The zeros on complex differential-difference polynomials of certain types

Abstract

In this paper, we consider the zeros distribution of $f(z)P(z,f) -q(z)$, where $P(z,f)$ is a linear differential-difference polynomial of a finite-order transcendental entire function $f(z)$, and $q(z)$ is a nonzero polynomial. To a certain extent, Theorem 1.1 generalizes the recent results (Latreuch and Belaïdi in Arab. J. Math. 7(1):27–37, 2018; Lü et al. in Kodai Math. J. 39(3):500–509, 2016) related to Hayman conjecture (Hayamn in Ann. Math. 70:9–42, 1959).

1 Introduction

We assume that the readers are familiar with the basic symbols and fundamental results of Nevanlinna theory [8, 9, 18]. A function $a(z)\not\equiv0, \infty$ is a small function with respect to $f(z)$ if $T(r,a)=S(r,f)$, where $S(r,f)=o(T(r,f))$ as $r\rightarrow \infty$ outside a possible exceptional set of finite logarithmic measure. We use $S(f)$ to denote the family of all small functions with respect to $f(z)$. In the paper, a linear differential-difference polynomial of a meromorphic function $f(z)$ is defined by

$$ P(z,f)=\sum_{i=1}^{n}\lambda_{i}(z)f^{(k_{i})}(z+c_{i}), $$

where $\lambda_{i}(z)\in S(f)$ and $c_{i}\in\mathbb{C}$, $k_{i}$ $(i=1,\ldots,n)$ are nonnegative integers.

In 1959, Hayman [7] considered the value distribution of the differential polynomial $f^{n}f'$ and obtained the following result.

Theorem A

Let $f(z)$ be a transcendental meromorphic function, and let $n\geq3$ be an integer. Then $f(z)^{n}f'(z)-d$ has infinitely many zeros, where d is a nonzero constant.

Since then, there were many studies on the zeros distribution of differential polynomials, such as [1, 2, 17]. Recently, some researchers considered the difference analogues of Theorem A, and many related results have been obtained. Laine and Yang [10] investigated the value distribution of $f(z)^{n}f(z+c)$ and proved the following result; for statement of this and others results, we recall the definition of the order of a meromorphic function $f(z)$:

$$ \rho(f)=\limsup_{r\rightarrow\infty}\frac{\log T(r,f)}{\log r}. $$

Theorem B

Let $f(z)$ be a transcendental entire function of finite order, and let c be a nonzero complex constant. If $n\geq2$, then $f(z)^{n}f(z+c)-a$ has infinitely many zeros, where a is a nonzero complex constant.

Some researchers improved Theorem B in different ways; for example, the constant a was replaced by a nonzero polynomial in [13]. In addition, the papers [12, 16, 19] are devoted to the cases of meromorphic functions f or more general difference products. Liu, Liu, and Zhou [14] obtained results related to Theorem B in differential-difference polynomials, which can be stated as follows.

Theorem C

Let $f(z)$ be a finite-order transcendental entire function, and let k be a positive integer. If $n\geq2$, then $f(z)^{n}f^{(k)}(z+c)-a(z)$ has infinitely many zeros, where $a(z)$ is an entire function with $\rho(a)<\rho(f)$.

Theorem D

Let $f(z)$ be a finite-order transcendental entire function with a Borel exceptional polynomial $d(z)$, and let k be a positive integer. If $n\geq1$, then $f(z)^{n}f^{(k)}(z+c)-b$ has infinitely many zeros, where b is a nonzero constant.

Remark 1

If $n=1$, then Theorem B is not true. For example, if $f(z)=e^{z}+1$ and $e^{c}=-1$, then $f(z)f(z+c)-1=-e^{2z}$ has no zeros. Chen, Huang, and Zheng [3] considered the case $n=1$ in Theorem B with $f(z)$ having a Borel exceptional value. In fact, the above function also shows that Theorem D happens: $f(z)f^{(k)}(z+c)-b=-e^{2z}-e^{z}-b$ has infinitely many zeros, and the value 1 is the Borel exceptional value of $e^{z}+1$.

However, it is still an open question whether Theorem D is true for a general transcendental entire function $f(z)$, that is, whether the condition that $f(z)$ has a Borel exceptional polynomial can be removed.

Question 1

Let $f(z)$ be a finite-order transcendental entire function, and let c be a nonzero constant and k be a positive integer. Have the differential-difference polynomials $f(z)f^{(k)}(z+c)-a(z)$ infinitely many zeros or not?

More generally, we can raise the following Question 2.

Question 2

What about the zeros distribution of $f(z)P(z,f)-a(z)$? Here $P(z,f)$ is a linear differential-difference polynomial in $f(z)$, which is a transcendental entire function of finite order, and $a(z)$ is a small function with respect to $f(z)$.

Two papers [11, 15] contribute greatly to this paper. In fact, assume that $f(z)$ is a transcendental entire function of finite order. Lü et al. [15] obtained that one of $f(z)f^{(k)}(z)-p(z)$ and $f(z)f^{(l)}(z)-p(z)$ must have infinitely many zeros, provided that k and l are nonzero distinct constants. Latreuch and Belaïdi [11] showed that one of $f(z)f(z+c_{1})-p(z)$ and $f(z)f(z+c_{2})-p(z)$ must have infinitely many zeros, where $f(z+c_{1})\not\equiv f(z+c_{2})$. We obtain the following result.

