Generalized q-Bessel function and its properties

Abstract

In this paper, the generalized q-Bessel function, which is a generalization of the known q-Bessel functions of kinds 1, 2, 3, and the new q-analogy of the modified Bessel function presented in (Mansour and Al-Shomarani in J. Comput. Anal. Appl. 15(4):655-664, 2013) is introduced. We deduced its generating function, recurrence relations and q-difference equation, which gives us the differential equation of each of the Bessel function and the modified Bessel function when q tends to 1. Finally, the quantum algebra $E_2(q)$ and its representations presented an algebraic derivation for the generating function of the generalized q-Bessel function.

MSC:33D45, 81R50, 22E70.

1 Introduction

The q-shifted factorials are defined by [1]

$$\begin{array}{r}{(a;q)}_0=1,\\ {(a;q)}_k=\prod _{i=o}^{k-1}(1-a{q}^k),\\ {(a_1,\dots ,a_r;q)}_k=\prod _{j=1}^r{(a_j;q)}_k;k=0,1,2,\dots ,\\ {(a;q)}_{\mathrm{\infty }}=\prod _{i=0}^{\mathrm{\infty }}(1-a{q}^i),\text{where }a,a_i\text{’s},q\in \mathbb{R}\text{ such that }0<q<1.\end{array}$$

The one-parameter family of q-exponential functions

$$E_q^{(\alpha )}(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{q^{\frac{\alpha n^2}{2}}}{{(q;q)}_n}{z}^n$$

with $\alpha \in \mathbb{R}$ has been considered in [2]. Consequently, in the limit when $q\to 1$, we have ${lim}_{q\to 1}{E}_q^{(\alpha )}((1-q)z)=e^z$. Exton [3] presented the following q-exponential functions:

$$E(\mu ,z;q)=\sum _{n=0}^{\mathrm{\infty }}\frac{q^{\mu n(n-1)}}{{[n]}_q!}{z}^n,$$

where ${[n]}_q!=\frac{{(q;q)}_n}{{(1-q)}^n}$. The relation between these two notations is given by

$$E(\lambda ,z;q)=E_q^{(2\lambda )}(q^{-\lambda }(1-q)z).$$

In Exton’s formula, if we replace z by $\frac{x}{1-q}$ and μ by 2a, we get the following q-exponential function:

$$E_q(x,a)=\sum _{n=0}^{\mathrm{\infty }}\frac{q^{a\left(\begin{array}{c}n\\ 2\end{array}\right)}}{{(q;q)}_n}{x}^n,$$

(1)

which satisfies the functional relation [4]

$$E_q(x,a)-E_q(qx,a)=x{E}_q(q^{a}x,a),$$

which can be rewritten by the formula

$$D_q{E}_q(x,a)=\frac{1}{1-q}{E}_q(q^{a}x,a),$$

(2)

where the Jackson q-difference operator $D_q$ is defined by [5]

$$D_{q}f(x)=\frac{f(x)-f(qx)}{(1-q)x}$$

(3)

and satisfies the product rule

$$D_q(f(x)g(x))=f(qx)D_{q}g(x)+g(x)D_{q}f(x).$$

There are two important special cases of the function $E_q(x,a)$

$$e_q(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{{(q,q)}_n},|x|<1$$

(4)

and

$$E_q(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{q^{\left(\begin{array}{c}n\\ 2\end{array}\right)}{x}^n}{{(q,q)}_n}.$$

(5)

The q-Bessel functions of kinds 1, 2 and 3 are defined by [6]

(6)

(7)

(8)

where ${}_r\phi _s$ is the basic hypergeometric function [1]

$${}_r\phi _s\left(\begin{array}{c}{a}_1,\dots ,a_r\\ b_1,\dots ,b_s\end{array}\right|q;z)=\sum _{k=0}^{\mathrm{\infty }}\frac{{(a_1,\dots ,a_r;q)}_k}{{(b_1,\dots ,b_s;q)}_k}{\left({(-1)}^k{q}^{\frac{k}{2}(k-1)}\right)}^{1+s-r}\frac{z^k}{{(q;q)}_k}.$$

(9)

The functions $J_n^{(i)}(x;q)$, $i=1,2$, are q-analogues of the Bessel function, and the function $J_n^{(3)}(x;q)$ is a q-analogue of the modified Bessel function.

