\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{amssymb}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 48, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/48\hfil Green function and Fourier transform]
{Green function and Fourier transform  for o-plus operators}

\author[W. Satsanit\hfil EJDE-2010/48\hfilneg]
{Wanchak Satsanit}

\address{Wanchak Satsanit \newline
Faculty of Science, Department of Mathematics, Maejo University \newline
Nong Han, San Sai,
Chiang Mai, 50290 Thailand}
\email{aunphue@live.com}


\thanks{Submitted January 8, 2010. Published April 6, 2010.}
\subjclass[2000]{46F10, 46F12}
\keywords{Fourier transform; diamond operator; tempered distribution}

\begin{abstract}
 In this article, we study the o-plus operator defined by
 \[
  \oplus^k =\Big(\Big(\sum^{p}_{i=1}\frac{\partial^2}{\partial
 x^2_i}\Big)^{4}-\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
 x^2_j}\Big)^{4}\Big)^k ,
 \]
 where $x=(x_1,x_2,\dots,x_n)\in \mathbb{R}^n$, $p+q=n$,
 and $k$ is a nonnegative integer. Firstly, we studied the
 elementary solution for the $\oplus^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.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks


 \section{Introduction}
Consider the ultra-hyperbolic operator iterated $k$ times,
\begin{equation}\label{1.1}
  \Box^k  = \Big(\frac{\partial^2}{\partial x_1^2}+
\frac{\partial^2}{\partial x_2^2}+\dots+\frac{\partial^2}{\partial
x_p^2}-\frac{\partial^2}{\partial
x_{p+1}^2}-\frac{\partial^2}{\partial
x_{p+2}^2}-\dots-\frac{\partial^2}{\partial x_{p+q}^2}\Big)^k .
\end{equation}
 Trione \cite{t1}  showed that the
generalized function $R^{H}_{2k}(x)$, defined by \eqref{2.1} below,
is the unique elementary solution for the $\Box^k $ operator, that is
$\Box^k R^{H}_{2k}(x)=\delta$ for $x\in \mathbb{R}^{n}$, the
$n$-dimensional Euclidian space.

Kananthai \cite{k1} studied the Diamond operator, iterated $k$
times,
\begin{equation}\label{1.2}
\diamondsuit^k =\Big(\Big(\sum^{p}_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^2 -\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^2\Big)^k ,
\end{equation}
for $x=(x_1,x_2,\dots,x_n)\in \mathbb{R}^n$, where $p+q=n$, $n$
is the dimension of Euclidean space $\mathbb{R}^n$,  and
$k$ is a nonnegative integer. The operator $\diamondsuit^k $
can be expressed in the form
\begin{equation}\label{1.3}
\diamondsuit^k =\triangle^k \Box^k =\Box^k \triangle^k
\end{equation}
 where $\triangle^k $ is the Laplacian operator iterated $k$ times,
\begin{equation}\label{1.4}
  \triangle^k  = \Big(\frac{\partial^2}{\partial
x^2_1}+\frac{\partial^2}{\partial
x^2_2}+\dots+\frac{\partial^2}{\partial x^2_n}\Big)^k\,.
\end{equation}

Kananthai \cite{k1}  showed that the function
$u(x)=(-1)^k R^{e}_{2k}(x)\ast R^{H}_{2k}(x)$ is the
unique elementary solution for the operator $\diamondsuit^k $, where
$\ast$ indicates convolution, and
$R^{e}_{2k}(x)$, $R^{H}_{2k}(x)$ are defined by \eqref{2.5} and
\eqref{2.2} with $\alpha=2k$ respectively; that is,
\begin{equation}\label{1.5}
\diamondsuit^k \left((-1)^k R^{e}_{2k}(x)\ast
R^{H}_{2k}(x)\right)=\delta\,.
\end{equation}
Furthermore, The operator $\oplus^k $ was first studied by
Kananthai,  Suantai and  Longani \cite{k3}. The $\oplus^k $ operator can
be expressed in the form
\begin{align*}
\oplus^k &=\Big[\Big(\sum^{p}_{i=1}\frac{\partial^2}{\partial
x_i^2}\Big)^2
            -\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
            x_j^2}\Big)^2\Big]^k
 \Big[\sum^{p}_{i=1}\frac{\partial^2}{\partial x_i^2}
            +i\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
            x_j^2}\Big]^k\\
&\quad \cdot \Big[\sum^{p}_{i=1}\frac{\partial^2}{\partial x_i^2}
            -i\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x_j^2}\Big]^k.
\end{align*}
The purpose of this work is to study the operator
\begin{equation} \label{1.6}
\begin{aligned}
\oplus^k &=\Big(\Big(\sum^{p}_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^4 -\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^4\Big)^k  \\
&= \Big(\Big(\sum^{p}_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^2 -\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^2\Big)^k \cdot \Big(\Big(\sum^{p}_{i=1}
\frac{\partial^2}{\partial x^2_i}\Big)^2
\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^2\Big)^k .
\end{aligned}
\end{equation}
Let us denote the operator
\[
\circledcirc^k =\Big(\Big(\sum^{p}_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^2 +\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^2\Big)^k .
\]
By \eqref{1.1} and \eqref{1.4} we obtain
\begin{equation} \label{1.7}
\begin{aligned}
\circledcirc^k
&=\Big(\Big(\sum^{p}_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^2 +\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^2\Big)^k  \\
&=\big[\big(\frac{\triangle+\Box}{2}\big)^{2}
+\big(\frac{\triangle-\Box}{2}\big)^{2}\big]^k  \\
&= \big(\frac{\triangle^2+\Box^2}{2}\big)^k .
\end{aligned}
\end{equation}
Thus,  \eqref{1.6} can be written as
\begin{equation}\label{1.8}
\oplus^k =\diamondsuit^k \circledcirc^k\,.
\end{equation}
For $k=1$ the operator $\diamondsuit$ can be expressed in the
form $\diamondsuit=\triangle\Box=\Box\triangle$ where $\Box$ is the
Ultra-hyperbolic operator
\begin{equation}\label{1.9}
  \Box = \frac{\partial^2}{\partial x_1^2}+
\frac{\partial^2}{\partial x_2^2}+\dots+\frac{\partial^2}{\partial
x_p^2}-\frac{\partial^2}{\partial
x_{p+1}^2}-\frac{\partial^2}{\partial
x_{p+2}^2}-\dots-\frac{\partial^2}{\partial x_{p+q}^2}.
\end{equation}
where $p+q=n$ and $\triangle$ is the Lapacian operator
\begin{equation}\label{1.10}
  \triangle = \frac{\partial^2}{\partial
x^2_1}+\frac{\partial^2}{\partial
x^2_2}+\dots+\frac{\partial^2}{\partial x^2_n}.
\end{equation}
By putting $p=1$ and $x_{1}=t(t=time)$ in \eqref{1.9}, we obtain
the wave operator
\begin{equation}\label{1.11}
  \Box = \frac{\partial^2}{\partial t_1^2}
-\frac{\partial^2}{\partial x_{2}^2}-\frac{\partial^2}{\partial
x_{3}^2}-\dots-\frac{\partial^2}{\partial x_{n}^2}\,.
\end{equation}
 From \eqref{1.6} with $q=0$ and $k=1$, we obtain
\begin{equation}\label{1.12}
\oplus=\triangle^{4}_{p}
\end{equation}
where
\begin{equation}\label{1.13}
  \triangle_{p} = \frac{\partial^2}{\partial
x^2_1}+\frac{\partial^2}{\partial
x^2_2}+\dots+\frac{\partial^2}{\partial x^2_p}.
\end{equation}

