Janvier Pesser Ntahomvukiye^*, Gaspard Bangerezako
University of Burundi, Faculty of Sciences, Department of Mathematics, Bujumbura, Burundi.
DOI: 10.4236/jamp.2026.141017   PDF    HTML   XML   43 Downloads   233 Views   Citations

Abstract

We first introduce the concepts of q-nonuniform exponential and trigonometric functions and then discuss the related q-nonuniform difference version of the integral Laplace transform and some of its applications.

Keywords

q-Nonuniform Difference Equations, q-Nonuniform Exponential and Trigonometric Functions, q-Nonuniform Laplace Transforms

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1. Introduction

Considering the most general divided difference derivative [1] [2],

$\mathcal{D}f\left(t\left(s\right)\right)=\frac{f\left(t\left(s+\frac{1}{2}\right)\right)-f\left(t\left(s-\frac{1}{2}\right)\right)}{t\left(s+\frac{1}{2}\right)-t\left(s-\frac{1}{2}\right)},$ (1)

admitting the property that if $f\left(t\right)=P_n\left(t\left(s\right)\right)$ is a polynomial of degree $n$ in $t\left(s\right)$ , then $\mathcal{D}f\left(t\left(s\right)\right)={\tilde{P}}_{n-1}\left(t\left(s\right)\right)$ is a polynomial in $t\left(s\right)$ of degree $n-1$ , one is led to the following most important canonical forms for $t\left(s\right)$ in order of increasing complexity:

$t\left(s\right)=t_0;$ (2)

$t\left(s\right)=s;$ (3)

$t\left(s\right)=q^s;$ (4)

$t\left(s\right)=q^s+q^{-s}2,q\in ℂ,s\in \mathbb{Z}.$ (5)

When the function $t\left(s\right)$ is given by (2)-(4), the divided difference derivative (1) leads to the ordinary differential derivative $Df\left(x\right)=\frac{\text{d}}{\text{d}x}f\left(x\right)$ , finite difference derivative $\text{Δ}f\left(t\right)=f\left(t+1\right)-f\left(t\right)=\left({\text{e}}^{\frac{\text{d}}{\text{d}t}}-1\right)f\left(t\right)$ and q-difference derivative (or Jackson derivative [3]) $D_{q}f\left(t\right)=\frac{f\left(qt\right)-f\left(t\right)}{qt-t}=\frac{q^{\frac{\text{d}}{\text{d}t}}-1}{qt-t}f\left(t\right)$ , respectively.

When $t\left(s\right)$ is given by (5), the corresponding derivative gives

$\begin{array}{c}\mathcal{D}f\left(t\left(s\right)\right)=\frac{f\left(t\left(s+\frac{1}{2}\right)\right)-f\left(t\left(s-\frac{1}{2}\right)\right)}{t\left(s+\frac{1}{2}\right)-t\left(s-\frac{1}{2}\right)}\\ =\frac{f\left(\frac{q^{\left(s+\frac{1}{2}\right)}+q^{\left(-s-\frac{1}{2}\right)}}{2}\right)-f\left(\frac{q^{\left(s-\frac{1}{2}\right)}+q^{\left(-s+\frac{1}{2}\right)}}{2}\right)}{\frac{q^{\left(s+\frac{1}{2}\right)}+q^{\left(-s-\frac{1}{2}\right)}}{2}-\frac{q^{\left(s-\frac{1}{2}\right)}+q^{\left(-s+\frac{1}{2}\right)}}{2}}.\end{array}$

It is usually referred to as the Askey-Wilson first order divided difference operator [4] that one can write:

$\mathcal{D}f\left(x\left(z\right)\right)=\frac{f\left(x\left(q^{\frac{1}{2}}z\right)\right)-f\left(x\left(q^{-\frac{1}{2}}z\right)\right)}{x\left(q^{\frac{1}{2}}z\right)-x\left(q^{-\frac{1}{2}}z\right)},$ (6)

where $x\left(z\right)=\frac{z+z^{-1}}{2}$ , having in mind that $z=q^s$ .

The Askey-Wilson polynomials [4], $P_n\left(x\left(z\right)\right),x\left(z\right)=\frac{z+z^{-1}}{2}$ are defined by:

$P_n\left(x\left(z\right)\right)=\frac{{\left(ab,ac,ad,q\right)}_n}{a^n}{4}^{{\varphi }_3}\left(\begin{array}{c}{q}^{-n},abcd{q}^{n-1},az,a{z}^{-1}\\ ab,ac,ad\end{array}|q;q\right)$ (7)

where the basic hypergeometric (or q-hypergeometric)series $r^{{\varphi }_s}$ read:

$r^{{\varphi }_s}\left(\begin{array}{c}{a}_1,\cdots ,a_r\\ b_1,\cdots ,b_s\end{array}|q;z\right):={\displaystyle \sum }_{k=0}^{\infty }\frac{{\left(a_1,\cdots ,a_r;q\right)}_k}{{\left(b_1,\cdots ,b_s;q\right)}_k}{\left(-1\right)}^{\left(1+s-r\right)k}{q}^{\left(1+s-r\right)\left(\begin{array}{l}k\hfill \\ 2\hfill \end{array}\right)}\frac{z^k}{{\left(q;q\right)}_k}$

with

${\left(a_1,\cdots ,a_r;q\right)}_k:={\left(a_1;q\right)}_k\cdots {\left(a_r;q\right)}_k,$

while

${\left(\sigma ;q\right)}_0:=1,{\left(\sigma ;q\right)}_k:={\displaystyle {\prod }_{i=0}^{k-1}\left(1-\sigma q^i\right)},k=1,2,3,\cdots$

They satisfy the following second order q-difference equation:

$\text{£}{P}_n\left(x\left(z\right)\right)=\lambda \left(n\right)P_n\left(x\left(z\right)\right)$ (8)

where

$\text{£}=\left[\frac{{\left(q-1\right)}^2}{4q}\frac{z^2-1}{z\omega \left(z\right)}{D}_q\frac{z^2-1}{z}v\left(q^{-\frac{1}{2}}z\right)\omega \left(q^{-\frac{1}{2}}z\right)\right]D_q,\text{with}\frac{\omega \left(qz\right)}{\omega \left(z\right)}=\frac{v\left(z\right)}{v\left(\left(qz{)}^{-1}\right)\right)}$

is the Askey-Wilson second order q-difference operator and

$\lambda \left(n\right)=-\left(1-q^{-n}\right)\left(1-abcd{q}^{n-1}\right).$

The analysis related to (2), (3), (4) and (5) are referred to as the differential, difference, q-difference and q-nonuniform difference analysis respectively.