Rogov [7, 8] introduced generalized modified q-Bessel functions, similarly to the classical case [9], as

$$I_n^i(x;q)=\frac{{(q^{n+1};q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}\left(\frac{x}{2}\right)^n{}_{\delta }{\phi }_1\left(\begin{array}{c}0,0,\dots ,0\\ q^{n+1}\end{array}\right|q;\frac{q^{\frac{(n+1)(2-\delta )}{2}}{x}^2}{4}),$$

where

$$\delta =\{\begin{array}{cc}2\hfill & \text{for }i=1,\hfill \\ 0\hfill & \text{for }i=2,\hfill \\ 1\hfill & \text{for }i=3.\hfill \end{array}$$

Recently, Mansour and et al. [10] studied the following q-Bessel function:

$$J_n^{(4)}(x;q)={\frac{{(x/2)}^n}{{(q;q)}_n}}_0{\phi }_2\left(\begin{array}{c}-\\ 0,q^{n+1}\end{array}\right|q;\frac{-q^{\frac{3(n+1)}{2}}{x}^2}{4}),$$

which is a q-analogy of the modified Bessel function.

In this paper, we define the generalized q-Bessel function and study some of its properties. Also, in analogy with the ordinary Lie theory [11, 12], we derive algebraically the generating function of the generalized q-Bessel function.

2 The generalized q-Bessel function and its generating function

Definition 2.1 The generalized q-Bessel function is defined by

$$J_n(x,a;q)=\frac{{(x/2)}^n}{{(q;q)}_n}\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}{q}^{\frac{ak}{2}(k+n)}}{{(q^{n+1};q)}_k}\frac{{(x^2/4)}^k}{{(q;q)}_k},$$

(10)

which converges absolutely for all x when $a\in {\mathbb{Z}}^{+}$ and for $|x|<2$ if $a=0$.

As special cases of $J_n(x,a;q)$, we get

$$\begin{array}{r}{J}_n^{(1)}(x;q)=J_n(x,0;q),\\ J_n^{(2)}(x;q)=J_n(x,2;q),\\ J_n^{(3)}(x;q)=J_n(x,1;q),\\ J_n^{(4)}(x;q)=J_n(x,3;q).\end{array}$$

Lemma 2.2 The function $J_n(x,a;q)$ is a q-analogy of each of the Bessel function and the modified Bessel function.

Proof

$$\begin{array}{rcl}\underset{q\to 1}{lim}{J}_n((1-q)x,a;q)& =& \underset{q\to 1}{lim}\left\{\frac{{(1-q)}^n}{{(q,q)}_n}{\left(\frac{x}{2}\right)}^n\sum _{k=0}^{\mathrm{\infty }}{(-1)}^{k(a+1)}{q}^{\frac{ak(k+n)}{2}}\frac{{(1-q)}^{2k}}{{(q^{n+1};q)}_k{(q;q)}_k}{\left(\frac{x}{2}\right)}^{2k}\right\}\\ =& \frac{1}{{(1)}_n}{\left(\frac{x}{2}\right)}^n\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}}{{(n+1)}_k{(1)}_k}{\left(\frac{x}{2}\right)}^{2k}\\ =& {\left(\frac{x}{2}\right)}^n\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}}{\mathrm{\Gamma }(n+k+1)\mathrm{\Gamma }(k+1)}{\left(\frac{x}{2}\right)}^{2k}.\end{array}$$

Hence, we get

$$\underset{q\to 1}{lim}{J}_n((1-q)x,a;q)=J_n(x);a=0,2,4,\dots$$

(11)

and

$$\underset{q\to 1}{lim}{J}_n((1-q)x,a;q)=I_n(x);a=1,3,5,\dots ,$$

(12)

where $J_n(x)$ is the Bessel function and $I_n(x)$ is the modified Bessel function. □

Lemma 2.3 The function $J_n(x,a;q)$ satisfies

$$J_{-n}(x,a;q)={(-1)}^{n(a+1)}{J}_n(x,a;q),n\in \mathbb{Z}.$$

(13)