 Firstly, we can find the elementary solution $G(x)$ of the
operator $\oplus^k $; that is,
\begin{equation}\label{1.14}
\oplus^k G(x)=\delta,
\end{equation}
 where $\delta$ is the Dirac-delta  distribution.
Moreover, we can find the relationship between $G(x)$
and the elementary solution of the wave operator defined by
\eqref{1.11} depending on the conditions of $p,q$ and $k$ of
\eqref{1.6} with $p=1,q=n-1,k=1$ and $x_{1}=t$(t is time) and also
we found that $G(x)$ relates to the elementary solution  the
Laplacian operator defined by \eqref{1.12} and \eqref{1.13}
depending on the conditions of $q$ and $k$ of \eqref{1.6} with $q=0$
and $k=1$. In finding the elementary solution of \eqref{1.6}, we use
the method of convolutions of the generalized function.
Finally, we study the Fourier transform of the elementary
solution of the $\oplus^k $  operator and also study their convolution.

\section{Preliminaries}

\begin{definition} \label{def2.1} \rm
Let $x = ( x_1, x_2, \dots , x_n )$ be a point   of the
$n$-dimensional Euclidean space $\mathbb{R}^n$.  Denoted by
\begin{equation}\label{2.1}
\upsilon = x_1^2 + x_2^2 + \dots  + x_p^2 - x_{p+1}^2- x_{p+2}^2 - \dots
- x_{p+q}^2
\end{equation}
the non-degenerated quadratic form, where  $p + q = n$ is the dimension
 the space $\mathbb{R}^n$.

Let  $\Gamma_+ = \{ x \in \mathbb{R}^n : x_1 > 0 $ and $u > 0 \}$ be the
interior of a forward cone and $\overline{\Gamma}_+$ denotes it
closure. For any complex number $\alpha$, define the function
\begin{equation}\label{2.2}
R_\alpha^H (\upsilon) = \begin{cases}
\frac{\upsilon^{\frac{\alpha - n}{2}}}{K_n (\alpha)},
&\text{for } x \in \Gamma_+, \\
0, &\text {for } x \not\in \Gamma_+,
\end{cases}
\end{equation}
where the constant $K_n (\alpha)$ is given by the formula
\begin{equation}\label{2.3}
K_n (\alpha) = \frac{\pi^{\frac{n-1}{2}} \Gamma ( \frac{2 + \alpha -
n}{2} ) \Gamma ( \frac{1 - \alpha}{2}) \Gamma (\alpha)}{\Gamma (
\frac{2 + \alpha - p}{2} ) \Gamma ( \frac{p - \alpha}{2})}.
\end{equation}
\end{definition}

 The function $R_{\alpha}^H(\upsilon)$ is called the Ultra-hyperbolic
kernel of Marcel
Riesz and was introduced by  Nozaki \cite[p.72]{n1}.
It is well known that $R_\alpha^H(\upsilon)$ is an ordinary function
if $Re (\alpha) \ge n$ and is a distribution of $\alpha$ if
$\mathop{\rm Re}(\alpha) < n$.
Let $\mathop{\rm supp }R_\alpha^H (\upsilon)$ denote the support
of $R_\alpha^H (\upsilon)$ and suppose
$\mathop{\rm supp}R_\alpha^H (\upsilon)\subset \bar{\Gamma}_+$, that
is $\mathop{\rm supp}R_\alpha^H (\upsilon)$ is compact.

If  $p=1$, then \eqref{2.2} reduces to the function
\begin{equation}\label{2.4}
M_\alpha^H (\upsilon) = \begin{cases}
\frac{\upsilon^{\frac{\alpha - n}{2}}}{H_n (\alpha)},
&\text {for } x \in \Gamma_+, \\
0, &\text{for } x \not\in \Gamma_+,
\end{cases}
\end{equation}
where $\upsilon=x^{2}_{1}-x^{2}_{2}-\dots -x^{2}_{n}$ and
$H_{n}{(\alpha)}=\pi^{\frac{n-1}{2}}2^{\alpha-1}
\Gamma(\frac{\alpha-n+2}{2})\Gamma(\frac{\alpha}{2})$.
The function $M^{H}_{\alpha}(\upsilon)$ is called the hyperbolic
kernel of Marcel Riesz.

\begin{definition} \label{def2.2} \rm
Let $x = ( x_1, x_2, \dots , x_n )$ be a point of $\mathbb{R}^n$
and $|x| = (x_1^2 + x_2^2 + \dots  + x_n^2)^{1/2}$.
The elliptic kernel of Marcel Riesz  is defined as
\begin{equation}\label{2.5}
R_\alpha^e (x) =\frac{ |x|^{\alpha-n}}{W_n(\alpha)},
\end{equation}
where
\begin{equation}\label{2.6}
W_n(\alpha) =\frac{
\pi^{\frac{n}{2}}2^{\alpha}\Gamma(\alpha/2)}
{\Gamma\big(\frac{n-\alpha}{2}\big)},
\end{equation}
with $\alpha$  a complex parameter and $n$  the dimension of
$\mathbb{R}^n$.
\end{definition}

 It can be shown that
$R^{e}_{-2k}(x)=(-1)^k \triangle^k \delta(x)$ where
$\triangle^k $ is defined by \eqref{1.4}. It follows that
$R^{e}_{0}(x)=\delta(x)$; see \cite{k2}. The function $R^{e}_{2k}(x) $ is
called the elliptic kernel of Marcel Riesz and is ordinary function
if $\mathop{\rm Re}(\alpha)\geq n$ and is a distribution of $\alpha$ for
$\mathop{\rm Re}(\alpha)<n$.

\begin{definition} \label{def2.3} \rm
Let $f(x) \in L_{1}(\mathbb{R}^n)$ (the space of integrable function in
$\mathbb{R}^n$).
The Fourier transform of $f(x)$ is defined as
\begin{equation}\label{2.7}
\widehat{f(\xi)}=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}
e^{-i\xi\cdot x}f(x)dx
\end{equation}
where $\xi=(\xi_1,\xi_2,\dots,\xi_n)$,
$x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$,
$\xi\cdot x=\xi_1x_1+\xi_2x_2+\dots+\xi_nx_n$ is the usual inner
product in $\mathbb{R}^n$ and $dx=dx_1dx_2 \dots dx_n$.
The inverse of Fourier transform is defined by
\begin{equation}\label{2.8}
f(x)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{i\xi\cdot x}
\widehat{f(\xi)}d\xi.
\end{equation}
If $f$ is a distribution with compact supports by
\cite[Theorem 7.4-3]{z1}, Equation \eqref{2.8} can be written as
\begin{equation}\label{2.9}
\widehat{f}(\xi)=\frac{1}{(2\pi)^{n/2}}\langle f(x),
e^{-i\xi\cdot x}\rangle.
\end{equation}
\end{definition}

\begin{lemma} \label{lem2.1}
The function $R^{H}_{2k}(\upsilon)$ and
$(-1)^k R^{e}_{2k}(x)$ are the elementary solutions of the operator
$\Box^k $ and $\triangle^k $ respectively, where $\Box^k $ and
$\triangle^k $  are  defined by
\eqref{1.4} and \eqref{1.3} respectively. The function
$R^{H}_{2k}(\upsilon)$
 defined by \eqref{2.2}with $\alpha=2k$  and $R^{e}_{2k}(x)$ defined by \eqref{2.5}
with $\alpha=2k$.
\end{lemma}

\begin{proof}
 We have to show that $ \Box^k R^{H}_{2k}(\upsilon)=\delta(x)$
and that $\triangle^k ((-1)^k R^{e}_{2k}(x)=\delta(x)$.
 The first part follows from \cite[Lemma 2.4]{t2}, and the
second part from  \cite[p.31]{k1}.
\end{proof}

\begin{lemma} \label{lem2.2}
The convolution $R^{H}_{2k}(\upsilon)\ast(-1)^k R^{e}_{2k}(x)$
is an elementary solution for the operator
$\diamondsuit^k $ iterated $k$ times and is defined by \eqref{1.1}
\end{lemma}

