On the weak solutions of the compound Bessel ultra-hyperbolic equation

Abstract

In this article, we have studied the compound Bessel ultra-hyperbolic equation of the form

$$\sum _{r=0}^m{C}_r{\square }_{\text{B}}^{r}u(x)=f(x)\text{,}$$

where ${\square }_{\text{B}}^r$ is the Bessel ultra-hyperbolic type operator iterated r-times, f is a given generalized function, u is an unknown function, $x\in {\mathbb{R}}_n^{+}$ and C_r is a constant. In this work we study the weak solution u(x) of above the equation which is of the form Bessel Diamond operator and moreover, such a solution is unique.

Introduction

Gelfand and Shilov [2] have first introduced the elementary solution of the n-dimensional classical diamond operator. Trione [13] has shown that the n-dimensional ultra-hyperbolic equation has $u(x)=R_{2k}(x)$ as a unique elementary solution. Later Tellez [14] has proved that $R_{2k}(x)$ exists only for case p is odd with $p+q=n$. Kananthai [3], [4], [5], [6] has proved that the distribution related to the n-dimensional ultra-hyperbolic equation, the solutions of n-dimensional classical diamond operator and weak solution of the compound ultra-hyperbolic equation, 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 ${◊}^{k}u(x)=\delta$. Furthermore, Yıldırım et al. [12] have introduced the Bessel diamond operator ${◊}_{\text{B}}^k$ and have studied the elementary solution of this operator and also the Fourier–Bessel transform of the elementary solution.

In this article, we will define the Bessel ultra-hyperbolic type operator iterated k times with $x\in {\mathbb{R}}_n^{+}=\{x:x=(x_1\text{,}\dots \text{,}{x}_n)\text{,}{x}_1>0\text{,}\dots \text{,}{x}_n>0\}$

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

Consider the equation

$${\square }_{\text{B}}^{k}u(x)=f(x)\text{,}$$

where $u(x)$ and f are some generalized functions.

We will develop the Eq. (2) to the form

$$\sum _{r=0}^m{C}_r{\square }_{\text{B}}^{r}u(x)=f(x)\text{,}$$

which is called the compound Bessel ultra-hyperbolic and by convention ${\square }_{\text{B}}^{0}u(x)=u(x)$. In finding the solutions of (3), we use the properties of B-convolutions for the generalized functions.

We give some notations and definitions.

The function $E(x)$ as defined by (8) is an elementary solution of the Laplace–Bessel operator

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

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

The operator ${◊}_{\text{B}}^k$ can be expressed as the product of the operators □_B and Δ_B, that is

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

Denoted by T^y the generalized shift operator acting according to the law [9]

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

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 [9]:

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

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

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

Convolution (6) 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

  $${\int }_0^{\infty }|f(x)|\left(\prod _{i=1}^n{x}_i^{2v_i}\right)\mathrm{d}x<\infty \text{,}$$

  then

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

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

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

• $(f\ast g)(x)=(g\ast f)(x)$.

The Fourier–Bessel transformation and its inverse transformation are defined as follows [10], [11]:

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

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 following equalities for Fourier–Bessel transformation [7], [8]:

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

Lemma 1

There is a following equality for Fourier–Bessel transformation

$$F_{\mathrm{B}}(|x{|}^{-\alpha })=2^{n+2|v|-2\alpha }\Gamma \left(\frac{n+2|v|-\alpha }{2}\right){\left[\Gamma \left(\frac{\alpha }{2}\right)\right]}^{-1}|x{|}^{\alpha -n-2|v|}\text{,}$$

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

The proof of this lemma is given in [10], [11].

Lemma 2

Given the equation ${\mathrm{\Delta }}_{\mathrm{B}}E(x)=\delta$ for $x\in R_n^{+}$, where Δ_B is the Laplace–Bessel operator defined by (4). Then

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

is an elementary solution of the operator Δ_B.

The proof of this lemma is given in [12].

Lemma 3

Given the equation ${\mathrm{\Delta }}_{\mathrm{B}}^{k}u(x)=\delta$ for $x\in R_n^{+}$, where ${\mathrm{\Delta }}_{\mathrm{B}}^k$ is the Laplace–Bessel operator iterated k times defined by (4). Then $u(x)=(-1{)}^k{S}_{2k}(x)$ is an elementary solution of the operator ${\mathrm{\Delta }}_{\mathrm{B}}^k$, where

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

The proof of Lemma 3 is similar to the proof of Lemma 2.

