We will first show that the following differential equation 𝐹(π‘˜)βˆ’π‘§=𝑒𝛼(πΉβˆ’π‘§) has transcendental entire solutions, where 𝐹=𝑓𝑛 and 𝛼 is an entire function, which improve what we have known and answer an open question raised in the work of Zhang and Yang (2009). And then, the examples are discussed.

1. Introduction and Main Results

In this paper, a meromorphic (resp., entire) function means meromorphic (resp., analytic) in the whole complex plane. We will adopt the standard notations in Nevanlinna's value distribution theory of meromorphic functions, such as the characteristic function 𝑇(π‘Ÿ,𝑓), the counting function of the poles 𝑁(π‘Ÿ,𝑓), the proximity function π‘š(π‘Ÿ,𝑓), and the reduction counting function 𝑁(π‘Ÿ,𝑓) (see [1–3]). In addition, 𝑆(π‘Ÿ,𝑓) denotes any quantity satisfying 𝑆(π‘Ÿ,𝑓)=π‘œ{𝑇(π‘Ÿ,𝑓)} as π‘Ÿβ†’βˆž, possibly outside a set of π‘Ÿ of finite linear measure.

Let 𝑓(𝑧) and 𝑔(𝑧) be two nonconstant meromorphic functions, β‹ƒπ‘ŽβˆˆπΆ{∞}. We say that 𝑓 and 𝑔 share the value π‘Ž IM if π‘“βˆ’π‘Ž and π‘”βˆ’π‘Ž have the same zeros. If π‘“βˆ’π‘Ž and π‘”βˆ’π‘Ž have the same zeros with the same multiplicities, we say that they share the value π‘Ž CM. Moreover, if π‘“βˆ’π‘§ and π‘”βˆ’π‘§ share 0 CM (resp., IM), we say that 𝑓 and 𝑔 share 𝑧 CM (resp., IM), or we say that 𝑓 and 𝑔 have the same fixed-points CM (resp., IM).

In 2008, Lei et al. [4] proved the follwing.

Theorem A. Suppose that 𝑓 is a nonconstant meromorphic function, 𝑛, π‘˜ are positive integers, and 𝑛β‰₯π‘˜+5,β€‰β€‰π‘Ž is a nonzero constant. If 𝑓𝑛 and (𝑓𝑛)(π‘˜) share π‘Ž CM, then 𝑓=𝑐e𝑀𝑧/𝑛, where 𝑐,𝑀 are nonzero constants such that π‘€π‘˜=1.

Recently, Zhang and Yang investigated the power of an entire function sharing one value with its derivative and obtained the following results.

Theorem B (see [5]). Let 𝑓 be a nonconstant entire function, and let 𝑛β‰₯7 be an integer. If 𝑓𝑛 and (𝑓𝑛)ξ…ž share 1 CM, then 𝑓𝑛=(𝑓𝑛)ξ…ž, and 𝑓=𝑐e𝑧/𝑛, where 𝑐 is a nonzero constant.

Theorem C (see [6]). Let 𝑓 be a non-constant entire function, and let 𝑛 be a positive integer. If 𝑓𝑛 and (𝑓𝑛)(π‘˜) share 1 CM, and 𝑛β‰₯π‘˜+1, then 𝑓𝑛=(𝑓𝑛)(π‘˜), and 𝑓=𝑐e𝑀𝑧/𝑛, where 𝑐,𝑀 are nonzero constants such that π‘€π‘˜=1.

According to the above theorems, one may ask an interesting question: what can be said β€œif 𝑓𝑛 and (𝑓𝑛)(π‘˜) share 𝑧 CM”? Our purpose of this paper is to solve this question by giving the transcendental entire solutions of the equation πΉξ…žβˆ’π‘§=e𝛼(πΉβˆ’π‘§),(1.1) where 𝐹=𝑓𝑛, and 𝛼 is an entire function. As an application, we now use a completely different method from that in [4–6] and give the following results.

Theorem 1.1. Let 𝑛(𝑛β‰₯2) be a positive integer, and 𝛼 be an entire function, then (1.1) has transcendental entire solutions 𝑓 and 𝑓=𝑐e𝑧/𝑛,(1.2) where c is a nonzero constant.

Obviously, from Theorem 1.1, we have the following result.

