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Solutions of several general quadratic partial differential-difference equations in ℂ2

  • Molla Basir Ahamed EMAIL logo and Sanju Mandal
Published/Copyright: June 18, 2025
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Abstract

In this article, we have introduced general transformation to solving the general quadratic equations. It is of interest to know about the existence and form of the solutions of general quadratic functional equations. By utilizing Nevanlinna theory for several complex variables and the transformation, we study the existence and form of the solutions to the general quadratic partial differential or partial differential-difference equations of the form a f 2 + 2 α f g + b g 2 + 2 β f + 2 γ g + C = 0 in C 2 . Consequently, we obtain certain corollaries of the main results of this article concerning trinomial equations which generalize many results. In addition, some examples relevant to the content of the article has been exhibited.

MSC 2010: 39A45; 30D35; 35M30; 32W50

1 Introduction

It is known that the Fermat equation x m + y m = 1 when m 3 does not admit nontrivial solutions in rational numbers by Fermat’s last theorem [1,2] the equation when m = 2 does admit nontrivial rational solutions. For positive integer m , the functional equation

(1.1) f m ( z ) + g m ( z ) = 1 for integer m 1

can be regarded as the Fermat-type equations over function fields. The study of equation goes back to Montel [3] and Gross [4]. In fact, the study of Fermat-type functional equations has a long history. In 1939 , Iyer [5] studied the entire solutions of (1.1) for m = 2 and showed that the only solution is of the form f ( z ) = cos ( φ ( z ) ) and g ( z ) = sin ( φ ( z ) ) , where φ is an entire function. The study of Fermat-type equations continue to be a subject of research and exploration for mathematicians. Investigating properties of solutions, generalizations, and related equations leads to new insights into number theory and related mathematical areas [69].

The study of general quadratic equations is motivated by its importance and relevance in various branches of mathematics and science. Such an equations often arise in problems related to optimization, control theory, physics, and engineering. This type of equations is essential for the development of modern technology and science. Moreover, the general quadratic equations has been an interesting subject in the field of complex analysis in connection with extensions of Nevanlinna’s theory. As we know that, partial differential equations are occurring in various areas of applied mathematics, such as fluid mechanics, nonlinear acoustic, gas dynamics and traffic flow [10,11]. In general, it is difficult to find entire and meromorphic solutions for a nonlinear PDE. By employing Nevanlinna theory and the method of complex analysis, there were a number of literature focusing on the solutions of the PDEs and theirs many variants.

Let us also recall here the definitions of difference equations and partial differential-difference equations (PDDEs) in C n .

Definition 1.1

[12] An equation is called a difference equation, if the equation includes shifts or difference of f ( z ) in C n , which can be called DE for short. An equation is called a complex PDDE, if this equation includes partial derivatives, shifts, or differences of f ( z ) in C n .

Due to the development of the difference analogues of Nevanlinna theory in C by Halburd and Korhonen [13], Chiang and Feng [14], as well as in C n [1520], researchers have devoted considerable attention to exploring various properties of the solutions to complex difference equations in both C and C n . Of particular note, in [21], Khavinson (and also Saleeby in [22]) derived that any entire solution of the partial differential equation

u z 1 2 + u z 2 2 = 1

must be linear, i.e., the solution takes the form u ( z 1 , z 2 ) = a z 1 + b z 2 + c , where a , b , c C , and a 2 + b 2 = 1 . In fact, Khavinson’s work inspired researchers on finding the entire solutions to the equations u z 1 2 + u z 2 2 = 1 in C 2 . For example, Li [2325] investigated on the partial differential equations with more general forms such as u z 1 2 + u z 2 2 = p , u z 1 2 + u z 2 2 = e q , etc., where p , q are polynomials in C 2 . In this article, with the help of the difference Nevanlinna theory for several complex variables [16,26], we obtain some results on the existence of the solutions for some Fermat-type PDDEs from one complex variable to several complex variables. There are few results in the study of solutions for complex PDDEs [2735].

In [36], Li showed that meromorphic solutions f and g of Fermat-type functional equation f 2 + g 2 = 1 in C 2 must be constant if, and only if, f z 2 and g z 1 have the same zeros (counting multiplicities). This shows the relationship between the existence of solutions and differential operators of solutions. Saleeby in [37] made elementary observations on right factors of meromorphic functions to describe complex analytic solutions to the quadratic functional equation f 2 + 2 α f g + g 2 = 1 , where α 2 1 is a constant in C . Saleeby’s work also associated with partial differential equations of the form u x 2 + 2 α u x u y + u y 2 = 1 . Liu and Yang [38] further investigated the existence and form of solutions for quadratic trinomial functional equations and proved that if α ± 1 , 0, then the equation f ( z ) 2 + 2 α f ( z ) f ( z ) + f ( z ) 2 = 1 has no transcendental meromorphic solutions. However, the equation f ( z ) 2 + 2 α f ( z ) f ( z + c ) + f ( z + c ) 2 = 1 must have transcendental entire solutions of order equal to one.

Recently, in [39], Xu et al. have investigated solutions of the following Fermat-type PDDEs. Our objective in this article is to explore the solutions of equations in [39] with a more general setting, specifically for general quadratic PDDEs. The study of finding solutions of general quadratic equations provide a broader perspective compared to solutions of these binomial and trinomial equations in one and several complex variables. We have seen that the equations in [39] are of the form F 2 + G 2 = 1 (binomial type). Additionally, it is evident that F 2 + 2 α F G + G 2 = 1 (trinomial type) represents a generalization of the binomial equation. Furthermore, the general quadratic equation of the form a F 2 + 2 α F G + b G 2 + 2 β F + 2 γ G + C = 0 serves as a natural extension encompassing both the binomial and trinomial equations.

Therefore, to further explore Fermat-type functional equations, it is natural to pose the following question.

Question 1.1

Does there exists solutions in C 2 when considering general quadratic equations instead of the equations in [39]? If so, can we determine the exact form of the solutions for such equations?

In this article, our main objective is to provide answer to Question 1.1. In other words, our main interest lies in finding the solutions of the functional equation P ( f , g ) = 0 , where P ( f , g ) = a f 2 + 2 α f g + b g 2 + 2 β f + 2 γ g + C , and ensuring that it is irreducible, which is equivalent to saying that the corresponding curve is nonsingular. Henceforth, we introduce the following notations:

L 1 ( f ( z ) ) = f ( z ) + f ( z ) z 1 , L 2 ( f ( z ) ) = f ( z ) + f ( z ) z 2 , L 3 ( f ( z ) ) = f ( z ) + 2 f ( z ) z 1 2 , L 4 ( f ( z ) ) = f ( z ) + 2 f ( z ) z 1 z 2 , M 1 ( f ( z ) ) = f ( z + c ) + f ( z ) z 1 , M 2 ( f ( z ) ) = f ( z + c ) + f ( z ) z 2 , M 3 ( f ( z ) ) = f ( z + c ) + 2 f ( z ) z 1 2 .

Specifically, we will utilize Nevanlinna theory to investigate the existence and precise form of the solutions to the following general quadratic PDDEs:

(1.2) a L 1 ( f ) 2 + 2 α L 1 ( f ) L 2 ( f ) + b L 2 ( f ) 2 + 2 β L 1 ( f ) + 2 γ L 2 ( f ) + C = 0 , a L 1 ( f ) 2 + 2 α L 1 ( f ) L 3 ( f ) + b L 3 ( f ) 2 + 2 β L 1 ( f ) + 2 γ L 3 ( f ) + C = 0 , a L 1 ( f ) 2 + 2 α L 1 ( f ) L 4 ( f ) + b L 4 ( f ) 2 + 2 β L 1 ( f ) + 2 γ L 4 ( f ) + C = 0 , a M 1 ( f ) 2 + 2 α M 1 ( f ) M 2 ( f ) + b M 2 ( f ) 2 + 2 β M 1 ( f ) + 2 γ M 2 ( f ) + C = 0 , a M 1 ( f ) 2 + 2 α M 1 ( f ) M 3 ( f ) + b M 3 ( f ) 2 + 2 β M 1 ( f ) + 2 γ M 3 ( f ) + C = 0 .

Motivated by Saleeby’s articles in [37,40], our main focus in this article is to establish results for the equations in (1.2), providing a comprehensive answer to Question 1.1. In these equations, we assume that the corresponding companion matrices are nonsingular, i.e., Δ 0 . If this condition is not met, each equation can be linearly factorized, which would demonstrate that the function f becomes a constant.

2 Method for solving the general quadratic equations

The most general quadratic first order partial differential equation P ( u x , u y ) = 0 over C can be written as follows:

(2.1) P ( u x , u y ) a u x 2 + 2 α u x u y + b u y 2 + 2 β u x + 2 γ u y + C ,

where a , α , b , β , γ , C are complex constants and ( x , y ) C 2 . It is easy to see that (2.1) is associated with the algebraic equation:

(2.2) a x 2 + 2 α x y + b y 2 + 2 β x + 2 γ y + C = 0 .

This will represent a pair of straight lines only if the L.H.S of (2.2) can be resolve into two linear factors. By arranging the equation as a quadratic in x , we have

a x 2 + 2 ( α y + β ) x + ( b y 2 + 2 γ y + C ) = 0 .

Therefore, if a 0 , then a simple computation shows that

x = ( α y + β ) ± ( α y + β ) 2 a ( b y 2 + 2 γ y + C ) a .

Hence, we see that left-hand side of (2.2) can be resolve into two linear factors, if ( α y + β ) 2 a ( b y 2 + 2 γ y + C ) , i.e., y 2 ( α 2 a b ) + 2 ( α β a γ ) y + ( β 2 a C ) would be a perfect square. This is a quadratic in y and hence is a perfect square, if its discriminant be zero, i.e., if 4 ( α β a γ ) 2 4 ( α 2 a b ) ( β 2 a C ) = 0 or if

(2.3) a ( a b C + 2 α β γ a γ 2 b β 2 C α 2 ) = 0 .

The companion matrix of the quadratic equation (2.2) is the following

A = a α β α b γ β γ C .

Let Δ det ( A ) , where Δ is known to be an invariant of the conic as Δ = 0 in invariant under linear transformations (see [41]). A simple computation shows that Δ = a b C + 2 α β γ a γ 2 b β 2 C α 2 . Since a 0 , from (2.3), we see that Δ = 0 . Therefore, when Δ = 0 , equation (2.2) represents a pair of straight lines. These pair of straight lines will be parallel if α 2 = a b and be intersecting if α 2 a b . In addition, when Δ 0 , equation (2.2) is transformed into a circle x 2 + y 2 = 1 , or, an ellipse x 2 a 2 + y 2 b 2 = 1 , or, a parabola, y 2 = x . Define, D = det a α α b = a b α 2 .

Let α 2 a b . Then under the transformation

(2.4) x = X + x 1 and y = Y + y 1 , where x 1 = α γ b β a b α 2 and y 1 = α β a γ a b α 2 ,

a straightforward computation shows that equation (2.2) reduces to

(2.5) a X 2 + 2 α X Y + b Y 2 + C 1 = 0 ,

where C 1 = a x 1 2 + 2 α x 1 y 1 + b y 1 2 + 2 β x 1 + 2 γ y 1 + C . In view of the values of x 1 and y 1 , we see that C 1 can be written as follows:

C 1 = a b C + 2 α β γ a γ 2 b β 2 C α 2 a b α 2 = Δ D .

Hence, equation (2.5) can be written as follows:

(2.6) a X 2 + 2 α X Y + b Y 2 + Δ D = 0 ,

where D 0 as α 2 a b .

