R E S E A R C H
Open Access
New concepts of Hahn calculus and
impulsive Hahn difference equations
Jessada Tariboon
1*, Sotiris K Ntouyas
2,3and Weerawat Sudsutad
1*Correspondence:
1Nonlinear Dynamic Analysis
Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok, 10800, Thailand Full list of author information is available at the end of the article
Abstract
In this paper, we introduce new concepts of Hahn difference operator, the
qk,
ω
k-Hahn difference operator. We aim to establish a calculus of differences basedon theqk,
ω
k-Hahn difference operator. We construct a right inverse of theqk,
ω
k-Hahn operator and study some of its properties. As applications, we establishexistence and uniqueness results for first- and second-order impulsiveqk,
ω
k-Hahndifference equations.
MSC: 34A08; 34A12; 34A60; 39A10; 39A13
Keywords: Hahn difference operator; Jacksonq-difference operator; Jackson q-integral; Nörlund sums; impulsive difference equations
1 Introduction and preliminaries
Many physical phenomena are described by equations involving nondifferentiable func-tions,e.g., generic trajectories of quantum mechanics []. Several different approaches to deal with nondifferentiable functions are followed in the literature, including the time scale approach, the fractional approach, and the quantum approach.
Quantum difference operators are receiving an increase of interest due to their appli-cations see,e.g., [–]. Roughly speaking, a quantum calculus substitutes the classical derivative by a difference operator, which allows one to deal with sets of nondifferentiable functions.
In [], Hahn introduced the quantum difference operator Dq,ω, whereq∈(, ) and
ω> are fixed. The Hahn operator unifies (in the limit) the two best-known and most-used quantum difference operators: the Jacksonq-difference derivativeDq, whereq∈(, ) (cf.[, , ]); and the forward differenceDωwhereω> (cf.[–]). The Hahn
differ-ence operator is a successful tool for constructing families of orthogonal polynomials and investigating some approximation problems (cf.[–]).
The aim of this paper is to introduce new concepts of Hahn’s difference operator, the
qk,ωk-Hahn difference operator, to establish a calculus based on this operator and to
con-struct the associated integral. The steps are parallel to []. While some properties are straightforward extensions of classical results, some others need special treatments. As applications of theqk,ωk-Hahn difference operator we establish existence and uniqueness
results for first- and second-order impulsive fractional differential equations.
Impulsive differential equations serve as basic models to study the dynamics of pro-cesses that are subject to sudden changes in their states. Recent development in this field
has been motivated by many applied problems, such as control theory, population dynam-ics, and medicine. For some recent works on the theory of impulsive differential equations, we refer the interested reader to the monographs [–]. Impulsive quantum difference equations have been established by Tariboon and Ntouyas in [] by improving the classi-cal quantum classi-calculus which does not work when there exists at least one impulsive point appearing between two different points in the definition of q-derivative. For recent re-sults on the topics of initial and boundary value problems of impulsive quantum difference equations, we refer the reader to [].
We organize this paper as follows. In Section , some basic formulas of Hahn’s differ-ence operator and the associated Jackson-Nörlund integral calculus are briefly reviewed. Our results are formulated and proved in Section . Applications to impulsive fractional difference equations are given in Section .
2 Preliminaries
Letq∈(, ) andω> . Define
ω:=
ω
–q (.)
and letIbe a real interval containingω.
Definition .(Hahn’s difference operator []) Letf:I→R. The Hahn difference oper-ator off is defined by
Dq,ωf(t) = ⎧ ⎨ ⎩
f(t)–f(qt+ω)
t(–q)–ω , t=ω, f(ω), t=ω,
(.)
provided thatf is differentiable atω.
The functionf is calledq,ω-differentiable onI, ifDq,ωf(t) exists for allt∈I.
Note that whenq→ we obtain the forwardω-difference operator
D,ωf(t) =
f(t+ω) –f(t)
ω , (.)
and whenω= we obtain the Jacksonq-difference operator
Dq,f(t) =
⎧ ⎨ ⎩
f(t)–f(qt)
t(–q) , t= ,
f(), t= , (.)
provided thatf() exists. Heref is supposed to be defined on aq-geometric setA⊂R, for whichqt∈Awhenevert∈A.
