Research article

Extremal orders and races between palindromes in different bases

  • Received: 23 August 2021 Accepted: 07 November 2021 Published: 10 November 2021
  • MSC : 11A63, 11N37, 11N56

  • Let b2 and n1 be integers. Then n is said to be a palindrome in base b (or b-adic palindrome) if the representation of n in base b reads the same backward as forward. Let Ab(n) be the number of b-adic palindromes less than or equal to n. In this article, we obtain extremal orders of Ab(n). We also study the comparison between the number of palindromes in different bases and prove that if bb1, then Ab(n)Ab1(n) changes signs infinitely often as n.

    Citation: Phakhinkon Phunphayap, Prapanpong Pongsriiam. Extremal orders and races between palindromes in different bases[J]. AIMS Mathematics, 2022, 7(2): 2237-2254. doi: 10.3934/math.2022127

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  • Let b2 and n1 be integers. Then n is said to be a palindrome in base b (or b-adic palindrome) if the representation of n in base b reads the same backward as forward. Let Ab(n) be the number of b-adic palindromes less than or equal to n. In this article, we obtain extremal orders of Ab(n). We also study the comparison between the number of palindromes in different bases and prove that if bb1, then Ab(n)Ab1(n) changes signs infinitely often as n.



    We first assume that the readers are familiar with the notations of the Nevanlinna theory such as T(r,f),m(r,f),N(r,f) and so on, which can be found, for instant, in Hayman [12], Yang [31], Yi and Yang [32]. As is known to all, Nevanlinna theory is a powerful tool in analysing the properties of meromorphic functions including meromorphic functions in arbitrary plane regions, algebroid functions, functions of several variables, holomorphic curves, uniqueness theory of meromorphic function and complex differential equations, and so on. Moreover, it is also used in lots of areas of mathematics including potential theory, measure theory, differential geometry, topology and others.

    Around 2006, Chiang and Feng [6], Halburd and Korhonen [8] established independently the difference analogues of Nevanlinna theory in C, respectively. In the past two decades, many scholars paid a lot of attention to a large number of interesting topics on complex difference of meromorphic functions by making use of the difference analogue of Nevanlinna theory, and complex difference results are in the rapid development.

    In all the previous articles, Korhonen, Halburd, Chen and his students investigated the properties of solutions for a series of linear difference equations, the difference Painlevé equations, the Riccati equations, the Pielou logistic equations, and obtained the conditions of the existence and estimations of growth order for the above difference equations [1,4,5,9,27,35]; Laine, Korhonen, Yang, Chen, etc. established the difference analogue of Clunie Lemma and Mohonko Lemma, and gave the existence theorems on solutions for some complex difference equations including difference Malmquist type and some nonlinear difference equations [16,17,21]; Korhonen, Zhang, Liu, Yang paid close attention to some complex differential difference equations [11,18,19,20,24,34], in special, Liu and his co-authors [18,19,20] investigated some types of complex differential difference equations of Fermat type such as

    f(z)2+f(z+c)2=1, (1.1)
    f(z)2+[f(z+c)f(z)]2=1. (1.2)

    They proved that the transcendental entire solutions with finite order of equation (1.1) must satisfy f(z)=sin(z±Bi), where B is a constant and c=2kπ or c=(2k+1)π, k is an integer, and the transcendental entire solutions with finite order of equation (1.2) must satisfy f(z)=12sin(2z+Bi), where c=(2k+1)π, k is an integer, and B is a constant. In 2016, Gao [7] discussed the solutions for the system of complex differential-difference equations

    {[f1(z)]2+f2(z+c)2=1,[f2(z)]2+f1(z+c)2=1, (1.3)

    and extend the results given by Liu and Cao.

    Theorem 1.1. (see [7,Theorem 1.1]). Suppose that (f1,f2) is a pair of transcendental entire solutions with finite order for the system of differential-difference equations (1.3). Then (f1,f2) satisfies

    (f1(z),f2(z))=(sin(zbi),sin(zb1i)),

    or

    (f1(z),f2(z))=(sin(z+bi),sin(z+b1i)),

    where b,b1 are constants, and c=kπ, k is a integer.

