# Nonlinear Fractionalq-Difference Equation with Fractional Hadamard and Quantum Integral Nonlocal Conditions

Authors: Athasit Wongcharoen; Arisa Thatsatian; Sotiris K. Ntouyas; Jessada Tariboon
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we establish existence and uniqueness results for a boundary value problem consisting by a nonlinear fractionalq-difference equation subject to a new type of boundary condition, combining the fractional Hadamard and quantum integrals. Our analysis is based on Banach’s fixed point theorem, a fixed point theorem for nonlinear contractions, Krasnosel’ski i˘’s fixed point theorem, and Leray-Schauder nonlinear alternative. Examples are given to illustrate our results.


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## Body

## 1. Introduction

The aim of this paper is to investigate the existence and uniqueness of solutions for a nonlinear fractionalq-difference equation subject to fractional Hadamard and quantum integral condition of the form:
(1)Dqαxt=ft,xt,1<α≤2,t∈0,T,x0=0,∑i=1nγiIpiμixξi=∑j=1mβjJσjxηj,where Dqα is the fractional q-derivative of order α, with a quantum number q∈0,1, f:0,T×ℝ⟶ℝ is a nonlinear continuous function, Ipiμi denotes the fractional quantum integral of order μi>0, with quantum number 0<pi<1, Jσj is the Hadamard fractional integral of order σj>0, γi and βj are given constants, and ξi,ηj∈0,T are fixed points, for i=1,⋯,n and j=1,⋯,m.The subject of fractional differential equations has recently evolved into an interesting subject for many researchers due to its multiple applications in economics, engineering, physics, chemistry, signal analysis, etc. Various types of fractional derivative and integral operator were studied: Riemann-Liouville, conformable fractional integral operators, Caputo, Hadamard, Erdelyi-Kober, Grünwald-Letnikov, Marchaud, and Riesz are just a few to name. The Hadamard-type fractional derivative differs from the preceding ones in the sense that the kernel of the integral and derivative contain logarithmic function of arbitrary exponent. Details and properties of Hadamard fractional derivatives and integrals can be found in Kilbas et al. [1]. Recently, there were some results on Hadamard-type fractional differential equations, see [2–11] and references cited therein.Nonlinear fractionalq-difference equations appear in the mathematical modeling of many phenomena in engineering and science and have attracted much attention by many researchers, see for example [12–21] and references therein.In the present paper, the novelty lies in the fact that we combine in boundary conditions both Hadamard and quantum integrals. To the best of our knowledge, this type of boundary condition appears for the first time in the literature. It is important to notice that we are combining in our work, fractional calculus, and quantum calculus. The key tool for this combination is the Property 2.25 of [1].Some special cases of the second condition of (1) can be seen by reducing m=n=1 as
(2)γ1Γp1μ1∫0+ξ1ξ1−p1sp1μ1−1xsdp1s=β1Γσ1∫0+η1logη1sσ1−1xssds,which is mixed quantum and Hadamard calculus. If p1=1, then we have
(3)γ1Γμ1∫0+ξ1ξ1−sμ1−1xsds=β1Γσ1∫0+η1logη1sσ1−1xssds,which is also mixed Riemann-Liouville and Hadamard fractional integral condition. If μ1=σ1=1, we have integral condition of the form:
(4)γ1∫0+ξ1xsds=β1∫0+η1xssds,which is a variety used in physical boundary value problems.We establish existence and uniqueness results by using standard fixed point theorems. We prove two existence and uniqueness results with the help of the Banach contraction mapping principle and a fixed point theorem on nonlinear contractions due to Boyd and Wong. Moreover, we prove two existence results, one via Leray-Schauder nonlinear alternative and another one via Krasnosel’skii˘’s fixed point theorem.The paper is organized as follows: in Section2, we recall some preliminary facts that we need in the sequel. In Section 3, we prove our main results. Some examples to illustrate our results are presented in Section 4.


