Properties of q-shift difference-differential polynomials of meromorphic functions

Abstract

In this paper, we deal with the zeros of the q-shift difference-differential polynomials ${[P(f){\prod }_{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}]}^{(k)}-\alpha (z)$ and ${(P(f){\prod }_{j=1}^d{[f(q_{j}z+c_j)-f(z)]}^{s_j})}^{(k)}-\alpha (z)$, where $P(f)$ is a nonzero polynomial of degree n, $q_j,c_j\in \mathbb{C}\setminus \{0\}$ ($j=1,\dots ,d$) are constants, $n,d,s_j(j=1,\dots ,d)\in {\mathbb{N}}_{+}$ and $\alpha (z)$ is a small function of f. The results of this paper are an extension of the previous theorems given by Chen and Chen and Qi. We also investigate the value sharing for q-shift difference polynomials of entire functions and obtain some results which extend the recent theorem given by Liu, Liu and Cao.

MSC:39A50, 30D35.

1 Introduction and main results

The purpose of this paper is to study some properties of zeros and uniqueness of complex q-shift difference polynomials of meromorphic functions. A polynomial $Q_q(z,f)$ can be called a q-shift difference-differential polynomial in f whenever $Q_q(z,f)$ is a polynomial in $f(z)$, its q-shift $f(qz+c)$ and their derivatives, with small functions of $f(z)$ as the coefficients. The fundamental results and the standard notations of the Nevanlinna value distribution theory of meromorphic functions will be used (see [1, 2]). A meromorphic function f means f is meromorphic in the complex plane. If no poles occur, then f reduces to an entire function. Given a meromorphic function $f(z)$, a meromorphic function $a(z)$ is called a small function with respect to f if $T(r,a(z))=S(r,f)$, where $S(r,f)$ is used to denote any quantity satisfying $S(r,f)=o(T(r,f))$ for all r outside a possible exceptional set E of finite logarithmic measure ${lim}_{r\to \mathrm{\infty }}{\int }_{[1,r)\cap E}\frac{dt}{t}<\mathrm{\infty }$. We also use $S_1(r,f)$ to denote any quantity satisfying $S_1(r,f)=o(T(r,f))$ for all r on a set F of logarithmic density 1; the logarithmic density of a set F is defined by

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{1}{logr}{\int }_{[1,r]\cap F}\frac{1}{t}dt.$$

In addition, for some $a\in \mathbb{C}\cup \{\mathrm{\infty }\}$, if the zeros of $f(z)-a$ and $g(z)-a$ (if $a=\mathrm{\infty }$, zeros of $f(z)-a$ and $g(z)-a$ are the poles of $f(z)$ and $g(z)$, respectively) coincide in locations and multiplicities we say that $f(z)$ and $g(z)$ share the value a CM (counting multiplicities) and if they coincide in locations only we say that $f(z)$ and $g(z)$ share a IM (ignoring multiplicities).

Definition 1.1 (see [3, 4])

Let l be a nonnegative integer or infinity. For $a\in \mathbb{C}\cup \{\mathrm{\infty }\}$, we denote by $E_l(a;f)$ the set of all a-points of f where an a-point of multiplicity k is counted k times if $k\le l$ and $l+1$ times if $k>l$. If $E_l(a;f)=E_l(a;g)$, we say that f, g share the value a with weight l.

Definition 1.2 (see [5])

When f and g share 1 IM, we denote by ${\overline{N}}_L(r,\frac{1}{f-1})$ the counting function of the 1-points of f whose multiplicities are greater than 1-points of g, where each zero is counted only once; similarly, we have ${\overline{N}}_L(r,\frac{1}{g-1})$. Let $z_0$ be a zero of $f-1$ of multiplicity p and a zero of $g-1$ of multiplicity q, we also denote by $N_{11}(r,\frac{1}{f-1})$ the counting function of those 1-points of f where $p=q=1$.

In recent years, there has been an increasing interest in studying difference equations, difference products and q-differences in the complex plane ℂ, and a number of papers (including [6–13]) have focused on the value distribution and uniqueness of differences and differences operator analogs of Nevanlinna theory.

For a transcendental meromorphic function f of finite order, herein and hereinafter, c is a nonzero complex constant and $a(z)$ is small function with respect to f, Liu et al. [14], Chen et al. [15], and Luo and Lin [16] studied the zeros distributions of difference polynomials of meromorphic functions and obtained: if $n\ge 6$, then $f{(z)}^{n}f(z+c)-\alpha (z)$ has infinitely many zeros [[14], Theorem 1.2]; if $n>m$, then $P(f)f(z+c)-a(z)$ has infinitely many zeros (see [[16], Theorem 1]), where $P(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0$ is a nonzero polynomial, where $a_0,a_1,\dots ,a_n$ (≠0) are complex constants, and m is the number of the distinct zeros of $P(z)$.

For transcendental meromorphic (resp. entire) function f of zero order and nonzero complex constant q, Zhang and Korhonen [17] studied the value distribution of q-difference polynomials of meromorphic functions and obtained the result that if $n\ge 6$ (resp. $n\ge 2$), then $f{(z)}^{n}f(qz)$ assumes every nonzero value $a\in \mathbb{C}$ infinitely often (see [[17], Theorem 4.1]).

Recently, Liu and Qi [18] firstly investigated the value distributions for q-shift of meromorphic function and obtained the following result.

Theorem A (see [[18], Theorem 3.6])

Let f be a zero-order transcendental meromorphic function, $n\ge 6$, $q\in \mathbb{C}\setminus \{0\}$, $\eta \in \mathbb{C}$, and let $R(z)$ be a rational function. Then the q-shift difference polynomial $f{(z)}^{n}f(qz+\eta )-R(z)$ has infinitely many zeros.

