Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 37, pp. 1–15. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.37 FORMS OF ENTIRE SOLUTIONS OF PARTIAL DIFFERENTIAL DIFFERENCE EQUATIONS WITH CONSTANT COEFFICIENTS XIN DING, XIU MIN ZHENG Abstract. The purpose of this article is to describe the transcendental en- tire solutions of quadratic trinomial partial differential equations (PDDEs) with constant coefficients. We establish theorems on the forms of finite order transcendental entire solutions for such PDDEs, which generalize and improve previous theorems. Some examples confirm the existence and the forms of transcendental entire solutions with finite order of such equations. 1. Introduction In this article, we consider transcendental entire solutions of certain quadratic trinomial partial differential difference equations (PDDEs) in C2, related to the Fermat type functional equations with constant coefficients. We begin with the Pythagorean functional equation f2 + g2 = 1, (1.1) which is frequently studied as analogue of Diophantine equation over number fields. In 1966, Gross [4] proved the classical result that entire solutions of (1.1) are f = cos a(z), g = sin a(z), where a(z) is an entire function. In fact, the study of these Fermat type functional equations (1.1) goes back to Montel [12] and Pólya [13]. The Pythagorean functional equations also include the eikonal equation u2 z1 + u2 z2 = 1, which was considered by Li [7] and by Khavinson [6]. They proved that entire solutions of u2 z1 + u2 z2 = 1 must be linear in C2. Clearly, the eikonal equation is a typical partial differential equation of Fermat type. As is known, partial differential equations (PDEs) occure in various areas of applied mathematics, such as nonlinear acoustic wave propagation, geometric op- tics, and traffic flow (see [2, 3]). In general, it is difficult for us to find entire and meromorphic solutions of nonlinear PDEs. By employing Nevanlinna theory and other methods of complex analysis, there are a number of publications focusing on the solutions of some PDEs and their variants, see [1]. For instance, Yuan [24] obtained all traveling meromorphic exact solutions of the modified Zakharov- Kuznetsov equation by using a method of complex analysis; Khavinson [6] in 1995 showed that each entire solution of the partial differential equation u2 z1 +u2 z2 = 1 in C2 is necessarily linear by using Nevanlinna theory. Later, Saleeby [15, 17] studied 2020 Mathematics Subject Classification. 30D35, 39A13. Key words and phrases. Nevanlinna theory; entire solution; existence. ©2024. This work is licensed under a CC BY 4.0 license. Submitted April 30, 2024. Published June 22, 2024. 1 2 X. DING, X. M. ZHENG EJDE-2024/37 the forms of entire and meromorphic solutions of some PDEs with several complex variables, and obtained the following result. Theorem 1.1 ([15, Theorem 1]). If f is an entire solution of(∂f(z1, z2) ∂z1 )2 + (∂f(z1, z2) ∂z2 )2 = 1 (1.2) in C2, then f(z1, z2) = c1z1 + c2z2 + c, where c, c1, c2 ∈ C and c21 + c22 = 1. Saleeby [16] further investigated entire and meromorphic solutions of the qua- dratic trinomial equation f2 + 2αfg + g2 = 1, α2 ̸= 1, α ∈ C, (1.3) and obtained the following result. Theorem 1.2 ([16, Theorem 2.1]). Entire and meromorphic solutions of equation (1.3) (respectively) have the forms f = 1√ 2 ( cosh√ 1 + α + sinh√ 1− α ) , g = 1√ 2 ( cosh√ 1 + α − sinh√ 1− α ) and f = α1 − α2β 2 (α1 − α2)β , g = 1− β2 (α1 − α2)β , where h is entire, β is mermorphic in C2 and α1 = −α + √ α2 − 1, α2 = −α −√ α2 − 1. Liu, Cao and et al.