# The Study of Solutions of Several Systems of Nonlinear Partial Differential Difference Equations

Authors: Hong Yan Xu; Meiying Yu; Keyu Zhang
Journal: Journal of Function Spaces
 (2022)

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## Abstract

Our main aim is to describe the entire solutions of several systems ofα1f1z2+α2f2z+c2=1,β1f2z2+β2f1z+c2=1,α1∂f1/∂z1n1+α2f2z+cm1=1,β1∂f2/∂z1n2+β2f1z+cm2=1, and α1∂f1/∂z12+α2f2z+c2=1,β1∂f2/∂z12+β2f1z+c2=1,α1∂f1/∂z12+α2f2z+c+α3f1z2=1,β1∂f2/∂z12+β2f1z+c+β3f2z2=1, where αj,βjj=1,2,3 are nonzero constants in ℂ and mj,njj=1,2 are positive integers. We obtain several theorems on the existence and the forms of solutions for these systems, which are some improvements and supplements of the previous theorems given by Xu and Cao, Gao, and Liu and Yang. Moreover, we give some examples to explain the existence of solutions for such systems.


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## Body

## 1. Introduction

As everyone knows, the study of the existence of solutions for Fermat type equations has always been an important and interesting problem. The famous Fermat’s Last Theorem has attracted the attention of many mathematical scholars [1, 2]. About 60 years ago or even earlier, Montel [3] and Gross [4] had considered the equation fm+gm=1 and obtained that the entire solutions of f2+g2=1 are f=cosζz,g=sinζz for the case m=2, where ζz is an entire function, and this equation does not admit any nonconstant entire solution for any positive integer m>2.With the establishment and rapid development of Nevanlinna value distribution theory for meromorphic functions and theirs difference [5–7], Liu [8] in 2009, Liu et al. [9] in 2012, and Liu and Yang [10] in 2013 studied some complex Fermat type difference and Fermat type differential difference equations and obtained some results.Theorem 1 (see [9], Theorem 1.1).
The transcendental entire solutions with finite order of(1)fz2+fz+c2=1must satisfy fz=sinAz+B, where B is a constant and A=4k+1π/2c, with k an integer.Theorem 2 (see [9], Theorem 1.3).
The transcendental entire solutions with finite order of(2)f′z2+fz+c2=1must satisfy fz=sinz±Bi, where B is a constant and c=2kπ or c=2k+1π, with k an integer.After that, Gao [11] in 2016 extended Theorem 2 from complex differential difference equation to the system of complex differential difference equations.Theorem 3 (see [11], Theorem 1.1).
Suppose thatf1,f2 is a pair of finite-order transcendental entire solutions for the system of differential difference equations
(3)f1′z2+f2z+c2=1,f2′z2+f1z+c2=1.
Then,f1,f2 satisfies
(4)f1,f2=sinz−bi,sinz−b1i,
or(5)f1,f2=sinz+bi,sinz+b1i,where b,b1 are constants and c=kπ, where k is an integer.In recent, Xu and Cao [12, 13] further discussed the solutions for some Fermat type PDDEs and obtained the following:Theorem 4 (see [13], Theorem 1.4).
Letc∈ℂn−0. Then, any nonconstant entire solution with finite order of the equation
(6)fz2+fz+c2=1has the form of fz=cosLz+B, where L is a linear function of the form Lz=a1z1+⋯+anzn on ℂn such that Lc=−π/2−2kπ,k∈ℤ, and B is a constant on ℂ.Theorem 5 (see [13], Theorem 1.1).
Letc=c1,c2∈ℂ2. Then,
(7)∂fz1,z2∂z1n+fz1+c1,z2+c2m=1does not have any transcendental entire solution with finite order, where m and n are two distinct positive integers.Theorem 6 (see [13], Theorem 1.2).
Letc=c1,c2∈ℂ2. Then, any transcendental entire solution with finite order of the PDDE
(8)∂f∂z12+fz1+c1,z2+c22=1has the form of fz1,z2=sinAz1+B, where A is a constant on ℂ satisfying AeiAc1=1 and B is a constant on ℂ; in the special case whenever c1=0, we have fz1,z2=sinz1+B.By analyzing Theorems3–6, a natural question is as follows: What will happen about the transcendental entire solutions for the system of the PDDEs of Fermat type? Although many scholars have paid considerable attention to the complex difference equation with a single variable and the complex Fermat type difference equation in recent years, a series of important and meaningful results (including [7, 14–22]) were obtained, however, to our knowledge, there were not much results about the complex difference equation in several complex variables. Of course, the references involving the results of systems of complex PDDEs are even less.This manuscript is aimed at studying the solutions of several Fermat type systems involving both difference operator and partial differential. We establish four theorems on the forms of solutions for several systems of Fermat type PDDEs, which are improvement of the previous theorems given by Liu et al., Gao, and Xu and Cao [8, 9, 11, 13]. We mainly employ the Nevanlinna value distribution theory and difference Nevanlinna theory of several complex variables in this article, and the readers can refer to [23, 24]. Now, we start to state our main results below.Theorem 7.
Letc=c1,c2∈ℂ2, αj,βjj=1,2∈ℂ−0, and mj,njj=1,2∈ℕ+. If the Fermat type PDDE system
(9)α1∂f1∂z1n1+α2f2z+cm1=1,β1∂f2∂z1n2+β2f1z+cm2=1satisfies one of the conditions
(i)
m1m2>n1n2(ii)
nj>mj/mj−1 and mj≥2, j=1,2
then system (9) does not exist any pair of finite-order transcendental entire solution.Remark 8.
Here, we say thatf,g is a pair of finite-order transcendental entire solution for
