# On Complex Differential-Difference Equations of Malmquist Type

Authors: Jianjun Zhang
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we shall give some properties of complex differential-difference equation of the form∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ/∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν=Rz,fz, where aijμμ=1,2,…,n and bijνν=1,2,…,mi=1,2,…,t;j=0,1,…,ki are non-negative integers, cii=1,2,…,t are distinct, nonzero complex numbers, αμzμ=1,2,…,n and βνzν=1,2,…,m are small functions relative to fz, and Rz,fz is a rational function in fz with coefficients, which are small functions of fz. We also consider related complex functional equations in the paper. These results are improvements of some several previous results.


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## Body

## 1. Introduction and Main Results

In this note, we assume that the reader is familiar with the standard notations and results about Nevanlinna’s value distribution theory of meromorphic functionfz, such as the characteristic function Tr,f, proximity function mr,f, and counting function Nr,f ( e.g., [2, 7, 9, 14]). The notation N¯r,f denotes the counting function of the poles of fz whose every pole is counted only once and Sr,f denotes any quantity that satisfies the condition: Sr,f=oTr,f as r⟶∞ possibly outside an exceptional set of r of finite logarithmic measure. A meromorphic function az is called a small function of fz if Tr,az=Sr,f.As one knows, Malmquist type problem is an important topic in complex differential equations. For the first-order complex differential equations, Malmquist [16] proved the following important result independent of Nevanlinna theory in 1913.Theorem 1.
LetRz,f be birational function. If a differential equation of the form(1)f′z=Rz,fz,admits a transcendental meromorphic solution, then the equation can be reduced into a Riccati differential equation(2)f′z=a0z+a1zfz+a2zf2z,where aizi=0,1,2 are rational functions.Later, Yosida [24] gave a proof of Theorem 1 by using Nevanlinna theory. Laine [13], Yang [23], Hille [11], Steinmetz [21], Rieth [19], He and Laine [8], Gackstatter and Laine [4], and Rieth [19] extended Theorem 1 to more general types. In particular, Gackstatter and Laine [4] gave a following generalized result of Theorem 1 in 1980.Theorem 2.
If the algebraic differential equation(3)∑λ∈Iαλzfi0f′i1⋯fnin=Rz,f=∑i=0paizfi∑j=0qbjzfj,where I=λ=i0,i1,…,in|ik∈ℕ∪0,k=0,1,…,n is a finite index set, coefficients αλzλ∈I, aizi=0,1,…,p, and bjzj=0,1,…,q are meromorphic and small functions relative to fz, possesses an transcendental meromorphic solution, then Rz,f reduces to a polynomial in f of degree ≤ΓP:=maxλ∈Ii0+2i1+⋯+n+1in.Recently, Zhang and Liao [26] gave a simple proof of Theorem 2, which is different from the methods that of Gackstatter and Laine [4] and Steinmetz [21]. However, Malmquist type problem even for special second-order algebraic differential equations remains open. One may refer to [14, p. 251].In the past fifteen or more years, people have already had great interest in complex difference equations. In particular, there are a large number of papers on Malmquist