[8-11] further studied entire solutions of some variants of Fermat type equations with more general forms than the difference equation f(z)2+ f(z + c)2 = 1 and obtained the following result. Theorem 1.3 ([10, Theorem 1.15]). Let a1, a2, a3, a4 be nonzero constants. If [a1f(z + c) + a2f(z)] 2 + [a3f(z + c) + a4f(z)] 2 = 1 admits transcendental entire solutions with finite order, then a21 + a23 = a22 + a24 and f(z) = a2 cos(aiz + bi) + a1 sin(aiz + bi) a2a3 − a1a4 , where a is a nonzero constant and b is a constant. Cao and Xu [1,19-23] investigated the existence of the solutions for some Fermat type partial differential difference equations with several variables by using the difference analogue of the logarithmic derivative lemma of several complex variables and obtained the following result. Theorem 1.4 ([22, Theorem 1.2]). Let c = (c1, c2) be a constant in C2. Then any transcendental entire solution with finite order of the partial differential difference equation (∂f(z1, z2) ∂z1 )2 + f(z1 + c1, z2 + c2) 2 = 1 (1.4) has the form of f(z1, z2) = sin(Az1 + B), where A is a constant in C satisfying AeiAc1 = 1, and B is a constant in C; in the special case whenever c1 = 0, we have f(z1, z2) = sin(z1 +B). EJDE-2024/?? FORMS OF ENTIRE SOLUTIONS 3 Recently, with the help of Nevanlinna theory and its difference analogues with several complex variables, Xu and et al. obtained some interesting results about Fermat type partial differential difference equations with several complex variables (see e.g.[21-24]). Especially, Zheng and Xu [25] in 2022 obtained the following result. When ∂f(z1,z2) ∂z1 and f(z1 + c1, z2 + c2) in equation (1.4) are replaced by the partial differential difference polynomials in C2. Theorem 1.5 ([25, Theorem 3.1]). Let c = (c1, c2) ∈ C2\{(0, 0)}, and a1, a2, a3, a4 be constants in C such that D := a1a4−a2a3 ̸= 0. Let f(z1, z2) be a transcendental entire solution with finite order of the partial differential difference equation[ a1f(z + c) + a2 ∂f ∂z1 ]2 + [ a3f(z + c) + a4 ∂f ∂z1 ]2 = 1. (1.5) Then f(z1, z2) is of the form f(z1, z2) = − 1 D (a3 + ia1 2α1 eL(z)+B − a3 − ia1 2α1 e−L(z)−B ) , where L(z) = α1z1 + α2z2, α1(̸= 0), α2, B ∈ C and L(z) satisfies α2 1 = −a21 + a23 a22 + a24 , e2L(c) = (ia2 + a4)(a3 − ia1) (ia2 − a4)(a3 + ia1) . Theorem 1.6 ([25, Theorem 3.3]). Let c = (c1, c2) ∈ C2\{(0, 0)}, a1, a2, a3, a4 be nonzero constants in C, and u0 = c2 + a1a4 a2a3 c1. Let f(z1, z2) be a transcendental entire solution with finite order of the partial differential difference equation[ a1f(z + c) + a2 ∂f ∂z1 ]2 + [ a3f(z + c) + a4 ∂f ∂z2 ]2 = 1. (1.6) Then f(z1, z2) is of one of the following two forms: (i) f(z1, z2) = ϕ(z2 + a1a4 a2a3 z1), where ϕ(u) is a transcendental entire function with finite order in u := z2 + a1a4 a2a3 z1 satisfying ϕ(u+ u0) + a4 a3 ϕ(u) = ± 1√ a21 + a23 ; (ii) f(z1, z2) = a3 + ia1 2(α1a2a3 − α2a1a4) eL(z)+B − a3 − ia1 2(α1a2a3 − α2a1a4) e−L(z)−B + φ(u), where L(z) = α1z1 + α2z2, α1, α2, B ∈ C and φ(u) satisfy (a2α1) 2 + (a4α2) 2 a21 + a23 = −1, eL(c) = −a4α2 + ia2α1 a3 + ia1 = − a3 − ia1 ia2α1 − a4α2 , a4 a3 φ′(u) + φ(u+ u0) = 0. Inspired by the above results, the following question can be raised naturally: How about replacing binomials in the left-hand sides of both equa- tions (1.5) and (1.6) with trinomials? 