# Results on Solutions for Severalq-Painlevé Difference Equations concerning Rational Solutions, Zeros, and Poles

Authors: Bu Sheng Li; Rui Ying; Xiu Min Zheng; Hong Yan Xu
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this article, we discuss the problem about the properties on solutions for several types ofq-difference equations and obtain some results on the exceptional values of transcendental meromorphic solutions fz with zero order, their q-differences Δqfz=fqz−fz, and divided differences Δqfz/fz. In addition, we also investigated the condition on the existence of rational solution for a class of q-difference equations. Our theorems are some extensions and supplement to those results given by Liu and Zhang and Qi and Yang.


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## Body

## 1. Introduction and Main Results

All the time, Painlevé equations have attracted much interest due to the reduction of solution equations, which are solvable by inverse scattering transformations, and they often occur in many physical situations: plasma physics, statistical mechanics, and nonlinear waves. The study of Painlevé equations has spanned more than one hundred years (see [1–3]).Around 2006, Halburd and Korhonen [4, 5] and Ronkainen [6] used Nevanlinna theory to discuss the following equations:(1)fz+1+fz−1=Rz,f=Pz,fQz,f,fz+1fz−1=Rz,f=Pz,fQz,f,where Rz,f is rational in f and meromorphic in z, respectively, and they singled out the following difference equations:(2)fz+1+fz−1=az+bfz+c,(3)fz+1+fz−1=az+bfz+c1−fz2,(4)fz+1fz−1=ηzfz2−λzfz+μzfz−1fz−υz,(5)fz+1fz−1=ηzfz2−λzfzfz−1,(6)fz+1fz−1=ηzfz−λzfz−1,(7)fz+1fz−1=hzfzm,where ηz,λz,andυz satisfy some conditions. In these equations, equation (2) is called as the difference Painlevé I equation, equation (3) is called as the difference Painlevé II equation, and the last four equations are called as the difference Painlevé III equations.In the last decade or so, there were a lot of papers focusing on the properties of solutions for difference PainlevéI–IV equations (see [7–11]). For example, Chen and Shon [12] in 2010 considered the difference Painlevé I equation (3) and obtained the following theorem.Theorem 1 (see [12], Theorem 4).
Leta,b,c be constants, where a,b are not both equal to zero. Then, the following holds:(i)
Ifa≠0, then (3) has no rational solution.(ii)
Ifa=0 and b≠0, then (3) has a nonzero constant solution wz=A, where A satisfies 2A2−cA−b=0.The other rational solutionwz satisfies wz=Pz/Qz+A, where Pz and Qz are relatively prime polynomials and satisfy degP<degQ.In 2013 and 2018, Zhang and Yi [11] and Du et al. [13] studied the difference Painlevé III equations with the constant coefficients and obtained the result as follows.Theorem 2 (see [11, 13]).
Iff is a transcendental finite-order meromorphic solution of(8)fz+1fz−1fz−12=fz2−λfz+μ,where λ and μ are constants, then the following holds:(i)
τf=σf.(ii)
Ifλμ≠0, then λf=σf.(iii)
For anyη∈C/0, τfz+η=σf.(iv)
λ1/Δf=λ1/Δf/f=σf.Ramani et al. [14] in 2003 investigated the existence of transcendental solution of equation(9)fz+1+fzfz+fz−1=Rz,f=Pz,fQz,f,which is called as difference Painlevé IV equations and obtained the result as follows.Theorem 3 (see [14]).
If the second-order difference equation (9) admits a nonrational meromorphic solution of finite order, then degzP≤4 and degzQ≤2.Of late, many mathematicians paid considerable attention to the value distribution of solutions for complexq-difference equations, which are formed by replacing the q-difference fqz, q∈C/0,1 with fz+c of meromorphic function in some expressions concerning complex difference equations, by utilizing the logarithmic derivative lemma on q-difference operators given by Barnett et al. [15] in 2007 (see [16–26]). For example, Qi and Yang [27] considered the following equation:(10)fqz+fzq=az+bfz+c,which can be seen as q-difference analogues of (2) and obtained the result as follows.Theorem 4 (see [27], Theorem 1).
