EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 4, 2021, 1306-1323
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On the Fourier Transform Related to the Diamond
Klein–Gordon Kernel

Sudprathai Bupasiri

Faculty of Education, Sakon Nakhon Rajabhat University, Sakon Nakhon 47000, Thailand

Abstract. In this article, we study the fundamental solution of the operator((
♢+m2

)(△2 +⊡2

2

))k

iterated k-times, which is defined by (10), where m is a non-negative real number, and k is a non-

negative integer. After that, we study the Fourier transform of the operator
((

♢+m2
) (△2+⊡2

2

))k
δ,

where δ is the Dirac delta function.

2020 Mathematics Subject Classifications: 46F10

Key Words and Phrases: Diamond Klein–Gordon kernel; Diamond operator; Laplace operator;
Fourier transform; wave equation

Introduction

The operator ⋄k has been first introduced by Kananthai [5], is named as the diamond
operator iterated k-times, and is defined by

♢k =

( p∑
r=1

∂2

∂x2r

)2

−

 p+q∑
j=p+1

∂2

∂x2j

2k

, p+ q = n, (1)

where n is the dimension of the space Rn, for x = (x1, x2, . . . , xn) ∈ Rn and k is a non-
negative integer. The operator ♢k can be expressed in the form ♢k = ⊡k△k = △k⊡k,
where the operator △k is Laplace operator iterated k-times, which is defined by

△k =

(
∂2

∂x21
+

∂2

∂x22
+ · · ·+ ∂2

∂x2n

)k

(2)

DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4101

Email address: sudprathai@gmail.com (S. Bupasiri)

http://www.ejpam.com 1306 © 2021 EJPAM All rights reserved.



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1307

and the operator ⊡k is the ultra-hyperbolic operator iterated k-times, which is defined by

⊡k =

(
∂2

∂x21
+

∂2

∂x22
+ · · ·+ ∂2

∂x2p
− ∂2

∂x2p+1

− ∂2

∂x2p+2

− · · · − ∂2

∂x2p+q

)k

. (3)

By putting p = 1 and x1 = t (time) in (3), then we obtain the wave operator

⊡ =
∂2

∂t2
−

n−1∑
j=1

∂2

∂x2j
. (4)

In 1997, Kananthai [5] showed that the convolution (−1)kRe
2k(x) ∗ RH

2k(x) is the funda-
mental solution of the operator ♢k, that is

♢k((−1)kRe
2k(x) ∗RH

2k(x)) = δ, (5)

where the function RH
2k(x) is defined by (20) and Re

2k(x) is defined by (19). The fundamen-
tal solution (−1)kRe

2k(x) ∗ RH
2k(x) is called the diamond kernel of Marcel Riesz. Satsanit

[20] showed that

⊙k =

( p∑
r=1

∂2

∂x2r

)2

+

 p+q∑
j=p+1

∂2

∂x2j

2k

=

((
△+⊡

2

)2

+

(
△−⊡

2

)2
)k

=

(
△2 +⊡2

2

)k

. (6)

Moreover, Kananthai, Suantai and Longani [7] studied the fundamental solution of the
operator ⊕k and the weak solution of the equation ⊕ku(x) = f(x), where the operator ⊕k

is defined by

⊕k =

( p∑
r=1

∂2

∂x2r

)4

−

 p+q∑
j=p+1

∂2

∂x2j

4k

=

( p∑
r=1

∂2

∂x2r

)2

−

 p+q∑
j=p+1

∂2

∂x2j

2k ( p∑
r=1

∂2

∂x2r

)2

+

 p+q∑
j=p+1

∂2

∂x2j

2k

= ⋄kLk
1L

k
2

= ⋄kLk (7)

where p+ q = n is the dimension of the Euclidean space Rn, k is a non-negative integer,
and f(x) is a generalized function.



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1308

Next, Kananthai, Suantai and Longani [6] studied the relationship between the oper-
ator ⊕k and the wave operator, and the relationship between the operator ⊕k and the
Laplace operator. Moreover, they studied equation ⊕kK(x) = δ and they showed that

K(x) = [RH
2k(x) ∗ (−1)kRe

2k(x)] ∗ S2k(x) ∗ T2k(x)

is the fundamental solution of the operator ⊕k. Later, Kananthai [3] studied the inversion
of the kernel Kα,β,γ,ν related to the operator ⊕k.