Theorem 1.1

Let $f(z)$ be a transcendental entire function of finite order, let $F_{1}(z,f)$ and $F_{2}(z,f)$ be two linear differential-difference polynomials in $f(z)$ with entire coefficients such that $F_{1}(z,f)\not \equiv F_{2}(z,f)$, and let $q(z)$ be a nonzero polynomial. Then at least one of $f(z)F_{1}(z,f)-q(z)$ and $f(z)F_{2}(z,f)-q(z)$ has infinitely many zeros except the case where only one of $F_{1}(z,f)$ or $F_{2}(z,f)$ is a small function with respect to $f(z)$.

Remark 2

Considering the case that all $c_{i}$ and $k_{i}$ are zeros, that is, $F_{1}(z,f)=\lambda_{1}(z)f(z)$ and $F_{2}(z,f)=\lambda_{2}(z)f(z)$, where $\lambda_{1}(z)\not\equiv\lambda_{2}(z)$ are small functions with respect to $f(z)$. In this case, it is easy to get that one of $f(z)F_{1}(z,f)-p(z)$ and $f(z)F_{2}(z,f)-p(z)$ has infinitely many zeros by the second main theorem for three small functions [8, Theorem 2.5]. In addition, if $F_{1}(z,f)=f^{(k)}(z)$ and $F_{2}(z,f)=f^{(l)}(z)$ in Theorem 1.1, then it is the result given in [15]. If $F_{1}(z,f)=f(z+c_{1})$ and $F_{2}(z,f)=f(z+c_{2})$ in Theorem 1.1, it is the result considered in [11]. If $F_{1}(z,f)=f^{(k)}(z+c_{1})$ and $F_{2}(z,f)=f^{(k)}(z+c_{2})$, Theorem 1.1 partially answers Question 1.

Remark 3

The condition that $q(z)\not\equiv0$ and $F_{1}(z,f)\not\equiv F_{2}(z,f)$ cannot be removed, which can be seen by taking $f(z)=e^{z}$ and $c_{1}=2\pi i$, $c_{2}=4\pi i$. In this case, we have that $f(z)f^{(k)}(z+c_{1})=f(z)f^{(k)}(z+c_{2})=e^{2z}$ has no zeros.

Remark 4

The exceptional case in Theorem 1.1 may happen. For example, consider $F_{1}(z,f)\equiv1$ and $F_{2}(z,f)=f(z+i\pi)$ with $f(z)=e^{z}+1$. Then $f(z)F_{1}(z,f)-1=e^{z}$ and $f(z)F_{2}(z,f)-1=-e^{2z}$ have no zeros.

The following corollary follows directly from Theorem 1.1.

Corollary 1.2

Let $\alpha, \beta, p_{1}, p_{2}$ and $q\not\equiv0$ be nonconstant polynomials. Then the system of equations

$$\begin{aligned} \textstyle\begin{cases} f(z)F_{1}(z,f)-q(z)=p_{1}(z)e^{\alpha(z)}, \\ f(z)F_{2}(z,f)-q(z)=p_{2}(z)e^{\beta(z)}, \end{cases}\displaystyle \end{aligned}$$

has no transcendental entire functions of finite order for every $F_{1}(z,f)\neq F_{2}(z,f)$ such that $F_{1}(z,f)$ and $F_{2}(z,f)$ are not small functions with respect to $f(z)$.

2 Some lemmas

The following lemma on the logarithmic derivative of meromorphic functions plays a crucial role in the paper.

Lemma 2.1

([8, 18])

Let f be a finite-order meromorphic function, and let $k\in\mathbb {N}$. Then

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

The difference analogue of the logarithmic derivative lemma, which is also very important in the proof of Theorem 1.1, was independently found by Halburd and Korhonen [6] and Chiang and Feng [4]. Let us state the result as follows.

Lemma 2.2

Let f be a transcendental meromorphic function of finite order, and let $c\in\mathbb{C}$. Then

$$m \biggl(r,\frac{f(z+c)}{f(z)} \biggr)=O \biggl(\frac{\log r}{r}{T(r,f)} \biggr)=S(r,f) $$

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

The following result is trivial by Lemma 2.1 and Lemma 2.2.

Lemma 2.3

Let f be a transcendental meromorphic function of finite order, and let $P(z,f)$ be a linear differential-difference polynomial. Then

$$m \biggl(r,\frac{P(z,f)}{f(z)} \biggr)=S(r,f) $$

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

Lemma 2.4

([5])

Let $f(z)$ be a transcendental meromorphic function of finite order $\rho(f)=\rho$, and let $\varepsilon>0$ be a given constant. Then there exists a set $E_{0}\subset(0,+\infty)$ of finite logarithmic measure such that, for all z satisfying $|z|\notin E_{0}\cup[0,1]$ and all $k,j$ such that $0\leq j\leq k$, we have

$$ \biggl\vert \frac{f^{(k)}(z)}{f^{(j)}(z)} \biggr\vert \leq \vert z \vert ^{(k-j)(\rho -1+\varepsilon)}. $$

We also need the following lemma to estimate the counting function and the characteristic function for transcendental meromorphic functions of finite order.

