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R E S E A R C H

Open Access

Impulsive fractional quantum Hahn

difference boundary value problems

Jessada Tariboon

1*

, Sotiris K. Ntouyas

2,3

and Bundit Sutthasin

1

*Correspondence:

jessada.t@sci.kmutnb.ac.th 1Intelligent and Nonlinear Dynamic

Innovations Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok, Thailand Full list of author information is available at the end of the article

Abstract

In this paper, we study the existence and uniqueness of solutions for two classes of boundary value problems for impulsive Caputo type fractional Hahn difference equations, by using the Banach contraction mapping principle and the nonlinear alternative of Leray–Schauder. The obtained results are well illustrated by examples.

MSC: Primary 39A10; 39A13; secondary 39A70

Keywords: Hahn difference equations; Boundary value problems; Existence; Uniqueness; Fixed point theorems

1 Introduction and preliminaries

Our purpose of this paper is to establish the existence and uniqueness results for two im-pulsive fractional Hanh difference boundary value problems. More precisely, we consider the first boundary value problem of orderνk, 0 <νk≤1,

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ c tkD

νk

qk,ωkx(t) =f(t,x(t)), tJk,k= 0, 1, 2, . . . ,m,

x(tk) =ϕk(x(tk)), k= 1, 2, . . . ,m, ξ1x(0) +ξ2x(T) =ξ3,

(1)

and the second boundary value problem of orderνk, 1 <νk≤2, ⎧

⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ c tkD

νk

qk,ωkx(t) =f(t,x(t)), tJk,k= 0, 1, 2, . . . ,m,

x(tk) =ϕk(x(tk)), k= 1, 2, . . . ,m, tkDqk,ωkx(t

+

k) –tk–1Dqk–1,ωk–1x(t

k) =ϕk∗(x(tk–)), k= 1, 2, . . . ,m,

x(0) =η1, tmDqm,ωmx(T) =η2,

(2)

wherec tkD

νk

qk,ωk is the fractional quantum Hahn difference operator of Caputo type, 0 <

qk< 1,ωk≥0,k= 0, 1, 2, . . . ,m,f :J×R→R,J= [0,T],ϕ,ϕk∗:R→R,k= 1, 2, . . . ,m, are given functions,x(tk) =x(tk) –x(tk–),tkDqk,ωk is the first order quantum Hahn difference

operator on intervalJk,k= 0, 1, 2, . . . ,m, and given constantsξ1,ξ2,ξ3,η1,η2∈R.

Theq-calculus appeared as a connection between mathematics and physics, especially, in elementary particle physics, which have used quantum numbers to present the discrete

(2)

values of energy levels in atoms. In 1910, Jackson [1] introduced the notion ofq-derivative as

Dqf(t) = ⎧ ⎨ ⎩

f(t)–f(qt) t(1–q) , t= 0,

f(0), t= 0, (3) provided thatf(0) exists. He also was the first to developq-calculus andq-difference equa-tions in a systematic way. The book by Kac and Cheung [2] covers many of the fundamental aspects of quantum calculus and alsoq-special functions. Theq-calculus has many ap-plications in mathematical areas such as orthogonal polynomials, basic hypergeometric functions, combinatorics, the calculus of variations, quantum mechanics, and the theory of relativity. Some recent results in quantum calculus can be found in [3–9] and the refer-ences cited therein.

Hahn [10] introduced his difference operatorDq,ωas

Dq,ωf(t) =

⎧ ⎨ ⎩

f(qt+ω)–f(t)

t(q–1)+ω , t=ω0,

f(ω0), t=ω0,

(4)

provided thatf is differentiable atω0, whereq∈(0, 1) andω≥0 are fixed. Heref is de-fined on an intervalI⊆Rcontainingω0:=ω/(1 –q). The Hahn difference operator unifies (in the limit) the two most well known and used quantum difference operators: the Jack-sonq-difference derivativeDq, whereq∈(0, 1), defined by (3), forω= 0, and the forward differenceforq→1, defined by

Dωf(t) =

f(t+ω) –f(t)

ω , (5)

where ω> 0 is a fixed constant. The Hahn difference operator is a successful tool for constructing families of orthogonal polynomials and investigating some approximation problems (cf. [11–14]). For some recent results on the boundary value problems of Hahn difference equations we refer to [15–19] and references therein.