To remove the X Y -term from equation (2.6), we take the transformation ( X , Y ) = ( U ξ 1 ± V η 1 ± , U η 1 ± + V ξ 1 ± ) , where ξ 1 ± and η 1 ± are given by

ξ 1 ± 2 α ( b a ) ± ( a b ) 2 + 4 α 2 2 + 4 α 2 , η 1 ± ( b a ) ± ( a b ) 2 + 4 α 2 ( b a ) ± ( a b ) 2 + 4 α 2 2 + 4 α 2 .

In light of this transformation, equation (2.6) is transformed to

(2.7) U 1 A ± 2 + V 1 B 2 + Δ D = 0 ,

where

A ± ( a + b ) ± ( a b ) 2 + 4 α 2 2 and B ( a + b ) ( a b ) 2 + 4 α 2 2 .

It is clear that (2.7) can be expressed as follows:

(2.8) D A ± Δ U 2 + D B Δ V 2 = 1 .

Clearly, if we consider the given general equation (2.1) as a functional equation in C 2 , then (2.8) is associated with a circle in C 2 . Therefore, in view of [37, Theorem 1.3], we must have

( U , V ) = cos h ( z ) D A ± Δ , sin h ( z ) D B Δ ,

where h : C 2 C is an entire function.

Thus, it follows from (2.4) that the solution ( f , g ) of the general quadratic functional equation a f 2 + 2 α f g + b g 2 + 2 β f + 2 γ g + C = 0 must be of the following form:

f ( z ) = D 11 cos h ( z ) D 12 sin h ( z ) + T 1 , g ( z ) = E 11 cos h ( z ) + E 12 sin h ( z ) + T 2 ,

where

D 11 = ξ 1 ± D A ± Δ , D 12 = η 1 ± D B Δ and T 1 = α γ b β a b α 2 , E 11 = η 1 ± D A ± Δ , E 12 = ξ 1 ± D B Δ and T 2 = α β a γ a b α 2 .

3 Main results and their consequences

In this article, we investigate the existence and form of solutions for general quadratic PDEs and PDDEs in C 2 . We present Theorems 3.13.5 for the general quadratic PDDEs (1.2) as discussed in Section 1. Thus, we completely answer Question 1.1. The first main result of the article is the following, which extends [39, Theorem 2.1] to general quadratic PDEs.

Theorem 3.1

Let a , b , C , α , β , γ C with α 2 0 , a b and Δ 0 , and f ( z 1 , z 2 ) be a finite order transcendental entire solutions for the general quadratic partial differential equation

(3.1) a L 1 ( f ) 2 + 2 α L 1 ( f ) L 2 ( f ) + b L 2 ( f ) 2 + 2 β L 1 ( f ) + 2 γ L 2 ( f ) + C = 0 .

Then f ( z 1 , z 2 ) takes one of the following forms:

  1. f ( z 1 , z 2 ) = A 1 + R 4 e ( z 1 + z 2 ) , where K 1 , K 2 , R 4 are constants with K 1 2 + K 2 2 = 1 ,

    A 1 = K 1 ξ 1 ± D A ± Δ K 2 η 1 ± D B Δ + α γ b β a b α 2 , K 2 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 , R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ a n d R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

  2. f ( z 1 , z 2 ) = E 11 + E 12 2 sin R 12 R 11 z 1 + z 2 + R 5 + E 11 E 12 2 cos R 12 R 11 z 1 + z 2 + R 5 + R 6 e ( z 1 + z 2 ) + T 2 .

    satisfying T 1 = T 2 ; where R 5 , R 6 are constants; R 11 = E 11 ( D 11 E 11 ) E 12 ( D 12 + E 12 ) , R 12 = D 11 ( D 11 E 11 ) + D 12 ( D 12 + E 12 ) .

By providing the following examples, we show that the form of transcendental entire solutions of (3.1) in Theorem 3.1 is precise.

Example 3.1

The finite order transcendental entire solutions for the general quadratic partial differential equation

L 1 ( f ) 2 + 4 L 1 ( f ) L 2 ( f ) + 3 L 2 ( f ) 2 2 L 1 ( f ) + 2 L 2 ( f ) = 1

is of the form

f ( z 1 , z 2 ) = 5 π i 7 e ( z 1 + z 2 ) + ± 59 53 5 4 2 ( 1 ± 5 ) 2 ( 5 ± 5 ) 2 5 ( 1 ± 5 ) { 4 7 ( 3 ± 5 ) 2 5 ± i 14 ( 1 5 ) 2 ± 5 } 8 i 7 ( 5 ± 5 ) 2 ± 5 7 .

Example 3.2

The finite order transcendental entire solutions for the general quadratic partial differential equation

3 L 1 ( f ) 2 4 L 1 ( f ) L 2 ( f ) + L 2 ( f ) 2 + 4 L 1 ( f ) 4 L 2 ( f ) = 1

is of the form

f ( z 1 , z 2 ) = ( 1 ± 5 ) 2 5 2 2 ± 5 2 i 10 2 5 sin z 1 + z 2 + π i 5 + ( 1 ± 5 ) 2 5 + 2 2 ± 5 2 i 10 2 5 cos z 1 + z 2 + π i 5 + π i 13 e ( z 1 + z 2 ) 2 .

Before we state some consequences of the main result, which provide the precise form of the solutions for trinomial equations, we introduce the following notations:

K 11 = ( a + b ) ± ( a b ) 2 + 4 α 2 , K 12 = ( ( b a ) ± ( a b ) 2 + 4 α 2 ) 2 + 4 α 2 , K 13 = ( a + b ) ( a b ) 2 + 4 α 2 , K 14 = ( b a ) ± ( a b ) 2 + 4 α 2 , A 11 = 2 2 α K 11 K 12 , A 12 = 2 K 14 K 12 K 13 , B 11 = 2 K 14 K 11 K 12 , B 12 = 2 2 α K 12 K 13 .

From Theorem 3.1, we obtain the following result, in which the form of the solutions for trinomial quadratic PDEs is characterized.

Corollary 3.1

Let a , b , α C with α 2 0 , a b and f ( z 1 , z 2 ) be a finite order transcendental entire solutions for the general quadratic partial differential equation:

a f ( z ) + f ( z ) z 1 2 + 2 α f ( z ) + f ( z ) z 1 f ( z ) + f ( z ) z 2 + b f ( z ) + f ( z ) z 2 2 = 1 .

Then f ( z 1 , z 2 ) takes the following forms:

  1. f ( z 1 , z 2 ) = A 31 + R 4 e ( z 1 + z 2 ) , where K 1 , K 2 , R 4 are constants with

    K 1 2 + K 2 2 = 1 , A 31 = A 11 K 1 A 12 K 2 , K 2 = ± S 11 S 11 2 + S 12 2 ,

    where S 11 = 2 ( 2 α K 14 ) K 11 K 12 a n d S 12 = 2 ( 2 α + K 14 ) K 11 K 12 .

  2. f ( z 1 , z 2 ) = B 11 + B 12 2 sin S 14 S 13 z 1 + z 2 + R 5 + B 11 B 12 2 cos S 14 S 13 z 1 + z 2 + R 5 + R 6 e ( z 1 + z 2 ) .

    where R 5 is a constant, S 13 = B 11 ( A 11 B 11 ) B 12 ( A 12 + B 12 ) and S 14 = A 11 ( A 11 B 11 ) + A 12 ( A 12 + B 12 ) .

Extending the scope of [39, Theorem 2.2], this article presents its second primary result, which is applicable to a broader range of quadratic PDEs.

Theorem 3.2

Let a , b , C , α , β , γ C with α 2 0 , a b and Δ 0 . The partial differential equation

(3.2) a L 1 ( f ) 2 + 2 α L 1 ( f ) L 3 ( f ) + b L 3 ( f ) 2 + 2 β L 1 ( f ) + 2 γ L 3 ( f ) + C = 0

does not admit any transcendental entire solution with finite order.

Remark 3.1

In view of the proof of Theorem 3.2, it can be easily shown that the trinomial quadratic PDE

a f ( z ) + f ( z ) z 1 2 + 2 α f ( z ) + f ( z ) z 1 f ( z ) + 2 f ( z ) z 1 2 + b f ( z ) + 2 f ( z ) z 1 2 2 = 1

does not admit any transcendental entire solutions. Consequently, we present a more general framework for the equation.

The following statement constitutes the third significant outcome of this article, involving the expansion of [39, Theorem 2.3] to address general quadratic partial differential equations.

Theorem 3.3

Let a , b , C , α , β , γ C with α 2 0 , a b and Δ 0 , and f ( z 1 , z 2 ) be a finite order transcendental entire solutions for the general quadratic partial differential equation

(3.3) a L 1 ( f ) 2 + 2 α L 1 ( f ) L 4 ( f ) + b L 4 ( f ) 2 + 2 β L 1 ( f ) + 2 γ L 4 ( f ) + C = 0 .

Then f ( z 1 , z 2 ) is of the form f ( z 1 , z 2 ) = A 3 + B 11 e ( z 1 z 2 ) , where K 5 , K 6 , and B 11 are constants with K 5 2 + K 6 2 = 1 ,

A 3 = K 5 ξ 1 ± D A ± Δ K 6 η 1 ± D B Δ + α γ b β a b α 2 , K 6 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 , R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ a n d R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

Example 3.3

The finite order transcendental entire solutions for the general quadratic partial differential equation

3 L 1 ( f ) 2 + 6 L 1 ( f ) L 4 ( f ) + 4 L 4 ( f ) 2 + 2 L 1 ( f ) 2 L 4 ( f ) = 1

of the form

f ( z 1 , z 2 ) = 7 π i 6 e ( z 1 z 2 ) ± 6 129215 118559 37 234 i 491 ( 1 ± 37 ) ( 37 ± 37 ) 3 ( 7 ± 37 ) ( 1 ± 37 ) { 39 ( 7 ± 37 ) 7 ± 37 ± i 1473 ( 5 37 ) 7 37 } 2 ( 37 ± 37 ) 3 ( 7 37 ) 7 3 .

In case of trinomial quadratic PDEs, we obtain the following result as a consequence of Theorem 3.3.

Corollary 3.2

Let a , b , α C with α 2 0 , a b , and f ( z 1 , z 2 ) be a finite order transcendental entire solutions for the general quadratic trinomial PDE

a f ( z ) + f ( z ) z 1 2 + 2 α f ( z ) + f ( z ) z 1 f ( z ) + 2 f ( z ) z 1 z 2 + b f ( z ) + 2 f ( z ) z 1 z 2 2 = 1 .

Then f ( z 1 , z 2 ) is of the forms f ( z 1 , z 2 ) = A 31 + B 11 e ( z 1 z 2 ) , where K 5 * , K 6 * , and B 11 are constants with K 5 * 2 + K 6 * 2 = 1 , A 31 = A 11 K 5 * A 12 K 6 * and K 6 * = ± S 11 S 11 2 + S 12 2 , where S 11 = 2 ( 2 α K 14 ) K 11 K 12 a n d S 12 = 2 ( 2 α + K 14 ) K 11 K 12 .

We obtain the next result as a generalization of [39, Theorem 2.4] for quadratic PDEs.

Theorem 3.4

Let a , b , C , α , β , γ C with α 2 0 , a b and Δ 0 , and c = ( c 1 , c 2 ) ( ( 0,0 ) ) C 2 , s 0 = c 1 + c 2 , and f ( z 1 , z 2 ) be a finite order transcendental entire solutions for the general quadratic partial differential equation

(3.4) a M 1 ( f ) 2 + 2 α M 1 ( f ) M 2 ( f ) + b M 2 ( f ) 2 + 2 β M 1 ( f ) + 2 γ M 2 ( f ) + C = 0 .

