On Meromorphic Solutions of Non-linear Differential-Difference Equations

Abstract

In this paper, we investigate the non-existence of transcendental entire solutions for non-linear differential-difference equations of the forms

$$\begin{aligned} f^{n}(z)+Q(z,f)=\beta _{1}e^{\alpha _{1}z}+\beta _{2}e^{\alpha _{2}z}+\cdots +\beta _{s}e^{\alpha _{s}z} \end{aligned}$$

and

$$\begin{aligned} f^{n}(z)f^{(k)}(z)+L_d(z,f)=\sum ^{s}_{i=1}p_i(z)e^{\alpha _i{(z)}}, \end{aligned}$$

where n,  s are positive integers, $n\ge s+2,$ Q(z, f) is a differential-difference polynomial in f of degree d.

1 Introduction and Main Results

In nonlinear science, the study of nonlinear mathematical physics equation is particularly meaningful. Nonlinear wave equation is an important mathematical model to describe natural phenomena. Examples include the Camassa–Holm equation [2] that describes the motion of water waves, the KdV–Burgers–Kuramoto equation [6] that describes unstable drift waves in plasma physics, the KdV equation [10] that describes the motion of shallow water waves, and etc. Similarly, the discrete nonlinear equation, especially the nonlinear differential-difference equation plays an important role in the study of nonlinear problems. Recently, many methods for studying the exact solutions of continuous nonlinear wave equations have been extended to nonlinear differential-difference equations. Tsuchida et al. [17] has extended the backscattering method to the study of differential-difference equations. Qian et al. [16] has extended the separation method of multilinear variables to solving differential-difference equations, so it is of great significance to study differential-difference equations in mathematical physics. In this paper, we will prove the existence of entire solutions of two kinds of differential-difference equations by using Nevanlinna’s value distribution theory.

Let f(z) be a meromorphic function in the complex plane ${\mathbb{C}}.$ We assume that the reader is familiar with standard notations in Nevanlinna’s value distribution theory, such as T(r, f),  m(r, f),  N(r, f),  as well as the main conclusions, for references see [8, 20]. We use the symbols $\sigma (f),$ $\lambda (f)$ and $\sigma _{2}(f)$ to denote the order of growth, the exponent of convergence of zeros and the hyper-order of f,  respectively. In addition, we use $N_{p)}(r,\frac{1}{f-a})$ to denote the counting function of the zeros of $f-a,$ whose multiplicities are not greater than p. For $a\in {\mathbb{C}},$ the deficiency $\delta (a,f)$ and $\delta _{p)}(a,f)$ are defined as

$$\begin{aligned} \delta (a,f)=1-\limsup \limits _{r\rightarrow \infty }\frac{N\left( r,\frac{1}{f-a}\right) }{T(r,f)},\quad \delta _{p)}(a,f)=1-\limsup \limits _{r\rightarrow \infty } \frac{N_{p)}\left( r,\frac{1}{f-a}\right) }{T(r,f)}. \end{aligned}$$

By S(r, f),  we denote any quantity satisfying $S(r,f)=o(T(r, f ))$ as $r\rightarrow \infty$ outside of an exceptional set E with finite logarithmic measure. A meromorphic function $\alpha (z)$ is said to be a small function of f if it satisfies $T(r, a) = S(r, f ).$

Nevanlinna’s value distribution theory of meromorphic functions has been widely applied to investigate the solvability and existence of entire or meromorphic solution of complex differential equations, difference equation and differential-difference equations. Proving the existence of entire or meromorphic solutions f of a given differential-difference equation and finding out the solutions, if any, are usually interesting and quite challenging problems.

Recently, many researchers focused on the non-existence of meromorphic solutions of nonlinear differential equations of the form

$$\begin{aligned} f^{n}(z)+Q_d(z,f)=g(z), \end{aligned}$$

(1)

where $Q_d(z,f)$ is a differential polynomial in f with degree d. In 2010, Yang and Laine [19] considered a special case of Eq. (1) regarding the existence of solutions, and proved the following result.

Theorem A

[19] Let p(z) be a non-vanishing polynomial, and let b,  c be nonzero complex numbers. If p is nonconstant, then the differential equation

$$\begin{aligned} f^3(z)+p(z)f''(z)=c\sin bz \end{aligned}$$

(2)

admits no transcendental entire solutions, while if p is constant, then Eq. (2) admits three distinct transcendental entire solutions, provided $(\frac{pb^2}{27})^3=\frac{1}{4}c^{2}.$

Remark 1.1

Notice that $c\sin {bz}=\frac{c}{2i}(e^{biz}-e^{-biz}),$ thus, $c\sin {bz}$ is a linear combination of the $e^{biz}$ and $e^{-biz}.$ Therefore, it is meaningful to investigate the existence of solutions of the differential equation

$$\begin{aligned} f^{n}(z)+Q_d(z,f)=p_{1}(z)e^{\alpha _{1}z}+p_{2}(z)e^{\alpha _{2}z}. \end{aligned}$$

(3)

Li and Yang [13] investigated meromorphic solutions of Eq. (3) and obtained the following result.

Theorem B

[13] Let $n\ge 4$ be an integer and $Q_d(z,f)$ denote an algebraic differential polynomial in f(z) of degree $d\le n-3.$ If $p_{1}(z),$ $p_{2}(z)$ are two non-zero polynomials and $\alpha _{1} , \alpha _{2}$ are two non-zero constant such that $\frac{\alpha _{1}}{\alpha _{2}}$ is not rational, then Eq. (3) has no transcendental entire solutions.

In view of the progress on the difference analogues of classical Nevanlinna theory of meromorphic functions [5, 7], it is quite natural to investigate difference analogues of Eq. (2). In [19], Yang and Laine showed that similar conclusions follow if restricting the solutions to be of finite order.

Theorem C

[19] A nonlinear difference equation

$$\begin{aligned} f^3(z)+q(z)f(z+1)=c\sin bz \end{aligned}$$

(4)

where q(z) is a nonconstant polynomial and $b,c\in {\mathbb{C}}$ are nonzero constants, does not admit entire solutions of finite order.

In 2011, Zhang and Liao discussed the case that $Q_d(z,f)$ in Eq. (3) is a difference polynomials instead of differential polynomial.

Theorem D

[21] Let $n\ge 4$ be an integer and $Q_d(z,f)$ denote an algebraic difference polynomial in f(z) of degree $d\le n-3.$ If $p_1(z), p_2(z)$ are two non-zero polynomials and $\alpha _1, \alpha _2$ are two non-zero constants with $\frac{\alpha _1}{\alpha _2}\ne (\frac{d}{n})^{\pm 1},1,$ then the Eq. (3) does not have any transcendental entire solution of finite order.

Remark 1.2

The proofs of the above theorems always need to use the logarithmic derivative lemma and its difference analogue. As we know, the logarithmic derivative lemma is valid for all meromorphic functions, while the difference analogue of the logarithmic derivative lemma is only valid for meromorphic functions with finite order or hyper-order less than one. Thus, for non-linear difference or differential-difference equations, only the entire (or meromorphic) solutions with finite order or hyper-order less than one were discussed.

Later on, some results were obtained when Q(z, f) is a differential-difference or difference polynomials, see [4]. In particular, Zhang and Huang [22] considered the case that the right-hand side of the Eq. (3) has $s\ge 2$ terms and Q(z, f) is a difference polynomials in f,  and they obtained the result as follow.

Theorem E

[22] Let $n\ge 2+s$ be a integer, $p(z)\not \equiv 0$ be a polynomial, $\eta$ be a constant, $\beta _{1}, \beta _{2},\ldots , \beta _{s}$ be nonzero constants and $\alpha _{1}, \alpha _{2},\ldots , \alpha _{s}$ be distinct non-zero constants. Suppose that $\frac{\alpha _{i}}{\alpha _{j}}\ne n$ for all $i,j \in {1,2,\ldots ,s}.$ And when $s\ge 5,$ suppose further that $n\alpha _k\ne l_{k1}\alpha _1+ l_{k2}\alpha _2 +\cdots + l_{ks}\alpha _s$ for $k = 5, 6,\ldots , s,$ where $l_{k1}, l_{k2},\ldots , l_{ks}\in \{0, 1,\ldots , n-1\}$ and $l_{k1}+ l_{k2}+ \cdots + l_{ks}= n.$ Then any meromorphic solution f(z) of the equation

$$\begin{aligned} f^{n}(z)+p(z)f(z+\eta )=\sum ^{s}_{i=1}\beta _ie^{\alpha _i{z}} \end{aligned}$$

(5)

must satisfy $\sigma _{2} (f)\ge 1.$

Remark 1.3

Clearly, we see that Eq. (5) has no meromorphic solutions with $\sigma _2(f)<1$ under the assumption of Theorem E. In addition, observing the above theorems, we notice that the existence of entire or meromorphic solutions of Eq. (1) is related to the number of terms on the right-hand of the equation and the form of Q(z, f). In particular, when Q(z, f) is a differential-difference or difference polynomial and the right hand of Eq. (5) has $s\ge 2$ terms, most of results discussed the simple case that Q(z, f) has only a monomial.

