q-Fractional calculus for Rubin’s q-difference operator

Abstract

In this paper we introduce a fractional q-integral operator and derivative as a generalization of Rubin’s q-difference operator. We also reformulate the definition of the $q^2$-Fourier transform and the q-analogue of the Fourier multiplier introduced by Rubin in (J. Math. Anal. Appl. 212(2):571-582, 1997; Proc. Am. Math. Soc. 135(3):777-785, 2007). As applications, we give summation formulas for ${}_2\varphi _1$ finite series, we also use the $q^2$-Fourier transform and Hahn q-Laplace transform to solve a fractional q-diffusion equation.

MSC:39A12, 33D15, 42A38, 35R11.

1 Introduction and preliminaries

Let q be a positive number, $0<q<1$. In the following, we follow the notations and notions of q-hypergeometric functions, the q-gamma function ${\mathrm{\Gamma }}_q(x)$, Jackson q-exponential functions $E_q(x)$, and the q-shifted factorial as in [1, 2]. The q-difference operator is defined by

$$D_{q}f(z):=\frac{f(z)-f(qz)}{z-qz}(z\ne 0).$$

(1.1)

Jackson [3] introduced an integral denoted by

$${\int }_a^{b}f(x)d_{q}x$$

as a right inverse of the q-derivative. It is defined by

$${\int }_a^{b}f(t)d_{q}t:={\int }_0^{b}f(t)d_{q}t-{\int }_0^{a}f(t)d_{q}t(a,b\in \mathbb{C}),$$

(1.2)

where

$${\int }_0^{x}f(t)d_{q}t:=(1-q)\sum _{n=0}^{\mathrm{\infty }}x{q}^{n}f\left(x{q}^n\right)(x\in \mathbb{C}),$$

(1.3)

provided that the series on the right-hand side of (1.3) converges at $x=a\text{ and }b$.

There is no unique canonical choice for the q-integration over $[0,\mathrm{\infty })$. In [4], Hahn defined the q-integration for a function f over $[0,\mathrm{\infty })$ by

$${\int }_0^{\mathrm{\infty }}f(t)d_{q}t=(1-q)\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{n}f\left(q^n\right),$$

while in [5] Matsuo defined q-integrations on the interval $[0,\mathrm{\infty })$ and $(-\mathrm{\infty },\mathrm{\infty })$ by

$${\int }_0^{\mathrm{\infty }/b}f(t)d_{q}t:=\frac{1-q}{b}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{n}f(q^n/b)(b>0),$$

(1.4)

$${\int }_{-\mathrm{\infty }/b}^{\mathrm{\infty }/b}f(t)d_{q}t=\frac{1-q}{b}\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{q}^n(f(q^n/b)+f(-q^n/b)),$$

(1.5)

respectively, provided that the series converges absolutely. For any $q\in (0,1)$ and $0<b<\mathrm{\infty }$, we define the spaces

$$\begin{array}{c}{L}^p({\mathbb{R}}_{b,q}):=\{f:{\int }_{-\mathrm{\infty }/b}^{\mathrm{\infty }/b}{|f(x)|}^p{d}_{q}x<\mathrm{\infty },p\ge 1\},{\mathbb{R}}_{b,q}:=\{\pm q^n/b:n\in \mathbb{Z}\},\hfill \\ L^{\mathrm{\infty }}({\mathbb{R}}_{b,q}):=\{f:{\parallel f\parallel }_{\mathrm{\infty }}:=sup\{f(\pm q^n/b)\mid n\in \mathbb{Z}\}<\mathrm{\infty }\}.\hfill \end{array}$$

We shall use the particular notation ${\mathbb{R}}_q$, ${\tilde{\mathbb{R}}}_q$ and ${\tilde{\mathbb{R}}}_{q,+}$ to denote ${\mathbb{R}}_{1,q}$, ${\mathbb{R}}_{\sqrt{1-q},q}$ and $\{\frac{q^k}{\sqrt{1-q}},k\in \mathbb{Z}\}$, respectively. One can verify that $L^2({\mathbb{R}}_{b,q})$ associated with the inner product

$$〈f,g〉:={\int }_{-\mathrm{\infty }/b}^{\mathrm{\infty }/b}f(t)\overline{g(t)}{d}_{q}t,f,g\in L^2({\mathbb{R}}_{b,q}),$$

is a Hilbert space. The Riemann-Liouville fractional q-integral operator is introduced by Al-Salam in [6] and later by Agarwal in [7] and defined by

$$I_q^{\alpha }f(x):=\frac{x^{\alpha -1}}{{\mathrm{\Gamma }}_q(\alpha )}{\int }_0^x{(qt/x;q)}_{\alpha -1}f(t)d_{q}t,\alpha \notin \{-1,-2,\dots \}.$$

(1.6)

Using (1.3), (1.6) reduces to

$$I_q^{\alpha }f(x)=x^{\alpha }{(1-q)}^{\alpha }\sum _{n=0}^{\mathrm{\infty }}{q}^n\frac{{(q^{\alpha };q)}_n}{{(q;q)}_n}f\left(x{q}^n\right),$$

(1.7)

which is valid for all α. The Riemann-Liouville fractional q-derivative of order α, $\alpha >0$, is defined by

$$D_q^{\alpha }=D_q^k{I}_q^{k-\alpha }(k=\lceil \alpha \rceil ).$$

Rubin in [8, 9] introduced the q-difference operator

$${\partial }_{q}f(z)=\frac{f(q^{-1}z)+f(-q^{-1}z)-f(qz)+f(-qz)-2f(-z)}{2(1-q)z}(z\ne 0).$$

(1.8)

It is straightforward to prove that if a function f is differentiable at a point z, then

$$\underset{q\to 1^{-}}{lim}{\partial }_{q}f(z)=f^{\prime }(z).$$

Also,

$${\delta }_{q}f(z)=\{\begin{array}{cc}{D}_{q}f(z)\hfill & \text{if }f\text{ is odd},\hfill \\ \frac{1}{q}{D}_{q^{-1}}f(z)\hfill & \text{if }f\text{ is even}.\hfill \end{array}$$

Let f and g be functions defined on a set A, where A satisfies

$$z\in A\to \pm q^{\pm 1}z\in A,$$

and let $f_e$ and $f_o$ be the even and odd parts of f, respectively. The following properties of the ${\partial }_q$ operator are from [9, 10] and hold for all $z\in A\setminus \{0\}$.

• (i) ${\partial }_{q}f(z)=\frac{1}{q}{D}_{q^{-1}}{f}_e(z)+D_q{f}_o(z)$.

• (ii) For two functions f and g,

• if f is even and g is odd, then

  $${\partial }_q(fg)(z)=qg(z)({\partial }_{q}f)(qz)+f(qz){\partial }_{q}g(z);$$

• if f and g are even, then

  $${\partial }_q(fg)(z)={\partial }_{q}f(z)g(z)+f(z/q){\partial }_{q}g(z);$$

• if f and g are odd, then

  $${\partial }_q(fg)(z)=\frac{1}{q}(f(z)({\partial }_{q}g)(z/q)+({\partial }_{q}f)(z/q)g(z/q)).$$

The q-translation ${\epsilon }^y$ is introduced by Ismail in [2] and is defined on monomials by

$${\epsilon }^y{x}^n:=x^n{(-y/x;q)}_n,$$

(1.9)

and it is extended to polynomials as a linear operator. Thus

$${\epsilon }^y\left(\sum _{n=0}^m{f}_n{x}^n\right):=\sum _{n=0}^m{f}_n{x}^n{(-y/x;q)}_n.$$

(1.10)

The q-translation operator is defined for $x^a$, $a>0$, to be

$${\epsilon }^y{x}^a:=x^a{(-y/x;q)}_a.$$

(1.11)

In [4], Hahn defined the following q-analogue of the Laplace transform:

$${}_{q}L_{s}f(x)=\varphi (s)=\frac{1}{1-q}{\int }_0^{s^{-1}}{E}_q(-qsx)f(x)d_{q}x(Re(s)>0).$$

(1.12)

Abdi [11] studied certain properties of these q-transforms. In [12], he used these analogues to solve linear q-difference equations with constant coefficients and certain allied equations. In [[4], equation (9.5)], Hahn defined the convolution of two functions F, G to be

$$(F\ast G)(x)=\frac{x}{1-q}{\int }_0^{1}F(tx)G[x-tqx]d_{q}t,$$

(1.13)

where $G[x-y]$, for

$$G(x):=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n,$$

is defined to be

$$G[x-y]:=\sum _{n=0}^{\mathrm{\infty }}{a}_n{[x-y]}_n,\text{with }{[x-y]}_n:=x^n{(y/x;q)}_n.$$

Using the definition of q-integration, $(F\ast G)$ is nothing but

$$(F\ast G)(x)=\frac{1}{1-q}{\int }_0^{x}F(t){\epsilon }^{-qt}G(x)d_{q}t,$$

(1.14)

where ε is the translation operator (1.10). It is remarked by Hahn [[4], p.373] that the convolution theorem

$${}_{q}L_s(F\ast G){=}_q{L}_s{F}_q{L}_{s}G$$

(1.15)

holds. One can verify that if $\mathrm{\Phi }(s):{=}_q{L}_{s}F(x)$ and $0<\alpha <1$, then

$${}_{q}L_s{D}_q^{\alpha }F(x)=\frac{s^{\alpha }}{{(1-q)}^{\alpha }}\mathrm{\Phi }(s)-I_q^{1-\alpha }F\left(0^{+}\right)\frac{1}{(1-q)};$$

(1.16)

see [13].