Letfz be a transcendental meromorphic solution with zero order of equation (10) and a,b,c be three constants such that a,b cannot vanish simultaneously. Then, the following holds:(i)
fz has infinitely many poles.(ii)
Ifa≠0 and any d∈C, then fz−d has infinitely many zeros.(iii)
Ifa=0 and fz takes a finite value A finitely often, then A is a solution of 2z2−cz−b=0.In 2018, Liu and Zhang [28] further investigated the following equation:(11)Yωz+Yz+Yzω=ξz+oYz+ν,and obtained the result as follows.Theorem 5 (see [28], Theorem 1).
LetYz be a transcendental meromorphic solution with zero order of (11) and ξ,o,ν be three constants such that ξ,o cannot vanish simultaneously. Then, the following holds:(i)
Yz has infinitely many poles.(ii)
For any finite valueB, if ξ=0, then Yz−B has infinitely many zeros.(iii)
Ifξ=0 and Yz−A has finite zeros, then A is a solution of 3z2−o−νz=0.Motivated by the idea [27, 28], a natural question is what is the result if we give q-difference analogues of (9). For this question, our main aim of this article is further to investigate some properties of meromorphic solutions for some q-Painlevé difference IV equations. It seems that this topic has never been treated before.In what follows, it should be assumed that the readers are familiar with the fundamental results and the standard notations in the theory of Nevanlinna value distribution (see Hayman [29], Yang [30], and Yi and Yang [31]). Let f be a meromorphic function, and we denote σf,λf, and λ1/f to be the order, the exponent of convergence of zeros, and the exponent of convergence of poles of fz, respectively, and denote τf to be the exponent of convergence of fixed points of fz, which is defined by(12)τf=limsupr⟶+∞logNr,1/fz−zlogr.In addition, we useSr,f denotes any quantity satisfying Sr,f=oTr,f for all r on a set F of logarithmic density 1, and the logarithmic density of a set F is defined by(13)limsupr⟶∞1logr∫1,r∩F1tdt.Now, our main results are listed as follows.Theorem 6.
LetRz=Az/Bz be an irreducible rational function, and let q≠0∈C and q≠1, and (14)fqz+fzfz+fzq=Rz=AzBz,where Az,Bz are polynomials with degzAz=a and degzBz=b.(i)
Suppose thata≥b and a−b are even numbers or zero. If equation (14) has an irreducible rational solution fz=μz/νz, where μz,νz are polynomials with degzμz=μ and degzνz=ν, then the following holds:(15)μ−ν=a−b2.(ii)
Suppose thatb≥a and b−a are even numbers or zero. If equation (14) has an irreducible rational solution fz=μz/νz, where μz,νz are polynomials with degzμz=μ and degzνz=ν, then(16)ν−μ=b−a2.(iii)
Ifa−b is an odd number, then equation (14) has no rational solution.Theorem 7.
Forq≠0∈C and q≠1, let fz be a transcendental meromorphic solution with zero order of equation(17)fqz+fzfz+fzq=afz2,where a≠0,4 is a constant. Let Δqf=fqz−fz. Then, the following holds:(i)
Bothf and Δqf have no nonzero finite Nevanlinna exceptional value.(ii)
Ifq+1/q≠a−2, then fηz and Δqfηz have infinitely many fixed points and τfηz=τΔqfηz=σf for any nonzero constant η.Theorem 8.
Forq≠0∈C and q≠1, and let fz be a transcendental meromorphic solution with zero order of equation(18)fqz+fzfz+fzq=dzfz,where dz is a nonconstant rational function satisfying that dqz/dz is not a constant. Then, the following holds:(i)
Bothf and Δqf/f have no Nevanlinna exceptional value.(ii)
Δqf has infinitely many poles and zeros, and λ1/Δqf=λΔqf=σf.(iii)
Δqf/f has infinitely many fixed points and τΔqf/f=σf.


