Electronic Journal of Differential Equations, Vol. 2010(2010), No. 48, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
GREEN FUNCTION AND FOURIER TRANSFORM FOR O-PLUS
OPERATORS
WANCHAK SATSANIT
Abstract. In this article, we study the o-plus operator defined by
⊕k =
p+q
p
““ X
∂ 2 ”4 “ X ∂ 2 ”4 ”k
−
,
2
∂xi
∂x2j
i=1
j=p+1
where x = (x1 , x2 , . . . , xn ) ∈ Rn , p + q = n, and k is a nonnegative integer.
Firstly, we studied the elementary solution for the ⊕k operator and then this
solution is related to the solution of the wave and the Laplacian equations.
Finally, we studied the Fourier transform of the elementary solution and also
the Fourier transform of its convolution.
1. Introduction
Consider the ultra-hyperbolic operator iterated k times,
∂2
∂2
∂2
∂2
∂2
∂ 2 k
−
.
+
+
·
·
·
+
−
−
·
·
·
−
k =
∂x21
∂x22
∂x2p
∂x2p+1
∂x2p+2
∂x2p+q
(1.1)
H
(x), defined by (2.1) below,
Trione [6] showed that the generalized function R2k
k
H
is the unique elementary solution for the operator, that is k R2k
(x) = δ for
n
x ∈ R , the n-dimensional Euclidian space.
Kananthai [2] studied the Diamond operator, iterated k times,
p
p+q
X
∂ 2 2 X ∂ 2 2 k
♦ =
−
,
∂x2i
∂x2j
i=1
j=p+1
k
(1.2)
for x = (x1 , x2 , . . . , xn ) ∈ Rn , where p + q = n, n is the dimension of Euclidean
space Rn , and k is a nonnegative integer. The operator ♦k can be expressed in the
form
♦k = 4k k = k 4k
(1.3)
where 4k is the Laplacian operator iterated k times,
∂2
∂2
∂ 2 k
4k =
+ 2 + ··· + 2
.
2
∂x1
∂x2
∂xn
2000 Mathematics Subject Classification. 46F10, 46F12.
Key words and phrases. Fourier transform; diamond operator; tempered distribution.
c
2010
Texas State University - San Marcos.
Submitted January 8, 2010. Published April 6, 2010.
1
(1.4)
2
W. SATSANIT
EJDE-2010/48
e
H
Kananthai [2] showed that the function u(x) = (−1)k R2k
(x) ∗ R2k
(x) is the
k
unique elementary solution for the operator ♦ , where ∗ indicates convolution, and
e
H
R2k
(x), R2k
(x) are defined by (2.5) and (2.2) with α = 2k respectively; that is,
e
H
♦k (−1)k R2k
(x) ∗ R2k
(x) = δ .
(1.5)
Furthermore, The operator ⊕k was first studied by Kananthai, Suantai and Longani
[4]. The ⊕k operator can be expressed in the form
⊕k =
p
p+q
p+q
p
h X
X
∂ 2 ik
∂ 2 2 X ∂ 2 2 ik h X ∂ 2
+
i
−
2
2
2
∂xi
∂xj
∂xi
∂x2j
i=1
j=p+1
j=p+1
i=1
p+q
p
hX
X
∂ 2 ik
∂2
−i
·
.
2
∂xi
∂x2j
j=p+1
i=1
The purpose of this work is to study the operator
p+q
p
X
∂ 2 4 X ∂ 2 4 k
k
−
⊕ =
∂x2i
∂x2j
j=p+1
i=1
(1.6)
p
p+q
p+q
p
X
∂ 2 2 X ∂ 2 2 k X ∂ 2 2 X ∂ 2 2 k
=
−
·
.
∂x2i
∂x2j
∂x2i
∂x2j
i=1
j=p+1
j=p+1
i=1
Let us denote the operator
}k =
p
p+q
X
∂ 2 2 X ∂ 2 2 k
+
.
∂x2i
∂x2j
i=1
j=p+1
By (1.1) and (1.4) we obtain
}k =
p
p+q
X
∂ 2 2 X ∂ 2 2 k
+
∂x2i
∂x2j
i=1
j=p+1
4 + 2
4 − 2 k
+
2
2
42 + 2 k
=
.
2
Thus, (1.6) can be written as
(1.7)
=
⊕k = ♦k }k .
(1.8)
For k = 1 the operator ♦ can be expressed in the form ♦ = 4 = 4 where is
the Ultra-hyperbolic operator
=
∂2
∂2
∂2
∂2
∂2
∂2
+
+
·
·
·
+
−
−
−
·
·
·
−
.
∂x21
∂x22
∂x2p
∂x2p+1
∂x2p+2
∂x2p+q
(1.9)
where p + q = n and 4 is the Lapacian operator
4=
∂2
∂2
∂2
+ 2 + ··· + 2 .
2
∂x1
∂x2
∂xn
(1.10)
By putting p = 1 and x1 = t(t = time) in (1.9), we obtain the wave operator
=
∂2
∂2
∂2
∂2
− 2 − 2 − ··· − 2 .
