Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 61, pp. 1–15. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.61 EXISTENCE AND FORMS OF ENTIRE SOLUTIONS TO SYSTEM OF NON-LINEAR PARTIAL DIFFERENTIAL EQUATIONS ABHIJIT BANERJEE, JHUMA SARKAR Abstract. The main objective of this article is to explore the existence and forms of transcendental entire solutions of some systems of non-linear partial differential equations. We obtain two results and illustrate the results with several examples. This article improves the results in [5, 15]. In the last section we discuss the differences between the solutions involving homogeneous and non-homogeneous operators, and state an open question for the sake of future research. 1. Introduction The development of the difference analogue of the Nevanlinna theory [4, 11] has greatly influenced the study of difference and difference-differential equations. Naturally, this topic has become a central focus for many researchers in the field. In 1966, Gross [6] studied the existence and form of transcendental entire solution of the equation f(z)m+g(z)m = 1, and settled the problem for m = 2 and pointed out that the equation does not possess any non-constant transcendental entire solution if m > 2. This significant result opened new avenues for further exploration about the existence and form of transcendental entire solutions for variants of classical Fermat- type equations. In course of time, this line of research has gained momentum, leading to a number of interesting results by many researchers, thereby enriching the field. Theorem 1.1 ([12]). For any two positive integers m and n with m ̸= n, the equation f ′(z)n + f(z + c)m = 1, has no transcendental entire solution with finite order. Theorem 1.2 ([12]). The finite order transcendental entire solution of f ′(z)2 + f(z + c)2 = 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. 2020 Mathematics Subject Classification. 32H30, 35M30. Key words and phrases. Fermat-type equation; transcendental entire solution; partial differential operator. ©2024. This work is licensed under a CC BY 4.0 license. Submitted June 12, 2024. Published October 15, 2024. 1 2 A. BANERJEE, J. SARKAR EJDE-2024/61 In 2017, Gao [5] investigated the existence and form of transcendental entire solutions, for the following systems of Fermat-type equations: f ′ 1(z) n1 + f2(z + c)m1 = Q1(z), f ′ 2(z) n2 + f1(z + c)m2 = Q2(z), (1.1) and f ′ 1(z) 2 + f2(z + c)2 = Q1(z), f ′ 2(z) 2 + f1(z + c)2 = Q2(z), (1.2) where Qj(z), j = 1, 2 are non-zero polynomials in C. For systems (1.1) and (1.2), the following results were obtained: Theorem 1.3 ([5]). There does not exist any finite order transcendental entire solutions (f1(z), f2(z)) of (1.1) if any of the following conditions is satisfied: (i) m1m2 > n1n2; (ii) mj > nj nj−1 , j = 1, 2. Theorem 1.4 ([5]). Let (f1(z), f2(z)) be a finite order transcendental entire solu- tion of (1.2) in C. Then Q1(z) = c11c12, Q2(z) = c21c22 and f1(z) = c11e az+b1 − c12e −az−b1 2a , f2(z) = c21e az+b2 − c21e −az−b2 2a , where a4 = 1, b1, b2, ckj( ̸= 0), k, j = 1, 2 are constants. In 2018, Xu-Cao [15] investigated the existence and form of transcendental entire solutions of shift-differential equation in C2 to obtain the following result. Theorem 1.5 ([15]). Let c = (c1, c2) be a non-zero constant in C2. Then the Fermat-type partial differential equation(∂f(z1, z2) ∂z1 )n + f(z1 + c1, z2 + c2) m = 1, does not have a finite order transcendental entire solution whenever m, n are two distinct positive integer. Theorem 