 For the proof of the above lemma, see \cite[p.33]{k1}.

\begin{lemma} \label{lem2.3}
The function $R^{H}_{\alpha}(x)$ and $R^{e}_{\alpha}(x)$, defined
by \eqref{2.2} and \eqref{2.5} respectively,
for $\mathop{\rm Re}(\alpha)$ are
homogeneous distribution of order $\alpha-n$ and also a tempered
distribution.
\end{lemma}

\begin{proof}
 Note taht $R^{H}_{\alpha}(x)$ and
$R^{e}_{\alpha}(x)$ satisfy the Euler equation; that is,
\[
(\alpha-n)R^{H}_{\alpha}(x)=\sum^{n}_{i=1}x_{i}\frac
{\partial}{\partial x_{i}}R^{H}_{\alpha}(x), \\
(\alpha-n)R^{e}_{\alpha}(x)=\sum^{n}_{i=1}x_{i}\frac
{\partial}{\partial x_{i}}R^{e}_{\alpha}(x).
\]
Then $R^{H}_{\alpha}(x)$ and $R^{e}_{\alpha}(x)$ are homogeneous
distributions of order $\alpha-n$.
Since Donoghue \cite[pp.154-155]{d1} proved the every homogeneous distribution
is a tempered distribution, the proof is complete.
\end{proof}

\begin{lemma}[Convolution of tempered distributions] \label{lem2.4}
 $R^{e}_{\alpha}(x)\ast R^{H}_{\alpha}(x)$ exists and is
a tempered distribution.
\end{lemma}

\begin{proof} Choose
$\mathop{\rm supp}R^{H}_{\alpha}(x)=K\subset\Gamma_{+}$ where $K$
is a compact set.
Then $R^{H}_{\alpha}(x)$ is a tempered distribution with compact
support. By Donoghue \cite[pp.156-159]{d1}, $R^{e}_{\alpha}(x)\ast
R^{H}_{\alpha}(x)$ exists and is a tempered distribution.
\end{proof}

\begin{lemma} \label{lem2.5}
The functions $R^{H}_{-2k}(x)$ and $(-1)^k R^{e}_{-2k}(x)$ are the
inverse in the convolution algebra of $R^{H}_{2k}(x)$ and
$(-1)^k R^{e}_{2k}(x)$, respectively. That is,
\begin{gather*}
R^{H}_{-2k}(x)\ast R^{H}_{2k}(x)=R^{H}_{-2k+2k}(x)=R^{H}_{0}(x)
=\delta(x), \\
(-1)^k R^{e}_{-2k}(x)\ast
(-1)^k R^{e}_{2k}(x)=(-1)^{2k}R^{e}_{-2k+2k}(x)=R^{e}_{0}(x)=\delta(x)
\end{gather*}
\end{lemma}

For the proof of the above lemma, see
\cite[p.123]{t2}, \cite[p.118, p.158]{d1}, \cite[p.10]{t1}.

\begin{lemma}[Convolution of $R^{e}_{\alpha}(x)$ and
$R^{H}_{\alpha}(x)$] \label{lem2.6}
 Let $R^{e}_{\alpha}(x)$ and
$R^{H}_{\alpha}(x)$ defined by \eqref{2.5} and \eqref{2.2}
respectively, then we obtain:
\begin{enumerate}
\item $R^{e}_{\alpha}(x)\ast R^{e}_{\beta}(x)=R^{e}_{\alpha+\beta}(x)$
              where $\alpha$ and $\beta$ are complex parameters;
\item $R^{H}_{\alpha}(x)\ast R^{H}_{\beta}(x)=R^{H}_{\alpha+\beta}(x)$
              for $\alpha$ and $\beta$ are both integers and except only the case
              both $\alpha$ and $\beta$ are both integers.
\end{enumerate}
\end{lemma}

\begin{proof} Part (1) can be found in \cite[p.158]{d1}.
For the second formula, when $\alpha$ and
$\beta$ are both even integers, see \cite{k2}.
When $\alpha$ is odd and $\beta$ is even, or $\alpha$ is even
and $\beta$ is odd,
from Trione \cite{t2} we have
\begin{gather}\label{2.10}
\Box^k R^{H}_{\alpha}(x)=R^{H}_{\alpha-2k}(x), \\
\label{2.11}
\Box^k R^{H}_{2k}(x)=\delta(x),\quad k=0,1,2,3,\dots
\end{gather}
where $\Box^k $ is the Ultra-hyperbolic operator iterated $k$-times
defined by
$$
\Box^k =\Big(\sum^{p}_{i=1}\frac {\partial^{2}}{\partial
x^{2}_{i}}- \sum^{p+q}_{j=p+1}\frac {\partial^{2}}{\partial
x^{2}_{j}}\Big)^k .
$$
Now let $m$ be an odd integer, we have
\begin{gather*}
\Box^k R^{H}_{m}(x)=R^{H}_{m-2k}(x), \\
 R^{H}_{2k}(x)\ast \Box^k R^{H}_{m}(x)=R^{H}_{2k}(x)\ast
R^{H}_{m-2k}(x)
\end{gather*}
 or
\begin{gather*}
(\Box^k R^{H}_{2k}(x))\ast R^{H}_{m}(x)= R^{H}_{2k}(x)\ast R^{H}_{m-2k}(x),
\delta \ast R^{H}_{m}(x)=R^{H}_{2k}(x)\ast R^{H}_{m-2k}(x).
\end{gather*}
Thus
$$
R^{H}_{m}(x)=R^{H}_{2k}(x)\ast R^{H}_{m-2k}(x).
$$
Since $m$ is odd, hence $m-2k$ is odd and $2k$ is a positive
even. Put $\alpha=2k,~\beta=m-2k$, we obtain
$$
R^{H}_{\alpha}(x)\ast R^{H}_{\beta}(x)=R^{H}_{\alpha+\beta}(x)
$$
for $\alpha$ is a nonnegative even and $\beta$ is odd.

