Skip to main content

Theory and Modern Applications

On basic Horn hypergeometric functions \(\mathbf{H}_{3}\) and \(\mathbf{H}_{4}\)

Abstract

The purpose of this work is to demonstrate several interesting contiguous function relations and q-differential formulas for basic Horn hypergeometric functions \(\mathbf{H}_{3}\) and \(\mathbf{H}_{4}\). Some properties of our main results are also constructed.

1 Introduction

Quantum calculus (or, q-calculus) is the study of calculus without limits. Many extensions of q-calculus have been developed and applied as one of the most active areas of research in mathematics and physics. These new extensions have proved to be very useful in various fields such as physics, engineering, statistics, actuarial sciences, economics, survival analysis, life checking out and telecommunications, and many others (see for example [17, 23, 24, 31, 33]). The applications have largely stimulated our present study. One of the most important branches of q-calculus is q-special functions. Jackson [19, 20], Andrews [4, 5], Gupta [16], Agarwal [2, 3], Ismail and Libis [18], Jain [21, 22], Jain and Vertna [23], Mishra [24], Sahai and Verma [25–27], Srivastava [30], Srivastava and Jain [31], Swarttouw [33], Verma and Sahai [34] introduced and discussed some interesting properties for various families of the basic Appell series, basic hypergeometric series, and q-Lauricella series by applying certain operators of q-calculus and its applications. Acikgoz et al. [1] and Araci et al. [6, 7] introduced a class of q-Euler, q-Frobenius–Euler, and q-Bernoulli polynomials based on q-exponential functions. Duran et al. [13, 14] introduced q-Bernoulli, q-Euler, and q-Genocchi polynomials and obtained the q-analogues of familiar earlier formulas and identities. In [8], Bagdasaryan et al. constructed Apostol q-Bernoulli, Apostol q-Genocchi, and Apostol q-Euler polynomials. Bansal and Choi [9], Bansal and Kumar [10], and Bansal et al. [11, 12] introduced and investigated the Pathway fractional integral formulas, fractional integral operators and integral transform of incomplete H-functions, incomplete ℵ-functions, incomplete I-functions, and S-generalized Gauss hypergeometric function. In [28, 29], Shehata investigated and discussed the generating functions for \((p,q)\)-Bessel and \((p,q)\)-Humbert functions.

Throughout this paper, we observe the following notations \(0<|q|<1\), \(q\in \mathbb{C}-\{1\}\). Let \(\mathbb{N}\) and \(\mathbb{C}\) be the sets of natural numbers and complex numbers, respectively.

The q-number (basic or quantum number) \([\beta ]_{q}\) is defined by [5, 15, 17]

$$\begin{aligned}{} [\beta ]_{q}=\frac{1-q^{\beta }}{1-q},\quad \beta \in \mathbb{C}, 0< \vert q \vert < 1, q \in \mathbb{C}-\{1\}. \end{aligned}$$
(1.1)

The q-number and q-factorial are given by

$$\begin{aligned}{} [n]_{q}=\textstyle\begin{cases} \frac{1-q^{n}}{1-q}, & n\in \mathbb{N}, 0< \vert q \vert < 1, q\in \mathbb{C}-\{1\}; \\ 1, & n=0, \end{cases}\displaystyle \end{aligned}$$
(1.2)

and

$$\begin{aligned} \begin{aligned}{}[n]_{q}!=\textstyle\begin{cases} \prod_{r=1}^{n}[r]_{q}=[n]_{q}[n-1]_{q}\ldots [2]_{q}[1]_{q}= \frac{(q;q)_{n}}{(1-q)^{n}}, \\ \hphantom{1, \quad\,} n\in \mathbb{N}, 0< \vert q \vert < 1, q\in \mathbb{C}-\{1\}; \\ 1, \quad n=0, 0< \vert q \vert < 1, q\in \mathbb{C}-\{1\}, \end{cases}\displaystyle \end{aligned} \end{aligned}$$
(1.3)

where \((\beta ;q)_{n}\) is the q-shifted factorial (q-Pochhammer symbol) which is definoted as follows: for \(n\in \mathbb{N}\), \(\beta \in \mathbb{C}\setminus \{1, q^{-1}, q^{-2},\ldots ,q^{1-n}\}\), \(0<|q|<1\), \(q\in \mathbb{C}-\{1\}\);

$$\begin{aligned}& \begin{aligned} &(\beta ;q)_{n}=\prod _{r=0}^{n-1}\bigl(1-\beta q^{r} \bigr)=(1- \beta ) (1-\beta q)\ldots \bigl(1-\beta q^{n-1}\bigr), \\ &(\beta ;q)_{0}=1,\quad \beta \in \mathbb{C}, 0< \vert q \vert < 1, q \in \mathbb{C}-\{1 \}. \end{aligned} \end{aligned}$$
(1.4)

Note that taking limit as tends to (1.1) in the above relations gives the shifted factorial \((\beta )_{n}\) (see [32])

$$ \begin{aligned} \lim_{q\longrightarrow 1^{-}} \frac{(q^{\beta };q)_{n}}{(1-q)^{n}}&=( \beta )_{n}. \end{aligned} $$

We recall some notations and defifinitions from q-calculus for \(\beta \in \mathbb{C}\), \(n\in \mathbb{N}\), \(0<|q|<1\), \(q\in \mathbb{C}-\{1\}\), which are essential in the sequel (see [15]):

$$\begin{aligned}& \begin{aligned} &(\beta q;q)_{n}=\frac{1-\beta q^{n}}{1-\beta }( \beta ;q)_{n} \\ &\hphantom{(\beta q;q)_{n}}=\bigl(1-\beta q^{n-1}\bigr) (\beta ;q)_{n-1}, \end{aligned} \end{aligned}$$
(1.5)
$$\begin{aligned}& \begin{aligned} &\bigl(\beta q^{-1};q\bigr)_{n}= \frac{1-\beta q^{-1}}{1-\beta q^{n-1}}( \beta ;q)_{n} \\ &\hphantom{\bigl(\beta q^{-1};q\bigr)_{n}}=\bigl(1-\beta q^{-1}\bigr) (\beta ;q)_{n-1}, \end{aligned} \end{aligned}$$
(1.6)

and

$$\begin{aligned} \begin{aligned} &(\beta ;q)_{n+k}=(\beta ;q)_{n} \bigl(\beta q^{n};q\bigr)_{k} \\ &\hphantom{(\beta ;q)_{n+k}}=(\beta ;q)_{k}\bigl(\beta q^{k};q \bigr)_{n}. \end{aligned} \end{aligned}$$
(1.7)

Definition 1.1

Based on q-Pochhammer’ symbol (1.4), we define the basic Horn hypergeometric functions \(\mathbf{H}_{3}\) and \(\mathbf{H}_{4}\) as follows:

$$\begin{aligned} \mathbf{H}_{3}(a,b;c;q,x,y)=\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n},\quad c \neq 1, q^{-1}, q^{-2},\ldots , \end{aligned}$$
(1.8)

and

$$\begin{aligned}& \mathbf{H}_{4}(a,b;c,d;q,x,y) \\& \quad =\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n},\quad c,d \neq 1, q^{-1}, q^{-2},\ldots . \end{aligned}$$
(1.9)

Remark 1.1

If \(q\rightarrow 1\), the basic Horn hypergeometric functions reduce to the Horn hypergeometric functions defined in [[32], p. 56, Eq. (27), p. 57, Eq. (28)].

To simplify the notation, we write \(\mathbf{H}_{3}\) for the series \(\mathbf{H}_{3}(a,b;c;q,x,y)\), \(\mathbf{H}_{3}(aq^{\pm 1})\) for the series \(\mathbf{H}_{3}(aq^{\pm 1},b;c;q,x,y)\), \(\mathbf{H}_{4}\) for the series \(\mathbf{H}_{4}\), … , and \(\mathbf{H}_{4}(cq^{\pm 1})\) stands for the series \(\mathbf{H}_{4}(a;b,cq^{\pm 1};q,x,y)\).

For a wide variety of other investigations involving basic Horn hypergeometric functions, see, for instance, [2, 3, 5, 30]. Motivated by the previous works in q-analysis (see [25, 26, 34]), in this paper we introduce a class of new extended forms of the basic Horn hypergeometric function. Our study can be detailed as follows: In Sect. 2, we introduce and study some contiguous functions relations and q-differential formulas for our considered basic Horn hypergeometric functions \(\mathbf{H}_{3}\) by permuting parameters. We discuss some family relations between basic Horn hypergeometric functions \(\mathbf{H}_{4}\) in Sect. 3. Finally, we discuss our main results and related results involving contiguous relations for \(\mathbf{H}_{3}\) and \(\mathbf{H}_{4}\) in Sect. 4.

2 Contiguous functions for \(\mathbf{H}_{3}(a,b;c;q,x,y)\)

Here we derive several properties as well as the contiguous function relations for \(\mathbf{H}_{3}\) with \(c\neq 1, q^{-1}, q^{-2},\ldots \) .

Theorem 2.1

For \(c\neq 1\), the contiguous relations of \(\mathbf{H}_{3}\) hold true for the numerator parameter a:

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(aq)=\mathbf{H}_{3}+ \frac{ax(1-aq)}{1-c} \mathbf{H}_{3}\bigl(aq^{2},b;cq;q,x,y \bigr) \\ &\hphantom{\mathbf{H}_{3}(aq)=}{}+\frac{axq(1-aq)}{1-c}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,xq,y \bigr) \\ &\hphantom{\mathbf{H}_{3}(aq)=}{}+ \frac{ay(1-b)}{1-c}\mathbf{H}_{3}\bigl(aq,bq;cq;q,xq^{2},y \bigr), \end{aligned} \end{aligned}$$
(2.1)
$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(aq)=\mathbf{H}_{3}+ \frac{ay(1-b)}{1-c} \mathbf{H}_{3}(aq,bq;cq;q,x,y) \\ &\hphantom{\mathbf{H}_{3}(aq)=}{}+\frac{ax(1-aq)}{1-c}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,x,yq \bigr) \\ &\hphantom{\mathbf{H}_{3}(aq)=}{}+ \frac{axq(1-aq)}{1-c}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,xq,yq \bigr), \end{aligned} \end{aligned}$$
(2.2)
$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}\bigl(aq^{-1} \bigr)=\mathbf{H}_{3}- \frac{ax(1-a)}{q(1-c)}\mathbf{H}_{3}(aq,b;cq;q,x,y) \\ &\hphantom{\mathbf{H}_{3}\bigl(aq^{-1} \bigr)=}{}-\frac{ax(1-a)}{1-c}\mathbf{H}_{3}(aq,b;cq;q,xq,y)- \frac{ay(1-b)}{q(1-c)}\mathbf{H}_{3}\bigl(a,bq;cq;q,xq^{2},y \bigr), \end{aligned} \end{aligned}$$
(2.3)

and

$$\begin{aligned} \begin{aligned} &\mathbf{H}_{3}\bigl(aq^{-1} \bigr)=\mathbf{H}_{3}- \frac{ay(1-b)}{q(1-c)}\mathbf{H}_{3}(a,bq;cq;q,x,y) \\ &\hphantom{\mathbf{H}_{3}\bigl(aq^{-1} \bigr)=}{}-\frac{ax(1-a)}{q(1-c)}\mathbf{H}_{3}(aq,b;cq;q,x,yq) \\ &\hphantom{\mathbf{H}_{3}\bigl(aq^{-1} \bigr)=}{}- \frac{ax(1-a)}{1-c}\mathbf{H}_{3}(aq,b;cq;q,xq,yq). \end{aligned} \end{aligned}$$
(2.4)

Proof

Replacing a by aq in (1.8), we get

$$\begin{aligned} \begin{aligned} &\mathbf{H}_{3}(aq)-\mathbf{H}_{3}= \sum_{m,n=0}^{\infty } \frac{(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}} { \bigl[}(aq;q)_{2m+n}-(a;q)_{2m+n} \bigr]x^{m}y^{n}. \end{aligned} \end{aligned}$$

