# Existence Results for Fractional Hahn Difference and Fractional Hahn Integral Boundary Value Problems

Authors: Nichaphat Patanarapeelert; Thanin Sitthiwirattham
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

The existence and uniqueness results of two fractional Hahn difference boundary value problems are studied. The first problem is a Riemann-Liouville fractional Hahn difference boundary value problem for fractional Hahn integrodifference equations. The second is a fractional Hahn integral boundary value problem for Caputo fractional Hahn difference equations. The Banach fixed-point theorem and the Schauder fixed-point theorem are used as tools to prove the existence and uniqueness of solution of the problems.


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## Body

## 1. Introduction

The quantum calculus is the subject of calculus without limits and deals with a set of nondifferentiable functions. The quantum operators are widely used in mathematic fields such as hypergeometric series, complex analysis, orthogonal polynomials, combinatorics, hypergeometric functions, and the calculus of variations. The quantum calculus is also found in many applications, such as quantum mechanics and particle physics [1–9].In 1949, the Hahn difference operator was introduced by Hahn [10]. This operator is a combination of two well-known operators, the forward difference operator and the Jackson q-difference operator. The Hahn difference operator is defined by (1)Dq,ωft=fqt+ω-fttq-1+ω,t≠ω0≔ω1-q.We note that (2)Dq,ωft=Δωftwheneverq=1,Dq,ωft=Dqftwheneverω=0,Dq,ωft=f′twheneverq=1,ω⟶0.The Hahn difference operator has been used in finding families of orthogonal polynomials as well to determine some approximation problems (see [11–13]).The right inverse of the Hahn difference operator was proposed by Aldwoah in 2009 [14, 15]. This operator is in the terms of the Jackson q-integral containing the right inverse of Dq [16] and Nörlund sum containing the right inverse of Δω [16].In 2010, Malinowska and Torres [17, 18] introduced the Hahn quantum variational calculus. In 2013, Malinowska and Martins [19] presented the generalized transversality conditions for the Hahn quantum variational calculus. Next, Hamza and Ahmed [20, 21] established the theory of linear Hahn difference equations. The authors also study the existence and uniqueness of solution for the initial value problems for Hahn difference equations by employing the method of successive approximations; in addition, they proved Gronwall’s and Bernoulli’s inequalities with respect to the Hahn difference operator and investigated the mean value theorems for this calculus. In 2016, Hamza and Makharesh [22] investigated Leibniz’s rule and Fubini’s theorem associated with Hahn difference operator. In the same year, Sitthiwirattham [23] studied the nonlocal boundary value problem for nonlinear Hahn difference equation as given by(3)Dq,ω2xt+ft,xt,Dp,θxpt+θ=0,t∈ω0,Tq,ω,xω0=φx,xT=λxη,η∈ω0,Tq,ω,where 0<q<1, 0<ω<T, ω0≔ω/1-q, 1≤λ<T-ω0/η-ω0, p=qm,m∈N, θ=ω(1-p)/(1-q), f:ω0,Tq,ω×R×R→R is a given function, and φ:Cω0,Tq,ω,R→R is a given functional.In 2010, Čermák and Nechvátal [24] established the fractional (q,h)-difference operator and the fractional (q,h)-integral for q>1. In 2011, Čermák et al. [25] studied linear fractional difference equations with discrete Mittag-Leffler functions for q>1 in 2011. Rahmat [26, 27] presented the (q,h)-Laplace transform and some (q,h)-analogues of integral inequalities on discrete time scales for q>1. In 2016, Du et al. [28] studied the monotonicity and convexity for nabla fractional (q,h)-difference for q>0, q≠1. Since fractional Hahn operators require the condition 0<q<1, we note that the operators mentioned above are not fractional Hahn operators. Recently, the fractional Hahn operators have been introduced by Brikshavana and Sitthiwirattham [29].In order to gain further insight into fractional Hahn operators, in this paper, we study the boundary value problem for fractional Hahn difference equation. Particularly, we consider a Riemann-Liouville fractional Hahn difference boundary value problem for a fractional Hahn integrodifference equation of the form(4)Dq,ωαut=Ft,ut,Ψr,ϕγut,t∈ω0,Tq,ω,uω0=uT,Dp,θβuω0=Dp,θβupT+θand a fractional Hahn integral boundary value problem for a Caputo fractional Hahn difference equation of the form(5)DCq,ωαut=Gt,ut,DCr,ϕγurt+ϕ,t∈ω0,Tq,ω,uω0=Au,Ip,θβuT=1Γpβ∫ω0TT-σp,ωsp,θβ-1_usdp,θs=Bu,where ω0,Tq,ω≔qkT+ωkq:k∈N0∪ω0; α∈(1,2], β,γ∈(0,1], ω>0, p,q,r∈(0,1), p=qm,r=qn,m,n∈N, θ=ω(1-p)/(1-q), and ϕ=ω(1-r)/(1-q); F,G∈C(ω0,Tq,ω×R×R,R) is given function; A,B:C(ω0,Tq,ω,R)→R are given functionals; and for φ∈C(ω0,Tq,ω×ω0,Tq,ω,[0,∞)), define (6)Ψr,ϕγut≔Ir,ϕγφut=1Γrγ∫ω0tt-σr,ϕsr,ϕγ-1_φt,susdr,ϕs.In Section2, we provide some basis definitions, properties, and lemma used as material for this work. In Sections 3 and 4, we prove the existence results of problems (4) and (5), respectively, and we prove the existence and uniqueness of a solution by using the Banach fixed-point theorem and the existence of at least one solution by using the Schauder fixed-point theorem. Finally, some examples are provided to illustrate our results in the last section.


