Abstract
In this paper, we investigate the distribution of the zeros of some high differential-difference polynomials and obtain some uniqueness theorems with regard to it.
MSC:30D35, 34M10.
Advances in Difference Equations volume 2012, Article number: 160 (2012)
In this paper, we investigate the distribution of the zeros of some high differential-difference polynomials and obtain some uniqueness theorems with regard to it.
MSC:30D35, 34M10.
In this paper, a meromorphic function always means it is meromorphic in the whole complex plane \mathbb{C}. We assume that the reader is familiar with the standard notations in the Nevanlinna theory. We use the following standard notations in value distribution theory (see [1, 4, 11, 14]):
And we denote by S(r,f) any quantity satisfying
possibly outside of a set E with finite linear measure, not necessarily the same at each occurrence. A meromorphic function a(z) is said to be a small function with respect to f(z) if T(r,a)=S(r,f). We say that two meromorphic functions f(z) and g(z) share the value a IM (ignoring multiplicities) if f(z)-a and g(z)-a have the same zeros. If f(z)-a and g(z)-a have the same zeros with the same multiplicities, then we say that they share the value a CM (counting multiplicities). A polynomial Q(z,f) is called a differential-difference polynomial in f if Q is a polynomial in f, its derivatives and shifts with meromorphic coefficients, say \{{a}_{\lambda}|\lambda \in I\}, such that T(r,{a}_{\lambda})=S(r,f) for all \lambda \in I. We define the difference operators \mathrm{\Delta}f=f(z+1)-f(z) and {\mathrm{\Delta}}^{n}f={\mathrm{\Delta}}^{n-1}(\mathrm{\Delta}f).
In 2007, Laine and Yang [8] considered zeros of one certain type of difference polynomials and obtained the following theorem.
Theorem A Let f be a transcendental entire function of finite order and c be a nonzero complex constant. If n\ge 2, then {f}^{n}(z)f(z+c)-a has infinitely many zeros, where a\in C\setminus 0.
Liu [13] considered the case of general difference products of a meromorphic function and made relevant improvements to Theorem A. Recently, a number of papers [5, 6] focusing on the distribution of zeros of some difference polynomials of differential types emerged. In this paper, we consider the general differential difference cases in some sense and obtain some results as follows.
Theorem 1 Let f be a transcendental entire function of finite order and \alpha (z)\not\equiv 0 be a small function with respect to f(z). If one of the following conditions holds:
n\ge k+2 when m\le k+1;
n\ge 2k-m+3 when m\ge k+2,
then the differential-difference polynomial {[{f}^{n}(z){f}^{m}(z+1)]}^{(k)}-\alpha (z) has infinitely many zeros.
Theorem 2 Let f be a transcendental entire function of finite order and {\mathrm{\Delta}}^{m}f\not\equiv 0, and \alpha (z)\not\equiv 0 be a small function with respect to f(z). If n\ge k+3, then the differential-difference polynomial {[{f}^{n}(z){\mathrm{\Delta}}^{m}f]}^{(k)}-\alpha (z) has infinitely many zeros.
Theorem 3 Let f be a transcendental meromorphic function of finite order, λ be a nonzero constant, and \alpha (z)\not\equiv 0 be a small function with respect to f(z). If n\ge 2m+2, then the differential-difference polynomial {f}^{n}(z)({f}^{m}(z+1)+\lambda )-\alpha (z) has infinitely many zeros.
Theorem 4 Let f, g be two transcendental entire functions of finite order, λ be a nonzero constant, and \alpha (z)\not\equiv 0 be a small function with respect to f(z). If n\ge 3m+5 and {f}^{n}(z)({f}^{m}(z+1)+\lambda ), {g}^{n}(z)({g}^{m}(z+1)+\lambda ) share \alpha (z) CM, then one of the following cases holds:
f=hg, where {h}^{n}={h}^{m}=1, i.e., if m, n are two prime integers, then f\equiv g;
{f}^{n}(z)({f}^{m}(z+1)+\lambda ){g}^{n}(z)({g}^{m}(z+1)+\lambda )={\alpha}^{2}(z).
