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Some results about a special nonlinear difference equation and uniqueness of difference polynomial
Journal of Inequalities and Applications volume 2011, Article number: 50 (2011)
Abstract
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.
1 Introduction
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 [3] proved:
Theorem A Let p be a non-vanishing polynomial, and let b, c be nonzero complex numbers. If p is nonconstant, then the differential equation
admits no transcendental entire solutions, while if p is constant, then the equation admits three distinct transcendental entire solutions, provided .
Remark 1. As an example of the case with p(z) constant, recall the nonlinear differential equation
As pointed out by Li and Yang [4], Equation (1.2) admits exactly three distinct transcendental
entire solutions: f1(z) = sinz, , . It is easy to see the condition given in Theorem A is satisfied.
Very recently, Yang and Laine [3] 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 [3] obtained:
Theorem B A nonlinear difference equation
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.
Theorem C Let p, q be polynomials. Then a nonlinear difference equation
has no transcendental entire solutions of finite order.
In this article, by the same method of [3], we replace f (z + 1) by Δ f (z) in Theorem B. We obtain:
Theorem 1 A nonlinear difference equation
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).
Example 1. In the special case of
a finite order entire solution is
The other two immediately follow from the conditions above:
where is a cubic of unity.
Theorem 2. A nonlinear difference equation
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:
Theorem 3 Let p, q be polynomials and let m, n be positive integers satisfying m ≥ 3. Then a nonlinear difference equation
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 [5] 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 [7] when n ≥ 2; Clunie [8] when n ≥ 1 and f is entire; Bergweiler and Eremenko [9] verified the case when n = 1 and f is of finite order, and finally by Chen and Fang [10] for the case n = 1. For an analogue result in difference, in 2007, Laine and Yang [11] 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 [12] 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. [13] obtained the following uniqueness theorem about the above results.
Theorem F [13] 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), gng(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.
If λ ≠ 0 and n ≥ 4m + 5 and f (z)n(μ f (z + c)m + λ), g(z)n(μ f (z + c)m + λ) share α(z) CM, then
-
1.
f (z) ≡ tg(z) (where t is a constant and tn = 1); or
-
2.
f n(z)(μ f m (z + c) + λ)gn(z)(μg m (z + c) + λ) ≡ α2(z).
If λ = 0 and n + m ≥ 9 and f (z)n(μ f (z + c)m), g(z)n (μ f (z + c)m) share α(z) CM, then
-
1.
f ≡ h1g(h1 is a constant and ); or
-
2.
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 [15], for the corresponding results on differential polynomials, see [16].
Lemma 2.1 Let f be a transcendental meromorphic solution of finite order ρ of a difference equation of the form
where H (z, f ), P(z, f ), Q(z, f ) are difference polynomials in f such that the total degree of H (z, f ) in f and its shifts is n, and the corresponding total degree of Q(z, f ) is ≤ n. If H (z, f ) contains just one term of maximal total degree, then for any ε > 0,
possibly outside of an exceptional set of finite logarithmic measure.
Lemma 2.2 [17, 18] Let f be a transcendental meromorphic function of finite order ρ. Then for any given complex numbers c1, c2, and for each ε > 0,
Lemma 2.3 [19] 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 ).
Lemma 2.4 [17] Let f be a meromorphic function of finite order ρ and c is a non-zero complex constant. Then, for each ε > 0, We have
It is evident that S(r, f (z + c)) = S(r, f ) from Lemma 2.4.
Lemma 2.5 [17] Let f be a meromorphic function with finite exponent of convergence of poles and c is a non-zero complex constant. Then, for each ε > 0, We have
Lemma 2.6 Let f (z), n, m, μ , λ and c be as in Theorem 5. Denote F (z) = f n (z)(μ f m (z)+c). Then
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.
Case 1. λ ≠ 0, Using Lemma 2.4, we have
On the other hand, using Lemma 2.2, we have
that is
So
Case 2. λ = 0 By the same method of case 1 and Lemma 2.4, we obtain
The proof of Theorem 5 is complete.
Lemma 2.7 [20] Let F and G be two nonconstant meromorphic functions. If F and G share 1 CM, then one of the following three cases holds:
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.