1.6 ([15, 16]). Let c = (c1, c2) be a non-zero constant in C2, then each finite order transcendental entire solution of the Fermat-type partial differential equation (∂f(z1, z2) ∂z1 )2 + f(z1 + c1, z2 + c2) 2 = 1, has the form f(z1, z2) = sin(Az1 + Bz2 + H(z2)), where A, B are constants in C satisfying A2 = 1, ei(Ac1+Bc2) = 1 and H(z2) is a polynomial in one variable z2 such that H(z2) = H(z2 + c2). In special case whenever c2 ̸= 0, we have f(z1, z2) = sin(Az1 +Bz2 + constant). Motivated by the results several authors made contribution to this field; see [1, 14], [17]-[21] and the references therein. EJDE-2024/61 EXISTENCE AND FORMS OF ENTIRE SOLUTIONS 3 2. Formulation of main problem and relevant examples To proceed further, we introduce the following differential-operator Definition 2.1. The partial differential operator PLn in Cn is defined as PLn = n∑ |J|=1 bj1...jn ∂|J| ∂zj11 . . . ∂zjnn , where bj1...jn ̸= 0 are constants in C and J = (j1, j2, . . . , jn), |J | = ∑n t=1 jt. From now onwards, we use z⃗n = (z1, z2, . . . , zn), c⃗n = (c1, c2, . . . , cn) and z⃗n + c⃗n = (z1 + c1, z2 + c2, . . . , zn + cn) and 0⃗ = (0, 0, . . . , 0). This article is based on exploring existence of finite order transcendental entire solutions in n (n ≥ 1) dimensional complex field of the equations (PLn (f1(z⃗n))) l1 + f2(z⃗n + c⃗n) k1 = Q1(z⃗n) (PLn (f2(z⃗n))) l2 + f1(z⃗n + c⃗n) k2 = Q2(z⃗n); (2.1) where Qj(z⃗n), j = 1, 2 are two non-zero polynomials in Cn and are of finite order transcendental entire solution for n = 2, i.e. in C2 of the equations (PL2 (f1(z⃗2))) 2 + f2(z⃗2 + c⃗2) 2 = 1 (PL2 (f2(z⃗2))) 2 + f1(z⃗2 + c⃗2) 2 = 1. (2.2) Theorem 2.2. Let c⃗ = (c1, c2, . . . , cn) be a non-zero constant in Cn. Then (2.1) can not have a finite order transcendental entire solution (f1(z⃗n), f2(z⃗n), . . . , fn(z⃗n)) if the exponents satisfy one of the following two conditions: (i) k1k2 > l1l2; (ii) kt > lt lt−1 for lt ≥ 2, t = 1, 2. The above theorem motivate us to explore the case lt = 1, and kt = 1; t = 1, 2. In this respect, the following example shows that the solution exists. Example 2.3. Let l1 = 1, l2 = 1, k1 = 1, k2 = 1, b10 = 1, b01 = 1, b11 = 1, b20 = 1, b02 = 1, Q1(z⃗2) = 1, Q2(z⃗2) = 1. Then f(z) = (f1(z⃗2), f2(z⃗2)), where fj(z⃗2) = ez1+z2 + 1, j = 1, 2 is a solution of (2.1) when ec1+c2 = −5. For the sake of convenience and to proceed further we us use the following ex- pressions A1(r, s) = −b10s+ b01r + b11(d2s− d1r) + 2b20d1s− 2b02d2r, A2(r, s) = b10s− b01r + b11(d2s− d1r) + 2b20d1s− 2b02d2r, B(r, s) = −b11rs+ b20s 2 + b02r 2, D1(r, s) = −b10r − b01s+ b11rs+ b20r 2 + b02s 2, D2(r, s) = b10r + b01s+ b11rs+ b20r 2 + b02s 2, L(r, s) = d1r + d2s, where r, s are parameters and d1, d2 are two constants in C. Theorem 2.4. Let (c1, c2) ̸= (0, 0) ∈ C2 be a constant and (f1(z⃗2), f2(z⃗2)) be a finite order transcendental entire solution of (2.2) in C2. Also let B(c1, c2), 4 A. BANERJEE, J. SARKAR EJDE-2024/61 A1(c1, c2) and B(c1, c2), A2(c1, c2) be nonzero simultaneously. Then (f1(z⃗2)f2(z⃗2)) takes one of the following form: (A) When D1(d1, d2)D2(d1, d2) = 1, e2L ⃗(c2) = D2(d1, d2) D1(d1, d2) , e2(W1+W2) = −1, eW1+W2 = i D1(d1, d2) e−L(c⃗2), we have f1 ⃗(z2) = e−L ⃗(z2)+L ⃗(c2)+W2 − eL ⃗(z2)−L ⃗(c2)−W2 2i , f2 ⃗(z2) = eL ⃗(z2)−L ⃗(c2)+W1 − e−L ⃗(z2)+L ⃗(c2)−W1 2i , where W1, W2 are two constants in C. (B) When D1(d1, d2)D2(d1, d2) = 1, e2L(c⃗2) = −D2(d1, d2) D1(d1, d2) , e2(W1−W2) = 1, eW1−W2 = − i D1(d1, d2) e−L(c⃗2), f1(z⃗2) = eL(z⃗2)−L(c⃗2)+W2 − e−L(z⃗2)+L(c⃗2)−W2 2i , f2(z⃗2) = eL(z⃗2)−L(c⃗2)+W1 − e−L(z⃗2)+L(c⃗2)−W1 2i , where W1, W2 are two constants in C. The following examples justify Theorem 2.4. Example 2.5. Let b10 = −2i, b01 = i, b11 = −2i, b20 = i, b02 = i, d1 = 1, d2 = 1, c1 = πi 4 , c2 = πi 4 , W1 = −πi 4 , W2 = −πi 4 . Then (f1(z⃗), f2(z⃗2)) = (−ie−z1−z2+ πi 4 + iez1+z2−πi 4 2 ,−e−z1−z2+ πi 4 + ez1+z2−πi 4 2 ) is a solution of (2.2). Example 2.6. Let b10 = 2i, b01 = i, b11 = 2i, b20 = i, b02 = i, d1 = 1, d2 = −1, c1 = πi 4 , c2 = −πi 4 , W1 = πi 4 , W2 = πi 4 . Then (f1(z⃗), f2(z⃗2)) = ( ie−z1+z2−πi 4 − iez1−z2+ πi 4 2 ,−e−z1+z2−πi 4 + ez1−z2+ πi 4 2 ) is a solution of (2.2). Example 2.7. Let b10 = 2i, b01 = i, b11 = 2i, b20 = i, b02 = i, d1 = 1, d2 = −1, c1 = 2πi, c2 = −2πi, W1 = 0, W2 = 0. Then (f1(z⃗2), f2(z⃗2)) = ( ie−z1+z2 − iez1−z2 2 , ie−z1+z2 − iez1−z2 2 ) is a solution of (2.2). EJDE-2024/61 EXISTENCE AND FORMS OF ENTIRE SOLUTIONS 5 Example 2.8. Let b10 = 2i, b01 = i, b11 = 2i, b20 = i, b02 = i, d1 = 1, d2 = −1, c1 = 1, c2 = 1, W1 = 0, W2 = 0. Then (f1(z⃗), f2(z⃗2)) = ( ie−z1+z2 − iez1−z2 2 , ie−z1+z2 − iez1−z2 2 ) is a solution of (2.2). Example 2.9. Let b10 = i, b01 = 2i, b11 = 2i, b20 = i, b02 = i, d1 = 1, d2 = −1, c1 = πi, c2 = −πi, W1 = πi 2 , W2 = −πi 2 . Then (f1(z⃗2), f2(z⃗2)) = (−ie−z1+z2 + iez1−z2 2 , ie−2z1+z2 − iez1−z2 2 ) is a solution of (2.2). Example 2.10. Let b10 = −2i, b01 = −i, b11 = 2i, b20 = i, b02 = i, d1 = 1, d2 = −1, c1 = 1, c2 = 1, W1 = 2πi, W2 = πi. Then (f1(z⃗), f2(z⃗2)) = (−ie−z1+z2 + iez1−z2 2 , ie−z1+z2 − iez1−z2 2 ) is a solution of (2.2). Example 2.11. Let b10 = 1, b01 = 1, b11 = 1, b20 = 1, b02 = 1, d1 = 1, d2 = −1, c1 = c2 = 1, W1 = πi 4 , W2 = πi 4 . Then (f1(z⃗), f2(z⃗2)) = (e−z1+z2−πi 4 + ez1−z2+ πi 4 2 , ie−z1+z2−πi 4 − iez1−z2+ πi 4 2 ) is a solution of (2.2). Example 2.12. Let b10 = 1, b01 = 2, b11 = 2, b20 = 1, b02 = 1, d1 = 2, d2 = −1, c1 = 2πi, c2 = 2πi, W1 = πi 4 , W2 = πi 4 . Then (f1(z⃗), f2(z⃗2)) = (e−2z1+z2−πi 4 + e2z1−z2+ πi 4 2 , ie−2z1+z2−πi 4 − ie2z1−z2+ πi 4 2 ) is a solution of (2.2). Example 2.13. Let b10 = 1, b01 = 2, b11 = 2, b20 = 1, b02 = 1, d1 = 2i, d2 = −i, c1 = i, c2 = 2i, W1 = −πi 4 , W2 = −πi 4 . Then (f1(z⃗), f2(z⃗2)) = ( − e−2iz1+iz2+ πi 4 + e2iz1−iz2−πi 4 2 , ie−2iz1+iz2+ πi 4 − ie2iz1−iz2−πi 4 2 ) is a solution of (2.2). Example 2.14. Let b10 = 1, b01 = 2, b11 = 2, b20 = 1, b02 = 1, d1 = 2i, d2 = −i, c1 = 2π, c2 = 2π, W1 = −πi 4 , W2 = −πi 4 , Then (f1(z⃗), f2(z⃗2)) = ( − e−2iz1+iz2+ πi 4 + e2iz1−iz2−πi 4 2 , ie−2iz1+iz2+ πi 4 − ie2iz1−iz2−πi 4 2 ) is a solution of (2.2). Example 2.15. Let b10 = 1, b01 = 2, b11 = 2, b20 = 1, b02 = 1, d1 = 2, d2 = −1, c1 = πi 2 , c2 = −πi 2 , W1 = πi 2 , W2 = πi 2 . Then (f1(z⃗), f2(z⃗2)) = (e−2z1+z2 + e2z1−z2 2 , e−2z1+z2 + e2z1−z2 2 ) is a solution of (2.2). 6 A. BANERJEE, J. SARKAR EJDE-2024/61 Example 2.16. Let b10 = b01 = − 6 5 , b11 = 1 5 , b20 = 8 5 , b02 = 4 5 , d1 = 1, d2 = 1, c1 = 1, c2 = log( 15 )− 1, W1 = W2 = πi 4 . Then (f1(z⃗), f2(z⃗2)) = ( 1 5e −z1−z2+ πi 4 − 5ez1+z2−πi 4 2i , 5ez1+z2+ πi 4 − 1 5e −z1−z2−πi 4 2i ) is a solution of (2.2). Corollary 2.17. Let (f1(z⃗2), f2(z⃗2)) be a transcendental entire function of order properly greater than one with B(c1, c2), A1(c1, c2) and A2(c1, c2) are not zero si- multaneously. Then (f1(z⃗2), f2(z⃗2)) can not be a solution of (2.2). 