# Entire Solutions for Complex Systems of the Second-Order Partial Differential Difference Equations of Fermat Type

Authors: Si Min Liu; Hong Yan Xu
Journal: Journal of Mathematics
 (2021)

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## Abstract

This article is mainly concerned with the existence and the forms of entire solutions for several systems of the second-order partial differential difference equations of Fermat typeα∂2f1z1,z2/∂z12+β∂2f1z1,z2/∂z22n1+f2z1+c1,z2+c2m1=1α∂2f2z1,z2/∂z12+β∂2f2z1,z2/∂z22n2+f1z1+c1,z2+c2m2=1 and ∂2f1z1,z2/∂z122+f2z1+c1,z2+c22=1∂2f2z1,z2/∂z122+f1z1+c1,z2+c22=1. Our results about the existence and the forms of solutions for these systems generalize the previous theorems given by Xu and Cao, Gao, Liu, and Yang. In addition, we give some examples to explain the existence of solutions of this system in each case.


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## Body

## 1. Introduction and Main Results

The issue on the existence and form of solutions for Fermat-type equationxm+yn=1 has attracted considerable attention from many scholars. Especially, Taylor and Wiles [1, 2] pointed out that this equation does not admit a nontrivial solution in rational numbers for m=n≥3, and this equation does admit a nontrivial rational solution for m=n=2. In fact, the study of this issue should go back to sixty years ago or even earlier; Montel [3] and Gross [4] had pointed out that the entire solutions of the functional equation fm+gn=1 for m=n=2 are f=cosaz,g=sinaz, where az is an entire function; for m=n>2, there are no nonconstant entire solutions.In 2004, Yang and Li [5] investigated a certain nonlinear differential equational of Malmquist type, by making use of Nevanlinna theory, and obtained the following.Theorem 1.
(See [5]). Let a1,a2, and a3 be nonzero meromorphic functions. Then, a necessary condition for the differential equation,(1)a1f2+a2f′2=a3,to have a transcendental meromorphic solution satisfying Tr,ak=Sr,f,k=1,2,3 is a1/a3≡constant.In the past ten years, Liu and his collaborators investigated the existence of solutions for a series of complex difference equations and complex differential difference equations of Fermat type, by using the difference Nevanlinna theory for meromorphic functions (see [6–8]), and obtained a lot of interesting original results (see [9–11]). In order to be consistent with the following text, here, we only list one of results are given by Liu.Theorem 2.
(See [10], Theorem 9). The transcendental entire solutions with finite order of(2)f′z2+fz+c2=1,must satisfy fz=sinz±Bi, where B is a constant and c=2kπ or c=2k+1π, where k is an integer.In 2019, Liu and Gao [12] further studied the entire solutions of the second-order differential and difference equation with single complex variable and obtained the following.Theorem 3.
(see [12], Theorem 2.1). Suppose that f is a transcendental entire solution with finite order of the complex differential difference equation(3)f″z2+fz+c2=Qz.Then,Qz=c1c2 is a constant, and fz satisfies(4)fz=c1eaz+b+c2e−az−b2a2,where a,b∈ℂ and a4=1,c=log−ia2+2kπi/a, where k∈ℤ.In 2016, Gao [13] further investigated the form of solutions for a class of system of differential difference equations corresponding to Theorem 2 and obtained the following.Theorem 4.
(See [13], Theorem 7). Suppose that f1,f2 is a pair of finite-order transcendental entire solutions for the system of differential difference equations(5)f1′z2+f2z+c2=1,f2′z2+f1z+c2=1.Then,f1,f2 satisfies(6)f1z,f2z=sinz−bi,sinz−b1i,orf1z,f2z=sinz+bi,sinz+b1i,where b,b1 are constants and c=kπ, where k is a integer.For the differential difference equations with several complex variables, Xu and Cao [14, 15] recently investigated the existence of the entire and meromorphic solutions for some Fermat-type partial differential difference equations by using the Nevanlinna theory in several complex variables and obtained the following theorems.Theorem 5.
(See [14], Theorem 7). Let c=c1,c2 be a constant in ℂ2. Then, the Fermat-type partial differential difference equation,(7)∂fz1,z2∂z1n+fz1+c1,z2+c2m=1,does not have any transcendental entire solution with finite order, where m and n are two distinct positive integers.Theorem 6.
(See [14], Theorem 8). Let c=c1,c2 be a constant in ℂ2. Then, any transcendental entire solution with finite order of the partial differential difference equation,(8)∂fz1,z2∂z12+fz1+c1,z2+c22=1,has 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.Inspired by the form of the abovementioned equations in Theorems3–6, a question naturally arises: What will happen about the existence and the form of the solutions when the equations are put into the system and the first-order partial differential is replaced by the second-order partial differentials? For nearly two decades, although there were a lot of important and meaningful results focusing on the solutions of the complex difference equation of single variable (including [8, 16–21]), as far as we all know, there are few literature about the system of the second-order partial differential difference equations of Fermat type in several complex variables. It seems that this topic has never been treated before.The purpose of this article is concerned with the properties of the solutions for some Fermat-type systems including both the difference operator and the second-order partial differential by making use of the (difference) Nevanlinna theory of several complex variables [22, 23]. We give the existence theorem and the forms of solutions for the Fermat-type systems of the second-order partial differential difference equations, which are generalization of the previous theorems given by Liu, Liu et al., Gao, Xu and Cao, and Xu et al. [9, 10, 13, 14, 24]. Here and below, let z+w=z1+w1,z2+w2 for any z=z1,z2,w=w1,w2, and c=c1,c2. Now, our main results of this paper are listed as follows.Theorem 7.
Letc=c1,c2∈ℂ2, and mj,njj=1,2 be positive integers, and α,β be constants in ℂ that are not zero at the same time. If the following system of Fermat-type partial differential difference equations,(9)α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22n1+f2z1+c1,z2+c2m1=1,α∂2f2z1,z2∂z12+β∂2f2z1,z2∂z22n2+f1z1+c1,z2+c2m2=1,satisfies one of the conditions(i)
m1m2>n1n2(ii)
mj>nj/nj−1 for nj≥2, j=1,2,then system (9) does not have any pair of transcendental entire solution with finite order.Remark 1.
Here,f,g is called as a pair of finite-order transcendental entire solutions for the system(10)fn1+gm1=1,fn2+gm2=1,if f,g are transcendental entire functions and ρ=maxρf,ρg<∞.Remark 2.
By observing the proof of Theorem7, it is easy to see that the conclusions of Theorem 7 also hold if ∂2fjz1,z2/∂z12 or ∂2fjz1,z2/∂z22 of system (9) is replaced by ∂2fjz1,z2/∂z1∂z2, j=1,2.Theorem 8.
Letc=c1,c2∈ℂ2. Then, any pair of transcendental entire solution f1,f2 with finite order for the system of Fermat-type difference equations,(11)∂2f1z1,z2∂z122+f2z1+c1,z2+c22=1,∂2f2z1,z2∂z122+f1z1+c1,z2+c22=1,is of the following forms:(12)f1z,f2z=±sinLz+B0,±cosLz+B0,or(13)f1z,f2z=±cosLz+B0,±cosLz+B0,where Lz=a1z1+a2z2, a14=1, e2iLc=±1, and B0 is a constant in ℂ.The following examples show the existence of solutions for system (11).Example 1.
Leta=a1,a2=1,π, c=c1,c2=π,1, Lz=z1+πz2, and(14)f1z,f2z=−sinLz+B0,−cosLz+B0,where B0 is a constant in ℂ. Thus, f1,f2 satisfies the following system:(15)∂2f1z1,z2∂z122+f2z1+π,z2+12=1,∂2f2z1,z2∂z122+f1z1+π,z2+12=1.Example 2.
Leta=a1,a2=−i,1, c=c1,c2=−π/2,π, Lz=−iz1+z2, and(16)f1z,f2z=−cosLz+B0,cosLz+B0,where B0 is a constant in ℂ. Thus, f1,f2 satisfies the following system:(17)∂2f1z1,z2∂z122+f2z1−π2,z2+π2=1,∂2f2z1,z2∂z122+f1z1−π2,z2+π2=1.Theorem 9.
Letc=c1,c2∈ℂ2. Then, any pair of transcendental entire solution f1,f2 with finite order for the system of Fermat-type partial differential difference equations,(18)∂2f1z1,z2∂z12+∂2f1z1,z2∂z1∂z22+f2z1+c1,z2+c22=1,∂2f2z1,z2∂z12+∂2f2z1,z2∂z1∂z22+f1z1+c1,z2+c22=1,is of the following forms:(19)f1z,f2z=±sinLz+B0,±cosLz+B0,or(20)f1z,f2z=±cosLz+B0,±cosLz+B0,where Lz=a1z1+a2z2, a1a1+a22=1, e2iLc=1, and B0 is a constant in ℂ.We also list two examples to exhibit the existence of solutions for system (18).Example 3.
Leta=a1,a2=2−1,2, c=c1,c2=−2+1π,π, Lz=2−1z1+2z2, and(21)f1z,f2z=sinLz+B0,−cosLz+B0,where B0 is a constant in ℂ. Thus, f1,f2 satisfies the following system:(22)∂2f1z1,z2∂z12+∂2f1z1,z2∂z1∂z22+f2z1−2+1π,z2+π2=1,∂2f2z1,z2∂z12+∂2f2z1,z2∂z1∂z22+f1z1−2+1π,z2+π2=1.Example 4.
Leta=a1,a2=−1,2, c=c1,c2=π,π/4, Lz=−z1+2z2, and(23)f1z,f2z=cosLz+B0,cosLz+B0,where B0 is a constant in ℂ. Thus, f1,f2 satisfies the following system:(24)∂2f1z1,z2∂z12+∂2f1z1,z2∂z1∂z22+f2z1+π,z2+π42=1,∂2f2z1,z2∂z12+∂2f2z1,z2∂z1∂z22+f1z1+π,z2+π42=1.Remark 3.
In fact, in view of the proofs of Theorems8 and 9, it is easy to get that the conclusions of Theorem 9 still holds, if the system (18) is replaced by the following systems:(25)∂2f1z1,z2∂z1∂z22+f2z1+c1,z2+c22=1,∂2f2z1,z2∂z1∂z22+f1z1+c1,z2+c22=1,or(26)∂2f1z1,z2∂z12+∂2f1z1,z2∂z1∂z22+f2z1+c1,z2+c22=1,∂2f2z1,z2∂z12+∂2f2z1,z2∂z1∂z22+f1z1+c1,z2+c22=1,respectively. The only thing that we need to do is to modify the condition a1a1+a22=1 to a1a22=1 or a2a1+a22=1, respectively.


