Entire Solutions of Certain Type of Non-Linear Difference Equations

Abstract

In this paper, we study the existence of entire solutions of finite-order of non-linear difference equations of the form

$$\begin{aligned} f^{n}(z)+q(z)\Delta _{c}f(z)=p_{1}\mathrm{e}^{\alpha _{1}z}+p_{2}\mathrm{e}^{\alpha _{2}z},\quad n\ge 2 \end{aligned}$$

and

$$\begin{aligned} f^{n}(z)+q(z)\mathrm{e}^{Q(z)}f(z+c)=p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z},\quad n\ge 3 \end{aligned}$$

where q, Q are non-zero polynomials, $c,\lambda ,p_{i},\alpha _{i}(i=1,2)$ are non-zero constants such that $\alpha _{1}\ne \alpha _{2}$ and $\Delta _{c}f(z)=f(z+c)-f(z)\not \equiv 0$. Our results are improvements and complements of Wen et al. (Acta Math Sin 28:1295–1306, 2012), Yang and Laine (Proc Jpn Acad Ser A Math Sci 86:10–14, 2010) and Zinelâabidine (Mediterr J Math 14:1–16, 2017).

1 Introduction and Main Results

In this paper, we assume that the reader is familiar with the fundamental results and standard notation of Nevanlinna theory [5, 7, 14]. In addition, we use $\rho (f)$ to denote the order of growth of f and $\lambda (f)$ to denote the exponent of convergence of zeros sequence of f. For simplicity, we denote by S(r, f) any quantify satisfying $S(r,f)=o(T(r,f)),$ as $r\rightarrow \infty$, outside of a possible exceptional set of finite logarithmic measure, we use S(f) to denote the family of all small functions with respect to f.

Recently, many scholars have investigated solvability and existence of solutions of non-linear differential equations or difference equations, see [3, 4, 8,9,10,11,12, 16].

Exponential polynomials are important in complex analysis as they have many interesting properties as mentioned, for example, in the paper [13] due to Wen, Heittokangas and Laine. In this paper, we mainly give exact expressions of exponential polynomial solutions of certain class of non-linear difference equations.

In [16], Yang and Laine proved the following result:

Theorem A

A non-linear difference equation

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

(1.1)

where q is a non-constant polynomial and b,  c are non-zero complex constants, Eq. (1.1) does not admit entire solutions of finite order. If q is a non-zero constant, then Eq. (1.1) possesses three district entire solutions of finite order, provided $b=3\pi n$ and $q^{3}=(-1)^{n+1}\frac{27}{4}c^{2}$ for a non-zero integer n.

Given Theorem A, it is natural to ask about the solutions of the following more general form

$$\begin{aligned} f^{n}(z)+q(z)\Delta _{c}f(z)=p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}, \end{aligned}$$

(1.2)

where q is a non-zero polynomial, $c,\lambda ,p_{i}(i=1,2)$ are non-zero constants such that $\Delta _{c}f(z)=f(z+c)-f(z)\not \equiv 0$ and $n\ge 2$ is an integer.

In this paper, we study this problem and obtain the following result.

Theorem 1.1

Let $n \ge 2$ be an integer, q be a non-zero polynomial, $c,\lambda ,p_{1},p_{2}$ be non-zero constants. If there exists some entire solution f of finite order to Eq. (1.2), such that $\Delta _{c}f(z)=f(z+c)-f(z)\not \equiv 0$, then q is a constant, and $n=2$ or $n=3$.

When $n=2$, then

$$\begin{aligned} f(z)=q+c_{1}\mathrm{e}^{\frac{\lambda }{2}z}+c_{2}\mathrm{e}^{-\frac{\lambda }{2}z}, \end{aligned}$$

where $q^{4}=4p_{1}p_{2}$, $c_{1}^{2}=p_{1},~c_{2}^{2}=p_{2}$, $\lambda c=2k\pi i$, $k\in {\mathbb {Z}}$ and k is an odd.

When $n=3$, then

$$\begin{aligned} f(z)=c_{1}\mathrm{e}^{\frac{\lambda }{3}z}+c_{2}\mathrm{e}^{-\frac{\lambda }{3}z}, \end{aligned}$$

where $q^{3}=\frac{27}{8}p_{1}p_{2}$, $c_{1}^{3}=p_{1},~c_{2}^{3}=p_{2}$, $\lambda c=3k\pi i$, $k\in {\mathbb {Z}}$ and k is an odd.

More recently, Zinelâabidine showed in [17]:

Theorem B

Let q be a polynomial, $p_{i},\alpha _{i}(i=1,2)$ be non-zero constants such that $\alpha _{1}\pm \alpha _{2}\ne 0$. If f is an entire solution of finite order of equation

$$\begin{aligned} f^{3}(z)+q(z)\Delta f(z)=p_{1}\mathrm{e}^{\alpha _{1}z}+p_{2}\mathrm{e}^{\alpha _{2}z}, \end{aligned}$$

(1.3)

such that $\Delta f(z)=f(z+1)-f(z)\not \equiv 0$, then q is a constant, and one of the following relations holds:

• 1. $f(z)=c_{1}\mathrm{e}^{\frac{\alpha _{1}}{3}z}$ and $c_{1}(\mathrm{e}^{\frac{\alpha _{1}}{3}}-1)q=p_{2}$, $\alpha _{1}=3\alpha _{2}$,

• 2. $f(z)=c_{2}\mathrm{e}^{\frac{\alpha _{2}}{3}z}$ and $c_{2}(\mathrm{e}^{\frac{\alpha _{2}}{3}}-1)q=p_{1}$, $\alpha _{2}=3\alpha _{1}$, where $c_{1},c_{2}$ are non-zero constants satisfying $c_{1}^{3}=p_{1},~c_{2}^{3}=p_{2}$.

The aim of this paper is to study the difference equation

$$\begin{aligned} f^{2}(z)+q(z)\Delta _{c} f(z)=p_{1}\mathrm{e}^{\alpha _{1}z}+p_{2}\mathrm{e}^{\alpha _{2}z}, \end{aligned}$$

(1.4)

where q is a non-zero polynomial, $c,p_{i},\alpha _{i}(i=1,2)$ are non-zero constants such that $\alpha _{1}\pm \alpha _{2}\ne 0$ and $\Delta _{c} f(z)=f(z+c)-f(z)\not \equiv 0$. In fact, we prove the following result.

Theorem 1.2

Let q be a non-zero polynomial, $c,p_{i},\alpha _{i}(i=1,2)$ be non-zero constants such that $\alpha _{1}\pm \alpha _{2}\ne 0$. If f is an entire solution of finite order of Eq. (1.4), such that $\Delta _{c} f(z)=f(z+c)-f(z)\not \equiv 0$, then q is a constant, $\rho (f)=1$ and one of the following conclusions holds:

• 1. $f(z)=c_{1}\mathrm{e}^{\frac{\alpha _{1}}{2}z}$, and $c_{1}(\mathrm{e}^{\frac{\alpha _{1}}{2}c}-1)q=p_{2}$, $\alpha _{1}=2\alpha _{2}$;

• 2. $f(z)=c_{2}\mathrm{e}^{\frac{\alpha _{2}}{2}z}$, and $c_{2}(\mathrm{e}^{\frac{\alpha _{2}}{2}c}-1)q=p_{1}$, $\alpha _{2}=2\alpha _{1}$, where $c_{1},c_{2}$ are non-zero constants satisfying $c_{1}^{2}=p_{1},~c_{2}^{2}=p_{2}$;

• 3.
  $$\begin{aligned} T(r,\varphi )+S(r,f)= & {} \kappa T(r,f),~0<\kappa \le 1,~\text{ and }\\ N\left( r,\frac{1}{f}\right) +S(r,f)= & {} \iota T(r,f),~1-\frac{\kappa }{2}\le \iota \le 1, \end{aligned}$$

  where $\varphi =\alpha _{1}\alpha _{2}f^{2}-2(\alpha _{1}+\alpha _{2})ff'+2(f')^{2}+2ff''$.

