Complex oscillation of a second-order linear differential equation with entire coefficients of $[p,q]-\phi$ order

Abstract

In this paper, the authors investigate the interaction between the growth, zeros of solutions with the coefficients of second-order linear differential equations in terms of $[p,q]-\phi$ order and obtain some results in general form.

MSC:30D35, 34A20.

1 Introduction and notations

In this paper, we shall assume that readers are familiar with the standard notations of Nevanlinna value distribution theory (see [1–3]). The theory of complex linear equations has been developed since 1960s. Many authors have investigated the second-order linear differential equation

$$f^{″}+A(z)f=0,$$

(1.1)

where $A(z)$ is an entire function or a meromorphic function of finite order or finite iterated order, and have obtained many results about the interaction between the solutions and the coefficient of (1.1) (see [4–7]). What about the case when $A(z)$ is an entire function of $[p,q]$-order or more general growth? In the following, we will introduce some notations about $[p,q]$-order, where p and q are two positive integers and satisfy $p\ge q\ge 1$ throughout this paper (see [8–11]). Firstly, for $r\in [0,+\mathrm{\infty })$, we define ${exp}_{1}r=e^r$ and ${exp}_{i+1}r=exp({exp}_{i}r)$, $i\in \mathbb{N}$, and for all sufficiently large r, we define ${log}_{1}r=logr$ and ${log}_{i+1}r=log({log}_{i}r)$, $i\in \mathbb{N}$. Especially, we have ${exp}_{0}r=r={log}_{0}r$ and ${exp}_{-1}r={log}_{1}r$. Secondly, we denote the linear measure and the logarithmic measure of a set $E\subset (1,+\mathrm{\infty })$ by $mE={\int }_{E}dt$ and $m_{l}E={\int }_E\frac{dt}{t}$.

Definition 1.1 ([10])

If $f(z)$ is a meromorphic function, the $[p,q]$-order of $f(z)$ is defined by

$${\sigma }_{[p,q]}(f)=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}T(r,f)}{{log}_{q}r}.$$

(1.2)

Especially, if $f(z)$ is an entire function, then the $[p,q]$-order of $f(z)$ is defined by (see [8, 9, 11, 12])

$${\sigma }_{[p,q]}(f)=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}T(r,f)}{{log}_{q}r}=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p+1}M(r,f)}{{log}_{q}r}.$$

(1.3)

Remark 1.1 We use ${\sigma }_{[1,1]}(f)=\sigma (f)$ and ${\sigma }_{[p,1]}(f)={\sigma }_p(f)$ to denote the order and the iterated order of a function $f(z)$.

Definition 1.2 ([10, 13])

The growth index (or the finiteness degree) of the iterated order of a meromorphic function $f(z)$ is defined by

$$i(f)=\{\begin{array}{ll}0& \text{if }f\text{ is rational},\\ min\{n\in \mathbb{N}:{\sigma }_n(f)<\mathrm{\infty }\}& \text{if }f\text{ is transcendental and }{\sigma }_n(f)<\mathrm{\infty }\text{ for some }n\in \mathbb{N},\\ \mathrm{\infty }& \text{if with }{\sigma }_n(f)=\mathrm{\infty }\text{ for all }n\in \mathbb{N}.\end{array}$$

Remark 1.2 By Definition 1.2, we can similarly give the definition of the growth index of the iterated exponent of convergence of the zero-sequence of a meromorphic function $f(z)$ by $i_{\lambda }(f,0)$.

Definition 1.3 ([10, 11])

The $[p,q]$ exponent of convergence of the (distinct) zero-sequence of a meromorphic function $f(z)$ is respectively defined by

$${\lambda }_{[p,q]}(f)=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}n(r,\frac{1}{f})}{{log}_{q}r}=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}N(r,\frac{1}{f})}{{log}_{q}r},$$

(1.4)

$${\overline{\lambda }}_{[p,q]}(f)=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p\overline{n}(r,\frac{1}{f})}{{log}_{q}r}=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p\overline{N}(r,\frac{1}{f})}{{log}_{q}r}.$$

(1.5)

Definition 1.4 ([10])

The $[p,q]$ exponent of convergence of the (distinct) pole-sequence of a meromorphic function $f(z)$ is respectively defined by

$${\lambda }_{[p,q]}\left(\frac{1}{f}\right)=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}n(r,f)}{{log}_{q}r},$$

(1.6)

$${\overline{\lambda }}_{[p,q]}\left(\frac{1}{f}\right)=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p\overline{n}(r,f)}{{log}_{q}r}.$$

(1.7)

Remark 1.3 We use ${\lambda }_{[1,1]}(f)=\lambda (f)$, ${\lambda }_{[p,1]}(f)={\lambda }_p(f)$ and ${\lambda }_{[1,1]}(\frac{1}{f})=\lambda (\frac{1}{f})$, ${\lambda }_{[p,1]}(\frac{1}{f})={\lambda }_p(\frac{1}{f})$ to denote the (iterated) exponent of convergence of the zero-sequence and pole-sequence of a meromorphic function $f(z)$.

Recently, some authors have investigated the exponent of convergence of the zero-sequence and pole-sequence of the solutions of second-order linear differential equations (see [13–15]) and have obtained the following results.

Theorem A ([5])

Let A be a transcendental meromorphic function of order $\sigma (A)$, where $0<\sigma (A)\le \mathrm{\infty }$, and assume that $\overline{\lambda }(A)<\sigma (A)$. Then, if $f\not\equiv 0$ is a meromorphic solution of (1.1), we have

$$\sigma (A)\le max\{\overline{\lambda }(f),\overline{\lambda }\left(\frac{1}{f}\right)\}.$$

Theorem B ([13])

Let $A(z)$ be an entire function with $i(A)=p\in {\mathbb{N}}_{+}$. Let $f_1$, $f_2$ be two linearly independent solutions of (1.1) and denote $F=f_1{f}_2$. Then $i_{\lambda }(F,0)\le p+1$ and

$${\lambda }_{p+1}(F,0)={\sigma }_{p+1}(F)=max\{{\lambda }_{p+1}(f_1,0),{\lambda }_{p+1}(f_2,0)\}\le {\sigma }_p(A).$$

If $i_{\lambda }(F,0)\le p$, then $i_{\lambda }(f,0)=p+1$ holds for all solutions of type $f=c_1{f}_1+c_2{f}_2$, where $c_1{c}_2\ne 0$.