In this paper, we assume $q_j,c_j\in \mathbb{C}\setminus \{0\}$ ($j=1,\dots ,d$) are constants, $n,d,s_j(j=1,\dots ,d)\in {\mathbb{N}}_{+}$, $\lambda =s_1+s_2+\cdots +s_d$, and $a(z)$ is a small function of f. We will study the value distribution of difference polynomials of more general form,

$$F(z)=P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j},$$

(1)

$$F_1(z)=P(f)\prod _{j=1}^d{[f(q_{j}z+c_j)-f(z)]}^{s_j},$$

(2)

where $P(f)$ is a nonzero polynomial of degree n, m is the number of the distinct zeros of $P(z)$, and we obtain the following results.

Theorem 1.1 Let f be a transcendental meromorphic (resp. entire) function of zero order and $F(z)$ be stated as in (1). If $k\in \mathbb{N}$ and $n>m(k+1)+2d+1+\lambda$ (resp. $n>m(k+1)+d-\lambda$). Then ${(F(z))}^{(k)}-a(z)$ has infinitely many zeros, where ${(F(z))}^{(k)}=F(z)$, if $k=0$.

Theorem 1.2 Let f be a transcendental meromorphic (resp. entire) function of zero order and $F_1(z)$ be stated as in (2). Assume $k\in \mathbb{N}$ and $n>(m+2d)(k+1)+\lambda +d+1$ (resp. $n>(m+2d)(k+1)$). Then ${(F_1(z))}^{(k)}-a(z)$ has infinitely many zeros, provided that $f(q_{j}z+c_j)\ne f(z)$, $j=1,2,\dots ,d$.

Recently, there were obtained some results on the existence and growth of solutions of difference-differential equations (see [19, 20]). Here, from Theorem 1.1 and Theorem 1.2, we get the following result on some nonlinear q-shift difference-differential equations.

Corollary 1.1 Let $p(z)$, $q(z)$ be nonzero polynomials and $F(z)$ be stated as in (1). Then the nonlinear q-shift difference-differential equation

$${[F(z)]}^{(k)}-p(z)=q(z)$$

(3)

has no transcendental meromorphic solution of zero order, provided that $n\ge \lambda +1$.

Corollary 1.2 Let $p(z)$, $q(z)$ be nonzero polynomials and $F_1(z)$ be stated as in (2). Then the nonlinear q-shift difference-differential equation

$${[F_1(z)]}^{(k)}-p(z)=q(z)$$

(4)

has no transcendental meromorphic solution of zero order, provided that $f(q_{j}z+c_j)\ne f(z)$, $j=1,2,\dots ,d$, and $n\ge \lambda +1$.

For the uniqueness of difference and q-difference of meromorphic functions, some results had been obtained (see [17, 21–23]). Here, we only state some of the latest theorems as follows.

Theorem B (see [[16], Theorem 2])

Let f and g be transcendental entire functions of finite order, c be a nonzero complex constant, $P(z)$ be stated as in Theorem  1.3, and let $n>2{\mathrm{\Gamma }}_0+1$ be an integer, where ${\mathrm{\Gamma }}_0=m_1+2m_2$, $m_1$ is the number of the simple zero of $P(z)$, and $m_2$ is the number of multiple zeros of $P(z)$. If $P(f)f(z+c)$ and $P(g)g(z+c)$ share 1 CM, then one of the following results holds:

• (i) $f\equiv tg$ for a constant t such that $t^l=1$, where $l=GCD\{{\lambda }_0,{\lambda }_1,\dots ,{\lambda }_n\}$ and

  $${\lambda }_i=\{\begin{array}{ll}i+1,& a_i\ne 0,\\ n+1,& a_i=0,\end{array}i=0,1,2,\dots ,n.$$

• (ii) f and g satisfy the algebraic equation $R(f,g)\equiv 0$, where $R({\omega }_1,{\omega }_2)=P({\omega }_1){\omega }_1(z+c)-P({\omega }_2){\omega }_2(z+c)$;

• (iii) $f(z)=e^{\alpha (z)}$, $g(z)=e^{\beta (z)}$, where $\alpha (z)$ and $\beta (z)$ are two polynomials, b is a constant satisfying $\alpha +\beta \equiv b$ and $a_n^2{e}^{(n+1)b}=1$.

In this paper, we will investigated the uniqueness problem of q-shifts of entire functions and obtain the following results.

Theorem 1.3 Let f, g be transcendental entire functions of zero order, $F(z)$ be stated as in (1) and

$$G(z)=P(g)\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j},$$

where ${\mathrm{\Gamma }}_0$ is stated as in Theorem B. If $F(z)$ and $G(z)$ share 1 CM and $n>max\{2({\mathrm{\Gamma }}_0+2d)-\lambda ,\lambda \}$, then one of the following cases holds:

• (i) $f\equiv tg$ for a constant t such that $t^{\kappa }=1$ where $\kappa =GCD\{{\lambda }_0+\lambda ,{\lambda }_1+\lambda ,\dots ,{\lambda }_n+\lambda \}$ and

  $${\lambda }_i=\{\begin{array}{ll}i+1,& a_i\ne 0,\\ n+1,& a_i=0,\end{array}i=0,1,2,\dots ,n.$$

• (ii) f and g satisfy the algebraic equation $R(f,g)\equiv 0$, where

  $$R({\omega }_1,{\omega }_2)=P({\omega }_1)\prod _{j=1}^d{\omega }_1{(q_{j}z+c_j)}^{s_j}-P({\omega }_2)\prod _{j=1}^d{\omega }_2{(q_{j}z+c_j)}^{s_j}.$$

Theorem 1.4 Under the assumptions of Theorem  1.3, if

$$E_l(1;P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j})=E_l(1;P(g)\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j})$$

and l, n, m are integers satisfying one of the following conditions:

• (I) $l\ge 3$, $n>max\{2{\mathrm{\Gamma }}_0+4d-\lambda ,\lambda \}$;

• (II) $l=2$, $n>max\{2{\mathrm{\Gamma }}_0+5d+m-\lambda -d\chi ,\lambda \}$;

• (III) $l=1$, $n>max\{2{\mathrm{\Gamma }}_0+6d+2m-\lambda -2d\chi ,\lambda \}$;

• (IV) $l=0$, $n>max\{2{\mathrm{\Gamma }}_0+7d+3m-\lambda -3d\chi ,\lambda \}$.

