Uniqueness of Meromorphic Functions Whose Differential Polynomials Share One Value

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

• 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 meromorphic 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 common 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 g CM share a. If f − a and g − a have the same zero point without counting the number, then f and g IM share a [2] . In addition, the following definitions are required: let p be a positive integer, and a ∈ C ∪ { ∞ } . Next

N p ) ( r , 1 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 function of the weight of the value point a within | z | < r . N ¯ ( p ( r , 1 f − a ) means corresponding reduced count function. Suppose k is a non-negative number. Mark N k ( r , 1 f − a ) defined as follows. See the literature [3] for details.

N k ( r , 1 f − a ) = N ¯ ( r , 1 f − a ) + N ¯ ( 2 ( r , 1 f − a ) + ⋅ ⋅ ⋅ + N ¯ ( k ( r , 1 f − a )

Before, Xiaomin Li and Zhitao Wen expanded Jilong Zhang’s theorem, where ( f n ) ′ changes to f n − 1 f ( k ) , so when k = 1 , that is Zhang’s theorem. Similarly, in this paper, we continuously change f n − 1 f ( k ) to f n − 1 ( f − 1 ) f ( k ) , which contained 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 positive integer, if f n and ( f n ) ′ CM share 1, then f n = ( f n ) ′ .

Theorem B [4] Suppose f is a non-constant meromorphic function, n ≥ 12 is a positive integer, if f n and ( f n ) ′ CM share 1, then f n = ( f n ) ′ .

Recently Zhang Jilong improved the above theorem. Get the following result:

Theorem C [5] Suppose f is a non-constant entire function, n ≥ 3 is a positive integer, if f n and ( f n ) ′ CM share 1, then f n = ( f n ) ′ .

Theorem D [5] Suppose f is a non-constant meromorphic function, n ≥ 4 is a positive integer, if f n and ( f n ) ′ CM share 1, then f n = ( f n ) ′ .

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 positive integer, n is a positive integer and satisfies 2 n ≥ k + 3 + k 2 + 2 k + 9 , if f n and ( f n ) ′ CM share 1, then f n = ( f n ) ′ .

Theorem F [5] Suppose f is a non-constant meromorphic function, k is a positive integer, n is a positive integer and satisfies 2 n ≥ 3 k + 8 + 9 k 2 + 40 k + 64 , if f n and ( f n ) ′ IM share 1, then f n = ( f n ) ′ .

Now we mainly improve the theorem of Li Xiaomin. Which that changes f n and f n − 1 f ( k ) to f n ( f − 1 ) and f n − 1 ( f − 1 ) f ( k ) . We get the following theorem:

Theorem 1 Suppose f is a non-constant meromorphic function, k is a positive integer, n is a positive integer and satisfies 2 n ≥ 3 k + 13 + 9 k 2 + 62 k + 129 , if f n ( f − 1 ) and f n − 1 ( f − 1 ) f ( k ) IM share 1, and the zeros of f − 1 with multiplicity 2 at least., then f = f ( k ) .

2. Some Lemmas

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

H = ( F ″ F ′ − 2 F ′ F − 1 ) − ( G ″ G ′ − 2 G ′ G − 1 )

and suppose H ≠ 0 , if F and G IM share 1, then

N E 1 ) ( r , 1 G − 1 ) ≤ N ( r , H ) + S ( r , F ) + S ( r , G )

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

P ( f ) = a n f n + a n − 1 f n − 1 + ⋅ ⋅ ⋅ + a 1 f

where a 1 , a 2 , ⋯ , a n − 1 , a n ( ≠ 0 ) are constants, then T ( r , P ( f ) ) = n T ( r , f ) + O ( 1 ) .

Lemma 3 Let f be a non-constant meromorphic function, k ( ≥ 1 ) and n ( ≥ 2 ) 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

T ( r , F ) = ( n + 1 ) T ( r , f ) + O ( 1 ) (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

T ( r , G ) ≤ ( n − 1 ) T ( r , f ) + T ( r , f − 1 ) + T ( r , f ( k ) ) ≤ ( n − 1 ) T ( r , f ) + T ( r , f ) + k N ¯ ( r , f ) + S ( r , f ) ≤ ( n + k ) T ( r , f ) + S ( r , f ) (3)

According to the second basic theorem and (2)