2 Orthogonality relations and completeness criteria

Koornwinder and Swarttouw introduced a q-analogue of the cosine and sine Fourier transform in [14] with the functions $Cos(z;q^2)$ and $Sin(z;q^2)$ defined by

$$\begin{array}{r}Cos(z;q^2)=\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^k{q}^{k(k+1)}{(z(1-q))}^{2k}}{{(q;q)}_{2k}}\\ \phantom{Cos(z;q^2)}{=}_1{\varphi }_1(0;q;q^2,q^2{z}^2{(1-q)}^2),\\ Sin(z;q^2)=\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^k{q}^{k(k+1)}{(z(1-q))}^{2k+1}}{{(q;q)}_{2k+1}}\\ \phantom{Sin(z;q^2)}=z_1{\varphi }_1(0;q^3;q^2,q^2{z}^2{(1-q)}^2).\end{array}$$

(2.1)

A q-analogue of the exponential function is introduced in [8, 9] and defined by

$$e(z;q^2):=Cos(iz;q^2)-iSin(iz;q^2).$$

Straightforward calculations give

$${\delta }_{q}Cos(\lambda x;q^2)=-\lambda Sin(\lambda x;q^2),{\delta }_{q}Sin(\lambda x;q^2)=\lambda Cos(\lambda x;q^2)$$

and

$${\delta }_{q}e(\lambda x;q^2)=\lambda e(\lambda x;q^2),$$

where $x\in \mathbb{C}$ and λ is a fixed complex number. Fitouhi et al. in [15] proved that

$$\left|{\frac{{(z;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}}_1{\varphi }_1(0;z,q,q^{1+n})\right|\le \frac{{(-|z|,-q;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}\{\begin{array}{cc}1\hfill & \text{if }n\ge 0,\hfill \\ {|z|}^{-n}{q}^{\frac{n(n+1)}{2}}\hfill & \text{if }n<0.\hfill \end{array}$$

Hence,

$$|Sin(\frac{q^n}{1-q};q^2)|\le \frac{{(-q^2;q^2)}_{\mathrm{\infty }}}{{(q;q^2)}_{\mathrm{\infty }}}\{\begin{array}{cc}1\hfill & \text{if }n\ge 0,\hfill \\ q^{n^2}\hfill & \text{if }n<0\hfill \end{array}$$

(2.2)

and

$$|Cos(\frac{q^n}{1-q};q^2)|\le \frac{{(-q^2;q^2)}_{\mathrm{\infty }}}{{(q;q^2)}_{\mathrm{\infty }}}\{\begin{array}{cc}1\hfill & \text{if }n\ge 0,\hfill \\ q^{n^2-2n}\hfill & \text{if }n<0.\hfill \end{array}$$

(2.3)

Consequently,

$$\left|e(\frac{q^n}{1-q};q^2)\right|\le 2\frac{{(-q^2;q^2)}_{\mathrm{\infty }}}{{(q;q^2)}_{\mathrm{\infty }}}\{\begin{array}{cc}1,\hfill & n\ge 0,\hfill \\ q^{n^2},\hfill & n<0.\hfill \end{array}$$

(2.4)

The following orthogonality relation is proved in [14].

Theorem 2.1 Let $|z|<1$ and n, m be integers. Then

$${\delta }_{mn}=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{z}^{k+n}{\frac{{(z^2;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}}_1{\varphi }_1(0;z^2;q,q^{n+k+1})z^{k+m}{\frac{{(z^2;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}}_1{\varphi }_1(0;z^2;q,q^{m+k+1}),$$

(2.5)

where the sum converges absolutely and uniformly on compact subsets of the open unit disc.

The following identity, which follows from (2.5) when we replace q by $q^2$ and z by $q^{\alpha }$, $\alpha >0$, is essential in our investigations.

$$\begin{array}{r}{q}^{-\alpha (n+m)}\frac{{(q^2;q^2)}_{\mathrm{\infty }}^2}{{(q^{2\alpha };q^2)}_{\mathrm{\infty }}^2}{\delta }_{mn}\\ =\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}q^{2k\alpha }{}_1{\varphi }_1{(0;q^{2\alpha };q^2,q^{2n+2k+2})}_1{\varphi }_1(0;q^{2\alpha };q^2,q^{2m+2k+2}).\end{array}$$

(2.6)

Theorem 2.2 For $0<q<1$,

$${\int }_0^{\mathrm{\infty }/\sqrt{1-q}}Sin(\frac{q^{n}z}{\sqrt{1-q}};q^2)Sin(\frac{q^{m}z}{\sqrt{1-q}};q^2)d_{q}z=\frac{\sqrt{1-q}{(q^2;q^2)}_{\mathrm{\infty }}^2}{{(q;q^2)}_{\mathrm{\infty }}^2}{q}^{-n}{\delta }_{n,m},$$

(2.7)

$${\int }_0^{\mathrm{\infty }/\sqrt{1-q}}Cos(\frac{q^{n}z}{\sqrt{1-q}};q^2)Cos(\frac{q^{m}z}{\sqrt{1-q}};q^2)d_{q}z=\frac{\sqrt{1-q}{(q^2;q^2)}_{\mathrm{\infty }}^2}{{(q;q^2)}_{\mathrm{\infty }}^2}{q}^{-n}{\delta }_{n,m}$$

(2.8)

and

$${\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}e(i\frac{q^{n}z}{\sqrt{1-q}};q^2)e(-i\frac{q^{m}z}{\sqrt{1-q}};q^2)d_{q}z=4\frac{\sqrt{1-q}{(q^2;q^2)}_{\mathrm{\infty }}^2}{{(q;q^2)}_{\mathrm{\infty }}^2}{q}^{-n}{\delta }_{n,m}.$$

(2.9)

Proof We start with proving (2.7). Since

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}Sin(\frac{q^n}{\sqrt{1-q}}z;q^2)Sin(\frac{q^m}{\sqrt{1-q}}z;q^2)d_{q}z\\ =\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^k\sqrt{1-q}Sin(\frac{q^{n+k}}{1-q};q^2)Sin(\frac{q^{m+k}}{1-q};q^2)\\ =\frac{q^{n+m}}{{(1-q)}^{3/2}}\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}q^{3k}{}_1{\varphi }_1{(0;q^3,q^2;q^{2+2n+2k})}_1{\varphi }_1(0;q^3,q^2;q^{2+2m+2k}).\end{array}$$

(2.10)

In (2.6), set $\alpha =3/2$ to obtain

$$\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}q^{3k}{}_1{\varphi }_1{(0;q^3;q^2,q^{2+2n+2k})}_1{\varphi }_1(0;q^3,q^2;q^{2+2m+2k})=q^{-3(n+m)/2}\frac{{(q^2;q^2)}_{\mathrm{\infty }}^2}{{(q^3;q^2)}_{\mathrm{\infty }}^2}{\delta }_{n,m}.$$

(2.11)

Combining (2.10) and (2.11) yields (2.7). The proof of (2.8) follows similarly and the proof of (2.9) follows by combining (2.7) and (2.8). □

Theorem 2.3 For any $q\in (0,1)$,

• (a) the set $\{e(\pm \frac{q^n}{\sqrt{1-q}}x;q^2),n\in \mathbb{Z}\}$ is a complete orthogonal set in $L_q^2({\tilde{\mathbb{R}}}_q)$,

• (b) both of the sets $\{Cos(\frac{x{q}^n}{\sqrt{1-q}};q^2),n\in \mathbb{Z}\}$ and $\{Sin(\frac{x{q}^n}{\sqrt{1-q}};q^2),n\in \mathbb{Z}\}$ are complete orthogonal sets in $L_q^2({\tilde{\mathbb{R}}}_{q,+})$.

Proof We only proove (a). The proof of (b) is similar and is omitted. From Theorem 2.2, it remains only to prove that the set $\{e(\pm \frac{q^n}{\sqrt{1-q}}x;q^2),n\in \mathbb{Z}\}$ is complete in $L_q^2({\tilde{\mathbb{R}}}_q)$. This is equivalent to proving that if there exists a function $f\in L^2({\tilde{\mathbb{R}}}_q)$ such that

$$〈f,e(\pm \frac{q^n}{\sqrt{1-q}}x;q^2)〉=0(n\in \mathbb{Z}),$$

(2.12)

then

$$f(\pm \frac{q^n}{1-q})=0(n\in \mathbb{Z}).$$

From (2.12) we deduce

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{f}_e(t)Cos(\frac{q^n}{\sqrt{1-q}}t;q^2)d_{q}t=0(n\in \mathbb{Z}),\\ {\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{f}_o(t)Sin(\frac{q^n}{\sqrt{1-q}}t;q^2)d_{q}t=0(n\in \mathbb{Z}),\end{array}$$

where $f_e$ and $f_o$ are the even and odd parts of the function f. Then from (3.8)-(3.9) we obtain $f_e(t)=f_o(t)=f(t)=0$ for all $t\in {\tilde{\mathbb{R}}}_q$. Hence $\{e(\pm \frac{q^n}{\sqrt{1-q}}x;q^2),n\in \mathbb{Z}\}$ is a complete orthogonal set in $L^2({\tilde{\mathbb{R}}}_q)$. □

Theorem 2.4

• (1) If $f\in L^2({\tilde{\mathbb{R}}}_q)$, then

  $$f(x)=\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}e(ix\frac{q^n}{\sqrt{1-q}};q^2)+\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{d}_{n}e(-ix\frac{q^n}{\sqrt{1-q}};q^2)(x\in {\tilde{\mathbb{R}}}_q),$$

  (2.13)

where

$$\begin{array}{c}{c}_n=\frac{q^n}{C}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t)e(-it\frac{q^n}{\sqrt{1-q}};q^2)d_{q}t(n\in \mathbb{Z}),\hfill \\ d_n=\frac{q^n}{C}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t)e(it\frac{q^n}{\sqrt{1-q}};q^2)d_{q}t(n\in \mathbb{Z}),\hfill \end{array}$$

and $C:=\frac{4\sqrt{1-q}{(q^2;q^2)}_{\mathrm{\infty }}^2}{{(q;q^2)}_{\mathrm{\infty }}^2}$.

• (2) If $f\in L^2({\tilde{\mathbb{R}}}_q)$, then

  $$f(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{n}Cos(\frac{x{q}^n}{\sqrt{1-q}};q^2)+\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{b}_{n}Sin(\frac{x{q}^n}{\sqrt{1-q}};q^2)(x\in {\tilde{\mathbb{R}}}_q),$$

  (2.14)

where

$$a_n=\frac{4}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{f}_e(t)Cos(\frac{t{q}^n}{\sqrt{1-q}};q^2)d_{q}t(n\in \mathbb{Z})$$

and

$$b_n=\frac{4}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{f}_o(t)Sin(\frac{t{q}^n}{\sqrt{1-q}};q^2)d_{q}t(n\in \mathbb{Z}).$$

Proof The proof of (1) follows directly from Theorem 2.3 and the orthogonality relations (2.9). In the following we give in detail the proof of (2). Let $f=f_e+f_o$ be any function in $L^2({\tilde{\mathbb{R}}}_q)$. Clearly both $f_e$ and $f_o$ belong to $L^2({\tilde{\mathbb{R}}}_q)$. The restriction of $f_e$ to ${\tilde{\mathbb{R}}}_{q,+}$ can be represented in the complete orthogonal set $\{Cos(\frac{x{q}^n}{\sqrt{1-q}}),n\in \mathbb{Z}\}$ as

$$f_e(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{n}Cos(\frac{x{q}^n}{\sqrt{1-q}};q^2)(x\in {\tilde{\mathbb{R}}}_{q,+}),$$

(2.15)

where

$$a_n=\frac{4}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{f}_e(t)Cos(\frac{t{q}^n}{\sqrt{1-q}};q^2)d_{q}t(n\in \mathbb{Z}).$$

The orthogonal set $\{Sin(\frac{x{q}^n}{\sqrt{1-q}}),n\in \mathbb{Z}\}$ also spans $L^2({\tilde{\mathbb{R}}}_{q,+})$, hence

$$f_o(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{b}_{n}Sin(\frac{x{q}^n}{\sqrt{1-q}};q^2)(x\in {\tilde{\mathbb{R}}}_{q,+}),$$

(2.16)

where

$$b_n=\frac{4}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{f}_o(t)Sin(\frac{t{q}^n}{\sqrt{1-q}};q^2)d_{q}t(n\in \mathbb{Z}).$$

