Three Results on the Nonlinear Differential Equations and Differential-Difference Equations

Abstract

We mainly study the transcendental entire solutions of the differential equation $f^n(z)+P(f)=p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z}$, where $p_1$, $p_2$, ${\alpha }_1$ and ${\alpha }_2$ are nonzero constants satisfying ${\alpha }_1\ne {\alpha }_2$ and $P(f)$ is a differential polynomial in f of degree $n-1$. We improve Chen and Gao’s results and partially answer a question proposed by Li (J. Math. Anal. Appl. 375 (2011), pp. 310–319).

1. Introduction and Main Results

In the past several decades, a great deal of mathematical effort in complex analysis has been devoted to studying differential equations, differential-difference equations and difference equations. The essential reason is penetration and application of Nevanlinna theory for the difference operator, see [1,2,3,4]. In this study, we assume readers are familiar with the standard notations and fundamental results used in the theory such as the characteristic function $T(r,f)$, the proximity function $m(r,f)$ and the counting function $N(r,f)$, see [5,6,7,8]. Moreover, we use the notations $\rho (f)$ and ${\rho }_2(f)$ to denote the order and the hyper-order of f, respectively.

Many scholars recently have had tremendous interest in developing solvability and existence of solutions of non-linear differential equations and differential-difference equations in the complex plane, see [9,10,11,12,13,14,15].

In 2011, Li [16] considered to find all entire solutions of the following nonlinear differential equation

$$f^n(z)+P(f)=p_1{e}^{\lambda z}+p_2{e}^{-\lambda z}$$

(1)

and obtained the following result.

Theorem 1.

(see [16]) Let $n\ge 2$ be an integer, $P(f)$ be a differential polynomial in f of degree at most $n-1$ and $\lambda ,p_1,p_2$ be three nonzero constants. If f is a meromorphic function of Equation (1) satisfying $N(r,f)=S(r,f)$, then there exist two nonzero constants $c_1,c_2(c_i^n=p_i)$ and a small function $c_0$ of f such that

$$f=c_0+c_1{e}^{\frac{\lambda z}{n}}+c_2{e}^{-\frac{\lambda z}{n}}.$$

Li [16] also investigated $p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z}$ for two distinct constants ${\alpha }_1$ and ${\alpha }_2$ instead of $p_1{e}^{\lambda z}+p_2{e}^{-\lambda z}$ in the right side of Equation (1) and obtained the following results.

Theorem 2.

(see [16]) Let $n\ge 2$ be an integer, $P(f)$ be a differential polynomial in $f(z)$ of degree at most $n-2$ and ${\alpha }_1,{\alpha }_2,p_1,p_2$ be nonzero constants satisfying ${\alpha }_1\ne {\alpha }_2$. If $f(z)$ is a transcendental meromorphic solution of the following equation

$$f^n(z)+P(f)=p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z}$$

(2)

satisfying $N(r,f)=S(r,f)$, then one of the following relations holds:

(1) 

$f(z)=c_0(z)+c_1{e}^{\frac{{\alpha }_{1}z}{n}};$

(2) 

$f(z)=c_0(z)+c_2{e}^{\frac{{\alpha }_{2}z}{n}};$

(3) 

$f(z)=c_1{e}^{\frac{{\alpha }_{1}z}{n}}+c_2{e}^{\frac{{\alpha }_{2}z}{n}}$ and ${\alpha }_1+{\alpha }_2=0$,

where $c_0(z)$ is a small function of f and constants $c_1$ and $c_2$ satisfy $c_1^n=p_1$ and $c_2^n=p_2$, respectively.

For further study, Li proposed a related question:

Question 1.How to find the solutions of Equation (2) if $degP(f)=n-1$?

The question was studied by Chen and Gao [17]. They partially answered it and obtained the following result.

Theorem 3.

(see [17]) Let $a(z)$ be a nonzero polynomial and $p_1,p_2,{\alpha }_1,{\alpha }_2$ be nonzero constants such that ${\alpha }_1\ne {\alpha }_2$. If $f(z)$ is a transcendental entire solution of finite order of the differential equation

$$f^2(z)+a(z)f^{\prime }(z)=p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z}$$

(3)

satisfying $N(r,\frac{1}{f})=S(r,f)$, then $a(z)$ must be a constant and one of the following relations holds:

(1) 

$f(z)=c_1{e}^{\frac{{\alpha }_{1}z}{2}}$, $a{c}_1{\alpha }_1=2p_2$ and ${\alpha }_1=2{\alpha }_2$;

(2) 

$f(z)=c_2{e}^{\frac{{\alpha }_{2}z}{2}}$, $a{c}_2{\alpha }_2=2p_1$ and ${\alpha }_2=2{\alpha }_1$,

where $c_1$ and $c_2$ are constants satisfying $c_1^2=p_1$ and $c_2^2=p_2$, respectively.

