© Wuhan University 2022

0 Introduction and Main Results

In this paper, Nevanlinna value distribution theory is a useful tool and its standard notations as well as well-known theorems can be found in Refs.[1,2]. Let $f(z)$ be a meromorphic function in the complex plane $\mathbb{C}$. The proximity function, counting function of poles and Nevanlinna characteristic function with regard to $f(z)$ are denoted by $m(r,f)$,$N(r,f)$ and $T(r,f)$, respectively. We also use $S(r,f)$ to denote any small quantity of $f(z)$ satisfying $S(r,f)=o(T(r,f))$, as $r\to \mathrm{\infty }$, possibly outside of a set of finite logarithmic measure. The order of growth of $f(z)$ and the exponent of convergence of zeros of $f(z)$ are defined by

$\sigma (f)=\overline{\underset{r\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}}\frac{\mathrm{l}\mathrm{o}{\mathrm{g}}^{+}T(r,f)}{\mathrm{l}\mathrm{o}\mathrm{g}r},\lambda (f)=\overline{\underset{r\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}}\frac{\mathrm{l}\mathrm{o}\mathrm{g}N(r,\frac{\mathrm{1}}{f})}{\mathrm{l}\mathrm{o}\mathrm{g}r}$

A large number of studies concerning the solvability and existence for entire and meromorphic solutions of nonlinear complex differential equations, difference equations and differential-difference equations can be found in Refs.[3-12].

The equation $\mathrm{4}{f}^{\mathrm{3}}+\mathrm{3}f″=-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{3}z$ has exactly three nonconstant entire solutions $f_{\mathrm{1}}(z)=\mathrm{s}\mathrm{i}\mathrm{n}z,f_{\mathrm{2}}(z)=-\frac{\sqrt[]{\mathrm{3}}}{\mathrm{2}}\mathrm{c}\mathrm{o}\mathrm{s}z-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{s}\mathrm{i}\mathrm{n}z,f_{\mathrm{3}}(z)=\frac{\sqrt[]{\mathrm{3}}}{\mathrm{2}}\mathrm{c}\mathrm{o}\mathrm{s}z-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{s}\mathrm{i}\mathrm{n}z$, according to Yang and Li ^[11]. It is worth noting that $-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{3}z=\frac{\mathrm{1}}{\mathrm{2}\mathrm{i}}({\mathrm{e}}^{-\mathrm{3}\mathrm{i}z}-{\mathrm{e}}^{\mathrm{3}\mathrm{i}z})$, that is, $-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{3}z$ is a linear combinations of ${\mathrm{e}}^{\mathrm{3}\mathrm{i}z}$ and ${\mathrm{e}}^{-\mathrm{3}\mathrm{i}z}$. As a result, it is meaningful to investigate the existence of the solutions of the differential equation

$f^n(z)+Q_d(z,f)=p_{\mathrm{1}}{\mathrm{e}}^{\lambda z}+p_{\mathrm{2}}{\mathrm{e}}^{-\lambda z}$(1)

where $p_{\mathrm{1}},p_{\mathrm{2}},\lambda$ are non-zero constants, and $Q_d(z,f)$ denotes a differential in $f$ with degree $d$.

For Eq.(1), Li ^[6] obtained the following theorem.

Theorem 1 ^[6] Let $n\ge \mathrm{2}$ be an integer, $Q_d(z,f)$ be a differential polynomial in $f$ of degree at most $n-\mathrm{2}$, and $p_{\mathrm{1}},p_{\mathrm{2}},\lambda$ are three non-zero constants. If $f$ is an entire solution of equation (1), then $f(z)=c_{\mathrm{1}}{\mathrm{e}}^{\lambda z/n}+c_{\mathrm{2}}{\mathrm{e}}^{-\lambda z/n}$, where $c_{\mathrm{1}}$ and $c_{\mathrm{2}}$ are constants and ${c_i}^n=p_i(i=\mathrm{1,2})$.

Later, Li ^[13] proved by weakening the requirement on the degree of $Q_d(z,f)$ in Theorem A.

