Research article

Existence results of sequential fractional Caputo sum-difference boundary value problem

  • Received: 10 March 2022 Revised: 13 May 2022 Accepted: 07 June 2022 Published: 15 June 2022
  • MSC : 39A10, 39A13, 39A70

  • In this article, we study the existence and uniqueness results for a sequential nonlinear Caputo fractional sum-difference equation with fractional difference boundary conditions by using the Banach contraction principle and Schaefer's fixed point theorem. Furthermore, we also show the existence of a positive solution. Our problem contains different orders and four fractional difference operators. Finally, we present an example to display the importance of these results.

    Citation: Chanisara Metpattarahiran, Thitiporn Linitda, Thanin Sitthiwirattham. Existence results of sequential fractional Caputo sum-difference boundary value problem[J]. AIMS Mathematics, 2022, 7(8): 15120-15137. doi: 10.3934/math.2022829

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  • In this article, we study the existence and uniqueness results for a sequential nonlinear Caputo fractional sum-difference equation with fractional difference boundary conditions by using the Banach contraction principle and Schaefer's fixed point theorem. Furthermore, we also show the existence of a positive solution. Our problem contains different orders and four fractional difference operators. Finally, we present an example to display the importance of these results.



    Fractional calculus is an emerging field drawing attention from both theoretical and applied disciplines. In particular, fractional calculus is a powerful tool for explaining problems in ecology, biology, chemistry, physics, mechanics, networks, flow in porous media, electricity, control systems, viscoelasticity, mathematical biology, fitting of experimental data, and so forth. One may see the papers [1,2,3,4,5] and the references therein.

    Fractional difference calculus or discrete fractional calculus is a very new field for mathematicians. Some real-world phenomena are being studied with the assistance of fractional difference operators. Basic definitions and properties of fractional difference calculus can be found in the book [6]. Fractional boundary value problems can be found in the books [7,8]. Now, the studies of boundary value problems for fractional difference equations are extended to be more complex. Excellent papers related to discrete fractional boundary value problems can be found in [9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35] and references cited therein. In particular, there are some recent papers that present the Caputo fractional difference calculus [36,37,38,39,40,41]. In the literature, there are apparently few research works studying boundary value problems for Caputo fractional difference-sum equations. For example, [42] studied a boundary value problem for pLaplacian Caputo fractional difference equations with fractional sum boundary conditions of the forms

    {ΔαC[ϕp(ΔβCx)](t)=f(t+α+β1,x(t+α+β1)),tN0,T:={0,1,,T},ΔβCx(α1)=0,x(α+β+T)=ρΔγx(η+γ). (1.1)

    In [43], investigated a nonlocal fractional sum boundary value problem for a Caputo fractional difference-sum equation of the form

    ΔαCu(t)=F[t+α1,ut+α1,ΔβCu(t+αβ)],tN0,T,ΔγCu(αγ1)=0,u(T+α)=ρΔωu(η+ω). (1.2)

    In addition, [44] considered a periodic boundary value problem for Caputo fractional difference-sum equations of the form

    ΔαCu(t)=F[t+α1,u(t+α1),Ψγu(t+α1)],tN0,T,t+α1tk,Δu(tk)=Ik(u(tk1)),k=1,2,...,p,Δ(Δβu(tk+β))=Jk(Δβu(tk+β1)),k=1,2,...,p,Au(α1)+BΔβu(α+β1)=Cu(T+α)+DΔβu(T+α+β). (1.3)

    We aim to fill the gaps related to the boundary value problem of Caputo fractional difference-sum equations. The goal of this paper is to enrich this new research area by using the unknown function of Caputo fractional difference and fractional sum in the problem. So, in this paper, we consider a sequential nonlinear Caputo fractional sum-difference equation with fractional difference boundary value conditions of the form

    CΔααCΔβα+β1x(t)=H[t+α+β1,x(t+α+β1),CΔνα+β1x(t+α+βν),Ψμ(t+α+β1,x(t+α+β1))],tN0,T,ρ1x(α+β2)=x(T+α+β),ρ2CΔγα+β2x(α+βγ1)=CΔγα+β2x(T+α+βγ+1), (1.4)

    where ρ1,ρ2R, 0<α,β,γ,ν,μ1, 1<α+β2 are given constants, HC(Nα+β2,T+α+β×R3,R), and for φC(×R2,[0,)), :={(t,r):t,rNα+β2,T+α+βandrt}. The operator Ψμ is defined by

    Ψμ(t,x(t)):=tμs=α+βμ2(tσ(s))μ1_Γ(μ)φ(t,s+μ,x(s+μ),CΔνα+β2x(s+μν+1)).

