Certain results related to the N-transform of a certain class of functions ...

Al-Omari Advances in Difference Equations (2018) 2018:7 https://doi.org/10.1186/s13662-017-1457-y

RESEARCH

Open Access

Certain results related to the N-transform of a certain class of functions and differential operators Shrideh K Al-Omari* *

Correspondence: [email protected] Department of Physics and Basic Sciences, Faculty of Engineering Technology, Al-Balqa Applied University, Amman, Jordan

Abstract In this paper, we aim to investigate q-analogues of the natural transform on various elementary functions of special type. We obtain results associated with classes of q-convolution products, Heaviside functions, q-exponential functions, q-hyperbolic functions and q-trigonometric functions as well. Further, we give definitions and derive results involving some q-differential operators. Keywords: q-differential operator; q-trigonometric function; q-convolution product; q-natural transform; q-hyperbolic function; Heaviside function

1 Introduction The subject of fractional calculus has gained noticeable importance and popularity due to its mainly demonstrated applications in diverse fields of science and engineering problems. Much of developed theories of fractional calculus were based upon the familiar Riemann-Liouville activity related to the q-calculus has been noticed due to applications of this theory in mathematics, statistics and physics. Jackson in [1] presented a precise definition and developed the q-calculus in a systematic way. Thereafter, some remarkable integral transforms were associated with different analogues in a q-calculus context. Among those examined integrals, we recall the q-Laplace integral operator [2–5], the q-Sumudu integral operator [6–9], the Weyl fractional q-integral operator [10], the q-wavelet integral operator [11], the q-Mellin type integral operator [12], Mangontarum transform [13], Widder potential transform [14], E2;1 transform [15] and a few others. In this paper, we give some q-analogues of some recently investigated integral transform, named the natural transform, and obtain some new desired q-properties. In the following section, we present some notations and terminologies from the qcalculus. In Section 3, we give the definition and properties of the natural transform. In Section 4, we give the definition of the analogue Nq of the natural transform and apply it to some functions of special type. In more details, Sections 5 to 7 are associated with the application of the analogue Nq to Heaviside functions, convolutions and differentiations. The remaining two sections investigate the N q analogue of the natural transform on some elementary functions with similar approach. © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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2 Definitions and preliminaries We present some usual notions and notations used in the q-calculus [16–18]. Throughout this paper, we assume q to be a fixed number satisfying 0 < q < 1. The q-calculus begins with the definition of the q-analogue dq f (x) of the differential of functions dq f (x) = f (qx) – f (x).

(1)

Having said this, we immediately get the q-analogue of the derivative of f (x), called its q-derivative, (Dq f )(x) :=

dq f (x) f (x) – f (qx) := , dq x (1 – q)x

x = 0,

(2)

(Dq f )(0) = f´ (0) provided f´ (0) exists. If f is differentiable, then (Dq f )(x) tends to f´ (0) as q tends to 1. It can be seen that the q-derivative satisfies the following q-analogue of the Leibniz rule:   Dq f (x)g(x) = g(x)Dq f (x) + f (qx)Dq g(x).

(3)

The q-Jackson integrals are defined by [1, 12] 

x

f (x) dq x = (1 – q)x

∞    f xqk qk ,

0



(4)

0 ∞

f (x) dq x = (1 – q)x 0

∞    f qk qk ,

(5)

–∞

provided the sum converges absolutely. The q-Jackson integral in a generic interval [a, b] is given by [1] 



b



b

f (x) dq x =

f (x) dq x – 0

a

a

f (x) dq x.

(6)

0

The improper integral is defined in the way that 

∞ A

f (x) dq x = (1 – q)

0

∞  k k  q q , f A A –∞

(7)

and, for n ∈ Z, we have 

∞ qn

0

 f (x) dq x =

∞

f (x) dq x.

(8)

0

The q-integration by parts is defined for functions f and g by 



b

g(x)Dq f (x) dq x = f (b)g(b) – f (a)g(a) – 0

b

f (qx)Dq g(x) dq x. a

(9)

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For x ∈ C, the q-shifted factorials are defined by (x; q)0 = 1; (x; q)t = n–1 k 0 (1 – xq ), and hence (x; q)∞ =

∞   1 – xqk .

