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Research article

On redundancy, separation and connectedness in multiset topological spaces

  • Received: 27 December 2019 Accepted: 27 February 2020 Published: 09 March 2020
  • MSC : 54A05, 54A10

  • This paper makes an attempt to study M-topology as a novel structure and emphasizes its importance by projecting how it differs from general topology. Primarily, the issue of redundancy in Mtopology is addressed by pointing out the importance of complementation with appropriate examples. Unlike general topology, M-topology induces two subspace M-topologies on a submset. In general topology, these two definitions of the subspace topologies coincide. The situations in which these two subspace M-topologies coincide are also analyzed for the purpose. Furthermore, two types of Mconnectedness and M-separations in M-topology are introduced and it is proved that neither of which implies the other.

    Citation: Rajish Kumar P, Sunil Jacob John. On redundancy, separation and connectedness in multiset topological spaces[J]. AIMS Mathematics, 2020, 5(3): 2484-2499. doi: 10.3934/math.2020164

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  • This paper makes an attempt to study M-topology as a novel structure and emphasizes its importance by projecting how it differs from general topology. Primarily, the issue of redundancy in Mtopology is addressed by pointing out the importance of complementation with appropriate examples. Unlike general topology, M-topology induces two subspace M-topologies on a submset. In general topology, these two definitions of the subspace topologies coincide. The situations in which these two subspace M-topologies coincide are also analyzed for the purpose. Furthermore, two types of Mconnectedness and M-separations in M-topology are introduced and it is proved that neither of which implies the other.


    Differential equations of arbitrary order have been shown to be useful in the study of models of many phenomena in various fields such as: Electrochemistry and material science, they are in fact described by differential equations of fractional order [9,10,15,16,25,26,27,28,29]. For more details, we refer the reader to the books of Hilfer [30], Podlubny [31], Kilbas et al. [34], Miller and Ross [2] and to the following research papers [1,2,3,4,5,6,7,8,11,12,14,16,17,19,20,24,31,35,36,37,38,39,40,41,42]. In this work, we discuss the existence and uniqueness of the solutions for multi-point boundary value problems of nonlinear fractional differential equations with two Riemann-Liouville fractionals:

    {Dαx(t)=mi=1fi(t,x(t),y(t),φ1x(t),ϕ1y(t)),α]1,2],t[0,T]Dβy(t)=mi=1gi(t,x(t),y(t),φ2x(t),ϕ2y(t)),β]1,2],t[0,T]I2αx(0)=0, Dα2x(T)=θIα1(x(η)), 0<η<T,I2βy(0)=0, Dβ2x(T)=ωIβ1(x(γ)), 0<γ<T, (1.1)

    where D(.), I(.) denote the Riemann-Liouville derivative and integral of fractional order (.), respectively, fi, gi:[0,T]×R4R, i=1,,m are continuous functions on [0,T] and

    (φ1x)(t)=t0A1(t,s)x(s)ds, (ϕ1y)(t)=t0B1(t,s)y(s)ds,
    (φ2x)(t)=t0A2(t,s)x(s)ds, (ϕ1y)(t)=t0B2(t,s)y(s)ds,

    with Ai and Bi being continuous functions on [0,1]×[0,1]. However, it is rare to find a work in nonlinear term fi depends on fractional derivative of unknown functions x(t),y(t),φ1x(t),ϕ1y(t) and solutions for multi-order fractional differential equations on the infinite interval [0,T). Motivated by [8,11,12,13,14] and the references therein, we consider the existence and unicity of solution for multi-order fractional differential equations on infinite interval [0,T).

    The rest of this paper is organized as follow. In section 2, we present some preliminaries and lemmas. Section 3 is dedicated to showing the existence of a solution for problem (1.1). Finally, section 4 illustrated the proposed results with two examples.

    Remark 1.1. This work generalizes the work of Houas and Benbachir [14] on different boundary conditions and for another type of integral.

    This section covers the basic concepts of Riemann-Liouville type fractional calculus that will be used throughout this paper.

    Definition 2.1. [31,32] The Riemann-Liouville fractional integral operator of order α0, of a function f:(0,)R is defined as

    {Jαf(t)=1Γ(α)t0(tτ)α1f(τ)dτ,J0f(t)=f(t),

    where Γ(α):=0euuα1du.

    Definition 2.2. [31,32] The Riemann-Liouville fractional derivative of order α>0, of a continuous function h:(0,)R is defined as

    Dαh(t)=1Γ(nα)(ddt)nt0(tτ)nα1h(τ)dτ=(ddt)nInαh(τ),

    where n=[α]+1.