Then f ( z 1 , z 2 ) takes one of the following forms:

  1. f ( z 1 , z 2 ) = ψ ( z 1 + z 2 ) , where ψ ( s ) is a transcendental entire function with finite order in s z 1 + z 2 , satisfying ψ ( s ) + ψ ( s + s 0 ) = A 4 , where K 7 , K 8 are constants with K 7 2 + K 8 2 = 1 , where

    A 4 = K 7 ξ 1 ± D A ± Δ K 8 η 1 ± D B Δ + α γ b β a b α 2 , K 8 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 , R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ a n d R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

  2. f ( z 1 , z 2 ) = ( D 11 E 11 ) ( α 1 α 2 ) sin ( L ( z ) + B ) + ( D 12 + E 12 ) ( α 1 α 2 ) cos ( L ( z ) + B ) + ψ 1 ( s ) ,

    where L ( z ) = α 1 z 1 + α 2 z 2 , α 1 , α 2 , B C and satisfying T 1 = T 2 ;

    e 2 i L ( c ) = { ( E 11 i E 12 ) α 1 ( D 11 + i D 12 ) α 2 } { ( i D 11 + D 12 ) ( i E 11 E 12 ) } { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } { ( E 11 + i E 12 ) α 1 ( D 11 i D 12 ) α 2 }

    and

    ( E 11 2 + E 12 2 ) α 1 2 + ( D 11 2 + D 12 2 ) α 2 2 2 ( D 11 E 11 D 12 E 12 ) α 1 α 2 = ( D 11 E 11 ) 2 + ( D 12 + E 12 ) 2 ;

    and also ψ 1 ( s ) satisfies

    ψ 1 ( s ) + ψ 1 ( s + s 0 ) = E 11 α 1 D 11 α 2 α 1 α 2 cos ( L ( z ) + B ) + E 12 α 1 + D 12 α 2 α 1 α 2 sin ( L ( z ) + B ) D 11 E 11 α 1 α 2 sin ( L ( z ) + L ( c ) + B ) D 12 + E 12 α 1 α 2 cos ( L ( z ) + L ( c ) + B ) + T 1 .

The following examples are exhibited to ensure the form of the transcendental entire solution of (3.4) in Theorem 3.4.

Example 3.4

Let c = ( c 1 , c 2 ) = ( 3 , 2 ) . The finite order transcendental entire solutions for the general quadratic partial differential equation

M 1 ( f ) 2 + 4 M 1 ( f ) M 2 ( f ) + 3 M 2 ( f ) 2 2 M 1 ( f ) + 2 M 2 ( f ) = 1

takes the form

f ( z 1 , z 2 ) = ψ ( z 1 + z 2 ) ,

where ψ ( s ) is a transcendental entire function with finite order in s z 1 + z 2 , satisfying

ψ ( s ) + ψ ( s + s 0 ) = ± 59 53 5 4 2 ( 1 ± 5 ) 2 ( 5 ± 5 ) 2 5 ( 1 ± 5 ) { 4 7 ( 3 ± 5 ) 2 5 ± i 14 ( 1 5 ) 2 ± 5 } 8 i 7 ( 5 ± 5 ) 2 ± 5 7 .

Example 3.5

Let c = ( c 1 , c 2 ) = ( 3 , m ) ; m = λ 1 λ 2 and n = 3 3 ± 2 8 10 3 , where

λ 1 = 3 ln 3 3 ( 4 10 ) 3 ( 2 + 10 ) ( 3 3 ± 2 8 10 ) 3 ( 8 + 10 ) ( 14 5 10 ) ( 3 3 ± 2 8 10 ) 6 i , λ 2 = 2 i ( 3 3 ± 2 8 10 ) .

The finite order transcendental entire solutions for the general quadratic partial differential equation

M 1 ( f ) 2 6 M 1 ( f ) M 2 ( f ) M 2 ( f ) 2 + 2 M 1 ( f ) + 4 M 2 ( f ) = 1

is of the form

f ( z 1 , z 2 ) = 3 ( 2 + 10 ) i 2 10 ( 4 3 2 8 10 ) 20 2 10 sin z 1 + n z 2 + π i 13 + 3 ( 4 10 ) 2 10 ( 4 3 2 8 10 ) 20 2 10 cos z 1 + n z 2 + π i 13 + ψ 1 ( s ) ,

where ψ 1 ( s ) satisfies the following:

ψ 1 ( s ) + ψ 1 ( s + s 0 ) = 30 10 3 ± 6 8 10 i 2 10 ( 4 3 2 8 10 ) 20 2 10 cos z 1 + n z 2 + π i 13 + 3 30 ± ( 1 + 10 ) 2 8 10 2 10 ( 4 3 2 8 10 ) 20 2 10 sin z 1 + n z 2 + π i 13 + 3 ( 2 + 10 ) i 2 10 ( 4 3 2 8 10 ) 20 2 10 sin z 1 + n z 2 + 3 + m n + π i 13 + 3 ( 4 10 ) 2 10 ( 4 3 2 8 10 ) 20 2 10 cos z 1 + n z 2 + 3 + m n + π i 13 + 1 2 .

In the case of trinomial quadratic PDEs, the following result can be obtained as a corollary of Theorem 3.4.

Corollary 3.3

Let a , b , α C with α 2 0 , a b and c = ( c 1 , c 2 ) ( ( 0,0 ) ) C 2 , s 0 = c 1 + c 2 , and f ( z 1 , z 2 ) be a finite order transcendental entire solutions for the general quadratic partial differential equation

a f ( z + c ) + f ( z ) z 1 2 + 2 α f ( z + c ) + f ( z ) z 1 f ( z + c ) + f ( z ) z 2 + b f ( z + c ) + f ( z ) z 2 2 = 1 .

The function f ( z 1 , z 2 ) takes one of the following forms:

  1. f ( z 1 , z 2 ) = ψ ( z 1 + z 2 ) , where ψ ( s ) is a transcendental entire function with finite order in s z 1 + z 2 , satisfying ψ ( s ) + ψ ( s + s 0 ) = A 31 , where K 7 * , K 8 * are constants with K 7 * 2 + K 8 * 2 = 1 , A 31 = A 11 K 7 * A 12 K 8 * , K 8 * = ± S 11 S 11 2 + S 12 2 , S 11 = 2 ( 2 α K 14 ) K 11 K 12 , and S 12 = 2 ( 2 α + K 14 ) K 11 K 12 .

  2. f ( z 1 , z 2 ) = ( A 11 B 11 ) ( α 1 α 2 ) sin ( L ( z ) + B ) + ( A 12 + B 12 ) ( α 1 α 2 ) cos ( L ( z ) + B ) + ψ 1 ( s ) .

    where L ( z ) = α 1 z 1 + α 2 z 2 , α 1 , α 2 , B C and satisfying

    e 2 i L ( c ) = { ( B 11 i B 12 ) α 1 ( A 11 + i A 12 ) α 2 } { ( i A 11 + A 12 ) ( i B 11 B 12 ) } { ( A 12 i A 11 ) + ( B 12 + i B 11 ) } { ( B 11 + i B 12 ) α 1 ( A 11 i A 12 ) α 2 }

    and

    ( B 11 2 + B 12 2 ) α 1 2 + ( A 11 2 + A 12 2 ) α 2 2 2 ( A 11 B 11 A 12 B 12 ) α 1 α 2 = ( A 11 B 11 ) 2 + ( A 12 + B 12 ) 2 ;

    and, also ψ 1 ( s ) satisfies

    ψ 1 ( s ) + ψ 1 ( s + s 0 ) = B 11 α 1 A 11 α 2 α 1 α 2 cos ( L ( z ) + B ) + B 12 α 1 + A 12 α 2 α 1 α 2 sin ( L ( z ) + B ) A 11 B 11 α 1 α 2 sin ( L ( z ) + L ( c ) + B ) A 12 + B 12 α 1 α 2 cos ( L ( z ) + L ( c ) + B ) .

We obtain the following result as a generalization [39, Theorem 2.5] for quadratic partial differential equations PDEs.

Theorem 3.5

Let a , b , C , α , β , γ C with α 2 0 , a b and Δ 0 , and c = ( c 1 , c 2 ) ( ( 0,0 ) ) C 2 , s 0 = c 1 + c 2 , and f ( z 1 , z 2 ) be a finite order transcendental entire solutions for the general quadratic partial differential equation

(3.5) a M 1 ( f ) 2 + 2 α M 1 ( f ) M 3 ( f ) + b M 3 ( f ) 2 + 2 β M 1 ( f ) + 2 γ M 3 ( f ) + C = 0 .

Then f ( z 1 , z 2 ) is one of the following forms

  1. f ( z 1 , z 2 ) = A 5 + e ( z 1 + A z 2 + B ) , where K 9 , K 10 , B are constant with K 9 2 + K 10 2 = 1 , where

    A 5 = K 9 ξ 1 ± D A ± Δ K 10 η 1 ± D B Δ + α γ b β a b α 2 , A = 2 k π i ± π i c 1 c 2 , K 10 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 , R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ a n d R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

  2. f ( z 1 , z 2 ) = ( e c 1 ) z 2 c 2 π ( z 2 ) e z 1 + R 11 sin [ L ( z ) L ( c ) + B ] + R 12 cos [ L ( z ) L ( c ) + B ] + T 2 ,

    where L ( z ) = α 1 z 1 + α 2 z 2 , α 1 , α 2 , B C , π ( z 2 + c 2 ) = π ( z 2 ) , R 11 = ( D 11 α 1 + E 12 ) α 1 ( D 12 α 1 + E 11 ) α 1 2 + 1 , R 12 = ( D 12 α 1 + E 11 ) + α 1 ( D 11 α 1 + E 12 ) α 1 2 + 1 and satisfying the following T 1 = T 2 ,

    e 2 i L ( c ) = { ( i D 11 D 12 ) α 1 ( E 11 i E 12 ) } { ( i D 11 + D 12 ) ( i E 11 E 12 ) } { ( i D 11 D 12 ) ( i E 11 + E 12 ) } { ( i D 11 + D 12 ) α 1 + ( i E 12 + E 11 ) } ,

    and

    ( D 11 2 + D 12 2 ) α 1 4 + 2 ( D 11 E 12 + D 12 E 11 ) α 1 3 + ( E 11 2 + E 12 2 ) α 1 2 = ( D 11 E 11 ) 2 + ( D 12 + E 12 ) 2 .

To support the accuracy of the solutions’ form in (3.5) of Theorem 3.5, we provide the following two examples.

Example 3.6

Let c = ( c 1 , c 2 ) = ( 7 , 3 ) . The finite order transcendental entire solutions for the general quadratic partial differential equation

3 M 1 ( f ) 2 + 6 M 1 ( f ) M 3 ( f ) + 4 M 3 ( f ) 2 + 2 M 1 ( f ) 2 M 3 ( f ) = 1

of the form

f ( z 1 , z 2 ) = e z 1 + ( 5 π i 7 ) 3 z 2 + 3 π i 7 + ± 6 129215 118559 37 234 i 491 ( 1 ± 37 ) ( 37 ± 37 ) 3 ( 7 ± 37 ) ( 1 ± 37 ) { 39 ( 7 ± 37 ) 7 ± 37 ± i 1473 ( 5 37 ) 7 37 } 2 ( 37 ± 37 ) 3 ( 7 37 ) 7 3 .