Theorem C ([13])

Let $A(z)$ be an entire function with $0<i(A)=p<\mathrm{\infty }$, let f be any non-trivial solution of (1.1), and assume ${\overline{\lambda }}_p(A,0)<{\sigma }_p(A)\ne 0$. Then ${\lambda }_{p+1}(f,0)\le {\sigma }_p(A)\le {\lambda }_p(f,0)$.

Theorem D ([13])

Let $A(z)$ be an entire function with $i(A)=p$ and ${\sigma }_p(A)=\sigma <\mathrm{\infty }$. Let $f_1$ and $f_2$ be two linearly independent solutions of (1.1) such that $max\{{\lambda }_p(f_1,0),{\lambda }_p(f_2,0)\}<\sigma$. Let $\mathrm{\Pi }(z)\not\equiv 0$ be any entire function for which either $i(\mathrm{\Pi })<p$ or $i(\mathrm{\Pi })=p$ and ${\sigma }_p(\mathrm{\Pi })<\sigma$. Then any two linearly independent solutions $g_1$ and $g_2$ of the differential equation $y^{″}+(A(z)+\mathrm{\Pi }(z))y=0$ satisfy $max\{{\lambda }_p(g_1),{\lambda }_p(g_2)\}\ge \sigma$.

Theorem E ([14])

Let A be a meromorphic function with $i(A)=p\in {\mathbb{N}}_{+}$, and assume that ${\overline{\lambda }}_p(A)<{\sigma }_p(A)$. Then, if f is a nonzero meromorphic solution of (1.1), we have

$${\sigma }_p(A)\le max\{{\overline{\lambda }}_p(f),{\overline{\lambda }}_p\left(\frac{1}{f}\right)\}.$$

In the special case where either $\delta (\mathrm{\infty },f)>0$ or the poles of f are of uniformly bounded multiplicities, we can conclude that

$$max\{{\lambda }_{p+1}(f),{\lambda }_{p+1}\left(\frac{1}{f}\right)\}\le {\sigma }_p(f)\le \{{\overline{\lambda }}_p(f),{\overline{\lambda }}_p\left(\frac{1}{f}\right)\}.$$

In [16], Chyzhykov and his co-authors introduced the definition of φ-order of $f(z)$, where $f(z)$ is a meromorphic function in the unit disc and used it to investigate the interaction between the analytic coefficients and solutions of

$$f^{(k)}+A_{k-1}(z)f^{(k-1)}+\cdots +A_0(z)f=0$$

in the unit disc, where the definition of φ-order of $f(z)$ is given as follows.

Definition 1.5 ([16])

Let $\phi :[0,1)\to (0,+\mathrm{\infty })$ be a non-decreasing unbounded function, the φ-order of a meromorphic function $f(z)$ in the unit disc is defined by

$$\sigma (f,\phi )=\underset{r\to 1^{-}}{\overline{lim}}\frac{{log}^{+}T(r,f)}{log\phi (r)}.$$

(1.8)

On the basis of Definition 1.5, it is natural for us to give the $[p,q]-\phi$ order of a meromorphic function $f(z)$ in the complex plane.

Definition 1.6 Let $\phi :[0,+\mathrm{\infty })\to (0,+\mathrm{\infty })$ be a non-decreasing unbounded function, the $[p,q]-\phi$ order and $[p,q]-\phi$ lower order of a meromorphic function $f(z)$ are respectively defined by

$${\sigma }_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}T(r,f)}{{log}_q\phi (r)},$$

(1.9)

$${\mu }_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_{p}T(r,f)}{{log}_q\phi (r)}.$$

(1.10)

Similar to Definition 1.6, we can also define the $[p,q]-\phi$ exponent of convergence of the (distinct) zero-sequence of a meromorphic function $f(z)$.

Definition 1.7 The $[p,q]-\phi$ exponent of convergence of the (distinct) zero-sequence of a meromorphic function $f(z)$ is respectively defined by

$${\lambda }_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}n(r,\frac{1}{f})}{{log}_q\phi (r)},$$

(1.11)

$${\overline{\lambda }}_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p\overline{n}(r,\frac{1}{f})}{{log}_q\phi (r)}.$$

(1.12)

Proposition 1.1 If $f_1(z)$, $f_2(z)$ are meromorphic functions satisfying ${\sigma }_{[p,q]}(f_1,\phi )=a$, ${\sigma }_{[p,q]}(f_2,\phi )=b$, then

• (i) ${\sigma }_{[p,q]}(f_1+f_2,\phi )\le max\{a,b\}$, ${\sigma }_{[p,q]}(f_1\cdot f_2,\phi )\le max\{a,b\}$;

• (ii) If $a\ne b$, ${\sigma }_{[p,q]}(f_1+f_2,\phi )=max\{a,b\}$, ${\sigma }_{[p,q]}(f_1\cdot f_2,\phi )=max\{a,b\}$.

In this paper, we add two conditions on $\phi (r)$ as follows: $\phi (r):[0,+\mathrm{\infty })\to (0,+\mathrm{\infty })$ is a non-decreasing unbounded function and satisfies (i) ${lim}_{r\to \mathrm{\infty }}\frac{{log}_{p+1}r}{{log}_q\phi (r)}=0$, (ii) ${lim}_{r\to \mathrm{\infty }}\frac{{log}_q\phi (\alpha r)}{{log}_q\phi (r)}=1$ for some $\alpha >1$. Throughout this paper, we assume that $\phi (r)$ always satisfies the above two conditions without special instruction.

Proposition 1.2 Let $\phi (r)$ satisfy the above two conditions (i)-(ii).