4 X. DING, X. M. ZHENG EJDE-2024/37 2. Main results and examples Motivated by the above question, our purpose of this paper is to explore the finite order transcendental entire solutions of the following quadratic trinomial partial differential difference equations D1(f) 2 + 2αD1(f)D2(f) +D2(f) 2 = 1 (2.1) and D1(f) 2 + 2αD1(f)D3(f) +D3(f) 2 = 1, (2.2) where α(̸= 0,±1), λj (j = 1, 2, 3, 4 and λ1λ4 − λ2λ3 ̸= 0) are constants in C, and D1(f) = λ1f(z+c)+λ2 ∂f ∂z1 ,D2(f) = λ3f(z+c)+λ4 ∂f ∂z1 ,D3(f) = λ3f(z+c)+λ4 ∂f ∂z2 . If α = 0, then equations (2.1) and (2.2) reduce to equations (1.5) and (1.6) respectively. If α = ±1, that is, equations (2.1) and (2.2) can be represented as [D1(f)±D2(f)] 2 = 1 and [D1(f)±D3(f)] 2 = 1 respectively, then we have (λ1 ± λ3)f(z + c) + (λ2 ± λ4) ∂f ∂z1 = ±1 (2.3) and (λ1 ± λ3)f(z + c) + λ2 ∂f ∂z1 ± λ4 ∂f ∂z2 = ±1, (2.4) respectively. Further, equation (2.3) has finite order transcendental entire solutions with the form f(z) = ± 1 λ1±λ3 + eβ1z1+β2z2+β0 , where (λ1 ±λ3)e β1c1+β2c2 = −(λ2 ± λ4)β1; equation (2.4) has finite order transcendental entire solutions with the form f(z) = ± 1 λ1±λ3 + eγ1z1+γ2z2+γ0 , where (λ1 ± λ3)e γ1c1+γ2c2 = −(λ2γ1 ± λ4γ2). In the following, we assume that α ̸= 0,±1 and denote A1 = 1√ 1 + α + 1 i √ 1− α , A2 = 1√ 1 + α − 1 i √ 1− α . The first main theorem is about the existence and the forms of transcendental entire solutions of the quadratic trinomial partial differential difference equation (2.1). Theorem 2.1. Let c = (c1, c2) ∈ C2\{(0, 0)}, and λj(j = 1, 2, 3, 4) be nonzero con- stants in C such that D := λ1λ4 −λ2λ3 ̸= 0. If equation (2.1) has a transcendental entire solution f(z1, z2) with finite order, then f(z1, z2) has the form f(z1, z2) = √ 2 4D [ λ1A2 − λ3A1 α1 eL(z)+B − λ1A1 − λ3A2 α1 e−L(z)−B ], where L(z) = α1z1 + α2z2, α1(̸= 0), α2, B ∈ C and L(z) satisfies α2 1 = − (λ1A1 − λ3A2)(λ1A2 − λ3A1) (λ4A2 − λ2A1)(λ4A1 − λ2A2) , e2L(c) = (λ1A1 − λ3A2)(λ4A1 − λ2A2) (λ3A1 − λ1A2)(λ4A2 − λ2A1) . The following examples confirm the conclusion about the form of transcendental entire solutions of equation (2.1). Example 2.2. Let f(z1, z2) = 3 √ 21− √ 7i 14 e √ 21 3 iz1+log √ 13+3 √ 3i 14e2 iz2+b0 EJDE-2024/?? FORMS OF ENTIRE SOLUTIONS 5 + 3 √ 21 + √ 7i 14 e− √ 21 3 iz1−log √ 13+3 √ 3i 14e2 iz2−b0 , where b0 ∈ C. Then f(z1, z2) is a transcendental entire solution of equation (2.1) with λ1 = 2, λ2 = λ3 = λ4 = 1, c1 = − √ 21i 7 , c2 = 1, α1 = √ 21i 3 , α2 = log √ 13+3 √ 3i 14e2 i, α = 1 2 and ρ(f) = 1. Example 2.3. Let f(z1, z2) = 21 √ 2− 7 √ 6i 6 √ 51 e √ 51i 14 z1+(9+ √ 3i)z2+b0+ 21 √ 2 + 7 √ 6i 6 √ 51 e− √ 51i 14 z1−(9+ √ 3i)z2−b0 , where b0 ∈ C. Then f(z1, z2) is a transcendental entire solution of equation (2.1) with λ1 = 2, λ2 = √ 2, λ3 = 1, λ4 = √ 2, c1 = 14√ 51i , c2 = 3 14 , α1 = √ 51i 14 , α2 = 9 + √ 3i, α = 1 2 , and ρ(f) = 1. When ∂f ∂z1 in equation (2.1) is replaced by ∂f ∂z1 + ∂f ∂z2 , we obtain the second theorem as follows. Theorem 2.4. Let c = (c1, c2) ∈ C2\{(0, 0)}, and λj(j = 1, 2, 3, 4) be nonzero constants in C such that D := λ1λ4 − λ2λ3 ̸= 0. Let f(z1, z2) be a transcendental entire solution with finite order of the partial differential difference equation D∗ 1(f) 2 + 2αD∗ 1(f)D∗ 2(f) +D∗ 2(f) 2 = 1, (2.5) where α(̸= 0,±1) ∈ C, and D∗ 1(f) = λ1f(z + c) + λ2( ∂f ∂z1 + ∂f ∂z2 ), D∗ 2(f) = λ3f(z + c) + λ4( ∂f ∂z1 + ∂f ∂z2 ). Then f(z1, z2) is of the form f(z1, z2) = √ 2 4D [ λ1A2 − λ3A1 α1 + α2 eL(z)+B − λ1A1 − λ3A2 α1 + α2 e−L(z)−B ], where L(z) = α1z1 + α2z2, α1(̸= 0), α2, B ∈ C and L(z) satisfies (α1 + α2) 2 = − (λ1A1 − λ3A2)(λ1A2 − λ3A1) (λ4A2 − λ2A1)(λ4A1 − λ2A2) , e2L(c) = (λ1A1 − λ3A2)(λ4A1 − λ2A2) (λ3A1 − λ1A2)(λ4A2 − λ2A1) . Since the proof of Theorem 2.4 is similar as the one of Theorem 2.1, we omit its proof. The following example confirms the conclusion about the forms of transcen- dental entire solutions of equation (2.5). Example 2.5. Let f(z1, z2) = 3 + 9 √ 3 2 √ 66 e √ 22iz1+2 √ 22iz2+b0 + 3− 9 √ 3 2 √ 66 e− √ 22iz1−2 √ 22iz2−b0 , where b0 ∈ C. Then f(z1, z2) is a transcendental entire solution of equation (2.5) with λ1 = 2, λ3 = 1, λ2 = λ4 = 1 4 , c1 = 1, c2 = 5 6 , α1 = √ 22i, α2 = 2 √ 22i, α = 1 2 , and ρ(f) = 1. When λ4 ∂f ∂z1 in equation (2.1) is replaced by λ4 ∂f ∂z2 , that is equation (2.2), we obtain the following theorem. 6 X. DING, X. M. ZHENG EJDE-2024/37 Theorem 2.6. Let c = (c1, c2) ∈ C2\(0, 0), and λj(j = 1, 2, 3, 4) be nonzero constants in C. If equation (2.2) has a transcendental entire solution f(z1, z2) with finite order, then f(z1, z2) is of one of the following two forms. (i) f(z1, z2) = ϕ ( z2 + λ1λ4 λ2λ3 z1 ) , where ϕ(u) is a transcendental entire solution with finite order in u := z2 + λ1λ4 λ2λ3 z1 satisfying ϕ(u+ u0) + λ4 λ3 ϕ′(u) = ± 1√ λ2 1 + λ2 3 + 2αλ1λ3 , u0 = c2 + λ1λ4 λ2λ3 c1. (ii) If c1λ4 ̸= ±c2λ2 and λ2λ3α1 − λ1λ4α2 = 0, then f(z1, z2) = √ 2 4λ2λ3 z1[(λ3A1 − λ1A2)e α2u+B + (λ3A2 − λ1A1)e −α2u−B ] + φ(u); if λ2λ3α1 − λ1λ4α2 ̸= 0, then f(z1, z2) = √ 2 4 [ λ3A1 − λ1A2 λ2λ3α1 − λ1λ4α2 eL(z)+B − λ3A2 − λ1A1 λ2λ3α1 − λ1λ4α2 e−L(z)−B ] + φ(u), where L(z) = α1z1 + α2z2, α1, α2, B ∈ C and L(z) satisfies eL(c) = λ3A2 − λ1A1 λ4A2α2 − λ2A1α1 = λ2A2α1 − λ4A1α2 λ3A1 − λ1A2 , and φ(u) is an entire function with finite order in u = z2 + λ1λ4 λ2λ3 z1 satisfying φ(u+ u0) + λ4 λ3 φ′(u) = 0, u0 = c2 + λ1λ4 λ2λ3 c1. We also give two examples to confirm the conclusion about the forms of tran- scendental entire solutions of equation (2.2). Example 2.7. Let f(z1, z2) = ± √ 7 7 + e2z1+z2 . Then f(z1, z2) is a transcendental entire solution of equation (2.2) with λ1 = 2, λ2 = λ3 = λ4 = 1, c1 = c2 = πi, α = 1/2. Example 2.8. Let f(z1, z2) = √ 6 12 eL(z)+B − √ 6 12 e−(L(z)+B) + e2z1+z2 , where L(z) = 1 2z1 + (2 − √ 3)z2, B ∈ C. Then f(z1, z2) is a transcendental entire solution of equation (2.2) with λ1 = λ3 = λ4 = 1, λ2 = 2, c1 = (1 + √ 3) log 2 √ 3 1 + 3 √ 3− (3 + √ 3)i + (1− √ 3)πi, c2 = −1 + √ 3 2 ( log 2 √ 3 1 + 3 √ 3− (3 + √ 3)i − πi ) , α1 = 1/2, α2 = 2− √ 3i, α = 2, and ρ(f) = 1. Next, we give some lemmas which play the key role in proving our results. EJDE-2024/?? FORMS OF ENTIRE SOLUTIONS 7 Lemma 2.9 ([14, 18]). For an entire function F in Cn, F (0) ̸= 0 put ρ(nF ) = ρ < ∞. Then there exist a canonical function fF and a function gF (z) ∈ Cn such that F (z) = fF (z)e gF (z). For