On the B-convolutions of the Bessel diamond kernel of Riesz

Abstract

In this article, the operator ${♢}_B^k$ is introduced and named as the Bessel diamond operator iterated k-times and is defined by

$${♢}_B^k=[(B_{x_1}+B_{x_2}+\cdots +B_{x_p}{)}^2-(B_{x_{p+1}}+\cdots +B_{x_{p+q}}{)}^2{]}^k$$

where $p+q=n\text{,}{B}_{x_i}=\frac{{\partial }^2}{\partial x_i^2}+\frac{2v_i}{x_i}\frac{\partial }{\partial x_i}$, $2v_i=2{\alpha }_i+1$, ${\alpha }_i>-\frac{1}{2}$ [8], $x_i>0\text{,}i=1\text{,}2\text{,}\dots \text{,}n\text{,}k$ is a nonnegative integer and n is the dimension of the ${\mathbb{R}}_n^{+}$. In this work, we study the elementary solution of the operator ${♢}_B^k$ and this elementary solution is called the Bessel diamond kernel of Riesz. Then, we study the B-convolution of this elementary solution.

Introduction

Gelfand and Shilov [2] have first introduced the elementary solution of the n-dimensional classical diamond operator. Trione [14] has shown that the n-dimensional ultra-hyperbolic equation has $u(x)=R_{2k}(x)$ as a unique elementary solution. Later, Tellez [15] has proved that $R_{2k}(x)$ exists only for case p is odd with $p+q=n$. Kananthai [3], [4], [5] has proved that the distribution related to the n-dimensional ultra-hyperbolic equation, the solutions of n-dimensional classical diamond operator and Fourier transformation of the diamond kernel of Marcel Riesz, and has showed that the solution of the convolution form $u(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)$ is an unique elementary solution of the ${♢}^{k}u(x)=\delta \text{.}$ Furthermore, Yıldırım et al. [13] have introduced the Bessel diamond operator and have proved that the convolution solution $u(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)$ is the elementary solution of ${♢}_B^k=\delta$ where ${♢}_B^k$ is the Bessel diamond operator iterated k times with $x\in {\mathbb{R}}_n^{+}=\{x:x=(x_1\text{,}\dots \text{,}{x}_n)\text{,}{x}_1>0\text{,}\dots \text{,}{x}_n>0\}$

$${♢}_B^k=[(B_{x_1}+B_{x_2}+\cdots +B_{x_p}{)}^2-(B_{x_{p+1}}+\cdots +B_{x_{p+q}}{)}^2{]}^k\text{,}p+q=n\text{.}$$

Here $S_{2k}(x)$ and $R_{2k}(x)$ are defined by (1.7), (1.8) respectively. Later, by considering the results in [13], Sarıkaya and Yıldırım have proved that the Bessel diamond operator and the nonlinear Bessel diamond operator related to the n-dimensional wave equation [10] and they have studied, in [9], the compound Bessel ultra-hyperbolic equation of the form

$$\sum _{k=0}^m{C}_k{\square }_B^{k}u(x)=f(x)\text{,}$$

where ${\square }_B^k$ is the Bessel ultra-hyperbolic type operator iterated k-times with $x\in {\mathbb{R}}_n^{+}$

$${\square }_B^k=(B_{x_1}+B_{x_2}+\cdots +B_{x_p}-B_{x_{p+1}}-\cdots -B_{x_{p+q}}{)}^k\text{,}p+q=n\text{.}$$

Now $u(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)$ is called the Bessel diamond kernel of Riesz and defined such a kernel by

$$T_k(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)\text{.}$$

In this paper, we study the existence of $T_k(x{)}^{\ast }{T}_m(x)$ and moreover the inverse $T_k^{\ast -1}$ of $T_k(x)$ in the B-convolution algebra $u^{\prime }$ is also considered. If $v_i=0$ for $i=1\text{,}\dots \text{,}n$, then the Bessel diamond operator is called the classical diamond operator iterated k-times. This case has been studied by Kananthai and others (see [2], [3], [4], [5], [14], [15]).

The function $E(x)=-S_2(x)$ as defined by (1.6) is an elementary solution of the Laplace–Bessel operator

$${\mathrm{\Delta }}_B=\sum _{i=1}^n{B}_{x_i}=\sum _{i=1}^n\left(\frac{{\partial }^2}{\partial x_i^2}+\frac{2v_i}{x_i}\frac{\partial }{\partial x_i}\right)$$

that is, ${\Delta }_{B}E(x)=\delta$ where $x\in {\mathbb{R}}_n^{+}$.

