R E S E A R C H
Open Access
Fractional
q
-symmetric calculus on a time
scale
Mingzhe Sun and Chengmin Hou
**Correspondence:
cmhou@foxmail.com Department of Mathematics, Yanbian University, Yanji, 133002, P.R. China
Abstract
In this paper, the definitions ofq-symmetric exponential function andq-symmetric gamma function are presented. By aq-symmetric exponential function, we shall illustrate the Laplace transform method and define and solve several families of linear fractionalq-symmetric difference equations with constant coefficients. We also introduce aq-symmetric analogue Mittag-Leffler function and studyq-symmetric Caputo fractional initial value problems. It is hoped that our work will provide foundation and motivation for further studying of fractionalq-symmetric difference systems.
MSC: 92B20; 68T05; 39A11; 34K13
Keywords: fractionalq-symmetric exponential function; fractionalq-symmetric gamma function; Laplace transform; initial value problem
1 Introduction
Theq-calculus is not of recent appearance. It was initiated in the twenties of the last cen-tury. However, it has gained considerable popularity and importance during the last three decades or so. This is due to its distinguished applications in numerous diverse fields of physics such as cosmic strings and black holes [], conformal quantum mechanics [], nu-clear and high energy physics [], just to name a few. Early developments forq-fractional calculus can be found in the work of Al-Salam and co-authors [, ] or Agarwal []. A q-Laplace transform method has been developed by Abdi [] and applied toq-difference equations [, ]. Moreover, there is currently much activity to reexamine and further de-velop theq-special functions. Notable early work includes the work of Jackson [–], Hahn [, ] and Agarwal []. We also refer the reader to more recent articles [–] and [, ].
Theq-symmetric quantum calculus has proven to be useful in several fields, in partic-ular in quantum mechanics []. As noticed in [], consistently with theq-deformed theory, the standardq-symmetric integral must be generalized to the basic integral de-fined. Recently, we introduced basic concepts of fractional q-symmetric integral and derivative operators in []. The basic theory ofq-symmetric quantum calculus opera-tors need be explored. The object of this paper is to further develop the theory of frac-tionalq-symmetric calculus. First, the definitions of q-symmetric exponential function andq-symmetric gamma function are presented. Second, we give a fractionalq-symmetric transform method. We then define someq-symmetric difference equations and apply
the q-symmetric transform method to obtain solutions. Finally, we study a Caputo q-fractional initial value problem and give aq-analogue Mittag-Leffler function.
2 Theq-symmetric gamma and theq-symmetric exponential functions Lett∈Rand define
Tt=
t:t=tqn,n∈N
∪t:t=tq–n,n∈N
∪ {}, <q< .
If there is no confusion concerningt, we shall denoteTtbyT.
The basicq-symmetric integrals off :T−→Rare defined through the relations
(Iq,f)(t) =
x
f(t)dqt=x
–q
∞
k=
qkfxqk+. (.)
Fora∈T,
(Iq,af)(t) =
x
a
f(t)dqt=
x
f(t)dqt–
a
f(t)dqt.
For a functionf :T−→R, theq-symmetric derivative is defined by
(Dqf)(x) =
f(qx) –f(q–x)
x(q–q–) , (Dqf)() :=f
(), (.)
and theq-symmetric derivatives of higher order are defined by Dqf(x) =f(x), Dqnf(x) =DqDnq–f
(x), n∈N+.
As forq-symmetric derivatives, we can define an operatorIqnby Iq,af(x) =f(x), Iqn,af(x) =Iq,aIqn,–af
(x), n∈N+. (.)
For operators defined in this manner, the following is valid:
(DqIq,)f(x) =f(x), (Iq,Dqf)(x) =f(x) –f(). (.)
The formula forq-symmetric integration by parts is b
a
u(qx)(Dqv)(x)dqx= u(x)v(x) b a–
b
a
vq–x(Dqu)(x)dqx. (.)
Definition . Theq-symmetric factorial is defined in the following way. Ifnis a
non-negative integer, then
(a–b)()= , (a–b)(k)=
k–
i=
Their natural expansions to reals are
(a–b)(α)=aα ∞
i=( –baq
i+) ∞
i=( –baq(i+α)+)
(α∈R,a= ). (.)
Next, for a real parameterq∈R+\ {}, we introduce aq-real number [a]
qby
[a]q=
–qa
–q (a∈R).
We shall state several properties of theq-symmetric factorial function, each property is verified using the definition and a straightforward calculation.