Laplace transform of an exponential type function $f\left(x\right)$ is given by

$F\left(p\right)=ℒ\left\{f\left(x\right)\right\}\stackrel{\text{def}}{=}{\displaystyle {\int }_0^{+\infty }{\text{e}}^{-px}f\left(x\right)\text{d}x},p=a+ib\in ℂ,$ (9)

and plays a major role in pure and applied analysis, especially in solving differential equations. If we consider $f\left(x\right)$ as a function of a discrete variable i.e. $t\in \mathbb{Z}$ , then the transformation (9) reads

$F\left(z\right)=Z\left\{f\left(x\right)\right\}\stackrel{\text{def}}{=}{\displaystyle \sum }_{j=0}^{+\infty }f\left(j\right)z^{-j},z={\text{e}}^{-p}.$ (10)

It is referred to as Z transform and plays similar role in difference analysis as Laplace transform in continuous analysis, especially in solving difference equations [5].

For a given function $f\left(x\right)$ on the lattice (4), we define its q-Laplace transform as the function [6]

$F\left(p\right)={ℒ}_q\left\{f\left(x\right)\right\}\stackrel{\text{def}}{=}{\displaystyle {\int }_0^{+\infty }{e}_{q^{-1}}^{-px}f\left(x\right){\text{d}}_{q}x},p=s+i\sigma \in ℂ,$ (11)

where the intégral is the Jackson intégral [3]

${\displaystyle {\int }_0^{x}f\left(x\right){\text{d}}_{q}x}=\left(1-q\right)x{\displaystyle \sum }_{i=0}^{\infty }{q}^{i}f\left(x{q}^i\right)$ (12)

and

${\displaystyle {\int }_0^{\infty }f\left(x\right){\text{d}}_{q}x}=\left(1-q\right){\displaystyle \sum }_{i=-\infty }^{\infty }{q}^{i}f\left(q^i\right).$ (13)

It plays similar role in q-difference analysis as Laplace transform in continuous analysis or Z transform in difference analysis, especially in solving q-difference equations [6].

In this work, we introduce q-nonuniform Laplace transform defined for functions $f\left(x\left(z\right)\right)$ given on the lattice (5), as

$\begin{array}{l}F\left(p\right)={ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z\right)\right)\right\}\\ \stackrel{\text{def}}{=}{\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;x_0;x\left(q^{-\frac{1}{2}}\right)\right)f\left(x\left(z\right)\right)\text{d}x\left(z\right),\\ p=s+i\sigma \in ℂ,0\le z_0,z\le \infty ,0<q<1,x_0=x\left(z_0\right),x\left(\infty \right)=\infty ,\end{array}$ (14)

where the intégral is defined by [7] [8]

${\displaystyle {\int }_{x\left(q^N\right)}^{x\left(q^s\right)}g\left(x\left(z\right)\right){\text{d}}_{q}x\left(z\right)}\stackrel{\text{def}}{=}\frac{1}{2}\left(1-q\right){\displaystyle {\sum }_{z=q^s}^{q^{N-1}}z\left(1-\frac{1}{q{z}^2}\right)g\left(x\left(q^{\frac{1}{2}}z\right)\right)}$ (15)

and

${\displaystyle {\int }_{x\left(z\right)}^{\infty }g\left(x\left(z\right)\right){\text{d}}_{q}x\left(z\right)}=\frac{1}{2}\left(1-q\right)z{\displaystyle {\sum }_{i=0}^{\infty }{q}^i}\left(\frac{1}{q^{2i+1}{z}^2}-1\right)g\left(x\left(q^{\frac{1}{2}+i}z\right)\right).$ (16)

It is expected to play similar role in q-nonuniform difference analysis as Laplace transform, Z transform and q-Laplace transform in differential, difference and q-difference analysis respectively, especially in solving q-nonuniform difference equations [9].

In the next section, we begin by introducing the concepts of q-nonuniform exponential and trigonometric functions as solutions of first or second order q-nonuniform difference equations.

2. q-Nonuniform Exponential and Trigonometric Functions

Before introducing the concept of q-non nuniform Laplace Transform, we clearly first study these of q-non exponential and trigonometric functions.

Consider first the following equation

$\mathcal{D}{y}_0\left(x\left(z\right)\right)=a\left(x\left(z\right)\right)y_0\left(x\left(z{q}^{-\frac{1}{2}}\right)\right).$ (17)

Developping it

$\frac{y_0(x\left(z{q}^{\frac{1}{2}}\right)-y_0(x\left(z{q}^{-\frac{1}{2}}\right)}{x\left(z{q}^{\frac{1}{2}}\right)-x\left(z{q}^{-\frac{1}{2}}\right)}=a\left(x\left(z\right)\right)y_0\left(x\left(z{q}^{-\frac{1}{2}}\right)\right),$

we get

$y_0\left(x\left(z{q}^{\frac{1}{2}}\right)\right)=p\left(x\left(z\right)\right)y_0\left(x\left(z{q}^{-\frac{1}{2}}\right)\right)$ (18)

where

$\begin{array}{c}p\left(x\left(z\right)\right)=1+\left(x\left(z{q}^{\frac{1}{2}}\right)-x\left(z{q}^{-\frac{1}{2}}\right)\right)a\left(x\left(z\right)\right)\\ =1+\frac{q\frac{1}{2}-q^{-\frac{1}{2}}}{2}\left(z-z^{-1}\right)a\left(x\left(z\right)\right).\end{array}$ (19)

By the recursion

$y_0\left(x\left(z\right)\right)={\left[p\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\right]}^{-1}{y}_0\left(x\left(zq\right)\right),$ (20)

we get

$y_0\left(x\left(z\right)\right)=y_0\left(x\left(z_0\right)\right){\displaystyle \prod }_{t=z}^{q^{-1}{z}_0}{\left[p\left(x\left(t{q}^{\frac{1}{2}}\right)\right)\right]}^{-1},$ (21)

or

$y_0\left(x\left(z\right)\right)=\left({\displaystyle \prod }_{i=0}^{N-1}{\left[p\left(x\left(z{q}^{\frac{1}{2}+i}\right)\right)\right]}^{-1}\right)y_0\left(x\left(z{q}^N\right)\right).$ (22)

Define the q-nonuniform exponential function

$E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\{\begin{array}{ll}{\displaystyle {\prod }_{t=z}^{q^{-1}{z}_0}{\left[p\left(x\left(t{q}^{\frac{1}{2}}\right)\right)\right]}^{-1}},\hfill & z>z_0,\hfill \\ 1,\hfill & z=z_0>0.\hfill \end{array}$ (23)

Clearly, we have the relation

$E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)={\left[p\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\right]}^{-1}{E}_{q^{-\frac{1}{2}}}\left(a;z_0;qz\right).$ (24)

This means that $E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)$ is a solution of (20) and consequently, that of (17). Moreover,

$E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)={\left[p\left(x\left(z_0{q}^{-\frac{1}{2}}\right)\right)\right]}^{-1}{E}_{q^{-\frac{1}{2}}}\left(a;q^{-1}{z}_0;z\right).$ (25)

From the definition (23), we can evaluate

$E_{q^{-\frac{1}{2}}}\left(a;0;z\right)={\displaystyle \prod }_{i=0}^{\infty }{\left[p\left(x\left(q^{i}z{q}^{\frac{1}{2}}\right)\right)\right]}^{-1},$ (26)

and

$E_{q^{-\frac{1}{2}}}\left(a;z_0;\infty \right)={\displaystyle \prod }_{i=0}^{\infty }{\left[p\left(x\left(q^{-i}{q}^{-1}{z}_0{q}^{\frac{1}{2}}\right)\right)\right]}^{-1}={\displaystyle \prod }_{i=-1}^{-\infty }{\left[p\left(x\left(q^i{z}_0{q}^{\frac{1}{2}}\right)\right)\right]}^{-1}.$ (27)

Also, one easily verifies that $E_{q^{-\frac{1}{2}}}\left(a;z_0;\infty \right)$ is a solution of (25) for $z=\infty$ .

