The properties of solutions of certain type of difference equations

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R E S E A R C H

Open Access

The properties of solutions of certain type of

difference equations

Xiaoguang Qi

1*

_{, Jia Dou}

2

_{and Lianzhong Yang}

3

*_{Correspondence:}

[email protected]

1_{School of Mathematics, University} of Jinan, Jinan, Shandong 250022, P.R. China

Full list of author information is available at the end of the article

Abstract

In this paper, we shall utilize Nevanlinna value distribution theory to study the solvability of the diﬀerence equations of the form:f(z)n+p(z)(

cf)m₌_r₍_z₎_eq(z)_and

f(z)n₊_p₍_z₎_eq(z)₍

cf)m=r(z), and we shall study the growth of their entire solutions.

Moreover, we will give a number of examples to show that the results in this paper are the best possible in certain senses. This article extends earlier results by Liuet al.

(Czechoslov. Math. J. 61:565-576, 2011; Ann. Pol. Math. 102:129-142, 2011).

MSC: Primary 39A05; secondary 30D35

Keywords: meromorphic functions; diﬀerence equation; growth; ﬁnite order

1 Introduction and main results

In this paper, we use the basic notions in Nevanlinna theory of meromorphic functions, as found in []. In addition, we useδ(f),λ(f), andλ(_f) to denote the order, and the exponents of the convergence of zeros and poles of a meromorphic functionf(z), respectively. We deﬁne diﬀerence operators ascf =f(z+c) –f(z), wherecis a non-zero constant.

In , Yang [] started to study the existence and uniqueness of ﬁnite order entire solutions of the following type of non-linear diﬀerential equation:

L(f) –p(z)f(z)n=h(z), ()

whereL(f) is a linear diﬀerential polynomial inf with polynomial coeﬃcients,p(z) is a non-vanishing polynomial,h(z) is an entire function, andnis an integer such thatn≥. Subsequently, several papers have appeared in which the solutions of () are studied. The reader is invited to see [–].

Recently, Yang and Laine [] considered the existence of the non-linear diﬀerential-diﬀerence equation of the form

f(z)n+L(z,f) =h(z), ()

whereL(z,f) is a ﬁnite sum of product off, derivatives off, and their shifts. They obtained the following result.

Theorem A Let n≥be an integer,L(z,f)be a linear differential-difference polynomial of f,with small meromorphic coefficients,and h(z)be a meromorphic function of finite order.

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Then ()possesses at most one admissible transcendental entire solution of ﬁnite order, unless L(z,f)vanishes identically. If such a solution f(z)exists,then f(z)is of the same order as h(z).

In particular, it is shown in [] that the equation

f(z)+P(z)f(z+ ) =Q(z)

has no transcendental entire solution of ﬁnite order, whereP(z),Q(z) are polynomials. Some improvements of Theorem A can be found in [–] as well. Replacingf(z+c) withcf, one of the present authors considered the existence of entire solutions of

f(z)n+P(z)(cf)m=Q(z), ()

we get the following result in [].

Theorem B Let P(z),Q(z)be polynomials,n and m be integers satisfying n>m≥.Then ()has no transcendental entire solution of ﬁnite order.

In fact, the special case of () withn=m, andp(z) =q(z) = , can be viewed as the Fermat type functional equation. It is well known that () has no transcendental entire solutions whenn≥, which can be seen in [].

It is natural to ask what happens ifP(z) orQ(z) is a transcendental entire function in (). Corresponding to this question, we will investigate the ﬁnite order entire (meromorphic) solution of

f(z)n+p(z)(cf)m=r(z)eq(z) and f(z)n+p(z)eq(z)(cf)m=r(z). Theorem . Consider the non-linear diﬀerence equation of the form

f(z)n+p(z)(cf)m=r(z)eq(z), ()

where p(z)≡,q(z),r(z)are polynomials,n and m are positive integers.Suppose that f(z) is a transcendental entire function of ﬁnite order,not of period c.If n>m,then f(z)cannot be a solution of().

Remarks () In casen≤m, Theorem . is not true. In the special case thatm=n= , the functionf(z) =ez_solves

f(z) +zlnf= (z+ )ez.

Moreover, the entire functionf(z) =ez+zis a solution of the following equation:

f(z) + z

π(πif)

₌_ez_.