type theorem of the complex difference equations, for instance [1, 3, 10, 12, 15, 25]. In 2000, Ablowitz et al. [1] first obtained two results on Malmquist type of complex difference equations by utilizing Nevanlinna theory. One year later, Heittokangas et al. [10] extended these two results to the case of more higher-order complex difference equations. Subsequently, Laine et al. [15] and Huang and Chen [12], respectively, generalized the preceding results. At the same time, Laine et al. [15] also obtained Tumura-Clunie theorem about complex difference equation of Malmquist type and considered the growth of meromorphic solutions of corresponding complex functional equations. In 2010, Zhang and Liao [25] generalized these results to more general complex difference equations and functional equations.A natural question is whether we can obtain some corresponding results for complex differential-difference equations of Malmquist type? The answer is positive. Now we give our main results as follows.Theorem 3.
Letc1,c2,…,ct be distinct, nonzero complex numbers and suppose that fz is a transcendental meromorphic solution of the differential-difference equation(4)∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν=∑i=0paizfi∑j=0qbjzfj,where aijμμ=1,2,…,n and bijνν=1,2,…,mi=1,2,…,t;j=0,1,…,ki are non-negative integers, αμzμ=1,2,…,n,βνzν=1,2,…,m,aizi=0,1,…,p and bjzj=0,1,…,q are small functions relative to fz. We denote(5)σi=maxμ,νai0μ+2ai1μ+⋯+ki+1aikiμ,bi0ν+2bi1ν+⋯+ki+1bikiν,i=1,2,…,t,σ=∑i=1tσi.If the orderρf:=ρ is finite, then d=maxp,q≤σ.According to Theorem 3, we can obtain the following two results easily.Corollary 1.
Letc1,c2,…,ct be distinct, nonzero complex numbers and suppose that fz is a transcendental meromorphic solution of the differential-difference equation(6)∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ=∑i=0paizfi∑j=0qbjzfj,where aijμμ=1,2,…,n;i=1,2,…,t;j=0,1,…,ki are non-negative integers, αμzμ=1,2,…,n,aizi=0,1,…,p, and bjzj=0,1,…,q are small functions relative to fz. We denote(7)σi=maxμ,νai0μ+2ai1μ+⋯+ki+1aikiμ,i=1,2,…,tσ=∑i=1tσi.If the orderρf:=ρ is finite, then d=maxp,q≤σ.Corollary 2.
Letc1,c2,…,cn be distinct, nonzero complex numbers and suppose that fz is a transcendental meromorphic solution of the differential-difference equation(8)∑λ∈Iαλz∏ν=1nfkλ,νz+cνlλ,ν∑μ∈Jβμz∏ν=1nfsμ,νz+cνmμ,ν=∑i=0paizfi∑j=0qbjzfj,with coefficients αλzλ∈I,βμzμ∈J,aizi=0,1,…,p and bjzj=0,1,…,q are small functions relative to fz, where I=λ=kλ,1,lλ,1,kλ,2,lλ,2,…,kλ,n,lλ,n|kλ,ν,lλ,ν∈ℕ∪0,ν=1,2,…,n and J=μ=sμ,1,mμ,1,sμ,2,mμ,2,…,sμ,n,mμ,n|sμ,ν,mμ,ν∈ℕ∪0,ν=1,2,…,n are two finite index sets, we denote(9)σν=maxλ,μkλ,ν+1lλ,ν,sμ,ν+1mμ,νν=1,2,…,n,σ=∑ν=1nσν.If the orderρf:=ρ is finite, then d=maxp,q≤σ.We are now giving the following two examples. Example 1 shows that the estimation in Theorem 3 is sharp and Example 2 shows that the conditionρf is finite in Theorem 3 and cannot be removed.Example 1.
Letc1=arctan2, c2=−π/4. Then, fz=tanz solves the following differential-difference equation:(10)f′z+c12fz+c22fz+c1+f″z+c2=25f+13f−142f−137f4−17f3+3f2−19f+2.
In Example1, we have d=σ=7 and ρf=1<+∞.Example 2.
