# The Existence of Meromorphic Solutions of Some Types of Systems of Complex Functional Equations

Authors: Hong-Yan Xu; Jin-Lian Wang; Hua Wang
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## Abstract

We investigate the existence of transcendental
meromorphic solutions of some types of systems of complex functional
equations and obtain some results about the existence of meromorphic
solutions of such systems. Our results are
improvement of the previous theorems given by Gao, Xu,
and Zheng, and our examples show that our results are sharp to some extent.


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## Body

## 1. Introduction and Main Results

Recently, with the establishment of the differences analogues of Nevanlinna’s theory (see [1–3]), people obtained many interesting theorems about the growth and existence of solutions of difference equations, q-difference equations, and so on (see [4–10]). To state some results, we should introduce some basic definition and standard notations. We firstly assume that readers are familiar with the fundamental results and the standard notations of the Nevanlinna value distribution theory of meromorphic functions such as m(r,f),N(r,f),T(r,f),… (see Hayman [11], Yang [12], and Yi and Yang [13]). In addition, we also use ρ(f), λ(f), and λ(1/f) to denote the order, the exponent of convergence of zeros, and the exponent of convergence of poles of f(z), respectively, and S(r,f) to denote any quantity satisfying S(r,f)=o(T(r,f)) for all r outside a possible exceptional set E of finite logarithmic measure limr→∞∫[1,r)∩Edt/t<∞, and a meromorphic function a(z) is called small function with respect to f, if T(r,a)=S(r,f)=o(T(r,f)).In 2001, Heittokangas et al. considered the existence of solutions of some difference equations and obtained the following results [14].Theorem 1 (see [14, Propositions 8 and 9]).
Letc1,…,cn∈C∖{0}. If the following equations(1)∑j=1nfz+cj=Rz,fz,(2)∏j=1nfz+cj=Rz,fz,where (3)Rz,fz≔Pz,fzQz,fz=a0z+a1zfz+⋯+aszfzsb0z+b1zfz+⋯+btzfzt,with small coefficients ai(z), bj(z) with respect to f, admit transcendental meromorphic solutions of finite order, then max⁡s,t≤n.Theorem 2 (see [14]).
Letc1,…,cn∈C∖{0}. If (2) with small coefficients ai(z), bi(z) of growth S(r,f) such that asbt≢0. If (4)max⁡λf,λ1f<ρf,then (2) is of the form(5)∏j=1nfz+cj=czfzk,where c(z) is meromorphic, T(r,c)=S(r,f), and k∈Z.In 2002, Gundersen et al. considered the solutions ofq-difference equation and obtained the following result.Theorem 3 (see [15]).
Suppose thatf is a transcendental meromorphic solution of a q-difference equation of the form(6)fqz=a0z+a1zfz+⋯+aszfzsb0z+b1zfz+⋯+btzfzt,where q∈C,|q|>1,bt(z)≡1, and meromorphic coefficients ai(z),bj(z) are of growth S(r,f). If (7)N¯r,f+N¯r,1f=Sr,f,then (6) is either of the form (8)fqz=aszfzsorfqz=a0zfz-t.In 2012, Zheng and Chen [16] further considered a q-difference equation more general than (2) under a condition similar to Theorem 2 on meromorphic solutions and obtained the following result.Theorem 4 (see [16, Theorem 1]).
Suppose thatf is a transcendental meromorphic solution of a q-difference equation of the form(9)∏j=1nfqjz=Rz,fz=a0z+a1zfz+⋯+aszfzsb0z+b1zfz+⋯+btzfzt,where qj∈C∖{0,1},j=1,…,n, and R(z,f) is an irreducible rational function in f with meromorphic coefficients ai(z)(i=0,…,s) and bj(z)(j=0,…,t) of growth S(r,f) such that bt(z)≡1,as(z)≢0. If (10)max⁡λf,λ1f<ρf,then (9) is reduced to the form (11)∏j=1nfqjz=aszfzsor∏j=1nfqjz=a0zfz-t.In 2012, Gao [17] studied the similar problem when (9) is replaced by the following system of functional equations,(12)f1pz=R1z,f2z=∑i=0s2aizf2zi∑j=0t2bjzf2zj,f2pz=R2z,f1z=∑i=0s1cizf1zi∑j=0t1djzf1zj,where p(z) is an entire function, R1(z,f2(z)),R2(z,f1(z)) are irreducible rational functions, and the coefficients are small functions, and obtained the following theorems.Theorem 5 (see [17, Theorem 1]).
Let(f1,f2) be a pair of nonconstant meromorphic solutions of system (12). Then, p(z) is a polynomial.Theorem 6 (see [17, Theorem 2]).
Let the polynomialp(z)=qzk+⋯ be of degree k≥2,(f1,f2) a pair of transcendental meromorphic solutions of system (12), and as2bt2≢0,cs1dt1≢0,ai,bj,ci,dj small functions. If (13)N¯r,ftpz+N¯r,1ftpz=Sr,f,t=1,2,then system (12) is of the form (14)f1pz=azf2zs,f2pz=bzf1zt,where a(z),b(z) are meromorphic, T(r,a(z))=S(r,fi),T(r,b(z))=S(r,fi), and s,t∈Z,s≠0,t≠0.Remark 7.
