Skip to main content

Advertisement

We’d like to understand how you use our websites in order to improve them. Register your interest.

Sharing values of q-difference-differential polynomials

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.

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

$$\begin{aligned} \limsup_{r\rightarrow\infty}\frac{1}{\log r} \int_{[1,r]\cap E}\frac{1}{t}\,dt. \end{aligned}$$

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 [46, 1114]. Zhang [12] obtained the following result.

Theorem A

([12])

Let\(f(z)\)and\(g(z)\)be transcendental entire functions of zero order, and letn, m, dbe 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 letmbe 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 andCM, 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 letkbe 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:

$$f(z)=\frac{1}{(z-z_{0})^{p}}f_{1}(z), \qquad g(z)=(z-z_{0})^{q}g_{1}(z), $$

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

$$\begin{aligned} L(z,f)=\prod_{i=1}^{d}f(q_{i}z+c_{i}), \end{aligned}$$

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 letmbe 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 letkbe a positive integer. If\(n \geq6\)and\((f(z)^{n}f(qz+c))^{(k)}\)and\((g(z)^{n}g(qz+c))^{(k)}\)share 1 andCM, 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 letsbe 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 andCM, 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 andCM, then

$$f(z)^{n}\bigl(f(qz+c)-f(z)\bigr)= g(z)^{n} \bigl(g(qz+c)-g(z)\bigr). $$

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

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

$$\begin{aligned} T\bigl(r,f(qz+c)\bigr)=T(r,f)+S(r,f) \end{aligned}$$

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

$$\begin{aligned} m \biggl(r,\frac{f(qz+c)}{f(z)} \biggr)=S(r,f) \end{aligned}$$

on a set of logarithmic density 1.

Lemma 2.3

Iffis a transcendental zero-order entire function, then

$$\begin{aligned} T\bigl(r, f(z)^{n}\bigl(f(z)^{m}-1\bigr)L(z,f) \bigr)=(m+n+d)T(r,f)+S(r,f) \end{aligned}$$

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

$$\begin{aligned} (n+m+d)T(r,f) =& T\bigl(r,f^{n+d}\bigl(f^{m}-1\bigr) \bigr)+S(r,f) \\ =& m\bigl(r,f^{n+d}\bigl(f^{m}-1\bigr)\bigr)+S(r,f) \\ \leq& m \biggl(r,\frac{f^{n+d}(f^{m}-1)}{f^{n}(f^{m}-1)L(z,f)} \biggr)+m(r,F)+S(r,f) \\ \leq& m \biggl(r,\frac{f^{d}}{L(z,f)} \biggr)+m(r,F)+S(r,f) \\ \leq& T(r,F)+S(r,f) \end{aligned}$$

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

$$\begin{aligned} T(r,F) \leq& T\bigl(r,f^{n}\bigl(f^{m}-1\bigr)\bigr)+T \bigl(r,L(z,f)\bigr) \\ \leq& (n+m+d)T(r,f)+S(r,f) \end{aligned}$$

on a set of logarithmic density 1. □

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,

$$\begin{aligned} F=BG+1-B, \end{aligned}$$
(1)

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

$$\begin{aligned} (n+m+d)T(r,f) =&T(r,F)+S(r,f) \\ \leq& \overline{N}(r,F)+\overline{N} \biggl(r,\frac{1}{F} \biggr) + \overline{N} \biggl(r,\frac{1}{F-1+B} \biggr)+S(r,f) \\ \leq& \overline{N} \biggl(r,\frac{1}{f^{n}} \biggr)+\overline{N} \biggl(r,\frac{1}{f^{m}-1} \biggr) +\overline{N} \biggl(r, \frac{1}{L(z,f)} \biggr) +\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& (m+d+1) \bigl(T(r,f)+T(r,g)\bigr)+S(r,f)+S(r,g). \end{aligned}$$
(2)

Using the same method, we have

$$ (n+m+d)T(r,g)\leq(m+d+1) \bigl(T(r,f)+T(r,g) \bigr)+S(r,f)+S(r,g). $$
(3)

