Existence of entire solutions of some non linear differential difference equations

R E S E A R C H

Open Access

Existence of entire solutions of some

non-linear differential-difference equations

Minfeng Chen

_{, Zongsheng Gao and Yunfei Du}

*_{Correspondence:}

[email protected] LMIB and School of Mathematics and Systems Science, Beihang University, Beijing, 100191, P.R. China

Abstract

In this paper, we investigate the admissible entire solutions of finite order of the differential-difference equations (f(z))2+P2(z)f2(z+c) =Q(z)eα(z)and

(f(z))2+ [f(z+c) –f(z)]2=Q(z)eα(z), whereP(z),Q(z) are two non-zero polynomials,

α

(f(z))n_+P(z)fm_{(z+c) =Q(z), whereP(z),Q(z) are two non-constant polynomials,} c∈C\{0},m,nare positive integers and satisfy_m1 +1_n< 2 except form= 1,n= 2.

MSC: 39B32; 39A10; 34M05

Keywords: differential-difference equation; admissible entire solution; finite order

1 Introduction and main results

In this paper, we assume that the reader is familiar with the standard symbols and funda-mental results of Nevanlinna theory [, ]. In addition, we denote byS(r,f) any quantify satisfyingS(r,f) =o(T(r,f)), asr→ ∞, outside of a possible exceptional set of finite log-arithmic measure. We define the loglog-arithmic measure ofEto belm(E) =_{E∩(,∞)}dr_r. A set E⊂(,∞) is said to have finite logarithmic measure iflm(E) <∞. Throughout this paper, all constants are complex constants unless otherwise specified.

Nevanlinna’s value theory of meromorphic functions has been used to study the proper-ties of entire or meromorphic solutions of differential equations and difference equations in complex plane, such as [–]. In [], Montel stated the following theorem.

Theorem A Let f(z), g(z)be two transcendental entire functions. Then if m and n are integers≥,the functional equation

fn(z) +gn(z) =  (.)

cannot hold.

However, whenn=  andg(z) has a specific relationship withf(z) in (.), the problem that whether we can obtain the accurate expressions of entire solutions or not is worth to be considered. Recently, many results focused on this problem that were obtained by using the Nevanlinna theory, such as [–].

In [], Liuet al.considered Fermat type differential-difference equation and obtained the following results.

Theorem B The transcendental entire solutions with finite order of

f(z)+f(z+c) =  (.)

must satisfy f(z) =sin(z±iB),where B is a constant and c= kπor c= kπ+π,k is an integer.

Theorem C The transcendental entire solutions with finite order of

f(z)+f(z+c) –f(z)=  (.)

must satisfy f(z) =_sin(z+iB),where c=kπ+π_,k is an integer,and B is a constant.

In [], Chen and Gao improved Theorem B and obtained the following result.

Theorem D Let P(z),Q(z)be two non-zero polynomials.If the differential-difference equa-tion

f(z)+P(z)f(z+c) =Q(z) (.)

admits a transcendental entire solution of finite order,then P(z),Q(z)reduce to constants, and

f(z) =pe

az+b_–qe–(az+b)

a ,

where a=±iA,A=(–)_ckkπ,k is an integer,b is a constant and p,q,c are non-zero constants.

Remark . Equation (.) is a special case of (.). Theorem D generalized Theorem B. From Theorem D, we see that ifP(z) andQ(z) are non-constant polynomials, then equa-tion (.) has no transcendental entire soluequa-tion of finite order.

In this paper, we generalize equations (.)-(.) and obtain the following results.

Theorem . Let P(z),Q(z)be two non-zero polynomials,c∈C\{}andα(z)be a polyno-mial.If the differential-difference equation

f(z)+P(z)f(z+c) =Q(z)eα(z) (.)

admits a transcendental entire solution of finite order,then f(z)must satisfy one of the following cases:

(i)P(z)and Q(z)reduce to constants,and

f(z) =qe

Az+B A +

where A=ieAc_{p,A= –ie}Ac_{p,B,Bare constants and A,A,q,q,p,c are non-zero} constants;

(ii)P(z)reduces to a constant,Q(z)is a polynomial with degree,and

f(z) =(az+a–

a

A)e

Az+B

A +

qeAz+B A ,

 A =c,

 A =c,

or

f(z) =qe

Az+B A +

(bz+b– b

A)e

Az+B

A ,

 A =c,

 A=c,

where A=ieAc_{p,A= –ie}Ac_{p,B,B,a,bare constants and A,A,q,q,a,b,p,c} are non-zero constants;

(iii)

f(z) =B(z)eAz, α(z) = Az+D,

where B(z)satisfies[B(z) +AB(z)]_+P_(z)B_(z+c)eAc_=Q(z)eD_{,A,c are non-zero}

con-stants,D is a constant.