Hence, we can state that the Dq,ω operator generalizes (in the limit) the forward
ω-difference and the Jacksonq-difference operators [, ]. Notice also that, under appropriate conditions,
lim
q→,ω→Dq,ωf(t) =f (t).
Lemma .([]) Let f,g:I→Rbe q,ω-differentiable at t∈I.Then the following state-ments are true:
(i) Dq,ω(f +g)(t) =Dq,ωf(t) +Dq,ωg(t),
(ii) Dq,ωfg(t) =g(t)Dq,ωf(t) +f(qt+ω)Dq,ωg(t),
(iii) Dq,ωcf(t) =cDq,ωf(t),for any constantc∈R,
(iv) Dq,ω(fg)(t) =
g(t)Dq,ωf(t)–f(t)Dq,ωg(t)
g(t)g(qt+ω) ,forg(t)g(qt+ω)= ,
(v) f(tq+ω) =f(t) + ((qt+ω) –t)Dq,ωf(t),t∈I.
Leth(t) =qt+ω,t∈I. Note thathis a contraction,h(I)⊆I,h(t) <tfort>ω,h(t) >t
fort<ω, andh(ω) =ω.
We use the standard notation of theq-number as [α]q=–q
α
–q forα∈R.
Lemma .([]) Let k∈Nand t∈I.Then
hk(t) =h◦h◦ · · · ◦ h(t)
i-times
=qkt+ω[k]q, t∈I. (.)
Next, we define the notion of aq,ω-integral, known as the Jackson-Nörlund integral.
Definition .([]) Letf :I→Rbe a function anda,b,ω∈I. Theq,ω-integral off
fromatobis defined by
b
a
f(s)dq,ωs= b
ω
f(s)dq,ωs– a
ω
f(s)dq,ωs, (.)
where
t
ω
f(s)dq,ωs=
t( –q) –ω
∞
k=
qkftqk+ω[k]q
, t∈I, (.)
provided that the series converges att=aandt=b.
The functionf isq,ω-integrable overIif it isq,ω-integrable over [a,b], for alla,b∈I. Note that in the integral formulas (.) and (.), whenω→, we obtain the Jackson
q-integral
b
a
f(s)dqs=
b
f(s)dqs–
a
f(s)dqs,
where
t
f(s)dqs=t( –q)
∞
k=
qkftqk, t∈I
(see,e.g., []); while ifq→ we obtain the Nörlund sum,
b
a
f(s)ωs= b
+∞f(s)ωs– a
where
t
+∞f(s)ωs= –ω +∞
k=
f(t+kω)
(see,e.g., [, , ]).
The following properties of Jackson-Nörlund integration can be found in [].
Lemma . Let f,g:I→Rbe q,ω-integrable on I,K∈R,and a,b,c∈I.Then the follow-ing formulas hold:
(i) aaf(t)dq,ωt= ,
(ii) abKf(t)dq,ωt=K b
a f(t)dq,ωt,
(iii) abf(t)dq,ωt= – a
b f(t)dq,ωt,
(iv) abf(t)dq,ωt= b
c f(t)dq,ωt+ c
af(t)dq,ωt,
(v) ab(f(t) +g(t))dq,ωt= b
a f(t)dq,ωt+ b
a g(t)dq,ωt,
(vi) abf(t)Dq,ωg(t)dq,ωt= [f(t)g(t)]ba– b
a Dq,ωf(t)g(qt+ω)dq,ωt.
Property (vi) of the above lemma is known asq,ω-integration by parts. The next result is thefundamental theorem of Hahn calculus.
Lemma .([]) Let f :I→Rbe continuous atωand define F(t) := t
ωf(s)dq,ωs.Then
F is continuous atω.In addition,Dq,ωF(t)exists for every t∈I and
Dq,ωF(t) =f(t). (.)
On the other hand,
b
a
Dq,ωf(s)dq,ωs=f(b) –f(a) for all a,b∈I. (.)
