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# Sharing values of *q*-difference-differential polynomials

*Advances in Difference Equations*
**volume 2020**, Article number: 212 (2020)

## Abstract

This paper is devoted to the uniqueness of *q*-difference-differential polynomials of different types. Using the idea of common zeros and common poles (Chin. Ann. Math., Ser. A 35:675–684, 2014), we improve the conditions of the former theorems and obtain some new results on the uniqueness of *q*-difference-differential polynomials of meromorphic functions.

## 1 Introduction and main results

In this paper, a meromorphic function is assumed meromorphic in the whole complex plane. We assume that the reader is familiar with the basic symbols and fundamental results of Nevanlinna theory; see, for example, [2, 3, 10]. We say that two meromorphic functions *f* and *g* share a point *a* CM (IM) if \(f(z)-a\) and \(g(z)-a\) have the same zeros counting multiplicities (ignoring multiplicities). The logarithmic density of the set *E* is defined by

Denote by \(S(r,f)\) a quantity of \(o(T(r,f))\) as \(r\rightarrow\infty\) outside a possible exceptional set *E* of logarithmic density 0.

Yang and Hua [9] obtained an important result on the uniqueness when the differential polynomials \(f^{n}f'\) and \(g^{n}g'\) share one value CM. Recently, many studies are devoted to the uniqueness of difference and *q*-difference polynomials; see [4–6, 11–14]. Zhang [12] obtained the following result.

### Theorem A

([12])

*Let*\(f(z)\)*and*\(g(z)\)*be transcendental entire functions of zero order*, *and let**n*, *m*, *d**be positive integers*. *If*\(n \geq m+5d\)*and*\(f(z)^{n}(f(z)^{m}-1)\prod_{i=1}^{d}f(q_{i}z)\)*and*\(g(z)^{n} (g(z)^{m}-1)\prod_{i=1}^{d}g(q_{i}z)\)*share* 1 *CM*, *then*\(f(z)\equiv tg(z)\), \(t^{n+d}=t^{m}=1\).

Liu, Liu, and Cao [4] and Zhang and Korhonen [11] obtained the following two theorems.

### Theorem B

([4, Theorem 1.5])

*Let*\(f(z)\)*and*\(g(z)\)*be transcendental zero*-*order entire functions*, *and let**m**be a positive integer*. *If*\(n \geq m+5\)*and*\(f(z)^{n}(f(z)^{m}-a)f(qz+c)\)*and*\(g(z)^{n}(g(z)^{m}-a)g(qz+c)\)*share a nonzero polynomial*\(p(z)\)*CM*, *then*\(f(z)\equiv g(z)\).

### Theorem C

([11, Theorem 5.1])

*Let*\(f(z)\)*and*\(g(z)\)*be transcendental zero*-*order meromorphic functions*. *If*\(n\geq8\)*and*\(f(z)^{n}f(qz)\)*and*\(g(z)^{n}g(qz)\)*share* 1 *and* ∞ *CM*, *then*\(f(z)\equiv tg(z)\), \(t^{n+1}=1\).

Zhao and Zhang [13] proved the following theorem.

### Theorem D

([13, Theorem 1.4])

*Let*\(f(z)\)*and*\(g(z)\)*be transcendental zero*-*order entire functions*, *and let**k**be a positive integer*. *If*\(n \geq2k+6\)*and*\((f(z)^{n}f(qz+c))^{(k)}\)*and*\((g(z)^{n}g(qz+c))^{(k)}\)*share* 1 *CM*, *then*\(f(z)\equiv tg(z)\), *where*\(t^{n+1}=1\).