The relations

$$\begin{aligned} \begin{aligned} (a;q)_{2m+n}= {(}1-a {)}(aq;q)_{2m+n-1} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} (aq;q)_{2m+n}= {\bigl(}1-aq^{2m+n} {\bigr)}(aq;q)_{2m+n-1}, \end{aligned} \end{aligned}$$

imply

$$\begin{aligned} &\mathbf{H}_{3}(aq)-\mathbf{H}_{3} \\ &\quad = \sum_{m,n=0}^{\infty } \frac{a(1-q^{2m+n})(aq;q)_{2m+n-1}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ &\quad =\sum_{m,n=0}^{\infty } \frac{a(1-q^{m}+q^{m}(1-q^{m})+q^{2m}(1-q^{n}))(aq;q)_{2m+n-1}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ &\quad =\sum_{m=1,n=0}^{\infty } \frac{a(aq;q)_{2m+n-1}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n}+ \sum_{m=1,n=0}^{\infty } \frac{aq^{m}(aq;q)_{2m+n-1}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\ &\qquad {}+\sum_{m=0,n=1}^{\infty } \frac{aq^{2m}(aq;q)_{2m+n-1}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n-1}}x^{m}y^{n} \\ &\quad =\frac{ax(1-aq)}{1-c}\sum_{m,n=0}^{\infty } \frac{(aq^{2};q)_{2m+n}(b;q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ &\qquad {}+ \frac{axq(1-aq)}{1-c} \sum_{m,n=0}^{\infty } \frac{q^{m}(aq^{2};q)_{2m+n}(b;q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ &\qquad {}+\frac{ay(1-b)}{1-c}\sum_{m,n=0}^{\infty } \frac{q^{2m}(aq;q)_{2m+n}(bq;q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ &\quad =\frac{ax(1-aq)}{1-c}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,x,y \bigr)+ \frac{axq(1-aq)}{1-c}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,xq,y \bigr) \\ &\qquad {}+ \frac{ay(1-b)}{1-c}\mathbf{H}_{3}\bigl(aq,bq;cq;q,xq^{2},y \bigr),\quad c\neq 1. \end{aligned}$$

By using the relation \(1-q^{2m+n}=1-q^{n}+q^{n}(1-q^{m})+q^{m+n}(1-q^{m})\), we get (2.2). Performing the replacement \(a\rightarrow aq^{-1}\) in the contiguous relations (2.1) and (2.2), we obtain (2.3) and (2.4). □

Theorem 2.2

For \(c\neq 1\), \(\mathbf{H}_{3}\) satisfies the derivative equations

$$\begin{aligned} \begin{aligned} &\theta _{x,q}\mathbf{H}_{3}=x \frac{(1-a)(1-aq)}{(1-c)(1-q)}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,x,y \bigr) \end{aligned} \end{aligned}$$
(2.5)

and

$$\begin{aligned} \begin{aligned} &\theta _{y,q}\mathbf{H}_{3}=y \frac{(1-a)(1-b)}{(1-c)(1-q)}\mathbf{H}_{3}(aq,bq;cq;q,x,y). \end{aligned} \end{aligned}$$
(2.6)

Proof

From (1.8), we consider the operators \(\theta _{x,q}=x\frac{\partial }{\partial x}=xD_{x,q}\) and \(\theta _{y,q}=y\frac{\partial }{\partial y}=yD_{y,q}\) to get

$$\begin{aligned} \theta _{x,q}\mathbf{H}_{3} =& \sum_{m,n=0}^{\infty } \biggl[ \frac{1-q^{m}}{1-q} \biggr] \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =&\sum_{m=1,n=0}^{\infty } \biggl[ \frac{1}{1-q} \biggr] \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\ =& \sum_{m,n=0}^{\infty } \biggl[ \frac{1}{1-q} \biggr] \frac{(a;q)_{2m+n+2}(b;q)_{n}}{(c;q)_{m+n+1}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\ =&x\frac{(1-a)(1-aq)}{(1-c)(1-q)}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,x,y \bigr),\quad c \neq 1, \end{aligned}$$

and

$$\begin{aligned} \theta _{y,q}\mathbf{H}_{3} =& \sum_{m,n=0}^{\infty } \biggl[ \frac{1-q^{n}}{1-q} \biggr] \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =&\sum_{m=0,n=1}^{\infty } \biggl[ \frac{1}{1-q} \biggr] \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n-1}}x^{m}y^{n} \\ =& \sum_{m,n=0}^{\infty } \biggl[ \frac{1}{1-q} \biggr] \frac{(a;q)_{2m+n+1}(b;q)_{n+1}}{(c;q)_{m+n+1}(q;q)_{m}(q;q)_{n}}x^{m}y^{n+1} \\ =&y\frac{(1-a)(1-b)}{(1-c)(1-q)}\mathbf{H}_{3}(aq,bq;cq;q,x,y),\quad c\neq 1. \end{aligned}$$

 □

Theorem 2.3

\(\mathbf{H}_{3}\) satisfies the difference equations

$$\begin{aligned}& \biggl[a\theta _{x,q}+ \frac{1-a}{1-q} \biggr] \mathbf{H}_{3}+a\theta _{x,q}\mathbf{H}_{3}(xq)+a\theta _{y,q} \mathbf{H}_{3}\bigl(xq^{2}\bigr)=\frac{1-a}{1-q} \mathbf{H}_{3}(aq), \end{aligned}$$
(2.7)
$$\begin{aligned}& \biggl[a\theta _{y,q}+ \frac{1-a}{1-q} \biggr] \mathbf{H}_{3}+a\theta _{x,q}\mathbf{H}_{3}(yq)+a\theta _{x,q} \mathbf{H}_{3}(xq,yq)=\frac{1-a}{1-q}\mathbf{H}_{3}(aq), \end{aligned}$$
(2.8)
$$\begin{aligned}& \biggl[aq^{-1}\theta _{x,q}+ \frac{1-aq^{-1}}{1-q} \biggr]\mathbf{H}_{3}\bigl(aq^{-1} \bigr)+aq^{-1}\theta _{y,q}\mathbf{H}_{3} \bigl(aq^{-1},xq\bigr)+aq^{-1} \theta _{y,q} \mathbf{H}_{3}\bigl(aq^{-1},xq,yq\bigr) \\& \quad = \frac{1-aq^{-1}}{1-q} \mathbf{H}_{3}, \end{aligned}$$
(2.9)

and

$$\begin{aligned}& \biggl[aq^{-1}\theta _{y,q}+ \frac{1-aq^{-1}}{1-q} \biggr]\mathbf{H}_{3}\bigl(aq^{-1} \bigr)+aq^{-1}\theta _{x,q}\mathbf{H}_{3} \bigl(aq^{-1},yq\bigr) \\& \qquad {}+aq^{-1} \theta _{x,q} \mathbf{H}_{3}\bigl(aq^{-1},xq,yq\bigr) \\& \quad = \frac{1-aq^{-1}}{1-q} \mathbf{H}_{3}. \end{aligned}$$
(2.10)

Proof

With the help of the above differential operators and using (2.5) and (2.6) for \(\mathbf{H}_{3}\), we get the results (2.7)–(2.10). □

Theorem 2.4

For \(c\neq 1\), the contiguous function relations of \(\mathbf{H}_{3}\) with the numerator parameter b give

$$\begin{aligned} \begin{aligned} &\mathbf{H}_{3}(bq)=\mathbf{H}_{3}+ \frac{by(1-a)}{1-c} \mathbf{H}_{3}(aq,bq;cq;q,x,y) \end{aligned} \end{aligned}$$
(2.11)

and

$$\begin{aligned} \begin{aligned} &\mathbf{H}_{3}\bigl(bq^{-1} \bigr)=\mathbf{H}_{3}- \frac{by(1-a)}{q(1-c)}\mathbf{H}_{3}(aq,b;cq;q,x,y). \end{aligned} \end{aligned}$$
(2.12)

Proof

If we replace b by bq in (1.8), we get

$$\begin{aligned} \begin{aligned} &\mathbf{H}_{3}(bq)-\mathbf{H}_{3}= \sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}} { \bigl[}(bq;q)_{n}-(b;q)_{n} \bigr]x^{m}y^{n}. \end{aligned} \end{aligned}$$

Using the relations

$$\begin{aligned} (b;q)_{n}= {(}1-b {)}(bq;q)_{n-1} \end{aligned}$$

and

$$\begin{aligned} (bq;q)_{n}= {\bigl(}1-bq^{n} {\bigr)}(bq;q)_{n-1}, \end{aligned}$$

we have

$$\begin{aligned} \mathbf{H}_{3}(bq)-\mathbf{H}_{3} =& \sum_{m,n=0}^{\infty } \frac{b(1-q^{n})(a;q)_{2m+n}(bq;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =&\sum_{m=0,n=1}^{\infty } \frac{b(a;q)_{2m+n}(bq;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n-1}}x^{m}y^{n} \\ =& \frac{by(1-a)}{1-c}\sum_{m,n=0}^{\infty } \frac{(aq;q)_{2m+n}(bq;q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =&\frac{by(1-a)}{1-c}\mathbf{H}_{3}(aq,bq;cq;q,x,y),\quad c\neq 1. \end{aligned}$$

Replacing \(b\rightarrow bq^{-1}\) in relation (2.11), we obtain (2.12). □

Theorem 2.5

The difference equations hold true for \(\mathbf{H}_{3}\):

$$\begin{aligned} \begin{aligned} & \biggl[b\theta _{y,q}+ \frac{1-b}{1-q} \biggr] \mathbf{H}_{3}=\frac{1-b}{1-q} \mathbf{H}_{3}(bq) \end{aligned} \end{aligned}$$
(2.13)

and

$$\begin{aligned} \begin{aligned} & \biggl[bq^{-1}\theta _{y,q}+ \frac{1-bq^{-1}}{1-q} \biggr]\mathbf{H}_{3}\bigl(bq^{-1} \bigr)=\frac{1-bq^{-1}}{1-q}\mathbf{H}_{3}. \end{aligned} \end{aligned}$$
(2.14)

Proof

From (2.5) and (2.6), we get (2.13) and (2.14). □

Theorem 2.6

For \(c\neq 1\), \(b\neq q\), the formulas hold true for \(\mathbf{H}_{3}\):

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}\bigl(aq,bq^{-1};c;q,x,y \bigr) \\ &\quad =\mathbf{H}_{3}+ \frac{ay}{1-c}\mathbf{H}_{3}(aq,b;c,dq;q,x,y) \\ &\qquad {}+\frac{ax(1-aq)}{1-bq^{-1}}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,x,yq \bigr) \\ &\qquad {}+ \frac{axq(1-aq)}{1-bq^{-1}}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,xq,yq \bigr) \\ &\qquad {}-\frac{abq^{-1}x(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,x,yq \bigr) \\ &\qquad {}- \frac{abx(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,xq,yq \bigr) \\ &\qquad {}-\frac{bq^{-1}y}{1-c}\mathbf{H}_{3}(aq,b;cq;q,x,y) \end{aligned} \end{aligned}$$
(2.15)

and

$$\begin{aligned}& \mathbf{H}_{3}\bigl(aq,bq^{-1};c;q,x,y \bigr) \\& \quad =\mathbf{H}_{3}+ \frac{ax(1-aq)}{(1-c)(1-bq^{-1})}\mathbf{H}_{3} \bigl(aq^{2},bq^{-1};cq;q,x,y\bigr) \\& \qquad {}+\frac{axq(1-aq)}{1-bq^{-1}}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,xq,y \bigr) \\& \qquad {}+ \frac{ay}{1-bq^{-1}}\mathbf{H}_{3}\bigl(aq^{2},bq;cq;q,xq^{2},y \bigr) \\& \qquad {}-\frac{abxq(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,x,yq \bigr) \\& \qquad {}- \frac{abx(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,xq,yq \bigr) \\& \qquad {}-\frac{bq^{-1}(1-aq)y}{1-c}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,x,y \bigr). \end{aligned}$$
(2.16)