In 1988, Trione [22] studied the fundamental solution of the ultra-hyperbolic Klein–
Gordon operator iterated k-times, which is defined by

(⊡+m2)k =

[
∂2

∂x21
+

∂2

∂x22
+ · · ·+ ∂2

∂x2p
− ∂2

∂x2p+1

− ∂2

∂x2p+2

− · · · − ∂2

∂x2p+q

+m2

]k
. (8)

Later, Lunnaree and Nonlaopon [11] introduced the operator (♢+m2)k, that is named
as the diamond Klein-Gordon operator iterated k-times, which is defined by

(♢+m2)k =

( p∑
r=1

∂2

∂x2r

)2

−

 p+q∑
j=p+1

∂2

∂x2j

2

+m2

k

, (9)

where p+q = n is the dimension of the space Rn, for x = (x1, x2, . . . , xn) ∈ Rn,m is a non-
negative real number and k is a non-negative integer, see [9, 10, 17, 18] for more details.
V.N. Mishra, K. Khatri and L.N. Mishra [15] studied the linear operators to approximate
signals of Lip (α, p), (p ≥ 1)-class, see [2, 12–14, 16] for more details.

Moreover, Kananthai [4] studied the fundamental solution for the (♢ + m4)k, which
related to the Klein-Gordon operator. From (7) the operator

( p∑
r=1

∂2

∂x2r

)2

+
m2

2

2

−

 p+q∑
j=p+1

∂2

∂x2j

2

− m2

2

2

k

can be expressed in the form
( p∑

r=1

∂2

∂x2r

)2

+
m2

2

2

−

 p+q∑
j=p+1

∂2

∂x2j

2

− m2

2

2

k

=

( p∑
r=1

∂2

∂x2r

)2

−

 p+q∑
j=p+1

∂2

∂x2j

2

+m2

k( p∑
r=1

∂2

∂x2r

)2

+

 p+q∑
j=p+1

∂2

∂x2j

2k

= (♢+m2)k
(
△2 +⊡2

2

)k



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1309

= (♢+m2)k ⊙k . (10)

From (10) with q = m = 0 and k = 1, we obtain Laplace operator △4
p of p-dimension,

where

△p =
∂2

∂x21
+

∂2

∂x22
+ · · ·+ ∂2

∂x2p
. (11)

In this article, we study the fundamental solution of the equation of the form
( p∑

r=1

∂2

∂x2r

)2

+
m2

2

2

−

 p+q∑
j=p+1

∂2

∂x2j

2

− m2

2

2


k

K(x,m) = δ,

or (
(♢+m2)

(
△2 +⊡2

2

))k

K(x,m) = δ,

where K(x,m) is the fundamental solution, δ is the Dirac delta function, k is a non-
negative integer, and m is a non-negative real number. Moreover, we study the Fourier

transform of the operator
(
(♢+m2)

(
△2+⊡2

2

))k
δ.

Preliminary Notes

Definition 1. Let x = (x1, x2, . . . , xn) be a point of the n-dimensional space Rn,

u = x21 + x22 + · · ·+ x2p − x2p+1 − x2p+2 − · · · − x2p+q, (12)

where p+ q = n.
Define Γ+ = {x ∈ Rn : x1 > 0 and u > 0}, which designates the interior of the forward

cone and Γ+ designates its closure and the following functions introduce by Nozaki [19,
Page 72], that

RH
α (x) =

{
u

α−n
2

Kn(α)
, if x ∈ Γ+;

0, if x ̸∈ Γ+

(13)

is called the ultra-hyperbolic kernel of Marcel Riesz. Here, α is a complex parameter and
n the dimension of the space. The constant Kn(α) is defined by

Kn(α) =
π

n−1
2 Γ

(
2+α−n

2

)
Γ
(
1−α
2

)
Γ(α)

Γ
(
2+α−p

2

)
Γ
(p−α

2

) (14)

and p is the number of positive terms of

u = x21 + x22 + · · ·+ x2p − x2p+1 − x2p+2 − · · · − x2p+q, p+ q = n



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1310

and let supp RH
α (x) ⊂ Γ+. Now, RH

α (x) is an ordinary function if Re α ≥ n and is a
distribution of α if Re α < n. Now, if p = 1 then (13) reduces to the function Mα(u), and
is defined by

Mα(u) =

{
u

α−n
2

Hn(α)
, if x ∈ Γ+;

0, if x ̸∈ Γ+,
(15)

where u = x21 − x22 − · · · − x2n and Hn(α) = π
(n−1)

2 2α−1Γ(α−n+2
2 ). The function Mα(u) is

called the hyperbolic kernel of Marcel Riesz.

