. The function x:ℝ+×ℝ⟶ℝ is the unique solution of the problem (19) if and only if
(20)xt=x0+1Γα∫0tt−sα−1fs,xsds,t∈ℝ+.Definition 14.
A functionx,y∈Cb is said to be a solution of (2) if and only if
(21)xt=x0+1Γα∫0tt−sα−1fs,xs,ysds,yt=y0+1Γβ∫0tt−sβ−1gs,xs,ysds,t∈J.In this section, we assume the following conditions.(H1).
There exists functionshi,α,hj,β∈L1J,ℝ+, i=1,2, j=3,4 such that
(22)fs,x,y−fs,x¯,y¯≤hi,αsx−x¯+h2,αsy−y¯forallx,x¯,y,y¯∈ℝ,gs,x,y−gs,x¯,y¯≤h3,βsx−x¯+h4,βsy−y¯forallx,x¯,y,y¯∈ℝ,where
(23)hi,α∞=supt∈J∫0tθαt−shi,αsds<∞,i=1,2,hj,β∞=supt∈J∫0tθβt−shj,βsds<∞,j=3,4,where for γ=α,β,
(24)θγt=tγ−1,t>0,0,t≤0.(H2).
The functionsfα,gβ:J⟶ℝ are defined by
(25)fαt=θαt−sfs,0,0,gβt=θβt−sgs,0,0,t∈J,satisfies
(26)fα∞=supt∈J∫0tθαt−sfs,0,0ds<∞,gβ∞=supt∈J∫0tθβt−sgs,0,0ds<∞.Now, we are in a position to prove our existence and uniqueness solution for the problem (2) using the Perov fixed point theorem and show that for each initial condition x0,y0, the solution is bounded.Theorem 15.
Assume that (H1)-(H2) are satisfied. If the matrix
(27)M=h1,α∞Γαh2,α∞Γαh3,β∞Γβh4,β∞Γβ∈M2×2ℝ+,converges to zero. Then, the problem (2) has a unique bounded solution.Proof.
Transform the problem (2) into a fixed point theorem of the operator N:Cb×Cb⟶Cb×Cb defined by Nx,y=N1x,y,N2x,y, where
(28)N1x,yt=x0+1Γα∫0tt−sα−1fs,xs,ysds,N2x,yt=y0+1Γβ∫0tt−sβ−1gs,xs,ysds,t∈0,∞.
First, we show that the operatorN is well-defined. Let x,y∈Cb×Cbandt∈0,∞, then we have
(29)N1x,yt≤x0+1Γα∫0tt−sα−1fs,xs,ysds≤x0+1Γα∫0tt−sα−1fs,xs,ys−fs,0,0ds+1Γα∫0tfs,0,0t−s1−αds.Then,
(30)N1x,ycb≤x0+h1,α∞Γαxcb+h2,α∞Γαycb+fα∞Γα,N2x,ycb≤y0+h3,β∞βxcb+h4,β∞βycb+gβ∞Γβ.
Hence, the operatorN is well-defined.
Clearly, the fixed points of operatorN are solutions of problem (2). Now, we show that N is a contraction. For all x,y,x¯,y¯∈Cb×Cb, we have
(31)N1x,yt−N1x¯,y¯t≤1Γα∫0tt−sα−1fs,xt,yt−fs,x¯t,y¯tds≤1Γα∫0th1,αsxt−x¯t+h2,αsyt−y¯tds.Then,
(32)N1x,y−N1x¯,y¯cb≤h1,α∞Γαx−x¯cb+h2,α∞Γαy−y¯cb.Similarly, we have
(33)N2x,y−N2x¯,y¯Cb≤h3,β∞Γβx−x¯Cb+h4,β∞Γβy−y¯Cb.Therefore,
(34)Nx,y−Nx¯,y¯Cb×Cb≤Mx−x¯Cby−y¯Cb.