## 2. Proof of Theorem 7

To prove Theorem7, we need the following lemmas.Lemma 1.
(See [25, 26]). Let f be a nonconstant meromorphic function on ℂn, and let I=i1,…,in be a multi-index with length I=∑j=1nij. Assume that Tr0,f≥e for some r0. Then,(27)mr,∂Iff=Sr,f,holds for all r≥r0 outside a set E⊂0,+∞ of finite logarithmic measure ∫Edt/t<∞, where ∂If=∂If/∂z1i1,…,∂znin.Lemma 2.
(See [22, 23]). Let f be a nonconstant meromorphic function with finite order on ℂn such that f0≠0,∞, and let ε>0. Then, for c∈ℂn,(28)mr,fzfz+c+mr,fz+cfz=Sr,f,holds for all r≥r0 outside a set E⊂0,+∞ of finite logarithmic measure ∫Edt/t<∞.Proof.
The proof of Theorem7: suppose that f1,f2 is a pair of transcendental entire functions with finite order, satisfying system (9); then, it follows that α∂2f1z1,z2/∂z12+β∂2f1z1,z2/∂z22 and α∂2f2z1,z2/∂z12+β∂2f2z1,z2/∂z22 are transcendental. Here, the following two cases will be considered:(i)
Case 1:m1m2>n1n2: in view of Lemma 2, it yields that(29)mr,fjz1,z2fjz1+c1,z2+c2=Sr,fj,j=1,2,hold for allr>0 outside of a possible exceptional set Ej⊂1,+∞ of finite logarithmic measure ∫Ejdt/t<∞. Thus, it follows from (29) that(30)Tr,fjz1,z2=mr,fjz1,z2≤mr,fjz1,z2fz1+c1,z2+c2+mr,fjz1+c1,z2+c2+log2,=mr,fjz1+c1,z2+c2+Sr,fj,=Tr,fjz1+c1,z2+c2+Sr,fj,j=1,2,for allr∉E=:E1∪E2. In view of (30), Lemma 1, and the Mokhon’ko theorem in several complex variables ([27], Theorem 3.4), it yields that(31)m1Tr,f2z1,z2≤m1Tr,f2z1+c1,z2+c2+Sr,f2,=Tr,f2z1+c1,z2+c2m1+Sr,f2,=Tr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22n1−1+Sr,f2,=n1Tr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22+Sr,f2+Sr,f1,=n1mr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22+Sr,f2+Sr,f1≤n1mr,α∂2f1z1,z2/∂z12+β∂2f1z1,z2/∂z22f1z1,z2+mr,f1z1,z2+Sr,f1+Sr,f2,=n1Tr,f1z1,z2+Sr,f1+Sr,f2,for allr∈E. Similarly, we have(32)m2Tr,f1z1,z2≤n2Tr,f2z1,z2+Sr,f1+Sr,f2,r∉E.In view of (31) and (32), it yields that(33)m1m2−n1n2Tr,fjz1,z2≤Sr,f1+Sr,f2,r∉E.Sincef1,f2 are transcendental and m1m2>n1n2, this is a contradiction.(ii)
Case 2:mj>nj/nj−1 for nj≥2, j=1,2: in view of the Nevanlinna second fundamental theorem, Lemma 2, and system (9), it follows that(34)n1−1Tr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22≤N¯r,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22+∑q=1n1N¯r,1α∂2f1z1,z2/∂z12+β∂2f1z1,z2/∂z22−wq+Sr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22≤N¯r,1α∂2f1z1,z2/∂z12+β∂2f1z1,z2/∂z22n1−1+Sr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22≤N¯r,1f2z1+c1,z2+c2+Sr,f1≤Tr,f2z1+c1,z2+c2+Sr,f1+Sr,f2,where wq is a root of wn1−1=0. Similarly, we have(35)n2−1Tr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22≤Tr,f1z1+c1,z2+c2+Sr,f1+Sr,f2.
On the other hand, by the Mokhon’ko theorem in several complex variables ([27], Theorem 3.4), it follows from system (9) that(36)m1Tr,f2z1+c1,z2+c2,=Tr,f2z1+c1,z2+c2m1+Sr,f2,=Tr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22n1−1+Sr,f2,=n1Tr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22+Sr,f1+Sr,f2.
Similarly, we have(37)m2Tr,f1z1+c1,z2+c2,=n2Tr,α∂2f1z1,z2∂z12+β∂2f1z1,z2∂z22+Sr,f1+Sr,f2.
In view of (34)–(37) and mj>nj/nj−1, it follows that(38)m1−n1n1−1Tr,f2z1+c1,z2+c2≤Sr,f1+Sr,f2,m2−n2n2−1Tr,f1z1+c1,z2+c2≤Sr,f1+Sr,f2.
It leads to a contradiction with the assumption thatf1,f2 are transcendental entire functions.
Therefore, this completes the proof of Theorem7.


