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On the meromorphic solutions of certain class of nonlinear differential equations


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., [811]).

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 [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

$$ f^{n}f^{(k)}+a_{n-1}f^{n-1}+ \cdots+a_{1}f+a_{0}=R\mathrm{e}^{\alpha}, $$

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

$$ P\bigl(z,f,f',\ldots,f^{(n)}\bigr)=0, $$

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: \(N(r,f)=S(r,f)\),

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

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

$$ \frac{nf^{n-1}f'f^{(k)}+f^{n}f^{(k+1)}+a'_{n-1}f^{n-1}+\cdots +a'_{1}f+a_{1}f'+a'_{0}}{ f^{n}f^{(k)}+a_{n-1}f^{n-1}+\cdots+a_{1}f+a_{0}}=\frac{R'}{R}+\alpha'. $$

It follows by (2.3) that

$$\begin{aligned}& -\biggl(\frac{R'}{R}+\alpha'\biggr)f^{n}f^{(k)}+nf^{n-1}f'f^{(k)}+f^{n}f^{(k+1)}+ \biggl\{ a'_{n-1}-\biggl(\frac{R'}{R}+ \alpha'\biggr)a_{n-1}\biggr\} f^{n-1} \\& \quad {}+(n-1)a_{n-1}f^{n-2}f'+\cdots+\biggl\{ a'_{1}-\biggl(\frac{R'}{R}+\alpha' \biggr)a_{1}\biggr\} f+a_{1}f' =\biggl( \frac{R'}{R}+\alpha'\biggr)a_{0}-a'_{0}. \end{aligned}$$

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

$$ f^{n}f^{(k)}+a_{n-1}f^{n-1}+ \cdots+a_{1}f=(A-1)a_{0}. $$

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.


$$\biggl(\frac{R'}{R}+\alpha'\biggr)a_{0}-a'_{0}:= \varphi\not\equiv0. $$

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

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

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

$$ h=\frac{a_{1}f'+a'_{0}-(\frac{R'}{R}+\alpha')a_{0}}{f}. $$

Then (2.7) gives \(T(r,h)=S(r,f)\). We have

$$ f'=\frac{1}{a_{1}}\biggl\{ hf-a'_{0}+ \biggl(\frac{R'}{R}+\alpha'\biggr)a_{0}\biggr\} := \mu_{1}f+\nu _{1}. $$

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

$$ f^{n-1}\psi=P_{n-1}(f), $$

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

$$ f^{(k)}=\mu_{k}f+\nu_{k}, $$

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

$$ \mu_{k}f^{n+1}+\nu_{k}f^{n}+a_{n-1}f^{n-1}+ \cdots+a_{1}f+a_{0}=R\mathrm{e}^{\alpha }. $$

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

$$ \frac{1}{n+1}\frac{\nu_{k}}{\mu_{k}}=(n+1)\frac{a_{0}}{a_{1}}\quad \mbox{and}\quad \mu _{k}= \biggl(\frac{1}{n+1}\frac{a_{1}}{a_{0}} \biggr)^{n+1}a_{0}. $$

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

$$ f^{(k)}= \biggl(\frac{1}{n+1}\frac{a_{1}}{a_{0}} \biggr)^{n+1}a_{0} \biggl[g\mathrm {e}^{\gamma}-(n+1)\frac{a_{0}}{a_{1}}\biggr]+(n+1) \biggl( \frac{1}{n+1}\frac {a_{1}}{a_{0}} \biggr)^{n}a_{0}. $$

In addition, by \(f=g\mathrm{e}^{\gamma}-(n+1)\frac{a_{0}}{a_{1}}\) we get

$$ f^{(k)}=Q\bigl(g,g',\ldots,g^{(k)}\bigr) \mathrm{e}^{\gamma}-(n+1) \biggl(\frac {a_{0}}{a_{1}}\biggr)^{(k)}, $$

where \(Q(g,g',\ldots,g^{(k)})\) is a differential polynomial of g.