Now, we remove the condition that $f(z)$ is a finite-order function, improve Theorem 3 and obtain the following result.

Theorem 4.

Let $a(z)$ be a nonzero polynomial and $p_1,p_2,{\alpha }_1,{\alpha }_2$ be nonzero constants such that ${\alpha }_1\ne {\alpha }_2$. Suppose that $f(z)$ is a transcendental entire solution of the differential Equation (3) satisfying $N(r,\frac{1}{f})=S(r,f)$. Then $a(z)$ must be a constant and one of the following relations holds:

(1) 

$f(z)=c_1{e}^{\frac{{\alpha }_{1}z}{2}}$, $a{c}_1{\alpha }_1=2p_2$ and ${\alpha }_1=2{\alpha }_2$;

(2) 

$f(z)=c_2{e}^{\frac{{\alpha }_{2}z}{2}}$, $a{c}_2{\alpha }_2=2p_1$ and ${\alpha }_2=2{\alpha }_1$,

where $c_1$ and $c_2$ are constants satisfying $c_1^2=p_1$ and $c_2^2=p_2$, respectively.

Next we consider the general case in Question 1 and obtain the following theorem.

Theorem 5.

Let $n\ge 2$ be an integer. Suppose that $P(f)$ is a differential polynomial in $f(z)$ of degree $n-1$ and that ${\alpha }_1$, ${\alpha }_2$, $p_1$ and $p_2$ are nonzero constants such that ${\alpha }_1\ne {\alpha }_2$. If $f(z)$ is a transcendental meromorphic solution of the differential Equation (2) satisfying $N(r,f)=S(r,f)$, then $\rho (f)=1$ and one of the following relations holds:

(1) 

$f(z)=c_1{e}^{\frac{{\alpha }_{1}z}{n}}$ and $c_1^n=p_1$;

(2) 

$f(z)=c_2{e}^{\frac{{\alpha }_{2}z}{n}}$ and $c_2^n=p_2$, where $c_1$ and $c_2$ are constants;

(3) 

$T(r,f)\le N_{1)}(r,\frac{1}{f})+T(r,\phi )+S(r,f)$, where $N_{1)}(r,\frac{1}{f})$ denotes the counting function corresponding to simple zeros of f and $\phi (\not\equiv 0)$ is equal to ${\alpha }_1{\alpha }_2{f}^2-n({\alpha }_1+{\alpha }_2)f{f}^{\prime }+n(n-1){(f^{\prime })}^2+nf{f}^{″}$.

Three examples are shown to illustrate the cases (1)–(3) of Theorem 5.

Example 1.

Let $f(z)=e^z$ be an entire solution of the differential equation

$$f^2(z)+f^{\prime }(z)=e^{2z}+e^z,$$

where $c_1=1$ and $p_1=1$. It implies the case (1) occurs.

Example 2.

Let $f(z)=2e^{2z}$ be an entire solution of the differential equation

$$f^2(z)+\frac{1}{8}{f}^{″}(z)=e^{2z}+4e^{4z},$$

where $c_2=2$ and $p_2=4$. It implies case (2) occurs.

Example 3.

Let $f(z)=e^z-1$ be an entire solution of the differential equation

$$f^2(z)+(f^{\prime }-1)=e^{2z}-e^z.$$

We can easily verify the inequality $T(r,f)\le N_{1)}(r,\frac{1}{f})+T(r,\phi )+S(r,f)$, where $\phi =2f^2-6f{f}^{\prime }+2{(f^{\prime })}^2+2f{f}^{″}=2$. It implies that case (3) occurs.

Remark 1.

From Theorem 4 and Example 3, we conjecture that case (3) in Theorem 5 can be removed if $N(r,1/f)=S(r,f)$.

In [18], Wang and Li investigated the following differential-difference equation

$$f^n(z)+q(z)f^{(k)}(z+c)=a{e}^{ibz}+d{e}^{-ibz}$$

(4)

and obtained the existence of entire solutions when $n\ge 3$.

In 2018, Chen and Gao went far to study Equation (4) with $n=2$. They obtained the following theorem.

Theorem 6.