Combining (2) with (3), we have

$$\begin{aligned} (n-m-d-2) \bigl(T(r,f)+T(r,g)\bigr)\leq S(r,f)+S(r,g), \end{aligned}$$

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

$$ f^{n}\bigl(f^{m}-1\bigr)L(z,f)=g^{n} \bigl(g^{m}-1\bigr)L(z,g). $$
(4)

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

$$ g^{m}\bigl(h^{n+m}L(z,h)-1 \bigr)=h^{n}L(z,h)-1. $$
(5)

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

$$T\bigl(r,h^{n+m}L(z,h)\bigr)\leq(n+m+d)T(r,h)+S(r,h). $$

We also have

$$\begin{aligned} (n+m)T(r,h) =&T\bigl(r,h^{n+m}\bigr) \\ \leq& T \biggl(r,\frac{1}{h^{n+m}L(z,h)} \biggr) +T\bigl(r,L(z,h)\bigr))+O(1) \\ \leq& T \bigl(r,h^{n+m}L(z,h) \bigr) +dT(r,h)+S(r,h). \end{aligned}$$

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

$$\overline{N}\bigl(r,h^{n+m}L(z,h)\bigr)\leq\overline{N}(r,h)+ \overline{N}_{1}(r). $$

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

$$\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)} \biggr)\leq\overline {N}\biggl(r, \frac{1}{h}\biggr)+\overline{N}_{0}(r). $$

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

$$\begin{aligned} T\bigl(r,h^{n+m}L(z,h)\bigr)&\leq\overline{N} \bigl(r,h^{n+m}L(z,h)\bigr) +\overline{N} \biggl(r, \frac{1}{h^{n+m}L(z,h)} \biggr) \\ &\quad+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+{S \bigl(r,h^{n+m}L(z,h)\bigr)} \\ &\leq\overline{N}(r,h)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{h} \biggr) +\overline{N}_{0}(r) \\ &\quad+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+S(r,h). \end{aligned}$$
(6)

On the other hand,

$$\begin{aligned}& (n+m)m(r,h)=m\bigl(r,h^{n+m}\bigr)\leq m \bigl(r,h^{n+m}L(z,h)\bigr)+m \biggl(r,\frac {1}{L(z,h)} \biggr)+O(1), \end{aligned}$$
(7)
$$\begin{aligned}& \begin{aligned}[b](n+m)N(r,h)&=N\bigl(r,h^{n+m}\bigr)= N \biggl( \frac{h^{n+m}L(z,h)}{L(z,h)} \biggr) \\ &\leq N\bigl(r,h^{n+m}L(z,h)\bigr)+N \biggl(r,\frac{1}{L(z,h)} \biggr)-\overline {N}_{1}(r)-\overline{N}_{0}(r).\end{aligned} \end{aligned}$$
(8)

From (7) and (8) we get

$$\begin{aligned} (n+m)T(r,h)&\leq T\bigl(r,h^{n+m}L(z,h)\bigr)+T \biggl(r,\frac{1}{L(z,h)} \biggr) \\ &\quad-\overline{N}_{1}(r)-\overline{N}_{0}(r)+O(1). \end{aligned}$$
(9)

From (6) and (9) we get

$$\begin{aligned} (n+m)T(r,h) \leq& \overline{N}(r,h)+\overline{N} \biggl(r, \frac {1}{h} \biggr)+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr) \\ &{}+T \biggl(r,\frac{1}{L(z,h)} \biggr)+S(r,h) \\ \leq& (d+2)T(r,h)+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+S(r,h). \end{aligned}$$
(10)

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

$$ \overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr) \leq \overline{N} \biggl(r,\frac{1}{h^{m}-1} \biggr) \leq mT(r,h)+S(r,h). $$
(11)

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

$$\begin{aligned} (n+m)T(r,h)=T\bigl(r,L(z,h)\bigr)\leq dT(r,h)+S(r,h), \end{aligned}$$

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

$$\begin{aligned} \frac{F^{(k)}-1}{G^{(k)}-1}=C, \end{aligned}$$

where C is a nonzero constant, that is,

$$ F^{(k)}=CG^{(k)}-C+1. $$
(12)