(10)fn1+gm1=1,fn2+gm2=1,if f,g are transcendental entire functions satisfying the above system and ρ=maxρf,ρg<∞.Remark 9.
We list an example to demonstrate that the conditionmj≥2 in Theorem 7 cannot be removed. Let
(11)f1=f2=1+14c12−14z12+c12c2z2+b+e2πi/c2z2z1−c1−c12c2z2−c2+b+e2πi/c2z22,where c1,b∈ℂ and c2≠0. Thus, f1,f2 satisfies the system (9) with n1=n2=2, m1=m2=1, and αj=βj=1, j=1,2.Theorem 10.
Letc=c1,c2∈ℂ2 and αj,βjj=1,2∈ℂ−0. If the system of Fermat type difference equations
(12)α1f1z2+α2f2z+c2=1,β1f2z2+β2f1z+c2=1
admits a pair of finite-order transcendental entire solutionf1,f2, then α12/β22=β12/α22=1, and f1,f2 have the following forms
(13)f1z,f2z=eLz+B0+e−Lz+B02α1,β2β1A21eLz+B0+A22e−Lz+B02α1,where Lz=a1z1+a2z2, a1,a2, B0∈ℂ, and A21,A22,c satisfy one of the following cases.
(i)
Lc=kπi, A21=−i and A22=i or A21=i and A22=−i, k is a integer(ii)
Lc=2k±1/2πi, A21=−1 and A22=−1, or A21=1 and A22=1.Remark 11.
From Theorem10, we can conclude that f1,f2 have the following relationships
(i)
f2=ηf1(ii)
f2=iηeLz+B1−e−Lz+B1/2α1, where η=±β2/β1 and f1z=eLz+B1+e−Lz+B1/2α1.Now, two examples can verify the existence of solutions for (12).Example 1.
Letc1,c2 and Lz=a1z1+a2z2 satisfy Lc=a1c1+a2c2=2k±1/2πi, and B0∈ℂ. Then, the function
(14)f1z,f2z=eLz+B0+e−Lz−B04,−eLz+B0+e−Lz−B02satisfies the system (12) with α1=2, α2=1, and β1=β2=1.Example 2.
Letc1,c2 and Lz=a1z1+a2z2 satisfy Lc=a1c1+a2c2=kπi, and B0∈ℂ. Then, the function
(15)f1z,f2z=eLz+B0+e−Lz−B02,13eLz+B0−e−Lz−B02isatisfies the system (12) with α1=1=β2 and α2=3=β1.Theorem 12.
Letc=c1,c2∈ℂ2 and αj,βjj=1,2∈ℂ−0. If the system of Fermat type PDDEs
(16)α1∂f1∂z12+α2f2z+c2=1,β1∂f2∂z12+β2f1z+c2=1
admits a pair of finite-order transcendental entire solutionf1,f2, then α1α22=β1β22 and f1,f2 is the form of
(17)f1,f2=A11eLz+B0+A12e−Lz+B02β2,ηα1a1α2eLz+B0−e−Lz+B02β2,where Lz=a1z1+a2z2, B0 is a constant in ℂ, and a1,c,A11,A12,η satisfy a12=−β22/α12=−α22/β12 and one of the following cases
(i)
Lc=2kπi, and η=−1, A11=−i, A12=i, or η=1, A11=i, A12=−i(ii)
Lc=2k+1πi, and η=−1, A11=i, A12=−i, or η=1, A11=−i, A12=i(iii)
Lc=2k+1/2πi, and η=1, A11=−1, A12=−1, or η=−1, A11=1, A12=1(iv)
Lc=2k−1/2πi, and η=1, A11=1, A12=1, or η=−1, A11=−1, A12=−1.
Here, two examples can verify the existence of solutions for (16).Example 3.
Leta1,a2=i,π, A11=−i, A12=i, η=−1, and B0∈ℂ. That is, Lz=iz1+πz2 and
(18)f1z,f2z=−ieLz+B0−e−Lz+B04,−ieLz+B0−e−Lz+B02.
Thus,f1,f2 satisfies the system (16) with c1,c2=π,i, α1=2, β1=1, α2=1, and β2=2.Example 4.
Leta1,a2=1,−πi, A11=−1, A12=−1, η=1, and B0∈ℂ. That is, Lz=z1−πiz2 and
(19)f1z,f2z=−eLz+B0+e−Lz+B04i,−eLz+B0−e−Lz+B02i.
Thus,f1,f2 satisfies the system (16) with c1,c2=πi,1/2, α1=2, β1=1, α2=i, and β2=2i.Example 5.
Leta1,a2=2i,i, A11=i, A12=−i, η=−1, and B0∈ℂ. That is, Lz=2iz1+iz2 and
(20)f1z,f2z=ieLz+B0−e−Lz+B04,−ieLz+B0−e−Lz+B08.
Thus,f1,f2 satisfies the system (16) with c1,c2=π,−π, α1=1, β1=2, α2=4, and β2=2.Example 6.
Leta1,a2=3,1, A11=1, A12=1, η=i, and B0∈ℂ. That is, Lz=3z1+z2 and
(21)f1z,f2z=eLz+B0+e−Lz+B06,ieLz+B0−e−Lz+B018.
Thus,f1,f2 satisfies the system (16) with c1,c2=π,−π, α1=i, α2=9, β1=3i, and β2=3.Theorem 13.
Letc=c1,c2∈ℂ2 and αj,βjj=1,2,3∈ℂ−0. Let f1,f2 be a pair of transcendental entire solutions of finite order for the system
(22)α1∂f1∂z12+α2f2z+c+α3f1z2=1,β1∂f2∂z12+β2f1z+c+β3f2z2=1.
Then,f1,f2 is one of the forms
(23)f1,f2=eiLz+B0−e−iLz−B02ia1α1+eηz2G1z2,±eiLz+B0−e−iLz−B02ia1β1+eηz2G2z2,or
(24)f1,f2=eiLz+B0−e−iLz−B02ia1α1+eηz2G1z2,±eiLz+B0+e−iLz−B02a1β1+eηz2G2z2,
whereLz=a1z1+a2z2, a1≠0,a2,B0∈ℂ, and G1z2,G2z2 are entire period functions of finite order with period 2c2, and a1,a2,αj,βj,η,c1,c2 satisfy e4iLc=1 and the following conditions
(C1) η=0 if α2β2=α3β3, and η=logα2β2−logα3β3/2c2 if α2β2≠α3β3
(C2) β1/α2a1−α3/α12=α1/β2a1−β3/β12=1, or β1/α2a1−α3/α12=α1/β2a1+β3/β12=1(25)f1z,f2z=elog−1/2c2z2G1z2+D1,elog−1/2c2z2G2z2+D2,where D1=α2ξ2−β3ξ1/2α2β2, D2=β2ξ1−α3ξ2/2α2β2, ξ1=±1, and ξ2=±1;
(26)f1z,f2z=b1z1+γ1z2+G1z2,−α3α2b1z1+γ2z2+G2z2,
whereG1z2,G2z2 are stated as in ((23)) and ((24)), and b1≠0,γ1,γ2 satisfy
(27)γ1=α2b3−β3b2−α2β2+α3β3b1c12α2β2c2,(28)γ2=β2b2−α3b3+α3β2+β3α2b1c12α2β2c2,(29)b22+α1b12=1,b32+β1α3α2b12=1.
Here, five examples can verify the existence of solutions for (22).Example 7.
LetB0∈ℂ, a1=1, a2=1, and
(30)f1z,f2z=eiz1+z2+B0−e−iz1+z2−B0/4i+eiz2,eiz1+z2+B0−e−iz1+z2−B0/2i−4eiz2.
Thus,f1,f2 satisfies system (22) with c1,c2=π,π, α1=2, β1=1, α2=−1, β2=4, α3=4, and β3=1.Example 8.
LetB0∈ℂ, a1=i, a2=1, and
(31)f1z1,z2=eiiz1+z2+B0−e−iiz1+z2−B0/−2+elog−1+2i/2πz2eiz2,f2z1,z2=eiiz1+z2+B0+e−iiz1+z2−B0−2+i2elog−1+2i/2πz2eiz2.
Thus,f1,f2 satisfies the system (22) with c1,c2=1/2π,π, α1=1, β1=i, α2=2, β2=i−2, α3=−i, and β3=2.Example 9.
Letα1∈ℂ and f1z,f2z=e3πi/2z2+1,e3πi/2z2. Thus, f1,f2 satisfies the system (22) with c1,c2=c1,1, c1∈ℂ, α2=2, α2=1, β2=1, and β3=−2.Example 10.
Letf1,f2 be of the forms
(32)f1z,f2z=eiz2+i−14πz2,2eiz2+1−i2πz2+1+i2.
Thus,f1,f2 satisfies the system (22) with c1,c2=c1,π, c1∈ℂ, α2=1, α3=2, β2=2i, β3=i, and α1,β1∈ℂ.Example 11.
Let(33)f1z1,z2=12z1+15−43−416z2+e2πiz2,f2z1,z2=−14z1+8−15+4332z2−12e2πiz2+43+1532.
Thus,f1,f2 satisfies system (22) with c=c1,c2=1,1, α1=1, β1=1, α2=2, β2=2, α3=1, and β3=4.