Wang and Ye [8] improved the conditions of Theorems B and C to \(n \geq m+4\) and \(n \geq6\), respectively, by using the idea of common zeros and common poles. Here we give the main idea of common zeros and common poles. Let *f*, *g* be two nonconstant meromorphic functions. Denote by \(\overline{n}_{0}(r)\) or \(\overline {n}_{1}(r)\) the numbers of common zeros or poles of *fg* and *g*, ignoring multiplicities. Let *p*, *q* be positive integers. We assume that the Laurent series of *f* and *g* at \(z_{0}\) are as follows:

where \(f_{1}(z)\) and \(g_{1}(z)\) are analytic functions at \(z_{0}\), and \(f_{1}(z_{0})\neq0\), \(g_{1}(z_{0})\neq0\); the other cases can be discussed in a similar way. So \(z_{0}\) is a zero of \(g(z)\) with multiplicities *q*. If \(q>p\), then \(z_{0}\) is a zero of \(f(z)g(z)\) with multiplicity \(q-p\), and thus the contribution to \(\overline{n}_{0}(r)\) is 1 at \(z_{0}\). If \(q\leq p\), then \(z_{0}\) is a pole of \(f(z)g(z)\) with multiplicity \(p-q\) or an analytic point of \(f(z)g(z)\), and thus the contribution to \(\overline{n}_{0}(r)\) is 0 at \(z_{0}\). A similar method can be discussed for \(\overline{n}_{1}(r)\). As usual, denote by \(\overline{N}_{0}(r)\) or \(\overline{N}_{1}(r)\) the counting functions of the common zeros or poles of *fg* and *g*, ignoring multiplicities. Thus we have \(\overline{N}(r,\frac{1}{fg})\leq\overline{N}(r,\frac {1}{f})+\overline{N}_{0}(r)\) and \(\overline{N}(r,fg)\leq\overline{N}(r,f)+\overline{N}_{1}(r)\). In this paper, we continue to consider the uniqueness of *q*-difference-differential polynomials. Firstly, we improve the condition \(n\geq m+5d\) in Theorem A to \(n\geq m+d+3\) in Theorem 1.1. Set

where \(c_{i}\) and \(q_{i}\neq0\) (\(i=1,\ldots, d\)) are constants, and *d* is a positive integer.

### Theorem 1.1

*Let*\(f(z)\)*and*\(g(z)\)*be transcendental zero*-*order entire functions*, *and let**m**be a positive integer*. *If*\(n \geq m+d+3\)*and*\(f(z)^{n}(f(z)^{m}-1)L(z,f)\)*and*\(g(z)^{n}(g(z)^{m}-1)L(z,g)\)*share* 1 *CM*, *then*\(f(z)\equiv c_{1}g(z)\), \(c_{1}^{n+d}=c_{1}^{m}=1\).

In the following theorem, we improve the condition \(n\geq2k+6\) in Theorem D to \(n\geq6\).

### Theorem 1.2

*Let*\(f(z)\)*and*\(g(z)\)*be transcendental zero*-*order meromorphic functions*, *and let**k**be a positive integer*. *If*\(n \geq6\)*and*\((f(z)^{n}f(qz+c))^{(k)}\)*and*\((g(z)^{n}g(qz+c))^{(k)}\)*share* 1 *and* ∞ *CM*, *then*\(f(z)\equiv c_{2}g(z)\), \(c_{2}^{n+1}=1\).

We also consider the following theorems for *q*-difference polynomials of different types. The following theorem is also an improvement of Theorem C.

### Theorem 1.3

*Let*\(f(z)\)*and*\(g(z)\)*be transcendental zero*-*order meromorphic functions*, *and let**s**be a positive integer*. *If*\(n\geq(d+1)s+4\)*and*\(f(z)^{n}L(z,f)^{s}\)*and*\(g(z)^{n}L(z,g)^{s}\)*share* 1 *and* ∞ *CM*, *then*\(f(z)\equiv c_{3}g(z)\), \(c_{3}^{n+sd}=1\).

### Theorem 1.4

*Let*\(f(z)\)*and*\(g(z)\)*be transcendental zero*-*order meromorphic functions*, \(q, c\in\mathbb{C}\), *and*\(q\neq0\). *If*\(n\geq7\), *and*\(f(z)^{n}(f(qz+c)-f(z))\)*and*\(g(z)^{n}(g(qz+c)-g(z))\)*share* 1 *and* ∞ *CM*, *then*

*If*\(\frac{g(z)}{g(qz+c)}\)*is transcendental with only finitely many zeros*, *then*\(f(z)\equiv c_{4}g(z)\), *where*\(c_{4}^{n+1}=1\).