 For the case $\alpha$ is a negative even and $\beta$ is odd,
by \eqref{2.8} we have
\begin{gather*}
\Box^k R^{H}_{0}(x)=R^{H}_{-2k}(x), \\
\Box^k \delta=R^{H}_{-2k}(x),
\end{gather*}
where $R^{H}_{0}(x)=\delta$. Now for $m$ is odd,
$$
R^{H}_{-2k}(x)\ast \Box^k R^{H}_{m}(x)=R^{H}_{-2k}(x)\ast
R^{H}_{m-2k}(x)
$$
 or
\begin{gather*}
\left(\Box^k \delta\right)\ast \Box^k R^{H}_{m}(x)= R^{H}_{-2k}(x)\ast
R^{H}_{m-2k}(x), \\
\delta \ast\Box^{2k}R^{H}_{m}(x)=R^{H}_{-2k}(x)\ast R^{H}_{m-2k}(x).
\end{gather*}
Thus
$$
R^{H}_{m-2(2k)}(x)=R^{H}_{-2k}(x)\ast R^{H}_{m-2k}(x).
$$
 Put $\alpha=-2k$ and $\beta=m-2k$, now $\alpha$ is a
negative even and $\beta$ is odd. Then we obtain
$$
R^{H}_{\alpha}(x)\ast R^{H}_{\beta}(x)=R^{H}_{\alpha+\beta}(x).
$$
This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.7}
Given the equation
\begin{equation}\label{2.12}
\circledcirc^k H(x)=\delta(x)
\end{equation}
for $x\in R^{n}$, where $\circledcirc^k $ is the  operator iterated $k-$ times is defined by
\eqref{1.7} and $\triangle^k $ is the Laplace operator iterated $k$
times defined by \eqref{1.4} and $\Box^k $ is Ultra-hyperbolic
operator  iterated $k-$ times is  defined by \eqref{1.1}. Then we obtain $H(x)$ is an
elementary solution of \eqref{2.12},  where
\begin{equation}\label{2.13}
H(x)=\left(R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)\right)
*\left(C^{*k}(x)\right)^{*-1}
\end{equation}
where
\begin{equation}\label{2.14}
C(x)=\frac{1}{2}R^{H}_{4}(x)+\frac{1}{2}(-1)^{2}R^{e}_{4}(x)\,.
\end{equation}
Here $C^{*k}(x)$ denotes the convolution of  $C(x)$ itself $k$ times,
$\left(C^{*k}(x)\right)^{*-1}$ denotes the inverse of $C^{*k}(x)$ in
the convolution algebra. Moreover $H(x)$ is a tempered distribution.
\end{lemma}

\begin{proof}
 We have
$$
\circledcirc^k H(x)=\big(\frac{\triangle^{2}+\Box^{2}}{2}\big)^k H(x)
=\delta(x)
$$
or we can write
$$
\big(\frac{1}{2}\triangle^{2}+\frac{1}{2}\Box^2\big)
\big(\frac{1}{2}\triangle^{2}+\frac{1}{2}\Box^2\big)^{k-1}H(x)=\delta(x).
$$
Convolving both sides of the above equation by
$R^{H}_{4}(x)*(-1)^{2}R^{e}_{4}(x)$,
\begin{align*}
& \big(\frac{1}{2}\triangle^{2}+\frac{1}{2}\Box^2\big)\ast
\big(R^{H}_{4}(x)\ast(-1)^{2}R^{e}_{4}(x)\big)
\big(\frac{1}{2}\triangle^{2} +\frac{1}{2}\Box^2\big)^{k-1}H(x)\\
&=\delta(x)*R^{H}_{4}(x)\ast (-1)^{2}R^{e}_{4}(x)
\end{align*}
or
\begin{align*}
&\big(\frac{1}{2}\triangle^{2}(R^{H}_{4}(x)\ast (-1)^{2}R^{e}_{4}(x))+\frac{1}{2}\Box^2(R_{4}(x)\ast
     (-1)^{2}S_{4}(x))\big)\ast
\big(\frac{1}{2}\triangle^{2} +\frac{1}{2}\Box^2\big)^{k-1}H(x)\\
&=\delta(x)*R^{H}_{4}(x)\ast (-1)^{2}R^{e}_{4}(x).
\end{align*}
By properties of convolutions,
\begin{align*}
&\big(\frac{1}{2}\triangle^{2}((-1)^{2}R^{e}_{4}(x))
 \ast R^{H}_{4}(x)+\frac{1}{2}\Box^2(R^{H}_{4}(x))\\
& \ast(-1)^{2}R^{e}_{4}(x)     \big)\ast
\big(\frac{1}{2}\triangle^{2} +\frac{1}{2}\Box^2\big)^{k-1}H(x)\\
&=\delta(x)*R^{H}_{4}(x)\ast (-1)^{2}R^{e}_{4}(x)
\end{align*}
By Lemmas \ref{lem2.1} and  \ref{lem2.2}, we obtain
\[
\Big(\frac{1}{2}\delta\ast
     R^{H}_{4}(x)+\frac{1}{2}\delta\ast(-1)^{2}R^{e}_{4}(x)\Big)\ast
\Big(\frac{1}{2}\triangle^{2} +\frac{1}{2}\Box^2\Big)^{k-1}H(x) =
R^{H}_{4}(x)\ast (-1)^{2}R^{e}_{4}(x)
\]
or
\[
\Big(\frac{1}{2}R^{H}_{4}(x)+\frac{1}{2}(-1)^{2}R^{e}_{4}(x)\Big)\ast
\Big(\frac{1}{2}\triangle^{2} +\frac{1}{2}\Box^2\Big)^{k-1}H(x) =
R^{H}_{4}(x)\ast (-1)^{2}R^{e}_{4}(x)
\]
keeping on convolving both sides of the above equation by
$R^{H}_{4}(x)*(-1)^{2}R^{e}_{4}(x)$ up to $k-1$ times, we obtain
\begin{equation}\label{2.15}
C^{*k}(x)*H(x)=\left(R^{H}_{4}(x)*(-1)^{2}R^{e}_{4}(x)\right)^{*k}
\end{equation}
 the symbol $*k$ denotes the convolution of itself $k-$times.
 By properties of $R^{H}_\alpha(x)$ and $R^{e}_{\alpha}(x)$ in
Lemma \ref{lem2.3}, we have
 $$
\left(R^{H}_{4}(x)*(-1)^2R^{e}_{4}(x)\right)^{*k}
=R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x).
$$
 Thus \eqref{2.15} becomes,
$$
C^{*k}(x)*H(x)=\left(R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)\right)
$$
or
\begin{equation}\label{2.16}
H(x)=\left(R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)\right)\ast ({C^{\ast
k}}(x))^{\ast-1}
\end{equation}
 is an elementary solution of \eqref{2.12}.
 We consider the function $C^{\ast k}(x)$, since
$R^{H}_{4}(x)*(-1)^{2}R^{e}_{4}(x)$ is a  tempered distribution.
Thus $C(x)$ defined by \eqref{2.14} is a tempered distribution, we
obtain
$C^{*k}(x)$ is a tempered distribution.

 Now, $R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)\in \mathcal{S'}$,
the space of tempered distribution. Choose $\mathcal{S'} \subset
\mathcal{D'_R}$ where $\mathcal{D'_R}$ is the right-side
distribution which is a
subspace of $\mathcal{D'}$ of distribution.
 Thus $R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)\in
\mathcal{D'_R}$. It follow that
$R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)$ is an element of convolution
algebra, since $\mathcal{D'_R}$ is a convolution algebra. Hence
Zemanian \cite{z1},  \eqref{2.16} has a unique solution
\[
H(x)=\left(R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)\right)*
\left(C^{*k}(x)\right)^{*-1},
\]
where $\left(C^{*k}(x)\right)^{*-1}$ is an inverse of $C^{*k}(x)$ in
the convolution algebra. $H(x)$ is called the Green function of the
operator $\circledcirc^k$.