Existence and uniqueness results for first-order abstract Hahn difference equations were studied in [], by using the method of successive approximation.
3 New concepts of Hahn calculus
Let there be a dense intervalJk= [tk,tk+]⊆Rand given constants <qk< ,ωk> and
θk=
ωk
–qk +tk. (.)
Note that iftk= ,qk=q, andωk=ω, thenθk=ω, whereωis defined in (.).
Definition . Letf be a function defined onJk. Theqk,ωk-Hahn difference operatoris
given by
tkDqk,ωkf(t) =
f(t) –f(qkt+ ( –qk)tk+ωk)
( –qk)(t–tk) –ωk
, t=θk, (.)
We say thatf isqk,ωk-differentiable onJkprovidedtkDqk,ωkf(t) exists for allt∈Jk. Note that ifωk= in (.), thentkDqk,f=tkDqkf, wheretkDqkis theqk-derivative of the function
f(t) which was first established in [] by
tkDqkf(t) =
f(t) –f(qkt+ ( –qk)tk) ( –qk)(t–tk)
. (.)
It is easy to see that iftk= andqk=q, then (.) is reduced to the Jacksonq-difference
operator in (.).
Example . Letf(t) =tfort∈J
k= [, ] and constantsqk= /,ωk= . Thenθk=
and theqk,ωk-Hahn derivative onJkis given by
D ,f(t) =
t– ( t+ )
(
)(t– ) –
=t
– t–
(t– ) , t= ,
andD
,f() = .
It is easy to prove the following results.
Theorem . Let f,g:Jk→Rbe qk,ωk-differentiable at t∈Jk.Then the following formu-las hold:
(i) tkDqk,ωk(f+g)(t) =tkDqk,ωkf(t) +tkDqk,ωkg(t),
(ii) tkDqk,ωkfg(t) =g(t)tkDqk,ωkf(t) +f(qkt+ ( –qk)tk+ωk)tkDqk,ωkg(t),
(iii) tkDqk,ωkcf(t) =ctkDqk,ωkf(t),for any constantc∈R,
(iv) tkDqk,ωk(
f g)(t) =
g(t)tkDqk,ωkf(t)–f(t)tkDqk,ωkg(t)
g(t)g(qkt+(–qk)tk+ωk) ,forg(t)g(qkt+ ( –qk)tk+ωk)= .
Next, we define the higher-orderqk,ωk-derivative of functions.
Definition . Letf be a function defined on Jk. We define the second-order qk,ωk
-derivative tkDqk,ωkf provided tkDqk,ωkf is qk,ωk-differentiable on Jk with tkDqk,ωkf =
tkDqk,ωk(tkDqk,ωkf) :Jk →R. In addition, we define the higher-order qk,ωk-derivative
tkD
n
qk,ωkf:Jk→R, withtkD
n
qk,ωkf =tkDqk,ωk(tkD
n–
qk,ωkf) andtkD
qk,ωkf =f.
The new definition ofqk,ωk-integral is given as follows.
Definition . Assumef :Jk→Ris a function anda,b∈Jk. We define theqk,ωk-integral
off fromatobby
b
a
f(s)tkdqk,ωks:=
b
θk
f(s)tkdqk,ωks–
a
θk
f(s)tkdqk,ωks, (.)
where
t
θk
f(s)tkdqk,ωks=
(t–tk)( –qk) –ωk ∞
i=
qikfqikt+ –qiktk+ωk[i]qk
fort∈Jk, provided that the series converge att=aandt=b. The functionf is called
qk,ωk-integrable onJkand we say thatf isqk,ωk-integrable over [a,b] for alla,b∈Jk.
Note that iftk= ,qk=q, andωk=ω, then (.) and (.) are reduced to (.) and (.),
respectively.
As customary, the following properties should be to stated. However, the proof is easy and we omit it.