## 2. Preliminaries

To present the preliminary, we suggest the basic quantum calculus in the book of Kac and Cheung [22], fractional quantum calculus in [23–25], and the Hadamard fractional calculus in [1]. Let a fixed constant q∈0,1 be a quantum number. The q-number is defined by
(5)aq=1−qa1−q,a∈ℝ.For example,3q=1+q+q2. The q-power function for any a,b∈ℝ, a≠0, is defined as
(6)a−bqγ=aγ∏i=0∞1−b/aqi1−b/aqγ+i.Ifγ=k∈ℕ0=0,1,2,⋯, then a−bqk=∏i=0k−1a−bqi and a−bq0≔1. For example, a−bq3=a−ba−qba−q2b. The notation of q-power function is appeared in kernels of fractional q-calculus as Definitions 1 and 2. Now, the q-gamma function Γqt is defined by
(7)Γqt=1−qqt−11−qt−1,fort∈ℝ\0,−1,−2,⋯.Now, we observe thatΓqt+1=tqΓqt. Next, we discuss about the q-derivative of a function f:0,∞⟶ℝ which is defined by
(8)Dqft=ft−fqt1−qt,t≠0,andDqf0=limt→0Dqft.Iff′t exists, then limq→1Dqft=f′t. The q-integral formula can be presented as
(9)Iqft=∫0tfsdqs=t1−q∑n=0∞qnftqn,t∈0,∞.The higher order ofq-derivative and q-integral operators is
(10)Dqkht=DqDqk−1ftandIqkht=IqIqk−1ft,k∈ℕ,with Dq0ft=ft and Iq0ft=ft. Next, the fundamental theorem of calculus for operators Dq and Iq can be stated as formulas
(11)DqIqft=ft,and if f is continuous at the point t=0, then
(12)IqDqft=ft−f0.Let us give the definitions of fractional quantum calculus of the Riemann-Liouville type fractional derivative and also integral operators.Definition 1 [24].
Let a constantα≥0 and f be the function on 0,∞. The Riemann-Liouville fractional q-integral of f order α is defined by
(13)Iqαft=1Γqα∫0+tt−qsqα−1fsdqs,α>0,t∈0,∞,and Iq0ft=ft.Definition 2 [24].
The Riemann-Liouville fractionalq-derivative of order α≥0 of a function f:0,∞⟶ℝ is given by
(14)Dqαft=DqnIqn−αft=1Γqn−αDqn∫0+tt−qsqn−α−1fsdqs,α>0,and Dq0ft=ft, where n is the smallest integer greater than or equal to α.Now, fort,s>0, the q-beta function is presented by
(15)Bqt,s=∫0+1ut−11−quqs−1dqu,which is related to the q-gamma function by
(16)Bqt,s=ΓqtΓqsΓqt+s.The fundamental formulas for fractional quantum calculus are in the following lemma.Lemma 3 [24, 26].
Letα,β≥0, n be a positive integer and f be a function defined in 0,∞. Then, the following formulas hold
(17)IqβIqαft=Iqα+βft,DqαIqαft=ft,IqβDqnft=DqnIqβft−∑k=0n−1tβ−n+kΓqβ+k−n+1Dqkf0.The fractionalq-integration of the two deferent quantum numbers is given by lemma.Lemma 4 [27].
Let constantsα,β>0 and 0<p,q<1 be quantum numbers. Then, for η∈ℝ+, we have