If we set $z=z_0=1$ in (26) and (27), we get

$E_{q^{-\frac{1}{2}}}\left(a;0;\infty \right)={\displaystyle \prod }_{i=-\infty }^{\infty }{\left[p\left(x\left(q^i{q}^{\frac{1}{2}}\right)\right)\right]}^{-1}.$ (28)

On the other side, since $0<q<1$ , then, for $a\left(x\left(z\right)\right)=a=const$ , $z_0,z\ne 0,\infty$ , one can verify that

$E_{q^{-\frac{1}{2}}}\left(a;0;z\right)=0=E_{q^{-\frac{1}{2}}}\left(a;z_0;\infty \right).$ (29)

Indeed,

$\begin{array}{c}p\left(x\left(z{q}^{\frac{1}{2}}\right)\right)=1+\frac{q^{\frac{1}{2}}-q^{-\frac{1}{2}}}{2}\left(q^{\frac{1}{2}}z-q^{-\frac{1}{2}}{z}^{-1}\right)a\\ =\frac{a\left(q-1\right)}{2}\frac{\left(z-z_1\right)\left(z-z_2\right)}{z};\end{array}$ (30)

with

$z_1=\frac{1-2q+2\sqrt{q^2-2a^2{q}^2+a^2{q}^3+a^{2}q}}{2\left(a{q}^2-aq\right)},$

$z_2=\frac{1-2q-2\sqrt{q^2-2a^2{q}^2+a^2{q}^3+a^{2}q}}{2\left(a{q}^2-aq\right)}.$

Hence

$\begin{array}{c}{E}_{q^{-\frac{1}{2}}}\left(a;0;z\right)={\displaystyle \prod }_{i=0}^{\infty }{\left[p\left(x\left(q^{i}z{q}^{\frac{1}{2}}\right)\right)\right]}^{-1}\\ =\frac{2z}{a\left(q-1\right)z_1{z}_2}\frac{1}{{\left(z_1^{-1}z;q\right)}_{\infty }{\left(z_2^{-1}z;q\right)}_{\infty }}\cdot {\displaystyle \prod }_{i=0}^{\infty }{q}^i=0,\end{array}$ (31)

where ${\left(u;q\right)}_{\infty }\stackrel{\text{def}}{=}{\displaystyle {\prod }_{i=0}^{\infty }\left(1-u{q}^i\right)}$. Similarly, one proves that $E_{q^{-\frac{1}{2}}}\left(a;z_0;\infty \right)=0$ .

Consider now the following equation

$\mathcal{D}\tilde{y}\left(x\left(z\right)\right)=\tilde{a}\left(x\left(z\right)\right)\tilde{y}\left(x\left(z{q}^{\frac{1}{2}}\right)\right).$ (32)

Developping, we get the relation

$\tilde{y}\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\tilde{p}\left(x\left(z\right)\right)=\tilde{y}\left(x\left(z{q}^{-\frac{1}{2}}\right)\right).$ (33)

where

$\begin{array}{c}\tilde{p}\left(x\left(z\right)\right)=1-\left(x\left(z{q}^{\frac{1}{2}}\right)-x\left(z{q}^{-\frac{1}{2}}\right)\right)a\left(x\left(z\right)\right)\\ =1-\frac{q^{\frac{1}{2}}-q^{-\frac{1}{2}}}{2}\left(z-z^{-1}\right)\tilde{a}\left(x\left(z\right)\right).\end{array}$ (34)

Using the recursion

${\tilde{y}}_0\left(x\left(z\right)\right)=\tilde{p}\left(x\left(z{q}^{\frac{1}{2}}\right)\right){\tilde{y}}_0\left(x\left(zq\right)\right),$ (35)

we get

${\tilde{y}}_0\left(x\left(z\right)\right)={\tilde{y}}_0\left(x\left(z_0\right)\right){\displaystyle \prod }_{t=z}^{q^{-1}{z}_0}\tilde{p}\left(x\left(t{q}^{\frac{1}{2}}\right)\right),$ (36)

or

${\tilde{y}}_0\left(x\left(z\right)\right)=\left({\displaystyle \prod }_{i=0}^{N-1}\tilde{p}\left(x\left(z{q}^{\frac{1}{2}+i}\right)\right)\right){\tilde{y}}_0\left(x\left(z{q}^N\right)\right).$ (37)

Define the other $q-$ nonuniform exponential function

$E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;z\right)\stackrel{\text{def}}{=}\{\begin{array}{ll}{\displaystyle {\prod }_{t=z}^{q^{-1}{z}_0}\tilde{p}\left(x\left(t{q}^{\frac{1}{2}}\right)\right)},\hfill & z>z_0,\hfill \\ 1,\hfill & z=z_0>0.\hfill \end{array}$ (38)

Here also, the function $E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;z\right)$ admet representations and properties similar to that of $E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)$ . In particular, $E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;z\right)$ verifies the relation

$E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;z\right)=\tilde{p}\left(x\left(z{q}^{\frac{1}{2}}\right)\right)E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;zq\right).$ (39)

This means that $E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;z\right)$ is a solution of (35) and consequently, that of (32). Moreover,

$E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;z\right)=\tilde{p}\left(x\left(z_0{q}^{-\frac{1}{2}}\right)\right)E_{q^{\frac{1}{2}}}\left(\tilde{a};q^{-1}{z}_0;z\right).$ (40)

Similarly, we evaluate

$E_{q^{\frac{1}{2}}}\left(\tilde{a};0;z\right)={\displaystyle \prod }_{i=0}^{\infty }p\left(x\left(q^{i}z{q}^{\frac{1}{2}}\right)\right),$ (41)

and

$E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;\infty \right)={\displaystyle \prod }_{i=-1}^{-\infty }\tilde{p}\left(x\left(q^i{z}_0{q}^{\frac{1}{2}}\right)\right).$ (42)

Also, one easily verifies that $E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;\infty \right)$ is a solution of (40) for $z=\infty$ .