Theorem . Let f,g:Jk→Rbe qk,ωk-integrable on Jk,K∈R,and a,b,c∈Jk.Then the following formulas hold:
(i) aaf(s)tkdqk,ωks= ,
(ii) abKf(s)tkdqk,ωks=K
b
a f(s)tkdqk,ωks,
(iii) abf(s)tkdqk,ωks= –
a
b f(s)tkdqk,ωks,
(iv) abf(s)tkdqk,ωks=
b
c f(s)tkdqk,ωks+
c
af(s)tkdqk,ωks,
(v) ab(f(s) +g(s))tkdqk,ωks=
b
a f(s)tkdqk,ωks+
b
a g(s)tkdqk,ωks.
Lemma . Let h be the transformation
h(t) :=qkt+ ( –qk)tk+ωk, t∈Jk, (.)
andθk∈Jkis defined by(.).Then the ith-order iteration of h is given by
hi(t) =h◦h◦ · · · ◦ h(t)
i-times
=qikt+ –qiktk+ωk[i]qk, t∈Jk. (.)
In addition,the sequence{hi(t)}∞
i=is an increasing(a decreasing)sequence in i when t<θk
(θk<t)with
lim i→∞h
i(t) =θ
k, t∈Jk. (.)
Proof By directly computation, it is easy to show that (.) holds. Fort∈Jkandi∈N, we have
hi+(t) –hi(t) =qki( –qk)(tk–t) +ωk
[i+ ]qk– [i]qk
=qik( –qk)(θk–t).
Ift<θk orθk<t, then we see that the sequence{hi(t)}∞i= is increasing or decreasing,
re-spectively. Therefore, equation (.) is true for allt∈Jk.
Now, we will state and prove the fundamental theorem ofqk,ωk-Hahn calculus.
Theorem . Suppose that the function f:Jk→Ris continuous atθk∈Jk.We define
F(t) :=
t
θk
f(s)tkdqk,ωks, t∈Jk. (.)
(i) tkDqk,ωkF(t) =f(t),
(ii) θt
ktkDqk,ωkf(s)tkdqk,ωks=f(t) –f(θk),
(iii) a tb kDqk,ωkf(s)tkdqk,ωks=f(b) –f(a).
Proof From (.), we observe that
Fqkt+ ( –qk)tk+ωk
=qkt+ ( –qk)tk+ωk
–tk( –qk) –ωk
× ∞
i=
qikfqkt+ ( –qk)tk+ωk
qik+ –qiktk+ωk[i]qk
=qk(t–tk) +ωk
( –qk) –ωk ∞
i=
qkifqki+t+ –qi+k tk+ωk[i+ ]qk
.
Then, by (.), we have
tkDqk,ωkF(t) =
F(t) –F(qkt+ ( –qk)tk+ωk)
( –qk)(t–tk) –ωk
=
∞
i= qik
fqikt+ –qkitk+ωk[i]qk
–(qk(t–tk) +ωk)( –q) –ωk ( –qk)(t–tk) –ωk
fqi+k t+ –qki+tk+ωk[i+ ]qk
=
∞
i=
qikfqikt+ –qiktk+ωk[i]qk
–qkfqi+k t+ –qi+k tk+ωk[i+ ]qk
=f(t).
This shows that (i) holds.
To prove (ii), by Definitions ., ., and Lemma ., we get
t
θk
tkDqk,ωkf(s)tkdqk,ωks
=(t–tk)( –qk) –ωk ∞
i=
qik(tkDqk,ωkf)
qikt+ –qiktk+ωk[i]qk
=(t–tk)( –qk) –ωk ∞
i= qik
×f(qikt+ ( –qik)tk+ωk[i]qk) –f(qk(q
i
kt+ ( –qik)tk+ωk[i]qk) + ( –qk)tk+ωk) ( –qk)(qikt+ ( –qik)tk+ωk[i]qk–tk) –ωk
=
∞
i=
fqikt+ –qkitk+ωk[i]qk
–fqi+k t+ –qi+k tk+ωk[i+ ]qk
Now, we show that (iii) holds. From (ii) for anya,b∈Jk, we obtain
b
a
tkDqk,ωkf(s)tkdqk,ωks=
b
θk
tkDqk,ωkf(s)tkdqk,ωks–
a
θk
tkDqk,ωkf(s)tkdqk,ωks
=f(b) –f(a).