Lemma 2.5

([4, Theorems 2.1, 2.2])

Let $f(z)$ be a transcendental meromorphic function with finite order $\rho(f)=\rho$, and let η be a fixed nonzero complex number. Then, for each $\varepsilon> 0$, we have

$$\begin{aligned} &T\bigl(r,f(z+\eta)\bigr)=T(r,f)+O\bigl(r^{\rho-1+\varepsilon}\bigr)+O(\log r)=T(r,f)+S(r,f), \\ &N\bigl(r,f(z+\eta)\bigr)=N(r,f)+O\bigl(r^{\rho-1+\varepsilon}\bigr)+O(\log r)=N(r,f)+S(r,f). \end{aligned}$$

Lemma 2.6

([18, Theorem 1.22])

Let $f(z)$ be a transcendental meromorphic function. Then

$$ T\bigl(r,f^{(n)}\bigr)\leq T(r,f)+n\overline{N}(r,f)+S(r,f). $$

Remark 5

Let $f(z)$ be a meromorphic function of finite order. Combining Lemma 2.5 and Lemma 2.6, for a linear differential-difference polynomial

$$ P(z,f)=\sum_{i=1}^{n} \lambda_{i}(z)f^{(k_{i})}(z+c_{i}), $$

we have

$$T\bigl(r,P(z,f)\bigr)\leq \Biggl(n+\sum_{i=1}^{n}k_{i} \Biggr)T(r,f)+S(r,f). $$

If $f(z)$ is a transcendental entire function of finite order, this inequality reduces to

$$T\bigl(r,P(z,f)\bigr)\leq T(r,f)+S(r,f). $$

Lemma 2.7

([9, Lemma 2.4.2])

Let $f(z)$ be a transcendental meromorphic solution of

$$f^{n}A(z,f)=B(z,f), $$

where $A(z,f)$ and $B(z,f)$ are differential polynomials in f and its derivatives with small meromorphic coefficients $a_{\lambda}$ in the sense of $m(r,a_{\lambda})=S(r,f)$ for all $\lambda\in I$ (I is a set of distinct complex numbers). If the total degree of $B(z,f)$ as a polynomial in f and its derivatives is less than or equal to n, then $m(r,A(z,f))=S(r,f)$.

Lemma 2.8

Let $f(z)$ be a transcendental meromorphic function of finite order, let $F(z)$ be a linear differential-difference polynomial of $f(z)$, and let and $a(z)$ and $b(z)$ be nonzero small functions with respect to $f(z)$. The equation

$$ a(z)f(z)F(z)-f'(z)F(z)-f(z)F'(z)=b(z) $$

(2.1)

gives

$$ T(r,f)=N \biggl(r,\frac{1}{f} \biggr)+S(r,f)=N_{1} \biggl(r, \frac {1}{f} \biggr)+S(r,f), $$

where $N_{1}$ denotes the counting function of the simple zeros of f.

Proof

Dividing both sides of (2.1) by $f^{2}$, since $a(z)$ and $b(z)$ are small functions of f, we get that

$$ \begin{aligned} 2m \biggl(r,\frac{1}{f} \biggr) & \leq m \biggl(r,\frac{F}{f} \biggr)+m \biggl(r,\frac{f'}{f} \frac{F}{f} \biggr) +m \biggl(r,\frac{F'}{f} \biggr)+S(r,f)=S(r,f) \end{aligned} $$

(2.2)

by Lemmas 2.1–2.3 and 2.5. By the Nevanlinna first main theorem we have

$$ T(r,f)=N \biggl(r,\frac{1}{f} \biggr)+S(r,f). $$

(2.3)

From (2.1) it is easy to see that

$$ N_{2} \biggl(r,\frac{1}{f} \biggr)\leq N \biggl(r,\frac{1}{b(z)} \biggr)=S(r,f), $$

(2.4)

where $N_{2}$ denotes the counting function of zeros of f with multiplicities not less than 2. Inequality (2.4) implies that the zeros of f are mainly simple zeros. Thus, by (2.3) and (2.4) we deduce that

$$ T(r,f)=N \biggl(r,\frac{1}{f} \biggr)+S(r,f)=N_{1} \biggl(r, \frac {1}{f} \biggr)+S(r,f). $$

 □

Remark 6

If $a(z)$ and $b(z)$ are rational functions in (2.1), then there are at most finitely many multiple zeros of $f(z)$.

3 Proof of Theorem 1.1

The proof utilizes the ideas of the papers [11, 15], although some details are different. Suppose contrary to the assertion that both $f(z)F_{1}(z,f)-q(z)$ and $f(z)F_{2}(z,f)-q(z)$ have finitely many zeros. Since $f(z)$ is of finite order, by the Hadamard factorization theorem we can write

$$ f(z)F_{1}(z,f)-q(z)=p_{1}(z)e^{\alpha(z)} $$

(3.1)

and

$$ f(z)F_{2}(z,f)-q(z)=p_{2}(z)e^{\beta(z)}, $$

(3.2)

where $\alpha(z)$, $\beta(z)$, $p_{1}(z)$, $p_{2}(z)$ are polynomials. From Remark 5 we know that $T(r,F_{1}(z,f))\leq T(r,f)+S(r,f)$ and $T(r,F_{2}(z,f))\leq T(r,f)+S(r,f)$. First, if $T(r,F_{1}(z,f))=S(r,f)$ and $T(r,F_{2}(z,f))=S(r,f)$, then from the second main theorem for three small functions [8, Theorem 2.5] we know that

$$\begin{aligned} T(r,f) \leq& \overline{N}(r,f)+\overline{N} \biggl(r,\frac{1}{f(z)-\frac {q(z)}{F_{1}(z,f)}} \biggr)+ \overline{N} \biggl(r,\frac{1}{f(z)-\frac {q(z)}{F_{2}(z,f)}} \biggr)+S(r,f) \\ =&\overline{N} \biggl(r,\frac{1}{\frac{p_{1}(z)}{F_{1}(z,f)}} \biggr)+\overline{N} \biggl(r, \frac{1}{\frac{p_{2}(z)}{F_{2}(z,f)}} \biggr)+S(r,f)=S(r,f), \end{aligned}$$

which is impossible.