## 2. Preliminaries

In this section, we suggest some notations, definitions, and lemmas which are used in the main results. Letq∈(0,1) and ω>0 and define (7)nq≔1-qn1-q=qn-1+⋯+q+1,nq!≔∏k=1n1-qk1-q,n∈R.The q-analogue of the power function a-bqn_ with n∈N0≔[0,1,2,…] is defined by (8)a-bq0_≔1,a-bqn_≔∏k=0n-1a-bqk,a,b∈R.The q,ω-analogue of the power function (a-b)q,ωn_ with n∈N0≔[0,1,2,…] is defined by (9)a-bq,ω0_≔1,a-bq,ωn_≔∏k=0n-1a-bqk+ωkq,a,b∈N.In general, for α∈R, we define (10)a-bqα_=aα∏n=0∞1-b/aqn1-b/aqα+n,a≠0,a-bq,ωα_=a-ω0α∏n=0∞1-b-ω0/a-ω0qn1-b-ω0/a-ω0qα+n=a-ω0-b-ω0qα_,a≠ω0.We note that aqα_=aα and a-ω0q,ωα_=a-ω0α and use the notation 0qα_=ω0q,ωα_=0 for α>0. The q-gamma and q-beta functions are defined by (11)Γqx≔1-qqx-1_1-qx-1,x∈R∖0,-1,-2,…,Bqx,s≔∫01tx-11-qtqs-1_dqt=ΓqxΓqsΓqx+s.Definition 1.
Forq∈(0,1), ω>0, and f defined on an interval I⊆R containing ω0≔ω/(1-q), the Hahn difference of f is defined by (12)Dq,ωft=fqt+ω-fttq-1+ωfort≠ω0,and Dq,ωf(ω0)=f′(ω0). Providing that f is differentiable at ω0, we call Dq,ωf the q,ω-derivative of f and say that f is q,ω-differentiable on I.Remarks 2.
We give some properties for the Hahn difference as follows:(1)
Dq,ω[f(t)+g(t)]=Dq,ωf(t)+Dq,ωg(t).(2)
Dq,ω[αf(t)]=αDq,ωf(t).(3)
Dq,ω[f(t)g(t)]=f(t)Dq,ωg(t)+g(qt+ω)Dq,ωf(t).(4)
Dq,ωf(t)/g(t)=(g(t)Dq,ωf(t)-f(t)Dq,ωg(t))/g(t)g(qt+ω).Lettinga,b∈I⊆R with a<ω0<b and kq=(1-qk)/(1-q),k∈N0≔N∪{0}, we define the q,ω-interval by (13)Iq,ωa,b=a,bq,ω≔qka+ωkq:k∈N0∪qkb+ωkq:k∈N0∪ω0=a,ω0q,ω∪ω0,bq,ω=a,bq,ω∪a,b=a,bq,ω∪b=a,bq,ω∪a,Iq,ωT≔Iq,ωω0,T=ω0,Tq,ω.Observe that, for each s∈a,bq,ω, the sequence σq,ωksk=0∞=qks+ωkqk=0∞ is uniformly convergent to ω0.We also define the forward jump operator asσq,ωk(t)≔qkt+ωkq and the backward jump operator as ρq,ωk(t)≔(t-ωkq)/qk for k∈N.Definition 3.
LetI be any closed interval of R containing a,b, and ω0. Assuming that f:I→R is a given function, we define q,ω-integral of f from a to b by (14)∫abftdq,ωt≔∫ω0bftdq,ωt-∫ω0aftdq,ωt,where (15)∫ω0xftdq,ωt≔x1-q-ω∑k=0∞qkfxqk+ωkq,x∈I.Providing that the series converges at x=a and x=b, we consider f as q,ω-integrable on [a,b] and the sum to the right-hand side of above equation will be called the Jackson-Nörlund sum.We note that the actual domain of functionf defined on a,bq,ω⊂I.We next introduce the fundamental theorem of Hahn calculus in the following lemma.Lemma 4 (see [14]).