## 2 Lemmas

Combining [11, Theorem 1.1] and [1, Theorem 2.1], we easily get the following lemma.

### Lemma 2.1

*Let*\(f(z)\)*be a transcendental zero*-*order meromorphic function*, \(q, c\in \mathbb{C}\), *and*\(q\neq0\). *Then*

*on a set of logarithmic density* 1.

### Lemma 2.2

([7])

*Let*\(f(z)\)*be a zero*-*order meromorphic function*\(q, c\in\mathbb{C}\), *and*\(q\neq0\). *Then*

*on a set of logarithmic density* 1.

### Lemma 2.3

*If**f**is a transcendental zero*-*order entire function*, *then*

*on a set of logarithmic density* 1.

### Proof

Set \(F(z)=f(z)^{n}(f(z)^{m}-1)L(z,f)\). By Lemma 2.2 and the standard Valiron–Mohon’ko theorem, if *f* is a transcendental zero-order entire function, then

on a set of logarithmic density 1. On the other hand, combining Lemma 2.1 with the fact that *f* is a transcendental zero-order function, we have

on a set of logarithmic density 1. □

## 3 Proofs of theorems

### Proof of Theorem 1.1

Let \(F(z)=f^{n}(f^{m}-1)L(z,f)\) and \(G(z)=g^{n}(g^{m}-1)L(z,g)\). Since \(F(z)\) and \(G(z)\) share 1 and ∞ CM, we have that \(\frac{F-1}{G-1}=B\), that is,

where *B* is a nonzero constant.

If \(B\neq1\), then from the second main theorem of Nevanlinna theory, Lemma 2.1, and Lemma 2.3 we obtain

Using the same method, we have

Combining (2) with (3), we have

which contradicts to \(n\geq m+d+3\). Thus \(B=1\), and from (1) we have

Let \(h(z)=\frac{f(z)}{g(z)}\). So \(\frac{L(z,f)}{L(z,g)}=\prod_{i=1}^{d}\frac{f(q_{i}z+c_{i})}{g(q_{i}z+c_{i})} =\prod_{i=1}^{d}h(q_{i}z+c_{i})=L(z,h)\), and then (4) can be written as

Next, we prove that \(h(z)\equiv c_{1}\) and \(c_{1}^{n+d}=c_{1}^{m}=1\), where \(c_{1}\) is a constant. Assume on the contrary that \(h(z)\) is not a constant. From Lemma 2.1 we have

We also have

Since \(n\geq m+d+3\), from the last two inequalities it follows that \(S(r,h^{n+m}L(z,h))=S(r,h)\). Denote by \(\overline{N}_{1}(r)\) the counting function of the common poles of \(h^{n+m}L(z,h)\) and \(L(z,h)\) ignoring multiplicities. Then

Here we should remark that the poles of \(L(z,h)\) may be the zeros of *h* and the zeros of \(L(z,h)\) may be the poles of *h*. Similarly, denote by \(\overline{N}_{0}(r)\) the counting function of the common zeros of \(h^{n+m}L(z,h)\) and \(L(z,h)\) ignoring multiplicities, and then

From the second main theorem of Nevanlinna theory and the last two inequalities we have

On the other hand,

Since \(n\geq m+d+3\), the value 1 is not the Picard exceptional value of \(h^{n+m}L(z,h)\) from (10). Furthermore, we prove \(h^{n+m}L(z,h)\equiv1\) and \(h(z)\equiv c_{1}\) is a nonzero constant. If \(h^{n+m}L(z,h)\not\equiv1\),then since 1 is not the Picard exceptional value of \(h^{n+m}L(z,h)\), there exists a point \(z_{0}\) satisfying \(h(z_{0})^{n+m}L(z,h(z_{0}))=1\). From the condition that \(g(z)\) is an entire function and (5) we have \(h(z_{0})^{m}=1\) and