    In 2012, Korhonen [14] firstly established the difference version of logarithmic derivative lemma (shortly, we may say logarithmic difference lemma) for meromorphic functions on Cm with hyper order strictly less than 23, and then used it to consider a class of partial difference equations in the same paper. Later, Cao and Korhonen [2] improved the logarithmic difference lemma to the case where the hyper order is strictly less than one. In 2019, Cao and Xu [3] further improved the logarithmic difference lemma in several variables under the condition of minimal type lim suprlogT(r,f)r=0. As far as we know, however, there are very little of results on solutions of complex partial difference equations by using Nevanlinna theory. By making use of the Nevanlinna theory with several complex variables (see [2,3,14]), Xu and Cao [3,29,30], Lu and Li [22] investigated the existence of the solutions for some Fermat type partial differential-difference equations with several variables, and obtained the following theorem.

    Theorem 1.2. (see [29,Theorem 1.2]). Let c=(c1,c2)C2. Then any transcendental entire solutions with finite order of the partial differential-difference equation

    (f(z1,z2)z1)2+f(z1+c1,z2+c2)2=1

    has the form of f(z1,z2)=sin(Az1+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 f(z1,z2)=sin(z1+B).

    In 2020, Xu, Liu and Li [28] further explored the existence and forms of several systems of complex partial differential difference equations and extend the previous results given by [29,30].

    Theorem 1.3. (see [28,Theorem 1.3]). Let c=(c1,c2)C2. Then any pair of transcendental entire solutions with finite order for the system of Fermat type partial differential-difference equations

    {(f1(z1,z2)z1)2+f2(z1+c1,z2+c2)2=1,(f2(z1,z2)z1)2+f1(z1+c1,z2+c2)2=1 (1.4)

    has the following form

    (f1(z),f2(z))=(eL(z)+B1+e(L(z)+B1)2,A21eL(z)+B1+A22e(L(z)+B1)2),

    where L(z)=a1z1+a2z2, B1 is a constant in C, and a1,c,A21,A22 satisfy one of the following cases

    (i) A21=i, A22=i, and a1=i, L(c)=(2k+12)πi, or a1=i, L(c)=(2k12)πi;

    (ii) A21=i, A22=i, and a1=i, L(c)=(2k12)πi, or a1=i, L(c)=(2k+12)πi;

    (iii) A21=1, A22=1, and a1=i, L(c)=2kπi, or a1=i, L(c)=(2k+1)πi;

    (iv) A21=1, A22=1, and a1=i, L(c)=(2k+1)πi, or a1=i, L(c)=2kπi.

    The above results suggest the following questions.

    Question 1.1. What will happen about the solutions when f(z), f(z+c), fz1 and fz2 are included in the equations in Theorem B?

    Question 1.2. What will happen about the solutions when the equations are turned into system under the hypothesis of Question 1.1?

    In view of the above questions, this article is concerned with the description of the existence and the forms of entire solutions for several partial differential difference of Fermat type. The main tools are used in this paper are the Nevanlinna theory and difference Nevanlinna theory with several complex variables. Our main results are obtained generalized the previous theorems given by Xu and Cao, Liu, Cao and Cao [20,29]. Throughout this paper, for convenience, we assume that z+w=(z1+w1,z2+w2) for any z=(z1,z2),w=(w1,w2). The main results of this paper are stated below.

    Theorem 2.1. Let c=(c1,c2)C2{(0,0)}. Then any transcendental entire solution f(z1,z2) with finite order of the partial differential equation

    f(z)2+[f(z+c)+fz1]2=1 (2.1)

    must be of the form

    f(z1,z2)=eL(z)+b+eL(z)b2,

    where L(z)=α1z1+α2z2, and L(c):=α1c1+α2c2, α1, α2,bC satisfy

    (i) L(c)=2kπi+π2i and α1=2i;

    (ii) L(c)=2kπi+3π2i and α1=0.

    The following examples show that the forms of transcendental entire solutions with finite order for equation (2.1) are precise.

