Growth of Meromorphic Function Sharing Functions and Some Uniqueness Problems

Qi, Jianming; Meng, Fanning; Yuan, Wenjun

doi:https://doi.org/10.1155/2016/6250867

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Research Article | Open Access

Volume 2016 | Article ID 6250867 | https://doi.org/10.1155/2016/6250867

Growth of Meromorphic Function Sharing Functions and Some Uniqueness Problems

Jianming Qi,^1,2Fanning Meng ,³and Wenjun Yuan³

Academic Editor: Kehe Zhu

Received29 Mar 2016

Accepted05 May 2016

Published05 Sept 2016

Abstract

Estimating the growth of meromorphic solutions has been an important topic of research in complex differential equations. In this paper, we devoted to considering uniqueness problems by estimating the growth of meromorphic functions. Further, some examples are given to show that the conclusions are meaningful.

1. Introduction and Main Results

Assuming that the reader is familiar with the notations and results on Nevanlinna theory [1] and the applications of normal family theory on estimating the growth of meromorphic functions (see [2–4]), it is an interesting attempt to consider the growth properties of meromorphic functions under the condition involved sharing value or some complex differential (or difference) equations (see [5–9]).

For a meromorphic function , the order and hyperorder of are defined as follows [1]:

Let and be two nonconstant meromorphic functions in the complex plane , and let be a meromorphic function or a finite complex number. If whenever , we write . If and , we write and say that and share IM (ignoring multiplicity). If whenever and the multiplicity of the zero of is greater than or equal to that of the zero of , then we denote this condition by . Let be a rational function which behaves asymptotically as , where are constants. The degree of at infinity is defined as .

In the following, for a linear differential polynomial of , we writewhere are constants and is an integer.

In 1986, Jank et al. [10] proved that, for an entire function , if and share a finite nonzero value IM and if , then . In 2006, Wang [11] replaced the value by a polynomial and obtained the following result: let be a nonconstant entire function, let be a polynomial of degree , and let be an integer. If and share CM and if , then .

In 2010, Lü and Yi [12] obtained the following results.

Theorem 1. Let be a nonconstant meromorphic function with finitely many poles, and let be a nonzero rational function. If where is a positive number, then is of order at most 1.

Theorem 2. Let be a nonconstant meromorphic function with finitely many poles, and let be a nonzero rational function such that and have no common poles. If then one of the following three cases holds:(i) and , where is a nonzero constant;(ii) reduces to a constant, say , and for some nonzero constant such that where is a nonzero constant;(iii) is a nonconstant polynomial with , , and where are two nonzero constants and is a polynomial such that

Problem 3. In Theorem 1, we see that , , and share one function with zero order. So it is natural to ask what will happen if they share a function of infinite order or positive finite order?

Considering Problem 3, we derive the following results.

Theorem 4 (main theorem). Let be a nonzero rational function and let be two entire functions. Let be an integer and let be defined as (2). If where , and if has at most finitely many zeros, then .

The following examples show that our conclusion really exists and is sharp.

Example 5. Let , where is a nonzero constant. Let . Noting that , we have Obviously, has no zeros. Thus it satisfies the assumptions of Theorem 4 and .

Example 6. Let and . Noting that , then It satisfies the assumptions of Theorem 4 and .

Example 7. Let , . Noting that , then , , and . It is easy to see and and . It satisfies the assumptions of Theorem 4 and .

Example 8. Let and , where is a constant. Differentiating twice yields and ; then . Thus , but .

Theorem 9 (main theorem). Let be a meromorphic function with at most finitely many poles, and let , where is a rational function and is a nonconstant polynomial. Let be an integer and let be defined as (2). Then , if

Example 10. Let , , so , and . Noting that then .

Remark 11. Example 10 illustrates that the conclusion of Theorem 9 really occurs.

Problem 12. If the rational function is replaced by a function (here is a polynomial) in Theorem 2, what will happen?

Investigating Problem 12, we obtain the following result.

Theorem 13. Let be a nonconstant transcendental meromorphic function with finitely many poles. Let (here is a polynomial and ) be a function and let be a nonzero rational function such that and have no common poles, and let be defined as (2). Let . If and , then the conclusions of Theorem 2 still hold and must be a constant.

2. Some Lemmas

In order to prove our theorems, we need the following lemmas.

Normal families, in particular, of holomorphic functions often appear in operator theory on spaces of analytic functions; for example, see in [13, Lemma ] and in [14, Lemma ]. Using the same method of the famous Pang-Zalcman Lemma [15, Lemma ] and the result of Lü et al. [8, Lemma , page 595], it is easy to obtain the following lemma. It plays an important role in the proof of Theorems 4 and 9.