Second, suppose that $T(r,F_{1}(z,f))\neq S(r,f)$ and $T(r,F_{2}(z,f))\neq S(r,f)$. We affirm that $e^{\alpha}$, $e^{\beta}$, and $e^{\alpha+\beta}$ are not small functions with respect to $f(z)$. Otherwise, if $e^{\alpha}$ is a small function with respect to $f(z)$, we have $f(z)F_{1}(z,f)=t(z)$ from (3.1), where $t(z)=q(z)+p_{1}(z)e^{\alpha(z)}$ is a small function with respect to $f(z)$. Lemma 2.7 and Lemma 2.3 imply that

$$ T(r,F_{1})=m(r,F_{1})=S(r,f), $$

and we get $T(r,f)=S(r,f)$, a contradiction. We can use a similar method to obtain that $e^{\beta}$ is not a small function of $f(z)$. From (3.1) and (3.2) we have

$$f(z)^{2}F_{1}(z,f)F_{2}(z,f)-q(z)f(z) \bigl[F_{1}(z,f)+F_{2}(z,f)\bigr]+q(z)^{2}=p_{1}(z)p_{2}(z)e^{\alpha +\beta}. $$

If $e^{\alpha+\beta}$ is a small function, by Lemma 2.7 and Lemma 2.3 we have $T(r, F_{1}F_{2})=S(r,f)$ and $T(r, \frac {F_{1}F_{2}}{f})=S(r,f)$, and hence $T(r,f)=S(r,f)$, which is impossible.

Since α is a polynomial, α and $\alpha'$ are small functions of f. Differentiating (3.1) and eliminating $e^{\alpha }$, we obtain

$$ a_{1}fF_{1}-f'F_{1}-fF'_{1}=b_{1}, $$

(3.3)

where $a_{1}=\frac{p_{1}'}{p_{1}}+\alpha'$ and $b_{1}=(\frac {p_{1}'}{p_{1}}+\alpha')q-q'$. Now, we will show that $a_{1}\not\equiv 0$. Indeed, if $a_{1}\equiv0$, then by a simple integration there exists a nonzero constant $C_{1}$ such that $C_{1}=p_{1}e^{\alpha}$, which implies a contradiction to the fact that $e^{\alpha}$ is not a small function with respect to $f(z)$. Similarly, we have $b_{1}\not \equiv0$.

By the same arguments as before, (3.2) gives

$$ a_{2}fF_{2}-f'F_{2}-fF'_{2}=b_{2}, $$

(3.4)

where $a_{2}=\frac{p_{2}'}{p_{2}}+\beta'$ and $b_{2}=(\frac {p_{2}'}{p_{2}}+\beta')q-q'$. We also obtain $a_{2}\not\equiv0$ and $b_{2}\not\equiv0$.

In view of Lemma 2.8 and Remark 6, we assume that $z_{0}$ is a simple zero of f such that $z_{0}$ is not a zero or pole of $b_{i}$ $(i=1,2)$. Equations (3.3) and (3.4) imply that

$$ \bigl(f'F_{1}+b_{1}\bigr) (z_{0})=0 $$

(3.5)

and

$$ \bigl(f'F_{2}+b_{2}\bigr) (z_{0})=0. $$

(3.6)

Hence we have

$$ (b_{2}F_{1}-b_{1}F_{2}) (z_{0})=0. $$

(3.7)

We will consider two cases depending on whether $b_{2}F_{1}-b_{1}F_{2}\equiv0$ or not.

Case 1. $b_{2}F_{1}-b_{1}F_{2}\not\equiv0$. Set

$$ h=\frac{b_{2}F_{1}-b_{1}F_{2}}{f}. $$

(3.8)

From Lemma 2.3 we have $m(r,h)=S(r,f)$. On the other hand, from (3.7) and Lemma 2.8 we have

$$\begin{aligned} N(r,h) =&N \biggl(r,\frac{b_{2}F_{1}-b_{1}F_{2}}{f} \biggr) \\ =&N_{1} \biggl(r,\frac{b_{2}F_{1}-b_{1}F_{2}}{f} \biggr)+S(r,f)=S(r,f). \end{aligned}$$

(3.9)

Thus $T(r,h)=S(r,f)$. Rewrite (3.8) in the form

$$ F_{1}=\frac{h}{b_{2}}f+\frac{b_{1}}{b_{2}}F_{2}. $$

(3.10)

By differentiating (3.10) we have

$$ F'_{1}= \biggl(\frac{h}{b_{2}} \biggr)'f+\frac{h}{b_{2}}f'+ \biggl( \frac {b_{1}}{b_{2}} \biggr)'F_{2}+\frac{b_{1}}{b_{2}}F'_{2}. $$

(3.11)

Substituting (3.10) and (3.11) into (3.3), we get

$$ \begin{aligned} \biggl[a_{1} \frac{h}{b_{2}}- \biggl(\frac{h}{b_{2}} \biggr)' \biggr]f^{2}-2\frac{h}{b_{2}}ff' + \biggl[a_{1}\frac{b_{1}}{b_{2}}- \biggl(\frac{b_{1}}{b_{2}} \biggr)' \biggr]fF_{2} &-\frac{b_{1}}{b_{2}}f'F_{2}- \frac{b_{1}}{b_{2}}fF'_{2}=b_{1}. \end{aligned} $$