    The plan of this paper is as follows. In Section 2, we recall some definitions and basic lemmas. Also, we derive the solution of (1.4) by converting the problem to an equivalent equation. In Section 3, we prove existence and uniqueness results of the problem (1.4) using the Banach contraction principle and Schaefer's theorem. Furthermore, we also show the existence of a positive solution to (1.4). An illustrative example is presented in Section 4.

    In the following, there are notations, definitions and lemmas which are used in the main results.

    Definition 2.1. [10] We define the generalized falling function by tα_:=Γ(t+1)Γ(t+1α), for any t and α for which the right-hand side is defined. If t+1α is a pole of the Gamma function and t+1 is not a pole, then tα_=0.

    Lemma 2.1. [9] Assume the following factorial functions are well defined. If tr, then tα_rα_ for any α>0.

    Definition 2.2. [10] For α>0 and f defined on Na:={a,a+1,}, the α-order fractional sum of f is defined by

    Δαaf(t)=Δαf(t):=1Γ(α)tαs=a(tσ(s))α1_f(s),

    where tNa+α and σ(s)=s+1.

    Definition 2.3. [11] For α>0 and f defined on Na, the α-order Caputo fractional difference of f is defined by

    CΔαaf(t)=ΔαCf(t):=Δ(Nα)aΔNf(t)=1Γ(Nα)t(Nα)s=a(tσ(s))Nα1_ΔNf(s),

    where tNa+Nα and NN is chosen so that 0N1<α<N. If α=N, then ΔαCf(t)=ΔNf(t).

    Lemma 2.2. [11] Assume that α>0 and 0N1<αN. Then,

    Δαa+NαCΔαay(t)=y(t)+C0+C1t1_+C2t2_+...+CN1tN1_,

    for some CiR, 0iN1.

    To study the solution of the boundary value problem (1.4), we need the following lemma that deals with a linear variant of the boundary value problem (1.4) and gives a representation of the solution.

    Lemma 2.3. Let Λ(ρ11)0, 0<α,β,γ,ν,μ1, 1<α+β2 and hC(Nα+β1,T+α+β1,R) be given. Then, the problem

    CΔααCΔβα+β1x(t)=h(t+α+β1),tN0,T (2.1)
    {ρ1x(α+β2)=x(T+α+β),ρ2CΔγα+β2x(α+βγ1)=CΔγα+β2x(T+α+βγ+1), (2.2)

    has the unique solution

    x(t)=T+(ρ11)(tαβ)+2ρ1Λ(ρ11)Γ(1γ)Γ(β1)Γ(α)T+α+βs=α+β1sβ+1r=αrαξ=0(T+α+βγ+1σ(s))γ_×(sσ(r))β2_(rσ(ξ))α1_h(ξ+α+β1)+1(ρ11)Γ(β)Γ(α)T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_h(ξ+α+β1)+1Γ(β)Γ(α)tβs=αsαξ=0(tσ(s))β1_(sσ(ξ))α1_h(ξ+α1), (2.3)

    where

    Λ=ρ2Γ(Tγ+4)Γ(2γ)Γ(T+3). (2.4)

    Proof. Using the fractional sum of order α(0,1] for (2.1) and from Lemma 2.2, we obtain

    CΔβα+β2x(t)=C1+1Γ(α)tαs=0(tσ(s))α1_h(s+α+β1), (2.5)

    for tNα1,T+α.

    Using the fractional sum of order 0<β1 for (2.5), we obtain

    x(t)=C2+C1t+1Γ(β)Γ(α)tβs=αsαξ=0(tσ(s))β1_(sσ(ξ))α1_h(ξ+α+β1), (2.6)

    for tNα+β2,T+α+β.

