# Solution of Several Functional Equations on Nonunital Semigroups Using Wilson’s Functional Equations with Involution

Authors: Jaeyoung Chung; Prasanna K. Sahoo
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

LetS be a nonunital commutative semigroup, σ
:
S
→
S an involution, and C the
set of complex numbers. In this paper, first we determine the general solutions f
,
g
:
S
→
C of
Wilson’s generalizations of  d’Alembert’s functional equations  f
x
+
y
+
f
x
+
σ
y
=
2
f
(
x
)
g
(
y
)
and f
x
+
y
+
f
x
+
σ
y
=
2
g
(
x
)
f
(
y
) on nonunital commutative semigroups, and then using
the solutions of these equations we solve a number of other functional equations on more general
domains.


---

## Body

## 1. Introduction

The functional equation(1)
f
(
x
+
y
)
+
f
(
x
-
y
)
=
2
f
(
x
)
f
(
y
)
is known as the d’Alembert’s functional equation. It has a long history going back to d’Alembert [1]. As the name suggests this functional equation was introduced by d’Alembert in connection with the composition of forces and plays a central role in determining the sum of two vectors in Euclidean and non-Euclidean geometries [2].Several authors have determined the general solutionf
:
S
→
F of the functional equation
(2)
f
(
x
+
y
)
+
f
(
x
+
σ
y
)
=
2
f
(
x
)
f
(
y
)
,
for all x
,
y
∈
S, where σ
:
S
→
S is an automorphism of order 2. Stetkær [3] studied (2) when F
=
C, S is an abelian topological group and f
:
S
→
C and σ
:
S
→
S are continuous. Sinopoulos [4] determined the general solution f
:
S
→
F of (2) when S is a commutative semigroup, F is a quadratically closed commutative field of characteristic different from 2, and σ is an endomorphism of S of order 2.Wilson’s functional equationf
(
x
+
y
)
+
f
(
x
-
y
)
=
2
f
(
x
)
g
(
y
) is a generalization of d’Alembert’s functional equation. Among others Wilson’s functional equation was studied by Wilson [5] and [6], Kaczmarz [7], van der Lyn [8], Fenyö [9], Chung et al. [10], and Aczél [11]. Stetkær (see [12] and [3]) studied Wilson’s equation with an involution; namely,
(3)
f
(
x
+
y
)
+
f
(
x
+
σ
y
)
=
2
f
(
x
)
g
(
y
)
,
∀
x
,
y
∈
S
,
where S is a topological monoid (i.e., S is a semigroup with an identity element), f, g are continuous functions defined on S and taking values on C, and σ
:
S
→
S is a continuous automorphism of order 2. He gave several properties of the solution (
f
,
g
) of Wilson’s equation (see [3, Chapter 11, Lemmas 11.3 and 11.4]). In Chapter 11 of [3] (see Corollary 11.7) a complete and satisfactory description of the solutions of Wilson’s equation (3) was given on abelian groups.The functional equationf
(
x
+
y
)
+
f
(
x
-
y
)
=
2
g
(
x
)
f
(
y
) was introduced in [6] (see also [3] Chapter 11). For more on known results related to Wilson’s functional equations we refer the reader to [13–20]. The above functional equation with an involution, namely,
(4)
f
(
x
+
y
)
+
f
(
x
+
σ
y
)
=
2
g
(
x
)
f
(
y
)
,
∀
x
,
y
∈
S
,
was studied by Stetkær (see [21]) on abelian groups.In this paper, first we present the solutions of Wilson’s functional equations (3) and (4) on nonunital commutative semigroups and then using the general solutions of these functional equations we determine the general solutions of several functional equations in two variables in more general domains. In particular, we determine the general solutions (
f
,
g
) of functional equations; namely,
(5)
f
(
p
r
,
q
s
)
+
f
(
p
s
,
q
r
)
=
2
f
(
p
,
q
)
g
(
r
,
s
)
kkkkkkkkkkkkkkkkkk
∀
p
,
q
,
r
,
s
∈
(
0,1
]
,
f
(
p
r
,
q
s
)
+
f
(
p
s
,
q
r
)
=
2
g
(
p
,
q
)
f
(
r
,
s
)
kkkkkkkkkkkkkkkkkkk
∀
p
,
q
,
r
,
s
∈
(
0,1
]
treated by Riedel and Sahoo in [22], on more general domains. We also determine the general solutions of functional equations:
(6)
f
(
z
w
)
+
f
(
z
w
-
)
=
2
f
(
z
)
g
(
w
)
∀
z
,
w
∈
C
,
f
(
z
w
)
+
f
(
z
w
-
)
=
2
g
(
z
)
f
(
w
)
∀
z
,
w
∈
C
,
f
(
u
x
-
v
y
,
u
y
+
v
x
)
+
f
(
u
x
+
v
y
,
u
y
-
v
x
)
=
2
f
(
x
,
y
)
g
(
u
,
v
)
∀
x
,
y
,
u
,
v
∈
R
,
f
(
u
x
+
v
y
,
u
y
+
v
x
)
+
f
(
u
x
-
v
y
,
u
y
-
v
x
)
=
2
f
(
x
,
y
)
g
(
u
,
v
)
∀
x
,
y
,
u
,
v
∈
R
.


