Electronic Journal of Differential Equations, Vol. 2023 (2023), No. 43, pp. 1–18.

ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu

DOI: https://doi.org/10.58997/ejde.2023.43

SOLUTIONS OF COMPLEX NONLINEAR FUNCTIONAL

EQUATIONS INCLUDING SECOND ORDER PARTIAL

DIFFERENTIAL AND DIFFERENCE IN C2

HONG YAN XU, GOUTAM HALDAR

Abstract. This article is devoted to exploring the existence and the form of

finite order transcendental entire solutions of Fermat-type second order partial

differential-difference equations(∂2f
∂z21

+ δ
∂2f

∂z22
+ η

∂2f

∂z1∂z2

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

and(∂2f
∂z21

+ δ
∂2f

∂z22
+ η

∂2f

∂z1∂z2

)2
+ (f(z1 + c1, z2 + c2)− f(z1, z2))2 = eg(z),

where δ, η ∈ C and g(z1, z2) is a polynomial in C2. Our results improve the
results of Liu and Dong [23], Liu et al. [24], and Liu and Yang [25]. Several

examples confirm that the form of transcendental entire solutions of finite order

in our results are precise.

1. Introduction

It is well known that for a positive integer m, the equation

fm + gm = 1 (1.1)

is regarded as Fermat type equation over function fields. With the help of Nevan-
linna theory [11, 16], Montel [27], Iyer [15], and Gross [5] studied the existence
and form of the solutions of the functional equation (1.1) and pointed out that for
m = 2, the entire solutions of (1.1) are f(z) = cos(ξ(z)) and g(z) = sin(ξ(z)), where
ξ is an entire function, and for m > 2, there are no non-constant entire solutions
of (1.1). In 2004, Yang and Li [42] investigated (1.1) by replacing g with f ′ when
m = 2, and proved that the transcendental entire solution of f(z)2 +f ′(z)2 = 1 has
the form f(z) = Aeαz/2 + e−αz/2A, where A,α are non-zero complex constants.

After the development of difference Nevanlinna theory (see [4, 6]), many re-
searcher began to study the existence and form of entire or meromorphic solutions
of Fermat-type difference and differential-difference equations (see [7, 21, 22, 23,
24, 25]). In 2012, Liu et al. [24] proved that the transcendental entire solutions
with finite order of the Fermat-type difference equation f(z)2 + f(z+ c)2 = 1 must

2020 Mathematics Subject Classification. 30D35, 35M30, 32W50, 39A45.
Key words and phrases. Functions of several complex variables; Fermat-type equations;

entire solutions; Nevanlinna theory.
©2023. This work is licensed under a CC BY 4.0 license.

Submitted April 17, 2023. Published June 26 2023.

1



2 H. Y. XU, G. HALDAR EJDE-2023/43

satisfy f(z) = sin(Az+B), where B is a constant and A = (4k+1)π/2c, where k is
an integer. In 2019, Han and Lü [10] investigated the more general complex differ-
ence equation f(z)2 + f(z + c)2 = eαz+β , α, β ∈ C, and proved that the nontrivial
meromorphic solutions of this equation are of the form de(αz+β)/2, where d ∈ C
such that d2(1 + eαc) = 1.

Hereinafter, we denote by z + w = (z1 + w1, z2 + w2) for any z = (z1, z2), w =
(w1, w2) ∈ C2. The study of several characteristics of the solutions to partial
differential equations in several complex variables is an important topic; see [1, 2,
3, 8, 9, 12, 14, 18, 26, 34, 35, 36, 37, 38]). It was Saleebly, who in 1999, first studied
the existence and form of entire and meromorphic solutions of Fermat-type partial
differential equations (see [30, 31, 32]). Most noticeably, Khavinson [14] proved that
any entire solution of the partial differential equation f2z1 + f2z2 = 1 must be linear,

i.e., f(z1, z2) = az1+bz2+c, where a, b, c ∈ C, and a2+b2 = 1. Here fz1 and fz2 are
the partial derivatives of f with respect to z1 and z2, respectively. Later, Li [19, 20]
investigated on the partial differential equations with more general forms such as
f2z1 + f2z2 = p, f2z1 + f2z2 = eq, etc, where p, q are polynomials in C2. Recently, Xu
and Cao [40] extended several results from one complex variable to several complex
variables. We recall some of them here.

Theorem 1.1 ([40]). Let c = (c1, c2, . . . , cn) ∈ Cn \ {(0, 0, . . . , 0)}. Then, any
non-constant entire solution f : Cn → P1(C) with finite order of the Fermat-type
difference equation

f(z)2 + f(z + c)2 = 1 (1.2)

has the form of f(z) = cos(L(z) + B), where L is a linear function of the form
L(z) = a1z1 + · · ·+ anzn on Cn such that L(c) = −π/2− 2kπ (k ∈ Z), and B is a
constant on C.

Theorem 1.2 ([40]). Let c = (c1, c2) be a constant in C2. Then any transcendental
entire solution with finite order of the Fermat-type partial differential-difference
equation (∂f(z1, z2)

∂z1

)2
+ f2(z1 + c1, z2 + c2) = 1 (1.3)

has the form f(z1, z2) = sin(Az1 + Bz2 + H(z2)), Where A,B are constant on C
satisfying A2 = 1 and Aei(Ac1+Bc2) = 1, and H(z2) is a polynomial in one variable
z2 such that H(z2) ≡ H(z2 + c2). In the special case whenever c2 6= 0, we have
f(z1, z2) = sin(Az1 +Bz2 + Constant).

In 2021, Zheng and Xu [43] obtained the following result.

Theorem 1.3 ([43]). Let c = (c1, c2) ∈ C2 \{(0, 0)}. Then there are no finite order
transcendental entire solutions of

f(z)2 + [f(z + c)− f(z)]2 = 1. (1.4)

In 2022, Xu et al. [41] extended Theorems 1.1 and 1.2 by replacing 1 with eg(z1,z2)

in the right-hand side of equations (1.2) and (1.3), and ∂f(z1,z2)
∂z1

with α∂f(z1,z2)∂z1
+

β ∂f(z1,z2)∂z2
in equation (1.3). We list some of the results here.

Theorem 1.4 ([41]). Let c = (c1, c2) ∈ C2, and let α, β be constants in C that are
not zero at the same time. If the partial differential-difference equation(

α
∂f(z1, z1)

∂z1
+ β

∂f(z1, z1)

∂z2

)2
+ f(z1 + c1, z2 + c2)2 = eg(z1,z1) (1.5)



EJDE-2023/43 COMPLEX NONLINEAR FUNCTIONAL EQUATIONS 3

admits a transcendental entire solution of finite order, then f and g must satisfy
one of the following cases:

(i) f(z1, z2) = ±e 1
2 g(z−c), where g(z) = φ(βz1 −αz2) and φ is a polynomial in

C;
(ii) g(z) must be of the form g(z) = L(z) + H(s1) + B, where L(z) is a linear

function of the form L(z) = A1z1 + A2z2, H(s1) is a polynomial in s1 :=
c2z1 − c1z2, A1, A2, B ∈ C and

f(z1, z2) =
ξ2 + 1

ξ(αA1 + βA2)
e

1
2 (L(z)+H(s1)+B),

where ξ(6= 0), A1, A2, B ∈ C satisfying

(αc2 − βc1)H ′ ≡ 0,
1

2i

ξ2 − 1

ξ2 + 1
(αA1 + βA2) = e

1
2 (A1c1+A2c2);

(iii)

f(z1, z2) =
eL1(z)+H1(s1)+B1

2(αA11 + βA12)
+
eL2(z)+H2(s1)+B2

2(αA21 + βA22)
,

where L1(z) = A11z1 + A12z2 +B1 and L2(z) = A21z1 + A22z2 +B2, with
Aj1, Aj2, Bj ∈ C( j =1, 2), satisfy

g(z) = L1(z) + L2(z) +H1(s1) +H2(s1) +B1 +B2,

L1(z) +H1(s1) 6= L2(z) +H2(s1), (αc2 − βc1)H ′j ≡ 0

−i(αA11 + βA12)e−L1(c) = i(αA21 + βA22)e−L2(c) = 1,

where Hj(s1) for j = 1, 2 are polynomials in s1 = c2z1 − c1z2.