    For α<0, we use the convention that Dαh=Jαh. Also for 0ρ<α, it is valid that DρJαh=hαρ. We note that for ε>1 and εα1,α2,...,αn, we have

    Dαtε=Γ(ε+1)Γ(εα+1)tεα,Dαtαi=0, i=1,2,...,n.

    In particular, for the constant function h(t)=1, we obtain

    Dα1=1Γ(1α)tα,αN.

    For αN, we obtain, of course, Dα1=0 because of the poles of the gamma function at the points 0,1,2,... For α>0, the general solution of the homgeneous equation Dαh(t)=0 in C(0,T)L(0,T) is

    h(t)=c0tαn+c1tαn1+......+cn2tα2+cn1tα1,

    where ci,i=1,2,....,n1, are arbitrary real constants. Further, we always have DαIαh=h, and

    DαIαh(t)=h(t)+c0tαn+c1tαn1+......+cn2tα2+cn1tα1.

    Lemma 2.1. [33] Let E be Banach space. Assume that T:EE is a completely continuous operator. If the set V={xE:x=μTx, 0<μ<1} is bounded, then T has a fixed point in E.

    To define the solution for problem (1.1). We consider the following lemma.

    Lemma 2.2. Suppose that (Hi)i=1,,mC([0,1],R), and consider the problem

    Dαh(t)mi=1Hi(t)=0, tj, 1<α<2, mN, (2.1)

    with the conditions

    I2αh(0)=0, Dα2h(T)=θIα1(h(η)), 0<η<T. (2.2)

    Then we have

    h(t)=1Γ(α)mi=1t0(tτ)α1Hi(τ)dτ+tα1ψ(mi=1T0(Tτ)Hi(τ)dτθΓ(2α)mi=1η0(ητ)2α2Hi(τ)dτ)

    with ψ=θΓ(α)Γ(2α1)η2α2Γ(α)T.

    Proof. We have

    h(t)=mi=1IαHi(t)+c0tα2+c1tα1,

    where ciR, i=0,1.

    We obtain

    I2αh(τ)=mi=1I2Hi(τ)+c0I2ατα2+c1I2ατα1=mi=1I2Hi(τ)+c0+c1τ,Iα1h(τ)=mi=1I2α1Hi(τ)+c0Iα1τα2+c1Iα1τα1=mi=1I2α1Hi(τ)+c0Γ(α1)Γ(2α2)τ2α3+c1Γ(α)Γ(2α1)τ2α2,Dα2h(τ)=mi=1I2Hi(τ)+c0Γ(α1)+c1Γ(α)τ.

    Using the given conditions: I2αh(0)=0, we find that c0=0, and since Dα2h(T)θIα1(h(η))=0, we have

    mi=1I2hi(T)+c1Γ(α)Tθ[mi=1I2α1hi(η)+c1Γ(α)Γ(2α1)η2α2]=0,

    then

    c1[Γ(α)Γ(2α1)η2α2Γ(α)T]=mi=1I2hi(T)θmi=1I2α1hi(η)

    and

    c1=1ψ(mi=1I2Hi(T)θmi=1I2α1Hi(η))=1ψ(mi=1T0(Tτ)Hi(τ)dτθΓ(2α)mi=1η0(ητ)2α2Hi(τ)dτ)

    with

    ψ=θΓ(α)Γ(2α1)η2α2Γ(α)T.

    Finally, the solution of (2.1) and (2.2) is

    h(t)=1Γ(α)mi=1t0(tτ)α1Hi(τ)dτ+tα1ψ(mi=1T0(Tτ)Hi(τ)dτθΓ(2α)mi=1η0(ητ)2α2Hi(τ)dτ).

    We denote by

    E={x,yC([0,T],R);φix,ϕiyC([0,T],R)  i=1,2},

    and the Banach space of all continuous functions from [0,T] to R endowed with a topology of uniform convergence with the norm defined by

    ||(x,y)||E=max(||x||,||y||,||φ1x||,||ϕ1y||,||φ2x||,||ϕ2y||),

    where

    ||x||=suptj|φix(t)|,||y||=suptj|y(t)|,||ϕix||=suptj|φix(t)|,||ϕiy||=suptj|ϕiy(t)|.

    In this section, we prove some existence and uniqueness results to the nonlinear fractional coupled system (1.1).