The operator ${♢}_B^k$ can be expressed as the product of the operators ${\square }_B^k$ and ${\mathrm{\Delta }}_B^k$, that is

$${♢}_B^k={\left[{\left(\sum _{i=1}^p{B}_{x_i}\right)}^2-{\left(\sum _{i=p+1}^{p+q}{B}_{x_i}\right)}^2\right]}^k={\left[\sum _{i=1}^p{B}_{x_i}-\sum _{i=p+1}^{p+q}{B}_{x_i}\right]}^k{\left[\sum _{i=1}^p{B}_{x_i}+\sum _{i=p+1}^{p+q}{B}_{x_i}\right]}^k={\square }_B^k{\mathrm{\Delta }}_B^k\text{.}$$

Denoted by $T_x^y$ the generalized shift operator acting according to the law [8]

$$T_x^y\phi (x)=C_v^{\ast }{\int }_0^{\pi }\cdots {\int }_0^{\pi }\phi \left(\sqrt{x_1^2+y_1^2-2x_1{y}_1\cos {\theta }_1}\text{,}\dots \text{,}\sqrt{x_n^2+y_n^2-2x_n{y}_n\cos {\theta }_n}\right)\left(\prod _{i=1}^n{\sin }^{2v_i-1}{\theta }_i\right)d{\theta }_1\cdots d{\theta }_n\text{,}$$

where $x\text{,}y\in {\mathbb{R}}_n^{+}$, $C_v^{\ast }={\prod }_{i=1}^n\frac{\Gamma (v_i+1)}{\Gamma (\frac{1}{2})\Gamma (v_i)}$. We remark that this shift operator is closely connected with the Bessel differential operator [8]

$$\begin{array}{cc}\hfill & \frac{d^{2}U}{{\mathit{dx}}^2}+\frac{2v}{x}\frac{\mathit{dU}}{\mathit{dx}}=\frac{d^{2}U}{{\mathit{dy}}^2}+\frac{2v}{y}\frac{\mathit{dU}}{\mathit{dy}}\text{,}\hfill \\ \hfill & U(x\text{,}0)=f(x)\text{,}\hfill \\ \hfill & U_y(x\text{,}0)=0\text{.}\hfill \end{array}$$

The convolution operator determined by the $T^y$ is as follows:

$$(f\ast \phi )(x)={\int }_{{\mathbb{R}}_n^{+}}f(y)T_x^y\phi (x)\left(\prod _{i=1}^n{y}_i^{2v_i}\right)\mathit{dy}\text{.}$$

Convolution (1.4) known as a B-convolution. We note the following properties of the B-convolution and the generalized shift operator.

• $T_x^y.1=1$;

• $T_x^0.f(x)=f(x)$;

• If $f(x)\text{,}g(x)\in C({\mathbb{R}}_n^{+})$, $g(x)$ is a bounded function all $x>0$ and

  $${\int }_{{\mathbb{R}}_n^{+}}\mid f(x)\mid \left(\prod _{i=1}^n{x}_i^{2v_i}\right)\mathit{dx}<\infty$$

  then

  $${\int }_{{\mathbb{R}}_n^{+}}{T}_x^{y}f(x)g(y)\left(\prod _{i=1}^n{y}_i^{2v_i}\right)\mathit{dy}={\int }_{{\mathbb{R}}_n^{+}}f(y)T_x^{y}g(x)\left(\prod _{i=1}^n{y}_i^{2v_i}\right)\mathit{dy}\text{.}$$

• From c, we have the following equality for $g(x)=1$.

  $${\int }_{{\mathbb{R}}_n^{+}}{T}_x^{y}f(x)\left(\prod _{i=1}^n{y}_i^{2v_i}\right)\mathit{dy}={\int }_{{\mathbb{R}}_n^{+}}f(y)\left(\prod _{i=1}^n{y}_i^{2v_i}\right)\mathit{dy}\text{.}$$

• $(f\ast g)(x)=(g\ast f)(x)$.

The Fourier–Bessel transformation and its inverse transformation are defined as follows [11], [12]

$$\begin{array}{cc}\hfill & (F_{B}f)(x)=C_v{\int }_{{\mathbb{R}}_n^{+}}f(y)\left(\prod _{i=1}^n{j}_{v_i-\frac{1}{2}}(x_i{y}_i)y_i^{2v_i}\right)\mathit{dy}\text{,}\hfill \\ \hfill & (F_B^{-1}f)(x)=(F_{B}f)(-x)\text{,}{C}_v={\left(\prod _{i=1}^n{2}^{v_i-\frac{1}{2}}\Gamma \left(v_i+\frac{1}{2}\right)\right)}^{-1}\text{,}\hfill \end{array}$$

where $j_{v_i-\frac{1}{2}}(x_i{y}_i)$ is the normalized Bessel function which is the eigenfunction of the Bessel differential operator. There are the following equalities for Fourier–Bessel transformation [6], [7]

$$\begin{array}{cc}\hfill & F_B\delta (x)=1\text{,}\hfill \\ \hfill & F_B(f\ast g)(x)=F_{B}f(x)·F_{B}g(x)\text{.}\hfill \end{array}$$

The proof of the following lemmas can be seen our paper in [11], [12], [13].

