Certain results related to the N-transform of a certain class of functions and differential operators

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.

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], $E_{2;1}$-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 q-calculus. In Section 3, we give the definition and properties of the natural transform. In Section 4, we give the definition of the analogue $N_{q}$ 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 $N_{q}$ 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.

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 $d_{q}f ( x )$ of the differential of functions

$$ d_{q}f ( x ) =f ( qx ) -f ( x ). $$

(1)

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

$$ ( D_{q}f ) ( x ):=\frac{d_{q}f ( x ) }{d_{q}x}:=\frac{f ( x ) -f ( qx ) }{ ( 1-q ) x},\quad x \neq0, $$

(2)

$( D_{q}f ) ( 0 ) =\acute{f} ( 0 )$ provided $\acute{f} ( 0 )$ exists. If f is differentiable, then $( D_{q}f ) ( x )$ tends to $\acute{f} ( 0 )$ as q tends to 1. It can be seen that the q-derivative satisfies the following q-analogue of the Leibniz rule:

$$ D_{q} \bigl( f ( x ) g ( x ) \bigr) =g ( x ) D_{q}f ( x ) +f ( qx ) D_{q}g ( x ). $$

(3)

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

$$\begin{aligned} &\int_{0}^{x}f ( x ) \,d_{q}x= ( 1-q ) x\sum _{0}^{\infty}f \bigl( xq^{k} \bigr) q^{k}, \end{aligned}$$

(4)

$$\begin{aligned} &\int_{0}^{\infty}f ( x ) \,d_{q}x= ( 1-q ) x\sum _{-\infty}^{\infty}f \bigl( q^{k} \bigr) q^{k}, \end{aligned}$$

(5)

provided the sum converges absolutely.

The q-Jackson integral in a generic interval $[ a,b ]$ is given by [1]

$$ \int_{a}^{b}f ( x ) \,d_{q}x= \int_{0}^{b}f ( x ) \,d_{q}x- \int_{0}^{a}f ( x ) \,d_{q}x. $$

(6)

The improper integral is defined in the way that

$$ \int_{0}^{\frac{\infty}{A}}f ( x ) \,d_{q}x= ( 1-q ) \sum _{-\infty}^{\infty}f \biggl( \frac{q^{k}}{A} \biggr) \frac {q^{k}}{A}, $$

(7)

and, for $n\in Z$, we have

$$ \int_{0}^{\frac{\infty}{q^{n}}}f ( x ) \,d_{q}x= \int_{0}^{\infty}f ( x ) \,d_{q}x. $$

(8)

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

$$ \int_{0}^{b}g ( x ) D_{q}f ( x ) \,d_{q}x=f ( b ) g ( b ) -f ( a ) g ( a ) - \int_{a}^{b}f ( qx ) D_{q}g ( x ) \,d_{q}x. $$

(9)

For $x\in C$, the q-shifted factorials are defined by $( x;q )_{0}=1; ( x;q )_{t}=\frac{ (x;q )_{\infty}}{ ( xq^{t};q ) _{\infty}}; ( x;q ) _{n}=\prod_{0}^{n-1} ( 1-xq^{k} )$, and hence

$$ ( x;q ) _{\infty}=\prod_{k=0}^{\infty} \bigl( 1-xq^{k} \bigr). $$

(10)

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

$$ [ x ] =\frac{1-q^{x}}{1-q},\qquad [ \infty ] =\frac{1}{1-q}. $$

(11)

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

$$ E_{q} ( x ) =\sum_{n=0}^{\infty}q^{\frac{n ( n-1 ) }{2}} \frac{x^{n}}{ ( [ n ] _{q} ) !}\quad ( x\in C ) $$

(12)

and

$$ e_{q} ( x ) =\sum_{n=0}^{\infty} \frac{x^{n}}{ ( [ n ]_{q} ) !} \quad\bigl( \vert x \vert < 1 \bigr), $$

(13)

where $( [ n ] _{q} ) != [ n ] _{q} [ n-1 ] _{q}\cdots [ 2 ] _{q} [ 1 ] _{q}$, $[ n ] =\frac{1-q^{n}}{1-q}=q^{n-1}+\cdots+q+1$.

However, in view of the product expansions, $( e_{q} ( x ) ) ^{-1}=E_{q} ( -x )$, which explains the need for both analogues of the exponential function.

The q-derivative of $E_{q} ( x )$ is $D_{q}E_{q} ( x ) =E_{q} ( qx )$, whereas the q-derivative of $e_{q} ( x )$ is $D_{q}e_{q} ( x ) =e_{q} ( x )$, $e_{q} ( 0 ) =1$.