    For the sake of convenience, we impose the following hypotheses:

    (H1) For each i=1,2,,m, the functions fi and gi  :[0,T]×R4R are continuous.

    (H2) There exist nonnegative real numbers ξik,φik,k=1,2,3,4,i=1,2,,m, such that for all t[0,T] and all (x1,x2,x3,x4), (y1,y2,y3,y4)R4, we have

    |fi(t,x1,x2,x3,x4)fi(t,y1,y2,y3,y4)|4k=1 ξik|xkyk|,

    and

    |gi(t,x1,x2,x3,x4)gi(t,y1,y2,y3,y4)|4k=1 χik|xkyk|.

    (H3) There exist nonnegative constants (Li) and  (Ki) i=1,...,m, such that: For each t[0,T] and all (x1,x2,x3,x4)R4,

    |fi(t,x1,x2,x3,x4)|Li,|gi(t,x1,x2,x3,x4)|Ki,i=1,...,m.

    We also consider the following quantities:

    A1=TαΓ(α+1)mi=1(ξi1+ξi2+ξi3+ξi4),A2=TβΓ(β+1)mi=1(χi1+χi2+χi3+χi4),A3=maxt,s[0,1]||A1(t,s)||×A1,A4=maxt,s[0,1]||A2(t,s)||×A1,A5=maxt,s[0,1]||B1(t,s)||×A2,A6=maxt,s[0,1]||B2(t,s)||×A2,ν1=[TαΓ(α+1)+1ψ(Tα+12+θT3α2(2α1)2Γ(2α1))],ν2=[TβΓ(β+1)+1ψ(Tβ+12+ωT3β2(2β1)2Γ(2β1))],ν3=maxt,s[0,1]|A1(t,s)|ν1,ν4=maxt,s[0,1]|A2(t,s)|ν1,ν5=maxt,s[0,1]|B1(t,s)|ν2,ν6=maxt,s[0,1]|B2(t,s)|ν2.

    The first result is based on Banach contraction principle. We have

    Theorem 3.1. Assume that (H2) holds. If the inequality

    max(A1,A2,A3,A4,A5,A6)<1, (3.1)

    is valid, then the system (1.1) has a unique solution on [0,T].

    Proof. We define the operator T:EE by

    T(x,y)(t)=(T1(x,y)(t),T2(x,y)(t)),t[0,T],

    such that

    T1(x,y)(t)=1Γ(α)mi=1t0(tτ)α1Hi(τ)dτ+tα1ψ(mi=1T0(Tτ)Hi(τ)dτθΓ(2α)mi=1η0(ητ)2α2Hi(τ)dτ) (3.2)

    and

    T2(x,y)(t)=1Γ(β)mi=1t0(tτ)β1Gi(τ)dτ+tβ1ψ(mi=1T0(Tτ)Gi(τ)dτωΓ(2β)mi=1γ0(γτ)2β2Gi(τ)dτ) (3.3)

    where

    Hi(τ)=fi(τ,x(τ),y(τ),φ1x(τ),ϕ1y(τ))

    and

    Gi(τ)=gi(τ,x(τ),y(τ),φ2x(τ),ϕ2y(τ)).

    We obtain

    φiT1(x,y)(t)=t0Ai(t,s)T1(x,y)(s)ds, ϕiT2(x,y)(t)=t0Bi(t,s)T2(x,y)(s)ds

    where i=1,2.

    We shall now prove that T is contractive.

    Let T1(x1,y1),T2(x2,y2)E. Then, for each t[0,T], we have

    |T1(x1,y1)T1(x2,y2)|[1Γ(α)mi=1t0(tτ)α1dτ+tα1ψ(mi=1T0(Tτ)dτθΓ(2α)mi=1η0(ητ)2α2dτ)]×maxτ[0,T]mi=1|(fi(τ,x1(τ),y1(τ),φ1x1(τ),ϕ1y1(τ))fi(τ,x2(τ),y2(τ),φ1x2(τ),ϕ1y2(τ)))|TαΓ(α+1)maxτ[0,T]mi=1|(fi(τ,x1(τ),y1(τ),φ1x1(τ),ϕ1y1(τ))fi(τ,x2(τ),y2(τ),φ1x2(τ),ϕ1y2(τ)))|.

    By (H2), it follows that

    ||T1(x1,y1)T1(x2,y2)||TαΓ(α+1)mi=1(ξi1+ξi2+ξi3+ξi4)×max(||x1x2||,||y1y2||,||φ1(x1x2)||,||φ2(x1x2)||,||ϕ1(y1y2)||,||ϕ2(y1y2)||).