Then the conclusions of Theorem  1.3 hold, where $\chi =min\{\mathrm{\Theta }(0,f),\mathrm{\Theta }(0,g)\}$.

2 Some lemmas

In the following, we explain some definitions and notations which are used in this paper. For $a\in \mathbb{C}\cup \{\mathrm{\infty }\}$, we define

$$\mathrm{\Theta }(a,f)=1-\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\overline{N}(r,\frac{1}{f-a})}{T(r,f)}.$$

For $a\in \mathbb{C}\cup \{\mathrm{\infty }\}$ and k is a positive integer, we denote by ${\overline{N}}_{(k}(r,\frac{1}{f-a})$ the counting function of those a-points of f whose multiplicities are not less than k in counting the a-points of f we ignore the multiplicities (see [1]) and $N_k(r,\frac{1}{f-a})=\overline{N}(r,\frac{1}{f-a})+{\overline{N}}_{(2}(r,\frac{1}{f-a})+\cdots +{\overline{N}}_{(k}(r,\frac{1}{f-a})$.

Lemma 2.1 (see [2])

Let f and g be two meromorphic functions. If f and g share 1 CM, then one of the following three cases holds:

• (i) $T(r,f)+T(r,g)\le 2N_2(r,f)+2N_2(r,g)+2N_2(r,\frac{1}{f})+2N_2(r,\frac{1}{g})+S(r,f)+S(r,g)$;

• (ii) $f\equiv g$;

• (iii) $f\cdot g=1$.

Lemma 2.2 (see [24])

Let f and g be two meromorphic functions, and let l be a positive integer. If $E_l(1;f)=E_l(1;g)$, then one of the following cases must occur:

• (i)
  $$\begin{array}{rl}T(r,f)+T(r,g)\le & N_2(r,f)+N_2(r,g)+N_2(r,\frac{1}{f})+N_2(r,\frac{1}{g})+\overline{N}(r,\frac{1}{f-1})\\ +\overline{N}(r,\frac{1}{g-1})-N_{11}(r,\frac{1}{f-1})+{\overline{N}}_{(l+1}(r,\frac{1}{f-1})\\ +{\overline{N}}_{(l+1}(r,\frac{1}{g-1})+S(r,f)+S(r,g);\end{array}$$

• (ii) $f=\frac{(b+1)g+(a-b-1)}{bg+(a-b)}$, where a (≠0), b are two constants.

Lemma 2.3 (see [24])

Let f and g be two meromorphic functions. If f and g share 1 IM, then one of the following cases must occur:

• (i)
  $$\begin{array}{rl}T(r,f)+T(r,g)\le & 2[N_2(r,f)+N_2(r,\frac{1}{f})+N_2(r,g)+N_2(r,\frac{1}{g})]\\ +3{\overline{N}}_L(r,\frac{1}{f-1})+3{\overline{N}}_L(r,\frac{1}{g-1})+S(r,f)+S(r,g);\end{array}$$

• (ii) $f=\frac{(b+1)g+(a-b-1)}{bg+(a-b)}$, where a (≠0), b are two constants.

By combining [7] and [17], we get the following lemma easily.

Lemma 2.4 Let $f(z)$ be a transcendental meromorphic function of zero order and q, η be two nonzero complex constants. Then

$$\begin{array}{r}T(r,f(qz+\eta ))=T(r,f(z))+S_1(r,f),N(r,\frac{1}{f(qz+\eta )})=N(r,\frac{1}{f})+S_1(r,f),\\ N(r,f(qz+\eta ))=N(r,f)+S_1(r,f),\overline{N}(r,\frac{1}{f(qz+\eta )})=\overline{N}(r,\frac{1}{f})+S_1(r,f),\\ \overline{N}(r,f(qz+\eta ))=\overline{N}(r,f)+S_1(r,f).\end{array}$$

Lemma 2.5 (see [[18], Theorem 2.1])

Let $f(z)$ be a nonconstant zero-order meromorphic function and $q\in \mathbb{C}\setminus \{0\}$. Then

$$m(r,\frac{f(qz+\eta )}{f(z)})=S(r,f),$$

on a set of logarithmic density 1.

Lemma 2.6 Let f be a transcendental meromorphic function of zero order, and $F(z)$ be stated as in (1). Then we have

$$(n-\lambda )T(r,f)+S_1(r,f)\le T(r,F(z))\le (n+\lambda )T(r,f)+S_1(r,f).$$

(5)

If f is a transcendental entire function of zero order, we have

$$T(r,F(z))=T(r,P(f)f^{\lambda })+S_1(r,f)=(n+\lambda )T(r,f)+S_1(r,f),$$

(6)

where $\lambda =s_1+s_2+\cdots +s_d$.

