Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 23, pp. 1–26. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.23 FORM OF SOLUTIONS TO QUADRATIC TRINOMIAL PARTIAL DIFFERENTIAL EQUATIONS WITH TWO COMPLEX VARIABLES JIN TU, HUIZHEN WEI Abstract. This article describes the from of entire solutions to quadratic trinomial partial differential equations (PDEs). By applying the Nevanlinna theory and the characteristic equation of PDEs, we extend some of the results obtained in [24] . Also we also provide examples that illustrate our results. 1. Introduction In 1995, Wiles and Taylor [18, 19] pointed out that the Fermat-type functional equation (also called Pythagorean functional equation) xm + ym = 1 (1.1) does not admit non-trivial solutions when m ≥ 3, but it admits non-trivial solutions when m = 2. Actually, the study of (1.1) can be tracked back to Montel [12] and Gross [1]. They proved that the equation fm + gm = 1 has entire solutions and pointed out that for m = 2, the equation has non-constant entire solutions f = cos p, g = sin p, where p is any non-constant entire function. Recently, with the evolution of Nevanlinna theory, many scholars gained plentiful results about these equations of Fermat-type. Liu, Cao and Cao [8] in 2012 investigated the existence of entire solutions with finite order of Fermat equations and obtained the following result. Theorem 1.1 ([8]). Suppose f is a transcendental entire solution of f ′(z)2 + f(z + c)2 = 1, (1.2) then f must satisfy f(z) = sin(z±Bi), where B ∈ C and c = 2kπ, or c = (2k+1)π with k an integer. In 2013, Saleeby [16] generalized the Pythagorean functional equation f2+g2 = 1 and studied the quadratic trinomial functional equation f2 + 2αfg + g2 = 1, α ∈ C− {1,−1} (1.3) and obtained the following result. 2020 Mathematics Subject Classification. 30D35, 32W50, 35M30, 39A45. Key words and phrases. Entire solution; meromorphic function; existence; partial differential equation. ©2024. This work is licensed under a CC BY 4.0 license. Submitted December 12, 2023. Published March 12, 2024. 1 2 J. TU, H. WEI EJDE-2024/23 Theorem 1.2 ([16]). If equation (1.3) has a transcendental entire solution, then f, g must satisfy f = 1√ 2 ( cosh√ 1 + α + sinh√ 1− α ) , g = 1√ 2 ( cosh√ 1 + α − sinh√ 1− α ) or f = α1 − α2β 2 (α1 − α2)β , g = 1− β2 (α1 − α2)β , where h is an entire function, β is a meromorphic function and α1 = −α+ √ α2 − 1, α2 = −α− √ α2 − 1. In 2016, Liu and Yang [9] researched the related properties on the meromorphic solutions of the following equations, for α2 ̸= 0, 1, f(z)2 + 2αf(z)f ′(z) + f ′(z)2 = 1, (1.4) f(z)2 + 2αf(z)f(z + c) + f(z + c)2 = 1. (1.5) If α2 ̸= 0, 1, then (1.4) has no transcendental meromorphic solutions but (1.5) has transcendental meromorphic solutions with finite order and the order must be equal to one. Now, let us mention some previous results about the Fermat-type PDEs with two complex variables. In 1995, Khavinson [3] pointed out that any entire solution of the partial differential equations( ∂u ∂z1 )2 + ( ∂u ∂z2 )2 = 1 (1.6) in C2 is necessarily linear. Later, Saleeby in [15] extended the result by exploring the solutions of Fermat-type functional equations (1.6) and obtain the following result. Theorem 1.3 ([15]). The entire solution of (1.6) must satisfy u(z1, z2) = c1z1 + c2z2 + c, where c, c1, c2 ∈ C and c21 + c22 = 1. Later, Li et al. discussed equations (1.6) with more general forms( ∂f ∂z1 )2 + ( ∂f ∂z2 )2 = fn, ( ∂f ∂z1 )2 + ( ∂f ∂z2 )2 = p, ( ∂f ∂z1 )2 + ( ∂f ∂z2 )2 = eg, where n ∈ N+, p, g are polynomials in C2 (see [4, 5, 6, 7]). Li in 2005 further investigated the functional equation of Fermat-type( ∂u ∂z1 )2 + ( ∂u ∂z2 )2 = eg (1.7) and obtained the following result. Theorem 1.4 ([6]). If equation (1.7) admits an entire solution of f(z) with finite order in C2, where g is a polynomial, then u is an entire solution of (1.7) if and only if (i) u = f(c1z1 + c2z2) or (ii) u = ϕ1(z1 + iz2) + ϕ2(z1 − iz2), where f is an entire solution and f ′(c1z1 + c2z2) = ±e 1 2 g(z), c1 and c2 are two constants satisfying c1 2 + c2 2 = 1, and ϕ′ 1(z1 + iz2) + ϕ′ 2(z1 − iz2) = 1 4e g(z). EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 3 Recently, Lü [10] studied the quadratic trinomial partial differential equation uz1 2 + 2Buz1uz2 + uz2 2 = eg(z), (1.8) where B is a constant and g is a polynomial or an entire function in C2, and obtained the following result. Theorem 1.5 ([10]). Let g be a polynomial in C2, let t1 and t2 be two different roots of the equation 1+2At+ t2 = 0(A ̸= ±1). Then u is an entire solution of the partial differential equation (1.8) if and only if (i) u(z1, z2) = G(z1 + t2z2 +B(z1 + t1z2)) or (ii) u(z1, z2) = ϕ1(z1 + t1z2) + ϕ2(z1 + t2z2), where G is an entire function in C satisfying BG′2(z1 + t2z2 +B(z1 + t1z2)) = eg−log τ , where B ∈ C−{0}, τ = 4(1−A2), ϕ1 and ϕ2 are entire functions in C and satisfy ϕ′ 1(z1 + t1z2) = eα(z1+t1z2), ϕ′ 2(z1 + t2z2) = eβ(z1+t2z2), where α and β are two polynomials such that α(z1 + t1z2) + β(z1 + t2z2) = g(z1, z2)− log τ. Theorem 1.6 ([10]). Let g be an entire function in C2. Then u is an entire solution of the partial differential equation (1.8) in C2 if and only if u(z1, z2) = F (z1, z2 ∓ z1) + f(z2 ∓ z1), where f is an entire function in C and F (t, s) = ∫ t 0 ±e g(t,±t+s) 2 dt. In 2020, Xu, Tu andWang [24] researched several Fermat-type PDEs and PDDEs with two complex variables( f(z) + ∂f ∂z1 )2 + ( f(z) + ∂f ∂z2 )2 = 1, (1.9)( f(z) + ∂f ∂z1 )2 + ( f(z) + ∂2f ∂z1∂z2 )2 = 1, (1.10) and obtained interesting results: Theorem 1.7 ([24]). If equation (1.9) has an entire solution in C2, then f(z1, z2) = ± √ 2 2 + ηe−(z1+z2) or f(z1, z2) = 1 2 sin(z2 − z1 + η1) + 1 2 cos(z2 − z1 + η1) + η2e −(z1+z2), where η, η1, η2 ∈ C. Theorem 1.8 ([24]). If equation (1.10) has an entire solution in C2, then f(z1, z2) = ± √ 2 2 + ηez2−z1 , where η ∈ C. 4 J. TU, H. WEI EJDE-2024/23 Other results on Fermat-type PDEs with two complex variables can be found in [2, 3, 20, 11, 21, 22, 23, 25, 26]). Inspired by the aforesaid theorems, the following question raises spontaneously. What will happen when uz1 is superseded by f(z)+ ∂f ∂z1 , and uz2 is superseded by f(z)+ ∂f ∂z2 or f(z)+ ∂2f ∂z2 1 , or f(z)+ ∂2f ∂z1∂z2 in question (1.8), where g(z) is a polynomial? 2. Results and examples Motivated by the above question, we study the solutions of the following qua- dratic trinomial partial differential equations, utilizing the Nevanlinna theory and the characteristic equation of partial differential equations:( f(z) + ∂f ∂z1 )2 + 2α ( f(z) + ∂f ∂z1 )( f(z) + ∂f ∂z2 ) + ( f(z) + ∂f ∂z2 )2 = eg(z), (2.1)( f(z) + ∂f ∂z1 )2 + 2α ( f(z) + ∂f ∂z1 )( f(z) + ∂2f ∂z21 ) + ( f(z) + ∂2f ∂z21 )2 = eg(z), (2.2)( f(z) + ∂f ∂z1 )2 + 2α ( f(z) + ∂f ∂z1 )( f(z) + ∂2f ∂z1∂z2 ) + ( f(z) + ∂2f ∂z1∂z2 )2 = eg(z), (2.3) where α2 ∈ C − {0, 1} and g(z) be a polynomial with the linear form g(z1, z2) = α1z1 + α2z2 + α0, where α1 ̸= 0, α2 ̸= 0, α0 ∈ C. For simplicity, let α2 ̸= 0, 1, and A1 := 1 2 √ 1 + α − i 2 √ 1− α , A2 := 1 2 √ 1 + α + i 2 √ 1− α . (2.4) Our main results read as follows. Theorem 2.1. Suppose equation (2.1) admits a transcendental entire solution f(z) of finite order. Then f(z1, z2) must satisfy the following: (i) f(z1, z2) = ζ1(β, α, α1, α2)e g(z)/2 + ηe−(z1+z2), where ζ1(β, α, α1, α2) =  √ 2(β2−1) iβ(α1−α2) √ 1−α , α1 ̸= α2, ± √ 2 (2+α1) √ 1+α , α1 = α2 ̸= −2, (β2+1)(z1+z2) 2β √ 2(1+α) + (β2−1)(z1−z2) 2iβ √ 2(1−α) , α1 = α2 = −2, and β, α, α1, α2 satisfy (4 + α1 + α2)(β 2 − 1) i √ 1− α = (α1 − α2)(β 2 + 1)√ 1 + α ; (ii) if B11 = B21, B12 = B22, then (2.1) has no transcendental entire solution f(z) of finite order, hence B11 = B21, B12 = B22 cannot coexist, then f(z1, z2) = 1√ 2 ϑ1(B11, B12, B21, B22) + ηe−(z1+z2), where ϑ1(B11, B12, B21, B22) EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 5 =  A1−A2 B11−B12 eγ1(z) + A2−A1 B21−B22 eγ2(z), B11 ̸= B12, B21 ̸= B22, (A1z1 +A2z2)e γ1(z) + A2−A1 B21−B22 eγ2(z), B11 = B12, B21 ̸= B22, A1−A2 B11−B12 eγ1(z) + (A2z1 +A1z2)e γ2(z), B11 ̸= B12, B21 = B22, moreover, η ̸= 0, γ1(z) = B11z1 + B12z2 + β1, γ2(z) = B21z1 + B22z2 + β2, Bj1, Bj2, βj ∈ C(j = 1, 2) satisfy g(z) = γ1(z) + γ2(z) = α1z1 + α2z2 + α0, and B11, B12, B21, B22 satisfy A2(B11 + 1) = A1(B12 + 1), A1(B21 + 1) = A2(B22 + 1). We list several examples to show the forms of solutions in Theorem 2.1 are precise. Example 2.2. f(z1, z2) = ± 2√ 37 e z1+2z2 2 + ie−(z1+z2) is a solution of (2.1) with g(z) = z1 + 2z2. Here, α = 1/2, α1 = 1, α2 = 2, α0 = 0, η = i. Example 2.3. f(z1, z2) = ± √ 6 3 e −z1−z2+2 2 + √ 2e−(z1+z2) is a solution of (2.1) with g(z) = −z1 − z2 + 2. Here, α = 2, α0 = 2, α1 = α2 = −1, η = √ 2. Example 2.4. f(z1, z2) = ± √ 6 6 (z1 + z2)e −2z1−2z2+1 2 + √ 2e−(z1+z2) is a solution of (2.1) with g(z) = −2z1 − 2z2 + 1. Here, α = 2, α0 = 1, α1 = α2 = −2, η = √ 2. Example 2.5. Let α = 1/2, γ1(z) = √ 6− 6− 3 √ 2i 6 z1 + √ 6− 6 + 3 √ 2i 6 z2, γ2(z) = √ 6− 6 + 3 √ 2i 6 z1 + √ 6− 6− 3 √ 2i 6 z2, β1 = β2 = 0, and η = √ 5 2 . Then f(z1, z2) = 1√ 2 ( e √ 6−6−3 √ 2i 6 z1+ √ 6−6+3 √ 2i 6 z2 + e √ 6−6+3 √ 2i 6 z1+ √ 6−6−3 √ 2i 6 z2 ) + √ 5 2 e−(z1+z2) is a transcendental entire solution of (2.1) with g(z) = √ 6−6 3 (z1 + z2). Example 2.6. Let α = 1/2, γ1(z) = −z1−z2, γ2(z) = √ 6−6+3 √ 2i 6 z1+ √ 6−6−3 √ 2i 6 z2, β1 = β2 = 0, η = i. Then f(z1, z2) = (√3− 3i 6 z1 + √ 3 + 3i 6 z2 + i ) e−z1−z2 + 1√ 2 e √ 6−6+3 √ 2i 6 z1+ √ 6−6−3 √ 2i 6 z2 is a transcendental entire solution of (2.1) with g(z) = √ 6+3 √ 2i−12 6 z1+ √ 6−3 √ 2i−12 6 z2. Example 2.7. Let α = 1/2, γ1(z) = √ 6−6−3 √ 2i 6 z1+ √ 6−6+3 √ 2i 6 z2, γ2(z) = −z1−z2, β1 = β2 = 0, η = √ 5. Then f(z1, z2) = 1√ 2 e √ 6−6−3 √ 2i 6 z1+ √ 6−6+3 √ 2i 6 z2 + (√3 + 3i 6 z1 + √ 3− 3i 6 z2 + √ 5 ) e−z1−z2 6 J. TU, H. WEI EJDE-2024/23 is a transcendental entire solution of (2.1) with g(z) = √ 6−3 √ 2i−12 6 z1+ √ 6+3 √ 2i−12 6 z2. Theorem 2.8. Suppose equation (2.2) admits a transcendental entire solution f(z) of finite order, then f(z1, z2) must satisfy the following: (i) f(z1, z2) = ζ2(β, α, α1, α2)e g(z)/2, where ζ2(β, α, α1, α2) =  2 √ 2(β2−1) iβα1(2−α1) √ 1−α , α1 ̸= 2, ± 1 2 √ 2(1+α) , α1 = 2, and β, α, α1, α2 satisfy (α1 2 + 2α1 + 8)(β2 − 1) i √ 1− α = (2α1 − α1 2)(β2 + 1)√ 1 + α ; (ii) f(z1, z2) = 1 2 √ 2 [ (A1 +A2 −A1B11)e γ1(z) + (A1 +A2 −A2B21)e γ2(z) ] , where η ̸= 0, γ1(z) = B11z1+H(z2)+B12z2+β1, γ2(z) = B21z1−H(z2)+B22z2+β2, Bj1, Bj2, βj ∈ C(j = 1, 2) satisfy g(z) = γ1(z) + γ2(z) = α1z1 + α2z2 + α0, A2(B11 + 1) = A1(B11 2 + 1), A1(B21 + 1) = A2(B21 2 + 1). We give several examples to show the results in Theorem 2.8 are precise to some extent. Example 2.9. f(z1, z2) = ± 1 2e −z1+2z2 is a solution of (2.2) with g(z) = −2z1+4z2. Here, α = 1/2, α1 = −2, α2 = 4, α0 = 0. Example 2.10. f(z1, z2) = ± 1 2 √ 3 ez1+2z2 is a solution of (2.2) with g(z) = 2z1+4z2. Here, α = 1/2, α1 = 2, α2 = 4, α0 = 0. Example 2.11. Let