We can see that the numerical values ofs,t are uncertain in the conclusion of Theorem 6.After his works, there were some papers focusing on the growth and existence of meromorphic solutions of some types of systems of complex functional equations (see [18–21]).In this paper, we will further investigate some properties of solutions of the system of complex functional equations. The main purpose of this paper is to find out the relationship between “s,t” and “si,tj” and further to confirm the numerical values of s, t in the conclusion of Theorem 6. Some results are obtained in this paper, and the first main result is about meromorphic solutions with few zeros and poles of a type of system of complex functional equations.Theorem 8.
Letqjt∈C∖0,1,jt=1,…,nt,t=1,2, and let (f1,f2) be a pair of nonrational meromorphic solutions of the system(15)∏j1=1n1f1qj1z=a0z+a1zf2z+⋯+as2zf2zs2b0z+b1zf2z+⋯+bt2zf2zt2,∏j2=1n2f2qj2z=e0z+e1zf1z+⋯+es1zf1zs1d0z+d1zf1z+⋯+dt1zf1zt1,with the meromorphic coefficients ai(z),bi(z),ei(z),di(z) of growth S(r,ft),t=1,2, and as2(z)es1(z)≢0,dt1(z)≡bt2(z)≡1. If(16)max⁡λft,λ1ft<ρft,t=1,2,then system (15) is one of the four following forms:(17a)∏j1=1n1f1qj1z=as2zf2zs2,∏j2=1n2f2qj2z=es1zf1zs1,(17b)∏j1=1n1f1qj1z=as2zf2zs2,∏j2=1n2f2qj2z=e0zf1z-t1,(18a)∏j1=1n1f1qj1z=a0zf2z-t2,∏j2=1n2f2qj2z=es1zf1zs1,(18b)∏j1=1n1f1qj1z=a0zf2z-t2,∏j2=1n2f2qj2z=e0zf1z-t1.Corollary 9.
Let(f1,f2) be a pair of transcendental solutions of (15) with finite order ρ(f1)=ρ1,ρ(f2)=ρ2,   ρ(f1),ρ(f2) are positive integers and all the other assumptions of Theorem 8 hold.(i)
If0<ρ1<ρ2<∞  and  ∑j2=1n2qj2ρ2≠0, then system (15) is reduced to (18b) only.(ii)
If0<ρ2<ρ1<∞  and  ∑j1=1n1qj1ρ1≠0, then system (15) is reduced to (18b) only.Corollary 10.
Let(f1,f2) be a pair of transcendental solutions of (15) with finite order ρ(f1)=ρ1,ρ(f2)=ρ2, ρ(f1),ρ(f2) are positive integers and all the other assumptions of Theorem 8 hold. If system (15) is reduced to (18b) only, and (19)∑j2=1n2qj2ρ2+t2∑j1=1n1qj1ρ1+t1≠0,then ρ1=ρ2.Furthermore, we obtain the following result when system (15) is replaced by the other system.Theorem 11.
Letqjt,cjt∈C∖0,1,jt=1,…,nt,t=1,2, and let (f1,f2) be a pair of nonrational meromorphic solutions of the system(20)∏j1=1n1f1qj1z+cj1=a0z+a1zf2∘p+⋯+as2zf2∘ps2b0z+b1zf2∘p+⋯+bt2zf2∘pt2,∏j2=1n2f2qj2z+cj2=e0z+e1zf1∘p+⋯+es1zf1∘ps1d0z+d1zf1∘p+⋯+dt1zf1∘pt1,with the meromorphic coefficients ai(z),bi(z),ei(z),di(z) of growth S(r,ft),t=1,2, and as2(z)es1(z)≢0,dt1(z)≡bt2(z)≡1. If f1, f2 satisfy (16) and f∘p=f(p(z)), where p(z)=qzk+⋯ is a polynomial in z of degree k≥2, then f1,f2 satisfy the following forms: (21)∏j1=1n1f1qj1z+cj1=as2zf2∘ps2or∏j1=1n1f1qj1z+cj1=a0zf2∘p-t2,∏j2=1n2f2qj2z+cj2=es1zf1∘ps1or∏j2=1n2f2qj2z+cj2=e0zf1∘p-t1.The researches on the properties of solutions of complex differential equations in the whole complex plane, disc, and angular domain are always interesting in the past several decades (see [22–26]). In 2014, Liu and Dong [27] considered the existence of solutions of some differential-difference equation, where the differential-difference equation is an equation including derivatives, shifts, or differences of f(z), and obtained the following result.Theorem 12 (see [27, Theorem 2.3]).
Letk be a positive integer. If the equation(22)∑k=1nfkz=Rz,fz+c=a0z+a1zfz+c+⋯+apzfz+cpb0z+b1zfz+c+⋯+bqzfz+cqadmits a transcendental meromorphic solution of finite order, then max⁡{p,q}≤n+1. If (22) admits a transcendental entire solution of finite order, then max⁡{p,q}≤1.Then, we will further consider the system of differential-difference equations with the analogue form of (22) and obtain the following theorem.Theorem 13.
Letk1,k2,n1,n2 be integers. If system(23)∑k1=1n1f1k1z=a0z+a1zf2z+c+⋯+as2zf2z+cs2b0z+b1zf2z+c+⋯+bt2zf2z+ct2,∑k2=1n2f2k2z=e0z+e1zf1z+c+⋯+es1zf1z+cs1d0z+d1zf1z+c+⋯+dt1zf1z+ct1,with the meromorphic coefficients ai(z),bi(z),ei(z),di(z) of growth S(r,ft),t=1,2, and as2(z)es1(z)≢0, admits a pair of transcendental meromorphic solutions of finite order, set σ1=max⁡{s1,t1}, σ2=max⁡{s2,t2}; then (24)σ1≤n1+1,σ2≤n2+1.If (23) admits a pair of transcendental entire solutions of finite order, then (25)σ1≤1,σ2≤1.We give the following example to show that Theorem13 is not valid for a pair of meromorphic solutions with infinite order of (23).Example 14.
Let(f1,f2)=(eez,e-ez); then (f1,f2) are a pair of solutions of the system (26)f1′z=ezf2z+cp,f2′z=-ezf1z+cp,where c is a complex constant and p is a positive integer satisfying ec=-1/p,p≥3.Theorem 15.
Under the assumptions of Theorem13, let (f1,f2) be a pair of transcendental meromorphic solutions of system(27)∑k1=1n1f1k1z=a0z+a1zf2∘p+⋯+as2zf2∘ps2b0z+b1zf2∘p+⋯+bt2zf2∘pt2,∑k2=1n2f2k2z=e0z+e1zf1∘p+⋯+es1zf1∘ps1d0z+d1zf1∘p+⋯+dt1zf1∘pt1,where p(z)=qzk+⋯(q(≠0)∈C) of degree k≥2. Then (28)Tr,f1=Olog⁡rς+ɛ,Tr,f2=Olog⁡rς+ɛ,k2σ1σ2≤n1+1n2+1,where ɛ>0 and (29)ς=log⁡n1+1n2+1-log⁡σ1σ22log⁡k.