This completes the proof.
Lemma . Let f,g:Jk→Rbe qk,ωk-integrable on Jk.Then the following integration by parts formula holds:
b
a
g(s)tkDqk,ωkf(s)tkdqk,ωks
=f(s)g(s)ba–
b
a
fqks+ ( –qk)tk+ωk
tkDqk,ωkg(s)tkdqk,ωks.
Proof By Theorem . we have
b
a tk Dqk,ωk
f(s)g(s)tkdqk,ωks= (fg)(b) – (fg)(a).
On the other hand, by (ii) of Theorem . and (v) of Theorem .,
b
a
tkDqk,ωk
f(s)g(s)tkdqk,ωks
=
b
a
g(s)tkDqk,ωkf(s)tkdqk,ωks
+
b
a
fqks+ ( –qk)tk+ωk
tkDqk,ωkg(s)tkdqk,ωks.
Combining these two equalities we get the desired formula.
Lemma . Letθk∈Jk,α∈R,andβ∈R\ {–}.Then for t∈Jkthe following formulas hold:
(i) tkDqk(t–θk)
α= [α]
qk(t–θk)
α–,
(ii) θt
k(s–θk)
β
tkdqk,ωks= (
–qk
–qβk+)(t–θk) β+.
Proof From Definition ., fort=θk, we have
tkDqk,ωk(t–θk)
α
=(t–θk)
α– (q
kt+ ( –qk)tk+ωk–θk)α
( –qk)(t–tk) –ωk
=(t–θk)
α–qα k(t–θk)α
( –qk)(t–θk)
= [α]qk(t–θk)
α–.
Now, we are going to prove (ii). Forβ∈R\ {–}, Definition . implies
The proof is completed.
Corollary . For a,b∈Jk,the following formula holds:
(i) and (ii) are obvious. To prove (iii), from (i) and (ii) we obtain
b
Theorem . Let f be the qk,ωk-integrable function on Jk.Then we have t
Proof By Definition ., we have
= ( –qk)
4 Impulsiveqk,
ω
k-Hahn difference equationsIn this section, we use our results on qk,ωk-Hahn calculus to establish existence and
uniqueness results for impulsiveqk,ωk-Hahn difference equations of the first and second
order. LetJ= [t,t],Jk= (tk,tk+] fork= , , . . . ,mbe subintervals ofJ= [,T] such that
θk∈Jkfork= , , , . . . ,m. LetPC(J,R) ={x:J→R:x(t) is continuous everywhere
ex-cept for sometkat whichx(t+k) andx(tk–) exist andx(t–k) =x(tk),k= , , . . . ,m}.PC(J,R) is
a Banach space with the normxPC=sup{|x(t)|:t∈J}.
4.1 First-order impulsiveqk,
ω
k-Hahn difference equationsIn this subsection, we study the existence and uniqueness of solutions for the following initial value problem for first-order impulsiveqk,ωk-Hahn difference equation
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
tkDqk,ωkx(t) =f(t,x(t)), t∈J,t=tk, x(tk) =ϕk(x(tk)), k= , , . . . ,m, x() =α,
(.)
whereα∈R, =t<t<t<· · ·<tk<· · ·<tm<tm+=T,f :J×R→Ris a continuous function, ϕk∈C(R,R),x(tk) =x(tk+) –x(tk),k= , , . . . ,m, and quantum numbers < qk< ,ωk> such thatθk∈Jkfork= , , , . . . ,m.
Lemma . Let x∈PC(J,R)satisfying(.).The impulsive qk,ωk-Hahn difference initial value problem(.)is equivalent to the integral equation
x(t) =α+
t<tk<t
tk
tk–
fs,x(s)tk–dqk–,ωk–s+
t<tk<t ϕk
x(tk)
+
t
tk
fs,x(s)tkdqk,ωks, (.)
witht<t= .