(18)IpαIqβ1η=Γpβ+1Γpα+β+1Γqβ+1ηα+β.The Hadamard fractional calculus is the subject of fractional derivative and integral which have logarithm kernels inside the singular integral formulas as in the definitions.Definition 5 [1].
The Hadamard derivative of fractional orderα for a function f:0,∞⟶ℝ is defined as
(19)HDαft=1Γn−αtddtn∫0+tlogtsn−α−1fssds,n=α+1,where the notation α denotes the integer part of the real number α, log·=loge·, and Γ is the usual Gamma function.Definition 6 [1].
The Hadamard fractional integral of orderα for a function f:0,∞⟶ℝ is defined by
(20)Jαft=1Γα∫0+tlogtsα−1fssds,α>0,provided the integral in right hand side exists.The key tool for combining the two type of fractional calculus in our work is the following lemma.Lemma 7 ([1], Property 2.25).
Letα>0 and β>0. The following formulas hold
(21)Jαtβ=β−αtβandHDαtβ=βαtβ.To accomplish our main purpose, we will use the fixed point theory for considering an operator equationx=Qx. For finding the operator Q, let us see the following lemma.Lemma 8.
Suppose that the pointsξi, ηj∈0,T and the constant
(22)Ω=∑i=1nγiΓpiαξiα+μi−1Γpiα+μi−∑j=1mβjα−1−σjηjα−1≠0,where α, μi, pi, γi, σj, βj, i=1,⋯,n, and j=1,⋯,m are defined in problem (1). Then, the linear fractional q-difference equation
(23)Dqαxt=ht,0<t<T,where h:0,T⟶ℝ, and subject to mixed fractional integrals of Hadamard and quantum boundary conditions
(24)x0=0,∑i=1nγiIpiμixξi=∑j=1mβjJσjxηjis equivalent to the linear integral equation
(25)xt=1Ω∑j=1mtα−1βjJσjIqαhηj−∑i=1ntα−1γiIpiμiIqαhξi+Iqαht.Proof.
Sinceα∈1,2, then (23) can be written as
(26)Dq2Iq2−αxt=ht,0<t<T.
Applying the fractionalq-integral of order α and using Lemma 3, we obtain
(27)IqαDq2Iq2−αxt=Dq2IqαIq2−αxt−k1tα−1−k2tα−2=xt−k1tα−1−k2tα−2=Iqαht,which yields
(28)xt=k1tα−1+k2tα−2+Iqαht,where k1,k2∈ℝ. The first boundary condition of (24) implies that k2=0. Then, (28) is reduced to
(29)xt=k1tα−1+Iqαht.
Now, we apply the fractional quantum integral of Riemann-Liouville of orderμi with quantum number pi to (29) as
(30)Ipiμixt=k1Γpiαtα+μi−1Γpiα+μi+IpiμiIqαht.
Using Lemma7 for taking the Hadamard fractional integral of order σj to (29), we get
(31)Jσjxt=k1α−1−σjtα−1+JσjIqαht.
From the second boundary condition of (24) and above two equations, it follows that
(32)k1∑i=1nγiΓpiαξiα+μi−1Γpiα+μi+∑i=1nγiIpiμiIqαhξi=k1∑j=1mβjα−1−σjηjα−1+∑j=1mβjJσjIqαhηj,and consequently
(33)k1=1Ω∑j=1mβjJσjIqαhηj−∑i=1nγiIpiμiIqαhξi,where the nonzero constant Ω is defined by (22). Substituting the constant k1 in (29), then, we obtain (25), which is the solution of BVP (23) and (24). The converse can be obtained by a direct computation. The proof is completed.