Proof Using the definition (10), we get

$$J_{-n}(x,a;q)=\sum _{k=n}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}{q}^{\frac{ak(k-n)}{2}}{(q^{-n+k+1};q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}{(q;q)}_k}{\left(\frac{x}{2}\right)}^{2k-n}.$$

For $s=k-n$, we obtain

$$J_{-n}(x,a;q)=\sum _{s=0}^{\mathrm{\infty }}\frac{{(-1)}^{(s+n)(a+1)}{q}^{\frac{as(s+n)}{2}}{(q^{s+1};q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}{(q;q)}_{s+n}}{\left(\frac{x}{2}\right)}^{2s+n},$$

and using the relations [1]

$$\begin{array}{r}{(q^{s+1};q)}_{\mathrm{\infty }}={(q^{n+s+1};q)}_{\mathrm{\infty }}{(q^{s+1};q)}_n,\\ {(q;q)}_{s+n}={(q;q)}_s{(q^{s+1};q)}_n,\end{array}$$

we obtain

$$\begin{array}{rcl}{J}_{-n}(x,a;q)& =& {(-1)}^{n(a+1)}\sum _{s=0}^{\mathrm{\infty }}\frac{{(-1)}^{s(a+1)}{q}^{\frac{as(s+n)}{2}}{(q^{n+s+1};q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}{(q;q)}_s}{\left(\frac{x}{2}\right)}^{2s+n}\\ =& {(-1)}^{n(a+1)}{J}_n(x,a;q).\end{array}$$

 □

Lemma 2.4 The function $J_n(x,a;q)$ satisfies the relation

$$J_n(-x,a;q)={(-1)}^n{J}_n(x,a;q),n\in \mathbb{Z},$$

(14)

and hence it is even (or odd) function if the integer n is even (or odd).

Now we will deduce the generating function of the generalized q-Bessel function $J_n(x,a;q)$.

Theorem 1 The generating function $g(x,t,a;q)$ of the function $J_n(x,a;q)$ is given by

$$g(x,t,a;q)=E_q(\frac{q^{\frac{a}{4}}xt}{2},\frac{a}{2})E_q(\frac{{(-1)}^{a+1}{q}^{\frac{a}{4}}x}{2t},\frac{a}{2})=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{\frac{a{n}^2}{4}}{J}_n(x,a;q)t^n.$$

(15)

Proof

Let

$$g(x,t,a;q)=E_q(\frac{q^{\frac{a}{4}}xt}{2},\frac{a}{2})E_q(\frac{{(-1)}^{a+1}{q}^{\frac{a}{4}}x}{2t},\frac{a}{2}),$$

then

$$g(x,t,a;q)=\sum _{r=0}^{\mathrm{\infty }}\sum _{s=0}^{\mathrm{\infty }}\frac{{(-1)}^{s(a+1)}{q}^{\frac{a}{2}[\left(\begin{array}{c}r\\ 2\end{array}\right)+\left(\begin{array}{c}s\\ 2\end{array}\right)]+\frac{a}{4}(s+r)}}{{(q;q)}_r{(q;q)}_s}{\left(\frac{x}{2}\right)}^{s+r}{t}^{r-s}.$$

For $s=r-n$, we get

$$g(x,t,a;q)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{r=0}^{\mathrm{\infty }}\frac{{(-1)}^{(r-n)(a+1)}{q}^{\frac{a}{2}[\left(\begin{array}{c}r\\ 2\end{array}\right)+\left(\begin{array}{c}r-n\\ 2\end{array}\right)]+\frac{a}{4}(2r-n)}}{{(q;q)}_r{(q;q)}_{r-n}}{\left(\frac{x}{2}\right)}^{2r-n}{t}^n.$$