Proof Fort∈J, applyingq,ω-integral fromttotin the first equation of (.) and using
Theorem .(iii), we obtain
x(t) =α+
t
t
fs,x(s)tdq,ωs.
Sinceθ∈J, we havet≥θand also, fort=t,
x(t) =α+ t
t
fs,x(s)tdq,ωs.
Fort∈J, taking theq,ω-integral to the first equation of (.) withk= and applying
Theorem .(iii) again, we have
x(t) =xt++
t
t
From the impulsive conditionx(t+) =x(t) +ϕ(x(t)), we get
x(t) =α+
t
t
fs,x(s)tdq,ωs+
t
t
fs,x(s)tdq,ωs+ϕ
x(t)
.
Fort∈J, theq,ω-integration and impulsive condition imply
x(t) =xt++
t
t
fs,x(s)tdq,ωs
=α+
t
t
fs,x(s)tdq,ωs+
t
t
fs,x(s)tdq,ωs+
t
t
fs,x(s)tdq,ωs
+ϕ
x(t)+ϕ
x(t).
From the above process, for anyt∈Jk,k= , , . . . ,m, we obtain the desired result in (.). Conversely, for anyt∈Jk,k= , , . . . ,m, applyingqk,ωk-derivative to (.) and using
Theorem .(i), we have
tkDqk,ωkx(t) =f
t,x(t).
By direct computation, we havex(tk) =ϕk(x(tk)) and alsox() =α. The proof is
com-pleted.
Now, we are in a position to prove an existence and uniqueness result for the problem (.), via Banach contraction mapping principle.
Theorem . Suppose that the following assumptions are fulfilled:
(H) the continuous functionf :J×R→Rsatisfies
f(t,x) –f(t,y)≤L|x–y|, L> ,∀t∈J,x,y∈R;
(H) the continuous functionsϕk:R→R,k= , , . . . ,msatisfy ϕk(x) –ϕk(y)≤L|x–y|, L> ,∀x,y∈R.
If
LT+mL< , (.)
then the impulsive qk,ωk-Hahn difference initial value problem(.)has a unique solution on J.
Proof Let us define an operatorA:PC(J,R)→PC(J,R) by
Ax(t) =α+
t<tk<t
tk
tk–
fs,x(s)tk–dqk–,ωk–s+
t<tk<t ϕk
x(tk)
+
t
tk
witht<t= . Letsupt∈J|f(t, )|=Mandmax{|ϕk()|:k= , , . . . ,m}=M. Choosing
a positive constantrsuch that
r≥|α|+MT+mM
– (LT+mL) ,
and setting a ballBr={x∈PC(J,R) :x ≤r}, we will show thatABr⊂Br. For anyx∈Br
andt∈J, we have
Ax(t)≤ |α|+
t<tk<t
tk
tk–
fs,x(s)tk–dqk–,ωk–s+
t<tk<t
ϕk
x(tk)
+
t
tk
fs,x(s)tkdqk,ωks
≤ |α|+
t<tk<T
tk
tk–
f
s,x(s)–f(t, )+f(t, )tk–dqk–,ωk–s
+
t<tk<T
ϕk
x(tk)–ϕk()+ϕk()
+
T
tm
fs,x(s)–f(t, )+f(t, )tmdqm,ωms
≤ |α|+ (Lr+M)
t<tk<T
tk
tk–
tk–dqk–,ωk–s
+m(Lr+M) + (Lr+M)
T
tm
tmdqm,ωms
=|α|+MT+mM+r(LT+mL)≤r.
This means thatAx ≤r, which yieldsABr⊂Br. Forx,y∈PC(J,R) and for eacht∈J, we have
Ax(t) –Ay(t)≤
t<tk<t
tk
tk–
fs,x(s)–fs,y(s)tk–dqk–,ωk–s
+
t<tk<t
ϕk
x(tk)–ϕk
y(tk)
+
t
tk
fs,x(s)–fs,y(s)tkdqk,ωks
≤Lx–y
t<tk<T
tk
tk–
tk–dqk–,ωk–s
+mLx–y+Lx–y T
tm
tmdqm,ωms
= (LT+mL)x–y,
which leads toAx–Ay ≤(LT+mL)x–y. AsLT +mL< , it follows from the
Example . Consider the first-order impulsiveqk,ωk-Hahn difference initial value
prob-Therefore, by Theorem ., we deduce that the problem (.) has a unique solution on [, ].