Lemma 4

If ${\square }_{\mathrm{B}}^{k}u(x)=\delta$ for $x\in {\Gamma }_{+}=\{x\in {\mathbb{R}}_n^{+}:x_1>0\text{,}{x}_2>0\text{,}\dots \text{,}{x}_n>0$ and $V>0\}$ , where ${\square }_{\mathrm{B}}^k$ is the Bessel-ultra-hyperbolic operator iterated k-times defined by (1). Then $u(x)=R_{2k}(x)$ is the unique elementary solution of the operator ${\square }_{\mathrm{B}}^k$, where

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

for

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

The proof of this lemma can be from Lemma 1, Lemma 2, Lemma 3.

Lemma 5

$R_{2k}(x)$ is homogeneous distributions of order $(2k-n-2|v|)$. In particular, it is a tempered distribution.

Proof

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

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

Now

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

Hence $R_{2k}(x)$ is a homogeneous distribution of order $(2k-n-2|v|)$. Now chosen $\text{supp}{R}_{2k}=K\subset {\overline{\Gamma }}_{+}$ where K is a compact set and ${\overline{\Gamma }}_{+}$ designates of closure Γ₊. Then R_2k is a tempered distribution with compact support. □

Lemma 6

• $({\square }_{\mathrm{B}}^k\delta )\ast u(x)={\square }_{\mathrm{B}}^{k}u(x)$ where u is any tempered distribution.

• Let $R_{2k}(x)$ and $R_{2m}(x)$ be defined by (10), then $R_{2k}(x)\ast R_{2m}(x)$ exists and is a tempered distribution.

• Let $R_{2k}(x)$ and $R_{2m}(x)$ be defined by (10), then $R_{2k}(x)\ast R_{2m}(x)=R_{2k+2m}(x)$, where k and m are a nonnegative integer.

• Let $R_{2k}(x)$ and $R_{2m}(x)$ be defined by (10) 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

• First, we consider the case $k=1$, now

  $${\square }_{\text{B}}\delta (x)=\sum _{i=1}^p\left(\frac{{\mathrm{\partial }}^2\delta }{\mathrm{\partial }{x}_i^2}+\frac{2v_i}{x_i}\frac{\mathrm{\partial }\delta }{\mathrm{\partial }{x}_i}\right)-\sum _{j=p+1}^{p+q}\left(\frac{{\mathrm{\partial }}^2\delta }{\mathrm{\partial }{x}_j^2}+\frac{2v_j}{x_j}\frac{\mathrm{\partial }\delta }{\mathrm{\partial }{x}_j}\right)\text{,}p+q=n$$

  and let $\phi (x)$ be a testing function in the Schwarts space S. By the definition odd B-convolution, we have

  $$〈{\square }_{\text{B}}\delta \ast u(x)\text{,}\phi (x)〉=〈u(x)\text{,}〈{\square }_{\text{B}}\delta (y)\text{,}\phi (x+y)〉〉=〈u(x)\text{,}〈\sum _{i=1}^p\left(\frac{{\mathrm{\partial }}^2\delta (y)}{\mathrm{\partial }{x}_i^2}+\frac{2v_i}{x_i}\frac{\mathrm{\partial }\delta (y)}{\mathrm{\partial }{x}_i}\right)-\sum _{j=p+1}^{p+q}\left(\frac{{\mathrm{\partial }}^2\delta (y)}{\mathrm{\partial }{x}_j^2}+\frac{2v_j}{x_j}\frac{\mathrm{\partial }\delta (y)}{\mathrm{\partial }{x}_j}\right)\text{,}\phi (x+y)〉〉=〈u(x)\text{,}〈\delta (y)\text{,}\sum _{i=1}^p\left(\frac{{\mathrm{\partial }}^2\phi (x+y)}{\mathrm{\partial }{x}_i^2}+\frac{2v_i}{x_i}\frac{\mathrm{\partial }\phi (x+y)}{\mathrm{\partial }{x}_i}\right)-\sum _{j=p+1}^{p+q}\left(\frac{{\mathrm{\partial }}^2\phi (x+y)}{\mathrm{\partial }{x}_j^2}+\frac{2v_j}{x_j}\frac{\mathrm{\partial }\phi (x+y)}{\mathrm{\partial }{x}_j}\right)〉〉=〈u(x)\text{,}\sum _{i=1}^p\left(\frac{{\mathrm{\partial }}^2\phi (x)}{\mathrm{\partial }{x}_i^2}+\frac{2v_i}{x_i}\frac{\mathrm{\partial }\phi (x)}{\mathrm{\partial }{x}_i}\right)-\sum _{j=p+1}^{p+q}\left(\frac{{\mathrm{\partial }}^2\phi (x)}{\mathrm{\partial }{x}_j^2}+\frac{2v_j}{x_j}\frac{\mathrm{\partial }\phi (x)}{\mathrm{\partial }{x}_j}\right)〉=〈\sum _{i=1}^p\left(\frac{{\mathrm{\partial }}^{2}u(x)}{\mathrm{\partial }{x}_i^2}+\frac{2v_i}{x_i}\frac{\mathrm{\partial }u(x)}{\mathrm{\partial }{x}_i}\right)-\sum _{j=p+1}^{p+q}\left(\frac{{\mathrm{\partial }}^{2}u(x)}{\mathrm{\partial }{x}_j^2}+\frac{2v_j}{x_j}\frac{\mathrm{\partial }u(x)}{\mathrm{\partial }{x}_j}\right)\text{,}\phi (x)〉=〈{\square }_{\text{B}}u(x)\text{,}\phi (x)〉\text{.}$$

