\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 $$\label{1.1} \frac{\partial}{\partial t} u(x,t)=c^2\Delta u(x,t),$$ with the initial condition $u(x,0)=f(x)$, has solution $$\label{1.2} u(x,t)=\frac{1}{(4c^2\pi t)^{n/2}}\int_{\mathbb{R}^n}\exp\Big(-\frac{|x-y|^2}{4c^2t} \Big)f(y)dy\,,$$ 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 $$\label{1.3} E(x,t)=\frac{1}{(4c^2\pi t)^{n/2}}\exp\big(-\frac{|x|^2}{4c^2t}\big),$$ 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 $$\label{1.4} \diamondsuit=\Big(\sum^p_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^2 -\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x^2_j}\Big)^2,\quad\text{with } p+q=n\,.$$ This operator can be written in the form $\diamondsuit=\Delta\Box=\Box\Delta$, where $$\label{1.5} \Delta = \frac{\partial^2}{\partial x^2_1}+\frac{\partial^2}{\partial x^2_2}+\dots+\frac{\partial^2}{\partial x^2_n}$$ is the Laplacian, and $$\label{1.6} \Box = \frac{\partial^2}{\partial x_1^2}+ \frac{\partial^2}{\partial x_2^2}+\dots+\frac{\partial^2}{\partial x_p^2}-\frac{\partial^2}{\partial x_{p+1}^2}-\frac{\partial^2}{\partial x_{p+2}^2}-\dots-\frac{\partial^2}{\partial x_{p+q}^2}$$ 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 $$\label{1.7} \frac{\partial}{\partial t}\,u(x,t)-c^2\circledast u(x,t)=0,$$ 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 $$\label{1.8} E(x,t)=\frac{1}{(2\pi)^n}\int_{\Omega} \exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3 +\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x) \Big]d\xi\,,$$ 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 $$\label{2.1} \widehat{f}(\xi)=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} e^{-i(\xi,x)}f(x)\,dx,$$ 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 $$\label{2.2} f(x)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{i(\xi,x)} \widehat{f}(\xi)\,d\xi.$$ If $f$ is a distribution with compact support by \cite[Theorem 7.4-3]{z1}, we can write $$\label{2.3} \widehat{f}(\xi)=\frac{1}{(2\pi)^{n/2}}\langle f(x), e^{-i(\xi,x)}\rangle.$$ \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 $$\label{2.4} \widehat{E(\xi,t)}= \begin{cases} \frac{1}{(2\pi)^{n/2}}\exp\Big[-c^2\Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3 +\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)t\Big]& \text{for }\xi\in\Gamma_+,\\ 0& \text{for }\xi\notin \Gamma_+. \end{cases}$$ \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 $$\label{2.5} L =\frac{\partial}{\partial t}-c^2\Big[\Big(\sum^p_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^3 +\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x^2_j}\Big)^3\Big],$$ 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 $$\label{2.6} E(x,t)=\frac{1}{(2\pi)^n}\int_{\Omega} \exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3 +\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x)\Big]d\xi.$$ 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 $$\label{3.1} \frac{\partial}{\partial t}\,u(x,t)-c^2\circledast u(x,t)=0$$ with initial condition $$\label{3.2} u(x,0)=f(x)$$ 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 $$\label{3.3} \widehat{u}(\xi,t)=K(\xi)\exp\Big[-c^2 \Big(\Big(\sum^p_{i=1}\xi^2_i\Big)^3 +\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big)t\Big]$$ where $K(\xi)$ is constant and $\widehat{u}(\xi,0)=K(\xi)$. By \eqref{3.2} we have $$\label{3.4} K(\xi)=\widehat{u}(\xi,0)=\widehat{f}(\xi) =\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{-i(\xi,x)}f(x)\,dx$$ 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 $$\label{3.5} u(x,t)=\frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3 +\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x-y) \Big]f(y)\,dy\,d\xi.$$ Set $$\label{3.6} E(x,t)=\frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n} \exp\Big[-c^2\Big[\Big(\sum^p_{i=1}\xi^2_i\Big)^3 +\Big(\sum^{p+q}_{j=p+1}\xi^2_j\Big)^3\Big]t+i(\xi,x) \Big]\,d\xi.$$ We choose $\Omega\subset\mathbb{R}^n$ be the spectrum of $E(x,t)$ and by \eqref{2.5}, we have \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} 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, $$\lim_{t\to 0} E(x,t) =\frac{1}{(2\pi)^n}\int_{\Omega}e^{i(\xi,x)}\,d\xi =\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i(\xi,x)}\,d\xi =\delta(x), \label{3.8}$$ 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. \begin{thebibliography}{00} \bibitem{h1} R. Haberman; \emph{Elementary Applied Partial Differential Equations}, 2-nd Edition, Prentice-Hall International, Inc. (1983). \bibitem{j1} F. John; \emph{Partial Differential Equations}, 4-th Edition, Springer-Verlag, New York, (1982). \bibitem{k1} A. Kananthai; \emph{On the Fourier Transform of the Diamond Kernel of Marcel Riesz}, Applied Mathematics and Computation 101:151-158 (1999). \bibitem{k2} A. Kananthai; \emph{On the Solution of the n-Dimensional Diamond Operator}, Applied Mathematics and Computational 88:27-37 (1997). \bibitem{n1} K. Nonlaopon, A. 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