Theorem . For a q-symmetric factorial function,we have
(i) (t–s)(β+ν)= (t–s)(β)t–qβs(ν), (ii) (at–as)(β)=aβ(t–s)(β),
(iii) tDq(t–s)
(β)
= [β]qq–t–s(β–), (iv) sDq
t–q–s(β)= –[β]
q(t–s)
(β–) .
Definition . Theq-symmetric exponential function is defined as
eq(t) =
∞
n=
–qn+t, eq() = .
Note thateq(q–) = and
Dqeq
q–t= –
–qeq(t). (.)
We are now in a position to give the integral representation of theq-symmetric gamma function. Letα∈R\ {. . . , –, –, }. Define theq-symmetric gamma function by
q(α) =
–q
q–t –q
α–
eq(t)dqt. (.)
For an integern, we denote
[]q! = , [n]q! = [n]q[n– ]q· · ·[]q=
( –qn)
( –q)
( –qn–) ( –q) · · ·
( –q) ( –q).
Lemma . Forα∈R,
q(α+ ) = [α]qq(α), q() = .
For any positive integer k,
Proof By (.), (.), (.), we can get
Remark . In [], authors introduced theq-symmetric gamma function as
hence
q(α) =
( –q)α–
∞
i=
( –q(i+)) ( –q(i+α))
= –q–α( –q)(α–).
3 Fractionalq-symmetric integral and derivative
We now introduce the fractionalq-symmetric integral operator (see []) Iα
q,f
(x) = q(α)
q α
x
(x–τ)(α–)fqα–τd
qτ
α∈R+, (.)
where
α k
= (α+ )
(k+ )(α–k+ ) (k∈N).
Lemma .([]) Letα,β∈R+.The fractional q-symmetric integration has the following
semigroup property:
Iqα,Iqβ,f(x) =Iqα+β, f(x).
Lemma .([]) Forα∈R+,the following identity is valid:
Iqα,f(x) =Iqα+,Dqf
(x) +f() q(α+ )
q α
xα.
We define the fractional q-symmetric derivative of Riemann-Liouville type of a function f(x)by
Dαq,f(x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
(I–α
q,f)(x), α< ,
f(x), α= ,
(D[α]
q I
[α]–α
q, f)(x), α> ,
(.)
where[α]denotes the smallest integer greater or equal toα.
Lemma .([]) Forα∈R\N,the following is valid:
DqDαq,f
(x) =Dα+q,f(x).
Lemma .([]) Forα∈R\N,the following is valid:
DqDαq,f
(x) =Dα
q,Dqf
(x) +f() q(–α)
q –(α+)
x–α–.
Lemma .([]) Forα∈R\N,the following is valid:
Lemma .([]) Letα∈(N– ,N].Then,for some constants ci∈R,i= , , . . . ,N,the
following equality holds:
Iqα,Dαq,f(x) =f(x) +cxα–+cxα–+· · ·+cNxα–N.
If we change the order of operators, we can introduce another type of fractional q-derivative.
The fractionalq-symmetric derivative of Caputo type is
c
Dαq,f(x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
(Iq–α,f)(x), α< ,
f(x), α= ,
(Iq[α]–α, D[α]q f)(x), α> ,
(.)
where [α] denotes the smallest integer greater or equal toα.
Lemma .([]) Forα∈R\N,and x> ,the following is valid:
c
Dα+q,f(x) –cDαq,Dqf
(x)
= ⎧ ⎨ ⎩
f()
q(–α)q
–(α+)
x–α–, α< –, , α≥–.
(.)
Lemma .([]) Forα∈R\Nand x> ,the following is valid:
DcqDα
q,f
(x) –cDα+q,f(x)
= ⎧ ⎨ ⎩
, α≤–,
f()
q(–α)q
–(α+)
x–α–, α> –. (.)
Lemma .([]) Letα∈(N– ,N].Then,for some constants ci∈R,i= , , . . . ,N– ,the
following equality holds:
Iqα,cDαq,f(x) =f(x) +c+cx+cx+· · ·+cN–xN–.
Lemma . Let f :T×T−→R,then the following identity holds:
Dq
t
f(t,s)dqs=
q–t t
Dqf(t,s)dqs+f(qt,t).
