$q$-Deformed conformable fractional Natural transform
UDC 517.9 We develop a new deformation and generalization of the natural integral transform based on the conformable fractional $q$-derivative. We obtain transformation of some deformed functions and apply the transform to solve linear differential equation with given initial conditions.
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| author | Herscovici, O. Mansour, T. Herscovici, O. Mansour, T. |
| author_facet | Herscovici, O. Mansour, T. Herscovici, O. Mansour, T. |
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| description | UDC 517.9
We develop a new deformation and generalization of the natural integral transform based on the conformable fractional $q$-derivative. We obtain transformation of some deformed functions and apply the transform to solve linear differential equation with given initial conditions. |
| doi_str_mv | 10.37863/umzh.v74i8.6099 |
| first_indexed | 2026-03-24T03:26:05Z |
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DOI: 10.37863/umzh.v74i8.6099
UDC 517.9
O. Herscovici1,2, T. Mansour (Univ. Haifa, Israel)
\bfitq -DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM
\bfitq -ДЕФОРМОВАНЕ КОНФОРМНЕ ДРОБОВЕ НАТУРАЛЬНЕ ПЕРЕТВОРЕННЯ
We develop a new deformation and generalization of the natural integral transform based on the conformable fractional
q-derivative. We obtain transformation of some deformed functions and apply the transform to solve linear differential
equation with given initial conditions.
Розроблено нову деформацiю та узагальнення натурального iнтегрального перетворення на основi конформної
дробової q-похiдної. Отримано перетворення деяких деформованих функцiй. Це перетворення застосовано до
розв’язування лiнiйного диференцiального рiвняння з заданими початковими умовами.
1. Introduction. Differential equations appear in many problems of physics, engineering, and
other sciences. So we need powerful mathematical tools to handle them. The integral transforms
are one of the widely used techniques applied for solving differential equations. Generalizations
of integral transforms turn them into more flexible tools to deal with complicated problems. Some
generalizations are based on an extension of transforms to multivariate cases. Other generalizations
can be done by deforming a differential operator and, consequently, an integral. Two leading kinds
of deformations are fractional calculus and q-calculus. Some applications of fractional calculus can
be found in [23], and of q-calculus and fractional q-calculus in [1].
In this paper, we develop a new deformation of the natural integral transform. We define new
extensions of some special functions and apply our deformed transform to them. Among other tools,
a new extension of q, \alpha -Taylor series is proposed. We start here by recalling three integral transforms,
namely the Laplace, the Sumudu, and the natural transform. Further, we develop a new deformation
of the natural transform and show its application to some deformed differential equations.
The Laplace transform is one of the most famous of the integral transforms. It is defined as
F (s) =
\infty \int
0
f(t)e - stdt.
This transform is very useful in solving differential equations with given initial and boundary condi-
tions. Moreover, it can be used for evaluating new identities for functions and integrals (see [18, 21]).
One of its important features is the transformation from the time domain to the frequency domain.
In 1993, Watugala [20] proposed a new integral transform, named Sumudu transform, which is
defined as
S\{ f(t)\} = Fs(u) =
\infty \int
0
1
u
e -
t
u f(t)dt.
1 Corresponding author, orli.herscovici@gmail.com.
2 The research of the first author was supported by the Ministry of Science and Technology, Israel.
c\bigcirc O. HERSCOVICI, T. MANSOUR, 2022
1128 ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
q-DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM 1129
Among its properties, Watugala remarks an easier visualization. This transform was further widely
studied by different researchers (see, for example, [4, 9, 14] and references therein). In his works
Belgacem (see [4] and references therein) denotes the Sumudu transform as an ideal tool for solving
many engineering problems. The Sumudu transform, unlike the Laplace transform, is not focusing
on transformation into the frequency domain. One of its main features is preserving units and scale
during transformation [4].
In 2008, Khan and Khan [12] defined a new transform, called N -transform and renamed later to
natural transform, as following:
R(u, s) = N(f(t)) =
\infty \int
0
f(ut)e - stdt.
It is easy to see that in the case u = 1 we obtain the Laplace transform, and in the case s = 1
we obtain the Sumudu transform. Due to its dual nature and close relationship with both, Laplace
and Sumudu, transforms, the natural transform is more flexible and lets easily to choose during
the problem solution, what way is preferable in each concrete case. In 2017, Kiliçman and Omran
generalized this transform for the two-dimensional case [13].
For both Sumudu and Laplace transforms their q-analogues were obtained and studied. The q-
analogues of the Sumudu transform based on Jackson q-derivative and q-integral were studied by
D. Albayrak and others (see [3] and references therein). The q-analogues of the Laplace transform
based some on the Jackson’s and some on the Tsallis q-derivative and q-integral were studied in
[6, 10, 15 – 17, 19]. Recently a q-analogue was proposed also for the natural transform [2].
In this paper, we define and study a deformation of the natural transform based on the conformable
fractional q-derivative defined by Chung [7]. This deformation is actually a generalization of the q-
deformation based on the Jackson q-derivative. In case when certain transform’s parameters equal
1, it proposes another definition for the q-Sumudu transform, different from [3]. Moreover, our
transform generalizes some results for q analogue of the natural transform defined in [13]. Finally,
we demonstrate some applications of the q-deformed conformable fractional natural transform.
2. Definitions and some properties of the conformable \bfitq -derivative. We start from a defini-
tion of a conformable fractional q-derivative D q,\alpha
x given by Chung in [7], namely,
D q,\alpha
x f(x) =
[\alpha ](f(x) - f(qx)
x\alpha (1 - q\alpha )
= x1 - \alpha D q
xf(x), (2.1)
where [\alpha ] =
1 - q\alpha
1 - q
is the q-number of \alpha , and D q
x is the Jackson q-derivative with respect to the
variable x. It is easy to check that the operator D q,\alpha
x defined by (2.1) is a linear operator. One can
see that in case \alpha = 1, this differential operator coincides with Jackson q-derivative. The following
notation is widely used in q-calculus (a+ b)nq =
\prod n - 1
j=0
(a+ qjb). Accordingly, with this definition,
we have
(a+ b)nq\alpha =
\left\{
\prod n - 1
j=0
(a+ q\alpha jb) for integer n > 0,
1 for n = 0.
(2.2)
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1130 O. HERSCOVICI, T. MANSOUR
Another notation closely related to q-calculus is a q-Pochhammer symbol
(a; q)n =
\left\{
\prod n - 1
j=0
(1 - aqj) for integer n > 0,
1 for n = 0
and
(a; q)\infty =
\infty \prod
j=0
(1 - aqj).
