Uniqueness of Meromorphic Functions Whose Differential Polynomials Share One Value

ISSN Online: 2327-4379 ISSN Print: 2327-4352

DOI: 10.4236/jamp.2018.611188 Nov. 19, 2018 2264 Journal of Applied Mathematics and Physics

Uniqueness of Meromorphic Functions Whose

Differential Polynomials Share One Value

Jin Tao, Xinli Wang

College of Science, University of Shanghai for Science and Technology, Shanghai, China

Abstract

In this paper, we prove a uniqueness theorem of meromorphic functions whose some nonlinear differential shares 1 IM with powers of the meromor-phic functions, where the degrees of the powers are equal to those of the non-linear differential polynomials. This result improves the corresponding one given by Zhang and Yang, and other authors.

Keywords

Differential Polynomials, Meromorphic Functions, Uniqueness Theorems

1. Introduction

The meromorphic function mentioned in this paper refers to the meromorphic function over the entire complex plane. Let f and g be two non-constant mero-morphic functions. E⊂

(

0,∞

)

_{means Linear measure finite set.} S r f

(

,

)

means S r f

(

,

)

=o T r f

{

(

,

)

}

,

(

r→ ∞ ∉,r E

)

_{. CM is the abbreviation of} com-mon multiplicities. And IM is the abbreviation of ignored multiplicities. These concepts can be found in the literature [1]. Let a _{be a finite complex number, if}

f a− _and g a− have the same zero point and the same number of weights, then f and gCM share a. If f a− _and g a− have the same zero point without counting the number, then f and gIM share a[2]. In addition, the following de-finitions are required: let p be a positive integer, and a C∈ ∪ ∞

{ }

_{. Next}

) , 1

p

N r f a

 

 ₋ 

  means f has a weight less than p count function of the weight of

the value point a within z r< . N_p) r, 1

f a

 

 ₋ 

  means corresponding reduced

count function; N_(p r, 1

f a

 

 ₋ 

  means the weight of f is not less than p count

How to cite this paper: Tao, J. and Wang, X.L. (2018) Uniqueness of Meromorphic Functions Whose Differential Polynomials Share One Value. Journal of Applied Ma-thematics and Physics, 6, 2264-2272.

https://doi.org/10.4236/jamp.2018.611188

Received: September 29, 2018 Accepted: November 16, 2018 Published: November 19, 2018

Copyright © 2018 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

DOI: 10.4236/jamp.2018.611188 2265 Journal of Applied Mathematics and Physics function of the weight of the value point a within z r< . N(p r, _{f a}1

 

 ₋ 

  means

corresponding reduced count function. Suppose k is a non-negative number. Mark N rk ,_{f a}1

 

 ₋ 

  defined as follows. See the literature [3] for details.

(2 (

1 1 1 1

, , , ,

k k

N r N r N r N r

f a f a f a f a

       

= + + ⋅⋅⋅ +

 ₋   ₋   ₋   ₋ 

       

Before, Xiaomin Li and Zhitao Wen expanded Jilong Zhang’s theorem, where

( )

_fn ′ _{changes to f f}n−1 ( )k _{, so when k=1, that is Zhang’s theorem. Similarly,} in this paper, we continuously change _{f f}n−1 ( )k _{to f}n−1

₍

_{f −1}

₎

_f( )k _{, which}

con-tained f −1_{. So we expended Xiaomin Li and Zhitao Wen’s theorem.}

In 2008, Lianzhong Yang and Jilong Zhang proved the following theorems: Theorem A [4] Suppose f is a non-constant entire function, n≥7 is a posi-tive integer, if _fn _and

( )

_fn ′ _{CM share 1, then f}n₌

( )

_fn ′_.

Theorem B [4] Suppose f is a non-constant meromorphic function, n≥12 is a positive integer, if _fn _and

( )

_fn ′ _{CM share 1, then f}n₌

( )

_fn ′_.

Recently Zhang Jilong improved the above theorem. Get the following result: TheoremC [5] Suppose f is a non-constant entire function, n≥3 is a posi-tive integer, if _fn _and

( )

_fn ′ _{CM share 1, then f}n₌

( )

_fn ′_.