Definition 2. Let f(x) ∈ L1(Rn) (the space of integrable function in Rn). The Fourier
transform of f(x) is defined as

f̂(ξ) =
1

(2π)n/2

∫
Rn

e−iξ·xf(x)dx, (16)

where ξ = (ξ1, ξ2, . . . , ξn), x = (x1, x2, . . . , xn) ∈ Rn, ξ · x = (ξ1x1, ξ2x2, . . . , ξnxn) is the
usual inner product in Rn and dx = dx1dx2 . . . dxn. The inverse of the Fourier transform
is defined by

f(x) =
1

(2π)n/2

∫
Rn

e−iξ·xf̂(ξ)dξ. (17)

If f is a distribution with compact supports, by [24, Theorem 7.4-3], Equation (17) can be
written as

f̂(ξ) = Ff(x) =
1

(2π)n/2

〈
f(x), e−iξ·x

〉
. (18)

Lemma 1. [5] Given the equation △ku(x) = δ for x ∈ Rn, where △k is the Laplace opera-
tor iterated k-times, which is defined by (2). Then u(x) = (−1)kRe

2k(x) is the fundamental
solution of the operator △k, where

Re
2k(x) =

Γ
(
n−2k

2

)
22kπ

n
2 Γ(k)

|x|2k−n. (19)

Lemma 2. [22] If ⊡ku(x) = δ for x ∈ Γ+ = {x ∈ Rn : x1 > 0 and u > 0}, where ⊡kis the
ultra-hyperbolic operator iterated k-times, which is defined by (3). Then u(x) = RH

2k(x) is
the unique fundamental solution of the operator ⊡k, where

RH
2k(x) =

u(
2k−n

2
)

Kn(2k)
=

(x21 + x22 + · · ·+ x2p − x2p+1 − · · · − x2p+q)
( 2k−n

2 )

Kn(2k)
(20)

and

Kn(2k) =
π

n−1
2 Γ

(
2+2k−n

2

)
Γ
(
1−2k
2

)
Γ(2k)

Γ
(
2+2k−p

2

)
Γ(p−2k

2 )
. (21)



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1311

Lemma 3. [5] Given the equation ♢ku(x) = δ for x ∈ Rn, then u(x) = (−1)kRe
2k(x) ∗

RH
2k(x) is the unique fundamental solution of the operator ♢k, where ♢k is the diamond

operator iterated k-times, which is defined by (1), Re
2k(x) and RH

2k(x) are defined by (19)
and (20), respectively. Moreover, (−1)kRe

2k(x) ∗RH
2k(x) is a tempered distribution.

It is not difficult to show that Re
−2k(x) ∗RH

−2k(x) = (−1)k♢kδ, for k is a non-negative
integer.

Definition 3. Let x = (x1, x2, . . . , xn) be a point of Rn, the function Pα(x,m) is defined
by

Pα(x,m) =

∞∑
r=0

(
−α/2

r

)
(m2)r(−1)α/2+rRe

α+2r(x) ∗RH
α+2r(x), (22)

where α is a complex parameter, m is a non-negative real number, RH
α+2r(x) and Re

α+2r(x)
are defined by (20) and (19), respectively.

From the definition of Pα(x,m) and by putting α = −2k, we have

P−2k(x,m) =
∞∑
r=0

(
k

r

)
(m2)r(−1)−k+rRe

2(−k+r)(x) ∗R
H
2(−k+r)(x).

Since the operator (♢+m2)k defined in equation (9) is a linearly continuous and has 1−1
mapping, then it has inverse. From Lemma 3, we obtain

P−2k(x,m) =

∞∑
r=0

(
−k

r

)
(m2)r♢−k−rδ

= (♢+m2)kδ. (23)

By putting k = 0 in (23), we have P0(x,m) = δ. By putting α = 2k into (22), we have

P2k(x,m) =

(
−k

0

)
(m2)0(−1)k+0Re

2k+0(x) ∗RH
2k+0(x)

+

∞∑
r=1

(
−k

r

)
(m2)r(−1)k+rRe

2k+2r(x) ∗RH
2k+2r(x). (24)

The second summand of the right-hand member of (24) vanishes for m = 0 and then, we
have

P2k(x,m = 0) = (−1)kRe
2k(x) ∗RH

2k(x) (25)

is the fundamental solution of the diamond operator ♢k.

Lemma 4. The function RH
−2k(x) and (−1)kRe

−2k(x) are the inverse in the convolution

algebra of RH
2k(x) and (−1)kRe

2k(x), respectively. That is,

RH
−2k(x) ∗RH

2k(x) = RH
−2k+2k(x) = RH

0 (x) = δ

and
(−1)kRe

−2k(x) ∗ (−1)kRe
2k(x) = (−1)2kRe

−2k+2k(x) = Re
0(x) = δ.