Theorem 2 ^[13] Let $n\ge \mathrm{2}$ be an integer, $Q_d(z,f)$ be a differential polynomial in $f$ of degree at most $n-\mathrm{1}$, and $p_{\mathrm{1}},p_{\mathrm{2}},\lambda$ are three non-zero constants. If $f$ is a meromorphic solution of equation (1) and $N(r,f)=S(r,f)$, then there exist two nonzero constants $c_{\mathrm{1}},c_{\mathrm{2}}({c_j}^n=p_j)$ and a small function $c_{\mathrm{0}}$ of $f$ such that $f(z)=c_{\mathrm{0}}+c_{\mathrm{1}}{\mathrm{e}}^{\frac{\lambda }{n}z}+c_{\mathrm{2}}{\mathrm{e}}^{-\frac{\lambda }{n}z}$.

Then it is natural to ask: is there a similar result if $p_{\mathrm{1}},p_{\mathrm{2}}$ are polynomials? The question was answered by the following theorem.

Theorem 3 ^[14] Let $n\ge \mathrm{4}$ be an integer, $Q_d(z,f)$ be a differential polynomial in $f$ of degree $d\le n-\mathrm{3}$. Let $p_{\mathrm{1}}(z)$ and $p_{\mathrm{2}}(z)$ be two nonzero polynomials, ${\alpha }_{\mathrm{1}},{\alpha }_{\mathrm{2}}$ be two nonzero constants with ${\alpha }_{\mathrm{1}}/{\alpha }_{\mathrm{2}}\ne$ rational. Then the differential equation

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

has no transcendental entire solutions.

The exponential polynomial plays a key role in the study of non-linear complex differential equations and many interesting properties have been mentioned, for example, in 2012, Wen, Heittokangas and Laine ^[15] investigated and classified the finite order entire solutions $f(z)$ of the following equation

$f^n(z)+q(z){\mathrm{e}}^{Q(z)}f(z+c)=P(z)$(3)

in terms of growth and zero distribution, where $n\ge \mathrm{2}$ is an integer, $c\in \mathbb{C}\backslash \{\mathrm{0}\}$, $q(z),Q(z),P(z)$ are polynomials.

Recall that an exponential polynomial of the form

$f(z)=P_{\mathrm{1}}(z){\mathrm{e}}^{Q_{\mathrm{1}}(z)}+P_{\mathrm{2}}(z){\mathrm{e}}^{Q_{\mathrm{2}}(z)}+\cdots +P_k(z){\mathrm{e}}^{Q_k(z)}$(4)

where $P_j(z),Q_j(z)(j=\mathrm{1,2},\cdots ,k)$ are polynomials in $z$ such that $\mathrm{m}\mathrm{a}\mathrm{x}\{\mathrm{d}\mathrm{e}\mathrm{g}{Q}_j(z):\mathrm{1}\le j\le k\}=s\ge \mathrm{1}$ is called an exponential polynomial of degree $s$.

In the following, we define two classes of transcendental entire functions:

${\mathrm{\Gamma }}_{\mathrm{0}}(z)=\{{\mathrm{e}}^{\alpha (z)}:\alpha (z)$is a non-constant polynomial$\}$

${\mathrm{\Gamma }}_{\mathrm{1}}(z)=$

$\{{\mathrm{e}}^{\alpha (z)}+d:\alpha (z)$ is a non-constant polynomial and $d\in \mathbb{C}\}.$

Theorem 4 ^[15] Let$n\ge \mathrm{2}$ be an integer, let $c\in \mathbb{C}\backslash \{\mathrm{0}\}$, and let $q(z),Q(z),P(z)$ be polynomials such that $Q(z)$ is not a constant and $q(z)\cancel{\equiv }\mathrm{0}$. Then we identify the finite order entire solutions $f$ of equation (3) as follows:

(i) Every solution $f$ satisfies $\sigma (f)=\mathrm{d}\mathrm{e}\mathrm{g}(Q)$ and is of mean type;

(ii) Every solution $f$ satisfies $\lambda (f)=\sigma (f)$ if and only if $P(z)\equiv \mathrm{0}$;

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

(iv) If a solution $f$ belongs to ${\mathrm{\Gamma }}_{\mathrm{0}}$ and if $g$ is any other finite order entire solution to (3), then $f=\eta g$, where ${\eta }^{n-\mathrm{1}}=\mathrm{1}$;

(v) If $f$ is an exponential polynomial solution of the form (4), then $f\in {\mathrm{\Gamma }}_{\mathrm{1}}$. Moreover, if $f\in {\mathrm{\Gamma }}_{\mathrm{1}}\backslash {\mathrm{\Gamma }}_{\mathrm{0}}$ , then $\sigma (f)=\mathrm{1}$.