Proof

Using (1.8), (1.5), (1.6), (1.7) and the relation \((bq^{-1};q)_{n}=(1-bq^{-1})(b;q)_{n-1}\) implies

$$\begin{aligned} &\mathbf{H}_{3}\bigl(aq,bq^{-1};c;q,x,y \bigr)-\mathbf{H}_{3} \\ &\quad =\sum_{m,n=0}^{\infty } \frac{(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}} \bigl[abq^{2m+n-1}-bq^{-1}-aq^{2m+n}+a-abq^{n-1}+bq^{n-1} \bigr]x^{m}y^{n} \\ &\quad =a\sum_{m=0,n=1}^{\infty } \frac{(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n-1}}x^{m}y^{n}+a \sum _{m=1,n=0}^{\infty } \frac{q^{n}(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\ &\qquad {}+a\sum_{m=1,n=0}^{\infty } \frac{q^{m+n}(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\ &\qquad {}-ab \sum _{m=1,n=0}^{\infty } \frac{q^{n-1}(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\ &\qquad {}-ab\sum_{m=1,n=0}^{\infty } \frac{q^{m+n-1}(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\ &\qquad {}-bq^{-1} \sum _{m=0,n=1}^{\infty } \frac{(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n-1}}x^{m}y^{n} \\ &\quad =\frac{a}{1-c}\sum_{m,n=0}^{\infty } \frac{(aq;q)_{2m+n}(b;q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n+1} \\ &\qquad {}+ \frac{a(1-aq)}{(1-bq^{-1})(1-c)} \sum_{m,n=0}^{\infty } \frac{q^{n}(aq^{2};q)_{2m+n}(bq^{-1};q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\ &\qquad {}+\frac{a(1-aq)}{(1-bq^{-1})(1-c)}\sum_{m,n=0}^{\infty } \frac{q^{m+n+1}(aq^{2};q)_{2m+n}(bq^{-1};q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\ &\qquad {}-\frac{ab}{(1-bq^{-1})(1-c)}\sum_{m,n=0}^{\infty } \frac{q^{n-1}(aq^{2};q)_{2m+n}(bq^{-1};q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\ &\qquad {}-\frac{ab(1-aq)}{(1-bq^{-1})(1-c)}\sum_{m,n=0}^{\infty } \frac{q^{m+n}(aq^{2};q)_{2m+n}(bq^{-1};q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\ &\qquad {}- \frac{bq^{-1}}{1-c} \sum_{m,n=0}^{\infty } \frac{(aq;q)_{2m+n}(b;q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n+1} \\ &\quad =\frac{ay}{1-c}\mathbf{H}_{3}(aq,b;cq;q,x,y)+ \frac{ax(1-aq)}{1-bq^{-1}}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,x,yq \bigr) \\ &\qquad {}+\frac{axq(1-aq)}{1-bq^{-1}}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,xq,yq \bigr) \\ &\qquad {}- \frac{abq^{-1}x(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,x,yq \bigr) \\ &\qquad {}-\frac{abx(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{3}\bigl(aq^{2},bq^{-1};cq;q,xq,yq \bigr) \\ &\qquad {}- \frac{bq^{-1}y}{1-c}\mathbf{H}_{3}(aq,b;cq;q,x,y),\quad c\neq 1,b \neq q. \end{aligned}$$

Using the relation \(a(1-q^{2m+n})-abq^{n-1}(1-q^{2m})-bq^{-1}(1-q^{n})=a(1-q^{m})+aq^{m}(1-q^{m})+aq^{2m}(1-q^{n}) -abq^{n-1}(1-q^{m})-abq^{m+n-1}(1-q^{m})-bq^{-1}(1-q^{n})\), we obtain (2.16). □

Theorem 2.7

Each of the following properties for \(\mathbf{H}_{3}\) in (1.8) holds true:

$$\begin{aligned}& b(1-a)\mathbf{H}_{3}(aq)-a(1-b) \mathbf{H}_{3}(bq) \\& \quad =(b-a)\mathbf{H}_{3}+\frac{abx(1-a)(1-aq)}{1-c}\mathbf{H}_{3} \bigl(aq^{2},b;cq;q,x,yq\bigr) \\& \qquad {}+\frac{abxq(1-a)(1-aq)}{1-c}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,xq,yq \bigr),\quad c \neq 1. \end{aligned}$$
(2.17)

Proof

From (1.8), we have

$$\begin{aligned}& b(1-a)\mathbf{H}_{3}(aq)-a(1-b) \mathbf{H}_{3}(bq) \\& \quad =\sum_{m,n=0}^{\infty } \frac{b(1-a)(aq;q)_{2m+n}(b;q)_{n}-a(1-b)(a;q)_{2m+n}(bq;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n}. \end{aligned}$$

By using the equation

$$\begin{aligned} (1-a) (aq;q)_{2m+n}=\bigl(1-aq^{2m+n}\bigr) (a;q)_{2m+n} \end{aligned}$$

and

$$\begin{aligned} (1-b) (bq;q)_{n}=\bigl(1-bq^{n}\bigr) (b;q)_{n}, \end{aligned}$$

we get

$$\begin{aligned}& b(1-a)\mathbf{H}_{3}(aq)-a(1-b) \mathbf{H}_{3}(bq) \\& \quad =\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(b;q)_{n}[b(1-aq^{2m+n})-a(1-bq^{n})]}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \quad =(b-a)\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}+ab \sum _{m,n=0}^{\infty } \frac{q^{n}(1-q^{m})(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(1-q^{m})(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}+ab\sum_{m,n=0}^{\infty } \frac{q^{n+m}(1-q^{m})(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(1-q^{m})(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\& \quad =(b-a)\mathbf{H}_{3}+ab\sum_{m,n=0}^{\infty } \frac{q^{n}(a;q)_{2m+n+2}(b;q)_{n}}{(c;q)_{m+n+1}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\& \qquad {}+ab\sum_{m,n=0}^{\infty } \frac{q^{n+m+1}(a;q)_{2m+n+2}(b;q)_{n}}{(c;q)_{m+n+1}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\& \quad =(b-a)\mathbf{H}_{3}+\frac{abx(1-a)(1-aq)}{1-c}\mathbf{H}_{3} \bigl(aq^{2},b;cq;q,x,yq\bigr) \\& \qquad {}+\frac{abxq(1-a)(1-aq)}{1-c}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,xq,yq \bigr),\quad c \neq 1. \end{aligned}$$

 □

Theorem 2.8

The contiguous function relations of \(\mathbf{H}_{3}\) with denominator parameter c are valid:

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}\bigl(cq^{-1} \bigr)=\mathbf{H}_{3}+ \frac{cx(1-a)(1-aq)}{(q-c)(1-c)}\mathbf{H}_{3} \bigl(aq^{2},b;cq;q,x,y\bigr) \\ &\hphantom{\mathbf{H}_{3}\bigl(cq^{-1} \bigr)=}{}+\frac{cy(1-a)(1-b)}{(q-c)(1-c)}\mathbf{H}_{3}(aq,bq;cq;q,xq,y),\quad c \neq 1, q, \end{aligned} \end{aligned}$$
(2.18)
$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}\bigl(cq^{-1} \bigr)=\mathbf{H}_{3}+ \frac{cy(1-a)(1-b)}{(q-c)(1-c)}\mathbf{H}_{3}(aq,bq;cq;q,x,y) \\ &\hphantom{\mathbf{H}_{3}\bigl(cq^{-1} \bigr)=}{}+\frac{cx(1-a)(1-aq)(1-b)}{(q-c)(1-c)}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,x,yq \bigr),\quad c \neq 1,q, \end{aligned} \end{aligned}$$
(2.19)
$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(cq)=\mathbf{H}_{3}- \frac{cx(1-a)(1-aq)}{(1-cq)(1-c)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr) \\ &\hphantom{\mathbf{H}_{3}(cq)=}{}-\frac{cy(1-a)(1-b)}{(1-cq)(1-c)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq,y \bigr),\quad c \neq 1,q^{-1}, \end{aligned} \end{aligned}$$
(2.20)

and

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(cq)=\mathbf{H}_{3}- \frac{cy(1-a)(1-b)}{(1-cq)(1-c)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,x,y \bigr) \\ &\hphantom{\mathbf{H}_{3}(cq)=}{}-\frac{cx(1-a)(1-aq)}{(1-cq)(1-c)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,yq \bigr),\quad c \neq 1,q^{-1}. \end{aligned} \end{aligned}$$
(2.21)

Proof

By the definition of basic Horn function, we get

$$\begin{aligned} \mathbf{H}_{3}\bigl(cq^{-1} \bigr)-\mathbf{H}_{3} =&\sum_{m,n=0}^{ \infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(q;q)_{m}(q;q)_{n}} \biggl[ \frac{1}{(cq^{-1};q)_{m+n}}-\frac{1}{(c;q)_{m+n}} { \biggr]}x^{m}y^{n} \\ =&\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(q;q)_{m}(q;q)_{n}} \frac{(c;q)_{m+n}-(cq^{-1};q)_{m+n}}{(cq^{-1};q)_{m+n}(c;q)_{m+n}}x^{m}y^{n}. \end{aligned}$$

Using

$$\begin{aligned} \bigl(cq^{-1};q\bigr)_{m+n}= {\bigl(}1-cq^{-1} {\bigr)}(c;q)_{m+n-1} \end{aligned}$$

and

$$\begin{aligned} (c;q)_{m+n}= {\bigl(}1-cq^{m+n-1} {\bigr)}(c;q)_{m+n-1}, \end{aligned}$$

we can rewrite the above equation as follows:

$$\begin{aligned}& \mathbf{H}_{3}\bigl(cq^{-1} \bigr)-\mathbf{H}_{3} \\& \quad =\sum_{m,n=0}^{ \infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(q;q)_{m}(q;q)_{n}} \frac{ {[}1-q^{m+n} {]}cq^{-1}(c;q)_{m+n-1}}{ {[}1-cq^{-1} {]}(c;q)_{m+n-1}(c;q)_{m+n}}x^{m}y^{n} \\& \quad =c\sum_{m,n=0}^{\infty }\frac{1-q^{m}}{q-c} \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}+c \sum _{m,n=0}^{\infty }q^{m}\frac{1-q^{n}}{q-c} \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \quad =\frac{c}{q-c}\sum_{m,n=0}^{\infty } \frac{(1-a)(1-aq)(aq^{2};q)_{2m+n}(b;q)_{n}}{(1-c)(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\& \qquad {}+ \frac{c}{q-c} \sum_{m,n=0}^{\infty }q^{m} \frac{(1-a)(a;q)_{2m+n}(1-b)(b;q)_{n}}{(1-c)(c;q)_{m+n+1}(q;q)_{m}(q;q)_{n}}x^{m}y^{n+1} \\& \quad =\frac{cx(1-a)(1-aq)}{(q-c)(1-c)}\mathbf{H}_{3}\bigl(aq^{2},b;cq;q,x,y \bigr) \\& \qquad {}+ \frac{cy(1-a)(1-b)}{(q-c)(1-c)}\mathbf{H}_{3}(aq,bq;cq;q,xq,y),\quad c\neq 1,q, \end{aligned}$$

which is the desired result. The proof of Eq. (2.19) can run parallel to Eq. (2.18), so details are omitted here.

