Some results about a special nonlinear difference equation and uniqueness of difference polynomial
© Qi et al; licensee Springer. 2011
Received: 25 February 2011
Accepted: 6 September 2011
Published: 6 September 2011
In this paper, we continue to study a special nonlinear difference equation solutions of finite order entire function. We also continue to investigate the value distribution and uniqueness of difference polynomials of meromorphic functions. Our results which improve the results of Yang and Laine [Proc. Jpn. Acad. Ser. A Math. Sci. 83:50-55 (2007)]; Qi et al. [Comput. Math. Appl. 60:1739-1746 (2010) ].
Mathematics Subject Classification (2000): 30D35, 39B32, 34M05.
Let f (z) be a meromorphic function in the whole complex plane ℂ. It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory such as the characteristic function T (r, f ), proximity function m(r, f ), counting function N(r, f ), the first and second main theorem etc.,(see [1, 2]). The notation S(r, f ) denotes any quantity that satisfies the condition: S(r, f ) = o(T (r, f )) as r → ∞ possibly outside an exceptional set of r of finite linear measure. A meromorphic a(z) is called a small function of f (z) if and only if T (r, a(z)) = S(r, f ). A polynomial Q(z, f ) is called a differential-difference polynomial in f whenever f is a polynomial in f (z), its derivatives and its shifts f (z+c), with small functions of f again as the coefficients. Denote Δ c f := f (z+c) - f (z), and for all n ∈ N, n ≥ 2, where c is nonzero complex constant.
Recently, Yang and Laine  proved:
admits no transcendental entire solutions, while if p is constant, then the equation admits three distinct transcendental entire solutions, provided .
entire solutions: f1(z) = sinz, , . It is easy to see the condition given in Theorem A is satisfied.
Very recently, Yang and Laine  present some studies on differential-difference analogues of Equation (1.2) showing that similar conclusions follow if one restricts the solutions to be of finite order. Yang and Laine  obtained:
where q(z) is a nonconstant polynomial and b, c ∈ ℂ are nonzero constants, does not admit entire solutions of finite order. If q(z) = q is a nonzero constant, then Equation (1.3) possesses three distinct entire solutions of finite order, provided b = 3πn and for a nonzero integer n.
has no transcendental entire solutions of finite order.
In this article, by the same method of , we replace f (z + 1) by Δ f (z) in Theorem B. We obtain:
where q(z) is a nonconstant polynomial and b, c ∈ ℂ are nonzero constants, does not admit entire solutions of finite order. If q(z) = q is a nonzero constant, then equation (1.3) possesses three distinct entire solutions of finite order, provided b = 3πn(n must be odd integer) and for a nonzero integer n. Without loss of generality, we may assume Δf (z) = f (z + 1) - f (z) in (1.4).
where is a cubic of unity.
where q(z) is a nonconstant polynomial and b, c ∈ ℂ are nonzero constants, does not admit entire solutions of finite order. If q(z) = q is a nonzero constant, then equation (1.5) possesses three distinct entire solutions of finite order, provided b = 3πn(n must be odd integer) and for a nonzero integer n. Without loss of generality, we may assume Δf (z) = f (z + 1) - f (z) in (1.5).
We also replace f (z + c) by Δ f (z) in the above Theorem C. We obtain:
has no transcendental entire solutions of finite order. With out loss of generality, we may assume Δ f (z) = f (z + 1) - f (z) in (1.6).
In 1959, Hayman  proposed:
Conjecture A If f is a transcendental meromorphic function, then f n f' assumes every finite non-zero complex number infinitely often for any positive integer n.
Hayman [5, 6] himself confirmed it for n ≥ 3 and for n ≥ 2 in the case of entire f. Further, it was proved by Mues  when n ≥ 2; Clunie  when n ≥ 1 and f is entire; Bergweiler and Eremenko  verified the case when n = 1 and f is of finite order, and finally by Chen and Fang  for the case n = 1. For an analogue result in difference, in 2007, Laine and Yang  proved:
Theorem D Let f be a transcendental entire function of finite order and c be non-zero complex constant. Then for n ≥ 2, f (z) n f (z + c) assumes every non-zero value a ∈ ℂ infinitely often.
Recently, Liu and Yang  improved Theorem D and obtained the next result.
Theorem E Let f be a transcendental entire function of finite order, and c be a non-zero complex constant. Then, for n ≥ 2; f (z) n f (z + c) - p(z) has infinitely many zeros, where p(z) is a non-zero polynomial.
Very recently, Qi et al.  obtained the following uniqueness theorem about the above results.
Theorem F  Let f and g be transcendental entire functions of finite order, and c be a non-zero complex constant; let n ≥ 6 be an integer. If f n f (z + c), g n g(z + c) share z CM, then f = t1g for a constant t1 that satisfies .
