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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011(2011), No. 04, 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 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/04\hfil Triharmonic heat equation]
{Solution to the triharmonic heat equation}

\author[W. Satsanit\hfil EJDE-2011/04\hfilneg]
{Wanchak Satsanit}

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

\thanks{Submitted June 14, 2010. Published January 7, 2011.}
\subjclass[2000]{46F10, 46F12}
\keywords{Fourier transform; tempered distribution; diamond operator}

\begin{abstract}
 In this article, we study the equation
 $$
 \frac{\partial}{\partial t}\,u(x,t)-c^2\circledast u(x,t)=0
 $$
 with initial condition $u(x,0)=f(x)$. Where
 $x$ is in the Euclidean space $\mathbb{R}^n$,
 $$
 \circledast=\Big(\sum^p_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^3
 +\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x^2_j}\Big)^3
 $$
 with $p+q=n$, $u(x,t)$ is an unknown function,
 $(x,t)=(x_1,x_2,\dots,x_n,t)\in \mathbb{R}^n\times (0,\infty)$,
 $f(x)$ is a generalized function, and $c$ is a positive
 constant.
 Under suitable conditions on $f$ and $u$, we obtain a
 unique solution. Note that for $q=0$, we
 have the triharmonic  heat equation
 $$
 \frac{\partial}{\partial t} u(x,t)-c^2\Delta^3 u(x,t)=0\,.
 $$
\end{abstract}

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

\section{Introduction}

It is well known that  the heat equation
\begin{equation}\label{1.1}
\frac{\partial}{\partial t} u(x,t)=c^2\Delta u(x,t),
\end{equation}
with the initial condition
$u(x,0)=f(x)$, has solution
\begin{equation}\label{1.2}
u(x,t)=\frac{1}{(4c^2\pi
t)^{n/2}}\int_{\mathbb{R}^n}\exp\Big(-\frac{|x-y|^2}{4c^2t}
\Big)f(y)dy\,,
\end{equation}
where $(x,t)=(x_1,x_2,\dots,x_n,t)\in \mathbb{R}^n\times (0,\infty)$,
and $\Delta=\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}$
is the Laplace operator.
It is also known that the solution
can be written as the convolution $u(x,t)=E(x,t)\ast f(x)$, where
\begin{equation}\label{1.3}
E(x,t)=\frac{1}{(4c^2\pi t)^{n/2}}\exp\big(-\frac{|x|^2}{4c^2t}\big),
\end{equation}
which is called {\it the heat kernel} \cite[pp. 208-209]{h1}.
Here $|x|^2=x_1^2+x_2^2+\dots+x_n^2$ and $t>0$.

In 1996, Kananthai \cite{k1} introduced the Diamond operator
\begin{equation}\label{1.4}
\diamondsuit=\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,\quad\text{with }  p+q=n\,.
\end{equation}
This operator can be written in the form
$\diamondsuit=\Delta\Box=\Box\Delta$, where
\begin{equation}\label{1.5}
  \Delta = \frac{\partial^2}{\partial
x^2_1}+\frac{\partial^2}{\partial
x^2_2}+\dots+\frac{\partial^2}{\partial
x^2_n}
\end{equation}
 is the Laplacian, and
 \begin{equation}\label{1.6}
  \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}
 is the  ultra-hyperbolic operator.
The Fourier transform and the elementary solution of
the Diamond operator has been studied; see for example \cite{k1}.
Nonlaopon and  Kananthai \cite{n1} studied the equation
$$
\frac{\partial}{\partial t} u(x,t)=c^2\Box u(x,t),
$$
and obtain the ultra-hyperbolic heat kernel
$$
E(x,t)=\frac{i^{q}}{(4c^2\pi t)^{n/2}}\exp\Big(-\frac{\sum
^p_{i=1}x^2_i-\sum^{p+q}_{j=p+1}x^2_j}{4c^2t}\Big),
$$
where $p+q=n$,  and $i=\sqrt{-1}$.