3. Lemmas We assume that the readers are familiar with the basic notations of the Nevan- linna theory such as N(r, f), N(r, 1 f ), m(r, f), T (r, f) in complex variable [7]. For several complex variables we refer to [10] and the references therein. By S(r, f) we will mean any quantity satisfying S(r, f) = o(T (r, f)), r → ∞, outside possibly an exceptional set of finite logarithmic measure. Based on the notations, the following lemmas will play important role in proving our theorems. Lemma 3.1 ([13]). For each entire function F in Cn, F (⃗0) ̸= 0⃗ and put ρ(nF ) = ρ < ∞. Then there exists a canonical function fF and a function gF ∈ Cn such that F (z) = fF (z)e gF (z⃗n) . For special case n = 1, fF is the canonical product of Weierstrass. Here ρ(nf ) denotes the order of the counting function of zeros of F . Lemma 3.2 ([22, 2]). Let f(z) be a non-constant meromorphic function in Cn and let I = (i1, . . . , in) be a multi index with length |I| = ∑n j=1 ij. Assume that T (r0, f) ≥ e for some r0. Then m ( r, ∂If f ) = S(r, f), holds for all r ≥ r0, outside a set E ⊂ (0,+∞) of finite logarithmic measure∫ E dt t < ∞, where ∂If = ∂If ∂z i1 1 ...∂zin n . Lemma 3.3 ([9]). Let fj (̸≡ 0), j = 1, 2, 3 be meromorphic function in Cn such that f1 is not constant, f1 + f2 + f3 = 1 and 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 positive number, then either f2 ≡ 1 or f3 ≡ 1. Lemma 3.4 ([3]). Let f(z⃗n) be a non-constant meromorphic function with finite order in Cn such that f (⃗0) ̸= 0,∞ and let ϵ > 0. Then, for c⃗n ∈ Cn, m ( r, f(z⃗n) f(z⃗n + c⃗n) ) +m ( r, f(z⃗n + c⃗n) f(z⃗n) ) = S(r, f), holds for all r ≥ r0, outside a set E ⊂ (0,+∞) of finite logarithmic measure∫ E dt t < ∞. EJDE-2024/61 EXISTENCE AND FORMS OF ENTIRE SOLUTIONS 7 Lemma 3.5 ([9, Lemma 3.1]). Suppose that a0(z⃗m), a1(z⃗m), . . . , an(z⃗m), n ≥ 1, are meromorphic in Cm and g0(z⃗m), g2(z⃗m), . . . , gn(z⃗m) are entire in Cm. gj(z⃗m) − gk(z⃗m) are non-constant for 0 ≤ j < k ≤ n. If n∑ j=0 aj(z⃗m)egj(z⃗m) ≡ 0 and T (r, aj) = o (T (r)), j = 0, 1, 2 . . . , n, T (r) = min 0≤j l1l2. Using Lemma 3.4, we have that m ( r, fj(z⃗n) fj(z⃗n + c⃗n) ) = S(r, fj), (4.1) holds for all r > 0 outside a possible set Ej ⊂ [1,+∞), j = 1, 2, . . . , n of finite logarithmic measure ∫ Ej dt t < ∞. Clearly, we have the following T (r, fj(z⃗n)) = m(r, fj(z⃗n)), ≤ m ( r, fj(z⃗n) fj(z⃗n + c⃗n) × fj(z⃗n + c⃗n) ) , ≤ m ( r, fj(z⃗n) fj(z⃗n + c⃗n) ) +m (r, fj(z⃗n + c⃗n)) + log 2, = m(r, fj(z⃗n + c⃗n)) + log 2 + S(r, fj), = T (r, fj(z⃗n + c⃗n)) + log 2 + S(r, fj), (4.2) for all r ̸∈ E1 ∪ E2. Applying Valliron Mohon’ko theorem in several complex variables [8] we have k1T (r, f2(z⃗n)) ≤ k1T (r, f2(z⃗n + c⃗n)) + S(r, f2) ≤ T (r, f2(z⃗n + c⃗n)) k1) + S(r, f2), = T ( r, (PLn(f1(z⃗n))) l1 −Q1(z⃗n) ) + S(r, f2), = l1T (r, PLn (f1(z⃗n))) + S(r, f1) + S(r, f2), = l1m (r, PLn (f1(z⃗n))) + S(r, f1) + S(r, f2), ≤ l1 [ m ( r, PLn (f1(z⃗n)) f1(z⃗n) ) +m(r, f1(z⃗n)) + log 2 ] + S(r, f1) + S(r, f2), = l1T (r, f1(z⃗n) + S(r, f1) + S(r, f2), (4.3) i.e. from (4.3) wee obtain (k1 + o(1))T (r, f2(z⃗n)) ≤ (l1 + o(1))T (r, f1(z⃗n)), r ̸∈ E1. (4.4) 8 A. BANERJEE, J. SARKAR EJDE-2024/61 Similarly,we obtain (k2 + o(1))T (r, f2(z⃗n)) ≤ (l2 + o(1))T (r, f1(z⃗n)), r ̸∈ E2. (4.5) From (4.4) and (4.5) clearly we have a contradiction. Case 2: Let kt > lt lt−1 , lt ≥ 2, t = 1, 2. Using the Nevanlinna second main theorem, from (2.1) we obtain (l1 − 1)T (r, PLn (f1(z⃗n))) ≤ N ( r, PLn (f1(z⃗n)) ) +N ( r, 1 (PLn (f1(z⃗n))) l1 −Q1(z⃗n) ) + S (r, PLn (f1)) , ≤ N ( r, 1 f2(z⃗n + c⃗n) ) + S(r, f1), ≤ T (r, f2(z⃗n + c⃗n)) + S(r, f1), ≤ T (r, f2(z⃗n)) + S(r, f1) + S(r, f2). (4.6) Proceeding with the similar arguments, from the second equation we obtain (l2 − 1)T (r, PLn f2((z⃗n )) ≤ T (r, f1(z⃗n)) + S(r, f1) + S(r, f2). (4.7) From the first equation of (2.1) and using Valliron Mohon’ko theorem in