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 p−Laplacian Caputo fractional difference equations with fractional sum boundary conditions of the forms
{ΔαC[ϕp(ΔβCx)](t)=f(t+α+β−1,x(t+α+β−1)),t∈N0,T:={0,1,…,T},ΔβCx(α−1)=0,x(α+β+T)=ρΔ−γx(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+α−1,ut+α−1,ΔβCu(t+α−β)],t∈N0,T,ΔγCu(α−γ−1)=0,u(T+α)=ρΔ−ωu(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+α−1,u(t+α−1),Ψγu(t+α−1)],t∈N0,T,t+α−1≠tk,Δu(tk)=Ik(u(tk−1)),k=1,2,...,p,Δ(Δ−βu(tk+β))=Jk(Δ−βu(tk+β−1)),k=1,2,...,p,Au(α−1)+BΔ−βu(α+β−1)=Cu(T+α)+DΔ−βu(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+β−1x(t)=H[t+α+β−1,x(t+α+β−1),CΔνα+β−1x(t+α+β−ν),Ψμ(t+α+β−1,x(t+α+β−1))],t∈N0,T,ρ1x(α+β−2)=x(T+α+β),ρ2CΔγα+β−2x(α+β−γ−1)=CΔγα+β−2x(T+α+β−γ+1), | (1.4) |
where ρ1,ρ2∈R, 0<α,β,γ,ν,μ≤1, 1<α+β≤2 are given constants, H∈C(Nα+β−2,T+α+β×R3,R), and for φ∈C(⊙×R2,[0,∞)), ⊙:={(t,r):t,r∈Nα+β−2,T+α+βandr≤t}. The operator Ψμ is defined by
Ψμ(t,x(t)):=t−μ∑s=α+β−μ−2(t−σ(s))μ−1_Γ(μ)φ(t,s+μ,x(s+μ),CΔνα+β−2x(s+μ−ν+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+1−α), for any t and α for which the right-hand side is defined. If t+1−α 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 t≤r, 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
Δ−αaf(t)=Δ−αf(t):=1Γ(α)t−α∑s=a(t−σ(s))α−1_f(s), |
where t∈Na+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−(N−α)aΔNf(t)=1Γ(N−α)t−(N−α)∑s=a(t−σ(s))N−α−1_ΔNf(s), |
where t∈Na+N−α and N∈N is chosen so that 0≤N−1<α<N. If α=N, then ΔαCf(t)=ΔNf(t).
Lemma 2.2. [11] Assume that α>0 and 0≤N−1<α≤N. Then,
Δ−αa+N−αCΔαay(t)=y(t)+C0+C1t1_+C2t2_+...+CN−1tN−1_, |
for some Ci∈R, 0≤i≤N−1.
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(ρ1−1)≠0, 0<α,β,γ,ν,μ≤1, 1<α+β≤2 and h∈C(Nα+β−1,T+α+β−1,R) be given. Then, the problem
CΔααCΔβα+β−1x(t)=h(t+α+β−1),t∈N0,T | (2.1) |
{ρ1x(α+β−2)=x(T+α+β),ρ2CΔγα+β−2x(α+β−γ−1)=CΔγα+β−2x(T+α+β−γ+1), | (2.2) |
has the unique solution
x(t)=T+(ρ1−1)(t−α−β)+2ρ1Λ(ρ1−1)Γ(1−γ)Γ(β−1)Γ(α)T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(T+α+β−γ+1−σ(s))−γ_×(s−σ(r))β−2_(r−σ(ξ))α−1_h(ξ+α+β−1)+1(ρ1−1)Γ(β)Γ(α)T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_h(ξ+α+β−1)+1Γ(β)Γ(α)t−β∑s=αs−α∑ξ=0(t−σ(s))β−1_(s−σ(ξ))α−1_h(ξ+α−1), | (2.3) |
where
Λ=ρ2−Γ(T−γ+4)Γ(2−γ)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+β−2x(t)=C1+1Γ(α)t−α∑s=0(t−σ(s))α−1_h(s+α+β−1), | (2.5) |
for t∈Nα−1,T+α.
Using the fractional sum of order 0<β≤1 for (2.5), we obtain
x(t)=C2+C1t+1Γ(β)Γ(α)t−β∑s=αs−α∑ξ=0(t−σ(s))β−1_(s−σ(ξ))α−1_h(ξ+α+β−1), | (2.6) |
for t∈Nα+β−2,T+α+β.