(x;q)∞ ; (x; q)n (xqt ;q)∞

=

(10)

k=0

The q-analogues of x and ∞ are respectively defined by [x] =

1 – qx , 1–q

[∞] =

1 . 1–q

(11)

The q-analogues of the exponential function of the first and second kind are respectively given as follows:

Eq (x) =

∞ 

q

n(n–1) 2

n=0

xn ([n]q )!

(x ∈ C)

(12)

and eq (x) =

∞  n=0

xn ([n]q )!



 |x| < 1 ,

(13) n

= qn–1 + · · · + q + 1. where ([n]q )! = [n]q [n – 1]q · · · [2]q [1]q , [n] = 1–q 1–q However, in view of the product expansions, (eq (x))–1 = Eq (–x), which explains the need for both analogues of the exponential function. The q-derivative of Eq (x) is Dq Eq (x) = Eq (qx), whereas the q-derivative of eq (x) is Dq eq (x) = eq (x), eq (0) = 1. The gamma and beta functions satisfy the q-integral representations 1

1–q q (t) = 0 xt–1 Eq (–qx) dq x

1 Bq (t; s) = 0 xt–1 (1 – qx)s–1 q dq x

 (14) (t, s > 0)  (s) (t)

q and are related to the identities Bq (t; s) = q q (s+t) and q (t + 1) = [t]q q (t). Due to (12) and (13), the q-analogues of sine and cosine functions of the second and first kind are respectively given as

sinq (at) = cosq (at) =

∞

2n+1

0

(at) (–1)n ([2n+1] ; q )!

0

(at) (–1)n ([2n] ; q )!

∞ ∞

2n

n(n+1) 2

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬

q (–1)n ([2n+1] a2n+1 t 2n+1 ; ⎪ ⎪ q )! ⎪ ⎪ ⎪ n(n–1) ⎪ ∞ ⎪ nq 2 2n 2n ⎭ cosq (at) = 0 (–1) ([2n]q )! a t .

sinq (at) =

(15)

0

3 The N-transform or the natural transform The natural transform of a function f (x), x ∈ (0, ∞), was proposed by Khan and Khan [19] as an extension to Laplace and Sumudu transforms to solve some fluid flow problems. Later, Silambarasan and Belgacem [20] derived certain electric field solutions to

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Maxwell’s equation in conducting media. In [21], the author applied the natural transform to some ordinary differential equations and some space of Boehmians. For further investigations of the natural transform, one can refer to [22] and [23]. Over the set A, where A = {f (t)|∃τ1 , τ2 , M > 0, |f (t)| < Met/τ j , t ∈ (–1)j × [0, ∞)|}, the N -transform of f (t) is given as [19, (1)], 

∞

f (ut) exp(–vt) dt

(Nf )(u; v) =

(u, v > 0).

(16)

0

Provided the integral above exists, it is so easy to see that (Nf )(1; v) = (Lf )(v),

(Nf )(u; 1) = (Sf )(u),

(17)

where Sf and Lf are respectively the Sumudu and Laplace transforms of f . Moreover, the natural-Laplace and natural-Sumudu dualities are given as [21, (5), (6)], 

∞

(Nf )(u; v) = 0

  vt 1 f (t) exp – dt u u

(18)

  1 ut f exp(–t) dt, v v

(19)

and 

∞

(Nf )(u; v) = 0

respectively. Further, from (18) and (19), it can be easily observed that (Nf )(u; v) =

  1 u (Lf ) , u v

  1 u (Nf )(u; v) = (Sf ) . v v

(20)

The natural transform of some known functions is given as follows [21, p. 731]: (i) N(a)(u; v) = 1v , where a is a constant; (ii) N(δ)(u; v) = 1v , where δ is the delta function; 1 , a is a constant. (iii) N(eat )(u; v) = v–au

4 The Nq analogue of the natural transform Hahn [24] and later Ucar and Albayrak [4] defined the q-analogues of the first and second type of Laplace transforms by means of the q-integrals

Lq (f ; s) = q L(f ; s) =

1 1–q 1 1–q



1 s

Eq (qst)f (t) dq t,

(21)

eq (–st)f (t) dq t.

