Growth of meromorphic solutions of certain types of q-difference differential equations

Abstract

In this paper, relying on Nevanlinna theory of the value distribution of meromorphic functions, we mainly study meromorphic solutions of certain types of q-difference differential equations, obtain estimates of the growth order of their meromorphic solutions, and give a number of examples to show what our results are the best possible in certain senses. Improvements and extensions of some results in the literature are presented.

1 Introduction and main results

In this paper, we assume that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory [1]. In addition, we use notations $\lambda (\frac{1}{f} )$ and $\rho(f)$ to denote the exponent of convergence of the pole-sequence and the order of growth of meromorphic function $f(z)$, respectively. We denote by $S(r,f)$ any quantify satisfying $S(r,f)=o(T(r,f))$ as $r\rightarrow\infty$ outside of a possible exceptional set of finite logarithmic measure. We define the logarithmic measure of E to be

$$\operatorname{lm}(E)= \int_{E\cap(1,\infty)}\frac{dr}{r}. $$

A set $E\subset(1,\infty)$ is said to have finite logarithmic measure if $\operatorname{lm}(E)<\infty$. Further, we recall the definitions of the truncated exponent of convergence of the pole-sequence and the lower order in complex plane:

$$\overline{\lambda} \biggl(\frac{1}{f} \biggr)=\limsup_{r\to\infty} \frac {\log^{+}\overline{N}(r,f)}{\log r}, \qquad\mu(f)=\liminf_{r \to\infty}\frac{\log^{+}T(r,f)}{\log r}. $$

There has been a lot of work on the growth order of meromorphic solution to certain types of complex differential equations and complex difference equations (or complex functional equations); see [2–10]. Malmquist [8] investigated the existence of transcendental meromorphic solutions of a complex differential equation and obtained the following result.

Theorem A

[8]

Let

$$ \frac{df(z)}{dz}=R\bigl(z,f(z)\bigr)=\frac{P(z,f(z))}{Q(z,f(z))}= \frac{\sum_{i=0}^{p}a_{i}(z)f^{i}(z)}{\sum_{j=0}^{q}b_{j}(z)f^{j}(z)}, $$

(1.1)

where $P(z,f(z))$ and $Q(z,f(z))$ are relatively prime polynomials in $f(z)$, the coefficients $a_{i}(z)$ ($i=0,\ldots,p$) and $b_{j}(z)$ ($j=0,\ldots,q$) are rational functions. If equation (1.1) admits a transcendental meromorphic solution, then $q=0$ and $p\leq2$.

Recently, Gundersen et al. [7] considered meromorphic solutions of a functional equation of the form

$$ f(qz)=R\bigl(z,f(z)\bigr)=\frac{\sum_{i=0}^{k}a_{i}(z)f^{i}(z)}{\sum_{j=0}^{l}b_{j}(z)f^{j}(z)}, $$

(1.2)

where the coefficients $a_{i}(z)$ ($i=0,\ldots,k$) and $b_{j}(z)$ ($j=0,\ldots,l$) are of growth $S(r,f)$, and q ($\vert q\vert > 1$) is a constant. In fact, they obtained the following theorem.

Theorem B

([7]) Suppose that $f(z)$ is a transcendental meromophic of equation (1.2) with $\vert q\vert > 1$. Assuming that $d:=\deg_{f}R(z,f(z))=\max\{k,l\}\geq1$, $a_{k}(z)\not\equiv0$, $b_{l}(z)\not\equiv0$, and that $R(z,f(z))$ is irreducible in $f(z)$. Then

$$\rho(f)=\frac{\log d}{\log q}. $$

In this paper, we continue to investigate the growth order of meromorphic solutions to certain types of complex q-difference differential equations and generalize Theorems A and B. Now, we state our results as follows.

Theorem 1.1

Suppose that $f(z)$ is a solution of the equation

$$ \bigl(f'(qz)\bigr)^{n}=R\bigl(z,f(z)\bigr)= \frac{\sum_{i=0}^{k}a_{i}(z)f^{i}(z)}{\sum_{j=0}^{l}b_{j}(z)f^{j}(z)} $$

(1.3)

with meromorphic coefficients $a_{i}(z)$ ($i=0,\ldots,k$) and $b_{j}(z)$ ($j=0,\ldots,l$) of growth $S(r,f)$ and a constant $q\in\mathbb{C}\backslash\{0\}$, assuming that $d:=\operatorname{deg}_{f}R(z,f(z))=\max\{k,l\}\geq1$, $a_{k}(z)\not\equiv0$, $b_{l}(z)\not\equiv0$, and that $R(z,f(z))$ is irreducible in $f(z)$. Then one of the following cases holds.

• (i) For $\vert q\vert >1$, if $f(z)$ is a transcendental meromorphic solution of equation (1.3) and $d>2n$, then

  $$\frac{\log d-\log2n}{\log \vert q\vert }\leq\mu(f)\leq\rho(f). $$

  If $f(z)$ is a transcendental entire solution of equation (1.3) and $d>n$, then

  $$\frac{\log d-\log n}{\log \vert q\vert }\leq\mu(f)\leq\rho(f). $$

• (ii) For $\vert q\vert <1$, if $f(z)$ is a transcendental meromorphic solution of equation (1.3), then $d\leq2n$, and

  $$\rho(f)\leq\frac{\log2n-\log d}{-\log \vert q\vert }. $$

  If $f(z)$ is a transcendental entire solution of equation (1.3), then $d\leq n$, and

  $$\rho(f)\leq\frac{\log n-\log d}{-\log \vert q\vert }. $$

• (iii) For $\vert q\vert =1$, if $f(z)$ is a transcendental meromorphic solution of equation (1.3), then $d\leq2n$. Furthermore, if $n< d\leq2n$, then $\overline{\lambda} (\frac{1}{f} )=\lambda (\frac {1}{f} )=\rho(f)$. If $f(z)$ is a transcendental entire solution of equation (1.3), then $d\leq n$.

Example 1.1

The function $f(z)=\frac{e^{z}+1}{e^{z}-1}$ is a solution to the q-difference differential equation

$$f'(3z)=-\frac{(f^{2}(z)-1)^{3}}{2(3f^{2}(z)+1)^{2}}, $$

where $d=6$, $n=1$, $q=3$, so that $d>2n$, $\vert q\vert >1$. Then $\frac{\log d-\log2n}{\log \vert q\vert }=1=\mu(f)=\rho(f)$.