Proof If f is a transcendental entire function of zero order, from the Valiron-Mohon’ko lemma and Lemma 2.5, we have

$$\begin{array}{rl}T(r,F(z))& =m(r,F(z))\le m(r,P(f)f^{\lambda }(z))+m(r,\frac{{\prod }_{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}}{f^{\lambda }(z)})\\ \le m(r,P(f)f^{\lambda }(z))+S_1(r,f)=T(r,P(f)f^{\lambda }(z))+S_1(r,f)\\ =(n+\lambda )T(r,f)+S_1(r,f).\end{array}$$

On the other hand, from Lemma 2.5, we have

$$\begin{array}{rl}(n+\lambda )T(r,f)& =T(r,P(f)f^{\lambda }(z))+S(r,f)=m(r,P(f)f^{\lambda }(z))+S(r,f)\\ \le m(r,F(z))+m(r,\frac{f^{\lambda }(z)}{{\prod }_{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}})\\ =T(r,F(z))+S_1(r,f).\end{array}$$

Thus, we get (6).

If f is a meromorphic function of zero order, from the Valiron-Mohon’ko lemma and Lemma 2.4, we have

$$T(r,P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j})\le T(r,P(f))+T(r,\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j})\le (n+\lambda )T(r,f)+S_1(r,f).$$

On the other hand, from the Valiron-Mo’honko lemma and Lemma 2.5, we have

$$\begin{array}{rl}(n+\lambda )T(r,f)=& T(r,P(f)f^{\lambda })+S(r,f)=m(r,P(f)f^{\lambda })+N(r,P(f)f^{\lambda })+S(r,f)\\ \le & m(r,F(z)\frac{f^{\lambda }(z)}{{\prod }_{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}})\\ +N(r,F(z)\frac{f^{\lambda }(z)}{{\prod }_{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}})+S(r,f)\\ \le & T(r,F(z))+2\lambda T(r,f)+S_1(r,f).\end{array}$$

Thus, we get (5). □

Using the same method as in Lemma 2.6, we get the following lemma easily.

Lemma 2.7 Let f be a transcendental meromorphic function of zero order, and $F_1(z)$ be stated as in (2). Then we have

$$(n-\lambda )T(r,f)+S_1(r,f)\le T(r,F_1(z))\le (n+2\lambda )T(r,f)+S_1(r,f).$$

If f is a transcendental entire function of zero order, we have

$$T(r,F_1(z))=T(r,P(f)\prod _{j=1}^d{[f(q_{j}z+c_j)-f(z)]}^{s_j})\ge nT(r,f)+S_1(r,f).$$

Lemma 2.8 (see [2] and [[25], Lemma 2.4])

Let f be a nonconstant meromorphic function, and p, k be positive integers. Then

$$\begin{array}{r}T(r,f^{(k)})\le T(r,f)+k\overline{N}(r,f)+S(r,f),\\ N_p(r,\frac{1}{f^{(k)}})\le T(r,f^{(k)})-T(r,f)+N_{p+k}(r,\frac{1}{f})+S(r,f),\\ N_p(r,\frac{1}{f^{(k)}})\le k\overline{N}(r,f)+N_{p+k}(r,\frac{1}{f})+S(r,f).\end{array}$$

Lemma 2.9 Let $f(z)$ and $g(z)$ be transcendental entire functions of zero order, $P(z)$ be stated as in Theorem  1.1. If $n>\lambda$, then for any complex constant $t\ne 0$, we have

$$P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}P(g)\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}\not\equiv t.$$

Proof For any complex constant $t\ne 0$, suppose that

$$P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}P(g)\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}\equiv t.$$

(7)

Suppose that the roots of $P(z)=0$ are $b_1,b_2,\dots ,b_m$ with multiplicities $l_1,l_2,\dots ,l_m$. Then we have $l_1+l_2+\cdots +l_m=n$. From (7), we have

$$\begin{array}{r}{(f-b_1)}^{l_1}{(f-b_2)}^{l_2}\cdots {(f-b_m)}^{l_m}\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}\\ \times {(g-b_1)}^{l_1}{(g-b_2)}^{l_2}\cdots {(g-b_m)}^{l_m}\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}\equiv t.\end{array}$$

(8)

Since f, g are nonconstant entire functions, from (8), we deduce that $b_1=b_2=\cdots =b_m=0$. If fact, from (8), we get that $b_1,b_2,\dots ,b_m$ are the Picard exceptional values. If $m\ge 2$ and $b_j\ne 0$ ($j=1,2,\dots ,m$), by Picard’s theorem of entire function, we can see that the Picard exceptional values of f are at least three. Thus, we get a contradiction. Hence, $m=1$ and $l_1=n$, that is, there exists a complex constant γ satisfying $P(f)=a_n{(f-\gamma )}^n$ and $P(g)=a_n{(g-\gamma )}^n$. Then

$$a_n{(f-\gamma )}^n\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}{a}_n{(g-\gamma )}^n\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}\equiv t.$$

(9)

Since f, g are transcendental entire functions, by the Picard theorem, we can see that $f-\gamma =0$ and $g-\gamma =0$ do not have zeros. Then we obtain $f(z)=e^{\alpha (z)}+\gamma$, $g(z)=e^{\beta (z)}+\gamma$ where $\alpha (z)$, $\beta (z)$ are two nonconstant functions. From (9), we see that $f(q_{j}z+c_j)\ne 0$ and $g(q_{j}z+c_j)\ne 0$. Thus, we get $\gamma =0$, that is,

$$a_n^{2}f{(z)}^n\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}g{(z)}^n\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}\equiv t.$$

(10)

Set $M(z)=f(z)g(z)$. If $M(z)$ is nonconstant, from (10), we have

$$a_n^{2}M{(z)}^n\prod _{j=1}^{d}M{(q_{j}z+c_j)}^{s_j}\equiv t,$$

that is,

$$a_n^{2}M{(z)}^n\equiv \frac{t}{{\prod }_{j=1}^{d}M{(q_{j}z+c_j)}^{s_j}}.$$

(11)

Since f, g are transcendental entire functions of zero order, from (11), Lemma 2.4 and $n>\lambda$, we get a contradiction.