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1312

For the proof of the this Lemma is given in [1, 21, 23].

Lemma 5. [20](Convolution of Re
α(x) and RH

α (x)). If Re
α(x) and RH

α (x) are defined by
(19) and (20), respectively, then

(i) Re
α(x) ∗Re

β(x) = Re
α+β(x), where α and β are complex parameters;

(ii) RH
α (x)∗RH

β (x) = RH
α+β(x), where α and β are both integers and except only the case

both α and β are both integers.

Lemma 6. [11] Given the equation (♢+m2)ku(x) = δ, where (♢+m2)k is the diamond
Klein-Gordon operator, which is defined by

(♢+m2)k =

( p∑
r=1

∂2

∂x2r

)2

−

 p+q∑
j=p+1

∂2

∂x2j

2

+m2

k

, (26)

where x = (x1, x2, . . . , xn) ∈ Rn, k is a non-negative integer, m is a non-negative real
number and δ is the Dirac delta function. Then, we obtain

P2k(x,m) =
∞∑
r=0

(
−k

r

)
m2r(−1)k+rRe

2k+2r(x) ∗RH
2k+2r(x) (27)

is the fundamental solution of the operator (♢ +m2)k, defined by (9), where RH
2k(x) and

Re
2k(x) are defined by (20) and (19), respectively. Moreover, u(x) = P2k(x,m) is tempered

distribution.

Lemma 7. [20] Given the equation

⊙kG(x) = δ, (28)

where ⊙k is the operator iterated k-times is defined by (6). Then, we obtain G(x) is the
fundamental solution of the equation (28), where

G(x) = (RH
4k(x) ∗ (−1)2kRe

4k(x)) ∗ (H∗k(x))∗−1 (29)

and

H(x) =
1

2
RH

4 (x) +
1

2
(−1)2Re

4(x). (30)

Here, H∗k(x) denotes the convolution of H(x) itself k-times, (H∗k(x))∗−1 denotes the
inverse of H∗k(x) in the convolution algebra. Moreover, G(x) is a tempered distribution.

Lemma 8. (The Fourier transform of
(
(♢+m2)

(
△2+⊡2

2

))k
δ.)

Let
||ξ|| =

(
ξ21 + ξ22 + · · ·+ ξ2n

)1/2



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1313

for ξ ∈ Rn. Then∣∣∣∣∣F
(
(♢+m2)

(
△2 +⊡2

2

))k

δ

∣∣∣∣∣ ≤ 1

(2π)n/2
(||ξ||4 +m2)k||ξ||4k.

That is, F
(
(♢+m2)

(
△2+⊡2

2

))k
δ is bounded and continuous on the space S ′

of the

tempered distribution. Moreover, by the inverse Fourier transformation(
(♢+m2)

(
△2 +⊡2

2

))k

δ = F−1 1

(2π)n/2

[(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 +
m2

2

)2

−
(
(ξ2p+1 + ξ2p+2 + · · ·+ ξ2p+q)

2 − m2

2

)2
]k

.

Proof. From the Fourier transform (16), we have

F
(
(♢+m2)

(
△2 +⊡2

2

))k

δ

=
1

(2π)n/2

〈
δ, (♢+m2)k

(
△2 +⊡2

2

)k

e−iξ·x

〉

=
1

(2π)n/2

〈
δ, (♢+m2)k

(−1)2k

2

(
(ξ21 + ξ22 + · · ·+ ξ2n)

2 + (ξ21 + ξ22 + · · ·+ ξ2p

−ξ2p+1 − ξ2p+2 − · · · − ξ2n)
2
)k

e−iξ·x
〉

=
1

(2π)n/2

〈
δ,
(−1)2k

2

(
(ξ21 + ξ22 + · · ·+ ξ2n)

2 + (ξ21 + ξ22 + · · ·+ ξ2p

−ξ2p+1 − ξ2p+2 − · · · − ξ2n)
2
)k

(♢+m2)ke−iξ·x
〉

=
1

(2π)n/2

〈
δ,

( p∑
i=1

ξ2i

)2

+

 p+q∑
j=p+1

ξ2j

2k ( p∑
i=1

ξ2i

)2

−

 p+q∑
j=p+1

ξ2j

2

+m2

k

e−iξ·x

〉

=
1

(2π)n/2

〈
δ,


( p∑

i=1

ξ2i

)2

+
m2

2

2

−

 p+q∑
j=p+1

ξ2j

2

− m2

2

2

k

e−iξ·x

〉

=
1

(2π)n/2


( p∑

i=1

ξ2i

)2

+
m2

2

2

−

 p+q∑
j=p+1

ξ2j

2

− m2

2

2

k

=
1

(2π)n/2

[(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 +
m2

2

)2

−
(
(ξ2p+1 + ξ2p+2 + · · ·+ ξ2p+q)

2 − m2

2

)2
]k

.