Proof
tDq
t
f(t,s)dqs=tDq
t –q
∞
k=
qkft,qk+t
= (q–q–)t
qt –q
∞
k=
–q–t –q
4 Theq-symmetric Laplace transform For convenience, we need some preliminaries.
Letq=q, then we have
q(x) =
∞
i=( –qi+) ∞
i=( –q(i+x–)+)
–q–x
= ( –q)(x–)( –q)–x=q(x)
x∈R\{, –, –, . . .}. (.)
The basicq-integrals are defined by
(Iq,f)(t) =
x
f(t)dqt=x( –q)
∞
k=
qkfxqk,
(Iq,af)(t) =
x
a
f(t)dqt=
x
f(t)dqt–
a
f(t)dqt.
(.)
Definition .([]q-beta function) For anyx,y> ,Bq(x,y) =
t
x–( –qt)(y–)d
qt.
Recall that
Bq(x,y) =
q(x)q(y)
q(x+y)
.
Therefore,
Bq(x,y) =
q(x)q(y)
q(x+y)
=q(x)q(y) q(x+y)
. (.)
Similar toq-beta function, we shall define theq-symmetric beta function.
Definition . (q-symmetric beta function) For any x,y> , Bq(x,y) =
(q–t)x–× ( –t)(y–)dqt.
Lemma . For any x,y> ,the following equality is valid:
Bq(x,y) =
q(x)q(y)
q(x+y)
.
Proof By (.), (.), (.) and Definition ., we have
Bq(x,y) =
q–tx–( –t)(y–)dqt
= –q
∞
k=
qkqkx– –qk+(y–)
= –q
∞
k=
qkqkx–
∞
n=
–qk+n+ –qk+n+y
=
tx– –qt(y–)dqt=Bq(x,y) =
q(x)q(y)
q(x+y)
Lemma .([]) Forα∈R+\N,λ∈(–,∞),the following is valid: (i)Iα
q,xλ=
q(λ+)
q(λ+α+)q
α
+λα xλ+α, (ii) Dα
q,xλ=
q(λ+)
q(λ–α+)q
α
–λα xλ–α.
We shall define aq-symmetric Laplace transform as follows:
Lq
f(t)(s) = –q
s
fq–teq(st)dqt.
Lemma . For anyα∈R\ {. . . , –, –, },
Lq
tα– ( –q)α–
(s) =q(α) sα .
Proof
Lq
tα– ( –q)α–
(s) = –q
s
q–t –q
α–
eq(st)dqt
= ( –q)sα
q–t –q
α– eq(t)dqt
=
sαq(α).
We now turn our attention to a shift theorem for theq-symmetric Laplace transform. First note the following identity.
Dqe–q (t) =Dq
∞
n=( –qn+t) =
(q–q–)t
∞
n=( –qn+t)
–∞
n=( –qnt)
= (q–q–)t
∞
n=( –qn+t)
– –t
= q ( –q)e
–
q
q–t.
Lemma . Let a be any real number.Then
Lq
e–q (at)(s) = s–aq.
Let n denote a positive integer.Then
Lq
tne–q (at)(s) = s–aq
n
j=
–qj
s–qj+a
=( –q )n[n]
q!
Proof Note that
so
Lq
tne–q (at)(s) = –q n
s–aqn+Lq
tn–e–q (at)(s)
= –q n
s–aqn+
–qn– s–aqn–Lq
tn–e–q (at)(s) =· · ·
=
n
j=
–qj
s–aqj+ =
( –q)n[n] q!
(s–a)(n+) .
Next, letF(t) =tμ,F(t) =tν–. DefineF[t] = (t–rt)(ν–)and the convolution
(F∗F)(t) = t –q
(t–rt)(ν–)Frq–tdqr,
whereν∈R\ {. . . , –, –, },μ∈(–, +∞). By the power rule, we have
(F∗F)(t) = q(ν)q(μ+ ) ( –q)
q(μ+ν+ )
tμ+ν.
In fact, by (.), Lemma ., we have
(F∗F)(t) = t –q
(t–rt)(ν–)Frq–tdqr
= –q
t
(t–s)(ν–)Fq–sdqs
= –q
t
(t–s)(ν–)q–sμdqs
= –q
t
(t–s)(ν–)qν–sμq–μνdqs
=q –μν
q(ν)
( –q) q –ν
Iq,sμ
= q(ν)q(μ+ ) ( –q)
q(μ+ν+ )
tμ+ν.
Now we simply apply Lemma . to each ofF,F, F∗F and obtain a convolution theorem.
Theorem .