Remark . In equation (.), whenα(z) is a constant, then equation (.) reduces to equa-tion (.). That is, Theorem . generalizes Theorems B and D.

Theorem . Let Q(z)be a non-zero polynomial,α(z)be a polynomial and c∈C\{}.If the differential-difference equation

f(z)+f(z+c) –f(z)=Q(z)eα(z) (.)

admits a transcendental entire solution of finite order,then f(z)must satisfy one of the following cases:

(i)

f(z) =B(z)eAz+c, α(z) = Az+D,

where B(z)satisfies[B(z) +AB(z)]_{+ [B(z+c)e}Ac_–B(z)]_=Q(z)eD_{,A,c are non-zero}

con-stants,c,D are constants;In particular,if A=±i,then

f(z) =qe

B_+q eB

–i e

–iz_{+c, qe}B_=qeB_eic_{– ,}

or

f(z) =be

B_{(iz+ ) +i(q}

eB+beB)

 e

–iz_{+c, b(ic+ ) = –iqe}B–B_,e–ic_{= ,}

or

f(z) =qe

B_+q eB

i e

or

f(z) =ae

B_{(–iz+ ) –i(ae}B_+qeB₎

 e

iz_+c

, a(ic– ) = –iqeB–B,eic= ,

where c,c,c,c,B,B,a,bare constants and a,b,q,q,c are non-zero constants; (ii)

f(z) =qe

B_z

 +

qeAz+B

A +c, e

Ac_{–  =iA} ,c= –i,

or

f(z) =qe

Az+B A +

qeBz

 +c, e

Ac_{–  = –iA,c=i,}

where B,B,c,care constants and A,A,q,qare non-zero constants; (iii)

f(z) =qe

Az+B A

+qe

Az+B A

+c,

or

f(z) =(az+a–

a

A)e

Az+B

A +

qeAz+B A +c,

 A+i=c,

or

f(z) =qe

Az+B A +

(bz+b– b

A)e

Az+B

A +c,

 A–i=c,

or

f(z) =(az+a–

a

A)e

Az+B

A +

(bz+b–b

A)e

Az+B

A +c,

 A+i =

 A–i=c,

where eAc_{–  = –iA,e}Ac_{–  =iA,B,B,a,bare constants and A,A,q,q,a,b,c} are non-zero constants.

Remark . In equation (.), whenQ(z) is non-zero constant andα(z) is a constant, then equation (.) reduces to equation (.). That is, Theorem . generalizes Theorem C.

Fermat type functional equations were investigated by Gross [, ] and many others. In [], Yang studied the Fermat type functional equation

a(z)fn(z) +b(z)gm(z) = , (.)

wherea(z),b(z) are small functions with respect tof(z) and obtained the following result.

When g(z) has a specific relationship with f(z) in (.), Liu et al. [] studied the differential-difference equation

f(z)n+fm(z+c) = , (.)

and obtained the following result.

Theorem F Equation(.)has no transcendental entire solutions with finite order, pro-vided that m=n,where m,n are positive integers.

Now, we generalize (.) and obtain the following result.

Theorem . Let P(z), Q(z)be two non-constant polynomials and c∈C\{},then the equation

f(z)n+P(z)fm(z+c) =Q(z) (.)

has no transcendental entire solutions with finite order,provided that _m +_n< except for m= ,n= ,where m,n are positive integers.

2 Some lemmas

In order to prove our conclusions, we need some lemmas.

Lemma .(see [, ]) Let f(z)be a transcendental meromorphic solution of

fnP(z,f) =Q(z,f),

where P(z,f)and Q(z,f)are polynomials in f(z)and its derivatives with meromorphic co-efficients,say{aλ|λ∈I},such that m(r,aλ) =S(r,f)for allλ∈I.If the total degree of Q(z,f)

as a polynomial in f(z)and its derivatives is≤n,then m(r,P(z,f)) =S(r,f).

Lemma .(see []) Suppose that f(z),f(z), . . . ,fn(z) (n≥)are meromorphic functions

and g(z),g(z), . . . ,gn(z)are entire functions satisfying the following conditions:

() n_j=fj(z)egj(z)≡.

() gj(z) –gk(z)are not constants for≤j<k≤n.

() For≤j≤n,≤h<k≤n,T(r,fj(z)) =o(T(r,egh(z)–gk(z)))(r→ ∞,r∈/E).

Then fj(z)≡ (j= , . . . ,n).

