Meromorphic solutions of $f^{n}+P_{d}(f) = p_{1}e^{\alpha_{1}z}+p_{2}e^{\alpha_{2}z}+p_{3}e^{\alpha_{3}z}$

1.   Introduction and main results

It is an important and quite difficult problem to prove the existence of solutions of complex differential equations. Recently, more and more people investigate the solutions of complex differential equations by using Nevanlinna theory. Throughout this paper, we assume that the reader is familiar with the standard notations and fundamental results in Nevanlinna theory, such as $T(r, f)$, $m(r, f)$, $N(r, f)$, $S(r, f)$, the first and second main theorems, lemma on the logarithmic derivatives etc. of Nevanlinna theory. For more details, see ^[3,4,14].

First of all, we give the following notations. For $\lambda\in\mathbb{C}\setminus\{0\}$, denote

$\Gamma_{0} = \{\lambda_{0}e^{\lambda z}$: $\lambda_{0}$ is a nonzero constant},

$\Gamma_{1} = \{\lambda_{1}e^{\lambda z}+\lambda_{0}$: $\lambda_{0}$ and $\lambda_{1}$ are nonzero constants},

$\Gamma_{2} = \{\lambda_{1}e^{\lambda z}+\lambda_{2}e^{-\lambda z}$: $\lambda_{1}$ and $\lambda_{2}$ are nonzero constants},

$\Gamma_{3} = \{\lambda_{1}e^{\lambda z}+\lambda_{2}e^{-2\lambda z}$: $\lambda_{1}$ and $\lambda_{2}$ are nonzero constants}.

In ^[13], the authors show that the differential equation $4f^{3}+3f'' = -\sin 3z$ has exactly three nonconstant entire solutions, namely $f_{1}(z) = \sin z$, $f_{2}(z) = \frac{\sqrt{3}}{2}\cos z-\frac{1}{2}\sin z$ and $f_{3}(z) = -\frac{\sqrt{3}}{2}\cos z-\frac{1}{2}\sin z$. Since $\sin 3z$ is a linear combination of $e^{3iz}$ and $e^{-3iz}$, this has attracted many scholars to study the following more general differential equation

where $p_{d}(f)$ is a differential polynomial in $f$ of degree $d$, see ^{[5,6,7,8,9,10,13,15,16]}. Now, we recall the following classic result due to Li ^[6].

Theorem A.^[6] Let $n\geq2$ be an integer and $P_{d}(f)$ be a differential polynomial in $f$ of degree at most $n-2$, and $p_{i}, \alpha_{i}(i = 1, 2)$ be nonzero constants and $\alpha_{1}\neq\alpha_{2}$. If $f(z)$ is a transcendental meromorphic solution of (1.1) and satisfying $N(r, f) = S(r, f)$, then one of the following holds:

${\rm (1)}$ $f(z) = c_{0}+c_{1}e^{\alpha_{1}z/n};$

${\rm (2)}$ $f(z) = c_{0}+c_{2}e^{\alpha_{2}z/n};$

${\rm (3)}$ $f(z) = c_{1}e^{\alpha_{1}z/n}+c_{2}e^{\alpha_{2}z/n},$ and $\alpha_{1}+\alpha_{2} = 0$,

where $c_{0}$ is a small function of $f(z)$ and $c_{1}, \; c_{2}$ are constants satisfying $c_{1}^{n} = p_{1}, c_{2}^{n} = p_{2}$.

If $d\leq n-2$, Theorem A shows that the entire solutions of (1.1) are exponential sums and they are in $\Gamma_{0}\cup\Gamma_{1}\cup\Gamma_{2}$. Firstly, we consider the existence of entire solutions of (1.1) under the condition $d = n-1$. We get the following result.

Theorem 1.1. Let $n\geq2$ be an integer and $P_{d}(f)$ be a differential polynomial in $f$ of degree $n-1$, and $p_{i}, \; \alpha_{i}(i = 1, 2)$ be nonzero constants and $\alpha_{1}\neq\alpha_{2}$. If $f$ is a transcendental entire solution of $(1.1)$, then $f\not\in\Gamma_{3}$.