 Since $R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)$ and
$\left(C^{*k}(x)\right)^{*-1}$ are lies in $\mathcal{S'}$, then by
\cite[p.152]{z1} again, we have
$\left(R^{H}_{4k}(x)*(-1)^{2k}R^{e}_{4k}(x)\right)
*\left(C^{*k}(x)\right)^{*-1}\in\mathcal{S'}$.
 Hence $H(x)$ is a tempered distribution.
\end{proof}

\begin{lemma} \label{lem2.8}
The Fourier transform of $\oplus^k \delta$ is
$$
\mathcal{F}\oplus^k \delta =\frac{1}{(2\pi)^{n/2}}
\left[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\right]^k
$$
where $\mathcal{F}$ is the Fourier transform defined by \eqref{2.7}
and if the norm of $\xi$ is given by
$\|\xi\|=\left(\xi_1^2+\xi_2^2+\dots+\xi_n^2\right)^{1/2}$ then
$$
\mathcal{F}\oplus^k  \delta\leq\frac{M}{(2\pi)^{n/2}}\|\xi\|^{8k}.
$$
Since $M$ is constant thus is $\mathcal{F}\oplus^k \delta$ is
bounded and continuous on the space $\mathcal{S'}$ of the tempered
distribution. Moreover, by \eqref{2.8},
$$
\oplus^k \delta=\mathcal{F}^{-1}\frac{1}{(2\pi)^{n/2}}
\left[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\right]^k\,.
$$
\end{lemma}

\begin{proof}  By  \eqref{2.10}
\begin{align*}
\mathcal{F}\oplus^k  \delta
  &=  \frac{1}{(2\pi)^{n/2}}\langle\oplus^k  \delta ,
e^{-i\xi\cdot x}\rangle \\
   &=  \frac{1}{(2\pi)^{n/2}}\langle \delta ,
\oplus^k  e^{-i\xi\cdot x}\rangle \\
 &=  \frac{1}{(2\pi)^{n/2}}\langle \delta ,
\diamondsuit^k \circledcirc^k  e^{-i\xi\cdot x}\rangle\quad
\text{by \eqref{1.6}}
\\
   &= \frac{1}{(2\pi)^{n/2}}\big\langle \delta ,
 \diamondsuit^k \Big(\frac{1}{2}\triangle^{2}+\frac{1}{2}\Box^2\Big)^k
e^{-i\xi\cdot x}\big\rangle\\
&= \frac{1}{(2\pi)^{n/2}}\big\langle \delta ,
 \diamondsuit^k \frac{(-1)^{2k}}{2}(\left(\xi^{2}_{1}+\dots\xi^{2}_{n}
\right)^{2}+\big(\xi^{2}_{1}+\dots+\xi^{2}_{p}-\xi^{2}_{p+1}
-\dots\\
&\quad -\xi^{2}_{n}\big)^{2})^k
 e^{-i\xi\cdot x}\big\rangle\\
&= \frac{1}{(2\pi)^{n/2}}\langle \delta ,
 \frac{(-1)^{2k}}{2}(\left(\xi^{2}_{1}+\dots+\xi^{2}_{n}\right)^{2}
 +\big(\xi^{2}_{1}+\dots+\xi^{2}_{p}-\xi^{2}_{p+1}-\dots\\
&\quad -\xi^{2}_{n}\big)^{2})^k
 \diamondsuit^k e^{-i\xi\cdot x}\rangle\\
&= \frac{1}{(2\pi)^{n/2}}\langle \delta ,
 (\left(\xi^{2}_{1}+\dots+\xi^{2}_{p}\right)^{2}+\left(\xi^{2}_{p+1}+\dots+\xi^{2}_{p+q}\right)^{2})
 \triangle^k \Box^k e^{-i\xi\cdot x}\rangle\\
&= \frac{1}{(2\pi)^{n/2}}\big\langle \delta ,
 \Big[\Big(\sum^{p}_{i=1}\xi^2_i\Big)^{2}
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^{2}\Big]^k (-1)^{2k}
\Big[\Big(\sum^{p}_{i=1}\xi^2_i\Big)^{2}\\
&\quad -\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^{2}\Big]^k
e^{-i\xi\cdot x}\big\rangle\\
&= \frac{1}{(2\pi)^{n/2}}\langle\delta,(-1)^{2k}
\Big[\Big(\sum^{p}_{i=1}\xi^2_i\Big)^{4}
-\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^{4}\Big]^k e^{-i\xi\cdot x}\rangle\\
&= \frac{1}{(2\pi)^{n/2}}\Big[\Big(\sum^{p}_{i=1}\xi^2_i\Big)^4
-\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^4\Big]^k .\\
&= \frac{1}{(2\pi)^{n/2}}\big[\big(\xi_1^2+\xi_2^2+\dots+\xi_p^2\big)^4-
\big(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\big)^4\big]^k .
\end{align*}
Next, we consider the boundness of $\mathcal{F}\oplus^k \delta$.
Since
\begin{align*}
  \oplus^k
&=  \left((\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\right)^k \\
&=  \big(\left(\xi_1^2+\dots+\xi_p^2\big)^{2}
 -\big(\xi_{p+1}^2+\dots+\xi_n^2\big)^{2}\right)^k
\Big( \big(\xi_{1}^2+\dots+\xi_{p}^2\big)^2\\
&\quad + \big(\xi_{p+1}^2
 +\dots+\xi_{n}^2\big)^2 \Big)^k  \\
&= \left(\big(\xi_1^2+\dots+\xi_n^2\big)
\left(\xi_{1}^2+\dots+\xi_{p}^{2}-\xi_{p+1}^{2}\dots
\xi_n^2\right)\right)^k \Big(
  \big(\xi_{1}^2+\dots+\xi_{p}^2\big)^2\\
&\quad + \big(\xi_{p+1}^2+\dots+\xi_{n}^2\big)^2 \Big)^k
\end{align*}
Thus
\begin{align*}
\mathcal{F}\oplus^k  \delta
&= \frac{1}{(2\pi)^{n/2}}\left(\left(\xi_1^2+\dots +\xi_n^2\right)
\left(\xi_{1}^2+\dots+\xi_{p}^{2}-\xi_{p+1}^{2}\dots
-\xi_n^2\right)\right)^k \\
&\quad \times\left( \big(\xi_{p+1}^2+\dots+\xi_{n}^2\big)^2
+ \big(\xi_{p+1}^2+\dots+\xi_{n}^2\big)^2 \right)^k,
\end{align*}
\begin{align*}
|\mathcal{F}\oplus^k  \delta|
&= \frac{1}{(2\pi)^{n/2}}|\big(\xi_1^2+\dots+\xi_n^2\big)
 \left(\xi_{1}^2+\dots +\xi_{p+1}^{2}\dots -\xi_n^2\right)|^k \\
&\quad \times\left|  \big(\xi_{p+1}^2+\dots+\xi_{n}^2\big)^2
+ \big(\xi_{p+1}^2+\dots+\xi_{n}^2\big)^2 \right|^k \\
&\leq \frac{M}{(2\pi)^{n/2}}
 \left|\big(\xi_1^2+\dots+\xi_n^2\big)\right|^k \left|
\left(\xi_{1}^2+\dots+\xi_n^2\right)\right|^k \left|
\big(\xi_1^2+\dots+\xi_n^2\big)\right|^{2k}.
\end{align*}
It follows that
$$
|\mathcal{F}\oplus^k  \delta|\leq \frac{M}{(2\pi)^{n/2}}\|\xi\|^{8k},
$$
 where $M$ is constant and
$\|\xi\|=\left(\xi_1^2+\xi_2^2+\dots+\xi_n^2\right)^{1/2}$,
$\xi_i(i=1,2,\dots,n)\in \mathbb{R}$. Hence we obtain
$\mathcal{F}\oplus^{k}\delta$ is bounded and continuous on the space
$\mathcal{S'}$ of the tempered distribution. Since $\mathcal{F}$ is
one-to-one transformation from the space $\mathcal{S'}$ of the tempered
distribution to the real space $\mathbb{R}$, then by \eqref{2.8}
$$
\oplus\delta=\mathcal{F}^{-1}\frac{1}{(2\pi)^{n/2}}
\left[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\right].
$$
That completes the proof.
\end{proof}