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1314

Next, we consider the boundedness of F
(
(♢+m2)

(
△2+⊡2

2

))k
δ. Since(

(♢+m2)

(
△2 +⊡2

2

))k

=

[(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 +
m2

2

)2

−
(
(ξ2p+1 + ξ2p+2 + · · ·+ ξ2p+q)

2 − m2

2

)2
]k

=
[(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 − (ξ2p+1 + ξ2p+2 + · · ·+ ξ2n)
2 +m2

)k
×
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 + (ξ2p+1 + ξ2p+2 + · · ·+ ξ2n)
2
)k]

=
[(
(ξ21 + ξ22 + · · ·+ ξ2n)(ξ

2
1 + · · ·+ ξ2p − ξ2p+1 − · · · − ξ2n) +m2

)k
×
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 + (ξ2p+1 + · · ·+ ξ2n)
2
)k]

.

Thus

F
(
(♢+m2)

(
△2 +⊡2

2

))k

δ

=
1

(2π)n/2

[(
(ξ21 + ξ22 + · · ·+ ξ2n)(ξ

2
1 + · · ·+ ξ2p − ξ2p+1 − · · · − ξ2n) +m2

)k
×
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 + (ξ2p+1 + · · ·+ ξ2n)
2
)k]

,∣∣∣∣∣F
(
(♢+m2)

(
△2 +⊡2

2

))k

δ

∣∣∣∣∣
=

1

(2π)n/2
(∣∣ξ21 + ξ22 + · · ·+ ξ2n

∣∣ ∣∣ξ21 + · · ·+ ξ2p − ξ2p+1 − · · · − ξ2n
∣∣+m2

)k
×
∣∣((ξ21 + ξ22 + · · ·+ ξ2p)

2 + (ξ2p+1 + · · ·+ ξ2n)
2
)∣∣k

≤ 1

(2π)n/2

(∣∣ξ21 + ξ22 + · · ·+ ξ2n
∣∣2 +m2

)k ∣∣ξ21 + ξ22 + · · ·+ ξ2n
∣∣2k

=
1

(2π)n/2
(||ξ||4 +m2)k||ξ||4k,

where ||ξ|| =
(
ξ21 + ξ22 + · · ·+ ξ2n

)1/2
, ξi(i = 1, 2, . . . , n) ∈ R. Hence, we obtain

F
(
(♢+m2)

(
△2 +⊡2

2

))k

δ

is bounded and continuous on the space S ′
of the tempered distribution.

Since F is 1− 1 transformation from the space S ′
of the tempered distribution to the

real space R, then by (17), we have(
(♢+m2)

(
△2 +⊡2

2

))k

δ



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1315

=
1

(2π)n/2
F−1

[(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 +
m2

2

)2

−
(
(ξ2p+1 + ξ2p+2 + · · ·+ ξ2p+q)

2 − m2

2

)2
]k

.

Main Results

Theorem 1.

(
The fundamental solution of

(
(♢+m2)

(
△2+⊡2

2

))k)
.

Given the equation (
(♢+m2)

(
△2 +⊡2

2

))k

K(x,m) = δ, (31)

where
(
(♢+m2)

(
△2+⊡2

2

))k
is the operator iterated k-times, which is defined by (10), δ is

the Dirac-delta function, x ∈ Rn, m is a non-negative real number and k is a non-negative
integer. Then, we obtain

K(x,m) =
(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m) (32)

is the fundamental solution for the operator iterated k-times, which is defined by (10). In
particular, for m = 0 then (31) becomes

⊕kK(x, 0) = δ, (33)

and we obtain

K(x, 0) =
(
RH

6k(x) ∗ (−1)3kRe
6k(x)

)
∗
(
(H∗k(x))∗−1

)
(34)

is the fundamental solution of the o-plus operator ⊕k, for q = m = 0 then (31) becomes

△4k
p K(x, 0) = δ, (35)

and we obtain
K(x, 0) = Re

8k(x) (36)

is the fundamental solution of (35), where △4k
p is the Laplace operator of p-dimension,

iterated 4k-times which is defined by (11).
Moreover, from (34), we obtain(

RH
−4k(x) ∗ (−1)3kRe

−6k(x)
)
∗
(
H∗k(x)

)
∗K(x, 0) = RH

2k(x) (37)

is the fundamental solution of the ultra-hyperbolic operator ⊡k iterated k-times, which
defined by (3), where Re

−6k(x) and RH
−4k(x) are inverse of Re

6k(x) and RH
4k(x), respectively.



