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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 76, pp. 1--9.\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-2010/76\hfil Solution of a PDE]
{Solutions of a partial differential equation related to the
oplus operator}

\author[W. Satsanit\hfil EJDE-2010/76\hfilneg]
{Wanchak Satsanit}

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

\thanks{Submitted April 8, 2010. Published June 8, 2010.}
\subjclass[2000]{46F10, 46F12}
\keywords{Ultra-hyperbolic kernel; diamond operator; tempered distribution}

\begin{abstract}
 In this article, we consider the equation
 $$
 \oplus^ku(x)=\sum^{m}_{r=0}c_{r}\oplus^{r}\delta
 $$
 where $\oplus^k$ is the operator iterated $k$ times and
 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 $p+q=n$, $x=(x_1,x_2,\dots,x_n)$ is in
 the $n$-dimensional  Euclidian space $\mathbb{R}^n$,
 $c_{r}$ is a constant, $\delta$ is the
 Dirac-delta distribution, $\oplus^{0}\delta=\delta$, and
 $k=0,1,2,3,\dots$. It is shown that, depending on the
 relationship between $k$ and $m$, the solution to this equation
 can be ordinary functions, tempered distributions,
 or singular distributions.
\end{abstract}

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

 \section{Introduction}

The diamond operator, iterated $k$ times, was studied by
Kananthai \cite{k1}, and is defined by
\begin{equation}\label{1.1}
\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,\quad p+q=n,
\end{equation}
where $n$ is the dimension of the space $\mathbb{R}^n$,
$x=(x_1,x_2,\dots,x_n)\in \mathbb{R}^n$, and $k$ is a nonnegative
integer. This operator  can be expressed as
\begin{equation}\label{1.2}
\diamondsuit^k=\Delta^k\square^k=\square^k\Delta^k
\end{equation}
where $\Delta^k$ is the Laplacian operator iterated $k$ times,
 defined by
\begin{equation}\label{1.3}
  \Delta^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}
and $\square^k$ is the Ultra-hyperbolic operator iterated $k$ times,
 defined by
\begin{equation}\label{1.4}
  \square^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}

Kananthai \cite{k1}  showed that  the
convolution
$$
u(x)=(-1)^kR^{e}_{2k}(x)\ast R^{H}_{2k}(x)
$$
is a unique elementary solution of the operator $\diamondsuit^k$,
where $R^{e}_{2k}(x)$ and $R^{H}_{2k}(x)$ 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)^kR^{e}_{2k}(x)\ast
R^{H}_{2k}(x)\right)=\delta\,.
\end{equation}
 Satsanit \cite{s2} introduced the $\circledcirc^k$ operator, defined by
\[
\circledcirc^k=\Big(\big(\sum^p_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^2 +\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^2\Big)^k.
\]
 From \eqref{1.3} and \eqref{1.4}, we obtain
\begin{equation} \label{1.6}
\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{\Delta+\square}{2}\Big)^{2}
  +\Big(\frac{\Delta-\square}{2}\Big)^{2}\Big)^k \\
  &= \Big(\frac{\Delta^2+\square^2}{2}\Big)^k .
\end{aligned}
\end{equation}
 The  $\oplus^k$ operator  has been studied by
Kananthai,  Suantai and  Longani \cite{k3}, and can
be expressed in the form
\begin{equation}\label{1.7}
\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\cdot\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
 \end{equation}
Thus,  \eqref{1.7} can be written as
\begin{equation}\label{1.8}
\oplus^k=\diamondsuit^k\circledcirc^k,
\end{equation}
where $\diamondsuit^k$ and $\circledcirc^k$ are defined by
\eqref{1.1}, \eqref{1.6} respectively.