\section{Main Results}

\begin{theorem} \label{thm3.1}
Given the equation
\begin{equation}\label{3.1}
\oplus^k G(x)=\delta(x),
\end{equation}
where $\oplus^k $ is the Oplus operator iterated $k$ times defined
by \eqref{1.8}, $\delta(x)$ is the dirac-delta distribution, $x\in
R^{n}$ and $k$ is a nonnegative integer. Then we obtain
\begin{equation}\label{3.2}
G(x)=\left(R^{H}_{2k}(\upsilon)\ast(-1)^k R^{e}_{2k}(x)\right)\ast
H(x)
\end{equation}
or by \eqref{2.13} and Lemma \ref{lem2.7}, we obtain
\begin{equation}\label{3.3}
G(x)=\left(R^{H}_{6k}(\upsilon)\ast(-1)^{3k}R^{e}_{6k}(x)\right)*
\left(C^{*k}(x)\right)^{*-1}
\end{equation}
 is a Green's function or an elementary solution for the operator
$\oplus^k$ iterated
$k-$times where $\oplus^k $ is defined by \eqref{1.8}, and $H(x)$
defined by \eqref{2.13}.

 For $q=0$, then\eqref{3.1}
becomes
\begin{equation}\label{3.4}
 \triangle^{4k}G(x)=\delta(x)
 \end{equation}
  By Lemma \ref{lem2.1}, we obtain
\[
G(x)=(-1)^{4k}R^{e}_{8k}(x)=R^{e}_{8k}(x)
\]
is an elementary solution of \eqref{3.4} where $\triangle^{4k}_{p}$
is the Laplacian of $p-$ dimension, iterated $4k-$ times and is
defined by \eqref{1.13}.
 Moreover, from \eqref{3.3}, we  obtain
\begin{align*}
&R^{H}_{-4k}(\upsilon)\ast (-1)^{3k}R^{e}_{-6k}(x)\ast\left(C^{\ast
k}(x)\right)\ast G(x)\\
&=\left(R^{H}_{4k}(\upsilon)\ast
R^{H}_{-4k}(\upsilon)\right)\ast\left((-1)^{3k}R^{e}_{6k}(x)\ast
(-1)^{3k}R^{e}_{-6k}(x)\right)\\
&\quad \ast\Big(\left(C^{\ast
k}(x)\right)\ast\left(C^{*k}(x)\right)^{*-1}\Big)\ast
R^{H}_{2k}(\upsilon).
\end{align*}
By \eqref{2.14}  the above equation becomes
\begin{equation}\label{3.5}
\Big(\frac{(-1)^{3k}}{2}\ast R^{e}_{-6k}(x)+R^{H}_{-4k}(\upsilon)\ast
\frac{(-1)^{5k}}{2}R^{e}_{-2k}(x)\Big)\ast G(x)=R^{H}_{2k}(\upsilon)
\end{equation}
as an elementary  solution  of the  operator  $k$  times is defined
by \eqref{1.4}
 In particular, if we put $p=1$, $q=n-1$, $k=1$ and $x_{1}=t$ in
\eqref{1.4},\eqref{3.3}, we obtain
\begin{equation}\label{3.6}
\Big(\frac{(-1)^{3}}{2} R^{e}_{-6}(x)+M^{H}_{-2}(\upsilon)\ast
\frac{(-1)^{5}}{2}R^{e}_{-6}(x)\Big)\ast G(x)=M^{H}_{2}(\upsilon),
\end{equation}
as an elementary solution of the wave operator defined by
\begin{equation}\label{3.7}
  \Box = \frac{\partial^2}{\partial t_1^2}
-\frac{\partial^2}{\partial x_{2}^2}-\frac{\partial^2}{\partial
x_{3}^2}-\dots-\frac{\partial^2}{\partial x_{n}^2},
\end{equation}
where $M^{H}_{2}(\upsilon)$ and $M^{H}_{-4}(\upsilon)$ defined by
\eqref{2.4} with $\alpha=2$ and $\alpha=-4$ respectively, and
$\upsilon=t^{2}_{1}-x^{2}_{2}-\dots-x^{2}_{n}$. The
function $R^{e}_{-6}(x)$ is defined by \eqref{2.5} with $\alpha=-6$.
\end{theorem}

\begin{proof}
  From \eqref{3.1} and \eqref{1.8}, we have
\begin{equation}\label{3.8}
\oplus^kG(x)=\left(\diamondsuit^k \circledcirc^k \right)G(x)=\delta(x).
\end{equation}
 Convolving both sides of \eqref{3.8} by
$\left(R^{H}_{2k}(\upsilon)\ast(-1)^k R^{e}_{2k}(x)\right)\ast
H(x)$, we obtain
$$
\left(R^{H}_{2k}(\upsilon)\ast (-1)^k R^{e}_{2k}(x)\right)\ast H(x)\ast
(\diamondsuit^k \circledcirc^k )G(x)=\delta\ast \left(
R^{H}_{2k}(\upsilon)\ast (-1)^k R^{e}_{2k}(x)\right)\ast H(x)
$$
By properties of convolution
$$
\diamondsuit^k \left(R^{H}_{2k}(\upsilon)\ast(-1)^k R^{e}_{2k}(x)\right)\ast \circledcirc^k (H(x))\ast
G(x)=\left(R^{H}_{2k}(\upsilon)\ast(-1)^k R^{e}_{2k}(x)\right)\ast
H(x).
$$
By Lemma \ref{lem2.2} and \ref{lem2.7}, we obtain,
\[
\delta \ast \delta\ast
G(x)=G(x)=\left(R^{H}_{2k}(x)\ast(-1)^k R^{e}_{2k}(\upsilon)\right)\ast
H(x).
\]
By Lemma \ref{lem2.6} and \eqref{2.13}, we obtain
 \begin{equation}\label{3.9}
 G(x)=R^{H}_{6k}(\upsilon)\ast(-1)^{3k}R^{e}_{6k}(x)*\left(C^{*k}(x)\right)^{*-1}
\end{equation}
is an elementary solution or Green's function of $\oplus^k $
operator.
 Now, for $q=0$ the \eqref{3.1} becomes
\begin{equation}\label{3.10}
\triangle^{4k}_{p}G(x)=\delta,.
\end{equation}
where $\triangle^{4k}_{p}$ is the Laplacian operator of p-dimension
iterated $4k$ times. By Lemma \ref{lem2.1}, we have
$$G(x)=(-1)^{4k}R^{e}_{8k}(x)$$
is an elementary solution of \eqref{3.10}.

