Some Results Concerning Meromorphic Solutions for the Pielou Logistic Equation

Abstract

In this paper, we investigate meromorphic solutions y(z) for the Pielou logistic equation and obtain some estimates of exponents of convergence of the sequence of zeros of $y(z+n)-a(z)$,$\Delta y(z)-a(z)=y(z+1)-y(z)-a(z)$ and $\frac{\Delta y(z)}{y(z)}-a(z)$, where $a(z)\,(\not \equiv 0)$ is an entire function such that $\sigma (a)<1$.

1 Introduction and Results

In this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna’s value distribution theory of meromorhpic functions (see [21, 24, 26]). In addition, we use the notations $\lambda (f)$ and $\lambda (\frac{1}{f})$ to denote, respectively, the exponent of convergence of the sequence of zeros and poles of a meromorphic function f and $\sigma (f)$ to denote the order growth of f. We also use the notation $\tau (f)$ to denote the exponent of convergence of fixed points of f that is defined as

$$\begin{aligned} \tau (f)=\mathop {\overline{\mathrm{lim}}}_{r\rightarrow \infty }\frac{\log N\left( r,\frac{1}{f(z)-z}\right) }{\log r}. \end{aligned}$$

For a meromorphic function f(z), we use S(f) to denote the family of all meromorphic functions $\alpha (z)$ that satisfy $T(r,\alpha )=S(r,f)$, where $S(r,f)=o(T(r,f)),$ as $r\rightarrow \infty$ outside of a possible exceptional set of finite logarithmic measure. Functions in the set S(f) are called small functions with respect to f(z).

Recently, a number of papers (including [1, 3,4,5,6,7,8,9,10,11,12,13,14, 18,19,20, 22, 23, 25]) focus on complex difference equations and difference analogues of Nevanlinna’s theory.

The Pielou logistic equation

$$\begin{aligned} y(z+1)=\frac{R(z)y(z)}{Q(z)+P(z)y(z)}, \end{aligned}$$

(1.1)

where P(z), Q(z), R(z) are nonzero polynomials, is an important difference equation because it is obtained by transform from the well-known Verhulst-Pearl equation (see [16])

$$\begin{aligned} x'(t)=x(t)[a-bx(t)] \quad (a, b>0), \end{aligned}$$

which is the most popular continuous model of growth of a population.

Chen [7] obtained the following theorem.

Theorem A

Let P(z), Q(z), R(z) be polynomials with $P(z)Q(z)R(z)\not \equiv 0$, and y(z) be a finite-order transcendental meromorphic solution of Eq. (1.1). Then

$$\begin{aligned} \lambda \left( \frac{1}{y}\right) =\sigma (y)\ge 1. \end{aligned}$$

We say a meromorphic function f(z) is oscillation if f(z) has infinitely many zeros.

Example 1.1

The function $y(z)=\frac{z3^z}{3^z-1}$ satisfies the Pielou logistic equation

$$\begin{aligned} y(z+1)=\frac{3(z+1)y(z)}{z+2y(z)}, \end{aligned}$$

where y(z) satisfies

$$\begin{aligned} \lambda (y)=0\quad \text{ and }\quad \lambda \left( \frac{1}{y}\right) =\sigma (y)=1. \end{aligned}$$

This example shows that the result of Theorem A is sharp.

The problems on zeros and fixed points of meromorphic functions are important topics in theory of meromorphic functions. Recently, several papers (including [3, 11, 12]) investigate the problem on zeros of difference $\Delta f(z)=f(z+1)-f(z)$ and divided difference $\frac{\Delta f(z)}{f(z)}$ of a meromorphic function f(z), particularly, for the meromorphic function f(z) of small growth, zeros of $\Delta f(z)$ and $\frac{\Delta f(z)}{f(z)}$ are investigated. Bergweiler and Langley obtained the following theorem.

Theorem B

(See [3]) There exists $\delta _0\in (0,1/2)$ with the following property. Let f(z) be a transcendental entire function with order

$$\begin{aligned} \sigma (f)\le \sigma<\frac{1}{2}+\delta _0<1. \end{aligned}$$

Then

$$\begin{aligned} G(z)=\frac{\Delta f(z)}{f(z)}=\frac{f(z+1)-f(z)}{f(z)} \end{aligned}$$

has infinitely many zeros.

But for an entire function of $\sigma (f)\ge 1$, the conclusion of Theorem B does not hold. For example, $f(z)=e^z$ satisfies $\frac{\Delta f(z)}{f(z)}=e-1$ which has only finitely many zeros.

When f(z) is meromorphic, Bergweiler and Langley consider the existence of zeros of the difference $g(z)=f(z+c)-f(z)$, also gave the following construction theorem to show that even if for a meromorphic function f(z) of lower order 0, g(z) may have only finitely many zeros.

Theorem C

(See [3]) Let $\phi (r)$ be a positive non-decreasing function on $[0,+\infty )$ which satisfies $\lim \limits _{r\rightarrow \infty }\phi (r)=\infty .$ Then there exists a function f(z) transcendental and meromorphic in the plane with

$$\begin{aligned} \mathop {\overline{\mathrm{lim}}}\limits _{r\rightarrow \infty }\frac{T(r,f)}{r}<\infty \quad \text{ and } \quad \mathop {\underline{\mathrm{lim}}}\limits _{r\rightarrow \infty }\frac{T(r,f)}{\phi (r)\log r}<\infty . \end{aligned}$$

such that $g(z)=\Delta f(z)=f(z+1)-f(z)$ has only one zero. Moreover, the function g(z) satisfies

$$\begin{aligned} \mathop {\overline{\mathrm{lim}}}\limits _{r\rightarrow \infty }\frac{T(r,g)}{\phi (r)\log r}<\infty . \end{aligned}$$

From Theorem A and Example 1.1, we see that although a transcendental meromorphic solution y(z) of (1.1) may have only finite many zeros, Chen discovered that $\Delta y(z)-a$ and $\frac{\Delta y(z)}{y(z)}-a$ are oscillation and obtained the following theorems.

Theorem D

(See [8]) Let P(z), Q(z), R(z) be nonzero polynomials, such that $\frac{R(z)-Q(z)}{P(z)}$ is a nonconstant. Set $\Delta y(z)=y(z+1)-y(z).$ Then for every finite-order transcendental meromorphic solution y(z) of (1.1), its difference $\Delta y(z)$ and divided difference $\frac{\Delta y(z)}{y(z)}$ are oscillation and satisfy

$$\begin{aligned} \lambda (\Delta y(z))=\lambda \left( \frac{\Delta y(z)}{y(z)}\right) =\sigma (y(z))\ge 1. \end{aligned}$$

Theorem E

(See [8]) Let $P_j(z), j=1,2,3$ be nonzero polynomials, such that

$$\begin{aligned} \deg P_3(z)>\max \{\deg P_j(z): j=1,\,2\}. \end{aligned}$$

(1.2)

If $a\in {\mathbb {C}}\backslash \{0\}$, then every finite-order transcendental meromorphic solution y(z) of equation

$$\begin{aligned} y(z+1)=\frac{P_1(z)y(z)}{P_2(z)+P_3(z)y(z)}, \end{aligned}$$

(1.3)

satisfies

• (i) $\lambda (y(z+n)-a)=\sigma (y(z))\ge 1\quad (n=0,1,2,\ldots );$

• (ii) if $\deg P_1\ne \deg P_2$, then $\lambda \left( \frac{\Delta y(z)}{y(z)}-a\right) =\sigma (y(z))$;

• (iii) if there is a polynomial h(z) satisfying

  $$\begin{aligned} (P_2(z)-P_1(z)+aP_3(z))^2-4aP_2(z)P_3(z)=h(z)^2, \end{aligned}$$

then $\lambda (\Delta y(z)-a)=\sigma (y(z))$.

Many papers and books (including [2, 15, 16]) investigate fixed points of meromorphic functions and their shifts, differences, divided differences. Chen and Shon obtained the following Theorem F.

