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# On the meromorphic solutions of certain class of nonlinear differential equations

Journal of Inequalities and Applications20152015:149

https://doi.org/10.1186/s13660-015-0672-x

• Received: 22 September 2014
• Accepted: 21 April 2015
• Published:

## Abstract

Let α be an entire function, $$a_{n-1},\ldots,a_{1},a_{0}$$, R be small functions of f, and let $$n\geq2$$ be an integer. Then, for any positive integer k, the differential equation $$f^{n}f^{(k)}+a_{n-1}f^{n-1}+\cdots+a_{1}f+a_{0}=R\mathrm{e}^{\alpha}$$ has transcendental meromorphic solutions under appropriate conditions on the coefficients. In addition, for $$n=1$$ and $$k=1$$, we have extended some well-known and relevant results obtained by others, by using different arguments.

## Keywords

• entire function
• differential polynomial
• Nevanlinna theory
• Hayman’s alternative
• differential equation

• 34M10
• 30D35

## 1 Introduction and main results

In this paper, a meromorphic function means meromorphic in the whole complex plane. We shall adopt the standard notations in Nevanlinna’s value distribution theory of meromorphic functions (see, e.g., [1, 2]).

Given a meromorphic function f, recall that $$\alpha\not\equiv0,\infty$$ is a small function with respect to f, if $$T (r,\alpha) = S(r, f )$$, where $$S(r,f)$$ denotes any quantity satisfying $$S(r,f)=o\{T(r,f)\}$$ as $$r\to\infty$$, possibly outside a set of r of finite linear measure.

### Theorem A

Let f be a transcendental meromorphic function, n (≥3) be an integer. Then $$F=f^{n}f'$$ assumes all finite values, except possibly zero, infinitely many times.

The above theorem was derived by Hayman  in 1959. Later, he conjectured  that Theorem A remains valid even if $$n=1$$ or $$n=2$$. Mues  proved the result for $$n=2$$ and the case $$n=1$$ was proved by Bergweiler and Eremenko  and independently by Chen and Fang . For entire functions and difference polynomials, similar results have been obtained by others earlier (see, e.g., ).

### Theorem B

()

If f is a transcendental meromorphic function of finite order and a (0) is a polynomial, then $$ff'-a$$ has infinitely many zeros.

Wang  obtained the following result.

### Theorem C

Let f be a transcendental entire function and n, k be positive integers, and let $$c(z)$$ (0) be a small function with respect to f. If $$T(r,f)\neq\tau N_{1)}(r,1/f)+S(r,f)$$, then $$f^{n}(z)f^{(k)}(z)-c(z)$$ has infinitely many zeros, where $$\tau=0$$ if $$n\geq2$$ or $$k=1$$; $$\tau=1$$ otherwise.

In this paper, by using methods different from that were used by others (see, e.g., [10, 14] and ), we shall extend and generalize the above results with $$f^{n}f^{(k)}$$ being replaced by a differential polynomial $$P_{n+1}(f)$$. Specifically, our main results can be stated as follows.

### Theorem 1.1

Let α be an entire function, R and $$a_{i}$$ ($$i=0,1, \ldots,n-1$$) be small functions of f with $$a_{0}\not\equiv0$$. If, for $$n\geq2$$, a transcendental meromorphic function f satisfies the differential equation
$$f^{n}f^{(k)}+a_{n-1}f^{n-1}+ \cdots+a_{1}f+a_{0}=R\mathrm{e}^{\alpha},$$
(1.1)
then, for any positive integer k, we have $$f=g\exp(\alpha /(n+1))-(n+1)\frac{a_{0}}{a_{1}}$$ with $$g^{n+1}=[\frac {(n+1)a_{0}}{a_{1}}]^{n+1}\frac{R}{a_{0}}$$, and $$(\frac{a_{0}}{a_{1}})^{(k)}+\frac {n}{n+1} (\frac{1}{n+1}\frac{a_{1}}{a_{0}} )^{n}a_{0}\equiv0$$.