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In the following, we will consider the properties of the solutions of the equation

f(z)n+p(z)eq(z)(cf)m=r(z), ()

wherep(z)≡,q(z),r(z) are polynomials. In fact, () may have solutions. For example, the functionf(z) =zez _{is a solution of the equation}_f₍_z₎_– z

πie

z

πif = . One solution

of the equationf(z) –ez 

πiπif =zis the functionf(z) =ez+z. In addition, the function f(z) =ez₊_z_{can solve}_f₍_z_{) +}_ez 

π(πif)=zas well. Hence, here we just study the order

of the growth of the solutions. We state our ﬁndings as follows.

Theorem . Let p(z)≡,q(z),r(z)be polynomials,n and m be positive integers satisfying n>m.Let f(z)be ﬁnite order entire solutions of(),thenδ(f) =degq(z).

Remark We will give the following examples to show that the assumption thatn>min Theorem . is sharp. Clearly,f(z) =ezis a solution of the equationf(z) –_lnf = . The

functionf(z) = eπiz_–_eπiz₊_z_solves_f₍_z_{) – (}

–_f)=z–. However,δ(f)=degq(z) in

the above two examples.

Corollary . Let p(z)≡,q(z),r(z)be polynomials,n> .Let f(z)be ﬁnite order entire solutions of the equation

f(z)n+p(z)eq(z)cf =r(z), ()

thenδ(f) =degq(z).

According to (), we will give a further discussion on the existence of meromorphic solutions. We get:

Theorem . Let p(z)≡,q(z),r(z)be polynomials.Assume one of the following assertions holds:

(i) n> ,f(z)is a ﬁnite order meromorphic function(not entire),

(ii) n> ,r(z)≡,andf(z)is a finite order entire function with infinitely many zeros, (iii) n≥,r(z)≡,andf(z)is a finite order entire function satisfyingλ(f) <δ(f). Then f(z)cannot be a solution of().

Remarks () From Theorem ., we know that the solution of the equation f(z)n₊ p(z)eq(z)_cf _{=  must be expressed as}_f₍_z_{) =}_α₍_z₎_eβ(z)_{, where}_α₍_z_{) and}_β₍_z_{) are polynomials.}

() The condition thatn>  in Theorem . is sharp. In fact, ifn= , then we know the functionf(z) = e_zz is a solution of the equationf(z) + (_z_π_i + )πif = . Moreover, the

functionf(z) =coszcan solvef(z) +_πf= . () Some ideas of this paper are from [].

2 Preliminary lemmas

Lemma .[, Theorem .] Let f(z)be a meromorphic function of ﬁnite order,and let c∈C,then

m

r,f(z+c) f(z)

+m

r, f(z)

f(z+c)

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Remark From Lemma ., we knowT(r,f) =T(r,f(z+c)) +S(r,f) whenf(z) is an entire function of ﬁnite order.

Lemma .[, Theorem ..] Let f(z)be a transcendental meromorphic solution of

fnA(z,f) =B(z,f),

where A(z,f),B(z,f)are diﬀerential polynomials in f and its derivatives with small mero-morphic coeﬃcients aλ,in the sense of m(r,aλ) =S(r,f)for allλ∈I.If d(B(z,f))≤n,then

m(r,A(z,f)) =S(r,f).

Lemma .[, Theorem .] Let fj(z) (j= , , )be meromorphic functions that satisfy



j=

fj(z)≡.

If f(z)is not a constant,and 

j=

N

r, fj

+ 



j=

N(r,fj) <

λ+o()T(r),

whereλ< and T(r) =max≤j≤{T(r,fj)},then either f(z)≡or f(z)≡.

Lemma . [, Theorem .] Suppose that fj(z) (j= , . . . ,n) (n≥)are meromorphic functions and gj(z) (j= , . . . ,n)are entire functions satisfying the following conditions.

() n_j_=fj(z)egj(z)≡__.

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

T(r,fj) =oTr,egh–gk _, _r→ ∞_,_r∈_/_E_,

whereE⊂(,∞)is of ﬁnite linear measure.

Then fj(z)≡.