several complex variables [8] we obtain k1T (r, f2(z⃗n + c⃗n)) = T (r, (PLn(f1(z⃗n))) l1 −Q1(z⃗n)) + S(r, f1) ≤ l1T (r, PLn (f1(z⃗n))) + S(r, f1) + S(r, f1). (4.8) Proceeding, in the similar way from the second equation of (2.1) we obtain k2T (r, f1(z⃗n + c⃗n)) ≤ l2T (r, PLn(f2(z⃗n)) + S(r, f1) + S(r, f2). (4.9) From (4.6)-(4.9) we obtain( k1 − l1 l1 − 1 + o(1) ) T (r, f2(z⃗n)) ≤ S(r, f1),( k2 − l2 l2 − 1 + o(1) ) T (r, f1(z⃗n)) ≤ S(r, f2). Since (f1(z⃗n), f2(z⃗n), . . . , fn(z⃗n)) is a transcendental entire function, we obtain( k1 − l1 l1 − 1 + o(1) )( k2 − l2 l2 − 1 + o(1) ) ≤ 0. Since kt > lt lt−1 , t = 1, 2, we have a contradiction. The proof of Theorem 2.2 is complete □ The following expression is used several times to prove the next theorem. Mm,u(p) = b10 ∂pm(z⃗2) ∂z1 + b01 ∂pm(z⃗2) ∂z2 + b11 {∂2pm(z⃗2) ∂z1∂z2 + (−1)u−1 ∂pm(z⃗2) ∂z1 ∂pm(z⃗2) ∂z2 } + b20 {∂2pm(z⃗2) ∂z21 + (−1)u−1 (∂pm(z⃗2) ∂z1 )2} + b02 {∂2pm(z⃗2) ∂z22 + (−1)u−1 (∂pm(z⃗2) ∂z2 )2} , for m,u = 1, 2. EJDE-2024/61 EXISTENCE AND FORMS OF ENTIRE SOLUTIONS 9 Proof of Theorem 2.4. Let (f1(z⃗2), f2(z⃗2)) be a pair of finite order transcendental entire solution of (2.2) in C2. Clearly, system (2.2) can be re-written as follows {PL2 (f1(z⃗2)) + if2(z⃗2 + c⃗2)}{PL2 (f1(z⃗2))− if2(z⃗2 + c⃗2)} = 1, {PL2 (f2(z⃗2) + if1(z⃗2 + c⃗2)}{PL2 (f2(z⃗2))− if1(z⃗22 + c⃗2)} = 1. (4.10) Now using Lemma 3.1, from (4.10) we obtain PL2(f1(z⃗2)) + if2(z⃗2 + c⃗2) = ep1(z⃗2), PL2 (f1(z⃗2))− if2(z⃗2 + c⃗2) = e−p1(z⃗2), PL2 (f2(z⃗2)) + if1(z⃗2 + c⃗2) = ep2(z⃗2), PL2(f2(z⃗2)− if1(z⃗2 + c⃗2) = e−p2(z⃗2), (4.11) where p1(z⃗2), p2(z⃗2) are two non-constant polynomials in C2. By an easy compu- tation from (4.11), we obtain PL2 (f1(z⃗2)) = ep1(z⃗2) + e−p1(z⃗2) 2 , f2(z⃗2 + c⃗2) = ep1(z⃗2) − e−p1(z⃗2) 2i , PL2 (f2(z⃗2)) = ep2(z⃗2) + e−p2(z⃗2) 2 , f1(z⃗2 + c⃗2) = ep2(z⃗2) − e−p2(z⃗2) 2i . (4.12) Combining the first and the last equations, and the second and the third equations of (4.12) we obtain respectively −iM2,1e p1(z⃗2+c⃗2)+p2(z⃗2) − iM2,2e p1(z⃗2+c⃗2)−p2(z⃗2) − e2p1(z⃗2+c⃗2) = 1, (4.13) and −iM1,1e p2(z⃗2+c⃗2)+p1(z⃗2) − iM1,2e p2(z⃗2+c⃗2)−p1(z⃗2) − e2p2(z⃗2+c⃗2) = 1. (4.14) Now taking into consideration equation (4.13), we discuss the following possibil- ities: (i) Let M2,1 ≡ 0, M2,2 ≡ 0. Then we have −e2p1(z⃗2+c⃗2) = 1, which shows that p1(z⃗2) is a constant polynomial, a contradiction. (ii) Let M2,1 ≡ 0 and M2,2 ̸≡ 0. Then we have −iM2,2e p1(z⃗2+c⃗2)−p2(z⃗2) − e2p1(z⃗2+c⃗2) = 1. (4.15) Since p1(z⃗2) is a non-constant polynomial, (4.15) implies that p1(z⃗2 + c⃗2)− p2(z⃗2) is also non-constant. We claim that −p2(z⃗2)− p1(z⃗2 + c⃗2) is also non-constant. On the contrary, let −p2(z⃗2)− p1(z⃗2 + c⃗2) = A′ 1, where A′ 1 is a constant in C. Then from (4.15) we obtain −iM2,2e A′ 1+2p1(z⃗2+c⃗2) − e2p1(z⃗2+c⃗2) = 1, i.e. (iM2,2e A′ 1 + 1)e2p1(z⃗2+c⃗2) = −1. Then we have p1(z⃗2) is a constant polynomial, a contradiction. Clearly, we can rewrite (4.15) as −iM2,2e −p2(z⃗2) − ep1(z⃗2+c⃗2) − e−p1(z⃗2+c⃗2) = 0. (4.16) Now applying Lemma 3.5 in (4.16) we have M2,2 ≡ 0, a contradiction. 