T ( r , F ) ≤ N ¯ ( r , 1 F − 1 ) + N ¯ ( r , 1 F ) + N ¯ ( r , F ) + S ( r , F ) = N ¯ ( r , 1 G − 1 ) + N ¯ ( r , 1 f ) + N ¯ ( r , f ) + S ( r , f ) ≤ T ( r , G ) + 2 T ( r , f ) + S ( r , f )

The above formula is combined with (1) to get

T ( r , G ) ≥ ( n − 1 ) 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

N p ( r , 1 f ( k ) ) ≤ k N ¯ ( r , f ) + N p + k ( r , 1 f ) + S ( r , f )

N ¯ ( r , 1 f − 1 ) ≤ N ( r , f f ′ ) ≤ N ¯ ( r , f ) + N ¯ ( r , 1 f ) + S ( r , 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 G = f n − 1 ( f − 1 ) f ( k ) , if F and G IM share 1, then

a) N ¯ ( r , 1 G ) ≤ ( k + 3 ) N ¯ ( r , 1 f ) + ( k + 1 ) N ¯ ( r , f ) + S ( r , f )

b) N ¯ L ( r , 1 F − 1 ) ≤ 2 N ¯ ( r , 1 f ) + 2 N ¯ ( r , f ) + S ( r , f ) ;

c) N ¯ L ( r , 1 G − 1 ) ≤ ( k + 3 ) N ¯ ( r , 1 f ) + ( k + 2 ) N ¯ ( r , f ) + S ( r , f ) .

Proof According to Lemma 4, we have

N ¯ ( r , 1 G ) ≤ N ¯ ( r , 1 f ) + N ¯ ( r , 1 f − 1 ) + N 1 ( r , 1 f ( k ) ) + S ( r , f ) ≤ N ¯ ( r , 1 f ) + N ¯ ( r , f ) + N ¯ ( r , 1 f ) + N k + 1 ( r , 1 f ) + k N ¯ ( r , f ) + S ( r , f ) ≤ ( k + 3 ) N ¯ ( r , 1 f ) + ( k + 1 ) N ¯ ( r , f ) + S ( r , f )

This leads to the conclusion (a), obtained from the definition of the N L ( r , 1 F − 1 ) and Lemma 3:

N L ( r , 1 F − 1 ) ≤ N ( r , F F ′ ) ≤ T ( r , F ′ F ) + O ( 1 ) = m ( r , F ′ F ) + N ( r , F ′ F ) + O ( 1 ) = N ¯ ( r , F ) + N ¯ ( r , 1 F ) + S ( r , F ) = 2 N ¯ ( r , f ) + 2 N ¯ ( r , 1 f ) + S ( r , f )

This leads to conclusions (b), the same reason

N ¯ L ( r , 1 G − 1 ) ≤ N ¯ ( r , 1 G ) + N ¯ ( r , G ) + S ( r , f )

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 satisfy N ¯ ( r , F ) + N ¯ ( r , 1 F ) = S ( r , F ) and N ¯ ( r , G ) + N ¯ ( r , 1 G ) = S ( r , G ) . If F and G IM share a non-zero constant a, then F = G or F G = 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

V = ( F ′ F − 1 − F ′ F ) − ( G ′ G − 1 − G ′ G ) (7)

If V = 0 ,

1 − 1 F = B − B G (8)

where B ≠ 0 is a constant, if N ¯ ( r , f ) ≠ S ( r , f ) .

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

So F = G , contradictory with the assumption of case 1.

Therefore, N ¯ ( r , f ) = S ( r , f ) .

So B ≠ 1 , by (8), we get

G = B F ( B − 1 ) F + 1 = B F F − 1 1 − B ( 1 B − 1 ) (9)

N ¯ ( r , 1 F − 1 B − 1 ) = S ( r , f ) (10)

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

T ( r , F ) ≤ N ¯ ( r , 1 F ) + N ¯ ( r , 1 F − 1 B − 1 ) + N ¯ ( r , F ) + S ( r , F ) ≤ N ¯ ( r , 1 f ) + N ¯ ( r , 1 f − 1 ) + S ( r , f ) ≤ 2 N ¯ ( r , f ) + S ( r , f ) (11)

By lemma 2, we have

T ( r , F ) = ( n + 1 ) T ( r , f ) + O ( 1 ) (12)

Then

( n + 1 ) T ( r , f ) − 3 T ( 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

V = F ′ F ( F − 1 ) − G ′ G ( G − 1 ) (14)

Suppose z 0 is a pole of f with multiplicity p, then z 0 is pole of F with multiplicity ( n + 1 ) p . and z 0 is zero of F ′ F ( F − 1 ) with multiplicity ( n + 1 ) p − 1 at least. z 0 is zero of G ′ G ( G − 1 ) with multiplicity ( n + 1 ) p + k − 1 at least.