Let

$$\begin{aligned} Q(z,f)=\sum ^{p}_{j=1}a_{j}(z)f^{(m_{j})}(z+\tau _{j})+a_0(z)\not \equiv 0, \end{aligned}$$

where $0\le m_{1}<\cdots <m_{p}$ and $\tau _{j}(1\le j\le p)$ are complex constants, and $a_{j}(0\le j\le p)\ne 0$ are small functions of f.

Inspired by the above results, our first result generalizes the result of Zhang and Huang and investigates the non-existence of entire solutions of the following differential-difference equations

$$\begin{aligned} f^{n}(z)+Q(z,f)=p_{1}e^{\alpha _{1}z}+p_{2}e^{\alpha _{2}z}+\cdots +p_{s}e^{\alpha _{s}z}, \end{aligned}$$

(6)

and obtain the following result.

Theorem 1.1

Let $n\ge 2+s$ be a integer, $Q(z,f)\not \equiv 0,$ and n, s be the positive integers and let $p_{1},\ldots ,p_{s},$ $\alpha _{1},\ldots ,\alpha _{s}$ be constants with $p_{1},\ldots ,p_{s}, \alpha _{1},\ldots \alpha _{s}\ne 0.$ If $\alpha _{1},\ldots ,\alpha _{s}$ are distinct constants such that $\frac{\alpha _{i}}{\alpha _{j}}\ne n$ for all $i,j\in {1,2,\ldots ,s},$ then Eq. (6) admits no finite order transcendental entire function with a finite Borel exceptional value.

Example 1.1

Let $n=4,$ $s=2$ and $Q(z,f)=-5(f')^2-f^2-2f'(z+2\pi i).$ Then, the following differential equation

$$\begin{aligned} f^4(z)-5(f')^2-f^2-2f'(z+2\pi i)=e^{4z}+4e^{3z} \end{aligned}$$

has a transcendental entire solution $f=e^z+1.$ Here f(z) has 1 as a Borel exceptional value.

Example 1.2

Let $n=5,$ $s=3$ and $Q(z,f)=ff'+f^2f'(z+2\pi i).$ Then, the following differential equation

$$\begin{aligned} f^5(z)+ff'+f^2f'(z+2\pi i)=e^{5z}+e^{2z}+e^{3z} \end{aligned}$$

has a transcendental entire solution $f=e^z.$ Here f(z) has zero as a Borel exceptional value.

Remark 1.4

The condition $\deg (Q(z,f))=1$ is necessary. If $\deg (Q(z,f))\ge 2,$ then Theorem 1.1 may not hold, as seen from Examples 1.1 and 1.2.

Observing the above results, one can find that every exponential part of two terms on the right side of the equations is the first order monomial. Naturally, a question arise: Can the solutions be obtained when every exponential part of two terms on right side of the equations is the k-th order monomial? Actually, the solutions of such equations can exist in this case. Below we provide a special example to illustrate this problem.

Example 1.3

Let $n=5,$ $s=3$ and $Q(z,f)=f'(z)+2zf(z+2).$ Then, the function $f(z)=ze^{z^{2}}$ is a transcendental entire solution of

$$\begin{aligned} f^5(z)+f'(z)+2zf(z+2)=z^5e^{5z^2}+(2z^2+1)e^{z^2}+2z(z+2)e^{(z+2)^2}. \end{aligned}$$

In 2013, Liao et al. [14] answered this question, and they obtained a result of the nonlinear differential equation of the following form

$$\begin{aligned} f^{n}(z)+Q_d(z,f)=p_{1}(z)e^{\alpha _{1}(z)}+p_{2}(z)e^{\alpha _{2}(z)}. \end{aligned}$$

(7)

Theorem E

[14] Let $n\ge 3$ and $Q_d(f)$ be a differential polynomial in f of degree d with rational functions as its coefficients. Suppose that $p_1(z), p_2(z)$ are rational functions and $\alpha _1(z), \alpha _2(z)$ are polynomials. If $d\le n-2,$ the differential equation (7) admits a meromorphic function f with finitely many poles, then $\frac{\alpha '_1}{\alpha '_2}$ is a rational number. Furthermore, only one of the following four cases holds:

• (1) $f(z) = q(z)e^{P (z)}$ and $\frac{\alpha '_1}{\alpha ' _2} = 1,$ where q(z) is a rational function and P(z) is a polynomial with $nP'(z) = \alpha '_1(z) = \alpha '_2(z);$

• (2) $f(z) = q(z)e^{P (z)}$ and either $\frac{\alpha '_1}{\alpha ' _2} = \frac{k}{n}$ or $\frac{\alpha '_1}{\alpha ' _2} = \frac{n}{k},$ where q(z) is a rational function, k is an integer with $1\le k \le d$ and P(z) is a polynomial with $nP'(z) = \alpha '_1(z)$ or $nP'(z) = \alpha '_2(z);$

• (3) f satisfies the first order linear differential equation $f'=\left( \frac{1}{n}\frac{p'_2}{p_2}+\frac{1}{n}\alpha '_2\right) f+\psi$ and $\frac{\alpha '_1}{\alpha ' _2}=\frac{n-1}{n}$ or f satisfies the first order linear differential equation $f'=\left( \frac{1}{n}\frac{p'_1}{p_1}+\frac{1}{n}\alpha '_1\right) f+\psi$ and $\frac{\alpha '_1}{\alpha ' _2}=\frac{n}{n-1},$ where $\psi$ is a rational function;

• (4) $f(z) = \gamma _1(z)e^{\beta _1(z)} + \gamma _2(z)e^{\beta _1(z)}$ and $\frac{\alpha '_1}{\alpha ' _2} = -1,$ where $\gamma _1,\gamma _2$ are rational functions and $\beta _1(z)$ is a polynomial with $n \beta _1'(z) = \alpha '_1$ or $n \beta _1'(z) = \alpha '_2.$

After that, Li and Yang [12] and Chen and Gui [3] continued to investigate this question, and got some results for certain types of non-linear differential equations.

Let

$$\begin{aligned} L_d(z,f)=\sum _{{\textbf{I}}\in {\mathbb{I}}}a_{{\textbf{I}}}(z) \prod ^{t}_{l=0}(f^{({\textbf{k}})}(z+c_{l}))^{I_l}\not \equiv 0, \end{aligned}$$

where ${\textbf{k}}=(0,1,\ldots ,k);$ ${\textbf{I}}=(I_0,I_1,\ldots ,I_{t}),$ $I_{l}=(i_{l0},i_{l1},\ldots ,i_{lk})$ are multiindices of nonnegative integers ${\mathbb{Z}}_{+};$ $d:=\deg (L_{d}(z,f))=\max \{\sum ^{t}_{i=l}I_{l}\}$ and $c_{i}$ be constants; ${\mathbb{I}}$ is a finite set of ${\mathbb{Z}}_{+}^{(t+1)(k+1)}$ and $a_{{\textbf{I}}}\ne 0$ are meromorphic functions. Our second result considers the following differential-difference equation

$$\begin{aligned} f^{n}(z)f^{(k)}(z)+L_d(z,f)=\sum ^{s}_{i=1}p_i(z)e^{\alpha _i{(z)}}. \end{aligned}$$

(8)

Theorem 1.2

Let $n\ge 2+s,$ $s\ge 3,$ $L_d(z)\not \equiv 0,$ $d\le n-s-1$ and n, s, d, k be positive integers. Suppose that $p_{1}(z),\ldots , p_{s}(z)$ are non-vanishing rational functions and $\alpha _{1}(z), \ldots ,\alpha _{s}(z)$ are nonconstant polynomials such that $\deg \{\alpha _{i}(z)-\alpha _{j}(z)\}\ge 1$ $(i,j=1,2,\ldots ,s).$ If (8) has a meromorphic solution f with finitely many poles, then f must be of the form:

$$\begin{aligned} f(z)=q(z)e^{P(z)} \end{aligned}$$

where q(z) is a non-vanishing rational function and P(z) is a non-constant polynomial. Furthermore, there must exist positive integers $l_0, l_1,\ldots l_s$ with $\{l_1, l_2,\ldots ls\} = \{1, 2, \ldots s\}$ and distinct integers $j_{1},j_{2},\ldots ,j_{s-1}$ with $0\le j_{1},j_{2}, \ldots , j_{s-1}\le d$ such that $\alpha '_{l_1}:\alpha '_{l_2}:\alpha '_{l_3}:\ldots : \alpha '_{l_s}=n+1:j_{1}:j_{2}:\ldots :j_{s-1},$ $(n+1)P'=\alpha '_{l_{1}},$ $L_d(z,f)=\sum ^{s}_{i=2}p_{l_i}(z)e^{\alpha _{l_i}{(z)}}.$

Remark 1.5

Theorem 1.2 is also an extension of [18], in which Wang and Wang discuss the expressions of Eq. (8) when $\alpha _1(z),\ldots ,\alpha _s(z)$ are polynomials with degree 1.