## 2. Some Lemmas

Lemma 16 (Valiron-Mohon’ko (see [24])).
Letf(z) be a meromorphic function. Then for all irreducible rational functions in f, (30)Rz,fz=∑i=0maizfzi∑j=0nbjzfzj,with meromorphic coefficients ai(z),bj(z), the characteristic function of R(z,f(z)) satisfies that (31)Tr,Rz,fz=dTr,f+OΨr,where d=max⁡{m,n} and Ψ(r)=maxi,j{T(r,ai),T(r,bj)}.Lemma 17 (see [14, p. 37]).
If a meromorphic functionf satisfies (32)max⁡λf,λ1f<ρf,then f is of regular growth.Lemma 18 (see [15, p. 127]).
The differential field(33)Lf=gismeromorphic∣Tr,g=Sr,fis algebraically closed in the field of meromorphic functions in the complex plane. That is, any meromorphic function satisfying an algebraic equation over the field Lf actually belongs to Lf.Remark 19 (see [28, p. 249] and [16, p. 728 Remark]).
The following observation(34)Nr,fqz=Nqr,f+O1holds for any meromorphic function f and any nonzero complex constant q. Clearly, we can immediately obtain (35)λ1fqz=λ1f.Similarly, we also have (36)Nr,1fqz=Nqr,1f+O1,λfqz=λf.Lemma 20 (see [29]).
Letf(z) be a transcendental meromorphic function and p(z)=qzk+⋯ a complex polynomial of degree k>0. For given 0<δ<q, let λ=q+δ,μ=q-δ; then for given ɛ>0 and for r large enough, (37)1-ɛTμrk,f≤Tr,f∘p≤1+ɛTλrk,f.Lemma 21.
Suppose thatf1, f2 are meromorphic functions of finite order ρ(f1),ρ(f2), ρ(f1),ρ(f2) are positive integers satisfying (16), and let qjt∈C∖{0},jt=1,…,nt,t=1,2.
(i) Ifρ1=ρ(f1)<ρ(f2)=ρ2 and (38)∑j2=1n2qj2ρ2≠0,then, H(z)=∏j1=1n1f1(qj1z)∏j2=1n2f2(qj2z) satisfies ρ(H)=ρ2.
(ii) Ifρ1=ρ(f1)>ρ(f2)=ρ2 and (39)∑j1=1n1qj1ρ1≠0,then, H(z)=∏j1=1n1f1(qj1z)∏j2=1n2f2(qj2z) satisfies ρ(H)=ρ1.Remark 22.
The following example shows that the conclusions are not valid if the conditionρ1≠ρ2 is removed in Lemma 21.Example 23.
Let(f1,f2)=(ez,e-z); then ρ1=ρ2=1 and 0=max⁡{λ(ft),λ(1/ft)}<1. If qj,j=1,2,…,n, are complex constants and n1=n2=n satisfy (40)∑j1=1nqj1=∑j1=1nqj1≠0,then we have H(z)=1; that is, ρ(H)=0≠1.Proof.
From the assumptions in Lemma21 and by Hadamard Theorem, we have (41)ftz=zmtQgtzQhtzePtz,t=1,2,where mt∈Z, P1(z),P1(z) are polynomials of degree ρ1,ρ2, respectively, and Qgt(z),Qht(z) are the canonical products formed with nonzero zeros and poles of ft,  t=1,2, respectively. So, we have (42)Hz=∏j1=1n1qj1zm1Qg1qj1zQh1qj1zeP1qj1z∏j2=1n2qj2zm2Qg2qj2zQh2qj2zeP2qj2z.
Denoting(43)Gjtz=qjtzmtQgtqjtzQhtqjtz,t=1,2,then we can rewrite H(z) as the following form:(44)Hz=exp⁡∑t=12∑jt=1ntqjtρtaρtzρt+Pρt-1z∏t=12∏jt=1ntGjtz,where aρt,t=1,2, are the leading coefficient of Pt(z) and Pρt-1(z) are polynomials of degree ≤ρt-1.
(i)
ρ
1
<
ρ
2. By Remark 19, it follows from (16) that(45)λH≤max1≤jt≤nt⁡λftqjtz,t=1,2=max⁡λf1,λf2<ρ2,(46)λ1H≤max1≤jt≤nt⁡λ1ftqjtz,t=1,2=max⁡λ1f1,λ1f2<ρ2.It follows from (44)–(46) that ρ(Gjt)<ρ2 for all jt=1,…,nt,t=1,2. Since ∑j2=1n2qj2n2≠0, we have ρ(H)=ρ2.
(ii)
ρ
1
>
ρ
2. By using the same argument as in (i), we can get ρ(H)=ρ1 easily.
Thus, this completes the proof of Lemma21.Lemma 24 (see [2]).
Letf(z) be a transcendental meromorphic function of finite order and c a nonzero complex constant. Then, we have (47)Tr,fz+c=Tr,fz+Sr,f.Lemma 25 (see [13, p. 37] or [12]).
Letf(z) be a nonconstant meromorphic function in the complex plane and l a positive integer. Then(48)Nr,fl=Nr,f+lN¯r,f,Tr,fl≤Tr,f+lN¯r,f+Sr,f.Lemma 26 (see [24, 30]).
Letg:(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 r0 such that g(r)≤h(αr) for all r≥r0.Lemma 27 (see [31]).
Letψ(r) be a function of r(r≥r0), positive, and bounded in every finite interval.
(i) Suppose thatψ(μrm)≤Aψ(r)+B(r≥r0), where μ(μ>0), m(m>1), A(A≥1), and B are constants. Then ψ(r)=O((log⁡r)α) with α=log⁡A/log⁡m, unless A=1 and B>0; and if A=1 and B>0, then, for any ɛ>0, ψ(r)=O((log⁡r)ɛ).
(ii) Suppose that (with the notation of (i))ψ(μrm)≥Aψ(r)(r≥r0). Then for all sufficiently large values of r, ψ(r)≥K(log⁡r)α with α=log⁡A/log⁡m, for some positive constant K.


