# Existence Results of Initial Value Problems for Hybrid Fractional Sum-Difference Equations

Authors: Saowaluck Chasreechai; Thanin Sitthiwirattham
Journal: Discrete Dynamics in Nature and Society
 (2018)

---

## Abstract

We consider a hybrid fractional sum-difference initial value problem and a hybrid fractional sequential sum-difference initial value problem. The existence results of these two problems are proved by using the hybrid fixed point theorem for three operators in a Banach algebra and the generalized Krasnoselskii’s fixed point theorem, respectively.


---

## Body

## 1. Introduction

As recognized that fractional difference calculus is a powerful tool used to describe many real world phenomena problems such as physics, chemistry, mechanics, control systems, flow in porous media, and electrical networks [1, 2], this is an impact on a researcher’s motivation to develop the research works in this area. Basic definitions and properties of fractional difference calculus were proposed by Goodrich and Peterson [3]. The developments of the theory related to discrete fractional boundary value problems were studied by many authors (see [4–48]). In particular Sitthiwirattham [41, 42] studied three-point Caputo fractional difference-fractional sum boundary value problem for sequential Caputo fractional difference equation of the form(1)ΔCαϕpΔCβxt=ft+α+β-1,xt+α+β-1,ΔCβxα-1=0,xα+β+T=ρΔ-γxη+γ,and three-point fractional sum boundary value problem for sequential Riemann-Liouville fractional difference equation of the form(2)ΔααΔα+β-1β+λEβxt=ft+α+β-1,xt+α+β-1,xα+β-2=0,xα+β+T=ρΔα+β-1-γxη+γ,where t∈N0,T, 0<α,β≤1, 1<α+β≤2, 0<γ≤1, η∈Nα+β-1,α+β+T-1, ρ is a constant, f:Nα+β-2,α+β+T×R→R is a continuous function, Eβx(t)=x(t+β-1), and ϕp is the p-Laplacian operator.Calculus which deals with derivatives and integrals of arbitrary orders is known as hybrid differential equations (i.e., quadratic perturbation of a nonlinear differential equation). Hybrid fractional differential equations are initialized to be used to model successfully several physical phenomena (see [49–51]). Apparently, this issue has found numerous miscellaneous applications connected with real world problems as they appear in many fields of engineering and science, including biology, chemistry, diffusion, control theory, electromagnetic theory, fluid flow, signal and image processing, fractals theory, fitting of experimental data, potential theory, and viscoelasticity. For some recent developments on the topic, see [52–57]. Recently, there are several research works related to boundary value problems for hybrid differential equations (see [58–65]). For example, Sun et al. [60] studied the existence of solutions for the boundary value problem of fractional hybrid differential equations(3)Dνxtft,xt+gt,xt=0,0<t<1,x0=x1=0,where Dν denotes the Riemann-Liouville fractional derivative of order ν, 1<ν≤2.Sitho et al. [64] studied existence results for the following initial value problem for hybrid fractional integrodifferential equations,(4)Dαxt-∑i=1mIβihit,xtft,xt=Ft,xt,t∈J:0,T,x0=0,and the initial value problem for hybrid fractional sequential integrodifferential equations,(5)DαDωxt-∑i=1mIβihit,xtft,xt=Gt,xt,Iγxt,t∈J,x0=Dωx0=0,where Dα denotes the Riemann-Liouville fractional derivative of order α, Iϕ is the Riemann-Liouville fractional integral of order ϕ>0, ϕ∈β1,β2,…,βm, 0<α,ω≤1, 1<α+ω≤2, and functions are F∈C(J×R,R), G∈C(J×R2),R, f∈C(J×R,R-0), and hi∈C(J×R,R) with hi(0,0)=0, i=1,2,…,m.While the boundary value problem for hybrid fractional difference equations has not been studied, to fill this gap, we study a hybrid fractional difference initial value problem of the form,(6)Δαut-Δ-γpt+γ,ut+γft,ut=Ft+α,ut+α,uα-1=0,and a hybrid fractional sequential sum-difference initial value problem of the form(7)ΔαΔβut-Δ-γgt+β+γ-1,ut+β+γ-1ht,ut=Ht+α+β-1,ut+α+β-1,Δ-ωut+α+β+ω-1,uα+β-2=Δβuα-1=0,where t∈N0,T≔0,1,…,T, α,β,γ,ω∈(0,1], 1<α+β≤2 are given constants, f∈C(Nα-1,T+α×R,R-{0}), h∈C(Nα+β-2,T+α+β×R,R-{0}), p∈C(Nα-1,T+α×R,R) with p(α-1,0)=0, g∈C(Nα+β-2,T+α+β×R,R) with g(α+β-2,0)=0, F∈C(Nα-1,T+α×R,R), and H∈C(Nα+β-2,T+α+β×R×R,R).The article is organized as follows. In Section2, we recall some definitions and basic lemmas used in this work. Then, we present the solutions of (6) and (7) by converting the problem to an equivalent summation equation. In Sections 3 and 4, we prove existence results of problems (6) and (7) by employing the hybrid fixed point theorem for three operators in a Banach algebra and the generalized Krasnoselskii’s fixed point theorem, respectively. We end with some examples to illustrate our results in the last section.