## 3. Proofs of Theorems 8 and 9

The following lemma plays the key role in proving Theorems8 and 9.Lemma 3.
(See [28, 29]). For an entire function F on ℂn, F0≠0 and put ρnF=ρ<∞. Then, there exist a canonical function fF and a function gF∈ℂn such that Fz=fFzegFz. For the special case n=1, fF is the canonical product of Weierstrass.Remark 4.
Here, we denoteρnF to be the order of the counting function of zeros of F.Lemma 4.
(See [30]). If g and h are entire functions on the complex plane ℂ and gh is an entire function of finite order, then there are only two possible cases: either(a)
the internal functionh is a polynomial, and the external function g is of finite order or else(b)
the internal functionh is not a polynomial but a function of finite order, and the external function g is of zero orderLemma 5.
(See [31], Lemma 3). Let fj≡0,j=1,2,3 be meromorphic functions on ℂm such that f1 is not constant and f1+f2+f3=1 and such that(39)∑j=13N2r,1fj+2N¯r,fj<λTr,f1+Olog+Tr,f1,for all r outside possibly a set with finite logarithmic measure, where λ<1 is a positive number. Then, either f2=1 or f3=1.


### 3.1. The Proof of Theorem8

Suppose thatf1,f2 is a pair of transcendental entire solutions with finite order of system (11). System (11) can be represented as follows:(40)∂2f1z1,z2∂z12+if2z1+c1,z2+c2∂2f1z1,z2∂z12−if2z1+c1,z2+c2=1,∂2f2z1,z2∂z12+if1z1+c1,z2+c2∂2f2z1,z2∂z12−if1z1+c1,z2+c2=1.Sincef1,f2 are transcendental entire functions with finite order, then by (40), we can see that the functions(41)∂2f1z1,z2∂z12+if2z1+c1,z2+c2,∂2f1z1,z2∂z12−if2z1+c1,z2+c2,∂2f2z1,z2∂z12+if1z1+c1,z2+c2,∂2f2z1,z2∂z12−if1z1+c1,z2+c2,have no any zeros and poles. Moreover, by Lemmas 3 and 4, we have that there exist two polynomials p1z,p2z such that(42)∂2f1z1,z2∂z12+if2z1+c1,z2+c2=eip1z1+c1,z2+c2,∂2f1z1,z2∂z12−if2z1+c1,z2+c2=e−ip1z1+c1,z2+c2,∂2f2z1,z2∂z12+if1z1+c1,z2+c2=eip2z1+c1,z2+c2,∂2f2z1,z2∂z12−if1z1+c1,z2+c2=e−ip2z1+c1,z2+c2.In view of (42), it yields that(43)∂2f1z1,z2∂z12=eip1z+e−ip1z2,f2z+c=eip1z−e−ip1z2i,∂2f2z1,z2∂z12=eip2z+e−ip2z2,f1z+c=eip2z−e−ip2z2i,which implies(44)∂2p2∂z12+i∂p2∂z12eip1z+c+p2z+∂2p2∂z12−i∂p2∂z12eip1z+c−p2z−e2ip1z+c≡1,(45)∂2p1∂z12+i∂p1∂z12eip2z+c+p1z+∂2p1∂z12−i∂p1∂z12eip2z+c−p1z−e2ip2z+c≡1.Now, we claim that∂2p2z1,z2/∂z12−i∂p2z1,z2/∂z12≡0. If ∂pz1,z2/∂z1≡0, then equation (44) becomes e2ip1z1+c1,z2+c2+1≡0, and this is impossible since p2z is a nonconstant polynomial. If ∂2p2z1,z2/∂z12−i∂p2z1,z2/∂z12≡0 and ∂p2z1,z2/∂z1≡0, then ∂u/∂z1=iu2, where u=∂p2z1,z2/∂z1. Solving this equation, we have −1/u=iz1+φ1z2, that is, ∂p2z1,z2/∂z1=u=−1/iz1+φ1z2, where φ1z2 is a polynomial in z2. Thus, it follows that p2z1,z2=ilogiz1+φ1z2+φ2z2, where φ2z2 is a polynomial in z2. This is a contradiction with the assumption of p2z1,z2 being a nonconstant polynomial. Hence, ∂2p2z1,z2/∂z12−i∂p2z1,z2/∂z12≡0. Similarly, we have ∂2p2z1,z2/∂z12+i∂p2z1,z2/∂z12≡0 and ∂2p2z1,z2/∂z12±i∂p2z1,z2/∂z12≡0.In view of Lemma5 and (44) and (45), it follows that(46)∂2p2∂z12+i∂p2∂z12eip1z+c+p2z≡1or∂2p2∂z12−i∂p2∂z12eip1z+c−p2z≡1,∂2p1∂z12+i∂p1∂z12eip2z+c+p1z≡1or∂2p1∂z12−i∂p1∂z12eip2z+c−p1z≡1.Now, the four cases will be taken into account below.(i)