Following that, Liu ^[16] and Chen et al ^[5] investigated the cases when $f(z+c)$ in Eq.(3) is replaced by $f^{(k)}(z+c)$ or ${\mathrm{\Delta }}_{c}f(z)$, respectively, and obtained a special form of solutions of the equations.

According to the foregoing results, all equations have only one main term $f^n$ on the left side.

Chen et al ^[17] proposed the following question based on whether a similar result can be obtained when there are two main terms.

In 2021, Chen, Hu and Wang ^[17] considered the following differential-difference equation

$f^n(z)+\omega f^{n-\mathrm{1}}(z)f\text{'}(z)+q(z){\mathrm{e}}^{Q(z)}f(z+c)=p_{\mathrm{1}}{\mathrm{e}}^{\lambda z}+p_{\mathrm{2}}{\mathrm{e}}^{-\lambda z}$(5)

and obtained the following result.

Theorem 5 ^[17] If $f$ is a transcendental entire solution with finite order of Eq.(5), then the following conclusions hold:

(i) If $n\ge \mathrm{4}$ for $\omega \ne \mathrm{0}$ and $n\ge \mathrm{3}$ for $\omega \ne \mathrm{0}$, then every solution $f$ satisfies$\sigma (f)=\mathrm{d}\mathrm{e}\mathrm{g}(Q)=\mathrm{1}$.

(ii) If $n\ge \mathrm{1}$ and $f$ is a solution of (5), which belongs to ${\mathrm{\Gamma }}_{\mathrm{0}}$, then

$f(z)={\mathrm{e}}^{\frac{\lambda }{n}z+B},Q(z)=-\frac{n+\mathrm{1}}{n}z+b$

or

$f(z)={\mathrm{e}}^{-\frac{\lambda }{n}z+B},Q(z)=\frac{n+\mathrm{1}}{n}z+b$

where $b,B\in \mathbb{C}$.

Inspired by the above results, we consider the following types of equations:

$f^n(z)+\omega f^{n-\mathrm{1}}(z)f\text{'}(z)+f^{\left(k\right)}(z+c)=p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(6)

and

$f^n(z)+\omega f^{n-\mathrm{1}}(z)f\text{'}(z)+q(z)f^{\left(k\right)}(z+c){\mathrm{e}}^{Q\left(z\right)}=p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(7)

Theorem 6   Let $n\ge \mathrm{5}$ and $k\ge \mathrm{0}$ be integers, $\omega ,p_{\mathrm{1}},p_{\mathrm{2}},{\alpha }_{\mathrm{1}}$ and ${\alpha }_{\mathrm{2}}$ be nonzero constants such that ${\alpha }_{\mathrm{1}}\ne {\alpha }_{\mathrm{2}}$. Suppose that ${\alpha }_{\mathrm{1}}/{\alpha }_{\mathrm{2}}\ne n$ and ${\alpha }_{\mathrm{2}}/{\alpha }_{\mathrm{1}}\ne n$. Then Eq.(6) has no transcendental entire solutions.

Theorem 7   Let $n\ge \mathrm{4}$ and $k\ge \mathrm{0}$ be integers, $q(z)$ be a non-vanishing polynomial and $Q(z)$ be a non-constant polynomial. Suppose that $\omega ,p_{\mathrm{1}},p_{\mathrm{2}},{\alpha }_{\mathrm{1}}$ and ${\alpha }_{\mathrm{2}}$ be nonzero constants such that ${\alpha }_{\mathrm{1}}\ne {\alpha }_{\mathrm{2}}$. If Eq.(7) admits a finite order transcendental entire function $f(z)$ with $\lambda (f)<\sigma (f)$, then the following conclusions hold:

(i) If $n\ge \mathrm{4}$ for $\omega \ne \mathrm{0}$, then every solution $f$ satisfies $\sigma (f)=\mathrm{d}\mathrm{e}\mathrm{g}(Q)=\mathrm{1}$.