Example 1.2

The function $f(z)=\cos z$ is a solution to the q-difference differential equation

$$\bigl(f'(2z)\bigr)^{2}=-4f^{4}(z)+4f^{2}(z), $$

where $d=4$, $n=2$, $q=2$, so that $d>n$, $\vert q\vert >1$. Then $\frac{\log d-\log n}{\log \vert q\vert }=1=\mu(f)=\rho(f)$.

Example 1.3

The function $f(z)=\frac{e^{z}}{z+1}$ is a solution to the q-difference differential equation

$$\bigl(f'(z/3)\bigr)^{3}=\frac{27(z+1)z^{3}}{(z+3)^{6}}f(z), $$

where $d=1$, $n=3$, $q=\frac{1}{3}$, so that $d<2n$, $\vert q\vert <1$. Then $1=\rho(f)<\frac{\log2n-\log d}{-\log \vert q\vert }=1+\frac{\log2}{\log3}$.

Example 1.4

The function $f(z)=e^{z^{2}}+1$ is a solution to the q-difference differential equation

$$\bigl(f'(z/2)\bigr)^{2}=z^{4}f(z)-z^{4}, $$

where $d=1$, $n=4$, $q=\frac{1}{2}$, so that $d< n$, $\vert q\vert <1$. Then $\rho(f)=2=\frac{\log n-\log d}{-\log \vert q\vert }$.

Example 1.5

The function $f(z)=\frac{e^{z}+1}{z}$ is a solution to the q-difference differential equation

$$f'(-z)=-\frac{f(z)+1}{z^{2}f(z)-z}, $$

where $d=1$, $n=1$, $q=-1$, so that $n=d<2n$, $\vert q\vert =1$.

Example 1.6

The function $f(z)=e^{z}+1$ is a solution to the q-difference differential equation

$$f'(z)=f(z)-1, $$

where $d=1$, $n=1$, $q=1$, so that $d=n$, $\vert q\vert =1$.

Theorem 1.2

Suppose that $f(z)$ is a solution of the equation

$$ \bigl(f'(qz)\bigr)^{n}=R\bigl(z,f\bigl(p(z)\bigr)\bigr)= \frac{\sum_{i=0}^{k}a_{i}(z)f^{i}(p(z))}{\sum_{j=0}^{l}b_{j}(z)f^{j}(p(z))} $$

(1.4)

with meromorphic coefficients $a_{i}(z)$ ($i=0,\ldots,k$) and $b_{j}(z)$ ($j=0,\ldots,l$) of growth $S(r,f)$, a constant $q\in\mathbb{C}\backslash\{0\}$, and $p(z)=c_{m}z^{m}+c_{m-1}z^{m-1}+\cdots+c_{0}$, where $c_{m}$ (≠0), $c_{m-1}$, … , $c_{0}$ are complex constants, and m (≥2) is an integer. Assume that $d:=\operatorname{deg}_{f}R(z,f(z))=\max\{k,l\}\geq1$, $a_{k}(z)\not\equiv0$, $b_{l}(z)\not\equiv0$, and that $R(z,f(z))$ is irreducible in $f(z)$. Then, if $f(z)$ is a transcendental meromorphic solution of equation (1.4) and $d\leq2n$, then

$$T\bigl(r,f(z)\bigr)=O\bigl((\log r)^{\alpha}\bigr), $$

where

$$\alpha=\frac{\log2n-\log d}{\log m}. $$

If $f(z)$ is a transcendental entire solution of equation (1.4) and $d\leq n$, then

$$T\bigl(r,f(z)\bigr)=O\bigl((\log r)^{\alpha}\bigr), $$

where

$$\alpha=\frac{\log n-\log d}{\log m}. $$

Theorem 1.3

Suppose that $f(z)$ is a solution of equation

$$ \sum_{s=1}^{n}\alpha_{s}(z)f^{(\lambda_{s})}(q_{s}z)=R \bigl(z,f(z)\bigr)=\frac {\sum_{i=0}^{k}a_{i}(z)f^{i}(z)}{\sum_{j=0}^{l}b_{j}(z)f^{j}(z)} $$

(1.5)

with meromorphic coefficients $\alpha_{s}(z)$ ($s=1,\ldots,n$), $a_{i}(z)$ ($i=0,\ldots,k$), and $b_{j}(z)$ ($j=0,\ldots,l$) of growth $S(r,f)$, distinct constants $q_{s}$ with $\vert q_{s}\vert \geq1$, and finite nonnegative integers $\lambda_{s}$. Suppose that $a_{k}(z)\not \equiv0$, $b_{l}(z)\not\equiv0$, and $R(z,f(z))$ is irreducible in $f(z)$. Denote

$$d:=\operatorname{deg}_{f}R\bigl(z,f(z)\bigr)=\max\{k,l\}\geq1;\qquad \lambda=\sum _{s=1}^{n}\lambda _{s};\qquad \vert q\vert = \max_{1\leq s\leq n}\bigl\{ \vert q_{s}\vert \bigr\} >1. $$

Then if $f(z)$ is a transcendental meromorphic solution of equation (1.5) and $d>\lambda+n$, then

$$\frac{\log d-\log(\lambda+n)}{\log \vert q\vert }\leq\mu(f)\leq\rho(f). $$

If $f(z)$ is a transcendental entire solution of equation (1.5) and $d>n$, then

$$\frac{\log d-\log n}{\log \vert q\vert }\leq\mu(f)\leq\rho(f). $$

Example 1.7

The function $f(z)=\frac{1}{e^{z}+1}$ is a solution to the q-difference differential equation

$$f''(2z)=\frac{-f^{2}(z)(f(z)-1)^{2}(2f(z)-1)}{(2f^{2}(z)-2f(z)+1)^{3}}, $$

where $d=6$, $n=1$, $\lambda=2$, so that $d>\lambda+n=3$, $\vert q\vert =2>1$. Then $1=\frac{\log d-\log(\lambda+n)}{\log \vert q\vert }=\mu(f)=\rho(f)$.

Example 1.8

The function $f(z)=e^{z}+1$ is a solution to the q-difference differential equation

$$f'(z)+f''(3z)=f^{3}(z)-3f^{2}(z)+4f(z)-2, $$

where $d=3$, $n=2$, so that $d>n$, $\vert q\vert =\max\{\vert q_{1}\vert , \vert q_{2}\vert \}=3>1$. Then $1-\frac{\log2}{\log3}=\frac{\log d-\log n}{\log \vert q\vert }<\mu(f)=\rho(f)=1$.