the special case n = 1, fF is the canonical product of Weierstrass. Here we denote by ρ(nF ) the order of the counting function of zeros of F . Lemma 2.10 ([13]). If g and h are entire functions in the complex plane and g(h) is an entire function of finite order, then there are only two possible cases: either (i) the internal function h is a polynomial and the external function g is of finite order; or (ii) the internal function h is not a polynomial but a function of finite order, and the external function g is of zero order. Lemma 2.11 ([5]). Let fj( ̸≡ 0), j = 1, 2, 3 be meromorphic functions in Cn such that f1 is not a constant, and f1 + f2 + f3 = 1, and such that 3∑ j=1 {N2(r, 1 fj ) + 2N(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 possible number. Then either f2 ≡ 1 or f3 ≡ 1. Here, N2(r, 1 f ) 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. 3. Proof of Theorem 2.1 Suppose that f is a transcendental entire solution with finite order of equation (2.1). Denote D1(f) = 1√ 2 (m+ n), D2(f) = 1√ 2 (m− n), (3.1) where m,n are entire functions in C2. Thus, equation (2.1) can be rewritten as (1 + α)m2 + (1− α)n2 = 1. (3.2) If √ 1 + αm is not a transcendental entire function, by (3.2), then √ 1− αn is not a transcendental entire function, which implies that f is not a transcendental entire function, a contradiction with the assumption that f is a transcendental entire function. Hence, √ 1 + αm and √ 1− αn are transcendental functions. We can rewrite equation (3.2) as ( √ 1 + αm+ i √ 1− αn)( √ 1 + αm− i √ 1− αn) = 1. (3.3) Noting that m,n are transcendental entire functions with finite order, we have that√ 1 + αm+i √ 1− αn and √ 1 + αm−i √ 1− αn have no zeros and poles. Therefore, by Lemmas 2.9 and 2.10, there exists a nonconstant polynomial p(z) in C2 such that √ 1 + αm+ i √ 1− αn = ep(z), √ 1 + αm− i √ 1− αn = e−p(z). (3.4) 8 X. DING, X. M. ZHENG EJDE-2024/37 In view of (3.1) and (3.4), it follows that D1(f) = λ1f(z + c) + λ2 ∂f ∂z1 = √ 2 4 [( 1√ 1 + α + 1 i √ 1− α )ep(z) + ( 1√ 1 + α − 1 i √ 1− α )e−p(z)] = √ 2 4 (A1e p(z) +A2e −p(z)) (3.5) and D2(f) = λ3f(z + c) + λ4 ∂f ∂z1 = √ 2 4 [( 1√ 1 + α − 1 i √ 1− α )ep(z) + ( 1√ 1 + α + 1 i √ 1− α )e−p(z)] = √ 2 4 (A2e p(z) +A1e −p(z)). (3.6) Noting that D = λ1λ4 − λ2λ3 ̸= 0, and solving the system consisting of (3.5) and (3.6), we deduce that f(z + c) = √ 2 4D [(λ4A1 − λ2A2)e p(z) + (λ4A2 − λ2A1)e −p(z)], (3.7) ∂f ∂z1 = √ 2 4D [(λ1A2 − λ3A1)e p(z) + (λ1A1 − λ3A2)e −p(z)]. (3.8) Then (3.7) and (3.8) yield ω1e p(z+c) + ω2e −p(z+c) = ω3 ∂p ∂z1 ep(z) − ω4 ∂p ∂z1 e−p(z), (3.9) where ω1 = λ1A2 − λ3A1, ω2 = λ1A1 − λ3A2, ω3 = λ4A1 − λ2A2, ω4 = λ4A2 − λ2A1. If ω2 = 0, then ω1 ̸= 0. Otherwise, A2 1 = A2 2, a contradiction. If ω3 ∂p ∂z1 = 0, then either ∂p ∂z1 = 0 or ω3 = 0. If ∂p ∂z1 = 0, then from (3.9), it follows that ω1e p(z+c) = 0, which implies ω1 = 0, a contradiction. If ω3 = 0, then ω4 ̸= 0. Otherwise, A2 1 = A2 2, a contradiction. Thus, (3.9) yieldS ep(z+c)+p(z) = −ω4 ω1 ∂p ∂z1 . Since p(z) is a nonconstant polynomial, p(z+c)+p(z) can not be a constant. Hence, the above equation implies a contradiction that the left-hand side is transcendental but the right-hand side is not transcendental. Thus, it follows that ω3 ̸= 0. Then, ω4 ̸= 0. Otherwise, ω2 = ω4 deduces a contradiction