Case 1.(47)∂2p2∂z12+i∂p2∂z12eip1z+c+p2z≡1,∂2p1∂z12+i∂p1∂z12eip2z+c+p1z≡1.Sincep1z,p2z are polynomials, from (47), it follows that p1z+c+p2z≡C1 and p2z+c+p1z≡C2, and here and below, C1,C2 are constants. Thus, it yields that p1z+2c−p1z≡C1−C2 and p2z+2c−p2z≡C2−C1. Hence, we have that p1z=Lz+Hz+B1,p2z=−Lz−Hz+B2, where L is a linear function of the form Lz=a1z1+a2z2, a1≠0,a2,B1,B2 are constants, and Hz:=Hs, Hs is a polynomial in s in ℂ, s=c2z1−c1z2.Here, we will prove thatHz≡0. If degsH=n, equation (47) implies(48)d2Hds2+idHds2≡ζ0,that is,(49)d2Hds2≡ζ0−idHds2,whereζ0∈ℂ. By comparing the degree of s in both sides of the abovementioned equation, we have n−2=2n−1, that is, n=0. Thus, the form of Lz+Hz+B is still the linear form of A1z1+A2z2+B, which means that Hz≡0. Thus, this means that ∂2p1/∂z12≡∂2p2/∂z12≡0. Substituting these into (47), we have(50)ia12eiLc+iB1+B2≡1,ia12e−iLc+iB1+B2≡1.In addition, in view of (44)–(47), it follows that(51)∂2p2∂z12−i∂p2∂z12ei−p1z+c−p2z≡1,∂2p1∂z12−i∂p1∂z12ei−p2z+c−p1z≡1,which means that(52)−ia12e−iLc−iB1+B2≡1,−ia12eiLc−iB1+B2≡1.Thus, we can deduce from (50) and (52) that(53)a14=1,e2iLc=1,a12eiLc+iB1+B2≡−i.In view of (43), f1,f2 are of the forms(54)f1z=ei−Lz+B2+iLc−eiLz−B2−iLc2i,(55)f2z=eiLz+B1−iLc−ei−Lz−B1+iLc2i.Ifa12=1 and eiLc=1, then Lc=2kπ and eiB1+B2=−i. Thus, it follows from (54) and (55) that(56)f1z1,z2=−eiLz−B2+ei−Lz+B22i=−sinLz+B0,whereB0=−B2, and(57)f2z1,z2=eiLz+B1−ei−Lz−B12i=eiLz−B2eiB1+B2−ei−Lz+B2e−iB1+B22i,=−eiLz−B2+ei−Lz+B22=−cosLz+B0.Ifa12=1 and eiLc=−1, then Lc=2k+1π and eiB1+B2=i. Thus, it follows from (54) and (55) that(58)f1z1,z2=eiLz−B2−ei−Lz+B22i=sinLz+B0,f2z1,z2=−eiLz+B1+ei−Lz−B12i=−eiLz−B2eiB1+B2+ei−Lz+B2e−iB1+B22i,=−eiLz−B2+ei−Lz+B22=−cosLz+B0.Ifa12=−1 and eiLc=1, then Lc=2kπ and eiB1+B2=i. Thus, it follows from (54) and (55) that(59)f1z1,z2=−sinLz+B0,f2z1,z2=cosLz+B0.Ifa12=−1 and eiLc=−1, then Lc=2k+1π and eiB1+B2=−i. Thus, it follows from (54) and (55) that(60)f1z1,z2=sinLz+B0,f2z1,z2=cosLz+B0.(ii)
Case 2.(61)∂2p2∂z12+i∂p2∂z12eip1z+c+p2z≡1,∂2p1∂z12−i∂p1∂z12eip2z+c−p1z≡1.Sincep1z,p2z are polynomials, from (61), it follows that p1z+c+p2z≡C1 and p2z+c−p1z≡C2, which imply that p2z+2c+p2z≡C1+C2, and this is a contradiction with the condition of p2z being a nonconstant polynomial.(iii)
Case 3.(62)∂2p2∂z12−i∂p2∂z12eip1z+c−p2z≡1,∂2p1∂z12+i∂p1∂z12eip2z+c+p1z≡1.Sincep1z,p2z are polynomials, from (62), it follows that p1z+c−p2z≡C1 and p2z+c+p1z≡C2, which imply that p1z+2c+p1z≡C1+C2, and this is also a contradiction.(iv)
Case 4.(63)∂2p2∂z12−i∂p2∂z12eip1z+c−p2z≡1,∂2p1∂z12−i∂p1∂z12eip2z+c−p1z≡1.Sincep1z,p2z are polynomials, then from (63), it follows that p1z+c−p2z≡C1 and p2z+c−p1z≡C2. This means that p1z+2c−p1z≡C1+C2 and p2z+2c−p2z≡C1+C2. Similar to the argument as in case 1 in Theorem 8, we can deduce that p1z=Lz+B1,p2z=Lz+B2, where L is a linear function of the form Lz=a1z1+a2z2, a1≠0,a2,B1,B2 are constants. Hence, it follows that ∂2p1/∂z12≡∂2p2/∂z12≡0. Substituting these into (63), we have(64)−ia12eiLc+iB1−B2≡1,−ia12eiLc+iB2−B1≡1.In addition, in view of (44)–(47), it follows that(65)∂2p2∂z12+i∂p2∂z12ei−p1z+c+p2z≡1,∂2p1∂z12+i∂p1∂z12ei−p2z+c+p1z≡1,which means that(66)ia12e−iLc+iB2−B1≡1,ia12e−iLc+iB1−B2≡1.Thus, we can deduce from (63) and (64) that(67)a14=1,e2iLc=−1,a12eiLc+iB1−B2≡i.In view of (43), f1,f2 are of the forms(68)f1z=eiLz+B2−iLc−ei−Lz−B2+iLc2i,(69)f2z=eiLz+B1−iLc−ei−Lz−B1+iLc2i.Ifa12=1 and eiLc=i, then Lc=2k+1/2π and eiB1−B2=1. Thus, it follows from (68) and (69) that(70)f1z1,z2=−ieiLz+B2−iei−Lz−B22i=−cosLz+B0,where B0=B2, and(71)f2z1,z2=−ieiLz+B1−iei−Lz−B12i=−ieiLz+B2eiB1−B2−iei−Lz−B2eiB2−B12i=−eiLz+B2+ei−Lz−B22=−cosLz+B0.Ifa12=1 and eiLc=−i, then Lc=2k−1/2π and eiB1−B2=−1. Thus, it follows from (68) and (69) that(72)f1z1,z2=cosLz+B0,f2z1,z2=−cosLz+B0.Ifa12=−1 and eiLc=i, then Lc=2k+1/2π and eiB1−B2=−1. Thus, it follows from (68) and (69) that(73)f1z1,z2=−cosLz+B0,f2z1,z2=cosLz+B0.Ifa12=−1 and eiLc=−i, then Lc=2k−1/2π and eiB1−B2=1. Thus, it follows from (68) and (69) that(74)f1z1,z2=cosLz+B0,f2z1,z2=cosLz+B0.Thus, in view of Cases 1–4, this completes the proof of Theorem8.