Let us emphasize that the definition (3) does not remain valid for impulse points tk,

k∈Z, such thattk∈(qt,t). For example, let [0,T],T> 4 be a dense interval andt= 2 be an impulsive point, i.e.,f(2+)=f(2). Then we haveD

1/2f(4+)=D1/2f(4–), which implies thatD1/2f(4) does not exist. On the other hand, this situation does not arise for impul-sive equations onq-time scales{0, . . . ,q2t,qt,t}, as the domains consist of isolated points covering the case of consecutive points oftandqtwith impulsive pointstk∈/(qt,t). Due to this reason, the subject of impulsive quantum difference equations on dense domains could not be studied. In [20], the authors modified the classical quantum calculus on [a,b] by defining

aDqf(t) = ⎧ ⎨ ⎩

f(t)–f(qt+(1–q)a)

(1–q)(t–a) , t=a, limt→a aDqf(t), t=a.

(6)

(3)

se-ries of impulsive quantum initial and boundary value problems were studied. We refer the interested reader to the recent monograph [21] for details.

In [22], the authors defined a quantum shifting operator by

aΦq(m) =qm+ (1 –q)a, (7)

m,a∈Rwithma. Then theq-derivative of a functionf on an interval [a,b] in (6) can be rewritten as

aDqf(t) = ⎧ ⎨ ⎩

f(t)–f(aΦq(t))

t–aΦq(t) , t=a,

limt→a aDqf(t), t=a.

(8)

Now, we consider the interval [a,b]⊆R, the quantum numbers 0 <q< 1,ω≥0, and

θ= ω

1 –q+a, (9)

with θ ∈[a,b]. The Hahn difference operator was generalized recently in [23] toaDq,ω

defined by

aDq,ωf(t) =

⎧ ⎨ ⎩

f(t)–f(qt+a(1–q)+ω)

(t–a)(1–q)–ω , t=θ,

f(θ), t=θ, (10)

provided thatf is differentiable atθ.

Next, we introduce a new quantum Hahn shifting operator by

θΦq(m) =qm+ (1 –q)θ. (11)

As a special case, ifω=a= 0, then (11) is reduced to classical quantum shifting in [1]. If ω= 0, then (11) is reduced to theq-shifting in (7) studied in [20], and ifa= 0, then (11) is reduced to Hahn shifting as appeared in [10]. In addition, the iteratedk-times of quantum shifting is defined by

θΦqk(m) =θΦqk–1

θΦq(m)

=qkm+1 –qkθ,

withθΦq0(m) =m.

Proposition 1 The following relations hold:

(i) (1 –q)(ta) –ω= (1 –q)(tθ) =tθΦq(t);

(ii) θΦq(θ) =θ;

(iii) (1 –qk)a+ω[k]

q= (1 –qk)θ,where[kq] = (1 –qk)/(1 –q),k= 0, 1, 2, . . ..

(4)

Definition 1 Letfbe a function defined on [a,b]. The quantum Hahn difference operator is defined by

aDq,ωf(t) =

⎧ ⎨ ⎩

f(t)–f(θΦq(t))

t–θΦq(t) , t=θ,

f(θ), t=θ,

(12)

provided thatf is differentiable atθ.

Definition 2 Assumef : [a,b]→Ris a given function and consider two pointsc,d

[a,b]. Theq,ω-quantum Hahn integral off fromctodis defined by d

c

f(s)adq,ωs:=

d

θ

f(s)adq,ωs

c

θ

f(s)adq,ωs, (13)

where t

θ

f(s)adq,ωs= tθΦq(t) ∞

i=0

qifθΦqi(t)

, (14)

fort∈[a,b], provided that the series converges att=candt=d. Let us define theθ-power function as

(nm)(0)θ = 1, (nm)

(k)

θ =

k–1

i=0

nθΦqi(m)

, k∈N∪ {∞}. (15)

For example, (nm)(4)θ = (nm)(nθΦq(m))(nθΦq2(m))(nθΦq3(m)). More generally, if γ ∈R, then

(nm)(θγ)=

i=0

(nθΦqi(m))

(nθΦ γ+i q (m))

,

withθΦqγ(m) =qγm+ (1 –)θ,γ ∈R. For example,

(nm)( 3 2)

θ =

(nm)(nθΦq(m))(nθΦq2(m))· · ·

(nθΦ

3 2

q(m))(nθΦ

5 2

q(m))(nθΦ

7 2 q(m))· · ·

.

