On the meromorphic solutions of certain class of nonlinear differential equations

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.

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 [3] in 1959. Later, he conjectured [4] that Theorem A remains valid even if $n=1$ or $n=2$. Mues [5] proved the result for $n=2$ and the case $n=1$ was proved by Bergweiler and Eremenko [6] and independently by Chen and Fang [7]. For entire functions and difference polynomials, similar results have been obtained by others earlier (see, e.g., [8–11]).

Theorem B

([12])

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

Wang [13] 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 [15]), 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 [16].

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 [20] and combining it with the proof of Proposition E in [21]. 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., [1], 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

$$T\bigl(r,f^{n}f^{(k)}+a_{n-1}f^{n-1}+ \cdots+a_{1}f\bigr)= S(r,f), $$

a contradiction.

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., [1], 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., [2], 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 [21]. 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),

$$N\biggl(r,\frac{1}{ff^{(k)}-a}\biggr)\sim2 T(r,f)+S(r,f). $$