Sharing of Borel Exceptional Values between Meromorphic Functions and Differential Polynomial involving Shift Function

Sharing of Borel Exceptional Values between Meromorphic Functions and Differential Polynomial involving Shift Function

Dr. Jayashri Pattar

Department of Mathematics, Maharanis Science College for Women, Maharani Cluster University, Palace Road, Bengaluru-560001, Karnataka, India.

Dr. Shilpa N

Department of Mathematics, School of Engineering, Presidency University, Bengaluru-560064, Karnataka, India.

Abstract – In this paper we establish a first result for a transcendental meromorphic function of finite order sharing two Borel exceptional values under two cases. In the first case and differential polynomial () share a non-zero complex number and as Borel exceptional values. In the second case they share and

as Borel exceptional values. We also prove a second result in which (())() and (())() share the value 1 counting multiplicities (CM), while (()) and (()) share ignoring multiplicities (IM).

Keywords: Uniqueness, Meromorphic function, linear, differential polynomial, shift function. Mathematics Subject Classification (2020): 30D20, 30D30 30D35.

1. INTRODUCTION

   Definition 1. Let us define a differential polynomial involving shift function

   where (), () are small functions of (), > 0 +, is a complex constant.

   In 2013, Chen [3] proved the relationships between Picard values of entire functions () and their forward differences ().

   Theorem 1.1. [3] Let be a transcendental entire function of finite order, let ( 0) be a constant, and let be a positive integer. If 0, 0, then () = +, where (

   0), are constants.

   In 2016, Chen et al., [2] proved difference analogue to theorem 1.1.

   Theorem 1.2. [2] Let ( ), be two distinct complex numbers ( may be ), let be a transcendental meromorphic function of finite order with two Borel exceptional values , and be a non zero constant such that 0. If and share , CM, then = 0, =

   and () = +, where ( 0), are constants.

   In 2021, M. Fang and Y. Wang [7] worked for higher order difference operators.

   Theorem 1.3. [7] Let ( ), be two distinct complex numbers and +, let be a transcendental meromorphic function of finite order with two Borel exceptional values , and is a non-zero constant such that 0. If and share , CM, then = 0,

   = and () = +, where ( 0), are constants.

   In the year 1998, W. Yuefei and F. Mingliang[14] proved the criteria for normality of families of meromorphic functions.

   Theorem 1.4. [14] Let () be a transcendental entire function, , with + 1. Then ()() = 1 has infinitely many solutions.

   In 2002, M-L Fang [8] obtained the below result corresponding to unicity theorem.

   Theorem 1.5. [8] Let and be two nonconstant entire functions, and let , be two positive integers with > 2 + 4. If ()() and ()() share 1 CM, then either () =

   1, () = 2 , where 1, 2 and are three constants satisfying

   (1) (12) ()2 = 1, or = for a constant such that = 1.

   J. Fan et al., [6] extended theorem 1.5 to prove the following.

   Theorem 1.6. [6] Let and be two nonconstant meromorphic functions, and let , be two

   positive integers with > 3 + 8

   , otherwise > 3 + 6. If

   ()() and ()() share 1 CM, and share IM, then either () = 1, () =

   2 , where 1, 2 and are three constants satisfying (1) (12) ()2 = 1, or =

   for a constant such that = 1.

2. LEMMAS

   Lemma 2.1. [9, 4] Let be a nonconstant meromorphic function of finite order, let be a nonzero finite complex number. Then

   and for any > 0, we have

   Lemma 2.2. [12, 7] Suppose that 1(), 2(),· · · , () are meromorphic functions satisfying the following identity

   If () 0 and

   where is a set of (0, ) with infinite linear measure, , = 1, 2,· · · , 1, < 1, then 1.

   Lemma 2.3. [5] Let be a meromorphic function of order () = < 1. Then for each given > 0, and a positive integer , there exists a set (1, ) that depends on , and it has finite logarithmic measure, such that for all satisfying || = [0, 1], we have

   Lemma 2.4. [7] Let be a meromorphic function, let be a positive integer, and let be a nonzero finite complex number. If 0, then either () 1 or is a polynomial with

   Lemma 2.5. [12, 6] Let be a meromorphic function such that () 0, and let be a positive integer. Then

   Lemma 2.6. [12, 11, 6] Let be a nonconstant meromorphic function, let be a positive integer, and let be a nonzero finite complex number. Then

   where 0

   ) is the counting function which only counts those points such that

   Lemma 2.7. [10, 12, 11, 6] If is a meromorphic function, . And then

   Lemma 2.8. [10, 1] Let f(z) be a meromorphic function and a be a finite complex number. Then

   where 1(), 2() are two meromorphic functions such that () = (, ), ( = 1, 2).