Let us state the definitions of Riemann–Liouville type of fractional derivative and inte-gral of quantum Hahn calculus and also Caputo type fractional derivative, which can be found in [24].

Definition 3 The fractional quantum Hahn difference of Riemann–Liouville type of order ν≥0 on an interval [a,b] is defined by (aDq,0ωf)(t) =f(t) and

aDνq,ωf

(t) = 1 Γq(lν)

aDlq,ω

t

a

tθΦq(s) (l–ν–1)

θ f(s)adq,ωs, ν> 0,

(5)

Definition 4 Letν≥0 andfbe a function defined on [a,b]. The fractional quantum Hahn integral of Riemann–Liouville type is given by (aI0q,ωf)(t) =f(t) and

aIq,νωf

(t) = 1 Γq(ν)

t

a

tθΦq(s) (ν–1)

θ f(s)adq,ωs, ν> 0,t∈[a,b].

Definition 5 The fractional quantum Hahn difference of Caputo typeν≥0 on an interval [a,b] is defined by (caDq,0ωf)(t) =f(t) and

c aDνq,ωf

(t) = 1 Γq(lν)

t

a

tθΦq(s) (l–ν–1)

θ aD

l

q,ωf(s)adq,ωs, ν> 0,

wherelis the smallest integer greater than or equal toν.

Ifω= 0, thenθ=aand the above fractional quantum Hahn calculus is reduced to frac-tional quantum calculus on the interval [a,b] as appeared in [22].

Theorem 1([24]) Letα,β∈R+,λ(–1,)andθ[a,b].The following formulas hold:

(i) (aIq,αω(xa)

(λ)

θ )(t) = Γq(λ+1)

Γq(α+λ+1)(ta) (α+λ)

θ ;

(ii) (aDαq,ω(xa)

(λ)

θ )(t) = Γq(λ+1) Γq(λα+1)(ta)

(λα)

θ .

Theorem 2([24]) Let f(t)be a function defined on an interval[a,b],β,ν∈R+,α∈(N– 1,N)andθ∈[a,b].Then,we have:

(i) (aIq,βωaIq,νωf)(t) = (aIq,νωaIq,βωf)(t) = (aIq,β+ωνf)(t); (ii) (aDβq,ωaIq,βωf)(t) = (acDβq,ωaIq,βωf) =f(t); (iii) (aIq,αωaDαq,ωf)(t) =f(t) +c1(ta)(

α–1)

θ +c2(ta)(

α–2)

θ +· · ·+cN(ta)(

α–N)

θ ;

(iv) (aIq,αωcaDαq,ωf)(t) =f(t) +d0+d1(ta)(1)θ +d2(ta)(2)θ +· · ·+dN–1(ta)(Nθ –1),for someci,dj∈R,i= 1, 2, . . . ,N,j= 0, 1, . . . ,N– 1.

The rest of the paper is organized as follows: In Sect.2.1we prove the existence and uniqueness results for the the impulsive Hahn difference boundary value problem (1), while the corresponding results for the impulsive Hahn difference boundary value prob-lem (2) are presented in Sect.2.2. Examples illustrating the obtained results are presented in Sect.3.

2 Impulsive fractional Hahn difference equations

To establish our results, we define intervalsJk= [tk,tk+1),k= 0, 1, 2, . . . ,m– 1,Jm= [tm,T] andJ= [0,T], with impulsive points 0 =t0<t1<· · ·<tk<tk+1<· · ·<tm<tm+1=T. In ad-dition, we define the spacePC(J,R) ={x:J→R:x(t) is continuous everywhere except for sometkat whichx(tk+) andx(tk) exist andx(t+k) =x(tk),k = 1, 2, . . . ,m}. Observe that

PC(J,R) is a Banach space equipped with the normx=sup{|x(t)|:tJ}. From Sect.1, we replace all parameters,a,q,ωandνof fractional quantum Hahn calculus in Defini-tions3–5bytk,qk,ωkandνk,k= 0, 1, 2, . . . ,m, respectively. Also we assume that

θk= ωk 1 –qk

(6)

In the next subsections, the fractional quantum Hahn calculus is used to establish the existence and uniqueness results for the impulsive fractional Hahn difference boundary value problems (1) and (2).

