Arab J Math 018 7:7 37 https://doiorg/101007/s40065-017-0189-x Arabian Journal o Mathematics Zinelâabidine Latreuch Benharrat Belaïdi On Picard value problem o some dierence polynomials Received: 4 April 017 / Accepted: 4 October 017 / Published online: 16 November 017 The Authors 017 This article is an open access publication Abstract In this paper, we study the value distribution o zeros o certain nonlinear dierence polynomials o entire unctions o inite order Mathematics Subject Classiication 30D35 39A05 1 Introduction results Throughout this paper, we assume that the reader is amiliar with the undamental results the stard notations o the Nevanlinna s value distribution theory 10,13] In addition, we will use ρ to denote the order o growth o, we say that a meromorphic unction a z is a small unction o z i T r, a = S r,, where S r, = o T r,, as r + outside o a possible exceptional set o inite logarithmic measure, we use S to denote the amily o all small unctions with respect to z For a meromorphic unction z, we deine its shit by c z = z + c In 1959, Hayman proved in 11] thati is a transcendental entire unction, then n assume every nonzero complex number ininitely many times, provided that n 3 Later, Hayman 1] conjectured that this result remains to be valid when n = 1n = Then Mues 18] conirmed the case when n = Bergweiler-Eremenko ] Chen-Fang 3] conirmed the case when n = 1,independentlySince then,there are many research publications see 17] regarding this type o Picard value problem In 1997, Bergweiler obtained the ollowing result Theorem 11 1] I is a transcendental meromorphic unction o inite order q is a not identically zero polynomial, then q has ininitely many zeros In 007, Laine Yang studied the dierence analogue o Hayman s theorem proved the ollowing result Theorem 1 14] Let z be a transcendental entire unction o inite order, c be a nonzero complex constant Then or n, n z z + c assume every nonzero value a C ininitely oten Z Latreuch B Belaïdi B Laboratory o Pure Applied Mathematics, Department o Mathematics, University o Mostaganem UMAB, B P 7, Mostaganem, Algeria E-mail: benharratbelaidi@univ-mostadz Z Latreuch E-mail: zlatreuch@gmailcom
8 Arab J Math 018 7:7 37 In the same paper, Laine Yang showed that Theorem 1 does not remain valid or the case n = 1 Indeed, take z = e z + 1 Then z z + πi 1 = 1 + e z 1 e z 1 =e z Ater their, a stream o studies on the value distribution o nonlinear dierence polynomials in has been launched many related results have been obtained, see, eg 5,14 16] For example, Liu Yang improved the previous result obtained the ollowing Theorem 13 15] Let z be a transcendental entire unction o inite order, c be a nonzero complex constant Then or n, n z z + c p z has ininitely many zeros, where p z 0 is a polynomial in z Hence, it is natural to ask: What can be said about the value distribution o z z + c qz; when is a transcendental meromorphic unction q be a not identically zero small unction o? In this paper, as an attempt in resolving this question, we obtain the ollowing results Theorem 14 Let be a transcendental entire unction o inite order, let c 1, c be two nonzero complex numbers such that z + c 1 z + c q be not identically zero polynomial Then z z + c 1 q z z z + c q z at least one o them has ininitely many zeros The ollowing corollary arises directly rom Theorems 14 13 