Theorem 1.4

Suppose that $f(z)$ is a solution of the equation

$$ \sum_{s=1}^{n}\alpha_{s}(z)f^{(\lambda_{s})}(q_{s}z)=R \bigl(z,f\bigl(p(z)\bigr)\bigr)=\frac {\sum_{i=0}^{k}a_{i}(z)f^{i}(p(z))}{\sum_{j=0}^{l}b_{j}(z)f^{j}(p(z))} $$

(1.6)

with meromorphic coefficients $\alpha_{s}(z)$ ($s=1,\ldots,n$), $a_{i}(z)$ ($i=0,\ldots,k$), and $b_{j}(z)$ ($j=0,\ldots,l$) of growth $S(r,f)$, distinct nonzero constants $q_{s}$, finite nonnegative integers $\lambda_{s}$, and $p(z)=c_{m}z^{m}+c_{m-1}z^{m-1}+\cdots+c_{0}$, where $c_{m}$ (≠0), $c_{m-1}$, … , $c_{0}$ are complex constants, and m (≥2) is an integer. Suppose that $a_{k}(z)\not\equiv0$, $b_{l}(z)\not\equiv0$, and $R(z,f(z))$ is irreducible in $f(z)$. Denote

$$d:=\operatorname{deg}_{f}R\bigl(z,f(z)\bigr)=\max\{k,l\}\geq1;\qquad \lambda=\sum _{s=1}^{n}\lambda _{s};\qquad \vert q\vert = \max_{1\leq s\leq n}\bigl\{ \vert q_{s}\vert \bigr\} >0. $$

Then if $f(z)$ is a transcendental meromorphic solution of equation (1.6) and $d\leq\lambda+n$, then

$$T\bigl(r,f(z)\bigr)=O\bigl((\log r)^{\alpha}\bigr), $$

where

$$\alpha=\frac{\log(\lambda+n)-\log d}{\log m}. $$

If $f(z)$ is a transcendental entire solution of equation (1.6) and $d\leq n$, then

$$T\bigl(r,f(z)\bigr)=O\bigl((\log r)^{\alpha}\bigr), $$

where

$$\alpha=\frac{\log n-\log d}{\log m}. $$

Beardon [3] studied entire solutions of the generalized function equation

$$ f(qz)=qf(z)f'(z), \quad f(0)=0, $$

(1.7)

where q is a nonzero complex number. First, we give some notations. The formal series $\mathcal{O}$ and $\mathcal{I}$ are defined by $\mathcal{O}:=0+0z+0z^{2}+\cdots$ and $\mathcal {I}:=0+1z+0z^{2}+0z^{3}+\cdots$ . We also introduce the sets $\mathcal{K}_{p}=\{z:z^{p}=p+2\}$ ($p=1,2,\ldots$) and $\mathcal{K}=\mathcal{K}_{1}\cup\mathcal{K}_{2}\cup\cdots$ . Clearly, $\mathcal{K}_{p}$ contains exactly p points, which are equally spaced around the circle $\vert z\vert =r_{p}$, where $r_{p}=(p+2)^{\frac{1}{p}}>1$ and $r_{p}\in\mathcal{K}_{p}$. Also, since $x^{-1}\log(x+2)$ is decreasing when $x>1$. we see that $r_{1}=3>r_{2}=2>\cdots>1$ and $r_{p}\to1$ as $p\to\infty$. Based on these notations, Beardon obtained the following two main theorems.

Theorem C

[3]

Any transcendental solution of (1.7) is of the form

$$f(z)=z+z\bigl(bz^{p}+\cdots\bigr), $$

where p is a positive integer, $b\neq0$, and $q\in\mathcal{K}_{p}$. In particular, if $q\notin\mathcal{K}$, then the only formal solutions of (1.7) are $\mathcal{O}$ and $\mathcal{I}$.

Theorem D

[3]

For each positive integer p, there is a unique real entire function

$$F_{p}=z\bigl(1+z^{p}+b_{2}z^{2p}+b_{3}z^{3p}+ \cdots\bigr) $$

that is a solution of (1.7) for each q in $\mathcal{K}_{p}$. Further, if $q\in\mathcal{K}_{p}$, then the only transcendental solutions of (1.7) are the linear conjugates of $F_{p}$.

More recently, Zhang [10] investigated the growth of solutions of (1.7) and obtained the following theorem.

Theorem E

[10]

Suppose that $f(z)$ is a transcendental solution of (1.7) for $k\in\mathcal{K}$. Then the order of growth $\rho(f)\leq\frac{\log2}{\log \vert q\vert }$.

In this paper, we generalize equation (1.7) and investigate the growth of solution of certain types of q-difference differential equations and obtain the following results.

Theorem 1.5

Let q be a complex constant satisfying $\vert q\vert >1$. Suppose that $f(z)$ is a solution to the equation

$$ \sum_{s=1}^{n}\alpha_{s}(z)f^{(\lambda_{s})}(z)= \frac {A(qz,f(qz))}{B(z,f(z))}, $$

(1.8)

where $A(z,y)$ and $B(z,y)$ are rational functions with meromorphic coefficients of growth $S(r,f)$ such that $A(z,y)$ and $B(z,y)$ are irreducible in y. Denote

$$\lambda=\sum_{s=1}^{n}\lambda_{s};\qquad 1\leq a:=\deg_{f}A\leq\deg_{f}B=:b. $$

Then,

• (i) if $f(z)$ is a transcendental meromorphic solution of equation (1.8), then

  $$\rho(f)\leq\frac{\log(b+\lambda+n)-\log a}{\log \vert q\vert }. $$

  Furthermore, if $b>a+\lambda+n$, then

  $$\frac{\log(b-\lambda-n)-\log a}{\log \vert q\vert }\leq\mu(f)\leq\rho(f)\leq\frac {\log(b+\lambda+n)-\log a}{\log \vert q\vert }. $$

• (ii) If $f(z)$ is a transcendental entire solution of equation (1.8), then

  $$\rho(f)\leq\frac{\log(b+n)-\log a}{\log \vert q\vert }. $$

  Furthermore, if $b>a+n$, then

  $$\frac{\log(b-n)-\log a}{\log \vert q\vert }\leq\mu(f)\leq\rho(f)\leq\frac{\log (b+n)-\log a}{\log \vert q\vert }. $$

Example 1.9

The function $f(z)=\tan z$ is a solution to the q-difference differential equation

$$f''(z)=\frac{f^{2}(2z)}{\frac{2f(z)}{f^{6}(z)-f^{4}(z)-f^{2}(z)+1}}, $$

where $a=2$, $b=6$, $n=1$, $q=2$, $\lambda=2$, so that $b>a+\lambda+n=5$. Then $\frac{\log3}{\log2}-1=\frac{\log(b-\lambda-n)-\log a}{\log \vert q\vert }<\mu (f)=\rho(f)=1< \frac{\log(b+\lambda+n)-\log a}{\log \vert q\vert }=2\frac{\log3}{\log2}-1$.

