# 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 , Yang  and Yi and Yang ). 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 )

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

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

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

In 1976, Rubel and Yang  proved the following result.

Theorem 1.1 

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

In 1996, Brück  gave the following conjecture.

Conjecture 1.1 

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  proved that Brück’s conjecture holds for entire functions of finite order and obtained the following result.

Theorem 1.2 [, 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 ).

In 2009, Chang and Zhu  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 [, 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.  got the following result.

Theorem 1.6 [, 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 

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 

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 

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 

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 [, 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 [, 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 [, 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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