## 3. Proofs of Theorem8, Corollaries 9 and 10

### 3.1. The Proof of Theorem8

LetGt(z)=∏jt=1ntft(qjtz),t=1,2. By applying Valiron-Mohon’ko theorem [24] to (15), we have(49)Tr,G1=max⁡s2,t2Tr,f2+Sr,f1+Sr,f2,Tr,G2=max⁡s1,t1Tr,f1+Sr,f1+Sr,f2.From (16), we can take constants ξt,δt such that (50)max⁡λft,λ1ft<ξt<δt<ρft,t=1,2,and then we have (51)Tr,ft′ft=N¯r,ft+N¯r,1ft+Sr,ft=Orξt+Sr,ft,t=1,2.From the definitions of Gt(t=1,2), similar to the above argument, we have (52)Tr,Gt′Gt=Nr,Gt′Gt+mr,Gt′Gt≤∑jt=1ntN¯r,ftqjtz+N¯r,1ftqjtz+Sr,f1+Sr,f2=∑jt=1ntN¯qjtr,ftz+N¯qjtr,1ftz+Sr,f1+Sr,f2=Orξt+Sr,f1+Sr,f2.From (16), we know that zeros and poles are Borel exceptions of ft(t=1,2), and from [32, Satz 13.4], we have that ft(t=1,2) is of regular growth. Hence, there exists r0>0 that T(r,ft)>rδt for r>r0. So, we can get that(53)Tr,Gt′Gt=Sr,f1+Sr,f2,Tr,ft′ft=Sr,f1+Sr,f2,t=1,2.Now, we rewrite system (15) as(54)bt2zas2zG1z=P2z,f2Q2z,f2=u2z,f2,dt1zes1zG2z=P1z,f1Q1z,f1=u1z,f1,without loss of generality, assume that Pt,Qt are monic polynomials in ft with coefficients of growth S(r,f1),S(r,f2), and set Ft≔ft′/ft,Ut≔ut′/ut,t=1,2; from (54), we have T(r,Ut)=S(r,f1)+S(r,f2). And because (55)P2′Q2-P2Q2′Q22=u2′=U2u2=U2P2Q2,P1′Q1-P1Q1′Q12=u1′=U1u1=U1P1Q1,it follows that (56)P2′Q2-P2Q2′=U2P2Q2,P1′Q1-P1Q1′=U1P1Q1.Substituting ft′=Ftft,t=1,2, to the above equalities and comparing the leading coefficients, we can get (57)s1-t1F1=U1,s2-t2F2=U2.Solving the above system, we get(58)u1=Γ2f1zs1-t1,u2=Γ1f2zs2-t2,where Γt∈C,t=1,2. From (54)–(58) and bt2(z)≡1,et1(z)≡1, it follows that (59)G1z=Γ2as2zf2zs2-t2,G2z=Γ1es1zf1zs1-t1.Substituting the above system into (15), we have (60)Γ2as2zf2zs2-t2=a0z+a1zf2z+⋯+as2zf2zs2b0z+b1zf2z+⋯+f2zt2,Γ1es1zf1zs1-t1=e0z+e1zf1z+⋯+es1zf1zs1d0z+d1zf1z+⋯+f1zt1,that is,(61)Γ2as2zf2zs2+⋯+Γ2as2zb1f2zs2-t2+1+Γ2as2zb0f2zs2-t2=a0z+a1zf2z+⋯+as2zf2zs2,where s2≥t2, or(62)Γ2as2zf2zt2+⋯+Γ2as2zb1f2z+Γ2as2zb0=a0zf2zt2-s2+a1zf2zt2-s2+1+⋯+as2zf2zt2,where t2>s2. By regarding (61) or (62) as an algebraic equation in f2 with coefficients of growth S(r,f2), then it follows by Lemma 18 that Γ2=1. Moreover, if s2≠0 and t2≠0, we can get a contradiction with the condition that R2(z,f2(z)) is irreducible in f2. Thus, we have s2=0 or t2=0. Then it follows from (61) that G1(z)=as2(z)f2(z)s2 or G1(z)=a0(z)f2(z)-t2.Similarly, we can getG2(z)=es1(z)f1(z)s1 or G2(z)=e0(z)f1(z)-t1.Thus, we complete the proof of Theorem8.