The gamma and beta functions satisfy the q-integral representations

$$ \left . \textstyle\begin{array}{l} \Gamma_{q} ( t ) =\int_{0}^{\frac{1}{1-q}}x^{t-1}E_{q} ( -qx ) \,d_{q}x \\ B_{q} ( t;s ) =\int_{0}^{1}x^{t-1} ( 1-qx ) _{q}^{s-1}\,d_{q}x\quad ( t,s>0 ) \end{array}\displaystyle \right \} $$

(14)

and are related to the identities $B_{q} ( t;s ) =\frac{\Gamma _{q} ( s ) \Gamma_{q} ( t ) }{\Gamma_{q} ( s+t ) }$ and $\Gamma_{q} ( t+1 ) = [ t ] _{q}\Gamma _{q} ( t )$. Due to (12) and (13), the q-analogues of sine and cosine functions of the second and first kind are respectively given as

$$ \left . \textstyle\begin{array}{l} \sin^{q} ( at ) =\sum_{0}^{\infty} ( -1 ) ^{n}\frac{ ( at ) ^{2n+1}}{ ( [ 2n+1 ] _{q} ) !}; \\ \cos^{q} ( at ) =\sum_{0}^{\infty} ( -1 ) ^{n}\frac{ ( at ) ^{2n}}{ ( [ 2n ] _{q} ) !}; \\ \sin_{q} ( at ) =\sum_{0}^{\infty} ( -1 ) ^{n}\frac {q^{\frac{n ( n+1 ) }{2}}}{ ( [ 2n+1 ] _{q} ) !} a^{2n+1}t^{2n+1}; \\ \cos_{q} ( at ) =\sum_{0}^{\infty} ( -1 ) ^{n}\frac {q^{\frac{n ( n-1 ) }{2}}}{ ( [ 2n ] _{q} ) !} a^{2n}t^{2n}. \end{array}\displaystyle \right \} $$

(15)

3 The N-transform or the natural transform

The natural transform of a function $f ( x )$, $x\in ( 0,\infty )$, 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 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 $A= \{ f ( t ) \vert \exists\tau_{1},\tau _{2},M>0, \vert f ( t ) \vert < Me^{t/\tau j}, t\in ( -1 ) ^{j}\times [ 0,\infty ) \vert \}$, the N-transform of $f ( t )$ is given as [19, (1)],

$$ ( Nf ) ( u;v ) = \int_{0}^{\infty}f (ut ) \exp ( -vt ) \,dt\quad ( u,v>0 ). $$

(16)

Provided the integral above exists, it is so easy to see that

$$ ( Nf ) ( 1;v ) = ( Lf ) ( v ),\qquad ( Nf ) ( u;1 ) = ( Sf ) ( u ), $$

(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)],

$$ ( Nf ) ( u;v ) = \int_{0}^{\infty } \frac{1}{u} f ( t ) \exp \biggl( -\frac{vt}{u} \biggr) \,dt $$

(18)

and

$$ ( Nf ) ( u;v ) = \int_{0}^{\infty } \frac{1}{v}f \biggl( \frac{ut}{v} \biggr) \exp ( -t ) \,dt, $$

(19)

respectively.

Further, from (18) and (19), it can be easily observed that

$$ ( Nf ) ( u;v ) =\frac{1}{u} ( Lf ) \biggl( \frac{u}{v} \biggr),\qquad ( Nf ) ( u;v ) =\frac{1}{v} ( Sf ) \biggl( \frac{u}{v} \biggr). $$

(20)

The natural transform of some known functions is given as follows [21, p. 731]:

• (i) $N ( a ) ( u;v ) =\frac{1}{v}$, where a is a constant;

• (ii) $N ( \delta ) ( u;v ) =\frac{1}{v}$, where δ is the delta function;

• (iii) $N ( e^{at} ) ( u;v ) =\frac{1}{v-au}$, a is a constant.

4 The $N_{q}$ 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

$$\begin{aligned} &L_{q} ( f;s ) =\frac{1}{1-q} \int_{0}^{\frac{1}{s}}E_{q} ( qst ) f ( t ) \,d_{q}t, \end{aligned}$$

(21)

$$\begin{aligned} & {}_{q}L ( f;s ) =\frac{1}{1-q} \int_{0}^{\infty}e_{q} ( -st ) f ( t ) \,d_{q}t. \end{aligned}$$

(22)

On the other hand, the q-analogues of the Sumudu transform of the first and second type were defined by [6, 7]

$$\begin{aligned} &S_{q} ( f;s ) =\frac{1}{ ( 1-q ) _{s}}\int _{0}^{s}E_{q} \biggl( \frac{q}{s}t \biggr) f ( t ) \,d_{q}t, \end{aligned}$$

(23)

$$\begin{aligned} &{}_{q}S ( f;s ) =\frac{1}{1-q} \int_{0}^{\infty}e_{q} \biggl( - \frac{t}{s} \biggr) f ( t ) \,d_{q}t, \end{aligned}$$