Note that the right-hand side of (1.1) is a linear combination of $e^{\alpha_{1}z}$ and $e^{\alpha_{2}z}$, it is natural and interesting to replace $p_{1}e^{\alpha_{1}z}+p_{2}e^{\alpha_{2}z}$ with $h(z): = p_{1}e^{\alpha_{1}z}+p_{2}e^{\alpha_{2}z}+\cdots+p_{k}e^{\alpha_{k}z}$. In other words, how to find the solutions of the following equation

where $k\geq3$ is an integer, $\alpha_{i}(i = 1, 2, \cdots, k)$ are distinct nonzero constants, $P_{d}(f)$ is a differential polynomial in $f$ with degree $d$ and $p_{i}, \alpha_{i}(i = 1, 2, \cdots, k)$ are nonzero constants?

In this perspective, Xue ^[12] investigated the entire solutions of (1.2) for $k = 3$ and he proved the following result.

Theorem B.^[12] Let $n\geq2$ and $P_{d}(f)$ be a differential polynomial in $f$ of degree $d\leq n-1$. Suppose that $p_{i}, \; \alpha_{i}(i = 1, 2, 3)$ are nonzero constants and $|\alpha_{1}|\geq|\alpha_{2}|\geq|\alpha_{3}|$. If $f(z)$ is a transcendental entire solution of of (1.2), then $f(z) = a_{1}e^{\alpha_{1}z/n}$, where $a_{1}$ is a non-zero constant such that $a_{1}^{n} = p_{1}$, and $\alpha_{j}$ are in one line for $j = 1, 2, 3$.

In fact, Examples 2–4 in this section show that (1.2) has some solutions different from $a_{1}e^{\alpha_{1}z/n}$ under the condition $k = 3$. In this paper, we further consider the existence of the transcendental meromorphic solutions of (1.2) under the conditions $d\leq n-3$ and $k = 3$. Without loss of generality, we assume that $|\alpha_{1}|\geq|\alpha_{2}|\geq|\alpha_{3}|$. We obtain the following result.

Theorem 1.2. Let $n\geq3$ be an integer, $k = 3$ and $P_{d}(f)$ be a differential polynomial in $f$ of degree $d\leq n-3$. If $f$ is a transcendental meromorphic solution of $(1.2)$ satisfying $N(r, f) = S(r, f)$, then one of the following holds:

${\rm (1)}$ $f(z) = c_{1}e^{\alpha_{1}z/n}$;

${\rm (2)}$ $f(z) = c_{1}e^{\alpha z}+c_{0}$ and $\alpha_{j} = \frac{n-j}{n}\alpha_{1}(j = 2, 3)$;

${\rm (3)}$ $f(z) = c_{1}e^{2\alpha z}+c_{3}e^{-\alpha z}$ and $\alpha_{2} = \frac{2n-3}{2n}\alpha_{1}$, $\alpha_{3} = -\frac{1}{2}\alpha_{1}$,

where $\alpha = \frac{(\alpha_{1}+\alpha_{2}+\alpha_{3})}{3(n-1)}$ and $c_{0}$, $c_{1}, \; c_{3}$ are nonzero constants satisfying $c_{1}^{n} = p_{1}, \; c_{3}^{n} = p_{3}$. Furthermore, if $d < n-3$, then $f(z) = c_{1}e^{\alpha_{1}z/n}$.

Remark. By the result due to Steinmetz ^[11], we have $T(r, h) = K\left(1+o(1)\right)\frac{r}{2\pi}(r\rightarrow \infty)$, where $K$ denote the perimeter of convex polygon which is formed by $\{0, \alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}\}$ (if $0, \alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are collinear, then $K = 2\max_{i, j\in\Lambda}|\alpha_{i}-\alpha_{j}|$, where $\Lambda = \{0, 1, 2, \cdots, k\}$ and $\alpha_{0} = 0$). This implies (1.2) has no rational solution.

By Theorem A and Theorem 1.1, we see that the entire solutions of (1.1) are not in $\Gamma_{3}$. Since the entire solutions of (1.2) are in $\Gamma_{0}\cup\Gamma_{1}\cup\Gamma_{3}$ under the conditions $k = 3$ and $d\leq n-3$, this implies that the solutions of (1.2) are different from (1.1). Below, we give some examples to show the existence of the solutions of (1.2).

