# On solvability of a boundary value problem for the Poisson equation with the boundary operator of a fractional order

## Abstract

In this work, we investigate the solvability of a boundary value problem for the Poisson equation. The considered problem is a generalization of the known Dirichlet and Neumann problems on operators of a fractional order. We obtain exact conditions for solvability of the studied problem.

MSC:35J05, 35J25, 26A33.

## 1 Introduction

Let $\mathrm{\Omega }=\left\{x\in {R}^{n}:|x|<1\right\}$ be the unit ball, $n\ge 3$, $\partial \mathrm{\Omega }=\left\{x\in {R}^{n}:|x|=1\right\}$ be the unit sphere, ${|x|}^{2}={x}_{1}^{2}+{x}_{2}^{2}+\cdots +{x}_{n}^{2}$. Further, let $u\left(x\right)$ be a smooth function in the domain Ω, $r=|x|$, $\theta =x/|x|$. For any $0<\alpha$, the expression

${J}^{\alpha }\left[u\right]\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(1-\alpha \right)}{\int }_{0}^{r}{\left(r-\tau \right)}^{\alpha -1}u\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau$

is called the operator of order α in the sense of Riemann-Liouville . From here on, we denote ${J}^{0}\left[u\right]\left(x\right)=u\left(x\right)$. Let $m-1<\alpha \le m$, $m=1,2,\dots$ ,

$\frac{\partial }{\partial r}=\sum _{j=1}^{n}\frac{{x}_{j}}{r}\frac{\partial }{\partial {x}_{j}},\phantom{\rule{2em}{0ex}}\frac{{\partial }^{k}u}{\partial {r}^{k}}=\frac{\partial }{\partial r}\left(\frac{{\partial }^{k-1}u}{\partial {r}^{k-1}}\right),\phantom{\rule{1em}{0ex}}k=1,2,\dots .$

The operators

$\begin{array}{c}{D}^{\alpha }\left[u\right]\left(x\right)=\frac{{\partial }^{m}}{\partial {r}^{m}}{J}^{m-\alpha }\left[u\right]\left(x\right),\hfill \\ {D}_{\ast }^{\alpha }\left[u\right]\left(x\right)={J}^{m-\alpha }\left[\frac{{\partial }^{m}}{\partial {r}^{m}}\left[u\right]\right]\left(x\right)\hfill \end{array}$

are called the derivative of order α in the sense of Riemann-Liouville and Caputo, respectively . Further, let $0\le \beta \le 1$, $0<\alpha \le 1$. Consider the operator

${D}^{\alpha ,\beta }\left[u\right]\left(x\right)={J}^{\beta \left(1-\alpha \right)}\frac{d}{dr}{J}^{\left(1-\beta \right)\left(1-\alpha \right)}u\left(x\right).$

${D}^{\alpha ,\beta }$ is said to be the derivative of the order α in the Riemann-Liouville sense and of type β.

We note that the operator ${D}^{\alpha ,\beta }$ was introduced in . Some questions concerning solvability for differential equations of a fractional order connected with ${D}^{\alpha ,\beta }$ were studied in [3, 4]. Introduce the notations:

$\begin{array}{c}{B}^{\alpha ,\beta }\left[u\right]\left(x\right)={r}^{\alpha }{D}^{\alpha ,\beta }\left[u\right]\left(x\right),\hfill \\ {B}^{-\alpha }\left[u\right]\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -1}{s}^{-\alpha }u\left(sx\right)\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$

In what follows, we denote

$\begin{array}{c}{B}^{\alpha ,0}={B}^{\alpha },\hfill \\ {B}^{\alpha ,1}={B}_{\ast }^{\alpha }.\hfill \end{array}$

## 2 Statement of the problem and formulation of the main result

Consider in the domain Ω the following problem:

$\mathrm{\Delta }u\left(x\right)=g\left(x\right),\phantom{\rule{1em}{0ex}}x\in \mathrm{\Omega },$
(2.1)
${D}^{\alpha ,\beta }\left[u\right]\left(x\right)=f\left(x\right),\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega }.$
(2.2)

We call a solution of the problem (2.1), (2.2) a function $u\left(x\right)\in {C}^{2}\left(\mathrm{\Omega }\right)\cap C\left(\overline{\mathrm{\Omega }}\right)$ such that ${B}^{\alpha ,\beta }\left[u\right]\left(x\right)\in C\left(\overline{\mathrm{\Omega }}\right)$, which satisfies the conditions (2.1) and (2.2) in the classical sense. If $\alpha =1$, then the equality

${B}^{1,\beta }=r\frac{\partial }{\partial r}=\frac{\partial }{\partial \nu }$

holds for all $x\in \partial \mathrm{\Omega }$, here ν is the vector of the external normal to Ω. Therefore, the problem (2.1), (2.2) is represented the Neumann problem in the case of $\alpha =1$, and the Dirichlet problem for the equation (2.1) in the case of $\alpha =0$.

It is known, the Dirichlet problem is undoubtedly solvable, and the Neumann problem is solvable if and only if the following condition is valid :

${\int }_{\partial \mathrm{\Omega }}f\left(x\right)\phantom{\rule{0.2em}{0ex}}d{s}_{x}={\int }_{\mathrm{\Omega }}g\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.$
(2.3)

Problems with boundary operators of a fractional order for elliptic equations are studied in [6, 7]. The problem (2.1), (2.2) is studied for the Riemann-Liouville and Caputo operators in the case of the Laplace equation, i.e., when $g\left(x\right)=0$, in the same works [8, 9]. It is established that the problem (2.1), (2.2) is undoubtedly solvable for the case of the Riemann-Liouville operator

${D}^{\alpha ,0}={D}^{\alpha },\phantom{\rule{1em}{0ex}}0<\alpha <1,$

and in the case of the Caputo operator

${D}^{\alpha ,1}={D}_{\ast }^{\alpha },\phantom{\rule{1em}{0ex}}0<\alpha <1,$

the problem (2.1), (2.2) is solvable if and only if the condition

${\int }_{\partial \mathrm{\Omega }}f\left(x\right)\phantom{\rule{0.2em}{0ex}}d{s}_{x}=0$

is valid, i.e., in this case, the condition for solvability of the problem (2.1), (2.2) coincides with the condition of the Neumann problem.

Let $v\left(x\right)$ be a solution of the Dirichlet problem

$\left\{\begin{array}{ll}\mathrm{\Delta }v\left(x\right)={g}_{1}\left(x\right),& x\in \mathrm{\Omega },\\ v\left(x\right)=f\left(x\right),& x\in \partial \mathrm{\Omega }.\end{array}$
(2.4)

The main result of the present work is the following.