Example 1.10

The function $f(z)=ze^{z}$ is a solution to the q-difference differential equation

$$f'(z)+f''(z)=\frac{f^{2}(3z)+f(3z)}{\frac{9f^{5}(z)}{(2z+3)z^{3}}+\frac {3f^{2}(z)}{(2z+3)z}}, $$

where $a=2$, $b=5$, $n=2$, $q=3$, so that $b>a+n=4$. Then $1-\frac{\log2}{\log3}=\frac{\log(b-n)-\log a}{\log \vert q\vert }<\mu(f)=\rho(f)=1< \frac{\log(b+n)-\log a}{\log \vert q\vert }=\frac{\log7-\log2}{\log3}$.

Theorem 1.6

Let q be a complex constant satisfying $\vert q\vert >1$. Suppose that $f(z)$ is a solution to the equation

$$ \bigl(f'(z)\bigr)^{n}=\frac{A(qz,f(qz))}{B(z,f(z))}, $$

(1.9)

where $A(z,y)$ and $B(z,y)$ are rational functions with meromorphic coefficients of growth $S(r,f)$ such that $A(z,y)$ and $B(z,y)$ are irreducible in y. Denote $1\leq a:=\deg_{f}A\leq\deg_{f}B=:b$. Then,

• (i) if $f(z)$ is a transcendental meromorphic solution of equation (1.9), then

  $$\rho(f)\leq\frac{\log(b+2n)-\log a}{\log \vert q\vert }. $$

  Furthermore, if $b>a+2n$, then

  $$\frac{\log(b-2n)-\log a}{\log \vert q\vert }\leq\mu(f)\leq\rho(f)\leq\frac{\log (b+2n)-\log a}{\log \vert q\vert }. $$

• (ii) If $f(z)$ is a transcendental entire solution of equation (1.9), then

  $$\rho(f)\leq\frac{\log(b+n)-\log a}{\log \vert q\vert }. $$

  Furthermore, if $b>a+n$, then

  $$\frac{\log(b-n)-\log a}{\log \vert q\vert }\leq\mu(f)\leq\rho(f)\leq\frac{\log (b+n)-\log a}{\log \vert q\vert }. $$

Example 1.11

The function $f(z)=\tan z$ is a solution to the q-difference differential equation

$$\bigl(f'(z)\bigr)^{2}=\frac{f(2z)+1}{\frac {f^{2}(z)-2f(z)-1}{f^{6}(z)+f^{4}(z)-f^{2}(z)-1}}, $$

where $a=1$, $b=6$, $n=2$, $q=2$, so that $b>a+2n=5$. Then $1=\frac{\log(b-2n)-\log a}{\log \vert q\vert }=\mu(f)=\rho(f)< \frac{\log(b+2n)-\log a}{\log \vert q\vert }=1+\frac{\log5}{\log2}$.

Example 1.12

The function $f(z)=ze^{z}$ is a solution to the q-difference differential equation

$$\bigl(f'(z)\bigr)^{2}=\frac{f^{2}(4z)+f(4z)}{\frac {16f^{6}(z)}{(z+1)^{2}z^{4}}+\frac{4f^{2}(z)}{z(z+1)^{2}}}, $$

where $a=2$, $b=6$, $n=2$, $q=4$, so that $b>a+n=4$. Then $\frac{1}{2}=\frac{\log(b-n)-\log a}{\log \vert q\vert }<\mu(f)=\rho(f)= \frac{\log(b+n)-\log a}{\log \vert q\vert }=1$.

2 Some lemmas

Lemma 2.1

See [4], Lemma 4

Let $f(z)$ be a transcendental meromorphic function, and $p(z)=a_{k}z^{k}+a_{k-1}z^{k-1}+\cdots+a_{1}z+a_{0}$ ($a_{k}\neq0$) be a nonconstant polynomial of degree k. Given $0<\delta<\vert a_{k}\vert$, let $\lambda=\vert a_{k}\vert +\delta$ and $\mu=\vert a_{k}\vert -\delta$, then, for any given $\varepsilon>0$,

$$(1-\varepsilon)T\bigl(\mu r^{k},f(z)\bigr)\leq T\bigl(r,f\bigl(p(z) \bigr)\bigr)\leq(1+\varepsilon )T\bigl(\lambda r^{k},f(z)\bigr) $$

for sufficiently large r.

Lemma 2.2

See [7], Lemma 3.1

Let $\Phi:(1,\infty)\to(0,\infty)$ be an increasing function, and let $f(z)$ be a nonconstant meromorphic function. If for some real constant $\alpha\in(0,1)$, there exist real constants $K_{1}>0$ and $K_{2}\geq1$ such that

$$T\bigl(r,f(z)\bigr)\leq K_{1}\Phi(r)+K_{2}T\bigl(\alpha r,f(z)\bigr)+S(\alpha r,f), $$

then

$$\rho(f)\leq\frac{\log K_{2}}{-\log\alpha}+\limsup_{r\to\infty}\frac {\log\Phi(r)}{\log r}. $$

Lemma 2.3

See [9], Lemma 2.2

Let $\Phi:(r_{0},\infty)\to(1,\infty)$, where $r_{0}\geq1$, be an increasing function. If for some real constant $\alpha>1$, there exists a real number $K>1$ such that $\Phi(\alpha r)>K\Phi(r)$, then

$$\liminf_{r\to\infty}\frac{\log\Phi(r)}{\log r}\geq\frac{\log K}{\log \alpha}. $$

Lemma 2.4

See [5], Lemma 3

Let $\Psi(r)$ be a function of r ($r\geq r_{0}$), positive and bounded in every finite interval. Suppose that $\Psi(\mu r^{m})\leq A\Psi (r)+B(r\geq r_{0})$, where μ (>0), m (>1), A (≥1), and B are constants. Then $\Psi(r)=O((\log r)^{\alpha})$ with $\alpha=\frac{\log A}{\log m}$, unless $A=1$ and $B>0$; and if $A=1$ and $B>0$, then, for any $\varepsilon>0$, $\Psi(r)=O((\log r)^{\varepsilon})$.