Letc1=log2, c2=log5. It is easy to check that fz=eez satisfies the following differential-difference equation:(11)f′z+c12fz+c22+f″z+c2fz+c1=4e2zf12+5ez+25e2zf3.
Clearly, we haved=12>7=σ and ρf=+∞.
The following result is the corresponding Tumura-Clunie theorem on complex differential-difference equation.Theorem 4.
Suppose thatc1,c2,…,ct are distinct, nonzero complex numbers and that fz is a transcendental meromorphic solutions of the following equation:(12)∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν=Pz,fzQz,fz,where aijμμ=1,2,…,n and bijνν=1,2,…,mi=1,2,…,t;j=0,1,…,ki are non-negative integers, and the coefficients αμzμ=1,2,…,n,βνzν=1,2,…,m are nonvanishing small functions relative to fz and Pz,fz and Qz,fz are relatively prime polynomials in fz over the field of small functions relative to fz. We denote(13)σi=maxμ,νai0μ+2ai1μ+⋯+ki+1aikiμ,bi0ν+2bi1ν+⋯+ki+1bikiν,i=1,2,…,t,σ=∑i=1tσi.Moreover, we assume thatq=degfQ>0,(14)σ=maxp,q≔=maxdegfP,degfQ,and that, without restricting generality, Q is a monic polynomial. If there exists α∈0,σ such that for all r sufficiently large,(15)N¯r,∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν≤αN¯r+C,fz+Sr,f,(16)∑i=1tσiN¯r,fz+ci≤αN¯r+C,fz+Sr,f,where C:=maxc1,c2,…,ct, then either the order ρf=+∞, or(17)Qz,fz≡fz+hzq,where hz is a small meromorphic function relative to fz.Note that by the fundamental property of counting function, we have(18)N¯r,∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ≤∑i=1tσiN¯r,fz+ci+Sr,f.Hence, if the left-hand side of (12) in Theorem 4 is replaced by the left-hand side of (6) in Corollary 1, then (16) implies (15). Therefore, we have the following result.Corollary 3.
Suppose thatc1,c2,…,ct are distinct, nonzero complex numbers and that fz is a transcendental meromorphic solutions of the following equation:(19)∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ=Pz,fzQz,fz,where aijμμ=1,2,…,n;i=1,2,…,t;j=0,1,…,ki are non-negative integers, and the coefficients αμzμ=1,2,…,n are nonvanishing small functions relative to fz and Pz,fz and Qz,fz are relatively prime polynomials in fz over the field of small functions relative to fz. We denote(20)σi=maxμ,νai0μ+2ai1μ+⋯+ki+1aikiμ,i=1,2,…,t,σ=∑i=1tσi.Moreover, we assume thatq=degfQ>0,(21)σ=maxp,q≔=maxdegfP,degfQ,and that, without restricting generality, Q is a monic polynomial. If there exists α∈0,σ such that for all r sufficiently large,(22)∑i=1tσiN¯r,fz+ci≤αN¯r+C,fz+Sr,f,where C:=maxc1,c2,…,ct, then either the order ρf=+∞, or(23)Qz,fz≡fz+hzq,where hz is a small meromorphic function relative to fz.Finally, we give a result about the growth of meromorphic solutions of complex functional equations.Theorem 5.
Letc1,c2,…,ct be distinct, nonzero complex numbers and suppose that f is a transcendental meromorphic solution of the equation,(24)∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν=fpz,where aijμμ=1,2,…,n;i=1,2,…,t;j=0,1,…,ki are non-negative integers and pz is a polynomial of degree k≥2. We denote(25)σi=maxμ,νai0μ+2ai1μ+⋯+ki+1aikiμ,bi0ν+2bi1ν+⋯+ki+1bikiν,i=1,2,…,t,σ=∑i=1tσi.Moreover, we assume that the coefficientsαμzμ=1,2,…,n and βνzν=1,2,…,m are small functions relative to f and that σ≥k. Then,(26)Tr,f=Ologrα+ε,where α=logσ/logk.