Thus, $M(z)$ is a constant. From (11), we get $f(z)g(z)\equiv \mu$, where μ is a complex constant satisfying $a_n^2{\mu }^{n+\lambda }\equiv t$. Since f, g are entire functions of zero order, then f, g are constants, which is a contradiction with f, g being transcendental. Hence,

$$P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}P(g)\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}\not\equiv t.$$

This completes the proof of Lemma 2.9. □

3 Proofs of Theorems 1.1 and 1.2

Proof of Theorem 1.1 From (1), by the Valiron-Mohon’ko lemma and Lemma 2.6, we find that $F(z)$ is not constant and $S_1(r,F^{(k)})=S_1(r,F)=S_1(r,f)$. Next, we will consider the following two cases when $k\ge 1$.

Case 1. If f is a transcendental meromorphic function of zero order, we first suppose that ${(P(f){\prod }_{j=1}^{d}f{(q_{j}z+c_j)}^{s_j})}^{(k)}=a(z)$ has finitely solutions. By the Second Fundamental Theorem for three small functions (see [[1], Theorem 2.25]) and the Valiron-Mohon’ko lemma, we have

$$\begin{array}{rl}T(r,F^{(k)})& \le \overline{N}(r,F^{(k)})+\overline{N}(r,\frac{1}{F^{(k)}})+\overline{N}(r,\frac{1}{F^{(k)}-a(z)})+S(r,F^{(k)})\\ \le \overline{N}(r,f)+\sum _{j=1}^d\overline{N}(r,f(q_{j}z+c_j))+N_1(r,\frac{1}{F^{(k)}})+S(r,F^{(k)})\\ \le (d+1)T(r,f)+T(r,F^{(k)})-T(r,F)+N_{k+1}(r,\frac{1}{F^{(k)}})+S(r,F^{(k)}).\end{array}$$

By Lemma 2.6 and Lemma 2.8, we obtain

$$\begin{array}{r}(n-\lambda )T(r,f)+S_1(r,f)\\ \le T(r,F)\le (d+1)T(r,f)+N_{k+1}(r,\frac{1}{F^{(k)}})+S(r,f)\\ \le (d+1)T(r,f)+m(k+1)\overline{N}(r,\frac{1}{f})+\sum _{j=1}^{d}N(r,\frac{1}{f(q_{j}z+c_j)})+S_1(r,f)+S(r,f)\\ \le [m(k+1)+2d+1]T(r,f)+S_1(r,f)+S(r,f).\end{array}$$

(12)

From the definitions of $S_1(r,f)$, $S(r,f)$ and $n>m(k+1)+2d+1+\lambda$, we get a contradiction to (12). Then $F{(z)}^{(k)}-a(z)$ has infinitely many zeros.

Case 2. If f is a transcendental entire function. Suppose that ${(P(f){\prod }_{j=1}^{d}f{(q_{j}z+c_j)}^{s_j})}^{(k)}=a(z)$ has finitely solutions. By using the same argument as in Case 1 and (4), we have

$$(n+\lambda )T(r,f)+S_1(r,f)\le [m(k+1)+d]T(r,f)+S_1(r,f)+S(r,f),$$

which is a contradiction with $n>m(k+1)+d-\lambda$.

For $k=0$, similar to the proofs of Case 1 and Case 2, and by the Second Fundamental Theorem and Lemma 2.6, we get the conclusions of Theorem 1.1.

Thus, we complete the proof of Theorem 1.1. □

Proof of Theorem 1.2 Similar to the proof of Theorem 1.1, and using Lemma 2.8, we can prove Theorem 1.2 easily. □

4 Proofs of Corollaries 1.1 and 1.2

The proofs of Corollaries 1.1 and 1.2 are similar. Here, we just give the proof of Corollary 1.2.

Proof of Corollary 1.2 Assume that f is a transcendental meromorphic solution of zero order of (4), provided that $f(q_{j}z+c_j)\ne f(z)$, $j=1,2,\dots ,d$, then

$${[P(f)\prod _{j=1}^d{[f(q_{j}z+c_j)-f(z)]}^{s_j}]}^{(k)}=p(z)+q(z).$$

Integrating above equation k times, it follows

$$\frac{H(z)}{P(f)f^{\lambda }}=\frac{{\prod }_{j=1}^d{[f(q_{j}z+c_j)-f(z)]}^{s_j}}{f^{\lambda }},$$

where $H(z)$ is a polynomial. From Lemma 2.5 and since f is a meromorphic function of zero order, we have

$$\begin{array}{rl}(n+\lambda )T(r,f)& =T(r,\frac{{\prod }_{j=1}^d{[f(q_{j}z+c_j)-f(z)]}^{s_j}}{f^{\lambda }})\\ =N(r,\frac{{\prod }_{j=1}^d{[f(q_{j}z+c_j)-f(z)]}^{s_j}}{f^{\lambda }})+S_1(r,f)\\ \le 2\lambda T(r,f)+S_1(r,f),\end{array}$$

which is contradiction with $n>\lambda$.

This completes the proof of Corollary 1.2. □

5 Proofs of Theorems 1.3 and 1.4

Proof of Theorem 1.3 From the assumptions of Theorem 1.3, we see that $F(z)$, $G(z)$ share 1 CM. Then the following three cases will be considered.