By substituting t=α+β−2,T+α+β into (2.6) and employing the first condition of (2.2), we obtain
−C2(ρ1−1)+C1[(T−(ρ1−1)(α+β)+2ρ1]=−1Γ(β)Γ(α)T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_h(ξ+α+β−1). | (2.7) |
Using the fractional Caputo difference of order 0<γ≤1 for (2.6), we obtain
CΔγα+β−2x(t)=C1Γ(1−γ)t+γ−1∑s=α+β−2(t−σ(s))−γ_+1Γ(1−γ)Γ(β)Γ(α)×t+γ−1∑s=α+β−2(t−σ(s))−γ_sΔ[s−β∑r=αr−α∑ξ=0(s−σ(r))β−1_(r−σ(ξ))α−1_h(ξ+α+β−1)]=C1Γ(1−γ)t+γ−1∑s=α+β−2(t−σ(s))−γ_+1Γ(1−γ)Γ(β−1)Γ(α)×t+γ−1∑s=α+β−2s−β+1∑r=αr−α∑ξ=0(t−σ(s))−γ_(s−σ(r))β−2_(r−σ(ξ))α−1_h(ξ+α+β−1), | (2.8) |
for Nα+β−γ−1,T+α+β−γ+1.
By substituting t=α+β−γ−1,T+α+β−γ+1 into (2.8) and employing the second condition of (2.2), it implies
C1=−1ΛΓ(1−γ)Γ(β−1)Γ(α)T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(T+α+β−γ+1−σ(s))−γ_×(s−σ(r))β−2_(r−σ(ξ))α−1_h(ξ+α+β−1). |
The constant C2 can be obtained by substituting C1 into (2.7). Then, we get
C2=T−(ρ1−1)(α+β)+2ρ1(ρ1−1)ΛΓ(1−γ)Γ(β−1)Γ(α)T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(T+α+β−γ+1−σ(s))−γ_×(s−σ(r))β−2_(r−σ(ξ))α−1_h(ξ+α+β−1)+1(ρ1−1)Γ(β)Γ(α)T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_h(ξ+α+β−1), |
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+β−2,T+α+β,R) as the Banach space of all functions x with the norm defined by
‖x‖C=‖x‖+‖ΔνCx‖, |
where ‖x‖=maxt∈Nα+β−2,T+α+β|x(t)| and ‖ΔνCx‖=maxt∈Nα+β−2,T+α+β|ΔνCx(t−ν+1)|.
The following assumptions are assumed:
(A1) H[t,x,y,z]:Nα+β−2,T+α+β×R3→R is a continuous function.
(A2) There exist constants K1,K2>0 such that for each t∈Nα+β−2,T+α+β and all xi,yi,zi∈R,i=1,2, we have
|H[t,x1,y1,z1]−H[t,x2,y2,z2]|≤K1[|x1−x2|+|y1−y2|+|z1−z2|], |
and
K2=maxt∈Nα+β−2,T+α+β|H[t,0,0,Ψμ(t,0)]|, |
where Ψμ(t,0):=1Γ(μ)t−μ∑s=α+β−μ−2(t−σ(s))μ−1_φ(t,s+μ,0,0).
(A3) φ:⊙×R2→R is continuious for (t,s)∈⊙, and there exists a constant L>0, such that for each (t,s)∈⊙ and all xi,yi∈C,i=1,2 we have
|φ(t,s+μ,x1,y1)−φ(t,s+μ,x2,y2)|≤L[|x1−x2|+|y1−y2|]. |
Let us define the operator ˜H[t,x(t)] by
˜H[t,x(t)]:=H[t,x(t),ΔνCx(t−ν+1),Ψμ(t,x(t))], | (3.1) |
for each t∈Nα+β−2,T+α+β and x∈C.
Note that Δ−βΔ−α˜H[t,x(t)] and ΔνCΔ−βΔ−α˜H[t,x(t)] exist when ν<α+β≤2.