(24)

where $s\in ( -\tau_{1},\tau_{2} )$ and f is a function of the set A, $A= \{ f ( t ) |\exists M,\tau_{1},\tau _{2}>0, \vert f ( t ) \vert < Me^{\frac{ \vert t \vert }{\tau_{j}}},t\in ( -1 ) ^{j}\times [ 0,\infty) \}$. Now, we are in a position to demonstrate our results as follows: the q-analogue of the natural transform of the first kind is

$$ ( N_{q}f ) ( u;v ):= \bigl( N_{q}f; ( u,v ) \bigr):= \int_{0}^{\infty} \frac{1}{u}f ( t )E_{q} \biggl( -q\frac{vt}{u} \biggr) \,d_{q}t, $$

(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

$$ ( N_{q}f ) ( u;v ) =\frac{1}{ ( 1-q ) u}\sum _{k\in Z }q^{k}f \bigl( q^{k} \bigr) E_{q} \biggl( -q^{k+1}\frac{v}{u} \biggr), $$

(26)

which, by (10), can be put into the form

$$ ( N_{q}f ) ( u;v ) =\frac{ ( q;q ) _{\infty}}{ ( 1-q ) ^{n}}\sum _{k\in Z}\frac{q^{k}f ( q^{k} ) }{ ( -\frac{v}{u};q ) _{k+1}}. $$

(27)

Let us now discuss some general applications of $N_{q}$ to some elementary functions.

Theorem 1

Let $\alpha\in R$. Then we have

$$ \bigl( N_{q}t^{\alpha} \bigr) ( u;v ) =\frac{u^{\alpha }}{v^{\alpha+1}}\Gamma ( \alpha+1 ). $$

(28)

Proof

By setting the variables, we get

$$\begin{aligned} \bigl( N_{q}t^{\alpha} \bigr) ( u;v ) =&\frac{u^{\alpha}}{ v^{\alpha+1}} \int_{0}^{\infty}t^{\alpha}E_{q} ( -qt ) \,d_{q}t =\frac{u^{\alpha}}{v^{\alpha+1}}\Gamma_{q} ( \alpha+1 ). \end{aligned}$$

The theorem hence follows. □

The direct consequence of (28) is that

$$ \bigl( N_{q}t^{n} \bigr) ( u;v ) =\frac{u^{n}}{v^{n+1}} \bigl( [ n ] _{q} \bigr)!. $$

(29)

Theorem 2

Let a be a positive real number. Then we have

$$ \bigl( N_{q}e_{q} ( at ) \bigr) ( u;v )= \frac {1}{v-au},\quad au< v. $$

(30)

Proof

By aid of the $e_{q}$ analogue of the q-exponential function, we write

$$ \bigl( N_{q}e_{q} ( at ) \bigr) ( u;v ) =\int_{0}^{\infty} \frac{1}{u} e_{q} ( at ) E_{q} \biggl( -q\frac{v}{u} t \biggr) \,d_{q}t. $$

(31)

In series representation, (31) can be written as

$$ \bigl( N_{q}e_{q} ( at ) \bigr) ( u;v ) =\sum _{0}^{\infty}\frac{a^{n}}{u ( [ n ] _{q} ) !} t^{n}E^{q} \biggl( -q\frac{v}{u}t \biggr) \,d_{q}t. $$

(32)

By (29), (32) gives a geometric series expansion, and hence we have $( N_{q}e_{q} ( at ) ) ( u;v ) =\frac {1}{v}\sum_{0}^{\infty} ( \frac{au}{v} ) ^{n}=\frac{1}{v-au}, au< v$.

This finishes the proof of the lemma. □

Theorem 3

Let a be a positive real number. Then we have

$$ \bigl( N_{q}E_{q} ( at ) \bigr) ( u;v ) =\frac{1}{v} \sum_{0}^{\infty}q\frac{n ( n-1 ) }{2} \biggl( \frac {au}{v} \biggr) ^{n}. $$

(33)

Proof

In view of the $E_{q}$ analogue of the exponential function, we indeed get

$$ \bigl( N_{q} ( at ) \bigr) ( u;v ) =\sum_{0}^{\infty} \frac{a_{n}q\frac{n ( n-1 ) }{2}}{ ( [ n ] _{q} ) !}\frac{1}{u} \int_{0}^{\infty}t^{n}E_{q} \biggl( -q \frac{v}{u}t \biggr) \,d_{q}t. $$

(34)

The parity of (29) gives $( N_{q}E_{q} ( at ) ) ( u;v ) =\frac {1}{v}\sum q\frac{n ( n-1 ) }{2} ( \frac{au}{v} ) ^{n}$.

This completes the proof of the theorem.

The hyperbolic q-cosine and q-sine functions are given as $\cosh^{q}t=\frac{e_{q} ( t ) +e_{q} ( -t ) }{2}$ and $\sinh^{q}t=\frac{e_{q} ( t ) -e_{q} ( -t ) }{2}$. Hence, as a corollary of Theorem 2, we have

$$\begin{aligned} &\bigl( N_{q}\cosh^{q}t \bigr) ( u;v ) =\frac {v}{v^{2}+a^{2}u^{2}},\quad au< v, \end{aligned}$$

(35)

$$\begin{aligned} &\bigl( N_{q}\sinh^{q}t \bigr) ( u;v ) =\frac {au}{v^{2}+a^{2}u^{2}},\quad au< v. \end{aligned}$$

(36)

 □

Theorem 4

Let a be a positive real number. Then we have $( N_{q}\cos^{q}at ) ( u;v ) =\frac {v}{v^{2}-a^{2}u^{2}}$ provided $au< v$.