Corollary 1.2. Let 𝑓 be a transcendental entire function, and let 𝑛(𝑛β‰₯2) be a positive integer. If 𝑓𝑛 and (𝑓𝑛)ξ…ž share 𝑧 CM, then the conclusion of Theorem 1.1 is valid.

Obviously, Corollary 1.2 improves the results in [6] and answers the open question in [7].

In order to illustrate our condition 𝑛β‰₯2 is sharp, we give examples as follows.

Example 1.3. Let 𝑓=eeπ‘§βˆ«π‘§0eβˆ’e𝑑(1βˆ’e𝑑)𝑑d𝑑, then 𝑓 is a non-constant solution of π‘“ξ…žβˆ’π‘§=e𝑧(π‘“βˆ’π‘§),(1.3) and 𝑓′, 𝑓 share 𝑧 CM, while π‘“β‰’π‘“ξ…ž.

Example 1.4. Let 𝑓=e3𝑧+2𝑧/3+2/9, then 𝑓 is a non-constant solution of π‘“ξ…žβˆ’π‘§=3(π‘“βˆ’π‘§),(1.4) and 𝑓, 𝑓′ share 𝑧 CM, but 𝑓≒𝑓′.

Theorem 1.5. Let 𝑓 be a transcendental entire function, 𝑛, π‘˜ positive integers, and 𝑛β‰₯π‘˜+2; if 𝑓𝑛, (𝑓𝑛)(π‘˜) share 𝑧 IM, then 𝑓=𝑐e𝑀𝑧/𝑛, where 𝑐, 𝑀 are nonzero constants and π‘€π‘˜=1.

2. Preliminaries

In order to prove the theorems above, we need some lemmas.

Lemma 2.1 (see [8, 9]). Suppose that 𝑓 is a meromorphic and transcendental in the whole complex plane and that 𝑓𝑛𝑃(𝑓)=𝑄(𝑓),(2.1) where 𝑃(𝑓) and 𝑄(𝑓) are differential polynomials in 𝑓 with functions of small proximity related to 𝑓 as the coefficients, and the degree of 𝑄(𝑓) is at most 𝑛, then π‘š(π‘Ÿ,𝑃(𝑓))=𝑆(π‘Ÿ,𝑓).(2.2)

Lemma 2.2 (see [1, 10]). Let 𝑓 be a transcendental meromorphic function, then π‘šξ‚΅π‘“π‘Ÿ,(𝑙)𝑓=𝑆(π‘Ÿ,𝑓),(2.3) for every positive integer 𝑙.

Lemma 2.3 (see [1, 3]). Let 𝑓 be a meromorphic function, and 𝛼𝑖(𝑧) with 𝑖=1,2,3 are three distinct small functions of 𝑓(𝑧), then 𝑇(π‘Ÿ,𝑓)≀𝑁1π‘Ÿ,π‘“βˆ’π›Ό1ξ‚Ά+𝑁1π‘Ÿ,π‘“βˆ’π›Ό2ξ‚Ά+𝑁1π‘Ÿ,π‘“βˆ’π›Ό3ξ‚Ά+𝑆(π‘Ÿ,𝑓).(2.4)