The following lemma is proved by Bergweiler et al. [2], p. 2.

Lemma 2.5

$$T\bigl(r,f(qz)\bigr)=T\bigl(\vert q\vert r,f(z)\bigr)+O(1),\qquad \overline{N} \bigl(r,f(qz)\bigr)=\overline {N}\bigl(\vert q\vert r,f(z)\bigr)+O(1) $$

for any meromorphic function $f(z)$ and any nonzero constant q.

3 Proof of Theorems 1.1-1.2

Proof of Theorem 1.1

If $\vert q\vert >1$ and $f(z)$ is a transcendental meromorphic solution of (1.3), then by applying the Valiron-Mohon’ko identity (see [11], Theorem 2.2.5), Lemma 2.5, and [1], Theorem 3.1, it follows from (1.3) that

$$\begin{aligned} T\bigl(r,R\bigl(z,f(z)\bigr)\bigr) =& T \biggl(r,\frac{\sum_{i=0}^{k}a_{i}(z)f^{i}(z)}{\sum_{j=0}^{l}b_{j}(z)f^{j}(z)} \biggr) \\ =& d T\bigl(r,f(z)\bigr)+S(r,f) \\ = & T\bigl(r,\bigl(f'(q z)\bigr)^{n}\bigr) \\ \leq& n\bigl[T\bigl(r,f(q z)\bigr)+\overline{N}\bigl(r,f(q z)\bigr)+S \bigl(r,f(q z)\bigr)\bigr] \\ = & n\bigl[T\bigl(\vert q\vert r,f(z)\bigr)+\overline{N}\bigl(\vert q \vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z)\bigr)\bigr] \\ \leq& 2nT\bigl(\vert q\vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z) \bigr), \end{aligned}$$

that is,

$$ d T\bigl(r,f(z)\bigr)+S(r,f)\leq2nT\bigl(\vert q\vert r,f(z)\bigr)+S\bigl( \vert q\vert r,f(z)\bigr). $$

(3.1)

By (3.1), for any small $\varepsilon>0$,

$$ d(1-\varepsilon)T\bigl(r,f(z)\bigr)\leq2n(1+\varepsilon)T\bigl(\vert q\vert r,f(z)\bigr) $$

(3.2)

for sufficiently large $r\notin E$, where $\operatorname{lm}(E)<\infty$. By an application of [6], Lemma 5, with $\beta>1$ and (3.2) we see that

$$ d(1-\varepsilon)T\bigl(r,f(z)\bigr)\leq2n(1+\varepsilon)T\bigl(\beta \vert q \vert r,f(z)\bigr) $$

(3.3)

for all $r\geq r_{0}$. If $d\leq2n$, then since $\beta \vert q\vert >1$, estimate (3.3) is trivial. So we only have to consider the case where $d>2n$. Then $\frac {d(1-\varepsilon)}{2n(1+\varepsilon)}>1$. It follows from Lemmas 2.3 and (3.3) that

$$\mu(f)=\liminf_{r\to\infty}\frac{\log^{+}T(r,f)}{\log r}\geq\frac{\log (d(1-\varepsilon))-\log(2n(1+\varepsilon))}{\log\beta \vert q\vert }. $$

As $\varepsilon\to0^{+}$ and $\beta\to1^{+}$, we have

$$\rho(f)\geq\mu(f)\geq\frac{\log d-\log2n}{\log \vert q\vert }. $$

If $\vert q\vert >1$ and $f(z)$ is a transcendental entire solution of (1.3), then similarly to (3.1), for any small $\varepsilon>0$, we have

$$ d(1-\varepsilon)T\bigl(r,f(z)\bigr)\leq n(1+\varepsilon)T\bigl(\vert q\vert r,f(z)\bigr) $$

(3.4)

for sufficiently large $r\notin E$, where $\operatorname{lm}(E)<\infty$. If $d>n$, similarly to the previous argument, we conclude that

$$\rho(f)\geq\mu(f)\geq\frac{\log d-\log n}{\log \vert q\vert }. $$

If $\vert q\vert <1$ and $f(z)$ is a transcendental meromorphic solution of (1.3), then applying [6], Lemma 5, and (3.2), we obtain that there exists $\alpha>1$ such that

$$ \alpha \vert q\vert < 1 \quad \mbox{and} \quad d(1-\varepsilon)T\bigl(r,f(z)\bigr) \leq 2n(1+\varepsilon)T\bigl(\alpha \vert q\vert r,f(z)\bigr) $$

(3.5)

for all $r\geq r_{0}$. Since $\alpha \vert q\vert <1$, if $d>2n$, then $\frac {2n(1+\varepsilon)}{d(1-\varepsilon)}<1$, a contradiction to (3.5). Thus, we have $d\leq2n$. Then $\frac {2n(1+\varepsilon)}{d(1-\varepsilon)}>1$, and from Lemma 2.2 we have that

$$\rho(f)\leq\frac{\log(2n(1+\varepsilon))-\log(d(1-\varepsilon))}{-\log \alpha \vert q\vert }. $$

As $\varepsilon\to0^{+}$ and $\alpha\to1^{+}$, we have

$$\rho(f)\leq\frac{\log2n-\log d}{-\log \vert q\vert }. $$

If $\vert q\vert <1$ and $f(z)$ is a transcendental entire solution of (1.3), then similarly to the previous argument, we have

$$d\leq n \quad\mbox{and} \quad\rho(f)\leq\frac{\log n-\log d}{-\log \vert q\vert }. $$

If $\vert q\vert =1$ and $f(z)$ is a transcendental meromorphic solution of (1.3), then by the proof of (3.1) we conclude that

$$\begin{aligned} d T\bigl(r,f(z)\bigr)+S(r,f) \leq& n\bigl[T\bigl(r,f(z)\bigr)+\overline{N} \bigl(r,f(z)\bigr)+S(r,f)\bigr] \\ \leq& 2nT\bigl(r,f(z)\bigr)+S(r,f). \end{aligned}$$