The conformable fractional q-integral is an inverse operation of the conformable fractional q-
derivative
I q,\alpha
x f(x) =
1
[\alpha ]
(1 - q\alpha )x\alpha
\sum
j\geq 0
q\alpha jf(qjx) = I q
x (x
\alpha - 1f(x)),
where I q
x is the Jackson q-integral. Then, for \alpha -monomial x\alpha n, we have
D q,\alpha
x x\alpha n = [n\alpha ]x\alpha (n - 1), I q,\alpha
x x\alpha n =
x\alpha (n+1)
[(n+ 1)\alpha ]
. (2.3)
It can be shown that the Leibniz rule for the conformable fractional q-derivative has the following
form:
D q,\alpha
x (f(x)g(x)) = f(qx)D q,\alpha
x g(x) + (D q,\alpha
x f(x)) g(x). (2.4)
Therefore, by integrating both sides of (2.4), we obtain a rule for integrating by parts\int
(D q,\alpha
x f(x)) g(x)dq,\alpha x = f(x)g(x) -
\int
f(qx)D q,\alpha
x g(x)dq,\alpha x. (2.5)
Chung defined also a conformable fractional q-exponential function
eq,\alpha (x) =
\sum
j\geq 0
x\alpha j
[j\alpha ]!
=
\bigl(
(1 - q)x\alpha ; q\alpha
\bigr)
\infty , (2.6)
where [n\alpha ]! = [\alpha ][2\alpha ] . . . [n\alpha ], with the property
D q,\alpha
x eq,\alpha (ax) = a\alpha eq,\alpha (ax). (2.7)
Note that, as usual, [0]! = 1.
Two new deformations of trigonometric functions were proposed in [7]
eq,\alpha (i
1
\alpha x) = cq,\alpha (x) + isq,\alpha (x), (2.8)
where
cq,\alpha (x) =
\sum
n\geq 0
( - 1)n
[2n\alpha ]!
x2\alpha n, sq,\alpha (x) =
\sum
n\geq 0
( - 1)n\bigl[
(2n+ 1)\alpha
\bigr]
!
x\alpha (2n+1).
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
q-DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM 1131
From (2.8) one can obtain
cq,\alpha (x) =
1
2
\Bigl(
eq,\alpha
\bigl(
i
1
\alpha x
\bigr)
+ eq,\alpha (( - i)
1
\alpha x)
\Bigr)
, (2.9)
sq,\alpha (x) =
1
2i
\Bigl(
eq,\alpha
\bigl(
i
1
\alpha x
\bigr)
- eq,\alpha (( - i)
1
\alpha x)
\Bigr)
. (2.10)
By applying (2.7), it is easy to show that
D q,\alpha
x cq,\alpha (x) = - sq,\alpha (x), D q,\alpha
x sq,\alpha (x) = cq,\alpha (x).
By using the definition of the deformed conformable fractional derivative (2.1), we evaluate con-
formable derivative of a function
1
eq,\alpha (ax)
:
D q,\alpha
x
1
eq,\alpha (ax)
= x1 - \alpha D q
x
1
eq,\alpha (ax)
= x1 - \alpha
1
eq,\alpha (ax)
- 1
eq,\alpha (qax)
x - qx
=
= x1 - \alpha eq,\alpha (qax) - eq,\alpha (ax)
eq,\alpha (qax)eq,\alpha (ax)(x - qx)
= - D q,\alpha
x eq,\alpha (ax)
eq,\alpha (qax)eq,\alpha (ax)
,
and, by applying (2.7), we obtain
D q,\alpha
x
1
eq,\alpha (ax)
= - a\alpha eq,\alpha (ax)
eq,\alpha (qax)eq,\alpha (ax)
= - a\alpha
eq,\alpha (qax)
. (2.11)
It is easy to see that D q,\alpha
x C = 0, where C is a constant (C does not depend on x). Indeed, from
(2.1), we have D q,\alpha
x C = x1 - \alpha D q
xC = 0.
We can see that D q,\alpha
x x\alpha = [\alpha ]. We would like to build a sequence of polynomials P0(x), P1(x), . . .
. . . , Pn(x) of degrees 0, \alpha , . . . , n\alpha , respectively, so that
D q,\alpha
x Pn(x) = Pn - 1(x),
Pn
\bigl(
a
1
\alpha
\bigr)
= 0
with initial condition P0(x) = 1. Therefore, the polynomial P1(x) has the form P1(x) = (x\alpha -
- a)/[\alpha ]. Obviously, P1
\bigl(
a
1
\alpha
\bigr)
=
\Bigl( \bigl(
a
1
\alpha
\bigr) \alpha - a
\Bigr)
/[\alpha ] = 0 and D q,\alpha
x P1(x) = 1 = P0(x).
Proposition 2.1. For all n \in \BbbN , we have
D q,\alpha
x (x\alpha - a)nq\alpha = [n\alpha ](x\alpha - a)n - 1
q\alpha .
Proof. We proceed the proof by induction on n. It is easy to see that, for n = 1, we have
D q,\alpha
x (x\alpha - a)1q\alpha = D q,\alpha
x (x\alpha - a) = [\alpha ],
and statement holds. Let us assume that statement holds for some integer k. We will prove it for
k + 1. By (2.2), we have (x\alpha - a)k+1
q\alpha = (x\alpha - a)kq\alpha (x
\alpha - q\alpha ka). By applying (2.4), we obtain
D q,\alpha
x (x\alpha - a)k+1
q\alpha = D q,\alpha
x
\Bigl(
(x\alpha - a)kq\alpha (x
\alpha - q\alpha ka)
\Bigr)
=
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1132 O. HERSCOVICI, T. MANSOUR
= (q\alpha x\alpha - q\alpha ka)[k\alpha ](x\alpha - a)k - 1
q\alpha + [\alpha ](x\alpha - a)kq\alpha =
= q\alpha (x\alpha - q\alpha (k - 1)a)[k\alpha ](x\alpha - a)k - 1
q\alpha + [\alpha ](x\alpha - a)kq\alpha =
= q\alpha [k\alpha ](x\alpha - a)kq\alpha + [\alpha ](x\alpha - a)kq\alpha =
= (x\alpha - a)kq\alpha
\biggl(
q\alpha
1 - qk\alpha
1 - q
+
1 - q\alpha
1 - q
\biggr)
=
= (x\alpha - a)kq\alpha
q\alpha - q\alpha (k+1) + 1 - q\alpha
1 - q
=
= (x\alpha - a)kq\alpha
1 - q\alpha (k+1)
1 - q
= [(k + 1)\alpha ](x\alpha - a)kq\alpha ,
which completes the induction.
Proposition 2.2. For all n \in \BbbN , we have
D q,\alpha
x (a - x\alpha )nq\alpha = - [n\alpha ](a - q\alpha x\alpha )n - 1
q\alpha .
Proof. By (2.2), we have
(a - x\alpha )nq\alpha = (a - x\alpha )(a - q\alpha x\alpha )(a - q2\alpha x\alpha ) . . . (a - q(n - 1)\alpha x\alpha ) =
= (a - x\alpha )q\alpha (q - \alpha a - x\alpha )q2\alpha (q - 2\alpha a - x\alpha ) . . . q(n - 1)\alpha (q - (n - 1)\alpha a - x\alpha ) =
= ( - 1)nq
\alpha n(n - 1)
2 (x\alpha - q - (n - 1)\alpha a) . . . (x\alpha - q - 2\alpha a)(x\alpha - q - \alpha a)(x\alpha - a) =
= ( - 1)nq
\alpha n(n - 1)
2 (x\alpha - q - (n - 1)\alpha a)nq\alpha .