## 2. Solution of Wilson’s Equations on Commutative Nonunital Semigroups

In this section, we present the solutions(
f
,
g
) of Wilson’s functional equations (3) and (4) on nonunital semigroups. We solve these equations on more general domains than the domains used in [3]. In particular we give the description of the solutions of Wilson’s equations (3) and (4) on nonunital semigroups whereas in [3] the description of the solutions of (3) is given on commutative topological monoids (i.e., semigroups with an identity element). Note that Lemmas 11.3 and 11.4 of Chapter  11 in [3] are not very useful for solving (3) and (4) on nonunital commutative semigroups due to the lack of identity element. Our method of proof is quite elementary and transparent.Throughout this paper we denote byS a 2-divisible commutative semigroup with no identity element and denote by C the set of complex numbers. A function σ
:
S
→
S is said to bean involution if σ
(
x
+
y
)
=
σ
(
y
)
+
σ
(
x
), for all x
,
y
∈
S, and σ
(
σ
(
x
)
)
=
x, for all x
∈
S. For simplicity we write σ
x instead of σ
(
x
). A function m
:
S
→
C is calledan exponential function provided that m
(
x
+
y
)
=
m
(
x
)
m
(
y
), for all x
,
y
∈
S, and a
:
S
→
C is calledan additive function provided that a
(
x
+
y
)
=
a
(
x
)
+
a
(
y
), for all x
,
y
∈
S.As a direct consequence of a theorem of Sinopoulos [4] we have the following lemma.Lemma 1.
Letg
:
S
→
C satisfy
(7)
g
(
x
+
y
)
+
g
(
x
+
σ
y
)
=
2
g
(
x
)
g
(
y
)
for all x
,
y
∈
S. Then there exists an exponential function m
:
S
→
C such that
(8)
g
(
x
)
=
m
(
x
)
+
m
(
σ
x
)
2
for all x
∈
S.In the following, we exclude the trivial cases whenf
(
x
)
≡
0 or g
(
x
)
≡
0.Theorem 2.
Letf
,
g
:
S
→
C satisfy the functional equation
(9)
f
(
x
+
y
)
+
f
(
x
+
σ
y
)
=
2
f
(
x
)
g
(
y
)
for all x
,
y
∈
S. Then either g, f are of the form
(10)
g
(
x
)
=
m
(
x
)
+
m
(
σ
x
)
2
,
f
(
x
)
=
α
1
m
(
x
)
+
α
2
m
(
σ
x
)
,
for all x
∈
S, where m
:
S
→
C is an exponential function satisfying m
≠
m
∘
σ and α
1
,
α
2
∈
C, or else
(11)
g
(
x
)
=
m
(
x
)
,
f
(
x
)
=
{
m
(
x
)
(
β
+
a
(
x
)
)
,
i
f
x
∈
S
∖
K
,
0
i
f
x
∈
K
,
where m
:
S
→
C is an exponential function satisfying m
=
m
∘
σ, K
=
{
x
∈
S
∣
m
(
x
)
=
0
} and a
:
S
∖
K
→
C is an additive function satisfying a
=
-
a
∘
σ and β
∈
C.Proof.
Choose anx
0
∈
S such that f
(
x
0
)
≠
0. Replacing x by x
0 in (9) and dividing the result by 2
f
(
x
0
) we have
(12)
g
(
y
)
=
f
(
x
0
+
y
)
+
f
(
x
0
+
σ
y
)
2
f
(
x
0
)
for all y
∈
S. Multiplying both sides of (12) by 2
g
(
x
) and using (9) we have
(13)
2
g
(
y
)
g
(
x
)
=
2
f
(
x
0
+
y
)
g
(
x
)
+
2
f
(
x
0
+
σ
y
)
g
(
x
)
2
f
(
x
0
)
=
(
f
(
x
0
+
y
+
x
)
+
f
(
x
0
+
y
+
σ
x
)
k
k
k
+
f
(
x
0
+
σ
y
+
x
)
+
f
(
x
0
+
σ
y
+
σ
x
)
)
×
(
2
f
(
x
0
)
)
-
1
=
f
(
x
0
+
y
+
x
)
+
f
(
x
0
+
σ
y
+
σ
x
)
2
f
(
x
0
)
+
f
(
x
0
+
y
+
σ
x
)
+
f
(
x
0
+
σ
y
+
x
)
2
f
(
x
0
)
=
f
(
x
0
+
y
+
x
)
+
f
(
x
0
+
σ
(
y
+
x
)
)
2
f
(
x
0
)
+
f
(
x
0
+
y
+
σ
x
)
+
f
(
x
0
+
σ
(
y
+
σ
x
)
)
2
f
(
x
0
)
=
g
(
y
+
x
)
+
g
(
y
+
σ
x
)
for all x
,
y
∈
S. By Lemma 1, g has the form
(14)
g
(
x
)
=
m
(
x
)
+
m
(
σ
x
)
2
for all x
∈
S. Now, we find f. Let f
e and f
o be the even part and the odd part of f with respect to σ; that is,
(15)
f
e
(
x
)
=
1
2
(
f
(
x
)
+
f
(
σ
x
)
)
,
f
o
(
x
)
=
1
2
(
f
(
x
)
-
f
(
σ
x
)
)
for all x
,
y
∈
S. Then we have
(16)
f
(
x
)
=
f
e
(
x
)
+
f
o
(
x
)
,
f
e
(
σ
x
)
=
f
e
(
x
)
,
f
o
(
σ
x
)
=
-
f
o
(
x
)
for all x
∈
S. Replacing x by σ
x in (14) we obtain
(17)
g
(
σ
x
)
=
g
(
x
)
for all x
∈
S. Replacing (
x
,
y
) by (
σ
x
,
σ
y
) in (9) and using (17) we have
(18)
f
(
σ
x
+
σ
y
)
+
f
(
σ
x
+
y
)
=
2
f
(
σ
x
)
g
(
y
)
for all x
,
y
∈
S. Adding (18) to (9) and dividing the result by 2 we have
(19)
f
e
(
x
+
y
)
+
f
e
(
x
+
σ
y
)
=
2
f
e
(
x
)
g
(
y
)
for all x
,
y
∈
S. Replacing (
x
,
y
) by (
y
,
x
) in (19) and using (16) we have
(20)
f
e
(
x
+
y
)
+
f
e
(
x
+
σ
y
)
=
2
f
e
(
y
)
g
(
x
)
.