Theorem 2.1 Let $0<\alpha \le 1$, $0\le \beta \le 1$, $0<\lambda <1$, $f\left(x\right)\in {C}^{\lambda +2}\left(\partial \mathrm{\Omega }\right)$, $g\left(x\right)\in {C}^{\lambda +1}\left(\overline{\mathrm{\Omega }}\right)$.

Then:

1. (1)

If $0<\alpha <1$, $0\le \beta <1$, then a solution of the problem (2.1), (2.2) exists, is unique and is represented in the form of

$u\left(x\right)={B}^{-\alpha }\left[v\right]\left(x\right),$
(2.5)

where $v\left(x\right)$ is the solution of the problem (2.4) with the function

${g}_{1}\left(x\right)={|x|}^{-2}{B}^{\alpha ,\beta }\left[{|x|}^{2}g\right]\left(x\right).$
1. (2)

If $0<\alpha \le 1$, $\beta =1$, then the problem (2.1), (2.2) is solvable if and only if the condition

${\int }_{\partial \mathrm{\Omega }}f\left(x\right)\phantom{\rule{0.2em}{0ex}}d{s}_{x}={\int }_{\mathrm{\Omega }}\frac{{|x|}^{2-n}-1}{n-2}{|x|}^{-2}{B}^{\alpha ,1}\left[{|x|}^{2}g\right]\left(x\right)\phantom{\rule{0.2em}{0ex}}dx$
(2.6)

is satisfied.

If a solution of the problem exists, then it is unique up to a constant summand and is represented in the form of (2.5), where $v\left(x\right)$ is the solution of the problem (2.4) with the function

${g}_{1}\left(x\right)={|x|}^{-2}{B}^{\alpha ,1}\left[{|x|}^{2}g\right]\left(x\right),$

satisfying to the condition $v\left(0\right)=0$.

1. (3)

If the solution of the problem exists, then it belongs to the class ${C}^{\lambda +2}\left(\overline{\mathrm{\Omega }}\right)$.

## 3 Properties of the operators ${B}^{\alpha ,\beta }$ and ${B}^{-\alpha }$

It should be noted that properties and applications of the operators ${B}^{\alpha }$, ${B}_{\ast }^{\alpha }$ and ${B}^{-\alpha }$ in the class of harmonic functions in the ball Ω are studied in . Later on, we assume that $u\left(x\right)$ is a smooth function in the domain Ω. The following proposition establishes a connection between operators ${B}^{\alpha ,\beta }$ and ${B}^{\alpha }$.

Lemma 3.1 Let $0<\alpha \le 1$, $0\le \beta \le 1$. Then the equalities

${B}^{\alpha ,\beta }\left[u\right]\left(x\right)=\left\{\begin{array}{ll}{B}^{\alpha }\left[u\right]\left(x\right),& 0\le \beta <1,0<\alpha <1,\\ {B}^{\alpha }\left[u\right]\left(x\right)-\frac{u\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)},& \beta =1,0<\alpha \le 1,\end{array}$
(3.1)

hold for any $x\in \mathrm{\Omega }$.

Proof

Denote

$\begin{array}{c}{\delta }_{1}=\beta \left(1-\alpha \right),\hfill \\ {\delta }_{2}=\left(1-\beta \right)\left(1-\alpha \right).\hfill \end{array}$

Let $0<\alpha <1$, $0\le \beta <1$. Using definition of the operator ${B}^{\alpha ,\beta }$, we obtain

$\begin{array}{rl}{B}^{\alpha ,\beta }\left[u\right]\left(x\right)& ={r}^{\alpha }{J}^{\beta \left(1-\alpha \right)}\frac{d}{dr}{J}^{\left(1-\beta \right)\left(1-\alpha \right)}u\left(x\right)\\ ={r}^{\alpha }\left\{\frac{1}{\mathrm{\Gamma }\left({\delta }_{1}\right)}{\int }_{0}^{r}{\left(r-\tau \right)}^{{\delta }_{1}-1}\frac{1}{\mathrm{\Gamma }\left({\delta }_{2}\right)}\frac{d}{d\tau }{\int }_{0}^{\tau }{\left(\tau -s\right)}^{{\delta }_{2}-1}u\left(s\theta \right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ ={r}^{\alpha }\left\{\frac{1}{\mathrm{\Gamma }\left({\delta }_{1}\right)}\frac{d}{dr}{\int }_{0}^{r}\frac{{\left(r-\tau \right)}^{{\delta }_{1}}}{{\delta }_{1}}\frac{1}{\mathrm{\Gamma }\left({\delta }_{2}\right)}\frac{d}{d\tau }{\int }_{0}^{\tau }{\left(\tau -s\right)}^{{\delta }_{2}-1}u\left(s\theta \right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ ={r}^{\alpha }\frac{1}{\mathrm{\Gamma }\left({\delta }_{1}\right)}\frac{1}{\mathrm{\Gamma }\left({\delta }_{2}\right)}\frac{d}{dr}\left\{{\int }_{0}^{r}{\left(r-\tau \right)}^{{\delta }_{1}-1}{\int }_{0}^{\tau }{\left(\tau -s\right)}^{{\delta }_{2}-1}u\left(s\theta \right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ ={r}^{\alpha }\frac{1}{\mathrm{\Gamma }\left({\delta }_{1}\right)}\frac{1}{\mathrm{\Gamma }\left({\delta }_{2}\right)}\frac{d}{dr}\left\{{\int }_{0}^{r}u\left(s\theta \right){\int }_{s}^{r}{\left(r-\tau \right)}^{{\delta }_{1}-1}{\left(\tau -s\right)}^{{\delta }_{2}-1}\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\right\}.\end{array}$

Consider the inner integral. If we change variables $\tau =r+\xi \left(s-r\right)$, this integral can be represented in the form of

${\int }_{s}^{r}{\left(r-\tau \right)}^{{\delta }_{1}-1}{\left(\tau -s\right)}^{{\delta }_{2}-1}\phantom{\rule{0.2em}{0ex}}d\tau ={\left(r-s\right)}^{{\delta }_{1}+{\delta }_{2}-1}{\int }_{0}^{1}{\left(1-\xi \right)}^{{\delta }_{1}-1}{\xi }^{{\delta }_{2}-1}\phantom{\rule{0.2em}{0ex}}d\xi .$