Lemma .(see []) Suppose that f(z),f(z), . . . ,fn(z) (n≥)are meromorphic functions

which are not constants except for fn(z).Furthermore,let

n

j=fj(z)≡.If fn(z) ≡and

n

j=

N r, 

fj(z)

+ (n– )

n

j=

Nr,fj(z)

<λ+o()Tr,fk(z)

,

Lemma . Let Q(z)be a non-zero polynomial and satisfy

Q(z+c) –Q(z)≡aQ(z) +b, (.)

where a,c are non-zero constants,b is a constant,then one of the following cases holds:

(i) ifb= anda=c,thenQ(z)reduces to a non-zero constant;

(ii) ifb= anda=c,thenQ(z)reduces to a non-zero constant orQ(z) =az+a,where

ais a non-zero constant,ais a constant;

(iii) ifb= anda=c,thenQ(z) =az+aandb=a(c–a),whereais a non-zero

constant,ais a constant;

(iv) ifb= anda=c,thenQ(z) =az_{+az+aandb=ac}_{,whereais a non-zero}

constant,a,aare constants.

Proof Denote

Q(z) =aszs+as–zs–+· · ·+a (as= ).

Then

Q(z) =saszs–+ (s– )as–zs–+· · ·+a,

Q(z+c) =as(z+c)s+as–(z+c)s–+· · ·+a,

Q(z+c) –Q(z) =sasczs–+

asCsc+as–Cs–c

zs–+· · ·.

(i) Ifb=  anda=c, comparing the coefficients ofzs–_{on both sides of (.), we see that} sasc=asas, it contradicts witha=candas= .

(ii) Ifb= ,a=cands≥, comparing the coefficients ofzs–on both sides of (.), we see thatasCsc+as–Cs–c=a(s– )as–, thenasCsc= , a contradiction. Thuss≤, that

is,Q(z) reduces to a non-zero constant orQ(z) is a non-constant polynomial with degree . (iii) Ifb=  anda=c, Q(z) is a non-zero constant, note thatb= , clearly (.) is a contradiction. Ifs≥, comparing the coefficients ofzs–on both sides of (.), we see that sasc=asas, a contradiction. Ifs= , by (.), we see thatb=a(c–a)= .

(iv) Ifb=  anda=c,Q(z) is a non-zero constant, note thatb= , clearly (.) is a contra-diction. Ifs= , by (.), we see thatb=a(c–a) = , a contradiction. Ifs≥, comparing the coefficients ofzs–_{on both sides of (.), we see thata}

sCsc+as–Cs–c=a(s– )as–, thenasCsc= , a contradiction. Ifs= , by (.), we see thatb=ac= .

Lemma .(see []) Letη,ηbe two complex numbers such thatη=ηand let f(z)be a

finite order meromorphic function.Then we have

m r,f(z+η) f(z+η)

=S(r,f).

3 Proof of Theorem 1.1

Suppose thatf(z) is a transcendental entire solution of finite order of (.), then

Thus, bothf(z) +iP(z)f(z+c) andf(z) –iP(z)f(z+c) are entire functions with finitely many zeros. Combining (.) with the Hadamard factorization theorem [], Theorem ., we assume that

f(z) +iP(z)f(z+c) =Q(z)eα(z)

and

f(z) –iP(z)f(z+c) =Q(z)eα(z)_,

whereQ(z),Q(z) are two non-zero polynomials,α(z),α(z) are two polynomials and cannot be constants simultaneously, otherwisef(z) is a polynomial. Thus, we have

f(z) =Q(z)e

α(z)_+Q(z)eα(z)

 (.)

and

f(z+c) = Q(z)e

α(z)_–Q(z)eα(z)

iP(z) . (.)

Differentiating (.) and shifting (.) by replacingzwithz+c, we have

[Q_(z) +Q(z)α_(z)]P(z) –Q(z)P(z) iP_(z)Q

(z+c)

eα(z)–α(z+c)

–[Q

(z) +Q(z)α(z)]P(z) –Q(z)P(z) iP_(z)Q(z+c) e

α(z)–α(z+c)

–Q(z+c) Q(z+c)

eα(z+c)–α(z+c)≡. (.)

We deduce that [Q(z) +Q(z)α(z)]P(z) –Q(z)P(z) ≡ and [Q(z) +Q(z)α(z)]P(z) – Q(z)P(z) ≡. If [Q(z) +Q(z)α(z)]P(z) –Q(z)P(z)≡, thenP(z)≡AQ(z)eα(z), where Ais a non-zero constant. Note thatP(z),Q(z) are non-zero polynomials, thenα(z) must be a constant. Letα(z)≡A. Sinceα(z) andα(z) cannot be constants simultaneously, thusα(z) cannot be a constant, then [Q(z) +Q(z)α(z)]P(z) –Q(z)P(z) ≡. Then (.) can be rewritten as

Q_(z) +Q(z)α(z)

P(z) –Q(z)P(z)

eα(z)_+iP_(z)Q

(z+c)eα(z+c)

+iP(z)Q(z+c)eA= . (.)