Theorem F

(See [12]) Let $c\in {\mathbb {C}}\backslash \{0\}$ be a constant and f(z) be a transcendental meromorphic function of order of growth $\sigma (f)=\sigma <1$ or of the form $f(z)=h(z)e^{az}$, where a is a nonzero constant, h(z) is a transcendental meromorphic function with $\sigma (h)<1$. Suppose that p(z) is a nonconstant polynomial. Then

$$\begin{aligned} G(z)=\frac{f(z+c)-f(z)}{f(z)}-p(z) \end{aligned}$$

has infinitely many zeros.

From Theorem F, we easily see that under conditions of Theorem F, the divided difference $G_1(z)=\frac{f(z+c)-f(z)}{f(z)}$ has infinitely many fixed points. Example $f(z)=ze^z$ shows that the result of Theorem F is sharp $\left( \frac{f(z+1)-f(z)}{f(z)}=\frac{(z+1)e-z}{z}\right)$.

Chen and Shon discovered that the properties on fixed points of meromorphic solutions of (1.3) are very well. They proved the following theorem.

Theorem G

(See [9]) Let $P_j(z), j=1,2,3$ be nonzero polynomials, such that

$$\begin{aligned} \deg P_3(z)\ge \max \{\deg P_j(z): j=1,\,2\}\quad \text{ and }\quad \deg P_3(z)\ge 1. \end{aligned}$$

Set $\Delta y(z)=y(z+1)-y(z)$. Then every finite-order transcendental meromorphic solution y(z) of Eq. (1.3) satisfies:

• (i) $\tau (y(z+n))=\sigma (y(z))\ge 1\;(n=0,1,2,\ldots );$

• (ii) if $P_1(z)-(z+1)P_2(z)\not \equiv 0$, then $\tau \left( \frac{\Delta y(z)}{y(z)}\right) =\sigma (y(z))$;

• (iii) if there is a polynomial h(z) satisfying

  $$\begin{aligned} (P_2(z)-P_1(z)+zP_3(z))^2-4zP_2(z)P_3(z)=h(z)^2, \end{aligned}$$

  then $\tau (\Delta y(z))=\sigma (y(z)).$

For a meromorphic function f(z) and a small function a(z), we see that there are many examples to show that either $f(z)-a(z)$ may have no zero, for example $f_1(z)=e^z+4z^3+2, a(z)=4z^3+2$, or $f(z+c)-a(z)$, or $\Delta _cf(z)-a(z)=f(z+c)-f(z)-a(z)$ may have only finitely many zeros. For example, for the function $f_2(z)=e^z+(z-c)^2, a(z)=z^2$, $f_2(z+c)-a(z)=e^{z+c}$, and $\Delta _{2\pi i}f_2(z)-a(z)=-\,(z-2\pi i)^2$ have only finitely many zeros. Even if for a meromorphic function of small growth, Chen and Shon show that there exists a meromorphic function $f_0(z)$ such that $\sigma (f_0)<1$ and $\Delta _cf_0(z)-a(z)=f_0(z+c)-f_0(z)-a(z)$ has only finitely many zeros, where $a(z)\equiv z$ (see Theorem 6 of [11]).

Furthermore, $\frac{f(z+c)-f(z)}{f(z)}-a(z)$ may also have only finitely many zeros. For example, the function $f(z)=z^2e^z$ satisfies that $\frac{f(z+1)-f(z)}{f(z)}-a(z)=\frac{e(z+1)^2-z^2}{z^2}-a(z)$ has only finitely many zeros, where $a(z)\equiv z^2$. This example also shows that the result of Theorem F is sharp.

The main purpose in this paper is to study meromorphic solutions of the Pielou logistic Eq. (1.3) and obtain some estimates of exponents of convergence of the sequence of zeros of $y(z+n)-a(z)$, $\Delta y(z)-a(z)=y(z+1)-y(z)-a(z)$ and $\frac{\Delta y(z)}{y(z)}-a(z)$. At the same time, we weaken some conditions of Theorem E and prove the following theorems.

Theorem 1.1

Let $P_j(z), j=1,2,3$ be nonzero polynomials, such that

$$\begin{aligned} \deg P_3(z)\ge \max \{\deg P_j(z): j=1,\,2\}. \end{aligned}$$

(1.4)

If a(z) is a nonconstant entire function such that $\sigma (a)<1$, then every finite-order transcendental meromorphic solution y(z) of Eq. (1.3) satisfies

• (i) $\lambda (y(z+n)-a(z))=\sigma (y(z))\ge 1\quad (n=0,1,2,\ldots );$

• (ii) if $P_1(z)-(a(z)+1)P_2(z)\not \equiv 0$, then $\lambda \left( \frac{\Delta y(z)}{y(z)}-a(z)\right) =\sigma (y(z))$;

• (iii) if there is an entire function h(z) satisfying

  $$\begin{aligned} (P_2(z)-P_1(z)+a(z)P_3(z))^2-4a(z)P_2(z)P_3(z)=h(z)^2, \end{aligned}$$

then $\lambda (\Delta y(z)-a(z))=\sigma (y(z)).$

Theorem 1.2

Let $P_j(z), j=1,2,3$ be nonzero polynomials and satisfy (1.4) in Theorem 1.1. If $a\in {\mathbb {C}}\backslash \{0\}$, then every finite-order transcendental meromorphic solution y(z) of Eq. (1.3) satisfies

• (i) if $\frac{P_1(z)-P_2(z)}{P_3(z)}\not \equiv a$, then $\lambda (y(z+n)-a)=\sigma (y(z))\ge 1\quad (n=0,1,2,\ldots );$

• (ii) if $\deg ((a+1)P_2-P_1)=\max \{\deg P_j:\,j=1,\,2\}$, then $\lambda \left( \frac{\Delta y(z)}{y(z)}-a\right) =\sigma (y(z))$;

• (iii) if there is a polynomial h(z) satisfying

  $$\begin{aligned}&(P_2(z)-P_1(z)+aP_3(z))^2-4aP_2(z)P_3(z)=h(z)^2,\\&\quad \deg (P_2-aP_3-h)=\deg h=\deg P_3, \end{aligned}$$

then $\lambda (\Delta y(z)-a)=\sigma (y(z)).$

Remark 1.1

In the special case, if we take $a(z)\equiv z$ in Theorem 1.1, we can also obtain Theorem G under more relaxed condition (1.4). In Theorem 1.2, if $P_j\,(j=1,2,3)$ satisfy (1.2), we see that $\frac{P_2-P_1}{P_3}\not \equiv a$ and by Theorem 1.2, we can obtain the conclusion (i) of Theorem E; if $\deg P_1\ne \deg P_2$, we see that $\deg ((a+1)P_2-P_1)=\max \{\deg P_j:\,j=1,\,2\}$ and obtain the conclusion (ii) of Theorem E; if $P_j\,(j=1,2,3)$ satisfy (1.2), then $\deg (P_2-aP_3-h)=\deg h=\deg P_3$ obviously holds and we can obtain the conclusion (iii) of Theorem E. Hence, Theorems 1.1 and 1.2 improve and generalize, respectively, Theorems G and E.

Remark 1.2

Generally, $\lambda (f(z)-a(z))\ne \lambda (f(z+c)-a(z))$ for a meromorphic function f(z) of finite order and a small function a(z). For example, the function $f_3(z)=e^z+4z+3, a(z)=4z+3$ satisfies

$$\begin{aligned} \lambda (f_3(z)-a(z))=0\ne \lambda (f_3(z+1)-a(z))=1. \end{aligned}$$

In what follows, we give Example 1.2 to illustrate that the existence condition of h(z) in Theorem 1.2 is attainable, and give Example 1.3 to show that there is a finite-order transcendental meromorphic function satisfying Eq. (1.3) when h(z) is a polynomial.