2.1 Impulsive problem of fractional quantum Hahn difference equation of order 0 <

ν

k≤1

In this subsection, we investigate the impulsive Hahn difference boundary value problem (1).

The following lemma deals with the linear variant of problem (1) and gives a represen-tation of the solution.

Lemma 1 Letξ1+ξ2= 0and hC(J,R)be a given function.Then,the function x is a

solution of the impulsive Hahn difference boundary value problem

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ c tkD

νk

qk,ωkx(t) =h(t), tJk,k= 0, 1, 2, . . . ,m,

x(tk) =ϕk(x(tk–)), k= 1, 2, . . . ,m, ξ1x(0) +ξ2x(T) =ξ3,

(16)

if and only if

x(t) = ξ3 ξ1+ξ2

ξ2 ξ1+ξ2

m

i=0

tiI νi

qi,ωih

ti+1+ m

j=1 ϕj

xtj

+ k–1

i=0

tiI νi

qi,ωih

ti+1– + k

j=1 ϕj

xtj–+tkI νk

qk,ωkh

(t), (17)

for tJk,k= 0, 1, 2, . . . ,m,with b

a(·) = 0,when b<a.

Proof Applying Theorem2(iv), fortJ0, we obtain

t0I

ν0 q0,ω0

c t0D

ν0 q0,ω0x

(t) =x(t) =c0+

t0I

ν0 q0,ω0h

(t),

for somec0∈R. In particular, whent=t1–, it follows that

xt1–=c0+

t0I

ν0 q0,ω0h

t1–.

FortJ1, using the same process, we have

x(t) =x(t1) +

t1I

ν1 q1,ω1h

(t).

The impulsive condition,x(t1) =x(t1–) +ϕ1(x(t1–)), yields

x(t) =xt1+ϕ1

xt1–+t1I

ν1 q1,ω1h

(t) =c0+

t0I

ν0 q0,ω0h

t1+ϕ1

xt1–+t1I

ν1 q1,ω1h

(7)

Repeating the above argument, fortJk,k= 0, 1, 2, . . . ,m, we get

x(t) =c0+ k–1

i=0

tiI νi

qi,ωih

ti+1– + k

j=1 ϕj

xtj–+tkI νk

qk,ωkh

(t), (18)

withba(·) = 0, whenb<a. Sincex(0) =c0and

x(T) =c0+ m

i=0

tiI νi

qi,ωih

ti+1– + m

j=1 ϕj

xtj,

witht

m+1=T, we can compute, with boundary condition in (16), that

c0= ξ3 ξ1+ξ2

ξ2 ξ1+ξ2

m

i=0

tiI νi

qi,ωih

ti+1+ m

j=1 ϕj

xtj

.

Substituting the constantc0in the integral equation (18), we obtain the desired result in (17). The converse follows by direct computation. The proof is completed.

In the following, for convenience we use the abbreviation

tkI

νk

qk,ωkf

s,x(s)(t) = 1 Γqk(νk)

t

tk

tθkΦqk(s)

(νk–1) θk f

s,x(s)tkdqk,ωks

=tkI νk

qk,ωkfx

(t), fork= 0, 1, 2, . . . ,m, and put

Λ1=

1 + |ξ2| |ξ1+ξ2|

L1 m

i=0

(ti+1ti)(θνii)

Γqi(νi+ 1)

+L2

2 m(m+ 1)

,

Λ2= | ξ3| |ξ1+ξ2|+

1 + |ξ2| |ξ1+ξ2|

M

m

i=0

(ti+1ti)(

νi) θi

Γqi(νi+ 1) +

N

2m(m+ 1)

.

Now, we are in the position to establish the existence of a unique solution of problem (1) by using the Banach contraction mapping principle.