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)]|≤Θ‖x1−x2‖C, |
for each t∈Nα+β−2,T+α+β and x1,x2∈C, where
Θ:=K1[1+LΓ(T+μ+3)Γ(μ+1)Γ(T+3)]. | (3.2) |
Proof. By (A3), for each t∈Nα+β−2,T+α+β and x1,x2∈C, we obtain
|(Ψμx1)(t)−(Ψμx2)(t)|≤1Γ(μ)t−μ∑s=α+β−μ−2(t−σ(s))μ−1_|φ(t,s+μ,x1(s+μ),ΔνCx1(s+μ−ν+1))−φ(t,s+μ,x2(s+μ),ΔνCx2(s+μ−ν+1))|≤1Γ(μ)T+α+β−μ∑s=α+β−μ−2(T+α+β−σ(s))μ−1_×L[|x1(s+μ)−x2(s+μ)|+|ΔνCx1(s+μ−ν+1)−ΔνCx2(s+μ−ν+1)|]≤LΓ(T+μ+3)Γ(μ+1)Γ(T+3){‖x1−x2‖+‖ΔνCx1−ΔνCx2‖}, |
and hence
|˜H[t,x1(t)]−˜H[t,x2(t)]|≤K1[|x1(t)−x2(t)|+|ΔνCx1(t−ν+1)−ΔνCx2(t−ν+1)|]+K1LΓ(T+μ+3)Γ(μ+1)Γ(T+3){‖x1−x2‖+‖ΔνCx1−ΔνCx2‖}=Θ‖x1−x2‖C. |
Next, we define the operator F:C⟶C by
(Fx)(t)=T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0Aα,β,γ,ρ1,ρ2(t,s,r,ξ)˜H[ξ+α+β−1,x(ξ+α+β−1)]+T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_(ρ1−1)Γ(β)Γ(α)˜H[ξ+α+β−1,x(ξ+α+β−1)]+t−β∑s=αs−α∑ξ=0(t−σ(s))β−1_(s−σ(ξ))α−1_Γ(β)Γ(α)˜H[ξ+α+β−1,x(ξ+α+β−1)], | (3.3) |
where
Aα,β,γ,ρ1,ρ2(t,s,r,ξ):=T+(ρ1−1)(t−α−β)+2ρ1Λ(ρ1−1)Γ(1−γ)Γ(β−1)Γ(α)×(T+α+β−γ+1−σ(s))−γ_(s−σ(r))β−2_(r−σ(ξ))α−1_. | (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+β−2,T+α+β×Nα+β−2,T+α+β×Nα−1,T+α+1×N0,T+2=:D, and there exist two constants Ω1,Ω2>0, such that
T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)|≤Ω1,T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)|≤Ω2, |
where
Ω1:=[(1+|ρ1−1|)T+2p1|ρ1−1||Λ|]Γ(T+γ+3)Γ(T+α+β+1)Γ(2−γ)Γ(α+β)[Γ(T+2)]2, | (3.5) |
Ω2:=|1−α−βΛ|Γ(T−γ+3)[Γ(T+α+β+1)]2Γ(2−ν)Γ(2−γ)Γ(α+β)[Γ(T+2)]2Γ(T+α+β−ν), | (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+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)|≤maxt∈Nα+β−2,T+α+βT+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)|=maxt∈Nα+β−2,T+α+β|T+(ρ1−1)(t−α−β)+2ρ1Λ(ρ1−1)Γ(1−γ)Γ(β−1)Γ(α)|×T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(T+α+β−γ+1−σ(s))−γ_(s−σ(r))β−2_(r−σ(ξ))α−1_≤[(1+|ρ1−1|)T+2p1|ρ1−1||Λ|]T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(T+α+β−γ+1−σ(s))−γ_(s−σ(r))β−2_(r−σ(ξ))α−1_≤[(1+|ρ1−1|)T+2p1|ρ1−1||Λ|]Γ(T+γ+3)Γ(T+α+β+1)Γ(T+2)Γ(T+2)Γ(2−γ)Γ(α+β)=Ω1, |
and
T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)|≤maxt∈Nα+β−2,T+α+βT+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)|=maxt∈Nα+β−2,T+α+β|t+ν−1∑s=α+β−2(t−σ(s))−ν_sΔ[T+(ρ1−1)(s−α−β)+2ρ1]ΛΓ(1−ν)Γ(1−γ)Γ(β−1)Γ(α)|×T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(T+α+β−γ+1−σ(s))−γ_(s−σ(r))β−2_(r−σ(ξ))α−1_≤|T+α+β+ν−1∑s=α+β−2(T+α+β−σ(s))−ν_(1−α−β)ΛΓ(1−ν)|Γ(T−γ+3)Γ(T+α+β+1)[Γ(T+2)]2Γ(2−γ)Γ(α+β)≤|1−α−βΛ|Γ(T−γ+3)[Γ(T+α+β+1)]2Γ(2−ν)Γ(2−γ)Γ(α+β)[Γ(T+2)]2Γ(T+α+β−ν)=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+|ρ1−1||ρ1−1|]Γ(T+α+β+1)Γ(T+1)Γ(α+β+1), | (3.8) |
ϕ2=Γ(T+2)Γ(T+α+β+ν)Γ(2−ν)Γ(α+β+ν+1)[Γ(T+ν+1)]2, | (3.9) |
then, the problem (1.4) has a unique solution in Nα+β−2,T+α+β.
