Results on meromorphic solutions of linear difference equations
- 1.7k Downloads
In this paper, we investigate meromorphic solutions of linear difference equations and prove a number of results. We give estimates for the growth of meromorphic solutions under some special cases and provide some examples to show that the answer to a question of Laine and Yang is not always positive. The zeros, poles and fixed points of finite order solutions are also studied.
MSC:39A13, 39A22, 30D35.
Keywordsdifference equations growth fixed points
1 Introduction and results
Recently, numbers of papers (see, e.g., [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]) are devoted to considering the complex difference equations and difference analogues of Nevanlinna theory. For the growth of meromorphic solutions of difference equations, Chiang and Feng [8, 9] considered the polynomial coefficients case and got
then we have .
The following result shows that the polynomial coefficients in Theorem A can be extended to rational functions.
then, for any meromorphic solution to (1.1), we have .
And hence Theorem 1.1 follows from Theorem A. We omit the details of its proof.
Theorem B 
If is a meromorphic solution to (1.1), then .
Theorem C 
Let be entire functions of finite order such that among those coefficients having the maximal order , exactly one has its type strictly greater than the others. Then, for any meromorphic solution to (1.1), we have .
In Theorems B and C, there is always some dominating coefficient such that . A natural question is what happens if the dominating coefficient is of order zero? Another question raised by Laine and Yang in  is whether all meromorphic solutions of (1.1) satisfy , even if there is no dominating coefficient. For the first question, we get the following result.
Theorem 1.2 Let be meromorphic functions such that there exists an integer l, , such that is a transcendental entire function, while , , are all rational functions. If is a meromorphic solution to (1.1), then .
Considering Laine and Yang’s question, we get the following example which indicates that the answer to their question is not always positive.
In this example, the relationship between and exactly depends on k.
However, the answer may be positive in some special case. In fact, we prove the following results, in which there is still some coefficient dominating in some angle.
where are all entire functions such that and . Then we have .
Theorem 1.4 Under exactly one of assumptions for the coefficients of (1.1) in Theorems A-C and Theorems 1.1 and 1.2, if is a finite order meromorphic solution to (1.1), then . What is more, either or , where .
Theorem 1.5 Under the assumption for the coefficients of (1.2) in Theorem 1.3, if is a finite order meromorphic solution to (1.2), then . What is more, either or .
The following examples show the sharpness of the estimates in Theorems 1.4 and 1.5.
which satisfies the assumptions in Theorem A and Theorem 1.1. We see that and .
which satisfies the assumptions in Theorems B and C and Theorem 1.2. We have , and , .
which satisfies the assumptions in Theorem 1.3. We have , and and .
2 Proof of Theorem 1.2
We first recall a key lemma used to prove Theorems A and B and the pointwise estimates for difference quotient which are counterparts to Gundersen’s logarithmic derivative estimates  (see , Corollary 2.6, Theorem 8.3).
Lemma 2.1 
Proof of Theorem 1.2 If , the assertion follows from Theorem B. We next consider the case that .
a contradiction. Our proof is thus finished. □
3 Proof of Theorem 1.3
for all r outside of a possible exceptional set with finite logarithmic measure.
for all r outside of a possible exceptional set with finite linear measure.
a contradiction. And hence we have . □
4 Proofs of Theorems 1.4 and 1.5
Lemma 4.1 
are not constant functions for .
(, ), where E is an exceptional set of finite linear measure, and .
Proofs of Theorems 1.4 and 1.5 In fact, we only give the proof of Theorem 1.5 since the proof of Theorem 1.4 is similar.
Obviously, we have .
Next, we assert that either or . If the assertion does not hold, we have .
where , are entire functions such that , and is a polynomial such that .
for . Notice that for and . Thus, Lemma 4.1 is valid for (4.1) and hence we get that for , a contradiction to our assumption. This completes our proof. □
The authors would like to thank the editor and the referees for their constructive comments to improve the readability of our paper. This work is supported by the National Natural Science Foundation of China (No. 11226091) and the Natural Science Research Projects of GDOU (No. 1212331).
This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.