## 2. Proof of Theorem6

Proof:.
assume that (14) has a rational solution fz=μz/vz and has poles z1,z2,…,zk. Then, fz can be represented in the following form:(19)fz=μzνz=∑j=1kcjλjz−zjλj+⋯+cj1z−zj+a0+a1z+⋯+aszs,where cjλj≠0,…,cj1j=1,2,…,k and a0,a1,…,as are constants; zjj=1,2,…,k are poles of fz with multiplicity λj, respectively.(i)
Suppose thata>b and a−b are even numbers. Then, in view of (14) and (19), it yields(20)μqzνqz+μzνzμzνz+μz/qνz/q=AzBz.Ifdegzμz=μ<ν=degzνz, then for z⟶∞, it follows(21)μqzνqz⟶0,μzνz⟶0,μz/qνz/q⟶0.However,Az/Bz⟶∞ as z⟶∞; thus, from (20), we can get a contradiction easily.Ifμ=ν, then let z⟶∞, and it leads to(22)μqzνqz⟶α,μzνz⟶α,μz/qνz/q⟶α,whereα is a nonzero constant. Thus, let z⟶+∞; in view of (20), we also get a contradiction. So, it follows μ>ν. Thus, assume that as≠0,s≥1, where s=μ−ν. As z⟶∞, it yields(23)fz=aszs1+o1,fqz=asqszs1+o1,fzq=asq−szs1+o1,AzBz=βza−b1+o1,whereβ≠0 is a constant, and it follows now in view of (20) that(24)2+qs+q−sas2z2s1+o1=βza−b1+o1,asz⟶∞. Since q≠1, then qs+2+q−s≠0. Hence, it follows from (24) that(25)μ−ν=s=a−b2.Next, assume thata=b. As z⟶∞, it follows(26)AzBz=β1+o1,whereβ≠0 is a constant. If μ<ν, then by using the same argument as above, we get a contradiction. If μ≥ν, then we assume that as≠0,s≥1. By using the same argument as above, we conclude(27)qs+2+q−sas2z2s=β1+o1,asz⟶∞. Thus, if μ>ν, then in view of (27), we can get a contradiction; if μ=ν, then we have(28)μ−ν=0=a−b2.(ii)
Suppose thatb>a and b−a are even numbers. Then, in view of (14) and (19), we get (20).Ifμ>ν, then for z⟶∞, it leads to(29)μqzνqz⟶∞,μzνz⟶∞,μz/qνz/q⟶∞.However,Az/Bz⟶0 as z⟶∞; thus, from (20), we can get a contradiction easily.Ifμ=ν, then let z⟶∞, it follows(30)μqzνqz⟶α,μzνz⟶α,μz/qνz/q⟶α,whereα is a nonzero constant. Thus, let z⟶+∞; in view of (20), we also get a contradiction. Thus, μ<ν. We rewrite (14) as the following form:(31)Bzμqzμzνzqνz+μqzμzqνz2+μz2νqzνzq+μzμzqνqzνz=Azνz2νqzνzq.Denote(32)Az=ξaza+⋯,Bz=δbzb+⋯,μz=γμzμ+⋯,νz=ζνzν+⋯,wherea≥1,b≥0, and μ≥0,ν≥1 are all nonnegative integers. Thus, in view of (31) and (32), we can deduce(33)qμ−ν+2+qν−μδbγμ2ζν2z2μ+ν+b+⋯=ξaζν4z4ν+a.Sinceq≠1, then qμ−ν+2+qν−μ≠0. Thus, by combining with this and (33), we have(34)2μ+ν+b=4ν+a,that isν−μ=b−a2,andζν2/γμ2=δb/ξaqμ−ν+2+qν−μ.(iii)
Ifa>b, then a−b=a−b is an odd number. Assume that fz=μz/νz is a rational solution of (14). In view of the conclusion of Theorem 6 (i), it follows μ−ν=a−b/2. This means a contradiction with the assumption that a−b is an odd number. Thus, (14) has no rational solution.
Ifa<b, then a−b=b−a is an odd number. Similar to the above argument, we also conclude that (14) has no rational solution.
Therefore, this completes the proof of Theorem6.