Recently, Yang and Laine [10] considered the following interesting differential equation and they proved
Theorem B Let p(z) be a nonvanishing polynomial and b,c\in C be two nonzero complex constants. If p(z) is nonconstant, then the differential equation
admits no transcendental entire solutions. If p(z) is a nonzero constant, then equation above admits three distinct transcendental entire solutions, provided {(\frac{p{b}^{2}}{27})}^{3}=\frac{{c}^{2}}{4}.
They also presented some results on difference analogues of the equation in Theorem B and obtained the following theorem.
Theorem C A nonlinear difference equation
where q(z) is a nonconstant polynomial and b,c\in C are two nonzero constants, has no transcendental entire function of finite order. If q(z) is a nonzero constant, then the difference equation above admits three distinct transcendental entire functions of finite order, provided b=3n\pi and {q}^{3}={(-1)}^{n+1}\frac{27{c}^{2}}{4} for a nonzero integer n.
In this paper, we also consider a more general class of difference equation related to Theorem B and Theorem C and obtain the following results, which generalize the above related results.
Theorem 5 Let p, q be two polynomials and m\ge 2, n be two positive integers. Then the difference equation
has no transcendental entire function of finite order.
Theorem 6 Let q(z) be a nonconstant polynomial and b, c be two nonzero constants. Then the difference equation
has no transcendental entire function of finite order. If q(z) is a nonzero constant, and the difference equation above admits a transcendental entire function of finite order, then
and
where n\in N, k=0,1,\dots ,m-1.
To prove our results, we need some lemmas as follows.
Lemma 1 (see [9])
Let f(z) be a transcendental meromorphic function with finite order. Then
Lemma 2 (see [9])
Let f(z) be a transcendental meromorphic function with finite order σ and η be a nonzero complex number, then for each \epsilon >0, we have
Lemma 3 Let f(z) be a transcendental entire function with finite order and F={f}^{n}(z)\times {f}^{m}(z+1). Then
Proof On the one hand, from Lemma 1, Lemma 2, and the standard Valiant-Mohon’ko theorem, we obtain
on the other hand, by Lemma 2 once again, we obtain
Combing the two equations above, we can get our conclusion immediately. □
Lemma 4 Let f(z) be a transcendental entire function with finite order and {\mathrm{\Delta}}^{m}f\not\equiv 0. Then
Proof It is obvious that the difference operator {\mathrm{\Delta}}^{m}f can be expressed in the following form:
where {\alpha}_{i} (i=1,\dots ,m-1) are some integers. By Lemma 1 and the equation above, we can deduce that
Then from the standard Valiant-Mohon’ko theorem and the equation above, we can obtain
The proof of Lemma 4 is completed. □
Lemma 5 (see [7])
Let f(z) be a transcendental meromorphic solution of finite order σ of a difference equation of the form
where H(z,f), P(z,f), Q(z,f) are difference polynomials in f(z) such that the total degree of H(z,f) in f(z) and its shifts is n and that the corresponding total degree of Q(z,f) is at most n. If H(z,f) just contains one term of maximal total degree, then for any \epsilon >0,
holds possibly outside of an exceptional set of finite logarithmic measure.
Lemma 6 (see [3])
Suppose c is a nonzero constant and α is a nonconstant meromorphic function. Then the differential equation
has no transcendental meromorphic solutions satisfying T(r,\alpha )=S(r,f).
Lemma 7 (see [11])
Suppose that {f}_{1}(z),{f}_{2}(z),\dots ,{f}_{n}(z) (n\ge 2) are meromorphic functions and {g}_{1}(z),{g}_{2}(z),\dots ,{g}_{n}(z) are entire functions satisfying the following conditions:
{\sum}_{j=1}^{n}{f}_{j}(z){e}^{{g}_{j}(z)}\equiv 0;
{g}_{j}(z)-{g}_{k}(z) are not constants for 1\le j<k\le n;
For 1\le j\le n, 1\le h<k\le n. T(r,{f}_{j})=o\{T(r,{e}^{{g}_{h}-{g}_{k}})\} (r\to \mathrm{\infty}, r\notin E).