(see [17]) Let $a(z)$ be a nonzero polynomial, $k\ge 0$ be an integer and $p_1,p_2,\lambda ,c$ be nonzero constants. If $f(z)$ is a transcendental entire solution of finite order of the differential-difference equation

$$f^2(z)+a(z)f^{(k)}(z+c)=p_1{e}^{\lambda z}+p_2{e}^{-\lambda z},$$

(5)

then $a(z)$ must be a constant and one of the following relations holds:

(1) 

$f(z)=\pm \frac{i}{2}a{(\frac{\lambda }{2})}^k+c_1{e}^{\frac{\lambda z}{2}}+c_2{e}^{-\frac{\lambda z}{2}}$ and $e^{\lambda c}=-1$, when k is odd;

(2) 

$f(z)=\pm \frac{1}{2}a{(\frac{\lambda }{2})}^k+c_1{e}^{\frac{\lambda z}{2}}+c_2{e}^{-\frac{\lambda z}{2}}$ and $e^{\lambda c}=1$, when k is even and $k>0$, where a, $c_1$ and $c_2$ are constants with $\frac{1}{64}{a}^4{(\frac{\lambda }{2})}^{4k}=p_1{p}_2$ and $c_i^2=p_i$$(i=1,2)$;

(3) 

$f(z)=\pm \frac{1}{2}a+c_1{e}^{\frac{\lambda z}{2}}+c_2{e}^{-\frac{\lambda z}{2}}$ and $e^{\lambda c}=1$, when $k=0$, where a, $c_1$ and $c_2$ are constants with $\frac{1}{64}{a}^4=p_1{p}_2$ or $\frac{9}{64}{a}^4=p_1{p}_2$ and $c_i^2=p_i$$(i=1,2)$.

For the right side of Equations (4) and (5), a question to be raised is how to find the existence of solutions if $e^{\lambda z}$ and $e^{-\lambda z}$ can be replaced by a linear combination of $e^{{\alpha }_{1}z}$ and $e^{{\alpha }_{2}z}$ for two distinct constants ${\alpha }_1$ and ${\alpha }_2$. We consider the question and obtain the following result.

Theorem 7.

Let ${\alpha }_1$, ${\alpha }_2$, $p_1$, $p_2$ and h be nonzero constants satisfying ${\alpha }_1\ne {\alpha }_2$. Suppose that $k\ge 0$ and $n\ge 2$ are integers and that $q(z)$ is a nonzero polynomial. If $f(z)$ is a transcendental entire solution with ${\rho }_2(f)<1$ of the differential-difference equation

$$f^n(z)+q(z)f^{(k)}(z+h)=p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z},$$

(6)

then we have $\rho (f)=1$, $q(z)$ must be a constant and one of the following relations holds:

(1) 

$f(z)=c_1{e}^{\frac{{\alpha }_{1}z}{n}},q{c}_1{\left(\frac{{\alpha }_1}{n}\right)}^k{e}^{\frac{{\alpha }_{1}h}{n}}=p_2,{\alpha }_1=n{\alpha }_{2}and{c}_1^n=p_1$;

(2) 

$f(z)=c_2{e}^{\frac{{\alpha }_{2}z}{n}},q{c}_2{\left(\frac{{\alpha }_2}{n}\right)}^k{e}^{\frac{{\alpha }_{2}h}{n}}=p_1,{\alpha }_2=n{\alpha }_{1}and{c}_2^n=p_2$;

(3) 

If $n=2$, we have $T(r,f)\le N_{1)}(r,1/f)+T(r,\phi )+S(r,f)$, where $N_{1)}(r,1/f)$ and φ are the same as defined in Theorem 5. If $n=3$, we have $T(r,f)=N_{1)}(r,1/f)+S(r,f)$. If $n\ge 4$, we only have the cases (1) and (2).

Next we give three examples to show existence of solutions of Equation (6).

Example 4.

Let $f(z)=e^z$. Then f is a transcendental entire solution of the following differential-difference equation

$$f^3(z)+f^{\prime }(z+2\pi i)=e^{3z}+e^z,$$

where ${\alpha }_1=3=3{\alpha }_2$, $c_1=1$, $q=1$ and $p_1=p_2=1$. Thus, case (1) occurs.

Example 5.

Let $f(z)=\sqrt{2}{e}^z$. Then f is a transcendental entire solution of the following differential-difference equation

$$f^2(z)+\sqrt{2}{f}^{(3)}(z+2\pi i)=2e^z+2e^{2z},$$

where ${\alpha }_2=2=2{\alpha }_1$, $c_2=\sqrt{2}$, $q=\sqrt{2}$ and $p_1=p_2=2$. Thus, case (2) occurs.