2 Preliminary Lemmas

In order to prove our results, we fist give some lemmas as follows.

Lemma 2.1

[19] Let f(z) be a transcendental meromorphic function with finite order $\rho ,$ and P(z, f), Q(z, f) be two differential-difference polynomials of f(z). If

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

holds and if the total degree of Q(z, f) in f(z) and its derivatives and their shifts is at most n,  then for any $\varepsilon >0,$

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

possible outside of an exception set of finite logarithmic measure.

Lemma 2.2

[1] Let $f_{k}(z)(1 \le k \le n)$ be meromorphic functions and $g_{k}(z)(1 \le k \le n)$ be entire functions satisfying

• (i) $\sum ^{n}_{k=1}f_{k}e^{g_{k}}\equiv 0;$

• (ii) $g_{k}-g_{l}$ are not constants for $1 \le k<l\le n;$

• (iii) $T(r, f_{j}) = o(T(r, e^{g_{k}-g_{l}} ))$ when $r\rightarrow \infty ,$ $r \notin E$ for $1 \le j \le n$ and $1 \le k<l \le n,$

where $E\subset [1,\infty )$ is a set of finite logarithmic measure.

Then $f_{j} (z) \equiv 0$ for $1 \le j \le n.$

Lemma 2.3

[11] Let f(z) be an entire function, and $a_{1}, a_{2}, \ldots$ denote all nonzero zeros of f(z) repeated according to multiplicity, suppose also that f(z) has a zero at $z = 0$ of multiplicity $m \ge 0.$ Then there exists an entire function g(z) and a sequence of nonnegative integers $p_{1}, p_{2}$ such that

$$\begin{aligned} f(z) = E(z)e^{g(z)} \end{aligned}$$

where $E(z) = z^{m}\prod ^{\infty }_{n=1}E_{pn}\frac{z}{a_{n}}$ is the canonical product formed by the zeros of f(z),  and $E_{n}(z)$ is given by $E_{0}(z)=1-z;$ $E_{n}(z) = (1-z) exp(z + z^{2}/2 +\cdots + z^{n}/n), n\ge 1.$ A well known fact about Lemma 2.3asserts that $\sigma (E) = \lambda (f)\le \sigma (f),$ and $\sigma (f) = \sigma (e^{g})$ when $\lambda (f) < \sigma (f).$

Lemma 2.4

[8] Let f be a non-constant meromorphic function on ${\mathbb{C}}$ and $k\ge 1.$ Then

$$\begin{aligned} m\left( r,\frac{f^{(k)}}{f}\right) =S(r,f) \end{aligned}$$

holds outside of a set of finite logarithmic measure.

Lemma 2.5

[9] The elementary symmetric function $E_{i}\equiv \sum \alpha _{1}\alpha _{2}\ldots \alpha _{i}$ of the n variables $\alpha _{1}, \alpha _{2},\ldots ,\alpha _{n}$ is equal to the quotient of the secondary Vandermondian $V_{n_i}$ by the principal Vandermondian $V_{n_0}.$

Lemma 2.6

[15] Consider the system of linear equation $AX=B,$ where

$$\begin{aligned} A= & {} \left( \begin{array}{cccc} a_{11} &{} a_{12} &{} \ldots &{} a_{1n} \\ a_{21} &{} a_{22} &{} \ldots &{} a_{2n} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ a_{n1} &{} a_{n2} &{} \ldots &{} a_{nn} \end{array}\right) ,\\ X= & {} \left( \begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array}\right) \end{aligned}$$

and

$$\begin{aligned} B=\left( \begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{n} \end{array}\right) \end{aligned}$$

If $\det (A)\ne 0,$ then the system has unique solution

$$\begin{aligned} (x_1,x_2,\ldots ,x_n)=\left( \frac{detA_1}{detA},\frac{detA_2}{detA},\ldots , \frac{detA_n}{detA}\right) , \end{aligned}$$

where

$$\begin{aligned} A_i=\left( \begin{array}{ccccccc} a_{11} &{} \cdots &{} a_{1,i-1}&{}b_1 &{} a_{1,i+1} &{} \cdots &{} a_{1n} \\ \vdots &{} \ddots &{} \vdots &{} \vdots &{} \ddots &{} \vdots &{} \\ a_{n1} &{} \cdots &{} a_{n,i-1}&{}b_n &{} a_{n,i+1} &{} \cdots &{} a_{nn} \end{array}\right) . \end{aligned}$$

Lemma 2.7

[20] Let f(z) be a nonconstant meromorphic function in the complex plane and k be a positive integer. Set

$$\begin{aligned} \Psi (z)=\sum ^{n}_{k=0}{a_{k}(z)f^{(k)}(z)}, \end{aligned}$$

where $a_{k}(z)(k=0,1,\ldots ,n)$ are all small functions of f(z). Then

$$\begin{aligned} T(r,\Psi )\le & {} T(r,f) +k{\overline{N}}(r,f)+S(r,f)\\\le & {} (k+1)T(r,f)+S(r,f),\\ N\left( r,\frac{1}{\Psi }\right)\le & {} N\left( r,\frac{1}{f}\right) +k{\overline{N}}(r,f)+S(r,f). \end{aligned}$$

Lemma 2.8

[5] Let f(z) be a nonconstant meromorphic function of finite order and $c\in {\mathbb{C}}.$ Then

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

and

$$\begin{aligned} {\overline{N}}(r,f(z+c))={\overline{N}}(r,f)+S(r,f), \quad {\overline{N}}\left( r,\frac{1}{f(z+c)}\right) ={\overline{N}} \left( r,\frac{1}{f(z)}\right) +S(r,f). \end{aligned}$$

3 Proof of Theorem 1.1

Suppose that Eq. (6) has an entire solution f(z) with a finite Borel exceptional value. We assert that f(z) is transcendental. If f(z) is a polynomial, it’s obvious that the order of growth of the left side is 0,  while the order of growth of the right side of Eq. (6) is 1. This is impossible. Thus it follows that f is transcendental, as asserted. We will distinguish two cases to deduce contradictions.

Case 1. $s=1.$

Then Eq. (6) becomes

$$\begin{aligned} f^{n}(z)+Q(z,f)=p_{1}e^{\alpha _{1}z}. \end{aligned}$$

(9)

Differentiating both sides of (9), we get

$$\begin{aligned} nf^{n-1}f'+Q'(z,f)=\alpha _{1}p_{1}e^{\alpha _{1}z}. \end{aligned}$$

(10)

Eliminating $e^{\alpha _1z}$ from (9) and (10), we can get

$$\begin{aligned} f^{n-1}(nf'-\alpha _{1}f)=\alpha _{1}Q(z,f)-Q'(z,f). \end{aligned}$$

(11)

If $nf'-\alpha _{1} f\not \equiv 0,$ note that $n\ge 2+s=3,$ then from (11) and Lemma 2.1, we obtain

$$\begin{aligned}{} & {} T(r,nf'-\alpha _{1}f)=m(r,nf'-\alpha _{1}f)=S(r,f). \end{aligned}$$

(12)

$$\begin{aligned}{} & {} T(r,f(nf'-\alpha _{1}f))=m(r,f(nf'-\alpha _{1}f))=S(r,f). \end{aligned}$$

(13)

Combining (12) and (13), we get

$$\begin{aligned} T(r,f)\le T(r,f(nf'-\alpha _{1}f))+T\left( r,\frac{1}{nf'-\alpha _{1}f}\right) =S(r,f). \end{aligned}$$

It is a contradiction.

If $nf'-\alpha _{1} f\equiv 0,$ then $f(z)=c_{1}e^{\frac{\alpha _{1}}{n}z},$ where $c_{1}$ is a constant.

From (11) we can deduce that $\alpha _{1}Q(z,f)-Q'(z,f)=0.$ Then

$$\begin{aligned} Q(z,f)=c_{2}e^{\alpha _{1}z}, \end{aligned}$$

where $c_{2}$ is a constant.

Substituting $f(z)=c_{1}e^{\frac{\alpha _{1}}{n}z}$ and $Q(z,f)=c_{2}e^{\alpha _{1}z}$ into (9), we can get

$$\begin{aligned} {c_{1}}^{n}e^{\alpha _{1}z}+c_{2}e^{\alpha _{1}z}-p_{1}e^{\alpha _{1}z}=0. \end{aligned}$$

By Lemma 2.2, we can deduce that

$$\begin{aligned} c_{1}^{n}+c_{2} -p_{1}=0. \end{aligned}$$

(14)

In another way, substituting $f(z)=c_{1}e^{\frac{\alpha _{1}}{n}z}$ into Eq. (9), we get

$$\begin{aligned} (c_{1}^{n}-p_1)e^{\alpha _{1}z}+\sum _{j=1}^{p}c_{1}a_{j}(z) \left( \frac{\alpha _{1}}{n}\right) ^{m_{j}}e^{\frac{\alpha _{1}}{n}(z+\tau _{j})}+a_0(z)=0. \end{aligned}$$

By Lemma 2.2, we can get $c_{1}^{n}-p_{1}=0.$ Combining (14) we have $c_{2} =0,$ that is $Q(z,f)\equiv 0,$ a contradiction.