3. Proof of Theorem 1.1

By differentiation to (1.1), we have πΉξ…žξ…žβˆ’1=π›Όξ…že𝛼(πΉβˆ’π‘§)+eπ›Όξ€·πΉξ…žξ€Έβˆ’1.(3.1) Combining (1.1) and (3.1) yields ξ€·πΉξ…žξ…žξ€Έβˆ’1(πΉβˆ’π‘§)=π›Όξ…žξ€·πΉξ…žξ€Έξ€·πΉβˆ’π‘§(πΉβˆ’π‘§)+ξ…žπΉβˆ’π‘§ξ€Έξ€·ξ…žξ€Έβˆ’1,(3.2) that is, πΉξ…žξ…žπΉβˆ’π›Όξ…žπΉπΉξ…žβˆ’ξ€·πΉξ…žξ€Έ2=ξ€·1βˆ’π›Όξ…žπ‘§ξ€ΈπΉ+π‘§πΉξ…žξ…žβˆ’ξ€·π›Όξ…žξ€ΈπΉπ‘§+𝑧+1ξ…ž+π›Όξ…žπ‘§2.(3.3) Substituting πΉξ…ž=π‘›π‘“π‘›βˆ’1π‘“ξ…ž,  𝐹=𝑓𝑛 into (3.3) results in π‘“π‘›π‘“π‘›βˆ’2𝑃=𝑄,(3.4) where 𝑃=π‘›π‘“π‘“ξ…žξ…žβˆ’π‘›π›Όξ…žπ‘“π‘“ξ…žξ€·π‘“βˆ’π‘›ξ…žξ€Έ2,𝑄=1βˆ’π›Όξ…žπ‘§ξ€Έπ‘“π‘›ξ‚†+𝑧𝑛(π‘›βˆ’1)π‘“π‘›βˆ’2ξ€·π‘“ξ…žξ€Έ2+π‘›π‘“π‘›βˆ’1π‘“ξ…žξ…žξ‚‡ξ€·π›Όβˆ’π‘›ξ…žξ€Έπ‘“π‘§+𝑧+1π‘›βˆ’1π‘“ξ…ž+π›Όξ…žπ‘§2(3.5) are differential polynomials in 𝑓, and the degree of 𝑄 is 𝑛. Lemma 2.1 gives π‘š(π‘Ÿ,π‘“π‘›βˆ’2𝑃)=𝑆(π‘Ÿ,𝑓) and π‘‡ξ€·π‘Ÿ,π‘“π‘›βˆ’2𝑃=𝑆(π‘Ÿ,𝑓).(3.6) We now prove 𝑃≑0. Suppose that 𝑃≒0, and 𝑛β‰₯3, then by (3.6) and Lemma 2.1, we obtain 𝑇(π‘Ÿ,𝑓)=𝑆(π‘Ÿ,𝑓), which is impossible. We may now assume that 𝑛=2. So, we get 𝑃=2π‘“π‘“ξ…žξ…žβˆ’2π›Όξ…žπ‘“π‘“ξ…žξ€·π‘“βˆ’2ξ…žξ€Έ2.(3.7) It follows from (3.4) and Lemma 2.2 that 𝑇(π‘Ÿ,𝑃)=𝑆(π‘Ÿ,𝑓), and π‘š(π‘Ÿ,𝑃/𝑓2)=𝑆(π‘Ÿ,𝑓). These equalities above show that π‘šξ‚΅1π‘Ÿ,𝑓=𝑆(π‘Ÿ,𝑓).(3.8) In addition, we can see from the expression of 𝑃 that the multiple zeros of 𝑓 must be the zeros of 𝑃; hence, 𝑁(π‘Ÿ,1/𝑓)=𝑁(π‘Ÿ,1/𝑓)+𝑆(π‘Ÿ,𝑓). This together with (3.8) and the first theorem (e.g., see [2, Theorem 1.2]) will result in 𝑇(π‘Ÿ,𝑓)=𝑁1π‘Ÿ,𝑓+𝑆(π‘Ÿ,𝑓).(3.9) By (3.7), we find π‘ƒξ…ž=2π‘“π‘“ξ…žξ…žξ…žβˆ’2π›Όξ…žξ…žπ‘“π‘“ξ…žβˆ’2π›Όξ…žπ‘“π‘“ξ…žξ…žβˆ’2π›Όξ…žξ€·π‘“ξ…žξ€Έ2βˆ’2π‘“ξ…žπ‘“ξ…žξ…ž.(3.10) Let 𝑧0 be a simple zero of 𝑓; it follows from (3.7) and (3.10) that 𝑃(𝑧0)+2{π‘“ξ…ž(𝑧0)}2=0 and π‘ƒξ…ž(𝑧0)=βˆ’2π›Όξ…ž(𝑧0){π‘“ξ…ž(𝑧0)}2βˆ’2π‘“ξ…ž(𝑧0)π‘“ξ…žξ…ž(𝑧0), which implies that 𝑧0 is a zero of π‘ƒπ‘“ξ…žξ…ž+(π›Όξ…žπ‘ƒβˆ’π‘ƒξ…ž)π‘“ξ…ž. Let 𝑔=π‘ƒπ‘“ξ…žξ…ž+ξ€·π›Όξ…žπ‘ƒβˆ’π‘ƒξ…žξ€Έπ‘“ξ…žπ‘“.