If we replace \(c\rightarrow cq\) in relation (2.18), we obtain

$$\begin{aligned} \mathbf{H}_{3}(cq) =&\mathbf{H}_{3}- \frac{cx(1-a)(1-aq)}{(1-cq)(1-c)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr) \\ &{}-\frac{cy(1-a)(1-b)}{(1-cq)(1-c)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq,y \bigr),\quad c \neq 1,q^{-1}. \end{aligned}$$

The proof of Eq. (2.21) would run parallel to Eq. (2.20), so we may skip the involved details. □

Theorem 2.9

The derivative formulas for \(\mathbf{H}_{3}\) are satisfied:

$$\begin{aligned}& \mathbf{H}_{3}\bigl(cq^{-1} \bigr)=\frac{c}{c-q}\mathbf{H}_{3}(a,b;c;q,xq,yq)- \frac{q}{c-q}\mathbf{H}_{3},\quad c\neq q, \end{aligned}$$
(2.22)
$$\begin{aligned}& \mathbf{H}_{3}(cq)+(c-1) \mathbf{H}_{3}=c\mathbf{H}_{3}(a,b;cq;q,xq,yq), \end{aligned}$$
(2.23)
$$\begin{aligned}& \biggl[cq^{-1}\theta _{x,q}+ \frac{1-cq^{-1}}{1-q} \biggr]\mathbf{H}_{3}+cq^{-1} \theta _{y,q}\mathbf{H}_{3}(xq)= \frac{1-cq^{-1}}{1-q} \mathbf{H}_{3}\bigl(cq^{-1}\bigr), \end{aligned}$$
(2.24)

and

$$\begin{aligned}& \biggl[cq^{-1}\theta _{y,q}+ \frac{1-cq^{-1}}{1-q} \biggr]\mathbf{H}_{3}+cq^{-1} \theta _{x,q}\mathbf{H}_{3}(yq)= \frac{1-cq^{-1}}{1-q} \mathbf{H}_{3}\bigl(cq^{-1}\bigr). \end{aligned}$$
(2.25)

Proof

Using the definition \(\mathbf{H}_{3}\) in (1.8) with the relation \(\frac{1}{(cq^{-1};q)_{m+n}}=\frac{1}{(c;q)_{m+n}} {[} \frac{c}{c-q}q^{m+n}-\frac{q}{c-q} {]}\), we get

$$\begin{aligned} \mathbf{H}_{3}\bigl(cq^{-1} \bigr) =&\sum_{m,n=0}^{\infty } \biggl[ \frac{c}{c-q}q^{m+n}-\frac{q}{c-q} \biggr] \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =&\frac{c}{c-q}\mathbf{H}_{3}(a,b;c;q,xq,yq)- \frac{q}{c-q}\mathbf{H}_{3},\quad c \neq q. \end{aligned}$$

Replacing \(c=cq\) in (2.22) implies the contiguous relation

$$\begin{aligned} \mathbf{H}_{3}(cq)+(c-1) \mathbf{H}_{3} =&c\mathbf{H}_{3}(a,b;cq;q,xq,yq). \end{aligned}$$

By using Eqs. (2.5) and (2.6), we obtain the required results (2.23) and (2.24). □

Theorem 2.10

For \(c\neq 1\) and \(cq\neq 1\), the following formulas are valid:

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(aq,b;cq;q,x,y)- \mathbf{H}_{3} \\ &\quad =\frac{ax(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr) \\ &\qquad {} + \frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr) \\ &\qquad {}+\frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,xq,y \bigr) \\ &\qquad {}+ \frac{cy(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq,y \bigr) \\ &\qquad {}+\frac{ay(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq^{2},y \bigr) \\ &\qquad {}- \frac{acxq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},bq;cq^{2};q,xq,yq \bigr), \end{aligned} \end{aligned}$$
(2.26)
$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(aq,b;cq;q,x,y)- \mathbf{H}_{3} \\ &\quad =\frac{ax(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr) \\ &\qquad {}+ \frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,xq,y \bigr) \\ &\qquad {}+\frac{cy(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq^{2},y \bigr) \\ &\qquad {}+ \frac{ay(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq^{2},y \bigr) \\ &\qquad {}+\frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,yq \bigr) \\ &\qquad {}- \frac{acxq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,xq,yq \bigr), \end{aligned} \end{aligned}$$
(2.27)
$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(aq,b;cq;q,x,y)- \mathbf{H}_{3} \\ &\quad =\frac{ax}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,x,y \bigr) \\ &\qquad {}+ \frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,yq \bigr) \\ &\qquad {}+\frac{ax(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,xq,yq \bigr) \\ &\qquad {}+ \frac{cy(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr) \\ &\qquad {}+\frac{ay(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq,y \bigr) \\ &\qquad {}- \frac{acxq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},bq;cq^{2};q,xq,yq \bigr), \end{aligned} \end{aligned}$$
(2.28)

and

$$\begin{aligned} \begin{aligned} &\mathbf{H}_{3}(aq,b;cq;q,x,y)- \mathbf{H}_{3} \\ &\quad =\frac{ay(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,x,y \bigr) \\ &\qquad {}+ \frac{ax(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,yq \bigr) \\ &\qquad {}+\frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,xq,yq \bigr) \\ &\qquad {}+ \frac{cy(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,x,y \bigr) \\ &\qquad {}+\frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,yq \bigr) \\ &\qquad {}- \frac{acxq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},bq;cq^{2};q,xq,yq \bigr). \end{aligned} \end{aligned}$$
(2.29)

Proof

Replacing a and c by aq and cq in (1.8), we get

$$\begin{aligned}& \mathbf{H}_{3}(aq,b;cq;q,x,y)- \mathbf{H}_{3} \\& \quad =\sum_{m,n=0}^{ \infty } \biggl[ \frac{(aq;q)_{2m+n}(b;q)_{n}}{(cq;q)_{m+n}(q;q)_{m}(q;q)_{n}}- \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m+n}(q;q)_{m}(q;q)_{n}} \biggr]x^{m}y^{n} \\& \quad =\sum_{m,n=0}^{\infty } \frac{(aq;q)_{2m+n-1}(cq;q)_{m+n-1}(b;q)_{n}}{(q;q)_{m}(q;q)_{n}} \\& \qquad {}\times \biggl[ \frac{(1-aq^{2m+n})(1-c)-(1-a)(1-cq^{m+n})}{(1-cq^{m+n})(cq;q)_{m+n-1}(c;q)_{m+n}} \biggr]x^{m}y^{n} \\& \quad =\frac{ax(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr)+ \frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,xq,y \bigr) \\& \qquad {}+\frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr)+ \frac{cy(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq,y \bigr) \\& \qquad {}+\frac{ay(1-b)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq^{2},y \bigr) \\& \qquad {}- \frac{acxq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},bq;cq^{2};q,xq,yq \bigr),\quad c \neq 1,q^{-1}. \end{aligned}$$

By using the relations

$$\begin{aligned}& a\bigl(1-q^{2m+n}\bigr)+c \bigl(1-q^{m+n}\bigr)-acq^{m+n}\bigl(1-q^{m} \bigr) \\& \quad =a\bigl(1-q^{m}\bigr)+aq^{m}\bigl(1-q^{m} \bigr)+aq^{2m}\bigl(1-q^{n}\bigr)+c\bigl(1-q^{n} \bigr) \\& \qquad {}+cq^{n}\bigl(1-q^{m}\bigr)-acq^{m+n} \bigl(1-q^{m}\bigr), \\& \quad =a\bigl(1-q^{n}\bigr)+aq^{n}\bigl(1-q^{m} \bigr)+aq^{m+n}\bigl(1-q^{m}\bigr)+c\bigl(1-q^{m} \bigr) \\& \qquad {}+cq^{m}\bigl(1-q^{n}\bigr)-acq^{m+n} \bigl(1-q^{m}\bigr), \\& \quad =a\bigl(1-q^{n}\bigr)+aq^{n}\bigl(1-q^{m} \bigr)+aq^{m+n}\bigl(1-q^{m}\bigr)+c\bigl(1-q^{n} \bigr) \\& \qquad {}+cq^{n}\bigl(1-q^{m}\bigr)-acq^{m+n} \bigl(1-q^{m}\bigr). \end{aligned}$$

The proof of Eqs. (2.27)–(2.29) would run parallel to Eq. (2.26) by using the above relations, so we omit the involved details. □

Theorem 2.11

For \(\mathbf{H}_{3}\) defined by (1.8), each of the formulas holds:

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(a,bq;cq;q,x,y) \\ &\quad = \mathbf{H}_{3}+\frac{by(1-a)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,x,y \bigr) \\ &\qquad {}+ \frac{cx(1-a)(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr) \\ &\qquad {}+\frac{cy(1-a)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq,y \bigr) \\ &\qquad {}+ \frac{cbx(1-a)(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,yq \bigr) \end{aligned} \end{aligned}$$
(2.30)

and

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(a,bq;cq;q,x,y) \\ &\quad = \mathbf{H}_{3}+\frac{(b+c)y(1-a)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,x,y \bigr) \\ &\qquad {}+ \frac{c(b+1)x(1-a)(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,yq \bigr) \end{aligned} \end{aligned}$$
(2.31)

for \(c\neq 1\) and \(cq\neq 1\).

Proof

From (1.8), we have

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{3}(a,bq;cq;q,x,y)- \mathbf{H}_{3} \\ &\quad =\sum_{m,n=0}^{ \infty } \frac{(a;q)_{2m+n}}{(q;q)_{m}(q;q)_{n}} \biggl[ \frac{(bq;q)_{n}}{(cq;q)_{m+n}}-\frac{(b;q)_{n}}{(c;q)_{m+n}} { \biggr]}x^{m}y^{n} \\ &\quad =\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(cq;q)_{m+n-1}(bq;q)_{n-1}}{(q;q)_{m}(q;q)_{n}} \biggl[ \frac{(1-bq^{n})(1-c)-(1-b)(1-cq^{m+n})}{(1-cq^{m+n})(cq;q)_{m+n-1}(c;q)_{m+n}} \biggr]x^{m}y^{n} \\ &\quad =\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(bq;q)_{n-1}}{(q;q)_{m}(q;q)_{n}} \biggl[ \frac{b(1-q^{n})+c(1-q^{m+n})+bcq^{n}(1-q^{m})}{(1-c)(cq;q)_{m+n}} \biggr]x^{m}y^{n} \\ &\quad =\frac{by(1-a)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,x,y \bigr)+ \frac{cx(1-a)(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,y \bigr) \\ &\qquad {}+\frac{cy(1-a)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq,bq;cq^{2};q,xq,y \bigr) \\ &\qquad {}+ \frac{cbx(1-a)(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{3}\bigl(aq^{2},b;cq^{2};q,x,yq \bigr),\quad c \neq 1,cq\neq 1. \end{aligned} \end{aligned}$$

Hence, we obtain (2.30), one can derive the result (2.31) by a similar way. □

3 Contiguous functions for \(\mathbf{H}_{4}\)

Relying on a similar procedure as the one used in the previous section, we obtain the following list of results for basic Horn function \(\mathbf{H}_{4}(a,b;c,d;q,x,y)\) with \(c,d\neq 1, q^{-1}, q^{-2},\ldots \) .