## 3. Main Results

At first, we denote byC=C0,T,ℝ the Banach space of all continuous functions from 0,T to ℝ endowed with the sup norm as x=supxt,t∈0,T. In view of Lemma 8 and replacing the function h by ft,xt, we define the operator Q:C⟶C by
(34)Qxt=tα−1Ω∑j=1mβjJσjIqαfxηj−∑i=1nγiIpiμiIqαfxξi+Iqαfxt,where Iqαfxv is denoted by
(35)Iqαfxv=1Γqα∫0+vv−qsqα−1fs,xsdqs≔gv,v∈t,ξi,ηj,while JσjIqαfxηj and IpiμiIqαfxξi are the Hadamard and quantum fractional integrals of a function g as
(36)JσjIqαfxηj=1Γσj∫0+ηjlogηjsσj−1gssds,IpiμiIqαfxξi=1Γpiμi∫0+ξiξi−pispiμi−1gsdpis,respectively. Now, we are going to prove the main results which are the existence criteria of solution for nonlocal mixed fractional integrals boundary value problem (1). The first, an existence and uniqueness result for (1), is given by using Banach’s fixed point theorem.Theorem 9.
Letf:0,T×ℝ⟶ℝ be a nonlinear continuous function satisfying the assumption.H1 There exists a positive constant L such that ∣ft,x−ft,y∣≤Lx−y, for each t∈0,T and x,y∈ℝ.If(37)LΦ<1,where Φ is given by
(38)Φ=TαΓqα+11TΩ∑j=1mβjα−σjηjα+1TΩ∑i=1nγiΓpiα+1Γpiα+μi+1ξiα+μi+1,then the boundary value problem (1) has a unique solution on 0,T.Proof.
The result allows from the operator equationx=Qx, where the operator Q is defined by (34). The Banach fixed point theorem is used to show that Q has a fixed point which is the unique solution of problem (1). Since the function f is continuous, then, we can set supft,0,t∈0,T=M<∞. After that, we define the radius r satisfying
(39)r≥ΦM1−ΦLof a ball Br=x∈C:x≤r. For any x∈Br, we see that
(40)Qxt≤supt∈0,Ttα−1Ω∑j=1mβjJσjIqαfxηj+tα−1Ω∑i=1nγiIpiμiIqαfxξi+Iqαfxt≤Tα−1Ω∑j=1mβjJσjIqαfx−f0+f0ηj+Iqαfx−f0+f0T+Tα−1∣Ω∣∑i=1nγiIpiμiIqαfx−f0+f0ξi≤Lr+MTα−1Ω∑j=1mβjJσjIqα1ηj+Tα−1Ω∑i=1nγiIpiμiIqα1ξi+Iqα1T,in which we used the following fact:
(41)fx−f0+f0=fv,xv−fv,0+fv,0≤Lx+M≤Lr+M,where v∈T,ξi,ηj. By applying Lemmas 4 and 2.3, we have
(42)IpiμiIqα1ξi=Γpiα+1Γpiα+μi+1Γqα+1ξiα+μi,JσjIqα1ηj=1Γqα+1Jσjtαηj=1Γqα+1α−σjηjα.
Then, we obtain(43)Qxt≤Lr+MTαΓqα+11T∣Ω∣∑j=1m∣βj∣α−σjηjα+1T∣Ω∣∑i=1n∣γi∣Γpiα+1Γpiα+μi+1ξiα+μi+1=Lr+MΦ≤r.
From this, we conclude that∥Qx∥≤r which yields QBr⊂Br.
Next, we will prove that the operatorQ is a contraction. Let x,y∈C, and for each t∈0,T, then, we have
(44)Qxt−Qyt≤Tα−1∣Ω∣∑j=1m∣βj∣JσjIqα∣fx−fy∣ηj+Tα−1∣Ω∣∑i=1n∣γi∣IpiμiIqα∣fx−fy∣ξi+Iqα∣fx−fy∣t≤Tα−1∣Ω∣∑j=1m∣βj∣JσjIqα1ηj+Tα−1∣Ω∣∑i=1n∣γi∣IpiμiIqα1ξi+Iqα1TL∥x−y∥=LΦ∥x−y∥.
Hence, we get the result that∥Qx−Qy∥≤LΦ∥x−y∥. As LΦ<1, from (37), the operator Q is a contraction. Applying the well known Banach fixed point theorem, it follows that Q has a fixed point which is the unique solution of the boundary value problem (1). This completes the proof.Next, the nonlinear contraction theorem will be used to prove a second existence and uniqueness result.Definition 10.
LetE be a Banach space and let A:E⟶E be a mapping. The operator A is said to be a nonlinear contraction if there exists a continuous nondecreasing function Ψ:ℝ+⟶ℝ+ such that Ψ0=0 and Ψt<t for all t>0 with the property:
(45)Ax−Ay≤Ψx−y,∀x,y∈E.Lemma 11 (see [28]).
LetE be a Banach space and let A:E⟶E be a nonlinear contraction. Then, A has a unique fixed point in E.Theorem 12.