## 2. Main Lemmas

In order to prove our results, we need the following lemmas.Lemma 1 (see [17]).
Letfz be a meromorphic function. Then, for all irreducible rational functions in f,(27)Rz,f=Pz,fQz,f=∑i=0paizfi∑j=0qbjzfj,such that the meromorphic coefficients aiz,bjz satisfy(28)Tr,ai=Sr,f,i=0,1,…,p,Tr,bj=Sr,f,j=0,1,…,q,we have(29)Tr,Rz,f=maxp,q⋅Tr,f+Sr,f.The estimation for the Nevanlinna characteristic of some classes of meromorphic functions plays a very important role in the proof of our theorems. According to the Proof of Theorem 3 in [18], we have the following.Lemma 2 (see [18]).
Letf1,f2,…,ft be distinct meromorphic functions and(30)Φz=∑μ=1nαμz∏i=1tfiai0μfi′ai1μ⋯fikiaikiμ∑ν=1mβνz∏i=1tfibi0νfi′bi1ν⋯fikibikiν,where aijμμ=1,2,…,n and bijνν=1,2,…,m are non-negative integers, αμzμ=1,2,…,n and βνzν=1,2,…,n are small functions of fizi=1,2,…,t. We set(31)σi=maxμ,νai0μ+2ai1μ+⋯+ki+1aikiμ,bi0ν+2bi1ν+⋯+ki+1bikiν,i=1,2,…,t.Then,(32)Tr,Φ≤∑i=1tσiTr,fi+oTr,fi,for r⟶∞ outside of a possible exceptional set of finite linear measure.Lemma 3 (see [3]).
Letfz be a meromorphic function with order ρ=ρf,ρ<+∞, and c be a fixed nonzero complex number, then for each ε>0, we have(33)Tr,fz+c=Tr,f+Orρ−1+ε+Ologr.Lemma 4 (see [22]).
Letfz be a meromorphic function and let ϕ be given by(34)ϕ=fn+an−1fn−1+⋯+a0,where aii=0,1,…,n−1 are small meromorphic functions relative to fz. Then, either(35)ϕ=f+an−1nn,or(36)Tr,f≤N¯r,1ϕ+N¯r,f+Sr,f.Lemma 5 (see [15, 20]).
Letfz be a nonconstant meromorphic function and Pz,f, Qz,f two polynomials in fz with meromorphic coefficients small relative to fz. If Pz,f and Qz,f have no common factors of positive degree in fz over the field of small functions relative to fz, then(37)N¯r,1Qz,f≤N¯r,Pz,fQz,f+Sz,f.Lemma 6 (see [5]).
Letf be a transcendental meromorphic function, and pz=akzk+ak−1zk−1+⋯+a1z+a0,ak≠0, be a nonconstant polynomial of degree k. Given 0<δ<ak, we denote λ=ak+δ and μ=ak−δ. Then, given ε>0 and a∈ℂ∪∞, we have(38)1−εTμrk,f≤Tr,fpz≤1+εTλrk,ffor all r large enough.Lemma 7 (see [6]).
Letϕ:r0,+∞⟶0,+∞ be positive and bounded in every finite interval, and suppose that ϕμrm≤Aϕr+B holds for all r large enough, where μ>0,m>1,A>1 and B are real constants. Then,(39)ϕr=Ologrα,where α=logA/logm.