α = 1/2, B11 = − √ 3i−1 2 , and B21 = √ 3i−1 2 . Then f(z1, z2) = √ 3 3 e − √ 3i−1 2 z1+z2 3+z2+1 + √ 3 3 e √ 3i−1 2 z1−z2 3+z2+3 is a transcendental entire solution of (2.2) with g(z) = −z1 + 2z2 + 4. Theorem 2.12. Suppose equation (2.3) admits a transcendental entire solution f(z) of finite order. Then f(z1, z2) must satisfy the following: (i) f(z1, z2) = ζ3(β, α, α1, α2)e g(z)/2 + ηe−z1+z2 , where ζ3(β, α, α1, α2) =  4(β2+1) β(α1α2+2α1+8) √ 2(1+α) , α1 + α2 ̸= 0, 4(β2−1) iβ(2+α1)α1 √ 2(1−α) , α1 + α2 = 0, and β, α, α1, α2 satisfy (α1α2 + 2α1 + 8)(β2 − 1) i √ 1− α = (2α1 − α1α2)(β 2 + 1)√ 1 + α ; (ii) if B11 = B21 and B12 = B22, then (2.3) does not admit any transcendental entire solution with finite order, if B11 = B21 and B12 = B22 do not coexist, and if B11 +B12 ̸= 0, B21 +B22 ̸= 0, then f(z1, z2) = 1√ 2 ϑ2(B11, B12, B21, B22) + ηe−z1+z2 , EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 7 where ϑ2(B11, B12, B21, B22) = A1 −A2 +A1B12 B11 +B12 eγ1(z) + A2 −A1 +A2B22 B21 +B22 eγ2(z); if B11 +B12 = 0, B21 +B22 ̸= 0, then f(z1, z2) = 1√ 2 ϑ3(B11, B12, B21, B22) + ηe−z1+z2 , where ϑ3(B11, B12, B21, B22) = { A2−A1+A2B22 B21+B22 eγ2(z) + A1 2 2A1−A2 eγ1(z), B11 = A1−A2 A1 ; (A1z2 −A2z2 +A1z1)e γ1(z) + A2−A1+A2B22 B21+B22 eγ2(z), B11 = −1; if B11 +B12 ̸= 0 and B21 +B22 = 0, then f(z1, z2) = 1√ 2 ϑ4(B11, B12, B21, B22) + ηe−z1+z2 , where ϑ4(B11, B12, B21, B22) = { A1−A2+A1B12 B11+B12 eγ1(z) + A2 2 2A2−A1 eγ2(z), B21 = A2−A1 A2 ; A1−A2+A1B12 B11+B12 eγ1(z) + (A2z2 −A1z2 +A2z1)e γ2(z), B21 = −1; if B11 +B12 = 0 and B21 +B22 = 0, then f(z1, z2) = 1√ 2 ϑ4(B11, B12, B21, B22) + ηe−z1+z2 , where ϑ4(B11, B12, B21, B22) =  A1 2 2A1−A2 eγ1(z) + A2 2 2A2−A1 eγ2(z), B11 = A1−A2 A1 , B21 = A2−A1 A2 ; (A1z2 −A2z2 +A1z1)e γ1(z) + A2 2 2A2−A1 eγ2(z), B11 = −1, B21 = A2−A1 A2 ; A1 2 2A1−A2 eγ1(z) + (A2z2 −A1z2 +A2z1)e γ2(z), B11 = A1−A2 A1 , B21 = −1; moreover, η ̸= 0, γ1(z) = B11z1 + B12z2 + β1, γ2(z) = B21z1 + B22z2 + β2, and Bj1, Bj2, βj ∈ C(j = 1, 2) satisfy g(z) = γ1(z) + γ2(z) = α1z1 + α2z2 + α0, A2(B11 + 1) = A1(B11B12 + 1), A1(B21 + 1) = A2(B21B22 + 1). Now we give some examples to show that the results in Theorem 2.12 are precise. Example 2.13. f(z1, z2) = 2 √ 3 9 e z1+2z2 2 +3e−z1+z2 is a solution of (2.3) with g(z) = z1 + 2z2. Here, α = 1/2, α1 = 1, α2 = 2, α0 = 0, η = 3. Example 2.14. f(z1, z2) = ± 1 2e z1−z2 +5e−z1+z2 is a solution of (2.3) with g(z) = 2z1 − 2z2. Here, α = 1/2, α1 = 2, α2 = −2, α0 = 0, η = 5. Example 2.15. Let α = 1/2, γ1(z) = z1 + (−2 + √ 3i)z2, γ2(z) = 3z1 + (−1 − 2 √ 3i 3 )z2, β1 = β2 = 0, and η = 5 2 . Then f(z1, z2) = −3i+ √ 3 12 ez1+(−2+ √ 3i)z2 + 3i+ √ 3 24 e3z1−(1+ 2 √ 3i 3 )z2 + 5 2 e−z1+z2 8 J. TU, H. WEI EJDE-2024/23 is a transcendental entire solution of (2.3) with g(z) = 4z1 + (−3 + √ 3i 3 )z2. Example 2.16. Let α = 1/2, γ1(z) = 3− √ 3i 2 z1 + −3+ √ 3i 2 z2, γ2(z) = 3z1 + (−1 − 2 √ 3i 3 )z2, β1 = β2 = 0, η = 0, then f(z1, z2) = 2 √ 3− 3i 21 e 3− √ 3i 2 z1+ −3+ √ 3i 2 z2 + 3i+ √ 3 24 e3z1−(1+ 2 √ 3i 3 )z2 is a transcendental entire solution of (2.3) with g(z) = 9− √ 3i 2 z1 + −15− √ 3i 6 z2. Example 2.17. Let α = 1/2, γ1(z) = −z1 + z2, γ2(z) = 3z1 + (−1 − 2 √ 3i 3 )z2, β1 = β2 = 0, η = 6, then f(z1, z2) = (√3− 3i 6 z1 − iz2 ) e−z1+z2 + 3i+ √ 3 24 e3z1−(1+ 2 √ 3i 3 )z2 + 6e−z1+z2 is a transcendental entire solution of (2.3) with g(z) = 2z1 − 2 √ 3i 3 z2. Example 2.18. Let α = 1/2, γ1(z) = z1 + (−2 + √ 3i)z2, γ2(z) = 3+ √ 3i 2 z1 + −3− √ 3i 2 z2, β1 = β2 = 0, η = 1, then f(z1, z2) = −3i+ √ 3 12 ez1+(−2+ √ 3i)z2 + 2 √ 3 + 3i 21 e 3+ √ 3i 2 z1+ −3− √ 3i 2 z2 + e−z1+z2 is a transcendental entire solution of (2.3) with g(z) = 5+ √ 3i 2 z1 + −7+ √ 3i 2 z2. Example 2.19. Let α = 1/2, γ1(z) = z1 + (−2 + √ 3i)z2, γ2(z) = −z1 + z2, β1 = β2 = 0, η = 4, then f(z1, z2) = −3i+ √ 3 12 ez1+(−2+ √ 3i)z2 + (√3 + 3i 6 z1 + iz2 ) e−z1+z2 + 4e−z1+z2 is a transcendental entire solution of (2.3) with g(z) = −1 + √ 3iz2. Example 2.20. Let α = 1/2, γ1(z) = 3− √ 3i 2 z1 + −3+ √ 3i 2 z2, γ2(z) = 3+ √ 3i 2 z1 + −3− √ 3i 2 z2, β1 = β2 = 0, η = 2i, then f(z1, z2) = 2 √ 3− 3i 21 e 3− √ 3i 2 z1+ −3+ √ 3i 2 z2 + 2 √ 3 + 3i 21 e 3+ √ 3i 2 z1+ −3− √ 3i 2 z2 + 2ie−z1+z2 is a transcendental entire solution of (2.3) with g(z) = 3z1 − 3z2. Example 2.21. Let α = 1/2, γ1(z) = −z1 + z2, γ2(z) = 3+ √ 3i 2 z1 + −3− √ 3i 2 z2, β1 = β2 = 0, η = √ 2, then f(z1, z2) = (√3− 3i 6 z1 − iz2 ) e−z1+z2 + 2 √ 3 + 3i 21 e 3+ √ 3i 2 z1+ −3− √ 3i 2 z2 + √ 2e−z1+z2 is a transcendental entire solution of (2.3) with g(z) = 1+ √ 3i 2 z1 + −1− √ 3i 2 z2. Example 2.22. Let α = 1/2, γ1(z) = 3− √ 3i 2 z1 + −3+ √ 3i 2 z2, γ2(z) = −z1 + z2, β1 = β2 = 0, η = √ 2i, then f(z1, z2) = 2 √ 3− 3i 21 e 3− √ 3i 2 z1+ −3+ √ 3i 2 z2 + (√3 + 3i 6 z1 + iz2 ) e−z1+z2 + √ 2ie−z1+z2 is a transcendental entire solution of (2.3) with g(z) = 1− √ 3i 2 z1 + −1+ √ 3i 2 z2. EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 9 3. Proof of Theorem 2.1 Proof. Suppose that f(z) is a transcendental entire solution of (2.1) with finite order. Let f(z) + ∂f ∂z1 = 1√ 2 (u+ v), f(z) + ∂f ∂z2 = 1√ 2 (u− v), where u and v are entire functions. Thus, we rewrite (2.1) in the form (1 + α)u2 + (1− α)v2 = eg(z). (3.1) Then it follows from (3.1) that(√1 + αu eg(z)/2 )2 + (√1− αv eg(z)/2 )2 = 1. This formula leads to(√1 + αu eg(z)/2 + i √ 1− αv eg(z)/2 )(√1 + αu eg(z)/2 − i √ 1− αv eg(z)/2 ) = 1, (3.2) which implies that both √ 1+αu eg(z)/2 + i √ 1−αv eg(z)/2 and √ 1+αu eg(z)/2 − i √ 1−αv eg(z)/2 have