10 A. BANERJEE, J. SARKAR EJDE-2024/61 (iii) Let M2,1 ̸≡ 0 and M2,2 ≡ 0. Then proceeding in the similar way as done in case (ii) we obtain a contradiction. So we must have M2,1 ̸≡ 0 and M2,2 ̸≡ 0. Using similar arguments from (4.14) we obtain M1,1 ̸≡ 0 and M1,2 ̸≡ 0. Hence using Lemma 3.3, in (4.13) and (4.14) we obtain −iM2,1e p1(z⃗2+c⃗2)+p2(z⃗2) ≡ 1 or − iM2,2e p1(z⃗2+c⃗2)−p2(z⃗2) ≡ 1; −iM1,1e p2(z⃗2+c⃗2)+p1(z⃗2) ≡ 1 or − iM1,2e p2(z⃗2+c⃗2)−p1(z⃗2) ≡ 1, respectively. Now we consider the following four cases: Case 1: −iM2,1e p1(z⃗2+c⃗2)+p2(z⃗2) ≡ 1, −iM1,1e p2(z⃗2+c⃗2)+p1(z⃗2) ≡ 1. Clearly we have p1(z⃗2 + c⃗2) + p2(z⃗2) ≡ η1, p2(z⃗2 + c⃗2) + p1(z) ≡ η2, where η1, η2 are two constants in C. Then we have p1(z⃗2) = L(z⃗2) + H(s) + W1, p2 ⃗(z2) = −L(z⃗2)−H(s) +W2, where W1, W2 are two constants in C, H(s) is a polynomial in s = c2z1 − c1z2. Now combining with (4.13) and (4.14) we obtain b10 (−d1 −H ′(s)c2) + b01 (−d2 +H ′(s)c1) + b11{H ′′(s)c1c2 + (−d1 −H ′(s)c2) (−d2 +H ′(s)c1)} + b20{−H ′′(s)c22 + (−d1 −H ′(s)c2) 2} + b02{−H ′′(s)c21 + (−d2 +H ′(s)c1) 2}eL(c⃗2)+W1+W2 ≡ i, b10 (d1 +H ′(s)c2) + b01 (d2 −H ′(s)c1) + b11{−H ′′(s)c1c2 + (d1 +H ′(s)c2) (d2 −H ′(s)c1)}+ b20{H ′′(s)c22 + (d1 +H ′(s)c2) 2}+ b02{H ′′(s)c21 + (d2 −H ′(s)c1) 2}e−L(c⃗2)+W1+W2 ≡ i, b10 (−d1 −H ′(s)c2) + b01 (−d2 +H ′(s)c1) + b11{H ′′(s)c1c2 − (−d1 −H ′(s)c2) (−d2 +H ′(s)c1)} + b20{−H ′′(s)c22 − (−d1 −H ′(s)c2) 2} + b02{−H ′′(s)c21 − (−d2 +H ′(s)c1) 2}e−L(c⃗2)−W1−W2 ≡ i, b10 (d1 +H ′(s)c2) + b01 (d2 −H ′(s)c1) + b11{−H ′′(s)c1c2 − (d1 +H ′(s)c2) (d2 −H ′(s)c1)} + b20{H ′′(s)c22 − (d1 +H ′(s)c2) 2} + b02{H ′′(s)c21 − (d2 −H ′(s)c1) 2}eL(c⃗2)−W1−W2 ≡ i. (4.17) We note that coefficient of H ′(s) of the first, second, third, and fourth equa- tions are A1(c1, c2), A2(c1, c2), −A2(c1, c2), and −A1(c1, c2) respectively. Also, coefficients of H ′(s)2 of the first, second, third, and fourth equations are B(c1, c2), B(c1, c2), −B(c1, c2), and −B(c1, c2) respectively. Further, the coefficients ofH ′′(s) of the first, second, third, and fourth equations are−B(c1, c2), B(c1, c2), −B(c1, c2), and B(c1, c2) respectively. EJDE-2024/61 EXISTENCE AND FORMS OF ENTIRE SOLUTIONS 11 Then (4.17) reduces to [D1(d1, d2) +A1(c1, c2)H ′(s) +B(c1, c2){H ′(s)2 −H ′′(s)}] × eL(c⃗2)+W1+W2 ≡ i, [D2(d1, d2) +A2(c1, c2)H ′(s) +B(c1, c2){H ′(s)2 +H ′′(s)}] × e−L(c⃗2)+W1+W2 ≡ i, [−D2(d1, d2)−A2(c1, c2)H ′(s)−B(c1, c2){H ′(s)2 +H ′′(s)}] × e−L(c⃗2)−W1−W2 ≡ i, [−D1(d1, d2)−A1(c1, c2)H ′(s)−B(c1, c2){H ′(s)2 −H ′′(s)}] × eL(c⃗2)−W1−W2 ≡ i. (4.18) Taking into consideration the first and fourth equations of (4.18), we have the following: (a) A1(c1, c2) ̸= 0, B(c1, c2) ̸= 0, then degree of H(s) ≤ 1. (b) A1(c1, c2) = 0, B(c1, c2) ̸= 0, then degree of H(s) ≤ 1. (c) A1(c1, c2) ̸= 0, B(c1, c2) = 0, then degree of H(s) ≤ 1. (d) A1(c1, c2) = 0, B(c1, c2) = 0, then degree of H(s) can be any finite number. Now using the first assumption of Theorem 2.4, i.e. B(c1, c2) and A1(c1, c2) are not zero simultaneously, we obtain degree of H(s) ≤ 1. Since under H(s) ≤ 1; p1(z⃗2), p2(z⃗2) both become linear polynomials, without loss of generality we can consider H(s) ≡ 0. Then from first and fourth equations of (4.18) we must have D1(d1, d2)e L(c⃗2)+W1+W2 = i, −D1(d1, d2)e L(c⃗2)−W1−W2 = i. (4.19) Let us consider the second and third equations of (4.18). We have the following 4 possibilities: (e) A2(c1, c2) ̸= 0, B(c1, c2) ̸= 0, degree of H(s) ≤ 1. (f) A2(c1, c2) = 0, B(c1, c2) ̸= 0, degree of H(s) ≤ 1. (g) A2(c1, c2) ̸= 0, B(c1, c2) = 0, degree of H(s) ≤ 1. (h) A2(c1, c2) = 0, B(c1, c2) = 0, degree of H(s) is arbitrary finite number. Using the assumption of Theorem 2.4, which is B(c1, c2) and A2(c1, c2) are not zero simultaneously, we must have deg(H(s) ≤ 1. Since p1(z⃗2), p2(z⃗2) becomes a linear polynomial, without any loss of generality we consider H(s) ≡ 0. Then from second and third equations of (4.18) we obtain D2(d1, d2)e −L(c⃗2)+W1+W2 = i, −D2(d1, d2)e −L(c⃗2)−W1−W2 = i. (4.20) Considering all conditions such that degree of H(s) ≤ 1 i.e. B(c1, c2), A1(c1, c2) and B(c1, c2), A2(c1, c2) are not zero simultaneously, from (4.19) and (4.20) we have D1(d1, d2)D2(d1, d2) = 1, e2L(c⃗2) = D2(d1, d2) D1(d1, d2) , e2(W1+W2) = −1, eW1+W2 = i D1(d1, d2) e−L(c⃗2). 