Integrating both sides of (12), we have

$$ F=CG+\frac{1-C}{k!}z^{k}+p_{1}(z), $$
(13)

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

$$\begin{aligned} T(r,F) \leq& \overline{N}(r,F)+\overline{N} \biggl(r, \frac{1}{F} \biggr) +\overline{N} \biggl(r,\frac{1}{F-p} \biggr)+S(r,f) \\ \leq& \overline{N}(r,f)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{f} \biggr) +\overline{N}_{0}(r)+\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& 2T(r,f)+2T(r,g)+\overline{N}_{1}(r)+\overline{N}_{0}(r)+S(r,f)+S(r,g), \end{aligned}$$
(14)

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,

$$\begin{aligned}& nm(r,f)=m\bigl(r,f^{n}\bigr)\leq m(r,F)+m \biggl(r, \frac{1}{f(qz+c)} \biggr)+O(1). \end{aligned}$$
(15)
$$\begin{aligned}& \begin{aligned}[b]nN(r,f)&=N\bigl(r,f^{n}\bigr)=N \biggl( \frac{F(z)}{f(qz+c)} \biggr) \\ &\leq N(r,F)+N \biggl(r,\frac{1}{f(qz+c)} \biggr)-\overline{N}_{1}(r)- \overline {N}_{0}(r).\end{aligned} \end{aligned}$$
(16)

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

$$ (n-1)T(r,f)\leq T(r,F)-\overline{N}_{1}(r)- \overline{N}_{0}(r)+O(1). $$
(17)

Substituting (14) into (17), we obtain

$$ (n-3)T(r,f)\leq2T(r,g)+S(r,f)+S(r,g). $$
(18)

Using the same method, we also get

$$ (n-3)T(r,g)\leq2T(r,f)+S(r,f)+S(r,g). $$
(19)

Combining (18) with (19), we have

$$\begin{aligned} (n-5) \bigl(T(r,g)+T(r,f)\bigr)\leq S(r,f)+S(r,g), \end{aligned}$$

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

$$\begin{aligned} f^{n}f(qz+c)=g^{n}g(qz+c). \end{aligned}$$

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

$$\begin{aligned} nT(r,h)=T\bigl(r,h(qz+c)\bigr)\leq T(r,h)+S(r,h), \end{aligned}$$

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

$$ \frac{f(z)^{n}L(z,f)^{s}-1}{g(z)^{n}L(z,g)^{s}-1}=E, $$
(20)

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

$$ Eg(z)^{n}L(z,g)^{s}=f(z)^{n}L(z,f)^{s}-1+E. $$
(21)

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

$$\begin{aligned} T(r,F) \leq& \overline{N}(r,F)+\overline{N} \biggl(r, \frac{1}{F} \biggr)+\overline{N} \biggl(r,\frac{1}{F-1+E} \biggr) +S(r,f) \\ \leq& \overline{N}(r,f)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{f} \biggr) +\overline{N}_{0}(r)+\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& 2T(r,f)+\overline{N} \biggl(r,\frac{1}{g} \biggr)+\overline{N} \biggl(r,\frac{1}{L(z,g)} \biggr) +\overline{N}_{1}(r) + \overline{N}_{0}(r)+S(r,f) \\ \leq& 2T(r,f)+(d+1)T(r,g)+\overline{N}_{1}(r)+\overline {N}_{0}(r)+S(r,f)+S(r,g), \end{aligned}$$
(22)

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

$$\begin{aligned}& nm(r,f)=m\bigl(r,f^{n}\bigr)\leq m(r,F)+sm \biggl(r, \frac{1}{L(z,f)} \biggr)+O(1), \\& \begin{aligned}nN(r,f)&=N\bigl(r,f^{n}\bigr)=N \biggl(r,F(z) \biggl[ \frac{1}{L(z,f)} \biggr]^{s} \biggr) \\ &\leq N(r,F)+sN \biggl(r,\frac{1}{L(z,f)} \biggr)-\overline{N}_{1}(r)- \overline{N}_{0}(r).\end{aligned} \end{aligned}$$