no zeros. Therefore, in view of [13]-[17], there exist a polynomial p(z) such that √ 1 + αu eg(z)/2 + i √ 1− αv eg(z)/2 = ep(z), √ 1 + αu eg(z)/2 − i √ 1− αv eg(z)/2 = e−p(z). (3.3) We denote γ1(z) = g(z) 2 + p(z), γ2(z) = g(z) 2 − p(z). (3.4) Then from (3.3), we have f(z) + ∂f ∂z1 = 1√ 2 ( A1e γ1(z) +A2e γ2(z) ) , (3.5) f(z) + ∂f ∂z2 = 1√ 2 ( A2e γ1(z) +A1e γ2(z) ) , (3.6) where A1, A2 are defined by (2.4). Thus, from (3.5) and (3.6) it follows that ∂f ∂z1 − ∂f ∂z2 = 1√ 2 [(A1 −A2)e γ1(z) + (A2 −A1)e γ2(z)]. (3.7) On the other hand, from (3.5) and (3.6), by combining with ∂2f ∂z1∂z2 = ∂2f ∂z2∂z1 , we have ∂f ∂z1 − ∂f ∂z2 = 1√ 2 [( A2 ∂γ1 ∂z1 −A1 ∂γ1 ∂z2 ) eγ1(z) + ( A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z2 ) eγ2(z) ] . (3.8) Thus, (3.7) and (3.8) yield( A1 −A2 −A2 ∂γ1 ∂z1 +A1 ∂γ1 ∂z2 ) eγ1(z)−γ2(z) = A1 −A2 +A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z2 . (3.9) Now we consider two cases. 10 J. TU, H. WEI EJDE-2024/23 Case 1. If eγ1(z)−γ2(z) is a constant, then γ1(z)−γ2(z) is a constant. By combining with γ1(z) − γ2(z) = 2p(z), it follows that p(z) is a constant. Let β = ep(z), then equations (3.5)-(3.6) can be represented as f(z) + ∂f ∂z1 = 1√ 2 ( k1√ 1 + α + k2√ 1− α ) eg(z)/2, (3.10) f(z) + ∂f ∂z2 = 1√ 2 ( k1√ 1 + α − k2√ 1− α ) eg(z)/2, (3.11) where k1 = β+β−1 2 , k2 = β−β−1 2i , and k21 + k22 = 1. This leads to ∂f ∂z1 − ∂f ∂z2 = √ 2k2√ 1− α eg(z)/2. (3.12) On the other hand, differentiating both sides of equations (3.10) and (3.11) with respect to z2 and z1, respectively, and combining this with ∂2f ∂z1∂z2 = ∂2f ∂z2∂z1 , we deduce that ∂f ∂z1 − ∂f ∂z2 = 1 2 √ 2 ( k1√ 1 + α ∂g ∂z1 − k2√ 1− α ∂g ∂z1 − k1√ 1 + α ∂g ∂z2 − k2√ 1− α ∂g ∂z2 ) eg(z)/2. (3.13) From (3.12) and (3.13) it follows that 4k2√ 1− α = k1√ 1 + α ∂g ∂z1 − k2√ 1− α ∂g ∂z1 − k1√ 1 + α ∂g ∂z2 − k2√ 1− α ∂g ∂z2 . (3.14) Since g(z) is a polynomial with the linear form g(z1, z2) = α1z1 +α2z2 +α0, where α1 ̸= 0, α2 ̸= 0, α0 ∈ C. Hence, from (3.14) we deduce that (4 + α1 + α2)k2√ 1− α = (α1 − α2)k1√ 1 + α . (3.15) The characteristic equations of (3.12) are dz1 dt = 1, dz2 dt = −1, df dt = √ 2k2√ 1− α eg(z)/2. Using the initial conditions: z1 = 0, z2 = s, and f(z1, z2) = f(0, s) := ϕ0(s) with a parameter s. Thus, we obtain the following parametric representation for the solutions of the characteristic equations: z1 = t, z2 = −t+ s, f(t, s) = ∫ t 0 √ 2k2√ 1− α e α1t−α2t+α2s+α0 2 dt+ ϕ0(s), (3.16) where ϕ0(s) is a finite order transcendental entire function in s = z1 + z2. Subcase 1.1. If α1 − α2 ̸= 0, it follows from (3.16) that f(t, s) = ∫ t 0 √ 2k2√ 1− α e α1t−α2t+α2s+α0 2 dt+ ϕ0(s) = 2 √ 2k2√ 1− α(α1 − α2) e α1t−α2t+α2s+α0 2 + ϕ1(s), (3.17) where ϕ1(s) = ϕ0(s)− 2 √ 2k2√ 1− α(α1 − α2) e α2s+α0 2 EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 11 is a finite order transcendental entire function in s. Thus, from (3.17), it follows that f(z1, z2) = 2 √ 2k2√ 1− α(α1 − α2) e α1z1+α2z2+α0 2 + ϕ1(z1 + z2). (3.18) Substituting (3.18) into (3.10) or (3.11), then by combining with (3.15), it yields that ϕ1(z1 + z2) + ϕ′ 1(z1 + z2) = 0, (3.19) which implies ϕ1(z1 + z2) = η1e −(z1+z2), η1 ∈ C\ {0}. Subcase 1.2. If α1 − α2 = 0, it follows from (3.16) that f(t, s) = ∫ t 0 √ 2k2√ 1− α e α1t−α2t+α2s+α0 2 dt+ ϕ0(s) = √ 2k2√ 1− α e α2s+α0 2 t+ ϕ2(s), (3.20) where ϕ2(s) = ϕ0(s) is a transcendental entire function with finite order in s. In view of (3.20), we have f(z1, z2) = √ 2k2√ 1− α e α2z1+α2z2+α0 2 z1 + ϕ2(z1 + z2). (3.21) From (3.15) and α1 − α2 = 0, we have α1 = α2 = −2 or k2 = 0. Subcase 1.2.1. If α1 = α2 ̸= −2, then we have k2 = 0, it follows from (3.21) that f(z1, z2) = ϕ3(z1 + z2). (3.22) Substituting (3.22) into the (3.10) or (3.11) yields ϕ3(z1 + z2) + ϕ′ 3(z1 + z2) = ± 1√ 2(1 + α) e α1z1+α2z2+α0 2 , which implies ϕ3(z1 + z2) = ± √ 2 (2 + α1) √ 1 + α eg(z)/2 + η3e −(z1+z2), η3 ∈ C. Subcase 1.2.2. If α1 = α2 = −2, it follows from (3.21) that f(z1, z2) = √ 2k2√ 1− α e −2z1−2z2+α0 2 z1 + ϕ4(z1 + z2). (3.23) Substituting (3.23) into the (3.10) or (3.11), yields ϕ4(z1 + z2) + ϕ′ 4(z1 + z2) = 1√ 2 ( k1√ 1 + α − k2√ 1− α )e −2z1−2z2+α0 2 , which implies ϕ4(z1 + z2) = 1√ 2 ( k1√ 1 + α − k2√ 1− α )(z1 + z2)e g(z)/2 + η4e −(z1+z2), η4 ∈ C. The proof of Theorem 2.1(i) is complete. 12 J. TU, H. WEI EJDE-2024/23 Case 2. If eγ1(z)−γ2(z) is not a constant, then p(z) is not a constant. It follows from (3.9) that A1 −A2 −A2 ∂γ1 ∂z1 +A1 ∂γ1 ∂z2 = 0, A1 −A2 +A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z2 = 0. (3.24) Otherwise, without loss of generality, if A1 −A2 −A2 ∂γ1 ∂z1 +A1 ∂γ1 ∂z2 ̸= 0, we have e2p(z) = A1 −A2 +A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z2 A1 −A2 −A2 ∂γ1 ∂z1 +A1 ∂γ1 ∂z2 . (3.25) Since p(z), g(z) are polynomials, the left-hand side of (3.25) is transcendental, which contradicts with the right-hand side of (3.25) is a rational function. Thus, in view of (3.24), we have A2 ∂γ1 ∂z1 −A1 ∂γ1 ∂z2 = A1 −A2, A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z2 = A2 −A1. (3.26) Next, we prove that γ1(z) and γ2(z) are linear forms of z1, z2. Similar calcula- tions to the ones in equation (3.12) can be used to (3.26); we can obtain γ1 = A1 −A2 A2 z1 + φ1(z2 + A1 A2 z1), γ2 = A2 −A1 A1 z1 + φ2(z2 + A2 A1 z1). Since g(z) is a polynomial with the linear form g(z1, z2) = γ1(z) + γ2(z) = α1z1 + α2z2 + α0, where α1 ̸= 0, α2 ̸= 0, α0 ∈ C, it follows that φ1 + φ2 = [ α1 − (A1 −A2) 2 A1A2 ] z1 + α2z2 + α0. Let φ1 = bms1 m + bm−1s1 