Theorem D [5] Suppose f is a non-constant meromorphic function, n≥4 _is a positive integer, if _fn _and

( )

_fn ′ _{CM share 1, then f}n₌

( )

_fn ′_.

Li Xiaomin and Wen Zhitao have improved on the basis of Zhang Jilong’s theorem, as follows.

Theorem E [5] Suppose f is a non-constant meromorphic function, k is a pos-itive integer, n is a positive integer and satisfies _{2n k≥ + +3 k}2_{+2k+9, if f}n and

( )

_fn ′ _{CM share 1, then f}n ₌

( )

_fn ′_.

Theorem F [5] Suppose f is a non-constant meromorphic function, k is a pos-itive integer, n is a positive integer and satisfies _{2n≥3k+ +8 9k}2_{+40k+64, if}

n

f and

( )

_fn ′ _{IM share 1, then f}n ₌

( )

_fn ′_.

Now we mainly improve the theorem of Li Xiaomin. Which that changes _fn and _{f f}n−1 ( )k _{to f}n

(

_{f −1}

)

_{and f}n−1

₍

_{f −1}

₎

_f( )k _{. We get the following}

theo-rem:

Theorem 1 Suppose f is a non-constant meromorphic function, k is a positive integer, n is a positive integer and satisfies _{2n≥3 13k+ + 9k}2_{+62k+129, if}

(

1

)

n

f f − _andfn−1

₍

_{f −1}

₎

_f( )k _{IM share 1, and the zeros of f −1 with}

mul-tiplicity 2 at least., then _{f = f}( )k _.

2. Some Lemmas

Lemma 1 [6] Suppose F and G are non-constant meromorphic functions, let

2 2

1 1

F F G G

H

F F G G

′′ ′ ′′ ′

   

=_ − _{ }− − _

′ − ′ −

   

DOI: 10.4236/jamp.2018.611188 2266 Journal of Applied Mathematics and Physics

(

)

(

)

(

)

1) _, 1 _{, , ,}

1 E

N r N r H S r F S r G G

 _{ ≤ + +}

 ₋ 

 

Lemma 2 [7] Let f be a non-constant meromorphic function, and

( )

1

1 1

n n

n n

P f a f a f − a f

−

= + + ⋅⋅⋅ +

where a a1, , ,2 an−1,an

( )

≠0 are constants, then

( )

(

,

)

(

,

)

( )

1

T r P f =nT r f +O _.

Lemma 3 Let f be a non-constant meromorphic function, k

( )

≥1 _and

( )

2

n ≥ _{are two positive integers. Let F = f}n

(

_{f −}1

)

_{and G f=} n−1

₍

_{f −1}

₎

_f( )k _.

If F and G IM share 1, then S r F

(

,

)

=S r G

(

,

)

=S r f

(

,

)

_.

Proof According to Lemma 2, we obtain

(

,

) (

1

) (

,

)

( )

1

T r F = n+ T r f +O ₍₁₎

It can be seen from the above formula,

(

,

)

(

,

)

S r F =S r f ₍₂₎

due to _{G f=} n−1

(

_{f −1}

)

_f( )k _{, we have}

(

) (

) (

)

(

)

(

( )

)

(

) (

)

(

)

(

) (

)

(

) (

) (

)

, 1 , , 1 ,

1 , , , ,

, ,

k

T r G n T r f T r f T r f

n T r f T r f kN r f S r f n k T r f S r f

≤ − + − +

≤ − + + +

≤ + +

(3)

According to the second basic theorem and (2)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1 1

, , , , ,

1

1 1

, , , ,

1

, 2 , ,

T r F N r N r N r F S r F

F F

N r N r N r f S r f

G f

T r G T r f S r f

   

≤ _{ }+ _{ }+ +

−

   

 

 

= _{ }+ _{ }+ +

−

   

≤ + +

The above formula is combined with (1) to get

(

,

) (

1

) (

,

) (

,

)

T r G ≥ n− T r f +S r f (4)

According to (3) and (4), we have

(

,

)

(

,

)

S r G =S r f ₍₅₎

According to (2) and (5), we can get the conclusion of Lemma 3.