Example 6.

Let $f(z)=e^z-1$. Then f is a transcendental entire solution of the following equation

$$f^2(z)+f(z+\pi i)=e^{2z}-3e^z.$$

A routine computation yields $T(r,f)\le N_{1)}(r,\frac{1}{f})+T(r,\phi )+S(r,f)$, where $\phi =2f^2-6f{f}^{\prime }+2{(f^{\prime })}^2+2f{f}^{″}=2$. Thus, case (3) occurs.

Example 7.

Let $f(z)=e^z+e^{-z}$. Then f is a transcendental entire solution of the following differential-difference equation

$$f^3(z)+f^{″}(z+\pi i)=e^{3z}+e^{-3z}.$$

A routine computation yields $T(r,f)=N_{1)}(r,\frac{1}{f})+S(r,f)$.

Remark 2.

From Examples 6 and 7, we conjecture that case (3) in Theorem 7 can be removed if $N(r,1/f)=S(r,f)$ for $n=2,3$.

Remark 3.

In Theorem 3, our result holds for ${\alpha }_1\ne {\alpha }_2$. However, if ${\alpha }_1+{\alpha }_2=0$, we just know the solutions satisfy case (3) for $n=2,3$. The expression of solutions can be obtained when $n=2$ in Theorem 6.

2. Some Lemmas

In this section, we introduce several lemmas to prove three theorems.

Lemma 1.

(see [5]) Let $f(z)$ be an entire function and k be a positive integer. Then

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

Lemma 2.

(see [3]) Let $c\in \mathbb{C}\backslash \left\{0\right\}$, $\epsilon >0$ and $f(z)$ be a meormorphic function of ${\rho }_2(f)<1$. Then

$$m\left(r,\frac{f(z+c)}{f(z)}\right)=o\left(\frac{T(r,f)}{r^{1-{\rho }_2(f)-\epsilon }}\right)$$

outside of an exceptional set of finite logarithmic measures.

Lemma 3.

(see [8]) Suppose that $f_1(z),f_2(z),\dots ,f_n(z)(n\ge 2)$ are meromorphic functions and that $g_1(z),g_2(z),\dots ,g_n(z)(n\ge 2)$ are entire functions satisfying the following conditions:

(1) 

$f_1(z)e^{g_1(z)}+f_2(z)e^{g_2(z)}+\dots +f_n(z)e^{g_n(z)}\equiv 0$;

(2) 

$g_j(z)-g_k(z)$ are not constants for $1\le j<k\le n$;

(3) 

For $1\le j\le n$ and $1\le h<k\le n$, $T(r,f_j(z))=o\left(T(r,e^{g_h(z)-g_k(z)})\right)(r\to \infty ,r\notin E)$, where $E\subset [1,\infty )$ is a finite linear measure or finite logarithmic measure.

Then $f_j(z)\equiv 0$$(j=1,2,\dots ,n)$.

Applying Lemmas 1 and 2 to Theorem 2.3 of [19], we get the following lemma, which is a version of the difference analogue of the Clunie lemma.

Lemma 4.

Let f be a transcendental meromorphic solution of ${\rho }_2(f)<1$ of a difference equation of the form

$$H(z,f)P(z,f)=Q(z,f),$$

where $H(z,f),P(z,f),Q(z,f)$ are difference polynomials in f such that the total degree of $H(z,f)$ in f and its shifts is n, and that the corresponding total degree of $Q(z,f)$ is $\le n$. If $H(z,f)$ contains just one term of maximal total degree, then for any $\epsilon >0$

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

possibly outside of an exceptional set of finite logarithmic measure.

3. Proof of Theorem 4

Proof. 

Denote $P_1(f)a(z)f^{\prime }(z)$. Suppose $f(z)$ be a transcendental entire solution of Equation (3).

