On the Growth of Solutions of Higher Order Linear Homogeneous and Non-homogeneous Differential Equations

Abstract

In this paper, we discuss the growth of meromorphic solutions of some classes of homogeneous and nonhomogeneous differential equations. We will prove that if meromorphic functions $F, A_j, D_j$ and polynomials $P_j$ with degree $n\ge 1$ $(j=0,1,\cdots ,k-1)$ satisfy some conditions, then the equation $f^{(k)}+(A_{k-1}e^{P_{k-1}}+D_{k-1})f^{(k-1)}+\cdots +(A_1e^{P_1}+D_1)f'+(A_0e^{P_0}+D_0)f=F(z) (k\ge 2)$ when $F\equiv 0$, all solutions $f\not \equiv 0$ have infinite order and hyper order $\sigma _2(f)\ge n$. This is a continue work of [Gan and Sun in Adv. Math. 36(1), 51–60 (2007)] and [Chen and Xu in Electron. J. Qual. Theory Differ. Equ. 1: 1–13 (2009)]

1 Introduction and Main Results

In this paper, we assume that the readers are familiar with the standard notations of the Nevanlinna theory of meromorphic functions, such as T(r, f), N(r, f), m(r, f)(for the detail, see [10, 12, 13]). The order and hyper order of f are defined as

$$\begin{aligned} \sigma (f)=\limsup \limits _{r\rightarrow \infty }\frac{\log T(r,f)}{\log r} \end{aligned}$$

and

$$\begin{aligned} \sigma _2(f)=\limsup \limits _{r\rightarrow \infty }\frac{\log \log T(r,f)}{\log r}. \end{aligned}$$

For a polynomial $P(z)=(\alpha +i\beta ) z^n+\cdots$ with degree n, $z=re^{i\theta }$, define

$$\begin{aligned} \delta (P,\theta )=\alpha \cos n\theta -\beta \sin n\theta . \end{aligned}$$

We can see that the sets $E^+=\{\theta \in [0,2\pi ): \delta (P,\theta )>0\}$ and $E^-=\{\theta \in [0,2\pi ): \delta (P,\theta )<0\}$ are both of nonzero measure ([8]).

For the differential equations

$$\begin{aligned} f^{(k)}+A_{k-1}e^{P_{k-1}}f^{(k-1)}+\cdots +A_1e^{P_1}f'+A_0e^{P_0}f=F(z) \end{aligned}$$

and

$$\begin{aligned} f^{(k)}+(A_{k-1}e^{P_{k-1}}+D_{k-1})f^{(k-1)}+\cdots +(A_1e^{P_1}+D_1)f'+(A_0e^{P_0}+D_0)f=F(z) \end{aligned}$$

with entire coefficients $A_j, D_j$, Gan and Sun [7] investigated the growth of the solutions. A natural question is that the growth of the solutions of the differential equations with meromorphic coefficients $A_j, D_j$. We discuss this problem in this paper and obtain three theorems as follows.

Theorem 1.1

Let $P_j(z)=a_jz^n+b_{j,1}z^{n-1}+\cdots +b_{j,n-1}z+b_{j,n}(j=0,1,\cdots ,k-1)$ be k polynomials with degree $n\ge 1$, $a_j\ne 0$, $A_j(z)$ be meromorphic functions with order $\sigma (A_j)<n(j=0,1,\cdots ,k-1)$, F(z) be a meromorphic function with finite order. If $a_j=\alpha _ja_1, \alpha _j>0, {\text {arg}}a_1\ne {\text {arg}}a_0(j=2,\cdots ,k-1)$ or $a_j=c_ja_0, 0<c_j<1, c_j<0 (j=1,\cdots ,k-1)$ and suppose that all the poles of f are uniformly bounded multiplicity. Consider the differential equation

$$\begin{aligned} f^{(k)}+A_{k-1}e^{P_{k-1}}f^{(k-1)}+\cdots +A_1e^{P_1}f'+A_0e^{P_0}f=F(z). \end{aligned}$$

(1.1)

(I) If $F\equiv 0$, then every transcendental meromorphic solution f of the Eq. (1.1) has infinite order and $\sigma _2(f)=n$.

