Abstract

Relying on the normal family theory, we mainly study uniqueness problems of meromorphic functions and their th derivatives and estimate sharply the growth order of their meromorphic functions. Our theorems improve some previous results.

1. Introduction and Main Results

In this paper, a meromorphic function means a function that is meromorphic in the whole complex plane . We will use the fundamental results and the standard notation of the Nevanlinna theory of meromorphic functions such as , , and as explained in [1].

Let be a rational function (or a polynomial) that belongs asymptotically to as , where are constants. The degree of at infinity is defined as .

Suppose that and are two nonconstant meromorphic functions. We define as a meromorphic function or a finite complex number. In this paper, we define whenever ; then we write . If and , then we write and say that and share IM (ignoring multiplicity). If and have the same zeros with the same multiplicities, we write and say that and share CM (counting multiplicity).

The subject on sharing values between entire functions and their derivatives was first studied by Rubel and Yang (cf. [2]). In 1977, they proved a result that if a nonconstant entire function and share two distinct finite numbers CM, then . Since then, sharing value problems have been studied by many authors and a number of profound results have been obtained (cf. [1, 3, 4]). To state our main results, we also need the following concepts.

Definition 1. The order and the hyperorder of a meromorphic function are both defined as follows: On estimating the order or hyperorder of meromorphic functions, there has been a lot of work on the growth order of meromorphic function to certain types of complex differential equations and complex difference equations (or complex functional equations) (cf. [3, 58]).
In 2011, Lü and Yi ([9]) investigated the estimating order of meromorphic functions with their derivatives about sharing values and obtained the following results.

Theorem A. Let be a nonconstant meromorphic function with finitely many poles, and let and , where and are three polynomials. Let be a positive integer. If all the zeros of have multiplicity at least and then .

Theorem B. Let and be two distinct polynomials, let be a positive integer, and let be a transcendental entire function with all the zeros of having multiplicity at least . If and , then one of the following cases must occur: (1)(2) and , where and are two nonzero constants.

In this paper, we continue to investigate the growth order and hyperorder of meromorphic functions with their derivatives about sharing functions and improve Theorems A and B. Now, we state our results as follows.

Theorem 2. Let be a nonconstant meromorphic function with finitely many poles, and let and , where are two rational functions and is an entire function. Let be a positive integer. If all the zeros of have multiplicity at least and then .

Remark 3. The following examples show that the conclusion is sharp.

Example 4. Let , where is a nonzero constant. Let and . Noting that , one has Thus, it satisfies the assumptions of Theorem 2 and .

Example 5. Let , , and , where is a nonzero constant. Differentiating yields , and then and . Thus, .

Theorem 6. Let be a nonconstant meromorphic function with finitely many poles, and let and , where , and are three polynomials with , , and , where are three nonnegative integers. Let be a positive integer. If all the zeros of have multiplicity at least and then .

Remark 7. The next two examples can illustrate that the conclusion of Theorem 6 is meaningful.

Example 8. Let , , and , where is a nonzero constant. Differentiating twice yields and , and then and . Thus, it satisfies the assumption of Theorem 6. Noting that , , and , we obtain that holds.

Example 9. Let , , and , where is a nonzero constant. Differentiating twice yields and , and then and . Thus, it satisfies the assumption of Theorem 6. We have . We obtain that holds.

Theorem 10. Let be a transcendental meromorphic function with finitely many poles, and let and , where , and are three polynomials. Let be a positive integer. If all the zeros of have multiplicity at least and and then the conclusions of Theorem B still hold, and must be a constant.

The next example will show if is rational; then Theorem 10 fails.

Example 11. Let , , and . Note that However, it does not satisfy any case of Theorem B.

Remark 12. The condition plays an important pole in the proof of Theorem 10. Unfortunately, we do not know whether the condition could be weakened.

2. Some Lemmas

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

A family of meromorphic functions defined on a domain (here ) is said to be normal in if every sequence of elements of contains a subsequence that converges locally uniformly in with respect to the spherical metric to a meromorphic function or (cf. [10]).

Lemma 13 (see [3, 11]). Let be a family of meromorphic (analytic) functions on the unit disc . If , , and , then there exist(a)a subsequence of (which we still write as );(b)points , ;(c)positive numbers , such that locally uniformly, where is a nonconstant meromorphic (entire) function on , such that where is a constant that is independent of .
Here, as usual, is the spherical derivative.

The next lemma is an extending result obtained by Lü and Qi in [12].

Lemma 14. 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 15 (cf. [5]). Let be an entire function; . Then, for each , there exist points , such that

Lemma 16 (cf. [1]). Suppose that and are two nonconstant meromorphic functions in the complex plane with and as their orders, respectively. Then

Lemma 17 (cf. [13]). Let and be meromorphic functions of finite order such that both of and have only finite many poles, and have no common poles, and the order of is less than the order of . If and share CM, then for some nonzero constant c.

3. The Proof of Theorem 2

By the method in [3, 12, 1416], we will present our proof as follows.

Note that ; then . It suffices to show .

By reduction to absurdity, we assume that . Set , and set ; then(i),(ii),

where and is a differential polynomial of and . Let We may set and get . By Lemma 14, for , there exists a sequence as such that Note that has at most finitely many zeros and poles and has at most finitely many poles. Then there exists a positive number such that has no poles in .