 On the other hand, we can also find $G(x)$ from \eqref{3.9}.
Since $q=0$, we have $R^{H}_{2k}(\upsilon)$ reduces to
$(-1)^k R^{e}_{2k}(x)$.
 Thus, by \eqref{3.9} for $q=0$, we obtain
\begin{align*}
G(x)&= \left((-1)^{6k}R^{e}_{6k}(x)\ast R^{e}_{6k}(x)\right)\ast
\left((-1)^{2k}R^{e}_{4k}(x)\right)^{\ast -1}\\
&=  (-1)^{6k}R^{e}_{6k+6k}(x)\left((-1)^{2k}R^{e}_{4k}(x)\right)^{\ast -1}\\
&=  R^{e}_{8k}(x).
\end{align*}
Now, consider the case of the wave equation. From \eqref{3.3}, we
have
\[
G(x)=\left(R^{H}_{6k}(\upsilon)\ast(-1)^{3k}R^{e}_{6k}(x)\right)*
\left(C^{*k}(x)\right)^{*-1}.
\]
Convolving the above equation by $R^{H}_{-4k}(\upsilon)\ast
(-1)^{3k}R^{e}_{-6k}(x)\ast\left(C^{\ast k}(x)\right)$ and by
Lemma \ref{lem2.5}, we obtain
\begin{align*}
&R^{H}_{-4k}(\upsilon)\ast (-1)^{3k}R^{e}_{-6k}(x)\ast\left(C^{\ast
k}(x)\right)\ast G(x)\\
&=\left(R^{H}_{4k}(\upsilon)\ast
R^{H}_{-4k}(\upsilon)\right)
\ast\left((-1)^{3k}R^{e}_{6k}(x)\ast
(-1)^{3k}R^{e}_{-6k}(x)\right)\\
&\quad \ast\Big(\left(C^{\ast
k}(x)\right)\ast\left(C^{*k}(x)\right)^{*-1}\Big)\ast
R^{H}_{2k}(\upsilon).
\end{align*}
By Lemma \ref{lem2.5}, we obtain
$$
R^{H}_{-4k}(\upsilon)\ast (-1)^{3k}R^{e}_{-6k}(x)\ast\left(C^{\ast
k}(x)\right)\ast G(x)=R^{H}_{0}(x)\ast R^{e}_{0}(x)\ast
\delta(x)\ast R^{H}_{2k}(\upsilon)
$$
or
\[
R^{H}_{-4k}(\upsilon)\ast (-1)^{3k}R^{e}_{-6k}(x)\ast\left(C^{\ast
k}(x)\right)\ast G(x)=\delta(x)\ast \delta(x)\ast \delta(x)\ast
R^{H}_{2k}(\upsilon).
\]
It follows that
\begin{equation}\label{3.11}
R^{H}_{-4k}(\upsilon)\ast (-1)^{3k}R^{e}_{-6k}(x)\ast\left(C^{\ast
k}(x)\right)\ast G(x)=R^{H}_{2k}(\upsilon)
\end{equation}
as an elementary solution of the operator $\Box^k $ iterated $k$
times defined by \eqref{1.4}. In particular , if we  put
$p=1$, $q=n-1$, $k=1$
and $x_{1}=t$ in \eqref{3.3} and \eqref{3.9} then $R^{H}_{-4}$
reduces to $M^{H}_{-4}(\upsilon)$ and $R^{H}_{2}(\upsilon)$  reduces to
$M^{H}_{2}(\upsilon)$where $M^{H}_{-4}(\upsilon)$ and
$M^{H}_{2}(\upsilon)$ is defined by \eqref{2.4} with
$\alpha=-4,~\alpha=2$ respectively. Thus \eqref{3.11} becomes
\begin{equation}\label{3.12}
M^{H}_{-4}(\upsilon)\ast (-1)^{3}R^{e}_{-6}(x)\ast\left(C^{\ast
1}(x)\right)\ast G(x)=M^{H}_{2}(\upsilon)
\end{equation}
by \eqref{2.14}, we obtain
\[
M^{H}_{-4}(\upsilon)\ast
(-1)^{3}R^{e}_{-6}(x)\ast
\Big(\frac{1}{2}M^{H}_{4}(\upsilon)+\frac{(-1)^{2}}{2}R^{e}_{4}(x)\Big)
\ast G(x)=M^{H}_{2}(\upsilon)
\]
or
\begin{align*}
&\Big(M^{H}_{-4}(\upsilon)\ast (-1)^{3}R^{e}_{-6}(x)\ast
\frac{1}{2}M^{H}_{4}(\upsilon)+M^{H}_{-4}(\upsilon)\\
&\ast (-1)^{3}R^{e}_{-6}(x)\ast
\frac{1}{2}(-1)^{2}R^{e}_{4}(x)\Big)
\ast G(x)\\
&=M^{H}_{2}(\upsilon).
\end{align*}
By Lemma \ref{lem2.5}, we obtain
\begin{equation}\label{3.13}
\Big(\frac{(-1)^{3}}{2} R^{e}_{-6}(x)+M^{H}_{-4}(\upsilon)\ast
\frac{(-1)^{5}}{2}R^{e}_{-2}(x)\Big)\ast G(x)=M^{H}_{2}(\upsilon)
\end{equation}
as an elementary solution of the wave operator defined by
\[
  \Box = \frac{\partial^2}{\partial t_1^2}
-\frac{\partial^2}{\partial x_{2}^2}-\frac{\partial^2}{\partial
x_{3}^2}-\dots-\frac{\partial^2}{\partial x_{n}^2},
\]
and  $R^{e}_{-6}(x)$ defined by \eqref{2.5} with $\alpha=-6$. This
completes the proof.
\end{proof}

\begin{theorem} \label{thm3.2}
\begin{align*}
& \mathcal{F}\left(\left(R^{H}_{6k}(x)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast
 -1}\right)\\
&= \frac{1}{(2\pi)^{n/2}}\big[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\big]^k
\end{align*}
and
\begin{equation}\label{3.14}
|\mathcal{F}\left(\left(R^{H}_{6k}(x)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast
 -1}\right)|\leq \frac{1}{(2\pi)^{\frac{n}{2}}}M
 \end{equation}
for a  large $\xi_{i}\in R$,
where $M$ is a constant  and  $C(x)$ is defined by \eqref{2.14}. That is,
 $\mathcal{F}$ is bounded and
continuous on the space $S'$ of the tempered distributions.
\end{theorem}