that D = 0. By combining this with (3.9), yields ω1e p(z+c)+p(z) = ω3 ∂p ∂z1 e2p(z) − ω4 ∂p ∂z1 . (3.10) Noting that N(r, 1 ep(z+c)+p(z) ) = 0, N(r, ep(z+c)+p(z)) = 0 and N(r, 1 e2p(z) ) = 0, by the Nevanlinna second main theorem in several complex variables, and in view of EJDE-2024/?? FORMS OF ENTIRE SOLUTIONS 9 (3.10), we conclude that T (r, ep(z+c)+p(z)) ≤ N(r, 1 ep(z+c)+p(z) ) +N(r, 1 ep(z+c)+p(z) − χ ) +N(r, ep(z+c)+p(z)) + S(r, ep(z+c)+p(z)) ≤ N(r, 1 ep(z+c)+p(z) − χ ) + S(r, ep(z+c)+p(z)) = N(r, 1 ω3 ω1 ∂p ∂z1 e2p(z) ) + S(r, ep(z+c)+p(z)) = S(r, ep(z+c)+p(z)), where χ = −ω4 ω1 ∂p ∂z1 . This is a contradiction. We conclude that ω2 ̸= 0. Similarly, we have ω1 ̸= 0, ω3 ̸= 0, ω4 ̸= 0. Thus, we rewrite (3.9) in the form ω3 ω2 ∂p ∂z1 ep(z+c)+p(z) − ω4 ω2 ∂p ∂z1 ep(z+c)−p(z) − ω1 ω2 e2p(z+c) = 1. (3.11) In view of ω1 ω2 e2p(z+c) ̸= 0 and ep(z+c)+p(z) is nonconstant, by Lemma 2.3, ω4 ω2 ∂p ∂z1 ep(z+c)−p(z) ≡ −1. (3.12) Thus, it follows from (3.11) that ω3 ω1 ∂p ∂z1 ep(z)−p(z+c) ≡ 1. (3.13) Since p(z) is a polynomial, (3.12) (or (3.13)) implies p(z+ c)− p(z) = ζ, where ζ is a constant in C. Thus, it follows that p(z) = L(z)+H(z)+B, where L(z) = α1z1+ α2z2, α1, α2 ∈ C, H(z) := H(s), H(s) is a polynomial in s = c2z1 − c1z2, c1, c2 ∈ C, and B ∈ C. Next, we prove that H(z) ≡ 0. (3.12) implies ω4 ω2 α1 + ω4 ω2 c2 dH ds ≡ −e−ζ , which also means that degs H ≤ 1. Thus, the form of L(z) +H(z) + B is still the linear form of α1z1 + α2z2 +B,α1, α2, B ∈ C, which means that H(z) ≡ 0. Hence, it follows that p(z) = L(z) +B = α1z1 + α2z2 +B,α1, α2, B ∈ C. By substituting this into (3.12) and (3.13), we deduce that −ω4 ω2 α1e L(c) ≡ 1, ω3 ω1 α1e −L(c) ≡ 1, (3.14) which implies that α2 1 = −ω1ω2 ω3ω4 , e2L(c) = −ω2ω3 ω1ω4 . (3.15) By applying (3.14) to (3.7), we have f(z) = √ 2 4D [ω3e L(z)+B−L(c) + ω4e −L(z)−B+L(c)] = √ 2 4D [ ω1 α1 eL(z)+B − ω2 α1 e−L(z)−B ]. The proof of Theorem 2.1 is complete. 10 X. DING, X. M. ZHENG EJDE-2024/37 4. Proof of Theorem 2.6 Suppose that f is a transcendental entire solution with finite order of equation (2.2). Denote D1(f) = 1√ 2 (u+ v), D3(f) = 1√ 2 (u− v), (4.1) where u, v are entire functions in C2. Thus, equation (2.2) can be rewritten as (1 + α)u2 + (1− α)v2 = 1. (4.2) As in Theorem 2.1, we discuss the folowing two cases. Case 1. √ 1 + αu is a constant. We denote √ 1 + αu = γ1, γ1 ∈ C. (4.3) In view of equation (4.2), it follows that √ 1− αv is also a constant. We denote √ 1− αv = γ2, γ2 ∈ C. (4.4) This leads to γ2 1 + γ2 2 = 1. Thus, we deduce from (4.1) that D1(f) = λ1f(z + c) + λ2 ∂f ∂z1 = 1√ 2 ( γ1√ 1 + α + γ2√ 1− α ), (4.5) D3(f) = λ3f(z + c) + λ4 ∂f ∂z2 = 1√ 2 ( γ1√ 1 + α − γ2√ 1− α ). (4.6) In view of (4.5) and (4.6), we have λ2λ3 ∂f ∂z1 −λ1λ4 ∂f ∂z2 = 1√ 2 [ λ3 ( γ1√ 1 + α + γ2√ 1− α ) −λ1 ( γ1√ 1 + α − γ2√ 1− α )] . (4.7) By differentiating both two sides of (4.5) and (4.6) for the variables z2 and z1 respectively, and by combining with the fact ∂2f ∂z1∂z2 = ∂2f ∂z2∂z1 , we deduce that λ2λ3 ∂f(z + c) ∂z1 − λ1λ4 ∂f(z + c) ∂z2 = 0, (4.8) which also implies by (4.7) that λ3 ( γ1√ 1 + α + γ2√ 1− α ) = λ1 ( γ1√ 1 + α − γ2√ 1− α ) . (4.9) Thus, from (4.9) and γ2 1 + γ2 2 = 1 it follows that γ1 = ± √ 1 + α(λ1 + λ3)√ 2λ2 1 + 2λ2 3 + 4αλ1λ3 , γ2 = ± √ 1− α(λ1 − λ3)√ 2λ2 1 + 2λ2 3 + 4αλ1λ3 . The characteristic