So

$$ \begin{aligned} nT(r,f)&\leq T(r,F)+sT \biggl(r, \frac{1}{L(z,f)} \biggr)-\overline {N}_{1}(r)- \overline{N}_{0}(r)+O(1) \\ &\leq T(r,F)+dsT(r,f)-\overline{N}_{1}(r)-\overline{N}_{0}(r)+S(r,f). \end{aligned} $$
(23)

Substituting (22) into (23), we obtain

$$ (n-ds-2)T(r,f)\leq(d+1)T(r,g)+S(r,f)+S(r,g). $$
(24)

Using the same method, we can get

$$ (n-ds-2)T(r,g)\leq(d+1)T(r,f)+S(r,f)+S(r,g). $$
(25)

Combining (24) with (25), it follows

$$\begin{aligned} (n-d-ds-3) \bigl(T(r,g)+T(r,f)\bigr)\leq S(r,f)+S(r,g), \end{aligned}$$

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

$$\begin{aligned} f(z)^{n}L(z,f)^{s}=g(z)^{n}L(z,g)^{s}. \end{aligned}$$

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

$$ h(z)^{n}\bigl[L(z,h)\bigr]^{s}=1. $$
(26)

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

$$\begin{aligned} f(z)^{n}\bigl(f(qz+c)-f(z)\bigr)=g(z)^{n} \bigl(g(qz+c)-g(z)\bigr). \end{aligned}$$

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

$$ g(qz+c) \bigl(H(z)-1 \bigr)=g(z) \bigl(h(z)^{n+1}-1 \bigr). $$
(27)

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

$$\begin{aligned} T\bigl(r,H(z)\bigr) \leq& \overline{N}\bigl(r,H(z)\bigr)+ \overline{N} \biggl(r,\frac {1}{H(z)} \biggr) +\overline{N} \biggl(r, \frac{1}{H(z)-1} \biggr)+S(r,H) \\ \leq& 2T\bigl(r,h(z)\bigr)+2T\bigl(r,h(qz+c)\bigr)+S(r,h) \\ \leq& 4T\bigl(r,h(z)\bigr)+S(r,h). \end{aligned}$$
(28)

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

$$\begin{aligned} nT(r,h)&=T\bigl(r,h(z)^{n}\bigr)\leq T\bigl(r,H(z) \bigr)+T \biggl(r,\frac{1}{h(qz+c)} \biggr)+S(r,h) \\ &\leq5T\bigl(r,h(z)\bigr)+S(r,h), \end{aligned}$$

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

$$\begin{aligned} \overline{N} \biggl(r,\frac{1}{H(z)-1} \biggr) \leq \overline{N} \biggl(r,\frac{1}{h(qz+c)-h(z)} \biggr)\leq2T\bigl(r,h(z) \bigr)+S(r,h). \end{aligned}$$
(29)

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

$$\begin{aligned} T\bigl(r,H(z)\bigr) \leq& \overline{N}\bigl(r,H(z)\bigr)+ \overline{N} \biggl(r,\frac{1}{H(z)} \biggr) +\overline{N} \biggl(r, \frac{1}{H(z)-1} \biggr)+S(r,h) \\ \leq& \overline{N}(r,h)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{h} \biggr)+\overline{N}_{0}(r) +\overline{N} \biggl(r,\frac{1}{H(z)-1} \biggr)+S(r,h) \\ \leq& 4T(r,h)+\overline{N}_{1}(r)+\overline{N}_{0}(r)+S(r,h), \end{aligned}$$
(30)

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

$$\begin{aligned}& nm(r,h)=m\bigl(r,h^{n}\bigr)\leq m\bigl(r,H(z) \bigr)+m \biggl(r,\frac{1}{h(qz+c)} \biggr)+O(1), \end{aligned}$$
(31)
$$\begin{aligned}& nN(r,h)=N\bigl(r,h^{n}\bigr)\leq N\bigl(r,H(z) \bigr)+N \biggl(r,\frac{1}{h(qz+c)} \biggr)-\overline{N}_{1}(r)- \overline{N}_{0}(r). \end{aligned}$$
(32)