Lemma 14 (see [8, 15]). Let be a family of meromorphic (analytic) functions in the unit disc . If , , and , and if there exists such that whenever , then there exist(i)a subsequence of (which we still write as ),(ii)points ,(iii)positive numbers ,such that locally uniformly, where is a nonconstant meromorphic (resp., entire) function on , such that (resp., ), , and where is a constant which is independent of .
Here, as usual, is the spherical derivative.

The next lemma is an extending result obtained by Lü and Qi in [16].

Lemma 15 (see [16]). Let be a meromorphic function of hyperorder . Then, for any , there exists a sequence such that for large enough , where if or is an arbitrary positive number.

Lemma 16 (see [5]). Let be an entire function with ; then for each , there exist points , such that

Lemma 17 (see [17]). Let be a nonconstant entire function with order , let be an integer, and let be a nonzero finite value. If and , then , where is a constant.

Lemma 18 (see [1]). Suppose that and are two nonconstant meromorphic functions in the complex plane with and as their orders, respectively. Then

The following lemma is from the proof of Theorem 2 in [18] (see pages 493–495 in [18]). It plays an important role in the proof of Theorem 13.

Lemma 19 (see [18]). Let and be two meromorphic functions of finite order such that both and have finitely many poles, and have no common poles, and the order of is less than the order of . Let and , let be a nonzero polynomial, and let Q be a polynomial. If is a solution of differential equation then reduces to a constant and reduces a constant.

3. Proof of Theorem 4

In the proof, we use some ideas of [8, 19–21]. The proof of Theorem 4 is as follows.

Noting that , thus . So we just need to obtain .

On the contrary, suppose that . Take such that , and set . Then(I),(II).

Put and . Here is a differential polynomial about and and is a positive number. Set . Obviously, . By Lemma 15, then for , there exists a sequence as such that

Noting that has at most finitely many zeros, then there exists a positive number such that has no poles in .

In view of as , without loss of generality, we may assume for all . Define and then every is analytic in . Now, fix . If , then . It is clear from (I) that . Hence (for large enough) Also as . It follows from Marty’s criterion that is not normal at .

Therefore, we can apply Lemma 14. Choosing an appropriate subsequence of if necessary, we may assume that there exist sequences and with and such that sequence is defined bylocally uniformly in , where is a nonconstant entire function of order at most 1. Moreover, for all andfor a positive number .

We claim thatlocally uniformly in .

Using the mathematical induction, we prove the claim as follows.

From (21), we have

Noting that and , we have and . In view of the definition of order, we havewhere is a positive constant and is an integer. Noting that , we have . Then, combining (22) and (25) yieldsFrom (24) and (26), we deduce thatlocally uniformly in , which implies that the claim is right when .

We assume that the claim is also right when ; that is,locally uniformly in .

Define . Thenlocally uniformly in . Note the fact that

We immediately derive thatlocally uniformly in , which finishes the proof of the claim.

Furthermore, we havelocally uniformly in .

We claim that(1),(2).

Suppose that ; then by Hurwitz’s theorem there exist , , such that (for sufficiently large) Thus ; by we have

By (27), we derive that which implies that .

To prove (2), suppose that . We know ; otherwise is a contradiction. Hence by (27) and Hurwitz’s theorem, there exist such that (for sufficiently large) It is obvious from (II) that . By (23), (32), and , similarly with (25) and (26), we obtainwhich yields (2). By Lemma 17, it is easy to deduce that , where is a constant; then , which is also a contradiction.

Thus, we complete the proof of Theorem 4.

4. Proof of Theorem 9

Set . Then(I),(II).

Put and . Here , and are two rational functions, , is a positive number. Set .

Noting that has at most finitely many zeros and has finitely many poles, then there exists a positive number such that has no poles in .

If , by Lemma 16, for every , there exist , such that, for ,

In view of as , without loss of generality, we may assume for all . Define and then every is analytic in . Now, fix . If , then . It is clear from that . Hence (for large enough) Also as . It follows from Marty’s criterion that is not normal at .

Therefore, we can apply Lemma 14. Choosing an appropriate subsequence of if necessary, we may assume that there exist sequences and with and such that sequences is defined bylocally uniformly in , where is a nonconstant entire function of order at most 1. Moreover, for all andfor a positive number .

We claim thatlocally uniformly in .

Using the mathematical induction, we prove the claim as follows.