Wen, Heittokangas and Laine [13] studied and classified the finite order entire solutions f of equation

$$\begin{aligned} f^{n}(z)+q(z)\mathrm{e}^{Q(z)}f(z+c)=P(z), \end{aligned}$$

(1.5)

where q, Q, P are polynomials, $n\ge 2$ is an integer and $c\in {\mathbb {C}}{\setminus }\{0\}$, and obtained the following Theorem C.

Recall that a function f of the form

$$\begin{aligned} f(z)=P_{1}(z)\mathrm{e}^{Q_{1}(z)}+\cdots +P_{k}(z)\mathrm{e}^{Q_{k}(z)}, \end{aligned}$$

(1.6)

where $P_{j}'s$ and $Q_{j}'s$ are polynomials in z is called an exponential polynomial. Furthermore, let

$$\begin{aligned} \Gamma _{1}= & {} \{\mathrm{e}^{\alpha (z)}+d: d\in {\mathbb {C}}~\text{ and }~\alpha ~\text{ is } \text{ a } \text{ non-constant } \text{ polynomial }\},\\ \Gamma _{0}= & {} \{\mathrm{e}^{\alpha (z)}: \alpha ~\text{ is } \text{ a } \text{ non-constant } \text{ polynomial }\}. \end{aligned}$$

Theorem C

(See [13])   Let $n\ge 2$ be an integer, let $c\in {\mathbb {C}}{\setminus }\{0\}$, and let q, Q, P be polynomials such that Q is not a constant and $q\not \equiv 0$. Then, we identify the finite order entire solutions f of equation (1.5) as follows:

• (a) Every solution f satisfies $\rho (f)=\deg Q$ and is of mean type.

• (b) Every solution f satisfies $\lambda (f)=\rho (f)$ if and only if $P\not \equiv 0$.

• (c) A solution belongs to $\Gamma _{0}$ if and only if $P\equiv 0$. In particular, this is the case if $n\ge 3$.

• (d) If a solution f belongs to $\Gamma _{0}$ and if g is any other finite-order entire solution to (1.5), then $f =\eta g$, where $\eta ^{n-1}=1$.

• (e) If f is an exponential polynomial solution of the form (1.6), then $f\in \Gamma _{1}$. Moreover, if $f\in \Gamma _{1}{\setminus }\Gamma _{0}$, then $\rho (f)=1$.

A natural question to ask is about $P(z)=p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}$ in (1.5), where $\lambda ,p_{1},p_{2}\in {\mathbb {C}}{\setminus }\{0\}$ are constants. We consider this question and obtain the following result.

Theorem 1.3

Let $n\ge 3$ be an integer, let $c,\lambda ,p_{1},p_{2}\in {\mathbb {C}}{\setminus }\{0\}$ be constants and let q, Q be polynomials such that Q is not a constant and $q\not \equiv 0$. If f is an entire solution of finite order of the equation

$$\begin{aligned} f^{n}(z)+q(z)\mathrm{e}^{Q(z)}f(z+c)=p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}, \end{aligned}$$

(1.7)

then the following conclusions hold.

• 1. Every solution f satisfies $\rho (f)=\deg Q=1$.

• 2. If a solution f belongs to $\Gamma _{0}$, then $f(z)=e^{\frac{\lambda }{n}z+B},~Q(z)=-\frac{n+1}{n}\lambda z+b$ or $f(z)=e^{-\frac{\lambda }{n}z+B},~Q(z)=\frac{n+1}{n}\lambda z+b$, where $b,B\in {\mathbb {C}}$.

Remark 1.1

We conjecture that if $n=2$, the conclusions of Theorem 1.3 are still valid, although we have not found a suitable method of proof yet. For example, $f(z)=\mathrm{e}^{z}$ is an entire solution of finite order of the difference equation

$$\begin{aligned} f^{2}(z)+2\mathrm{e}^{-3z}f(z-\log 2)=\mathrm{e}^{2z}+\mathrm{e}^{-2z}, \end{aligned}$$

and $f(z)=\mathrm{e}^{z}+\mathrm{e}^{-z}$ is an entire solution of finite order of the difference equation

$$\begin{aligned} f^{2}(z)+2\mathrm{e}^{z}f(z+\pi i)=-\mathrm{e}^{2z}+\mathrm{e}^{-2z}. \end{aligned}$$

The following example shows that our estimates in Theorem 1.3 are accurate.

Example 1.1

If $f(z)=\mathrm{e}^{z}$ is an entire solution of finite order of the difference equation

$$\begin{aligned} f^{3}(z)+\frac{1}{2}\mathrm{e}^{-4z}f(z+\log 2)=\mathrm{e}^{3z}+\mathrm{e}^{-3z}, \end{aligned}$$

where $\lambda =3,~n=3,~b=B=0$, then $f(z)=\mathrm{e}^{\frac{\lambda }{n}z+B}=\mathrm{e}^{z}$, $Q(z)=-\frac{n+1}{n}\lambda z+b=-4z$ and $\rho (f)=\deg Q=1$.

2 Some Lemmas

Lemma 2.1

$(Clunie's~Lemma)$ (See [1, 7, Lem. 2.4.2])   Let f be a transcendental meromorphic solution of

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

where P(z, f) and Q(z, f) are polynomials in f and its derivatives with meromorphic coefficients, say $\{a_{\lambda }|\lambda \in I\}$, such that $m(r,a_{\lambda })=S(r,f)$ for all $\lambda \in I$. If the total degree of Q(z, f) as a polynomial in f and its derivatives is at most n, then

$$\begin{aligned} m(r,P(z,f))=S(r,f). \end{aligned}$$

(2.1)

Lemma 2.2

(See [6, Cor. 3.3]) Let f be a non-constant finite order meromorphic solution of

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

where P(z, f) and Q(z, f) are difference polynomials in f with small meromorphic coefficients, and let $c\in {\mathbb {C}}$, $\delta <1$. If the total degree of Q(z, f) as a polynomial in f and its shifts at most n, then

$$\begin{aligned} m(r,P(z,f))=o\left( \frac{T(r+|c|,f)}{r^{\delta }}\right) +o(T(r,f)) \end{aligned}$$

(2.2)

for all r outside of a possible exceptional set with finite logarithmic measure.

Remark 2.1

In Lemma 2.2, if f is a transcendental meromorphic function with finite order $\rho$, and P(z, f), Q(z, f) are differential-difference polynomials in f, then by the same reasoning as in the proof of Lemma 2.1, we also obtain the conclusion (2.2). Furthermore, if the coefficients of P(z, f) and Q(z, f) are polynomials $A_{j},~j=1,\ldots ,n$, for each $\varepsilon >0$, then (2.2) can be written as:

$$\begin{aligned} m(r,P(z,f))=O(r^{\rho -1+\varepsilon }) +O\left( \sum _{j=1}^{n}m(r,A_{j})\right) , \end{aligned}$$

(2.3)

where r is sufficiently large.

Lemma 2.3

(See [15, Thm. 1.51])   Suppose that $f_{1}, f_{2},\ldots ,f_{n}(n\ge 2)$ are meromorphic functions and $g_{1}, g_{2},\ldots ,g_{n}$ are entire functions satisfying the following conditions:

• 1. ${\sum \limits _{j=1}^{n} f_{j}\mathrm{e}^{g_{j}}\equiv 0.}$

• 2. $g_{j}-g_{k}$ are not constants for $1\le j<k\le n$.

• 3. For $1\le j\le n, 1\le h<k\le n$,

  $$\begin{aligned} T(r,f_{j})=o(T(r,e^{g_{h}-g_{k}}))~(r\rightarrow \infty , r\not \in E), \end{aligned}$$

where $E\subset [1,\infty )$ is finite linear measure or finite logarithmic measure.

Then $f_{j}\equiv 0~(j=1,\ldots , n)$.

