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## Complex Differences and Difference Equations

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Volume 2014 |Article ID 584969 | https://doi.org/10.1155/2014/584969

Zhiqiang Mao, Huifang Liu, "Difference Equations and Sharing Values Concerning Entire Functions and Their Difference", Abstract and Applied Analysis, vol. 2014, Article ID 584969, 6 pages, 2014. https://doi.org/10.1155/2014/584969

# Difference Equations and Sharing Values Concerning Entire Functions and Their Difference

Accepted22 Mar 2014
Published07 Apr 2014

#### Abstract

The value distribution of solutions of certain difference equations is investigated. As its applications, we investigate the difference analogue of the Brück conjecture. We obtain some results on entire functions sharing a finite value with their difference operators. Examples are provided to show that our results are the best possible.

#### 1. Introduction and Main Results

In this paper, the term meromorphic function will mean being meromorphic in the whole complex plane . It is assumed that the reader is familiar with the standard notations and the fundamental results of the Nevanlinna theory; see, for example, . In addition, we use notations to denote the order and the exponent of convergence of the sequence of zeros of a meromorphic function , respectively. The notation is defined to be any quantity satisfying as , possibly outside a set of of finite logarithmic measure.

Let and be two nonconstant meromorphic functions, and let . We say that and share CM, provided that and have the same zeros with the same multiplicities. Similarly, we say that and share IM, provided that and have the same zeros ignoring multiplicities.

The famous results in the uniqueness theory of meromorphic functions are the 5 IM and 4 CM shared values theorems due to Nevanlinna . It shows that if two nonconstant meromorphic functions and share five different values IM or four different values CM, then or is a linear fractional transformation of . Condition 4 CM shared values have been improved to 2 CM + 2 IM by Gundersen , while the case 1 CM + 3 IM still remains an open problem. Specifically, Brück posed the following conjecture.

Conjecture 1 (see ). Let be a nonconstant entire function satisfying the hyperorder , where is not a positive integer. If and share a finite value CM, then for some nonzero constant .

In , Brück proved that the conjecture is true provided that or . He also gave counterexamples to show that the restriction on the growth of is necessary.

In recent years, as the research on the difference analogues of Nevanlinna theory is becoming active, lots of authors  started to consider the uniqueness of meromorphic functions sharing values with their shifts or their difference operators.

Heittokangas et al. proved the following result which is a shifted analogue of Brück’s conjecture.

Theorem A (see ). Let be a meromorphic function of and a nonzero complex number. If and share a finite value and CM, then for some constant .

In , Heittokangas et al. gave the example which shows that cannot be relaxed to .

For a nonzero complex number , we define difference operators as Regarding the difference analogue of Brück’s conjecture, we mention the following results.

Theorem B (see ). Let be a finite order transcendental entire function which has a finite Borel exceptional value , and let be a constant such that . If and share CM, then for some nonzero constant .

Theorem C (see ). Let be a nonperiodic transcendental entire function of finite order. If and share a nonzero finite value CM, then ; that is, where is an entire function with and is a polynomial with .

Let be a nonperiodic transcendental entire function of finite order. Theorem B shows that if a nonzero finite value is shared by and , then . It is obvious that the result in Theorem B is sharper than Theorem C for . In this paper, we continue to investigate the difference analogue of Brück’s conjecture and obtain the following result.

Theorem 2. Let be a finite order entire function, an integer, and a constant such that . If and share a finite value CM, then ; that is, where is an entire function with and is a polynomial with .

Remark 3. It is obvious that Theorem 2 is sharper than Theorem C and a supplement of Theorem B for .

The discussions in Theorems C and 2 are concerning the case that shared value . When , we obtain the following result.

Theorem 4. Let be a finite order entire function, a positive integer, and a constant such that . If and share CM, then ; that is, where is an entire function with and is a polynomial with .

It is well known that if a finite order entire function shares CM with , then satisfies the difference equation where is a polynomial. Hence in order to prove the above results, we consider the value distribution of entire solutions of the difference equation and obtain the following result.

Theorem 5. Let be polynomials, and let be a polynomial with degree . Then every entire solution of finite order of (8) satisfies and (i)if , then ;(ii)if , then or has only finitely many zeros.

#### 2. Lemmas

Lemma 6 (see ). Let be a nondecreasing continuous function, , and let be the set of all such that . If the logarithmic measure of is infinite, then

Lemma 7 (see ). Let be a nonconstant meromorphic function of finite order, . Then for all outside a possible exceptional set with finite logarithmic measure .

Remark 8. By Lemmas 6 and 7, we know that, for a nonconstant meromorphic function of finite order,

Lemma 9 (see ). Let and be entire functions such that (i),(ii)the order of is less than the order of for ; and furthermore, the order of is less than the order of for and .Then .