Because both sides of (2.15) are even functions on ${\tilde{\mathbb{R}}}_q$, the equality extends on ${\tilde{\mathbb{R}}}_q$; and similarly the two sides of (2.16). Hence we have the representation (2.14) of any $f\in L^2({\tilde{\mathbb{R}}}_q)$. □

3 Rubin’s $q^2$-Fourier transform

Koornwinder and Swarttouw [14] introduced the pair of q-transforms

$$\begin{array}{r}g\left(q^n\right)=\frac{{(q;q^2)}_{\mathrm{\infty }}}{{(q^2;q^2)}_{\mathrm{\infty }}}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^k\{\begin{array}{c}Cos(\frac{q^{k+n}}{1-q};q^2)\hfill \\ \text{or}\hfill \\ Sin(\frac{q^{k+n}}{1-q};q^2)\hfill \end{array}f\left(q^k\right),\\ f\left(q^k\right)=\frac{{(q;q^2)}_{\mathrm{\infty }}}{{(q^2;q^2)}_{\mathrm{\infty }}}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^n\{\begin{array}{c}Cos(\frac{q^{k+n}}{1-q};q^2)\hfill \\ \text{or}\hfill \\ Sin(\frac{q^{k+n}}{1-q};q^2)\hfill \end{array}g\left(q^n\right),\end{array}$$

(3.1)

where $0<q<1$ and f, g are in the space $L^2({\mathbb{R}}_q)$. Now assume that $\frac{log(1-q)}{logq}\in 2\mathbb{Z}$ or, equivalently,

$$q\in \{q\in (0,1):1-q=q^{2m}\text{ for some integer }m\}.$$

(3.2)

Then, by replacing $q^k$ and $q^n$ in (3.1) by $q^k\sqrt{1-q}$ and $q^n\sqrt{1-q}$, and then $f(q^k\sqrt{1-q})$ and $g(q^n\sqrt{1-q})$ by $f(q^k)$ and $g(q^k)$, Koornwinder and Swarttouw obtained the following q-analogue of the cosine and sine Fourier transforms:

$$\begin{array}{r}g(\lambda )=\frac{\sqrt{1+q}}{{\mathrm{\Gamma }}_q^2(1/2)}{\int }_0^{\mathrm{\infty }}f(t)\{\begin{array}{c}Cos(t\lambda ;q^2)\hfill \\ \text{or}\hfill \\ Sin(t\lambda ;q^2)\hfill \end{array}{d}_{q}t,\\ f(x)=\frac{\sqrt{1+q}}{{\mathrm{\Gamma }}_q^2(1/2)}{\int }_0^{\mathrm{\infty }}f(t)\{\begin{array}{c}Cos(x\lambda ;q^2)\hfill \\ \text{or}\hfill \\ Sin(x\lambda ;q^2)\hfill \end{array}{d}_q\lambda .\end{array}$$

(3.3)

Therefore, if we let $q\to 1^{-}$ for such q’s that satisfy (3.2), we obtain the cosine and sine Fourier transforms

$$g(\lambda )=\sqrt{\frac{2}{\pi }}{\int }_0^{\mathrm{\infty }}f(t)cos(\lambda t)dt,f(t)=\sqrt{\frac{2}{\pi }}{\int }_0^{\mathrm{\infty }}g(\lambda )cos(t\lambda )d\lambda ,$$

(3.4)

$$g(\lambda )=\sqrt{\frac{2}{\pi }}{\int }_0^{\mathrm{\infty }}f(t)sin(\lambda t)dt,f(t)=\sqrt{\frac{2}{\pi }}{\int }_0^{\mathrm{\infty }}g(\lambda )sin(t\lambda )d\lambda .$$

(3.5)

The pair of functions $Cos(\lambda x;q^2)$ and $Sin(\lambda x;q^2)$ satisfy

$$-\frac{1}{q}{D}_{q^{-1}}{D}_{q}y(x)=\{\begin{array}{cc}{\lambda }^{2}y(x)\hfill & \text{if }y(x)=Sin(\lambda x;q^2),\hfill \\ q{\lambda }^{2}y(x)\hfill & \text{if }y(x)=Cos(\lambda x;q^2).\hfill \end{array}$$

Therefore, the eigenfunctions $\{Cos(\lambda x;q^2),Sin(\lambda x;q^2)\}$ have two different eigenvalues. Consequently, as remarked by Koornwinder and Swarttouw in [14], no q-exponential functions built from $\{Cos(x;q^2),Sin(x;q^2)\}$ will satisfy an eigenfunction problem. This motivated Rubin [8] to define the q-difference operator (1.8) since for this operator, the functions $\{Cos(\lambda x;q^2),Sin(\lambda x;q^2)\}$ are solutions of the eigenvalue problem

$$-{\delta }_q^{2}y(x)={\lambda }^{2}y(x).$$

Rubin [8] introduced a $q^2$-analogue of the Fourier transform in the form

$$\hat{f}(x;q^2):={\mathcal{F}}_q(f)(x)=\frac{\sqrt{1+q}}{2{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)e(-itx;q^2)d_{q}t,$$

(3.6)

where $f\in L^1({\mathbb{R}}_q)$ and q satisfies condition (3.2).

Remark 3.1 Rubin [9] proved that

• (1) the $q^2$-Fourier transform defines a bounded linear operator from $L^1({\mathbb{R}}_q)$ to $L^{\mathrm{\infty }}({\mathbb{R}}_q)$,

• (2) the $q^2$-Fourier transform is defined and bounded on $L^1({\mathbb{R}}_q)\cap L^2({\mathbb{R}}_q)$,

• (3) $L^1({\mathbb{R}}_q)\cap L^2({\mathbb{R}}_q)$ is dense in $L^2({\mathbb{R}}_q)$ (consider the functions with finite support).

Consequently, the $q^2$-Fourier transform defines a bounded extension to $L^2({\mathbb{R}}_q)$.

Koornwinder and Swarttouw introduced the q-Hankel transforms (3.1) which can be written in the form (3.3) only if q satisfies condition (3.2). In fact, we can write the q-transforms in (3.1) as q-integral on $(-\mathrm{\infty },\mathrm{\infty })$ by using Matsuo definition (1.4) as in the following. Rewrite the transform pair in (3.1) as

$$\begin{array}{r}g\left(\frac{q^n}{\sqrt{1-q}}\right)=\frac{{(q;q^2)}_{\mathrm{\infty }}}{{(q^2;q^2)}_{\mathrm{\infty }}}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^k\{\begin{array}{c}Cos(\frac{q^{k+n}}{1-q};q^2)\hfill \\ \text{or}\hfill \\ Sin(\frac{q^{k+n}}{1-q};q^2)\hfill \end{array}f\left(\frac{q^k}{\sqrt{1-q}}\right),\\ f\left(\frac{q^k}{\sqrt{1-q}}\right)=\frac{{(q;q^2)}_{\mathrm{\infty }}}{{(q^2;q^2)}_{\mathrm{\infty }}}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^n\{\begin{array}{c}Cos(\frac{q^{k+n}}{1-q};q^2)\hfill \\ \text{or}\hfill \\ Sin(\frac{q^{k+n}}{1-q};q^2)\hfill \end{array}g\left(\frac{q^n}{\sqrt{1-q}}\right),\end{array}$$

(3.7)

where we assume that the functions f and g are in the space $L^1({\tilde{\mathbb{R}}}_q)\cap L^2({\tilde{\mathbb{R}}}_q)$. Using Matsuo definition of the q-integration on $(0,\mathrm{\infty })$, (1.4) with $b=\sqrt{1-q}$, the transformations in (3.7) can be written as

$$\begin{array}{r}g(x)=\frac{\sqrt{1+q}}{{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}f(t)Cos(xt;q^2)d_{q}t,\\ f(t)=\frac{\sqrt{1+q}}{{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}g(x)Cos(xt;q^2)d_{q}x\end{array}$$

(3.8)

and

$$\begin{array}{r}g(x)=\frac{\sqrt{1+q}}{{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}f(t)Sin(xt;q^2)d_{q}t,\\ f(t)=\frac{\sqrt{1+q}}{{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}g(x)Sin(xt;q^2)d_{q}x,\end{array}$$

(3.9)

where $x,t\in {\tilde{\mathbb{R}}}_q$ and f, g are in $L^1({\tilde{\mathbb{R}}}_q)\cap L^2({\tilde{\mathbb{R}}}_q)$. This is similar to Rubin’s work in [9]. Consequently, we set the following reformulation of Rubin’s definition of the $q^2$-Fourier transform (3.6).

Definition 3.2 Let $0<q<1$. We define the $q^2$-Fourier transform for any function $f\in L^1({\tilde{\mathbb{R}}}_q)$ to be

$$\hat{f}(x;q^2):={\mathcal{F}}_q(f)(x)=\frac{\sqrt{1+q}}{2{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t)e(-itx;q^2)d_{q}t.$$

(3.10)

It is clear that Rubin’s definition of the $q^2$-Fourier transform is a special case of (3.2) because if $1-q=q^{2m}$ for some $m\in \mathbb{Z}$, then

$${\int }_0^{\mathrm{\infty }/\sqrt{1-q}}f(t)d_{q}t={\int }_0^{\mathrm{\infty }/q^m}f(t)d_{q}t={\int }_0^{\mathrm{\infty }}f(t)d_{q}t.$$

However, we get the classical Fourier transform only when $q\to 1^{-}$ and q satisfies (3.2). Similar to Rubin’s results mentioned in Remark 3.1, we can prove that the $q^2$-Fourier transform defines a bounded linear operator from $L^1({\tilde{\mathbb{R}}}_q)$ to $L^{\mathrm{\infty }}({\tilde{\mathbb{R}}}_q)$, and $L^1({\tilde{\mathbb{R}}}_q)\cap L^2({\tilde{\mathbb{R}}}_q)$ is dense in $L^2({\tilde{\mathbb{R}}}_q)$. Therefore, the $q^2$-Fourier transform in (3.2) defines a bounded extension to $L^2({\tilde{\mathbb{R}}}_q)$.

The proofs of the following results, which are valid for any $q\in (0,1)$, are similar to the proofs in [8]. Therefore, we state them without proofs.

• (1) If $f\in L^2({\tilde{\mathbb{R}}}_q)$, then

  $$f(t)=\frac{\sqrt{1+q}}{2{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}\mathcal{F}(f)(x)e(itx(1-q);q^2)d_{q}x,t\in {\tilde{\mathbb{R}}}_q.$$

  (3.11)

• (2) If $f(u)$ and $uf(u)\in L_q^1({\tilde{\mathbb{R}}}_q)$, then

  $${\partial }_q({\mathcal{F}}_{q}f)(x)={\mathcal{F}}_q(-iuf(u))(x).$$

• (3) If f and ${\partial }_{q}f\in L_q^1({\tilde{\mathbb{R}}}_q)$, then

  $${\mathcal{F}}_q({\partial }_{q}f)(x)=ix{\mathcal{F}}_q(f)(x).$$

  (3.12)

We reformulate the definitions of $q^2$-Fourier multiplier and the $q^2$-Fourier convolution formula introduced by Rubin in [9] with the restriction (3.2) to any $q\in (0,1)$.