Case 1. Suppose that $F(z)$, $G(z)$ satisfy Lemma 2.1(i). Since $f(z)$, $g(z)$ are entire functions of zero order, from the Valiron-Mohon’ko lemma and Lemma 2.6, we have $S_1(r,F)=S_1(r,f)$, $S_1(r,G)=S_1(r,g)$,

$$\begin{array}{rl}{N}_2(r,\frac{1}{F})& \le N_2(r,\frac{1}{P(f)})+\sum _{j=1}^d{N}_2(r,\frac{1}{f{(q_{j}z+c_j)}^{s_j}})\\ \le {\mathrm{\Gamma }}_{0}T(r,f)+2\sum _{j=1}^d\overline{N}(r,\frac{1}{f(q_{j}z+c_j)})\le ({\mathrm{\Gamma }}_0+2d)T(r,f)+S_1(r,f),\end{array}$$

(13)

and

$$N_2(r,\frac{1}{G})\le ({\mathrm{\Gamma }}_0+2d)T(r,g)+S_1(r,g).$$

(14)

Then, from Lemma 2.1(i) and Lemma 2.7, since f, g are entire, we have

$$\begin{array}{rl}T(r,F(z))+T(r,G(z))& \le 2N_2(r,\frac{1}{F})+2N_2(r,\frac{1}{G})+S(r,f)+S(r,g)\\ \le 2({\mathrm{\Gamma }}_0+2d)[T(r,f)+T(r,g)]+S_1(r,f)+S_1(r,g).\end{array}$$

(15)

From Lemma 2.6 and (12), we have

$$(n+\lambda )[T(r,f)+T(r,g)]\le 2({\mathrm{\Gamma }}_0+2d)[T(r,f)+T(r,g)]+S_1(r,f)+S_1(r,g),$$

that is,

$$(n+\lambda -2{\mathrm{\Gamma }}_0-4d)[T(r,f)+T(r,g)]\le S_1(r,f)+S_1(r,g).$$

(16)

Since $n>2({\mathrm{\Gamma }}_0+2d)-\lambda$ and f, g are transcendental functions, we get a contradiction.

Case 2. If $F(z)\equiv G(z)$, that is,

$$P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}\equiv P(g)\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}.$$

(17)

Set $h=\frac{f}{g}$. If h is not a constant, from (17), we find that f and g satisfy the algebraic equation $R(f,g)\equiv 0$, where

$$R({\omega }_1,{\omega }_2)=P({\omega }_1)\prod _{j=1}^d{\omega }_1{(q_{j}z+c_j)}^{s_j}-P({\omega }_2)\prod _{j=1}^d{\omega }_2{(q_{j}z+c_j)}^{s_j}.$$

If h is a constant. Substituting $f=gh$ into (17), we get

$$\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}[a_n{g}^n(h^{n+\lambda }-1)+a_{n-1}{g}^{n-1}(h^{n+\lambda -1}-1)+\cdots +a_0(h^{\lambda }-1)]\equiv 0,$$

(18)

where $a_n(\ne 0),a_{n-1},\dots ,a_0$ are constants.

Since g is transcendental entire function, we have ${\prod }_{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}\not\equiv 0$. Then, from (18), we have

$$a_n{g}^n(h^{n+\lambda }-1)+a_{n-1}{g}^{n-1}(h^{n+\lambda -1}-1)+\cdots +a_0(h^{\lambda }-1)\equiv 0.$$

(19)

If $a_n\ne 0$ and $a_{n-1}=a_{n-2}=\cdots =a_0=0$, then from (19) and g is transcendental function, we get $h^{n+\lambda }=1$.

If $a_n\ne 0$ and there exists $a_i\ne 0$ ($i\in \{0,1,2,\dots ,n-1\}$). Suppose that $h^{n+\lambda }\ne 1$, from (19), we have $T(r,g)=S(r,g)$ which is contradiction with transcendental function g. Then $h^{n+\lambda }=1$. Similar to this discussion, we can see that $h^{j+\lambda }=1$ when $a_j\ne 0$ for some $j=0,1,\dots ,n$.

Thus, from the definition of l, we can see that $f\equiv tg$ where t is a constant such that $t^{\kappa }=1$, $\kappa =GCD\{{\lambda }_0+\lambda ,{\lambda }_1+\lambda ,\dots ,{\lambda }_n+\lambda \}$.

Case 3. If $F(z)G(z)\equiv 1$. From Lemma 2.9, we get that $fg=\mu$ for a constant μ such that $a_n^2{\mu }^{n+\lambda }\equiv 1$.

Thus, this completes the proof of Theorem 1.3. □

Proof of Theorem 1.4 From the assumptions of Theorem 1.4, we have $E_l(1;F(z))=E_l(1;G(z))$.

• (I) $l\ge 3$. Since

  $$\begin{array}{r}\overline{N}(r,\frac{1}{F(z)-1})+\overline{N}(r,\frac{1}{G(z)-1})+{\overline{N}}_{(l+1}(r,\frac{1}{F(z)-1})\\ +{\overline{N}}_{(l+1}(r,\frac{1}{G(z)-1})-N_{11}(r,\frac{1}{F(z)-1})\\ \le \frac{1}{2}N(r,\frac{1}{F(z)-1})+\frac{1}{2}(r,\frac{1}{G(z)-1})+S(r,F)+S(r,G)\\ \le \frac{1}{2}T(r,F)+\frac{1}{2}T(r,G)+S(r,F)+S(r,G).\end{array}$$

Case 1. Suppose that $F(z)$, $G(z)$ satisfy Lemma 2.2(i). From (13), (14), and Lemma 2.6, we have

$$(n+\lambda )[T(r,f)+T(r,g)]\le 2({\mathrm{\Gamma }}_0+2d)[T(r,f)+T(r,g)]+S_1(r,f)+S_1(r,g),$$

that is,

$$(n+\lambda -2{\mathrm{\Gamma }}_0-4d)[T(r,f)+T(r,g)]\le +S_1(r,f)+S_1(r,g).$$

Since $n>2{\mathrm{\Gamma }}_0+4d-\lambda$ and f, g are transcendental, a contradiction is obtained.