Letf:I→R be continuous at ω0. Define (16)Fx≔∫ω0xftdq,ωt,x∈I.Then, F is continuous at ω0. Furthermore, Dq,ωF(x) exists for every x∈I and (17)Dq,ωFx=fx.Conversely, one has (18)∫abDq,ωFtdq,ωt=Fb-Fa∀a,b∈I.Lemma 5 (see [23]).
Letq∈(0,1), ω>0, and f:I→R be continuous at ω0. Then, (19)∫ω0t∫ω0rfsdq,ωsdq,ωr=∫ω0t∫qs+ωtfsdq,ωrdq,ωs.Lemma 6 (see [23]).
Letq∈(0,1) and ω>0. Then, (20)∫ω0tdq,ωs=t-ω0,∫ω0tt-σq,ωsdq,ωs=t-ω021+q.We next introduce fractional Hahn integral and fractional Hahn difference of Riemann-Liouville and Caputo types as follows.Definition 7.
Forα,ω>0,q∈(0,1), and f defined on ω0,Tq,ω, the fractional Hahn integral is defined by(21)Iq,ωαft≔1Γqα∫ω0tt-σq,ωsq,ωα-1_fsdq,ωs=t1-q-ωΓqα∑n=0∞qnt-σq,ωn+1tq,ωα-1_fσq,ωnt,and (Iq,ω0f)(t)=f(t).Definition 8.
Forα,ω>0, q∈(0,1), N-1<α<N,N∈N, and f defined on ω0,Tq,ω, the fractional Hahn difference of the Riemann-Liouville type of order α is defined by (22)Dq,ωαft≔Dq,ωNIq,ωN-αft=1Γq-α∫ω0tt-σq,ωsq,ω-α-1_fsdq,ωs.The fractional Hahn difference of the Caputo type of order α is defined by (23)DCq,ωαft≔Iq,ωN-αDq,ωNft=1ΓqN-α∫ω0tt-σq,ωsq,ωN-α-1_Dq,ωNfsdq,ωs,and Dq,ω0f(t)=DCq,ω0f(t)=f(t).Lemma 9 (see [29]).
Letα>0, q∈(0,1), ω>0, and f:Iq,ωT→R. Then, (24)Iq,ωαDq,ωαft=ft+C1t-ω0α-1+⋯+CNt-ω0α-N,for some Ci∈R,i=N1,N, and N-1<α≤N,N∈N.Lemma 10 (see [29]).
Letα>0, q∈(0,1), ω>0, and f:Iq,ωT→R. Then, (25)Iq,ωαDCq,ωαft=ft+C0+C1t-ω0+⋯+CN-1t-ω0N-1,for some Ci∈R,i=N0,N-1, and N-1<α≤N,N∈N.Next, we give some auxiliary lemmas for simplifying our calculations.Lemma 11 (see [29]).
Letα,β>0, p,q∈(0,1), and ω>0. Then, (26)∫ω0tt-σq,ωsq,ωα-1_s-ω0q,ωβ_dq,ωs=t-ω0α+βBqβ+1,α,∫ω0t∫ω0xt-σp,ωxp,ωα-1_x-σq,ωsq,ωβ-1_dq,ωsdp,ωx=t-ω0α+ββqBpβ+1,α.In the following, we give a lemma that deals with the linear variant of problem (4) and gives a representation of the solution.Lemma 12.
LetΩ≠0, α∈(1,2], β∈(0,1], ω>0, p,q∈(0,1), p=qm,m∈N, and θ=ω(1-p)/(1-q) and let h∈C([ω0,T]q,ω,R) be a given function. Then the problem(27)Dq,ωαut=ht,t∈ω0,Tq,ω,uω0=uT,Dp,θβuω0=Dp,θβupT+θhas the unique solution(28)ut=t-ω0α-1ΩT-ω0α-2Ph-Λ2Qh-t-ω0α-2ΩT-ω0α-1Ph-Λ1Qh+1Γqα∫ω0tt-σq,ωsq,ωα-1_hsdq,ωs,where the functions P[h] and Q[h] are defined by (29)Ph≔1ΓqαΓp-β∫ω0pT+θ∫ω0xpT+θ-σp,θxp,θ-β-1_x-σq,ωsq,ωα-1_hsdp,θsdq,ωx,(30)Qh≔1Γqα∫ω0TT-σq,ωsq,ωα-1_hsdq,ωs,respectively. The constants Ω, Λ1, and Λ2 are defined by(31)Ω≔T-ω0α-1Λ2-T-ω0α-2Λ1,(32)Λ1≔1Γp-β∫ω0pT+θpT+θ-σp,θsp,θ-β-1_s-ω0α-1dp,θs=ΓpαpT-ω0α-β-1Γpα-β,(33)Λ2≔1Γp-β∫ω0pT+θpT+θ-σp,θsp,θ-β-1_s-ω0α-2dp,θs=Γpα-1pT-ω0α-β-2Γpα-β-1,respectively.Proof.