## 2. Preliminaries

In what follows are the notations, definitions, and lemmas which are used in the main results.Definition 1 (see [6]).
Define the generalized falling function bytα_≔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 (see [4]).
Assume that the factorial functions are well defined. Ift≤r, then tα_≤rα_ for any α>0.Definition 3 (see [6]).
Forα>0 and f defined on Na≔a,a+1,…, the α-order fractional sum of f is defined by(8)Δ-αft≔1Γα∑s=at-αt-σsα-1_fs,where t∈Na+α and σ(s)=s+1.Definition 4 (see [6]).
Forα>0 and f defined on Na, the α-order Riemann-Liouville fractional difference of f is defined by(9)Δαft≔ΔNΔ-N-αft=1Γ-α∑s=at+αt-σs-α-1_fs,where t∈Na+N-α and N∈N is satisfied with 0≤N-1<α<N.Lemma 5 (see [5]).
For any real numberν and any positive integer p, the following equality holds:(10)Δ-νΔpft=ΔpΔ-νft-∑k=0p-1t-aν-p+kΓν-p+k+1Δkfa,or f is defined on Na.Lemma 6 (see [4]).
Let0≤N-1<α≤N. Then,(11)Δ-αΔαyt=yt+C1tα-1_+C2tα-2_+⋯+CNtα-N_,for some Ci∈R, with 1≤i≤N.We provide the following lemma dealing with linear variant of the boundary value problems (6) and (7) and give a representation of the solution.Lemma 7.
Letα,γ∈(0,1], p∈C(Nα-1,T+α×R,R) with p(α-1,0)=0, y∈C(Nα-1,T+α,R-{0}), and h∈C(Nα-1,T+α,R). Then, for t∈N0,T, the problem(12)Δαut-Δ-γpt+γ,ut+γyt=ht+α,uα-1=0,has the unique solution(13)ut=ytΓα∑s=0t-αt-σsα-1_hs+α+1Γγ∑s=α-γ-1t-γt-σsγ-1_ps+γ,us+γ,for t∈Nα-1,T+α.Proof.
Using the fractional sum of orderα:Δ-α for (12) and Lemmas 5 and 6, we obtain(14)Δ-αht+α=ut-Δ-γpt+γ,ut+γyt-tα-1_ΓαΔ-1-αuα-1-Δ-γpα-1+γ,uα-1+γyα-1=ut-Δ-γpt+γ,ut+γyt-tα-1_ΓαΔ-1-α1yα-1×uα-1-1Γγ∑s=α-1α+γ-1-γα+γ-1-σsγ-1_ps,us=ut-Δ-γpt+γ,ut+γyt-tα-1_ΓαΔ-1-α1yα-1uα-1-γpα-1,uα-1,for t∈Nα-1,T+α.
Sinceu(α-1)=0, p(α-1,0)=0, and y(α-1)≠0, it follows that(15)ut-Δ-γpt+γ,ut+γyt=Δ-αht+α.Thus, (13) holds. The proof is completed.Lemma 8.
Letα,β,γ∈(0,1], 1<α+β≤2, g∈C(Nα+β-2,T+α+β×R,R) with g(α+β-2,0)=0, x∈C(Nα+β-2,T+α+β,R-{0}), and k∈C(Nα+β-2,T+α+β,R) be given. Then, for t∈N0,T, the problem(16)ΔαΔβut-Δ-γgt+β+γ-1,ut+β+γ-1xt=kt+α+β-1,uα+β-2=Δβuα-1=0,has the unique solution(17)ut=1ΓαΓβ∑s=α-1t-βt-σsβ-1_xs+β∑ξ=0s-αs-σξα-1_kξ+α+β-1+1Γγ∑s=α+β-γ-2t-γt-σsγ-1_gs+β+γ,us+β+γ,for t∈Nα+β-2,T+α+β.Proof.
Using the fractional sum of orderα:Δ-α for (16) and Lemmas 5 and 6, we obtain(18)Δ-αkt+α+β-1=Δβut-Δ-γgt+β+γ-1,ut+β+γ-1xt-tα-1_ΓαΔ-1-αΔβuα-1-Δ-γgα-1+β+γ-1,uα-1+β+γ-1xα-1=Δβut-Δ-γgt+β+γ-1,ut+β+γ-1xt-tα-1_ΓαΔ-1-α1xα-1×Δβuα-1-1Γγ∑s=α+β-2α+β+γ-2-γα+β+γ-2-σsγ-1_gs,us=Δβut-Δ-γgt+β+γ-1,ut+β+γ-1xt-tα-1_ΓαΔ-1-α1xα-1Δβuα-1-γgα+β-2,uα+β-2,for t∈Nα-1,T+α.
Sinceu(α+β-2)=Δβuα-1=0, g(α+β-2,0)=0, and x(α-1)≠0, we have(19)Δβut-Δ-γgt+β+γ-1,ut+β+γ-1xt=Δ-αkt+α+β-1.That can be arranged in the form(20)Δβut=xtΔ-αkt+α+β-1+Δ-γgt+β+γ-1,ut+β+γ-1.Using the fractional sum of order β for (20), we obtain(21)ut=Δ-βxtΔ-αkt+α+β-1+Δ-β-γgt+β+γ-1,ut+β+γ-1,for t∈Nα+β-2,T+α+β.
Thus (17) holds. Our proof is completed.Lemma 9 (Arzelá-Ascoli theorem [66]).
A set of functions inC[a,b] with the sup norm is relatively compact if and only if it is uniformly bounded and equicontinuous on [a,b].Lemma 10 (see [66]).
If a set is closed and relatively compact, then it is compact.