Differentiating Equation (3), we obtain

$$2f{f}^{\prime }+P_1^{\prime }={\alpha }_1{p}_1{e}^{{\alpha }_{1}z}+{\alpha }_2{p}_2{e}^{{\alpha }_{2}z}.$$

(7)

Eliminating $e^{{\alpha }_{2}z}$ from Equations (3) and (7) gives

$${\alpha }_2{f}^2-2f{f}^{\prime }+{\alpha }_2{P}_1-P_1^{\prime }=({\alpha }_2-{\alpha }_1)p_1{e}^{{\alpha }_{1}z}.$$

(8)

Differentiating Equation (8) yields

$$2{\alpha }_{2}f{f}^{\prime }-2{(f^{\prime })}^2-2f^{\prime }{f}^{″}+{\alpha }_2{P}_1^{\prime }-P_1^{″}={\alpha }_1({\alpha }_2-{\alpha }_1)p_1{e}^{{\alpha }_{1}z}.$$

(9)

It follows from Equations (8) and (9) that

$$\phi =Q,$$

where

$$\phi ={\alpha }_1{\alpha }_2{f}^2-2({\alpha }_1+{\alpha }_2)f{f}^{\prime }+2{(f^{\prime })}^2+2f{f}^{″}$$

and

$$Q=-{\alpha }_1{\alpha }_2{P}_1+({\alpha }_1+{\alpha }_2)P_1^{\prime }-P_1^{″}.$$

Here we distinguish two cases below.

Case 1.$\phi \not\equiv 0$.

Similar to the proof of Theorem 3 [17], we can obtain a contradiction.

Case 2.$\phi \equiv 0$.

By taking $n=2$, we use the method of Case 1 of Theorem 5 to obtain $t_1=\frac{{\alpha }_1}{2}$ and $t_2=\frac{{\alpha }_2}{2}$, where $t_i=\frac{f^{\prime }}{f}(i=1,2)$.

Now if $t_1=\frac{{\alpha }_1}{2}$, then $f(z)=c_1{e}^{\frac{{\alpha }_{1}z}{2}}$, where $c_1$ is a constant satisfying $c_1^2=p_1$. Substituting these formulas into Equation (3), we have $a(z)c_1{\alpha }_1=2p_2$ and ${\alpha }_1=2{\alpha }_2$, where $a(z)$ must be a constant. Set $aa(z)$.

Similarly, if $t_2=\frac{{\alpha }_2}{2}$, then we have $f(z)=c_2{e}^{\frac{{\alpha }_{2}z}{2}}$, $a{c}_2{\alpha }_2=2p_1$ and ${\alpha }_2=2{\alpha }_1$, where $c_2$ is a constant satisfying $c_2^2=p_2$. □

4. Proof of Theorem 5

Proof. 

Assume that $f(z)$ is a transcendental meromorphic solution of Equation (2) with $N(r,f)=S(r,f)$.

A differential polynomial $P(f)$ with $degP(f)=n-1$ can be written in the following form

$$P(f)=\sum _{i=1}^{n-1}{a}_i{M}_i(f)=a_1{M}_1(f)+a_2{M}_2(f)+\dots +a_{n-1}{M}_{n-1}(f),$$

where $a_i$ are the small functions of f and $M_i(f)=f^{n_{0i}}{(f^{\prime })}^{n_{1i}}\dots {(f^{(k)})}^{n_{ki}}$ are the differential monomials such that $\mathrm{deg}{M}_i(f)=n_{0i}+n_{1i}+\dots +n_{ki}=i\le n-1$.

We can represent $P(f)$ as

$$P(f)=\frac{a_1{M}_1(f)}{f}f+\frac{a_2{M}_2(f)}{f^2}{f}^2+\dots +\frac{a_{n-1}{M}_{n-1}(f)}{f^{n-1}}{f}^{n-1}.$$

By Lemma 1, we derive

$$m\left(r,\frac{a_i{M}_i(f)}{f^i}\right)=m\left(r,\frac{a_i{f}^{n_{0i}}{(f^{\prime })}^{n_{1i}}\dots {(f^{(k)})}^{n_{ki}}}{f^i}\right)=S(r,f)$$

for $1\le i\le n-1$. Furthermore, we have

$$m(r,P(f))\le (n-1)m(r,f)+S(r,f).$$

Since $N(r,f)=S(r,f)$

$$T(r,P(f))\le (n-1)T(r,f)+S(r,f)$$

(10)

holds.

By Equation (10), we obtain

$$\begin{array}{cc}\hfill & T(r,p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z})=T(r,f^n(z)+P(f))\hfill \\ \le \hfill & T(r,f^n(z))+T(r,P(f))+O(1)\hfill \\ \le \hfill & nT(r,f)+(n-1)T(r,f)+S(r,f)\hfill \\ =\hfill & (2n-1)T(r,f)+S(r,f)\hfill \end{array}$$

(11)

and

$$\begin{array}{cc}\hfill & T(r,p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z})=T(r,f^n(z)+P(f))\hfill \\ \ge \hfill & T(r,f^n(z))-T(r,P(f))+O(1)\hfill \\ \ge \hfill & nT(r,f)-(n-1)T(r,f)+S(r,f)\hfill \\ =\hfill & T(r,f)+S(r,f).\hfill \end{array}$$

(12)

It follow from Equations (11) and (12) that

$$T(r,f)+S(r,f)\le T(r,p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z})\le (2n-1)T(r,f)+S(r,f),$$

which implies $\rho (f)=1$.