If degα(z)≥, thendegα(z) =degα(z+c)≥,deg(α(z+c) –α(z))≥, andeα(z), eα(z+c)_,eα(z+c)–α(z)_{are of regular growth, by Lemma ., we have}

Q_(z) +Q(z)α_(z)P(z) –Q(z)P(z)≡iP(z)Q(z+c)≡iP(z)Q(z+c)≡,

a contradiction. Thusdegα(z)≤, note that α(z) cannot be a constant, then α(z) = Az+B, whereAis a non-zero constant. Rewriting (.) as

whereH(z) =iP_{(z)Q(z+c)e}Ac+B_{+ [(Q}

(z) +Q(z)α(z))P(z) –Q(z)P(z)]eB. IfH(z)≡, sinceiP_{(z)Q(z+c)e}A ≡_{, clearly (.) is a contradiction. IfH(z)} ≡_{, we can see that the} left side of (.) is a transcendental entire function, and the right side of (.) is a non-zero polynomial, a contradiction.

Similarly, we can prove that [Q_(z) +Q(z)α_(z)]P(z) –Q(z)P(z) ≡.

Thus, [Q_(z)+Q(z)α_(z)]P(z)–Q(z)P(z) ≡ and [Q_(z)+Q(z)α_(z)]P(z)–Q(z)P(z) ≡ , by (.) and Lemma ., we see that if any two of eα(z)–α(z+c)_{, e}α(z)–α(z+c) _and eα(z+c)–α(z+c)_{are not constants, then the third term must be constant. If any two of them} are constants, then the third term also must be constant. In what follows, we discuss four cases: Case ,eα(z)–α(z+c)_andeα(z)–α(z+c)_{are not constants; Case ,e}α(z)–α(z+c)_and eα(z+c)–α(z+c)_{are not constants; Case , e}α(z)–α(z+c)_andeα(z+c)–α(z+c) _{are not constants;} Case ,eα(z)–α(z+c)_,eα(z)–α(z+c)_andeα(z+c)–α(z+c)_{are all constants.}

Case,eα(z)–α(z+c)_andeα(z)–α(z+c)_{are not constants, by (.) and Lemma ., we have}

–Q(z+c) Q(z+c)

eα(z+c)–α(z+c)≡, (.)

which implies thatα(z+c) –α(z+c) is a constant, and

[Q(z) +Q(z)α(z)]P(z) –Q(z)P(z) [Q_(z) +Q(z)α_(z)]P(z) –Q(z)P(z)e

α(z)–α(z)_{≡, (.)}

which implies thatα(z) –α(z) is a constant.

Denoteeα(z+c)–α(z+c)_=eα(z)–α(z)_{=k(= ), by (.), we getQ(z) = –kQ(z), substituting} it into (.) yields

P(z)Q(z) –P(z)Q_(z)≡P(z)Q(z)α_(z) +α_(z).

SinceP(z) andQ(z) are non-zero polynomials,α(z) andα(z) are polynomials, from the above identity, we getα_(z) +α_(z)≡, that is,α(z) +α(z) is a constant. Note thatα(z) –

α(z) is a constant, then bothα(z) andα(z) are constants, a contradiction.

Case,eα(z)–α(z+c)_andeα(z+c)–α(z+c)_{are not constants, by (.) and Lemma ., we have}

–[Q

(z) +Q(z)α(z)]P(z) –Q(z)P(z) iP_(z)Q(z+c) e

α(z)–α(z+c)_≡,

which implies thatα(z) –α(z+c) is a constant, thenα(z+c) –α(z+ c) is also a constant, and

[Q_(z) +Q(z)α_(z)]P(z) –Q(z)P(z) iP_(z)Q

(z+c)

eα(z)–α(z+c)≡_,

which implies thatα(z) –α(z+c) is a constant.

Byα(z)–α(z+c) = [α(z)–α(z+c)]+[α(z+c)–α(z+c)], we see thatα(z)–α(z+c) is a constant, thenα(z) is a constant or a polynomial with degree , which implies that

α(z) –α(z+c) is also a constant, a contradiction.