Also

$E_{q^{\frac{1}{2}}}\left(\tilde{a};0;\infty \right)={\displaystyle \prod }_{i=-\infty }^{\infty }\tilde{p}\left(x\left(q^i{z}_0{q}^{\frac{1}{2}}\right)\right),$ (43)

and

$E_{q^{\frac{1}{2}}}\left(\tilde{a};0;z\right)=\infty =E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;\infty \right).$ (44)

Consider now the product $h\left(z\right)=E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)\cdot E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;z\right)$ . Define $K_{a,\tilde{a}}$ by

$K_{a,\tilde{a}}=h\left(\infty \right)=E_{q^{-\frac{1}{2}}}\left(a;z_0;\infty \right)\cdot E_{q^{\frac{1}{2}}}\left(\tilde{a};z_0;\infty \right)={\displaystyle \prod }_{i=1}^{\infty }\frac{\tilde{p}\left(x\left(q^{-i}{z}_0{q}^{\frac{1}{2}}\right)\right)}{p\left(x\left(q^{-i}{z}_0{q}^{\frac{1}{2}}\right)\right)}.$ (45)

Simple computations give

$K_{a,\tilde{a}}=\frac{-\tilde{a}}{a}\frac{\left(1-z_0^{-1}{z}_1\right)\left(1-z_0^{-1}{z}_2\right)}{\left(1-z_0^{-1}{\tilde{z}}_1\right)\left(1-z_0^{-1}{\tilde{z}}_2\right)}\frac{{\left(z_0^{-1}{\tilde{z}}_1;q\right)}_{\infty }{\left(z_0^{-1}{\tilde{z}}_2;q\right)}_{\infty }}{{\left(z_0^{-1}{z}_1;q\right)}_{\infty }{\left(z_0^{-1}{z}_2;q\right)}_{\infty }},$ (46)

where $z_1,z_2$ and ${\tilde{z}}_1,{\tilde{z}}_2$ are finite roots of $p\left(q^{\frac{1}{2}}z\right)=0$ and $\tilde{p}\left(q^{\frac{1}{2}}z\right)=0$ , respectively, knowing that $\tilde{p}\left(x\left(z\right)\right)$ is obtained from $p\left(x\left(z\right)\right)$ by replacing $a$ by $-\tilde{a}$ .

From (19), (23), (34) and (38), we get the following

Theorem 2.1

$E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)\cdot E_{q^{\frac{1}{2}}}\left(-a;z_0;z\right)=1=E_{q^{-\frac{1}{2}}}\left(-a;z_0;z\right)\cdot E_{q^{\frac{1}{2}}}\left(a;z_0;z\right).$ (47)

More generally, we have the following

Theorem 2.2 If $y\left(x\left(z\right)\right)$ and $\tilde{y}\left(x\left(z\right)\right)$ are respective solutions of the equations

$\mathcal{D}y\left(x\left(z\right)\right)=a\left(x\left(z\right)\right)y\left(x\left(z{q}^{-\frac{1}{2}}\right)\right),$

$\mathcal{D}\tilde{y}\left(x\left(z\right)\right)=-a\left(x\left(z\right)\right)\tilde{y}\left(x\left(z{q}^{-\frac{1}{2}}\right)\right),$ (48)

satifying the conditions

$y\left(x\left(z_0\right)\right)\tilde{y}\left(x\left(z_0\right)\right)=1,$ (49)

then

$y\left(x\left(z\right)\right)\tilde{y}\left(x\left(z\right)\right)=1.$ (50)

Proof.

$\begin{array}{c}\mathcal{D}\left(y\tilde{y}\right)=y\left(x\left(q^{-\frac{1}{2}}z\right)\right)\mathcal{D}\tilde{y}+\tilde{y}\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\mathcal{D}y\\ =y\left(x\left(q^{-\frac{1}{2}}z\right)\right)\left(-a\right)\tilde{y}\left(x\left(z{q}^{\frac{1}{2}}\right)\right)+\tilde{y}\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\left(a\right)y\left(x\left(q^{-\frac{1}{2}}z\right)\right)\\ =0.\end{array}$

This means that $y\tilde{y}=const$ , which, by (49) gives (50).

Consider next the following definitions

1)

${\text{cos}}_{q^{\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{\frac{1}{2}}}\left(ai;z_0;z\right)+E_{q^{\frac{1}{2}}}\left(-ai;z_0;z\right)}{2};$

${\text{sin}}_{q^{\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{\frac{1}{2}}}\left(ai;z_0;z\right)-E_{q^{\frac{1}{2}}}\left(-ai;z_0;z\right)}{2i};$

${\text{cosh}}_{q^{\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{\frac{1}{2}}}\left(a;z_0;z\right)+E_{q^{\frac{1}{2}}}\left(-a;z_0;z\right)}{2};$

${\text{sinh}}_{q^{\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{\frac{1}{2}}}\left(a;z_0;z\right)-E_{q^{\frac{1}{2}}}\left(-a;z_0;z\right)}{2}.$ (51)

2)

${\text{cos}}_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{-\frac{1}{2}}}\left(ai;z_0;z\right)+E_{q^{-\frac{1}{2}}}\left(-ai;z_0;z\right)}{2};$

${\text{sin}}_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{-\frac{1}{2}}}\left(ai;z_0;z\right)-E_{q^{-\frac{1}{2}}}\left(-ai;z_0;z\right)}{2i};$

${\text{cosh}}_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)+E_{q^{-\frac{1}{2}}}\left(-a;z_0;z\right)}{2};$

${\text{sinh}}_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)-E_{q^{-\frac{1}{2}}}\left(-a;z_0;z\right)}{2}.$ (52)

3)

${\text{cos}}_{q^{\frac{1}{2}}{q}^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{\frac{1}{2}}}\left(ai;z_0;z\right)+E_{q^{-\frac{1}{2}}}\left(-ai;z_0;z\right)}{2};$

${\text{sin}}_{q^{\frac{1}{2}}{q}^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{\frac{1}{2}}}\left(ai;z_0;z\right)-E_{q^{-\frac{1}{2}}}\left(-ai;z_0;z\right)}{2i};$

${\text{cosh}}_{q^{\frac{1}{2}}{q}^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{\frac{1}{2}}}\left(a;z_0;z\right)+E_{q^{-\frac{1}{2}}}\left(-a;z_0;z\right)}{2};$

${\text{sinh}}_{q^{\frac{1}{2}}{q}^{-\frac{1}{2}}}\left(a;z_0;z\right)\stackrel{\text{def}}{=}\frac{E_{q^{\frac{1}{2}}}\left(a;z_0;z\right)-E_{q^{-\frac{1}{2}}}\left(-a;z_0;z\right)}{2}.$ (53)

We get

$\mathcal{D}{\cos }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=-a{\sin }_{q^{\frac{1}{2}}}\left(a;z_0;z{q}^{\frac{1}{2}}\right);$

${\mathcal{D}}^{-}{\cos }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=-a{\sin }_{q^{\frac{1}{2}}}\left(a;z_0;z\right),$