In the same paper [41], they also explored the existence and the forms of entire
and meromorphic solutions of the partial differential difference equation(

α
∂f

∂z1
+ β

∂f

∂z2

)2
+ (f(z + c)− f(z))2 = eg(z), (1.6)

where g(z) is a polynomial in C2 and α, β are constants in C and obtained the
following result.

Theorem 1.5. [41] Let c = (c1, c2) ∈ C2, α( 6= 0), β constants in C, and αc2−βc1 6=
0. Let f be a finite order transcendental entire solution of the partial differential-
difference equation (1.6), then f must satisfy one of the following cases:

(i) f(z1, z2) = φ1(βz1 − αz2), where φ1 is a finite order transcendental entire
function such that

±e 1
2 g(z) = φ1(βz1 − αz2 + βc1 − αc2)− φ1(βz1 − αz2),

(ii) g(z) = A1z1 +A2z2 +H(s1) +B and

f(z) = ± 1

α

∫ z1/α

0

e
1
2 (A1z1+A2z2+H(s1)+B)dz1 +G

(αz2 − βz1
α

)
,

where e
1
2 (A1c1+A2c2) = 1, H(s1) is a polynomial in s1 = c2z1 − c1z2, G is a

finite order period entire function with period (αc2 − βc1)/α, and A1, A2 ∈
C;



4 H. Y. XU, G. HALDAR EJDE-2023/43

(iii) g(z) = A1z1 +A2z2 +B and

f(z) =
(
ξ +

1

ξ

)e 1
2 (A1z1+A2z2+B)

αA1 + βA2
+G

(αz2 − βz1
α

)
,

where A1, A2, B ∈ C, G is a finite order entire period function with period
(αc2 − βc1)/α and ξ( 6= 0), A1, A2, B ∈ C satisfying

1

2i

ξ2 − 1

ξ2 + 1
(αA1 + βA2) + 1 = e

1
2 (A1c1+A2c2);

(iv) g(z) = A1z1 +A2z2 and

f(z1, z2) =
eL1(z)+B1

2(αA11 + βA12)
+

eL2(z)+B2

2(αA21 + βA22)
+G

(αz2 − βz1
α

)
,

where A1, A2, B ∈ C, G is a finite order entire period function with period
(αc2−βc1)/α and L1(z) = A11z1+A12z2+B1 and L2(z) = A21z1+A22z2+
B2, with Aj1, Aj2, Bj ∈ C( j =1, 2), satisfy

L1(z) 6= L2(z), g(z) = L1(z) + L2(z) +B1 +B2,

[1− i(αA11 + βA12)]e−L1(c) = [1 + i(αA21 + βA22)]e−L2(c) = 1.

r For the second-order partial differential-difference equations of Fermat type in
C2, Xu et al. [39] obtained the following important results.

Theorem 1.6. [39] Let c = (c1, c2) ∈ C2 and c2 6= 0. If the difference equation(∂2f(z1, z2)

∂z21

)2
+ f(z1 + c1, z2 + c2)2 = eg(z1,z2) (1.7)

admits a finite order transcendental entire solution, then g(z) must be of the form
g(z) = L(z) + H(s1) + B, where L(z) = A1z1 + A2z2, H(s1) is a polynomial in
s1 := c2z1 − c1z2, and A1, A2 ∈ C. Further, f(z) must satisfy one of the following
cases:

(i)

f(z1, z2) =
4(ξ2 + 1)

A2
1ξ

e
1
2 [A1z1+A2z2+B],

where ξ is a non-zero complex number in C and e(A1c1+A2c2)/2 = A2
1(ξ2 −

1)/4i(ξ2 + 1).
(ii)

f(z1, z2) =
A2

21e
L1(z)+B1 +A2

11e
L2(z)+B2

2
,

where L1(z) = A11z1 + A12z2 +B1 and L2(z) = A21z1 + A22z2 +B2, with
Aj1, Aj2, Bj ∈ C( j =1, 2), satisfy

g(z) = L1(z) + L2(z) +B1 +B2,

L1(z) 6= L2(z), −iA2
21e

L1(c) = iA2
21e

L2(c) = 1.

Theorems 1.3–1.6 suggest the following questions as open problems.

(1) What can be said about the existence and forms of solutions of the equation
(1.4) when the constant 1 is replaced by a function eg(z1,z2) in Theorem 1.3?



EJDE-2023/43 COMPLEX NONLINEAR FUNCTIONAL EQUATIONS 5

(2) What can be said about the existence and forms of solutions of the equation

(1.7) when ∂2f(z1,z2)
∂z21

is replaced by more general operator ∂2f
∂z21

+ δ ∂
2f
∂z22

+

η ∂2f
∂z1∂z2

in Theorem 1.6?

(3) What can be said about the existence and forms of solutions of (1.5) and

(1.6) when α ∂f
∂z1

+ β ∂f
∂z2

is replace by second order homogeneous linear

partial differential operator ∂2f
∂z21

+ δ ∂
2f
∂z22

+ η ∂2f
∂z1∂z2

in Theorems 1.5 and 1.6?

2. Results

Motivated by the above questions and utilizing difference analogues of Nevan-
linna theory of several complex variables [1, 2, 3], we obtain Theorems 2.1, 2.6,
and 2.10. Theorem 2.1 is an extension and generalization of Theorems 1.4 and 1.6.
Theorem 2.6 is an extension of Theorem 1.5. And Theorem 2.10 is the extension
of Theorem 1.3. Now we consider the second-order partial differential difference
equations (∂2f

∂z21
+ δ

∂2f

∂z22
+ η

∂2f

∂z1∂z2

)2
+ f(z1 + c1, z2 + c2)2 = eg(z1,z2), (2.1)(∂2f

∂z21
+ δ

∂2f

∂z22
+ η

∂2f

∂z1∂z2

)2
+Big(f(z1 + c1, z2 + c2)− f(z1, z2)

)2
= eg(z), (2.2)

and the difference equation

f(z)2 + [f(z + c)− f(z)]2 = eg(z1,z2), (2.3)

where δ, η ∈ C, c = (c1, c2) ∈ C2 and g(z1, z2) is a polynomial in C2.
Before we state our main results, let us first set the following.

A1 = a1 +
1

2
ηa2, A2 = δa2 +

1

2
ηa1, A3 = c22 + δc21 − ηc1c2,

A4 =
1

2
(a21 + δa22 + ηa1a2), Aj5 = 2aj1 + ηaj2, Aj6 = 2δaj2 + ηaj1,

Aj7 = a2j1 + δa2j2 + ηaj1aj2, j = 1, 2.

(2.4)

Now we state our results as follows.

Theorem 2.1. Let c = (c1, c2) ∈ C2 and g(z1, z2) be a polynomial in C2. If f(z)
be a finite order transcendental entire solution of (2.1), then one of the following
cases occurs.