Now, by using Proposition 2.1, we obtain
D q,\alpha
x (a - x\alpha )nq\alpha = D q,\alpha
x
\Bigl(
( - 1)nq
\alpha n(n - 1)
2 (x\alpha - q - (n - 1)\alpha a)nq\alpha
\Bigr)
=
= [n\alpha ]( - 1)nq
\alpha n(n - 1)
2 (x\alpha - q - (n - 1)\alpha a)n - 1
q\alpha =
= ( - 1)n[n\alpha ]q\alpha q2\alpha . . . q(n - 1)\alpha (x\alpha - q - (n - 1)\alpha a) . . . (x\alpha - q - 2\alpha a)(x\alpha - q - \alpha a) =
= - [n\alpha ]q\alpha q2\alpha . . . q(n - 1)\alpha (q - (n - 1)\alpha a - x\alpha ) . . . (q - 2\alpha a - x\alpha )(q - \alpha a - x\alpha ) =
= - [n\alpha ](a - q(n - 1)\alpha x\alpha ) . . . (a - q2\alpha x\alpha )(a - q\alpha x\alpha ) =
= - [n\alpha ](a - q\alpha x\alpha )n - 1
q\alpha ,
and the proof is complete.
Now, we can state that Pn(x) =
(x\alpha - a)nq\alpha
[n\alpha ]!
. Indeed, Pn
\bigl(
a
1
\alpha
\bigr)
= 0 and
D q,\alpha
x Pn(x) = D q,\alpha
x
(x\alpha - a)nq\alpha
[n\alpha ]!
=
[n\alpha ](x\alpha - a)n - 1
q\alpha
[n\alpha ]!
=
(x\alpha - a)n - 1
q\alpha
[(n - 1)\alpha ]!
= Pn - 1(x).
Therefore, by using the [11] (Theorems 2.1 and 8.1), we can state the following result.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
q-DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM 1133
Theorem 2.1. Any polynomial or formal power series function f(x) can be expressed via the
generalized conformable fractional q-Taylor expansion about x = a
1
\alpha as
f(x) =
\sum
n\geq 0
\bigl(
D q,\alpha
x
\bigr) n
f
\bigl(
a
1
\alpha
\bigr) (x\alpha - a)nq\alpha
[n\alpha ]!
.
Let us define one more q-deformed conformable fractional exponential function
Eq,\alpha (x) =
\sum
j\geq 0
q\alpha j(j - 1)/2 x\alpha j
[j\alpha ]!
= (1 + (1 - q)x\alpha )\infty q\alpha . (2.12)
It is easy to see that Eq,\alpha (0) = 1. By using (2.3), we obtain
D q,\alpha
x Eq,\alpha (ax) =
\sum
j\geq 1
q\alpha j(j - 1)/2a
\alpha jx\alpha (j - 1)
[(j - 1)\alpha ]!
=
\sum
j\geq 0
q\alpha (j+1)j/2a
\alpha (j+1)x\alpha j
[j\alpha ]!
=
= a\alpha
\sum
j\geq 0
q\alpha j(j - 1)/2 q
\alpha ja\alpha jx\alpha j
[j\alpha ]!
= a\alpha Eq,\alpha (qax).
Let us define the q-deformed conformable fractional Gamma function for some n \geq 1:
\Gamma q,\alpha (n+ 1) =
\infty \int
0
x\alpha n
1
eq,\alpha (qx)
dq,\alpha x. (2.13)
Proposition 2.3. For all n > 0, the function \Gamma q,\alpha (n + 1) defined by (2.13) satisfies the recur-
rence relation
\Gamma q,\alpha (n+ 1) = [n\alpha ]\Gamma q,\alpha (n)
with the initial condition \Gamma q,\alpha (1) = 1.
Proof. We proceed the proof by induction on n. For n = 0, we have
\Gamma q,\alpha (1) =
\infty \int
0
1
eq,\alpha (qx)
dq,\alpha x = -
\infty \int
0
D q,\alpha
x
1
eq,\alpha (x)
= - 1
eq,\alpha (x)
\bigm| \bigm| \bigm| \bigm| \infty
0
= 1.
Let us assume that the claim holds for k - 1 and let us prove it for k. Let us consider now the
function \Gamma q,\alpha (k + 1) for some k. By (2.13), we have
\Gamma q,\alpha (k + 1) =
\infty \int
0
x\alpha k
1
eq,\alpha (qx)
dq,\alpha x,
from where, by rearranging and using (2.4), we get
\Gamma q,\alpha (k + 1) = -
\infty \int
0
x\alpha k
\biggl(
D q,\alpha
x
1
eq,\alpha (x)
\biggr)
dq,\alpha x =
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1134 O. HERSCOVICI, T. MANSOUR
= - x\alpha k
1
eq,\alpha (x)
\bigm| \bigm| \bigm| \infty
0
+
\infty \int
0
\Bigl(
D q,\alpha
x x\alpha k
\Bigr) 1
eq,\alpha (qx)
dq,\alpha x =
= [k\alpha ]
\infty \int
0
x\alpha (k - 1) 1
eq,\alpha (qx)
dq,\alpha x = [k\alpha ]\Gamma q,\alpha (k),
which completes the proof.
One can immediately obtain from the last proposition the following result.
Corollary 2.1. For all n \in \BbbN , it holds that
\Gamma q,\alpha (n+ 1) = [n\alpha ]!.
The function \Gamma q,\alpha (n) defined as (2.13) is a q-deformed conformable fractional extension of the
\Gamma -function. It is well-known that \Gamma -function is closely related to the B-function, that is, for the
B-function, defined as
B(m,n) =
1\int
0
xm - 1(1 - x)n - 1dx,
holds that
B(m,n) =
\Gamma (m)\Gamma (n)
\Gamma (m+ n)
. (2.14)
Let us define the function
Bq,\alpha (m,n) =
1\int
0
x\alpha (m - 1)(1 - q\alpha x\alpha )n - 1
q\alpha dq,\alpha x.
Proposition 2.4. For all natural m, n, it holds that
Bq,\alpha (m,n) =
\Gamma q,\alpha (m)\Gamma q,\alpha (n)
\Gamma q,\alpha (m+ n)
.
