Growth properties at infinity for solutions of modified Laplace equations

Transcription

1 Sun et al. Journal of Inequalities and Applications (2015) 2015:256 DOI /s R E S E A R C H Open Access Growth properties at infinity for solutions of modified Laplace equations Jianguo Sun 1,BinghangHe 2* and Corchado Peixoto-de-Büyükkurt 3 * Correspondence: binghanghe@qq.com 2 Mathematics Institute of Jilin University, Jinlin, , China Full list of author information is available at the end of the article Abstract Let F be a family of solutions of Laplace equations in a domain D and for each f F, f has only zeros of multiplicity at least k.letnbe a positive integer and such that n k(k+1) 2.Leta be a complex number such that a 0. If for each pair of functions f and g in F, f n f (k) and g n g (k) share a value in D,thenF is normal in D. Keywords: growth property; modified Laplace equation; normal family 1 Introduction Let D be a domain in C. LetF be a solution of certain Laplace equations defined in the domain D. F is said to be normal in D, in the sense of Montel, if for any sequence {f n } F, there exists a subsequence {f n } such that f n converges spherically locally uniformly in D to a meromorphic function or. Let g(z) be a solution of certain Laplace equations and a be a finite complex number. If f (z)andg(z) have the same zeros, then we say that they share a IM (ignoring multiplicity) (see [1]). In 1998, Wang and Fang [2] proved the following result. Theorem A Let f be a transcendental meromorphic function in the complex plane. Let n and k be two positive integers such that n k +1,then (f n ) (k) assumes every finite non-zero value infinitely often. Corresponding to Theorem A, there are the following theorems about normal families in [3]. Theorem B Let F be a family of meromorphic functions in D, n, k be two positive integers such that n k +3.If (f n ) (k) 1for each function f F, then F is normal in D. Recently, corresponding to Theorem B, Yang[4] provedthefollowingresult. Theorem C Let F be a family of meromorphic functions in D. Let n, kbetwopositive integers such that n k +2.Let a 0beafinitecomplexnumber. If (f n ) (k) and (g n ) (k) share aindforeachpairoffunctionsf andginf, then F is normal in D. Recently, Zhang and Li[5] proved the following theorem Sun et al. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Sun et al. Journal of Inequalities and Applications (2015) 2015:256 Page 2 of 7 Theorem D Let f be a transcendental meromorphic function in the complex plane. Let k be a positive integer. Let L[f ]=a k f (k) + a k 1 f (k 1) + + a 0 f, where a 0, a 1,...,a k are small functions and a ( 0) ( =1,2,...,k). For c 0,, let F = f n L[f ] c, where n is a positive integer. Then, for n 2, F = f n L[f ] c has infinitely many zeros. From Theorem D, we immediately obtain the followingresult. Corollary D Let f be a transcendental meromorphic function in the complex plane. Let c be a finite complex number such that c 0.Let n, k be two positive integers. Then, for n k(k+1) 2, f n f (k) chasinfinitelymanyzeros. It is natural to ask whether Corollary D can be improved by the idea of sharing values similarly with Theorem C. In this paper we investigate the problem and obtain the following result. Theorem 1 Let F be a family of meromorphic functions in D. Let n, kbetwopositive integers such that n 1+ 1+(k+1) 2. Let a be a complex number such that a 0.For each f F, f has only zeros of multiplicity at least k. If f n f (k) and g n g (k) share a in D for every pair of functions f, g F, then F is normal in D. Remark 1 From Theorem 1,itiseasytosee k(k+1) 2 2 for any positive integer k. Example 1 Let D = {z : z <1}, n, k N with n 1+ 1+(k+1) 2 and n be a positive integer; for k =2,let F = { f m (z)=mz k 1, z D, m =1,2,... }. Obviously, for any functions f m and g m in F,wehavefm nf m (k) share any a 0inD.ButF is not normal in D. =0,obviouslyfm nf m (k) and gm n g(k) m Example 2 Let D = {z : z <1}, n, k N with n 1+ 1+(k+1) 2 and n is a positive integer, and let F = { f m (z)=e mz, z D, m =1,2,... }. Obviously, for any f m and g m in F,wehavefm nf m (k) share 0 in D.ButF is not normal in D. Example 3 Let D = {z : z <1}, n, k N with n 1+ let { F = f m (z)= ( m z + 1 ) }, z D, m =1,2,.... m = m k e (mn+m)z,obviouslyf n m f (k) m 1+(k+1) 2 and g n m g(k) m,and n be a positive integer, For functions f m and g m in F,wehavef m f m = mz +1.Obviouslyf mf m and g mg m share 1 in D.ButF is not normal in D.