In view of as , without loss of generality, we assume that for all . Define and and then every is analytic in . Also as . It follows from Marty’s criterion that is not normal at .

By Lemma 13 and choosing an appropriate subsequence of if necessary, we may assume that there exist sequences and with and such that the sequence is defined bylocally uniformly in , where is a nonconstant entire function of order at most 1, whose zeros have multiplicity at least andfor a positive number .

We assert that

We prove the assertion by induction. By (15), the assertion holds for . We assume that the assertion is right for ; that is, Now we will prove that the assertion is also right for . Let and thenRecall the above definition of degree of rational function; we haveand here is a positive constant and is an integer. Noting that , we have . Then, combining (16) and (21), we conclude thatThen Hence, we complete the proof of the assertion.

Obviously, . Otherwise would be a polynomial of degree less than and could not have zeros of multiplicity at least . In the following, we will prove that(a),(b).

Firstly, we prove (a). Suppose that . By Hurwitz’s theorem, there exists a sequence , , such that (for being sufficiently large)Thus, and

By (16), (17), (21), and (22), we derive thatThus, . So (a) holds.

Similarly, we can prove . Now we prove From (16) and (17), we haveSuppose that . Then . Hence, by (27) and Hurwitz’s theorem, there exists a sequence , , such that (for being sufficiently large) Then, it follows from the assumption that we get . It implies that Thus, we prove that and obtain (b).

From (a) and (b), it is easy to deduce that . Thus, ; it contradicts with . So . Therefore, we complete the proof of Theorem 2.

4. The Proof of Theorem 6

Using some ideas of [3, 14], we give the following proof of Theorem 6.

Let . Then all the zeros of have multiplicities at least and where and is a polynomial. A careful calculation leads us to obtain . Let where is a polynomial. Set .

If , by Lemma 15, for each , there exist such that, for ,In view of as , without loss of generality, we may assume that for all . Define and Note that has at most finitely many zeros and has at most finitely poles; then there exists a positive number such that has no poles in , so is analytic in . Therefore, all are analytic in . Thus, we have structured a family of holomorphic functions.

Also as . It follows from Marty’s criterion that is not normal at . Therefore, by applying Lemma 13 and choosing an appropriate subsequence of if necessary, we may assume that there exist subsequences and with and such that the sequence is defined bylocally uniformly in , where is a nonconstant entire function of order at most 1, whose zeros have multiplicity at least andfor a positive number .

We assert that

We will use the mathematical induction to prove the assertion. By (34), the assertion is right for . We assume that the assertion is right for ; that is, Now we prove that the assertion is also right when . Let and thenSince , we have In view of the degree of a polynomial function, one haswhere is a nonnegative integer. Then, combining (35) and (41) yieldsFurthermore, So we complete the proof of the assertion.

Obviously, . Otherwise, would be a polynomial of degree less than and so could not have zeros of multiplicity at least . In the following, we will prove

(c) ,

(d) .

Firstly, we prove (c). Suppose that . Then, by Hurwitz’s theorem, we get that there exists a sequence , , such that (for being sufficiently large)Thus, and . By (36) and (39), we derive thatAlso,where is a nonnegative integer. Then, combining (35) and (46) yieldsSo,Thus, and (c) holds.

Next, we will prove that .

Similar to what was stated above, suppose that . Then, by Hurwitz’s theorem, there exists a sequence , , such that (for being sufficiently large)Thus, and By (35) and (36), we deduce that

By calculation, where is a nonnegative integer. Then, combining (35) and (51) shows thatSo,It implies that . Now, we prove From (35), (36), (51), and (52), we haveSuppose that . Since , by (55) and Hurwitz’s theorem, there exists a sequence , , such that (for being sufficiently large) By the assumption, we get and Thus, we prove and obtain (d).

From (c) and (d), it is easy to deduce that . Thus, . But it contradicts with . So .

Next, we will prove .

If , noting the fact that and Lemma 16, then . Rewrite the function as . Thus, by Lemma 16, we conclude that . Furthermore, it follows from that .

If , noting that , then .

If , noting that and is a rational function, then and . So, .

Therefore, holds. This completes the proof of Theorem 6.

5. The Proof of Theorem 10

Based on some ideas in [9], we will prove the theorem.

By Theorem 6, Lemma 17, and the assumption of Theorem 10, we havewhere is a nonzero constant. It follows thatIf , then ; this is result (1) in Theorem B. Assuming that and is a zero of , we have . Substituting into (59), we haveNoting that , we know that all the zeros of are the zeros of , so has only finitely many zeros. Set , where is a nonzero polynomial and is a polynomial. Then , where is a polynomial with and is a polynomial with . Substituting them into (59), we have Then,From (63), we get and . Thus is a constant, say . So and , where is a constant. Furthermore,We deduce that . Since , we get that is a constant. Set as a constant. Due to the fact that and (64), must be a constant. Hence, we complete the proof of Theorem 10.

Conflicts of Interest

The authors declare that there are no conflicts of interest.

Acknowledgments

This work was supported by NNSF of China (11501367 and 1701111), NSF of Guangdong Province (2016A030310257), and Leading Academic Discipline Project of Shanghai Dianji University (16JCXK02).