Lemma 4 [8] Let f be a non-constant meromorphic function, k p, are two positive integers. The zero point of f −1 _{is at least 2, then}

( )

( )

( )

1 1

, , , ,

p k p k

N r kN r f N r S r f

f

f +

 _{≤ +}  ₊

   

  _{ }

 

(

)

(

)

1 1

, , , , ,

1

f

N r N r N r f N r S r f

f f f

     

≤ ≤ + +

 ₋   _′  

     

Lemma 5 Let f be a non-constant meromorphic function, n

( )

≥2 _{, p are two}

positive integers. The zero point of f −1 _{is at least 2. Let F f=} n

(

_{f −}1

)

_and

(

)

( )

1 ₁ k

n

G f₌ − f ₋ f _{, if F}

DOI: 10.4236/jamp.2018.611188 2267 Journal of Applied Mathematics and Physics a) N r,1

(

k 3

)

N r,1

(

k 1

) (

N r f,

)

S r f

(

,

)

G f

 

 _{ ≤ + + + +}

 

 

   

b) , 1 2 ,1 2

(

,

)

(

,

)

1 L

N r N r N r f S r f

F f

 

 _{ ≤ + +}

 

 ₋ 

    ;

c) , 1

(

3

)

,1

(

2

) (

,

)

(

,

)

1 L

N r k N r k N r f S r f

G f

 

 _{ ≤ + + + +}

 

 ₋ 

    .

Proof According to Lemma 4, we have

( )

(

)

(

)

(

)

(

)

(

)

(

) (

)

(

)

1

1

1 1 1 1

, , , , ,

1

1 1 1

, , , , , ,

1

3 , 1 , ,

k

k

N r N r N r N r S r f

G f f f

N r N r f N r N r kN r f S r f

f f f

k N r k N r f S r f f

+

 

   

 _{ ≤ + + +}

 

   

  _{−  }

       

     

≤ _{ }+ + _{ }+ _{ }+ +

     

 

≤ + _{ }+ + +

 

This leads to the conclusion (a), obtained from the definition of the

1 ,

1 L

N r F

 

 ₋ 

  and Lemma 3:

( )

( )

(

)

(

)

(

)

(

)

1

, , , 1 , , 1

1

1 1

, , , 2 , 2 , ,

L F F F F

N r N r T r O m r N r O

F F F F F

N r F N r S r F N r f N r S r f

F f

′ ′ ′

 _≤  _≤  _{+ =}  ₊  ₊

 ₋   _′      

         

 

 

= + _{ }+ = + _{ }+

   

This leads to conclusions (b), the same reason

(

)

(

)

1 1

, , , ,

1 L

N r N r N r G S r f

G G

 _≤  _{+ +}

 ₋   

   

Combine _{G f=} n−1

(

_{f −1}

)

_f( )k

and the q form in Lemma 5, we can get (c). Lemma 6 Suppose F and G are non-constant meromorphic functions, and sa-tisfy N r F

(

,

)

N r,1 S r F

(

,

)

F

 

+ _{ }=

  and

(

)

(

)

1

, , ,

N r G N r S r G G

 

+ _{ }=

  . If F and

G IM share a non-zero constant a, then F G= _or FG=1_.

Proof

Suppose _{F f=} n

(

_{f −}1

)

_{, G f=} n−1

₍

_{f −1}

₎

_f( )k _{. (6)}

Let H be defined by Lemma 1. The following two discussions, Case 1 Suppose H≠0_{, then} F G≠ _{, let}

1 1

F F G G

V

F F G G

′ ′ ′ ′

   

=_ − _{ }− − _

− −

    (7)

If V =0_,

1

1 B B

F G

− = − ₍₈₎

where B≠0 _{is a constant, if} N r f

(

,

)

≠S r f

(

, .

)

By (6) and (8), we get B=1_.