3 Sun et al. Journal of Inequalities and Applications (2015) 2015:256 Page 3 of 7 Remark 2 Example 1 shows that the condition that f has only zeros of multiplicity at least k is necessary in Theorem 1. Example 2 shows that the condition a 0inTheorem1 is inevitable. Example 3 shows thattheorem 1 is not truefor n =1. 2 Lemmas In order to prove our theorem, we need the following lemmas. Lemma 2.1 (Zalcman s lemma, see [6]) Let F be a family of meromorphic functions in the unit disc with the property that, for each f F, all zeros of multiplicity are at least k. Suppose that there exists a number A 1 such that f (k) (z) Awheneverf F and f =0. If F is not normal in, then for 0 α k, there exist: 1. anumberr (0, 1); 2. asequenceofcomplexnumbersz n, z n < r; 3. asequenceoffunctionsf n F; 4. asequenceofpositivenumbersρ n 0 + such that g n (ξ)=ρ n αf n(z n + ρ n ξ) locally uniformly converges (with respect to the spherical metric) to a non-constant meromorphic function g(ξ) on C, and, moreover, the zeros of g(ξ) are of multiplicity at least k, g (ξ) g (0) = ka +1.In particular, ghasorderatmost2. Lemma 2.2 Let n, k be two positive integers such that n k(k+1) 2, and let a 0be a finite complex number. If f is a rational but not a polynomial meromorphic function and f has only zeros of multiplicity at least k, then f n f (k) a has at least two distinct zeros. Proof If f n f (k) a has zeros and has exactly one zero. We set f = A(z α 1) m 1(z α 2 ) m 2 (z α s ) m s, (2.1) (z β 1 ) n 1(z β 2 ) n 2 (z β t ) n t where A is a non-zero constant. Because the zeros of f are at least k, weobtainm i k (i =1,2,...,s), n 1( =1,2,...,t). For simplicity, we denote m 1 + m m s = m ks, (2.2) n 1 + n n t = n t. (2.3) From (2.1), we obtain f (k) = (z α 1) m1 k (z α 2 ) m 2 k (z α s ) ms k g(z), (2.4) (z β 1 ) n 1+k (z β 2 ) n 2+k (z β t ) n t+k where g is a polynomial of degree at most k(s + t 1). From (2.1)and(2.4), we obtain f n f (k) = An (z α 1 ) M 1(z α 2 ) M 2 (z α s ) M s g(z) (z β 1 ) N 1(z β 2 ) N 2 (z β t ) N t = p q. (2.5) Here p and q are polynomials of degree M and N, respectively.alsop and q have no common factor, where M i =(n +1)m i k and N =(n +1)n + k. By(2.2) and(2.3), we

4 Sun et al. Journal of Inequalities and Applications (2015) 2015:256 Page 4 of 7 deduce M i =(n +1)m i k k(n +1) k = nk, N =(n +1)n + k n + k +1.Forsimplicity, we denote deg P = M = deg q = N = s M i + deg(g) nks + k(s + t 1) i=1 =(nks + ks)+k(t 1) (nk + k)s, (2.6) t N (k +1+n)t. (2.7) =1 Since f n f (k) a = 0 has ust a unique zero z 0,from(2.5)weobtain f n f (k) B(z z 0 ) l = a + (z β 1 ) N 1(z β 2 ) N 2 (z β t ) N t = p q. (2.8) By a 0,weobtainz 0 α i (i =1,...,s), where B is a non-zero constant. From (2.5), we obtain [ f n f (k)] (z α 1 ) M1 1 (z α 2 ) M 2 1 (z α s ) Ms 1 g 1 (ξ) =, (2.9) (z β 1 ) N 1+1 (z β t ) N t+1 where g 1 (ξ) is a polynomial of degree at most (k +1)(s + t 1). From (2.8), we obtain [ f n f (k)] = (z z 0 ) l 1 g 2 (z) (z β 1 ) N 1+1 +(z β t ) N t+1, (2.10) where g 2 (ξ)=b(l N)z t + B 1 z t B t is a polynomial (B 1,...,B t are constants). Now we distinguish two cases. Case 1. If l N,by(2.8), then we obtain deg p deg q.som N.By(2.9)and(2.10), we obtain s i=1 (M i 1) deg g 2 = t.som s deg(g) t,andm s + t + deg(g) (k +1)(s + t) k <(k +1)(s + t). By (2.6)and(2.7), we obtain [ M M <(k +1)(s + t) (k +1) nk + k + ] [ N (k +1) n + k +1 1 nk + k + 1 n + k +1 ] M. By n k(k+1) 2,wededuceM < M, which is impossible. Case 2. If l = N, then we distinguish two subcases. Subcase 2.1. If M N, by(2.9) and(2.10), we obtain s i=1 (M i 1) deg g 2 = t. SoM s deg(g) t,andm s + t + deg(g) (k +1)(s + t) k <(k +1)(s + t), then we can proceed similarly to Case 1. This is impossible. Subcase 2.2. If M < N,by(2.9)and(2.10), we obtain l 1 deg g 1 (s + t 1)(k +1),and then N = l deg g 1 +1 (k +1)(s + t) k <(k +1)(s + t) [ ] 1 (k +1) nk + k + 1 N N. n + k +1 By n k(k+1) 2,wededuceN < N. This is impossible.

