# Existence Results for Fractional Difference Equations with Three-Point Fractional Sum Boundary Conditions

Authors: Thanin Sitthiwirattham; Jessada Tariboon; Sotiris K. Ntouyas
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We consider a discrete fractional boundary value problem of the formΔαu(t)=f(t+α-1,u(t+α-1)),t∈[0,T]ℕ0:={0,1,…,T},u(α-2)=0,u(α+T)=Δ-βu(η+β), where 1<α≤2, β>0, η∈[α-2,α+T-1]ℕα-2:={α-2,α-1,…,α+T-1}, and f:[α-1,α,…,α+T-1]ℕα-1×ℝ→ℝ is a continuous function. The existence of at least one solution is proved by using Krasnoselskii's fixed point theorem and Leray-Schauder's nonlinear alternative. Some illustrative examples are also presented.


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## Body

## 1. Introduction

Fractional differential and difference equations have received increasing attention during recent years since fractional derivatives provide an excellent instrument for the description of memory and hereditary properties of various materials and processes. For a reader interested in the systematic development of the topic, a good account of papers dealing with fractional or discrete fractional boundary value problems can be found in [1–25] and references cited therein.In this paper we consider a nonlinear discrete fractional boundary value problem of the form(1)Δαu(t)=f(t+α-1,u(t+α-1)),t∈[0,T]ℕ0:={0,1,…,T},u(α-2)=0,u(α+T)=Δ-βu(η+β),
where1<α≤2,β>0,η∈[α-2,α+T-1]ℕα-2:={α-2,α-1,…,α+T-1}, andf:[α-1,…,α+T-1]ℕα-1×ℝ→ℝis a continuous function. Letℕbe the nonnegative integers. Denote[i,j]ℕi={k∈ℕ∣i≤k≤j}.Problem (1) was studied recently by authors in [26] where existence and uniqueness results are obtained by using Banach’s contraction mapping principle, the nonlinear contraction, and Schaefer’s fixed point theorem. Here we continue this study by proving some new existence results by using Krasnoselskii’s fixed point theorem and Leray-Schauder’s nonlinear alternative. Thus the results of this paper complement the results of [26].The plan of this paper is as follows. In Section2 we recall some definitions and basic lemmas. Also we give a representation for the solution to (1) by converting the problem to an equivalent summation equation. In Section 3, using this representation, we prove our main results. Some illustrative examples are presented in Section 4.


## 2. Preliminaries

In this section, we introduce notations, definitions, and lemmas needed in the following.Definition 1.
One defines the generalized falling function bytα_:=Γ(t+1)/Γ(t+1-α), for anytandαfor which the right-hand side is defined. Ift+1-αis a pole of the Gamma function andt+1is not a pole, thentα_=0.Definition 2 (see [17]).
Theαth fractional sum of a functionf, forα>0, is defined by
(2)Δ-αf(t)=Δ-αf(t;a):=1Γ(α)∑s=at-α‍(t-σ(s))α-1_f(s),
fort∈{a+α,a+α+1,…}:=ℕa+αandσ(s)=s+1. One also defines theαth fractional difference forα>0byΔαf(t):=ΔNΔα-Nf(t), wheret∈ℕa+αandN∈ℕare chosen so that0≤N-1<α≤N.Lemma 3.
Lettandαbe any numbers for whichtα_andtα-1_are defined. ThenΔtα_=αtα-1_.Lemma 4 (see [17]).
Let0≤N-1<α≤N.Then
(3)Δ-αΔαy(t)=y(t)+C1tα-1_+C2tα-2_+⋯+CNtα-N_,
for someCi∈ℝ, with1≤i≤N.Theorem 5 (Arzelá-Ascoli theorem, see [27]).
LetXbe a Banach space andS={s(t)}a function family of continuous mappingss:[a,b]→X. If S is uniformly bounded and equicontinuous, and, for anyt*∈[a,b], the set{s(t*)}is relatively compact, then there exists a uniformly convergent function sequence{sn(t)}forn=1,2,…,t∈[a,b]inS.Lemma 6 (Mazur Lemma, see [28]).
IfSis a compact subset of Banach spaceX, then its convex closure conv¯  S is compact.To define the solution of the boundary value problem (1) we need the following lemma, proved in [26] which deals with linear variant of the boundary value problem (1) and gives a representation of the solution.Lemma 7.
Let(4)Λ=∑s=1η-α+2‍(η+β-s-α+1)β-1_Γ(s+α-1)Γ(β)Γ(s)-Γ(α+T+1)Γ(T+2)≠0
and let h:[α-1,…,α+T-1]ℕα-1→ℝbe given. Then the problem
(5)Δαu(t)=h(t+α-1),t∈[0,T]ℕ0,α∈(1,2],u(α-2)=0,u(α+T)=Δ-βu(η+β),η∈[α-2,α+T-1]ℕα-2,β>0,
has a unique solution
(6)u(t)=-tα-1_ΛΓ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_h(ξ+α-1)-∑s=0T‍(T+α-σ(s))α-1_h(s+α-1)1Γ(β)∑s=αη‍∑ξ=0s-α‍]+1Γ(α)∑s=0t-α‍(t-σ(s))α-1_h(s+α-1).