So z 0 is a zero of V with multiplicity ≥ n at least.

n N ¯ ( r , f ) ≤ N ( r , 1 V ) + S ( r , f ) ≤ N ( r , V ) + S ( r , f ) ≤ N ¯ ( r , 1 G ) + N ¯ L ( r , 1 F − 1 ) + N ¯ L ( r , 1 G − 1 ) + S ( r , f ) ≤ ( k + 3 ) N ¯ ( r , 1 f ) + ( k + 1 ) N ¯ ( r , f ) + 2 N ¯ ( r , f )     + 2 N ¯ ( r , 1 f ) + ( k + 3 ) N ¯ ( r , 1 f ) + ( k + 2 ) N ¯ ( r , f ) + S ( r , f ) ≤ ( 2 k + 8 ) N ¯ ( r , 1 f ) + ( 2 k + 5 ) N ¯ ( r , f ) + S ( r , f )

Then

( n − 2 k − 5 ) N ¯ ( r , f ) ≤ ( 2 k + 8 ) N ¯ ( r , 1 f ) + S ( r , f ) (15)

The following two sub-cases are discussed:

Sub-case 1.1 suppose

U = F ′ F − 1 − G ′ G − 1 (16)

If U = 0 , we have

F = D G + 1 − D

where D ≠ 0 is a constant. Then

N ¯ ( r , f ) = S ( r , f ) . (17)

Suppose N ¯ ( r , 1 f ) ≠ S ( r , f ) , D = 1 , F = G , contradiction;

Suppose N ¯ ( r , 1 f − 1 ) ≠ S ( r , f ) , D = 1 , F = G , contradiction.

So N ¯ ( r , 1 f ) = S ( r , f ) , N ¯ ( r , 1 f − 1 ) = S ( r , f ) . (18)

If D ≠ 1 , N ¯ ( r , 1 G ) = N ¯ ( r , 1 F + D − 1 ) ,

N ¯ ( r , 1 G ) = N ¯ ( r , 1 f ) + N ¯ ( r , 1 f ( k ) ) + N ¯ ( r , 1 f − 1 ) = S ( r , f )

So N ¯ ( r , 1 F + D − 1 ) = S ( r , f ) .

Then

( n + 1 ) T ( r , f ) = T ( r , F ) + S ( r , f ) ≤ N ¯ ( r , 1 F ) + N ¯ ( r , 1 F + D − 1 ) + N ¯ ( r , F ) + S ( r , f ) = S ( r , f )

Obviously impossible.

Suppose U is not always equal to 0, let z 1 be a zero of f with multiplicity q, then z 1 is a zero F with multiplicity nq and z 1 is zero of F ′ F − 1 with multiplicity n q − 1 at least. z 0 is zero of G ′ G − 1 with multiplicity ( n − 1 ) q − 1 at least.

So z 1 is a zero of U with multiplicity n − 2 at least.

( n − 2 ) N ¯ ( r , 1 f ) ≤ N ( r , 1 U ) + S ( r , f ) ≤ N ( r , U ) + S ( r , f ) ≤ N ¯ ( r , f ) + N ¯ L ( r , 1 F − 1 ) + N ¯ L ( r , 1 G − 1 ) + S ( r , f ) ≤ ( k + 5 ) N ¯ ( r , 1 f ) + ( k + 4 ) N ¯ ( r , f ) + S ( r , f )

So

( n − k − 7 ) N ¯ ( r , 1 f ) ≤ ( k + 5 ) N ¯ ( r , f ) + S ( r , f ) (19)

Also

T ( r , F ) ≤ N ¯ ( r , 1 F ) + N ¯ ( r , 1 F − 1 ) + N ¯ ( r , F ) + S ( r , F ) ≤ 2 N ¯ ( r , 1 f ) + 2 N ¯ ( r , f ) + N ¯ ( r , 1 F − 1 ) + S ( r , f )