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

$$ \begin{aligned}[b] nT(r,h)&\leq T\bigl(r,H(z)\bigr)+T \biggl(r,\frac{1}{h(qz+c)} \biggr)-\overline {N}_{1}(r)- \overline{N}_{0}(r)+O(1) \\ &\leq T\bigl(r,H(z)\bigr)+T(r,h)-\overline{N}_{1}(r)- \overline{N}_{0}(r)+S(r,h). \end{aligned} $$
(33)

Substituting (30) into (33), we obtain

$$\begin{aligned} nT(r,h)\leq5T(r,h)+S(r,h), \end{aligned}$$

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

$$\begin{aligned} nT(r,h)=T\bigl(r,h(qz+c)\bigr)\leq T(r,h)+S(r,h), \end{aligned}$$

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. □

References

  1. 1.

    Chiang, Y.M., Feng, S.J.: On the Nevanlinna characteristic of \(f(z+\eta)\) and difference equations in the complex plane. Ramanujan J. 16(1), 105–129 (2008)

  2. 2.

    Hayman, W.K.: Meromorphic Functions. Clarendon Press, Oxford (1964)

  3. 3.

    Laine, I.: Nevanlinna Theory and Complex Differential Equations. de Gruyter, Berlin (1993)

  4. 4.

    Liu, K., Liu, X.L., Cao, T.B.: Uniqueness and zeros of q-shift difference polynomials. Proc. Indian Acad. Sci. Math. Sci. 121(3), 301–310 (2011)

  5. 5.

    Liu, K., Liu, X.L., Cao, T.B.: Value distribution and uniqueness of difference polynomials. Adv. Differ. Equ. 2011, Article ID 234215 (2011)

  6. 6.

    Liu, K., Liu, X.L., Cao, T.B.: Some difference results on Hayman conjecture and uniqueness. Bull. Iran. Math. Soc. 38(4), 1007–1020 (2012)

  7. 7.

    Liu, K., Qi, X.G.: Meromorphic solutions of q-shift difference equations. Ann. Pol. Math. 101, 215–225 (2011)

  8. 8.

    Wang, Q.Y., Ye, Y.S.: Value distribution and uniqueness of difference polynomials for meromorphic functions. Chin. Ann. Math., Ser. A 35, 675–684 (2014) (in Chinese)

  9. 9.

    Yang, C.C., Hua, X.H.: Uniqueness and value sharing of meromorphic functions. Ann. Acad. Sci. Fenn., Math. 22(2), 395–406 (1997)

  10. 10.

    Yang, C.C., Yi, H.X.: Uniqueness Theory of Meromorphic Functions. Kluwer Academic, Dordrecht (2003)

  11. 11.

    Zhang, J.L., Korhonen, R.J.: On the Nevanlinna characteristic of \(f(qz)\) and its applications. J. Math. Anal. Appl. 369, 537–544 (2010)

  12. 12.

    Zhang, K.Y.: Uniqueness of q-difference polynomials of meromorphic functions. Adv. Mater. Res. 756–759, 2948–2951 (2013)

  13. 13.

    Zhao, Q.X., Zhang, J.L.: Zeros and shared one value of q-shift difference polynomials. J. Contemp. Math. Anal. 50(2), 63–69 (2015)

  14. 14.

    Zheng, X.M., Xu, H.Y.: On value distribution and uniqueness of meromorphic function with finite logarithmic order concerning its derivative and q-shift difference. J. Inequal. Appl. 2014, Article ID 295 (2014)

Download references

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).

Author information

Affiliations

Authors

Contributions

Both authors drafted the manuscript, read, and approved the final manuscript.

Corresponding author

Correspondence to Kai Liu.

Ethics declarations

Competing interests

The authors declare 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

Verify currency and authenticity via CrossMark

Cite this article

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

Download citation

MSC

  • 30D35

Keywords

  • Meromorphic functions
  • q-Difference-differential polynomials
  • Sharing values
  • Common zeros
  • Common poles