Since

${\int }_{0}^{1}{\left(1-\xi \right)}^{{\delta }_{1}-1}{\xi }^{{\delta }_{2}-1}\phantom{\rule{0.2em}{0ex}}d\xi =\frac{\mathrm{\Gamma }\left({\delta }_{1}\right)\mathrm{\Gamma }\left({\delta }_{2}\right)}{\mathrm{\Gamma }\left({\delta }_{1}+{\delta }_{2}\right)},\phantom{\rule{1em}{0ex}}{\delta }_{1}+{\delta }_{2}=1-\alpha ,$

we have

$\begin{array}{rl}{B}^{\alpha ,\beta }\left[u\right]\left(x\right)& =\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left({\delta }_{1}+{\delta }_{2}\right)}\frac{d}{dr}\left\{{\int }_{0}^{r}{\left(r-s\right)}^{{\delta }_{1}+{\delta }_{2}-1}u\left(s\theta \right)\phantom{\rule{0.2em}{0ex}}ds\right\}\\ =\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}\left\{{\int }_{0}^{r}{\left(r-s\right)}^{-\alpha }u\left(s\theta \right)\phantom{\rule{0.2em}{0ex}}ds\right\}={B}^{\alpha }\left[u\right]\left(x\right).\end{array}$

The first equality from (3.1) is proved.

If $\beta =1$, then

$\begin{array}{rl}{B}^{\alpha ,1}\left[u\right]\left(x\right)& =\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\left\{{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }\frac{d}{d\tau }u\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ =\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}\left\{{\int }_{0}^{r}\frac{{\left(r-\tau \right)}^{1-\alpha }}{1-\alpha }\frac{d}{d\tau }u\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ =\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}\left\{-\frac{{r}^{1-\alpha }}{1-\alpha }u\left(0\right)+{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }u\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ =\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\left\{-{r}^{-\alpha }u\left(0\right)+\frac{d}{dr}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }u\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ ={B}^{\alpha }\left[u\right]\left(x\right)-\frac{u\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)}.\end{array}$

The lemma is proved. □

This lemma implies that for any $0\le \beta \le 1$, the problem (2.1), (2.2) can be always reduced to the problems with the boundary Riemann-Liouville or Caputo operators.

Corollary 3.2 If $0<\alpha <1$, $\beta =1$, then the equality

${B}_{\ast }^{\alpha }\left[u\right]\left(x\right)={B}^{\alpha }\left[u\right]\left(x\right)-\frac{u\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)}$
(3.2)

is correct.

Lemma 3.3 If $0<\alpha <1$, $\beta =1$, then the equality

${B}_{\ast }^{\alpha }\left[u\right]\left(0\right)=0$

holds.

Proof

Since

$\begin{array}{rl}{B}^{\alpha }\left[u\right]\left(x\right)=& \frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{r}{\left(r-s\right)}^{-\alpha }u\left(s\theta \right)\phantom{\rule{0.2em}{0ex}}ds\\ =& \frac{1-\alpha }{\mathrm{\Gamma }\left(1-\alpha \right)}{\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi \\ +\frac{r}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi ,\end{array}$

by virtue of smoothness of the function $u\left(x\right)$ at $x\to 0$, the second integral converges to zero.

Then the equality (3.2) implies

$\begin{array}{rl}\underset{x\to 0}{lim}{B}_{\ast }^{\alpha }\left[u\right]\left(x\right)& =\underset{x\to 0}{lim}\left[{B}^{\alpha }\left[u\right]\left(x\right)-\frac{u\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)}\right]\\ =\frac{1-\alpha }{\mathrm{\Gamma }\left(1-\alpha \right)}\underset{x\to 0}{lim}{\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi -\frac{u\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)}\\ =\frac{1-\alpha }{\mathrm{\Gamma }\left(1-\alpha \right)}u\left(0\right){\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }\phantom{\rule{0.2em}{0ex}}d\xi -\frac{u\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)}=0.\end{array}$

The lemma is proved. □

Lemma 3.4 Let $0<\alpha <1$. Then the equality

$u\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau$
(3.3)

holds for any $x\in \mathrm{\Omega }$.

Proof Let $x\in \mathrm{\Omega }$ and $t\in \left(0,1\right]$. Consider the function

${\mathrm{\Im }}_{t}\left[u\right]\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{t}{\left(t-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau .$

Represent ${\mathrm{\Im }}_{t}\left[u\right]\left(x\right)$ in the form of

${\mathrm{\Im }}_{t}\left[u\right]\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}\frac{d}{dt}\left\{{\int }_{0}^{t}\frac{{\left(t-\tau \right)}^{\alpha }}{\alpha }{\tau }^{-\alpha }{B}^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau \right\}.$

Further, using definition of the operator ${B}^{\alpha }$, we have

$\begin{array}{r}{\mathrm{\Im }}_{t}\left[u\right]\left(x\right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}\left\{{\int }_{0}^{t}\frac{{\left(t-\tau \right)}^{\alpha }}{\alpha \mathrm{\Gamma }\left(\alpha \right)}{\tau }^{-\alpha }{\tau }^{\alpha }\frac{d}{d\tau }{\int }_{0}^{\tau }{\left(\tau -\xi \right)}^{-\alpha }u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}\left\{\frac{{\left(t-\tau \right)}^{\alpha }}{\alpha }{\int }_{0}^{\tau }{\left(\tau -\xi \right)}^{-\alpha }u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi {|}_{\tau =0}^{\tau =t}\\ \phantom{\rule{2em}{0ex}}+{\int }_{0}^{t}{\left(t-\tau \right)}^{\alpha -1}{\int }_{0}^{\tau }{\left(\tau -\xi \right)}^{-\alpha }u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}\left\{{\int }_{0}^{t}{\left(t-\tau \right)}^{\alpha -1}{\int }_{0}^{\tau }{\left(\tau -\xi \right)}^{-\alpha }u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right\}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}\left\{{\int }_{0}^{t}u\left(\xi x\right){\int }_{\xi }^{t}{\left(t-\tau \right)}^{\alpha -1}{\left(\tau -\xi \right)}^{-\alpha }\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}d\xi \right\}.\end{array}$

It is easy to show that

${\int }_{\xi }^{t}{\left(t-\tau \right)}^{\alpha -1}{\left(\tau -\xi \right)}^{-\alpha }\phantom{\rule{0.2em}{0ex}}d\tau =\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(1-\alpha \right).$

Then

${\mathrm{\Im }}_{t}\left[u\right]\left(x\right)=\frac{d}{dt}{\int }_{0}^{t}u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi =u\left(tx\right).$

If now we suppose $t=1$, then

$u\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau .$

The lemma is proved. □

Using connection between operators ${B}^{\alpha }$ and ${B}_{\ast }^{\alpha }$, one can prove the following.