## 2. Proof of Theorem7

Proof.
Letf1,f2 be a pair of finite-order transcendental entire functions satisfying (9). Here, let us consider two cases below.Case 1.
m1m2>n1n2. Owing to Refs. [23, 24], we have the following facts that
(34)mr,fjzfjz+c=Sr,fj,j=1,2
hold for allr>0 outside of a possible exceptional set Ej⊂1,+∞ of finite logarithmic measure ∫Ejdt/t<∞. Due to the above fact, we have
(35)Tr,fj=mr,fj≤mr,fjzfz+c+mr,fjz+c+log2=mr,fjz+c+Sr,fj=Tr,fjz+c+Sr,fj,j=1,2,
for allr∈E≕E1∪E2. By the Mokhon’ko theorem ([25], Theorem 3.4) and the Logarithmic Derivative Lemma [26], it yields from (35) that
(36)m1Tr,f2≤m1Tr,f2z+c+Sr,f2=Tr,α2f2z+cm1+Sr,f2=Tr,α1∂f1∂z1n1−1+Sr,f2=n1Tr,∂f1∂z1+Sr,f1+Sr,f2=n1mr,∂f1∂z1+Sr,f1+Sr,f2≤n1mr,∂f1/∂z1f1+mr,f1+Sr,f1+Sr,f2=n1Tr,f1+Sr,f1+Sr,f2,
for allr∈E. Similarly, we also get
(37)m2Tr,f1≤n2Tr,f2+Sr,f1+Sr,f2,r∈E.
Thus, we conclude from (36) and (37) that
(38)m1m2−n1n2Tr,fj≤Sr,f1+Sr,f2,r∈E.
By combining with the condition thatm1m2>n1n2 and f1,f2 being transcendental functions, we obtain a contradiction.Case 2.
nj>mj/mj−1 and mj≥2, j=1,2. Thus, it is easy to get that mj>nj/nj−1. In view of the Nevanlinna second fundamental theorem, the difference logarithmic derivative lemma in several complex variables [23, 24], we thus obtain from (9) that
(39)n1−1Tr,∂f1∂z1≤N¯r,∂f1∂z1+∑q=1n1‍N¯r,1∂f1/∂z1−wq/α1+Sr,∂f1∂z1≤N¯r,1α1∂f1/∂z1n1−1+Sr,∂f1∂z1≤N¯r,1f2z+c+Sr,f1≤Tr,f2z+c+Sr,f1+Sr,f2,
wherewq is a roots of wn1−1=0. Similarly, we also have
(40)n2−1Tr,∂f2∂z1≤Tr,f1z+c+Sr,f1+Sr,f2.
In addition, by applying the Mokhon’ko theorem in several complex variables ([25], Theorem 3.4) for (9), we can conclude
(41)m1Tr,f2z+c=Tr,α2f2z+cm1+Sr,f2=Tr,α1∂f1∂z1n1−1+Sr,f2=n1Tr,∂f1∂z1+Sr,f1+Sr,f2.
Similarly, we also get(42)m2Tr,f1z+c=n2Tr,∂f2∂z1+Sr,f1+Sr,f2.
Due tomj>nj/nj−1, it follows from (39)–(42) that
(43)m1−n1n1−1Tr,f2z+c≤Sr,f1+Sr,f2,m2−n2n2−1Tr,f1z+c≤Sr,f1+Sr,f2,and this is a contradiction with f1,f2 being transcendental functions.Therefore, Theorem7 is proved.