Example 1.2

Suppose that $P_1(z)\equiv P_2(z)\equiv P_3(z)$ and $\deg P_3(z)\ge 0$. Then nonzero polynomial $h(z):=\pm \sqrt{a(a-4)}P_3(z)\,(a\in {\mathbb {C}}\backslash \{0,4,-\frac{1}{2}\})$, satisfy the condition appearing in (iii) of Theorem 1.2. In fact, by simple calculation, we see that $P_i(z)\,(i=1,2,3)$ satisfy (1.4) and h(z) satisfy

$$\begin{aligned} (P_2(z)-P_1(z)+aP_3(z))^2-4aP_2(z)P_3(z)=a(a-4)[P_3(z)]^2=h(z)^2, \end{aligned}$$

and

$$\begin{aligned} \deg (P_2-aP_3-h)=\deg \left[ (1-a\mp \sqrt{a(a-4)})P_3(z)\right] =\deg h=\deg P_3, \end{aligned}$$

since $\sqrt{a(a-4)}\ne 0$ and $1-a\mp \sqrt{a(a-4)}\ne 0$.

Example 1.3

Set $y(z)=\frac{1}{e^{2\pi iz}+z}$. Then y(z) solves equation

$$\begin{aligned}y(z+1)=\frac{P_1(z)y(z)}{P_2(z)+P_3(z)y(z)}, \end{aligned}$$

where $P_j(z)$ are nonzero polynomials and satisfy $P_1(z)\equiv P_2(z)\equiv P_3(z)$. In fact, by $\frac{1}{y(z+1)}=e^{2\pi iz}+z+1$ and $\frac{1}{y(z)}=e^{2\pi iz}+z$, we have

$$\begin{aligned} \frac{1}{y(z+1)}-\frac{1}{y(z)}=1. \end{aligned}$$

Hence

$$\begin{aligned} y(z+1)=\frac{y(z)}{1+y(z)}=\frac{P_1(z)y(z)}{P_2(z)+P_3(z)y(z)}. \end{aligned}$$

Thus, by the conclusion (iii) of Theorem 1.2 and Example 1.2, we have, for every $a\in {\mathbb {C}}\backslash \{0,4,-\frac{1}{2}\}$,

$$\begin{aligned} \lambda \left( \Delta y(z)-a\right) =\lambda \left( \frac{-1}{(e^{2\pi iz}+z+1)(e^{2\pi iz}+z)}-a\right) =\sigma \left( \frac{1}{e^{2\pi iz}+z}\right) =1. \end{aligned}$$

2 Proof of Theorem 1.1

We need the following lemmas for the proof of Theorem 1.1.

Lemma 2.1

(See [7]) Let $h_2(z)\not \equiv 0, h_1(z), F(z)$ be polynomials, $c_2, c_1(\ne c_2)$ be constants. Suppose that f(z) is a finite-order transcendental meromorphic solution of difference equation

$$\begin{aligned} h_2(z)f(z+c_2)+h_1(z)f(z+c_1)=F(z). \end{aligned}$$

Then $\sigma (f(z))\ge 1$.

Lemma 2.2

(See [18, 25]) Let w(z) be a nonconstant finite-order meromorphic solution of

$$\begin{aligned} P(z,w)=0, \end{aligned}$$

where P(z, w) is a difference polynomial in w(z). If $P(z,a)\not \equiv 0$ for a meromorphic function a(z) satisfying $T(r,a)=S(r,w)$, then

$$\begin{aligned} m\left( r,\frac{1}{w(z)-a(z)}\right) =S(r,w(z)) \end{aligned}$$

holds for all r outside of a possible exceptional set with finite logarithmic measure.

Remark 2.1

Use the same method as in the proof of Lemma 2.2 (see [18]), we can prove that in Lemma 2.2, if all coefficients $b_{\lambda }(z)$ of P(z, w) satisfy $\sigma (b_{\lambda }(z))=\sigma _1<\sigma (w(z))=\sigma$, and if $P(z,a)\not \equiv 0$ for a meromorphic function a(z) satisfying $T(r,a)=S(r,w),$ then for a given $\varepsilon \, (0<\varepsilon <\sigma -\sigma _1)$

$$\begin{aligned} m\left( r,\frac{1}{w(z)-a(z)}\right) =S(r,w(z))+O(r^{\sigma _1+\varepsilon }) \end{aligned}$$

holds for all r outside of a possible exceptional set with finite logarithmic measure.

Lemma 2.3

Suppose that $P_j(z),\, j=1,2,3$ satisfy condition (1.4) in Theorem 1.1, y(z) is a nonconstant meromorphic function, and $a(z)(\not \equiv 0)$ is an entire function such that $\sigma (a)<1$. Then

$$\begin{aligned} f_1(z)= & {} -(1+a(z))P_3(z)y(z)+P_1(z)-(1+a(z))P_2(z)\quad \text{ and }\\ f_2(z)= & {} P_2(z)+P_3(z)y(z) \end{aligned}$$

have at most finitely many common zeros.

Proof

Suppose that $z_0$ is a common zero of $f_1(z)$ and $f_2(z)$. Then $f_2(z_0)=P_2(z_0)+P_3(z_0)y(z_0)=0$. Thus, $y(z_0)=-\frac{P_2(z_0)}{P_3(z_0)}$. Substituting $y(z_0)=-\frac{P_2(z_0)}{P_3(z_0)}$ into $f_1(z)$, we obtain

$$\begin{aligned} f_1(z_0)= & {} -(1+a(z_0))P_3(z_0)\left( -\frac{P_2(z_0)}{P_3(z_0)}\right) +P_1(z_0)-(1+a(z_0))P_2(z_0)\\= & {} P_1(z_0)\\= & {} 0. \end{aligned}$$

Since $P_1(z)$ has only finitely many zeros, we see that $f_1(z)$ and $f_2(z)$ have at most finitely many common zeros. $\square$

Lemma 2.4

(See [20]) Let f(z) be a nonconstant finite-order meromorphic function. Then

$$\begin{aligned} T(r+1,f(z))=T(r,f(z))+S(r,f(z)). \end{aligned}$$

Gol’dberg [17] (p.66) or [13] gives that for any constant b,

$$\begin{aligned} (1+o(1))T(r-|b|,f(z))\le T(r,f(z+b))\le (1+o(1))T(r+|b|,f(z)). \end{aligned}$$

The above equation and Lemma 2.4 give the following lemma.

Lemma 2.5

Let f(z) be a nonconstant finite-order meromorphic function. Then

$$\begin{aligned} T(r,f(z+1))=T(r,f(z))+S(r,f(z)). \end{aligned}$$

Lemma 2.6

Let f(z) be a meromorphic function of finite order, and let c be a nonzero complex constant. Then

$$\begin{aligned} m\left( r,\frac{f(z+c)}{f(z)}\right) +m\left( r,\frac{f(z)}{f(z+c)}\right) =S(r,f). \end{aligned}$$

Lemma 2.7

Suppose that $P_j(z),\,j=1,2,3$ satisfy condition (1.4) in Theorem 1.1, and y(z) is a nonconstant meromorphic function, $a(z)(\not \equiv 0)$ is an entire function such that $\sigma (a)<1$. Then

$$\begin{aligned} f_1(z)= & {} -(1+a(z))P_3(z)\quad \text{ and }\\ f_2(z)= & {} -(1+a(z))y(z)+P_1(z)-(1+a(z))P_2(z) \end{aligned}$$

have at most finitely many common zeros.

Proof

Suppose that $z_0$ is a common zero of $f_1(z)$ and $f_2(z)$. Then $f_1(z_0)=-\,(1+a(z_0))P_3(z_0)=0$. Thus, $1+a(z_0)=0$ or $P_3(z_0)=0$. If $1+a(z_0)=0$, substituting $1+a(z_0)=0$ into $f_2(z)$, we obtain

$$\begin{aligned} f_2(z_0)= & {} -(1+a(z_0))y(z_0)+P_1(z_0)-(1+a(z_0))P_2(z_0) \\= & {} P_1(z_0)\\= & {} 0. \end{aligned}$$

Since $P_1(z)$ has only finitely many zeros, we see that $f_1(z)$ and $f_2(z)$ have at most finitely many common zeros. If $P_3(z_0)=0$, since $P_3(z)$ has only finitely many zeros, we see that $f_1(z)$ and $f_2(z)$ have at most finitely many common zeros. $\square$

Using the same method as in the proof of Lemmas 2.3 and 2.7, we can prove the following Lemma 2.8.