The purpose of this work is to study the equation
\begin{equation}\label{1.7}
\frac{\partial}{\partial t}\,u(x,t)-c^2\circledast u(x,t)=0,
\end{equation}
with the initial condition
$u(x,0)=f(x)$, for $x\in\mathbb{R}^n$. The operator is
\begin{align*}
  \circledast
&=\Big(\sum^p_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^3 +\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^3\\
&= \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)
\Big[\Big(\sum^p_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^2- \Big(\sum^p_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)\\
&\quad +\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^2\Big] \\
&=\Delta(\Delta^{2}-\frac{3}{4}(\Delta+\Box)(\Delta-\Box))\\
&=\frac{3}{4}\diamondsuit\Box+\frac{1}{4}\Delta^3
\end{align*}
where $\Delta, \Box, \diamondsuit$ are defined by
\eqref{1.5}, \eqref{1.6} and \eqref{1.4}  respectively.

Here, $p+q=n$, $u(x,t)$ is an unknown function,
$(x,t)=(x_1,x_2,\dots,x_n,t)$ is in $\mathbb{R}^n\times (0,\infty)$,
$f(x)$ is a  generalized
function, and $c$ is a positive constant. We obtain
a solution $u(x,t)=E(x,t)\ast f(x)$, where
\begin{equation}\label{1.8}
E(x,t)=\frac{1}{(2\pi)^n}\int_{\Omega}
\exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)
\Big]d\xi\,,
\end{equation}
and $\Omega \subset \mathbb{R}^n$ is the spectrum of $E(x,t)$ for
any fixed $t>0$.

Here $E(x,t)$ is the elementary solution
of \eqref{1.7}, whose properties will be studied in this article.
If we put $q=0$, then \eqref{1.7} reduces to
the equation
$$
\frac{\partial}{\partial t}\,u(x,t)-c^2\Delta^3u(x,t)=0
$$
which is related to the triharmaoic heat equation.

\section{Preliminaries}

\begin{definition} \label{def2.1} \rm
Let $f(x)\in L _1(\mathbb{R}^n)$, the space of
integrable function in $\mathbb{R}^n$. Then the Fourier transform of
$f(x)$ is
\begin{equation}\label{2.1}
\widehat{f}(\xi)=\frac{1}{(2\pi)^{n/2}}
\int_{\mathbb{R}^n} e^{-i(\xi,x)}f(x)\,dx,
\end{equation}
where $\xi=(\xi_1,\xi_2,\dots,\xi_n)$,
$x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$,
$(\xi,x)=\xi_1 x_1+\xi_2 x_2+\dots+\xi_n x_n$, and
$dx=dx_1\,dx_2\dots dx_n$.
The inverse of Fourier transform is defined as
\begin{equation}\label{2.2}
f(x)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{i(\xi,x)}
\widehat{f}(\xi)\,d\xi.
\end{equation}
If $f$ is a distribution with compact support by
\cite[Theorem 7.4-3]{z1}, we  can write
\begin{equation}\label{2.3}
\widehat{f}(\xi)=\frac{1}{(2\pi)^{n/2}}\langle f(x),
e^{-i(\xi,x)}\rangle.
\end{equation}
\end{definition}

\begin{definition} \label{def2.2} \rm
The spectrum of the kernel $E(x,t)$ in \eqref{1.6} is the bounded
support of the Fourier transform $\widehat{E(\xi,t)}$ for
any fixed $t>0$.
\end{definition}

\begin{definition} \label{def2.3} \rm
 Let $\xi=(\xi_1,\xi_2,\dots,\xi_n)$ be a point in $\mathbb{R}^n$ and
let $$
\Gamma_+=\{\xi\in\mathbb{R}^n :
\xi_1^2+\xi_2^2+\dots+\xi_p^2-\xi_{p+1}^2-\xi_{p+2}^2-
\dots-\xi_{p+q}^2>0 \textrm{ and }\xi_1>0\}
$$
be the interior of the forward cone, and $\overline{\Gamma}_+$
denote the closure of $\Gamma_+$.
\end{definition}