Lemma 3.5 Let $0<\alpha <1$. Then the representation

$u\left(x\right)=u\left(0\right)+\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}_{\ast }^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau$
(3.4)

is valid for any $x\in \mathrm{\Omega }$.

Proof Using the equality (3.3), taking into account (3.2), we obtain

$\begin{array}{rl}u\left(x\right)& =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }\left[\frac{u\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)}+{B}_{\ast }^{\alpha }\left[u\right]\left(\tau x\right)\right]\phantom{\rule{0.2em}{0ex}}d\tau \\ =u\left(0\right)+\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}_{\ast }^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau .\end{array}$

The lemma is proved. □

Lemma 3.6 Let $0<\alpha <1$. Then the equalities

${B}^{-\alpha }\left[{B}^{\alpha }\left[u\right]\right]\left(x\right)={B}^{\alpha }\left[{B}^{-\alpha }\left[u\right]\right]\left(x\right)=u\left(x\right)$
(3.5)

hold for any $x\in \mathrm{\Omega }$.

Proof Let us prove the first equality. Apply to the function ${B}^{\alpha }\left[u\right]$, the operator ${B}^{-\alpha }$. By definition of ${B}^{-\alpha }$, we have

${B}^{-\alpha }\left[{B}^{\alpha }\left[u\right]\right]\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau .$

But by virtue of the equality (3.3), the last integral is equal to $u\left(x\right)$, i.e.,

${B}^{-\alpha }\left[{B}^{\alpha }\left[u\right]\right]\left(x\right)=u\left(x\right).$

Now let us prove the second equality. Applying the operator ${B}^{\alpha }$ to the function ${B}^{-\alpha }\left[u\right]\left(x\right)$, we obtain

$\begin{array}{rl}{B}^{\alpha }\left[{B}^{-\alpha }\left[u\right]\right]\left(x\right)& =\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{B}^{-\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}\left\{{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -1}{s}^{-\alpha }u\left(s\tau \theta \right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \right\}.\end{array}$

Further, it is not difficult to verify correctness of the following equalities:

$\begin{array}{r}\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }u\left(s\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{1em}{0ex}}\underset{s\tau =\xi }{=}\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{rs}{\left(r-\frac{\xi }{s}\right)}^{-\alpha }u\left(\xi \theta \right)\frac{d\xi }{s}\\ \phantom{\rule{1em}{0ex}}=\frac{{r}^{\alpha }{s}^{\alpha -1}}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{rs}{\left(sr-\xi \right)}^{-\alpha }u\left(\xi \theta \right)\phantom{\rule{0.2em}{0ex}}d\xi \\ \phantom{\rule{1em}{0ex}}=\frac{{\left(sr\right)}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{d\left(sr\right)}{\int }_{0}^{rs}{\left(sr-\xi \right)}^{-\alpha }u\left(\xi \theta \right)\phantom{\rule{0.2em}{0ex}}d\xi ={B}^{\alpha }\left[u\right]\left(sx\right).\end{array}$

Here, it is taken into account $\theta =\frac{x}{|x|}=\frac{sx}{|sx|}$. Therefore,

${B}^{\alpha }\left[{B}^{-\alpha }\left[u\right]\right]\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -1}{s}^{-\alpha }{B}^{\alpha }\left[u\right]\left(sx\right)\phantom{\rule{0.2em}{0ex}}ds.$

Hence, using the equality (3.3), we obtain

${B}^{\alpha }\left[{B}^{-\alpha }\left[u\right]\right]\left(x\right)=u\left(x\right).$

The lemma is proved. □

In , the following is proved.

Lemma 3.7 Let $u\left(x\right)\in {C}^{1}\left(\mathrm{\Omega }\right)$. Then the equality

$u\left(x\right)=u\left(0\right)+{\int }_{0}^{1}\left[\sum _{k=1}^{n}{x}_{k}\frac{\partial u\left(sx\right)}{\partial {x}_{k}}\right]\phantom{\rule{0.2em}{0ex}}ds$
(3.6)

holds for any $x\in \mathrm{\Omega }$.

Since

$\sum _{k=1}^{n}{x}_{k}\frac{\partial u\left(x\right)}{\partial {x}_{k}}=r\frac{\partial u\left(x\right)}{\partial r}={B}^{1}\left[u\right]\left(x\right)={B}_{\ast }^{1}\left[u\right]\left(x\right),$

the equality (3.5) can be represented in the form of

$u\left(x\right)=u\left(0\right)+{\int }_{0}^{1}{s}^{-1}{B}^{1}\left[u\right]\left(sx\right)\phantom{\rule{0.2em}{0ex}}ds.$
(3.7)

Then Lemma 3.5 and the equality (3.6) imply the following.

Corollary 3.8 Let $0<\alpha \le 1$. Then for any $x\in \mathrm{\Omega }$, the representation

$u\left(x\right)=u\left(0\right)+\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}_{\ast }^{\alpha }\left[u\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau$

is valid. Hence, as the inverse operator to ${B}^{1}$, we can consider the following operator:

${B}^{-1}\left[u\right]\left(x\right)={\int }_{0}^{1}{\tau }^{-1}u\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau .$

Note that if

$u\left(0\right)\ne 0,$

then the operator ${B}^{-1}$ is not defined in such functions. Let $u\left(x\right)$ be a smooth function. Obviously,

${B}^{1}\left[u\right]\left(0\right)=0.$

Consider action of the operator ${B}^{-1}$ to the function ${B}^{1}\left[u\right]\left(x\right)$. By definition of the operator ${B}^{-1}$, we have

${B}^{-1}\left[{B}^{1}\left[u\right]\right]\left(x\right)={\int }_{0}^{1}{s}^{-1}{B}^{1}\left[u\right]\left(tx\right)\phantom{\rule{0.2em}{0ex}}dt.$

By virtue of (3.7), the value of the last integral is equal to $u\left(x\right)-u\left(0\right)$. Thus, the equality

${B}^{-1}\left[{B}^{1}\left[u\right]\right]\left(x\right)=u\left(x\right)-u\left(0\right)$

holds.

Conversely, let $u\left(0\right)=0$. Then the operator ${B}^{-1}$ is defined for such functions, and

${B}^{1}\left[{B}^{-1}\left[u\right]\right]\left(x\right)=r\frac{\partial }{\partial r}\left[{\int }_{0}^{1}{s}^{-1}u\left(sx\right)\phantom{\rule{0.2em}{0ex}}ds\right]=r\frac{\partial }{\partial r}\left[{\int }_{0}^{r}{\xi }^{-1}u\left(\xi \theta \right)\phantom{\rule{0.2em}{0ex}}d\xi \right]=u\left(x\right).$

It means that

${B}^{1}\left[{B}^{-1}\left[u\right]\right]\left(x\right)=u\left(x\right).$

Thus, we prove the following.