We next turn to proving conclusions (1)–(3).

Differentiating Equation (2), we have

$$n{f}^{n-1}{f}^{\prime }+P^{\prime }={\alpha }_1{p}_1{e}^{{\alpha }_{1}z}+{\alpha }_2{p}_2{e}^{{\alpha }_{2}z}.$$

(13)

Eliminating $e^{{\alpha }_{2}z}$ from Equations (2) and (13) gives

$${\alpha }_2{f}^n-n{f}^{n-1}{f}^{\prime }+{\alpha }_{2}P-P^{\prime }=({\alpha }_2-{\alpha }_1)p_1{e}^{{\alpha }_{1}z}.$$

(14)

Differentiating Equation (14) yields

$$n{\alpha }_2{f}^{n-1}{f}^{\prime }-n(n-1)f^{n-2}{(f^{\prime })}^2-n{f}^{n-1}{f}^{″}+{\alpha }_2{P}^{\prime }-P^{″}={\alpha }_1({\alpha }_2-{\alpha }_1)p_1{e}^{{\alpha }_{1}z}.$$

(15)

By Equations (14) and (15), we have

$$f^{n-2}\phi =Q,$$

where

$$\phi ={\alpha }_1{\alpha }_2{f}^2-n({\alpha }_1+{\alpha }_2)f{f}^{\prime }+n(n-1){(f^{\prime })}^2+nf{f}^{″}$$

(16)

and

$$Q=-{\alpha }_1{\alpha }_{2}P+({\alpha }_1+{\alpha }_2)P^{\prime }-P^{″}.$$

We still consider two cases below.

Case 1.$\phi \equiv 0$.

Dividing with $f^2$ on both sides in Equation (16) and recalling $\frac{f^{″}}{f}={\left(\frac{f^{\prime }}{f}\right)}^{\prime }+{\left(\frac{f^{\prime }}{f}\right)}^2$, we get a Riccati equation

$$t^{\prime }+n{t}^2-({\alpha }_1+{\alpha }_2)t+\frac{{\alpha }_1{\alpha }_2}{n}=0$$

where $t=\frac{f^{\prime }}{f}$. A routine computation yields two constant solutions $t_1=\frac{{\alpha }_1}{n}$ and $t_2=\frac{{\alpha }_2}{n}$.

Given that $t\ne t_1$ and $t\ne t_2$ hold, we have

$$\frac{1}{t_1-t_2}\left(\frac{t^{\prime }}{t-t_1}-\frac{t^{\prime }}{t-t_2}\right)=-n.$$

Integrating it on both sides gives

$$ln\frac{t-t_1}{t-t_2}=n(t_2-t_1)z+C,C\in \mathbb{C},$$

which is equivalent to

$$\frac{t-t_1}{t-t_2}=e^{n(t_2-t_1)z+C}.$$

It immediately yields

$$t=t_2+\frac{t_2-t_1}{e^{n(t_2-t_1)z+C}-1}=\frac{f^{\prime }}{f}.$$

Note that zeros of $e^{n(t_2-t_1)z+C}-1$ are the zeros of f. If $z_0$ is a zero of f with multiplicity k, then

$$k=\mathrm{Res}\left[\frac{f^{\prime }}{f},z_0\right]=\mathrm{Res}\left[t_2+\frac{t_2-t_1}{e^{n(t_2-t_1)z+C}-1},z_0\right]=\frac{1}{n}$$

is a contradiction.

If $t_1=\frac{{\alpha }_1}{2}$, then $f(z)=c_1{e}^{\frac{{\alpha }_{1}z}{2}}$, where $c_1$ is a constant satisfying $c_1^2=p_1$.

Similarly, if $t_2=\frac{{\alpha }_2}{2}$, then we have $f(z)=c_2{e}^{\frac{{\alpha }_{2}z}{2}}$, where $c_2$ is a constant satisfying $c_2^2=p_2$.

Case 2.$\phi \not\equiv 0$.