Let $\Omega$ be spectrum of $E(x,t)$ defined by Definition
\ref{def2.2} for any fixed $t>0$, and
$\Omega\subset\overline{\Gamma}_+$. Let
the Fourier transform of $E(x,t)$ be
\begin{equation}\label{2.4}
\widehat{E(\xi,t)}=
\begin{cases}
\frac{1}{(2\pi)^{n/2}}\exp\Big[-c^2\Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)t\Big]&
\text{for }\xi\in\Gamma_+,\\
0& \text{for }\xi\notin \Gamma_+.
\end{cases}
\end{equation}

\begin{lemma} \label{lem2.1}
The Fourier transform of $\circledast \delta$ is
$$
\mathcal{F}\circledast\delta
=\frac{(-1)^3}{(2\pi)^{n/2}}[(\xi_1^2+\xi_2^2+\dots+\xi_p^2\Big)^3+
(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2)^3]
$$
where $\mathcal{F}$ is defined by \eqref{2.1}.
Let the norm of $\xi$ be
$\|\xi\|=(\xi_1^2+\xi_2^2+\dots+\xi_n^2)^{1/2}$. Then
$$
|\mathcal{F}\circledast\delta|\leq\frac{M}{(2\pi)^{n/2}}\|\xi\|^6,
$$
where $M$ is a positive constant.
That is, $\mathcal{F}\circledast$ is bounded and continuous on the
space $\mathcal{S'}$ of the tempered distribution. Moreover, by
\eqref{2.2},
$$
\circledast\delta=\mathcal{F}^{-1}\frac{1}{(2\pi)^{n/2}}
[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^3+
(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2)^3]
$$
\end{lemma}

\begin{proof}  By \eqref{2.3},
\begin{align*}
  \mathcal{F}\circledast\delta
&= \frac{1}{(2\pi)^{n/2}}\langle\circledast\delta ,
 e^{-i(\xi,x)}\rangle \\
&= \frac{1}{(2\pi)^{n/2}}\langle \delta ,
 \circledast e^{-i(\xi,x)}\rangle \\
&=\frac{1}{(2\pi)^{n/2}}\langle \delta ,
 \Big(\frac{3}{4}\diamondsuit\Box+\frac{1}{4}\Delta^3\Big)e^{-i(\xi,x)}\rangle\\
&=\frac{1}{(2\pi)^{n/2}}\langle \delta ,
 \frac{3}{4}\diamondsuit\Box e^{-i(\xi,x)}\rangle+
 \frac{1}{(2\pi)^{n/2}}\langle\delta,\frac{1}{4}\Delta^3e^{-i(\xi,x)}\rangle\\
&=\frac{1}{(2\pi)^{n/2}}\Big\langle \delta ,
 \frac{3}{4}(-1)^3\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^{2}
-\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^{2}\Big]
\\
&\quad\times \Big[\Big(\sum^p_{i=1}\xi^2_i\Big)
 -\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)\Big]e^{-i(\xi,x)}\Big\rangle\\
&\quad + \frac{1}{(2\pi)^{n/2}}\Big\langle\delta,\frac{1}{4}(-1)^3
 \Big[\Big(\sum^{n}_{i=1}\xi^2_i\Big)
 \Big]^3e^{-i(\xi,x)}\Big\rangle\\
&=\frac{1}{(2\pi)^{n/2}}\Big[\frac{3}{4}(-1)^3\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^{2}
 -\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^{2}\Big]\Big]
 \Big[\Big(\sum^p_{i=1}\xi^2_i\Big)-
 \Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)\Big]\\
&\quad +\frac{1}{(2\pi)^{n/2}}\Big(\frac{1}{4}(-1)^3\Big[\Big(\sum^{n}_{i=1}\xi^2_i\Big)
 \Big]^3\Big)\\
&=\frac{(-1)^3}{(2\pi)^{n/2}}\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]\\
&=\frac{(-1)^3}{(2\pi)^{n/2}}
 \Big[\Big(\xi_1^2+\xi_2^2+\dots+\xi_p^2\Big)^3+
\Big(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\Big)^3\Big].
\end{align*}
Then
\begin{align*}
&|\mathcal{F}\circledast\delta|\\
&= \frac{1}{(2\pi)^{n/2}}\big|
 \big(\xi_1^2+\xi_2^2+\dots+\xi_p^2\big)^3+
 \big(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2\big)^3\big| \\
&\leq  \frac{1}{(2\pi)^{n/2}}|\xi_1^2+\dots+\xi_n^2|\big|
  (\xi_1^2+\dots+\xi_n^2)^2+ (\xi_1^2+\dots+\xi_n^2)^2
 + (\xi_1^2+\dots+\xi_n^2)^2 \big| \\
&\leq \frac{M}{(2\pi)^{n/2}}\|\xi\|^6,
\end{align*}
where $\|\xi\|=\big(\xi_1^2+\xi_2^2+\dots+\xi_n^2\big)^{1/2}$,
$\xi_i\in \mathbb{R}$ $(i=1,2,\dots,n)$. Hence we obtain
$\mathcal{F}\circledast\delta$ is bounded and continuous on the
space $\mathcal{S'}$ of the tempered distribution.