Theorem 3.1

For \(c\neq 1\) and \(d\neq 1\), the contiguous function relations of \(\mathbf{H}_{4}\) with numerator parameter a hold true:

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{4}(aq)=\mathbf{H}_{4}+ \frac{ax(1-aq)}{1-c} \mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,x,y \bigr) \\ &\hphantom{\mathbf{H}_{4}(aq)=}{}+\frac{axq(1-aq)}{1-c}\mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,xq,y \bigr) \\ &\hphantom{\mathbf{H}_{4}(aq)=}{}+ \frac{ay(1-b)}{1-d}\mathbf{H}_{4}\bigl(aq,bq;c,dq;q,xq^{2},y \bigr), \end{aligned} \end{aligned}$$
(3.1)
$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{4}(aq)=\mathbf{H}_{4}+ \frac{ay(1-b)}{1-d} \mathbf{H}_{4}(aq,bq;c,dq;q,x,y) \\ &\hphantom{\mathbf{H}_{4}(aq)=}{}+\frac{ax(1-aq)}{1-c}\mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,x,yq \bigr) \\ &\hphantom{\mathbf{H}_{4}(aq)=}{}+ \frac{axq(1-aq)}{1-c}\mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,xq,yq \bigr), \end{aligned} \end{aligned}$$
(3.2)
$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{4}\bigl(aq^{-1} \bigr)=\mathbf{H}_{4}- \frac{aq^{-1}x(1-a)}{1-c}\mathbf{H}_{4}(aq,b;cq,d;q,x,y) \\ &\hphantom{\mathbf{H}_{4}\bigl(aq^{-1} \bigr)=}{}-\frac{ax(1-a)}{1-c}\mathbf{H}_{4}(aq,b;cq,d;q,xq,y) \\ &\hphantom{\mathbf{H}_{4}\bigl(aq^{-1} \bigr)=}{}- \frac{aq^{-1}y(1-b)}{1-d}\mathbf{H}_{4}\bigl(a,bq;c,dq;q,xq^{2},y \bigr), \end{aligned} \end{aligned}$$
(3.3)

and

$$\begin{aligned} \begin{aligned} &\mathbf{H}_{4}\bigl(aq^{-1} \bigr)=\mathbf{H}_{4}- \frac{ay(1-b)}{1-d}\mathbf{H}_{4}(a,bq;c,dq;q,x,y) \\ &\hphantom{\mathbf{H}_{4}\bigl(aq^{-1} \bigr)=}{}-\frac{aq^{-1}x(1-a)}{1-c}\mathbf{H}_{4}(aq,b;cq,d;q,x,yq) \\ &\hphantom{\mathbf{H}_{4}\bigl(aq^{-1} \bigr)=}{}- \frac{ax(1-a)}{1-c}\mathbf{H}_{4}(aq,b;cq,d;q,xq,yq). \end{aligned} \end{aligned}$$
(3.4)

Proof

In (1.9), replacing a by aq, we get

$$\begin{aligned} \mathbf{H}_{4}(aq)-\mathbf{H}_{4} =& \sum_{m,n=0}^{\infty } \frac{(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}} { \bigl[}(aq;q)_{2m+n}-(a;q)_{2m+n} \bigr]x^{m}y^{n}. \end{aligned}$$

Using the relations

$$\begin{aligned} (a;q)_{2m+n}= {(}1-a {)}(aq;q)_{2m+n-1} \end{aligned}$$

and

$$\begin{aligned} (aq;q)_{2m+n}= {\bigl(}1-aq^{2m+n} {\bigr)}(aq;q)_{2m+n-1}, \end{aligned}$$

we get

$$\begin{aligned} \mathbf{H}_{4}(aq)-\mathbf{H}_{4} =& \sum_{m,n=0}^{\infty } \frac{a(1-q^{2m+n})(aq;q)_{2m+n-1}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =&\sum_{m,n=0}^{\infty } \frac{a(aq;q)_{2m+n+1}(b;q)_{n}}{(c;q)_{m+1}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\ &{}+ \sum_{m,n=0}^{\infty } \frac{aq^{m+1}(aq;q)_{2m+n+1}(b;q)_{n}}{(c;q)_{m+1}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m+1}y^{n} \\ &{}+\sum_{m,n=0}^{\infty } \frac{aq^{2m}(aq;q)_{2m+n}(b;q)_{n+1}}{(c;q)_{m}(d;q)_{n+1}(q;q)_{m}(q;q)_{n}}x^{m}y^{n+1} \\ =&\frac{ax(1-aq)}{1-c}\mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,x,y \bigr) \\ &{}+ \frac{axq(1-aq)}{1-c}\mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,xq,y \bigr) \\ &{}+\frac{ay(1-b)}{1-d}\mathbf{H}_{4}\bigl(aq,bq;c,dq;q,xq^{2},y \bigr),\quad c\neq 1,d \neq 1. \end{aligned}$$

The proof of Eq. (3.2) would run parallel to Eq. (3.2), so details are omitted here. Performing the replacement \(a\rightarrow aq^{-1}\) in the contiguous relations (3.1) and (3.2), we get (3.3) and (3.4). □

Theorem 3.2

The q-derivatives of \(\mathbf{H}_{4}\) defined in (1.9) are valid:

$$\begin{aligned} \begin{aligned} &D^{r}_{x,q} \mathbf{H}_{4}= \frac{(a;q)_{2r}}{(c;q)_{r}(1-q)^{r}}\mathbf{H}_{4} \bigl(aq^{2r},b;cq^{r},d;q,x,y\bigr),\quad c \neq 1, q^{-1}, q^{-2},\ldots \end{aligned} \end{aligned}$$
(3.5)

and

$$\begin{aligned} \begin{aligned} &D^{r}_{y,q} \mathbf{H}_{4}= \frac{(a;q)_{r}(b;q)_{r}}{(d;q)_{r}(1-q)^{r}}\mathbf{H}_{4} \bigl(aq^{r},bq^{r};c,dq^{r};q,x,y\bigr),\quad d \neq 1, q^{-1}, q^{-2},\ldots . \end{aligned} \end{aligned}$$
(3.6)

Proof

In (1.9), we apply the operators \(\frac{\partial }{\partial x}=D_{x,q}\) and \(\frac{\partial }{\partial y}=D_{y,q}\) to get

$$\begin{aligned} D_{x,q}\mathbf{H}_{4} =&\sum _{m,n=0}^{\infty } \biggl[ \frac{1-q^{m}}{1-q} \biggr] \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m-1}y^{n} \\ =&\frac{(1-a)(1-aq)}{(1-q)(1-c)}\sum_{m,n=0}^{\infty } \frac{(aq^{2};q)_{2m+n}(b;q)_{n}}{(cq;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =& \frac{(1-a)(1-aq)}{(1-c)(1-q)} \mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,x,y\bigr),\quad c \neq 1 \end{aligned}$$

and

$$\begin{aligned} D_{y,q}\mathbf{H}_{4} =&\sum _{m,n=0}^{\infty } \biggl[ \frac{1-q^{n}}{1-q} \biggr] \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n-1} \\ =&\frac{(1-a)(1-b)}{(1-q)(1-c)}\sum_{m,n=0}^{\infty } \frac{(aq;q)_{2m+n}(bq;q)_{n}}{(c;q)_{m}(dq;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =& \frac{(1-a)(1-b)}{(1-d)(1-q)} \mathbf{H}_{4}(aq,bq;cq,d;q,x,y),\quad d\neq 1. \end{aligned}$$

Iterating this technique n times on \(\mathbf{H}_{4}\), we obtain (3.5) and (3.6). □

Theorem 3.3

For \(\mathbf{H}_{4}\) defined in (1.9), we have

$$\begin{aligned}& \biggl[a\theta _{x,q}+ \frac{1-a}{1-q} \biggr] \mathbf{H}_{4}+a\theta _{x,q}\mathbf{H}_{4}(xq)+a\theta _{y,q} \mathbf{H}_{4}\bigl(xq^{2}\bigr)=\frac{1-a}{1-q} \mathbf{H}_{4}(aq), \end{aligned}$$
(3.7)
$$\begin{aligned}& \biggl[a\theta _{y,q}+ \frac{1-a}{1-q} \biggr] \mathbf{H}_{4}+a\theta _{x,q}\mathbf{H}_{4}(yq)+a\theta _{x,q} \mathbf{H}_{4}(xq,yq)=\frac{1-a}{1-q}\mathbf{H}_{4}(aq), \end{aligned}$$
(3.8)
$$\begin{aligned}& \biggl[aq^{-1}\theta _{x,q}+ \frac{1-aq^{-1}}{1-q} \biggr]\mathbf{H}_{4}\bigl(aq^{-1} \bigr)+aq^{-1}\theta _{y,q}\mathbf{H}_{4} \bigl(aq^{-1},xq\bigr)+aq^{-1} \theta _{y,q} \mathbf{H}_{4}\bigl(aq^{-1},xq,yq\bigr) \\& \quad = \frac{1-aq^{-1}}{1-q} \mathbf{H}_{4}, \end{aligned}$$
(3.9)

and

$$\begin{aligned}& \biggl[aq^{-1}\theta _{y,q}+ \frac{1-aq^{-1}}{1-q} \biggr]\mathbf{H}_{4}\bigl(aq^{-1} \bigr)+aq^{-1}\theta _{x,q}\mathbf{H}_{4} \bigl(aq^{-1},yq\bigr)+aq^{-1} \theta _{x,q} \mathbf{H}_{4}\bigl(aq^{-1},xq,yq\bigr) \\& \quad = \frac{1-aq^{-1}}{1-q} \mathbf{H}_{4}. \end{aligned}$$
(3.10)

Proof

By using these q-derivatives of \(\mathbf{H}_{4}\) in (3.5) and (3.6), we get the recursion formulas (3.7) and (3.8).

Using (3.5), (3.7), and (3.8) for \(\mathbf{H}_{4}\), we obtain (3.9) and (3.10). □

Theorem 3.4

For \(d\neq 1\), the contiguous function relations of \(\mathbf{H}_{4}\) with the numerator parameter b hold true:

$$\begin{aligned} \mathbf{H}_{4}(bq) =&\mathbf{H}_{4}+ \frac{by(1-a)}{1-d} \mathbf{H}_{4}(aq,bq;c,dq;q,x,y) \end{aligned}$$
(3.11)

and

$$\begin{aligned} \mathbf{H}_{4}\bigl(bq^{-1} \bigr) =&\mathbf{H}_{4}- \frac{by(1-a)}{q(1-d)}\mathbf{H}_{4}(aq,b;c,dq;q,x,y). \end{aligned}$$
(3.12)

Proof

In (1.9), we replace b by bq to obtain

$$\begin{aligned} \begin{aligned} &\mathbf{H}_{4}(bq)-\mathbf{H}_{4}= \sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}} { \bigl[}(bq;q)_{n}-(b;q)_{n} \bigr]x^{m}y^{n}. \end{aligned} \end{aligned}$$

By means of the relations

$$\begin{aligned} (b;q)_{n}= {(}1-b {)}(bq;q)_{n-1} \end{aligned}$$

and

$$\begin{aligned} (bq;q)_{n}= {\bigl(}1-bq^{n} {\bigr)}(bq;q)_{n-1}, \end{aligned}$$

we have

$$\begin{aligned} \mathbf{H}_{4}(bq)-\mathbf{H}_{4} =& \sum_{m,n=0}^{\infty } \frac{b(1-q^{n})(a;q)_{2m+n}(bq;q)_{n-1}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =&\sum_{m,n=0}^{\infty } \frac{b(a;q)_{2m+n+1}(bq;q)_{n}}{(c;q)_{m+1}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n+1} \\ =& \frac{by(1-a)}{1-d}\mathbf{H}_{4}(aq,bq;c,dq;q,x,y),\quad d\neq 1. \end{aligned}$$

Replacing \(b\rightarrow bq^{-1}\) in relation (3.12), we get

$$\begin{aligned} \mathbf{H}_{4}\bigl(a,bq^{-1};c,d;q,x,y\bigr)= \mathbf{H}_{4}- \frac{by(1-a)}{q(1-d)}\mathbf{H}_{4}(aq,b;c,dq;q,x,y),\quad d \neq 1. \end{aligned}$$

 □

Theorem 3.5

The q-differential formulas are valid:

$$\begin{aligned} \biggl[b\theta _{y,q}+\frac{1-b}{1-q} \biggr] \mathbf{H}_{4}= \frac{1-b}{1-q}\mathbf{H}_{4}(bq) \end{aligned}$$
(3.13)

and

$$\begin{aligned} \biggl[bq^{-1}\theta _{y,q}+\frac{1-bq^{-1}}{1-q} { \biggr]}\mathbf{H}_{4}\bigl(bq^{-1}\bigr)= \frac{1-bq^{-1}}{1-q}\mathbf{H}_{4}. \end{aligned}$$
(3.14)

Proof

From (3.5) and (3.6), we obtain (3.13) and (3.14). □

Theorem 3.6

For \(c\neq 1\), \(d\neq 1\), and \(b\neq q\), the contiguous function relations hold true for the numerator parameters a and b of \(\mathbf{H}_{4}\):