    Example 2.1. Let α1=2i, α2=3π2 and bC, that is,

    f(z1,z2)=e2z2+b+e2z2b2.

    Then f(z1,z2) satisfies equation (2.1) with (c1,c2)=(c1,3πi4), where c1C.

    Example 2.2. Let α1=0, α2=2 and b=0, that is,

    f(z1,z2)=e2iz1+3π2+e2iz13π22.

    Then f(z1,z2) satisfies equation (2.1) with (c1,c2)=(π2,i).

    If equation (2.1) contains both two partial differentials and difference, we obtain

    Theorem 2.2. Let c=(c1,c2)C2{(0,0)} and s=c2z1c1z2. Then any transcendental entire solution f(z1,z2) with finite order of the partial differential equation

    f(z)2+[f(z+c)+fz1+fz2]2=1 (2.2)

    must be of the form

    f(z1,z2)=eL(z)+H(s)+b+eL(z)H(s)b2,

    where L(z)=α1z1+α2z2, H(s) is a polynomial in s, and L(c), α1, α2,bC satisfy

    (i) c1=c2=kπ+34π, α1+α2=2i and degsH2;

    (ii) c1c2, H(s)0, and L(c)=2kπi+3π2i, α1+α2=0 or L(c)=2kπi+π2i, α1+α2=2i.

    We list the following examples to show that the forms of transcendental entire solutions with finite order for equation (2.2) are precise.

    Example 2.3. Let L(z)=3iz1+iz2, H(s)=3πi4(z1z2)n, nN+ and b=0, that is,

    f(z1,z2)=eL(z)+H(s)+eL(z)H(s)2=cos[3z1+z2+3π4(z1z2)n].

    Then f(z1,z2) satisfies equation (2.2) with (c1,c2)=(3π4,3π4).

    Example 2.4. Let L(z)=3iz1+iz2, H(s)0, and b=0, that is,

    f(z1,z2)=eL(z)+eL(z)2=cos(3z1+z2).

    Then f(z1,z2) satisfies equation (2.2) with (c1,c2)=(π2,π).

    Example 2.5. Let L(z)=iz1iz2, and b=0, that is, α1+α2=0 and

    f(z1,z2)=eL(z)+eL(z)2=cos(z1z2).

    Then f(z1,z2) satisfies equation (2.2) with (c1,c2)=(2π,π2).

    When equation (2.1) is turned to system of functional equations, we have

    Theorem 2.3. Let c=(c1,c2)C2{(0,0)}. Then any pair of finite order transcendental entire solution (f1,f2) for the system of the partial differential equations

    {f1(z)2+[f2(z+c)+f1z1]2=1,f2(z)2+[f1(z+c)+f2z1]2=1 (2.3)

    must be of the form

    (f1(z),f2(z))=(eL(z)+b+eL(z)b2,A1eL(z)+b+A2eL(z)b2),

    where L(z)=α1z1+α2z2, and α1,α2,b,A1,A2C satisfy one of the following cases:

    (i) if α1=0, then α2c2=2kπi+π2i, A1=A2=1 or α2c2=2kπi+3π2i, A1=A2=1 or α2c2=2kπi, A1=i,A2=i or α2c2=(2k+1)πi, A1=i,A2=i, kZ;

    (ii) if α1=2i, then α1c1+α2c2=2kπi+π2i, A1=A2=1 or α1c1+α2c2=2kπi+3π2i, A1=A2=1, kZ.

    Some examples show the existence of transcendental entire solutions with finite order for system (2.3).

    Example 2.6. Let L(z)=iz2 and b=0, that is, α1=0, α2=i and

    (f1,f2)=(eiz2+eiz22,eiz2+eiz22)=(cosz2,cosz2).

    Then (f1,f2) satisfies system (2.3) with (c1,c2)=(c1,π2), where c1C.

    Example 2.7. Let L(z)=iz2 and b=0, that is, α1=0, α2=i and

    (f1,f2)=(eiz2+eiz22,ieiz2+ieiz22)=(cosz2,sinz2).