From this inequality we have $d\leq2n$. If $n< d\leq2n$, then

$$\begin{aligned} \frac{d-n}{n}T\bigl(r,f(z)\bigr)+S(r,f) \leq& \overline{N}\bigl(r,f(z) \bigr)+S(r,f) \\ \leq& N\bigl(r,f(z)\bigr)+S(r,f) \\ \leq& T\bigl(r,f(z)\bigr)+S(r,f), \end{aligned}$$

that is, $\overline{\lambda} (\frac{1}{f} )=\lambda (\frac {1}{f} )=\rho(f)$. If $\vert q\vert =1$ and $f(z)$ is a transcendental entire solution of (1.3), then we similarly obtain that $d\leq n$. This completes the proof of Theorem 1.1. □

Proof of Theorem 1.2

If $f(z)$ is a transcendental meromorphic solution of (1.4), then by the Valiron-Mohon’ko identity ([11], Theorem 2.2.5), Lemma 2.5, and [1], Theorem 3.1, it follows from (1.4) that

$$\begin{aligned} T\bigl(r,R\bigl(z,f\bigl(p(z)\bigr)\bigr)\bigr) =& T \biggl(r,\frac{\sum_{i=0}^{k}a_{i}(z)f^{i}(p(z))}{\sum_{j=0}^{l}b_{j}(z)f^{j}(p(z))} \biggr) \\ =& d T\bigl(r,f\bigl(p(z)\bigr)\bigr)+S\bigl(r,f\bigl(p(z)\bigr)\bigr) \\ = & T\bigl(r,\bigl(f'(q z)\bigr)^{n}\bigr) \\ \leq& n\bigl[T\bigl(r,f(q z)\bigr)+\overline{N}\bigl(r,f(q z)\bigr)+S \bigl(r,f(q z)\bigr)\bigr] \\ = & n\bigl[T\bigl(\vert q\vert r,f(z)\bigr)+\overline{N}\bigl(\vert q \vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z)\bigr)\bigr] \\ \leq& 2nT\bigl(\vert q\vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z) \bigr), \end{aligned}$$

that is,

$$ d T\bigl(r,f\bigl(p(z)\bigr)\bigr)+S\bigl(r,f\bigl(p(z)\bigr)\bigr)\leq2nT\bigl( \vert q\vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z)\bigr). $$

(3.6)

By Lemma 2.1, for given $0<\delta<\vert c_{m}\vert$ and $\mu =\vert c_{m}\vert -\delta$ and for any small $\varepsilon>0$,

$$ d(1-\varepsilon)T\bigl(\mu r^{m},f(z)\bigr)\leq2n(1+\varepsilon)T \bigl(\vert q\vert r,f(z)\bigr) $$

(3.7)

for sufficiently large $r\notin E$, where $\operatorname{lm}(E)<\infty$. An application of [6], Lemma 5, with $\beta>1$ and (3.7) yields

$$d(1-\varepsilon)T\bigl(\mu r^{m},f(z)\bigr)\leq2n(1+\varepsilon)T \bigl(\beta \vert q\vert r,f(z)\bigr) $$

for $r\geq r_{0}$. Put $R=\beta \vert q\vert r$. Then the last inequality can be rewritten as

$$ T \biggl(\frac{\mu R^{m}}{\beta^{m}\vert q\vert ^{m}},f(z) \biggr)\leq \frac{2n(1+\varepsilon)}{d(1-\varepsilon)}T\bigl(R,f(z) \bigr). $$

(3.8)

If $d\leq2n$, then $\frac{2n(1+\varepsilon)}{d(1-\varepsilon)}\geq1$. Since $\frac{\mu}{\beta^{m}\vert q\vert ^{m}}>0$, $m\geq2$, by Lemma 2.4 we get that

$$T\bigl(r,f(z)\bigr)=O\bigl((\log r)^{\alpha}\bigr), $$

where

$$\begin{aligned} \alpha =& \frac{\log(2n(1+\varepsilon))-\log(d(1-\varepsilon))}{\log m} \\ =& \frac{\log2n-\log d}{\log m}+\frac{\log(1+\varepsilon)-\log (1-\varepsilon)}{\log m} \\ \rightarrow& \frac{\log2n-\log d}{\log m}\quad (\varepsilon\rightarrow0). \end{aligned}$$

If $f(z)$ is a transcendental entire solution of (1.4) and $d\leq n$, then we similarly have

$$T\bigl(r,f(z)\bigr)=O\bigl((\log r)^{\alpha}\bigr), $$

where

$$\alpha=\frac{\log n-\log d}{\log m}. $$

This completes the proof of Theorem 1.2. □

4 Proof of Theorems 1.3-1.4

Proof of Theorem 1.3

If $f(z)$ is a transcendental meromorphic solution of (1.5), then by applying the Valiron-Mohon’ko identity ([11], Theorem 2.2.5), Lemma 2.5, and [1], Theorem 3.1, it follows from (1.5), $\vert q_{s}\vert >1$, and $\vert q\vert =\max_{1\leq s\leq n}\{\vert q_{s}\vert \}>1$ that

$$\begin{aligned} T\bigl(r,R\bigl(z,f(z)\bigr)\bigr) =& T \biggl(r,\frac{\sum_{i=0}^{k}a_{i}(z)f^{i}(z)}{\sum_{j=0}^{l}b_{j}(z)f^{j}(z)} \biggr) \\ =& d T\bigl(r,f(z)\bigr)+S(r,f) \\ =& T \Biggl(r,\sum_{s=1}^{n} \alpha_{s}(z)f^{(\lambda_{s})}(q_{s}z) \Biggr) \\ \leq& \sum_{s=1}^{n}T\bigl(r,f^{(\lambda_{s})}(q_{s}z) \bigr)+S(r,f) \\ \leq& \sum_{s=1}^{n}\bigl[T \bigl(r,f(q_{s}z)\bigr)+\lambda_{s}\overline {N} \bigl(r,f(q_{s}z)\bigr)+S\bigl(r,f(q_{s}z)\bigr) \bigr]+S(r,f) \\ =& \sum_{s=1}^{n}\bigl[T\bigl(\vert q_{s}\vert r,f(z)\bigr)+\lambda_{s}\overline {N}\bigl( \vert q_{s}\vert r,f(z)\bigr)+S\bigl(\vert q_{s} \vert r,f(z)\bigr)\bigr]+S(r,f) \\ \leq& \sum_{s=1}^{n}\bigl[(1+\lambda _{s})T\bigl(\vert q_{s}\vert r,f(z)\bigr)+S\bigl( \vert q_{s}\vert r,f(z)\bigr)\bigr]+S(r,f) \\ \leq& \sum_{s=1}^{n}(1+ \lambda_{s})T\bigl(\vert q\vert r,f(z)\bigr)+S\bigl(\vert q \vert r,f(z)\bigr)+S(r,f) \\ =& (\lambda+n)T\bigl(\vert q\vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z)\bigr)+S(r,f), \end{aligned}$$

that is,

$$ d T\bigl(r,f(z)\bigr)+S(r,f)\leq(\lambda+n)T\bigl(\vert q\vert r,f(z) \bigr)+S\bigl(\vert q\vert r,f(z)\bigr) $$

