Some results on transcendental entire solutions to certain nonlinear differential-difference equations

1.   Introduction

Let $f(z)$ be a nonconstant function meromorphic on the complex plane $\mathbb{C}$. We assume that the reader is familiar with the fundamental results and standard notations of Nevanlinna theory (see ^[10,12,21]). For simplicity, we denote by $S(r, f)$ any quantity satisfying $S(r, f) = o(T(r, f))$ as $r\to \infty$, outside of a possible exceptional set of finite logarithmic measure.

Nevanlinna theory and its difference analogues have been used to study the growth, oscillation, solvability and existence of entire or meromorphic solutions to differential equations, difference equations and differential-difference equations, see, e.g., ^{[1,2,4,6,12,13,14,15,16,18,19,20]}.

In 1964, Hayman ^[10] considered the following non-linear differential equation

where $Q_{d}(z, f)$ is a differential polynomial in $f$ with degree $d$ and obtained the following result which is an extension of Tumura–Clunie theory.

Theorem A. Suppose that $f(z)$ is a nonconstant meromorphic function, $d\leq n-1$, and $f, g$ satisfy $N(r, f)+N(r, 1/g) = S(r, f)$ in (1.1). Then we have $g(z) = (f(z)+\gamma(z))^{n}$, where $\gamma(z)$ is meromorphic and a small function of $f(z)$.

Since then, the non-linear differential equation (1.1) has been studied extensively over the years, see ^[16,19,20] etc.

In 2006, Li and Yang ^[13] studied the particular case that $g(z)$ has the form $p_{1}e^{\alpha_{1}z}+p_{2}e^{\alpha_{2}z}$ in Eq (1.1) and proved the following result.

Theorem B. Let $n\geq 4$ be an integer and $Q_{d}(z, f)$ denote an algebraic differential polynomial in $f$ of degree $d\leq n-3$. Let $p_{1}, p_{2}$ be two nonzero polynomials, $\alpha_{1}$ and $\alpha_{2}$ be two nonzero constants with $\alpha_{1}/\alpha_{2}\neq$ rational. Then the differential equation

has no transcendental entire solutions.

Moreover, Yang and Li ^[13] also studied the case when $n = 3$, and found the exact forms of solutions to Eq (1.2) under some related conditions.

In 2014, Liao and Ye ^[14] investigated the structure of solutions to the following differential equation

with non-zero rational function $u$ and non-constant polynomial $v$ and obtained the following result.

Theorem C. Suppose that $f$ is a meromorphic solution of (1.3) which has finitely many poles. Then

for $n\geq d+1$ and $s$ is a rational function satisfying $s^{n}[(n+1)s'+v's] = (n+1)u$.

In 2012, Wen et al. ^[18] classified the finite order entire solutions of the equation

where $q, Q, P$ are polynomials, $n\geq 2$ is an integer, and $c\in \mathbb{C}\setminus\{0\}$. In 2019, Chen et al. ^[2] derived some conclusions when the term $P(z)$ on the right-hand side of Eq (1.4) is replaced by $p_{1}e^{\lambda z}+p_{2}e^{-\lambda z}$, where $p_{1}, p_{2}, \lambda$ are non-zero constants.

Based on the above results, one can see that there exists only one dominant term $f^{n}$ or $f^{n}f'$ on the left-hand side of these equations. In 2020, Chen, Hu and Wang ^[4] investigated the following non-linear differential-difference equation which has two dominated terms

where $n$ is a positive integer, $c\neq 0$, $\omega$ are constants, $q, Q, u, v$ are polynomials such that $Q, v$ are not constants and $q, u \not\equiv0$, and obtained the following result.

Theorem D. Let $n$ be an integer satisfying $n\geq 3$ for $\omega\neq0$ and $n\geq2$ for $\omega = 0$. Suppose that $f$ is a non-vanishing transcendental entire solution of finite order of (1.5). Then every solution $f$ satisfies one of the following results:

(1) $\rho(f) < \deg v = \deg Q$ and $f = Ce^{-z/\omega}$, where $C$ is a constant.

(2) $\rho(f) = \deg Q\geq \deg v$.

In the meantime, Chen et al. ^[4] proposed a conjecture that the conclusions of Theorem D are still valid when $n = 2$ and $\omega\neq0$.

In this paper, we consider the above conjecture and obtain the following result, which is a complement of Theorem D.