    Then (f1,f2) satisfies system (2.3) with (c1,c2)=(c1,2π), where c1C.

    Example 2.8. Let L(z)=2iz1+iz2 and b=0, that is, α1=2i, α2=i and

    (f1,f2)=(eL(z)+eL(z)2,eL(z)+eL(z)2)=(cos(2z1+z2),cos(2z1+z2)).

    Then (f1,f2) satisfies system (2.3) with (c1,c2)=(π2,π2).

    Similar to the argument as in the proof of Theorem 2.3, we can get the following result easily.

    Theorem 2.4. Let c=(c1,c2)C2 and c1c2. Then any pair of transcendental entire solution (f1,f2) with finite order for the system of the partial differential equations

    {f1(z)2+[f2(z+c)+f1z1+f1z2]2=1,f2(z)2+[f1(z+c)+f2z1+f2z2]2=1 (2.4)

    must be of the form

    (f1(z),f2(z))=(eL(z)+b+eL(z)b2,A1eL(z)+b+A2eL(z)b2),

    where L(z)=α1z1+α2z2, and α1,α2,b,A1,A2C satisfy one of the following cases:

    (i) if α1+α2=0, then α1c1+α2c2=2kπi+π2i, A1=A2=1 or α1c1+α2c2=2kπi+3π2i, A1=A2=1 or α1c1+α2c2=2kπi, A1=i,A2=i or α1c1+α2c2=(2k+1)πi, A1=i,A2=i, kZ;

    (ii) if α1+α2=2i, then α1c1+α2c2=2kπi+π2i, A1=A2=1 or α1c1+α2c2=2kπi+3π2i, A1=A2=1, kZ.

    The following example shows that the condition c1c2 in Theorem 2.4 can not be removed.

    Example 2.9. Let c10 and

    (f1,f2)=(eg(z1,z2)+eg(z1,z2)2,ieg(z1,z2)+ieg(z1,z2)2),

    where g(z1,z2)=z1z2+β(z1z2)n, β0 and nN+. Then (f1,f2) is a pair of transcendental entire solutions with finite order for system (2.4) with c=(c1,c1).

    The following lemmas play the key roles in proving our results.

    Lemma 3.1. ([25,26]). For an entire function F on Cn, F(0)0 and put ρ(nF)=ρ<. Then there exist a canonical function fF and a function gFCn such that F(z)=fF(z)egF(z). For the special case n=1, fF is the canonical product of Weierstrass.

    Remark 3.1. Here, denote ρ(nF) to be the order of the counting function of zeros of F.

    Lemma 3.2. ([23]). If g and h are entire functions on the complex plane C and g(h) is an entire function of finite order, then there are only two possible cases: either

    (a) the internal function h is a polynomial and the external function g is of finite order; or else

    (b) the internal function h is not a polynomial but a function of finite order, and the external function g is of zero order.

    Lemma 3.3. ([13,Lemma 3.1]). 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

    3j=1{N2(r,1fj)+2¯N(r,fj)}<λT(r,f1)+O(log+T(r,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, where N2(r,1f) is the counting function of the zeros of f in |z|r, where the simple zero is counted once, and the multiple zero is counted twice.

    Suppose that f(z) is a transcendental entire solution with finite order of equation (2.1). Thus, we rewrite (2.1) as the following form

    [f(z)+i(f(z+c)+fz1)][f(z)i(f(z+c)+fz1)]=1, (3.1)

    which implies that both f(z)+i(f(z+c)+fz1) and f(z)i(f(z+c)+fz1) have no poles and zeros. Thus, by Lemmas 3.1 and 3.2, there thus exists a polynomial p(z) such that

    f(z)+i(f(z+c)+fz1)=ep(z),f(z)i(f(z+c)+fz1)=ep(z).