Example 3.7

Let c = ( c 1 , c 2 ) = ( 7 , 2 π ) , where

α 1 = i 5 2 + 1 2 1 3 28 73 m m + 56 3 + 73 m + m 3 18 i 1 30 28 73 m m ,

m = 3025 18 27042 3 , α 2 = 1 2 π 1 2 i ln ( 8 + 10 ) α 1 + i ( 14 5 10 ) ( 3 2 12 ) α 1 + i ( 6 + 3 10 ) 7 α 1 ,

and

L ( c ) = 1 2 i ln ( 8 + 10 ) α 1 + i ( 14 5 10 ) ( 3 2 12 ) α 1 + i ( 6 + 3 10 ) .

The finite order transcendental entire solutions for the general quadratic partial differential equation

M 1 ( f ) 2 + 6 M 1 ( f ) M 3 ( f ) M 3 ( f ) 2 + 4 M 1 ( f ) + 2 M 3 ( f ) = 1

is of the form

f ( z 1 , z 2 ) = ( e 7 ) z 2 2 π sin ( z 2 ) e z 1 1 2 + { ( 4 10 ) α 1 + i ( α 1 2 ( 10 1 ) 3 ) } 5 2 10 ( α 1 2 + 1 ) 20 2 10 sin α 1 z 1 + α 2 z 2 L ( c ) + 5 π i 11 + { ( 3 α 1 2 + 10 1 ) i α 1 ( 2 + 10 ) } 5 2 10 ( α 1 2 + 1 ) 20 2 10 cos α 1 z 1 + α 2 z 2 L ( c ) + 5 π i 11 .

As a consequence of Theorem 3.5, we establish the existence and form of trinomial quadratic PDEs and obtain the following result.

Corollary 3.4

Let a , b , α C with α 2 0 , a b and c = ( c 1 , c 2 ) ( ( 0,0 ) ) C 2 , s 0 = c 1 + c 2 , and f ( z 1 , z 2 ) be a finite order transcendental entire solutions for the general quadratic partial differential equation

a f ( z + c ) + f ( z ) z 1 2 + 2 α f ( z + c ) + f ( z ) z 1 f ( z + c ) + 2 f ( z ) z 1 2 + b f ( z + c ) + 2 f ( z ) z 1 2 2 = 1 .

Then f ( z 1 , z 2 ) is one of the following forms

  1. f ( z 1 , z 2 ) = A 31 + e ( z 1 + A z 2 + B ) , where K 9 * , K 10 * , and B are constants with K 9 * 2 + K 10 * 2 = 1 , where

    A 31 = A 11 K 9 * A 12 K 10 * , A = 2 k π i ± π i c 1 c 2 a n d K 10 * = ± S 11 S 11 2 + S 12 2 , S 11 = 2 ( 2 α K 14 ) K 11 K 12 a n d S 12 = 2 ( 2 α + K 14 ) K 11 K 12 .

  2. f ( z 1 , z 2 ) = ( e c 1 ) z 2 c 2 π ( z 2 ) e z 1 + S 15 sin [ L ( z ) L ( c ) + B ] + S 16 cos [ L ( z ) L ( c ) + B ] ,

    where L ( z ) = α 1 z 1 + α 2 z 2 , α 1 , α 2 , B C , π ( z 2 + c 2 ) = π ( z 2 ) , S 15 = ( A 11 α 1 + B 12 ) α 1 ( A 12 α 1 + B 11 ) α 1 2 + 1 , S 16 = ( A 12 α 1 + B 11 ) + α 1 ( A 11 α 1 + B 12 ) α 1 2 + 1 and satisfying

    e 2 i L ( c ) = { ( i A 11 A 12 ) α 1 ( B 11 i B 12 ) } { ( i A 11 + A 12 ) ( i B 11 B 12 ) } { ( i A 11 A 12 ) ( i B 11 + B 12 ) } { ( i A 11 + A 12 ) α 1 + ( i B 12 + B 11 ) } ,

    and

    ( A 11 2 + A 12 2 ) α 1 4 + 2 ( A 11 B 12 + A 12 B 11 ) α 1 3 + ( B 11 2 + B 12 2 ) α 1 2 = ( A 11 B 11 ) 2 + ( A 12 + B 12 ) 2 .

4 Proof of the main results

Let C denote the complex plane and let f be a nonconstant meromorphic function in C . We will utilize the following standard notation of value distribution theory (for further details, refer to [4244]): T ( r , f ) , m ( r , f ) , N ( r , f ) , N ¯ ( r , f ) , ; We denote by S ( r , f ) , any function satisfying S ( r , f ) = { T ( r , f ) } as r , possibly outside a set of finite measure. Define the order of f by

ρ ( f ) lim r ¯ log + T ( r , f ) log r .

First, we recall here some necessary lemmas, which will play a key role in proving the main results.

Lemma 4.1

[45,46] For any entire function F on C n , F ( 0 ) 0 and put ρ ( n F ) = ρ < . Then there exist a canonical function f F and a function g F C n such that F ( z ) = f F ( z ) e g F ( z ) . For the special case n = 1 , f F is the canonical product of Weierstrass.

Remark 4.1

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

Lemma 4.2

[47] Let f j ( 0 ) , j = 1, 2, 3, be meromorphic functions on C m such that f 1 is not constant and f 1 + f 2 + f 3 = 1 such that

j = 1 3 N 2 r , 1 f j + 2 N ¯ ( r , f j ) < λ T ( r , f 1 ) + O ( log + T ( r , f 1 ) ) ,

for all r outside possibly a set with finite logarithmic measure, where λ < 1 is a positive number. Then either f 2 = 1 or f 3 = 1 .

Remark 4.2

Here, N 2 ( 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.

Lemma 4.3

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

  1. the internal function h is a polynomial and the external function g is of finite order or

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

In the following section, we will discuss proof of the main results.

Proof of Theorem 3.1

Assuming that f is a transcendental entire solution of (3.1) with a finite order, we can express equation (3.1) in the following form using the transformations discussed in Section 2:

(4.1) D A ± Δ u 2 + D B Δ v 2 = 1 .

The proof is divided into two distinct cases for clarity.

Case 1: If D A ± Δ u is a constant, then it follows from (4.1) that D B Δ v is also constant. Denoting

D A ± Δ u = K 1 and D B Δ v = K 2 ,

in view of (4.1), it is easy to see that K 1 2 + K 2 2 = 1 . This evidently leads to

(4.2) L 1 ( f ) = f ( z ) + f z 1 = K 1 ξ 1 ± D A ± Δ K 2 η 1 ± D B Δ + α γ b β a b α 2 A 1 ,

(4.3) L 2 ( f ) = f ( z ) + f z 2 = K 1 η 1 ± D A ± Δ + K 2 ξ 1 ± D B Δ + α β a γ a b α 2 B 1 .

Considering (4.2) and (4.3), it is evident that

(4.4) f z 1 f z 2 = A 1 B 1 .

The characteristic equations of (4.4) are as follows:

d z 1 d t = 1 , d z 2 d t = 1 , d f d t = A 1 B 1 .

Using the initial conditions z 1 = 0 and z 2 = s , we can express f as, f = f ( 0 , s ) ϕ ( s ) , where ϕ ( s ) represents a function that relies on the parameter s . Accordingly, a straightforward calculation shows that z 1 = t , z 2 = t + s , and

f ( t , s ) = 0 t ( A 1 B 1 ) d t + ϕ ( s ) = ( A 1 B 1 ) t + ϕ ( s ) ,

where ϕ ( s ) is a transcendental entire function with finite order in s . Then we obtain z 1 = t and s = z 1 + z 2 , the solution of equation (4.4) is of the form

(4.5) f ( z 1 , z 2 ) = ( A 1 B 1 ) z 1 + ϕ ( z 1 + z 2 ) .

Since f is an entire function, we have 2 f z 1 z 2 = 2 f z 2 z 1 , hence from (4.2) and (4.3), we obtain f z 1 = f z 2 . Therefore, from (4.4), it follows that A 1 = B 1 . Furthermore, from (4.2), (4.3) and using A 1 = B 1 , K 1 2 + K 2 2 = 1 , a simple computation shows that

K 2 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 ,

where

R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ and R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

Consequently, equation (4.5) is transformed into

(4.6) f ( z 1 , z 2 ) = ϕ ( z 1 + z 2 ) .

By substituting (4.6) into (4.3), we obtain ϕ ( z 1 + z 2 ) + ϕ ( z 1 + z 2 ) = A 1 , and an easy computation shows that

(4.7) ϕ ( z 1 + z 2 ) = A 1 + R 4 e ( z 1 + z 2 ) , R 4 C .

Thus considering (4.6) and (4.7), we can conclude that the solution takes the precise form of f ( z 1 , z 2 ) = A 1 + R 4 e ( z 1 + z 2 ) .

Case 2: Assuming that D A ± Δ u is not constant, we can apply the knowledge that the solutions to the functional equation f 2 + g 2 = 1 are given by f ( z ) = cos h ( z ) and g ( z ) = sin h ( z ) , with h representing an entire function. With this understanding, we can deduce from (4.1) that

(4.8) L 1 ( f ) = f ( z ) + f z 1 = D 11 cos h ( z ) D 12 sin h ( z ) + T 1 ,

(4.9) L 2 ( f ) = f ( z ) + f z 2 = E 11 cos h ( z ) + E 12 sin h ( z ) + T 2 .

By subtracting (4.9) from (4.8), we obtain

(4.10) f z 1 f z 2 = ( D 11 E 11 ) cos h ( z ) ( D 12 + E 12 ) sin h ( z ) + ( T 1 T 2 ) .

Using the fact that 2 f z 1 z 2 = 2 f z 2 z 1 , from (4.8) and (4.9), a routine computation shows that

(4.11) f z 1 f z 2 = D 11 h z 2 E 11 h z 1 sin h ( z ) + D 12 h z 2 + E 12 h z 1 cos h ( z ) .

From (4.10) and (4.11), we obtain

(4.12) D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) sin h ( z ) + D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) cos h ( z ) = ( T 1 T 2 ) .

Sub-case 2.1: Suppose that

D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) 0 , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) 0 and ( T 1 T 2 ) 0 .

It follows from (4.12) that

tan h ( z ) = D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) + ( T 1 T 2 ) cos h ( z ) D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) .

By considering the Nevanlinna characteristic function and applying the “First Fundamental Theorem of Nevanlinna,” we derive

T ( r , tan h ) T r , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) + 2 T r , D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) + 2 T ( r , cos h ) + S ( r , h ) ,

which shows that

(4.13) T ( r , tan h ) 3 T r , h z 2 + 3 T r , h z 1 + 2 T ( r , e i h ) + S ( r , h ) .

Given that f is a finite order transcendental entire solution of equation (3.1), we can deduce from Lemmas 4.1 and 4.3 in combination with (4.8) and (4.9) that h is a nonconstant polynomial in C 2 . Assuming that h can be expressed as h ( z ) = k = 0 n P k ( z ) of degree n , where P k ( z ) is a homogeneous polynomial of degree k , a simple computation shows that

T ( r , e i h ( z ) ) T ( r , e i P n ( z ) ) = T ( r , e i a n m , m z 1 n m z 2 m ) = a n m , m π r n ( as r ) = a n m , m π r deg ( h ( z ) ) .

Therefore, from (4.13), we see that

T ( r , tan h ) = O { T ( r , h ) + log r + r deg h ( z ) }

outside possibly a set of finite Lebesgue measure in view of the results (see [18,46]) that T ( r , G z j ) = O ( T ( r , G ) ) for any meromorphic function G outside a set of finite Lebesgue measure and that T ( r , P ) = O ( log r ) for any polynomial P . However, it is easy to see that

lim r T ( r , tan h ) T ( r , h ) + log r + r deg h ( z ) = ,

when h is a nonconstant polynomial. Therefore, h must be a constant, a contradiction.