### Remark 1.1

Let $$a_{0}$$ and $$a_{1}$$ be non-zero constants in Theorem 1.1. Then (1.1) has no transcendental meromorphic solutions.

A meromorphic solution f of (1.1) is called admissible, if $$T (r, \alpha_{j}) = S(r, f )$$ holds for all coefficients $$\alpha_{j}$$ ($$j=0,\ldots,n-1$$) and $$T(r,R)=S(r,f)$$.

### Remark 1.2

If $$a_{0}\equiv 0$$ and $$n\geq2$$, $$k \geq 1$$, then the other coefficients $$a_{1},\ldots, a_{n-1}$$ must be identically zero. In this case, (1.1) becomes $$f^{n}f^{(k)}=R\mathrm{e}^{\alpha}$$ and f has the form $$f=u\exp(\alpha/(n+1))$$ as the only possible admissible solution of (1.1), where u is a small function of f.

We have the following corollary by Theorem 1.1.

### Corollary 1.1

Let f be a transcendental meromorphic function with $$N(r,f)=S(r,f)$$, $$n\geq2$$ be an integer. If $$(\frac {a_{0}}{a_{1}})^{(k)}+\frac{n}{n+1} (\frac{1}{n+1}\frac{a_{1}}{a_{0}} )^{n}a_{0}\not\equiv0$$, then $$F=f^{n}f^{(k)}+a_{n-1}f^{n-1}+\cdots+a_{1}f+a_{0}$$ has infinitely many zeros, where $$a_{i}$$ ($$i=0,1, \ldots,n-1$$) are small functions of f such that $$a_{0}\not\equiv0$$.

Note that in Theorem 1.1, it is assumed that $$n\geq2$$ and $$k\geq1$$. However, for $$n=1$$ and $$k=1$$, we can derive the following result.

### Theorem 1.2

Let p, q, and R be non-zero polynomials, α be an entire function. Then the differential equation $$pff'-q=R\mathrm{e}^{\alpha}$$ has no transcendental meromorphic solutions, where p, q, and R are small functions of f with $$pq\not \equiv0$$.

### Remark 1.3

From the proof of Theorem 1.2, we see that the restriction in Theorem 1.2 to p, q, and R may extend to small functions. In fact, it is easy to find that the conclusion is valid provided that p, q, and R are non-vanishing small functions of f. The following corollary arises directly from an immediate consequence of Theorem 1.2.

### Corollary 1.2

Let f be a transcendental meromorphic function with $$N(r,f)=S(r,f)$$, p and q be non-vanishing small functions of f. Then $$F=pff'-q$$ has infinitely many zeros.

## 2 Some lemmas and proofs of theorems

In order to prove our conclusions, we need some lemmas. The following lemma is fundamental to Clunie’s theorem .

### Lemma 2.1

([17, 18])

Let f be a transcendental meromorphic solution of
$$f^{n}P(z, f) = Q(z, f),$$
where $$P(z, f)$$ and $$Q(z, f)$$ are polynomials in f and its derivatives with meromorphic coefficients $$\{a_{\lambda}|\lambda\in I\}$$ such that $$m(r, a_{\lambda }) = S(r, f)$$ for all $$r\in I$$. If the total degree of $$Q(z, f)$$ as a polynomial in f and its derivatives is less than or equal to n, then $$m(r, P(r, f)) = S(r, f)$$.

The following lemma is crucial to the proof of our theorems.

### Lemma 2.2

([18, 19])

Let f be a meromorphic solution of an algebraic equation
$$P\bigl(z,f,f',\ldots,f^{(n)}\bigr)=0,$$
(2.1)
where P is a polynomial in $$f,f', \ldots,f^{(n)}$$ with meromorphic coefficients small with respect to f. If a complex constant c does not satisfy (2.1), then
$$m\biggl(r,\frac{1}{f-c}\biggr)=S(r,f).$$

### Proof of Theorem 1.1

Let f be a transcendental meromorphic function that satisfies (1.1). Then two cases are to be treated, namely case 1: $$N(r,f)\neq S(r,f)$$, and case 2: $$N(r,f)=S(r,f)$$. For case 1, it is impossible as α is an entire function and R, $$a_{1},\ldots,a_{n}$$ are small functions of f.