Lemma 2.8

Suppose that $P_j(z),\,j=1,2,3$ satisfy condition (1.4) in Theorem 1.1, y(z) is a nonconstant meromorphic function, and $a(z)(\not \equiv 0)$ is an entire function such that $\sigma (a)<1$. Then

$$\begin{aligned} f_1(z)=P_3(z)y(z)^2+(P_2(z)-P_1(z)+P_3(z)a(z))y(z)+P_2(z)a(z) \end{aligned}$$

and

$$\begin{aligned} f_2(z)=P_2(z)+P_3(z)y(z) \end{aligned}$$

have at most finitely many common zeros.

Lemma 2.9

(See [25]) Let f(z) be a transcendental meromorphic solution of finite-order $\rho$ of a difference equation of the form

$$\begin{aligned} U(z,f)P(z,f)=Q(z,f), \end{aligned}$$

where $U(z,f),\,P(z,f),\,Q(z,f)$ are difference polynomials such that the total degree $\deg U(z,f)=n$ in f(z) and its shifts, and $\deg Q(z,f)\le n.$ Moreover, we assume that U(z, f) contains just one term of maximal total degree in f(z) and its shifts. Then, for each $\varepsilon >0$,

$$\begin{aligned} m\left( r,P(z,f)\right) =O(r^{\rho -1+\varepsilon })+S(r,f), \end{aligned}$$

possibly outside of an exceptional set of finite logarithmic measure.

Remark 2.2

From the proof of Lemma 2.5 in [25], we can see that if the coefficients of $U(z,f),\,P(z,f),\,Q(z,f)$, namely $\alpha _{\lambda }(z)$, satisfy $m(r,\alpha _{\lambda })=S(r,f),$ then the same conclusion still holds.

Proof of Theorem 1.1

Suppose that y(z) is a finite-order transcendental meromorphic solution of (1.3).

• (i) We prove that $\lambda (y(z+n)-a(z))=\sigma (y(z))\ge 1\quad (n=0,\,1,\ldots )$. Suppose that $n=0$. Set $y(z)-a(z)=g(z)$. By Theorem A, we obtain $\sigma (y(z))\ge 1$. Since $\sigma (a)<1$, we see that g(z) is transcendental and satisfies

  $$\begin{aligned} \sigma (g(z))=\sigma (y(z))\ge 1\quad \text{ and }\quad S(r,g)=S(r,y). \end{aligned}$$

  (2.1)

  Substituting $y(z)=g(z)+a(z)$ into (1.3), we obtain

  $$\begin{aligned} H_0(z,g):= & {} P_3(z)[g(z+1)+a(z+1)][g(z)+a(z)]+P_2(z)[g(z+1)+a(z+1)]\\&-P_1(z)[g(z)+a(z)]\\= & {} 0. \end{aligned}$$

  Thus,

  $$\begin{aligned} H_0(z,0)=P_3(z)a(z+1)a(z)+P_2(z)a(z+1)-P_1(z)a(z). \end{aligned}$$

  (2.2)

  We assert that $H_0(z,0)\not \equiv 0$. If a(z) is a nonconstant polynomial, by (1.4), we know that in the right side of (2.2), there exists only one term $P_3(z)a(z+1)a(z)$ being of the highest degree $\deg P_3(z)+2\deg a(z)$. Hence, $H_0(z,0)\not \equiv 0$. If a(z) is a transcendental entire function, on the contrary, we suppose that $H_0(z,0)\equiv 0$. Thus, we have

  $$\begin{aligned} P_3(z)a(z+1)a(z)+P_2(z)a(z+1)-P_1(z)a(z)=0, \end{aligned}$$

  that is,

  $$\begin{aligned} -\frac{P_2(z)}{a(z)}+\frac{P_1(z)}{a(z+1)}=P_3(z). \end{aligned}$$

  By Lemma 2.1, we have $\sigma \left( \frac{1}{a(z)}\right) =\sigma (a(z))\ge 1$, which contradicts our hypothesis $\sigma (a(z))<1$. Hence, $H_0(z,0)\not \equiv 0$. Thus, by Lemma 2.2 and $H_0(z,0)\not \equiv 0$, we obtain

  $$\begin{aligned} N\left( r,\frac{1}{g(z)}\right) =T(r,g)+S(r,g) \end{aligned}$$

  (2.3)

  for all r outside of a possible exceptional set with finite logarithmic measure. So that, by (2.1) and (2.3), we obtain $\lambda (g(z))=\sigma (g(z))=\sigma (y(z))$. Combining this equality and the conclusion of Theorem A, we have

  $$\begin{aligned} \lambda (y(z)-a(z))=\sigma (y(z))\ge 1. \end{aligned}$$

  Now suppose that $n=1$. By (1.3), we obtain

  $$\begin{aligned} y(z+1)-a(z)= & {} \frac{\left[ P_1(z)-P_3(z)a(z)\right] y(z)-P_2(z)a(z)}{P_2(z)+P_3(z)y(z)}. \end{aligned}$$

  (2.4)

  Set $y(z+1)-a(z)=g_1(z)$. Then $g_1(z)$ is transcendental. By Lemma 2.5 and $\sigma (y(z))\ge 1$, we have

  $$\begin{aligned} \sigma (g_1(z))=\sigma (y(z+1))=\sigma (y(z))\ge 1\quad \text{ and }\quad S(r,g_1)=S(r,y). \end{aligned}$$

  (2.5)

  Substituting $y(z)=g_1(z-1)+a(z-1)$ into (2.4), we obtain

  $$\begin{aligned} H_1(z,g_1):= & {} g_1(z)\left\{ P_2(z)+P_3(z)\left[ g_1(z-1)+a(z-1)\right] \right\} \\&-\left[ P_1(z)-P_3(z)a(z)\right] \left[ g_1(z-1)\right. \\&\left. +a(z-1)\right] +P_2(z)a(z)\\= & {} 0. \end{aligned}$$

  Thus,

  $$\begin{aligned} H_1(z,0)=P_3(z)a(z)a(z-1)-P_1(z)a(z-1)+P_2(z)a(z). \end{aligned}$$

  Using the same method as above, we can prove that $H_1(z,0)\not \equiv 0$ and (2.3) holds, thus, we can get

  $$\begin{aligned} \lambda (y(z+1)-a(z))=\lambda (g_1(z))=\sigma (g_1(z))=\sigma (y(z))\ge 1. \end{aligned}$$

  Now suppose that $n=2$. By (1.3), we have

  $$\begin{aligned} y(z+2)-a(z)= & {} \frac{\left[ P_1(z+1)-P_3(z+1)a(z)\right] y(z+1)-P_2(z+1)a(z)}{P_2(z+1)+P_3(z+1)y(z+1)}. \end{aligned}$$

  (2.6)

  Set $f(z)=y(z+1)$. Thus, (2.6) is transformed as

  $$\begin{aligned} f(z+1)-a(z)= & {} \frac{\left[ P_1(z+1)-P_3(z+1)a(z)\right] f(z)-P_2(z+1)a(z)}{P_2(z+1)+P_3(z+1)f(z)}. \end{aligned}$$

  (2.7)

  Since $P_j(z+1)\,(j=1,2,3)$ satisfy (1.4) for (2.7), applying the conclusion for $n=1$ above, we obtain

  $$\begin{aligned} \lambda (y(z+2)-a(z))=\lambda (f(z+1)-a(z))=\sigma (f(z))=\sigma (y(z))\ge 1. \end{aligned}$$

  Continuing to use the same method as above, we obtain

  $$\begin{aligned} \lambda (y(z+n)-a(z))=\sigma (y(z))\ge 1\quad n=1,\,2,\ldots . \end{aligned}$$

• (ii) Suppose that $P_1(z)-(a(z)+1)P_2(z)\not \equiv 0$. We prove

  $$\begin{aligned} \lambda \left( \frac{\Delta y(z)}{y(z)}-a(z)\right) =\sigma (y(z)). \end{aligned}$$

  (2.8)

  By (1.3), we obtain

  $$\begin{aligned} \frac{\Delta y(z)}{y(z)}-a(z)=\frac{-(1+a(z))P_3(z)y(z)+P_1(z)-(1+a(z))P_2(z)}{P_2(z)+P_3(z)y(z)}. \end{aligned}$$

  (2.9)

  We divide this proof into two cases: (1)  a(z) is transcendental;   (2)  a(z) is a nonconstant polynomial.