Lemma 2.4

(See [11, Lem. 6])   Suppose that f is a transcendental meromorphic function, a, b, c, d are small functions with respect to f and $acd\not \equiv 0$. If

$$\begin{aligned} af^{2}+bff'+c(f')^{2}=d, \end{aligned}$$

(2.4)

then

$$\begin{aligned} c(b^{2}-4ac)\frac{d'}{d}+b(b^{2}-4ac)-c(b^{2}-4ac)'+(b^{2}-4ac)c'=0. \end{aligned}$$

(2.5)

Lemma 2.5

(See [2, Cor. 2.6])   Let $\eta _{1},\eta _{2}$ be two complex numbers such that $\eta _{1}\ne \eta _{2}$ and let f be a finite order meromorphic function. Let $\rho$ be the order of f. Then for each $\varepsilon >0$, we have

$$\begin{aligned} m\left( r,\frac{f(z+\eta _{1})}{f(z+\eta _{2})}\right) =O(r^{\rho -1+\varepsilon }). \end{aligned}$$

(2.6)

Using an argument similar to that used in [3, Lem. 2.3], we get the following result.

Lemma 2.6

Let $n\ge 1$ be an integer, $\lambda$ be a non-zero constant and H be a non-zero polynomial. Then, the differential equation

$$\begin{aligned} \lambda ^{2}f-n^{2}f''=H \end{aligned}$$

(2.7)

has a special solution $c_{0}$ which is a non-zero polynomial.

3 Proof of Theorem 1.1

Denote $P=q\Delta _{c}f$. Suppose that f is a transcendental entire solution of finite order of Eq. (1.2). Differentiating (1.2), we have

$$\begin{aligned} nf^{n-1}f'+P'=\lambda (p_{1}\mathrm{e}^{\lambda z}-p_{2}\mathrm{e}^{-\lambda z}). \end{aligned}$$

(3.1)

Differentiating (3.1) yields

$$\begin{aligned} n(n-1)f^{n-2}(f')^{2}+nf^{n-1}f''+P''=\lambda ^{2}(p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}). \end{aligned}$$

(3.2)

It follows from (1.2) and (3.1) that

$$\begin{aligned}&\lambda ^{2}f^{2n}-n^{2}f^{2(n-1)}(f')^{2}\nonumber \\&\quad +2\lambda ^{2}Pf^{n}-2nP'f^{n-1}f'+\lambda ^{2}P^{2}-(P')^{2}-4\lambda ^{2}p_{1}p_{2}=0. \end{aligned}$$

(3.3)

It follows from (1.2) and (3.2) that

$$\begin{aligned} \lambda ^{2}f^{n}-n(n-1)f^{n-2}(f')^{2}-nf^{n-1}f''+\lambda ^{2}P-P''=0. \end{aligned}$$

(3.4)

Eliminating $(f')^{2}$ from (3.3) and (3.4) yields

$$\begin{aligned} f^{2n-1}\varphi =Q(z,f), \end{aligned}$$

(3.5)

where

$$\begin{aligned} \varphi =\lambda ^{2}f-n^{2}f'' \end{aligned}$$

(3.6)

and

$$\begin{aligned} Q(z,f)= & {} [(n-2)\lambda ^{2}P+nP'']f^{n}-2n(n-1)P'f^{n-1}f'\nonumber \\&+(n-1) [\lambda ^{2}P^{2}-(P')^{2}-4\lambda ^{2}p_{1}p_{2}], \end{aligned}$$

(3.7)

Q(z, f) is a differential-difference polynomial in f and the total degree is at most $n+1$. Note that when $n\ge 2$, then $2n-1\ge n+1$, by Lemma 2.2 and Remark 2.1, we have $m(r,\varphi )=S(r,f)$; therefore, $T(r,\varphi )=S(r,f)$. We distinguish two cases below:

Case 1 If $\varphi \equiv 0$, i.e. $\lambda ^{2}f-n^{2}f''\equiv 0$. Every entire solution $f(\not \equiv 0)$ of this equation can be expressed as:

$$\begin{aligned} f(z)=c_{1}\mathrm{e}^{\frac{\lambda }{n} z}+c_{2}\mathrm{e}^{-\frac{\lambda }{n} z}, \end{aligned}$$

(3.8)

where $c_{1},c_{2}$ are non-zero constants. Otherwise, if one of $c_{1},c_{2}$ is equal to zero, substituting (3.8) into (1.2) and using Lemma 2.3, we obtain a contradiction.

When $n=3$, then $f(z)=c_{1}\mathrm{e}^{\frac{\lambda }{3} z}+c_{2}\mathrm{e}^{-\frac{\lambda }{3} z}$, substituting this into (1.2) yields

$$\begin{aligned}&(c_{1}^{3}-p_{1})\mathrm{e}^{\lambda z}+(c_{2}^{3}-p_{2})\mathrm{e}^{-\lambda z}+c_{1}\left[ q(z)\left( \mathrm{e}^{\frac{\lambda }{3} c}-1\right) +3c_{1}c_{2}\right] \mathrm{e}^{\frac{\lambda }{3} z} \nonumber \\&\quad +\,c_{2}\left[ q(z)\left( \mathrm{e}^{-\frac{\lambda }{3} c}-1\right) +3c_{1}c_{2}\right] \mathrm{e}^{-\frac{\lambda }{3} z}=0. \end{aligned}$$

(3.9)

It follows from (3.9) and Lemma 2.3 that

$$\begin{aligned} \left\{ \begin{array}{ll} c_{1}^{3}=p_{1}, c_{2}^{3}=p_{2};\\ c_{1}\left[ q(z)\left( \mathrm{e}^{\frac{\lambda }{3} c}-1\right) +3c_{1}c_{2}\right] \equiv 0;\\ c_{2}\left[ q(z)\left( \mathrm{e}^{-\frac{\lambda }{3} c}-1\right) +3c_{1}c_{2}\right] \equiv 0. \end{array}\right. \end{aligned}$$

Note that $c_{1},c_{2}\in {\mathbb {C}}{\setminus }\{0\}$, then $q(z)(\mathrm{e}^{\frac{\lambda }{3} c}-1)=q(z)(\mathrm{e}^{-\frac{\lambda }{3} c}-1)=-3c_{1}c_{2}\ne 0$, q(z) reduces to a constant q and $\lambda c=3k\pi i$, $k\in {\mathbb {Z}}{\setminus }\{0\}$ and k is an odd, $q^{3}=\frac{27}{8}p_{1}p_{2}$.

When $n=2l(l\ge 1)$ is even, $f(z)=c_{1}\mathrm{e}^{\frac{\lambda }{2l} z}+c_{2}\mathrm{e}^{-\frac{\lambda }{2l} z}$. Substituting this into (1.2) yields

$$\begin{aligned}&(c_{1}^{2l}-p_{1})\mathrm{e}^{\lambda z}+(c_{2}^{2l}-p_{2})\mathrm{e}^{-\lambda z}+\sum _{k=1}^{2l-1}\left( {\begin{array}{c}k\\ 2l\end{array}}\right) c_{1}^{2l-k}c_{2}^{k}\mathrm{e}^{\frac{2l-2k}{2l}\lambda z}\nonumber \\&\quad +c_{1}q(z)(\mathrm{e}^{\frac{\lambda c}{2l}}-1)\mathrm{e}^{\frac{\lambda z}{2l}} +c_{2}q(z)\left( \mathrm{e}^{-\frac{\lambda c}{2l}}-1\right) \mathrm{e}^{-\frac{\lambda z}{2l}}=0. \end{aligned}$$

(3.10)

If $k=l$, then $\sum _{k=1}^{2l-1}\left( {\begin{array}{c}k\\ 2l\end{array}}\right) c_{1}^{2l-k}c_{2}^{k}e^{\frac{2l-2k}{2l}\lambda z}$ must have a constant term. That is $\left( {\begin{array}{c}l\\ 2l\end{array}}\right) c_{1}^{l}c_{2}^{l}=\frac{(2l)!}{l!l!}(c_{1}c_{2})^{l}$. By Lemma 2.3, we obtain $c_{1}c_{2}=0$, a contradiction.