• (i) If $f(z)$ is an entire function, then

  $$\begin{array}{r}{\sigma }_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}T(r,f)}{{log}_q\phi (r)}=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p+1}M(r,f)}{{log}_q\phi (r)},\\ {\mu }_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_{p}T(r,f)}{{log}_q\phi (r)}=\underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_{p+1}M(r,f)}{{log}_q\phi (r)}.\end{array}$$

• (ii) If $f(z)$ is a meromorphic function, then

  $$\begin{array}{r}{\lambda }_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}n(r,\frac{1}{f})}{{log}_q\phi (r)}=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}N(r,\frac{1}{f})}{{log}_q\phi (r)},\\ {\overline{\lambda }}_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p\overline{n}(r,\frac{1}{f})}{{log}_q\phi (r)}=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p\overline{N}(r,\frac{1}{f})}{{log}_q\phi (r)}.\end{array}$$

Proof (i) By the inequality $T(r,f)\le {log}^{+}M(r,f)\le \frac{R+r}{R-r}T(R,f)$ ($0<r<R$), set $R=\alpha r$ ($\alpha >1$), we have

$$T(r,f)\le {log}^{+}M(r,f)\le \frac{\alpha +1}{\alpha -1}T(\alpha r,f).$$

(1.13)

By (1.13) and ${lim}_{r\to \mathrm{\infty }}\frac{{log}_q\phi (\alpha r)}{{log}_q\phi (r)}=1$, it is easy to see that conclusion (i) holds.

• (ii) Without loss of generality, assume that $f(0)\ne 0$, then $N(r,\frac{1}{f})={\int }_0^r\frac{n(t,\frac{1}{f})}{t}dt$. Since

  $$N(r,\frac{1}{f})-N(r_0,\frac{1}{f})={\int }_{r_0}^r\frac{n(t,\frac{1}{f})}{t}dt\le n(r,\frac{1}{f})log\frac{r}{r_0}(0<r_0<r),$$

  (1.14)

then by (1.14) and ${lim}_{r\to \mathrm{\infty }}\frac{{log}_{p+1}r}{{log}_q\phi (r)}=0$, we have

$$\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}N(r,\frac{1}{f})}{{log}_q\phi (r)}\le max\{\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}n(r,\frac{1}{f})}{{log}_q\phi (r)},\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p+1}r}{{log}_q\phi (r)}\}=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}n(r,\frac{1}{f})}{{log}_q\phi (r)}.$$

(1.15)

On the other hand, since $\alpha >1$, we have

$$N(\alpha r,\frac{1}{f})={\int }_0^{\alpha r}\frac{n(t,\frac{1}{f})}{t}dt\ge {\int }_r^{\alpha r}\frac{n(t,\frac{1}{f})}{t}dt\ge n(r,\frac{1}{f})log\alpha .$$

(1.16)

By (1.16) and ${lim}_{r\to \mathrm{\infty }}\frac{{log}_q\phi (\alpha r)}{{log}_q\phi (r)}=1$, we have

$$\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}N(r,\frac{1}{f})}{{log}_q\phi (r)}\ge \underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p}n(r,\frac{1}{f})}{{log}_q\phi (r)}.$$

(1.17)

By (1.15) and (1.17), it is easy to see that ${\lambda }_{[p,q]}(f,\phi )={\overline{lim}}_{r\to \mathrm{\infty }}\frac{{log}_{p}n(r,\frac{1}{f})}{{log}_q\phi (r)}={\overline{lim}}_{r\to \mathrm{\infty }}\frac{{log}_{p}N(r,\frac{1}{f})}{{log}_q\phi (r)}$. By the same proof above, we can obtain the conclusion ${\overline{\lambda }}_{[p,q]}(f,\phi )={\overline{lim}}_{r\to \mathrm{\infty }}\frac{{log}_p\overline{n}(r,\frac{1}{f})}{{log}_q\phi (r)}={\overline{lim}}_{r\to \mathrm{\infty }}\frac{{log}_p\overline{N}(r,\frac{1}{f})}{{log}_q\phi (r)}$. □

Remark 1.4 If $\phi (r)=r$, Definitions 1.1 and 1.3 are special cases of Definitions 1.6 and 1.7.

2 Main results

In this paper, our aim is to make use of the concept of $[p,q]-\phi$ order of entire functions to investigate the growth, zeros of the solutions of equation (1.1).

Theorem 2.1 Let $A(z)$ be an entire function satisfying ${\sigma }_{[p,q]}(A,\phi )>0$. Then ${\sigma }_{[p+1,q]}(f,\phi )={\sigma }_{[p,q]}(A,\phi )$ holds for all non-trivial solutions of (1.1).

Theorem 2.2 Let $A(z)$ be an entire function satisfying ${\sigma }_{[p,q]}(A,\phi )>0$, let $f_1$, $f_2$ be two linearly independent solutions of (1.1) and denote $F=f_1{f}_2$. Then $max\{{\lambda }_{[p+1,q]}(f_1,\phi ),{\lambda }_{[p+1,q]}(f_2,\phi )\}={\lambda }_{[p+1,q]}(F,\phi )={\sigma }_{[p+1,q]}(F,\phi )\le {\sigma }_{[p,q]}(A,\phi )$. If ${\sigma }_{[p+1,q]}(F,\phi )<{\sigma }_{[p,q]}(A,\phi )$, then ${\lambda }_{[p+1,q]}(f,\phi )={\sigma }_{[p,q]}(A,\phi )$ holds for all solutions of type $f=c_1{f}_1+c_2{f}_2$, where $c_1{c}_2\ne 0$.

Theorem 2.3 Let $A(z)$ be an entire function satisfying ${\overline{\lambda }}_{[p,q]}(A,\phi )<{\sigma }_{[p,q]}(A,\phi )$. Then ${\lambda }_{[p+1,q]}(f,\phi )\le {\sigma }_{[p,q]}(A,\phi )\le {\lambda }_{[p,q]}(f,\phi )$ holds for all non-trivial solutions of (1.1).