2
∂t1
∂x2
∂x3
∂xn
(1.11)
EJDE-2010/48
GREEN FUNCTION AND FOURIER TRANSFORM
3
From (1.6) with q = 0 and k = 1, we obtain
⊕ = 44p
(1.12)
where
4p =
∂2
∂2
∂2
+ 2 + ··· + 2.
2
∂x1
∂x2
∂xp
(1.13)
Firstly, we can find the elementary solution G(x) of the operator ⊕k ; that is,
⊕k G(x) = δ,
(1.14)
where δ is the Dirac-delta distribution. Moreover, we can find the relationship
between G(x) and the elementary solution of the wave operator defined by (1.11)
depending on the conditions of p, q and k of (1.6) with p = 1, q = n − 1, k = 1 and
x1 = t(t is time) and also we found that G(x) relates to the elementary solution
the Laplacian operator defined by (1.12) and (1.13) depending on the conditions of
q and k of (1.6) with q = 0 and k = 1. In finding the elementary solution of (1.6),
we use the method of convolutions of the generalized function. Finally, we study
the Fourier transform of the elementary solution of the ⊕k operator and also study
their convolution.
2. Preliminaries
Definition 2.1. Let x = (x1 , x2 , . . . , xn ) be a point of the n-dimensional Euclidean
space Rn . Denoted by
υ = x21 + x22 + · · · + x2p − x2p+1 − x2p+2 − · · · − x2p+q
(2.1)
the non-degenerated quadratic form, where p + q = n is the dimension the space
Rn .
Let Γ+ = {x ∈ Rn : x1 > 0 and u > 0} be the interior of a forward cone and Γ+
denotes it closure. For any complex number α, define the function
( α−n
υ 2
, for x ∈ Γ+ ,
H
(2.2)
Rα (υ) = Kn (α)
0,
for x 6∈ Γ+ ,
where the constant Kn (α) is given by the formula
Kn (α) =
π
n−1
2
Γ( 2+α−n
)Γ( 1−α
2
2 )Γ(α)
)Γ( p−α
Γ( 2+α−p
2
2 )
.
(2.3)
H
The function Rα
(υ) is called the Ultra-hyperbolic kernel of Marcel Riesz and
H
was introduced by Nozaki [5, p.72]. It is well known that Rα
(υ) is an ordinary
H
function if Re(α) ≥ n and is a distribution of α if Re(α) < n. Let supp Rα
(υ)
H
H
H
denote the support of Rα (υ) and suppose supp Rα (υ) ⊂ Γ̄+ , that is supp Rα (υ) is
compact.
If p = 1, then (2.2) reduces to the function
( α−n
υ 2
, for x ∈ Γ+ ,
H
Mα (υ) = Hn (α)
(2.4)
0,
for x 6∈ Γ+ ,
n−1
where υ = x21 − x22 − · · · − x2n and Hn (α) = π 2 2α−1 Γ( α−n+2
)Γ( α2 ). The function
2
H
Mα (υ) is called the hyperbolic kernel of Marcel Riesz.
4
W. SATSANIT
EJDE-2010/48
Definition 2.2. Let x = (x1 , x2 , . . . , xn ) be a point of Rn and |x| = (x21 + x22 +
· · · + x2n )1/2 . The elliptic kernel of Marcel Riesz is defined as
e
Rα
(x) =
where
|x|α−n
,
Wn (α)
(2.5)
n
Wn (α) =
π 2 2α Γ(α/2)
,
Γ n−α
2
(2.6)
with α a complex parameter and n the dimension of Rn .
e
It can be shown that R−2k
(x) = (−1)k 4k δ(x) where 4k is defined by (1.4). It
e
e
follows that R0 (x) = δ(x); see [3]. The function R2k
(x) is called the elliptic kernel
of Marcel Riesz and is ordinary function if Re(α) ≥ n and is a distribution of α for
Re(α) < n.
Definition 2.3. Let f (x) ∈ L1 (Rn ) (the space of integrable function in Rn ). The
Fourier transform of f (x) is defined as
Z
1
fd
(ξ) =
e−iξ·x f (x)dx
(2.7)
(2π)n/2 Rn
where ξ = (ξ1 , ξ2 , . . . , ξn ), x = (x1 , x2 , . . . , xn ) ∈ Rn , ξ · x = ξ1 x1 + ξ2 x2 + · · · + ξn xn
is the usual inner product in Rn and dx = dx1 dx2 . . . dxn . The inverse of Fourier
transform is defined by
Z
1
eiξ·x fd
(ξ)dξ.
(2.8)
f (x) =
(2π)n/2 Rn
If f is a distribution with compact supports by [8, Theorem 7.4-3], Equation (2.8)
can be written as
1
fb(ξ) =
hf (x), e−iξ·x i.
(2.9)
(2π)n/2
H
e
Lemma 2.4. The function R2k
(υ) and (−1)k R2k
(x) are the elementary solutions
k
k
of the operator and 4 respectively, where k and 4k are defined by (1.4) and
H
e
(1.3) respectively. The function R2k
(υ) defined by (2.2)with α = 2k and R2k
(x)
defined by (2.5) with α = 2k.