(i) f(z1, z2) = φ1(z2 − αz1) + φ2(z2 − βz1), where φ1, φ2 are finite order tran-
scendental entire functions in C2 such that

φ1(z2 − αz1 + c2 − αc1) + φ2(z2 − βz1 + c2 − βc1) = ±e 1
2 g(z1,z2),

α, β ∈ C with α+ β = η, αβ = δ.
(ii) g(z1, z2) is of the form g(z1, z2) = L(z) +H(s1) +B, where L(z) = a1z1 +

a2z2, H(s1) is a polynomial in s1 := c2z1 − c1z2, a1, a2, B ∈ C, and the
form of the solution is

f(z1, z2) =
ξ2 − 1

2iξ
e

1
2 [L(z)+H(s1)−L(c)+B],



6 H. Y. XU, G. HALDAR EJDE-2023/43

where ξ 6= 0,±1,±i and L(z) satisfies the relation

e
1
2 [a1c1+a2c2] =

ξ2 − 1

2i(ξ2 + 1)
[A4 + (A1c2 −A2c1)a0 +

1

2
A3a

2
0],

where a0 is the coefficient of linear term of the polynomial H(s1) and Aj’s
are defined in (2.4). In particular, if A1c2 − A2c1 6= 0 or A3 6= 0, then
H(s1) becomes linear in s1.

(iii) g(z1, z2) is of the form g(z1, z2) = L(z) +H(s1) +B, where L(z) = L1(z) +
L2(z), H(s1) = H1(s1) + H2(s1) with L1(z) + H1(s1) 6= L2(z) + H2(s1),
Lj(z) = aj1z1 + aj2z2, B = B1 + B2, Hj(s1) is a polynomial in s1 =
c2z1 − c1z2 for j = 1, 2, B1, B2, aji are constants in C, and the form of the
solution is

f(z1, z2) =
1

2i
[A2e

(L1(z)+H1(s1)−L1(c)+B1) +A1e
(L2(z)+H2(s1)−L2(c)+B2)],

where L1(z) and L2(z), respectively satisfy the relations

eL1(c) = −i[A17+(A15c2−A16c1)a0+A3a
2
0]eL2(c) = i[A27+(A25c2−A26c1)a00+A3a

2
00],

a0 and a00, respectively the coefficients of the linear term of the polynomials
H1(s1) and H2(s1), and Aij’s are defined in (2.4) In particular, if A15c2−
A16c1 6= 0 or A3 6= 0, then H1 becomes linear in s1. Similarly, if A25c2 −
A26c1 6= 0 or A3 6= 0, then H2 becomes linear in s1.

Next, we exhibit some examples in support of the Theorem 2.1.

Example 2.2. Let α = β = 1, c1 = 2, c2 = 3 and g(z) = 4(z2 − z1 + 1)2. Then, in

view of Theorem 2.1(i), it can be easily seen that f(z1, z2) = e(z2−z1)
2

is a solution
of (2.1).

Example 2.3. Let c1 = c2 = 1, ξ = 3, δ = 1, η = 2 and g(z1, z2) = z1 + z2 + (z1 −
z2)n + 10, n ∈ N. Then in view of of Theorem 2.1(ii), one can easily verify that
f(z) = 5

3e
[z1+z2+(z1−z2)n+10]/2 is a solution of (2.1).

Example 2.4. Let δ = η = 4, ξ = 5, c1 = 2, c2 = 3, a0 = 1, L(z) = z1 − z2 and
g(z1, z2) = 4z1 − 3z2. Then in view of Theorem 2.1(ii), we can easily verify that

f(z1, z2) = − 12i
25 e

1
2 (4z1−3z2) is a solution of (2.1).

Example 2.5. Let c = (c1, c2) ∈ C such that c1 6= c2, δ = 1, η = 2, L1(z) = z1+z2,
L2(z) = z1 + 2z2, H1(s1) = H2(s2) = 0 and B1 = B2 = 1, and g(z1, z2) =
2z1 + 3z2 + 2. Then in view of Theorem 2.1(iii), it can be easily verified that
f(z1, z2) = 1

2 [ 14e
z1+z2+1 + 1

9e
z1+2z2+1] is a solution of (2.1).

Theorem 2.6. Let c = (c1, c2) ∈ C2, δ, η ∈ C and g(z) is a polynomial in C2.
Let f(z) be a finite order transcendental entire solution of (2.2). Then, one of the
following cases must occur.

(i) f(z1, z2) = φ1(z2 − αz1) + φ2(z2 − βz1), where φ1, φ2 are finite order tran-
scendental entire functions in C2 satisfying

φ1(z2 − αz1 + c2 − αc1) + φ2(z2 − βz1 + c2 − βc1)

− φ1(z2 − αz1)− φ2(z2 − βz1) = ±eg(z)/2,

with α, β ∈ C such that α+ β = η and αβ = δ.



EJDE-2023/43 COMPLEX NONLINEAR FUNCTIONAL EQUATIONS 7

(ii) g(z1, z2) = a1z1 + a2z2 + H(c2z1 − c1z2) + B, where H is a polynomial in
c2z1 − c1z2 and a1c1 + a2c2 = 4kπi, k ∈ Z,

f(z1, z2) = ±
∫ z1

0

∫ z1

0

e
1
2 [a1z1+a2z2+H(c2z1−c1z2)+B]dz1dz1

+

∫ z1/α2

0

G0(z2 − βz1)dz1 +G1(z2 − αz1),

where G0, G1 are finite order transcendental entire functions in C2 satisfy-
ing ∫ z1

0

[G0(z2 − βz1 + c2 − βc1)−G0(z2 − βz1)]dz1

+G1(z2 − αz1 + c2 − αc1)−G1(z2 − αz1) = 0.

(iii) If γc22 + δc21 6= ηc1c2, then g(z) must be of the form g(z) = a1z1 +a2z2 +B,
a1, a2, B ∈ C, and the solution has the form

f(z1, z2) = φ1(z2 − αz1) + φ2(z2 − βz1)

+
2(ξ + ξ−1)

a21 + δa22 + ηa1a2
e

1
2 [a1z1+a2z2+B],

where ξ( 6= 0) ∈ C, a21 + δa22 + ηa1a2 6= 0, α, β are same as in (i), φ1, φ2 are
finite order transcendental entire functions in C2 such that

φ1(z2 − αz1 + c2 − αc1) + φ2(z2 − βz1 + c2 − βc1)

= φ1(z2 − αz1) + φ2(z2 − βz1)

and

e
1
2 [a1c1+a2c2] =

(ξ − ξ−1)(a21 + δa22 + ηa1a2)

4i(ξ + ξ−1)
+ 1.

(iv) If γc22 + δc21 6= ηc1c2, then g(z) must be of the form g(z) = L1(z) +L2(z) +
B1 +B2, where Lj(z) = aj1z1 + aj2z2 with L1(z) 6= L2(z), aij , Bj ∈ C and
the form of the solution is

f(z1, z2) = φ1(z2 − αz1) + φ2(z2 − βz1) +
eL1(z)+B1

2(a211 + δa212 + ηa11a12)

+
eL2(z)+B2

2(a221 + δa222 + ηa21a22)
,

where a221 + δa222 + ηa21a22 6= 0, a211 + δa212 + ηa11a12 6= 0, α, β are same
as in (i), φ1, φ2 are finite order transcendental entire functions in C2 such
that

φ1(z2 − αz1 + c2 − αc1) + φ2(z2 − βz1 + c2 − βc1)

= φ1(z2 − αz1) + φ2(z2 − βz1)

and L1(z), L2(z) satisfy the relations

eL1(c) = −i(a211 + δa212 + ηa11a12) + 1,

eL2(c) = i(a221 + δa222 + ηa21a22) + 1.

The following examples show that the forms of solutions are precise.



8 H. Y. XU, G. HALDAR EJDE-2023/43

Example 2.7. Let α = β = −1. Choose c = (c1, c2) ∈ C2 such that c1+c2 = 2kπi,
k ∈ C. Then in view of Theorem 2.6(i), we can easily deduce that f(z1, z2) = ez1+z2

is a solution of (2.2) with g(z1, z2) = 2(z1 + z2).

Example 2.8. Let α = β = 1 and ξ = 2. Choose c = (c1, c2) ∈ C2 such that
c1 6= c2 and c2−c1 = 2kπi, k ∈ Z. Let ψ(z2−z1) = φ1(z2−z1)+φ2(z2−z1) = ez2−z1

and g(z) = L(z)+1 = z1 +2z2 +1. Then, in view of Theorem 2.6(iii), we can easily
verify that f(z1, z2) = ez1−z2 + 5

9e
(z1+z2+1)/2 is a solution of (2.2).