### 3.2. The Proof  of  Theorem 9

Suppose thatf1,f2 is a pair of transcendental entire solutions with finite order of system (18). System (18) can be represented as follows:(75)∂2f1z∂z12+∂2f1z∂z1∂z2+if2z+c∂2f1z∂z12+∂2f1z∂z1∂z2−if2z+c=1,∂2f2z∂z12+∂2f2z∂z1∂z2+if1z+c∂2f2z∂z12+∂2f2z∂z1∂z2−if1z+c=1.Sincef1,f2 are transcendental entire functions with finite order, then by (75), we can see that the functions(76)∂2f1z∂z12+∂2f1z∂z1∂z2+if2z+c,∂2f1z∂z12+∂2f1z∂z1∂z2−if2z+c,∂2f2z∂z12+∂2f2z∂z1∂z2+if1z+c,∂2f2z∂z12+∂2f2z∂z1∂z2−if1z+c,have no any zeros and poles. Moreover, by Lemmas 3 and 4, we have that there exist two polynomials p1z,p2z such that(77)∂2f1z1,z2∂z12+∂2f1z1,z2∂z1∂z2+if2z1+c1,z2+c2=eip1z1+c1,z2+c2,∂2f1z1,z2∂z12+∂2f1z1,z2∂z1∂z2−if2z1+c1,z2+c2=e−ip1z1+c1,z2+c2,∂2f2z1,z2∂z12+∂2f2z1,z2∂z1∂z2+if1z1+c1,z2+c2=eip2z1+c1,z2+c2,∂2f2z1,z2∂z12+∂2f2z1,z2∂z1∂z2−if1z1+c1,z2+c2=e−ip2z1+c1,z2+c2.In view of (77), it yields that(78)∂2f1z∂z12+∂2f1z∂z1∂z2=eip1z+e−ip1z2,f2z+c=eip1z−e−ip1z2i,∂2f2z∂z12+∂2f2z∂z1∂z2=eip2z+e−ip2z2,f1z+c=eip2z−e−ip2z2i,which implies(79)P21+iP22eip1z+c+p2z+P21−iP22eip1z+c−p2z−e2ip1z+c≡1,(80)P11+iP12eip2z+c+p1z+P11−iP12eip2z+c−p1z−e2ip2z+c≡1,where(81)P11=∂2p1∂z12+∂2p1∂z1∂z2,P12=∂p1∂z12+∂p1∂z1∂p1∂z2,P21=∂2p2∂z12+∂2p2∂z1∂z2,P22=∂p2∂z12+∂p2∂z1∂p2∂z2.Now, we claim thatP21−iP22≡0. If P21−iP22≡0; then, equation (79) becomes P21+iP22eip1z+c+p2z−e2ip1z1+c1,z2+c2≡1. If P21+iP22≡0, then it yields e2ip1z1+c1,z2+c2≡−1, and this is a contradiction with the condition of p1 being a nonconstant polynomial. If P21+iP22≡0, we have(82)P21+iP22eip1z+c+p2z≡e2ip1z1+c1,z2+c2+1.By making use of the Mokhon’ko theorem in several complex variables ([27], Theorem 3.4), in view of (82), it follows that(83)Tr,e2ip1z1+c1,z2+c2=Tr,P21+iP22eip1z+c+p2z+O1.In view of the Nevanlinna second fundamental theorem, (82) and (83), it follows that(84)Tr,e2ip2z1+c1,z2+c2≤Nr,1e2ip2z1+c1,z2+c2+Nr,1e2ip2z1+c1,z2+c2+1+Sr,e2ip2z1+c1,z2+c2≤Nr,1P21+iP22eip1z+c+p2z+Sr,e2ip2z1+c1,z2+c2≤Nr,1P21+iP22+Sr,e2ip2z1+c1,z2+c2≤OTr,p2+Sr,e2ip2z1+c1,z2+c2,outside possibly a set of finite Lebesgue measure. This is a contradiction with the fact that(85)limr⟶+∞Tr,e2ip2Tr,p2=+∞,for p2z being a nonconstant polynomial. Hence, P21−iP22≡0. Similarly, we have P21+iP22≡0 and P11+iP12≡0.In view of Lemma5 and (79) and (80), it follows that(86)P21+iP22eip1z+c+p2z≡1orP21−iP22eip1z+c−p2z≡1,P11+iP12eip2z+c+p1z≡1orP11−iP12eip2z+c−p1z≡1.Now, we will consider the four cases below.Case 1.(87)P21+iP22eip1z+c+p2z≡1,P11+iP12eip2z+c+p1z≡1.Sincep1z,p2z are polynomials, from (87), it follows that p1z+c+p2z≡C1 and p2z+c+p1z≡C2. Thus, it yields that p1z+2c−p1z≡C1−C2 and p2z+2c−p2z≡C2−C1. Hence, similar to the argument as in case 1 of Theorem 8, we have that p1z=Lz+B1,p2z=−Lz+B2, where L is a linear function of the form Lz=a1z1+a2z2, a1≠0,a1≠−a2,B1,B2 are constants, which means that ∂2p1/∂z12≡∂2p2/∂z12≡∂2p2/∂z1∂z2≡0. Substituting these into (87), we have(88)ia1a1+a2eiLc+iB1+B2≡1,ia1a1+a2e−iLc+iB1+B2≡1.In addition, in view of (79)–(87), it follows that(89)P21−iP22ei−p1z+c−p2z≡1,P11−iP12ei−p2z+c−p1z≡1,which means that(90)−ia1a1+a2e−iLc−iB1+B2≡1,−ia1a1+a2eiLc−iB1+B2≡1.Thus, we can deduce from (88) and (90) that(91)a12a1+a22=1,e2iLc=1,a1a1+a2eiLc+iB1+B2≡−i.Similar to the argument as in the proof of Theorem8 and by combining with (91), we have that f1,f2 is of the form(92)f1,f2=±sinLz+B0,±cosLz+B0.Case 2.(93)P21+iP22eip1z+c+p2z≡1,P11−iP12eip2z+c−p1z≡1.Sincep1z,p2z are polynomials, from (93), it follows that p1z+c+p2z≡C1 and p2z+c−p1z≡C2, which imply that p2z+2c+p2z≡C1+C2, and this is a contradiction with the condition of p2z being a nonconstant polynomial.Case 3.(94)P21−iP22eip1z+c−p2z≡1,P11+iP12eip2z+c+p1z≡1.Sincep1z,p2z are polynomials, then from (94), it follows that p1z+c−p2z≡C1 and p2z+c+p1z≡C2, which imply that p1z+2c+p1z≡C1+C2, and this is also a contradiction.Case 4.(95)P21−iP22eip1z+c−p2z≡1,P11−iP12eip2z+c−p1z≡1.Sincep1z,p2z are polynomials, from (95), it follows that p1z+c−p2z≡C1 and p2z+c−p1z≡C2. This means that p1z+2c−p1z≡C1+C2 and p2z+2c−p2z≡C1+C2. Thus, similar to the argument as in case 1 of Theorem 2, we can deduce that p1z=Lz+B1,p2z=Lz+B2, where L is a linear function of the form Lz=a1z1+a2z2, a1≠0,a1≠−a2,B1,B2 are constants. Hence, it follows that ∂2p1/∂z12≡∂2p2/∂z12≡∂2p2/∂z1∂z2≡0. Substituting these into (95), we have(96)−ia1a1+a2eiLc+iB1−B2≡1,−ia1a1+a2eiLc+iB2−B1≡1.In addition, in view of (80)–(87), it follows that(97)P21+iP22ei−p1z+c+p2z≡1,P11+iP12ei−p2z+c+p1z≡1,which means that(98)ia1a1+a2e−iLc+iB2−B1≡1,ia1a1+a2e−iLc+iB1−B2≡1.Thus, we can deduce from (96) and (98) that(99)a12a1+a22=1,e2iLc=−1,a1a1+a2eiLc+iB1−B2≡i.Similar to the argument as in the proof of Theorem8, and by combining with (99), we can deduce that f1,f2 is of the form(100)f1,f2=±cosLz+B0,±cosLz+B0.Thus, in view of cases 1–4, this completes the proof of Theorem8.


