Abstract and Applied Analysis

Volume 2014, Article ID 306360, 5 pages

http://dx.doi.org/10.1155/2014/306360

## Admissible Solutions of the Schwarzian Type Difference Equation

College of Science, Guangdong Ocean University, Zhanjiang 524088, China

Received 14 January 2014; Accepted 20 March 2014; Published 7 April 2014

Academic Editor: Zong-Xuan Chen

Copyright © 2014 Baoqin Chen and Sheng Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

This paper is to investigate the Schwarzian type difference equation where is a rational function in with polynomial coefficients, , respectively are two irreducible polynomials in of degree , respectively . Relationship between and is studied for some special case. Denote . Let be an admissible solution of such that ; then for (≥2) distinct complex constants , In particular, if , then

#### 1. Introduction and Results

Throughout this paper, a meromorphic function always means being meromorphic in the whole complex plane, and always means a nonzero constant. For a meromorphic function , we define its shift by and define its difference operators by In particular, for the case . We use standard notations of the Nevanlinna theory of meromorphic functions such as , , and and as stated in [1–3]. For a constant , we define the Nevanlinna deficiency by

Recently, numbers of papers (see, e.g., [4–12]) are devoted to considering the complex difference equations and difference analogues of Nevanlinna theory. Due to some idea of [13], we consider the admissible solution of the Schwarzian type difference equation: where is a rational function in with polynomial coefficients, , respectively , are two irreducible polynomials in of degree , respectively, . Here and in the following, “admissible” always means “transcendental.” And we denote from now on. For the existence of solutions of (3), we give some examples below.

*Examples. *(1) is an admissible solution of the Schwarzian type difference equation:

(2) is an admissible solution of the Schwarzian type difference equation

(3) Let , then solves the Schwarzian type difference equation:
This example shows that (3) may admit polynomial solutions.

Considering the relationship between and in those examples above, we prove the following result.

Theorem 1. *For the Schwarzian type difference equation (3) with polynomial coefficients, note the following.*(i)*If it admits an admissible solution such that , then
* *In particular, if , then .*(ii)*If its coefficients are all constants and it admits a polynomial solution with degree , then and .*

*Remark 2. *From examples (1) and (2), we conjecture that in Theorem 1(i). However, we cannot prove it currently. From example (3) given before, we see that the restriction on the coefficients in Theorem 1(ii) cannot be omitted.

For the Schwarzian differential equation,
where , , and are as stated before; Ishizaki [13] proved the following result (see also Theorem in [2]).

*Theorem A (see [2, 13]). Let be an admissible solution of (8) with polynomial coefficients, and let be ≥2 distinct complex constants. Then
*

*For the Schwarzian type difference equation (3), we prove the following result.*

*Theorem 3. Let be an admissible solution of (3) with polynomial coefficients such that , and let be ≥2 distinct complex constants. Then
In particular, if , then
*

*Remark 4. *From Theorem 1, under the condition in Theorem 3, we have in (11). The behavior of the zeros and the poles of in is essentially different from that in the . We wonder whether the restriction can be omitted or not.

*2. Lemmas*

*2. Lemmas**The following lemma plays a very important role in the theory of complex differential equations and difference equations. It can be found in Mohon’ko [14] and Valiron [15] (see also Theorem in the book of Laine and Yang [2]).*

*Lemma 5 (see [14, 15]). Let be a meromorphic function. Then, for all irreducible rational functions in ,
with meromorphic coefficients such that
and the characteristic function of satisfies
where .*

*The following two results can be found in [10]. In fact, Lemma 6 is a special case of Lemma 8.3 in [10].*

*Lemma 6 (see [10]). Let be a meromorphic function of hyper order , and . Then
possibly outside of a set of with finite logarithmic measure.*

*Lemma 7 (see [10]). Let be a meromorphic function of hyper order , and . Then
possibly outside of a set of with finite logarithmic measure.*

*From Lemma 7, we can easily get the following conclusion.*

*Lemma 8. Let be a meromorphic function of hyper order , and . Then
possibly outside of a set of with finite logarithmic measure.*

*Lemma 9. Let be an admissible solution of (3) with coefficients. Then, using the notation ,
In particular, if , then
*

*Proof. *We use the idea by Ishizaki [13] (see also [2]) to prove Lemma 9. It follows from Lemma 8 that
From this and Lemma 5, we get
and hence

If , since all coefficients of and are polynomials, there are at the most finitely many poles of , neither the poles of nor the zeros of . Therefore, we see that
We obtain (18) from this and (22) immediately.

If , there are at most finitely many poles of , not the zeros of , then
Now (18) follows from (22) and (24).

Notice that if , then (24) always holds. This finishes the proof of Lemma 9.

*3. Proof of Theorem 1*

*3. Proof of Theorem 1**Case 1. *
Equation (3) admits an admissible solution such that . Since all coefficients of and are polynomials, there are at the most finitely many poles of that are not the poles of and . This implies that

*From Lemma 5, we get
*

*We can deduce from (3), (25), (26), and Lemma 8 that
It follows from this that
What is more is that if , then we obtain from (28) that *

*Case 2. *The coefficients of (3) are all constants and it admits a polynomial solution with degree . Set
then
where
From (29) and (30), we obtain that

*If , then , which yields that . That is a contradiction to our assumption. Thus, .*

*If , then , , and . Now from (3), we get
Considering degrees of both sides of the equation above, we can see that .*

*If , we can deduce similarly that
where are polynomials such that .*

*Rewrite (3) as follows:
*

*From (34), we find that the leading coefficient of is
*

*Considering degrees of both sides of (35), we prove that .*

*4. Proof of Theorem 3*

*4. Proof of Theorem 3**Firstly, we consider the general case. As mentioned in Remark 1 in [13], due to Jank and Volkmann [16], if (3) admits an admissible solution, then there are at most common zeros of and . Since all coefficients of are polynomials, there are at the most finitely many poles of that are the zeros of . Therefore, from (3), we have
Combining this and Lemma 9, applying the second main theorem, we get
Thus, we prove that (10) holds.*

*Secondly, we consider the case that . From (3) and Lemma 8, we similarly get that
From this and applying Lemma 9 with (19), as arguing before, we can prove that (11) holds.*

*Conflict of Interests*

*Conflict of Interests**The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgments*

*Acknowledgments**The authors would like to thank the referees for their valuable suggestions. This work is supported by the NNSFC (nos. 11226091 and 11301091), the Guangdong Natural Science Foundation (no. S2013040014347), and the Foundation for Distinguished Young Talents in Higher Education of Guangdong (no. 2013LYM_0037).*

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