# On some complex differential and difference equations concerning sharing function

## Abstract

By using the theory of complex differential equations, the purpose of this paper is to investigate a conjecture of Brück concerning an entire function f and its differential polynomial $L(f)= a k (z) f ( k ) +⋯+ a 0 (z)f$ sharing a function $α(z)$ and a constant β. We also study the problem on entire function and its difference polynomials sharing a function.

MSC:39A50, 30D35.

## 1 Introduction and main results

Let f be a nonconstant meromorphic function in the whole complex plane . We shall use the following standard notations of the value distribution theory:

$T(r,f),m(r,f),N(r,f), N ¯ (r,f),…$

(see Hayman [1], Yang [2] and Yi and Yang [3]). We denote by $S(r,f)$ any quantity satisfying $S(r,f)=o(T(r,f))$, as $r→+∞$, possibly outside of a set with finite measure. A meromorphic function $a(z)$ is called a small function with respect to f if $T(r,a)=S(r,f)$. In addition, we will use the notation $σ(f)$ to denote the order of meromorphic function $f(z)$, and $τ(f)$ to denote the type of an entire function $f(z)$ with $0<σ(f)=σ<+∞$, which are defined to be (see [1])

$σ(f)= lim sup r → ∞ log T ( r , f ) log r ,τ(f)= lim sup r → ∞ log M ( r , f ) r σ .$

We use $σ 2 (f)$ to denote the hyper-order of $f(z)$, $σ 2 (f)$ is defined to be (see [3])

$σ 2 (f)= lim sup r → ∞ log log T ( r , f ) log r .$

In 1976, Rubel and Yang [4] proved the following result.

Theorem 1.1 [4]

Let f be a nonconstant entire function. If f and $f ′$ share two finite distinct values CM, then $f≡ f ′$.

In 1996, Brück [5] gave the following conjecture.

Conjecture 1.1 [5]

Let f be a nonconstant entire function. Suppose that $σ(f)$ is not a positive integer or infinite, if f and $f ′$ share one finite value a CM, then

$f ′ − a f − a =c$

for some nonzero constant c.

In 1998, Gundersen and Yang [6] proved that Brück’s conjecture holds for entire functions of finite order and obtained the following result.

Theorem 1.2 [[6], Theorem 1]

Let f be a nonconstant entire function of finite order. If f and $f ′$ share one finite value a CM, then $f ′ − a f − a =c$ for some nonzero constant c.

The shared value problems relative to a meromorphic function f and its derivative $f ( k )$ have been a more widely studied subtopic of the uniqueness theory of entire and meromorphic functions in the field of complex analysis (see [712]).

In 2009, Chang and Zhu [13] further investigated the problem related to Brück’s conjecture and proved that Theorem 1.2 remains valid if the value a is replaced by a function.

Theorem 1.3 [[13], Theorem 1]

Let f be an entire function of finite order and $a(z)$ be a function such that $σ(a)<σ(f)<∞$. If f and $f ′$ share $a(z)$ CM, then $f ′ − a f − a =c$ for some nonzero constant c.

Thus, there are natural questions to ask:

1. (i)

What would happen when $σ(a)<σ(f)<∞$ is replaced by $0<σ(a)=σ(f)<∞$ in Theorem 1.3?

2. (ii)

For Theorems 1.1-1.3, what would happen when $f ′$ is replaced by differential polynomial

$L(f)= a k (z) f ( k ) + a k − 1 (z) f ( k − 1 ) +⋯+ a 1 (z) f ′ + a 0 (z)f,$
(1)

where $a k (z)(≢0),…, a 0 (z)$ are polynomials?

Theorem 1.4 Let $f(z)$ and $α(z)$ be two nonconstant entire functions and satisfy $0<σ(α)=σ(f)<∞$ and $τ(f)>τ(α)$, and $L(f)$ be stated as in (1) such that

$σ(f)>1+max { deg z a j − deg z a k k − j , 0 } .$

If $f(z)$ and $L(f(z))$ share $α(z)$ CM, then

$L ( f ( z ) ) − α ( z ) f ( z ) − α ( z ) =c$

for some nonzero constant c.

Theorem 1.5 Let $f(z)$ be a nonconstant transcendental entire function with $σ 2 (f)<∞$, let $σ 2 (f)$ be not an integer, and let $L(f)$ be stated as in (1). If f and $L(f)$ share a nonzero constant a CM and $δ(0,f)>0$, then

$L ( f ( z ) ) − a f ( z ) − a =c$

for some nonzero constant c.