Lemma 3.9 For any $x\in \mathrm{\Omega }$, the following equalities are valid:

1. (1)

${B}^{-1}\left[{B}^{1}\left[u\right]\right]\left(x\right)=u\left(x\right)-u\left(0\right)$;

2. (2)

if $u\left(0\right)=0$, then

${B}^{1}\left[{B}^{-1}\left[u\right]\right]\left(x\right)=u\left(x\right).$

Using Lemma 3.9 and connection between operators ${B}^{\alpha }$ and ${B}_{\ast }^{\alpha }$, we get the following.

Corollary 3.10 Let $0<\alpha \le 1$. Then for any $x\in \mathrm{\Omega }$, the following equalities hold: if $0<\alpha <1$, then

$\begin{array}{c}{B}^{-\alpha }\left[{B}_{\ast }^{\alpha }\left[u\right]\right]\left(x\right)=u\left(x\right)-\frac{u\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)};\hfill \\ {B}^{-1}\left[{B}^{1}\left[u\right]\right]\left(x\right)=u\left(x\right)-u\left(0\right);\hfill \end{array}$

if $0<\alpha \le 1$ and $u\left(0\right)=0$, then

${B}_{\ast }^{\alpha }\left[{B}^{-\alpha }\left[u\right]\right]\left(x\right)=u\left(x\right).$

Lemma 3.11 Let

$\mathrm{\Delta }u\left(x\right)=g\left(x\right),\phantom{\rule{1em}{0ex}}x\in \mathrm{\Omega },0<\alpha \le 1.$

Then for any $x\in \mathrm{\Omega }$, the equality

$\mathrm{\Delta }{B}^{\alpha }\left[u\right]\left(x\right)=|x{|}^{-2}{B}^{\alpha }\left[|x{|}^{2}g\right]\left(x\right)$
(3.8)

holds.

Proof Let $0<\alpha <1$. After changing of variables, the function ${B}^{\alpha }\left[u\right]\left(x\right)$ can be represented in the form of

${B}^{\alpha }\left[u\right]\left(x\right)=\frac{1-\alpha }{\mathrm{\Gamma }\left(1-\alpha \right)}{\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi +r\frac{d}{dr}{\int }_{0}^{1}\frac{{\left(1-\xi \right)}^{-\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}u\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi ={I}_{1}\left(x\right)+{I}_{2}\left(x\right).$

Since

$\mathrm{\Delta }u\left(x\right)=g\left(x\right),$

it is easy to show that

$\mathrm{\Delta }{I}_{1}\left(x\right)=\frac{1-\alpha }{\mathrm{\Gamma }\left(1-\alpha \right)}{\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }{\xi }^{2}g\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi .$

Further, if $v\left(x\right)$ is a smooth function, then obviously,

$\mathrm{\Delta }\left[r\frac{\partial }{\partial r}v\left(x\right)\right]=r\frac{\partial }{\partial r}\mathrm{\Delta }v\left(x\right)+2\mathrm{\Delta }v\left(x\right).$

That is why

$\mathrm{\Delta }{I}_{2}\left(x\right)=r\frac{d}{dr}{\int }_{0}^{1}\frac{{\left(1-\xi \right)}^{-\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}{\xi }^{2}g\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi +2{\int }_{0}^{1}\frac{{\left(1-\xi \right)}^{-\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}{\xi }^{2}g\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi .$

Consider the integral

${\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }{\xi }^{2}g\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi .$

After changing of variables $\xi r=\tau$, $\xi ={r}^{-1}\tau$, the integral can be transformed to the following form:

${\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }{\xi }^{2}g\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi ={r}^{\alpha -3}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau .$

Then

$\begin{array}{rl}r\frac{d}{dr}{\int }_{0}^{1}\frac{{\left(1-\xi \right)}^{-\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}{\xi }^{2}g\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi =& r\frac{d}{dr}\left[{r}^{\alpha -3}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \right]\\ =& \left(\alpha -3\right){r}^{\alpha -3}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \\ +{r}^{\alpha -2}\frac{d}{dr}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau .\end{array}$

Hence,

$\begin{array}{rl}\mathrm{\Delta }{I}_{1}\left(x\right)+\mathrm{\Delta }{I}_{2}\left(x\right)=& \frac{1-\alpha }{\mathrm{\Gamma }\left(1-\alpha \right)}{r}^{\alpha -3}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{\left(\alpha -3\right)}{\mathrm{\Gamma }\left(1-\alpha \right)}{r}^{\alpha -3}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{{r}^{\alpha -2}}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{2}{\mathrm{\Gamma }\left(1-\alpha \right)}{r}^{\alpha -3}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =& \frac{{r}^{\alpha -2}}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =& {r}^{-2}\frac{{r}^{\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dr}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2}g\left(\tau \theta \right)\phantom{\rule{0.2em}{0ex}}d\tau ={r}^{-2}{B}^{\alpha }\left[|x{|}^{2}g\right]\left(x\right).\end{array}$

Let now $\alpha =1$. In this case,

${B}^{1}\left[u\right]\left(x\right)=r\frac{\partial u\left(x\right)}{\partial r},$

and therefore.

$\mathrm{\Delta }{B}^{1}\left[u\right]\left(x\right)=\mathrm{\Delta }\left[r\frac{\partial u\left(x\right)}{\partial r}\right]=r\frac{\partial \mathrm{\Delta }u\left(x\right)}{\partial r}+2\mathrm{\Delta }u\left(x\right)=r\frac{\partial g\left(x\right)}{\partial r}+2g\left(x\right).$

On the other hand,

$\left(r\frac{\partial }{\partial r}+2\right)g\left(x\right)={|x|}^{-2}r\frac{\partial }{\partial r}\left[{r}^{2}g\left(x\right)\right]={|x|}^{-2}{B}^{1}\left[{|x|}^{2}g\right]\left(x\right).$

The lemma is proved. □

## 4 Some properties of a solution of the Dirichlet problem

Let $v\left(x\right)$ be a solution of the problem (2.4). It is known (see ), if functions$f\left(x\right)$ and ${g}_{1}\left(x\right)$ are sufficiently smooth, then a solution of the problem (2.4) exists and is represented in the form of

$v\left(x\right)=-\frac{1}{{\omega }_{n}}{\int }_{\mathrm{\Omega }}G\left(x,y\right){g}_{1}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy+\frac{1}{{\omega }_{n}}{\int }_{\partial \mathrm{\Omega }}P\left(x,y\right)f\left(y\right)\phantom{\rule{0.2em}{0ex}}d{s}_{y},$
(4.1)

here ${\omega }_{n}$ is the area of the unit sphere, $G\left(x,y\right)$ is the Green function of the Dirichlet problem for the Laplace equation, and $P\left(x,y\right)$ is the Poisson kernel.