To prove Theorem 1.1, we now suppose that $$N(r,f)= S(r,f)$$.

Denoting $$\phi:= f^{n}f^{(k)} + a_{n-1}f^{n-1} +\cdots+ a_{1}f$$, and assuming that $$T(r,\phi) = S(r, f)$$, then by Lemma 2.1, we get $$m(r, f^{(k)}) = S(r, f)$$ and then $$T(r, f^{(k)}) = S(r, f)$$, since $$N(r, f) = S(r, f)$$ by the assumption. The contradiction $$T(r, f) = S(r, f)$$ now follows by the theorem in  and combining it with the proof of Proposition E in . Thus, for any transcendental meromorphic function f under the condition: $$N(r,f)=S(r,f)$$,
$$T\bigl(r,f^{n}f^{(k)}+a_{n-1}f^{n-1}+ \cdots+a_{1}f\bigr)\neq S(r,f).$$
(2.2)
From (1.1) and the result of Milloux (see, e.g., , Theorem 3.1), one can easily show that
$$T\bigl(r,\mathrm{e}^{\alpha}\bigr)\leq(n+1)T(r,f)+S(r,f),$$
which leads to $$T(r,\alpha)+T(r,\alpha')=S(r,f)$$.
By taking the logarithmic derivative on both sides of (1.1), we have
$$\frac{nf^{n-1}f'f^{(k)}+f^{n}f^{(k+1)}+a'_{n-1}f^{n-1}+\cdots +a'_{1}f+a_{1}f'+a'_{0}}{ f^{n}f^{(k)}+a_{n-1}f^{n-1}+\cdots+a_{1}f+a_{0}}=\frac{R'}{R}+\alpha'.$$
(2.3)
It follows by (2.3) that
\begin{aligned}& -\biggl(\frac{R'}{R}+\alpha'\biggr)f^{n}f^{(k)}+nf^{n-1}f'f^{(k)}+f^{n}f^{(k+1)}+ \biggl\{ a'_{n-1}-\biggl(\frac{R'}{R}+ \alpha'\biggr)a_{n-1}\biggr\} f^{n-1} \\& \quad {}+(n-1)a_{n-1}f^{n-2}f'+\cdots+\biggl\{ a'_{1}-\biggl(\frac{R'}{R}+\alpha' \biggr)a_{1}\biggr\} f+a_{1}f' =\biggl( \frac{R'}{R}+\alpha'\biggr)a_{0}-a'_{0}. \end{aligned}
(2.4)
If $$(\frac{R'}{R}+\alpha')a_{0}-a'_{0}\equiv0$$, then $$Aa_{0}=Re^{\alpha}$$, where A is a non-zero constant. From (1.1), we get
$$f^{n}f^{(k)}+a_{n-1}f^{n-1}+ \cdots+a_{1}f=(A-1)a_{0}.$$
(2.5)
If $$A=1$$, then from (2.5), we obtain
$$f^{n}f^{(k)}+a_{n-1}f^{n-1}+ \cdots+a_{1}f\equiv0,$$
which contradicts (2.2). However, if $$A\neq1$$, then again from (2.5), we would derive
$$T\bigl(r,f^{n}f^{(k)}+a_{n-1}f^{n-1}+ \cdots+a_{1}f\bigr)= S(r,f),$$
Thus
$$\biggl(\frac{R'}{R}+\alpha'\biggr)a_{0}-a'_{0}:= \varphi\not\equiv0.$$
In this case, from (2.4), we have
$$N_{(2}\biggl(r,\frac{1}{f}\biggr)\leq N\biggl(r, \frac{1}{\varphi}\biggr)+S(r,f)\leq T(r,\varphi )+S(r,f)=S(r,f),$$