## 3. Main Results

LetC([α-2,α+T]ℕα-2,ℝ)be the Banach space of all continuous functionsxwith the norm∥x∥=max{|x(t)|,t∈[α-2,α+T]ℕα-2}.Define an operatorF:C([α-2,α+T]ℕα-2,ℝ)→C([α-2,α+T]ℕα-2,ℝ)by
(7)Fx(t)=-tα-1_ΛΓ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_f(x(ξ+α-1))ξ+α-1,x(ξ+α-1))-∑s=0T‍(T+α-σ(s))α-1_×f(s+α-1,x(s+α-1))1Γ(β)∑s=αη‍∑ξ=0s-α‍]+1Γ(α)∑s=0t-α‍(t-σ(s))α-1_×f(s+α-1,x(s+α-1)),
fort∈[α-2,α+T]ℕα-2, whereΛ≠0is defined by (4). It is easy to see that the problem (1) has solutions if and only if the operatorFhas fixed points.In the following, for the sake of convenience, we set a constant(8)Ω=Γ(T+α+1)|Λ|Γ(α)Γ(T+2)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]+Γ(T+α+1)Γ(T+1)Γ(α+1).Our first result is based on the Krasnoselskii’s fixed point theorem [29].Theorem 8.
LetKbe a bounded closed convex and nonempty subset of a Banach spaceX. LetA,Bbe operators such that: (i)
Ax+By∈Kwheneverx,y∈K(ii)
Ais compact and continuous(iii)
Bis a contraction mapping.Then there existsz∈Ksuch thatz=Az+Bz.Theorem 9.
Assume that(H1)
there exists a constantL>0such that|f(t,x)-f(t,y)|≤L|x-y|for eacht∈[α-2,α+T]ℕα-2andx,y∈ℝ;(H2)
|f(t,u)|≤μ(t), for all(t,u)∈[α-2,T+α]ℕα-2×ℝ, withμ∈C([α-2,T+α]ℕα-2,ℝ+);(H3)
the setS1={(t-σ(s))α-1_f(s+α-1,u(s+α-1))∣u∈C([α-2,T+α]ℕα-2,ℝ),s∈{-1,…,T+1}}is relatively compact for anyt∈[0,T]ℕ0.
If(9)LΩ<1,
whereΩis given by (8), then the boundary value problem (1) has at least one solution on[α-2,T+α]ℕα-2.Proof.
By settingmaxt∈[α-2,T+α]ℕα-2|μ(t)|=∥μ∥and choosing a constant
(10)R≥∥μ∥Ω,
we now consider a suitable ballB¯R={u∈C([α-2,T+α]ℕα-2):∥u∥≤R}.
In view of Lemma7, we define the operatorsℱ1andℱ2on the ballB¯Ras
(11)(ℱ1u)(t)=1Γ(α)∑s=0t-α‍(t-σ(s))α-1_×f(ξ+α-1,u(ξ+α-1)),(ℱ2u)(t)=-tα-1_ΛΓ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_×f(ξ+α-1,u(ξ+α-1))-∑s=0T‍(T+α-σ(s))α-1_×f(s+α-1,u(s+α-1))1Γ(β)∑s=αη‍∑ξ=0s-α‍].
Foru,v∈B¯R, we have
(12)|(ℱ1u)(t)+(ℱ2v)(t)|≤maxt∈[α-2,T+α]ℕα-2{+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]}∥μ∥Γ(α)∑s=0t-α‍(t-σ(s))α-1_+∥μ∥tα-1_|Λ|Γ(α)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]}=∥μ∥Γ(α)maxt∈[α-2,α+T]ℕα-2∑s=0t-α‍(t-σ(s))α-1_+∥μ∥|Λ|Γ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]×maxt∈[α-2,α+T]ℕα-2tα-1_=∥μ∥maxt∈[α-2,α+T]ℕα-2Γ(t+1)Γ(t-α+1)Γ(α+1)+∥μ∥|Λ|Γ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]×maxt∈[α-2,α+T]ℕα-2Γ(t+1)Γ(t-α+2)≤∥μ∥Γ(T+α+1)Γ(T+1)Γ(α+1)+∥μ∥|Λ|Γ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]×Γ(T+α+1)Γ(T+2)=∥μ∥Ω≤R.