Example 2.9. Let δ = 1, η = 2, c1 = log(10−8i)/4, and c2 = [log(1−9i)− log(1+
i)]/2. Let L1(z) = z1 + 2z2, L2(z) = −z1 + 2z2. Then in view of Theorem 2.6(iv),
we can easily deduce that f(z1, z2) = eL1(z)+1/18 + eL2(z)+2/2 is a solution of (2.2)
with g(z1, z2) = L1(z) + L2(z) + 2.

Theorem 2.10. Let c = (c1, c2) ∈ C2 and g(z1, z2) be a polynomial in C2. If f
be a finite order transcendental entire solution of (2.3), then one of the following
cases must occur.

(i) g(z1, z2) must be of the form g(z1, z2) = L(z) + H(s) + B, where L(z) =
a1z1 + a2z2, H(s) is a polynomial in s := c2z1 − c1z2, a1, a2, B ∈ C and

f(z1, z2) = ±e 1
2 [L(z)+H(s)+B], where e

1
2L(c) = 1.

(ii) g(z1, z2) must be of the form g(z1, z2) = L(z) +H(s) + B, L(z), H(s) and
B are same as (i) and

f(z1, z2) =
ξ2 + 1

2ξ
e

1
2 [L(z)+H(s)+B],

where ξ 6= 0,±i,±1 and L(z) satisfies the relation

e
1
2L(c) =

(1− i)ξ2 + 1 + i

ξ2 + 1
.

(iii)

f(z1, z2) =
eL1(z)+H1(s1)+B1 + eL2(z)+H2(s1)+B2

2
,

where L1(z) = a11z1 + a12z2 + B1 and L2(z) = a21z1 + a22z2 + B2, with
aj1, aj2, Bj ∈ C( j =1, 2), satisfy

g(z1, z2) = L1(z) + L2(z) +H1(s1) +H2(s1) +B1 +B2,

L1(z) +H1(s1) 6= L2(z) +H2(s1), eL1(c) = 1− i, eL2(c) = 1 + i.

Example 2.11. Let L(z) = z1 + 2z2, H(s) = −π2(z1 − 2z2)2, B = 1, c1 = 2πi
and c2 = πi. Then in view of Theorem 2.10(i), it can be shown that f(z1, z2) =

e
1
2 [z1+2z2−π2(z1−2z2)2+1] is a solution of (2.3), where g(z) = z1+2z2−π2(z1−2z2)2+

1.

Example 2.12. Let L(z) = z1 − z2 and c = (c1, c2) ∈ C such that c1 − c2 =
(5− 3i)/5. Let H(s) = (c2z1 − c1z2)n, n ∈ N. Then in view of Theorem 2.10(ii), it

can be shown that f(z1, z2) = e
5
4 [z1−z2+(c2z1−c1z2)n+2] is a solution of (2.3), where

g(z) = z1 − z2 + (c2z1 − c1z2)n + 2.



EJDE-2023/43 COMPLEX NONLINEAR FUNCTIONAL EQUATIONS 9

3. Proofs of main results

Before we starting, we present some necessary lemmas which will play key role
to prove the main results.

Lemma 3.1 ([13]). Let fj 6≡ 0 (j = 1, 2, 3) be meromorphic functions on Cn such
that f1 are not constant, f1 + f2 + f3 = 1, and such that

3∑
j=1

{
N2

(
r,

1

fj

)
+ 2N(r, fj)

}
< λT (r, fj) +O(log+ T (r, fj))

holds 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 3.2 ([17, 29, 33]). For an entire function F on Cn, F (0) 6≡ 0 and put
ρ(nF ) = ρ <∞. Then there exist a canonical function fF and a function gF ∈ Cn
such that F (z) = fF (z)egF (z). For the special case n = 1, fF is the canonical
product of Weierstrass.

Lemma 3.3 ([28]). 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

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

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

Lemma 3.4 ([13]). Let a0(z), a1(z), . . . , an(z) (n ≥ 1) be meromorphic functions
on Cm and g0(z), g1(z), . . . , gn(z) are entire functions on Cm such that gj(z)−gk(z)

are not constants for 0 ≤ j < k ≤ n. If
∑n
j=0 aj(z)e

gj(z) ≡ 0, and ||T (r, aj) =

o(T (r)), where T (r) = min0≤j<k≤n T (r, egj−gk) for j = 0, 1, . . . , n, then aj(z) ≡ 0
for each j = 0, 1, . . . , n.

Proof of Theorem 2.1. Let f(z) be a transcendental entire solution of (2.1). First
rewrite (2.1) as( P (f)

eg(z)/2
+ i

f(z1 + c1, z2 + c2)

eg(z)/2

)( P (f)

eg(z)/2
− if(z1 + c1, z2 + c2)

eg(z)/2

)
= 1, (3.1)

where P (f) = ∂2f
∂z21

+ δ ∂
2f
∂z22

+ η ∂2f
∂z1∂z2

.

Since f is a transcendental entire function of finite order, in view of (3.1), we
conclude that (P (f) + if(z1 + c1, z2 + c2))/eg(z)/2 and (P (f) − if(z1 + c1, z2 +
c2))/eg(z)/2 have no zeros and poles. Thus, by Lemmas 3.2 and 3.3, there exists a
non-constant polynomial h(z) in C2 such that

∂2f
∂z21

+ δ ∂
2f
∂z22

+ η ∂2f
∂z1∂z2

eg(z)/2
+ i

f(z1 + c1, z2 + c2)

eg(z)/2
= eh(z),

∂2f
∂z21

+ δ ∂
2f
∂z22

+ η ∂2f
∂z1∂z2

eg(z)/2
− if(z1 + c1, z2 + c2)

eg(z)/2
= e−h(z).

(3.2)

We set

γ1(z) =
g(z)

2
+ h(z), γ2(z) =

g(z)

2
− h(z). (3.3)



10 H. Y. XU, G. HALDAR EJDE-2023/43

Therefore, in view of (3.2) and (3.3), we obtain that

∂2f

∂z21
+ δ

∂2f

∂z22
+ η

∂2f

∂z1∂z2
=
eγ1(z) + eγ2(z)

2
,

f(z1 + c1, z2 + c2) =
eγ1(z) − eγ2(z)

2i
.

(3.4)

After simple computations, it follows from the two equations of (3.4) that

− iQ1(z)eγ1(z)−γ1(z+c)) + iQ2(z)eγ2(z)−γ1(z+c) − eγ2(z+c)−γ1(z+c) = 1, (3.5)

where

Qj(z) =
(∂γj
∂z1

)2
+
∂2γj
∂z21

+ δ
((∂γj

∂z2

)2
+
∂2γj
∂z22

)
+ η
(∂γj
∂z1

∂γj
∂z2

+
∂2γj
∂z1∂z2

)
, j = 1, 2.

(3.6)

Now, we discuss two possible cases.

Case 1. Let γ2(z + c) − γ1(z + c) be a constant, say k ∈ C. In view of (3.3), we
conclude that h(z) is constant. Let ξ = eh(z) ∈ C. Then, (3.4) yields that

∂2f

∂z21
+ δ

∂2f

∂z22
+ η

∂2f

∂z1∂z2
= D1e

g(z)/2, f(z + c) = D2e
g(z)/2, (3.7)

where D1 =
(
ξ + ξ−1

)
/2, D2 = −i

(
ξ − ξ−1

)
/2. Note that D2 6= 0 and D2

1 +D2
2 =

1.
If D1 = 0, in view of (3.7), it follows that

∂2f(z)

∂z21
+ δ

∂2f(z)

∂z22
+ η

∂2f(z)

∂z1∂z2
= 0,

f(z1 + c1, z2 + c2) = ±eg(z)/2.
(3.8)

From the first equation of (3.8), we obtain f(z1, z2) = φ1(z2−αz1) +φ2(z2− βz1),
where φ1, φ2 are finite order transcendental entire functions in C2, and α, β ∈ C
such that α + β = η, αβ = δ. Hence, from the second equation of (3.8), it follows
that

φ1(z2 − αz1 + c2 − αc1) + φ2(z2 − βz1 + c2 − βc1) = ±e 1
2 g(z1,z2).