Therefore, Eq. (6) does not have transcendental entire solution when $s=1.$

Case 2. $s>1.$ For convenience, let

$$\begin{aligned} h=f^{n}(z)+Q(z,f). \end{aligned}$$

(15)

Differentiating Eq. (15) $s-1$ times yield the system of equations

$$\begin{aligned} \left\{ \begin{array}{l} h=p_{1}e^{\alpha _{1}z}+p_{2}e^{\alpha _{2}z}+\cdots +p_{s}e^{\alpha _{s}z},\\ h'=\alpha _{1}p_{1}e^{\alpha _{1}z}+\alpha _{2}p_{2}e^{\alpha _{2}z}+\cdots + \alpha _{s}p_{s}e^{\alpha _{s}z},\\ h''=\alpha _{1}^{2}p_{1}e^{\alpha _{1}z}+\alpha _{2}^{2}p_{2}e^{\alpha _{2}z}+ \cdots +\alpha _{s}^{2}p_{s}e^{\alpha _{s}z},\\ \cdots \\ h^{(s-1)}=\alpha _{1}^{s-1}p_{1}e^{\alpha _{1}z}+\alpha _{2}^{s-1}p_{2}e^{\alpha _{2}z}+\cdots +\alpha _{s}^{s-1}p_{s}e^{\alpha _{s}z}. \end{array}\right. \end{aligned}$$

(16)

By Lemma 2.6, we obtain

$$\begin{aligned} p_{1}e^{\alpha _{1}z}=\frac{D_{1}}{D}, \end{aligned}$$

(17)

where

$$\begin{aligned} D=\left| \begin{array}{cccc} 1&{}1&{} \cdots &{}1\\ \alpha _{1}&{}\alpha _{2}&{}\cdots &{}\alpha _{s}\\ \alpha _{1}^{2}&{}\alpha _{2}^{2}&{}\cdots &{}\alpha _{s}^{2}\\ \vdots &{} \vdots &{} \ddots &{}\vdots \\ \alpha _{1}^{s-1}&{}\alpha _{2}^{s-1}&{}\cdots &{}\alpha _{s}^{s-1} \end{array}\right| \end{aligned}$$

(18)

is a principal Vandermondian with order s, 

$$\begin{aligned} D_{1}=\left| \begin{array}{cccc} h&{}1&{} \cdots &{}1\\ h'&{}\alpha _{2}&{}\cdots &{}\alpha _{s}\\ \vdots &{} \vdots &{} \ddots &{}\vdots \\ h^{(s-1)}&{}\alpha _{2}^{s-1}&{}\cdots &{}\alpha _{s}^{s-1} \end{array}\right| . \end{aligned}$$

(19)

Noting that $\alpha _{1},\alpha _{2},\ldots \alpha _{s}$ are distinct constants, hence,

$$\begin{aligned} D=\prod _{1\le j\le i\le s}(\alpha _{i}-\alpha _{j})\not \equiv 0. \end{aligned}$$

By (17), (18)and (19), we obtain

$$\begin{aligned}{} & {} p_{1}e^{\alpha _{1}z}=\frac{D_1}{D}=\frac{1}{D}((-1)^{s+1}M_{s1}h^{(s-1)} \\{} & {} \quad +\cdots +(-1)^{s-j+1}M_{s-j,1}h^{(s-j-1)}+\cdots +M_{11}h), \end{aligned}$$

(20)

where

$$\begin{aligned} M_{s1}= & {} \left| \begin{array}{cccc} 1&{}1&{} \cdots &{}1\\ \alpha _{2}&{}\alpha _{3}&{}\cdots &{}\alpha _{s}\\ \alpha _{2}^{2}&{}\alpha _{3}^{2}&{}\cdots &{}\alpha _{s}^{2}\\ \vdots &{} \vdots &{} \ddots &{}\vdots \\ \alpha _{2}^{s-2}&{}\alpha _{3}^{s-2}&{}\cdots &{}\alpha _{s}^{s-2} \end{array}\right| =\prod _{2\le j\le i\le s}(\alpha _{i}-\alpha _{j})\not \equiv 0, \end{aligned}$$

(21)

$$\begin{aligned} M_{s-j,1}= & {} \left| \begin{array}{cccc} 1&{}1&{} \cdots &{} 1\\ \alpha _{2}&{}\alpha _{3}&{}\cdots &{}\alpha _{s}\\ \vdots &{} \vdots &{} \ddots &{}\vdots \\ \alpha _{2}^{s-j-2}&{}\alpha _{3}^{s-j-2}&{}\cdots &{}\alpha _{s}^{s-j-2}\\ \alpha _{2}^{s-j}&{}\alpha _{3}^{s-j}&{}\cdots &{}\alpha _{s}^{s-j}\\ \vdots &{} \vdots &{} \ddots &{}\vdots \\ \alpha _{2}^{s-1}&{}\alpha _{3}^{s-1}&{}\cdots &{}\alpha _{s}^{s-1} \end{array}\right| ,\quad (j=1,\ldots ,s-1). \end{aligned}$$

(22)

Differentiating both sides of (20), we get

$$\begin{aligned}{} & {} \alpha _{1}p_1e^{\alpha _{1}z}=\frac{D_1'}{D}=\frac{1}{D}((-1)^{s+1}M_{s1}h^{(s)} \\{} & {} \quad +\cdots +(-1)^{s-j+1}M_{s-j,1}h^{(s-j)}+\cdots +M_{11}h'). \end{aligned}$$

(23)

By eliminating $e^{\alpha _{1}z}$ from (20) and (23), we get

$$\begin{aligned}{} & {} ((-1)^{s+1}M_{s1}h^{(s)}+((-1)^{s}M_{s-1,1}-(-1)^{s+1}\alpha _{1}M_{s1})h^{(s-1)} \\{} & {} \quad +\cdots +((-1)^{s-j+1}M_{s-j,1}-(-1)^{s-j+2}\alpha _{1}M_{s-j+1,1})h^{(s-j)} \\{} & {} \quad +\cdots -\alpha _{1}M_{11}h)=0. \end{aligned}$$

(24)

Let H(w) be a linear differential operator defined by

$$\begin{aligned}{} & {} H(w)=w^{(s)}+\frac{(-1)^{s}M_{s-1,1}-(-1)^{s+1}\alpha _{1}M_{s1}}{(-1)^{s+1}M_{s1}}w^{(s-1)}+\cdots \\{} & {} \frac{(-1)^{s-j+1}M_{s-j,1}-(-1)^{s-j+2}\alpha _{1}M_{s-j+1,1}}{(-1)^{s+1}M_{s1}}w^{(s-j)} \\{} & {} \quad +\cdots -\frac{\alpha _{1}M_{11}}{(-1)^{s+1}M_{s1}}w. \end{aligned}$$

(25)

We deduce from (21), (22), (25) and Lemma 2.5 that

$$\begin{aligned}{} & {} \frac{(-1)^{s}M_{s-1,1}-(-1)^{s+1}\alpha _{1}M_{s1}}{(-1)^{s+1}M_{s1}} =-(\alpha _{1}+\alpha _{2}+\cdots +\alpha _{s}), \end{aligned}$$

(26)

$$\begin{aligned}{} & {} \frac{(-1)^{s-j+1}M_{s-j,1}-(-1)^{s-j+2}\alpha _{1}M_{s-j+1,1}}{(-1)^{s+1}M_{s1}}=(-1)^{j}\sum \alpha _{2}\alpha _{3}\cdots \alpha _{j+1} \\{} & {} \quad +(-1)^{j}\alpha _{1}\sum \alpha _{2}\alpha _{3}\cdots \alpha _{j}=(-1)^{j}\sum \alpha _{1}\alpha _{2}\alpha _{3}\cdots \alpha _{j},\quad (j=2,3,\ldots ,s-1), \end{aligned}$$

(27)

and

$$\begin{aligned} -\frac{\alpha _{1}M_{11}}{(-1)^{s+1}M_{s1}}=(-1)^{s}(\alpha _{1}\alpha _{2}\cdots \alpha _{s}). \end{aligned}$$

(28)

Denote

$$\begin{aligned} \tau _{j}\equiv \sum \alpha _{1}\alpha _{2}\cdots \alpha _{j},\quad (j=1, 2,\ldots ,s-1). \end{aligned}$$

(29)

Here, $\tau _{j}$ $(j=1, 2,\ldots ,s-1)$ are the elementary symmetric function of s variables $\alpha _{1}, \alpha _{2},\ldots , \alpha _{s}.$ So H(w) can be rewritten as

$$\begin{aligned} H(w)=w^{(s)}-\tau _{1}w^{(s-1)}+\cdots (-1)^{j}\tau _{j}w^{(s-1)}+\cdots (-1)^{s}\tau _{s}w. \end{aligned}$$

(30)

Since $h=f^{n}(z)+Q(z,f),$ it follows from (24) that

$$\begin{aligned} H(f^{n}(z))=-H(Q(z,f)). \end{aligned}$$

(31)