When $n=2l+1(l\ge 2)$ is odd, $f(z)=c_{1}\mathrm{e}^{\frac{\lambda }{2l+1} z}+c_{2}\mathrm{e}^{-\frac{\lambda }{2l+1} z}$. Substituting this into (1.2) yields

$$\begin{aligned}&\sum _{k=0}^{2l+1}\left( {\begin{array}{c}k\\ 2l+1\end{array}}\right) c_{1}^{2l+1-k}c_{2}^{k}\mathrm{e}^{\frac{2l+1-2k}{2l+1}\lambda z} +c_{1}q(z)\left( \mathrm{e}^{\frac{\lambda c}{2l+1}}-1\right) \mathrm{e}^{\frac{\lambda z}{2l+1}}\\&\quad +\,c_{2}q(z)\left( \mathrm{e}^{-\frac{\lambda c}{2l+1}}-1\right) \mathrm{e}^{-\frac{\lambda z}{2l+1}}=p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}, \end{aligned}$$

i.e.

$$\begin{aligned} \begin{aligned}&\left( c_{1}^{2l+1}-p_{1}\right) \mathrm{e}^{\lambda z}+\left( c_{2}^{2l+1}-p_{2}\right) \mathrm{e}^{-\lambda z}+\sum _{\begin{array}{c} k=1~\\ {k\ne l,l+1} \end{array}}^{2l}\left( {\begin{array}{c}k\\ 2l+1\end{array}}\right) c_{1}^{2l+1-k}c_{2}^{k}\mathrm{e}^{\frac{2l+1-2k}{2l+1}\lambda z}\\&\quad +\left[ c_{1}q(z)\left( \mathrm{e}^{\frac{\lambda c}{2l+1}}-1\right) +\left( {\begin{array}{c}l\\ 2l+1\end{array}}\right) c_{1}^{l+1}c_{2}^{l}\right] \mathrm{e}^{\frac{\lambda z}{2l+1}}\\&\quad +\left[ c_{2}q(z)\left( \mathrm{e}^{-\frac{\lambda c}{2l+1}}-1\right) +\left( {\begin{array}{c}l+1\\ 2l+1\end{array}}\right) c_{1}^{l}c_{2}^{l+1}\right] \mathrm{e}^{-\frac{\lambda z}{2l+1}}=0. \end{aligned} \end{aligned}$$

(3.11)

Since $l\ge 2$, then $\sum \nolimits _{\begin{array}{c} k=1\\ k\ne l,l+1 \end{array}}^{2l}\left( {\begin{array}{c}k\\ 2l+1\end{array}}\right) c_{1}^{2l+1-k}c_{2}^{k} \mathrm{e}^{\frac{2l+1-2k}{2l+1}\lambda z}$ contains at least two terms. By Lemma 2.3, we have $\left( {\begin{array}{c}k\\ 2l+1\end{array}}\right) c_{1}^{2l+1-k}c_{2}^{k}=0,~k\ne l,l+1,~k=1,\ldots ,2l$. Then, $c_{1}c_{2}=0$, a contradiction.

Case 2 As in the beginning of the proof of Theorem 1.2 below, we obtain $\rho (f)=1$. If $\varphi \not \equiv 0$, since f is a transcendental entire function with order $\rho (f)=1$, we see that (3.5) satisfies the conditions of Lemma 2.2 and Remark 2.1. Thus, we have

$$\begin{aligned} m(r,\lambda ^{2}f-n^{2}f'')=S(r,f)+O(m(r,q))=O(\log r), \end{aligned}$$

which implies that $\lambda ^{2}f-n^{2}f''$ is a polynomial. Denote

$$\begin{aligned} \lambda ^{2}f-n^{2}f''=H, \end{aligned}$$

(3.12)

where H is a non-zero polynomial. By Lemma 2.6, we see that (3.12) must have a non-zero polynomial solution, say, $c_{0}(z)$. Since the differential equation

$$\begin{aligned} \lambda ^{2}f-n^{2}f''=0, \end{aligned}$$

has two fundamental solutions

$$\begin{aligned} f_{1}(z)=\mathrm{e}^{\frac{\lambda }{n}z},\quad f_{2}(z)=\mathrm{e}^{-\frac{\lambda }{n}z}, \end{aligned}$$

the general entire solution $f(\not \equiv 0)$ of (3.12) can be expressed as:

$$\begin{aligned} f(z)=c_{0}(z)+c_{1}\mathrm{e}^{\frac{\lambda }{n}z}+c_{2}\mathrm{e}^{-\frac{\lambda }{n}z}, \end{aligned}$$

(3.13)

where $c_{1},c_{2}$ are non-zero constants, $c_{0}(z)$ is a non-zero polynomial. Otherwise, if one of $c_{1},c_{2}$ is equal to zero, substituting (3.13) into (1.2) and using Lemma 2.3, we obtain a contradiction.

When $n=2$, then $f(z)=c_{0}(z)+c_{1}\mathrm{e}^{\frac{\lambda }{2}z}+c_{2}\mathrm{e}^{-\frac{\lambda }{2}z}$. Substituting this into (1.2) yields

$$\begin{aligned}&(c_{1}^{2}-p_{1})\mathrm{e}^{\lambda z}+(c_{2}^{2}-p_{2})\mathrm{e}^{-\lambda z}+c_{0}^{2}(z)+q(z)[c_{0}(z+c)-c_{0}(z)]+2c_{1}c_{2}\nonumber \\&\quad +c_{1}[q(z)(\mathrm{e}^{\frac{\lambda c}{2}}-1)+2c_{0}(z)]\mathrm{e}^{\frac{\lambda }{2}z} +c_{2}\left[ q(z)\left( \mathrm{e}^{-\frac{\lambda c}{2}}-1\right) +2c_{0}(z)\right] \mathrm{e}^{-\frac{\lambda }{2}z}=0.\nonumber \\ \end{aligned}$$

(3.14)

It follows from (3.14) and Lemma 2.3 that

$$\begin{aligned} \left\{ \begin{array}{ll} c_{1}^{2}=p_{1}, c_{2}^{2}=p_{2};\\ c_{0}^{2}(z)+q(z)[c_{0}(z+c)-c_{0}(z)]+2c_{1}c_{2}\equiv 0;\\ c_{1}[q(z)(\mathrm{e}^{\frac{\lambda c}{2}}-1)+2c_{0}(z)]\equiv 0;\\ c_{2}[q(z)(\mathrm{e}^{-\frac{\lambda c}{2}}-1)+2c_{0}(z)]\equiv 0. \end{array}\right. \end{aligned}$$

Note that $c_{1},c_{2}\in {\mathbb {C}}{\setminus }\{0\}$, then $q(z)(\mathrm{e}^{\frac{\lambda c}{2}}-1)\equiv q(z)(\mathrm{e}^{-\frac{\lambda c}{2}}-1)\equiv -2c_{0}(z)\not \equiv 0$, that is $q(z)\equiv c_{0}(z)$ and $\lambda c=2k\pi i$, $k\in {\mathbb {Z}}{\setminus }\{0\}$ and k is odd. Substituting $q(z)\equiv c_{0}(z)$ into $c_{0}^{2}(z)+q(z)[c_{0}(z+c)-c_{0}(z)]+2c_{1}c_{2}\equiv 0$ yields $c_{0}(z+c)c_{0}(z)+2c_{1}c_{2}\equiv 0$. Then, $c_{0}(z)$ must be a non-zero constant. Therefore, $c_{0}^{4}=q^{4}=4c_{1}^{2}c_{2}^{2}=4p_{1}p_{2}$.

When $n\ge 3$, then $2n-2\ge n+1$, it follows from (3.5) that $f^{2n-2}(f\varphi )=Q(z,f)$. By Lemma 2.2 and Remark 2.1, we have $m(r,f\varphi )=S(r,f)$. Therefore, $T(r,f\varphi )=S(r,f)$. Since $\varphi \not \equiv 0$, then $T(r,f)=m(r,f)\le m(r,f\varphi )+m\left( r,\frac{1}{\varphi }\right) \le T(r,\varphi )+S(r,f) =S(r,f)$, which gives a contradiction.

This completes the proof of Theorem 1.1.