Example 1. For any $n\geq5$, $f^{n}+ff''+f' = e^{nz}+e^{2z}+e^{z}$ has exactly one entire solution $f(z) = e^{z}$.

Example 2. For $n = 5$ and $d = 2$, $f^{5}-10ff'+5f'-1 = 10e^{3z}+5e^{4z}+e^{5z}$ has exactly one entire solution $f(z) = e^{z}+1$.

Example 3. For $n = 5$ and $d = 2$, $f^{5}-\frac{50}{9}f^{2}-\frac{5}{9}f'f'' = e^{-5z}+5e^{7z}+e^{10z}$ has exactly one entire solution $f(z) = e^{-z}+e^{2z}$.

Example 4. Let $n = 3$ and $d = 0$, then

has three entire solutions $f_{j}(z) = \omega_{j}(e^{z}+1)$, where $\omega_{j} = e^{\frac{2j\pi i}{3}}(j = 1, 2, 3)$.

And

has three solutions $f_{j}(z) = \omega_{j}(e^{-z}+e^{2z})$, where $\omega_{j} = e^{\frac{2j\pi i}{3}}(j = 1, 2, 3)$.

By Theorem 1.2, we can prove the following two corollaries.

Corollary 1. Let $b_{j}(j = 0, 1, \cdots, k)$ be constants and $\{\lambda, p_{1}, p_{2}, p_{3}\}\subset\mathbb{C}\setminus\{0\}$. If $f$ is a transcendental merommorphic solution of the following nonlinear differential equation

with $N(r, f) = S(r, f)$. Then $\lambda$ is a root of polynomials $-3b_{0}+b_{1}z+b_{2}z^{2}+\cdots+b_{k}z^{k}$. In particularly, if $b_{0} = 0$, then $(1.3)$ has no transcendental merommorphic solution satisfying $N(r, f) = S(r, f)$.

Let $b_{0} = 1$, $b_{1} = 2$, $b_{2} = 1$, then $-3b_{0}+b_{1}z+b_{2}z^{2}$ has two roots $1$ and $-3$. If $\lambda = 1$ or $\lambda = -3$, it can be verified that equation $f^{4}+f+2f'+f'' = e^{4\lambda z}+4e^{3\lambda z}+6e^{2\lambda z}$ has a solution $f(z) = -e^{\lambda z}-1$.

Corollary 1.2. Let $d$, $b_{j}(j = 0, 1, \cdots, k)$ be constants and $\{\lambda, p_{1}, p_{2}, p_{3}\}\subset\mathbb{C}\setminus\{0\}$. If $f$ is a transcendental meromorphic solution of the following nonlinear differential equation

with $N(r, f) = S(r, f)$. Then $d = 0$ and $\lambda$ is a root of polynomials $b_{0}+7b_{1}z-5b_{2}z^{2}+\cdots+$ $\left(3+(-2)^{k+1}\right)b_{k}z^{k}$.

Let $b_{0} = -70$, $b_{1} = 5$, $b_{2} = -7$, then $b_{0}+7b_{1}z-5b_{2}z^{2}$ has two roots $1$ and $-2$. It can be verified that $f^{4}-70f+5f'-7f'' = 16e^{-8z}+96e^{-5z}+81e^{4z}$ has a solution $f(z) = 2e^{-2z}+3e^{z}$ and $f^{4}-70f+5f'-7f'' = 81e^{16z}+324e^{10z}+81e^{-8z}$ has a solution $f(z) = 3e^{-4z}+3e^{2z}$.

The paper is devoted to investigate the solutions of (1.2) under the conditions $k = 3$ and $d\leq n-3$ and we obtain some new results. There are still many problems to be solved. For further study, we arise the following questions.

Question 1. How to find the solutions of (1.2) under the condition $k > 3$?

Question 2. How to find the solutions of (1.2) under the conditions $k = 3$ and $d\leq n-2$?

2.   Lemmas

To prove the Theorems, we need the following lemmas.