Taking fractional Hahnq,ω-integral of order α for (27), we obtain(34)ut=Iq,ωαht+C1t-ω0α-1+C2t-ω0α-2=1Γqα∫ω0tt-σq,ωsq,ωα-1_hsdq,ωs+C1t-ω0α-1+C2t-ω0α-2.Taking fractional Hahn p,θ-difference of order β for (34), we obtain(35)Dp,θβut=1ΓqαΓp-β∫ω0t∫ω0xt-σp,θxp,θ-β-1_x-σq,ωsq,ωα-1_hsdq,ωsdp,θx+C1Γq-β∫ω0tt-σp,θsp,θ-β-1_s-ω0α-1dp,θs+C2Γq-β∫ω0tt-σp,θsp,θ-β-1_s-ω0α-2dp,θs.Substituting t=ω0,T into (34) and using the first condition of (27), we have(36)C1T-ω0α-1+C2T-ω0α-2=-1Γqα∫ω0TT-σq,ωsq,ωα-1_hsdq,ωs.By letting t=ω0, pT+θ into (35) and employing the second condition of (27), we obtain(37)C1Γp-β∫ω0pT+θpT+θ-σp,θsp,θ-β-1_s-ω0α-1dp,θs+C2Γp-β∫ω0pT+θpT+θ-σp,θsp,θ-β-1_s-ω0α-2dp,θs=-1ΓqαΓp-β∫ω0pT+θ∫ω0xpT+θ-σp,θxp,θ-β-1_x-σq,ωsq,ωα-1_hsdq,ωsdp,θx.To find C1 and C2, we solve the system of equations (36) and (37). Then, we obtain (38)C1=1ΩT-ω0α-2Ph-Λ2Qh,C2=-1ΩT-ω0α-1Ph-Λ1Qh,where P[h], Q[h], Ω, Λ1, and Λ2 are defined as (29)–(33), respectively.
Substituting the constantsC1 and C2 into (34) and by Cramer’s rule, it is implied that (28) is the uniqueness solution.We next prove a lemma that deals with the linear variant of problem (5) and gives a representation of the solution.Lemma 13.
Letα∈(1,2], β∈(0,1], ω>0, p,q∈(0,1), p=qm,m∈N, and θ=ω(1-p)/(1-q) and let h∈C(ω0,Tq,ω,R) be a given function and A,B:C(ω0,Tq,ω,R)→R be given functionals. Then, the problem(39)DCq,ωαut=ht,t∈ω0,Tq,ω,uω0=Au,Ip,θβuT=1Γpβ∫ω0TT-σp,θsp,θβ-1_usdp,θs=Buhas the unique solution(40)ut=Au+t-ω0Γqβ+2T-ω0β+1Bu-T-ω0βΓqβ+1Au-1ΓqαΓpβ×∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_hsdq,ωsdp,θx+1Γqα∫ω0tt-σq,ωsq,ωα-1_hsdq,ωs.Proof.
Taking fractional Hahnq,ω-integral of order α for (39), we obtain(41)ut=Iq,ωαht+C1t-ω0+C2.Taking fractional Hahn p,θ-integral of order β for (41), we have(42)Ip,θβut=Ip,θβIq,ωαht+1Γqβ∫ω0tt-σp,ωsp,ωβ-1_C1s-ω0+C2dp,ωs=1ΓqαΓpβ∫ω0t∫ω0xt-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_hsdq,ωsdp,θx+t-ω0β+1Γpβ+2C1+t-ω0βΓpβ+1C2.Letting t=ω0 into (41) and employing the first condition of (39), we get (43)C2=Au.Substituting t=T into (42) and employing the second condition of (39), we obtain (44)C1=Γqβ+2T-ω0β+1Bu-T-ω0βΓqβ+1Au-1ΓqαΓpβ×∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_hsdq,ωsdp,θx.Substituting the constants C1 and C2 into (41) and by Cramer’s rule, it is implied that (40) is the uniqueness solution.We next introduce Schauder’s fixed-point theorem used to prove the existence of a solution to (4) and (5).Lemma 14 ([30] (Arzelá-Ascoli theorem)).
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 15 (see [30]).
If a set is closed and relatively compact, then it is compact.Lemma 16 ([31] (Schauder’s fixed-point theorem)).
Let(D,d) be a complete metric space, let U be a closed convex subset of D, and let T:D→D be the map such that the set Tu:u∈U is relatively compact in D. Then the operator T has at least one fixed point u∗∈U:Tu∗=u∗.