Recently, some papers have studied Brück’s conjecture related to difference of entire function (including [14, 15]). In 2009, Heittokangas et al. [14] got the following result.

Theorem 1.6 [[14], Theorem 1]

Let f be a nonconstant meromorphic function of finite order $σ(f)<2$, and let η be a nonzero complex number. If $f(z+η)$ and $f(z)$ share a finite complex value a CM, then $f(z+η)−a=c(f(z)−a)$ for all $z∈C$, where c is some nonzero complex number.

In this paper, we further investigate Brück’s conjecture related to entire function and its difference polynomial and obtain the following result.

Theorem 1.7 Let $f(z)$ be a nonconstant entire function of finite order $0<σ(f)<∞$, $L 1 (f)$ be difference polynomial of f of the form

$L 1 ( f ( z ) ) =f(z+ η k )+f(z+ η k − 1 )+⋯+f(z+ η 1 ),$

where $η k , η k − 1 ,…, η 1$ are nonzero complex numbers. If $L 1 (f(z))=cf(z)$ and ξ (≠0) is a Borel exceptional value of $f(z)$, then $L 1 (f)=kf(z)$.

## 2 Some lemmas

To prove our theorems, we will require some lemmas as follows.

Lemma 2.1 [16]

Let $f(z)$ be a transcendental entire function, $ν(r,f)$ be the central index of $f(z)$. Then there exists a set $E⊂(1,+∞)$ with finite logarithmic measure, we choose z satisfying $|z|=r∉[0,1]∪E$ and $|f(z)|=M(r,f)$, we get

Lemma 2.2 [17]

Let $f(z)$ be an entire function of finite order $σ(f)=σ<∞$, and let $ν(r,f)$ be the central index of f. Then, for any ε (>0), we have

$lim sup r → ∞ log ν ( r , f ) log r =σ.$

Lemma 2.3 [18]

Let f be a transcendental entire function, and let $E⊂[1,+∞)$ be a set having finite logarithmic measure. Then there exists ${ z n = r n e i θ n }$ such that $|f( z n )|=M( r n ,f)$, $θ n ∈[0,2π)$, $lim n → ∞ θ n = θ 0 ∈[0,2π)$, $r n ∉E$ and if $0<σ(f)<∞$, then, for any given $ε>0$ and sufficiently large $r n$,

$r n σ ( f ) − ε <ν( r n ,f)< r n σ ( f ) + ε .$

Lemma 2.4 [16]

Let $P(z)= b n z n + b n − 1 z n − 1 +⋯+ b 0$ with $b n ≠0$ be a polynomial. Then, for every $ε>0$, there exists $r 0 >0$ such that for all $r=|z|> r 0$ the inequalities

$(1−ε)| b n | r n ≤|P(z)|≤(1+ε)| b n | r n$

hold.

Lemma 2.5 Let $f(z)$ and $A(z)$ be two entire functions with $0<σ(f)=σ(A)=σ<∞$, $0<τ(A)<τ(f)<∞$, then there exists a set $E⊂[1,+∞)$ that has infinite logarithmic measure such that for all $r∈E$ and a positive number $κ>0$, we have

$M ( r , A ) M ( r , f )

Proof By definition, there exists an increasing sequence ${ r m }→∞$ satisfying $(1+ 1 m ) r m < r m + 1$ and

$lim m → ∞ log M ( r m , f ) r m σ =τ(f).$
(2)

For any given β ($τ(A)<β<τ(f)$), there exists some positive integer $m 0$ such that for all $m≥ m 0$ and for any given ε ($0<ε<τ(f)−β$), we have

$logM( r m ,f)> ( τ ( f ) − ε ) r m σ .$
(3)

Thus, there exists some positive integer $m 1$ such that for all $m≥ m 1$, we have

$( m m + 1 ) σ > β τ ( f ) − ε .$
(4)

From (2)-(4), for all $m≥ m 2 =max{ m 0 , m 1 }$ and for any $r∈[ r m ,(1+ 1 m ) r m ]$, we have

$M ( r , f ) ≥ M ( r m , f ) > exp { ( τ ( f ) − ε ) r m σ } ≥ exp { ( τ ( f ) − ε ) ( m m + 1 r ) σ } > exp { β r σ } .$
(5)