Proof

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

$$\begin{aligned} \bigl( N_{q}\cos^{q}at \bigr) ( u;v ) &=\sum _{0}^{\infty }\frac{a^{2n}}{u ( [ 2n ] _{q} ) !} \int_{0}^{\infty }t^{2n}E_{q} \biggl( -q\frac{v}{u}t \biggr) \,d_{q}t \\ &=\frac{1}{v}\sum_{0}^{\infty} \biggl( \frac{au}{v} \biggr) ^{2n}. \end{aligned}$$

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

$$ \bigl( N_{q}\cos^{q}at \bigr) ( u;v ) =\frac {v}{v^{2}-a^{2}u^{2}} $$

(37)

provided $au< v$.

This finishes the proof. □

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

$$ \bigl( N_{q}\sin^{q}at \bigr) ( u;v ) =\frac {au}{v^{2}-a^{2}u^{2}},\quad au< v. $$

(38)

5 $N_{q}$ 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

$$ D_{q}E_{q} \biggl( -q\frac{u}{v}t \biggr) = \frac{v}{u}\sum_{0}^{\infty } ( -1 ) ^{n+1}q^{\frac{ ( n+1 ) ( n+2 ) }{2}}\frac {v^{n}}{u^{n}}t^{n}. $$

(39)

Proof

By using the q-representation of $E_{q}$ in (12), we write

$$\begin{aligned} D_{q}E_{q} \biggl( -q\frac{v}{u}t \biggr) &= D_{q}\sum_{0}^{\infty} \frac{ ( -1 ) ^{\frac{n ( n-1 ) }{2}}}{ ( [ n ] _{q} ) !} \biggl( \frac{qv}{u} \biggr) ^{n}t^{n} \\ &= \sum_{1}^{\infty} ( -1 ) ^{n} \frac{q^{\frac{ ( n+1 ) ( n+2 ) }{2}}}{ ( [ n-1 ] _{q} ) !}q^{n}\frac{v^{n}}{u^{n}}t^{n-1} \\ &=\sum_{1}^{\infty} ( -1 ) ^{n+1}q^{\frac{ ( n+1 ) ( n+2 ) }{2}}\frac{v^{n+1}}{u^{n+1}}t^{n}. \end{aligned}$$

Hence, it follows that $D_{q}E_{q} ( -q\frac{v}{u}t ) =\frac{v}{u}\sum_{0}^{\infty } ( -1 ) ^{n+1}q^{\frac{ ( n+1 ) ( n+2 ) }{2}}\frac {v^{n}}{u^{n}}t^{n}$. This finishes the proof. □

The natural transform of the q-derivative $D_{q}f$ can be written as follows.

Theorem 6

Let $u,v>0$, then we have

$$ N_{q} \bigl( D_{q}f ( t ) \bigr) ( u;v ) =-f ( 0 ) + \frac{v}{u}N_{q} ( f ) ( u;v ). $$

(40)

Proof

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

$$\begin{aligned} N_{q} \bigl( D_{q}f ( t ) \bigr) ( u;v ) &= \int_{0}^{\infty}D_{q}f ( t ) E_{q} \biggl( -q\frac{v}{u }t \biggr) \,d_{q}t \\ &=f ( t )D_{q}E_{q} \biggl( -q\frac{v}{u}t \biggr) \Big| ^{\infty} _{0}- \int_{0}^{\infty}f ( qt ) D_{q}E_{q} \biggl( -q\frac{v}{u}t \biggr) \,d_{q}t. \end{aligned}$$

The parity of Lemma 5, gives

$$\begin{aligned} N_{q} \bigl( D_{q}f ( t ) \bigr) ( u;v ) &=-f ( 0 ) - \int_{0}^{\infty}f ( qt ) \frac{v}{u}\sum_{0}^{\infty}\frac{ ( -1 ) ^{n+1}}{ ( [ n ] _{q} ) !} \\ &=-f ( 0 ) +\frac{v}{u} \int_{0}^{\infty}f ( qt ) \sum _{0}^{\infty}\frac{ ( -1 ) ^{n}}{ ( [ n ] _{q} ) !}q^{\frac{ ( n+1 ) ( n+2 ) }{2}} \frac{v^{n}}{u^{n}}t^{n}\,d_{q}t. \end{aligned}$$