## 3. Existence Results for Problem (4)

In this section, we prove the existence results for problem (4). Let X=CIq,ωT,R be a Banach space of all function u with the norm defined by (45)uX=maxt∈Iq,ωT⁡ut,where ω>0, p,q,r∈(0,1), p=qm,r=qn,m,n∈N, θ=ω(1-p)/(1-q), and ϕ=ω(1-r)/(1-q). Define an operator F:X→X by(46)Fut≔t-ω0α-1ΩT-ω0α-2P∗Fu-Λ2Q∗Fu-t-ω0α-2ΩT-ω0α-1P∗Fu-Λ1Q∗Fu+1Γqα∫ω0tt-σq,ωsq,ωα-1_Fs,us,Ψr,ϕγusdq,ωs,where Ω, Λ1, and Λ2 are defined as (31)–(33), respectively, and the functions P∗[Fu] and Q∗[Fu] are defined by(47)P∗Fu≔1ΓqαΓp-β∫ω0pT+θ∫ω0xpT+θ-σp,θxp,θ-β-1_x-σq,ωsq,ωα-1_×Fs,us,Ψr,ϕγusdq,ωxdp,θs,Q∗Fu≔1Γqα∫ω0TT-σq,ωsq,ωα-1_Fs,us,Ψr,ϕγusdq,ωs.Obviously, problem (4) has solutions if and only if the operator F has fixed points.Theorem 17.
Assume thatF:Iq,ωT×R×R→R is continuous and φ:Iq,ωT×Iq,ωT→[0,∞) is continuous with φ0=max⁡φt,s:t,s∈Iq,ωT×Iq,ωT. In addition, suppose that the following conditions hold:
(
H
1
) There exist constants λ1,λ2>0 such that, for each t∈Iq,ωT and u,v∈R,(48)Ft,u,Ψr,ϕγu-Ft,v,Ψr,ϕγv≤λ1u-v+λ2Ψr,ϕγu-Ψr,ϕγv.(H2)LΘ<1, where(49)L≔λ1+λ2φ0T-ω0γΓrγ+1,(50)Θ≔T-ω02α-1Ωpα+βΓq-βT-ω0α-β-2Γp-βΓqα-β+1-Λ2Γqα+1+T-ω02α-2Ωpα+βΓq-βT-ω0α-β-1Γp-βΓqα-β+1-Λ1Γqα+1+T-ω0αΓqα+1.Then, problem (4) has a unique solution.Proof.
To prove thatF is contraction, we consider(51)Hu-vt≔Ft,ut,Ψr,ϕγut-Ft,vt,Ψr,ϕγvt≤Ft,u,Ψr,ϕγu-Ft,v,Ψr,ϕγv≕Hu-v,for each t∈Iq,ωT and u,v∈X. We find that(52)P∗Fu-P∗Fv=1ΓqαΓp-β∫ω0pT+θ∫ω0xpT+θ-σp,θxp,θ-β-1_×x-σq,ωsq,ωα-1_Hu-vsdq,ωsdp,θx≤λ1u-v+λ2Ψr,ϕγu-Ψr,ϕγvΓq-βpT-ω0α-βΓp-βΓqα-β+1,Q∗Fu-Q∗Fv=1Γqα∫ω0TT-σq,ωsq,ωα-1_Hu-vsdq,ωs≤λ1u-v+λ2Ψr,ϕγu-Ψr,ϕγvT-ω0αΓqα+1.Finally, for each t∈Iq,ωT and u,v∈X, (53)Fut-Fvt≤λ1u-v+λ2Ψr,ϕγu-Ψr,ϕγvt-ω0α-1T-ω0αΩ×pα+βΓq-βT-ω0α-β-2Γp-βΓqα-β+1-Λ2Γqα+1+t-ω0α-2T-ω0αΩ×pα+βΓq-βT-ω0α-β-1Γp-βΓqα-β+1-Λ1Γqα+1+t-ω0αΓqα+1≤λ1+λ2φ0T-ω0γΓrγ+1Θu-vX=LΘu-vX,where L and Θ are defined on (49)-(50), respectively.
(
H
2
) implies that F is a contraction. Therefore, by using Banach fixed-point theorem, F has a fixed point which is a unique solution of problem (4).We next show the existence of a solution to (4) by the following Schauder’s fixed-point theorem.Theorem 18.
Suppose that(H1) holds. Then, problem (4) has at least one solution.Proof.
We organize the proof into three steps.