\begin{proof}
 By Theorem \ref{thm3.1}, we have
$$
\oplus^k (\left(\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast
 -1}\right)=\delta(x)
$$
or
$$
\left(\oplus^k \delta\right)\ast(\left(\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast
 -1}\right)=\delta(x).
$$
Taking the Fourier transform on both sides of the above equation, we
obtain
$$
\mathcal{F}(\left(\oplus^k \delta\right)\ast[\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast
 -1}])=\mathcal{F}\delta=\frac{1}{(2\pi)^{n/2}}.
$$
By \eqref{2.10}
$$
\frac{1}{(2\pi)^{n/2}}\langle(\left(\oplus ^k \delta\right)\ast[\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast
 -1}]),e^{-i(\xi\cdot x)}\rangle=\frac{1}{(2\pi)^{n/2}}.
$$
By the definition of convolution
\begin{align*}
&\frac{1}{(2\pi)^{n/2}}\langle(\left(\oplus ^k \delta\right)\ast[\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast
 -1}]),e^{-i\xi.(x+r)}\rangle\\
&=\frac{1}{(2\pi)^{n/2}},
\end{align*}
\begin{align*}
&\frac{1}{(2\pi)^{n/2}}\langle[\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast
 -1}]),e^{-i(\xi.r)}\rangle\langle(\left(\oplus ^k
\delta\right),e^{-i\xi\cdot x}\rangle\\
&=\frac{1}{(2\pi)^{n/2}},
\end{align*}
\[
\mathcal{F}([\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast -1}])
\left( 2\pi\right)^{\frac{n}{2}}\mathcal{F}\left(\oplus^k \delta\right)
=\frac{1}{\left (2\pi\right)^{n/2}},
\]
\begin{align*}
&\mathcal{F}([\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(S^{\ast k}(x))^{\ast -1}])\\
& \cdot (-1)^{3k}\left[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\right]^k \\
&=\frac{1}{\left (2\pi\right)^{n/2}}.
\end{align*}
It follows that
\begin{align*}
&\mathcal{F}([\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast -1}])\\
&=\frac{1}{(-1)^{4k}\left
(2\pi\right)^{n/2}\left[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\right]^k }.
\end{align*}
Since
\begin{equation} \label{3.15}
\begin{aligned}
&\frac{1}{\left(\xi_1^2+\dots+\xi_p^2\right)^{4}-\left(\xi^{2}_{p+1}
+\dots+\xi^{2}_{p+q}\right) ^{4}}\\
&=\frac{1}{\left(\left(\xi_1^2+\dots+\xi_p^2\right)^{2}
 +\left(\xi^{2}_{p+1}+\dots+\xi^{2}_{p+q}\right)^{2}\right)}\\
&\quad \times\frac{1}{\left(\left(\xi_1^2+\dots
+\xi_n^2\right)\left(\xi_{1}^{2}+\dots+\xi_{p}^{2}-\xi_{p+1}^{2}
-\dots-\xi_{p+q}^{2}\right)\right)}.
\end{aligned}
\end{equation}
 Let $\xi=\left(\xi_{1},\xi_{2},\dots,\xi_{n}\right)\in\Gamma_{+}$ with
$\Gamma_{+}$ defined by Definition \ref{def2.1}. Then
$(\xi_1^2+\dots+\xi_p^2$$+\xi^{2}_{p+1}+\dots+\xi^{2}_{p+q})>0$
and for a large $k$, the right-hand side of \eqref{3.15} tend to
zero. It follows that it is bounded by a positive constant $M$ say,
that is we obtain \eqref{3.15} as required and also by \eqref{3.13}
$\mathcal{F} $ is continuous on the space $S'$ of the tempered
distribution.
\end{proof}

\begin{theorem} \label{thm3.3}
\begin{align*}
&\mathcal{F}\Big([\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast -1}]\\
& \ast[\left(R^{H}_{6m}(\upsilon)\ast
 (-1)^{3m}R^{e}_{6m}(x)\right)\ast(C^{\ast m}(x))^{\ast -1}]\Big)
\\
&=(2\pi)^{n/2}\mathcal{F}\left([\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast -1}\right])\\
&\quad \times\mathcal{F}([\left(R^{H}_{6m}(\upsilon)\ast
 (-1)^{3m}R^{e}_{6m}(x)\right)\ast(C^{\ast m}(x))^{\ast -1}])\\
&=\frac{1}{(2\pi)^{n/2}\left[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\right]^{k+m}},
\end{align*}
where $k$ and $m$ are nonnegative integer and $\mathcal{F}$ is
bounded and continuous on the space $S'$ of the tempered
distribution.
\end{theorem}

\begin{proof} Since $R^{H}_{6k}(\upsilon)$ and
$R^{e}_{4k}(x)$ are tempered distribution with compact support,
\begin{align*}
&\left([\left(R^{H}_{6k}(\upsilon)\ast (-1)^{3k}R^{e}_{6k}(x)\right)
\ast(C^{\ast k}(x))^{\ast -1}\right])\\
&\ast\Big([\left(R^{H}_{6m}(\upsilon)\ast (-1)^{3m}R^{e}_{6m}(x)\right)
\ast(C^{\ast m}(x))^{\ast -1}]\Big)\\
&=\left(R^{H}_{6k}(\upsilon)\ast R^{H}_{6m}(\upsilon)\right)\ast
\left((-1)^{3(k+m)}R^{e}_{6k}(x)\ast R^{e}_{6m}(x)\right)\\
&\quad \ast\left(\left(C^{\ast k}(x)\right)^{\ast -1}\ast (\left(C^{\ast
m}(x)\right)^{\ast -1}\right)\\
&=\Big(R^{H}_{6(k+m)}(\upsilon)\ast
(-1)^{3(k+m)}R^{e}_{6(k+m)}(x)\Big)\ast
\big(C^{\ast(k+m)}(x)\big)^{\ast -1}
\end{align*}
by \cite[pp.156-159]{t2} and \cite[Lemma 2.5]{k2}. Taking the Fourier
transform on both sides and using Theorem \ref{thm3.2}, we obtain
\begin{align*}
&\mathcal{F}\left[\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x)^{\ast -1}\right])\\
&\ast\Big([\left(R^{H}_{6m}(\upsilon)\ast
 (-1)^{3m}R^{e}_{6m}(x)\right)\ast(C^{\ast m}(x))^{\ast -1}]\Big)\\
&=\frac{1}{\left
(2\pi\right)^{n/2}\left[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\right)^4\right]^{k+m}}\\
&=\frac{1}{\left
(2\pi\right)^{n/2}\left[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^4-
\left(\xi_{p+1}^2+\xi_{p+2}^{2}+\dots+\xi_{p+q}^2\right)^4\right]^k }\\
&\quad \times\frac{(2\pi)^{n/2}}{\left
(2\pi\right)^{n/2}\left[\left(\xi_1^2+\dots+\xi_p^2\right)^4-
\left(\xi_{p+1}^2+\dots+\xi_{p+q}^2\right)^4\right]^{m}}\\
&=(2\pi)^{n/2}\mathcal{F}\left([\left(R^{H}_{6k}(\upsilon)\ast
 (-1)^{3k}R^{e}_{6k}(x)\right)\ast(C^{\ast k}(x))^{\ast -1}\right])\\
&\quad\times \mathcal{F}([\left(R^{H}_{6m}(\upsilon)\ast
 (-1)^{3m}R^{e}_{6m}(x)\right)\ast(C^{\ast m}(x))^{\ast -1}]).
\end{align*}
Since $(R^{H}_{6(k+m)}(\upsilon)\ast
(-1)^{3(k+m)}R^{e}_{6(k+m)}(x))\ast
(C^{\ast(k+m)}(x))^{\ast -1}\in S'$, the space of
tempered distribution and by Theorem \ref{thm3.2}, we obtain that
$\mathcal{F}$ is bounded and continuous on $S'$.
\end{proof}

\subsection*{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.

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\end{document}