 The purpose of this article, is finding the solution to the
equation
\begin{equation}\label{1.9}
\oplus^ku(x)=\sum^{m}_{r=0}c_{r}\oplus^{r}\delta
\end{equation}
by using convolutions of the generalized function.
It is also shown that
the type of solution to \eqref{1.9}  depends on the
relationship  between  $k$ and $m$, according to the following
cases:
\begin{itemize}
\item[(1)] If $m<k$ and $m=0$, then \eqref{1.9} has the solution
$$
    u(x)=c_{0}\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))
    \ast (C^{\ast k}(x))^{\ast-1}\right)
$$
which is an elementary solution of the $\oplus^k$ operator
in Theorem \ref{thm3.1}, is an ordinary function when  $6k\geq n$, and
     is a tempered distribution when $6k<n$.

\item[(2)] If $0<m<k$ then the solution of \eqref{1.9} is
$$
u(x)=\sum^{m}_{r=1}c_{r}\left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)
 \ast R^{H}_{6(k-r)}(x))\ast (C^{\ast (k-r)}(x))^{\ast-1}\right)
$$
which is an ordinary function when
        $6k-6r\geq n$ and is tempered distribution when $6k-6r<n$.

\item[(3)] If $m\geq k $ and  $k\leq m\leq M$, then
    \eqref{1.9} has the solution
 $$
u(x)= \sum^{M}_{r=k}c_{r}\oplus^{r-k}\delta
$$
which is only a singular distribution.
\end{itemize}

 Before going that point, the following definitions and some
concepts are needed.

\section{ Preliminaries}
\begin{definition} \label{def2.1} \rm
Let $x = ( x_1, x_2, \dots, x_n )$ be a point in $\mathbb{R}^n$.
Define
\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}
where $p + q = n$ is the dimension of 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 let
$\overline{\Gamma}_+$ denote its closure.
For any complex number $\alpha$, define the function
\begin{equation}\label{2.2}
R_\alpha^H (\upsilon) = \begin{cases}
\frac{\upsilon^{(\alpha - n)/2}}{K_n (\alpha)},
&\text{for } x \in \Gamma_+, \\
0, &\text{for }x \not\in \Gamma_+,
\end{cases}
\end{equation}
where
\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}
 The function $R_{\alpha}^H(\upsilon)$
was introduced by  Nozaki \cite[p. 72]{n1} and
is called the Ultra-hyperbolic kernel of Marcel Riesz.
\end{definition}

It is well known that $R_\alpha^H(\upsilon)$ is an ordinary function
if $\mathop{\rm Re }(\alpha) \ge n$ 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.

 From Trione \cite[p. 11]{t2}, $R^{H}_{2k}(\upsilon)$ is an elementary
 solution of the operator $\square^k$; that is,
 \begin{equation}\label{2.4}
 \square^kR^{H}_{2k}(\upsilon)=\delta(x)\,.
 \end{equation}

\begin{definition} \label{def2.2} \rm
Let $x = ( x_1, x_2, \dots, x_n )$
and $|x| = (x_1^2 + x_2^2 + \dots + x_n^2)^{1/2}$.
The elliptic kernel of Marcel Riesz and 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\left(\frac{\alpha}{2}\right)}
{\Gamma\left(\frac{n-\alpha}{2}\right)},
\end{equation}
$\alpha$ is a complex parameter, and $n$ is the dimension of
$\mathbb{R}^n$.
\end{definition}

 It can be shown that
$R^{e}_{-2k}(x)=(-1)^k\Delta^k\delta(x)$ where
$\Delta^k$ is defined by \eqref{1.3}. It follows that
$R^{e}_{0}(x)=\delta(x)$, \cite[p. 118]{k1}.
Moreover,  $(-1)^kR^{e}_{2k}(x)$ is an elementary
solution of the operator $\Delta^k$ \cite[Lemma 2.4]{k1};
that is,
 \begin{equation}\label{2.7}
 \Delta^k((-1)^kR^{e}_{2k}(x)=\delta(x)\,.
 \end{equation}