Proof. With the notations f(qx) = (1 - q\alpha x\alpha )n - 1
q\alpha and D q,\alpha
x g(x) = x\alpha (m - 1)dq,\alpha x, we obtain
f(x) = (1 - x\alpha )n - 1
q\alpha and g(x) =
x\alpha m
[\alpha m]
. Therefore, by using Proposition 2.2, we get D q,\alpha
x f(x) =
= - [(n - 1)\alpha ](1 - q\alpha x\alpha )n - 2
q\alpha . Applying (2.5) with our notations yields
Bq,\alpha (m,n) = (1 - x\alpha )n - 1
q\alpha
x\alpha m
[\alpha m]
\bigm| \bigm| \bigm| \bigm| 1
0
+
1\int
0
[(n - 1)\alpha ](1 - q\alpha x\alpha )n - 2
q\alpha
x\alpha m
[\alpha m]
dq,\alpha x =
=
[(n - 1)\alpha ]
[\alpha m]
1\int
0
x\alpha m(1 - q\alpha x\alpha )n - 2
q\alpha dq,\alpha x.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
q-DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM 1135
Thus, by assuming m and n are natural numbers, we obtain
Bq,\alpha (m,n) =
1\int
0
x\alpha (m - 1)(1 - q\alpha x\alpha )n - 1
q\alpha dq,\alpha x =
=
[(n - 1)\alpha ]
[\alpha m]
1\int
0
x\alpha m(1 - q\alpha x\alpha )n - 2
q\alpha dq,\alpha x =
=
[(n - 1)\alpha ]
[\alpha m]
[(n - 2)\alpha ]
[\alpha (m+ 1)]
1\int
0
x\alpha (m+1)(1 - q\alpha x\alpha )n - 3
q\alpha dq,\alpha x = . . .
. . . =
[(n - 1)\alpha ]
[m\alpha ]
[(n - 2)\alpha ]
[(m+ 1)\alpha ]
. . .
[2\alpha ]
[(m+ n - 3)\alpha ]
1\int
0
x\alpha (m+n - 3)(1 - q\alpha x\alpha )1q\alpha dq,\alpha x =
=
[(n - 1)\alpha ]
[m\alpha ]
[(n - 2)\alpha ]
[(m+ 1)\alpha ]
. . .
[2\alpha ]
[(m+ n - 3)\alpha ]
\times
\times
\left( x\alpha (m+n - 2)
[(m+ n - 2)\alpha ]
(1 - x\alpha )
\bigm| \bigm| \bigm| 1
0
+
1\int
0
[\alpha ]
x\alpha (m+n - 2)
[(m+ n - 2)\alpha ]
dq,\alpha x
\right) =
=
[(n - 1)\alpha ][(n - 2)\alpha ] . . . [2\alpha ][\alpha ]
[m\alpha ][(m+ 1)\alpha ] . . . [(m+ n - 3)\alpha ][(m+ n - 2)\alpha ]
x\alpha (m+n - 1)
[\alpha (m+ n - 1)]
\bigm| \bigm| \bigm| \bigm| 1
0
=
=
[(n - 1)\alpha ]! [(m - 1)\alpha ]!
[(m+ n - 1)\alpha ]!
=
\Gamma q,\alpha (n)\Gamma q,\alpha (m)
\Gamma q,\alpha (m+ n)
. (2.15)
Proposition 2.4 is proved.
Remark 2.1. It is easy to see that for \alpha = q = 1, (2.15) turns into (2.14). Thus, functions \Gamma q,\alpha
and Bq,\alpha are q-deformed conformable fractional extensions of the well-known \Gamma - and B-functions,
respectively. This proposition may be extended for all positive m, n.
3. \bfitq -Deformed conformable fractional natural transform. We define now a q-deformed
conformable fractional natural transform as
Nq,\alpha (f(t)) =
\infty \int
0
f(ut)
1
eq,\alpha (qst)
dq,\alpha t, s > 0. (3.1)
Then we have
Nq,\alpha (1) =
\infty \int
0
1
eq,\alpha (qst)
dq,\alpha t =
= - 1
s\alpha
\infty \int
0
D q,\alpha
t
1
eq,\alpha (st)
dq,\alpha t = - 1
s\alpha
1
eq,\alpha (st)
\bigm| \bigm| \bigm| \bigm| \infty
0
=
1
s\alpha
.
Let us now to obtain a transform of \alpha -monomial.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1136 O. HERSCOVICI, T. MANSOUR
Proposition 3.1. For all integer N \geq 0,
Nq,\alpha (t
\alpha N ) =
u\alpha N
s\alpha (N+1)
\Gamma q,\alpha (N + 1).
Proof. By definition (3.1), we have
Nq,\alpha (t
\alpha N ) =
\infty \int
0
u\alpha N t\alpha N
1
eq,\alpha (qst)
dq,\alpha t =
= - 1
s\alpha
u\alpha N
\infty \int
0
\biggl(
D q,\alpha
t
1
eq,\alpha (st)
\biggr)
t\alpha Ndq,\alpha t.
Integrating by parts of the last equation leads to
Nq,\alpha (t
\alpha N ) = - u\alpha N
s\alpha
\left\{ 1
eq,\alpha (st)
t\alpha N
\bigm| \bigm| \bigm| \infty
0
-
\infty \int
0
1
eq,\alpha (qst)
\bigl(
D q,\alpha
t t\alpha N
\bigr)
dq,\alpha t
\right\} =
=
u\alpha N
s\alpha
\infty \int
0
[N\alpha ] t\alpha (N - 1) 1
eq,\alpha (qst)
dq,\alpha t =
=
u\alpha
s\alpha
[N\alpha ]
\infty \int
0
u\alpha (N - 1)t\alpha (N - 1) 1
eq,\alpha (qst)
dq,\alpha t =
=
u\alpha
s\alpha
[N\alpha ]Nq,\alpha (t
\alpha (N - 1)). (3.2)
Thus, by (3.2), we obtain
Nq,\alpha (t
\alpha N ) =
u\alpha
s\alpha
[N\alpha ]
u\alpha
s\alpha
[(N - 1)\alpha ] . . .
u\alpha
s\alpha
[\alpha ]
1
s\alpha
=
u\alpha N
s\alpha (N+1)
[N\alpha ]!. (3.3)
Applying Corollary 2.1 completes the proof.
Now let us consider the transform of two deformed exponential functions.
Proposition 3.2. The q-deformed conformable natural transforms of the q-deformed conformable
exponential functions are given by
Nq,\alpha (eq,\alpha (at)) =
1
s\alpha - a\alpha u\alpha
,
Nq,\alpha (Eq,\alpha (at)) =
\infty \sum
n=0
q
\alpha n(n - 1)
2
(ua)\alpha n
s\alpha (n+1)
.
Proof. By applying transform (3.1) to the deformed exponential funciton (2.6), we obtain
Nq,\alpha (eq,\alpha (at)) =
\infty \int
0
eq,\alpha (aut)
1
eq,\alpha (qst)
dq,\alpha t =
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
q-DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM 1137
=
\infty \int
0
\infty \sum
n=0
a\alpha nu\alpha nt\alpha n
[n\alpha ]!
1
eq,\alpha (qst)
dq,\alpha t =
=
\infty \sum
n=0
a\alpha n
[n\alpha ]!