N ¯ ( r , 1 F − 1 ) ≤ N ¯ ( r , 1 f n − 1 ( f − 1 ) f ( k ) f n ( f − 1 ) − 1 ) ≤ T ( r , f ( k ) f ) + S ( r , f ) = m ( r , f ( k ) f ) + N ¯ ( r , f ( k ) f ) + S ( r , f ) ≤ k N ¯ ( r , f ) + k N ¯ ( r , 1 f ) + S ( r , f )

Then

T ( r , F ) ≤ ( k + 2 ) N ¯ ( r , 1 f ) + ( k + 2 ) N ¯ ( r , f ) + S ( r , f ) (20)

If n − 2 k − 5 > 0 , and N ¯ ( r , 1 f ) = S ( r , f ) , N ¯ ( r , f ) = S ( r , f ) one of the two forms is established. Then N ¯ ( r , 1 f ) + N ¯ ( r , f ) = S ( r , f ) , substituting the above formula is obviously impossible.

Or T ( r , F ) = S ( r , f ) , contradiction.

So N ¯ ( r , 1 f ) + N ¯ ( r , f ) ≠ S ( r , f ) , we get

N ¯ ( r , 1 f ) ≠ S ( r , f ) , N ¯ ( r , f ) ≠ S ( r , f ) , then

( n − k − 7 ) N ¯ ( r , 1 f ) ≤ ( k + 4 ) ( 2 k + 8 ) n − 2 k − 6 N ¯ ( r , 1 f ) + S ( r , f )

( n − k − 7 ) ≤ ( k + 4 ) ( 2 k + 8 ) n − 2 k − 6

3 k + 13 − 9 k 2 + 62 k + 129 2 ≤ n ≤ 3 k + 13 + 9 k 2 + 62 k + 129 2 (21)

Case 2

Situation 2.1

If N ¯ ( r , 1 f ) + N ¯ ( r , f ) = S ( r , f ) , we get

N ¯ ( r , 1 F ) + N ¯ ( r , F ) = S ( r , F ) , N ¯ ( r , 1 G ) + N ¯ ( r , G ) = S ( r , F )

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

Firstly, if F = G , f n ( f − 1 ) = f n − 1 ( f − 1 ) f ( k ) , f = f ( k ) , conclusion established.

Secondly, if F G = 1 , f n ( f − 1 ) ⋅ f n − 1 ( f − 1 ) f ( k ) = 1 , f 2 n − 1 ( f − 1 ) 2 f ( k ) = 1 .

Obviously f is entire function.

And

T ( r , f 2 n − 1 ) ≤ N ¯ ( r , 1 f ( k ) ) + N ¯ ( r , 1 ( f − 1 ) 2 ) + S ( r , f ) = T ( r , f ) + 2 T ( r , f ) + S ( r , f ) = 3 T ( r , f ) + S ( r , f )

( 2 n − 1 ) T ( r , f ) ≤ 3 T ( 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 ) ,

T ( r , F ) ≤ N ¯ ( r , 1 f ) + N ¯ ( r , f ) + N ¯ ( r , 1 f − 1 ) + N ¯ ( r , 1 F − 1 ) + S ( r , f ) ≤ ( k + 2 ) 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 2 n ≥ 3 k + 13 + 9 k 2 + 62 k + 129 .

Suppose N ¯ ( r , f ) = S ( r , f ) ,

If N ¯ ( r , 1 f ) ≠ S ( r , f ) we get B = A − 1 .

If A = 1 , we get B = 0 , then F = G .

If A ≠ 1 , we get

F − 1 = G − 1 G − 1 A − 1 ( 1 A − 1 )

N ¯ ( r , f ) = N ¯ ( r , G ) = N ¯ ( r , 1 G − 1 A − 1 ) = S ( r , f )

And T ( r , F ) = T ( r , G ) + O (1)

( n + 1 ) T ( r , f ) = T ( r , F ) + O ( 1 ) = T ( r , G ) ≤ N ¯ ( r , G ) + N ¯ ( r , 1 G ) + N ¯ ( r , 1 G − 1 A − 1 ) + S ( r , f ) ≤ N ¯ ( r , 1 G ) + S ( r , f ) ≤ ( k + 3 ) N ¯ ( r , 1 f ) + S ( r , f )

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

Therefore, Theorem 1 is proved.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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