This is the conclusion (i).
If D1 6= 0, then from (3.7), we obtain that

1

2

( ∂g
∂z1

)2
+
∂2g

∂z21
+ δ
(1

2

( ∂g
∂z2

)2
+
∂2g

∂z22

)
+ η
(1

2

∂g

∂z1

∂g

∂z2
+

∂2g

∂z1∂z2

)
=

2D1

D2
e

1
2 [g(z+c)−g(z)].

(3.9)

Since g(z) is a polynomial in C2, it follows from (3.9) that g(z + c) − g(z)
must be constant. Then, g(z) can be written as g(z) = L(z) + H(s1) + B, where
L(z) = a1z1+a2z2, H(s1) is a polynomial in s1 = c2z1−c1z2, a1, a2, B are constants
in C. Hence, it follows from (3.9) that

(A1c2 −A2c1)H ′ +A3

(1

2
H ′2 +H ′′

)
=

2(ξ + ξ−1)

ξ − ξ−1
e

1
2L(c) −A4, (3.10)

where Aj ’s are defined in (2.4).



EJDE-2023/43 COMPLEX NONLINEAR FUNCTIONAL EQUATIONS 11

If A1c2 −A2c1 = 0 = A3, then from (3.10), we obtain that

e
1
2L(c) =

ξ − ξ−1

2(ξ + ξ−1)
A4.

If A1c2 − A2c1 6= 0 or A3 6= 0, then it follows from (3.10) that H ′ must be
constant, say a0, which is the coefficient of s1 in the polynomial H(s1).

Therefore, from (3.10), we obtain that

e
1
2L(c) =

ξ2 − 1

2i(ξ2 + 1)
[A4 + (A1c2 −A2c1)a0 +

1

2
A3a

2
0]. (3.11)

Hence, in either case L(z) satisfies the relation (3.11).
Therefore, in view of the second equation of (3.7), we obtain the form of the

solution as

f(z) =
ξ2 − 1

2iξ
e

1
2 [L(z)+H(s1)−L(c)+B].

This is conclusion (ii).

Case 2 Let γ2(z+ c)− γ1(z+ c) be non-constant. Then in view of (3.5), it follows
that Q1(z) and Q2(z) both can not be zero at the same time.

If Q1(z) ≡ 0 and Q2(z) 6≡ 0, then (3.5) yields that

iQ2(z)eγ2(z)−γ1(z+c) − eγ2(z+c)−γ1(z+c) = 1.

In view of the above equation, it follows that

N
(
r,

1

eγ2(z+c)−γ1(z+c) + 1

)
= N

(
r,

1

Q2(z)eγ2(z)−γ1(z+c)

)
= S

(
r, eγ2(z)−γ1(z+c)

)
.

Also, notice that

N
(
r, eγ2(z+c)−γ1(z+c)

)
= S

(
r, eγ2(z+c)−γ1(z+c)

)
,

N
(
r,

1

eγ2(z+c)−γ1(z+c)

)
= S

(
r, eγ2(z+c)−γ1(z+c)

)
.

By the second main theorem of Nevanlinna for several complex variables, we obtain

T
(
r, eγ2(z+c)−γ1(z+c)

)
≤ N

(
r, eγ2(z+c)−γ1(z+c)

)
+N

(
r,

1

eγ2(z+c)−γ1(z+c)

)
+N

(
r,

1

eγ2(z+c)−γ1(z+c) + 1

)
+ S

(
r, eγ2(z+c)−γ1(z+c)

)
≤ S

(
r, eγ2(z+c)−γ1(z+c)

)
+ S

(
r, eγ2(z)−γ1(z+c)

)
.

This implies that γ2(z + c)− γ1(z + c) is constant, which is a contradiction.
Similarly, we can get a contradiction for the case Q1(z) 6≡ 0 and Q1(z) ≡ 0.

Hence, Q1(z) 6≡ 0 and Q2(z) 6≡ 0.
Since γ1(z) and γ2(z) are polynomials in C2 and γ2(z + c) − γ1(z + c) is non-

constant, applying Lemma 3.1 to the equation (3.5), we obtain that either−iQ1(z)eγ1(z)−γ1(z+c) =
1, or iQ2(z)eγ2(z)−γ1(z+c) = 1.

If

− iQ1(z)eγ1(z)−γ1(z+c) = 1, (3.12)

then from (3.5), it follows that

iQ2(z)eγ2(z)−γ2(z+c) = 1. (3.13)



12 H. Y. XU, G. HALDAR EJDE-2023/43

As γ1(z) and γ2(z) are polynomials, in view of (3.12) and (3.13), we conclude that
γ1(z) − γ1(z + c) and γ2(z) − γ2(z + c) both are constants in C, and hence we
obtain that γ1(z) = L1(z) +H1(s1) +B1 and γ2(z) = L2(z) +H2(s1) +B2, where
Lj(z) = aj1z1 + aj2z2, Hj(s1) is a polynomial in s1 = c2z1 − c1z2, aj1, aj2, B1, B2

are constants in C for j = 1, 2. Note that L1(z) + H1(s1) 6= L2(z) + H2(s1).
Otherwise, γ2(z + c) − γ1(z + c) would become constant, a contradiction to our
assumption. Hence, the form of the polynomial g(z) is g(z) = L(z) + H(s1) + B,
where L(z) = L1(z) + L2(z), H(s1) = H1(s1) +H2(s1) and B = B1 +B2.

Therefore, in view of (3.12) and (3.13), we obtain that

(A15c2 −A16c1)H ′1 +A3(H ′21 +H ′′1 ) = ieL1(c) −A17,

(A25c2 −A26c1)H ′2 +A3(H ′22 +H ′′2 ) = −ieL2(c) −A27,
(3.14)

where Aij ’s are defined in (2.4).
Then, by similar arguments as in Case 1, we obtain from (3.14) that

eL1(c) = −i
[
A17 + (A15c2 −A16c1)a0 +A3a

2
0

]
,

eL2(c) = i[A27 + (A25c2 −A26c1)a00 +A3a
2
00],

(3.15)

where a0 and a00, respectively the coefficients of the linear term of the polynomials
H1(s1) and H2(s1).

Therefore, in view of the second equation of (3.4), we obtain

f(z) =
1

2i

(
eL1(z)+H1(s1)−L1(c)+B1 − eL2(z)+H2(s1)−L2(c)+B2

)
,

where L1(c) and L2(c) can be found from (3.15). This is conclusion (iii).
If iQ2(z)eγ2(z)−γ1(z+c) = 1, then it follows from equation (3.5) that

−iQ1(z)eγ1(z)−γ2(z+c) = 1.

Since γ1(z) and γ2(z) are both polynomials in C2, it follows that γ2(z)−γ1(z+c) =
η1 and γ1(z)−γ2(z+c) = η2, where η1, η2 ∈ C. This implies that γ1(z)−γ1(z+c) =
γ2(z)− γ2(z+ c) = η1 + η2. Therefore, we can write γ1(z) = L(z) +H(s1) + ζ1 and
γ2(z) = L(z) +H(s1) + ζ2. But, then we obtain γ2(z + c)− γ1(z + c) = ζ2 − ζ1, a
constants, which is a contradiction. �

Proof of Theorem 2.6. Let f(z) be a transcendental entire solution of the equation
(2.2). First rewrite (2.2) as( P (f)

eg(z)/2
+ i

f(z + c)− f(z)

eg(z)/2

)( P (f)

eg(z)/2
− if(z + c)− f(z)

eg(z)/2

)
= 1, (3.16)

where

P (f) =
∂2f

∂z21
+ δ

∂2f

∂z22
+ η

∂2f

∂z1∂z2
.