### 3.2. The Proof of Corollary9

Let(f1,f2) be a pair of transcendental solutions of (15) of finite order ρ(f1)=ρ1,ρ(f2)=ρ2.(i) Ifρ1<ρ2, we can exclude (17a) and (17b) easily by comparing the growth order of two sides of the equation (63)∏j1=1n1f1qj1z=as2zf2zs2.Next, we will exclude (18a). Suppose that(64)∏j2=1n2f2qj2z=e0zf1zs1.From the assumptions of Lemma 21 and by Hadamard Theorem, we can write(65)ftz=zmtQgtzQhtzePtz,t=1,2,where mt∈Z, P1(z),P2(z) are polynomials of degree ρ1,ρ2, respectively, and Qgt(z),Qht(z) are the canonical products formed with nonzero zeros and poles of ft, respectively. Substituting (65) into (64), we have (66)exp⁡∑j2=1n2qj2ρ2aρ2zρ2+Pρ2-1z∏j2=1n2qj2zm2Qg2qj2zQh2qj2z=zs1m1Qg1zs1Qh1zs1es1P1z.Since ∑j2=1n2qj2ρ2≠0 and ρ1<ρ2, we can get a contradiction easily.(ii) Ifρ1>ρ2, we can exclude (17a) and (18a). By using the same argument as in (i) and combining the condition ∑j1=1n1qj1ρ1≠0 and max⁡ρas2,ρes1<ρ1, we can exclude (17b).Therefore, we complete the proof of Corollary9.


### 3.3. The Proof of Corollary10

From the assumptions of Corollary10 and (18b), we have(67)f1zt1f2zt2∏j2=1n2f2qj2z∏j1=1n1f1qj1z=a0ze0z.We can rewrite (67) as the following form:(68)∏j2=1n2+t2f2qj2z∏j1=1n1+t1f1qj1z=a0ze0z,where qn1+1=⋯=qn1+t1=qn2+1=⋯=qn2+t2=1.Suppose thatρ1≠ρ2. Since (69)∑j2=1n2qj2ρ2+t2∑j1=1n1qj1ρ1+t1≠0,by using the similar method as in Lemma 21, from (68), we can get a contradiction with the assumption maxρa0,ρe0<min⁡ρ1,ρ2. Hence ρ1=ρ2.Thus, we complete the proof of Corollary10.