Equation (16) can be written as

$$\frac{1}{f^2}=\frac{1}{\phi }\left[{\alpha }_1{\alpha }_2-n({\alpha }_1+{\alpha }_2)\left(\frac{f^{\prime }}{f}\right)+n(n-1){\left(\frac{f^{\prime }}{f}\right)}^2+n\left(\frac{f^{″}}{f}\right)\right].$$

Using Lemma 1, we have

$$2m\left(r,\frac{1}{f}\right)=m\left(r,\frac{1}{f^2}\right)\le m\left(r,\frac{1}{\phi }\right)+S(r,f).$$

(17)

From Equation (16), if $z_0$ is a multiple zero of f, then $z_0$ must be a zero of $\phi$. Thus, it follows that

$$N_{(2}\left(r,\frac{1}{f}\right)\le N\left(r,\frac{1}{\phi }\right)+S(r,f),$$

(18)

where $N_{(2}(r,\frac{1}{f})$ denotes the counting function of multiple zeros of f. Equations (17) and (18) and the first fundamental theorem give

$$T(r,f)\le N_{1)}\left(r,\frac{1}{f}\right)+T(r,\phi )+S(r,f).$$

(19)

□

5. Proof of Theorem 7

Proof. 

Assume that $f(z)$ is a transcendental entire solution with ${\rho }_2(f)<1$ of Equation (6). Applying Lemmas 1 and 2 to Equation (6), we have

$$\begin{array}{cc}\hfill & T(r,p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z})=T(r,f^n(z)+q(z)f^{(k)}(z+h))\hfill \\ \le \hfill & T(r,f^n)+T(r,q(z)f^{(k)}(z+h))+O(1)\hfill \\ \le \hfill & T(r,f^n)+m\left(r,\frac{q(z)f^{(k)}(z+h)}{f(z)}\right)+m(r,f)+O(1)\hfill \\ \le \hfill & T(r,f^n)+m\left(r,q(z)\frac{f(z+h)}{f(z)}\right)+m\left(r,\frac{f^{(k)}(z+h)}{f(z+h)}\right)+m(r,f)+O(1)\hfill \\ \le \hfill & (n+1)T(r,f)+S(r,f).\hfill \end{array}$$

(20)

On the other hand, we deduce

$$\begin{array}{cc}\hfill & T(r,p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z})=T(r,f^n(z)+q(z)f^{(k)}(z+h))\hfill \\ \ge \hfill & T(r,f^n)-T(r,q(z)f^{(k)}(z+h))+O(1)\hfill \\ \ge \hfill & nT(r,f)-m\left(r,\frac{q(z)f^{(k)}(z+h)}{f(z)}\right)-m(r,f)+O(1)\hfill \\ \ge \hfill & nT(r,f)-m\left(r,\frac{q(z)f(z+h)}{f(z)}\right)-m\left(r,\frac{f^{(k)}(z+h)}{f(z+h)}\right)-m(r,f)+O(1)\hfill \\ \ge \hfill & nT(r,f)-T(r,f)+S(r,f)\hfill \\ =\hfill & (n-1)T(r,f)+S(r,f).\hfill \end{array}$$

(21)

Combining Equations (20) and (21), it follows that

$$(n-1)T(r,f)+S(r,f)\le T(r,p_1{e}^{{\alpha }_{1}z}+p_2{e}^{{\alpha }_{2}z})\le (n+1)T(r,f)+S(r,f),$$

which implies $\rho (f)=1$.

Denoting $P_2(f):=q(z)f^{(k)}(z+h)$ and differentiating Equation (6), we have

$$n{f}^{n-1}{f}^{\prime }+P_2^{\prime }={\alpha }_1{p}_1{e}^{{\alpha }_{1}z}+{\alpha }_2{p}_2{e}^{{\alpha }_{2}z}.$$

(22)

Eliminating $e^{{\alpha }_{2}z}$ from Equations (6) and (22) gives

$${\alpha }_2{f}^n-n{f}^{n-1}{f}^{\prime }+{\alpha }_2{P}_2-P_2^{\prime }=({\alpha }_2-{\alpha }_1)p_1{e}^{{\alpha }_{1}z}.$$

(23)

Differentiating Equation (23) yields

$$n{\alpha }_2{f}^{n-1}{f}^{\prime }-n(n-1)f^{n-2}{(f^{\prime })}^2-n{f}^{n-1}{f}^{″}+{\alpha }_2{P}_2^{\prime }-P_2^{″}={\alpha }_1({\alpha }_2-{\alpha }_1)p_1{e}^{{\alpha }_{1}z}.$$

(24)

It follows from Equations (23) and (24) that

$$f^{n-2}\phi =Q,$$

(25)

where

$$\phi ={\alpha }_1{\alpha }_2{f}^2-n({\alpha }_1+{\alpha }_2)f{f}^{\prime }+n(n-1){(f^{\prime })}^2+nf{f}^{″}$$

and

$$Q=-{\alpha }_1{\alpha }_2{P}_2+({\alpha }_1+{\alpha }_2)P_2^{\prime }-P_2^{″}.$$

Next we discuss two cases below.

