On the meromorphic solutions of certain class of nonlinear differential equations
- Nana Liu^{1},
- Weiran Lü^{1}Email author and
- Chungchun Yang^{2}
https://doi.org/10.1186/s13660-015-0672-x
© Liu et al.; licensee Springer. 2015
Received: 22 September 2014
Accepted: 21 April 2015
Published: 1 May 2015
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 equationMSC
34M10 30D351 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
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
The following lemma is crucial to the proof of our theorems.
Lemma 2.2
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 \(N(r,f)= S(r,f)\).
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.
This also completes the proof of Theorem 1.2. □
3 Remarks and a conjecture
Remark 3.1
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
Finally, we present the following more general and quantitative conjecture.
Conjecture 3.1
Notes
Declarations
Acknowledgements
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 (http://creativecommons.org/licenses/by/4.0/), 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
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