    This leads to

    f(z)=ep(z)+ep(z)2, (3.2)
    f(z+c)+f(z)z1=ep(z)ep(z)2i. (3.3)

    Substituting (3.2) into (3.3), it yields that

    (pz1+i)ep(z+c)+p(z)+(pz1+i)ep(z+c)p(z)e2p(z+c)=1. (3.4)

    Obviously, pz1+i0. Otherwise, e2p(z+c)1, thus p(z) is a constant. Then f(z) is a constant, this a contradiction. By Lemma 3.3, it follows from (3.4) that

    (pz1+i)ep(z+c)p(z)=1. (3.5)

    Thus, in view of (3.4) and (3.5), we conclude that

    (pz1+i)ep(z)p(z+c)=1. (3.6)

    Equations (3.5) and (3.6) mean that e2[p(z+c)p(z)]=1. Thus, it yields that p(z)=L(z)+H(s)+b, where L is a linear function as the form L(z)=α1z1+α2z2, e2L(c)=1, and H(s) is a polynomial in s:=c2z1c1z2, α1,α2,bC.

    On the other hand, it follows from (3.5) and (3.6) that (pz1+i)2=1, that is, pz1=0 or pz1=2i. In view of (3.5) or (3.6), it yields that eL(c)=i and α1=2i or eL(c)=i and α1=0, that is, L(c)=2kπi+π2i and α1=2i or L(c)=2kπi+3π2i and α1=0. Moreover, by combining with p(z)=L(z)+H(s)+b, we have that Hc2=η, where η is a constant. If c2=0, then H(s)=H(c1z2), and combining with e2(p(z+c)p(z))=1, then it follows that degsH1. If c20, it is easy to get that H(s) is a polynomial in s with degsH1. Hence, we can conclude that L(z)+H(s)+b is a linear form in z1,z2, w.l.o.g, we denote that the form is α1z1+α2z2+b, where α1,α2,bC. Thus, by combining with (3.5) and (3.6), we have

    f(z1,z2)=eL(z)+b+eL(z)b2,

    where L(z)=α1z1+α2z2, and L(c), α1, α2,bC satisfy L(c)=2kπi+π2i and α1=2i or L(c)=2kπi+3π2i and α1=0, kZ.

    Therefore, this completes the proof of Theorem 2.1.

    Suppose that f(z) is a transcendental entire solution with finite order of equation (2.2). Similar to the argument as in the proof of Theorem 2.1, there exists a nonconstant polynomial p(z) such that (3.2) and

    (pz1+pz2+i)ep(z+c)p(z)=1, (3.7)
    (pz1+pz2+i)ep(z)p(z+c)=1. (3.8)

    In view of (3.7) and (3.8), we have e2(p(z+c)p(z))=1. Thus, it yields that p(z)=L(z)+H(s)+b, where L is a linear function as the form L(z)=α1z1+α2z2, H(s) is a polynomial in s, and α1,α2,bC, and e2L(c)=1.

    On the other hand, from (3.7) and (3.8), we have (pz1+pz2+i)2=1. Noting that p(z)=L(z)+H(s)+b, it follow that

    [α1+α2+H(c2c1)+i]2=1. (3.9)

    If c1=c2, then H(c2c1)0. This shows that H(s) can be any degree polynomial in s. In fact, we can assume that degsH2. If degsH1, then L(z)+H(s)+b is still a linear function of the form a1z1+a2z2, this means that H(s)0. In view of (3.9), it yields that α1+α2=0 or α1+α2=2i. If α1+α2=0, that is, α1=α2, combining with c1=c2, then it leads to L(c)=α1c1α1c1=0, that is, e2L(c)=1, this is a contradiction with e2L(c)=1. If α1+α2=2i, then L(c)=(α1+α2)c1=2ic1. By combining with eL(c)=i, we have c1=c2=kπ+34π.

    If c1c2, then H must be a constant, that is, degsH1. Hence, we can conclude that L(z)+H(s)+b is a linear form in z1,z2. Thus, this means that H(s)0 and H(c2c1)0. Hence, we have p(z)=α1z1+α2z2+b, where α1,α2,bC. So, in view of (3.9), we also have α1+α2=0 or α1+α2=2i. Further, In view of (3.7) or (3.8), we conclude that L(c)=2kπi+π2i and α1+α2=2i or L(c)=2kπi+3π2i and α1+α2=0, kZ.