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1316

From (34) and (37) with p = 1, q = n− 1, k = 1,m = 0 and x1 = t (time), we obtain(
(−1)3

2
Re

−6(x) +MH
−4(u) ∗

(−1)5

2
Re

−2(x)

)
∗K(x, 0) = MH

2 (u) (38)

or ((
−1

2
Re

−6(x)

)
+MH

−4(u) ∗
(
−1

2
Re

−2(x)

))
∗K(x, 0) = MH

2 (u) (39)

is the fundamental solution of the wave operator is defined by (4), where M2(u) is defined
by (15) with α = 2.

Proof. From (10) and (31), we have(
(♢+m2)

(
△2 +⊡2

2

))k

K(x,m) = (♢+m2)k
(
△2 +⊡2

2

)k

K(x,m) = δ. (40)

Convolving both sides of (40) by
(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m), we

obtain[(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m)

]
∗ (♢+m2)k

(
△2 +⊡2

2

)k

K(x,m)

=
[(

RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1
)
∗ P2k(x,m)

]
∗ δ.

By properties of the convolution, we have

(♢+m2)k(P2k(x,m)) ∗
(
△2 +⊡2

2

)k ((
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

))
∗K(x,m)

=
(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m).

By Lemma 6 and Lemma 7, we obtain,

δ ∗ δ ∗K(x,m) = K(x,m) =
(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m) (41)

is the fundamental solution of
(
(♢+m2)

(
△2+⊡2

2

))k
operator.

In particular, for m = 0 then (31) becomes

⊕kK(x, 0) = δ. (42)

From Lemma 5, (22) and (41), we obtain

K(x, 0) =
(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x, 0)

=
(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ ((−1)kRe

2k(x) ∗RH
2k(x))



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1317

=
(
RH

6k(x) ∗ (−1)3kRe
6k(x)

)
∗
(
(H∗k(x))∗−1

)
(43)

is the fundamental solution of the o-plus operator ⊕k.
Putting q = m = 0, then (31) becomes

△4k
p K(x, 0) = δ, (44)

where △4k
p is Laplace operator of p-dimension iterated 4k-times. By Lemma 1, we have

K(x, 0) = (−1)4kRe
8k(x) = Re

8k(x)

is the fundamental solution of (44).
On the other hand, we can also find K(x,m) from (41). Since q = 0, we have RH

2k(x)
reduces to (−1)kRe

2k(x). Thus, by (41) for q = m = 0, we obtain

K(x, 0) =
(
(−1)2kRe

4k(x) ∗ (−1)2kRe
4k(x)

)
∗
(
(−1)2kRe

4k(x)
)∗−1

∗ P2k(x, 0)

= (−1)4kRe
4k+4k(x) ∗

(
(−1)2kRe

4k(x)
)∗−1

∗ ((−1)kRe
2k(x) ∗ (−1)kRe

2k(x))

= (−1)8kRe
8k(x) = Re

8k(x),

where (Re
4k(x))

∗−1 is the inverse of Re
4k(x) in the convolution algebra.

From (41), we have

K(x, 0) =
(
RH

4k(x) ∗ (−1)2kRe
4k(x)

)
∗
(
H∗k(x)

)∗−1
∗ P2k(x, 0).

Convolving the above equation by
(
RH

−4k(x) ∗ (−1)3kRe
−6k(x)

)
∗
(
H∗k(x)

)
. By Lemma 4,

Lemma 5, and (25), we obtain(
RH

−4k(x) ∗ (−1)3kRe
−6k(x)

)
∗
(
H∗k(x)

)
∗K(x, 0)

=
(
RH

4k(x) ∗RH
−4k(x)) ∗ ((−1)2kRe

4k(x) ∗ (−1)3kRe
−6k(x)

)
)

∗ (
(
H∗k(x)

)
∗
(
H∗k(x)

)∗−1
) ∗ P2k(x, 0)

or (
RH

−4k(x) ∗ (−1)3kRe
−6k(x)

)
∗
(
H∗k(x)

)
∗K(x, 0)

= δ(x) ∗ (−1)5kRe
−2k(x) ∗ δ(x) ∗ P2k(x, 0)

= δ(x) ∗ (−1)5kRe
−2k(x) ∗ δ(x) ∗ ((−1)kRe

2k(x) ∗RH
2k(x))

= δ(x) ∗ δ(x) ∗ δ(x) ∗RH
2k(x) = RH

2k(x).