Since f is a transcendental entire function of finite order, in view of (3.16), we
conclude that (P (f)+i(f(z+c)−f(z)))/eg(z)/2 and (P (f)−i(f(z+c)−f(z)))/eg(z)/2

have no zeros and poles. Thus, by Lemmas 3.2 and 3.3, there exists a non-constant



EJDE-2023/43 COMPLEX NONLINEAR FUNCTIONAL EQUATIONS 13

polynomial h(z) in C2 such that

∂2f
∂z21

+ δ ∂
2f
∂z22

+ η ∂2f
∂z1∂z2

eg(z)/2
+ i

f(z + c)− f(z)

eg(z)/2
= eh(z),

∂2f
∂z21

+ δ ∂
2f
∂z22

+ η ∂2f
∂z1∂z2

eg(z)/2
− if(z + c)− f(z)

eg(z)/2
= e−h(z).

(3.17)

We set

γ1(z) =
g(z)

2
+ h(z), γ2(z) =

g(z)

2
− h(z). (3.18)

Then, in view of (3.17) and (3.18), we obtain that

∂2f

∂z21
+ δ

∂2f

∂z22
+ η

∂2f

∂z1∂z2
=

1

2
[eγ1(z) + eγ2(z)]

f(z + c)− f(z) =
1

2i
[eγ1(z) − eγ2(z)]

. (3.19)

After simple computations, it follows from the two equations of (3.19) that

[1−iQ1(z)]eγ1(z)−γ1(z+c))+[1+iQ2(z)]eγ2(z)−γ1(z+c)−eγ2(z+c)−γ1(z+c) = 1, (3.20)

where Q1(z) and Q2(z) are defined in (3.6). Now we consider two possible cases.

Case 1. Let γ2(z + c) − γ1(z + c) = k ∈ C. In view of (3.18), we conclude that
h(z) is constant. Set eh = ξ ∈ C. Then (3.19) yields that

∂2f

∂z21
+ δ

∂2f

∂z22
+ η

∂2f

∂z1∂z2
= D1e

g(z)/2, f(z + c)− f(z) = D2e
g(z)/2, (3.21)

where D1 = 1
2 (ξ + ξ−1), D2 = 1

2i (ξ − ξ
−1t). Note that D2

1 +D2
2 = 1.

Subcase 1.1. Let D1 = 0. Therefore, it follows from (3.21) that

∂2f(z)

∂z21
+ δ

∂2f(z)

∂z22
+ η

∂2f(z)

∂z1∂z2
= 0,

f(z + c)− f(z) = ±eg(z)/2.
(3.22)

Now, in view of the first equation of (3.22), we obtain that

f(z1, z2) = φ1(z2 − αz1) + φ2(z2 − βz1),

where φ1, φ2 are finite order transcendental entire functions in C2, and α, β are
constants in C such that α+ β = η and αβ = δ.

In view of the second equation of (3.22), we obtain that

φ1(z2 − αz1 + c2 − αc1) + φ2(z2 − βz1 + c2 − βc1)− φ1(z2 − αz1)− φ2(z2 − βz1)

= ±e 1
2 g(z1,z2).

This is conclusion (i).

Subcase 1.2. Let D2 = 0. Therefore, it follows from (3.21) that

∂2f(z)

∂z21
+ δ

∂2f(z)

∂z22
+ η

∂2f(z)

∂z1∂z2
= ±e 1

2 g(z),

f(z + c)− f(z) = 0.

(3.23)

Clearly, the second equation of (3.23) shows that f is a periodic function of period

c. In view of the two equations in (3.23), it follows that e
1
2 (g(z+c)−g(z)) = 1. This



14 H. Y. XU, G. HALDAR EJDE-2023/43

implies that g(z1, z2) = a1z1 + a2z2 +H(c2z1− c1z2) +B, where H is a polynomial
in c2z1 − c1z2 and a1c1 + a2c2 = 4kπi, k ∈ Z.

Now, in view of the results in [41, page 2178, Line 21], the first equation of (3.23)
can be written as

(D + αD′)(D + βD′)f(z) = ±e 1
2 g(z), (3.24)

where D ≡ ∂
∂z1

, D′ ≡ ∂
∂z2

, α, β ∈ C such that α+ β = η and αβ = δ.

Let (D + βD′)f(z) = u(z). Then (3.24) yields that

∂u

∂z1
+ β

∂u

∂z2
= ±e 1

2 g(z1,z2). (3.25)

The characteristic equations of (3.25) are

dz1
dt

= 1,
dz2
dt

= β,
du

dt
= e

1
2 g(z1,z2).

Using the initial conditions: t = 0, z1 = 0, z2 = s, and u = u(0, s) := G0(s), with
a parameter s, we obtain the following parametric representation for the solutions
of the characteristic equations: z1 = t, z2 = βt+ s,

u(z1, z2) = ±
∫ z1

0

e
1
2 g(z)dz1 +G0(z2 − βz1),

where G0 is a finite order transcendental entire function in C2.
Since, we have assumed that (D+ βD′)f(z) = u(z), in view of (3.24), it follows

that
∂f(z)

∂z1
+ α

∂f(z)

∂z2
= ±

∫ z1

0

e
1
2 g(z)dz1 +G0(z2 − βz1). (3.26)

By similar arguments as above, we obtain from (3.26) that

f(z1, z2) = ±
∫ z1

0

∫ z1

0

e
1
2 [a1z1+a2z2+H(c2z1−c1z2)+B]dz1dz1

+

∫ z1

0

G0(z2 − βz1)dz1 +G1(z2 − αz1),

where G1 is a finite order transcendental entire function in C2.
In view of the fact that a1c1 + a2c2 = 4kπi, k ∈ Z, it follows from the second

equation of (3.23) that∫ z1

0

[G0(z2 − βz1 + c2 − βc1)−G0(z2 − βz1)]dz1

+G1(z2 − αz1 + c2 − αc1)−G1(z2 − αz1) = 0.

This is the conclusion (ii).

Subcase 1.3. Let D1 6= 0 and D2 6= 0. Then after simple calculations, (3.21)
yields that(∂2g

∂z21
+

1

2

( ∂g
∂z1

)2)
+ δ
(∂2g
∂z22

+
1

2

( ∂g
∂z2

)2)
+ η
( ∂2g

∂z1∂z2
+

1

2

∂g

∂z1

∂g

∂z2

)
=

2D1

D2
[e

1
2 [g(z+c)−g(z)] − 1.

(3.27)

Since g(z) is a polynomial in C2, in view of (3.27) we conclude that g(z+c)−g(z) =
ξ, ξ ∈ C. This implies that g(z) = L1(z)+H(s1)+B1, where L1(z) = a11z1+a12z2,



EJDE-2023/43 COMPLEX NONLINEAR FUNCTIONAL EQUATIONS 15

H(s1) is a polynomial in s1 := c2z1− c1z2, a11, a12, B1 ∈ C. Hence, we obtain from
(3.27) that [(

a11 +
1

2
ηa12

)
c2 −

(
δa12 +

1

2
ηa11

)
c1
]
H ′

+
(
c22 + δc21 − ηc1c2

)(1

2
H ′

2
+H ′′

)
=

2D1

D2
[e

1
2L1(c) − 1].