Case 2. If $F(z)$, $G(z)$ satisfy Lemma 2.2(ii), that is,

$$F=\frac{(b+1)G+(a-b-1)}{bG+(a-b)},$$

(20)

where a (≠0), b are two constants.

We now will consider three subcases as follows.

Subcase 2.1. $b\ne 0,-1$. If $a-b-1\ne 0$, then by (20) we know

$$\overline{N}(r,\frac{1}{G+\frac{a-b-1}{b+1}})=\overline{N}(r,\frac{1}{F}).$$

Since f, g are entire functions of zero order, by the Second Fundamental Theorem, we have

$$\begin{array}{rl}T(r,G)& \le \overline{N}(r,\frac{1}{G})+\overline{N}(r,\frac{1}{G+\frac{a-b-1}{b+1}})+S(r,g)\le \overline{N}(r,\frac{1}{G})+\overline{N}(r,\frac{1}{F})+S(r,g)\\ \le (m+d)T(r,g)+(m+d)T(r,f)+S_1(r,f)+S_1(r,g).\end{array}$$

Then from Lemma 2.6, we have

$$(n+\lambda -m-d)T(r,g)\le (m+d)T(r,f)+S_1(r,f)+S_1(r,g).$$

Similarly, we have

$$(n+\lambda -m-d)T(r,f)\le (m+d)T(r,g)+S_1(r,f)+S_1(r,g).$$

From the definitions of m and ${\mathrm{\Gamma }}_0$, we have $m=m_1+m_2$. Since $n>2{\mathrm{\Gamma }}_0+4d-\lambda$, we have $n+{\lambda }_{2}m-2d>2{\mathrm{\Gamma }}_0+4d-\lambda +\lambda -2m-2d=2{\mathrm{\Gamma }}_0+2d-2m>0$. From the above two inequalities, for any ε ($0<\epsilon <2{\mathrm{\Gamma }}_0+2d-2m$), we have

$$(n+\lambda -2m-2d-\epsilon )[T(r,f)+T(r,g)]\le S_1(r,f)+S_1(r,g),$$

which is a contradiction with f, g are transcendental.

If $a-b-1=0$, then by (20) we know $F=((b+1)G)/(bG+1)$. Since f, g are entire functions, we see that $-\frac{1}{b}$ is a Picard’s exceptional value of $G(z)$. By the Second Fundamental Theorem, we have

$$T(r,G)\le \overline{N}(r,\frac{1}{G})+S(r,G)\le (m+d)T(r,g)+S_1(r,g).$$

Then, from Lemma 2.6 and $n>2{\mathrm{\Gamma }}_0+4d-\lambda$, we know $T(r,g)\le S_1(r,g)$, a contradiction.

Subcase 2.2. $b=-1$. Then (20) becomes $F=a/(a+1-G)$.

If $a+1\ne 0$, then $a+1$ is a Picard exceptional value of G. Similar to the discussion as in Subcase 2.1, we can deduce a contradiction again.

If $a+1=0$, then $FG\equiv 1$, that is,

$$P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}P(g)\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}\equiv 1.$$

Since $n>max\{2{\mathrm{\Gamma }}_0+4d-\lambda ,\lambda \}$, by Lemma 2.9, we see that $fg=\mu$ for a constant μ such that $a_n^2{\mu }^{n+\lambda }\equiv 1$.

Subcase 2.3. $b=0$. Then (20) becomes $F=(G+a-1)/a$.

If $a-1\ne 0$, then $\overline{N}(r,\frac{1}{G+a-1})=\overline{N}(r,\frac{1}{F})$. Similar to the discussion as in Subcase 2.1, we can deduce a contradiction again.

If $a-1=0$, then $F\equiv G$, that is,

$$P(f)\prod _{j=1}^{d}f{(q_{j}z+c_j)}^{s_j}\equiv P(g)\prod _{j=1}^{d}g{(q_{j}z+c_j)}^{s_j}.$$

Using the same argument as in the proof of Case 2 in Theorem 1.3, we can see that f, g satisfy Theorem 1.3(ii).

• (II) $l=2$. Since

  $$\begin{array}{r}\overline{N}(r,\frac{1}{F-1})+\overline{N}(r,\frac{1}{G-1})-N_{11}(r,\frac{1}{F-1})\\ +\frac{1}{2}{\overline{N}}_{(l+1}(r,\frac{1}{F-1})+\frac{1}{2}{\overline{N}}_{(l+1}(r,\frac{1}{G-1})\\ \le \frac{1}{2}N(r,\frac{1}{F-1})+\frac{1}{2}N(r,\frac{1}{G-1})\\ \le \frac{1}{2}T(r,F)+\frac{1}{2}T(r,G)+S(r,F)+S(r,G),\end{array}$$

  (21)

$$\begin{array}{rl}{\overline{N}}_{(l+1}(r,\frac{1}{F-1})& \le \frac{1}{2}N(r,\frac{F}{F^{\prime }})=\frac{1}{2}N(r,\frac{F^{\prime }}{F})+S(r,F)\\ \le \frac{1}{2}\overline{N}(r,\frac{1}{F})+S(r,F)\\ \le \frac{m}{2}T(r,f)+\frac{d}{2}\overline{N}(r,\frac{1}{f})+S_1(r,f),\end{array}$$

(22)

and

$${\overline{N}}_{(l+1}(r,\frac{1}{G-1})\le \frac{m}{2}T(r,g)+\frac{d}{2}\overline{N}(r,\frac{1}{g})+S_1(r,g).$$