H
e
Proof. We have to show that k R2k
(υ) = δ(x) and that 4k ((−1)k R2k
(x) = δ(x).
The first part follows from [7, Lemma 2.4], and the second part from [2, p.31]. H
e
Lemma 2.5. The convolution R2k
(υ) ∗ (−1)k R2k
(x) is an elementary solution for
k
the operator ♦ iterated k times and is defined by (1.1)
For the proof of the above lemma, see [2, p.33].
H
e
Lemma 2.6. The function Rα
(x) and Rα
(x), defined by (2.2) and (2.5) respectively, for Re(α) are homogeneous distribution of order α − n and also a tempered
distribution.
H
e
Proof. Note taht Rα
(x) and Rα
(x) satisfy the Euler equation; that is,
H
(α − n)Rα
(x) =
n
X
i=1
n
xi
X
∂ e
∂ H
e
Rα (x), (α − n)Rα
(x) =
xi
Rα (x).
∂xi
∂x
i
i=1
EJDE-2010/48
GREEN FUNCTION AND FOURIER TRANSFORM
5
H
e
Then Rα
(x) and Rα
(x) are homogeneous distributions of order α − n. Since
Donoghue [1, pp.154-155] proved the every homogeneous distribution is a tempered
distribution, the proof is complete.
e
H
Lemma 2.7 (Convolution of tempered distributions). Rα
(x) ∗ Rα
(x) exists and is
a tempered distribution.
H
H
Proof. Choose supp Rα
(x) = K ⊂ Γ+ where K is a compact set. Then Rα
(x)
is a tempered distribution with compact support. By Donoghue [1, pp.156-159],
e
H
Rα
(x) ∗ Rα
(x) exists and is a tempered distribution.
H
e
Lemma 2.8. The functions R−2k
(x) and (−1)k R−2k
(x) are the inverse in the
H
k e
convolution algebra of R2k (x) and (−1) R2k (x), respectively. That is,
H
H
H
R−2k
(x) ∗ R2k
(x) = R−2k+2k
(x) = R0H (x) = δ(x),
e
e
e
(−1)k R−2k
(x) ∗ (−1)k R2k
(x) = (−1)2k R−2k+2k
(x) = R0e (x) = δ(x)
For the proof of the above lemma, see [7, p.123], [1, p.118, p.158], [6, p.10].
H
e
H
e
(x) defined by
(x) and Rα
(x)). Let Rα
(x) and Rα
Lemma 2.9 (Convolution of Rα
(2.5) and (2.2) respectively, then we obtain:
e
e
(x) ∗ Rβe (x) = Rα+β
(x) where α and β are complex parameters;
(1) Rα
H
H
(2) Rα
(x) ∗ RβH (x) = Rα+β
(x) for α and β are both integers and except only
the case both α and β are both integers.
Proof. Part (1) can be found in [1, p.158]. For the second formula, when α and β
are both even integers, see [3]. When α is odd and β is even, or α is even and β is
odd, from Trione [7] we have
H
H
k Rα
(x) = Rα−2k
(x),
k
H
R2k
(x)
= δ(x),
k = 0, 1, 2, 3, . . .
(2.10)
(2.11)
where k is the Ultra-hyperbolic operator iterated k-times defined by
p
p+q
X
X
∂2
∂ 2 k
k =
−
.
2
∂xi
∂x2j
i=1
j=p+1
Now let m be an odd integer, we have
H
H
k Rm
(x) = Rm−2k
(x),
H
H
H
H
R2k
(x) ∗ k Rm
(x) = R2k
(x) ∗ Rm−2k
(x)
or
H
H
H
H
H
H
H
(k R2k
(x)) ∗ Rm
(x) = R2k
(x) ∗ Rm−2k
(x), δ ∗ Rm
(x) = R2k
(x) ∗ Rm−2k
(x).
Thus
H
H
H
Rm
(x) = R2k
(x) ∗ Rm−2k
(x).
Since m is odd, hence m − 2k is odd and 2k is a positive even. Put α = 2k, β =
m − 2k, we obtain
H
H
Rα
(x) ∗ RβH (x) = Rα+β
(x)
for α is a nonnegative even and β is odd.
For the case α is a negative even and β is odd, by (2.8) we have
H
k R0H (x) = R−2k
(x),
6
W. SATSANIT
EJDE-2010/48
H
k δ = R−2k
(x),
where R0H (x) = δ. Now for m is odd,
H
H
H
H
R−2k
(x) ∗ k Rm
(x) = R−2k
(x) ∗ Rm−2k
(x)
or
H
H
H
k δ ∗ k Rm
(x) = R−2k
(x) ∗ Rm−2k
(x),
H
H
H
δ ∗ 2k Rm
(x) = R−2k
(x) ∗ Rm−2k
(x).
Thus
H
H
H
(x) = R−2k
(x) ∗ Rm−2k
(x).
Rm−2(2k)
Put α = −2k and β = m − 2k, now α is a negative even and β is odd. Then we
obtain
H
H
Rα
(x) ∗ RβH (x) = Rα+β
(x).
This completes the proof.