Proof. Choose a constant R satisfying
R≥K2(Ω1+Ω2+ϕ1+ϕ2)1−Θ(Ω1+Ω2+ϕ1+ϕ2). |
We will show that F(BR)⊂BR, where BR={x∈C:‖x‖C≤R}. For all x∈BR, we have
|(Fx)(t)|≤T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)|(|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,0]|+|˜H[ξ+α+β−1,0]|)+|1|ρ1−1|Γ(β)Γ(α)T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_×(|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,0]|+|˜H[ξ+α+β−1,0]|)+1Γ(β)Γ(α)t−β∑s=αs−α∑ξ=0(t−σ(s))β−1_(s−σ(ξ))α−1_(|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,0]|+|˜H[ξ+α+β−1,0]|)|≤(Θ‖x‖C+K2)Ω1+(Θ‖x‖C+K2)(1+|ρ1−1|Γ(β)Γ(α)|ρ1−1|)×T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_≤(Θ‖x‖C+K2){Ω1+(1+|ρ1−1||ρ1−1|)Γ(T+α+β+1)Γ(T+1)Γ(α+β+1)}≤(ΘR+K2)[Ω1+ϕ1], |
and
|(ΔνCFx)(t−ν+1)|≤T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)|(|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,0]|+|˜H[ξ+α+β−1,0]|)+1Γ(1−ν)Γ(β)Γ(α)t+ν−1∑s=α+β−2(t−σ(s))−ν_Δs[s−β∑r=αr−α∑ξ=0(s−σ(r))β−1_×(r−σ(ξ))α−1_(|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,0]|+|˜H[ξ+α+β−1,0]|)]≤(Θ‖x‖C+K2)Ω2+(Θ‖x‖C+K2)Γ(1−ν)Γ(β)Γ(α)×T+α+β+ν−1∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(T+α+β−σ(s))−ν_(s−σ(r))β−2_(r−σ(ξ))α−1_≤(Θ‖x‖C+K2){Ω2+Γ(T+2)Γ(T+α+β+ν)Γ(2−ν)Γ(α+β+ν+1)[Γ(T+ν+1)]2}≤(ΘR+K2)[Ω2+ϕ2]. |
Thus,
‖Fx‖C≤(ΘR+K2)[Ω1+Ω2+ϕ1+ϕ2]≤R, |
and hence, F(BR)⊂BR.
We next show that F is a contraction. For all x,y∈C and for each t∈Nα+β−2,T+α+β, we have
|(Fx)(t)−(Fy)(t)|≤T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,y(ξ+α+β−1)]|+1|ρ1−1|Γ(β)Γ(α)T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_×|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,y(ξ+α+β−1)]|+1Γ(β)Γ(α)t−β∑s=αs−α∑ξ=0(t−σ(s))β−1_(s−σ(ξ))α−1_|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,y(ξ+α+β−1)]|≤T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,y(ξ+α+β−1)]|+(1+|ρ1−1||ρ1−1|Γ(β)Γ(α))T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_×|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,y(ξ+α+β−1)]|≤Θ‖x−y‖CΩ1+Θ‖x−y‖C(1+|ρ1−1||ρ1−1|Γ(β)Γ(α))×T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_≤Θ[Ω1+ϕ1]‖x−y‖C, |
and
|(ΔνCFx)(t−ν+1)|≤T+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,y(ξ+α+β−1)]|+1Γ(1−ν)Γ(β)Γ(α)t+ν−1∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(t−σ(s))−ν_(s−σ(r))β−2_(r−σ(ξ))α−1_×|˜H[ξ+α+β−1,x(ξ+α+β−1)]−˜H[ξ+α+β−1,y(ξ+α+β−1)]|≤Θ‖x−y‖CΩ2+Θ‖x−y‖C1Γ(1−ν)Γ(β)Γ(α)×T+α+β+ν−1∑s=α+β−1s−β+1∑r=αr−α∑ξ=0(T+α+β−σ(s))−ν_(s−σ(r))β−2_(r−σ(ξ))α−1_≤Θ[Ω2+ϕ2]‖x−y‖C. |
Thus,
‖Fx−Fy‖C≤Θ[Ω1+Ω2+ϕ1+ϕ2]‖x−y‖C≤‖x−y‖C. |
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:X→X be a continuous and compact mapping. If the set