Case 1   Suppose that a(z) is a transcendental entire function. Then, we see that $P_1(z)-(1+a(z))P_2(z)\not \equiv 0$ and

$$\begin{aligned} \frac{\Delta y(z)}{y(z)}-a(z)= & {} \frac{-(1+a(z))P_3(z)y(z)+P_1(z)-(1+a(z))P_2(z)}{P_2(z)+P_3(z)y(z)}\nonumber \\= & {} \frac{Q(z)}{P_2(z)+P_3(z)y(z)}, \end{aligned}$$

(2.10)

where $Q(z)=-\,(1+a(z))P_3(z)y(z)+P_1(z)-(1+a(z))P_2(z)$. Since $P_j(z)\,(j=1,2,3)$ are polynomials, $\sigma (a)<1\le \sigma (y)$, we have $\sigma (Q)=\sigma (y)\ge 1$. By Lemma 2.3, we see that Q(z) and $P_2(z)+P_3(z)y(z)$ have at most finitely many common zeros. Hence, by (2.10), we have

$$\begin{aligned} N\left( r,\frac{1}{\frac{\Delta y(z)}{y(z)}-a(z)}\right) \ge N\left( r,\frac{1}{Q(z)}\right) +O(\log r). \end{aligned}$$

(2.11)

Hence, in order to prove (2.8), we only need to prove that

$$\begin{aligned} \lambda (Q(z))=\sigma (y(z)). \end{aligned}$$

(2.12)

On the contrary, we suppose that $\lambda (Q(z))<\sigma (y(z))=\sigma (Q(z))$. Then, Q(z) can be rewritten as

$$\begin{aligned}&-(1+a(z))P_3(z)y(z)+P_1(z)-(1+a(z))P_2(z)=Q(z)\nonumber \\&\quad =z^s\frac{b_0(z)}{H_0(z)}e^{h(z)}=\frac{b(z)}{H_1(z)}. \end{aligned}$$

(2.13)

where h(z) is a polynomial with $\deg h(z)\le \sigma (y(z)),\,b_0(z)$ and $H_0(z)$ are canonical products ($b_0(z)$ may be a polynomial) formed by nonzero zeros and poles of $-(1+a(z))P_3(z)y(z)+P_1(z)-(1+a(z))P_2(z)$, respectively, s is an integer, if $s\ge 0$, then $b(z)=z^sb_0(z),\,H_1(z)=H_0(z)e^{-h(z)}$; if $s<0$, then $b(z)=b_0(z),\,H_1(z)=z^{-s}H_0(z)e^{-h(z)}$. Combining Theorem A with property of canonical product, we see that

$$\begin{aligned} \left\{ \begin{array}{l} \lambda (b(z))=\sigma (b(z))= \lambda (Q(z))<\sigma (y(z))=\sigma (Q(z));\\ \lambda (H_1(z))=\sigma (H_1(z))=\sigma (y(z))=\sigma (Q(z)).\end{array}\right. \end{aligned}$$

(2.14)

By (2.13), we see that

$$\begin{aligned} y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}=\frac{b(z)}{-(1+a(z))P_3(z)H_1(z)}=\frac{b(z)}{H(z)}, \end{aligned}$$

(2.15)

where $H(z)=-\,(1+a(z))P_3(z)H_1(z)$. Thus, by (2.14) and $\lambda (a(z)+1)\le \sigma (a(z))<1\le \sigma (y(z))=\sigma \left( \frac{1}{y(z)}\right)$, we have

$$\begin{aligned} \left\{ \begin{array}{l}\lambda (b(z))=\sigma (b(z))= \lambda (Q(z))<\sigma (y(z))=\sigma (Q(z));\\ \lambda (H(z))=\lambda (H_1(z))=\sigma (H_1(z))=\sigma (H(z))=\sigma (y(z))=\sigma (Q(z)).\end{array}\right. \end{aligned}$$

(2.16)

By (2.15), we obtain

$$\begin{aligned} \left\{ \begin{array}{l}y(z)=\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}+b(z)f(z),\\ y(z+1)=\frac{P_1(z+1)-(1+a(z+1))P_2(z+1)}{(1+a(z+1))P_3(z+1)}+b(z+1)f(z+1).\end{array}\right. \end{aligned}$$

(2.17)

where $f(z)=\frac{1}{H(z)}$. Thus, by (2.16) and Lemma 2.5, we have

$$\begin{aligned} \sigma (f(z))=\sigma (H(z))=\sigma (y(z)),\quad \sigma (b(z+1))=\sigma (b(z))<\sigma (f(z)). \end{aligned}$$

(2.18)

Substituting (2.17) into (1.3), we obtain

$$\begin{aligned} H_2(z,f):= & {} \left\{ \frac{P_1(z+1)-(1+a(z+1))P_2(z+1)}{(1+a(z+1))P_3(z+1)}+b(z+1)f(z+1)\right\} \nonumber \\&\cdot \left\{ P_2(z)+P_3(z)\left[ \frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}+b(z)f(z) \right] \right\} \nonumber \\&-P_1(z)\left[ \frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}+b(z)f(z) \right] \nonumber \\= & {} 0. \end{aligned}$$

(2.19)

By (2.19), we obtain

$$\begin{aligned} H_2(z,0)= & {} \frac{P_1(z)}{1+a(z)}\left\{ \frac{P_1(z+1)-P_2(z+1)(1+a(z+1))}{P_3(z+1)(1+a(z+1))}\right. \nonumber \\&\left. -\frac{P_1(z)-P_2(z)(1+a(z))}{P_3(z)}\right\} . \end{aligned}$$

(2.20)

In what follows, we prove $H_2(z,0)\not \equiv 0$. On the contrary, we suppose that $H_2(z,0)\equiv 0$. By (2.20), we have

$$\begin{aligned} \frac{P_1(z+1)-P_2(z+1)(1+a(z+1))}{P_3(z+1)(1+a(z+1))}-\frac{P_1(z)-P_2(z)(1+a(z))}{P_3(z)}=0. \end{aligned}$$

(2.21)

Set $K(z)=\frac{P_1(z)-P_2(z)(1+a(z))}{P_3(z)}$. Thus, we have

$$\begin{aligned} T(r,K(z))=T(r,a(z))+O(\log r),\quad S(r,K)=S(r,a). \end{aligned}$$

By (2.21), we have

$$\begin{aligned} a(z+1)+1=\frac{K(z+1)}{K(z)}. \end{aligned}$$

(2.22)

Obviously, K(z) is a transcendental meromorphic function and satisfies $\sigma (K(z))=\sigma (a(z))<1$. From (2.22), Lemmas 2.5 and 2.6, we have

$$\begin{aligned} T(r,a(z))= & {} T(r,a(z+1)+1)+S(r,a)\\= & {} m(r,a(z+1)+1)+S(r,a)\\= & {} m\left( r,\frac{K(z+1)}{K(z)}\right) +S(r,a)\\= & {} S(r,K)+S(r,a)\\= & {} S(r,a), \end{aligned}$$

this is a contradiction. Hence, $H_2(z,0)\not \equiv 0$.