## 3. The Proof of Theorem10

Letf1,f2 be a pair of finite-order transcendental entire functions satisfying (12). We firstly rewrite the system (12) as
(44)α1f1+iα2f2z+cα1f1−iα2f2z+c=1,β1f2+iβ2f1z+cβ1f2−iβ2f1z+c=1.By applying the Hadamard factorization theorem (can be found in [27, 28]), then there exist two polynomials p1,p2 such that
(45)α1f1+iα2f2z+c=ep1,α1f1−iα2f2z+c=e−p1,β1f2+iβ2f1z+c=ep2,β1f2−iβ2f1z+c=e−p2.Thus, we have from (45) that
(46)α1f1=ep1+e−p12,α2f2z+c=ep1−e−p12i,β1f2=ep2+e−p22,β2f1z+c=ep2−e−p22i,which implies
(47)α1β2iep1z+c+p2+α1β2iep1z+c−p2−e2p1z+c≡1,(48)β1α2iep2z+c+p1+β1α2iep2z+c−p1−e2p2z+c≡1.By applying [29], Lemma 3.1 (can be found in [30]), for (47) and (48), we have that
(49)α1iep1z+c−p2≡β2,orα1ep1z+c+p2≡β2i,β1iep2z+c−p1≡α2,orβ1ep2z+c+p1≡α2i.Here, four cases will be discussed below.Case 1.
(50)α1iep1z+c−p2≡β2,β1iep2z+c−p1≡α2.
Thus, we can conclude from (50) that p1z+c−p2z≡C1 and p2z+c−p1z≡C2; here and below, C1,C2 are constants. So, this leads to p1z=Lz+B1,p2z=Lz+B2, where Lz=a1z1+a2z2 and a1,a2,B1,B2 are constants. Thus, by virtue of (47)–(50)), it yields that
(51)α1ieLc+B1−B2≡β2,β1ieLc−B1+B2≡α2,α1e−Lc−B1+B2≡β2i,β1e−Lc+B1−B2≡α2i,which implies
(52)α12β22=β12α22=1,e4Lc=1,eB1−B2=β2α1ie−Lc.
In view of (46), let
(53)f1z=eLz+B1+e−Lz−B12α1,f2z=eLz+B2+e−Lz−B22β1.
IfeLc=1, i.e., Lc=2kπi, k∈ℤ, then eB2−B1=α1/β2i. Thus,
(54)f2z=eLz+B2+e−Lz−B22β1=eLz+B1eB2−B1+e−Lz−B1eB1−B22β1=β2β1ieLz+B1−ie−Lz−B12α1.
IfeLc=−1, i.e., Lc=2k+1πi, k∈ℤ, then eB2−B1=−α1/β2i. Thus,
(55)f2z=eLz+B1eB2−B1+e−Lz−B1eB1−B22β1=−β2β1ieLz+B1−ie−Lz−B12α1.
IfeLc=i, i.e., Lc=2k+1/2πi, k∈ℤ, then eB2−B1=−α1/β2. Thus,
(56)f2z=−β2β1eLz+B1+e−Lz−B12α1=−β2β1f1z.
IfeLc=−i, i.e., Lc=2k−1/2πi, k∈ℤ, then eB2−B1=α1/β2. Thus,
(57)f2z=β2β1eLz+B1+e−Lz−B12α1=β2β1f1z.Case 2.
(58)α1iep1z+c−p2≡β2,β1ep1−p2z+c≡α2i.
Thus, it yields from (58) that p1z+c−p2z≡C1 and p1z+p2z+c≡C2. Hence, we obtain that p1z+2c+p1z≡C1+C2, and this is a contradiction with p1 is not a constant.Case 3.
(59)α1ep1z+c+p2≡β2i,β1iep2z+c−p1≡α2.
Thus, it yields from (59) that p1z+c+p2z≡C1 and p2z+c−p1z≡C2. Hence, we obtain that p2z+2c+p2z≡C1+C2, and this is a contradiction with p2 is not a constant.Case 4.
(60)α1ep1z+c+p2≡β2i,β1iep2z+c+p1≡α2.
Thus, it yields from (60) that p2z−p1z+c≡C1 and p1z−p2z+c≡C2. Hence, we obtian that p1z=Lz+B1,p2z=−Lz+B2, where Lz=a1z1+a2z2, a1,a2,B1,B2 are constants. By virtue of (47),(48), (60), it yields that
(61)α1eLc+B1+B2≡β2i,β1e−Lc+B1+B2≡α2i,α1e−Lc−B1−B2≡β2i,β1eLc−B1−B2≡α2i,which implies
(62)α12β22=β12α22=1,e4Lc=1,eB1+B2=β2α1ie−Lc=α2iβ1eLc.
In view of (46), let
(63)f1z=eLz+B1+e−Lz−B12α1,f2z=e−Lz+B2+eLz−B22β1.
IfeLc=1, i.e., Lc=2kπi, k∈ℤ, then eB1+B2=β2/α1i=α2/β1i. Thus,
(64)f2z=eLz+B1e−B2−B1+e−Lz−B1eB1+B22β1=β2β1−ieLz+B1+ie−Lz−B12α1.
IfeLc=−1, i.e., Lc=2k+1πi, k∈ℤ, then eB1+B2=−β2/α1i=−α2/β1i. Thus,
(65)f2z=β2β1ieLz+B1−ie−Lz−B12α1.
IfeLc=i, i.e., Lc=2k+1/2πi, k∈ℤ, then eB1+B2=β2/α1=−α2/β1. Thus,
(66)f2z=β2β1eLz+B1+e−Lz−B12α1=β2β1f1z.
IfeLc=−i, i.e., Lc=2k−1/2πi, k∈ℤ, then eB1−B2=−β2/α1=α2/β1. Thus,
(67)f2z=−β2β1eLz+B1+e−Lz−B12α1=−β2β1f1z.
Therefore, this completes the proof of Theorem10.