equations of (4.8) are dz1 dt = λ2λ3, dz2 dt = −λ1λ4, df dt = 0. By using the initial conditions z1 = 0, z2 = u, and f = f(0, u) := ϕ(u) with a parameter u, we obtain the following parametric representation for the solutions of the characteristic equations: z1 = λ2λ3t, z2 = −λ1λ4t+ u, f(t, u) = ∫ t 0 0dt+ ϕ(u) = ϕ(u), EJDE-2024/?? FORMS OF ENTIRE SOLUTIONS 11 where ϕ(u) is a transcendental entire function with finite order in u = z2 + λ1λ4 λ2λ3 z1. Then, by combining this with t = 1 λ2λ3 z1 and u = z2 + λ1λ4 λ2λ3 z1, the solution of equation (4.8) has the form f(z1, z2) = ϕ(z2 + λ1λ4 λ2λ3 z1). (4.10) By substituting (4.10) into (4.5), we obtain that ϕ(u+ u0) + λ4 λ3 ϕ′(u) = ± 1√ λ2 1 + λ2 3 + 2αλ1λ3 , (4.11) where u0 = c2 + λ1λ4 λ2λ3 c1. Case 2. √ 1 + αu is not a constant. Then we can rewrite equation (4.2) as ( √ 1 + αu+ i √ 1− αv)( √ 1 + αu− i √ 1− αv) = 1. (4.12) Noting that u, v are transcendental entire functions with finite order, we have that√ 1 + αu + i √ 1− αv and √ 1 + αu − i √ 1− αv have no zeros and poles. Thus, by Lemmas 2.9 and 2.10, there exists a nonconstant polynomial q(z) in C2 such that √ 1 + αu+ i √ 1− αv = eq(z), √ 1 + αu− i √ 1− αv = e−q(z). (4.13) In view of (4.1) and (4.13), similar to Theorem 2.1, it follows that D1(f) = λ1f(z + c) + λ2 ∂f ∂z1 = √ 2 4 (A1e q(z) +A2e −q(z)), (4.14) D3(f) = λ3f(z + c) + λ4 ∂f ∂z2 = √ 2 4 (A2e q(z) +A1e −q(z)), (4.15) which means λ2λ3 ∂f ∂z1 − λ1λ4 ∂f ∂z2 = √ 2 4 [(λ3A1 − λ1A2)e q(z) + (λ3A2 − λ1A1)e −q(z)]. (4.16) By differentiating both two sides of (4.14) and (4.15) for the variables z2 and z1 respectively, and by combining with the fact ∂2f ∂z1∂z2 = ∂2f ∂z2∂z1 , we conclude that λ2λ3 ∂f(z + c) ∂z1 − λ1λ4 ∂f(z + c) ∂z2 = √ 2 4 [(λ2A2 ∂q ∂z1 − λ4A1 ∂q ∂z2 )eq + (λ4A2 ∂q ∂z2 − λ2A1 ∂q ∂z1 )e−q]. (4.17) It follows from (4.16) and (4.17) that υ1e q(z+c) + υ2e −q(z+c) = υ3e q(z) + υ4e −q(z). (4.18) where υ1 = λ3A1 − λ1A2, υ2 = λ3A2 − λ1A1, υ3 = λ2A2 ∂q ∂z1 − λ4A1 ∂q ∂z2 , υ4 = λ4A2 ∂q ∂z2 − λ2A1 ∂q ∂z1 . If υ2 = 0, then υ1 ̸= 0. Otherwise, A2 1 = A2 2, a contradiction. If υ3 = 0, then υ4 ̸= 0. Otherwise, (4.18) leads to υ1e q(z+c) = 0, a contradiction. Thus, it follows that υ3 ̸= 0. Similarly, υ4 ̸= 0. By combining this with (4.18), we have eq(z+c)+q(z) = υ4 υ1 . 12 X. DING, X. M. ZHENG EJDE-2024/37 Since q(z) is a nonconstant polynomial, q(z+c)+q(z) can not be a constant. Hence, the above equation implies a contradiction that the left-hand side is transcendental but the right-hand side is not transcendental. Therefore, by (4.18), we obtain that υ1e q(z+c)+q(z) = υ3e 2q(z) + υ4. (4.19) Similar to Case 2 in Theorem 2.1, by the Nevanlinna second main theorem in several complex variables, and in view of (4.19), we conclude that it is a contradiction. We conclude that υ2 ̸= 0. Similarly, we have υ1 ̸= 0. Thus, (4.18) leads to υ3 υ2 eq(z+c)+q(z) + υ4 υ2 eq(z+c)−q(z) − υ1 υ2 e2q(z+c) = 1, (4.20) If υ3 = 0, then υ4 ̸= 0. Otherwise, (4.20) leads to −υ1 υ2 e2q(z+c) = 1, a contradiction. Then (4.20) becomes e2q(z+c) = υ4 υ1 eq(z+c)−q(z) − υ2 υ1 . (4.21) Noting that N(r, 1 e2q(z+c) ) = 0, N(r, e2q(z+c))) = 0 and N(r, 1 υ4 υ1 eq(z+c)−q(z) ) = 0, by the Nevanlinna second main theorem in several complex variables, and in view of (4.21), we conclude that T (r, e2q(z+c)) ≤ N(r, 1 e2q(z+c) ) +N(r, 1 υ2 υ1 + e2q(z+c) ) +N(r, e2q(z+c)) + S(r, e2q(z+c)) ≤ N(r, 1 υ4 υ1 eq(z+c)−q(z) ) + S(r, e2q(z+c)) = S(r, e2q(z+c)), which leads to a contradiction. Hence υ3 ̸= 0. Similarly, we have υ4 ̸= 0. By Lemma 2.11, and the fact that eq(z+c)+q(z) is nonconstant yields υ4 υ2 eq(z+c)−q(z) ≡ 1. (4.22) Thus, in view of (4.20), υ3 υ1 eq(z)−q(z+c) ≡ 1. (4.23) Since q(z) is a polynomial, (4.22) (or (4.23)) implies q(z+ c)− q(z) = η, where η is a constant in C. Thus, it follows that q(z) = L(z)+H(z)+B, where L(z) = α1z1+ α2z2, α1, α2 ∈ C, H(z) := H(s), H(s) is a polynomial in s = c2z1− c1z2, c1, c2 ∈ C, and B ∈ C. Next, we prove that H(z) ≡ 0. In view of (4.22) and (4.23), we have 1 υ2 [σ1 − (c1λ4A2 + c2λ2A1) dH ds ] = A−1, (4.24) 1 υ1 [σ2 + (c2λ2A2 + c1λ4A1) dH ds ] = A, (4.25) where A = eη = eL(c), σ1 = λ4A2α2 − λ2A1α1, σ2 = λ2A2α1 − λ4A1α2. This implies that both (c1λ4A2 + c2λ2A1) dH ds and (c2λ2A2 + c1λ4A1) dH ds are con- stants. By combining this with the fact c1λ4 ̸= ±c2λ2, it follows that dH ds is a constant; that is, degs H ≤ 1. Thus, the form of L(z) +H(z) +B is still the linear EJDE-2024/?? FORMS OF ENTIRE SOLUTIONS 13 form of α1z1 + α2z2 + B,α1, α2, B ∈ C, which means that H(z) ≡ 0. It follows that q(z) = L(z) +B = α1z1 +α2z2 +B,α1, α2, B ∈ C. Thus, we can deduce from (4.24) and (4.25) that σ1 υ2 eL(c) = 1, σ2 υ1 e−L(c) = 1, that is, σ1σ2 υ2υ1 = 1, eL(c) = υ2 σ1 = σ2 υ1 . (4.26) On the other hand, in view of (4.16), we can deduce that λ2λ3 ∂f(z) ∂z1 − λ1λ4 ∂f(z) ∂z2 = √ 2 4 [υ1e α1z1+α2z2+B + υ2e −(α1z1+α2z2+B)]. (4.27) The characteristic equations of (4.27) are dz1 dt = λ2λ3, dz2 dt = −λ1λ4, df dt = √ 2 4 [υ1e α1z1+α2z2+B + υ2e −(α1z1+α2z2+B)]. By using the initial comditions z1 = 0, z2 = u, and f = f(0, u) := φ0(u) with a parameter u, we obtain the following parametric representation for the solutions of the characteristic equations: z1 = λ2λ3t, z2 = −λ1λ4t+ u, f(t, u) = √ 2 4 ∫ t 0 [υ1e α1z1+α2z2+B + υ2e −(α1z1+α2z2+B)]dt+ φ(u), where φ(u) is an entire function with finite order in u such that φ(u) = { φ0(u), λ2λ3α1 − λ1λ4α2 = 0; φ0(u)− √ 2 4 ( υ1e α2u+B λ2λ3α1−λ1λ4α2 − υ2e −(α2u+B) λ2λ3α1−λ1λ4α2 ), λ2λ3α1 − λ1λ4α2 ̸= 0. If λ2λ3α1 − λ1λ4α2 = 0, then f(z1, z2) = √ 2 4λ2λ3 z1[υ1e α2u+B + υ2e −α2u−B ] + φ0(u). If λ2λ3α1 − λ1λ4α2 ̸= 0, then f(z1, z2) = √ 2 4 [ υ1 λ2λ3α1 − λ1λ4α2 eL(z)+B − υ2 λ2λ3α1 − λ1λ4α2 e−L(z)−B ] + φ(u). 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Comput., 219 (11) (2013), 6212-6223. [25] Zheng, X. M., Xu, H. Y.; Entire solutions for some Fermat type functional equations con- cerning difference and partial differntial in C2, Anal. Math., 48 (1) (2022), 199-226. Xin Ding School of Mathematics and Statistics, Jiangxi Normal University, Nanchang, 330022, China. Jiangxi Provincial center for Applied Mathematics, Jiangxi Normal University, Nan- chang, 330022, China Email address: dingxin2001@126.com EJDE-2024/?? FORMS OF ENTIRE SOLUTIONS 15 Xiu Min Zheng (corresponding author) School of Mathematics and Statistics, Jiangxi Normal University, Nanchang, 330022, China. Jiangxi Provincial center for Applied Mathematics, Jiangxi Normal University, Nan- chang, 330022, China Email address: zhengxiumin2008@sina.com 1. Introduction 2. Main results and examples 3. Proof of Theorem 2.1 4. Proof of Theorem 2.6 Acknowledgments References