    Therefore, this completes the proof of Theorem 2.2.

    Suppose that (f1(z),f2(z)) is a pair of transcendental entire solutions with finite order of (2.3). By using the same argument as in the proof of Theorem 2.1, there exist two polynomials p(z),q(z) in C2 such that

    {f1(z)=ep(z)+ep(z)2,f2(z+c)+f1z1=ep(z)ep(z)2i,f2(z)=eq(z)+eq(z)2,f1(z+c)+f2z1=eq(z)eq(z)2i. (4.1)

    In view of (4.1), it follows that

    (i+pz1)ep(z)+q(z+c)+(i+pz1)ep(z)+q(z+c)e2q(z+c)1, (4.2)

    and

    (i+qz1)eq(z)+p(z+c)+(i+qz1)eq(z)+p(z+c)e2p(z+c)1. (4.3)

    Thus, by applying Lemma 3.3, we can deduce from (4.2) and (4.3) that

    (i+pz1)ep(z)+q(z+c)=1,or(i+pz1)ep(z)+q(z+c)=1

    and

    (i+qz1)eq(z)+p(z+c)=1,or(i+qz1)eq(z)+p(z+c)=1.

    Now, we will discuss four cases below.

    Case 1.

    {(i+pz1)ep(z)+q(z+c)=1,(i+qz1)eq(z)+p(z+c)=1. (4.4)

    Thus, it follows that p(z)+q(z+c)=η1 and q(z)+p(z+c)=η2, that is, p(z+2c)p(z)=η2+η1 and q(z+2c)q(z)=η2+η1, here and below η1,η2 are constants in C, which each occurrence can be inconsistent. So, we can deduce that p(z)=L(z)+H(s)+B1 and q(z)=L(z)+H(s)+B2, where L(z) is a linear function of the form L(z)=α1z1+α2z2, H(s) is a polynomial in s:=c1z2c2z1, α1,α2,B1,B2 are constants in C. Noting that (i+pz1) and (i+qz1) are constants, it follows that p(z)=L(z)+B1 and q(z)=L(z)+B2. Thus, by combining with (4.2)-(4.4), it yields that

    {(i+α1)eL(c)+B2B1=1,(i+α1)eL(c)+B1B2=1,(i+α1)eL(c)+B1B2=1,(i+α1)eL(c)+B2B1=1, (4.5)

    where L(c):=α1c1+α2c2, this thus leads to

    (i+α1)2=1,e2L(c)=1,e2(B1B2)=1. (4.6)

    Then we have α1=0 or α1=2i.

    Assume that α1=0. If eL(c)=i, that is, L(c)=α2c2=2kπi+π2i, then it follows from (4.5) that eB2B1=1. Noting that system (4.1), we have

    f1(z)=ep(z)+ep(z)2=eL(z)+B1+eL(z)B12,

    and

    f2(z)=eL(z)+B2+eL(z)B22=eL(z)+B1+B2B1+eL(z)B1+B1B22=eL(z)+B1+eL(z)B12=f1(z).

    If eL(c)=i, that is, L(c)=α2c2=2kπi+3π2i, then it follows from (4.5) that eB2B1=1. Thus, it yields from (4.1) that

    f2(z)=eL(z)+B2+eL(z)B22=eL(z)+B1+B2B1+eL(z)B1+B1B22=eL(z)+B1+eL(z)B12=f1(z).

    Assume that α1=2i. If eL(c)=i, that is, L(c)=2ic1+α2c2=2kπi+π2i, then it follows from (4.5) that eB2B1=1. Noting that system (4.1), we have

    f2(z)=eL(z)+B2+eL(z)B22=eL(z)+B1+B2B1+eL(z)B1+B1B22=eL(z)+B1+eL(z)B12=f1(z).

    If eL(c)=i, that is, L(c)=2ic1+α2c2=2kπi+3π2i, then it follows from (4.5) that eB2B1=1. Thus, it yields from (4.1) that

    f2(z)=eL(z)+B2+eL(z)B22=eL(z)+B1+B2B1+eL(z)B1+B1B22=eL(z)+B1+eL(z)B12=f1(z).