This implies that∥ℱ1u+ℱ2v∥≤R. Therefore,ℱ1u+ℱ2v∈B¯R. Now, we will prove thatℱ2is a contraction mapping. For anyx,y∈C([α-2,α+T]ℕα-2,ℝ), we have
(13)|(ℱ2x)(t)-(ℱ2y)(t)|≤1|Λ|Γ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_×|f(ξ+α-1,x(ξ+α-1))-f(ξ+α-1,y(ξ+α-1))|+∑s=0T‍(T+α-σ(s))α-1_×|f(s+α-1,x(s+α-1))-f(s+α-1,y(s+α-1))|1Γ(β)∑s=αη‍∑ξ=0s-α‍]×maxt∈[α-2,α+T]ℕα-2tα-1_≤L∥x-y∥|Λ|Γ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_[1Γ(β)∑s=αη‍∑ξ=0s-α‍]×maxt∈[α-2,α+T]ℕα-2Γ(t+1)Γ(t-α+2)≤L∥x-y∥|Λ|Γ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍]×Γ(T+α+1)Γ(T+2)≤LΩ∥x-y∥.
By (9), we get thatℱ2is a contraction mapping.
Next, we will show thatℱ1is compact and continuous. Continuity offcoupled with the assumption(H2)implies that the operatorℱ1is continuous and uniformly bounded onB¯Ras
(14)∥ℱ1u∥≤∥u∥Γ(T+α+1)Γ(α)Γ(T+1)Γ(α+1).
From(H2), there exists a constantN>0such thatsup(t,u)∈[α-2,T+α]ℕα-2×BR|f(t,u)|≤N. For anyε>0, there exists a neighborhoodUoft,t∈[α-2,α+T]ℕα-2, such that, fort1,t2∈U,
(15)|(t2)(t2-1)⋯(t1+1)(t2-α)(t2-α-1)⋯(t1-α+1)-1|<αΓ(α)Γ(T-α+1)NΓ(T+1)ε.
Then, we have(16)|ℱ1u(t2)-ℱ1u(t1)|=1Γ(α)|∑s=0t2-α‍(t2-σ(s))α-1_×f(ξ+α-1,u(ξ+α-1))-1Γ(α)∑s=0t1-α‍(t1-σ(s))α-1_×f(ξ+α-1,u(ξ+α-1))∑s=0t2-α‍|≤1Γ(α)|∑s=0t1-α‍((t2-σ(s))α-1_-(t1-σ(s))α-1_)×f(s+α-1,u(s+α-1))∑s=0t1-α‍|+1Γ(α)|∑s=t1-α+1t2-α‍(t2-σ(s))α-1_×f(s+α-1,u(s+α-1))∑s=t1-α+1t2-α‍|≤NΓ(α)|∑s=t1-α+1t2-α‍∑s=0t1-α‍(t2-σ(s))α-1_-∑s=0t1-α‍(t1-σ(s))α-1_+∑s=t1-α+1t2-α‍(t2-σ(s))α-1_|=NαΓ(α)|Γ(t2+1)Γ(t2-α+1)-Γ(t1+1)Γ(t1-α+1)|=NαΓ(α)Γ(t1+1)Γ(t1-α+1)×|Γ(t2+1)Γ(t1-α+1)Γ(t2-α+1)Γ(t1+1)-1|=NαΓ(α)Γ(T+1)Γ(T-α+1)×|(t2)(t2-1)⋯(t1+1)(t2-α)(t2-α-1)⋯(t1-α+1)-1|<ε.
This means that the set ℱ1(B¯R) is an equicontinuous set.
In view of Lemma6 and the condition thatS1is relatively compact, we have thatconv¯S1is compact. For anyt*∈[0,T]ℕ0,
(17)(F1un)(t*)=1Γ(α)∑s=0t*-α‍(t*-σ(s))α-1_×f(s+α-1,u(s+α-1))=1Γ(α)θn,
where
(18)θn=∑s=0t*-α‍(t*-σ(s))α-1_f(s+α-1,u(s+α-1)).
Sinceconv¯S1is convex and compact, we have thatθn∈conv¯S1. Therefore, for anyt*∈[0,T]ℕ0, the set(F1un)(t*), forn=1,2,…, is relatively compact. From Theorem 5, every(F1un)(t)contains a uniformly convergent subsequence(F1unk)(t), fork=1,2,…, on∈[0,T]ℕ0which means that the set is relatively compact. Sinceℱ1(B¯R)is a bounded, equicontinuous, and relatively compact set, we have thatF1is compact onB¯R. Therefore, all the assumptions of Theorem 8 are satisfied, and the conclusion of Theorem 8 implies that boundary value problem (1) has at least one solution on[α-2,T+α]ℕα-2. This completes the proof.Corollary 10.