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

\begin{proof} We need to show that
$\square^kR^{H}_{2k}(\upsilon)=\delta(x)$
which is done in \cite[Lemma 2.4]{t2}.
Also we need to show that
 $\Delta^k((-1)^kR^{e}_{2k}(x)=\delta(x)$.
which is done in \cite[p. 31]{k1}.
\end{proof}

\begin{lemma} \label{lem2.2}
The convolution $R^{H}_{2k}(\upsilon)\ast
(-1)^kR^{e}_{2k}(x)$ is an elementary solution of the operator
$\diamondsuit^k$ iterated $k$ as 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 functions $R^{H}_{\alpha}(x)$ and $R^{e}_{\alpha}(x)$  defined
by \eqref{2.2} and \eqref{2.5} respectively, for $Re(\alpha)$, are
homogeneous distributions of order $\alpha-n$ and also a tempered
distributions.
\end{lemma}

\begin{proof} Since $R^{H}_{\alpha}(x)$ and
$R^{e}_{\alpha}(x)$ satisfy the Euler equation,
\begin{gather*}
(\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),
\end{gather*}
we have that
$R^{H}_{\alpha}(x)$ and $R^{e}_{\alpha}(x)$ are homogeneous
distributions of order $\alpha-n$.
 Donoghue \cite[pp. 154-155]{d1}
 proved that the every homogeneous distribution is a tempered
distribution. This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.4}
The convolution $R^{e}_{\alpha}(x)\ast R^{H}_{\alpha}(x)$ 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}[Convolution of $R^{e}_{\alpha}(x)$ and
$R^{H}_{\alpha}(x)$] \label{lemma2.5}
Let $R^{e}_{\alpha}(x)$ and
$R^{H}_{\alpha}(x)$ defined by \eqref{2.5} and \eqref{2.2}
respectively, then we obtain the following:
\begin{itemize}
\item[(1)] $R^{e}_{\alpha}(x)\ast R^{e}_{\beta}(x)
=R^{e}_{\alpha+\beta}(x)$
when $\alpha$ and $\beta$ are complex parameters;

\item[(2)] $R^{H}_{\alpha}(x)\ast R^{H}_{\beta}(x)
=R^{H}_{\alpha+\beta}(x)$ when $\alpha$ and $\beta$ are
integers, except when both $\alpha$ and $\beta$
are odd.
\end{itemize}
\end{lemma}

\begin{proof} For the first formula, see \cite[p. 158]{d1}.
For  the second formula, when $\alpha$ and
$\beta$ are both even integers; see \cite{k2}.
For the case $\alpha$ is odd and $\beta$ is even
or $\alpha$ is even and $\beta$ is odd,
by Trione \cite{t1}, we have
\begin{equation}\label{2.8}
\square^kR^{H}_{\alpha}(x)=R^{H}_{\alpha-2k}(x)
\end{equation}
and
\begin{equation}\label{2.9}
\square^kR^{H}_{2k}(x)=\delta(x),\quad k=0,1,2,3,\dots
\end{equation}
where $\square^k$ is the Ultra-hyperbolic operator iterated $k$-times
defined by
$$
\square^k=\Big(\sum^p_{i=1}\frac {\partial^{2}}{\partial
x^{2}_{i}}- \sum^{p+q}_{j=p+1}\frac {\partial^{2}}{\partial
x^{2}_{j}}\Big)^k.
$$
Now let $m$ be an odd integer. We have
$\square^kR^{H}_{m}(x)=R^{H}_{m-2k}(x)$
and
$$
R^{H}_{2k}(x)\ast \square^kR^{H}_{m}(x)=R^{H}_{2k}(x)\ast
R^{H}_{m-2k}(x)
$$
 or
\begin{gather*}
\left(\square^kR^{H}_{2k}(x)\right)\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
$$
R^{H}_{\alpha}(x)\ast R^{H}_{\beta}(x)=R^{H}_{\alpha+\beta}(x)
$$
when $\alpha$ is  nonnegative even and $\beta$ is odd.