(22)

0



∞

0

On the other hand, the q-analogues of the Sumudu transform of the first and second type were defined by [6, 7] Sq (f ; s) =

1 (1 – q)s





s

Eq 0

 q t f (t) dq t, s

(23)

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1 q S(f ; s) = 1–q



∞ 0

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  t eq – f (t) dq t, s

(24) |t|

where s ∈ (–τ1 , τ2 ) and f is a function of the set A, A = {f (t)|∃M, τ1 , τ2 > 0, |f (t)| < Me τj , t ∈ (–1)j × [0, ∞)}. Now, we are in a position to demonstrate our results as follows: the qanalogue of the natural transform of the first kind is   (Nq f )(u; v) := Nq f ; (u, v) :=



∞

0

  vt 1 f (t)Eq –q dq t, u u

(25)

provided the function f (t) is defined on A and u and v are the transform variables. The q-series representation of (25) can be written as     k 1 k k+1 v q f q Eq –q , (Nq f )(u; v) = (1 – q)u u

(26)

k∈Z

which, by (10), can be put into the form (Nq f )(u; v) =

(q; q)∞  qk f (qk ) . (1 – q)n (– uv ; q)k+1

(27)

k∈Z

Let us now discuss some general applications of Nq to some elementary functions. Theorem 1 Let α ∈ R. Then we have 

 uα Nq t α (u; v) = α+1 (α + 1). v

(28)

Proof By setting the variables, we get 

 uα Nq t α (u; v) = α+1 v



∞

t α Eq (–qt) dq t =

0

uα q (α + 1). vα+1 

The theorem hence follows. The direct consequence of (28) is that 

  un  Nq t n (u; v) = n+1 [n]q !. v

(29)

Theorem 2 Let a be a positive real number. Then we have 

 Nq eq (at) (u; v) =

1 , v – au

au < v.

(30)

Proof By aid of the eq analogue of the q-exponential function, we write 

 Nq eq (at) (u; v) =



∞ 0

  v 1 eq (at)Eq –q t dq t. u u

(31)

In series representation, (31) can be written as 

∞   Nq eq (at) (u; v) = 0

  v an t n Eq –q t dq t. u([n]q )! u

(32)

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By (29), (32) gives a geometric series expansion, and hence we have (Nq eq (at))(u; v) = 1 ∞ au n 1 0 ( v ) = v–au , au < v. v This finishes the proof of the lemma.  Theorem 3 Let a be a positive real number. Then we have 

  ∞  1  n(n – 1) au n Nq Eq (at) (u; v) = q . v 0 2 v

(33)

Proof In view of the Eq analogue of the exponential function, we indeed get 

∞   an q n(n–1) 1 2 Nq (at) (u; v) = ([n]q )! u 0



∞

0

  v t n Eq –q t dq t. u

The parity of (29) gives (Nq Eq (at))(u; v) = 1v q n(n–1) ( au )n . 2 v This completes the proof of the theorem. The hyperbolic q-cosine and q-sine functions are given as coshq t = e (t)–e (–t) sinhq t = q 2 q . Hence, as a corollary of Theorem 2, we have 

(34)

eq (t)+eq (–t) 2

 Nq coshq t (u; v) =

v , au < v, v2 + a2 u2   au Nq sinhq t (u; v) = 2 , au < v. v + a2 u2

Theorem 4 Let a be a positive real number. Then we have (Nq cosq at)(u; v) = vided au < v.

and

(35) (36)  v v2 –a2 u2

pro-

Proof On account of (15) and (29), we obtain 

∞   Nq cos at (u; v) = q

0

=

a2n u([2n]q )!



 ∞  1  au 2n v

0

v

0

∞

  v t Eq –q t dq t u 2n

.

For au < v, the geometric series converges to the sum 

 Nq cosq at (u; v) =

v v2 – a2 u2

(37)

provided au < v. This finishes the proof.



Similarly, by (15) and (29), we deduce that 

 Nq sinq at (u; v) =

v2

au , – a2 u2

au < v.