## 3. Proof of Theorem7

We first introduce some notations and some basic results about Nevanlinna theory, which can be used in Section3 and Section 4. Let f be a meromorphic function in C, the Nevanlinna characteristic Tr,f, which encodes information about the distribution of values of f on the disk z≤r, is defined by(35)Tr,f=mr,f+Nr,f.The proximity functionmr,f is defined by(36)mr,f=mr,∞,f=12π∫02πlog+freiθdθ,where log+x=max0,logx and(37)Nr,f=Nr,∞,f=∫0rnt,f−n0,ftdt+n0,flogr,where nr,f is the number of poles of f in the circle z=r, counted according to multiplicities.Leta∈C∪∞, and the deficiency of a with respect to fz is defined by(38)δa,f=liminfr⟶+∞mr,1/f−aTr,f=1−limsupr⟶+∞Nr,1/f−aTr,f.Ifδa,f>0, then the complex number a is called the Nevanlinna exceptional value. And the order σf, the exponent of convergence of zeros λf, and the exponent of convergence of poles λ1/f of fz are defined by(39)σf=limsupr⟶+∞log+Tr,flogr,λf=limsupr⟶+∞log+Nr,1/flogr,λ1f=limsupr⟶+∞log+Nr,flogr.Besides, we also use some properties ofTr,f,mr,f,andNr,f such as(40)Tr,1f−a=Tr,f+O1,mr,∏j=1pfj≤∑j=1pmr,fj,mr,∑j=1pfj≤∑j=1pNr,fj+logp,Nr,∏j=1pfj≤∑j=1pNr,fj,Nr,∑j=1pfj≤∑j=1pNr,fj,Tr,∏j=1pfj≤∑j=1pTr,fj,Tr,∑j=1pfj≤∑j=1pTr,fj+logp,where fjzj=1,2,…,p are p meromorphic functions and a∈C and require some lemmas as follows.Lemma 1 (see [15], Theorem 2).
Letf be a nonconstant zero-order meromorphic solution of Pz,f=0, where Pz,f is a q-difference polynomial in fz. If Pz,a≢0 for slowly moving target az, then(41)mr,1f−a=Sr,f,where Sr,f denotes any quantity satisfying Sr,f=oTr,f for all r on a set F of logarithmic density 1.Remark 1.
Forq∈C/0,1, a polynomial in fz and finitely many of its q-shifts fqz,…,fqnz with meromorphic coefficients in the sense that their Nevanlinna characteristic functions are oTr,f on a set F of logarithmic density 1 and can be called as a q-difference polynomial of f.Lemma 2 (see [24], Theorems 1 and 3).
Letfz be a nonconstant zero-order meromorphic function and q∈C/0. Then,(42)Tr,fqz=1+o1Tr,fz,Nr,fqz=1+o1Nr,fz,on a set of lower logarithmic density 1.The proof of Theorem7:.
(i) suppose thatfz is a transcendental meromorphic solution of equation (17), then in view of (17), let(43)P1z,f≔fqz+fzfz+fzq−afz2≡0.
For any given constantd∈C/0 and with a view of a≠4, it follows(44)P1z,d=4d2−ad2≠0.
In view ofP1z,d≢0 and by Lemma 1, we conclude that mr,1/f−d=Sr,f. This leads to(45)Nr,1f−d=Tr,f+Sr,f,which implies δd,f=0. Thus, fz has no nonzero finite Nevanlinna exceptional value.
Sincefz is of zero order and Δqf=fqz−fz, then by Lemma 2, it follows Tr,Δqf≤2Tr,f+Sr,f, which means that Δqf if of zero order. In view of (17), it follows(46)fq2z+fqzfqz+fz=afqz2.
With (17) subtraction, it leads to(47)fqz+fzfq2z+fqz−fz−fzq=afqz+fzfqz−fz.
From (17), we see that fqz+fz≡0. Otherwise, it leads to fz≡0, a contradiction. Thus, the above equality means(48)fq2z+fqz−fz−fzq=afqz−fz,that is,(49)Δqfqz+2−aΔqfz+Δqfzq=0.
Denote(50)P2z,Δqf≔Δqfqz+2−aΔqfz+Δqfzq≡0.
For any given constantd∈C/0, then from (50), we have P2z,d=4−ad. Hence, with a view of a≠4, it follows P2z,d≡0. Thus, by Lemma 1, we have mr,1/Δqf−d=Sr,Δqf, and this leads to(51)Nr,1Δqf−d=Tr,Δqf+Sr,Δqf,which implies that δd,Δqf=0. Thus, Δqf has no nonzero finite Nevanlinna exceptional value:(ii)
Replacingz by ηz in (17), we have(52)fqηz+fηzfηz+fηzq=afηz2.
Letg1z=fηz, it yields(53)g1qz+g1zg1z+g1zq=ag1z2.
Set(54)P3z,g1≔g1qz+g1zg1z+g1zq−ag1z2≡0.Thus, it follows P3z,z=q+12/q−az2, and with a view of q+1/q≠a−2, we have P3z,z≡0. By applying Lemma 1, it yields mr,1/g1z−z=Sr,g1. Thus, in view of Lemma 2, this leads to(55)Nr,1fηz−z=Nr,1g1z−z=Tr,g1z+Sr,g1=Tr,fηz+Sr,fηz=Tr,f+Sr,f,which implies that fηz has infinitely many fixed points and τfηz=σf.
In view of (48), set g2z=Δqfηz and(56)P4z,g2≔g2qz+2−ag2z+g2zq≡0,then P4z,z=q+2−a+1/qz. Since q+1/q≠a−2, then P4z,z≡0. By applying Lemma 1, we have mr,1/g2z−z=Sr,g2. Thus, in view of Lemma 2, this leads to(57)Nr,1Δqfηz−z=Nr,1g2z−z=Tr,g2z+Sr,g1=Tr,Δqfηz+Sr,fηz,which implies that Δqfηz has infinitely many fixed points and τΔqfηz=σf.
Therefore, the proof of Theorem7 is completed.