(II) If $F\not \equiv 0$, then all solutions f of non-homogeneous linear differential Eq. (1.1) with at most one exceptional solution $f_0$ of finite order satisfy

$$\begin{aligned} \lambda (f)=\overline{\lambda }(f)=\sigma (f)=\infty , \lambda _2(f)=\overline{\lambda }_2(f)=\sigma _2(f). \end{aligned}$$

Theorem 1.2

Let $P_j(z)=a_jz^n+b_{j,1}z^{n-1}+\cdots +b_{j,n-1}z+b_{j,n}(j=0,1,\cdots ,k-1)$ be k polynomials with degree $n\ge 1$, $a_j\ne 0$, $A_j(z), D_j(z)$ be meromorphic functions with order $\sigma (A_j)<n, \sigma (D_j)<1(j=0,1,\cdots ,k-1)$, F(z) be a meromorphic function with finite order. If $a_j=\alpha _ja_1, \alpha _j>0, {\text {arg}}a_1\ne {\text {arg}}a_0(j=2,\cdots ,k-1)$ or $a_j=c_ja_0, 0<c_j<1, c_j<0 (j=1,\cdots ,k-1)$ and suppose that all poles of f are uniformly bounded multiplicity. Consider the differential equation

$$\begin{aligned} f^{(k)}+(A_{k-1}e^{P_{k-1}}+D_{k-1})f^{(k-1)}+\cdots +(A_1e^{P_1}+D_1)f'+(A_0e^{P_0}+D_0)f=F(z). \end{aligned}$$

(1.2)

(I) If $F\equiv 0$, then every transcendental meromorphic solution f of the Eq. (1.2) has infinite order and $\sigma _2(f)=n$.

(II) If $F\not \equiv 0$, then all solutions f of non-homogeneous linear differential Eq. (1.2) with at most one exceptional solution $f_0$ of finite order satisfy

$$\begin{aligned} \lambda (f)=\overline{\lambda }(f)=\sigma (f)=\infty , \lambda _2(f)=\overline{\lambda }_2(f)=\sigma _2(f). \end{aligned}$$

Theorem 1.3

Let $P_j(z)=b_{j,0}z^n+b_{j,1}z^{n-1}+\cdots +b_{j,n-1}z+b_{j,n}(j=0,1,\cdots ,k-1)$ be k polynomials with degree $n\ge 1$, $a_j\ne 0$, $A_j(z)$ be meromorphic functions with order $\sigma (A_j)<n (j=0,1,\cdots ,k-1)$, F(z) be a meromorphic function with finite order. If $b_{j,0}$ are distinct complex numbers and suppose that all poles of f are uniformly bounded multiplicity and $A_0(z)\not \equiv 0$. Consider the differential equation

$$\begin{aligned} f^{(k)}+A_{k-1}e^{P_{k-1}}f^{(k-1)}+\cdots +A_1e^{P_1}f'+A_0e^{P_0}f=F(z). \end{aligned}$$

(1.3)

(I) If $F\equiv 0$ and the measure of the set of $E(\theta )=\{\theta \in [0,2\pi ):\delta (P_0,\theta )>\delta (P_j,\theta ), j=1,\cdots ,k-1\}$ is positive, then every transcendental meromorphic solution f of the Eq. (1.3) has infinite order and $\sigma _2(f)=n$.

(II) If $F\not \equiv 0$, then all solutions f of non-homogeneous linear differential Eq. (1.3) with at most one exceptional solution $f_0$ of finite order satisfy

$$\begin{aligned} \lambda (f)=\overline{\lambda }(f)=\sigma (f)=\infty , \lambda _2(f)=\overline{\lambda }_2(f)=\sigma _2(f). \end{aligned}$$

2 Some Lemmas

In this section, we will give some lemmas which are crucial to the proof of our theorems.

The following lemma is used widely in the discussing of growth order of differential equations.

Lemma 2.1

[11] Let F(r) and G(r) be monotone nondecreasing functions on $(0,\infty )$ such that (i) $F(r)\le G(r)$ n.e. or (ii) for $r\ne H\cup [0,1]$ having finite logarithmic measure. Then for any constant $\alpha >1$, there exists $r_0>0$ such that $F(r)\le G(\alpha r)$ for all $r>r_0$.

The next lemma was proved by Gundersen [9] and will be used to obtain the hyper order $\sigma _2(f)$ of the meromorphic solution f.