where $$N_{(2}(r,\frac{1}{f})$$, as usually, denotes the counting function of zeros of f whose multiplicities are not less than 2, which implies that the zeros of f are mainly simple zeros. Again, from (2.4), the fact that $$\alpha'$$ is a small function of f and Lemma 2.2 (where $$c=0$$ is used), we conclude $$m(r,\frac{1}{f})=S(r,f)$$. This together with Nevanlinna’s first theorem will result in
$$T(r,f)=N\biggl(r,\frac{1}{f}\biggr)+S(r,f)=N_{1)}\biggl(r, \frac{1}{f}\biggr)+S(r,f),$$
(2.6)
where in $$N_{1)}(r,1/f)$$ only the simple zeros of f are to be considered.
Assume that $$a_{1}\equiv0$$. It follows by (2.4) and $$n\geq2$$ that $$N_{1)}(r,1/f)=S(r,f)$$, which contradicts (2.6). Thus $$a_{1}\not\equiv 0$$. Let $$z_{0}$$ be a simple zero of f, and $$z_{0}$$ be not a pole of one of the coefficients $$a_{i}$$, $$(\frac{R'}{R}+\alpha')a_{i}-a'_{i}$$ ($$i=1,2,\ldots,n-1$$). From (2.4), we see that $$z_{0}$$ is a zero of $$a_{1}f'+a'_{0}-(\frac{R'}{R}+\alpha')a_{0}$$. Set
$$h=\frac{a_{1}f'+a'_{0}-(\frac{R'}{R}+\alpha')a_{0}}{f}.$$
(2.7)
Then (2.7) gives $$T(r,h)=S(r,f)$$. We have
$$f'=\frac{1}{a_{1}}\biggl\{ hf-a'_{0}+ \biggl(\frac{R'}{R}+\alpha'\biggr)a_{0}\biggr\} := \mu_{1}f+\nu _{1}.$$
(2.8)
Clearly, it follows from (2.6) and $$T(r,\mu_{1})+T(r,\nu_{1})=S(r,f)$$ that $$\mu_{1}\nu_{1}\not\equiv0$$. By (2.3), we obtain
$$f^{n-1}\psi=P_{n-1}(f),$$
(2.9)
where $$\psi=-(\frac{R'}{R}+\alpha')ff^{(k)}+nf'f^{(k)}+ff^{(k+1)}$$, $$P_{n-1}(f)=(\frac{R'}{R}+\alpha')(a_{n-1}f^{n-1}+\cdots +a_{1}f+a_{0})-(a_{n-1}f^{n-1}+\cdots+a_{1}f+a_{0})'$$. It follows by (2.2) that $$P_{n-1}(f)\not\equiv0$$. Thus $$\psi\not \equiv0$$. Moreover, by applying Lemma 2.1 to (2.9), we get $$m(r,\psi)=S(r,f)$$. It is easy to see by $$N(r,f)=S(r,f)$$ that $$T(r,\psi)=S(r,f)$$.
From (2.8) and induction, we have $$f''=(\mu'_{1}+\mu^{2}_{1})f+\mu_{1}\nu _{1}+\nu '_{1}:=\mu_{2}f+\nu_{2}$$, and
$$f^{(k)}=\mu_{k}f+\nu_{k},$$
(2.10)