Theorem 1.1. Let $c, \, \widetilde{\omega}\neq 0$ be constants, $q, Q, u, v$ be polynomials such that $Q, v$ are not constants and $q, u\not\equiv0$. Suppose that $f$ is a transcendental entire solution with finite order to

satisfying $\lambda(f) < \rho(f)$, then $\deg Q = \deg v$, and one of the following relations holds:

(1) $\sigma(f) < \deg Q = \deg v$, and $f = Ce^{-z/\widetilde{\omega}}$

(2) $\sigma(f) = \deg Q = \deg v$.

The following two examples given by Chen et al. ^[4] can illustrate the Conclusions (1) and (2) of Theorem 1.1, respectively.

Example 1.2. $f_{0}(z) = 2e^{-z}$ is a transcendental entire solution to the following differential-difference equation

where $\widetilde{\omega} = 1\neq0$, $Q = z^{2}+z+1$, $v = z^{2}$ and $0 = \lambda(f_{0}) < \sigma(f_{0}) = 1$. Then we have $\sigma(f_{0}) = 1 < 2 = \deg Q = \deg v$, and $f_{0} = Ce^{-z/\widetilde{\omega}}$, where $C = 2$. This shows that the Conclusion (1) of Theorem 1.1 occurs.

Example 1.3. $f_{1}(z) = e^{z^{2}}$ is a transcendental entire solution to the following differential-difference equation

where $\widetilde{\omega} = 1\neq0$, $Q = z^{2}-2z-1$, $v = 2z^{2}$ and $0 = \lambda(f_{1}) < \sigma(f_{1}) = 2$. Then we have $\sigma(f_{1}) = 2 = \deg Q = \deg v$. This illustrates that the Conclusion (2) of Theorem 1.1 also exits.

In ^[4], Chen, Hu and Wang also investigated the entire solutions with finite order to the following differential-difference equation

where $n$ is an integer, $c, \lambda, p_{1}, p_{2}$ are non-zero constants and $\omega$ is a constant, and $q\not\equiv 0$, $Q$ are polynomials such that $Q$ is not a constant. They obtained the following result.

Theorem E. If $f$ is a transcendental entire solution with finite order to Eq. (1.7), then the following conclusions hold:

(i) If $n\geq4$ for $\omega\neq0$ and $n\geq3$ for $\omega = 0$, then every solution $f$ satisfies $\rho(f) = \deg Q = 1$.

(ii) If $n\geq1$ and $f$ is a solution to Eq.(1.7) which belongs to $\Gamma_{0}$, then

or

where $b, B\in \mathbb{C}$, and $\Gamma_{0} = \{e^{\alpha(z)}:\alpha(z)$ is a non-constant polynomial$\}$.

Given Theorem E, it is natural to ask: how about the solutions to the following more general form

where $n$ is a positive integer, $\omega$ is a constant and $c, \lambda_{1}, \lambda_{2}, p_{1}, p_{2}$ are non-zero constants, $q, Q$ are polynomials such that $Q$ is not a constant and $q\not\equiv0$.

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

Theorem 1.4. If $f$ is a transcendental entire solution with finite order to Eq (1.8), then the following conclusions hold:

(1) If $n\geq4$ for $\omega\neq0$ and $n\geq 3$ for $\omega = 0$, then every solution $f$ satisfies $\sigma(f) = \deg Q = 1$.

(2) If $n\geq1$ and $f$ is a solution to Eq (1.8) with $\lambda(f) < \sigma(f)$, then

or

where $b_{1}, b_{2}\in \mathbb{C}$ satisfy $p_{1} = q\left(\frac{p_{2}n}{n+\omega\lambda_{2}} \right)^{\frac{1}{n}}e^{\frac{\lambda_{2}c}{n}+b_{1}}$ and $p_{2} = q\left(\frac{p_{1}n}{n+\omega\lambda_{1}} \right)^{\frac{1}{n}}e^{\frac{\lambda_{1}c}{n}+b_{2}}$, respectively.

Remark 1. It is easy to see that Theorem 1.4 generalizes and improves the Theorem E of Chen et al. ^[4]. In addition, we conjecture that the Conclusion (1) is still true for $n = 2$, and $n = 3$ when $\omega\neq0$ in Eq (1.8).

For the case when $n = 3$ and $\omega\neq0$ in Eq (1.8), we prove the following result under certain assumption.

Theorem 1.5. Let $\omega, c, \lambda_{1}, \lambda_{2}, p_{1}, p_{2}$ be non-zero constants, $q, Q$ be polynomials such that $Q$ is not a constant and $q\not\equiv0$. If $f$ is a transcendental entire solution with finite order to

satisfying $N_{1)}(r, 1/f) < (\kappa+o(1)) T(r, f)$, where $0\leq \kappa < 1$ and $N_{1)}(r, 1/f)$ denotes the counting functions corresponding to simple zeros of $f$, then $\sigma(f) = \deg Q = 1$.