m−1 + · · ·+ b0, s1 = z2 + A1 A2 z1, φ2 = dns2 n + dn−1s2 n−1 + · · ·+ d0, s2 = z2 + A2 A1 z1. If m ≥ 2, we have n = m and bj = −dj , j = 2, . . . , n. Furthermore, if bj ̸= 0 for j = 2, . . . , n, we need consider the coefficient of zm−1 2 z1 in φ1, φ2, then it yields that C1 mbm A1 A2 zm−1 2 z1 + C1 mdm A2 A1 zm−1 2 z1 = 0, further, we can obtain that A1 A2 = A2 A1 , this is a contradiction with the required condition of theorems. Thus, we deduce that m = 1, and the γ1, γ2 are linear forms of z1, z2. Without loss of generality, we set γ1(z) = B11z1 +B12z2 + β1, γ2(z) = B21z1 +B22z2 + β2. EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 13 According to equation (3.7), we obtain that the characteristic equations of (3.7) are dz1 dt = 1, dz2 dt = −1, df dt = 1√ 2 [(A1 −A2)e γ1 + (A2 −A1)e γ2 ]. Similarly, we obtain f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2)e (B11−B12)t+B12s+β1 + (A2 −A1)e (B21−B22)t+B22s+β2 ] dt+ ϕ0(s), (3.27) where ϕ0(s) is a transcendental entire function with finite order in s = z1 + z2. Subcase 2.1. If B11 −B12 ̸= 0, B21 −B22 ̸= 0, it follows from (3.27) that f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2)e (B11−B12)t+B12s+β1 + (A2 −A1)e (B21−B22)t+B22s+β2 ] dt+ ϕ0(s) = 1√ 2 [ A1 −A2 B11 −B12 e(B11−B12)t+B12s+β1 + A2 −A1 B21 −B22 e(B21−B22)t+B22s+β2 ] + ϕ5(s). (3.28) where ϕ5(s) = ϕ0(s)− 1√ 2 [ A1 −A2 B11 −B12 eB12s+β1 + A2 −A1 B21 −B22 eB22s+β2 ] is a finite order transcendental entire function in s. Thus, it follows (3.28) that f(z1, z2) = 1√ 2 [ A1 −A2 B11 −B12 eγ1(z) + A2 −A1 B21 −B22 eγ2(z) ] + ϕ5(z1 + z2). (3.29) Since g(z) is a polynomial with the linear form g(z1, z2) = γ1(z) + γ2(z) = α1z1 + α2z2 + α0, where α1 ̸= 0, α2 ̸= 0, α0 ∈ C. Hence, we deduce from (3.26) that A2(B11 + 1) = A1(B12 + 1), A1(B21 + 1) = A2(B22 + 1). (3.30) Substituting (3.29) into (3.5) or (3.6), then combining this with (3.30) yields that ϕ5(z1 + z2) + ϕ′ 5(z1 + z2) = 0, (3.31) which implies ϕ5(z1 + z2) = η5e −(z1+z2), η5 ∈ C. Subcase 2.2. If B11 −B12 = 0 and B21 −B22 ̸= 0, from (3.27) it follows that f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2)e B11s+β1 + (A2 −A1)e (B21−B22)t+B22s+β2 ] dt+ ϕ0(s) = 1√ 2 [ (A1 −A2)e B11s+β1t+ A2 −A1 B21 −B22 e(B21−B22)t+B22s+β2 ] + ϕ6(s), (3.32) where ϕ6(s) = ϕ0(s)− 1√ 2 A2 −A1 B21 −B22 eB22s+β2 is a transcendental entire function in s. Thus, in view of (3.32), we obtain f(z1, z2) = 1√ 2 [ (A1 −A2)e γ1(z)z1 + A2 −A1 B21 −B22 eγ2(z) ] + ϕ6(z1 + z2). (3.33) 14 J. TU, H. WEI EJDE-2024/23 Also, we deduce from (3.26) that B11 = B12 = −1, A1(B21 + 1) = A2(B22 + 1). (3.34) Substituting (3.33) into (3.5) or (3.6), then by combining this with (3.34), it yields that ϕ6(z1 + z2) + ϕ′ 6(z1 + z2) = 1√ 2 A2e γ1(z), (3.35) which implies ϕ6(z1 + z2) = 1√ 2 A2e γ1(z)(z1 + z2) + η6e −(z1+z2), η6 ∈ C. Subcase 2.3. If B11 −B12 ̸= 0 and B21 −B22 = 0, it follows from (3.27) that f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2)e (B11−B12)t+B12s+β1 + (A2 −A1)e B21s+β2 ] dt+ ϕ0(s) = 1√ 2 [ A1 −A2 B11 −B12 e(B11−B12)t+B12s+β1 + (A2 −A1)e B21s+β2t ] + ϕ7(s), (3.36) where ϕ7(s) = ϕ0(s)− 1√ 2 A1 −A2 B11 −B12 eB12s+β1 is a finite order transcendental entire function in s. Thus, from (3.36), it follows that f(z1, z2) = 1√ 2 [ A1 −A2 B11 −B12 eγ1(z) + (A2 −A1)e γ2(z)z1 ] + ϕ7(z1 + z2). (3.37) Also, we deduce from (3.26) that A2(B11 + 1) = A1(B12 + 1), B21 = B22 = −1. (3.38) Substituting (3.37) into (3.10) or (3.11), then by combining this with (3.38), it yields that ϕ7(z1 + z2) + ϕ′ 7(z1 + z2) = 1√ 2 A1e γ2(z), (3.39) which implies ϕ7(z1 + z2) = 1√ 2 A1e γ2(z)(z1 + z2) + η7e −(z1+z2), η7 ∈ C. Subcase 2.4. If B11 −B12 = 0 and B21 −B22 = 0, it follows from (3.27) that f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2)e B12s+β1 + (A2 −A1)e B21s+β2 ] dt+ ϕ0(s) = 1√ 2 [ (A1 −A2)e B12s+β1 + (A2 −A1)e B12s+β2 ] t+ ϕ8(s), (3.40) where ϕ8(s) = ϕ0(s) is a transcendental entire function with finite order in s. Then from (3.40) we obtain f(z1, z2) = 1√ 2 [ (A1 −A2)e γ1(z) + (A2 −A1)e γ2(z) ] z1 + ϕ8(z1 + z2). (3.41) Also, we deduce from (3.26) that B11 = B12 = −1, B21 = B22 = −1, which leads to p(z) being a constant. By the assumption at the begin of Case 2, we obtain a contradiction. Thus, the proof of Theorem 2.1(ii) is complete. □ EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 15 4. Proof of Theorem 2.8 Proof. Suppose that f(z) is a transcendental entire solution of (2.2) with finite order. By using the same discussion in the proof of Theorem 2.1, we obtain that f(z) + ∂f ∂z1 = 1√ 2 ( A1e γ1(z) +A2e γ2(z) ) , (4.1) f(z) + ∂2f ∂z21 = 1√ 2 ( A2e γ1(z) +A1e γ2(z) ) , (4.2) where A1, A2 are defined by (2.4). Thus, it follows from (4.1) and (4.2) that ∂f ∂z1 − ∂2f ∂z21 = 1√ 2 [ (A1 −A2)e γ1(z) + (A2 −A1)e γ2(z) ] . (4.3) On the other hand, differentiating with respect to z1 on equation (4.1), in accor- dance with (4.2), we have f(z)− ∂f ∂z1 = 1√ 2 [ (A2 −A1 ∂γ1 ∂z1 )eγ1(z) + (A1 −A2 ∂γ2 ∂z1 )eγ2(z) ] . (4.4) Differentiating with respect to z1 on equation (4.4) yields ∂f ∂z1 − ∂2f ∂z21 = 1√ 2 [ (A2 ∂γ1 ∂z1 −A1( ∂γ1 ∂z1 )2 −A1 ∂2γ1 ∂z21 )eγ1(z) + (A1 ∂γ2 ∂z1 −A2( ∂γ2 ∂z1 )2 −A2 ∂2γ2 ∂z21 )eγ2(z) ] . (4.5) Thus, in line with (4.3) and (4.5), it follows that( A1 −A2 −A2 ∂γ1 ∂z1 +A1( ∂γ1 ∂z1 )2 +A1 ∂2γ1 ∂z21 ) e2p(z) = A1 −A2 +A1 ∂γ2 ∂z1 −A2( ∂γ2 ∂z1 )2 −A2 ∂2γ2 ∂z21 . (4.6) Now, we consider two cases. Case 1: p(z) is a constant. Let β = ep(z), then equations (4.1)-(4.2) can be