Corollary 15 Let n 1 be an integer let c 1, c c 1 c = 0 be two distinct complex numbers Let α, β, p 1, p q 0 be nonconstant polynomials I is a inite order transcendental entire solution o { n z z + c 1 q z = p 1 z e αz n z z + c q z = p z e βz, then n = 1 must be a periodic unction o period c 1 c Some lemmas The ollowing lemma is an extension o the dierence analogue o the Clunie lemma obtained by Halburd Korhonen 8] Lemma 1 4] Let z be a nonconstant, inite order meromorphic solution o n P z, = Q z,, where P z,, Q z, are the dierence polynomials in z with meromorphic coeicients a j z j = 1,,s, let δ<1 I the degree o Q z, as a polynomial in z its shits is at most n, then T r + c, s m r, P z, = o + o T r, + O m r, a j, or all r outside an exceptional set o inite logarithmic measure r δ Lemma 6] Let z be a nonconstant, inite order meromorphic unction let c = 0 be an arbitrary complex number Then T r, z + c = T r, z + S r, Lemma 3 7] Let z be a transcendental meromorphic unction o inite order ρ, let ε>0be a given constant Then, there exists a set E 0 1, + that has inite logarithmic measure, such that or all z satisying z / E 0 0, 1], or all k, j, 0 j < k, we have k z j z z k jρ1+ε The ollowing lemma is the lemma o the logarithmic derivative j=1
Arab J Math 018 7:7 37 9 Lemma 4 10] Let be a meromorphic unction let k N Then m r, k = S r,, where S r, = O log T r, + log r, possibly outside a set E 1 0, + o a inite linear measure I is o inite order o growth, then m r, k = O log r The ollowing lemma is a dierence analogue o the lemma o the logarithmic derivative or inite order meromorphic unctions Lemma 5 6,8,9] Let η 1,η be two arbitrary complex numbers such that η 1 = η let z be a inite order meromorphic unction Let σ be the order o z Then or each ε>0, we have m r, z + η 1 = O r σ 1+ε z + η Lemma 6 Let z be a transcendental meromorphic solution o the system { z z + c1 q z = p 1 z e αz, z z + c q z = p z e βz, 1 where α, β are polynomials p 1, p, q are not identically zero rational unctions I N r, = S r,, then deg α = deg β = deg α + β = ρ > 0 Proo First, we prove that deg α = ρ by the same we can deduce that deg β = ρ It is clear rom 1thatdegα ρ Suppose that deg α<ρ, this means that z z + c 1 := F = q z + p 1 z e αz S Applying Lemmas 1 into, we obtain T r, c = T r, = S r, which is a contradiction Assume now that deg α + β <ρ, this leads to p 1 p e α+β S From this 1, we have where P z, = p 1 p e α+β + q, a = c 1 P z, = a z b z It is clear that P z, 0, using Lemma 1,weget which leads to T r, = m r, c c1, b = q + c m r, P z, = S r, b z + P z, = S r, a z which is a contradiction Hence, deg α + β = deg α = deg β Finally, using Lemma 1, it is easy to see that both o α β are nonconstant polynomials
30 Arab J Math 018 7:7 37 3 Proo o Theorem 14 We shall prove this theorem by contradiction Suppose contrary to our assertion that both o z z + c 1 q z z z + c q z have initely many zeros Then, there exist our polynomials α, β, p 1 p such that z z + c 1 q z = p 1 z e αz 31 z z + c q z = p z e βz 3 By dierentiating 31 eliminating e α, we get A 1 c1 c1 c 1 = B 1, 33 where A 1 = p 1 p 1 + α p, B 1 = 1 p 1 + α q q By Lemma 6, wehave deg α = deg β = deg α + β = ρ > 0 Now, we prove that A 1 0 To show this, we suppose the contrary Then, there exists a constant A such that A = p 1 z e α, which implies the contradiction deg α = ρ = 0 By the