## 3.1. The Proof of Theorem8

Suppose thatf1,f2 is a pair of transcendental entire solutions with finite order of system (11). System (11) can be represented as follows:(40)∂2f1z1,z2∂z12+if2z1+c1,z2+c2∂2f1z1,z2∂z12−if2z1+c1,z2+c2=1,∂2f2z1,z2∂z12+if1z1+c1,z2+c2∂2f2z1,z2∂z12−if1z1+c1,z2+c2=1.Sincef1,f2 are transcendental entire functions with finite order, then by (40), we can see that the functions(41)∂2f1z1,z2∂z12+if2z1+c1,z2+c2,∂2f1z1,z2∂z12−if2z1+c1,z2+c2,∂2f2z1,z2∂z12+if1z1+c1,z2+c2,∂2f2z1,z2∂z12−if1z1+c1,z2+c2,have no any zeros and poles. Moreover, by Lemmas 3 and 4, we have that there exist two polynomials p1z,p2z such that(42)∂2f1z1,z2∂z12+if2z1+c1,z2+c2=eip1z1+c1,z2+c2,∂2f1z1,z2∂z12−if2z1+c1,z2+c2=e−ip1z1+c1,z2+c2,∂2f2z1,z2∂z12+if1z1+c1,z2+c2=eip2z1+c1,z2+c2,∂2f2z1,z2∂z12−if1z1+c1,z2+c2=e−ip2z1+c1,z2+c2.In view of (42), it yields that(43)∂2f1z1,z2∂z12=eip1z+e−ip1z2,f2z+c=eip1z−e−ip1z2i,∂2f2z1,z2∂z12=eip2z+e−ip2z2,f1z+c=eip2z−e−ip2z2i,which implies(44)∂2p2∂z12+i∂p2∂z12eip1z+c+p2z+∂2p2∂z12−i∂p2∂z12eip1z+c−p2z−e2ip1z+c≡1,(45)∂2p1∂z12+i∂p1∂z12eip2z+c+p1z+∂2p1∂z12−i∂p1∂z12eip2z+c−p1z−e2ip2z+c≡1.Now, we claim that∂2p2z1,z2/∂z12−i∂p2z1,z2/∂z12≡0. If ∂pz1,z2/∂z1≡0, then equation (44) becomes e2ip1z1+c1,z2+c2+1≡0, and this is impossible since p2z is a nonconstant polynomial. If ∂2p2z1,z2/∂z12−i∂p2z1,z2/∂z12≡0 and ∂p2z1,z2/∂z1≡0, then ∂u/∂z1=iu2, where u=∂p2z1,z2/∂z1. Solving this equation, we have −1/u=iz1+φ1z2, that is, ∂p2z1,z2/∂z1=u=−1/iz1+φ1z2, where φ1z2 is a polynomial in z2. Thus, it follows that p2z1,z2=ilogiz1+φ1z2+φ2z2, where φ2z2 is a polynomial in z2. This is a contradiction with the assumption of p2z1,z2 being a nonconstant polynomial. Hence, ∂2p2z1,z2/∂z12−i∂p2z1,z2/∂z12≡0. Similarly, we have ∂2p2z1,z2/∂z12+i∂p2z1,z2/∂z12≡0 and ∂2p2z1,z2/∂z12±i∂p2z1,z2/∂z12≡0.In view of Lemma5 and (44) and (45), it follows that(46)∂2p2∂z12+i∂p2∂z12eip1z+c+p2z≡1or∂2p2∂z12−i∂p2∂z12eip1z+c−p2z≡1,∂2p1∂z12+i∂p1∂z12eip2z+c+p1z≡1or∂2p1∂z12−i∂p1∂z12eip2z+c−p1z≡1.Now, the four cases will be taken into account below.(i)
Case 1.(47)∂2p2∂z12+i∂p2∂z12eip1z+c+p2z≡1,∂2p1∂z12+i∂p1∂z12eip2z+c+p1z≡1.Sincep1z,p2z are polynomials, from (47), it follows that p1z+c+p2z≡C1 and p2z+c+p1z≡C2, and here and below, C1,C2 are constants. Thus, it yields that p1z+2c−p1z≡C1−C2 and p2z+2c−p2z≡C2−C1. Hence, we have that p1z=Lz+Hz+B1,p2z=−Lz−Hz+B2, where L is a linear function of the form Lz=a1z1+a2z2, a1≠0,a2,B1,B2 are constants, and Hz:=Hs, Hs is a polynomial in s in ℂ, s=c2z1−c1z2.Here, we will prove thatHz≡0. If degsH=n, equation (47) implies(48)d2Hds2+idHds2≡ζ0,that is,(49)d2Hds2≡ζ0−idHds2,whereζ0∈ℂ. By comparing the degree of s in both sides of the abovementioned equation, we have n−2=2n−1, that is, n=0. Thus, the form of Lz+Hz+B is still the linear form of A1z1+A2z2+B, which means that Hz≡0. Thus, this means that ∂2p1/∂z12≡∂2p2/∂z12≡0. Substituting these into (47), we have(50)ia12eiLc+iB1+B2≡1,ia12e−iLc+iB1+B2≡1.In addition, in view of (44)–(47), it follows that(51)∂2p2∂z12−i∂p2∂z12ei−p1z+c−p2z≡1,∂2p1∂z12−i∂p1∂z12ei−p2z+c−p1z≡1,which means