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{4}\bigl(aq,bq^{-1};c,d;q,x,y \bigr) \\ &\quad =\mathbf{H}_{4}+ \frac{ay}{1-d}\mathbf{H}_{4}(aq,b;c,dq;q,x,y) \\ &\qquad {}+\frac{ax(1-aq)}{1-bq^{-1}}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq;q,x,yq \bigr) \\ &\qquad {}+ \frac{axq(1-aq)}{1-bq^{-1}}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq;q,xq,yq \bigr) \\ &\qquad {}-\frac{abq^{-1}x(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq;q,x,yq \bigr) \\ &\qquad {}- \frac{abx(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq;q,xq,yq \bigr) \\ &\qquad {}-\frac{bq^{-1}y}{1-d}\mathbf{H}_{4}(aq,b;c,dq;q,x,y) \end{aligned} \end{aligned}$$
(3.15)

and

$$\begin{aligned}& \mathbf{H}_{4}\bigl(aq,bq^{-1};c,d;q,x,y \bigr) \\& \quad =\mathbf{H}_{4}+ \frac{ax(1-aq)}{(1-c)(1-bq^{-1})}\mathbf{H}_{4} \bigl(aq^{2},bq^{-1};cq;q,x,y\bigr) \\& \begin{aligned} &\qquad {}+\frac{axq(1-aq)}{1-bq^{-1}}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq;q,xq,y \bigr) \\ &\qquad {}+ \frac{ay}{1-d}\mathbf{H}_{4}\bigl(aq,bq;c,dq;q,xq^{2},y \bigr) \\ &\qquad {}-\frac{abxq(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq;q,x,yq \bigr) \end{aligned} \\& \qquad {}- \frac{abx(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq;q,xq,yq \bigr) \\& \qquad {}-\frac{bq^{-1}y}{1-d}\mathbf{H}_{4}(aq,b;c,dq;q,x,y). \end{aligned}$$
(3.16)

Proof

Using the definition of \(\mathbf{H}_{4}\) with the relation \((bq^{-1};q)_{n}=(1-bq^{-1})(b;q)_{n-1}\), we get

$$\begin{aligned}& \mathbf{H}_{4}\bigl(aq,bq^{-1};c,d;q,x,y \bigr)-\mathbf{H}_{4} \\& \quad =\sum_{m,n=0}^{\infty } \frac{(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}} \bigl[\bigl(1-aq^{2m+n}\bigr) \bigl(1-bq^{-1}\bigr)-(1-a) \bigl(1-bq^{n-1}\bigr) \bigr]x^{m}y^{n} \\& \quad =\sum_{m,n=0}^{\infty } \frac{a(1-q^{n})(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q) _{n}}x^{m}y^{n} \\& \qquad {}+ \sum_{m,n=0}^{\infty } \frac{aq^{n}(1-q^{m})(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}+\sum_{m,n=0}^{\infty } \frac{aq^{m+n}(1-q^{m})(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m}(d;q)_{n}(q;q) _{m}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}- \sum_{m,n=0}^{\infty } \frac{abq^{n-1}(1-q^{m})(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}-\sum_{m,n=0}^{\infty } \frac{abq^{m+n-1}(1-q^{m})(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}- \sum_{m,n=0}^{\infty } \frac{bq^{-1}(1-q^{n})(aq;q)_{2m+n-1}(b;q)_{n-1}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \quad =\frac{ay}{1-d}\mathbf{H}_{4}(aq,b;c,dq;q,x,y) \\& \qquad {}+ \frac{ax(1-aq)}{1-bq^{-1}}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq,d;q,x,yq \bigr) \\& \qquad {}+\frac{axq(1-aq)}{1-bq^{-1}}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq,d;q,xq,yq \bigr) \\& \qquad {}- \frac{abq^{-1}x(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq,d;q,x,yq \bigr) \\& \qquad {}-\frac{abx(1-aq)}{(1-bq^{-1})(1-c)}\mathbf{H}_{4}\bigl(aq^{2},bq^{-1};cq,d;q,xq,yq \bigr) \\& \qquad {}- \frac{bq^{-1}y}{1-d}\mathbf{H}_{4}(aq,b;c,dq;q,x,y),\quad c\neq 1,d\neq 1,b \neq q. \end{aligned}$$

Proceeding as above, one can get the proof of Eq. (3.16) that would run parallel to Eq. (3.15), so details are omitted here. □

Theorem 3.7

For \(c\neq 1\), the recursion formulas hold true for the numerator parameter b of \(\mathbf{H}_{4}\):

$$\begin{aligned}& \begin{aligned} &b(1-a)\mathbf{H}_{4}(aq)-a(1-b) \mathbf{H}_{4}(bq) \\ &\quad =(b-a)\mathbf{H}_{4}+\frac{abx(1-a)(1-aq)}{1-c}\mathbf{H}_{4} \bigl(aq^{2},b;cq,d;q,x,yq\bigr) \\ &\qquad {}+\frac{abxq(1-a)(1-aq)}{1-c}\mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,xq,yq \bigr). \end{aligned} \end{aligned}$$
(3.17)

Proof

From (1.9) we have

$$\begin{aligned}& b(1-a)\mathbf{H}_{4}(aq)-a(1-b) \mathbf{H}_{4}(bq) \\& \quad =\sum_{m,n=0}^{\infty } \frac{b(1-a)(aq;q)_{2m+n}(b;q)_{n}-a(1-b)(a;q)_{2m+n}(bq;q)_{n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n}. \end{aligned}$$

By using the equation

$$\begin{aligned} (1-a) (aq;q)_{2m+n}=\bigl(1-aq^{2m+n}\bigr) (a;q)_{2m+n} \end{aligned}$$

and

$$\begin{aligned} (1-b) (bq;q)_{n}=\bigl(1-bq^{n}\bigr) (b;q)_{n}, \end{aligned}$$

we get

$$\begin{aligned}& b(1-a)\mathbf{H}_{4}(aq)-a(1-b) \mathbf{H}_{4}(bq) \\& \quad =\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(b;q)_{n}[b(1-aq^{2m+n})-a(1-bq^{n})]}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \quad =(b-a)\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}+ab \sum _{m,n=0}^{\infty } \frac{q^{n}(1-q^{m})(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(1-q^{m})(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\& \qquad {}+ab\sum_{m,n=0}^{\infty } \frac{q^{n+m}(1-q^{m})(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(1-q^{m})(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\& \quad =(b-a)\mathbf{H}_{4}+\frac{abx(1-a)(1-aq)}{1-c}\mathbf{H}_{4} \bigl(aq^{2},b;cq;q,x,yq\bigr) \\& \qquad {}+\frac{abxq(1-a)(1-aq)}{1-c}\mathbf{H}_{4}\bigl(aq^{2},b;cq;q,xq,yq \bigr),\quad c \neq 1. \end{aligned}$$

 □

Theorem 3.8

The contiguous function relations of \(\mathbf{H}_{4}\) with denominator parameters c and d hold true:

$$\begin{aligned} \mathbf{H}_{4}\bigl(cq^{-1}\bigr)=\mathbf{H}_{4}+ \frac{cx(1-a)(1-aq)}{(q-c)(1-c)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,d;q,x,y \bigr),\quad c \neq 1,q \end{aligned}$$
(3.18)

and

$$\begin{aligned} \mathbf{H}_{4}\bigl(dq^{-1}\bigr)=\mathbf{H}_{4}+ \frac{dy(1-a)(1-b)}{(q-d)(1-d)}\mathbf{H}_{4}(aq,bq;c,dq;q,x,y),\quad d \neq 1,q. \end{aligned}$$
(3.19)

Proof

By the definition of \(\mathbf{H}_{4}\), we have

$$\begin{aligned} \mathbf{H}_{4}\bigl(a,b;cq^{-1};q,x,y \bigr)-\mathbf{H}_{4} =&\sum_{m,n=0}^{ \infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(d;q)_{n}(q;q)_{m}(q;q)_{n}} \frac{(c;q)_{m}(d;q)_{n}-(cq^{-1};q)_{m}}{(cq^{-1};q)_{m}(c;q)_{m}}x^{m}y^{n}. \end{aligned}$$

Using

$$\begin{aligned} \bigl(cq^{-1};q\bigr)_{m}= {\bigl(}1-cq^{-1} {\bigr)}(c;q)_{m-1} \end{aligned}$$

and

$$\begin{aligned} (c;q)_{m} =& {\bigl(}1-cq^{m-1} { \bigr)}(c;q)_{m-1}, \end{aligned}$$

we get

$$\begin{aligned}& \mathbf{H}_{4}\bigl(a,b;cq^{-1};q,x,y \bigr)-\mathbf{H}_{4} \\& \quad =\sum_{m,n=0}^{ \infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(q;q)_{m}(q;q)_{n}} \frac{ {[}1-q^{m} {]}cq^{-1}(c;q)_{m-1}}{ {[}1-cq^{-1} {]}(c;q)_{m-1}(c;q)_{m}(d;q)_{n}}x^{m}y^{n} \\& \quad =c\sum_{m,n=0}^{\infty }\frac{1-q^{m}}{q-c} \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(1-q^{m})(q;q)_{m-1}(q;q)_{n}}x^{m}y^{n} \\& \quad =\frac{cx(1-a)(1-aq)}{(q-c)(1-c)}\mathbf{H}_{4}\bigl(aq^{2},b;cq;q,x,y \bigr),\quad c \neq 1,q, \end{aligned}$$

which is the desired result. The recursion formula (3.19) can be proved in a similar manner. □

Theorem 3.9

The contiguous function relations hold true for the denominator parameter c of \(\mathbf{H}_{4}\):

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{4}(a,b;cq,d;q,x,y) \\ &\quad = \mathbf{H}_{4}- \frac{cx(1-a)(1-aq)}{(1-cq)(1-c)}\mathbf{H}_{4} \bigl(aq^{2},b;cq^{2},d;q,x,y\bigr),\quad c \neq 1,cq\neq 1, \\ &\mathbf{H}_{4}(a,b;c,dq;q,x,y) \\ &\quad =\mathbf{H}_{4}- \frac{dy(1-a)(1-b)}{(1-dq)(1-d)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,x,y \bigr),\quad d \neq 1,dq\neq 1. \end{aligned} \end{aligned}$$
(3.20)

Proof

Replacing \(c\rightarrow cq\) in (3.18) and \(d\rightarrow dq\) in (3.19), we obtain (3.20). □

Theorem 3.10

The following results of \(\mathbf{H}_{4}\) hold well:

$$\begin{aligned}& \mathbf{H}_{4}\bigl(cq^{-1} \bigr)=\frac{c}{c-q}\mathbf{H}_{4}(a,b;c,d;q,xq,y)- \frac{q}{c-q}\mathbf{H}_{4},\quad c\neq q, \end{aligned}$$
(3.21)
$$\begin{aligned}& \mathbf{H}_{4}\bigl(dq^{-1} \bigr)=\frac{d}{d-q}\mathbf{H}_{4}(a,b;c,d;q,x,yq)- \frac{q}{d-q}\mathbf{H}_{4},\quad d\neq q, \end{aligned}$$
(3.22)
$$\begin{aligned}& \mathbf{H}_{4}(cq)+(c-1) \mathbf{H}_{4}=c\mathbf{H}_{4}(a,b,cq,xq,y), \end{aligned}$$
(3.23)
$$\begin{aligned}& \mathbf{H}_{4}(dq)+(d-1) \mathbf{H}_{4}=d\mathbf{H}_{4}(a,b,c,dq,x,yq), \end{aligned}$$
(3.24)
$$\begin{aligned}& \biggl[cq^{-1}\theta _{x,q}+ \frac{1-cq^{-1}}{1-q} \biggr]\mathbf{H}_{4}=\frac{1-cq^{-1}}{1-q} \mathbf{H}_{4}\bigl(cq^{-1}\bigr), \end{aligned}$$
(3.25)

and

$$\begin{aligned} \biggl[dq^{-1}\theta _{y,q}+ \frac{1-dq^{-1}}{1-q} \biggr]\mathbf{H}_{4} =&\frac{1-dq^{-1}}{1-q} \mathbf{H}_{4}\bigl(dq^{-1}\bigr). \end{aligned}$$
(3.26)

Proof

We replace \(\frac{1}{(cq^{-1};q)_{m}}=\frac{1}{(c;q)_{m}} {[}\frac{c}{c-q}q^{m}- \frac{q}{c-q} {]}\) in relation (1.9) to get

$$\begin{aligned} \mathbf{H}_{4}\bigl(cq^{-1} \bigr) =&\sum_{m,n=0}^{\infty } \biggl[ \frac{c}{c-q}q^{m}-\frac{q}{c-q} \biggr] \frac{(a;q)_{2m+n}(b;q)_{n}}{(c;q)_{m}(d;q)_{n}(q;q)_{m}(q;q)_{n}}x^{m}y^{n} \\ =&\frac{c}{c-q}\mathbf{H}_{4}(a,b;c,d;q,xq,y)- \frac{q}{c-q}\mathbf{H}_{4};\quad c \neq q. \end{aligned}$$