Case,eα(z)–α(z+c)_andeα(z+c)–α(z+c)_{are not constants, by (.) and Lemma ., we have}

[Q_(z) +Q(z)α_(z)]P(z) –Q(z)P(z) iP_(z)Q(z+c) e

which implies that α(z) –α(z+c) is a constant, note that [Q(z) +Q(z)α(z)]P(z) – Q(z)P(z) ≡, then α(z) cannot be a constant, therefore,α(z) can only be a polyno-mial with degree . Denoteα(z) =Az+B, whereAis a non-zero constant, andBis a constant. Rewriting (.) as

AP(z)Q(z) +P(z)Q_(z) –P(z)Q(z)≡ieAc_P_{(z)Q(z+c), (.)}

P(z) must be a constant, denotedP(z)≡p(= ). Then (.) can be rewritten as

Q(z+c) –Q(z)≡  AQ

(z) and A=ieAcp.

By Lemma ., we have (i) if_A

=c, thenQ(z)≡q(constant); (ii) if 

A =c, thenQ(z)≡q (constant) orQ(z) =az+a, whereais a non-zero constant andais a constant.

By (.) and (.), we have

–[Q

(z) +Q(z)α(z)]P(z) –Q(z)P(z) iP_(z)Q

(z+c)

eα(z)–α(z+c)≡, (.)

which implies that α(z) –α(z+c) is a constant, note that [Q_(z) +Q(z)α_(z)]P(z) – Q(z)P(z) ≡, soα(z) cannot be a constant, thenα(z) can only be a polynomial with degree , denoteα(z) =Az+B, whereA is a non-zero constant andBis a constant. Note thatP(z)≡p(= ), then (.) can be rewritten as

Q(z+c) –Q(z)≡ 

AQ

(z) and A= –ieAcp.

By Lemma ., we have (i) if_A

 =c, thenQ(z)≡q(constant); (ii) if 

A =c, thenQ(z)≡q (constant) orQ(z) =bz+b, wherebis a non-zero constant andbis a constant.

Note thatA=ieAc_{pandA= –ie}Ac_{p, we see thatA=A, that is,} 

A = 

A. In what follows, we discuss three subcases: Subcase ., _A

=cand 

A =c; Subcase ., 

A=cand 

A =c; Subcase ., 

A =cand 

A =c. Subcase., _A

=cand 

A =c, thenQ(z)≡qandQ(z)≡q, by (.), (.) and (.), we obtain

f(z) =qe

Az+B_+qeAz+B 

and

f(z) =qe

Az+B A +

qeAz+B A .

Subcase.,_A

 =cand 

A =c, thenQ(z)≡qandQ(z)≡qorQ(z) =az+aand Q(z)≡q. IfQ(z)≡qandQ(z)≡q, the same as Subcase .. IfQ(z) =az+aand Q(z)≡q, by (.), (.) and (.), we obtain

f(z) =(az+a)e

and

f(z) =(az+a–

a

A)e

Az+B

A +

qeAz+B A .

Subcase.,_A

=cand 

A =c, thenQ(z)≡qandQ(z)≡qorQ(z)≡qandQ(z) = bz+b. IfQ(z)≡qandQ(z)≡q, the same as Subcase .. IfQ(z)≡qandQ(z) = bz+b, by (.), (.) and (.), we obtain

f(z) =qe

Az+B_{+ (b}

z+b)eAz+B 

and

f(z) =qe

Az+B A +

(bz+b– b

A)e

Az+B

A .

Case,eα(z)–α(z+c)_,eα(z)–α(z+c)_andeα(z+c)–α(z+c)_{are all constants, that is,α}

(z)–α(z+c),

α(z) –α(z+c) andα(z+c) –α(z+c) are all constants. Note thatα(z) andα(z) are not constants simultaneously, thenα(z) =Az+B,α(z) =Az+Bandα(z) = Az+D, where Ais non-zero constant andB,B,D(=B+B) are constants. Therefore, by (.), (.) and (.), we havef(z) =B(z)eAz, whereB(z) satisfies [B(z) +AB(z)]+P(z)B(z+c)eAc= Q(z)eD_.

This completes the proof of Theorem ..

4 Proof of Theorem 1.2

As in the beginning of the proof of Theorem ., from (.), we have

f(z) =Q(z)e

α(z)_+Q(z)eα(z)

 (.)

and

f(z+c) –f(z) =Q(z)e

α(z)_–Q(z)eα(z)

i . (.)

whereQ(z),Q(z) are two non-zero polynomials,α(z),α(z) are two polynomials and cannot be constants simultaneously, otherwisef(z) is a polynomial. Differentiating (.), shifting (.) by replacingzwithz+cand combining (.), we have

Q_(z) +Q(z)α_(z) +iQ(z) iQ(z+c)

eα(z)–α(z+c)

–Q

(z) +Q(z)α(z) –iQ(z)

iQ(z+c) e

α(z)–α(z+c)_–Q(z+c) Q(z+c)e

α(z+c)–α(z+c)_{≡. (.)}

IfQ_(z) +Q(z)α_(z) +iQ(z)≡, that is, –Q_(z) = (α_(z) +i)Q(z), then we haveα_(z) +i≡ andQ(z)≡q(constant). Thusα(z) = –iz+B, whereBis a constant, substitute it into (.) yields