According to Theorem6, we deduce that the operator N has unique fixed point which is a solution of problem (2). Now, we will prove that the solution x,y of problem (2) is bounded. For all t∈0,∞, we have
(35)xt≤x0+1Γα∫0tt−sα−1fs,xs,ysds≤x0+1Γα∫0th1,αsxs+h2,αsysds+1Γα∫0tfαs,0,0ds≤x0+1Γα∫0th1,αsxs+h2,αsysds+fα∞Γα,yt≤y0+1Γβ∫0tt−sβ−1gs,xs,ysds≤y0+1Γβ∫0th3,βsxs+h4,βsysds+1Γβ∫0tgβs,0,0ds≤y0+1Γβ∫0th3,βsxs+h4,αsysds+gβ∞Γβ.Therefore,
(36)xCb≤x0+h1,α∞ΓαxCb+h2,α∞ΓαyCb+fα∞Γα,yCb≤y0+h3,β∞ΓβxCb+h4,β∞ΓβyCb+gβ∞Γβ.Hence,
(37)xCbyCb≤MxCbyCb+σ1σ2,where
(38)σ1=x0+fα∞Γα,σ2=y0+gβ∞Γβ.
Then,(39)xCbyCb≤I−M−1σ1σ2.
From (H1) and (H2), we deduce that the solution (x,y) is bounded.For the next result, we prove the continuous dependence of solutions on initial conditions.Theorem 16.
Assume that (H1) and (H2) hold. If ft,0,0=gt,0,0=0, t∈J and the matrix M defined in (27) converges to zero.
For everyx0,y0∈ℝ×ℝ, we denoted by xt,x0,yt,y0 the solution of problem (2). Then, the map x0,y0⟶x·,x0,y·,y0 is continuous.Proof.
From Theorem15, for each initial condition x0,y0,x¯0,y¯0∈ℝ×ℝ, there exists unique solution x⋅,x0,y⋅,y0, x¯⋅,x¯0,y¯⋅,y¯0, then we get
(40)xt,x0=x0+1Γα∫0tt−sα−1fs,xs,ysds,yt,y0=y0+1Γβ∫0tt−sβ−1gs,xs,ysds,x¯t,x¯0=x¯0+1Γα∫0tt−sα−1fs,x¯s,y¯sds,y¯t,y¯0=y¯0+1Γβ∫0tt−sβ−1gs,xs,y¯sds.
Therefore,(41)xt,x0−x¯t,x¯0≤x0−x¯0+1Γα∫0th1,αsxs−x¯s+h2,αsys−y¯sds,yt,y0−y¯t,y¯0≤y0−y¯0+1Γβ∫0th3,βsxs−x¯s+h4,βsys−y¯sds.
Hence,(42)x⋅,x0−x¯⋅,x¯0Cby⋅,y0−y¯⋅,y¯0Cb≤I−M−1x0−x¯0y0−y¯0⟶0,asx0,y0⟶x¯0,y¯0.


## 4. Existence and Compactness of Solution Sets

For the existence and compactness result of problem (2), we consider the following Banach space:
(43)BCbJ,ℝ=u∈CJ,ℝ:supt∈Jut1+t<∞,with norm
(44)uB=supt∈Jut1+t.It is evident thatBCbJ,ℝ is a Banach space. The following compactness criterion on unbounded domains is called Corduneanu compactness criterion in which the proof is easy and similar to the classical one in Cbℝ+,ℝ (see [38]).Lemma 17.
LetH⊂BCbℝ+,ℝ. Then, H is relatively compact if the following conditions hold:
(a)
H is uniformly bounded in BCbℝ+,ℝn(b)
The functions belonging toH are almost equicontinuous on ℝ+, i.e., for all I⊂ℝ+ compact interval, for any ϵ>0, there exists δϵ>0 such that for every t1,t2∈J with t1−t2<δϵ, we have for all u∈H(45)ut21+t2−ut11+t1<∈(c)
The functions fromH are equiconvergent, that is, given ϵ>0, there corresponds Tϵ>0 such that
(46)ut11+t1−ut21+t2<ϵforanyt1,t2≥ϵandu∈HIn the sequel of this section, we will consider the following assumption.(H3).