Sub-case 2.2: Suppose that

D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) = 0 , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) 0 , T 1 T 2 .

Thus, (4.12) can be written as follows:

tan h ( z ) = D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) sin h ( z ) T 1 T 2 .

Given that f is an entire transcendental solution of equation (3.1) with finite order, we can utilize Lemmas 4.1, 4.3, (4.8), and (4.9) to deduce that h is a nonconstant polynomial in C . By the similar argument as in Sub-case 2.1, we see that

lim r T ( r , tan h ) T ( r , h ) + log r + r deg h ( z ) = ,

where h is a nonconstant polynomial. This shows that h must be a constant, a contradiction.

Sub-case 2.3: Let us assume

D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) 0 , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) = 0 , T 1 T 2 .

By the similar argument used in sub-case 1 , we arrive a contradiction.

Sub-case 2.4: Suppose that

D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) 0 , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) 0 , T 1 = T 2 .

By employing the same reasoning as in sub-case 1, we reach a contradictory conclusion.

Sub-case 2.5: Suppose that

D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) = 0 , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) = 0 , T 1 T 2 .

In view of (4.12), we see that T 1 = T 2 , which contradicts that T 1 T 2 .

Sub-case 2.6: Assume that

D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) = 0 , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) 0 , T 1 = T 2 .

By employing equation (4.12), we can establish that cos h ( z ) = 0 , which implies that h ( z ) must be a constant. This contradicts our initial reasoning.

Sub-case 2.7: Let

D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) 0 , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) = 0 , T 1 = T 2 .

By observing equation (4.12), it becomes clear that sin h ( z ) = 0 . This indicates that h ( z ) takes on a constant value, leading to a contradiction.

Sub-case 2.8: Suppose that

D 11 h z 2 E 11 h z 1 + ( D 12 + E 12 ) = 0 , D 12 h z 2 + E 12 h z 1 ( D 11 E 11 ) = 0 , T 1 = T 2 .

Since T 1 = T 2 , a simple computation shows that

(4.14) R 11 h z 1 R 12 h z 2 = 0 ,

where R 11 = E 11 ( D 11 E 11 ) E 12 ( D 12 + E 12 ) , R 12 = D 11 ( D 11 E 11 ) + D 12 ( D 12 + E 12 ) . As a result, it can be derived from equation (4.14) that

(4.15) h ( z ) = z 2 + R 12 R 11 z 1 + R 5 , where R 5 C .

By substituting this h ( z ) into (4.10), we obtain

(4.16) f z 1 f z 2 = ( D 11 E 11 ) cos z 2 + R 12 R 11 z 1 + R 5 ( D 12 + E 12 ) sin z 2 + R 12 R 11 z 1 + R 5 .

The characteristic equations of (4.16) are as follows:

d z 1 d t = 1 , d z 2 d t = 1 ,

and hence,

d f d t = ( D 11 E 11 ) cos z 2 + R 12 R 11 z 1 + R 5 ( D 12 + E 12 ) sin z 2 + R 12 R 11 z 1 + R 5 .

By using the initial conditions: z 1 = 0 , z 2 = s , and f = f ( 0 , s ) ϕ 1 ( s ) with a parameter s , wee see that z 1 = t , z 2 = t + s , and

f ( t , s ) = ϕ 1 ( s ) + 0 t ( D 11 E 11 ) cos z 2 + R 12 R 11 z 1 + R 5 ( D 12 + E 12 ) sin z 2 + R 12 R 11 z 1 + R 5 d t = ϕ 1 ( s ) + ( D 11 E 11 ) 0 t cos R 12 R 11 R 11 t + s + R 5 d t ( D 12 + E 12 ) 0 t sin R 12 R 11 R 11 t + s + R 5 d t = ϕ 2 ( s ) + ( D 11 E 11 ) R 11 ( R 12 R 11 ) sin R 12 R 11 R 11 t + s + R 5 + ( D 12 + E 12 ) R 11 ( R 12 R 11 ) cos R 12 R 11 R 11 t + s + R 5 ,

where

ϕ 2 ( s ) = ϕ 1 ( s ) ( D 11 E 11 ) R 11 ( R 12 R 11 ) sin ( s + R 5 ) ( D 12 + E 12 ) R 11 ( R 12 R 11 ) cos ( s + R 5 )

is a transcendental entire function with finite order in s . By using t = z 1 , s = z 1 + z 2 , we easily obtain

(4.17) f ( z 1 , z 2 ) = ( D 11 E 11 ) R 11 ( R 12 R 11 ) sin R 12 R 11 z 1 + z 2 + R 5 + ( D 12 + E 12 ) R 11 ( R 12 R 11 ) cos R 12 R 11 z 1 + z 2 + R 5 + ϕ 2 ( z 1 + z 2 ) .

Substituting (4.17) into (4.9) gives that

ϕ 2 ( z 1 + z 2 ) + ϕ 2 ( z 1 + z 2 ) = R 21 sin R 12 R 11 z 1 + z 2 + R 5 + R 22 cos R 12 R 11 z 1 + z 2 + R 5 + T 2 ,

where

R 21 = E 12 R 12 + ( E 11 D 11 + D 12 ) R 11 ( R 12 R 11 ) and R 22 = E 11 R 12 ( E 12 + D 11 + D 12 ) R 11 ( R 12 R 11 ) .

It follows easily that

(4.18) ϕ 2 ( z 1 + z 2 ) = ( R 21 + R 22 ) 2 sin R 12 R 11 z 1 + z 2 + R 5 ( R 21 R 22 ) 2 cos R 12 R 11 z 1 + z 2 + R 5 + R 6 e ( z 1 + z 2 ) + T 2 .

Consequently, in view of (4.17) and (4.18), we obtain form of the solution f as follows:

f ( z 1 , z 2 ) = E 11 + E 12 2 sin R 12 R 11 z 1 + z 2 + R 5 + E 11 E 12 2 cos R 12 R 11 z 1 + z 2 + R 5 + R 6 e ( z 1 + z 2 ) + T 2 .

This completes the proof.□

Proof of Theorem 3.2

Let us assume that f is a transcendental entire solution of equation (3.2) with finite order. The equation (3.2) can be expressed as follows:

(4.19) D A ± Δ u 2 + D B Δ v 2 = 1 .

We discuss the following two possible cases.

Case 1: If D A ± Δ u is a constant, then it follows from (4.19) that D B Δ v is also constant. Denote

D A ± Δ u = K 3 and D B Δ v = K 4 .

In view of (4.19), it is easy to see that K 3 2 + K 4 2 = 1 , hence

(4.20) L 1 ( f ) = f ( z ) + f z 1 = K 3 ξ 1 ± D A ± Δ K 4 η 1 ± D B Δ + α γ b β a b α 2 A 2 ,

(4.21) L 3 ( f ) = f ( z ) + 2 f z 1 2 = K 3 η 1 ± D A ± Δ + K 4 ξ 1 ± D B Δ + α β a γ a b α 2 B 2 .

By subtracting (4.21) from (4.20), we see that

(4.22) f z 1 2 f z 1 2 = A 2 B 2 .

By differentiating (4.20) with respect to z 1 , we obtain

(4.23) f z 1 + 2 f z 1 2 = 0 .

Consequently, it can be deduced from equations (4.21) and (4.23) that

(4.24) f ( z ) f z 1 = B 2 .

By differentiating (4.24) with respect to z 1 , we obtain

(4.25) f z 1 2 f z 1 2 = 0 .

In view of (4.22) and (4.25), it is clear that A 2 = B 2 . With the help of K 3 2 + K 4 2 = 1 , A 2 = B 2 , using (4.2) and (4.3), a simple computation shows that

K 4 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 ,

where

R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ , and R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

Consequently, from (4.25), it follows that

(4.26) f ( z 1 , z 2 ) = e z 1 + ϕ 1 ( z 2 ) + ϕ 2 ( z 2 ) ,

where ϕ 1 ( z 2 ) and ϕ 2 ( z 2 ) are two functions in z 2 . By substituting (4.26) into (4.20), we obtain 2 e z 1 + ϕ 1 ( z 2 ) + ϕ 2 ( z 2 ) = A 2 . Thus, it follows that ϕ 2 ( z 2 ) = A 2 and 2 e z 1 + ϕ 1 ( z 2 ) = 0 , which is impossible.

Case 2: If D A ± Δ u is not a constant. In a manner analogous to the argument applied in the proof of Theorem 3.1, we see that there exists an entire function h such that

(4.27) L 1 ( f ) = f ( z ) + f z 1 = D 11 cos h ( z ) D 12 sin h ( z ) + T 1 ,

(4.28) L 3 ( f ) = f ( z ) + 2 f z 1 2 = E 11 cos h ( z ) + E 12 sin h ( z ) + T 2 .

By the similar reasoning applied previously, we obtain

(4.29) f z 1 2 f z 1 2 = ( D 11 E 11 ) cos h ( z ) ( D 12 + E 12 ) sin h ( z ) + ( T 1 T 2 ) .

Differentiating (4.27) with respect to z 1 , it follows that

(4.30) f z 1 + 2 f z 1 2 = D 11 sin h ( z ) h z 1 D 12 cos h ( z ) h z 1 .

By considering equations (4.28) and (4.30), it becomes evident that

(4.31) f ( z ) f z 1 = E 11 + D 12 h z 1 cos h ( z ) + E 12 + D 11 h z 1 sin h ( z ) + T 2 .

By differentiating (4.31) with respect to z 1 and using (4.29), we obtain

( D 12 + E 12 ) E 11 h z 1 D 12 h z 1 2 + D 11 2 h z 1 2 sin h ( z ) = ( D 11 E 11 ) E 12 h z 1 D 11 h z 1 2 D 12 2 h z 1 2 cos h ( z ) + ( T 1 T 2 ) .

Since f is a finite order transcendental entire solution of equation (3.2), we can conclude, based on Lemmas 4.1 and 4.3, (4.27), and (4.28), that h is a nonconstant polynomial in C 2 . Similar arguments as in the proof of Theorem 3.1 gives that

(4.32) ( D 12 + E 12 ) E 11 h z 1 D 12 h z 1 2 + D 11 2 h z 1 2 = 0 ,

(4.33) ( D 11 E 11 ) E 12 h z 1 D 11 h z 1 2 D 12 2 h z 1 2 = 0 ,

(4.34) and ( T 1 T 2 ) = 0 .

In view of (4.32) and (4.33), we see that

(4.35) 2 h z 1 2 S 1 h z 1 = S 2 ,

where S 1 = D 11 E 11 D 12 E 12 D 11 2 + D 12 2 and S 2 = D 11 E 12 + D 12 E 11 D 11 2 + D 12 2 , which implies that

h ( z 1 , z 2 ) = e S 1 z 1 + ϕ 1 ( z 2 ) + e ϕ 2 ( z 2 ) S 2 S 1 z 1

is a transcendental, which is a contradiction as h is a nonconstant polynomial. This completes the proof.□

Proof of Theorem 3.3

Suppose that f is a transcendental entire solution of (3.3) with finite order. The equation (3.3) can be written as follows:

(4.36) D A ± Δ u 2 + D B Δ v 2 = 1 .

Now we will discuss two cases below.

Case 1: If D A ± Δ u is a constant, then it follows from (4.36) that D B Δ v is also constant. Denote

D A ± Δ u = K 5 and D B Δ v = K 6 .