where $$\mu_{k}$$, $$\nu_{k}$$ are small functions of f. By the expression of ψ and (2.6), we get $$\nu_{k}\not\equiv0$$. If $$\mu_{k}\equiv0$$, then (2.10) gives $$T(r,f^{(k)})=S(r,f)$$, which is impossible. Therefore, $$\mu _{k}\not\equiv0$$.
By (2.10), (1.1) becomes
$$\mu_{k}f^{n+1}+\nu_{k}f^{n}+a_{n-1}f^{n-1}+ \cdots+a_{1}f+a_{0}=R\mathrm{e}^{\alpha }.$$
(2.11)
By applying the Tumura-Clunie lemma (see, e.g., , Theorem 3.9) to the left-hand side of (2.11), we have $$\mu_{k}[f+\frac{\nu _{k}}{(n+1)\mu_{k}}]^{n+1}=R\mathrm{e}^{\alpha}$$ and $$f=g\mathrm{e}^{\alpha /(n+1)}-\frac{\nu_{k}}{(n+1)\mu_{k}}$$ with $$g^{n+1}=\frac{R}{\mu_{k}}$$.
In view of (2.11), we have
$$\mu_{k}f^{n+1}+\nu_{k}f^{n}+a_{n-1}f^{n-1}+ \cdots+a_{1}f+a_{0}=\mu_{k}\biggl[f+ \frac{\nu _{k}}{(n+1)\mu_{k}}\biggr]^{n+1}.$$
Thus, we have
$$\frac{1}{n+1}\frac{\nu_{k}}{\mu_{k}}=(n+1)\frac{a_{0}}{a_{1}}\quad \mbox{and}\quad \mu _{k}= \biggl(\frac{1}{n+1}\frac{a_{1}}{a_{0}} \biggr)^{n+1}a_{0}.$$
(2.12)
By (2.12), we obtain $$\nu_{k}=(n+1) (\frac{1}{n+1}\frac{a_{1}}{a_{0}} )^{n}a_{0}$$ and $$g^{n+1}=[\frac{(n+1)a_{0}}{a_{1}}]^{n+1}\frac{R}{a_{0}}$$.
Set $$(n+1)\gamma=\alpha$$. It follows by (2.10) and $$f=g\mathrm {e}^{\gamma}-(n+1)\frac{a_{0}}{a_{1}}$$ that
$$f^{(k)}= \biggl(\frac{1}{n+1}\frac{a_{1}}{a_{0}} \biggr)^{n+1}a_{0} \biggl[g\mathrm {e}^{\gamma}-(n+1)\frac{a_{0}}{a_{1}}\biggr]+(n+1) \biggl( \frac{1}{n+1}\frac {a_{1}}{a_{0}} \biggr)^{n}a_{0}.$$
(2.13)
In addition, by $$f=g\mathrm{e}^{\gamma}-(n+1)\frac{a_{0}}{a_{1}}$$ we get
$$f^{(k)}=Q\bigl(g,g',\ldots,g^{(k)}\bigr) \mathrm{e}^{\gamma}-(n+1) \biggl(\frac {a_{0}}{a_{1}}\biggr)^{(k)},$$
(2.14)
where $$Q(g,g',\ldots,g^{(k)})$$ is a differential polynomial of g.
Thus, (2.13) and (2.14) imply
$$Q\bigl(g,g',\ldots,g^{(k)}\bigr)= \biggl(\frac{1}{n+1} \frac{a_{1}}{a_{0}} \biggr)^{n+1}a_{0}g$$
and
$$(n+1) \biggl(\frac{a_{0}}{a_{1}}\biggr)^{(k)}= \biggl(\frac{1}{n+1} \frac{a_{1}}{a_{0}} \biggr)^{n+1}a_{0}(n+1)\frac{a_{0}}{a_{1}}-(n+1) \biggl(\frac{1}{n+1}\frac {a_{1}}{a_{0}} \biggr)^{n}a_{0}.$$
(2.15)
It follows by (2.15) that
$$\biggl(\frac{a_{0}}{a_{1}}\biggr)^{(k)}+\frac{n}{n+1} \biggl( \frac{1}{n+1}\frac {a_{1}}{a_{0}} \biggr)^{n}a_{0}=0.$$

