\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, $$\label{1.1} \Box^k = \Big(\frac{\partial^2}{\partial x_1^2}+ \frac{\partial^2}{\partial x_2^2}+\dots+\frac{\partial^2}{\partial x_p^2}-\frac{\partial^2}{\partial x_{p+1}^2}-\frac{\partial^2}{\partial x_{p+2}^2}-\dots-\frac{\partial^2}{\partial x_{p+q}^2}\Big)^k .$$ 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, $$\label{1.2} \diamondsuit^k =\Big(\Big(\sum^{p}_{i=1}\frac{\partial^2}{\partial x^2_i}\Big)^2 -\Big(\sum^{p+q}_{j=p+1}\frac{\partial^2}{\partial x^2_j}\Big)^2\Big)^k ,$$ 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 $$\label{1.3} \diamondsuit^k =\triangle^k \Box^k =\Box^k \triangle^k$$ where $\triangle^k$ is the Laplacian operator iterated $k$ times, $$\label{1.4} \triangle^k = \Big(\frac{\partial^2}{\partial x^2_1}+\frac{\partial^2}{\partial x^2_2}+\dots+\frac{\partial^2}{\partial x^2_n}\Big)^k\,.$$ 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, $$\label{1.5} \diamondsuit^k \left((-1)^k R^{e}_{2k}(x)\ast R^{H}_{2k}(x)\right)=\delta\,.$$ 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 \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} 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 \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} Thus, \eqref{1.6} can be written as $$\label{1.8} \oplus^k =\diamondsuit^k \circledcirc^k\,.$$ For $k=1$ the operator $\diamondsuit$ can be expressed in the form $\diamondsuit=\triangle\Box=\Box\triangle$ where $\Box$ is the Ultra-hyperbolic operator $$\label{1.9} \Box = \frac{\partial^2}{\partial x_1^2}+ \frac{\partial^2}{\partial x_2^2}+\dots+\frac{\partial^2}{\partial x_p^2}-\frac{\partial^2}{\partial x_{p+1}^2}-\frac{\partial^2}{\partial x_{p+2}^2}-\dots-\frac{\partial^2}{\partial x_{p+q}^2}.$$ where $p+q=n$ and $\triangle$ is the Lapacian operator $$\label{1.10} \triangle = \frac{\partial^2}{\partial x^2_1}+\frac{\partial^2}{\partial x^2_2}+\dots+\frac{\partial^2}{\partial x^2_n}.$$ By putting $p=1$ and $x_{1}=t(t=time)$ in \eqref{1.9}, we obtain the wave operator $$\label{1.11} \Box = \frac{\partial^2}{\partial t_1^2} -\frac{\partial^2}{\partial x_{2}^2}-\frac{\partial^2}{\partial x_{3}^2}-\dots-\frac{\partial^2}{\partial x_{n}^2}\,.$$ From \eqref{1.6} with $q=0$ and $k=1$, we obtain $$\label{1.12} \oplus=\triangle^{4}_{p}$$ where $$\label{1.13} \triangle_{p} = \frac{\partial^2}{\partial x^2_1}+\frac{\partial^2}{\partial x^2_2}+\dots+\frac{\partial^2}{\partial x^2_p}.$$ Firstly, we can find the elementary solution $G(x)$ of the operator $\oplus^k$; that is, $$\label{1.14} \oplus^k G(x)=\delta,$$ 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 $$\label{2.1} \upsilon = x_1^2 + x_2^2 + \dots + x_p^2 - x_{p+1}^2- x_{p+2}^2 - \dots - x_{p+q}^2$$ 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 $$\label{2.2} R_\alpha^H (\upsilon) = \begin{cases} \frac{\upsilon^{\frac{\alpha - n}{2}}}{K_n (\alpha)}, &\text{for } x \in \Gamma_+, \\ 0, &\text {for } x \not\in \Gamma_+, \end{cases}$$ where the constant $K_n (\alpha)$ is given by the formula $$\label{2.3} K_n (\alpha) = \frac{\pi^{\frac{n-1}{2}} \Gamma ( \frac{2 + \alpha - n}{2} ) \Gamma ( \frac{1 - \alpha}{2}) \Gamma (\alpha)}{\Gamma ( \frac{2 + \alpha - p}{2} ) \Gamma ( \frac{p - \alpha}{2})}.$$ \end{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 $$\label{2.4} M_\alpha^H (\upsilon) = \begin{cases} \frac{\upsilon^{\frac{\alpha - n}{2}}}{H_n (\alpha)}, &\text {for } x \in \Gamma_+, \\ 0, &\text{for } x \not\in \Gamma_+, \end{cases}$$ 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 $$\label{2.5} R_\alpha^e (x) =\frac{ |x|^{\alpha-n}}{W_n(\alpha)},$$ where $$\label{2.6} W_n(\alpha) =\frac{ \pi^{\frac{n}{2}}2^{\alpha}\Gamma(\alpha/2)} {\Gamma\big(\frac{n-\alpha}{2}\big)},$$ 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 $\mathop{\rm Re}(\alpha)0$ 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. \begin{thebibliography}{00} \bibitem{d1} W. 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