 For the case when $\alpha$ is negative even and $\beta$ is odd,
 by \eqref{2.8} we have
$$
\square^kR^{H}_{0}(x)=R^{H}_{-2k}(x)
$$
or
$\square^k\delta=R^{H}_{-2k}(x)$,
where $R^{H}_{0}(x)=\delta$. Now when $m$ is odd,
$$
R^{H}_{-2k}(x)\ast \square^kR^{H}_{m}(x)=R^{H}_{-2k}(x)\ast
R^{H}_{m-2k}(x)$$
 or
\begin{gather*}
\left(\square^k\delta\right)\ast \square^kR^{H}_{m}(x)
= R^{H}_{-2k}(x)\ast R^{H}_{m-2k}(x),\\
\delta \ast\square^{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
negative even and $\beta$ is odd. Then we obtain
$$
R^{H}_{\alpha}(x)\ast R^{H}_{\beta}(x)=R^{H}_{\alpha+\beta}(x).
$$
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,
$x\in \mathbb{R}^{n}$, and $k$ is a nonnegative integer.
Then
\begin{equation}\label{3.2}
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
$\oplus^k$, where
\begin{equation}\label{3.3}
C(x)=\frac{1}{2}R^H_{4}(x)+\frac{1}{2}(-1)^2R^e_{4}(x),
\end{equation}
where $C^{*k}(x)$ denotes the convolution of  $C$ with itself $k$ times,
$\left(C^{*k}(x)\right)^{*-1}$ denotes the inverse of $C^{*k}(x)$ in
the convolution algebra. Moreover $G(x)$ is a tempered distribution.
\end{theorem}

For a proof of the above theorem, see  \cite{s1}.

\begin{theorem} \label{thm3.2} For $0<r<k$,
\begin{equation} \label{3.4}
\begin{aligned}
&\oplus^{r}\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))\ast(C^{\ast
k}(x))^{\ast-1}\right)\\
&= \left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)\ast
R^{H}_{6(k-r)}(x))\ast(C^{\ast (k-r)}(x))^{\ast-1}\right)
\end{aligned}
\end{equation}
and for $k\leq m$,
\begin{equation}\label{3.5}
\oplus^{m}\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))\ast
(C^{\ast k}(x))^{\ast-1}\right)=\oplus^{m-k}\delta.
\end{equation}
\end{theorem}

\begin{proof}  For $0<r<k$, from \eqref{3.1},
$$
\oplus^k\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))\ast (C^{\ast
k}(x))^{\ast-1}\right)=\delta.
$$
Thus,
\[
\oplus^{k-r}\oplus^{r}\left(((-1)^{3k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (C^{\ast k}(x))^{\ast-1}\right)=\delta
\]
or
$$
\oplus^{k-r}\delta\ast\oplus^{r}\left(((-1)^{3k}R^{e}_{6k}(x)
\ast R^{H}_{6k}(x))\ast (C^{\ast
k}(x))^{\ast-1}\right)=\delta.
$$
 Convolving both sides by
$\left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)\ast R^{H}_{6(k-r)}(x))\ast
(C^{\ast k}(x))^{\ast-1}\right)$, we obtain
\begin{align*}
&\oplus^{k-r}\left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)\ast
R^{H}_{6(k-r)}(x))\ast (C^{\ast
k}(x))^{\ast-1}\right)\\
&\ast \oplus^{r}\left(((-1)^{3k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (C^{\ast k}(x))^{\ast-1}\right)\\
&=\left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)\ast R^{H}_{6(k-r)}(x))\ast (C^{\ast
(k-r)}(x))^{\ast-1}\right)\ast \delta.
\end{align*}
By theorem \ref{thm3.1},
\begin{align*}
&\delta\ast \oplus^{r}\left(((-1)^{3k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (C^{\ast k}(x))^{\ast-1}\right)\\
&=\left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)\ast R^{H}_{6(k-r)}(x))\ast (C^{\ast
(k-r)}(x))^{\ast-1}\right)\ast \delta.
\end{align*}
It follows that
\begin{align*}
&\oplus^{r}\left(((-1)^{3k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (C^{\ast k}(x))^{\ast-1}\right)\\
 &=\left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)\ast R^{H}_{6(k-r)}(x))\ast (C^{\ast
(k-r)}(x))^{\ast-1}\right)
\end{align*}
as required. For $k\leq m$
\begin{align*}
&\oplus^{m}\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))\ast
(C^{\ast k}(x))^{\ast-1}\right)\\
&=\oplus^{m-k}\oplus^k\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))\ast (C^{\ast
k}(x))^{\ast-1}\right).
\end{align*}
It follows that
\[
\oplus^{m}\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))\ast
(C^{\ast k}(x))^{\ast-1}\right)=\oplus^{m-k}\delta
\]
 by Theorem \ref{thm3.1}. This completes the proof.
\end{proof}