In addition, for $m=1,2,\ldots ,s,$

$$\begin{aligned}{} & {} (f^{n}(z))^{(m)}=n(n-1)\cdots (n-(m-1))f(z)^{n-m}(f'(z))^{m} \\{} & {} \quad +\sum _{j=2}^{m-1}\sum _{\lambda }\gamma _{j\lambda }f(z)^{n-j}(f'(z))^{\lambda _{j1}}(f''(z))^{\lambda _{j2}}\cdots (f^{(m-1)}(z))^{\lambda _{j,m-1}} \\{} & {} \quad +nf(z)^{n-1}f^{(m)}(z), \end{aligned}$$

(32)

where $\gamma _{j\lambda }$ are the positive integers, $\lambda _{j1},\lambda _{j2},\ldots \lambda _{jm-1}$ are the non-negative integers and the sum $\sum _{\lambda }$ is carried out such that $\lambda _{j1}+\lambda _{j2}+\cdots +\lambda _{j,m-1}=j$ and $\lambda _{j1}+2\lambda _{j2}+\cdots +(m-1)\lambda _{j,m-1}=m.$

By (25), (30), (31) and (32), we get

$$\begin{aligned} H(f^{n}(z))=f^{n-s}(z)\Phi =-H(Q(z,f)), \end{aligned}$$

(33)

where $\Phi$ is a differential polynomial in f(z) of degree s with constant coefficients and H(Q(z, f)) is a differential-difference polynomial in f(z) of degree 1 with small function coefficients of f.

If $\Phi =0,$ which yields $H(f^{n}(z))\equiv 0$ and $H(Q(z,f))\equiv 0.$

By $H(Q(z,f))\equiv 0,$ we get

$$\begin{aligned}{} & {} (Q(z,f))^{(s)}-\tau _{1}(Q(z,f))^{(s-1)}+\cdots (-1)^{j}\tau _{j}(Q(z,f))^{(s-j)}\\{} & {} \quad +\cdots +(-1)^{s}\tau _{s}(Q(z,f))\equiv 0. \end{aligned}$$

The characteristic equation for this equation is as follows:

$$\begin{aligned} \lambda ^{s}-\tau _{1}\lambda ^{s-1}+\cdots +(-1)^{j}\tau _{j}\lambda ^{s-j}+\cdots (-1)^{s}\tau _{s}=0. \end{aligned}$$

(34)

Due to the fact that $\alpha _{1},\alpha _{2},\ldots ,\alpha _{s}$ are s distinct roots of (34), we can obtain that

$$\begin{aligned} Q(z,f)=\tilde{c_{1}}e^{\alpha _{1}z}+\tilde{c_{2}}e^{\alpha _{2}z}+\cdots +\tilde{c_{s}}e^{\alpha _{s}z}, \end{aligned}$$

where $\tilde{c_{j}}(j=1,2\cdots ,s)$ are constants. Accordingly, we deduce from $H(f^{n}(z))\equiv 0$ that

$$\begin{aligned} f^{n}(z)=d_{1}e^{\alpha _{1}z}+d_{2}e^{\alpha _{2}z}+\cdots +d_{s}e^{\alpha _{s}z}, \end{aligned}$$

(35)

where $d_{j}(j=1,2\ldots ,s)$ are constants.

Since f(z) is a finite order transcendental entire function with a finite Borel exceptional value a,  using Weierstrass factorisation theorem, we can factorize f(z) as

$$\begin{aligned} f(z)=d(z)e^{g(z)}+a, \end{aligned}$$

(36)

where d(z) is the canonical product of $f-a$ formed with it’s zeros and d(z) is an entire function such that $\sigma (d(z))<\sigma (f(z)).$ We can get that g(z) is a polynomial of degree $\ge 1.$ By (36), we see that

$$\begin{aligned} f^{(m_{j})}(z)=G_{j}(z)e^{g(z)},\quad j=1,2,\ldots ,p, \end{aligned}$$

(37)

where $G_{1}(z),G_{2}(z),\ldots ,G_{p}(z)$ are polynomials in d(z), g(z) and their derivatives. Clearly, $T(r,G_{j})=o(T(r,f))$ and $G_{j}\ne 0$ for $j=1,2,3\ldots ,p.$ Then we can get

$$\begin{aligned} Q(z,f)=\sum _{j=1}^{p}A_{j}(z)e^{g(z+\tau _{j})}+a_0(z)= \sum _{k=1}^{s}\tilde{c_{k}}e^{\alpha _{k}z}, \end{aligned}$$

(38)

where $A_{j}=a_{j}(z)G_{j}(z+\tau _{j}).$ Since $T(r,a_{j}(z))=o(T(r,f))$ $(1\le j\le p),$ we obtain $T(r,A_{j})=o(T(r,f)), j=1,2\ldots p.$

Now we claim that g(z) is a polynomial of degree 1.

Otherwise, we suppose that g(z) is a polynomial of degree $\ge 2.$ Denote $g(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots +a_{0}$ with $a_{n}\ne 0,$ then we can rewrite (38) as

$$\begin{aligned} e^{a_{n}z^{n}}\sum _{j=1}^{p}\tilde{A_{j}}(z)+a_0(z)-\sum _{k=1}^{s}\tilde{c_{k}}e^{\alpha _{k}z}=0, \end{aligned}$$

(39)

where $\tilde{A_{j}}(z)=A_{j}(z)e^{g(z+\tau _{j})-a_{n}z^{n}},$ $\sigma (\tilde{A_{j}}(z))\le n-1$ and $T(r,\tilde{A_{j}}(z))= o(T(e^{a_{n}z^{n}-\alpha _{k}z})).$ Applying Lemma 2.2 to (39), it follows that $a_0(z)\equiv 0,$ which is a contradiction .

Then we can get

$$\begin{aligned} f=d(z)e^{\beta z}+a, \end{aligned}$$

(40)

where $\beta$ is a nonzero constant, and hence

$$\begin{aligned} f^{(m_{j})}(z+\tau _{j})=e^{\beta \tau _{j}}e^{\beta z}\sum _{i=0}^{m_{j}}\frac{m_{j}!\beta ^{m_{j}-i}}{i!(m_{j}-i)!}d^{(i)}(z+\tau _{j}). \end{aligned}$$

Therefore, we can rewrite (38) as

$$\begin{aligned} Q(z,f)=e^{\beta z}\sum _{j=1}^{p}e^{\beta \tau _{j}}a_{j}(z) \sum _{i=0}^{m_{j}}\frac{m_{j}!\beta ^{m_{j}-i}}{i!(m_{j}-i)!}d^{(i)}(z+\tau _{j})+a_0(z)=\sum _{k=1}^{s}\tilde{c_{k}}e^{\alpha _{k}z}. \end{aligned}$$

(41)

If $\beta \ne \alpha _{k}(k=1,2,\ldots ,s),$ then by Lemma 2.2, we can get

$$\begin{aligned} R(z)=\tilde{c_{k}}=0(k=1,2,\ldots ,s), \end{aligned}$$

where

$$\begin{aligned} R(z)=\sum _{j=1}^{p}e^{\beta \tau _{j}}a_{j}(z)\sum _{i=0}^{m_{j}}\frac{m_{j}!\beta ^{m_{j}-i}}{i!m_{j}-i}d^{(i)}(z+\tau _{j}), \end{aligned}$$

It is a contradiction.

On the other hand, if $\beta =\alpha _{r}$ for some $r\in \{1,2,\ldots ,s\},$ then by (40) we can get

$$\begin{aligned} f^{n}(z)=d^{n}(z)e^{n\beta z}+C_{n}^{1}ad^{n-1}(z)e^{(n-1)\beta z}+\cdots +C_{n}^{n-1}a^{n-1}d(z)e^{\beta z}+a^{n}. \end{aligned}$$

From (35) it finally concludes that

$$\begin{aligned}{} & {} d^{n}(z)e^{n\beta z}+C_{n}^{1}ad^{n-1}(z)e^{(n-1)\beta z}+\cdots +C_{n}^{n-1}a^{n-1}d(z)e^{\beta z}+a^{n} \\{} & {} \quad =d_{1}e^{\alpha _{1}z}+d_{2}e^{\alpha _{2}z}+\cdots +d_{s}e^{\alpha _{s}z}. \end{aligned}$$

(42)

We note that $\frac{\alpha _{i}}{\alpha _{j}}\ne n,$ so $n\alpha _{r}\ne \alpha _{k}(k=1,\ldots ,s).$ By Lemma 2.2, it yields that $d^{n}(z)=0.$

Then $f(z)\equiv a,$ a contradiction.

If $\Phi \ne 0,$ note that $n\ge s+2,$ then by (33) and Lemma 2.2 we get

$$\begin{aligned} T(r,\Phi )=m(r,\Phi )=S(r,\Phi ),\quad T(r,f\Phi )=m(r,f\Phi )=S(r,f). \end{aligned}$$

(43)

Then, we have

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

a contradiction.