Differentiating (1.4) results in
Combining (3.1) and (1.4), we get
This means that
where T4(z, f ) is a differential-difference polynomial of f , of total degree at most 4. If now T4(z, f ) vanishes identically, then or , and therefore,
Otherwise, the Clunie lemma applied to a differential-difference equation, see Remark after Lemma 2.1, implies that
Therefore, α := b2 f 2 + 9(f')2 is a small function of f , not vanishing identically. By Lemma 2.3, α must be a constant. Differentiating b2 f 2 + 9(f')2 = α, we immediately conclude that (3.3) holds in this case as well. Solving (3.3) shows that f must be of the form
Substituting the preceding expression of f into the original difference equation (1.4), expressing sin bz in the terms of exponential function, and denoting , an elementary computation results in
where
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
Suppose that f is a transcendental entire solution of finite order ρ to Equation (1.6). Without loss of generality, we may assume that q(z) does not vanish identically. From (1.6), we readily conclude by Lemma 2.2 that
and then
By the m ≥ 3, hence ρ(f) < ρ, a contradiction.
6 Proof of Theorem 4
Case 1. If λ ≠ 0, then we denote F (z) = f (z)n(μ f (z + c)m + λ). From Lemma 2.6, we have (2.1) and F (z) is not a constant. Since f (z) is a transcendental entire function of finite order, we get T (r, f (z + c)) = T (r, f (z)) + S(r, f ) from Lemma 2.4. By the second main theorem, we have
Thus
The assertion follows by n ≥ 3m + 2.
Case 2. If λ = 0, then we denote F (z) = f (z)n (μ f (z + c)m). From Lemma 2.6, we have (2.2) and F (z) is not a constant,
Thus
The assertion follows by n + m ≥ 3. The Proof of Theorem 5 is complete.
7 Proof of Theorem 5
Case 1 λ ≠ 0. Denote
Then F (z) and G(z) share 1 CM except the zeros or poles of α(z), and
from Lemma 2.6. By the definition of F , we get N2(r, F (z)) = S(r, f ) and
Then
Similarly,
Suppose that (2.3) hold. Substituting (7.4) and (7.5) into (2.3), we obtain
Then
Substituting (7.2) and (7.3) into the last inequality, yields
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.
Subcase 1.1. Suppose that F (z) ≡ G(z). That f (z)n(μ f (z + c)m + λ) ≡ g(z)n(μg (z + c)m + λ). Let h(z) = f (z)/g(z). If h(z)n hm(z + c) ≢1, we have
Then h is a transcendental meromorphic function with finite order since g is transcendental. By Lemma 2.4, we have
By the condition n ≥ 4m + 5, it is easily to show that h(z)n hm(z + c) is not a constant from (7.7).
Suppose that there exists a point z0 such that h(z0 )nhm(z0 + c) = 1. Then h(z0 )n = 1 from (7.6) since g(z) is entire function. Hence hm(z0 + c) = 1 and
Denote H := h(z)nhm(z + c), from the above inequality and (7.7), we apply the second main theorem to H , resulting in
Noting this, we have
and then
which is a contradiction since n ≥ 4m + 5. Therefore, h(z)n h(z + c)m ≡ 1. Thus hn(z) ≡ 1. Hence f (z) ≡ tg(z), where t is a constant and tn = 1.
Subcase 1.2. Suppose that F (z)G(z) ≡ α(z)2 . That is
Case 2. λ = 0. Denote
Then F (z) and G(z) share 1 CM except the zeros or poles of α(z), and
from Lemma 2.6. By the definition of F , we get N2(r, F (z)) = S(r, f ) and
Then
Similarly,
Suppose that (2.3) hold. Substituting (7.11) and (7.12) into (2.3), we obtain
Then
Substituting (7.9) and (7.10) into the last inequality, yields
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.
subcase 2.1. Suppose that F (z) ≡ G(z). That f (z)n (μ f (z + c)m ) ≡ g(z)n (μg (z + c)m ). Let h1(z) = f (z)/g(z). If h1(z) is not a constant, we have
Then h1 is a meromorphic function with finite order since g and f are of finite order. By Lemma 2.4 and (7.13), we have
and then
a contradiction. So h1(z) must be a constant, then f ≡ h1g(h1 is a constant and hn+m= 1).
subcase 2.2. Suppose that F (z)G(z) ≡ α(z)2. That is
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.
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Acknowledgements
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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JQ drafted the manuscript and have made outstanding contributions to this paper. JD and TZ participated in the sequence alignment.
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Qi, J., Ding, J. & Zhu, T. Some results about a special nonlinear difference equation and uniqueness of difference polynomial. J Inequal Appl 2011, 50 (2011). https://doi.org/10.1186/1029-242X-2011-50
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DOI: https://doi.org/10.1186/1029-242X-2011-50