    By substituting t=α+β2,T+α+β into (2.6) and employing the first condition of (2.2), we obtain

    C2(ρ11)+C1[(T(ρ11)(α+β)+2ρ1]=1Γ(β)Γ(α)T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_h(ξ+α+β1). (2.7)

    Using the fractional Caputo difference of order 0<γ1 for (2.6), we obtain

    CΔγα+β2x(t)=C1Γ(1γ)t+γ1s=α+β2(tσ(s))γ_+1Γ(1γ)Γ(β)Γ(α)×t+γ1s=α+β2(tσ(s))γ_sΔ[sβr=αrαξ=0(sσ(r))β1_(rσ(ξ))α1_h(ξ+α+β1)]=C1Γ(1γ)t+γ1s=α+β2(tσ(s))γ_+1Γ(1γ)Γ(β1)Γ(α)×t+γ1s=α+β2sβ+1r=αrαξ=0(tσ(s))γ_(sσ(r))β2_(rσ(ξ))α1_h(ξ+α+β1), (2.8)

    for Nα+βγ1,T+α+βγ+1.

    By substituting t=α+βγ1,T+α+βγ+1 into (2.8) and employing the second condition of (2.2), it implies

    C1=1ΛΓ(1γ)Γ(β1)Γ(α)T+α+βs=α+β1sβ+1r=αrαξ=0(T+α+βγ+1σ(s))γ_×(sσ(r))β2_(rσ(ξ))α1_h(ξ+α+β1).

    The constant C2 can be obtained by substituting C1 into (2.7). Then, we get

    C2=T(ρ11)(α+β)+2ρ1(ρ11)ΛΓ(1γ)Γ(β1)Γ(α)T+α+βs=α+β1sβ+1r=αrαξ=0(T+α+βγ+1σ(s))γ_×(sσ(r))β2_(rσ(ξ))α1_h(ξ+α+β1)+1(ρ11)Γ(β)Γ(α)T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_h(ξ+α+β1),

    where Λ is defined by (2.4). Substituting the constants C1 and C2 into (2.6), we obtain (2.3).

    In this section, we wish to establish the existence results for the problem (1.4). We denote C=C(Nα+β2,T+α+β,R) as the Banach space of all functions x with the norm defined by

    xC=x+ΔνCx,

    where x=maxtNα+β2,T+α+β|x(t)| and ΔνCx=maxtNα+β2,T+α+β|ΔνCx(tν+1)|.

    The following assumptions are assumed:

    (A1) H[t,x,y,z]:Nα+β2,T+α+β×R3R is a continuous function.

    (A2) There exist constants K1,K2>0 such that for each tNα+β2,T+α+β and all xi,yi,ziR,i=1,2, we have

    |H[t,x1,y1,z1]H[t,x2,y2,z2]|K1[|x1x2|+|y1y2|+|z1z2|],

    and

    K2=maxtNα+β2,T+α+β|H[t,0,0,Ψμ(t,0)]|,

    where Ψμ(t,0):=1Γ(μ)tμs=α+βμ2(tσ(s))μ1_φ(t,s+μ,0,0).

    (A3) φ:×R2R is continuious for (t,s), and there exists a constant L>0, such that for each (t,s) and all xi,yiC,i=1,2 we have

    |φ(t,s+μ,x1,y1)φ(t,s+μ,x2,y2)|L[|x1x2|+|y1y2|].

    Let us define the operator ˜H[t,x(t)] by

    ˜H[t,x(t)]:=H[t,x(t),ΔνCx(tν+1),Ψμ(t,x(t))], (3.1)

    for each tNα+β2,T+α+β and xC.

    Note that ΔβΔα˜H[t,x(t)] and ΔνCΔβΔα˜H[t,x(t)] exist when ν<α+β2.

    Lemma 3.1. Assume that (A1)(A3) hold. Then, the following property holds:

    (A4) There exits a positive constant Θ such that

    |˜H[t,x1(t)]˜H[t,x2(t)]|Θx1x2C,

    for each tNα+β2,T+α+β and x1,x2C, where

    Θ:=K1[1+LΓ(T+μ+3)Γ(μ+1)Γ(T+3)]. (3.2)

    Proof. By (A3), for each tNα+β2,T+α+β and x1,x2C, we obtain

    |(Ψμx1)(t)(Ψμx2)(t)|1Γ(μ)tμs=α+βμ2(tσ(s))μ1_|φ(t,s+μ,x1(s+μ),ΔνCx1(s+μν+1))φ(t,s+μ,x2(s+μ),ΔνCx2(s+μν+1))|1Γ(μ)T+α+βμs=α+βμ2(T+α+βσ(s))μ1_×L[|x1(s+μ)x2(s+μ)|+|ΔνCx1(s+μν+1)ΔνCx2(s+μν+1)|]LΓ(T+μ+3)Γ(μ+1)Γ(T+3){x1x2+ΔνCx1ΔνCx2},

    and hence

    |˜H[t,x1(t)]˜H[t,x2(t)]|K1[|x1(t)x2(t)|+|ΔνCx1(tν+1)ΔνCx2(tν+1)|]+K1LΓ(T+μ+3)Γ(μ+1)Γ(T+3){x1x2+ΔνCx1ΔνCx2}=Θx1x2C.