(3.12)

In addition, equation (3.4) can be written as

$$ a_{2}\frac{b_{1}}{b_{2}}fF_{2}- \frac{b_{1}}{b_{2}}f'F_{2}-\frac {b_{1}}{b_{2}}fF'_{2}=b_{1}. $$

(3.13)

Combining (3.12) and (3.13), we get

$$ \biggl[a_{1}\frac{h}{b_{2}}- \biggl( \frac{h}{b_{2}} \biggr)' \biggr]f-2\frac {h}{b_{2}}f'+ \biggl[a_{1}\frac{b_{1}}{b_{2}}- \biggl(\frac{b_{1}}{b_{2}} \biggr)'-a_{2}\frac{b_{1}}{b_{2}} \biggr]F_{2}=0. $$

(3.14)

Note that $-2\frac{h}{b_{2}}\not\equiv0$. We proceed to prove that

$$ a_{1}\frac{h}{b_{2}}- \biggl(\frac{h}{b_{2}} \biggr)'\not\equiv0\quad \mbox{and} \quad a_{1} \frac{b_{1}}{b_{2}}- \biggl(\frac {b_{1}}{b_{2}} \biggr)'-a_{2} \frac{b_{1}}{b_{2}}\not\equiv0. $$

(3.15)

Otherwise, if $a_{1}\frac{h}{b_{2}}-(\frac{h}{b_{2}})'\equiv0$, then by the definition of $a_{1}$ and a simple integration we have

$$ p_{1}e^{\alpha}=C_{2} \frac{h}{b_{2}}, $$

(3.16)

where $C_{2}$ is a nonzero constant. Since $T (r,\frac {h}{b_{2}} )=S(r,f)$, $p_{1}e^{\alpha}$ is a small function of f, a contradiction. If $a_{1}\frac{b_{1}}{b_{2}}-(\frac {b_{1}}{b_{2}})'-a_{2}\frac{b_{1}}{b_{2}}\equiv0$, then we have $a_{1}-a_{2}\equiv\frac{b_{1}'}{b_{1}}-\frac{b_{2}'}{b_{2}}$. A simple integration yields that

$$ \frac{p_{1}}{p_{2}}e^{\alpha-\beta}=C_{3}\frac{b_{1}}{b_{2}}:=\gamma, $$

where $C_{3}$ is a nonzero constant, and γ is a small function of f. From (3.1) and (3.2) we have

$$ f(F_{1}-\gamma F_{2})=q(1-\gamma). $$

(3.17)

If $\gamma\not\equiv1$, then by Lemma 2.7 we get

$$ m(r,F_{1}-\gamma F_{2})+S(r,f)=T(r,F_{1}-\gamma F_{2})=S(r,f). $$

From this and from (3.17) we have

$$ T(r,f)=T \biggl(r,\frac{q(1-\gamma)}{F_{1}-\gamma F_{2}} \biggr)=S(r,f), $$

a contradiction. If $\gamma\equiv1$, then we have $F_{1}\equiv F_{2}$, which contradicts the hypothesis of Theorem 1.1. Thus $a_{1}\frac {b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'-a_{2}\frac{b_{1}}{b_{2}}\not\equiv0$.

From these discussions we can rewrite (3.14) as

$$ F_{2}=mf+nf', $$

(3.18)

where

$$m=\frac{(\frac{h}{b_{2}})'-a_{1}\frac{h}{b_{2}}}{(a_{1}-a_{2})\frac {b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'},\qquad n=\frac{\frac {2h}{b_{2}}}{(a_{1}-a_{2})\frac{b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'}. $$

Furthermore, (3.10) and (3.18) give

$$ F_{1}=sf+tf', $$

(3.19)

where

$$ s=\frac{b_{1}m+h}{b_{2}},\qquad t=\frac{b_{1}}{b_{2}}n. $$

(3.20)

Differentiating (3.18), we have

$$ F'_{2}=m'f+ \bigl(m+n'\bigr)f'+nf''. $$

(3.21)

Substituting (3.18) and (3.21) into (3.4), we obtain

$$ \bigl(a_{2}m-m'\bigr)f^{2}+ \bigl(a_{2}n-2m-n'\bigr)ff'-n \bigl(f'\bigr)^{2}-nff''=b_{2}. $$

(3.22)

Differentiating (3.22), we have

$$ \begin{aligned}[b] & \bigl(a_{2}m-m' \bigr)'f^{2}+ \bigl[2\bigl(a_{2}m-m' \bigr)+\bigl(a_{2}n-2m-n'\bigr) \bigr]ff' \\ &\quad {}+\bigl(a_{2}n-2m-2n'\bigr) \bigl[\bigl(f' \bigr)^{2}+ff'' \bigr]-n\bigl(3f'f''+ff''' \bigr)=b_{2}'. \end{aligned} $$

(3.23)

Since we have supposed that $f(z_{0})=0$, $f'(z_{0})\neq0$, and $b_{2}(z_{0})\neq0$, from (3.22) and (3.23), respectively, we obtain

$$ \bigl(n\bigl(f'\bigr)^{2}+b_{2} \bigr) (z_{0})=0 $$

and

$$ \bigl[\bigl(a_{2}n-2m-2n'\bigr) \bigl(f' \bigr)^{2}-3nf'f''-b_{2}' \bigr](z_{0})=0, $$

which leads to

$$ \bigl(\bigl[b_{2}\bigl(a_{2}n-2m-2n' \bigr)+b_{2}'n\bigr]f'-3b_{2}nf'' \bigr) (z_{0})=0. $$