4 Proof of Theorem 1.2

Clearly, $\rho (p_{1}\mathrm{e}^{\alpha _{1}z}+p_{2}\mathrm{e}^{\alpha _{2}z})=1$, where $\alpha _{1}\pm \alpha _{2}\ne 0$. From (1.4) and Lemma 2.5, we have

$$\begin{aligned} T(r,p_{1}\mathrm{e}^{\alpha _{1}z}+p_{2}\mathrm{e}^{\alpha _{2}z})= & {} T(r,f^{2}(z)+q(z)\Delta _{c}f(z))\\\le & {} T(r,f^{2})+m\left( r,\frac{q(z)\Delta _{c}f(z)}{f}\right) +m(r,f)+O(1)\\\le & {} 3T(r,f)+S(r,f), \end{aligned}$$

and

$$\begin{aligned} T(r,f^{2}(z)+q(z)\Delta _{c}f(z))\ge & {} T(r,f^{2})-T(r,q(z)\Delta _{c}f(z))+O(1)\\\ge & {} 2T(r,f)-\left[ m\left( r,\frac{q(z)\Delta _{c}f(z)}{f}\right) +m(r,f)\right] +O(1)\\\ge & {} 2T(r,f)-T(r,f)+S(r,f)=T(r,f)+S(r,f), \end{aligned}$$

i.e.

$$\begin{aligned} T(r,f)+S(r,f)\le T(r,p_{1}\mathrm{e}^{\alpha _{1}z}+p_{2}\mathrm{e}^{\alpha _{2}z})\le 3T(r,f)+S(r,f), \end{aligned}$$

thus, $\rho (f)=1$. Denote $P=q\Delta _{c}f$. Suppose that f is a transcendental entire solution of finite order of equation (1.4). By differentiating (1.4), we have

$$\begin{aligned} 2ff'+P'=\alpha _{1}p_{1}\mathrm{e}^{\alpha _{1}z}+\alpha _{2}p_{2}\mathrm{e}^{\alpha _{2}z}. \end{aligned}$$

(4.1)

Eliminating $\mathrm{e}^{\alpha _{2}z}$ from (1.4) and (4.1), we have

$$\begin{aligned} \alpha _{2}f^{2}-2ff'+\alpha _{2}P-P'=(\alpha _{2}-\alpha _{1})p_{1}\mathrm{e}^{\alpha _{1}z}. \end{aligned}$$

(4.2)

Differentiating (4.2) yields

$$\begin{aligned} 2\alpha _{2}ff'-2(f')^{2}-2ff''+\alpha _{2}P'-P''=\alpha _{1}(\alpha _{2}-\alpha _{1})p_{1}\mathrm{e}^{\alpha _{1}z}. \end{aligned}$$

(4.3)

It follows from (4.2) and (4.3) that

$$\begin{aligned} \varphi (z)=Q(z,f), \end{aligned}$$

(4.4)

where

$$\begin{aligned} \varphi (z)=\alpha _{1}\alpha _{2}f^{2}-2(\alpha _{1}+\alpha _{2})ff'+2(f')^{2}+2ff'' \end{aligned}$$

(4.5)

and

$$\begin{aligned} Q(z,f)=-\alpha _{1}\alpha _{2}P+(\alpha _{1}+\alpha _{2})P'-P''. \end{aligned}$$

(4.6)

We distinguish two cases below:

Case 1 If $\varphi \equiv 0$, then $Q(z,f)\equiv 0$. Since $\alpha _{1}\ne \alpha _{2}$, we see that $\alpha _{1}P-P'\equiv 0$ and $\alpha _{2}P-P'\equiv 0$ cannot hold simultaneously. Suppose that $\alpha _{2}P-P'\not \equiv 0$. By (4.6), we have

$$\begin{aligned} \alpha _{2}P-P'=A\mathrm{e}^{\alpha _{1}z}, \end{aligned}$$

(4.7)

where A is a non-zero constant. Substituting (4.7) into (4.2), we have

$$\begin{aligned} f(\alpha _{2}f-2f')=\frac{[(\alpha _{2}-\alpha _{1})p_{1}-A]\alpha _{2}}{A}P-\frac{(\alpha _{2}-\alpha _{1})p_{1}-A}{A}P'. \end{aligned}$$

(4.8)

Since the right-hand side of (4.8) is a differential-difference polynomial in f of degree at most 1, by Lemma 2.2 and Remark 2.1, we have

$$\begin{aligned} m(r,\alpha _{2}f-2f')=S(r,f). \end{aligned}$$

Denote $\psi =\alpha _{2}f-2f'$. We consider two subcases as follows.

Subcase 1.1 If $\psi \equiv 0$, that is $\alpha _{2}f-2f'\equiv 0$, then $f^{2}={\widetilde{p}}_{2}\mathrm{e}^{\alpha _{2}z},~{\widetilde{p}}_{2}\in {\mathbb {C}}{\setminus }\{0\}$. Substituting this and (4.7) into (1.4) yields

$$\begin{aligned} \left( 1-\frac{p_{2}}{{\widetilde{p}}_{2}}\right) f^{2}=\frac{\alpha _{2}p_{1}-A}{A}P-\frac{p_{1}}{A}P'. \end{aligned}$$

If $p_{2}\ne {\widetilde{p}}_{2}$, by Lemma 2.2 and Remark 2.1, we have $T(r,f)=m(r,f)=S(r,f)$, a contradiction. Therefore, $p_{2}={\widetilde{p}}_{2}$, $f(z)=c_{2}\mathrm{e}^{\frac{\alpha _{2}}{2}z}$, $c_{2}(\mathrm{e}^{\frac{\alpha _{2}}{2}c}-1)q=p_{1}$, $c_{2}^{2}=p_{2}$, $\alpha _{2}=2\alpha _{1}$.

Subcase 1.2 If $\psi \not \equiv 0$, then $\psi '=\alpha _{2}f'-2f''$, $f'=\frac{\alpha _{2}}{2}f-\frac{\psi }{2}$, $f''=\frac{\alpha _{2}^{2}}{4}f-\frac{\alpha _{2}}{4}\psi -\frac{\psi '}{2}$. Note that $\varphi \equiv 0$ and substitute this into (4.5) gives

$$\begin{aligned} \left[ \left( \alpha _{1}-\frac{\alpha _{2}}{2}\right) \psi -\psi '\right] f=-\frac{\psi ^{2}}{2}. \end{aligned}$$

Since $\psi \not \equiv 0$, then $\left( \alpha _{1}-\frac{\alpha _{2}}{2}\right) \psi -\psi '\not \equiv 0$. From the above equality, we have $T(r,f)=S(r,f)$, which implies a contradiction.

Similarly, if $\alpha _{1}P-P'\not \equiv 0$, then we obtain $f(z)=c_{1}\mathrm{e}^{\frac{\alpha _{1}}{2}z}$, $c_{1}(\mathrm{e}^{\frac{\alpha _{1}}{2}c}-1)q=p_{2}$, $c_{1}^{2}=p_{1}$, $\alpha _{1}=2\alpha _{2}$.

Case 2 If $\varphi \not \equiv 0$, by applying the lemma of logarithmic derivative and Lemma 2.2, from (4.4)–(4.6), we have

$$\begin{aligned} m\left( r,\frac{\varphi }{f}\right) =m\left( r,\frac{Q}{f}\right) =S(r,f)\quad and \quad m\left( r,\frac{\varphi }{f^{2}}\right) =S(r,f). \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} 2m\left( r,\frac{1}{f}\right)&=m\left( r,\frac{1}{f^{2}}\right) \le m\left( r,\frac{\varphi }{f^{2}}\right) +m\left( r,\frac{1}{\varphi }\right) \\&\le T(r,\varphi )+S(r,f)=T(r,Q)+S(r,f)\\&\le m\left( r,\frac{Q}{f}\right) +m(r,f)+S(r,f)\\&\le T(r,f)+S(r,f). \end{aligned} \end{aligned}$$

(4.9)

Suppose that there exist $\kappa ,\iota \ge 0$ such that

$$\begin{aligned} T(r,\varphi )+S(r,f)=\kappa T(r,f),\quad N\left( r,\frac{1}{f}\right) +S(r,f)=\iota T(r,f), \end{aligned}$$

where $0\le \kappa ,~\iota \le 1$. It follows from (4.9) that

$$\begin{aligned} 2m\left( r,\frac{1}{f}\right) \le T(r,\varphi )+S(r,f)=\kappa T(r,f) \end{aligned}$$

and

$$\begin{aligned} T(r,f)\ge & {} N\left( r,\frac{1}{f}\right) +O(1)=T\left( r,\frac{1}{f}\right) -m\left( r,\frac{1}{f}\right) +O(1)\\\ge & {} T(r,f)-\frac{\kappa }{2}T(r,f)+O(1)=\left( 1-\frac{\kappa }{2}\right) T(r,f)+O(1), \end{aligned}$$

then we have $1-\frac{\kappa }{2}\le \iota \le 1$ and $0\le \kappa \le 1$.