    Next, we define the operator F:CC by

    (Fx)(t)=T+α+βs=α+β1sβ+1r=αrαξ=0Aα,β,γ,ρ1,ρ2(t,s,r,ξ)˜H[ξ+α+β1,x(ξ+α+β1)]+T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_(ρ11)Γ(β)Γ(α)˜H[ξ+α+β1,x(ξ+α+β1)]+tβs=αsαξ=0(tσ(s))β1_(sσ(ξ))α1_Γ(β)Γ(α)˜H[ξ+α+β1,x(ξ+α+β1)], (3.3)

    where

    Aα,β,γ,ρ1,ρ2(t,s,r,ξ):=T+(ρ11)(tαβ)+2ρ1Λ(ρ11)Γ(1γ)Γ(β1)Γ(α)×(T+α+βγ+1σ(s))γ_(sσ(r))β2_(rσ(ξ))α1_. (3.4)

    By Lemma 2.3, we find that any solution of the problem (1.4) is the fixed point of the operator F.

    Lemma 3.2. Assume that the function Aα,β,γ,ρ1,ρ2(t,s,r,ξ) satisfies the following properties:

    (A5) Aα,β,γ,ρ1,ρ2(t,s,r,ξ) is a continuous function for all (t,s,r,ξ)Nα+β2,T+α+β×Nα+β2,T+α+β×Nα1,T+α+1×N0,T+2=:D, and there exist two constants Ω1,Ω2>0, such that

    T+α+βs=α+β1sβ+1r=αrαξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)|Ω1,T+α+βs=α+β1sβ+1r=αrαξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)|Ω2,

    where

    Ω1:=[(1+|ρ11|)T+2p1|ρ11||Λ|]Γ(T+γ+3)Γ(T+α+β+1)Γ(2γ)Γ(α+β)[Γ(T+2)]2, (3.5)
    Ω2:=|1αβΛ|Γ(Tγ+3)[Γ(T+α+β+1)]2Γ(2ν)Γ(2γ)Γ(α+β)[Γ(T+2)]2Γ(T+α+βν), (3.6)

    and tΔνC is the Caputo fractional difference with respect to t.

    Proof. It is obvious that Aα,β,γ,ρ1,ρ2(t,s,r,ξ) is a continuous function for all (t,s,r,ξ)D. Next, we consider

    T+α+βs=α+β1sβ+1r=αrαξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)|maxtNα+β2,T+α+βT+α+βs=α+β1sβ+1r=αrαξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)|=maxtNα+β2,T+α+β|T+(ρ11)(tαβ)+2ρ1Λ(ρ11)Γ(1γ)Γ(β1)Γ(α)|×T+α+βs=α+β1sβ+1r=αrαξ=0(T+α+βγ+1σ(s))γ_(sσ(r))β2_(rσ(ξ))α1_[(1+|ρ11|)T+2p1|ρ11||Λ|]T+α+βs=α+β1sβ+1r=αrαξ=0(T+α+βγ+1σ(s))γ_(sσ(r))β2_(rσ(ξ))α1_[(1+|ρ11|)T+2p1|ρ11||Λ|]Γ(T+γ+3)Γ(T+α+β+1)Γ(T+2)Γ(T+2)Γ(2γ)Γ(α+β)=Ω1,

    and

    T+α+βs=α+β1sβ+1r=αrαξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)|maxtNα+β2,T+α+βT+α+βs=α+β1sβ+1r=αrαξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)|=maxtNα+β2,T+α+β|t+ν1s=α+β2(tσ(s))ν_sΔ[T+(ρ11)(sαβ)+2ρ1]ΛΓ(1ν)Γ(1γ)Γ(β1)Γ(α)|×T+α+βs=α+β1sβ+1r=αrαξ=0(T+α+βγ+1σ(s))γ_(sσ(r))β2_(rσ(ξ))α1_|T+α+β+ν1s=α+β2(T+α+βσ(s))ν_(1αβ)ΛΓ(1ν)|Γ(Tγ+3)Γ(T+α+β+1)[Γ(T+2)]2Γ(2γ)Γ(α+β)|1αβΛ|Γ(Tγ+3)[Γ(T+α+β+1)]2Γ(2ν)Γ(2γ)Γ(α+β)[Γ(T+2)]2Γ(T+α+βν)=Ω2.