DOI: 10.4236/jamp.2018.611188 2268 Journal of Applied Mathematics and Physics Therefore, N r f

(

,

)

=S r f

(

,

)

_.

So B≠1_{, by (8), we get}

(

1

)

1 1 11

1

BF BF

G

B F _F B

B

 

= = _{ }

− + ₋  − 

−

(9)

(

)

1

, ₁ , 1

N r S r f

F B

 

 

=

 

 ₋ 

 

 − 

(10)

According to the second basic theorem and (6) (8) (9) (10) we get

(

)

(

)

(

)

(

)

(

)

(

)

1 1

, , , ₁ , ,

1

1 1

, , ,

1

2 , ,

T r F N r N r N r F S r F

F _F

B

N r N r S r f

f f

N r f S r f

 

 

 

≤ _{ }+ _{ }+ +

  _{ − }

 − 

   

≤ _{ }+ _{ }+

−

   

≤ +

(11)

By lemma 2, we have

(

,

) (

1

) (

,

)

( )

1

T r F = n+ T r f +O ₍₁₂₎

Then

(

n+1

) (

T r f,

)

−3T r f

(

,

)

=S r f

(

,

)

(13) If B=1_{, then} F G= , contradiction.

If V was not always equal to 0, (7) can be rewritten into

(

1

)

(

1

)

F G

V

F F G G

′ ′

= −

− − (14)

Suppose z0 is a pole of f with multiplicity p, then z0 is pole of F with

mul-tiplicity

(

n+1

)

p. and z0 is zero of _{F F}

₍

F ₁

₎

′

− with multiplicity

(

n+1

)

p−1

at least. z0 is zero of _{G G}

₍

G ₁

₎

′

− with multiplicity

(

n+1

)

p k+ −1 at least.

So z0 is a zero of V with multiplicity ≥n at least.

(

)

(

)

(

)

(

)

(

)

(

)

(

) (

)

(

)

(

)

(

) (

)

(

)

(

)

(

) (

)

(

)

1

, , , , ,

1 1 1

, , , ,

1 1

1

3 , 1 , 2 ,

1 1

2 , 3 , 2 , , 1

2 8 , 2 5 , ,

L L

nN r f N r S r f N r V S r f V

N r N r N r S r f

G F G

k N r k N r f N r f f

N r k N r k N r f S r f

f f

k N r k N r f S r f f

 

≤ _{ }+ ≤ +

 

     

≤ _{ }+ _{ }+ _{ }+

− −

     

 

≤ + _{ }+ + +

 

   

+ _{ }+ + _{ }+ + +

   

 

≤ + _{ }+ + +

DOI: 10.4236/jamp.2018.611188 2269 Journal of Applied Mathematics and Physics Then

(

n 2k 5

) (

N r f,

) (

2k 8

)

N r,1 S r f

(

,

)

f

 

− − ≤ + _{ }+

  (15)

The following two sub-cases are discussed: Sub-case 1.1 suppose

1 1

F G

U

F G

′ ′

= −

− − (16) If U=0_{, we have}

1 F DG= + −D where D≠0 _{is a constant. Then}

(

,

)

(

,

)

N r f =S r f . (17)

Suppose N r,1 S r f

(

,

)

f

 

≠

 

  , D=1, F G= , contradiction;

Suppose , 1

(

,

)

1

N r S r f f

 

≠

 ₋ 

  , D=1, F G= , contradiction.

So N r,1 S r f

(

,

)

f

 

=

 

  ,

(

)

1

, ,

1

N r S r f f

 

=

 ₋ 

  . (18)

If D≠1_, ,1 , 1

1

N r N r

G F D

 ₌  

   _{+ −} 

   ,

( )

(

)

1 1 1 1

, , , , ,

1 k

N r N r N r N r S r f

G f f f

 

   

 _{ = + + =}

 

   

  _{  −}

       

So , 1

(

,

)

1

N r S r f F D

 _{ =}

 _{+ −} 

  .