Suppose that a continuous functionf:0,T×ℝ⟶ℝ satisfies the condition:
H2ft,x−ft,y≤htx−y/H∗+x−y, t∈0,T, x,y∈ℝ, where the function h:0,T⟶ℝ+ is continuous, and a positive constant H∗ is defined by
(46)H∗=Tα−1Ω∑j=1mβjJσjIqαhηj+Tα−1Ω∑i=1nγiIpiμiIqαhξi+IqαhT.Then, the mixed fractional Hadamard and quantum integrals nonlocal problem (1) has a unique solution on 0,T.Proof.
Let us consider the operatorQ:C⟶C defined in (34) and define a continuous nondecreasing function Ψ:ℝ+⟶ℝ+ by
(47)Ψλ=H∗λH∗+λ,∀λ≥0.
Then, we see that the functionΨ satisfies Ψ0=0 and Ψλ<λ for all λ>0.
Next, for anyx,y∈C and for each t∈0,T, we obtain
(48)Qxt−Qyt≤tα−1Ω∑j=1mβjJσjIqαfx−fyηj+tα−1Ω∑i=1nγiIpiμiIqαfx−fyξi+Iqαfx−fyt≤Tα−1Ω∑j=1mβjJσjIqαhx−yH∗+x−yηj+Tα−1Ω∑i=1nγiIpiμiIqαhx−yH∗+x−yξi+Iqαhx−yH∗+x−yT≤Ψx−yH∗Tα−1Ω∑j=1mβjJσjIqαhηj+Tα−1∣Ω∣∑i=1nγiIpiμiIqαhξi+IqαhT=Ψx−y,which implies that Qx−Qy≤Ψx−y and also satisfies Definition 10. Therefore, Q is a nonlinear contraction. Thus, by applying Lemma 11, the operator Q has a unique fixed point which is the unique solution of the boundary value problem (1). The proof is finished.Next, the first existence result will be obtained by applying the following theorem.Theorem 13 (Nonlinear alternative for single valued maps) [29].
LetE be a Banach space, C a closed, convex subset of E,U be an open subset of C, and 0∈U. Suppose that A:U¯⟶C is a continuous, compact (that is, AU¯ is a relatively compact subset ofC) map. Then, either
(i)
A has a fixed point in U¯, or(ii)
There is ax∈∂U (the boundary of U in C) and λ∈0,1 with x=λAx.Theorem 14.
Suppose thatf:0,T×ℝ⟶ℝ is a nonlinear continuous function which satisfies the following conditions:
H3 there exists a continuous nondecreasing function ψ:0,∞⟶0,∞ and also a function p∈C0,T,ℝ+ such that
(49)ft,x≤ptψxforeacht,x∈0,T×ℝ
H4 there exists a positive constant N such that
(50)NψNpΦ>1,where Φ defined by (38). Then, the problem (1) has at least one solution on 0,T.Proof.
For a positive numberρ, we let Bρ=x∈C:x≤ρ be a bounded ball in C. Now, we will prove that the set QBρ is uniformly bounded. For t∈0,T, we can compute that
(51)Qxt≤Tα−1Ω∑j=1mβjJσjIqαfxηj+∑i=1nγiIpiμiIqα∣fx∣ξi+Iqα∣fx∣T≤pψxTαΓqα+11TΩ∑j=1mβjα−σjηj+1TΩ∑i=1nγiΓpiα+1Γpiα+μi+1ξiα+μi+1≤pψρΦ,which can be deduced that
(52)Qx≤pψρΦ.
Then, the setQBρ is uniformly bounded. Next, we will show that the set QBρ is equicontinuous set of C. For any two points τ1,τ2∈0,T with τ1<τ2 and x∈Bρ, we have
(53)Qxτ2−Qxτ1≤τ2α−1−τ1α−1∣Ω∣∑j=1mβjJσjIqαfxηj+∑i=1nγiIpiμiIqαfxξi+1Γqα∫0τ1t2−qsqα−1−t1−sqα−1fs,xsdqs+1Γqα∫τ1τ2t2−qsqα−1fs,xsdqs≤pψρτ2α−1−τ1α−1∣Ω∣∑j=1m∣βj∣α−σjηjαΓqα+1+pψρτ2α−1−τ1α−1∣Ω∣∑i=1n∣γi∣Γpiα+1Γpiα+μi+1Γqα+1ξiα+μi+pψρΓqα+12τ2−τ1α+τ2α−τ1α.
Asτ2−τ1⟶0, the right hand side of the above inequality converses to zero, independently of x∈Bρ. Then, the set QBρ is equicontinuous. Thus, we conclude that the set QBρ is relatively compact. Therefore, by the Arzel a′-Ascoli theorem, the operator Q:C⟶C is completely continuous.
Finally, we show that the operatorQ cannot be fulfilled the condition ii in Theorem 13. Then, we have to claim that there exists an open set U⊂Bρ with x≠λQx for λ∈0,1 and x∈∂U. Then, for each t∈0,T, we apply the computation in the first step, that is
