Abstract and Applied Analysis

Volume 2013 (2013), Article ID 496096, 7 pages

http://dx.doi.org/10.1155/2013/496096

## Fixed Points of Meromorphic Solutions for Some Difference Equations

^{1}School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China^{2}Department of Mathematics, College of Natural Sciences, Pusan National University, Pusan 609-735, Republic of Korea

Received 1 December 2012; Accepted 17 April 2013

Academic Editor: Patricia J. Y. Wong

Copyright © 2013 Zong-Xuan Chen and Kwang Ho Shon. 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

We investigate fixed points of meromorphic solutions for the Pielou logistic equation and obtain some estimates of exponents of convergence of fixed points of and its shifts , differences , and divided differences .

#### 1. Introduction and Results

In this paper, we assume the reader is familiar with basic notions of Nevanlinna’s value distribution theory (see [1–3]). In addition, we use the notation to denote the order of growth of a meromorphic function and and to denote, respectively, the exponents of convergence of zeros and poles of . We also use the notation to denote the exponent of convergence of fixed points of that is defined as

Recently, a number of papers (including [4–17]) focus on complex difference equations and difference analogues of Nevanlinna’s theory.

The Pielou logistic equation where and are nonzero polynomials, is an important difference equation, because it is obtained by transform from the well-known Verhulst Pearl equation (see [18, page 99]) which is the most popular continuous model of growth of a population.

Chen [7] obtained the following theorem.

Theorem A. *Let and be polynomials with and let be a finite order transcendental meromorphic solution of (2). Then
*

*Example 1. *The function satisfies the Pielou logistic equation
where satisfies
This example shows that the result of Theorem A is sharp.

One of the main purposes in this paper is to study fixed points of meromorphic solutions of the Pielou logistic equation (2).

The problem of fixed points of meromorphic functions is an important one in the theory of meromorphic functions. Many papers and books (including [18–20]) investigate fixed points of meromorphic functions.

Now we consider fixed points of meromorphic functions and their shifts, differences, and divided differences. We see that there are many examples to show that either may have no fixed point, for example, or the shift of , or the difference of may have only finitely many fixed points; for example, for the function , its shift , and its difference have only finitely many fixed points. Even if for a meromorphic function of small growth, Chen and Shon show that there exists a meromorphic function such that and has only finitely many fixed points (see Theorem 6 of [9]).

A divided difference may also have only finitely many fixed points; for example, the function satisfies that its divided difference has only finitely many fixed points. Chen and Shon obtained Theorem B.

Theorem B (see [9]). *Let be a constant and let be a transcendental meromorphic function of order of growth or of the form , where is a constant and is a transcendental meromorphic function with . Suppose that is a nonconstant polynomial. Then
**
has infinitely many zeros.*

From Theorem B, we easily see that under conditions of Theorem B, the divided difference has infinitely many fixed points. The previous example shows that result of Theorem B is sharp.

However, we discover that the properties on fixed points of meromorphic solutions of (2) are very good. We prove the following theorem.

Theorem 2. *Let and be nonzero polynomials such that
**
Set . Then every finite order transcendental meromorphic solution of (2) satisfies the following:*(i)*;
*(ii)* if , then ; *(iii)* if there is a polynomial satisfying
**then .*

*Remark 3. *Generally, for a meromorphic function of finite order. For example, the function satisfies

#### 2. Proof of Theorem 2

We need the following lemmas for the proof of Theorem 2.

Lemma 4 (see [12, 17]). *Let be a nonconstant finite order meromorphic solution of
**
where is a difference polynomial in . If for a meromorphic function satisfying , then
**
holds for all outside of a possible exceptional set with finite logarithmic measure.*

*Remark 5. *Using the same method as in the proof of Lemma 4 (see [12]), we can prove that in Lemma 4, if all coefficients of satisfy and if for a meromorphic function satisfying , then for a given ,
holds for all outside of a possible exceptional set with finite logarithmic measure.