written as f(z) + ∂f ∂z1 = 1√ 2 ( k1√ 1 + α + k2√ 1− α ) eg(z)/2, (4.7) f(z) + ∂2f ∂z21 = 1√ 2 ( k1√ 1 + α − k2√ 1− α ) eg(z)/2, (4.8) where k1 = β+β−1 2 , k2 = β−β−1 2i , and k21 + k22 = 1. This leads to ∂f ∂z1 − ∂2f ∂z21 = √ 2k2√ 1− α eg(z)/2. (4.9) On the other hand, differentiating with respect to z1 on equation (4.7), and combining this with (4.8), we have f(z)− ∂f ∂z1 = 1√ 2 [ (1− α1 2 ) k1√ 1 + α − (1 + α1 2 ) k2√ 1− α ] eg(z)/2. (4.10) 16 J. TU, H. WEI EJDE-2024/23 Differentiating with respect to z1 on equation (4.10), combining this with (4.9) yields ∂f ∂z1 − ∂2f ∂z21 = 1√ 2 [ ( α1 2 − α2 1 4 ) k1√ 1 + α − ( α1 2 + α2 1 4 ) k2√ 1− α ] eg(z)/2. (4.11) Since g(z) is a polynomial with the linear form g(z1, z2) = α1z1 +α2z2 +α0, where α1 ̸= 0, α2 ̸= 0, α0 ∈ C. Hence, we deduce from (4.9) and (4.11) that 8 + 2α1 + α2 1√ 1− α k2 = 2α1 − α2 1√ 1 + α k1. (4.12) According to (4.7) and (4.10), we deduce that f(z1, z2) = 1 2 √ 2 [ (2− α1 2 ) k1√ 1 + α − α1 2 k2√ 1− α ] eg(z)/2. (4.13) Subcase 1.1. If α1 ̸= 2, combining (4.12) and (4.13) yields f(z1, z2) = 4 √ 2k2 α1(2− α1) √ 1− α eg(z)/2. (4.14) Subcase 1.2. If α1 = 2, combining (4.12) and (4.13) yields f(z1, z2) = ± k1 2 √ 2(1 + α) eg(z)/2, (4.15) where k1 = ±1, k2 = 0. Thus, the proof of Theorem 2.8(i) is complete. Case 2. If p(z) is a non-constant, it follows from (4.6) that A1 −A2 −A2 ∂γ1 ∂z1 +A1( ∂γ1 ∂z1 )2 +A1 ∂2γ1 ∂z21 = 0, A1 −A2 +A1 ∂γ2 ∂z1 −A2( ∂γ2 ∂z1 )2 −A2 ∂2γ2 ∂z21 = 0. (4.16) Otherwise, without loss of generality, if A1 −A2 −A2 ∂γ1 ∂z1 +A1( ∂γ1 ∂z1 )2 +A1 ∂2γ1 ∂z2 1 ̸= 0, we have e2p(z) = A1 −A2 +A1 ∂γ2 ∂z1 −A2( ∂γ2 ∂z1 )2 −A2 ∂2γ2 ∂z2 1 A1 −A2 −A2 ∂γ1 ∂z1 +A1( ∂γ1 ∂z1 )2 +A1 ∂2γ1 ∂z2 1 . (4.17) Since p(z) and g(z) are polynomials, the left-hand side of (4.17) is transcendental, which contradicts with the right-hand side of (4.17) being a rational function. Thus, we have A2 ∂γ1 ∂z1 −A1( ∂γ1 ∂z1 )2 −A1 ∂2γ1 ∂z21 = A1 −A2, A1 ∂γ2 ∂z1 −A2( ∂γ2 ∂z1 )2 −A2 ∂2γ2 ∂z21 = A2 −A1. (4.18) In view of (4.18), we obtain that γ1 and γ2 are of the form γ1(z) = B11z1 +H(z2) +B12z2 + β1, γ2(z) = B21z1 −H(z2) +B22z2 + β2, where H(z2) = dnz n 2 + dn−1z n−1 2 + · · ·+ d2z 2 2 . Then (4.18) can be rewritten as A2B11 −A1B 2 11 = A1 −A2, A1B21 −A2B 2 21 = A2 −A1. (4.19) EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 17 By (4.1) and (4.4) we deduce that f(z1, z2) = 1 2 √ 2 [ (A1 +A2 −A1B11)e γ1(z) + (A1 +A2 −A2B21)e γ2(z) ] . (4.20) Thus, the proof of Theorem 2.8(ii) is complete. □ 5. Proof of Theorem 2.12 Proof. Suppose that f(z) is a transcendental entire solution of (2.3) with finite order. By using the same argument in the proof of Theorem 2.1, we obtain f(z) + ∂f ∂z1 = 1√ 2 ( A1e γ1(z) +A2e γ2(z) ) , (5.1) f(z) + ∂2f ∂z1∂z2 = 1√ 2 ( A2e γ1(z) +A1e γ2(z) ) , (5.2) where A1 and A2 are defined by (2.4). Thus, it follows from (5.1) and (5.2) that ∂f ∂z1 − ∂2f ∂z1∂z2 = 1√ 2 [ (A1 −A2)e γ1(z) + (A2 −A1)e γ2(z) ] . (5.3) On the other hand, differentiating with respect to z2 on equation (5.1), and combining this with (5.2), we obtain f(z)− ∂f ∂z2 = 1√ 2 [( A2 −A1 ∂γ1 ∂z2 ) eγ1(z) + ( A1 −A2 ∂γ2 ∂z2 ) eγ2(z) ] . (5.4) Differentiating with respect to z1 on equation (5.4), then combining this with ∂2f ∂z1∂z2 = ∂2f ∂z2∂z1 yields ∂f ∂z1 − ∂2f ∂z1∂z2 = 1√ 2 ( Γ1e γ1(z) + Γ2e γ2(z) ) , (5.5) where Γ1 = A2 ∂γ1 ∂z1 −A1 ∂γ1 ∂z1 ∂γ1 ∂z2 −A1 ∂2γ1 ∂z2∂z1 , Γ2 = A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z1 ∂γ2 ∂z2 −A2 ∂2γ2 ∂z2∂z1 . Thus, on the basis of (5.3) and (5.5), we have (A1 −A2 − Γ1)e 2p(z) = A1 −A2 + Γ2. (5.6) Now, we consider two cases. Case 1.=: p(z) is a constant. Let β = ep(z), then equations (5.1)-(5.2) can be written as f(z) + ∂f ∂z1 = 1√ 2 ( k1√ 1 + α + k2√ 1− α ) eg(z)/2, (5.7) f(z) + ∂2f ∂z1∂z2 = 1√ 2 ( k1√ 1 + α − k2√ 1− α ) eg(z)/2, (5.8) where k1 = β+β−1 2 , k2 = β−β−1 2i , and k21 + k22 = 1. This leads to ∂f ∂z1 − ∂2f ∂z1∂z2 = √ 2k2√ 1− α eg(z)/2. (5.9) 18 J. TU, H. WEI EJDE-2024/23 On the other hand, differentiating with respect to z2 on equation (5.7), then in line with (5.8), we have f(z)− ∂f ∂z2 = 1√ 2 [ (1− α2 2 ) k1√ 1 + α − (1 + α2 2 ) k2√ 1− α ] eg(z)/2. (5.10) Differentiating with respect to z1 on equation (5.10), then combining this with (5.9) yields ∂f ∂z1 − ∂2f ∂z1∂z2 = 1√ 2 [ ( α1 2 − α1α2 4 ) k1√ 1 + α − ( α1 2 + α1α2 4 ) k2√ 1− α ] eg(z)/2. (5.11) Since g(z) is a polynomial with the linear form g(z1, z2) = α1z1 +α2z2 +α0, where α1 ̸= 0, α2 ̸= 0, α0 ∈ C. Hence, we deduce from (5.9) and (5.11) that 8 + 2α1 + α1α2√ 1− α k2 = 2α1 − α1α2√ 1 + α k1. (5.12) From (5.7) and (5.10), we deduce that ∂f ∂z1 + ∂f ∂z2 = [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] eg(z)/2. (5.13) The characteristic equations of (5.13) are dz1 dt = 1, dz2 dt = 1, df dt = [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] eg(z)/2. Similarly, we obtain f(t, s) = ∫ t 0 [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] eg(z)/2dt+ ϕ0(s), (5.14) where ϕ0(s) is a finite order transcendental entire function in s = z2 − z1. Subcase 1.1. If α1 + α2 ̸= 0, it follows from (5.14) that f(t, s) = ∫ t 0 [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] e (α1+α2)t+α2s+α0 2 dt+ ϕ0(s) = 2 α1 + α2 [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] e (α1+α2)t+α2s+α0 2 + ϕ9(s), (5.15) where ϕ9(s) = ϕ0(s)− 2 α1 + α2 [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] e α2s+α0 2 . It follows (5.15) that f(z1, z2) = 1 α1 + α2 [ α2k1√ 2(1 + α) + (α2 + 4)k2√ 2(1− α) ] e