(3.11) Thus, 𝑇(π‘Ÿ,𝑔)=𝑆(π‘Ÿ,𝑓).(3.12) By (3.11), we obtain π‘“ξ…žξ…ž=𝛼1𝑓+𝛽1π‘“ξ…ž,(3.13) where 𝛼1=𝑔/𝑃, 𝛽1=(π‘ƒβ€²βˆ’π›Όβ€²π‘ƒ)/𝑃,π‘‡ξ€·π‘Ÿ,𝛼1ξ€Έξ€·=𝑆(π‘Ÿ,𝑓),π‘‡π‘Ÿ,𝛽1ξ€Έ=𝑆(π‘Ÿ,𝑓).(3.14) Substituting (3.13) into (3.7) will yield 𝑃=2𝛼1𝑓2𝛽+21βˆ’π›Όξ…žξ€Έπ‘“π‘“ξ…žξ€·π‘“βˆ’2ξ…žξ€Έ2.(3.15) Also from (3.13), we find π‘“ξ…žξ…žξ…ž=𝛼2𝑓+𝛽2π‘“ξ…ž,(3.16) where 𝛼2=π›Όξ…ž1+𝛼1𝛽1, 𝛽2=𝛼1+π›½ξ…ž1+𝛽21,π‘‡ξ€·π‘Ÿ,𝛼2ξ€Έξ€·=𝑆(π‘Ÿ,𝑓),π‘‡π‘Ÿ,𝛽2ξ€Έ=𝑆(π‘Ÿ,𝑓).(3.17) It follows from (3.16), (3.13), and (3.10) that π‘ƒξ…žξ€·π›Ό=22βˆ’π›Όξ…žπ›Ό1𝑓2𝛽+22βˆ’π›Όξ…žπ›½1βˆ’π›Όξ…žξ…žβˆ’π›Ό1ξ€Έπ‘“π‘“ξ…žξ€·π›Όβˆ’2ξ…ž+𝛽1π‘“ξ€Έξ€·ξ…žξ€Έ2.(3.18) From (3.15) and (3.18), we get π‘ƒξ…žξ€·π›Ό=22βˆ’π›Όξ…žπ›Ό1𝑓2𝛽+22βˆ’π›Όξ…žπ›½1βˆ’π›Όξ…žξ…žβˆ’π›Ό1ξ€Έπ‘“π‘“ξ…žξ€·π›Όβˆ’2ξ…ž+𝛽1𝛼1𝑓2+𝛽1βˆ’π›Όξ…žξ€Έπ‘“π‘“ξ…žβˆ’π‘ƒ2.(3.19) As noted from 𝑃′=(𝛼′+𝛽1)𝑃 and the definitions of 𝛼2 and 𝛽2, we can deduce from (3.19) that ξ€·π›Όξ…ž1βˆ’2𝛼1π›Όξ…žξ€Έξ‚†π›½π‘“+ξ…ž1βˆ’π›Όξ…žξ…žβˆ’π›Όξ…žπ›½1+ξ€·π›Όξ…žξ€Έ2ξ‚‡π‘“ξ…ž=0.(3.20) Suppose that π›Όξ…ž1βˆ’2𝛼1𝛼′≑0, then 𝛼1=𝑐e2𝛼, where 𝑐 is a nonzero constant. If e𝛼≑1, then (1.1) gives 𝑓=𝑐e𝑧/𝑛, which is the desired result. If e𝛼≒1, this together with (1.1) will lead to 𝑁(π‘Ÿ,1/𝑓)=𝑆(π‘Ÿ,𝑓), which contradicts with (3.9). We now assume that π›Όξ…ž1βˆ’2𝛼1𝛼′≒0, in the same way, (3.20) shows that 𝑁(π‘Ÿ,1/𝑓)=𝑆(π‘Ÿ,𝑓), which also contradicts with (3.9). Hence, 𝑃≑0. Then, it follows from (3.4) that 𝑄≑0 and from (3.3) that πΉξ…žξ…žπΉβˆ’π›Όξ…žπΉξ…žξ€·πΉπΉβˆ’ξ…žξ€Έ2=0.(3.21) Obviously, we obtain from (3.21) that πΉξ…žξ…žπΉξ…ž=π›Όξ…ž+πΉξ…žπΉ.(3.22) By integration, we have 𝐹′=𝑑𝐹e𝛼, where 𝑑 is a nonzero constant. Substituting this into (1.1) will yield (π‘‘βˆ’1)𝑓𝑛=𝑧(1βˆ’e𝛼)e𝛼.(3.23) If 𝑑=1, by this and (1.1), we have πΉξ…ž=𝐹, that is, 𝑓=𝑐e𝑧/𝑛, where 𝑐 is a nonzero constant.