Then {f}_{j}(z)\equiv 0 (j=1,2,\dots ,n).
Lemma 8 (see [2])
Let f be a nonconstant meromorphic function and p, k be positive integers. Then
Remark {N}_{p}(r,\frac{1}{f}) denotes the counting function of zeros of f where an m-fold zero is counted m times if m\le p and p times if m>p.
Lemma 9 (see [12])
Suppose F and G are two nonconstant meromorphic functions. If F and G share 1 CM, then one of the following three cases holds:
max\{T(r,F),T(r,G)\}\le {N}_{2}(r,\frac{1}{G})+{N}_{2}(r,G)+{N}_{2}(r,\frac{1}{F})+{N}_{2}(r,F)+S(r,F)+S(r,G);
F\equiv G;
F\cdot G\equiv 1.
The proof of Theorem 1 Denote F(z)={f}^{n}(z){f}^{m}(z+1). From the definition of F and Lemma 3, we obtain
which means F is also a transcendental entire function. On the contrary, we suppose that {F}^{(k)}-\alpha (z) has only finitely many zeros, then we obtain
Applying the second main theorem for three small functions to the entire function {F}^{(k)}, we can obtain
And by applying Lemma 8 to the right side of the equation above, we can obtain
That is to say
Combing Equations (2)-(3), we obtain
Next, we discuss the following two cases separately.
Case 1. If m\le k+1, then from the definition of F again, we can obtain
Then, based on the equation above, Equation (4) becomes
which contradicts our assumption n\ge k+2.
Case 2. If m\ge k+2, then in the similar way as in Case 1, we can obtain
Then, based on the equation above, Equation (4) becomes
which also contradicts our assumption n\ge 2k+3-m. Therefore, {F}^{(k)}-\alpha (z) has infinitely many zeros. The proof of Theorem 1 is completed. □
The proof of Theorem 2 Denote F(z)={f}^{n}(z){\mathrm{\Delta}}^{m}f. From the definition of F and Lemma 4, we get
which means F is also a transcendental entire function. On the contrary, suppose that {F}^{(k)}-\alpha (z) has only finitely many zeros. Noting the following fact in Lemma 4 again
we can obtain
Then in the similar way as in Theorem 1, we can obtain
Thus, we can deduce
which contradicts our assumption n\ge k+3. Thus we proved {F}^{(k)}-\alpha (z) has infinitely many zeros. The proof of Theorem 2 is completed. □
The proof of Theorem 3 Denote F(z)={f}^{n}(z)({f}^{m}(z+1)+\lambda ). On the contrary, suppose that F(z)-\alpha (z) has only finitely many zeros. Then from the definition of F, we can obtain obviously that
which means F is also a transcendental entire function. Applying the second main theorem for three small functions to function F, we obtain
In other words,
which contradicts our assumption n\ge 2m+2. So we proved F(z)-\alpha (z) has infinitely many zeros. The proof of Theorem 3 is completed. □
The proof of Theorem 4 Set
Then F and G share 1 CM. From the definition of F, G and Lemma 2, we obtain
and
From the definition of {N}_{2}(r,\frac{1}{F}) and Lemma 2, we obtain
In other words,
In the similar way, we can obtain
Next, we consider three cases according to Lemma 9.
Case 1. Suppose the first equation in Lemma 9 holds. Then by Equations (7)-(8), we obtain
But by Equations (5), (6), and (9), we obtain
which is impossible when n\ge 3m+5.
Case 2. Suppose F\equiv G holds. That is to say
Set h=\frac{f}{g}, and suppose it is a nonconstant meromorphic function.