Definition 3.3 Let $q\in (0,1)$. We define the $q^2$-Fourier multiplier operator corresponding to translation by y to be

$$(T_{y}f)(x)=\frac{\sqrt{1+q}}{2{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}e(-ity;q^2)({\mathcal{F}}_{q}f)(t)e(itx;q^2)d_{q}t,$$

(3.13)

whenever the q-integral makes sense. If $f\in L^2({\tilde{\mathbb{R}}}_q)$ and $g\in L^1({\tilde{\mathbb{R}}}_q)$, we define the multiplier corresponding to Fourier convolution of f with g to be

$$(f\ast g)(z)={\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}[T_{\lambda }f](z)g(\lambda )d_q\lambda .$$

(3.14)

Theorem 3.4 Let f and g be two functions in $(L^1\cap L^2)({\tilde{\mathbb{R}}}_q)$. Then

$${\mathcal{F}}_q(f\ast g)(x)={\mathcal{F}}_q(f)(x){\mathcal{F}}_q(g)(x)(x\in {\tilde{\mathbb{R}}}_q).$$

(3.15)

Proof The proof of (3.15) is completely similar to the proof of [[9], Theorem 8] and is omitted. □

4 Fractional q-operator as a generalization of a q-difference operator

Let f be an integrable function of period 2π. Weyl, see Zygmund’s book [16], introduced a fractional operator which is more convenient for trigonometric series than the Riemann-Liouville fractional operator. This operator is defined by

$$(I_{\alpha }f)(x)\sim \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n\frac{e^{inx}}{{(in)}^{\alpha }}\text{if }f(x)\sim \sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{e}^{inx},c_0=0,$$

(4.1)

where $i^{\alpha }=e^{i\alpha \pi /2}$. Zygmund [[16], p.133] pointed out that

$$I_{\alpha }f(x)=\frac{1}{2\pi }{\int }_0^{2\pi }f(t){\mathrm{\Psi }}_{\alpha }(x-t)dt,{\mathrm{\Psi }}_{\alpha }(x)=\sum _{n\ne 0}\frac{e^{inx}}{{(in)}^{\alpha }}.$$

He also proved the semigroup identity

$$I_{\alpha }{I}_{\beta }=I_{\alpha +\beta },\alpha ,\beta >0.$$

In [17], Ismail and Rahman defined a q-analogue of the fractional operator $I_{\alpha }$, so that $\alpha =1$ represents a right inverse of the Askey-Wilson operator ${\mathcal{D}}_q$ which is defined by

$$({\mathcal{D}}_{q}f)(x):=\frac{\breve{f}(q^{1/2}{e}^{i\theta })-\breve{f}(q^{-1/2}{e}^{i\theta })}{(q^{1/2}-q^{-1/2})sin\theta },x=cos\theta ,$$

where $f(x)=\breve{f}(z)$ with $x=(z+1/z)/2$.

In this section, we introduce a q-analogue of the fractional operator (4.1) as a generalization of the q-difference operator defined by Rubin in [8]. From Theorem 2.3, consequently,

$$f(x)={\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t){\mathrm{\Psi }}_0(x,t)d_{q}t,$$

where

$$\begin{array}{rl}{\mathrm{\Psi }}_0(x,t)=& \sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{n}e\left(ix\frac{q^n}{\sqrt{1-q}}\right)e(-it\frac{q^n}{\sqrt{1-q}})+q^{n}e(-ix\frac{q^n}{\sqrt{1-q}})e\left(it\frac{q^n}{\sqrt{1-q}}\right)\\ =& \sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{n}Cos(x\frac{q^n}{\sqrt{1-q}};q^2)Cos\left(t\frac{q^n}{\sqrt{1-q}}\right)\\ +q^{n}Sin(x\frac{q^n}{\sqrt{1-q}};q^2)Sin(t\frac{q^n}{\sqrt{1-q}};q^2).\end{array}$$

Lemma 4.1 The series

$$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{n(1-\alpha )}e(ix\frac{q^n}{\sqrt{1-q}};q^2)e(-it\frac{q^n}{\sqrt{1-q}};q^2)(x,t\in {\tilde{\mathbb{R}}}_q)$$

(4.2)

is absolutely convergent only when $Re\alpha <1$.

Proof The series in (4.2) can be written as

$$(\sum _{n=-\mathrm{\infty }}^{-1}+\sum _{n=0}^{\mathrm{\infty }})q^{n(1-\alpha )}e(ix\frac{q^n}{\sqrt{1-q}};q^2)e(-it\frac{q^n}{\sqrt{1-q}};q^2).$$

From (2.4), the series ${\sum }_{n=0}^{\mathrm{\infty }}{q}^{n(1-\alpha )}e(ix\frac{q^n}{\sqrt{1-q}};q^2)e(-it\frac{q^n}{\sqrt{1-q}};q^2)$ is absolutely convergent for $Re\alpha <1$ and diverges for $Re\alpha \ge 1$, while the series ${\sum }_{n=-\mathrm{\infty }}^{-1}{q}^{n(1-\alpha )}e(ix\frac{q^n}{\sqrt{1-q}};q^2)e(-it\frac{q^n}{\sqrt{1-q}};q^2)$ is absolutely convergent for all $\alpha \in \mathbb{C}$. □

Set

$$\begin{array}{rl}{\mathrm{\Psi }}_{\alpha }(x,t):=& {(1-q)}^{\alpha /2}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^n\frac{e(ix\frac{q^n}{\sqrt{1-q}};q^2)e(-it\frac{q^n}{\sqrt{1-q}};q^2)}{{(i{q}^n)}^{\alpha }}\\ +{(1-q)}^{\alpha /2}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^n\frac{e(-ix\frac{q^n}{\sqrt{1-q}};q^2)e(it\frac{q^n}{\sqrt{1-q}};q^2)}{{(-i{q}^n)}^{\alpha }},\end{array}$$

where $x,t\in {\tilde{\mathbb{R}}}_q$ and $i^{\alpha }$ is defined with respect to the principal branch, i.e., $i^{\alpha }=e^{i\frac{\pi }{2}\alpha }$.

Lemma 4.2 For $k\in \mathbb{N}$ and $Re\alpha <1$,

$${\delta }_{q,x}^k{\mathrm{\Psi }}_{\alpha }(x,t)={\mathrm{\Psi }}_{\alpha -k}(x,t).$$

Proof The proof follows directly by using that

$${\delta }_{q,x}^{k}e(ix\frac{q^n}{\sqrt{1-q}};q^2)={\left(\frac{i{q}^n}{\sqrt{1-q}}\right)}^{k}e(ix\frac{q^n}{\sqrt{1-q}};q^2).$$

 □

A direct calculation yields the following identity, which holds for $Re\alpha <1$,

$${(1-q)}^{-\alpha /2}{\mathrm{\Psi }}_{\alpha }(x,t)=2cos\left(\frac{\pi }{2}\alpha \right)A_{\alpha }(x,t)+2sin\left(\frac{\pi }{2}\alpha \right)B_{\alpha }(x,t),$$

(4.3)

where

$$\begin{array}{rl}{A}_{\alpha }(x,t):=& \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{k(1-\alpha )}Cos(\frac{q^{n+k}}{1-q};q^2)Cos(\frac{q^{m+k}}{1-q};q^2)\\ +q^{k(1-\alpha )}Sin(\frac{q^{n+k}}{1-q};q^2)Sin(\frac{q^{m+k}}{1-q};q^2)\end{array}$$

(4.4)

and

$$\begin{array}{rl}{B}_{\alpha }(x,t):=& \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{k(1-\alpha )}Cos(\frac{q^{m+k}}{1-q};q^2)Sin(\frac{q^{n+k}}{1-q};q^2)\\ -q^{k(1-\alpha )}Cos(\frac{q^{n+k}}{1-q};q^2)Sin(\frac{q^{m+k}}{1-q};q^2).\end{array}$$

(4.5)

Theorem 4.3 For $Re(\alpha )<1$,

$$\begin{array}{r}{(1-q)}^{-\alpha /2}{\mathrm{\Psi }}_{\alpha }(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ =\{\begin{array}{c}2cos(\frac{\pi }{2}\alpha )q^{-m(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r}\hfill \\ -2sin(\frac{\pi }{2}\alpha )q^{-m(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r}\hfill \\ \mathit{\text{if}}n>m+[\frac{1-Re\alpha }{2}],\hfill \\ 2cos(\frac{\pi }{2}\alpha )q^{-n(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r}\hfill \\ -2sin(\frac{\pi }{2}\alpha )q^{-n(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r}\hfill \\ \mathit{\text{if}}m>n+[\frac{1-Re\alpha }{2}],\hfill \end{array}\end{array}$$

(4.6)

where ${\tau }_r=\{\begin{array}{ll}{q}^{-\frac{r}{2}}{q}^{-\frac{1-\alpha }{2}},& r\mathit{\text{is odd}},\\ q^{\frac{r}{2}},& r\mathit{\text{is even}}.\end{array}$

Moreover,

$$\begin{array}{r}{(1-q)}^{-\alpha /2}{\mathrm{\Psi }}_{\alpha }(\frac{q^n}{\sqrt{1-q}},-\frac{q^m}{\sqrt{1-q}})\\ =\{\begin{array}{c}2cos(\frac{\pi }{2}\alpha )q^{-m(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r}\hfill \\ +2sin(\frac{\pi }{2}\alpha )q^{-m(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r},\hfill \\ n>m+[\frac{1-Re\alpha }{2}],\hfill \\ 2cos(\frac{\pi }{2}\alpha )q^{-n(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r}\hfill \\ +2sin(\frac{\pi }{2}\alpha )q^{-n(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r},\hfill \\ m>n+[\frac{1-Re\alpha }{2}],\hfill \end{array}\end{array}$$

(4.7)

$$\begin{array}{r}{(1-q)}^{-\alpha /2}{\mathrm{\Psi }}_{\alpha }(-\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ =\{\begin{array}{c}2cos(\frac{\pi }{2}\alpha )q^{-m(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r}\hfill \\ -2sin(\frac{\pi }{2}\alpha )q^{-m(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r}\hfill \\ \mathit{\text{if}}n>m+[\frac{1-Re\alpha }{2}],\hfill \\ 2cos(\frac{\pi }{2}\alpha )q^{-n(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r}\hfill \\ -2sin(\frac{\pi }{2}\alpha )q^{-n(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r}\hfill \\ \mathit{\text{if}}m>n+[\frac{1-Re\alpha }{2}],\hfill \end{array}\end{array}$$

(4.8)

$$\begin{array}{r}{(1-q)}^{-\alpha /2}{\mathrm{\Psi }}_{\alpha }(-\frac{q^n}{\sqrt{1-q}},-\frac{q^m}{\sqrt{1-q}})\\ =\{\begin{array}{c}2cos(\frac{\pi }{2}\alpha )q^{-m(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r}\hfill \\ +2sin(\frac{\pi }{2}\alpha )q^{-m(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r}\hfill \\ \mathit{\text{if}}n>m+[\frac{1-Re\alpha }{2}],\hfill \\ 2cos(\frac{\pi }{2}\alpha )q^{-n(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r}\hfill \\ +2sin(\frac{\pi }{2}\alpha )q^{-n(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r}\hfill \\ \mathit{\text{if}}m>n+[\frac{1-Re\alpha }{2}].\hfill \end{array}\end{array}$$