Replacing \(c=cq\) in (3.21), we get

$$\begin{aligned} \mathbf{H}_{4}(cq)+(c-1)\mathbf{H}_{4}=c \mathbf{H}_{4}(a,b,cq,xq,y). \end{aligned}$$

The results (3.25), (3.26), (3.22), and (3.24) are along the same lines as those of Eqs. (2.22)–(2.25). □

Theorem 3.11

For \(c, d, cq \neq 1\), the results of \(\mathbf{H}_{4}\) hold:

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{4}(aq,b;cq,d;q,x,y)- \mathbf{H}_{4} \\ &\quad =\frac{ax(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr) \\ &\qquad {}+ \frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,xq,y \bigr) \\ &\qquad {}+\frac{ay(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq,bq;cq,dq;q,xq^{2},y \bigr) \\ &\qquad {}+ \frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr) \\ &\qquad {}-\frac{acy(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}(aq,bq;cq,dq;q,xq,y) \\ &\qquad {}- \frac{acxq(1-aq}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,xq,yq \bigr) \end{aligned} \end{aligned}$$
(3.27)

and

$$\begin{aligned}& \mathbf{H}_{4}(aq,b;cq,d;q,x,y)- \mathbf{H}_{4} \\& \quad =\frac{ax(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr) \\& \qquad {}+ \frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,xq,y \bigr) \\& \qquad {}+\frac{ay(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq,bq;cq,dq;q,xq^{2},y \bigr) \\& \qquad {}+ \frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr) \\& \qquad {}-\frac{acxq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},bq;cq^{2},d;q,xq,y \bigr) \\& \qquad {}- \frac{acy(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},bq;cq,dq;q,xq^{2},y \bigr). \end{aligned}$$
(3.28)

Proof

In (1.9), replacing a and c by aq and cq, we have

$$\begin{aligned}& \mathbf{H}_{4}(aq,b;cq,d;q,x,y)- \mathbf{H}_{4} \\& \quad =\sum_{m,n=0}^{ \infty } \frac{(b;q)_{n}}{(q;q)_{m}(q;q)_{n}} \biggl[ \frac{(aq;q)_{2m+n}(c;q)_{m}(d;q)_{n}-(a;q)_{2m+n}(cq;q)_{m}}{(cq;q)_{m}(c;q)_{m}(d;q)_{n}} \biggr]x^{m}y^{n} \\& \quad =\sum_{m,n=0}^{\infty } \frac{(aq;q)_{2m+n-1}(cq;q)_{m-1}(b;q)_{n}}{(q;q)_{m}(q;q)_{n}} \biggl[ \frac{(1-aq^{2m+n})(1-c)-(1-a)(1-cq^{m})}{(1-cq^{m})(cq;q)_{m-1}(c;q)_{m}(d;q)_{n}} \biggr]x^{m}y^{n} \\& \quad =\frac{ax(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr) \\& \qquad {}+ \frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,xq,y \bigr) \\& \qquad {}+\frac{ay(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq,bq;cq,dq;q,xq^{2},y \bigr) \\& \qquad {}+ \frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr) \\& \qquad {}-\frac{acxq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},bq;cq^{2},d;q,xq,y \bigr) \\& \qquad {}- \frac{acy(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},bq;cq,dq;q,xq,yq \bigr),\quad c, d, cq \neq 1. \end{aligned}$$

By using the relation

$$\begin{aligned}& a\bigl(1-q^{2m+n}\bigr)+c \bigl(1-q^{m}\bigr)-acq^{m}\bigl(1-q^{m+n} \bigr) \\& \quad =a\bigl(1-q^{m}\bigr)+aq^{m}\bigl(1-q^{m} \bigr) \\& \qquad {}+aq^{2m}\bigl(1-q^{n}\bigr)+c\bigl(1-q^{m} \bigr)-acq^{m}\bigl(1-q^{m}\bigr)-acq^{2m} \bigl(1-q^{n}\bigr). \end{aligned}$$

The proof of Eq. (3.27) is similar as that of Eq. (3.28). □

Theorem 3.12

For \(c, d, cq \neq 1\), the following contiguous function relations hold:

$$\begin{aligned}& \mathbf{H}_{4}(aq,b;cq,d;q,x,y)- \mathbf{H}_{4} \\& \quad =\frac{ay(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}(aq,bq;cq,dq;q,x,y) \\& \begin{aligned}&\qquad {}+ \frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,yq \bigr) \\ &\qquad {}+\frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,xq,yq \bigr) \\ &\qquad {}+ \frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr) \end{aligned} \\& \qquad {}-\frac{acy(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}(aq,bq;cq,dq;q,xq,y) \\& \qquad {}- \frac{acxq(1-aq}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,xq,yq \bigr) \end{aligned}$$
(3.29)

and

$$\begin{aligned}& \mathbf{H}_{4}(aq,b;cq,d;q,x,y)- \mathbf{H}_{4} \\& \quad =\frac{ay(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}(aq,bq;cq,dq;q,x,y) \\& \qquad {}+ \frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,yq \bigr) \\& \qquad {}+\frac{axq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,xq,yq \bigr) \\& \qquad {}+ \frac{cx(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr) \\& \qquad {}-\frac{acxq(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},bq;cq^{2},d;q,xq,y \bigr) \\& \qquad {}- \frac{acy(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},bq;cq,dq;q,xq^{2},y \bigr). \end{aligned}$$
(3.30)

Proof

By virtue of our calculation in Theorem 3.11, and with the help of the relations

$$\begin{aligned}& a\bigl(1-q^{2m+n}\bigr)+c \bigl(1-q^{m}\bigr)-acq^{m}\bigl(1-q^{m+n} \bigr) \\& \quad =a\bigl(1-q^{n}\bigr)+aq^{n}\bigl(1-q^{m} \bigr) \\& \qquad {}+aq^{m+n}\bigl(1-q^{m}\bigr)+c\bigl(1-q^{m} \bigr)-acq^{m}\bigl(1-q^{n}\bigr)-acq^{m+n} \bigl(1-q^{m}\bigr) \end{aligned}$$

and

$$\begin{aligned}& a\bigl(1-q^{2m+n}\bigr)+c \bigl(1-q^{m}\bigr)-acq^{m}\bigl(1-q^{m+n} \bigr) \\& \quad =a\bigl(1-q^{n}\bigr)+aq^{n}\bigl(1-q^{m} \bigr) \\& \qquad {}+aq^{m+n}\bigl(1-q^{m}\bigr)+c\bigl(1-q^{m} \bigr)-acq^{m}\bigl(1-q^{m}\bigr)-acq^{2m} \bigl(1-q^{n}\bigr). \end{aligned}$$

To simplify the above relationships, we get (3.29) and (3.30). □

Theorem 3.13

For \(c, d, dq \neq 1\), the contiguous relations hold true for the parameters of \(\mathbf{H}_{4}\):

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{4}(aq,b;c,dq;q,x,y)- \mathbf{H}_{4} \\ &\quad =\frac{ax(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,x,y \bigr) \\ &\qquad {}+ \frac{axq(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,xq,y \bigr) \\ &\qquad {}+\frac{ax(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}\bigl(aq,b;c,dq^{2};q,xq^{2},y \bigr) \\ &\qquad {}+ \frac{d(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}(aq,bq;c,dq;q,x,y) \\ &\qquad {}-\frac{adx(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,x,yq \bigr) \\ &\qquad {}- \frac{adyq(1-b}{(1-d)(1-dq)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,xq,yq \bigr), \end{aligned} \end{aligned}$$
(3.31)
$$\begin{aligned}& \mathbf{H}_{4}(aq,b;c,dq;q,x,y)- \mathbf{H}_{4} \\& \quad =\frac{ax(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,x,y \bigr) \\& \qquad {}+ \frac{axq(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,xq,y \bigr) \\& \qquad {}+\frac{ax(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}\bigl(aq,b;c,dq^{2};q,xq^{2},y \bigr) \end{aligned}$$
(3.32)
$$\begin{aligned}& \qquad {}+ \frac{d(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}(aq,bq;c,dq;q,x,y) \\& \qquad {}-\frac{adyq(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,x,yq \bigr) \\& \qquad {}- \frac{adx(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,x,yq^{2} \bigr), \\& \begin{aligned} &\mathbf{H}_{4}(aq,b;c,dq;q,x,y)- \mathbf{H}_{4} \\ &\quad =\frac{ay(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,x,y \bigr) \\ &\qquad {}+ \frac{axq(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,x,yq \bigr) \\ &\qquad {}+\frac{dy(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,xq,yq \bigr) \\ &\qquad {}+ \frac{dy(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,x,y \bigr) \\ &\qquad {}-\frac{adx(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq,d;q,x,yq \bigr) \\ &\qquad {}- \frac{ady(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,xq,yq \bigr), \end{aligned} \end{aligned}$$
(3.33)

and

$$\begin{aligned}& \mathbf{H}_{4}(aq,b;c,dq;q,x,y)- \mathbf{H}_{4} \\& \quad =\frac{ay(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,x,y \bigr) \\& \qquad {}+ \frac{axq(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,x,yq \bigr) \\& \qquad {}+\frac{dy(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,xq,yq \bigr) \\& \qquad {}+ \frac{dy(1-b)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,x,y \bigr) \\& \qquad {}-\frac{adyq(1-b)}{(1-d)(1-dq)}\mathbf{H}_{4}\bigl(aq,bq;c,dq^{2};q,x,yq \bigr) \\& \qquad {}- \frac{adx(1-aq)}{(1-c)(1-d)}\mathbf{H}_{4}\bigl(aq^{2},b;cq,dq;q,x,yq^{2} \bigr). \end{aligned}$$
(3.34)

Proof

Using the relations

$$\begin{aligned}& a\bigl(1-q^{2m+n}\bigr)+d \bigl(1-q^{n}\bigr)-adq^{n}\bigl(1-q^{m+n} \bigr) \\& \quad =a\bigl(1-q^{m}\bigr)+aq^{m}\bigl(1-q^{m} \bigr) \\& \qquad {}+aq^{2m}\bigl(1-q^{n}\bigr)+d\bigl(1-q^{n} \bigr)-adq^{n}\bigl(1-q^{m}\bigr)-adq^{m+n} \bigl(1-q^{n}\bigr), \\& a\bigl(1-q^{2m+n}\bigr)+d \bigl(1-q^{n}\bigr)-adq^{n}\bigl(1-q^{m+n} \bigr) \\& \quad =a\bigl(1-q^{m}\bigr)+aq^{m}\bigl(1-q^{m} \bigr) \\& \qquad {}+aq^{2m}\bigl(1-q^{n}\bigr)+d\bigl(1-q^{n} \bigr)-adq^{n}\bigl(1-q^{n}\bigr)-adq^{2n} \bigl(1-q^{m}\bigr), \\& a\bigl(1-q^{2m+n}\bigr)+d \bigl(1-q^{n}\bigr)-adq^{n}\bigl(1-q^{m+n} \bigr) \\& \quad =a\bigl(1-q^{n}\bigr)+aq^{n}\bigl(1-q^{m} \bigr) \\& \qquad {}+aq^{m+n}\bigl(1-q^{m}\bigr)+d\bigl(1-q^{n} \bigr)-adq^{n}\bigl(1-q^{m}\bigr)-adq^{m+n} \bigl(1-q^{n}\bigr), \end{aligned}$$

and

$$\begin{aligned}& a\bigl(1-q^{2m+n}\bigr)+d \bigl(1-q^{n}\bigr)-adq^{n}\bigl(1-q^{m+n} \bigr) \\& \quad =a\bigl(1-q^{n}\bigr)+aq^{n}\bigl(1-q^{m} \bigr) \\& \qquad {}+aq^{m+n}\bigl(1-q^{m}\bigr)+d\bigl(1-q^{n} \bigr)-adq^{n}\bigl(1-q^{n}\bigr)-adq^{2n} \bigl(1-q^{m}\bigr). \end{aligned}$$