If {h}^{n}(z){h}^{m}(z+1)\equiv 1, then {h}^{n}(z)=\frac{1}{{h}^{m}(z+1)}. By Lemma 2, we obtain
which is also impossible when n\ge 3m+5. So {h}^{n}(z){h}^{m}(z+1)\not\equiv 1.
Substituting h=\frac{f}{g} into Equation (11), we obtain
From Equation (12), it is obvious that if {h}^{n}(z){h}^{m}(z+1)-1=0, then 1-{h}^{n}(z)=0 and {h}^{m}(z+1)-1=0. Thus, we can obtain
Applying the second main theorem to H:={h}^{n}(z){h}^{m}(z+1) and noting the equation above, we obtain
That is to say,
which is impossible when n\ge 3m+5. Thus, we obtain h is a constant such that {h}^{n}={h}^{m}=1. Furthermore, if (m,n)=1, then h=1 and f\equiv g.
Case 3. Suppose F\cdot G\equiv 1 holds. Then
The proof of Theorem 4 is completed. □
The Proof of Theorem 5 Suppose the difference equation admits a transcendental entire function of finite order and q(z) does not vanish identically without loss of generality. Then, by Lemma 1, we obtain
which is impossible when m>1. The proof of Theorem 5 is completed. □
The Proof of Theorem 6 Suppose the difference equation (1) admits a transcendental entire function of finite order. First of all, we notice that
is a difference polynomial of f with total degree at most one. Differentiating Equation (1), we obtain
where Q={\mathrm{\Delta}}^{m}f. Combining the squares of Equation (1) and Equation (13), we can obtain
where
is a differential-difference polynomial of f with total degree at most four. Then by Lemma 5, we obtain
From Equation (15) and Lemma 6, we obtain {b}^{2}{f}^{2}+9{{f}^{\prime}}^{2} must be a constant. To put it another way,
It is clear that the solutions of homogeneous and linear Equation (16) with constant coefficients must be of the form
where {c}_{1}, {c}_{2} are two constants. Next, we claim {c}_{1}, {c}_{2} are two nonzero constants.
If {c}_{2}=0, then f(z)={c}_{1}{e}^{\frac{biz}{3}} and \mathrm{\Delta}f=Af, {\mathrm{\Delta}}^{m}f={A}^{m}f, where A={e}^{\frac{bi}{3}}-1. Substituting these into Equation (1), we obtain
By Lemma 7, we obtain c={c}_{1}=0, which is impossible. Thus {c}_{2}\ne 0, and we can obtain {c}_{1}\ne 0 in a similar way. Thus, our claim holds.
From Equation (17), we obtain
where A={e}^{\frac{bi}{3}}-1 and B={e}^{\frac{-bi}{3}}-1. Substituting Equation (19) into Equation (1), we can obtain
By Lemma 7 again, we obtain
and
If q(z) is a nonconstant polynomial, then we can obtain a contradiction from Equation (22) obviously.
If q(z) is a nonzero constant, then we can set A=Bx according to Equation (22), where x={e}^{\frac{2k\pi i}{m}}, k=0,1,\dots ,m-1. It is easy to see {e}^{\frac{bi}{3}}=1 or −x.
If {e}^{\frac{bi}{3}}=1, then A=0 and we obtain {c}_{1}{c}_{2}=0 from Equation (22), which is a contradiction. So {e}^{\frac{bi}{3}}\ne 1.
If {e}^{\frac{bi}{3}}=-x, then
where n is an integer. From Equation (22), we obtain
The proof of Theorem 6 is completed. □
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The author would like to thank the main editor and anonymous referees for their valuable comments and suggestions leading to improvement of this paper. This research was supported by the Fundamental Research Funds for the Central Universities (No. 2011QNA25).
The author declares that they have no competing interests.
The author carried out the proof of the theorems and approved the final manuscript.
Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Zhang, J. Some results on zeros and the uniqueness of one certain type of high difference polynomials. Adv Differ Equ 2012, 160 (2012). https://doi.org/10.1186/1687-1847-2012-160
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DOI: https://doi.org/10.1186/1687-1847-2012-160