This completes the proof of Theorem 1.1. □

### Proof of Remark 1.2

Let f be a transcendental meromorphic solution of (1.1). Since $$a_{0}\equiv0$$, we have $$N(r,1/f)\leq N(r,1/R)+S(r,f)=S(r,f)$$. Obviously, $$N(r,f)=S(r,f)$$. In this case, there exist a meromorphic function u and an entire function v such that $$f=u\mathrm{e}^{v}$$, and $$N(r,1/u)+N(r,u)=S(r,f)$$. Clearly, from the expressions of f and the Borel lemma (see, e.g., , Theorem 1.52), all the $$a_{j}$$ ($$j=1,2,\ldots,n-1$$) must be identically zero. Thus, Remark 1.2 follows. □

### Proof of Theorem 1.2

Now we proceed to prove the theorem by contradiction. Let f be a transcendental meromorphic function that satisfies $$pff'-q=R\mathrm {e}^{\alpha}$$. Then two cases are to be retreated, namely $$N(r,f)\neq S(r,f)$$ and $$N(r,f)=S(r,f)$$. For $$N(r,f)\neq S(r,f)$$, this is impossible as α is an entire function and R, p, q are non-zero polynomials.

To prove Theorem 1.2, we now suppose that $$N(r,f)= S(r,f)$$. We differentiate $$pff'-q=R\mathrm{e}^{\alpha}$$ and eliminate $$\mathrm{e}^{\alpha}$$,
$$t_{1}ff'+p\bigl(f'\bigr)^{2}+pff''=t_{2},$$
(2.16)
where $$t_{1}=p'-(\frac{R'}{R}+\alpha')p$$, $$t_{2}=q'-(\frac{R'}{R}+\alpha')q$$.
If $$t_{2}\equiv0$$, then, by integrating the definition of $$t_{2}$$, α must be a constant, hence $$ff'$$ is rational, and then, by Lemma 2.1, $$m(r, f') = S(r, f)$$. Hence $$T(r, f') = S(r, f)$$. This is a contradiction by Proposition E in . Thus, $$t_{2}\not\equiv0$$, and then by (2.16), we get (2.6). By differentiating both sides of (2.16), we have
$$t'_{1}ff'+\bigl(t_{1}+p' \bigr) \bigl(f'\bigr)^{2}+\bigl(t_{1}+p' \bigr)ff''+3pf'f''+pff'''=t'_{2}.$$
(2.17)
Letting $$z_{0}$$ be a simple zero of f, (2.16) and (2.17) imply
$$\bigl(p\bigl(f'\bigr)^{2}-t_{2}\bigr) (z_{0})=0$$
(2.18)
and
$$\bigl\{ \bigl(t_{1}+p'\bigr) \bigl(f' \bigr)^{2}+3pf'f''-t'_{2} \bigr\} (z_{0})=0.$$
(2.19)
Let
$$g=\frac{3pt_{2}f''+[t_{2}(t_{1}+p')-t'_{2}p]f'}{f}.$$
(2.20)
From (2.6), (2.18), and (2.19), we get
$$T(r,g)=S(r,f).$$
By (2.20), we obtain
$$f''=\alpha_{1}f+\beta_{1}f',$$
(2.21)
where
$$\alpha_{1}=\frac{g}{3pt_{2}}, \qquad \beta_{1}= \frac{t'_{2}p-t_{2}(t_{1}+p')}{3pt_{2}}$$
and
$$T(r,\alpha_{1})=S(r,f), \qquad T(r,\beta_{1})=S(r,f).$$
Substituting (2.21) into (2.16) yields
$$(t_{1}+p\beta_{1})ff'+p\bigl(f' \bigr)^{2}+\alpha_{1}pf^{2}=t_{2}.$$
(2.22)
On the other hand, from (2.21), we have
$$f'''=\alpha_{2}f+ \beta_{2}f',$$
(2.23)
where $$\alpha_{2}=\alpha'_{1}+\alpha_{1}\beta_{1}$$, $$\beta_{2}=\alpha _{1}+\beta'_{1}+\beta^{2}_{1}$$, and
$$T(r,\alpha_{2})=S(r,f),\qquad T(r,\beta_{2})=S(r,f).$$
Substituting (2.23) into (2.17), we have