## 3. Hybrid Fractional Sum-Difference Initial Value Problem (6)

In this section, we aim to show the existence results for problem (6). To accomplish this, we let E=C(Nα-1,T+α,R) be a space of all functions u and defined a norm and a multiplication in E by(22)u=maxt∈Nα-1,T+α⁡ut,uvt=utvt.In addition, we define operator F:E→E by(23)Fut=ft,utΓα∑s=0t-αt-σsα-1_Fs+α,us+α+1Γγ∑s=α-γ-1t-γt-σsγ-1_ps+γ,us+γ.Clearly, problem (6) has solutions if and only if operator F has fixed points. The first shows the existence and uniqueness of a solution to problem (6) by using the Banach contraction principle.Theorem 11.
Assume thatf∈C(Nα-1,T+α×R,R-{0}), p∈C(Nα-1,T+α×R,R) with p(α-1,0)=0, and H∈C(Nα+β-2,T+α+β×R×R,R). In addition, suppose that(A1)
there exist constantsL1,L2>0 such that for each t∈Nα-1,T+α and u,v∈R(24)Ft,u-Ft,v≤L1u-v,pt,u-pt,v≤L2u-v,(A2)
there exists a positive functionθ with bound θ such that for each t∈Nα-1,T+α and u,v∈R(25)ft,u-ft,v≤θt.If L1θTα_/Γ(α+1)+L2(T+α)γ_/Γ(γ+1)<1, then problem (6) has a unique solution.Proof.
We shall show thatF is a contraction. For any u,v∈C and for each t∈Nα-1,T+α, we have(26)Fut-Fvt≤fΓα∑s=0t-αt-σsα-1_Fs+α,us+α-Fs+α,vs+α+1Γγ∑s=α-γ-1t-γt-σsγ-1_ps+γ,us+γ-ps+γ,vs+γ≤θΓαL1u-v∑s=0TT+α-σsα-1_+1ΓγL2u-v∑s=α-γ-1T+α-γT+α-σsγ-1_≤u-vL1θTα_Γα+1+L2T+αγ_Γγ+1.Thus, we have (Fu)-(Fv)<u-v.
Consequently,F is a contraction. Therefore, by the Banach fixed point theorem, we get that F has a fixed point which is a unique solution of problem (6).In the second result, we deduce the existence of at least one solution of the initial value problem (6) by using the hybrid fixed point theorem for three operators in a Banach algebra. Clearly, E is a Banach algebra with respect to the above norm and multiplication in it.Theorem 12 (hybrid fixed point theorem for three operators in a Banach algebra [67]).
LetS be a nonempty, closed convex, and bounded subset of the Banach algebra E, and let A,C:E→E and B:S→E be three operators such that(i)
A and C are Lipschitzian with Lipschitz constants δ and ρ, respectively,(ii)
B is completely continuous,(iii)
x=AxBy+Cx⇒x∈S for all y∈S,(iv)
δM+ρ<1, where M=B(S).
Then, the operator equationAxBx+Cx=x has a solution.Theorem 13.
Assume thatf∈C(Nα-1,T+α×R,R-{0}), p∈C(Nα-1,T+α×R,R) with p(α-1,0)=0, and H∈C(Nα+β-2,T+α+β×R×R,R). In addition, suppose that(H1)
there exist two positive functionsϕ and ψ with bound ϕ and ψ, respectively, such that for each t∈Nα-1,T+α and u,v∈R(27)ft,u-ft,v≤ϕtu-v,pt,u-pt,v≤ψtu-v,(H2)
there exists a functionω∈C(Nα-1,T+α,R+) and a continuous nondecreasing function Ψ:0,∞→(0,∞) such that for each (t,u)∈Nα-1,T+α×R(28)Ft,u≤ωtΨu,H3