\infty \int
0
u\alpha nt\alpha n
1
eq,\alpha (qst)
dq,\alpha t =
=
\infty \sum
n=0
a\alpha n
[n\alpha ]!
Nq,\alpha (t
\alpha n) =
\infty \sum
n=0
a\alpha n
[n\alpha ]!
u\alpha n
s\alpha (n+1)
[n\alpha ]! =
=
1
s\alpha
\infty \sum
n=0
(au)\alpha n
s\alpha n
=
1
s\alpha - a\alpha u\alpha
.
By applying transform (3.1) to the deformed exponential funciton (2.12), we obtain
Nq,\alpha (Eq,\alpha (at)) =
\infty \int
0
Eq,\alpha (aut)
1
eq,\alpha (qst)
dq,\alpha t =
=
\infty \int
0
\infty \sum
n=0
q
\alpha n(n - 1)
2
a\alpha nu\alpha nt\alpha n
[n\alpha ]!
1
eq,\alpha (qst)
dq,\alpha t =
=
\infty \sum
n=0
q
\alpha n(n - 1)
2
a\alpha n
[n\alpha ]!
\infty \int
0
u\alpha nt\alpha n
1
eq,\alpha (qst)
dq,\alpha t =
=
\infty \sum
n=0
q
\alpha n(n - 1)
2
a\alpha n
[n\alpha ]!
u\alpha n
s\alpha (n+1)
[n\alpha ]! =
=
\infty \sum
n=0
q
\alpha n(n - 1)
2
(ua)\alpha n
s\alpha (n+1)
,
and the proof is complete.
Let us consider now the transform of the deformed trigonometric functions (2.9), (2.10).
Proposition 3.3. The deformed conformable fractional natural transform of deformed trigono-
metric functions, defined by (2.9) and (2.10), is given by
Nq,\alpha (cq,\alpha (t)) =
s\alpha
s2\alpha + u2\alpha
,
Nq,\alpha (sq,\alpha (t)) =
u\alpha
s2\alpha + u2\alpha
.
Proof. It follows from the definition (2.9) and the linearity of the transform Nq,\alpha that
Nq,\alpha (cq,\alpha (at)) =
1
2
\Bigl(
Nq,\alpha
\Bigl(
eq,\alpha
\bigl(
i
1
\alpha at
\bigr)
+Nq,\alpha
\bigl(
eq,\alpha
\bigl(
( - i)
1
\alpha at
\bigr) \bigr) \Bigr) \Bigr)
=
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1138 O. HERSCOVICI, T. MANSOUR
=
1
2
\Biggl(
1
s\alpha - (i
1
\alpha a)\alpha u\alpha
+
1
s\alpha - (( - i)
1
\alpha a)\alpha u\alpha
\Biggr)
=
=
1
2
\biggl(
1
s\alpha - i(au)\alpha
+
1
s\alpha + i(au)\alpha
\biggr)
=
=
1
2
s\alpha + i(au)\alpha + s\alpha - i(au)\alpha
(s\alpha - i(au)\alpha )(s\alpha + i(au)\alpha )
=
s\alpha
s2\alpha + (au)2\alpha
.
In the same way, from the definition (2.10) and the linearity of the transform Nq,\alpha , we obtain
Nq,\alpha (sq,\alpha (at)) =
1
2i
\Bigl(
Nq,\alpha
\Bigl(
eq,\alpha
\bigl(
i
1
\alpha at
\bigr)
- Nq,\alpha
\bigl(
eq,\alpha
\bigl(
( - i)
1
\alpha at
\bigr) \bigr) \Bigr) \Bigr)
=
=
1
2i
\Biggl(
1
s\alpha - (i
1
\alpha a)\alpha u\alpha
- 1
(s\alpha - (( - i)
1
\alpha a)\alpha u\alpha
\Biggr)
=
=
1
2i
\biggl(
1
s\alpha - i(au)\alpha
- 1
s\alpha + i(au)\alpha
\biggr)
=
=
1
2i
s\alpha + i(au)\alpha - s\alpha + i(au)\alpha
(s\alpha - i(au)\alpha )(s\alpha + i(au)\alpha )
=
=
(au)\alpha
s2\alpha + (au)2\alpha
,
and the proof is complete.
Suppose that function f(t) has a polynomial or formal power series expansion in \alpha -monomials
t\alpha n. Let us denote such function f(t) by f\alpha (t). We consider now the transform of a derivative
D q,\alpha
t f\alpha (t):
Nq,\alpha
\bigl(
D q,\alpha
t f\alpha (t)
\bigr)
=
\infty \int
0
(D q,\alpha
t f\alpha ) (ut)
1
eq,\alpha (qst)
dq,\alpha t =
=
1
u\alpha
\infty \int
0
(D q,\alpha
t f\alpha ) (y)
1
eq,\alpha
\Bigl(
qs
y
u
\Bigr) dq,\alpha y =
=
1
u\alpha
f\alpha (y)
1
eq,\alpha
\Bigl(
qs
y
u
\Bigr)
\bigm| \bigm| \bigm| \bigm| \bigm| \bigm|
\infty
0
+
1
u\alpha
s\alpha
u\alpha
\infty \int
0
f\alpha (y)
1
eq,\alpha
\Bigl(
qs
y
u
\Bigr) dq,\alpha y =
= - 1
u\alpha
f\alpha (0) +
s\alpha
u\alpha
\infty \int
0
f\alpha (ut)
1
eq,\alpha (qst)
dq,\alpha t =
= - 1
u\alpha
f\alpha (0) +
s\alpha
u\alpha
Nq,\alpha (f\alpha (t)).
Let us rewrite it as
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
q-DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM 1139
Nq,\alpha (D
q,\alpha
t f\alpha (t)) =
s\alpha
u\alpha
Nq,\alpha (f\alpha (t)) -
1
u\alpha
f\alpha (0). (3.4)
Therefore,
Nq,\alpha
\bigl( \bigl(
D q,\alpha
t
\bigr) 2
f\alpha (t)
\bigr)
=
s\alpha
u\alpha
Nq,\alpha
\bigl(
D q,\alpha
t f\alpha (t)
\bigr)
- 1
u\alpha
\bigl(
D q,\alpha
t f\alpha
\bigr)
(0) =
=
s\alpha
u\alpha
\biggl(
s\alpha
u\alpha
Nq,\alpha (f\alpha (t)) -
1
u\alpha
f\alpha (0)
\biggr)
- 1
u\alpha
\bigl(
D q,\alpha
t f\alpha
\bigr)
(0) =
=
\biggl(
s\alpha
u\alpha
\biggr) 2
Nq,\alpha (f\alpha (t)) -
1
u\alpha
s\alpha
u\alpha
f\alpha (0) -
1
u\alpha
\bigl(
D q,\alpha
t f\alpha
\bigr)
(0).
Thus, we can state the following result.