Then

(

) (

)

(

)

(

)

(

)

(

)

(

)

1 , , ,

1 1

, , , ,

1 ,

n T r f T r F S r f

N r N r N r F S r f

F F D

S r f

+ = +

   

≤ _{ }+ _{ }+ +

+ −

   

=

Obviously impossible.

Suppose U is not always equal to 0, let z1 be a zero of f with multiplicity q,

then z1 is a zero F with multiplicity nq and z1 is zero of _FF ₁

′

− with

multip-licity nq−1 _{at least.} z₀ is zero of _GG₋′₁ with multiplicity

(

n−1

)

q−1 _{at least.}

So z1 is a zero of U with multiplicity n−2 at least.

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

) (

)

(

)

1 1

2 , , , , ,

1 1

, , , ,

1 1

1

5 , 4 , ,

L L

n N r N r S r f N r U S r f

f U

N r f N r N r S r f

F G

k N r k N r f S r f f

   

− _{ }≤ _{ }+ ≤ +

 

 

   

≤ + _{ }+ _{ }+

− −

   

 

≤ + _{ }+ + +

DOI: 10.4236/jamp.2018.611188 2270 Journal of Applied Mathematics and Physics So

(

n k 7

)

N r,1

(

k 5

) (

N r f,

)

S r f

(

,

)

f

 

− − _{ }≤ + +

  (19)

Also

(

)

(

)

(

)

(

)

(

)

1 1

, , , , ,

1

1 1

2 , 2 , , ,

1

T r F N r N r N r F S r F

F F

N r N r f N r S r f

f F

   

≤ _{ }+ _{ }+ +

−

   

   

≤ _{ }+ + _{ }+

−

 

 

(

)

( )

(

)

( )

(

)

( ) ( )

(

)

(

)

(

)

1

1 1

, , , ,

1 1

1 1

, , ,

1

, , ,

k

k n

n

k k

f

N r N r T r S r f

F f f f f

f f

f f

m r N r S r f

f f

kN r f kN r S r f f

−

 

 

 

 _≤  _{≤ +}

 

 ₋     

  _ − _{  }

−

 ₋ 

 

   

= _ _+ _ _+

   

 

≤ + _{ }+

 

Then

(

,

) (

2

)

,1

(

2

) (

,

)

(

,

)

T r F k N r k N r f S r f f

 

≤ + _{ }+ + +

  (20)

If n−2k− >5 0_{, and} N r,1 S r f

(

,

)

f

 

=

 

  , N r f

(

,

)

=S r f

(

,

)

one of the two

forms is established. Then N r,1 N r f

(

,

)

S r f

(

,

)

f

 

+ =

 

  , substituting the above

formula is obviously impossible. Or T r F

(

,

)

=S r f

(

,

)

_{, contradiction.}

So N r,1 N r f

(

,

)

S r f

(

,

)

f

 

+ ≠

 

  , we get

(

) (

)

(

)

1

, , , , ,

N r S r f N r f S r f f

 

≠ ≠

 

  , then

(

7

)

,1

(

4 2

)(

8

)

,1

(

,

)

2 6

k k

n k N r N r S r f

f n k f

+ +

   

− − _{ }≤ _{ }+

− −

   

(

7

) (

4 2

)(

8

)

2 6

k k

n k

n k

+ +

− − ≤

− −

2 2

3 13 9 62 129 3 13 9 62 129

2 2

k+ − k + k+ _n k+ + k + k+

≤ ≤ ₍₂₁₎

DOI: 10.4236/jamp.2018.611188 2271 Journal of Applied Mathematics and Physics If N r,1 N r f

(

,

)

S r f

(

,

)

f

 

+ =

 

  , we get

(

)

(

)

1

, , ,

N r N r F S r F F

 _{ + =}

 

  ,

(

)

(

)

1

, , ,

N r N r G S r F G

 _{ + =}

 

 

According to Lemma 6, we get F G= or FG=1.

Firstly, if F G= _{, f}n

(

_{f − =1}

)

_fn−1

(

_{f −1}

)

_f( )k _{, f = f}( )k _{, conclusion} estab-lished.