The two examples below exhibit the occurence of Theorem 1.5.

Example 1.6. ^[4] $f_{2}(z) = e^{z}$ is a transcendental entire solution to the nonlinear differential-difference equation

where $\omega = 1\neq0$, $Q = -4z$, $N_{1)}(r, 1/f_{2}) = 0$ from the fact that $0$ is a Picard exceptional value of $f_{2}$. Thus, the conclusion $\sigma(f_{2}) = 1 = \deg Q$ holds.

Example 1.7. $f_{3}(z) = e^{2z}-e^{z}$ is a transcendental entire solution to the nonlinear differential-difference equation

where $\omega = -1\neq0$, $Q = 3z$, $\kappa = 1/2$ since $N_{1)}(r, 1/f_{3}) = N(r, 1/f_{3}) = r/\pi+o(r)$ and $T(r, f_{3}) = 2r/\pi+o(r)$ by using the following Lemma 2.5. Thus, the conclusion $\sigma(f_{3}) = 1 = \deg Q$ holds.

2.   Preliminary Lemmas

In order to prove our results, we need the following lemmas. Lemmas 2.1 and 2.2 are useful tools to solve differential-difference equations.

Lemma 2.1 (^[21]). Let $f_{j}(z)\, (j = 1, \ldots, n)\, (n\geq2)$ be meromorphic functions, and let $g_{j}(z)\, (j = 1, \ldots, n)$ be entire functions satisfying

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

(ii) when $1\leq j < k \leq n,$ then $g_{j}(z)-g_{k}(z)$ is not a constant;

(iii) when $1\leq j \leq n, 1\leq h < k \leq n$, then

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

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

Lemma 2.2 (^[21]). Let $f_{j}(z), \, j = 1, 2, 3$ be meromorphic functions and $f_{1}(z)$ is not a constant. If

and

where $\lambda < 1$, $T(r) = \max_{1\leq j\leq3}\{T(r, f_{j})\}$ and $I$ represents a set of $r\in (0, \infty)$ with infinite linear measure. Then $f_{2}\equiv 1$ or $f_{3}\equiv 1$.

The difference analogues of the logarithmic derivative lemma (see ^[3,7,8,9,11]) are of great importance in the study of complex difference equations. The following version is a particular case of [11,Lemma 2.2].

Lemma 2.3 (^[11]). Let $f$ be a non-constant meromorphic function, let $c, h$ be two complex numbers such that $c\neq h$. If he hyper-order of $T(r, f)$ i.e. $\sigma_{2}(f) < 1$, then

for all $r$ outside of a set of finite logarithmic measure.

The following lemma, which is a special case of [11,Theorem 3.1], gives a relationship for the Nevanlinna characteristic of a meromorphic function with its shift.

Lemma 2.4 (^[12]). Let $f(z)$ be a meromorphic function with the hyper-order less that one, and $c\in \mathbb{C}\setminus \{0\}$. Then we have

The following lemma gives the Nevanlinna characteristic function, proximity function and counting function of a given exponential polynomial. For convenience of the readers, we give the definition of an exponential polynomial with the following form:

where $P_{j}$ and $Q_{j}$ are polynomials in $z$ for $1\leq j\leq k$. Denote $q = \max\{\deg Q_{j}:Q_{j}\not\equiv 0\}$, and let $\omega_{1}, \ldots, \omega_{m}$ be pairwise distinct leading coefficients of polynomials that attain the maximum degree $q$. Thus $f$ in (2.1) can be written in the normalized form

where $H_{j}$ are either exponential polynomials of order $< q$ or ordinary polynomials in $z$, and $m\leq k$. In addition, we denote the convex hull of a finite set $W(\subset \mathbb{C})$ by $co(W)$, and the circumference of $co(W)$ by $C(co(W))$, refer to ^[17,18] for more details.

Lemma 2.5 (^[17]). Let $f(z)$ be given by (2.2), $W = \{\overline{\omega_{1}}, \ldots, \overline{\omega_{m}}\}$ and $W_{0} = W\cup \{0\}$. Then

If $H_{0}(z)\not\equiv 0$, then

while if $H_{0}(z)\equiv0$, then

The following lemma is a revised version of [12,Lemma 2.4.2].