Lemma 2.10. Given the equation
}k H(x) = δ(x)
n
(2.12)
k
for x ∈ R , where } is the operator iterated k− times is defined by (1.7) and 4k is
the Laplace operator iterated k times defined by (1.4) and k is Ultra-hyperbolic operator iterated k− times is defined by (1.1). Then we obtain H(x) is an elementary
solution of (2.12), where
∗−1
H
e
H(x) = R4k
(x) ∗ (−1)2k R4k
(x) ∗ C ∗k (x)
(2.13)
where
1 H
1
R4 (x) + (−1)2 R4e (x) .
(2.14)
2
2
∗−1
Here C ∗k (x) denotes the convolution of C(x) itself k times, C ∗k (x)
denotes
∗k
the inverse of C (x) in the convolution algebra. Moreover H(x) is a tempered
distribution.
C(x) =
Proof. We have
}k H(x) =
42 + 2 k
H(x) = δ(x)
2
or we can write
1 2 1 2 1 2 1 2 k−1
4 + 4 + H(x) = δ(x).
2
2
2
2
Convolving both sides of the above equation by R4H (x) ∗ (−1)2 R4e (x),
1 2 1 2 k−1
1 2 1 2
4 + ∗ R4H (x) ∗ (−1)2 R4e (x)
4 + H(x)
2
2
2
2
H
2 e
= δ(x) ∗ R4 (x) ∗ (−1) R4 (x)
or
1 2 H
1
1
1 k−1
4 (R4 (x) ∗ (−1)2 R4e (x)) + 2 (R4 (x) ∗ (−1)2 S4 (x)) ∗ 42 + 2
H(x)
2
2
2
2
= δ(x) ∗ R4H (x) ∗ (−1)2 R4e (x).
By properties of convolutions,
1 2
1
4 ((−1)2 R4e (x)) ∗ R4H (x) + 2 (R4H (x))
2
2
EJDE-2010/48
GREEN FUNCTION AND FOURIER TRANSFORM
7
1
1 k−1
∗ (−1)2 R4e (x) ∗ 42 + 2
H(x)
2
2
= δ(x) ∗ R4H (x) ∗ (−1)2 R4e (x)
By Lemmas 2.4 and 2.5, we obtain
1
1
1
1 k−1
δ ∗ R4H (x) + δ ∗ (−1)2 R4e (x) ∗
42 + 2
H(x) = R4H (x) ∗ (−1)2 R4e (x)
2
2
2
2
or
1
1
1
1 k−1
R4H (x) + (−1)2 R4e (x) ∗
42 + 2
H(x) = R4H (x) ∗ (−1)2 R4e (x)
2
2
2
2
keeping on convolving both sides of the above equation by R4H (x) ∗ (−1)2 R4e (x) up
to k − 1 times, we obtain
∗k
C ∗k (x) ∗ H(x) = R4H (x) ∗ (−1)2 R4e (x)
(2.15)
H
(x)
the symbol ∗k denotes the convolution of itself k−times. By properties of Rα
e
and Rα (x) in Lemma 2.6, we have
∗k
H
e
R4H (x) ∗ (−1)2 R4e (x)
= R4k
(x) ∗ (−1)2k R4k
(x).
Thus (2.15) becomes,
H
e
C ∗k (x) ∗ H(x) = R4k
(x) ∗ (−1)2k R4k
(x)
or
H
e
H(x) = R4k
(x) ∗ (−1)2k R4k
(x) ∗ (C ∗k (x))∗−1
(2.16)
∗k
R4H (x) ∗
is an elementary solution of (2.12). We consider the function C (x), since
(−1)2 R4e (x) is a tempered distribution. Thus C(x) defined by (2.14) is a tempered
distribution, we obtain C ∗k (x) is a tempered distribution.
H
e
Now, R4k
(x) ∗ (−1)2k R4k
(x) ∈ S 0 , the space of tempered distribution. Choose
0
0
0
S ⊂ DR where DR is the right-side distribution which is a subspace of D0 of disH
e
H
e
0
. It follow that R4k
(x) ∗ (−1)2k R4k
(x)
tribution. Thus R4k
(x) ∗ (−1)2k R4k
(x) ∈ DR
0
is an element of convolution algebra, since DR is a convolution algebra. Hence
Zemanian [8], (2.16) has a unique solution
∗−1
H
e
H(x) = R4k
(x) ∗ (−1)2k R4k
(x) ∗ C ∗k (x)
,
∗−1
where C ∗k (x)
is an inverse of C ∗k (x) in the convolution algebra. H(x) is
called the Green function of the operator }k .
∗−1
H
e
Since R4k
(x) ∗ (−1)2k R4k
(x) and C ∗k (x)
are lies in S 0 , then by [8, p.152]
∗−1
H
e
again, we have R4k
∈ S 0 . Hence H(x) is a
(x) ∗ (−1)2k R4k
(x) ∗ C ∗k (x)
tempered distribution.