## 3.1. The Proof of Theorem8

LetGt(z)=∏jt=1ntft(qjtz),t=1,2. By applying Valiron-Mohon’ko theorem [24] to (15), we have(49)Tr,G1=max⁡s2,t2Tr,f2+Sr,f1+Sr,f2,Tr,G2=max⁡s1,t1Tr,f1+Sr,f1+Sr,f2.From (16), we can take constants ξt,δt such that (50)max⁡λft,λ1ft<ξt<δt<ρft,t=1,2,and then we have (51)Tr,ft′ft=N¯r,ft+N¯r,1ft+Sr,ft=Orξt+Sr,ft,t=1,2.From the definitions of Gt(t=1,2), similar to the above argument, we have (52)Tr,Gt′Gt=Nr,Gt′Gt+mr,Gt′Gt≤∑jt=1ntN¯r,ftqjtz+N¯r,1ftqjtz+Sr,f1+Sr,f2=∑jt=1ntN¯qjtr,ftz+N¯qjtr,1ftz+Sr,f1+Sr,f2=Orξt+Sr,f1+Sr,f2.From (16), we know that zeros and poles are Borel exceptions of ft(t=1,2), and from [32, Satz 13.4], we have that ft(t=1,2) is of regular growth. Hence, there exists r0>0 that T(r,ft)>rδt for r>r0. So, we can get that(53)Tr,Gt′Gt=Sr,f1+Sr,f2,Tr,ft′ft=Sr,f1+Sr,f2,t=1,2.Now, we rewrite system (15) as(54)bt2zas2zG1z=P2z,f2Q2z,f2=u2z,f2,dt1zes1zG2z=P1z,f1Q1z,f1=u1z,f1,without loss of generality, assume that Pt,Qt are monic polynomials in ft with coefficients of growth S(r,f1),S(r,f2), and set Ft≔ft′/ft,Ut≔ut′/ut,t=1,2; from (54), we have T(r,Ut)=S(r,f1)+S(r,f2). And because (55)P2′Q2-P2Q2′Q22=u2′=U2u2=U2P2Q2,P1′Q1-P1Q1′Q12=u1′=U1u1=U1P1Q1,it follows that (56)P2′Q2-P2Q2′=U2P2Q2,P1′Q1-P1Q1′=U1P1Q1.Substituting ft′=Ftft,t=1,2, to the above equalities and comparing the leading coefficients, we can get (57)s1-t1F1=U1,s2-t2F2=U2.Solving the above system, we get(58)u1=Γ2f1zs1-t1,u2=Γ1f2zs2-t2,where Γt∈C,t=1,2. From (54)–(58) and bt2(z)≡1,et1(z)≡1, it follows that (59)G1z=Γ2as2zf2zs2-t2,G2z=Γ1es1zf1zs1-t1.Substituting the above system into (15), we have (60)Γ2as2zf2zs2-t2=a0z+a1zf2z+⋯+as2zf2zs2b0z+b1zf2z+⋯+f2zt2,Γ1es1zf1zs1-t1=e0z+e1zf1z+⋯+es1zf1zs1d0z+d1zf1z+⋯+f1zt1,that is,(61)Γ2as2zf2zs2+⋯+Γ2as2zb1f2zs2-t2+1+Γ2as2zb0f2zs2-t2=a0z+a1zf2z+⋯+as2zf2zs2,where s2≥t2, or(62)Γ2as2zf2zt2+⋯+Γ2as2zb1f2z+Γ2as2zb0=a0zf2zt2-s2+a1zf2zt2-s2+1+⋯+as2zf2zt2,where t2>s2. By regarding (61) or (62) as an algebraic equation in f2 with coefficients of growth S(r,f2), then it follows by Lemma 18 that Γ2=1. Moreover, if s2≠0 and t2≠0, we can get a contradiction with the condition that R2(z,f2(z)) is irreducible in f2. Thus, we have s2=0 or t2=0. Then it follows from (61) that G1(z)=as2(z)f2(z)s2 or G1(z)=a0(z)f2(z)-t2.Similarly, we can getG2(z)=es1(z)f1(z)s1 or G2(z)=e0(z)f1(z)-t1.Thus, we complete the proof of Theorem8.


## 3.2. The Proof of Corollary9

Let(f1,f2) be a pair of transcendental solutions of (15) of finite order ρ(f1)=ρ1,ρ(f2)=ρ2.(i) Ifρ1<ρ2, we can exclude (17a) and (17b) easily by comparing the growth order of two sides of the equation (63)∏j1=1n1f1qj1z=as2zf2zs2.Next, we will exclude (18a). Suppose that(64)∏j2=1n2f2qj2z=e0zf1zs1.From the assumptions of Lemma 21 and by Hadamard Theorem, we can write(65)ftz=zmtQgtzQhtzePtz,t=1,2,where mt∈Z, P1(z),P2(z) are polynomials of degree ρ1,ρ2, respectively, and Qgt(z),Qht(z) are the canonical products formed with nonzero zeros and poles of ft, respectively. Substituting (65) into (64), we have (66)exp⁡∑j2=1n2qj2ρ2aρ2zρ2+Pρ2-1z∏j2=1n2qj2zm2Qg2qj2zQh2qj2z=zs1m1Qg1zs1Qh1zs1es1P1z.Since ∑j2=1n2qj2ρ2≠0 and ρ1<ρ2, we can get a contradiction easily.(ii) Ifρ1>ρ2, we can exclude (17a) and (18a). By using the same argument as in (i) and combining the condition ∑j1=1n1qj1ρ1≠0 and max⁡ρas2,ρes1<ρ1, we can exclude (17b).Therefore, we complete the proof of Corollary9.


## 3.3. The Proof of Corollary10

From the assumptions of Corollary10 and (18b), we have(67)f1zt1f2zt2∏j2=1n2f2qj2z∏j1=1n1f1qj1z=a0ze0z.We can rewrite (67) as the following form:(68)∏j2=1n2+t2f2qj2z∏j1=1n1+t1f1qj1z=a0ze0z,where qn1+1=⋯=qn1+t1=qn2+1=⋯=qn2+t2=1.Suppose thatρ1≠ρ2. Since (69)∑j2=1n2qj2ρ2+t2∑j1=1n1qj1ρ1+t1≠0,by using the similar method as in Lemma 21, from (68), we can get a contradiction with the assumption maxρa0,ρe0<min⁡ρ1,ρ2. Hence ρ1=ρ2.Thus, we complete the proof of Corollary10.