Lemma 2.6. Let $f(z)$ be a transcendental meromorphic solution to the equation:

where $P(z, f)$ and $Q(z, f)$ are polynomials in $f$ and its derivatives with meromorhphic coefficients, say $\{a_{\lambda}|\lambda\in I\}$, $n$ be a positive integer. If the total degree of $Q(z, f)$ as a polynomial in $f$ and its derivatives is at most $n$, then

Lemma 2.7. (Hadamard factorization theorem [21,Theorem 2.7] or [5,Theorem 1.9]) Let $f$ be a meromorphic function of finite order $\sigma(f)$. Write

near $z = 0$ and let $\{a_{1}, a_{2}, \ldots\}$ and $\{b_{1}, b_{2}, \ldots\}$ be the zeros and poles of $f$ in $\mathbb{C}\backslash\{0\}$, respectively. Then

where $P_{1}(z)$ and $P_{2}(z)$ are the canonical products of $f$ formed with the non-null zeros and poles of $f(z)$, respectively, and $Q(z)$ is a polynomial of degree $\leq \sigma(f)$.

Remark 2. A well known fact about Lemma 2.7 asserts that $\lambda(f) = \lambda(z^{k}P_{1}) = \sigma(z^{k}P_{1})\leq \sigma(f)$, $\lambda(1/f) = \lambda(P_{2}) = \sigma(P_{2})\leq \sigma(f)$ if $k\geq 0$; and $\lambda(f) = \lambda(P_{1}) = \sigma(P_{1})\leq \sigma(f)$, $\lambda(1/f) = \lambda(z^{-k}P_{2}) = \sigma(z^{-k}P_{2})\leq \sigma(f)$ if $k < 0$. So we have $\sigma(f) = \sigma(e^{Q})$ when $\max\{\lambda(f), \lambda(1/f)\} < \sigma(f)$.

By combinig [21,Theorem 1.42] and [21,Theorem 1.44], we have the following lemma.

Lemma 2.8 (^[21]). Let $f(z)$ be a non-constant meromorphic function in the complex plane. If $0, \infty$ are Picard exceptional values of $f(z)$, then $f(z) = e^{h(z)}$, where $h(z)$ is a non-constant entire function. Moreover, $f(z)$ is of normal growth, and

(i) if $h(z)$ is a polynomial of degree $p$, then $\sigma(f) = p$;

(ii) if $h(z)$ is a transcendental entire function, then $\sigma(f) = \infty$.

The following lemma gives a relationship between the growth order of a meromorphic function with its derivative.

Lemma 2.9 (^[21]). Suppose that $f(z)$ is meromorphic in the complex plane and $n$ is a positive integer. Then $f(z)$ and $f^{(n)}(z)$ have the same order.

3.   Proof of Theorem 1.1.

Suppose that $f(z)$ is a finite-order transcendental entire solution to Eq (1.6) and satisfies $\lambda(f) < \sigma(f)$. Then, by Lemma 2.7 and Remark 2, we have

where $d$ is the canonical product formed by zeros of $f$ such that $\sigma(d) = \lambda(f) < \sigma(f)$, and $h = a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{0}$ is a non-constant polynomial, where $a_{m}(\neq 0), \ldots, a_{0}$ are constants, $\deg h = m = \sigma(f)$ is a positive integer.

Set $f_{c} = f(z+c)$, we rewrite (1.6) as

By Lemmas 2.4 and 2.9, we have $\sigma(f_{c}) = \sigma(f) = \sigma(f')$. From (3.2), by the order property, we get

By substituting (3.1) into (3.2), we obtain that

Next, we consider the following three cases.

Case 1. $\sigma(f) > \deg Q$. Then $\deg h > \deg Q\geq 1$, and by (3.3) we have $\deg v\leq \deg h$.

Subcase 1.1. $\deg h > \deg v$. From (3.4) we have

where $h_{1} = 2a_{m-1}z^{m-1}+\cdots$ and $h_{2} = Q+(a_{m}mc+a_{m-1})z^{m-1}+\cdots$ are polynomials with degree $\leq m-1$. So, noting $\sigma(d') = \sigma(d) = \sigma(d_{c}) < m$, by applying Lemma 2.1 to (3.5), it follows that

which yields a contradiction.

Subcase 1.2. $\deg h = \deg v$. Denote $v(z) = v_{m}z^{m}+v_{m-1}z^{m-1}+\cdots+v_{0}$, where $v_{m}(\neq 0), \ldots, v_{0}$ are constants. From (3.4) we have

where $h_{1}$ and $h_{2}$ are defined as in Subcase 1.1, and $h_{3} = v_{m-1}z^{m-1}+\cdots+v_{0}$ is a polynomial with $\deg h_{3}\leq m-1$.

If $v_{m}\neq 2a_{m}$ and $v_{m}\neq a_{m}$, since $\sigma(d') = \sigma(d) = \sigma(d_{c}) < m$, by applying Lemma 2.1 to (3.6), we get $u\equiv 0$, which is a contradiction.