there exists a number(29)R≥f^0ωΨRTα/Γα+1+p^0T+αγ/Γγ+11-ϕωΨRTα/Γα+1-ψT+αγ/Γγ+1,wheref^0=maxt∈Nα-1,T+α|f(t,0)|, p^0=maxt∈Nα-1,T+α|p(t,0)|, and(30)ϕωΨRTαΓα+1+ψT+αγΓγ+1<1.Then problem (6) has a unique solution on Nα-1,T+α.Proof.
Define subsetS of E as(31)S=u∈E:u≤R.We see that S is closed, convex, and bounded subset of the Banach space C. By Lemma 7, we define three operators A:E→E, B:S→E, and C:E→E by(32)Aut≔ft,ut,t∈Nα-1,T+α,But≔1Γα∑s=0t-αt-σsα-1_Fs+α,us+α,t∈Nα-1,T+α,Cut≔1Γγ∑s=α-γ-1t-γt-σsγ-1_ps+γ,us+γ,t∈Nα-1,T+α.Note that problem (6) has solutions if and only if the operator u=AuBv+Cu has fixed points.
To show that all operators satisfy all the conditions of Theorem12, we proceed with the following steps.
Step  1. Prove that A and C are Lipschitzian on E.
For anyu,v∈R and for each t∈Nα-1,T+α, then by (H1), we have(33)Aut-Avt=ft,ut-ft,vt≤ϕtu-v≤ϕu-v,Cut-Cvt=1Γγ∑s=α-γ-1t-γt-σsγ-1_ps+γ,us+γ-ps+γ,vs+γ≤1Γγ∑s=α-γ-1T+α-γT+α-σsγ-1_ψsu-v≤ψsT+αγΓγ+1u-v.This implies that, for all u,v∈R,(34)Aut-Avt≤ϕu-v,Cut-Cvt≤ψsT+αγΓγ+1u-v.Therefore, Aand C are Lipschitzian on E with Lipschitz constants ϕ and ψ(s)(T+α)γ/Γ(γ+1).
Step  2. We prove that B is completely continuous on S.
SinceF is continuous, the operator B is continuous on S. Next, we will prove that the set B(S) is uniformly bounded in S. For any u∈S, we find that(35)But≤1Γα∑s=0t-αt-σsα-1_ωsΨR≤ωΨRTαΓα+1≕K,for all t∈Nα-1,T+α. Therefore, B≤K, which shows that B is uniformly bounded on S.
Next we show thatB(S) is an equicontinuous set in E. For any ϵ>0, there exists a positive constant δ such that, for t1,t2∈Nα-1,T+α,(36)t2α-t1α<Γα+1ωΨR,whenevert2-t1<δ.Then we obtain(37)Bxt2-Bxt1≤ωΨRΓα∑s=0t2-αt2-σsα-1_-∑s=0t1-αt1-σsα-1_≤ωΨRΓα+1t2α-t1α<ϵ.This implies that the set B(S) is an equicontinuous set. From the Arzelá-Ascoli theorem, we find that B is completely continuous.
Step  3. u=AuBv+Cu⇒u∈S for all v∈S.
Letu∈E and v∈S be arbitrary elements such that u=AuBv+Cu. Then,(38)ut≤AutBvt+Cut≤ft,utΓα∑s=0t-αt-σsα-1_Fs+α,us+α+1Γγ∑s=α-1tt+γ-σsγ-1_ps,us≤ft,ut-ft,0+ft,0Γα∑s=0TT+α-σsα-1_ωΨR+1Γγ∑s=α-1T+αT+α+γ-σsγ-1_ps,us-pt,0+pt,0≤Rϕ+f^0ωΨRTαΓα+1+Rψ+p^0T+αγΓγ+1.We find that(39)u≤Rϕ+f^0ωΨRTαΓα+1+Rψ+p^0T+αγΓγ+1≤R.Therefore, u∈S.
Step  4. We prove that δM+ρ<1. Since(40)M=BS=supu∈Smaxt∈Nα-1,T+αBut≤ωΨRTαΓα+1,and by (H3), we have(41)ϕM+ψT+αγΓγ+1<1,with δ=ϕ and ρ=ψ(T+α)γ/Γ(γ+1).
We see that all the conditions of Theorem12 are satisfied. Hence, the operator equation u=AuBv+Cu has a solution in S. In consequence, problem (6) has a solution on Nα-1,T+α. This completes the proof.