Next, we deduce that $\kappa \ne 0$.

If $\kappa =0$, then $T(r,\varphi )=S(r,f)$. It follows from (4.9) that

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

By (4.5), if $z_{0}$ is a multiple zero of f, then $z_{0}$ must be a zero of $\varphi$. Hence, $N_{(2}\left( r,\frac{1}{f}\right) =S(r,f)$. Differentiating (4.5) gives

$$\begin{aligned} \varphi '=2\alpha _{1}\alpha _{2}ff'-2(\alpha _{1}+\alpha _{2})ff''-2(\alpha _{1}+\alpha _{2})(f')^{2}+6f'f''+2ff'''. \end{aligned}$$

(4.10)

If $z_{0}$ is a simple zero of f, it follows from (4.5) and (4.10) that $z_{0}$ is a zero of $3\varphi f''-[\varphi '+(\alpha _{1}+\alpha _{2})\varphi ]f'$. Define

$$\begin{aligned} \alpha :=\frac{3\varphi f''-[\varphi '+(\alpha _{1}+\alpha _{2})\varphi ]f'}{f}, \end{aligned}$$

(4.11)

then we have $T(r,\alpha )=S(r,f)$. It follows that

$$\begin{aligned} f''=\frac{1}{3}\left( \frac{\varphi '}{\varphi }+\alpha _{1}+\alpha _{2}\right) f'+\frac{\alpha }{3\varphi }f. \end{aligned}$$

(4.12)

Substituting (4.12) into (4.5) yields

$$\begin{aligned} af^{2}+bff'+2(f')^{2}=\varphi , \end{aligned}$$

(4.13)

where $a=\alpha _{1}\alpha _{2}+\frac{2\alpha }{3\varphi }$, $b=\frac{2}{3}\left[ \frac{\varphi '}{\varphi }-2(\alpha _{1}+\alpha _{2})\right]$. By Lemma 2.4, we have

$$\begin{aligned} 2(b^{2}-8a)\frac{\varphi '}{\varphi }=2(b^{2}-8a)'-b(b^{2}-8a). \end{aligned}$$

(4.14)

Now, we distinguish two subcases below.

Subcase 2.1 Suppose that $b^{2}-8a\not \equiv 0$. It follows from (4.14) that

$$\begin{aligned} 4\frac{\varphi '}{\varphi }=2(\alpha _{1}+\alpha _{2})+3\frac{(b^{2}-8a)'}{b^{2}-8a}. \end{aligned}$$

(4.15)

By integration, we see that there exists a $B\in {\mathbb {C}}{\setminus }\{0\}$ such that

$$\begin{aligned} \mathrm{e}^{2(\alpha _{1}+\alpha _{2})z}=B\varphi ^{4}(b^{2}-8a)^{-3}, \end{aligned}$$

(4.16)

which implies $\mathrm{e}^{2(\alpha _{1}+\alpha _{2})z}\in S(f)$, then $\alpha _{1}+\alpha _{2}=0$, a contradiction.

Subcase 2.2 Suppose that $b^{2}-8a\equiv 0$. Differentiating (4.13) yields

$$\begin{aligned} \varphi '=a'f^{2}+(2a+b')ff'+b(f')^{2}+bff''+4f'f''. \end{aligned}$$

(4.17)

Suppose $z_{0}$ is a simple zero of f which is not the zero of a, b. It follows from (4.13) and (4.17) that $z_{0}$ is a zero of $2\varphi f''-\left( \varphi '-\frac{b}{2}\varphi \right) f'$. Putting

$$\begin{aligned} \beta :=\frac{2\varphi f''-\left( \varphi '-\frac{b}{2}\varphi \right) f'}{f}, \end{aligned}$$

(4.18)

we have $T(r,\beta )=S(r,f)$. It follows that

$$\begin{aligned} f''=\left( \frac{1}{2}\frac{\varphi '}{\varphi }-\frac{b}{4}\right) f'+\frac{\beta }{2\varphi }f. \end{aligned}$$

(4.19)

Substituting (4.19) into (4.17) yields

$$\begin{aligned} \varphi '=cf^{2}+dff'+2\frac{\varphi '}{\varphi }(f')^{2}, \end{aligned}$$

(4.20)

where $c=a'+\frac{b\beta }{2\varphi }$, $d=2a+b'+\frac{b}{2}\frac{\varphi '}{\varphi }-\frac{b^{2}}{4}+\frac{2\beta }{\varphi }$. Eliminating $(f')^{2}$ from (4.13) and (4.20), we have

$$\begin{aligned} A(z)f(z)+B(z)f'(z)\equiv 0, \end{aligned}$$

(4.21)

where

$$\begin{aligned} A(z)= & {} c-a\frac{\varphi '}{\varphi }=a'+\frac{b\beta }{2\varphi }-a\frac{\varphi '}{\varphi },\\ B(z)= & {} d-b\frac{\varphi '}{\varphi }=2a+b'-\frac{b}{2}\frac{\varphi '}{\varphi }-\frac{b^{2}}{4}+\frac{2\beta }{\varphi }. \end{aligned}$$

Note that A(z) and B(z) are small functions of f. If $z_{0}$ is a simple zero of f and not the zero of B(z), it follows from (4.21) that $A(z)=B(z)\equiv 0$. By (4.19), we have

$$\begin{aligned} f''=\left( \frac{1}{2}\frac{\varphi '}{\varphi }-\frac{b}{4}\right) f'-\frac{1}{b}\left( a'-a\frac{\varphi '}{\varphi }\right) f, \end{aligned}$$

(4.22)

where $b=\frac{2}{3}\left[ \frac{\varphi '}{\varphi }-2(\alpha _{1}+\alpha _{2})\right] \not \equiv 0$. Otherwise, $e^{2(\alpha _{1}+\alpha _{2})z}=C\varphi \in S(f)$, then $\alpha _{1}+\alpha _{2}=0$, a contradiction. Substituting $b^{2}-8a\equiv 0$ into (4.22) yields

$$\begin{aligned} f''=\frac{1}{3}\left( \frac{\varphi '}{\varphi }+\alpha _{1}+\alpha _{2}\right) f' -\frac{1}{6}\left[ \left( \frac{\varphi '}{\varphi }\right) '-\frac{1}{2}\left( \frac{\varphi '}{\varphi }\right) ^{2} +(\alpha _{1}+\alpha _{2})\frac{\varphi '}{\varphi }\right] f.\qquad \quad \end{aligned}$$

(4.23)

It follows from (4.12) and (4.23) that

$$\begin{aligned} \frac{\alpha }{\varphi }=-\frac{1}{2}\left[ \left( \frac{\varphi '}{\varphi }\right) '-\frac{1}{2}\left( \frac{\varphi '}{\varphi }\right) ^{2} +(\alpha _{1}+\alpha _{2})\frac{\varphi '}{\varphi }\right] . \end{aligned}$$

(4.24)

We deduce that $\varphi '\not \equiv 0$. Otherwise, we assume that $\varphi '\equiv 0$, then $\frac{\alpha }{\varphi }\equiv 0$. Substituting this into $b^{2}-8a\equiv 0$ yields

$$\begin{aligned} 2\left( \frac{\alpha _{1}}{\alpha _{2}}\right) ^{2}-5\frac{\alpha _{1}}{\alpha _{2}}+2=0, \end{aligned}$$

which implies that $\frac{\alpha _{1}}{\alpha _{2}}=2$ or $\frac{\alpha _{1}}{\alpha _{2}}=\frac{1}{2}$. By substituting $\varphi '\equiv 0$ into (4.23), we obtain

$$\begin{aligned} f''=\frac{1}{3}(\alpha _{1}+\alpha _{2})f', \end{aligned}$$

then

$$\begin{aligned} f=\frac{3C_{1}}{\alpha _{1}+\alpha _{2}}\mathrm{e}^{\frac{1}{3}(\alpha _{1}+\alpha _{2})z}+C_{2}, \end{aligned}$$

where $C_{1},C_{2}\in {\mathbb {C}}{\setminus }\{0\}$. Without loss of generality, when $\frac{\alpha _{1}}{\alpha _{2}}=2$, then

$$\begin{aligned} f=\frac{C_{1}}{\alpha _{2}}\mathrm{e}^{\alpha _{2}z}+C_{2}. \end{aligned}$$

Substituting this into (1.4) and using Lemma 2.3, we can obtain a contradiction.