  It follows that

  $${\square }_{\text{B}}\delta \ast u(x)={\square }_{\text{B}}u(x)\text{.}$$

  Similarly for any k we can show that

  $${\square }_{\text{B}}^k\delta \ast u(x)={\square }_{\text{B}}^{k}u(x)\text{.}$$

• Since R_2k and R_2m are tempered distribution by Lemma 5. Now chosen $\text{supp}{R}_{2k}=K\subset {\overline{\Gamma }}_{+}$ where K is a compact set and ${\overline{\Gamma }}_{+}$ designates of ${\Gamma }_{+}$ closure. Then R_2k is a tempered distribution with compact support by [1] $R_{2k}(x)\ast R_{2m}(x)$ exists and is a tempered distribution.

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

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

  then by Lemma 4 we have the following equality:

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

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

  $$R_{2m}(x)\ast {\square }_{\text{B}}^{m}u(x)=R_{2k}(x)\ast R_{2m}(x)\text{,}$$

  or

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

  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′ of distribution. Now $R_{2k}(x)\ast R_{2m}(x)=\delta$ then by Zemanian [15] show that $R_{2k}(x)=R_{2m}^{\ast -1}(x)$ is a unique inverse. □

Lemma 7

Let $R_{2k}(x)$ and $K_{n\text{,}v}(2k)$ be defined by (10), (11). Then

• $K_{n\text{,}v}(2k+2)=2k(2k+2-n-2|v|)K_{n\text{,}v}(2k)$,

• ${\square }_{\mathrm{B}}^k{R}_{2m}(x)=R_{2m-2k}(x)$, where k and m are a nonnegative integer,

• $R_{-2k}(x)={\square }_{\mathrm{B}}^k\delta (x)$, where k is a nonnegative integer.

Proof

• From (11) we have

  $$K_{n\text{,}v}(2k+2)=\frac{{\pi }^{\frac{n+2|v|-1}{2}}\Gamma \left(\frac{4+2k-n-2|v|}{2}\right)\Gamma \left(\frac{-1-2k}{2}\right)\Gamma (2k+2)}{\Gamma \left(\frac{4+2k-p-2|v|}{2}\right)\Gamma \left(\frac{p+2|v|-2k-2}{2}\right)}=\frac{{\pi }^{\frac{n+2|v|-1}{2}}\frac{2+2k-n-2|v|}{2}\Gamma \left(\frac{2+2k-n-2|v|}{2}\right)\frac{-2}{1+2k}\Gamma \left(\frac{1-2k}{2}\right)2k(2k+1)\Gamma (2k)}{\frac{2+2k-p-2|v|}{2}\Gamma \left(\frac{2+2k-p-2|v|}{2}\right)\frac{2}{p+2|v|-2k-2}\Gamma \left(\frac{p+2|v|-2k}{2}\right)}=2k(2k+2-n-2|v|)K_{n\text{,}v}(2k)\text{.}$$

• By Lemma 6(c) we have

  $$\begin{array}{cc}\hfill & \delta \ast R_{2m}(x)=R_{2k}(x)\ast R_{2m-2k}\text{,}\hfill \\ \hfill & {\square }_{\text{B}}^k{R}_{2k}(x)\ast R_{2m}(x)=R_{2k}(x)\ast R_{2m-2k}\text{,}\hfill \\ \hfill & R_{2k}(x)\ast {\square }_{\text{B}}^k{R}_{2m}(x)=R_{2k}(x)\ast R_{2m-2k}\hfill \end{array}$$

  and

  $${\square }_{\text{B}}^k{R}_{2m}(x)=R_{2m-2k}\text{.}$$

• For m = k, by Lemma 7(b) we have

  $${\square }_{\text{B}}^k{R}_{2k}(x)=R_0\text{,}\delta =R_0\text{.}$$

  For m = 0, by Lemma 7(b) we have

  $${\square }_{\text{B}}^k{R}_0=R_{-2k}(x)\text{,}$$

  or

  $${\square }_{\text{B}}^k\delta =R_{-2k}(x).\square$$

Theorem 1

Given the compound Bessel ultra-hyperbolic equation

$$\sum _{r=0}^m{C}_r{\square }_{\mathrm{B}}^{r}u(x)=f(x)\text{,}$$

where ${\square }_{\mathrm{B}}^r$ is the Bessel ultra-hyperbolic operator iterated r-times defined by (1), f is a tempered distribution, $x\in {\mathbb{R}}_n^{+}$ and n is odd and C_r is a constant. Then (14) has a unique weak solution