## 4. Hybrid Fractional Sequential Sum-Difference Initial Value Problem (7)

In this section, we prove existence results of problem (7). To accomplish this, we denote that X=C(Nα+β-2,T+α+β,R) and define the space of all functions u with the norm by(42)u=maxt∈Nα+β-2,T+α+β⁡ut.In addition, we define the operator H:X→X by(43)Hut=1ΓαΓβ∑s=α-1t-βt-σsβ-1_hs+β,us+β×∑ξ=0s-αs-σξα-1_Hξ+α+β-1,uξ+α+β-1+1Γγ∑s=α+β-γ-2t-γt-σsγ-1_gs+β+γ,us+β+γ.Clearly, problem (7) has solutions if and only if the operator H has fixed points. The first shows the existence and uniqueness of a solution to problem (7) by using the Banach contraction principle.Theorem 14.
Assume thath∈C(Nα+β-2,T+α+β×R,R-0), g∈C(Nα+β-2,T+α+β×R,R) with g(α+β-2,0)=0, and H∈C(Nα+β-2,T+α+β×R×R,R). In addition, suppose that(B1)
there exist constantsM1,M2,N>0 such that for each t∈Nα+β-2,T+α+β and u,v,u∗,v∗∈R(44)Ht,u,u∗-Ht,v,v∗≤M1u-v+M2u∗-v∗,gt,u-gt,v≤Nu-v,(B2)
there exists a positive functionϑ with bound ϑ such that for each t∈Nα+β-2,T+α+β and u,v∈R(45)ht,u-ht,v≤ϑt.If Ω<1 then problem (7) has a unique solution, where(46)Ω≔ϑΓαΓβT+ββ-1_T+α+βα-1_M1+M2T+α+βω_Γω+1+NT+α+βγ_Γγ+1.Proof.
We shall show thatH is a contraction. For any u,v∈C and for each t∈Nα-1,T+α, we have(47)Hut-Hvt≤hΓαΓβ∑s=α-1t-βt-σsβ-1_∑ξ=0s-αs-σξα-1_Hξ+α+β-1,uξ+α+β-1-Hξ+α+β-1,vξ+α+β-1+1Γγ∑s=α+β-γ-2t-γt-σsγ-1_gs+β+γ,us+β+γ-gs+β+γ,vs+β+γ≤ϑΓαΓβ∑s=α-1T+α∑ξ=0s-αT+α+β-σsβ-1_s-σξα-1_M1u-v+M2Δ-ωu-Δ-ωv+NΓγu-v∑s=α+β-γ-2T+α+β-γT+α+β-σsγ-1_≤ϑΓαΓβM1+M2T+α+βω_Γω+1u-vT+ββ-1_T+α+βα-1_+NΓγu-vT+α+βγ_≤u-vϑΓαΓβT+ββ-1_T+α+βα-1_M1+M2T+α+βω_Γω+1+NT+α+βγ_Γγ+1=u-vΩ.Thus, we have (Hu)-(Hv)<u-v.
Consequently,H is a contraction. Therefore, by the Banach fixed point theorem, we get that H has a fixed point which is a unique solution of problem (7).In the second result, we deduce the existence of at least one solution of the initial value problem (7) by using the generalized Krasnoselskii’s fixed point theorem.Theorem 15 (generalized Krasnoselskii’s fixed point theorem [68]).
LetM be a nonempty, closed convex, and bounded subset of the Banach space X. Let A:X→X and B:M→X be two operators such that(i)
A is a contraction,(ii)
B is completely continuous,(iii)
x=Ax+By for all y∈M⇒x∈M.
Then, the operator equationAx+Bx=x has a solution.Theorem 16.
Assume thath∈C(Nα+β-2,T+α+β×R,R-{0}), g∈C(Nα+β-2,T+α+β×R,R) with g(α+β-2,0)=0, and H∈C(Nα+β-2,T+α+β×R×R,R). In addition, suppose that(K1)