(ii) If a solution $f$ belongs to ${\mathrm{\Gamma }}_{\mathrm{0}}$ , then

$f(z)={\mathrm{e}}^{\frac{{\alpha }_{\mathrm{2}}}{n}z+B},Q(z)=({\alpha }_{\mathrm{1}}-\frac{{\alpha }_{\mathrm{2}}}{n})z+b$

or

$f(z)={\mathrm{e}}^{\frac{{\alpha }_{\mathrm{1}}}{n}z+B},Q(z)=({\alpha }_{\mathrm{2}}-\frac{{\alpha }_{\mathrm{1}}}{n})z+b$

where $b,B\in \mathbb{C}$.

1 Preliminary Lemmas

In order to prove our results, we introduce the following Lemmas.

Lamma 1 ^[18] Let $f(z)$ be a non-constant meromorphic function and let $k\ge \mathrm{1}$. Then if the growth order of $f(z)$ is finite, we have

$m(r,\frac{f^{(k)}}{f})=O(\mathrm{l}\mathrm{o}\mathrm{g}r)$

And if the growth order of $f(z)$ is infinite, we have

$m(r,\frac{f^{(k)}}{f})=O(\mathrm{l}\mathrm{o}\mathrm{g}T(r,f)+\mathrm{l}\mathrm{o}\mathrm{g}r),\mathrm{a}\mathrm{s}r\to \mathrm{\infty }$

possibly outside a set of finite linear measure.

Lamma 2 ^[19] Let $f(z)$ be a transcendental meromorphic function of finite order. Then

$T(r,f(z+c))=T(r,f(z))+S(r,f(z))$

where $S(r,f(z))$ denotes any quantity $S(r,f(z))=o(T(r,f(z)))$ as $r\to \mathrm{\infty }$, possible outside of a set of finite logarithmic measure.

In the study of complex differential-difference equations, the Clunie lemma^[10] plays an important part. Let's recall the Clunie lemma's precise definition.

Lamma 3 (Clunie's lemma) ^[10] Let $f(z)$ be a transcendental meromorphic solution of finite order of differential-difference equation

$f^{n}P(z,f)=Q(z,f)$

where $P(z,f),Q(z,f)$ are polynomials in $f(z)$ and its derivatives and its shifts with small meromorphic coefficients. If the total degree of $Q(z,f)$ as a polynomial in $f(z)$ and its derivatives and its shifts is $n$, then

$m(r,P(z,f))=S(r,f)$

for all $r$ out of a possible exceptional set of finite logarithmic measure.

Lamma 4 ^[18] Let $f_j(z)(j=\mathrm{1,2},\cdots ,n,n\ge \mathrm{2})$ be meromorphic functions, and let $g_j(z)(j=\mathrm{1,2},\cdots ,n,n\ge \mathrm{2})$ be entire functions satisfying

(i) ${\sum }_{j=\mathrm{1}}^n{f}_j(z){\mathrm{e}}^{g_j(z)}\equiv \mathrm{0}$;

(ii) when $\mathrm{1}<j<k\le n$, then $g_j(z)-g_k(z)$ is not a constant;

(iii) when $\mathrm{1}\le j\le n,\mathrm{1}\le h<k\le n$, then

$T(r,f_j(z))=o(T(r,{\mathrm{e}}^{g_h-g_j}))(r\to \mathrm{\infty },r\notin E)$

where $E\subset (\mathrm{1},\mathrm{\infty })$ is of finite linear measure or logarithmic measure.

Then $f_j(z)\equiv \mathrm{0}(j=\mathrm{1},\cdots ,n)$.

2 Proof of Theorem 6

Let $f$ be a transcendental entire solution of Eq.(6).

Set $L=f^n+\omega f^{n-\mathrm{1}}f\text{'}$,$H=f^{(k)}(z+c)$. Then we write Eq.(6) as

$L+H=p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(8)

Differentiating (8) yields

$L\text{'}+H\text{'}=p_{\mathrm{1}}{\alpha }_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\alpha }_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(9)

Eliminating ${\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}$ from (8) and (9), we get

$({\alpha }_{\mathrm{1}}L-L\text{'})+({\alpha }_{\mathrm{1}}H-H\text{'})=p_{\mathrm{2}}({\alpha }_{\mathrm{1}}-{\alpha }_{\mathrm{2}}){\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(10)

Differentiating (10) yields

$({\alpha }_{\mathrm{1}}L\text{'}-L″)+({\alpha }_{\mathrm{1}}H\text{'}-H″)=p_{\mathrm{2}}({\alpha }_{\mathrm{1}}-{\alpha }_{\mathrm{2}}){\alpha }_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(11)