It follows that (
RH

−4k(x) ∗ (−1)3kRe
−6k(x)

)
∗
(
H∗k(x)

)
∗K(x, 0) = RH

2k(x) (45)



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1318

as the fundamental solution of the ultra-hyperbolic operator iterated k-times defined by
(3).

In particular, if we put p = 1, q = n − 1, k = 1,m = 0 and x1 = t (time) in (41) then
RH

−4(x) reduces to MH
−4(u) and RH

2 (x) reduce to MH
2 (u), where MH

4 (u) and MH
2 (u) are

defined by (15) with α = −4, α = 2, respectively. Thus, (45) becomes

(
MH

−4(u) ∗ (−1)3Re
−6(x)

)
∗
(
1

2
MH

4 (x) +
(−1)2

2
Re

4(x)

)
∗K(x, 0) = MH

2 (u). (46)

By Lemma 7, we obtain(
(−1)3

2
Re

−6(x) +MH
−4(u) ∗

(−1)5

2
Re

−2(x)

)
∗K(x, 0) = MH

2 (u) (47)

or ((
−1

2
Re

−6(x)

)
+MH

−4(u) ∗
(
−1

2
Re

−2(x)

))
∗K(x, 0) = MH

2 (u) (48)

as the fundamental solution of the wave operator defined by

⊡ =
∂2

∂t2
−

n−1∑
j=1

∂2

∂x2j
, (49)

where Re
−6(x) defined by (19). This completes the proof.

Theorem 2.

F
[(

RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1
)
∗ P2k(x,m)

]
=

1

(2π)n/2
[(

(ξ21 + ξ22 + · · ·+ ξ2p)
2 + m2

2

)2
−
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 − m2

2

)2]k
=
∣∣∣F [(RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m)

]∣∣∣ ≤ 1

(2π)
n
2

M (50)

for a large ξi ∈ R, where m is a non-negative real number and M is a constant. That is,
F is bounded and continuous on the space S ′

of the tempered distributions.

Proof. By Theorem 1, we obtain(
(♢+m2)

(
△2 +⊡2

2

))k ((
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m)

)
= δ

or((
(♢+m2)

(
△2 +⊡2

2

))k

δ

)
∗
((

RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1
)
∗ P2k(x,m)

)
= δ.



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1319

Taking the Fourier transform on both sides of the above equation, we obtain

F

(((
(♢+m2)

(
△2 +⊡2

2

))k

δ

)
∗
((

RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1
)

∗P2k(x,m))) = Fδ =
1

(2π)n/2
.

By (18), we have

1

(2π)n/2

〈((
(♢+m2)

(
△2 +⊡2

2

))k

δ

)
∗
((

RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1
)

∗P2k(x,m)) , e−i(ξ·x)
〉
=

1

(2π)n/2
.

By the definition of convolution

1

(2π)n/2

〈((
(♢+m2)

(
△2 +⊡2

2

))k

δ

)
∗
((

RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1
)

∗P2k(x,m)) , e−iξ·(x+r)
〉
=

1

(2π)n/2
,

1

(2π)n/2

〈
(RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1) ∗ P2k(x,m), e−i(ξ·r)

〉
×

〈(
(♢+m2)

(
△2 +⊡2

2

))k

δ, e−i(ξ·x)

〉
=

1

(2π)n/2
,

F((RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1) ∗ P2k(x,m))(2π)
n
2 F

((
(♢+m2)

(
△2 +⊡2

2

))k

δ

)
=

1

(2π)n/2
.

By Lemma 8, we obtain

F((RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1) ∗ P2k(x,m))

×

[(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 +
m2

2

)2

−
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 − m2

2

)2
]k

=
1

(2π)n/2
.