Putting y
=
y
0 in (19) and (20) such that g
(
y
0
)
≠
0 and equating the right hand sides of the results we have
(21)
f
e
(
x
)
=
f
e
(
y
0
)
g
(
y
0
)
g
(
x
)
:
=
β
g
(
x
)
for all x
∈
S. From now on, we determine f
o. Subtracting (18) from (9) and dividing the result by 2 we have
(22)
f
o
(
x
+
y
)
+
f
o
(
x
+
σ
y
)
=
2
f
o
(
x
)
g
(
y
)
for all x
,
y
∈
S. Replacing (
x
,
y
) by (
y
,
x
) and using (16) we have
(23)
f
o
(
x
+
y
)
-
f
o
(
x
+
σ
y
)
=
2
f
o
(
y
)
g
(
x
)
for all x
,
y
∈
S. Subtracting (23) from (22) and dividing the result by 2 we have
(24)
f
o
(
x
+
σ
y
)
=
f
o
(
x
)
g
(
y
)
-
f
o
(
y
)
g
(
x
)
for all x
,
y
∈
S. We divide into two cases when (i) m
(
σ
x
)
≠
m
(
x
), for some x
∈
S, and (ii) m
(
σ
x
)
=
m
(
x
), for all x
∈
S.
Case1. First, we consider the case when m
(
σ
x
)
≠
m
(
x
), for some x
∈
S. We show that if m
≠
m
∘
σ, then g is not an exponential function. Indeed, from (10) we have
(25)
g
(
x
+
y
)
=
1
2
(
m
(
x
+
y
)
+
m
(
σ
x
+
σ
y
)
)
=
1
2
(
m
(
x
)
m
(
y
)
+
m
(
σ
x
)
m
(
σ
y
)
)
for all x
,
y
∈
S, and
(26)
g
(
x
)
g
(
y
)
=
1
4
(
m
(
x
)
+
m
(
σ
x
)
)
(
m
(
y
)
+
m
(
σ
y
)
)
for all x
,
y
∈
S. If g is exponential, equating (25) and (26) and multiplying the result by 4 we have
(27)
(
m
(
x
)
-
m
(
σ
x
)
)
(
m
(
y
)
-
m
(
σ
y
)
)
=
0
for all x
,
y
∈
S, which implies m
(
x
)
=
m
(
σ
x
), for all x
∈
S, contradicting our assumption that m
(
x
)
≠
m
(
σ
x
), for some x
∈
S. Thus, g is not exponential. Let
(28)
f
o
p
(
x
)
=
1
2
(
m
(
x
)
-
m
(
σ
x
)
)
.
Then it is easy to see that (
f
p
,
g
) satisfies the functional equation (22). Since g is not exponential, we can choose x
0
,
y
0
∈
S such that g
(
x
0
+
y
0
)
≠
g
(
x
0
)
g
(
y
0
). Let f
o be an arbitrary solution of the functional equation (22). We can choose c
1
,
c
2
∈
C (not both zero) such that
(29)
c
1
f
o
(
x
0
)
+
c
2
f
o
p
(
x
0
)
=
0
.
Let
(30)
h
o
(
x
)
=
c
1
f
o
(
x
)
+
c
2
f
o
p
(
x
)
for all x
∈
S. Then h
0
(
x
0
)
=
0 and h
o satisfies the functional equation (22) and satisfies
(31)
h
o
(
x
+
σ
y
)
=
h
o
(
x
)
g
(
y
)
-
h
o
(
y
)
g
(
x
)
for all x
,
y
∈
S. We claim that h
0
(
x
)
≡
0. If not, that is, there exists α
∈
S such that h
0
(
α
)
≠
0; then putting y
=
α in (31) and dividing the result by h
0
(
α
) we have
(32)
g
(
x
)
=
h
o
(
x
)
g
(
α
)
-
h
o
(
x
+
σ
α
)
h
o
(
α
)
for all x
∈
S. Putting y
=
σ
x
0 in (31) and using (17) and (30) we have
(33)
h
o
(
x
+
x
0
)
=
h
o
(
x
)
g
(
σ
x
0
)
-
h
0
(
σ
x
0
)
g
(
x
)
=
h
o
(
x
)
g
(
x
0
)
+
h
0
(
x
0
)
g
(
x
)
=
h
o
(
x
)
g
(
x
0
)
for all x
∈
S. Now, from (32) and (33) we have
(34)
g
(
x
0
)
g
(
x
)
=
-
h
o
(
x
0
+
σ
α
)
(
h
o
(
x
)
g
(
α
)
-
h
o
(
x
+
σ
α
)
)
h
o
(
α
)
2
=
(
h
o
(
α
)
g
(
x
0
)
h
0
(
x
)
g
(
α
)
kk
-
h
o
(
α
)
g
(
x
0
)
h
o
(
x
+
σ
α
)
)
×
(
h
o
(
α
)
2
)
-
1
=
g
(
x
0
)
h
0
(
x
)
g
(
α
)
-
g
(
x
0
)
h
o
(
x
+
σ
α
)
h
o
(
α
)
=
h
0
(
x
+
x
0
)
g
(
α
)
-
h
o
(
x
+
x
0
+
σ
α
)
h
o
(
α
)
=
g
(
x
+
x
0
)
for all x
∈
S, which contradicts the fact that g
(
x
0
+
y
0
)
≠
g
(
x
0
)
g
(
y
0
). Thus, we conclude that h
0
(
x
)
≡
0 and it follows from (30) that
(35)
c
1
f
o
(
x
)
+
c
2
f
o
p
(
x
)
=
0
for all x
∈
S. If c
1
=
0, then c
2
≠
0 and hence
(36)
f
o
p
(
x
)
=
1
2
(
m
(
x
)
-
m
(
σ
x
)
)
=
0
for all x
∈
S, which contracts the assumption m
(
x
)
≠
m
(
σ
x
) for some x
∈
S. Thus, c
1
≠
0 and from (35) we have
(37)
f
o
(
x
)
=
-
c
2
c
1
f
o
p
(
x
)
for all x
∈
S. From (21), (28), and (37) we have
(38)
f
(
x
)
=
f
e
(
x
)
+
f
o
(
x
)
=
β
g
(
x
)
-
c
2
c
1
f
o
p
(
x
)
=
α
1
m
(
x
)
+
α
2
m
(
σ
x
)
for all x
∈
S, where α
1
=
(
1
/
2
)
(
β
-
(
c
2
/
c
1
)
)
,
α
2
=
(
1
/
2
)
(
β
+
(
c
2
/
c
1
)
). Thus, we get (10).