(38)

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5 Nq and q-differentiation In this section, we discuss some q-differentiation formulae. On account of (12), we derive the following differentiation result. Lemma 5 Let u, v > 0. Then we have   ∞ n (n+1)(n+2) v v u Dq Eq –q t = (–1)n+1 q 2 tn . v u 0 un

(39)

Proof By using the q-representation of Eq in (12), we write n(n–1)     ∞  qv n n v (–1) 2 Dq Eq –q t = Dq t u ([n]q )! u 0

=

∞  1

=

∞ 

(n+1)(n+2)

q 2 vn qn n t n–1 (–1) ([n – 1]q )! u n

(–1)n+1 q

(n+1)(n+2) 2

1

Hence, it follows that Dq Eq (–q uv t) =

v u

vn+1 n t . un+1

∞ 0

(–1)n+1 q

(n+1)(n+2) n v n 2 un

t . This finishes the proof. 

The natural transform of the q-derivative Dq f can be written as follows. Theorem 6 Let u, v > 0, then we have   v Nq Dq f (t) (u; v) = –f (0) + Nq (f )(u; v). u

(40)

Proof Using the idea of q-integration by parts and the formula in (9), we write 

  v Dq f (t)Eq –q t dq t u 0      ∞ v ∞ v = f (t)Dq Eq –q t  – f (qt)Dq Eq –q t dq t. u 0 u 0

 Nq Dq f (t) (u; v) =



∞

The parity of Lemma 5, gives   Nq Dq f (t) (u; v) = –f (0) –



∞

∞

f (qt) 0

= –f (0) +

v u



v  (–1)n+1 u 0 ([n]q )!

∞

f (qt) 0

∞  (–1)n (n+1)(n+2) vn n q 2 t dq t. ([n]q )! un 0

Changing the variables as qt = y and t n = q–n yn implies   v Nq Dq f (t) (u; v) = –f (0) + u



∞

f (t) 0

∞  0

n(n–1)

(–1)n

q 2 vn n t dq t. ([n]q )! un

(41)

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∞ By virtue of (12), (41) yields Nq (Dq f (t))(u; v) = –f (0) + uv 0 f (t)Eq (–q uv t) dq t. Hence, Nq (Dq f (t))(u; v) = –f (0) + uv Nq (f )(u; v). This finishes the proof.  Now we extend Theorem 6 to the nth derivative as in the following theorem. Theorem 7 Let u, v > 0 and n ∈ Z+ . Then we have n–1  n–1–i     vn  u Diq f (0). Nq Dnq f (t) (u; v) = n Nq (f ) (u, v) – u v i=0

(42)

Proof On account of Theorem 6, we can write Nq (D2q f (t))(u; v) = –Dq f (0) + uv (–f (0) + v (Nq f )(u; v)). Hence we have obtained u   v2 v Nq D2q f (t) (u; v) = –Dq f (0) – f (0) + 2 (Nq f )(u; v). u u Proceeding as in (43), we obtain Nq (Dnq f (t))(u; v) = This finishes the proof.

(43)

v2 v n–1–i i N (f )(u; v) – n–1 Dq f (0). i=0 ( u ) u2 q



6 Nq of q-convolutions Let functions f and g be given in the form f (t) = t α and g(t) = t β–1 for α, β > 0. We define the q-convolution of f and g as  (f ∗ g)q (t) =

t

f (τ )g(t – qτ ) dq t.

(44)

0

The q-convolution theorem can be read as follows. Theorem 8 Let α, β > 0. Then we have       Nq (f ∗ g)q (u; v) = u2 Nq t α (u, v)Nq t β–1 (u; v). Proof By aid of (44) and (14), we get Nq ((f ∗ g)q )(u; v) = Hence, by (28) we obtain

(45) Bq (α+1,β) u

∞ 0

t α+β Eq (–q vtu ) dq t.

  uα+β+1 Nq (f ∗ g)q (u; v) = q (α + 1)q (β) α+β+1 . v

(46)

Motivation on (46) gives Nq ((f ∗ g)q )(u; v) = u2 (Nq t α )(u, v)(Nq t β–1 )(u; v). The proof is therefore completed.  In a similar way, we extend the q-convolution to functions of a power series form. Theorem 9 Let f (t) = u2 (Nq f )(u; v)(Nq g)(u; v).