Step  1. Verify F map bounded sets into bounded sets in BR=u∈X:uX≤R. We set maxt∈Iq,ωT⁡Ft,0,0=K and choose a constant(54)R≥KΘ1-LΘ,where L and Θ are defined on (49)-(50), respectively. Denote that (55)St,u,0=Ft,ut,Ψr,ϕγut-Ft,0,0+Ft,0,0.For each t∈Iq,ωT and u∈BR, we obtain(56)P∗Fu=1ΓqαΓp-β∫ω0pT+θ∫ω0xpT+θ-σp,θxp,θ-β-1_×x-σq,ωsq,ωα-1_St,u,0dq,ωsdp,θx≤λ1+λ2φ0Γrγ+1T-ω0r,ϕγ+1_uX+KΓq-βpT-ω0α-βΓp-βΓqα-β+1,Q∗Fu=1Γqα∫ω0TT-σq,ωsq,ωα-1_St,u,0dq,ωs≤λ1+λ2φ0Γrγ+1T-ω0r,ϕγ+1_uX+KT-ω0αΓqα+1.Similarly with Theorem 17, we obtain (57)Fut≤λ1+λ2φ0Γrγ+1T-ω0r,ϕγ+1_uX+KΘ=LuX+KΘ≤LR+KΘ≤R.Therefore FuX≤R. This implies that F is uniformly bounded.
Step  2. Show that F is continuous on BR.
Lettingϵ>0, there exists δ>0 such that, for each t∈Iq,ωT and for all u,v∈BR, (58)Hu-v<ϵΘwheneveru-v<δ.Then, we have(59)Fu-FvX≤Hu-vΘ<ϵ.This means that the operator F is continuous on BR.
Step  3. Examine that F is equicontinuous with BR. For any t1,t2∈Iq,ωT with t1<t2, we have(60)Fut2-Fut1≤t2-ω0α-1Ω-t1-ω0α-1ΩT-ω0α-2P∗Fu-Λ2Q∗Fu+t2-ω0α-2Ω-t1-ω0α-2ΩT-ω0α-1P∗Fu-Λ1Q∗Fu+FΓqα∫ω0t2t2-σq,ωsq,ωα-1_dq,ωs-∫ω0t1t1-σq,ωsq,ωα-1_dq,ωs.If |t2-t1|→0, the right-hand side of the above inequality tends to be zero. So F is relatively compact on BR.
This means that the setF(BR) is an equicontinuous set. As a consequence of Steps I to III together with the Arzelá-Ascoli theorem, we get that F:X→X is completely continuous. By Schauder’s fixed-point theorem, we can conclude that problem (4) has at least one solution. The proof is done.


## 4. Existence Results for Problem (5)

In this section, we present the existence results for problem (5). Let Y=CIq,ωT,R be a Banach space of all function u with the norm defined by(61)uY=maxt∈Iq,ωT⁡ut,DCr,ϕγurt+ϕ,where ω>0, p,q,r∈0,1, p=qm,r=qn,m,n∈N, θ=ω(1-p)/(1-q), ϕ=ω(1-r)/(1-q), and A,B:CIq,ωT,R→R. Define an operator G:Y→Y by(62)Gut≔Au+t-ω0Γqβ+2T-ω0β+1Bu-T-ω0βΓqβ+1Au-1ΓqαΓpβ×∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_Gt,ut,DCr,ϕγurt+ϕdq,ωsdp,θx+1Γqα∫ω0tt-σq,ωsq,ωα-1_Gt,ut,DCr,ϕγurt+ϕdq,ωs.Obviously, problem (5) has solutions if and only if the operator G has fixed points.Theorem 19.
Assume thatG:Iq,ωT×R×R→R is continuous and A,B:C(Iq,ωT,R)→R. In addition, suppose that the following conditions hold:
(
H
3
) There exist constants μ1,μ2>0 such that, for each t∈Iq,ωT and u,v∈R, (63)Gt,u,DCr,ϕγu-Gt,v,DCr,ϕγv≤μ1u-v+μ2DCr,ϕγu-DCr,ϕγv.(H4) There exist constants l1,l2>0 such that, for each u,v∈Y, (64)Au-Av≤l1u-vY,Bu-Bv≤l2u-vY.(H5)Φ<1, where(65)Φ≔1+β+1ql1+Γqβ+2T-ω0βl2+μ1+μ2T-ω0αΓqβΓqβ+2ΓpβΓqα+β+1+1Γqα+1.Then, problem (5) has a unique solution.Proof.
To show thatF is contraction, we consider (66)Ku-vt≔Gt,ut,DCr,ϕγurt+ϕ-Gt,vt,DCr,ϕγvrt+ϕ≤Gt,ut,DCr,ϕγurt+ϕ-Gt,vt,DCr,ϕγvrt+ϕ≕Ku-v,for each t∈Iq,ωT and u,v∈Y.