(4.9)

Proof Using the following formula from [[14], p.455]

$$\begin{array}{r}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{s}^k{y}^{n+k}{\frac{{(y^2;q^2)}_{\mathrm{\infty }}}{{(q^2;q^2)}_{\mathrm{\infty }}}}_1{\varphi }_1(0;y^2;q^2,q^{2n+2k+2})x^{m+k}{\frac{{(x^2;q^2)}_{\mathrm{\infty }}}{{(q^2;q^2)}_{\mathrm{\infty }}}}_1{\varphi }_1(0;x^2;q^2,q^{2m+2k+2})\\ =s^{-m}{y}^{n-m}{\frac{{(s^{-1}x{y}^{-1},y^2;q^2)}_{\mathrm{\infty }}}{{(sxy,q^2;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2}s{x}^{-1}y,sxy;y^2;q^2,q^{2n-2m}{s}^{-1}x{y}^{-1})\\ =s^{-n}{x}^{m-n}{\frac{{(s^{-1}y{x}^{-1},x^2;q^2)}_{\mathrm{\infty }}}{{(sxy,q^2;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2}sx{y}^{-1},sxy;x^2;q^2,q^{2m-2n}{s}^{-1}y{x}^{-1}),\end{array}$$

where $|sxy|<1$, we can prove that

$$\begin{array}{r}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{k(1-\alpha )}Cos(\frac{q^{m+k}}{1-q};q^2)Cos(\frac{q^{n+k}}{1-q};q^2)\\ =\{\begin{array}{cc}{q}^{-m(1-\alpha )}{\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2-\alpha },q^{1-\alpha };q;q^2,q^{2n-2m+\alpha }),\hfill & n\ge m+[Re\alpha /2],\hfill \\ q^{-n(1-\alpha )}{\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2-\alpha },q^{1-\alpha };q;q^2,q^{2m-2n+\alpha }),\hfill & m\ge n+[Re\alpha /2],\hfill \end{array}\end{array}$$

where $Re(1-\alpha )>0$ and

$$\begin{array}{r}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{k(1-\alpha )}Sin(\frac{q^{m+k}}{1-q};q^2)Sin(\frac{q^{n+k}}{1-q};q^2)\\ =\{\begin{array}{cc}{q}^{n-m}\frac{q^{-m(1-\alpha )}}{1-q}{\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{3-\alpha },q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2-\alpha },q^{3-\alpha };q^3;q^2,q^{2n-2m+\alpha }),\hfill & n>m+[Re\alpha /2],\hfill \\ q^{m-n}\frac{q^{-n(1-\alpha )}}{1-q}{\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{3-\alpha },q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2-\alpha },q^{3-\alpha };q^3;q^2,q^{2m-2n+\alpha }),\hfill & m>n+[Re\alpha /2].\hfill \end{array}\end{array}$$

Also,

$$\begin{array}{r}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{k(1-\alpha )}Sin(\frac{q^{m+k}}{1-q};q^2)Cos(\frac{q^{n+k}}{1-q};q^2)\\ =\{\begin{array}{c}{q}^{-m(1-\alpha )}{\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{1-\alpha },q^{2-\alpha };q;q^2,q^{2n-2m+\alpha +1}),\hfill \\ q^{m-n}\frac{q^{-n(1-\alpha )}}{1-q}{\frac{{(q^{\alpha -1},q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2-\alpha },q^{3-\alpha };q^3;q^2,q^{2m-2n+\alpha -1}).\hfill \end{array}\end{array}$$

Hence, if $x:=\frac{q^n}{\sqrt{1-q}}$ and $t:=\frac{q^m}{\sqrt{1-q}}$, then

$$\begin{array}{r}{A}_{\alpha }(x,t)\\ =\{\begin{array}{c}{t}^{-(1-\alpha )}{(1-q)}^{-(1-\alpha )/2}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{(\frac{x}{t})}^r,\hfill \\ n>m+[-Re\alpha /2],\hfill \\ x^{-(1-\alpha )}{(1-q)}^{-(1-\alpha )/2}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{(\frac{t}{x})}^r,\hfill \\ m>n+[-Re\alpha /2].\hfill \end{array}\end{array}$$

(4.10)

Also,

$$\begin{array}{r}{A}_{\alpha }(x,-t)\\ =\{\begin{array}{c}{t}^{-(1-\alpha )}{(1-q)}^{-(1-\alpha )/2}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{(\frac{x}{t})}^r,\hfill \\ n>m+[-Re\alpha /2],\hfill \\ x^{-(1-\alpha )}{(1-q)}^{-(1-\alpha )/2}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\alpha [\frac{r}{2}]}\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{(\frac{t}{x})}^r,\hfill \\ m>n+[-Re\alpha /2],\hfill \end{array}\end{array}$$

(4.11)

$$\begin{array}{r}{B}_{\alpha }(x,t)\\ =-\{\begin{array}{c}{q}^{-m(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r},\hfill \\ n\ge m+[\frac{1-Re\alpha }{2}],\hfill \\ q^{-n(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r},\hfill \\ m>n+[\frac{1-Re\alpha }{2}],\hfill \end{array}\end{array}$$

(4.12)

$$\begin{array}{r}{B}_{\alpha }(x,-t)\\ =\{\begin{array}{c}{q}^{-m(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(n-m)r},\hfill \\ n\ge m+[\frac{1-Re\alpha }{2}],\hfill \\ q^{-n(1-\alpha )}\frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{r=0}^{\mathrm{\infty }}{(-1)}^r{q}^{\frac{\alpha r}{2}}{\tau }_r\frac{{(q^{1-\alpha };q)}_r}{{(q;q)}_r}{q}^{(m-n)r},\hfill \\ m>n+[\frac{1-Re\alpha }{2}].\hfill \end{array}\end{array}$$

(4.13)

Substituting from (4.10)-(4.13) into (4.3) yields the values ${\mathrm{\Psi }}_{\alpha }(\pm x,\pm t)$ and the theorem follows. □

Remark 4.4 In the previous theorem, we calculated the value of ${\mathrm{\Psi }}_{\alpha }(x,t)$, $x,t\in {\tilde{\mathbb{R}}}_q$ and for specific values of x, t. We can calculate the values of ${\mathrm{\Psi }}_{\alpha }(x,t)$ for all x, t by using the identity

$$\begin{array}{r}\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{s}^{k+m}{y}^{k+n}{\frac{{(y^2;q^2)}_{\mathrm{\infty }}}{{(q^2;q^2)}_{\mathrm{\infty }}}}_1{\varphi }_1(0;y^2;q^2,q^{2k+2n+2})x^{k+m}{\frac{{(x^2;q^2)}_{\mathrm{\infty }}}{{(q^2;q^2)}_{\mathrm{\infty }}}}_1{\varphi }_1(0;x^2;q^2,q^{2k+2m+2})\\ =y^{n-m}{\frac{{(s^{-1}x{y}^{-1},q^{2n-2m+2};q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m}{s}^{-1}x{y}^{-1},q^2;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2n-2m}{s}^{-1}x{y}^{-1},s^{-1}y{x}^{-1};q^{2n-2m+2};q^2,sxy)\end{array}$$

for $|sxy|<1$. See Proposition 4.1 of [14].

In this case we have

$$\begin{array}{r}{\mathrm{CC}}_{\alpha }(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ :=\sum _{r=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{r(1-\alpha )}Cos(\frac{q^{n+r}}{1-q};q^2)Cos(\frac{q^{m+r}}{1-q};q^2)\\ =q^{-m(1-\alpha )}{\frac{{(q^{\alpha },q^{2n-2m+2},q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m+\alpha },q,q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2n-2m+\alpha },q^{\alpha };q^{2n-2m+2};q^2,q^{1-\alpha }),\end{array}$$

(4.14)

$$\begin{array}{r}\begin{array}{r}{\mathrm{SS}}_{\alpha }(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ :=\sum _{r=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{r(1-\alpha )}Sin(\frac{q^{n+r}}{1-q};q^2)Sin(\frac{q^{m+r}}{1-q};q^2)\\ =q^{n-m}{q}^{-m(1-\alpha )}{\frac{{(q^{\alpha },q^{2n-2m+2},q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m+\alpha },q,q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2n-2m+\alpha },q^{\alpha };q^{2n-2m+2};q^2,q^{3-\alpha }),\end{array}\\ \begin{array}{r}{\mathrm{SC}}_{\alpha }(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ :=\sum _{r=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{r(1-\alpha )}Sin(\frac{q^{n+r}}{1-q};q^2)Cos(\frac{q^{m+r}}{1-q};q^2)\\ =q^{n-m}{q}^{-m(1-\alpha )}{\frac{{(q^{\alpha -1},q^{2n-2m+2},q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m+\alpha -1},q,q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2n-2m+\alpha -1},q^{\alpha +1};q^{2n-2m+2};q^2,q^{2-\alpha }),\end{array}\\ \begin{array}{r}{\mathrm{CS}}_{\alpha }(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ :=\sum _{r=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{r(1-\alpha )}Sin(\frac{q^{m+r}}{1-q};q^2)Cos(\frac{q^{n+r}}{1-q};q^2)\\ =q^{-m(1-\alpha )}{\frac{{(q^{\alpha +1},q^{2n-2m+2},q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m+\alpha +1},q,q;q^2)}_{\mathrm{\infty }}}}_2{\varphi }_1(q^{2n-2m+\alpha +1},q^{\alpha -1};q^{2n-2m+2};q^2,q^{2-\alpha }).\end{array}\end{array}$$

(4.15)

Corollary 4.5 For each fixed $x,t\in {\tilde{\mathbb{R}}}_q$, the function ${\mathrm{\Psi }}_{\alpha }(x,t)$ as a function of α can be extended to an entire function on ℂ.