Theorem 2.4 Let $A(z)$ be an entire function satisfying ${\sigma }_{[p,q]}(A,\phi )={\sigma }_1>0$, let $f_1$ and $f_2$ be two linearly independent solutions of (1.1) such that $max\{{\lambda }_{[p,q]}(f_1,\phi ),{\lambda }_{[p,q]}(f_2,\phi )\}<{\sigma }_1$. Let $\mathrm{\Pi }(z)\not\equiv 0$ be any entire function satisfying ${\sigma }_{[p,q]}(\mathrm{\Pi },\phi )<{\sigma }_1$. Then any two linearly independent solutions $g_1$ and $g_2$ of the differential equation $f^{″}+(A(z)+\mathrm{\Pi }(z))f=0$ satisfy $max\{{\lambda }_{[p,q]}(g_1,\phi ),{\lambda }_{[p,q]}(g_2,\phi )\}\ge {\sigma }_1$.

3 Some lemmas

Lemma 3.1 ([17–19])

Let $f(z)$ be a transcendental entire function, and let z be a point with $|z|=r$ at which $|f(z)|=M(r,f)$. Then, for all $|z|$ outside a set $E_1$ of r of finite logarithmic measure, we have

$$\frac{f^{(j)}(z)}{f(z)}={\left(\frac{v_f(r)}{z}\right)}^j(1+o(1))(j\in \mathbb{N}),$$

(3.1)

where $v_f(r)$ is the central index of $f(z)$.

Lemma 3.2 ([7, 19, 20])

Let $g:[0,+\mathrm{\infty })⟶\mathbb{R}$ and $h:[0,+\mathrm{\infty })⟶\mathbb{R}$ be monotone non-decreasing functions such that $g(r)\le h(r)$ outside of an exceptional set $E_2$ of finite linear measure or finite logarithmic measure. Then, for any $d>1$, there exists $r_0>0$ such that $g(r)\le h(dr)$ for all $r>r_0$.

Lemma 3.3 ([18, 21])

Let $f(z)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{z}^n$ be an entire function, $\mu (r)$ be the maximum term, i.e., $\mu (r)=max\{|a_n|r^n;n=0,1,\dots \}$, and let $v_f(r)$ be the central index of f.

• (i) If $|a_0|\ne 0$, then

  $$log\mu (r)=log|a_0|+{\int }_0^r\frac{v_f(t)}{t}dt.$$

  (3.2)

• (ii) For $r<R$, we have

  $$M(r,f)<\mu (r)\{v_f(R)+\frac{R}{R-r}\}.$$

  (3.3)

Lemma 3.4 Let $f(z)$ be an entire function satisfying ${\sigma }_{[p,q]}(f,\phi )={\sigma }_2$ and ${\mu }_{[p,q]}(f,\phi )={\mu }_1$, and let $v_f(r)$ be the central index of f, then

$$\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p{v}_f(r)}{{log}_q\phi (r)}={\sigma }_2,\underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_p{v}_f(r)}{{log}_q\phi (r)}={\mu }_1.$$

Proof Let $f(z)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{z}^n$. Without loss of generality, we can assume that $|a_0|\ne 0$. From (3.2), for any $1<{\alpha }_1<\alpha$, we have

$$log\mu ({\alpha }_{1}r)=log|a_0|+{\int }_0^{{\alpha }_{1}r}\frac{v_f(t)}{t}dt\ge log|a_0|+{\int }_r^{{\alpha }_{1}r}\frac{v_f(t)}{t}dt\ge log|a_0|+v_f(r)log{\alpha }_1.$$

By the Cauchy inequality, it is easy to see $\mu ({\alpha }_{1}r)\le M({\alpha }_{1}r,f)$, hence

$$v_f(r)log{\alpha }_1\le logM({\alpha }_{1}r,f)+c_3,$$

(3.4)

where $c_3>0$ is a constant. By Proposition 1.2, (3.4) and ${lim}_{r\to \mathrm{\infty }}\frac{{log}_q\phi ({\alpha }_{1}r)}{{log}_q\phi (r)}=1$ ($1<{\alpha }_1<\alpha$), we have

$$\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p{v}_f(r)}{{log}_q\phi (r)}\le \underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p+1}M({\alpha }_{1}r,f)}{{log}_q\phi ({\alpha }_{1}r)}\cdot \underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_q\phi ({\alpha }_{1}r)}{{log}_q\phi (r)}={\sigma }_{[p,q]}(f,\phi ),$$

(3.5)

$$\underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_p{v}_f(r)}{{log}_q\phi (r)}\le \underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_{p+1}M({\alpha }_{1}r,f)}{{log}_q\phi ({\alpha }_{1}r)}\cdot \underset{r\to \mathrm{\infty }}{lim}\frac{{log}_q\phi ({\alpha }_{1}r)}{{log}_q\phi (r)}={\mu }_{[p,q]}(f,\phi ).$$

(3.6)

On the other hand, set $R={\alpha }_{1}r$, by (3.3), we have

$$M(r,f)<\mu (r)(v_f({\alpha }_{1}r)+\frac{{\alpha }_1}{{\alpha }_1-1})=|a_{v_f({\alpha }_{1}r)}|r^{v_f({\alpha }_{1}r)}(v_f({\alpha }_{1}r)+\frac{{\alpha }_1}{{\alpha }_1-1}).$$

(3.7)

Since ${\{|a_n|\}}_{n=1}^{\mathrm{\infty }}$ is a bounded sequence, by (3.7), we have

$${log}_{p+1}M(r,f)\le {log}_p{v}_f({\alpha }_{1}r)[1+\frac{{log}_{p+1}{v}_f({\alpha }_{1}r)}{{log}_p{v}_f({\alpha }_{1}r)}]+{log}_{p+1}r+c_4,$$

(3.8)

where $c_4>0$ is a constant. By Proposition 1.2, (3.8), ${lim}_{r\to \mathrm{\infty }}\frac{{log}_q\phi ({\alpha }_{1}r)}{{log}_q\phi (r)}=1$ ($1<{\alpha }_1<\alpha$) and ${lim}_{r\to \mathrm{\infty }}\frac{{log}_{p+1}r}{{log}_q\phi (r)}=0$, we have

$${\sigma }_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p+1}M(r,f)}{{log}_q\phi (r)}\le \underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p{v}_f({\alpha }_{1}r)}{{log}_q\phi ({\alpha }_{1}r)}=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_p{v}_f(r)}{{log}_q\phi (r)},$$