Theorem 3.1. Suppose that function f\alpha (t) has polynomials or formal power series expansion
in \alpha -monomials t\alpha . Then, for all integer n > 0, it holds that
Nq,\alpha
\bigl( \bigl(
D q,\alpha
t
\bigr) n
f\alpha (t)
\bigr)
=
\biggl(
s\alpha
u\alpha
\biggr) n
Nq,\alpha (f\alpha (t)) -
1
u\alpha
n - 1\sum
j=0
\biggl(
s\alpha
u\alpha
\biggr) n - 1 - j\bigl(
D q,\alpha
t
\bigr) j
f\alpha (0).
Proof. The proof is by induction on n. The theorem statement holds for n = 1 as it shown in
(3.4). Let us assume the formula holds for k, and let us prove it for k + 1. By using (3.4), we have
Nq,\alpha
\Bigl( \bigl(
D q,\alpha
t
\bigr) k+1
f\alpha (t)
\Bigr)
=
s\alpha
u\alpha
Nq,\alpha
\Bigl( \bigl(
D q,\alpha
t
\bigr) k
f\alpha (t)
\Bigr)
- 1
u\alpha
\bigl( \bigl(
D q,\alpha
t
\bigr) k
f\alpha
\bigr)
(0),
and, by induction’s assumption, we get
Nq,\alpha
\Bigl( \bigl(
D q,\alpha
t
\bigr) k+1
f\alpha (t)
\Bigr)
= - 1
u\alpha
\bigl( \bigl(
D q,\alpha
t
\bigr) k
f\alpha
\bigr)
(0)+
+
s\alpha
u\alpha
\left( \biggl( s\alpha
u\alpha
\biggr) k
Nq,\alpha (f\alpha (t)) -
1
u\alpha
k - 1\sum
j=0
\biggl(
s\alpha
u\alpha
\biggr) k - 1 - j \bigl(
D q,\alpha
t
\bigr) j
f\alpha (t)
\right) =
=
\biggl(
s\alpha
u\alpha
\biggr) k+1
Nq,\alpha (f\alpha (t)) -
1
u\alpha
k - 1\sum
j=0
\biggl(
s\alpha
u\alpha
\biggr) k - j \bigl(
D q,\alpha
t
\bigr) j
f\alpha (t) -
1
u\alpha
\bigl( \bigl(
D q,\alpha
t
\bigr) k
f\alpha
\bigr)
(0) =
=
\biggl(
s\alpha
u\alpha
\biggr) k+1
Nq,\alpha (f\alpha (t)) -
1
u\alpha
k\sum
j=0
\biggl(
s\alpha
u\alpha
\biggr) k - j \bigl(
D q,\alpha
t
\bigr) j
f\alpha (t),
which completes the proof.
Let us consider now two examples of applying the conformable fractional q-deformed natural
transform for solving differential equations.
Example 3.1. This example is an extension of the Example 4.2.4 in [8]. We have a differential
equation \Bigl( \bigl(
D q,\alpha
t
\bigr) 3
+
\bigl(
D q,\alpha
t
\bigr) 2 - 6D q,\alpha
t
\Bigr)
f(t) = 0,
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1140 O. HERSCOVICI, T. MANSOUR
with the initial condition
f(0) = 1, D q,\alpha
t f(0) = 0,
\bigl(
D q,\alpha
t
\bigr) 2
f(0) = 5.
We apply our conformable fractional natural q-transform to the differential equation, and, by using
the Theorem 3.1, obtain\biggl(
s\alpha
u\alpha
\biggr) 3
\=f - 1
u\alpha
2\sum
j=0
\biggl(
s\alpha
u\alpha
\biggr) 2 - j \bigl(
D q,\alpha
t
\bigr) j
f(0) +
\biggl(
s\alpha
u\alpha
\biggr) 2
\=f -
- 1
u\alpha
1\sum
j=0
\biggl(
s\alpha
u\alpha
\biggr) 1 - j \bigl(
D q,\alpha
t
\bigr) j
f(0) - 6 \cdot s
\alpha
u\alpha
\=f +
6
u\alpha
f(0) = 0,
where \=f = Nq,\alpha (f(t)). Let w =
s\alpha
u\alpha
. Then we get
w3 \=f - w2
u\alpha
f(0) - w
u\alpha
D q,\alpha
t f(0) - 1
u\alpha
\bigl(
D q,\alpha
t
\bigr) 2
f(0) + w2 \=f -
- w
u\alpha
f(0) - 1
u\alpha
D q,\alpha
t f(0) - 6w \=f +
6
u\alpha
f(0) = 0,
and, by applying the initial conditions, we obtain
\=f =
1
u\alpha
w2 + w - 1
w(w2 + w - 6)
=
=
1
u\alpha
\biggl(
1
6w
+
1
3(w + 3)
+
1
2(w - 2)
\biggr)
=
=
1
6
1
s\alpha
+
1
3
1
s\alpha + 3u\alpha
+
1
2
1
s\alpha - 2u\alpha
=
=
1
6
1
s\alpha
+
1
3
1
s\alpha -
\bigl(
( - 3)
1
\alpha
\bigr) \alpha
u\alpha
+
1
2
1
s\alpha -
\bigl(
2
1
\alpha
\bigr) \alpha
u\alpha
.
Now, by using the results of the Proposition 3.2, we can find the original function f(t) as following:
f(t) =
1
6
+
1
3
eq,\alpha
\bigl(
( - 3)
1
\alpha t
\bigr)
+
1
2
eq,\alpha
\bigl(
2
1
\alpha t
\bigr)
.
One can easily see that this solution for q = 1, \alpha = 1 becomes f(t) =
1
6
+
1
3
e - 3t +
1
2
e2t, which
coincides with the solution of [8].
Example 3.2. Let us consider now an extension of the differential equation appearing in Example
4 of [22]:
D q,\alpha
t f(t) + 3f(t) = 13sq,\alpha
\bigl(
2
1
\alpha t
\bigr)
,
with the initial condition f(0) = 6. Again, let \=f = Nq,\alpha (f(t)). By applying the integral transform
to this differential equation, we obtain the equation
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
q-DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM 1141
s\alpha
u\alpha
\=f - 1
u\alpha
f(0) + 3 \=f = 13 \cdot 2u\alpha
s2\alpha + 4u2\alpha
,
which, by applying the initial condition, can be rewritten as
s\alpha + 3u\alpha
u\alpha
\=f =
26u\alpha
s2\alpha + 4u2\alpha
+
6
u\alpha
.