For everyl>0, the functions x,y⟶ft,x,y,gt,x,y are uniformly continuous on the sets −l,l×−l,l uniformly with respect to t∈ℝ+, i.e., f and g satisfied the following condition: for all l>0, ϵ>0, there exists δϵ>0 such that for all x,y,x¯,y¯∈−l,l×−l,l and for all t∈ℝ+ with x/1+t−x¯/1+t<δϵ, y/1+t−y¯/1+t<δϵ, we have
(47)ft,x,y−ft,x¯,y¯<ϵ,gt,x,y−gt,x¯,y¯<ϵ.(H4).
There existai,bi,ci>0, i=1,2 such that
(48)ft,x,y≤a1x1+t+b1y1+t+c1,gt,x,y≤a2x1+t+b2y1+t+c2,∀x,y∈ℝ,t∈ℝ+.Theorem 18.
Assume that (H3) and (H4) hold. Then, the problem (2) has at least one bounded solution. Moreover, the solution set
(49)Sx0,y0=x,y∈BCbJ,ℝ×BCbJ,ℝ:x,yissolutionof2,is compact and the multivalued map S:x0,y0⟶Sx0,y0 is u.s.c.Proof.
LetN=N1,N2 is defined in the proof of Theorem 15.Step 1.
N is well defined. Let x,y∈BCbJ,ℝ×BCbJ,ℝ, then
(50)N1xt,yt1+l≤x01+l+11+lΓα∫0tt−sα−1fs,xs,ysds≤x01+l11+lΓα∫0tt−sα−1a1xs1+s+b1ys1+s+c1ds≤x01+l+c1tαΓα+11+t+11+tΓα∫0tt−sα−1a1xs1+s+b1ys1+s+ds≤x0+c1Γα+1+11+tΓα∫0tt−sα−1dsa1xB+b1yB.Hence,(51)supt∈JN1xt,yt1+t≤x0+c1Γα+1+a1xB+b1yBΓα+1,supt∈JN2xt,yt1+t≤y0+c1Γβ+1+a2xB+b2yBΓβ+1.Step 2.
N is continuous.Letxn,yn⟶x,y in BCbJ,ℝ×BCbJ,ℝ. Then, there exists l>0 such that for any t∈ℝ+ and n∈ℕ, we have
(52)xnt1+t,ynt1+t,xt1+t,yt1+t∈−l,l.By (H3), for every ϵ>0, there exists δϵ>0 such that for all x,y,x¯,y¯∈−l,l and for all t∈ℝ+ with x/1+t−x¯/1+t<δϵ, y/1+t−y¯/1+t<δϵ we have
(53)ft,x,y−ft,x¯,y¯<ϵΓα+1,gt,x,y−gt,x¯,y¯<ϵΓβ+1.Sincexn,yn converge to x,y, then there exists n0∈ℕ such that for all n>n0,
(54)xnt1+t−xt1+t<δϵ,ynt1+t−yt1+t<δϵ,∀t∈ℝ+.Hence,(55)N1xn,ynt1+t−N1x,yt1+t≤1Γα∫0tt−sα−11+tfs,xns,yns−fs,xs,ysds≤1Γα∫0tt−sα−11+tdsΓα+1ϵ≤tα1+tϵ≤ϵ.Thus(56)N1xn,yn−N1x,yB⟶0,N2xn,yn−N2x,yB⟶0,asn⟶∞.Step 3.
We will show thatN maps bounded sets into bounded sets in BCbJ,R×BCbJ,R.LetBr≔x,y∈BCbJ,ℝ×BCbJ,ℝ:x,y∞≤r, where r=r1,r2 and if x,y∈Br, then we obtain
(57)N1xt,yt1+t≤1Γα∫0tt−sα−11+tfs,xs,ysds≤1Γαa1r1+b1r2+c1.Similarly, we have(58)N2xt,yt1+t≤1Γβa2r1+r2b2+c2.Hence,(59)N1x,y,N2x,y≤l1,l2,where
(60)l1=1Γαa1r1+b1r2+c1,l2=1Γβa2r1+r2b2+c2.Step 4.