Hence, for $n\le 0$, the coefficient of $t^n$ is given by

$$c_n={(-1)}^{-n(a+1)}{q}^{\frac{a{n}^2}{4}}\frac{{(x/2)}^{-n}}{{(q;q)}_{-n}}\sum _{r=0}^{\mathrm{\infty }}\frac{{(-1)}^{r(a+1)}{q}^{\frac{ar}{2}(r-n)}}{{(q;q)}_r{(q^{-n+1};q)}_r}{\left(\frac{x}{2}\right)}^{2r},$$

where ${(q,q)}_{r-n}={(q;q)}_{-n}{(q^{-n+1};q)}_r$ for $n\le 0$. Then

$$c_n={(-1)}^{-n(a+1)}{q}^{\frac{a{n}^2}{4}}{J}_{-n}(x,a;q)=q^{\frac{a{n}^2}{4}}{J}_n(x,a;q).$$

Similarly, for $n\ge 0$. □

As special cases of $g_n(x,t,a;q)$, we obtain

$$g_n(x,t,0;q)=E_q(\frac{xt}{2},0)E_q(\frac{-x}{2t},0)=e_q\left(\frac{xt}{2}\right)e_q\left(\frac{-x}{2t}\right),$$

(16)

which is a generating function of the q-Bessel function $J_n^{(1)}(x;q)$ [13],

$$g_n(x,\frac{t}{\sqrt{q}},2;q)=E_q(\frac{xt}{2},1)E_q(\frac{-qx}{2t},1)=E_q\left(\frac{xt}{2}\right)E_q\left(\frac{-qx}{2t}\right),$$

(17)

which is a generating function of the q-Bessel function $J_n^{(2)}(x;q)$ [13],

$$g_n(x,t,1;q)=E_q(\frac{\sqrt[4]{q}xt}{2},\frac{1}{2})E_q(\frac{\sqrt[4]{q}x}{2t},\frac{1}{2}),$$

(18)

which is a generating function of the q-Bessel function $J_n^{(3)}(x;q)$ and

$$g_n(x,t,3;q)=E_q(\frac{q^{3/4}xt}{2},\frac{3}{2})E_q(\frac{q^{3/4}x}{2t},\frac{3}{2}),$$

(19)

which is a generating function of the q-Bessel function $J_n^{(4)}(x;q)$ [10].

3 The q-difference equation of the function $J_n(x,a;q)$

Now the generating function method [13] will be used to deduce the q-difference equation of the generalized q-Bessel function. Using equation (15), we have

$$E_q(\frac{q^{\frac{a}{4}}xth}{2},\frac{a}{2})E_q(\frac{{(-1)}^{a+1}{q}^{\frac{a}{4}}xh}{2t},\frac{a}{2})=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{\frac{a{n}^2}{4}}{J}_n(xh,a;q)t^n;h\in \mathbb{R}-\{0\}.$$

(20)

By applying the operator $D_q$, we get

$$\begin{array}{r}\frac{{(-1)}^{a+1}{q}^{\frac{a}{4}}h}{2(1-q)t}{E}_q(\frac{q^{\frac{a+4}{4}}xth}{2},\frac{a}{2})E_q(\frac{{(-1)}^{a+1}{q}^{\frac{3a}{4}}xh}{2t},\frac{a}{2})\\ +\frac{q^{\frac{a}{4}}th}{2(1-q)}{E}_q(\frac{q^{\frac{3a}{4}}xth}{2},\frac{a}{2})E_q(\frac{{(-1)}^{a+1}{q}^{\frac{a}{4}}xh}{2t},\frac{a}{2})\\ =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{\frac{a{n}^2}{4}}{D}_q{J}_n(xh,a;q)t^n.\end{array}$$

Using equation (20), we obtain

$$\begin{array}{r}\frac{{(-1)}^{a+1}{q}^{\frac{a}{4}+\frac{a}{2}\left(\begin{array}{c}n+1\\ 2\end{array}\right)+\frac{n+1}{2}}}{2(1-q)}{J}_{n+1}(q^{\frac{a+2}{4}}xh,a;q)+\frac{q^{\frac{a}{4}+\frac{an(n-1)}{4}}}{2(1-q)}{J}_{n-1}(q^{\frac{a}{4}}xh,a;q)\\ =\frac{q^{\frac{a{n}^2}{4}}}{h}{D}_q{J}_n(xh,a;q).\end{array}$$

Hence

(21)