$\begin{array}{c}G\left(x,y\right)=\frac{1}{n-2}\left[|x-y{|}^{2-n}-||y|x-\frac{y}{|y|}{|}^{2-n}\right],\hfill \\ P\left(x,y\right)=\frac{1-|x{|}^{2}}{|x-y{|}^{n}}\hfill \end{array}$

take place.

Lemma 4.1 Let $v\left(x\right)$ be a solution of the problem (2.4).

Then

1. (1)

if $v\left(0\right)=0$, then

${\int }_{\partial \mathrm{\Omega }}f\left(y\right)\phantom{\rule{0.2em}{0ex}}d{s}_{y}={\int }_{\mathrm{\Omega }}\frac{|y{|}^{2-n}-1}{n-2}{g}_{1}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy;$
(4.2)
2. (2)

if the equality (4.2) is valid, then the condition $v\left(0\right)=0$ is fulfilled for a solution of the problem (2.4).

Proof Let a solution of the problem (2.4) exist. Represent it in the form of (4.1). We have from the representation of the function $G\left(x,y\right)$

$G\left(0,y\right)=\frac{1}{n-2}\left[|y{|}^{2-n}-1\right]$

and

$P\left(0,y\right)=1.$

Then

$0=v\left(0\right)=-\frac{1}{{\omega }_{n}}{\int }_{\mathrm{\Omega }}G\left(0,y\right){g}_{1}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy+\frac{1}{{\omega }_{n}}{\int }_{\partial \mathrm{\Omega }}P\left(0,y\right)f\left(y\right)\phantom{\rule{0.2em}{0ex}}d{s}_{y}.$

Hence,

${\int }_{\partial \mathrm{\Omega }}f\left(y\right)\phantom{\rule{0.2em}{0ex}}d{s}_{y}={\int }_{\mathrm{\Omega }}\frac{|y{|}^{2-n}-1}{n-2}{g}_{1}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy.$

The equality (4.2) is proved. The second assertion of the lemma is proved in the inverse order. The lemma is proved. □

Lemma 4.2 Let $v\left(x\right)$ be a solution of the problem (2.4), and the function ${g}_{1}\left(x\right)$ be represented in the form of

${g}_{1}\left(y\right)=\left(\rho \frac{\partial }{\partial \rho }+2\right)g\left(y\right),\phantom{\rule{1em}{0ex}}\rho =|y|.$

Then the condition (4.2) can be represented in the form of

${\int }_{\partial \mathrm{\Omega }}f\left(y\right)\phantom{\rule{0.2em}{0ex}}d{s}_{y}={\int }_{\mathrm{\Omega }}g\left(y\right)\phantom{\rule{0.2em}{0ex}}dy.$
(4.3)

Proof Using representation of the function ${g}_{1}\left(y\right)$, we have

$\begin{array}{r}{\int }_{\mathrm{\Omega }}\frac{|y{|}^{2-n}-1}{n-2}{g}_{1}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{1}{\rho }^{n-1}{\int }_{|\xi |=1}\frac{{\rho }^{2-n}-1}{n-2}\left(\rho \frac{\partial }{\partial \rho }+2\right)g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\rho .\end{array}$

Then

$\begin{array}{rl}{\int }_{0}^{1}{\rho }^{n-1}\frac{{\rho }^{2-n}-1}{n-2}\left(\rho \frac{\partial }{\partial \rho }+2\right)g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\rho =& \frac{1}{n-2}{\int }_{0}^{1}\left[{\rho }^{2}-{\rho }^{n}\right]\frac{\partial }{\partial \rho }\left[g\right]\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\rho \\ +\frac{2}{n-2}{\int }_{0}^{1}\left[\rho -{\rho }^{n-1}\right]g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\rho ={I}_{1}+{I}_{2}.\end{array}$

Consider ${I}_{1}$. After integrating by parts, we get

${I}_{1}=\frac{1}{n-2}{\int }_{0}^{1}\left[n{\rho }^{n-1}-2\rho \right]g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\rho ,$

what follows

$\begin{array}{r}{I}_{1}+{I}_{2}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n-2}\left\{{\int }_{0}^{1}\left(n-2\right){\rho }^{n-1}g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\rho -2{\int }_{0}^{1}\rho g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\rho +2{\int }_{0}^{1}\rho g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\rho \right\}\\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{1}{\rho }^{n-1}g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\rho .\end{array}$

Hence,

${\int }_{\mathrm{\Omega }}\frac{|y{|}^{2-n}-1}{n-2}{g}_{1}\left(y\right)\phantom{\rule{0.2em}{0ex}}dy={\int }_{0}^{1}{\rho }^{n-1}{\int }_{|\xi |=1}g\left(\rho \xi \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\rho ={\int }_{\mathrm{\Omega }}g\left(y\right)\phantom{\rule{0.2em}{0ex}}dy.$

The lemma is proved. □

## 5 The proof of the main proposition

Let $0<\alpha <1$, $0\le \beta <1$, and $u\left(x\right)$ be a solution of the problem (2.1), (2.2). In this case, by Lemma 3.1, ${B}^{\alpha ,\beta }={B}^{\alpha }$. Apply to the function $u\left(x\right)$ the operator ${B}^{\alpha }$, and denote

$v\left(x\right)={B}^{\alpha }\left[u\right]\left(x\right).$

Then, using the equality (3.8), we obtain

$\mathrm{\Delta }v\left(x\right)=\mathrm{\Delta }{B}^{\alpha }\left[u\right]\left(x\right)={|x|}^{-2}{B}^{\alpha }\left[{|x|}^{2}g\right]\left(x\right)\equiv {g}_{1}\left(x\right).$

Since ${B}^{\alpha ,\beta }={B}^{\alpha }$, it is obviously, Thus, if $u\left(x\right)$ is a solution of the problem (2.1), (2.2), then we obtain for the function

$v\left(x\right)={B}^{\alpha }\left[u\right]\left(x\right)$

the problem (2.4) with

${g}_{1}\left(x\right)={|x|}^{-2}{B}^{\alpha }\left[{|x|}^{2}g\right]\left(x\right).$

Further, since

${g}_{1}\left(x\right)={|x|}^{-2}{B}^{\alpha }\left[{|x|}^{2}g\right]\left(x\right)=\left[r\frac{d}{dr}+3-\alpha \right]{\int }_{0}^{1}\frac{{\left(1-\xi \right)}^{-\alpha }}{\mathrm{\Gamma }\left(1-\alpha \right)}{\xi }^{2}g\left(\xi x\right)\phantom{\rule{0.2em}{0ex}}d\xi ,$

for $g\left(x\right)\in {C}^{\lambda +1}\left(\overline{\mathrm{\Omega }}\right)$, we have ${g}_{1}\left(x\right)\in {C}^{\lambda }\left(\overline{\mathrm{\Omega }}\right)$.