(3.9)

$${\mu }_{[p,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_{p+1}M(r,f)}{{log}_q\phi (r)}\le \underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_p{v}_f({\alpha }_{1}r)}{{log}_q\phi ({\alpha }_{1}r)}=\underset{r\to \mathrm{\infty }}{\underline{lim}}\frac{{log}_p{v}_f(r)}{{log}_q\phi (r)}.$$

(3.10)

By (3.5), (3.6), (3.9) and (3.10), we obtain the conclusion of Lemma 3.4. □

Lemma 3.5 Let $f_1(z)$ and $f_2(z)$ be entire functions of $[p,q]-\phi$ order and denote $F=f_1{f}_2$. Then

$${\lambda }_{[p,q]}(F,\phi )=max\{{\lambda }_{[p,q]}(f_1,\phi ),{\lambda }_{[p,q]}(f_2,\phi )\}.$$

Proof Let $n(r,F)$, $n(r,f_1)$ and $n(r,f_2)$ be unintegrated counting functions for the number of zeros of $F(z)$, $f_1(z)$ and $f_2(z)$. For any $r>0$, it is easy to see

$$n(r,F)\ge max\{n(r,f_1),n(r,f_2)\}.$$

(3.11)

By Definition 1.7 and (3.11), we have

$${\lambda }_{[p,q]}(F,\phi )\ge max\{{\lambda }_{[p,q]}(f_1,\phi ),{\lambda }_{[p,q]}(f_2,\phi )\}.$$

(3.12)

On the other hand, since the zeros of $F(z)$ must be the zeros of $f_1(z)$ or the zeros of $f_2(z)$, for any $r>0$, we have

$$n(r,F)\le n(r,f_1)+n(r,f_2)\le 2max\{n(r,f_1),n(r,f_2)\}.$$

(3.13)

By Definition 1.7 and (3.13), we have

$${\lambda }_{[p,q]}(F,\phi )\le max\{{\lambda }_{[p,q]}(f_1,\phi ),{\lambda }_{[p,q]}(f_2,\phi )\}.$$

(3.14)

Therefore, by (3.12) and (3.14), we have ${\lambda }_{[p,q]}(F,\phi )=\{{\lambda }_{[p,q]}(f_1,\phi ),{\lambda }_{[p,q]}(f_2,\phi )\}$. □

Lemma 3.6 Let $f(z)$ be a transcendental meromorphic function satisfying ${\sigma }_{[p,q]}(f,\phi )={\sigma }_3$, where $\phi (r)$ only satisfies $\frac{{log}_{p+1}r}{{log}_q\phi (r)}=0$, and let k be any positive integer. Then, for any $\epsilon >0$, there exists a set $E_3$ having finite linear measure such that for all $r\notin E_3$, we have

$$m(r,\frac{f^{(k)}}{f})=O\left\{{exp}_{p-1}\{({\sigma }_3+\epsilon ){log}_q\phi (r)\}\right\}.$$

Proof Set $k=1$, since ${\sigma }_{[p,q]}(f,\phi )={\sigma }_3<\mathrm{\infty }$, for sufficiently large r and for any given $\epsilon >0$, we have

$$T(r,f)<{exp}_p\{({\sigma }_3+\epsilon ){log}_q\phi (r)\}.$$

(3.15)

By the lemma of logarithmic derivative, we have

$$m(r,\frac{f^{\prime }}{f})=O\{logT(r,f)+logr\}(r\notin E_3),$$

(3.16)

where $E_3\subset [0,+\mathrm{\infty })$ is a set of finite linear measure, not necessarily the same at each occurrence. By (3.15), (3.16) and $\frac{{log}_{p+1}r}{{log}_q\phi (r)}=0$, we have $m(r,\frac{f^{\prime }}{f})=O\{{exp}_{p-1}\{(\sigma +\epsilon ){log}_q\phi (r)\}\}$ ($r\notin E_3$).

We assume that $m(r,\frac{f^{(k)}}{f})=O\{{exp}_{p-1}\{({\sigma }_3+\epsilon ){log}_q\phi (r)\}\}$ ($r\notin E_3$) holds for any positive integer k. By $N(r,f^{(k)})\le (k+1)N(r,f)$, for all $r\notin E_3$, we have

$$\begin{array}{rl}T(r,f^{(k)})& =m(r,f^{(k)})+N(r,f^{(k)})\le m(r,\frac{f^{(k)}}{f})+m(r,f)+(k+1)N(r,f)\\ \le (k+1)T(r,f)+O\left\{{exp}_{p-1}\{({\sigma }_3+\epsilon ){log}_q\phi (r)\}\right\}.\end{array}$$

(3.17)

By (3.16) and (3.17), for $r\notin E_3$, we have

$$m(r,\frac{f^{(k+1)}}{f})\le m(r,\frac{f^{(k+1)}}{f^{(k)}})+m(r,\frac{f^{(k)}}{f})=O\left\{{exp}_{p-1}\{({\sigma }_3+\epsilon ){log}_q\phi (r)\}\right\}.$$

 □

Lemma 3.7 ([19])

Let $f(z)$ be an entire function of $[p,q]$-order, and $f(z)$ can be represented by the form

$$f(z)=U(z)e^{V(z)},$$

where $U(z)$ and $V(z)$ are entire functions such that

$${\lambda }_{[p,q]}(f)={\lambda }_{[p,q]}(U)={\sigma }_{[p,q]}(U),{\sigma }_{[p,q]}(f)=max\{{\sigma }_{[p,q]}(U),{\sigma }_{[p,q]}\left(e^V\right)\}.$$

If $f(z)$ is an entire function of $[p,q]-\phi$ order, we have a similar result as follows.