Lemma 2.11. The Fourier transform of ⊕k δ is
h
4 ik
1
2
2
2 4
2
2
2
(ξ
+
ξ
+
·
·
·
+
ξ
)
−
ξ
+
ξ
+
·
·
·
+
ξ
F ⊕k δ =
1
2
p
p+1
p+2
p+q
(2π)n/2
where F is the Fourier transform defined by (2.7) and if the norm of ξ is given by
1/2
kξk = ξ12 + ξ22 + · · · + ξn2
then
F ⊕k δ ≤
M
kξk8k .
(2π)n/2
8
W. SATSANIT
EJDE-2010/48
Since M is constant thus is F ⊕k δ is bounded and continuous on the space S 0 of
the tempered distribution. Moreover, by (2.8),
h
4 ik
1
2
2
2 4
2
2
2
⊕k δ = F −1
(ξ
.
+
ξ
+
·
·
·
+
ξ
)
−
ξ
+
ξ
+
·
·
·
+
ξ
1
2
p
p+1
p+2
p+q
(2π)n/2
Proof. By (2.10)
1
h⊕k δ, e−iξ·x i
(2π)n/2
1
=
hδ, ⊕k e−iξ·x i
(2π)n/2
1
hδ, ♦k }k e−iξ·x i by (1.6)
=
(2π)n/2
1 1 2 k −iξ·x k 1 2
=
δ,
♦
4
+
e
2
2
(2π)n/2
2
(−1)2k
1 2
( ξ12 + . . . ξn2 + ξ12 + · · · + ξp2 − ξp+1
− ...
δ, ♦k
=
n/2
2
(2π)
2
− ξn2 )k e−iξ·x
F ⊕k δ =
2
1
(−1)2k
2
hδ,
( ξ12 + · · · + ξn2 + ξ12 + · · · + ξp2 − ξp+1
− ...
n/2
2
(2π)
2
− ξn2 )k ♦k e−iξ·x i
2
2 k k −iξ·x
1
2
2
hδ, ( ξ12 + · · · + ξp2 + ξp+1
+ · · · + ξp+q
)4 e
i
=
n/2
(2π)
p
p+q
p
h X
2
1 h X 2 2 X 2 2 ik
2k
ξ
δ,
+
ξ
(−1)
ξi2
=
i
j
n/2
(2π)
i=1
j=p+1
i=1
=
−
p+q
X
ξj2
2 ik
e−iξ·x
j=p+1
=
p
p+q
h X
4 X
4 ik
1
2k
2
2
hδ,
(−1)
ξ
−
ξ
e−iξ·x i
i
j
(2π)n/2
i=1
j=p+1
p
p+q
1 h X 2 4 X 2 4 ik
ξ
−
ξj
.
=
(2π)n/2 i=1 i
j=p+1
=
4
4 k
1 2
2
2
2
ξ1 + ξ22 + · · · + ξp2 − ξp+1
+ ξp+2
+ · · · + ξp+q
.
n/2
(2π)
Next, we consider the boundness of F ⊕k δ. Since
4 k
2
2
2
⊕k = (ξ12 + ξ22 + · · · + ξp2 )4 − ξp+1
+ ξp+2
+ · · · + ξp+q
2 k 2
2
2
2
= ξ12 + · · · + ξp2 − ξp+1
+ · · · + ξn2
ξ1 + · · · + ξp2
2 k
2
+ ξp+1
+ · · · + ξn2
k 2
2
2
= ξ12 + · · · + ξn2 ξ12 + · · · + ξp2 − ξp+1
. . . ξn2
ξ1 + · · · + ξp2
EJDE-2010/48
GREEN FUNCTION AND FOURIER TRANSFORM
2
+ ξp+1
+ · · · + ξn2
9
2 k
Thus
F ⊕k δ =
|F ⊕k δ| =
≤
k
1
2
ξ12 + · · · + ξn2 ξ12 + · · · + ξp2 − ξp+1
· · · − ξn2
(2π)n/2
2 k
2
2
2
× ξp+1
+ · · · + ξn2
,
+ · · · + ξn2 + ξp+1
1
2
| ξ12 + · · · + ξn2 ξ12 + · · · + ξp+1
· · · − ξn2 |k
n/2
(2π)
2 k
2
2
2
+ · · · + ξn2 × ξp+1
+ · · · + ξn2 + ξp+1
k k 2k
M 2
ξ1 + · · · + ξn2 ξ12 + · · · + ξn2 ξ12 + · · · + ξn2 .
n/2
(2π)
It follows that
M
kξk8k ,
(2π)n/2
1/2
, ξi (i = 1, 2, . . . , n) ∈ R.
where M is constant and kξk = ξ12 + ξ22 + · · · + ξn2
k
Hence we obtain F ⊕ δ is bounded and continuous on the space S 0 of the tempered distribution. Since F is one-to-one transformation from the space S 0 of the
tempered distribution to the real space R, then by (2.8)
h
4 i
1
2
2
2 4
2
2
2
⊕δ = F −1
(ξ
+
ξ
+
·
·
·
+
ξ
)
−
ξ
+
ξ
+
·
·
·
+
ξ
.
1
2
p
p+1
p+2
p+q
(2π)n/2
|F ⊕k δ| ≤
That completes the proof.