It follows that

F((RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1) ∗ P2k(x,m))



S. Bupasiri / Eur. J. Pure Appl. Math, 14 (4) (2021), 1306-1323 1320

=
1

(2π)n/2
[(

(ξ21 + ξ22 + · · ·+ ξ2p)
2 + m2

2

)2
−
(
(ξ2p+1 + ξ2p+2 + · · ·+ ξ2p+q)

2 − m2

2

)2]k .
Since

1[(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 + m2

2

)2
−
(
(ξ2p+1 + ξ2p+2 + · · ·+ ξ2p+q)

2 − m2

2

)2]
=

1[
(ξ21 + ξ22 + · · ·+ ξ2p)

2 + (ξ2p+1 + ξ2p+2 + · · ·+ ξ2p+q)
2
]

× 1[
(ξ21 + ξ22 + · · ·+ ξ2n)(ξ

2
1 + ξ22 + · · ·+ ξ2p − ξ2p+1 − · · · − ξ2p+q) +m2

] . (51)

Let ξ = (ξ1, ξ2, . . . , ξn) ∈ Γ+ with Γ+ defined by Definition 1. Then (ξ21 + ξ22 + · · ·+ ξ2p +
ξ2p+1 + ξ2p+2 + · · ·+ ξ2p+q) > 0 and for a large k, the right-hand side of (51) tend to zero. It
follows that it is bounded by a positive constant M say, that is we obtain (50) as required
and also by (50) F is continuous on the space S ′

of the tempered distribution.

Theorem 3.

F
([(

RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1
)
∗ P2k(x,m)

]
∗
[(

RH
4l (x) ∗ (−1)2lRe

4l(x) ∗ (H∗l(x))∗−1
)
∗ P2l(x,m)

])
= (2π)n/2F

[(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m)

]
×F

[(
RH

4l (x) ∗ (−1)2lRe
4l(x) ∗ (H∗l(x))∗−1

)
∗ P2l(x,m)

]
=

1

(2π)n/2
[(

(ξ21 + ξ22 + · · ·+ ξ2p)
2 + m2

2

)2
−
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 − m2

2

)2]k+l
,

where k and l are non-negative integers and F is bounded and continuous on the space S ′

of tempered distribution.

Proof. Since RH
4k(x), R

e
4k(x) and P2k(x,m) are tempered distribution with compact

support,([(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m)

]
∗
[(

RH
4l (x) ∗ (−1)2lRe

4l(x) ∗ (H∗l(x))∗−1
)
∗ P2l(x,m)

])
=
[
RH

4k(x) ∗RH
4l (x)

]
∗
[
(−1)2k+2lRe

4k(x) ∗Re
4l(x)

]
∗
[
(H∗k(x))∗−1(H∗l(x))∗−1

]
∗ [P2k(x,m) ∗ P2l(x,m)]



REFERENCES 1321

=
[
RH

4(k+l)(x)
]
∗
[
(−1)2(k+l)Re

4(k+l)(x)
]
∗
[
(H∗(k+l)(x))∗−1

]
∗
[
P2(k+l)(x,m)

]
by [8, Pages 156–159] and [21, Lemma 2.45]. Taking the Fourier transform on both sides
and using Theorem 2, we obtain

F
([(

RH
4k(x) ∗ (−1)2kRe

4k(x) ∗ (H∗k(x))∗−1
)
∗ P2k(x,m)

]
∗
[(

RH
4l (x) ∗ (−1)2lRe

4l(x) ∗ (H∗l(x))∗−1
)
∗ P2l(x,m)

])
=

1

(2π)n/2
[(

(ξ21 + ξ22 + · · ·+ ξ2p)
2 + m2

2

)2
−
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 − m2

2

)2]k+l

=
1

(2π)n/2
[(

(ξ21 + ξ22 + · · ·+ ξ2p)
2 + m2

2

)2
−
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 − m2

2

)2]k
× (2π)n/2

(2π)n/2
[(

(ξ21 + ξ22 + · · ·+ ξ2p)
2 + m2

2

)2
−
(
(ξ21 + ξ22 + · · ·+ ξ2p)

2 − m2

2

)2]l
= (2π)n/2F

[(
RH

4k(x) ∗ (−1)2kRe
4k(x) ∗ (H∗k(x))∗−1

)
∗ P2k(x,m)

]
×F

[(
RH

4l (x) ∗ (−1)2lRe
4l(x) ∗ (H∗l(x))∗−1

)
∗ P2l(x,m)

]
.

Since
(
RH

4(k+l)(x) ∗ (−1)2(k+l)Re
4(k+l)(x) ∗ (H

∗(k+l)(x))∗−1
)
∗
(
P2(k+l)(x,m)

)
∈ S ′

, the space

of tempered distribution and by Theorem 2, we obtain that F is bounded and continuous
on S ′

.

Acknowledgements

The author would like to thank the referees for their suggestions which enhanced
the presentation of the paper. The author was supported by Sakon Nakhon Rajabhat
University, Thailand.

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