Theorem 3 Let f :J×R→Randϕk:R→R,k= 1, 2, . . . ,m,be given functions satisfying f(t,x) –f(t,y)≤L1|xy|, L1> 0,∀tJ,x,y∈R, (19)

and

ϕk(x) –ϕk(y)≤L2|xy|, L2> 0,∀x,y∈R. (20)

If

Λ1< 1, (21)

(8)
(9)

+tkI νk

qk,ωk|fxfy|

(t) ≤L1xy

|ξ2| |ξ1+ξ2|

m

i=0

tiI νi

qi,ωi(1)

ti+1– + m

i=0

tiI νi

qi,ωi(1)

ti+1

+L2xy

|

ξ2| |ξ1+ξ2|

+ 1 m

j=1 (1)

=Λ1xy,

which yields AxAyΛ1xy. From (21), we conclude that the operatorAis a contraction onBr. By the Banach contraction mapping principle, therefore, the impulsive boundary value problem of fractional quantum Hahn difference equation (1) has a unique solutionxonJsuch thatxr. The proof is complete.

Corollary 1 Let constantsξ1= 0andξ2= 0in(1),then we have the impulsive initial value

problem

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ c tkD

νk

qk,ωkx(t) =f(t,x(t)), tJk,k= 0, 1, 2, . . . ,m,

x(tk) =ϕk(x(tk–)), k= 1, 2, . . . ,m,

x(0) =ξ3

ξ1.

(22)

If the functions f andϕi,i= 1, 2, . . . ,m,satisfy(19)and(20),respectively,and if

L1 m

i=0

(ti+1ti)(θνii)

Γqi(νi+ 1)

+L2

2m(m+ 1) < 1,

then the impulsive initial value problem of fractional quantum Hahn difference equation

(22)has a unique solution on J.

2.2 Impulsive problem of fractional quantum Hahn difference equation of order 1 <

ν

k≤2

Consider now the impulsive fractional Hahn difference boundary value problem (2).

Lemma 2 Let gC(J,R).Then,the function x is a solution of the impulsive Hahn differ-ence boundary value problem

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ c tkD

νk

qk,ωkx(t) =g(t), tJk,k= 0, 1, 2, . . . ,m,

x(tk) =ϕk(x(tk–)), k= 1, 2, . . . ,m, tkDqk,ωkx(t

+

k) –tk–1Dqk–1,ωk–1x(t

k) =ϕk∗(x(tk–)), k= 1, 2, . . . ,m,

x(0) =η1, tmDqm,ωmx(T) =η2,

(10)

if and only if

(11)

then (27) can be written as

Therefore, the result in (24) holds when substituting the constantc1in (28). The converse follows by direct computation. This competes the proof.

To accomplish our goal, we define the operatorG:PC(J,R)→PC(J,R) by

(12)

value problem (2). Now, we set

Λ3=T

m

i=0

(ti+1ti)(θνii–1)

Γqi(νi)

+ m

i=0

(ti+1ti)(θνii)

Γqi(νi+ 1)

+ m

i=1

(ti+1ti) i–1

j=0

(tj+1tj) (νj–1) θj

Γqj(νj)

,

Λ4=mT+ m

i=1

(ti+1ti)i, Λ5=L1Λ3+L2m+L3Λ4, Λ6=|η1|+|η2|T.

Theorem 4 Let f andϕkbe given functions satisfying(19)and(20),respectively,for all

k= 1, 2, . . . ,m.Assume theϕk∗:R→Rsuch that

ϕk∗(x) –ϕk∗(y)≤L3|xy|, L3> 0,∀x,y∈R. (30)

IfΛ5< 1,then the impulsive fractional quantum Hahn difference boundary value problem (2)has a unique solution on J.

Proof The existence of a unique solution for the problem (2) will be proved by consider-ing an operator equationx=Gx, whereGis defined by (29). Consider the setBR={x

PC(J,R) :xR}, where a positive constantRsatisfies

R>Λ6+Λ3M+mN+Λ4K 1 –Λ5

,

andK=maxj|ϕj∗(0)|and constantsM,Nare defined in the proof of Theorem3. We claim thatGBRBR. Since

Gx(t)

≤ |η1|+

|η2|+ m

j=0

tjI νj–1

qj,ωj|fx|

tj+1– + m

i=1

ϕixti

m

i=0

(ti+1ti)