Now, we prove thatN maps bounded sets in BCbJ,ℝ×BCbJ,ℝ into almost equicontinuous sets of BCbJ,ℝ×BCbJ,ℝ.Letx,y∈Br. Then, for all t1<t2,
(61)N1xt2,yt21+t2−N1xt1,yt11+t1≤1Γα∫0t1t2−sα−11+t2−t1−sα−11+t1fs,xs,ysds+1Γα∫t1t2t2−sα−11+t2fs,xs,ysds≤1Γα∫0t1t2−sα−11+t2−t1−sα−11+t1a1r1+b1r2+c1ds+1Γα∫t1t2t2−sα−11+t2a1r1+b1r2+c1ds≤a1r1+b1r2+c1Γα+1t2α1+t2−t1α1+t1+2t2−t1α1+t2.Thus,(62)N1xt2,tt21+t2−N1xt1,tt11+t1⟶0ast1⟶t2.Similarly, we have(63)N2xt2,yt21+t2−N2xt1,yt11+t1⟶0ast1⟶t2.Step 5.
The setNB¯0,r is equiconvergent, i.e., for every ϵ>0, there exists Tϵ>0 such that Nxt2,yt2−Nxt1,yt1<ϵ, for every t1,t2>Tϵ and each x,y∈B¯0,r.
(64)N1xt2,yt21+t2−N1xt1,yt11+t1≤a1r1+b1r2+c1Γα+13t2α1+t2+t1α1+t1,N2xt2,yt21+t2−N2xt1,yt11+t1≤a2r1+b1r2+c1Γα+13t2β1+t2+t1β1+t1.It is clear that(65)3t2β1+t2⟶0,t1β1+t1⟶0,ast2,t1⟶∞.Then, for anyϵ>0, there exists Tϵ>0 such that for all t1,t2>Tϵ, we have
(66)N1xt2,yt21+t2−N1xt1,yt11+t1≤ϵ,forallx,y∈B¯0,r,N2xt2,yt21+t2−N2xt1,yt11+t1≤ϵ,foranyx,y∈B¯0,r.Step 6.
The set is defined as follows:
M
=
x
,
y
∈
BC
b
J
,
ℝ
×
BC
b
J
,
ℝ
:
x
,
y
=
μ
N
x
,
y
,
μ
∈
0
,
1 is bounded. Let x,y∈M, then
(67)xt1+t≤x01+t+μ1+tΓα∫0tt−sα−1fs,xs,ysds≤x01+t+μ1+tΓα∫0tt−sα−1a1xs1+s+b1ys1+s+c1ds≤x01+t+c1tαΓα+11+t+11+tΓα∫0tt−sα−1a1xs1+s+b1ys1+sds.Similarly, we get(68)yt1+t+y01+t+c2tβΓβ+11+t+11+tΓβ∫0tt−sβ−1a2xs1+s+b2ys1+sds.Thus,(69)xt1+t+yt1+t≤X01+t+Ctγ1+t+C∗1+t∫0tt−sγ−1xs1+s+b2ys1+sds,where
(70)X0=x0+y0,C=c1Γα+1+c2Γβ+1,C∗=a1+b1Γα+a2+b2Γβ,γ=maxα,β.Hence,(71)xt1+t+yt1+t≤X0+C+C∗1+t∫0tt−sγ−1xt1+s+ys1+sds.By the Gronwall-Bellman Lemma12, we have
(72)xt1+t+yt1+t≤X0+C+∫0t∑n=1∞C∗ΓγnΓnγt−snγ−1ds≤X0+C+∑n=1∞tγC∗Γγn1+tnΓnγ+1X0+C≤X0+C+∑n=0∞C∗ΓγnΓnγ+1X0+C.Then, for everyt∈0,∞, we have
(73)xt1+t+yt1+t≤X0+C1+EγC∗Γγ≔M,where Eγ is the Mittag-Leffler function defined by