Set $E= ⋃ m = m 2 ∞ [ r m ,(1+ 1 m ) r m ]$, then

$m l E= ∑ m = m 2 ∞ ∫ r m ( 1 + 1 m ) r m d t t = ∑ m = m 2 ∞ log ( 1 + 1 m ) =∞.$

From the definition of type of entire function, for any sufficiently small $ε>0$, we have

$M(r,A)
(6)

By (5) and (6), set $κ=β−τ(A)−ε$, for all $r∈E$, we have

$M ( r , A ) M ( r , f )

Thus, this completes the proof of this lemma. □

Lemma 2.6 [[19], Theorem 2.1]

Let $f(z)$ be a meromorphic function of finite order σ, and let η be a fixed nonzero complex number, then, for each $ε>0$, we have

$m ( r , f ( z + c ) f ( z ) ) +m ( r , f ( z ) f ( z + c ) ) =O ( r σ − 1 + ε ) .$

Lemma 2.7 [[19], Corollary 2.5]

Let $f(z)$ be a meromorphic function with order $σ=σ(f)$, $σ<+∞$, and let η be a fixed nonzero complex number, then, for each $ε>0$, we have

$T ( r , f ( z + η ) ) =T(r,f)+O ( r σ − 1 + ε ) +O(logr).$

Lemma 2.8 [1, 20]

Let $g:(0,+∞)→R$, $h:(0,+∞)→R$ be monotone increasing functions such that $g(r)≤h(r)$ outside of an exceptional set E with finite linear measure, or $g(r)≤h(r)$, $r∉H∪(0,1]$, where $H⊂(1,∞)$ is a set of finite logarithmic measure. Then, for any $α>1$, there exists $r 0$ such that $g(r)≤h(αr)$ for all $r≥ r 0$.

## 3 The proof of Theorem 1.4

Since $f(z)$ is an entire function, and $f(z)$ and $L(f(z))$ share $α(z)$ CM, then there is an entire function $γ(z)$ such that

$L ( f ( z ) ) − α ( z ) f ( z ) − α ( z ) = e γ ( z ) .$
(7)

Next, we will claim that $γ(z)$ is a constant.

Suppose that $γ(z)$ is transcendental. It follows that $σ( e γ ( z ) )=∞$. However, since $0<σ(f)=σ(α)<∞$, it follows from the left-hand side of (7) that $σ( L ( f ( z ) ) − α ( z ) f ( z ) − α ( z ) )<∞$, a contradiction. Thus, $γ(z)$ is not transcendental.

Suppose that $γ(z)$ is a nonconstant polynomial, let

$γ(z)= b m z m + b m − 1 z m − 1 +⋯+ b 0 ,$

where $b m ,…, b 0$ are constants and $b m ≠0$, $m≥1$. Thus, it follows from (7) and Lemma 2.4 that

$| b m | r m ( 1 + o ( 1 ) ) =|γ(z)|=|log L ( f ( z ) ) f ( z ) − α ( z ) f ( z ) 1 − α ( z ) f ( z ) |.$
(8)

Since $L(f)= a k f k + a k − 1 f ( k − 1 ) +⋯+ a 0 f$, from Lemma 2.1, then there exists a subset $E 1 ⊂(1,+∞)$ with finite logarithmic measure such that for some point $z=r e i θ$ ($θ∈[0,2π)$), $r∉ E 1$ and $M(r,f)=|f(z)|$, we have

$f ( j ) ( z ) f ( z ) = { ν ( r , f ) z } j ( 1 + o ( 1 ) ) ,1≤j≤k.$

Thus, it follows that

$L ( f ( z ) ) f ( z ) = a k { ν ( r , f ) z } k ( 1 + o ( 1 ) ) + ⋯ + a 1 { ν ( r , f ) z } ( 1 + o ( 1 ) ) + a 0 = a k z k ( 1 + o ( 1 ) ) [ ν ( r , f ) k + ∑ j = 1 k a k − j a k z j ν ( r , f ) k − j ( 1 + o ( 1 ) ) ] .$
(9)

From Lemma 2.3, there exists ${ z n = r n e i θ n }$ such that $|f( z n )|=M( r n ,f)$, $θ n ∈[0,2π)$, $lim n → ∞ θ n = θ 0 ∈[0,2π)$, $r n ∉ E 1$, then, for any given ε satisfying