Lemma 3.8 Let $f(z)$ be an entire function of $[p,q]-\phi$ order, and $f(z)$ can be represented by the form

$$f(z)=U(z)e^{V(z)},$$

where $U(z)$ and $V(z)$ are entire functions of $[p,q]-\phi$ order such that

$$\begin{array}{c}{\lambda }_{[p,q]}(f,\phi )={\lambda }_{[p,q]}(U,\phi )={\sigma }_{[p,q]}(U,\phi ),\hfill \\ {\sigma }_{[p,q]}(f,\phi )=max\{{\sigma }_{[p,q]}(U,\phi ),{\sigma }_{[p,q]}(e^V,\phi )\}.\hfill \end{array}$$

4 Proofs of Theorems 2.1-2.4

Proof of Theorem 2.1 Set ${\sigma }_{[p,q]}(A,\phi )={\sigma }_4>0$. First, we prove that every solution of (1.1) satisfies ${\sigma }_{[p+1,q]}(f,\phi )\le {\sigma }_4$. If $f(z)$ is a polynomial solution of (1.1), it is easy to know that ${\sigma }_{[p+1,q]}(f,\phi )=0\le {\sigma }_4$ holds. If $f(z)$ is a transcendental solution of (1.1), by (1.1) and Lemma 3.1, there exists a set $E_1\subset (1,+\mathrm{\infty })$ having finite logarithmic measure such that for all z satisfying $|z|=r\notin [0,1]\cup E_1$ and $|f(z)|=M(r,f)$, we have

$${\left(\frac{v_f(r)}{r}\right)}^2(1+o(1))\le {exp}_{p+1}\{({\sigma }_4+\frac{\epsilon }{2}){log}_q\phi (r)\}.$$

And hence, we have

$$v_f(r)\le r{exp}_{p+1}\{({\sigma }_4+\epsilon ){log}_q\phi (r)\}(r\notin E_1).$$

(4.1)

By (4.1) and Lemma 3.2, there exists some ${\alpha }_1$ ($1<{\alpha }_1<\alpha$) such that for all $r\ge r_0$, we have

$$v_f(r)\le {\alpha }_{1}r{exp}_{p+1}\{({\sigma }_4+\epsilon ){log}_q\phi ({\alpha }_{1}r)\}.$$

(4.2)

By Lemma 3.4, (4.2) and the two conditions on $\phi (r)$, we have

$${\sigma }_{[p+1,q]}(f,\phi )=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{{log}_{p+1}{v}_f(r)}{{log}_q\phi (r)}\le {\sigma }_4.$$

(4.3)

On the other hand, by (1.1), we have

$$m(r,A)=m(r,-\frac{f^{″}}{f})=O\{logrT(r,f)\}.$$

(4.4)

By (4.4), we have ${\sigma }_{[p,q]}(A,\phi )\le {\sigma }_{[p+1,q]}(f,\phi )$. Therefore, we have that ${\sigma }_{[p+1,q]}(f,\phi )={\sigma }_{[p,q]}(A,\phi )$ holds for all non-trivial solutions of (1.1). □

Proof of Theorem 2.2 Set ${\sigma }_{[p,q]}(A,\phi )={\sigma }_5>0$, by Theorem 2.1, we have ${\sigma }_{[p+1,q]}(f_1,\phi )={\sigma }_{[p+1,q]}(f_2,\phi )={\sigma }_{[p,q]}(A,\phi )={\sigma }_5$. Hence, we have

$${\lambda }_{[p+1,q]}(F,\phi )\le {\sigma }_{[p+1,q]}(F,\phi )\le max\{{\sigma }_{[p+1,q]}(f_1,\phi ),{\sigma }_{[p+1,q]}(f_2,\phi )\}={\sigma }_{[p,q]}(A,\phi ).$$

(4.5)

By Lemma 3.5 and (4.5), we have

$$max\{{\lambda }_{[p+1,q]}(f_1,\phi ),{\lambda }_{[p+1,q]}(f_2,\phi )\}={\lambda }_{[p+1,q]}(F,\phi )\le {\sigma }_{[p+1,q]}(F,\phi )\le {\sigma }_{[p,q]}(A,\phi ).$$

(4.6)

It remains to show that ${\lambda }_{[p+1,q]}(F,\phi )={\sigma }_{[p+1,q]}(F,\phi )$. By (1.1), we have (see [[13], pp.76-77]) that all zeros of $F(z)$ are simple and that

$$F^2=C^2{({\left(\frac{F^{\prime }}{F}\right)}^2-2\left(\frac{F^{″}}{F}\right)-4A)}^{-1},$$

(4.7)

where $C\ne 0$ is a constant. Hence,

$$\begin{array}{rl}2T(r,F)& =T(r,{\left(\frac{F^{\prime }}{F}\right)}^2-2\left(\frac{F^{″}}{F}\right)-4A)+O(1)\\ \le O(\overline{N}(r,\frac{1}{F})+m(r,\frac{F^{\prime }}{F})+m(r,\frac{F^{″}}{F})+m(r,A)).\end{array}$$

(4.8)

By Lemma 3.6, for all $r\notin E_3$, we have $m(r,A)=m(r,\frac{f^{″}}{f})=O\{{exp}_p\{({\sigma }_5+\epsilon ){log}_q\phi (r)\}\}$, $m(r,\frac{F^{\prime }}{F})=O\{{exp}_p\{({\sigma }_5+\epsilon ){log}_q\phi (r)\}\}$ and $m(r,\frac{F^{″}}{F})=O\{{exp}_p\{({\sigma }_5+\epsilon ){log}_q\phi (r)\}\}$. By (4.8), for all $r\notin E_3$, we have

$$T(r,F)=O\{\overline{N}(r,\frac{1}{F})+{exp}_p\{({\sigma }_5+\epsilon ){log}_q\phi (r)\}\}.$$

(4.9)

Let us assume ${\lambda }_{[p+1,q]}(F,\phi )<\beta <{\sigma }_{[p+1,q]}(F,\phi )$. Since all zeros of $F(z)$ are simple, we have

$$\overline{N}(r,\frac{1}{F})=N(r,\frac{1}{F})=O\left\{{exp}_{p+1}\{\beta {log}_q\phi (r)\}\right\}.$$

(4.10)

By (4.9) and (4.10), for all $r\notin E_3$, we have

$$T(r,F)=O\left\{{exp}_{p+1}\{\beta {log}_q\phi (r)\}\right\}.$$

By Definition 1.6 and Lemma 3.2, we have ${\sigma }_{[p+1,q]}(F,\phi )\le \beta <{\sigma }_{[p+1,q]}(F,\phi )$, this is a contradiction. Therefore, the first assertion is proved.