Since $\mathcal{F}$ is a one-to-one transformation from the space
$\mathcal{S'}$ of the tempered distribution to the real space
$\mathbb{R}$,  by \eqref{2.2}, we have
$$
\circledast\delta=\mathcal{F}^{-1}\frac{1}{(2\pi)^{n/2}}
[(\xi_1^2+\xi_2^2+\dots+\xi_p^2)^3+
(\xi_{p+1}^2+\xi_{p+2}^2+\dots+\xi_{p+q}^2)^3].
$$
This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.2}
Let
\begin{equation}\label{2.5}
L =\frac{\partial}{\partial
t}-c^2\Big[\Big(\sum^p_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x^2_j}\Big)^3\Big],
\end{equation}
where
$$
\Big(\sum^p_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x^2_j}\Big)^3
=\frac{3}{4}\diamondsuit\Box+\frac{1}{4}\Delta^3,
$$
 $p+q=n$, $(x,t)=(x_1,x_2,\dots,x_n,t)\in \mathbb{R}^n\times (0,\infty)$,
and $c$ is a positive constant.
Then
\begin{equation}\label{2.6}
E(x,t)=\frac{1}{(2\pi)^n}\int_{\Omega}
\exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)\Big]d\xi.
\end{equation}
is an elementary solution of \eqref{2.5}.
\end{lemma}

\begin{proof}
Let $E(x,t)$ be an elementary solution of operator $L$. Then
$$
L E(x,t)=\delta(x,t),
$$
where $\delta$ is the Dirac-delta distribution. Thus
$$
\frac{\partial }{\partial t}E(x,t)-c^2
\Big[\Big(\sum^p_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x^2_j}\Big)^3\Big]
E(x,t)=\delta(x)\delta(t).
$$
Taking the Fourier transform on both sides of the
equation, we obtain
$$
\frac{\partial }{\partial t}\widehat{E(\xi,t)}
 +c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]
\widehat{E(\xi,t)}=\frac{1}{(2\pi)^{n/2}}\delta(t).
$$
Thus
$$
\widehat{E(\xi,t)}=\frac{H(t)}{(2\pi)^{n/2}}\exp
\Big[-c^2\Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)t\Big]
$$
where $H(t)$ is the Heaviside function. Since $H(t)=1$ for $t>0$.
Therefore,
$$
\widehat{E(\xi,t)}=\frac{1}{(2\pi)^{n/2}}
\exp\Big[-c^2\Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)t\Big]
$$
which by \eqref{2.3}, we obtain
\[
E(x,t)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{i(\xi,x)}
 \widehat{E(\xi,t)}\,d\xi
=\frac{1}{(2\pi)^{n/2}}\int_{\Omega}e^{i(\xi,x)}
\widehat{E(\xi,t)}\,d\xi
\]
where $\Omega$ is the spectrum of $E(x,t)$.
Thus from \eqref{2.2},
$$
E(x,t)=\frac{1}{(2\pi)^n}\int_{\Omega}
\exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)
\Big]d\xi.
$$
This completes the proof.
\end{proof}