In view of (4.36), we have K 5 2 + K 6 2 = 1 . Consequently,

(4.37) L 1 ( f ) = f ( z ) + f z 1 = K 5 ξ 1 ± D A ± Δ K 6 η 1 ± D B Δ + α γ b β a b α 2 A 3 ,

(4.38) L 4 ( f ) = f ( z ) + 2 f z 1 z 2 = K 5 η 1 ± D A ± Δ + K 6 ξ 1 ± D B Δ + α β a γ a b α 2 B 3 .

An easy computation using (4.37) and (4.38) shows that

(4.39) f z 1 2 f z 1 z 2 = A 3 B 3 .

By differentiating (4.37) with respect to z 2 , we obtain

(4.40) f z 2 + 2 f z 2 z 1 = 0 .

From (4.38) and (4.40), it follows that

(4.41) f ( z ) f z 2 = B 3 .

By differentiating (4.41) with respect to z 1 , we obtain

(4.42) f z 1 2 f z 1 z 2 = 0 .

In view of (4.40) and (4.42), it is easy to see that A 3 = B 3 . Using K 5 2 + K 6 2 = 1 , A 3 = B 3 and from (4.37), (4.38) a simple computation shows that

K 6 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 ,

where

R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ and R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

Through the combination of equations (4.40) and (4.42), it follows that

(4.43) f z 1 + f z 2 = 0 ,

which shows that f ( z 1 , z 2 ) = ϕ ( z 2 z 1 ) . By substituting f ( z 1 , z 2 ) into (4.37), we obtain ϕ ( z 2 z 1 ) ϕ ( z 2 z 1 ) = A 3 . Of course, ϕ ( z 2 z 1 ) = A 3 + B 11 e ( z 1 z 2 ) , B 11 C . Hence, f takes the form f ( z 1 , z 2 ) = A 3 + B 11 e ( z 1 z 2 ) .

Case 2: If D A ± Δ u is not a constant, employing a similar reasoning as in the proof of Theorem 3.1, we can conclude the existence of an entire function h ( z ) such that

(4.44) L 1 ( f ) = f ( z ) + f z 1 = D 11 cos h ( z ) D 12 sin h ( z ) + T 1 ,

(4.45) L 4 ( f ) = f ( z ) + 2 f z 1 z 2 = E 11 cos h ( z ) + E 12 sin h ( z ) + T 2 .

Considering equations (4.44) and (4.45), a direct calculation shows that

(4.46) f z 1 2 f z 1 z 2 = ( D 11 E 11 ) cos h ( z ) ( D 12 + E 12 ) sin h ( z ) + ( T 1 T 2 ) .

By differentiating (4.44) with respect to z 2 , we obtain

(4.47) f z 2 + 2 f z 2 z 1 = D 11 sin h ( z ) h z 2 D 12 cos h ( z ) h z 2 .

From (4.45) and (4.47), it follows that

(4.48) f ( z ) f z 2 = D 11 h z 2 + E 12 sin h ( z ) + D 12 h z 2 + E 11 cos h ( z ) + T 2 .

Differentiating (4.48) with respect to z 1 , and using (4.46), the following can be obtained

(4.49) D 11 2 h z 1 z 2 D 12 h z 1 h z 2 E 11 h z 1 + ( D 12 + E 12 ) sin h ( z ) = ( D 11 E 11 ) D 12 2 h z 1 z 2 D 11 h z 1 h z 2 E 12 h z 1 cos h ( z ) + ( T 1 T 2 ) .

However, since f is a finite order transcendental entire solution of equation (3.3), by Lemmas 4.1, 4.3, (4.44), and (4.45), we conclude that h is a nonconstant polynomial in C 2 . The similar argument being used in the proof of Theorem 3.1 will help us to obtain the following:

(4.50) D 11 2 h z 1 z 2 D 12 h z 1 h z 2 E 11 h z 1 + ( D 12 + E 12 ) = 0 ,

(4.51) ( D 11 E 11 ) D 12 2 h z 1 z 2 D 11 h z 1 h z 2 E 12 h z 1 = 0

(4.52) and ( T 1 T 2 ) = 0 .

In view of (4.50) and (4.51), a simple computation shows that h z 1 h z 2 + S 3 h z 1 = S 4 , where

S 3 = D 11 E 12 + D 12 E 11 D 11 2 + D 12 2 and S 4 = D 11 ( D 11 E 11 ) + D 12 ( D 12 + E 12 ) D 11 2 + D 12 2 ,

which implies that

h ( z 1 , z 2 ) = ϕ ( z 1 ) e S 3 z 2 + ϕ ( z 2 ) + S 4 S 3 z 1

is a transcendental, which is a contradiction as h is a nonconstant polynomial. This completes the proof.□

Proof of Theorem 3.4

Suppose that f is a transcendental entire solution of (3.4) with finite order. Under the transformation mentioned in Section 2, (3.4) can be written as follows:

(4.53) D A ± Δ u 2 + D B Δ v 2 = 1 .

To complete the proof, it is sufficient to discuss the following cases.

Case 1: If D A ± Δ u is a constant, then it follows from (4.53) that D B Δ v is also constant. Let us denote

D A ± Δ u = K 7 and D B Δ v = K 8 .

From (4.53), it is clear that K 7 2 + K 8 2 = 1 . In fact, we observe that

(4.54) M 1 ( f ) = f ( z + c ) + f z 1 = K 7 ξ 1 ± D A ± Δ K 8 η 1 ± D B Δ + α γ b β a b α 2 A 4 ,

(4.55) M 2 ( f ) = f ( z + c ) + f z 2 = K 7 η 1 ± D A ± Δ + K 8 ξ 1 ± D B Δ + α β a γ a b α 2 B 4 .

From (4.54) and (4.55), we obtain

(4.56) f z 1 f z 2 = A 4 B 4 .

The characteristic equations of (4.56) are as follows:

d z 1 d t = 1 , d z 2 d t = 1 , d f d t = A 4 B 4 .

Using the initial conditions: z 1 = 0 , z 2 = s and f = f ( 0 , s ) ψ ( s ) with a parameter s . Therefore, a simple computation shows that z 1 = t , z 2 = t + s , and

f ( t , s ) = 0 t ( A 4 B 4 ) d t + ψ ( s ) = ( A 4 B 4 ) t + ψ ( s ) ,

where ψ ( s ) is a transcendental entire function with finite order in s . Then we obtain z 1 = t and s = z 1 + z 2 , the solution of equation (4.56) is of the form

(4.57) f ( z 1 , z 2 ) = ( A 4 B 4 ) z 1 + ψ ( z 1 + z 2 ) .

In view of 2 f z 1 z 2 = 2 f z 2 z 1 , using (4.2) and (4.3), we obtain that f ( z + c ) z 1 = f ( z + c ) z 2 . Thus, it follows from (4.56) that A 4 = B 4 . Also, from (4.54), (4.55) and using A 4 = B 4 , K 7 2 + K 8 2 = 1 , a basic calculation gives that K 8 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 , where

R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ and R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

From (4.57) and using A 4 = B 4 , we obtain f ( z 1 , z 2 ) = ψ ( z 1 + z 2 ) , where ψ ( s ) is a transcendental entire function with finite order in s . Also ψ ( s ) satisfying ψ ( s ) + ψ ( s + s 0 ) = A 4 for s 0 = c 1 + c 2 .

Case 2: If D A ± Δ u is not a constant, then using a similar argument as before, we obtain

(4.58) M 1 ( f ) = f ( z + c ) + f z 1 = D 11 cos h ( z ) D 12 sin h ( z ) + T 1 ,

(4.59) M 2 ( f ) = f ( z + c ) + f z 2 = E 11 cos h ( z ) + E 12 sin h ( z ) + T 2 .

Subtracting (4.59) from (4.58), we see that

(4.60) f z 1 f z 2 = ( D 11 E 11 ) cos h ( z ) ( D 12 + E 12 ) sin h ( z ) + ( T 1 T 2 ) .

Since 2 f z 1 z 2 = 2 f z 2 z 1 , using (4.58) and (4.59), a simple computation shows that

(4.61) f ( z + c ) z 1 f ( z + c ) z 2 = D 11 h z 2 E 11 h z 1 sin h ( z ) + D 12 h z 2 + E 12 h z 1 cos h ( z ) .

Therefore, it follows from (4.60) and (4.61) that

(4.62) { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } e i h ( z + c ) + 2 i ( T 2 T 1 ) ( i D 11 + D 12 ) ( i E 11 E 12 ) e i h ( z + c ) + ( D 11 + i D 12 ) h z 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) + h ( z ) ] + ( E 11 + i E 12 ) h z 1 ( D 11 i D 12 ) h z 2 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) h ( z ) ] = 1 .

The equation (4.62) can be written as f 11 + f 12 + f 13 = 1 . It is easy to see that f 11 0 , i.e.,

{ ( D 12 i D 11 ) + ( E 12 + i E 11 ) } e i h ( z + c ) + 2 i ( T 2 T 1 ) ( i D 11 + D 12 ) ( i E 11 E 12 ) 0 .

Otherwise, we obtain

e i h ( z + c ) = 2 i ( T 1 T 2 ) ( D 12 i D 11 ) + ( E 12 + i E 11 ) ,

and this shows that the polynomial h ( z + c ) is a constant, i.e., the polynomial h ( z ) is a constant, which is not possible.

If

( D 11 + i D 12 ) h z 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) = 0 ,

then we have

( E 11 + i E 12 ) h z 1 ( D 11 i D 12 ) h z 2 ( i D 11 + D 12 ) ( i E 11 E 12 ) 0 .

Otherwise, we have

{ ( D 12 i D 11 ) + ( E 12 + i E 11 ) } e i h ( z + c ) + 2 i ( T 2 T 1 ) ( i D 11 + D 12 ) ( i E 11 E 12 ) e i h ( z + c ) = 1 ,

which implies that

e i h ( z + c ) = i ( T 1 T 2 ) ± i T ( D 12 i D 11 ) + ( E 12 + i E 11 ) ,

where T ( T 2 T 1 ) 2 + [ ( D 12 i D 11 ) + ( E 12 + i E 11 ) ] [ ( i E 11 E 12 ) ( i D 11 + D 12 ) ] , and hence, the polynomial h must be a constant. This shows that D A ± Δ u is a constant, which is a contradiction. Therefore, (4.62) is reduced to

{ ( D 12 i D 11 ) + ( E 12 + i E 11 ) } e i h ( z + c ) + 2 i ( T 2 T 1 ) ( i D 11 + D 12 ) ( i E 11 E 12 ) e i h ( z + c ) + ( E 11 + i E 12 ) h z 1 ( D 11 i D 12 ) h z 2 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) h ( z ) ] = 1 .

By the similar argument being used in [39], using second fundamental theorem of Nevanlinna for several complex variables, we obtain T ( r , e i h ( z + c ) ) = o ( T ( r , e h ) ) , which is a contradiction. Therefore, we have

( D 11 + i D 12 ) h z 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) 0 .

Similarly, it can be shown that

( E 11 + i E 12 ) h z 1 ( D 11 i D 12 ) h z 2 ( i D 11 + D 12 ) ( i E 11 E 12 ) 0 .

That is, f 12 0 and f 13 0 . Because f 11 , f 12 , f 13 are all entire functions, they have no poles. Consequently, we see that the following condition in Lemma 4.2

j = 1 3 N 2 r , 1 f 1 j + 2 N ¯ ( r , f 1 j ) = 0 < λ T ( r , f 11 ) + O ( log + T ( r , f 11 ) )

is satisfied for all r outside possibly a set with finite logarithmic measure, where λ < 1 is a positive number. Thus, in view of Lemma 4.2, we have

( D 11 + i D 12 ) h z 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) + h ( z ) ] 1 ,

or,

( E 11 + i E 12 ) h z 1 ( D 11 i D 12 ) h z 2 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) h ( z ) ] 1 .