If ${\sigma }_{[p+1,q]}(F,\phi )<{\sigma }_{[p,q]}(A,\phi )$, let us assume that ${\lambda }_{[p+1,q]}(f,\phi )<{\sigma }_{[p,q]}(A,\phi )$ holds for any solution of type $f=c_1{f}_1+c_2{f}_2$ ($c_1{c}_2\ne 0$). We denote $F=f_1{f}_2$ and $F_1=f{f}_1$, then we have ${\lambda }_{[p+1,q]}(F,\phi )<{\sigma }_{[p,q]}(A,\phi )$ and ${\lambda }_{[p+1,q]}(F_1,\phi )<{\sigma }_{[p,q]}(A,\phi )$. Since (4.9) holds for $F(z)$ and $F_1(z)$ and $F_1=f{f}_1=(c_1{f}_1+c_2{f}_2)f_1=c_1{f}_1^2+c_{2}F$, we have

$$\begin{array}{rl}T(r,f_1)& =O(T(r,F_1)+T(r,F))\\ =O\{\overline{N}(r,\frac{1}{F_1})+\overline{N}(r,\frac{1}{F})+{exp}_p\{({\sigma }_5+\epsilon ){log}_q\phi (r)\}\}.\end{array}$$

(4.11)

By ${\lambda }_{[p+1,q]}(F,\phi )<{\sigma }_{[p,q]}(A,\phi )$, ${\lambda }_{[p+1,q]}(F_1,\phi )<{\sigma }_{[p,q]}(A,\phi )$ and (4.10), for some $\beta <{\sigma }_{[p,q]}(A,\phi )$, we have

$$T(r,f_1)=O\left\{{exp}_{p+1}\{\beta {log}_q\phi (r)\}\right\}.$$

(4.12)

By Definition 1.6 and (4.12), we have ${\sigma }_{[p+1,q]}(f_1,\phi )\le \beta <{\sigma }_{[p,q]}(A,\phi )$, this is a contradiction with Theorem 2.1. Therefore, we have that ${\lambda }_{[p+1,q]}(f,\phi )={\sigma }_{[p,q]}(A,\phi )$ holds for all solutions of type $f=c_1{f}_1+c_2{f}_2$, where $c_1{c}_2\ne 0$. □

Proof of Theorem 2.3 By Theorem 2.1 and ${\lambda }_{[p+1,q]}(f,\phi )\le {\sigma }_{[p+1,q]}(f,\phi )$, it is easy to know that ${\lambda }_{[p+1,q]}(f,\phi )\le {\sigma }_{[p,q]}(A,\phi )$ holds. It remains to show that ${\sigma }_{[p,q]}(A,\phi )\le {\lambda }_{[p,q]}(f,\phi )$. Let us assume ${\sigma }_{[p,q]}(A,\phi )>{\lambda }_{[p,q]}(f,\phi )$. By (1.1) and a similar proof of Theorem 5.6 in [[13], p.82], we have

$$T(r,\frac{f}{f^{\prime }})=O\{\overline{N}(r,\frac{1}{f})+\overline{N}(r,\frac{1}{A})\}(r\notin E_3).$$

(4.13)

By (4.13), the assumption ${\sigma }_{[p,q]}(A,\phi )>{\lambda }_{[p,q]}(f,\phi )$ and ${\overline{\lambda }}_{[p,q]}(A,\phi )\le {\sigma }_{[p,q]}(A,\phi )$, for some $\beta <{\sigma }_{[p,q]}(A,\phi )$, we have

$$T(r,\frac{f}{f^{\prime }})=O\left\{{exp}_p\{\beta {log}_q\phi (r)\}\right\}.$$

(4.14)

By Definition 1.6 and (4.14), we have ${\sigma }_{[p,q]}(\frac{f}{f^{\prime }},\phi )={\sigma }_{[p,q]}(\frac{f^{\prime }}{f},\phi )\le \beta <{\sigma }_{[p,q]}(A,\phi )$. By

$$-A(z)={\left(\frac{f^{\prime }}{f}\right)}^{\prime }+{\left(\frac{f^{\prime }}{f}\right)}^2,$$

we have ${\sigma }_{[p,q]}(A,\phi )\le {\sigma }_{[p,q]}(\frac{f^{\prime }}{f},\phi )<{\sigma }_{[p,q]}(A,\phi )$, this is a contradiction. Therefore, we have that ${\lambda }_{[p+1,q]}(f,\phi )\le {\sigma }_{[p,q]}(A,\phi )\le {\lambda }_{[p,q]}(f,\phi )$ holds for all non-trivial solutions of (1.1). □

Proof of Theorem 2.4 As a similar proof of Theorem 3.1 in [6], we denote $F=f_1{f}_2$ and $F_2=g_1{g}_2$. Let us assume

$${\lambda }_{[p,q]}(F_2,\phi )=max\{{\lambda }_{[p,q]}(g_1,\phi ),{\lambda }_{[p,q]}(g_2,\phi )\}<{\sigma }_1.$$

By Theorem 2.1, we have ${\sigma }_{[p+1,q]}(F,\phi )\le max\{{\sigma }_{[p+1,q]}(f_1,\phi ),{\sigma }_{[p+1,q]}(f_2,\phi )\}={\sigma }_1$, and hence, by Lemma 3.6, for any integer $k\ge 1$ and for any $\epsilon >0$, we have

$$m(r,\frac{F^{(k)}}{F})=O\left\{{exp}_p\{({\sigma }_1+\epsilon ){log}_q\phi (r)\}\right\}(r\notin E_3).$$