Lemma 10 (see ). Let be a meromorphic function with finite order , . Then for any given and integers , there exists a set of finite logarithmic measure, so that, for all , we have

Lemma 11 (see ). Let be entire functions with finite order. If there exists an integer such that holds, then every meromorphic solution of the difference equation satisfies .

#### 3. Proofs of Results

Proof of Theorem 5. Let be an entire solution of finite order of (8). By Remark 8 and (8), we get By (15) we get .
Case  1 (). Suppose that , by the Weierstrass factorization; we get , where is an entire function and is a polynomial such that Substituting into (8), we get If , then by (16) we know that the order of the right side of (17) is , and the order of the left side of (17) is less than . This is a contradiction. Hence . Set where are complex numbers. By (17) we get Next we discuss the following two subcases.
Subcase 1 (). Then by Lemma 9, (16), and (19), we get . This is impossible.
Subcase  2 (). Suppose that Then . By , we obtain that is a nonzero constant. Hence is a nonzero polynomial. By (19) we get Since for , then by Lemma 9 and (21), we get This is impossible. Hence we have . Then from the order consideration, we know that the order of the right side of (19) is , and the order of the left side of (19) is less than . This is a contradiction. Hence .
Case  2 (). Then by , we get . Suppose that has infinitely many zeros and ; by the Weierstrass factorization, we get where is a complex number and is an entire function such that Let , where are complex numbers. Substituting into (8), we get Note that ; otherwise has only finitely many zeros. If, then the order of the right side of (25) is 1, but the order of the left side of (25) is less than 1. This is absurd. If, then by Lemma 9, (24), and (25), we get . This is impossible. Hence . Theorem 5 is thus completely proved.

Proof  of Theorem 2. Since and share CM and is of finite order, then where is a polynomial with . Now we will take two steps to complete the proof.
Step  1. We prove that .
Let ; then and . By this and (26), we get Next we discuss the following three cases.
Case 1 (  and  ). Then . By Theorem 5(i), (27), and (28), we get .
Case 2 ( and  ). If , then by Theorem 5(i), (27), and (28), we get . If , then by Theorem 5(ii) and (27), we obtain that , or has only finitely many zeros.
If has only finitely many zeros, set where is a polynomial and is a complex number; then substituting (29) into (28), we get By (30) and , we know that the order of the left side of (30) is 0 and the order of the right side of (30) is 1 unless and . In this case, take it into the left side of (30); we have . Since all , and are not zero, it is impossible. Hence we get .
Case  3. is a complex constant. Then by (28) we get where is a complex number. Suppose that . Let , where is an entire function and is a polynomial such that Substituting into (31), we get Since , by (32) we obtain that the order of the left side of (33) is less than and the order of the right side of (33) is . This is absurd. Hence we get .
Step  2. We prove that .
Suppose that . Since and share CM, then where is a nonzero constant. Let ; then by (34) we get Differentiating (35), we get Note that and . So by Lemma 10 and (36), we get This is absurd. So . Theorem 2 is thus completely proved.

Proof of Theorem 4. Since and share CM and is of finite order, then where is a polynomial. By and (38), we get We discuss the following two cases.
Case  1. is a polynomial with . Then by Lemma 11 and (39), we get . Now we prove . Suppose that ; then by the Weierstrass factorization, we get , where is an entire function and is a polynomial such that Substituting into (39), we get
If , then by (40), (41), and , we obtain that the order of the left side of (41) is and the order of the right side of (41) is less than . This is absurd.
If , then by (41) we get Set where are complex numbers. Then Now we discuss the following two subcases.
Subcase  1. holds for every . Then by (40), (42), , and Lemma 9, we get . This is absurd.
Subcase  2. There exist some such that . Then by (44) we have for . Merging the term into , by (42) we get or where satisfying . If , then by (40), (45), , and Lemma 9, we get . This is absurd. If , then by (46) and , we get . By this we know that the order of the left side of (46) is and the order of the right side of (46) is less than . This is absurd. Hence we get .
Case  2. is a complex constant. Then by Lemma 10 and (38), we get . Now we prove . Suppose that . If , then by the similar argument to that of case 1, we get . This is absurd. If , then by (40) we get . Since , then . Theorem 4 is thus completely proved.

#### 4. Some Examples

The following examples show the existence of such entire functions which satisfy Theorems 25. Moreover, Example 2 shows that the result in Theorem 4 is the best possible.

Example 1. Let , and , where are constants. Then and share CM and .

Example 2. Let , and . Then and share CM and .

Example 3. Let , and , where is an entire function with period such that and . Then and share CM and . (Ozawa  proved that for any there exists a period entire function of order .)

Example 4. The entire function satisfies the difference equation where . Here and has only finitely many zeros.

#### Conflict of Interests

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

#### Acknowledgments

This work is supported by the National Natural Science Foundation of China (nos. 11201195, 11171119) and the Natural Science Foundation of Jiangxi, China (nos. 20122BAB201012, 20132BAB201008.).

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