If $v_{m} = 2a_{m}$, then (3.6) can be written as

By applying Lemma 2.1, we have $qd_{c}\equiv 0$, which implies a contradiction.

If $v_{m} = a_{m}$, then (3.6) can be written as

Similarly as above, by Lemma 2.1, we get

Noting $d\not\equiv 0$, it follows that

By integrating the above equation, we have

Noting that $\deg h > \deg Q\geq 1$, we obtain $\sigma(d) = \deg h = \sigma(f)$, which contradicts with $\sigma(d) < \sigma(f)$.

Case 2. $\sigma(f) = \deg Q$. Then from (3.3), we have $\deg v\leq \deg h = \deg Q$. Next, we deduce that $\deg v = \deg h = \deg Q$. Otherwise, if $\deg v < \deg h = \deg Q$, we will get a contradiction by the following. We suppose that $Q(z) = b_{m}z^{m}+b_{m-1}z^{m-1}+\cdots+b_{0}$, where $b_{m}(\neq0), \ldots, b_{0}$ are constants. Then Eq (3.4) can be written as

where $h_{1}$ is defined as in Subcase 1.1, and $\widetilde{h_{2}} = (a_{m}mc+a_{m-1}+b_{m-1})z^{m-1}+\cdots$ with $\deg\widetilde{h_{2}} \leq m-1$.

If $b_{m}\neq \pm a_{m}$, since $\sigma(d') = \sigma(d) = \sigma(d_{c}) < m$, by applying Lemma 2.1 to (3.7), we get $u\equiv0$, which is a contradiction.

If $b_{m} = a_{m}$, then Eq (3.7) can be rewritten as

Thus, by using Lemma 2.1, we have $u\equiv 0$, a contradiction.

If $b_{m} = -a_{m}$, then Eq (3.7) can be rewritten as

So by Lemma 2.1, we get

which implies

Noting that $\deg h > \deg v\geq 1$, we have $\sigma(d) = \deg h = \sigma(f)$, which contradicts with $\sigma(d) < \sigma(f)$. Therefore, $\deg v = \deg h = \deg Q$, which implies that the Conclusion (2) holds.

Case 3. $\sigma(f) < \deg Q$. Then we have $T(r, f) = S(r, e^{Q})$. By Milloux's theorem and lemma 4, we have $T(r, f') = S(r, e^{Q})$ and $T(r, f_{c}) = S(r, e^{Q})$. Therefore, it follows from (3.2) that

Thus,

Differentiating (3.2) yields

where $A = q'f_{c}+qf_{c}'+qf_{c}Q'$.

Eliminating $e^{v}$ from (3.2) and (3.8), we have

where

Note that $\sigma(f'') = \sigma(f') = \sigma(f) = \sigma(f_{c}) < \deg Q$, then by applying Lemma 2.1 to (3.9), we get $B_{1}\equiv B_{2}\equiv 0$. It follows from $B_{1}\equiv 0$ that

by integration, we have $qf_{c}e^{Q} = c_{3}ue^{v}$, where $c_{3}$ is a non-zero constant.

If $c_{3} = 1$, then substituting $qf_{c}e^{Q} = ue^{v}$ into (3.2), we see that $f^{2}+\widetilde{\omega} f f' = 0$. Thus we can easily get $f = c_{4}e^{-z/\widetilde{\omega}}$, where $c_{4}\in \mathbb{C}\setminus \{0\}$, which implies that the Conclusion (1) holds.

If $c_{3}\neq1$, we have $f = c_{3}u_{-c}/q_{-c}e^{v_{-c}-Q_{-c}}$. By substituting this expression into (3.2), we obtain

Since $1\leq\sigma(f) = \deg(v_{-c}-Q_{-c}) < \deg Q = \deg v$, then by Lemma 2.1 and $(1-c_{3})u\not\equiv 0$ we can easily deduce a contradiction.

4.   Proof of Theorem 1.4.

Suppose that $f$ is a finite-order transcendental entire solution to Eq (1.8). Denote $f_{c} = f(z+c)$, then Eq (1.8) can be written as

By Lemma 2.4, we have $\sigma(f) = \sigma(f_{c})$.

Differentiating both sides of (4.1), we have

where $A_{1} = q'f_{c}+qf_{c}'+qf_{c}Q'$.

Eliminating $e^{\lambda_{2} z}$ from (4.1) and (4.2), we get

where $A_{2} = \lambda_{2}qf_{c}-A_{1}$.