   Lemma 2.9. [6] Let be a nonconstant entire function, and let ( 2) be a positive integer. If ()()() 0, then () = + , where ( 0), are two constants.

   Lemma 2.10. [13, 6] Let and be two nonconstant entire functions, and let ( 1) be a positive integer. If 1, then () = 1, () = 2, where 1, 2 and are three constants satisfying (12)+12 = 1.

3. MAIN RESULTS

   Theorem 3.1. Let 1( ), 2 be two distinct complex numbers and +, let be a transcendental meromorphic function of finite order with two Borel exceptional values 1,

   2 and is a non-zero constant such that () 0. If and () share 1, 2 CM, then

   1 = 0, 2 = .

   Proof. Case 1. 1 is a nonzero finite complex number, 2 = . Since 1, are two distinct Borel exceptional values of and is of finite order, by Hadamards factorization theorem, we have

   where ( 0, ) is a meromorphic function such that () < () and is a non constant polynomial with () = (). Hence we have

   follows that () 0. Thus ( 0) is a meromorphic function with () < ().

   Hence is a small function of . By second fundamental theorem and (3.3) we have

   Thus, we deduce from (3.2) and (3.5) that ( 1) = (), this contradicts that 1 is a Borel exceptional value of . Hence this is absurd.

   Case 2. 1 = 0, 2 = . Since 0, are two distinct Borel exceptional values of and is of finite order, by Hadamards factorization theorem we have

   where ( 0, ) is a meromorphic function such that () < () and is a non constant polynomial with () = () 1. Hence we have

   Since and () share 0, CM, there exists a polynomial satisfying

   It follows from (3.9) and Lemma (2.1) that

   We consider two subcases.

   Case 2.1. () 2. Here again we have two subcases.

   Case 2.1.1 1 () () 1.

   Thus by (3.9) we obtain

   a contradiction.

   If 2 then by (3.12) – (3.14) we know that 1, 2,· · · , are nonconstant. +1 0 and (2.1) is valid, thus by Lemma (2.2) we obtain that +1 1 a contradiction.

   Case 2.1.2. () = 0. If 1 () 0, then by using the same argument as used in case

   2.1.1 we obtain a contradiction.

   If 1 () = 0, then by (3.11) we have

   By second fundamental theorem and Lemma (2.2) and using the same argument as used in case 2.1.1 we obtain a contradiction.

   Case 2.2. () = 1. Thus by (3.6) we have

   where ( 0, ) is a meromorphic function such that () < 1 and is a nonzero constant. By (3.9) and (3.15) we obtain

   where , are two nonzero constants. We now write equation (3.16) in the form

   where = (), 1,· · · , 1 are constants.

   We choose such that 0 < < 1 (). Lemma (2.3) asserts that there exists a set (1, +) of finite logarithmic measure, such that for all || = [0, 1]

   Let || = [0, 1] and || , then it follows from (3.17) and (3.18) that = 0.

   Thus, we have

   ()(()) + 1(1 ()) + · · · + 1(1 ()) = 0. (3.19) If (1 ()) = 0 then by Lemma (2.4) we know that is a nonzero constant and () =

   +. If (1 ()) 0 then it follows from ((1 ())) () < 1, (3.19) and Lemma (2.3) that 1 = 0. Now suppose that 0, 1 = 0,· · · , 1 = 0, 2 . Thus, we have

   ()( ()) + 1( ()) + · · · + ( ()) = 0, 0. (3.20) We claim that (1 ()) = 0. Otherwise, we have

   By Lemma (2.3) and (3.21) we deduce that = 0 a contradiction. Thus, we prove that

   (1 ()) = 0. Hence, we have from (3.17) that

   where 1,1,· · · , 1,1, 1,0, are constants and 0, (0 ()) = .

   Now using the same argument as for proving (()) = 0 we obtain that ( ()) = 0, 1 1 1.

   By taking 1 = 1 we have (1 ()) = 0 and by Lemma (2.4) we deduce that () is a nonzero constant. Hence the theorem is proved.