Similarly, we can prove the following relation:

(22)

By using equations (21) and (22), we obtain

(23)

But

$$\begin{array}{rcl}{J}_{n-1}(\sqrt{q}xh,a;q)& =& \frac{{(\frac{\sqrt{q}xh}{2})}^{n-1}}{{(q;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}{q}^{\frac{ak}{2}(k+n-1)}{(q^{n+k};q)}_{\mathrm{\infty }}}{{(q;q)}_k}{\left(\frac{\sqrt{q}xh}{2}\right)}^{2k}\\ =& \frac{q^{\frac{n-1}{2}}{(\frac{xh}{2})}^{n-1}}{{(q;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}{q}^{\frac{ak}{2}(k+n-1)}{(q^{n+k};q)}_{\mathrm{\infty }}}{{(q;q)}_k}(q^k-1+1){\left(\frac{xh}{2}\right)}^{2k}\\ =& q^{\frac{n-1}{2}}{J}_{n-1}(xh,a;q)-\frac{q^{\frac{n-1}{2}}{(\frac{xh}{2})}^{n-1}}{{(q;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{(k+1)(a+1)}{q}^{\frac{a(k+1)(k+n)}{2}}}{{(q;q)}_k}\\ \times {(q^{n+k+1};q)}_{\mathrm{\infty }}{\left(\frac{xh}{2}\right)}^{2k+2}\\ =& q^{\frac{n-1}{2}}\{J_{n-1}(xh,a;q)+\frac{{(-1)}^a{q}^{\frac{an}{2}}xh}{2}{J}_n(q^{\frac{a}{4}}xh,a;q)\}.\end{array}$$

Then

$$q^{\frac{1-n}{2}}{J}_{n-1}(\sqrt{q}xh,a;q)-J_{n-1}(xh,a;q)=\frac{{(-1)}^a{q}^{\frac{an}{2}}xh}{2}{J}_n(q^{\frac{a}{4}}xh,a;q).$$

(24)

Equations (23) and (24) give the relation

$$q^{\frac{1+n}{2}}{J}_{n+1}(\sqrt{q}xh,a;q)-J_{n+1}(xh,a;q)=\frac{-xh}{2}{J}_n(q^{\frac{a}{4}}xh,a;q).$$

(25)

Equations (21) and (25) give the relation

$$\{D_q+\frac{q^{\frac{a(1-n)}{4}}}{(1-q)x}[q^{\frac{n}{2}}{\delta }_q-1]\}{J}_n(xh,a;q)=\frac{{(-1)}^{a+1}h{q}^{\frac{(a+2)(n+1)}{4}}}{2(1-q)}{J}_{n+1}(q^{\frac{a+2}{4}}xh,a;q),$$

(26)

where the operator ${\delta }_q$ is given by ${\delta }_{q}f(x)=f(\sqrt{q}x)$.

Now consider the following operator:

$$M_{n,q}=\{D_q+\frac{q^{\frac{a(1-n)}{4}}}{(1-q)x}[q^{\frac{n}{2}}{\delta }_q-1]\},$$

(27)

then we can rewrite equation (26) by the formula

$$M_{n,q}{J}_n(xh,a;q)=\frac{{(-1)}^{a+1}h{q}^{\frac{(a+2)(n+1)}{4}}}{2(1-q)}{J}_{n+1}(q^{\frac{a+2}{4}}xh,a;q).$$

(28)

Also, equations (22) and (24) give the relation

$$\{D_q+\frac{q^{\frac{-a(1+n)}{4}}}{(1-q)x}[q^{\frac{-n}{2}}{\delta }_q-1]\}{J}_n(xh,a;q)=\frac{h{q}^{\frac{(a+2)(1-n)}{4}}}{2(1-q)}{J}_{n-1}(q^{\frac{a+2}{4}}xh,a;q).$$

(29)

If we consider the operator

$$N_{n,q}=\{D_q+\frac{q^{\frac{-a(1+n)}{4}}}{(1-q)x}[q^{\frac{-n}{2}}{\delta }_q-1]\},$$