3. Main Results
Theorem 3.1. Given the equation
⊕k G(x) = δ(x),
(3.1)
k
where ⊕ is the Oplus operator iterated k times defined by (1.8), δ(x) is the diracdelta distribution, x ∈ Rn and k is a nonnegative integer. Then we obtain
H
e
G(x) = R2k
(υ) ∗ (−1)k R2k
(x) ∗ H(x)
(3.2)
or by (2.13) and Lemma 2.10, we obtain
∗−1
H
e
G(x) = R6k
(υ) ∗ (−1)3k R6k
(x) ∗ C ∗k (x)
(3.3)
is a Green’s function or an elementary solution for the operator ⊕k iterated k−times
where ⊕k is defined by (1.8), and H(x) defined by (2.13).
For q = 0, then(3.1) becomes
44k G(x) = δ(x)
(3.4)
By Lemma 2.4, we obtain
e
e
G(x) = (−1)4k R8k
(x) = R8k
(x)
is an elementary solution of (3.4) where 44k
p is the Laplacian of p− dimension,
iterated 4k− times and is defined by (1.13). Moreover, from (3.3), we obtain
H
e
R−4k
(υ) ∗ (−1)3k R−6k
(x) ∗ C ∗k (x) ∗ G(x)
10
W. SATSANIT
EJDE-2010/48
H
H
e
e
= R4k
(υ) ∗ R−4k
(υ) ∗ (−1)3k R6k
(x) ∗ (−1)3k R−6k
(x)
∗−1 H
∗ C ∗k (x) ∗ C ∗k (x)
∗ R2k
(υ).
By (2.14) the above equation becomes
(−1)3k
(−1)5k e
e
H
H
∗ R−6k
(x) + R−4k
(υ) ∗
R−2k (x) ∗ G(x) = R2k
(υ)
(3.5)
2
2
as an elementary solution of the operator k times is defined by (1.4) In particular,
if we put p = 1, q = n − 1, k = 1 and x1 = t in (1.4),(3.3), we obtain
(−1)3
(−1)5 e
e
H
R−6
R−6 (x) ∗ G(x) = M2H (υ),
(x) + M−2
(υ) ∗
(3.6)
2
2
as an elementary solution of the wave operator defined by
=
∂2
∂2
∂2
∂2
− 2 − 2 − ··· − 2 ,
2
∂t1
∂x2
∂x3
∂xn
(3.7)
H
where M2H (υ) and M−4
(υ) defined by (2.4) with α = 2 and α = −4 respectively,
e
2
2
(x) is defined by (2.5) with α = −6.
and υ = t1 − x2 − · · · − x2n . The function R−6
Proof. From (3.1) and (1.8), we have
⊕k G(x) = ♦k }k G(x) = δ(x).
(3.8)
H
k e
Convolving both sides of (3.8) by R2k (υ) ∗ (−1) R2k (x) ∗ H(x), we obtain
H
e
H
e
R2k
(υ) ∗ (−1)k R2k
(x) ∗ H(x) ∗ (♦k }k )G(x) = δ ∗ R2k
(υ) ∗ (−1)k R2k
(x) ∗ H(x)
By properties of convolution
H
e
H
e
♦k R2k
(υ) ∗ (−1)k R2k
(x) ∗ }k (H(x)) ∗ G(x) = R2k
(υ) ∗ (−1)k R2k
(x) ∗ H(x).
By Lemma 2.5 and 2.10, we obtain,
H
e
δ ∗ δ ∗ G(x) = G(x) = R2k
(x) ∗ (−1)k R2k
(υ) ∗ H(x).
By Lemma 2.9 and (2.13), we obtain
∗−1
H
e
G(x) = R6k
(υ) ∗ (−1)3k R6k
(x) ∗ C ∗k (x)
(3.9)
k
is an elementary solution or Green’s function of ⊕ operator. Now, for q = 0 the
(3.1) becomes
44k
(3.10)
p G(x) = δ, .
where 44k
p is the Laplacian operator of p-dimension iterated 4k times. By Lemma
2.4, we have
e
G(x) = (−1)4k R8k
(x)
is an elementary solution of (3.10).
On the other hand, we can also find G(x) from (3.9). Since q = 0, we have
H
e
R2k
(υ) reduces to (−1)k R2k
(x). Thus, by (3.9) for q = 0, we obtain
∗−1
6k e
e
e
G(x) = (−1) R6k (x) ∗ R6k
(x) ∗ (−1)2k R4k
(x)
∗−1
e
e
= (−1)6k R6k+6k
(x) (−1)2k R4k
(x)
e
= R8k
(x).
Now, consider the case of the wave equation. From (3.3), we have
∗−1
H
e
G(x) = R6k
(υ) ∗ (−1)3k R6k
(x) ∗ C ∗k (x)
.
EJDE-2010/48
GREEN FUNCTION AND FOURIER TRANSFORM
11
H
e
Convolving the above equation by R−4k
(υ) ∗ (−1)3k R−6k
(x) ∗ C ∗k (x) and by
Lemma 2.8, we obtain
H
e
R−4k
(υ) ∗ (−1)3k R−6k
(x) ∗ C ∗k (x) ∗ G(x)
H
H
e
e
= R4k
(υ) ∗ R−4k
(υ) ∗ (−1)3k R6k
(x) ∗ (−1)3k R−6k
(x)
∗−1 H
∗ C ∗k (x) ∗ C ∗k (x)
∗ R2k
(υ).