Differentiating (4.24) gives

$$\begin{aligned} \left( \frac{\alpha }{\varphi }\right) '=-\frac{1}{2}\left[ \left( \frac{\varphi '}{\varphi }\right) '' -\left( \frac{\varphi '}{\varphi }\right) '\frac{\varphi '}{\varphi } +(\alpha _{1}+\alpha _{2})\left( \frac{\varphi '}{\varphi }\right) '\right] . \end{aligned}$$

It follows from $b^{2}-8a\equiv 0$ that $bb'=4a'$, that is

$$\begin{aligned} \left( \frac{\alpha }{\varphi }\right) '=\frac{1}{6}\left( \frac{\varphi '}{\varphi }\right) '\left[ \frac{\varphi '}{\varphi }-2(\alpha _{1}+\alpha _{2})\right] . \end{aligned}$$

Putting $\gamma :=\frac{\varphi '}{\varphi }$ and combining the above two equality yields

$$\begin{aligned} (\alpha _{1}+\alpha _{2})\gamma '=2\gamma \gamma '-3\gamma ''. \end{aligned}$$

(4.25)

If $\gamma '\equiv 0$, then $\varphi =C_{3}e^{C_{4}z},~C_{3},C_{4}\in {\mathbb {C}}$. It follows from $\varphi '\not \equiv 0$ that $C_{3},C_{4}\ne 0$, which implies that $\varphi \not \in S(f)$, a contradiction. If $\gamma '\not \equiv 0$, it follows from (4.25) that

$$\begin{aligned} e^{(\alpha _{1}+\alpha _{2})z}=C_{5}\varphi ^{2}\left( \left( \frac{\varphi '}{\varphi }\right) '\right) ^{-3}, ~C_{5}\in {\mathbb {C}}{\setminus }\{0\}, \end{aligned}$$

which implies that $e^{(\alpha _{1}+\alpha _{2})z}\in S(f)$, then $\alpha _{1}+\alpha _{2}=0$, a contradiction.

This completes the proof of Theorem 1.2.

5 Proof of Theorem 1.3

Suppose that f is a transcendental entire solution of finite order of Eq. (1.7). In what follows, we consider three cases.

Case 1 If $\rho (f)<1$, using Lemma 2.5, from (1.7) we have

$$\begin{aligned} T(r,e^{Q(z)})= & {} m(r,\mathrm{e}^{Q(z)})=m\left( \frac{p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}-f^{n}(z)}{q(z)f(z+c)}\right) \\\le & {} m\left( r,\frac{1}{q(z)f(z+c)}\right) +m(r,p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z})\\&+m(r,f^{n}(z))+O(1)\\\le & {} m\left( r,\frac{f(z)}{q(z)f(z+c)}\right) +m\left( r,\frac{1}{f(z)}\right) +n T(r,f)\\&+2T(r,\mathrm{e}^{\lambda z})+S(r,\mathrm{e}^{\lambda z})\\\le & {} 2T(r,\mathrm{e}^{\lambda z})+S(r,\mathrm{e}^{\lambda z})+(n+1)T(r,f)+S(r,f)\\\le & {} 2T(r,\mathrm{e}^{\lambda z})+S(r,\mathrm{e}^{\lambda z}), \end{aligned}$$

then $\deg Q\le 1$, note that $\deg Q\ge 1$, therefore $\deg Q=1$. Denote $Q(z)=az+b,~a\in {\mathbb {C}}{\setminus }\{0\},~b\in {\mathbb {C}}$. Rewriting (1.7) in the following form:

$$\begin{aligned} f^{n}(z)+q(z)\mathrm{e}^{az+b}f(z+c)=p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z} \end{aligned}$$

(5.1)

and differentiating (5.1) we get

$$\begin{aligned} nf^{n-1}f'+A(z)\mathrm{e}^{az+b}=\lambda (p_{1}\mathrm{e}^{\lambda z}-p_{2}\mathrm{e}^{-\lambda z}), \end{aligned}$$

(5.2)

where

$$\begin{aligned} A(z)=q'(z)f(z+c)+aq(z)f(z+c)+q(z)f'(z+c). \end{aligned}$$

Eliminating $\mathrm{e}^{\lambda z}$ and $\mathrm{e}^{-\lambda z}$ from (5.1) and (5.2) yields

$$\begin{aligned} B(z)\mathrm{e}^{2(az+b)}+C(z)\mathrm{e}^{az+b}+D(z)\equiv 0, \end{aligned}$$

(5.3)

where

$$\begin{aligned} \left\{ \begin{array}{l} B(z)=\lambda ^{2}q^{2}(z)f^{2}(z+c)-A^{2}(z);\\ C(z)=2\lambda ^{2}q(z)f^{n}(z)f(z+c)-2nA(z)f^{n-1}(z)f'(z);\\ D(z)=\lambda ^{2}f^{2n}(z)-n^{2}f^{2(n-1)}(z)(f'(z))^{2}-4\lambda ^{2}p_{1}p_{2}. \end{array}\right. \end{aligned}$$

(5.4)

Thus, from Lemma 2.3, we have

$$\begin{aligned} B(z)\equiv C(z)\equiv D(z)\equiv 0. \end{aligned}$$

It follows from $D(z)\equiv 0$ that

$$\begin{aligned} f^{2(n-1)}(z)(\lambda ^{2}f^{2}(z)-n^{2}(f'(z))^{2})\equiv 4\lambda ^{2}p_{1}p_{2},~n\ge 3. \end{aligned}$$

Using Lemma 2.1, we have

$$\begin{aligned} m(r,\lambda ^{2}f^{2}(z)-n^{2}(f'(z))^{2})=S(r,f) \end{aligned}$$

and

$$\begin{aligned} m(r,f(z)(\lambda ^{2}f^{2}(z)-n^{2}(f'(z))^{2}))=S(r,f). \end{aligned}$$

We deduce that $\lambda ^{2}f^{2}(z)-n^{2}(f'(z))^{2}\not \equiv 0$. Otherwise, $4\lambda ^{2}p_{1}p_{2}=0$, a contradiction. Since f is entire,

$$\begin{aligned} T(r,f)= & {} m(r,f)\le m(r,f(z)(\lambda ^{2}f^{2}(z)-n^{2}(f'(z))^{2}))\\&+m\left( r,\frac{1}{\lambda ^{2}f^{2}(z)-n^{2}(f'(z))^{2}}\right) \\\le & {} T(r,\lambda ^{2}f^{2}(z)-n^{2}(f'(z))^{2})+S(r,f)=S(r,f), \end{aligned}$$

a contradiction.