Lemma 2.2

[9] Let f be a transcendental meromorphic function. Let $\alpha >1$ be a constant, k, j be two integers satisfying $k>j\ge 0$. Then exists a set $E_1\subset (1,\infty )$ which has finite logarithmic measure, and a constant $C>0$, such that for all z satisfying $|z|=r\notin E_1\cup [0,1]$, we have

$$\begin{aligned} \left| \frac{f^{(k)}(z)}{f^{(j)}(z)}\right| \le C\left[ \frac{T(\alpha r,f)}{r}(\log r)^\alpha \log T(\alpha r,f)\right] ^{k-j}. \end{aligned}$$

(2.1)

The following lemma was first established by Chen [1], which is useful for our study. We will use it for the case of $A(z)\equiv 1$.

Lemma 2.3

[1] Let $P(z)=(\alpha +i\beta )z^n+\cdots (|\alpha |+|\beta |\ne 0)$ be a polynomial with degree $n\ge 1$, A(z) be an entire function with $\sigma (A)<n$. Set $g(z)=A(z)e^{P(z)}, z=re^{i\theta }, \delta (P,\theta )=\alpha \cos n\theta -\beta \sin n\theta$, then for any $\varepsilon >0$, there exists a set $H_1\subset [0,2\pi )$, with zero measure, for any $\theta \in [0,2\pi )\backslash (H_1\cup H_2)$, there exists a number $R>0$, for any $|z|=r>R$, we have

if $\delta (P,\theta )>0$, then

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P,\theta )r^n]\le |g(re^{i\theta })|\le \exp [(1+\varepsilon )\delta (P,\theta )r^n] \end{aligned}$$

if $\delta (P,\theta )<0$, then

$$\begin{aligned} \exp [(1+\varepsilon )\delta (P,\theta )r^n]\le |g(re^{i\theta })|\le \exp [(1-\varepsilon )\delta (P,\theta )r^n] \end{aligned}$$

where $H_2=\{\theta \in [0,2\pi ):\delta (P,\theta )=0\}$ is a finite set.

Lemma 2.4

[6] Let $k\ge 2$ and $A_0, A_1, \cdots , A_{k-1}$ be meromorphic functions. Suppose that $\sigma =\max \{\sigma (A_j),0\le j\le k-1\}$ and all the poles of f are of uniformly bounded multiplicity. Then every transcendental meromorphic solution of the differential equation

$$\begin{aligned} f^{(k)}+A_{k-1}f^{(k-1)}+\cdots +A_0f=0 \end{aligned}$$

satisfies $\sigma _2(f)\le \sigma$.

We will use Lemma 2.4 to estimate the hyper order of f.

Lemma 2.5

[5] Let $A_j, F(j=0,1,2,\cdots ,k-1)$ be finite order meromorphic functions. If f is an infinite order meromorphic solution of the equation

$$\begin{aligned} f^{(k)}+A_{k-1}f^{(k-1)}+\cdots +A_0f=F, \end{aligned}$$

then $\overline{\lambda }(f)=\lambda (f)=\sigma (f)=\infty$.

Lemma 2.6

[4] Let g(z) be a meromorphic function with $\sigma (g)=\beta <+\infty$. Then for any $\varepsilon >0$, there exists a set $E\subset (1,\infty )$ with $lm(E)=\int _1^\infty \chi _E(t)/tdt<+\infty$, such that for all z with $|z|=r\notin [0,1]\cup E, r\rightarrow \infty$, then

$$\begin{aligned} \exp (-r^{\beta +\varepsilon })\le |g(z)|\le \exp (r^{\beta +\varepsilon }). \end{aligned}$$

3 Proof of the Theorems

From the conditions of the theorems and by Lemma 2.4, we can obtain that $\sigma _2(f)\le n$. What we need to do is to prove that $\sigma _2(f)\ge n$.

Proof of Theorem 1.1

(I) Let us treat it as three cases.

Case i. When $a_j=\alpha _ja_1, \alpha _j>0, {\text {arg}}a_1\ne {\text {arg}}a_0 (j=2, 3, \cdots , k-1)$.