Case 2   Suppose that a(z) is a nonconstant polynomial. Then, by $P_1(z)-(a(z)+1)P_2(z)\not \equiv 0$ and (2.9), we have

$$\begin{aligned} \frac{\Delta y(z)}{y(z)}-a(z)= & {} \frac{y(z+1)-y(z)}{y(z)}-a(z)\nonumber \\= & {} \frac{-(1+a(z))P_3(z)y(z)+P_1(z)-(1+a(z))P_2(z)}{P_2(z)+P_3(z)y(z)}\nonumber \\= & {} \frac{-(1+a(z))P_3(z)\left\{ y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}\right\} }{P_2(z)+P_3(z)y(z)}. \end{aligned}$$

(2.23)

Since $a(z),\,P_j(z)\,(j=1,2,3)$ are polynomials, by (2.23), we see that $y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}$ and $P_2(z)+P_3(z)y(z)$ have the same poles, except possibly finitely many. By Lemma 2.3, we see that $-(1+a(z))P_3(z)y(z)+P_1(z)-(1+a(z))P_2(z)$ and $P_2(z)+P_3(z)y(z)$ have at most finitely many common zeros. Hence, in order to prove (2.8), we only need to prove that

$$\begin{aligned} \lambda \left( y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}\right) =\sigma (y(z)). \end{aligned}$$

(2.24)

On the contrary, we suppose that $\lambda \left( y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}\right) <\sigma (y(z))$. From $\lambda \left( \frac{1}{y(z)}\right) =\sigma (y(z))\ge 1$, we know that

$$\begin{aligned} \lambda \left( \frac{1}{y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}}\right) =\lambda \left( \frac{1}{y(z)}\right) =\sigma (y(z)). \end{aligned}$$

Thus, $y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}$ can be rewritten as

$$\begin{aligned} y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}=z^s\frac{b_0(z)}{H_0(z)}e^{h(z)}=\frac{b(z)}{H(z)}. \end{aligned}$$

(2.25)

where h(z) is a polynomial with $\deg h(z)\le \sigma (y(z)),\,b_0(z)$ and $H_0(z)$ are canonical products ($b_0(z)$ may be a polynomial) formed by nonzero zeros and poles of $y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}$, respectively, s is an integer, if $s\ge 0$, then $b(z)=z^sb_0(z),\,H(z)=H_0(z)e^{-h(z)}$; if $s<0$, then $b(z)=b_0(z),\,H(z)=z^{-s}H_0(z)e^{-h(z)}$. Combining Theorem A with property of canonical product, we see that

$$\begin{aligned} \left\{ \begin{array}{l}\lambda (b(z))=\sigma (b(z))= \lambda \left( y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}\right) <\sigma (y(z));\\ \lambda (H(z))=\sigma (H(z))=\sigma (y(z)).\end{array}\right. \end{aligned}$$

(2.26)

Moreover, we see that (2.17)–(2.19) still hold and

$$\begin{aligned} H_2(z,0)= & {} \frac{P_1(z)}{1+a(z)}\left\{ \frac{P_1(z+1)-P_2(z+1)(1+a(z+1))}{P_3(z+1)(1+a(z+1))}\right. \nonumber \\&\left. -\frac{P_1(z)-P_2(z)(1+a(z))}{P_3(z)}\right\} . \end{aligned}$$

(2.27)

Set

$$\begin{aligned} h_1(z)=P_1(z)-P_2(z)(1+a(z)),\quad h_2(z)=P_3(z)(1+a(z)). \end{aligned}$$

Thus, (2.27) can be rewritten as

$$\begin{aligned} H_2(z,0)= & {} \frac{P_1(z)}{1+a(z)}\frac{h_1(z+1)}{h_2(z+1)}-P_1(z)\frac{h_1(z)}{h_2(z)}\\= & {} P_1(z)\frac{h_1(z+1)h_2(z)-(1+a(z))h_1(z)h_2(z+1)}{(1+a(z))h_2(z+1)h_2(z)}. \end{aligned}$$

Since $h_1(z),\,h_2(z)$ and a(z) are nonzero polynomials, we see that

$$\begin{aligned} h_1(z+1)h_2(z)-(1+a(z))h_1(z)h_2(z+1)\not \equiv 0, \end{aligned}$$

that is, $H_2(z,0)\not \equiv 0.$

Hence, for two cases above, we have

$$\begin{aligned} H_2(z,0)\not \equiv 0. \end{aligned}$$

(2.28)

Thus, by (2.18), (2.28), Lemma 2.2 and Remark 2.1, we obtain for any given $\varepsilon \,(0<\varepsilon <\sigma (y(z))-\sigma (b(z))=\sigma (f(z))-\sigma (b(z)))$

$$\begin{aligned} N\left( r,\frac{1}{f(z)}\right) =T(r,f(z))+S(r,f(z))+O(r^{\sigma (b(z))+\varepsilon }) \end{aligned}$$

(2.29)

holds for all r outside of a possible exceptional set with finite logarithmic measure.

On the other hand, by $f(z)=\frac{1}{H(z)}$ and the fact that H(z) is an entire function, we see that

$$\begin{aligned} N\left( r,\frac{1}{f(z)}\right) =N(r,H(z))=0. \end{aligned}$$

Thus, (2.29) is a contradiction. Hence, $\lambda \left( y(z)-\frac{P_1(z)-(1+a(z))P_2(z)}{(1+a(z))P_3(z)}\right) =\sigma (y(z)).$

(iii)   Suppose that there is an entire function h(z) satisfying

$$\begin{aligned} (P_2(z)-P_1(z)+a(z)P_3(z))^2-4a(z)P_2(z)P_3(z)=h(z)^2. \end{aligned}$$

(2.30)

In what follows, we prove

$$\begin{aligned} \lambda (\Delta y(z)-a(z))=\sigma (y(z)). \end{aligned}$$

(2.31)

By (1.3), we obtain

$$\begin{aligned}&y(z+1)-y(z)-a(z)\nonumber \\&\quad =-\frac{P_3(z)y(z)^2+[P_2(z)-P_1(z)+P_3(z)a(z)]y(z)+P(z)a(z)}{P_2(z)+P_3(z)y(z)}. \end{aligned}$$

(2.32)

By (2.30) and (2.32), we obtain

$$\begin{aligned}&y(z+1)-y(z)-a(z)\nonumber \\&\quad =-\frac{P_3(z)\left\{ y(z)-\frac{[-P_2(z)+P_1(z)-P_3(z)a(z)]+h(z)}{2P_3(z)}\right\} \left\{ y(z)-\frac{[-P_2(z)+P_1(z)-P_3(z)a(z)]-h(z)}{2P_3(z)}\right\} }{P_2(z)+P_3(z)y(z)}. \nonumber \\ \end{aligned}$$

(2.33)

Now we divide this proof into two cases: (1) a(z) is transcendental; (2) a(z) is a nonzero polynomial. $\square$

Case 1   Suppose that a(z) is transcendental. From (2.30), we know that h(z) is also a transcendental entire function and satisfies $\sigma (h(z))=\sigma (a(z))<1$ and

$$\begin{aligned} T(r,h(z))=T(r,a(z))+O(\log r). \end{aligned}$$

(2.34)

In addition, we have $T(r,P_3(z)a(z))=T(r,a(z))+O(\log r)$. Combining this with (2.34), we see that there is at least one of $-P_3(z)a(z)+h(z)$ and $-P_3(z)a(z)-h(z)$, say, $-P_3(z)a(z)-h(z)$, such that

$$\begin{aligned} T(r,-P_3(z)a(z)-h(z))=T(r,a(z))+O(\log r). \end{aligned}$$

(2.35)

Set $G(z)=-\,P_3(z)a(z)-h(z)$. Then, we see that G(z) is a transcendental entire function of order $\sigma (G(z))=\sigma (a(z))<1$ and satisfies

$$\begin{aligned} T(r,G(z))=T(r,a(z))+O(\log r). \end{aligned}$$

(2.36)

Since $P_j(z)\,(j=1,2,3)$ are polynomials, we see that poles of $P_2(z)+P_3(z)y(z)$ must be poles of $P_3(z)y(z)^2+[P_2(z)-P_1(z)+P_3(z)a(z)]y(z)+P_2(z)a(z)$. Thus, poles of $P_2(z)+P_3(z)y(z)$ are not zeros of $y(z+1)-y(z)-a(z)$. By Lemma 2.8, we see that the numerator and the denominator of the right side of (2.32) have at most finitely many common zeros. Thus, in order to prove (2.31), by (2.33), we only need to prove that

$$\begin{aligned} \lambda \left( y(z)-\frac{[-P_2(z)+P_1(z)-P_3(z)a(z)]+h(z)}{2P_3(z)}\right) =\sigma (y(z)), \end{aligned}$$