Suppose degα(z)≥. Clearly, Q_(z) +Q(z)(α_(z) –i) ≡ . According to degα(z) = degα(z+c)≥,degα(z+c) = ,deg(α(z+c) –α(z+c)) =deg(α(z) –α(z+c))≥ and deg(α(z+c) –α(z))≥, andeα(z),eα(z+c),eα(z+c),eα(z+c)–α(z+c),eα(z)–α(z+c),eα(z+c)–α(z) are of regular growth, by Lemma ., we have

Q_(z) +Q(z)α_(z) –i≡iQ(z+c)≡iQ(z+c)≡,

a contradiction. Thusdegα(z)≤, that is,α(z) =B(constant) orα(z) =Az+B, where Ais a non-zero constant. Ifα(z)≡B, by (.), we haveeα(z+c)_≡–eB[Q(z)–iQ(z)+iQ(z+c)]

iQ(z+c) , the left side of this identity is a transcendental entire function, and the right side of this identity is a rational function, a contradiction. Hence,α(z) =Az+B. Rewriting (.) as

–iqe(–i–A)z–ic+B–B≡Q(z) + (A–i)Q(z) +ieAcQ(z+c).

If –i–A= , clearly the above identity is a contradiction. ThenA= –i. The above identity can be rewritten as

iQ(z) –Q_(z) –iqe–ic+B–B≡_ie–ic_{Q(z+c). (.)}

IfQ(z)≡q(constant), by (.), we getqeB=qeB(eic– ). By (.), (.) and (.), we have

f(z) =qe

B_+q eB

 e

–iz

and

f(z) =qe

B_+qeB

–i e

–iz_+c.

IfQ(z) is a non-constant polynomial, by (.), we obtain

iQ(z+c) –Q(z)= –Q_(z) –iqe–ic+B–B_{, i=ie}–ic_.

Note that –iqe–ic+B–B_{=  and –}

i=c, by Lemma ., we see thatQ(z) =bz+b, where

bis a non-zero constant andbis a constant. From (.), we getb(ic+ ) = –iqeB–B_, by (.), (.) and (.), we have

f(z) =(bz+b)e

B_+q eB

 e

–iz

and

f(z) =be

B_{(iz+ ) +i(qe}B_+beB₎

 e

–iz_+c.

Similarly, ifQ_(z) +Q(z)α_(z) –iQ(z)≡, then we have

f(z) =qe

B_+q eB

i e

or

f(z) =ae

B_{(–iz+ ) –i(ae}B_+qeB₎

 e

iz_{+c, a(ic– ) = –iqe}B–B_,eic_{= ,}

wherec,c,B,B,aare constants anda,q,qare non-zero constants.

According to the above proof, we can see that Q_(z) +Q(z)α_(z) +iQ(z)≡ and Q_(z) +Q(z)α(z) –iQ(z)≡ cannot be valid simultaneously. In what follows, we as-sume thatQ_(z) +Q(z)α_(z) +iQ(z) ≡ andQ_(z) +Q(z)α_(z) –iQ(z) ≡. By (.) and Lemma ., we see that if any two ofeα(z)–α(z+c)_,eα(z)–α(z+c) _andeα(z+c)–α(z+c) _are not constants, then the third term must be constant. If any two of them are constants, then the third term also must be constant. In the following, we discuss four cases: Case ,eα(z)–α(z+c)_andeα(z)–α(z+c)_{are not constants; Case ,e}α(z)–α(z+c)_andeα(z+c)–α(z+c)_are not constants; Case ,eα(z)–α(z+c)_andeα(z+c)–α(z+c)_{are not constants; Case ,e}α(z)–α(z+c)_, eα(z)–α(z+c)_andeα(z+c)–α(z+c)_{are all constants.}

Case andCase, similarly to the proof of Case  and Case  of Theorem ., we can obtain a contradiction.

Case,eα(z)–α(z+c)_andeα(z+c)–α(z+c)_{are not constants, by (.) and Lemma ., we have}

Q_(z) +Q(z)α_(z) +iQ(z)

iQ(z+c) e

α(z)–α(z+c)_{≡, (.)}

which implies thatα(z) –α(z+c) is a constant, thenα(z) =Az+Borα(z)≡B, where Ais a non-zero constant andBis a constant. By (.) and (.), we also have

–Q

(z) +Q(z)α(z) –iQ(z)

iQ(z+c) e

α(z)–α(z+c)_{≡, (.)}

which implies that α(z) –α(z+c) is a constant, then α(z) =Az+B orα(z)≡B, whereAis a non-zero constant andBis a constant. Note thatα(z) andα(z) cannot be constants simultaneously, In what follows, we discuss three subcases: Subcase .,α(z)≡ Bandα(z) =Az+B; Subcase .,α(z) =Az+Bandα(z)≡B; Subcase .,α(z) = Az+Bandα(z) =Az+B.