Case2. Next, we assume that m
(
σ
x
)
=
m
(
x
), for all x
∈
S. Then from (14) we have
(39)
g
(
x
)
=
m
(
x
)
,
m
(
σ
x
)
=
m
(
x
)
for all x
∈
S. Thus, from (24) and (39) we have
(40)
f
o
(
x
+
σ
y
)
=
f
o
(
x
)
m
(
y
)
-
f
o
(
y
)
m
(
x
)
for all x
,
y
∈
S.
LetK
=
{
x
∈
S
∣
m
(
x
)
=
0
}. Then K is an ideal in S and S
∖
K is a subsemigroup of S. Dividing (40) by m
(
x
+
σ
y
) and using (39) we have
(41)
f
o
(
x
+
σ
y
)
m
(
x
+
σ
y
)
=
f
o
(
x
)
m
(
x
)
-
f
o
(
y
)
m
(
y
)
for all x
,
y
∈
S
∖
K. Let
(42)
a
(
x
)
=
f
o
(
x
)
m
(
x
)
for all x
∈
S
∖
K. Then we have
(43)
a
(
x
+
σ
y
)
=
a
(
x
)
-
a
(
y
)
for all x
,
y
∈
S
∖
K. Replacing x by y in (43) we have
(44)
a
(
y
+
σ
y
)
=
a
(
y
)
-
a
(
y
)
=
0
.
Replacing x by x
+
y in (43) we have
(45)
a
(
x
+
y
+
σ
y
)
=
a
(
x
+
y
)
-
a
(
y
)
,
and replacing y by y
+
σ
y in (43) and using (44) we have
(46)
a
(
x
+
y
+
σ
y
)
=
a
(
x
)
-
a
(
y
+
σ
y
)
=
a
(
x
)
.
From (45) and (46) we have
(47)
a
(
x
+
y
)
=
a
(
x
)
+
a
(
y
)
for all x
,
y
∈
S
∖
K. Replacing y by σ
y in (47) and using (43) we have
(48)
a
(
σ
y
)
=
-
a
(
y
)
for all y
∈
S
∖
K. From (21), (39), (42), and (48) we have
(49)
f
(
x
)
=
f
e
(
x
)
+
f
o
(
x
)
=
β
g
(
x
)
+
a
(
x
)
m
(
x
)
=
β
m
(
x
)
+
a
(
x
)
m
(
x
)
=
m
(
x
)
(
β
+
a
(
x
)
)
for all x
∈
S
∖
K. Let x
∈
K. Then we have m
(
x
/
2
)
=
m
(
σ
(
x
/
2
)
)
=
0. Thus, replacing both x and σ
(
y
) by (
x
/
2
) in (40) we have f
o
(
x
)
=
0 and also by (21) f
e
(
x
)
=
0. Hence f
(
x
)
=
0 for any x
∈
K. Thus, from (39), (48), and (49) we get (11). This completes the proof of the theorem.Remark 3.
The solution(
g
,
f
) in (11) can be written in the form
(50)
g
(
x
)
=
m
(
x
+
σ
x
)
,
f
(
x
)
=
{
m
(
x
+
σ
x
)
(
β
+
a
(
x
)
-
a
(
σ
x
)
)
,
if
x
∈
S
∖
K
,
0
,
if
x
∈
K
,
where m is an exponential function and a is an additive function. Also, let us assume that S has the property:for any
x
,
y
∈
S,there exists a positive integer
k
and
z
∈
S
such that
(51)
x
+
z
=
k
y
,
which is satisfied by every group and also most of well known semigroups such as S
=
(
0,1
)
,
(
0
,
∞
). Then, if m
(
x
0
)
=
0 for some x
0
∈
S, then for any y
∈
S we can choose a positive integer k and z
∈
S such that x
0
+
z
=
k
y. Thus, we have
(52)
m
(
y
)
k
=
m
(
k
y
)
=
m
(
x
0
+
z
)
=
m
(
x
0
)
m
(
z
)
=
0
,
which implies m
(
y
)
=
0 for all y
∈
S. Thus, K
=
{
x
∈
S
∣
m
(
x
)
=
0
}
=
∅ and g
,
f in (11) are given by
(53)
g
(
x
)
=
m
(
x
+
σ
x
)
,
f
(
x
)
=
m
(
x
+
σ
x
)
(
β
+
a
(
x
)
-
a
(
σ
x
)
)
for all x
∈
S.In view of the proof of Theorem2, we can see that 2-divisibility of S is necessary only in showing that f
(
x
)
=
0, for all x
∈
K. Thus, if S is a commutative group possibly without 2-divisibility, we have the following.Corollary 4.
LetG be a commutative group and f
,
g
:
G
→
C satisfy the functional equation
(54)
f
(
x
+
y
)
+
f
(
x
+
σ
y
)
=
2
f
(
x
)
g
(
y
)
for all x
,
y
∈
G. Then either g, f are of the form
(55)
g
(
x
)
=
m
(
x
)
+
m
(
σ
x
)
2
,
f
(
x
)
=
α
1
m
(
x
)
+
α
2
m
(
σ
x
)
for all x
∈
G, where m is an exponential function satisfying m
≠
m
∘
σ and α
1
,
α
1
∈
C, or else
(56)
g
(
x
)
=
m
(
x
)
,
f
(
x
)
=
m
(
x
)
(
β
+
a
(
x
)
)
for all x
∈
G, where m
:
G
→
C is an exponential function satisfying m
=
m
∘
σ, a
:
G
→
C is an additive function satisfying a
=
-
a
∘
σ, and β
∈
C.Remark 5.