Let

$$ H=\frac{[b_{2}(a_{2}n-2m-2n')+b_{2}'n]f'-3b_{2}nf''}{f}. $$

It is obvious that H is a small function of f and

$$ f''=\frac{b_{2}(a_{2}n-2m-2n')+b_{2}'n}{3b_{2}n}f'- \frac{H}{3b_{2}n}f. $$

(3.24)

Substituting (3.24) into (3.22), we have

$$ g_{1}f^{2}+g_{2}ff'+g_{3} \bigl(f'\bigr)^{2}=b_{2}, $$

(3.25)

where

$$\begin{aligned} &g_{1}=a_{2}m-m'+\frac{H}{3b_{2}}, \\ &g_{2}=\frac{1}{3} \biggl(2a_{2}-\frac{b_{2}'}{b_{2}} \biggr)n-\frac {4}{3}m-\frac{1}{3}n',\qquad g_{3}=-n. \end{aligned}$$

It is easy to see that $g_{3}\not\equiv0$. We proceed to prove $g_{2}\not\equiv0$. Otherwise, we get

$$ \frac{g_{2}}{g_{3}}=\frac{2}{3}\frac{h'}{h}+ \frac{1}{3}\frac {n'}{n}-\frac{1}{3}\frac{b_{2}'}{b_{2}}- \frac{2}{3}(a_{1}+a_{2})=0, $$

(3.26)

since, by the definition of $a_{1}$ and $a_{2}$,

$$ \alpha'+\beta'+\frac{p_{1}'}{p_{1}}+\frac{p_{2}'}{p_{2}}= \frac {h'}{h}+\frac{1}{2} \biggl(\frac{n'}{n}-\frac{b_{2}'}{b_{2}} \biggr). $$

By integration we have

$$ p_{1}p_{2}e^{\alpha+\beta}=C_{3}h\biggl( \frac{n}{b_{2}}\biggr)^{\frac{1}{2}}, $$

where $C_{3}$ is a nonzero constant. Since $T (r,C_{3}h(\frac {n}{b_{2}})^{\frac{1}{2}} )=S(r,f)$, we can deduce that $e^{\alpha +\beta}$ is a small function of f, a contradiction. Thus $g_{2}\not \equiv0$. Differentiating (3.25), we have

$$ g_{1}'f^{2}+ \bigl(2g_{1}+g_{2}'\bigr)ff'+ \bigl(g_{2}+g_{3}'\bigr) \bigl(f' \bigr)^{2}+g_{2}ff''+2g_{3}f'f''=b_{2}'. $$

(3.27)

By the same method used to deal with (3.22) and (3.23) we have

$$ f''=\frac{b_{2}'g_{3}-b_{2}(g_{2}+g_{3}')}{2b_{2}g_{3}}f'+ \frac {R}{2b_{2}g_{3}}f, $$

(3.28)

where

$$ R=\frac{[b_{2}(g_{2}+g_{3}')-b_{2}'g_{3}]f'+2b_{2}g_{3}f''}{f} \quad \mbox{and}\quad T(r,R)=S(r,f). $$

Substituting (3.28) into (3.27), we get

$$ \begin{aligned}[b] &\biggl(g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}} \biggr)f^{2}+ \biggl(2g_{1}+g_{2}'+\frac{1}{2} \frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2}\bigl(g_{2}+g_{3}' \bigr)\frac{g_{2}}{g_{3}} +\frac{R}{b_{2}} \biggr)f'f \\ &\quad {}+\frac{b_{2}'g_{3}}{b_{2}}\bigl(f'\bigr)^{2}=b_{2}'. \end{aligned} $$

(3.29)

Combining (3.29) and (3.25), we have

$$ \biggl(g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}}- \frac{b_{2}}{b_{2}}'g_{1} \biggr)f+ \biggl(2g_{1}+g_{2}'- \frac{1}{2}\frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2} \bigl(g_{2}+g_{3}'\bigr)\frac{g_{2}}{g_{3}}+ \frac{R}{b_{2}} \biggr)f'=0. $$

(3.30)

Let

$$ q_{1}=g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}}- \frac{b_{2}}{b_{2}}'g_{1} $$

and

$$ q_{2}=2g_{1}+g_{2}'- \frac{1}{2}\frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2} \bigl(g_{2}+g_{3}'\bigr)\frac{g_{2}}{g_{3}}+ \frac{R}{b_{2}}. $$

Now we claim that $q_{1}$ and $q_{2}$ vanish identically. If $q_{2}\not \equiv0$, then by the previous analysis we can get that $q_{2}$ is a small function of f. From (3.30) we have $q_{2}(z_{0})=0$. Thus

$$ T(r,f)=N_{1} \biggl(r,\frac{1}{f} \biggr)+S(r,f)\leq N \biggl(r,\frac {1}{q_{2}} \biggr)+S(r,f)=S(r,f), $$

a contradiction. Again by (3.30) we get $q_{1}\equiv0$. Eliminating R from $q_{1}\equiv0$ and $q_{2}\equiv0$, we have

$$ \begin{aligned}[b] &g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)\frac {b_{2}'}{b_{2}}+g_{2}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)+2g_{2}g_{2}'g_{3}-g_{2}^{2}g_{3}'-4g_{1}'g_{3}^{2} \\ &\quad =g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)\frac {b_{2}'}{b_{2}}+g_{2}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)-g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)'+g_{3}'\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)=0. \end{aligned} $$

(3.31)

We continue by discussing two subcases depending on whether $4g_{1}g_{3}-g_{2}^{2}$ vanishes identically or not.