$0<ε< min 1 ≤ j ≤ k j σ ( f ) − j − d k − j 3 k − j ,$

where $d k − j = deg z a k − j − deg z a k$, and sufficiently large $r n$, we have

$r n σ ( f ) − ε <ν( r n ,f)< r n σ ( f ) + ε .$
(10)

Since $a j (z)$, $j=0,1,…,k$, are polynomials, let $a j (z)= ∑ t = 0 s j l j t z t$, where $s j = deg z a j$, $j=0,1,…,k$. Then, from Lemma 2.4 and (10), we have

$| a k − j a k z j ν ( r , f ) k − j ( 1 + o ( 1 ) ) | ≤ M | l k − j , s k − j | r n s k − j | l k , s k | r n s k r n j r n ( σ ( f ) + ε ) ( k − j ) = M | l k − j , s k − j | | l k , s k | r n d k − j + j + ( σ ( f ) + ε ) ( k − j ) ≤ M | l k − j , s k − j | | l k , s k | r n k σ ( f ) − j σ ( f ) + d k − j + j + ( k − j ) ε ,$
(11)

where $d k − j = s k − j − s k$ and M is a positive constant. Since $−jσ(f)+ d k − j +j+(k−j)ε<−2kε<0$, it follows from (11) that

$| a k − j a k z j ν ( r , f ) k − j ( 1 + o ( 1 ) ) |
(12)

Since $0<σ(α)=σ(f)<∞$ and $τ(α)<τ(f)<∞$, from Lemma 2.5, there exists a set $E⊂[1,+∞)$ that has infinite logarithmic measure such that for a sequence ${ r n } 1 ∞ ∈ E 2 =E− E 1$, we have

(13)

From (8), (9), (12), (13) and Lemma 2.2, we can get that

$| b m | r n m ( 1 + o ( 1 ) ) =|γ(z)|=O(log r n ),$
(14)

which is impossible. Thus, $γ(z)$ is not a polynomial.

Therefore, $γ(z)$ is a constant, that is, there exists some nonzero constant c such that

$L ( f ( z ) ) − α ( z ) f ( z ) − α ( z ) =c.$

Thus, this completes the proof of Theorem 1.4.

## 4 The proof of Theorem 1.5

Since $L(f)$ and f share the constant a CM, then there exists an entire function $φ(z)$ such that

$L ( f ) − a f − a = e φ .$
(15)

We will consider two cases as follows.

Case 1. If $a=0$, it follows from (15) that

$L ( f ( z ) ) f ( z ) = e φ ( z ) .$
(16)

Since $L(f(z))= a k (z) f ( k ) (z)+⋯+ a 1 (z) f ′ (z)+ a 0 (z)$ and $a j (z)$, $j=0,1,…,k$, are polynomials, it follows from (16) that

$T ( r , e φ ) =m ( r , e φ ) =m ( r , L ( f ) f ) ≤ ∑ j = 1 k m ( r , f ( j ) f ) + ∑ i = 0 k m(r, a i )=O ( log r T ( r , f ) ) ,$

outside of an exceptional set $E 3$ with finite linear measure. Thus, there exists a constant K such that

By Lemma 2.8, there exists an $r 0 >0$, and for all $r≥ r 0$, we have

(17)

Thus, we can deduce from (17) that $σ( e φ )≤ σ 2 (f)<∞$, that is, $φ(z)$ is a polynomial.

By using the same argument as in [[21], Theorem 1.1], we can get that $σ 2 (f)= deg z φ$, which is a contradiction to $σ 2 (f)$ is not a positive integer. Thus, $φ(z)$ is only a constant, it follows from (15) that $L(f(z))=cf(z)$, where c is a nonzero constant.

Case 2. If $a≠0$, from the derivation of (15) and eliminating $e φ$, we can get

$φ ′ (z)= L ′ ( f ( z ) ) L ( f ( z ) ) − a − f ′ ( z ) f ( z ) − a .$
(18)

If $φ ′ (z)≡0$, that is, $φ(z)≡c$, c is a constant. Thus, we can prove the conclusion of Theorem 1.5 easily.