## 4. Proof of Theorem8

Lemma 3 (see [15], Theorem 1).
Letfz be a nonconstant zero-order meromorphic function and q∈C/0. Then,(58)mr,fqzfz=Sr,f.Lemma 4 (see [17], Theorem 2.5).
Letf be a transcendental meromorphic solution of order zero of a q-difference equation of the form(59)Uqz,fPqz,f=Qqz,f,where Uqz,f,Pqz,f, and Qqz,f are q-difference polynomials such that the total degree degUqz,f=n in fz and its q-shifts, whereas degQqz,f≤n. Moreover, we assume that Uqz,f contains just one term of maximal total degree in fz and its q-shifts. Then,(60)mr,Pqz,f=Sr,f.Proof of Theorem8:.
(i) suppose thatfz is a transcendental meromorphic solution of equation (18). We firstly prove that Δqf/f has no Nevanlinna exceptional value. Equation (18) can be rewritten as(61)fqzfz+1fz/qfz+1=dzfz.
Setg3z=fqz/fz. In view of (61) and Lemma 2, it follows(62)Tr,fz=Tr,dzfz+Ologr≤Tr,fqzfz+Tr,fz/qfz+Ologr=Tr,g3z+Tr,g3zq+Ologr=2Tr,g3+Sr,g3=2Tr,Δqff+Sr,f.
And with a view ofTr,Δqf/f≤2Tr,f+Sr,f, we thus conclude that Δqf/f is transcendental and of order zero, and dz is small with respect to g3z.
Replacingz by qz in (18), it follows(63)fq2z+fqzfqz+fz=dqzfqz.
By combining with (18), we have(64)fq2z+fqzfz+fz/q=dqzfqzdzfz.
Sinceg3z=fqz/fz, then it yields(65)fqz=g3zfz,fq2z=g3qzfqz=g3zg3qzfz,fzq=fzg3z/q.
Substituting (65) into (64), we obtain(66)g3zg3qzfz+g3zfzfz+fz/g3z/q=dqzdzg3z,that is,(67)g3zqg3qz+1=dqzdzgzq+1.
By applying Lemma4 for (67), it follows mr,g3z/q=Sr,g3. This leads to(68)Nr,g3zq=Tr,g3+Sr,g3.
Thus, in view of Lemma2, it yields(69)1+o1Nr,g3=Nr,g3zq=Tr,g3zq+Sr,g3=1+o1Tr,g3+Sr,g3.
This shows(70)Nr,Δqff=Nr,g3=Tr,g3+Sr,g3=Tr,Δqff+Sr,Δqff,which implies δ∞,Δqf/f=0.
Set(71)P5z,g3≔g3zqg3qz+1−dqzdzgzq+1≡0.
For any constantϱ∈C/−2, we have(72)P5z,ϱ+1=ϱ+1ϱ+1+1−dqzdzϱ+1+1=ϱ+2ϱ+1−dqzdz.
Sincedqz/dz is not a constant, then P5z,ϱ+1≢0. By Lemma 1, it follows(73)mr,1Δqf/f−ϱ=mr,1g3−ϱ−1=Sr,g3=Sr,Δqff.
This means(74)Nr,1Δqf/f−ϱ=Tr,Δqff+Sr,Δqff,which implies that δϱ,Δqf/f=0 for any constant ϱ∈C/−2.
In view of (18) and Lemma 3, we have(75)mr,1fqz+fz=mr,fz+fz/qdzfz≤mr,1dz+mr,fz+fz/qfz+O1≤Sr,f.
Thus, we can conclude from (18), (75), and Lemma 2 that(76)Nr,1fqz+fz=Tr,1fqz+fz+Sr,f=Tr,1dzfz+fz/qfz+Sr,f=Tr,fz/qfz+Ologr+Sr,f=Tr,fqzfz+Sr,f=Tr,Δqff+Sr,f.
On the contrary, we can see that the zero ofΔqf/f+2 is the zero of fqz+fz, and the zero of fqz+fz is also the zero of Δqf/f+2. Indeed, if z0 is a zero of Δqf/f+2, that is, fqz0−fz0/fz0+2=0, then it follows fqz0+fz0=0, and this shows that z0 is a zero of fqz+fz; if z1 is a zero of fqz+fz, that is, fqz1+fz1=0, then it follows Δqfqz1/fz1+2=0, and this shows that z1 is a zero of Δqf/f+2. Thus, by combining with (76), we conclude(77)Nr,1Δqf/f+2=Nr,1fqz+fz=Tr,Δqff+Sr,Δqff.