4.2 Second-order impulsiveqk,
ω
k-Hahn difference equationsIn this subsection, we consider the second-order initial value problem of the impulsive
Proof Fort∈J, takingq,ω-integral for the first equation of (.) and using the second
Applying the first impulsive condition of (.), that is,x(t+
=α+βt+
Repeating the above method, fort∈J, we obtain (.) as desired.
Conversely, it can easily be shown by direct computation that the integral equation (.) satisfies the impulsive initial value problem (.). This completes the proof.
From Example .(iii) withb=tk+, we set the notation
(k) =(tk+–tk)
–ω
k(tk+–tk)
+qk .
Also, we use the notations
(U) =U
witht<t= . By transforming the impulsive initial value problem (.) into a fixed point problemx=Qx, we will show that the operatorQhas a fixed point which is a unique solution of problem (.) via the Banach contraction mapping principle.
Setting supt∈J|f(t, )|=N, max{|ϕk()|: k= , , . . . ,m}=N, and max{|ϕk∗()|: k=
, , . . . ,m}=N, we will prove thatQBR⊂BR, whereBR={x∈PC(J,R) :x ≤R}and
the positive constantRsatisfies
R≥|α|+|β|T+ (N)
– (L) . (.)
Forx∈BR, taking into account Example .(iii), we get
Qx(t)≤ |α|+|β|T
+
t<tk<T
tk
tk–
s
tk–
fu,x(u)–f(u, )
+f(u, )tk–dqk–,ωk–utk–dqk–,ωk–s+ϕk
x(tk)
–ϕk()+ϕk()
+T t<tk<T
tk
tk–
f
s,x(s)–f(s, )+f(s, )tk–dqk–,ωk–s
+ϕk∗x(tk)–ϕk∗()+ϕ∗k()
+
t<tk<T
tk tk
tk–
fs,x(s)–f(s, )+f(s, )tk–dqk–,ωk–s
+ϕk∗x(tk)
–ϕk∗()+ϕ∗k()
+
T
tm
s
tm
f
u,x(u)–f(u, )+f(u, )tmdqm,ωmutmdqm,ωms
≤ |α|+|β|T+
m
k=
(LR+N)(k– ) +LR+N
+T m
k=
(tk–tk–)(LR+N) +LR+N
+
m
k= tk
(tk–tk–)(LR+N) +LR+N
+ (LR+N)(m)
=|α|+|β|T+R (L) + (N)≤R.
Then we haveQx ≤R, which impliesQBR⊂BR.
Finally, forx,y∈PC(J,R) and for eacht∈J, we get
Qx(t) –Qy(t)≤
t<tk<T
tk
tk–
s
tk–
fu,x(u)–fu,y(u)tk–dqk–,ωk–utk–dqk–,ωk–s
+ϕk
x(tk)
–ϕk
+T traction mapping principle thatQis a contraction. Therefore, we see that the operatorQ has a fixed point which is a unique solution of the impulsiveqk,ωk-Hahn difference initial
value problem (.) onJ. The proof is completed.
Example . Consider the second-order impulsiveqk,ωk-Hahn difference initial value
problem of the form
respectively. From the above information, we find that
Therefore, by Theorem ., we deduce that the problem (.) has a unique solution on [, ].
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally in this article. They read and approved the final manuscript.
Author details
1Nonlinear Dynamic Analysis Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s
University of Technology North Bangkok, Bangkok, 10800, Thailand.2Department of Mathematics, University of Ioannina,
Ioannina, 451 10, Greece.3Nonlinear Analysis and Applied Mathematics (NAAM) - Research Group, Department of
Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia.
Acknowledgements
This research was funded by Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Thailand. Contract No. 5942107.
Received: 25 February 2016 Accepted: 22 September 2016
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