## 4. The Proof of Theorem12

Proof.
Letf1,f2 be a pair of finite-order transcendental entire functions satisfying (16). Firstly, (16) may be represented as the following form:
(68)α1∂f1∂z1+iα2f2z+cα1∂f1∂z1−iα2f2z+c=1,β1∂f2∂z1+iβ2f1z+cβ1∂f2∂z1−iβ2f1z+c=1.
By the Hadamard factorization theorem (can be found in [27, 28]), there are two nonconstant polynomials p1,p2 satisfying
(69)α1∂f1∂z1+iα2f2z+c=ep1,α1∂f1∂z1−iα2f2z+c=e−p1,β1∂f2∂z1+iβ2f1z+c=ep2,β1∂f2∂z1−iβ2f1z+c=e−p2.
In view of (69), it yields that
(70)α1∂f1∂z1=ep1+e−p12,α2f2z+c=ep1−e−p12i,β1∂f2∂z1=ep2+e−p22,β2f1z+c=ep2−e−p22i,which implies
(71)α1β2i∂p2∂z1ep1z+c+p2+α1β2i∂p2∂z1ep1z+c−p2−e2p1z+c≡1,(72)β1α2i∂p1∂z1ep2z+c+p1+β1α2i∂p1∂z1ep2z+c−p1−e2p2z+c≡1.
Obviously,∂p1/∂z1≡0. Otherwise, e2p2z+c≡1. This leads to a contradiction with p1 is not a constant. Similarly, ∂p2/∂z1≡0. Thus, due to [29], Lemma 3.1 (can be found in [30]), (71), and (72), we obtain that
(73)α1β2i∂p2∂z1ep1z+c+p2≡1orα1β2i∂p2∂z1ep1z+c−p2z≡1,β1α2i∂p1∂z1ep2z+c+p1≡1orβ1α2i∂p1∂z1ep2z+c−p1≡1.Hence, four cases will be discussed below.Case 1.
(74)α1β2i∂p2∂z1ep1z+c+p2≡1,β1α2i∂p1∂z1ep2z+c+p1≡1.
Thus, it follows from (74) that p1z+c+p2≡C1 and p2z+c+p1≡C2. These lead to p1z+2c−p1≡C1−C2 and p2z+c−p2≡C2−C1. Hence, we obtain that p1z=Lz+B1,p2z=−Lz+B2, where Lz=a1z1+a2z2, a1≠0,a2,B1,B2 are constants. By combining with (71)–(74), we have
(75)α1a1β2ieLc+B1+B2≡1,β1a1α2ie−Lc+B1+B2≡1,α1a1β2ie−Lc−B1−B2≡1,β1a1α2ieLc−B1−B2≡1,and this leads to
(76)a12=−β22α12=−α22β12,e4Lc=1,eB1+B2=β2α1a1ie−Lc=α2iβ1a1eLc.Subcase 1.
IfeLc=1, then Lc=2kπi and eB1+B2=β2/α1a1i=α2i/β1a1. Due to (70), we have that
(77)f1=e−Lz+B2+Lc−eLz−B2−Lc2β2i=ieLz−B2−e−Lz+B22β2,f2=eLz+B1−Lc−e−Lz−B1+Lc2α2i=eLz+B1−e−Lz−B12α2i=eLz−B2eB1+B2−e−Lz+B2e−B1−B22α2i=−β2i/α1a1eLz−B2−β2i/−α1a1e−Lz+B22α2i=α1a1α2eLz−B2−e−Lz+B22β2.Subcase 2.
IfeLc=−1, then Lc=2k+1πi, k∈ℤ and eB1+B2=−β2/α1a1i=−α2i/β1a1. Due to (70), we have that
(78)f1=e−Lz+B2+Lc−eLz−B2−Lc2β2i=−ieLz−B2−e−Lz+B22β2,f2=eLz+B1−Lc−e−Lz−B1+Lc2α2i=−eLz+B1+e−Lz−B12α2i=−eLz−B2eB1+B2+e−Lz+B2e−B1−B22α2i=α1a1α2eLz−B2−e−Lz+B22β2.Subcase 3.
IfeLc=i, then Lc=2k+1/2πi, k∈ℤ, and eB1+B2=−β2/α1a1=−α2/β1a1. Due to (70), we have that
(79)f1=e−Lz+B2+Lc−eLz−B2−Lc2β2i=eLz−B2+e−Lz+B22β2,f2=eLz+B1−Lc−e−Lz−B1+Lc2α2i=−eLz+B1−e−Lz−B12α2=−eLz−B2eB1+B2−e−Lz+B2e−B1−B22α2=−α1a1α2eLz−B2−e−Lz+B22β2.Subcase 4.