+ m–1

i=0 tiI

νi

qi,ωi|fx|

ti+1+ϕi+1

xti+1

+ m

i=1

(ti+1ti) i–1

j=0 tjI

νj–1

qj,ωj|fx|

tj+1– +ϕj+1xtj+1

+tmI

νm

qm,ωm|fx|

(T),

and|fx| ≤ |fxf0|+|f0| ≤L1R+M,|ϕj(x)| ≤ |ϕj(x) –ϕj(0)|+|ϕj(0)| ≤L2R+N,|ϕj∗(x)| ≤ |ϕj∗(x) –ϕj∗(0)|+|ϕj(0)| ≤L3R+K,j= 1, 2, 3, . . . ,m, for anyxBR, we have

Gx(t)≤Λ6+5Λ3M+mN+Λ4K<R,

which yieldsGxR. To prove the contraction property of operatorG, for anyx,yBR, we consider the inequalities

Gx(t) –Gy(t)

T

m

j=0

tjI νj–1

qj,ωj|fxfy|

tj+1+T

m

i=1

(13)

+ m–1

i=0 tiI

νi

qi,ωi|fxfy|

ti+1– +ϕi+1

xti+1– –ϕi+1

yti+1

+ m

i=1

(ti+1ti) i–1

j=0 tjI

νj–1

qj,ωj|fxfy|

tj+1

+ϕj+1xtj+1– –ϕj+1ytj+1

+tmI

νm

qm,ωm|fxfy|

(T)

L1xyT

m

j=0

tjI νj–1

qj,ωj(1)

tj+1+L3xyT

m

i=1 (1)

+L1xy

m–1

i=0

tiI νi

qi,ωi(1)

ti+1+L2xy

m–1

i=0 (1)

+L1xy

m

i=1

(ti+1ti) i–1

j=0

tjI νj–1

qj,ωj(1)

tj+1

+L3xy

m

i=1

(ti+1ti) i–1

j=0

(1) +L1xytmI νm

qm,ωm(1)

(T)

=L1Λ3xy+L2mxy+L3Λ4xy =Λ5xy.

Then we get GxGyΛ5xywhich implies thatGis a contraction operator as Λ5< 1. Therefore problem (2) has a unique solutionxonJ.

Lemma 3(Nonlinear alternative for single valued maps, [25]) Let E be a Banach space,

C a closed,convex subset of E,U an open subset of C and0∈U.Suppose that F:UC is a continuous,compact(that is, F(U)is a relatively compact subset of C)map.Then either

(i) Fhas a fixed point inU,or

(ii) There is au∂U(the boundary ofUinC)andλ∈(0, 1)withu=λF(u).

Theorem 5 Assume that the functions f :J×R→R,ϕk,ϕk:R→R,k= 1, 2, . . . ,m,are

continuous.In addition,we suppose that:

(H1) There exist a continuous nondecreasing functionψ1: [0,∞)→(0,∞)and a

contin-uous functionp:J→R+such that

f(t,x)≤p(t)ψ1

|x| for each(t,x)∈J×R.

(H2) There exist continuous nondecreasing functionsψ2,ψ3: [0,∞)→(0,∞)such that ϕk(x)≤ψ2

|x| and ϕk(x)≤ψ3

|x|,

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(H3) There exists a constantQ> 0such that

Q

Λ6+pψ1(Q)Λ3+ψ2(Q)m+ψ3(Q)Λ4 > 1,

wherep∗=sup{p(t) :tJ}.

Then the impulsive fractional quantum Hahn difference boundary value problem(2)has at least one solution on J.

Proof Let us prove the theorem by applying Lemma3. For a positive numberρ, we define the set={xPC(J,R) :xρ}. Clearly,is a closed, convex subset ofPC(J,R). Let

Hence the operatorGis continuous which is one of assumptions in Lemma3. In the next step, we will prove the compactness of operatorG.

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+

The right-hand side of the above inequality tends to zero asτ1→τ2(independently ofx). This shows that the setGBρis an equicontinuous set. Therefore the setGBρis relatively

compact. From the above and Arzelá–Ascoli theorem, the operatorGis completely con-tinuous or compact. Hence one more of assumptions of Lemma3holds.

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which yields

x

Λ6+pψ1(x)Λ3+ψ2(x)m+ψ3(x)Λ4 ≤1.

By assumption (H3), there exists a positive constant Qsuch that x =Q. Let us de-fineU={x:x<Q}. It is obvious thatG:UPC(J,R) is continuous and

com-pletely continuous. Therefore, there is nox∂Usuch thatx=λGxfor someλ∈(0, 1). By Lemma3, thus, we get the result thatGhas a fixed pointxUwhich is a solution of

problem (2) onJ. The proof is completed.