∞ 0

ai t αi and g(t) = t β–1 . Then we have Nq ((f ∗ g)q )(u; v) =

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Proof Under the hypothesis of the theorem and Theorem 8, we write ∞       Nq (f ∗ g)q (u; v) = ai Nq t αi ∗ t β–1 q (u; v) 0

=

∞      ai Nq t αi (u; v) Nq t β–1 (u; v) 0

= u2 (Nq f )(u; v)(Nq g)(u; v). 

Hence, the proof of the theorem is finished.

7 Nq and Heaviside functions The Heaviside function is defined by ⎧ ⎨1, t ≥ a, ´ – a) = u(t ⎩0, 0 ≤ t < a,

(47)

where a is a real number. In this part of the paper, we establish the following theorem. Theorem 10 If u´ denotes the Heaviside function and u, v > 0, then we have     1 v ´ – a) (u; v) = Eq – a . Nq u(t v u

(48)

´ – a))(u; v) = Proof By (47) we have Nq (u(t 

 1 1 ´ – a) (u; v) = – Nq u(t v u =

1 1 – v u



∞ a

0

0

∞ 

∞ a

Eq (–q uv t) dq t. On account of (21), we get

n(n–1)   v n q 2 –q t dq t ([n]q )! u n(n–1)

(–1)n

0

1 u

q 2 vn qn n ([n]q )! u



a

t n dq t. 0

Upon integrating and using simple calculation, the above equation reveals ∞  1 1  q 2 vn an+1 ´ – a) (u; v) = – Nq u(t v u 0 [n]q ! un [n + 1]q n(n–1)



∞

n

∞

(n+1)n

1 1 qn+1 2 vn n+1 = + (–1)n+1 a v u 0 [n + 1]q ! un =

vn+1 n+1 1 1 q 2 + (–1)n+1 a . v v 0 [n + 1]q ! un+1

´ – a))(u; v) = This can be written as Nq (u(t mation from 0 gives

1 v

+

1 v

∞ 1

m(m–1)

(–1)m q([m]q )!

m(m–1)   ∞   1 –v q 2 vm m 1 ´ – a) (u; v) = E a . (–1)m a = Nq u(t q v 1 ([m]q )! um v u

vm m a . um

Starting the sum-



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8 The q-analogue of the second kind Over the set A = {f (t)|∃M, τ1 , τ2 > 0, |f (t)| < Met/τj , t ∈ (–1)j × [0, ∞)}, the q-analogue of the N -transform of the second type is expressed as 

q





  v 1 f (t)eq – t dq t. u u

∞

N f (u; v) = 0

(49)

As the q-analogue of the gamma function of the second kind is defined as  γq (t) =

∞

xt–1 eq (–x) dq x,

(50)

0

it follows that γq (1) = 1, γq (t + 1) = q–t [t]q γq (t),

γq (n) = q

n(n–1) 2

q (n),

(51)

q being the q-analogue of the gamma function of the first kind. In the remainder of this section, we derive certain results analogous to the ones we have obtained in the previous chapters. Lemma 11 Let α > –1. Then we have 

 uα N q t α (u; v) = α+1 γq (α + 1). v

(52)

In particular, 

  un –n(n–1)  N q t n (u; v) = n+1 q 2 [nq ] !. v

(53)

Proof Let α > –1. Then, by the change of variables, we have 



1 N t (u; v) = u q α





∞

α

t eq 0

  ∞ –vt uα dq t = α+1 t α eq (–t) dq t. u v 0 α

By aid of (50), we get (N q t α )(u; v) = vuα+1 γq (α + 1). Proof of the second part of the theorem follows from (51). Hence, we have completed the proof of the theorem.  Theorem 12 Let a ∈ R, a > 0, then we have 

∞  1  an un –n(n–1) q 2 . N q eq (at) (u; v) = uv 0 vn

(54)

Proof By (13) we write 

 1 N q eq (at) (u; v) = u

 0

∞

     ∞ ∞ v v 1  an eq (at)eq – t dq t = t n eq – t dq t. u u 0 ([n]q )! 0 u

By aid of Theorem 2, we get (N q eq (at))(u; v) = This completes the proof of the theorem.