(4.1)

for sufficiently large $r\notin E$, where $\operatorname{lm}(E)<\infty$. By using (4.1) and [6], Lemma 5, with $\beta>1$, for any given $\varepsilon>0$, we have

$$ d(1-\varepsilon)T\bigl(r,f(z)\bigr)\leq(\lambda+n) (1+\varepsilon)T\bigl(\beta \vert q\vert r,f(z)\bigr) $$

(4.2)

for all $r\geq r_{0}$. Since $\beta \vert q\vert >1$, if $d\leq\lambda+n$, then estimate (4.2) is trivial. So we only have to consider the case where $d>\lambda+n$. Then $\frac {d(1-\varepsilon)}{(\lambda+n)(1+\varepsilon)}>1$. It follows from Lemma 2.3 and (4.2) that

$$\mu(f)=\liminf_{r\to\infty}\frac{\log^{+}T(r,f)}{\log r}\geq \frac{\log(d(1-\varepsilon))-\log((\lambda+n)(1+\varepsilon))}{\log\beta \vert q\vert }. $$

As $\varepsilon\to0^{+}$ and $\beta\to1^{+}$, we have

$$\rho(f)\geq\mu(f)\geq\frac{\log d-\log(\lambda+n)}{\log \vert q\vert }. $$

Similarly, if $f(z)$ is a transcendental entire solution of (1.5) and $d>n$, we have

$$\rho(f)\geq\mu(f)\geq\frac{\log d-\log n}{\log \vert q\vert }. $$

This completes the proof of Theorem 1.3. □

Proof of Theorem 1.4

If $f(z)$ is a transcendental meromorphic solution of (1.6), similarly to (4.1), by applying the Valiron-Mohon’ko identity ([11], Theorem 2.2.5), Lemma 2.5, and [1], Theorem 3.1, it follows from (1.6) and $\vert q\vert =\max_{1\leq s\leq n}\{\vert q_{s}\vert \}>0$ that

$$ d T\bigl(r,f\bigl(p(z)\bigr)\bigr)+S\bigl(r,f\bigl(p(z)\bigr)\bigr)\leq(\lambda +n)T\bigl(\vert q\vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z) \bigr). $$

(4.3)

By Lemma 2.1, for given $0<\delta<\vert c_{m}\vert$ and $\mu =\vert c_{m}\vert -\delta$ and for any small $\varepsilon>0$,

$$ d(1-\varepsilon)T\bigl(\mu r^{m},f(z)\bigr)\leq(\lambda+n) (1+ \varepsilon )T\bigl(\vert q\vert r,f(z)\bigr) $$

(4.4)

for sufficiently large $r\notin E$, where $\operatorname{lm}(E)<\infty$. An application of [6], Lemma 5, with $\beta>1$ and (4.4) yields

$$ d(1-\varepsilon)T\bigl(\mu r^{m},f(z)\bigr)\leq(\lambda+n) (1+ \varepsilon)T\bigl(\beta \vert q\vert r,f(z)\bigr) $$

(4.5)

for $r\geq r_{0}$. Set $R=\beta \vert q\vert r$. Then (4.5) can be rewritten as

$$ T \biggl(\frac{\mu R^{m}}{\beta^{m}\vert q\vert ^{m}},f(z) \biggr)\leq \frac{(\lambda+n)(1+\varepsilon)}{d(1-\varepsilon)}T\bigl(R,f(z) \bigr). $$

(4.6)

If $d\leq\lambda+n$, then $\frac{(\lambda+n)(1+\varepsilon )}{d(1-\varepsilon)}\geq1$. Since $\frac{\mu}{\beta^{m}\vert q\vert ^{m}}>0$, $m\geq2$, from Lemma 2.4 we get that

$$T\bigl(r,f(z)\bigr)=O\bigl((\log r)^{\alpha}\bigr), $$

where

$$\begin{aligned} \alpha =& \frac{\log((\lambda+n)(1+\varepsilon))-\log(d(1-\varepsilon ))}{\log m} \\ =& \frac{\log(\lambda+n)-\log d}{\log m}+\frac{\log(1+\varepsilon)-\log (1-\varepsilon)}{\log m} \\ \rightarrow& \frac{\log(\lambda+n)-\log d}{\log m}\quad (\varepsilon \rightarrow0). \end{aligned}$$

If $f(z)$ is a transcendental entire solution of (1.6) and $d\leq n$, we similarly have

$$T\bigl(r,f(z)\bigr)=O\bigl((\log r)^{\alpha}\bigr), $$

where

$$\alpha=\frac{\log n-\log d}{\log m}. $$

This completes the proof of Theorem 1.4. □

5 Proof of Theorems 1.5-1.6

Proof of Theorem 1.5

If $f(z)$ is a transcendental meromorphic solution of (1.8), by applying the Valiron-Mohon’ko identity ([11], Theorem 2.2.5), Lemma 2.5, and [1], Theorem 3.1, it follows from (1.8) that

$$\begin{aligned} T\bigl(r,A\bigl(q z,f(q z)\bigr)\bigr) =& a T\bigl(r,f(q z)\bigr)+S\bigl(r,f(q z)\bigr) \\ =& a T\bigl(\vert q\vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z)\bigr) \\ =& T \Biggl(r,B\bigl(z,f(z)\bigr)\sum_{s=1}^{n} \alpha_{s}(z)f^{(\lambda _{s})}(z) \Biggr) \\ \leq& T(r,B\bigl(z,f(z)\bigr)+T \Biggl(r,\sum_{s=1}^{n}f^{(\lambda_{s})}(z) \Biggr)+S(r,f) \\ \leq& b T\bigl(r,f(z)\bigr)+\sum_{s=1}^{n} \bigl[T\bigl(r,f(z)\bigr)+\lambda_{s}\overline {N}\bigl(r,f(z) \bigr)+S(r,f)\bigr]+S(r,f) \\ \leq& b T\bigl(r,f(z)\bigr)+\sum_{s=1}^{n}(1+ \lambda_{s})T\bigl(r,f(z)\bigr)+S(r,f) \\ =& b T\bigl(r,f(z)\bigr)+(\lambda+n)T\bigl(r,f(z)\bigr)+S(r,f) \\ =& (b+\lambda+n)T\bigl(r,f(z)\bigr)+S(r,f), \end{aligned}$$