Changing the variables as $qt=y$ and $t^{n}=q^{-n}y^{n}$ implies

$$ N_{q} \bigl( D_{q}f ( t ) \bigr) ( u;v ) =-f ( 0 ) + \frac{v}{u} \int_{0}^{\infty}f ( t ) \sum _{0}^{\infty} ( -1 ) ^{n}\frac{q^{\frac{n ( n-1 ) }{2}}}{ ( [ n ] _{q} ) !} \frac{v^{n}}{u^{n}} t^{n}\,d_{q}t. $$

(41)

By virtue of (12), (41) yields $N_{q} ( D_{q}f ( t ) ) ( u;v ) =-f ( 0 ) +\frac{v}{u}\int_{0}^{\infty}f ( t ) E_{q} ( -q\frac{v}{u}t ) \,d_{q}t$. Hence, $N_{q} ( D_{q}f ( t ) ) ( u;v ) =-f ( 0 ) +\frac{v}{u}N_{q} ( f ) ( u;v )$. 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 $n\in Z^{+}$. Then we have

$$ N_{q} \bigl( D_{q}^{n}f ( t ) \bigr) ( u;v ) = \frac {v^{n}}{u^{n}} \bigl( N_{q} ( f ) \bigr) ( u,v ) -\sum _{i=0}^{n-1} \biggl( \frac{u}{v} \biggr) ^{n-1-i}D_{q}^{i}f ( 0 ). $$

(42)

Proof

On account of Theorem 6, we can write $N_{q} ( D_{q}^{2}f ( t ) ) ( u;v )=-D_{q}f ( 0 ) +\frac{v}{u} ( -f ( 0 ) +\frac {v}{u} ( N_{q}f ) ( u;v ) )$. Hence we have obtained

$$ N_{q} \bigl( D_{q}^{2}f ( t ) \bigr) ( u;v ) =-D_{q}f ( 0 ) -\frac{v}{u}f ( 0 ) +\frac{v^{2}}{u^{2}} ( N_{q}f ) ( u;v ). $$

(43)

Proceeding as in (43), we obtain $N_{q} ( D_{q}^{n}f ( t ) ) ( u;v ) =\frac {v^{2}}{u^{2}}N_{q} ( f ) ( u;v ) -\sum_{i=0}^{n-1} ( \frac{v}{u} ) ^{n-1-i}D_{q}^{i}f ( 0 )$.

This finishes the proof. □

6 $N_{q}$ of q-convolutions

Let functions f and g be given in the form $f ( t ) =t^{\alpha}$ and $g ( t ) =t^{\beta-1}$ for $\alpha,\beta>0$. We define the q-convolution of f and g as

$$ ( f\ast g ) _{q} ( t ) = \int_{0}^{t}f ( \tau ) g ( t-q\tau ) \,d_{q}t. $$

(44)

The q-convolution theorem can be read as follows.

Theorem 8

Let $\alpha,\beta>0$. Then we have

$$ N_{q} \bigl( ( f\ast g ) _{q} \bigr) ( u;v ) =u^{2}N_{q} \bigl( t^{\alpha} \bigr) ( u,v ) N_{q} \bigl( t^{\beta -1} \bigr) ( u;v ). $$

(45)

Proof

By aid of (44) and (14), we get $N_{q} ( ( f\ast g ) _{q} ) ( u;v ) =\frac{B_{q} ( \alpha+1,\beta ) }{u}\int_{0}^{\infty }t^{\alpha+\beta}E_{q} ( -q\frac{vt}{u} ) \,d_{q}t$. Hence, by (28) we obtain

$$ N_{q} \bigl( ( f\ast g ) _{q} \bigr) ( u;v ) = \Gamma_{q} ( \alpha+1 ) \Gamma_{q} ( \beta ) \frac{ u^{\alpha+\beta+1}}{v^{\alpha+\beta+1}}. $$

(46)

Motivation on (46) gives $N_{q} ( ( f\ast g ) _{q} ) ( u;v ) =u^{2} ( N_{q}t^{\alpha} ) ( u,v ) ( N_{q}t^{\beta-1} ) ( u;v )$. The proof is therefore completed. □

In a similar way, we extend the q-convolution to functions of a power series form.

Theorem 9

Let $f ( t ) =\sum_{0}^{\infty}a_{i}t^{\alpha i}$ and $g ( t ) =t^{\beta -1}$. Then we have $N_{q} ( ( f\ast g ) _{q} ) ( u;v ) =u^{2} ( N_{q}f ) ( u;v ) ( N_{q}g ) ( u;v )$.

Proof

Under the hypothesis of the theorem and Theorem 8, we write

$$\begin{aligned} N_{q} \bigl( ( f\ast g ) _{q} \bigr) ( u;v ) &=\sum _{0}^{\infty}a_{i}N_{q} \bigl( \bigl( t^{\alpha i}\ast t^{\beta-1} \bigr) _{q} \bigr) ( u;v ) \\ &=\sum_{0}^{\infty}a_{i} \bigl( N_{q}t^{\alpha i} \bigr) ( u;v ) \bigl( N_{q}t^{\beta-1} \bigr) ( u;v ) \\ &=u^{2} ( N_{q}f ) ( u;v ) ( N_{q}g ) ( u;v ). \end{aligned}$$

Hence, the proof of the theorem is finished. □

7 $N_{q}$ and Heaviside functions

The Heaviside function is defined by

$$ \acute{u} ( t-a )= \textstyle\begin{cases} 1, &t\geq a, \\ 0, &0\leq t< a,\end{cases} $$

(47)

where a is a real number.