(2.37)

or

$$\begin{aligned} \lambda \left( y(z)-\frac{[-P_2(z)+P_1(z)-P_3(z)a(z)]-h(z)}{2P_3(z)}\right) =\sigma (y(z)). \end{aligned}$$

(2.38)

Now we prove that (2.38) holds. On the contrary, we suppose that

$$\begin{aligned} \lambda \left( y(z)-\frac{[-P_2(z)+P_1(z)-P_3(z)a(z)]-h(z)}{2P_3(z)}\right) <\sigma (y(z)). \end{aligned}$$

Thus, $y(z)-\frac{[-P_2(z)+P_1(z)-P_3(z)a(z)]-h(z)}{2P_3(z)}$ can be rewritten as

$$\begin{aligned} y(z)=\frac{[-P_2(z)+P_1(z)-P_3(z)a(z)]-h(z)}{2P_3(z)}+b_2(z)f_2(z), \end{aligned}$$

(2.39)

where $f_2(z)=\frac{1}{H_2(z)},\,b_2(z)$ and $H_2(z)$ are nonzero entire functions, such that

$$\begin{aligned} \lambda (b_2(z))=\sigma (b_2(z))<\sigma (y(z))\quad \text{ and }\quad \lambda (H_2(z))=\sigma (H_2(z))=\sigma (y(z)). \end{aligned}$$

Using the same method as in the proof of (2.17)–(2.20), we can obtain

$$\begin{aligned} H_3(z,f_2):= & {} \left\{ \frac{P_1(z+1)-P_2(z+1)-P_3(z+1)a(z+1)-h(z+1)}{2P_3(z+1)}\right. \\&\left. +\,b_2(z+1)f_2(z+1)\right\} \\&\cdot \left\{ \frac{P_1(z)+P_2(z)-P_3(z)a(z)-h(z)}{2}+b_2(z)f_2(z)\right\} \\&-\,P_1(z)\left\{ \frac{P_1(z)-P_2(z)-P_3(z)a(z)-h(z)}{2P_3(z)}+b_2(z)f_2(z)\right\} \\= & {} 0, \end{aligned}$$

and

$$\begin{aligned} H_3(z,0)=\frac{W_1(z)-W_2(z)}{4P_3(z)P_3(z+1)}, \end{aligned}$$

(2.40)

where

$$\begin{aligned} \left\{ \begin{array}{l}W_1(z)=P_3(z)\{G(z+1)+P_1(z+1)-P_2(z+1)\}\{G(z)+P_1(z)+P_2(z)\};\\ W_2(z)=2P_1(z)P_3(z+1)\{G(z)+P_1(z)-P_2(z)\}.\end{array}\right. \nonumber \\ \end{aligned}$$

(2.41)

We assert that $W_1(z)-W_2(z)\not \equiv 0$. On the contrary, we suppose that $W_1(z)-W_2(z)\equiv 0$. That is,

$$\begin{aligned}&P_3(z)[G(z+1)+P_1(z+1)-P_2(z+1)][G(z)+P_1(z)+P_2(z)]\nonumber \\&\quad =2P_1(z)P_3(z+1)\left[ G(z)+P_1(z)-P_2(z)\right] . \end{aligned}$$

(2.42)

Noting that $P_j(z)\,(j=1,2,3)$ are polynomials and G(z) is a transcendental entire function of finite order, applying Lemma 2.9 to (2.42), we have

$$\begin{aligned} T(r,G(z))= & {} T(r,G(z)+P_1(z)+P_2(z))+O(\log r)\\= & {} m(r,G(z)+P_1(z)+P_2(z))+O(\log r)\\= & {} S(r,G(z))+O(\log r)\\= & {} S(r,G(z)). \end{aligned}$$

This is a contradiction. Hence, $W_1(z)-W_2(z)\not \equiv 0$. So that $H_2(z,0)\not \equiv 0.$

Thus, by Lemma 2.2 and using the same method as in the proof of (ii), we see that (2.31) holds.

Case 2   Suppose that a(z) is a nonconstant polynomial. Then, by (1.4) and (2.30), we see that

$$\begin{aligned} \deg h(z)=\deg P_3(z)+\deg a(z). \end{aligned}$$

(2.43)

By Lemma 2.8 and (2.33), we see that to prove (2.31), we only need to prove (2.37) or (2.38), where a(z) is a nonconstant polynomial. By (2.43), we have $\deg (P_3(z)a(z))=\deg h(z)=\deg P_3(z)+\deg a(z)$. Combining this with (1.4), we see that there exists at least one of

$$\begin{aligned}{}[-P_2(z)+P_1(z)-P_3(z)a(z)]+h(z)\quad \text{ and }\quad [-P_2(z)+P_1(z)-P_3(z)a(z)]-h(z), \end{aligned}$$

say, $[-P_2(z)+P_1(z)-P_3(z)a(z)]-h(z)$, such that

$$\begin{aligned} \deg \left( [-P_2(z)+P_1(z)-P_3(z)a(z)]-h(z)\right) =\deg P_3(z)+\deg a(z). \end{aligned}$$

(2.44)

Set $a(z)=az^s+\cdots$ and $P_j(z)=b_jz^{d_j}+\cdots (j=1,2,3)$, where $a,\,b_j$ are nonzero constants, $s\,(\ge 1),\,d_1,\,d_2,\,d_3\,(\ge 0)$ are nonnegative integers. Then, from (1.4) and (2.44), we have

$$\begin{aligned} P_3(z)a(z)+h(z)=dz^{d_3+s}+\cdots , \end{aligned}$$

(2.45)

where d is a nonzero constant. Thus, from (2.41) and (2.45), we have

$$\begin{aligned} \left\{ \begin{array}{l}W_1(z)=\left( b_3z^{d_3}+\cdots \right) \left( -dz^{d_3+s}+\cdots \right) \left( -dz^{d_3+s}+\cdots \right) =b_3d^2z^{3d_3+2s}+\cdots ,\\ W_2(z)=\left( 2b_1z^{d_1}+\cdots \right) \left( b_3z^{d_3}+\cdots \right) \left( -dz^{d_3+s}+\cdots \right) =-\,2b_1b_3dz^{2d_3+d_1+s}+\cdots .\end{array}\right. \nonumber \\ \end{aligned}$$

(2.46)

By (1.4), we see that $3d_3+2s\ge 2d_3+d_1+2s>2d_3+d_1+s$, that is, $W_1(z)-W_2(z)\not \equiv 0$, so that $H_2(z,0)\not \equiv 0$.

Thus, by Lemma 2.2 and using the same method as in the proof of (ii), we see that (2.31) holds.

Thus, Theorem 1.1 is proved.

3 Proof of Theorem 1.2

Using the following Lemma 3.1 and the same method as in the proof of Theorem 1.1, we can obtain Theorem 1.2.

