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On the meromorphic solutions of certain class of nonlinear differential equations

Journal of Inequalities and Applications20152015:149

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


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.


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




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.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

Department of Mathematics, China University of Petroleum, Qingdao, 266580, P.R. China
Department of Mathematics, Nanjing University, Nanjing, 210093, P.R. China


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