1 uv

∞ 0

an un –n(n–1) q 2 . vn



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Theorem 13 Let a > 0, a ∈ R, then we have 

 N q Eq (at) (u; v) =

1 , u(v – au)

au < v.

(55)

Proof After simple calculations and using Theorem 2, we obtain 

  v Eq (at)eq – t dq t u 0 n(n–1)    ∞ ∞ v 1q 2 t n eq – t dq t an = u 0 ([n]q )! u 0

 1 N q eq (at) (u; v) = u



∞

∞

=

1  n un a n. uv 0 v

Since the above series determines a geometric series, it reads (N q eq (at))(u; v) = 1 , au < v. u(v–au)

1 1 uv 1– au v

= 

Hence the theorem is proved. The N q transform of cosq and sinq is given as follows. Theorem 14 Let a > 0. Then we have 

∞  (2n–1) 1 u2n N q cosq (at) (u; v) = (–1)n a2n 2n q–2n 2 . u 0 v

(56)

Proof Using the definition of cosq , we write 

  v cos (at)eq – t dq t u 0    ∞ ∞ 2n v a 1 = (–1)n t 2n eq – t dq t u 0 ([2n]q )! 0 u

 1 N cos (at) (u; v) = u q

q



∞

q

∞

=

(2n–1) u2n 1 (–1)n a2n 2n q–2n 2 . u 0 v

This completes the proof of the theorem.



The N q transform of sinq (at) is given as follows. Theorem 15 Let a > 0, then we have 

∞  (2n–1) 1  u2n+1 N q sinq (at) (u; v) = (–1)n a2n+1 2n+1 q–2n 2 . uv 0 v

The proof of Theorem 15 follows from a proof similar to that of Theorem 14.

(57)

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9 Nq of q-differentiation Before we start investigations, we first assert that     v v v Dq eq – t = – eq – t . u u u

(58)

In detail, this can be interpreted as    ∞ v (–1)n vn D tn Dq eq – t = n q u ([n] )! u q 0 =

∞  1

=

∞  0

=–

(–1)n vn n–1 t ([n – 1]q )! un (–1)n+1 vn+1 n t ([n]q )! un+1

  ∞ v v  (–1)n vn n v – e t . t = – q u 0 ([n]q )! un u u

This proves the above assertion. Hence we prove the following theorem. Theorem 16 Let u, v > 0. Then we have 

 N q Dq f (t) (u; v) = –f (0) –

 0

∞

  v f (qt)Dq eq – t dq t. u

Proof By (58), the above equation gives 

  v Dq f (t)eq – t dq t u 0    ∞ v v f (qt)eq – t dq t = –f (0) + u 0 u    ∞ v –1 v –1 = –f (0) + q f (t)eq – q t dq t. u u 0

 N q Dq f (t) (u; v) =



∞

By setting variables, we have (N q Dq f (t))(u; v) = –f (0) + uv q–1 (N q f )(q–1 v; u). This completes the proof. Now, we extend (59) to have 

   N q D2q f (u; v) = N q Dq (Dq f ) (u; v)   v –1  q q N Dq f q–1 v; u u   v –1  q  –2  v –1 –f (0) + q N f q v; u = –Dq f (0) + q u u     v –1 v 2–2  q  –2  = –Dq f (0) – N f q v; u . q f (0) + q u u

= –Dq f (0) +

(59)

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Proceeding to the nth derivative, we get (N q Dnq f )(u; v) = ( uv )n q–n (N q f )(q–n v; u) v n–1–i i – n–1 Dq f (0). i=0 ( u ) This completes the proof of the theorem. 

10 Conclusion The q-analogues of the N -transform have been applied to various classes of exponential functions, trigonometric functions, hyperbolic functions and certain differential operators as well. As the N -transform defines an extension to Laplace and Sumudu transforms, our results generalize the results existing in the literature. Acknowledgements The author would like to express many thanks to the anonymous referees for their corrections and comments on this manuscript. Funding Not applicable. Availability of data and materials Not applicable. Competing interests The author declares that he has no competing interests. Authors’ contributions The author read and approved the final manuscript.

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