Eliminating ${\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$ from (10) and (11), we get

${\alpha }_{\mathrm{1}}{\alpha }_{\mathrm{2}}L-({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})L\text{'}+L″=-[{\alpha }_{\mathrm{1}}{\alpha }_{\mathrm{2}}H-({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})H\text{'}+H″]$(12)

Note that

$L\text{'}=n{f}^{n-\mathrm{1}}f\text{'}+(n-\mathrm{1})\omega f^{n-\mathrm{2}}{(f\text{'})}^{\mathrm{2}}+\omega f^{n-\mathrm{1}}f″$

and

$\begin{array}{l}L\text{'}=n(n-\mathrm{1})f^{n-\mathrm{2}}{(f\text{'})}^{\mathrm{2}}+n{f}^{n-\mathrm{1}}f″+(n-\mathrm{1})(n-\mathrm{2})\omega f^{n-\mathrm{3}}{(f\text{'})}^{\mathrm{3}}\\ +\mathrm{3}(n-\mathrm{1})\omega f^{n-\mathrm{2}}f\text{'}f″+\omega f^{n-\mathrm{1}}f‴\end{array}$

Substituting $L,L\text{'},L″$ into (12), we have

$f^{n-\mathrm{3}}\phi =-[{\alpha }_{\mathrm{1}}{\alpha }_{\mathrm{2}}H-({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})H\text{'}+H″]$(13)

where

$\phi ={\alpha }_{\mathrm{1}}{\alpha }_{\mathrm{2}}{f}^{\mathrm{3}}+[\omega {\alpha }_{\mathrm{1}}{\alpha }_{\mathrm{2}}-n({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})]f^{\mathrm{2}}f\text{'}$

$+[n-\omega ({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})]f^{\mathrm{2}}f″+\omega f^{\mathrm{2}}f‴$

$+(n-\mathrm{1})[n-\omega ({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})]f{(f\text{'})}^{\mathrm{2}}+\mathrm{3}\omega (n-\mathrm{1})ff\text{'}f‴$

$+\omega (n-\mathrm{1})(n-\mathrm{2}){(f\text{'})}^{\mathrm{3}}$(14)

Note that $n-\mathrm{3}\ge \mathrm{2}$ and ${\alpha }_{\mathrm{1}}{\alpha }_{\mathrm{2}}H-({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})H\text{'}+H″$ is a differential-difference polynomial in $f$ and the total degree is at most 1. By Lemma 3, we obtain

$m(r,\phi )=S(r,f),m(r,f\phi )=S(r,f)$

If $\phi \cancel{\equiv }\mathrm{0}$, then

$T(r,f)=m(r,f)\le m(r,f\phi )+m(r,\frac{\mathrm{1}}{\phi })$

$\le T(r,\phi )+S(r,f)=S(r,f)$(15)

which yields a contradiction.

Thus, $\phi \equiv \mathrm{0}$. From (13), we get ${\alpha }_{\mathrm{1}}{\alpha }_{\mathrm{2}}H-({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})H\text{'}+H″\equiv \mathrm{0}$. Then $H$ has the form

$H=f^{(k)}(z+c)=a_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+a_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(16)

where $a_{\mathrm{1}}$ and $a_{\mathrm{2}}$ are constants.

By integration, we have

$f(z)=t_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+t_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(17)

where $t_{\mathrm{1}}=\frac{a_{\mathrm{1}}{\mathrm{e}}^{-{\alpha }_{\mathrm{1}}c}}{{{\alpha }_{\mathrm{1}}}^k}$ and $t_{\mathrm{2}}=\frac{a_{\mathrm{2}}{\mathrm{e}}^{-{\alpha }_{\mathrm{2}}c}}{{{\alpha }_{\mathrm{2}}}^k}$ are constants.