Assume that there exists a constantG>0such that|f(t,u)|≤Gfor anyt∈[0,T]ℕ0and u∈C([α-2,T+α]ℕα-2×ℝ), and the set S={(t-σ(s))α-1_f(s+α-1,u(s+α-1))∣u∈C([α-2,T+α]ℕα-2,ℝ), and s∈{1,…,T+1}} is relatively compact for everyt∈[0,T]ℕ0; then there exists at least one solution of boundary value problem (1) on t∈[0,T]ℕ0.Proof.
Lettingμ(t)≡Gfort∈[α-2,T+α]ℕα-2, we get the result by using Theorem 9.The second result is based on Leray-Schauder nonlinear alternative [30].Theorem 11.
LetEbe a Banach space,Ca closed, convex subset ofE,Uan open subset of C, and0∈U. Suppose thatF:U¯→Cis a continuous, compact (i.e.,F(U)is a relatively compact subset ofC) map. Then either(i)
Fhas a fixed point inU¯ or(ii)
there areu∈∂U(the boundary ofUinC) andλ∈(0,1)withu=λF(u).Theorem 12.
Letf:[α-2,T+α]ℕα-2×ℝ→ℝbe a continuous function. Assume that(H4)
there exist a functionp∈C([α-2,T+α]ℕα-2,ℝ+) and a continuous nondecreasing functionΨ:ℝ+→ℝ+such that|f(t,u)|≤p(t)Ψ(∥u∥), for all(t,u)∈[α-2,T+α]ℕα-2×ℝ;(H5)
the setS~=S1∩S2is relatively compact for anyt∈[0,T]ℕ0, whereS1is defined as in(H3)andS2={tα-1_f(s+α-1,u(s+α-1))∣u∈C([α-2,T+α]ℕα-2,ℝ),s∈{-1,…,T+1}};(H6)
there exists a constantM>0such that
(19)MΓ(α)Ψ(M)∥p∥Φ>1,where
(20)Φ=Γ(T+α+1)|Λ|Γ(T+2)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_v×(s-σ(ξ))α-1_v+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]+∑s=0T‍(T+α-σ(s))α-1_.
Then the boundary value problem (1) has at least one solution on[α-2,T+α]ℕα-2.Proof.
We consider the operator equationu=F(u), whereF:C([α-2,T+α]ℕα-2)→C([α-2,T+α]ℕα-2)defined by (7).
Now, we will show thatFmaps bounded sets into bounded sets inC([α-2,T+α]ℕα-2,ℝ). For a positive numberr, letB¯r={u∈C([α-2,T+α]ℕα-2,ℝ)∣∥u∥≤r}be a bounded set inC([α-2,T+α]ℕα-2,ℝ). Then
(21)|Fu(t)|≤tα-1_|Λ|Γ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_×|f(ξ+α-1,u(ξ+α-1))|+∑s=0T‍(T+α-σ(s))α-1_×|f(s+α-1,u(s+α-1))|1Γ(β)∑s=αη‍∑ξ=0s-α‍]+1Γ(α)∑s=0t-α‍(t-σ(s))α-1_×|f(s+α-1,u(s+α-1))|≤tα-1_|Λ|Γ(α)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_p(ξ+α-1)Ψ(∥u∥)+∑s=0T‍(T+α-σ(s))α-1_×p(s+α-1)Ψ(∥u∥)1Γ(β)∑s=αη‍∑ξ=0s-α‍]+1Γ(α)∑s=0t-α‍(t-σ(s))α-1_×p(s+α-1)Ψ(∥u∥)≤Ψ(∥u∥)(T+α)α-1_|Λ|Γ(α)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_×p(ξ+α-1)+∑s=0T‍(T+α-σ(s))α-1_p(s+α-1)1Γ(β)∑s=αη‍∑ξ=0s-α‍]+Ψ(∥u∥)Γ(α)∑s=0T‍(T+α-σ(s))α-1_p(s+α-1)=Ψ(∥u∥)Γ(α){[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_∑s=0T‍Γ(T+α+1)|Λ|Γ(T+2)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_×p(s+α-1)+∑s=0T‍(T+α-σ(s))α-1_p(s+α-1)1Γ(β)∑s=αη‍∑ξ=0s-α‍]+∑s=0T‍(T+α-σ(s))α-1_p(s+α-1)[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_}.