same, we can prove that B 1 0 By the same arguments as above, 3gives A c c c = B, 34 where A = p p + β p B = p + β q q Obviously, A 0B 0 Dividing both sides o 33 34 by, we get or each ε>0 m r, 1 m r, c i + m r, ci + m r, c i ci + O log r ci = O r ρ1+ε + O log r = S r, So, by the irst undamental theorem, we deduce that T r, = N r, 1 + O r ρ1+ε + O log r 35 It is clear rom 3334 that any multiple zero o is a zero o B i i = 1, Hence, N r, 1 N r, 1Bi = O log r, where N r, 1 denotes the counting unction o zeros o whose multiplicities are not less than It ollows by this 35that T r, = N 1 r, 1 + O r ρ1+ε + O log r, 36 where N 1 r, 1 is the counting unction o zeros, where only the simple zeros are considered From 33 34, or every zero z 0 such that z 0 = 0 which is not zero or pole o B 1 B, we have c1 + B 1 z0 = 0 37 c + B z0 = 0 38 By 3738, we obtain B c1 B 1 c z0 = 0, 39 which means that the unction B c1 B 1 c has at most a inite number o simple poles We consider two cases:
Arab J Math 018 7:7 37 31 Case 1 B c1 B 1 c 0 Set h z = B c1 B 1 c 310 z Then, rom the lemma o logarithmic dierences, we have m r, h = O r ρ1+ε + Olog r On the other h, N r, h = N r, B c1 B 1 c = N 1 r, B c1 B 1 c + O r ρ1+ε + O log r = S r, Thus, T r, h = O r ρ1+ε + O log r = Sr, From the Eq 310, we have By dierentiating 311, we get c1 z = B 1 B c z + h B z 311 c 1 z = h B B z + h B z + Substituting 31131into33 A1 h h ] B B A1 B ] 1 + Equation 34, can be rewritten as B + h B B ] c z + B 1 B c z 31 c B 1 B c B 1 B c = B 1 313 B 1 A c + B 1 c + B 1 c B B B =B 1 By adding this to 313, we get A1 h h ] + h ] A1 B 1 + B ] 1 A c = 0 314 B B B B B B Its clear that h B 0 To complete the proo o our theorem, we need to prove A 1 h B h 0 B A 1 B 1 B B B 1 A B 0 Suppose contrary to our assertion that A 1h B hb 0 Then, by the deinition o A1 by simple integration, we get p 1 e α h = C 1, B where C 1 is a nonzero constant This implies that deg α = ρ 1, which is a contradiction Hence, A 1 h B hb 0 Next, we shall prove A 1 B 1 B B B 1 A B 0 Suppose that A 1 B 1 B B B 1 A B 0 Then, we obtain p 1 e αβ B 1 = C := γ, p B where C is a nonzero constant γ is a small unction o From 313, we get c1 γ c = 1 γ q 315
3 Arab J Math 018 7:7 37 I γ 1, then by applying Clunie s lemma to 315, we obtain m r, c1 γ c = T r, c1 γ c = S r, By this 315, we have T r, = T r, 1 γ q = S r, c1 γ c which is a contradiction I γ 1, then we obtain the contradiction c1 z c z Thus, A 1 B 1 B 1 A B where 0 From the above discussion 314, we have M = B B c z = M z z + N z z 316 c1 z = ϕ z z + ψ z z, 317 hb h A1 B, N = A 1 A B 1 B B h B A 1 A B 1 B B ϕ z = B 1 M + h, ψ = B 1 N B B B Dierentiation o 316gives c = M + M + N + N 318 Substituting 316318into34, we get M A M ] + N A N + M ] + N + =B 319 Dierentiating 319, we get M A M ] + M A M ] + N A N + M ] + N A N + M + + N 3 + =B 30 Suppose z 0 is a simple zero o not a zero or pole o B Then rom 31930, we have N + B z 0 = 0, N A N + M + 3N + B ] z 0 = 0 It ollows that z 0 is a zero o B N A N + M B N] + 3B N Thereore, the unction B N B A N + B M B H = N] + 3B N satisies T r, H = S r, = H B 3B N + N + B A N B M + B N] 31 3B N Substituting 31into319, we get q 1 + q + q 3 =B, 3