Simplifying the above relations, we obtain (3.31)–(3.34). □

Theorem 3.14

The formulas hold true of \(\mathbf{H}_{4}\):

$$\begin{aligned}& \mathbf{H}_{4}(a,bq;cq,d;q,x,y) \\& \quad = \mathbf{H}_{4}+ \frac{by(1-a)}{(1-d)(1-cq)}\mathbf{H}_{4} \bigl(aq,bq;cq^{2},dq;q,x,y\bigr) \\& \qquad {}-\frac{cx(1-a)(1-aq)(1-b)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr),\quad c, d, cq \neq 1 \end{aligned}$$
(3.35)

and

$$\begin{aligned}& \begin{aligned} &\mathbf{H}_{4}(a,bq;c,dq;q,x,y) \\ &\quad = \mathbf{H}_{4}+ \frac{(b-d)y(1-a)}{(1-d)(1-dq)}\mathbf{H}_{4} \bigl(aq,bq;c,dq^{2};q,x,y\bigr),\quad d, dq \neq 1. \end{aligned} \end{aligned}$$
(3.36)

Proof

By using the definition of \(\mathbf{H}_{4}\), we get

$$\begin{aligned}& \mathbf{H}_{4}(a,bq;cq;q,x,y)- \mathbf{H}_{4} \\& \quad =\sum_{m,n=0}^{ \infty } \frac{(a;q)_{2m+n}(b;q)_{n}}{(d;q)_{n}(q;q)_{m}(q;q)_{n}} \biggl[ \frac{(bq;q)_{n}(c;q)_{m}-(b;q)_{n}(cq;q)_{m}}{(cq;q)_{m}(c;q)_{m}} \biggr]x^{m}y^{n} \\& \quad =\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(cq;q)_{m-1}(bq;q)_{n-1}}{(d;q)_{n}(q;q)_{m}(q;q)_{n}} \biggl[ \frac{(1-bq^{n})(1-c)-(1-b)(1-cq^{m})}{(1-cq^{m})(cq;q)_{m-1}(c;q)_{m}} \biggr]x^{m}y^{n} \\& \quad =\sum_{m,n=0}^{\infty } \frac{(a;q)_{2m+n}(bq;q)_{n-1}}{(d;q)_{n}(q;q)_{m}(q;q)_{n}} \biggl[ \frac{b(1-c)(1-q^{n})+c(b-1)(1-q^{m})}{(1-c)(cq;q)_{m}} \biggr]x^{m}y^{n} \\& \quad =\frac{by(1-a)}{(1-d)(1-cq)}\mathbf{H}_{4}\bigl(aq,bq;cq^{2},dq;q,x,y \bigr) \\& \qquad {}+ \frac{c(b-1)x(1-a)(1-aq)}{(1-c)(1-cq)}\mathbf{H}_{4}\bigl(aq^{2},b;cq^{2},d;q,x,y \bigr),\quad c, d, cq \neq 1. \end{aligned}$$

We prove relation (3.36) in a similar way as relation (3.35). □

4 Conclusion

We conclude by the remark that the results established in this paper are general forms and one can deduce several contiguous function relations and q-differential relations of basic Horns hypergeometric functions \(\mathbf{H}_{3}\) and \(\mathbf{H}_{4}\) as different cases of our main findings. Also, other types of these extensions are recommended for a parallel study of this work. More work will be carried out in the coming future results in other fields of interest for fractional quantum calculus.

Availability of data and materials

All data are included within this paper.

References

  1. Acikgoz, M., Araci, S., Duran, U.: New extensions of some known special polynomials under the theory of multiple q-calculus. Turk. J. Anal. Number Theory 3, 128–139 (2015)

    Article  Google Scholar 

  2. Agarwal, R.P.: Some relations between basic hypergeometric functions of two variables. Rend. Circ. Mat. Palermo (2) 3, 1–7 (1954)

    Article  MathSciNet  Google Scholar 

  3. Agarwal, R.P.: Contiguous relations for bilateral basic hypergeometric series. Int. J. Math. Sci. 3(2), 375–388 (2004)

    MathSciNet  MATH  Google Scholar 

  4. Andrews, G.E.: On q-difference equations for certain well-poised basic hypergeometric series. Quart. J. Math. 19(2), 433–447 (1968)

    Article  MathSciNet  Google Scholar 

  5. Andrews, G.E.: Problems and Prospects for Basic Hypergeometric Functions. In: Askey, R.A. (ed.) Theory and Applications of Special Functions. Academic Press, San Diego (1975)

    Google Scholar 

  6. Araci, S., Acikgoz, M., Park, K.H., Jolany, H.: On the unification of two families of multiple twisted type polynomials by using q-adic q-integral at \(q=-1\). Bull. Malays. Math. Sci. Soc. 37, 17–41 (2014)

    MathSciNet  MATH  Google Scholar 

  7. Araci, S., Duran, U., Acikgoz, M.: Symmetric identities involving q-Frobenius-Euler polynomials under Sym (4). Turk. J. Anal. Number Theory 3, 90–93 (2015)

    Article  Google Scholar 

  8. Bagdasaryan, A., Araci, S., Acikgoz, M., He, Y.: Some new identities on the Apostol–Bernoulli polynomials of higher order derived from Bernoulli basis. J. Nonlinear Sci. Appl. 9, 2697–2704 (2016)

    Article  MathSciNet  Google Scholar 

  9. Bansal, M.K., Choi, J.: A note on pathway fractional integral formulas associated with the incomplete H-functions. Int. J. Appl. Comput. Math. 5(5), 133 (2019)

    Article  MathSciNet  Google Scholar 

  10. Bansal, M.K., Kumar, D.: On the integral operators pertaining to a family of incomplete I-functions. AIMS Math. 5(2), 1247–1259 (2020)

    Article  MathSciNet  Google Scholar 

  11. Bansal, M.K., Kumar, D., Jain, R.: A study of Marichev–Saigo–Maeda fractional integral operators associated with S-generalized Gauss hypergeometric function. Kyungpook Math. J. 59(3), 433–443 (2019)

    MathSciNet  MATH  Google Scholar 

  12. Bansal, M.K., Kumar, D., Nisar, K.S., Singh, J.: Certain fractional calculus and integral transform results of incomplete ℵ-functions with applications. Math. Methods Appl. Sci. 43(8), 5602–5614 (2020)

    Article  MathSciNet  Google Scholar 

  13. Duran, U., Acikgoz, M., Araci, S.: On a class of multiple q-Bernoulli, multiple q-Euler and multiple q-Genocchi polynomials of order α. Adv. Appl. Math. Sci. 15(8), 229–254 (2016)

    MathSciNet  MATH  Google Scholar 

  14. Duran, U., Acikgoz, M., Esi, A., Araci, S.: Some new symmetric identities involving q-Genocchi polynomials under \(S_{4}\). J. Math. Anal. 6(5), 22–31 (2015)

    MathSciNet  MATH  Google Scholar 

  15. Gasper, G., Rahman, M.: Basic Hypergeometric Series, 2nd edn. Encyclopedia of Mathematics and Its Applications, vol. 96. Cambridge University Press, Cambridge (2004)

    Book  Google Scholar 

  16. Gupta, D.: Contiguous relations, basic hypergeometric functions, and orthogonal polynomials. III. Associated continuous dual q-Hahn polynomials. J. Comput. Appl. Math. 68(1–2), 115–149 (1996)

    Article  MathSciNet  Google Scholar 

  17. Ismail, M.E.H.: Classical and Quantum Orthogonal Polynomials in One Variable. Encyclopedia of Mathematics and Its Applications, vol. 98. Cambridge University Press, Cambridge (2009)

    MATH  Google Scholar 

  18. Ismail, M.E.H., Libis, C.A.: Contiguous relations, basic hypergeometric functions, and orthogonal polynomials I. J. Math. Anal. Appl. 141(2), 349–372 (1989)

    Article  MathSciNet  Google Scholar 

  19. Jackson, F.H.: On basic double hypergeometric functions. Quart. J. Math. 13, 69–82 (1942)

    Article  MathSciNet  Google Scholar 

  20. Jackson, F.H.: On some formulae in partition theory, and bilateral basic hypergeometric series. J. Lond. Math. Soc. 24, 233–237 (1949)

    Article  MathSciNet  Google Scholar 

  21. Jain, V.K.: Some transformations of basic hypergeometric series and their applications. Proc. Am. Math. Soc. 78, 375–384 (1980)

    Article  MathSciNet  Google Scholar 

  22. Jain, V.K.: Some transformations of basic hypergeometric functions. Part II. SIAM J. Math. Anal. 12, 957–961 (1981)

    Article  MathSciNet  Google Scholar 

  23. Jain, V.K., Vertna, A.: Transformations between basic hypergeometric series on different bases and identities of Rogers–Ramanujan type. J. Math. Anal. Appl. 76, 230–269 (1980)

    Article  MathSciNet  Google Scholar 

  24. Mishra, B.P.: On certain transformation formulae involving basic hypergeometric functions. J. Ramanujan Soc. Math. Math. Sci. 2(2), 9–16 (2014)

    MathSciNet  MATH  Google Scholar 

  25. Sahai, V., Verma, A.: nth-order q-derivatives of multivariable q-hypergeometric series with respect to parameters. Asian-Eur. J. Math. 7(2), 1450019 (2014)

    Article  MathSciNet  Google Scholar 

  26. Sahai, V., Verma, A.: nth-order q-derivatives of Srivastava’s general triple q-hypergeometric series with respect to parameters. Kyungpook Math. J. 56, 911–925 (2016)

    Article  MathSciNet  Google Scholar 

  27. Sahai, V., Verma, A.: Recursion formulas for q-hypergeometric and q-Appell series. Commun. Korean Math. Soc. 33(1), 207–236 (2018)

    MathSciNet  MATH  Google Scholar 

  28. Shehata, A.: On the \((p,q)\)-Humbert functions from the view point of the generating functions method. J. Funct. Spaces 2020, Article ID 4794571 (2020)

    MathSciNet  MATH  Google Scholar 

  29. Shehata, A.: On the \((p,q)\)-Bessel functions from the view point of the generating functions method. J. Interdiscip. Math., 1–14 (2020)

  30. Srivastava, B.: A note on certain bibasic q-Appell and Lauricella series. Glas. Mat. 30(1), 29–36 (1995)

    MathSciNet  MATH  Google Scholar 

  31. Srivastava, H.M., Jain, V.K.: q-series identities and reducibility of basic double hypergeometric functions. Can. J. Math. 38(1), 215–231 (1986)

    Article  MathSciNet  Google Scholar 

  32. Srivastava, H.M., Manocha, H.L.: A Treatise on Generating Functions. Halsted, New York (1984)

    MATH  Google Scholar 

  33. Swarttouw, R.F.: The contiguous function relations for basic hypergeometric function. J. Math. Anal. Appl. 149(1), 151–159 (1990)

    Article  MathSciNet  Google Scholar 

  34. Verma, A., Sahai, V.: Some recursion formulas for q-Lauricella series. Afr. Math. 30(9), 1–44 (2019)

    MATH  Google Scholar 

Download references

Acknowledgements

The author would like to thank the anonymous reviewers for the suggestions that helped to improve the presentation of the paper.

Funding

The author received no specific funding for this work.

Author information

Authors and Affiliations

Authors

Contributions

All of this work was done by AS. The author read and approved the final manuscript.

Corresponding author

Correspondence to Ayman Shehata.

Ethics declarations

Competing interests

The author declares that they have no competing interests.

Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Reprints and permissions

About this article

Check for updates. Verify currency and authenticity via CrossMark

Cite this article

Shehata, A. On basic Horn hypergeometric functions \(\mathbf{H}_{3}\) and \(\mathbf{H}_{4}\). Adv Differ Equ 2020, 595 (2020). https://doi.org/10.1186/s13662-020-03056-3

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/s13662-020-03056-3

MSC

Keywords