Consequently,(22)∥Fu∥≤Ψ(r)∥p∥Γ(α)×{[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_Γ(T+α+1)|Λ|Γ(T+2)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]+∑s=0T‍(T+α-σ(s))α-1_[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_}.
Next we show thatFmaps bounded sets into equicontinuous sets ofC([α-2,T+α]ℕα-2,ℝ). Letu∈B¯r, whereB¯ris a bounded set ofC([α-2,T+α]ℕα-2,ℝ). For anyε>0, there exists a neighborhoodUoft,t∈[α-2,T+α]ℕα-2, such that, fort1,t2∈U,
(23)|t2α-1_-t1α-1_|∥p∥Ψ(r)|Λ|Γ(α)×|1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍|+∥p∥Ψ(r)Γ(α)Γ(T+1)Γ(T-α+1)×|(t2)(t2-1)⋯(t1+1)(t2-α)(t2-α-1)⋯(t1-α+1)-1|<ε.
Then, we have(24)|Fu(t2)-Fu(t1)|=|-t2α-1_-t1α-1_ΛΓ(α)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_×f(ξ+α-1,u(ξ+α-1))-∑s=0T‍(T+α-σ(s))α-1_×f(s+α-1,u(s+α-1))1Γ(β)∑s=αη‍∑ξ=0s-α‍]+1Γ(α)[∑s=0t1-α‍((t2-σ(s))α-1_-(t1-σ(s))α-1_)×f(s+α-1,u(s+α-1))+∑s=t1-α+1t2-α‍(t2-σ(s))α-1_×f(s+α-1,u(s+α-1))∑s=0t1-α‍]|≤|t2α-1_-t1α-1_|∥p∥Ψ(r)|Λ|Γ(α)×|1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍|+∥p∥Ψ(r)Γ(α)×|+∑s=t1-α+1t2-α‍(t2-σ(s))α-1_|∑s=0t1-α‍((t2-σ(s))α-1_-(t1-σ(s))α-1_)+∑s=t1-α+1t2-α‍(t2-σ(s))α-1_|=|t2α-1_-t1α-1_|∥p∥Ψ(r)|Λ|Γ(α)×|1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_|1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_|+∥p∥Ψ(r)Γ(α)Γ(T+1)Γ(T-α+1)×|(t2)(t2-1)⋯(t1+1)(t2-α)(t2-α-1)⋯(t1-α+1)-1|<ε.
This means that the setF(B¯r)is an equicontinuous set. By(H5), Lemma 6, and Theorem 5, it follows thatF:C([α-2,T+α]ℕα-2,ℝ)is completely continuous.
Letu=Fu. Then, fort∈[α-2,T+α]ℕα-2 and using the computations as in the first step, we get
(25)|u(t)|≤tα-1_|Λ|Γ(α)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_×|f(ξ+α-1,x(ξ+α-1))|+∑s=0T‍(T+α-σ(s))α-1_×|f(s+α-1,x(s+α-1))|1Γ(β)∑s=αη‍∑ξ=0s-α‍]+1Γ(α)∑s=0t-α‍(t-σ(s))α-1_×|f(s+α-1,x(s+α-1))|≤Ψ(∥u∥)∥p∥Γ(α)×{∑s=0T‍Γ(T+α+1)|Λ|Γ(T+2)×[1Γ(β)∑s=αη‍∑ξ=0s-α‍(η+β-σ(s))β-1_×(s-σ(ξ))α-1_+∑s=0T‍(T+α-σ(s))α-1_1Γ(β)∑s=αη‍∑ξ=0s-α‍]+∑s=0T‍(T+α-σ(s))α-1_}.
Consequently,(26)∥u∥Γ(α)Ψ(∥u∥)∥p∥Φ≤1.
In view of(H4), there existsMsuch that∥u∥<M. Let us set
(27)U={u∈C([α-2,T+α]ℕα-2,ℝ)∣∥u∥<M}.
Note that the operatorF:U¯→C([α-2,T+α]ℕα-2,ℝ)is continuous and completely continuous. From the choice ofU, there is nou∈∂Usuch thatu=λFufor someλ∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Theorem 11), we deduce thatFhas a fixed pointu∈U¯which is a solution of the problem (1). This completes the proof.