    Case 2.

    {(i+pz1)ep(z)+q(z+c)=1,(i+qz1)eq(z)+p(z+c)=1.

    Thus, it follows that p(z)+q(z+c)=η1 and q(z)+p(z+c)=η2, that is, q(z+2c)+q(z)=η1η2. Noting that q(z) is a nonconstant polynomial in C2, we can get a contradiction.

    Case 3.

    {(i+pz1)ep(z)+q(z+c)=1,(i+qz1)eq(z)+p(z+c)=1.

    Thus, it follows that p(z)+q(z+c)=η1 and q(z)+p(z+c)=η2, that is, p(z+2c)+p(z)=η2+η1. Noting that p(z) is a nonconstant polynomial in C2, we can get a contradiction.

    Case 4.

    {(i+pz1)ep(z)+q(z+c)=1,(i+qz1)eq(z)+p(z+c)=1. (4.7)

    Thus, it follows that p(z)+q(z+c)=η1 and q(z)+p(z+c)=η2, that is, p(z+2c)p(z)=η2η1 and q(z+2c)q(z)=η1η2. Similar to the same argument as in Case 1, we get that p(z)=L(z)+B1 and q(z)=L(z)+B2, where L(z)=α1z1+α2z2, α1,α2,B1,B2 are constants in C. Thus, by combining with (4.2)-(4.3) and (4.7), it yields that

    {(i+α1)eL(c)B1B2=1,(iα1)eL(c)B1B2=1,(i+α1)eL(c)+B1+B2=1,(iα1)eL(c)+B1+B2=1, (4.8)

    which leads to (iα1)2=(i+α1)2=1, that is, α1=0. Then, it leads to

    e2L(c)=1,e2(B1+B2)=1. (4.9)

    If eL(c)=1, that is, L(c)=2kπi, then it follows from (4.8) that eB1+B2=i. Noting that system (4.1), we have

    f1(z)=ep(z)+ep(z)2=eL(z)+B1+eL(z)B12,

    and

    f2(z)=eL(z)+B2+eL(z)B22=eL(z)+B1B1B2+eL(z)B1+B1+B22=ieL(z)+B1+ieL(z)B12.

    If eL(c)=1, that is, L(c)=(2k+1)πi, then it follows from (4.8) that eB1+B2=i. Noting that system (4.1), we have

    f2(z)=eL(z)+B2+eL(z)B22=eL(z)+B1B1B2+eL(z)B1+B1+B22=ieL(z)+B1ieL(z)B12.

    Therefore, this completes the proof of Theorem 2.3.

    By observing the argument as in Theorems 2.1 and 2.2, when f(z) is a function with one complex variable, we can obtain the following conclusions easily.

    Theorem 5.1. Then the following complex differential equation

    f(z)2+[f(z)+f(z)]2=1 (5.1)

    has no any transcendental entire solutions with finite order.

    Theorem 5.2. Let cC{0}. Suppose that the complex differential difference equation

    f(z)2+[f(z+c)+f(z)]2=1 (5.2)

    admits a transcendental entire solution with finite order. Then f must be of the form

    f(z)=e2iz+b+e2izb2,

    and c=π4±kπ, kZ, where bC.

    Theorem 5.3. The following system of complex differential equations

    {f1(z)2+[f2(z)+f2(z)]2=1,f2(z)2+[f1(z)+f1(z)]2=1 (5.3)

    has no any transcendental entire solutions with finite order.

    For further studying the solutions of (2.1)-(2.4), we raise the following question which can not be solved now.

    Question 5.1. How to describe the transcendental meromorphic solutions of (2.1)–(2.4)?

    We thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments which improved the presentation.

    This work was supported by the National Natural Science Foundation of China (11561033), the Natural Science Foundation of Jiangxi Province in China (20181BAB201001), and the Foundation of Education Department of Jiangxi (GJJ190876, GJJ191042, GJJ190895) of China.

    The authors declare that none of the authors have any competing interests in the manuscript.



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