Case 1.$\phi \equiv 0$.

This case can be completed by the same method as employed in Case 1 of Theorem 5. We obtain $f(z)=c_2{e}^{\frac{{\alpha }_{2}z}{n}}$, where $c_2$ is a constant satisfying $c_2^n=p_2$. Substituting these formulas into Equation (6), we have

$$q(z)c_2{\left(\frac{{\alpha }_2}{n}\right)}^k{e}^{\frac{{\alpha }_{2}h}{n}}{e}^{\frac{{\alpha }_2}{n}z}-p_1{e}^{{\alpha }_{1}z}=0.$$

According to ${\alpha }_1\ne {\alpha }_2$ and Lemma 3, we have

$${\alpha }_2=n{\alpha }_1\mathrm{and}q(z)c_2{\left(\frac{{\alpha }_2}{n}\right)}^k{e}^{\frac{{\alpha }_{2}h}{n}}=p_1,$$

which implies that $q(z)$ is a constant. Set $q:=q(z)$.

Similarly, we proceed to obtain $f(z)=c_1{e}^{\frac{{\alpha }_{1}z}{n}}$, $q{c}_1{\left(\frac{{\alpha }_1}{n}\right)}^k{e}^{\frac{{\alpha }_{1}h}{n}}=p_2$, ${\alpha }_1=n{\alpha }_2$ and $c_1^n=p_1$.

Case 2.$\phi \not\equiv 0$.

For $n\ge 4$, we shall derive a contradiction. In fact, Q is a difference-differential polynomial in f and its degree at most is 1. By Equation (25) and Lemma 4, we have $m(r,\phi )=S(r,f)$ and $T(r,\phi )=S(r,f)$. On the other hand, we can rewrite Equation (25) as $f^{n-3}(f\phi )=Q$, which implies $m(r,f\phi )=S(r,f)$ and $T(r,f\phi )=S(r,f)$. If $\phi \not\equiv 0$, then $T(r,f)=T(r,\frac{f\phi }{\phi })=S(r,f)$ and this is impossible.

For $n=3$, since Q is a difference-differential polynomial in f and its degree at most is 1, it follows from Equation (25) and Lemma 4 that $m(r,\phi )=S(r,f)$ and

$$T(r,\phi )=S(r,f).$$

(26)

We still use the same method in Case 2 of Theorem 5 to obtain the inequality of Equation (19). Equations (19) and (26) and the first fundamental theorem result in

$$T(r,f)=N_{1)}\left(r,\frac{1}{f}\right)+S(r,f).$$

For $n=2$, we just obtain the inequality of Equation (19). □

6. Conclusions

In this study, we consider two questions. Firstly, the first question posed by Li in [16] is how to find the solutions of Equation (2) if $degP(f)=n-1$. Since the degree of $P(f)$ is bigger than $n-2$, one cannot use Clunie’s lemma which is a key in the proof in Theorem 2. It is very difficult to resolve the question. Chen and Gao considered the entire solution f of Equation (2) with the order $\rho (f)<\infty$ and $N(r,1/f)=S(r,f)$ when $n=2$ and partially answered the question. We remove the condition that the order $\rho (f)<\infty$ by a different method and improve the result of Chen and Gao in Theorem 4. For the general case of Li’s question, we use the method of Theorem 4 and give a partial answer in Theorem 5.

Secondly, motived by Theorem 2, a question to be raised is how to find the existence of solutions to Equation (5) if $e^{\lambda z}$ and $e^{-\lambda z}$ can be replaced by a linear combination of $e^{{\alpha }_{1}z}$ and $e^{{\alpha }_{2}z}$ for two distinct constants ${\alpha }_1$ and ${\alpha }_2$. We consider the general case by the similar method with Theorem 5 and give the partial solutions of Equation (6).

For further study, we conjecture that the inequality $T(r,f)\le N_{1)}(r,\frac{1}{f})+T(r,\phi )+S(r,f)$ or $T(r,f)=N_{1)}(r,\frac{1}{f})+S(r,f)$ can be removed if $N(r,1/f)=S(r,f)$ in Theorems 5 and 7.