Furthermore, by Theorem 2.1, we have ${\lambda }_{[p,q]}(F,\phi )=max\{{\lambda }_{[p,q]}(f_1,\phi ),{\lambda }_{[p,q]}(f_2,\phi )\}<{\sigma }_1$, and hence we have $\overline{N}(r,\frac{1}{F})=O\{{exp}_p\{\beta {log}_q\phi (r)\}\}$ for some $\beta <{\sigma }_1$. And the $[p,q]-\phi$ order of the function $A(z)$ implies that

$$T(r,A)=O\left\{{exp}_p\{({\sigma }_1+\epsilon ){log}_q\phi (r)\}\right\}(r\to \mathrm{\infty }).$$

By (4.9), we obtain

$$T(r,F)=O\{\overline{N}(r,\frac{1}{F})+{exp}_p\{({\sigma }_1+\epsilon ){log}_q\phi (r)\}\}=O\left\{{exp}_p\{(\beta {log}_q\phi (r)\}\right\}.$$

(4.15)

By Definition 1.6 and (4.15), we have ${\sigma }_{[p,q]}(F,\phi )\le {\sigma }_1$. On the other hand, by

$$4A={\left(\frac{F^{\prime }}{F}\right)}^2-2\frac{F^{″}}{F}-\frac{1}{F^2},$$

(4.16)

we have ${\sigma }_{[p,q]}(A,\phi )={\sigma }_1\le {\sigma }_{[p,q]}(F,\phi )$, hence ${\sigma }_{[p,q]}(F,\phi )={\sigma }_1$. The same reasoning is valid for the function $F_2$, we have

$$4(A+\mathrm{\Pi })={\left(\frac{F_2^{\prime }}{F_2}\right)}^2-2\frac{F_2^{″}}{F_2}-\frac{1}{F_2^2},$$

(4.17)

and ${\sigma }_{[p,q]}(F_2,\phi )={\sigma }_1$. Since ${\lambda }_{[p,q]}(F,\phi )<{\sigma }_1$ and ${\lambda }_{[p,q]}(F_2,\phi )<{\sigma }_1$, by Lemma 3.8, we may write

$$F=Q{e}^P,F_2=R{e}^S,$$

(4.18)

where P, Q, R, S are entire functions satisfying ${\sigma }_{[p,q]}(Q,\phi )={\lambda }_{[p,q]}(F,\phi )<{\sigma }_1$, ${\sigma }_{[p,q]}(R,\phi )={\lambda }_{[p,q]}(F_2,\phi )<{\sigma }_1$ and ${\sigma }_{[p,q]}(e^P,\phi )={\sigma }_{[p,q]}(e^S,\phi )={\sigma }_1$. Substituting (4.18) into (4.16) and (4.17), we have

$$4A=-\frac{1}{Q^2{e}^{2P}}+G_1(z),$$

(4.19)

$$4(A+\pi )=-\frac{1}{R^2{e}^{2S}}+G_2(z),$$

(4.20)

where $G_1(z)$ and $G_2(z)$ are meromorphic functions satisfying ${\sigma }_{[p,q]}(G_j,\phi )<{\sigma }_1$ ($j=1,2$). Equation (4.19) subtracting (4.20), we have

$$\frac{1}{R^2{e}^{2S}}-\frac{1}{Q^2{e}^{2P}}=G_3(z),$$

(4.21)

where $G_3(z)$ is a meromorphic function satisfying ${\sigma }_{[p,q]}(G_3,\phi )<{\sigma }_1$. From (4.21), we have

$$e^{-2S}+H_1{e}^{-2P}=H_2,$$

(4.22)

where $H_1(z)$ and $H_2(z)$ are meromorphic functions satisfying ${\sigma }_{[p,q]}(H_j,\phi )<{\sigma }_1$ ($j=1,2$), and $H_1=-\frac{R^2}{Q^2}$. Deriving (4.22), we have

$$-2S^{\prime }{e}^{-2S}+(H_1^{\prime }-2P^{\prime }{H}_1)e^{-2P}=H_3,$$

(4.23)

where $H_3(z)$ is a meromorphic function satisfying ${\sigma }_{[p,q]}(H_3,\phi )<{\sigma }_1$. Eliminating $e^{-2S}$ by (4.22) and (4.23), we have

$$(H_1^{\prime }-2(P^{\prime }-S^{\prime })H_1)e^{-2P}=H_4,$$

(4.24)

where $H_4(z)$ is a meromorphic function satisfying ${\sigma }_{[p,q]}(H_4,\phi )<{\sigma }_1$. Since ${\sigma }_{[p,q]}(e^P,\phi )={\sigma }_1$, therefore by (4.24), we have $H_1^{\prime }-2(P^{\prime }-S^{\prime })H_1\equiv 0$, thus we have $H_1=c{e}^{2(P-S)}$, $c\ne 0$. Hence

$$\frac{F^2}{F_2^2}=\frac{Q^2}{R^2}{e}^{2(P-S)}=-\frac{1}{c}.$$

(4.25)

From (4.16), (4.17) and (4.25), we have

$$4(A+\mathrm{\Pi }+\frac{1}{c}A)={\left(\frac{F_2^{\prime }}{F_2}\right)}^2-2\frac{F_2^{″}}{F_2}+\frac{1}{c}{\left(\frac{F^{\prime }}{F}\right)}^2-\frac{2}{c}\frac{F^{″}}{F}.$$

By Lemma 3.6, we obtain

$$\begin{array}{rl}T(r,(1+\frac{1}{c})A+\mathrm{\Pi })& =m(r,(1+\frac{1}{c})A+\mathrm{\Pi })\\ =O\left\{{exp}_{p-1}\{({\sigma }_1+\epsilon ){log}_q\phi (r)\}\right\}(r\to \mathrm{\infty }).\end{array}$$

This implies

$${\sigma }_{[p,q]}((1+\frac{1}{c})A+\mathrm{\Pi },\phi )=0.$$

Hence, by Proposition 1.1, we have $c=-1$. Since $F^2=F_2^2$, we have

$$\frac{F^{\prime }}{F}=\frac{F_2^{\prime }}{F_2},\frac{F^{″}}{F}=\frac{F_2^{″}}{F_2}.$$

From (4.13) and (4.17), we have $\mathrm{\Pi }\equiv 0$, this is a contradiction. Therefore, we obtain the conclusion of Theorem 2.4. □