Lemma 1

There is a following equality for Fourier–Bessel transformation

$$F_B(\mid x{\mid }^{-\alpha })=2^{n+2\mid v\mid -2\alpha }\Gamma \left(\frac{n+2\mid v\mid -\alpha }{2}\right){\left[\Gamma \left(\frac{\alpha }{2}\right)\right]}^{-1}\mid x{\mid }^{\alpha -n-2\mid v\mid }$$

where $\mid v\mid =v_1+\cdots +v_n$.

Lemma 2

Given the equation ${\mathrm{\Delta }}_{B}E(x)=\delta$ for $x\in R_n^{+}$, where ${\mathrm{\Delta }}_B$ is the Laplace–Bessel operator defined by (1.3). $E(x)=-S_2(x)$ is an elementary solution of the operator ${\mathrm{\Delta }}_B$ where

$$S_2(x)=\frac{2^{n+2\mid v\mid -4}\Gamma \left(\frac{n+2\mid v\mid -2}{2}\right)}{{\prod }_{i=1}^n{2}^{v_i-\frac{1}{2}}\Gamma \left(v_i+\frac{1}{2}\right)}\mid x{\mid }^{2-n-2\mid v\mid }\text{.}$$

Lemma 3

Given the equation ${\mathrm{\Delta }}_B^{k}u(x)=\delta$ for $x\in R_n^{+}$, where ${\mathrm{\Delta }}_B^k$ is the Laplace–Bessel operator iterated k times defined by (1.3). Then $u(x)=(-1{)}^k{S}_{2k}(x)$ is an elementary solution of the operator ${\mathrm{\Delta }}_B^k$ where

$$S_{2k}(x)=\frac{2^{n+2\mid v\mid -4k}\Gamma \left(\frac{n+2\mid v\mid -2k}{2}\right)}{{\prod }_{i=1}^n{2}^{v_i-\frac{1}{2}}\Gamma \left(v_i+\frac{1}{2}\right)\Gamma (k)}\mid x{\mid }^{2k-n-2\mid v\mid }\text{.}$$

Lemma 4

If ${\square }_B^{k}u(x)=\delta$ for $x\in {\Gamma }_{+}=\{x\in {\mathbb{R}}_n^{+}:x_1>0\text{,}{x}_2>0\text{,}\dots \text{,}{x}_n>0\mathit{and}V>0\}$, where ${\square }_B^k$ is the Bessel ultra-hyperbolic operator iterated k-times defined by (1.3). Then $u(x)=R_{2k}(x)$ is the unique elementary solution of the operator ${\square }_B^k$ where

$$R_{2k}(x)=\frac{V^{\left(\frac{2k-n-2\mid v\mid }{2}\right)}}{K_{n\text{,}v}(2k)}=\frac{(x_1^2+x_2^2+\cdots +x_p^2-x_{p+1}^2-\cdots -x_{p+q}^2{)}^{\left(\frac{2k-n-2\mid v\mid }{2}\right)}}{K_{n\text{,}v}(2k)}$$

for

$$K_{n\text{,}v}(2k)=\frac{{\pi }^{\frac{n+2\mid v\mid -1}{2}}\Gamma \left(\frac{2+2k-n-2\mid v\mid }{2}\right)\Gamma \left(\frac{1-2k}{2}\right)\Gamma (2k)}{\Gamma \left(\frac{2+2k-p}{2}\right)\Gamma \left(\frac{p-2k}{2}\right)}\text{.}$$

Lemma 5

The functions $S_{2k}(x)$ and $R_{2k}(x)$ are homogeneous distributions of order $(2k-n-2\mid v\mid )$ for $\mathit{Re}2k<n+2\mid v\mid$. In particular, the B-convolution $S_{2k}(x)\ast R_{2k}(x)$ exists and is a tempered distribution.