    Thus, the condition (A5) holds.

    In what follows, we consider the existence and uniqueness of a solution to the problem (1.4) using the Banach contraction principle.

    Theorem 3.1. Assume that (A1)(A5) hold. If

    Θ[Ω1+Ω2+ϕ1+ϕ2]<1, (3.7)

    where Ω1,Ω2 are defined as (3.5)–(3.6), and

    ϕ1=[1+|ρ11||ρ11|]Γ(T+α+β+1)Γ(T+1)Γ(α+β+1), (3.8)
    ϕ2=Γ(T+2)Γ(T+α+β+ν)Γ(2ν)Γ(α+β+ν+1)[Γ(T+ν+1)]2, (3.9)

    then, the problem (1.4) has a unique solution in Nα+β2,T+α+β.

    Proof. Choose a constant R satisfying

    RK2(Ω1+Ω2+ϕ1+ϕ2)1Θ(Ω1+Ω2+ϕ1+ϕ2).

    We will show that F(BR)BR, where BR={xC:xCR}. For all xBR, we have

    |(Fx)(t)|T+α+βs=α+β1sβ+1r=αrαξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)|(|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,0]|+|˜H[ξ+α+β1,0]|)+|1|ρ11|Γ(β)Γ(α)T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_×(|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,0]|+|˜H[ξ+α+β1,0]|)+1Γ(β)Γ(α)tβs=αsαξ=0(tσ(s))β1_(sσ(ξ))α1_(|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,0]|+|˜H[ξ+α+β1,0]|)|(ΘxC+K2)Ω1+(ΘxC+K2)(1+|ρ11|Γ(β)Γ(α)|ρ11|)×T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_(ΘxC+K2){Ω1+(1+|ρ11||ρ11|)Γ(T+α+β+1)Γ(T+1)Γ(α+β+1)}(ΘR+K2)[Ω1+ϕ1],

    and

    |(ΔνCFx)(tν+1)|T+α+βs=α+β1sβ+1r=αrαξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)|(|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,0]|+|˜H[ξ+α+β1,0]|)+1Γ(1ν)Γ(β)Γ(α)t+ν1s=α+β2(tσ(s))ν_Δs[sβr=αrαξ=0(sσ(r))β1_×(rσ(ξ))α1_(|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,0]|+|˜H[ξ+α+β1,0]|)](ΘxC+K2)Ω2+(ΘxC+K2)Γ(1ν)Γ(β)Γ(α)×T+α+β+ν1s=α+β1sβ+1r=αrαξ=0(T+α+βσ(s))ν_(sσ(r))β2_(rσ(ξ))α1_(ΘxC+K2){Ω2+Γ(T+2)Γ(T+α+β+ν)Γ(2ν)Γ(α+β+ν+1)[Γ(T+ν+1)]2}(ΘR+K2)[Ω2+ϕ2].

    Thus,

    FxC(ΘR+K2)[Ω1+Ω2+ϕ1+ϕ2]R,

    and hence, F(BR)BR.

    We next show that F is a contraction. For all x,yC and for each tNα+β2,T+α+β, we have

    |(Fx)(t)(Fy)(t)|T+α+βs=α+β1sβ+1r=αrαξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,y(ξ+α+β1)]|+1|ρ11|Γ(β)Γ(α)T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_×|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,y(ξ+α+β1)]|+1Γ(β)Γ(α)tβs=αsαξ=0(tσ(s))β1_(sσ(ξ))α1_|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,y(ξ+α+β1)]|T+α+βs=α+β1sβ+1r=αrαξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,y(ξ+α+β1)]|+(1+|ρ11||ρ11|Γ(β)Γ(α))T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_×|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,y(ξ+α+β1)]|ΘxyCΩ1+ΘxyC(1+|ρ11||ρ11|Γ(β)Γ(α))×T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_Θ[Ω1+ϕ1]xyC,