Put $z=re^{i\theta }$. Then

$$\begin{aligned} \Re (a_0z^n)= & {} \delta (P_0,\theta )r^n|a_0|=|a_0|r^n\cos ({\text {arg}}a_0+n\theta ), \nonumber \\ \Re (a_jz^n)= & {} \delta (P_j,\theta )r^n|a_j|=|a_j|r^n\cos ({\text {arg}}a_j+n\theta )=\alpha _j|a_1|r^n\cos ({\text {arg}}a_1+n\theta ).\nonumber \\ \delta (P_j,\theta )= & {} \alpha _j\delta (P_1,\theta ), \ j=2,3,\cdots , k-1. \ \end{aligned}$$

(3.1)

By (3.1) and Lemma 2.3, for any $0<\varepsilon <1$, there exists an ${\text {arg}}z=\theta$, satisfying $\theta \in [0,2\pi )\backslash (H_1\cup H_2)$, such that

$$\begin{aligned} \delta (P_0,\theta )>0,\delta (P_j,\theta )=\alpha _j\delta (P_1,\theta )<0, \ j=2,\cdots ,k-1. \end{aligned}$$

By (1.1), we have

$$\begin{aligned} -e^{P_0}=\frac{1}{A_0}\frac{f^{(k)}}{f}+\frac{A_{k-1}}{A_0} e^{P_{k-1}}\frac{f^{(k-1)}}{f}+\cdots +\frac{A_1}{A_0}e^{P_1}\frac{f'}{f}. \end{aligned}$$

(3.2)

By Lemma 2.3 ($A(z)\equiv 1$), for r large enough,

$$\begin{aligned} |e^{P_0}|\ge & {} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n],\\ |e^{P_j}|\le & {} \exp [(1-\varepsilon )\delta (P_j,\theta )r^n]=\exp [(1-\varepsilon )\alpha _j\delta (P_1,\theta )r^n](j=1,2,\cdots ,k-1). \end{aligned}$$

Since $\sigma =\max \{\sigma (A_j), j=0,1,2,\cdots ,n-1\}$, by Lemma 2.6, for any $\varepsilon >0$ small enough, there exists a set $E\subset (1,\infty )$ with $lm(E)<+\infty$, such that for all z with $|z|=r\notin [0,1]\cup E, r\rightarrow \infty$, we have $(j=0,1,\cdots ,k-1)$

$$\begin{aligned} \left| \frac{A_j}{A_0}\right| \le \exp (r^{\frac{\sigma +n}{2}}). \end{aligned}$$

(3.3)

For $|z|=r\notin E\cup E_1\cup [0,1]$, we have

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n]\le Ck\exp [(1-\varepsilon )\alpha \delta (P_1,\theta )r^n]\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}, \end{aligned}$$

where $\alpha =\max \{1, \alpha _j,j=2,\cdots ,k-1\}$. Hence, we obtain $\sigma (f)=\infty$ and $\sigma _2(f)\ge n.$

Case ii. When $a_j=c_ja_0, 0<c_j<1(j=1,\cdots ,k-1)$.

Let $c=\max \{c_j,j=1,\cdots ,k-1\}$ and $0<c<1$. For any $0<\varepsilon <\frac{1-c}{2}$, there exists an ${\text {arg}}z=\theta$, satisfying $\theta \in [0,2\pi )\backslash (H_1\cup H_2)$, such that $\delta (P_j,\theta )=c_j\delta (P_0,\theta )>0, j=1,2,\cdots ,k-1$. By Lemma 2.3 ($A(z)\equiv 1$), for r large enough,

$$\begin{aligned} |e^{P_0}|\ge & {} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n],\\ |e^{P_j}|\le & {} \exp [(1-\varepsilon )\delta (P_j,\theta )r^n]= \exp [(1-\varepsilon )c_j\delta (P_0,\theta )r^n](j=1,2,\cdots ,k-1). \end{aligned}$$

In view of Lemma 2.2 and (3.2), (3.3) for $|z|=r\notin E\cup E_1\cup [0,1],$ we have

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n]\le Ck\exp [(1-\varepsilon )c\delta (P_0,\theta )r^n]\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}. \end{aligned}$$

Thus

$$\begin{aligned} \exp \left[ \frac{(1-c)}{2}\delta (P_0,\theta )r^n\right] \le Ck\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}. \end{aligned}$$

We obtain $\sigma (f)=\infty$ and $\sigma _2(f)\ge n$.

Case iii. When $a_j=c_ja_0, c_j<0(j=1,\cdots ,k-1)$.