From (41), we have

Noting that and , we have

By (38) and (42), combining with and , we deduce thatwhere = = is a fixed constant. Then, combining (38), (42), (45), and (46) yieldsFrom (44) and (47), we deduce thatlocally uniformly in , which implies that the claim is right when .

We assume that the claim is also right when ; that is,locally uniformly in .

Define . Thenlocally uniformly in . Note the fact that

We immediately derive thatlocally uniformly in , which finishes the proof of the claim.

Furthermore, we havelocally uniformly in .

We claim that

(1) ,

(2) .

The proof of (1) is exactly the same as in the proof of Theorem 4. To prove (2), just replace (53) in the previous proof byHere is also a fixed constant.

By (38) and (42), combining with and , we deduce thatwhich yields (2). From Lemma 17, it is easy to deduce that , where is a constant; then , which is also a contradiction.

So . Next we will prove .

If , noting that , by Lemma 18, then . Due to , we have . Thus, by Lemma 18, . Then, it follows from that .

If , noting that , we obtain .

If , due to , hence . Because is a rational function, we get . By Lemma 18, we may obtain , a contradiction.

At last, we obtain .

Thus, we complete the proof of Theorem 9.

5. Proof of Theorem 13

Now we distinguish two cases.

Case 1 (if is a nonconstant polynomial). By the assumption of Theorem 13 and the result of Theorem 9, we easily deduce that is of order at most . DefineThe fact that is transcendental and implies .
Because of , it is easy to obtain that has no zeros. With the assumption , we derive that has finitely many multiple zeros. We know that the possible poles of are from the multiple zeros of and the poles of ; thus has finitely many poles. Moreover, from (56), we have . Therefore, we can setwhere is a nonzero polynomial and is at most .
Set and . We obtainBy Lemma 19, we get and ; both are constants. Thus is a nonzero constant. Set , rewriting (56) aswhere .
If , then and , where is a nonzero constant. So . Since is a nonconstant polynomial, we get . It is a contradiction with the condition .
Next, we consider . Differentiating (59) times yields Furthermore, we getSetObviously, is a rational function.
Suppose is a zero of . Substituting and into (61), we get ; this shows that is also a zero of .
Next, we will prove that . Otherwise, we assume that . Combining the above discussion and the fact that has finitely many multiple zeros yieldsand it implies that has finitely many zeros. Thus we can set , where is a rational function and is a polynomial and degree at most . Differentiating the above equation leads toBy the fact that has finitely many zeros and the assumption , we deduce that also has finitely many zeros. Noting thatthus or . By the assumptions of Theorem 13, , this is impossible.
If , then . By the definition, we get and also . So we may set or , where is a nonzero constant and are two constants.
If , then . So . It is a contradiction about the condition .
If , then . So (64) changes toand (65) turns intoThus or . Also is impossible. If , then , a contradiction.
At last, we complete the proof ofNext, we distinguish the following subcases.
Subcase 1.1 ( is not a polynomial). Suppose that is a pole of with multiplicity . By (62) and the fact that and is a pole of with multiplicity , it contradicts (68).
Subcase 1.2 ( is nonconstant polynomial). From (62), is a polynomial. By simple calculation, we get . Combining with (68), we deduce . So and . Here is a nonzero constant, and is a constant.

By integrating (59), we havewhere is a nonzero constant and is a polynomial.

So , a contradiction.

Subcase 1.3 ( is a constant). Set . So . As subcase 1.2, we also get . Here is a nonzero constant, and is a constant. Integrating (59) yieldswhere is a constant.

If , by calculation, (here and is an arbitrary constant), and , a contradiction.

So , by calculation; (here and is an arbitrary constant), and , a contradiction.

Case 2. If is a constant, because is a transcendental meromorphic function, obviously the condition and hold.

So Theorem 13 turns into Theorem 2, and the results of Theorem 2 still hold.

The proof is completed.

Competing Interests

The authors declare that they have no competing interests.

Acknowledgments

This work was supported by the NSF of China (11271090), Innovation Program of Shanghai Municipal Education Commission (14YZ164), Projects 13XKJC01 from the Leading Academic Discipline Project of Shanghai Dianji University, NSF of Guangdong Province (2015A030313346 and S2012010010121), Young Talents in Educational Commission of Guangdong Province (2015KQNCX116), National Natural Science Foundation of Guangdong Province (2016A030310257), and Shanxi Scholarship Council of China (no. 2013-045).

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Copyright © 2016 Jianming Qi et al. 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.

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