$$u(x)=f(x)\ast R_{2m}(x)\ast (C_m{R}_0(x)\ast w(x)R_2(x){)}^{\ast -1}\text{,}$$

where

$$w(x)=C_{m-1}+C_{m-2}\frac{V}{2(4-n-2|v|)}+C_{m-3}\frac{V^2}{2·4(4-n-2|v|)(6-n-2|v|)}+\cdots +C_0\frac{V^{m-1}}{2·4·6\cdots 2(m-1)(4-n-2|v|)(6-n-2|v|)\cdots (2m-n-2|v|)}$$

and V defined by (10) and $(C_m{R}_0(x)\ast w(x)R_2(x){)}^{\ast -1}$ is an inverse of $C_m{R}_0(x)\ast w(x)R_2(x)$.

Proof

By Lemma 6(a), Eq. (14) can be written as

$$(C_m{\square }_{\text{B}}^m\delta +C_{m-1}{\square }_{\text{B}}^{m-1}\delta +\cdots +C_1{\square }_{\text{B}}\delta +C_0\delta )\ast u=f(x)\text{.}$$

B-convolving both sides by $R_{2m}(x)$ defined by (10) we obtain

$$(C_m{\square }_{\text{B}}^m{R}_{2m}+C_{m-1}{\square }_{\text{B}}^{m-1}{R}_{2m}+\cdots +C_1{\square }_{\text{B}}{R}_{2m}+C_0{R}_{2m})\ast u=f(x)\ast R_{2m}\text{.}$$

By Lemma 4, Lemma 7(c) we obtain

$$(C_m\delta +C_{m-1}{R}_2(x)+C_{m-2}{R}_4(x)+\cdots +C_1{R}_{2(m-1)}(x)+C_0{R}_{2m}(x))\ast u=f(x)\ast R_{2m}\text{.}$$

By Lemma 7(a) we obtain

$$R_4(x)=\frac{V^{\frac{4-n-2|v|}{2}}}{K_n(4)}=R_2(x)\frac{V}{2(4-n-2|v|)K_n(2)}\text{,}$$

similarly

$$\begin{array}{cc}\hfill & R_6(x)=\frac{V^{\frac{6-n-2|v|}{2}}}{K_n(6)}=R_4(x)\frac{V}{4(6-n-2|v|)}=R_2(x)\frac{V^2}{2·4(4-n-2|v|)(6-n-2|v|)}\text{,}\hfill \\ \hfill & R_8(x)=R_2(x)\frac{V^3}{2·4·6(4-n-2|v|)(6-n-2|v|)(8-n-2|v|)}\text{,}\hfill \\ \hfill & \cdots \hfill \\ \hfill & R_{2m}(x)=R_2(x)\frac{V^{m-1}}{2·4·6(4-n-2|v|)(6-n-2|v|)(8-n-2|v|)\cdots (2m-n-2|v|)}\text{.}\hfill \end{array}$$

Thus we obtain the function $w(x)$ of (16). Now $w(x)$ is continuous and infinitely differentiable in classical sense for n is odd. Since $R_2(x)$ is a tempered distribution with compact support, hence $w(x)R_2(x)$ also is tempered distribution with compact support and so $C_m{R}_0(x)\ast w(x)R_2(x)$. By Lemma 6(d), $C_m{R}_0(x)\ast w(x)R_2(x)$ has a unique inverse denote by $(C_m{R}_0(x)\ast w(x)R_2(x){)}^{\ast -1}$. Now (17) can be written as

$$(C_m{R}_0(x)\ast w(x)R_2(x))\ast u(x)=f(x)\ast R_{2m}(x)\text{,}{R}_0=\delta \text{.}$$

B-convolving both sides by $(C_m{R}_0(x)\ast w(x)R_2(x){)}^{\ast -1}$, we have

$$u(x)=f(x)\ast R_{2m}\ast (C_m{R}_0(x)\ast w(x)R_2(x){)}^{\ast -1}\text{.}$$

Since $R_{2m}(x)$ is a unique by Lemma 4 and $(C_m{R}_0(x)\ast w(x)R_2(x){)}^{\ast -1}$ also is a unique by Lemma 6(d), it follows that $u(x)$ is a unique weak solution of (14) with odd-dimensional n.

This completes the proof. □