For any $\varepsilon >0$ small enough, there exists an ${\text {arg}}z=\theta$, satisfying $\theta \in [0,2\pi )\backslash (H_1\cup H_2)$, such that $\delta (P_0,\theta )>0, \delta (P_j,\theta )=c_j\delta (P_0,\theta )<0, j=1,2,\cdots ,k-1$. By Lemma 2.3 ($A(z)\equiv 1$), for r large enough,

$$\begin{aligned} |e^{P_0}|\ge & {} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n],\\ |e^{P_j}|\le & {} \exp [(1-\varepsilon )\delta (P_j,\theta )r^n]\le M(j=1,2,\cdots ,k-1). \end{aligned}$$

In view of Lemma 2.2 and (3.2), (3.3) for $|z|=r\notin E\cup E_1\cup [0,1],$ we have

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n]\le CkM\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}. \end{aligned}$$

Thus

$$\begin{aligned} \exp \left[ \frac{(1-\varepsilon )}{2}\delta (P_0,\theta )r^n\right] \le CkM[rT(2r,f)]^{k+1}. \end{aligned}$$

We obtain $\sigma (f)=\infty$ and $\sigma _2(f)\ge n$.

(II) Assume $f_0$ is a solution of finite order of (1.1). If there exists another solution $f_1$ of finite order of (1.1), then $\sigma (f_1-f_0)<\infty$, and $f_1-f_0$ is a solution of the corresponding homogeneous differential equation. However, we get that $\sigma (f_1-f_0)=\infty$, which contradicts to $\sigma (f_1-f_0)<\infty$. Therefore all solutions f of non-homogeneous linear differential Eq. (1.1), with at most one exceptional solution $f_0$ of finite order, satisfy $\sigma (f)=\infty$.

Now suppose that f is a solution of infinite order of (1.1), then by Lemma 2.5, we obtain

$$\begin{aligned} \lambda (f)=\overline{\lambda }(f)=\sigma (f)=\infty . \end{aligned}$$

Next, we will show that every solution f of infinite order of (1.1) satisfy $\overline{\lambda }_2(f)=\sigma _2(f)$. It is easy to see that the zeros of f occur at the poles of $A_j(z)(j=1,2,\cdots ,k-1)$ or the zeros of F(z). If f has a zero at $z_0$ of order n, $n>k$, then F(z) must have a zero at $z_0$ of order $n-k$. Therefore, we get

$$\begin{aligned} N\left( r,\frac{1}{f}\right) \le k\overline{N}\left( r,\frac{1}{f}\right) + N\left( r,\frac{1}{F}\right) +\sum \limits _{j=1}^{k-1}N(r,A_j). \end{aligned}$$

On the other hand, (1.1) may be rewritten as

$$\begin{aligned} \frac{1}{f}=\frac{1}{F}\left[ \frac{f^{(k)}}{f}+A_{k-1} e^{P_{k-1}}(z)\frac{f^{(k)}}{f}+\cdots +A_0e^{P_0}\right] . \end{aligned}$$

Hence

$$\begin{aligned} m\left( r,\frac{1}{f}\right) \le m\left( r,\frac{1}{F}\right) +\sum \limits _{j=1}^{k-1}[m(r,A_j)+m(r,e^{P_j})] +\sum \limits _{j=1}^{k-1}m\left( r,\frac{f^{(j)}}{f}\right) +O(1). \end{aligned}$$

Hence, by the logarithmic derivative lemma, there exists a set E having finite linear measure such that for all $r\notin E$, we have

$$\begin{aligned} \begin{aligned} T(r,f)&=T\left( r,\frac{1}{f}\right) +O(1)\\&\le T\left( r,\frac{1}{F}\right) +k\overline{N}\left( r,\frac{1}{f}\right) +\sum \limits _{j=1}^{k-1}[T(r,A_j) +m(r,e^{P_j})]+\sum \limits _{j=1}^{k-1}m\left( r,\frac{f^{(j)}}{f}\right) \\&\le T(r,F)+\sum \limits _{j=1}^{k-1}[T(r,A_j)+T(r,e^{P_j})]+k\overline{N}\left( r,\frac{1}{f}\right) +O(\log T(r,f)+\log r). \end{aligned} \end{aligned}$$

(3.4)