Then for ${g}_{1}\left(x\right)\in {C}^{\lambda }\left(\overline{\mathrm{\Omega }}\right)$, $f\left(x\right)\in {C}^{\lambda +2}\left(\partial \mathrm{\Omega }\right)$, a solution of the problem (2.4) exists and belongs to the class ${C}^{\lambda +2}\left(\overline{\mathrm{\Omega }}\right)$ (see, for example, ).

Further, applying to the equality

$v\left(x\right)={B}^{\alpha }\left[u\right]\left(x\right)$

the operator ${B}^{-\alpha }$, by virtue of the first equality of the formula (3.5), we obtain

$u\left(x\right)={B}^{-\alpha }\left[v\right]\left(x\right).$

The last function satisfies to all the conditions of the problem (2.1), (2.2).

Really,

$\begin{array}{rl}\mathrm{\Delta }u\left(x\right)=\mathrm{\Delta }{B}^{-\alpha }\left[v\right]\left(x\right)& =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{2-\alpha }\mathrm{\Delta }v\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{2-\alpha }{g}_{1}\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{2-\alpha }{\tau }^{-2}{|x|}^{-2}{B}^{\alpha }\left[{|x|}^{2}g\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{|x|}^{-2}{B}^{\alpha }\left[{|x|}^{2}g\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =\frac{{|x|}^{-2}}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-\tau \right)}^{\alpha -1}{\tau }^{-\alpha }{B}^{\alpha }\left[{|x|}^{2}g\right]\left(\tau x\right)\phantom{\rule{0.2em}{0ex}}d\tau ={|x|}^{-2}{|x|}^{2}g\left(x\right)=g\left(x\right).\end{array}$

Now, using the second equality from (3.5), we obtain So, the function $u\left(x\right)={B}^{-\alpha }\left[v\right]\left(x\right)$ satisfies equation (2.1) and the boundary condition (2.2). Let now $0<\alpha <1$, $\beta =1$, and $u\left(x\right)$ be a solution of the problem (2.1), (2.2). Apply to the function $u\left(x\right)$ the operator ${B}^{\alpha ,1}={B}_{\ast }^{\alpha }$, and denote $v\left(x\right)={B}_{\ast }^{\alpha }\left[u\right]\left(x\right)$. In this case, we obtain for the function $v\left(x\right)$ the problem (2.4) with the function

${g}_{1}\left(x\right)={|x|}^{-2}{B}_{\ast }^{\alpha }\left[{|x|}^{2}g\right]\left(x\right).$

Since ${B}_{\ast }^{\alpha }\left[u\right]\left(0\right)=0$, the function $v\left(x\right)$ must satisfy in addition to the condition $v\left(0\right)=0$.

Arbitrary solution of the problem (2.4) at smooth $f\left(x\right)$ and $g\left(x\right)$ is represented in the form of (4.1). And in order that this solution satisfies to the condition $v\left(0\right)=0$, according to Lemma 3.11, it is necessary and sufficient fulfillment of the condition (4.2).

In our case, the condition (4.2) has the form

${\int }_{\partial \mathrm{\Omega }}f\left(y\right)\phantom{\rule{0.2em}{0ex}}d{s}_{y}={\int }_{\mathrm{\Omega }}\frac{{|y|}^{2-n}-1}{n-2}{|y|}^{-2}{B}_{\ast }^{\alpha }\left[{|y|}^{2}g\right]\left(y\right)\phantom{\rule{0.2em}{0ex}}dy.$

In this case, ${B}^{\alpha ,1}={B}_{\ast }^{\alpha }$ and, therefore, the condition (4.2) coincides with the condition (2.6).

Thus, necessity of (2.6) is proved. This condition is also sufficient condition for existence of a solution for the problem (2.1), (2.2).

In fact, if the condition (2.6) holds, then $v\left(0\right)=0$, and the function

$u\left(x\right)={B}^{-\alpha }\left[v\right]\left(x\right)+C$

satisfies to all conditions of the problem (2.1), (2.2). Let us check these conditions. Fulfillment of the condition

$\mathrm{\Delta }u\left(x\right)=g\left(x\right)$

can be checked similarly as in the case of the proof of the first part of the theorem. Further, using the equality (3.5) and connection between operators ${B}^{\alpha }$ and ${B}_{\ast }^{\alpha }$, we get

$\begin{array}{rl}{B}_{\ast }^{\alpha }\left[u\right]\left(x\right)& ={B}_{\ast }^{\alpha }\left[{B}^{-\alpha }\left[v\right]+C\right]\left(x\right)\\ ={B}_{\ast }^{\alpha }\left[{B}^{-\alpha }\left[v\right]\right]\left(x\right)+{B}_{\ast }^{\alpha }\left[C\right]\\ ={B}^{\alpha }\left[{B}^{-\alpha }\left[v\right]\right]\left(x\right)+\frac{{B}^{-\alpha }\left[v\right]\left(0\right)}{\mathrm{\Gamma }\left(1-\alpha \right)}=v\left(x\right).\end{array}$

Hence, If $\alpha =1$, then

${D}^{\alpha ,1}={D}_{\ast }^{\alpha }=\frac{\partial }{\partial r}$

and

${B}^{\alpha ,1}=r\frac{\partial }{\partial r}.$

In this case,

${|x|}^{-2}{B}^{\alpha ,1}\left[{|x|}^{2}g\right]\left(x\right)=\left(r\frac{\partial }{\partial r}+2\right)g\left(x\right).$

Then by virtue of Lemma 4.1, the solvability condition of the problem (2.1), (2.2) can be rewritten in the form of

${\int }_{\partial \mathrm{\Omega }}f\left(x\right)\phantom{\rule{0.2em}{0ex}}d{s}_{x}={\int }_{\mathrm{\Omega }}g\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.$

It is the solvability condition for the Neumann problem. Further, since $v\left(x\right)\in {C}^{\lambda +2}\left(\overline{\mathrm{\Omega }}\right)$, the function $u\left(x\right)={B}^{-\alpha }\left[v\right]\left(x\right)$ also belongs to the class ${C}^{\lambda +2}\left(\overline{\mathrm{\Omega }}\right)$. The theorem is proved.