4 Proof of Theorem 1.2

Let

$$\begin{aligned} h(z)=f^nf^{(k)}+L_{d}(z,f). \end{aligned}$$

(44)

Suppose that f is a rational solution of Eq. (6). Obviously, h(z) is a small function of $p_{i}(z)e^{\alpha _{i}(z)} (i=1,2,\ldots ,s).$ Since deg$\{\alpha _{i}(z)-\alpha _{j}(z)\}\geqslant 1,$ by Lemma 2.2 we deduce that $p_{i}(z)=0 (i=1,2,\ldots ,s),$ which contradicts $p_{i}(z)(i=1,2,\ldots ,s)$ are non-vanishing rational functions. Hence f must be transcendental.

Suppose that f is a meromorphic solution with finitely poles of Eq. (6). Now we prove that f has finite order. Clearly,

$$\begin{aligned}{} & {} (n+1)T(r,f)=T(r,f^{n+1})=T\left( r,\frac{1}{f^{n+1}}\right) +S(r,f) \\{} & {} \quad \le m\left( r,\frac{1}{f^nf^{(k)}}\right) +m\left( r,\frac{f^{(k)}}{f} \right) +N\left( r,\frac{1}{f^nf^{(k)}}\right) \\{} & {} \qquad +N\left( r,\frac{1}{f}\right) +S(r,f) \\{} & {} \quad \le T(r,f^nf^{(k)})+N\left( r,\frac{1}{f}\right) +S(r,f) \\{} & {} \quad =m(r,f^nf^{(k)})+N\left( r,\frac{1}{f}\right) +S(r,f) \\{} & {} \quad \le T\left( r,\sum ^s_{i=1}p_i(z)e^{\alpha _i(z)}\right) +m(r,L_d(z,f))+S(r,f) +N\left( r,\frac{1}{f}\right) \\{} & {} \quad \le Ar^{\eta }+(d+1)T(r,f)+S(r,f), \end{aligned}$$

(45)

where

$$\begin{aligned} A=\frac{sum\ of\ the\ leading\ coefficients\ of\ \alpha _i(z)}{\pi } \end{aligned}$$

and

$$\begin{aligned} \eta =\max \{\deg \alpha _i(z)\} \quad (i=1,2,\ldots ,s). \end{aligned}$$

Hence $(n-d)T(r,f)\le A r^{\eta }+S(r,f)$ and f is of finite order.

Differentiating Eq. (44) $s-1$ $(s\ge 3)$ times yields the system of differential equations

$$\begin{aligned} \left\{ \begin{array}{l} h=p_1(z)e^{\alpha _1(z)}+p_2(z)e^{\alpha _2(z)}+\cdots +p_s(z)e^{\alpha _s(z)},\\ h'=({p_1}^{\prime }(z)+p_1(z){\alpha _1}^{\prime }(z))e^{\alpha _1(z)}+({p_2}^{\prime }(z) +p_2(z){\alpha _2}^{\prime }(z))e^{\alpha _2(z)}+\cdots \\ \quad +({p_s}^{\prime }(z)+p_s(z){\alpha _s}^{\prime }(z))e^{\alpha _s(z)},\\ h''=\sum _{i=1}^{s}({p_i}^{\prime \prime }(z)+2{p_i}^{\prime }(z){\alpha _i}^{\prime }(z) +p_i(z){\alpha _i}^{\prime \prime }(z)+p_i(z)({{\alpha _i}^{\prime }(z)})^2)e^{\alpha _i(z)},\\ \cdots \\ h^{(s-1)}=\sum _{i=1}^{s}[p_i^{(s-1)}(z)+ p_i^{(s-2)}(z)H_1(\alpha _i'(z))+p_i^{(s-3)}(z)H_2(\alpha _i'(z), \alpha _i''(z))+\cdots \\ \quad +p_i(z)H_{s-1}(\alpha _i'(z),\alpha _i''(z),\ldots ,\alpha _i^{(s-1)}(z))]e^{\alpha _i(z)}, \end{array}\right. \end{aligned}$$

(46)

where $H_{j}(\alpha _i',\alpha _i'',\ldots ,\alpha _i^{(j)})$ are differential polynomials of $\alpha _i^{(j)}$ with degree j $(j=1,2,\ldots ,s-1; i=1,2,\ldots ,s).$ By Lemma 2.6, we obtain

$$\begin{aligned} De^{\alpha _1(z)}=D_1, \end{aligned}$$

(47)

where

$$\begin{aligned} D= & {} \left| \begin{array}{ccc} p_1&{} \cdots &{}p_s \\ {p_1}^{\prime }+p_1{\alpha _1}^{\prime }&{}\cdots &{}{p_s}^{\prime }+p_s{\alpha _s}^{\prime } \\ \vdots &{}\ddots &{}\vdots \\ \begin{aligned}&{}p_1^{(s-1)}+p_1^{(s-2)}L_1(\alpha _1^{\prime })+\cdots \\ {} &{}p_1L_{s-1}(\alpha _1^{\prime },\alpha _1^{\prime \prime },\ldots ,\alpha _1^{(s-1)}) \\ \end{aligned} &{}\cdots &{}\begin{aligned} &{}p_s^{(s-1)}+p_s^{(s-2)}L_1(\alpha _s^{\prime })+\cdots \\ {} &{}+p_sL_{s-1}(\alpha _s',\alpha _s'',\ldots ,{\alpha _s}^{(s-1)})\\ \end{aligned} \end{array}\right| , \end{aligned}$$

(48)

$$\begin{aligned} D_1= & {} \left| \begin{array}{cccc} h&{}p_2&{}\cdots &{}p_s \\ h'&{}{p_2}^{\prime }+p_2{\alpha _2}^{\prime }&{}\cdots &{}{p_s}^{\prime }+p_s{\alpha _s}^{\prime }\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ h^{(s-1)}&{} \begin{aligned}&{}p_2^{(s-1)}+p_2^{(s-2)}L_1(\alpha _2^{\prime })+\cdots \\ {} &{}+p_2L_{s-1}(\alpha _2^{\prime },\alpha _2^{\prime \prime },\ldots ,\alpha _2^{(s-1)})\\ \end{aligned}&{}\cdots &{} \begin{aligned}&{}p_s^{(s-1)}+p_s^{(s-2)}L_1(\alpha _s^{\prime })+\cdots \\ {} &{}+p_sL_{s-1}(\alpha _s',\alpha _s'',\ldots ,{\alpha _s}^{(s-1)})\\ \end{aligned} \end{array}\right| . \end{aligned}$$

(49)

We will distinguish the following two cases:

Case 1. If $D\equiv 0,$ then by (47) we have $D_1\equiv 0.$ So,

$$\begin{aligned} M_{11}h-M_{21}h'+\cdots +(-1)^{s-1}M_{s1}h^{(s-1)}\equiv 0 , \end{aligned}$$

(50)

where

$$\begin{aligned} M_{11}= & {} \left| \begin{array}{ccc} {p_2}^{\prime }+p_2{\alpha _1}^{\prime }&{}\cdots &{}{p_s}^{\prime }+p_s{\alpha _s}^{\prime } \\ \vdots &{}\ddots &{}\vdots \\ \begin{aligned}&{}p_2^{(s-1)}+p_2^{(s-2)}L_1(\alpha _2^{\prime })+\cdots \\ {} &{}+p_2L_{s-1}(\alpha _2^{\prime },\alpha _2^{\prime \prime },\ldots ,\alpha _2^{(s-1)})\\ \end{aligned}&{}\cdots &{} \begin{aligned}&{}p_s^{(s-1)}+p_s^{(s-2)}L_1(\alpha _s^{\prime })+\cdots \\ {} &{}+p_sL_{s-1}(\alpha _s',\alpha _s'',\ldots ,{\alpha _s}^{(s-1)})\\ \end{aligned} \end{array}\right| ,\\ M_{21}= & {} \left| \begin{array}{ccc} p_2&{}\cdots &{}p_s\\ {p_2}''+2p_2'{\alpha _2}^{\prime }+p_2\alpha _2''+p_2(\alpha _2')^2&{}\cdots &{}{p_s}''+2p_s'{\alpha _s}^{\prime }+p_s\alpha _s''+p_s(\alpha _s')^2 \\ \vdots &{}\ddots &{}\vdots \\ \begin{aligned}&{}p_2^{(s-1)}+p_2^{(s-2)}L_1(\alpha _2^{\prime })+\cdots \\ {} &{}+p_2L_{s-1}(\alpha _2^{\prime },\alpha _2^{\prime \prime },\ldots ,\alpha _2^{(s-1)})\\ \end{aligned}&{}\cdots &{} \begin{aligned}&{}p_s^{(s-1)}+p_s^{(s-2)}L_1(\alpha _s^{\prime })+\cdots \\ {} &{}+p_sL_{s-1}(\alpha _s',\alpha _s'',\ldots ,{\alpha _s}^{(s-1)})\\ \end{aligned} \end{array}\right| \end{aligned}$$

and

$$\begin{aligned} M_{s1}= & {} \left| \begin{array}{ccc} p_2 &{}\cdots &{} p_s \\ {p_2}^{\prime }+p_2{\alpha _1}^{\prime }&{}\cdots &{}{p_s}^{\prime }+p_s{\alpha _s}^{\prime } \\ \vdots &{}\ddots &{}\vdots \\ p_2^{(s-2)}+p_2^{(s-3)}L_1(\alpha _2^{\prime }) &{} &{} p_s^{(s-2)}+p_s^{(s-3)}L_1(\alpha _{s}^{\prime })+\cdots \\ +\cdots p_2L_{s-2}(\alpha _2^{\prime },\alpha _2^{\prime \prime },\ldots ,\alpha _2^{(s-2)})&{}\cdots &{}+p_sL_{s-2}(\alpha _s',\alpha _s'',\ldots ,{\alpha _s}^{(s-2)}) \end{array}\right| \end{aligned}$$

are rational functions.