In this part of the paper, we establish the following theorem.

Theorem 10

If ú denotes the Heaviside function and $u,v>0$, then we have

$$ N_{q} \bigl( \acute{u} ( t-a ) \bigr) ( u;v ) =\frac {1}{v}E_{q} \biggl( -\frac{v}{u}a \biggr). $$

(48)

Proof

By (47) we have $N_{q} ( \acute{u} ( t-a ) ) ( u;v ) =\frac {1}{u}\int_{a}^{\infty}E_{q} ( -q\frac{v}{u}t ) \,d_{q}t$. On account of (21), we get

$$\begin{aligned} N_{q} \bigl( \acute{u} ( t-a ) \bigr) ( u;v ) &=\frac{1}{ v}- \frac{1}{u} \int_{0}^{a}\sum_{0}^{\infty} \frac{q^{\frac{ n ( n-1 ) }{2}}}{ ( [ n ] _{q} ) !} \biggl( -q \frac{v}{u}t \biggr) ^{n}\,d_{q}t \\ &=\frac{1}{v}-\frac{1}{u}\sum_{0}^{\infty} ( -1 ) ^{n}\frac{q^{\frac{n ( n-1 ) }{2}}}{ ( [ n ] _{q} ) ! }q^{n} \frac{v^{n}}{u^{n}} \int_{0}^{a}t^{n}\,d_{q}t. \end{aligned}$$

Upon integrating and using simple calculation, the above equation reveals

$$\begin{aligned} N_{q} \bigl( \acute{u} ( t-a ) \bigr) ( u;v ) &=\frac{1}{ v}- \frac{1}{u}\sum_{0}^{\infty} \frac{q^{\frac{n ( n-1 ) }{2} }}{ [ n ] _{q}!}\frac{v^{n}}{u^{n}}\frac{a^{n+1}}{ [ n+1 ] _{q}} \\ &=\frac{1}{v}+\frac{1}{u}\sum_{0}^{\infty} ( -1 ) ^{n+1}\frac{q^{n+1\frac{n}{2}}}{ [ n+1 ] _{q}!}\frac{v^{n}}{u^{n}}a^{n+1} \\ &=\frac{1}{v}+\frac{1}{v}\sum_{0}^{\infty} ( -1 ) ^{n+1} \frac{q^{\frac{ ( n+1 ) n}{2}}}{ [ n+1 ] _{q}!}\frac{ v^{n+1}}{u^{n+1}}a^{n+1}. \end{aligned}$$

This can be written as $N_{q} ( \acute{u} ( t-a ) ) ( u;v ) =\frac{1}{v} +\frac{1}{v}\sum_{1}^{\infty} ( -1 ) ^{m}\frac{q^{m ( m-1 ) }}{ ( [ m ] _{q} ) !}\frac{v^{m}}{u^{m}}a^{m}$. Starting the summation from 0 gives

$$\begin{aligned} N_{q} \bigl( \acute{u} ( t-a ) \bigr) ( u;v ) =\frac{1}{ v}\sum _{1}^{\infty} ( -1 ) ^{m} \frac{q^{\frac{m ( m-1 ) }{2}}}{ ( [ m ] _{q} ) !}\frac{v^{m}}{u^{m}} a^{m} =\frac{1}{v}E_{q} \biggl( \frac{-v}{u}a \biggr). \end{aligned}$$

 □

8 The q-analogue of the second kind

Over the set $A= \{ f ( t ) | \exists M,\tau_{1},\tau _{2}>0, \vert f ( t ) \vert < Me^{t/\tau_{j}},t\in ( -1 ) ^{j}\times [ 0,\infty ) \}$, the q-analogue of the N-transform of the second type is expressed as

$$ \bigl( N^{q}f \bigr) ( u;v ) = \int_{0}^{\infty} \frac{1}{u}f ( t ) e_{q} \biggl(- \frac{v}{u}t \biggr) \,d_{q}t. $$

(49)

As the q-analogue of the gamma function of the second kind is defined as

$$ \gamma_{q} ( t ) = \int_{0}^{\infty}x^{t-1}e_{q} ( -x ) \,d_{q}x, $$

(50)

it follows that

$$ \gamma_{q} ( 1 ) =1,\gamma_{q} ( t+1 ) =q^{-t} [ t ] _{q}\gamma_{q} ( t ),\quad \gamma_{q} ( n ) =q^{\frac{n ( n-1 ) }{2}}\Gamma_{q} ( n ), $$

(51)