Note that in general2-divisible commutative semigroup, the solution (
g
,
f
) can not be written in the simple form as above (e.g., S
=
(
R
,
×
): see Theorem 18).Example 6.
LetG
=
〈
R
2
,
+
〉 in Corollary 4 and let a
,
b
∈
R with b
≠
0 and let
(57)
σ
=
(
a
b
1
-
a
2
b
-
a
)
.
Then σ is an involution on R
2 and we can consider the functional equation
(58)
f
(
t
+
u
,
s
+
v
)
+
f
(
t
+
a
u
+
b
v
,
s
+
1
b
(
1
-
a
2
)
u
-
a
v
)
=
2
f
(
t
,
s
)
g
(
u
,
v
)
for all t
,
s
,
u
,
v
∈
R. As a direct consequence of Corollary 4 we can exhibit all regular solutions of (58) in a transparent form. By Corollary 4, g has the form
(59)
g
(
t
,
s
)
=
m
(
t
,
s
)
+
(
m
∘
σ
)
(
t
,
s
)
2
=
m
(
t
,
s
)
+
m
(
a
t
+
b
s
,
(
1
/
b
)
(
1
-
a
2
)
t
-
a
s
)
2
for all t
,
s
∈
R, where m
:
R
2
→
C is an exponential function. In view of the proof in [4], m is given by m
(
x
)
=
g
(
x
)
-
α
(
g
(
x
+
x
0
)
-
g
(
x
+
σ
x
0
)
), for all x
∈
R
2 and for some α
∈
C
,
x
0
∈
R
2. Thus, if g is Lebesgue measurable, so is m and has the form m
(
t
,
s
)
=
e
c
1
t
+
c
2
s for some c
1
,
c
2
∈
C. Thus, g has the form
(60)
g
(
t
,
s
)
=
1
2
(
e
c
1
t
+
c
2
s
+
e
(
1
/
b
)
(
a
b
c
1
+
c
2
-
a
2
c
2
)
t
+
(
b
c
1
-
a
c
2
)
s
)
for all t
,
s
∈
R. Also, by Corollary 4, if m
≠
m
∘
σ, that is, c
1
≠
(
1
/
b
)
(
1
+
a
)
c
2, then f has the form
(61)
f
(
t
,
s
)
=
α
1
e
c
1
t
+
c
2
s
+
α
2
e
(
1
/
b
)
(
a
b
c
1
+
c
2
-
a
2
c
2
)
t
+
(
b
c
1
-
a
c
2
)
s
for all t
,
s
∈
R and for some α
1
,
α
2
∈
C, and if c
1
=
(
1
/
b
)
(
1
+
a
)
c
2, then f has the form
(62)
f
(
t
,
s
)
=
e
d
1
(
(
1
+
a
)
t
+
b
s
)
(
β
+
d
2
(
(
1
-
a
)
t
+
b
s
)
)
for all t
,
s
∈
R and for some d
1
,
d
2
,
β
∈
C.In the following theorem, we present the general solution of the second Wilson’s functional equation.Theorem 7.
Letf
,
g
:
S
→
C satisfy
(63)
f
(
x
+
y
)
+
f
(
x
+
σ
(
y
)
)
=
2
g
(
x
)
f
(
y
)
for all x
,
y
∈
S. Then there exists an exponential function m and β
∈
C with β
≠
0 such that
(64)
f
(
x
)
=
m
(
x
)
+
m
(
σ
x
)
2
β
,
g
(
x
)
=
m
(
x
)
+
m
(
σ
x
)
2
for all x
∈
S.Proof.
Replacingy by σ
y in (63) we see that
(65)
f
(
y
)
=
f
(
σ
y
)
for all y
∈
S. Replacing (
x
,
y
) by (
y
,
x
) in (63) and using (65) we have
(66)
f
(
x
+
y
)
+
f
(
x
+
σ
y
)
=
2
g
(
y
)
f
(
x
)
for all x
,
y
∈
S. Choosing y
0
∈
S such that g
(
y
0
)
≠
0, putting y
=
y
0 in (63) and (66), and equating the right hand sides of the results we have
(67)
f
(
x
)
=
f
(
y
0
)
g
(
y
0
)
g
(
x
)
:
=
1
β
g
(
x
)
for all x
∈
S. Replacing f
(
x
) by (
1
/
β
)
g
(
x
) in (63) and multiplying the result by β we have
(68)
g
(
x
+
y
)
+
g
(
x
+
σ
y
)
=
2
g
(
x
)
g
(
y
)
for all x
,
y
∈
S. By Lemma 1 and (67) we get (64). This completes the proof.


## 3. Applications of Wilson’s Equations in Solving Several Other Equations

Let(
H
,
∘
) be a commutative semigroup, and f
,
g
:
H
×
H
→
C. As an application of Theorems 2 and 7, we determine all general solutions of the functional equations
(69)
f
(
p
∘
r
,
q
∘
s
)
+
f
(
p
∘
s
,
q
∘
r
)
=
2
f
(
p
,
q
)
g
(
r
,
s
)
,
(70)
f
(
p
∘
r
,
q
∘
s
)
+
f
(
p
∘
s
,
q
∘
r
)
=
2
g
(
p
,
q
)
f
(
r
,
s
)
for all p
,
q
∈
H. For the case H
=
(
0,1
], (69) was treated by Riedel and Sahoo [22]. In the following we exclude the trivial cases when f
=
0 or g
=
0.Theorem 8 (c.f. [22]).