By taking the derivative of (4.3), we get

Then eliminating $e^{\lambda_{1} z}$ from (4.3) and (4.4) gives

Case 1. $\sigma(f) < 1$. By applying the logarithmic derivative lemma, Lemmas 2.3 and 2.5 to Eq (4.1), we obtain

where $W_{0} = \{0, \overline{\lambda_{1}}, \overline{\lambda_{2}}\}$. Thus we have $\deg Q \leq 1$. Noting $\deg Q\geq 1$, we know that $\deg Q = 1$. Let $Q = az+b$, $a\in \mathbb{C}\setminus \{0\}$, $b\in \mathbb{C}$.

Thus, by applying Lemma 2.1 to (4.5), we have

Subcase 1.1.$A_{2}\equiv 0$. That is $\lambda_{2}qf_{c}-q'f_{c}-qf_{c}'-qf_{c}a \equiv 0.$ Noting $qf_{c}\not\equiv 0$, it follows that

By integration, we have

If $a\neq\lambda_{2}$, then $\sigma(f) = \sigma(f_{c}) = 1$, which contradicts with $\sigma(f) < 1$. Thus we have $a = \lambda_{2}$, and $f_{c} = c_{1}/q$. Then $f(z) = c_{1}/q(z-c)$, which is impossible for a transcendental function $f$.

Subcase 1.2.$A_{2}\not\equiv 0$. From (4.6), we get

It follows that

Since $\lambda_{1}\neq \lambda_{2}$, we consider three subcases as follows.

Subcase 1.2.1. $a = \lambda_{2}$. Then (4.7) can be written as $(qf_{c})' = -c_{2}e^{(\lambda_{1}-\lambda_{2})z}$. By integration, we obtain $qf_{c} = \frac{c_{2}}{\lambda_{2}-\lambda_{1}} e^{(\lambda_{1}-\lambda_{2})z}+c_{3}$, where $c_{3}\in \mathbb{C}$. By Lemma 2.5 we have

Since $f$ is transcendental, by Lemma 2.4, it follows that

which contradicts with $\sigma(f) < 1$.

Subcase 1.2.2. $a = \lambda_{1}$. Then (4.7) can be written as $(qf_{c})'+(\lambda_{1}-\lambda_{2})(qf_{c}) = -c_{2}$. Thus, we obtain $qf_{c} = \frac{c_{2}}{\lambda_{2}-\lambda_{1}} +c_{4}e^{(\lambda_{2}-\lambda_{1})z}$, where $c_{4}\in \mathbb{C}$. We assert that $c_{4}\neq 0$. Otherwise, we have $f(z) = \frac{c_{2}}{\lambda_{2}-\lambda_{1}}\frac{1}{q(z-c)}$, which contradicts with the assumption that $f$ is transcendental. Therefore, $c_{4}\neq 0$, similarly as in Subcase 1.2.1, by combining Lemmas 2.4, 2.5, and $\sigma(f) < 1$, we can get a contradiction.

Subcase 1.2.3. $a\neq\lambda_{1}$ and $a\neq\lambda_{2}$. Then by (4.7), we get that

Since $c_{2}\neq 0$, $a\neq\lambda_{1}$, and $\lambda_{1}\neq \lambda_{2}$, similarly as in Subcase 1.2.1, by combining Lemmas 2.4, 2.5, and $\sigma(f) < 1$, we also get a contradiction.

Case 2. $\sigma(f) > 1$. Denote $P = p_{1}e^{\lambda_{1} z}+p_{2}e^{\lambda_{2} z}$, then by Lemma 2.5 we have $\sigma(P) = 1$. We rewrite Eq (4.1) as

Differentiating (4.8) yields

where $L = (qf_{c})'+Q'(qf_{c})$.

Subcase 2.1. $\omega\neq 0$ and $n\geq4$. Eliminating $e^{Q}$ from (4.8) and (4.9), we have

where

Subcase 2.1.1. $H\not\equiv 0$. We rewrite $PL-P'(qf_{c})$ as

and $H$ as

thus both $PL-P'(qf_{c})$ and $H$ are differential polynomials with meromorphic coefficients. By the logarithmic derivative lemma and Lemma 2.4, we have $m(r, f_{c}'/f_{c}) = S(r, f_{c}) = S(r, f)$; by Lemma 2.3, we have $m(r, f_{c}/f) = S(r, f)$; and by Lemma 2.5, we have $m(r, P) = O(r)$. Note that $n\geq 4$ and $H$, $fH$ are entire, by applying Lemma 2.6 to (4.10), it follows that

and

Thus, by $H\not\equiv 0$ we deduce that

which contradicts with $\sigma(f) > 1$.