For eacht∈Iq,ωT and u,v,DCr,ϕγu,DCr,ϕγv∈Y, we find that(67)Gut-Gvt≤1+t-ω0β+1qT-ω0Au-Av+t-ω0Γqβ+2T-ω0β+1Bu-Bv+t-ω0Γqβ+2T-ω0β+1ΓqαΓpβ∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_Ku-vsdq,ωsdp,θx+1Γqα∫ω0tt-σq,ωsq,ωα-1_Ku-vsdq,ωs≤1+β+1ql1u-vY+Γqβ+2T-ω0βl2u-vY+μ1ut-vt+μ2Dr,ϕγut-Dr,ϕγvtΓqβ+2T-ω0βΓqαΓpβ×∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_dq,ωsdp,θx+1Γqα∫ω0TT-σq,ωsq,ωα-1_dq,ωs≤u-vY1+β+1ql1+Γqβ+2T-ω0βl2+u-vYμ1+μ2T-ω0αΓqβΓqβ+2ΓpβΓqα+β+1+1Γqα+1≤u-vYΦ,DCr,ϕγGurt+ϕ-DCr,ϕγGvrt+ϕ≤1+β+1qT-ω0DCr,ϕγrt-ω0Au-Av+Γqβ+2T-ω0β+1Bu-BvDCr,ϕγrt-ω0+Γqβ+2DCr,ϕγrt-ω0T-ω0β+1ΓqαΓpβ∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_Ku-vsdq,ωsdp,θx+1ΓqαDCr,ϕγ∫ω0rt+ϕrt+ϕ-σq,ωsq,ωα-1_Ku-vsdq,ωs≤1+r1-γβ+1qT-ω0-γΓr2-γl1u-vY+r1-γΓqβ+2T-ω0β-γΓr2-γl2u-vY+r1-γΓqβΓqβ+22T-ω0α+β-γΓqαΓpβ2Γqα+β+1Γr2-γ·u-vYμ1+μ2+rα-γΓrα+1T-ω0α-γ1-γrΓqα+1Γrα-γ+1·u-vYμ1+μ2<u-vYΦ.Hence, from (67), it is implied that (Gu)(t)-(Gv)(t)Y≤u-vYΦ.
By(H5), it is implied that G is a contraction. Therefore, by using Banach fixed-point theorem, G has a fixed point which is a unique solution of problem (5).We also deduce the existence of a solution to (5) by the following Schauder’s fixed-point theorem.Theorem 20.
Suppose that(H3)-(H4) hold. Then, problem (5) has at least one solution.Proof.
We divide the proof into three steps.
Step  1. Verify G map bounded sets into bounded sets in BL={u∈Y:uY≤L}. We set maxt∈Iq,ωT⁡Gt,0,0=K, supu∈Y⁡|A(u)|=M, and supu∈Y⁡|B(u)|=N and choose a constant(68)L≤1+β+1qM+Γqβ+2/T-ω0βN+Kμ1+μ2T-ω0αΓqβΓqβ+2/ΓpβΓqα+β+1+1/Γqα+11-Φ,where Φ are defined on (65). Denote that (69)Tt,u,0=Gt,ut,DCr,ϕγurt+ϕ-Gt,0,0+Gt,0,0.For each t∈Iq,ωT and u∈BL, we obtain(70)Gut≤1+t-ω0β+1qT-ω0Au+t-ω0Γqβ+2T-ω0β+1Bu+t-ω0Γqβ+2T-ω0β+1ΓqαΓpβ∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_Tu-vsdq,ωsdp,θx+1Γqα∫ω0tt-σq,ωsq,ωα-1_Tu-vsdq,ωs≤1+β+1ql1uY+M+Γqβ+2T-ω0βl2uY+N+μ1ut+μ2Dr,ϕγut+KΓqβ+2T-ω0βΓqαΓpβ×∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_dq,ωsdp,θx+1Γqα∫ω0TT-σq,ωsq,ωα-1_dq,ωs≤uY1+β+1ql1+Γqβ+2T-ω0βl2+1+β+1qM+Γqβ+2T-ω0βN+uY+Kμ1+μ2T-ω0αΓqβΓqβ+2ΓpβΓqα+β+1+1Γqα+1≤LΦ+1+β+1qM+Γqβ+2T-ω0βN+Kμ1+μ2T-ω0αΓqβΓqβ+2ΓpβΓqα+β+1+1Γqα+1≤L.Similar to Theorem 19, we obtain (71)DCr,ϕγGurt+ϕ<L.So, we have GuY≤L. This implies that G is uniformly bounded.
Step  2. Show that G is continuous on BL.
Lettingϵ>0, there exist δ=max⁡{δ1,δ2,δ3}>0 such that, for each t∈Iq,ωT and for all u,v∈BL,(72)Ku-v<ϵ3T-ω0αΓqβΓqβ+2/ΓpβΓqα+β+1+1/Γqα+1wheneveru-v<δ1,Au-Av<ϵ31+β+1qwheneveru-v<δ2,Bu-Bv<ϵT-ω0β3Γqβ+2wheneveru-v<δ3.Then, we have (73)Gu-Gv≤1+β+1qAu-Av+Γqβ+2T-ω0βBu-Bv+T-ω0αΓqβΓqβ+2ΓpβΓqα+β+1+1Γqα+1Ku-v<ϵ3+ϵ3+ϵ3=ϵ.By a similar proof to the above, we obtain (74)DCr,ϕγGu-DCr,ϕγGv<ϵ.Hence, Gu-GvY<ϵ. This means that the operator G is continuous on BL.