Subcase.,α(z)≡Bandα(z) =Az+B, by (.), we have

Q(z+c) –Q(z)≡–iQ_(z).

By Lemma ., we see that ifc= –i, thenQ(z)≡q(constant); Ifc= –i, thenQ(z)≡q (constant) orQ(z) =az+a, whereais a non-zero constant andais a constant.

By (.), we have

Q(z+c) –Q(z)≡ 

A–i

Q_(z) and A–i= –ieAc_.

By Lemma ., we see that if_A–_i=c, thenQ(z) =q(constant); If_A–_i=c, thenQ(z)≡ q(constant) orQ(z) =bz+b, wherebis a non-zero constant andbis a constant.

IfQ(z)≡q andQ(z)≡q, by (.), (.) and (.), we getc= –i. Note thatA–i= –ieAc_{, then} 



A–i=c(= –i),Q(z)≡qandQ(z)≡q; Subcase ..,



A–i=c(= –i),Q(z) =az+a andQ(z)≡q; Subcase .., _A

–i=c(= –i),Q(z)≡qandQ(z) =bz+b.

Subcase..,_A– _i=c(= –i),Q(z)≡qandQ(z)≡q, by (.), (.) and (.), we get

f(z) =qe

B_+q eAz+B 

and

f(z) =qe

B_z

 +

qeAz+B A +c.

Subcase..,_A– _i=c(= –i),Q(z) =az+aandQ(z)≡q, by (.), (.) and (.), we have

f(z) =(az+a)e

B_+qeAz+B

 ,

f(z) =ae

B_z

 +

aeB_z

 +

qeAz+B A

+c_,

and

f(z+c) –f(z) =(az+a)e

B_–q eAz+B

i –

aeB 

=(az+a)e

B_–q eAz+B

i ,

thenamust be zero, a contradiction.

Subcase..,_A–_i=c(= –i),Q(z)≡qandQ(z) =bz+b, by (.), (.) and (.), we have

f(z) =qe

B_{+ (b}

z+b)eAz+B

 ,

f(z) =qe

B_z

 +

(bz+b–_Ab_)eAz+B

A +c

,

and

f(z+c) –f(z) =qe

B_c

 –

(bz+b)eAz+B i

=qe

B_{– (bz+b)e}Az+B

i ,

thenc= –i, note thatc= –i, a contradiction.

Subcase.,α(z) =Az+Bandα(z)≡B, similarly to the proof of Subcase ., (.) admits a transcendental entire solution, if and only if_A+ _i=c(=i),Q(z)≡qandQ(z)≡ q. By (.), (.) and (.), we get

f(z) =qe

and

f(z) =qe

Az+B A +

qeB_z

 +c, e

Ac_{–  = –iA,c=i,}

whereB,B,care constants andq,qare non-zero constants.

Subcase.,α(z) =Az+Bandα(z) =Az+B, by (.) and (.), we obtain, respec-tively,

Q(z+c) –Q(z)≡ 

A+i

Q_(z) and A+i=ieAc _(.)

and

Q(z+c) –Q(z)≡ 

A–iQ

(z) and A–i= –ieAc. (.)

Clearly,A =A. In what follows, we discuss four subcases: Subcase .., _A+ _i=cand 

A–i=c; Subcase ..,



A+i =cand



A–i =c; Subcase ..,



A+i =cand



A–i =c;

Sub-case .., _A+ _i=_A– _i=c.

Subcase.., 

A+i=cand



A–i=c, by (.), (.) and Lemma ., we haveQ(z)≡q andQ(z)≡q, whereqandqare non-zero constants. By (.), (.) and (.), we get

f(z) =qe

Az+B_+q eAz+B 

and

f(z) =qe

Az+B A +

qeAz+B A +c.

Subcase..,_A+ _i=cand_A– _i=c. By_A+_i=c, (.) and Lemma ., we haveQ(z)≡q orQ(z) =az+a, wherea,qare non-zero constants andais a constant. By _A–_i=c, (.) and Lemma ., we haveQ(z)≡q, whereqis a non-zero constant. IfQ(z)≡q andQ(z)≡q, the same as Subcase ... IfQ(z) =az+aandQ(z)≡q, by (.), (.) and (.), we get

f(z) =(az+a)e

Az+B_+qeAz+B 

and

f(z) =(az+a–

a

A)e

Az+B

A +

qeAz+B A +c.