Since for any $\varepsilon >0$ and sufficiently large r, we have

$$\begin{aligned} O(\log (rT(r,f)))\le & {} \frac{1}{2}T(r,f), T(r,F)\le r^{\sigma (F)+\varepsilon }, T(r,A_j)\le r^{\sigma +\varepsilon }(j=0,1,\cdots ,k-1); \\ T(r,e^{P_j})\le & {} r^{n+\varepsilon }(j=0,1,\cdots ,k-1). \end{aligned}$$

so that for $r\notin E$ and r large enough, we have

$$\begin{aligned} T(r,f)\le 2k\overline{N}\left( r,\frac{1}{f}\right) +2r^{\sigma (F)+\varepsilon }+(4k+5)(r^{\sigma +\varepsilon }+r^{n+\varepsilon }). \end{aligned}$$

Hence by Lemma 2.4, we get $\sigma _2(f)\le \overline{\lambda }_2(f)$. On the other hand, we can easily get $\sigma _2(f)\ge \lambda _2(f)\ge \overline{\lambda }_2(f)$. Therefore, we complete the proof.

Proof of Theorem 1.2

(I) Suppose $F\equiv 0$, then we have

$$\begin{aligned} f^{(k)}+(A_{k-1}(z)e^{P_{k-1}(z)}+D_{k-1}(z))f^{(k-1)}+\cdots +(A_0(z)e^{P_0(z)}+D_0(z))f=0. \end{aligned}$$

(3.5)

Case i. When $a_j=\alpha _ja_1, \alpha _j>0, {\text {arg}}a_1\ne {\text {arg}}a_0$.

Put $z=re^{i\theta }$. Then

$$\begin{aligned} \Re (a_0z^n)= & {} \delta (P_0,\theta )r^n|a_0|=|a_0|r^n\cos ({\text {arg}}a_0+n\theta ),\nonumber \\ \Re (a_jz^n)= & {} \delta (P_j,\theta )r^n|a_j|=|a_j|r^n\cos ({\text {arg}}a_j+n\theta )=\alpha _j|a_1|r^n\cos ({\text {arg}}a_1+n\theta ).\nonumber \\ \delta (P_j,\theta )= & {} \alpha _j\delta (P_1,\theta ) \ j=2,3,\cdots , k-1. \ \end{aligned}$$

(3.6)

By (3.6) and Lemma 2.3, for any $0<\varepsilon <1$, there exists an ${\text {arg}}z=\theta$, satisfying $\theta \in [0,2\pi )\backslash (H_1\cup H_2)$, such that

$$\begin{aligned} \delta (P_0,\theta )>0,\delta (P_j,\theta )<0,(j=1,2,\cdots ,k-1). \end{aligned}$$

By (3.5),

$$\begin{aligned} -e^{P_0}-\frac{D_0}{A_0}=\frac{1}{A_0}\frac{f^{(k)}}{f}+ \left( \frac{A_{k-1}}{A_0}e^{P_{k-1}}+\frac{D_{k-1}}{A_0}\right) \frac{f^{(k-1)}}{f}+\cdots +\left( \frac{A_1}{A_0}e^{P_1}+\frac{D_1}{A_0}\right) \frac{f'}{f}. \end{aligned}$$

(3.7)

By Lemma 2.3, for r large enough,

$$\begin{aligned}&\left| e^{P_0(re^{i\theta })}+\frac{D_0(re^{i\theta })}{A_0(re^{i\theta })} \right| \ge \exp [(1-\varepsilon )\delta (P_0,\theta )r^n](1+o(1)),\\&\begin{aligned} \left| e^{P_j}+\frac{D_j}{A_j}\right|&\le \exp [(1-\varepsilon )\delta (P_j,\theta )r^n]\exp (r^{\max \{\sigma (D_j),\sigma (A_j)\}+\frac{\varepsilon }{2}})\\&\le \exp [r^{\max \{\sigma (D_j),\sigma (A_j)\}+\varepsilon }](j=1,2,\cdots ,k-1). \end{aligned} \end{aligned}$$

For $|z|=r\notin E\cup E_1\cup [0,1],$ we have

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n](1+o(1))\le Ck\exp [r^{\max \{\sigma (D_j),\sigma (A_j)\}+\varepsilon }]\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}. \end{aligned}$$

By noting that $\sigma _D+\varepsilon \le \frac{1}{2}+\frac{1}{2}\sigma _D\le 1, r^{\frac{\sigma +n}{2}}<\frac{1}{2}(1-\varepsilon )\delta (P_0,\theta )r^n , \sigma +\varepsilon <\frac{\sigma +n}{2}$, we obtain $\sigma (f)=\infty$ and $\sigma _2(f)\ge n.$

Case ii. When $a_j=c_ja_0, 0<c_j<1(j=1,\cdots ,k-1)$.