\section{Main Results}

\begin{theorem} \label{thm3.1}
Consider the equation
\begin{equation}\label{3.1}
\frac{\partial}{\partial t}\,u(x,t)-c^2\circledast u(x,t)=0
\end{equation}
with initial condition
\begin{equation}\label{3.2}
u(x,0)=f(x)
\end{equation}
and the operator
\begin{align*}
  \circledast
&=\Big(\sum^p_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x^2_j}\Big)^3\\
&= \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)
\Big[\Big(\sum^p_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)^2- \Big(\sum^p_{i=1}\frac{\partial^2}{\partial
x^2_i}\Big)\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big) \\
&\quad +\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^2\Big] \\
&=\frac{3}{4}\diamondsuit\Box+\frac{1}{4}\Delta^3
\end{align*}
where $p+q=n$, $k$ is a positive integer, $u(x,t)$ is an
unknown function for $(x,t)=(x_1,x_2,\dots,x_n,t)\in
\mathbb{R}^n\times (0,\infty)$, $f(x)$ is a given generalized
function, and $c$ is a positive constant.
Then
\[
u(x,t)=E(x,t)\ast f(x)
\]
as a solution of \eqref{3.1}-\eqref{3.2}, where
$E(x,t)$ is given by \eqref{2.5}.
\end{theorem}

\begin{proof}
Taking the Fourier transform  on both sides
of \eqref{3.1}, we obtain
$$
\frac{\partial}{\partial t}\widehat{u}(\xi,t)+
c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]\widehat{u}(\xi,t)=0,
$$
(see Lemma \ref{lem2.1}). Thus
\begin{equation}\label{3.3}
\widehat{u}(\xi,t)=K(\xi)\exp\Big[-c^2
\Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)t\Big]
\end{equation}
where $K(\xi)$ is constant and $\widehat{u}(\xi,0)=K(\xi)$.
By \eqref{3.2} we have
\begin{equation}\label{3.4}
K(\xi)=\widehat{u}(\xi,0)=\widehat{f}(\xi)
=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{-i(\xi,x)}f(x)\,dx
\end{equation}
and by the inversion in \eqref{2.2}, \eqref{3.3} and \eqref{3.4}
we obtain
\begin{align*}
&u(x,t)\\
&=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{i(\xi,x)}
 \widehat{u}(\xi,t)\,d\xi\\
&=\frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}
 e^{i(\xi,x)}e^{-i(\xi,y)}f(y)
\exp\Big[-c^2\Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)t\Big]\,dy\,d\xi.
\end{align*}
Thus
$$
u(x,t)=\frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}
\int_{\mathbb{R}^n}e^{i(\xi,x-y)}
\exp\Big[-c^2\Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)t\Big]f(y)\,dy\,d\xi
$$
or
\begin{equation}\label{3.5}
u(x,t)=\frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}
\exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x-y)
\Big]f(y)\,dy\,d\xi.
\end{equation}
 Set
\begin{equation}\label{3.6}
E(x,t)=\frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}
\exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)
\Big]\,d\xi.
\end{equation}
We choose $\Omega\subset\mathbb{R}^n$ be the spectrum of $E(x,t)$
and by \eqref{2.5}, we have
\begin{equation} \label{3.7}
\begin{aligned}
E(x,t)&=\frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}
\exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)
\Big]\,d\xi \\
&=\frac{1}{(2\pi)^{n}}\int_{\Omega}
\exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)
\Big]\,d\xi.
\end{aligned}
\end{equation}
Thus \eqref{3.5} can be written in the convolution form
$$
u(x,t)=E(x,t)\ast f(x).
$$
Since $E(x,t)$ exists,
\begin{equation}
\lim_{t\to  0} E(x,t)
=\frac{1}{(2\pi)^n}\int_{\Omega}e^{i(\xi,x)}\,d\xi
=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i(\xi,x)}\,d\xi
=\delta(x), \label{3.8}
\end{equation}
for $x\in\mathbb{R}^n$;
see \cite[Eq. (10.2.19b)]{k1}.
Thus for the solution $u(x,t)=E(x,t)\ast f(x)$ of \eqref{3.1},
$$
\lim_{t\to  0} u(x,t)=u(x,0)=\delta\ast f(x)=f(x)
$$
which satisfies \eqref{3.2}.
\end{proof}

\begin{theorem} \label{thm3.2}
The kernel $E(x,t)$ defined by \eqref{3.7} has the following
properties:
\begin{itemize}
\item[(1)] $E(x,t)\in\mathcal{C}^\infty$ for $x\in\mathbb{R}^n$ and
$t>0$, the space of function with infinitely many continuous
derivatives.

\item[(2)] For $t>0$,
$$
\Big(\frac{\partial}{\partial t}
-c^2\Big[\Big(\sum^p_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial
x^2_j}\Big)^3\Big]\Big)E(x,t)=0\,.
$$

\item[(3)]  $E(x,t)>0$ for $t>0$.