\begin{theorem} \label{thm3.3}
Consider the linear differential equation
\begin{equation}\label{3.6}
\oplus^k u(x)=\sum^{m}_{r=0}c_{r}\oplus^{r}\delta,
\end{equation}
where
\[
 \oplus^k = \Big(\Big(\sum^p_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^{4}
-\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^{4}\Big)^k,\\
\]
 $p+q=n$, $n$ is  odd with $p$ odd and $q$ even,
 or $n$ is  even with $p$ odd and $q$ odd,
$x\in \mathbb{R}^n$, $c_{r}$ is a constant, $\delta$ is the Dirac-delta
distribution, and $\oplus^{0}\delta=\delta$.
 Then the type of solution to \eqref{3.6} depends on the
relationship  between  $k$ and $m$, according to the
following cases:
\begin{itemize}
    \item[(1)]
 If $m<k$ and $m=0$, then \eqref{3.6} has  solution
    $$
u(x)=c_{0}\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))
\ast (C^{\ast k}(x))^{\ast-1}\right)
$$
which is an elementary solution of the $\oplus^k$ operator
in Theorem \ref{thm3.1},  when  $6k\geq n$, and
     is a tempered distribution when  $6k<n$.

\item[(2)] If $0<m<k$, then the solution of \eqref{3.6} is
             $$
u(x)=\sum^{m}_{r=1}c_{r}\left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)
\ast R^{H}_{6(k-r)}(x))\ast (C^{\ast (k-r)}(x))^{\ast-1}\right)
$$
which is an ordinary function when $6k-6r\geq n$, and is a
tempered distribution when  $6k-6r<n$.

\item[(3)] If $m\geq k $ and  $k\leq m\leq M$, then
    \eqref{3.6} has  solution
$$
u(x)= \sum^{M}_{r=k}c_{r}\oplus^{r-k}\delta
$$
 which is only a singular distribution.
\end{itemize}
\end{theorem}

\begin{proof}
  (1) For $m=0$, we have
$\oplus^ku(x)=c_{0}\delta$, and by Theorem \ref{thm3.1} we obtain
$$
u(x)=\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))
\ast (C^{\ast k}(x))^{\ast-1}\right)
$$
 Now, $(-1)^{3k}R^{e}_{6k}(x)$ and $R^{H}_{6k}(x)$ are the
analytic function for $6k\geq n$ and also
$(-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x)\ast (C^{\ast k}(x))^{-1}$
exists and is an analytic function by \eqref{3.2}. It follows that
$(-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x)\ast (S^{\ast k}(x))^{-1}$
is an ordinary function for  $6k\geq n$. By Lemma \ref{lem2.1}
with $\alpha =6k$, $(-1)^{3k}R^{e}_{6k}(x)$ and with
$\alpha =6k,~R^{H}_{6k}(x)$ are tempered distribution with $6k<n$,
 we obtain $(-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x)\ast
(C^{\ast k}(x))^{-1}$
exists and is a tempered distribution.