IfeLc=−i, then Lc=2k−1/2πi, k∈ℤ, and eB1+B2β2/α1a1=α2/β1a1. Due to (70), we have that
(80)f1=e−Lz+B2+Lc−eLz−B2−Lc2β2i=−eLz−B2+e−Lz+B22β2,f2=eLz+B1−Lc−e−Lz−B1+Lc2α2i=eLz+B1+e−Lz−B12α2=eLz−B2eB1+B2+e−Lz+B2e−B1−B22α2=−α1a1α2eLz−B2−e−Lz+B22β2.Case 2.
(81)α1β2i∂p2∂z1ep1z+c+p2≡1,β1α2i∂p1∂z1ep2z+c−p1≡1.
Thus, it yields from (81) that p1z+c+p2≡C1 and p2z+c−p1≡C2. We have that p2z+2c+p2≡C1+C2, and this leads to a contradiction with p2 being not constant.Case 3.
(82)α1β2i∂p2∂z1ep1z+c−p2≡1,β1α2i∂p1∂z1ep2z+c+p1≡1.
Sincep1z,p2z are polynomials, then from (82), it follows that p1z+c−p2z≡C1 and p2z+c+p1z≡C2. This means p1z+2c+p1z≡C1+C2, and this is a contradiction because p1z is not a constant.Case 4.
(83)α1β2i∂p2∂z1ep1z+c−p2≡1,β1α2i∂p1∂z1ep2z+c−p1≡1.
Then, from (83), it yields that p1z+c−p2≡C1 and p2z+c−p1≡C2, and this leads to p1z+2c−p1≡C1+C2 and p2z+2c−p2≡C2+C1. Thus, it follows that p1z=Lz+B1,p2z=Lz+B2, where Lz=a1z1+a2z2, a1≠0,a2,B1,B2 are constants in ℂ. In view of (71), (72), and (83), we have
(84)α1β2ia1eLc+B1−B2≡1,β1α2ia1eLc−B1+B2≡1,α1β2ia1e−Lc−B1+B2≡1,β1α2ia1e−Lc+B1−B2≡1,which implies
(85)a12=−β22α12=−α22β12,e4Lc=1,eB1−B2=β2iα1a1e−Lc=β1a1α2ieLc.Subcase 4.1. IfeLc=1, then Lc=2kπi, k∈ℤ, and eB1−B2=β2i/α1a1=β1a1/α2i. By virtue of (70), it follows that
(86)f1=eLz+B2−Lc−e−Lz−B2+Lc2β2i=−ieLz+B2−e−Lz−B22β2,f2=eLz+B1−Lc−e−Lz−B1+Lc2α2i=eLz+B1−e−Lz−B12α2i=eLz+B2eB1−B2−e−Lz−B2e−B1+B22α2i=β2i/α1a1eLz+B2−β2i/α1a1e−Lz−B22α2i=−α1a1α2eLz+B2−e−Lz−B22β2.Subcase 4.2. IfeLc=−1, then Lc=2k+1πi, k∈ℤ, and eB1−B2=−β2i/α1a1=−β1a1/α2i. By virtue of (70), it follows that
(87)f1=eLz+B2−Lc−e−Lz−B2+Lc2β2i=ieLz+B2−e−Lz−B22β2,f2=eLz+B1−Lc−e−Lz−B1+Lc2α2i=−eLz+B1+e−Lz−B12α2i=−eLz+B2eB1−B2+e−Lz−B2e−B1+B22α2i=−α1a1α2eLz+B2−e−Lz−B22β2.Subcase 4.3. IfeLc=i, then Lc=2k+1/2πi and eB1−B2=β2/α1a1=α2/β1a1. By virtue of (70), we have that
(88)f1=eLz+B2−Lc−e−Lz−B2+Lc2β2i=−eLz+B2+e−Lz−B22β2,f2=eLz+B1−Lc−e−Lz−B1+Lc2α2i=−eLz+B1+e−Lz−B12α2=−eLz+B2eB1−B2+e−Lz−B2e−B1+B22α2=α1a1α2eLz+B2−e−Lz−B22β2.Subcase 4.4. IfeLc=−i, then Lc=2k−1/2πi and eB1−B2=−β2/α1a1=−α2/β1a1. By virtue of (70), we have that
(89)f1=eLz+B2−Lc−e−Lz−B2+Lc2β2i=eLz−B2+e−Lz+B22β2,f2=eLz+B1−Lc−e−Lz−B1+Lc2α2i=eLz+B1+e−Lz−B12α2=eLz+B2eB1−B2+e−Lz−B2e−B1+B22α2=α1a1α2eLz+B2−e−Lz−B22β2.Hence, the proof of Theorem12 is completed.