Substituting the expression (44) of h(z) and $h'(z)$ $h''(z)\cdots$ $h^{(s-1)}$ into (50), we get

$$\begin{aligned} M_{11}f^nf^{(k)}-M_{21}(f^nf^{(k)})'+\cdots +(-1)^{s-1}M_{s1}(f^nf^{(k)})^{(s-1)}=H_{1}(z), \end{aligned}$$

(51)

where

$$\begin{aligned} H_1(z)=-[M_{11}L_d(z,f)-M_{21}L_d'(z,f)+\cdots +(-1)^{s-1}M_{s1}L_d^{(s-1)}(z,f)] \end{aligned}$$

is a differential polynomial in f with rational functions as its coefficients and degree of $H_1\le d.$ In addition,

$$\begin{aligned}{} & {} (f(z)^{n})^{(m)}=n(n-1)\cdots (n-(m-1))f(z)^{n-m}(f'(z))^{m} \\{} & {} \quad +\sum _{j=2}^{m-1}\sum _{\lambda }\gamma _{j\lambda }f(z)^{n-j} (f'(z))^{\lambda _{j1}}(f''(z))^{\lambda _{j2}}\cdots (f^{(m-1)}(z))^{\lambda _{j,m-1}} \\{} & {} \quad +nf(z)^{n-1}f^{(m)}(z), \end{aligned}$$

(52)

where $\gamma _{j\lambda }$ are the positive integers, $\lambda _{j1},\lambda _{j2},\ldots \lambda _{jm-1}$ are the non-negative integers and the sum $\sum _{\lambda }$ is carried out such that $\lambda _{j1}+\lambda _{j2}+\cdots +\lambda _{j,m-1}=j$ and $\lambda _{j1}+2\lambda _{j2}+\cdots +(m-1)\lambda _{j,m-1}=m.$

Since

$$\begin{aligned} (f^n(z)f(z)^{(k)})'=nf^{n-1}(z)f'(z)f(z)^{(k)}+f^n(z)f(z)^{(k+1)} \end{aligned}$$

and

$$\begin{aligned}{} & {} (f^n(z)f(z)^{(k)})''=nf^{n-1}(z)f(z)''f(z)^{(k)}+2nf^{n-1}(z)f(z)'f(z)^{(k+1)}\\{} & {} \quad +n(n-1)f^{n-2}(z)(f(z)')^{2}f(z)^{(k)}+f^{n}(z)f(z)^{(k+2)}. \end{aligned}$$

We deduce inductively that,

$$\begin{aligned}&(f^nf^{(k)})^{(t)}=\sum ^t_{i=0}\left( {\begin{array}{c}t\\ i\end{array}}\right) (f^n)^{(i)}(f^{(k)})^{(t-i)}\\ &\quad =\sum ^t_{i=1}\left( {\begin{array}{c}t\\ i\end{array}}\right) (f^{(k)})^{(t-i)}\cdot [nf^{n-1}f^{(i)}+\sum _{j=2}^{i-1}\sum _{\lambda }\gamma _{j\lambda }f^{n-j}(f')^{\lambda _{j1}}(f'')^{\lambda _{j2}}\cdots (f^{(i-1)})^{\lambda _{j,i-1}}\\ &\qquad +n(n-1)\cdots (n-(i-1))f^{n-i}(f')^i]+f^nf^{(k+t)} \end{aligned}$$

(53)

for $t=0, 1,\ldots ,s ,$ where $\gamma _{j\lambda }$ are positive integers, $\lambda _{j1},\lambda _{j2},\ldots \lambda _{j,i-1}$ are nonnegative integers, and $\lambda _{j1}+\lambda _{j2}+\cdots +\lambda _{j,i-1}=j$ and $\lambda _{j1}+2\lambda _{j2}+\cdots +(i-1)\lambda _{j,i-1}=i.$

Now we define

$$\begin{aligned} \varphi _t=\frac{(f^nf^{(k)})^{(t)}}{f^{n-s+1}}, \end{aligned}$$

(54)

for $t=1,2,\ldots ,s-1.$

It follows from (54) and (51), we can rewrite (51) as

$$\begin{aligned} f^{n-s+1}R_1=H_1, \end{aligned}$$

(55)

where

$$\begin{aligned} R_1=M_{11}f^{s-1}f^{(k)}-M_{21}\varphi _1+\cdots (-1)^{s-1}M_{s1}\varphi _{s-1}. \end{aligned}$$

(56)

Note that $d\le n-s-1$ and f is finite order, so combining (55) with Lemma 2.1 we get

$$\begin{aligned} m(r,R_1)=O(\log r). \end{aligned}$$

On the other hand, since f has finitely many poles, we have

$$\begin{aligned} T(r,R_1)=m(r,R_1)+N(r,R_1)=O(\log r), \end{aligned}$$

which means that $R_1$ is a rational function.

Next we discuss the following two subcases:

Case 1.1. If $R_1(z)\equiv 0,$ then from (55) and (51), we have

$$\begin{aligned} M_{11}f^{s-1}f^{(k)}=-[-M_{21}\varphi _1+\cdots (-1)^{s-1}M_{s1}\varphi _{s-1}]. \end{aligned}$$

(57)

We will show that f has at most finitely many zeros. Suppose that f has infinitely many zeros. Then there exists a point $z_1$ such that $f(z_1) = 0$ and $z_1$ is not the zero and pole of $M_{j1} (j = 1, 2,\ldots , s).$

Take $z=z_1.$ First, assume that $M_{s1}\not \equiv 0.$ Then by (54) and (53), we know $\psi _1(z_1)=\psi _2(z_1)\cdots =\psi _{s-2}(z_1)=0.$ Then by (57), we have $\psi _{s-1}(z_1)=0.$ Noting that

$$\begin{aligned} \varphi _{s-1}&=\frac{(f^nf^{(k)})^{(s-1)}}{f^{n-s+1}}=\frac{\sum ^{s-1}_{i=0}\left( {\begin{array}{c}s-1\\ i\end{array}}\right) (f^n)^{(i)}(f^{(k)})^{(s-1-i)}}{f^{n-s+1}}\\ &=\frac{\sum ^{s-1}_{i=1}\left( {\begin{array}{c}s-1\\ i\end{array}}\right) (f^{(k)})^{(s-1-i)}}{f^{n-s+1}}\cdot \Bigg [nf^{n-1}f^{(i)}+\sum _{j=2}^{i-1}\sum _{\lambda }\gamma _{j\lambda }f^{n-j}(f')^{\lambda _{j1}}\\ {}&\quad (f'')^{\lambda _{j2}}\cdots (f^{(i-1)})^{\lambda _{j,i-1}} +n(n-1)\cdots (n-(i-1))f^{n-i}(f')^i\Bigg ]+\frac{f^nf^{(k+s-1)}}{f^{n-s+1}}, \end{aligned}$$

(58)

we can deduce that $(f'(z_1))^{s-1}f^{(k)}(z_1) = 0.$ Now we will deduce a contradiction by comparing the zero multiplicity of (57) with respect to $z_1.$ We need to treat three cases:

Subcase 1. $f'(z_{1})=0$ and$f^{(k)}(z_{1})=0.$ Suppose that $z_1$ is a zero of $f^{(k)}(z)$ with multiplicity h $(h\ge 1)$ and $z_1$ is a zero of f with multiplicity m $(m\ge 2).$ Then we can deduce form (58), (52) and (57) that $z_1$ is a zero of the right hand side of (57) with multiplicity $(m-1)(s-1)+h ,$ while the multiplicity of the left hand side of (57) is $(s-1)m+h.$ It is a contradiction.

Subcase 2. $f'(z_{1})=0$ and$f^{(k)}(z_{1})\ne 0.$ Suppose that $z_1$ is a multiple zero of f which multiplicity is m $(k\ge m\ge 2).$ Then we can deduce form (58), (52) and (57) that $z_1$ is a zero of the right hand side of (57) with multiplicity $(m-1)(s-1) ,$ while the multiplicity of the left hand side of (57) is $(s-1)m,$ a contradiction.