Lemma 2.1. ^[1,2]Suppose that $f(z)$ is meromorphic and transcendental in the plane and that

where $P(f)$ and $Q(f)$ are differential polynomials in $f$ with functions of small proximity related to $f$ as the coefficients and the degree of $Q(f)$ is at most n. Then

Lemma 2.2. If $f$ is a transcendental meromorphic solution of the following equation

where $a, b, c, d, E$ are constants and $adE\neq0$, then $f$ satisfies one of the following cases:

${\rm (i)}$ $f(z) = Ae^{\frac{-c}{3d} z}+B$;

${\rm (ii)}$ $f(z) = \frac{Dd}{c}e^{\frac{c}{3d}z}-\frac{E}{D^{2}c}e^{-\frac{2c}{3d}z}$,

where $A, B, D$ are nonzero constants satisfying $B^{3} = E$.

Proof. Suppose that $f$ is a transcendental meromorphic solution of (2.1). Since $E\neq0$, it's obviously that $f$ has no pole and all zeros of $f$ with multiplicity $1$. By (2.1), we have

By the lemma of logarithmic derivative and (2.2), we get $m(r, 1/f) = S(r, f)$. This implies

Differentiating (2.1), we obtain

Let $\omega(z) = \frac{cf'+3df''}{f}$, by a similar analysis as in Lemma $6$ in ^[6], we can deduce $\omega(z)$ has no pole. So we have $T(r, \omega) = m(r, \omega) = S(r, f)$. Let

Substituting (2.5) into (2.4), we obtain

By combining (2.3) and (2.6), we conclude that

Now, we distinguish the following two cases to discuss.

Case 1. If $\omega = 0$. By (2.5), we have $f' = c_{0}e^{\frac{-c}{3d} z}$, this implies $f(z) = Ae^{\frac{-c}{3d} z}+B$ with $A, B\in\mathbb{C}$. By substituting $f(z) = Ae^{\frac{-c}{3d} z}+B$ into (2.1), we have $B^{3} = E$. This gives $(i)$.

Case 2. If $\omega\neq0$. By (2.7), it is easy to see that $b = 0$ and $4c^{3} = -27ad^{2}$. Therefore (2.1) is equivalent to the following equation.

Since $f$ has no pole, we obtain by (2.8) that

where $\alpha(z)$ is an entire function and $C\in\mathbb{C}\backslash\{0\}$. By (2.9), we can deduce

It follows from (2.10) that $\alpha(z) = \frac{3c}{d}z+c_{1}$, where $c_{1}\in\mathbb{C}$. Hence $f = \frac{Dd}{c}e^{\frac{c}{3d}z}-\frac{E}{D^{2}c}e^{-\frac{2c}{3d}z}$ with $D\in\mathbb{C}\backslash\{0\}$. This gives $(ii)$.

Remark. If $f$ is a transcendental meromorphic solution of (2.7), it follows from Lemma 2.2 that $f\in\Gamma_{1}\cup\Gamma_{3}$.

$\bullet$ If $f\in\Gamma_{1}$, then $a, b, c, d$ satisfy $bc = 9ad$ and $3bd = c^{2}$ by (2.7). Suppose that $b = c = 3$ and $a = d = 1$. Then for any $A\in\mathbb{C}\backslash\{0\}$ and $B^{3} = E$, it can be verified that $f(z) = Ae^{-z}+B$ is a solution of

$\bullet$ If $f\in\Gamma_{3}$, then $a, b, c, d$ satisfy $b = 0$ and $4c^{3}+27ad = 0$ by (2.7). Let $a = -4$, $b = 0$, $c = 3$ and $d = 1$, then $f(z) = \frac{D}{3}e^{z}-\frac{E}{3D^{2}}e^{-2z}$ with $D\in\mathbb{C}\backslash\{0\}$ is a solution of

Lemma 2.3. Let $a_{i}(i = 1, 2, 3, 5)$ be constants, $a_{4} = -n$, $a_{6} = -3n(n-1)$, $a_{7} = -n(n-1)(n-2)$ with $n\geq3$ and $\psi\not\equiv0$ is a small function of $e^{z}$. If $f$ is a transcendental meromorphic solution of the following equation

with $\rho(f)\geq1$, then one of the following holds:

${\rm (i)}$ $f(z) = c_{1}e^{\alpha z}+c_{2}$;

${\rm (ii)}$ $f(z) = \frac{ c_{3}}{3\alpha}e^{-\alpha z}-\frac{(n-1)\psi}{a_{5}(n-2)c_{3}^{2}}e^{2\alpha z}$,

where $\alpha = \frac{a_{5}}{3n(n-1)^{2}}$ and $c_{1}, c_{2}, c_{3}$ are nonzero constants satisfying $c_{2}^{3} = \psi$.