(30)

then we can rewrite equation (29) by the formula

$$N_{n,q}{J}_n(xh,a;q)=\frac{h{q}^{\frac{(a+2)(1-n)}{4}}}{2(1-q)}{J}_{n-1}(q^{\frac{a+2}{4}}xh,a;q).$$

(31)

Hence, the q-difference equation of the function $J_n(x,a;q)$ takes the formula

$$M_{n-1,q}{N}_{n,q}{J}_n(xh,a;q)=\frac{{(-1)}^{a+1}{q}^{\frac{a+2}{4}}{h}^2}{4{(1-q)}^2}{J}_n(q^{\frac{a+2}{2}}xh,a;q).$$

(32)

If we replace h by $1-q$ and consider the limit as q tends to 1, then we obtain

$$(\frac{1}{2}\frac{d}{dx}-\frac{n-1}{2x})(\frac{1}{2}\frac{d}{dx}+\frac{n}{2x})y(x)=\frac{{(-1)}^{a+1}}{4}y(x),$$

(33)

or

$$x^2{y}^{″}(x)+x{y}^{\prime }-(n^2+{(-1)}^{a+1}{x}^2)y=0.$$

(34)

The differential equation (34) gives the Bessel function at $a=0,2,4,\dots$ and the modified Bessel function at $a=1,3,5,\dots$ , which proves again that $J_n(x,a;q)$ is a q-analogy of each of them.

4 The recurrence relations of the function $J_n(x,a;q)$

Lemma 4.1

$$J_n(x,a;q)=\frac{2}{x}(1-q^{n+1})J_{n+1}(q^{-a/4}x,a;q)+{(-1)}^{a+1}{q}^{\frac{(a+2)(n+1)}{2}}{J}_{n+2}(x,a;q).$$

(35)

Proof

$$\begin{array}{rcl}{J}_n(x,a;q)& =& \frac{{(x/2)}^n}{{(q;q)}_n}\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}{q}^{\frac{ak}{2}(k+n)}}{{(q;q)}_k{(q^{n+1};q)}_{k+1}}(1-q^{n+1}+q^{n+1}-q^{n+k-1}){\left(\frac{x^2}{4}\right)}^k\\ =& \frac{2}{x}(1-q^{n+1})\frac{{(x/2)}^{n+1}}{{(q;q)}_{n+1}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}{q}^{\frac{ak}{2}(k+n+1)}}{{(q;q)}_k{(q^{n+2};q)}_k}{\left(\frac{{(q^{-\frac{a}{4}}x)}^2}{4}\right)}^k\\ +q^{n+1}\frac{{(x/2)}^{n+2}}{{(q;q)}_{n+2}}\sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}{q}^{\frac{ak}{2}(k+n)}}{{(q;q)}_{k-1}{(q^{n+3};q)}_{k-1}}{\left(\frac{x^2}{4}\right)}^{k-1}\\ =& \frac{2}{x}(1-q^{n+1})J_{n+1}(q^{-a/4}x,a;q)\\ +{(-1)}^{a+1}{q}^{\frac{(a+2)(n+1)}{2}}\frac{{(x/2)}^{n+2}}{{(q;q)}_{n+2}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^{k(a+1)}{q}^{\frac{ak}{2}(k+n+2)}}{{(q;q)}_k{(q^{n+3};q)}_k}{\left(\frac{x^2}{4}\right)}^k\\ =& \frac{2}{x}(1-q^{n+1})J_{n+1}(q^{-a/4}x,a;q)+{(-1)}^{a+1}{q}^{\frac{(a+2)(n+1)}{2}}{J}_{n+2}(x,a;q).\end{array}$$

 □

Similarly, if we write $(1-q^k+q^k-q^{n+k-1})$ instead of $(1-q^{n+1}+q^{n+1}-q^{n+k-1})$, we can prove the following lemma.

Lemma 4.2

$$J_n(x,a;q)=\frac{2}{x}(1-q^{n+1})J_{n+1}(q^{\frac{2-a}{4}}x,a;q)+{(-1)}^{a+1}{q}^{\frac{a(n+1)}{2}}{J}_{n+2}(x,a;q).$$

(36)

Now, if we replace a by $a+2$ in the recurrence relation (36), we get the recurrence relation (35). Then we have the following lemma.