Proof If α is a positive integer, then the series on the right-hand sides of (4.6)-(4.9) are finite sums and hence are convergent. Since the zeros of the function $cos(\frac{\pi }{2}\alpha )$ are the poles of the function ${(q^{1-\alpha };q^2)}_{\mathrm{\infty }}$ with the same orders. In fact

$$\underset{\alpha \to (2j+1)}{lim}\frac{cos\frac{\pi }{2}\alpha }{{(q^{1-\alpha };q^2)}_{\mathrm{\infty }}}=-\frac{\pi }{2lnq}\frac{q^{j^2+j}}{{(q^2;q^2)}_j{(q^2;q^2)}_{\mathrm{\infty }}}(j\in {\mathbb{N}}_0).$$

Similarly, the zeros of the function $sin(\frac{\pi }{2}\alpha )$ are the poles of the function ${(q^{2-\alpha };q^2)}_{\mathrm{\infty }}$ with the same orders and

$$\underset{\alpha \to (2j)}{lim}\frac{sin\frac{\pi }{2}\alpha }{{(q^{2-\alpha };q^2)}_{\mathrm{\infty }}}=-\frac{\pi }{2lnq}\frac{q^{j^2-j}}{{(q^2;q^2)}_{j-1}{(q^2;q^2)}_{\mathrm{\infty }}}(j\in \mathbb{N}).$$

Then the left-hand sides of equations (4.6)-(4.9) are entire functions. Hence, as a function of α, the functions ${\mathrm{\Psi }}_{-\alpha }(x,t)$ ($x,t\in {\tilde{\mathbb{R}}}_q$) can be analytically extended by defining its values when $\mathrm{\Re }\alpha \ge 1$ by the left-hand sides of (4.6)-(4.9). □

It also should be noted that for $Re(\alpha )\ge 1$, the left-hand sides of (4.6) and (4.7) determine ${\mathrm{\Psi }}_{-\alpha }(x,t)$ for all $x,t\in {\tilde{\mathbb{R}}}_q$, which is different from the case of $Re(\alpha )<1$; see Remark 4.4.

Definition 4.6 For $Re\alpha >0$, we define a fractional q-integral operator $J_q^{\alpha }$ on $L^2({\tilde{\mathbb{R}}}_q)\cap L^1({\tilde{\mathbb{R}}}_q)$ by

$$J_q^{\alpha }f(x):=\frac{1}{C}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t){\mathrm{\Psi }}_{\alpha }(x,t)d_{q}t.$$

The following properties follow at once from (4.3) and their analytic continuation on ℂ.

• If $f\in L^2({\tilde{\mathbb{R}}}_q)\cap L({\tilde{\mathbb{R}}}_q)$ and f is even, then

  $$J_q^{\alpha }f(x)=2{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}f(t){\phi }_{\alpha }(x,t)d_{q}t,$$

where

$$\begin{array}{r}{(1-q)}^{-\alpha /2}{\phi }_{\alpha }(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ =\{\begin{array}{c}2q^{-m(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}cos\frac{\pi }{2}\alpha {\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j}}{{(q;q)}_{2j}}{q}^{j(2n-2m+\alpha )}\hfill \\ -2q^{n-m-(m+1)(1-\alpha )}sin\frac{\pi }{2}\alpha \frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j+1}}{{(q;q)}_{2j+1}}{q}^{j(2n-2m+\alpha -1)},\hfill \\ 2q^{-n(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}cos\frac{\pi }{2}\alpha {\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j}}{{(q;q)}_{2j}}{q}^{j(2m-2n+\alpha )}\hfill \\ +2q^{-n(1-\alpha )}sin\frac{\pi }{2}\alpha \frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j}}{{(q;q)}_{2j}}{q}^{j(2m-2n+\alpha +1)}\hfill \end{array}\end{array}$$

and

$$\begin{array}{r}{(1-q)}^{-\alpha /2}{\phi }_{\alpha }(-\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ =\{\begin{array}{c}2q^{-m(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}cos\frac{\pi }{2}\alpha {\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{\alpha +1};q)}_{2j}}{{(q;q)}_{2j}}{q}^{i(2n-2m+\alpha )}\hfill \\ +2q^{n-m-m(1-\alpha )}sin\frac{\pi }{2}\alpha \frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j+1}}{{(q;q)}_{2j+1}}{q}^{j(2n-2m+\alpha -1)},\hfill \\ 2q^{-n(1-\alpha )}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}cos\frac{\pi }{2}\alpha {\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j}}{{(q;q)}_{2j}}{q}^{j(2m-2n+\alpha )}\hfill \\ -2q^{-n(1-\alpha )}sin\frac{\pi }{2}\alpha \frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j}}{{(q;q)}_{2j}}{q}^{j(2m-2n+\alpha +1)}.\hfill \end{array}\end{array}$$

Hence, if f is an even function, then ${\delta }_q^{2k}f(x)$ is an even function and ${\delta }_q^{2k+1}f(x)$ is an odd function for all $k\in {\mathbb{N}}_0$.

• If $f\in L^2({\tilde{\mathbb{R}}}_q)\cap L({\tilde{\mathbb{R}}}_q)$ and f is odd, then

  $$J_q^{\alpha }f(x)=2{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}f(t){\psi }_{\alpha }(x,t)d_{q}t,$$

where

$$\begin{array}{r}{(1-q)}^{-\alpha /2}{\psi }_{\alpha }(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ =\{\begin{array}{c}2q^{-m(1-\alpha )}{q}^{n-m}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}cos\frac{\pi }{2}\alpha {\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j+1}}{{(q;q)}_{2j+1}}{q}^{j(2n-2m+\alpha )}\hfill \\ -2q^{-m(1-\alpha )}sin\frac{\pi }{2}\alpha \frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j}}{{(q;q)}_{2j}}{q}^{j(2n-2m+\alpha +1)},\hfill \\ 2q^{-n(1-\alpha )}{q}^{m-n}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}cos\frac{\pi }{2}\alpha {\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j+1}}{{(q;q)}_{2j+1}}{q}^{j(2m-2n+\alpha )}\hfill \\ +2q^{-(n+1)(1-\alpha )}{q}^{m-n}sin\frac{\pi }{2}\alpha \frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j+1}}{{(q;q)}_{2j+1}}{q}^{j(2m-2n+\alpha -1)}\hfill \end{array}\end{array}$$

and

$$\begin{array}{r}{(1-q)}^{-\alpha /2}{\psi }_{\alpha }(-\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})\\ =\{\begin{array}{c}-2q^{-m(1-\alpha )}{q}^{n-m}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}cos\frac{\pi }{2}\alpha {\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j+1}}{{(q;q)}_{2j+1}}{q}^{j(2n-2m+\alpha )}\hfill \\ -2q^{-m(1-\alpha )}sin\frac{\pi }{2}\alpha \frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j}}{{(q;q)}_{2j}}{q}^{j(2n-2m+\alpha +1)},\hfill \\ 2q^{-n(1-\alpha )}{q}^{m-n}\frac{{(q^{\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{1-\alpha },q;q^2)}_{\mathrm{\infty }}}cos\frac{\pi }{2}\alpha {\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j+1}}{{(q;q)}_{2j+1}}{q}^{j(2m-2n+\alpha )}\hfill \\ -2q^{-(n+1)(1+\alpha )}{q}^{m-n}sin\frac{\pi }{2}\alpha \frac{{(q^{1+\alpha },q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2-\alpha },q;q^2)}_{\mathrm{\infty }}}{\sum }_{j=0}^{\mathrm{\infty }}\frac{{(q^{1-\alpha };q)}_{2j+1}}{{(q;q)}_{2j+1}}{q}^{j(2m-2n+\alpha -1)}.\hfill \end{array}\end{array}$$

Hence, if f is an odd function, then ${\delta }_q^{2k}f(x)$ is an odd function and ${\delta }_q^{2k+1}f(x)$ is an even function for all $k\in {\mathbb{N}}_0$.

Theorem 4.7 Let $f\in L^2({\tilde{\mathbb{R}}}_q)$. Then

$${\delta }_q(J_{q}f)(x)=f(x)\mathit{\text{for all}}x\in {\tilde{\mathbb{R}}}_q.$$

(4.16)

Moreover, if f is q-regular at zero, then

$$J_q({\delta }_{q}f)(x)=f(x)\mathit{\text{for all}}x\in {\tilde{\mathbb{R}}}_q.$$

(4.17)

Proof If $f\in L^2({\tilde{\mathbb{R}}}_q)$, then

$$J_{q}f(x)=\frac{2}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{f}_e(t){\phi }_1(x,t)d_{q}t+\frac{2}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{f}_e(t){\psi }_1(x,t)d_{q}t,$$

where

$${\phi }_1(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})=\{\begin{array}{cc}\frac{\pi \sqrt{1-q}}{lnq},\hfill & n>m,\hfill \\ \frac{\pi \sqrt{1-q}}{lnq}+\frac{C}{2},\hfill & m\ge n\hfill \end{array}$$

and

$${\psi }_1(\frac{q^n}{\sqrt{1-q}},\frac{q^m}{\sqrt{1-q}})=\{\begin{array}{cc}-\frac{C}{2},\hfill & n>m-1,\hfill \\ 0,\hfill & m>n.\hfill \end{array}$$

Hence,

$$\begin{array}{rl}{J}_{q}f(x)=& (1+\frac{2\pi \sqrt{1-q}}{Clnq}){\int }_0^x{f}_e(t)d_{q}t+\frac{2\pi \sqrt{1-q}}{Clnq}{\int }_x^{\mathrm{\infty }}{f}_e(t)d_{q}t\\ -{\int }_{qx}^{\mathrm{\infty }}{f}_0(t)d_{q}t.\end{array}$$

(4.18)

One can verify that the first and second q-integrals of (4.18) are odd functions, while the last q-integral is an even function. Consequently,

$$\begin{array}{rl}{\delta }_q{J}_{q}f(x)=& (1+\frac{2\pi \sqrt{1-q}}{Clnq})D_{q,x}{\int }_0^x{f}_e(t)d_{q}t+\frac{2\pi \sqrt{1-q}}{Clnq}{D}_{q,x}{\int }_x^{\mathrm{\infty }}{f}_e(t)d_{q}t\\ -D_{q,x}{\int }_x^{\mathrm{\infty }}{f}_0(t)d_{q}t.\end{array}$$

(4.19)

Using the fundamental theorem of q-calculus, see [13], we obtain (4.16). To prove (4.17), we assume that f is q-regular at zero,

$$\begin{array}{rl}{J}_q{\delta }_{q}f(x)& =J_q({\delta }_q{f}_e)(x)+J_q({\delta }_q{f}_0)(x)\\ =\frac{2}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{\delta }_q{f}_e(t){\psi }_1(x,t)d_{q}t+\frac{2}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{\delta }_q{f}_0(t){\phi }_1(x,t)d_{q}t.\end{array}$$