Now we can express the transformation \=f as
\=f =
1
s\alpha + 3u\alpha
5 - u2\alpha + 6s2\alpha
s2\alpha + 4u2\alpha
=
=
A
s\alpha + 3u\alpha
+
Bs\alpha + Cu\alpha
s2\alpha + 4u2\alpha
. (3.5)
The unknown constants A, B, C can be found by comparing two expressions for \=f. One can easily
check that A = 8, B = - 2, and C = 6. Therefore (3.5) can be rewritten as
\=f =
8
s\alpha + 3u\alpha
- 2
s\alpha
s2\alpha + 4u2\alpha
+ 3
2u\alpha
s2\alpha + 4u2\alpha
,
where, by Propositions 3.2 and 3.3, we can obtain the original function f(t) as following:
f(t) = 8eq,\alpha
\bigl(
( - 3)
1
\alpha t
\bigr)
- 2cq,\alpha
\bigl(
2
1
\alpha t
\bigr)
+ 3sq,\alpha
\bigl(
2
1
\alpha t
\bigr)
.
Note that for q = 1, \alpha = 1 we obtain the solution of [22].
We have considered transforms of functions and their derivatives. Let us consider now derivatives
of the transform. We would like to emphasize that a function f(t) is, actually, polynomial or formal
power series in t\alpha -monomials. Let us denote by Rq,\alpha (u, s) the conformable fractional q-deformed
natural transform (3.1). With the notation e - 1
q,\alpha (t) =
1
eq,\alpha (t)
we can state the following lemma.
Lemma 3.1. For all integer n > 0,\bigl(
D q,\alpha
s
\bigr) n
e - 1
q,\alpha
\bigl(
q - (n - 1)st
\bigr)
= ( - 1)nt\alpha nq - (
n
2)\alpha e - 1
q,\alpha (qst).
Proof. By applying the operator D q,\alpha with respect to s consequently n - 1 times to the function
e - 1
q,\alpha
\bigl(
q - (n - 1)st
\bigr)
and using (2.11), we obtain\bigl(
D q,\alpha
s
\bigr) n
e - 1
q,\alpha
\bigl(
q - (n - 1)st
\bigr)
=
\bigl(
- (q - (n - 1)t)
\alpha \bigr)
. . .
\bigl(
- (q - 1t)\alpha
\bigr)
D q,\alpha
s e - 1
q,\alpha (st).
Applying (2.11) one more time, we have\bigl(
D q,\alpha
s
\bigr) n
e - 1
q,\alpha
\bigl(
q - (n - 1)st
\bigr)
=
\bigl(
- (q - (n - 1)t)
\alpha \bigr)
. . .
\bigl(
- (q - 1t)
\alpha \bigr)
( - t\alpha )e - 1
q,\alpha (qst) =
= ( - 1)nt\alpha nq - (
n
2)\alpha e - 1
q,\alpha (qst),
where from the lemma’s statement follows.
Proposition 3.4. Suppose that function f\alpha (t) has polynomials or formal power series expansion
in \alpha -monomials t\alpha . Then, for all integer n > 0, it holds that
Nq,\alpha
\bigl(
t\alpha nf\alpha (t)
\bigr)
= ( - 1)nq(
n
2)\alpha u\alpha n
\bigl(
D q,\alpha
s
\bigr) n
Rq,\alpha (u, q
- ns),
where Rq,\alpha (u, s) = Nq,\alpha (f\alpha (t)).
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1142 O. HERSCOVICI, T. MANSOUR
Proof. We have
\bigl(
D q,\alpha
s
\bigr) n
Rq,\alpha (u, q
- ns) =
\bigl(
D q,\alpha
s
\bigr) n \infty \int
0
f\alpha (ut)e
- 1
q,\alpha
\bigl(
q - (n - 1)st
\bigr)
dq,\alpha t =
=
\infty \int
0
f\alpha (ut)
\bigl(
D q,\alpha
s
\bigr) n
e - 1
q,\alpha
\bigl(
q - (n - 1)st
\bigr)
dq,\alpha t.
By using Lemma 3.1, we get
\bigl(
D q,\alpha
s
\bigr) n
Rq,\alpha (u, q
- ns) =
\infty \int
0
f\alpha (ut)( - 1)nt\alpha nq - (
n
2)\alpha e - 1
q,\alpha (qst)dq,\alpha t =
= ( - 1)n
q - (
n
2)\alpha
u\alpha n
\infty \int
0
(ut)\alpha nf\alpha (ut)e
- 1
q,\alpha
\bigl(
q - (n - 1)st
\bigr)
dq,\alpha t =
= ( - 1)n
q - (
n
2)\alpha
u\alpha n
Nq,\alpha
\bigl(
t\alpha nf\alpha (t)
\bigr)
.
The rearrangement of the last equation completes the proof.
The natural transform is a function of two variables, namely u and s. The previous proposition
establishes a connection between the transform of product of f\alpha (t) with a positive power of \alpha -
monomials t\alpha and q, \alpha -deformed derivative with respect to one of the variables, namely s, of
q, \alpha -transform. Let us consider now a derivative of deformed transform with respect to its another
variable u.
Proposition 3.5. Suppose that function f\alpha (t) has the following expansion:
f\alpha (t) =
\infty \sum
m=0
amt\alpha m.
Then, for all integer n > 0, it holds that
Nq,\alpha
\bigl(
t\alpha nf\alpha (t)
\bigr)
=
u\alpha n
s\alpha n
(D q,\alpha
u )nu\alpha nRq,\alpha (u, s).
Proof. If f\alpha (t) =
\sum \infty
m=0
amt\alpha m, then
Nq,\alpha
\bigl(
t\alpha nf\alpha (t)
\bigr)
= Nq,\alpha
\Biggl( \infty \sum
m=0
amt\alpha (m+n)
\Biggr)
.
By using the linearity of the transform and applying (3.3), we get
Nq,\alpha
\bigl(
t\alpha nf\alpha (t)
\bigr)
=
\infty \sum
m=0
u\alpha (n+m)
s\alpha (n+m+1)
[(n+m)\alpha ]!am =
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
q-DEFORMED CONFORMABLE FRACTIONAL NATURAL TRANSFORM 1143
=
u\alpha n
s\alpha n
\infty \sum
m=0
u\alpha m
s\alpha (m+1)
[(n+m)\alpha ]!am =
=
u\alpha n
s\alpha n
\infty \sum
m=0
(D q,\alpha
u )n
[m\alpha ]!amu\alpha (n+m)
s\alpha (m+1)
=
=
u\alpha n
s\alpha n
\infty \sum
m=0
(D q,\alpha
u )n
u\alpha n \cdot [m\alpha ]!amu\alpha m
s\alpha (m+1)
=
=
u\alpha n
s\alpha n
(D q,\alpha
u )nu\alpha nNq,\alpha (f\alpha (t)).
The replacement Nq,\alpha (f\alpha (t)) by Rq,\alpha (u, s) in the last equation completes the proof.
These results are in complete agreement with those obtained for non-deformed Sumudu and
natural transform investigated by Belgacem and others (see [5] and references therein). Now we
will give another representation of the q, \alpha -deformed natural transform of the product of f\alpha (t) with
positive degree of \alpha -monomial t\alpha . This proposition extends the [5] (Theorem 4.2).