Similarly, by $\phi \equiv \mathrm{0}$ and (12) we can also obtain ${\alpha }_{\mathrm{1}}{\alpha }_{\mathrm{2}}L-({\alpha }_{\mathrm{1}}+{\alpha }_{\mathrm{2}})L\text{'}+L″\equiv \mathrm{0}$, then we have

$L=f^n+\omega f^{n-\mathrm{1}}f\text{'}=m_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+m_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(18)

where $m_{\mathrm{1}}$ and $m_{\mathrm{2}}$ are constants. From (17) and (18), we obtain

$\begin{array}{l}(t_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+t_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}{)}^{n-\mathrm{1}}[t_{\mathrm{1}}(\omega {\alpha }_{\mathrm{1}}+\mathrm{1}){\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+t_{\mathrm{2}}(\omega {\alpha }_{\mathrm{2}}+\mathrm{1}){\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}]\\ =m_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+m_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}\end{array}$(19)

Dividing both sides of (19) by ${\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$, we get

$\begin{array}{l}{t}_{\mathrm{1}}(\omega {\alpha }_{\mathrm{1}}+\mathrm{1})[{\displaystyle \sum _{i=\mathrm{0}}^{n-\mathrm{2}}}{C}_{n-\mathrm{1}}^i(t_{\mathrm{1}}{)}^i(t_{\mathrm{2}}{)}^{n-i-\mathrm{1}}{\mathrm{e}}^{(i+\mathrm{1}){\alpha }_{\mathrm{1}}z+(n-i-\mathrm{2}){\alpha }_{\mathrm{2}}z}\\ +(t_{\mathrm{1}}{)}^{n-\mathrm{1}}{\mathrm{e}}^{n{\alpha }_{\mathrm{1}}z-{\alpha }_{\mathrm{2}}z}]\\ +t_{\mathrm{2}}(\omega {\alpha }_{\mathrm{2}}+\mathrm{1}){\displaystyle \sum _{i=\mathrm{0}}^{n-\mathrm{1}}}{C}_{n-\mathrm{1}}^i(t_{\mathrm{1}}{)}^i(t_{\mathrm{2}}{)}^{n-i-\mathrm{1}}{\mathrm{e}}^{i{\alpha }_{\mathrm{1}}z+(n-i-\mathrm{1}){\alpha }_{\mathrm{2}}z}\\ =m_{\mathrm{1}}{\mathrm{e}}^{({\alpha }_{\mathrm{1}}-{\alpha }_{\mathrm{2}})z}+m_{\mathrm{2}}\end{array}$(20)

Since ${\alpha }_{\mathrm{1}}/{\alpha }_{\mathrm{2}}\ne n$, ${\alpha }_{\mathrm{2}}/{\alpha }_{\mathrm{1}}\ne n$ and ${\alpha }_{\mathrm{1}}\ne {\alpha }_{\mathrm{2}}$, by (20) and Lemma 4, we obtain $(t_{\mathrm{1}}{)}^{n-\mathrm{1}}\equiv \mathrm{0}$, that is $t_{\mathrm{1}}\equiv \mathrm{0}$. Similarly, dividing both sides of (19) by ${\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}$, we can get $t_{\mathrm{2}}\equiv \mathrm{0}$. So $f(z)\equiv \mathrm{0}$, this is a contradiction.

Therefore, Eq.(7) has no transcendental entire solutions.

3 Proof of Theorem 7

Suppose that $f$ is a transcendental entire solution of finite order of Eq.(7). In what follows, we firstly prove that $\sigma (f)=\mathrm{1}$.

Case A Suppose that $\sigma (f)<\mathrm{1}$. By Lemma 1, Lemma 2 and (7), we have

$\begin{array}{l}T(r,{\mathrm{e}}^{Q(z)})=m(r,{\mathrm{e}}^{Q(z)})\\ =m(r,\frac{p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}-f^n-\omega f^{n-\mathrm{1}}f\text{'}}{q(z)f^{(k)}(z+c)})\\ \le m(r,p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z})+m(r,f^n+\omega f^{n-\mathrm{1}}f\text{'})\\ +m(r,\frac{\mathrm{1}}{q(z)f^{(k)}(z+c)})\\ \le m(r,p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{e}^{{\alpha }_{\mathrm{2}}z})+m(r,f^n)+m(r,\frac{f\text{'}}{f})\\ \le (n+k+\mathrm{1})T(r,f)+T(r,p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}\\ +S(r,p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z})\end{array}$

$\le \mathrm{m}\mathrm{a}\mathrm{x}\{\sigma (p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}),\sigma (f)\}$(21)

then $\mathrm{d}\mathrm{e}\mathrm{g}(Q)\le \mathrm{1}$.