    and

    |(ΔνCFx)(tν+1)|T+α+βs=α+β1sβ+1r=αrαξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,y(ξ+α+β1)]|+1Γ(1ν)Γ(β)Γ(α)t+ν1s=α+β1sβ+1r=αrαξ=0(tσ(s))ν_(sσ(r))β2_(rσ(ξ))α1_×|˜H[ξ+α+β1,x(ξ+α+β1)]˜H[ξ+α+β1,y(ξ+α+β1)]|ΘxyCΩ2+ΘxyC1Γ(1ν)Γ(β)Γ(α)×T+α+β+ν1s=α+β1sβ+1r=αrαξ=0(T+α+βσ(s))ν_(sσ(r))β2_(rσ(ξ))α1_Θ[Ω2+ϕ2]xyC.

    Thus,

    FxFyCΘ[Ω1+Ω2+ϕ1+ϕ2]xyCxyC.

    Therefore, F is a contraction. Hence, by using Banach fixed point theorem, we get that F has a fixed point which is a unique solution of the problem (1.4).

    We next deduce the existence of a solution to (1.4) by using the following Schaefer's fixed point theorem.

    Theorem 3.2. [45] (Arzelá-Ascoli Theorem) A set of functions in C[a,b] with the sup norm is relatively compact if and only if it is uniformly bounded and equicontinuous on [a,b].

    Theorem 3.3. [45] If a set is closed and relatively compact, then it is compact.

    Theorem 3.4. [46] Let X be a Banach space and T:XX be a continuous and compact mapping. If the set

    {xX:x=λT(x),forsomeλ(0,1)}

    is bounded, then T has a fixed point.

    Theorem 3.5. Suppose that (A1)(A5) hold. Then, the problem (1.4) has at least one solution on Nα+β2,T+α+β.

    Proof. We shall use Schaefer's fixed point theorem to prove that the operator F defined by (3.3) has a fixed point. It is clear that F:CC is completely continuous. So, it remains to show that the set

    E={uC(Nα+β2,T+α+β):u=λFuforsome0<λ<1}isbounded.

    Let uE. Then,

    u(t)=λ(Fu)(t)forsome0<λ<1.

    Thus, for each tNα+β2,T+α+β, we have

    |λ(Fx)(t)|λT+α+βs=α+β1sβ+1r=αrαξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β1,x(ξ+α+β1)]|+λ|ρ11|Γ(β)Γ(α)T+αs=αsαξ=0(T+α+βσ(s))β1_(sσ(ξ))α1_×|˜H[ξ+α+β1,x(ξ+α+β1)]|+λΓ(β)Γ(α)tβs=αsαξ=0(tσ(s))β1_(sσ(ξ))α1_|˜H[ξ+α+β1,x(ξ+α+β1)]|<(ΘR+K2)[Ω1+ϕ1],

    and

    |λ(ΔνCFx)(tν+1)|λT+α+βs=α+β1sβ+1r=αrαξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β1,x(ξ+α+β1)]|+λΓ(1ν)Γ(β)Γ(α)t+ν1s=α+β2(tσ(s))ν_Δs[sβr=αrαξ=0(sσ(r))β1_×(rσ(ξ))α1_|˜H[ξ+α+β1,x(ξ+α+β1)]|]<(ΘR+K2)[Ω2+ϕ2].

    Hence,

    λ(Fx)(t)<˜ΘR[Ω1+Ω2+ϕ1+ϕ2]<R.

    This shows that E is bounded. By Schaefer's fixed point theorem, we conclude that the problem (1.4) has at least one solution.

    In the sequel, we discuss the positivity of the obtained solution xC. To this end, we add adequate assumptions and provide the following theorem.

    We note that a positive solution of (1.4) in C is a function x(t)>0 which has ΔνCx(tν+1)>0 for all tNα+β2,T+α+β.

    Theorem 3.6. Suppose that (A1)(A5) are fulfilled in R+, where HC(Nα+β2,T+α+β×R+×R+×R+,R+) and φC(×R+×R+,R+). If condition (3.7) is satisfied, for α,β,γ,ν,μ(0,1), and in addition

    ρ1>1andρ2>Γ(T+γ+4)Γ(2γ)Γ(T+3),

    then a solution in C of the problem (1.4) is positive.