Proof. If $f$ is a transcendental meromorphic solution of (2.11), we can deduce $N(r, f) = N(r, \psi) = S(r, f)$ and

Differentiating (2.11) and it yields

Let

It follows from (2.11)–(2.14) that $N(r, \varphi)\leq N_{(2}\left(r, \frac{1}{f}\right)+N(r, f)+N(r, \psi) = S(r, f)$. This implies $T(r, \varphi) = S(r, f)$. Now we rewrite (2.14) as the form

where

Differentiating (2.15), we get

and

By substituting (2.15) and (2.17) into (2.11), we have

By substituting (2.15), (2.17) and (2.18) into (2.13), we obtain

where

and

By Combining (2.19) and (2.20), we have

Applying the lemma of logarithmic derivative to (2.11), we get $m\left(r, \frac{1}{f}\right) = S(r, f)$. This implies $N\left(r, \frac{1}{f}\right) = T(r, f)+S(r, f)$. We then obtain by (2.21) that

Now, we claim that $\psi$ has no zero and pole.

If not, firstly, we assume that $\psi$ has a pole and denote it by $z_{0}$. Obviously, $z_{0}$ is also a pole of $f$ by (2.11). Hence, there is a sufficiently small neighborhood $U(z_{0})$ of $z_{0}$, for any $z$ in $U(z_{0})$, by (2.11), we have

where $\mu\nu\neq0$ and $m\in\mathbb{N^{+}}$. Note that $a_{6} = -3n(n-1)$, $a_{7} = -n(n-1)(n-2)$.

By substituting (2.23) into (2.14) and (2.16), we can deduce

where $A_{1} = -\frac{(n-2)(m+1)}{n-1}$ and $B_{1} = \frac{m(m+1)}{n-1}$. By substituting (2.23) and (2.24) into (2.22), we have

where $K_{1} = \left((\frac{n-2}{n-1})^{2}(m+1)+4\frac{n-2}{n-1}+3m-6\right)(m+1)$. If $n\geq4$, it is easy to see that $K_{1}\geq\frac{5}{9} > 0$. By (2.22) and (2.25), we can get a contradiction. So $n = 3$. This shows that

Note that $K_{1} > 0$ if $m = n = 3$. We can easily get a contradiction by (2.22), (2.25) and (2.26). This means $\psi$ has no pole.

Secondly, if $\psi$ has a zero and denote it by $z_{1}$. It is easy to see that $z_{1}$ is not a zero of $f$ with multiplicity $1$ by (2.11). Next, we distinguish the following two cases to discuss.

Case i). Suppose that $z_{1}$ is a zero of $f$ with multiplicity $t(t\geq2)$. Hence, there is a sufficiently small neighborhood $U(z_{1})$ of $z_{1}$, and for any $z$ in $U(z_{1})$, by (2.11), we have

where $\mu_{1}\nu_{1}\neq0$. Furthermore, we can deduce $s = 3t-3$.

By substituting (2.27) into (2.14) and (2.16), we get

where $A_{2} = \frac{(n-2)(t-1)}{n-1}$ and $B_{2} = \frac{t(t-1)}{n-1}$. By substituting (2.27) and (2.28) into (2.22), we have

where $K_{2} = \left(3t+6-\frac{4(n-2)}{n-1}\right)(t-1) > 0$ for $n\geq3$. This implies $\Gamma(z)\not\equiv0$, which contradicts to (2.22). Hence $z_{1}$ is not a zero of $f$ with multiplicity $\geq2$.

Case ii). Suppose that $z_{1}$ is not a zero of $f$. Note that $z_{1}$ is a pole of $A(z)$ with multiplicity $1$ by (2.16), therefore $z_{1}$ is a pole of $B(z)$ with multiplicity $1$ or it is not a pole of $B$ by (2.17). Suppose that $\psi$ satisfies (2.27), we then deduce $A(z) = \frac{(n-2)s}{3(n-1)}(z-z_{1})^{-1}+O(1)$. By a calculation, we have

where $A_{3} = \frac{n-2}{3(n-1)}$. It's easy to see that $(6n-4)A_{3}-3(n-1) < 0$ for $n\geq3$. So $\Lambda(z)\not\equiv0$, which also contradicts to (2.22).