Let $c=\max \{c_j,j=1,\cdots ,k-1\}$ and $0<c<1$. For any $0<\varepsilon <\frac{1-c}{2}$, there exists an ${\text {arg}}z=\theta \in [0,2\pi )\backslash (H_1\cup H_2)$, satisfying $\delta (P_j,\theta )=c_j\delta (P_0,\theta )>0, j=1,2,\cdots ,k-1$. By Lemma 2.3, for r large enough,

$$\begin{aligned}&\left| e^{P_0}+\frac{D_0}{A_0}\right| \ge \exp [(1-\varepsilon )\delta (P_0,\theta )r^n](1+o(1)),\\&\begin{aligned} \left| e^{P_j}+\frac{D_j}{A_j}\right|&\le \exp [(1+\varepsilon )\delta (P_j,\theta )r^n]\exp (r^{\max \{\sigma (D_j),\sigma (A_j)\}+\frac{\varepsilon }{2}})\\&=\exp [(1+\varepsilon )c_j\delta (P_0,\theta )r^n]\exp (r^{\max \{\sigma (D_j),\sigma (A_j)\}+\frac{\varepsilon }{2}})\\&\le \exp [(1+\varepsilon )c\delta (P_0,\theta )r^n](1+o(1)) \ (j=1,2,\cdots ,k-1). \ \end{aligned} \end{aligned}$$

In view of Lemma 2.2 and (3.7), for $|z|=r\notin E\cup E_1\cup [0,1],$ we have

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n](1+o(1))\le Ck\exp [(1+\varepsilon )c\delta (P_0,\theta )r^n](1+o(1))\exp (r^\frac{\sigma +n}{2})[rT(2r,f)]^{k+1}. \end{aligned}$$

Thus

$$\begin{aligned} \exp \left[ \frac{(1-c)}{2}\delta (P_0,\theta )r^n\right] \le Ck\exp (r^\frac{\sigma +n}{2})[rT(2r,f)]^{k+1}, \end{aligned}$$

we obtain $\sigma (f)=\infty$ and $\sigma _2(f)\ge n$.

Case iii. When $a_j=c_ja_0, c_j<0(j=1,\cdots ,k-1)$.

For any $\varepsilon >0$ small enough, there exists an ${\text {arg}}z=\theta$, satisfying $\theta \in [0,2\pi )\backslash (H_1\cup H_2)$, such that $\delta (P_0,\theta )>0, \delta (P_j,\theta )=c_j\delta (P_0,\theta )<0, j=1,2,\cdots ,k-1$. By Lemma 2.3($A(z)\equiv 1$), for r large enough,

$$\begin{aligned}&\left| e^{P_0}+\frac{D_0}{A_0}\right| \ge \exp [(1-\varepsilon )\delta (P_0,\theta )r^n](1+o(1)),\\&\begin{aligned} \left| e^{P_j}+\frac{D_j}{A_j}\right|&\le \exp [(1-\varepsilon )\delta (P_j,\theta )r^n]\exp (r^{\max \{\sigma (D_j),\sigma (A_j)\}+\frac{\varepsilon }{2}})\\&\le M\exp (r^{\max \{\sigma (D_j),\sigma (A_j)\}+\frac{\varepsilon }{2}})(j=1,2,\cdots ,k-1). \end{aligned} \end{aligned}$$

In view of Lemma 2.2 and (3.2), (3.3) for $|z|=r\notin E\cup E_1\cup [0,1],$ we have

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n]\le CkM\exp (r^{\max \{\sigma (D_j),\sigma (A_j)\}+\frac{\varepsilon }{2}}) \exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}. \end{aligned}$$

Thus

$$\begin{aligned} \exp \left[ \frac{(1-\varepsilon )}{2}\delta (P_0,\theta )r^n\right] \le CkM[rT(2r,f)]^{k+1}, \end{aligned}$$

we obtain $\sigma (f)=\infty$ and $\sigma _2(f)\ge n$.