$\mathcal{D}{\sin }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=a{\cos }_{q^{\frac{1}{2}}}\left(a;z_0;z{q}^{\frac{1}{2}}\right);$

${\mathcal{D}}^{-}{\sin }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=a{\cos }_{q^{\frac{1}{2}}}\left(a;z_0;z\right),$ (54)

where ${\mathcal{D}}^{-}f\left(x\left(z\right)\right)={Df\left(x\left(z\right)\right)|}_{z:=z{q}^{-\frac{1}{2}}}$ . Hence

${\left[{\mathcal{D}}^{-}\right]}^2{\cos }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=-a^2{\cos }_{q^{\frac{1}{2}}}\left(a;z_0;z\right),$

${\left[{\mathcal{D}}^{-}\right]}^2{\sin }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=-a^2{\sin }_{q^{\frac{1}{2}}}\left(a;z_0;z\right).$ (55)

Similarly

$\mathcal{D}{\cosh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=a{\sinh }_{q^{\frac{1}{2}}}\left(a;z_0;z{q}^{\frac{1}{2}}\right);$

${\mathcal{D}}^{-}{\cosh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=a{\sinh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right),$

$\mathcal{D}{\sinh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=a{\cosh }_{q^{\frac{1}{2}}}\left(a;z_0;z{q}^{\frac{1}{2}}\right);$

${\mathcal{D}}^{-}{\sinh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=a{\cosh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right),$ (56)

and consequently

${\left[{\mathcal{D}}^{-}\right]}^2{\cosh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=a^2{\cosh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right),$

${\left[{\mathcal{D}}^{-}\right]}^2{\sinh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right)=a^2{\sinh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right).$ (57)

On the other side, we have

$\mathcal{D}{\cos }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=-a{\sin }_{q^{-\frac{1}{2}}}\left(a;z_0;z{q}^{-\frac{1}{2}}\right);$

${\mathcal{D}}^{+}{\cos }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=-a{\sin }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right),$

$\mathcal{D}{\sin }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=a{\cos }_{q^{-\frac{1}{2}}}\left(a;z_0;z{q}^{-\frac{1}{2}}\right);$

${\mathcal{D}}^{+}{\sin }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=a{\cos }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right),$ (58)

where ${\mathcal{D}}^{+}f\left(x\left(z\right)\right)={Df\left(x\left(z\right)\right)|}_{z:=z{q}^{\frac{1}{2}}}$ . Hence

${\left[{\mathcal{D}}^{+}\right]}^2{\cos }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=-a^2{\cos }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right),$

${\left[{\mathcal{D}}^{+}\right]}^2{\sin }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=-a^2{\sin }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right).$ (59)

Similarly

$\mathcal{D}{\cosh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=a{\sinh }_{q^{-\frac{1}{2}}}\left(a;z_0;z{q}^{-\frac{1}{2}}\right);$

${\mathcal{D}}^{-}{\cosh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=a{\sinh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right),$

$\mathcal{D}{\sinh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=a{\cosh }_{q^{-\frac{1}{2}}}\left(a;z_0;z{q}^{-\frac{1}{2}}\right);$

${\mathcal{D}}^{-}{\sinh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=a{\cosh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right),$

and consequently

${\left[{\mathcal{D}}^{+}\right]}^2{\cosh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=a^2{\cosh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right),$

${\left[{\mathcal{D}}^{+}\right]}^2{\sinh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=a^2{\sinh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right).$ (60)

From the preceding, one easily verifies that

a)

${\text{cos}}_{q^{\frac{1}{2}}{q}^{-\frac{1}{2}}}^2\left(a;z_0;z\right)+{\text{sin}}_{q^{\frac{1}{2}}{q}^{-\frac{1}{2}}}^2\left(a;z_0;z\right)=1,$ (61)

b)

${\text{cosh}}_{q^{\frac{1}{2}}{q}^{-\frac{1}{2}}}^2\left(a;z_0;z\right)-{\text{sinh}}_{q^{\frac{1}{2}}{q}^{-\frac{1}{2}}}^2\left(a;z_0;z\right)=1,$ (62)

c)

${\text{cos}}_{q^{\frac{1}{2}}}\left(a;z_0;z\right){\text{cos}}_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)+{\text{sin}}_{q^{\frac{1}{2}}}\left(a;z_0;z\right){\text{sin}}_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=1,$ (63)

d)

${\cosh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right){\cosh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)-{\sinh }_{q^{\frac{1}{2}}}\left(a;z_0;z\right){\sinh }_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)=1.$ (64)

3. q-Nonuniform Laplace Transform

For a given function $f\left(x\left(z\right)\right)$ , we define its q-nonuniform Laplace transform or Laplace transform on the q-nonuniform lattice $x\left(z\right)=\left(z+z^{-1}\right)/2$ , $z=q^s$ , $0<q<1$ , $-\infty \le s\le +\infty$ , as the function

$\begin{array}{l}F\left(p\right)={ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z\right)\right)\right\}\\ \stackrel{\text{def}}{=}{\displaystyle {\int }_{x\left(z_0\right)=x_0}^{x\left(\infty \right)=\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;z{q}^{-\frac{1}{2}}\right)f\left(x\left(z\right)\right)\text{d}x\left(z\right);p=s+i\sigma \in ℂ.\end{array}$

We denote $f\left(x\left(z\right)\right){\rightleftharpoons }_{q^{\frac{1}{2}}}F\left(p\right)$ , and we say that $f\left(x\left(z\right)\right)$ is the original of $F\left(p\right)$ while $F\left(p\right)$ is the image of $f\left(x\left(z\right)\right)$ by the Laplace transform on q-nonuniform lattices.

3.1. Properties of the q-Nonuniform Laplace Transform

1) Linearity. By the linearity of the integral, it becomes clear that

${ℒ}_{q^{\frac{1}{2}}}\left\{\alpha f\left(x\left(z\right)\right)+\beta g\left(x\left(z\right)\right)\right\}=\alpha {ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z\right)\right)\right\}+\beta {ℒ}_{q^{\frac{1}{2}}}\left\{g\left(x\left(z\right)\right)\right\}.$

2) Transform of derivatives. We have

$\begin{array}{c}{ℒ}_{q^{\frac{1}{2}}}\left\{Df\left(x\left(z\right)\right)\right\}={\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;z{q}^{-\frac{1}{2}}\right)\mathcal{D}f\left(x\left(z\right)\right)\text{d}x\left(z\right)}\\ ={\left[E_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)f\left(x\left(z\right)\right)\right]}_{z_0}^{\infty }\\ -{\displaystyle {\int }_{x_0}^{\infty }f\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\mathcal{D}{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\text{d}x\left(z\right)}\\ =-f\left(x\left(z_0\right)\right)+p{\displaystyle {\int }_{x_0}^{\infty }f\left(x\left(z{q}^{\frac{1}{2}}\right)\right)E_{q^{-\frac{1}{2}}}\left(-p;z_0;z{q}^{-\frac{1}{2}}\right)\text{d}x\left(z\right)}\\ =p{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\right\}-f\left(x\left(z_0\right)\right),\end{array}$ (65)

where we used the q-nonuniform integration by parts [8]:

$\begin{array}{l}{\displaystyle {\int }_{x_0}^{x\left(z\right)}g}\left(x\left(z{q}^{-\frac{1}{2}}\right)\right)\mathcal{D}f\left(x\left(z\right)\right)\text{d}x\left(z\right)\\ ={\left[g\left(x\left(z\right)\right)f\left(x\left(z\right)\right)\right]}_{x_0}^{x\left(z\right)}-{\displaystyle {\int }_{x_0}^{x\left(z\right)}f}\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\mathcal{D}g\left(x\left(z\right)\right)\text{d}x\left(z\right),\end{array}$ (66)

and (29). Thus

i)

${ℒ}_{q^{\frac{1}{2}}}\left\{\mathcal{D}f\left(x\left(z\right)\right)\right\}=p{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\right\}-f\left(x\left(z_0\right)\right),$ (67)

and

${ℒ}_{q^{\frac{1}{2}}}\left\{\mathcal{D}f\left(x\left(z{q}^{\alpha }\right)\right)\right\}=p{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z{q}^{\alpha +\frac{1}{2}}\right)\right)\right\}-f\left(x\left(q^{\alpha }{z}_0\right)\right).$ (68)

Hence

$\mathcal{D}f\left(x\left(z{q}^{-\frac{1}{2}}\right)\right){\rightleftharpoons }_{q^{\frac{1}{2}}}pF\left(p\right)-f\left(x\left(z_0{q}^{-\frac{1}{2}}\right)\right),$ (69)

and

$\mathcal{D}f\left(x\left(z{q}^{\frac{1}{2}}\right)\right){\rightleftharpoons }_{q^{\frac{1}{2}}}p{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(zq\right)\right)\right\}-f\left(x\left(z_0{q}^{\frac{1}{2}}\right)\right).$ (70)

Using (67), one easily obtains

ii)

$\begin{array}{l}{ℒ}_{q^{\frac{1}{2}}}\left\{D^{2}f\left(x\left(z\right)\right)\right\}={ℒ}_{q^{\frac{1}{2}}}\left\{\mathcal{D}\left(\mathcal{D}f\left(x\left(z\right)\right)\right)\right\}\\ =p^2{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(zq\right)\right)\right\}-pf\left(x\left(q^{\frac{1}{2}}{z}_0\right)\right)-\mathcal{D}f\left(x\left(z_0\right)\right).\end{array}$ (71)

Hence

$\begin{array}{l}{ℒ}_{q^{\frac{1}{2}}}\left\{D^{2}f\left(x\left(z{q}^{\frac{1}{2}}\right)\right)\right\}\\ =p^2{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z{q}^{\frac{3}{2}}\right)\right)\right\}-pf\left(x\left(q{z}_0\right)\right)-\mathcal{D}f\left(x\left(q^{\frac{1}{2}}{z}_0\right)\right).\end{array}$ (72)

iii)

$\begin{array}{l}{ℒ}_{q^{\frac{1}{2}}}\left\{{\mathcal{D}}^{3}f\left(x\left(z\right)\right)\right\}={ℒ}_{q^{\frac{1}{2}}}\left\{\mathcal{D}\left({\mathcal{D}}^{2}f\left(x\left(z\right)\right)\right)\right\}\\ =p^3{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z{q}^{\frac{3}{2}}\right)\right)\right\}-\left[p^{2}f\left(x\left(q{z}_0\right)\right)+p\mathcal{D}f\left(x\left(q^{\frac{1}{2}}{z}_0\right)\right)+{\mathcal{D}}^{2}f\left(x\left(z_0\right)\right)\right].\end{array}$ (73)

$\cdots$

n)

$\begin{array}{l}{ℒ}_{q^{\frac{1}{2}}}\left\{D^{n}f\left(x\left(z\right)\right)\right\}=p^n{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z{q}^{\frac{n}{2}}\right)\right)\right\}\\ -\left[p^{n-1}f\left(x\left(q^{\frac{n-1}{2}}{z}_0\right)\right)+p^{n-2}\mathcal{D}f\left(x\left(q^{\frac{n-2}{2}}{z}_0\right)\right)+\cdots +{\mathcal{D}}^{n-1}f\left(x\left(z_0\right)\right)\right],\end{array}$ (74)

or equivalently

$\begin{array}{l}{ℒ}_{q^{\frac{1}{2}}}\left\{D^{n}f\left(x\left(q^{-\frac{n}{2}}z\right)\right)\right\}=p^n{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z\right)\right)\right\}\\ -\left[p^{n-1}f\left(x\left(q^{-\frac{1}{2}}{z}_0\right)\right)+p^{n-2}\mathcal{D}f\left(x\left(q^{-1}{z}_0\right)\right)+\cdots +{\mathcal{D}}^{n-1}f\left(x\left(q^{-\frac{n}{2}}{z}_0\right)\right)\right].\end{array}$ (75)

For example, for $n=2$ , we have

${ℒ}_{q^{\frac{1}{2}}}\left\{D^{2}f\left(x\left(z{q}^{-1}\right)\right)\right\}=p^2{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(z\right)\right)\right\}-\left[pf\left(x\left(q^{-\frac{1}{2}}{z}_0\right)\right)+\mathcal{D}f\left(x\left(q^{-1}{z}_0\right)\right)\right].$ (76)

3) Transform of integrals. In the equation

${ℒ}_{q^{\frac{1}{2}}}\left\{\mathcal{D}f\left(x\left(z\right)\right)\right\}=p{ℒ}_{q^{\frac{1}{2}}}\left\{f\left(x\left(q^{\frac{1}{2}}z\right)\right)\right\}-f\left(x\left(z_0\right)\right),$

set

$f\left(x\left(z\right)\right)={\displaystyle {\int }_{x_0}^{x\left(z\right)}g}\left(x\left(z\right)\right)\text{d}x\left(z\right).$ (77)

We have

${ℒ}_{q^{\frac{1}{2}}}\left\{g\left(x\left(z\right)\right)\right\}=p{ℒ}_{q^{\frac{1}{2}}}\left\{{\displaystyle {\int }_{x_0}^{x\left(z{q}^{\frac{1}{2}}\right)}g}\left(x\left(z\right)\right)\text{d}x\left(z\right)\right\}.$ (78)

Hence

${ℒ}_{q^{\frac{1}{2}}}\left\{{\displaystyle {\int }_{x_0}^{x\left(z{q}^{\frac{1}{2}}\right)}g}\left(x\left(z\right)\right)\text{d}x\left(z\right)\right\}=\frac{1}{p}{ℒ}_{q^{\frac{1}{2}}}\left\{g\left(x\left(z\right)\right)\right\}.$ (79)