But

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{\delta }_q{f}_e(t){\psi }_1(x,t)d_{q}t& ={\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{D}_{q,t}{f}_e(t/q){\psi }_1(x,t)d_{q}t\\ =-\frac{C}{2}{\int }_{qx}^{\mathrm{\infty }/\sqrt{1-q}}{D}_{q,t}{f}_e(t/q)d_{q}t=\frac{C}{2}{f}_e(x)\end{array}$$

and

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{\delta }_q{f}_0(t){\phi }_1(x,t)d_{q}t\\ ={\int }_0^{\mathrm{\infty }/\sqrt{1-q}}{D}_{q,t}{f}_0(t){\phi }_1(x,t)d_{q}t\\ =(\frac{\pi \sqrt{1-q}}{lnq}+\frac{C}{2}){\int }_0^x{D}_{q,t}{f}_0(t)d_{q}t+\frac{\pi \sqrt{1-q}}{lnq}{\int }_x^{\mathrm{\infty }}{D}_{q,t}{f}_0(t)d_{q}t\\ =\frac{C}{2}{f}_o(x)-(\frac{\pi \sqrt{1-q}}{lnq}+\frac{C}{2})f_0(0)=\frac{C}{2}{f}_o(x)\end{array}$$

since $f_0(0)=0$. This proves (4.17) and completes the proof of the theorem. □

We can also prove that

$$\begin{array}{c}{\psi }_2(x,t):=\{\begin{array}{cc}-\frac{c}{2}(x-qt)+(x-t)\frac{\pi \sqrt{1-q}}{lnq},\hfill & x\le q^{-1}t,\hfill \\ -\frac{c}{2}(t-qx)+(x-t)\frac{\pi \sqrt{1-q}}{lnq},\hfill & x\ge qt,\hfill \end{array}\hfill \\ {\psi }_2(x,-t):=\{\begin{array}{cc}\frac{c}{2}(x+qt)+(x+t)\frac{\pi \sqrt{1-q}}{lnq},\hfill & x\le q^{-1}t,\hfill \\ \frac{c}{2}(t+qx)+(x+t)\frac{\pi \sqrt{1-q}}{lnq},\hfill & x\ge qt\hfill \end{array}\hfill \end{array}$$

and

$${\psi }_2(-x,t):=\{\begin{array}{cc}\frac{c}{2}(x+qt)-(x+t)\frac{\pi \sqrt{1-q}}{lnq},\hfill & x\le q^{-1}t,\hfill \\ \frac{c}{2}(t+qx)-(x+t)\frac{\pi \sqrt{1-q}}{lnq},\hfill & x\ge qt.\hfill \end{array}$$

Hence,

$$\begin{array}{rl}{J}_q^{2}f(x)=& \frac{\pi \sqrt{1-q}}{clnq}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}(x-t)f(t)d_{q}t\\ +{\int }_0^{q^{-1}x}(qx{f}_e(t)-t{f}_0(t))d_{q}t+{\int }_{q^{-1}x}^{\mathrm{\infty }/\sqrt{1-q}}(x{f}_e(t)-qt{f}_0(t))d_{q}t.\end{array}$$

Definition 4.8 For $\alpha >0$, we define a fractional q-difference operator ${\delta }_q^{\alpha }$ on $L^2({\tilde{\mathbb{R}}}_q)\cap L^1({\tilde{\mathbb{R}}}_q)$ by

$${\delta }_q^{\alpha }f(x):=J_q^{-\alpha }f(x)=\frac{1}{C}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t){\mathrm{\Psi }}_{-\alpha }(x,t)d_{q}t.$$

(4.20)

Lemma 4.9 The operator ${\delta }_q^{\alpha }$ coincides with Rubin’s q-difference operator when α is a positive integer.

Proof Let $\alpha =k$ for some $k\in \mathbb{N}$, and let $f\in L^2({\tilde{\mathbb{R}}}_q)\cap L^1({\tilde{\mathbb{R}}}_q)$. Using (2.7), we conclude

$$f(x)=\frac{1}{C}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t){\mathrm{\Psi }}_0(x,t)d_{q}t.$$

Then from Lemma 4.2 we obtain

$${\delta }_q^{k}f(x)=\frac{1}{C}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t){\mathrm{\Psi }}_{-k}(x,t)d_{q}t,$$

and the lemma follows. □

Lemma 4.10 If $\alpha >0$ and $f\in L^2({\tilde{\mathbb{R}}}_q)$, then

$${\delta }_q^{\alpha }f(x)={\delta }_q^k{J}_q^{k-\alpha }f(x)(\alpha >0;k=\lceil \alpha \rceil ;x\in {\mathbb{R}}_q).$$

(4.21)

Now, if α is a positive integer, then $\lceil \alpha \rceil =\alpha$ and from (4.21)

$${\partial }_q^{\alpha }={\partial }_q^k.$$

Proof Since

$${\delta }_q^k{J}_q^{k-\alpha }f(x)=\frac{1}{C}{\delta }_q^k{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(t){\mathrm{\Psi }}_{k-\alpha }(x,t)d_{q}t$$

and ${\delta }_{q,x}^k{\mathrm{\Psi }}_{k-\alpha }(x,t)={\mathrm{\Psi }}_{-\alpha }(x,t)$, the proof follows. □

Theorem 4.11 If $f\in L^1({\tilde{\mathbb{R}}}_q)\cap L^2({\tilde{\mathbb{R}}}_q)$ and α, β are complex numbers such that $Re(\alpha )<1$ and $Re(\beta )<1$ such that $Re(\alpha +\beta )<1$, then

$$J_q^{\alpha }{J}_q^{\beta }f=J_q^{\beta }{J}_q^{\alpha }f=J_q^{\alpha +\beta }f.$$

Proof Let $x\in {\tilde{\mathbb{R}}}_q$. Since

$$\begin{array}{rcl}{J}_q^{\alpha }\left(J_q^{\beta }f\right)(x)& =& \frac{1}{C^2}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(u){\mathrm{\Psi }}_{\alpha }(x,t){\mathrm{\Psi }}_{\beta }(t,u)d_{q}u{d}_{q}t\\ =& \frac{1}{C^2}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(u){\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}{\mathrm{\Psi }}_{\alpha }(x,t){\mathrm{\Psi }}_{\beta }(t,u)d_{q}t{d}_{q}u,\end{array}$$

using the orthogonality relation (2.9), we obtain

$${\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}{\mathrm{\Psi }}_{\alpha }(x,t){\mathrm{\Psi }}_{\beta }(t,u)d_{q}t=C{\mathrm{\Psi }}_{\alpha +\beta }(x;u).$$

Consequently,

$$J_q^{\alpha }\left(J_q^{\beta }f\right)(x)=\frac{1}{C}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}f(u){\mathrm{\Psi }}_{\alpha +\beta }(x;u)d_{q}u=J_q^{\alpha +\beta }f(x).$$

 □

Example 4.12 Let $n\in \mathbb{Z}$ and let $f_n(x)$ be the even function defined on ${\tilde{\mathbb{R}}}_q$ by

$$f_n(x):=\{\begin{array}{cc}1,\hfill & x=\pm x_n,x_n:=\frac{q^n}{\sqrt{1-q}},\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

Hence,

$$f_n(x)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{k}Cos(x\frac{q^k}{\sqrt{1-q}};q^2)+\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{b}_{k}Sin(x\frac{q^k}{\sqrt{1-q}};q^2).$$

Since $f_n$ is an even function, then $b_k=0$ for all $k\in \mathbb{Z}$ and

$$a_k=\frac{q^k}{C}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}{f}_n(t)Cos(t\frac{q^k}{\sqrt{1-q}};q^2)d_{q}t=\frac{2q^{n+k}\sqrt{1-q}}{C}Cos(\frac{q^{n+k}}{1-q};q^2).$$

Consequently,

$$f_n(x)=\frac{2q^n\sqrt{1-q}}{C}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{k}Cos(\frac{q^{n+k}}{1-q};q^2)Cos(\frac{q^{k}x}{\sqrt{1-q}};q^2).$$

Hence,

$$\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{k}Cos(\frac{q^{n+k}}{1-q};q^2)Cos(\frac{q^{k}x}{\sqrt{1-q}};q^2)=\{\begin{array}{cc}\frac{C{q}^{-n}}{2\sqrt{1-q}},\hfill & x=x_n,\hfill \\ \text{zero},\hfill & \text{otherwise}.\hfill \end{array}$$

Since

$${\delta }_q{f}_n(x)=\frac{1}{q}{D}_{q^{-1}}{f}_n(x):=\{\begin{array}{cc}\frac{\pm 1}{x_n(1-q)},\hfill & x=\pm x_n,\hfill \\ \frac{\mp 1}{q{x}_n(1-q)},\hfill & x=\pm q{x}_n,\hfill \\ \text{zero},\hfill & \text{otherwise}\hfill \end{array}$$

and

$${\delta }_q{f}_n(x)=-\frac{2q^n}{C}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{2k}Cos(\frac{q^{n+k}}{1-q};q^2)Sin(\frac{q^{k}x}{\sqrt{1-q}};q^2)(x\in {\tilde{\mathbb{R}}}_q).$$

We obtain

$$\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{2k}Sin(\frac{x{q}^k}{\sqrt{1-q}};q^2)Cos(\frac{q^{n+k}}{1-q};q^2)=\{\begin{array}{cc}\text{zero},\hfill & x\notin \{\pm x_n,\pm q{x}_n\},\hfill \\ \pm \frac{C{q}^{-n}}{2x_n(1-q)},\hfill & x=\pm x_n,\hfill \\ \pm \frac{C{q}^{-n}}{2q{x}_n(1-q)},\hfill & x=\pm q{x}_n.\hfill \end{array}$$

(4.22)

Similarly,

$${\delta }_q^2{f}_n(x)=\{\begin{array}{cc}\frac{-1}{q{x}_n^2(1-q)},\hfill & x=\pm x_n,\hfill \\ \frac{-1}{q^2{x}_n^2{(1-q)}^2},\hfill & x=\pm q{x}_n,\hfill \\ \frac{-q}{x_n^2{(1-q)}^2},\hfill & x=\pm q^{-1}{x}_n,\hfill \\ \text{zero},\hfill & \text{otherwise}.\hfill \end{array}$$

But

$${\delta }_q^2{f}_n(x)=-2\frac{q^n}{C\sqrt{1-q}}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{3k}Cos(\frac{q^{n+k}}{1-q};q^2)Cos\left(\frac{q^{k}x}{\sqrt{1-q}}\right)(x\in {\tilde{\mathbb{R}}}_q).$$

Hence,

$$\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{3k}Cos(\frac{q^{n+k}}{1-q};q^2)Cos(\frac{q^{k}x}{\sqrt{1-q}};q^2)=\{\begin{array}{cc}\frac{C{q}^{-n-1}}{2x_n^2\sqrt{1-q}},\hfill & x=\pm x_n,\hfill \\ \frac{C{q}^{-n-2}}{2x_n^2{(\sqrt{1-q})}^3},\hfill & x=\pm q{x}_n,\hfill \\ \frac{C{q}^{-n+1}}{2x_n^2{(\sqrt{1-q})}^3},\hfill & x=\pm q^{-1}{x}_n,\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

(4.23)

In the following two examples, we show how we can use (4.20) when α is a positive integer to obtain new summation formulae.