\item[(4)] For $t>0$,
$$
|E(x,t)|\leq \frac{2^{2-n}}{\pi^{n/2}}
\frac{M(t)}{\Gamma(\frac{p}{2})\Gamma(\frac{q}{2})},
$$
where $M(t)$ is a function of $t$ in the spectrum $\Omega$, and
$\Gamma$ denotes the Gamma function.
Thus $E(x,t)$ is bounded for any fixed $t>0$.

\item[(5)] $\lim_{t\to  0} E(x,t)=\delta$.
\end{itemize}
\end{theorem}

\begin{proof}
(1) From \eqref{3.7},
$$
\frac{\partial^n}{\partial x^n} E(x,t)=
\frac{1}{(2\pi)^{n}}\int_{\Omega}\frac{\partial^n}{\partial
x^n} \exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)
\Big]\,d\xi.
$$
Thus $E(x,t)\in\mathcal{C}^\infty$ for $x\in\mathbb{R}^n$ and
$t>0$.

(2) By a computation,
$$
\Big(\frac{\partial}{\partial t}-c^2
 \Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]\Big)E(x,t)=0.
$$

(3) $E(x,t)>0$ for $t>0$ is obvious by \eqref{3.7}.

(4) We have
\begin{gather*}
E(x,t)=\frac{1}{(2\pi)^{n}}\int_{\Omega}\exp
\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)
\Big]\,d\xi,
\\
|E(x,t)|\leq\frac{1}{(2\pi)^{n}}\int_{\Omega}\exp
\Big[-c^2\Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3
+\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)\Big]\,d\xi.
\end{gather*}
By changing to bipolar coordinates
\begin{gather*}
\xi_1=r\omega_1,\quad \xi_2=r\omega_2,\quad \dots,
\quad \xi_p=r\omega_p,\\
\xi_{p+1}=s\omega_{p+1},\quad \xi_{p+2}=s\omega_{p+2}, \quad
\dots,\quad \xi_{p+q}=s\omega_{p+q},
\end{gather*}
where $\sum_{i=1}^p\omega_i^2=1$ and
$\sum_{j=p+1}^{p+q}\omega_j^2=1$. Thus
$$
|E(x,t)|\leq\frac{1}{(2\pi)^{n}}\int_{\Omega}\exp
[-c^2(s^6+r^6)t]r^{p-1}s^{q-1}\,dr\,ds\,
d\Omega_p\,d\Omega_q
$$
where
$d\xi=r^{p-1}s^{q-1}\,dr\,ds\,d\Omega_p\,d\Omega_q$, $d\Omega_p$ and
$\Omega_q$ are the elements of surface area of the unit sphere in
$\mathbb{R}^p$ and $\mathbb{R}^q$ respectively. Since
$\Omega\subset\mathbb{R}^n$ is the spectrum of $E(x,t)$ and we
suppose $0\leq r\leq R$ and $0\leq s \leq T$ where $R$ and $T$ are
constants. Thus we obtain
\begin{align*}
|E(x,t)|&\leq\frac{\Omega_p\,\Omega_q}{(2\pi)^{n}}\int_0^R\int_0^T\exp
[-c^2(s^6+r^6\Big)t]r^{p-1}s^{q-1}\,ds\,dr\\
&=\frac{\Omega_p\,\Omega_q}{(2\pi)^{n}}M(t)\\
&=\frac{2^{2-n}}{\pi^{n/2}}
\frac{M(t)}{\Gamma(\frac{p}{2})\Gamma(\frac{q}{2})}
\end{align*}
for any fixed $t>0$ in the spectrum $\Omega$,
where
$$
M(t)=\int_0^R\int_0^T\exp[-c^2(s^6+r^6)t]r^{p-1}s^{q-1}\,ds\,dr
$$
is a function of $t$,
$\Omega_p=2\pi^{p/2}/\Gamma(\frac{p}{2})$
and
$\Omega_q=2\pi^{p/2}/\Gamma(\frac{q}{2})$.
Thus, for any fixed $t>0$, $E(x,t)$ is bounded.

(5) This statement follows from \eqref{3.8}.
\end{proof}

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


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\end{document}