## 5. The Proof of Theorem13

Proof.
Assume thatf1,f2 is a pair of finite-order transcendental entire functions satisfying (22). Thus, let us discuss two following cases.
(i)
Suppose that∂f1/∂z1 is transcendental, then α2f2z+c+α3f1 is transcendental. Noting that αj,βj are nonzero constants, we next prove that β2f1z+c+β3f2 and β1∂f2/∂z1 are transcendental
Suppose thatα2∂f2z+c/∂z1+α3∂f1/∂z1 is not transcendental. Since ∂f1/∂z1 is transcendental, then ∂f2z+c/∂z1 and ∂f2z/∂z1 are transcendental. By observing the second equation of (22), we can conclude that β2f1z+c+β3f2 is transcendental.
Suppose thatα2∂f2z+c/∂z1+α3∂f1/∂z1 is transcendental. If ∂f2z+c/∂z1 is transcendental, similar to the above argument, β2f1z+c+β3f2 and ∂f2/∂z1 are transcendental. If ∂f2z+c/∂z1 is not transcendental, it thus leads to that ∂f2/∂z1 is not transcendental. From (22), we thus get that β2f1z+c+β3f2 is not transcendental. Thus, it yields that β2∂f1z+c/∂z1+β3∂f2/∂z1 is not transcendental. This is a contradiction with ∂f1z+c/∂z1 is transcendental and ∂f2/∂z1 is not transcendental.
Hence, if∂f1/∂z1 is transcendental, then α2f2z+c+α3f1, β2f1z+c+β3f2, and ∂f2/∂z1 are transcendental. Hence, system (22) can be represented as
(90)α1∂f1∂z1+iα2f2z+c+α3f1α1∂f1∂z1−iα2f2z+c+α3f1=1,β1∂f2∂z1+iβ2f1z+c+β3f2β1∂f2∂z1−iβ2f1z+c+β3f2=1.
Thus, by the Hadamard factorization theorem (can be found in [27, 28]), there are two nonconstant polynomials p,q such that
(91)α1∂f1∂z1+iα2f2z+c+α3f1=eip,α1∂f1∂z1−iα2f2z+c+α3f1=e−ip,β1∂f2∂z1+iβ2f1z+c+β3f2=eiq,β1∂f2∂z1−iβ2f1z+c+β3f2=e−iq.
In view of (91), it yields that
(92)α1∂f1∂z1=eip+e−ip2,α2f2z+c+α3f1=eip−e−ip2i,β1∂f2∂z1=eiq+e−iq2,β2f1z+c+β3f2=eiq−e−iq2i,which implies
(93)β1α2∂p∂z1−α3α1eip+qz+c+β1α2∂p∂z1−α3α1eiqz+c−p−e2iqz+c≡1,(94)α1β2∂q∂z1−β3β1eiq+pz+c+α1β2∂q∂z1−β3β1eipz+c−q−e2ipz+c≡1.
Obviously,∂p/∂z1≠α3/α1. Otherwise, we have that −e2iqz+c≡1, and this leads to a contradiction since q is not a constant. Similarly, ∂q/∂z1≠β3/β1. Thus, due to [29], Lemma 3.1 (can be found in [30]), and in view of (93) and (94), we can deduce that
(95)β1α2∂p∂z1−α3α1eiqz+c−p≡1,orβ1α2∂p∂z1−α3α1eip+qz+c≡1,α1β2∂q∂z1−β3β1eipz+c−q≡1,orα1β2∂q∂z1−β3β1eiq+pz+c≡1.Now, let us consider the following four cases.Case 1.
(96)β1α2∂p∂z1−α3α1eiqz+c−p≡1,α1β2∂q∂z1−β3β1eipz+c−q≡1.
Then, (96) can lead to that qz+c−p≡C1 and pz+c−q≡C2. Thus, we obtain that pz+2c−p≡C2+C1 and qz+2c−q≡C1+C2. Hence, we can conclude that pz=Lz+B1,qz=Lz+B2, where Lz=a1z1+a2z2, a1,a2,B1,B2 are constants. By combining with (93)–(96), we have
(97)β1α2a1−α3α1eiLc+B2−B1≡1,α1β2a1−β3β1eiLc+B1−B2≡1,β1α2a1−α3α1e−iLc+B2−B1≡1,α1β2a1−β3β1e−iLc+B1−B2≡1.
This means that(98)β1α2a1−α3α12=α1β2a1−β3β12=1,e4iLc=1,eiB1−B2=β1α2a1−α3α1eiLc.
By combining with (92), f1,f2 have the following forms:
(99)f1z=eiLz+B1−e−iLz−B12ia1α1+φ1z2,f2z=eiLz+B2−e−iLz+B22ia1β1+φ2z2,where φ1z2,φ2z2 are entire functions of finite order in z2. Substituting the above expressions into (92), we can deduce that
(100)α2φ2z2+c2+α3φ1z2=0,β2φ1z2+c2+β3φ2z2=0.
This leads to(101)φ1z2+2c2=α3β3α2β2φ1z2,φ2z2+2c2=α3β3α2β2φ2z2.
Due to (101), we have
(102)φ1z2=eηz2G1z2,φ2z2=eηz2G2z2,where G1z2,G2z2 are entire period functions of finite order with period 2c2, and in (102), η=0, if α2β2=α3β3, and η=logα2β2−logα3β3/2c2, if α2β2≠α3β3. Further, in view of (100) and (102), we have G2z2=−α3/α2G1z2; if α2β2≠α3β3, we have G2z2=−α3/α2G1z2.IfeiLc=1, it follows from (97) that eiB1−B2=±1. Thus, it yields that
(103)f2z=eiLz+B2−e−iLz+B22ia1β1+φ2z2=α1β2eiLz+B1eiB2−B1−e−iLz+B1eiB1−B22ia1α1+φ2z2=±α1β2eiLz+B1−e−iLz−B12ia1α1+φ2z2.IfeiLc=−1, it follows from (97) that e2iB1−B2=1. Thus, similar to the above argument, we obtain that
(104)f2z=±α1β2eiLz+B1−e−iLz−B12ia1α1+φ2z2.IfeiLc=i, it follows from (97) that e2iB1−B2=−1. Thus, we obtain that
(105)f2z=±α1β2eiLz+B1+e−iLz−B12a1α1+φ2z2.IfeiLc=−i, it follows from (97) that e2iB1−B2=−1. Thus, we obtain that