α1z1+α2z2+α0 2 + ϕ9(z2 − z1). (5.16) Substituting (5.16) into the (5.7) or (5.8), then combining this with (5.12) yields ϕ9(z2 − z1)− ϕ′ 9(z2 − z1) = 0, (5.17) which implies ϕ9(z2 − z1) = η9e −z1+z2 , η9 ∈ C. EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 19 Subcase 1.2. If α1 + α2 = 0, then from (5.14) it follows that f(t, s) = ∫ t 0 [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] e (α1+α2)t+α2s+α0 2 dt+ ϕ0(s) = [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] e α2s+α0 2 t+ ϕ10(s), (5.18) where ϕ10(s) = ϕ0(s). In view of (5.18), we have f(z1, z2) = [ α2k1 2 √ 2(1 + α) + (α2 + 4)k2 2 √ 2(1− α) ] e α2(z2−z1)+α0 2 z1 + ϕ10(z2 − z1). (5.19) Substituting (5.19) into the (5.7) or (5.8), then combining this with (5.12) yields ϕ10(z2 − z1)− ϕ′ 10(z2 − z1) = 4k2 α1 √ 2(1− α) e α2(z2−z1)+α0 2 , (5.20) which implies ϕ10(z2 − z1) = 8k2 (2+α1)α1 √ 2(1−α) eg(z)/2 + η10e −z1+z2 , η10 ∈ C. The proof of Theorem 2.12(i) is complete. Case 2. If p(z) is a non-constant, it follows from (5.6) that A1 −A2 −A2 ∂γ1 ∂z1 +A1 ∂γ1 ∂z1 ∂γ1 ∂z2 +A1 ∂2γ1 ∂z2∂z1 = 0, A1 −A2 +A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z1 ∂γ2 ∂z2 −A2 ∂2γ2 ∂z2∂z1 = 0. (5.21) Otherwise, without loss of generality, if A1 −A2 −A2 ∂γ1 ∂z1 +A1 ∂γ1 ∂z1 ∂γ1 ∂z2 +A1 ∂2γ1 ∂z2∂z1 ̸= 0, we have e2p(z) = A1 −A2 +A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z1 ∂γ2 ∂z2 −A2 ∂2γ2 ∂z2∂z1 A1 −A2 −A2 ∂γ1 ∂z1 +A1 ∂γ1 ∂z1 ∂γ1 ∂z2 +A1 ∂2γ1 ∂z2∂z1 . (5.22) Since p(z) and g(z) are polynomials, the left-hand side of (5.22) is transcendental, which contradicts with the right-hand side of (5.22) being a rational function. In view of (5.21), we have A2 ∂γ1 ∂z1 −A1 ∂γ1 ∂z1 ∂γ1 ∂z2 −A1 ∂2γ1 ∂z2∂z1 = A1 −A2, A1 ∂γ2 ∂z1 −A2 ∂γ2 ∂z1 ∂γ2 ∂z2 −A2 ∂2γ2 ∂z2∂z1 = A2 −A1. (5.23) Next, we discuss the forms of γ1 and γ2. Set γ1 = n∑ k=0 αk(z2)z k 1 = αn(z2)z k 1 + αn−1(z2)z k−1 1 + · · ·+ α0(z2), where αn(z2), αn−1(z2), . . . , α0(z2) are polynomials is z2 and notice that z2 does not have a degree n. Differentiating with respect to z1 and z2 on γ1 respectively, 20 J. TU, H. WEI EJDE-2024/23 and substituting ∂γ1 ∂z1 , ∂γ1 ∂z2 and ∂2γ1 ∂z2∂z1 into (5.23), we have A2 n∑ k=1 kαk(z2)z k−1 1 −A1 n∑ k=1 kαk(z2)z k−1 1 n∑ k=0 α′ k(z2)z k 1 −A1 n∑ k=1 kα′ k(z2)z k−1 1 = A1 −A2. (5.24) Considering the highest degree of z1, if k ≥ 1, then α′ k(z2) ≡ 0, αk(z2) is a constant. Otherwise, the left of (5.24) is a non-constant polynomial, which contradicts the right-hand side of (5.24) being a constant. Hence, equation (5.24) can be rewritten as A2 n∑ k=1 kαkz1 k−1 −A1 n∑ k=1 kαkz1 k−1α′ 0(z2) = A1 −A2. (5.25) Obviously, α0 ′(z2) is a constant. Otherwise, considering the coefficients on both sides of z2 leads to a contradiction. Hence, we let α0 ′(z2) = c, where c is a constant. Further, if k ≥ 2, then we have A2kαk −A1ckαk = 0, A2α1 −A1α1c = A1 −A2. The formula above yields A1 = A2 which is a contradiction. If k = 0, the left-hand side of (5.23) is zero, which contradicts with the right-hand side t of (5.23) being a nonzero constant. Hence, k = 1. Whereupon, γ1 = α0(z2) + α1z1, where α0 ′(z2) is a constant. Similar to the arguments in γ2, we have the same form for γ1. Without loss of generality, we set γ1(z) = B11z1 +B12z2 + β1, γ2(z) = B21z1 +B22z2 + β2. In view of (5.23), this can be rewritten as A2B11 −A1B11B12 = A1 −A2, A1B21 −A2B21B22 = A2 −A1. (5.26) According to equation (5.1) and (5.4), we deduce that ∂f ∂z1 + ∂f ∂z2 = 1√ 2 [ (A1 −A2 +A1B12)e γ1 + (A2 −A1 +A2B22)e γ2 ] . (5.27) The characteristic equations of (5.27) are dz1 dt = 1, dz2 dt = 1, df dt = 1√ 2 [(A1 −A2 +A1B12)e γ1 + (A2 −A1 +A2B22)e γ2 ]. Similarly, we obtain f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2 +A1B12)e (B11+B12)t+B12s+β1 + (A2 −A1 +A2B22)e (B21+B22)t+B22s+β2 ] dt+ ϕ0(s), (5.28) where ϕ0(s) is a transcendental entire function with finite order in s = z2 − z1. EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 21 Subcase 2.1. If B11 +B12 ̸= 0, B21 +B22 ̸= 0, it follows from (5.28) that f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2 +A1B12)e (B11+B12)t+B12s+β1 + (A2 −A1 +A2B22)e (B21+B22)t+B22s+β2 ] dt+ ϕ0(s) = 1√ 2 [A1 −A2 +A1B12 B11 +B12 e(B11+B12)t+B12s+β1 + A2 −A1 +A2B22 B21 +B22 e(B21+B22)t+B22s+β2 ] + ϕ11(s), (5.29) where phi11(s) = ϕ0(s)− 1√ 2 [A1 −A2 +A1B12 B11 +B12 eB12s+β1 + A2 −A1 +A2B22 B21 +B22 eB22s+β2 ] . Thus, from (5.29), it follows that f(z1, z2) = 1√ 2 [A1 −A2 +A1B12 B11 +B12 eγ1(z) + A2 −A1 +A2B22 B21 +B22 eγ2(z) ] + ϕ11(z2 − z1). (5.30) Substituting (5.30) into (5.1) or (5.2), then combining this with (5.26) yields ϕ11(z2 − z1)− ϕ′ 11(z2 − z1) = 0, (5.31) which implies ϕ11(z2 − z1) = η11e −z1+z2 , η11 ∈ C. Subcase 2.2. If B11 +B12 = 0, B21 +B22 ̸= 0, then it follows from (5.28) that f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2 +A1B12)e B12s+β1 + (A2 −A1 +A2B22)e (B21+B22)t+B22s+β2 ] dt+ ϕ0(s) = 1√ 2 [ (A1 −A2 +A1B12)e B12s+β1t + A2 −A1 +A2B22 B21 +B22 e(B21+B22)t+B22s+β2 ] + ϕ12(s), (5.32) where phi12(s) = ϕ0(s)− 1√ 2 A2 −A1 +A2B22 B21 +B22 eB22s+β2 . Thus, in view of (5.32), it follows that f(z1, z2) = 1√ 2 [ (A1 −A2 +A1B12)e γ1(z)z1 + A2 −A1 +A2B22 B21 +B22 eγ2(z) ] + ϕ12(z2 − z1). (5.33) Since B11 +B12 = 0, B21 +B22 ̸= 0, in view of (5.26), we can deduce that B11 = A1 −A2 A1 , B12 = A2 −A1 A1 , or B11 = −1, B12 = 1. (5.34) Thus, there exist several cases as follows. Subcase 2.2.1. If B11 = A1−A2 A1 , it follows from (5.33) that f(z1, z2) = 1√ 2 A2 −A1 +A2B22 B21 +B22 eγ2(z) + ϕ12(z2 − z1). (5.35) 22 J. TU, H. WEI EJDE-2024/23 Substituting (5.35) into the (5.1) or (5.2), then combining this with (5.34) yields ϕ12(z2 − z1)− ϕ′ 12(z2 − z1) = 1√ 2 A1e γ1(z), ϕ12(z2 − z1)− ϕ′′ 12(z2 − z1) = 1√ 2 A2e γ1(z), which implies ϕ12(z2 − z1) = 1√ 2 A1 2 2A1 −A2 eγ1(z) + η12e −z1+z2 , η12 ∈ C. Subcase 2.2.2. If B11 = −1, it follows from (5.33) that f(z1, z2) = 1√ 2 [ (2A1 −A2)e γ1(z)z1 + A2 −A1 +A2B22 B21 +B22 eγ2(z) ] + ϕ13(z2 − z1). (5.36) Substituting (5.36) into the (5.1) or (5.2), thencombining this with (5.34) yields ϕ13(z2 − z1)− ϕ′ 13(z2 − z1) = 1√ 2 (A2 −A1)e γ1(z), ϕ13(z2 − z1)− ϕ′′ 13(z2 − z1) = 1√ 2 (2A2 − 2A1)e γ1(z), which implies ϕ13(z2 − z1) = 1√ 2 (A2 −A1)e γ1(z)(z1 − z2) + η13e −z1+z2 , η13 ∈ C. Subcase 2.3. If B11+B12 ̸= 0 and B21+B22 = 0, then it follows from (5.28) that f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2 +A1B12)e (B11+B12)t+B12s+β1 + (A2 −A1 +A2B22)e B22s+β2 ] dt+ ϕ0(s) = 1√ 2 [A1 −A2 +A1B12 B11 +B12 e(B11+B12)t+B12s+β1 + (A2 −A1 +A2B22)e B22s+β2t ] + ϕ14(s), (5.37) where ϕ14(s) = ϕ0(s)− 1√ 2 A1 −A2 +A1B12 B11 +B12 eB12s+β1 . Thus, in view of (5.37), it follows that f(z1, z2) = 1√ 2 [A1 −A2 +A1B12 B11 +B12 eγ1(z) + (A2 −A1 +A2B22)e γ2(z)z1 ] + ϕ14(z2 − z1). (5.38) Since B11 +B12 ̸= 0 and B21 +B22 = 0, in view of (5.26), we deduce that B21 = A2 −A1 A2 , B22 = A1 −A2 A2 or B21 = −1, B22 = 1. There exists several cases as follows. EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 23 Subcase 2.3.1. If B21 = A2−A1 A2 , then it follows from (5.38) that f(z1, z2) = 1√ 2 A1 −A2 +A1B12 B11 +B12 eγ1(z) + ϕ14(z2 − z1). (5.39) Substituting (5.39) into (5.1) or (5.2), then combining this with (5.26) yields ϕ14(z2 − z1)− ϕ′ 14(z2 − z1) = 1√ 2 A2e γ2(z), ϕ14(z2 − z1)− ϕ′′ 14(z2 − z1) = 1√ 2 A1e γ2(z), which implies ϕ14(z2 − z1) = 1√ 2 A2 2 2A2 −A1 eγ2(z) + η14e −z1+z2η14 ∈ C. Subcase 2.3.2. If B21 = −1, then it follows from (5.33) that f(z1, z2) = 1√ 2 [ A1 −A2 +A1B12 B11 +B12 eγ1(z) + (2A2 −A1)e γ2(z)z1] + ϕ15(z2 − z1). (5.40) Substituting (5.40) into the (5.1) or (5.2), then combining this with (5.26) yields ϕ15(z2 − z1)− ϕ′ 15(z2 − z1) = 1√ 2 (A1 −A2)e γ2(z), ϕ15(z2 − z1)− ϕ′′ 15(z2 − z1) = 1√ 2 (2A1 − 2A2)e γ2(z), which implies ϕ15(z2 − z1) = 1√ 2 (A1 −A2)e γ2(z)(z1 − z2) + η15e −z1+z2 , η15 ∈ C. Subcase 2.4. If B11+B12 = 0 and B21+B22 = 0, then it follows from (5.28) that f(t, s) = ∫ t 0 1√ 2 [ (A1 −A2 +A1B12)e B12s+β1 + (A2 −A1 +A2B22)e B22s+β2 ] dt+ ϕ0(s) = 1√ 2 [ (A1 −A2 +A1B12)e B12s+β1t + (A2 −A1 +A2B22)e B22s+β2t ] + ϕ16(s), (5.41) where ϕ16(s) = ϕ0(s) is a transcendental entire function with finite order in s. Thus, in view of (5.41), it follows that f(z1, z2) = 1√ 2 [ (A1 −A2 +A1B12)e γ1(z) + (A2 −A1 +A2B22)e γ2(z) ] z1 + ϕ16(z2 − z1). (5.42) Since B11 +B12 = 0, B21 +B22 = 0, in view of (5.26), we deduce that B11 = A1 −A2 A1 , B12 = A2 −A1 A1 or B11 = −1, B12 = 1; 24 J. TU, H. WEI EJDE-2024/23 B21 = A2 −A1 A2 , B22 = A1 −A2 A2 or B21 = −1, B22 = 1. Thus, there exist several cases, as follows/. Subcase 2.4.1. If B11 = A1−A2 A1 and B21 = A2−A1 A2 , it follows from (5.42) that f(z1, z2) = ϕ16(z2 − z1). (5.43) Substituting (5.43) into (5.1) or (5.2), and then combining this with (5.26) yields ϕ16(z2 − z1)− ϕ′ 16(z2 − z1) = 1√ 2 (A1e γ1(z) +A2e γ2(z)), ϕ16(z2 − z1)− ϕ′′ 16(z2 − z1) = 1√ 2 (A2e γ1(z) +A1e γ2(z)), which implies phi16(z2 − z1) = 1√ 2 [ A1 2 2A1 −A2 eγ1(z) + A2 2 2A2 −A1 eγ2(z) ] + η16e −z1+z2 , with η16 ∈ C. Subcase 2.4.2. If B11 = −1 and B21 = A2−A1 A2 , it follows from (5.42) that f(z1, z2) = 1√ 2 (2A1 −A2)e γ1(z)z1 + ϕ17(z2 − z1). (5.44) Substituting (5.44) into (5.1) or (5.2), then combining this with (5.26) yields ϕ17(z2 − z1)− ϕ′ 17(z2 − z1) = 1√ 2 [ (A2 −A1)e γ1(z) +A2e γ2(z) ] , ϕ17(z2 − z1)− ϕ′′ 17(z2 − z1) = 1√ 2 [ (2A2 − 2A1)e γ1(z) +A1e γ2(z) ] , which implies ϕ17(z2 − z1) = 1√ 2 [ (A2 −A1)e γ1(z)(z1 − z2) + A2 2 2A2 −A1 eγ2(z) ] + η17e −z1+z2 , with η17 ∈ C. Subcase 2.4.3. If B11 = A1−A2 A1 and B21 = −1, it follows from (5.28) that f(z1, z2) = 1√ 2 (2A2 −A1)e γ2(z)z1 + ϕ18(z2 − z1). (5.45) Substituting (5.45) into (5.1) or (5.2), then combining this with (5.26) yields ϕ18(z2 − z1)− ϕ′ 18(z2 − z1) = 1√ 2 [ A1e γ1(z) + (A1 −A2)e γ2(z) ] , ϕ18(z2 − z1)− ϕ′′ 18(z2 − z1) = 1√ 2 [ A2e γ1(z) + (2A1 − 2A2)e γ2(z) ] , which implies ϕ18(z2 − z1) = 1√ 2 [ A1 2 2A1 −A2 eγ1(z) + (A1 −A2)e γ2(z)(z1 − z2) ] + η18e −z1+z2 , with η18 ∈ C. Subcase 2.4.4. If B11 = −1 and B21 = −1, lead to p(z) being a constant. By the assumption at the begin of Case 2, we obtain a contradiction. The proof of Theorem 2.12(ii) is complete. □ EJDE-2024/23 SOLUTIONS TO QUADRATIC TRINOMIAL PDES 25 Acknowledgements. This work was supported by the National Natural Science Foundation of China (11561031, 12161074, 11761035). We want to thank the ref- erees for their suggestions which improved this article. References [1] F. Gross; On the equation fn + gn = 1, Bull. Am. Math. Soc., 72 (1966), 86-88. [2] P. C. Hu, B. Q. 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Jin Tu School of Mathematics and Statistics, Jiangxi Normal University, Nanchang Jiangxi 330022, China Email address: tujin2008@sina.com Huizhen Wei School of Mathematics and Statistics, Jiangxi Normal University, Nanchang Jiangxi 330022, China Email address: whz991013@126.com 1. Introduction 2. Results and examples 3. Proof of Theorem 2.1 4. Proof of Theorem 2.8 5. Proof of Theorem 2.12 Acknowledgements References