(II) Assume $f_0$ is a solution of finite order of (1.2). Using the same method from proof of Theorem 1.1, we can prove that all solutions f of non-homogeneous linear differential Eq. (1.2), with at most one exceptional solution $f_0$ of finite order, satisfy $\sigma (f)=\infty$.

Suppose that f be an infinite order solution of (1.2), we can rewrite (1.2) as

$$\begin{aligned} \frac{1}{f}=\frac{1}{F}\left[ \frac{f^{(k)}}{f}+B_{k-1}\frac{f^{(k)}}{f}+\cdots +B_0\right] , \end{aligned}$$

where $B_j=A_je^{P_j}(z)+D_j(z)(j=0,1,\cdots ,k-1)$. Since $\sigma (D_j)<1(j=0,1,\cdots ,k-1), n\ge 1$, $\sigma (B_j)<n(j=0,1,\cdots ,k-1)$. Using the same method from Theorem 1.1, we can complete the proof.

Proof of Theorem 1.3

If $F(z)\equiv 0$, suppose that f(z) is a transcendental solution of (1.3). Since the measure of the set of $E(\theta )=\{\theta \in [0,2\pi ):\delta (P_0,\theta )>\delta (P_j,\theta ), j=1,\cdots ,k-1\}$ is bigger than zero, there exists a ${\text {arg}}z=\theta \in [0,2\pi )\backslash (H_1\cup H_2)$ such that $\delta (P_0,\theta )>\delta (P_j,\theta ), (j=1,\cdots ,k-1)$ and $\delta (P_0,\theta )>0$.

By Lemma 2.3, for $\varepsilon >0$, we obtain for sufficiently large r,

$$\begin{aligned} |e^{P_0}|\ge & {} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n],\\ |e^{P_j}|\le & {} \exp [(1-\varepsilon )\delta (P_j,\theta )r^n](j=1,2,\cdots ,k-1). \end{aligned}$$

Set $\delta _1=\max _{1\le j\le k-1}\{\delta (P_j,\theta )\}$. Since $\sigma =\max \{\sigma (A_j), j=0,1,2,\cdots ,n-1\}$, by Lemma 2.6, for any $\varepsilon >0$ small enough, there exists a set $E\subset (1,\infty )$ with $lm(E)<+\infty$, such that for all z with $|z|=r\notin [0,1]\cup E, r\rightarrow \infty$, we have $(j=0,1,\cdots ,k-1)$

$$\begin{aligned} \left| \frac{A_j}{A_0}\right| \le \exp (r^{\frac{\sigma +n}{2}}). \end{aligned}$$

We treat it as two cases.

Case i. $\delta _1>0$. For $|z|=r\notin E\cup E_1\cup [0,1]$, for r large enough, we have

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n]\le Ck\exp [(1-\varepsilon )\delta _1r^n]\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}, \end{aligned}$$

i.e.

$$\begin{aligned} \begin{aligned}\exp [(1-\varepsilon )(\delta (P_0,\theta )-\delta _1)r^n]&\le Ck\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}\\&\le Ck\exp (\frac{1}{2}(1-\varepsilon )(\delta (P_0,\theta )-\delta _1)r^n)[rT(2r,f)]^{k+1}.\\ \end{aligned} \end{aligned}$$

Case ii. $\delta _1<0$. For $|z|=r\notin E\cup E_1\cup [0,1]$, for r large enough, we have

$$\begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n]\le Ck\exp [(1+\varepsilon )\delta _1r^n]\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}, \end{aligned}$$

i.e.

$$\begin{aligned} \begin{aligned} \exp [(1-\varepsilon )\delta (P_0,\theta )r^n]&\le Ck M\exp (r^{\frac{\sigma +n}{2}})[rT(2r,f)]^{k+1}\\&\le Ck M\exp (\frac{1}{2}(1-\varepsilon )\delta (P_0,\theta )r^n)[rT(2r,f)]^{k+1}.\\ \end{aligned} \end{aligned}$$

Combining Case i and ii, we obtain $\sigma (f)=\infty$ and $\sigma _2(f)\ge n.$

The proof of (II) is similar to Proof of Theorem 1.1, we omit it.

4 Conclusions

We discuss the growth of meromorphic solutions of some class of homogeneous and nonhomogeneous differential equations.