{x∈X: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+β−2,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:C⟶C is completely continuous. So, it remains to show that the set
E={u∈C(Nα+β−2,T+α+β):u=λFuforsome0<λ<1}isbounded. |
Let u∈E. Then,
u(t)=λ(Fu)(t)forsome0<λ<1. |
Thus, for each t∈Nα+β−2,T+α+β, we have
|λ(Fx)(t)|≤λT+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|Aα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β−1,x(ξ+α+β−1)]|+λ|ρ1−1|Γ(β)Γ(α)T+α∑s=αs−α∑ξ=0(T+α+β−σ(s))β−1_(s−σ(ξ))α−1_×|˜H[ξ+α+β−1,x(ξ+α+β−1)]|+λΓ(β)Γ(α)t−β∑s=αs−α∑ξ=0(t−σ(s))β−1_(s−σ(ξ))α−1_|˜H[ξ+α+β−1,x(ξ+α+β−1)]|<(ΘR+K2)[Ω1+ϕ1], |
and
|λ(ΔνCFx)(t−ν+1)|≤λT+α+β∑s=α+β−1s−β+1∑r=αr−α∑ξ=0|tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)||˜H[ξ+α+β−1,x(ξ+α+β−1)]|+λΓ(1−ν)Γ(β)Γ(α)t+ν−1∑s=α+β−2(t−σ(s))−ν_Δs[s−β∑r=αr−α∑ξ=0(s−σ(r))β−1_×(r−σ(ξ))α−1_|˜H[ξ+α+β−1,x(ξ+α+β−1)]|]<(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 x∈C. 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(t−ν+1)>0 for all t∈Nα+β−2,T+α+β.
Theorem 3.6. Suppose that (A1)–(A5) are fulfilled in R+, where H∈C(Nα+β−2,T+α+β×R+×R+×R+,R+) and φ∈C(⊙×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(2−γ)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(2−γ)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+(ρ1−1)(t−α−β)+2ρ1(ρ1−1)(ρ2−Γ(T+γ+4)Γ(2−γ)Γ(T+3))Γ(1−γ)Γ(β)Γ(α)]×(T+α+β−γ+1−σ(s))−γ_(s−σ(r))β−2_(r−σ(ξ))α−1_>0, |
and
tΔνCAα,β,γ,ρ1,ρ2(t,s,r,ξ)=t+ν−1∑s=α+β−2[(t−σ(s))−ν_(s−α−β)(ρ2−Γ(T+γ+4)Γ(2−γ)Γ(T+3))Γ(1−ν)Γ(1−γ)Γ(β)Γ(α)]×(T+α+β−γ+1−σ(s))−γ_(s−σ(r))β−2_(r−σ(ξ))α−1_>0. |
It results that the unique solution x(t) of problem (1.4) which satisfies with (3.3) is positive for each t∈Nα+β−2,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(t−1)((t+12)+5)5[1+|CΔ1212x(t−1)|]+Ψ14(t+12,x(t+12)),t∈N0,4,2x(−12)=x(112),20Δ13x(16)=Δ13x(376), | (4.1) |
whereΨ14(t+12,x(t+12))=t−14∑s=−34(t−σ(s))−34Γ(14)[e−(s+14)[x(s+14)+1]((t+12)+5)2[3+|x(s+14)|]+e−(s+14)[CΔ12−12x(s+34)+1]((t+12)+5)2[3+|CΔ12−12x(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]=e−s(t+5)2[x+13+|x|+y+13+|y|], we can show that
Λ≈16.0098,Θ≈0.000199,Ω1≈35.0489,Ω2≈19.7664,Φ1≈18.0469andΦ2≈2.9653. |
Observe that (A1)–(A5) hold for all xi,yi,zi∈R,i=1,2, and for each t∈N−12,112, we obtain
|H[t,x1,y1,z1]−H[t,x2,y2,z2]|≤1(t+5)5[|x1−x2|+|y1−y2|+|z1−z2|]. |
So, K1=(211)5≈0.000199, and K2=maxt∈N−12,112˜H[t,0]≈0.0000394.
Next, for all xi,yi∈R,i=1,2, and each (t,s)∈N−12,112×N−12,112, we obtain
|φ[t,s+34,x1,y1]−H[t,s+34,x2,y2]|≤e−s(t+5)5[|x1−x2|+|y1−y2|]. |
So, K2=e−12(211)5≈0.000121.
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 N−12,112.
Furthermore, H,φ∈R+, 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 N−12,112 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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