By the above discussion, we conclude that $\psi$ has no zero and pole. It follows from $T(r, \psi) = S(r, e^{z})$ that $\psi$ is a constant. We then obtain by (2.22) that $b_{1} = b_{2} = b_{3}\equiv0$. Furthermore, it is easy to see that $A(z)\equiv-\frac{a_{5}}{a_{6}+3a_{7}}$ by (2.16), we then deduce $B(z)$ is a constant by the fact $b_{3}\equiv0$. Now (2.20) becomes

By Lemma 2.2, we have $f(z) = c_{1}e^{\alpha z}+c_{2}$ or $f(z) = \frac{ c_{3}}{3\alpha}e^{-\alpha z}-\frac{(n-1)\psi}{a_{5}(n-2)c_{3}^{2}}e^{2\alpha z}$, where $\alpha = \frac{a_{5}}{3n(n-1)^{2}}$ and $c_{1}, c_{2}, c_{3}$ are nonzero constants satisfying $c_{2}^{3} = \psi$.

Lemma 2.4. [14, Corollary of Theorem 1.52] If $f_{j}(z)$ $(j = 1, 2, \cdots, n+1) (n\geq2)$, $g_{j}(z) (j = 1, 2, \cdots, n)(n\geq1)$ are entire functions satisfying the following two conditions:

(i) $\sum_{j = 1}^{n}f_{j}(z)e^{g_{j}(z)}\equiv f_{n+1}(z);$

(ii) When $1\leq j\leq n+1, 1\leq k \leq n$, the order of $f_{j}$ is less than that of $e^{g_{k}(z)}$. When $n\geq 2$ and $1\leq j\leq n+1, 1\leq h\leq k \leq n$, the orders of $f_{j}$ are also less than that of $e^{g_{h}(z)-g_{k}(z)}.$

Then $f_{j}(z)\equiv0 (j = 1, 2, \cdots, n+1).$

By Lemma 2.4. we have

Lemma 2.5. Let $a_{0}, \cdots, a_{n}$ be constant, and let $b_{1}, \cdots, b_{n}\in\mathbb{C}\backslash \{0\}$ be distinct constants. Then

holds only when $a_{0} = a_{1} = \cdots = a_{n} = 0$.

3.   Proof of theorems

Proof of Theorem 1.1. Suppose that $f = \lambda_{1}e^{\lambda z}+\lambda_{2}e^{-2\lambda z}$ is a solution of (1.1). By substituting $f$ into (1.1), we have

where $\Xi = \{n, -2n+3, -2n\}$. By Lemma 2.5 and (3.1), we can get a contradiction. Hence $f\not\in\Gamma_{3}$. This completes the proof of Theorem 1.1.

Proof of Theorem 1.2. Let $f$ be a transcendental meromorphic solution of (1.2) satisfying $N(r, f) = S(r, f)$. By differentiating (1.2), we have

and

where $q_{i} = p_{i}\alpha_{i}$ and $r_{i} = p_{i}\alpha_{i}^{2}$$(i = 1, 2, 3)$. Notice that (1.2), (3.2), (3.3) can be written as inhomogeneous linear systems of equations form as $AX = b$, where

By a calculation, $D = |A| = p_{1}p_{2}p_{3}(\alpha_{3}-\alpha_{2})(\alpha_{3}-\alpha_{1})(\alpha_{2}-\alpha_{1})\neq0$. So, there is a unique solution $X^{T} = (e^{\alpha_{1}z}, e^{\alpha_{2}z}, e^{\alpha_{3}z})$ of $AX = b$. It can be verified that $e^{\alpha_{1}z}$ satisfies

where

By differentiating (3.4) and eliminate $e^{\alpha_{1}z}$, we can deduce

where

which is a differential polynomial in $f$ with degree $\leq n-3$ and

By (3.6) and Lemma 2.1, we have $T(r, \Psi) = m(r, \Psi)+S(r, f) = S(r, f)$. Below, we distinguish two cases to discuss.