(106)f2z=±α1β2eiLz+B1+e−iLz−B12a1α1+φ2z2.Case 2.
(107)β1α2∂p∂z1−α3α1eiqz+c−pz≡1,α1β2∂q∂z1−β3β1eiqz+pz+c≡1.
We thus get from (107) that pz+c+qz≡C2 and qz+c−p≡C1. This means qz+2c+qz≡C1+C2, and this yields a contradiction with q being not a constant.Case 3.
(108)β1α2∂p∂z1−α3α1eip+qz+c≡1,α1β2∂q∂z1−β3β1eipz+c−q≡1.
We thus get from (108) that qz+c+pz≡C1 and pz+c−qz≡C2. So, we conclude that pz+2c+pz≡C1+C2, and this leads to a contradiction with p being not a constant.Case 4.
(109)β1α2∂p∂z1−α3α1eip+qz+c≡1,α1β2∂q∂z1−β3β1eiq+pz+c≡1.
Then, it follows from (109) that p+qz+c≡C1 and q+pz+c≡C2. These yield that pz+2c−p≡C1+C2 and qz+2c−q≡C2+C1, which leads to p=Lz+B1,q=−Lz+B2, where Lz=a1z1+a2z2, a1,a2,B1,B2 are constants. In view of (93), (94), and (109), we have
(110)β1α2a1−α3α1e−iLc−B1−B2≡1,α1β2−a1−β3β1eiLc+B1+B2≡1,β1α2a1−α3α1eiLc−B1−B2≡1,α1β2−a1−β3β1e−iLc+B1+B2≡1.
In view of (110), it follows that
(111)β1α2a1−α3α12=α1β2a1+β3β12=1,e4iLc=1,eiB1+B2=β1α2a1−α3α1eiLc.
By combining with (92), f1,f2 are of the following forms
(112)f1z=eiLz+B1−e−iLz−B12ia1α1+φ1z2,f2z=ei−Lz+B2−eiLz−B22ia1β1+φ2z2,where φ1z2,φ2z2 are finite-order entire functions in z2. By using the same argument as in Case 1, we have (102).IfeiLc=1, it follows from (110) that e2iB1+B2=1. Thus, we can deduce that
(113)f2z=ei−Lz+B2−eiLz−B22ia1β1+φ2z2=α1β2−eiLz+B1e−iB1+B2+e−iLz+B1eiB1+B22ia1α1+φ2z2=±α1β2eiLz+B1−e−iLz−B12ia1α1+φ2z2.IfeiLc=−1, it follows from (110) that e2iB1+B2=1. We have that
(114)f2z=±α1β2eiLz+B1−e−iLz−B12ia1α1+φ2z2.IfeiLc=i, it follows from (110) that e2iB1+B2=−1. Thus, we obtain that
(115)f2z=±α1β2eiLz+B1+e−iLz−B12a1α1+φ2z2.IfeiLc=−i, it follows from (110) that e2iB1+B2=−1. Thus, we obtain that
(116)f2z=±α1β2eiLz+B1+e−iLz−B12a1α1+φ2z2.Therefore, from (102)–(106) and (113)–(116), we can prove the conclusions (23) and (24) of Theorem 13.(i)Assume that∂f1/∂z1=0. Thus, from (22), it follows that
(117)f1z=ϕ1z2,α2f2z+c+α3f1z≡ξ1,ξ1=±1.This leads to∂f2/∂z1=0. We thus get from (22) that
(118)f2z=ϕ2z2,β2f1z+c+β3f2z≡ξ2,ξ2=±1.By combining with (117) and (118), it yields
(119)α2ϕ2z2+c2+α3ϕ1z2≡ξ1,β2ϕ1z2+c2+β3ϕ2z2≡ξ2,which implies that
(120)ϕ1z2+2c2=α3β3α2β2ϕ1z2+α2ξ2−β3ξ1α2β2,ϕ2z2+2c2=α3β3α2β2ϕ2z2+β2ξ1−α3ξ2α2β2.Ifα2β2=α3β3, then from (120), it follows that
(121)ϕ1z2+2c2=ϕ1z2+α2ξ2−β3ξ1α2β2,ϕ2z2+2c2=ϕ2z2+β2ξ1−α3ξ2α2β2,which implies that
(122)ϕ1z2=G1z2+γ1z2,ϕ2z2=G2z2+γ2z2,whereG1z2,G2z2 are entire period functions of finite order with period 2c2, and
(123)γ1=α2ξ2−β3ξ12c2α2β2,γ2=β2ξ1−α3ξ22c2α2β2.Ifα2β2≠α3β3, then from (120), it follows that
(124)ϕ1z2=elogα2β2−logα3β3/2c2z2G1z2+D1,ϕ2z2=elogα2β2−logα3β3/2c2z2G2z2+D2,whereD1=α2ξ2−β3ξ1/α2β2−α3β3 and D2=β2ξ1−α3ξ2/α2β2−α3β3. Substituting (124) into (119), it follows that α2β2=−α3β3, G1z2=β3/β2iG2z2−c2 and G2z2=α3/α2iG1z2−c2. Thus, we have
(125)ϕ1z2=elog−1/2c2z2G1z2+α2ξ2−β3ξ12α2β2,ϕ2z2=elog−1/2c2z2G2z2+β2ξ1−α3ξ22α2β2.(ii)
Suppose that∂f1z1,z2/∂z1=b1≠0. Then, it yields in view of (22) that(126)f1z=b1z1+ψ1z2,α2f2z+c+α3f1z=b2,b22+α1b12=1,whereψ1z2 is a transcendental entire function of finite order in z2. Equation (126) leads to ∂f2/∂z1=−α3/α2b1. Thus, due to the second equation in (22), we have
(127)f2z=−α3α2b1z1+ψ2z2,β2f1z+c+β3f2z=b3,b32+β1α3α2b12=1,whereψ2z2 is a transcendental entire function of finite order in z2. Combining with (126) and (127), we can deduce that α2β2=α3β3 and
(128)α2ψ2z2+c2+α3ψ1z2=b2+α3b1c1,β2ψ1z2+c2+β3ψ2z2=b3−β2b1c1,This means that(129)ψ1z2=G1z2+γ1z2,ψ2z2=G2z2+γ2z2,whereG1z2,G2z2 are entire period functions of finite order with period 2c2 satisfying
(130)G2z2+c2+α3α2G1z2=β2b2+α3b32α2β2,G1z2+c2+β3β2G2z2=β3b2+α2b32α2β2,γ1=α2b3−β3b2−2α2β2b1c12α2β2c2,γ2=β2b2−α3b3+2α3β2b1c12α2β2c2.Hence, from (126)–(129), it is easy to get the cases ((26)) and ((27)) of Theorem 13.Therefore, the proof of Theorem13 is completed.



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*Source: 4449502-2022-07-12.xml*