By Lemma 2.8, we obtain
H
e
H
R−4k
(υ) ∗ (−1)3k R−6k
(x) ∗ C ∗k (x) ∗ G(x) = R0H (x) ∗ R0e (x) ∗ δ(x) ∗ R2k
(υ)
or
H
e
H
R−4k
(υ) ∗ (−1)3k R−6k
(x) ∗ C ∗k (x) ∗ G(x) = δ(x) ∗ δ(x) ∗ δ(x) ∗ R2k
(υ).
It follows that
H
e
H
R−4k
(υ) ∗ (−1)3k R−6k
(x) ∗ C ∗k (x) ∗ G(x) = R2k
(υ)
(3.11)
k
as an elementary solution of the operator iterated k times defined by (1.4). In
particular , if we put p = 1, q = n − 1, k = 1 and x1 = t in (3.3) and (3.9) then
H
H
H
R−4
reduces to M−4
(υ) and R2H (υ) reduces to M2H (υ)where M−4
(υ) and M2H (υ)
is defined by (2.4) with α = −4, α = 2 respectively. Thus (3.11) becomes
H
e
M−4
(υ) ∗ (−1)3 R−6
(x) ∗ C ∗1 (x) ∗ G(x) = M2H (υ)
(3.12)
by (2.14), we obtain
H
e
M−4
(υ) ∗ (−1)3 R−6
(x) ∗
1
2
M4H (υ) +
(−1)2 e R4 (x) ∗ G(x) = M2H (υ)
2
or
1
H
e
H
M−4
(υ) ∗ (−1)3 R−6
(x) ∗ M4H (υ) + M−4
(υ)
2
1
e
∗ (−1)3 R−6
(x) ∗ (−1)2 R4e (x) ∗ G(x)
2
= M2H (υ).
By Lemma 2.8, we obtain
(−1)3
(−1)5 e
e
H
R−6
(x) + M−4
(υ) ∗
R−2 (x) ∗ G(x) = M2H (υ)
2
2
as an elementary solution of the wave operator defined by
=
(3.13)
∂2
∂2
∂2
∂2
−
−
−
·
·
·
−
,
∂t21
∂x22
∂x23
∂x2n
e
and R−6
(x) defined by (2.5) with α = −6. This completes the proof.
Theorem 3.2.
H
e
R6k
(x) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1
4 k
1 2
2
2
2
=
(ξ1 + ξ22 + · · · + ξp2 )4 − ξp+1
+ ξp+2
+ · · · + ξp+q
n/2
(2π)
F
and
|F
H
e
R6k
(x) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 | ≤
1
n M
(2π) 2
(3.14)
12
W. SATSANIT
EJDE-2010/48
for a large ξi ∈ R, where M is a constant and C(x) is defined by (2.14). That is,
F is bounded and continuous on the space S 0 of the tempered distributions.
Proof. By Theorem 3.1, we have
H
e
⊕k ( R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 = δ(x)
or
H
e
⊕k δ ∗ ( R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 = δ(x).
Taking the Fourier transform on both sides of the above equation, we obtain
1
H
e
F( ⊕k δ ∗ [ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 ]) = Fδ =
.
(2π)n/2
By (2.10)
1
1
H
e
h( ⊕k δ ∗ [ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 ]), e−i(ξ·x) i =
.
n/2
(2π)
(2π)n/2
By the definition of convolution
1
H
e
h( ⊕k δ ∗ [ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 ]), e−iξ.(x+r) i
n/2
(2π)
1
,
=
(2π)n/2
1
H
e
h[ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 ]), e−i(ξ.r) ih( ⊕k δ , e−iξ·x i
n/2
(2π)
1
,
=
(2π)n/2
n
1
H
e
,
F([ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 ]) (2π) 2 F ⊕k δ =
n/2
(2π)
H
e
F([ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (S ∗k (x))∗−1 ])
h
4 ik
2
2
2
· (−1)3k (ξ12 + ξ22 + · · · + ξp2 )4 − ξp+1
+ ξp+2
+ · · · + ξp+q
1
=
n/2
.
(2π)
It follows that
H
e
F([ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 ])
1
=
h
4 ik .
n/2
2
2
2
(−1)4k (2π)
(ξ12 + ξ22 + · · · + ξp2 )4 − ξp+1
+ ξp+2
+ · · · + ξp+q
Since
1
4
2
2
− ξp+1
+ · · · + ξp+q
1
=
2
2 2
2
2
2
ξ1 + · · · + ξp + ξp+1
+ · · · + ξp+q
ξ12
+ · · · + ξp2
×
(ξ12
4
+ ··· +
ξn2 )
ξ12
1
.