## 4. The Proof of Theorem11

LetGt(z)=∏jt=1ntftqjtz+cjt and gt(z)=ft∘p,t=1,2. By applying Valiron-Mohon’ko theorem [24] to (20), we have(70)Tr,G1=max⁡s2,t2Tr,g2+Sr,g1+Sr,g2,Tr,G2=max⁡s1,t1Tr,g1+Sr,g1+Sr,g2.From (70) and the definitions of Gt(t=1,2), similar to the above argument, we have(71)Tr,Gt′Gt=Nr,Gt′Gt+mr,Gt′Gt≤∑jt=1ntN¯r,ftqjtz+cjt+N¯r,1ftqjtz+cjt+Sr,g1+Sr,g2≤∑jt=1ntN¯qjtr+cjt,ft+N¯qjtr+cjt,1ft+Sr,g1+Sr,g2.Since T(r,f) is an increasing function for r and p(z)=qzk+⋯, hence, there exists r0>0 such that for all r>r0>0(72)Tq-δtrk,ft≥Tqt∗r+ct∗,ft,t=1,2,where qt∗=max⁡qjt,jt=1,2,…,nt,ct∗=max⁡cjt,jt=1,2,…,nt, and 0<δt<q,t=1,2. Thus, it follows from the above inequality and Lemma 20 that (73)N¯qt∗r+ct∗,ftTqt∗r+ct∗,ft=N¯qt∗r+ct∗,ftTr,gtTr,gtTqt∗r+ct∗,ft≥N¯qt∗r+ct∗,ftTr,gtTq-δtrk,ftTqt∗r+ct∗,ft,t=1,2.From (16), we have N¯(qt∗r+ct∗,ft)/T(qt∗r+ct∗,ft)→0 as r→∞ for t=1,2. Thus, we can get N¯(qt∗r+ct∗,ft)/T(r,gt)→0 as r→∞ for t=1,2; that is,(74)N¯qt∗r+ct∗,ft=Sr,gt,t=1,2.Similarly, we have(75)N¯qt∗r+ct∗,1ft=Sr,gt,t=1,2.From (71)–(75), we have(76)Tr,Gt′Gt=Sr,gt,t=1,2.Now, we rewrite system (20) as (77)bt2zas2zG1z=P2z,g2Q2z,g2=u2z,g2,dt1zes1zG2z=P1z,g1Q1z,g1=u1z,g1,and, without loss of generality, assume that Pt,Qt are monic polynomials in gt with coefficients of growth S(r,g1),S(r,g2). Thus, by using the same argument as in Theorem 8, we can get the conclusion of Theorem 11 easily.Therefore, this completes the proof of Theorem11.


## 5. Proofs of Theorems13 and 15

### 5.1. The Proof of Theorem13

Let(f1,f2) be a pair of transcendental meromorphic solutions of (23); by using Lemmas 16 and 25, it follows from (23) that (78)σ2Tr,f2z+c=Tr,∑k1=1n1f1k1z+Sr,f1+Sr,f2≤mr,f1+n1+1Nr,f1+Sr,f1+Sr,f2≤n1+1Tr,f1+Sr,f1+Sr,f2,σ1Tr,f1z+c=Tr,∑k2=1n2f2k2z+Sr,f1+Sr,f2≤mr,f2+n2+1Nr,f2+Sr,f1+Sr,f2≤n2+1Tr,f2+Sr,f1+Sr,f2.By Lemma 24, it follows that (79)σ2Tr,f2≤n1+1Tr,f1+Sr,f1+Sr,f2,σ1Tr,f1≤n2+1Tr,f2+Sr,f1+Sr,f2.From the above inequalities and the fact that f1,f2 are transcendental functions, then we have σ1≤n1+1 and σ2≤n2+1. If f1,f2 are entire functions, we can get σ1≤1 and σ2≤1, easily.Hence, this completes the proof of Theorem13.