   Theorem 3.2. Let and be two meromorphic functions and , be two positive integers

   with > 3 + 8

   , otherwise > 3 + 6. If (())() and

   (())() share 1 CM; () and () share IM and

   then either () = 1, () = 2 , where 1, 2 and are constants satisfying

   (1) (12) ()2 = 1, or = for a constant such that = 1. Proof. Set = [()](), = [()]().

   Since [()]() and [()]() share 1 C M then and share 1 C M. By Lemma (2.5) we obtain

   It follows (, ) = (, ). Similarly, we get (, ) = (, ).

   Set =

   Next we consider two cases.

   Case 1. = 0 then by (3.24)

   where is a finite complex constant.

   In the following we consider two subcases.

   Case 1.1 = 1. It follows from (3.25) that = , that is [()]() = [()](). Which implies () = () + , where is a polynomial with () 1.

   If 0, then we have

   Since and are two nonconstant meromorphic functions, then

   (, ) + (1), (, ) + (1). (3.27) By second fundamental theorem and (3.27) we obtain

   Hence, we get

   By > 2 + 4, Nevanlinna second fundamental theorem and (3.26) – (3.28) we have

   Which implies

   Similarly

   By either > 3 + 6 or > 3 + 8 ( + 4) 2 + 4 we get

   a contradiction.

   Hence 0. It follows = where is a constant such that = 1.

   Case 1.2 1. Then by (3.25) we obtain

   Since and share IM, it follows from (3.32) that and . Hence 1 0 and

   then by (3.32) we deduce that . By Lemma (2.6) we obtain

   It follows from either > 3 + 6 or > 3 + 8 ( + 4) 2 + 4 that

   (, ) (, ) a contradiction.

   Case 2. 0. Let 0 be a pole of [()] with multiplicity 1. Then by [()]and [()]share IM we know that 0 is a pole of [()] with multiplicity 2.

   Set = {1, 2} by (3.24) we deduce that 0 is a zero of with multiplicity + 1.

   Hence by Lemma (2.7) we have

   Similarly,

   Suppose 1 0. Let 0 be a common simple zero of () 1 and () 1, by a simple computation we see that 1(0) = 0. Thus by first fundamental theorem and Lemma (2.7) we have

   where (, 1

   ) is the counting function of simple zeros of () 1. It follows from

   and share 1 CM and (3.37) that

   ) is the counting function for which (())(+1) = 0 and

   ()[() 1] 0. Since and share 1 CM, then we get

   By Lemma (2.6) we have

   It follows from (3.38) – (3.42) that

   Since, (, 1

   We obtain from (3.43) that

   Without loss of generality, we suppose that there exist a set I with infinite measure such that

   That is

   Which implies

   which is a contradiction to our hypothesis > 6 + + . Hence we get 1 0.

   That is

   Integrating this equation

   where ( 0), are two finite complex numbers. Next, we consider two subcases.

   Case 2.1 0. Since () and () share IM. We know that and share IM. It

   follows from (3.46) that , . Hence 1

   0 thus by (3.46) we deduce

   Now we consider two subcases.

   Case 2.1.1 = . It follows from

   0 and (3.46) that

   In the following, we consider two subcases.

   Case 2.1.1.1 1. Then we have

   0. By Lemma (2.6) we obtain

   From (3.46) we can write

   By Lemma (2.8) we obtain the following inequality

   Hence

   Hence by (3.23) and (3.48) we deduce that

   (, ()) (, ) a contradiction.

   Case 2.1.1.2 = 1. Thus = 1 by (3.46) we deduce that 1. That is

   Since () and () share IM then by (3.49) we deduce that () , () . It follows from (3.49) that (())() 0, (())() 0, () 0, () 0. If 2, then by Lemma (2.9) we get () = 1, () = 2 , where 1, 2 and are constants satisfying (1) (12) ()2 = 1. If = 1 then by Lemma (2.10) we get

   () = 1, () = 2 , where 1, 2 and are constants satisfying

   (12)+1 = 1.

   Case 2.1.2 . Hence we have 0, 0. In this case by using the same

   argument as in 2.1.1.1 we get a contradiction.

   Case 2.2 = 0 then by (3.46)

   If = 1 then by (3.50) we have . That is (())() = (())(), by using the same argument as in case 1.1, we get , where is a constant such that = 1. If 1 then by (3.50) we get (())() (())() + 1. That is (())() (())() =

   1. Thus, we obtain ()

   ) + where is a polynomial of degree . Then by

   using the same argument as in case 1.1 we get a contradiction. Hence the proof.

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