Substituting (11) into (10), we get a contradiction to \(n\geq m+d+3\), so \(h^{n+m}L(z,h)\equiv1\), that is, \(h^{n+m}=\frac {1}{L(z,h)}\). From Lemma 2.1 we have

which also contradicts to \(n\geq m+d+3\), so \(h(z)\equiv c_{1}\), where \(c_{1}\) is a nonzero constant, that is, \(f(z)\equiv c_{1}g(z)\), and \(L(z,h)=c_{1}^{d}\). From (5) we can get \(c_{1}^{m}=c_{1}^{n+d}=1\). Thus the theorem is proved. □

### Proof of Theorem 1.2

Let \(F(z)=f(z)^{n}f(qz+c)\) and \(G(z)=g(z)^{n}g(qz+c)\). From the condition in Theorem 1.2 we know that \(F^{(k)}\) and \(G^{(k)}\) share 1 and ∞ CM, so

where *C* is a nonzero constant, that is,

Integrating both sides of (12), we have

where \(p_{1}(z)\) is a polynomial of degree at most \(k-1\). Denote \(\frac {1-C}{k!}z^{k}+p_{1}(z)\) by \(p(z)\). If \(p(z)\not\equiv0\), then by the second main theorem of Nevanlinna theory, Lemma 2.1, and (13) we obtain

where \(\overline{N}_{0}(r)\) denotes the counting function ignoring multiplicities of the common zeros of \(F(z)\) and \(f(qz+c)\), and \(\overline{N}_{1}(r)\) denotes the counting function ignoring multiplicities of the common poles of \(F(z)\) and \(f(qz+c)\). On the other hand,

From (15), (16), and Lemma 2.1 we have

Substituting (14) into (17), we obtain

Using the same method, we also get

Combining (18) with (19), we have

which contradicts to \(n\geq6\), and thus \(p(z)\equiv0\). Since the degree of \(p_{1}(z)\) is at most \(k-1\), we have \(C=1\) and \(p_{1}(z)\equiv 0\). From (13) we get

Assume that \(h(z)=\frac{f(z)}{g(z)}\). Then \(h(qz+c)h(z)^{n}=1\), that is, \(h(z)^{n}=\frac{1}{h(qz+c)}\), and from Lemma 2.1 we have

which also contradicts to \(n\geq6\), so \(h(z)\) is a nonzero constant, say \(c_{2}\). So \(f(z)\equiv c_{2}g(z)\), and \(c_{2}^{n+1}=1\). Thus the theorem is proved. □

### Proof of Theorem 1.3

Since \(f(z)\) and \(g(z)\) are transcendental zero-order meromorphic functions and \(f(z)^{n}L(z,f)^{s}\) and \(g(z)^{n}L(z,g)^{s}\) share 1 and ∞ CM, we have

where *E* is a nonzero constant. Then (20) can be rewritten as

Let \(F(z)=f(z)^{n}L(z,f)^{s}\) and \(G(z)=g(z)^{n}L(z,g)^{s}\). We affirm that \(E=1\). On the contrary, assume that \(E\neq1\). Using the second main theorem of Nevanlinna theory and Lemma 2.1 for (21), we get

where \(\overline{N}_{0}(r)\) denotes the counting function ignoring multiplicities of the common zeros of \(F(z)\) and \(L(z,f)\), and \(\overline{N}_{1}(r)\) denotes the counting function ignoring multiplicities of the common poles of \(F(z)\) and \(L(z,f)\).