Letf
,
g
:
H
×
H
→
C satisfy the functional equation (69) for all p
,
q
,
r
,
s
∈
H. Then either g, f are of the form
(71)
g
(
p
,
q
)
=
1
2
(
m
1
(
p
)
m
2
(
q
)
+
m
2
(
p
)
m
1
(
q
)
)
,
f
(
p
,
q
)
=
α
1
m
1
(
p
)
m
2
(
q
)
+
α
2
m
2
(
p
)
m
1
(
q
)
for all p
,
q
∈
H, where m
1
,
m
2
:
H
→
C are exponential functions such that m
1
≠
m
2 and α
1
,
α
2
∈
C, or else
(72)
g
(
p
,
q
)
=
m
*
(
p
∘
q
)
,
p
,
q
∈
H
,
f
(
p
,
q
)
=
{
m
*
(
p
∘
q
)
(
β
+
a
*
(
p
)
-
a
*
(
q
)
)
,
i
f
p
,
q
∈
H
∖
I
,
0
,
i
f
p
∈
I
o
r
q
∈
I
,
where I
=
{
p
∈
H
∣
m
*
(
p
)
=
0
}, a
*
:
H
∖
I
→
C is an additive function, m
*
:
H
→
C is an exponential function, and β
∈
C.Proof.
Letσ
(
p
,
q
)
=
(
q
,
p
) for all (
p
,
q
)
∈
S. Then by Theorem 2 for the case when m
∘
σ
≠
m, all general solutions (
f
,
g
) of (69) are given by
(73)
g
(
p
,
q
)
=
m
(
p
,
q
)
+
m
(
q
,
p
)
2
,
f
(
p
,
q
)
=
α
1
m
(
p
,
q
)
+
α
2
m
(
q
,
p
)
for all p
,
q
∈
H, where m
:
S
→
C is an exponential function and α
1
,
α
2
∈
C. Since m can be written in the form m
(
p
,
q
)
=
m
1
(
p
)
m
2
(
q
), for some exponential functions m
1
,
m
2
:
H
→
C, from (73) we get (71). Assume that m
=
m
∘
σ; that is, m
(
p
,
q
)
=
m
(
q
,
p
), for all p
,
q
∈
H. Then we have
(74)
m
(
p
,
q
)
=
m
1
(
p
)
m
2
(
q
)
=
m
1
(
q
)
m
2
(
p
)
for all p
,
q
∈
H. Let I
1
=
{
p
∈
H
∣
m
1
(
p
)
=
0
}
,
I
2
=
{
p
∈
H
∣
m
2
(
p
)
=
0
}. If there exists q
0
∈
I
1
∖
I
2, putting q
=
q
0 in (74) we have m
1
(
p
)
=
0, for all ∈
H, and hence m
=
g
=
0. Since we exclude the case f
=
0 or g
=
0, we have I
1
⊂
I
2. Similarly, we have I
2
⊂
I
1 and I
1
=
I
2
:
=
I. From (74), we have
(75)
m
1
(
p
)
m
2
(
p
)
=
m
1
(
q
)
m
2
(
q
)
for all p
,
q
∉
I, which implies
(76)
m
1
(
p
)
=
α
m
2
(
p
)
for all p
∉
I and for some α
∈
C. Thus, we have
(77)
m
1
(
p
)
2
=
α
2
m
2
(
p
)
2
=
α
·
α
m
2
(
p
2
)
=
α
m
1
(
p
2
)
=
α
m
1
(
p
)
2
for all p
∉
I, which implies α
=
1 and hence m
1
=
m
2
:
=
m
*. Thus, from (73) and (74) we have
(78)
g
(
p
,
q
)
=
m
(
p
,
q
)
+
m
(
q
,
p
)
2
=
m
(
p
,
q
)
=
m
1
(
p
)
m
2
(
q
)
=
m
*
(
p
∘
q
)
for all p
,
q
∈
H. Since m
(
p
,
q
)
=
0 if and only if p
∈
I or q
∈
I, by Remark 3 we have
(79)
f
(
p
,
q
)
=
m
*
(
p
∘
q
)
(
β
+
a
(
p
,
q
)
-
a
(
q
,
p
)
)
=
m
*
(
p
∘
q
)
(
β
+
a
1
(
p
)
+
a
2
(
q
)
-
a
1
(
q
)
-
a
2
(
p
)
)
=
m
*
(
p
∘
q
)
(
β
+
(
a
1
-
a
2
)
(
p
)
-
(
a
1
-
a
2
)
(
q
)
)
:
=
m
*
(
p
∘
q
)
(
β
+
a
*
(
p
)
-
a
*
(
q
)
)
for all p
,
q
∈
H
∖
I, where a
*
:
H
∖
I
→
C is an additive function and β
∈
C. Thus, we get (72). This completes the proof.Remark 9.
SinceH
=
(
0,1
] has the property stated in Remark 3, we have I
=
{
p
∈
H
∣
m
*
(
p
)
=
0
}
=
∅. Thus, Theorem 8 includes the result of Riedel and Sahoo [22].Using Theorem7, we have the following.Theorem 10.