12 A. BANERJEE, J. SARKAR EJDE-2024/61 The form of the solution is f1 ⃗(z2) = e−L ⃗(z2)+L ⃗(c2)+W2 − eL ⃗(z2)−L ⃗(c2)−W2 2i , f2 ⃗(z2) = eL ⃗(z2)−L ⃗(c2)+W1 − e−L ⃗(z2)+L ⃗(c2)−W1 2i . Case 2: Let −iM2,1e p1(z⃗2+c⃗2)+p2(z⃗2) ≡ 1, −iM1,2e p2(z⃗2+c⃗2)−p1(z⃗2) ≡ 1. Clearly we have p1(z⃗2+ c⃗2)+p2(z⃗2) ≡ η1, p2(z⃗2+ c⃗2)−p1(z⃗2) ≡ η2, where η1, η2 are two constants in C. Then by easy computation we obtain p1(z⃗2 + 2c⃗2) + p1(z⃗2) ≡ η1 − η2, which contradicts that p1(z⃗2) is a non-constant polynomial. Case 3: Let −iM2,2e p1(z⃗2+c⃗2)−p2(z⃗2) ≡ 1, −iM1,1e p2(z⃗2+c⃗2)+p1(z⃗2) ≡ 1. Then by using similar arguments as in Case 2, we obtain a contradiction. Case 4: Let −iM2,2e p1(z⃗2+c⃗2)−p2(z⃗2) ≡ 1, −iM1,2e p2(z⃗2+c⃗2)−p1(z⃗2) ≡ 1. Then clearly we have p1(z⃗2 + c⃗2) − p2(z⃗2) ≡ η1, p2(z⃗2 + c⃗2) − p1(z⃗2) ≡ η2, where η1, η2 be two constants in C. Let us take p1(z⃗2) = L(z⃗2) + H(s) + W1, p2(z⃗2) = L(z⃗2) + H(s) + W2, where W1, W2 are constants in C, H(s) is a polynomial in s = c2z1 − c1z2. Then combining this with (4.13) and (4.14), we obtain b10 (d1 +H ′(s)c2) + b01 (d2 −H ′(s)c1) + b11{−H ′′(s)c1c2 − (d1 +H ′(s)c2) (d2 −H ′(s)c1)} + b20{H ′′(s)c22 − (d1 +H ′(s)c2) 2} + b02{H ′′(s)c21 − (d2 −H ′(s)c1) 2}eL(c⃗2)+W1−W2 ≡ i, b10 (d1 +H ′(s)c2) + b01 (d2 −H ′(s)c1) + b11{−H ′′(s)c1c2 − (d1 +H ′(s)c2) (d2 −H ′(s)c1)} + b20{H ′′(s)c22 − (d1 +H ′(s)c2) 2} + b02{H ′′(s)c21 − (d2 −H ′(s)c1) 2}eL(c⃗2)−W1+W2 ≡ i, (4.21) b10 (d1 +H ′(s)c2) + b01 (d2 −H ′(s)c1) + b11{−H ′′(s)c1c2 + (d1 +H ′(s)c2) (d2 −H ′(s)c1)} + b20{H ′′(s)c22 + (d1 +H ′(s)c2) 2} + b02{H ′′(s)c21 + (d2 −H ′(s)c1) 2}e−L(c⃗2)−W1+W2 ≡ i, EJDE-2024/61 EXISTENCE AND FORMS OF ENTIRE SOLUTIONS 13 b10 (d1 +H ′(s)c2) + b01 (d2 −H ′(s)c1) + b11{−H ′′(s)c1c2 + (d1 +H ′(s)c2) (d2 −H ′(s)c1)} + b20{H ′′(s)c22 + (d1 +H ′(s)c2) 2} + b02{H ′′(s)c21 + (d2 −H ′(s)c1) 2}e−L(c⃗2)+W1−W2 ≡ i. Proceeding with the similar methods as done in Case 1 we conclude that H(s) ≡ 0. Then from (4.21) we obtain −D1(d1, d2)e L(c⃗2)+W1−W2 = i, −D1(d1, d2)e L(c⃗2)−W1+W2 = i, D2(d1, d2)e −L(c⃗2)−W1+W2 = i, D2(d1, d2)e −L(c⃗2)+W1−W2 = i. Clearly we have D1(d1, d2)D2(d1, d2) = 1, e2L(c⃗2) = −D2(d1, d2) D1(d1, d2) , e2(W1−W2) = 1, eW1−W2 = − i D1(d1, d2) e−L(c⃗2). In this case the form of the solution is f1(z⃗2) = eL(z⃗2)−L(c⃗2)+W2 − e−L(z⃗2)+L(c⃗2)−W2 2i , f2(z⃗2) = eL(z⃗2)−L(c⃗2)+W1 − e−L(z⃗2)+L(c⃗2)−W1 2i . □ 5. Discussion related to Theorem 2.4 and an open question From the expressions of D1(d1, d2), D2(d1, d2) we see that they are related in a certain way. More elaborately, when we consider only the second degree homo- geneous differential operator, then D1(d1, d2) = D2(d1, d2) and when we consider only the first order differential operator, then D1(d1, d2) = −D2(d1, d2). Now we discuss the following cases: Case 1: Let D1(d1, d2) = D2(d1, d2) = D(d1, d2). Then from Case 1 and Case 4 in Theorem 2.4, we obtain D2(d1, d2) = 1, that is D(d1, d2) = ±1. Case 2: Let D1(d1, d2) = −D2(d1, d2) = D(d1, d2). Then from Case 1 and Case 4 in Theorem 2.4, we obtain D2(d1, d2) = −1, that is D(d1, d2) = ±i. Combining Case 1 and Case 2 we clearly see that D1(d1, d2) and D2(d1, d2) can take the values {1,−1, i,−i} with D1(d1, d2)D2(d1, d2) = 1. In particular, we can write the solution of equation (2.2) as the follows: Let f1(z⃗) = S11e −L(z⃗2)−W1 + S12e L(z⃗2)+W1 2 , f2(z⃗) = S21e −L(z⃗2)−W1 + S22e L(z⃗2)+W1 2 , where