Example 4.13 Let f be an even function. Then, for each $k\in {\mathbb{N}}_0$, we have

$${\delta }_q^{2k}f(x)=q^{k(k-1)}{D}_{q,x}^{2k}f(x/q^k),{\delta }_q^{2k+1}f(x)=q^{k(k+1)}{D}_{q,x}^{2k+1}f(x/q^{k+1}).$$

Hence,

$${\delta }_q^{2k}f(x)=q^{k(k-1)}{(1-q)}^{-2k}{x}^{-2k}\sum _{r=0}^{2k}{q}^r\frac{{(q^{-2k};q)}_r}{{(q;q)}_r}f\left(x{q}^{r-k}\right).$$

On the other hand,

$${\delta }_q^{2k}f(x)=4\frac{{(-1)}^k}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}f(t){\mathrm{CC}}_{-2k}(x,t)d_{q}t.$$

Let $m\in \mathbb{Z}$ and let $f_m(x)$ be the even function defined on ${\tilde{\mathbb{R}}}_q$ by

$$f_m(x):=\{\begin{array}{cc}1,\hfill & x=\pm x_m,x_m:=\frac{q^n}{\sqrt{1-q}},\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

Set $x=\frac{q^n}{\sqrt{1-q}}$. Hence,

$$\begin{array}{rl}{\delta }_q^{2k}f(x)=& 4\frac{{(-1)}^k\sqrt{1-q}}{C}{q}^{-2mk}{(1-q)}^{-k}\frac{{(q^{-2k},q^{2n-2m+2},q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m-2k},q,q;q^2)}_{\mathrm{\infty }}}\\ {\times }_2{\varphi }_1(q^{2n-2m-2k},q^{-2k};q^{2n-2m+2};q^2,q^{2k+1})\\ =& q^{k^2+m-n-2nk}{(1-q)}^{-k}\frac{{(q^{-2k};q)}_{m-n+k}}{{(q;q)}_{m-n+k}}.\end{array}$$

Since

$$\frac{{(q^{-2k},q^{2n-2m+2};q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m-2k};q^2)}_{\mathrm{\infty }}}=0\text{if }n\ge m+k\text{ or }n<m.$$

Hence, if $m\le n\le m+k-1$, we obtain

$$\begin{array}{r}{}_2\varphi _1(q^{2n-2m-2k},q^{-2k};q^{2n-2m+2};q^2,q^{2k+1})\\ ={(-1)}^k{q}^{k^2+(m-n)(2k+1)}\frac{{(q^{-2k};q)}_{m-n+k}}{{(q;q)}_{m-n+k}}\frac{{(q^2,q^{2n-2m-2k};q^2)}_{\mathrm{\infty }}}{{(q^{-2k},q^{2n-2m+2};q^2)}_{\mathrm{\infty }}}.\end{array}$$

Example 4.14 Let f be an odd function. Then, for each $k\in {\mathbb{N}}_0$, we have

$${\delta }_q^{2k}f(x)=q^{k^2}{D}_{q,x}^{2k}f(x/q^k),{\delta }_q^{2k+1}f(x)=q^{k^2}{D}_{q,x}^{2k+1}f(x/q^k).$$

Hence,

$${\delta }_q^{2k}f(x)=q^{k^2}{(1-q)}^{-2k}{x}^{-2k}\sum _{r=0}^{2k}{q}^r\frac{{(q^{-2k};q)}_r}{{(q;q)}_r}f\left(x{q}^{r-k}\right).$$

On the other hand,

$${\delta }_q^{2k}f(x)=4\frac{{(-1)}^k}{C}{\int }_0^{\mathrm{\infty }/\sqrt{1-q}}f(t){\mathrm{CS}}_{-2k}(x,t)d_{q}t.$$

Let $m\in \mathbb{Z}$ and let $f_m(x)$ be the odd function defined on ${\tilde{\mathbb{R}}}_q$ by

$$f_m(x):=\{\begin{array}{cc}\pm 1,\hfill & x=\pm x_m,x_m:=\frac{q^m}{\sqrt{1-q}},\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

Set $x=\frac{q^n}{\sqrt{1-q}}$. Hence,

$$\begin{array}{rl}{\delta }_q^{2k}f(x)=& 4\frac{{(-1)}^k\sqrt{1-q}}{C}{q}^{-2mk}{(1-q)}^{-k-\frac{1}{2}}\frac{{(q^{-2k},q^{2n-2m+2},q^2;q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m-2k},q,q;q^2)}_{\mathrm{\infty }}}\\ {\times }_2{\varphi }_1(q^{2n-2m-2k},q^{-2k-2};q^{2n-2m+2};q^2,q^{2k+3})\\ =& q^{k^2+k+m-n}{(1-q)}^{-k-\frac{1}{2}}\frac{{(q^{-2k-1};q)}_{m-n+k}}{{(q;q)}_{m-n+k}}.\end{array}$$

Since

$$\frac{{(q^{-2k},q^{2n-2m+2};q^2)}_{\mathrm{\infty }}}{{(q^{2n-2m-2k};q^2)}_{\mathrm{\infty }}}=0\text{if }n\ge m+k\text{ or }n<m.$$

Hence, if $m\le n\le m+k-1$, we obtain

$$\begin{array}{r}{}_2\varphi _1(q^{2n-2m-2k},q^{-2k-2};q^{2n-2m+2};q^2,q^{2k+3})\\ ={(-1)}^k{q}^{k^2+k+(m-n)(2k+1)}\frac{{(q^{-2k-1};q)}_{m-n+k}}{{(q;q)}_{m-n+k}}\frac{{(q^2,q^{2n-2m-2k};q^2)}_{\mathrm{\infty }}}{{(q^{-2k},q^{2n-2m+2};q^2)}_{\mathrm{\infty }}}.\end{array}$$

5 Application of the $q^2$-analogue of the Fourier transform to solve q-fractional difference equations

In [18], Ho explored the possibility of using the classical Fourier and Mellin integral transforms to solve the class of q-difference differential equations

$$D_{q,t}^{n}u(x,t)=\frac{{\partial }^2}{\partial x^2}u(x,t),x\in \mathbb{R},t>0,n\ge 1,$$

(5.1)

with the initial conditions

$$y(x,0)=f(x),D_{q,t}^{k}y(x,t){|}_{t=0^{+}}=g_k(x)(k=1,\dots ,n-1),$$

where the functions $f(x)$ and $g_k(x)$ are assumed to vanish as $x\to \pm \mathrm{\infty }$. In [19] Brahim and Quanes used the $q^2$-Fourier transform and the q-Mellin transform to solve equation (5.1) in case of $n=1,2$, and only for q satisfying condition (3.2). In this section, we use the $q^2$-Fourier transform with the ${}_{q}L_s$ transform to solve the q-fractional diffusion equation

$$\begin{array}{r}{D}_{q,t}^{\alpha }u(x,t)=\lambda {\partial }_{q,x}^{2}u(x,t),\\ x\in {\tilde{\mathbb{R}}}_q,t\in \mathbb{R},0\le \alpha <1,0<q<1,\end{array}$$

(5.2)

with the initial conditions

$$I_{q,t}^{1-\alpha }u(x,t){|}_{t=0^{+}}=\varphi (x),{\partial }_{q,x}^{k}u(x,t)\in L_q^1({\tilde{\mathbb{R}}}_q)(k=0,1),$$

(5.3)

$$\varphi \in (L_q^1\cap L_q^2)({\tilde{\mathbb{R}}}_q).$$

(5.4)

Theorem 5.1 The solution of q-fractional diffusion equation (5.2) subject to the initial conditions (5.3)-(5.4) is given by

$$u(x,t)={\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}[T_{y}G(x,t)](x)\varphi (y)d_{q}y,$$

where

$$[T_{y}G(x,t)](x)=\frac{{(1+q)}^{1/2}}{2{\mathrm{\Gamma }}_{q^2}(\frac{1}{2})}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}e(-iy\xi ;q^2)g(\xi ,t)e(ix\xi ;q^2)d_q\xi$$

and

$$g(\xi ,t):=\{\begin{array}{cc}{t}^{\alpha -1}{e}_{\alpha ,\alpha }(-\lambda {\xi }^2{t}^{\alpha };q),\hfill & |\lambda {\xi }^2{t}^{\alpha }|<\frac{1}{{(1-q)}^{\alpha }},\hfill \\ 0,\hfill & \mathit{\text{otherwise}},\hfill \end{array}$$

where, in general, $e_{\alpha ,\beta }(x;q)$ is the q-analogue of the q-Mittag-Leffler function defined for $Re(\alpha >0)$ and $\beta \in \mathbb{C}$ by

$$e_{\alpha ,\beta }(x;q)=\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{{\mathrm{\Gamma }}_q(\alpha k+\beta )},\left|x{(1-q)}^{\alpha }\right|<1.$$

Proof First we calculate the $q^2$-Fourier transform of (5.2) with respect to the variable x. Hence, applying (3.12) yields

$$D_{q,t}^{\alpha }U(\xi ,t)=-\lambda {\xi }^{2}U(\xi ,t),$$

(5.5)

where

$$U(\xi ,t):={\mathcal{F}}_{q,x}(u(x,t))(\xi ).$$

Now we calculate the ${}_{q}L_s$ transform of (5.5) with respect to the variable t. Using (1.16) we obtain

$$(P^{\alpha }+\lambda {\xi }^2)V(\xi ,s)=\frac{{\mathcal{F}}_q(\varphi )(\xi )}{1-q},$$

(5.6)

where

$$V(\xi ,s){=}_{q,t}{L}_s(U(\xi ,t))(s),p=\frac{s}{1-q}.$$

One can verify that

$${}_{q,t}L_s\left(t^{\alpha -1}{e}_{\alpha ,\alpha }(-\lambda {\xi }^2{t}^{\alpha };q)\right)=\frac{1}{1-q}\frac{1}{p^{\alpha }+\lambda {\xi }^2}.$$

Consequently,

$$U(\xi ,t)={\mathcal{F}}_q(\varphi )(\xi )t^{\alpha -1}{e}_{\alpha ,\alpha }(-\lambda {\xi }^2{t}^{\alpha }),\left|\lambda {\xi }^2{t}^{\alpha }\right|<\frac{1}{{(1-q)}^{\alpha }}.$$

It follows from the inversion formula of the $q^2$-Fourier transform that

$$\begin{array}{c}{t}^{\alpha -1}{e}_{\alpha ,\alpha }(-\lambda {\xi }^2{t}^{\alpha };q)={\mathcal{F}}_{q,x}(G(x,t))(\xi ),\hfill \\ G(x,t)=\frac{\sqrt{1+q}}{2{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}{t}^{\alpha -1}{e}_{\alpha ,\alpha }(-\lambda {\xi }^2{t}^{\alpha };q)e(i\xi x;q^2)d_q\xi ,\hfill \end{array}$$

where the variable of the q-integration ξ runs only over all $\xi \in {\tilde{\mathbb{R}}}_q$ such that

$$\left|\lambda {\xi }^2{t}^{\alpha }\right|<\frac{1}{{(1-q)}^{\alpha }}.$$

Consequently,

$$u(x,t)=\varphi (x)\ast G(x,t).$$

Applying the $q^2$-Fourier convolution formula gives

$$u(x,t)={\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}[T_{y}G(x,t)](x)\varphi (y)d_{q}y,$$

where

$$[T_{y}G(x,t)](x)=\frac{\sqrt{1+q}}{2{\mathrm{\Gamma }}_{q^2}(1/2)}{\int }_{-\mathrm{\infty }/\sqrt{1-q}}^{\mathrm{\infty }/\sqrt{1-q}}e(-iy\xi ;q^2)t^{\alpha -1}{e}_{\alpha ,\alpha }(-\lambda {\xi }^2{t}^{\alpha };q)e(ix\xi ;q^2)d_q\xi .$$

 □