$\Gamma_{q}$ 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 $\alpha>-1$. Then we have

$$ \bigl( N^{q}t^{\alpha} \bigr) ( u;v ) =\frac{u^{\alpha}}{v^{\alpha+1}} \gamma_{q} ( \alpha+1 ). $$

(52)

In particular,

$$ \bigl( N^{q}t^{n} \bigr) ( u;v ) =\frac {u^{n}}{v^{n+1}}q^{\frac{-n ( n-1 ) }{2}} \bigl( [ n_{q} ] \bigr) !. $$

(53)

Proof

Let $\alpha>-1$. Then, by the change of variables, we have

$$\begin{aligned} \bigl( N^{q}t^{\alpha} \bigr) ( u;v ) =\frac{1}{u}\int_{0}^{\infty}t^{\alpha}e_{q} \biggl( \frac{-vt}{u} \biggr) \,d_{q}t =\frac{u^{\alpha}}{v^{\alpha+1}} \int_{0}^{\infty}t^{\alpha }e_{q} ( -t ) \,d_{q}t. \end{aligned}$$

By aid of (50), we get $( N^{q}t^{\alpha} ) ( u;v ) =\frac{u^{\alpha}}{v^{\alpha+1}}\gamma_{q} ( \alpha+1 )$. Proof of the second part of the theorem follows from (51). Hence, we have completed the proof of the theorem. □

Theorem 12

Let $a\in R,a>0$, then we have

$$ \bigl( N^{q}e_{q} ( at ) \bigr) ( u;v ) = \frac {1}{uv}\sum_{0}^{\infty} \frac{a^{n}u^{n}}{v^{n}}q^{\frac{-n ( n-1 ) }{2}}. $$

(54)

Proof

By (13) we write

$$\begin{aligned} \bigl( N^{q}e_{q} ( at ) \bigr) ( u;v ) =\frac{1}{u} \int_{0}^{\infty}e_{q} ( at ) e_{q} \biggl( -\frac{v}{u}t \biggr) \,d_{q}t = \frac{1}{u}\sum_{0}^{\infty} \frac{a^{n}}{ ( [ n ] _{q} ) !} \int_{0}^{\infty}t^{n}e_{q} \biggl( -\frac{v}{u}t \biggr) \,d_{q}t. \end{aligned}$$

By aid of Theorem 2, we get $( N^{q}e_{q} ( at ) ) ( u;v ) =\frac {1}{uv}\sum_{0}^{\infty}\frac{a^{n}u^{n}}{v^{n}}q^{\frac{-n ( n-1 ) }{2}}$.

This completes the proof of the theorem. □

Theorem 13

Let $a>0,a\in R$, then we have

$$ \bigl( N^{q}E_{q} ( at ) \bigr) ( u;v ) = \frac{1}{u ( v-au ) },\quad au< v. $$

(55)

Proof

After simple calculations and using Theorem 2, we obtain

$$\begin{aligned} \bigl( N^{q}e_{q} ( at ) \bigr) ( u;v ) &= \frac {1}{u}\int_{0}^{\infty}E_{q} ( at ) e_{q} \biggl( -\frac{v}{u}t \biggr) \,d_{q}t \\ &=\frac{1}{u}\sum_{0}^{\infty} \frac{q^{\frac{n ( n-1 ) }{2}}}{ ( [ n ] _{q} ) !}a^{n} \int_{0}^{\infty }t^{n}e_{q} \biggl( -\frac{v}{u}t \biggr) \,d_{q}t \\ &=\frac{1}{uv}\sum_{0}^{\infty}a^{n} \frac{u^{n}}{v^{n}}. \end{aligned}$$

Since the above series determines a geometric series, it reads $( N^{q}e_{q} ( at ) ) ( u;v ) =\frac {1}{uv}\frac{1}{1-\frac{au}{v}}=\frac{1}{u ( v-au ) },au< v$.

Hence the theorem is proved. □

The $N^{q}$ transform of $\cos^{q}$ and $\sin^{q}$ is given as follows.

Theorem 14

Let $a>0$. Then we have

$$ \bigl( N^{q}\cos^{q} ( at ) \bigr) ( u;v ) = \frac {1}{u}\sum_{0}^{\infty} ( -1 ) ^{n}a^{2n}\frac{u^{2n}}{v^{2n}}q^{-2n\frac{ ( 2n-1 ) }{2}}. $$

(56)

Proof

Using the definition of $\cos^{q}$, we write

$$\begin{aligned} \bigl( N^{q}\cos^{q} ( at ) \bigr) ( u;v ) &= \frac{1}{u } \int_{0}^{\infty}\cos^{q} ( at ) e_{q} \biggl( -\frac{v}{ u}t \biggr) \,d_{q}t \\ &=\frac{1}{u}\sum_{0}^{\infty} ( -1 ) ^{n}\frac{a^{2n}}{ ( [ 2n ] _{q} ) !} \int_{0}^{\infty }t^{2n}e^{q} \biggl( -\frac{v}{u}t \biggr) \,d_{q}t \\ &=\frac{1}{u}\sum_{0}^{\infty} ( -1 ) ^{n}a^{2n}\frac{ u^{2n}}{v^{2n}}q^{-2n\frac{ ( 2n-1 ) }{2}}. \end{aligned}$$

This completes the proof of the theorem. □

The $N^{q}$ transform of $\sin^{q} ( at )$ is given as follows.