Lemma 4.3 The two functions $J_n(x,a;q)$ and $J_n(x,a+2;q)$ have the same recurrence relation.

Then we have two cases of the recurrence relation.

Case (1): The function $J_n(x,a;q)$ has the recurrence relation

$$J_n(x,a;q)=\frac{2}{x}(1-q^{n+1})J_{n+1}(x,a;q)-q^{n+1}{J}_{n+2}(x,a;q);\mathrm{\forall }a=0,2,4,\dots ,$$

(37)

which is the recurrence relation of each of $J_n^{(1)}(x;q)$ and $J_n^{(2)}(x;q)$.

Case (2): The function $J_n(x,a;q)$ has the recurrence relation

$$\begin{array}{rcl}{J}_n(x,a;q)& =& q^{\frac{n}{4}}[\frac{2}{x}(1-q^{n+1})-q^{\frac{n}{2}}\frac{x}{2}]J_{n+1}(x,a;q)\\ +q^{n+1/2}{J}_{n+2}(x,a;q);\mathrm{\forall }a=1,3,5,\dots ,\end{array}$$

(38)

which is the recurrence relation of each of $J_n^{(3)}(x;q)$ and $J_n^{(4)}(x;q)$.

5 The quantum algebra approach to $J_n(x,a;q)$

The quantum algebra $E_q(2)$ is determined by generators H, $E_{+}$ and $E_{-}$ with the commutation relations

$$[H,E_{+}]=E_{+},[H,E_{-}]=-E_{-},[E_{-},E_{+}]=0.$$

(39)

By considering the irreducible representations $(\omega )$ of $E_q(2)$ characterized by $\omega \in \mathbb{C}$, then the spectrum of the operator H will be the set of integers ℤ, and the basis vectors $f_m$, $m\in \mathbb{Z}$, satisfy

$$E_{\pm }{f}_m=\omega f_{m\pm 1},H{f}_m=m{f}_m,E_{+}{E}_{-}{f}_m={\omega }^2{f}_m,$$

(40)

where $C=E_{+}{E}_{-}$ is the Casimir operator which commutes with the generators H, $E_{+}$ and $E_{-}$. The following differential operators presented a simple realization of $(\omega )$

$$H=z\frac{d}{dz},E_{+}=\omega z,E_{-}=\frac{\omega }{z}$$

(41)

acting on the space of all linear combinations of the functions $z^m$, z a complex variable, $m\in \mathbb{Z}$, with basis vectors $f_m(z)=z^m$.

In the ordinary Lie theory, matrix elements $T_{sm}$ of the complex motion group in the representation $(\omega )$ are typically defined by the expansions [7–9]

$$e^{\alpha E_{+}}{e}^{\beta E_{-}}{e}^{\tau H}{f}_m=\sum _{s=-\mathrm{\infty }}^{\mathrm{\infty }}{T}_{sm}(\alpha ,\beta ,\tau )f_s.$$

(42)

If we replace the mapping $e^x$ by the mapping $E_q(x,a/2)$ from the Lie algebra to the Lie group with putting $\tau =0$ in equation (42), we can use the model (41) to find the following q-analog of matrix elements of $(\omega )$:

$$E_q(\alpha E_{+},a/2)E_q(\beta E_{-},a/2)f_m=\sum _{s=-\mathrm{\infty }}^{\mathrm{\infty }}{T}_{sm}(\alpha ,\beta )f_s,$$

(43)

and hence

$$E_q(\alpha \omega z,\frac{a}{2})E_q(\beta \frac{\omega }{z},\frac{a}{2})z^m=\sum _{r,t=0}^{\mathrm{\infty }}\frac{q^{\frac{a}{2}[\left(\begin{array}{c}r\\ 2\end{array}\right)+\left(\begin{array}{c}t\\ 2\end{array}\right)]}}{{(q;q)}_r{(q;q)}_t}{\omega }^{r+t}{\alpha }^r{\beta }^t{z}^{m-t+r}.$$