Subcase 1. If $4g_{1}g_{3}-g_{2}^{2}\not\equiv0$, then from (3.31) we have

$$ \frac{g_{2}}{g_{3}}=\frac {(4g_{1}g_{3}-g_{2}^{2})'}{4g_{1}g_{3}-g_{2}^{2}}-\frac {b_{2}'}{b_{2}}-\frac{g_{3}'}{g_{3}}. $$

Combining this equation with (3.26), we get that

$$ 2(a_{1}+a_{2})=2\frac{h'}{h}+\frac{n'}{n}+2 \frac{b_{2}'}{b_{2}}-3\frac {(4g_{1}g_{3}-g_{2}^{2})'}{4g_{1}g_{3}-g_{2}^{2}}. $$

Similarly as in the proof of $g_{2}\not\equiv0$, we can deduce that $e^{\alpha+\beta}$ is a small function of f, a contradiction.

Subcase 2. If $4g_{1}g_{3}-g_{2}^{2}\equiv0$, then, on the one hand, by using $q_{1}\equiv0$ we can rewrite equation (3.28) as

$$ f''= \biggl(\frac{b_{2}'g_{1}}{b_{2}g_{2}}-\frac{g_{1}'}{g_{2}} \biggr)f +\frac{1}{2} \biggl(\frac{b_{2}'}{b_{2}}-\frac{g_{2}}{g_{3}}- \frac {g_{3}'}{g_{3}} \biggr)f', $$

and then from this equation and from (3.24) we have

$$ \frac{g_{1}'}{g_{2}}-\frac{b_{2}'}{b_{2}}\frac{g_{1}}{g_{2}}=\frac{H}{3b_{2}n}. $$

On the other hand, by the definitions of $g_{1}$ and $g_{3}$ we have

$$ \frac{ g_{1}}{g_{3}}=\frac{m'}{n}-a_{2}\frac{m}{n}- \frac{H}{3b_{2}n}. $$

Therefore we get

$$ \frac{g_{1}'}{g_{2}}-\frac{b_{2}'}{b_{2}}\frac{g_{1}}{g_{2}}+ \frac{ g_{1}}{g_{3}}-\frac{m'}{n}+a_{2}\frac{m}{n}=0. $$

(3.32)

For brevity, we denote

$$ D=(a_{1}-a_{2})\frac{b_{1}}{b_{2}}-\biggl( \frac{b_{1}}{b_{2}}\biggr)'. $$

By calculation we have

$$ \frac{ g_{1}}{g_{3}}=\frac{1}{4} \biggl(\frac{g_{2}}{g_{3}} \biggr)^{2},\qquad \frac{g_{1}}{g_{2}}=\frac{1}{4} \frac{g_{2}}{g_{3}},\qquad \frac{ g_{1}'}{g_{2}}=\frac{1}{2} \biggl( \frac{g_{2}}{g_{3}} \biggr)'+\frac{1}{4}\frac{g_{2}}{g_{3}} \frac{g_{3}'}{g_{3}}, $$

and

$$ \frac{m'}{n}= \biggl(\frac{m}{n} \biggr)'+ \frac{m}{n}\frac{n'}{n},\qquad \frac{m}{n}=\frac{1}{2} \biggl(\frac{h'}{h}-\frac {b_{2}'}{b_{2}}-a_{1} \biggr),\qquad \frac{n'}{n}=\frac{h'}{h}-\frac {b_{2}'}{b_{2}}-\frac{D'}{D}. $$

From (3.26) we have

$$ \frac{g_{2}}{g_{3}}=\frac{h'}{h}-\frac{2}{3}\frac{b_{2}'}{b_{2}}- \frac {1}{3}\frac{D'}{D}-\frac{2}{3}(a_{1}+a_{2}). $$

Substituting these identities into (3.32), we have

$$ \begin{aligned} &\frac{1}{4} \biggl(\frac{h'}{h}- \frac{2}{3}\frac{b_{2}'}{b_{2}}-\frac {1}{3}\frac{D'}{D}- \frac{2}{3}(a_{1}+a_{2}) \biggr) \biggl( \frac{8}{3}\frac{b_{2}'}{b_{2}}-2\frac{h'}{h}+\frac{4}{3} \frac {D'}{D}+\frac{2}{3}(a_{1}+a_{2}) \biggr) \\ &\quad {}+\frac{1}{2} \biggl(\biggl(\frac{h'}{h}\biggr)'- \biggl(\frac{b_{2}'}{b_{2}}\biggr)'-a_{1}' \biggr)+ \frac{1}{2} \biggl(\frac{h'}{h}-\frac{b_{2}'}{b_{2}}- \frac {D'}{D}-a_{2} \biggr) \biggl(\frac{h'}{h}- \frac{b_{2}'}{b_{2}}-a_{1} \biggr) \\ &\quad {}-\frac{1}{2} \biggl(\biggl(\frac{h'}{h}\biggr)'- \frac{2}{3}\biggl(\frac {b_{2}'}{b_{2}}\biggr)'- \frac{1}{3} \biggl(\frac{D'}{D} \biggr)'- \frac {2}{3}(a_{1}+a_{2})' \biggr)=0, \end{aligned} $$