5 Sun et al. Journal of Inequalities and Applications (2015) 2015:256 Page 5 of 7 If f n f (k) a 0andweknowf is rational but not a polynomial, then f n f (k) also is rational but not a polynomial. At this moment, l =0for(2.8),andproceedingasinCase1,wehave a contradiction. Lemma 2.2 is proved. 3 Proof of Theorem 1 We may assume that D = { z <1}. SupposethatF is not normal in D. Withoutlossof generality, we assume that F is not normal at z 0 = 0. Then, by Lemma 2.1,thereexist: 1. a number r (0, 1); 2. asequenceofcomplexnumbersz, z 0 ( ); 3. asequenceoffunctionsf F; 4. asequenceofpositivenumbersρ 0 + such that g (ξ)=ρ n+1 k f (z + ρ ξ) converges uniformly with respect to the spherical metric to a non-constant meromorphic function g(ξ)inc.moreover,g(ξ)isoforderatmost2. By Hurwitz s theorem, the zeros of g(ξ)areatleastk multiple. On every compact subset of C which contains no poles of g,weseethat (z + ρ ξ)f (k) (z + ρ ξ) a = g n (ξ)( g (k) (ξ) ) a (3.1) f n converges uniformly with respect to the spherical metric to g n (ξ)(g (k) (ξ)) a. If g n (ξ)(g (k) (ξ)) a (a 0)andg has only zeros of multiplicity at least k, theng has no zeros. From the g n g (k) having no zeros and the g n (ξ)(g (k) (ξ)) a,weknowg has no poles. Because the g(ξ) is a non-constant meromorphic function in C and g has order at most 2. We obtain g(ξ) =e dξ 2 +hξ+c,whered, h, c are constants and dh 0.Sog n (ξ)(g (k) (ξ)) a, which is a contradiction. When g n (ξ)(g (k) (ξ)) a 0,(a 0), we distinguish three cases. Case 1. If g is a transcendental meromorphic function, by Corollary D,thisisacontradiction. Case 2. If g is a polynomial and the zeros of g(ξ) areatleastk multiple, and n k(k+1) 2,theng n (ξ)(g (k) (ξ)) a = 0 must have zeros, which is a contradiction. Case 3. If g is a non-polynomial ration function, by Lemma 2.2,thisisacontradiction. Next we will prove that g n g (k) a has ust a unique zero. To the contrary, let ξ 0 and ξ0 be two distinct solutions of g n g (k) a,andchooseδ (> 0) small enough such that D(ξ 0, δ) D(ξ0, δ)= where D(ξ 0, δ)={ξ : ξ ξ 0 < δ} and D(ξ0, δ)={ξ : ξ ξ 0 < δ}.from(3.1), by Hurwitz s theorem, there exist points ξ D(ξ 0, δ), ξ large, f n (z + ρ ξ ) ( f (k) (z + ρ ξ ) ) a =0, f n (z + ρ ξ ) ( f (k) (z + ρ ξ ) ) a =0. D(ξ0, δ) suchthatforsufficiently By the hypothesis that for each pair of functions f and g in F, f n f (k) and g n g (k) share a in D, we know that for any positive integer m f n m (z + ρ ξ ) ( f (k) m (z + ρ ξ ) ) a =0, f n m (z + ρ ξ ) ( f (k) m (z + ρ ξ ) ) a =0.