If

( D 11 + i D 12 ) h z 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) + h ( z ) ] 1 ,

then we have h ( z + c ) + h ( z ) = λ 1 , where λ 1 is a constant in C . This shows that the polynomial h is a constant, which is a contradiction. Hence,

(4.63) ( E 11 + i E 12 ) h z 1 ( D 11 i D 12 ) h z 2 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) h ( z ) ] 1 .

Since h is a polynomial, from (4.63), it follows that h ( z + c ) h ( z ) = ξ , where ξ C . Hence, we see that h ( z ) = L ( z ) + H ( z ) + B , where L ( z ) = α 1 z 1 + α 2 z 2 , H ( z ) H ( s 1 ) , H ( s 1 ) is a polynomial in s 1 = c 2 z 1 c 1 z 2 .

Now, we claim that T 1 = T 2 . However, if T 1 T 2 , then using (4.62) and (4.63), a simple computation shows that

(4.64) 2 i ( T 2 T 1 ) e i h ( z + c ) { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } = 1 + { ( D 11 + i D 12 ) ( α 2 c 1 H ( z ) ) ( E 11 i E 12 ) ( α 1 + c 2 H ( z ) ) } e i L ( c ) ( D 12 i D 11 ) + ( E 12 + i E 11 ) .

It follows that h ( z + c ) must be a constant, i.e., h ( z ) is a constant, which is not possible. Consequently, the claim T 1 = T 2 is established. Thus, from (4.64), we see that

(4.65) { ( D 11 + i D 12 ) ( α 2 c 1 H ) ( E 11 i E 12 ) ( α 1 + c 2 H ) } e i L ( c ) { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } = 1 .

Next, we aim to show that H ( z ) 0 . Using (4.63) and (4.65), a simple computation shows that

(4.66) ( E 11 + i E 12 ) α 1 ( D 11 i D 12 ) α 2 + [ ( D 11 i D 12 ) c 1 + ( E 11 + i E 12 ) c 2 ] H ( i D 11 + D 12 ) ( i E 11 E 12 ) = e i L ( c )

and

(4.67) ( D 11 + i D 12 ) α 2 ( E 11 i E 12 ) α 1 [ ( D 11 + i D 12 ) c 1 + ( E 11 i E 12 ) c 2 ] H { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } = e i L ( c ) .

In view of (4.66) and (4.67), it follows that [ ( D 11 i D 12 ) c 1 + ( E 11 + i E 12 ) c 2 ] H and [ ( D 11 + i D 12 ) c 1 + ( E 11 i E 12 ) c 2 ] H are constants. Since ( c 1 , c 2 ) ( 0,0 ) , we see that H is a constant, that is, deg s 1 H 1 . Hence, it is easy to see that L ( z ) + H ( z ) + B is still linear form of α 1 z 1 + α 2 z 2 + B , which means that H ( z ) 0 . Therefore, it is easy to see that h ( z ) = L ( z ) + B = α 1 z 1 + α 2 z 2 + B . By using (4.66) and (4.67), we obtain

( E 11 + i E 12 ) α 1 ( D 11 i D 12 ) α 2 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i L ( c ) = 1 ; ( E 11 i E 12 ) α 1 ( D 11 + i D 12 ) α 2 ( D 12 i D 11 ) + ( E 12 + i E 11 ) e i L ( c ) = 1 ,

which implies that

e 2 i L ( c ) = { ( E 11 i E 12 ) α 1 ( D 11 + i D 12 ) α 2 } { ( i D 11 + D 12 ) ( i E 11 E 12 ) } { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } { ( E 11 + i E 12 ) α 1 ( D 11 i D 12 ) α 2 }

and

( E 11 2 + E 12 2 ) α 1 2 + ( D 11 2 + D 12 2 ) α 2 2 2 ( D 11 E 11 D 12 E 12 ) α 1 α 2 = ( D 11 E 11 ) 2 + ( D 12 + E 12 ) 2 .

By substituting h ( z ) = α 1 z 1 + α 2 z 2 + B and T 1 = T 2 into (4.60), we easily obtain

(4.68) f z 1 f z 2 = ( D 11 E 11 ) cos ( α 1 z 1 + α 2 z 2 + B ) ( D 12 + E 12 ) sin ( α 1 z 1 + α 2 z 2 + B ) .

The characteristic equations of (4.68) are d z 1 d t = 1 , d z 2 d t = 1 . Thus, we have

(4.69) d f d t = ( D 11 E 11 ) cos ( α 1 z 1 + α 2 z 2 + B ) ( D 12 + E 12 ) sin ( α 1 z 1 + α 2 z 2 + B ) .

By using the initial conditions: z 1 = 0 , z 2 = s and f = f ( 0 , s ) ψ 0 ( s ) with a parameter s , we see that z 1 = t , z 2 = t + s , and

f ( t , s ) = 0 t [ ( D 11 E 11 ) cos ( α 1 z 1 + α 2 z 2 + B ) ( D 12 + E 12 ) sin ( α 1 z 1 + α 2 z 2 + B ) ] d t + ψ 0 ( s ) = 0 t ( ( D 11 E 11 ) cos [ ( α 1 α 2 ) t + α 2 s + B ] ( D 12 + E 12 ) sin [ ( α 1 α 2 ) t + α 2 s + B ] ) d t + ψ 0 ( s ) = ( D 11 E 11 ) ( α 1 α 2 ) sin [ ( α 1 α 2 ) t + α 2 s + B ] + ( D 12 + E 12 ) ( α 1 α 2 ) cos [ ( α 1 α 2 ) t + α 2 s + B ] + ψ 1 ( s ) ,

where ψ 1 ( s ) is a transcendental entire function with finite order in s such that

ψ 1 ( s ) = ψ 0 ( s ) ( D 11 E 11 ) ( α 1 α 2 ) sin ( α 2 s + B ) ( D 12 + E 12 ) ( α 1 α 2 ) cos ( α 2 s + B ) .

Consequently, we have

(4.70) f ( z 1 , z 2 ) = ( D 11 E 11 ) ( α 1 α 2 ) sin ( α 1 z 1 + α 2 z 2 + B ) + ( D 12 + E 12 ) ( α 1 α 2 ) cos ( α 1 z 1 + α 2 z 2 + B ) + ψ 1 ( s ) .

which can be written as follows:

f ( z 1 , z 2 ) = ( D 11 E 11 ) ( α 1 α 2 ) sin ( L ( z ) + B ) + ( D 12 + E 12 ) ( α 1 α 2 ) cos ( L ( z ) + B ) + ψ 1 ( s ) .

Substituting (4.70) into (4.58), it is easy to verify that ψ 1 ( s ) satisfies

ψ 1 ( s ) + ψ 1 ( s + s 0 ) = E 11 α 1 D 11 α 2 α 1 α 2 cos ( L ( z ) + B ) + E 12 α 1 + D 12 α 2 α 1 α 2 sin ( L ( z ) + B ) D 11 E 11 α 1 α 2 sin ( L ( z ) + L ( c ) + B ) D 12 + E 12 α 1 α 2 cos ( L ( z ) + L ( c ) + B ) + T 1 .

This completes the proof.□

Proof of Theorem 3.5

Suppose that f is a transcendental entire solution of (3.5) with finite order. The equation (3.5) can be written as follows:

(4.71) D A ± Δ u 2 + D B Δ v 2 = 1 .

Below, we will analyze two cases.

Case 1: If D A ± Δ u is a constant, then it follows from (4.71) that D B Δ v is also constant. Denote

D A ± Δ u = K 9 , D B Δ v = K 10 .

In view of (4.71), it is easy to see that K 9 2 + K 10 2 = 1 . Consequently, we have

(4.72) M 1 ( f ) = f ( z + c ) + f z 1 = K 9 ξ 1 ± D A ± Δ K 10 η 1 ± D B Δ + α γ b β a b α 2 A 5 ,

(4.73) M 3 ( f ) = f ( z + c ) + 2 f z 1 2 = K 9 η 1 ± D A ± Δ + K 10 ξ 1 ± D B Δ + α β a γ a b α 2 B 5 .

A simple computation using (4.72) and (4.73) shows that

(4.74) f z 1 2 f z 1 2 = A 5 B 5 .

Differentiating (4.72) with respect to z 1 , we obtain

(4.75) f ( z + c ) z 1 + 2 f z 1 2 = 0 .

Thus, it follows from (4.73) and (4.75) that

(4.76) f ( z + c ) f ( z + c ) z 1 = B 5 .

By differentiating (4.76) with respect to z 1 , we obtain

(4.77) f ( z + c ) z 1 2 f ( z + c ) z 1 2 = 0 .

In view of (4.74) and (4.77), it is easy to see that A 5 = B 5 . Using K 9 2 + K 10 2 = 1 , A 5 = B 5 and from (4.72), (4.73), a simple computation shows that K 10 = R 2 R 3 ± R 1 R 1 2 + R 2 2 R 3 2 R 1 2 + R 2 2 , where

R 1 = ξ 1 ± η 1 ± D A ± Δ , R 2 = ξ 1 ± + η 1 ± D B Δ and R 3 = β ( α + b ) γ ( α + a ) a b α 2 .

In view of (4.74), using A 5 = B 5 , we obtain

(4.78) f ( z 1 , z 2 ) = e z 1 + ψ 1 ( z 2 ) + ψ 2 ( z 2 ) .

where ψ 1 ( z 2 ) , ψ 2 ( z 2 ) are two function in z 2 . From (4.72) and (4.78), it follows that e ( z 1 + c 1 ) + ψ 1 ( z 2 + c 2 ) + e z 1 + ψ 1 ( z 2 ) + ψ 2 ( z 2 + c 2 ) = A 5 , which implies that ψ 2 ( z 2 ) = A 5 , ψ 1 ( z 2 ) = A z 2 + B , where A = 2 k π i ± π i c 1 c 2 Therefore, from (4.78), we obtain f ( z 1 , z 2 ) = A 5 + e ( z 1 + A z 2 + B ) .

Case 2: If D A ± Δ u is not a constant. By the similar arguments being used in the proof of Theorem 3.1, there exists an nonconstant entire function h ( z ) such that

(4.79) M 1 ( f ) = f ( z + c ) + f z 1 = D 11 cos h ( z ) D 12 sin h ( z ) + T 1 ,

(4.80) M 3 ( f ) = f ( z + c ) + 2 f z 1 2 = E 11 cos h ( z ) + E 12 sin h ( z ) + T 2 .

In view of (4.79) and (4.80), we obtain

(4.81) f z 1 2 f z 1 2 = ( D 11 E 11 ) cos h ( z ) ( D 12 + E 12 ) sin h ( z ) + ( T 1 T 2 ) .

By differentiating (4.79) with respect to z 1 , we obtain

(4.82) f ( z + c ) z 1 + 2 f z 1 2 = D 11 sin h ( z ) h z 1 D 12 cos h ( z ) h z 1 .

Thus, it follows from (4.80) and (4.82) that

(4.83) f ( z + c ) f ( z + c ) z 1 = D 12 h z 1 + E 11 cos h ( z ) + D 11 h z 1 + E 12 sin h ( z ) + T 2 .

By differentiating (4.83) with respect to z 1 and using (4.81), we obtain

D 11 2 h z 1 2 D 12 h z 1 2 E 11 h z 1 sin h ( z ) + D 12 2 h z 1 2 + D 11 h z 1 2 + E 12 h z 1 cos h ( z ) = ( D 11 E 11 ) cos h ( z + c ) ( D 12 + E 12 ) sin h ( z + c ) + ( T 1 T 2 ) .