that is,

$$a T\bigl(\vert q\vert r,f(z)\bigr)+S\bigl(\vert q\vert r,f(z)\bigr) \leq(b+\lambda+n)T\bigl(r,f(z)\bigr)+S(r,f), $$

that is,

$$ a T\bigl(r,f(z)\bigr)+S(r,f)\leq(b+\lambda+n)T \biggl(\frac{r}{\vert q\vert },f(z) \biggr)+S \biggl(\frac{r}{\vert q\vert },f \biggr) $$

(5.1)

for sufficiently large $r\notin E$, where $\operatorname{lm}(E)<\infty$. By using [6], Lemma 5, and $\frac{1}{\vert q\vert }<1$, from (5.1) we obtain that there exists $\beta>1$ such that

$$\frac{\beta}{ \vert q\vert }< 1 \quad\mbox{and} \quad a(1-\varepsilon )T\bigl(r,f(z)\bigr) \leq(b+\lambda+n) (1+\varepsilon)T \biggl(\frac{\beta r}{\vert q\vert },f(z) \biggr) $$

for all $r\geq r_{0}$. Since $1\leq a\leq b$, then $\frac{(b+\lambda+n)(1+\varepsilon )}{a(1-\varepsilon)}\geq1$, and from Lemma 2.2 we get that

$$\rho(f)\leq\frac{\log((b+\lambda+n)(1+\varepsilon))-\log (a(1-\varepsilon))}{-\log\frac{\beta}{ \vert q\vert }}. $$

As $\varepsilon\to0^{+}$ and $\beta\to1^{+}$, we have

$$\rho(f)\leq\frac{\log(b+\lambda+n)-\log a}{\log \vert q\vert }. $$

On the other hand, by applying the Valiron-Mohon’ko identity ([11], Theorem 2.2.5), Lemma 2.5, and [1], Theorem 3.1, it follows from (1.8) that

$$\begin{aligned} T\bigl(r,B\bigl(z,f(z)\bigr)\bigr) =& b T\bigl(r,f(z)\bigr)+S(r,f) \\ =& T \biggl(r,\frac{A(q z,f(q z))}{\sum_{s=1}^{n}\alpha_{s}(z)f^{(\lambda _{s})}(z)} \biggr) \\ \leq& T\bigl(r,A\bigl(q z,f(q z)\bigr)\bigr)+T \Biggl(r,\sum _{s=1}^{n}\alpha _{s}(z)f^{(\lambda_{s})}(z) \Biggr)+O(1) \\ \leq& a T\bigl(r,f(q z)\bigr)+T \Biggl(r,\sum_{s=1}^{n}f^{(\lambda_{s})}(z) \Biggr)+S\bigl(r,f(q z)\bigr)+S(r,f) \\ \leq& a T\bigl(\vert q\vert r,f(z)\bigr)+\sum_{s=1}^{n} \bigl[T\bigl(r,f(z)\bigr)+\lambda_{s}\overline {N}\bigl(r,f(z) \bigr)+S(r,f)\bigr]+S\bigl(\vert q\vert r,f\bigr) \\ \leq& a T\bigl(\vert q\vert r,f(z)\bigr)+\sum_{s=1}^{n}(1+ \lambda _{s})T\bigl(r,f(z)\bigr)+S\bigl(\vert q\vert r,f \bigr)+S(r,f) \\ \leq& a T\bigl(\vert q\vert r,f(z)\bigr)+(\lambda+n)T\bigl(r,f(z)\bigr)+S \bigl(\vert q\vert r,f\bigr)+S(r,f), \end{aligned}$$

that is,

$$b T\bigl(r,f(z)\bigr)+S(r,f)\leq a T\bigl(\vert q\vert r,f(z)\bigr)+( \lambda+n)T\bigl(r,f(z)\bigr)+S\bigl(\vert q\vert r,f\bigr). $$

If $b>a+\lambda+n$, then for any given $\varepsilon>0$, this inequality can be rewritten as

$$ (b-\lambda-n) (1-\varepsilon)T\bigl(r,f(z)\bigr)\leq a(1+\varepsilon )T\bigl( \vert q\vert r,f(z)\bigr) $$

(5.2)

for sufficiently large $r\notin E$, where $\operatorname{lm}(E)<\infty$. By using [6], Lemma 5, with $\beta>1$ from (5.2) we have that

$$ (b-\lambda-n) (1-\varepsilon)T\bigl(r,f(z)\bigr)\leq a(1+\varepsilon)T \bigl( \beta \vert q\vert r,f(z) \bigr) $$

(5.3)

for all $r\geq r_{0}$. Since $\beta \vert q\vert >1$, $\frac{(b-\lambda-n)(1-\varepsilon)}{a(1+\varepsilon)}>1$, and it follows from Lemma 2.3 and (5.3) that

$$\mu(f)=\liminf_{r\to\infty}\frac{\log^{+}T(r,f)}{\log r}\geq \frac{\log((b-\lambda-n)(1-\varepsilon))-\log(a(1+\varepsilon))}{\log \beta \vert q\vert }. $$

As $\varepsilon\to0^{+}$ and $\beta\to1^{+}$, we have

$$\rho(f)\geq\mu(f)\geq\frac{\log(b-\lambda-n)-\log a}{\log \vert q\vert }. $$

Similarly, if $f(z)$ is a transcendental entire solution of (1.8) and $b>a+n$, we have

$$\frac{\log(b-n)-\log a}{\log \vert q\vert }\leq\mu(f)\leq\rho(f)\leq\frac{\log (b+n)-\log a}{\log \vert q\vert }. $$

This completes the proof of Theorem 1.5. □

Proof of Theorem 1.6

Using the same method as in the proof of Theorem 1.5, the conclusion of Theorem 1.6 follows immediately. We omit the proof here. □