Subcase 2.1.2. $H\equiv 0$. Then from (4.10), we have

Noting $qf_{c}P\not\equiv0$, it follows that

By integration, we have

Substituting (4.11) into (4.8), we get that

If $c_{6} = 1$, then from (4.12) we have $f+\omega f' = 0$. By integration, we get $f = c_{7}e^{-\frac{1}{\omega}z}$, $c_{7}\in \mathbb{C}\setminus\{0\}$, which contradicts with $\sigma(f) > 1$. Thus, $c_{6}\neq1$. It follows from (4.11) that

Then $\deg Q = \deg Q_{-c} = \sigma(f) > 1$ since $\sigma(P_{-c}) = 1$ by Lemma 2.5.

By Substituting (4.13) into (4.12), we have

Thus, from $\deg Q > 1$, $\sigma(P_{-c}) = \sigma(P) = 1$ and Lemma 2.1, we get $P(z)\equiv0$, which is impossible.

Subcase 2.2. $\omega = 0$ and $n\geq 3$. Eliminating $e^{Q}$ from (4.8) and (4.9), we obtain

Subcase 2.2.1. $Lf-nqf_{c} f'\not\equiv 0$. Since $n\geq 3$ and $\omega = 0$, similarly as in Subcase 2.1.1, we have

and

By $Lf-nqf_{c} f'\not\equiv 0$, we deduce that

which contradicts with $\sigma(f) > 1$.

Subcase 2.2.2. $Lf-nqf_{c} f'\equiv 0$. Then

By integration, we obtain

Substituting (4.14) into (4.8) yields

If $c_{8}\neq -1$, then we have $nT(r, f)+S(r, f) = T(r, (1+c_{8})f^{n}) = T(r, P) = O(r)$, which contradicts with $\sigma(f) > 1$. Thus, $c_{8} = -1$, and then $p_{1}e^{\lambda_{1} z}+p_{2}e^{\lambda_{2} z} = P = (1+c_{8})f^{n}\equiv 0$, which is impossible.

Case 3. $\sigma(f) = 1$. By (4.1), Lemma 2.3, and the logarithmic derivative lemma, we obtain

Note that $\deg Q \geq 1$, then by combining Lemma 2.5, we get

that is $\sigma(f) = \deg Q = 1$. Thus, the Conclusion (1) is proved.

Next, we will prove the Conclusion (2). Suppose that $f$ is a finite-order transcendental entire solution to Eq (1.8) and satisfies $\lambda(f) < \sigma(f)$. Then, similarly as in the proof of Theorem 1.1, by Lemma 2.7 and Remark 2, we have

where $d$ is the canonical product formed by zeros of $f$ such that $\sigma(d) = \lambda(f) < \sigma(f)$, and $h$ is a polynomial with $\deg h = \sigma(f)\geq1$.

By substituting (4.15) into (1.8), we get

Dividing both sides of (4.16) by $p_{2}e^{\lambda_{2}z}$, we obtain

where

Obviously, $f_{1}$ is not a constant by the fact that $\lambda_{1}\neq\lambda_{2}$. Let $T(r) = \max\{T(r, f_{1}), T(r, f_{2}), T(r, f_{3})\}$. Next we distinguish two cases below.

Case 1. $\sigma(f) > 1$. It follows from $\max\{\sigma(d') = \sigma(d), 1\} < \deg h$ that $d^{n-1}(d+\omega(d'+dh'))/p_{2}e^{-\lambda_{2}z}$ is a small function of $e^{h}$. Then $T(r)\geq T(r, f_{2}) = nT(r, e^{h})+S(r, e^{h})$. Thus, by Milloux's theorem and Lemma 2.4, we get

and

as $r\to \infty$.

Therefore, by applying Lemma 2.2, we can deduce that $f_{2}\equiv 1$ or $f_{3}\equiv 1$.

If $f_{2}\equiv 1$, then $d^{n-1}(d+\omega(d'+dh'))e^{nh-\lambda_{2}z}\equiv p_{2}$. We deduce that $d^{n-1}(d+\omega(d'+dh'))\not\equiv0$. Otherwise, $p_{2}\equiv0$, which is a contradiction. So, by Lemma 2.5 and Milloux's theorem, we obtain

which contradicts with $\deg h = \sigma(f) > \max\{\sigma(d), 1\}$.

If $f_{3}\equiv 1$, then by (4.17), we have $f_{1}+f_{2}\equiv 0$. It follows that

By a similar discussion as above, we can also get a contradiction.

Case 2. $\sigma(f) = 1$. Then we have $\sigma(d) < 1 = \deg h = \sigma(e^{(\lambda_{1}-\lambda_{2})z})$ and $T(r)\geq T(r, f_{1}) = T\left(r, e^{(\lambda_{1}-\lambda_{2})z}\right)+S\left(r, e^{(\lambda_{1}-\lambda_{2})z}\right)$. Thus, by Milloux's theorem and Lemma 2.4, we obtain

and

as $r\to \infty$.