### 5.2. The Proof of Theorem15

By using the same argument as in Theorem13, we can get (80)σ2Tr,f2∘p≤n1+1Tr,f1+Sr,f1+Sr,f2,σ1Tr,f1∘p≤n2+1Tr,f2+Sr,f1+Sr,f2.Since p(z)=qzk+⋯, by Lemma 20, we can get that, for ϑt=q-δt>0,t=1,2 and sufficiently large r, (81)σ11-ɛTϑ1rk,f1≤n1+11+o1Tr,f2,r∉E1,σ21-ɛTϑ2rk,f2≤n2+11+o1Tr,f1,r∉E2,where E1,E2 are two sets of finite linear measure. By Lemma 26, we have that, for any given γ1>1,γ2>1 and for sufficiently large r, (82)σ11-ɛTϑ1rk,f1≤n1+11+o1Tγ1r,f2,σ21-ɛTϑ2rk,f2≤n2+11+o1Tγ2r,f1;that is,(83)Tϑ1γ1kR1k,f1≤n1+11+o1σ11-ɛTR1,f2,(84)Tϑ2γ2kR2k,f2≤n2+11+o1σ21-ɛTR2,f1,where R1=γ1r and R2=γ2r. From (83) and (84), we have(85)Tϑ1ϑ2kγ1kγ22kr2k,f1≤n1+11+o1σ11-ɛn2+11+o1σ21-ɛTr,f1,Tϑ2ϑ1kγ2kγ12kr2k,f2≤n2+11+o1σ21-ɛn1+11+o1σ11-ɛTr,f2.And since k≥2, we get that σ1σ2≤(n1+1)(n2+1). Then it follows from (85) and by Lemma 27 that (86)Tr,f1=Olog⁡rς1,Tr,f2=Olog⁡rς1,where (87)ς1=log⁡n1+1n2+1-log⁡σ1σ2+2log1+o1-2log⁡1-ɛ2log⁡k=log⁡n1+1n2+1-log⁡σ1σ22log⁡k+ε1.Set ς=log⁡(n1+1)(n2+1)-log⁡σ1σ2/2log⁡k; then we have (88)Tr,f1=Olog⁡rς+ε1,Tr,f2=Olog⁡rς+ε1.Next, we will prove thatk2σ1σ2≤(n1+1)(n2+1). Suppose that k2σ1σ2>(n1+1)(n2+1); then we can get that ς=log⁡(n1+1)(n2+1)-log⁡σ1σ2/2log⁡k<1. For sufficiently small ε1>0, we have ς1=ς+ε1<1. This contradicts the condition on the transcendence of f1,f2.Thus, the proof of Theorem15 is completed.


## 5.1. The Proof of Theorem13

Let(f1,f2) be a pair of transcendental meromorphic solutions of (23); by using Lemmas 16 and 25, it follows from (23) that (78)σ2Tr,f2z+c=Tr,∑k1=1n1f1k1z+Sr,f1+Sr,f2≤mr,f1+n1+1Nr,f1+Sr,f1+Sr,f2≤n1+1Tr,f1+Sr,f1+Sr,f2,σ1Tr,f1z+c=Tr,∑k2=1n2f2k2z+Sr,f1+Sr,f2≤mr,f2+n2+1Nr,f2+Sr,f1+Sr,f2≤n2+1Tr,f2+Sr,f1+Sr,f2.By Lemma 24, it follows that (79)σ2Tr,f2≤n1+1Tr,f1+Sr,f1+Sr,f2,σ1Tr,f1≤n2+1Tr,f2+Sr,f1+Sr,f2.From the above inequalities and the fact that f1,f2 are transcendental functions, then we have σ1≤n1+1 and σ2≤n2+1. If f1,f2 are entire functions, we can get σ1≤1 and σ2≤1, easily.Hence, this completes the proof of Theorem13.


## 5.2. The Proof of Theorem15

By using the same argument as in Theorem13, we can get (80)σ2Tr,f2∘p≤n1+1Tr,f1+Sr,f1+Sr,f2,σ1Tr,f1∘p≤n2+1Tr,f2+Sr,f1+Sr,f2.Since p(z)=qzk+⋯, by Lemma 20, we can get that, for ϑt=q-δt>0,t=1,2 and sufficiently large r, (81)σ11-ɛTϑ1rk,f1≤n1+11+o1Tr,f2,r∉E1,σ21-ɛTϑ2rk,f2≤n2+11+o1Tr,f1,r∉E2,where E1,E2 are two sets of finite linear measure. By Lemma 26, we have that, for any given γ1>1,γ2>1 and for sufficiently large r, (82)σ11-ɛTϑ1rk,f1≤n1+11+o1Tγ1r,f2,σ21-ɛTϑ2rk,f2≤n2+11+o1Tγ2r,f1;that is,(83)Tϑ1γ1kR1k,f1≤n1+11+o1σ11-ɛTR1,f2,(84)Tϑ2γ2kR2k,f2≤n2+11+o1σ21-ɛTR2,f1,where R1=γ1r and R2=γ2r. From (83) and (84), we have(85)Tϑ1ϑ2kγ1kγ22kr2k,f1≤n1+11+o1σ11-ɛn2+11+o1σ21-ɛTr,f1,Tϑ2ϑ1kγ2kγ12kr2k,f2≤n2+11+o1σ21-ɛn1+11+o1σ11-ɛTr,f2.And since k≥2, we get that σ1σ2≤(n1+1)(n2+1). Then it follows from (85) and by Lemma 27 that (86)Tr,f1=Olog⁡rς1,Tr,f2=Olog⁡rς1,where (87)ς1=log⁡n1+1n2+1-log⁡σ1σ2+2log1+o1-2log⁡1-ɛ2log⁡k=log⁡n1+1n2+1-log⁡σ1σ22log⁡k+ε1.Set ς=log⁡(n1+1)(n2+1)-log⁡σ1σ2/2log⁡k; then we have (88)Tr,f1=Olog⁡rς+ε1,Tr,f2=Olog⁡rς+ε1.Next, we will prove thatk2σ1σ2≤(n1+1)(n2+1). Suppose that k2σ1σ2>(n1+1)(n2+1); then we can get that ς=log⁡(n1+1)(n2+1)-log⁡σ1σ2/2log⁡k<1. For sufficiently small ε1>0, we have ς1=ς+ε1<1. This contradicts the condition on the transcendence of f1,f2.Thus, the proof of Theorem15 is completed.



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*Source: 723025-2015-10-15.xml*