3 Examples

In this section we give examples to illustrate the usefulness of our main results.

Example1 Consider the following impulsive fractional quantum Hahn difference bound-ary value problem:

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ c tkD

k+2

k+3

k+1

k+2,k2+51k+6

x(t) =1 2(

et t+17)(

x2(t)+2|x(t)|

1+|x(t)| ) +27, tJk,J= [0, 4], x(tk) = k+181 |sinx(tk)|, tk=k,k= 1, 2, 3,

1 2x(0) +

2 3x(4) =

4 5.

(31)

Hereνk= (k+ 2)/(k+ 3) < 1,qk= (k+ 1)/(k+ 2),ωk= 1/(k2+ 5k+ 6),k= 0, 1, 2, 3,m= 3,

T= 4,ξ1= 1/2,ξ2= 2/3,ξ3= 4/5.

In addition, we observe thatθk= (1/(k+ 3)) +kJk,k= 0, 1, 2, 3. By using a mathematical program, we can find that

1 + |ξ2| |ξ1+ξ2|

m

i=0

(ti+1ti)(

νi) θi

Γqi(νi+ 1) ≈8.357863592111553,

1 2

1 + |ξ2| |ξ1+ξ2|

m(m+ 1)≈9.428571428571431.

Setting

f(t,x) = 1 2

e–t

t+ 17

x2+ 2|x| 1 +|x|

+2

7 and ϕk(x) = 1

k+ 18|sinx|, we compute that

f(t,x) –f(t,y)≤(1/17)|xy| and ϕk(x) –ϕk(y)≤(1/19)|xy|.

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Example2 Consider the impulsive fractional quantum Hahn difference boundary value problem

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ c tkD

2k+3

k+2

k+3

k+5,k1+5

x(t) = (t+10)1 2( x 8(t)

|x7(t)|+1+ 1), tJk,J= [0, 4], x(tk) = k

2

10(k2+1)|x(tk)|+ 2, k= 1, 2, 3,tk=k, tkDk+3

k+5, 1

k+5x(t +

k) –tk–1Dk+2

k+4, 1

k+4x(tk) =

sin2k 100 |x(t

k)|+ 3, k= 1, 2, 3,tk=k,

x(0) =47, t3D43,18x(4) =59.

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Hereνk= (2k+ 3)/(k+ 2), 1 <νk≤2,qk= (k+ 3)/(k+ 5),ωk= 1/(k+ 5),k= 0, 1, 2, 3,

m= 3,T= 4,η1= 4/7,η2= 5/9. We find thatθk= (1/2) +kJk,k= 0, 1, 2, 3. By using a mathematical program, we obtain constants as

Λ3≈40.771217238380451, Λ4= 18, Λ6≈2.793650793650793.

Setting

f(t,x) = 1 (t+ 10)2

x8 |x7|+ 1+ 1

,

ϕk(x) =

k2

10(k2+ 1)|x|+ 2 and ϕk(x) =

sin2k

100 |x|+ 3, and choosingp(t) = (1/(t+10)2),ψ

1(u) =u+1,ψ2(u) = (1/10)u+2 andψ3(u) = (1/100)u+3, it follows that

f(t,u)≤p(t)ψ1|u|, ϕk(u)≤ψ2|u|, ϕk∗(u)≤ψ3|u|,

for allu∈R,k= 1, 2, 3, which imply that conditions (H1)–(H2) hold. Forp∗= (1/100), by direct computation, there exists a constantQ> 562.8514177160808 satisfying the inequal-ity in (H3). Applying Theorem5, we deduce that the impulsive fractional quantum Hahn difference boundary value problem (32) has at least one solution on [0, 4].

Funding

This research was funded by King Mongkut’s University of Technology North Bangkok. Contract no. KMUTNB-61-GOV-B-22.

Availability of data and materials

Data sharing not applicable to this article as no data sets were generated or analyzed during the current study.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to this work. All authors read and approved the final manuscript.

Author details

1Intelligent and Nonlinear Dynamic Innovations Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok, Thailand.2Department of Mathematics, University of Ioannina, Ioannina, Greece.3Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah, Saudi Arabia.

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