which leads to

$$\begin{aligned} &\frac{7}{18}\frac{b_{2}'}{b_{2}}\frac{D'}{D}+\frac{1}{6} \biggl(\frac {D'}{D} \biggr)' -\frac{5}{18}(a_{1}+a_{2}) \frac{D'}{D}-\frac{1}{6}(a_{1}+a_{2}) \frac {h'}{h}+\frac{1}{18} \biggl(\frac{b_{2}'}{b_{2}} \biggr)^{2} \\ &\qquad {}-\frac{1}{18}(a_{1}+a_{2})\frac{b_{2}'}{b_{2}}+ \frac{1}{2}a_{1}\frac{D'}{D} +\frac{1}{2}a_{1}a_{2}- \frac{1}{6}\frac{b_{2}'}{b_{2}}-\frac {1}{2}a_{1}'+ \frac{1}{3}(a_{1}+a_{2})' \\ &\quad =\frac{1}{9}(a_{1}+a_{2})^{2}. \end{aligned}$$

Since

$$\lim_{z\rightarrow\infty}\frac{R'(z)}{R(z)}=0 $$

if $R(z)$ is a nonzero rational function, dividing both sides of the above equation by $\frac{(a_{1}+a_{2})^{2}}{2}$ and taking the limit, we have

$$ \lim_{z\rightarrow\infty}\frac{a_{1}a_{2}}{(a_{1}+a_{2})^{2}}=\frac {2}{9}+\lim _{z\rightarrow\infty}\frac{1}{3}\frac{h'}{h}\frac{1}{(a_{1}+a_{2})}. $$

It follows by Lemma 2.4 that

$$ \lim_{z\rightarrow\infty}\frac{a_{1}a_{2}}{(a_{1}+a_{2})^{2}}=\lim_{z\rightarrow\infty} \frac{\alpha'\beta'}{(\alpha'+\beta')^{2}}=\frac{2}{9}. $$

By setting $\alpha(z)=a_{m}z^{m}+\cdots+a_{0}$ and $\beta (z)=b_{m}z^{m}+\cdots+b_{0}$ we have

$$ \lim_{z\rightarrow\infty}\frac{\alpha'\beta'}{(\alpha'+\beta ')^{2}}=\frac{a_{m}b_{m}}{(a_{m}+b_{m})^{2}}= \frac{2}{9}, $$

which implies that $\frac{a_{m}}{b_{m}}=2$ or $\frac{a_{m}}{b_{m}}=\frac {1}{2}$. We first consider the case $\frac{a_{m}}{b_{m}}=2$. From (3.1), (3.2), (3.18), and (3.19) we have

$$ sf^{2}+tff'-q=A(z)e^{2b_{m}z^{m}} $$

and

$$ mf^{2}+nff'-q=B(z)e^{b_{m}z^{m}}, $$

where

$$ A(z)=p_{1}e^{a_{m-1}z^{m-1}+\cdots+a_{0}} \quad \mbox{and} \quad B(z)=p_{2}e^{b_{m-1}z^{m-1}+\cdots+b_{0}}. $$

Then

$$ mf^{2}+nff'=q+B \biggl(\frac{sf^{2}+tff'-q}{A} \biggr)^{\frac{1}{2}}. $$

Combining (2.2) with the expressions of A and B, we obtain

$$ \begin{aligned}T(r,F_{2})&=T\bigl(r, mf+nf' \bigr) \\ &=m \biggl(r,\frac{q}{f}+B \biggl(\frac{sf^{2}+tff'-q}{Af^{2}} \biggr)^{\frac{1}{2}} \biggr)+S(r,f) \\ &\leq m\biggl(r,\frac{q}{f}\biggr)+m(r,B)+\frac{1}{2}m \biggl(r, \frac {sf^{2}+tff'-q}{Af^{2}} \biggr)+S(r,f) \\ &\leq m\biggl(r,\frac{s}{A}\biggr)+m\biggl(r,\frac{t}{A}\biggr)+m \biggl(r,\frac{f'}{f}\biggr)+m\biggl(r,\frac {q}{Af^{2}}\biggr)+S(r,f) \\ &=S(r,f). \end{aligned} $$

Hence we get $T(r,F_{2})=S(r,f)$, a contradiction. For the case $\frac{a_{m}}{b_{m}}=\frac{1}{2}$, by the same argument we can also deduce $T(r,F_{1})=S(r,f)$, a contradiction.

Case 2. $b_{2}F_{1}-b_{1}F_{2}\equiv0$.

Using the same arguments as in the proof of (3.14), we have

$$ a_{1}\frac{b_{1}}{b_{2}}-\biggl(\frac{b_{1}}{b_{2}} \biggr)'-a_{2}\frac {b_{1}}{b_{2}}\equiv0, $$

which leads to

$$ \frac{p_{1}}{p_{2}}e^{\alpha-\beta}=C_{3}\frac{b_{1}}{b_{2}}, $$

where $C_{3}$ is a nonzero constant. Then by (3.1) and (3.2), if $\frac{C_{3}b_{1}}{b_{2}}\equiv1$, then $F_{1}\equiv F_{2}$, which contradicts the hypothesis of Theorem 1.1. If $\frac{C_{3}b_{1}}{b_{2}}\not\equiv1$, then from

$$fF_{1}-\frac{C_{3}b_{1}}{b_{2}}fF_{2}=q-q\frac{C_{3}b_{1}}{b_{2}} $$

we have $T(r,f)=S(r,f)$, which is impossible.

Thus we obtain that at least one of $f(z)F_{1}(z,f)-q(z)$ and $f(z)F_{2}(z,f)-q(z)$ has infinitely many zeros.