## 4. Some Examples

In this section, in order to illustrate our result, we consider two examples.Example1. Consider the following three-point fractional sum boundary value problem:
(28)Δ3/2x(t)=e-sin2t(t+7/2)2·|x||x|+1,t∈[12,92]ℕ1/2,x(-12)=0,x(112)=Δ-4/3x(236).Hereα=3/2,β=4/3,η=5/2,T=4, and f(t,x)=e-sin2t/(t+3)2·|x|/(|x|+1).Also we find
(29)Λ=∑s=13‍(10/3-s)1/3_Γ(s+1/2)(1/3)Γ(1/3)Γ(s)-Γ(13/2)Γ(6)=-283≠0,Ω=Γ(13/2)|-283|Γ(3/2)Γ(6)×[1Γ(4/3)∑s=3/25/2‍∑ξ=0s-3/2‍(52+43-σ(s))4/3-1_×(s-σ(ξ))3/2-1_+∑s=04‍(4+32-σ(s))3/2-1_1Γ(4/3)∑s=3/25/2‍∑ξ=0s-3/2‍]+Γ(4+3/2+1)Γ(4+1)Γ(3/2+1)≈10.29.Since|f(t,x)-f(t,y)|≤(4/49)|x-y|is satisfied withγ=4/49, we can show that
(30)γΩ≈449×10.29=0.84<1.Hence, by Theorem9, the boundary value problem (28) has at least one solution on[1/2,9/2]ℕ1/2.Example2. Consider the following three-point fractional sum boundary value problem:
(31)Δ3/2x(t)=t+1/2100πsin(πx)+(t+3/2)x21+x2,t∈[12,92]ℕ1/2,x(-12)=0,x(112)=Δ-4/3x(236).Hereα=3/2,β=4/3,η=5/2,T=4,f(t,x)=(t/100π)sin(πx)+(t+1)x2/(1+x2),Λ=-283≠0, and Φ=8.168. It is clear that |f(t,x)|=|(t/100π)sin(πx)+(t+1)x2/(1+x2)|≤(t+1)∥x∥/100+1.We choosep(t)=t+1,Ψ(∥x∥)=∥x∥/100+1and find that
(32)MΓ(α)∥p∥Φ>M100+1
which impliesM>103.27. Hence, by Theorem 12, the boundary value problem (31) has at least one solution on [1/2,9/2]ℕ1/2.



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*Source: 104276-2013-09-11.xml*