If 𝑑≠1, suppose that 𝑓 has infinitely many zeros, and e𝛼 is not a constant, then the zeros of 𝑓 are the zeros of 1βˆ’e𝛼, with multiplicities at least 𝑛. It follows from the equation above and the second fundamental theorem (or Lemma 2.3) that 𝑇(π‘Ÿ,e𝛼)≀𝑁(π‘Ÿ,e𝛼)+𝑁1π‘Ÿ,e𝛼+𝑁1π‘Ÿ,eπ›Όξ‚βˆ’1+𝑆(π‘Ÿ,e𝛼),(3.24) which with 𝑛β‰₯2 may lead to a contradiction. Now, we suppose that 𝑓 has infinitely many zeros, and e𝛼 is a constant. We suppose that e𝛼=𝐡, then we can get a contradiction.

If 𝑑≠1, suppose that 𝑓 has finitely many zeros, and let 𝑓=π‘š(𝑧)e𝛽,(3.25) where π‘š(𝑧) is a polynomial and 𝛽 is an entire function. In this case, (3.23) can be written as (1βˆ’π‘‘)π‘šπ‘›π‘§e𝑛𝛽+eβˆ’π›Ό=1.(3.26)

Applying the second fundamental theorem (or Lemma 2.3) to (3.26), we have 𝑛(𝑇(π‘Ÿ,𝑓)=π‘‡π‘Ÿ,e𝑛𝛽+𝑆(π‘Ÿ,𝑓)=𝑆(π‘Ÿ,𝑓),(3.27) which is impossible. This also proves Theorem 1.1.

Remark 3.1. By examining the proof of Theorem 1.1 more carefully, here as an extension of Theorem 1.1, we would like to pose the following.

Conjecture 3.2. Let 𝑓 be a transcendental entire function, and let 𝑛 be a positive integer. If 𝑓𝑛 and (𝑓𝑛)(π‘˜) share 𝑧 CM, and 𝑛β‰₯π‘˜+1, then 𝑓𝑛=(𝑓𝑛)(π‘˜), 𝑓=𝑐e𝑀𝑧/𝑛,(3.28) where 𝑐, 𝑀 are nonzero constants such that π‘€π‘˜=1.

4. Proof of Theorem 1.5

Let 𝐹=𝑓𝑛, if 𝐹≒𝐹(π‘˜), Lemma 2.3 implies 𝑇(π‘Ÿ,𝐹)≀𝑁(π‘Ÿ,𝐹)+𝑁1π‘Ÿ,𝐹+𝑁1π‘Ÿ,ξ‚β‰€πΉβˆ’π‘§+𝑆(π‘Ÿ,𝐹)𝑁1π‘Ÿ,𝑓+𝑁1π‘Ÿ,𝐹(π‘˜)≀/πΉβˆ’1+𝑆(π‘Ÿ,𝑓)𝑁1π‘Ÿ,𝑓𝐹+π‘‡π‘Ÿ,(π‘˜)𝐹≀+𝑆(π‘Ÿ,𝑓)𝑁1π‘Ÿ,𝑓𝐹+π‘π‘Ÿ,(π‘˜)𝐹≀+𝑆(π‘Ÿ,𝑓)𝑁1π‘Ÿ,𝑓+π‘˜π‘ξ‚΅1π‘Ÿ,𝑓≀+𝑆(π‘Ÿ,𝑓)(π‘˜+1)𝑇(π‘Ÿ,𝑓)+𝑆(π‘Ÿ,𝑓),(4.1) which contradicts with 𝑛β‰₯π‘˜+2, and hence 𝐹≑𝐹(π‘˜), and our desired result follows.

Finally, we conclude the paper with the following.

Question 1. Let 𝑅 be a rational function, and let 𝛼 be an entire function, then for two integers 𝑛, π‘˜ such that 𝑛β‰₯π‘˜+1, what can be said about the transcendental entire solution of (𝑓𝑛)(π‘˜)βˆ’π‘§=𝑅e𝛼(π‘“π‘›βˆ’π‘§), can 𝑓 be expressed as 𝑓=𝑐e𝑀𝑧/𝑛 for some constants 𝑐(β‰ 0), π‘€π‘˜=1?

Question 2. What can be said if the condition in Theorem 1.5 β€œπ‘›β‰₯π‘˜+2” is replaced by β€œπ‘›β‰₯π‘˜+1”?


This works was supported by the Fundamental Research Funds for the Central Universities (no. 10CX04038A).