(3.28)

Since c22 + δc21 6= ηc1c2, in view of (3.28), we conclude that H ′ is constant. This
implies that H(s1) = a0s1 + b0. Hence, g(z) reduces to the form

g(z) = L(z) +B = a1z1 + a2z2 +B, (3.29)

where a1 = a11 + a0c2, a2 = a12 − a0c1 and B = B1 + b0.
Therefore, in view of (3.27) and (3.29) we obtain that

e
1
2 [a1c1+a2c2] =

D2

4D1

(
a21 + δa22 + ηa1a2

)
+ 1. (3.30)

Now, in view of the results in [41, page 2178, Line 21], the first equation of (3.21)
can be written as

(D2 + δD′2 + ηDD′)f(z) = D1e
1
2 [a1z1+a2z2+B], (3.31)

where D ≡ ∂
∂z1

and D′ ≡ ∂
∂z2

. Therefore, complementary function of (3.31) is

C.F. = φ1(z2 − αz1) + φ2(z1 − βz1), where φ1, φ2 are finite order transcendental
entire functions in C2, α, β ∈ C such that α+β = η and αβ = δ. Particular integral
of (3.31) is

P.I. =
4D1e

B/2

a21 + δa22 + ηa1a2

∫ ∫
evdvdv =

4D1

a21 + δa22 + ηa1a2
e

1
2 [a1z1+a2z2+B],

where v = a1z1 + a2z2. Hence, from (3.21), we obtain

f(z1, z2) = φ1(z2 − αz1) + φ2(z2 − βz1)

+
2(ξ + ξ−1)

a21 + δa22 + ηa1a2
e

1
2 [a1z1+a2z2+B].

Substituting f(z1, z2) into the second equation of (3.21) and combining with (3.30),
we obtain that

φ1(z2 − αz1 + c2 − αc1) + φ2(z2 − βz1 + c2 − βc1) = φ1(z2 − αz1) + φ2(z2 − βz1).

This is the conclusion (iii).

Case 2. Let γ2(z+ c)−γ1(z+ c) be non-constant. Then, obviously 1− iQ1(z) and
1+iQ2(z) can not be identically zero at the same time. Otherwise, in view of (3.20),
it follows that eγ2(z+c)−γ1(z+c) is a constant, which implies that γ2(z+c)−γ1(z+c)
is a constant. This is a contradiction to our assumption.

If 1− iQ1(z) ≡ 0 and 1 + iQ2(z) 6≡ 0, the (3.20) it yields that

(1 + iQ2(z))eγ2(z) − eγ2(z+c) − eγ1(z+c) ≡ 0. (3.32)

Note that γ2(z) − γ2(z + c) is non-constant. Otherwise, if γ2(z) − γ2(z + c) =
ζ ∈ C, then (3.32) yields that [(1 + iQ2(z))eζ − 1]eγ1(z+c)−γ2(+c) = 1. But, then
γ1(z + c) − γ2(+c) becomes a constant, which is a contradiction. Also, note that
γ2(z) − γ1(z + c) is non-constant. Otherwise, in view of (3.32), we obtain that
γ1(z + c) − γ2(+c) is constant, which is a contradiction. Hence, in view of (3.32)
and the Lemma 3.4, we can easily get a contradiction. Similarly, we can get a



16 H. Y. XU, G. HALDAR EJDE-2023/43

contradiction for the case 1− iQ1(z) 6≡ 0 and 1 + iQ2(z) ≡ 0. Therefore, we must
have 1− iQ1(z) 6≡ 0 and 1 + iQ2(z) 6≡ 0.

Now, in view of Lemma 3.1, we obtain from (3.20) that either

[1− iQ1(z)]eγ1(z)−γ1(z+c) ≡ 1, or [1 + iQ2(z)]eγ2(z)−γ1(z+c) ≡ 1.

If [1 + iQ2(z)]eγ2(z)−γ1(z+c) ≡ 1, then in view of (3.20), it follows that [1 −
iQ1(z)]eγ1(z)−γ2(z+c) ≡ 1. Therefore, we must obtain that γ2(z) − γ1(z + c) = ξ1
and γ1(z) − γ2(z + c) = ξ2, ξ1, ξ2 ∈ C. Thus, it follows that γ1(z) − γ1(z + 2c) =
γ2(z) − γ2(z + 2c) = ξ1 + ξ2. This implies that γ1(z) = L(z) + H(s1) + B1

and γ2(z) = L(z) + H(s1) + B2, where L(z) = a1z1 + a2z2 and H(s1) is a
polynomial in s1 := c2z1 − c1z2, a1, a2, B1, B2 ∈ C. Hence, we must have that
γ2(z + c) − γ1(z + c) = B2 − B1, a constant in C, which is a contradiction to the
assumption. Therefore, we must have

[1− iQ1(z)]eγ1(z)−γ1(z+c) ≡ 1. (3.33)

In view of (3.20) and (3.33), we obtain that

[1 + iQ2(z)]eγ2(z)−γ2(z+c) ≡ 1. (3.34)

Since γ1(z) and γ2(z) are polynomials in C2, from (3.33) and (3.34), we can conclude
that γ1(z)− γ1(z + c) = η1 and γ2(z)− γ2(z + c) = η2, η1, η2 ∈ C. Thus, we have
γ1(z) = L1(z) + H1(s1) + B1 and γ2(z) = L2(z) + H2(s1) + B2, where Lj(z) =
aj1z1 + aj2z2 and Hj(s1) is a polynomial in s1 := c2z1 − c1z2, aj1, aj2, Bj ∈ C for
j = 1, 2. Therefore, in view of (3.5), (3.33), we obtain that

[(2a11 + ηa12)c2 − (2δa12 + ηa11)c1]H ′1 + (c22 + δc21 − ηc1c2)(H ′21 +H ′′1 )

= i[eL1(c) − 1]− (a211 + δa212 + ηa11a12).

Since c22 + δc21 − ηc1c2 6= 0, in view of the above equation, we conclude that H1(s1)
is a linear polynomial in s1, and thus L1(z) +H1(s1) becomes linear in C. For the
sake of convenience, we still denote that γ1(z) = L1(z) +B1. In a similar manner,
from (3.5) and (3.34), we can conclude that γ2(z) = L2(z) +B2. Therefore, in view
of (3.5), it follows from (3.33) and (3.34) that

eL1(c) = −i(a211 + δa212 + ηa11a12) + 1,

eL2(c) = i(a221 + δa222 + ηa21a22) + 1.
(3.35)

Now, in view of the results in [41, page 2178, Line 21], and the form of γ1(z) and
γ2(z), the first equation of (3.19) can be written as

(D2 + δD′2 + ηDD′)f(z) =
1

2
[eL1(z)+B1 + eL2(z)+B2 ], (3.36)

where D ≡ ∂
∂z1

and D′ ≡ ∂
∂z2

.

The complementary function of (3.36) is φ1(z2 − αz1) + φ2(z1 − βz1), where
φ1, φ2 are finite order transcendental entire functions in C2, α, β ∈ C such that
α+ β = η and αβ = δ, and the particular integral is

P.I. =
eL1(z)+B1

2(a211 + δa212 + ηa11a12)
+

eL2(z)+B2

2(a221 + δa222 + ηa21a22)
.



EJDE-2023/43 COMPLEX NONLINEAR FUNCTIONAL EQUATIONS 17

Hence, the form of the solution of (3.36) is

f(z1, z2) = φ1(z2 − αz1) + φ2(z2 − βz1) +
eL1(z)+B1

2(a211 + δa212 + ηa11a12)

+
eL2(z)+B2

2(a221 + δa222 + ηa21a22)
.

(3.37)

Substituting (3.37) into the second equation of (3.19) and combining with (3.35),
we obtain that

φ1(z2 − αz1 + c2 − αc1) + φ2(z2 − βz1 + c2 − βc1) = φ1(z2 − αz1) + φ2(z2 − βz1).

This is conclusion (iv). �

Theorem 2.10 can be proved by similar arguments as in Theorem 2.1, Therefore,
we omit its proof.

References

[1] T. B. Cao; The growth, oscillation and fixed points of solutions of complex linear differential
equations in the unit disc, J. Math. Anal. Appl., 352(2) (2009), 739–748.