Now replace s by $m-t+r$ to get

$$E_q(\alpha \omega z,\frac{a}{2})E_q(\beta \frac{\omega }{z},\frac{a}{2})z^m=\sum _{s=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{r=0}^{\mathrm{\infty }}\frac{q^{\frac{a}{2}[\left(\begin{array}{c}r\\ 2\end{array}\right)+\left(\begin{array}{c}m+r-s\\ 2\end{array}\right)]}}{{(q;q)}_r{(q;q)}_{m+r-s}}{\omega }^{m+2r-s}{\alpha }^r{\beta }^{m+r-s}{z}^s$$

and by equating the coefficient of $z^s$ for $m\ge s$ on both sides, we get

$$\begin{array}{rcl}{T}_{sm}(\alpha ,\beta )& =& \frac{q^{\frac{a}{2}\left(\begin{array}{c}m-s\\ 2\end{array}\right)}}{{(q;q)}_{m-s}}{(\omega \beta )}^{m-s}\sum _{r=0}^{\mathrm{\infty }}\frac{q^{\frac{a}{2}r(r+m-s-1)}}{{(q;q)}_r{(q^{m-s+1};q)}_r}{\left({\omega }^2\alpha \beta \right)}^r,m\ge s\\ =& q^{\frac{a}{4}{(m-s)}^2}{\left({(-1)}^{-(a+1)}\frac{\beta }{\alpha }\right)}^{\frac{m-s}{2}}{(-1)}^{(m-s)(a+1)}{J}_{m-s}(2q^{\frac{-a}{4}}\omega \sqrt{{(-1)}^{a+1}\alpha \beta },a;q),\\ {(-1)}^{a+1}\alpha \beta >0;m\ge s\\ =& q^{\frac{a}{4}{(s-m)}^2}{\left({(-1)}^{(a+1)}\frac{\alpha }{\beta }\right)}^{\frac{s-m}{2}}{J}_{s-m}(2q^{\frac{-a}{4}}\omega \sqrt{{(-1)}^{a+1}\alpha \beta },a;q),\\ {(-1)}^{a+1}\alpha \beta >0;m\ge s,\end{array}$$

where $J_{-n}(x,a;q)={(-1)}^{n(a+1)}{J}_n(x,a;q)$.

Similarly,

$$\begin{array}{r}{T}_{sm}(\alpha ,\beta )=q^{\frac{a}{4}{(s-m)}^2}{\left({(-1)}^{(a+1)}\frac{\alpha }{\beta }\right)}^{\frac{s-m}{2}}{J}_{s-m}(2q^{\frac{-a}{4}}\omega \sqrt{{(-1)}^{a+1}\alpha \beta },a;q),\\ {(-1)}^{a+1}\alpha \beta >0;s\ge m.\end{array}$$

The combination between the two cases gives us the following expression:

$$\begin{array}{r}{T}_{sm}(\alpha ,\beta )=q^{\frac{a}{4}{(s-m)}^2}{\left({(-1)}^{(a+1)}\frac{\alpha }{\beta }\right)}^{\frac{s-m}{2}}{J}_{s-m}(2q^{\frac{-a}{4}}\omega \sqrt{{(-1)}^{a+1}\alpha \beta },a;q),\\ {(-1)}^{a+1}\alpha \beta >0,\end{array}$$

(44)

which is valid for all $m,s\in \mathbb{Z}$. Then we get the following result.

Lemma 5.1

(45)

where ${(-1)}^{a+1}\alpha \beta >0$.

As special cases:

Considering (45) with $a=0$, $\alpha =-\beta =1$, $z=t$ and $\omega =\frac{x}{2}$, we obtain the relation (16).

Considering (45) with $a=2$, $\alpha =-\beta =1$, $z=\frac{t}{\sqrt{q}}$ and $\omega =\frac{\sqrt{q}x}{2}$, we obtain the relation (17).

Considering (45) with $a=1$, $\alpha =\beta =1$, $z=t$ and $\omega =\frac{\sqrt[4]{q}x}{2}$, we obtain the relation (18).

Considering (45) with $a=3$, $\alpha =\beta =1$, $z=t$ and $\omega =\frac{q^{3/4}x}{2}$, we obtain the relation (19).