Secondly, if FG=1_{, f}n

(

_{f − ⋅1}

)

_fn−1

(

_{f −1}

)

_f( )k _{=1, f}2 1n−

(

_{f −1}

)

2_f( )k _=1.

Obviously f is entire function. And

(

)

( )

₍

₎

(

)

(

)

(

)

(

)

(

)

(

)

2 1

2

1 1

, , , ,

1

, 2 , ,

3 , ,

n

k

T r f N r N r S r f

f f

T r f T r f S r f T r f S r f

− _≤ _ _+ _ _+

   ₋ 

  _{ }

= + +

= +

(

2 1n−

) (

T r f,

)

≤3T r f

(

,

) (

+S r f,

)

(

,

)

(

,

)

T r f =S r f _{, contradiction.}

Situation 2.2

If N r f

(

,

)

≠S r f

(

,

)

, we get B=0. So G− =1 A F

(

−1

)

_.

If N r f

(

,

)

≠S r f

(

,

)

, we get A=1_{, then} F G= . And if N r f

(

,

)

=S r f

(

,

)

_,

(

)

(

)

(

)

(

) (

)

(

)

1 1 1

, , , , , ,

1 1

2 , ,

T r F N r N r f N r N r S r f

f f F

k N r f S r f

     

≤ _{ }+ + _{ }+ _{ }+

−  − 

   

≤ + +

(

n+1

) (

T r f,

) (

≤ k+2

) (

N r f,

) (

+S r f,

)

We get n k≤ +1_{, that contradict with 2n≥3 13k+ + 9k}2_+62k+129

. Suppose N r f

(

,

)

=S r f

(

,

)

_,

If N r,1 S r f

(

,

)

f

 

≠

 

  we get B A= −1.

If A=1_{, we get} B=0, then F G= . If A≠1_{, we get}

1 1

1 ₁

1 1

G F

A G

A

−  

− = _{ }

−

 

− −

(

,

)

(

,

)

, 1₁

(

,

)

1

N r f N r G N r S r f

G A

 

 

= =  =

 ₋ 

 

 − 

DOI: 10.4236/jamp.2018.611188 2272 Journal of Applied Mathematics and Physics

(

) (

)

(

)

( )

(

)

(

)

(

)

(

) (

)

(

)

1 , , 1 ,

1 1

, , , ₁ ,

1

1 1

, , 3 , ,

n T r f T r F O T r G

N r G N r N r S r f

G _G

A

N r S r f k N r S r f

G f

+ = + =

 

 

 

≤ + _ _+  +

 ₋ 

 

 − 

 

 

≤ _{ }+ ≤ + _{ }+

   

We get n k≤ +2_{,which contradicts with 2n≥3 13k+ + 9k}2_+62k+129.

Therefore, Theorem 1 is proved.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this pa-per.

References

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[3] Alzahary, T.C. and Yi, H.X. (2004) Weighted Sharing Three Values and Uniqueness of Meromorphic Functions. Journal of Mathematical Analysis and Applications, 295, 247-257. https://doi.org/10.1016/j.jmaa.2004.03.040

[4] Yang, L.Z. and Zhang, J.L. (2008) Non-Existence of Meromorphic Solusions of Fermat Type Functional Equation. Aequationes Mathematicae, 76, 140-150.

https://doi.org/10.1007/s00010-007-2913-7

[5] Yang, L.Z. and Zhang, J.L. (2009) A Power of a Meromorphic Function Sharing a Small Function Sharing a Small Function with Its Derivative. Annales Academiae Scientiarum Fennicae Mathematica, 34, 249-260.

[6] Yi, H.X. (1997) Uniqueness Theorems for Meromorphic Functions Whose n-th De-rivatives Share the Same 1-Points. Complex Variables, Theory and Application, 34, 421-436.https://doi.org/10.1080/17476939708815064

[7] Yang, C.C. (1972) On Deficiencies of Differential Polynomial. Mathematische Zeit-schrift, 125, 107-112.https://doi.org/10.1007/BF01110921