[2] T. B. Cao, R. J. Korhonen; A new version of the second main theorem for meromorphic

mappings intersecting hyperplanes in several complex variables, J. Math. Anal. Appl., 444(2)
(2016), 1114–1132.

[3] T. B. Cao, L. Xu; Logarithmic difference lemma in several complex variables and partial

difference equations, Ann. Math. Pure Appl., 199 (2020), 767–794.
[4] Y. M. Chiang, S. J. Feng; On the Nevanlinna characteristic of f(z + η) and difference

equations in the complex plane, Ramanujan J., 16 (2008), 105–129.
[5] F. Gross; On the equation fn(z) + gn(z) = 1, Bull. Amer. Math. Soc., 72 (1966), 86–88.

[6] R. G. Halburd, R. J. Korhonen; Difference analogue of the lemma on the logarithmic deriv-

ative with applications to difference equations, J. Math. Anal. Appl., 314 (2006), 477–487.
[7] R. G. Halburd, R. J. Korhonen; Finite-order meromorphic solutions and the discrete Painleve

equations, Proc. Lond. Math. Soc. 94(2) (2007), 443–474.

[8] G. Haldar; Solutions of Fermat-type partial differential difference equations in C2, Mediterr.
J. Math., 20 (2023), 50. https://doi.org/10.1007/s00009-022-02180-6

[9] G. Haldar, M. B. Ahamed; Entire solutions of several quadratic binomial and trinomial

partial differential-difference equations in C2, Anal. Math. Phys. 12 (2022), Article number:
113. https:10.1007/s13324-022-00722-5

[10] Q. Han, F. Lü; On the equation fn(z)+gn(z) = eαz+β , J. Contemp. Math. Anal., 54 (2019),

98–102.
[11] W. K. Hayman; Meromorphic Functions, The Clarendon Press, Oxford, 1964.

[12] P. C. Hu, B. Q. Li; On meromorphic solutions of nonlinear partial differential equations of

first order, J. Math. Anal. Appl., 377 (2011), 881–888.
[13] P. C. Hu, P. Li, C. C Yang; Unicity of Meromorphic Mappings, Advances in Complex Analysis

and its Applications, vol. 1. Kluwer Academic Publishers, Dordrecht, Boston, London, 2003.
[14] D. Khavinson; A note on entire solutions of the eiconal equation, Amer. Math. Mon., 102

(1995), 159–161.

[15] G. Iyer; On certain functional equations, J. Indian. Math. Soc., 3 (1939), 312–315.
[16] I. Laine; Nevanlinna Theory and Complex Differential Equations, Walter de Gruyter

Berlin/Newyork, 1993.
[17] P. Lelong; Fonctionnelles Analytiques et Fonctions Enti‘eres (n variables), Presses de

L’Université de Montreal, 1968.
[18] B. Q. Li; On entire solutions of Fermat type partial differential equations, Int. J. Math., 15

(2004), 473–485.
[19] B. Q. Li; Entire solutions of (uz1 )m + (uz2 )n = eg , Nagoya Math. J., 178 (2005), 151–162.
[20] B. Q. Li; Entire solutions of eiconal type equations, Arch. Math., 89 (2007), 350–357.

[21] P. Li, C. C. Yang; On the nonexistence of entire solutions of certain type of nonlinear
differential equations, J. Math. Anal. Appl., 320 (2006), 827–835.



18 H. Y. XU, G. HALDAR EJDE-2023/43

[22] L. W. Liao, C. C. Yang, J. J. Zhang; On meromorphic solutions of certain type of non-linear

differential equations, Ann. Acad. Sci. Fenn. Math., 38 (2013), 581–593.

[23] K. Liu, X. J. Dong; Fermat type differential and difference equations, Electron. J. Differential.
Equ., 2015 (159) (2015), 1–10.

[24] K. Liu, T. B. Cao, H. Z. Cao; Entire solutions of Fermat type differential-difference equations,

Arch. Math., 99 (2012), 147–155.
[25] K. Liu, L. Z. Yang; A note on meromorphic solutions of Fermat types equations, An. Stiint.

Univ. Al. I. Cuza Lasi Mat. (N. S.), 1 (2016), 317–325.

[26] F. Lü, Z. Li; Meromorphic solutions of Fermat type partial differential equations, J. Math.
Anal. Appl., 478 (2) (2019), 864–873.

[27] P. Montel; Lecons sur les familles de nomales fonctions analytiques et leurs applications,

Gauthier-Viuars Paris, (1927), 135–136.
[28] G. Pòlya; On an integral function of an integral function, J. Lond. Math. Soc., 1 (1926),

12–15.
[29] L. I. Ronkin; Introduction to the Theory of Entire Functions of Several Variables, Moscow:

Nauka 1971 (Russian), American Mathematical Society, Providence, 1974.

[30] E. G. Saleebly; Entire and meromorphic solutions of Fermat type partial differential equa-
tions, Analysis (Munich), 19 (1999), 369–376.

[31] E. G. Saleeby; On entire and meromorphic solutions of λuk +
∑n
i=1 u

m
zi

= 1, Complex Var.

Theory Appl., 49 (2004), 101–107.

[32] E. G. Saleeby; On complex analytic solutions of certain trinomial functional and partial

differential equations, Aequationes Math., 85 (2013), 553–562.
[33] W. Stoll; Holomorphic Functions of Finite Order in Several Complex Variables, American

Mathematical Society, Providence, 1974.

[34] H. Y. Xu, Z. Xuan; Some inequalities on the convergent abscissas of Laplace-Stieltjes trans-
forms, J. Math. Inequal. 17 (1) (2023), 163–183.

[35] H. Y. Xu, H. Li, Z. Xuan ; Some new inequalities on Laplace-Stieltjes transforms involving
logarithmic growth, Fractal Fract., (2022), 6, 233.

[36] H. Y. Xu, L. Xu; Transcendental entire solutions for several quadratic binomial and trinomial

PDEs with constant coefficients, Anal. Math. Phys., 12 (2022), 64.
[37] H. Y. Xu, X. L. Liu, Y. H. Xu; On solutions for several systems of complex nonlinear partial

differential equations with two variables, Anal. Math. Phys. 10.1007/s13324-023-00811-z.

[38] H. Y. Xu, Y. Y. Jiang; Results on entire and meromorphic solutions for several systems of
quadratic trinomial functional equations with two complex variables, RACSAM (2022) 116:8,

https://doi.org/10.1007/s13398-021-01154-9.

[39] H. Y. Xu, D. W. Meng, S. Y. Liu, H. Wang; Entire solutions for several second-order partial
differential-difference equations of Fermat type with two complex variables, Adv. Differ. Equ.,

2021, Article number 52. https://doi.org/10.1186/s13662-020-03201-y

[40] L. Xu and T. B. Cao; Solutions of complex Fermat-type partial difference and differential-
difference equations, Mediterr. J. Math. 15 (2018), 1–14.

[41] H. Y. Xu, K. Y. Zhang and X. M. Zheng; Entire and meromorphic solutions for several

Fermat type partial differential difference equations in C2, Rocky Mt. J. Math. 52 (6) (2022),
2169–2187.

[42] C. C. Yang, P. Li; On the transcendental solutions of a certain type of non-linear differential
equations, Arch. Math., 82 (2004), 442–448.

[43] X. M. Zheng, H. Y. Xu; Entire solutions of some Fermat type functional equations concerning
difference and partial differential in C2, Anal. Math., 48 (2022), 199–226.

Hong Yan Xu
School of Arts and Sciences, Suqian University, Suqian, Jiangsu 223800, China

Email address: xhyhhh@126.com

Goutam Haldar (corresponding author)

Department of Mathematics, Malda College - 732101, West Bengal, India

Email address: goutamiit1986@gmail.com, goutamiitm@gmail.com


	1. Introduction
	2. Results
	3. Proofs of main results
	References