## 3. Proofs of Theorems

Proof of Theorem 3.
We assume thatfz is a transcendental meromorphic solution with finite order of equation (4). It follows from Lemma 1 to Lemma 3 that for each ε>0,(40)maxp,qTr,f=Tr,∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν+Sr,f≤∑i=1tσiTr,fz+ci+Sr,f=∑i=1tσiTr,fz+Orρ−1+ε+Ologr+Sr,f=∑i=1tσiTr,fz+Orρ−1+ε+Ologr+Sr,f.=σTr,fz+Orρ−1+ε+Ologr+Sr,f.
Therefore, we haved=maxp,q≤σ.Proof of Theorem 4.
Supposefz is a transcendental meromorphic solution of equation (12) and(41)Qz,fz≠fz+hzq.
Then, by Lemma4, Lemma 5 and condition (15), we have(42)Tr,f≤N¯r,1Qz,f+N¯r,f+Sr,f≤N¯r,Pz,fQz,f+N¯r,f+Sr,f=N¯r,∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν+N¯r,f+Sr,f≤αN¯r+C,fz+N¯r,f+Sr,f.
Thus,(43)Tr,f−N¯r,f≤αN¯r+C,fz+Sr,f.
Now we assume that the orderρf<+∞, then we have Sr,fz+ci=Sr,f and(44)Tr,fz+ci−N¯r,fz+ci≤αN¯r+C,fz+ci+Sr,f,for i=1,2,…,t. By using Lemma 1, Lemma 2 and condition (16), we conclude that,(45)σTr,f=Tr,∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν+Sr,f≤∑i=1tσiTr,fz+ci+Sr,f=∑i=1tσiTr,fz+ci−N¯r,fz+ci+∑i=1tσiN¯r,fz+ci+Sr,f≤∑i=1tσiαN¯r+C,fz+ci+αN¯r+C,fz+Sr,f≤∑i=1tσiαN¯r+2C,fz+αN¯r+C,fz+Sr,f≤∑i=1tσiαN¯r+2C,fz+αN¯r+2C,fz+Sr,f=σ+1αN¯r+2C,fz+Sr,f.
Thus, we have(46)Tr,f−N¯r,f≤σ+1σαN¯r+2C,f−N¯r,f+Sr,f.
Next we shall prove the following inequality by induction,(47)Tr,f−N¯r,f≤σ+mσαN¯r+2mC,f−mN¯r,f+Sr,f.
The casem=1 has been proved. We assume inequality (47) holds when m=k. Now we are going to prove that the inequality (47) holds for m=k+1. From (16) and (45), we have(48)σTr,f≤∑i=1tσiTr,fz+ci−N¯r,fz+ci+αN¯r+C,fz+Sr,f≤∑i=1tσiσ+kσαN¯r+2kC,fz+ci−kN¯r,fz+ci+αN¯r+C,fz+Sr,f≤∑i=1tσiσ+kσαN¯r+2kC+C,fz−kN¯r−C,fz+αN¯r+C,fz+Sr,f≤∑i=1tσiσ+kσαN¯r+2kC+C,fz−kN¯r−C,fz+αN¯r+2kC+C,fz+Sr,f=σ+k+1αN¯r+2kC+C,fz−σkN¯r−C,fz+Sr,f.
SinceTr,fz≤Tr+C,fz, we have(49)σTr,fz≤σTr+C,fz≤σ+k+1αN¯r+2kC+2C,fz−σkN¯r,fz+Sr,f.
Hence,(50)Tr,fz≤σ+k+1σαN¯r+2k+1C,fz−kN¯r,fz+Sr,f.
This implies that(51)Tr,fz−N¯r,fz≤σ+k+1σαN¯r+2k+1C,fz−k+1N¯r,fz+Sr,f.
It follows from (47) that(52)N¯r,fz≤σ+mσmαN¯r+2mC,f+Sr,f.
Letm be large enough such that(53)1γ≔=σ+mσmα=1m+1σα<1.
Since(54)N¯r,fz≤1γN¯r+2mC,fz+Sr,f,
We have for anys∈ℕ,(55)N¯r,fz≤1γsN¯r+2smC,f+Sr+s−1mC,f.
Thus, for eachε>0,(56)γsN¯r,fz≤N¯r+2smC,f+Sr+s−1mC,f≤1+εTr+2smC,fz,for r+2smC large enough holds. Now fixing r=r0 and setting r0+2smC=t. Then,(57)γt−r0/2mCN¯r0,fz≤1+εTt,f,logTt,flogt+log1+εlogt≥tlogγ2mClogt−r0logγ2mClogt+logN¯r0,flogt.
Finally, lett⟶∞, and we conclude that the order ρf=∞. Therefore, we get a contradiction and the assertion follows.Proof of Theorem 5.
We assumefz is a transcendental meromorphic solution of equation (24). Denoting again C=maxc1,c2,…,cn. By Lemma 6 and Lemma 2, we have(58)1−εTμrk,f≤Tr,fpz=Tr,∑μ=1nαμz∏i=1tfz+ciai0μf′z+ciai1μ⋯fkiz+ciaikiμ∑ν=1mβνz∏i=1tfz+cibi0νf′z+cibi1ν⋯fkiz+cibikiν,≤∑i=1tσiTr,fz+ci+Sr,f≤∑i=1tσiTr+C,fz+Sr,f=∑i=1tσiTr+C,fz+Sr,f=σTr+C,fz+Sr,f.
SinceTr+C,f≤Tβr,f holds for r large enough and β>1, we may assume r to be large enough such that(59)1−εTμrk,f≤σ1+εTβr,f.
Outside a possible exceptional set of finite linear measure. By the standard idea of removing the exceptional set (see [14], p5), we know that whenever γ>1,(60)1−εTμrk,f≤σ1+εTγβr,f,holds for all r large enough. We denote t=γβr, then (60) may be written as follows:(61)Tμγβktk,f≤σ1+ε1−εTt,f.
By Lemma7, we have(62)Tr,f=Ologrs,where(63)s=logσ1+ε/1−εlogk=logσlogk+o1,
Denoting nowα=logσ/logk, thus we obtain the required form. Thus, we complete the Proof 3.



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