Case 2 If $\rho (f)>1$. Denote $P(z)=p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}$, $H(z)=q(z)f(z+c)$. It is clear that $\rho (P)=1$, then $T(r,P)=S(r,f)$. Equation (1.7) can be written as:

$$\begin{aligned} f^{n}(z)+H(z)\mathrm{e}^{Q(z)}=P(z). \end{aligned}$$

(5.5)

Differentiating (5.5) yields

$$\begin{aligned} nf^{n-1}(z)f'(z)+L(z)\mathrm{e}^{Q(z)}=P'(z), \end{aligned}$$

(5.6)

where $L(z)=H'(z)+Q'(z)H(z)$. Eliminating $\mathrm{e}^{Q(z)}$ from (5.5) and (5.6), we have

$$\begin{aligned} f^{n-1}(z)(L(z)f(z)-nH(z)f'(z))=P(z)L(z)-P'(z)H(z). \end{aligned}$$

(5.7)

Note that $n-1\ge 2$ and $P(z)L(z)-P'(z)H(z)$ is a differential-difference polynomial in f and the total degree is at most 1. By Lemma 2.2 and Remark 2.1, we obtain

$$\begin{aligned} m(r,L(z)f(z)-nH(z)f'(z))=S(r,f) \end{aligned}$$

and

$$\begin{aligned} m(r,f(z)(L(z)f(z)-nH(z)f'(z)))=S(r,f). \end{aligned}$$

If $L(z)f(z)-nH(z)f'(z)\not \equiv 0$, then

$$\begin{aligned} T(r,f)= & {} m(r,f)\le m(r,f(z)(L(z)f(z)-n H(z)f'(z)))\\&+\,m\left( r,\frac{1}{L(z)f(z)-n H(z)f'(z)}\right) \\\le & {} T(r,L(z)f(z)-n H(z)f'(z))+S(r,f)=S(r,f), \end{aligned}$$

which yields a contradiction. If $L(z)f(z)-nH(z)f'(z)\equiv 0$, then

$$\begin{aligned} \frac{q'(z)}{q(z)}+Q'(z)+\frac{f'(z+c)}{f(z+c)}=n\frac{f'(z)}{f(z)}. \end{aligned}$$

By integration, we see that there exists a $C\in {\mathbb {C}}{\setminus }\{0\}$ such that

$$\begin{aligned} q(z)f(z+c)\mathrm{e}^{Q(z)}=Cf^{n}(z),~C\in {\mathbb {C}}{\setminus }\{0\}. \end{aligned}$$

(5.8)

Substituting (5.8) into (1.7) gives

$$\begin{aligned} (1+C)f^{n}(z)=P. \end{aligned}$$

If $1+C\ne 0$, we have $T(r,f)=m(r,f)=S(r,f)$, a contradiction. If $1+C=0$, then $P(z)=p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}\equiv 0$, a contradiction.

Case 3 If $\rho (f)=1$, by applying Lemma 2.5, from (1.7) we obtain

$$\begin{aligned} T(r,\mathrm{e}^{Q(z)})= & {} m(r,\mathrm{e}^{Q(z)})=m\left( \frac{p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z}-f^{n}(z)}{q(z)f(z+c)}\right) \\\le & {} m\left( r,\frac{1}{q(z)f(z+c)}\right) +m(r,p_{1}\mathrm{e}^{\lambda z}+p_{2}\mathrm{e}^{-\lambda z})+m(r,f^{n}(z))+O(1)\\\le & {} m\left( r,\frac{f(z)}{q(z)f(z+c)}\right) +m\left( r,\frac{1}{f(z)}\right) +n T(r,f)\\&+2T(r,\mathrm{e}^{\lambda z})+S(r,\mathrm{e}^{\lambda z})\\\le & {} (n+1)T(r,f)+S(r,f)+2T(r,\mathrm{e}^{\lambda z})+S(r,\mathrm{e}^{\lambda z}). \end{aligned}$$

Note that $\deg Q\ge 1$, then $1\le \deg Q=\sigma (\mathrm{e}^{Q(z)})\le \max \{\rho (\mathrm{e}^{\lambda z}),\rho (f)\}=1$, that is $\rho (f)=\deg Q=1$.

If f belongs to $\Gamma _{0}$, and noting that $\rho (f)=\deg Q=1$, we define $f=\mathrm{e}^{Az+B}$ and $Q(z)=az+b$, where $a,A\in {\mathbb {C}}{\setminus }\{0\}$ and $b,B\in {\mathbb {C}}$. Substituting these into (1.7) yields

$$\begin{aligned} \mathrm{e}^{nB}\mathrm{e}^{(nA+\lambda )z}+\mathrm{e}^{Ac+b+B}q(z)\mathrm{e}^{(A+a+\lambda )z}-p_{1}\mathrm{e}^{2\lambda z}-p_{2}=0. \end{aligned}$$

(5.9)

We distinguish four cases below.

Case 1 If $nA+\lambda =0$ and $A+a+\lambda =0$. By Lemma 2.3 and (5.9), we obtain $p_{1}=0$, a contradiction.

Case 2 If $nA+\lambda =0$ and $A+a+\lambda \ne 0$. If $A+a+\lambda \ne 2\lambda$, by Lemma 2.3 and (5.9), we have $p_{1}=q(z)\equiv 0$, a contradiction. If $A+a+\lambda =2\lambda$, then $A=-\frac{\lambda }{n},~a=\frac{n+1}{n}\lambda$, substituting these into (5.9) yields

$$\begin{aligned} \left( q(z)\mathrm{e}^{Ac+b+B}-p_{1}\right) \mathrm{e}^{2\lambda z}=p_{2}-\mathrm{e}^{nB}, \end{aligned}$$

by Lemma 2.3, we have $p_{2}-\mathrm{e}^{nB}\equiv q(z)\mathrm{e}^{Ac+b+B}-p_{1}\equiv 0$, then q(z) reduces to a non-zero constant, and $f(z)=\mathrm{e}^{-\frac{\lambda }{n}z+B},~Q(z)=\frac{n+1}{n}\lambda z+b$.

Case 3 If $nA+\lambda \ne 0$ and $A+a+\lambda =0$. If $nA+\lambda \ne 2\lambda$, it follows from Lemma 2.3 that $p_{1}=0$, a contradiction. If $nA+\lambda =2\lambda$, then $A=\frac{\lambda }{n},~a=-\frac{n+1}{n}\lambda$. Substituting these into (5.9) yields

$$\begin{aligned} (\mathrm{e}^{nB}-p_{1})\mathrm{e}^{2\lambda z}=p_{2}-q(z)\mathrm{e}^{Ac+b+B}, \end{aligned}$$

by Lemma 2.3, we have $\mathrm{e}^{nB}-p_{1}\equiv p_{2}-q(z)\mathrm{e}^{Ac+b+B}\equiv 0$, then q(z) reduces to a non-zero constant, and $f(z)=\mathrm{e}^{\frac{\lambda }{n}z+B},~Q(z)=-\frac{n+1}{n}\lambda z+b$.

Case 4 If $nA+\lambda \ne 0$ and $A+a+\lambda \ne 0$. If $nA+\lambda$, $A+a+\lambda$ and $2\lambda$ are pairwise distinct from each other, by Lemma 2.3 and (5.9), we have $p_{1}=p_{2}\equiv q(z)\equiv 0$, a contradiction. If only two of $nA+\lambda$, $A+a+\lambda$ and $2\lambda$ coincide, without loss of generality, suppose that $nA+\lambda =A+a+\lambda \ne 2\lambda$, then (5.9) can be written as:

$$\begin{aligned} \left( \mathrm{e}^{nB}+\mathrm{e}^{Ac+b+B}q(z)\right) \mathrm{e}^{(nA+\lambda )z}-p_{1}\mathrm{e}^{2\lambda z}-p_{2}=0. \end{aligned}$$

From the above equality and using Lemma 2.3, we have $p_{1}=p_{2}=0$, which implies a contradiction. If $nA+\lambda =A+a+\lambda =2\lambda$, then we write (5.9) as:

$$\begin{aligned} \left( \mathrm{e}^{nB}+\mathrm{e}^{Ac+b+B}q(z)-p_{1}\right) \mathrm{e}^{2\lambda z}-p_{2}=0. \end{aligned}$$

It follows from Lemma 2.3 that $p_{2}=0$, a contradiction.

This completes the proof of Theorem 1.3.