Proposition 3.6. Suppose that function f\alpha (t) has polynomial or formal power series expansion
in \alpha -monomials t\alpha . Then, for all integer n > 0, we have
Nq,\alpha
\bigl(
t\alpha nf\alpha (t)
\bigr)
=
u\alpha n
s\alpha n
n\sum
k=0
bn,ku
\alpha k(D q,\alpha
u )kRq,\alpha (u, s),
where the coefficients bn,k satisfy the recurrence relationship
bn,k =
\left\{
[n\alpha ]bn - 1,0, k = 0,
[(n+ k)\alpha ]bn - 1,k + q\alpha (n - 1+k)bn - 1,k - 1, 0 < k < n,
q\alpha (2n - 1)bn - 1,n - 1, k = n,
with initial condition b0,0 = 1.
Proof. We proceed the proof by induction on n. For n = 0, we have Nq,\alpha (f\alpha (t)) = Rq,\alpha (u, s),
so that b0,0 = 1. By previous proposition, for n = 1 we get
Nq,\alpha (t
\alpha f\alpha (t)) =
u\alpha
s\alpha
D q,\alpha
u u\alpha Rq,\alpha (u, s),
which, by applying deformed Leibniz rule (2.4), can be rewritten as
Nq,\alpha (t
\alpha f\alpha (t)) =
u\alpha
s\alpha
\bigl(
[\alpha ]Rq,\alpha (u, s) + q\alpha u\alpha D q,\alpha
u Rq,\alpha (u, s)
\bigr)
.
Thus, b1,0 = [\alpha ] = [\alpha ]b0,0, b1,1 = q\alpha = q\alpha b0,0, and the claim holds. Assuming that the claim holds
for m \leq n, we will prove it, for m = n+ 1,
Nq,\alpha
\bigl(
t\alpha (n+1)f\alpha (t)
\bigr)
= Nq,\alpha
\bigl(
t\alpha
\bigl(
t\alpha nf\alpha (t)
\bigr) \bigr)
=
=
u\alpha
s\alpha
D q,\alpha
u u\alpha
u\alpha n
s\alpha n
n\sum
k=0
bn,ku
\alpha k(D q,\alpha
u )kRq,\alpha (u, s) =
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1144 O. HERSCOVICI, T. MANSOUR
=
u\alpha
s\alpha
D q,\alpha
u
n\sum
k=0
bn,k
u\alpha (1+n+k)
s\alpha n
(D q,\alpha
u )kRq,\alpha (u, s) =
=
u\alpha
s\alpha
n\sum
k=0
bn,k
[(n+ k + 1)\alpha ]u\alpha (n+k)
s\alpha n
(D q,\alpha
u )kRq,\alpha (u, s)+
+
u\alpha
s\alpha
n\sum
k=0
bn,k
q\alpha (1+n+k)u\alpha (1+n+k)
s\alpha n
(D q,\alpha
u )k+1Rq,\alpha (u, s) =
=
u\alpha (n+1)
s\alpha (n+1)
n\sum
k=0
bn,k[(n+ k + 1)\alpha ]u\alpha k(D q,\alpha
u )kRq,\alpha (u, s)+
+
u\alpha (n+1)
s\alpha (n+1)
n - 1\sum
k=1
bn,k - 1q
\alpha (n+k)u\alpha k(D q,\alpha
u )kRq,\alpha (u, s) =
=
u\alpha (n+1)
s\alpha (n+1)
n+1\sum
k=0
bn+1,ku
\alpha k(D q,\alpha
u )kRq,\alpha (u, s),
where
bn+1,0 = bn,0[(n+ 1)\alpha ],
bn+1,k = bn,k[(n+ k + 1)\alpha ] + bn,k - 1q
\alpha (n+k), 1 \leq k \leq n,
bn+1,n+1 = bn,nq
\alpha (2n+1),
which completes the proof.
We end this paper by the following conclusion. Our new generalization of the natural trans-
form proposes also new generalizations of other widely used integral transforms. By applying the
techniques described here, one can solve a k-order linear q-differential equation with constant co-
efficients. There is no need to find separately homogeneous solution and a particular solution. In
order to solve a differential equation by applying the integral transform one need to know the integral
transform of the right-hand side function of the differential equation\sum
0\leq j\leq k
aj
\bigl(
D q,\alpha
x
\bigr) k - j
f(x) = b(x)
and the initial conditions
\bigl(
D q,\alpha
x
\bigr) j
y(0) = yj for j = 0, . . . , k - 1.
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Received 02.05.20,
after revision — 21.05.21
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
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| id | umjimathkievua-article-6099 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:26:05Z |
| publishDate | 2022 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/50/a32ff26219d95c1fa9cd5c67e9e37850.pdf |
| spelling | umjimathkievua-article-60992022-10-24T17:56:02Z $q$-Deformed conformable fractional Natural transform $q$-Deformed conformable fractional natural transform Herscovici, O. Mansour, T. Herscovici, O. Mansour, T. Laplace transform Sumudu transform q-deformation Jackson q-derivative conformable fractional q-derivative q-Leibniz rule UDC 517.9 We develop a new deformation and generalization of the natural integral transform based on the conformable fractional $q$-derivative. We obtain transformation of some deformed functions and apply the transform to solve linear differential equation with given initial conditions. УДК 517.9$q$ -Деформоване конформне дробове натуральне перетворенняРозроблено нову деформацiю та узагальнення натурального iнтегрального перетворення на основi конформної дробової $q$-похiдної. Отримано перетворення деяких деформованих функцiй. Це перетворення застосовано до розв’язування лiнiйного диференцiального рiвняння з заданими початковими умовами. Institute of Mathematics, NAS of Ukraine 2022-10-04 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6099 10.37863/umzh.v74i8.6099 Ukrains’kyi Matematychnyi Zhurnal; Vol. 74 No. 8 (2022); 1128- 1145 Український математичний журнал; Том 74 № 8 (2022); 1128- 1145 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6099/9291 Copyright (c) 2022 Орлі Гершкович |
| spellingShingle | Herscovici, O. Mansour, T. Herscovici, O. Mansour, T. $q$-Deformed conformable fractional Natural transform |
| title | $q$-Deformed conformable fractional Natural transform |
| title_alt | $q$-Deformed conformable fractional natural transform |
| title_full | $q$-Deformed conformable fractional Natural transform |
| title_fullStr | $q$-Deformed conformable fractional Natural transform |
| title_full_unstemmed | $q$-Deformed conformable fractional Natural transform |
| title_short | $q$-Deformed conformable fractional Natural transform |
| title_sort | $q$-deformed conformable fractional natural transform |
| topic_facet | Laplace transform Sumudu transform q-deformation Jackson q-derivative conformable fractional q-derivative q-Leibniz rule |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6099 |
| work_keys_str_mv | AT herscovicio qdeformedconformablefractionalnaturaltransform AT mansourt qdeformedconformablefractionalnaturaltransform AT herscovicio qdeformedconformablefractionalnaturaltransform AT mansourt qdeformedconformablefractionalnaturaltransform |