2
2
+ · · · + ξp2 − ξp+1
− · · · − ξp+q
(3.15)
EJDE-2010/48
GREEN FUNCTION AND FOURIER TRANSFORM
13
Let ξ = (ξ1 , ξ2 , . . . , ξn ) ∈ Γ+ with Γ+ defined by Definition 2.1. Then (ξ12 + · · · +
2
2
ξp2 +ξp+1
+ · · · + ξp+q
) > 0 and for a large k, the right-hand side of (3.15) tend
to zero. It follows that it is bounded by a positive constant M say, that is we
obtain (3.15) as required and also by (3.13) F is continuous on the space S 0 of the
tempered distribution.
Theorem 3.3.
H
e
F [ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 ]
H
e
∗ [ R6m
(υ) ∗ (−1)3m R6m
(x) ∗ (C ∗m (x))∗−1 ]
H
e
= (2π)n/2 F [ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 )
H
e
× F([ R6m
(υ) ∗ (−1)3m R6m
(x) ∗ (C ∗m (x))∗−1 ])
1
=
h
4 ik+m ,
2
2
2
(2π)n/2 (ξ12 + ξ22 + · · · + ξp2 )4 − ξp+1
+ ξp+2
+ · · · + ξp+q
where k and m are nonnegative integer and F is bounded and continuous on the
space S 0 of the tempered distribution.
H
e
Proof. Since R6k
(υ) and R4k
(x) are tempered distribution with compact support,
H
e
[ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 )
H
e
∗ [ R6m
(υ) ∗ (−1)3m R6m
(x) ∗ (C ∗m (x))∗−1 ]
H
H
e
e
= R6k
(υ) ∗ R6m
(υ) ∗ (−1)3(k+m) R6k
(x) ∗ R6m
(x)
∗−1
∗−1
∗ ((C ∗m (x))
∗ C ∗k (x)
∗−1
H
e
= R6(k+m)
(υ) ∗ (−1)3(k+m) R6(k+m)
(x) ∗ C ∗(k+m) (x)
by [7, pp.156-159] and [3, Lemma 2.5]. Taking the Fourier transform on both sides
and using Theorem 3.2, we obtain
H
e
F R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x)∗−1 )
H
e
∗ [ R6m
(υ) ∗ (−1)3m R6m
(x) ∗ (C ∗m (x))∗−1 ]
1
=
n/2
h
n/2
h
(2π)
2
2
2
+ ξp+2
+ · · · + ξp+q
(ξ12 + ξ22 + · · · + ξp2 )4 − ξp+1
4 ik+m
1
=
(2π)
2
2
2
(ξ12 + ξ22 + · · · + ξp2 )4 − ξp+1
+ ξp+2
+ · · · + ξp+q
(2π)n/2
4
4 im
n/2
2
2
(2π)
ξ12 + · · · + ξp2 − ξp+1
+ · · · + ξp+q
H
e
= (2π)n/2 F [ R6k
(υ) ∗ (−1)3k R6k
(x) ∗ (C ∗k (x))∗−1 )
H
e
× F([ R6m
(υ) ∗ (−1)3m R6m
(x) ∗ (C ∗m (x))∗−1 ]).
×
h
4 ik
14
W. SATSANIT
EJDE-2010/48
H
e
Since (R6(k+m)
(υ) ∗ (−1)3(k+m) R6(k+m)
(x)) ∗ (C ∗(k+m) (x))∗−1 ∈ S 0 , the space of
tempered distribution and by Theorem 3.2, we obtain that F is bounded and continuous on S 0 .
Acknowledgements. The authors would like to thank Prof. Amnuay Kananthai, Department of Mathematics, Faculty of Science, Chiang Mai University, for
many helpful discussion and also to the Graduate School, Maejo University for the
financial support.
References
[1] W. F. Donoghue; Distributions and Fourier Transform, Academic Press, New York,(1969)
[2] A. Kananthai; On the Solution of the n-Dimensional Diamond Operator, Applied Mathematics and Computational, Elsevier Science Inc., New York, 1997, pp 27-37
[3] A. Kananthai; On the distribution related to the Ultra-hyperbolic equation, Applied Mathematics and Computational, Elsevier Science Inc., New York, 1997, pp. 101-106.
[4] A. Kananthai, S. Suantai, V. Longani; On the operator ⊕K related to the wave equation and
Laplacian, Applied Mathematics and Computational,Elsevier Science Inc., New York, (2002),
pp 219-229
[5] Y. Nozaki; On Riemann-Liouville integral of Ultra-hyperbolic type, Kodai Math. Sem. Rep.
6(2)(1964)69-87.
[6] S. E. Trione; On the Ultra-hyperbolic kernel, Trabajos de Mathematica, 116 preprint (1987).
[7] S. E. Trione, M. Aguirre Tellez; The distribution convolution products of Marcel Riesz’s
Ultra-hyperbolic kernel, Ravista de la Union Mathematica Argentina 39(1995) 115-124.
preprint(1987).
[8] A. H. Zemanian; Distribution Theory and Transform Analysis, McGraw-Hill, New York,
(1965).
Wanchak Satsanit
Faculty of Science, Department of Mathematics, Maejo University
Nong Han, San Sai, Chiang Mai, 50290 Thailand
E-mail address: aunphue@live.com