6 Sun et al. Journal of Inequalities and Applications (2015) 2015:256 Page 6 of 7 Fix m,take, and note z + ρ ξ 0, z + ρ ξ f n m (0)( f (k) m (0)) a =0. 0, then Since the zeros of fm n (k) (0)(f m (0)) a have no accumulation point, so z + ρ ξ =0,z + ρ ξ =0. Hence ξ = z ρ, ξ = z ρ. This contradicts with ξ D(ξ 0, δ), ξ D(ξ0, δ), and D(ξ 0, δ) D(ξ0, δ)=. Sogn g (k) a has ust a unique zero, which can be denoted by ξ 0. Fromtheabove,weknowg n g (k) a has ust a unique zero. If g is a transcendental meromorphic function, by Corollary D,theng n g (k) a = 0 has infinitely many solutions, which is a contradiction. Fromtheabove,weknowg n g (k) a has ust a unique zero. If g is a polynomial, then we set g n g (k) a = K(z z 0 ) l,wherek is a non-zero constant, l is a positive integer. Because the zeros of g(ξ)areatleastk multiple, and n 1+ 1+(k+1) 2,weobtainl 3. Then [g n g (k) ] = Kl(z z 0 ) l 1 (l 1 2). But [g n g (k) ] has exactly one zero, so g n g (k) has the same zero z 0 too. Hence g n g (k) (z 0 ) = 0, which contradicts g n g (k) (z 0 )=a 0. If g is a rational function but not a polynomial, by Lemma 2.2,theng n g (k) a =0atleast has two distinct zeros, which is a contradiction. Theorem 1 is proved. 4 Discussion In 2013, Yang and Nevo [4] has proved the following. Theorem E Let F be a family of meromorphic functions in D, n be a positive integer and a, b be two constants such that a 0, and b. If n 3 and for each function f F, f af n b, then F is normal in D. Recently, Zhang improved Theorem E by the idea of shared values. Meanwhile, Zhang [7] has proved the following. Theorem F Let F be a family of meromorphic functions in D, n be a positive integer and a, b be two constants such that a 0, and b. If n 4 and for each pair of functions fandginf, f af n and g ag n share the value b, then F is normal in D. By Theorem 1, we immediately obtain the following result. Corollary 1 Let F be a family of meromorphic functions in a domain D and each f has only zeros of multiplicity at least k +1.Let n, k be positive integers and n k(k+1) 2 and let a 0, be a complex number. If f (k) af n and g (k) ag n share 0 for each pair function of f and g in F, then F is normal in D. Remark 4.1 Obviously, for k =1 and b =0, Corollary 1 occasionally investigates the situationwhenthepoweroff is negative in Theorem F.

7 Sun et al. Journal of Inequalities and Applications (2015) 2015:256 Page 7 of 7 Recently, Zhang [8] proved the following. Theorem G Let F be a family of meromorphic functions in the plane domain D. Let n, be a positive integer such that n 2. Let a be a finite complex number such that a 0.If f n f and g n g share a in D for every pair of functions f, g F, then F is normal in D. Question 1 It is natural to ask if the conclusion of Theorems G and 1 still holds for n 1. Competing interests The authors declare that they have no competing interests. Authors contributions BH and CP made the main contribution in conceiving the presented research. JS, BH, and CP worked ointly on each section, while ML and JS drafted the manuscript. All authors read and approved the final manuscript. Author details 1 Department of Computer Science and Technology, Harbin Engineering University, Harbin, , China. 2 Mathematics Institute of Jilin University, Jinlin, , China. 3 Department of Mathematics, University of Delaware, 501 Ewing Hall, Newark, DE 19716, USA. Acknowledgements This work was written while the corresponding author was at the Department of Mathematics of the University of Delaware as a visiting professor, and he is grateful to the department for their support. This work was supported by the National Science Foundation of China under Grant Nos , ; the Doctoral Program of Higher Education Foundation of China under Grant No ; the National Science Foundation of Heilongiang Province under Grant No. F201306; the Specialized Foundation for the Basic Research Operating expenses Program of Central College No. HEUCF100612, HEUCFT1202. Received: 9 April 2015 Accepted: 2 August 2015 References 1. Drasin, D: Normal families and Nevanlinna theory. Acta Math. 122, (1969) 2. Wang, YF, Fang, ML: Picard values and normal families of meromorphic functions with multiple zeros. Acta Math. Sin. 41, (1998) 3. Schiff, J: Normal Families. Springer, Berlin (1993) 4. Yang, P, Nevo, S: Derivatives of meromorphic functions with multiple zeros and elliptic functions. Acta Math. Sin. Engl. Ser. 29(7), (2013) 5. Zhang, ZL, Li, W: Picard exceptional values for two class differential polynomials. Acta Math. Sin. 34(6), (1994) 6. Zalcman, L: Normal families: new perspectives. Bull. Am. Math. Soc. 35, (1998) 7. Zhang, QC: Meromorphic functions sharing values in an angular domain. J. Math. Anal. Appl. 34(1), (2013) 8. Zhang, QC: Some normality criteria of meromorphic functions. Complex Var. Elliptic Equ. 53(8), (2013)