Theory and Modern Applications

# 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 . 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 . Malmquist  investigated the existence of transcendental meromorphic solutions of a complex differential equation and obtained the following result.

### Theorem A



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.  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

() 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.

1. (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

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

$$\rho(f)\leq\frac{\log n-\log d}{-\log \vert q\vert }.$$
3. (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  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



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



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  investigated the growth of solutions of (1.7) and obtained the following theorem.

### Theorem E



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,

1. (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

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

1. (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

$$\frac{\log(b-2n)-\log a}{\log \vert q\vert }\leq\mu(f)\leq\rho(f)\leq\frac{\log (b+2n)-\log a}{\log \vert q\vert }.$$
2. (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 , 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 , 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 , 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 , 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. , 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 , Theorem 2.2.5), Lemma 2.5, and , 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 , 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 , 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 (, Theorem 2.2.5), Lemma 2.5, and , 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 , 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 (, Theorem 2.2.5), Lemma 2.5, and , 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 , 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 (, Theorem 2.2.5), Lemma 2.5, and , 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 , 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 (, Theorem 2.2.5), Lemma 2.5, and , 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 , 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 (, Theorem 2.2.5), Lemma 2.5, and , 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 , 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. □

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## Acknowledgements

The authors thank the referee for his/her valuable suggestions to improve the present article. The first author was supported by the NNSF of China (11371225). The second author was supported by PhD research startup foundation of Jiangxi Science and Technology Normal University, the Natural Science Foundation of Jiangxi (No. 20132BAB201008), the Science and Technology Project Founded by the Education Department of Jiangxi Province (No. GJJ150817), and the NNSF of China (11661044).

## Author information

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### Corresponding author

Correspondence to Yeyang Jiang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

M-FC, Y-YJ, and Z-SG completed the main part of this article, M-FC, Z-SG corrected the main theorems. All authors read and approved the final manuscript.

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