Lemma 3.1

Let $P_j(z),\,j=1,2,3$ be nonzero polynomials and satisfy (1.4) in Theorem 1.1, a is a nonzero constant, and h(z) is a polynomial satisfying

$$\begin{aligned}&(P_2-P_1+aP_3)^2-4aP_2P_3=h^2,\quad \deg (P_2-aP_3-h)\nonumber \\&\quad =\deg h=\deg P_3. \end{aligned}$$

(3.1)

Suppose that

$$\begin{aligned} \left\{ \begin{array}{l} W_1(z)= P_3(z)\left\{ P_1(z+1)-aP_3(z+1)-h(z+1)-P_2(z+1)\right\} \\ \quad \cdot \left\{ P_1(z)-aP_3(z)-h(z)+P_2(z)\right\} ;\\ W_2(z) = 2P_1(z)P_3(z+1)\left\{ P_1(z)-aP_3(z)-h(z)-P_2(z) \right\} .\end{array}\right. \end{aligned}$$

(3.2)

Then

$$\begin{aligned} W_1(z)-W_2(z)\not \equiv 0. \end{aligned}$$

Proof

Suppose that $P_j(z)=b_jz^{d_j}+\cdots (j=1,2,3)$, where $b_j$ are nonzero constants, $d_j$ are nonnegative integers. By (1.4), we have

$$\begin{aligned} d_3\ge \max \{d_1,\,d_2\}. \end{aligned}$$

(3.3)

From (3.1) and (3.3), we see that $\deg h(z)=\deg P_3(z)=d_3$. Set $h(z)=h_*z^{d_3}+\cdots$, where $h_*$ is a nonzero constant. By (3.1), we have

$$\begin{aligned}&(P_2(z)+aP_3(z)-P_1(z)-h(z))(P_2(z)+aP_3(z)-P_1(z)+h(z))\nonumber \\&\quad =4aP_2(z)P_3(z). \end{aligned}$$

(3.4)

We divide this proof into two cases: (1) $d_3>\max \{d_1,\, d_2\}$; (2) $d_3=\max \{d_1,\,d_2\}$.

Case 1   Suppose that $d_3>\max \{d_1,\, d_2\}$. Then, by $\deg h(z)=\deg P_3(z)=d_3$, we see that there exists at least one of $-aP_3(z)-h(z)$ and $-aP_3(z)+h(z)$, say, $-aP_3(z)-h(z)$, such that $\deg [-aP_3(z)-h(z)]=\deg P_3(z)$. Thus, we have

$$\begin{aligned} aP_3(z)+h(z)=dz^{d_3}+\cdots , \end{aligned}$$

(3.5)

where d is a nonzero constant. Noting that $d_3>\max \{d_1,\, d_2\}$ and (3.5), we have

$$\begin{aligned} P_1(z)-aP_3(z)-h(z)\pm P_2(z)=-\,dz^{d_3}+\cdots . \end{aligned}$$

(3.6)

Thus, from (3.2) and (3.6), we have

$$\begin{aligned} \left\{ \begin{array}{l}W_1(z)=\left( b_3z^{d_3}+\cdots \right) \left( -dz^{d_3}+\cdots \right) \left( -dz^{d_3}+\cdots \right) =b_3d^2z^{3d_3}+\cdots ,\\ W_2(z)=\left( 2b_1z^{d_1}+\cdots \right) \left( b_3z^{d_3}+\cdots \right) \left( -dz^{d_3}+\cdots \right) =-\,2b_1b_3dz^{2d_3+d_1}+\cdots .\end{array}\right. \end{aligned}$$

(3.7)

Since $3d_3>2d_3+d_1$, from (3.7), we see that $W_1(z)-W_2(z)\not \equiv 0.$

Case 2   Suppose that $d_3=\max \{d_1,\,d_2\}$. We divide this proof into three subcases: (1)  $d_3=d_1>d_2$; (2) $d_3=d_1=d_2$; (2) $d_3=d_2>d_1$.

Subcase 2.1    Suppose that $d_3=d_1>d_2$. Then, by $\deg h(z)=d_3$, we see that there exists at least one of $aP_3(z)-P_1(z)-h(z)$ and $aP_3(z)-P_1(z)+h(z)$, say, $aP_3(z)-P_1(z)+h(z)$, such that $\deg (aP_3(z)-P_1(z)+h(z))=d_3$. Since $d_3>d_2$, comparing the degrees of both sides of (3.4), we have

$$\begin{aligned} \deg (P_2(z)+aP_3(z)-P_1(z)+h(z))=d_3\quad \text{ and }\quad ab_3-b_1+h_*\ne 0. \end{aligned}$$

(3.8)

By (3.4) and $\deg (4aP_2(z)P_3(z))=d_3+d_2$, we have

$$\begin{aligned} \deg (P_2(z)+aP_3(z)-P_1(z)-h(z))=d_2. \end{aligned}$$

(3.9)

Hence, by (3.9) and $d_2<d_1=d_3=\deg h(z)$, we see that

$$\begin{aligned} \deg (aP_3(z)-P_1(z)-h(z))<d_3\quad \text{ and }\quad ab_3-b_1-h_*=0. \end{aligned}$$

(3.10)

From (3.2), (3.8) and $d_2<d_1=d_3=\deg h(z)$, we see that

$$\begin{aligned} \left\{ \begin{array}{l}W_1(z)=b_3(b_1-ab_3-h_*)^2z^{3d_3}+\cdots ,\\ W_2(z)=2b_1b_3(b_1-ab_3-h_*)z^{2d_3+d_1}+\cdots .\end{array}\right. \end{aligned}$$

(3.11)

From (3.10), (3.11) and $d_3=d_1$, we see that

$$\begin{aligned} W_1(z)-W_2(z)= & {} -b_3(b_1-ab_3-h_*)(b_1+ab_3+h_*)z^{3d_3}+\cdots \nonumber \\= & {} -2ab_3^2(b_1-ab_3-h_*)z^{3d_3}+\cdots . \end{aligned}$$

(3.12)

By (3.8) and (3.12), we see that $W_1(z)-W_2(z)\not \equiv 0.$

Subcase 2.2   Suppose that $d_3=d_1=d_2$. Then, by $\deg h(z)=d_3$ and (3.4), we see that

$$\begin{aligned} (b_2+ab_3-b_1)^2-h_*^2=4ab_2b_3\quad \text{ and }\quad b_2+ab_3-b_1+h_*\ne 0. \end{aligned}$$

(3.13)

Thus, from (3.13), we have

$$\begin{aligned} (b_2-ab_3-b_1)^2-h_*^2=4ab_1b_3\quad \text{ and }\quad b_2-ab_3-b_1-h_*\ne 0. \end{aligned}$$

(3.14)

From (3.2), (3.13), (3.14) and $d_3=d_1=d_2$, we see that

$$\begin{aligned} \left\{ \begin{array}{l}W_1(z)=b_3(b_1-ab_3-h_*-b_2)(b_1-ab_3-h_*+b_2)z^{3d_3}+\cdots ,\\ W_2(z)=2b_1b_3(b_1-ab_3-h_*-b_2)z^{3d_3}+\cdots .\end{array}\right. \end{aligned}$$

(3.15)

Hence, from (3.15), we have

$$\begin{aligned} W_1(z)-W_2(z)= & {} b_3(b_1-ab_3-h_*-b_2)(b_2-ab_3-h_*-b_1)z^{3d_3}+\cdots .\nonumber \\ \end{aligned}$$

(3.16)

By (3.13), (3.14) and (3.16), we see that $W_1(z)-W_2(z)\not \equiv 0.$

Subcase 2.3   Suppose that $d_3=d_2>d_1$. Then, by $\deg h(z)=d_3$, we see that there exists at least one of $P_2(z)+aP_3(z)-h(z)$ and $P_2(z)+aP_3(z)+h(z)$, say, $P_2(z)+aP_3(z)+h(z)$, such that $\deg (P_2(z)+aP_3(z)+h(z))=d_3$. Since $d_3>d_1$, we have

$$\begin{aligned} \deg (P_2(z)+aP_3(z)-P_1(z)+h(z))=d_3\quad \text{ and }\quad b_2+ab_3+h_*\ne 0. \end{aligned}$$

(3.17)

By the condition $\deg (P_2-aP_3-h)=\deg h$, we see that

$$\begin{aligned} ab_3+h_*-b_2\ne 0. \end{aligned}$$

(3.18)

From (3.2) and $d_1<d_2=d_3=\deg h(z)$, we see that

$$\begin{aligned} \left\{ \begin{array}{l}W_1(z)=b_3(ab_3+h_*+b_2)(ab_3+h_*-b_2)z^{3d_3}+\cdots ,\\ W_2(z)=-\,2b_1b_3(ab_3+h_*+b_2)z^{2d_3+d_1}+\cdots .\end{array}\right. \end{aligned}$$

(3.19)

By (3.17)–(3.19) and $d_3>d_1$, we see that $W_1(z)-W_2(z)\not \equiv 0.$$\square$

Thus, Lemma 3.1 is proved.