In the present paper, we get analogue results in difference, along with the following.
Theorem 4 Let f be a transcendental meromorphic function of finite order ρ, and α(z) be a small function with respect to f (z). Suppose that c is a nonzero complex constant and λ, μ are constants, n, m are positive integers.
If λ ≠ 0 and n ≥ 3m + 2, then f (z) n (μ f m (z + c) + λ) - α(z) has infinitely many zeros.
If λ = 0 and n + m ≥ 3, then f (z) n (μ f m (z + c)) - α(z) has infinitely many zeros.
Theorem 5 Let f and g be transcendental entire functions of finite order, and c be a non-zero complex constant. Suppose that and λ, μ are constants, n, m are distinct positive integers.
f (z) ≡ tg(z) (where t is a constant and t n = 1); or
f n (z)(μ f m (z + c) + λ)g n (z)(μg m (z + c) + λ) ≡ α2(z).
f ≡ h1g(h1 is a constant and ); or
f g = h2(h2 is a constant).
2 Some Lemmas
In order to prove our theorem, we need the following Lemmas:
As far as Clunie type lemmas are concerned, same conclusions hold as long as the proximity functions of the coefficients α(z) satisfy m(r, α) = S(r, f ). The next lemma is a rather general variant of difference counterpart of the Clunie Lemma, see , for the corresponding results on differential polynomials, see .
possibly outside of an exceptional set of finite logarithmic measure.
Lemma 2.3  Suppose c is a nonzero constant and α is a nonconstant meromorphic function. Then the differential equation f 2 + (c f (n))2 = α has no transcendental meromorphic solutions satisfying T (r, α) = S(r, f ).
It is evident that S(r, f (z + c)) = S(r, f ) from Lemma 2.4.
where k is a real constant and k ∈ [-2m, m].
Proof. It is easy to see F (z) is not a constant. Otherwise, we may set d = f n (z)(μ f m (z + c) + λ) (d is a constant). So nT (r, f ) = T (r, μ f m (z + c) + λ) + O(1) = mT (r, f ) + S(r, f ). It is contradict with m, n are distinct.
The proof of Theorem 5 is complete.
where denotes the counting function of zeros of F such that simple zeros are counted once and multiple zero twice.
3 Proof of Theorem 1
Let f be an entire solution of Equation (1.4). Without loss of generality, we may assume that f is transcendental entire.
Since w(z) is transcendental, we must have a0 = a2 = a4 = a6 = 0. Therefore, c1 ≠ 0, c2 ≠ 0, and the condition a4 = 0 implies that q(z) is a constant, say q ≠ 0. Combing now the conditions a4 = 0 and a2 = 0 we conclude that , and then , hence b = 3πn. The connection between q and c now follows from 3c1c2 + (-1) n q - q = 0 and . We obtain, , so n must be an odd. Finally, . The proof of Theorem 1 is complete.
4 Proof of Theorem 2
Due to the same idea of Proof Theorem 1, we omit the proof.
5 Proof of Theorem 3
By the m ≥ 3, hence ρ(f) < ρ, a contradiction.
6 Proof of Theorem 4
The assertion follows by n ≥ 3m + 2.
The assertion follows by n + m ≥ 3. The Proof of Theorem 5 is complete.
7 Proof of Theorem 5
contradicting with n ≥ 4m + 5. Hence F (z) ≡ G(z) or F (z)G(z) ≡ α(z)2 by Lemma 2.7, We discuss the following two subcases.
By the condition n ≥ 4m + 5, it is easily to show that h(z) n h m (z + c) is not a constant from (7.7).
which is a contradiction since n ≥ 4m + 5. Therefore, h(z) n h(z + c) m ≡ 1. Thus h n (z) ≡ 1. Hence f (z) ≡ tg(z), where t is a constant and t n = 1.
contradicting with n + m ≥ 9. Hence F (z) ≡ G(z) or F (z)G(z) ≡ α(z)2 by Lemma 2.7. We discuss the following two subcases.
a contradiction. So h1(z) must be a constant, then f ≡ h1g(h1 is a constant and hn+m= 1).
Let f (z)g(z) = h2(z), By a reasoning similar to that mentioned at the end of the proof of Case2.1, we know that h2(z) must be a constant, then f g = h2(h2 is a constant). The proof of Theorem 6 is complete.
The authors are grateful to the referees for their valuable suggestions and comments. The authors would like to express their hearty thanks to Professor Hongxun Yi for his valuable advice and helpful information. Supported by project 10XKJ01 from Leading Academic Discipline Project of Shanghai Dianji University and also supported by the NSFC (No.10771121, No.10871130), the NSF of Shandong (No. Z2008A01) and the RFDP (No.20060422049), and The National Natural Science Youth Fund Project (51008190).
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