(74)Eγz=∑n=0∞znΓnγ+1.Then,(75)xCb≤M,yCb≤M.According to Theorem7, problem (2) has at least one solution.Now, we show that the set(76)Sx0,y0=x,y∈BCbJ,ℝ×BCbJ,ℝ:x,yissolutionof2,is compact. It is clear that Sx0,y0⊂NSx0,y0. From (75), we deduced that Sx0,y0 is bounded sets in BCbJ,ℝ. Since N is compact, then Sx0,y0 is compact if and only if Sx0,y0 is closed. Let xn,ynn≥1⊂Sx0,y0 be a sequence converge to x,y. Thus,
(77)xnt=x0+1Γαt−sα−1fs,xns,ynsds,ynt=y0+1Γβt−sβ−1gs,xns,ynsds,t∈J.Similarly to Step2, we can prove that
(78)xt=x0+1Γα∫0tt−sα−1fs,xs,ysds,yt=y0+1Γβ∫0tt−sβ−1gs,xs,ysds,t∈J.This implies thatSx0,y0 is compact.The solution operatorS is u.s.c.
(a)
S has a closed graph:To see this, first note that the graph ofS is the set
(79)GS=a,b,x,y∈ℝ×ℝ×BCbJ,ℝ×BCbJ,ℝ∣x,y∈Sa,b.Letaq,bq,xq,yqq be a sequence in GS, and let aq,bq,xq,yqq⟶a,b,x,yasq⟶∞.Sincexq,yq∈Saq,bq, then we have
(80)xqt=aq+1Γα∫0tt−sα−1fs,xqs,yqsds,yqt=bq+1Γβ∫0tt−sβ−1gs,xqs,yqsds,t∈J.Let(81)Zt=Z1t,Z2t=a+1Γα∫0tt−sα−1fs,xs,ysds,b+1Γβ∫0tt−sβ−1gs,xs,ysdst∈J.As in Step2, we can prove that
(82)xq,yq⟶x,yasq⟶∞.(b)
Using the same method as Steps3 to 4, we find that for each bounded set C⊂BCbJ,ℝ×BCbJ,ℝ, we can show that SC is compact.From (a) and (b), we concluded that,S is u.s.c.


## 5. Applications

In this section, we show the applicability of our main result. We begin by illustrating by Theorem15.Example 19.
Consider the problem:(83)Dc1/2x=h1t1+k1sinx+k2cosx,Dc1/2y=h2t1+k3x+k4y1/2,t∈ℝ,x0=x0,y0=y0,where
(84)h1t=y∫0tse−s2ds1001+t2andh2s=tt2001+t2,t∈ℝ+,ki≥0,i=1,2,3,4,ft,x,y=h1t1+k1sinx+k2cos∣y∣,gt,x,y=h2t1+k3x+k4∣y∣.
It is clear that(85)∫0th1st−s1/2ds=∫0ts∫0sre−r2dr1001+s2t−s1/2≤∫0∞e−r2dr1001+s2t−s1/2∫0tss1+s2t−s1/2ds=∫0∞e−r2dr100∫0tt2ll1+tl21−l1/2dl≤∫0∞e−r2dr100∫0t1−l−1/2dl≤∫0∞e−r2dr50=π100,∫0th2st−s1/2ds=∫0tss2001+s2t−s1/2ds=≤1100.
Then,(86)supt∈ℝ+∫0thist−s1/2ds<∞,i=1,2.
Observe that everyx,y,x¯,y¯∈ℝ and t∈ℝ+, we have
(87)ft,x,y−ft,x¯,y¯≤k1h1tx−x¯+k2h1ty−y¯,gt,x,y−gt,x¯,y¯≤k3h2tx−x¯+k4h2ty−y¯.
Hence, the condition (H1) holds.