## 6 Example

Example Let $0<\alpha <1$, $\beta =1$ and

$g\left(x\right)={|x|}^{2k},\phantom{\rule{1em}{0ex}}k=0,1,\dots .$

Then

$\begin{array}{rl}{|x|}^{-2}{B}^{\alpha ,1}\left[{|x|}^{2}g\right]\left(x\right)& =\frac{{|x|}^{\alpha -2}}{\mathrm{\Gamma }\left(1-\alpha \right)}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }\frac{\partial }{\partial \tau }{\tau }^{2k+2}\phantom{\rule{0.2em}{0ex}}d\tau \\ =\frac{\left(2k+2\right){|x|}^{\alpha -2}}{\mathrm{\Gamma }\left(1-\alpha \right)}{\int }_{0}^{r}{\left(r-\tau \right)}^{-\alpha }{\tau }^{2k+2-1}\phantom{\rule{0.2em}{0ex}}d\tau \\ =\frac{\left(2k+2\right){|x|}^{\alpha -2}}{\mathrm{\Gamma }\left(1-\alpha \right)}{|x|}^{2k+2-\alpha }{\int }_{0}^{1}{\left(1-\xi \right)}^{-\alpha }{\xi }^{2k+1}\phantom{\rule{0.2em}{0ex}}d\xi \\ =\frac{\left(2k+2\right){|x|}^{2k}}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{\mathrm{\Gamma }\left(1-\alpha \right)\mathrm{\Gamma }\left(2k+2\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}=\frac{\mathrm{\Gamma }\left(2k+3\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}{|x|}^{2k}.\end{array}$

Since

$\begin{array}{rl}{\int }_{0}^{1}{r}^{2k+n-1}\left({r}^{2-n}-1\right)\phantom{\rule{0.2em}{0ex}}dr& ={\int }_{0}^{1}\left({r}^{2k+1}-{r}^{2k+n-1}\right)\phantom{\rule{0.2em}{0ex}}dr\\ =\frac{1}{2k+2}-\frac{1}{2k+n}=\frac{n-2}{\left(2k+2\right)\left(2k+n\right)},\end{array}$

we have

$\begin{array}{rl}{\int }_{\mathrm{\Omega }}\frac{{|y|}^{2-n}-1}{n-2}{|y|}^{-2}{B}_{\ast }^{\alpha }\left[{|y|}^{2}g\right]\left(y\right)\phantom{\rule{0.2em}{0ex}}dy& ={\int }_{|\xi |=1}{\int }_{0}^{1}\frac{{r}^{2-n}-1}{n-2}{r}^{-2}{B}_{\ast }^{\alpha }\left[{|y|}^{2}g\right]\left(r\xi \right)\phantom{\rule{0.2em}{0ex}}dr\phantom{\rule{0.2em}{0ex}}d\xi \\ =\frac{\mathrm{\Gamma }\left(2k+2\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}\frac{{\omega }_{n}}{\left(2k+n\right)}.\end{array}$

Then the solvability condition for the problem (2.1), (2.2) has in this case the form

${\int }_{\partial \mathrm{\Omega }}f\left(y\right)\phantom{\rule{0.2em}{0ex}}d{s}_{y}=\frac{\mathrm{\Gamma }\left(2k+2\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}\frac{{\omega }_{n}}{\left(2k+n\right)}.$

For example, if $f\left(x\right)=1$, this condition is not fulfilled. If

$f\left(x\right)=\frac{\mathrm{\Gamma }\left(2k+2\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}\frac{1}{\left(2k+n\right)},$

then the solvability condition of the problem is carried out. In this case, solving the Dirichlet problem (2.4) with the functions

$\begin{array}{c}{g}_{1}\left(x\right)\equiv {|x|}^{-2}{B}^{\alpha ,1}\left[{|x|}^{2}g\right]\left(x\right)=\frac{\mathrm{\Gamma }\left(2k+3\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}{|x|}^{2k},\hfill \\ f\left(x\right)=\frac{\mathrm{\Gamma }\left(2k+2\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}\frac{1}{\left(2k+n\right)},\hfill \end{array}$

we obtain (see )

$v\left(x\right)=\frac{\mathrm{\Gamma }\left(2k+3\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}\frac{{|x|}^{2k+2}}{\left(2k+2\right)\left(2k+n\right)}=\frac{\mathrm{\Gamma }\left(2k+2\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}\frac{{|x|}^{2k+2}}{2k+n}.$

Using the formula (2.5), we obtain the solution of the problem (2.1), (2.2)

$\begin{array}{rl}u\left(x\right)& ={B}^{-\alpha }\left[v\right]\left(x\right)\\ =\frac{\mathrm{\Gamma }\left(2k+2\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}\frac{{|x|}^{2k+2}}{2k+n}\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -1}{s}^{2k+2-\alpha }\phantom{\rule{0.2em}{0ex}}ds\\ =\frac{\mathrm{\Gamma }\left(2k+2\right)}{\mathrm{\Gamma }\left(2k+3-\alpha \right)}\frac{{|x|}^{2k+2}}{2k+n}\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(2k+3-\alpha \right)}{\mathrm{\Gamma }\left(2k+3\right)}=\frac{{|x|}^{2k+2}}{\left(2k+2\right)\left(2k+n\right)}.\end{array}$

Thus, the solution of the problem (2.1), (2.2) has the form

$u\left(x\right)=\frac{{|x|}^{2k+2}}{\left(2k+2\right)\left(2k+n\right)}+C.$

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## Acknowledgements

This paper is financially supported by the grant of the Ministry of Science and Education of the Republic of Kazakhstan (Grant No. 0830/GF2). The authors would like to thank the editor and referees for their valuable comments and remarks, which led to a great improvement of the article.

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Correspondence to Berikbol T Torebek.

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The authors declare that they have no competing interests.

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All authors completed the paper together. All authors read and approved the final manuscript.

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Torebek, B.T., Turmetov, B.K. On solvability of a boundary value problem for the Poisson equation with the boundary operator of a fractional order. Bound Value Probl 2013, 93 (2013). https://doi.org/10.1186/1687-2770-2013-93

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• DOI: https://doi.org/10.1186/1687-2770-2013-93

### Keywords

• the Poison equation
• boundary value problem
• fractional derivative
• the Riemann-Liouville operator
• the Caputo operator 