Theorem 15

Let $a>0$, then we have

$$ \bigl( N^{q}\sin^{q} ( at ) \bigr) ( u;v ) = \frac {1}{uv}\sum_{0}^{\infty} ( -1 ) ^{n}a^{2n+1}\frac{u^{2n+1}}{v^{2n+1}}q^{-2n\frac{ ( 2n-1 ) }{2}}. $$

(57)

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

9 $N^{q}$ of q-differentiation

Before we start investigations, we first assert that

$$ D_{q}e_{q} \biggl( -\frac{v}{u}t \biggr) =- \frac{v}{u}e_{q} \biggl( -\frac {v}{u}t \biggr) . $$

(58)

In detail, this can be interpreted as

$$\begin{aligned} D_{q}e_{q} \biggl( -\frac{v}{u}t \biggr) &=\sum _{0}^{\infty}\frac{ ( -1 ) ^{n}}{ ( [ n ] _{q} ) !} \frac {v^{n}}{u^{n}}D_{q}t^{n} \\ &=\sum_{1}^{\infty}\frac{ ( -1 ) ^{n}}{ ( [ n-1 ] _{q} ) !} \frac{v^{n}}{u^{n}}t^{n-1} \\ &=\sum_{0}^{\infty}\frac{ ( -1 ) ^{n+1}}{ ( [ n ] _{q} ) !} \frac{v^{n+1}}{u^{n+1}}t^{n} \\ &=-\frac{v}{u}\sum_{0}^{\infty} \frac{ ( -1 ) ^{n}}{ ( [ n ] _{q} ) !}\frac{v^{n}}{u^{n}}t^{n} =-\frac{v}{u}e_{q} \biggl( -\frac{v}{u}t \biggr). \end{aligned}$$

This proves the above assertion.

Hence we prove the following theorem.

Theorem 16

Let $u,v>0$. Then we have

$$ \bigl( N^{q}D_{q}f ( t ) \bigr) ( u;v ) =-f ( 0 ) - \int_{0}^{\infty}f ( qt ) D_{q}e_{q} \biggl( -\frac{v}{u}t \biggr) \,d_{q}t. $$

(59)

Proof

By (58), the above equation gives

$$\begin{aligned} \bigl( N^{q}D_{q}f ( t ) \bigr) ( u;v ) &= \int_{0}^{\infty}D_{q}f ( t ) e_{q} \biggl( -\frac{v}{u}t \biggr) \,d_{q}t \\ &=-f ( 0 ) +\frac{v}{u} \int_{0}^{\infty}f ( qt ) e_{q} \biggl( - \frac{v}{u}t \biggr) \,d_{q}t \\ &=-f ( 0 ) +\frac{v}{u}q^{-1} \int_{0}^{\infty}f ( t ) e_{q} \biggl( - \frac{v}{u}q^{-1}t \biggr) \,d_{q}t. \end{aligned}$$

By setting variables, we have $( N^{q}D_{q}f ( t ) ) ( u;v ) =-f ( 0 ) +\frac{v}{u}q^{-1} ( N^{q}f ) ( q^{-1}v;u )$.

This completes the proof.

Now, we extend (59) to have

$$\begin{aligned} \bigl( N^{q}D_{q}^{2}f \bigr) ( u;v ) &= \bigl( N^{q}D_{q} ( D_{q}f ) \bigr) ( u;v ) \\ &=-D_{q}f ( 0 ) +\frac{v}{u}q^{-1} \bigl( N^{q}D_{q}f \bigr) \bigl( q^{-1}v;u \bigr) \\ &=-D_{q}f ( 0 ) +\frac{v}{u}q^{-1} \biggl( -f ( 0 ) +\frac{v }{u}q^{-1} \bigl( N^{q}f \bigr) \bigl( q^{-2}v;u \bigr) \biggr) \\ &=-D_{q}f ( 0 ) - \biggl( \frac{v}{u} \biggr) q^{-1}f ( 0 ) + \biggl( \frac{v}{u} \biggr) q^{2-2} \bigl( N^{q}f \bigr) \bigl( q^{-2}v;u \bigr). \end{aligned}$$

Proceeding to the nth derivative, we get $( N^{q}D_{q}^{n}f ) ( u;v ) = ( \frac {v}{u} ) ^{n}q^{-n} ( N^{q}f ) ( q^{-n}v;u ) -\sum_{i=0}^{n-1} ( \frac{v}{u} ) ^{n-1-i}D_{q}^{i}f ( 0 )$.

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.