Case 1. If $\Psi\equiv0$. Then $R_{d}(f)\equiv0$ by (3.6). We now claim that $E_{d}(f)\equiv0$, if not, by (3.7), we have

where $A$ is a non-zero constant. Therefore, by (3.4), we get

If $A\neq p_{1}(\alpha_{3}-\alpha_{1})(\alpha_{2}-\alpha_{1})$, it follows from (3.10) that $f$ has no pole and zero. Therefore $f = B_{1}e^{az}$. By substituting $f = B_{1}e^{az}$ into (3.10), we can deduce $a = \frac{\alpha_{1}}{n}$. Note that $P_{d}(f)$ is an differential polynomial in $f$ of degree $d\leq n-3$ and $|\alpha_{1}|\geq|\alpha_{2}|\geq|\alpha_{3}|$, it can be verified that $E_{d}(f)\neq Ae^{\alpha_{1}z}$ by (3.5). This contradicts (3.9).

So we have $A = p_{1}(\alpha_{3}-\alpha_{1})(\alpha_{2}-\alpha_{1})$. By (3.10), we obtain

It can be verified that $f$ also has no pole and zero by (3.11), this means $f = B_{2}e^{bz}$. By substituting $f = B_{2}e^{bz}$ into (3.10), we get $b = \frac{\alpha_{2}}{n}$ or $b = \frac{\alpha_{3}}{n}$. This implies that $E_{d}(f)\neq Ae^{\alpha_{1}z}$ by $|\alpha_{1}|\geq |\alpha_{j}|(j = 2, 3)$ and (3.5).

Hence $E_{d}(f)\equiv0$. It follows from (3.5) that

By (3.12), we see that $f$ has no pole and zero. By a similar discussion as above, we have $f^{n} = p_{1}e^{\alpha_{1}z}$. This is $(i)$ of Theorem 1.2.

Case 2. If $\Psi\not\equiv0$. Applying the lemma of logarithmic derivative to (3.8), we get $m(r, \frac{\psi}{f^{3}}) = \frac{1}{3}m(, \frac{\psi}{f^{3}})+S(r, f) = S(r, f)$. Therefore,

Since $N(r, f) = S(r, f)$, by the above equation, we have

By (1.2) and (3.13), we get $\rho(f) = \rho(p_{1}e^{\alpha_{1}z}+p_{2}e^{\alpha_{2}z}+p_{3}e^{\alpha_{3}z}) = 1$. It follows from Lemma 3 that $f(z) = c_{1}e^{\beta z}+c_{0}$ or $f(z) = c_{1}e^{2\beta z}+c_{3}e^{-\beta z}$, where $c_{i}(i = 0, 1, 2, 3)$ are nonzero constants and $\beta = \frac{\alpha_{1}+\alpha_{2}+\alpha_{3}}{3(n-1)}$.

If $f = c_{1}e^{\beta z}+c_{0}$, by substituting $f$ into (1.2), we have

Note that $|\alpha_{1}|\geq|\alpha_{2}|\geq|\alpha_{3}|$ and $p_{1}p_{2}p_{3}c_{1}c_{0}\neq0$, then by (3.14) and Lemma 2.5, we get

By the above equations, we can deduce $\alpha_{j} = \frac{n-j}{n}\alpha_{1}(j = 2, 3)$, which satisfies the $(ii)$ of Theorem 1.2.

If $f = c_{2}e^{2\beta z}+c_{3}e^{-\beta z}$, substitute $f$ into (1.2), we have

By (3.15) and Lemma 2.5, we have

Since $\beta = \frac{\alpha_{1}+\alpha_{2}+\alpha_{3}}{3(n-1)}$, we can deduce $\alpha_{2} = \frac{2n-3}{2n}\alpha_{1}$ and $\alpha_{3} = -\frac{1}{2}\alpha_{1}$ by (3.16). So, we have $(iii)$ of Theorem 1.2.

Furthermore, if $d < n-3$, then by (3.6) and lemma 2.1, we have $T(r, \Psi) = S(r, f) = T(r, f\Psi)$. This means $\Psi\equiv0$. It follows that $f^{n} = p_{1}e^{\alpha_{1}z}$ by the above discussion of Case 1. Hence we complete the proof of Theorem 1.2.