We see that(88)ft,0,0=h1t,gt,0,0=h2t,t∈ℝ+.
Hence, the condition (H2) holds. Assume that k1k4−k2k3≠0. For this, we have
(89)M~=k1πΓ1/2100k2πΓ1/2100k3πΓ1/2100k4πΓ1/2100.
det
M
~
≠
0. If we add that k2=0 or k3 and 0<k1<Γ1/2100/π,0<k4<Γ1/2100≤1. The ρM¯<1. By Theorem 15, it follows that problem (83) has a unique solution.In this last example, we illustrate the applicability of Theorem18.Lemma 20.
Let us consider the functionf:ℝ+×ℝ×ℝ⟶ℝ defined by
(90)ft,x,y=atϕx,y,where a:ℝ+⟶ℝ,ϕ:ℝ×ℝ⟶ℝ satisfy the following assumptions:
(S1)a∈Cbℝ.
(S2) The functionφ is continuous and bounded. Then, the function
(91)f∗t,x,y=atϕx1+t,y1+t,t∈ℝ+,x,y∈ℝ×ℝ,satisfied the condition (H3).Proof.
Sincea is bounded, then there exists A>0 such that
(92)at<A,∀t∈ℝ+.
Letl>0, then for any x,y,x¯,y¯∈−l,l×−l,l, we obtain
(93)ft,x,y−ft,x¯,y¯=atϕx,y−ϕx¯,y¯.
Hence,(94)ft,x,y−ft,x¯,y¯≤Aϕx,y−ϕx¯,y¯.
It is clear that form (S2)ϕ is uniformly continuous on the −l,l×−l,l. Thus, for any ϵ>0 there exists δϵ>0 such that for all x,y,x¯,y¯∈−l,l×−l,l with x−x¯<δϵ,y−y¯<δϵ then,
(95)ft,x,y−ft,x¯,y¯≤Aϵ.
This implies that for everyt∈ℝ+ such that for all x,y,x¯,y¯∈−l,l×−l,l with x/1+t−x¯/1+t<δϵ, y/1+t−y¯/1+t<δϵ, we get
(96)ft,x,y−ft,x¯,y¯≤Aϵ.Theorem 21.
Leta1,a2:ℝ+⟶ℝ,ϕ1,ϕ2:ℝ×ℝ⟶ℝ be functions that satisfy the conditions (S1) and (S2).
Assume
(S3) There existsdi>0,i=1,4 such that
(97)ϕ1x,y≤d1x+d2y+d3,∀x,y∈ℝ,t∈ℝ+,ϕ2x,y≤d4x+d5y+d6,∀x,y∈ℝ,t∈ℝ+.
Then, the following problem(98)Dcαxt=ft,x,t,Dcβxt=gt,x,t,t∈J,x0=x0,y0=y0,where
(99)ft,x,y=a1tϕ1x1+t,y1+t,t∈ℝ+,x,y∈ℝ,gt,x,y=a2tϕ2x1+t,y1+t,t∈ℝ+,x,y∈ℝ.Proof.
From Lemma20, we deduce that f and g satisfies (H3). By (S1) and (S3), there exist A1,A2>0 such that
(100)ft,x,y≤A1x1+t,y1+t,gt,x,y≤A2x1+t,y1+t,t∈ℝ+,x,y∈ℝ.
Therefore, all the conditions of Theorem18 hold. Then, problem (98) has at least one solution in BCb.Remark 22.
For another application, we can replace the condition (S2) with (S4). The functionsφ1 and φ2 are uniformly continuous and bounded.


## 6. Conclusion

In this paper, we investigated a system of fractional differential equations under various assumptions on the right-hand-side nonlinearity and we obtain a number of results regarding the existence and uniqueness of solutions in an appropriate space of continuous functions. In this paper, we have focused on the dependence continuity of a solution, compactness of solution sets, and upper semicontinuity of operator solutions. We hope this paper can provide some contribution to the questions of existence and topological structure for the system of fractional differential equations on unbounded domains.



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