    Hence,

    ||T1(x1,y1)T1(x2,y2)||A1||x1x2,y1y2||E. (3.4)

    With the same arguments as before, we can show that

    ||T2(x1,y1)T2(x2,y2)||A2||x1x2,y1y2||E. (3.5)

    On the other hand, we have

    ||φ1(T1(x1,y1)T1(x2,y2))||t0||A1(t,s)||||T1(x1,y1)T1(x2,y2)||dsmaxt,s[0,1]||A1(t,s)||×A1||x1x2,y1y2||E.

    Hence,

    ||φ1(T1(x1,y1)T1(x2,y2))||A3||x1x2,y1y2||E (3.6)

    and

    ||φ2(T1(x1,y1)T1(x2,y2))||A4||x1x2,y1y2||E. (3.7)

    Also, we have

    ||ϕ1(T2(x1,y1)T2(x2,y2))||A5||x1x2,y1y2||E (3.8)

    and

    ||ϕ2(T2(x1,y1)T2(x2,y2))||A6||x1x2,y1y2||E. (3.9)

    Thanks to (3.4)–(3.9), we get

    ||T(x1,y1)T(x2,y2)||max(A1,A2,A3,A4,A5,A6)×||(x1x2,y1y2)||E. (3.10)

    Thanks to (3.10), we conclude that T is a contractive operator. Therefore, by Banach fixed point theorem, T has a unique fixed point which is the solution of the system (1.1).

    Our second main result is based on Lemma 2.1. We have

    Theorem 3.2. Assume that the hypotheses (H1) and (H3) are satisfied. Then, system (1.1) has at least a solution on [0,T].

    Proof. The operator T is continuous on E in view of the continuity of fi and gi (hypothesis (H1)).

    Now, we show that T is completely continuous:

    (i) First, we prove that T maps bounded sets of E into bounded sets of E. Taking λ>0, and (x,y)Ωλ={(x,y)E;||(x,y)||λ}, then for each t[0,T], we have:

    |T1(x,y)|[1Γ(α)t0(tτ)α1dτ+tα1ψ(T0(Tτ)dτθΓ(2α)η0(ητ)2α2dτ)]×supt[0,T]mi=1|fi(t,x(t),y(t),φ1x(t),ϕ1y(t))|[TαΓ(α+1)+1ψ(Tα+12+θT3α2(2α1)2Γ(2α1))]×supt[0,T]mi=1|fi(t,x(t),y(t),φ1x(t),ϕ1y(t))|,

    Thanks to (H3), we can write

    ||T1(x,y)||[TαΓ(α+1)+1ψ(Tα+12+θT3α2(2α1)2Γ(2α1))]mi=1Li.

    Thus,

    ||T1(x,y)||ν1mi=1Li. (3.11)

    As before, we have

    ||T2(x,y)||ν2mi=1Ki. (3.12)

    On the other hand, for all j=1,2, we get

    |ϕjT1(x,y)(t)|=|t0Aj(t,s)T1(x,y)(s)ds|maxt,s[0,1]|Aj(t,s)|ν1mi=1Li.

    This implies that

    ||ϕ1T1(x,y)(t)||ν3mi=1Li, (3.13)
    ||ϕ2T1(x,y)(t)||ν4mi=1Li. (3.14)

    Similarly, we have

    ||φ1T2(x,y)(t)||ν5mi=1Ki, (3.15)
    ||φ2T2(x,y)(t)||ν6mi=1Ki. (3.16)

    It follows from (3.11)–(3.16) that:

    ||T(x,y)||Emax(ν1mi=1Li,ν2mi=1Ki,ν3mi=1Li,ν4mi=1Li,,ν5mi=1,ν6mi=1).

    Thus,

    ||T(x,y)||E<.

    (ii) Second, we prove that T is equi-continuous:

    For any 0t1<t2T and (x,y)Ωλ, we have

    |T1(x,y)(t2)T1(x,y)(t1)|[1Γ(α)t10(t2τ)α1(t1τ)α1dτ+1Γ(α)t2t1(t2τ)α1dτ+tα12tα11ψ(T22θη2α1Γ(2α1)2Γ(2α1))]×supt[0,T]mi=1|fi(t,x(t),y(t),φ1x(t),ϕ1y(t))|[2Γ(α+1)(t2t1)α1+(tα12tα11)[T22ψθη2α1ψΓ(2α1)2Γ(2α1)+1Γ(α+1)]]×mi=1Li.