Lemma 6. *Suppose that and satisfy the condition (8) in Theorem 2 and that is a nonconstant meromorphic function. Then
**
have at most finitely many common zeros.*

*Proof. *Suppose that is a common zero of and . Then . Thus, . Substituting into , we obtain
Since has only finitely many zeros, we see that and have at most finitely many common zeros.

Lemma 7 (see [14]). *Let be a nonconstant finite order meromorphic function. Then
*

Gol’dberg and Ostrovskii [21, page 66] give that for any constant , This and Lemma 7 give the following lemma.

Lemma 8. *Let be a nonconstant finite order meromorphic function. Then
*

Using the same method as in the proof of Lemma 6, we can prove Lemmas 9 and 10.

Lemma 9. *Suppose that and satisfy the condition (8) in Theorem 2 and that is a nonconstant meromorphic function. Then
**
have at most finitely many common zeros.*

Lemma 10. *Suppose that and satisfy the condition (8) in Theorem 2 and is a nonconstant meromorphic function. Then
**
have at most finitely many common zeros.*

*Proof of Theorem 2. *(i) We prove that . Suppose that . Set . So, is transcendental, , and . Substituting into (2), we obtain
Thus,
By (8) and (22), we see that . Thus, by Lemma 4 and , we obtain
for all outside of a possible exceptional set with finite logarithmic measure. Thus,
for all outside of a possible exceptional set with finite logarithmic measure. So, by Theorem A and (24), we obtain .

Now suppose that . By (2), we obtain
By (8), we see that . Since and are polynomials, by (25), we see that and have the same poles, except possibly finitely many poles. By Lemma 6, we see that and have at most finitely many common zeros. Hence, by (25), we have that
Suppose that . Thus, can be rewritten as the following form:
where is a polynomial with , and are canonical products ( may be a polynomial) formed by nonzero zeros and poles of , respectively, and is an integer; if , then ; if , then . Combining Theorem A with properties of canonical product, we see that
By (27), we obtain
where . Thus, by (28) and Lemma 8, we have that
Substituting (29) into (2), we obtain
By (31), we obtain
By (8), we see that in the numerator of the right side of (32), there exists only one term being of the highest degree. So,
Thus, by (28), (33), Lemma 4, and its Remark 5, we obtain that for any given
holds for all outside of a possible exceptional set with finite logarithmic measure.

On the other hand, by and the fact that is an entire function, we see that
Thus, by this and (28), we see that (34) is a contradiction. Hence, . By (26), we obtain

Now suppose that . By (2), we obtain
where . By Lemma 8, we have that . By (8), we have
Thus, for (37), applying the conclusion of above, we obtain
Continuing to use the same method as above, we can obtain

(ii) Suppose that . We prove that . By (2), we obtain
Since and have the same poles, except possibly finitely many poles, by Lemma 9 and (41), we only need to prove that

Set
Suppose that . Using the same method as in the proof of (i), can be rewritten as the following form:
where and and are nonzero entire functions, such that
Substituting (44) into (2), we obtain
Since and are polynomials, we see that
that is,
Using the same method as in the proof of (i), we obtain a contradiction. Hence, (42) holds; that is, .

(iii) Suppose that there is a polynomial satisfying
Thus, by (8) and (49), we see that

Now we prove that . By (2), we obtain
By (49) and (51), we obtain
Since and are polynomials, we see that poles of must be poles of . Thus, poles of are not zeros of . By Lemma 10, we see that the numerator and the denominator of the right side of (51) have at most finitely many common zeros. Thus, in order to prove , by (52), we only need to prove that
or
By (50), we have . Combining this with (8), we see that there exists at least one of
such that its degree is equal to . Without loss of generality, we may suppose that

Now we prove that (53) holds. Suppose that
Using a similar method as in the proof of (i), we see that can be rewritten as the following form:
where and and are nonzero entire functions, such that
Substituting (58) into (2), we obtain
where
By (8), (50), and (56), we see that
Thus, we obtain
So, by (60) and (63), we see that .

Using the same method as in the proof of (i), we see that (53) holds.

Thus, Theorem 2 is proved.

#### Acknowledgments

The first author was supported by the National Natural Science Foundation of China (no. 11171119). The second author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2010-0009646).

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