W1, W2 and S11, S12, S21, S22 are constants in C. Now under the conclusion (A) in Theorem 2.4, we have the following: (i) D1(d1, d2) = i, D2(d1, d2) = −i and eL(c⃗2) = i, eW1+W2 = −i, then S11 = −i, S12 = i, S21 = −1, S22 = −1 or eL(c⃗2) = −i, eW1+W2 = i, then S11 = −i, S12 = i, S21 = 1, S22 = 1; or 14 A. BANERJEE, J. SARKAR EJDE-2024/61 (ii) D1(d1, d2) = −i, D2(d1, d2) = i and eL(c⃗2) = i, eW1+W2 = i, then S11 = i, S12 = −i, S21 = −1, S22 = −1 or eL(c⃗2) = −i, eW1+W2 = −i, then S11 = i, S12 = −i, S21 = 1, S22 = 1; or (iii) D1(d1, d2) = 1, D2(d1, d2) = 1 and eL(c⃗2) = 1, eW1+W2 = i, then S11 = 1, S12 = 1, then S21 = i, S22 = −i; eL(c⃗2) = −1, eW1+W2 = −i, then S11 = 1, S12 = 1, then S21 = i, S22 = −i; or (iv) D1(d1, d2) = −1, D2(d1, d2) = −1 and eL(c⃗2) = 1, eW1+W2 = −i, then S11 = −1, S12 = −1, S21 = i, S22 = −i or eL(c⃗2) = −1, eW1+W2 = i, then S11 = −1, S12 = −1, S21 = −i, S22 = i. Similarly, under conclusion (B) in Theorem 2.4, we have the following (i) D1(d1, d2) = −i, D2(d1, d2) = i, and eL(c⃗2) = 1, eW1−W2 = 1, then S11 = i, S12 = −i, S21 = i, S22 = −i; or eL(c⃗2) = −1, eW1−W2 = −1, then S11 = i, S12 = −i, S21 = −i, S22 = i; o (ii) D1(d1, d2) = i, D2(d1, d2) = −i and eL(c⃗2) = 1, eW1−W2 = −1, then S11 = −i, S12 = i, S21 = i, S22 = −i; or eL(c⃗2) = −1, eW1−W2 = 1, then S11 = −i, S12 = i, S21 = −i, S22 = i; or (iii) D1(d1, d2) = −1, D2(d1, d2) = −1 and eL(c⃗2) = i, eW1−W2 = 1, then S11 = −1, S12 = −1, S21 = −1, S22 = −1, or eL(c⃗2) = −i, eW1−W2 = −1, then S11 = −1, S12 = −1, S21 = 1, S22 = 1; or (iv) D1(d1, d2) = 1, D2(d1, d2) = 1, and eL(c⃗2) = i, eW1−W2 = −1, then S11 = 1, S12 = 1, S21 = −1, S22 = −1; eL(c⃗2) = −i, eW1−W2 = 1, then S11 = 1, S12 = 1, S21 = 1, S22 = 1. In view of (2.1) and (2.2) the following question is inevitable: What will be the possible form of transcendental entire solution of the following system of equation in Cn (PL2 (f1(z⃗2))) 2 + f2(z⃗2 + c⃗2) 2 = Q1(z⃗2), (PL2 (f2(z⃗2)) 2 + f1(z⃗2 + c⃗2) 2 = Q2(z⃗2); where Qj(z⃗2), j = 1, 2 are two non-zero polynomials in Cn? Acknowledgements. The authors would like to thank the anonymous referee for his/her valuable suggestions taht improved the overall presentation of the manu- script. J. Sarkar wishes to thank the Council of Scientific and Industrial Research (India), for their financial help under File No.-09/0106(13572)/2022-EMR-I. References [1] A. Banerjee, J. Sarkar; On the solutions of Fermat-type quadratic trinomial equations in C2 generated by first order linear c-shift and partial differential operators, Advances. Pure. App. Math., 15(1) (2024), 44-69. [2] A. Biancofiore, W. 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Haldar; Entire solutions to Fermat-type difference and partial differential- difference equations in Cn, Electronic J. Diff. Equa., 2024 (26) (2024), 1-21. [21] H. Y. Xu, K. Liu, Z. X. Xuan; Results on solutions of several product type nonlinear partial differential equations in C3, J. Math. Anal. Appl. In press (2025), no. 128885, 1-20. [22] Z. Ye; On Nevanlinna’s second main theorem in projective space, Invent. Math., 122(1) (1995), 475-507. Abhijit Banerjee Department of Mathematics, University of Kalyani, West Bengal 741235, India Email address: abanerjee kal@yahoo.co.in Jhuma Sarkar Department of Mathematics, University of Kalyani, West Bengal 741235, India Email address: jhumasarkar928@gmail.com 1. Introduction 2. Formulation of main problem and relevant examples 3. Lemmas 4. Proof of the main results 5. Discussion related to Theorem 2.4 and an open question Acknowledgements References