 (2) For the case $0<m<k$, we have
$$
\oplus^ku(x)=c_{1}\oplus \delta +c_{2}\oplus^{2}\delta +\dots
+c_{m}\oplus^{m}\delta.
$$
We convolved both sides of the above
equation  by $(-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x)\ast (C^{\ast
k}(x))^{-1}$
 to obtain
\begin{align*}
&\oplus^k\left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))\ast
(C^{\ast k}(x))^{-1}\right)\ast u(x)\\
&=c_{1}\oplus \left(((-1)^{3k}R^{e}_{6k}(x)\ast R^{H}_{6k}(x))\ast
(C^{\ast k}(x))^{-1}\right)\\
&\quad +c_{2}\oplus^{2}\left(((-1)^{3k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (C^{\ast k}(x))^{-1}\right)\\
&\quad +\dots +c_{m}\oplus^{m}\left(((-1)^{3k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (C^{\ast k}(x))^{-1}\right).
\end{align*}
By Theorems \ref{thm3.1} and  \ref{thm3.2}, we obtain
\begin{align*}
u(x)&=c_{1}\left(((-1)^{3(k-1)}R^{e}_{6(k-1)}(x)\ast
R^{H}_{6(k-1)}(x))\ast (C^{\ast
(k-1)}(x))^{\ast-1}\right)\\
&\quad +c_{2}\left(((-1)^{4(k-2)}R^{e}_{6(k-2)}(x)\ast
R^{H}_{6(k-2)}(x))\ast (C^{\ast
(k-2)}(x))^{\ast-1}\right)\\
&\quad +\dots+c_{m}\left(((-1)^{3(k-m)}R^{e}_{6(k-m)}(x)\ast
R^{H}_{6(k-m)}(x))\ast (C^{\ast (k-m)}(x))^{\ast-1}\right)
\end{align*}
or
$$
u(x)=\sum^{m}_{r=1}c_{r}\left(((-1)^{3(k-r)}R^{e}_{6(k-r)}(x)\ast
R^{H}_{6(k-r)}(x))\ast (C^{\ast (k-r)}(x))^{\ast-1}\right).
$$
Similarly, as in the case(1), $u(x)$ is an ordinary function for
$6k-6r\geq n$ and is a tempered distribution for
 and  $6k-6r<n$.

 (3) For the case $m\geq k$ and  $k\leq m \leq M$, we
have
$$
\oplus^ku(x)=c_{k}\oplus^k\delta+c_{k+1}\oplus^{k+1}\delta
+\dots+c_{M}\oplus^{M}\delta.
$$
Convolved  both sides of the above equation  by
$(-1)^{3k}R^{e}_{6k}(x)\ast R_{6k}(x)\ast (C^{\ast k}(x))^{\ast-1}$
to obtain
\begin{align*}
&\oplus^k\left(((-1)^{3k}R^{e}_{4k}(x)\ast R^{H}_{6k}(x))\ast
(S^{\ast
k}(x))^{\ast-1}\right)\ast u(x)\\
&=c_{k}\oplus^k \left(((-1)^{2k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (S^{\ast
k}(x))^{-1}\right)\\
&\quad +c_{k+1}\oplus^{k+1}\left(((-1)^{3k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (C^{\ast k}(x))^{\ast-1}\right)\\
&\quad +\dots +c_{M}\oplus^{M}\left(((-1)^{3k}R^{e}_{6k}(x)\ast
R^{H}_{6k}(x))\ast (C^{\ast k}(x))^{\ast-1}\right).
\end{align*}
By Theorems \ref{thm3.1} and  \ref{thm3.2} again, we obtain
\[
u(x)=  c_{k}\delta +c_{k+1}\oplus
\delta+c_{k+2}\oplus^{2}\delta+\dots + c_{M}\oplus^{M-k}\delta
=  \sum^{M}_{r=k}c_{r}\oplus^{r-k}\delta.
\]
Since $\oplus^{r-k}\delta$ is a singular distribution, hence $u(x)$
is only the singular distribution. This completes the proofs.
\end{proof}

\subsection*{Acknowledgements}
The authors would like to thank Prof. Amnuay Kananthai, Department
of Mathematics, at  Chiang Mai University, for many
helpful discussions. The authors want to thank  also the Thailand
Research Fund and Graduate School, Maejo  University,
Thailand for its financial support.

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\end{document}