    Proof. By Theorem 3.1 and the fact that, for ρ1>1andρ2>Γ(T+γ+4)Γ(2γ)Γ(T+3), the condition (3.7) is a particular case, the problem (1.4) admits a unique solution in C.

    Moreover, since α,β,γ,ν,μ(0,1), we obtain for each (t,s,r,ξ)D,

    Aα,β,γ,ρ1,ρ2(t,s,r,ξ)=[T+(ρ11)(tαβ)+2ρ1(ρ11)(ρ2Γ(T+γ+4)Γ(2γ)Γ(T+3))Γ(1γ)Γ(β)Γ(α)]×(T+α+βγ+1σ(s))γ_(sσ(r))β2_(rσ(ξ))α1_>0,

    and

    tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)=t+ν1s=α+β2[(tσ(s))ν_(sαβ)(ρ2Γ(T+γ+4)Γ(2γ)Γ(T+3))Γ(1ν)Γ(1γ)Γ(β)Γ(α)]×(T+α+βγ+1σ(s))γ_(sσ(r))β2_(rσ(ξ))α1_>0.

    It results that the unique solution x(t) of problem (1.4) which satisfies with (3.3) is positive for each tNα+β2,T+α+β.

    In this section, we present an example to illustrate our results.

    Example. Consider the following fractional difference boundary value problem:

    CΔ2323CΔ5612x(t)=x(t+12)((t+12)+5)5[1+|x(t+12)|]+CΔ1212x(t1)((t+12)+5)5[1+|CΔ1212x(t1)|]+Ψ14(t+12,x(t+12)),tN0,4,2x(12)=x(112),20Δ13x(16)=Δ13x(376), (4.1)
    whereΨ14(t+12,x(t+12))=t14s=34(tσ(s))34Γ(14)[e(s+14)[x(s+14)+1]((t+12)+5)2[3+|x(s+14)|]+e(s+14)[CΔ1212x(s+34)+1]((t+12)+5)2[3+|CΔ1212x(s+34)|]].

    By letting α=23,β=56,γ=13,ν=12,μ=14,T=4,ρ1=2,ρ2=20, H[t,x,y,z]=1(t+5)5[x1+|x|+y1+|y|+z]andφ[t+12,s+14,x,y]=es(t+5)2[x+13+|x|+y+13+|y|], we can show that

    Λ16.0098,Θ0.000199,Ω135.0489,Ω219.7664,Φ118.0469andΦ22.9653.

    Observe that (A1)(A5) hold for all xi,yi,ziR,i=1,2, and for each tN12,112, we obtain

    |H[t,x1,y1,z1]H[t,x2,y2,z2]|1(t+5)5[|x1x2|+|y1y2|+|z1z2|].

    So, K1=(211)50.000199, and K2=maxtN12,112˜H[t,0]0.0000394.

    Next, for all xi,yiR,i=1,2, and each (t,s)N12,112×N12,112, we obtain

    |φ[t,s+34,x1,y1]H[t,s+34,x2,y2]|es(t+5)5[|x1x2|+|y1y2|].

    So, K2=e12(211)50.000121.

    Finally, we can show that

    Θ[Ω1+Ω2+Φ1+Φ2]0.0151<1.

    Hence, by Theorem 3.1, the problem (4.1) has a unique solution.

    Moreover, by Theorem 3.5, the problem (4.1) has at least one solution on N12,112.

    Furthermore, H,φR+, and

    ρ1=2>1,ρ2=20>Γ(253)Γ(53)Γ(7)15.293.

    Therefore, the solution of the problem (4.1) is positive on N12,112 by Theorem 3.6.

    In the present research, we considered a sequential nonlinear Caputo fractional sum-difference equation with fractional difference boundary conditions. Notice that the unknown function of this problem is in the form of Caputo fractional difference and fractional sum with different orders, which expands the research scope of the problems in [42,43,44]. Existence results are established by a Banach contraction principle and Schaefer's fixed point theorem. {The results of the paper are new and enrich the subject of boundary value problems for Caputo fractional difference-sum equations. In future work, we may extend this work by considering new boundary value problems.

    This research was funded by the National Science, Research and Innovation Fund (NSRF) and Suan Dusit University with Contract no. 64-FF-06.

    The authors declare no conflict of interest.



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