Since \(F(z)=f(z)^{n}L(z,f)^{s}\), we have

So

Substituting (22) into (23), we obtain

Using the same method, we can get

Combining (24) with (25), it follows

which contradicts to \(n\geq(d+1)s+4\), and thus \(E=1\). From (21) we get

Let \(h(z)=\frac{f(z)}{g(z)}\). So \(\frac{L(z,f)}{L(z,g)}=L(z,h)\). Then

So \(nT(r,h)=sT(r,L(z,h))\leq sdT(r,h)+S(r,h)\). Since \(n\geq(d+1)s+4\), \(h(z)\) must be a constant, say \(c_{3}\), that is, \(f(z)\equiv c_{3}g(z)\). Then from (26) it follows that \(c_{3}^{n+sd}=1\). □

### Proof of Theorem 1.4

Letting \(L(z,f)=f(qz+c)-f(z)\) and \(L(z,g)=g(qz+c)-g(z)\), \(s=1\) in Theorem 1.3, we obtain that if \(n\geq7\), then

Let \(h(z)=\frac{f(z)}{g(z)}\) and \(H(z)=h(qz+c)h(z)^{n}\). The last equation implies that

We know that \(T(r,H)\leq(n+1)T(r,h)+S(r,h)\) from the expression of \(H(z)\) and Lemma 2.1. Thus \(S(r,H)=S(r,h)\). Next, we prove that \(h(z)\equiv c_{4}\), where \(c_{4}^{n+1}=1\), when \(\frac {g(z)}{g(qz+c)}\) is transcendental with only finitely many zeros. Obviously, \(h(z)\) is neither a constant except \(c_{4}\) nor a rational function from (27) since \(\frac{g(z)}{g(qz+c)}\) is transcendental. Thus we assume that \(h(z)\) is a transcendental meromorphic function.

First, we affirm that \(H(z)-1\) has infinitely many zeros. Otherwise, by the second main theorem of Nevanlinna theory and Lemma 2.1

From the Valiron–Mohon’ko theorem, Lemma 2.1, and (28) we obtain

which contradicts to \(n\geq7\). Thus \(H(z)-1\) has infinitely many zeros.

Then we prove that \(H(z)\equiv1\) and \(h(z)\equiv c_{4}\) is a nonzero constant, that is, \(c_{4}^{n+1}=1\). If \(H(z)\not\equiv1\), then since \(H(z)-1\) has infinitely many zeros, we can choose a point \(z_{0}\) satisfying \(H(z_{0})=1\), and \(z_{0}\) is not the zero of \(\frac{g(z)}{g(qz+c)}\). From (27) we have \(h(qz_{0}+c)=h(z_{0})\). By Lemma 2.1

Using the second main theorem of Nevanlinna theory, Lemma 2.1, and (29), we obtain

where \(\overline{N}_{0}(r)\) denotes the counting function ignoring multiplicities of the common zeros of \(H(z)\) and \(h(qz+c)\), and \(\overline{N}_{1}(r)\) denotes the counting function ignoring multiplicities of the common poles of \(H(z)\) and \(h(qz+c)\).

On the other hand, from \(H(z)=h(qz+c)h(z)^{n}\) we have

From (31), (32), and Lemma 2.1 we have

Substituting (30) into (33), we obtain

which contradicts to \(n\geq7\), so \(H(z)=h(qz+c)h(z)^{n}\equiv1\), that is, \(h(z)^{n}=\frac{1}{h(qz+c)}\). From Lemma 2.1 we have

which also contradicts to \(n\geq7\), so \(h(z)\) is a nonzero constant, say \(c_{4}\), and from (27) we get \(c_{4}^{n+1}=1\). Thus the theorem is proved. □

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### Acknowledgements

The authors thanks the anonymous referees for their valuable comments and suggestions, which improved the presentation of this manuscript.

### Availability of data and materials

Data sharing not applicable to this paper as no data sets were generated or analyzed during the current study.

## Funding

This work was supported by the NSFC (No. 11661052) and the outstanding youth scientist foundation plan of Jiangxi (No. 20171BCB23003).

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Li, J., Liu, K. Sharing values of *q*-difference-differential polynomials.
*Adv Differ Equ* **2020**, 212 (2020). https://doi.org/10.1186/s13662-020-02668-z

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DOI: https://doi.org/10.1186/s13662-020-02668-z