Therefore, by applying Lemma 2.2, we can deduce that $f_{2}\equiv 1$ or $f_{3}\equiv 1$.

If $f_{2}\equiv 1$, then

We assert that $h' = \lambda_{2}/n$. Otherwise, since $\sigma(d') = \sigma(d) < 1 = \deg (nh-\lambda_{2}z)$, by applying Lemma 2.1 to (4.18), we get $p_{2}\equiv0$, which is a contradiction. Thus $h' = \lambda_{2}/n$. We set $h = \lambda_{2}z/n+B$, where $B$ is a constant. Substituting this into (4.18), we have

Next, we deduce that $d$ is a constant. Otherwise, if $d$ is a non-constant entire function, then it follows from (4.19) that $0$ is a Picard exceptional value of $d$. Thus by Lemma 2.8, we have $d = e^{\alpha}$, where $\alpha$ is a non-constant polynomial, which contradicts with $\sigma(d) < 1$. So $d$ is a non-zero constant, and (4.19) can be written as

Therefore,

Moreover, by $f_{2}\equiv 1$ and (4.17), we also have $f_{1}+f_{3}\equiv0$. That is

which implies that

where $b_{1}$ satisfies $p_{1} = q\left(\frac{p_{2}n}{n+\omega\lambda_{2}} \right)^{\frac{1}{n}}e^{\frac{\lambda_{2}c}{n}+b_{1}}$.

If $f_{3}\equiv 1$, then by (4.17) we have $f_{1}+f_{2} = 0$. It follows that

By using a similar method as in the case $f_{2}\equiv 1$, we get

Furthermore, it follows from $f_{3}\equiv 1$ that $qd_{c}e^{Q+h_{c}-\lambda_{2}z}\equiv p_{2}$. Then we can deduce

where $b_{2}$ satisfies $p_{2} = q\left(\frac{p_{1}n}{n+\omega\lambda_{1}} \right)^{\frac{1}{n}}e^{\frac{\lambda_{1}c}{n}+b_{2}}$. From the above discussion, the proof of the Conclusion (2) is complete.

5.   Proof of Theorem 1.5.

Suppose that $f$ is a finite-order transcendental entire solution to Eq (1.9).

If $\sigma(f) < 1$, then by a similar method as in Theorem 1.4 (Case 1), we can get a contradiction.

If $\sigma(f) > 1$, we denote $P = p_{1}e^{\lambda_{1} z}+p_{2}e^{\lambda_{2} z}$. Then Eq (1.9) can be written as

Differentiating (5.1) yields

where $L = (qf_{c})'+Q'(qf_{c})$.

Eliminating $e^{Q}$ from (5.1) and (5.2), we obtain

where

If $H\equiv 0$, then from (5.3) we have $PL-P'(qf_{c})\equiv0$. By a similar reasoning as in Theorem 1.4 (Subcase 2.1.2), we get a contradiction. Therefore, $H\not\equiv 0$. Noting that $H$ is entire and $PL-P'(qf_{c})$, $H/f$ are differential polynomials with meromorphic coefficients, similarly as in Theorem 1.4 (Subcase 2.1.1), by applying Lemma 2.6 to (5.3), we obtain

and

Obviously, the poles of $H/f$ arise from the poles of $(f')^{2}/f$. Suppose that $z_{0}$ is a zero of $f$ with multiplicity $p$, then it is a simple pole of $(f')^{2}/f$ when $p = 1$, and a zero of $(f')^{2}/f$ with multiplicity $p-2$ when $p\geq2$. Noting that $f$ is entire, we obtain

Therefore, by $H\not\equiv 0$, it follows that

Thus we have

which contradicts with $\sigma(f) > 1$.

If $\sigma(f) = 1$, then by a similar method as in Theorem 1.4(Case 3), we can get that $\sigma(f) = \deg Q = 1$.

6.   Conclusions

By using the Nevanlinna theory and its difference analogues, we study the transcendental entire solutions for two types of nonlinear differential-difference equations, and obtain three main theorems, which are improvements and complements of some previous results. Meanwhile, some examples are given to illustrate the conclusions.

Acknowledgments

The authors would like to thank the referee for his/her thorough review with constructive suggestions and valuable comments. This work was supported by NNSF of China (No.11801215), and the NSF of Shandong Province, P. R. China (No.ZR2016AQ20 & No. ZR2018MA021).

Conflict of interest

The authors declare no conflict of interest.