Proof

We need to show that $R_{2k}(x)$ satisfies the Euler equation

$$(2k-n-2\mid v\mid )R_{2k}(x)=\sum _{i=1}^n{x}_i\frac{\partial R_{2k}(x)}{\partial x_i}\text{.}$$

Now

$$\sum _{i=1}^n{x}_i\frac{\partial R_{2k}(x)}{\partial x_i}=\frac{1}{K_{n\text{,}v}(2k)}\sum _{i=1}^n{x}_i\frac{\partial }{\partial x_i}(x_1^2+x_2^2+\cdots +x_p^2-x_{p+1}^2-\cdots -x_{p+q}^2{)}^{\left(\frac{2k-n-2\mid v\mid }{2}\right)}=\frac{(2k-n-2\mid v\mid )}{K_{n\text{,}v}(2k)}(x_1^2+x_2^2+\cdots +x_p^2-x_{p+1}^2-\cdots -x_{p+q}^2{)}^{\left(\frac{2k-n-2\mid v\mid }{2}-1\right)}\times (x_1^2+x_2^2+\cdots +x_p^2-x_{p+1}^2-\cdots -x_{p+q}^2)=\frac{(2k-n-2\mid v\mid )}{K_{n\text{,}v}(2k)}(x_1^2+x_2^2+\cdots +x_p^2-x_{p+1}^2-\cdots -x_{p+q}^2{)}^{\left(\frac{2k-n-2\mid v\mid }{2}\right)}=(2k-n-2\mid v\mid )R_{2k}(x)\text{.}$$

Hence $R_{2k}(x)$ is a homogeneous distribution of order $(2k-n-2\mid v\mid )$ as required and similarly $S_{2k}(x)$ is also a homogeneous distribution of order $(2k-n-2\mid v\mid )$. Now choose ${\mathit{suppR}}_{2k}=K\subset {\overline{\Gamma }}_{+}$ where K is a compact set and ${\overline{\Gamma }}_{+}$ designates of closure ${\Gamma }_{+}$. Then, $R_{2k}$ is a tempered distribution with compact support and by [1], $S_{2k}(x)\ast R_{2k}(x)$ exists and is a tempered distribution. □

Lemma 6

• Let $S_{2k}(x)$ and $S_{2m}(x)$ be defined by (1.7), then $S_{2k}(x)\ast S_{2m}(x)=S_{2k+2m}$, where k and m are nonnegative integer,

• Let $R_{2k}(x)$ and $R_{2m}(x)$ be defined by (1.8), then $R_{2k}(x)\ast R_{2m}(x)=R_{2k+2m}(x)$, where k and m are nonnegative integer,

• Let $R_{2k}(x)$ and $R_{2m}(x)$ be defined by (1.8) and if $R_{2k}(x{)}^{\ast }{R}_{2m}(x)=\delta$, then $R_{2k}(x)$ is an inverse of $R_{2m}$ in the B-convolution algebra, denoted by $R_{2k}(x)=R_{2m}^{\ast -1}(x)$, moreover $R_{2m}^{\ast -1}(x)$ is unique.

Proof

• The proof of Lemma 6(a) can be seen in [13].

• From equation ${\square }_B^{k+m}u(x)=\delta$, we obtain $u(x)=R_{2k+2m}(x)$ by Lemma 4. For any m is a nonnegative integer, we write

  $${\square }_B^{k+m}u(x)={\square }_B^k{\square }_B^{m}u(x)=\delta \text{,}$$

  then by Lemma 4 we have the following equality:

  $${\square }_B^{m}u(x)=R_{2k}(x)\text{.}$$

  B-convolving both sides by $R_{2m}(x)$ we obtain

  $$R_{2m}(x{)}^{\ast }{\square }_B^{m}u(x)=R_{2k}(x{)}^{\ast }{R}_{2m}(x)$$

  or

  $${\square }_B^m{R}_{2m}(x{)}^{\ast }u(x)=R_{2k}(x{)}^{\ast }{R}_{2m}(x)\text{.}$$

  Then from Lemma 4 we have the following equality:

  $${\delta }^{\ast }u(x)=R_{2k}(x{)}^{\ast }{R}_{2m}(x)\text{.}$$

  It follows that:

  $$u(x)=R_{2k}(x{)}^{\ast }{R}_{2m}(x)\text{.}$$

  From the fact that $u(x)=R_{2k+2m}(x)$, we obtain

  $$R_{2k}(x{)}^{\ast }{R}_{2m}(x)=R_{2k+2m}(x)\text{.}$$

• Since $R_{2k}(x)$ and $R_{2m}(x)$ are tempered distributions with compact supports, thus $R_{2k}(x)$ and $R_{2m}(x)$ are the elements of space of B-convolution algebra $u\prime$ of distribution. If $R_{2k}(x)\ast R_{2m}(x)=\delta$, then $R_{2k}(x)=R_{2m}^{\ast -1}(x)$ is a unique inverse of $R_{2m}(x)$ [16]. □