Arab J Math 018 7:7 37 33 where q 1 = M A M + q = 1 3 N + 1 3 We prove irst q 0 Suppose the contrary Then which leads to H 3B, B B A N + 4 3 M, q 3 = N q = N q 3 3 N 1 B 3 B 3 A 1 + A + h 3 h = 0 α + β = N N B + h B h p 1 p p 1 p By simple integration o both sides o the above equation, we get p 1 p e α+β = c N B h, 33 where c is a nonzero constant, this leads to the contradiction deg α + β < deg α = deg β Hence, q 0 Dierentiating 3, we obtain q 1 + q 1 + q + q + q 3 + q + q 3 =B 34 Let z 0 be a simple zero o which is not a zero or pole o B Then rom 334, we have q 3 + B z 0 = 0, q + q 3 + q 3 + B ] z 0 = 0 Thereore, z 0 is a zero o B q + q 3 B q 3 + B q 3 Hence, the unction B q + q 3 B R = q 3 + B q 3 satisies T r, R = S r, = R + B q 3 B q + q 3 35 B q 3 B q 3 Substituting 35into34 q 1 + q ] R + q 1 + q B q + 1 B q 1 q + q q 3 B 3 + R ] q 3 B + B q 3 B =B 36 Combining 363, we obtain q 1 + q ] R B q 1 + q 1 + q B q 3 B 1 B q 1 q + q q B 3 + R ] = 0 37 q 3 B From 37, we deduce that q 1 + q R B q 3 B B q 1 = 0
34 Arab J Math 018 7:7 37 q 1 + q 1 By eliminating R rom the above two equations, we obtain B q 1 q + q q B 3 + R = 0 q 3 B q 3 4q1 q 3 q B + q 4q1 q 3 q q3 4q1 q 3 q + q B 3 4q1 q 3 q = 0 38 Thus, Eq 35 can be rewritten as = B q 1 q 1 + 1 B q N 39 B q q B q 3 N Subcase 1 I 4q 1 q 3 q 0, then rom 38, we have q 4q1 q 3 q = q 3 4q1 q 3 q B q 3 B q 3 On the other h, q = 1 N q 3 3 N + 1 B 3 B 3 A 1 + A + hb 3 h B Hence, 4q1 q 3 q A 1 + A =3 4q1 q 3 q + 4 N hb N + 4 B + B By the deinition o A i i = 1, simple integration, we deduce that h B deg α + β < deg α = deg β which is a contradiction Subcase I 4q 1 q 3 q, then rom 3931, we have B B q 1 q q 1 q = H 3B N 330 On the other h, q 1 M A M q 3 N = H 3B N 331
Arab J Math 018 7:7 37 35 Combining 330331, we obtain A 1 A B 1 B B 5 B 4 B A 1 A B 1 + 1 6 5 h 3 h B A 1 A B 1 B B B A 1 A B 1 B B A 1 A B 1 B B 1 9 A 1 + A + 9 1 6 A 1 A B 1 B B B A 1 + A 7 9 B B = 1 9 A 1 + A B A 1 A B 1 B A 1 A B 1 1 B A 1 + A B A 1 A B 1 B B 5 6 A 1 + A h h + 3 B h 1 B h 5 4 B A 1 A B 1 B B B 19 36 + B 1 B 9 B h h A 1 A B 1 B B A 1 A B 1 A 1 A B 1 B B B A 1 A B 1 B B 1 A 1 + 1 A 3 1 + A 1 + 3 A 1 + A B + 1 B A A 1 B Dividing both sides o the above equation by A 1+A R since lim z z Rz = 0iR is a nonzero rational unction, we obtain A A 1 A 1 + A 9 5 h h 3 A 1 + A + 3 B h h 6 B A 1 + A + 5 h h + o 1 33 A 1 + A On the other h, since ρ h ρ 1 by Lemma 3 h z h z z ρ +ε, 333 or all z satisying z / E 0 0, 1], where E 0 1, is a set o inite logarithmic measure By combining 33333, we deduce lim z z / E 0 0,1] A A 1 A 1 + A = lim z z / E 0 0,1] α β α + β = 9 By setting α z = a m z m + +a 0 β z = b m z m + +b 0, we deduce which implies that a m bm lim z z / E 0 0,1] α β α + β = a m b m a m + b m = 9 = or 1 We consider irst the case a m bm = 1, we get rom 31317 ϕ + ψ q = Ae 1 b m z m 334
36 Arab J Math 018 7:7 37 M + N q = Be b m z m, 335 where A = p 1 e a m1z m1 + +a 0 B = p e b m1z m1 + +b 0 From 334335,weget Hence, Thereore, M ϕ + ψ + N q = q + A B 1 ϕ + ψ = q M + A + N q 1 B T r,ϕ + ψ = m r,ϕ + ψ + S r, = 1 log + ϕ re iθ re iθ + ψ re iθ re iθ dθ π E 1 + 1 log + ϕ re iθ re iθ + ψ re iθ re iθ dθ + S r,, π E where E 1 = { θ : re iθ 1 } E = { θ : re iθ > 1 } Now 1 log + ϕ re iθ re iθ + ψ re iθ re iθ dθ π E 1 1 log + re iθ dθ + S r, π E 1 1 log + re iθ π E 1 re iθ dθ + S r, = S r, On the other h, 1 log + ϕ re iθ re iθ + ψ re iθ re iθ dθ π E = 1 log + q re iθ π E re iθ dθ + 1 log + M re iθ 4π E B re iθ + N re iθ B re iθ re iθ re iθ q re iθ re iθ dθ + S r, = S r, Hence, T r, c1 = T r,ϕ + ψ = S r, which is a contradiction I a m bm =, then by the same argument, we have which implies the contradiction ϕ M + N + ψ q = q + B A T r, c = T r, M + N = S r, Case B c1 B 1 c 0, using the same arguments as in the proo o 314, we obtain that A 1 B 1 B 1 A 0 B B B 1