3 Proof of Theorem 1.1

Ifq(z) is a constant orr(z)≡, then the conclusion follows from Theorem B. It remains to consider the caseq(z) is a non-constant polynomial andr(z)≡. Assumef(z) is a tran-scendental entire solution of (), which is ﬁnite order, not of periodc. Diﬀerentiating () and eliminatingeq(z), we have

f(z)n–

nf(z) –

q(z) +r ₍_z₎

r(z)

f(z)

=

q(z) +r ₍_z₎

r(z)

p(z)(cf)m–p(z)(cf)m–mp(z)(cf)m–(cf). ()

Ifnf(z) – (q(z) +r_r₍(_zz₎))f(z)≡, then we havef(z)n=Ar(z)eq(z). Writingf(z) =h(z)eq(nz)_,

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we get

(A– )r(z)eq(z)+p(z)(cf)m≡. ()

Clearly, ifA= , thenf(z) is a period function with periodc, which contradicts the assump-tion. Hence,A= . Letg=eqn(z)_{, then (}_cf₎m_{can be expressed as}m

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Hence

Sincef(z) is entire, applying Lemma . and the lemma on the logarithmic derivative to the above equation, we conclude that

T(r,H) =m(r,H) +S(r,f) =S(r,f). ()

DiﬀerentiatingH(z), we have

(m+ )f–

The above equation can be rewritten in the following form:

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Subcase . Supposezis a zero ofH(z) with multiplicityt. From (), by simple

calcula-tions, we know thatk=  +t≤t, which means thatzis a contribution ofS(r,f) toN(r,_f)

by ().

Subcase . Supposezis not a zero ofH(z). By (), we getk–k= , then such a zero

off(z) must be simple and we ﬁnd thatq+r_r + (m+ )H_H must vanish atz. That implies

thatzmakes a contribution ofS(r,f) toN(r,_f) by ().

In a word, N(r,_f) =S(r,f). Therefore, we can assumef(z) to be of the form f(z) = G(z)es(z)_{, where}_s₍_z_{) is a non-constant polynomial, and}_G₍_z_{) is an entire function}

satis-fyingN(r,_G) =T(r,G) =S(r,f) by the Hadamard factorization theorem. Substituting this expression into (), we get

G(z)m+e(m+)s(z)+p(z)G(z+c)es(z+c)–G(z)es(z)m=r(z)eq(z),

and so

G(z)m+e(m+)s(z)+p(z) _m_–

i=

m

i

(–)iG(z+c)m–iG(z)ie(m–i)s(z+c)+is(z)

+p(z)(–)mG(z)mems(z)=r(z)eq(z),

which may be written in the form

G(z)es(z)

p(z)(–)m–+

r(z)eq(z)

p(z)(–)m_G₍_z₎m_ems(z)

+ m–

i=

_m i

(–)iG(z+c)m–iG(z)ie(m–i)s(z+c)+is(z)

(–)m–_G₍_z₎m_ems(z) = . ()

From Lemma ., we have either

f(z) =

r(z)eq(z)

p(z)(–)m_G₍_z₎m_ems(z) ≡

or

f(z) =

m–

i=

m i

(–)i_G₍_z₊_c₎m–i_G₍_z₎i_e(m–i)s(z+c)+is(z)

(–)m–_G₍_z₎m_ems(z) ≡.

Iff(z)≡, then we get

r(z)eq(z)=p(z)(–)mG(z)mems(z)=p(z)(–)mf(z)m.

Plugging the above equation into (), similar to (), we getT(r,req_–_p₍

cf)m)≤mT(r,f) + S(r,f). Comparing both sides of (), we get a contradiction. If f(z)≡, then we get

f(z)m+_≡_r₍_z₎_eq(z)_{, which means that}

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Applying Lemma . to (), we get

nm(r,f) =mr,r–peq(cf)m

≤mr,eq+m

r,(cf) m

fm

+mr,fm+S(r,f)

≤mr,eq+mm(r,f) +S(r,f),

which means that (n–m)m(r,f)≤m(r,eq_{) +}_S₍_r_,_f_{). From the assumption that}_n_>_m_and the above inequality, we conclude thatδ(f)≤degq(z).

Now we show thatδ(f) =degq(z). Suppose to the contrary thatδ(f) <degq(z). Then

δ(fn₊_peq₍

cf)m) =degq(z) >  from Remark below Lemma ., andδ(r) = . This is a contradiction by ().

First, we conﬁrm that one period function with periodccannot be a solution of (). Actu-ally, iff(z) is a period function, then the order off(z) satisﬁesδ(f)≥. At the same time, we getf(z)n=r(z), which is impossible. It remains to consider the function which is not of periodc.

(i) Suppose thatf(z)is a ﬁnite order meromorphic solution of (), andzis a pole of

f(z)with multiplicityt. Then we getz+cis a pole off(z)with multiplicity≥nt. By

calculation, we see thatz+kcis a pole off(z)with multiplicitynkt, wherekis a

positive integer. Sincen> , we conclude thatλ(f) =∞,i.e.,δ(f) =∞. This is a contradiction.