that(52)−ia12e−iLc−iB1+B2≡1,−ia12eiLc−iB1+B2≡1.Thus, we can deduce from (50) and (52) that(53)a14=1,e2iLc=1,a12eiLc+iB1+B2≡−i.In view of (43), f1,f2 are of the forms(54)f1z=ei−Lz+B2+iLc−eiLz−B2−iLc2i,(55)f2z=eiLz+B1−iLc−ei−Lz−B1+iLc2i.Ifa12=1 and eiLc=1, then Lc=2kπ and eiB1+B2=−i. Thus, it follows from (54) and (55) that(56)f1z1,z2=−eiLz−B2+ei−Lz+B22i=−sinLz+B0,whereB0=−B2, and(57)f2z1,z2=eiLz+B1−ei−Lz−B12i=eiLz−B2eiB1+B2−ei−Lz+B2e−iB1+B22i,=−eiLz−B2+ei−Lz+B22=−cosLz+B0.Ifa12=1 and eiLc=−1, then Lc=2k+1π and eiB1+B2=i. Thus, it follows from (54) and (55) that(58)f1z1,z2=eiLz−B2−ei−Lz+B22i=sinLz+B0,f2z1,z2=−eiLz+B1+ei−Lz−B12i=−eiLz−B2eiB1+B2+ei−Lz+B2e−iB1+B22i,=−eiLz−B2+ei−Lz+B22=−cosLz+B0.Ifa12=−1 and eiLc=1, then Lc=2kπ and eiB1+B2=i. Thus, it follows from (54) and (55) that(59)f1z1,z2=−sinLz+B0,f2z1,z2=cosLz+B0.Ifa12=−1 and eiLc=−1, then Lc=2k+1π and eiB1+B2=−i. Thus, it follows from (54) and (55) that(60)f1z1,z2=sinLz+B0,f2z1,z2=cosLz+B0.(ii)
Case 2.(61)∂2p2∂z12+i∂p2∂z12eip1z+c+p2z≡1,∂2p1∂z12−i∂p1∂z12eip2z+c−p1z≡1.Sincep1z,p2z are polynomials, from (61), it follows that p1z+c+p2z≡C1 and p2z+c−p1z≡C2, which imply that p2z+2c+p2z≡C1+C2, and this is a contradiction with the condition of p2z being a nonconstant polynomial.(iii)
Case 3.(62)∂2p2∂z12−i∂p2∂z12eip1z+c−p2z≡1,∂2p1∂z12+i∂p1∂z12eip2z+c+p1z≡1.Sincep1z,p2z are polynomials, from (62), it follows that p1z+c−p2z≡C1 and p2z+c+p1z≡C2, which imply that p1z+2c+p1z≡C1+C2, and this is also a contradiction.(iv)
Case 4.(63)∂2p2∂z12−i∂p2∂z12eip1z+c−p2z≡1,∂2p1∂z12−i∂p1∂z12eip2z+c−p1z≡1.Sincep1z,p2z are polynomials, then from (63), it follows that p1z+c−p2z≡C1 and p2z+c−p1z≡C2. This means that p1z+2c−p1z≡C1+C2 and p2z+2c−p2z≡C1+C2. Similar to the argument as in case 1 in Theorem 8, we can deduce that p1z=Lz+B1,p2z=Lz+B2, where L is a linear function of the form Lz=a1z1+a2z2, a1≠0,a2,B1,B2 are constants. Hence, it follows that ∂2p1/∂z12≡∂2p2/∂z12≡0. Substituting these into (63), we have(64)−ia12eiLc+iB1−B2≡1,−ia12eiLc+iB2−B1≡1.In addition, in view of (44)–(47), it follows that(65)∂2p2∂z12+i∂p2∂z12ei−p1z+c+p2z≡1,∂2p1∂z12+i∂p1∂z12ei−p2z+c+p1z≡1,which means that(66)ia12e−iLc+iB2−B1≡1,ia12e−iLc+iB1−B2≡1.Thus, we can deduce from (63) and (64) that(67)a14=1,e2iLc=−1,a12eiLc+iB1−B2≡i.In view of (43), f1,f2 are of the forms(68)f1z=eiLz+B2−iLc−ei−Lz−B2+iLc2i,(69)f2z=eiLz+B1−iLc−ei−Lz−B1+iLc2i.Ifa12=1 and eiLc=i, then Lc=2k+1/2π and eiB1−B2=1. Thus, it follows from (68) and (69) that(70)f1z1,z2=−ieiLz+B2−iei−Lz−B22i=−cosLz+B0,where B0=B2, and(71)f2z1,z2=−ieiLz+B1−iei−Lz−B12i=−ieiLz+B2eiB1−B2−iei−Lz−B2eiB2−B12i=−eiLz+B2+ei−Lz−B22=−cosLz+B0.Ifa12=1 and eiLc=−i, then Lc=2k−1/2π and eiB1−B2=−1. Thus, it follows from (68) and (69) that(72)f1z1,z2=cosLz+B0,f2z1,z2=−cosLz+B0.Ifa12=−1 and eiLc=i, then Lc=2k+1/2π and eiB1−B2=−1. Thus, it follows from (68) and (69) that(73)f1z1,z2=−cosLz+B0,f2z1,z2=cosLz+B0.Ifa12=−1 and eiLc=−i, then Lc=2k−1/2π and eiB1−B2=1. Thus, it follows from (68) and (69) that(74)f1z1,z2=cosLz+B0,f2z1,z2=cosLz+B0.Thus, in view of Cases 1–4, this completes the proof of Theorem8.