This can be formulated as follows:

(4.84) { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } e i h ( z + c ) + 2 i ( T 2 T 1 ) ( i D 11 + D 12 ) ( i E 11 E 12 ) e i h ( z + c ) + ( D 11 + i D 12 ) 2 h z 1 2 + ( i D 11 D 12 ) h z 1 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) + h ( z ) ] + ( i D 12 D 11 ) 2 h z 1 2 + ( i D 11 + D 12 ) h z 1 2 + ( i E 12 + E 11 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) h ( z ) ] = 1 .

Equation (4.84) can be written as g 11 + g 12 + g 13 = 1 . It is easy to see that g 11 0 , i.e.,

{ ( D 12 i D 11 ) + ( E 12 + i E 11 ) } e i h ( z + c ) + 2 i ( T 2 T 1 ) ( i D 11 + D 12 ) ( i E 11 E 12 ) 0 .

Otherwise, we obtain

e i h ( z + c ) = 2 i ( T 1 T 2 ) ( D 12 i D 11 ) + ( E 12 + i E 11 ) .

This shows that h ( z + c ) is a constant, i.e., h ( z ) is a constant, which is a contradiction.

If

( D 11 + i D 12 ) 2 h z 1 2 + ( i D 11 D 12 ) h z 1 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) = 0 ,

then from (4.84) it is easy to see that

( i D 12 D 11 ) 2 h z 1 2 + ( i D 11 + D 12 ) h z 1 2 + ( i E 12 + E 11 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) 0 .

Otherwise, it follows that

{ ( D 12 i D 11 ) + ( E 12 + i E 11 ) } e i h ( z + c ) + 2 i ( T 2 T 1 ) ( i D 11 + D 12 ) ( i E 11 E 12 ) e i h ( z + c ) = 1 ,

and a simple computation shows that

e i h ( z + c ) = i ( T 1 T 2 ) ± i T ( D 12 i D 11 ) + ( E 12 + i E 11 ) ,

where T ( T 2 T 1 ) 2 + [ ( D 12 i D 11 ) + ( E 12 + i E 11 ) ] [ ( i E 11 E 12 ) ( i D 11 + D 12 ) ] . Clearly, it follows that h ( z ) is a constant, which is a contradiction. Thus, equation (4.84) reduces to

{ ( D 12 i D 11 ) + ( E 12 + i E 11 ) } e i h ( z + c ) + 2 i ( T 2 T 1 ) ( i D 11 + D 12 ) ( i E 11 E 12 ) e i h ( z + c ) + ( i D 12 D 11 ) 2 h z 1 2 + ( i D 11 + D 12 ) h z 1 2 + ( i E 12 + E 11 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) h ( z ) ] = 1 .

By the similar argument being used previously, in view of the second fundamental theorem of Nevanlinna for several complex variables, we obtain T ( r , e 2 i h ( z + c ) ) = o ( T ( r , e h ) ) , which is a contradiction. Hence,

( D 11 + i D 12 ) 2 h z 1 2 + ( i D 11 D 12 ) h z 1 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) 0 .

Similarly, we obtain

( i D 12 D 11 ) 2 h z 1 2 + ( i D 11 + D 12 ) h z 1 2 + ( i E 12 + E 11 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) 0 .

That is, g 12 0 and g 13 0 . Moreover, since g 11 , g 12 , g 13 are all entire functions, we see that

j = 1 3 N 2 r , 1 g 1 j + 2 N ¯ ( r , g 1 j ) = 0 < λ T ( r , g 11 ) + O ( log + T ( r , g 11 ) )

for all r outside possibly a set with finite logarithmic measure, where λ < 1 is a positive number. Therefore, in view of the Lemma 4.2 using (4.84), we obtain

( D 11 + i D 12 ) 2 h z 1 2 + ( i D 11 D 12 ) h z 1 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) + h ( z ) ] 1

or

( i D 12 D 11 ) 2 h z 1 2 + ( i D 11 + D 12 ) h z 1 2 + ( i E 12 + E 11 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) h ( z ) ] 1 .

If

( D 11 + i D 12 ) 2 h z 1 2 + ( i D 11 D 12 ) h z 1 2 ( E 11 i E 12 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) + h ( z ) ] 1 ,

then it is easy to see that h ( z + c ) + h ( z ) = λ 2 , where λ 2 is a constant in C . This shows that h ( z ) is a constant, which is not possible. Therefore,

(4.85) ( i D 12 D 11 ) 2 h z 1 2 + ( i D 11 + D 12 ) h z 1 2 + ( i E 12 + E 11 ) h z 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i [ h ( z + c ) h ( z ) ] 1 .

Since h ( z ) is a polynomial, then it follows from (4.85) that h ( z + c ) h ( z ) = η , where η is a constant in C . Therefore, it follows that h ( z ) = L ( z ) + H ( z ) + B , where L ( z ) = α 1 z 1 + α 2 z 2 , H ( z ) H ( s 1 ) , H ( s 1 ) is a polynomial in s 1 = c 2 z 1 c 1 z 2 . We claim that T 1 = T 2 . On contrary, we suppose that T 1 T 2 . Then in view of (4.84) and (4.85), a simple computation shows that

(4.86) 2 i ( T 2 T 1 ) e i h ( z + c ) { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } = 1 + { ( i D 11 D 12 ) ( α 1 + c 2 H ( z ) ) 2 ( E 11 i E 12 ) ( α 1 + c 2 H ( z ) ) } e i L ( c ) ( D 12 i D 11 ) + ( E 12 + i E 11 ) .

It follows that the polynomial h ( z + c ) must be a constant. In other words, the polynomial h ( z ) becomes a constant, which is a contradiction. Consequently, we have T 1 = T 2 . Thus, from (4.86), it is easy to see that

(4.87) { ( i D 11 D 12 ) ( α 1 + c 2 H ) 2 ( E 11 i E 12 ) ( α 1 + c 2 H ) } e i L ( c ) { ( D 12 i D 11 ) + ( E 12 + i E 11 ) } = 1 .

We claim that H ( z ) 0 . From (4.85) and (4.87), it follows that

(4.88) ( i D 11 + D 12 ) α 1 2 + ( i E 12 + E 11 ) α 1 + c 2 [ 2 α 1 ( i D 11 + D 12 ) + ( i E 12 + E 11 ) ] H + c 2 2 ( i D 11 + D 12 ) ( H ) 2 = e i L ( c ) [ ( i D 11 + D 12 ) ( i E 11 E 12 ) ]

and

(4.89) ( i D 11 D 12 ) α 1 2 ( E 11 i E 12 ) α 1 + c 2 [ 2 α 1 ( i D 11 D 12 ) ( E 11 i E 12 ) ] H + c 2 2 ( i D 11 D 12 ) ( H ) 2 = e i L ( c ) [ ( i D 11 D 12 ) ( i E 11 + E 12 ) ] .

Combining with the fact that ( c 1 , c 2 ) ( 0,0 ) , it follows that H is a constant, that is deg s 1 H 1 . Hence, L ( z ) + H ( z ) + B is still linear form of α 1 z 1 + α 2 z 2 + B , which means that H ( z ) 0 . Therefore, it is easy to see that h ( z ) = L ( z ) + B = α 1 z 1 + α 2 z 2 + B . Thus, equations (4.88) and (4.89) become

(4.90) ( i D 11 + D 12 ) α 1 2 + ( i E 12 + E 11 ) α 1 ( i D 11 + D 12 ) ( i E 11 E 12 ) e i L ( c ) = 1

and

(4.91) ( i D 11 D 12 ) α 1 2 ( E 11 i E 12 ) α 1 ( i D 11 D 12 ) ( i E 11 + E 12 ) e i L ( c ) = 1 .

From (4.90) and (4.91), a simple computation shows that

e 2 i L ( c ) = { ( i D 11 D 12 ) α 1 ( E 11 i E 12 ) } { ( i D 11 + D 12 ) ( i E 11 E 12 ) } { ( i D 11 D 12 ) ( i E 11 + E 12 ) } { ( i D 11 + D 12 ) α 1 + ( i E 12 + E 11 ) }

and

( D 11 2 + D 12 2 ) α 1 4 + 2 ( D 11 E 12 + D 12 E 11 ) α 1 3 + ( E 11 2 + E 12 2 ) α 1 2 = ( D 11 E 11 ) 2 + ( D 12 + E 12 ) 2 .

Furthermore, we observe that equation (4.83) can be formulated as follows:

f ( z + c ) f ( z + c ) z 1 = ( D 12 α 1 + E 11 ) cos ( α 1 z 1 + α 2 z 2 + B ) + ( D 11 α 1 + E 12 ) sin ( α 1 z 1 + α 2 z 2 + B ) + T 2 ,

that is,

(4.92) f z 1 f ( z ) = ( D 12 α 1 + E 11 ) cos [ L ( z ) L ( c ) + B ] ( D 11 α 1 + E 12 ) sin [ L ( z ) L ( c ) + B ] T 2 ,

Therefore, a simple computation shows that

(4.93) f ( z 1 , z 2 ) = ϕ ( z 2 ) e z 1 + R 11 sin [ L ( z ) L ( c ) + B ] + R 12 cos [ L ( z ) L ( c ) + B ] + T 2 ,

where ϕ ( z 2 ) is an entire function in z 2 and

R 11 = ( D 11 α 1 + E 12 ) α 1 ( D 12 α 1 + E 11 ) α 1 2 + 1 , R 12 = ( D 12 α 1 + E 11 ) + α 1 ( D 11 α 1 + E 12 ) α 1 2 + 1 .

Through the substitution of (4.93) into (4.79), a simple computation leads to

ϕ ( z 2 + c 2 ) e c 1 e z 1 + ( R 11 + D 12 ) sin ( L ( z ) + B ) + ( R 12 D 11 ) cos ( L ( z ) + B ) + R 11 α 1 cos [ L ( z ) L ( c ) + B ] R 12 α 1 sin [ L ( z ) L ( c ) + B ] = ϕ ( z 2 ) e z 1 .

By comparing the coefficients of e z 1 on both sides, we derive the equation ϕ ( z 2 + c 2 ) = e c 1 ϕ ( z 2 ) , from which we deduce that ϕ ( z 2 ) = ( e c 1 ) z 2 c 2 π ( z 2 ) , where π ( z 2 + c 2 ) = π ( z 2 ) . Consequently, by examining (4.93), we deduce that form of the solution is

f ( z 1 , z 2 ) = ( e c 1 ) z 2 c 2 π ( z 2 ) e z 1 + R 11 sin ( L ( z ) L ( c ) + B ) + R 12 cos ( L ( z ) L ( c ) + B ) + T 2 .

This completes the proof.□



Acknowledgement

The authors express their gratitude to the referee for the helpful suggestions and insightful comments aimed at improving the exposition of the article. The first named author is supported by the DST FIST (SR/FST/MS-II/2021/101(C)), Department of Mathematics, Jadavpur University. The second named author is supported by CSIR-SRF (File No: 09/0096(12546)/2021-EMR-I, dated: 08/10/2024), Government of India, New Delhi.

  1. Funding information: Authors state no funding involved.

  2. Author contributions: Both authors have contributed equally to the manuscript.

  3. Conflict of interest: The authors declare that there is no conflict of interest regarding the publication of this article.

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2024-02-16
Revised: 2024-10-16
Accepted: 2024-11-26
Published Online: 2025-06-18

© 2025 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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