Lq
(F∗F)(t)(s) =Lq
(F)(t)(s)Lq
Proof
Lq
(F∗F)(t)
(s) = q(ν)q(μ+ ) ( –q)
q(μ+ν+ )
s
q–tμ+νeq(st)dqt
=q(ν)q(μ+ )( –q )μ+ν
( –q)
q(μ+ν+ )
s
q–t –q
μ+ν
eq(st)dqt
=q(ν)q(μ+ )( –q )μ+ν
( –q)
q(μ+ν+ )
q(μ+ν+ )
sμ+ν+
=q(μ+ )q(ν)( –q )μ+ν–
sμ+ν+ , Lq
(F)(t)(s) =( –q )μ
–q
s
q–t –q
μ
eq(st)dqt
= –qμq(μ+ ) sμ+ . Similarly,
Lq
(F)(t)(s) = –qν–q(ν) sν .
Thus (.) holds.
The convolution theorem will be valid for functionsFrepresenting linear sums of func-tions of the formtμ. Clearly,μis not necessary an integer.
Corollary . Let Fbe an analytic function,and let F=tν–onT\{}.Then Theorem.
holds.
We now obtain some of the standard properties for theLq-transform.
Lemma . Assume f =Fis of the type such that(.)is valid.Then
(i) LqIqνf
q–νt(s) =q
–ν
(–q)ν
sν Lq
f(t)(s). (ii) Lq{Dνqf(qνt)}(s) =
q
ν
(–q)ν
sν Lq{f(t)}(s).
Proof (i) Note that
Iν
qf
q–νt= q(ν)
q ν
q–νt
q–νt–s(ν–)fqν–sd
qs
= q(ν)
q ν
–ν t
(t–s)(ν–)fq–sdqs
=t q(ν)
q –ν
(t–ts)(ν–)fq–tsdqs
= –q q(ν)
q –ν
we have
Similarly, by Theorem ., we may easily see that (ii) holds.
For the next set of properties, first note that
It follows by induction that ifmdenotes a positive integer, then
LqDmqf
By Lemma . and (.), we may easily obtain the following Theorem ..
Example . Consider the following fractionalq-symmetric difference equations: Note that each equation given in each of (a), (b) and (c) is not equivalent since our method requires knowledge of the fractional derivatives or integrals of the solution de-fined at zero as well.
We search for analytic solutions onT\ {}for each equation by using aq-symmetric Laplace transform.
For part (a), if we take the Laplace transform of each side of the equation, then by The-orem . we havey(t) = .
For part (b), if we take the Laplace transform of each side of the equation and use the properties of theq-symmetric Laplace transform, we obtain
Lq
By using Theorem ., we have
y(t) = q
For part (c), if we take the Laplace transform of each side of the equation and use the properties of theq-symmetric Laplace transform, we obtain
Lq
Example . Consider the following fractionalq-symmetric difference equation:
DqIq/y
q/t=tμ forT\ {}.
Applying the Laplace transform to each side of the equation, we can get
sq
Example . Consider the problem
Applying the Laplace transform to each side of the equation, we can get
5 Aq-symmetric fractional initial problem andq-symmetric Mittag-Leffler function
The following identity, which is useful for transformingq-symmetric fractional difference equations intoq-symmetric fractional integrals, will be our key in this section.
Example . Let <α≤ and consider theq-symmetric fractional difference equation
cDα
To obtain an explicit clear solution, we apply the method of successive approximation. Sety(t) =aand
ym(t) =a+λIqα,ym–
Form= , we have by the power formula Theorem .
If we proceed inductively and letm−→ ∞, we obtain the solution.
y(t) =a
Definition . Forz,z∈CandR(α) > , theq-symmetric Mittag-Leffler function is defined by
qEα,β(λ,z–z) =
∞
k= λkqk
α
(z–z)(αk)
q
q(αk+β)
.
Whenβ= , we simply useqEα(λ,z–z) :=qEα,(λ,z–z).
According to Definition . above, the solution of theq-symmetric difference equation in Example . is expressed by
y(t) =aqEα(λ,t) +
t
(t–s)(α–)qEα,α
λ,t–q(α–)sfq–sdqs.
Acknowledgements
The authors would like to thank the referee for invaluable comments and insightful suggestions. This work was supported by NSFC project (No. 11161049).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
CH and MS worked together in the derivation of the mathematical results. All authors read and approved the final manuscript.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Received: 6 December 2016 Accepted: 25 May 2017
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