Letf
,
g
:
S
→
C satisfy the functional equation (70) for all p
,
q
,
r
,
s
∈
H. Then g, f are of the form
(80)
g
(
p
,
q
)
=
m
1
(
p
)
m
2
(
q
)
+
m
2
(
p
)
m
1
(
q
)
2
,
f
(
p
,
q
)
=
m
1
(
p
)
m
2
(
q
)
+
m
2
(
p
)
m
1
(
q
)
2
β
for all p
,
q
∈
H, where m
1
,
m
2
:
H
→
C are exponential functions and β
∈
C with β
≠
0.Using polar form of complex numbers, we have the following two lemmas. In the following we definearg
(
z
)
=
0 for z
=
0.Lemma 11.
Letm
:
C
→
C satisfy the functional equation
(81)
m
(
z
w
)
=
m
(
z
)
m
(
w
)
for all z
,
w
∈
C. Then m has the form
(82)
m
(
z
)
=
M
(
|
z
|
)
E
(
arg
z
)
for all z
∈
C, where M
:
R
→
C is a multiplicative function, that is, M
(
x
y
)
=
M
(
x
)
M
(
y
), for all x
,
y
∈
R, and E
:
R
→
C is an exponential function such that E
(
x
+
2
π
)
=
E
(
x
) for all x
∈
R.Lemma 12.
Leta
:
C
→
C satisfy the functional equation
(83)
a
(
z
w
)
=
a
(
z
)
+
a
(
w
)
for all z
,
w
∈
C. Then a has the form
(84)
a
(
z
)
=
M
(
|
z
|
)
+
A
(
arg
z
)
for all z
∈
C, where M
:
R
→
C is a multiplicative function and A
:
R
→
C is an additive function such that A
(
x
+
2
π
)
=
A
(
x
) for all x
∈
R.Using Theorem2 and Lemma 11 we have the following.Theorem 13.
Letf
,
g
:
C
→
C satisfy the functional equation
(85)
f
(
z
w
)
+
f
(
z
w
-
)
=
2
f
(
z
)
g
(
w
)
for all z
,
w
∈
C. Then f, g are of the form
(86)
f
(
z
)
=
M
(
|
z
|
)
(
α
1
E
(
arg
z
)
+
α
2
E
(
arg
z
-
)
)
,
g
(
z
)
=
1
2
M
(
|
z
|
)
(
E
(
arg
z
)
+
E
(
arg
z
-
)
)
for all z
∈
C, or
(87)
g
(
z
)
=
M
(
|
z
|
)
,
f
(
z
)
=
M
(
|
z
|
)
(
β
+
A
(
arg
z
)
)
for all z
∈
C, where M
:
R
→
C is a multiplicative function, E
:
R
→
C is an exponential function such that E
(
x
+
2
π
)
=
E
(
x
), for all x
∈
R, and A
:
R
→
C is an additive function such that A
(
x
+
2
π
)
=
A
(
x
), for all x
∈
R, and α
1
,
α
2
,
β
∈
C.Remark 14.
If some regularities onm are assumed in Theorem 13, it can be verified that E
=
1 and A
=
0. Thus, all regular solutions (
g
,
f
) (e.g., continuous solutions) of (85) are given by
(88)
g
(
z
)
=
|
z
|
c
,
f
(
z
)
=
β
|
z
|
c
for all z
∈
C, where c
,
β
∈
C.Using Theorem7 and Lemma 11 we have the following.Theorem 15.
Letf
,
g
:
C
→
C satisfy the functional equation
(89)
f
(
z
w
)
+
f
(
z
w
-
)
=
2
g
(
z
)
f
(
w
)
for all z
,
w
∈
C. Then f, g are of the form
(90)
f
(
z
)
=
α
M
(
|
z
|
)
(
E
(
arg
z
)
+
E
(
arg
z
-
)
)
,
g
(
z
)
=
1
2
M
(
|
z
|
)
(
E
(
arg
z
)
+
E
(
arg
z
-
)
)
for all z
∈
C, where M
:
R
→
C is a multiplicative function and α
∈
C.As consequence of Theorems13 and 15, we also find general solutions of the following functional equations:
(91)
f
(
u
x
-
v
y
,
u
y
+
v
x
)
+
f
(
u
x
+
v
y
,
u
y
-
v
x
)
-
2
f
(
x
,
y
)
g
(
u
,
v
)
=
0
,
(92)
f
(
u
x
-
v
y
,
u
y
+
v
x
)
+
f
(
u
x
+
v
y
,
u
y
-
v
x
)
-
2
g
(
x
,
y
)
f
(
u
,
v
)
=
0
,
(93)
f
(
u
x
+
v
y
,
u
y
+
v
x
)
+
f
(
u
x
-
v
y
,
u
y
-
v
x
)
-
2
f
(
x
,
y
)
g
(
u
,
v
)
=
0
,
(94)
f
(
u
x
+
v
y
,
u
y
+
v
x
)
+
f
(
u
x
-
v
y
,
u
y
-
v
x
)
-
2
g
(
x
,
y
)
f
(
u
,
v
)
=
0
,
for all x
,
y
,
u
,
v
∈
R, where f
,
g
:
R
2
→
R.Theorem 16.
Letf
,
g
:
R
2
→
C satisfy the functional equation (91). Then f, g are of the form
(95)
f
(
x
,
y
)
=
M
(
x
2
+
y
2
)
×
(
α
1
E
(
arctan
(
y
x
)
)
+
α
2
E
(
-
arctan
(
y
x
)
)
)
,
g
(
x
,
y
)
=
1
2
M
(
x
2
+
y
2
)
×
(
E
(
arctan
(
y
x
)
)
+
E
(
-
arctan
(
y
x
)
)
)
for all (
x
,
y
)
∈
R
2, or
(96)
g
(
x
,
y
)
=
M
(
x
2
+
y
2
)
,
f
(
x
,
y
)
=
M
(
x
2
+
y
2
)
(
β
+
A
(
arctan
(
y
x
)
)
)
for all (
x
,
y
)
∈
R
2, where M
:
[
0
,
∞
)
→
C is a multiplicative function, E
:
R
→
C is an exponential function such that E
(
x
+
2
π
)
=
E
(
x
), for all x
∈
R, and A
:
R
→
C is an additive function such that A
(
x
+
2
π
)
=
A
(
x
), for all x
∈
R, and α
1
,
α
2
,
β
∈
C.Proof.