Case 1. If $F(z)$, $G(z)$ satisfy Lemma 2.2(i), from the fact that $f(z)$, $g(z)$ are transcendental entire functions and (21)-(22), we have

$$\begin{array}{rl}T(r,F(z))+T(r,G(z))\le & 2N_2(r,\frac{1}{F})+2N_2(r,\frac{1}{G})+mT(r,f)+mT(r,g)\\ +d\overline{N}(r,\frac{1}{f})+d\overline{N}(r,\frac{1}{g})+S_1(r,f)+S_1(r,g).\end{array}$$

From (13), (14), Lemma 2.6, and $\chi =min\{\mathrm{\Theta }(0,f),\mathrm{\Theta }(0,g)\}$, for any ε ($0<\epsilon <n+\lambda -m-2{\mathrm{\Gamma }}_0-5d+d\chi$), we have

$$(n+\lambda -m-2{\mathrm{\Gamma }}_0-5d+d\chi -\epsilon )[T(r,f)+T(r,g)]\le S_1(r,f)+S_1(r,g).$$

(23)

Since $n>2{\mathrm{\Gamma }}_0+5d+m-\lambda -d\chi$ and f, g are transcendental functions, we get a contradiction.

Case 2. If $F(z)$, $G(z)$ satisfy Lemma 2.2(ii). Similar to the proof of Case 2 in (I), we get the conclusions of Theorem 1.4.

• (III) $l=1$. Since

  $$\begin{array}{rl}\overline{N}& (r,\frac{1}{F-1})+\overline{N}(r,\frac{1}{G-1})-N_{11}(r,\frac{1}{F-1})\\ \le \frac{1}{2}N(r,\frac{1}{F-1})+\frac{1}{2}N(r,\frac{1}{G-1})\\ \le \frac{1}{2}T(r,F)+\frac{1}{2}T(r,G)+S(r,F)+S(r,G),\end{array}$$

  (24)

we have

$$\begin{array}{rl}{\overline{N}}_{(2}(r,\frac{1}{F})& \le N(r,\frac{F}{F^{\prime }})=N(r,\frac{F^{\prime }}{F})+S(r,f)\le \overline{N}(r,\frac{1}{F})+S(r,f)\\ \le mT(r,f)+d\overline{N}(r,\frac{1}{f})+S_1(r,f),\end{array}$$

(25)

and

$${\overline{N}}_{(2}(r,\frac{1}{G})\le mT(r,g)+d\overline{N}(r,\frac{1}{g})+S_1(r,g).$$

(26)

Case 1. If $F(z)$, $G(z)$ satisfy Lemma 2.2(i), from f, g are entire functions, (13), (14), (21), (22), (23), and (24), we have

$$\begin{array}{rl}T(r,F)+T(r,G)\le & 2({\mathrm{\Gamma }}_0+2d+m)[T(r,f)+T(r,g)]+2d\overline{N}(r,\frac{1}{f})\\ +2d\overline{N}(r,\frac{1}{g})+S_1(r,f)+S_1(r,g).\end{array}$$

From Lemma 2.6 and $\chi =min\{\mathrm{\Theta }(0,f),\mathrm{\Theta }(0,g)\}$, for any ε ($0<\epsilon <n+\lambda -2{\mathrm{\Gamma }}_0-6d-2m+2d\chi$), we have

$$[n+\lambda -2{\mathrm{\Gamma }}_0-6d-2m+2d\chi -\epsilon ][T(r,f)+T(r,g)]\le S_1(r,f)+S_1(r,g).$$

(27)

Since $n>2{\mathrm{\Gamma }}_0+6d+2m-\lambda -2d\chi$, from (27) and f, g are transcendental, we get a contradiction.

Case 2. If $F(z)$, $G(z)$ satisfy Lemma 2.2(ii). Similar to the proof of Case 2 in (I), we get the conclusions of Theorem 1.4.

• (IV) $l=0$, that is, $F(z)$, $G(z)$ share 1 IM. From the definitions of $F(z)$, $G(z)$, we have

  $$\begin{array}{rl}{\overline{N}}_L(r,\frac{1}{F-1})& \le N(r,\frac{F}{F^{\prime }})=N(r,\frac{F^{\prime }}{F})+S(r,F)\le \overline{N}(r,\frac{1}{F})+S(r,F)\\ \le mT(r,f)+d\overline{N}(r,\frac{1}{f})+S_1(r,f),\end{array}$$

  (28)

similarly, we have

$${\overline{N}}_L(r,\frac{1}{G-1})\le mT(r,g)+d\overline{N}(r,\frac{1}{g})+S_1(r,f).$$

(29)

Case 1. Suppose that $F(z)$, $G(z)$ satisfy Lemma 2.3(i). From (28) and (29), we have

$$\begin{array}{rl}T(r,F(z))+T(r,G(z))\le & 2N_2(r,\frac{1}{F})+2N_2(r,\frac{1}{G})+3mT(r,f)+3mT(r,g)\\ +3d\overline{N}(r,\frac{1}{f})+3d\overline{N}(r,\frac{1}{g})+S_1(r,f)+S_1(r,g).\end{array}$$

From Lemma 2.6 and (13)-(14), for any ε ($0<\epsilon <n+\lambda -2{\mathrm{\Gamma }}_0-7d-3m+3d\chi$), we get

$$(n+\lambda -2{\mathrm{\Gamma }}_0-7d-3m+3d\chi -\epsilon )[T(r,f)+T(r,g)]\le S_1(r,f)+S_1(r,g).$$

(30)

Since $n>2{\mathrm{\Gamma }}_0+7d+3m-\lambda -3d\chi$, we get a contradiction.

Case 2. Suppose that $F(z)$, $G(z)$ satisfy Lemma 2.3(ii). Similar to the proof of Case 2 in (I), we get the conclusions of Theorem 1.4 easily.

Thus, the proof of Theorem 1.4 is completed. □