Proof of Corollary 1.1. Suppose that $f$ is a transcendental meromorphic solution of (1.3) satisfying $N(r, f) = S(r, f)$. By Theorem 1.2, we have

where $\lambda_{0}$ and $\lambda_{1}$ are non-zero constant satisfying $\lambda_{1}^{n} = p_{1}$. By substituting (3.17) into (1.3), we can deduce

This means $\lambda$ is a root of polynomials $-3b_{0}+b_{1}z+b_{2}z^{2}+\cdots+b_{k}z^{k}$.

In particularly, if $a_{0} = 0$, it follows from (3.18) that $\lambda_{0} = 0$. It is a contradiction. Hence, (1.3) has no transcendental merommorphic solution satisfying $N(r, f) = S(r, f)$. This completes the proof of Corollary 1.

Proof of Corollary 1.2. Suppose that $f$ is a transcendental meromorphic solution of (1.4) satisfying $N(r, f) = S(r, f)$. By Theorem 1.2, we have

where $\lambda_{0}$ and $\lambda_{1}$ are non-zero constant satisfying $\lambda_{j}^{n} = p_{j}(j = 1, 3)$. By substituting (3.19) into (1.4), we can deduce that $d = 0$ and

By (3.20) and (3.21), we can deduce $\lambda$ is a root of polynomials $b_{0}+7b_{1}z-5b_{2}z^{2}+\cdots+\left(3+(-2)^{k+1}\right)b_{k}z^{k}$. We complete the proof of Corollary 1.2.

4.   Conclusions

In this paper, we consider the meromorphic solutions of (1.2) with few poles under the conditions $k = 3$, $n\geq3$ and $d\leq n-3$. We proved that all of the solutions of (1.2) are in $\Gamma_{0}\cup\Gamma_{1}\cup\Gamma_{3}$. In particular, for $n-3 = d = 1$, if the following differential equation

has a transcendental meromorphic solution $f$ satisfying $N(r, f) = S(r, f)$, then $\{\alpha_{1}, \; \alpha_{2}, \; \alpha_{3}\} = \{4\lambda, \; 3\lambda, \; 2\lambda\}$ or $\{\alpha_{1}, \; \alpha_{2}, \; \alpha_{3}\} = \{-8\lambda, \; -5\lambda, \; 4\lambda\}$. Moreover

${\rm (1)}$ $\lambda$ is a root of polynomials $-3b_{0}+b_{1}z+b_{2}z^{2}+\cdots+b_{k}z^{k}$ if $\{\alpha_{1}, \; \alpha_{2}, \; \alpha_{3}\} = \{4\lambda, \; 3\lambda, \; 2\lambda\}$;

${\rm (2)}$ $\lambda$ is a root of polynomials $b_{0}+7b_{1}z-5b_{2}z^{2}+\cdots+\left(3+(-2)^{k+1}\right)b_{k}z^{k}$ if $\{\alpha_{1}, \; \alpha_{2}, \; \alpha_{3}\} = \{-8\lambda, \; -5\lambda, \; 4\lambda\}$.

It is a natural idea to investigate the solutions of (1.2) for $k = 3$, $n\geq2$ and $d\leq n-2$. Are all entire solutions of (1.2) in $\Gamma_{0}\cup\Gamma_{1}\cup\Gamma_{3}$ under this conditions? In fact, the following examples show that (1.2) has other forms of entire solutions.

$\bullet$ Let $n = 2$ and $d = 1$, then $f(z) = e^{z}+6e^{-z}+6e^{-2z}$ is a solution of

$\bullet$ Let $n = 3, 4$ and $d = n-2$, then $f(z) = e^{-z}+e^{z}$ is a solution of

and

$\bullet$ Let $n = 3$ and $d = 1$, then $f(z) = e^{z}+e^{3z}$ is a solution of

Acknowledgments

We would like to thank the editors and reviewers for their valuable suggestions which has significantly improve the paper. The research was supported by the National Natural Science Foundation of China under Grant (No.12071047, No.12171264).

Conflict of interest

The authors declare that there is no conflict of interest regarding the publication of this paper.