there exist two positive functionsϑ and θ with bound ϑ and θ, respectively, such that, for each t∈Nα+β-2,T+α+β and u,v,Δ-ωu,Δ-ωv∈R,(48)ht,u-ht,v≤ϑtu-v,Ht,u,Δ-ωu-Ht,u,Δ-ωv≤θtu-v-Δ-ωu-Δ-ωv,(K2)
there exist functionsμ,ν,κ∈C(Nα+β-2,T+α+β,R+) such that(49)ht,u≤μt∀t,u∈Nα+β-2,T+α+β×R,Ht,u,Δ-ωu≤νt∀t,u,Δ-ωu∈Nα+β-2,T+α+β×R×R,gt,u≤κt∀t,u∈Nα+β-2,T+α+β×R.
If(50)T+α+ββΓβ+1ϑνT+α+βαΓα+1+μθ1+T+α+βωΓω+1<1,then problem (7) has a unique solution on Nα+β-2,T+α+β.Proof.
Letmaxt∈Nα+β-2,T+α+βμt=μ, maxt∈Nα+β-2,T+α+βνt=ν, and maxt∈Nα+β-2,T+α+βκt=κ, and choose a constant(51)R≥κT+α+βγΓγ+1+μνT+α+βα+βΓα+1Γβ+1.We consider BR=u∈X:u≤R. Define four operators P:X→X, Q:BR→X, and S:X→X by(52)Put≔ht,ut,Qut≔1Γα∑s=0t-αt-σsα-1_Hs+α+β-1,us+α+β-1,Sut≔1Γγ∑s=α+β-γ-2t-γt-σsγ-1_gs+β+γ,us+β+γ,for t∈Nα+β-2,T+α+β, and(53)Tut≔1Γβ∑s=0t-βt-σsβ-1_PusQus.Problem (7) has solutions if and only if the operator u=Su+Tu has fixed points.
The proof is divided into three steps as follows.
Step  1. Verify u=Su+Tu map bounded sets into bounded sets in BR.
For eachv∈BR, we obtain(54)ut=Sut+Tut≤1Γγ∑s=α+β-γ-2t-γt-σsγ-1_gs+β+γ,us+β+γ+1Γβ∑s=0t-βt-σsβ-1_PusQus≤κΓγ∑s=α+β-γ-2T+α+β-γT+α+β-σsγ-1_+μνΓαΓβ∑s=0T+αT+α+β-σsβ-1_∑ξ=0s-αξ-σξα-1_≤κT+α+βγΓγ+1+μνT+αα+βΓα+1Γβ+1≤R.Thus, u≤R. This implies that Su+Tu is uniformly bounded.
Hence, condition (iii) of Theorem15 holds.
Step  2. Check that T is contraction mapping.
For anyu,v∈R and for each t∈Nα+β-2,T+α+β, by (K1), we have(55)Tut-Tvt≤1Γβ∑s=0t-βt-σsβ-1_PusQus-PvsQvs=1Γβ∑s=0t-βt-σsβ-1_QusPus-Pvs+PvsQus-Qvs=1Γβ∑s=0t-βt-σsβ-1_ϑνT+α+βαΓα+1u-v+μθu-v+u-v1Γω∑s=α+β-2tt-σsω-1≤u-vT+α+ββΓβ+1ϑνT+α+βαΓα+1+μθ1+T+α+βωΓω+1.By (50), T is a contraction mapping. Hence, condition (i) of Theorem 15 holds.
Step  3. Check that S is completely continuous on BR.
The operatorS is obviously continuous on BR. Furthermore, S is uniformly bounded on BR since(56)S≤κT+α+βγΓγ+1.For any ϵ>0, there exists a positive constant δ such that for t1,t2∈Nα+α+β,T+α+β(57)t2γ-t1γ<Γγ+1κ,whenevert2-t1<δ.Then,(58)Sxt2-Sxt1≤1Γγ∑s=α+β-γ-2t2-γt2-σsγ-1_gs+β+γ,us+β+γ-∑s=α+β-γ-2t1-γt1-σsγ-1_gs+β+γ,us+β+γ≤κΓγ+1t2γ-t1γ<ϵ.This implies that the set S(BR) is an equicontinuous set. Therefore, by the Arzelá-Ascoli theorem, we find that S is completely continuous. Hence, condition (ii) of Theorem 15 holds.
We see that all the assumptions of Theorem15 are satisfied. Therefore, we can conclude that problem (7) has at least one solution. The proof is completed.