(54)xt≤pψxΦwhich yields inequality
(55)xpψxΦ≤1.
The conditionH4 implies that there exists a constant N such that x≠N. Now, we define the set
(56)U=x∈Bρ:x<N.
From the previous results, we obtain that the operatorQ~U¯⟶C is continuous and completely continuous. Then, there is no x∈∂U such that x=λQx for some λ∈0,1. By applying the nonlinear alternative of the Leray-Schauder type, we get that the operator Q has a fixed point x∈U¯ which is a solution of the nonlinear fractional q-difference equation with fractional Hadamard and quantum integral nonlocal conditions. This finishes the proof.The next existence result is based on Krasnosel’skii˘’s fixed point theorem which can be used to relax the condition in Theorem 9.Theorem 15 (Krasnosel’skii˘’s fixed point theorem) [30].
LetC be a closed, bounded, convex, and nonempty subset of a Banach space E. Let A,B be the operators such that (a) Ax+By∈C whenever x,y∈C; (b) A is compact and continuous; (c) B is a contraction mapping. Then, there exists z∈C such that z=Az+Bz.Theorem 16.
Assume that a continuous functionf:0,T×ℝ⟶ℝ is satisfied condition H1 in Theorem 9 and is bounded as the following condition:
(i)
H5ft,x≤κt, ∀t,x∈0,T×ℝ, and κ∈C0,T,ℝ+.If inequality(57)LTαΓqα+1<1holds, then the nonlocal problem (1) has at least one solution on 0,T.Proof.
Now, we definesup∣κt∣:t∈0,T=κ and choose a positive constant r¯ such that
(58)r¯≥κΦ,where Φ is defined by (38), to be a radius of the ball Br¯=x∈C:x≤r¯. Furthermore, we set the operators Q1 and Q2 on Br¯ as A and B in Theorem 15, respectively, by
(59)Q1xt=tα−1Ω∑j=1mβjJσjIqαfxηj−∑i=1nγiIpiμiIqαfxξi,t∈0,T,Q2xt=Iqαfxtt∈0,T.
The combination of two operators showsQ1+Q2=Q. We have
(60)Q1x+Q2y≤κTα−1Ω∑j=1mβjα−σjηjαΓqα+1+Tα−1Ω∑i=1nγiΓpiα+1Γpiα+μi+1Γqα+1ξiα+μi+TαΓqα+1=κΦ≤r¯.
Therefore, we haveQ1x+Q2y∈Br¯, and thus condition a of Theorem 15 is satisfied. Since the function f is fulfilled by condition H1 in Theorem 9, then the operator Q2 is a contraction mapping with inequality (57).
Finally, we will show that the operatorQ1 should satisfy condition b in Theorem 15. Using the continuity of f, we can show that the operator Q1 is continuous. The uniformly boundedness of the set Q1Br¯ can be shown by
(61)Q1x≤κTα−1∣Ω∣Γqα+1∑j=1mβjα−σjηjα+∑i=1nγiΓpiα+1Γpiα+μi+1ξiα+μi.
To proveQ1Br¯ is equicontinuous set, we let two points t1,t2∈0,T, t2<t1. For any x∈Br¯, we have
(62)Q1xt1−Q1xt2≤κt1α−1−t2α−1∣Ω∣∑j=1mβjα−σjηjαΓqα+1+κt1α−1−t2α−1Ω∑i=1nγiΓpiα+1Γpiα+μi+1Γqα+1ξiα+μi,which converses to zero independently of x as t1−t2⟶0. So, Q1Br¯ is an equicontinuous set. Therefore, Q1Br¯ is a relative compact and by the Arzelá-Ascoli theorem, Q1 is compact on Br¯. Thus, the assumptions a,b, and c of Krasnosel’ski i˘’s fixed point theorem are satisfied. Then, the nonlinear fractional q-difference equation with fractional Hadamard and quantum integral nonlocal conditions (1) has at least one solution on 0,T. The proof is completed.Remark 17.
The interchanging of operatorsQ1 and Q2 gives another result by replacing inequality (57) by the following condition:
(63)LTα−1ΩΓqα+1∑j=1mβjα−σjηjα+∑i=1nγiΓpiα+1Γpiα+μi+1ξiα+μi<1.