Subcase.., _A+ _i=cand _A– _i =c, similarly to the proof of Subcase .., we can ob-tain

f(z) =qe

Az+B A +

or

f(z) =qe

Az+B A +

(bz+b–_Ab_)eAz+B

A +c,

where B, B, b, c, c are constants and A, A, q, q, b, c are non-zero con-stants.

Subcase.., _A+_i =_A– _i=c. By _A+_i =c, (.) and Lemma ., we haveQ(z)≡q or Q(z) =az+a, wherea,q are non-zero constants andais a constant. By _A– _i =c, (.) and Lemma ., we haveQ(z)≡q orQ(z) =bz+b, whereb,qare non-zero constants andbis a constant. IfQ(z)≡qandQ(z)≡q, the same as Subcase ... IfQ(z) =az+a andQ(z)≡q, the same as Subcase ... IfQ(z)≡q andQ(z) = bz+b, the same as Subcase ... IfQ(z) =az+aandQ(z) =bz+b, by (.), (.) and (.), we get

f(z) =(az+a)e

Az+B_{+ (bz+b)e}Az+B 

and

f(z) =(az+a–

a

A)e

Az+B

A +

(bz+b–b

A)e

Az+B

A +c.

Case,eα(z)–α(z+c)_,eα(z)–α(z+c)_andeα(z+c)–α(z+c)_{are all constants, that is,α}

(z)–α(z+c),

α(z) –α(z+c) andα(z+c) –α(z+c) are all constants. Note thatα(z) andα(z) are not constants simultaneously, thenα(z) =Az+B,α(z) =Az+Bandα(z) = Az+D, where Ais non-zero constant andB,B,D(=B+B) are constants. Therefore, by (.), (.) and (.), we havef(z) =B(z)eAz_{+c, whereB(z) satisfies [B(z) +AB(z)]}_{+ [B(z+c)e}Ac_–B(z)]₌ Q(z)eD_.

This completes the proof of Theorem ..

5 Proof of Theorem 1.3

Suppose thatf(z) is a transcendental entire function of finite order satisfying (.). In what follows, we will discuss four cases: Case ,m=n≥; Case ,m>n; Case ,n>m≥; Case ,n≥,m= .

Case,m=n≥. Ifm=n= , note thatP(z) andQ(z) are non-constant polynomials, by TheoremD, we see that (.) has no transcendental entire solutions of finite order. If m=n≥, rewriting (.) as _Q_(z)(f(z))n+_QP(₍z_z)₎fm(z+c) = , by TheoremE, we see that (.) has no transcendental entire solutions of finite order.

Case andCase, similarly to the proof of the Case  and Case  of [], Theorem ., we can also obtain (.) has no transcendental entire solutions of finite order.

Case,n≥,m= . Differentiating (.), we get

nf(z)n–f(z) +P(z)f(z+c) +P(z)f(z+c) =Q(z).

Substituting (.) into the above equation yields

f(z)n–

nf(z) –P _(z)

P(z)f

_{(z)= –P(z)f(z+c) +Q(z) –Q(z)}P(z)

DenoteF(z) =f(z),ϕ(z) =nf(z) –P_P((_zz₎)f(z) =nF(z) –P_P((_zz₎)F(z). Then (.) can be rewrit-ten as

Fn–(z)ϕ(z) = –P(z)F(z+c)

F(z) F(z) +Q

_{(z) –Q(z)}P(z)

P(z). (.)

By Lemma ., we see thatm(r,F_F(z₍+_z)c)) =S(r,F) andm(r,Q(z) –Q(z)P_P((_zz₎)) =S(r,F), note thatn– ≥, by Lemma ., we have

mr,ϕ(z)=S(r,F) and mr,F(z)ϕ(z)=S(r,F).

We see thatϕ(z) ≡, otherwise (f(z))n_=Fn_{(z)≡AP(z), whereAis a non-zero constant,}

a contradiction. Note thatf(z) is a transcendental entire function, thenN(r,ϕ(z)) =S(r,F) and

Tr,F(z)=mr,F(z)≤mr,F(z)ϕ(z)+m r, 

ϕ(z)

≤mr,ϕ(z)+Nr,ϕ(z)+S(r,F) =S(r,F),

that is,T(r,f(z)) =T(r,F(z))≤S(r,F) =S(r,f), a contradiction. This completes the proof of Theorem ..

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors drafted the manuscript, and they read and approved the final manuscript.

Acknowledgements

The authors would like to thank the referee for his/her reading of the original version of the manuscript with valuable suggestions and comments. This work was supported by the National Natural Science Foundation of China (11371225).

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Received: 15 November 2016 Accepted: 18 April 2017 References

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