Step  3. Examine that G is equicontinuous with BL. For any t1,t2∈Iq,ωT with t1<t2,(75)Gut2-Gut1≤t2-t1Γqβ+2T-ω0β+1Bu-T-ω0βΓqβ+1Au-GΓqαΓpβ×∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_dq,ωsdp,θx+GΓqα∫ω0t2t2-σq,ωsq,ωα-1_dq,ωs-∫ω0t1t1-σq,ωsq,ωα-1_dq,ωs,DCr,ϕγGurt2+ϕ-DCr,ϕγGurt1+ϕ≤DCr,ϕγt2-t1Γqβ+2T-ω0β+1Bu-T-ω0βΓqβ+1Au-GΓqαΓpβ×∫ω0T∫ω0xT-σp,θxp,θβ-1_x-σq,ωsq,ωα-1_dq,ωsdp,θx+GΓqαDCr,ϕγ∫ω0t2t2-σq,ωsq,ωα-1_dq,ωs-∫ω0t1t1-σq,ωsq,ωα-1_dq,ωs.If |t2-t1|→0, the right-hand side of the above inequalities (75) tends to be zero. So G is relatively compact on BL.
Therefore the setG(BL) is an equicontinuous set. As a consequence of Steps I to III together with the Arzelá-Ascoli theorem, we get that G:Y→Y is completely continuous. By Schauder’s fixed-point theorem, we can conclude that problem (5) has at least one solution. The proof is completed.


## 5. Some Examples

In this section, we provide some examples to illustrate our results.Example 1.
Consider the following fractional Hahn boundary value problem:(76)D1/2,24/3ut=e-2tut+23u+1+arctan⁡sin2⁡πt10etΓ1/81/3∫4tt-σ1/8,7/2s1/8,7/2-2/3_×e-3s-t100+cos2⁡πtusd1/8,7/2s,t∈4,101/2,2,u4=u10,D1/4,31/2u4=D1/4,31/2u112.
Hereα=4/3, β=1/2, γ=1/3, q=1/2, p=1/4, r=1/8, ω=2, θ=3, ϕ=7/2, ω0=ω/(1-q)=4, T=10, φ(t,s)=e-3s-t/(100+cos2⁡πt), φ0=1/200, and Ft,u,Ψ1/8,7/21/3u=e-2tu/t+23(u+1)+(arctan⁡(sin2⁡πt)/10etΓ1/81/3)∫4tt-σ1/8,7/2s1/8,7/2-2/3_φ(t,s)u(s)d1/8,7/2s.
Since(77)Ft,u,Ψ1/8,7/21/3u-Ft,v,Ψ1/8,7/21/3v≤1216e8x-y+π400e4Ψ1/8,7/21/3u-Ψ1/8,7/21/3v,then H1 is satisfied with λ1=1/216e8 and λ2=π/400e4.
Also, we have(78)Λ1=0.00402,Λ2=-0.1387,Ω=0.00184.We can show that (79)L=0.00002858,Θ=64.270.Therefore, we find that (80)LΘ≈0.00184<1.Hence, by Theorem 17, problem (76) has a unique solution on 4,101/2,2.Example 2.
Consider the following fractional Hahn boundary value problem:(81)DC1/2,24/3ut=e-sin2⁡πtu100+ecos2πt/2·utt-2+DC1/8,7/21/3ut1+ut,t∈4,101/2,2,u4=u125e3cos2⁡πu,I1/4,31/2u10=u100π2sin2⁡πu.
Hereα=4/3, β=1/2, γ=1/3, q=1/2, p=1/4, r=1/8, ω=2, θ=3, ϕ=7/2, ω0=ω/(1-q)=4, T=10, A(u)=(|u|/125e3)cos2⁡πu, B(u)=(|u|/100π2)sin2⁡πu, and Ft,u,DC1/8,7/21/3u=(e-sin2⁡πtu/100+ecos2πt/2)·((|u(t)|t-2+DC1/8,7/21/3u(t))/(1+|u(t)|)).
Since(82)Ft,u,DC1/8,7/21/3u-Ft,v,DC1/8,7/21/3v≤11616u-v+1101DC1/8,7/21/3u-DC1/8,7/21/3v,(83)Au-Av≤1125e3u-vY,(84)Bu-Bv≤1100π2u-vY,then (H3) and (H4) are satisfied with μ1=1/1616 and μ2=1/101 and l1=1/125e3 and l2=1/100π2.
Therefore, we get(85)Φ≈0.2135<1.Hence, by Theorem 19, problem (81) has a unique solution on [4,10]1/2,2.



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*Source: 7895186-2017-12-17.xml*