This shows thatδ−2,Δqf/f=0; thus, by combining with δ∞,Δqf/f=0 and δϱ,Δqf/f=0 for any ϱ∈C/−2, we have δϱ,Δqf/f=0 for any ϱ∈C. So, Δqf/f has no Nevanlinna exceptional value.
Next, we prove thatfz has no Nevanlinna exceptional value.
Firstly, in view of (61) and Lemma 3, we have(78)mr,1f≤mr,1dz+mr,fqzfz+mr,fz/qfz=Sr,f,which implies(79)Nr,1f=Tr,f+Sr,f.
This meansδ0,f=0.
Secondly, in view of (18), we denote(80)P6z,f≔fqz+fzfz+fzq−dzfz≡0.
Sincedz is a nonconstant function, then for any constant ϱ∈C/0, it yields(81)P6z,ϱ=4ϱ2−ϱdz=ϱ4ϱ−dz≢0.
Thus, from Lemma1, we conclude mr,1/f−ϱ=Sr,f, which implies(82)Nr,1f−ϱ=Tr,f+Sr,f.
This means thatδϱ,f=0 for any constant ϱ∈C/0.
Finally, in view of (77), we have(83)mr,1Δqf/f+2=Sr,Δqff=Sr,f.
Thus, from (18) and (83), it yields(84)mr,fz+fzq=mr,dzfzfqz+fz=mr,1fqz/fz+1+Ologr=mr,1Δqf/f+2+Ologr=Sr,f.
Hence, by the above equality and in view of (18) and Lemma 3, it leads to(85)mr,f≤mr,fz+fzq+mr,fqz+fz+Ologr≤2mr,fz+fzq+mr,fqz+fzfz+fz/q+Ologr=Sr,f,which implies(86)Nr,f=Tr,f+Sr,f.
Hence,δ∞,f=0. Together with δ0,f=0 and δϱ,f=0 for any ϱ∈C/0, we obtain that fz has no Nevanlinna exceptional value.(ii)
Sincefqz=Δqfz+fz and fz/q=fz−Δqfz/q, then by substituting these into (18), it follows(87)2fz+Δqfz2fz−Δqfzq=dzfz.Ifz0 is a zero of fz and is not a pole of dz, then in view of (87), we conclude that z0 is a zero of Δqfz or a zero of Δqfz/q. Thus, it follows from Lemma 2 that(88)Nr,1f≤Nr,1Δqfz+Nr,1Δqfz/q+Nr,d=2Nr,1Δqf+Sr,f.Thus, by combining with (79), it yields that Δqfz has infinitely many zeros and λΔqf=σf.In view of (85) and Lemma 3, we can deduce(89)mr,Δqf≤mr,f+mr,Δqff=Sr,f=Sr,Δqf,which implies(90)Nr,Δqf=Tr,Δqf+Sr,Δqf=Tr,1Δqf+Sr,Δqf.Thus, by combining with (88) and (79), we have that Δqfz has infinitely many poles and λ1/Δqf=σf. This proves the conclusions of Theorem 8 (ii).(iii)
In view of (71), it follows(91)P5z,z+1=zq+1qz+1+1−dqzdzzq+1+1=z+qqz+2q−dqzdzz+2qq.
Sincedz is a nonconstant rational function, then let z⟶∞, and we have dqz/dz⟶qκ, where κ=degzd. But, for q≠0, it follows z+qqz+2/z+2q⟶∞ as z⟶∞. Thus, we can deduce P5z,z+1≡0. By Lemma 1, we conclude that(92)mr1Δqf/f−z=mr,1g3z−1−z=Sr,g3=Sr,Δqff.
This leads to(93)Nr1Δqf/f−z=Tr,Δqff+Sr,Δqff,which implies that Δqf/f has infinitely many fixed points and τΔqf/f=σf.
Therefore, this completes the proof of Theorem8.



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*Source: 3781942-2020-09-01.xml*