    Therefore,

    ||T1(x,y)(t2)T1(x,y)(t1)||E[2Γ(α+1)(t2t1)α1+(tα12tα11)[T22ψ+1Γ(α+1)]]×mi=1Li. (3.17)

    We also have

    ||T2(x,y)(t2)T2(x,y)(t1)||E[2Γ(β+1)(t2t1)β1+(tβ12tβ11)[T22ψ+1Γ(β+1)]]×mi=1Ki. (3.18)

    On the other hand,

    |ϕiT1(x,y)(t2)ϕiT1(x,y)(t1)|[maxs[0,1]|Ai(t2,s)Ai(t1,s)|+(t2t1)maxs[0,1]|Ai(t1,s)|]×sups[0,1]|T1(x,y)(s)|.

    Consequently, for all i=1,2, we obtain

    ||ϕiT1(x,y)(t2)ϕiT1(x,y)(t1)||[maxs[0,1]|Ai(t2,s)Ai(t1,s)|+(t2t1)maxs[0,1]|Ai(t1,s)|]ν1mi=1Li. (3.19)

    Similarly,

    ||φiT1(x,y)(t2)φiT1(x,y)(t1)||[maxs[0,1]|Bi(t2,s)Bi(t1,s)|+(t2t1)maxs[0,1]|Bi(t1,s)|]ν2mi=1Ki. (3.20)

    where i=1,2. Using (3.17)–(3.20), we deduce that

    ||T(x,y)(t2)T(x,y)(t1)||E0

    as t2t1.

    Combining (i) and (ii), we conclude that T is completely continuous.

    (iii) Finally, we shall prove that the set F defined by

    F={(x,y)E,(x,y)=ρT(x,y), 0<ρ<1}

    is bounded.

    Let (x,y)F, then (x,y)=ρT(x,y), for some 0<ρ<1. Thus, for each t[0,T], we have:

    x(t)=ρT1(x,y)(t), y(t)=ρT2(x,y)(t). (3.21)

    Thanks to (H3) and using (3.11) and (3.12), we deduce that

    ||x||ρν1mi=1Li, ||y||ρν2mi=1Ki. (3.22)

    Using (3.13)–(3.16), it yields that

    {||ϕ1x||ρν3mi=1Li||ϕ2x||ρν4mi=1Li||φ1y||ρν5mi=1Ki||φ2y||ρν6mi=1Ki. (3.23)

    It follows from (3.22) and (3.23) that

    ||T(x,y)||Eρmax(ν1mi=1Li,ν2mi=1Ki,ν3mi=1Li,ν4mi=1Li,,ν5mi=1,ν6mi=1).

    Consequently,

    ||(x,y)||E<.

    This shows that F is bounded. By Lemma (2.1), we deduce that T has a fixed point, which is a solution of (1.1).

    To illustrate our main results, we treat the following examples.

    Example 4.1. Consider the following system:

    {D32x(t)=cos(πt)(x+y+φ1x(t)+ϕ1y(t))10π(x+y+φ1x(t)+ϕ1y(t))+132π2e(cosx(t)+cosy(t)+φ1x(t)+ϕ1y(t)4π),D32y(t)=18π3(t+1)(x+y+φ2x(t)+ϕ2y(t)3+x+y+φ2x(t)+ϕ2y(t))+1(10π+et)e(t+1)(sinx(t)+siny(t)+cosφ2x(t)+cosϕ2y(t)2+sinx(t)+siny(t)+cosφ2x(t)+cosϕ2y(t)),I12x(0)=0, D12x(T)=I12(x(1)),I12y(0)=0, D12y(T)=I12(y(1)). (4.1)

    We have

    α=32, β=32, T=1, θ=1, ω=1, γ=1, m=2, η=1.

    Also,

    f1(t,x(t),y(t),φ1x(t),ϕ1y(t))=cos(πt)(x+y+φ1x(t)+ϕ1y(t))10π(1+x+y+φ1x(t)+ϕ1y(t)), (4.2)
    f2(t,x(t),y(t),φ1x(t),ϕ1y(t))=132π2e(cosx(t)+cosy(t)+φ1x(t)+ϕ1y(t)4π). (4.3)