(ii) Suppose that a finite order entire solutionf(z)has infinitely many zeros and r(z)≡. We assume that all zeros ofp(z)are inD={z:|Rez| ≤M,|Imz| ≤M}, whereM> is some constant. From the assumption, we know thatf(z)has infinitely many zeros which are not inD. Ifz∈/Dsatisfiesf(z) = , then from ()

we know thatz+kcare zeros off(z). Moreover, we havez+kcare not inD, as

k→ ∞. In the same way as (i), we getλ(_f) =∞,i.e.,δ(f) =∞, which contradicts the assumption thatδ(f) <∞.

(iii) Supposef(z)is an entire solution of ﬁnite order satisfyingλ(f) <δ(f), then we know

δ(f)≥by the conclusion of Corollary .. Hence, from the Hadamard factorization theorem,f(z)can be written asf(z) =α(z)eβ(z)_{, where}_β₍_z₎_{is a}

non-constant polynomial.

Case . n> . From the conclusion of Corollary ., we know λ(f) = δ(α) <δ(f) = degq(z) =q≥. So we can rewritef(z) in the following form:

f(z) =γ(z)eaqzq_, _()

where aq is a non-zero constant,δ(γ) <q, andT(r,γ) =S(r,f). From the conclusion of Corollary ., we conclude that

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wherebqis a non-zero constant,δ(μ) <q, andT(r,μ) =S(r,f). By (), (), and (), we have

γ(z)nenaqzq₊_p₍_z₎_μ₍_z₎_γ₍_z₊_c₎_eν(z)_–_γ₍_z₎_e(aq+bq)zq₌_r₍_z_), _() whereν(z) is a polynomial such thatdegν(z)≤q– .

Subcase . Ifnaq=bq+aq, then () can be expressed as

γ(z)n+p(z)μ(z)γ(z+c)eν(z)_–_γ₍_z₎₌ r(z)

enaqzq.

Sincer(z)≡, we getγ(z)n₊_p₍_z₎_μ₍_z₎₍_γ₍_z₊_c₎_eν(z)_–_γ₍_z₎₎_≡_{. Comparing the}

character-istics function of both sides of the above equation, we get a contradiction. Subcase . Ifnaq=bq+aq, then we rewrite () in the form

γ(z)nenaqzq₊_p₍_z₎_μ₍_z₎_γ₍_z₊_c₎_eν(z)_–_γ₍_z₎_e(aq+bq)zq_–_r₍_z_{) = .}

Applying Lemma . to the above equation, we get r(z)≡, which contradicts the as-sumption.

Case . n= . Comparing both sides of (), we obtainδ(f)≥degq(z) =q by Remark below Lemma .. Ifδ(f) =q, then in the same way as Case , the conclusion follows. If k=δ(f) >q, thenf(z) can be written asf(z) =φ(z)eakzk_{, where}_δ₍_φ_{) <}_k_{, and}_T₍_r_,_φ_{) =}_S₍_r_,_f_).

Substituting this expression into (), we get

φ(z) +p(z)eq(z)φ(z+c)ϕ(z) –φ(z)eakzk₌_r₍_z_), _()

whereδ(ϕ) <k, andT(r,ϕ) =S(r,f). Sincer(z)≡, we getφ(z) +p(z)eq(z)₍_φ₍_z₊_c₎_ϕ₍_z_{) –}

φ(z))≡. Comparing the characteristics function of both sides of the above equation, we get a contradiction.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

XQ completed the main part of this paper, JD and LY corrected the main theorems. All authors read and approved the ﬁnal manuscript.

Author details

1_{School of Mathematics, University of Jinan, Jinan, Shandong 250022, P.R. China.}2_{Quancheng Middle School, Jinan,}

Shandong 250000, P.R. China.3_{School of Mathematics, Shandong University, Jinan, Shandong 250100, P.R. China.}

Acknowledgements

The authors thank the referee for his/her valuable suggestions to improve the present paper. This work was supported by the National Natural Science Foundation of China (No. 11301220 and No. 11371225) and the Tianyuan Fund for Mathematics (No. 11226094), the NSF of Shandong Province, China (No. ZR2012AQ020) and the Fund of Doctoral Program Research of University of Jinan (XBS1211).

Received: 23 April 2014 Accepted: 17 September 2014 Published:03 Oct 2014

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