Thus, (2.13) and (2.14) imply

$$Q\bigl(g,g',\ldots,g^{(k)}\bigr)= \biggl(\frac{1}{n+1} \frac{a_{1}}{a_{0}} \biggr)^{n+1}a_{0}g $$


$$ (n+1) \biggl(\frac{a_{0}}{a_{1}}\biggr)^{(k)}= \biggl(\frac{1}{n+1} \frac{a_{1}}{a_{0}} \biggr)^{n+1}a_{0}(n+1)\frac{a_{0}}{a_{1}}-(n+1) \biggl(\frac{1}{n+1}\frac {a_{1}}{a_{0}} \biggr)^{n}a_{0}. $$

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 \(\mathrm{e}^{\alpha}\),

$$ t_{1}ff'+p\bigl(f'\bigr)^{2}+pff''=t_{2}, $$

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

$$ t'_{1}ff'+\bigl(t_{1}+p' \bigr) \bigl(f'\bigr)^{2}+\bigl(t_{1}+p' \bigr)ff''+3pf'f''+pff'''=t'_{2}. $$

Letting \(z_{0}\) be a simple zero of f, (2.16) and (2.17) imply

$$ \bigl(p\bigl(f'\bigr)^{2}-t_{2}\bigr) (z_{0})=0 $$


$$ \bigl\{ \bigl(t_{1}+p'\bigr) \bigl(f' \bigr)^{2}+3pf'f''-t'_{2} \bigr\} (z_{0})=0. $$


$$ g=\frac{3pt_{2}f''+[t_{2}(t_{1}+p')-t'_{2}p]f'}{f}. $$

From (2.6), (2.18), and (2.19), we get

$$T(r,g)=S(r,f). $$

By (2.20), we obtain

$$ f''=\alpha_{1}f+\beta_{1}f', $$


$$\alpha_{1}=\frac{g}{3pt_{2}}, \qquad \beta_{1}= \frac{t'_{2}p-t_{2}(t_{1}+p')}{3pt_{2}} $$


$$T(r,\alpha_{1})=S(r,f), \qquad T(r,\beta_{1})=S(r,f). $$

Substituting (2.21) into (2.16) yields

$$ (t_{1}+p\beta_{1})ff'+p\bigl(f' \bigr)^{2}+\alpha_{1}pf^{2}=t_{2}. $$

On the other hand, from (2.21), we have

$$ f'''=\alpha_{2}f+ \beta_{2}f', $$

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

$$ \bigl[t'_{1}+\beta_{1}\bigl(t_{1}+p' \bigr)+3p\alpha_{1}+p\beta_{2}\bigr]ff' + \bigl(t_{1}+p'+3p\beta_{1}\bigr) \bigl(f'\bigr)^{2}+\bigl[\alpha_{1} \bigl(t_{1}+p'\bigr)+\alpha _{2}p \bigr]f^{2}=t'_{2}. $$

It follows by (2.22) and (2.24) that

$$\begin{aligned}& \bigl\{ p\bigl[t'_{1}+\beta_{1} \bigl(t_{1}+p'\bigr)+3p\alpha_{1}+p \beta_{2}\bigr]-\bigl(t_{1}+p' +3p \beta_{1}\bigr) (t_{1}+p\beta_{1})\bigr\} ff' \\& \quad {}+\bigl\{ p\bigl[\alpha_{1}\bigl(t_{1}+p' \bigr)+\alpha_{2}p\bigr]-\alpha_{1}p\bigl(t_{1}+p'+3p \beta _{1}\bigr)\bigr\} f^{2} =t'_{2}p-t_{2} \bigl(t_{1}+p'+3p\beta_{1}\bigr). \end{aligned}$$

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

$$ t'_{2}p-t_{2}\bigl(t_{1}+p'+3p \beta_{1}\bigr)\equiv0. $$

Now, (2.25) and (2.26) lead to

$$ p\bigl[\alpha_{1}\bigl(t_{1}+p'\bigr)+ \alpha_{2}p\bigr]-\alpha_{1}p\bigl(t_{1}+p'+3p \beta _{1}\bigr)\equiv0. $$

From the definition of \(\alpha_{2}\) and (2.27), we deduce

$$ \alpha'_{1}\equiv2\alpha_{1} \beta_{1}. $$

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),

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


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The authors would like to thank the referee for his/her several important suggestions and valuable comments to our original manuscript which enabled us to improve greatly the quality and readability of the paper.

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Correspondence to Weiran Lü.

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Dedicated to Professor George Csordas on the occasion of his retirement.

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Liu, N., Lü, W. & Yang, C. On the meromorphic solutions of certain class of nonlinear differential equations. J Inequal Appl 2015, 149 (2015).

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