Subcase 3.$f'(z_{1})\ne 0$ and$f^{(k)}(z_{1})=0.$ Suppose that $z_1$ is a multiple zero of $f^{(k)}$ which multiplicity is h $(h\ge 1)$ and $z_1$ is a multiple zero of $f'$ which multiplicity is $m=1$ . Then we can deduce form (58), (52) and (57) that $z_1$ is a zero of the right hand side of (57) with multiplicity h,  while the multiplicity of the left hand side of (57) is $s-1+h,$ a contradiction.

If not, assume that $M_{s1} \equiv 0.$ If $z_1$ is a simple zero of f(z),  then $f'(z_1)\ne 0.$ Suppose that $z_1$ is a zero of $f^{(k)}(z)$ whose multiplicity is h $(h\ge 0).$ So we get that $z_1$ is a zero with multiplicity $s-1+h$ of left hand side of (57) and a zero with multiplicity $1+h$ of right hand side of (57). Since $s\ge 3,$ it is a contradiction.

If $z_1$ is a multiple zero of f(z) with multiplicity $m\ge 2,$ then $z_1$ is a zero with multiplicity $(s-1)m+h$ of left hand side of (57) and a zero with multiplicity $(s-2)(m-1)+m+h$ of right hand side of (57), where $h\ge 0$ is the multiplicity of $f^{(k)}(z).$ A contradiction. Thus, f has at most finitely many zeros.

Case 1.2. If $R_1(z)\not \equiv 0,$ then (55) leads to

$$\begin{aligned} f^{n-s}(z)(fR_1)=H_1. \end{aligned}$$

(59)

By Lemma 2.1, we get

$$\begin{aligned} m(r,fR_1)=O(\log r). \end{aligned}$$

Since $fR_1$ has finitely many poles, then we have

$$\begin{aligned} T(r,fR_1)=m(r,fR_1)+N(r,fR_1)=O(\log r). \end{aligned}$$

Therefore, $fR_1$ is a rational function, which contradicts that f is transcendental.

Case 2. If $D\not \equiv 0,$ then differentiating (47), we have

$$\begin{aligned} (D'+D\alpha _1')e^{\alpha _1}=D_1'. \end{aligned}$$

(60)

Eliminating $e^{\alpha _1}$ from (47) and (60), we get

$$\begin{aligned} (D_1'D-D_1D')=\alpha _1'D_1D. \end{aligned}$$

(61)

Substituting $D_1=M_{11}h-M_{21}h'+\cdots +(-1)^{s-1}M_{s1}h^{(s-1)}$ and $D'_1=M_{11}'h+(M_{11}-M_{21}')h'+\cdots +(-1)^{s-2}(M_{s-1 1}-M_{s1}')h^{(s-1)}+(-1)^{(s-1)}M_{s1}h^{(s)}$ into (61), we obtain that

$$\begin{aligned} A_{1}h+A_{2}h'+\cdots +A_{s+1}h^{(s)}=0, \end{aligned}$$

(62)

where

$$\begin{aligned} \left\{ \begin{array}{l} A_1=M_{11}'D_0-M_{11}(D_0'+\alpha _1'D_0),\\ A_2=(M_{11}-M_{21}')D_0+M_{21}(D_0'+\alpha _1'D_0),\\ \vdots \\ A_s=(-1)^{s-2}(M_{s-1 1}-M_{s1}')D_0-(-1)^{s-1}M_{s1}(D_0'+\alpha _1'D_0),\\ A_{s+1}=(-1)^{s-1}M_{s1}D_0, \end{array}\right. \end{aligned}$$

(63)

are rational functions. Substituting the expression (44) of h(z) and $h'(z)$ $h''(z)\cdots h^{(s)}$ into (62), we get

$$\begin{aligned} A_{1}f^nf^{(k)}+A_{2}(f^nf^{(k)})'+\cdots +A_{s+1}(f^nf^{(k)})^{(s)}=H_2, \end{aligned}$$

(64)

where

$$\begin{aligned} H_2=-[A_{1}L_d(z,f)+A_{2}L_d'(z,f)+\cdots +A_{s+1}L_d^{(s)}(z,f)]. \end{aligned}$$

Clearly, $H_2$ is a differential-difference polynomial in f with degree $\le d$ and rational functions as its coefficients. By (53), we set

$$\begin{aligned} \psi _t=\frac{(f^nf^{(k)})^{(t)}}{f^{n-s}}, \end{aligned}$$

(65)

for $t=1,2,\ldots ,s.$ Combining (65) and (64), we have

$$\begin{aligned} f^{n-s}R_2=H_2, \end{aligned}$$

(66)

where

$$\begin{aligned} R_2=A_{1}f^{s}f^{(k)}+A_{2}\psi _1+\cdots A_{s+1}\psi _{s}. \end{aligned}$$

(67)

Since $d\le n-s-1$ and f is finite order, by Lemma 2.1 we get

$$\begin{aligned} m(r,R_2)=O(\log r). \end{aligned}$$

On the other hand, we know that f has finitely many poles, and hence

$$\begin{aligned} T(r,R_2)=m(r,R_2)+N(r,R_2)=O(\log r). \end{aligned}$$

Then, $R_2$ is a rational function. Next we consider the following two cases:

Case 2.1. If $R_2(z)\equiv 0,$ then from (67), we have

$$\begin{aligned} A_{1}f^{s}f^{(k)}=-[-A_{2}\psi _1+\cdots A_{s+1}\psi _{s}]. \end{aligned}$$

(68)

Now counting zero multiplicity of the left hand side of (68) and right hand side of (68) in the same way as done in Case 1.1, we can find that the multiplicity of zeros is different between the left sides and the right sides of (68). It is a contradiction. So we can obtain that f has finitely many zeros.

Case 2.2. If $R_2(z)\not \equiv 0,$ then from (67) we have

$$\begin{aligned} f^{n-s-1}(z)(fR_2)=H_2(z,f). \end{aligned}$$

(69)

It follows from Lemma 2.1 that

$$\begin{aligned} m(r,fR_2)=O(\log r). \end{aligned}$$

Since $fR_2$ has finitely many poles,

$$\begin{aligned} T(r,fR_2)=m(r,fR_2)+N(r,fR_2)=O(\log r). \end{aligned}$$

Therefore, $fR_2$ is a rational function, which contradicts that f is transcendental. So from the above discussion we conclude that f is a transcendental meromorphic function with finite many zeros and poles. Now by Lemma 2.3, we can say that

$$\begin{aligned} f(z)=q(z)e^{P(z)} \end{aligned}$$

(70)

and

$$\begin{aligned} f^{(k)}(z)=R(z)e^{P(z)}, \end{aligned}$$

(71)

where q(z) is a non-vanishing rational function, P(z) is a non-constant polynomial, R(z) is a differential polynomial in q(z) and P(z). Substituting (70) and (71) into (8) yields

$$\begin{aligned} q^nRe^{(n+1)P}+\sum ^{d}_{j=0}\beta _{j}e^{P_{c_{j}}}e^{jP}=\sum ^{s}_{i=1}p_ie^{\alpha _i} \end{aligned}$$

(72)

where $\beta _{j}$ are rational functions and $P_{c_{j}}$ are polynomials with $\deg (P_{c_{j}})<\deg (P(z)).$ Since $\deg \{\alpha _{i}(z)-\alpha _{j}(z)\}\ge 1$ and $p_i(z)$ are non-vanishing rational functions, by Lemma 2.2 we can deduce that there must exist positive integers $l_0, l_1,\ldots l_s$ with $\{l_1, l_2,\ldots ls\} = \{1, 2, \ldots s\}$ and distinct integers $j_{1},$ $j_{2},\ldots , j_{s-1}$ with $0\le j_{1},j_{2}, \ldots , j_{s-1}\le d$ such that

$$\begin{aligned} \left\{ \begin{array}{l} (n+1)P=\alpha _{l_1}+\gamma _1\\ j_{1}P=\alpha _{l_2}+\gamma _2\\ \cdots \\ j_{s-1}P=\alpha _{l_s}+\gamma _s\\ \end{array}\right. \end{aligned}$$

(73)

where $\gamma _1,$ $\gamma _2,\ldots ,\gamma _s$ are constants, and

$$\begin{aligned} \left\{ \begin{array}{l} q^{n}R=p_{l_1}e^{-\gamma _1}\\ \beta _{j_{1}}Pe^{P_{c_{j_{1}}}}=p_{l_2}e^{-\gamma _2}\\ \cdots \\ \beta _{j_{s-1}}Pe^{P_{c_{j_{s-1}}}}=p_{l_s}e^{-\gamma _s}\\ \end{array}\right. \end{aligned}$$

(74)

Hence, $\beta _{j}\equiv 0$ hold for all $j\ne j_{i}(i=1,2,3,\ldots ,s-1),$ $\alpha '_{l_1}:\alpha '_{l_2}:\alpha '_{l_3}:\ldots :\alpha '_{l_s}=n+1: j_{1}:j_{2}:\ldots :j_{s-1},$ $(n+1)P'=\alpha '_{l_{1}},$ $L_d(z,f)=\sum ^{s}_{i=2}p_{l_i}(z)e^{\alpha _{l_i}{(z)}}.$