If $φ ′ (z)≢0$, then it follows from (18) that

$m ( r , φ ′ ) =S(r,f).$
(19)

We can rewrite (18) in the following form:

$φ ′ = f [ L ( f ) f 1 L ( f ) L ′ ( f ) L ( f ) − a − 1 f f ′ f − a ] = f a [ L ( f ) f L ′ ( f ) L ( f ) − a − L ′ ( f ) f − f ′ f − a + f ′ f ] .$
(20)

Since $φ ′ ≢0$ and f is transcendental, set

$Ψ:= L ( f ) f L ′ ( f ) L ( f ) − a − L ′ ( f ) f − f ′ f − a + f ′ f ,$
(21)

then we have $m(r,Ψ)=S(r,f)$. Thus, it follows from (20) and (21) that

$a f ( z ) = Ψ ( z ) φ ′ ( z ) .$
(22)

Since $φ(z)$ is an entire function, from (18)-(22), then we have

$m ( r , 1 f ) ≤ m ( r , Ψ ) + m ( r , 1 φ ′ ) ≤ S ( r , f ) + T ( r , φ ′ ) = S ( r , f ) + m ( r , φ ′ ) = S ( r , f ) .$

It follows that

$δ(0,f)= lim inf r → ∞ m ( r , 1 f ) T ( r , f ) =0,$

which is a contradiction to the assumption of Theorem 1.5.

Thus, from Case 1 and Case 2, we complete the proof of Theorem 1.5.

## 5 The proof of Theorem 1.7

Since $f(z)$ is an entire function of finite order $0<σ(f)<∞$ and ξ (≠0) is a Borel exceptional value of $f(z)$, then $f(z)$ can be written in the form

$f(z)=ξ+p(z) e h ( z ) ,$
(23)

where $h(z)$ is a polynomial of degree l and $p(z)$ is an entire function satisfying $σ(p(z))<σ(f(z))= deg z h(z)=l$. Thus, we have

$f(z+ η j )=ξ+p(z+ η j ) e h ( z + η j ) ,j=1,2,…,k.$
(24)

From Lemma 2.7, we have $σ(p(z+ η j ))<σ(f(z+ η j ))=σ(f(z))$ and $deg z h(z+ η j )= deg z h(z)=l$ for $j=1,2,…,k$. Since $L 1 (f(z))=cf(z)$, it follows from (23) and (24) that

$∑ j = 1 k p(z+ η j ) e h ( z + η j ) =(c−k)d+cp(z) e h ( z ) .$
(25)

Set $h(z)= μ l z l +⋯$ and $μ l ≠0$, then we can deduce from (25) that

$∑ j = 1 k p(z+ η j ) e μ m − 1 j z m − 1 + ⋯ = ( c − k ) d + c p ( z ) e h ( z ) e μ l z l .$
(26)

Let $Φ:= ∑ j = 1 k p(z+ η j ) e μ m − 1 j z m − 1 + ⋯$, it is easy to see that $Φ≢0$ and $σ(Φ)<σ(f)$, that is, $T(r,Φ)=o(T(r,f))=o(T(r, e h ( z ) ))$.

Suppose that $c≠k$. Since $ξ≠0$, it follows from (26) that

$N ( r , 1 e h ( z ) − ( c − k ) ξ c p ( z ) ) =N ( r , 1 Φ ) ≤T(r,Φ)=S ( r , e h ( z ) ) .$

By the second fundamental theorem concerning small functions, for any $ε>0$, we have

$T ( r , e h ( z ) ) ≤ N ( r , 1 e h ( z ) − ( c − k ) ξ c p ( z ) ) + ε T ( r , e h ( z ) ) + S ( r , e h ( z ) ) = ε T ( r , e h ( z ) ) + S ( r , e h ( z ) ) .$

Since ε is arbitrary, we can get a contradiction from the above inequality. Thus, we can get that $c=k$.

Therefore, we prove that $L 1 (f(z))=kf(z)$, that is, the conclusion of Theorem 1.7 holds.

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## Acknowledgements

The authors thank the referee for his/her valuable suggestions to improve the present article. The first author was supported by the NSF of China (11301233, 61202313), the Natural Science Foundation of Jiang-Xi Province in China (20132BAB211001), and the Foundation of Education Department of Jiangxi (GJJ14644) of China. The second author was supported by the NSF of China (11371225, 11171013).

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Correspondence to Hua Wang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

HW, L-ZY and H-YX completed the main part of this article, HW, H-YX corrected the main theorems. All authors read and approved the final manuscript.

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Wang, H., Yang, LZ. & Xu, HY. On some complex differential and difference equations concerning sharing function. Adv Differ Equ 2014, 274 (2014). https://doi.org/10.1186/1687-1847-2014-274