Theorem 1

Given the equation ${♢}_B^{k}u(x)=\delta$ for $x\in R_n^{+}\text{,}$, then $u(x)=(-1{)}^k{S}_{2k}(x{)}^{\ast }{R}_{2k}(x)$ is the unique elementary solution of the operator ${♢}_B^k$ where ${♢}_B^k$ is the diamond Bessel operator iterated k-times defined by (1.1), $S_{2k}(x)$ and $R_{2k}(x)$ are defined by (1.7), (1.8), respectively. Moreover $u(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)$ is a tempered distribution.

The proof of this Theorem can be seen in [13].

Theorem 2

Let $T_k(x)$ the Bessel diamond kernel of Riesz which is defined by (1.2), then $T_k(x)$ is a tempered distribution and can be expressed by

$$T_k(x)=T_{k-r}(x{)}^{\ast }{T}_r(x)$$

where r is a nonnegative integer and $r<k$. Moreover if we put $\ell =k-r\text{,}n=r$, then we obtain

$$T_{\ell +n}(x)=T_{\ell }(x{)}^{\ast }{T}_n(x)\text{<mi mathvariant="italic" is="true">for</mi>}\ell +n=k\text{.}$$

Proof

The $T_k(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)$ is a tempered distribution from Lemma 5. Then, we have ${♢}_B^r{♢}_B^{k-r}{T}_k(x)=\delta$ for $r<k$ from ${♢}_B^k{T}_k(x)=\delta$. From Theorem 1 we have the following equality:

$${♢}_B^{k-r}{T}_k(x)=(-1{)}^r{S}_{2r}(x)\ast R_{2r}(x)\text{.}$$

B-convolving both sides by $(-1{)}^{k-r}{S}_{2k-2r}(x)\ast R_{2k-2r}(x)$, we have

$$[(-1{)}^{k-r}{S}_{2k-2r}(x)\ast R_{2k-2r}(x)]\ast {♢}_B^{k-r}{T}_k(x)=[(-1{)}^{k-r}{S}_{2k-2r}(x{)}^{\ast }{R}_{2k-2r}(x){]}^{\ast }[(-1{)}^r{S}_{2r}(x{)}^{\ast }{R}_{2r}(x)]$$

or

$${♢}_B^{k-r}[(-1{)}^{k-r}{S}_{2k-2r}(x)\ast R_{2k-2r}(x){]}^{\ast }{T}_k(x)=(-1{)}^k[S_{2k-2r}(x{)}^{\ast }{S}_{2r}(x)]\ast [R_{2k-2r}(x{)}^{\ast }{R}_{2r}(x)]\text{.}$$

Since $S_{2k}(x)$ and $R_{2k}(x)$ are tempered distributions and are the elements of the space of B-convolution algebra $u^{\prime }$. From Lemma 6 and Theorem 1, we have

$$\begin{array}{cc}\hfill & \delta \ast T_k(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)\hfill \\ \hfill & T_k(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)\text{.}\hfill \end{array}$$

From (1.9) there is the following equality:

$$T_k(x)=T_{k-r}(x{)}^{\ast }{T}_r(x)\text{.}$$

Now put $\ell =k-r\text{,}n=r$ it follows that:

$$T_{\ell }(x)\ast T_n(x)=T_{\ell +n}(x)=T_k(x)\text{,}$$

as required. □

Theorem 3

Let $T_k(x)$ the Bessel diamond kernel of Riesz which is defined by (1.2), then $T_m(x)$ is an element of the space $u\prime$ of B-convolution algebra and there exist an inverse $T_m^{\ast -1}$ of $T_m(x)$ such that

$$T_m(x)\ast T_m^{\ast -1}=T_m^{\ast -1}\ast T_m(x)=\delta \text{.}$$

Proof

The $T_k(x)=(-1{)}^k{S}_{2k}(x)\ast R_{2k}(x)$ is a tempered distribution from Lemma 5. Therefore the supports of $S_{2k}(x)$ and $R_{2k}(x)$ are compact. Then they are the elements of the space of B-convolution algebra $u\prime$ of distribution. By Lemma 6(c) there exist a unique inverse $T_m^{\ast -1}$ such that

$$T_m(x)\ast T_m^{\ast -1}=T_m^{\ast -1}\ast T_m(x)=\delta \text{.}$$

This completes the proof. □