Arab J Math 018 7:7 37 37 which leads to p 1 e αβ = k B 1 = k c 1, 336 p B c where k is a nonzero complex constant By this 313, we have 1 c c1 c = q c k c1 337 I k = 1, then by applying Clunie lemma to 337, we deduce the contradiction T r, ci = S r, Hence, k = 1 rom the equation 336, we conclude that c1 c which exclude the hypothesis o our theorem This shows that at least one o z z + c 1 q z z z + c q z has ininitely many zeros Acknowledgements The authors are grateul to the anonymous reeree or his/her valuable comments suggestions which lead to the improvement o this paper Open Access This article is distributed under the terms o the Creative Commons Attribution 40 International License http:// creativecommonsorg/licenses/by/40/, which permits unrestricted use, distribution, reproduction in any medium, provided you give appropriate credit to the original authors the source, provide a link to the Creative Commons license, indicate i changes were made Reerences 1 Bergweiler, W: On the product o a meromorphic unction its derivative Bull Hong Kong Math Soc 1, 97 101 1997 Bergweiler, W; Eremenko, A: On the singularities o the inverse to a meromorphic unction o inite order Rev Mat Iberoam 11, 355 373 1995 3 Chen, HH; Fang, ML: The value distribution o n Chin Ser A 387, 789 798 1995 4 Chen, ZX: Complex Dierences Dierence Equations, vol 9 Mathematics Monograph SeriesScience Press, Beijing 014 5 Chen, ZX: On value distribution o dierence polynomials o meromorphic unctions Abstr Appl Anal Art ID 39853, pp 9 011 6 Chiang, YM; Feng, SJ: On the Nevanlinna characteristic o z + η dierence equations in the complex plane Ramanujan J 161, 105 19 008 7 Gundersen, GG: Estimates or the logarithmic derivative o a meromorphic unction, plus similar estimates J Lond Math Soc 371, 88 104 1988 8 Halburd, RG; Korhonen, RJ: Dierence analogue o the lemma on the logarithmic derivative with applications to dierence equations J Math Anal Appl 314, 477 487 006 9 Halburd, RG; Korhonen, RJ: Nevanlinna theory or the dierence operator Ann Acad Sci Fenn Math 31, 463 478 006 10 Hayman, WK: Meromorphic Functions Oxord Mathematical Monographs, Clarendon Press, Oxord 1964 11 Hayman, WK: Picard values o meromorphic unctions their derivatives Ann Math 70, 9 4 1959 1 Hayman, WK: Research Problems in Function Theory The Athlone Press University o London, London 1967 13 Laine, I: Nevanlinna Theory Complex Dierential Equations, de Gruyter Studies in Mathematics, vol 15 Walter de Gruyter & Co, Berlin 1993 14 Laine, I; Yang, CC: Value distribution o dierence polynomials Proc Jpn Acad Ser A Math Sci 838, 148 151 007 15 Liu, K; Yang, LZ: Value distribution o the dierence operator Arch Math Basel 93, 70 78 009 16 Li, N, Yang, L: Value distribution o certain type o dierence polynomials Abstr Appl Anal Art ID 78786, pp 6 014 17 Lü, W; Liu, N; Yang, C; Zhuo, C: Notes on value distributions o k b Kodai Math J 393, 500 509 016 18 Mues, E: Über ein problem von Hayman Math Z 1643, 39 59 1979 in German Publisher s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps institutional ailiations