DefineF
,
G
:
C
→
R by
(97)
F
(
x
+
i
y
)
=
f
(
x
,
y
)
,
G
(
x
+
i
y
)
=
g
(
x
,
y
)
for all x
,
y
∈
R. Then, (91) is reduced to
(98)
F
(
z
w
)
+
F
(
z
w
-
)
-
2
F
(
z
)
G
(
w
)
=
0
for all z
,
w
∈
C. Using Theorem 13 we get the result.Similarly, using Theorem15 we have the following.Theorem 17.
Letf
,
g
:
R
2
→
C satisfy the functional equation (92). Then f, g are of the form
(99)
f
(
x
,
y
)
=
α
M
(
x
2
+
y
2
)
×
(
E
(
arctan
(
y
x
)
)
+
E
(
-
arctan
(
y
x
)
)
)
,
g
(
x
,
y
)
=
1
2
M
(
x
2
+
y
2
)
×
(
E
(
arctan
⁡
(
y
x
)
)
+
E
(
-
arctan
⁡
(
y
x
)
)
)
,
for all (
x
,
y
)
∈
R
2, where M
:
[
0
,
∞
)
→
C is a multiplicative function, E
:
R
→
C is an exponential function such that E
(
x
+
2
π
)
=
E
(
x
), for all x
∈
R, and α
∈
C.Theorem 18.
Letf
,
g
:
R
2
→
C satisfy the functional equation (93). Then either g, f are of the form
(100)
g
(
p
,
q
)
=
1
2
(
M
1
(
p
+
q
)
M
2
(
p
-
q
)
k
k
k
k
k
+
M
2
(
p
+
q
)
M
1
(
p
-
q
)
)
,
f
(
p
,
q
)
=
α
1
M
1
(
p
+
q
)
M
2
(
p
-
q
)
+
α
2
M
2
(
p
+
q
)
M
1
(
p
-
q
)
for all p
,
q
∈
R, where M
1
,
M
2
:
R
→
C are multiplicative functions such that M
1
≠
M
2 and α
1
,
α
2
∈
C, or
(101)
g
(
p
,
q
)
=
M
(
p
2
-
q
2
)
,
p
,
q
∈
R
,
f
(
p
,
q
)
=
{
M
(
p
2
-
q
2
)
(
β
+
L
(
p
+
q
)
-
L
(
p
-
q
)
)
,
i
f
p
2
≠
q
2
,
0
,
i
f
p
2
=
q
2
,
where L
:
R
∖
{
0
}
→
C is a logarithmic function, that is, L
(
x
y
)
=
L
(
x
)
+
L
(
y
), M
:
R
→
C is a multiplicative function, and β
∈
C, or
(102)
g
(
p
,
q
)
=
1
,
f
(
p
,
q
)
=
β
for all p
,
q
∈
R.Proof.
DefineF
,
G
:
R
2
→
R by
(103)
F
(
x
,
y
)
=
f
(
x
+
y
2
,
x
-
y
2
)
,
G
(
x
,
y
)
=
g
(
x
+
y
2
,
x
-
y
2
)
for all x
,
y
∈
R. Then, (93) is reduced to
(104)
F
(
p
r
,
q
s
)
+
F
(
p
s
,
q
r
)
=
2
F
(
p
,
q
)
G
(
r
,
s
)
for all p
,
q
,
r
,
s
∈
R. Applying Theorem 8 with H
=
〈
R
,
×
〉 for m
≠
m
σ, we get (100). It is easy to see that I
:
=
{
x
∈
R
:
m
*
(
x
)
=
0
}
=
{
0
} or I
=
∅ in the case when m
=
m
σ of Theorem 8. If I
=
{
0
}, from (71) we get (101) with M
=
m
*
,
L
=
a
*. If I
=
∅, then m
*
(
0
)
≠
0 and it follows that g
=
m
*
=
1. In this case, a
* is a logarithmic function on R; that is, a
* satisfies
(105)
a
*
(
p
q
)
=
a
*
(
p
)
+
a
*
(
q
)
for all p
,
q
∈
R. Putting q
=
0 in (105) we have a
*
=
0. Thus, from (71) we get (102). This completes the proof.DefineF
,
G
:
R
2
→
R by
(106)
F
(
x
,
y
)
=
f
(
x
+
y
2
,
x
-
y
2
)
,
G
(
x
,
y
)
=
g
(
x
+
y
2
,
x
-
y
2
)
for all x
,
y
∈
R. Then, (94) is reduced to
(107)
F
(
p
r
,
q
s
)
+
F
(
p
s
,
q
r
)
=
2
G
(
p
,
q
)
F
(
r
,
s
)
for all p
,
q
,
r
,
s
∈
R. Using Theorem 10 we have the following.Theorem 19.
Letf
,
g
:
R
2
→
C satisfy the functional equation (94). Then either g or f has the form
(108)
g
(
p
,
q
)
=
M
1
(
p
+
q
)
M
2
(
p
-
q
)
+
M
2
(
p
+
q
)
M
1
(
p
-
q
)
2
,
f
(
p
,
q
)
=
M
1
(
p
+
q
)
M
2
(
p
-
q
)
+
M
2
(
p
+
q
)
M
1
(
p
-
q
)
2
β
,
for all p
,
q
∈
R, where M
1
,
M
2
:
R
→
C are multiplicative functions and β
∈
C with β
≠
0.



---

*Source: 463918-2014-08-27.xml*