## 4. Examples

Example 18.
Consider the nonlinear fractionalq-difference equation with fractional Hadamard and quantum integral nonlocal conditions of the form:
(64)D1/23/2xt=ft,xt,t∈0,2,x0=0,38I1/61/2x14+25I1/33/2x12+19I1/25/2x32=13J1/3x15+49J2/3x35+712J4/3x75+815J5/3x95.Here,α=3/2, q=1/2, T=2, γ1=3/8, γ2=2/5, γ3=1/9, μ1=1/2, μ2=3/2, μ3=5/2, p1=1/6, p2=1/3, p3=1/2, ξ1=1/4, ξ2=1/2, ξ3=3/2, n=3, β1=1/3, β2=4/9, β3=7/12, β4=8/15, σ1=1/3, σ2=2/3, σ3=4/3, σ4=5/3, η1=1/5, η2=3/5, η3=7/5, η4=9/5, m=4. Then, we can compute constants as Ω≈2.51852 and Φ≈3.27524.i Let the nonlinear function f be defined by
(65)ft,x=e−cos2tt+23x2+2∣x∣1+∣x∣+t24+1.Then, by direct computation, we getft,x−ft,y≤1/4x−y, which satisfies condition H1 in Theorem 9 with L=1/4. Therefore, we have
(66)LΦ≈0.81881<1.By the conclusion of Theorem9, the boundary value problem (64) with (65) has a unique solution on 0,2.ii Consider now the function f by
(67)ft,x=1t+23x18x16+1+1.Then, we can see that(68)∣ft,x∣=1t+23x18x16+1+1≤1t+23x2+1.Settingpt=1/t+23 and ψx=x2+1, we have p=1/8, and there exists a constant N∈0.52019,1.92238 satisfying inequality in H4. Hence, all assumptions in Theorem 14 are completed. Thus, the problem (64) with (67) has at least one solution on 0,2.iii If the function f is
(69)ft,x=sin2tmxx+1+14,m∈ℝ+,then, we have ft,x−ft,y≤1/mx−y with L=1/m. If m≤Φ≈3.27524, then Theorem 9 cannot be used to apply for the problem (64) with (69). For example, if m=2, then LΦ≈1.63762>1. But the inequality in Remark 17 is satisfied as
(70)12·Tα−1ΩΓqα+1∑j=1mβjα−σjηjα+∑i=1nγiΓpiα+1Γpiα+μi+1ξiα+μi=0.44979<1.Hence, by applying Theorem16 and Remark 17, the problem (64) with (69) has at least one solution on 0,2.


## 5. Conclusion

We investigated the existence and uniqueness of solutions for a nonlocal boundary value problem involving aq-difference equation, supplemented with a new type of boundary condition, including both Hadamard fractional and quantum integrals. In our first two results, we establish the existence and uniqueness of solutions by using Banach’s fixed point theorem and a fixed point theorem for nonlinear contractions due to Boyd and Wong. Then, we used the Leray-Schauder nonlinear alternative and Krasnosel’ski i˘’s fixed point theorem to derive two existence results. Examples are also presented to illustrate our results. It is worthwhile to point out that the results presented in this paper are new and significantly contribute to the existing literature on the topic.



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