Note that $\mathrm{d}\mathrm{e}\mathrm{g}(Q)\ge \mathrm{1}$, therefore $\mathrm{d}\mathrm{e}\mathrm{g}Q=\mathrm{1}$. Denote $Q(z)=az+b,a\in \mathbb{C}\backslash \{\mathrm{0}\},b\in \mathbb{C}$. Rewriting (7) in the following form:

$f^n+\omega f^{n-\mathrm{1}}f\text{'}+q(z){\mathrm{e}}^{az+b}{f}^{(k)}(z+c)=p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$(22)

Differentiating (22) yields

$\begin{array}{l}n{f}^{n-\mathrm{1}}f\text{'}+\omega (n-\mathrm{1})f^{n-\mathrm{2}}{(f\text{'})}^{\mathrm{2}}+\omega f^{n-\mathrm{1}}f″+A(z){\mathrm{e}}^{az+b}\\ ={\alpha }_{\mathrm{1}}{p}_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+{\alpha }_{\mathrm{2}}{p}_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}\end{array}$(23)

where

$A(z)=q\text{'}(z)f^{(k)}(z+c)+q(z)f^{(k+\mathrm{1})}(z+c)+aq(z)f^{(k)}(z+c)$

Eliminating ${\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}$ from (22) and (23), we obtain

$\begin{array}{l}(n-{\alpha }_{\mathrm{2}}\omega )f^{n-\mathrm{1}}f\text{'}+(n-\mathrm{1})\omega f^{n-\mathrm{2}}{(f\text{'})}^{\mathrm{2}}+\omega f^{n-\mathrm{1}}f″-{\alpha }_{\mathrm{2}}{f}^n\\ +[A(z)-{\alpha }_{\mathrm{2}}q(z)f^{(k)}(z+c)]{\mathrm{e}}^{az+b}=({\alpha }_{\mathrm{1}}-{\alpha }_{\mathrm{2}})p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}\end{array}$(24)

Subcase A1 If $a\ne {\alpha }_{\mathrm{1}}$, by (24) and Lemma 4, we have

${\alpha }_{\mathrm{1}}-{\alpha }_{\mathrm{2}}\equiv \mathrm{0}$

which contradicts with ${\alpha }_{\mathrm{1}}\ne {\alpha }_{\mathrm{2}}$.

Subcase A2 If $a={\alpha }_{\mathrm{1}}$, by (24), we can get

$\begin{array}{l}(n-{\alpha }_{\mathrm{2}}\omega )f^{n-\mathrm{1}}f\text{'}+(n-\mathrm{1})\omega f^{n-\mathrm{2}}{(f\text{'})}^{\mathrm{2}}+\omega f^{n-\mathrm{1}}f″-{\alpha }_{\mathrm{2}}{f}^n\\ +[(A(z)-{\alpha }_{\mathrm{2}}q(z)f^{(k)}(z+c)){\mathrm{e}}^b-({\alpha }_{\mathrm{1}}-{\alpha }_{\mathrm{2}})p_{\mathrm{1}}]{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}=\mathrm{0}\end{array}$(25)

From (25) and Lemma 4, we have

$(n-{\alpha }_{\mathrm{2}}\omega )f^{n-\mathrm{1}}f\text{'}+(n-\mathrm{1})\omega f^{n-\mathrm{2}}{(f\text{'})}^{\mathrm{2}}+\omega f^{n-\mathrm{1}}f″-{\alpha }_{\mathrm{2}}{f}^n\equiv \mathrm{0}$(26)

Dividing with $f^n$ on both sides of (26), we obtain

$(n-{\alpha }_{\mathrm{2}}\omega )\frac{f\text{'}}{f}+(n-\mathrm{1})\omega {(\frac{f\text{'}}{f})}^{\mathrm{2}}+\omega \frac{f″}{f}-{\alpha }_{\mathrm{2}}=\mathrm{0}$(27)

Since $\frac{f″}{f}=(\frac{f\text{'}}{f})\text{'}+{(\frac{f\text{'}}{f})}^{\mathrm{2}}$, which yields a Riccati differential equation:

$(n-{\alpha }_{\mathrm{2}}\omega )t+\omega t\text{'}+n\omega t^{\mathrm{2}}-{\alpha }_{\mathrm{2}}=\mathrm{0}$(28)

where $t=\frac{f\text{'}}{f}$.