    For t[0,1] and (x1,y1,φ1x1,ϕ1y1),(x2,y2,φ1x2,ϕ1y2)R4, we have

    |f1(t,x1,y1,φ1x1,ϕ1y1)f1(t,x2,y2,φ1x2,ϕ1y2)||cos(πt)|10π|x1+y1+φ1x1+ϕ1y11+x1+y1+φ1x1+ϕ1y1x2+y2+φ1x2+ϕ1y2)1+x2+y2+φ1x2+ϕ1y2)|110π(|x1x2|+|y1y2|+|φ1x1φ1x2|+|ϕ1y1ϕ1y2|) (4.4)

    and

    |f2(t,x1,y1,φ1x1,ϕ1y1)f2(t,x2,y2,φ1x2,ϕ1y2)|132πe(|x1x2|+|y1y2|+|φ1x1φ1x2|+|ϕ1y1ϕ1y2|). (4.5)

    So, we can take

    ξ11=ξ12=ξ13=ξ14=110π,
    ξ21=ξ22=ξ23=ξ24=132πe.

    We also have

    g1(t,x(t),y(t),φ2x(t),ϕ2y(t))=18π3(t+1)(x+y+φ2x(t)+ϕ2y(t)3+x+y+φ2x(t)+ϕ2y(t))

    and

    g2(t,x(t),y(t),φ2x(t),ϕ2y(t))=1(10π+et)et+1(sinx(t)+siny(t)+cosφ2x(t)+cosϕ2y(t)2+sinx(t)+siny(t)+cosφ2x(t)+cosϕ2y(t)) (4.6)

    For t[0,1] and (x1,y1,φ2x1,ϕ2y1),(x2,y2,φ2x2,ϕ2y2)R4, we can write

    |g1(t,x1,y1,φ2x1,ϕ2y1)g1(t,x2,y2,φ2x2,ϕ2y2)|18π3(|x1x2|+|y1y2|+|φ2x1φ2x2|+|ϕ2y1ϕ2y2|), (4.7)

    and

    |g2(t,x1,y1,φ2x1,ϕ2y1)g2(t,x2,y2,φ2x2,ϕ2y2)|110πe2(|x1x2|+|y1y2|+|φ2x1φ2x2|+|ϕ2y1ϕ2y2|). (4.8)

    Hence,

    χ11=χ12=χ13=χ14=18π3,
    χ21=χ22=χ23=χ24=110πe2.

    Therefore,

    A1=0.0589009676,A2=0.0250930393.

    Suppose

    Ai=Bi=1, i=1,2,

    so,

    A1=A3=A4,A2=A5=A6.

    Thus,

    max(A1,A2,A3,A4,A5,A6)<1, (4.9)

    and by Theorem 3.1, we conclude that the system (4.1) has a unique solution on [0,1].

    Example 4.2.

    {D32x(t)=π(t+1)sin(φ1x(t)+ϕ1y(t))2cos(x(t)+y(t))+et2π+cos(x(t)+φ1x(t))+sin(sin(y(t)+ϕ1y(t)), t[0,1],D43y(t)=e2sin(x(t)+y(t))2π+cos(φ2x(t)+ϕ2y(t))+3t2cosy(t)et3+1cos(x(t)+y(t)φ2x(t)ϕ2y(t)), t[0,1],I12x(0)=0, D12x(T)=I12(x(1)),I23y(0)=0, D23y(T)=I13(y(1)). (4.10)

    We have

    α=32, β=43, T=1, θ=1, ω=1, γ=1, m=2, η=1.

    Since

    |f1(t,x(t),y(t),φ1x(t),ϕ1y(t))|=|π(t+1)sin(φ1x(t)+ϕ1y(t))2cos(x(t)+y(t))|2π,|f2(t,x(t),y(t),φ1x(t),ϕ1y(t))|=|et2π+cos(x(t)+φ1x(t))+sin(sin(y(t)+ϕ1y(t))|e2π+2,|g1(t,x(t),y(t),φ2x(t),ϕ2y(t))|=|e2sin(x(t)+y(t))2π+cos(φ2x(t)+ϕ2y(t))|e22π+1,|g2(t,x(t),y(t),φ2x(t),ϕ2y(t))|=|3t2cosy(t)et3+1cos(x(t)+y(t)φ2x(t)ϕ2y(t))|3e1.

    The functions f1, f2, g1 and g2 are continuous and bounded on [0,1]×R4. So, by Theorem 3.2, the system (4.10) has at least one solution on [0,1].

    We have proved the existence of solutions for fractional differential equations with integral and multi-point boundary conditions. The problem is solved by applying some fixed point theorems. We also provide examples to make our results clear.

    The authors declare that they have no conflicts of interest in this paper.



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