3.2. q-Nonuniform Laplace Transform for Some Elementary Functions

1) $f\left(x\left(z\right)\right)=1$ . We have

$\begin{array}{c}{ℒ}_{q^{\frac{1}{2}}}\left\{1\right\}={\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)\text{d}x\left(z\right)\\ =-\frac{1}{p}{\displaystyle {\int }_{x_0}^{\infty }\mathcal{D}{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\text{d}x\left(z\right)}\\ =-{\frac{1}{p}{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)|}_{z_0}^{\infty }\\ =\frac{1}{p}.\end{array}$ (80)

2) $f\left(x\left(z\right)\right)=x\left(z{q}^{\frac{1}{2}}\right)$ . We have

$\begin{array}{c}{ℒ}_{q^{\frac{1}{2}}}\left\{x\left(z{q}^{\frac{1}{2}}\right)\right\}={\displaystyle {\int }_{x_0}^{\infty }x}\left(z{q}^{\frac{1}{2}}\right)E_{q^{-\frac{1}{2}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)\text{d}x\left(z\right)\\ =-\frac{1}{p}{\displaystyle {\int }_{x_0}^{\infty }x}\left(z{q}^{\frac{1}{2}}\right)\mathcal{D}{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\text{d}x\left(z\right)\\ =-\frac{1}{p}{\left[x\left(z\right)E_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\right]}_{z_0}^{\infty }+\frac{1}{p}{\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)\text{d}x\left(z\right)\\ =\frac{x\left(z_0\right)}{p}-\frac{1}{p^2}{\displaystyle {\int }_{x_0}^{\infty }\mathcal{D}{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;z\right)\text{d}x\left(z\right)\\ =\frac{x\left(z_0\right)}{p}-\frac{1}{p^2}{\left[E_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\right]}_{z_0}^{\infty }\\ =\frac{x\left(z_0\right)}{p}+\frac{1}{p^2}.\end{array}$ (81)

3) $f\left(x\left(z\right)\right)=E_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)$ . We calculate

$\begin{array}{c}I={ℒ}_{q^{\frac{1}{2}}}\left\{E_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}\\ ={\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)E_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\text{d}x\left(z\right)\\ =-\frac{1}{p}{\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\mathcal{D}{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\text{d}x\left(z\right)\\ =-\frac{1}{p}{\left[E_{q^{-\frac{1}{2}}}\left(a;z_0;z\right)E_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\right]}_{z_0}^{\infty }\\ +\frac{1}{p}{\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)\mathcal{D}{E}_{q^{\frac{1}{2}}}\left(a;x_0;z\right)\text{d}x\left(z\right)\\ =\frac{1}{p}+\frac{a}{p}{\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)E_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\text{d}x\left(z\right).\\ =\frac{1}{p}+\frac{a}{p}I.\end{array}$

Hence

$I=\frac{1}{p-a}.$ (82)

4) $f\left(x\left(z\right)\right)={\text{cos}}_{q^{-\frac{1}{2}}}\left(ai;z_0;q^{\frac{1}{2}}z\right)=\frac{E_{q^{-\frac{1}{2}}}\left(ai;z_0;q^{\frac{1}{2}}z\right)+E_{q^{-\frac{1}{2}}}\left(-ai;z_0;q^{\frac{1}{2}}z\right)}{2}$ .

Using the preceding case, we obtain

${ℒ}_{q^{\frac{1}{2}}}\left\{{\text{cos}}_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}=\frac{1}{2}\left(\frac{1}{p-ai}+\frac{1}{p+ai}\right)=\frac{p}{p^2+a^2}.$ (83)

5) $f\left(x\left(z\right)\right)={\text{sin}}_{q^{-\frac{1}{2}}}\left(ai;z_0;q^{\frac{1}{2}}z\right)=\frac{E_{q^{-\frac{1}{2}}}\left(ai;z_0;q^{\frac{1}{2}}z\right)-E_{q^{-\frac{1}{2}}}\left(-ai;z_0;q^{\frac{1}{2}}z\right)}{2i}$ .

Similarly, we get

${ℒ}_{q^{\frac{1}{2}}}\left\{{\text{sin}}_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}=\frac{1}{2i}\left(\frac{1}{p-ai}-\frac{1}{p+ai}\right)=\frac{a}{p^2+a^2}.$ (84)

6) $f\left(x\left(z\right)\right)={\text{cosh}}_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)=\frac{E_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)+E_{q^{-\frac{1}{2}}}\left(-a;z_0;q^{\frac{1}{2}}z\right)}{2}$ . Here

also, we obtain

${ℒ}_{q^{\frac{1}{2}}}\left\{{\text{cosh}}_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}=\frac{1}{2}\left(\frac{1}{p-a}+\frac{1}{p+a}\right)=\frac{p}{p^2-a^2}.$ (85)

7) $f\left(x\left(z\right)\right)={\text{sinh}}_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)=\frac{E_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)-E_{q^{-\frac{1}{2}}}\left(-a;z_0;q^{\frac{1}{2}}z\right)}{2}$ .

Similarly,

${ℒ}_{q^{\frac{1}{2}}}\left\{{\text{sinh}}_{q^{-\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}=\frac{1}{2}\left(\frac{1}{p-a}-\frac{1}{p+a}\right)=\frac{a}{p^2-a^2}.$ (86)

8) $f\left(x\left(z\right)\right)=E_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)$ . We calculate

$\begin{array}{c}I={ℒ}_{q^{\frac{1}{2}}}\left\{E_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}\\ ={\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)E_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\text{d}x\left(z\right)\\ =-\frac{1}{p}{\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{\frac{1}{2}}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\mathcal{D}{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\text{d}x\left(z\right)\\ =-\frac{1}{p}{\left[E_{q^{\frac{1}{2}}}\left(a;z_0;z\right)E_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\right]}_{z_0}^{\infty }\\ +\frac{1}{p}{\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)\mathcal{D}{E}_{q^{\frac{1}{2}}}\left(a;x_0;z\right)\text{d}x\left(z\right)\\ =-\frac{1}{p}{\left[E_{q^{\frac{1}{2}}}\left(a;z_0;z\right)E_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\right]}_{z_0}^{\infty }\\ +\frac{a}{p}{\displaystyle {\int }_{x_0}^{\infty }{E}_{q^{-\frac{1}{2}}}}\left(-p;z_0;q^{-\frac{1}{2}}z\right)E_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\text{d}x\left(z\right).\\ =-\frac{1}{p}{\left[E_{q^{\frac{1}{2}}}\left(a;z_0;z\right)E_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)\right]}_{z_0}^{\infty }+\frac{a}{p}I.\end{array}$ (87)

Hence

$\begin{array}{c}\left(1-\frac{a}{p}\right)I=-\frac{1}{p}{\left[E_{q^{-\frac{1}{2}}}\left(-p;z_0;z\right)E_{q^{\frac{1}{2}}}\left(a;x_0;z\right)\right]}_{z_0}^{\infty }\\ =\frac{1}{p}-\frac{1}{p}{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;\infty \right)E_{q^{\frac{1}{2}}}\left(a;z_0;\infty \right).\end{array}$

Consequently

$I=\frac{1}{p-a}-\frac{1}{p-a}{E}_{q^{-\frac{1}{2}}}\left(-p;z_0;\infty \right)E_{q^{\frac{1}{2}}}\left(a;z_0;\infty \right)=\frac{1-K_{-p,a}}{p-a},$ (88)

where $K_{-p,a}=E_{q^{-\frac{1}{2}}}\left(-p;z_0;\infty \right)E_{q^{\frac{1}{2}}}\left(a;z_0;\infty \right)$ (see (45)).