By a simple calculation, we can get

$t=\frac{\mathrm{1}}{n}\frac{{\left(-\frac{n}{{\alpha }_{\mathrm{2}}+\frac{n}{\omega }}{\mathrm{e}}^{-({\alpha }_{\mathrm{2}}+\frac{n}{\omega })z}+C_{\mathrm{1}}\right)}^{\text{'}}}{-\frac{n}{{\alpha }_{\mathrm{2}}+\frac{n}{\omega }}{\mathrm{e}}^{-({\alpha }_{\mathrm{2}}+\frac{n}{\omega })z}+C_{\mathrm{1}}}+\frac{{\alpha }_{\mathrm{2}}}{n}$

then

$f^n=C_{\mathrm{2}}\left[-\frac{n}{{\alpha }_{\mathrm{2}}+\frac{n}{\omega }}{\mathrm{e}}^{-({\alpha }_{\mathrm{2}}+\frac{n}{\omega })z}+C_{\mathrm{1}}\right]{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$

where $C_{\mathrm{1}},C_{\mathrm{2}}$ are constants.

If $C_{\mathrm{2}}=\mathrm{0}$, then $f^n=\mathrm{0}$, this is a contradiction.

If $C_{\mathrm{2}}\ne \mathrm{0}$, then $\sigma (f)=\mathrm{1}$, which contradicts with $\sigma (f)<\mathrm{1}$.

Case B Suppose that $\sigma (f)>\mathrm{1}$. Denote $P(z)=p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z}$,$G(z)=q(z)f^{(k)}(z+c)$. Equation (7) can be written as:

$f^n+\omega f^{n-\mathrm{1}}f\text{'}+G(z){\mathrm{e}}^{Q(z)}=P(z)$(29)

Differentiating (29) yields

$n{f}^{n-\mathrm{1}}f\text{'}+(n-\mathrm{1})\omega f^{n-\mathrm{2}}{(f\text{'})}^{\mathrm{2}}+\omega f^{n-\mathrm{1}}f″+L(z){\mathrm{e}}^{Q(z)}=P\text{'}(z)$(30)

where $L(z)=G\text{'}(z)+G(z)Q\text{'}(z)$.

Eliminating ${\mathrm{e}}^{Q(z)}$ from (29) and (30), we get

$f^{n-\mathrm{2}}\psi =PL-P\text{'}G$(31)

where $\psi =L{f}^{\mathrm{2}}+(\omega L-nG)ff\text{'}-\omega Gff″-(n-\mathrm{1})\omega G{(f\text{'})}^{\mathrm{2}}$.

Note that $n-\mathrm{2}\ge \mathrm{2}$ and $PL-P\text{'}G$ is a differential-difference polynomial in $f$ and the total degree is at most 1. By Lemma 3, we obtain

$m(r,\psi )=S(r,f)$and $m(r,f\psi )=S(r,f)$.

If $\psi \cancel{\equiv }\mathrm{0}$, since $f$ is a transcendental entire function, then

$\begin{array}{l}T(r,f)=m(r,f)\le m(r,f\psi )+m(r,\frac{\mathrm{1}}{\psi })\\ \le T(r,\psi )+S(r,f)=S(r,f)\end{array}$

which is a contradiction.

If $\psi \equiv \mathrm{0}$, from (31) yields $PL-P\text{'}G\equiv \mathrm{0}$, then

$\frac{P\text{'}}{P}=\frac{q\text{'}}{q}+\frac{f^{(k+\mathrm{1})}(z+c)}{f^{(k)}(z+c)}+Q\text{'}$

By integration, we see that there exists a constant $C_{\mathrm{3}}\in \mathbb{C}\backslash \{\mathrm{0}\}$ such that

$P=C_{\mathrm{3}}q{f}^{(k)}(z+c){\mathrm{e}}^Q$(32)

Substituting (32) into (7) yields

$f^n+\omega f^{n-\mathrm{1}}f\text{'}=(\mathrm{1}-\frac{\mathrm{1}}{C_{\mathrm{3}}})(p_{\mathrm{1}}{\mathrm{e}}^{{\alpha }_{\mathrm{1}}z}+p_{\mathrm{2}}{\mathrm{e}}^{{\alpha }_{\mathrm{2}}z})$(33)