$$\bigl[t'_{1}+\beta_{1}\bigl(t_{1}+p' \bigr)+3p\alpha_{1}+p\beta_{2}\bigr]ff' + \bigl(t_{1}+p'+3p\beta_{1}\bigr) \bigl(f'\bigr)^{2}+\bigl[\alpha_{1} \bigl(t_{1}+p'\bigr)+\alpha _{2}p \bigr]f^{2}=t'_{2}.$$
(2.24)
It follows by (2.22) and (2.24) that
\begin{aligned}& \bigl\{ p\bigl[t'_{1}+\beta_{1} \bigl(t_{1}+p'\bigr)+3p\alpha_{1}+p \beta_{2}\bigr]-\bigl(t_{1}+p' +3p \beta_{1}\bigr) (t_{1}+p\beta_{1})\bigr\} ff' \\& \quad {}+\bigl\{ p\bigl[\alpha_{1}\bigl(t_{1}+p' \bigr)+\alpha_{2}p\bigr]-\alpha_{1}p\bigl(t_{1}+p'+3p \beta _{1}\bigr)\bigr\} f^{2} =t'_{2}p-t_{2} \bigl(t_{1}+p'+3p\beta_{1}\bigr). \end{aligned}
(2.25)
From the definition of $$\beta_{1}$$, we now claim $$t'_{2}p-t_{2}(t_{1}+p'+3p\beta_{1})\equiv0$$. To show this, we assume the contrary, that is, $$t'_{2}p-t_{2}(t_{1}+p'+3p\beta_{1})\not\equiv0$$. Then from the fact that $$t'_{2}p-t_{2}(t_{1}+p'+3p\beta_{1})$$ is a small function of f and (2.25), we get
\begin{aligned} N_{1)}\biggl(r,\frac{1}{f}\biggr)&\leq N\biggl(r, \frac{1}{t'_{2}p-t_{2}(t_{1}+p'+3p\beta_{1})}\biggr) \\ &\leq T\bigl(r,t'_{2}p-t_{2} \bigl(t_{1}+p'+3p\beta_{1}\bigr) \bigr)+S(r,f)=S(r,f), \end{aligned}
and from this and (2.6) we deduce $$T(r,f)=S(r,f)$$, a contradiction. Thus, we have
$$t'_{2}p-t_{2}\bigl(t_{1}+p'+3p \beta_{1}\bigr)\equiv0.$$
(2.26)
Now, (2.25) and (2.26) lead to
$$p\bigl[\alpha_{1}\bigl(t_{1}+p'\bigr)+ \alpha_{2}p\bigr]-\alpha_{1}p\bigl(t_{1}+p'+3p \beta _{1}\bigr)\equiv0.$$
(2.27)
From the definition of $$\alpha_{2}$$ and (2.27), we deduce
$$\alpha'_{1}\equiv2\alpha_{1} \beta_{1}.$$
(2.28)
It follows from (2.28) and the definitions of $$t_{1}$$, $$\beta_{1}$$ that
$$\alpha^{3}_{1} p^{4}\equiv t^{2}_{2} \mathrm{e}^{2\alpha}.$$
In the beginning of the proof it was already shown that $$t_{2}\not\equiv0$$. Hence, the contradiction here is immediate.

This also completes the proof of Theorem 1.2. □

## 3 Remarks and a conjecture

### Remark 3.1

Corollary 1.1 or Corollary 1.2 can be strengthened to
$$N\biggl(r,\frac{1}{F}\biggr)\neq S(r,f).$$

### Remark 3.2

What can be said if ‘$$pff' -q$$’ is replaced by ‘$$pff^{(k)} -q$$’, for any integer $$k\geq2$$, in Theorem 1.2?

### Remark 3.3

Taking $$f(z)=\mathrm{e}^{z}$$, we have
$$N\biggl(r,\frac{1}{ff^{(k)}-a}\biggr)\sim2T(r,f)+S(r,f),$$
where k is a positive integer, and a is a non-zero constant.

Finally, we present the following more general and quantitative conjecture.

### Conjecture 3.1

Let f be a transcendental entire function. Then for any integer $$k\geq1$$, and any small function a (0),
$$N\biggl(r,\frac{1}{ff^{(k)}-a}\biggr)\sim2 T(r,f)+S(r,f).$$

## Authors’ Affiliations

(1)
Department of Mathematics, China University of Petroleum, Qingdao, 266580, P.R. China
(2)
Department of Mathematics, Nanjing University, Nanjing, 210093, P.R. China

## References 