9) $f\left(x\left(z\right)\right)={\text{cos}}_{q^{\frac{1}{2}}}\left(ai;z_0;q^{\frac{1}{2}}z\right)=\frac{E_{q^{\frac{1}{2}}}\left(ai;z_0;q^{\frac{1}{2}}z\right)+E_{q^{\frac{1}{2}}}\left(-ai;z_0;q^{\frac{1}{2}}z\right)}{2}$ . Using the preceding case, we obtain

${ℒ}_{q^{\frac{1}{2}}}\left\{{\text{cos}}_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}=\frac{1}{2}\left(\frac{1-K_{-p,ai}}{p-ai}+\frac{1-K_{-p,-ai}}{p+ai}\right).$ (89)

10) $f\left(x\left(z\right)\right)={\text{sin}}_{q^{\frac{1}{2}}}\left(ai;z_0;q^{\frac{1}{2}}z\right)=\frac{E_{q^{\frac{1}{2}}}\left(ai;z_0;q^{\frac{1}{2}}z\right)-E_{q^{\frac{1}{2}}}\left(-ai;z_0;q^{\frac{1}{2}}z\right)}{2i}$ .

Similarly, we get

${ℒ}_{q^{\frac{1}{2}}}\left\{{\text{sin}}_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}=\frac{1}{2i}\left(\frac{1-K_{-p,ai}}{p-ai}-\frac{1-K_{-p,-ai}}{p+ai}\right).$ (90)

11) $f\left(x\left(z\right)\right)={\text{cosh}}_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)=\frac{E_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)+E_{q^{\frac{1}{2}}}\left(-a;z_0;q^{\frac{1}{2}}z\right)}{2}$ . Here also, we obtain

${ℒ}_{q^{\frac{1}{2}}}\left\{{\text{cosh}}_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}=\frac{1}{2}\left(\frac{1-K_{-p,a}}{p-a}+\frac{1-K_{-p,-a}}{p+a}\right).$ (91)

12) $f\left(x\left(z\right)\right)={\text{sinh}}_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)=\frac{E_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)-E_{q^{\frac{1}{2}}}\left(-a;z_0;q^{\frac{1}{2}}z\right)}{2}$ .

Similarly,

${ℒ}_{q^{\frac{1}{2}}}\left\{{\text{sinh}}_{q^{\frac{1}{2}}}\left(a;z_0;q^{\frac{1}{2}}z\right)\right\}=\frac{1}{2}\left(\frac{1-K_{-p,a}}{p-a}-\frac{1-K_{-p,-a}}{p+a}\right).$ (92)

3.3. Inverse of the q-Nonuniform Laplace Transform

For searching the original function from its image, we have to use, as in other cases of Laplace transforms, the q-nonuniform Laplace transform properties of the subsection 3.1 and the q-nonuniform Laplace transform of elementary functions of the subsection 3.2.

3.4. Applications of q-Nonuniform Laplace Transform on q-Nonuniform Difference Equations

In the same way that the Laplace transfrom, Z-transform and q-Laplace transfrom are applied respectively in differential, difference [5] and q-difference equations [6], the q-nonuniform Laplace transform is expected to be applied in q-nonuniform difference equations [9].

For question of simplicity, we suppose that the order of the equation is $n=2$ .

So, consider the second order q-nonuniform difference equation

$a_0{\mathcal{D}}^{2}y\left(x\left(q^{-1}z\right)\right)+a_1\mathcal{D}y\left(x\left(q^{-\frac{1}{2}}z\right)\right)+a_{2}y\left(x\left(z\right)\right)=g\left(x\left(z\right)\right),$ (93)

with the initial conditions

$y\left(x\left(q^{-\frac{1}{2}}{z}_0\right)\right)=y_0;\mathcal{D}y\left(x\left(q^{-1}{z}_0\right)\right)=y_1.$ (94)

Applying the q-nonuniform Laplace transform as in the subsection 3.1, we obtain

${ℒ}_{q^{\frac{1}{2}}}\left\{{\mathcal{D}}^{2}y\left(x\left(q^{-1}z\right)\right)\right\}=p^2{ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}-py\left(x\left(q^{-\frac{1}{2}}{z}_0\right)\right)-\mathcal{D}y\left(x\left(q^{-1}{z}_0\right)\right),$ (95)

${ℒ}_{q^{\frac{1}{2}}}\left\{Dy\left(x\left(q^{-\frac{1}{2}}z\right)\right)\right\}=p{ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}-y\left(x\left(q^{-\frac{1}{2}}{z}_0\right)\right).$ (96)

Applying the q-nonuniform Laplace transform on both sides of the equation (93) and using (94), (95) and (96), we get

$\begin{array}{l}{a}_0{p}^2{ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}-a_0{y}_{0}p-a_0{y}_1+a_{1}p{ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}\\ -a_1{y}_0+a_2{ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}=G\left(p\right)\end{array}$

$\begin{array}{l}\iff a_0{p}^2{ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}+a_{1}p{ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}+a_2{ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}\\ =G\left(p\right)+a_0{y}_{0}p+a_0{y}_1+a_1{y}_0.\end{array}$

Setting ${ℒ}_{q^{\frac{1}{2}}}\left\{y\left(x\left(z\right)\right)\right\}=Y\left(p\right)$ , we obtain

$Y\left(p\right)\left(a_0{p}^2+a_{1}p+a_2\right)=G\left(p\right)+a_0{y}_{0}p+a_0{y}_1+a_1{y}_0.$

Consequently, $Y\left(p\right)=\frac{G\left(p\right)+a_0{y}_{0}p+a_0{y}_1+a_1{y}_0}{a_0{p}^2+a_{1}p+a_2}$ and $y\left(x\left(z\right)\right)={ℒ}_{q^{\frac{1}{2}}}^{-1}\left\{Y\left(p\right)\right\}$ .

4. Conclusion

In this work, basic concepts of exponential and trigonometric functions on q-nonuniform Lattices were introduced and q-nonuniform difference version of the integral Laplace tranform and also some of its applications were given. Clearly, more applications of q-nonuniform Laplace Transform for solving q-nonuniform difference equations are expected.

Acknowledgements

Avoid the stilted expression, “One of us (R. B. G.) thanks...” Instead, try “R. B. G. thanks”. Do NOT put sponsor acknowledgements in the unnumbered footnote on the first page, but at here.