## 3.2. The Proof  of  Theorem 9

Suppose thatf1,f2 is a pair of transcendental entire solutions with finite order of system (18). System (18) can be represented as follows:(75)∂2f1z∂z12+∂2f1z∂z1∂z2+if2z+c∂2f1z∂z12+∂2f1z∂z1∂z2−if2z+c=1,∂2f2z∂z12+∂2f2z∂z1∂z2+if1z+c∂2f2z∂z12+∂2f2z∂z1∂z2−if1z+c=1.Sincef1,f2 are transcendental entire functions with finite order, then by (75), we can see that the functions(76)∂2f1z∂z12+∂2f1z∂z1∂z2+if2z+c,∂2f1z∂z12+∂2f1z∂z1∂z2−if2z+c,∂2f2z∂z12+∂2f2z∂z1∂z2+if1z+c,∂2f2z∂z12+∂2f2z∂z1∂z2−if1z+c,have no any zeros and poles. Moreover, by Lemmas 3 and 4, we have that there exist two polynomials p1z,p2z such that(77)∂2f1z1,z2∂z12+∂2f1z1,z2∂z1∂z2+if2z1+c1,z2+c2=eip1z1+c1,z2+c2,∂2f1z1,z2∂z12+∂2f1z1,z2∂z1∂z2−if2z1+c1,z2+c2=e−ip1z1+c1,z2+c2,∂2f2z1,z2∂z12+∂2f2z1,z2∂z1∂z2+if1z1+c1,z2+c2=eip2z1+c1,z2+c2,∂2f2z1,z2∂z12+∂2f2z1,z2∂z1∂z2−if1z1+c1,z2+c2=e−ip2z1+c1,z2+c2.In view of (77), it yields that(78)∂2f1z∂z12+∂2f1z∂z1∂z2=eip1z+e−ip1z2,f2z+c=eip1z−e−ip1z2i,∂2f2z∂z12+∂2f2z∂z1∂z2=eip2z+e−ip2z2,f1z+c=eip2z−e−ip2z2i,which implies(79)P21+iP22eip1z+c+p2z+P21−iP22eip1z+c−p2z−e2ip1z+c≡1,(80)P11+iP12eip2z+c+p1z+P11−iP12eip2z+c−p1z−e2ip2z+c≡1,where(81)P11=∂2p1∂z12+∂2p1∂z1∂z2,P12=∂p1∂z12+∂p1∂z1∂p1∂z2,P21=∂2p2∂z12+∂2p2∂z1∂z2,P22=∂p2∂z12+∂p2∂z1∂p2∂z2.Now, we claim thatP21−iP22≡0. If P21−iP22≡0; then, equation (79) becomes P21+iP22eip1z+c+p2z−e2ip1z1+c1,z2+c2≡1. If P21+iP22≡0, then it yields e2ip1z1+c1,z2+c2≡−1, and this is a contradiction with the condition of p1 being a nonconstant polynomial. If P21+iP22≡0, we have(82)P21+iP22eip1z+c+p2z≡e2ip1z1+c1,z2+c2+1.By making use of the Mokhon’ko theorem in several complex variables ([27], Theorem 3.4), in view of (82), it follows that(83)Tr,e2ip1z1+c1,z2+c2=Tr,P21+iP22eip1z+c+p2z+O1.In view of the Nevanlinna second fundamental theorem, (82) and (83), it follows that(84)Tr,e2ip2z1+c1,z2+c2≤Nr,1e2ip2z1+c1,z2+c2+Nr,1e2ip2z1+c1,z2+c2+1+Sr,e2ip2z1+c1,z2+c2≤Nr,1P21+iP22eip1z+c+p2z+Sr,e2ip2z1+c1,z2+c2≤Nr,1P21+iP22+Sr,e2ip2z1+c1,z2+c2≤OTr,p2+Sr,e2ip2z1+c1,z2+c2,outside possibly a set of finite Lebesgue measure. This is a contradiction with the fact that(85)limr⟶+∞Tr,e2ip2Tr,p2=+∞,for p2z being a nonconstant polynomial. Hence, P21−iP22≡0. Similarly, we have P21+iP22≡0 and P11+iP12≡0.In view of Lemma5 and (79) and (80), it follows that(86)P21+iP22eip1z+c+p2z≡1orP21−iP22eip1z+c−p2z≡1,P11+iP12eip2z+c+p1z≡1orP11−iP12eip2z+c−p1z≡1.Now, we will consider the four cases below.Case 1.(87)P21+iP22eip1z+c+p2z≡1,P11+iP12eip2z+c+p1z≡1.Sincep1z,p2z are polynomials, from (87), it follows that p1z+c+p2z≡C1 and p2z+c+p1z≡C2. Thus, it yields that p1z+2c−p1z≡C1−C2 and p2z+2c−p2z≡C2−C1. Hence, similar to the argument as in case 1 of Theorem 8, we have that p1z=Lz+B1,p2z=−Lz+B2, where L is a linear function of the form Lz=a1z1+a2z2, a1≠0,a1≠−a2,B1,B2 are constants, which means that ∂2p1/∂z12≡∂2p2/∂z12≡∂2p2/∂z1∂z2≡0. Substituting these into (87), we have(88)ia1a1+a2eiLc+iB1+B2≡1,ia1a1+a2e−iLc+iB1+B2≡1.In addition, in view of (79)–(87), it follows that(89)P21−iP22ei−p1z+c−p2z≡1,P11−iP12ei−p2z+c−p1z≡1,which means that(90)−ia1a1+a2e−iLc−iB1+B2≡1,−ia1a1+a2eiLc−iB1+B2≡1.Thus, we can deduce from (88) and (90) that(91)a12a1+a22=1,e2iLc=1,a1a1+a2eiLc+iB1+B2≡−i.Similar to the argument as in the proof of Theorem8 and by combining with (91), we have that f1,f2 is of the form(92)f1,f2=±sinLz+B0,±cosLz+B0.Case 2.(93)P21+iP22eip1z+c+p2z≡1,P11−iP12eip2z+c−p1z≡1.Sincep1z,p2z are polynomials, from (93), it follows that p1z+c+p2z≡C1 and p2z+c−p1z≡C2, which imply that p2z+2c+p2z≡C1+C2, and this is a contradiction with the condition of p2z being a nonconstant polynomial.Case 3.(94)P21−iP22eip1z+c−p2z≡1,P11+iP12eip2z+c+p1z≡1.Sincep1z,p2z are polynomials, then from (94), it follows that p1z+c−p2z≡C1 and p2z+c+p1z≡C2, which imply that p1z+2c+p1z≡C1+C2, and this is also a contradiction.Case 4.(95)P21−iP22eip1z+c−p2z≡1,P11−iP12eip2z+c−p1z≡1.Sincep1z,p2z are polynomials, from (95), it follows that p1z+c−p2z≡C1 and p2z+c−p1z≡C2. This means that p1z+2c−p1z≡C1+C2 and p2z+2c−p2z≡C1+C2. Thus, similar to the argument as in case 1 of Theorem 2, we can deduce that p1z=Lz+B1,p2z=Lz+B2, where L is a linear function of the form Lz=a1z1+a2z2, a1≠0,a1≠−a2,B1,B2 are constants. Hence, it follows that ∂2p1/∂z12≡∂2p2/∂z12≡∂2p2/∂z1∂z2≡0. Substituting these into (95), we have(96)−ia1a1+a2eiLc+iB1−B2≡1,−ia1a1+a2eiLc+iB2−B1≡1.In addition, in view of (80)–(87), it follows that(97)P21+iP22ei−p1z+c+p2z≡1,P11+iP12ei−p2z+c+p1z≡1,which means that(98)ia1a1+a2e−iLc+iB2−B1≡1,ia1a1+a2e−iLc+iB1−B2≡1.Thus, we can deduce from (96) and (98) that(99)a12a1+a22=1,e2iLc=−1,a1a1+a2eiLc+iB1−B2≡i.Similar to the argument as in the proof of Theorem8, and by combining with (99), we can deduce that f1,f2 is of the form(100)f1,f2=±cosLz+B0,±cosLz+B0.Thus, in view of cases 1–4, this completes the proof of Theorem8.



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*Source: 4207579-2021-01-22.xml*