## 5. Examples

In this section, we provide some examples to illustrate our results.Example 1.
Consider the following fractional difference initial value problem:(59)Δ1/2ut-Δ-1/3pt+1/3,ut+1/3ft,ut=Ft+12,ut+12,t∈N0,10,u-12=0,where(60)pt,ut=u2t+4ut3+ut·tsin2⁡2πt50002+e-t,ft,ut=102cos2⁡2πt+3t3100π-2t2·u2t+3utut+10,Ft,ut=t2+58ut+10π80100π-t.We set α=1/2, γ=1/3, and T=10.
Noticing that(H1)-(H2) hold, for each t∈N-1/2,21/2, we have(61)pt,u-pt,v=4t150002+e-tu-v,ft,u-ft,v=2+3t100π-2t2u-v,Ft,u-Ft,v=t2+5100π-tu10+π8.Thus,(62)ϕ=0.000389,ψ=0.00139,ω=0.441.Finally, we find that(63)f^0=0.00389,p^0=0.00105.So, (H3) holds with a number R∈[730.5887,892.9406].
Hence, by Theorem13, problem (59) has a unique solution on N-1/2,21/2.Example 2.
Consider the following fractional difference boundary value problem:(64)Δ1/2Δ2/5ut-Δ-1/3gt-4/15,ut-4/15ht,ut=Ht-110,ut-110,Δ-3/4ut+1320,u-1110=Δ2/5u-12=0,for t∈N0,10, where(65)gt,ut=log⁡1+utetsin⁡5πt1+ut,ht,ut=ut+1ut+2e-sin2πt100e-t,Ht,ut,Δ-3/4ut+34=13arctan⁡3utsin25πt10+ut-13arctan⁡3Δ-3/4ut+3/4sin25πt10+Δ-3/4ut+3/4.We let α=1/2, β=2/5, γ=1/3, ω=3/4, and T=10.
Noticing that(K1)-(K2) hold, for each t∈N-11/10,109/10, we obtain(66)ht,u-ht,v=e-sin2πt400e-4tu-v,Ht,u,Δ-3/4u-Ht,v,Δ-3/4v=110sin25πtu-v+Δ-3/4u-Δ-3/4v,ht,u=e-sin2πt100e-t+e-πt,Ht,u,Δ-3/4u=3sin2⁡5πt.Thus, we obtain(67)ϑ=0.000923,θ=0.1,μ=0.00369,ν=3.Finally, we find that(68)T+α+ββΓβ+1ϑνT+α+βαΓα+1+μθ1+T+α+βωΓω+1≈0.2277<1.Hence, by Theorem 16, problem (64) has a unique solution on N-11/10,109/10.



---

*Source: 5268528-2018-01-30.xml*